Mathematics Applied to Continuum Mechanics (Classics in Applied Mathematics 52) -poor quality-

Mathematics Applied to Continuum Mechanics

SIAM'S, Classics in Applied Mathematics series «rr;ists of books that were previously allowed to Lcr out of- print. These books are republished by SIAM as a professional service because they continue to be important ramrurces for mathematical scientists. F.d.itor-inlChief R+7ben E. t)'Matlle*y, Jr., University of Washington Editorial Board Richard A. 13nGtkiï, University of Wiuorrsin ,Maclist}fl Leah Edelsrcin-Veshct, University of British Columbia Nicholas J. High,-tlsti, University of h1 2, is formed by setting two of its indices equal and performing the resulting sum. An example of bow contraction leads to a new tensor whose order is 2 less than the original tensor is provided by the Following theorem.

Car#esiw7 Tensors [Ch. 2

38 Theorem 4. Let T be a fourth order tensor, where

=

L^^+imn

TOR r -

Define f1 by ^ iR

in

—

—

Tim.

(15)

Then Li is a second order tensor, Proof The rule for obtaining the components of U is given by (15). Since T is a fourth order tensor, ^ ^p G j4 eJxr° J,_, TpgFi•

^4^JRR

so, setting

m j,

= But 1 fig

^

1R

(16)

c5-11 yr

as in (1.1.9), so e [pf R7 7;1, 1

Thin

-

By (15), the above equation is e quivalent to FF r which is the required transformation law. p Definition 7. Let A be an nth order tensor and B be an mth order tensor, where [AL i R = A i ,... 4 ,

[8 ],, ^ — Br,

Then the tensor product AB is defined by [ABI,...

D = A; ... ,nBi.,, ,

J...

(I7)

Note that the right side of (17) is the product of two real numbers. Example 4. if v and w are first order tensors with components u ; and w then

[vw],i — IJ i The ÿth component of vw, the product of the ith component of a with the jth component of w, can be found in the ith row and jihî column of the following array: CJ w 1

VI. W2

l} 1 14'3

NW) — 31;44' J

112 .19d2

r..12 w3 -

OE)

V3 WI 113 w1 31 3 W3

Two vectors have a scalar product, a vector product, and a tensor product These products arc, respectively, a scalar, a vector, and a special second order tensor called a dyad (see Exercise 24). NOTE-

S e t. 2.1) Tensor Algebru

39

Theorem 5. Let W = AB, where A is an ,nth order tenser and B is an mth order tensor. Then W is an (n + m)th order tensor. Proof for n I, in = 2. Suppose that A, and BA , and An, and }3 , are the components of A and ft in two different coordinate systems. By definition of W, A r j,

,

Werra

= Am Bp

(19)

Since A and B are tensors, A[

Bjk = /

irnA+n E jp i 4 Bpe .

Therefore, from (19), we obtain the desired transformation law,

Prooffor Me general rase.

Left to the reader. Q

Definition 8. Let A and B be second order tensors with components A 11 and B. Then the contraction product A • B is defined by [A B] — AuBig. The contraction product of tensors of arbitrary order is defined similarly, by summing ever adjacent indices .

Theorem 6. Let A be an !nth order tensor and B be an nth order tensor. Then A • Bis a tensor of order rn + n 2. Proof. A • B a contraction of the (m + n)th order tensor AB. 11 —

Easily proved results (Exercise 16), and useful ones, arc

rod[T - e ^1i^ = T . e M. T. e (1x — 1-

- ,,

(20a, b) (20c)

These equations assign interpretations to each individual tensor component. In Section 1.3 we showed in essence that if t = n • T holds for arbitrary t and n, then T must satisfy the tensor transformation law. it is useful here to restate this result in a general context. To make this section's presentation more complete, we shall repeat the main points of the proof. Theorem 7. Suppose that in any Cartesian coordinate system, there is a rule associating a unique ordered set of nine quantities with T. If, for an arbitrary vector a, a • T is a vector, then T is a second order tensor.* Tilts theorem brings ow

a sect1rtd order censor's characterization as a mapping of a vector

pasx into] itself (compare Section I.4).

s

2

Cartesian Tensors

40 Proof.

Let a •

T = v. We must show the following. Suppose that

(2la, b)

and o; T: ; _ 14,

a, Ti = t?

and suppose further that a and v satisfy the vector transformation law, so = /ad*. and t' 1

(22a. b)

I 11 04.

Then it must be true that Tpq — / ipf

But upon substituting (21a) and (21b) into (22a) and then using (226), we find that — { JP t

F

/ jig T2p-

To employ (16) for the purpose of moving all !'s to one side of the equation, we multiply by f 1,: / 1,/ jr„ a: Tip =

,a;Tlp

Tu.

or -

f.k,] = O.

Since a, is arbitrary, an argument used in Section 1.3 shows that the terms within the square brackets must vanish, which is equivalent to the desired conclusion. Q REMARK.

If a and b are real numbers, b # 0, their quotient alb is

defined by

a

=.c=.a = be.. By analogy, v = a • T could be interpreted as expressing T as the quotient of r and a_ Theorem 7 is called a quotient rule because it can be regarded as asserting that the quotient of two first order tensors is a second order tensor. While suggestive, the "quotient” aspect of tensor equations need not be taken too seriously. The following is another quotient rule of a rather general character. Theorem 8. Suppose that in any coordinate system, there is a rule associating a unique set of 3h quantities with T Suppose that for an arbitrary frith order tensor A, TA is a tensor* of order ni + n. Then T is an nth order tensor. The proof of this theorem and of other quotient rules will be given as

exercises _ • Sink T is opt known to be a knsut, strictly speaking TA is not defined. In any courdinase sysielll, by TA we mean the analog of t 17l.

See. 2. If Tensor Algebra

41

SPECIAL RESULTS FOR SECOND ORDER TENSORS

Definition 9. Let A be a second order tensor with components A. Then the transpose of A is denoted by A T and is defined by [AT%]IJ = A1,.

(23)

Theorem 9. If A is a second order tensor, then A' is a second order tensor. Proof. Left to the reader Definition 10. If

A — A 1 ` or A ll = A

^; ,

then A is said to be symmetric. If

A=

or A,1 = - A l; ,

then A is said to be antisymmetric (or skew-symmetric). For example, in a nonpolar material, the stress tensor is symmetric. A second order tensor may be neither symmetric nor antisymmetric, but it can be written as the sum of a symmetric and an antisyrnmetric tensor, as is shown in the Following theorem_ Theorem 10. Let T be a second order tensor. Then one can write T — S + A,

(24)

where S is a symmetric second order tensor and A is an antisymmetric second cider tensor. This decomposition is unique. Outline of proof T=+{T+ 1.T`)+

1(T —

T T`)-

( 25 )

On the right side of (25), the first term is symmetric and the second is antisymmetric. The uniqueness proof is requested in Exercise 9. G Matrices are an aid in performing certain manipulations with the components of second order tensors_ For example, in a given coordinate system [A], 1 can be written in the ith row and jth column of a 3 by 3 matrix. We shall denote this matrix by (A). Thus A l2

A,

A2z

A23

Ati2

A33

3

Note that A has a meaning independent of any coordinate system, but the matrix (A) is defined with reference to a particular coordinate system. In (18) we displayed the matrix corresponding to the second order tensor vw.

Car#eswn Tensors [Ch_ 2

42

Theorem 11_

Fer second order tensors A and B, (26)

(A - B) = (A)(B).

In words, the matrix of the contraction product of and B is the product of the A and B matrices. Proof.. The component in the pth row and qth column of the matrix (A • B) is A ; Bi . A F,Bk is found by taking the"dot product" of the pth row of (A) and the qth column of (Bl. By definition of matrix multiplication, this is the clement in the plh row and qth column of the matrix product (A)(B). f] Theorem 12. The vector transformation laws U^

Z2^

=

Duli uj,

can be written, using matrix notation, as l^ r

I' 2

(EH = /21

/l2 ^ i3

^i

/22 /23

^2

i 31

/ 33

^3

!^1

/ 32

and / 12 1 ;',

1'2

6) =

/22

(r1

/ 32

respectively As before, lei L be the transformation matrix, and let (v) be the column vector with components 07,, v 2 , t] 3 ). If a prime indicates use of a primed coordinate system and if a superscript (Tr) denotes matrix transpose, then we can w rite (v) • rr = (v) rr L (v) = L(vj', (27a) .

Prou[

Left to the reader.

Theorem 11 The transformation law for second order tensors, A ',„ni

can be written (A) = L(A)1.».

(27b)

Left to the reader. REMARK. Since tensor equations are valid if they are demonstrated in any single set of components, there is a theory for second order tensors that completely parallels the corresponding matrix theory. We have seen this in Theorem 10; the decomposition into the sum of symmetric and antisymmetric Proof.

Sr c _ 2 1]

Tc•usor Algebra

43

tensors is the same as the corresponding matrix decomposition. To give another example, one can term a second order tensor nonsingular if any of its component matrices has a nonzero determinant. If R is a nonsingular tensor. then [Exercise 25(c)J there exists an inverse tensor, which we denote by R with the property

R• R '' = R - ' i R = f_

[28)

ISOTROPIC TENSORS

There are certain materials whose structure singles out one or more directions. Examples are the direction of the grain in a piece of wood, and various symmetry axes in a crystal lattice. Other materials, like water. appear to have no internal preferred direction. The constitutive equations for materials of the latter kind must be expressed in terms of special tensors whose components are always the same. We shall now study this class of" isotropic" tensors. Definition 11_ Tensors, such as I and E, whose components are given by the same set of numbers in all coordinate systems, are called isotropic. ll follows immediately from the definition of zero order tensors that they are isotropic. Are t here any isotropic tensors of order I ?There is at least one, the zero tensor of order 1. Excluding this case ]which we shall term trivia°}, are there any others? There are not, as we shall now show. Theorem it There are no nontrivial isotropic tensors of order 1. Proof (We proceed in a way that makes up in conceptual simplicity what it lacks in elegance. We rule out the possibility that there is a nontrivial vector which has the same components in oil coordinate systems by showing that there is no such vector even if only three special coordinate systems are considered) Suppose that v is an isotropic vector and that its components are y, relative to a certain basis e l n. Consider another set of basis vectors satisfying

e

tx _

^

^z^

^

i zp

=e` , a, e" }- =

(29)

It is clear from Figure 2.1 that the components v: satisfy t•' l -

= fi 3 ,

l'2 ,

Z'3 = VI.

(30)

This cyclic change of subscripts, symbolized by l'

►

2,

2'

3,

3' -' I.

(31)

can also be verified by calculating the appropriatef and applying the vector transformation law (1.1.6). (The calculations are almost the same as those requested in Exercise I,17.)

44

Cartesian

Tensors [Ch_ 2

r fJl 0 cll

FICiURE 2.1, A

rülulron of axes

ecjuivulenf fo a cyclic exchange of subscript.

Since Y 1S isotropic, = rL,

11 2 ,

t' 2

(32)

118 = V3-

Combining (30) and (32), we Find that (33)

=112 =V3 = 1 3 =l1 } • et`l., Consider a third set of basis vectors satisfying V1

cox

y e{21, 1121" =

_

t1}

ear = ea}

(34)

f rom Figure 2.2, = i VI,

but, by isotropy, VZ = v2 .

Thus o2 : — L . This is consistent with (33) if and only if v / = v2 Consequently, the only isotropic vector is the zero vector. U

v3 = O.

Theorem 15. Any isotropic tensor of order 2 can be written 21 for some scalar A. Proof, Suppose that T is an isotropic tensor of second order and suppose that its components are T i relative to a certain basis ems. As in the proof of the previous theorem, consider a new set of basis vectors satisfying (29). By (31),

T12

— T22,

7-22 =

= T23,

T23

= 13

T3Z

- 1132

T 21 = T32

,

T33, ,,

T33` T11,

(35a)

T31 = T12 ,

(35b)

T13 =

T21

.

(35c)

See. 2.11 Tensor Algebra

45 t

[ll

, P

fll

/

F Lc; t I R E 2.2. The cJnubk-prrn:rd "

mtenr Is r o ta t ed 90° about the e' as axis. -

Since T is isotropic, . T IL

T11.

T22 - ^=2 T 33

- T33 -

(36)

From (35a) we deduce that Ti 1 - T22 - T3 3=

(37)

for some number A. Similarly, we can deduce Irons (3Sb) and (35c) that T1 2 =t T2 3 = Ti

,

•

T1

= T32

-

i3-

(38)

New consider the components ❑ f T relative to a third set of basis vectors satisfying (34). Applying the transformation law, one finds, as in Exercise I.3,8, that 1'12 =

—

f21, T31 = T32 .

But, by isotropy, 11 . ^13

M T12, T 31 — T31-

Thus T1 2 - -Tn.,

1-3 l

= T3 z -

(39)

For (38) and (39) to he consistent, we must have 712 =

T23 = ;1 = T21

—

T32

- f13 = O.

(40)

The equations of (37) and (40) show that if T has the same set of components for the three sets of axes related by (29) and (34), then T necessarily has the

Curiesian Tinsors [Ch . Z

46

form 21. But )i is a tensor by Example 1 and part (c) of Theorem 1, and Al certainly has the same set of components for all axes. [] Theorem 16. Any isotropic tensor of order 3 can be written 1.E for some scalar 1. (If reflections are allov.ed, E is not a tensor, so the only isotropic tensor of order 3 is O.) Proof. The argument is entirely analogous to that required for Theorem 15. Details are left to the reader. Theorem 17. If T is an isotropic tensor of order 4, then y qr) [T]pgrs = okr5 Urs ± ;ii(15Fr ya ± ^

+ i { b pr b gS

—

cPs qr)

(41)

for some scalars 1., /4, and Pti. Proof. Again the same ideas suffice, but the details are now somewhat lengthy. The reader is asked to supply a proof in Exercise 19. THE VECTOR ASSOCIATED WITH AN ANTISYMMETRIC TENSOR

To conclude this section, we shall present some formulas involving the alternating tensor that will be seen to be useful in our discussion of kinematics, To motivate our discussion, we start with the fact (see Exercise 21) that ifa rigid material is in uniform rotation about the origin, then the velocity v at the point x is given by (42)

r = ül A x,

where w is the angular velocity vector. in terms of the alternator, (42) can be written e,;k toi xk .

ft;

(43)

Since {43) holds for arbitrary x, the quotient rule implies that the quantities r ft cw} are the components of a second order tensor. Not surprisingly, it proves useful to regard v as given by the contraction product of this tensor and x. As will be seen in the formal development that starts in the next paragraph, it is conventional to deal with a tensor whose components are —E, ik to . With a vector CO we shall associate an antisymmetrk tensor Q defined by ^

=

^ Py r

^^

ir

^ = F_ - w.

or

(44)

Thus 4

(t2) _

— €^ 3 Wg

ca 3

^l 2

0

c), .

— (.1)1

0

(45)

Sec 2.1] T eruur Algebra

47

An explicit formula for w in terms of can be obtained by multiplying both and using the th rule: sides of (44) by

Eip4r/pq = Ei pq € pgr C+J

gipE

çç= f^ (60e5 î, - U9,üi9)(0,. =

3ui —

That is U li

= 8 i ^q ^i ^4 -

(46)

Similarly, (46) implies (44) when 12 is antisymmetric [Exercise 20(b)]. Sometimes (46) is written w = i113 _ (47 ) What we have shown can be summed up as Follows.

Theorem 18.

Corresponding to any antisymmetric tensor 13 there exists a vector to [given by (46)] such that 12 = - w. Here w is called the vector of the anuisymmetric tensor O. Corresponding to any vector w, there exists an antisymmetric tensor 12 [given by (44)] such that (46) holds_ We motivated our present discussion with some remarks concerning rigid body rotation. In this connection we can now draw the following useful

conclusion. —12 • x fct some Theorem l9. If a velocity vector v satisfies v = x • it antisymmetric tensor Q that is independent of the position vector x, then the motion is a uniform rigid body rotation about the origin with angular flpt . velocity w given by ta i = Proof. From Theorem 18, do) is defined as above, then Up =

xg f2gp = Xi? çpr w r —

'eprQ tt: r

X4

re)

A x]p.

Comparing with (42), we see that the velocity is that of a rigid body rotation

with angular velocity w. D REMARK. Later [see (4_1_33)] we use a result analogous to Theorem 19 to show, in essence, that infinitesimal displacements of the form u = S2 • x arc rotations. E XERCISES

1,

Prove Theorem I, parts (b) and (c).

2. Prove (9). 3. Verify (10), (11), (12), (13), and (14)_ is a 4. Prove from first principles that if T is a second order tensor, then scalar. 5. Prove Theorem 5: (a) for n = m = 1; (b) for n 1, rn = 3; (c) in general_ 6. Prove Theorem 8: (a) when m = 1, n = 2; (b) in general. 7. Prove Theorem 9.

Corleswn Tensors [Ch. 2

4^

relative to 8. Let T be a second order tensor with components T. and two different hases. Prove that (a) Td : T, implies that 7'1 = (b) 1; = — 7; ; implies that r ; 9. Show that the decomposition (24) is unique. 10. Prove (a) Theorem 12; (b) Theorem 13. 11. Suppose that for any Cartesian coordinate system there is a rule for associating T with 81 components 'C ity Suppose that for arbitrary vectors v and w, v • T w is a second order tensor. Prove that T is a fourth order tensor. 12. In a certain Cartesian coordinate system, the components of the vectors v and w are (1, 2, 3) and (4, 5, b). Find the components of vw. 13. Give a geometric interpretation of Example 3 in terms of the triple scalar product (see Example I.2.1). In particular, state why reflections must be prohibited. 14. in any Cartesian coordinate system there is a rule for associating T with components Ti such that T ; = Tfl . For arbitrary vectors a, a • T • a is a scalar. Prove that T is a symmetric tensor. 15. Let L be an entity whose components with respect to any unit vectors e[ 'l are given by [L]1 = ut`I• etrr. Here in'L,u[ 21, and u13 l are anorthonormal set of vectors that are regarded as fixed, while components of L are computed with respect to various coordinate systems_ Is L a second order tensor? 16. (a) Demonstrate (20). (b) Specialize (2Db) to the case where T is the stress tensor. Show that the original interpretation (Section 14.2 of i) of the components of T is recovered. 17. Prove Theorem 16. 18. What theorems and definitions show that the right side of (41) is an isotropic tensor'? 19. Prove Theorem 17 in the following manner (Jeffreys and Jeffreys. 1962, Sec. 3.031). (a) By considering special rotations, as in the text, show that ,

14 1111 — U2222 = 11 3333

^

CI.

W1122 = "2211 = 11 4233

11 3322 ! 14 33[1 — "1L33 " t 2+

"2323 = "1313 = "3L31

u2121 = 11 L212. = U 3232 — C3,

"2112 = W2332

1.1 3223 — 11 3113 = 1411331 a c4 ►

"1221

—

so that one can write

[T]

s = C 2 va

rs

+ C 3 e p, b y +

+ (C l — L.2 — C3 —

C 4)L•'pgr,,

where laps = 1 if all subscripts are equal and equals zero otherwise. „

Set. 211

The l:r,yerwakee Problem

49

(b) From the fact that T „x v x4,Xr Xs must be a scalar if x is a vector with components x , (why?), deduce that c, — r 2 — e 3 e4 = D. 20. (a] Verify (45). (b) Show that (46) implies (44) when .0 is antisymmctric. 21. Deduce (42) from Exercise I _ I.6(a). $22. Let To and £k be components of second order tensors_ Suppose that for arbitrary rpm iij

ilkmFkm-

(a) Show that if 7 is symmetric, then CI?,, Cjam. (b) Suppose that the E„ are the components of a symmetric tensor and that the CIA ., are the components of an isotropic tensor. Show that ;.=

2

E0

+ ft kk & i

for certain coefficients 1.4 and A. 23. True or false: If ecri r gives the components of et'k in a primed coordinate system, then (20c) implies that T- = etr}' T ,pg e rr

-r

9

Explain your answer. 24. The tensor product of two vectors is called a dyad_ A sum of such products is called a dyadic. 1(a) Write the dyad ab as the sum of symmetric and asymmetric terms, t(b) Let e( ') be a set of mutually orthogonal unit vectors. Show that (c) Try to state and prove some little theorems involving dyads. (d) Show by construction that every second order tensor is a dyadic. 25. (a) If U is a nonsingular symmetric tensor, show that U - t is symmetric _ (b) Show that (A B} = BT- AT . (c) Prove that a nonsingular second order tensor R possesses an inverse so that (28) holds. (d) Show that (C • D) - 1 = D - ' - C - ' `

2.2 The Eigenvalue Problem In this section we shall be concerned primarily with the problem of determining the eigenvalucs associated with a symmetric second order tensor.* We shall thereby discover a coordinate system in which the components of such a tensor assume an especially simple form. The eigenvalue problem turns out to be one of the most fundamental in all mathematics, but it is at first not

•

Many readers will he familiar with the eigenvalue problem fur square matrices andlor linear transformations. They wits Find tittle new in this section.

Kb, 2

Cartesian rr'nsurs

50

clear why one ought to consider it at ail. We can, however, find some motivation in the sort of investigation that moved Cauchy to begin the study of eigenvalue problems, namely a deeper study of stress. The relation between the stress vector t and the unit normal n is typically that shown in Figure 2.3(a), (We consider a particular point x and time t throughout, so we do not explicitly indicate the dependence oft on x and t.) It is not unreasonable to anticipate that there are directions n such that the corresponding stress vectors t(n) point in the n-direction, as in Figure 2.4(a). In such a case t(n) — do for some scalar A. Because tl — n) _ — t(n), a small •' flake" whose relatively large flat surfaces are perpendicular to n is subjected to a particularly simple set of normal stresses [Figure 2.4(b]]. The stresses simply act to extend the flake, or to compress it if A is negative. Compare Figure 2.4(b) with Figure 2.3(b); in the former there is a shear stress component of t perpendicular to n. This can perhaps be seen more clearly in Figure 2.3(c) in which the stresses have been resolved into components that are respectively normal and tangential to the flake. ti

tal

1

tb}

k)

2.3. (a) Typical refarrorPslirp between th e unit exterior aorr+ta l n and the stress rector t(n). (h) The stress rectors on opposite /ares aj' a small thin "flake" are equal in magnitude and point in opposite directions. (c) Stres.s vectors (dashed arrows) are resolved into components of shear (along the surface of the N1► e) und r.vtension (normal to the flake). FIGiJRe

Pure extension is a relatively simple state of stress. It is worth investigating whether such states generally de exist_ We thus ask ourselves the question: "Are there any directions n such that 1(n) = An for some scalar A?" Using the relation t(n) = n • T we can pose the question in terms of the stress tensor T: Given T, find n such that n - T = An. Indeed, if we can find any vector v satisfying v • T = ;ry

( 1)

we shall have virtually achieved our objective. We merely need to divide both sides of (l) by lid to see that r/Jv) is the desired unit vector n.

Sec. 2.2] The Eigenualue Problem

51

tl+tij - hn

1(n)

1( (J)

F1€;t3RE

2:4. (a)

(b)

Th e 3urftwe element with direction n suffers u purely normal

stress.. (t}) A flake that suffers purely extension, not to shear_

normal Stress will be subject only to

Let us select a particular set of base vectors 0 1 , In ternis of the associated components of v and T, CO becomes

o; T = "iv) or

v; }; _

(2a, l})

,

Fquurioris (1) and (2) are alternative sraterrterirs of the rnathernaticai problem that wewisii todiscriss Note that throughout this section by tensor" we shall understand "second order tensor." Writing (2b) as .

26,j ) = O,

(3)

we see that it can he considered as a set of three equations for the three unknowns r; - These equations clearly have the trivial solution y 1 = O. This solution is unique unless the determinant of coefficients vanishes. For a nontrivial solution to exist, therefore, it is necessary that the determinant of coefficients be iero, i.e., li

fJ^.`t

CI; —

--

A

TL 3

T12

7 :21

T22 -

T3]

T32

A

723

T3 3

;

O.

(4)

A

Expanding the determinant in (4), we find that --J. 3 + 1,1 2 - 1 +1 3 =a,

(5)

where l i — Ti.. z

T11 12 1

(6a)

T1 2 ^22

1 3 = dot (74).

+

Tl 1

T13

431

T33

+

T.2 2

3

T32

T3 3

(6b)

()

Cartesian Tensors [Ch. 2

52

Equation (5) is called the characteristic equation. Its roots (the zeros of the characteristic polynomial) are the eifienvalues or characteristic values of T. We shall denote them by A" ), i = I, 2, 3. The cigenvalues satisfy (1), a coordinate-free equation, so that they must have the same value no matter what coordinate system is used. The coefficients in (5) must therefore be scalars. This can be verified directly. For example, consider the first principal invariaat 1 , also termed the trace and denoted by tr T. This is the contraction of a second order tensor, and hence is a scalar (Exercise I.4). The coefficients of the characteristic equation can be economically characterized. To do this, we need two definitions. The principal dingoes' of (Ti j) consists of T11 , T22, and T33. An mth order principal minor of a third order matrix is an rn by m determinant formed by deleting the row and column of 3 — m terms in the determinant's principal diagonal. Thus the first, second, and third order principal minors of the matrix l 4 7

2 5 8

3 6 —9

are the determinants

^ll = 1, 1 5 1= 5, 1 - 91= —9, 3 7 1 —9 '

IS8 —9 '

1 4

2 5 '

and I 4 7

2

5 8

3 6 —9

respectively.

Theorem 1. The principal invariants I I , 12, and

equal, respectively. the sum of all different principal minors of orders 1, 2, and 3. Proof. We merely verify by inspection that the determinants in (6) are principal minors of the required order. p 13

Corresponding to each eigenvalue An», we can determine a nonzero eigeovector or characteristic vector v that satisfies (2). We denote the eigenvector corresponding to Al°1 by vtrl. With this notation, (2) reads t,4P 7 = xiri vl rybij (no sum on p). vrTii = Ali4v114 or (7a, b) Because our discussion was motivated by means of the stress tensor, we began with the eigenvalue problem (1) REMARK.

v •T= kw,

Sec. 211 The Eigenraahse Problem

53

but frequently there arises the related problem

T•w= .^w. It is easily seen that the characteristic equation corresponding to the latter problem is the same as (4), so one can speak of the eigcnvalucs of T But the equations that give the components of the eigenvectors are not the same in the two cases, so one must speak of Left and right eigenvectors of T. In the important special case wherein T is symmetric, one refers to the eigenvectors, for then the right and left eigenvectors are identical. Exemple I. Find the eigenva Ries and eigenvectors for the symmetric second order tensor T whose components in a certain coordinate system are I

0

0

^ 4

9

0 0

Solution.

-

4

In this case (4) becomes l

0

0

—A Il 0 4

A da 4

t]

4

=0

(8)

9 — ►i 4

or

0—(l —A)(

^

—h)^4 —

—6 1 (l—^l=(i— i^)(2—).1t3 —dl,

SO

Am=1, A''=2, A''-3. (Each individual can choose for himself which eigenvalue he chooses to designate kto, which ,1(21, and which dr31 .) When A _ At') = 1, the three equations of (7) become 0=0,

+ ^^^' t^' =o

.

-

3 r = 0,

^ _ at

(42"

Equations (9h) and (9c) have the unique solution 4" = t` ` = 0, so

1,

for any nonzero constant a

ut

( 9 a, b, c}

Cartesian Tensurs

54

[Ch. 2

When 2. = AI" = 2, (7) becomes

- n^r > _ {),

_

o, (43)42]

2f + (1») = 0. (1 a, b. C)

(-%(2)tl'

j

[The rosiest way to obtain the matrix of coefficients in (10) is to make the stibstiIUtian A -= Al21 in the matrix whose determinant appears in (8).1 Multiplication of (10c) by -

gives (10b), so (t0b) and (10c.) are dependent. We find that vS21 =

=0,

b, ll21 fb,

for any nonzero constant b. Thus 1.{21 = be {21 +

I31

When A = 1131 = 3, (7) becomes - 2 0»

T

U,

-

1 u,] f

4

—e^-)v^]} — (-)0311 = 0.

= 0,

—

which yields ^

{]I

= — V eeS^1 # C4 1]1 ,

for any nonzero cons t ant c. (By deflnition,eigenvectors arc nonzero, so e = O is un accepta bte.} If we want our eigenveclors le be of unit length, we can take V II)

=

e I^1 .

v121 =

^1^1

2

^t

+^

Observe that the unit vectors

2

{31,

^

1]^ - – 23 ^

– —

^^1 + 2 e1!}

(li ^

of (11) are mutually perpendicular: l,t11.

Yo

This is not an accident, but it is connected with the symmetric character of the particular matrix whose eigenvalues and eigcnvectors we just determined. For comparison, here are the results of a second example. Ex*mp!e 2. Consider the second order tensor whose components in a certain coordinate system are 1 t}

0

fl

0

1

t)

–4

U

Determine a set of eigcnvectors and discuss their orthogonality_ Partial solution. As the reader is asked to show [Exercise 3(a)1, the characteristic equation is (1 —2.)(A.' 4 4) = 0.

Sec. 2.2! The Eïye►rralue Problem

55

so ;he cigenvalues are a. 0 = t,

„p i]

2i,

dit3i =

with corresponding right eigenvcctors ue 1L

vr2t

= b&2

f 2jbe''f,

11°1

T

real

—

i^! ZiC+C^ .

and L are any nonzero constants_ Note that there are complex eigenvalucs and that the corresponding eigentectors have complex components. When complex components are possible, one usually takes the scalar product of vectors to be where a, h,

{r, w> - L

,

(12)

where an uverbar denotes complex conjugate_ This definition preserves the property of nonnegativc length: (13) Bul

n ow (w,") = (w. i>.

(14)

(Fer real vectors, v •w = w• v_) Whatever scalar product we use, v (21 and v'" are no longer or ihogonal, i.e., their scalar product is not zero For

v42 '. N 3 ' = 6e• + 4hr. # C; 4 on the fluid in the domain x 2 < O. This stress component. 72 1 , is the same all akwig the plane s 2 = I]: for definiteness we consider the w1 rests at the origin.

In Figure 3.2(a) this stress is denoted by a heavy arrow. Let us now consider a new flow wherein the velocity field is rotated counterclockwise by 9G [Figure 3.2(b)]. if we perforrn clockwise rotation of the tarok, Figurc 3.2(b) appears identical to Figure 3.2(4 Because the material is assumed to he isotropic, the correspondingly rotated stress [heavy arrow in Figure 3 2(b) I roust be identical with the original stress. The reason is that since there is no preferred orientation in the fluid, one cannot distinguish between an urtrotated look at the vciocity held of Figure 3.2(a) and a rotated look ai the (different} velocity field of Figure 3.2(b). Equivalent to rotating the book, in order to view Figure 3.2(b) sideways, is a 94" clockwise rotation of aptes Thus, using primed axes, the stress component T'2, in Figure 3.2(b) must be identical with the cotrtponent 72 1 in Figure 3.2(a). Similarly, the velocity gradient v'1 , 2 in Figure 3.2(b) must be identical with v 1 _ 2 in Figure 3.2(a). But from the respective constitutive equations for the two flows 1(13) and its primed counterpart] 721 —

T21 — C sI2e 2

C31t1v1.2•

,

(17)

Since we have lust seen that ^21 =

%t,

ut.t =

it follows from (1 7) that • 2112

= C2112.

5incc (13) must hold for arbitrary velocity fields, repetition of the argument we have just given for various special cases gives the result that for all , j, k, and 1,

Cffia =

That is, the tensor - in (13) with components

Cud must be isotropic

DERIVATION OF THE FINAL. EQUATIONS

According to Theorem 2.1.17, since the C,,kt isotropic fourth order tensor, we can write

are the components of an

Mfrs = ^%ij ars + /41:5ar45js + a;s ( jr) + 1 f (6 ir ^ js y 451s 6 jr)

for some scalars it, ti, and

K. It is

(18)

easy to see that the symmetry requirement

(14) holds if and only if K = 0 [Exercise 1(a)]. Now, using the decomposition

The Nai•ter Siokrs F"quarirms

Ser. 3. 1]

83 t2

= ritjt ^j

t2 .

Ob

!

3.2. (a l With the given velocity profile, t h e heavy arrow represents at utiyiR. on lo n e r (shaded) material du e to action or upper iurt,sltaded) material. (b) A 90' cYrunterr•Irki a•zge rotational the vault ypro frle hit ia )yit -es xhe profile here. The effec t cdi th e unshaded region on rire shaded results in a eorrresfJr+RdiruJi}• rotated .stress ( lrerr r ► arrow). This is because the material is Isotropic : orfiervise, resistance to dejerrmarinan would in general depend an errr eerteuton, so that a rotated velocity field would not far rise to a rotated stre ss. FlcURL

Strum

viscous fluids [Ch. 3

84

of Lr into its symmetric and antisymmetric parts Dr_, and — 4-ilrs Drs +

COrs Wr, ,

But with K = Q, (18) shows that C ur, is symmetric in its last two indices and we know that F'rs is antisymmetric — so the second term in the preceding equation is zero [Exercise 1(b)]. Therefore, in its final form the constitutive equation shows that the stress tensor is determined by just the symmetric part D of the velocity gradient tensor. This final form is [Exercise 1(c)]

1 îj

(—p + h11rr)"i1 + 2pDi J •

( 1 9)

The viscosity coefficients ). and p depend on the density p and temperature 0 (or any other convenient pair of thermodynamic variables). In direct notation, since D r, = v, , we have T = [--p

A(V - v)]! + 211)-

(20)

When our constitutive equations are substituted into the equation for linear momentum conservation (A2.l.2), we obtain [Exercise 3(a)] the ltiavier —Stakes equations P

If d1 and

p

Dr = pp — Vp + Di

• v.) + V • (2p]) !

(21)

-

can he regarded as constants, (19) simplifies to Dv

+ PL+ p !)1 _ pf + V[- ` p

v7 +

+',

(22)

where [Exercise 3(b)] ^ EV V/s

^

ü{

2 [^ t •' i

ex.

Dx 2

Pxj

[^;^,

(23)

In the incompressible case, when V • v — 0, we obtain simply p

f

= pf

Vp +

^, ►

(24)

while (20) becomes T = —pl + 2pD.

(25)

It is worth commenting on the fact that our analysis [which follows Batchelor (1967)] has shown that the stress tensor does not depend on the rotation tensor W If a more general relation for the stress tensor were sought (for fluids exhibiting more complex behavior than water and air), then this lack of dependence of W could not be deduced. But W still does not enter the

Sec. 3.11 7he11avier StokesEquemiruu

165

constitutive relations that are traditionally postulated. This is because of the assumptive ui material indifference-- that constitutive equations are the same for all observers. Observers who are rotating with respect to one another would see different solid body rotations but the same deformation, so the assumption of material indifference will not permit dependence of T on U hut will permit dependence on D. Written down in a precise mathematical manner-, the observer invariance' requirement that underlies the material indifference assumption is used to restrict more complicated constitutive equations. See Jaunzcmis (1967) for an exposition of this type of analysis. "Sufficiently small" deformation rates were required for the validity of the linearity assumption (13), but it can be argued that this means merely that strain should he small over the very short time that is characteristic of molecular motion. Thus it is perhaps not surprising• but in any case it certainly is true that there is no reason to doubt the validity of the Navier Stokes equations for common fluids such as water and air except under the most extreme conditions.* aoUNüArcY CONDITIONS

Equation (24) differs from the in viscid hauler equation 02.1 .6), owing to the in the former of the term /W . Addition of this term has the important mathematical consequence of increasing from one to two the highest order of spatial derivatives present. 1f experience with ordinary differential equations is any guide, a new boundary condition is called for (beside the requirement that fluid does not penetrate the boundary[. This view is perhaps reinforced by a feeling that the adjacent viscous fluid adheres to the knife blade as it slices through the cold molasses, Surely, the shear stresses present in a real fluid must somehow restrict the free slip along a boundary allowed by inviscid flow theory. But might not the fluid slip over a very thin stagnant layer next to a fixed wall? Could it be that the tangential velocity is proportional, via a very small constant, to the normal velocity gradient? Maybe molasses does not slip past the knife, but what about a fluid like mercury in contact with a glass wall? (Mercury does not wet glass.) What about violently eddying turbulent flow? What about gases, particularly gases of low density? Such questions have been a subject of research for over 150 years. To name only prominent nineteenth-century scientists, the investigators included Coulomb, mb, Navier, Poisson, Poiseuille, Couette, Stokes, and Maxwell. presence

• It has, in Iaet. recently been shown that an appropr.atc approach to viscous equations through kinctx theory to subtler matter than had previously been supposed) flocs not yield the Neuter Stokes equations in t e strictly two-dimensional raise [l. Opperthcim and T Keyes, Phys. Rev. A7,1384 (1972j. and ë, 100 (1973)]. There are no such unexpected results in three dimensions, however: thus even normal two-dimensional hydrodynamies is "sa4ed." roc such calculations are meant to provide a first appru,imatwn 10 three-dimensional problems x bun venation in one direction is small.

86

Viscous Fluids

[CPI 3

It is now generally accepted that for liquids and gases under "ordinary" conditions it is appropriate to assume that fluid adheres to a rigid wall. To be precise, let x be any point on the boundary of a rigid body immersed in fluid, and let x be a point in the fluid. Then the velocity v(x) will be assumed to satisfy the adherence boundary condition .

]lrn w(x) = v(x w)

,

(26)

where v(x,,) is the known wail velocity. Evidence for acceptance of this boundary condition includes direct microscopic measurements of the extremely small velocity of particles near rigid walls. Unless they are interpreted with some subtlety, however, direct measurements tend to answer questions about the molecular model of fluid flow. What interests us here is not molecular questions, and certainly not the question (assuming it is meaningful) of whether fluid "really" adheres to the boundary. We wish to he assured that the adherence boundary condition is an assumption in accord with the rest of the assumptions involved in the continuum mode! of fluid flow. Such assurance comes indirectly but nonetheless convincingly from the ever-increasing number of instances wherein viscous flow theory, with [he adherence boundary condition, provides results in accord with experiment. Nonetheless, under "extreme" conditions the correct boundary condition remains in doubt.* IIVCOMPRESS1t3I,E VISCOUS FLOW

Except for a few remarks, in our subsequent discussion of viscous fluid flow we shall assume the fluid to be incompressible. (Indeed, we shall usually take the density to be uniform_) For an incompressible fluid of specific volume i. - '

Dn

Dt =

1 Dp Di =

so the term pD(p - ')/Dr does not appear in the fundamental thermodynamical equation (14.4.12) of L Thus the pressure never arises in thermodynamic considerations. Re-salts concerning the pressure which are traditionally deduced from considerations of equilibrium thermodynamics do not necessarily apply to the hydrodynamic pressure. For example, there is no a priori reason to assert that the hydrodynamic pressure in a moving viscous incompressible fluid can be interpreted as the mean normal stress on a stationary wall. Nevertheless, that this interpretation is, in fact, valid * For djsc;ussinn and further references, see Serrin 11959, p.240). An mterestatg account of earlier work can be found ita CïoIdslcin 11938, Vol. 2, pp. 676-80). An argument from kinetic theory that yicids t he no-slip boundary co=idil ion, plus further intalesling comment. is pm, on pp. 83-84 of K. E. Meyer, bur odurrion io Me)hema rim! Fluid Dynamics (New York : ury,1971).

Sert. 3. f ]

7 he Novier-Stokes Equations

87

can be deduced from the constitutive equation (25) and the adherence boundary condition (26) (Exercise 4). Consider a viscous fltaid of uniform density p. Let us divide the Navier-Stokes equation (24) by p and make the definition

v4_

(27)

p The resulting momentum equation, and the appropriate equation of mass conservation, are

Dv

1 f — p gp -

—

D1

V-V

vVY,

(28)

0-

(29)

These equations will form the basis of our discussion of viscous fluid flow. We can regard vas a constant, so we have four equations for the pressure and the three velocity components.` The dimensions of the viscosity p and the kinematic viscosity y are as

follows (Exercise 6): dimensions of p = mass/length-time -,

(30a)

dimensions of v = length L ftime.

(30b)

Typical values for p and y are (at 1 atm pressure and 20 °CJt

:

p (g/cm-s): glycerine = 9, water = 10 -2 , air = 2 x 10 v (cm 2 /s): 2 , air = 15 x 1(3 water = 1

a2

Note that the effect of viscosity is represented by the term proportional to v in (28) and that yis 15 times as large for air as it is for water. The qualitative variation of p and v with temperature is depicted in Figure 3.3 for water and air. (Since the density of water is very nearly l gJcm` over the entire range of temperatures considered, v 12 for water when cgs units • One might be icinplcd to pose as a fundamental mathematical problem: "Given initial values of p and r and given the morion or any solid bvuneur ies, solar US l and 129) subjetl to thr boundary condition ( 26)... A comparable "fundamental problem - of purr rnathrrnai cs would be "characterize the conseslucnt:es of the group axioms," fn both c u.c the incredible richrre55 of phenomena emt]raced by the cr{uatiuns or axioms precludes the solution of such pt claims. In both cases, taste and judgment must be used to select trilerestrng or wolthwhrle phenornrna and to formulate and coke problem, or to prove theorems that will elucidate such phrnom€na. The more "applied" an applied mathematician is, the more he will rely on relevance to a sr.:lei:Idle matter as a criterion for dcterminirt$ which phenomena arc inletcstuil;. t An essential ingredient of the physical "feel required to docffrciive applied mathematic-5 is u knowledge of the rough magnitudes of the various parameters Char appear in the ovations_

68

t'isruus Fluids 1C i. 3

,u on (tags LIMOS.

I

atmosphere pressure'

FIGURE kine»kitit

3 3.

Appro.—tit—tame rYlriutluN 1+•02 temperature rrf viscosity (0) und for air and w a t er- Data from C'v1rL►tNrn (1938}, pp, 5-1,

are used.) There is very little variation with pressure. See Goldstein (1938, Chap. 1) for further details and additional references. At the moment, the viscosity coefficients pc and v may seem to lack a clear connection with our intuitive feeling for viscosity. This connection will be more firmly established below when we indicate how these coeftïcierits can be measured by studying viscous flow in a pipe [compare (23)] or the passage of a small sphere through viscous fluid [compare (6.11)]. Further clarification comes via an exact solution (2.2a, b. c) to the Navier Stokes ouations, a shear flow in which the only nonzero velocity compnnent u is proportional to the nnrrnal direction j':

u = ky,

k

a constant.

In this case the sheaf- force per unit area T} , is proportional to the velocity gradient via the viscosity:

Historically, the viscosity coefficients first appeared when such a rather natural proportionality was assumed. The tensorial Navier-.Stokes assumption (25) came much later, EX ER CI5F:5

L. (a) Show that (14) implies that rt — 0 in (18). (C,,u and l rs -=-14 implies that C ars Ws — O. (b) Show that C# (c) Duce (19) from (13) and (IS), with x = 0_

Sec_

1.11 The !Warier-Stokes Equations

Kq

I Show that (20) implies that the principal axes of T and D coincide. Would you expect this to remain true if the linearity assumption were dropped? If the isotropy assumption were dropped? Back up your answers with physical reasoning. 3. (a) Verify (21). (b) Verify (22). 4. From (25) and the adherence boundary condition, deduce that the magnitude of the normal stress exerted by a moving viscous incompressible fluid on a stationary waIl is equal to the hydrodynamic pressure p_ 5, Let q i denote the ith component of the heat flux vector q. Give reasons for assuming the constitutive equation gf i cii 0 j , where 0 is the temperature and the quantities ri, may depend on thermodynamic variables. From an isotropy requirement deduce the Newton-Fourier law:

rcV[i.

q^

ti, Derive (30a) and (30b) from (2 5) and (27). .

The object of the next set of problems is to deduce certain facts about energy changes, particularly as these changes are affected by viscosity. A knowledge of Cartesian tensors is assumed. 7. From the momentum balance equation (A2.1.2) deduce the kinetic: energy equation (14.4.32) of I:

fff, ,„ d _ JJj'pr. t

R(i)

cit +

-vdd— JJJT:D ctr.

RIO

BO)

where (using the summation convention)

T:D=

f 1:1{ .

Section 14.4 of 1 contains interpretation of this equation and several deductions from it. [We assume that T,, _ TA ; R(r) is a material volume.] S. (a) Modify the equation of Exercise 7 in the case where the body force f is given by f = — VP and the function P is time-inde pendent_ P is called the potential energy per unit mass or specific potential energy. Give a justification for this terminology by interpreting the equation just derived. (b) For an incompressible viscous fluid, show that

T: D = 2nD : D.

(31)

The quantity = 2AiD : D is called the viscous dissipation (per unit volume) (compare Section 14.4 of I). Note that this dissipation is nonnegative.

Viscvi13 Rinds

ye

(['h. 3

$9. The viscous dissipation vanishes if and only if D , U. Show by integrating the resulting partial differentia! equations that D = U if and only if W loY lot certain constants cuj and Ir. T what motion does this velocity field correspond? is it reasonable that it is the only viscous incompressible motion with no dissipation of energy? 10. (a) Prove that

D(V • v) Dr

V-a =

(h)

l

D:D

2 1w l ^ ,

-

(32)

where st E DvfDt and to is the vorticity vector V A v. Use (12) to show that for a viscous incompressible fluid.

fJf

(I)

dr

ff a - n eta_

riwI l dr +

(33)

If the boundary dR is composed of rigid walls on which•v vanishes, deduce the following remarkable relation between the dissipation and the magnitude of the vorticity vector: J]JtI) dT = J.L Jf

(34)

= /21E1)1 2 . LWe have here E. It is not, in general, true that followed Serrin's (1959) presentation of results obtained by Hobyleff and Forsythe toward the end of the nineteenth century.]

N n -I

The next three exercises outline a derivation of the equations of twodimensional viscous flow that makes no explicit use of tensors_ 11. (a)

Verify that the following equations describe a two dimensional velocity field at any point y rear a fixed reference point X. -

2

vAy) — VAX) =

E{Y# - x k )

[^Li

+ higher order terms

2

;

E Ûr„ — xkP ,k t

2 + ^=

I

(}'k

-

-.1q)44fk

+...^

1 — 1, 2,

where 1c? 4^

D ^`

eV,

2 dx ; + r^x r ^'

I^;,

=

t —

^^k

2 ax,

^^i

^ c^x k

S'a'c". â_r J The Nai ler Stokes Equations

9 1.

In the equations above, rau,{Ox k , Dtk , and I^I; at-e Evaluated at the fixed reference point x. (b) Suppose that higher order terms are neglected. Show that if D a, tl, then the velocity field is that of a material that rigidly rotates with constant angular velocity while it simultaneously translates with velocity v(x). (Compare Exercises 9 and 4.1.5.) 12. The object of this exercise is to determine expressions for the components corresponding to the Die, in a primed coordinate system rotated by an angle t} with respect to an original unprimed system. so that

x a =x; cos[ --x r2 sin B, x; =x, cos (a)

0 t- x 2 sin 0,

x 2 =x', sin f1+.tie cos 0; xz— —x, sin t)+x 2 cos O.

If velocity components in the unprimed and primed systems are denoted by (t;,, r: 1 ) and (r;l, r/ 2 ), respectively, verify that

=

u' r.x i

cos 0

(X 1

(e l cos tl +

12 2

sin 0)

+ sin 0- •(--t sin 1 0+r•2 cos 0) 1 —^

J (D 1 1 +D1a)+- -(D rl — D22 } cos 2U+D 13 sin 20.

2

(h) Note that the result in (a) is identical in structure to the transformation law for T1 1 which was derived in Exercise 14.2.8 of I. Show that the identity in structure continues to hold for the transformation laws giving the components of D; 2 = D21 and D. (Section 1.3 reveals the basis for the coincidence" in the formulas.) 13. In an inviscid fluid Ti =_ -p ; - For a viscous fluid, additional terms are needed on the right side to represent a local stress corresponding to a local shearing motion. Exercise 11 shows that the WI, terms in the local description correspond to a rigid rotation, so it is sensible to assume that there is no shear contribution from these terms. As for the remaining terms, the D ; ,, the simplest assumption (presumably appropriate if derivates are small enough) is a linear one. Thus we assume that

7^ r = — p+A1D1r 7;2 —

T22 — —p

A 2 D 13

A ri D12 ,

(35a)

B 1 D I1 + 133D,2 f HiD11,

(35b)

+ C~2Dr2 -i- C73D22.

(35c)

This exercise outlines the consequences of the isotropy assumption, which maintains that in fluids with no preferred direction the relations (35a, b, c) should be the same whatever coordinate system is used.

Viscous Fluids [(_h.

92

3

(a) Using the results of Exercise 12, substitute for D', ,, ❑ ', 3, and D P22 in r12

= B1Di1 + B2 D i e + B 3 D '22

[8,, B2, B 3 as in (35b)]

and deduce that

( 36a)

{D1 s + D22)( 8 1 4 B3) = 0,

1(7'22

—

T11) = iB2(D22 — D11) + (B,

B3 )0 12=

(36b) T12 ' B2 D12 f I(D11 — 1)22)(B1 — B3).

(36c) (b) (c)

Combining (35) and (36), deduce several relations among the A's, B's, and C's. Repeat the above steps, starting with

fi l = - p +

41011 f A2 1712 + A 3 D'22 .

Duce that A2 — 81 = 83 = C2 = 0 ,

and that one can write C 1 = A3 = A,

8 2 =2 1,

C 3 —A2—Â

+ 2p.

(d) Assuming that  and p are constants, use the results just obtained concerning the dependence of the stress on velocity gradients to deduce the Navier-Stokes equations (22) 14. Consider the following steady flow (already investigated to some extent in Exercise 1114 of I). L'1

(a) (b)

=

2x 1 + 3x 2i

V2

=0

Find the rates of deformation and rotation. Determine the principal axes of the deformation tensor and ils diagonal form. Consider a circular disk centered around the origin, of radius 1What becomes of the shape of this disk after a time interval 0.1 unit? Plot the curve of the new boundary. (Assume that the time interval is sufficiently small so that one may calculate the displacements by Ei[ 1 = u, ❑f,

(c)

X1 — Xi,

Ax e = v 2 ❑I.

Consider points on the circle bounding the disk.) Describe the relationship between the results obtained in (a) and (b):

State clearly the deformation and the rotation involved. Verify explicitly for at least one boundary point.

Sec . . 3.21 Exact Solutions

93

15. In a certain Cartesian coordinate system, a two-dimensicjrial steady flow has the velocity components

v, = 2x t + 3x 2 , n2 = x, — x 2 . (a)

Find the components of the following three tensors: velocity gradient, defntmation, and rotation. (b) Consider a set of material points (x,, x 2 ) that at time r = 0 lie on the unit circle x; + x = I. Find the locus of these points at time r =F 0 <E L Proceed as follows. Let (x,. x 2 ) —, (t. ,, 3 .2) in the time interval (0, r:). (1) Justify the assumption ,

y, = x, + (2x, + 3x 2 )e,

y 2 = x, 4 (x1 — x 2 )c.

(ii)

Use perturbation theory to solve for x, and x, in terms of r, and y 2 . Neglect OW) terms. (iii) Impose x; + 4 = I. Neglect OW) terms. (c) Using elementary methods, one can find a new rotated coordinate system in which the result of (iii) becomes revealed as an ellipse in standard form. Write down this standard form Manna fulrcrtcitwurr using part (a). Explain your reasoning.

3.2 Exact Solutions We begin our study of t he Navier-Stokes equations for a viscous fluid of uniform density by displaying three exact solutions (obtained without approximating the equations). We shall briefly discuss the use of such solutions in determining the value of the viscosity p. Since no terms are discarded as "negligible," there is no question that exact solutions represent one possible behavior allowed by the equations. It is natural to be skeptical about whether this behavior is typical. In response to this skepticism, one can only say that experience indicates that exact solutions which can be interpreted as solving "reasonable" physical problems generally illustrate behavior that is broadly typical of other physical situations which cannot be exactly analyzed. SOLUTION t: PLANE COUETTE FLOW

Consider viscous incompressible fluid contained between infinite parallel plates a distance d apart. Suppose that the bottom plate is stationary and that the tip plate is moving with a uniform speed U. Solution of this problem should give us an idea of the effects of the adherence boundary condition and of how tangential viscous stresses drag the fluid along. We choose an x-v-z Cartesian coordinate system. The components of the velocity vector will be denoted by (u, Thus the situation is as depicted

Viscous Fluids ICh. 3

94

Y

=d

F

r

u=

^

4 I

Y

=o =D

t

f z

Fit; v H is 14. The configuration that generates plane Couertefi"ow of u viscous fluid. The fluid is contained between two infinite parallel plates. One of these is stationary and dhe other »ores with speed U.

in Figure 14. Seeking the simplest possible solution, we ignore the effect of body forces and set f 0_ (Gravity, the most important body force, only adds a hydrostatic pressure in any case; see Section 7.2.) Since the plates are infinite, it seems reasonable to look for a solution in which the only nonzero velocity component is u. It is also reasonable to guess that u is a function only of the vertical component y.. With the assumptions u = w = 0 = h(y), mass conservation is automatically satisfied. With f = O the t^ momentum equation (1.28) implies that

_P s -

p+ vhrr= 0,

pr = 0+

A:= 4.

(la, b, c)

From (lb) and (1c) it follows that the pressure can depend only on x and t. Since h is assumed to depend only on y, we deduce from (lb) that p x and her must be constants. [Our reasoning is the same as that used in separation of variables; compare the material following (15.2.3) of I.] We consider a flow due entirely to the effects of tangential viscous stresses and, therefore, look for a solution with pi, = O. Then h,,,, = O. From the adherence boundary conditions

h= 0aty=0,

h=U tit y= d,

we obtain the final exact solution u

Uy = d,

t7

=

= 0,

p — constant.

(2a, b, c)

The linear increase of horizontal speed is shown in Figure 3.5. As in the figure it is customary to position the graph between the bounding walls and to draw inafew velocity vectors to make the flow easier to visualize. Such a diagram is called a velocity profile.

Se a- . 12]

Fxorr S^Jurrons

95 d

L

=0

F t G LP R E 3. 5. The ratted ►• profile for plane (`ouerfe j1uH 1_ et uscomputethe tangential stress exerted by the moving flail on the lower plate, Reverting to subscripts for a moment. we obtain the desired stress, using (1.25) and (1.6),

cr r

T,, = D# 1

c_r 2

r

t•2 --

ex,

In our current notation. the above equation is written ^

CU['t ) -r Cy ex —

Since u = fl} d and r – C. we find that L

(2d)

The stress is directly proportional to i,' and inversely proportional to d. if we could move a plate parallel to another plate at constant speed, then we presumably could measure p b} finding the proportionalil constant. The plates would have to he large enough so that end and side effects would he neglib_ Note that the force on the lower plate increases with p.m conformity wit h our intuitive discussion of the relative fonces required to slice a knife through molasses and through water. SOLUTION 2: PLANE POISEUILLE FLAW, To see how viscosity impedes flow forced by pressure gradients. let us consider horizontal flow between motionless parallel plates at y = ±d forced by a constant horizontal pressure gradient pz

–

C,

C a constant.

(Positive C corresponds to pressure decreasing with increasing x. Accordingly, we expect the fluid to flow in the direction of increasing x when C > û.) (Ti + constant, all With the assumptions u = h( '^, t _ +^ = ü, p = equations are satisfied, provided that –

p -1 C +i'IJ j ^ = 0 or

C -t - Ph„

s ^-

Viscous Fluids ICÏt_ 3

96

Adherence requires that

h =U

at

d,

So

C t1 =

(i} 2 —

(3)

The parabolic velocity profile is depicted in Figure 3.6. ^-^

-

-d

3.6_ /he t rlrrt ir; 1►roifile fur plane Paistti.rilfr flow TTxe jirm• is driven by if { tmsruur pressure gradient that LE dlrrr-rt'd parallel fr, flhr hrteurdin0 fnre F tCï ll s L

pfrunt•.ti_

Of chief physical interest here is the urmRss flow rtde per unit width, i.e.. the mass of fluid per unit time, per unit distance in the 2-direction. that passes a given plane _r = constant. This is given by

f

/}!! d}' _ 1t Cr t d 3. -

(4)

d

The pressure can be measured_ Cnnsequently, comparison of a measured mass flow with (4) could in principle provide an alternative way of measuring viscosity. In fact, viscosity is measured with the aid of solutions similar to but not identical with, the ones that we have just derived_ The first solution we considered (Figure 3.5) is called plane Couette flow. (Ordinary) Couette now is the tangential flow of a viscous hind in the annulus between infinitely long concentric rotating cylinders (Figure 3.7). If the gap between the cylinders is small compared to the mean radius (R, t R2)/2, then plane Couette fl ow should be a good approximation to ordinary Couette flow [Exercise 1(b)]. But the latter is easier to approximate experimentally and can be described in its own right by an exact solution [Exercise 1(a)] _ Indeed, an important method of measuring viscosity is to suspend the inner cylinder from a wire whose twisting is proportional to the torque exerted by stresses transmitted by the viscous fluid located between the inner cylinder and a rotating outer cylinder.

Sec. 3.2] Exert Solutions

97

^-^-^

--

yR -^-Rr

FIGURE 3.7. In Cnuette flow the fluid is confined bemaeetw infinite coaxial

cylinders_ The outer (inner) cyirr<der rotates with angular

velocity n,(12, ).

The exact solution illustrated in Figure 3.6 is called plane Poiseuille flow_ A more useful solution is that for (ordinary) Poiseuille flow in which fluid flows along a circular pipe because of an axial pressure gradient (sec Figure 3.4 The solution is considered in Exercise 2. Instead of (4), the following formula relates mass flow rate through the pipe E with an imposed pressure gradient of magnitude D; E

—8

Dv -1R4.

(5)

A practical method of measuring viscosity consists of determining the pressure gradient D and mass efflux E and solving for y from (5). Plane Couette flow and plane Poiseuille flow show the qualitative features or ordinary Couette and Poiseuille flow_ The former have been studied here so that we would not need to introduce cylindrical conrdinates and thereby divert attention from the issues we wish to emphasize at this point. It might appear that we can sum up by noting that the viscosity p or the kinematic viscosity ' can be determined by measuring the mass efflux in

Fee taR E 3.8. Poiseuille flow in o circular pipe, like plane Poiseuille flow. is driven by, a constant pressure gradient. The parabolic uxwl r ek eity profile is depicted.

Viseurs Fluais (Ch. 3

9$

Poiseuille flow and that the Navier- Stokes equations themselves can be checked by using this viscnsiLy to compare measured and predicted torques in Couette flow. While broadly correct, such a summary conceals certain difficulties. I. Measurements require theories for their interpretation, For example, confident measurement of torque by examining torsion of a wire requires secure theoretical understanding of the relation between torque and torsion (see Section 5.3). As another example, to measure pressure one needs a device and a theory to explain its operation_ In practice, a degree of empiricism is usually necessary. This is well illustrated by the lengthy discussion in Goldstein (1938, pp. 2481t,) concerning a pressure-measuring device called the Pilot tube. 2. Given an understanding of measuring devices, there remains the problem of whether the exact solutions involved are appropriate. Cylinders cannot be made infinitely Icing, nor can they be made perfectly circular. Measuring devices disturb the flow whose properties are being subjected to measurement. Failure to reconcile theory with experiment could be due to the fact that inevitable small deviations from theoretical ideality may mean that the appropriate mathematical solutions are entirely different in character from the simple exact solutions. One ought to examine the stability of the exact solutions to small perturbations (see Section 152 of I and Exercise 4). To illustrate this point, we remark that Stokes derived the Poiseuille flow solution from his equations and found complete disagreement with the experiments available to him. He delayed publication of his solution. Now we know that the experiments involved the violently eddying turbulent flow which replaces Poiseuille flow at sufficiently high imposed pressure gradients. SOLUTION 3:

RAYLEIC:M IMPULSIVE FLOW

We turn La a problem that illustrates how viscous effects develop with time. Consider fluid in the domain y > O which for r < O is motionless and is bounded below by a motionless horizontal plate. Suppose that at time t = 0 the plate instantaneously begins to move, in its own plane, with speed U. Presumably, viscous effects will bring the entire body of fluid into motion. How does this happen'? 0 (sec Figure 3.9). As above, this We assume that u = of y, t), r' requires that p r = ps = 0, p. = constant. We take p. = 0 so that the motion is due entirely to viscous effects. The governing equations then require that —

Ur

= v2 I1.1r,

14

-0

_

f

> 13,

for t

O.

while, by assumption,

(6)

Sec_ 3.2] Exact So luerulis

99 ^

I/I1II/INI/IIfIIiI7II7l7ÎÏÏTT speed = 6 tor r Q: tl(U, rl

U.

(71

A person with considerable insight might realize that in this simple problem the speed U plays only a limited role. All that the solutions will tell us is how u/U passes from its initial value of zero to its ultimate value of unity, at a given spatial position and I -or a fluid of a given viscosity. We thus expect that E4

(8)

= ![t', r,

the point being that l:` does nor! appear on the right side of (8)_ As is frequently the case, mathematics can be used to verity physical insight. With the change of variable it' = Nit/. (61 and (7) become = uu},

LA (}, f) — 1

> U).

(9E41:1

Since these equations do not involve U, re cannot depend on I1, from which (8) follows. The ratio uXU is dimensionless, It must, therefore, depend only on a dimensionless combination of y, t, and v. Since the dimensions of r are length/time, y, t, and w can only occur in the combination y 2 1 - Vv. More precisely, u/U = g(y 2 fvt) for some fu nct ion g. It is cnnvenient to write this last equation in the equivalent form:

u'

-

Lr

w hc°re7 = 2 }.`{vt}

,tz

(10)

for then the final answer will be expressible in terms of a certain tabulated function.

too

Visenu s Raids

tCh. 3

To test the conclusion (10), obtained by physical reasoning,* we substitute into the governing equation (9). By the chain Hile, (9a) implies that —

iv - 1 12t - 3f3 3y' _ v . iv - ] - l^n

or

F" -# 2r1F' = 0. For any fixed t, y --• 0 implies that ri - O. sc (9b) implies that (12)

F(0) = 1.

Note that for fixed y, rl cx:, so (12) requires that at any fixed location Oas r as r of if1 : -• I -• oo. This is a reasonable result_ Equation (11) is of second order, so there must be a boundary condition in addition to (12). We require that at any fixed time the velocity far from the plate approaches zero: Hat u(y, $) ---- 0,

t >0.

(13)

Trouble threatens, however, when further thought indicates that it is just as reasonable to require that at any fixed position the velocity approaches zero as r ,l, 0 (i.e., t > O, r - 0): lini u(y, r

s) -

0,

y > 0.

(14)

,o

It i s reassuring that both (13) and (14) amount to the same condition on yy r namely, } F('(),

11m F(/) = U

(15)

0,2

However arrived

at, the final problem, given by (11), (12), and (15), is a simple one. Regarded as an equation for F', (1 1) is first order. Its solution is

F' = C I exp ( - r7 2 ) T h us

F(rJ) =C, f e ^d^+

C2.

(16)

• Scc Scc`iiun 3,4 and Exercises 3.4.$-tO for other material concerning - similarity' solutions tu partial differentiae equations in which dependent .iambics appear only in certain combination t In [Eats and papers it is customary to vrgaruxe the subject matter its a "straight line" manner in which the mathematical problem is completely formutticd and then is solved. From time to time, however. it is useful to remind [he reader that creative scientific thinking mitre often than nul results in an interaction between the formulation of a problem and its solution-

See. 3.21 Exact Solutions

aoI

t'trrl - err, I

FIGURE 3. W. in tQuyfer`gh in,pulswr flaw, u! (i = Ftg), where ij = bl vt) "' A graph of F is presented in thF,r figurer.

Since the Leal value of Pi is of special interest, we use zero as the lower limit of the indefinite integral in (161. Because of this, (12) simply requires that C2 = 1, whence (IS) requires that

C 1 f e_

:

d + l = 0.

Thedefinit e integral in (17) hay the val tic ec

F(er),

where F (r)

( I 7)

[See Exercise 3.2(d)of 11. Taus = I

—

e

R

In 0

d.

(IS)

The right side of (18) happens to be a tabulated function, called the complementary error function and denoted by erf L (ri). The error /unction itself, erf (rr), is just the integral term in (18), so F(rt) =

err, 01) — l — erf (t,).

(19)

We can ascertain the main features of our solution without resorting 10 tables. Merely from the fact that the solution is a function of q 1}4,11) -- " 1 see that the place where n f U — constant is at a certain value of v(v r) - t, 2.*we Thus the portion of fluid that moves with less than a given per cent of its final velocity U is farther from the plate than a certain plane whose height is proportional to (vt')" Since exp ( x) decreases rapidly with increasing 4', the graph of F must have the qualitative features of Figure 3,10. From a table (ear by numerical —

- constant has a unique * Physical intuition would kid one to believe that the equation sululion, this k confirmed by IN, which shows that Fl(s) is continuous and monotonically

decreases from I to 0 asp/ goes horn 0

Le r[,.

:

o2

Viscous Fluids (Ch. 3

integration) we ascertain that F(17) = 0_005 when r1 _ 2. As a consequence, when 71 > 2 or y > 4(v0" the fluid speed has not reached even half of l per cent of its final value. Since erf, (1) = 0.16, when y > 2(%s)u 2 the fluid speed does not exceed 16 per cent of its final value_ From the highly simplified situation typical of an exact solution, we wish only to draw the most general type of conclusion. Such a conclusion in the present case is that the purely diffusive cruse of setting fluid into motion tykes effect in r region whose thickness is of order (vl)'". This spreading at a rate proportional to r''' is the single most important characteristic of diffusion. (Compare the remarks in 1 at the end of Section 4.1 .)

EXERCISES

In the following exercises, u, u, and w denote the velocity components when Cartesian coordinates x, y, and z are used When cylindrical coordinates r, O. and z are employed (sec Appendix 3.1 for relevant equations), the corresponding velocity components are denoted by Or ) , r.{e) , and w_ 1. (a) Solve for the Couette flow of Figure 3.7 by looking for a solution of the form Gtr} _ w Ü, ut'M = V(r), p = [Sec Exercise 4(b) P(r). for the answer.] (h) Find the stress 7. e exerted by the fluid on the inner cylinder_ Show that if the gap is small compared to the mean radius, then the result is dose to that of (2d). Explain why. 2. Solve for the Poiseuill e flow of Figure 3.8 by looking for a solution of D. Derive (5). the form i( = t = Q, w = f (r), p = = in as much detail as possible the similarities and differences 3. Discuss between (4) and (5). 4. In this problem we outline the beginnings of an investigation into the stability of Couette flow. Knowledge of stability theory, as in Section 15.2 of I, is assumed. (a) Show that the Couette flow of Exercise 1(a) satisfies the inviscid equatiionsof motion, so that viscosity enters only by means of the adherence boundary condition. In a first attempt to find out something about its stability, we shall regard Couette flow as inviscid. (b) Examine the stability of the following exact solution (Couette flow) of the inviscid equations: V(r) = Ar + Er - ^

p = P(r) = i{2 rT L ,

!] ( ri = w =

(2 ntfi A = 2R1 R1— R^ T

0;

8^

(Dt

112)RiM

T Ri`Ri

.

Sec.

.I21 Exarl Solutions

103

Assume disturbances that writing

are

c;[r 1 X, z, l4'

=

1 ,^ '^'t r,

axisymmetric (independent of 0),

Zt^^ _

I, f),

/(r) + L` B1 (r, z, 1),

P = P(r) + P(r,

Z+

I).

Linearize. Write ^,{r}^itoi+

i , z,} t Crl(r

As)

and make analogous assumptions For the forms of 0 19, iv, and p. Eliminate ° ',ti3, and p. To simplify notation, write WI X. Deduce -

dr

(1 ;

a

dr

rX) + 1, 2 (a

— 1)X

0,

X(R I ) — X(R 1 ) —

where

(I3(r) =

r

d

rlr

(r

(c) Note the similarity between this equation and (152.21) in 1. Exploit this similarity and draw conclusions such as Deductions I and 2 of 1, Section 15.2. (The required theorems are still valid, with substituted for dR jdz.) The mathematical similarity evidenced in our investigations of the stability of Couette now and of a stratified fluid is due to the physical similarity of the destabilizing effects of gravitational force and centrifugal force_ 5. In deriving the flow of Figure 3.5,we set px = O. Keep the sameboundary conditions, but now set p = — C, Ca constant. Show that the resulting fl ow is a superposition of the flows of Figures 3.5 and 3.6_ Sketch a typical velocity profile. 6. (a) Verify (11). (b) If u/U — y(y V /vt), find the equation satisfied by q. Recover (I1) by a change of variable. (c) Show that the assumption of (10) would be invalid if (9a) were 24; = vurrs 7. Show that 1„` exp (— e) 8. Instead of (16), take

do

is a convergent improper integral.

F(t7) — CI J ed4 +

C2.

Find Ci and C2 (a) when c = aç (b) for arbitrary finite c_ Verify that the solution is the same as (18). 9. Put a filled teacup on the center of a record turntable and suddenly turn the turntable on to 33 revolutions per minute. About how long does ,

L04

Visci us Fluids ICh 3

it take before the fluid is in solid body motion? Show that the estimate obtained from the diffusion distance (4 02 is much too large. This discrepancy has been explained only recently [see l -1. Greenspan's monograph The Theory taf Rotatin g Fluids New York : Cambridge University Press, 1968)]. 10. if the variable Li is regarded as corresponding to temperature, what problem in heat flow is described by exactly the sanie equations as those appropriate to the Rayleigh impulsive flow? Discuss the physical plausibility of the solution in the context of heat flow. +11. A viscous incompressible fluid fills the region z > O. It is bounded below by an oscillating plane, so :r{x, y, U e) = U exp (iw1),

U and co constants.

.

Find the velocity field. (Take p = constant.) 12. Advantages accrue in certain conditions if portions of an airplane wing are madeof porous material and a small amount of air is continually sucked into the wing. This relatively simple problem gives a first idea of the effect of suction. Consider two - dimensional flow of a viscous incompressible fluid in the half-space z > 0. Let the x-wise velocity component is approach a constant U as z # oc, . Suppose that at the porous plane z — 0 there is a perpendicular suction with speed W, so that at z : 0, w = — W where IV is a positive constant. Find a solution of the form u = u(z),

w = ff ,

u = 0,

p = constant.

14. Consider viscous steady incompressible flow

u = t? = i)

,

iv — f (x,

y)

in a pipe of uniform cross section whose generators are parallel to the z-axis. (a) Show that the axial pressure gradient satisfies ;1_p = — l,

constant, rdr so that the problem reduces to that of finding a solution of the equation ) 0 -? ♦

p

which vanishes on the curve in which the pipe cuts the x, y-plane. (b) The lines — 1 =0,

x+2--y,/3=4,

x+ 2+y

=0

bound an equilateral triangle whose sides have length 20 and whose centroid is the origin. For the pipe whose cross section is

Set_ 3.3] Orr Bnundur}' Lr^}r^rs

1 05

this triangle, show that f (x, y) is a constant multiple of (x — 1)x+2—y )(x+2+

yam)

and find the constant. $(c) Find the tangential force per unit length on the triangular pipe. NOTE. The mathematical problem that arises in viscous flaw down a pipe is identical to a problem that emerges in the study of the torsion of an elastic beam (see Section 5.2)_

3.3 On Boundary Layers* Books on science are full of plausible but noncompelling assumptions that lead to theories whose predictions are eventually confirmed with stunning accuracy. It is refreshing and instructive to contemplate instances where plausible reasoning led to trouble. One such instance conciûrrts the extensive nineteenth century study of the flow of air past a solid body_ There "the trouble had been by no means the lack of a theory, but rather the existence of an almost overwhelmingly large body of theory, constructed by many of the best mathematical physicists of the nineteenth century, according to the most respectable physical principles. This theory gave, for the motion of a wide variety of shapes through the atmosphere, which is treated as `a perfect the fullest information, none of which accorded with the most Rued' elementary observation of the facts. The failure of the theory was particularly disturbing because the only known mechanical properties of the air which had been neglected, namely compressibility and viscosity, could reasonably be supposed to produce small effects in the type of problem considered."t This section is concerned with the steady flow of fluid past obstacles. ft can be shown (as in the work of Lighthill just cited) that compressibility can be neglected in such flows provided that typical speeds are small compared to the speed of sound, that there are no large, imposed temperature differences, and that there are no motions of atmospheric dimensions. It is the aim of this section to explore the subtle effect of small viscosity when such conditions are satisfied. -

COMPARATIVE MAGNITUDES OF VISCOUS AND INVISCID TERMS

Suppose that you run at a speed of 15 kilometers per hour directly into a wind blowing at a speed of 25 km/h. If viscosity is neglected, theory predicts an absence of wind resistance (D'Alembert's paradox). Apparently, viscosity should be retained in equations that describe the air flow. But let us estimate * Appreciation of this Section will be greatly enhanced by ptior study of the maierial in I on sealing (Section 6.3) and on singular perturhalions (Chapter 9). t Quotation from M_ J. Light hill's introductory chapter in Rnsenhead (1963)_

[Ch_ 3

Viscous

106

the relative size of the inviscid and viscous terms. We assume that the wind blows along the x axis. Consider the Navier-Stokes equations for viscous flow ( 1.25), now written with an asterisk to emphasize that the variables are dimensional. With ajar* = 0 (steady flow), these equations are -

du* ^r +^ 0x*ôy*

1 Op* Mu* alai. p c7x* + 1 4-012 + ^Ÿ * ou•d * tr• r7f t^* ^p* 020 ^ 4 1R- e^y +u { # t - : #[^y* OX* [ lx Y e(y*) 2

c7u* (]D* ^^* + a^ ^ *—°'

,

( la. b, c)

A typical inviscid term is u* ate*/dx* and a typical viscous term is v 132 14*/21x*) 2 . For flow disturbed by the runner, it would appear than u* will be of the magnitude U0 (25 + 15) km/h or U0 x cm's. Significant changes in u* would be expected to take place over a distance of the order of some representative body size such as the width of a chest or a leg; say, over a distance of L — 10 cm if so, we would expect that

u

#

e?u* a2u* v *)Z ex* r3lx*)^

UO2 1.- -1 ,,U0 L

-x

Ut, f.

where

U0 L v

^

(4 x 104 crrt/s)(1#1 an) 15 x lU - ^ crn 2 js

10^.

(2)

We have used the value of v for air given at the end of Section 3.1. According to our estimates, the inviscid terms are 100,000 times as large as the viscous. The paradox seems sharper: Neglect of the viscous terms leads to the physically unreasonable result that the runner feels no resistance from the wind, but the viscous term truly seems negligible. Ludwig Yrandtl,at the begin ning of this century, had the insight required to resolve the paradox and point the way toward a procedure for calculating the effect of viscosity in many flows_ The essence of his ideas lies in the recognition that there is an extremely thin region, close to a solid body, where the fluid velocity undergoes great changes. Just outside this boundary layer the fluid "slips" along as predicted by inviscid theory. The velocity changes rapidly as the boundary layer is traversed, and becomes zero on the boundary as required by the boundary conditions for a viscous fluid_ The above estimates are not valid in the thin zone of rapid change, where the viscous terms are important no matter how small the viscosity.

-ei-Nr-

3.31

Chi

REYNOLDS

ro^

l'.r1Ye•fs

rw UMBER

Further progress is facilitated by the use of scaling. To permit some appreciation of this process for those who arc not familiar with it.? we summarize its main elements- The object is to introduce dimensionless variables in such a way that the maximum order of magnitude of the various terms is correctly estimated by the parameters that precede them. Roughly speaking, to do this one must nondimensionalize the dependent variables with constants estimating their maximum values. The dimensional independent variables x* and ys must be nondtntensionalized with lengths that are estimates of distances in which one observes significant changes in the dependent variables. The constants or sr ales that are appropriate in this nondimensionalization may differ from region to region_ The dimensionless group that we computed in (2) is an example of the Reynolds number, Re. The concept of scales is basic to the dehnitiuri of Re: Re — (overall velocity scale)(overall length scale) kinematic viscosity

LW_ t•

(3)

The word "overall" is used in (3) because there can he more than one scale of variation in viscous flow past a body. Thus the Reynolds number is an estimate of the ratio between the convective acceleration terms§ such as rr*clu*/(x* and viscous terms such as y r '- re*fi(x*)' except ni possible - special" regions of "unusual" variation. As we shall see, when the Reynolds n u miter is an order of magnitude or more greater than unity. it ceases to provide meaningful estimates near the solid boundaries, Prandtl recognized that the f'Alernbert prediction "no drag in nonviscous flows" is only slightly wrong in some slightly viscous flows. For flow parr streamlined bodies one can say that there is "slight drag in slightly viscous bows." The reason is that for such flows the itwi cc`td float' panern But when an obstacle is is only altered in the thin boundary layer region. bluff(i.e., blunt), the presence of viscosity in minute amounts alters the entire flow pattern. BOUNDARY LAYER EQUATIONS FOR STEADY FLAM: PAST A FLAT PLATE

We shall return later to some of these general ideas- Now we shall discuss in some detail the effects of a small amount of viscosity on a simple but representative "streamline" flow_ To this end, let us consider the case of a flat plate immersed in some steady two-dimensional flow_ Denote the (dimensional) velocity components by rc*(x*, y*), u*[x*, y*) and the pressure

by p*(x*, y*)- l'he flow could be disturbed by a number of objects, but we di,eussed in Section .63 or Rcrnernber ihat this discussion mals with steady flows_ The term PORN' may provide the

t Scaling is

dominant acccleration firms where the

vclocity at a pnirti varies with lime.

to8

Viscous Fluids (Ch. 3

shall fix attention on the fiat plate, which we imagine to be positioned along the x*-axis. The y*-axis is normal to the plate. We shall assume that the viscous flow does not differ significantly from the inviscid, except near the objects (Figure 3.11). We have already denoted the overall length and velocity scales by L and Zia, respectively. In the mainstream, then, a typical convective acceleration term such as u* Ott*fOx* should be of magnitude U /L. We expect a pressure gradient term like -- p - t 'p*/cox* to be the same magnitude, so we select pUt as a pressure scab:_ Thus the following are appropriate scaled dimensionless variables in the rnainstreamt :

X=

x*

y*

:a*

L. Y=U = U

Y=

v*

P=

o

u

p* P

^, (4)

With (4), the equations of steady viscous motion (1) are

f'L r x

-

UVx +

VU y _ —P x + Re t (Lf xx + Urr), VVy = `Pr ± Re - '

U x +. Vie = 0

(Vxx

( 5a)

4- Vier),

( 5b) ( 5 c)

-

Let U(X, Y), V(X, Y), and P(X, Y) denote the solution to the inviscid problem. These functions satisfy the above equations when the viscous terms are entirely ignored (formally, Re oo). Also satisfied are the boundary conditions of vanishing normal velocity. In particular. V vanishes on the plate, so from (5a) with Re oc we obtain the following result, to be used below: U(X, O)11x(X, 0) -- . F(X , O)

( 5 d)

-

-

Let us reiterate in present terms a point that we have already made. We are considering flows where Re may be 10 5 , 106 , or even larger. Nevertheless, the terms in (5a, b) preceded by Re r cannot be entirely disregarded. The f Here is more detail on the scaling isce also Section 7.3). Lei us use (4) Ow keep the pressure scale. P o unspecified for the moment, so lhal p p'/ Po . The Navicr--Stokes equation (la) becomes —

Uâ t U U^ r'll _ -— L U d } i. v Y

P4

—

op

— pL — ÔX

+ viscous terms.

Since Ua , P. and L are scales, by the definition of - scale" the approximate magnitudes of the three terms that appear explicitly in the preceding equation art ti /L, Uall. and PdpL. respectively. Away from obstacles, when Re > 1 we expect an essentially inviscid fl ow with all nonviscous terms in rough balance. This requires that U6IL = P01pL or Po pUo, determining the pressure scale P. NOTE, From the viewpaini of kinetic theory, pressure can be regarded as molecular momentum transfer per unit area. The pressure is of magnitude pti because momentum per unit volume of magnitude MI D is typically transported by bulk convection at a velocity of magnitude [1, D. ,

-

—

e _ 331 On Sc

Boundary

Lavers

109 Y.

l C[l[7i1:1rlt

& r15 a 3f

rirm5LIr[ =

P.

ja'

-

F ai tJ R E 3.11. A fiat

plate af' length 1.

and rrrher objet r.► émnreraiil tit a r-i-tc'rrWs

fluid.

reason is that the change of variables (4) introduces a scaling which is tint valid near the obstacles placed in the flow. Only away from these obstacles does this change of variables ensure (for example) that UU x and U }- r are approximately of unit magnitude, so that the ratio of Re -- 1 U ri, to UUx is the negligible quantity Re - I. We now turn our attention to the selection of scales that are appropriate within the boundary layer region. We expect the tangential velocity component ce*, the component parallel to the plate, to vary from its inviscid value outside the boundary layer to zero on the plate. Consequently, the maximum value of u* should be of magnitude Li e , the scale of the inviscid component. Variations along the plate have a scale that should be about the same as the scale L of inviscid variation, but the relatively rapid variations normal to the plate will have some scale e5 that is smalt compared to L. What about the normal velocity component v*'P We know that u* vanishes on the boundary. We shall estimate it at a representative distance 6 into the layer. We do this with the aid of the continuity equation ôu*/fix* + (70/4* 0, which shows that the normal derivative 4 t,*ley* has the same magnitude as 490/49x*, namely. U 0/L. Consequently, using Taylor's formula, We find that u*(x*, y*) x t.*fx*, 6)

v*(x*, O) + o * (x*, 0) Y

0

+

-

L

.

(6)

For a moment, let us take the pressure scale to be the unspecified constant Pl. Using the various scales to nondimensionalize, we write

v-

Z1 *

Lf^'

v=

^? *

u*

I.J 0bJL,

p=

Pi

, x r

x• L

,Y =

Y* ^

(7)

L_

110

Vr xc-r)uS Fluids [CA, 3

With (7) the steady Navier-Stokes equations (1) hecorne U,2,

eu

L

{fix

- 14 [

[UP] Ût + ^X

[

+

eu

fil L t' ; b ^-^

P,

ey

Fr ^1^

_

_

L

C^`}^

tip

rvU0

ax +

[

021,11

^ex2

vU o r' 2 ir + ^ 62 ^

oy2, (sa)

Op

(^y +

vU0B ^x v [vt.I.1P2 L3 Dx 2 + 31. (11, Y 2 (8 h)

U 01 iU u U eü a Q #. ['x 4 L [^y

(8c)

The object of our scaling was to choose variables so that the squarebracketed parameters in the preceding equations give the magnitudes of the terms of which they are a factor. 'These magnitudes must he such that the viscous terms are retained in the boundary layer- As we now show, this requirement allows us to determine b an d P 1 . Since ES is small, the largest viscous term in (8a) is proportional to a ufdy 2. This term will be retained, however small v is, only if b is such that the magnitude of this term is the same as the magnitude of the convective acceleration terms. Consequently, we require Uâ/f. = vU0/b2, SO

b = vrf

or

U0

b = Re - " 2 L.

(9 )

Knowing that 5 is proportional to v", we see from (8b) that i p/ra y is proportional to v_ Consequently, the change in p across the boundary layer of thickness b is of order V312 . TO first approximation, certainly, the pressure does not Mary ar~ross the boundary layer. This is an important qualitative conclusion of our analysis. In particular, it means that the pressure scale can he taken to be the same as the inviscid pressure scale: P, = p1426.

(10)

Using the expressions for the boundary layer thickness and the pressure scale given in (9) and (10), we can rewrite the governing equations (8) in the form + rlti y =

-4 rr,, y + Re - 1 m kt, -

pr = Re -1 (^ urn„ — LL J + t ») f- Re -2 uxx,

u + !Jy ^ 0. „

(11a, b, c)

We have assumed that Re > 1 and have introduced certain scaled dimensionless variables. We did this in such a way that in the boundary layer the relative magnitudes of the various terms in (11) should be approximately given by the Reynolds number factors that appear. [Compare the different dimensionless version (5) of the equations, the latter version being scaled

Sec. 33] On Bounrkorl Layer s for the mainstream.] When Re f, then we deduce from ( I I l that to a first approximation the following equations hold in the boundary Iayer:

tux + illy = - P x + u }y.,

py

tl,

u x -1- U}, = (i

(12a. h, e)

Equation (12h) states the previously noted fact that the pressure does not vary across the boundary layer. in (12a) we thus may replace p x(x, y) by its value just outside the boundary layer region. `phis is sometimes determined by measurement and is sometimes calculated by inviscid flow theory_ In the latter case, to first approximation we may use the inviscid equations evaluated right on the plate (rather than a thin boundary layer width away)_ Employing (5d), then, we see that the equations of (t 2) reduce to uu, +r)u

=D£1' +U) ,

fi(x, 0)V x (x, 0)

u+z'. = U, assumed known.

(131

The final steady boundary layer equations (13) have been derived for the flat plate. Nevertheless, it can be shown that they a re essentially valid for boundary layers near arbitrary two dimensional bodies, provided that x and 1' are interpreted as orthogonal coordinates measured along and perpendicularly to the wall, respectively (Rosenheati, 1963, pp. 201f1.). The only restriction is that wall curvature should not be so great that the centrifugal effect of the resulting curved flow requires a nonnegligible normal pressure gradient to balance it (That is, the wall's curvature radius must be large compared to the boundary layer thickness.) We shall restrict ourselves to flow past flat plates, but the reader should keep in mind that most of oui discussion applies much more widely. -

B c, N DAKI' CONDITIONS

We turn now to the question of boundary conditions. Since the most obvious defect of inviscid theory is that it permits fluid to slip along solid boundaries, we surely will want to require that both velocity components vanish at y — O. Next we must somehow link the boundary layer solution with the mainstream_ A way to accomplish this is called patching_ This method is used for a particular flow problem at a particular large Reynolds number. One requires that the boundary layer velocity component u(x, y) join with the corresponding exterior component U(x, Y) at some definite distance from the boundary y = O. Suppose for definiteness that Re 10 6 . Calculations show that the various possible solutions rl(x, y) cease varying appreciably as y increases beyond y = 5 or y = 10, say. Since y!Re" = Y, y = 10 corresponds to Y — 0.01. One can then patch together an approximate solution by using boundary layer flow for Y S 0.01 and inviscid flow for Y 0.01, where the boundary layer solution is chosen to satisfy u(x, 10) — 11'(x, 0.01) (see Figure 112). Since (typically) results do not differ much if patching is applied

112

Viscous Hair's

[Ch 3

Irtwiuid now: CFO. Yl, film, Y}

Patch here {i=re 1-T -

-

-

liaund ar y rayer Ilow ulr. kl. ntr.}'r

TfT TT^ ^ ^TT T^ ^TTTTTT F I U u R +:

3.12. Pinching toyerlrer lxurndurr• laver and !nr •isrid flow.

at y = 20 and Y 0.02, one has confidence that the general nature of one's results is independent of the somewhat arbitrary nature of the manner in which the boundary layer flow is linked to the mainstream_ We shall adopt a somewhat deeper philosophy of linkage called matching.* The idea here is to produce results that are correct in the limit as Re ar . One must drop the notion that the linkage is to be enforced at some tangible value of a normal coordinate. The boundary layer scale ô is proportional to Re - ' r2 , so certainly as Re all the boundary layer variation has occurred far closer to the wall than any fixed (i.e., Re-independent) positive value of the mainstream coordinate Y. On the other hand, any fixed value of the boundary layer coordinate y is at a definite position within the boundary layer and cannot be an entirely suitable place to Link with the mainstream. Therefore, we assert that the inviscid component LF(X Y) at Y = 0 should he equated with the boundary layer component ar(x, y) at y — oz;. From what we have just written. what eke could be appropriate?t The numerical results will not differ much from those obtained in the patching approach, but the matching approach is more readily capable of proof and extension. The previous paragraph still has the spirit of seeking a !oration For matching Better understanding of the boundary condition emerges if one frees oneself as much as possible from the patching frame of mind and seeks an overlap region where boundary layer and inviscid flow merge_ This region should be composed of values that, as Re ; a;, approach the wall, on the one hand, and the far extremities of the boundary layer, on the other- Such a set is given in terms of a function e(Re) by ,

^r

rr

i.e., y

— Re -

we.

(l4)

• We proceed in a manner that is else 311 ial ly self-cnnlained, but the reader who has studied Elementary singular perturbation theory, as in Chapter 9 of t will have a much more Secure understanding of the issLcs inwolrcdt One care also argue that ilk matching of {, at Y = O with a al y = cc Ô the most natural mathematical formulation of the fact that as Re ne, any tixcd 4alus oft lc free-stream variable Y. however small. corresponds to larger and larger 4alues of the boundary layer variable y. ,

Sec. 131 Ott Boundary Layers

113

provided that lim Re -I12 19(Re)

1im Q(Re) = Re

Re +^

^^

Using (14), we substitute for Y and y in terms of

O.

(15a, b)

the intermediate variable ri

and write our matching requirement as dim U [ x, Its-•oc \

JJJ

= dim Re-t

^ r;

i1(x,

R

((6) s)

We stress that ri is held constant in the above limits. Here we work only to the lowest approximation wherein the mainstream equations (5) and the boundary layer equations (11) are simplified by taking them in the limit Re x. The limiting equations contain no trace of Re, so (16) reduces to U(x, 0) = u(x, CO, the requirement previously arrived at. The notion of an overlap region gives this requirement added plausibility. More important, it provides the foundation of an approach that is more easily extended when higher approximations are desired. We have derived matching conditions for the horizontal velocity component 14, but what about the vertical component u? The structure of the equations (13) indicates that u should not he specified as y x : second order y-derivatives appear on u but only first derivatives on a. Thus two boundary conditions are anticipated on u (at y = O and y cc) but only one on u (the vanishing of y at y = 0, which is certainly required by the physical problem). Turning to the streamwise coordinate x, similar considerations indicate that we should require one condition on u and none on u, for (13) contains terms proportional to u„ but u is never differentiated with respect to x. Consequently, if we prescribe the entering horizontal velocity u(x, y) at a certain location (say x = 0), then we expect to be able to compute the flow downstream of that location. Summarizing, we find that the boundary layer problem for flow above* a flat plate of length I. has the following mathematical formulation, to first approximation: ux + vy = 0,

1411, + Vtl y ` U U ' ury ,

O < y< cc, O<x< L. u(x,0) =u(x,0)=0, u(0, y) = uo(y),

0<x< L,

y > O.

lln7 u(x, y) ; U(x, 0), 0< x < L . U(x, 0) and u o(y) are prescribed functions.

'

Flow below is determined by symmetry.

Vis-cous Ftyids

11 4

(C }I. 3

Formulation of the boundary layer problem is hardly a straightforward mailer, and the critical reader may well fed uneasy at this point. Can one trust "physical intuition" when one has no genuine experience of the flow inside the slivery thin boundary layer region? Awareness that a mathematical model is never identical with a physical situation is all the sharper when, as here, rather bold approximations are being attempted. Only further study of partial differential equations and of fluid mechanics can build lull confidence in boundary layer theory_ Nevertheless, the reader may find some reassurance in the discussion of Appendix 3.2, where we show that there appears to be just enough information in (17) to (201 to produce a unique numerical solution_ Moreover, it may be reassuring that a simplified "caricature" of the boundary layer problem seems to have a unique solution (Exercise 5). since most of the The "thickness .' of the boundary layer is of order Re variation across the layer takes place in a multiple or two of the normal length scale = Re - " L. For the runner mentioned at the beginning of this section, L = 10 cm, Re = l0', so the boundary layer thickness is of magnitude 113 mm, It is noteworthy that although the boundary layer is exceedingly thin, yet (as we shall see more clearly below) developments within it have vital and even dominant roles to play in the formation of the entire flow pattern. ' +3 ,

EXERCISES I. Estimate the overall Reynolds number: (a) For a fast-moving automobile. (b) For a hard-thrown baseball or hard-kicked soccer hall. (c) For a large jet airliner at cruising speed, (d) For a dust speck falling in air. le) For the motion of a whiplike cilium that propels a paramecium. (The cilia beat about 30 limes per second with an amplitude that is a sizable fraction of their length of about 10 pm = 10 - 3 cm.) 2. Estimate the boundary layer thickness in cases(a),(b), and (c) in Exercise 1. }3. la) A definite displacenient thickness 6 1 is sometimes used to characterize the extent of the boundary layer. By definition

45,(x)r FI _—^ j

1') dy. a(x LI( X , 01 1

Justify this definition. (b) A momei tum thickness is also useful on occasion. Give an appropriate definition. Explain your answer_ 4. This exercise is concerned with a change of variables, due to R. von Mises, that transforms the boundary layer equations (17) into a single partial differential equation that is well adapted for further analytical

Sri- 3 41 Boundary L f}, y > O.

= X)Y• tfi =(x, 0) _ ,Jr,,(x, 4) °- O,

y > 0.

/)(0, :) = 1, litre rfr)(x, y)

x > 0.

1,

x > 0.

(4a) (4b, c) (4d) (4e)

We expect the flow just at the entrance to the boundary layer to be undisturbed by the plate, so we have imposed the condition u* = U0 at x* = O. The dimensionless form of this condition is u1 =or = 1, because the velocity scale of the inviscid flow is its constant horizontal component U o . BI.ASIUS SIMILARITY SOLUTION

The above problem has a solution of the special similarity type in which the independent variables appear only in a certain combination. Experienced f

Further discussion of the legitimacy and uscfu Irless of the stream fund ion may be found in 15.1.3 and 15.1.4 of I. We have delayed the introduction of the stream function to maximize experience with systems of partial differential equations and to minimizt the appearance of Concepts that are somewhat special. albeit very useful.

Um-61,n

5rr. 3.4] Boundurt• Layer Firm. Prj.sr a .Semi-rrf nice

Flrtr Harr

117

analysis anticipate such solutions in relatively simple problems involving infinite or semi-infinite boundaries, whence determination of [heir precise form is a relatively simple matter [Exercise 2(b)7. Moreover, similarity solutions are the subject of a still-developing theory based on the invariance of problems under continuous groups of transformations (see Exercises 8 to 10). A way to guess the character of the solution front general principles stems from the observation that no geometric length scale L remains in our semi-infinite problem. This suggests that the form of the function 0 in (3b) must be such that I. does not appear. One possibility is iJ(x, p)

(2x)tr2 f(+11.

ry

ü

(2x) -

(5)

For then, from (3), * !Mx * y*) = (U 0 L;')l(3 2x*

-1

t+

= (2110 ux*}t^2 f ^,*

^

^ 2Vx*

2

_.. 3

.,

1; 2

In (5) the constant 2 has been inserted for cnnvenicnce; the alternatives to (5) lead to essentially the same final results [Exercise 2(a)],t Using (5), we find (Exercise ( (b)] ll -

(I?),

c _ (2_ )'

P " f•(ij LL l

— Pia J

d ter)

(7)

It is straightforward to verify that the problem (4) now becomes

1" t fir= t) , f(0) = f '(0) = 0, j '( -r )= I.

8la d y

There is indeed a solution of the form (5). If an x appeared explicitly in (8a), then f could not be a funclinn of r7 nnly, but this did not happen. Both the limits x — ► 0 and y —0 correspond to Ft x. so the two boundary conditions (4d) and (4e) cor respond to the single condition (8d), giving the three boundary conditions expected in a third order differential equation. 'the problem (8j could be solved numerically by assuming f "(0) = A, and computing j at = P1)1, rr = 0, 1, 2, .... Results of various computations can be used to select a good approximation to that value of A which provides a Anoi her way to Ira u ii t he form of

is to reason that in the absence oratternaiis'cs a multiple

of Y• itself must he the c€7rrect Icngih scats i, al a Indeed. if

L

distance sc. downstream from ittc leading edge

.

is replaced by 7r* in (h), then one Aims t

011% ..

4 21-T [j VA * ) " 0

t.^

k 2 . INA* US' .2 )

which is equivalent to tbl. We shall see seticral limes in what follows how the .x' plays the role of the overall length scab_

distance

riB

liiscrr:3s fluids (C7r. 3

solution satisfyingf'(cx;) = I (see Exercise 3). i-i. Blasius, the first to discuss this similarity solution, solved (8) by a series method in his 1908 paper. But the following observations give a more elegant and economical way of proceeding (see Exercise 14 for another approach). 1f F() is a solution of (8a), then so is uF(ori), for any constant a. Choosing F"(Ü) = 1 for convenience, one solves the initial value problem# F' + FF" = O. E(C) ` F'(4) = 0, F10) = L

(9)

if u is chosen to satisfy a

lim

(lo)

then aF(ini) will be the desired solution (Exercise 4). This solution is tahulated, for example, in Rosenhead (1963, p. 224)_ There is good agreement between the predicted velocities and measurements made in a number of experiments_ Of considerable interest is the predicted drag. This is obtained by integrating the wall shearing stress T x . at location x*, where [Exercise 1(c)1 t7u * L x'

_P

}^

*

= f• =

o

" 2 pilo(R x •} 112f10].

(llj

[T he answer is expressed in terms of a local Reynold; number R.. x* iv.] We find that the drag on one side of a unit- width portion of the plate which extends a distance L downstream is 21l2119W,LR,7 (12 f"(0). A customarily defined dimensionless drag coefficient C D is given by ,1 -

2312/ 10)R "

(12)

The numerical solution provides f "(0), with the result C.' = 1.3282I "2

(13)

An important general conclusion is that the viscous drag on a fixed area decreases to zero as the viscosity decreases to zero. DEFECTS IN THE BLAS1US SOLUTION

We note that the predicted shearing stress act ually becomes infinite at x 0 like x - "2. The integral of the stress remains finite but doubt is certainly cast on the validity of the theory for small values of x. Iii pursuing this matter, we recall that the boundary layer approximation is derived for flows whose overall Reynolds number is large compared to unity. For the flat plate, we (ft9) is much easier than chat of (13t, for only a single - march t Numerical initial values is required_ outward from attic given

.S+•r . 3.41 Boundary Layer Ffulr Pair o .Semr-infinite fur Harr

119

have seen that the correct choice for the length scale is the downstream distance x*; i_e., the flow is characterized by the local Reynolds number LF a x*/v. We must thus expect failure of our theory close enough to the Leading edge, so that {l a x*/v < 1, say. [Note that the theoretical defect reveals itself in the discontinuity of Oy at the origin, a discnntinuity required by (4c11 and 044) An incorrect picture of the flow over the short distance vfl a ' should have only a small effect on overall drag predictions, but nonetheless improvement of the theory is of considerable interest_ It can be shown that

1.22(2x)

"

as pc-

cr2 i

(14)

(see Exercise 6). As x ; 0 for fixed y, the vertical velocity component is thus seen to become unbounded. Consequently, the boundary layer solution is defective not only at the leading edge of the plate, but all along the line x : O. Except near y = 0, there seems to he no reason for this defect Indeed, except near the leading edge, the defect disappears if parabolic rather than Cartesian coordinates are employed. Formula (14) also applies for fixed positive x at the" edge" of the boundary layer (y ) and shows that the flow there is directed slightly away from Further, there is erificatioii of our assertion that ex, the plate. unlike 1411x, xij, will be predicted, so that no attempt should he made to match z to the -

inviscid flow. From the very form of the solution (7) we can deduce the qualitative result that the boundary layer thickness increases parabolically with downstream distance. Whatever criterion on u is used to define this thickness, the deduction follows from the fact that u is constant when q is constant, je., when x is a constant multiple of y 2 . Tabulation off' gives quantitative results_ it turns out that f is more than 90 per cent of the way to its ultimate value of unity when ti exceeds 2.4, and more than 99 per cent of the way when a exceeds 16. Thus if boundary thickness is defined as the location ' y , where the horizontal velocity component reaches 90 per cent of its inviscid value, then y2 g — 2.4(2 x * vilie) 1 ". if one employs the displacement thickness 0;,the dimensional version of the quantity defined in Exercise 3.3(a). it is found that = 12(2x. ilLf0) t r In principle the boundary layer approximation should become increasingly accurate as the speed U U is increased, but in practice steady boundary layer flow develops an instability when L!c x*/v exceeds a value of around 600. (See Section 15.2 of I on stability, particularly its concluding cxample.t) Once instability sets in, oscillations and irregularities become more prominent as x* increases and the flow soon exhibits the random eddying that is characteristic of turbulence. Local drag is much larger in turbulent flow than in See also Chapter SoFt:' C. Lin, Theory uf ffOrodynurrar Str3hoiir}' I New York Cambridge University Press, [955), lo supplement i he general stability coFesidrt atinns nt I wlth an account ,

o r rescarcir a=] this pariiCU[ar prvblcm.

120

t'i.sCUUS

M. irl.s

1C:h_ 3

/Laminar (nonturhulent). Roughly speaking, this is because the eddies are more efficient at mixing than molecular diffusion,so that a larger,"effec Live diffusion coefficient" must replace p in the computation of viscous stress, BOUNDARY LAYE R SEPARATION

Another deviation between theory and practice can be illustrated by considering uniform flow past spheres and cylinders. We shall restrict ourselves to two-dimensional flow past cylinders in which the circulation is zero. (Compare Section 15.E of I.) In both situations, therefore, the standard inviscid solutions are symmetric fore and aft_ Pressure decreases from the Forwardrnost point to the corresponding side point or equator, and then increases symmetrically. If boundary layer theory is used to improve on the inviscid results, profound consequences are seen to arise from the increase in pressure over the rear of the body. (The terni adverse pressure gradient is used when pressure increases with downstream distance. See Figure 3.13.) Since the pressure within the boundary layer is the same as that outside (to first approximation), the boundary layer flow is progressively slowed by the adverse pressure gradient. Beyond some point P the boundary

0

F tG t,tit t 3. # 3. Symmetrical inriscid flow pusr cylinder or _spfrerre. S.- stagnation is :OW. Both in the graph giving lrre.ssure und o n the cylinder. point,whervlcy the hoa ry line singles out the region of adverse pressure gradient n,he re in ?Alen > O.

Syr. 3.4]

l3rrurnfur}• Layer

Paw a Semi-infinite Flat Phare

12 I

layer is predicted to contain a region of upstream flow. But in essence this means that upstream fluid is deflected by a zone emaciating from P and extending outward into the flow. The boundary layer is said to have separated from the body at (or near) P (Figure 3.14). Obstacles in uniform flow can he divided into two classes, bluff nr streamlined, depending upon whether nr not significant separation occurs. As we shall now indicate very briefly, the character of the flow and the degree of Our understanding are very different in the two classes.

P

F1auR E 3. 1 4. Successive tangential velocity profil es in a separating boundary Slaw bac'kftr3 ► } . omits in rlae .shaded region, elownsire'urn of the separation layer. point P. SLIGHTLY VISCOUS UNIFORM FLOW PAST STREAMLINED BODIES

Separation is a negligible factor in uniform flow past a slender body with a pointed trailing portion, provided that t he body is not badly misaligned with the flow_ A first approximation to such streamline flows is provided by inviscid theory, and a correction by boundary layer methods_ In principle, a more accurate solution can be obtained by modifying the inviscid outer flow, so that it corresponds to flow past a body with a normal "blowing" given by r.(x, coo) as calculated by boundary layer theory. Passing beyond the first term of boundary layer theory has turned out, however, to require remarkably subtle analysis. This can be illustrated by considering the boundary layer downstream of the trailing edge of - a flat plaie of length L . The stream function must still satisfy (4a), but now the solution must be matched to the Blasius solution of(7) and (S) upstream of the trailing edge. The no-slip conditions on the plate must be replaced by symmetry conditions v = U and dujdy = O. A solution to this problem was published by S_ Goldstein in 1930. This solution had a relatively thin " wake layer" within the ordinary boundary layer. Only in 1969 and 1970, respectively, did K. Stewartsori and A. Messiter find that Goldstein's solution gave a defective description of the pressure. To fix this up, a layer had to be introduced that was thin compared to the

Viscous FJuirii [C'h. 3

J 22

wake and whose length shrank to zero as Re co. This may well seem an academic matter, but it is not. An accurate first correction to the Blasius drag formula (i3) is only obtained if the "triple deck" structure of the boundary layer at the trailing edge is taken into account. (And, indeed, aerodynamicists are aware of the fact that flow details at the trailing edge of wings have a remarkably large effect upon drag. l Thus a presumnably accurate cor rection to the drag coefficient for a flat plate only appeared six decades after Blasius's original calculation! It almost goes without saying that there is no proof that the formal calculations provide an appropriate approximation to the exact solution as Re —k OD. Streamline flows at large Re contain regions of turbulence in both boundary layer and wake.° Although much of value is known about turbulence, and its effect can be estimated in many practical situations, there is to this day no satisfactory foundation on which to base investigations of this phenomenon_ The combined statistical and nonlinear nature of turbulent flow gives rise to theoretical problems that have so far resisted all efforts to construct a theory of generality and power. Si.IGHTI.Y VISCOI:s [:NIFORNi FLOW PAST 81-11_, IF1 BODIES

Boundary layer separation will be present when the conditions cited above for streamline flow are not met. In particular, in the phenomenon of stall separation occurs even for flow past a streamlined airfoil if there is sufficient misalignment with the mainstream. Separation generally brings about an overall flow pattern which is markedly different from that predicted by straightlbrward inviscid theory; its main feature is a large and unsteady wake region. The idea that boundary layer theory can be straightforwardly used to correct a relatively simple inviscid flow must now be abandoned. Yet this theory remains useful for bluff bodies if ii is employed in conjunction with measurements of the mainstream flow. Also, the combination of inviscid and boundary layer theory can give worthwhile results somewhat upstream of the separation point-t Special note must be made of the fact that in spite of separation, the inviscid solution For the cylinder (studied in Section 15.4 of I) is still of great use in calculating genuine flows. With the aid of conformal mapping, this solution can he transformed so that it provides a uniform two-dimensional inviscid flow past an airfoil. Excluding stall, separation effects are negligible in such flows. An outstanding theoretical problem is to calculate the flow pattern around a bluff body, say a sphere or a cylinder, in the limit Re -• a;. Because of Al sufficiently high speeds compressibility effects become important. but these hate been excluded from our considerations here a circular cylinder al Reynolds numbers of order Qf magnitude I0 4 . separation occurs .

t For

ahoul 80° downstream from the forwardrnost tsiagnation) point. At higher tpxd their: is turbulence in the boundary layer forward or the separation pOint.and separation is delayed.

Sec. 3,411 Baum/elf I. Luger Flan Past rt Sears-urr intre flat F`Harr

123

separation, a pronounced lack of symmetry between upstream and downstream flow is anticipated. The limiting inviscrd solution can and certainly will contain discontinuities in velocity between certain adjacent streamlines. The general characteristics of this limiting solution remain unknown, although slow progress seems to be forthcoming from a combination of advances in asymptotic theory, numerical calculation, and experiment. Calculations and experiments cannot provide definitive results because the actual flow becomes unsteady at sufficiently high Re. The limiting solution is still of interest, however. even though it must be unstable to perturbations, The mathematical challenge rrf providing an asymptotic analysis of a prototype nonlinearity is perhaps sufficient motivation for continued study of tlhe problem, but the limiting solution should also be of use in approximating flows where Re is moderately large. For the flat plate, as we have seen, the drag approaches zero as the viscosity decreases. The same is true in general for streamlined bodies. But for bluff bodies the radically different nature of upstream and downstream velocity and pressure fields in large Reynolds number flows is responsible for the significant drag actually found, in contrast to the small drag expected from a slight alteration of D'Alembert's classical "no-drag" result for inviscid flow. The dramatic difference between the drag for streamlined and nonstreamlined bodies is illustrated in Figure 3.15. l

•

streamlined turfed au u l a rtrc•rtlur = 4 x IO ° r L. .{ Bak hrlrrr. I 967. e;rfsrda•r ►t itlr he ►urne dray at speed p.336.) The appeared ert hug.. .a.. Frrlktk'r. t'. M . and Walker, 11b'_ S. Aeronaut. Research Council. R ep . and Mern,, rrrr, l2.41 . Re•drut+ri tt all t"iiiLlltt

3 15 4c•Irt7nutre• crass sct'rie+tr rtJ .

fit

pernis.Ycm,

Inviscid flow, boundary layer, separation, turbulence the development of these concepts and others like them has made it possible to understand the general outlines of results as complicated as the cylinder drag plotted in Figure 3_16_ Moi ivaied by the desire to build better aircraft and ships, the study of uniform flow past obstacles forms one of the most developed parts of fluid mechanics. As we have briefly indicated, much has been accomplished but much remains to be explained. The following survey articles provide recent discussions of orne of the advanced topics that we have mentioned: M. Van Dyke, "Higher Order

Viscous Fluids [Ch, .ï

12 4

1.5

i .o

05

o lD° ee''

?atip

R. —

3. 16. Qrealitatire befutr•ior of the dray per unit axial length as u junction of Reynolds number for a circular cylinder of radius a_ The cross-hatched region represents a drop in dr a y due to bnuexdeary layer tur b elP n ce and corrsegreenr rnot Yrrur n t of rhe .separarir err mind from e i br l ur 80 to 120'. }lurelriny is tutcessary because results depe n d ro some . extent o n th e amount of turbulence in the wind tu ►trtel. (Botrlrekxr, 1 967, p. 34 i , or Ro.senhrad, 1963, p. 106.) FIGURE

Boundary Layer theory," and S. Brown and K. Stewartson " Laminar Separation," Ann. Rev. Fluid Mech. I W. Sears and M. Van Dyke, eds. (Palo Alla, Calif.: Annual Reviews, inc., 1969), pp. 265-92 and 45-72, respectively; P_ A_ Lagerstrom, "Solutions of the Navier-Stokes Equation at Large Reynolds Number," SIAM J. App!. Math_ 28, 202-14 (1975); K. Stewartson, "Multistructured Boundary Layers on Flat Plates and Related Bodies," Advances in Applied McwhaMics 14, 145--239 (1974), also "On the Asymptotic Theory of Separated and Unseparated Fluid Motions," SIAM J. App[. Math. 28,501-18(1975). ,

E X E R CIS E 5

1. (a) If Oi* and W are, respectively, dimensional and dimensionless stream functions, then ;11 = *IC, where C is a constant. This constant must be such that (2) is satisfied. Verify the text's result C U 0 15. Check that C is dimensionally correct. (b) Verify (7) and (8). (c) Verify (11). 2. (a) Suppose that = (2x) :11g(0), 0 = xjy2 . This accomplishes the goal of eliminating dependence on L just as well as assumption (5). Comment.

Sec_ 3.41

Boundary Layer Flow Pusf uSr m i-infinif e flu Haw

l'S

By substituting into (4), (5), and (6), discuss the process of determining whether these equations have a solution of the form = _eig(yx/ for some constants a and b. (c) Modify the differential equation so that the problem does not have a solution of the assumed form_ Modify the boundary conditions to the same end. 3. (a) Set up a difference scheme that could be used to solve the initial value problem consisting of (8a) to (Sc) plus the condition #'"(ü) = A. A a parameter. lb) Let f(x) = NIA); the notation makes explicit the fact that the limiting value of ' depends on the initial condition .4. revise an algorithm that in principle could be used to determine to arbitrary accuracy the value A which ensures f(x•) = 1. 4. (a) Verify that if (l C) holds, then aF(arl) will satisfy the equations of l8), as asserted in the text. Op) (Project) Using the method of (a(, write a computer program and use it to determine the solution to (s). 5. As was shown in Exercise 15.1.7 of I, in steady inviscid incompressible flows without body forces. Bernoulli's equation guarantees that (b)

Ps + ip[(u') 2 + (r"*)2 J = p + where p x is the pressure far from the obstacle. Use this to interpret the definition of the drag coefficient (12). 6. (a) To obtain the behavior of the solution to (8) for large Values of 17. one assumes that . f{.f )

_

-

rr

B + 4(O.

OW small,

B a constant.

Why is the assumption made? (b) Using the above assumption. deduce the approximation

B + A(ri — BY 2 exp [ - 40l — B1 2 ].

A a constant.

[Equation (14) follows at once, using the value of B obtained from the numerical solution.] 7. Separation occurs for a circular cylinder before the region of ads Erse pressure gradient begins, according to Figure 113_ Discuss. In the present section and elsewhere, we have studied problems for partial differential equations that have similarity solutions, wherein the dependent variables essentially appear only in a certain combination. The next three exercises explore this matter further. Further references are G. Birkhof,, Hydrodynamics, A Study in Logit, Fact, and Similitude (Princeton. N.J.: Princeton University Press, 1960) [also issued by Dover Publications] and G. W. Bluman and J. D. Cole, Shniturit y Methods for Constructing Solutions to Ordinary and Partial Differential Equations (New York: Springer-Verlag, 1974).

c a6

4'i.vcnus Fluids

(Ch. 3

8. (a) Consider the transformations {1i Ay'', x + Bx, and y -1 Cy_ Show that the boundary layer problem (4) is invariant under these transformations if and only if C = A and S = AC. (b) if the problem has a unique solution, the solution must also be invariant under the above set of transformations. From this, deduce the form of the similarity solution (5). NOTE. The idea that invariance to a group of stretching transformations requires a solution in which variables our in certain combinations can be regarded as underlying the reduction of parameter number obtained by the use of dimensionless variables. 9. To afford further practice in the use of similarity solutions, consider the fl ow of a thin two dimensional jet into fluid at rest. It can be shown [e.g., Rosenhead (1963, p. 75410 that the flow is governed by the boundary in layer equation (4a) but with the boundary conditions -

0

as y

O„ = 0

at y

► y,

cc, 0,

x>0

(symmetry in x-axis);

x > {]

1l► ^ dy = M, M a constant (a)

(flow dies out far from jet);

(same flow of momentum across each section).

Considering the same class of transformations as in Exercise S, deduce that there is a solution of the form xi3{•(7)

r=

yx- 213

Show that the problem has a solution which is a numerical constant times cc tank arl, where ct is determined by the momentum condition. 10. Consider diffusion of heat in an unbounded homogeneous medium wherein the temperature depends only on the distance [rani a plane, line, and point, respectively. The governing equation for temperature T, (lD)

T

= KV 2 T,

thermal conductivity,

then takes the form

7;= x[T„ +(n—l)r - 'T, where n = i , 2, 3, respectively. (a) We are interested in the diffusion of a fixed quantity of heat, initially at the origin. Show that the corresponding restriction on Tis

fc Jo

Tr" dr = constant.

.Ser. 3.4]

Boundary Layer A ra . I'u.0 a Semi-irlfrrlite flat plurr

127

lb) Show that search for a similarity solution leads to an ordinary differential equation which can be written f' -

fl" + i ► i(f '

f)

0

r 2 /4hr.

for a certain function f f), where =

:AO Solve the equation of (b) subject to the boundary conditions

f

as r

and f ` ,- ► iÏ

+(d) Use superposition (with integrals) to generalize the

solution of

(c)

so that you can deal with the diffusion of a fixed quantity of heat which is initially distributed in an arbitrary fashion. le) What is the solution for a line source at r — 0 (two-dimensional problem) producing heat at a constant rate? The next three exercises give a glimpse of a subject that is the focus of considerable current interest, the interconnectton between chemical reaction and flow. The material for these exercises is taken from Chapter 2 of Levich 11962), 11. Consider boundary layer flow past a semi infinite flat plate. -

(al Show that deep within the boundary layer (t.e. "very" near the plate) the horizontal velocity component u* satisfies ,

u* ^

k}^* I-14 vx • '

(15)

where k is a constant. In fact, k is about (b) Show that (15) is reasonable in view of the following fact. lithe thickness b of th e boundary layer is defined as the distance from the plate then numerical calculations show where u' is 90 per cent of that e5 5.2(vx*/Uoli l 12. Suppose that a chemical is present to a fluid !hawing past an object, and that the chemical reacts very quickly with the material of which the object is composed. It is assumed that the chemical concentration C*(x*, v) is governed by the following problem, where u' and v* are given functions. ,

u* C'* -- ►

la )

Co

r)C* cx*

a.,,.

re. [' c*

,1L'*

+ ^* Cy * =-

D

far from the object.

+

r7(y*)2

f 16a)

C* = O on the object. (16h)

What physical assumptions are implied by (16)'' (b) Suppose that both u• and i.* have a typical magnitude U 0 , and that u* and u* typically undergo significant changes in a distance L. Introduce scaled dimensionless variables rnto (16). Gomment

i28

Viscous Fluids [C'1,. 3

briefly nn the advantages of this procedure. Discuss in general terms the type of solution expected when the Peclet number Pe ï fi e L jD satisfies Pe i> I (a common occurrence). 13. We wish to examine WI) in the case where there is a "concentration boundary layer" close to the semi-infinite flat plate y* O. 0, x* Flow far from the plate satisfies = h 0 ^a* = t]. We expect that if D is small enough, the concentration boundary layer thickness d will be small, even compared with the ordinary viscous boundary layer thickness b. Let us introduce scaled dimensionless variables that are apprnpriate within such a Concentration boundary layer, set up an apprnpriate mathematical problem, and draw a general conclusion without solving. (a) Choose scales for x* and y*. (h) Formula (IS) can be employed for n*. Why? Use it and (a) to obtain a scale for ü*. (c) We assume that Ou*/ex* + De1*f y* = O. (What does this imply?) Consequently, the two terms on the left of (liôa) should have the same magnitude. Why? The scaling for v should now be apparent. (d) Find a formula for the order of magnitude of d in terms of the other parameters. Use it to show that the concentration boundary layer is about one-tenth as thick as the viscous boundary layer in those eases (which occur frequently) when Pe 1000 Re. (e) Set up a dimensionless mathematical problem for determining the leading approximation to the concentration, within the concent ration boundary layer. Remember: u* and u* are regarded as given, u* by (1S) and us by a r:nrresponding formula. (f) As a rough approximation, the rate at which chemical is being absorbed at the boundary, per unit area, is taken to be DCad. Justify this approximation. 14. This exercise is concerned with a methnd for obtaining the Blasius Nat!. Acad. Sci. 27, similarity solution that is due to H. Weyl 578 (1941)]. (a) Let 9(q) = F"(7). w here F is defined by (9). Show that (9) is equivalent to an equation of the form g = (Dig), where .

(Ng) = ex (b) The above equation can be treated by means of the following successive approximations scheme (compare Section 7.2 of I): fa=U, 1,2,....

Show that (z — C) 3 exp (—

g 3(2) = exp 0

^

dt;

-

Sec. 3.5} Vortiriry (7ruriyes in Viscous Fluid Mork*: (c)

(d)

129

Demonstrate that O(g) O, (g) (Dig*) if g s y, Show that all odd approximants (i.e., all g with odd subscripts) are greater than or equal than even approximants. Show further that as n increases, the odd approximants decrease and the even ones increase. Prove that r

g,(z)]

-

,

and thereby deduce the existence of a unique solution q to which the approximants converge.

3.5 Vorticity Changes in Viscous Fluid Motion If a Muid moves with velocity v(x, t], the fortieity trl(x, t) is defined by ru(x. ti = curl v(x, r)

(1)

A major result of Section 3.1 allows us to interpret to as twice the angular velocity vector of the sotid body rotation, which, together with translation and elongation, makes up the local fluid motion. It turns out that great insight into the nature ol- fluid motion is attained by studying the mechanisms by which vorticity changes, in general and in particular instances_ An introduction to these matters is the subject of this sectton-

In a viscous fluid of uniform density, subject to a conservative body force, an equation for the evolution of vorticity can be obtained by taking the curl of the Navier-Stokes equations (1.28). By judicious use of various identities (Exercise l) one arrives at the following result:

r

+ V • VW

w •''v + V

2 (1.1.

(2)

The pressure does not appear in (2), and this is one reason why, it is advantageous to study the development of vorticity. VORTICITY CONVECTION AND STRETCHING

Let us first consider how vorticity changes in the absence of viscosity. With v = D, (2) becomes Po.)

at

^-

v • Vw = w - Vv,

i.e.,

Dw

Dr —

op V v.

(3)

As discussed in Section 13.3 of 1, the term ' Vt.) represents a change in vorticity due to convection, in which voracity is bodily carried from place to place by the fluid motion.

Viscutis Fluids [Ch. 3

As a start toward understanding the fa. Vv terms consider at time r two material points P and Q (points fixed in the fluid). Suppose that P is located at x and Q at x + En, a distance r. along a unit vector n (Figure 3. i 7). A short time Ar later, P will have been displaced by 4'(x, r)Ar + ()UAW] while the corresponding approximate displacement for Q will her(x + nr, r)At O[(Ar) 2 ]. Neglecting 01Arl 1 terms, we see that the original vector in which -t joins P to Q will have been transformed into a new vector given by un + [+.(x + un) - v(x) Via --- un i (n V+)[:Ar + O(t: 2 ). -

(4)

Subtracting the original vector en and dividing by Ar, we conclude that to lowest order in a the quantity tai[ Vi is the rate of change per unit time of a {short) material vector En emanating from x. n

E t. 1.)^

1

[Ie -pYipr

1 ,. - [ Î A t

F u te r 1.11 A short line of parades PQ rs riirrte+f in sink. At [utrf? the line P'Q' by flow with t•elueirr v -

We can write D(Fn)

Dr

= [:n• V r

(as i— O).

(5)

Using (3) and (5), we sex. that if a—cn is initially zero, then it remains zero, for D(m — un) = tl. f)r

(6)

[The result holds if the differential equation has a unique solution, for w — Fla ; 0 is certainly one so lution of (6) that satisfies the initial conditions_ Uniquenesswill obtain if l Vr is bounded-j What we have shown is this. Units can be chosen so that to can initially be identified with a material vector to of small length F. The smaller t is the more nearly it is true that to continues to be identified with, and to move like, the • Our discussion hut! will be intuitive. A rtg,ornus appruaCh Section 2-3 on tensor caluul«s-

ta ou[li=3rd in trie cxEfeLSCS [or

Sec-. 3.5] Ifurtii ihti

n ges in Viscous Changes

Fluid h#otdnn

11

material vector i.n. Thus (31 can be interpreted as describing evolution of the vorticity vector equivalent to that which would recur if that vector were an (infinitesimally long) material vector. Relative motion of the fluid at the two ends of the vector causes it to undergo a rigid rotation (relative velocity normal to +o) and a contraction or elongation (relative velocity along (,))Rotation merely changes the orientation of the vorticity. But velocity changes along u} effectively alter the extent of the spinning region and bring about annihilation or production of vorticity, just as a twirling ice skater can change his rotation speed by laterally extending or pulling in his arms. One can think of vortex lines throughout the fluid passing from one vorticity vector to the "adjacent - one, and vorticity being generated by the stretching of these linesYtsCOUS VORTICITY DIFFUSION AND BOUNDARY GENERATION The introduction of 'viscosity brings into [he picture two new ways for vorticity to change. First, the vino term in (2) is responsible for vorticity diffusion; local spin al one location will, through shear stresses, engender spin at an adjacent location. etc. Second, the necessity for the fluid to adhere at the boundary brings about a shear. Thus in a viscous fluid the boundary is a source (or sink) of vorticity so that an initially irrotational vise iw fluid will not remain irrotational. To obtain formal confirmation ofthe boundary's role as a vorticity source, note from the vorticity equation (2) that the diffusion lean is the negative divergence of a (tensor) vorticity flux — vVw. Consider a fiat plate along z = a: here

so

rrwu=$1.=D

!t

i

=0

c+r rit C'L' Fv =[),--= — - + ,— =0. 0. y' f.: fly i'.t

where we have used the mass conservation condition V-v = O. The %vorticity generated by the plate is given by corrlputing the following expression at z =O: ^^•

iz

= r

l p fi l l) z = vV 2 =- ^ - . i7

p f. y.

We see that vorticity is generated even in the limit r O. It can be shown that there is a unique velocity field v(x, r) that corresponds to a given vorticity field ra(x, t). subject only to the restriction that the normal velocity component of the fluid equals that of an adjacent boundary (with

suitable conditions al infinity). In a short time interval (I, + ,)the vorticity (0(x, t) is convected, stretched, and diffused according to (2) but additional vorticity appears near the boundary (from the boundary source). This provides the unique w(x, r + At) that gives identical normal and tangential fluid and boundary velocities, as required by [he no-slip boundary condition for viscous fluids.

[

Viscous Fluids [Ch_ 3

32

Pressure perturbations propagate with the speed of sound (infinitely fast in an incompressible fluid). Vorticity changes are far slower; this is a Fundamental reason why study of vorticity is so revealing. For further information, start with M. J. Lighthill's article in Rosenhead (1964 In particular, Lighthill discusses how the ideas of the previous paragraph cart be made the basis for a numerical calculation of the flow. VOR7rCITY IN TWO-DIMENSIONAL FLOWS

Great tangling of vortex lines with concomitant vortex stretching is a major feature of turbulence. But such matters are beyond our scope here, and we shall restrict ourselves in what follows to two-dimensional flows [wherein w = (u(x, y), r(x y), 0)]. For such flows the vorticity vector is ,

eu

' ( 0r 0, cu),

^

^u (7)

The stretching term vanishes, since dw

w Vv =

= Q.

(8)

i.I

Vorticity changes only from convection and diffusion, according to the following simplified version of (2): c7rxJ —

+ w • 17r13 — v P r,3.

(9)

at

Integration of (9) over an area

II

Or J 1

w

.

A

bounded by a curve C gives (Exercise 2)

A

-

,

J

u„{ + v

-

-

an

ds,

(l0)

C

where v„ is the component of w in the direction of the exterior normal n, and ira / n is the derivative oral in the direction of n. We see the roles of convection and diffusion again, for the net vorticity within A is increased by the net amount of w convected and diffused across the boundaries. As expected, the rate of diffusion is proportional, via the viscosity y, to the normal derivative of w. Let Us examine how these ideas can be used to interpret the exact solutions of Section 3.2. VOitT'ICITY IN PARTICULAR YISCOL:S FLOWS

From (12.3), the downstream velocity component u in flow between parallel plates a distance d apart is u=

(d 2 - y1 )

(- C = pressure gradient).

Ser. 3.51

VI)rrirrrl. Changes in I['r.uwus Mad Afert?o/rr

^

33

The vorticity

is of maximum magnitude at the walls y = ±d and decreases to zero at the center. Convection ducs not play a role here, as there is no variation along streamlines. There is continual upward flow of vorficity(into the fluid from the lower boundary, through the fluid, and out at the upper boundary) at the Constant rate —tiI!

_

(gyp-

r1Y

=C

.

(I2)

The pressure gradient forces fluid past the boundary 3' = —d, but the fluid must adhere at the wall, so a spin is imparted to the fluid layer adjacent to the wall. This spin sets fluid in the "nex!" Dyer spinning, and so on (vorticity diffuses). Exactly the same process proceeds from the boundary at y d hut the spin is in the opposite direction. At the center of the fluid. opposite spins meet and annihilate each other, giving risc to the absence of vorticity noted above. When thehorizontal plate ofSolution 3,Section 3?, is snatched into motion at speed U, there instantaneously develops a discontinuity in the horizontal velocity component u- Th e horizontal speed is U at the plane and zero just above it. At the initial instant, the vorticity cry can be regarded as infinite at this discontinuity and zero elsewhere. A plane of velocity discontinuities is called a vortex shot here the instantaneous start of the motion from rest creates a vortex sheet, and diffusion of vorticity from this sheet accounts for the entire supply of vorticity in the now thereafter. This picture is affirmed by the exact solution 12.I8. which yields ,

CO

all

Oy

=—

c

^

=

Lf(?rid)

= - }1i1 (4TCV 3 r 1 ) - 1 ''3 exp {

1 ' 2 c°xp(— re ),

— rp 3 J .

r >ü

t

>

[11 = ( 4 l,r) i fz • ^

03) (14)

Upon substituting y = ü into (14), we obtain zero. Thus there is no flux of vorticity through the boundary after the motion starts. But (13) shows that vorticity is relatively appreciable when and only when is of order of magnitude unity or less, that is, when y is no greater in order of magnitude than (wt) I ' 2 . The spieading of an effect across a distance of magnitude (a)U 2 in time r is characteristic of diffusion with diffusivity v. (Compare the end of Section 4.1 in L) Here we see, in addition, an overall decrease in intensity t 312 , as the fixed amount of vorticity is spread over an ever-widening like area

134

Ili semis f kids [Ch_ 3

Suppose that a second plate were parallel to the plate which instantaneously begins to move with speed U. Diffusion would rearrange the fixed amount of vorticity that is brought into being until ultimately the vorticity was evenly spread between the two plates. This is precisely what we see in Solution 1 of Section 3.2, the ultimate steady flow with constant vorticity throughout (see Exercise 3). As a final illustration of the explanatory power of the vorticity concept, consider boundary layer flow. Since downstream speed is of the order of the mainstream speed U S„ a parcel of fluid at downstream location x* will have spent a time of order of magnitude x*1I} o near the plate (which commences at x* = 0). In this time, vorticity will have diffused a distance of order of magnitude (vr) 1 ' 2 = (vx* f U o )" But the region of extensive vorticity is coincident with the boundary layer, so this physical argument gives a boundary layer thickness of order (%x*/[J 0 )' 12 , a result obtained mathematically in Section 3.4. In two dimensional flow past an obstacle, diffusion is responsible for lateral spread of the vorticity generated at the boundary while convection sweeps the vorticity downstream. Bee.ause organized bodily transfer of vorticity by convection is so much more efficient than random diffusive transfer, flow speeds do not have to be very large before vorticity will be confined to a thin boundary layer region near an obstacle. -

SUMMARY OF THE ROLE OF VOMTICITY

As we have seen, vorticity in the interior of an incompressible fluid of uniform density is changed by stretching, convection, and viscous diffusion. None of these mechanisms can create vorticity where it is initially absent_ We have examined how vorticity can be generated at a solid boundary in viscous flow. But no such possibility exists in the absence of viscosity. This leads to the first great fact about vorticity. For a flow that can he idealized as (a) inviscid, (h) of uniform density or having a pressure that depends only on density [sec Exercise 1(h)1, and (c) devoid of body force or having a conservative body force—for such flows, if motion starts from rest or otherwise can be ascertained to be free of vorticity at some time, then vorticity is always zero_ Mathematically, this means that V A r = O, This, in turn (given certain smoothness conditions), guarantees the existence of a function 0. such that r = V4_ Profound consequences result, of which we mention only Bernoulli's theorem (Exercise 15.1.7 of I) and the fact, exploited for example in Section 15.4 01 1, that is harmonic if the flow is incompressible. As for the generation of vorticity, we have examined one major method-the spin imparted by solid boundaries to a viscous fluid. To mention one more, the pressure term drops out of the vorticity equation because it has been assumed to have the form of a gradient and, therefore, to vanish when the curl operator is applied to it. More generally, a contribution from the

Sec. 3.6] Slow Viscous Hoot Past a Strut" Sphere

1

35

pressure term will remain in the vorticity equation, with concomitant thermodynamic generation of vorticity. An important example occurs in shock waves (e.g., aircraft sonic booms). There, discontinuities in thermodynamic variables across the shock are accompanied by a production of vorticity. Expertise in fluid mechanics requires a much deeper study of vorticity than is appropriate here. But we have introduced some of the key concepts that are required, and have given samples of the general principles and their particular implementation.

EXERCISES

1. (a) (b)

Derive (2). Begin by employing the identity found in Exercise 15.1.7 of I or in Theorem 23.7. Show that the text's assumption of constant density is not necessary and that (2) holds for a barotropic fluid subject to a conservative body force.

2. Derive (10). 3. Show that the net amount of vorticity per unit length is independent of time in the exact solution (2.18). Show that this same amount of vorticity is found in the first exact solution of Section 3.2. (This bears out remarks made a few paragraphs above. Note that the vorticity which is instantaneously generated when a plate is impuis ivel ystartcd is the same whet tier or not a second plate is present.) 4. Discuss the distribution of vorticity (a) in the solution of Exercise 2.1; (h) in the solution of Exercise 22 5. Establish a connection between the inviscid version of (9) and the differential equation of Exercise 15.1.4 of I. What is the viscous version of the latter equation?

3.6 Slow Viscous Flow Past a Small Sphere If a small sphere falls steadily in a viscous fluid, the me, all Reynolds number is small. The scientific importance of this one "slow-flow" problem is attested to by the fact that its solution has been used (i) as the basis for accurate determination of fluid viscosities, (ii) as an essential element in Millikan's oil-drop experiment for measuring the charge of the electron, and (iii) as a necessary part of the Einstein-Perrin method for precise determination of Avogadro's number. [Item (iii) is discussed at the end of Section 3.1 of I.] To mention just one application of more general small Reynolds number flows, such flows accompany the motion of flagellae and cilia, the tiny whiplike organs that propel microorganisms through fluid and that also propel fluid through channels in various creatures, including man.

r3 6

Viscoua Fluids [Ch. 3

We shall now outline the calculations that provide a first approximation to slow flow past a sphere. The original work can be found in a famous paper of Stokes published in 1851. Useful formulas for the divergence and curl in spherical coordinates are quoted in passing_ Exercises ask the reader to fill in many of the omitted details. FORMULATION

As we stated below (3.3), the Reynolds number estimates the ratio between (dimensional) convective acceleration terms like u duii?x and viscous terms like r i u2/13x2. When the Reynolds number is small. the convective acceleration terms should he negligible to a first approximation. (it is to be expected that nonlinear terms are negligible in a slow flow.) Theis the pressure-gradient terms should be of the same magnitude as the viscous terms. The dimensional equations governing uniform density low Reynolds number flow therefore are, to lowest approximation, V • v - 0,

V

A

ca — _ fit - 'VP

(La , b)

Here we have used the following relation between the vorticity ea and the

Laplacian of the velocity vector v (given in Theorem 2.3-6)_ V 2111 = V(V • v) — V A [u

[U = V A

,

-

In (1), p is the pressure and p is the viscosity coefficient. From (1 b) we obtain VA

(2)

(Vhw)= Q.

We use spherical coordinates (p. ch, 0), where

x=

p Sin

¢

COS

El,

y

=

p sin ch sin 0,

z = p COS ch.

Unit vectors a {`'', el', and point along the directions p, rp, and U increasing. The corresponding velocity components are t. l", 6}, and 06} . From references such as Rosenhead (1963, p. 133), we obtain the following general formulas, where subscripts denote partial derivatives: p = Pn e" + p i p, e + (p sin O)- pe e'.

V - v = p 1 (p 2 010 + (p sin 4i pV

A v -

(sin 0) - 1(061 sin + [(sin 0 1(of -

(3a)

' (LC'bi sin 0)0 + (p sin 0) - '(v'°')e,

(3b)

— (041) e aervi -

(pu

ro})r] e i E

[(pi's %

—

( »)^etet_

(4)

We take the origin at the center of the sphere and assume that far from the sphere, flow of speed U is directed along the positive z-axis. There should be no swirl, so aim LP) = O. Consequently, from the requirement V • x = O, using (3b), we deduce the presence of a stream function Jr(p, 0) such that t] tPi

(P2 sin /) 14/

v"' _ —(p stn 0)-1 fi r,.

(5)

See. 3.45] 51o11' YisCaus

NOW PUS' a

Snug} Splaerp

137

Only the third component of the vorticity will be nonzero, so the problem can be reduced to

1 2

+

—

P2

c

a a2 2 where c : cos 0,

{'c 2

(6)

plus suitable boundary conditions (Exercise 3) . SOLUTION

It is not difficult to verify that the appropriate solution for flow past a sphere of radius a is ^

2

^^ s'n^

^ - l) (l ^ ü

^[3

^

^,

(7)

^

From (lb), the pressure is therefore given by

(8 )

P = P«L — ip Gra ff 2 cos cfr-

In (8), ? is used to denote density. Remember, p ordinate in this sect lor'_

used for a spherical co-

In general orthogonal coordinates, formula (1.25),

T, 1 = -- {^ii + 2pD ;. remains true, provided that, for example, 7, 2 is now interpreted as the stress vector on the surface element with normal in the direction of the first coordinate vector, projected in the direction of the second coordinate vector. To compute the stress on the sphere, we need the following formulas for selected components of the rate ofdeformation tensor in spherical coordinates : Don _ 2(21'),,

{}^ — ({a sin Or ' (0 ), -4_ p(p

D,m = p(p- rrn n + p -1 (0 nl ) 4

(9)

Using these, alter some calculation one can obtain the final rather simple result that the drag force Fd is given by Fd = 6npLra.

(10)

If the sphere has density 6', then its predicted terminal speed of fall V through a fluid of density (5 in a gravitational field of acceleration g is (Exercise 6) V = a 1g

'(Zr — L5)_

(11)

We note that full understanding of the equations in spherical coordinates requires a tensor calculus which is not restricted to Cartesian coordinates_ This is beyond our scope here, so we have just quoted various needed results. Aris (1962), among others, presents derivations of the required formulas.

L

(Ch 3

vJSC{,us

^^

The low Reynolds number approximation obtained above (Stokes flow) looks considerably more straightforward than its high Reynolds number counterpart, but it should be mentioned that there are actually strong similarities between the two. When improvements are considered, it becomes apparent that the Stokes now is actually an "inner" approximation which matches the "outer" uniform flow. Further progress requires modification of the outer flow. For additional material on these matters, see Rosenhead (1963, Chap. IV).

EXERCISES

I. fa) For what physical reason does the density not appear in the governing equations ( 1)? (b) Verify (2). (c) Verify (8). (d) Deduce from ru (l) a differential equation that in vol ves o n l y the pressure. 2. Justify the introduction of the stream function, as in (5). 3. Derive (6), and provide suitable boundary conditions. 4. Verify that (7) is the appropriate solution to (6). 5. Carry out the somewhat lengthy calculations necessary to obtain (10). 6. Verify (i I )_

Appendix 3.1 Navier--Stokes Equations in Cylindrical Coordinates A study of general tensors, which is beyond the scope of this hook, makes routine the transformation of the Navier-Stokes equations into cylindrical coordinates [or any other coordinates; see Aris (1962)]. We shall only state the results. Thus the principal equations for a viscous fluid of uniform density p are as follows, where subscripts denote partial derivatives. (Body forces are neglected.) Conservation of mass: V - Y = r

1 (^L;^^ M ^ r +

r

-F ri' ` - f).

(l}

Balance of linear momentum: Di]I

Dr D LP' )

D1

►1

- r

^

[Z!

Ld}]^ = —P

+ r - ' ^{ " L1 1ak = —

J — 1 }r + 4'[V ^[r t! r ^

( pr)_

D

Dr

l

pe + 1,[p^L''e' -Jr 2r

= _ p - 1 p,

vV214,.

^ trj t!

— 2r

-2(01)0

2 (r' !

— r

j

^ )e^ +

(2)

2r'Le j ^.

(3)

(4)

.1ftp0lr.6 j?l

et?

r

biro. Li/rrrrrrrue+

AS? rltr.

Here it". 1 ,`rt`, and w are velocity components in the directions r. O. and increasing, i.e., along the unit coordinate vectors eI". et"r and et".

D Dr

V'

^

r'

►

—+at"'_

t'r e12 r

-

r'r

+ r

l 51

f 112 rr'a_; . + r r'r + r'`' r 1)-' + I r'

(6)

The relation between stress and rate of strain is still when p # fir.

1=

P +

(7]

In f7) p and if represent all possi ble combinations of r. q, and k, For example. Tt, is the stress component along e l "' on a surface element whose exterior normal points along 0". The components of the rate-of-strain or deformation tensor are given by 1Drr = (1 ," 1 )r , iDBlr = r I (tnrr 4 r Y". } =

Dv) = r lit' + It `I,, ^ w

-

,

=w

} _r =

1^1

The vurttcfty vector is given by ILO = ❑ AV

t1 .101ld irF +

[ ( O rFf_

e [r tti4'u + r -I [rr [1^ , - (0' 1 0

wr l e

rü^

l`)

Note that the major vector operators have been expressed in polar coordinates: div in (I ), curl in (9), and grad in the first terms on the right of (2k (3), and (4} (the components of p ' Vp).

Appendix 3.2 Generation of Confidence in the Boundary

Layer Equations by Construction of a Finite Difference Scheme for Their Solution Part of the subject of partial differential equations deals with the question of what initial and boundary conditions are appropriate in various situations. It is usually a major task to show that a problem is well posed. i.e., that there is precisely one solution and that small chances in the problem cause small changes in the solution. Here we illustrate on the boundary layer equations a relatively simple way to generate confidence that a mathematical problem makes sense. The idea is to see whether one can establish a finite difference procedure which in principle coutd he used to determine an approximate solution to the problem, and which requires exactly the given boundary and initial conditions. This procedure has independent value as the possible basis of a numerical solution.

Viscous Fluids IClt. 3

: 4n

In contemplating a numerical analysis of the problem (117-20), we first observe that the semi-infinite interval in y trust be replaced by (0, y, ), where y h. is large enough so that no significant variation in the solution occurs. We then limit our calculations to a grid of points, as in Figure 3.18, where u and v. are prescribed as shown. For simplicity in exposition we consider both the horizontal and vertical distances between adjacent points to be the same constant h, but this requirement can easily be dropped.

—

w

e e t

t

ü —r

rp

o a o o o o

O^

R+

D

a D

o o

D

�

O 0 û U

Q

d

�

O

ü

V

O

o

o

d

40 D Q

a

O

C.

# • • •

i

•

O d

O O

• •

•

u - r=4 FIGURE 3. lb. Velocity CoxNporrenls u and r are ta be dererrn inert at the unshaded

grid points. Only u is known at the fruif-shaded points.

We wish to replace derivatives by difference quotients in such a way that we can determine u and a at the various grid po i nts. We shall use the notation r O) " f,. u(th,ih) = ll i , Cf(iit, O) ^Ux(Jh, (1) One possible way of proceeding uses (3.I7b) and the equation obtained by substituting (3.17b) into (3.17a)_ The. a equations are u ; + r+ = Q,

—

rt, , r + r' , = L F i

"

+ 11 .y .

(2a, b)

The first step is to use (2b) to compute :r along x = O. In doing so, we make the approximations vy(i{i, jh) = h -1 (v+! — Jh) —

h -! (sdt .1

(3)

—

fi i (t++,1+! -

1- s.),

— 214

1.j +

141,j- L }.

Upon evaluating (2a) at (x, y) = (ih, jet), using these approximations, and then solving for v,, f , we find that (u1,1

- ^C i j^1 j-1 .

.

— if — h ^(n1 .1 -^ `^ 21414 + ui.i-1)]

(4)

Appe'ndi_v 3.2] t,errrrurrl,+r uj Cilrr/rrierue in doe-

Bind! trklxl'

Le1te'r hyler^frrul ►

141

z r^

F Iii u ti r_ 3.19. firjurrrrrrrrrrar used w obtain a Mk' IYrlree u j r-. Thur sire required twines of u and r• I1re k 1lenrrr at er pone $3 i►rdie used ht' sltudins{ doe Hi r1rra1 rr,ylf halves. resfu'ru!-rh', 4.0 rhe r•Urreivrrefirrrl circle

Ism Figure 3.19). Specializing to i = 0. j = I, we see that t' ;,., is determined by an expression involving the known quantities $1 0 0 , ]f ,+ 8,+0 . 2 (Is is given along x = 0),A, and t' #, p (1 is zero when t --= Ûj_ Repeated use of (4) gives t) all the way along the initial line, up to but not including r e .N . (Note that c , is used in determining t'n.! _ 1 .) r1 The next step is to determine u alongx II. This we do by replacing 12a) by — r1; ) ♦ -11} = 0. h` l { i — ^' i.1

which yields trr^ 1.1 = Ir i.}

i' e

-^

+ r ";

-1

15)

Figure 3.20). Specializing to i — 0, j = I we observe that r1 r , is determined in terms du o. , (given), D o. , just computed). and re.. = 0. Repeated use of (5) allows computation of u along x h up to and including uI N, Since ti t. r , is given, rr is known on all grid points of _v' — h. The use of (5) with i = 0, j = N provides the missing value of r>, r a• N- By repeating the entire calculational process, we can successively determine vertical lines or ti's and 1.1 's. AU the given information has been exploited, and no more seems necessary, so added confidence is derived in the appropriateness of our boundary conditions. In particular, we see that additional prescription of ()along y = hi or x = 0 would be inappropriate, for r at these locations can be computed from the data given. The scheme we have outlined is the general type that is suited for use with an electronic computer. Untouched, and beyond our scope, is the question of convergence as h . [3 of the exact solution of the difference scheme to the actual solution of the differential equations_ Similarly left to other sources is determination of the accuracy of the approximate solution to the difference scheme computed by the machine at fixed h and with inevitable roundoff error. (see

—

Viscous Fluids [Ch 3

42

u

=

?

_

^+

ftu"L'Q E 124. lirformatriu ►t used f[a obtain a new value of it. known information. us ïrr Figure J. I9_

rxldrCYifr's

SJladiary

Of course, other difference schemes can be constructed. For example, one could use (3.17a) rather than (2a) to march rightward in the computation of u. Selection of the scheme that is best in some sense is another question which has to be left to a study of numerical analysis. See, for example, R. Richtmyer and K_ Morton's Difference Meih +ds for Initial Value Problems (New York: Inter'science, 1967) for valuable discussions of the type of proh!erns we have mentioned, We must noie the possibility of a "scheme" that does not require the value of u nn y = O. This seems to cr]ntradrel our statement that ai! given data are necessary. The scheme proceeds as before except that (5) is replaced by h - 1 (u,i l

li .!

J l + ÎI

1 fU

^ 1 — U ..) ]

—

t}

or

By taking = O,) = 0, L 2,_..,u can be computed onx = h star bug with y — 0, without knowledge of u(ii, O) = u I O . But u 1.s _, cannot be computed by this scheme, nor can uz,_ 2 and u 3 _ i when it is repeated, and so on. The whole process of finding evidence that our problem is appropriately forrnulatcd rests on the demonstration that u and tt al each grid point iS uniquely determined. If one or more of the unknowns is arbitrary, the scheme is inadmissible. Now it might be thought that u l - i can be computed from yy I.", I J uo.iv-11 ♦ 12- VD. k -1 - VO.'h'-21 =

h

}

but in using (6) we are indefensibly employing the same expression for t^ , at (0, N -- 2) and at (Q, N — 1), In the limit N —. co, h -6 Q, Nh • } , this means that ra},fO, 1 O, which is generally faise ,

.

Appendix

Generation of Canfid[rrCN in rhi'

BoülrifClfi' 1.-L11 -*V' {:i

M ürNN.'S

1 43

Use ofa finite difference scheme to verify the appropriateness of boundary conditions is not an approach that is free of problems. The total number of unknowns at the various grid points is always far exceeded by the number of equations that could be regarded as approximations to the given set of differential equations. One wishes to devise a set of difference equations that give a unique set of values for all these unknowns, These equations should reduce to the original differential equations in the limit 11 • 0, and no further relationships among the variables should result. Any '*sensible" explicit difference scheme satisfies these requirements, provided that and r grid distances are appropriately related. To prove that a solution to the original problem exists, it remains to show that the sn to the difference equation approaches a limit, and that this limit is a solution to the original differential equations. This may not he so, and proofs in any case are not easy,s But experience teaches that sensible formal procedures are justifiable more often than not"; this is why the ability to construct a sensible difference scheme gives confidence that a solution exists to the original problem_ In the present context there arc serious extra difficulties because of the necessity of passing from the finite domain treated in a difference calculation to the infinite domain specified in the original problem. Indeed, questions of existence and uniqueness for boundary layer problems are definitely not trivial, as is illustrated, for example, in the discussion of classes of similar solutions in Rosenhead (1963, Chap. V, Sec. 21). In some cases, for example, uniqueness is only obtained if decay as r — x is required to be as fast as possible. It is hard to see how the necessity of such extra conditions can be anticipated by the present approach_ But the very difficulty of a rigorous treatment increases the value of a heuristic approach, provided that the limitations of heurism are not forgotten.

* We have pointed nut that the process is not without difficulty, nonetheless, formulating appropriate difference equations and Mien passing to the limit proi.idcsanelIcctiveway to obtain existence and other theorems for partial differential equations. For the wave, heat, and Laplace equations, a classical reference is the paper al- Courant, Friedrichs. and Lewy tired ore p. 12t of 1 [IBM J. i 1. 125 (1967)]. Major works concerned with the existence of solutions to the boundary layer equations (3.I7-20) have employed the method of tines wherein difFerencing with respect to one variable transforms the original prnhlem unto a system of ordinary differential er;4ations_ Q_ A. 01einik [Sager Math. 4, 583 (19631] uses the rrari.sverse line rnethod in which the "limclike" x variable is discrttized_ W. Walter [Arch. Rarional Mech. Areal. 39, 169 (1970)] proves an cxislcnce theorem by means of distretixing the other, spacelike, variable after making a preliminary transformation (see Exercise 3.41

CHAPTER

4

F ound a tions of Ela sticity

I

chapter we shall lay the foundations of the classical threedimensional theory of elasticity. We shall restrict ourselves to the assumption of very small deformations and use as our stress-strain or constitutive relation the mathematical representation of an isotropic, homogeneous, linearly elastic solid. Simple exact solutions will be shown to provide relations between the various coefficients that appear in our linearity assumptions. Considerations of energy and virtual work will provide uniqueness theorems. The chapter concludes with a brief introduction to solid mechanics when the assumption of very small deformations is dropped. It should be noted that the derivation of the equations governing linear elastic solids bears strong similarities with the corresponding derivation (in Section 3. I ) of the Navier Stokes equations for a viscous fluid. N THIS

4.1 Analysis of Local Motion The ftcld equations that apply to all continuous media must be supplemented by constitutive equations that describe the particular nature of a given material_ This section contains a prerequisite to the postulation of constitutive equations in solid mechanics, a local mathematical description of material deformation. "Compatibility equations" are also discussed. Cartesian tensors are used throughout. STRAIN TENSOR IN MATFttIAt. COORDINATES

Consider a particle located at the initial time t o at Po , a point with the coordinates A. The medium, of which this is a generic particle, is deformed. At time t, the particle is located at P, which has the coordinates x(A, 1). The continuous medium is said to be strained when the relative position of two particles is altered. We shall now develop a quantitative measure of such a

strain. We introduce the displacement vector U defined by U=x—A

or U ; =x ; — A.

(la)

With explicit indication of dependent variables. (la) is written L',(A , t) = x{A, s) — A ; .

(lb)

Here we are referring the variables to material coordinates and thus are using the material or I,sgrangian description. 144

See. 4 !]

A nr11f•sis of

Lora! Moti o n

Let us write (lb) in the

1 45

forrni + i^^(A , ti•

x,{A. EP =

(2)

[In words, (2) states that the position x ; of the particle that was initially at A is now (at time t) equal to its initial position plus its displacement from that position.] Let us compute the derivative of x,(A, r) a+ii h respect to A. Using index notation_ we find that

xi.i _ o3i

(3)

Ut.j•

where the argument (A, r) is understood in x,., and L'

. Equation (31 will prove useful in a moment. At time t — 0 consider a vector dA whose initial and terminal coordinates are A, and A; + dA,, respectively. (For the present, regard dA 1 dA 2 , and dA 3 as arbitrary numbers.) If dA is embedded in deforming material, at time t its end points occupy the positions x(A, I) and MA + dA, r) (see F rgure 4 I) ,

,

We can write

x,-(A # dA, 9l — Y,(A. t) = x,, 4(A, t) dA ; + - - • ,

(4)

where - - indicates further terms in a Taylor expansion and the summation convention is used (as it will be throughout our discussion of elasticity). The smaller I dA I is, the closer to zero is the contribution made by these additional terms (assuming that x i. j is continuous in a neighborhood of A). A convenient way of stating this fact is to say that if dA is infinitesimal, and if dx(A. dA, r) is defined by dx ; =x i(A+d A, t)— xpA, t),

(SI

then we have

(6)

dA,_

The ellipses (• • -1 are absent from 16) because (6) is "exact" when dA is infinitesimal.

xrA*dA tl

1 1

f.0

1 ^+iA n

1

^I

t / dA

A

F to u RE 4.1. Defarmarra.3 dx

Jrne

^^f; tA + dA. ti

^

1

f

//

A + dA

changes rhe t nfi nftesrma l l i ne Element dA emu t1 new elmnt

Foundations of E%ISlteit} (Ch. 4

1 46

If we combine (6) and (3), w e obtain dx ;

_ [64 + UJA, r )] dA ;

Or

dx ; = del ; + f.+ i.i(A, t) dA ) .

(7)

Equation (7) has an important geometric interpretation. A straight line is generally deformed into a curve_ By contrast, (7) implies that an infinitesimal straight line (or Fine dement) is deformed into another infinitesimal straight line (Exercise l). Consider two line elements ds„ and ds4 through the point P 0 . After deformation, these become the line elements do and ds' passing through the point P- In component farm, the various line elements may be written as

dsa = dA ; et^M ds = dx ; e" I ,

dtir,

ds' = d)610 1 .

Here el is, as usual, a unit vector pointing out along the x ;-axis. The length of the line element ds will be denoted by ds and a corresponding notation will be used for the other line elements. We are now ready to introduce a measure of local deformation. It is not obvious how this measure should be chosen; indeed, more than one choice can be made-Trialand error indicate that one appropriate measure is the change in the scalar product ds„ • ds4 due to deformation. Thus we consider ds • ds' — ds„ - dsQ = ds ds' cos 0 — ds„ ds' o cos 0 c ,

(8)

where the angles are shown in Figure 4.2. Employing (3), we may write this difference in scalar products as dx ; dx; --

dA, dA; = (hi, + 1r1 }(o,k + U4 . 0 dA, dAx — dA 4 LC, = dA i dA: + (1i ,, + l`l j , k + U,,,Ui.x) dA } tfA k — dA i dA t (9) (U'k,i + I;F J k + [1;. }Ui, k j dA) dAi. ,

1

We designate the quantity in parentheses as twice the material [or Langian] strain tensor,* which we denote by 2 thi . Thus F

k1 = ^ k ^

j. k +

Ui.

(1U)

4 k-

Note that the strain tensor is symmetric: rl = difference in scalar products can now be written as ,

k

ds • ds' — ds o • ds#, = 2;4 1 dA ' The word "tcosur" is appropriate since ^ is Farmed by addition tives of a vector.

(11.) and contraction

ofideriva-

sec _ ¢_11 Analysis of Loc•ai Motion

1 47

A,

5

dst,

« A2 fil)

A1

FI C;URE 4.2. r]c}{rrrrrrarirsn changes the

angle between two line dements from

Igo ro A. Frequently, the strain is so small that I fi r J ! 4 1. In such cases it is natural to replace the strain tensor by its first nrder approximation, writing rhh i — (j +

;ark

(12)

When the strain tensor is given by the linear relation ( 12), we shall speak ref

infinitesimal strain . GEOMETRICAL. INTERPRETATION OF STRAIN COMPONF:NISti

now show huw changes in ienjths, angles, and volumes can bu described in terms of the components rh) . We consider first the special case when the two line elements ds o and ds4, coincide and lie along e": d , = dA; =ds o ,dA 2= d4 =dA 3 — dA — 0. In this case dx ; =dx; and [from We

(11)]

rfs 2 --- dsô = 21 1 , dsb.

(13)

Let us define p 1 by the relation dw = dsa( l 4 - m i )

-

(14)

Thus

1.11 =

ds — ds" ds n

(15)

in other words, p is the change in length per unit original Length parallel to the A s -axis. Substituting (14) into (13), we find 2a

1

= 2 fc, +

(16)

i48

Foundations of Fla_sririty ICh. 4

The solutions of this quadratic equation in j1 are Iii =

±11

+ 2/11) 112 — 1 -

Only the plus sign yields a meaningful root, since p 1 must vanish whenever 11 = O. Thus ra (17) p, — (1 + 2 7i 1 )' — 1 and (18) ds =, ds o(l + 2 ) `/2 1, and (17) implies that

For infinitesimal straits, I m i l

p 1 = 1 + 3.(27 11 ) + 0( 71i1) — 1

Thus ii +,

or

t7t

^

^

Nt•

(L9)

12 2 , and +Iaa represent the change in length per unit original length

el), e'2 t, and em, respectively_ Let us consider a second example. Here we examine the deformation of line elements which are initially orthogonal, so that

parallel to

dA t ds d , d4 2 —dA 3 = 0,

dAï=dif 3 - 0,

dA; — dso.

From (8) and (11) we find that ds ds' cos i} = 21 1 r ds L dso _ ,

(20)

Let y 2 = n — O. Thus y . 2 represents the decrement in the right angle that was orig i nally formed by elements in the directions of c' and ec2. We have ,

cos $= cos (-n — y t2 ) i sin y i 2, ds = ds (,(1 4 2ri l I )" and ds` ds'a(1 + 211 22 ) 1 ' 2 so that, from (20),

sin Yi

2/ 12 _ (1 + 410 1/2 (1 + 2g22)" 2.

(21)

For infinitesimal strains, (21) reduces to Y12

2 1 12

(22)

'

when i # j, rl rt represents one-half the decrement in the right angle that was originally formed by elements parallel to &' 1 and e1r1. A further interesting interpretation is found by considering the Jacobian which represents the ratio of a volume element after deformation to a 1, volume element before deformation (see Section 13.4 of I). The Jacobian is given by

Thus we see that

fur infinitesimal strains

=

Nx1, x#* xA

00 ] , #1.# r A3)

_ det

r At ♦ (id aAj

see ¢- f( Analysis

Loral MraliUrl

1 49

Expanding this determinant for the case of infiniteNtrrtaII 3t rains, we find (Exercise 3) that

J^ 1irhi .

(23)

In this ease, then, the trace 1.5f r),,, namely q„. represents the change in iwrlunie

after delimiiution, per unit original whittle. At this point it is worthwhile looking hack at what we have done in this section, in older to emphasize that there are two independent levels of approximation involved in arriving at the infinitesimal strain tensor (l2). Using the word loosely, "strain - at point 1' may be characterized by examining the deformation of various lines that pass through P. As is customary, we confined our attention to the fate of such lines in the limit as t h eir length approaches Zero. It can be said that in characterizing deformation we only examined tenpoints that are "nearest neighbors" of P. This restricted the precision or our characterization but was clearly the appropriate first step_ Having decided to limit our attention to the deformation of anfinitcstmal straight lines, we derived an expression (9) that can he used to provide information such as the increase in length of such lines, the change in angle between a pair of them, eic. Equation 19) provides this information even in cases n f the most extensive distortion. For a large class of important problems, however, we can limit ourselves to very small distortions. It is this rel,rlively simple but very useful linear theory. wherein )10) is simplified to (14 that concerns us almost exclusively in this volume. Rut Section 43 gives a taste of the theory when the strain is chararterifed by arbitrary distortions of infinitesimal line elements, STRAIT TENSOR IN SPATIAl. ; OORI}INA"iES

An argument entirely similar to that just used can be applied to develop a measure of strain in spatial coordinates For this purpose, we write - A,(1i, 1l Note the following connections between velocities r and V and displacements u and 11 in spatial and material coordinates. respectively: V-

six s1!

_

r''1A + U)

-

r'!

r' ll

Du

ci

Dr

We can develop the spatial (or F ulerianl strain tensor, c l ((x, [j, defined by t:r! —2(4r. i

-I

( 25 )

where. ; = i1j0x,. The individual compiments can he studied through examples similar to those used in the previous section (Exercise 4) For small strains. WC find that E..1 ` 1(14r. r -

I

ii ) , ,).

I

(2

i)

Foundations of EÏustirity

J5

[Ch. 4

For t — j, tt is the change in length per unit length in the deformed state parallel to the x ;-axis. For i j i: o is one-half the decrement in the angle between two directions that end up in the deformed state parallel to c'i and est, respectively. Comparing (12) and (26), we see that for infinitesimal strains the two strain tensors have the same form except that a material description is used in the former case and the differentiation is with respect to the material variables - while a spatial description is used in the latter case, and the differentiation is carried out with respect to the spatial variables. It should be noted that the difference between differentiating with respect to spatial variables rather than material variables is of second order. To be explicit, ira vertical line denotes a substitution 01- material variables for spatial, we have, by the chain rule, that

_

BfJ k _ k ex,,, Ûut C^(A,,, OA p = ax. 8A p cx^ DA ^U

&k

x ,n

^ Frrz

4-

^^^

t^F 16U„,

ôxp

ax,„

a

(27)

p

Neglecting second order terms, we find that t) _ ô14(x, i)

cOA ,

exp

- A. ffi

(2R)

and also

duk(x, [) Oxp

,au ,,(A, 1) P

(29) Mk.

o

Thus the difference between the two derivatives can be neglected when we are dealing with infinitesimal strains In this and the next three sections w e shall restrict ourselves to infinitesimal strains. In many classical texts, the infinitesimal strain tensor is denoted by t o , although each component is interpreted as if one were considering the material strain tensor_ Also, no distinction is made between :d ; and U 1 . Although this is not strictly consistent with the notation used up to now, we shall continue the practice of failing to make a sharp distinction between material and spatial coordinates in the remainder of the chapter except for the Final section (4.5) on nonlinear elasticity. THE ROTATION TENSOR

We shall now employ the strain tensor to show that an infinitesimal piece of material, alter deformation, is translated and rotated (a rigid motion) and also elongated along three mutually perpendicular axes. We begin by noting that (30) d u ; = u r {A + dA) - N ; (A) = ui . f(A) dAi.

Sec. 4. ! l

Analysis Of Laced MorioR

151

Thus points near A are displaced a common amount 1i ; (A) and arc subject to a dA i . According to Theorem 2. I.10, Ili 1 can further alteration given by u1. or be written as the sum a symmetric tensor and an antisymmetric tensor,

namely, (31)

coo ,

where €ii = ^( 1^c + cir1. i ) and wu =

l.j

.^

— ^i)

[32a, b]

The antisyrnmetric tensor w 1f is known as the rotation tensor. As has been shown at the end of Section 2.1, we may associate with such an antisymmetric tensor the rotadon vector fit , where [analogously to the fluid mechanical case in (3.1 .8)]

flk

""

ifL' n u)k •

ij (2) if

(33a)

As in (2.1.44), Cmr.i[ 1-2k

— WmA •

(33b)

Now [using (7)], dx ; — dA, _c = dA f +w dA i ,

and [using (33b)], we obtain co if dA i = Efl f2k dA = [dA Iti III.

Thus cow dA i is a small rotation through the angle I5li, about an axis parallel to Sk through the tail of dA. 1f eti 0, then the whole deformation reduces to a small rigid body translation and rotation.* We note that the components of the strain tensor c ;; for i = j arc known as the normal strains and the components for i # j are known as the shearing strains_ We have shown that normal strains represent elongations [see (19)] and shearing strains represent angle changes [see (22), and remember the equivalence of e ii and 11 i; in the case of infinitesimal strain], PRINCIPAL AXES OF STRAIN

Since the strain tensor €, 1 is symmetric, it has three mutually orthogonal characteristic vectors which are known as the principal axes of strain. The characteristic values are called the principal strains and are frequently denoted by, , c 2 , and r, . When referred to principal axes, the strain tensor takes the form t 1f

=

Ei

❑

❑

et

0

❑

❑

0.

• This result can also be shown cilrecLly (Exercise 5) ,,

t5z

Foundations of Elasticity

ICI?. 4

It follows that eki dA j corresponds to a stretching, by a percentage equal to the principal strains, along the principal axes of strain. If we take as our system of axes the principal axes of strain, a rectangular parallelepiped with edges d r , dA 2, and fill deforms into the parallelepiped shown in Figure 4.3. It is seen from Figure 4.3 that the change in volume is given by (1 + E 1 )(1 4 €2)(I + € 3 )

dA r dA 2 dA 3 — dA, dA 2 dA 3 .

Recalling that we have confined ourselves to small strains, we find that the change in volume is given by

(c, + t: 2 + E 3 ) dA, dA

Cl/13-

Asis Section 13.4 of 1, the dilatation A is defined as the change in volume per unit original volume We see from the above equation that -

(34a)

d=e: l +f: } + E,.

We note that this expression is the first invariant of the strain tensor evaluated in a coordinate system that coincides with the principal axes. Thus we can conclude that A = e.;; _

(34b)

This result coincides with the analysis of the Jacobian found above_ It can be shown (see Exercise 11.1-20) that the planes on which the maximum shearing strain will occur arc those planes that cut the principal planes (i.e., the planes normal to the principal axes) at 45°. The maximum shear strain will be the maximum of the three quantities Itl — t:1 , ^

E{

I

2 1t- 2

— 1.

3,

(35)

EF3 — ErI-

drt x

FIGURE 4.3. Deformation changes one infinitesimal rectartyulur

parallel-

epiped to another, provided that the original parallelepiped is oriented along the principal axes of straw. (For clarity of presentation, the defèrrtuitxur pictured is much larger than it should he

SET_ 41] Analysis of Loral Milian

1

53

COMPATIBILITY EQUATIONS Suppose that we are given the nine components of the strain tensor, t:,3(x, z), co . Can we guarantee that there will be a displacement vector, u,(x. i), that will satisfy the following equation? =

'PP i l ,

+

(36)

j, i)'

That is, are there three (single-valued) displacement functions u i , u z , and u that satisfy the six distinct equations provided by (36) when c, j is a given symmetric tensor? As we shall see, sometimes one can determine an expression for the strains in a given situation, but the integrability question we have posed must be answered before one is assured that this expression is appropriate. To be more precise, some problems in elasticity are naturally posed entirely in terms of the stresses: The stresses are given on the boundary of an object and are to be determined in its interior. We shall see that once the stresses are known, it is a simple matter to determine the strains. Knowing the strains, can one now find the displacements? (This question does not arise in fluid mechanics, for it is of virtually no interest to consider situations where stress is prescribed on the boundary of a fluid region) The integrability problem can be restated more conveniently in the following equivalent form. Given a strain tensor ar ; and an antisyrnmetric rotation tensor 0.) ;i , under what circumstances can we guarantee that there is a displacement vector u, satisfying the following equation? (37)

u.^ — e,) + cu .

The two forms (36) and (37) are equivalent because the heart of the matter is to relate deformation and displacement ; any required solid body rotation can easily be added. (A formal proof is requested in Exercise 12.) ' Because the problem has been changed from (36) to (37). we can regard it (for fixed i) as a question to which we can apply Theorem 2.3_10. This states that if a continuously differentiable vector field B; = Box} is defined over a simply connected region in space, then there is a single-valued function 4 _ ¢(x) such that

B;0,,

081

if and only if 41-, Bi.

(39)

0_

For our particular problem, we are seeking three functions such that ea,. = + a. As stated in (33h), we may write

[I,, 142,

and u

w J = ZIA i 4 .

We observe (Exercise 6) that nk, k — 1

(1)r1.

_ 0.

(0}

Found/J tiorls nf EIra.slrri 1• tC"h_ 4

154

We can thus formulate our problem once more as follows: Given the symmetric tensor E; and the vector 11. 1, with 1-4 .1, O, find necessary and sufficient conditions for the existence of a vector u ; such that + c;ik il_*

(41)

We may nnw apply our integrability theorem to each component of given in (41). Consequently, necessary and sufficient conditions are esri(Et!

+

€ig, n ).,, =

as

0.

Using the "ed" rule of Theorem 1.2.11,

( 15s145ph — skiFif' k P SI.1

si

+

r25 ^.

The integrabilility condition now reduces to —

t

sr^

c si. p

(42)

We seek a condition that is expressed solely in terms of the c o . Because of the introduction of II, this is easily accomplished by another application of the integrability theorem, this time to (42). The necessary and sufficient conditions for the existence of a displacement vector become Qrs = Û

where

Qrs

(43)

rini Espj E ll. -

The integrability conditions (43) are known as compatibility conditions, since they are the requirements that a given strain tensor be compatible with a sing Ze-valued displacement vector_ Since Q,# = s, [Exercise 7(a)], there are six independent equations. These can be taken to be the following [Exercise 7(b) or Exercise 14(a)] : ,

1.23 = ( - 23.1 + £31.2 + 6 12.3),1, 122.31 = ( — '6 31.2 + £12.3 + E23, X1

4 33,12 = ( -0 12.3

12, 12

+ E23+l

+'6 31.2).3*

(44)

1 11.22 + 1122, i I*

2E23, 23 = 6. 22.33 + C 33.223

2 &31.31

F33. t 1 + 6 11.33'

A multiply connected region M can be made simply connected as required by the integrability theorem by making appropriate cuts. if displacements are to be single-valued in Ai, they must certainly have this property in the cut region. Thus the conditions of (44j are still necessary, even in multiply • It is conventional lu use similar notation for the alternator symbol p, fl. Tbc dangcr of confusion should no be great_ c1, __ L i

E iji,

and the strain

Sir_

4_ Ij

Anufi'ses

Lut

at Morion

155

connected regions. in addition, one must require that the same displacements result at every point on each cut, no matter how these points are approached. (The additional condition has to be verified only on any nne set of cuts that makes M simply connected.) SOME EXAMPLES OF STRAIN

order to illustrate in concrete terms some of the general ideas just discussed, we present three elementary worked examples_ In each case the reader should be sure that he agrees with the given information; (a)a description or the strain, (b) the principal axes of strain and the principal strains, (c) the dilatation, and (d) the maximum shearing stress. In

Example 1. Consider uniform dilatation in which the sirain tensor is given by

Solution. in the particular coordinate system chosen, the strain tensor represents a change in length per unit length of the amount a in each of the x,-dire Lions_ The s hear strains vanish in this coordinate system_ On the other hand, we note that the tensor is isotropic; hence the strain tensor will be the same for every coordinate system_ The principal strains are e; every coordinate system is a set of principal axes; and the maximum shearing strain is zero. The cubic dilation is given by =t = 3e.

or a=

Example 2. Consider simple extension given by the strain tensor e, = e,

i = j = I.

t;r = 0 forail other i and j .

S lulion. In this state cf strai n, the material undergoes a change nie per unit length in the x i -direction only. Since the tensor is already in diagonal form, the given sel. or coordinates form a system of principal axes and the principal strates arc e. 0, and O. The dilatation equals e_ The maximum shearlrtg strain is le and planes of maximum shearing strain are shown in Figure 4.4. x

rr

^

XI (j}

(bl

F! c: u R E 4.4. Planes of maximum sfrenrinrg .srrain in sim ple exrensrnn along r!u' x, -{ixis_

156

Fouxulrarir3rl.s of Elastic- iry [Ch_ 4

Example 3. Consider simple shear in which the sale deformation is a change in the angle between the x,- and x,-directions of 2s. The strain tensor For this case is U

=

s

0

s ü C:). 0 0

(4s) []

The before and after pictures are shown in Figure 4.5, for a section that was a unit square before deformation_ The reader shuuld use this figure to check his results. N o r E. Rotation is not specified, since is not given. ln Figure 4_5(b), sufficient rotation is assumed to align the bottom of ltie figure with its position prior to deformation_ x,

= x,

!h! F tci u K r 4.5 Srrrrin

sfef'r1r. (a ) Be frrre deformation. ( h) After difor-

rrrfrrian

Solution. It can be shc'vn [Exercise 9(a)] that the principal axes of strain are the x 3 -axis and two mutually perpendicular vectors that make angles of 45 0 to the x ! - and x2 -axes. The principal strains are s, - s, and O. Figut 4.5 enables us to confirm these facts frorri the geometry_ It is apparent that the diagonals l'o and undergo the greatest changes in length. 1 fence they must lie along the principal axes of strain, which we thus sec make angles of 45° to the x,- and x 2 -coordinates. The principal strains can be computed from the geometry of the deformed parallelepiped_ We note that t = Io = - According to the law of cosines, the diagonal C is given by 2 ['

= [ I - 1 -- 2 cos (inn — 2s)]" 2 = (2 — 2 sin 2s)' ` 2 — 44112 23130

where advantage has been taken of the s mall strains to linearize the equations. Thus one principal strain is given by

^ ^n = -s. —

lU

Similarly,

I" = [I + 1 — 2 cos (in + 2s)]+r2 = 2 i + 2 0. -I- sj_ Consequently, the other principal strain is

r fu

set-. 4.11 Arscrlj•,ws fit Local Motion

157

The orientation assumed in drawing Figure 4.51bj is assured if u 2 (0, t), 1 ) = u3(0, t7, 1 ) = 0.

ia ;(ü, D, oj = 0,

(46)

Under these assumptions one can show [Exercise 9(b)] that thedisplacerrrents are given by H 1 - 2sx2,

11 = ar = O.

(47)

,

W e conclude with a mention of'the important case of plane strain_ Here the displacement vector is given by

u # - u Q(x t , x 2 ),

or ,

u s - 0,

-

I, 2.

The strain tensor is given by Evd

=

t: o {x

i: e3 = 0,

i

,, X 2 } — 2(u

3.^

a,

+ z^ ,) ^

— l, 2;

1,2,3. EXERCISES

1. Strictly speaking, (7) deals only with the end points of the vector dA. Show that our discussion implies that all points on dA are "deformed'" into a new straight line. 2. Use the quotient rule to show that the Lagrangian strain tensor is indeed a tensor_ 3. Verify (23), and deduce the text's interpretation of the trace of the strain tensor, for small strains_ 4. Carry out the development of the Fuleriari strain tensor, including the interpretation of the components for both general strains and infinitesimal strains. 5. If u; + tr ; - D for all i and j, show by direct integration that the deformation can only be a small rigid rotation plus a rigid body translation. (This exercise is very similar to Exercise 11.}.) 6. Verify (40). 7. (a) Show that (43) implies that Q,. Qom' (h) Show that (44) follows from (43). 8. Sketch the following displacement fields. Calculate the extensions, shears, dilatation, and rotation. (a)

u,

(h) ex 1 , ii i = u 3 Here e is a constant.

D-

9. (a) Find, by algebraic calculation, the principal axes of strain and the principal strains for simple shear_ (b)Demonstrate that Figure 4.5(b) is correctly drawn. 10. A cylindrical bar (not necessarily circular) with generators parallel to the x 3 axis undergoes the displacement -

u1

- Ox2x31 I4 = OXIx3

,

4i 3 =

xi x2), ,

ftxurda rior,s of nawiriry [C h . 4

l 5b

where H is a small angle. Clive a geometric description of the displacement. Find the strain tensor, the principal strains, and maximum shear strains. 1 L Let w = w(x 1* x 3 )and let Greek subscripts range over the values 1 and 2. The displacement vector is given by IJa =

UV a

• ü3 = w } -- — -

,

2(1 — 2v)

X 3 w a^ ti

where v is constant. Find the strain tensor. Evaluate T33

+v C33 ]

^1'

1—

^ik

1 2_ (a) Show that if u satisfies (37), then it satisfies (36). f.(b) Show that if a function u can be found that satisfies (36), then one can always find a (perhaps different) function u that satisfies (37) as well. 113. The compatibility conditions (43) represent six conditions on the six different components of the symmetric strain tensor. There is thus only a single tensor that can satisfy these conditions." Discuss. 14. (a) With the aid of (1.2.11), show that the compatibility conditions (43) can be written

Q rs

—

O.

, = tes.iii

— rr,.,I =

0.

(48)

$(b) Show that (43) is identically satisfied if €; = ^; + 14i.; for a smooth function n_ Why is this result not surprising? 15. Let V be a simply connected region with boundary V. Consider the incompatibility tensor Q defined in (43)_ (a) Show that Q„ . ,= O. $(b) Suppose that Qr,4 0in V r#s; ,

121

r =

Q2I = Q33

^

{]

on ôV.

Show that Q„= Oin V Thus if three of the compatibility equations hold in general and three others hold only on the boundary, then in fact all six independent equations hold generally.' * This exercise is adapted fro m K. Wa hixu , "A NINE on the Coniiiiiuris of Compatibility,” Mai, ph ys , 36, 306-312 (1957). Comments on the significance of the result are grade an the end of Section 43.

Se c 4_2] Nno4t•`:1- f ûnslifutrt•e Equation and Sot»r Exact .awtfitlotLs

t

59

4.2 Hooke's Constitutive Equation and Sonic Exact Solutions In this section we supplement the field ,equations, which apply to any continuum, by a suitably generalized version of Ilooke's "stress is proportional to strain" assumption for cylindrical bars under tension. The generalization involves an isotropic tensor of order 4, and hence two coefficients. (The third coefficient, normally found in the expression for a general fourth order isotropic tensor, is zero by symmetry.) Certain exact solutions will he shown to relate these "Lame coefficients" to the more down-to-earth elastic constants such as Young's modulus. This modulus was introduced in Chapter 12 of I, where a one-dimensional theory of elasticity was presented. The discussion here is independent of the earlier one, but of course bears many similarities to it GENERALIZED HOOKE'S LAW

Let us now state the basic equations governing the motion of a continuous medium. Anticipating a fuller discussion at the beginning of the next section, we exploit the assumed smallness of the displacement and its derivatives to replace the (material) velocity by the partial derivative with respect to time of the displacement: au, Du, (Jul aJ1 i - n t^ i ^ Di ^t ^

Similarly, we shall make the approximation 3 ;n; {Fl 1

Derivations of the basic equations can be found in Chapter 14 of l; the equations are listed in Appendix 2.1. With the above approximations the mass-conservation requirement takes the form

Dp

_ 0.

(1)

The three equations derived from the balance of linear momentum become

P t

! cox t-

{Tii ) +

fi .

(2)

Consonant with its conventional use in elasticity theory, we use ; ro denote the body force per unit volume. The notation pf, is used for this quantity in I, and in the present volumeexcepl for the chapters on elasticity.) Balance of angular momentum gives (3) = Energy equations will not be required in

our discussion.

I643

Foundations of £tusrit- 4y [Ch. 4

The body force per unit volume.. .6(x, r), is normally considered to be known. Thus (1) and (2) comprise four equations; and there are 10 unknowns - the density, the 3 components of the displacement vector (used in preference to velocity in elasticity), and the 6 components of the symmetric stress tensor. 1f we were to introduce the strains, the discrepancy between the number of equations and unknowns would not be altered, since we would he adding 6 components of the (symmetric) strain tensor and 6 relations between the strains and the displacement vector_ Constitutive equations, the stress - strain relations still remain to be postulated. In Chapter 12 of I. on the one-dimensional motion of a bar, we used as our stress- strain relation l llooke's law, which states that stress is proportional to strain. Let us consider this case in somewhat more detail_ Suppose that we apply a tension test to a cylindrical har of structural metal, such as steel. That is, suppose we measure the elongation of the bar as a function of an applied axial force. The. results turn out to be of the form shown in Figure 4.6_ The abscissa, e, is the uniform axial strain, i.e., the change in length per unit original length. The ordinate is the stress T computed by dividing the applied force F by the original cross-sectional area a.

T

T

u tr

__

f ! J ! I

r t 1

1 {}

k' t c u k E 4.6. SfrPs.s T eErso- strain e in a tension test on a cylindrical b u r

We find that up to the point P, the proportional limit, the stress T is proportional to the strain e. Between P and Y, the yield point, the relation is no longer proportional_ lithe load is released, however, the unloading curve will duplicate the loading curve; and, when the load is reduced to zero, the strain will likewise vanish. Beyond Y, the material becomes plastic and unloading is indicated by the dashed line. We observe in this case that when the load is reduced to zero, there is a permanent strain left in the specimen. if the material is subsequently reloaded, the stress strain curve will follow the unloading tine, which is approximately parallel to OP. After the reloading portion meets the original loading curve, the stress-strain diagram continues as though no unloading - reloading cycle had taken place. The stress corre-

Sec. 4_2j Hookr's Consfivaire E.quainori and Soin Lxucs Solutions

rbr

spending to the point U, a local maximum of the graph, is usually referred to as the ultimate load ; point B represents rupture of the specimen. Depending on the particular material, portions of this stress. strain curve may he missing or somewhat altered_ In some materials, the stress-strain curve beyond the yield point may he almost fl at. Moreover, the unloading reloading cycle may look something like a hysteresis loop. One should also realize that the apparent drop in stress after U is somewhat illusory. We recall that the stress plotted was defined as applied force divided by the original cross-sectional area At the point U. the area of the specimen has contracted. If- one were to calculate the stress by dividing the applied force by the instantaneous area, the curve would not drop. For low-carbon steel, P is about 35,000 pounds per square inch (psi) and 11 is about 60,000 psi. For iron, P is about 25,000 psi and Y is approximately 30,000 psi. The initial slope of the stress strain diagram is known as Young's modulus and designated by E. For steel, E 30 x 106 psi, and for aluE = 10 x 10' psi_ (I psi = 69,870 dynesjcrn minum, Classical elasticity is concerned with the linear portion of the stress-strain curve from O to P. It should be remembered that not all materials display this behavior_ For example, rubberlike materials usually have only a curved stress strain law that is approximated by a straight line through the origin only for extremely small stresses. We shall assume that the material is initially stress free, so that all components of stress vanish when the components of strain are zero. Furthermore, extrapolating the tension test results, we assume that the stresses are linear functions of the strains. Consequently, we may write -)

(4)

Ci1hmEkrn-

The quotient rule shows that the 81 coefficients C oin, form the components of a fourth order tensor C_ For most of our applications, we shall assume that the material is homogeneous, which means that the components of C11 ,, are constants independent of both x and t. There will be a few cases in which we shall want wallow these coefficients to be functions of x, but this will be particularly noted at the appropriate time. The components of C_'^ can be found by removing a representative sample from the material and applying a series of tests wherein both the stress and strain components can be measured_ A wide class 01 materials in which we shall be interested are devoid of "grain" or crystal orientation. They thus have the property that the values of the coefficients, so determined, will be independent of the orientation of the sample tested. Such materials are called isotropic. Consider (4) and its counterpart in another Cartesian coordinate system ^i 1 — ^, ijkm Ek

•

t62

Frrundarrrxrrs of F laxslirïr}.

lC'!1. 4

In an isotropic material the contribution, say, of £ 2.3 try T 2 , must be the same as the contribution of E3 3 to T, 2 . Th115 C 1223 =

0 1223

or, in general, C'ukra That is, C must be a fourth order isotropic tensor. The general form of such a tensor is given in (2-1.40. The scalar found there must be zero, however, since Tu is invariant to an interchange of i and ), but the terms multiplied by K are not. [See Exercise 2.1.22 or the derivation of (31.49).] Consequently, (4) takes the form` T,-; = 4E1:4 + 11-41‘ 15 i;-

(5)

Equations (5) are known as the generalized Hooke's law, and the constants . and p are known as the Lamé constants. INTERPRETATION OF THE ELASTIC COEFFICIENTS VIA EXACT SOLUTIONS

We shall consider a number of simple, "armchair" experiments that will give us some physical insight into the two Lam& constants, as well as certain important combinations ct the two. To do so, we must first solve the linear equations (5) for the strains as functions of the stresses. In (5), let us put i and j equal to a common value k. Performing the indicated sum, we find that Tkk -

(2p + 3,110: kk

or f.k k

= --

k

-_

21.i + 3A7

(6)

Thus we obtain the strain-stress equations

2u

2p(2p + 3,1) '

(7)

• The saint final result is obtained From the seemingly more general assumption Ti r To see this, one employs the decomposition ^1.^.

= i.F,. +

i 06w

of (1,31). Then one can make the assumption that an observer rotation will not affect the constitutive equation (principle of material indifference). Alternatively, one can prove that t ic contribution associated with w. in fact, vanishes. - flic discussion is precisely parallel to that for the corresponding case that arises in fluid mechanics. as presented above (3.1.19) and also bckrw {3.1.25].

Sec_ 4_1] Hoake'.s

Curpsrrlurree•

&yurclrrxi und Satin, Exact Su}urrm.ti.

t63

We shall now consider a sequence of situations that are so simple that exact solutions to the governing equations can be guessed. In all these situations the (symmetric) stress tensor is identically constant, there are no body forces, and there are no variations with time, so the linear momentum equation (2) and the angular momentum equation (3) are identically satisfied. The density p is constant, so the mass-conservation requirement (1) is also identically satisfied. The various situation s a re specified byprescribingstresses on theboundarics of bodies having given shapes. From the guessed form of 71 1 we compute the strains from (7). In all our examples the strain tensor is constant, so the compatibility conditions (1.43) are certainly satisfied. TENSION OF A CYLINDRICAL BAR

As our first example, we consider a cylindrical bar under tension, as illustrated in Figure 4.7. The magnitude of the force per unit area is prescribed as N at the ends of the bar, and zero along the sides.

FIGURE 4.7. A c 'li ufrrt ul brrr Wider rrrnwlr

The stress tensor in this case is given by

f) {l 7;o — 0 U 0 , 0UU N

for then both the end and side forces have their assigned values (Exercise 1). The example under study is precisely that of the tension test_ Therefore, we expect the axial stress and the axial strain to be in a ratio given by Young's modulus E:

N Elt

(R)

r64

Foundarrons of F.Ja.siiriry

1C.h. 4

Furthermore, it has king been known that lateral contraction occurs in the tension test. The ratio of lateral contraction to longitudinal extension is designated by Poisson's ratio v. That is, we expect that —

f:22

^1L

_

— ^3 3

=

(9)

V.

^li

From the known values of I we compute the t: i f from (7) and find accord with (8) and (9), provided [Exercise 2(a)] that ^

=^(2^+3^. ) + $.1'

ti:

2(..1-^ ^)

.

(10a, h)

Solving for the Lamé constants, we find [Exercise 2(b)] that

A

_ E l,), ^ ^ 2(1 +

Ev

(1 + v)(1 — 2 ► j '

(lla, b}

and that

v_ E

d

(12)

2,u{31 + 2}^j

The strain-stress equations ofMean now he written very simply in terms of the "engineering constants," Young's modulus and Poisson's ratio: = (1 + ►+)7; f - u 3rkk

^

.

(13)

These equations have a rather simple interpretation_ We note that (i 1 —

TrLr r2 2T33 (14)

and similarly for the other axial strains. Equation (14) states that the tensile strain in the x 1 '-direction is the elongation produced by the tensile stress in that direction minus the contraction produced in that direction by the tensile stresses in the x 2- and x 3 -directions. One could, in fact, develop the stress strain laws by starting with this fact as a hypothesis, together with the assumption that the principal axes of stress and strain coincide. From this it follows that the stress strain Laws in principal axes are given by (14) with corresponding expressions for the other two components. Since the constitutive relations must be tensorial, they must be given by (13) in any other Cartesian coordinate system. -

SHEAR OF A RECTANGULAR BAR

The second case we shall examine is that of pure shear stress. Imagine an infinite cylindrical bar of rectangular cross section that stretches along the x3 -axis. Subject the lateral faces of the bar to shear stress having magnitude s

See 4_4 Huokr's Cotrsatufirr Equation and Same FA ucv Sr, fuftr,ns

165

-r7 -

5

+

^

I

?

i

/ ^

/ ^ / ^

Fi[;t7 sE 4.8. !lie arrows represent Shear farces applied rrl u rectangular hur

.

The miss section delimits into Me dashed parallelogram

and orientation as in Figure 4.8, Then (Exercise 4) the stress tensor is given by F tiJ

s

7:i — s 0 {] .

(15)

t]

0 0

We see from (7) or (13) that the only nonvanishing strain component will be ' l 2 _ If we write E, 2 = ' , then (according to the interpretation of the nffdiagonal strain components found in Section 4.1), the cross section deforms into a parallelogram, as shown in Figure 4.8. The stress-strain law (5)yields T1 — 21"F 2

(i6a)

or, alternatively, my.

(16h)

will be proIn linear elasticity the (correct) expectation that the angle has led engineers to introduce a proportionto the applied shear s portional ality constant called the shear modulus G:

From (I fib) and (I lb) we sec that

E We expect that, for reasonable materials, G > 0 and E > O. Thus I +y>O

or w> —1.

We expect, additionally, that longitudinal extension is accompanied by lateral contraction. Consequently, the inequality stated above will be replaced by the assumption (17) v >p,

Foun dations of Elasticity [Ch. 4

166

COMPRESSION OF A RECTANGULAR PARALLELEPIPED

The final case that we shall consider involves pure compression. Consider a rectangular parallelepiped with faces normal to the coordinate axes. Impose on each face an inward normal stress of magnitude p, p > 0_ Then p =

0

—

0

0

p

U. p

0 0

—

The shear strains vanish and the normal strains satisfy

ECE l _ EE22 = 1;€.33 = (2); — 1)p.

The dilatation, A, is given

by

FA = E kk = — 3(1 — 2v)p. (18) We expect that the dilatation will be proportional to p. Thus the bulk modulus or compression, k, is defined by

= —

(19)

A

(The minus sign is present because dilatation is negative in compression.) From (18) and (19), k

_ 3(1 — 2v)

.(20)

In terms of the Lame constants, (20) can easily be shown (Exercise 5) to become k = À + p.

(21)

It is expected that material will not expand under the influence of a purely compressive stress. For this reason we assume that k 0, i.e., [from (20)] that y< To summarize the inequalities concerning Poisson's ratio, expectations that elongation will not lead to lateral expansion and that compression forces will not cause expansion give rise to the assumption .

0.

(28)

Therefore, it must be that the equilibrium equations are satisfied: 1.1+

f—

(29)

to the contrary, that for sonie i

1.70 at a point and therefore* by continuity, in a sphere about this point_ (Reversal of the above inequality is handled similarly.) Pick a smooth virtual displacement, proportional to a factors, that is positive within thissphereand vanishes outside it. A contradiction can now easily be generated from (28). The reason is that the first term is proportional to E. but the quadratic second term is proportional to £ 2 and thus cannot cancel the first term if t is small enough. CI There are a number of other minimum principles in the theory of elasticity ., for example, the true stress minimizes a certain "complementary energy_" (See Sokolnikoff 1956.) Extensions can be made to dynamic problems by a version of Hamilton's principle (see Exercises I l .3.2 and 1 1.3.2U). Aside from their general theoretical interest, such principles provide practical means of calculation. We shall see examples of this in our study of variational methods in Part D. We note in conclusion that minimum principles play a vital role in the field of elastic stability_ A glimpse of the issues involved is provided at the end of Section 5.1.

I.

Fourzdrsriont of Elasticity [Ch. 4

82

ExEitCI5E5

1. (a) Derive (3a). (b) Derive (3b) by writing 441 as the sum of a symmetric and an antisymmetric tensor. (c) Show further that a(ct,E0) _ ^ t?[ ^t

J.

(d) Use the above results to demonstrate (3d). 2. (a) Derive (9a). (b) Derive (l0). 3. A cylindrical bar of length L, with generators parallel to the x 3 -axis, is oluriiform density p. Gravity acts in the direction of the negative x 3-axis. The bar is supported by a uniform vertical stress on the top surface x 3 = L and all other sin faces are free of stress. (a) Formulate an appropriate mathematical problem that should yield the appropriate stress tensor under equilibrium conditions. (h) Show that the stress tensor given by 0

0

0

U

0

(]

0 0

(e)

p9x 3

satisfies all conditions of (a). Find the corresponding strains and show that the displacements a re given by —

p9x3xr E

_ pg(xl

—ppx 3 x 2 11 2

^- y'x r

+

2E

}'xi

— L2

)3

There will be some arbitrariness in the answer to (c). Why does this not contradict the uniqueness theorem? Remove the arbitrariness by assuming that there is neither displacement nor rotation at {0.0, L). (e) To what shape is the originally flat top surface warped under the action of gravity? 4. Consider two equilibrium states of an elastic body R. The body forces h and surface stresses z, produce displacements u ; . The body forces f and surface stresses t' prcxluce the displacements u. (d)

Si e'_ 4.4]

blend Caneeps und Me l'rini'ipre nf Virruur WOrk

1E3

(a) Prove the reciprocal theorem of Betti and Rayleigh:

fIf' tittdc JJJJu

d

T=

dû-ieR

f51 f?w

d

(h) ff T, and Eli are the stresses and strains corresponding (of, and t, and Tr, and arehe stresses and strains corresponding Wirt and rr, , develop the following alternative form of the reciprocal theorem: fir dr = ffJT 'r° j€} x ^

ij dr.

5. The reciprocal theorem can serve as a computational device in a number of cases. Let us consider one such example. Let

f * = 0 and (a) (h)

1-fi

bir .

Find the corresponding surface stresses r*. Show, from the reciprocal theorem, that the total change in volume AV

- JiikidT

is given by

1 ^ 2v

AV ^ =

x

7k►

d=.

(c) Conclude that I 2v ❑ V = -- ^

,x, du f

—

OR

fIl

fx;d2

6. Derive (20) and (21). 7. By following the lines of the argument given in the text, but proceeding from first principles, show that the equilibrium displacement of a simple Nookean spring provides an absolute minimum to the potential energy. N. (a) Verify (22). (b) Verily (24) and (25)_ 9. (a) Derive (27). (h) Following the line of argument given in the text, provide a format proof that (28) implies (29) In particular, provide an explicit formula for a thrice continuously differentiable virtual displacement that vanishes outside the sphere. DO. Show that if body forces vanish and if displacements are prescribed over the whole boundary, then it is the strain energy itself that is minimized at equilibrium. -

FoUrtdarions of Elasticity [Ch.

I H4

4

4.5 Some Effects of Finite Deformation In this section we take a brief look at some effects that were exeludcd from consideration heretofore, owing to the neglect of nonlinear terms in our characterization of deformation_ We begin with another examination of kinematics, based on the algebraic "polar decomposition theorem." Our deeper understanding of kinematics is used in formulating a rather general constitutive equation. We use this to reinvestigate simple shear. Although consideration of this one elementary case obviously can only scratch the surface of the subject, nonetheless we can discern some important effects of finite deformation and we have laid the groundwork for further study. Our approach is based on the book of Jaunzemis (1967), which is a good source for some of the reasoning that we shall omit, and which provides a useful bridge to modern research in finite deformation theory. KINEMATICS

We showed in Section 4.1 that in the linearized case local deformation can he regarded as a superposition of a solid body rotation and stretching along mutually perpendicular axes. We now show that the same type of result is true in the nonlinear case_ Later we demonstrate that the linear result can be recovered as a special case of the nonlinear theory. We begin with (1.6) dx• = x ; JA) d A ; .

(I)

which we write in the lams dx, = f {A) dA,

or

dx = F- dA.

(2)

Here F is a common notation for the position-gradient tensor Vx. We assume that F is nonsingular, so that each initial line element dA is transformed into a unique line element dx, and conversely_ tinder these circumstances the transformation is characterized by the following theorem. Polar Decomposition Theorem. if F is nonsingular, then it has a unique right decomposition of the form

F = R •tI -

(3)

R'r = R ' ;

(4)

1 Pere R is orthogonal,

and U is symmetric and positive. (A symmetric tensor is termed positive if all its eigenvalues are positive-) There is a unique left decomposition of the form F= V •R,

(5)

where R is the same tensor that appears in (3) and if is positive and symmetric.

Sec. 4.5J Srarrre Etr•Crs of FtntiY iJefnrmarron

185

A transformation associated with an orthogonal matrix' such as R preserves length, for y t = R rj x j implies that

= RraR^k^xr = x k x

(6)

since Rrjpfk =

(R1.R)

Thus the transformation induced by R is a rotation and R is called a rotation tensor, Moreover. (I and V induce stretching (or contraction) along their three mutually perpendicular eigenvectors and are termed the right and left stretch tensors, respectively. Proof of the polar decomposition theorem. Define w by

iv;—F

.1 1.

W e k n o w that ►

= 0 if and only if Y = 0,

f 7)

since F is nonsingular. Thus the right side of 14'î 11 r

t, I r ik t: 1 l k

(8)

is a positive definite quadratic form, since (7) guarantees that the form vanishes if and only if L. — v t = r 3, = O. Associated with this form is a (symmetric) matrix C:

i.e., C = F T `F•

(F r ' )k

(9)

Since the form is positive definite, C has positive eigenvalues. In principal axes, C is thus represented by a diagonal matrix with positive entries. Let U be the tensor whose representation with respect to these same axes is a diagonal matrix consisting of the positive square routs of the corresponding entries for C. In these axes, and therefore in all axes. U has the properties

U T ' = 11.

(10a, El

u• I) =C.

Since U has positive eigenvalues, it is nonsingular. We can thus define a tensor R by

(I 1l

1,1 = F-

Clearly_ 13) is satisfied. Also R t::

R lr,R T (1) Y

}

i

r-

urthcug❑ nal.

for [using Exercise ? t 251 h11

FTF-F-Ü t =

,- fi As here, we Swatch frçcl} bciwcCn Ytrtso

U

,u (12

and the inalricci that represcnl them in s[io -ne

coordinate system. Thus R is a main. that gi c Lorrmpo11 r of R. and (H T ' R,, is t he fifth component of the matrix product formed by the transpose of R and R uself Carte55an

i'•iaurrdafions nf EfusriRtily

lC'h. 4

Suppose that there were two right decompositions,

F = R• U

F R- U.

and

(13)

We have FTr •

F = i1 i r

• R Tr •

R. lrl = U • U and similarly F i r- F = iJ - U.

By considering the equal symmetric tensors U. U and U F in principal axes, we see from the uniqueness of the positive square root that U = V. Now R = R follows from (13). Thus the right decomposition is unique. Suppose that F — V - P provides a left decomposition. Since P• P T` = 1, we have

= P • (p Tr

J'

.

I/

Fs)

( 14)

-

Because of the uniqueness of the right decomposition F = R • [:, (14) that imples

P IT• V - P

P=R,

ii.

(15)

Thus there is a decomposition of the form (5), and the required properties of V now follow at once from the known properties of i1 [Exercise I (a)]. L C is called the right Cauchy-Grew deformation tensor_ Comparison with

(1.10) shows that it is related to the material or Lagrangian strain tensor if as follows [Exercise 1(b)] : =

1(C — 1.

(1 b)

An analogous development can easily be carried out using spatial coordinates. We write

dx , where f-1 = A .,. ,

,

(17)

A tensor e is defined by _ { T Lr 1 cik = (1 1 )jk = JLl

{ ik

(IS)

-

This tensor is related to the spatial or Lulerian strain tensor z by E—

4I

I

-

C)-

(19)

But, using the chain rule, we find that

_ fixa

.2

_

cox ;

G * xLAt A —A 0 side will be contracted (u 3 < a). We designate the coordinates of a point after displacement by x;, ,

x;—x,+ u,i(x)The x 3 -axis (x 1 = x2 = Ü) will be called the neutral axis. After displacement, we see from (14) that points on the neutral axis will occupy the positions x^

Mxi =

xz = f},

2E{

(15)

x r3 = x3.

Since the strain component E3 = u 3.3 vanishes when x 1 = L} we see that (according to our linear theory) the neutral axis is neither elongated nor compressed. Equation (15) shows that it is bent into a parabola that remains in the {x 1 , x 1 ) plane. Along the neutral axis, we also have ,

-

dxi _ Mx3

dx3

El

(16)

Some Exarnpfe s raf Static- Problems ïa Elasticity [Ch. 3

zoo

-

and

d 2x . dx 3

_

M

(17)

Eaf

From {14) and (16) we see that u3

_

--.xi

8xi(O,0,X3)

ex

(IR)

Throughout our whole discussion, it has been assumed that the displacements are sufficiently small that linear theory can he used. In this approximation the second derivative d 2x',{dx3 equals the curvature h of the neutral axis_ When it is desired to extend our ideas to include nonlinear effects, it tarns out that (17) can still be employed, provided that it is expressed in the form M Ï

1 R

(19)

where R is the radius of curvature. This is the Eider--Bernoulli rule for the bending of beams. (Note that the introduction aril allows this rule to beexpressed in acoordinate-fre,ernanner-) In the form M = EIR - ,the Euler -Bernoulli rule states that the applied moment M is directly proportional to the curvature R - ,with proportionality constant the flexural rigidity 'f. As expected, the rigidity increases with the stiffness (measured by Young's modulus F). The dependence on the moment of area I is explained by the observation that a bar will be more rigid if its material is concentrated tar from the center line, where the strain is greatest. LNTRC1Ut.[:TL(]N Ti] THE: ENGINEERING TLiE.URY O F BENDING— RAMC ASSUMPTIONS

We are now in a position to derive an approximate theory of the bending of beams in which the applied forces that produce the deformation are not purely terminal couples. We shall examine bending by lateral forces. (Our approach is in the spirit of the discussion of the lateral motion of a bar found in I Chapter 12.) A key assumption in the development is the use of the Euler-Bernoulli bending law (19). We recognize that this is a hypothesis that is somehow more restrictive than those already used in 0115- development of three-dimensional linear elasticity, since it has been derived for bending by couples rather than the more general forces considered here. A more complete theory of bending, which is technically known as flexure, is considerably more difficult mathematically. We shall not consider this more complicated problem but rather refer the reader to Sokolnikoff (1956). It should be noted, however, that one approach to flexure leads to the Dirichlet problem in two dimensions. This is essentially the same mathematical problem that arises in the next section when we analyze torsion. ,

See. 5. 1 1 Bending of Beams

2)L

We shall consider a uniform prismatic bar with the x 3 -axis passing through [he centraid of each cross section. The x 1 and r 2 axes are directed along the axes of principal moments of inertia of the cross section as shown in Figure 5.2, where ara I-beam section has been chosen For definiteness. For simplicity, we shall take the top surface to be horizontal, hut this is not necessary for our argument_ The cross-section area will be designated by A, the mass per unit volume by cr, and the mass per unit length, d 4, by p_ .

FIGURE

5.2. Di irihution a/ Alive cm the tap q an 1./war o

The problems that we have chosen to emphasize in this chapter conic from the literature of elastic equilibrium; that is ; they are independent of time. We shall depart from this program for the present example and consider time dependence. The analysis is not made much more difficult by this extension, interesting wave-propagation problems can then be considered, and the static equations will, of course, arise as a special case. We suppose [hat to the upper surface is applied a force that acts in the vertical direction only and varies only along the length of the beam. This force may then be specified in terms of its distribution per unit ierigrh, fix, r), where

fix,

= f (x 3 , I^[.

(20a)

The internal forces will be specified in terms of resultant forces and resultant moments acting on a section of the beam. Consider a portion of the

Some Examples of Static Problems in Elasticity [Ch. 5

202

f

.

1-

MO; r, +}

tit

r1l17. f+ r FIGURE 5.3. Force', and bending truurlrtttt arum; un arhitrur# +t•ynre+ft of a lu■rattt. 4surrt•iatexl uptll tln ttrtlsHle7iI rxrr she k i t (right} rnd of the bur ii v dut (rrrx.ti.ti) rr•/tre:sertriny head (mil) off to: rrrrlHl -. trr-r-rrrditict tr+ fllr rrgltrhuurl ride Am rr< trrr repres;krnarrr,.l u { raf+tmYlt ►

beam between x 3 � a and x 3 = b. as shown in Figure 5.3. The forces acting on the ends of the bar can be resolved into vertical and horizontal components. It then becomes clear that these forces sum up to a shear force Q, a bending moment M. and a tangential Force P. We assume that the latter is tern as we are restricting consideration to bending only. The shear force Q is defined in the following manner: Q(b, 1; + ) is the total vertical force exerted by the material on the side x 3 > h on the material on the side x3 < h across the cross section x3 = b;Q(a. t; --) is the total vertical force exerted by the material on the side x ^ < a on the material on the side x 3 >a across the cross section x 3 — a. Since vertical force acts in the x r direction, we may write (in terms caf the vertical unit vector ï shown in figure 5.2) -

l_'tlbt

Qih.r. +)— Ob. t)i_•

It can be shown IF.xercisc 7) that. for arbitrary b, Q(h. r: —1 = —QV?, r: +) _ —Q[h, In similar fashion we can define the bending ftputreeni M as follows: M(h, t; •1 ) is the total moment exerted by material x 3 > h (In the material x3 < h across the cross section x 3 -= b; M(a, t: ) is the total moment exerted by material x 3 a on the material x 3 > a across the cross section x 3 a. All the internal and external forces are assumed to act in the (x r , x3)- plane. —

• The (*uanlily Q(b, t) introduced in (206} is the Magnrtudc al the force (IOC tO the stress ttx 1 , .x 3 , b, i;

i2(h. r) _

JJ T1(xi.x.k iidx 1 di. .4

net

or resealwii vrrItcal

s .rl 8rnr1"fry

13rurna

201

Cooseyuently, we may write ; + j = M(6,

taj.

(1a)

Moreover (Exercise 7). m(/), r; - )

- Mih. t; +)

^bl(h tij.

(21b)

The various shear forces and bending nieinents are shown in Figure 5.1 We shall new assume that the displacements fulfill the following assumptions. (al The center line moves only vertically with the displacement n 3 , r (ln particular, u 3 — Q for points on the center line.) (b) All vertical displacements are the same as that of the center line_ (c)There is no displacement in the x ; -direction. (d) Plane sections that are normal to the conic' line before deformation remain plane and normal to the deformed center line. We shall shortly provide a rationale for these assumptions based on what we have previously learned about pure bending by terminal couples. For the moment, lei us observe that the first assumption is a consequence of the pure 'bending result WI. Assumptions (h) and (c) arise from the "thin" geometry of the heanu since its cross section dimm:nsions are small compared with its length, lateral contractions produced by longitudinal extension (Poisson ratio effect) can be neglected. Assumption (d) can be interpreted as saying that the deformed surface which corresponds to the cross section x 3 = constant is replaced by its tangent plane at the point of intersection with the deformed center line. It is left as an exercise to show that in pure bending the normal to this tangent plane is parallel to the tangent vector of the deformed center line (Exercise 5). The nature of the first two displacement components can immediately be determined from assumptions (a), (b). and (c): -

u,

]'(x3 • r),

it s = O.

Let us turn to the slightly more difficult task of characterizing u 3 . To this end, consider a point Q on the undeformed fiber BQ with equation x i = x ? . jSce Figure 5-4. Note that since the x,-axis points downward, the fiber depicted must be associated with a negative value of 4'".) The displacement of Q can be described as Follows_ (ii) By assumption lb), Q suffers a vertical displacement QS that is equal in magnitude to PP From this we deduce a result that we shall need in a moment : The length of SP' is — x;°j. (ii) The horizontal component of displacement, SQ', must be such as to transfer Q to a point Q' on the straight line PQ' that is the image of PQ [ by assumption (d)]. If the angle at P' of the triangle SP'Q' is denoted by AO, it follows that u 3 (.ti 1 . x 3 . r) = — x, tan All .

20 4

Some Exam ples of Static Problerns rn Elurticity [C'#+.

5

i+ j(Q)

—

x a)

Flu u K r 5 .4. fle}rirmatrrur of a small srymen r of u hnarrr ac•cr ► rdir# to the —

engineering erürrJ rheorti•_ ••

The minus sign arises in a situation wherein we have chosen the xi -axis to point downward and have drawn the deformed center line with positive curvature. Other choice4, however, consistently analyzed, will lead to the same result [Exercise 6(a)]. Since AP'' IS a right angle [assumption (d)], we see that tan Ail Summarizing our

ay(X, , t) ^

rx3

information concerning the displacement vector, we write

= y[k, , r),

11 3 =

tr 3 = — Y 1

r1yfx3 if

.

t1

(22

)

Now we can give further support to assumptions (a) to td). We argue that the qualitative Features of pure bending should apply locally in thin bars, for here the detailed distribution al forces should not matter. And indeed, equations (22) are the same as those which would be obtained by assuming the displacements of pure bending (14) and setting the Poisson ratio ratio v' equal to zero [Exercise 6(b)]. EQUATIONS OF ENGINEERING BENDING

THE a

Having described the displacements and the stress resultants, we are in a. position to write down the equations of motion for a portion of the beam taken arbitrarily between x 3 = u and x 3 = b, as shown in Figure 5.3_ Imposing linear momentum balance, we have t;

^

) + 11(xj, r) dx 3 + Q(h, r; +) =

^

rff

dx, dxi dxxQ(a. r3r2

A

(23)

Sec. 5.11

Benriin,y ui #rrurrrs

205

From (22) we see that u 2 vanishes while u 3 is an odd function of x i , Thus the choice of the x 1 - and x 2 -axes passing through the centraid leads to 2u

3

ylx 3. r1

dx z dx 2 = A

1r

, u.

124a)

A

Consequently, using 120). we find that

f (x 3 , r) rix a

(.4,

+^, ^( I)

QUA, 1) = a ri

r^ 2 }d-^ ^, 1) #ry r

.0

v

dx 3

t

(24b)

This can be rewritten as

dQ(x 3 . r) . /'[x^,r)+ rx

r12 ).lx,

p

l} â

{).

3

w

Since the section between x 3 = a and x # = b has been arbitrarily chosen, we may apply the IDubois Reyrnond lemma* to obtain the partial differential equation describing Ow balance of linear momentum (70.x

r)

{x t) = p

3,

xa

(J t JAX3, rl f,+2

(2i)

Next we write down the conscqucncesof the balance of angular momentum, where the moments are taken about the point x 1 = x2 = 0 x 3 - u- In our calculations, we shall designate by h the width in the x,-direction of the flat, top ; udace on which the external load fix 3 fi is applicd.t Equating the rate of change ill- angular momentum to the torques acting, we thus have. ,

,

f

M(u, r; —) +-

4

(bk

NO, r; f ) f

[lk } A

2

lx - ak) A h

Q(b, r: +- ) =

(x

dx 3 dx 2

ak) n rr

[^U

u

I e^x t dx 2 [!Ji 3 . 1

A

{27) W e now have to calculate the v ector products in (27) and simplify the integrals. W e find that L^ df (x -[dc) n

f^

l!

try

_ ^ ^ x^ r^t ^ r^xs

d 3y ^ 03yy0 ) 3 + x r r7t 2 Px + (x } y u dt' ^

02, — kx2 ^ -

• In the prrSencc

of dascon ITrlulitcs, ;he procedures must he altered. Soc per unir kngxh, so that

t Remember that I wds defined iu be then force

area_

(28) Exercise 12 the force per

Wilt

Some Examples of Static Problems il Elasticity (Ch. 5

2o6

Since the x,- and x 2-axes pass through the centroid of the cross section and are directed along the principal axes of inertia,

fJ

f

J'

x i xi dx, dx 2 = 5x 2 dx, dx 2 = 0,

Jx dx, dx 2 = 1.

A A One can then show that only the j component in (28) does not vanish [Exercise $(b)], and we obtain

— M(a, t ) + M(b, r) + f(x 3 — a) f (x 3 , t) dx3 + (b — a)Q(b, 1) a) 1 at ^^x+A(x3— x3

t

(29)

2 dx3-

As in the development of (25), (29) can be rewritten in terms of a single integral: [(

Q(x3,

x3

r)]

y —d 1

de e x

3

+A(x3—a) -

dx 3 =0.

(30)

The Dubois- Reymond lemma may be applied again, with the result that

â49Q (x3 — a)f(x3 , t) + (x3 --a) + Q(x3, t) 3 =a1 r

x +P(x3—a) d3

(31)

Multiplying (26) by (x 3 — a) and subtracting the result from (31), one obtains DM ^x

pl

^3y

+ Q(x3^ r) — A &2 ôx 3^

(32)

Among other consequences, (32) shows that the shear force Q can be found from the bending moment M and the deflection y. The shear force can be eliminated completely by differentiating (32) with respect to x 3 and subtracting (26), with the result

em

P

2 a4Y 1Or 2 - PK dt 2 3X 2—_ ex2+

49 2y

(33)

Here x is the radius of gyration defined by x2 - 1/A. Our problem has now been reduced to a single partial differential equation (33) connecting the bending moment and the deflection, i.e., one equation with two unknowns.

Ser' 5.1 j

f3e•ndinr) e}J

Beano

207

As in the case of the lateral motion of a bar, we are missing a constitutive equation connecting t h e stresses with the strains. We shall assume for this relation the linearized Euler Bernoulli law (17) developed for the problemil of' pure bending. For small displacements and slopes, we thus impose the condition -

hi = ^1 { ^ PA

134)

Substituting (341 into (33), we obt ekrr the final fourth order partial difterential equation for the displacement y(x 5 , r)

E

(35)

We can obtain the equation for static deflection of a beam by suppressing the time dependence in (35). For this case the displacement itx j ) depends on the force distribution f (x3) according to the equation

y El

(3)

ds 43 = J:

The second term in (35), due to Lord Rayleigh, represents the effect of rotary inertia, as can be traced from the derivation [Exercise Hlcl]. In many applications, its effect is small and it is usually neglected. Under these circumstances, (35) is replaced by f^d }'

^] 4 + C^X3

t•

y

(37)

p ^' — — ' 1. i i

Looking back, we can obtain more insight into cur approximation procedure_ From the footnote to (20b) we see that A ()(k, t) can be regarded as the average vertical force acting at the section x. Similarly, A - 1 M can be regarded as the average j-component of the moment vector. Nonvertical stress resultants and non-j components of moment resultants are completely . ignored. We wish to utilize only the average quantities Q and M. Thus in (23) and (27) we can impose linear and angular momentum balance only on arbitrary vertical slices cf the bar, not on completely arbitrary regions as required by an "exact" theory. It should be a good approximation to replace a function that does riot vary appreciably by its average, so the use of Q (fur example) should be appropriate if stresses do not vary significantly across a section of the beam. Such arguments again lead to the confusion that our approximations should provide useful answers for beams whose cross-sectional dimensions are sufficiently small. 'L

208

SOFTre f.xurr2ples uif S[u1 w

Problems in Elasticity [Ch. 5

BOUNDARY CONDITIONS

If the beam is of finite length, we must append approximate boundary conditions to the differential equation (37). Since we have reduced the geometrical freedom to the deflection of the center Tine only, and since the precise stress distribution is replaced by stress resultants, we cannot expect to formulate boundary conditions by prescribing components of displacement or components of traction at points of the boundary surface. In place of these, our boundary conditions are given in terms of the deflection or the stress resultants. Because the governing equation contains fourth partial Jeri •atives. we shall need two conditions at each end of the bears. We shall consider several of the standard boundary conditions that might occur at the right end of the bears, x 3 = L. It will be clear that in most cases exactly the same boundary conditions hold at the left-hand end of the bar. Possible exceptions are the subject of Exercise 13(a). The end x 3 — L is said to he built in lilt is so restrained that there is no deflection and the slope remains zero (Figure 5.5). Thus the boundary conditions become

built-in :

y(L, r)

ay (Z ' f)

r FIGURE

,

r,'x a

l^.

( 3 8)

.

A buifr -in encl.

A simply supported end is one in which both the deflection and bending moment vanish, as would occur in the case of a hinge [Figure 5.6(a)]. In terms of the deflection, this boundary condition is described by simple support: y(L, r)

d2

t) '

= 0.

(39)

[The vanishing of the second derivative follows from the Euler Bernoulli law (34).] It should he noted that both boundary conditions (38) and (39) remain the same whether the problem is static or dynamic. The diagram depicting the simple support is frequently simplified to the form shown in Figure 5.6(b). There are situations that arise in practice in which the applied loads are not transverse to the beam and hence have an axial component. -

Sec. 5.!)

Bending n/ beams

209

^.x

r^^S^^^r^rr^ #d)

FIGURE i,6. Various typ es of simply step forre1! end (a) .4 hinged end. Fihereui and bending moment r-ettti.sf+. th) A stntpfrftecf representation rif the hinged ord. (C) A support f1ie'tr ran srefrfiih• an axial reaction. (d) support Mai freely permits r1-vial mutton deffec•tiem

The simple support may be capable of opposing an axial force, as indicated in Figure 5. 6(c), or nor, as shown in Figure 5.6((1.)_ The third important boundary condition is the free end. In this case, the bending moment and the shear force vanish, or

PAL, r) = Q(L, t)

—

U.

From (32) and (34),

3'

t) __

N

13 y

A ôr2

f

x3

D'y Y^

(gai

Thus the conditions foi a free end may be written in terms of the displacement t) as -

y(x ; ,

c2y(L, :i)= 03C1

t,

a3 t) A Di ox 3 p

E

Oa, t)

(41)

r7x3 — U.

Boundary conditions (41) take into account the effect of rotary inertia. if this is neglected and (37) is used as the governing partial differential equation, then the corresponding boundary conditions, replacing (41), would be (12 y(L, t) = û , dx1

0 3 y(L, t)

it should be noted that boundary conditions

0.

(42)

(42) also represent a free end in

the case of static deflection. Another useful boundary condition is the one that corresponds to a concentrated force F at the end x 3 = L (Figure 5.7). ln this case the bending moment will vanish and the shear resultant at the end mist be in equilibrium

Some Examples of Static Problems in Elasticity [Ch. 5

210

i

{

FIG U RE 5.7_ .4 free end sichfeet to a concentrated force

F.

with the applied force. More precisely, by examining the forces on a slice of the bar between L — A and L and noting that momentum and external load terms are negligible in the limit A --> O. we see [Exercise 9(a)] that the force boundary condition requires that (43)

Q(L,t;—)+ F =O.

Since F = Fi and Q(L, r ; —) _ --42(L, r)i, (43) can be replaced by the scalar equation

Q(L, r) = F.

(44)

In terms of the displacement. the force and moment boundary conditions are

fil y(L, t_) ^ A [^1 2 ex 3

a3 AL, r}

F

e2AL' r^ ,

eC3

=

^.

(45)

If either the effect of ro tary inertia is ignored or the problem is independent of time, the boundary conditions are simplified to 2 e3 t, t) (46) — El - >L" t) = F. = Q, ex3 ex; Another way to obtain (44) is as follows. Consider (24b) for the right end section, for which b = L. We must set Q(L, t) = O, for there is no material to the right of x = L. On the other hand, we must add the concentrated force term F to the left side of the equation. Upon taking the limit a L, we obtain (44). In like manner, if there were a concentrated force G at x 3 = K, the left end of the beam, the corresponding boundary condition would be Q(K, t) = —G. The condition of vanishing moment similarly follows from (29). TRAVELING-WAVE

SOLUTIONS

We can gain some insight into the nature of the beam motion by examining a simple harmonic wave solution to the governing equation (35), for the homogenous case that occurs when f is set equal to zero. Following the standard procedure (as in 1, pp. 37881), we consider y(x3, t) = exp

`(x3 — cr)].

(47)

Sec 5.1 .1 Besdt,k of Beams

2i1

Here c• is the velocity of propagation of the wave and A is the wave length. Note that the wave propagates along the axis tithe bar, just as in the case al longitudinal motion. On the other hand, the displacement is in the i- direction normal to the axis of p ro pagation. This wave is said to he transverse (in contrast to the longitudinal wave of I, Chapter 12). Substituting (47} into the equation of motion (351, with f set equal to zero, WE note that there is a Cornrnon exponential factor in every terni. 11 this factor is suppressed, the following algebraic equation remains: 4rt 2 EIJ -; — pn 1 47*. 2

2 c-2

--

p; 3 .

is

08)

Solving for the speed c, we find that

4rr2E! c•2

p( J. 2 + 47( 2

(49) }

Equation (49) includes the effect of rotary inertia. In the event that this is neglected, the wave speed is given by

r =

2n^:'! A p

—.

Observe that the influence of rotary inertia depends upon the ratio of the radius of gyration of the cross section to the wavelength of the transverse wave. Unless this ratio is large, rotary inertia is of little significance in determining the wave speed_ On the other hand. rotary inertia must be included for short wares whose length is comparable to the radius ofgyration or smaller. Lt should be pointed out that in those situations where the Rayleigh rotary inertia term is of some significance, there is ail additional term that has not been included which is of the same order of magnitude. The further term takes into account the fact that in flexure there is a distribution of shear stress across the cross section; the assumption of plane sections remaining plane must be changed. This effect has been analyzed by Timoshenko, who found that it can be represented by a modification of the coefficient in the rotary inertia term together with an additional term in the differential equation proportional to 0 .1 .4 /01 4 , See S. Timoshenko. Vihruta'n Problems in €nqureeririy, 3rd ed. (New York: Van Nostrand. 1955. pp. 32e1 31. 334 35), A final observation applies whether or not rotary inertia is considered. We recall from I. p. 377, that the speed of propagation of longitudinal waves depended on the physical constants of the bar only- In the case of transverse waves, we see that the spedd of propagation r depends not only on the properties of the beam but also on the wavelength of the particular wave, 1. Such dependence of the wave speed on wavelength is called dispersion. To examine dispersion more closely, we differentiate e 2 in l49i with respect

12

Some

Examples of Static Probkms in Elasticity [Ch.

5

to A and find

dc 2

— 4rr 2 E}

dA $

p(^ $ + 47z 1 rc 1 ) 2-

(51)

Th u s

and the dispersion is anomalous in that long waves travel slower than short waves. The effects of dispersion are given detailed treatment in Part C. Calculations are largely carried out for water waves, but sec Exercise 9.1.8. Additional interesting solutions to the equations we have derived are the subject of exercises. BUCKLING O F A BEAM

If a long, straight elastic beam is gradually subjected to a greater and greater axial load P. a critical load P = P1. will arise at which the column suddenly will deform into a bowed state. This phenomenon is a prototype of elastic buckling. One expects that the governing equations of the problem will always be satisfied by an equilibrium statecOrresponding to a straight, slightly shortened, column. For P . P1 , however, it is anticipated that this state will become unstable to any small perturbation, resulting in the appearance of the bowed equilibrium state. It is therefore the case that buckling cannot be described by the classical linear theory of elasticity; in that theory, as was shown in Section 4.4, there is only a single possible state of the column at equilibrium. To give some idea of the issues involved, we now give one formulation of the problem that gives rise to a prediction of the critical load Pr at which buckling ensues_ For a far more comprehensive introduction to the important topic of elastic stability, the reader is referred to C. L_ Dym, "Stability Theory and Its Applications to Structural Mechanics" (Leyden: Noordhoff International Publishing, 1974). Although we have proved it (in Section 4.4) only for linear elasticity, we shall take as the basis of our presentation the assumption that equilibrium is characterized by an extremum of potential energy. In our analysis we shall continue to employ the approximate theory of bending that we have developed in this section, As mentioned under (22), our approximate theory gives results that are identical with those that would be obtained if Poisson's ratio is set equal to zero. With such an assumption, equation (4.4.6) for the strain energy gives

E. =

J)Lj

f f

:

d

i

.

(52)

Sec. 5.1]

blending oJ

2t3

We suppose that the buckling is associated with bending in the x i -direction. Using the expressions in (22) for displacemen ts, we thus find that the only nonzero strain component isr — —x i s". Consequently,

E, = 1E1

(}"j' ra?.x 3

(53)

c3

where, as before, I is the second area moment of the cross section about the x2 -axis. Suppose that there is a normal compressive force P at each end of the beam but that the sides are stress-free- Body forces are neglected. Then. according to 14.4.18), to obtain the total potential energy we. need only supplement the strain energy by the term r .,. Pr+ 3^

- ff rtr da =

P V

s= ^ i

^ k3

I.

( 54 )

R

Here we use the facts that

t(i i ,

=

 ^

t.(x i ,x 3 ,L,i1= — ^i ,

where A is the area of the cross section. We now utilize a nonlinear result to rewrite (54) in terms ay. In doing so, we make an additional constitutive assumption, namely, that the column is of such a material that it suffers no change in length during the deformations under consideration. Our variational principle concerns a given substance, so we use material coordinates and obtain from (4.1-10) that on the centerline 2 /13 3 — , 1.e.„ 2f13. 3 + (U 3.3 )2 ± (iLI s 3 ) = Û. (55)

❑

3y 3 , we obtain to first

Solving for the small quantity

b3.3

— i(^Ial; _

approxinisition

-102 -

(56)

Note that we are indeed taking nonlinear effects into account here, for if nonlinear terms were neglected entirely in (55) we would obtain U3 O. In the purely linear case, then, the contribution (54) from the compressive force to the potential energy is negligible. Employing (53), (54), and (56), we find the following expression for the potential energy V:

V(y) =

J

r.

[EI(y,.)3

—

P(y)1 dx 3 .

( 5 7

For definiteness, let us also assume that the column is pinned at both ends, so that

},{0} — Y ,}( 0) = JAL) = y.,( f,) = f1-

(58)

214

Some Examples of Static Problems in Elastievly [Ch. S

One can then show (Exercise 4 or Exercise 11.3.1) that the condition of extrerial potential energy leads to

El

^*

_ ^ 0.

(59)

The eigenvalue problem formed by (59) and boundary conditions (58) has only the trivial solution unless P=P

„

-

n'Pc ,

Pc = ir 2 EIL - '

(60)

in which case w

(tut);

== Cr sin - — , c.„ arbitrary. f

(61)

It appears that a natural candidate for the critical buckling load is P 1 Pc , for the first sign of nonuniqueness appea.rs here_ In considering how to proceed further, one might conjecture by analogy with point mechanics that equilibria correspond with statienary points of potential energy, but only energy minima correspond to stable equilibria. One might also conjecture that the arbitrariness in amplitude found in the buckled solution at P I would disappear in a more accurate theory. These conjectures are correct, as can be seen by consulting p]ym (op. cit.) or the literature cited therein. Some idea of the issues involved can be gleaned from the beginning of Section 12.4. Here arguments are given to demonstrate that, for a single particle subject to a potential force, an equilibrium point that corresponds to a minimum in potential energy is stable in the sense of l.iapunov. YAR1ATlUNAt.

METHODS IN ELASTICITY

In Chapter 11 we introduce the calculus of variations, a subject concerned primarily with the extrcmalization of integrals under various constraints. By applying variational techniques, one can show that the virtual displacement which minimizes the potential energy satisfies the correct equilibrium equations and boundary conditions (Exercise 11.3.26). Moreover, the same techniques can be used (Exercise 11.3.2 7) to demonstrate a generalization of Hamilton's principle of particle mechanics: The virtual displacement that cxtremalizcs the time integral of the difference between the kinetic and potential energy in fact satisfies the differential equations (4.3.4) of dynamic elasticity. The variational formulation of elasticity provides a useful method for obtaining appropriately simplified problems in special cases. This is illustrated in Section I 2.4 where the differential equation governing a flexible membrane is derived by assuming a sensibly approximated form of the potential energy_ Variational formulations of elasticity find another important application as a computational tool. If the correct displacements are known to minimize

Svc_ 5

- (

I

/knifing

of Bums

215

some integral, then a finite linear combination of suitable functions can be inserted into the integral and the constants of combination can then besought that provide the smallest possible value to the integral over the class of functions under consideration. The problem is thus reduced to an algebraic one that is well-suited to modern computers. This approach is illustrated at the end of Section 12.4, in connection with the torsion problem. For further discussions of variational methods in elasticity, see Weinstock (1952, Chapter 10) or Sokolnik off (1956, Chapter 7). EXERCISES

1. (a) Show that the strain tensor given by (5) satisfies the compatibility equations. (b) Carry out the detailed calculations that lead from (7) to ( 13)2. It was remarked in the text, following (13), that the linear terms in these equations represent a rigid body motion. Show this to be the case. 3. Consider a beam of square cross section bent by terminal couples_ Utilizingthe displacements given by (14), sketch the deformed middle surface, x , ï 0, Also sketch the boundary of the deformed cross section given by 3 = constant. 4. In (57), let y = yo + Ey, satisfy the boundary conditions (58) for all L. Show, using integration by parts, that the condition

d

V(yo -4- Ey

z

implies that y o must satisfy (59). t5. In our discussion of the geometrical approximation used in the approximate theory of bending, we stated that in pure bending by terminal couples, the normal to the deformed cross section at the neutral axis was parallel to the deformed neutral axis. Prove this statement. 6. (a) Suppose that the deformed center line has a curvature opposite to that shown in Figure 5.4. Show that it continues to be true that rr j = , (1 34:3, ilat- 3,(h) Prove the statement made after (22) that the displacements given there are the same as those for pure bending if the Poisson ratio is set equal to zero in the latter case. 7. Establish (as in I, p. 361, or from properties of the stress tensor) that for arbitrary b, Qlh.z: +) _ — Q(b,r; $. (a) Show the validity

(b) (c)

—

),

M(b,i; +)

—

-- M(b,i;

—

}-

of (24a). Complete the calculations needed to derive (29) from (27). By tracing its origin, show that the second term in (35) represents the effect of rotary inertia.

2115

Same Lxa»rpits of Static Problems in Elostrkiry

[Ch. 5

9. (a) Derive boundary condition (43) by the first method suggested in the text_ (see F igure 5.8). (b) A beam carries a heavy particle of mass in at x,

—L

.t 3

L FI[;I:RI.

5_8 A n end carrying a heavy lrurfir'1r r+ f rrraas en.

Show that the boundary conditions at x 3 = L are given by ni (C)

t

=

El

x3

—

ph.2

c•

Ô

+ In

2 x3

(62)

How would (62) change if we had to take into account the moment of inertia of [he dumbbell-shaped end mass shown in Figure 5.9?

F i t:UIeE 5.9. An earl carrying ar Jrmrhfletl - .siipprsrf nurss, r {•yuirrng crur ► rde ration erf die moment 4 iitPYrrrr_

10. Find the static deflection y(x 3 ) of a beam (as a Figure 5.10) that is clamped at x 3 = i, free at x 3 = L, and subject to the vertical loading

f _ (x 3 — L)F 0

.

x;

FI GU R t 5.10. Acantilever beam subject to a frneark decreasing toad-

Sec 5.1 I ldendiny uj Beams

217

11. Determine the natural frequencies of vibration of a beam (as in Figure 5.11) that is simply supported at x 3 = 0 and x3 = L. To carry out this consider solutions written in the form progam, ytx 3 , r) = ri(x 3 ) exp Irtt) of differential equation (35), with f = 0, and boundary conditions 139). -

I

r3

^

o

t, FIGURE 5. 1 1 . A beam simply setpporred tit bo th ends

(a)

Show

that u(x 3 } must satisfy d4u + ptc 22 2 dx1 dx 3 3 2 du(0) = d2 Lj 14(0) = u(L) = dx3 3

E!

p

ate

=

0,

—

(b)

— O.

Show that the only solution of this boundary value problem is, to within a multiplicative constant, u(x3) = sin acx 3 .

How a re a and , related? What are the permissible values of a and hence the natural frequencies or eigenvalues 1.? I2. We shall study the static deflection of a simply supported beam of length L carrying a concentrated unit load at an interior point .x 3 — 5 (see Figure 5.12). Thus the deflection y(x 3 ) will satisfy (c)

dxa d4 -2 Y

_ = 0, 0<x 3 < ,

<x3
0 and = - In r. Show that

2

(h)

fji(I

z} -

In '4), dc f

j' I (. 0 ,l OR

f^

ln r ï^ - - in r i c rt ds. an

By taking R to be a circle of radius p centered at (. ^ l , 2 ). prove the mean value inequality 4( 1 ,X71s

1

2n pJeRç

ds.

(c) To prove that there is no interior maximum unless is constant, 2 ) = 1 , x assume the contrary and Iet Cx M, the maximum value, where (x i , x 2 ) is an interior point of R. As shown in Figure 5.20, there is another interior point (y i , y 2 ) in R, where 0(y r , y 2 ) < M. Let (z 1 , z 2 ) be the first point on a curve C in R going from (x 1 , x 2 ) to (y i , y,), where iP(z r , z 7 ) = M. Let r be a circle centered at (z,, z 2 ) lying in R. Show that there must be an arc of F on which < M. Hence show that there can he no interior maximum_ Show that the only cross section that will admit a warping function which vanishes identically is a circle.

Set . . 5.2]

Sr. Venom.

Tar NUM

Problem

239

F [tir1RF 5.20, Construction s!?unvrrm Mar u cunrrudirirnrr results trim? the drutytolita on inferior maximum rrt ( -► i . x :).

ussur?aP r

9. We shall develop here a very useful alternative Formula for [he applied moment, using Green's theorem in the plane. (a) Starting from (2b), show that

L (b)

r 4 x a n ii ç{i d5,

—

r'V

Show that =J

(c)

where ^'t

A

- f 43 ` -- if s_

an

Finally, obtain the Dice-Weinsrein formula,

M3 = J -

J5

ctii fix ] clx 2 -

(am)

Using (60), prove that of all cross sections having the same polar moments of area .1, the Circle has the largest torsional rigidity. 10. (a) Provide all details omitted in the text's separation c1l` variables approach to finding a harmonic function f that satisfies 140) Also verify (47) and (48). (b) Prove that for a rectangular cross section the maximum shear stress is at (±u p) and verify (50). 11I. This exercise develops an approximate formula for the maximum shear stress on the boundary of a rectangle by starting with (50) written in the (d)

.

,

form

T=Ba 1

8

—

^2 sech

1 rib

sech

^^ + ^ — R =^

-

+ 1)t^i3j2a)] • 2 (2r^ + 1)

240

Some Exumptes of S7nrie Problems iM Elasticity

[Ch. S

(al Show that the series appearing in this expression can be bounded above by

9 (b)

sech (Zr+II

^^

.

^=i

Show that this expression cari he further bounded above by -n

x

e?4f+ —^ ^ n=

(c)

-

5

exp nil 2 ü

Sum this series and obtain a numerical estimate for the worst case,

which occurs when b = a. 12. For the elliptical cross section, find [he stress function `!', the lins of constant shearing stress, and the tordue angle relationship (54). 13. Show that the text's formulation of the torsion problem does indeed provide a zero net force on each of the two ends of the cylinder. (a) lise [he stress function. (b) lise the warping function.

5.3 Some Mane Problems An extensive literature has been developed on [he solution of static elasticity problems that are essentially two-dimensional in formulation. Concern with two dimensional static elasticity arises from two main motives: (I) there are many important technical problems that are two-dimensional in nature; tii) there is independent interest in the powerful mathematical methods that have been developed for the solution of the biharmonic equation, which plays a central role in the theory. It should also be noted that the experimental methods of photoelasticity are particularly well adapted to such two-dimensional problems, so that many experimental verifications of the mathematical theory have been made. In our brief introductory discussion of these matters,wc shall first consider the problem of plane strain and show how this can be formulated as a boundary value problem involving the biharmenic equation. A garticular example, the Kirsch problem, will be solved_ Although the mathematical techniques required are elementary, the solution is important as an introduction to the technically important problem of stress concentrations. We shall also consider the problem of plane stress, which is of somewhat limited applicability, and a more useful approximate theory of generalized plane stress. The resulting boundary value problems will be shown to be the same as those occurring in plane strain, and thus solutions obtained for one case can readily be translated into those for the other_ -

Set'_ 5.3)

Some Plane Prohlemt..

24 1

EQUATIONS FOR PLANE STRAIN

Let us consider a cylindrical body whose axis is parallel to the x 3-direction. We include multiply connected regions as well as infinite regions bounded by cylindrical holes. The surface tractions and body forces, if any, are assumed to be independent of the coordinate x 3 . The basic assumption of plane strain is that there is no displacement in the x 3 direction and the ot her two displacement components are independent of x 3 . As usual, Greek subscripts will take on only the values I and 2, whereas Latin subscripts range from I to 3. With this, the assumption of plane strain may be stated as -

Jig

= 4.1C1,x2 j,

u3 = O.

(1

)

The strain-displacement relations become =

^.

0

+

Up, g),

E;y

= O.

From Hooke's law,

T1 _ ?let,1 + .ich x ki a we have Te

T

T,r 3=0,

T0(Xt• Y2),

(2a)

T33 =

Carrying out a contraction in Hooke's law, we see that 1-33+

T33 =

or T33

=

(2p + 31)c,

'+' f

(2b)

.

Thus (2a) and (2b) show that the complete stress system can be determined onceTe is known. To simplify our discussion, we shall also assume that the body forces vanish.' The equations of equilibrium, To.) 0, become, for our stress system, —

Tt 1.t +Tt2.2 = 0,

T21.1 +

T22.2 =

(3a, b)

^•

AIRY'S STRESS FUNCTION

Equations (3a) and (313) can be regarded as integrability conditions. We can thus deduce that there exist functions V 1 (x i , x 2 ) and V2(x t , x2 ) with the properties Tt l = V1,2 , T12 =

—

V1,1, T21 = V2,2 , T22 =

-

V2.1•

• In the important cases where body forces are derivable from a potential, the resulting complications are not too serious. We shall not consider this case, however.

242

Some Examples u/ Slade Problems in £lastïrirr

[Ch. 5

Since the stress tensor is symmetric, we must have

V1.1 + F2. 2 =

(4)

^-

Equation (4) is again an integrability condition, and yields the conclusion that there exists a function U(x 1 , x 2 ) with the property II, = U , 2 and 1,2 — ff . We see that the components of the stress tensor can be found from U by ,

7', r = (1 32 , Ti2 =

U.129

1 a2

—

(5)

.11-

In terms of the two-dimensional alternating tensor eo we have (6) tiQr Cre U, fP' The function ti(x r , x 2 ) is known as Airy's stress function The next task is the determination of the partial differential equation and the boundary conditions that must be satisfied by this function. We recall that in view of (6), the stresses will satisfy the equations of equilibrium. In order that these stresses lead to compatible strains, the Beltrami-Michell equations (4.3.15) must be satisfied. For our purposes it is necessary and sufficient to use just one of these equations, namely {in the absence of body forces)

7;.. pp = U. In terms of the stress function, this becomes A= e., e,r, il mg* ` ï v 1) ,. n66 or

t 'r ,au00

(7)

Thus the stress function satisfies the hiharmonic equation in the region D of the Ix , x 2 )-plane occupied by the cross section of the cylindrical elastic body. ,

BOUNDARY CONDITIONS

Let us now turn to calculating tractions on a cylindrical surface in terms or the stress function. The trace of such a surface in the (x 1 , xj plane will be denoted by C. as shown in Figure 5.21(a). The normal to the surface will have ri, U. In addition, we shall make use of rp , the unit tangent vector to C. The traction t ; satisfies -

Therefore, t 3 = 0, and

In terms of It by (6)

Ser.

5.3]

Some Plum. Prvfr fe m.%

2

43

^rU

la I

FIG ufil: 5.21. (al Crrr C re frre,sivrrt parr erf a cylindrical boundary ime•r:u•e•red hr the (.r, ,x 2 )-plaire, which is perpendirrdur w the directrices s }f the cylinder. Depic•red also are the normal te ^{rrl. the rurryent (ro ), and die srress ree•rvr (ii ). t hl Normal (t `) and Irnr,ycwriuJ (il ) components of the stress recto; (0. ]-

Since the unit tangent vector and unit normal vector obey —

n. = e'ae z$3,

we may write r t — e#4,1 L or r y .

(10)

Thus we see that if the stresses are given along all or part of a bounding curve C, the derivatives of Li must satisfy (l0) along C'_ In the classical formulation of boundary value problems governed by the hiharmonic equation, the value of the Function and its normal derivative are specified along the boundary. We shall show that this is the appropriate formulation here when the stresses are given along the bounding curve C. To do so, we must investigate (10) somewhat more closely.

2

44

Some Examples r.J Static Problems in Elasticity [Ch_ 5

Let us rewrite (10) in terms of the arc length s along C as t = etfu

au ^s

or

0} =

rj.2

t2{s) —

'

au ,

Integrating with respect to s, we see that the first partial derivatives of U are determined on the boundary by

fr2(s}ds + c t =_ F,(s) + c

#.l,i(s] 11 .2(s) =

J

1 (s)ds :

+ c x - Fx(s) + C 2 .

Furthermore, I-'.1n + U, 2 1i 2 —

0 t1 [^fr

= (F 1 + cj }n r + (F2 + r'Anx

and

c? I1 _ U.rtz + 1.). 2 r 2 = Os — (F r -+- c s }t 1 4- (F 2 -I-c 2 )t 2

.

Integrating once more with respect to s, we have, for some function ii, )(s) = H(s, cr, CO 4 r 3 he traction is known along a boundary curve C, U and O i t iefi We see that Kale are also known along C and the resulting boundary value problem is of the classical type. It should be noted that this formulation introduces three constants or integration. Just as in the torsion problem, these can be neglected for simply connected regions since the stresses will satisfy compatibility conditions guaranteeing single valued strains and displacements. In the case of multiply connected regions, additional conditions to guarantee global singlevalued behavior must be added, and these will suffice to determine the unknown constants. These conditions were first developed in a paper by J. H. Michell_• For future purposes, it is convenient to consider a decomposition of the traction t into a component tangent to a curve C and a component normal to C. The former will be denoted by IT and the latter by e N [see Figure 5.21(b)]. Since -

au t ^ ^ ^^^ ^s - • On the Direct Determination of Str esa in an Elastic Solid, with Application inrheThcory of Plates." Proc.

Land. Morfl. Sur. 3(1, tOO 24 L 1899)

Ser 5 3] -

-

Some Plane Problems

2 45

we have r

r ^

r P rr _~°N, i, as•P

DU ix., I T = — n -- r! as .

^

In similar fashion,

r^ = rfinP _ ePpU.p^ t, na = G.ow t;,to =

._ rp-

( 12 )

In the event that the plane strain problem is stated in terms of stress boundary conditions, we see that the mathematical problem requires solving the biharrnonic equation for U with the boundary conditions either given in the form (10), or (1 l) and (12), or, after preliminary integration, given in terms of U and on the boundary_ Solutions can be found in a number of ways. For appropriately shaped boundaries, separation of variables provides the simplest and most direct approach. A much more powerful approach is possible, using complex variables. Discussion of these methods would take us somewhat beyond the scope of this chapter_ The interested reader will find a very good account in Sokolnikollf; (1956, pp_ 262 324 PfÎl.Ali COORDINATES

Ire order to consider a certain example, which is naturally handled with polar coordinates, we have to make some preliminary calculations. We shall see that these rather straightforward results have some interesting features in themselves. From (15.4.4) of I or other sources, the biharrnonic V 2 V 2 ü = 0 can be written in polar coordinates as a2 U

t 011

I 02U

ir e + rS• +-r (10 2 are t r Pr + r3 002 =°

(1 3)

If we look for product solutions of the Form 14r, Û) = el sin aU or

U(r, 0) = r' cos 20,

we find (Exercise I) that such solutions fall into the following groups: to o +b o Inr + ca r i +da r2 Inr,

(14)

(ea + Ïo In r + gür2 -t h o r 2 In r•)1J,

115)

(a i r + b i r - i + c i r ln r + d i r 3 ) sin L,

( 1f )

0,

(17)

(e z r + fir- i + gir In

r + h i d) cas

(A i r + B i r - i + C i r ln

r +

D i r ; )0 sin 0,

(l Ii}

(L i :- -4- F i r - ' + G i r ln

r +- If i r 3 }0 cos 0.

(19)

(a.

+

-

b q r ' + cr' +2 + dr - x •2 ) sin

(e.ra +J f

20,

r -' • 9.e+2 + ir o r' 42 ) COS `JCfr

Here a 0 0, 1, arid, for the moment, need not be an integer.

(20) (21)

246

Some Examples of Static Problems in Elasticity [Ch_ 5

Let us assume that the elastic region contains at least a portion which covers the full range in 0,0 w 2 /c 2 then f = A exp(hx 2 ) + B exp(

-

6x,),

(5)

where

c,3 2)112

b=

—

? .

C T,

On the other hand, if c2) e /4 3 s 2, then f = M sin

(axa

+ D cos (ax 2 j,

(G)

where 02r ii 52 a= (.7 r Equation (4) yields a differential equation for the unknown function g(x 2) in the displacement (2). On the other hand, we are seeking displacements in the semi-infinite substrate that do not penetrate far into the region. Thus the only physically possible solution is easily seen to he g = E exp(--kx 2 ),

(7)

where (0 2

k=

si—

112

*z ^r

Boundary conditions must be given on the free surface x 2 = 0 and at the interface x 2 = h. The traction vanishes on the free surface. Thus t i =T2 =0 The only nonvanishing component of 32 , so that Hooke's law yields

at x 2 = O.

To foi the given set of displacements is

u 3. 2(x1, t], () = O.

(8)

29 2

introduction to Dynamic Problems in Elasticity

Kit_ 45

It is assumed that the displacement as well as the traction is continuous at the interface. Consequently, u3(x1 , h , r) T u31 1, h, r)

(9)

and 1 32(x1, h, r) = T32(x1+ h, 1)

ar pi3,2(x1+ h , t) — p *4 3,2(x1, h, t).

(1O)

Suppose that cu e/ei < s 2 . Using (1), (5), and (#l), we find that A = B, and the full displacement can be written in the form u 3 = .el cosh bx 1 exp i(sx 1 — cot).

(11)

Substituting (2). (7). and Ill) into boundary conditions (9) and (10), we find ,sat cosh bh = E exp (—kb), pbsit sink bh = — kte*E exp (—kh). (12) In order for (12) to give nontrivial solutions for sif and E, the determinant of the coefficients must vanish. This yields I.01 — —

tanh bh.

( 1 3)

Since bJ, is positive, tanh bh is positive. Then the right-hand side of (13) is negative, whereas the Left-hand side is positive, which is impossible. It must be that

w2

> s=, 2 cT with the amplitude f (x2) given by (6). This requirement together with the decay requirement in the semi-infinite space implies that

co * C3

<s
`-n, reference to (I 6) shows that there will be nodal planes in

the layer for these higher modes. Denoting the wavelength corresponding to a; by ).;, it is seen from (22) that since the speed of transmission is the same in both cases, ap:ir = az

^^

Thus

It has been found in seismological problems that the most important case is the one without nodal planes, since the energy transmitted by the longer waves is greater than that associated with the shorter waves. CONCLUDING REMARKS

the general discussion as well as the special examples considered in this chapter, the important fact that there are two speeds of propagation in an elastic medium has been emphasized. We have noted that the basic field equations can be written as scalar and vector wave equations, each with a different speed of propagation. in many cases, however, we find that the boundary conditions require that both types of waves be present. This is one of the major reasons why elastic wave propagation problems are frequently more difficult mathematically than the corresponding problems in either In

acoustic or electromagnetic wave propagation_ We also observe that in our study of propagation in infinite and semiinfinite media, no dispersion took place_ (For example, the speed of Rayleigh waves is independent of their length.) This is in contrast with the results that we developed in the preceding chapter on the engineering theory of transverse wave motion in beams_ We can recognize a duc to the origin of dispersion from its appearance in Love waves, in the presence of boundaries a finite distance apart. indeed, dimensional reasoning shows that there cannot be dispersion in the absence of a length scale (such as the width of the dust or the beam), for otherwise how could the wave "know" how long it was and hence how fast it should travel? It seems that dispersion should be regarded as a normal state of affairs,. but dispersion cannot be present in problems lacking an intrinsic length scale_ Drspersrvc waLrür waves arc diuusscd al Icnglh ,n what Follows. slatting in Section 8.1.

296

hir+awr#rciior: ro

Dynamic Problems in Elasticity (Ch . 6

EXERCISES

Study Love waves for the case in which the substrate is infinitely rigid, ii-e-, u 3 = 0 for x 2 = h. lithe motion is dispersive, find the relation between the speed and the wavelength. 2. Study Love waves for a layer that has free surfaces both at x 2 0 and x 3 4 h. 3. The variation of h/A as a function of cir e- can be found in the following manner. Choose a value of Or e- in the range !.

I

^

• CT ^ _ CT CT

(27)

^

Calculate u; and solve for h/. from l7bl written in the form

(9(k)= tan

[u^]

1 .

Carry out the calculation for various valuesofc/c T in (27) for p*/p = I and the case te fir = 2, and 3. In addition, calculate the corresponding values of exp ( kA). (This can be done without undue effort using a programmable pocket calculator.) 4. We have seen that solutions of the form (1) and (2), with the restriction that the displacement dies out in the substrate, are only possible if c r < cr. Suppose that this inequality is reversed and the restriction of decay in the substrate is replaced by boundedness. Examine what can be said about the solution to such a problem. S. Are there alternatives to the text's assumption that in Love waves the displacement and the traction must both be continuous at the interface? Discuss. ,

—

PART C Water Waves

CH APTER 7 For m uIation of the Theory of S urface W aves in an Inviscid F luid

T

IE REAMER will surely grant that the wave concept is one of the most fruitful in theoretical science. Our senses continually record the arrival of sound and light ; we learn in elementary physics courses that these phenomena are due to pressure and electrodynamic waves. Quantum mechanics tells us that the objects around us are "really" light wave packets of probability densities. Other examples abound. As the theoretical physicist Sommerfeld has written, "Ever since waves were studied, water waves have served the natural scientist as a model for wave theory in general: . Part C is concerned with waves on a water surface. These are a paradigm for wave phenomena because of their simultaneous generality and accessibility to observation. In referring to generality, we have in mind the fact that the appearance of water waves is usually influenced by dispersion (dependence of wave speed on wavelength) and nonlinear effects. in special cases. water waves are essentially nondispersive (long waves) or linear (small-amplitude waves)_ Now, "ordinary „ sound waves are nondispersive (so that we may speak of the" speed of sound) and little affected by nonlincarities, but nonlinear effects in "strong" sound waves cause shocks or sonic booms. Light waves are dispersive (hence the rainbow) but are not ordinarily affected by nonlinearities_ With the advent of lasers, however, the new field of nonlinear optics was born. Thus an expert in the theory of nonlinear dispersive water wave is prepared for complications that may emerge in other wave systems. Special instruments are necessary to record even rudimentary facts ..

concerning most wave systems, but visual inspection suffices to reveal many rather subtle aspects of water waves. This accessibility to c'bçervatio i means that readers will possess the basic familiarity with the phenomenon that is a prerequisite to theoretical analysis, Elastic waves were discussed in the previous chapter_ The reader who becomes familiar with both elastic and Fluid waves will find the contrast enlightening. There exist many similarities, for example, the presence of dispersion except in the absence of a length scale. Another similarity resides in the important effects of density stratification. We considered one prototype problem of this type in elasticity when we treated hove, waves in Section 6.6. There are some basic results for the fluid mechanics case in Section 15.2 290

'oc

Formulation of the Theory

of Surface Wares arr hiri rid Fluid [C h. 7

of I, particularly in Exercise 8. See also C. S. Yih, D of Nonhortiogeneotis Fluids (New York : Macmillan, 1965). Understanding of density effects in one

medium is, of course, helpful when such effects are to be studied in another medium. Although similar in many respects, elastic waves also differ importantly from water waves. Perhaps the primary difference is that the a cisterice in elastic media of both transverse and longitudinal waves brings about interactions not found in fluidlike media that do not support shear stresses. By contrast, ar adequate study of water waves must concern itself with nonlinear effects, while these effects (at least for the usual applications) are normally negligible for elastic systems. In I we have derived field equations that suffice for an initial study of water waves, namely, the equations for an inviscid fluid. Complete mathematical characterization of the problem requires more thought, however, for it is necessary to seek appropriate boundary conditions at the air-water interface. A careful derivation of these boundary conditions is the subject of Section 7.1 In Section 7.2 we Formulate a mathematical problem that can serve as a model for a wide variety of situations involving water waves. This problem is rather difficult. so we conclude Section 7.2 by writing down a simplified problem, wherein nonlinear terms are deleted and boundary conditions are applied at the mean position of the interface. A justification of this simplification is the subject of Section 7.3. The linear problem is studied intensively in Chapters 8 and 9. Part C concludes with Chapter 10, an introduction to a study of' nonlinear effects. NoTt: to INSTRUCTORS. The standard treatment of water waves is based on Laplace's equation for the velocity potential. with appropriate boundary conditions obtained by the use of Bernoulli's equation. Here we proceed directly from the Fuler equations, thereby skirting certain strictly hydrodynamical topics such as Bernoulli's equation and the justification of the irrolationality assumption. The d isadvantage of slightly more complicated calculations is more than offset, we believe, by the advantage of giving experience in dealing with systems of partial differential equations. Moreover. in a viscous motion starting from rest, nonlinear effects lead to a second order vorticity throughout the fluid. There is a resulting mass transport, even in the limit of vanishing viscosity.* Weak turbulence initiated and maintained by the waves, or stemming from the wind, is responsible for the appearance of vorticity that affects the nonlinear properties of the waves.t It thus is likely that much future research on water waves will not employ the irrotationality assumption. • tier M S. Longuei-tiwir,s, "Masb TransPurs ru tksr liaunddry Layer ata FrceGscrttatmg Surface," J Flu i d .4fech. 8. 293 3O5 119b0) t O. M. P h illips. "A Nate an the Turbuicrsex Generated by Gravity Waves," J. Geaphys. .

R e s, â6, 2t389-21s93 t tS ► fi!

) .

Sec. 7,1 ] 8oeundar.1 C'cuWlrtrrrtvi

3a I

7.1 Boundary Conditions In the first part of this secuun we shall derive the "kinematie boundary conditions- a mathematical expression of the fact that the surface F(x, t) = 0 is the boundary between two immiscible substances. No knowledge of the nature of the substances will be necessary the discussion is entirely geometric. Logically, then, this topic should perhaps have been presented with the kinematic material of Section 13.2 in 1. li is presented here because there would seem to be higher motivation for studying the kinematic boundary condition in a familiar and tangible selling, that of water waves. Since our treatment olfthis condition is entirely independent of the succeeding treatment of water waves. it can be read al any 1 ime. In the latter part of this section we define and use the notion of surface tension in deriving a forte balance condition that holds at the interface between immiscible inviscid fluids,

kINL!1f1A'ftC - Bt)UNl)Aiiii ('[1Nili7iO N li is an idealization to say that a surface Fix, t1 = O separates two immiscible substances. According to the molecular point of view, the separation is not abrupt. but rather there is a mixed transition zone joining two regions of almost pure composition. Idealizing this transition zone by a surface is permissible and sensible when one is interested in phenomena which vary signilicantly only over distanc:.es that are large compared with the width of the transition zone. Suppose, for definiteness, that a surface Fix, r) = O is deemed to separate water and air. (Replacing a transition zone by a surface iscertainly permissible here, for even centimeter-long surface ripples have a scale that is perhaps 1 million times greater than typical transition widths.) We assume that all the water remains on one side of the surface and all the air on the other, i.e., that the bounding surface is impermeable. If nu fluid is to pass through fix. tI = 0, then ur any point x„ N this .+rrrfue -e the crmrtpoilettt of the /ljtiti reiu( it ►' v(x,,. 11 than is tnsrutauneuasl ' normal to F = f3 at x e must equal i (x c, , t), the normal speed t / F U ar x{,. The italicized requirement just stated, while roughly correct, must be modified because the fluid velocity al a boundary point may not be defined. The reason is that the tangential velocity component may suffer a discontinuity across the boundary (since the boundary is impermeable). Suppose, therefore, that the impermeable surface l-ix, r} - O separates one substance in region l2 from another in region R 2 . Let n be a unit normal to at x 0 : ,

VF'Ix ü , t)

n = IVF(xe,, tl^ ^

11)

302

Formulation of the theory of Swjw a Naves in an lrrriscid Fluid [Ch. 7

Then the boundary condition that we shall impose is line v{x , i) - n = VR(x 0 , r) ^

(2)

—x w

;Er,

for x€, on F=O;i— 1, 2. We now seek an expression for i.„ in terms of F.* Let x = %At) describe the trajectory of a point P on the surface F = O. Then F'Lxdt), t] = O.

(3)

Differentiating (3) with res pe ct to r, we obtain

dx f.

F,

O,

(4)

But dx Pfrlr is just the velocity of point l'. To obtain the normal speed V, we project this velocity vector on the unit normal n of I!). Thus, from (4). ` VF IVFI

•d

F,

x^ !

dr

(5)

G VF

Employing (5), we see that boundary condition (2) becomes limv(x, r). VF(x 0 , r) = — F,(xo. r).

(6)

11 -*So

ti*

This is usually

R,

written

F, + v -F — O.

(7a)

Equivalently, the kinematic boundary condition is l] J-

Dr

= 0

( l b)

with the understanding that the condition holds as the surface F = 0 is approached from either side, When the interface equation is written z = ((x, y, r) and v = (u, y. w), then (7) takes the form w =Cr + u t; x + ttç as the reader can show [Exercise l(b)]

on z=C( x, y, t),

(8)

-

• This wrss obn inc l in (6.2.3)- We repast the brref required argument hezc Ioz cornpletcness and because we wish tarts C and [] to be accessible wrthout the knowledge of Cartesian tensors that was dcvclnped in part A and applied Iii Part B.

Set - . 7.1]

Boundary C onuliii oris

303

BASIC F A C TS ABOUT SURFACE TENSION

As one learns in elementary physics courses, observations of the properties of a liquid-air interface suggest the need to take into account an elasticlike properly of the surface. One revealing experiment involves a soap film on a wire ring. A loop of thread in the film will quickly assume a circular shape if its interior is punctured (Figure Ti). Presumably, in Figure 7.l(ak each element of thread is motionless because of a balance between forces resulting from the presence of surface film on either side of the element. When the interior film is destroyed, surface forces act only on one side of each dcment of the string, pulling it out. Note that in the final state of Figure 7.1(b), symmetry shows that the forge associated with a portion of surface acts normally to the bounding string.

()

W)

7.1. (ai A wire rira! .► upporiini a soup film ni which rests u loup of thread. (b) The confiquraiton afier the film in the interior of the loop is punctured FIGURE

ln another common experiment, fluid in a thin tube rises until its weight is balanced by surface forces (Figure 7.2). `l o explain this, one postulates a surface tension force per unit length of magnitude T that acts at each point P on the bounding curve of the surface, in a direction that is normal to this curve and in the tangent plane to the surface at P. The horizontal components of this force add to zero. A vertical force balance on the column gives (in the notation of Figure 7.2)

2ttrT cos a = rrr 2 hpg,

rigig i.e., 7' _ — - 2 cos a

The surface tension coefficient T is thus expressed in terms of readily measurable quantities. On a molecular level, surface tension can be understood as a manifestation of the significant attractive forces that exist between fluid molecules. The molecular viewpoint is outside our scope here, but good discussions can be found in most basic physics texts.

lai

Formulation of the Theury of Surface I ailles in an frniisrid Muid [Ch 7

F I G u k E 7.2. A tlrrn rube of against yrar+r r r magnitude' y J.

u/ °W #rr I p r.w held up h 4 swr;lfue•e rerr irurr

QUICK DERIVATION OF A DYNAM1( BOUNDARY CONDITION

We now present a rapid derivation of a form of the dynamic boundary condition that is sufficiently general to suffice for nearl y all future needs in the present work. Then we shall retrace our steps more carefully. We shall restrict consideration to inviscid fluids. Figure 7.3 presents an edge C of a (cylindrical) surface demerit that does not vary in the y-direction, and that departs only slightly from the plane z = constant_ It can be shown that body and inertia forces on the element arc of higher order, so that the surface fore must add to zero. (As will be shown in the following discussion, the argument is similar to that used in Section 14.2 of 1 to demonstrate the principle of local stress equilibrium.) We impose the condition that the net upward forces, due to surface tension and to a pressure

a'

.

rlal

r

I

^ 01 I i3 ^ r

I k

I

izi i

I

1vi .

uKIE 7 -3. S►rrJure fernier: rir ► rrrrrr

r r acting on ri

^^

+ ❑x

rrr•arl.r planar cylindrical .surfrare •

Net- . 7.1 )

Boundary ['urrdi r i^ns

30 5

difference across the surface, add to zero.* the net upward pressure force (per unit distance in the y-direction) is approximately

(PB — Max + O(Ax) 2 .

(9)

Here p a is the pressure acting on the bottom of the surface and p r is the pressure acting on the top. The O(Ax)' error term in (9) arises from estimating the length of C try Ax. Let 1,1 denote the angle that the tangent to C makes with the positive x-axis, O S tfi < i (figure 7.3). Assuming a constant tangential tension of magnitude T, we obtain the following expression for the upward force contribution from this source: -

T stn [rr — fi(x)] + T sin Jr(x -+ ❑ .).

(IU)

Since the surface is nearly flat, rr — i/i(x) is small, so that sin [rr—+k(x)]- tan [rr

—

Vi(x)]_

tan (x)_

(Lk) --

^

Because sin tti(x + Ax) x tan tÿ(x f Ax), (JO) can be replaced by

—

z T C^.K

dz (x + ^x) + T dx

Expanding the second term in (1 l), we obtain the following expression {evaluated at v}:

— T dz + dx

2z dz T €t A.x IL- O(Ak) 3 = T ^ Ax T f dx` dx dx 2 ^

,

0(11vj 1 _ ( l2}

Equating the sum of (9) and (12) to zero, dividing by Ax, and then tak irtg the limit Ax # 0, we finally obtain

d2 .7

PT — F ►s = f dx z •

(13)

This is the desired dynamic boundary condition for a surface that varies only in one dimension. It is a linearized condition, restricted to surfaces that deviate only Slightly from a plane_ DETAILED DERIVATION OF THE DYNAMIC BOUNDARY CONDITION FOR AN tNYI4{'ill FLUID

We shall now reconsider the dynamic boundary condition, proceeding in a much more careful manner than before. The r€.triction to nearly flat boundaries will be dropped. Our discussion will contain some detailed arguments, which do not make for rapid reading. Time spent in careful `

Our more dead ed dtscussron iatertri this scctron shows that the x-components of the various

forces add to zero if we assume that T is cnnstant.

306

Fornrukrer un nj the Theory eery

of Surkier Wares in on Inriscid Fluid jCh_ 7

perusal of these arguments, however, will help the reader toward a genuine understanding of the iSSLtcs involved. h turns out to be helpful to regard an interface vii as a special state of matter, confined to a surface. For an ordinary material region, the influence of adjacent exterior substance is represented by a certain force per unit urea given by the stress t, a vector defined an t he boundary of the region in question. Analogously, we hypothesize that the influence of outside adjacent rnatenal on a patch of surface material P can be regarded as equivalent to the influence of a force per unit length I acting around the curve C that forms the boundary of P (Figure 7.4).* The total force of this kind would thus be given by ( 14 )

where dN is an element of arc length_

ds

FIGURE 7 4. A force p er emit letu y th 1 is hy7rrrlhesi_er.f to ere -i nri the boundary

porch p

raJ

surface Mai ,seppreale.v two rrollmiuis,

of'ex

Uwe consider a material region R(r) that intersects the interface .0 between two different substances, then the influence of outside material will require a body-force integral and a surface-stress integral, as in Section I4.2 of 1, plies a line integral such as (14). Here C will be the curve formed by the intersection of .0 and the houridary of R(r} (Figure 7-5). Thus instead of (14.2.5) of 1, we have the momentum balance equation

dr,,, #t

pv

=

fff

p

dr+

Jf t

pig

du+

f

i d-5.

(15)

(,i

There is an appealing symmetry in (15). Linear momentum can be represented as changing due to distributed sources per unit volume given by pF, • A full analogy would requirc irltroduciion of an " intcrfacc body force," hut these seems no compelling physical mason ro inteOdticr such grim-alit).-

Sec. 7.11

Boundary Co rirfil rorts

307

Ft ci.' R E 7_5. A rtrctrrrat region R(r) rhat mraddhs she

iMrerfac•e .1 be tw een

two d'rlieremt materials. sources per unit area (acting on the boundary) given by t, and sources per unit length (acting on a certain boundary curve) given by 1. Although we shall not pursue it. one can derive consequences of (15) which are analogous te the consequences of (14.2.5) that were developed in Section 14.2 of I [sec Aris (1962}1 What we wish to emphasize is that (15) shows that the fundamental equation (14.2.5) of I a generalization F = ma for continua, cannot be "obviously correct," because sometimes (when surface tension is present) it is not correct at all We shall at once adopt the simplest reasonable hypothesis concerning the nature of 1. Let n(x, r) be a unit normal to a small piece P of the interface at point x and time I. Let ulx, r) be a unit tangent to the boundary C at point x, time r. Choose n and u so that the unit vector N = u A n points outward from P (Figure 7.6). We shall assume the consri:wive equation ,

I(x, r) = T(x

r)N(x, r)

for x on C.

(1 6) Equation (16) states that the "rnernbranelike" force per unit length i acts in the direction of N with a magnitude that is independent of N; T, the magnitude of the membrane stress, is the surface tension.

>"• !CURE

,

7.6. Tangent cold rtorrrtp l rectors at a perirri on

surface P.

the boundary C

308

Formulation of thp

Afar),

of Surface. Wa v es in an ltaaisrid Fluid [Ch. 7

What justification is there for the constitutive assumption that surface effects are representable by a stress acting in the direction of N? First of all, it is the simplest sensible assumption and one that at least explains the static phenomena of Figures 7.1 and 7.2_ It is reasonable to adopt this assumption and sec if it leads to sensible answers. Moreover, there is an analogy between tension in the surface phase and stress in the bulk phase (the interior of the fluid)_ It is frequently useful to postulate that a fluid has zero viscosity; in this case the internal stress acts normally. One hopes, not ir1 vain as we shall see, that it will similarly be useful to ignore the possible presence of"surface viscosity" and thus to postulate normal-acting stresses in an interface as well as in the bulk phase. It should be pointed out that there is considerable interest in physicochemical problems where the standard surface tension postulate is inappropriate. Ark ( 19ô2) discusses some of the theoretical generalizations that are required when "surface viscosities" are considered. Nonetheless, it is clear that the hypothesis of normally acting surface tension rs at least as important for an analysis of surface effects as is the {i nviscidl hypothesis of normally acting stress in bulk phase fluid mechanics. FORCES ON A SURFACE ELEMENT

Having postulated a certain kind of interfacial stress, we shall derive the desired dynamic boundary condition for uni invirc•id fiuidt by requiring conservation of momentum to hold for a fluid element "straddling" the in terface_ Consider a portion of the interface cut out by planes x — constant and y = constant, where each pair of planes is a small d istance ki apart (Figure 7.71. Lei us locus our attention on a material element R whose midsurface is this portion of the interface and whose thickness has the uniform value r. . We shall apply the conservation of linear momentum requirement (15) at the instant t. It is not hard to show that the i nertia term and the body force term will be hounded by a multiple of r.j. . For example, suppose that the mid surface has the equation z = Mix, y) for (x, y) in the square S,,. 11 l' denotes the volume of R, then

JfJPF

di

!+

ax (pf ]

VR }

where the maximum is taken over some fixed domain containing R. But k,t • Surface viscosity seems to

haVC

no detectable cficx:t in pure liquids, but may be imponanl in

solutions_ t Effects of viscosity are usually secundary in water wave problems [sec Bachelor (1457, 370A-

Sec. 7.1] Boundary Conditions

309

I

'

I

I

^ I

I ^

+ I I t

I I I ^

-- — —'— — —

^

—r ^^

^r

u

Fm

u sc a=_ 7.7 A

.sr1r}urr " flake -

R.

can be found by multiplying the area of the rnidsurface by the thickness cp . Thus lfI< = efa f f(1 4 Ms

)"' dx dy Ç

-

F1.e' max if I t M .

-

4 M )" 2

sY

Similarly, the forces acting on the vertical sides of R will be bounded by the product of ell' and a constant independent of f: and p. Since x can be made arbitrarily small, the remaining terms of (15) have the property

Zi

rn 2 1f

, i]

.1;

+ f5tar + ^Ir ds

O.

(L 7)

i

Here C is the intersection of the interface .a and the boundary OR of R, and . is the portion of ,i that is interior to C. A plus or minus superscript is affixed to .^ depending on whether the surface is approached from the top (z > O) or the bottom (z < O). Equation (17) is a local equilibrium principle of the type derived in Chapter 14 of 1 - ln the present case the stress and surface tension forces on a surface element .o are in equilibrium. We now restrict ourselves. for simplicity, to a two-dimensional situation in which there is no variation in the y-direction. In particular, the interface height does not vary with y, so the interface is a cylindrical surface with generators parallel to the y - axis (Figure 7.8). The surface tension forces acting on the front edge of the interface and its back (curves cut by planes y = constant) are exactly equal and opposite and so add to zero. The surface tension force per unit length that :acts on each of the Iwo straight lines

310

Formulation of she Theory of Surface Waves in an #nviuid Fluid (Ch. T T{si.j N(s L j

th mugs 7.$. Edge stresses on a surJur fr. The surfil €e fs o cylinder wiih arr equation of 'he form f ( x, 2J = ^.

forming the side boundaries of the cylindrical interface does not vary along one such line. These force per unir length vectors are denoted by T(sJN(s L) and T(s n )N(sR ), respectively (Figure 7.8). The subscripts L and f refer to the left and right for one looking in the direction y increasing. The reason for the use of the letter s will appear in a moment. The surface tension_ forces acting on the left and right sides of the interface, each side having length , are pT(sL)N(sj and pT(sR )N(sR ). A "head-on" view of the from edge of the interface, as seen by someone looking in the direction of increasing y, appears in Figure 7.9. We designate by e the curve making up this front edge, The total length of r will be denoted by A__ Its arc length, taken increasing as x increases, will be denoted by s. At the left

Y EOuKF 7.9. The frorrr edge of t he interface dPpirre,d in tï:qure 7.$. fop and t ^ bottom ^7rYS]71 rf} magnitudes Cire {) r f S) and Sll , respectively.

Sec,

7.11

Boundary Conditions

3 11

TABLE 7.1 Forces on the iriterfaee. Upward

Rightward

tin directio n of z increasing)

(in dircction of s increasing)

Top: Bottom: Left:

Right:

< 5 '1( 5 R ,

1

= I,

3,4-

and right ends of c, s will be given the values st_ and s R , so — By 0(s) we shall mean the positive acute angle between the x-axis and the tangent to c at s We indicate in Figure 7.9 typical pressures p ris) and Nis) acting normally to the top and bottom surfaces. Note that in this view Ms ') and N(s,) are tangent to c at its left and right ends. We are now in a position to obtain the information needed for determining the consequences of the local equilibrium equation (17). In evaluating the line integral, we need only consider the contributions from the left and right edges of the interface, since the contributions from the front and back edges add to zero. We list the required information in Table 7. I . In the table, we have employed for c the parametric equations x = x(s), z =

z(s); s i < s < sx,

( 18 )

where the dependence on t has not been explicitly shown. We have also used the relations dz

Sin 0= ±—, fI' ï

-

dx

c[ts$ —

(19)

d5

The "Wngen[ angle" ti is defined somewhat differently from its counscrpan that was

used above, Which unambiguous do inition one adopts is essentrilly irrirrratereal. Of importance is the care with signs that we emphasize here and that we glossad over earlier by employing the .

'usual" formula tan

kir dzidx.

312

rF o rmulal ion of tlrp Theory of Surface Wares in an litriscid fluid

[Ch. 7

r

4x FIGURE'

7.10. Diagram used to rabtain the relationships ( C 9) between Me

rariyrnrx:l anyJr° El pfird dLrirorrres kith respect to arc feriylh s.

it,i2; therefore, the signs must be (sec Figure 7.10). By definition, 0 5 il chosen to make sin nonnegative. Let us carefully derive the first entry in the first column of the table Tri begin with, in Figure 7.9 the arrow marked p i correctly denotes the stress exerted by the top fluid on the inter face .0 since, because we are regarding the Fluid as inviscir1. R stress of triagnitude p r. acts in the direction of the exterior normal (which points out of .0 into the top fluid). Equivalently, a stress of magnitude +pr aCts in the[lirection of the interior normal (as shown). The corresponding force associated with an element of c having length ds is p 3 ds. The "top" vertical force upward on .0 is thus -

—

-

J

sR

r

pt(s) cos O(s)]p ds = —

fl_

FA p1{s) ds ds (s)p s^

dx pr(5 r)

( 01. ;

ds

(e haw used the integral mean value theorem in obtaining the above equation.) Let us turn to the third entry of column one The force upward due to the surface tension acting on the left-hand end of c is

oi5t) s i n vi s it } .

= J^^l^

)

f

dz —

Ts [tiL)l

L

The negative sign appears since, as can be seen from Figure 7.9, for s near S L , z decreases as s increases; so dzjds is negative. But suppose that the Figure were di awn differently; would the result be the same? consider Figure 7.11.

Here, since N(s L ) points downward, the upward force is

—117'(a L ) in 0 (sL) =

—

reT(S, p g . Negative K implies a picture like bulges as one would expect with Figure 7.11; PT < p g , which is again as one would expect. EXERCISES

The first four exercises involve two-dimensional flows, sa that the interface reduces to a curve in the xy-plane_

Sec. 7.1 ] Boundary C ündisians

315

1. (a) Show that any curve xy = Constant can be though[ of as an interface dividing two immiscible materials whose motion is described by the velocity vector v = xi — yi(b) Derive (8). 2. (a) In a two-dimensional steady motion any streamline can be considered as a boundary between immiscible materials. Why? (b) Verify the statement of Exercise 2(a) for the particular example worked out in Section 112 of 1. In that section the velocity vector is given in Equation (17). the streamlines in Equation 12 I). 3. (a) In two-dimensional unsteady motion a streamline generally cannot be considered as the boundary between immiscible materials. Why not? (h) Show that the streamlines of Exercise 13,2.4 of 1 cannot be considered as boundaries separating immiscible materials moving with the give velocity field. +4. h can he shown that the following two equations Carl he considered as giving particle paths for the flow of a fluid of uniform density: x = a + e6 sin la -f tj,

y =5

—

cos la

r).

00) (31)

Various values of (u, b) correspond to various fluid particles. Thus the spatial velocity field is i^

= exp fb(x, y, 11 cos [a(x, y, z) + r], = exp [b(x,y, t)] sin [a(x, y. r) -I- r], ]

where a(x, y, r] and b(x, y, t) are given implicitly by (30) and (31). Show that the kinematic houndarycondition is satisfied along the curve derived from (3Oland (31) by setting h - constant and regarding casa parameter. This is GerSrtier's rrochuitial wave, the single known exact solution to the problem of wave motion in a semi - infinite medium with no surface tension (that we formulate below) [see Milne - Thompson (1968)]. 1-5. Show [hat if the surface F(x, r) = 0 separates water and air, then it always consists of the same fluid particles. }6. Static problems involving surface tension are expressed in terms of the surface tension coefficient T. the gravitational constant g, and the density of the medium p. What conclusions can be drawn using dimensional analysis? 7. (a) Derive those entries in Table 7 that are not derived in the text_ (b) In the limit used to obtain (20), to what quantity does )/ i tend? 8. Derive (22)9. $(a) Derive (28). (h) Derive 129).

316

Formulation of the Theory of Surface Waves hi [ar lniiseid Fluid [Ch. 7

10. Derive the dynamic boundary condition without the restriction 0/0y = O.

Proceed by the following steps. *(a) Consider the interface element .fir of Figure 7.12. with equation = C(x, y). whose projection on the xy-plane is the square whose z corners are

(ye, yo -4 J1)J (A + /1, }'o + Fe)Write down an integral relationship requiring that the pressure forces and surface forces add to zero. (xo, yo), (xo + Jz, yo),

I

FIGURE 7.12. An interface element .0 whose projection on the -.. Y-plarrr square.

Is

a

(b) Combining the integrals and using the mean value theorem and the integral mean value theorem. deduce the following relationships:

+- C ÿ) I n

ôy [TCxC,,ll + —

^ [} • (1 # ^ ){1 + Ÿ

C.

ApC }, W — ^ [ T{1 -i- c!}[1 + + ^^ [

A P = 6 [ l c,(1 + +

^x

[Tc x(1 -3-

ÿ)

Cr) - ;], "

+

J ,zi

+

1,2

J.

(32)

+ cr) +

7- c. c,(1 +

+- ^

''/]-

(33)

Sec.

7.J1

Boundary Condilions

317

(c) Use 134) to eliminate Ap from (32). carry out the differentiations. and ohtain T. = O. Similarly, show that T,. = 0- Hence derive from (34) Ap ° T[

I + C) — 2Cz4,C., + C yy(l + CŸ)1(1 + ; + rÿ21

- ]" 2 (35)

The coefficient of Tmust have a geometric interpretation that can be described without reference to a particular coordinate system_ One such description is ic 1 + k:2, where !Ki and lc 1 are obtained as follows. Construct the tangent plane 2: to the surface 5 at point P_ Constructtwo planes r, and rt 3 perpendicular to each other and to I (Figure 7.13). Define c 1 as the cu rvature al P of the plane curve r ; formed by the intersection of ri and .0, 1 = 1, 2. The planes n, and n, are determined by the above description only up tri a rotation about their line of intcrsectior]. But it can he shown that xi f- rc2 has the same value however the planes are rotated; 3{rc ti # li 2 ) is called the mean curvature- For further details, see accounts r3f differential geometry. such as that in Chapter 2 of 1J. Struik, Di prentiai Geometry (Reading, Mass.: Addison-Wesley, 195U). REMARK.

FIG [1 R E

7- I 3. Diagram us ed to ex plain Ihe -

er,rrr eRr -

of

orrura ['urralure

11_ Improve the derivation of the dynamic boundary condition (26). This could be the subject of a brief essay. Here are some suggestions. (a) Give stronger physical reasons For assuming the constitutive equation (16). (h) Justify the statements above (17) concerning the forces on the vertical sides of R. (c) List precise smoothness conditions that guarantee the validity of the various mathematical steps. tZ. (This exercise requires knowledge of regular perturbation theory; see Chapter 7 of 1.) $(a) Let : = /ix) be the interface between fluid and air. Let the fluid be motionless, inviscid, and subject to constant atmospheric

318

Formula iton of the Theory of Surface Wares if! un Irwi..ccW Fluid

10i.

7

pressure and the force of gravity (acting in the direction of —z}. Let p be the density, assumed constant. Show that the dynamic boundary condition becomes —T - p9' + ((1 + CD-3r2 = C,

C a constant. (36)

(h) Figure 7.14 represents a fluid hounded on one side by a vertical plane wall. Show that with the coordinate system given in this figure, C must be zero_

FIGURE 7_ 1 4. 5raiiortary ntittiscus Own a fluid with sur farE• re•nsrun is in c•orriur•r

With a y e,riicuf u411.

Make the change of variables rÿ= h ,x= xh cot O.

(37)

Then show that (36), with C = 0, can be written =1 2 C[1 -#

'- d

')T",

dR

(38) '

where 2=

^9f?s

E—

r 7. ,

cat 3

O

(39a, b)

.

Show that appropriate boundary conditions are C(0) — 1,

(

0

as

-c -. oc

,

(40)

and also C '(0)

-1

.

(41)

If you are familiar with scaling procedures (Section 6.3 of i),

justify (37).

See. 7.2] Formul

t(d)

lion

319

and Simplification

For certain liquids. r. is small compared with unity. Therefore. we look for a series solution (►r)_Ço( )+zrg)+cÇÇ(7+ ... to (38) and 140). We regard the parameter r i* as fixed when we calculate this solution. Later we shall impose 141) and find a value for 7 2 . Show that (38) and 14C) imply that —

'i

-r/C0

= 0;

CD(0) = 1,

z i CL = ilr 2Ca(C1) 2 ;

Co(cc) = f-

C i (o} — C 1 (cc} = U.

‘_.

Find the equations satisfied by *(e) Show that i_ o = exp (— M. and find findr_ 2 . 1(1) Show that (41) implies that

L,.

if you are courageous.

1 =- rl + r:r; + O(r: 3 ). Assume that 7 2 = I + ar. + 0(r 2 ) and find a. [Note that although there is an f: in the denominator of (39a). nevertheless. r 2 = 0(1). as was implicitly assumed above.] ff your courage persists. find the O(c) Correction to -r. +(g) Consider fluid contained between two parallel vertical walls_ About how far apart would the walls have be before we could use the Le estimate the height of the meniscus? (h) This problem can be solved exactly, See for example. Landau and Lifschita (1959. p. 2351. For sufficiently small r:. show [ha' the expansion given above agrees with an expansion of the exact answer (assuming convergence of the various series).

to

result of lfl

,

7.2 Formulation and Simplification We shall formulate a problem involving surface waves of an inviscid fluid (which we shall call "water") whose density can be regarded as uniform. "Air," a relatively light fluid whose influence is negligible. will be assumed to occupy the space above the water surface. Although a number of simplifying assumptions will be made, the mathematical problem that emerges will still be quite difficult. Thus the section concludes with a further simplification, neglect of nonlinear terms. We use an asterisk (*) to show that variables are dimensional. As usual, subscripts denote partial derivatives.

320

Formulation of

the

Theory of Surface Waves in an lrmisrid Fluid [Ch. 7

EQUATIONS FOR TWO-DIMENSIONAL WAVES IN AN INFINITELY WIDE. LAYER OF JN VISCID FWD

We shall restrict ourselves to two-dimensional motions (u* = e3fe3y* The body force will be that corresponding to a uniform vertical gravitational acceleration of constant magnitude g_ The lower boundary of the Fluid will be assumed to be an impermeable horizontal plane (therefore, we cannot cQstsider such phenomena as the breaking of waves on a sloping beach). We choose a coordinate system so that the lower boundary has the equation z* — — H (H a constant) anti so that gravity acts in the direction °I - decreasing z* (Figure 7.15).

FIGURE

7_ i S. Natation

for urresiryae+art ol water NY]r e'.s in fluid al mean dept h H.

We first write down the requirement of linear momentum balance. From (A2.1.6) with f — gk, taking into account definition (A2 1_1b) of the material derivative, we have u* -t If*wx. + V0* u'. — — pi

w + u*w:. -1- x•*w=. = —p

x. ,

t p r: . —

(l )

g.

(2)

Here te* and I4 are velocity components in the directions x* Increasing and z * increasing, respectively_ The constant p is the density and p* denotes the pressure1r or a fluid of uniform density the mass-eonservatiorr requirement (A.2.1.1) is u =. + ti4 1. = o. (3) The condition that boundary condition

nn fluid permeates the bottom gives rise to the

at z' =

—

H,

w* _ 0.

bottom (4)

We shall assume that "side" boundaries are so far apart that their influence can be Neglected_ Consequently, the horizontal variable x• has the domain — cx < x* < co. Once such an infinite domain is allowed, experience teaches that there may be solutions which behave badly aslx* I —, c:. There is — really" rio such thing as an infinite domain in nature, so it is not a straightforward

Sec- 7.2] For w/ariun and Srtrplrfieution

32]

matter to assign boundary conditions 'al infinity." !Nonetheless, there are some types of behavior (e.g., exponential "blowup") that are almost certainly intolerable in a mathematical mode!. We deal with the situation for the present by imposing the following somewhat vague side boundary condition: solution "well behaved" as lx' l

(51

oc.

— C*(x*, r*]_ When Iet the water-air interface have the equation have, from (111), the surface kinematic boundary condition at r* = C * (x * , t#),

'' = Cr. + u*4*..

i,

•=0 (6}

For reasons that will be discussed shortly, we shall assume that [he pressure above the water surface is a constant p . With this assurruption s (' _26) and (1.29) give the surface dynamic boundary condition T(*.iotl •+ (f *•)3]

at z* = C • (x', l'1,

312 M

pA - p*.

(71

Here Tis the surface [ension. As is the case even with the simplest mechanical problems, initial conditions must be imposed to obtain a fully specified situation_ For, all other things being equal, the initial state of the water determines its subsequent history. Later we shall give examples of appropriate conditions; now it will suffice to make a formal statement that such conditions need in be imposed. We shall start measuring time from the initial instant, which will thus be designated r* = O. Hence we require the initial condition at t• = 0,

suitable conditions prescribed_

(g)

We have been thinking of a water layer having "depth" H , but we must make precise what we mean by this. The depth varies when surface irregularities are present. We can, however, speak ofa layer whose mean depth is H. Since the layer bottom is located ël ÿ* = —11, this means that 2* = O must be the mean position ni - the interface_f Hence the function Mx*, two must haven spatial mean value of zero. if * has period L, we therefore impose the mean Might requirement ([or a periodic interface) 1

i-

1, ^o

c.v.,t *) dx* = O.

(9a)

if C* is not periodic, we must make the more general requirement

Urn

L

J-14*(x*, t*}dx* = O.

(9b)

# Although the definition ]c prohahl}- known to most readers. for rnmpteteness she mean value of a function is defined in Exercise 1.

322

Formulation of the Theor y of Surface Waves in an h+riseulflu rd (Ch. 7

Possible changes in mean height could be permitted by imposing (9) only al t* = O. The mathematical problem posed by (1) to (9) seems likely to have a solution. lr, (1) to (3) we have three equations for the three unknowns u*, w', and p*. Also, en each part of the boundary equations (4) to (7) provide a N o T E.

boundary condition on a*, w*, and p*. [We are temporarily interpreting (6) and (7) as a sine implicit condition on p* and w*. This can be obtained in theory" by eliminating L'_J Initial conditions are provided by (8). while (9) is an extra normalization condition that pins down the average location of the bounding surface. Even if we could prove that the system of equations (I) to (9) has a unique solution, it would be quite another matter to characterize this solution in any detail. The system is nonlinear, and some of the boundary conditions are to be applied onacurve z' _ *(x', t* ) whose location will only be known when the problem is solved. These features cause considerable technical difficulty, so we shall soon restrict our attention to a simplified system. Before simplifying the equations, however, it is advantageous to reformulate there somewhat, by employing a slightly different form of the pressure which arises when a possible mot ionless state is "subtracted out:" STATIC AND DYNAMIC PRESSURE

A steady-state (Nat* = 0), exact solution to (1) to (9), cur responding to a motionless layer, can be found at once. The surface bounding a motionless layer should be plane, so we look for a solution wherein

u*= IA* —ÿ* = Equations (3) to (6) are identically satisfied, while (1) shows that p* is independent of x'_ Equations (2) and (7) require that p* = po(z*). where

pa(z*) _ — Paz * + pa ,

— H S z* < 0.

(1%

Initial condition (S) is irrelevant for our steady-state problem. Finally the mean height condition (9) is certainly satisfied. 11 is not difficult 10 discern the physical meaning of the solution just found_ The static pressurept, a distance — z* units below the surface of a motionless layer is due 10 the weight of a fluid column —z* unfits in height and one unit in cross section, plus the weight of the atmosphere. Returning now to the original problem ( I) to (9), we find that it is convenient to introduce the difference between the total pressure p' and the static pressure p" rather than the pressure itself. We accordingly define the dynamic pressure

p* by p * (X *, I*. t * ) = p* x * , z*. V') (

no(z').

(1 l)

Sec. 7.21

Fornrulariocr und Simplification

323

In terms of the dynamic pressure rather than the, total pressure, the governing equations [(1) to (9)] become

+ w*W# = —p 'i^•

,

wt*, + u*wx. + w*w= = — pp= ,

u*. + w= = 0. At x• _ — N,

x,* _ O.

As I x*

solution well behaved.

cc,

(15)

(16)

w* = Cr. + u*Cr..

At z* = 4*(x*. r*)

^^^•f •tl + (C1•)2]-312 = — P* + p9‘*-

is

= 0.

(13) (14)

At z* _ C*(x*, r*),

At

( 12)

(17)

suitable initial conditions prescribed.

(18)

(19)

For waves o f period L in x *,

^

f

L

x*. r*} dx* =

a

U,

(20a)

and for an aperiodic interface

1 i:, 2K

urn —

x

_

C*(x#, t*) dx* - 0_

(20a)

,^

Comparison of (12) to (20) with (1) to (9) shows that only two essential changes occur as a result of introducing the dynamic pressure i•)*. (i) The body force terms g is dropped in passing from (2) to (13), (ii) In passing from 7) to (18), we drop the constant term pa and add the term pgC . If there were no free surface [and hence no dynamic boundary condition (18)), the equations involving the dynamic pressure p * would thus be exactly the same as the equal ions involving the total pressure p* except that the gravity term would be absent in the former. Consequently, for an inviscid fluid of uniform density, in the absence of free surfaces one can always assume that gravity is nor present. The pressure obtained as part of the solution will be the dynamic pressure. The effect of gravity can be taken into account merely by adding the static pressure and the dynamic pressure to determine the total pressure. When free surfaces are present. the static pressure might better be termed the mean static pressure, as it gives the force on a unit area due to the (temporal) average weight of a column of fluid whose actual height fluctuates because of surface deflection. Consequently, as can be seen from the dynamic boundary condition ( le), in the presence of a free surface gravity must be taken into account throughout the calculations. [

324

Formulation of th e Theory of Surface Wares in an fm ist id Fluid [Ch.)

Note that the constant atmospheric pressure does not appear anywhere in the governing equations (12) TO (20). This is because motions are driven by pressure derences, so that a uniform pressure has no dynamical effect. (The situation would be different if thermodynamic effects entered, but recall that we are considering a fluid of uniform density.) In particular, the motion of the hounding surface is entirely unaffected by a constant atmospheric pressure_ The surface moves just as if it were entirely free of a constraining pressure force, and is thus often called a free surface_ In restricting consideration to fluids bounded by a free surface, we are tacitly putting to one side the important problem of wave generation by wind—for here air-pressure variations are all important. Subtler effects are

also being neglected. For any motion of the water will drive a motion of the adjacent air, producing air-pressure variations. These, in turn, will affect the motion of the water. Since the density of air is much less than that of water, the effect of the latter motion will be negligible except when it continues for too long a time. The exception must be noted, because small effects can accumulate to noticeable levels if they act over long periods. NON D1M ENSIONALIZATION

We shall now introduce dimensionless variables to reduce t he number of parameters on which the solution explicitly depends_ (Compare Section 61 of I_) For the time being we shall restrict our attention to "fairly regular" wave trains, which have the property that the distance between one crest and the next does not vary much. Any one of these distances can be chosen as a typical wavelength L (Figure 7.16). "Fairly regular" wave trains also have the property that the maximum heights of the individual waves do not differ appreciably; consequently, any one of these heights can be chosen as a typical amplitude A. The time it takes any selected wave crest m to travel a distance L will be called a typical period P. (In the special case of an exactly periodic wave train that propagates at constant speed C without changing shape, P will be the temporal period of the waves and C = LIP.) 1 n dealing with fairly regular waves, the quotient of a typical wavelength L and a typical period Pis sometimes referred tc3 as a typical wave speed. We now

1

„,

L and typical umpliructe R. !he ware cr est m 1rai5els a distance approximately equal to L in the typical per i od P. Ffoi1KE 7.I6. A

"

fairly regular" wave train of typical wavelength

Sec. 7,21 Formulation and Simplification

introduce new (unstarred) dimensionless variables, in terms of L, the density p:

la

x`

P,

x — L

w* w= ^^^,,

u*

u= A ^ , /

p

,

^

z

=

A,

325

P, and

Z^

f.

p* pLY^3

,

2'

^ — C` . A

(21)

in terms of these variables, (12) to (2C) can he written Exercise 6)

x + sm.} =

!1, + +

14*

+ rtx

At

z_

(24) ( 25)

w = CF 4- EuC Y .

At z — ES

p

L

031

+ w= =

At z = EC,

For waves of period

= — p_+

solution well hehaved_

As 1x1 i ac',

At t = 0,

(221

w = 0.

— ^J ,

,

— pY .

(26) (27)

BC„(1 + v2 (x ) -3t z ).

suitable initial conditions prescribed.

(28)

(29)

in x* (and hence of period unity in x),

f

C(x, t) dx = O.

(30a)

For aperiodic waves, lim A^

Here

t—

A L'

ce

1 21L

^

_} , H

‘, (x,

t } dx — O.

(30b)

- k

l

T p L. 2 '

qP 2 I

(3 l )

are dimensionless combinations of parameters that appear in the dimensionless equations (22) to (30)_ We see that the severe dimensional parameters A, L, H T, p, y, and P can appear only in the four dimensionless combinations of (31). The first two parameters of (31) are geometric and provide, respectively, the ratios of typical wave amplitude and mean water height in typical wavelength. The parameter B, sometimes called a Bond number provides a measure of the importance of surface tension. The Froude number Frrieasures the importance of gravity. It is thesquare of the ratio of two times, Pond WO". The former is a typical wave period and the latter is the trmc it takes a particle, starting from rest, to fall a distance equal to half a typical wavelength.

326

Fcar m ufari lm of fhe Th eory of Surface Waves

i1ltHS['id fluid [Ch. 7

LINEARIZATION

The mathematical problem posed by (22) to (30) is difficult_ There arc two main reasons for this. First, the equations contain many nonlinear terms, and there is no general theory for nonlinear partial differential equations. Second, the boundary conditions (27) and (28) must be imposed on the curve z = but {(x, t) is one of the unknown variables of the problem. If we restrict our considerations for the time being to waves of small amplitude, then both difficulties disappear. We expect that such waves accompany flows wherein velocities, pressures, surface deflections, and their rates of changes are small—so that all nonlinear terms can be neglected to a first approximation. The assumed smallness of surface deflection has another consequence: surface boundary conditions can safely be evaluated at the mean position of the interface. With nonlinear terms neglected and surface boundary conditions imposed at the mean water level z = 0, (22) to (30) simplify to the following sels of equations [with a superscript (I) denoting a first approximation]: =

_

f

w it {

t) x

(32)

(JJ

=

(33)

O. (34)

Ate — — h, w4 = D As 1x1 At

a= 0,

At z = 0, At

i

as,

(35)

solution well behaved.

wO,

P(

iJ

=_ ^+ , ^

(36)

(37)

^ F(Cl t t — 8CV2)-

— 0, suitable initial conditions prescribed.

(38)

(39)

For waves of unit period ,

J

^{ ^ ^[x,

^}

âx = 0.

(40a)

0

For aperiodic waves, li ril .0

I 2K

II: ^t IX, t} dx = O. _a

(40h)

The two succeeding chapters are devoted to a study of the linearized equations (32) to (40). The reader can begin this study at once if he wishes. But sooner or later he should peruse the next section , where the linearization process is examined more carefully. Let us mention here the most important point covered by this reexamination. Equations (32) to (mil) can be obtained

Sec.

7 3 Order-af Magniiiide Esiimares, Narjdimeresichalization, and Scaling

327

from (22) to (30) by setting c equal to zero. but we have provided no firm justi fication for such a formal step. EXERCISES

1. Let f be a continuous function defined on the interval [i . b]. Let x o = a, x1, x„ = b divide [a, bl into n equally spaced intervals of width Ax. Write down an expression for the average value of the pi + I quantities f (xi), i = 0, I, 2, ... , n. Multiply numerator and denominator by Ax and pass to the limit, thereby motivating the following definition: average value off (x)

1 -

b

f (x) dx.

2. Show that for a function of period L, (9a) holds if and only if (9h) holds. 3. (a) Formula tea mathematical problem that is relevant to wavegeneration by wind. (h) How would the Formulation in the text be altered if the bottom were irregular? In particular, how would (9) be changed? 4. In Section 15.4 of 1, gravity was neglected in finding the force on a body immersed in a uniform stream. Show that the presence of gravity adds a buoyancy force. 5. True or false : Even though a free surface is present, if surface-tension effects are dominant, then the effect of gravity can be determined in the same way as recommended in the text for situations in which free surfaces are absent. Explain your answer. 6. (a) Verify that the variables of (21) and the parameters of (31) are dimensionless_ (b) Verily (22) La (30). 7. If you are familiar with the concept of stability (see Sections 11. l and 152 of I), show that (32) to (40) would he obtained in analyzing the stability of a motionless layer of fluid to small perturbations.

7.3 Order-of-Magnitude Estimates, Nondimensionalimation, and Scaling In Section 2 we passed in a rather formal way from the dimensional set of equations (2.12 -20) to the linearized dimensionless set (132 -40). We shall now try to make this passage more convincing. For full appreciation of the ensuing discussion, the reader should be familiar with the content of Sections 6i and 6.3 of I on dimensional analysis and scaling, but the present section can be read independently. It will suffice to know that a rough definition of the concept of the niter- of magnitude of a function in some region is "an estimate of the function's maximum absolute value, for independent variables in the given region_"

328

Formulation of the Theory of Surface Waves or an }txviscirf Fluid (Ch. 7

ESTIMATING THE SIZE OF TERMS IN EQUATIONS GOVERNING WATER WAVES

In Section 2 we decided to neglect the nonlinear terms in (2.22) and (2.23). Using language introduced in Section 13.3 of 1, we are evidently assuming that the convective acceleration terms (preceded by r) are negligible compared with the local acceleration terms w and w. To gain more insight into this matter, let us estimate the order of magnitude of the acceleration terms for a train of "fairly regular" waves of typical wavelength L, amplitude A, and temporal period P (see Figure 7.16). The first step is to determine the order of magnitude of the velocity v* possessed by the fluid composing the wave train of Figure 7.16. Is it AL, or A/P or LJP, or an entirely different quantity? The first possibility can be ruled out on dimensional grounds, since AL has dimensions (length) 2 . Such quantities as A/P, LJP, (LIP) - (AIL)" all have the correct dimensions, so the requirement of correct dimensionality is of limited usefulness. More helpful is a knowledge of the general features of water waves, which yields the fact that during u period P, particles near the surface of the water travel a distance roughly equal to the amplitude A. The farther below the surface a particle is, the less extcnsive is its motion. Thus, since it is maximum values of Ive1 that we arc concerned with, v* has order of magnitude A/P. There are at least three ways that a person could acquire the knowledge to estimate v*. (i) He could have observed water waves.f Watching a piece of seaweed tossed about by ocean waves, he would see that the seaweed moves hack and forth over a distance approximately equal to the amplitude A, not to the wavelength L, in a period P. (ii) He could have drawn upon a knowledge of wave motion obtained from a study, either theoretical or experimental, of wave motion in other media- (iii) He could have solved the linearized problem (2.32 40) and computed the path of a particle (see Exercise 8.1.5). This last way might seem dishonest. We are attempting to provide convincing motivation for an approximation to be used in solving a problem. We there turn around and say that lithe problem is salved, then we can supply the motivation sought_ This is a typical and legitimate procedure, however Uncertain of t he validity of what one is doing, one makes an approximation for a formal mathematical reason. One solves the approximate problem, and hopes that the physical insight thereby gained will clarify the physical significance of the original approximation. We have argued that v* is of order of magnitude A/P. What about the local acceleration vt? To determine its magnitude. we begin by recalling that to a first approximation v* has period Pin t * . In a time Pf2. then, Iv* I rises from a ,

t Theoretical instruction 1n fluid mechanics. can b.e enhanced Ihruugh Trie use of films precd by the 1J.5. National C'onuncücc for 1-lurd Mechanics Films e nd disrrrbutrd by Encyiopacdia Eirilairniea Edllcalionik Corporation, 425 Miehigarti Ave.,

Chicago. Ill

-

6&61I- here

portions of A. l3ryson's him " waves in F1uicis" could certainly be shown to advantage. See the book of film notes, éf ssrrorrd Experiments ut Fluid Mechanics (C'amhtrdgc, Mass.: MIT Press,

1972).

Se c. 1.31

O rder - of-Magrrrfude Essinmtes, Nondlr ►rensio+rafizariori, and Scaling

32 9

value of zero, more or less levels off around its maximum value, and then falls back to zero. If we assign equal time intervals to each of these three behaviors,# we are led to the conclusion that 1 v* rises from zero to its maximum value of A/P in about P/6 time units. Suppose that I v* I = 0 at time r0 . As an estimate for I rh.*Jot * j, we have

r * (ic * P 6)1 ^v [ a l A/P – 0 6A

11

P/6

-

– – P/6

px

(l]

Similarly, keeping in mind that to a first approximation v* has a spatial period of L, we find that

j at,* j

AjP

(2)

ex";

We further assert that derivatives with respect to z* also reduce I v* I by a factor roughly equal to L/6. (This, too, requires some advance knowledge of water waves, such as direct observation that – unless the water is very shallow velocities seem to change no more rapidly in the horizontal direction than in the vertical) Thus it seems legitimate to assume that all spatial derivatives multiply orders of magnitude by a factor that is roughly equal to 6/L.§ As a first application of our order-of-magnitude estimates, we educe that

* ` Vsid

is of order of magnitude

Y*Jc''i *

(AM kl6ïL) A/Pl _ A IA/Pk(6 P) L

—

[1

Thus our neglect of the nonlinear comer ire acceleration terms means that

we ►viith their length ., are considering wades whose urnpilitade is smart compared i.e., AIL C L There is more that we can accomplish, now that we have an idea of the magnitudes of the various terms- As will be iliustratrd al once, we can choose dimensionless variables in such a way that each term is preceded by a combination of parameters that explicitly reveals the term's magnitude. Appropriate approximations are thereby suggested. SCALING

We could define a nondimensional horizontal variable x in a number of ways, such as x = x*{L or x = "A or x = Y' L) or x = x* /6A. The quantities L, A, A 2/ L, and 6A are called intrinsic reference lengths, since they t The three inter vats may well not be equal, so we Cannot be confident Ilya' our estimates are at ail precise. But ii is rough orders of rnrtgnilude that we seek, for these are usuafIystill:cornt for the purposes of comparing terrn sIn making order cif magnitude estimates, we have nni considered the efeci of the bottom. This is appropriate unless the water layer is shallow Thc kcy ED a proper scaling foi shallow water waves is i he understanding That the depth is, in fact. I he appropriate length scale to use in farmin g the scaled vertical independcni variable and velocity compnneni lser Stoker 11957. -

Sec. 2.4)1.

-

330

Formulation

of the Theory of Surface Waves un an lmiacid Mud [CIS_ 7

are standards of length that are expressed in terms of parameters appearing in the particular problem under consideration. Intrinsic reference velocities, intrinsic reference times. and intrinsic reference pressures are defined by analogy with intrinsic reference lengths_ They are specific velocities, times, and pressures that are formed from the parameters that appear in the formulation of a problem, and to which the velocities, times, and pressures of the problem can be compared in defining dimensionless variables. Examples of possible intrinsic reference velocities for our water wave problem are AIP and LIP. As we have seen, the choice of an intrinsic reference quantity is not unique_ We recommend that intrinsic reference quantities be chosen so that when they are used in forming dimensionless variables, then each term in she equations governing a problem is the product of a constant factor giving 'hose terms' order of magnitude and a function of unit order of magnitude_ The variables of that problem will then be termed scaled. The particular intrinsic reference quantity associated with each dependent and indcpcndunt variable will be called the scale of that variable. As we shall see, when variables are scaled (generally not an easy task), then semiautomatic perturbation procedures will often simplify the mathematical problem_ The concept of scale needs some discussion before its meaning becomes clear. To give one example, we assert that asa velocity scale in the water wave problem we have been discussing we should lake AIP, the order of magnitude of v* We would then define a dimensionless velocity v by (4)

so that A v* _ (-13)v_

(5)

We further assert that a reasonable choice for a length scale is L/6, so that dimensionless horizontal and vertical variables are defined by x*

x —

116

,

_*

a _ L{6

(6)

We have

0 0x*

d

6 L 8x'

ax*

60 L

( 7 a, b)

Consequently, using (4), v+-V

*Y*

— rriC} %i y ,

PLP

Vv.

(8)

Srr. 7.31 Order-oj-Maanin d

E_sJimmies.

dime+rsroncii 3 ion,

and Sca/i, i

33 I

Our choice of length scale has automatically led to results obtained above. That is.17) is equivaleiit to the assertion in the paragraph Following (21 that "all spatial derivatives multiply orders of magnitude by a factor that is roughly equal to b/L" Moreover, (Pam omat ically gives the numerator of the quotient in (3) as the order of magnitude of Or- V*v*_ The discussion of the preceding paragraph illustrates the following point. In scaling independent var iables one chooses those particular reference quantities which ensure that upon nondimensionaliairrg. terms involving derivatives appear preceded by a factor giving their order of magnitude f give another example, an appropriate time scale is P/6, for if

t* P/6'

(9)

then

ev* 11 4

A / P av P16 Ot

6A e?v 132 el

(IO1

Again, the correct order of magnitude appears expliciily on the right side [compare (1 )] while the dimensionless factor iv/r t is of unit order of magnitude. The surface deflection * has not yet been scaled, but this presents no difficulty. By definition, A is the maximum value of so that the scaled dimensionless surface deflection 5 should be defined as follows:

—— • A USE:

lll}

OF DtMENS1()NLES5 SCALED VARIABLES

A number of poilus of interest will emerge if we insert the dimensionless variables defined in (4), (6), (9), and (l l) into the kinematic boundary condition (2.61 We obtain the following:

at z = 6EC,

tit = 65, + 6^u

(12)

Y,

where

A

(13)

L

With the emergence of the parameter z, we are led naturally to the study of the subclass of water waves wherein I c 14 1. Such waves will be said to have small amplitudes, with the implicit understanding that amplitudes are being compared with wavelengths. For small-amplitude waves, a natural first approximation to the boundary condition (12) is obtained by retaining 0(1) tcrmst and by neglecting 0(e) t

Set Appendix 3 i ,

of 1

for definitions concerned wish the ••oh

natation used here.

f ïxmularian of Theory of Surface Wares in an &wsCid Fluid Irh. 7

33 2

terms. We obtain at z =ü w= ,

her .

"Automatically" we have emerged with a linearized boundary condition, applied at the mean position of the interface. Moreover, power-series expansions in c naturally suggest themselves as a means to improve upon the first approximation_ It now appears wise to drop the "6" that appears in (6) and (9), the definitions of scaled independent variables. We have seen that it is relative orders of magnitude O() compared with 0(l) that matter in our discussion. These are not affected by the "6." Moreover, scaling proceeds via order-of-magnitude es:imaies, and the appearance of precisely the number "6" might lull us into the illusion of an accuracy that we cannot legitimately claim. Keeping in mind that we can use a somewhat more accurate scaling if sharp numerical guesses are ever rLuired, we shall therefore employ the following dimensionless variables: t ;

is

^, x =

x` ^

,

__

z* ^,

u

u• ^^^,.

w•

a'^ = ^, {,

^' Ciis

These are the same variables that we used i n (2.24 With them, (12) is replaced by the following

at z =t:C. w—Sr - + This is identical, as it should be, with the corresponding equation (2.27). PRESSURE SCALE

We have avoided for as long as we could the question of the pressure scale. As often seems to be the case, this scale is the most difficult one to select. Perhaps the difficulty occurs because pressure measurements—in contrast to measurements of lengths, times, velocities, etc.—require instruments not possessed by the casual observer. In a first approach to the scaling we can argue as follows. We have seen above that when IE.I C I the convective acceleration is small compared with the local acceleration_ Then the acceleration of a fluid particle is of the magnitude of eviler*, i-e-, AP -2 . Consequently, the inertia of a unit volume of fluid has magnitude pAP -2 . How many such blocks of fluid are pushing in a coordinated fashion on a given unit area? As the scale of velocity changes is L, we would expect fluid in a column of length roughly comparable with L to be moving in the same direction. Consequently, the dynamic pressure, which is due to the inertia of moving fluid, should have magnitude pA P -2 L. This is the pressure scale used in (2.2!), so the justification of the choice of variables in Section 2 can be regarded as reasonably complete.

sec_ 7.3I Order-of Marirtiterdr Esti+rrot!'.s,

NondimeAsjorializiitnn, wid Scaling

333

Since the pressure scaling is subtle, it is worth looking at it from another point of view. The approach that follows is rather lengthy. but il is typical of the sort of argument that is sometimes necessary. For the moment, let us denote the unknown pressure scale by rt; the dimensionless (dynamic) pressure p is therefore defined by

A

(15)

In

The dynamic boundary condition (2.18) is the single must charauertstic feature of our water wave problem. so that it is perhaps not surprising that this condition turns out (after trial and error) to provide the key to scaling the pressure. In terms of our scaled dimensionless variables, (2_lli) can he written z = F.ÿ, Tfsr - 2 Cx,1 (I + F:2- C!) 3r] : —1rp + pgAly. 116) Pressure, as a force per unii area, has dimensions (mass)) length) (time) - 1 . The only parameter that has a mass dimension is the density. Consequently, it is convenient to write '

pressure scale n = (densï'ty)(length) 2 (tinie) r _

(17)

It is a ratio of lengths that forms the small parameter in our problem. so let

us concentrate on selecting the correct lengths in (17). Suppose that n V. If so, since the other terms in (16) are 0(A ) or smaller, the leading approximation to (16) For small I; = AIL. is p = O. This is a contradiction. We assume that the pressure is order unity: then we deduce that the order-unity term is zero, set that the magnitude of the pressure is small compared with unity. Suppose that in ^ A l . Then [he leading approximation ru (16) would not contain the pressure, being -

TAI -2 r,, = = pgAC.

(18)

From ( 18), when is positive, iâ xx is positive and the curve z = 1:(x, t) is concave upward. When 1Ç is negative, this curve is concave downward. Such concavity relationships are exactly opposite to the ones that roust occur for wavelike solutions (Figure 7_l7)_ We therefore roust reject the possibility that Concave down ward !

Concave

upward

i' luuxc 1117_

Ce+reial•rrr refaximshipl ferr o llpt;d

wart- mole

334

Formulaliom of !he Theory of Surface Wares in an a*nviscrd fluid (Ch. 7

the dynamic pressure is of smaller order than the 0(A) velocities and surface deflection. The other possibilities having been shown to lead to contradictions, it must be that the pressure is of order A. That is, from (17), Zr = pA[(length}(time] - }.

(19)

The crucial part of the scaling is now finished, for the leading approximation to the dynamic boundary condition has been established to be, in general, a balance among a pressure terra, a surface tension. term proportional to Ç , and a gravitational term proportional to C_ It remains to specify the square-bracketed contribution to (19). This contribution must be 0(1) if p is to be O(A). Both g and TJpL have the right dimensions, but each has the possibility of failing to be 0(1) if, respectively, gravity and surface tension are relatively unimportant. Moreover, it is L and P that have determined our scales thus far, so we anticipate that these

parameters will also enter here. Thus as a final version of (19) we lake it

t pALP -2

.

This is in agreement with our first approach to the scaling and hence with the scaling used in (2.21). Having struggled to understand the implications of the linearized water wave problem given at the end of the previous section, we are now ready to try to solve the problem. We begin a discussion of the solution in the next chapter, EXERCISE*

1. Six methods for obtaining order-of-magnitude estimates were listed at the end of Section 6.3 off_ Which of these were used at the beginning of the present section?

" Also sec Chapter 6 of 1. which cm-Rains exercises on dimensional analysis and scaling

C H APTE R 8

Solution in the Linear Theory

I

7 we formulated a system of linear partial differential equaaccompanying initial and boundary conditions, to describe the tions, with motion of small-amplitude water waves. In the present chapter we solve these equations. Section 8.l exhibits particular solutions composed of sinusoidal functions. in Section 8.2 it is shown that an infinite series of these functions will provide a solution that satisfies given spatially periodic initial conditions_ Section 83 treats the general initial value problem with a "smeared-out" integral superposition of the sinusoidal solutions. Our treatment of initial value problems illustrates the use or Fourier series (Section 8,2) and Fourier integral (Section 8.31. These topics have already been discussed in Chapters 4 anti 5 of f. [In I. Fourier analysis was used to solve the partial differential equation that describes diffusion, Here, a .s}•su ni of partial differential equations is being studied.) No knowledge of the material in I is assumed in the present chapter, but those who are familiar with this material cari rapidly skim part of the discussion in Sections 8.2 and 8.3. N CHAPTER

8.1 A Solution of the Linearized Equations In this section we shall find a particular periodic solution to the linearized equations for small- amplitude water waves_ in the next section we shall see that this solution is not so special as it may appear at first, since general periodic waves can be synthesized from it and solutions very similar to il, In the text we shall discuss the simplified case h — H}f, ac, which is applicable when the typical wavelength L is small compared with the mean depth H. The reader is asked to carry through the more complicated calculations required when this simplification is not made (Exercise 4). A,SSuMPTtrtN OF

A SOLUTION OF EXPONEN`i

IAL TYPE

The equations to he considered are those given at the end of Section 7.2, with it aC

= O,

+ As

z

-.

—

Ask! At z = 0,

oo cc,

,

wr 1 '

+pcir= 0, a » #

= O.

-' 0.

(2) (3)

solution well behaved.

w in = City*

(la, b,c)

p

ut = f .• IC"'

—

BrSS[]•

Solution periodic in space, with zero mean value. 335

(4a, b) (5)

Soiuiionire the Linear Theory ICS. 8

336

To find a solution that satisfies these conditions, we employ the following generalization of the well-known result concerning the exponential nature of solutions to ordinary differential equations with constant coefficients, Systems of linear homogeneous partial differential equations and !wundciry conditions with constant coefficients Lech as II) to (4)] laemr yo /ions in which each dependent variable is a different constant multiple of the same product of exponential factors, one factor for each independent variable.* We shall now consider a solution of this type and seek special choices For the various constants so that all required conditions are satisfied_ In particular, we take ut 't(x, z, r)

û

a'tt ttx, z, 1) i F ^ '(x, r. tJ#

-- Re

i„.

exp (cwt} exp (earx) exp (kz),

(6a)

h

and

5r 1 (x, t) = Re ÿ exp hew) exp where ü, +;.,

(tax).

(6h)

1i, C, w, a, and k ate constants_

R I:MARKS. (1) There is. of course. no factor exp (k:) in f6h) because the surface height tÿr" does not involve z. (ill Since the solution to our physical problem cannot be imaginary, we have prefixed the right side of (6) by "Re," indi sting our intention to take the Feat part of the final answer. (iii) We do not at first restrict w, a, and k to be real, so the presence or absence of i in the exponents involves no loss of generality. But the choice of exponents does reflect art anticipation of the final form of the solution. VERIFICATION THAT ALL CONDITIONS ARE SATISFIED

We note that a must be real. Otherwise, the solutions will become exponentially large either as _v —# * or._ or as x — — ac , which would violate 13) (Fxercise 2). Since a is real. wit satisfies the condition 151 Substitution of (6a) into (l) yields the following, on cancellation of the exponential factors that are common tea each term:

arrUw eafe

—f,

(7a)

k ji = 0,

(7h)

= O.

(7cl

+it

ic+ r

♦ i4tt•

Thesystemofhomogeneotisequationshasthetrivialsolutionü = t4 — p _ 0, corresponding to the motionless state_ For a nontrivial solution to exist, the determinant of the coefficients in (7) must vanish. so that W(2 — k 2 ) — ü. • [.-oarpyrc Stttion 15.2 of I, whetfi thrt abrrit appro ach is uscd.

(8)

. J]

Sec 8-

A Solution of Me Linearized F.gaarion.s

337

Suppose that ua = Q Then the only nontrivial solution to (I) to (5) of the form (6) is

_ 0,

= Fr r r =

C13

1

=

0'

1/ I J

=

h exp (h 2),

where Ce is arbitrary [Exercise 1(a)]. In this flow the horizontal speeds of adjacent horizontal planes of fluid are related by a'r' Y û exp (kz). But one can verify by inspection that (L } to (5) is satisfied by ti ll) = 1(4

w tI J

=

pri]

criJ

=

0.

Here f is an arbitrary function; it need not even be continuous. Such a general shearing motion is possible because planes of inviscid fluid can move over one another without hindrance_ Were ire to include the nontrivial solutions corresponding to w — 0, we would therefore be considering a combination of waves and at somewhat artificial shear flow. We wish to consider waves in the simplest context, so Met shall uhruti's ignore possible solotions a frrre°,cfxfridrng tb w — O.* For the remainder of t h is section we a swine that a > 0- The case of nega ( i vc a will be studied in Section 82) Since to # O. (X) implies that k = # a. Boundary condition (2] cannot he satisfied if < O. so

Since k is positive, the solutions (6) decay as z t_ The determinant of the coefficients in the homogeneous equations (7) is zero, so their solution will not he unique. In face, there turns out to be a one parameter family of Solutions_ Anticiipal ing this, we denote this parameter by a anis write —} —

-

=

(LO)

a,

where a is a complex constant, nonzero (to avoid the trivial solution). but otherwise arbitrary. From boundary condition (4a) plus Equations (7b) and (7c), we find —

fi — r,a 2 a - `a, ii = =utc (a it 0).

(l la, b, cJ

Equation (7a) is a consequence of (lb) and (7c) and provides no new information. [- the only nontrivial solutions corresponding to a T {) have w = 0 The full nonlinear equations (7.2.22-28) ate satisfied fur arbitrary 1 by u — fir1, k' l' C - D. This leads to the realiL1tion that even if rnilial cundiiruns arc prescribed. uniqueness requires a more prrcic condition ihan t71161 as r-Z 1 — x _ For example, if the initial disturbance is confined to a bounded region. Il seems approprraie to demand that their be no

inward

flow of energy ai sufficiently large values of 1x1 Another possibility is to prescribe some' inward new " as' x and to rei uire an ourward flow of energy as x —0 x,_ Such a " radiation — —

condition" was discussed at the end of Section 12.2 of l in connection with a problem involving clastic waves.

SoJurion in rhe Linear Thelon; [Ch. t3

338

(Exercise lb) and will be ignored, as stated above.] The remaining boundary condition (4b) implies that r.4a2a - ' = 1.(1 + Hal } or

ro 2 = F(a + Ba 3 )

(12)

e2 ; F(tx i ^ Bah

(13)

or

where t'=

—

a

(14)

.

IEt E M A R K. Since R is nnnnegative, ill is real and the solution (6) is oscillatory in time_ From (12). given a. there are two possible values of w. They are equal in magnitude and opposite in sign. 1t might avoid a little confusion if we write the two possibilities as ru = w, and to = — eo 1 . ri t > 0; but we choose to simplify the notation by considering the two roots to be ru and — w. where by vi we now mein the posiitve root of (12). With this change to notation. till is valid for solutions proportional to exp Innl), lout for solutions proportional to exp (— iieit) we most use

=

rrvu,

p=

iw z a - ' a .

oat?.

(15)

Using (l 11 and 04), we see that one set of solutions of the type IN is as follows: 11 4111 X. 2, r) r*"1it(x, z, t)

p`tt(x, 2,1)

—

=

Re

t^

ir4i 0a 7 a

exp (az)

exp [r'a( ( + ri}]

'

;11 1x, t) = R e exp [ia{ x +

This set represents leftward-traveling waves with dimensionless spatial wave number a and dimensionless temporal frequency w_• The dimensionless spatial and temporal periods are 2n/a and 2n/co. respectively, and the dimensionless wave speed is c. Equation (i 3) gives r as a function of a and the dimensionless parameters F and B. • Those unfamiliar with the basa facts concernsng traveling pcnodwt waves can consult Section 12.2 of I.

Sec_ 8.1] A Sohiiion of they Li eearezed Equations

339

Using (15), we see that the - negative w" set of solutions is 14" )(x, }(x, z. r)

w - in)

wt' qx, z, t) = Re

exp (az) exp [t 4x - (t)].

to 2 ,2 - l

plrl (x, r, r) CI] (x, r} =

Re exp 1icc(x - ci)].

This set represents rightward-traveling waves of wave number a and frequency r1). RETURN TO DIMENSIONA1. VARIABLES

Let us reintroduce dimensional variables. Then the Factor giving the periodic dependence cif nur scilution becomes exp

ax*

±

f * = exp [f W(xe ± Cr')],

where a

W

j and

rL

C

=

^.

From (16a) we see that the dimensional s pa tial and temporal periods are 2ni_jcc and 2KPicu. Equations (16h) and (I 6c) relate the dimensional wave speed C and the dimensional wave number W t o their dime nsionless counterparts c and ;x. Let us consider a class of waves of various dimensional wavelengths i = 1, 4.... We wish to compare the properties of these waves, so we choose a single representative length scale L and time scale P to serve for all. We have just seen that the spatial period (i.e., the wavelength) is given in general by 2nLja, where a is the dimensionless wave number. Consequently, the dimensionless wave number r ; of the ith wave will satisfy

2ati_ -- Lt .

(17)

a;

Recall that

9L

F

PilL

2

(18)

Upon substitution of (16e) and (18) into the dimensionless " speed equation" (13), we see that the ith wave has a dimensional speed C, given by

C?

^

T

-

ur

1

C1 , y1^ `' -1- TK ; . J ,

(19a)

Satdion

34e

in Me Lamar Theory (Ch. 8

Here

r- T

and K ; - a`= 2

L

(19b,

Li

where we have used (17) in obtaining the second equation of (19c). There are two limiting types of water waves in which, respectively, gravity and surface tension effects dominate. To bring this out, we write (19a) in the form

CC;°r]= + [C;``]'.

(20a)

where

[ cry

!

2rt f '

[

^"

= 2rtr ]

^

(20b, c)

INTERPRETATION OF THE SOLUTION

We see from (19a) that the speed C ; of a wave of length L4 = 27111(1 depends nra irs 1nglh. Such waves are called dispersive. As we shall see, a great deal about the wave's qualitative behavior can be deduced from dispersion relations* like (l9a), which provide the quantitative dependence of speed 2n Î; f.; then, from (20), C? [Cr]'. Thus when upon length. If gLif 2n gravitational effects dominate, waves move with a speed approximately equal to Coy, the speed of gravity waves of length L i _ if gL ;12nt 2n17L1 . then surface terisic,n, effects dominate and the waves move with a speed approximately equal to 0", the speed of capillary waves of length L i . As expected, surface tension effects dominate when waves have relatively small lengths. For then the surface possesses relatively high curvature and, as seen from the dynamic boundary condition (7.1.26), high curvature is a concomitant of large surface tension effects. Dispersion of gravity waves is said to be iwmull because longer waves move faster, just as in the case of light waves moving through air, where the longer, red waves move faster than the shorter. blue waves, Dispersion of capillary waves is called artumalous because shorter waves move faster. Figure 8.1 contains a graph of Cl when both capillary and gravitational effects are important. The graph's most interesting feature is a minimum wave speed er . For waves on water,

p = 1 gjcm a . T = 72 g/s 2 .

g = 980 cm/s 2 ,

sa [Exercise 3(b )]

e; = 23 emirs,

Lc = 1.7 cm.

(21)

In water, no waves move at a speed slower than 23 cm/s_ At speeds higher than this, two types of wave are possible: relatively long waves dominated by • A case of dispersive waves in elasiicily is discussed in Section 6.6.

Ser. s.lj A Solution of the Linearized Equations

341

SquBre or WE'VE speed

L,

^

Fra1 kE 8.1, Graph yn•hy l dimensional) wcu•r speed C, Karr u! lrrrettir Speeds Cr atul Cr lesr pure ,yrar-et y and rmrflur.r wares are also shown.

L

-

.

Note the rtisiettre (3:1 a mammon ware speed.

gravitational effects, and relatively short waves dominated by surface 1ensinn effects. To see one physical consequence of the existence ofa minim urn wage speed, consider the steady flow of water past a fixed obstacle say, mo emertt of a gentle stream past a fishing limo_ Lei the speed of the fluid far from the obstacle be a constant Y One anticipates that the presence of the obstacle will result in the formation of a wave system that is stationary when viewed from the obstacle.* Equivalently, such a wave system must move along with an obstacle that passes al speed V through stationary water. But our results tell IS that a train of stationary waves cannot, in fact, appear unless V is greater than 23 cm/s. The same cnnsidcral ions apply to the steady motion of a ship, but the speed ofa ship would of course normally be far greater than 23 cm/s. An important physical result to be gleaned from our solution is that fluid speed and pressure decay rapidly with depth. When = i — r/cc (at a depth of half a wavelength. speed and dynamic pressure are reduced by a factor of e - x 0.04. That is, motions half a wavelength below the surface are about 4 per cent as fast as they are at the surface_ At one wavelength below the surface, the velocities and dynamic pressures have about one-fifth of I per cent of their surface magnitudes. These results are in accord with the common experience that one has to swim only a surprisingly small distance underwater to avoid the effect of surface waves. EXERCISES

1. (a)

Fill in the details omitted under (8) by

examining the consequences

of satisfying 48) with w = 0. (b) Find all nontrivial solutions to (l l to 15) of the form (6) with

x = O.

• Such wave systems are discussed rather fully in Section 9 4 and more briefly al the cnd Section 4.1_

of

Solution is

34z

the L.bwar Theory [Ch. 8

2. Show that (3) is violated if Im (a) # D in (6). 3. (a) (consider solutions proportional to exp l— ja i x)exp (iwt) exp (ix i z)Show that the relationship between frequency and wavelength is still given by (12). (b) Verify (21). $4. Find periodic solutions to (7.2.32-40) without the simplifications h = co. In particular, find the dimensional wave speed_ 5. (a) Show that, to a first approximation, particle paths for gravitydriven periodic waves in a fluid of infinite depth are circles whose radius near the surface is nearly the wave amplitude A. Note that the radius decreases rapidly with depth. The particle path equation dxfdt = v(x, tj is, in general, nonlinear and difficult to salve. Approximate as follows. Let a be a fixed point near the initial particle position x 0 . No malter what the particle path is, for a sufficiently short time all points on it will be near a (Figure $.2). For this length of time we can solve the (simple) approximate equation (kids = v(a, I), x(0) = x4 . For our water waves, the resulting particle paths are such that a particle which starts near xa will always remain near xp (a result that ceases to be correct after a "long" tinge)

^ t

r

a

FIGURE 8.2.

The pad; elf o panicle Meat %fairs at same pwni x

o near

a.

The

solid portion of the path is near sr. so thor the particle path equation can he appropriately sitnplf%d.

(h) In Section 7.3 we assumed that particles move a distance A, approximately, in time P, so that the velocity scale is AJP. Is this assumption verified? 6. Show that for waves in a fluid of finite depth the particle paths are ellipses, to a first approximation. Note that the interface] distances are independent of depth.

Se,['. 8,11 A Solution vi the Linearized Equations

343

7. The a im of this exercise is to sh ow that the basic " separa t io n of variables" solution for radially symmetric linear waves on a fluid of infinite depth is

u

— iwJ 1(ar)

P' n

-.

C1

iwJ 0 (ar) Off-' J0(ar)

exp (leur) exp (az^

iw1 t (ar) + C2 iwJ0(crr) exp ( (i)2Œ - 'J0(ar) —

—

iwt ) exp (az),

C = C r Ja(cxr] exp (iwt) + Cr Mari exp (

—

it a),

where al and a are still related by (12). Here r (x 2 + y 2) 112 and u is the radial velocity (in the direction r increasing). (a) Show that the linearized equations tl, + f ryr -=0 ,

W^ _ 4 Pr

O.

(ru }Jr

± rw, = 0,

and the dynamic boundary condition is rr P = F — m],r+r- 'Cri).

(For material on polar coordinates, see Appendix 3.1 and Appendix 15.2 of l.) (h) Since r and r appear only as derivatives, assume that i and z dependence is exponential and look For a solution of the form u w= p

ù(r) w( r) exp (+ iwr) exp (az) 15(r)

C = C(r) exp (i irnt). Show that the problem reduces to the following equation for w(r}.

r(W }' + rac2 tiv = a. (c)

Complete the calculations_ Possibly of help is the material on Reesel functions in Appendix R.I.

The next three problems introduce king small-amplitude waves, which turn out to be governed by the classical wave equation. Tidal waves (i.e.. tsunamis) are an important example of such waves_ 8. Imagine a channel along the x-axis with the z axis pointing upward. Consider long (compared with the fluid depth), two-dimensional (no y-variation), small-amplitude (ignore all nonlinear terms) waves in an inviscid incompressible fluid under the action of gravity. Since the waves -

Solution in the Linear Theory [Ch. 8

344

are long, it is plausible that the vertical velocity terms are negligible. ignore surface tension and assume that the pressure at the free surface must take the constant value p a (a) Deduce

(z• _ C#)

.

pc•

r^?u*

X

(

22

)

^

from the requirement that linear momentum be conserved [see (721) and (7.2.2)]. Start by deleting terms containing w' and its derivatives in (7.2.2). Then integrate with respect to z* and obtain (23)

p* = PA + P9(C' — z*).

which stales that the pressure at a point is due entirely to the weight of the water above that point. [Our neglect of vertical acceleration has led us to (23) - It might be more convincing to begin with (23)1 (h) Suppose that the mean depth of the channel has the constant value H and the channel has constant width b (Figure 8.3). By using the fact that influx of mass into a given fixed volume must be compensated for by an increase in fluid height, derive the continuity equation

(bN+brâ*)

+a

(24)

#*) = 0,

where small nonlinear terms are again ignored. Side View

z

'

End View

-

0

z • — —fi

FIGURE

(c)

II ^6 - --i

8.3. l'lms uf awar r in a M [inr+pi.

Combining (22) and (24), derive (25)

City — Oa.* = 0, In the situation described by Exercise 8, consider an isolated wave

f(x —

/i ii);

f(x)=—Q when x< A,x>B.

=

(We drop asterisks for the remainder of these exercises.) Suppose that this wave strikes a rigid vertical wall which forms a cross section of the channel.

Sec.

g.11 A Solution GI the Liarired Equations

345

(a)

Show that the wave is reflected without change of shape_ (b) Show that during the impact of the wave on the wall, the water rises to twice the normal height of the isolated wave. (c) Briefly compare the results of (b) with the related discussion in Section 12.2 of I. 10. Show that Che derivation of Exercise 8 can readily be extended Co channels for which the depth H is variable: H H(x). 11. Suppose that the depths of a channel for x < 0 and x > [l are if, and H 2. A progressive wave =

ci

Cos cox — e1 1)

,

c 1 = gH 1 ,

travels along the portion of depth H 1 . Determine the amplitudes of the reflected and transmitted waves, as multiples of the amplitude of the incident wave. To obtain conditions to be applied at the discontinuity x D, first integrate (22) and (24) [without asterisks] with respect to x from x u Y e Co x = }. g , where 1. 1 < 0,1. 2 > O. Then let y,, 3 —, O. Show that must be continuous at x = 0 while s(x, 1) must suffer a prescribed discontinuity there. (What is the physical meaning of the latter condition?) Use (22) to prescribe the discontinuity in the free-surface slope (x . Finally, assume that the entire solution is the sum of the incident wave (displacement Ç i ) plus a reflected wave and a transmitted wave_ Show that the incident and reflected waves must have the same wavelength but that the wavelength of the transmitted wave differs From this value. Discuss your results. Do they depend in a sensible way on the parameters of the problem'? If these parameters are assigned typical values, are the results reasonable? Compare your results with those for the analogous elastic problem, as discussed in Sec. 12.2 of I. 112. Expand in powers of ; h the expression for waive speed that was obtained in Exercise 4. Thereby show that for a depth If satisfying H = 3T7pg, the speed is nearly constant for waves whose length is moderate or large compared with H r,. What is He for water? [Such nearly nondispersive water waves can he used to visualize the development and interaction of (nondispersiv e) SOU nd waves in air. See "Shallow Water Wave Generation by Unsteady Flow,"J. F_ Ffuwcs-Williams and

D. Hawkings, J. Fluid Mech. 31, 779 (1968)] I1 (a) Defend the following equations as a model for tides in large flat basins: 1711

^t (

— ^f1)2J =

0 Jx 4 l4 — ^}, ^ ^x

^t

eV al

+2Wtr = — y

ax ou)-- $y (ha].

d

0y

l^ — a

346

Sofurion in

the 1.irrear Theory (Ch . 8

Here a and y are velocity components, C the tidal elevation, and h the height of the basin (assumed TO be constant in what follows)_ What are and w? (b) Assume a time factor of exp (kit), solve for u and ❑ in terms of t:, and derive an equation of the form VC + k 2 C = O. Show that vanishing of the normal velocity component at a vertical boundary implies that

°C-- ïr^0, where the derivatives of C are, respectively, normal and tangential. N O T E: Solution of this problem is the subject of Exercises 12.2-16-20-

8.2 Initial Vaine Problems: Periodic Cases In the previous section we ignored initial conditions in the process of exhibiting some particular solutions to the linearized water wave equations. Now we begin our study of the initial value problem. Since the position and the velocity of the free surface is more readily observable than interior velocities and pressures, we shall consider conditions on the surface, namely, C( ` i(x,O)_ f(x),

al ix, O)

9(x),

( la, b)

where f and g are given functions. In this section we shall consider the important special case where f and g are periodic. The aperiodic case will be discussed in the next section. SUPERPOSING SOLUTIONS OF EXPONENTIAL TYPE

In mathematical terms, we are faced with an initial boundary value problem for a system of linear partial differential equations with constant coefficients. Our procedure is an illustration of a general approach to such problems that is of wide applicability. A fundamental fact upon which this procedure is based was illustrated in Section 8.1: Systems of linear differential equations and boundary conditions with constant coefficients always have solutions of exponential type. A second basis is the lemma that a linear combination of solutions to a linear homogeneous problem is also a solution. (See, for example, the concluding portion of Section 2.1 of I.) Therefore, one can attempt to satisfy the initial conditions by taking a suitable linear combination of solutions to the equations and boundary conditions. 'lb do this superposition successfully, it is necessary (but not always sufficient) to have available all linearly independent exponential-type solutions of the given equations aril boundary conditions (Exercise 12). in dealing with the initial value problem for small-amplitude water waves, then, our first task is to find all these solutions for Equations (IA 5). -

S„. S.l]

)r- rnm

lnirial

Vcrlue Prnhle n,s:

Periodic eases

347

(1.10) and (1.1 1 ), one exponential solution for

pin

will .0

ci1 1^

is the real part of -- L!k

Cxp (22)

1{.0

exp (2z) exp (a^)

0,12^

exp ( iacx) exp ( i t^f ) -. J 0 as discussed above (1.15). Also, in the factor exp (± ianc) the presence of the + symbol means that all possibilities are covered if a is restricted to positive values. [As noted under (1.11), the case a = U can be ignored.] Solutions proportional to both exp(iwt) and exp(—in.t) appear explicitly in (2) and (3) and implicit ly in (4) and (5). It remains only to consider solutions proportional to exp (—iax). The dependence of (o ors a is the sanie for solutions proportional to exp (ia.x) as it is for solutions proportional to exp 1—iax) (Exercise 1.3). Therefore, a complex linear combination of solutions proportional to exp (— lax) is C cos (Jr exp ( — fax) + D sin wt exp (— fax),

where C and D are arbitrary complex constants. Taking the real part, we obtain Ct' cos tot cos ax + C.

cos an sin ax

+ D"1 sin ira cos GLx + Dr'' sin we sin ax.

(9)

Expressions (7) and (9) are linear combinations of the same four functions, so if (7) is included in a list of solutions LC be superposed there is rio need to include (9). In the remainder of this section We shall consider situations in which the initial position C and "velocity" C r of the interface are periodic. We shall use as a length scale the spatial period that we require the solutions to have. Consequently, the initial functions of (1) have the periodicity property f(x + I) = f(x),

y(x + I) = g(x)-

(10)

The solutions of (4) and (5) will initially have a unit spatial period if and only if they always have such a period. For a given term to repeat itself in

• Since the cxpresSionS

in (8)

by means of complex constants.

are complex. real linear combinations of course arc composed

SEC.

8.21

Value Problems: Periodic Cases

349

grit distance, that term must have a least spatial period of l/n, n stn integer_ We can thus only accept solutions for which 2rrlct = l/rt or a = 2rt1, rr = 1, 2, 3, .... We shall denote the corresponding frequencies by w,,. From the dispersion relation (1-12), the W are given by the following formula: ,

F[2nn + Bl2rrrle],

at!

r1

= l, 2, 3, ..

I l 1)

--

In the climactic step of the separation of variables method, we attempt to satisfy the initial conditions by superposing all linearly independent solutions cal the equations and boundary conditions. Therefore, we assume that ![o,,

l ^I

=- Re

.^^

^

L1i n l{ ^ Ii l'C

h=I

^

^Ir1 5

sin ion ('

exp (2nrrzl

sin to,/

exp (2nrrz)

cos (Uh l

CxFi 12111iz)

exp 12:Ixi.r)

COS tt^ n f

-

+ Re ^

A°n

-

n=

1

cos t o „ t

exp

cos it),, t

exp (:_117iY )

s in rr1 r

exp (2tltrz)

(2rIR.7 )

exp (2>*reix).

(12)

sin frJ^r

ANOTHER WAY TO WRITS: THE REAL PART OF

COMPLEX slims

We shall soon want to multiply both sides of equations derived from (I 2) by a complex constant, hut [he operations of multiplication by a complex constant and taking the real part do not commute. Fortunately, we need not abandon compact complex notation, since difficulties caused by the lack of commulativity can be circumvented if we rewrite ( 12) with the aid of the relation Re.:=

ÿ *1,

(13)

where * denotes complex conjugate. Using I 0), we can write the first sum in (12) as sin --

) L n -1

tr7„

r43rt —fin

rt}„

t exp (2iln21

sin a.)„1 r:xp (2r1r})

„

['0 d n

exp (2rrrit)

exp (? ► rrixl

cos to„ r exp (2/177.z) [•W„ exp 12nitiix) COS

rJn r

exp ( —2nrli-r}]'

exp i

(

—

?•lrtit)]

.i1,* exp ( — 2rlrrix)]

[a n exp (2r17ir.C) + .4^* exp (

711n1 l )1

.

(14) There is a similar expression for the second sum.

Solution ira the Linear Theory

35 0

[Ch. 8

We can write each line of (14) as a single sum if we redefine the constants so that the second terms have the same form as the first terms, except that n appears instead of n. This notational symmetry is attained if we make the definitions —

[13-11

A,r

1st —

,,, A

- „

(15)

we.

i- , B„ —

—

fa,,. B-„=

. (16a,h,c,d)

(We have also inserted a factor j so that this factor will disappear in the final form of the sum) We must examine in detail how (15) and (16) lead to an elegant notation. First, let us consider the surface displacement. We see from the last line of (14) that the assumed form for the cos w,t contribution to VI becomes A„ cos w„ r exp (2rlrrix) f n=1

yA

,

cos co_„ t exp (

2ntzix).

„=

This can be compactly written as A„ cos w„ t exp (2inzix),

provided that we make the definition Ala

=û.

(16e)

We next turn our attention to the horizontal velocity component_ Using (15) and (16), we see from the first line of (l4) that the assumed form for the sin w„t terms in ur 11 is sin w„t exp (2rtrtz) exp(2rtrzlx) w =1

+ X KO „A_, in r^_t exp (2nnz) exp (-2rinlx). -

(17)

=t

Further juggling is required to achieve notational symmetry_ We write exp (2nttz) as exp (I 2 ► trr 1 z) in the first term and as exp (I — 2r,tz I x) in the second term. To take care of the opposing signs of the first and second sums, we use the signum function sgn where sgn x — 1, x > 0;

sgn x = —1, x < O.

We can now write (17) ast

stt rr = E -irr^ „ sgn N

t sgn (2nn)

Section 8..

(2rrtt)A„ exp (I2r171Izj exp arirtix) sin w„ f _

— CC

sgn (n), but uh.c of sgn (inn) permits readier generalization of our results rn

Sec.

8.21 Indio! Value Problems: Periodic Cares

35 1

Additional manipulations of the kind just performed (Exercise 1 ) allow one to transform (12) into the final assumed form,

- it „ sgn (2rnn) —

(A„ sin cot — B. cos tf„t) exp (I2rin

(A„

w„

sin w„ r

—

IZ)

B,, cos û,,r) exp (1 2r7a'tIa) exp (2rlrcix)

(A„ cos cu.: -+ $„ in r „r) exp (I2i1ar[0

bu I

A. cos L11 t f B„ sin r.iJ.n l ,

(IS) Having arrived at the solution (1 ), we no longer need concern ourselves with the constants .d, and at, that were used earlier. We must, however, adhere to the restriction

AA

A_,,,

B* = B_,,.

(19).

This follows from the definitions (1t) of A„ and BA in terms of ,nt. and At.. Without (l9}, the sums in (l8) will not represent real quantities. Example 1. As a check, show directly that w` i ' Is real!.

SFvfruoinm. (w ill

•=

- c0„(A:

Tn make the last factor look right, we j - —iI, Then

p.,,r19* _ y _ w

ra„!}exp(I2mtlZ)cxp( - 2n rr

sin Lud —

define a new dummy summailor, index

sin ())._ 4 : — B'_ cosa, t r1exp(I2inl-)expl2jnrx}. ,

j- a

l'rDm (19),

A# i = AT* = A} and

B*_ f = B,.

Using this and (151, we obtain

y —[op ; sinrrl f t - fi r COS ay) cxpf121^I-1cxp (2:12ix )

[+nsrr]* = 1 ^

--^

d 11

w

Y^.

j by

351

xiokaian in the Linear 77reary [Cif. 8

Example 2.

As a check, show formally that the continuity equation (i -Ic) is satis-

fied by (J 8).

Sofrafiun. Since our calculations are to be formal, we assume that tiffe order of drfïerentiation and summation cati he interchanged, so that wvl

s

E{^rlJIG1n sgn [2110

I2nrrIcu,](14, sin w,r

J3,cos eu,r]

..-^

cxp ll 201n lz) imp f2nnixj n sgn

(tr) ;

^

Q

!PI .

SATISFYING INITIAL CONDITIONS

We now substitute fit'', as given in (18), into the initial conditions (1). Assuming that term-by-term differentiation with respect to t is permissible, we obtain the requirements a

f {x) _

A„ exp (2rirrtx),

9(x) =

t.1„B„ exp (2nitix)- 120a, h)

The right sides of (20a) and (20b) are complex } ourier series. To determine A,,, we proud in a manner analogous to I he procedure used to determine coefficients in a real Fourier series (sec Section 41 of I). We multiply both sides of 120a) by exp (— 2knix), k an integer, and integrate over the interval [D, 11 Since

exp [2rti(rt — kJ dx

r+ rx=

^°`

1,

0

k,

(21)

we find that ^k

—

{ J (x} C7[^j +

2 ^i ^r i %]

C^x

,

Ifî = ^ 1 ^2 . . . . ,

^^ ^^G)

Si rr^ i larly, wk

Bk = f g(x) imp ( — 2itit[1x) dx,

k

= ± l . + 2. . . , .

(23)

Equations (18), (22), and (23) provide a formal solution to the initial endue problem. (i) Asa check, formulas (22) and ( 23 ) do satisfy (19). (ii) A REMARKS. considerable advantage of writing Fourier series in complex form is that the integration (2 C), unlike the corresponding integral for Fourier series involving sines and cosines, does not require manipulation of trigonometric identities. Although we shall not go into details, we also remark that the complex notation simplifies any nonlinear manipulations that one might choose to do. (iii) The coefficients A k involve only [he initial surface displacement f (x) and

Sec. 8.2] Iniliol Value Problems: Periodic Cases

353

the coefficients B* involve only the initial surface velocity g(x)_ Had we used exponential rather than trigonometric forms for the time dependence, the relation among f, g. and the various undetermined constants would have been more involved (Exercise 5). (iv) if we had used the solution (l2) instead of (20a), we would have obtained f(x) —

y .s^

Re w

=

exp (2rrrrex),

,r^ = sir +

t

or

(.. cos 2nlrx — .S.i1,1f stn 2nnx)_

} (x) =

(24)

w, l

The Fourier coefficients .s. „' and .c1 can the advantage of complex notation is lost.

he obtained in the usual

way, but

EXERCISES

L Complete the calculations, begun in the text, that are necessary to obtain (18). 2. Complete the verification, started in Example I, that the sums in (la) represent real quanti ties_ 3. Complete the Formal verification. started in Example 2, that the alleged solutions of (18) actually satisfy (1.1-5). 4. Extend t he fo rm u las of this section to the case of fi n i to H (see Exercise L4). 5. Verify remark (iii) following (23) by repeating the calculations without making the shift from (2) and (3) to (4) and (5). In particular, show that the final answer is the sank as that obtained in the text. 6. Replace (15) by w„ = w and find the counterpart of (18)_ Which convention For defining w,, n < O do you prefer? 7. Satisfy initial conditions starting from (24) rather than from (2Clal. Verily that the final answers are identical. 8. As a check, verify that formulas (22) and (23) do provide a solution that satisfies (l). +9. Find Ak and BIG when the water surface is initially motionless and initially has the shape pictured in Figure 8.4. —

It i I I.

^

i r—

!

x

^

I

] —

^

E - ,

^

I I

11 ia

I'

r^

5

FIG LIItE 8.4. Initial displacement needed /Or Exercise 9. Only one period is 5I1nK-n via Ant-lion ► i•irfl period unify.

Solution the Lin ear The ory

354

[Ch. 11

10. (a) Solve the initial valu: problem for the following special case of (l) :

Ox, ü) = cos

(2Trx

C,(x,

0) = O.

)

The resulting motion is given the name

srariding wait's. Why is the

name appropriate? (b) Find the particle paths and sketch some typical ones. I L la) Consider PE, the potential energy of gravity waves (B - 0) in one spatial perry]. Take z = 0 as The zero point of potential energy. If, as in this section, the spatial period is used as a length scale show that

PE = iP9A 1 L

f 1 {2 dx.

lb) Show that KE, the kinetic energy par spatial period, is given by (, KE _ "piA31.F (cl

-

'

j

J Iii + i4. 1 )r1k d:_ o

Using (18), find expressions for K E and PE. Show that

KE + PE = ifr9A 21.. ^ ^^A^ 1 + n= — aS

Note that energy is conserved, as expected. i12. Prove the validity of the italicized statement precuisng (2)

.

83 Aperiodic Initial Values In this section we shall consider the initia conditions Cm p) = f(x).

Lix,0l = g(x),

(lab)

as in the previous section, but we shall no longer require f and g to be periodic. This means that solution of the initial value problem involves Fourier integrals rather than the Fourier series that are relevant for the periodic case. We shall provide a formal treatment of the problem. As a first deduction from the mathematical manipulation, we shall then show that a symmetric initially motionless surface hump splits into congruent left- and rightmoving segments_ The section concludes with a perspective of the initial value problem in terms of the delta function_ ABANDONING DIMENSIONLESS VARIABLES

Two rativenitons. valid kir the remainder of our discussion of the linear

theory of water waves, will be made before we proceed. (al We shall drop the superscripts on utli tv"°, prat, and ts'r'

.+tprriodir India! Values

Sec. 83] (b)

We shall take A

355

L. = I cm and P = Is.By(7.2.21)

a* = Pi, x* Lx, i' =tLP - r v, C*=ELC,

p* =

0P-2p.

where c = AIL Thus no notational difference remains between dimensionless and dimensional distances and tunes, hut the latter have the units centimeters and seconds. Similarly, dimensional velocities and displacements have units of cm/s and cm. The dimensional dynamic pressure 15* divided by the density p has the units crn 2 /s 2 . Convention (b) can be viewed as using the centimeter is a reference length and the second as a reference time. We need no longer continue to employ carefully chosen length and time scales because our previous effort in This direction has served its purpose of helping us formulate a simplified version of our original problem. In many instances, even at this stage, careful choice of reference quantities makes inessential but nonetheless welcome simplifications in the answer to a given problem. We make convention (b) lx:cause we shall be cnnsidering -a wide class of problems and do not wish to keep changing reference quantities. SOLUTION VIA FOURIER INTEGRALS

Solution of the general initial value problem proceeds in a manner that is closely analogous to the procedure used in solving the problem when the initial surface displacement and velocity were periodic. Again we supLrpose all possible solutions of exponential type. Before, the periodicity condition allowed only those solutions whose wave number ar satisfied a = 2nn, r' an integer Without a periodicity requirement, a can be any real number. Since we have a continuum of wave numbers to "sum over,,, we shall superpose by means of an integral. Just as in Section 8.2, it is advantageous to write the solutions with their x-depdence expressed as a complex exponential. We therefore superpose the solutions of (2.18) by replacing 2nn by w„ by w(a), A n by A(a), and B„ by B(a). Integrating over a, we are led to the assumptions —

r^ sgn a[A(a) sin on — H(a) i^

cc3s

r u] exp (1 a 12 ± iax) da,

^i

(2a)

.a.i[A(a} sin cm' — R(a) cos rar] exp ( Ia I z+ iccx} r^a,

142(x, 2, 1) =

(2b)

^

G7 2 ( a l'tA(a:} cos cur + B(a) sin (in] exp (1 a 1 a+ iacx)

p[x, x, r}__ a

(2c)

ax, 1}

f

[A (a) cos lug -+

B(a) sin rllt] exp (fax) d a.

(2d)

Autun in the Linear Theory [Ch. 8

356

ln these equations. w is a function of a defined for nonnegative values o r as the nonnegative root of the dispersion relation (1.12). With L = P = 1, we have from (7.2.31) that B TIp9. F = g. Thus (1.12) becomes

0.4ct)

T ° T f p, a

= 9a + î'a 3,

0.

(2e)

For negative values of a, w is defined by (4 — cf) = ca(a),

(3a)

-

in analogy with the relation between m and co_ given in (2.15); A(a) and Bloc) are unknown functions that satisfy

A#(a) = A(

—

a),

Bs(a) = B(

—

a),

(3b, c)

in analogy with (2.19). Example 1. Asa partial check on our work so far, show that C is real. Safruiarr.

From (2d),

*(x, [)

[A' la) ces c a)t +

=

B•(a) in w(a)r] exp (— lax)

da.

B y (3b, c).

tA(—a) cos w(a)t + G1(

Z*(x,t) =

With the chan g e of variahlc a= ^ * (x, t) =

^

^

—

—

a)sincv(a)1]exp(

—

iax)da-

{^ this becomes ,

[ AV) cos { o( — P t + 1:3 O) sin (01-13)i] exp (ifix)1 - d^}.

or, using (3a), 4•(x, l) ;

sin L^(^^r] exp {r^ix) d{s = ^(x, t). 1 l^J^{^[Cos+-u^^^t + BO ^

h)]. E xnmpk 2. As a check, show formally that Hi +- P_ — 0 [Equation 11. ft)]. Sohuiwr. Differentiating under integral signs, from (2h) and (2e). w, + Pi — f {— tU s + cil' NA cos wt + B sin cm)

ex p{ E a l z+ iax) da. = O.

REMARK. The reader shouki be confident that the quantities

given in (2) are

formal solutions of the governing equations (1.I). To obtain (2) we superposed a continuum of solutions (and did some harmless but valuable notational juggling). The basic idea of the superposition is as follows. Symbolically, if a vector s(x, I. ml is a solution to a set of linear homogeneous equations for any real value of a. then the linear combination

r{x, t • a,jd.; i- 1

(da; - ar — oer - r 1

ffperirhfir India) Values

S. 831

357

with q

tr = f^ •

° h.

is also a solution. Assuming that +t exists. we cannot be sure that the limn ^

>w

lim rr

y s(x, t aoA2, _^ s{x

r, a) d a

I

is a solution only because we cannot be sure !hat ditfcren tiaiion under the integral sign permissible. nut (2) is a formal solution because in format c:alcuta[ions we assume that is such limit inicrchanges are possible. Moreover,

is a formal solution if it exists Formai eaiculatuons are meaningful because they are correct if appropriate restrichomi arc made on the entities involved. i is often appropriate ta defer the task of determining a sufficient set of restriciiois.

Our task now is to obtain an expression for the unknown functions A(a) and B(ar) from the integral equations resulting from substitution of (2) into the 'initial conditions 1 I) A(^j exp (irfxj dix

f (x) =

g(x) = f 23(^)r^^(^) exp (i^cx) da. (4a, ^)

,

cc

To find Alla), we mimic the derivation of (2.22) as closely as possible* and write

r

^

lirn Jexp( -

(x)dx = lim f exp ( - !p_rl

i.

r Ala} exp (iaxJ da dx.

r

Interchanging the order of integration. we Fuld that

exp(

Jim

i lrx)f(x)dx =

r-.cJJ r ^ r

A(a) I ^ c;xp [^(a - tr}.t] d. Jr1 a x j r

lirn

,•^JJJJ

-

flic integral in the square brackets is easily evaluated. When a # , I J

^

r

exp

[i(^

[I(bC - Rt)i] - cxp [- i(ü il jf^ — ox] d x = QX^ — — -- --

_

i(a -

2 2—

- sin (a -

so lirn r-^ '

f

I

a

exp( —i}tx)f(x)dx = r

lim ^ 2A(a) 1,^;

sin (a a—

p)i

da.

l- hc unknown functions A and B can also be dercrmtned from (5.3.1) and (5.3 8] of I.

(5}

Sokiioti iii the Linear Theory [Ch. 8

358

The right side of(5) has no readily apparent formal limit as it stands but with the change of variable (a — p)i = v it becomes* lim 2 r oc

A(^ + p

^

-

sin ^ ^

riv _=

where

C

u -1 sin v dr.

-

(6)

-m

We can determine the numerical constant C to any desired degree of accuracy by numerical methods, but in fact it can be shown by various analytic methods that C — n (Exercises 2 and 3). Putting all our results together, we see that (5) becomes l

2n

^

exp (

iux)f (x) dx

A(p)•

(7)

^ .3D

Similarly,

g(x) dx = ^)^(N^ 2Irr ^ eXP ( — iJ^)

(8)

In an important special case, the free surface is initially motionless and its initial position is an even function. That is, f(x)

f(

x),

(9a, b)

9(x) = O.

—

With (9a), equation (7) can be written A(^ll

2

cos px f (x) dx n

fo cas px f (x} dx,

(10)

since

f^

s ill f!X f (x) [SX = 0, ^

•

Equation (10) shows that A is an even function. Since B

r) f

A(a) COS wt cos ax da = 2 f4(a)ccswi!

0, (2d) becomes

cosaxdcc. (1 2)

^

where we have used the relation w(--a) = w(a).

•

We have assumed Thal

r-ao

A{ l

+ + tJ= A(2l. 11

This is certainly true for fixed finite v if A is continuous at FA., but w ranges from — oc to + ec. Min integrals con verge, however, contributions for Targe M must be small. Th is gives confidence that whit we have donc is valid- For proof. see Section 5,3 of 1_

Sec. 8.3]

.periodic Initroi Vuhies

359

Exampk 3. Suppose that the surface deflection is initially given by a rectangular bump.

(1x. 0)= 0,

1x1 >R;i

CÇx.0)

— Q. xi

R:

C,{x.g1= 0

(13)

(sec Figure 8.5). Find the surface deflection for later times. L

FIGURE 5 5. Surface deflection ui time - r= D. in the form taf a rerIOngulur hump_ ti,rrnglrr (15) giees the del: ecir`vn at later +rmes-

alutian. li is not dear that the given initial displacement will result in a motion whose convective acceleration is negligible as assumed by our linearized ihcory Only if such a motion is predicted will the results of the theory be consistent with the assumptions made ilt deriving it. Keeping this in mind, and noting that (9) holds, we use (tUM to find

I

ALA = -

t`

ccas

^ ü

Q sin pH

dx -- -

ln`

(14)

horn (12). 2Q

;(x, r) = -

-

sin all? I- — cos to(a)1 cos cue da.

( 15 )

G

However complicated a function at is of a, the integral could be evaluated numerically. The job would be difficult, as the interval of integration is infinite and the integrand is oscillatory. As we shall see in the next chapter, however, it will be possible ED simplify (15) so that no complicated calculations need be made. Note from (15) that as could have been anticipated, the surface deflection and its derivatives will indeed be small lithe initial height Q is small. A QUALITATIVE FEATURE OF

THE SOLUTION

We cannot emphasize loo strongly that although in a sense we have solved a wide class of problems, our work is almost worthless unless we can proceed further. Our purpose is to acquire a better understanding of water waves, but this does not yet emerge from the "solution" provided by (2), (7), and (S). Further effort is necessary, A discussion of our results forms the subject matter of the next chapter. Here we prescrit the derivation of but one general conclusion, when the free

Sviutia ►r in the Linear Theory [Ch. 8

360

surface is initially symmetric and motionless. In this case, we have by a slight modification of (12),

‘(x, r) = 2 Re

f

A(a) cds

CI1t exl7 (lx)

dm

Writing cos oil in terms of exponentials, we obtain

+sx(x, t) + tâ L(x, t},

(16)

where C R (x, t) — Re E J A(a) exp Max — wt)] (17)

Ci( x, r) = Re

fa

A(a)exp [i(ax +

r^t)] da.

We have CA— lc, t) = Re ^A(a)exp [i(— ax — oil)] da -- a

= Re

.a

A *(a) exp [i(ax + tot)] da,

because Re z = Re z. But A is real [by (10)], so that

Thus an originally motionless surface deflection splits into two parts, a portion ICR traveling to the right, and a symmetrically placed congruent portion CL traveling to the left. Since there is no distinguished direction in .a homogeneous fluid layer of infinite extent, this result could have been anticipated. Consider nondiypersiue wrwes for which the phase speed wfa is a constant c (independent of a). For such waves we deduce from (17) that (Exercise 8) CR(x, =

(x — ct, )

,

1( , t) — R(x + et, C),

(18)

where r = wfa. The original motionless deflection splits into a pair of congruent disturbances, of half the initial amplitude, that travel to right and left unaltered in shape with a constant speed c. Fly contrast, since their constituent sinusoidal waves of different lengths travel at different speeds, dispersive waves generally change their shape as they move. SUPERPOSITION AND THE DELTA FUNCTION The following considerations give a better understanding of how complicated linear phenomena can be regarded as the superposition of the effects of simple "causes." Example 3 can be regarded as giving the effect (surface displacement) caused by an initially motionless, rectangular surface bump,

Srr_ 13.31 Aperiadir lrririuf irufuea

;6e

centered at the origin, of width 2R and height Q_ In the solution given in (15), x is the distance from the original center of the bump_ If the bump were centered at x = 4„ then (15) would give the correct later "effect, - provided fi Thus tut initial motionless hump of width that x were replaced by x centered at 4; , would give a surface displacement of height f and —

❑

,

2

.I

Jo cc

sir[

Ilk

2

cos co(a)1 cos cr(x —

da.

( 19 )

Because the governing equations are linear, an initial motionless "step-bump" like that of Figure K.b(a) would cause a surface displacement which was the sum of the surface displacements due to the bumps of Figure 8.6(b) to (d). We should be able, therefore, to find the effect of a general motionless initial displacement ax, û) = f (x) by approximating f(x) by a step function, using (19). and taking the appropriate limit_ Multiplication of the numerator and

°tl

I

r I r I I

I 1 I I I

i

1

I

Ji

b

h

1 I I I I

(2}

i

❑tl

Rh]

d( 3

[r)

t) 6 t1

R h!

(d)

E} FIGURE 8.6. Th e three re*rfuagles of (:t) are separately depiered i,r tb), Cc). and id), in the !incur theory of water wa v es, the surftx'f deflection din' IC an initial configuration (a) inn b e ob tai ne d by summing the rlefieerions du e 10 irriti[e! cvrrfigurafiuns of the form (b), (c l, and {d).

362

Solution in the Linear Theory [Ch_ 8

the denominator by At la facilitates determination of the limit. We obtain C(x, r) —

f

2 iirn x

ef, -0 r a

f(

d2) cos (loos cos a(x ti) Air der 2 at6 2

aJr^ ( r

or

r) —

f() cos co(a)t cos ac(x — rÿ) da f:M

^

(20)

^ a

where we have used the well-known result Sin a

Inn

a

a-•0

=

1.

As it should he, (20) is the same as one form of the original solution (Exercise 13), A similar way to obtain the same final answer begins by considering (15) in the limit Il Oy 212R . The result is ^ sin aR cos cu(a)f cos ax ria o aR

2QH

lirn

C(x, [) =

Ti

R^o

WA - I

Or C(x, t) — -

l^

cos tot C(7s ax da _

(2l)

a

^

Here g(x, t) can be regarded as the displacement at time t due to a unit volume of water (per unit distance in the y direction) initially arranged in a tower of infinite height above the point x = O. That is, the original displacement is —

C(x, 0) = b[xj, where the delta function 6(x) satisfies (formally) ,51x ) _ 0, x

Cl;

I)dx — 1 ^.

(22)

Suppose that the motionless tower were initially located above x = e; and con tained f() units of wafer, i.e.,

«(x, 0)—Z(x — )I (0, Then the resultant displacement would be 1

-7r

CE

if ( a j cos cog CPIs ac(x — Oda, 0

Se e. 8J1

Aperrrxlfr.' lniriui Values

163

, as before_ Finally, it is plausible to where we have replaced x by x regard a motionless initial displacement as a composite of delta functions. ;led to obtain the resultant equation for by Integration. This yields -

1 jr(x r) -

. f(f,.

cos tat eels o(x — ) do! kit, a

123)

which indeed agrees with (20). R r M A R K. When NI) = 6(0, then (23) must reduce to (21). This is implied by the fermi] property

glo3(0 4 = ,((}). a property that is slightly more palatable than (22), and, in fact, is the basis of a well-founded definition of the delta function. Note that the integral in (211 does not exist in the conventional SLnSC_ Our recent manipulations have been circular, in that we started with a general solution, we specialized to the case of an initial rectangular bump, then we recovered our original result by superposing infinitely tall and thin initial bumps. Such formal experience with the delta function is a worthwhile preliminary to later study (not possible within the confines of this volume) of the delta function in its proper setting within the theory of "distributions" or "generalized functions. -

E7► 6RCI

S

1. Show that the improper integral in (6) converges. 2. (a) if you are familiar with contour integration, use that method to show that the constant (' defined in (6) has the value i. (b) Use a computing machine to evaluate C In three significant figures. 3. Certain formal manipulations with the Laplace transform yield C = 7E, where C is defined in (6)_ The Laplace transform F{s} of a function [(t). sometimes denoted by L( {tr)]. is defined by L[fltl] = 1 (s) = -

J,a

C]{p (

-

titlf

eh

if the integral converges. (a) Show that formal interchange of the order of integration gives F(r) de =

erg (- stji - ' f (OW.

+(b} Formally set ti = U in the above equation and use the result to obtain C. 4. Verify that u, w, and p, as defined in (2), are real.

Soh aia ►r in the Linear Theory [Ch. 8

3454

S. Show formally that the governing equations (1.1-4) are satisfied by the solution defined in (2). 6. Show that if f(x) is odd [f ( x) _ f(xi], then –

A( 43 .4) sin µx dµ f(x)1

–

provided that A(p) _ f f(x) sin ,ux dx_ rf p

7. Show that the formulas for A and B in (7) and (8) satisfy the conditions on A and B given in (3h, c). S. Verify (18). 9. Specialize the formulas of this section to the case when the initial conditions are C(x, 0) – (bfrr b 2 + x 2 ). Cr(x, 0} = O, where b is a positive constant. Consider infinitely deep water and assume that surface tension effects dominate gravity effects. Do either (a) or (h). (a) Work out A(a) from (7) by contour integration. (h) If you are unfamiliar with contour integration. verify that (4a) is satisfied if A(a) (2n) - ` exp (–bled). 10. Suppose that water waves were nondispersive (like waves in a string). Sketch the appearance of the free surface which is initially deformed as in Figure 8.5. Is this in accord with your physical expectations concerning water waves? (If so, revise your expectations.) 11. Extend the results of this section to the case of finite II. 12. Generalize the remarks about left and right moving waxes that were made after (17), when the restriction (9) is not assumed 13. Using(2d)and (7), show that when Cr(x, 0) = 0, C(x, t) can be written .

^

C(x, t) =

nI

^

^

^a

cos a(x ^ ^)f (+;) cos c.u(a)r d« 4g.

14. Find C(x, t) when

C(x, 0 ) _ exp ^

I

I— )

,

(x, 0 ) = 0-

R

15. Find C(x, t) when

C(x, D) = Q sin (.0x, Ix I

R;

((x,1)) =

-

0

, 1x1 > R,

1,(x, O) = 0.

Discuss the limiting case R ac. Obtain (19) by using the result of Exercise 13 when 16.

((x, O) = fad' , ix –

C IN , : p)=0.

C(x, (]) = 0 otherwise,

17. Make precise and verify the statement that the velocity, pressure, and surface displacement resulting from the initial conditions (1) can be

Appendix 8.11

Besse) Flint-lions

365

determined by adding the solutions corresponding to initia! conditions Clx, Oj =

f

Lk, D} = 0;

C(x, O) O,

(x O) = g(x).

18. Generalize Exercise 2_11 to show that. assuming convergence of the integrals, the sum of kinetic energy (KE) and potential energy (FE) for the waves given by (2) is KE + PE = >rpgA'

JT

I 4 1J1 2 +

1 B(P)I 2 ] d#.

(24)

Do this by writing down expressions for dx, etc.

r

and then using the following version of 14a) and 171'

exp [rxfx — p)lAla1 tfx dot — 2rAW).

• f.

(25)

^

19. +(a) Show that (2d) can he written = Re y, where

yfx, r) =

F(a) exp

—

F(a) _ 4(2) -

^x

so that, from (24), KE + PE

ra j,y 1'

f.,

F(012 da

126)

(b) Use (25) to show that KE + PE = .pyA'

127(

^

21). With the aid of several graphs. characterize the behavior of the transform A(p) Found in Example 3 as a function of Q and R. Given A(p), what can be ascertained concerning the parameters Q and R that characterize the

initial disturbance?

Appendix 8.1 Bessel Functions The following equation often arises, particularly when cylindrical wordinates are employed:

r 2 F"(r) 1- rF'(r) + (a ) r 2

—

p2 )F(r) -

Here p and a are parameters; the degenerate case

from the discussion.

f_

(1)

= 0 is often excluded

Solution in the Linear Theory [Ch. 8

366

Solutions of (I) are called Bessel functions. Asa result of over a century of analytical and numerical work, there exists a most extensive body of knowledge about these functions. [The classical reference, which only covers work up to 1922, is G. N, Watson, Treatise on the 'Theory of Besse! Functions (New York: Cambridge University Press, 1966).] Work on the Bessel equation is representative of the subject special functions, a large part of which can be regarded as a study of functions defined as solutions to certain important ordinary differential equations. Since so many of their properties are known, it is almost as meaningful to say that the solution of sonic problem is a Bessel function or a Legendre function, For example, as it is to say that it is a trigonometric function. (Indeed, sines and cosines can profitably be defined as solutions to the differential equation d2p/Jix 2 + y = U.) It cannot be denied, however, that mastering the properties of typical special functions is a Far more extensive task than the comparable task for trigonometric functions. Here we wish only to list some of the most essential properties of Besse! functions. A kw more results are given in Exercise 3.1.4 of I. The Bessel equation (I) has a regular singular point at the origin. The method of Frobenius provides the following Maclaurin expansion of one solution to the equation, the Bessel function of the first kind, of order p:

^r^ ^ `^ ( - 1 j"'{r/2)^ k2 )„,z_-0 m { r(p + m + 1 )'

J ^r)

r

U.

(2 )

lip is not zero or an integer, J_(x) is a second linearly independent solution. Otherwise, we have the "exceptional case" of the Frobenius method [see, e.g., Boyce and DiPrima (1969), or E. Kreyszig, Advanced Engineering Mathematics (New York: Wiley, 1967)]. Then the appropriate solution is Y (x], the Bessel function of the second kind:

Yr(r)

2 -.l

r 2 +

^

+

l

p^ ( (;)

^

y)

1r

r^

n

m =a

a

'(

p

^

-

I^" 1 (h. + hm+ p)(+'12e^ m !(rn+ 1^) !

tr^ —

—

r > O,p = O. t ,2, ----

in?

(3)

Here h c =Ü

,

h= ^ i !-

I

1

y = lir n '

— lii m) = 0.577 ....

m— Cr.

For small r, the first terms of the above series provide good approximations to alp and Y. To obtain approximations for larger,. we must recognize that r = co is an irregular singular point of (1)- Appropriate methods (Coddington and Levinson, 1955) yield series expansions for 1 F and Y. which are asympto-

Appendix 8_1] Besse/ f wrrricros

367

tic ally valid as r i cc. The first terms of these expansions provide the following approximations -

lAr) ^' x 1;(r)--

f cc's

Jii sm

r — ^ nr — ^ px ^ ,

( 4 a)

(r —4 ,- 2 RR).

(lb)

These expressions are exact when p = - . Although not at all evident from the Frobenius expansions (2) and (3). qualitative behavior of Besse! functions (oscillation with decreasing magnitude as r increases) is evident fmm (4), a single term of the expansions about the irregular singular point at r = x _ This is perhaps to be expected. for more should emerge from an analysis if we confront the function head on at its worst singularity. From (2) and (3) one can derive the following derivative formulas, where Fp stands for either Jp n r Yr : dr ^r New)]^ err"F_

dr

Aar),

i] _ —ar FF p + „W.

[r -PFp(

(5a)

(5b)

Using (5), we can deduce

d; [1.. P(ar)j

+ 1~ p(ar)

! ^IF p - r(ar) — pr —

af' F+ r (ar) f pr - rF p{ar) {

=-- ^af^_

Jar)

Ep + I(ar)]

(6a) 16h1 56c1

EXERCISE

1. (a) Show that formal differentiation of(2) and (3) yields (5a) and (5h). (b) Derive (6a) and (6h)_

C HAPTER 9 Group Speed and G roup V elocity

I

N THE previous chapter we exhibited explicit formulas for the solution of rather general water wave problems. It might seem that this would be the occasion For considerable satisfaction. but this is not the case. We seek understanding of natural phenomena; no such understanding emerges from passive contemplation of complicated formulas_ In the present context, and a great many others, insight does appear when solution formulas are simplified by means of asymptotic approximation methods. Roughly speaking, this is because "understanding" a complicated matter requires discernment of its essential aspects and relegation of details into the background_ AsyrupWlic approximations do just this. for they generally ignore a function's regular points and even its regular singularities and concentrate on revealing behavior in the neighborhood of an essential singularity.* For water waves, the asymptotic approximation is obtained by the method of stationary phase," which can be regarded as a variant of the more general method of steepest descent_ The former method allows one to simplify the Fourier integrals so that the behavior of the solution can he well characterized by introducing the notion of wimp .spred.t Group speed is a central unifying concept in wave propagation. It arose from observations on water waves by keen observers such as Scott Russell {cited on p. 31SO of Lamb (1932)]. The mathematical basis for the phenomenon was established by the great nineteenth-century British applied mathematicians, Stokes, Kelvin, and Rayleigh§ ; contributions of great value have also been made by a number of contemporary applied mathematicians. In this chapter we apply t he method of stationary phase to water waves and then compare some of our predictions with experiment. We also briefly. of the theory of water waves, particularly surveyompactil ns in connection with "tidal waves." We next turn to a kinematic approach to group selncity. which more clearly reveal; the physical basis of the • This poor is developed in L. A. Segel. "The importance olAsyimptour Analysis in Applied Mathematics." Amer. Malt. Morirhli 73.7 -14 (1966). t Group speed hears the same relation to group veloemy as ordinary speed does to oidinary

vciociiy. Many authors do not deem it wurthwhiit is make this distinction cxpthcrtiy and always refer to "group velocity_ .. The Irish mathematical physicist (or applied Enathrrnatician) Hamilton artiliriparcd some of lhcworl; on group speed during hm mrd-nirctcenth-t. mur' Ntudy of optics_ See Chapter 9 of Lamb (4932) for htsiorical notes_

368

Sec_ 9.1] Group Speed ria the Method of Siorionnr}• Phose

3t9

results and has the further advantage of being capable of extension to wave systems for which slight inhomogeneities or nonlinearilies must he taken Into account. Using this approach. we predict the striking wave pat tern associated with a ship—or a water beetle.

9.1

Group Speed via the Method of Stationary Phase

Pursuing the gnat of deriving physical understanding of water waves from thesolution formufas(8.3.21, (8.3.7). and (8.181. we obtain an approximation to these formulas for large values of x and r that permits a clear physical description of our rultx. NEEID FOR AN ASYMPTOTIC

APPROXIMATION

We assert that it should be informative to approximate our formulas in the limit r CC. Simplification would also he expected from approximations valid near r = 0, but the results should he of less interest than approximations valid for large t. One reason is that the behavior at small time is expected to be dominated by the particular initial conditions imposed, while for large times it is likely that effects arising from the intrinsic nature of water waves

will be prominent. Those versed in the theory of complex functions can apprcciale another reason. It appears from (8.3.2) that for fixed x the surface deflection C(x, r) [as well as u, w, and p] is analytic at t - 0, or any other finite value of t, but has an essential singularity as (We arc temporarily regarding r as a complex variable_ It is a central result of function theory that this is necessary to understand the behavior of C, even though we are ultimately interested only in real values of the time r.) One expects that a better understanding of the nature of iÿ will be gained by examining near its - worst " point, al the essential singularity r = x , rather than near points of good behavior Such as r = t) Turning to approximations based on the spatial variable, we assert that simplifications tinder die assumption of large x should be particularly informative. Cierierally, the initial disturbance is to some extent concentrated near x =- O. Thus approximations for small x, like approximations for small t, should be strongly influenced by the initial disturbance. Approximations valid for large x . , however, like approximations valid for large 1, should reveal more about the intrinsic nature of water waves. In addition, for fixed t the surface deflection appears to be analytic at x = 0 but appears to have an essential singularity at x = x. As above, therefore. complex function theory prompts us to examine the solution for large x. STATtONARY-t'FtASF. APPROXIMATION

Once the idea of looking at approximate behavior for large x and t has occurred to us, we are faced with the technical problem of finding suitable approximation formulas. Kelvin faced just this problem in his work on

Group Sp ee d and Grua!) Velocity (Cu.I 4

water waves; he solved it by initiating the method of stationary phase. The basic problem solved by this method is the approximation for Targe real A of the integral

1(a, b, 1.) =

r_ ^ U_

f(a ) ex p[if g(a)] (la,

(

1

)

^^

Suppose T hat g'(a) =0 fora ac < h if and only if a — cc ; , and that Oaf) 0, I = 1,..- MIN 1), a < a; < b. Under suitable conditions the following basic stationary phrase theorem can be demonstrated: ,

f(u, h,ï.}

^•

rr 2 { (a i)

1

i= r

t

Cxp [i Â9(Œ) +- ^ rn sgrl g'.(7^) ^-

l

(2) No

E.

The results arc still valid, with obvious slight changes, if a =

—

or h = ay. Since the error decreases as A — cc, the approximation (2) is

asymp to t ic_ A generalized stationary phase result, slightly vague but very useful, is this. Ifexp [(Mall oscillates rapidly compared withf (a), we have the approximation -

^

-

f (a) exp [ih {a)]

da

_ ,^2n >1: I it"(a !) I ; -

"2s f^a}

exp [i!I(a lj

sgn h"(a))], (3)

^

where

hlac)=C] if and only if* =ay. h'

. j).)-P6

D,

}= 1,---. N, N> I-

We discuss the derivation of (2) and (3) in Appendix 9.1. APPLICATION üF TIM APPROXIMATION

Consider the right-moving wave of (8.3.17): .,

R OC,

r) = R e

f A (a) 1.!xp [+rtX — iciga]r } da.

(4)

We can put (4) into the form (1) by moving from the origin at a fixed speed V so that x can be replaced by Vt. This gives

A(a) exp {r![a V -- co(a)1} da,

cÿx = Re

V =+,

(5)

a

By regarding Vas fixed and lettings —# co, we can now use the basic st at t ionaryphase theorem (2) to find an approximation to (S) that is valid for large t. Hut this theorem guarantees the validity of the approximation only for the single

see. 4 f 1 Group Speed ria the Merhod of Stationary

Phase

371

-

value of x- satisfying x = Vr. We desire results that hold for a range of values, so w e turn to the generalized version of the theorem, The exponential factor in the iritegrand of (4) should vary rapidly compared with A(a), as required, rovided that either x or r has sufficiently large values_ Comparing the integrals in (4) and (3), we set h(a) = ax — cel(a)i.

Points of stationary phase satisfy Ji'(a) .= 0 or —

aa'(tr)t — [}-

(6)

We anticipate that (6) will have a unique solution to the examples with ,t,iic.h we shall he dealing. We denote this solution by a(x, rj, so that x

—

Olt = 0-

(7)

This notation explicitly indicates the dependence of the stationary - phase wave number on x and r (which are parameters as far as the integration with respect to a is concerned). Equation uatian (7) can he regarded as an implicit definition of DIX, t}, the wave number giving the principal contribution to the surface deflection at (x, r)_ (We shall discuss this last poin t in more detail shortly.) In (3) we must substitute the value of h(s) at the stationary - phase point. This is simply h[a(x, 1)] i a{x, r}x — w[rx(x, ills

(g)

Other portions of (3) are handled similarly. In taking the real part of the integral as required by (4), we must bear in mind that A is real [by (8.3.10)]. With all this, application of (3) to (4) gives

isdx, t} ^

f fi, (x, *

A(x, r) cos [a(x, i)x —

(Aix, i )r — ()sgn u^ "(x,

r)], (9a}

where' fl(x, r)

—

A[a(x, f}] f^(x, r) .

°to[ot(x, t}], w"(x, t) =

d2 co(a) j 2 ^a

alx.+l

(4h, c, d) ncprnding on whether one nr two indepcndcnt variables arc displayed. the 5amc letters for different functions on opposite sidrs of the rçuatit ns in (9b- c. d). For example, alx.il is the valut al and t for thr function '1(, 1. whilr 44Œ) is the valut al a of the different (unahion .1i k

arc

used

Group Speed arid Group Velocity [Ch. 9

3 72

Example 1. Apply the method of stationary phase to simplify the expression for the surface deflection, due to an initial rectangular bump. that was Found in Example 8.3.3. For simplicity. ignore the effects of Finite depth and of surface tension. Sabrrïon. Using (8.3.2e), when depth and surface tension effects are ignored we have w=(

„al"].

w — ^^il 2 a -1l2 .

,

^

=

rl3

`

- 3ra

110a, ", C}

From (8.3.14), the function A(a) appropriate to this problem is Q site aR an

A(a)

( l Od)

The contribution of the right-moving wave is given by (9) where, from (1). a(x, t) is determined by

This equation has a solution if and only ifx O. (Similarly. the left-moving wave will make a nonncgligible contribution to the asymptotic expression we sack if and only if x < O. This is to be expected.) For x > 0, then, using gi 4x2 ^

u{x, r)

.

we find from (9) that for sufficiently large values of x and r the nght-moving wave is given by Cp

1 os 4x- c 4x 1

41:) ix ng

sin

(l2)

4

In Section 9.2 we shall show that (12) compares favorably with experiment.

Recall that the essence of the stationary-phase approximation lies in (6), which states that at point x and time t the surface deflection is mainly due to those wave numbers Œ(x, t), which are such that (13) For values of x and r satisfying x = VI, V a constant, (13) is independent of x and t and yields (assuming a unique solution) a = av ,

where cola,) = V,

a r a constant.

(14)

In more ponderous but explicit notation, we can write 4x, 1)1r Ye -

w'(x,

( Vt,

(15a)

r) = û y .

of x(x, toix=v, = '(ay)

V•

(15b)

5er. 9.1]

Group Speed um the M e thod of Siationary Ph ase

373

If the "wave number a(x, r) that explicitly appears in (9) is evaluated when Vt, then, by (15a), the constant U y is obtained. It thus seems that ifyou start from the origin and move with constant speed I", then at your location x = Vi there always appears the same wave number a,.. t he one given by (15a). This is not an entirely satisfactory statement, however, for the wave number (or spatial period) ar u polar has no obvious geometrical meaning- It turns ou t, nonetheless, that our basic idea about a v is correct. To clarify the situation, let us examine the appearance of the waves at time t near the point x which satisfies x = Vt. This requires a Taylor expansion doll constituents of (9). We start by considering the amplitude AIx, rl. Near x = Vr, Taylor's formula gives x

A(x, ii = 111V I, r) i- A,(Vr, t)(x

Vr) - 4 ;AYL(Vt. t)(x — V

-

+-

(16)

From the definition (9b), A(x, r) we find, by the chain rule, that Ax =

We can determine ax by implicit differentiation of (13): — ua"oc = 0,

so a x(x, t) = [rw"(x, r)] ' .

(l7a)

Using (15a), we see that the first two terms of (1 b) give A(x, r)

A(a ,) +- 'gay) (x — V;) + - - • . rm"far)

(17b)

Equation (17b) indicates that if x is restricted to be within a fixed distance of Vt, then we can replace the function A(x, t) by the constant A(a i.) if we are willing to make an error proportional to t We next expand

`.

h(x, :) - a(x, r)x — o{x, r)r about x = Vr. Since hx(x, t) =

4x

—

W"t)oc, + ü.

we find, using ( 15h), that

h, Vt, I)

= ay .

Differentiation of (19( gives. with the aid of (17a).

h, Vt, t) = [ruy"'(izr.)] -

374

Group Speed and Group Velocity [Ch. 9

Therefore, it appears that by allowing another 00 -i ) error we can replace h(x, t) by the first two terms in its Taylor series, i.e. , by h(VI,r)+h,(Vt

,

r)(x- Vt)=a t.Vr —(4a jt+a v(x— Vr) av x -

Similarly, offx, t) can be replaced by w"(a v ). We finally have for x near Vr, C.

^

^ (arj cos [ax — w[a Y)t — 4 n sgn m +'(°^r) - (20)

^

Example 2. For large t, what does the deflection C i, of Example 1 look like near x = Vi, V a given constant? Solution. From (I l }and (10), =

2,

e4aYl = 40wl ' 12 V 2

, S113

to°Iciw1 = -i4 - 'I/ 3 ,

ayR

(21)

sn 1 4

r sin I ^ gR V -') cos ( - gl^ - ^x - ^ g ^t +

CR . 4Q1 9

1

4

(22)

INTERPRETATfON: CROUP SPEED

Equation (20) follows immediately From (9) if the formal substitution V: is made in a, cry, and of. This substitution would be unconvincing without an argument such as the one just given, however. In the term ct(x, r)x, for example, why should one be able to replace x by Vt in the factor a(x, t) but not in the factor x? Such a replacement is legitimate because a(x, t) is a slowly varying function of x —a feature that was not evident before we made the calculations that led to (20). [The slow variation of a with x for large times is explicitly shown in (l 7a).] Equation (9) is an approximation valid for all sufficiently large values of x and s. Equation (20) is a fiirTher approximurion of (9) that is valid for x near Vt when f is large enough to justify our neglect of O(t - ' j terms. In order to cover" all large values ofx with (20), one must take all permissible values of the Constant I' Personification of the formulas may help. Equation (20) provides the surface deflection seen at large time by a man moving with uniform speed V, starting at the origin at time zero, in a magic boat that does not disturb the water surface as it moves. To get a complete picture of the surface deflection at large time, one must initially send out many men in many boats moving at various speeds. The appearance of (9) seems difficult to characterize generally. Only in (20). a special case of (9), do we see a function having manifest wavelike character.

See_ Q.!]

Group 5peerl via tleP MI hod ❑f 5ialronary Phase

375

This is illustrated by the results of Examples l and 2. The asymptotic approximation ( l 2) does not have the form of a slowly varying wave. hut later, when (22) can be used. this form emerges. Since(20) describes the surface deflection near x — Vt, for sufficiently large time an observer who is moving at speed V (starting at the origin at t = 0) will continually see waves of wave number « Y . This can only happen if waves of wave number a. move with this speed. The term group speed is applied to V. From (14), the group speed corresponding to the wave number at, is given by the formula V = '(cc v)- We conclude that whatever its initial

shape, al large times an initial disturbance sorts itself out into a wave Train consisting of oscillations of slowly varying wavelength_ A section of the wave train where the oscillation has a wavelength approximately equal to 1 y - 2n fa y, mores at a speed approximately equal to the group speed V = ri (a v ). The approximation improves as time increases. PHASE SPEED Recall that a purely sinusoidal wave, proportional to sin (cce x -- cap, r), say, moves with the phase speed wda e . We are considering a wave train formed of a continuous superposition of such waves. The above-mentioned role of phase speed is therefore inaccessible to direct observation, because without preliminary Fourier analysis it is impossible to isolate one of the continuum of harmonics that make up the wave train. As we shall now show, the phase speed can be observed by focusing attention on a point of the free surface at a certain fixed distance d above the mean height. The progress of such a point is given by the function x x(t) that is defined implicitly by a solution of C R(x, t) = d. Implicit differentiation will give x the speed of the point. Once again it is the approximation 120) that allows us to obtain a clear-rut result_ Using (20 ), implicit differentiation of bR = d gives ,

— it - 311 cos H - I -

"(

sin #! H = O,

(23)

where H

cofar)t — *7I sgn

W "(a Fr).

For large r, the first term on the left side of (23) is negligible compared to the second, so that the speed of the point x can be estimated from Ham O or ay x —(.u(a,,)=0.

Thus for large time points near x 4 Vt that are at a fixed distance S above the mean height move with speed w{av}fav, the local phase spy_ In particular, points at the mean height (d = 0) move with this speed. Further, it can he shown that maxima, minima, and points of inflection also move with the local phase speed (Exercise 3).

Group Speed and Group Velocity [Ch. 9

n6

Example 3. Discuss phase and group speed for water waves, ignoring surface tension and depth effects. Solution.

The frequency (Ai is given in terms of the wave number a by rat = 'Jga-

The phase speed is wfa = gJa. The group speed is LAO = 1 gJa. The phase speed is twice the group speed. Both speeds are decreasing functions of the wave number, so that both are increasing functions of wavelength.

F iG tr R t 9.1. successive "snapFirols" Al a,rloa -ly raw Ow ware tre7in. The lines RR' enclose wares of approximate leng th The speed of such a ware group is less than that of the point of tY ►mmon phase labeled P I . The lines QQ' enclose a ,group of longer wares, whir* more ..faster than those associated w ith RK The local maximum 1 3 mores with as phasr , speed toward the n ight of the ,

QQ . grouP.

The roles of phase and group speeds are shown schematically in Figure 9.1 wherein the free surface is depicted at the equally spaced instants t o , t p -f ❑f, r a + 2/it, .... We consider for definiteness a situation, like that of Example 3, in which both group and phase speed increase with wavelength and the phase speed at a given wavelength exceeds the group speed at that wavelength. On the left of the figure, between the pair of lines marked RR', are waves of approximate wavelength A t .* On the right, between the pair of lines marked A,. There is a transition region QQ', are waves of longer wavelength There is an intrinsic imprecision in the nolion of wavelength. since the distance b€Iwœn successive crests (or troughs) is slowly varying_

Sec. 9.1)

Group Speed via ihr Method fJJ Sialionar}' Phase

377

between the two sets of waves. The transition region is actually long, so that there is a continual gradual change of wavelength. We have shortened this region in order to he able to depict more clearly groups of waves moving at different speeds. The speed of the waves of length A L is given by the slope of the line RR', The longer waves of length 1.2 move with the greater speed given by the slope of QV. (To compute the slope in the conventional manner, one must look at the figures sideways so that the independent variable r increases to one's right. Although units are not given on our schematic diagram, we assume that they have been chosen so that the slopes are numerically equal to the relevant speeds.) A point where the free surface always has the mean height has been labeled P 1 . This point moves with the local phase speed, that is, with the phase speed associated with waves of length A L . As this is larger than the local group speed, P 1 moves forward relative to the group of waves having wavelength J L. Eventually, the point P I will be more accuntidy associated with waves of wavelength larger than A i and hence with a larger phase speed, Also depicted is a local maximum P 2 moving with the phase speed, larger than that of P 1 , associated with waves of length )r. SPECIAL. CUl*lt]ITIUNS NEAR EXTREMA O F GROUP SPEED When applying the method of stationary phase in (3). we replaced Pita) in the neighborhood of a stationary-phase point a, by what we took to be its first two nonvanishing terms. h(;) + }h"(Œ1)(Œ

—

ai)2 .

Our approximation is invalid when h"(a) D and will be poor when h"(a ;) is nearly zero. For waves, hi) = ax — coop. so [as is clear From (9a)] the usual stationary-phase approximation wilt fail al wave numbers where the graph of group speed (Oa) versus wave number a possesses a horizontal Ionyeni. For water waves. provided that the depth is not too small, the group speed has a local minimum V,,, at a certain wave number a,,, [Exercise 5(a)]. Consider an observer located at point VI at time f. If V < V„,, the observer should report a negligible surface displacement after a time. since C .:: f) according to the stationary - phase approximation_ If V is slightly larger than I/,„, on the other hand, the observer should report waves of wave number approximately equal to a n,. If V = V,,, wc can use the results of Exercise A9.1.5, but to describe the transition between the smooth-water region reported by the first observer and the waves reported by the second, it is necessary to use an approximation that is valid, for large time, uniformly for VL V 5 V 2 where Vi < V < V2. Such a uniformly valid approximation can he found in terms of the Airy function Ai, the simplest function whose behavior alters from monotonic to oscillatory as its argument changes. For details, see Jeffreys and Jeffreys (1962, pp. 517-18). These authors modify the

37 8

Group Speed and Group Vorocrry ((k 9

usual stationary-phase result by a clever device. A more straightforward approach can be made using heavier machinery.* A similar uniformly valid approximation must be used to obtain an adequate description of the waves near x ` t H. The reason is that (even in the presence of depth and surface tension effects) the group speed also has an extremum at V y yH [Exercise 5{b)]. It turns out (Jeffreys and Jeffreys, 1962, p. 51 7) that the disturbance becomes mere and more prominent near

x

;r. There are capillary waves ahead of this point, but they die out

relatively rapidly [Exercise 4(a)]. Moreover, these short waves, with their accompanying large velocity gradients. are more strongly damped by the viscous effects not considered in our analysis. Except for shallow layers that will not concern Lis here, 9H > V !, and the discussion of this paragraph applies to the leading edge of the wave train_ To summarize, the principal surface disturbance is a slowly varying wave train whci ,e "front" and "back „ boundaries move with speeds 1 1,„ and

gH, respectively_

SOME: APPLICATIONS TO FLOW PAST OBSTACLES

We shall now apply some of our conclusions to the understanding of flow past obstacles. In doing so, we anticipate to some extent more detailed discussions to follow. By making some applications of our conclusions now, however, we indicate their power to those who cannot at the moment pursue the subject further_ Moreover, the tentative nature ci f some of our arguments provides good motivation for further study. We have seen that lines of constant phase, notably crests and troughs, travel with the phase speed co/a. Sets of waves having roughly the same length, by contrast, have been shown to travel with the group speed drr*fda. 11 can also be shown (see Section 9.3) [hat energy travels with group speed, for waves of a given approximate length retain their energy as they move. The conclusions we have reached have been in the context of an initial value problem, hut they will he shown its Section 9.3 to hold whenever a slowly varying wave train is present_ In particular, the description in the previous paragraphs of the roles played by phase and group velocity holds in the case of steady flow past an obstacle. This should not he surprising, for the effect of an obstacle in a steady linear flow is the same as the superposed effect of appropriately chosen initial disturbances. The obstacle can be regarded as sending out identical disturbances at each instant. At any moment of time, the effect of these disturbance can be obtained by adding the effects of each individual disturbance, with appropriate delays to account for the varying times at which each disturbance has occurred. Sec C. Chester. B. Friedman, and F Utsdl "An Extension of the Method of Steepest Descents," Prat_ Cambridge Phil. Soc. 53, 599-îi l (1957). Sec also our Figure 9.5 and the refermez cited u> >ts caption. ,

Sly. 9_1]

(irotep Spud

rua ihe Myrhod of 51a1ionary Phase

379

Since energy travels with group speed, there is a significant difference between situations wherein group speed exceeds phase speed and situations wherein the opposite situation obtains. In the former (latter) case, wave groups and energy travel forward (backward) with respect to the patte rn of crests and troughs. In this connection it is noteworthy that group speed exceeds phase speed if and only if the latter is an increasing function of wave dumber. Thus the tvavellerigrh (if any} at which the difference between group speed and phase speed changes sign is the some us the Waveleriyth (ij' any) ut which phase speed is a minimum (Exercise 2). For water waves when depth effects can he neglected, this wavelength is 1.7 cm see 18.1.21)]. Solutions of numerous special problems bear out one's intuition that when the surface of flowing water is disturbed by some object (or when the object moves through the water). most of the wave energy will generally be found in waves whose length is of the same magnitude as a typical obstacle dimension.' It is no surprise, therefore, that the wave pattern forms a wake behind a moving ship. To be stationary with respect to the ship, a crest must have a (phase) velocity that is balanced by the component el- stream velocity (as se en from the ship) ahai is normal to the given crest (compare Exercise 4.211. All significant wave groups are gravity waves, with lengths considerably longer than 1..7 cm. Consequently, energy moves backward with respect to the pattern. Each wave group disperses. and this spreading causes adecrease in amplitude and a gradual disappearance of pattern with increasing distance aft of the ship. Water beetles, by contrast, are typically less than 1 cm long, so much of the wave energy they generate will be in the form of capillary waves (lengths less than 1.7 cm'. ]f the beetles move steadily at less than the minimum phase speed of about 23 cmis, no wave pattern can be produced (for no set of crests and troughs can move slowly enough to maintain their position with respect to the beetle). In this case one expects only a highly localized disturbance of the water surface, which in fact is jusi what is observed Beetles can move as fast as 50 to l&) cmis, so there should be instances in which wave patterns appear. Indeed, such a pattern is shown in the frontispiece. Although there a dead beetle is held stationary in a smoothly flowing stream, the insect is properly oriented and the waves are virtually the same as in nature - except that the beetles in question normally move in a jerky manner and so produce a superposition of the patterns obtained for steady motion. Striking in the frontispiece is the fact that all the noticeable waves precede the beetle Our present understanding of the situation would lead us to anticipate this, • In aIJ rases, short waves may be an irnporiant source of energy disswarron. t Thr tengrh of hie brute is in the inrrrrner}rete runtçc. where both capillary and gravity effecis are expected to he impnrtarti. Thus it is not surprising that gravity waves Can a [so be sccn behind beetles. although their relatively great length and srnatl arnphrude make detection difficult_

380

Group Speed and Group

[Ch. 9

however: Since group velocity exceeds phase velocity in capillary waves, the energy moves forward with respect to the pattern. Beetles seem to have an intuitive knowledge of the subtleties of group velocity, for there is some evidence that they use reflections of the waves they generate to warn them of objects ahead of them. (See the article by Tucker cited in the frontispiece caption, and the references listed therein.) Physicists, too, have reason to be concerned with deeper aspects of group velocity and related matters; otherwise paradoxes arise in connection with the velocity of light. An excellent reference here is L. Brillouin, Ware Propagation and Group Velocity (New York: Academic Press, 1960). The principal mathematical issue in Brillouin's book is the asymptotic evaluation of integrals, a topic that we have touched upon briefly in our own development of the group velocity concept. Further development of this concept in the present volume (Section 9.3) is based an an entirely different kinematic approach that is fruitful even in contexts where it seems impossible to express resu lts in the fern3 of integrals. EXERCISES

L Find asymptotic results, as in this section, when there is no initial displacement of the free surface but there is an initial velocity. 1. (a) Show that in general the difference between phase and group speed changes sign precisely at the wave number at which the phase velocity is a minimum. (h) For water of infinite depth, find (as a function of wave number) the ratio of group speed to phase speed. Sketch the graph. Note particularly the result for pu re capillary waves (or, equivalently, the infinite wave-number limit). (c) Find the ratio of group speed to phase speed for water of depth I!, ignoring surface tension. Graph. Do your results suggest any modifications for the concludèngdiscussion of the section in the case of waves that are long compared to water depth? 3. Show that according to {0), maxima, minima, and points of inflection move with the local phase speed. This involves defining x as a function of t by the equation dC R /r?x = [i and then finding ac. In determining x, show that only fi(x. t) need be considered time-dependent, since time derivatives of other terms are small_ 4. (a) When capillary effects dominate, (8_31e) gives (for fluid of infinite depth) 0=

`43;

T_

?

p

,

a > U.

(24)

Find the counterpart of (12). Deduce that capillary wave amplitude, in contrast to gravity waves,decreases relatively rapidly as x increases.

See_ 4_l] F.xperirnenrs and Prorl,rcrt Applications

3b

^

(b) If you keep your eye on a wave of a given length, will a maximumheight point for that wave move forward or backward relative to the wave as a whole? At a given instant, are the longer or the shorter waves nearer to the point of the initial disturbance? Explain. 5. When depth and surface tension effects are both included. the dimensional dispersion relationship is (see Exercise 8.1.4)

ro' — (ga + T c3 ) tanh fl

.

(25)

where H is the depth of the layer, 7 - = Tlp, etc. la) Set tanh xH = I in (25). (This is appropriate except for waves that are long compared to the depth 1/.1 Shot, that the group speed V12) has a minimum 4;,, when x — m„,. where x, w = {1.4ti y T. Find V,, and the corresponding wavelength for water waves [7- = 72 g s 2 . p — 1 g/cm 3 . ti = 980 ems']_ +(b) For small a.

=Wi[a—

w 3 a 3 +••-]

Find co 3 lc) Under what circumstances are the regions of validity of the calculations in (a) and (b) expected to overlap'' For water waves. about how large does f! have to be to ensure this? Sketch the graph tif l'i in this case. Show that a graphical construction for group velocity can be obtained as 6. follows_ Plot phase velocity e verses wavelength .. Construct the tangent line to this curve at the wavelength d o of interest. The dinireti group velocity is the intersection of the tangent line with the vertical axis. 7. Using (25), show that the phase velocity no longer has an interior minimum cm for water waves What if H ^ I7 £ , 17g - t. Show that modifications are suggested to the concluding discussion of this section? 8. (a) Find the relation between group and phase velocity in the anomalous dispersion case provided by transverse elastic waves in a beam [use the dispersion relations (5.I.50)1 Discuss the significance of your results. (b) Repeat part (a) using (5.1.49) and thereby determine sonic effects of rotary dispersion.

f!, 3

H, _

-

9.2 Experiments and Practical Applications A good theoretician should be familiar with all relevant experimental results. Although it is not appropriate here to present an extensive account of water wave experiments, to forbear entirely from mentioning them might foster an inappropriate disregard for reality. Thus we shall compare our theory to a certain set of careful experimental measurements. Moreover, we

Group Speed and Group Velocity [Ch. 9

382

shall provide a brief discussion of practical application of lite theory, particularly to rsummis (tidal waves). EXPERIMENTS ON THE COLLAPSE OF A RECTANGULAR DUMP OF WATER

J. E. Prins* performed experiments to test the validity of the theoretical results we have been discussing. He set up the equivalent (sec Exercise I) of an initial rectangular bump or depression of width 2R and height Q on a motionless channel of water about 18 m long and 30 cm wide (60 ft by 1 ft)# (see Figure 9.2). The depth H was typically 67 cm (2.3 ft), although shallower

FtG u lr

9.2_ Experimental setup used by J. Prins_ The -• bump" of water

un the kit is released w time t = O. Waves propagate to the right and ore prosily absorbed by maternal p(uced on the sloping heath, Thus there is negligible reflection. and merJs -urr ►rtrrits carrt.spund hell to the idealized ruse of an un-

bounded fluid layer.

depths were also used. Prins measured the vertical movement of the free surface at five places along the channel and found the oscillatory waves predicted by our theory when QM < 0.2 and R/H < 1. Our theory does not predict phenomena found by Prins at larger values of Q/ f and R/11, such as "a leading wave being a single wave with solitary wave§ characteristics, separated from the dispersive wave pattern by a more or less flat part at the still wafer level." This is not surprising, as the larger are (2/H and RIB, the larger is the initial bump and the more likely it is that our assumption of small-amplitude waves is inappropriate. -

• "Characteristics of Waves Generated by a Local Disturbance," Trues. Arne r. Geopiiys Union 39, 865-74 (t95). t This was done by placing an airtight box, apcii side down, into the water at the head of the channel_ (The box WAS the same width a5 the cha nnel.) The air in the box was par l ial ly evacuated to obiain a rectangular elevation, in other experiments, compre -ssed air was introduced to obtain an initial recta ngular depression_ At the initial imtant the downstream side of the box was quickly removed. k A solitary mare is long compared with the depth H otthew undisturbed fluid layer. Its maximum height above the undisturbed level, A, need not he small_ It moves without deforming, ai speed % f! i A) (sec Sicker (195711

▪

Sec. 1.2]

Experiments und Practical Applications

383

In Figure 93{a) and (c) we display some of Prins's measurements of the time interval P between successive relative maxima, due to an initial rectangular devation. These are plotted versus r, the time at which the intervening. relative minimum passed one of the fixed locations, x — 4 r . at which measurements were made. ln Figure 9_3(b) and (d), P refers to the time 3.0

1

{1

{1

11 tx ^ I.

r=a ^

I0

r^

^

g

It Ili

HP-1 )^^ r1.%%

E^ r

^

^

^

4

•

?4

115

1f

4

(A)

30

.0 l

0

Idl

tcl

F t (i Li x E 9, 3. Time interval P tzs function of tame z, as defined ln the schematic ware recordings depleted in (a) and (h). In (a) and (c), elet ►ation—Q = 0.3 f! (9 cm). fer lb} and (di, depression-1Q ` —0.3 fr. fer lai and (b(, hump extent R ; fi. In (c) and (d), R = I ft. Depth H ; 23 fr in all cases. Each graph contains four curries, giving successively larger takers of P. Experimentaipoints A. x. 0, are recorded at slalioris x = 5, 15, 25. 35 ft, resper 41474y (5 fi ft 150 con I _ Dashed laies. infinite depth theory [Equation (1) of Solid lines: lines: Kranzer-Keller finite depth theory_ f Redrawn with perrrrisiion from Fig. 3 of J. E. Prins, Trans. Amer. Geophys. Uni art 39, 865 874 (I 958). copyrighted by American Geophysical [brion.] .

,

Group Speed

384

and Group

Velocity [Ch- 9

between succeeding minima, since [in contrast to 9.3(a) and (c)], the waves were caused by a depression. COMPARISON OF THEORY AND EXPERIMENT To obtain a theoretical prediction of P, consider the asymptotic formula (1.22). Choose V = x ; /r so that the successive maxima in question are near x ; - VT when t is close to T. According to (1.22), successive maxima should occur in a time interval ofduration 2rr/(gV t). An observer at x x ; should therefore record a value of P given by —

P=

4rtx i gt •

(1)

As Figure 9-3 shows, agreement between experiment and theory (dashed lines) stems good when f' is less than about I but is only fair for larger values of P. Many factors might he responsible for the partial discrepancy between theory and experiment. The discrepancy occurs for relatively small values of r, so one possibility is that measurements were made before enough time had elapsed to allow use of (1.22). It turns out, however, that most of the discrepancy is due to the effect of finite depth. One can easily provide indications that when P begins to exceed a quantity with magnitude unity, then the waves begin to be long enough to he affected by the bottom's presence (Exercise 2). Further as Prins shows- see the solid line of Figure 9.3), predictions of depth theory arc in gond agreement wish all measurements made_ fi+^[r Figure 9.4 shows traces of the surface deflection when Q = —6 cm, R 10 cm. A noteworthy feature is the continual increase in surface deflection maxima for the first portion of the record obtained at each station. This increase is nearly linear as the reader can check with a straightedge. Just such a linear behavior at fixed x might be expected from (1.12) when s IR

gt z R — 2 eh

gt2R 3. 4x

IZ}

In this case t' R is the product ofa sinusoidal function and a linear function of t

fi)cosr —

CR t

.

(3)

Indeed (Exercise 5), except for a relatively short initiai period, the cosine Factor in (3) varies rapidly compared to t, and the maxima Of Ci (as functions of time) occur if t = t,,, where

gr H

^

^

4x

^

n

r2n,

(),

1, 2,

-•-•

(4)

When (2) holds, then, the maximum amplitudes do, in fact, increase linearly: max CR

x

!

X

/) _r<x

,ri

' 0,1, 2, •

()

Time, r(seç1 14

15

25

FIGURE 9.4. 1aPc heights as a function of rime, recorded at various fixed slolions. Q = — fi, R = } fr, H =re 2.3 ft. 'Redrawn with permission from Fig. 5 of J. E. Prins, Trans. Amer. Geophys. Union 39, 865'874 (1958). copyrighted by American Geophysical Union.]

Group Speed und Group I{frlorir y[C'h.

10

40

r

4

5o

^

rt

hr depth H) versus dimensionless time r ,yfhf, Corrrpurixem of 14ran:r'r Keller (op eft ) ewfrerlareor'rs with measured values of J. Prins," Water Wares Due to a Local Disturbance," F i C+ tr RlE 9.5. Values of the ware hetirthr ti (dirided

Proc. 6th Conc. Coastal Eng. 1Eng. Council Warr Res.), 1958. (f!

R Jt, N = 2} fit, observation at x = 35 ft.! [Redrawn from R. Wiekrel er nl,, "Warrr Wares Generated hy landslides in Recserru}r'rs.° ilnirersiry oJ California ff rdreur#e• Bnrtirrerrtrrrt Laboratory. Berkeley. Calif.. Wave Res. Project Tech. Rcpt. HEL 19 -1, Apr. 196c, Fig. 10: with permission,'

The approximation (2) is valid while r i is less than about 4x 2My ?. Using R 10 cm, q = 10 i cm/s 2 : for the first observation post (x = 150 ern) this condition is t
1 _ .. , there is a wedge with half-angle sin -1 (r,,,iUI with the (b)

4 32

Group Speed and Group Velocity [Ch. 9

property that capillary waves appear only inside the wedge and gravity waves only outside? 11. According to the original caption of Figure 9.15, the ratio of the wavelengths of the two types of waves where their crests are perpendicular to the direction of water flow uniquely determines wavelengths and water speed. Since the gravity waves are seventeen times as long as the capillary waves, the wavelengths of the capillary and gravity waves must be 0.042 and 0.71 m, respectively, and the water speed must he 034 m/sec. ' Verify these statements,

Appendix 9.1 The Method of Stationary Phase—An Informal Discussion In discussing wave dispersion, and in many other contexts, one is led Co seek approximations to integrals of the form 1(a, b, ).) = If(x) ex p [i4-1(x)) dx

(I)

The approximations should be valid for large values of the real parameter ;.. MOTIVATION

Instead of (1), it is easier at first to consider its real part, j f (4 cos [(x)] dx.

(2)

Ill. is very large, then cos ).g(x) will be a very rapidly oscillating function. The situation will be like that depicted in Figure 9.16: f (x) will change slowly

FIG Li R E 9.16. As shown, if ;. is large, ens P .g(x)] wilt in genFral eQry rapidly compared ro fix).

App endix 9.1]

The Mt^s h ud rtf Sturronurl- Phase An lnfnrrrro! Drsc -u.ssron

413

Kit

xt 1.u1 l

I

I

I I

I I I

I 11

^e

^i

Ir

I^

FIGURE 9_17. A function ,y with a ronsiunr rryrort No mazer h n x faro' i i s,

there

will be nr>'oscillation of cos [i.g(x)] lor

.x o < r < x I

compared with the change of cos Ag(x). Because of this, the contributions to the integral from intervals where cos Ag(x) is positive nearly cancel the contributions where cos Ag(x) is negative. On the other hand, suppose there is an interval where g(x) is constant. In that interval cos :tg(x) would also be constant, so that there would be no oscillations and no cancellation. It is thus plausible that, for large ;t, the contribution to (2) from an interval where g(x) is constant far outweighs contributions from intervals where g(x) varies. In Figure 9.17, for example, we expect the contribution from 1 3 to outweigh the contributions from 1 1 and 1 3 . Moreover, for very large •1 we expect that the !s contribution will dominate, even though I t is very narrow. In the limit A o) it is plausible that if g(x) has a horizontal tangent at x; (figure 9.18), then the contribution to the integral from the interval [x ; — d, x ; + 6], however small b is, dominates contributions from intervals in which y does not contain a horizontal tangent.

go1

I I t a

FIGURE

9.18. For sufficiently large

arise when x is

rc^^nt.

s,

i. the major contribution M ,

in a narrow region about x , the location of the ,

( 2) will

lrurrcrnirnl

Group Speed cowl Group Velocity [fit.

4 14

9

In the present context, points w here g'(x ; ) = 0 are cal led points of stationary phase. We assert that for large A the principal contributions to (1) do indeed come from these points, and that this fact allows (1) to be approximated by a relatively simple expression. DEVELOPMENT O F A THEOREM

In the first part of the demonstration of our assertion, we suppose that the closed interval [a, f] does not contain a point of stationary phase. Then, since g'(x) 0 for a S x S 13, we can write f(a P, a.) = f (x) exp [a9(x)] dx ` f .

t^g'

dx [exp

dx.

Integrating by parts, we find that i^.!(a, 13,) = _ exp (i).g)

. ' 4l)3Î9„ exp (ia.g) dx.

--

(3)

a

We next employ two forms of the triangle inequality, Ix + 'I

ru

and

Ix1 + IYI

F(X}dX S ^IF(x)I dx. o

I

(4)

Using (4) and the fact that I exp (iA.rg)I ° 1, we see from (3) that I Al(a,

P. ^) I ^ 10) + I (a) s'(1)

+

f'g' — fe" dx ^1

(9') 3

We can conclude that if there is no stationary-phase point in [a, fl], then certainly I(a. fi. AIM = O(A_ ').as A --• x, for IAI(afi. APB is bounded by a constant which is independent of di. We slate this result in the form of a lemma, which contains conditions guaranteeing that the various integrals exist. Lena.

If f (x) and y(x) have, respectively, one and two continuous derivatives for a < x S b, and if y`(x) # C for a < x < b. then I(x) exp [0.9(x)] dx

)

as A.

(5)

We turn now to the case where the interval [a, /3] does contain a point of stationary phase. Call such a point x„ so that g'(x ; } = C_ Suppose further that g"(x;) # O. Let 6 be such that x, is the only zero of g'(x) for I x — x i I ç S. We assert that J +a !^ = f (x) exp [ikg(x)] dx

^ Y a i

(6)

Appendix 9.1] The Method of Swrrnnary Phase—An lnformpf Discussion

can be approximated for S i = ( r) '

Lis

large positive A ) I — 112 f(xi)

4lS

by S i , where

exp [f 4(x,) + ;7Zi sgn g"(xi)]_*

(7)

To be more precise, we assert that under appropriate hypotheses =Si +O(). -

').

(8)

In considering a general case of the integral l(r^ b, A) of (1), let us suppose that there are N points of stationary phase in (a, b), N 1, and that all such points occur in the interior of the interval of integration. Then we can split up [a, b] into subintervals that contain stationary-phase points, where we can use (8), and into other subintervals that do not contain such points and hence, by (3), that make negligible OP, - I ) contribution to I. Such reasoning leads to the basic stationary-phase theorem (12). If one or both of the end points a and b are points of stationary phase; i.e_, C or = 0, then there are corresponding contributions to the if g`(a) approximating sum of 41.2). Not surprisingly, it can be shown that these contributions are precisely half the contribution of an interior point of stationary phase. If g"(x,) — 0 but q "(x i) V- 0, the corresponding contribution to (i.2) is a relatively large 0(AA " I ) term, and the first term neglected now is 0(A -2 °). See Exercise 5_ ,

HEURISTIC DERIVATION OF THE KEY APPROXIMATION

It remains to give evidence for our assertion that (6) can be approximated by (7). We shall give only a heuristic derivation of this result. For a rigorous discussion, see A. Erdelyi, Asymptotic Expansions (N.Y.: Dover Publications Inc., 1954) and also A. Wintner, "Remarks on the Method of Stationary Phases," J. Muth. Phys. 24, 127 30 (1945). In (6) we can make as small as we wish. t# is reasonable to suppose that in the narrow interval [x i --- 5, x i + 6], we can approximate f (x) by f (xi). We must be more accurate in considering g(x), since it is multiplied by the large quantity A. Remembering that g'(x,) = C, we approximate Mx) by the first two nonvanishing terms in its Taylor series about x i . Thus -

x, +d

.,

}l .I (Yi}exp (i+^Cg(xi) - f- I9 (xi)( j/!^ ` ^i) 2 J! dx

Q (` )

An - 6

or

, l exp [l4(x i)1

+

a exp [rrl,L(x — x i ) 1 ] dx,

2f(xi)

s,

• Here sgn is Itic siynung (r ^nr -rron: sgn x

x

=

1xl

^

1.

1.

>0, x c 0. ^

(10)

4 16

Group Speed and Group Velocity [Ch. 9

where

( II)

A — iy"(x ; ). If A > 0, we make the change of variable

^lrl(x — .x

= 42 ,

(I2)

with which the integral in (10) becomes 64.14 1 112

(d1,q)- 112

f0

exp

(13)

For large a., (13) can be approximated by ^«

exp (g2) fig = 0

cos (el d a

- i

J (2 n

}d .

(14)

Numerical methods could be used to evaluate (14), but those having some familiarity with Contour integration should have no difficulty in completing Exercise 4, which shows that its value is it'". Exercise 2 deals with the case A z 0. Putting our results together, we obtain the desired formula (7). CENERAI.IZA'r ION

The presence of the parameter A in (9)ensures that there will be a situation when A is large) wherein one factor of the integrant] performs many oscillations about a zero mean value while the other factor changes only slightly. Whenever such a situation occurs, the same heuristic reasoning that we have just presented suggests that we can approximate the integral by considering only those portions of the interval of integration where the oscillatory factor is stationary. Thus when exp [iii(x)1 oscillates rapidly compared with jr (x), we expect the generalized stationary-phase result of (I.3). What our hypotheses for this result lack in precision, they gain in emphasizing the essential

point. The condition h(x) 4(x), . co, is sufficient but not necessary to guarantee the relatively rapid oscillation of h, and hence to ensure that (1.3) is a good approximation. Often a theorem is proved in the literature in a form different from that desired for a particular application, and the sufficient conditions used for the proof may not be met in practice. This is illustrated in Section 9.1, where we apply the generalized stationary-phase approximation (1.3) without reducing our integrals to the clearly stated form of (1.2). Having some confidence that we could produce the required proofs if necessary, but feeling that the main issues do not lie in that direction, we are content to rely on our intuition that the necessary extensions of the basic theorem can be obtained.

Appendix 9.1]

The Method of Stationary Phase- An Informal Discussion

417

EX [FtC IS ES

1. To acquire some notion of how the stationary-phase concept is relevant even though no large parameter explicitly appears, show that if there is no stationary phase point in [a, l3], then under suitable hypotheses on f and h (which you should state) -

f

f(x) exp cih(x)] dx I < k max p

h ()

,

k is a constant.

2. Obtain (7) when A < O, i.e., when g"(x + ) < O. 3. (Project) If you have access to a digital computer, use it to evaluate the integrals of (14) to three significant figures. By considering the integral of exp (iz 9 over the contour depicted in Figure 4. 9.19, show that the evaluation of 114) reduces to the evaluation of exp ( — t9 dr. The final result can then be obtained with the aid of Exercise 3.2.2(d) of 1.

FIGURE

9.19. A contour in the :-plane

5. Consider 1(a, bd.) of (1). Suppose that in addition to the ordinary stationary-phase points x ; , there are also M points y, satisfying g'(yi) = 0,

g"1,y,) = Q, 9"1v1)

a;

a < yt < b, 1i = 1, 2,... , M.

Show formally that the series in (1.2) must be augmented by l'(,,) (0)1 !3 ,1-

1/3

L i

Ig (y)I - 1/3f( a)exp f

[idlAy i)]•

-

Under suitable hypotheses it can he shown (Stoker, 1957) that the error is now OP.-2/3).

C HAPTER 10 Nonlinear Effects

IJ

this point, our treatment of water waves has concentrated on the extensive classical developments of the linear theory. Now we turn our attention to nonlinear effects. Grappling with nonlinearity is one of the major tasks of the modern applied mathematician, so this is certainly not misplaced effort. Moreover, much understanding of nonlinear effects has emerged in the last few years from studies of prototype problems in water waves. The rather classical material that we present in some detail is of considerable value in itself, and also provides a necessary Foundation for more recent developments. Our discussion of recent results is brief, yet it will afford an appreciation of current research_ NTIL

10.1 Formation of Perturbation Equations for Traveling Waves in Chapters 8 and 9 we studied water waves under the assumption that wave amplitudes were so small that nonlinear effects could be entirely ignored. We first obtained sinusoidal traveling wave solutions to the governing equations. These solutions, although special, exhibited important properties--particularly the dependence of wave speed on wavelength. Moreover, the sinusoidal solutions could be superposed to solve a general initial value problem. Now we shall investigate how a sinusoidal traveling wave solution to the linearized equations is altered if nonlinear effects are taken into account An important new property will arise dependence of wave speed on amplitude. But superposition is no longer possible, so that, in contrast with linear theory, the obtaining of near-sinusoidal solutions is not a dominant step toward full understanding of the problem. "Adding" nearly sinusoidal nonlinear wavesgi%es rise to new effects that we can only describe briefly here. Useful insights into nonlinearity are nevertheless afforded. Our calculations are an example of regular perturbation theory. This subject was introduced in Chapter 7 of I, mostly in the context of a single ordinary differential equation. A system of two ordinary differential equations was discussed in Chapter 8 of I. Here we deal with a system of three prarririi differential equations_ There are two further complications: Boundary conditions are applied on an unknown curve and a series expansion of a certain parameter in the problem is necessary. Although knowledge equivalent to the elements of regular perturbation theory, as discussed in Chapter 4 [8

Sec. 10.11

Formation of Perzurharron Equnl(uns for Traueltmy Waves

419

7 of 1, would doubtless give much more confidence in our procedures, no such knowledge is required here. The reader need only accept the following idea: If various dependent variables and a parameter are known to be functions of a certain other parameter l:, it is reasonable to hope that expansions of all quantities in powers ofc will provide valid results when I k I is sufficiently small. in this section we shall first introduce a coordinate system that "moves with the wave_" Then we assume expansions in the amplitude parameter r.. The original nonlinear system is thereby reduced to a sequence of linear systems. Solutions of these systems are discussed in Section 10.2_ CtiA NICE OF VARIABLES

Consider small - amplitude water waves that retain their shape as they move with constant speed C. We look for waves t hat are spatially periadicand hence, since they do not deform, are temporally periodic_ {Later we shall restrict ourselves to "nearly sinusoidal waves," but now our discussion is more general.) We shall take 27rL to he the spatial period (wavelength) of these waves, 27rP to be their temporal period, and A to be their amplitude_ In terms of the dimensionless variables x = x*/L and r = t*/P, both spatial and temporal periods have the value 27t.f Recall that the pattern rnoves a wavelength 2 rL in a period 27zP, so that

L

_

2nL —

2711)

or C.

_ LP .

We shall look for right -moving waves, so that all variables will be assumed to be functions of x* — Ct* rather than x* and t* separately. Dependence on the second horizontal variable y* will be assumed to be negligible. Using the scaling of (7.2.21), we therefore assume dimensionless dependent variables '"

It .s

u ! A/f' ,

w

* {'

=

— Co A

pALP 2' -

A la d)

that are functions only of the dimensionless independent variables

x —

x*

—

Ct*

-*

r = — L

and

.

(

3e, f}

(Note that x can be regarded as the horizontal variable in coordinates that "move with the wave.") if ti* = Li*

(x* --

L

Cr* z* .

, L

t This is a hetter choice than unify for the dimensionless pct pd - il avoids the appearance of tir in Many later formulas_

Nonlinear Effects [Ch. JO

42

we have

au *_

c'e^•

*

€7x*

.

Analogous relations hold for r*-derivatives of rte* and (; thus the governing dimensional equations (7.2.12-15), (7.2.17-18), and (7.2.20) become -Eu x + E 2 (IA , + w14,) = —Ew1

(21

+ ex (u ' + ww,) — — ep,. Eu

At z = —

— epz,

(3)

+Ew,=0.

(4)

Ew — 0.

(5)

Atz= EC,

r.w = —EC„+E 2 riCz .

At z T Er,

Fp = F[E4

f

(6)

BECz„(1 + F 2 !) -3/']

(7)

x^ c^(x, t) dx = 0_

(8)

n

Instead of boundary condition (7.2.16) and initial condition (7.2.19), we impose a periodicity condition. We seek functions of period 27EI_ in x* and 2mrP = Dr1,VC in is. From (le) this implies that all functions have period 2r in x.

(9)

For a reason that will be explained shortly, we have chosen not to perform a division by E that would simplify the appearance of the equations. CONSEQUENCE OF TRAVELING-WAVE ASSUMPTION

that we have assumed that the nonlinear equations have solutions in the form of waves that travel at a constant speed C without It must be emphasized

changing shape. (Mathematically, we have assumed that all dependent variables are functions of x* and r* only in the combination x• — CO.) The validity of this assumption is yet to be established. Four parameters appear in the dimensionless governing equations (2) to (9)_ As in (7.2.31) these parameters are E

A L•

_ L

_ pg L2 .

rL C2

We have seen that according to linear water wave theory, periodic waves travel with a speed C that depends upon the wavelength 2rtL, the surface tension 7, and the depth H . Only the parameters h. B, and F enter in linear theory, and C occurs only in the Froude number F. Thus we would expect

&c. an Farmaiion of Perturbarion Equations for Traveling Wares

42

this dependence to be expressible in the form F = F(B, h).

(10)

Indeed [Exercise 1(a)], the dispersion relation (9. I .25) can be written as F

[(I + B) tanh h]

(11)

When nonlinear effects are taken into account, we expect that [10) will be generalized to F — F(B, k F).

(12)

In other words, traveling waves are expected to he possible only if the dimensionless speed parameter F depends in a specific way upon the other parameters of the problem. SERIES SOLUTION

We have seen that analysis of water waves can be considerably simplified if it is assumed that the amplitude-to-wavelength ratio i isa small parameter. The most drastic simplification is to retain only terms of lowest order to F. This results in a linearized theory that has been extensively discussed in Chapters 8 and 9. A more comprehensive approach is to assume that the dependent variables can be expressed as power series in e. Since it is reasonable to hope that such series will converge when z is small enough, we shall proceed with the formal calculations. We assume power-series expansions for all dependent kariables. Thus for the horizontal velocity component we posit the expansion

tu(x, z, t, E) = Ew'"(x, z, rj + elu` kx, z, r) - r. 3 ui 3 (x, z, r) +

- - - -

(13)

Note that the O(E) left side of (13) is matched by an OC') dominant term on the right. Superscripts match the power of e; and the lowest superscript is unity, in conformity with the notations of earlier chapters. (This last desirable feature emerges due to the fact that we did not divide the equations by r..) For the other variables we write (no longer indicating independent variables)

+ E3w(3' + - - - ,

Ew = t:w"'

+ E3p{3' f- . . . ,

( 15)

F- t1 + r. 2 cm + f;3 {31 + ...,

( 16)

t:p = Ep"i Ec, —

(14)

-}- E 2 /PP

F(B, h, i O. If f"(x 0 ) = 0, further sufficient conditions can be obtained by considering more terms in the Taylor expansion. Continuing the litany or elementary results, we recall that an extremum is either a maximum or a minimum. Using subtler tools than Taylor's theorem

_

• 111(x) b2>cz, which are closest to and farthest from the origin. a2

(12)

• For example. sec pp- 431 f. in Vol. 1 of V. L Smirnov. A Course of Higher Marhemaiks (Readins, Mass_: Addison-Wesley, 1964)_

See. 11 _ 1]

Fxrre ►na

Furnitrin -- layrarrye Jllultiplif•rs

453

Sohrtian. As usual, it is easier to cxtrernalize the square of the distance. To exy2 + Z 2 subject to (i l}, treinalixe f(x, y, z) = 12 we use (9) and 1itid candidates by simultaneously solving { I I) and ,

x2

yr

/x 2 +

t2

+ 2 + r .2 - )] -a 11 1

y2 +z2 -a.l

y

(13)

^.

Equation (I 3) gives 2x - 2.+ -2x = 0.

23 , - 0.

2y -

2: - 2Ae - :z = 1]_

(14)

These three linear equations for x, y, and z have nontrivial solutions only for certain values of d- The SO lurions are

â

▪ x=e t . 1'=0. _ • x = 0. y T c2 .

^•

0:

(LSa)

r = 0;

(l 5h)

c3;

( t SCl

.;, x = O, l' = 0, z

where c,, c 2 . and

-

c l are arbitary so far- Substitution of(IS 5i mue U I) gives ± b , c^ = ±C-

e t = # a. r- 2

(16)

The functions f and g have continuous derivatives and Vg is nonzero for points on y = 0. z) given by (15) and (1.6)_• (These points he a1 so the only candidates are the points the ends of the axes of the ellipsoids.) To determine the nature of the extremum provided by the candidates (a, 0, 0) and - u, 0, 0), we wril e = Fa+x',

y

(

z

the primed variables measure the extent of the departure of a typical point (^ y, rj from a candidate point. Subtracting the square of the distance in the origin of the candidate points (± a, 0, 0) kern the corresponding quantity for a typical point, and using the fact th;li trie latter point must lie on the ellipsoid, we find the following: ,

f (x, j, z)

11-1-d2,0,0) = (±a ±

But

t'1 ? +

(ty` + (c'} ? — a2

tijx + ( h ^ z + l rz2 s 1

}a ^ Or

{^Q

♦ ^ ) 2 a2

=

-

b2

(ÿ { }^

uz {r 'lz,

so fix.

a= y, z) - fl±a, U, 0) = (y') 2(II - - ) + (z') 2(1 -

6^

[

ur

(17(

2

y' = z' = 0. so both candidate By the inequalities 112j, this difference is negative unless points yield the same abswlure maximum distance from the origin_ Testing of the other two pairs of candidate points is left no Exercise I.

• Note that the num}wrs1in(15) g ivethe vaLuc of + y2 -4r' for the ascocinjied candid a[cs. The permitted values of t he Lagrange multiplier frequently have some such Ecumenic or physical interpretation_

Calculus of Variations [ CJr . 11

454

To test whether the stationary point x 103 is an extremurn of f, we have tried to determine whether the difference f(x"" 4 x') f(x 1°) ) is positive (minimum)or negative (maximum) for all x' which are such that x = xI°t + x' satisfies the constraints of the problem. For local maxima and minima, this -

—

difference will be of one sign only if Ix; I IxzI, and Jx3I are sufficiently small, and this can be taken into account by neglecting all but the largest terms in the difference. Because xt °1 is a stationary point, the difference will not contain any linear terms. Neglect of higher order terms will usually yield a quadratic expression [as in (17) and Exercise 2] whose character can be determined by various methods such as those noted in the footnote to the line following (6). ,

Example 2. {This example shows the importance of the exceptional case (9).1 Find the minimum distance from 10, 0, - I) to the surface A(x. }', z) = x z +

(18)

— z s = O.

From Figure 11-2 the answer is clearly unity. But let us see how our machinery operates_ r

FIGURE 11.2. The analytic determination of the minimum distance from (0, {) — I) ro the surface depicted tuas an exceptional character herause the surface has a cusp at the closest point ro (0, Q , — Ij .

Solution. To extremalire f(x, y, 2) = xt + y2 + (z + I )i subject to (I 8), we attempt to find candidate* by simultaneously solving (I8) arid F[xe +y2 + (z + 1) 2 —

+y2 —z 3 )]=4

or

2x(1 — d) = 0,

2y(I -- d) = D,

2(z + 1) + 5 rl.z` = ü-

(19a, b, c)

If A = 1, r + 1 S O from (19c), which contradicts z Q, a consequence of (18). If û contradicts (19c). The four # 1, (19a, h) and (18) give x = y == z = D. but z i

Sec. !!.!]

ExtrPrrra

of a Function—Lagrange Multipliers

455

equations (1$) and 119) therefore have no real solution, so no "successful candidates" have yet emerged, even though f and g have continuous derivatives everywhere. But Fg = Oat (O. 0, 0) and q(0. 0, 0) — 0. so the origin is the "candidate"for which we were looking_

Exercise 20 provides an application of Lagrange multipliers that is at once more physically significant and more subtle than the examples treated so far. INEQUALITY CONSTRAINTS

Lagrange multipliers can also be employed in extremalixation problems where the constraints are inequalities. Suppose, for example, that it is required to maximize a smooth function f(x 1 , )(,) subject to the constraint h(x 1 , x 2 , x 3 ) 5 fl_ The maximizing point is thus restricted to be in the closed region R bounded by the surface h = 0 (Figure 11.3). (Think of the problem of determining the maximum temperature_f among all points of R, whether intenor points or boundary points.)

The problem. is to maximize a funrtion f utttony all point's in the region R defined by h O. Suppose that the maximizing point is at x12y. on the boundary b= 0 of It Since rhe gradient points in the direction of maximum kJcrease, ;be normal ŸJr(x{ painrs away from negative txelueS of h and toward positive values. Hence, tf x rs any point interior to R and sufficiently close to 121 ) is x 421, then the angle 9 between x -- x^ 1 and the exterior normal C h(x obtuse. This fact is used in the derivation of 125)_ FIGURE 11.3_

Calculus of Variations

4515

[Ch. II

We can treat the present problem by combining our previous approaches. In the open set of points x satisfying h(x) < 0, we search for local maxima of where Vf = O. Possible maximizing points on the boundary are subject to the constraint h = 0, so a Lagrange multiplier can be employed_ But a maximum attained at x(2) on h 0 will not be a true local maximum for this problem unless f (x12) )is larger than the values of f at neighboring points in R. (Of course, this complication was not present when we were dealing solely with the constraint h = 0.) We shall find an additional necessary condition ensuring that a candidate point on the boundary will, in fact, be a iota) minimum off in the region H. The various conditions can be neatly combined, as the following detailed discussion shows. _f a maximizing point x;° is in the interior of R, then of course Vf (x" I) = O. In this case the usual condition Vf (0°) = ).Vh(xt 1)) holds for 1 = 0. If a maximizing point x(2 ° is on the boundary of R, then Mx" )) = 0,50 by Theorem 1 [if the exceptional case (9) is ruled out], Vf (x 121 ) -= AVh(x 12 '),

/!(x 121) = 0-

(20)

Since x# r} is by hypothesis a local maximum off, then (21)

_f ix) - f(x"l) < 0,

fer any point x that is in R and that is near x(21. But, to a first approximation [by Taylor's formula (4)].

fix) Since Vf

—

f( x(21) = [1(x 121 )] - (x

—

x121 ).

AWE, by (20), for boundary maxima.the above equation gives f (x)

—

f(x 14) _ A[Vh(x1")] . ( x

—

x 12 ').

(22)

Let U be the angle between x — x12F and Vh(xW W W). This angle is obtuse, so cos e < C. (See Figure 11.3--the convexity of R at x (21 is not necessary.) But (22) can he written

{ ix) — ( x i2)) — % I

VJ!(x(2I)I I x — x [21 I cos e,

which is consistent with (21) and a negative cos ( if and only if di is positive. If x12 ' were a minimum, ,f would have to be negative. Let us sum up. Assuming smooth functions, to extremalize f (x) subject to the constraint h(x) 0, we search for points x and constants ) which satisfy

Vf = Wh.

(23)

Solutions to (23) must further satisfy either h0{?. E. > 0• y2 > 0. xi c x28. Curves = }r(x), y ? 0, join (x 1 , y ^) to (x 2 , Y2). The curves are rotated about the x-axis. We wish to find that twice continuously differentiable curve which minimizes the area. Find and solve the relevant Euler equation to obtain the "candidate" curves (a)

(b)

y

C L cosh {C 1 - `(x

—

C r )],

C E and

C2 constants.

9. (Project) We refer here to Exercise 8. (a) Consider the family of curves passing through (x i , y t )- Show that these curves have an envelope (Franklin 1940, pp- 525OE). Use this fact and a sketch to deduce that there are two, one, or zero members of the family which pass through (x 2 , y 2 ), depending on the location of (x2 , Yi)-

C'Qfrulus

474

rit

Variations [CJ:, ! I

(b) Suppose that the conditions for admissibility are relaxed so that curves need not be expressed in the form y = y(x). Suppose further that continuous piecewise differentiable curves are admissible. There is now a curve composed of three line segments that always provides a relative minimum and often "provides" an absolute minimum. Can you guess what h is? 10. Consider spherical Coordinates, wherein x

_ p sal tb cos U,

y

= p sin 4) sin 0,

z=

p cos O .

Let curves on the sphere p M2 constant be given in the form 4 = f(0), #(a} Set up the calculus-of-variations problem whose solution is a curve of the above form that cxtrcmalizes the distance between two points on the sphere. (b) Use Theorem 2 tc reduce the problem to a first order differential equation. Snow that if the note after Theorem 2 is disregarded, one will be led to the erroneous conclusion that small circles on the sphere are extrenializing curves [Pars (1960. p. 43]. 11. Derive (23) by writing dx ' (h ` dy F x y. — dx = G x, y, - )dy^. dx — J.() dy } ,

(a)

Verify that dx (.;(x, ^x, y, ^

]'

= F x, y,

dx ^ Y -

r dx . dy

(b) Use the Euler equation d Gp — — G3 = i)

dy

and the hypothesis that F(x, y, y') is independent of x. (This formal derivation is more natural than the one given in the text, but its justification requires more restrictive hypotheses.) 12. In finding the shortest distance between (U, U) and (n, b), one obtains the right answer by trying to minimize S [I -- (y') 2 ] dx rather than rl + (y`) 3 ] " 2 dx. Is this trick always possible? Formulate and prove a theorem. 13. Give the natural boundary conditions for the following: (a) For (2) (comment on the answer). 1(b) For the F of Exercise I. (c) For the F of Exercise 3(a). $(d) For the F of Exercise 3(b).

See. l' 1_21 introduction rn riae f ïzkrrlu.x

!r(trruliorrA

475

Further exercises concerning natural boundary conditions, and more interesting ones, will be found in the next section 14. (a) Formally derive the Euler-Lagrange equation ( I ?) by choosing }•n, _P h to minimize N O T E.

}5'3, - - -

n ^} J j +1 ^a J E F ^* ff11 • 7^^ t^ (^ç+ l i=0 xj+i —

x;

-

where x u = a < x, ‹ x 2 • -

-tex„ + 1 -= b,

and

j(a) =

î

5^(x 1 ),

a. 11b) = P.

Then take the limit as tor; x ;# — xi -s Q.* N T E. In the limit, it is immaterial whether }}, t or}y,is used as the second argument in F. But the choice made here simplifies the calculations, (b) In (a) the function values at the end points x U a and x.. 1 = b were fixed. Allow arbitrary function values at these points and thereby derive natural boundary conditions_ 15. This problem yields Legendre's necessary condition for a relative minimum (maximum). that F33 = Fy. must he nonnegative (nonpositive). (a) Carry one step further the passage from (13) to (14) and derive

d 2fir cle 2 (b)

i

=

j"

fi

f 1: 22 52 -t- 2}•2I55i

+ F33(sr )2 ]

a

Suppose that F 33 (x, y, y' } is negative when x = . for some point in the interval (a, 11). Assuming that the second partials of F are continuous in a domain containing all points of interest, show that y could net provide a relative minimum to 1. Do this by considering an appropriate particular function s. for example, s(x) = C, i -x — I ? 6;

s(x) ^ I —

- 'fix— I

,

Ix —f

6; 6

C 1.

16. James and Johannes Bernoulli solved the brachistoehrone problem in 1697. Johannes wrote: "With justice we may admire Huygens because he first discovered that a heavy particle falls on a cycloid in the same time always, no matter what thestarting point may he. But you will he petrified with astonishment when I say that exactly this same cycloid, the tautochrone of lluygens, is the brachistochrone we are seeking" [quoted in E T. Bell's, Men of Matheniatica (London: Penguin, 1953), p_ l46. show that the cycloid is a tautochrone. • This is the approach used by luter, Lagrange -5 procedure is employed in the cexl_

Calculus of Variations [Ch. 11

476

11.3 Calculus of Variations—Generalizations Experience soon makes evident the need to extrerna]ize more general functionals than those treated in the previous section. If the material that has been presented so far is thoroughly understood, howeNer, its extension should present few difficulties_ A relatively terse presentation of the calculations required to obtain generalized Euler equations is the business of this section. We shall usually denote partial derivatives by numerical subscripts during the course of the derivation, and by appropriate subscript letters in writing the final answer. In part F it is more convenient to use the latter notation throughout. The first six exercises below are, in order, examples of problems requiring the six generalizations that we shall discuss. For proper motivation, these exercises- should be read over before proceeding. A.

MOR E I)ERtVA'rIVES

We wish to determine a function y that extremalizes F(x,!'. ft,

fLY)

n fix

(1

)

among all sufficiently smooth functions .f.J that satisfy _0(a)

Ao,

V(a) = A L , P(P) = B o , l''(f) = B1,

(2)

for given constants AD, AL, Bo , and B 1 . We introduce a smooth function s that satisfies

s(a)=

s'(a )=

49)= 5 '0)—

^.

Considering the admissible functions y + cs, we require that the following function 5(c) have an extremum at e O.

5(e) = fF(x , y + cs, }' +

rs, y" + r") dx.

= 0 now implies that

j:(F a s + F3 s' + F{ s ") dx -= Q _ Upon integration by parts and use of the boundary conditions satisfied by s, we obtain [Exercise 11(a)]

r.

(F2 —

d

dx

a F3 +

dx

; F4 c!x = D.

(3)

e _ 11 _3] Sc

['ufrulas of Variations--GPrterra&2orions

477

From (3) and Lemma A (Appendix 1 1.i) we deduce that it is necessary that the extremalizing function y satisfy the fourth order ordinary differential equation Fr

d dd2 F.+ -F^r a dx ^ dx#

(4)

and the four houndary conditions (2). Generalization to functions F involving the first n derivatives of y is straightforward. B - MORE FUNCTIONS

We wish to find two functions y, and y 2 that extremalize

rF(x,y1.54,y2,y2)dx

/Uri, Ÿs) among all sufficiently

(5)

smooth functions p i and $ that satisfy

.171(a) = AI , .f 1(P) _ R 1 ,

(h)

2(v) = A2, j32(() = Hz,

for given constants A ] , B 1 , A 2 , and B2- We consider admissible functions

k,

of the form ŸA ) !1x) + z;5.dx),

( 7)

where the si are smooth functions that satisfy the boundary conditions

st(a) - si(ii) = 0,

i=1,2

We make the definition

li .5 t , Z ) _F(x. yi +

J

r. tst , y;

+ r: is', , Yz + tz'z,y'3 +t z s'z )dx.

For fixed s i (x) and s i (x), .1(F 1* c 2 ) must be extremai when ri Necessary conditions are

afi ac,

when L I = O , f: 2

GE 2

=

r2

= 0.

= p-

These lead by a short familiar path [Exercise 11(h)] to the simultaneous ordinary differential equations d

F,,, - - dx

F = 0,

1,

2-

The generalization to n functions should be evident.

(8)

CRk uh1S üJ Variations

47S C.

[Ch. I 1

MO RE INDEPENDENT VARIABLES

We wish to determine a function z(x, y) that extremalizes F(x, y, f, 2,,, 2,) dx

}(f)

dy

(9)

R

among all sufficiently smooth functions of two variables 7 which are prescribed on S, the boundary of R. Consider admissible functions f of the form s = z + Es, where s is sufficiently smooth and s =0

on S.

( 10)

Our usual argument requires that .51"(c) vanish at r = C, where (^}

1

fF (x, y, z + cs, z i + rs, z + E.5,) di(dy.

(li)

Thus it must be that

1J

5F3

+ sF 4 + ,F 3)dx dy 0.

(12)

Transferral of the derivatives on s to derivatives on F, and F 3 is required if we are to proceed as before, To accomplish this, we define a vector function V by V = Fo i + F j.

(13)

Using V, the terms of interest in (12) can be written Jjs F4 + s y F 3 )dx dy = jf(V.s.V)dxdy. R

(14)

f^

But because of the identity [Exercise Mc)), V•(sV) = Vs• V + sV V,

(15)

we find, using the divergence theorem, that

ff

(Vs • V)dx dy =

—

f(s0 - V)dxdy +

(n sV)da, -

(16)

R

where n is the unit exterior normal to R. The line integral in (16) vanishes because s 0 on S. With all this, (12) becomes

Jfs[F3 ^

aF s

,

-

^

-

^ [ix dy

=

D.

(17)

Ser. f 1.3]

[ ahuJas of Variations —Generalizations

479

The two-dimensional version of Lemma A shows that the square-bracketed terms in (17) must vanish (if they are continuous), so that—switching to letter subscripts— d —F = — =0 . (18) Ox FY ^F zti The boundary condition (10) must also be satisfied. [It should he clear that FZM denotes the partial derivative of F(x, y z, z x , z y ) w ith respect to z. keeping x, y,,z, and z, constant. In taking a/ex the variable y is held constant. Compare Exercise 8.] Conditions sufficient for the validity of the formal calculations leading to (18) not only limit the class of admissible functions and functions F, but also require the boundary of R to be smooth enough so that all the integrals exist and may be suitably manipulated. If Jl is an integral of a function z of three variables x,, x 2 , x 3 and its first partials z,, z r , z 3 , then [Exercise 12(b}] (18) generalizes to ,

3

F#

a {

r= ` x^

F

,

}

=0

,

zf =

^

c7 z +dx

D. 1NTFC,RAt. CONSTRAINT

We wish to extremalize f

(v} _

^a

F(x, y, 37) dx

(19}

among all sufficiently smooth functions j) that satisfy the houndary conditions !'(Œ( = A,

Yî^^) = B,

(20)

and the constraint -

(l') _

a

4 (x, 37, j7) fix = C,

Ca given constant.

(21)

Denote the extremalizing function by y_ We can no longer proceed by considering admissible functions of the form y + Es, for if s is a fixed function we cannot expect y + Es to satisfy My + Es) = C for an interval of E va lues containing O. Only when such an interval exists can we proceed as before to require a vanishing derivative with respect to ti at E = 0_ Instead, we write y(x) + E l s,(x) ♦ E2 (x), where the s,{x) are smooth and satisfy s,(a) _ $0) 0, i = 1, 2. Regarding s, and s2 as fixed, our problem is to write a condition implied by the fact that .^(E I ,E Z ) _

j F(x,1' + E

o t + E2 S2 .y . + F t S, + F2 s3 )dx

Calculus of Variations [Ch. II

4 8O

has an extreme value at 1 i = E 2 = 0 among all nearby (s,, e2) satisfying j(E,, s z ) = C, where

=G(x, y +

+ Ea sa, Y' + c,si + s2sfd dx.

f

Using a Lagrange multiplier A, the procedure of Section 11,1 leads to 0

0 (.I — Ai) = (.o — Ai) = 0 as t -s 3

at t 1 =-

E2

= 0,

(22)

unless j,,( 0, 0) = /rr(O 0) = 0. ,

(23)

From (22) we can deduce in the usual way (Exercise 11(d)] that

f [

a

^ F. dx

F2

^ Gs

--

—

--

^d G3 x

,^jx) dx = 0,

i

= l, 2-

(24)

We must now be a little careful Equation (24) was derived under the supposition that s i and s2 were fixed. A may thus depend on s i and s 2 . But (24) with I = I yields F2 —

d dx

F3 , (x) dx

G2

d dx

fr3 t(x) dx•

so A is independent of s 2 . (The possibility of a vanishing denominator in the above equation is essentially disposed of in Exercise 13.) Equation (24) with I = 2 is now seen to be of a form allowing application of Lemma A. Thus [Exercise 11(400 y(x) and a. must satisfy

Fy --

dF

^ . — dj G ,, —

d^ G ,

= 0

(25)

plus the boundary conditions (20) and the constraint (21). The exceptional case (23) is the subject of Exercise 13. Note that (25) is appropriate to the unconstrained extremalization of I (F — ).G) dx. By straightforward generalization of the argument used here (Exercise 14), one can show that if Jâ F(x, , ÿ) dx is to be extremalixed among smooth functions 5 satisfying the boundary conditions (20) and the N constraints I:0 ` 1(x, }y, ÿ') dx — 0,

i = 1, 2, ... , 1'If, (26a)

then 1y and the N Lagrange multipliers A; must satisfy Fy —

dx

F—

^- 3

^ C;'^ — d G;) = 0 dx

(26b)

Set. 11.31 Calculus of Variations—Generalizations

q8 i

plus the constraints (26a) and the given boundary conditions y(13) = B.

y(a) = A,

(26c)

Moreover, by combining previous arguments, we may readily generalize to integrands that contain more derivatives, more functions, or depend on more independent variables. E. FUNCTIONAL CONSTRAINT*

We seek to extremalize

FO. xi, x3,x3, x t,z2,xs] dt

(27)

^o

among smooth functions that have prescribed boundary values xAto) = a1+

x4(11 ) =

= 1, 2, 3,

(28)

and satisfy the constraint G(t,

(29)

X1, X2, X31= 0.

Because the constraint does not involve integration, our previous line of attack is inappropriate, for it will not reduce the problem to one involving the extremalization of an ordinary function [Exercise 14(c)]_ We expect a result involving a Lagrange multiplier, and we obtain it by the approach often used in the derivation of such results, use of the constraint to eliminate a variable. (Compare Exercise 11.1.15.) Assuming that the hypotheses of the relevant implicit function theorem are satisfied, in particular that c GJr3x 3 # D, we employ (29) to express x 3 in terms of the other variables: (30)

x3 = f(r, x,, x 2 ),

Using this, our problem becomes that of determining the unconstrained extrema of f,

fl

0(tx1, x2 , xI , x 2)di,

(31 )

u

where it)(t, x 1 , x 2 ,

12) �

F t , X 1, X2+

1 (^, xl, x2>^ xt+ ^2t ^ i ( i ,

Xls

X2)

•

From Part B we know that the x ; must necessarily satisfy the Euler equations é

= 1, 2.

(32)

In this case it clutters the notation unduly to make explicit the distinction between a general admissible function and a particular exurentalizing function. `

482

Calculus of Variations

[Ch_ i

In computing (,b Y,, we must remember that d dt

+

f(t, x t , x2)

When i = 1, the Euler equation (32) gives, in terms of F,

F= , + F f + Fx,

d (Fr. + F= 1 [ 1 ) .=0 — dr Ox t dt

or d d f.. (F„, -- Fx , -}- f Y , — F;^, dr dt

= O.

(33)

But, using (30), we have that G[t, x i , x 2 , f{t, x t , x 2 )] = 0 identically in t, x t , and x 2 . Taking the partial derivative with respect to x i , we find

G„, + G I , f= , = O.

(34)

Consider (33) minus times (34):

di

, ti - AG xti + F x, ` dt x , — D. F x' T AG

(35)

Using our assumption that c'G fr?x 3 # 0, we can define a fe ncrion A(r) by

F — (dirdr)F i ,

(36a)

Equation (35) now implies that

F.

de

F^ — ,

l^Gf

(36b)

x,

'The equation

F.

d dt

F x2 — AG, L3 = 0

(36e)

can be obtained similarly, by considering {32) when i = 2 [Exercise 11(e)]_ Combining (36a), (36b), and (364 we can summarize our results by stating that necessary conditions on the extrernalizing functions and on the Lagrange multiplier function ,i(r) are the Euler equations Fx ,

d F'x, —C x,, di

i

— 1 , 2, 3,

(37)

plus the constraint (29) and the boundary conditions (28). Since Gx - 0, these results can be expressed in a manner that we have corne to expect' the equations (36) are appropriate to the unconstrained extrernalization of F — W. But remember that here A is a function, not a constant.

Sec. J 1 .31 Calculus of F'arrarions

Generalizations

48 3

The above derivation assumed that 060x 3 A C. if this were not the case, but if either r G/r x i 0 or r3G/ x x # Q, we could still obtain the same results by eliminating x i or x 2+ respectively. Thus an alternative to (37) is the possibility that

i = 1, 2, 3. Note that (38) implies that raG/i t

(38)

0 on the extremalizing curve since

G[r, x 1 (1), x 2 (t), x 3 (t}] — 0 implies that

. G r — 0_

+

F. END POINT FREE TO MOVE. ON A GIVEN CURVE

We wish to find a curve extremalizing a certain functional when that curve is required to begin at (a, b) and to terminate on a given curve G(x I , x2) = 0. Let all smooth curves satisfying these requirements be given parametrically by

xi =

x, r), (

c

1, 2, 0 < r

1.

Given that

G[. 1(l), x 2 (l}0 = 0,

Xi(0) — a, x 2 (0) = b,

(39)

consider the problem of extremalizing

x2) _

FEE, x1(t), ,gi(t), xx(t). rx(t)i dr,

where '

:1

di

(40)

REMARK. There is no loss of generality in assuming that the parameter r has the special values 0 and 1 at the beginning and end, respectively, of each admissible curve (Exercise 16). Denote the functions extremalizing (40) by x i and x 2 . Let s and 5 2 be smooth functions defined on [0, I] that satisfy

5,(0) = 0,

(41)

1 = 1, 2.

Then for any such functions s i and s 2 , regarded as fixed, it must be that

e2 ) _

J

F[t, x i

+ e i s,, x 'i + c,s xa +

has an extremum at E1 = E2 = 0 among all

El

E=sx, x 3

+ & zsi] di (42)

and E2 that satisfy the constraint

J(E t , E 2 ) r G[x.(1) + f i s t (1), x 2 (1) + E2 S2 (1)) = 0.

If (0,

0) — jr ,,(d, 0 ) ° 0 ,

(43)

Cîikulus of Variations [03_ i i

484

that is, if Gf:[x1(1), x 2(1)) _ 0,

G=.[xi( 1 ), x2(1)1 = 0,

(44a, b)

we have seen that the usual Lagrange multiplier scheme does not work. Assuming for the moment that neither equation of (44) holds, we have

r?€i

[- (EL, r2) — J(r^

,

e2

)J = 0 at E I = £ 2 = 0,

i=

1, 2.

(45)

On integrating by parts in the usual manner [Exercise 11(f)], we deduce that

^

= f st(f)(F

f,

!

O

d

Fxi di ♦ 3i(1 )1* si

LÎx i

— ).G,Sxt(I), x 2 (1)1s,(1),

1= L

t = 1, 2.

(4.6)

We proceed as in our earlier discussion of natural boundary conditions. We know that (46) holds for any of the class of smooth functions s i and s 2 which satisfy the boundary conditions (41). We investigate, in turn, the consequences of (46) for particular subclasses of such functions, Employing an argument of Part D, we first note that, (46) with i = 1 shows that À is equal to a ratio that is independent of s 2 . Equation (46) with i = 2 certainly holds for those functions s 2 which satisfy s 2 (1) = 0, and from this we deduce that d Fx2

dx z

F — 0.

(47)

Equation (46} with i = 2 also holds for functions s 2 that do not vanish at i — 1, so [using (47)1 we infer that at ! = 1,

Fxi — AG =2 = 0.

(48)

Similarly,

F

d

dx Fr = 0,

(49)

L

and at t = 1, F L..— ÀGx = 0.

(54)

If Gs , and G,r , do not vanish air = 1, we can combine (48) and (50) to obtain Fx, G.

FL . att=1. Gx,

(51)

The two unknown functions x i and x 2 must therefore satisfy the two second order equations (47) and (49) and the four boundary conditions (39) and (51).

Sec. 1 1.3] Calculus of Voriariotls --Generalizations

485

The transversaüty condition (51), in relating x t and x 2 to the components of VG, fixes the direction in which the extremalizing curve approaches the given terminal curve G = O. (In other words, the exact nature of the transverse approach is specified.) This condition was derived under the supposition that neither (44a) nor (44b) is valid. If just one of these equations holds, however, we know that at the end of the extremalizing curve, the end-point curve G = 0 may be, respectively, horizontal or vertical_ If both hold at a point (or if G does not have continuous partial derivatives), G may not have a well defined direction when it is met by the extremal. In all these exceptional cases the location of the possible meeting point is known. -

Example_ For the brachistochrone problem (where now - = didt) F = (if + .rÎ1 1 1 (2Y12) 1 2

so at t = 1, Gx ,

F,u

.ft

VG =

or

Gx

^ ix t • tea)

02 0 ü).

(521

Thus when the particle reaches G = 0 its direction is normal to that curve. in agreement with the remarks made in Example 22 in connection with the natural boundary condition for this problem. EXERCISES

+I. According to the theory of elasticity, a thin homogeneous cylindrical rod compressed by a constant longitudinal force P al both ends is in equilibrium if its potential energy is an extremum. Suppose that the rod lies along the x-axis for 0 < .x S L and let y(x) denote lateral displacement (see Figure 11.7). It can then be shown, as at the end of Section 5.1, that the potential energy corresponding to a deflection ÿ(x) is L

[E!0")2 - P02 ] dx,

(53)

where E is Young's modulus for the material of the rod (a stiffness factor) and ! is the moment of inertia of a cross section. In this problem

L

^

FIGURE 1 L7. A rod under compression.

486

Caicufus af Variations [Ch_ 1 1

we shall assume that L and I are constant and will use K as an abbreviation for P f F#_ (a) Find the Euler equation necessarily satisfied by a function y which extremalizes (53) subject to the imposed boundary conditions y(L) = y'(D) = y'(L) = 0.

y(0)

(54)

[Conditions (54) are appropriate to a bar with ends "built in" so they cannot move or bend. Note that the slight compression due to P is ignored in the boundary conditions] (b) Suppose that both ends of the bar are hinged, so that although their position is fixed [y(0) = y(L) = 0], they arc free to bend. Find the appropriate natural boundary conditions. Solve the resulting problem. Show that as K increases, a critical value is reached at which a nontrivial solution of the problem is possible. (The connection of such critical values with the phenomenon of buckling is discussed in Section 5.1.) (c) Suppose that the end of the bar at x = L is completely free_ What are the appropriate natural boundary conditions? 2, This problem concerns the motion of a particle of mass m due• to a conservative force F(x 1 , x 2 , x i ). We posit the existence of a potentialenergy function V(x 1 , x2 , x3) such that the ith component of F satisfies F; = Let T denote the kinetic energy of a motion with position components x i at time t, so that T`

?TOE,2,

d dt

The difference between the kinetic and potential energies is called the Lagrangian L: f r L(x 1 , x 2 , x 3 , x 1 , x 2 x3) = T(xl, X2 x3) ` V(x l, x2, x3). ,

Let xj = ,(r) describe the trajectory of a point that leaves a at time t i

an d arrives at p when t = t 3 so that x(t, ) = txi 41 2 ) = /30 Hamilton's ,

principle asserts that the particle will move on that path which renders stationary the quantity

Jr:r, ^(xl^

g2,173,

:Xi, X-2, k3)

[^t .

That is, from all conceivable trajectories which start at a when t = t 1 and end at p when t 1 2 , nature selects the trajectory x i = xi(t) which renders stationary the integrated excess of kinetic over potential energy_ (a) Show that Hamilton's principle implies Newton's laws of motion. (b) Generalize to a system involving many particles. (c) Show the converse of (a),

Sec_ H .3] Calculus of tfariunof,—Gen ralizalions

4E7

3. (a) Suppose that a fluid of constant density p is confined to a region R. and that a velocity potential 0 exists, so that the velocity v satisfies v = V4. Suppose further that the potential 0 is prescribed on the boundary S(R). It is desired to determine 01 in such a manner that the net kinetic energy f f l R pv • V - dr is minimized. Show that V 2 ç = 0 is a necessary condition. +(b) Suppose that the kinetic energy is to be minimized without restriction of 4) on the boundary_ Find the natural boundary condition and give a physical interpretation of your results_ $4. A flexible cable of length Land constant mass per unit length p hangs so as to minimize its potential energy. Let the ends of the cable be fixed at (O, O) and (a, b), where a > O. We must minimize

J

o

p9^l+(^`)^^

tl3 dx

among all smooth functions lv that satisfy 5'(0) = C,

jlu)

= b.

and

f [1 + ^ÿ,Yr2 dx —

L

Here g is the acceleration due to gravity and y = fix} is the equation of the cable. Use a Lagrange multiplier to find the shape of the cable. That is solve the relevant differential equation but do not attempt to determine the arbitrary constants. That matter is treated in Exercise 24. $5. Given two points on a surface, the geodesics joining the points are curves whose lengths are extrernal compared to nearby surface curves joining the points. Thus the geodesics of the surface G(x i , x 1 , x 3 } = 0 are found by extremalizing ,

[xi(t) + .q(i) + x3(t)]iri dt, subject to the constraint G(x t, and to the boundary conditions xi(tc) — a-,

x ; (i t) = b„

E = 1, 2, 3.

Show that the principal normal to the geodesic curve at a point coincides with the normal to the surface at that point. (Recall From differential geometry that the principal normal n satisfied n = + ' aids, where dxfds is the unit K is the curvature. s denotes arc length, and t tangent vector.)

Calraus of Varial io ns

4 86

[C IL 1 I

6. (a) Show that the shortest plane curve starting at (0, O) and ending on the given curve Glx, y) = 0 intersects G(x. y) = O at right angles. (b) Solve the problem completely when G(x, y) xy 1_ Sketch. $7. Write out (4) a (a) When F(x, (b) When F(x, y, y' yu ) = (y")2 m x(y') .2 + y 3 . $5l, (a) Given that —

,

F(tz, f3, y, ^ ^:) = ally + exp (ô) + s 2 , ,

if z is a function of x and y, find F3(x, y, z, z+ z)

a

F4(x, y, z, zx* $ y ) —

a

f Fs4x, y, z, Tx,

zy).

[Compare (17).] (b) Write out (18) when F(x, y, r, zx , z,,) = xz sin (zx). 9. Generalize part A to the case where the function F of ( l) depends on and (a) its first three derivatives, (b) its first n derivatives. 10. Generalize part B to the case where instead of (5), one has

f(}',.}2 Y3)_

fa

F(x, }i.}LIYi..92,Y2.yi.}'u, A. }+3}dx.

11. Firl in the details necessary to obtain the following equations: (a) (3); (b) (8); (c) (15); (d) (24) and (25); (e) (36c); (1) (46). 12. Generalize part C to the case where the integrand of (9) depends: (a) On z(x, y) and all its first and second partial derivatives. (b) On a function of three variables and its first partial derivatives. $13. Show that (23) implies G y — (dfdx)G„. _ O. Discuss. 14. (a) Generalize the results of part D to obtain the conditions (26). (b) Generalize the results of part D to the case where the integrals depend on a function of two variables and its first derivatives. (c) Show that previous methods will not work in part E by writing the analog of (22) and attempting to obtain the analog of (24). 15. What are the appropriate natural boundary conditions $(1)) In part D? +(a) In part 13? (c) In part E? +16 l .et x,— x,t), octe t

fi,

i = 1,2,3,

be the parametric equations of a curve. Find a new parameterization that preserves all differentiability properties and for which the parameter runs between zero and one. 17. Recall (Exercise 3) that Laplace's equation V20=0

Sec.. !!.3] C â1adiu.s of Variations— Grnerufizariu ►rs

489

is the Euler equation corresponding to the extremalization of the Dirichlet integral, F

5J

dA —

t

^V^^^ dA.

A

A

Write Laplace's equation in polar coordinates by making the change of variables in the integral and computing the new Euler equation. X18. Show directly that the harmonic function of Exercise 17 actually minimizes the Dirichlet integral. 19. Consider a particle performing planar motion under the influence of a central force field. Lei the particle's position at time t be given by the polar coordinates r(t) and 0(t). (a) Show that the Lagrangian of Exercise 2 is now L = m(^z + r 2192 ) — W(r).

(b)

20. (a)

Show that Hamilton's principle (Exercise 2) yields Euler equations which represent the radial equation of planetary motion plus Kepler's second law that the radius vector sweeps out equal areas in equal times. Use Hamilton's principle (Exercise 2) to determine the equation governing u(x, r), the perpendicular deflection of a point x at time t, of a string of constant linear density p vibrating between Fixed end points at x = 0 and x = L. Here the potential energy V is assumed to be proportional (with proportionality constant p, the tension) to the increase in length compared with the length at rest. For small deflections ( f u,, I C 1, p x constant) show that V — 4,0

^

u Y dx.

^

Find T and derive Purr

—

=

(b) Generalize to the vibrating membrane where V is proportional to the change in area 21. This problem is concerned with a variational principle that yields the differential equation and boundary conditions for the two-dimensional irrotationai fl ow of an inviscid incompressible fluid above a rigid plane at y = 0 and below the free surface y — 10, t). [See C, Luke, J. Fluid Mech. 27, 395-97 (1967).1 I1 x, y, t) is the velocity potential of the two-dimensional flow, we wish to extremalize ^.

Cafct,fus of Variations [ Ch. 1

490 v

FiuuRt 1.1.8. The two-dimensional mariun of a fluid can be described by a rleloeiry potential 4)(x, y, r). The upper boundary y ; 14x, r) isdepicted here in a space time plot.

among all smooth functions and Ii, where has prescribed values when x = x r , x = x 2 , i = t i , and t = 1 2 . Here g is a constant, the gravitational acceleration (see Figure 11.8). (a) Suppose that and h are the extremaiizing functions. Consider J40 + 01, h + s 2 H], where $ and H are such that + tï 1 $ and h + E2 H are admissible. Deduce g,^/ ,j, ^ (^^_ + 3Wy + 4`r + 9^ )y =h r2

x

= (x j

x,fh,J U

0,

(58)

d3- dx dt = 0.

(59)

H dx

di =

l

(4x^, + cr +()

Explain your procedure. (b)

To modify (59), use the formula

JJJv. VeD dr = —fff(V•r}0dr + JJain - y ckr, R

R

OR

where

V ^i ^^+j^t f k ^

Y

(60)

sE'r. 1 1.3] Ca !r•ulus of Vuriutir++u --CrNnerali_ari:+ns

49 1

and R is the three-dimensional region depicted_ (Observe that x, t, and y play the roles usually played by x, y, and z; otherwise (60) is just a three-dimensional version of (16).] Deduce from (59) xs

^ -fr ,

—

J

r1.3^,

(C.0)3,= D

fix di 4

^,

r.

Jx

[v( — ^i : ^ r "

,} ^y y hr + W1W JS h (IA dt

^

ra

r, ^si

^

^^

(^__ + ^Yr^ d 3° fix (If = 0,

ti^ ^ (1 + h!,+ h? ) (61)

(c) From (58) and (61), derive a differential equation and three boundary conditions_ Explicitly and carefully state your reasoning_ 22. +(a) Show that of all plane closed curves of given perimeter, the circle encloses the greatest area. (I'or our purposes, it will he sufficient to show that the circle extremalires the area and to argue geometrically that this extremum must be a maximum.) If A is the area of a curve bounded by C. Green's formula

c(^ _

^^ Q dy) _

fc(P

cY 1r1x-+ ^idrl

c x ay f

A

can easily he specialized to give the result A ï ;4(x dy -- y dx). Use the parametric form of this equation_ (b) Using (a), consider any smooth closed curve of length L which encloses an area A. Deduce the isoperimetric inequality

A
y, respectively, while Snell's law and the value of y emerge from natural boundary conditions. (d) Compare the advantages and disadvantages of the derivation of Snell's law presented here with that presented in elementary physics texts. fe) Consider the simpler problem of extremalizing J when • is a given fixed constant_ Derive a condition on the discontinuity of y'(x4. This is called the Weierstrass Erdrnwin vertex condition. 24. Exercise 4 reduces the hanging-cable problem to the following question: Is there a unique member of the three-parameter family of catenaries (b)

y — C' 1 =C_; cosh

—

C 3C'

)

that passes through (0, 0) and (a, b) and has length L. L u 2 +-62 ? Following Section 6.7 of Pars (1962)* we outline a proof that the answer is affirmative. We ask the reader to fill in the details. (a) Consider chords of the curve yt = cosh _x that make an angle = tan - I (b/u) with the x 1 -axis. Denote the length of a chord and the length of the arc below it, by 2P and S. respectively. Let R = S/21'_ Show that R varies monotonically from one to infinity as P varies from zero to infinity by deriving the following four equations (the notation is supplied by Figure 11.10)_

w•i = cosh

FIG u R E 1 1.10. The pant (4,g) bisects the chord of the cuterlrlry y t = coshx i •

CQ1rWfus of Varia

494

(1) (ii) (iii)

i f P sin a = cosh ( + P cos a). P sin a sinh sinh (P cos a). S = sinh ( + P cos a) — sinh ( — P S y a ---=sin + 2P

(iv)

r

(ORS

[Ch. 11

cas et).

sinh (P cos al s cos 2 a. f cos a

(h) By (a) there is a single chord for which R has the correct valut L/(u 2 + b 2 ) 1 '2. Show that this implies that one can then solve the problem by a change of scale and a translation. 25. (a) Consider the problem of extremalizing

_

_

fifE

^t

^t Ni r^tiz

u r , Lï^, u^, x1 ^ , dx2 , ^x3,

^û3^

xl ,.. ., r^x3 axis

(LX ]

dx2

dx3

A

among all smooth functions i L û 2 ie 3 having given values on the boundary of R. By slightly generalizing Part C show that the extremalizing funct ions u i necessarily satisfy ,

,

3

P., -- y (' )• = 0, •= 1 x J (b)

where

OU;

u

"

^

^x}

Suppose that u 1 , u 2 , and u 3 are prescribed on a part OR of the boundary of R but are not prescribed on the remaining part aR 2 Suppose further that one wishes to extremalize the integral of part (a) minus

1{4 uk da, t'R

where t is given on the boundary. Show that the equation of part (a) still holds, but in addition there is the natural boundary condition 3

tr =

}=L

Eu„ rti on OR 2

26. Consider the function u that extremalizes the potential energy [given in (4.4.18)1: 3 V(E) V ()=

^

R

moi dx1 dxa dx

3 }^r/ + ^3 (^ ^^ rr2 ^

I.

L

where e rj(u) — -'{u r

+ ai. d,

f and t are given.

3

dcr, y liur,

set. 11.31 Calculus of Variations—Generali=urio,ts

495

Suppose that u is prescribed on a portion OR ] of the boundary. lise the results of Exercise 25 to derive the equations of elastic equilibrium

3 dT,

ox;

L

(

wher

and the boundary conditions

a

e Tif -—^;^ ^ + .^^;^

^

^^)

t=1

3

EI Tej— t j on ÔR 2 .

u, given on aR 1 ,

j=

27. The aim of this exercise is to show that Hamilton's principle for particles (see Exercise 2) extends to continuous elastic media To this end, consider the extrernalication of

$12

i (u) Y{u)l df,

,

where the potential energy V is given in (64) and the kinetic energy T is defined by 3

T(û) = l p fff

3 rtx , dx 2 dx 3

auk)

R

(p = density, a given constant). Using the same boundary conditions as in Exercise 26, show that the same results are obtained except that the equilibrium equations are replaced by the dynamic equations 0214 ; 3 aT . + 2 — 1 ôt ax, P f-

28. Let A ; and y; denote eigen' alues and corresponding eigenfunctions of 28. the Sturm- Liouville problem, Ly—,y,

10) = .013) = O.

Here

L= --d p —d +g dx dx p and q are positive and smooth, +l t

Jn

F(x, y, y') dx,

)2

,

).3 - • • . Let y minimize

where F = p15702 + q$5 .

0

subject to the constraints

IFI6 G{x, j , $5 ) dx = 1 and '

a

J

G,{x, K $') dx

O

,

i = 1,... , N — 1,

Calculus u/ !luriwio ►ns Ef'h_

496

r

^

where G' and G, = YYr (a) (b)

Using Lagrange multipliers show that y is an eigenfunction of the Sturm Liouville problem. Employ the fact that eigenfunctions are mutually orthogonal (I, Chapter 5) to show that y is the Nth eigenfunction. (The variational characterization of eigenvalues is extensively considered in Chapter 12.)

Appendix 11.1 Lemma A Lemma A. Suppose that G =

(1)

G (x)s(x) dx 0

!

P

^

for all functions s possessing n continuous derivatives (n any positive integer) and satisfying

(a) = 0,

s(0) = O.

(2)

If G(x) is continuous for i < x < 13, then G(x) M O. Proof. If G(x) # U for a ' x z j , there exists a point x 0 , Œ ç xu 5 13, such that COO # O. Suppose that G(x 0 ) > O. Then (with slight modifications if x 0 is an endpoint of the interval) if G is continuous, there exists a positive number p such that G(x) >0

fer Ix —

< p.

As we shall show in a moment, it is not hard to construct a function s* which has rt continuous derivatives, satisfies ( 2 ), is identically zero for l x — x 0 1 p, and is positive for rx — x o l < p. But then [taking p small enough so that (— p, p) is contained in (a, e)t, dx

^ G(x)s * (x.) dx > 0,

J xt - p

contradicting (1). The possibility G(x e ) < Q is handled in the same way, Thus G(x) 0_ A suitable function s* is given by s*(x) s'(x)

0,

I x— Xi] f o p [133 — (x -- x 0 ) 2 _11' ,

Ix — x0 1 < p.

If m > n, at least n derivatives of s* vanish at x = x e ± p, ensuring the required smoothness_ G

Appendix 11.21

Variational ,Natation

497

The lemma and the proof generalize in an obviousway to many dimensions. Note the similarity between Lemma A and Dubois-Reyniond lemma that was used so often in I. Lemma A is also associated with the name of DuboisReymond—but here, such attribution would obviously make for ambiguity of reference.

Appendix 11.2 Variational Nutatian In common use is a notation that emphasizes the analogy between extrernalizing a function and extremalizing a functional- First consider a function y. Recall that the differential dy(x o , dx) is by definition the first order change in y(x) at x o due to an arbitrary change dx in x, i.e., dy y(x 0 ) dx. Thus a necessary condition that y have an extremum at x o is 0 at x - x 0 for arbitrary dx. dy(x, dx) To extremalize the functional I(y) - jâ F(x, y, y') rlx cf (11.2. 5), we make a change in notation and introduce the twice continuously differentiable function by instead of the function s of (11-2.9 j and (11.2-10), where by -= s.

(1)

so by(a) = bu(1) 0.

as the lowest order approximation

We define the first variation bl(y, by) to A/, where

AI : I(y + by) - I(y). "Lowest order" means that only terms linear in by and its derivatives are retained. But using by = s and regarding ❑ I as a function of E for fixed s, we have

AI= I(y-}- Es) -I^):

E

df(y + rs) de

=0

The required first derivative is precisely what was computed in our original approach to Problem A of Section 112 Using the result (11.2.14), we therefore obtain ^

I(y + t:s) - 1(y) =

E

(Fr s + F,. s') dx +

-

Thai is

bI

y r)by + F, .(x, y, Y + 1hy'] dx. .

= ^a

( 2)

Necessary for 1 to have an extremum at y yo is the condition bl y, by) = O at y = yo for arbitrary by; and the Euler equation (11.2.21) follows from this condition as in Section 11.2. Hence if one makes a correspondence between

Calculus of rariu+iuns IC h. 11 dx and y, and between dy and 5i, then extremalizing a functional is seen to be closely analogous to extremalizing a function. Formal manipulations are easy with variational notation, because of the close link with ordinary differentials. For example. if f , and I 2 are functionals, then the following rules hold: ± 1 2 ) = ô#, + 61 3 ,

pi)_

6(1 1 1 2 ) — l i I a + 12 61 1 ,

l 3 bU t t 1051 2

(3a,b,c)

Iz

f2 To prove (3b), we define

ff', ', } r (F.} = { [LY OE) + Es)/ 20)+ Fes'} — ) I(Y) 1 A(Y) and note that to lowest order ^ (E) ^ F 1 1(Y) ^

t

Go '{0)

(E)

120 + FS}

s- 0

or

+ } z(y)

dE

1^(Y + a:-s}

Proofs of the remaining formulas are left to the exercises. As another example of the natural character of the formal ruses, we observe that from (2) we can deduce such formulas as

t+ f

y" dx =

a

bf

%1yR - 1 by dx, Q

fi

sin (y') dx = f cos (y') b y' dx.

a

(4a, b)

Q

Example J. F ind a necessary condition that A he a stationary value of f i1ir2, where

ii=

fPly'3 1 +

=

dx.

rr r}' 3 dx.

lsl

Here p, q, and r are given functions caf x, r > O. The extremalizing function p must not be identically zero hut must equal zero at x = a and x Q. Solutiurr. The First variation must vanish. Thus, since .i = i,tj'.,'11i 1t .), we have

r

f ____ — 161 2 )

r 61 1 —

Since I, 34 0, we infer that r5i , — ,lfll = [1. Hence

j {2pyy'

4

20,4

—

22r OA dy = 0

L^l

for arbitrary srnoni h functions by that vanish at a and fl. Performing an integration by parts on the first term, we deduce in the usual manner that —(py'l' + lq

—

.lrjy = [l,

i(a) = ytjf1 -• O.

l7l

Appendix f }.2]

Variational Aü,rurton

499

Note that the eigenvalue problem (7) has been shown to be equivalent to the problem of rendering stationary the quotient of integrals] 1 tf 2 . This provides an example of a matter to he extensively discussed below. It is useful to know something about the variational notation, since it is used by many authors. On the other hand, the formal elegance of this notation can seduce the unwary into an unfounded belief that they understand the elements of the calculus of variations when, in fact, they possess only manipulative skill. We therefore believe that the more basic approach of Section 11.2 should be used at least until the subject is thoroughly understood.

EXERCISES

1. Prove (3b) and (3c) 2. (a) Verify (4a) and (4h). (b) Find c f! exp 3. (Chandrasekhar, 1961, Chap. 2.) Let F - (D 2 — kIG z (D 2 — k 2 )2 I1 where k is a constant and D ; d/dz. Show using variational notation that (D 2 — k2)F = — Rk 2 W

is the Euler equation which provides the required condition if R is to be a stationary value of 1 i /u 2 1 2 , where ,/,( W)_ Jr i_(DF) 2 + k 2 F2 1 dz, o

1 2(W) =

Je G

2 d2.

The function W that renders R stationary is required to satisfy these boundary conditions: at z = 0 and at z = 1, W = 0, F = D, either DW — 0 or D2 W = O. 4. Show that the first variation of the integral I defined in (2.5) is unchanged if a total derivative of the form d

dx Ox,

^]

is added to the integrand_ [This nonuniqueness of the functional whose extremalizati ❑ n leads to a given Euler equation can sometimes be used to advantage. See, for example, p. 9 and elsewhere in L. Landau and E Lifshitz, Mechanics (Oxford: Pergamon Press, 1960).]

CHAPT ER 12

Characterization of Eigenvalues an d Eq uilibrium S tate s a s Extrema

I

N CHAPTER l 1 we showed how variational problems could be reduced to the solution of boundary value problems in differential equations. In this chapter we proceed in exactly the opposite direction by showing that there are often advantages to casting problems in differential equations into variational form. A similar duality is present in the relation between differential and difference equations: Numerical analysts make their living by approximating differential equations by difference equations, but we saw in Chapter 3 of I that it can be useful to approximate difference equations by differential equations. For certain classes of variational problems, determining the EulerLagrange equations leads to a solution, since these equations can be readily solved, either analytically or numerically. In other cases the reverse procedure is useful, for an extremalizalion problem can be handled directly, without resort to a translation to differential equations_ After an introductory section the famous Ritz method for direct solutions to variational problems is thus described in Section 12.2. A variational formulation also permits ready deduction of certain qualitative conclusions. To illustrate this, the Courant max rein principle is used in Section 12.3 to demonstrate several qualitative features of vibration problems (as well as to permit quick proof of the essential feature of the Ritz method)_ Section 12.4 shows how variational formulation of certain positive problems in partial differential equations leads to simple uniqueness and stability proofs as well as to efficient calculation methods. The chapter concludes with a lengthy appendix on certain basic ideas from functional analysis that are useful, although not essential, in pursuing the topics under investigation here. -

12.1

Eigenvalues and Stationary Points

THREE STATIONARY VALUE PROBLEMS

To begin our considerations, it will he worthwhile to consider three instances in which stationary value problems lead to eigenvalue equations. (By definition, functions suffer no first order change al stationary points, i.e., their first derivative vanishes.) In the next section we show that for an impo rtant class of positive problems, the eigenvalues can profitably be characterized as minima, not just stationary points. 500

Sec. 17.1] Er9envalues und Sraiwnury

Puinrs

501

Problem A Determination of points on the quadric surface 3

E Ark r ac ; = 1,

r.,-

where A u = A i

(la. h)

I

that render stationary the squared distance to the origin i: 21 } x-- 2z + V-3 _ (Compare the particular case treated in Example 11.1.1_) Employing a Lagrange multiplier A 1 , we find that at the stationary point (x i , x 2 , x 3 ), we must have 3

0X t

—A

'LA; = p,

'

i= I

k

1, 2, 3.

r.j - t

( 2)

If we use the symmetry condition (lb) and divide by 2, we obtain 3

3 ^

;=1

A,,, xf = ;xk

y A dl x# x 1 =

Also,

j.j =

I

,

k = 1, 2, 3.

(3)

t

This is the much studied problem of determining the eigenvalues of the symmetric matrix with components A r (see Section 2.2), The eigcnvectors of A id , with lengths chosen to satisfy (lak give the stationary points. Problem B Determination of the unit vectors n = (n 1 , n 2 , n 3 ) giving stationary values to n • On), the normal component or the stress I. In terms of the symmetric stress tensor* with components To we must render stationary 3

X I1 t TJ fi,

i. j=

Oa)

l

subject to the constraint

(4b1 Upon introducing a Lagrange multiplier )1., we see that (4) requires finding components rii that satisfy a

3

drz k i, i=E1

3 \

i1r TfF} — EF; = 4 i= I

or 3

E Tioni = A n.% and -1

E

= 1.

(5)

4= 1

Readers who are unfamiliar with the stress tensor can just observe that (4) linplics (Sj and (b)-

502

ChQracterizalion oj Eiqenr:atues

and Equiitbrieoh Stales as Extrema

[Ch. 12

This is the already studied problem of obtaining the principal axes of the stress tensor (Section 2.2). Formally it is, of course, virtually the same as Problem A. Note that (as in Example 11.1.1) the Lagrange multiplier A is numerically equal to the desired stationery nunnal stress, for 3

E

3

nk

ni = Â

f,f°1

= Â.

(6)

1.4

Because symmetric matrices have mutually orthogonal eigenvectors, the stationary values are associated with three orthonormal vectors n. Problem C Determination of a function p(x) that renders stationary

V - J[EI(xHjY) 2 — o

fn23 dx.

(7)

We seek here stationary values of the potential energy V appropriate to the buckling problem for the bending of a bar that is hinged at both ends. [The expression for V can be determined from Exercise 113.1 by replacing the deflection slope y' used there by p and by allowing the moment of inertia I to vary. Of course, I(x) > O.] We are faced with an elementary problem in the calculus of variations. It is easily seen (Exercise 1) that it is necessary that a stationary function p satisfies the following Euler equation and natural boundary conditions:

—d

!{x) dx = Ap,

p'(0) = 0,

p'(L) = O.

(8)

Here A has been used as an abbreviation for the ratio of the compressing force f to Young's constant E. The lowest value of A for which (8) has a nontrivial solution is proportional to the critical buckling load. Mathematically speaking, (8) is an eigenvalue problem for a differential equation of a rather general Sturm-!_.iouville type (compare Section 5.2 of I). One can regard Problems A, B, and C, as illustrating what at timesalmost seems to be an unreasonable importance of eigenvalues. Three rather different stationary value problems lead to the necessity of computing eigenvalues. But here we emphasize a complementary point of view which regards Problems A, B, and C as evidence that eigenvalues can be characterized as stationary t;alnes of some quantity in a variety of ways. We shall explore this matter, whose importance is evidenced by the many times that eigenvalue problems have been encountered in this volume and in I. A PARTICULAR SEt.F-ADJOINT POSITIVF. PROBLEM

The material that we wish to present lends itself well to an abstract formulation. In the context of self-adjoint operators on vector spaces, the basic

Sec. 12_1l

Eigrnratues and Stationary Points

503

simplicity and generality of the ideas is clearly revealed, and an excellent opportunity is provided to illustrate some useful ideas of functional analysis_ Theabstract prerequisites are presented in Appendix 12.1_ The reader who wishes to obtain rapidly the main ideas on eigenvalues and extremalization can, however, omit this appendix and read the succeeding sections with the following particular case in mind_ Consider the eigenvalue problem

— dx

[

px1 dx

+ q(x)u(x) _ ,1u(x

u(a) = u(b) = 0.

),

(9)

We prefer to regard this problem as seeking certain particular functions (or generalized " vectors") that are members of the set S of all twice continuously differentiable, real-valued functions which vanish at a and b.* We take the set S as the domain of the linear differential operator d d L _— dx p

q

(10)

and look for nonzero elements of S ("cigenvectors") which satisfy Lu (The zero element of S is the function z, where z(x) = 0_ Although Lz z is as usual specifically excluded from being an eigcnfunction.) We discussed the eigenvalue problem (9) in Chapter 5 of I (where we included an additional bit of generality that is not needed at present). There, by an integration by parts, we showed the self-sdjointness property

Az,

(Lzi, = (u, Lv)

when u E S

and

G) e S,

where

(f, g)

f (x)9(x) d

—

(12)

^-

r

^

Equation (12) provides an illustration of the generalized "scalar product" ( , ), which will be repeatedly referred to. Furthermore, integration by parts shows that L also provides an example of a positive operator in the sense that if u E S, then

(Lu, u) ? 0, (Lu, = 0

(13)

if and only if u = O.

Those who wish to confine themselves to the special case we have just outlined should continually specialize the abbreviated notation we use to this case. For example, we shall show that the lowest eigenvalue is the greatest lower bound over S of (Lu, u)f(u, u).f In the present case, using the definition The results also hold for other boundary conditions. such as the conditions

kilo) -

ii(b) =C

cf (8).

t We have already shown, in Example I of Appendix 1 1.2 (with r = I), that eigenvalues of (9) render (Lu. ul{(u, u) stationary.

50 4

Charar -rerixarion of Et .yenrahres and Equilibrium Srorrs as Exrrernu [ eh_ 1 2

of ( ,

and an integration by parts,

(Lu, (u, u)

[—(pue)' +

(lulu dx

_ ^ô CP{u1 2

A iJ^ dx

te' dx

7a

qu 2) dx

EXERCISE

1. Verify that the conditions of (8) are appropriate. 12.2 Eigenvalues as Minima and the Ritz Method Using material introduced in the previous Section as motivation, we now characterize eigenvalues as the minima of certain functionats, i.e., of certain scalar-valued operations whose arguments arc elements of an appropriate vector spate of functions. It is then natural to obtain upper bounds to eigenvalues by "doing the best you can" to minimize the functionals by inserting the partial sum of a series that could converge to the desired eigenvector, and choosing the coefficients to make the functional as small as possible. This is the Ritz method,' a technique of considerable practical importance whose presentation will be the subject of the latter part of this section. MOTIVATION

motivate our characterization of eigenvalues, we focus attention on Problem 13 in Section 12.1. Leaving aside its interpretation in terms of stresses, the mathematical problem is to find stationary values among all vectors of unit length, of To

3

given that T1) = 7j. / . =t As we have seen, the stationary values are the eigenvalues 2 defined by T(x) = 71x 1 , x 2 , x a} _—

3

E

E x ; di x; ,

gl F x, — ;0c,,

t = 1 , 2 , 3.

(I)

= 1 where the stationary values are The points on the unit sphere E i3= taken ont are given by the three mutually perpendicular eigenvectors of T 1 . • The method is also called the Rayleigh Ritz method. In his 1877 bank The Theory of Sound (in Sec , 89. for example) Lord Rayleigh characterized as minima certain frequencies that roc in his study of vibrations. He then obtained useful estimates of these frequencies by guessing simple forms of the corresponding eiget+function, perhaps minimizing with respect W a single pa rameter. Although Rayleigh doubtless knew its potentialities, it is to the papers by the German scientist W. Ritz, beginning in 1908. that we owe the systematic exploitation of the method_ The "modern." somewhat abstract version is presented well by Mikhlie (1964), a Soviet contributor to the study and application of variational methods. t A function is said to rake on a given value if the function has this value for some point in the domain under consideration.

Sec. 12.2] Figenvedues -s Mjnanu rued the Ritz Method

505

For reasons that will become clear, it is of particular interest to study the positive definite case where 3

E x t Ti xj > Û Here the level surfaces

E xi To x1 =

K.

Li-I

K a positive constant,

( 3)

are ellipsoids. To fix ideas, let us make a particular selection of constants so that

L = , f2

—

I

1.

A3

(4)

Choosing principal axes, we have 3

Tj1 x,x^ = X^ + -tx3 + x3. i,

}= L

The unit-length eigenvectors associated with the Ai x t11 4 (1, 0, 0), x{2M - (0, 1,

arc ±xt' where

0), x 43} - (0, Q, 1).

Figure 12.1 shows sections by the coordinate planes of the level surface ellipsoids that correspond to stationary values of T, namely

x^ + txx + x

dt

i = 1, 2, 3.

Let us examine the geometric interpretation of the algebraic results. The lowest eigenvalue A l corresponds to the smallest ellipsoid that just touches the unit sphere. This is, x

i

+ 4x

+

9x; - 1,

and it touches the sphere at i xli As can be seen from Figure 12.1(a) and (h), all points on the sphere that are near +x { " can only be touched by larger level-surface ellipsoids of the given class (3). Among points on the

sphere, therefore, T has local and absolute minima at + xl L ^. As shown in Figure 12.1(d) and (f), the ellipsoid (3) for A _ 02.2 touches the unit sphere at + x {2}. In the (x i , x 2 )-plane, nearby points on the unit sphere correspond to smaller ellipsoids, but in the (x 2 , x 3 )-plane, nearby points Correspond to larger ellipsoids. The stationary points +x (21 are thus neither local minima nor local maxima of T(x). The same type of considerations shows that the third pair of stationary points ± x 111 provides local and absolute maxima for T1x) among all nearby

points on the unit sphere. It would he advantageous if the eigenvalues could all be characterized as minima, or as maxima, not merely stationary points. 1f the former were true,

gatr

Characterization of Eigenralues and Equilibrium Stares as Exrremu

[Ch. 12

Yr

t+r

ry

3

{i)

tri

{dM r^

X

sy

sl

rd!

01 i

J~

r G t; R E 12.1. Points on the unit sphere are sought that give stationary values ro T(x) -x + #;x2 + x}. Since level surfaces 7'(x1 = constant are ellipsoids. the problem is equivalent tu finding which of this family of ellipsoids is iwiger+r ro rire sphere. Depicted here as heuty crows are secrions (in one quadrant) mode by the coordinate planes through each of the three tangem ellipses_ Points of rangerrcy, the stationary points. are indicated by heavy dots.

for example, the approximate methods that did "as good a job as possible" in finding a minimum would give an upper bound to the desired eigenvalue. In the simple example just considered, however, no such uniform characterization has emerged from our naive considerations. But we shall now show that a slightly different point of view allows us to characterize all eigenvalues in our example as absolute minima of T over suitably chosen sets. Since T1 is a positive definite matrix, T(x) > 0 when Ix I = I_ Since 1(x) is bounded below (by zero) for x on a closed bounded set (the unit sphere), we know that T(x) has a greatest lower bound b and that there exists a point on the unit sphere where T b. An absolute minimum is a fortiori a stationary point, so the lowest eigenvalue A(t) of 7; in fact, provides an absolute minimum to T(x) among points on the unit sphere. Indeed, T(+ x {, }i = =

.r

b,

and Ttx) takes on its greatest lower bound at ± xu }. Thus the smallest stationary value provides an absolute minimum for T on I x I = 1, as we have already seen geometrically.

See- 12.2] Eiqene•afrrrs us Minima a n d the Ritz Method

507

Because Ti is symmetric, it has mutually orthogonal eigenvectors. Consequently, the remaining stationary points must lie on the great circle x3 -i x3 =1, in the plane x i = 0 perpendicular to x"/ These stationary points of T(x,, X2, X3) necessarily are stationary points of 710, x2„ x3). There are only two pairs of the latter [determined by a two-dimensional eigcnvalue problem that is obtained by setting x 3 = f} in (1)1. Thus the two stationary values of 71x 1 , x2 , 0) on the unit circle x; + x= = 1 will provide the two remaining eigenvalues. But by the same argument as in the three - dimensional case, one of these stationary values is, in fact, the greatest lower bound of the nonnegative function T(x t , x 2 , û):

= min T(x) for x such that x x — 1,x - x'" = O.

(5)

As shown in Figure 12.1(d) and (f), this bound is taken on ai the points ± x' 3 } Repetition of the above argument shows that the third eigenvalue is the greatest lower bound of T(x i x 2 , x 3) among points on the unit sphere that are orthogonal both to xir' and to x(2 . The only such points are s x(3 ', so that this greatest lower bound can only be 71 4 x 13) ) ,

— min T1x)

for x such that x • x'"

—

x x"' = U

(6)

-

To summarize, we knew that the various eigenvalues were smar ionary values of 7'(x) for x on the unit sphere. We demonstrated that, in fact, the eigenvalues are all minima of T1x) for x on subsets of the sphere that are selected in conformity with the fact that successive eigenvectors must be orthogonal to all previous eigenvectors. Our demonstration relied upon the symmetry of T o , to ensure the orthogonality. It also relied on the knowledge that (x) was bounded below (because of its positive definiteness), to ensure that T indeed took on a minimum when x was restricted to the various subsets of the unit sphere. It is certainly not evident from the discussion just given that the characterization we have made fora particular eigenvalue problem is of general utility. Nevertheless, experience with a number of particular problems-- such as Problems A, B, and C of Section 12.1-- gradually brought mathematicians to the realization of an underlying similarity, To gather the fruits of this Au, experience, we write the original eigenvalue problem (1 ) in the form Lu written [where the normalized eigenvectors are denoted by el u = l^r a^ u]^ Y = ,

,

,

(vi,

[t 3 ),

then v — Lu

l o ut

= J-1

The operator L is self-adjoint and positive if the scalar product a lu, Yti r I a i Li(

(7)

508

Characterization of Er genralue► and Equilibrium States as Exfre+Tms [(.'h. 12 ;

is employed (Appendix 12.1). In this notation our previous results can be written [where the normalized eigenvectors are denoted by nil = min (Lu, u), ). r = (Ln11), n u) ), (n`", nu)) — 1;

). 2 =

min (Lu, u), 2 2 = {Ln[2 ', um), {n12 , nu} ) = 0 (n12 ', n (2)) = 1; c..0= ,

^

(8) and so forth. In point of fact, it proves convenient to alter slightly the above characterizations of the eigenvalues. To do this, we recall that a scalar multiple of an eigenvector is also an eigenvector. If (u, u) = 1 and if w = cue, # 0, then (w, w) = (au, au) = a 2 (u, u) = x 2 ,

(9)

so (Lu, u) = (L[a - 1 w], a - r w)

fit)

a - 2 {Lw, w,) (

i

w)

(1 0) J

Thus the unnorrnalized eigenvector corresponding to the smallest eigenvalue can be thought of as providing a minimum of (Lw, w)/(w, w). The condition (u, u) = 1 merely provides the information (w, w) a 2 and need not be taken into account during the search for the minimum. To obtain the second eigenvalue, the search must be restricted to vectors that are orthogonal to the first eigenvector, etc. We turn now to the characterization of eigenvalues in a general class of problems wherein a reliable guide is provided by the results that we have found for a particular case. SPECIFICATION OF A LINEAR OPERATOR

The succeeding theorems and discussions of this section all involve a linear self-adjoint operator L with domain S and range T, i.e., L : S T We assume that Tis a real vector space with a scalar product ( , ) and that S is a subspace of 1 -(in which case the same scalar product will serve for 51. Finally, we assume that if t e T and (t, s) = 0 for arbitrary s e S,

then t = z,

(11)

where z is the zero vector in T. In other words, only the zero vector is orthogonal to all other vectors in T. in the important special case when S = T (L maps the vector space into itself), assumption (11) obviously holds. The reason is that we can take s wand apply the last of the scalar product axioms to deduce: = z from {t, 0=-0. Suppose that T is the space of continuous functions and S is the subspace of functions that vanish on the boundary of the region under consideration and that possess n continuous derivatives. Then (11) is nothing more than Lemma

ser. f 2.2]

Eiyent•aa'ues as Minima an d the Ritz Method

509

A (Appendix 11.1. Finally, it can be shown (Exercise 3) that (11) holds IFS is dense in T This property requires that

if t t= T, then s„ —. r fora sequence {s„}, s n c S n = 1, 2, .. . (12) The approach of s„ to t indicated by "s„ -* t” is specified in terms of the norm ,

.

IL If (a generalized length) defined by II I(

(s, s 1 ' R .

By definiti on, sn

^

if and only if

t

liln II S n —

(131

ill = 0-•

THE LOWEST EIGFNVAL(JE

Theorem 1. Suppose that the set of real numbers (Lu, u)/(u, u) has a greatest lower bound (glb) ). 1 and that this minimum is actually attaincdt when u — u t : gib

(Lu, u)

uE5 0i, II)

J. 1 —

(Lit ' , u1)

^

u^ t S.

,

(14a, b)

W3,140

Then ;., is the lowest eigenvalue of L and u 1 is the corresponding eigenvet;tor. Proof Let s be an element of S and let F. be a real number- Regarding s as fixed and E as varying, we can assert that the following function "f(1:) has a minimum al r: = 0: M(E) —

it/4u t + Es], t + i s) (u t + ES, u, -}- ES)

(15)

But straightforward calculation shows that ME !M = (Lu

i , u1) + 2i(Lu 1 , s) + c2(Ls, s)

`

`(e) (16)

(u1, u1) + 241, s] + $ 2 S,

s)

c)

Since (E) is minimal when e — 0, it must be that Afé`(o) = 0 or

0=

D( 0 ) N'(0) — N(0)D'(0)

[D(0)] 2

"

(a) — , 1 D`(0) U(0)

(17)

We have used the result A l N(0)/D(D), which is an alternative way of writing (14b). Using (16), we deduce from (17) that

(Lu1 — ,^ 1 u 1 s) = O. ,

It can easily be shown [Exercise

(18)

3(d)] that S is dense in T if and only if every clement of

I has an element of S arbitrarily close to (t. t If L is a positive operator. as in the example discusscd at the beginning of this section. then (Cu. ti)/(u, u) is bounded below by uro, and therefore possesses a (nnnnegative) greatest lower hound, Whether or not the bound is attained is a rather deep question that is beyond Our present scope-

510

[. Iraracrerirariun oj L{genvatues and Equilibrium States as Exrrenur [C'h. 12

Since s is an arbitrary member of S, (1 1) i mplies that Lu1 -- AO' = z or Lu =

(19)

The above equation shows that / I is an eigenvalue. To show that it is the lowest, note that if u is any eigenvector with corresponding eigenvalue A, then Lu

).u,

so (Lu, u) _ Mu, u) or 2 =

(Lu, (u, u)

(20)

From (14a), 2. HIGHER EIGFlNVALUES

Theorem 2. L, so that

Let fir , i = 1, 2, ... N — 1 be the first N — 1 eigenvalues of ,

2 1 X 1 2 < .,. ç Let the corresponding orthonormal eigenvectors be denoted by u ; , i 2, ... , N — 1. Thus

(ur,

1 ;)

—

i,J —

•

t

, 2

,-.-, N — L

1, (21)

[We know from Appendix 11 1 that eigenvectors can be chosen so that (21) holds.] Suppose that there exists a real number J.N and an element of S, uN , such that (u N , u ;) = 4 and .

(Lu, u) — --(u, u)

^ N = gla

.fs

fu, NJ-

i — 1, . . . , N — 1,

(Lu s , aN)

(22a)

(22b)

(uh. uN)

is the Nth eigenvalue of L, and uN is the corresponding eigenvector. Proof. Let s be any member of S, the domain of L, and define r by

Then

;.N

h- t f

X ( s. u ïlui ^= 1

(23 )

rr— I (S) u) — E (s, umif = 4,

(24)

=

s

33

Note that by (21) ( I , u)

4

r =1

so t is admissible to the "competition" of minimizing (Lu, u)/(u, u). Thus the

following function (E) is minimal when r

61(z) _

0:

(L[re,i + ^ , UN + ]

Li)

( U N + et, uN + e t )

(25)

Sec. 12.2]

Eigen{salues as Mrrttm ❑ and the Rirr Method

Slt

We deduce (Ltii , — 1r,u. , r) = D

(26)

,

in exactly the same way as we deduced (18). We cannot reach the desired conclusion at once, since r is not arbitrary.* But from the definition oft, IV -I (LAIN

r)

—

=

(Lu x 2NUr+, s) — t

E -

—

ÂNUU,

i

(s, uj4

( 27 )

Using the self adjointness of L and various properties of the scalar product, we see that the ith term of the sum in the above equation equals (S, u1)(Luu, le i)

(s, ur)14ur ur)

(s. Uj)(uri , Lu i ) 0 • (5, OWN, 2rui ) 1 - (s, ui1^4luhi, u 1 ) — O.

Since the sum in (27) is zero, as before we can

(Lu re — Xup 5) = 0, ,

—

deduce that

■ so UN

Thus i, is an eigenvalue. Suppose that 2 is any eigenvalue distinct from A 1 , a.3 , .. _, and 4,1 _ I . Since eigenvectors corresponding to distinct eigenvalues are orthogonal, the eigenveetor ti corresponding to 2 must satisfy (y,

i — 1,2, ... , N — 1_

-=

(28)

As in (20), 2 w (Lu, Ujj(u, c), so u is an element of the set of numbers of which _ . is the glh. Therefore, ). ? ;. N . d R E M A R K. Scalar multiplication of u by a nonzero constant does not change the value of the ratio (L14, &)/(u, er). instead of evaluating this ratio for u, then, we can use r = of II u li. As (t, r} — 1, we can write

---

gib (Lar, z)),

( 29 )

ui!

!p.m = I

with a corresponding statement for the higher eigenvalues. The converse of this remark, aS in the discussion of (9) and (10), shows the equivalence of the Iwo characterizations of the eigenvalues, with or without the auxiliary condition (r, r) = 1. Extension of the above results to the problem

Lu

=Ar#ee,

M positive and self- adjoint,

(30)

A little knowledge of vector space propertiieS permits a more elegant argument [hark that given in the main teat. Equation (26) holds for an arbrtrary vector in the subspace P composed of vectors that are orthogonal to the fixed vectors u,. r = I, ... , N — I. It is not hard to show that Lap,. — f,. "rid e P. Hence.. as before. L& . equals the zero of A but the afro of a subspace is identical wish the zero of the original space. —

5 12

t1

Charuc+rrizulivrt of E+;aernuah ►esartd Equilibrium Mates us Extrema [ Ch.

is straightforward (Exercise 4). For example, the lowest eigenvalue is now characterized by (La, u) r = gib f3 ^

(31)

YcS ^u, t1^

where, as in (Al2.1.24), (u, u) = (Mu, a)_ Example. As in Exercise 11.3.20, one can show that the defection i]fx, y, t) of a

vibrating membrane satisfies pv2 ^^ =-

2^ ^

(32)

^i

where the constant p denotes tension and p is the density of the membrane, which is assumed to vary with spatial position (x, y). We assume that p is continuous in R, the domain covered by the membrane. Suppose that the membrane is fixed along its boundary B. There are time-harmonic Solutions of the form U = cos tor u(x, y) or L' = sin ait u(x, y)

(33)

if pal } u = puI 2 u ,

u = Don 11.

(34)

The problem of (34) can be regarded as the search for elements 14 of the set S = (w lw has two continuous partial derivatives with respect [al and y , ^+ = 0 o B)

(35)

that satisfy Lu =

(36)

wher e Lu =_

❑2 u, Mu

-

-

pu, A =

w'

.

(37)

Both L and M are self- adjoint under the scalar product 91 =

(

J

ffc.})(xy)dcT.

(38i

R

Furthermore, Land M are positive operators, so all cigenvalues are positive_ in particular, using (3l) we see that the lowest eigeuvatue ; t satisfies

at

wV r w du i[ f gib — js wts

]J* p w do

(39)

Using Green's theorem to modify the numerator, and using the idea expressed in (29), we Can also write .l i — glb fJivwi 2IVH.I da,

(40)

where admissible functions must be elements of S and must also satisfy I = {w,w) =px''da.

(4 1)

Sec_ 12.2] Eigenvafues as Minima und she Rirz Meihod

5 13

THE RITZ METHOD

We have expressed eigenvalues as minima of certain functionals. The elements that are possible arguments of these functionals must be selected from a certain well-defined set S, the domain of L. A practical procedure for obtaining approximations to eigenvalues consists of selecting a few elements from S and making the functional as small as possible by using a linear combination of these trial elements. To illustrate the procedure, consider the eigenvalue defined by where u a S, (ee, u) = 1.

=- gib (Lu, ie),

(42)

Let w i , w 2 , ... • w* he a set of trial elements selected from the vector space S_ It is convenient to select trial elements that satisfy the orth ❑ normality condition (w1,

wi) _ c5 ;j•

(43)

We now consider the particular element of S given by the linear combination 14 =

E ïR; w;

( 4)

t=1

and we choose the real numbers a ; to make (Lee, u) as small as possible_ Of course, the /i must be such that the constraint (u, u) 1 is satisfied. This means that h'

1

=

E i= L

N

N

j= L

^j ^ !

N

^ ^ ^i^ J^^i a x°) } = L j' i. 1 = 1 i-L

(45)

where we have employed (43). But N

(Lu, u) =(L[

N

!M

y cL, w; , y x i

i =1

11 ^

= ^ m; a; a i,j

1- 1

=

,

(46)

1

Here

(Lw i , w,).

(47)

Note that = (Lw j , w1 ) = (w) , Lw ; ) = (Lw; , wi ) — m il , by the self-adjointness of L. Consequently, among all 2; satisfying ^

i

EL =

we must choose those that minimize N

2

1

,

(48)

C'haraclerizoiiun r f Eigenvalues a n d Equilibrium Stales as Exrrema

514

[Ch. 12

Using the approach of Section 11.1, we introduce a Lagrange multiplier A and deduce that the minimizing 2, must necessarily satisfy ^

N

y mjlat a i —

Oa{

N

A ^ at =-- 0.

i,1-1

(49)

i= l

Upon carrying out the differentiation and dividing by 2, we find from the above equation that

Er

+77 i O 4 — Act r - 0 or ic= l

i — 1, . .

— AS,k)cck

. ,

N.

( 50)

h

To obtain nontrivial solutions to the eigenvalue problem (50), we require that 0 = determinant

— A A ) _ (— 1)r`A N + - --

,

(51)

an Nth order characteristic equation for A. We have icduced our problem to the solution of the algebraic equation (51), even though the original problem certainly need not be algebraic. Now we must establish a connection between the roots of { 51) and the eigenvalues

of the operator L. To this end it will prove useful also to pose the algebraic eigenvalue problem (50) in the framework of linear self-adjoint operators. {This problem is merely a slight generalization of an illustrative example treated in the previous introductory section.) Consider the Euclidean vector space R N of N-tuplcs a — ([,, al# , _ , j, with inner product - -

N

a• d =

-

where d

x ;; ,

, hti).

(52)

1

The eigenvalue problem (50) can now he written Ma = Aa. The linear operator defined by

d

=

Ma ifb r

(53) ;-t

is self-adjoint, as shown in (Al2.1.13). By Theorem Al2.1.1 the eigenvalues are real. We denote the ith lowest eigenvalue by A ; , so that A, < Ax

...

AN,

( 54)

The corresponding eigenvectors irrr', i = 1, ... N, are determined only up to an arbitrary constant by the requirement that Ma Aa, but this constant must be selected so that the constraint (45) is satisfied. That is, in the present ,

notation, a{'M • ar"t = 1.

Using (55), we see that Mar ie

/4,o° implies that au) • Ma" ) = A i .

(55)

Sec. 12,211 Eigenvalues as Minim and the

Ritz Method

5t5

Now the extreme values of the quantiy (46), whose gib we seek, are

Ece N

F.4 =

{,)^ atri = ^n . p A4

1

4

Mau)

. -

Thus the lowest eigenvalue A t should be the desired approximation to the lowest eigenvalue 1 1 of L. As we have minimized (Lu, u) only over a restricted set of functions, our approximation ought to be an upper bound. Indeed, it is a natural conjecture that A, We shall prove in the

next

1,2 ... N ,

(56)

,

section that this conjecture is correct.

VALIDITY AND C]T- 1LJTY

OF THE RITZ

MET 110f)

We shall briefly discuss conditions for the convergence cf the Ritz method. Then we shall touch on some practical matters connected with the use of this method. As more and more trial elements are taken, the lowest eigenvalue cannot increase. Lei us assume that L is bounded below in the sense a(u, u),

(Lu,

2 a scalar.

(57)

Then the various approximations to the lowest eigenvalue are certainly bounded below by a, so they must converge to some value- It is natural to hope that this value is the correct answer, ,1 1 . To discuss sufflcientconditions for the justification of this hope, we first note that we can assume with no loss of generality that a is positive (and hence L is positives). For if L is not positive, we observe that if L i u + fin. /1 a scalar, then A + is an eigenvalue of L, if and only if À is an cigen slue of L. (57) and {1 > Eat , then J. is bounded below by a positive con- IfLsatie stant. (Lu, y). if L is positive (and self-adjoint) we make the definition Lu, t'] As in Exercise A 12.1.6(c), this provides a functional of two elements [ that satisfies all the axioms of a scalar product. Now a set of elements w i is said to be complete in S if for any u F S, ,

Lim E a i w; = u

for some set of scalars a l , ar e ,

...

]

.

The limit is in the sense of (13); i.e., the norm of the difference between the partial sum and u must approach zero_ if the elements w l arc complete in S when the norm is formed from [ , ] (so that Dug 2 = [u, u]). then convergence * 1[(57) ho lds for rr > O, then 1964).

L

positive 6ul the convcrse l5 not necessarily true (Mikhlin, ,

516

Charactert_atrün of Eigeni alues and Equ[libriurn

Blares ac EYrrerna [Ch. 11

of the Ritz method can be demonstrated for the lowest eigenvalue, Convergence for higher eigenvalues can also be demonstrated under certain conditions (Mikhlin, 1964, Secs. 32 and 78). Proofs of convergence are obviously of high theoretical interest. For practical purposes, however, computations often become formidable when N becomes even moderately large. Typically, this is because calculation of the ni d involves complicated integrals and because it is not easy to determine the eigenvalues of a large matrix. In many cases what one wishes to know is whether a good approximation can be simply obtained. And indeed the Ritz method often provides an excellent approximation to At even if only one or two trial elements are employed. One illustration of this is found in the calculation of I, Section 12.4, where a one-term approximation gives an upper bound to the fundamental frequency of a longitudinally vibrating wedge-shaped bar that differs from the exact value by less than 3 per cent. A further illustration of the method will be given below. There are three main reasons for the outstanding practical usefulness of the Ritz method. 1. The method is most powerful for approximating the lowest eigenvalue, but it is just this quantity that is often of highest importance. To give some examples, the lowest frequency determines the character of a vibration more than the higher frequency overtones, the lowest eigenvalue usually corresponds to the most unstable perturbation (see Section 15.2 of I), the lowest eigenvalue of the Schrôdinger equation* gives the ground state, etc. 2 Because an eigenvalue is the stationary value of a certain functional, a first order error in the eigetttaector will give °lily a second order error in the eigenvalue (see Figure 12.2). 3. In applications, one often has some idea of the general "shape" of solutions. This can be used to guide the selection of trial elements. A common "rule of thumb" in using the Ritz procedure is to compare the eigenvalue estimated when, say, one trial element is used with that obtained with two trial elements. if the answers are close, one feels satisfied with one's approximation. But it could he that convergence is to the second eigenvalue (say), not the first, because both trial elements are virtually orthogonal to the lowest eigenvect or. A theoretical landmark WAS achieved with the aid of a Ritz calculation by Hyllcraas alï t#ie ionization potential of h€Buret 1t is worth quoting in emenso from the account on p. 347 of E. U. Condon and G. H. Shortley's The Theory of Atomic Specira {New York: Cambridge University Press, 19571. "An eighth approximation led to 1,80749tthc = 198322 cm t for the iortiie.ation potential as compared with an experimental value al! 1 48298 ± 6 cm - L. It will be noticed that the theoretical value txeeeds the experimental value by 24 cm - ' , which appears to contradict our statement that the Ritz method always gives Too high an energy value. The discrepancy is duc to the neglect of the finite mass of the helium nucleus and to relativity effects. When these are included the theoretical value becomes 198307 cm ', which is in agreement with the experimental value_ This is an important accomplishment of quantum mechanics since it is known that the older quantized-orbit theories led definitely to the wrong value."

Sec_ 12.2J Eigenualue.e as Minima and the Ritz Methad

517

f

Y

F1 i; ri RE 1 2.2. A schematic illustration of why an D{E] error in rhe eigrrt:.recrar, represented by x a , leads to only an 042 ) error in the eigFnrralue A. The quantity R(A) is minimal when r = xQ, Mid 1 =— R(x o l. Since R'(x,j = O. if we nusr.stimure xa to obttrin x U + E, using R(x 0 4 El win give [r , a good approximation fo A since .

.

R(x r, -f r) — A = R(x U + Fj — Rim ^ R"{xoiE r-

GENERALIZATION

The Ritz method can be used foi - the eigenvalue problem Lu = ;Mu, where M is a positive self-adjoint operator. In this case, as in (31), if (u, y) i (Mgr, ti), (Lu, u]

A, = glb -- -- or urs (u, U}

A, =

gib (Lu, u),

(58a, b)

I,cs

while for higher eigenvalues the minimization is subject to orthogonality conditions of the form A (12) > AV ). Note that (77) can be used as a first approximation to the lowest root of the characteristic equation for a higher order approximation, to be improved by some perturbation or iteration scheme. From the second-lowest roots, Mikhlin obtains the following two approximations to the corresponding exact eigerivalue A` } = 1.8605,

/VP

= 1.7718.

It appears that accurate results are being achieved. A computer program could be set up to supply results for any given value of R, r, and a. RATURAI. BOUNDARY

CONDITIONS

Successful abstract theories reveal the unifying concepts that underlie seemingly disparate problems. At the same time, the very generality of these theories sometimes makes them incapable of dealing with important special features of a particular case_ An illustration of this fact in the present context stems from the observation that for differential operators, but not for matrix operators or integral operators, boundary conditions are usually a part of the definition of the domain over which the operators act. Consequently, the general theories cannot be expected to bring out the subtleties connected with the fact that under certain circumstances it is advantageous to omit specification of Some of the boundary conditions in describing the domain of a differential operator. We close the present section with an illustration of this point. Miktilin did nut introduce dimensionless variables, so we have had ID carry our a translation

of his resulrs,

Iltrle

522

Characterization cif Eigenvalues and Equilibrium Slates as Exart'tna [C'h. 12

Consider the eigenvalue problem (pu')' + qu = Ani,

u(o} — O,

(78a) (78b, c)

p(b)u"fh) + Ou(b) — O.

We regard this problem as an instance of L14 = Pilo. Here k is a positive constant,. As usual we assume that p, q and r are positive, that pis continuously differentiable on [4, 1], and that q and r are continuous on [0, 1]. A special case of this problem appears in 12.4.10 of I, where equations are given that describe longitudinal vibrations of a bar. The bar is secured at x = h by a spring, and k2 is the relevant spring constant. 13y Exercise 13(a) the operator L is self- adjoint and positive over a space whose elements must satisfy the boundary conditions (78h) and (784 Consequently, the eigenvalue problem can be treated by the methods that we have already presented. If it is, however, Ritz trial functions must satisfy the complicated boundary condition (78c). An alternative approach, as we now demonstrate, avoids this complication. We observe that on integration by parts, ,

j7,

1 — (pie')'

+

+ qu]G dx --= —pu'r ,

(pu'd 4 que) dx.

^

(79)

u

G

Let us define a vector space T consisting of twice continuously differentiable functions that vanish at a. ff u and y are elements of T, then (79} can be written (80a) where now (SÛb)

and (80c) Since p and q are positive, r product ( Exercise l3(b)J. Let

A L = gib R 1 (14), uc T

,

1

possesses all the attributes of a scalar

R 1 (uj

[14,

u]

k2u2(b)

(81)

r/+ L^, u)

where (u, re} i (Mu, u) as above. There is reason to expect that . exists, for R 1 (u) > Û; nonetheless, the lower bound may not be achieved for a function in T. ,

Theorem 3. Suppose that there exists an element u 1 of T such that 2 1 = R02 1 ). Then 2 is the lowest eigenvalue of (7 8 ) and u, is the corresponding eigenvector.

Sec _ 1 2_21 '

Eir#enr•afr i es Qs Minimu and alir R I t- ME' Mimi

Outline of Proof s E T and define

523

As in the proof of Theorem 1 we select an arbitrary 4 3 (0.mDi

= 1{ 1 (u 1 + r..$),

The requirement jr(0) = Q again yields N'i (0) — ). 1 19'1 (0) = O or [Exercise 13( c)]

2[u 1 ,

s]

-- 22 1 (u 1 , .$) + 21ï 2 I11(hh(h) = D.

We now employ (80) and the definition of
. Canceling a 2, v,c lind that

(Cll r — £ 1 Mu r . s) -}- [p(b)rri(10 + k 2 u r (h)]ssh) The conclusions that Lit t = ï i Mtt i ,

((h)ra r (h) +

1i r ((t) = b,

u.(h) = ()_

and that f. r is the smallest eigenvalue, now follow by the standard line of reasoning used in problems involving natural boundary conditions [Exercise 13(c)]. d 4

EX

FRCISES

1. Using geometric reasoning, the text provides two different characterizations of the stationary values in Problem B (for all admissible: x, on the one hand, and for increasingly restricted sets of x values. on the other)_ Provide such a discussion for Problem A. 2. Show algebraically that i., and are correctly glen by (5) and (6). (Take the additional constraints into account by means of additional Lagrange multipliers.) 3. #(a) From the identity .

(ii„,.

G„) ` (u, U) = (14, — te, V M — r) + ( ii —

prove that an

fortiori, (fa, um)

al.

r) +

it, zn y implies that (ir,,. 1.) (es. r). (tr ,,. i') (a. ri.

trr.

(u, r) and a

Prove that (I I) holds if S is dense in T. Show that with the inner product (7), l'Nil gives the usual length of of a vector in 3-space. (d) Show that S is dense in Tin the sense of (12) if and only if for every positive c and for every element t of T, there exists an element s of S such that Its — < 4. Extend Theorems 1 and 2 to the problem Lea = AMu. where M is positive and self-adjoint. 5. Extend the Ritz method to the problem LIE = i_Mu by verifying (6O), (61), (b) (c)

and (62).

514

Chwrueterisatiatt of E{genn•atues and Equilibrium Stairs as Extrema (Ch.

f

6. Let L be a positive self-adjoint operator with domain a vector space S possessing the scalar product ( , ). Let v i , ... , v, be elements of S. Similarly to (Al2.1.24), the definition [i i , (Lv i , vi ) introduces [ ], which satisfies the scalar product axioms. (a) Prove that if the r i are linearly dependent, then the following G ram determinant vanishes; ,

[t., 11 v i ] "' D., + vn] [^„, ^ :1 [ v., v„] ::: '

^

(b)

Show that if the Gram determinant vanishes, then there is a vector v = = t a1 eat with not all y = 0, which satisfies [u, u ; ] = 0, i = t, . . n. Deduce [v, u] = 0 and thereby prove the linear dependence of the v i . 7. (a) Verify that (64a) is the dimensionless version of (63). (b) Fill in the details necessary to obtain (65). H. Verify {a) (68); (b) (69); {c) (70); (d) (71); (e) (72). 9. (a)-(e) Verily the five formulas of {75); (f) verify (76). 10. (Project) Find at least a rough approximation to the lowest root of (69) and use it to check (77). 11. (Project) Find fairly accurate approximations to the first two roots of (69) using hand techniques. 12. {Project) (a) Outline a computer program that will successively find the roots of (69). (b) Write and run the program_ (c) Write a discussion of the issues involved in preparing a computer program to obtain the lowest three eigenvalues for the beam of Figure 12.3. Any sensible value of R, r, and a should be accepted by the program. 13. (a) Show the self-adjointness and positivity of the operator described in (78) and the sentences that follow (78). (b) Verify that (80b) defines a scalar product_ (c) Fill in the details required in the proof of Theorem 3_ 14. To afford practice with the Ritz method, consider ,

d2n dx2 (a)

(b)

(c)

+ i_u=0,

u(-1)=4(1)=0.

If 1 0 denotes the lowest eigenvalue, show that 4.4 0 ft2. Find an upper bound for 4/1 0 by evaluating (Lw, w)j(w, w) when w — 1 — x 2 . Compare with x 2 9.87, a result correct to two decimal places. To improve on the approximation alb), use the fact that the lowest eigenfunction is even. This permits the integrals to be evaluated between zero and 1, and suggests as an improved trial function w = 1 — x2 + a(1 — x 2 )x'. Use this to obtain an approximation

Se(' . 12.2] Eigenvalhes as Minima and the Ritz Aleihod

52 5

is

to 410 that is accurate to about one-tenth of 1 per cent. [It probably quickest to employ (62).] (d) If (1 - x 2 ) o a„x' is used as a trial function, to what number do you expect the second lowest root of the Ritz eigenvalue equation to converge as N ac.? 15. The bending oscillations of a plate that is rigidly fixed at its boundary B (a plane curve) are governed by the following problem. which is a two-dimensional version of the beam vibration equation (63) and the "built-in" boundary conditions (5.1.381 '

Vo w f k w - 4,

w=

Crt

O on

B.

182)

Here w is the vertical deflection, 1. is a dimensionless frequency of oscillation, and denotes a normal derivative. Use the Ritz method to find an approximation for the lowest eigenvalue when B is an isosceles right triangle. (a) Reduce the problem to the evaluation of definite integrate_ Defend your choice of trial functions. (b) Complete the calculation (project),

The following five exercises concern an eigenvalue problem that is the subject of "Free tidal oscillations in rotating fiat basins of the form of rectangles and of sectors of- circles,- by A. Pnueli and C. L. Pekeris, Phil. Trans RGy. Sr c. (London) A 263, 149-71 (1968)_ This paper seeks to gain insight into the effect of jagged coastline on tides. The wavelength cf tidal oscillations is so large compared to the depth of the region which contains the water that vertical variations can be ignored. Thus the wave height only depends on the horizontal coordinates x and and the time t. For wave heights of the form ax, y) exp (it7t) it can be shown (Exercise 8. t _13) that

N2 4 k 2 ) _ 0 for (t, y) in R,

en

- it

ids

= 0 for (_x, y) on B.

(83)

Here B is the boundary of the region R that contains the water; ltix (17 2 - 4w 2 )/gh - (o 2 fgh)(1 - r 2 ), where w is the constant angular speed at which R rotates, g is the (constant) gravitational acceleration. h is the (constant) average depth of the water: ?/rtt denotes differentiation along the outward normal to B, and c?/r's denotes differentiation along B_ Note that C is a complex - valued function. One purpose of this series of exercises is to afford practice in manipulating the ideas that we have presented by asking for the extensions required in the complex domain.` (No • It mensions to vector -valued functions arc also useful. For an example. see "A computertrnplerocr3led vector variational solution of loaded rectangular waveguides," by W. J. English ISh,iM 1. Appi, Math. 21, 461-68 ( 1971 }].

526

Characterization el - Eigen ahres and Equilibrium Slates as Ewsemna KA

U

true complex analysis is necessary for this, only an understanding of complex numbers.) 16. First consider the case of no rotation, where t = O. Show that if iÿ makes the integral

1_

J

J(VZ. Ÿ2 — k 2 Z2) dx dy,

E complex conjugate,

stationary among all complex-valued functions Zfx, y) which are twice continuously differentiable, then is an eigenfunction. Use the results of Exercise 17_ 17. (a) The formula

J

fv4 . VII' dx dy =

e n ds ff0 / 4,1.0 dx dy O

R

R

is well known when 4) and 1/ are real - valued: Prove that it remains true when and arc complex-valued functions of x and y. (b) Suppose that Re JJ f (x. y)g(x, y) dx dy = 0,

"Hey "real part of,"

for arbitrary twice continuously differentiable complex-valued functions f. Prove that if g is continuous, then g = O. NOTE. Exercise 16 also requires a one-dimensional version of the above lemma, but the proof of this is essentially the same as that for the

two-dimensional version. +18. Now let rotation be present ('r 0). Show that a correct variational formulation results if one adds the line integral I' to the integral 1 of Exercise 16, where =

c?2 — d5$ F^5

Also show that P is real. 19. One possible way to obtain an approximate solution to the problem of extremalizing 1 -4- 1' begins by assuming

Z=

ay complex constants, p=1

where ^^

Cp

+ kp ^e = 0,

^^

= 0 on B,

k; s

-

• . -

sec. 12.21 Eifiestralues as Minima and the Ric: Merlod

527

(As trial functions, this approach uses "no-rotation" eigenfunctions. It should therefore be useful for sufficiently small values of r) Demonstrate that the above assumption for Z leads to the problem +

r

1

—0)Niai =Q,

=

1, 2,

N.

m-t

/^ Nrm _

N? =JJ? dx dy,

— [ S! , ds = PMrg

L'S

R

Part of the demonstration requires extension of the Ritz method to the complex case.

For small z, a perturbation method is useful in obtaining the approximate eigenvalues k. Nothing further is required here, but the reader may wish to glance at Exercise Al2.1.7 to obtain an idea of how such calculations proceed. 20. Pnueli and Pekeris found that, except for small values of r, the following method was more successful than that of Exercise 19. Following a suggestion of Trefftz, one can consider trial functions of the form NOTE.

1a„ „(x, y; k) + i^i 7,1(x, y; k)], app and fl„ real constants, n-

1

where (V' + k e g„ =(V 2 + k l kl„ - 0 and the real functions and ri„ are complete on the boundary. (a) Show that surface integrals drop out so that one obtains

n

-

fH Langn D! +

1

Lg.) + ifin( ` +1 n ^^^ + tir Drin)] da = 0,

L ^ [1^^^^^7t - g1 DO + 13„(qaD►h + rledring ds =

n=1

8

where

p

=

- it i^5

t^i1

(b) Show that these equations are equivalent to

E qtr a + rP

ds = 0,

n=l •

a

°^s

1I

^ ^

6.

f("a‘n

=1 •

as

+

# ^^ âds = o. n

0,

52$

Chararterizalion of Fryrn, whirs mud Equilibrium Susses as F. .renia loir.

12

(The lutter" represent simultaneous equations for the determination of the coefficients a„ and f1„ and the vanishing of the determinant of these equations yields the cigenvalues r,, for a given k." To furnish a bolier understanding of the method, we note that for a hasin wherein kI a and Kyi s 1x, the following are suitable trial functions fer waves that arc symmetric with respect to the center of the rectangle: = CUh I. ti A Li)ti

]^ n

-

sin it x sin ^i y

-

I)ivl s iiili by !rn ensures that rf„ is real n:YCti when if. is imaginary.)

12.3 The Courant Maximum—Minimum Principle A PRont.l'. M1i 1N ► 'IBRA l ION ttlt,ï11t1'

Consider the longitudinal vibrations of itihotnogeneous bars, i.e.. bars whose composition and shape may vary along their length. Suppose that two bars are identical in every resp e i except that Ilse lirsi (surperscript I) is tihide of a slilfet material limit the second fsuperscript 2j. to other words, the respective Young's moduli satisfy 1.1I 1 0. > L12100. )

where f) is the length of the bar. What can one say about 11w frequency of vibrations? Thai the stiller bar possesses a higher fundamental frequency is casily shown from the variational charactcrir,.ition of the lowcsi eigenvalue. From Section 12.4 of 1. for a bar that is fixed at both ends this is Evi.,1'i02 rf x 10 (w).

1(H')= J

J

t = 1,2.

The set S consists of functions that have certain smoothness properties and that vanish at ii and D. For every admissible function, the qunttetii i is larger for the first bar than the second, ln particular, it nl" provides 111C iiiiliiiituui of #x"`lu•1, ibeii

lin = Rwitu 3 ")

min W 2 110 _

^t]

we* s

> r.}`' = Dut this approach sloes not lead to the desired statement the functions th:it must be tested to obtain the minimum 2, 3 For are orthogonal lo a i ` white those for 411 must be orthogonal to 0 1 Tliese arc different sets of fonctions, so we cannot carry through our previous arguaient. A characterization that does allow us to provide the desired result foc a wide class of problems, and that is also of considerable theoretical interest

n

Sec. il_j]

The Courant' Maximum --Minirrrtrrn Prirrc - rpk•

519

and elegance, is the Caurwu* maximum--minimum pri ncipl e . To this we now turn. The reader will not be surprised to find that we motivate the general principle by turning to the example that we discussed before, the determination of points on the unit sphere that minimize L - t 7;x i xf Previously, We characterized the second eigenvalue of the positive definite and symmetric matrix with components 1; ; by restricling attention Co the plane through the origin that is perpendicular to the first rigenvector, Now we wish to avoid reference to the first eigen vector and still formulaic a problem that gives the second eigenvalue. As Courant proved, this can be done by first considering an arbitrary plane through the origin. Such a plane cuts the unit sphere in a circle of course, and cuts a given level surface in au ellipsc t As before, try to find the ellipse of minimum area that just touches the Circle. Now try to choose the plane so as to maximize this minimum area If your geometric intuition is good, you will see that the plane should be rotated until it is perpendicular to the first eigenvector, so that the maximum minimum characterization leads to the same second eigenvalue as before. In any case, this result follows from the general formulation of the maximum minimum principle, which follows, . 1

-

.

THE MAX—MIN PRINCIPLE

Theorem 1. Consider a real linear self-adjoint operator L with domain S. as described in the material just preceding 12-111. Let the eigenvalues of L be f 1 < ). - • • with correspondmng orth ❑ n ❑ rural a igenveçtors to I . ii ... . I arbitrary vectors r, as follows§: Define a function lit of p —

rnilïv,._.- , rp_ tl =

min RI(irl, we s

f = t,-.-, p — 1,

(2)

where

Mu) • Richard Courant was

head ache

— (L^^_ u) (u, re)

(3)

Mathematical Institute m Gottingen when it waS a world

center For mathematical research. Driven our by the Nazis, he started all over again in the United Stairs and built up at New York University the intcrnaii wliy renowned tnatNUte tttdt bears his name.

through as center. a closed curve obviously results. it is apparent from the form of the attrehraic suhstttutrotn required obtain chip curve that its equation contains only linear and quadratic terse- Corlsequcrrtly, to the curve is an ellipse, t The level surfaces are ellipsoids. When such a surface is cut by a plane

The argurnents arc disirr to follow if we use the more intuitive -- rnux"' and "min -- in preference td lub and gib. To make sure that ihtr notation is understood, we remind the l eader that in (2) m denotes the minimum value of R(rrl among all elements of S that arc orthogonal to the vectors r,. i = I. _ _ _. p t_ As usual, we canon as well minimize (La, 4,11 if we add the additional restriction (di. rrI = I. in conformity with previous praciice. we are assuming the —

existence of all minima and mamma.

Characterization of EiyetrraJ:ws and Equilibrium States a-s Extrema [['h- 12

53D

Then max m = m(u t ,..., Ur— t) =

(4 )

hJ,I

Proof

According to Theorem 2-2, r(ut, u2, - - • u - t ) _ ). . so a fortiori .

,

,

max m> (r11 1 -o complete the proof, we show that the last inequality is also truc in reverse by demonstrating that for arbitrary ^; m(uy, .... v p _ t ) S Ar To provide this demonstration, we exhibit a vector u in S which is orthogonal to the v, and which is such that R(u) ç ). p . We write ,

-.

p

P

u =

E b; ti ; and require (u, u) = 1,

i.e.,

y bi2 — 1.

(Sa, b!

There is at least one vector u in the p-dimensional subspace spanned by that is orthogonal to L t, - • • , _ I and that is of unit length as required by (5b).* But for this vector u ❑F

A

P

R(u) _ (Lu, u) _ Xb ; I.u ; , s= 1

^

[,i =1

F

^^^► ^l^+i

P

—

},b?

Ç

^.

(6)

J1 _ . e =t

t= t

Because R(u) < 2 certainly mir t , _ < the ti's are arbitrary.

bjuj) i,j =I

J

P —

A

hj u; =

- -

, up 1 )

i! p , and the max of rn is also

Example. The pth eigenvalue of the self-adjotnt problem Mc = Ac that arasc in method can he chanicterized using the max mitt principle as Follows:

the Rit?

c Me pic) = C^c •

ii(a tlr ,

. ai'-CI] --

min PM'. i6 R"

A,,=

max p.

(7)

ia 1 Qt

p

Using the max-min principle, one can readily show that all frequencies of vibration are raised in a longitudinally vibrating beam if the beam is made stiffer in the sense of W. For, as in our argument concerning the lowest eigenvalue, Nii21(l}I,..

• In concrete terms, thcp

vp-

-,

t'y— l )

(8)

I linear homogeneous equations

y Nui. t'i = o, —

Cl}Il itlu l+

j — I•- - •P —

I-

1

have al least a one-parameter fancily of solutions for h1 , _ . , b p , dru} a particular solution cats be selected se that (5h ) is satisfied-

Sec. 123] The Coures zt Maximum-Minimum Principle

Sat

Lei the maximum value of the rrzi'' be provided by I, 2.

Then J. A2t _ mt2)(ui2is ..-.

uF} i l C ^tti{ut2w ._.. ^

1)

max ^Tr^^'1U • -

rt. I) ; 4".

^.

ia•, I

(9 )

APPl.1CATIDN TO THE RITZ !METHOD

Let us now prove our earlier assertion that eigenvalues obtained by the Ritz method are bounded below by the exa ct eigenvalues- This is done in an elegant manner by using the max-min principle to characterize the eigenvalues of both the original problem and the problem generated by the Ritz method. The latter case has been dealt with in the example above, and we shall employ the notation used there. Theorem 2. The pth smallest root of the algebraic equation (2-511 obtained in the Ritz method is an upper bound for the pth eigenvaluc of the original problem. Proof Let u 1 , - . - , up - 1 denote p — I arbitrary elements of S. As before, we are considering as trial functions a linear combination du orthonorrnal elements of S, namely, w 1 , . - i . Define p - - 1 n-tuplesa'"; i = I , p — 1; by f Rai t+x where alp = (n•,, v ;). a4') =lt,•^ sulk .

,

,

..

,

Note that the general trial element

C

!! = ;=

t

i

satisfies n

E ^`^w1^ ^} =

(u, vi) = y

Fl

i E c1a lp = e' a' ', 1=

i = t,

, p — 1.

( 12 )

I

Furthermore, using (7), we see that RN)

_

(1-14. u)

c • - ` R —

(u, u) T c • e

^(c)

(13)

Equation (13)just shows that R(u) reduces to p(c) when the trial function (l 1) is used. But we view (13) as an identity between the expressions whose maximum- minimum over suitable sets, respectively, characterizethe original eigenvalue problem and the Ritz approximation to it. Lxp]icitly noting the dependence on the a; '' of (1O), we fix our attention on the vector c = c(a { ", a ( v - 't that minimizes p among all vec[ors which are orthogonal to these si t '', i = 1, ... , p -- 1. Consider the function u of i l 1) with the components c; of this particular c as coefficients- By (12), since ...

}

Characterization of Eigetwafues and Equilibrium

532

Suites as Extrema [Ch. 12

= 0, this u is orthogonal to the ta; . Consequently,

e•

min p(c) = R(u)

for a certain u such that (u„ i4) = 0.

cc

(14)

But the left side of (14) is /t(tt{ tr, ... ,alp L) ) in the notation of (7), so that AR (

Ly

min

. ► air - I );

R(q) = rn(U1, - - -, up-1}-

ii

i4.w} =i

•Then max p(a t oo i

{ L},

} a tr- Lt

m(v i • , vr-

,^

In other words, by (7), A

?

m

(v

1,

- • • Lp

-

L).

But since the i: are arbitrary, Ap

max m(VL , ..-.ep-t) = J.p ,

which is what we wished to prove.

El

E X FRCISES

1. (a) Use the max-min principle to prove a result concerning the qualitative effect of increasing the density of a longitudinally vibrating inhomogencous beam. (b) What is the effect of increasing the moment of inertia I upon transverse beam oscillations? (See Exercise 12.4.4 of [.) 2. Using (12.4.10) of I, discuss the effect of the spring constant k on the longitudinal vibration frequency of a bar_ In particular, consider the limiting cases k = 0 and k = rx . 3. (a) Show that preventing a portion of a vibrating membrane from moving cannot lower the frequency of vibration, while slitting the membrane cannot raise the frequency. t(b) What will be the effect on the frequency of freeing part of the boundary of a vibrating membrane? t4. Regarding the buckling of a bar, consider three situations in which, respectively, the boundary conditions at an end of the bar are

(i) y=y'=O(ii) y =y"= O.

" + K y " T Oy t (iii) 1 _ 0Imagine the other end of the bar to be "built in" (y J y' = 0) in all cases. Referring to Exercise 11.3.1 for notation and results, discuss the relative magnitudes of longitudinal force that are required for buckling in the three situations.

sec. 12.41

Minima! Clrururrerizurrrgr

of linear

positive >prphtt•r11t

533

12.4 Minimal Characterization of Linear Positive Problems A surprisingly large proportion of natural laws can be viewed from either of two perspectives. In one, equilibrium configurations or motions are determined by imposing a necessary balance between various vector quantities. The other formulates the laws as extrenial principles. Since a single world of phenomena is under scrutiny, both points of view must. cf course, he equivalent. Sometimes the formulation of a natural law is bound up with extremum principles from the very beginning of theoretical work. An example can be found in optics- Willebrord Snell (1591- [6261 experimentally discovered the law of refraction that bears his name. Later, in 1657, Pierre de Fermat asserted that "nature always acts by the straightest course"; in particular, light travels between two points so as to minimize* the time of travel_ Assuming a speed of light propagation that varied from material to material, Fermat deduced Snell's law, More often than not, the formulation of a natural law as an extremum principle followed its original enunciation by some time. (An example is provided by Hamilton's variational formulation of Newtonian mechanics, the far-reaching and precise culmination of Maupertuis's grand but ill-supported proposition that nature always acts to minimize sonic "action.") Thus it seems that riot only should a theoretician be capable o f solving extremalization problems that arise "naturally" in the course of his work, but he also should be aware of the fact that natural (or social) behavior often, and with advantage, ran be regarded as being the consequences of an extremalization principle_ Possible advantages arc suggestive theoretical elegance and computational convenience. An illustration of the first advantage lies in the unification of optics and mechanics through the principles of Fermat and Hamilton- this was influential in forming the foundations of quantum mechanics. The second advantage was illustrated in the previous section, where we showed how a variational formulation of certain classes of eigenvalue problems led to the powerful Ritz. met hod of approximate computation_ In classical fields of "hard" science, al least, these days most problems are formulated as a set of equations or inequalities: most often differential but also integral, difference, delay-differential, and so on. One must make an extra effort not to lose sight of the possible advantages of an extrental formulation. In this section we wish to illustrate the possibility of reformulating as minimization problems certain equations that involve positive operators. We begin with a discussion of the fact that in particle mechanics stable equilibrium can he characterized as the state that minimizes potential energy. • It is now known rhai only m a restrictive formulation of

Fermai's

iruc

rosay

tl^xl the time of travel is minirnixcd. In general. all rifle Can asscrl is that the travel trine along the actual path

is stationary with respect to a suitable class of admissible path% Sue pp- 127fr ot' M

Born and E.

-

Wolf's classic,

Principk.s n} ()pries (London: Pergamon Press. 1959)

534

Characterization

EfgOnrrnlues a nd Equilibrium States as Extrema

(Ch. 12

An attempted generalization of this idea to the problem of a weighted membrane leads to a conjectured equivalence between the Poisson problem in partial differential equations and a certain minimization problem. The equivalence is proved. Indeed, a whole class of such equivalences is demonstrated by using functional analysis. (This approach emphasizes a structural similarity that was ultimately perceived to unify a number of particular problems) Again, a Ritz approximation procedure is formulated. Its strengths and weaknesses are illustrated in the context of the classical torsion problem of linear elasticity. PARTICLE EQUILIBRIUM AS A MINIMUM OF PDTENT1Al. ENERGY

The ideas we wish to present can be illustrated by considering the motion along the x-axis of a particle of unit mass that is subject to a force component f (along the x-axis). Let there be a force potential 0, so that f (x) = — ç'(x). Then x =— . (l) '

Upon multiplying by x and integrating with respect to time t, we find that

+00 2 + =

C,

C a constant,

(2)

the well - known expression for energy conservation in this conservative system. By definition, at equilibrium the particle is at rest (dx/dt = ü}. From (1) this means that if x = .x 0 is an equilibrium solution then 4 1(x0) = O. Thus at equilibrium the potential energy is a stationary function of position. We now show that the equilibrium is stable if the potenliui energy is u minimum. As was mentioned in Section 11.3 of I, the precise meaning of "stability" varies with context. This is not due to a mere lack of a standardized terminology but reflects the fact that the appropriate stability question genuinely differs from situation to situation. Thus the downward-hanging equilibrium of a damped pendulum was shown in Section 11.3 of I to be asymptotically stable in that perturbations from equilibrium eventually die out completely. There is no damping in the conservative system under present consideration. All one can hope for is that departure From equilibrium can be made as small as desired by keeping the level of disturbance sufficiently low. More precisely, one wishes that for all time the state of the system will differ by an arbitrarily small amount from the equilibrium state, provided that the initial state departs from equilibrium by a sufficiently small amount. This is the requirement of liapltnor Stability. * In other words, Liapunov stability requires that the state remains close to equilibrium if it starts close enough. * The Russian. scientist Liapunov introduced this definition in a paper published in 1893liapunov's early and profound contributions to siability theory were the forerunners of a Russian exptrtise in this field that Continues strongly to this by

Sec. 12.4] Mirrrrnut C'hararrerrzarir3rr clot Lirrear Posirit•e Problem

535

FIGURE 124 Liapttnov stability of the origin in the phase line. Given

«)

circle of radius E, there fnusl be a corresponding circle of radius point which starts in the lamer circle never mores our of rte farmer.

arl}'

such that a

Asymptotic stability, on the other hand, requires that the state gradually return to equilibrium, not just remain close to il. In the present case, the state of the system is given by the position and speed of the particle_ As we have found in several instances, it is convenient to measure displacement in terms of departure from equilibrium. Thus we introduce x—x0 ,

+x 0 ).

O()

In terms of the potential () is stationary at the equilibrium point i± = 4_ The Liapunov stability requirement is as follows (see Figure 12.4)_ The equilibrium point = Ois Liapunov-stable if for any given positive E, ,

[OW

+

whenever, for some positive

42 ( , )]nr 2 C 6 = (5(0,

for

7

0

[CM + 4 2 ( 0)]' < b.

(3)

(4)

That is one must be able to arrange that a state point will stay within any given circle about the equilibrium poinl (in the phase plane). This must he done by starting the slate point sufficiently close to equilibrium. We shall provide a demonstration of Liapunov stability under the assumptions that d"(0) > 0 and "'i) is continuous when I I is sufficiently small. Taylor's theorem with remainder then implies that

rrld =

(NO)

+ tat' + Oie],

where a = (110}.

(5)

Since 1 has a minimum when = 0, the constant a is positive. Keeping only the two lowest order terms in , we see that the energy-conservation requirement (2) yields (in terms of )

42+ ad z=2[G

-

0)1.

( harurreriru:iun of Eigcnralues and Equilibrium

536

Suie.`

as Euremu [(h- 12

Since the left side of the above equation is nonnegative, we can write

= K',

}a'

K a positive constant.

Thus the state point Os), (t)) moves in the 4 - 4 plane (the phase plane) in trajectories that have the form of ellipses near the equilibrium point O. U). The particular elliptical path is determined by the initial conditions. By ascertaining the major axis of the ellipses, we deduce tirai Ka - ]!2 [ n t}

1-(1)1 112

{

(6)

K,

Thus, as is geometrically obvious, the point ( , 4) will remain close to the origin if K is small enough. But

_

+ a42(0)

(7)

Now if 2

452,

(0) +

e5 > i},

we have

V(02 2 ^

herefore,

from

6 2,

1_4(0)] 2 < tl' K2 - 4"(0) -},

ai;'(0)

+ (4 ).

(6) and (7), L42 (t) + ^(t)]

:^^

^ E5g(a),

(8)

where

g(u) =

/() + R p,xo - rrz, (t + a)ia2

f2

< 1,

u ? 1.

We see, then, that the "nearness" requirement (3) holds if o = k/g(a). We have finished the prerequisites for the remainder of this section. We point out, however, that our demonstration of stability was not complete. Further effort is required to justify the neglect of the 0(tÿ 3 ) term in (5) (Exercise 1). We also point out that the ideas presented here can be generalized to form the second or direct method of Liapunon. To generalize to systems of high order is relatively simple: the ellipses of constant energy become hyperellipsoids, since the energy in the neighborhood of stable equilibrium will be given by a positive definite quadratic form. A more important generalization involves equations for which a positive definite energylike Liapunov function can be found that can be shown to decrease when evaluated at the coordinates of the moving state point. Then the equilibrium point can be shown to he asymptotically stable. For further reading see, for example, a treatise by W. Hahn, Stability of Motion (New York: Springer-Verlag, 1967) or a more elementary paperback, by D. Sanchez, Ordinary D erentictl Equations and Stability Theory (San Francisco: W. H. Freeman, 1 . 968),

Sec. 12.411 Minimal Characterization of Linear Pastore Prohkms

537

THE LOADED MEMBRANE

An important scientific idea, one of those which lay at the foundations of the simultaneous rise of calculus and mechanics, is that definitions and results for discrete systems can be extended to continuous systems by a process of "chopping up," summation, and passage to limiting integrals. With this now-familiar idea in mind, it is natural to conjecture that for continuous conservative systems, too, stable equilibrium is marked by a minimum of potential energy.* Rather than attempt to verify this conjecture, let us assume that it is true and see what some of the consequences are. We choose to study a problem involving two spatial variables so that a partial differential equation will emerge. Experience indicates that it is of more interest than may at first be apparent to select the problem of calculating the deflection of a loaded membrane. imagine, then, a drumhead that is loaded with a weight per unit area described by a continuous function f (x, y). Here x and y are Cartesian coordinates, and the undeflected drum will be regarded as being fixed to a curve B bounding a region A in the xy plane. Assume that the drum is stretched tautly, so that a small deflection w(x, y) will result from the loading. We assign w a positive value if it is downward (Figure 12.5). We proceed under the assumption (validated in Section 4.4) that the actual deflection will minimize the potential energy among all smooth functions that vanish on the boundary. A deflection will be regarded as conceivable (or admissible in the terminology we have been using) if it is smooth and if it vanishes on the boundary B where the membrane is fixed_

FIG U ttE

12,5. Deflec[iurr of a loaded membrane _

By definition, a membrane is a stretchable material which is so flexible that negligible force is required to bend it. Deflecting a membrane by a displaccment ►► (x, y), then, results in a storage of potential energy due solely to the stretching. One can proceed to calculate the potential energy from the standard linear constitutive assumption. However, we shall adopt a fresh (but equivalent) approach. We shall assume that for small deflections at least, the potential energy of stretching is a function of the difference between the

}s

• We have tried to make the chapters on variational methods as self-contained as possible, so we merely point out here that material in Section 4,4 has already shown that classes of Continuous elastic systems do indeed extremalize or minimize a suitable energy functional at equilibrium,

Characterization

538

of Eigentafue ► and Equilibrium Stoles as Exirema [Ch. 12

deformed and undeformed areas. This difference D is

JJ

+2,)`' 2 da—

D = 11(1 +

{9)

dc.

A

A

For sufficiently small deflections,

wx w^

l,

-}

so D :

if-1

w; + ++F^) d a ,

(10)

A

where we have used the approximation (1 + c)' 12 x i + ic. We shall assign a potential energy of zero to the undeforrncd membrane, so that 1 1,(0) = O. For small D fin a familiar process) we keep the first nonvanishing term in a Maclaurin expansion of K(D) about D — O. e thus have for some constant T that

TO

or

Vs 4 T JJ(

+ ► ÿ) da.

(11)

A

An additional contribution to potential energy, V. results from the sag due

to the weight V„,

placed on the membrane. We have

— (work necessary to lift membrane back to a flat position)

Of (x i y) d er . A

Since total potential energy V is the sum of V, and VV,, a conjecture that the deflection minimizes potential energy leads to the following characterization of w. Consider the sel of smooth functions w which vanish on B_ 'The actual deflection w is that member of the set which solves the problem minimize IT if _ + wÿ} der — A

We know, however, that to these conditions' on w: —

T Ÿ2w

---

f,

J

fx1)f(x. y) da.

(12)

A

a

force-balance approach to our problem leads

w—

0 on 8 (F 2

3 -

exz

+ dyz

.

(13)

Here T is the tension in the membrane. * A detailed derivation can be obtained by a slight modification of the derivation of the surface tension boundary condition, found in Section 7. I - In that section, as here, there is no resistaace to bending and a tension resists stretching- In water waves an interface is subject to a net pressure p T p, wihich here should be replaced by the weight density J. Since we are —

.

limiting ourselves to small deflections, the surface curvature can be represented by its leading term- 1 his was ❑ ' in Section 7-1. Since we take deflection to be positive downward, it is —y e w' here.

Sec. 12.4] Minima! Characterization of Linear Pasture Problerni

539

EQUIVALENCE OF AN INHOMOCENEOUS EQUATION AND A MINIMIZATION PROBLEM

We could proceed by showing that, as in particle mechanics so in appropriate problems of continuum mechanics, the minimum potential-energy characterization of stable equilibrium is equivalent to that obtained from a balance of forces. it is of more interest here, however, to provide a direct demonstration of the equivalence of (12) and (13). For the main purpose of this section is to show the equivalence of ce rt ain problems in differential equations and certain minimization problems. Some rearrangement and introduction of new notation will allow us to obtain results of considerable generality with little extra effort. The "rearrangement" is actually an application of Green's theorem, which shows that

J5

(wx f w)

do = JJi

Jiv

vwi da _ J

A

A

vi w do,

(14)

A

since w vanishes on B. We introduce the notation (w, f ) ^`

f

f wx. y) f (x, y) dc.

(15)

We define two sets of functions as follows:

R = {ii l>ti is continuous in AL

(16)

S = fw r is twice continuously differentiable in A, K• = 0 on B}.

(17)

Finally, we make the abbreviation

L-

TV2 .

118)

Note that L takes functions in S into functions in R:

(l9)

L : S —' R.

With all this, our task becomes that of proving the equivalence of the following two problems. Problem A Find an element w e S that satisfies Lw =

f

Problem B

Minimize F(R)= (Lii, ►t•) -- 2(i . f) among smooth functions tit- E S. [ln obtaining Problem B we have doubled the quantity to be minimized in (12), and have employed (141 In proving the equivalence of the two problems, we shall use the following properties of the entities we have defined.

54 0

Characrérizarion of Eigcntalucs and Equilibrium Slates

as Exert ma [Ch_ )2 -

(1)

S and R are vector spaces under the "usual" definitions of addition and scalar multiplication, as in the two-dimensional version of Example IV, Appendix 12.1. (ii) The expression { , ) of (15) defines a real-valued function on pairs of elements of S and R, and this function satisfies all the axioms of a scalar product. {Compare Appendix 12.1, Definition 2.) L is a linear operator {with domain S and range RI and it satisfies the self-adjointness property (Lu, (iv)

if u, r E S.

(u, Lr)

(20)

L is a positive operator, in that if w E S, (Lw, w)

0,

(Lw, w) = 0 if and only if w is the zero element of S. (21a, b)

This follows from (14) and the vanishing of w on B. NOTE. Both (iii) and {iv) are special cases of Exercise Al2. 1.10. (v) If it is true that

(NJ) = 0 for every s in S. then if f E R,f is the zero element. (This isa restatement of Lemma A. Appendix f 1.1.) The remaining results of this section hold whenever the properties (i) to (v) are applicable. This is the case for the original problem that motivated us. but it is also true in many ether instances. (Some of these are the subject of exercises.) The reader who seeks no added generality can simply verify that the various properties which we shall use actually hold for the particular definitions of ( , ), R, S, and L given in (15) to (1$). As in exercise A 12.1.ôc, it can be shown that (iv) implies that if [u, e}

(LRu, v);

u,

€ S;

(22)

then [ , J has the properties of a scalar product. In particular, if ( 23 )

then II J can be regarded as a generalized length or norm in the sense of (Al2.1.8). Theorem 1. if Problem A has a solution, the solution is unique. Proof. if Lw, = f and Lw 2 = f, then Lk, — w 2 ] = z (z being the Zero element of S). Consequently, by either (2 lb) or (Al2.1.8a), ^Iwl

— WI N

(L[x'i — w1]. w, — w 2 ) = 0, so w 1 = w 2 . 0

(24)

Theorem 2. If w is a solution of Problem A, it is a solution of Problem B. and conversely.

See. 1 2 .41 Prooaf.

Min imul Characterization of Linear Positive ProhJerf! w -

i4 i

Suppose that w is a solution of Problem A. so that L w — f. Then w} - 2[^r titi'] + [w, Iv] - [ ► {', iv] = EH' - Si' ^^ [w, +S'^ ,

.

or fliv)

Il çi

' w II 3 - W +• it -

125)

To solve Problem B, we must choose an clement of that minimizes Fits ). From (25) it is obvious that the choice is IT _ w. Note that Min F(i) = - 1I4 2 ras

{26)

To prove the converse, suppose that w minimizes Fi+'). Let s be an arbitrary clement al S. Then if Az) = F(w -}. cs), the condition .i'(0) = 0 is necessary. But (r.) = (Lk. + as]. w + rs) - 2(w + 4s, fl

(Lw, w)

—

2(w. f) + r. (Ls. w) + (Ln. s) - 2(a,1)] + 0(1: 2 ).

Using the self-adjointness property, we find .410) = (L.s. tr) + (Lw, }) - 2(+. !1 so that Lw = f by (r).

2(l.ws - f. ,)

(271

11

By demonstrating the equivalence of Problems A and B, we have in particular proved the equivalence of the minimization pro blcm(12) and the Poisson problem 04 In terms of the membrane deflection interpretation we have shown the equivalence of a force-balance approach and an approach that characterizes equilibrium as the state that minimizes potential energy. Note that for the membrane (Lit.. is•) is the potential energy of stretching. Thus the the positive definite character of L can be regarded as expressing the fact that any deflection requires work to accomplish or (equivalently in this conservative system) Stores energy in the membrane. Many other minimum energy characterizations of mechanical problems can be fitted into the framework of our general theory, Particularl) for problems in elasticity, one expects repeated occurrence of positive definite operators to express the fact that a positive amount of work is required to distort a system from its unstressed "natural - state. -

THE RITZ METHOD APPLIED Ti] THE TORSION PROBt.EM1

To obtain an approximate solution of Problem B (and hence Problem A). we make F(0) as small as possible by choosing appropriate constants 2 in the trial function 1►

=

E 7C j t1•, .

f281

54 2

C.ltaracierrzafion of Eagentalues and Equilibrium Suites as Exirema [Ch. 12

In (28) the w ; are preselected in the hope that a linear combination of them

ca n prov ide a good approximation to the solution w. This is the Ritz procedure again, and it proceeds much as it did in the eigenvalue application of Section 122 With (28) we have N

F(w) (Lw, w)

2(w, f ) =

y ^; ^ 1 rt; ►

M ,

— 2 1 a ; b; ,

4.;= t

(29)

r=

where

(Lw ; , w a),

fa = (w';. f )•

(30)

To find how to choose the coefficients a ; in order to minimize F, we differentiate with respect to ot s and obtain

y "la ak

bf .

(31)

A sufficient condition for theexistence of a unique solution to (31) is the linear independence of the w, and the positivity of L (Exercise 12.2.6). Because it shows the strengths and weaknesses of the Ritz method in an important but relatively easy problem, we now turn our attention to the torsion problem of linear elasticity. This concerns the twisting by stresses on its ends of a cylindrical bar whose generators we imagine to be parallel to the z-axis. Let the cross section A that is intersected by the xy-plane be bounded by the closed smooth curve B, as in Figure 12.6_ As we have shown in Section 5.2, there is a solution in which the cross section located a distance z from the xy - plane rotates through the angle Oz, where 0 is the constant rotation per unit length. The tangential stresses T_ and T) on any section z = constant are independent of z, and can be written

7

^

Ow =^ y . ^=

t?w

(32)

where the stress function w(x, y) satisfies

— V 2 w = 2GO,

w = Q on B.*

(33)

(Here G is the shear modulus, a constant that characterizes the material's resistance to shear.) The sides of the bar are free of stress and the resultant of the forces distributed over any section is zero. Rut these forces provide a

• Equation (33) occurs in several other physical contexts such as viscous fluid flow down a cylindrical pipe with cross section B (Exercise 3.2.l4) and random walk in a region bcundr l bY 811. f quation (14 20)].

Sec. 12.4] Minimal Characterization of - Linear Positive Pfnhieme

543

F l C uRr. 1 2.6. 'Iarsion of a cylindrical beam by u couple of R ►ktyniftKle

couple whose magnitude Al is given by

M T JJ(x T:,

(3 4)

^

so [ Exercise 3(b)3 Ilf = 2 flit} c .

( 1 S)

d

Equation (33) can be put into the form of Problem A merely by making the identification f 2Gfi. Let us test the Ritz principle when A is the square lying inside the lines x — ± 1, y = # 1. From symmetry considerations. the simplest sensible trial function is = otti

(x

x — l )(y 2 — 1).

(36)

From (3l) one finds [Exercise 4(a)] that ;" = Using this value of xi", one u', can substitute the approximation (36) into (35) and find that M

544

Uh uracierimrian of Eigeni lues and Equilibrium Srates

as Extre na In_ 12

where Mt" = 2.22G0. To improve the approximation, it is natural to try 1)[ iz, + a21(x2 ws^i = (x 2 T 1)(y2 + y 2)1 (37) —

.

Here and below, using results given, for example, by Mikhlirt (1964, Sec 65), the results turn out to be = (35)(31)

(8j(277j

GO, ^ {2ai ^

(5^(lU5) GOM (8)(554) >

:

A1,112 /

where 14!12 ' = 2.24660.

(38)

The close agreement between the two approximations Mt" and M"i encourages us to think that our results are accurate_ Indeed, M is actually a bit smaller than 2.250, so the error in MI ' is less than one-fifth of 1 per cent_ is it a coincidence that the approximate values of decreased as more terms were taken, and that both lay above the true value? No, it is not, as is shown in Exercise 6. Of principal interest in deciding whether or not the beam will fail when subject In torsion is the maximum magnitude of the shear stress_ We thus seek to approximate f

= ma?[ ( T'=„ + T}) oz Lx. Yjf A

(39)

The maximum occurs on the boundary of the square, in the middle of each side [Exercise 5(a)]. Thus its value can be determined from (40) r = Iw y( 0. 1 )1. Using the two trial functions IV° and yti.^2', we obtain approximate values = 1.25G0 and 1.41 GO, while the true value can be shown to be 1.35GO. We see that in contrast to the situation for the moment hi, here the Ritz method does not provide a sequence of nondecreasing upper bounds for the correct answer. Moreover, the approximations are not nearly as accurate. For these facts we can give the following reasons. The moment M requires evaluation of an integral. As is usual in such instances, small errors in the approximation to the integrand 1 should not have too large an effect_ Indeed, Exercise 6 shows that M itself can be characterized by an extremal property, so that first order errors in estimating should give rise only to second order errors in M. By contrast, t 1 " and rt e ' were obtained by evaluating, at a particular point, a derivative of an approximate function. This is a notoriously inaccurate procedure: A spurious twitch in the approximation can give rise to a large error. Whaiever method is used, reliable approximations to T arc difficult to obtain. In concluding, we point out that the torsion problem (33) is formally identical to the membrane problem (1 3) when the weight (or pressure) func-

Sc. e 12_4]

Almond Characterization af Linear Posirrioe Pruddcros

545

lion is the constant 2Gf7JT. In one case w is the stress function, in the other w is the deflection_ This membrane analogy was pointed out by Prandtl in 1903. The analogy furnishes a practical procedure for determining the shear stresses in a twisted beam of complicated section. One examines the deflection of a soap film that is fixed to a curve having the same shape as the boundary of the beam, and that is subjected to a uniform pressure. [Sec Sections 93 and 99 of Timoshenko and Goodier (1970). These sections are part of a chapter devoted entirely to the torsion problem.] Of several other scientific questions that reduce to (33), we make particular mention of the random-walk problem discussed in Section 3.4 of 1. There we pointed out how a probabilistic question that arose in a biological context was answered quantitatively by referring to classical calculations on the torsion problem.

EXERCISES L (Project) Justify the neglect of the O( e) term in (5)_ 2. Generalize to three dimensions the text's discussion of Liapunov instability at a minimum of potential energy_ 3. Verify: (a) equations (29) and (31); (b) equation 135). 4, Verify the text's results lai for x; 1 ' [as defined in (36)] and M 11 : (b) for oi412 j, ott22 °* and M i 2 I . 5. (a) Verify (40I. (b) Verify the text's values for r 0 ' and rte'_ 6. (a) Use (26) to characterize the couple M as a minimum, (b) Use (31) to show that the Ritz method provides an upper bound for M. 7. Referring to the example at the beginning of this section, show that equilibrium is unstable if the potential energyis(a) maximum, (b)stationary, but neither a maximum nor a minimum. As part of the problem, formulate an appropriate definition of instability. 8. By sketching relevant circles and ellipses, give an entirely geometric demonstration of Liapunov stability for the example considered in the text. Start from 42 + a a = K 2 . 9. Discuss the stability of equilibrium for the undamped pendulum using the Liapunov approach. (Compare Section 11.3 of I_) 10. The deflection w of a bent plate can be shown to he governed by the equation Vow = qjD, where D is a combination of elastic constants and q(x, y}) is the density of loading_ if the plate is rigidly fixed along its edge B, both w and its normal derivative must vanish at S. [The problem is a two-dimensional, time-independent version of (5.1.37) and (5.1.38).] (a) Discuss the variational characterization of this problem. (Proofs of self-adjoin tness a n d positivity are reg uestcd in the Exercise Al2.1.13.) (b) When q is a constant. find a one-term approximation to w. Assume that the plate is square, and use the Ritz method.

546

Characterization of Eiiernwdurs and Equilibrium 5;ores as

E_Yrrmw

[Ch. 12

11. Show that a change of variable reveals an analogy between planar membranes subjected to a varying loading and unloaded membranes fixed to a nonplanar boundary. 12. The electrostatic potential w[x, y, 2) exterior to a conductor bounded by a closed bounded surface A satisfies for x (x, y, z) exterior to A ; w = wo far x lim lx [wand lir ix I bounded.

E

A;

Here w0 is a constant. (a) Set the problem up to conform with the abstract formulation discussed in this section, proving the requisite self-adjointness and positivity. Begin by considering a region exterior to A and interior to a sphere of large radius. (b) Discuss in general terms the Ritz method and its use in determining the capacity C, where

A

In this definition, a/On denotes the exterior normal derivative.

Appendix 12.1 Self-adjoint Operators on Vector Spaces in Chapter 5 of I we showed that eigenvalues of Sturm--Liouville problems for second order ordinary differential equations were real, and that eigenfunctions corresponding to distinct eigenvalues were orthogonal in a certain defined sense. In Section 2.2 we showed that eigenvalues of real symmetric tensors were real, and that cigenvectors corresponding to distinct eigenvalues were orthogonal. In this appendix we formulate an abstract symmetric eigenvalue problem that embraces the two cases just mentioned and many nacre_ Indeed, we go further with our abstraction. We list axioms that define an abstract "vector" in such a way that operations of addition and scalar multiplication such as one carries out for directed line segments are equally valid, for example, when carried out on certain sets of functions. We also present an abstract version of the usual scalar or "dot" product. This is needed to define sell adjoint and positive linear operators and to derive properties of equations that involve such operators. A few more concepts are mentioned in what is essentially a brief introduction to functional analysis. Some prefer to learn powerful unifying ideas first, and then have them illustrated in a number of concrete cases. They can read this appendix early_ Possibly, an example or two will have to be omitted, but the material presented here is essentially self-contained. Indeed, although the material is

Appendix 12_ f ] Self-udjorrr; Operators on Vector Spaces

547

important, we have placed it in an appendix to emphasize the fact that it can be inserted at any of several points. It is more in the spirit of This book, however, to observe how the abstract grows out of the concrete, If two or three particular concrete instances a re presented first, at the price of some "inefficiency" one gains powerful motivation for studying the appropriate abstraction_ Often the abstract procedure can he obtained by selecting what comparison shows to be the common features of the particular arguments. Applied mathematicians must be vigilant about keeping in mind the concrete manifestations of the abstract ideas. More than that, they must keep in mind that in any particular instance, more can often be accomplished than is encompassable by a general theory. Yet one cannot deny the elegance of the abstract ideas nor the utility of separating the particular aspects of a situation from the common structure that underlies a number of related situations. Our plan in this appendix is to present a number of abstract definitions and a few important theorems. Several concrete examples of the abstractions are given, most of which are discussed elsewhere in ibis book. Many readers will be familiar with at least the beginning of the succeeding discussions, and it would he beyond our scope to convey a detailed picture to those who are not. Thus many proofs will largely be left to the exercises, particularly those that require a more-or-less routine verification of certain properties.

REAL VECTOR SPACES Definition 1. Let S be a set on which is defined the following two operations: (1) addition, which associates an entity !; + tv with any two elements z and w of S; and (ii) (real) scalar multiplication, which associates an entity au with every real number a and element y of S. If the following rules are satisfied for any elements u, y, and w of S, and any real numbers ar and then S is called a vector space over the real numbers or just a real vector space.• (a) r + w is an element of S. (h) {u+ :a)+ w=u+(a+w). (c) There exists an clement z of S (zero element) such that c + z = u for every u in S. -I(d) To every v in S there corresponds an element of S, denoted by -- e, such that u + ( u) -- z. [Notation: By r~i - w is meant to + (—w}1

?,

—

• 11 in the LO rector-space postulates taj to 011, Greek letters can be complex numbers, then we have a rector space over floe complex numbers, Abstraction of the reyutccd properties of real and compkx numbers is shown in algebra texts to give rise to the concept of a field. and thence to the corresponding notion of a vector space over an arbitrary Field. f Many authors denote the zero element by'' but we employ '`z" to emphasize that this element 15 not necessarily a number_

[haracrerizaRiun ufEiyent;afues and Equilibrium Stales as Exirema

548

[Ch_ 1 2

(e)

u } y = y + 14_ [Properties (a) to (e) mean that S is a commutative group under addition.] (f) arty is an element of S. +cow. (g) a(v + w) (h) (a+/3)u (i) (Nfi)v = OEM). ke = V. (j) The following sets S; , with accompanying definitions of addition and scalar multiplication, are examples of real vector spaces. Verification is left to the reader. Example 1. S t is the set of directed line segmen ts in ordinary titrer-dimcnsi nual space. Scalar multiplication and addition are defined in the usual geometrical way (prolongation of the segment, parallelogram law). . a„), where the a are Example 2. S2 is the sel of ordered n-tuplcs a (a 02 . real numbers. if and b =_ (b c , b 3 , _ _ _ hAj are elements of S3 and a is real number, by definition, a + h is the n-tuplc whose ith element is a ! 4 b.; as is the n-tuple whose ith element is aa1 . Also . .

,

z is the n-tuple whose ith element is zero; — a is the n-tuple whose ith clement is — ai .

Example 3. Si is the set of nth order Cartesian tensors. Definitions of addition and scalar multiplication are given in Section 2.!,asare elements of the proofthat the tensors form a vector space with these definitions. Example 4. S4 is the sel of real-valued functions defined on an interval [a, h] and possessing n continuous derivatives. if f and g are elements of S. by definition

f + p is the function whose value ai x isf (x) f

g(x);

of is the function whose value at x is of (xj.

Example S. Ss is the subset of S4 wherein the functions vanish at a and at b. The definitions of addition and scalar multiplication are "inherited" from Example 4_ [A theorem of algebra, applicable here, states that to show that a subspace ofa vector space is itself a vector space one only has to verify the closure properties (a) and (f) of Definition 1 .] A s an illustration of vector space* manipulation we prove that t

= z fur every i) in S.

(1)

(Recall that r is the zero element of S.) This follows from our postulates. since IL , can be "canceled" from both sides of the following equation! in +Da= (I I-(1)u= I

•

Vector spaces arc also called linear spans.

+ r.

Appendix 12_1] Neff-ttdjoint Operutor_ti un Vector Spo c r.►

549

SCAI.AR PRODUCTS

Definition 2. Let S he a vector space over the reals. Suppose that with every pair of elements n and w of S there is associated a real number (u, w) with

the following properties: (a) (air, w) = a(u, w), ce any real number. (L + t), w) = (u w) -F (y, w), where a is also an element of S. (b) ,

(ci (w. e) = (e, w). (d) (v, v) k 0 and (e, e) = û if and only if e = z. Then (ta, w) is called the (real) scalar product (or sometimes inner product) of V and w. 1f (ti, w) = 0, then y and w are termed orthogonal. As the reader can verify, the following are examples (Areal scalar products. Examples 6 to 9 refer to scalar products defined on the vector spaces that were given as Examples I to 4 , respectively. Example 6. For directed line segments r and w, the ordinary "clot" product it formed by multiplying the rnagniLudcsofi a.id Lt byihecosineof the angle hetweenthem. In the present context, the dot product is also correctly called a scalar product. Example 7. if a and hare ti-tuples, a scalar product is defined by (A,b) =

Er u1 b ;

(2)

-

^ -

Example S. If a and b are first order Cartesian tensors, a scalar product is formed by their contraction. Example 9. Lei j and y he elements of the vcaor space of functions defined and continuous on the interval [u. bi. A possible scalar product is given by (j. 9) =

fl.x)g(x)rixl dx,

1.3)

where r is a (fixed) continuous function Chas satisfies rlx) > 0 for u < x b. An important special case is r -= 1. tote for instance t hat under the scalar product

(f. = f f(x)g(. ) dx,

(4)

if f is even and g is odd, then f and g are orthogonal if a = b. As an illustration of manipulation with scalar products, we prove that = ti for any r

tri S

,

(5)

This follows by means of postulate (h) For scalar products. since (t,

tr,„

—(v.v) — tu,y)= 0

(^f

When performing manipulations with a scalar product, one roust he sure that all the appropriate axioms are saiislied for every possible element that can appear in the product In what follows, we shall always assume ;hut this is the case, without further explicit -

indications.

5S0

Charrscter[zuium of Eiyenvalues and Equilibrium Stales as Extrenua lCh. 12

Just as with "ordinary" vectors, elements v, of a vector space S (i called linearly independent provided that

• y a, v, = z

I, ... , n) arc

only if ere = O.

(7)

I

Example 10. Using (1) and (5), show that mutually orthogonal nonzero vectors are linearly independent _ Salado& If the nonzero vectors v, are mutually orthogonal, then (v ; , v1) = Ab u for some constants p,. The constants must be positive. (Why?) But upon taking the solar product of (7) with vj , we find that

E abb. u3) = C or y ct c p, 6 4 = ü.

i^1

r^l

Since

pi > 0, a l = 0. 1 = 1. -..,n. Kali a, are zero. then (I) implies (7). Vectors u in spaces with an inner product can be assigned a generalized length or norm Ilvi by means of the following definition:

u)" 2 .

411 =

[Here and below the nonnegative root is implied by ( ) I J 2 .1 It can be verified [Exercise 22] that the following properties hold:

I vII II III —

Ilvii = 0 if and only if :a

0,

(8a)

z.

(Rb)

IaIIIvII-

I(U, w)I s IIvII IIAII

if + +III s II v II + II + Il

(Schwarz inequality)_

(8c)

(triangle inequality).

(8d)

Example 11. Consider inequality (8c)—which is attributed by various authors to one or more of the mathematicians Cauchy. Schwarz, and Bunyakovsky For continuous functions un Erg b] subject to the scalar product .

,

f

f (x)g(x) dx.

a

this gives

Ç fxx

dx

r^2 Sfxdx^ixdx+r2 ■

■

while the triangle inequality (8d) implies that 1r2 1f2 fu(x) + gx)] dx 5f 3 (x)dx + Çy(x) dx • f •

L,2

Appendix 12.]]

55i

Self-adjoin, Operators on Vector Spaces

The degree to which the directions of ordinary vectorsr 1 and r 2 coincide can be measured by the angle between them_ The cosine of the angle, or (VI ' v211f ► 1 I1bs l) is perhaps an even better measure, for it ranges from —1 (when the vectors point in the opposite direction) to zero (when the vectors are orthogonal) to 1 (when the vectors have the same direction). The Schwarz inequality (8c) permits the same measure 0i- coincidence or correlation to be used for functions, for irx ffixJix ) dx^ f i(x) dx r1 f^(x ) dx

will also range from — I to 1. Such a measure of correlation is frequently used. in the theory of fluid turbulence, for example, the intensity of randomness might be gauged by measuring the correlation between pressure traces taken over a period of time at two different points. LINEAR SELF-ADJOINT OPERATORS Definition 3. Suppose that there is a rule L which associates with every element y in a (real) vector space S precisely one element w in a vector space T Then L is an uperazr r with domain S and range T(or a mapping fro n S info T). We write L : S —, T or w = L[v] or iv = Le. L is a linear operator if and only if for any elements n and li of S and any real numbers a and /3, we have L[oty + flw] = aL[r] + (r.[, J

(9)

or, equivalently [Exercise 3(h)], L[rxv] = rlL[4],

L[o + w] = L[r] + L[151

(14)

Examples 12 to 15 refer to the vector spaces of Examples 1 to 4, respectively. (Proofs are largely left to the reader.) Example 12. ifv is a directed line segment. L(u) = 5i defines a linear operator, for

L[cw + Pw] = 5(au /Ivy) = alSrl t {5w) = irL[r] + fJL[i ]

.

Exttntple 13. Let a be an n-tuple and M a fixed matrix with elements m,,. Let

c = L[a] be defined by c3

E

M4/. x .

j'• 1

Then L is a linear operator. Of a is regarded as a column vector, we can use matrix notation to write L[a] = Mai Example 14. If T is a Cartesian tensor of order n, and Q is a fixed Cartesian tensor of order n, then

L[T] = T • Q defines a linear operator mapping nth order tensors into tensors of order n — 2.

(1 2)

Characterization of Eir enesalurs an d Equilibrium Stairs as Extremes ([_h. 12

55 2

Example 15. Let if- be an element of the vector space of twice continuously differentiable functions_ Then L[ f] = d ifidxr defines a linear operator mapping this vector space into the vector space of continuous functions_ The operators

Ne(f) = f 2 and Ni(f) —

d

dx + 7

are not linear because

N 1 (af + jig) :ü hf 2

+ 2ixffy+ (11 g 1 but

aN 1 (f)

f

l!iN101 = af' +

while

N,(a^ + ^) =

+ ^ ^^ + 7

but

at iV= 1 f ) + ^i1^` 1 ^4)

T

a

df

+ ^i ^x + 14.

Definition 4. Let L be defined on a real vector space S with inner product. L is termed self-adjloint (or se/far/joint ewer S) if (Lu, y) = (u. Lu) for every 14 and u in S. Example 16. If L is a second order differential operator (dfdx)(pd/dx) and S is the vector space of twice continuously diflkrentiable functions that vanish at x = a and x =. b, then the self adjointness of L is asserted in i, Equation (5.2.7)_ The inner product is that of (4).

Example 17. Linear operators associated with symmetric second order tensors are proved to be self adjoint in Lemma 1 of Section 2.2. The " vectors " here are first order tensors, and the inner product is formed by contraction.

Example 18. Let S he the vector space of rr-tuplcs with scalar product (2) and let the linear operator L be defined as in (I l)- if the coefficients rt+ t1 satisfy the symmetry condition M ; j = mg. then L is self- adjoint_ The proof of this statement is virtually identical to that of Lemma I. Section 2.2. One uses definitions, subscript interchange, and the hypothesized symmetry; as follows: R

(b, La)

=

11

X hi 1=

rfJiJa l

1 1- 1

y

b,rtJi,ut

j - J i•

ai E rn„b — (a' Lb). ^

(13)

EIGF.NVAI.IlF. PROBLEMS Let u be a nonzero element of a

vector space S. and let Al be an operator (not

necessarily linear) whose domain is S_ If Mu = for some scalar A, then di is M, called an eigenvttlue of and u is the corresponding eigenveclor.

Appendix 12.1] Seff-adjr ►inr Operators on Vector Spaces

553

Example 19. Consider the vector space of functions that are defined and twice continuously differentiable on ftl, rt] and that vanish at the end points of the interval_ Then the linear operator

L

d^ —

(14

-^ x#

)

has as eigenvalues the squares of the positive integers (i.e.. the nth smallest eigenvalue equals rily The corresponding eigenvectors are any constant multiple of sin ax. „

The question arises: Are the eigenvalues real if the problem "looks" real? The answer in general is negative. (An example is given in Section 2.2— there the complex character of the eigenvalues LS a consequence of the fact that a real polynomial may have complex zeros.) As we now show, however. excursions into the complex domain can be avoided in a restricted but important class of eigenvalue problems_ Consider linear operators on column matrices defined via multiplication by a square matrix with real coefficients, or linear operators on functions defined by differential operators with real coefficients. We have in these cases L[u + iv] = L[u] + iL[r]

,

L [u ] real.

(15a)

Here we are expanding our paint of view to consider a new vector space C with elements u + it where u and :a are in the original vector space S. In general, if a linear operator satisfies (15a),wecall it real. Note that (15a) implies L[u + iv] = L[u] + :L[u] = L[u] — iL[t.] = L[14 + rv].

( I 5b)

We now rule out the possibility of complex eigenvalues and eigenfunctions for real self adjoin: operators L. That is. we show that there cannot be a number 2 2(r) 4- #O. and some u+i A 0 such that

L[u + i c] = J.{u + i v )_

(16)

Theorem 1. Eigenvalues of real self-adjoint linear operators are real. The corresponding eigenvector can always be taken to be real. Proof With ). _ {'' +tit'' (16) is equivalent to tr} L!1 =- ;t u — Lu = /4tr`t,' -F- P'ie. (17a, b) ,

Using (l7) and the self-adjointness of L, we have

0 = (ü, Li)) — (Lu, t)) = (u, A mu -F- A" u) — {. {r1u — = P'[IlL4 11 2 4 110 21

u) ( 1 8)

Since u + it; is an eigenvector of L, it must be nonzero. We can therefore

conclude from (18) that At' ) = 0, so the eigenvalue 1 is real_ Moreover, Lu = 1. {'fu and = J.{"U, so either u or L. cart serve as a real eigenvector associated with the real eigen value 2 = A'''. d

554

characterization of Elyerrt+alues and Equilibrium Stages as 6xlre+rru

Pe. 12

Theorem I means that we lose no eigenvalues or eigenveetors by remaining in the real domain. This is guaranteed to be the case, let us emphasize, only when L is Self-adjoins. We have seen special instances of this result for real self-adjoint Sturm-Liouville operators in Section 5.2 of l and for real symmetric second order tensors in Section 22. We now state the abstract version of the orthogonality results that were proved in these instances. Theorem 2. Eigenvectors corresponding to different eigenvalues of a real self-adjoins operator are orthogonal. if n linearly independent eigenvectors correspond to a single eigen value, n > 1, then one can find n orr hogon l eigenvectors corresponding to that eigenvalue. Proof. Suppose that

Lu= 2.0

Lta

and

—

ttu,

2#

Then, on taking the scalar product of the first equation with u and the second with u and subtracting, we find that (,

Lu) — (u, Lc) — ){t

,

u) — µ{u, u) _ (• — 1.4(u,

(19)

u).

(We have used the symmetry property of the scalar product_) But the selfadjointness of L implies that the left side of (19) is zero_ Since À # the orthogonality relation (u, a) = D is established. To construct mutually orthogonal eigenvectors from a number of linearly independent eigenvectors corresponding to one eigenvalue, one employs the Gram-Schmidt process—just as in Section 5.2 of I. 0 ,

Example N. The self-adjointness proved in (13) implies that the eigenvalrtes of a symmetric n by n matrix are real, and that the eigerivecturs can he regarded as

real

mutually orthogonal.

POSiT1VE OPERATORS

Let M be defined on a vector space S with a scalar product An operator WI is called positive (or positive over S) if (Mu, u)

0 for u in S,

(Mu, u) _ 0 if and only if u

z.

(20)

Example 2l. let She the space of functions that are defined and twice contiguously differentiable on [0, 1] and that vanish at O and L Then L = —d 2 /dx' is positive over S, using the "usual" scalar product (4). For if u E S, integration by parts yields t^

(u, L.u) = ^ Li( — rd") n

fix

^

JP dx o

(1-

(21)

Equality holds if and only in' = 0, or u = constant; and the constant must be zero because of the boundary conditions.

Pierwr Spaces

Appendix f 2.11 Sr}I-rit!'uinr Operutr ► rs

555

A useful generalization of the above material on eigenvalue problems concerns ;Mir,

Lit

(22

where L and M are self- adjoint, and M is positive. By following the line of proof used in Theorem I, one can show that the eigenvalues 2 are real [Exercise 6la)]. if we attempt to repeal the calculations used in Theorem 2 [Exercise 6(b)], we are led immediately to the relation lie', — p)(v, Mu) _- 0 or tr, M u ) = 0,

123a, b)

where tr and r are cigenvectors corresponding to distinct eigenvalues and pt. To interpret (23b) as an orthogonality ri u1t, we define a new scalar product < , ) by

n.

( 60 )

A Cauchy sequence {xj, x ; M, Las the property that for every positive e, there exists an integer n such that d(x; , x2 ) < c

whenever t > n

and

j > n.

(61)

A metric space is complete if every Cauchy sequence is a Convergent sequence. (Do not confuse the concepts of a complete metric space and a set of elements that are complete in a given vector space.) The rational numbers under the usual absolute value metric are an example of a metric space that is not complete, for a Cauchy sequence of rational numbers can, of course, converge to an irrational number A normed vector space (or normed linear space) is a vector space with elements x ; on which is defined a real-valued function II I with the properties

Ilx;ll

0, 11;11 = û if and only if x i is the zero element; ( 62 a, b, c) Ilax;i1 = 1( I IIx ; II; + 11)0. (Ixt + x3 11 ?

A Wormed vector space can always be regarded as a metric space if rune empl oys thenatural metric generated by the norm and writesd{u, tr) = 11 14 — 1'11 ABanchspeiormdvctspaehiomltnsaur metric. A scalar product space is a vector space with a scalar product defined as in Definition 2. Such a space has a natural norm x i II (x ; , x )` r z and hence a natural metric. It is a normed vector space, since equations (8a, b, d) of Definition 2 are the same as equations (62a, h, c). A Hilbert space is a scalar product space that is complete in its natural metric_ A Hilbert space, with its scalar product and norm, has more structure than a Banach space. A Hilbert space is a Banach space, but the converse is not necessarily true_ ,

,

COMPLETELY CONTINUOUS OPERATORS

The abstractions of functional analysis establish useful analogies between structured classes of functions and ordinary geometry_ Proofs in Hilbert space theory, for example, can often be constructed by drawing appropriate

564

C. Iwuruclerizurion of Eigrnrulue.s and EguilibritJrrt Surres as Extrema

[Ch. 12

diagrams and then translating the geometric steps into manipulations of the abstract lengths, scalar products, and angles that are defined by the axioms. The analogy between Hilbert space and ordinary three-dimensional Euclidean space breaks down in connection with compactness. A set of Hilbert space elements is bounded if each has a norm that is less than some constant; the set is compact 'revery sequence of elements chosen from it contains a convergent subsequence. In three (or n-}dimensional Euclidean space, the Bolrano-Weierstrass theorem shows that a closed bounded set is compact. But this connection does not hold generally in Hilbert spaces, as is shown by the Following example. Example mple 25. Consider the closed bounded set S of Hilbert space elements f that satisfy fI f Il 5 I. Let f, be a sequence of orthonorrnal elements in S:

f)

=

No subsequence converges, for the "distance" between any pair of elements is

. as is

shown by the following calculation:

6

-- fJ11 3 - 1Jf - I1 , ^ fjl f IL) ( - ^^++r 1 +{ + IIJ^ +^f^. = lfi• fJ^ = 1+ 6 +I -2. -

Again and again we have encountered examples of Sturm-Lionville problems. These involve differential operators, which "unsmooth," in contrast to "smoothing" integral operators. Thus an advantageous reformulation can be obtained by inverting the Sturm-Liouville problem Lu Au by means of the inverse operator L' (assuming that it exists)_ We have L14 =

Au

implies that

14 =

AL - 'u or L - 'u = 1 - ' il.

We see that the nonzero eigenvalues of L- t and the nonzero eigenvalues of L are mutually reciprocal, and the corresponding eigenvectors of the two operators are identical. For Sturm-Liouville problems, Green's functions can be used to invert the differential operators and thereby to obtain corresponding eigenvalue problems involving integral operators_ The self-adjointness property is retained by the inverse operator (Exercise 27). If we write the inverted problem in the form Au = pie, we expect that the operator A has a property that reflects its smoothing character. Such is the case: The sequence Ax, turns out to be compact whenever x„ is bounded. Operators with this property are said to be completely continuous. Theorems I and 2 of Section 12.2 characterize the eigenvalues of a Iinear self adjoint operator L under the assumption that (Lu, u)f(u, u) has a greatest lower bound and that this bound is attained for an element in the domain of L. The inversion that is necessary to permit the development of Sturm-

Appendix 12_11 Sr4f adjarni Operators on Vector Spaces

565

Liouville theory in terms of completely continuous operators prepares us for the fact that the eigenvalues of completely continuous self-adjoint operators arc characterized by a raximutn property. Let A be a completely continuous self-adjoint operator whose domain is a Hilbert space S. One can prove the existence of a solution to the problem of maximizing I (Au, 01/(u. 4 The maximum gives the modulus of a nonzero eigenvalue, and the maximizing vector is an eigenvector. A sequence of maximum problems with orthogonality constraints (as in Theorem 12.2.2) yields all nonzero eigenvalues and the correspondingcigenveclors. Moreover, if A does not have a zero eigenvalue, the eigenvectors are complete in S. It is fitting to illustrate the theorems of functional analysis with results that show the existence of the complete sets ofeigenvectors that we have been manipulating. Yet it can be argued that we have just changed the nature of the problem, so that the task is now one of proving complete continuity (or some other sufficient condition). There is some truth in this, but the functional analytic approach has the advantage of breaking the problem into a portion where detailed analysis is required to establish the characterization of operators and spaces, and another portion where one derives strong and clean geometric characterizations of these operators and spaces_ Proofs of the detailed assertions that we have made, and an introduction to the literature, will be found in such references as Friedman (1956h Stakgold (1967), and Riesz-Sz.-Nagy (1955).

EXERCISES

I. Verify that illustrations of Definition 1 are provided by (a) Example l ; (h) Example 2; (c) Example 3; (d) Example 4_ L Verify that illustrations of Definition 2 are provided by (a) Example 6; (b) Example 7; (c) Example 8; (d) Example 9. 3. (a) Verify that the L of Example 13 is linear. (h) Show the equivalence of (9) and (10). (c) Verify that illustrations of Definition 3 are provided by Examples 13 and 14. 4. A complex vector space C is introduced in the discussion of eigenvalue problems. Fill out the definition of C and verify that it is indeed a vector space. S. Verify that (15a) is satisfied in the instances mentioned in the sentence preceding this equation. 6. (a) Show that the eigenvalues of {22) are real. (h) Verify (23a, b). (c) Show that < , > as defined in (24) fulfills the conditions of Definition 2.

C hararrerizalion of Eiyertralues and hodarbwr, S1ure.s as Exrrerara (Ch _ 12

566

7. The object of this exercise is to solve via power series the eigenvalue problem

£L lt1(u) = Au, u E vector space S, I c l

Ll° l(u) -

We assume that a solution is known when e = 0. Here L,t °' and i. 111 are self- adjoint linear operators mapping S into R. We assume that the known eigenvalues of Lt ° E are simple; that is, no more than one linearly independent eigenvector corresponds to each eigenvalue. Let u} °w be the orthonorrnal set of eigenvectors corresponding to the eigenvalues of L 1°1 . Assume that this set is complete in S. Let us look for solutions in the form of power series in c: ^ r

(a)

(b)

=u;°)

+Elf

(')

+ 2 ui 2 '+-..,

d1.j = AID) +r.t1' ) +•

.

From equating the OW) terms to zero, find an inhomogeneous equation for le. Determine API by the orthogonality condition necessarily satisfied if a solution to such an equation exists. [See the paragraph following (46)J Assume that [4 1 ° has a series expansion in terms of the u1" } and find the coefficients. To obtain a unique solution, require that te satisfy the normalization condition ( ut,) , uln) f 1.

Carry the calculations one step further, and find Al21 and an expansion for 42). NOTE The formal calculations give the perturbation of any particular simple eigenvalue 2; ° ' even if some or all or the remaining eigenvalues are not simple. (It must be assumed that the unperturbed eigenvectors have been selected so as to form an orthnnormal set_) Calculations giving the perturbation of a multiple eigenvalue are somewhat more complicated. See Courant-Hilbert, Vol. 1 (1953, p. 346). 8. Use the approach of Exercise 7 to approximate the lowest eigenvalue of (c)

z dx 1

+ Exu =

Au,

u'(0) = 1410= O.

9. (a) Show that

d[ du p(x) rx + q(x )u dx

'

!i(a) -}-

nl1'(u)

— 0,

u(b) MAN = 0, can be regarded as an eigenvalue problem involving a self-adjoint operator. • Fora particular closely related problem. see Exercise 7.2.1 I of

Append 12.11 Se11-witurru Opercutors

Prutor Spaces

567

(b) Give conditions sufficient to guarantee that the operator is positive as well as self-adjoint. 10. Prove that

a -); Ox

3

E_

L=

Lit=

= i,t2

x 3)i

is a positive self-adjoint operator if ,4 04x) is positive definite for all x in R. Let the domain of L be the space of twice continuously differentiable functions that vanish on the boundary of R and use the "natural" scalar product. 11. Let L act on functions defined and continuous on a bounded threedimensional region R. Here if v = Lu, then s Y MY)

1ff

142 4'3,

where a given kernel IC(x, y) is defined and continuous in R. Suppose that K(x, y) K(y, x). Prove that L is a linear self- adjoint operator. 12. Suppose that L and L2 map a vector space S into itself. Let L I and L2 be self-aidjoint. Is their product (i.e., their composition) self-adjoint? 13. Show that if V 4 = VW) acts on smooth functions in R that vanish on the boundary B of R. together with their normal derivative, then V' is self-adjoint and positive, 14. Lei the domain of L be the set of functions f that are defined and fourtimes continuously differentiable on [a, bj, where f ) f f(b)

0. 11 L

d d2 d2 p(x) dx 2 dx 2 dx

)

d d—x

ri

prove that L is self-adjoint. Use the scalar product (11 1 f2)

=

J f,(x)f2

(x)dx

and assume that p, q, and r are sufficiently. smooth. 15. (a) Suppose that

d2 L = a 2(x)-dx 2

z (x) dx

adx).

Show that the equation Lu = Au can be put into the form considered in Exercise 9. (b) Show that an eigenvalue problem for a general fourth order ordinary differential operator can be put into the self-adjoint form of Exercise 14 only under special circumstances.

568

Chararreri arion of Eisuennyitue ► and Equilibrium Slates a.s E.rirema ICh. 12

16. Consider (38) in the case where L requires multiplication of an n-tuple by a given symmetric matrix. Compare the text's statements with standard results concerning solutions of linear equations. [Use the scalar product (2).] 17. If L, L 1 . and L2 are linear operators, find alternative expressions for the following: (a) (L+} + ; (b) (L k + L) F c) (L ,L 2 )+ 1$. (a) Verify that (53) is an appropriate complex scalar product. (h) Show that if the conjugate is omitted in (53), there are nonzero elements a(r) and d 2 i whose scalar product vanishes. 19. Extend the text's discussion of (38) to the case where L is not self-adjoint. ; (

.

In the next two exercises L will be a linear operator over a space S composed of real-valued functions that are defined and twice continuously differentiable over [u, b]. Further conditions on S will be stated_ The scalar product used will be that of Exercise 14.

20. if functions in S must satisfy u(0) = u(1) = 0

,

and L =

[Ix

(Fix) —d

x)

dx + q(

.

find L+ and the dual space T 21. Suppose that functions in S must satisfy u(1) = an(0), u1(1) _ (3u'(0),

x and fi constants_ Let Lu = u". Show that L' is formally self-adjoint but is not selfadjoint unless a special relationship exists between a and ji. 22. Consider the norm defined by IILII = (I, O w. (a) Verify (8a) and (8h). (b) Prove (8c), starting with the fact that 0 < (r + 2w, L + 2w) for any real number 2. (c) Use the Schwarz inequality to establish (8d). (d) When does equality hold in (Sc)? in (8d)? 23, Let the elements x d belong to a metric space M. Use the triangle inequality to show that if {x i } is a convergent sequence, then it is a Cauchy sequence. 24. Let {u i } be a complete orthonormal set in the scalar product space S and let u be an element of S_ z (the zero element). (a) Prove that if (u, ui) = 0, i = 1, 2, ... , then u (b) Suppose that S is a Hilbert space. Prove that E7=' , (u, u 1)u j must converge to an element of S. (This is part of the Riesz-Fischer theorem. in the proof, use the convergence guaranteed by Parsevai's theorem to show that the partial sums form a Cauchy sequence.) (c) Use (b) to prove the converse of (a) -.i.e., show that if only the zero element is orthogonal to each of an orthonormal set, then that set is complete.

Appendix 12.11 Self-adfnlrri ()pertnurs on Vector Spat -PA

5

25. Show that the Sturm- Liouville operator — d/dx(pd jdx) + q is selfadjoint when p, q, and the boundary conditions are appropriate to the following equations of Table 12.1: (a) Bessel; (h) Legendre; (c) Tchebycheff; (d) Hermite; (e) Laguerre_ NOTE. Either the equations have a regular singular point at one or both of the end points, or the intervals over which the equations hold are unbounded. The complete theory for these singular Sturm-Liauville equations is rather complex, but it can be found in advanced treatments of differential equations, for example, Coddington and Levinson (1955). 26. Let S be the vector space of functions defined and continuous on [0, 11 The linear operator D = dfdx is defined only on a subspace of S, the set of functions with continuous derivatives. Show, by contrast, that if a linear operator is defined on a single element of ordinary threedimensional Euclidean space, then it is defined on the whsle space. REMARK. The fact that linear operators are often only defined on subspaces causes difficulties in functional analysis, compared to the theory of finite-dimensional spaces_ Marry of these difficulties can be overcome, however, if (as is very often the case) the subspace is dense in the full space. 27. Let a linear operator L provide a one-to-one mapping of the whole vector space S onto itself. Show that L is self-adjoint if and only if L is self-adjoins

Bibliography General Applied Mathematics R., and D. HILBERT. (1953) Methods of Mathemcrrrcof Physics. Vol. 1. New York: Interscience Publishers. a division of John Wiley & Sons, Inc. 561 pp. "Mathematical methods originating in problems of physics arc developed and the attempt is made to shape results into unified mathematical theories." (Volume 2 does not have the classical style of Vol. 1 hut is a compendium DI - important results in the theory of partial differential equations.) JEFFREYS, H., and B. J FFREYS. (1962) Methods of Aiiithematicol Physics_ New York: Cambridge University Press_ 716 pp_ "This book is intended to provide an account of those parts of pure mathematics that arc most frequently needed in physics . _ .. Abundant applications to special problems are given as illustrations_" COURANT,

Ad v anc ed Calculus FRANKLIN, P. (1 940

A Treatise on Advanced Calculus_ New York.: John Wiley & Sons. Inc. Reprinted by Dover Publications Inc_, New York, 1964. 595 pp. 1i11DEBR AND. F. B. (1962) Arkunced Calculus Ji)r Applications. F:elglewood Cliffs. N.J. 7 Prentice-Hall, Inc- 646 pp. KAPtAN, W. (1952) Advanced Cakuhis, Reading, Mass. Addison-Wesley Publishing Company. Inc. 678 pp.

Differential Equations BOYCE, W„ and R. ()iliumA_ (1969) Elementary .r irrential

Equations crud Boundary

Value Problems, 2nd ed. New York: John Wiley & Sons, hic. 583 pp. "This book is written from the . _ viewpoint ttl the appliedr r^dttternattctan whose interests in differential equations may at the same time be quite theoretical as well as intensely practical." COI)DINCiT❑ N. F_, and N_ LEv1NSON. (1955) neon' of Ordinary Diflerenriaf Equations, New York : McGraw-Hill Book Company. 429 pp. Authoritative advanced work. GARAEïFI]IAN_ P_ (1964) Partial Dt_ ere„tïal Equations. New York; John Wiley & Sons, Inc. 672 pp. A graduate text "written for engineers and physicists as well as mathematicians. — Ordinary Difierentiuf Equations. Essex. England: Longman Group Ltd. INcE,.(1927) Reprinted by Dover Publications, Inc., New York. 1956. 558 pp. An excellent source for the classical theory_ ,

5,7 1.

572

Bibliography

KELLOGG, i,

0. (1929) Faurrda[ions of Potential Theory. Berlin: J. Springer. Reprinted by Dover Publications, Inc_, New York, 1953. 384 pp. "It is inherent in the nature of the subject that physical intuition and illustration be appealed to freely, and this has been done. However, in order that the book may present sound ideals to the student, and also serve the mathematician, both for purposes of reference and as a basis for further developments. the proofs have been given by rigorous methods." TYCKONOV, A., and A. SAMuutSKl_ (1964) Partial Differential Equations of Muthernrtfrcaf Physics, Vol_ L San Francisco: Holden-Day, Inc. 390 pp. Vol, ❑ 1967. 250 pp. Representative of a large class of books, this is relatively brief and has good .

illustrations of how the mathematics is used.

Tensor s ABRAM, J. (1965) Tensor Calculus Through Diffrrenriaf Geometry. metry. London: Butterworth & Company Ltd. 170 pp. About 25 per cent of the book is devoted to applications its dynamics and continuum mechanics.

ARts. R. (1962) Fectors, Teri.sors, acrd tiro flask- &guotions of Fluid Mechanics. Englewood

Cliffs, N.J.: Prentice-Hall, hic. 286 pp. "Sets out to show that the calculus of t ensors is the languagc most appropriate to the rational examination of physical field theories," The argument is illustrated by detailed consideration of fluid mechanical equations. KARAMCRETI, K. (1967) Vector Analysis and Cartesian Tensors with Selected Applications. San Francisco: Holden-Day, lac. 255 pp. Begins with vector algebra and calculus, and therefore provides u particularly good reference for readers who hick experienix with these tools_

G. (1960) Cartesian Tensors. London: Methuen & Company 1.td_ R9 pp. Tensors are defined as multilinear functions of direction in this introduction to the subject for applied mathematicians and physicists.

TEMPLE,

Continuum Mechanics HooGE, P. G. Jr. (1970) Continuum Mechanics Ncw York: McGraw-Hill Hook Company. 251 pp. The essentially modern innovation consists in viewing the subject o fall continuous media as a whole and concentrating on those properties and concepts which arc common to all of them. The ever-increasing body of knowledge which must be mustered by the engineer, and, in particular, the use of increasingly complex materials in modern technology has made this new approach necessary." Au introductory graduate text for engineers, this Ilk provides an alternative reference for the derivation of the Navier and Navier Stokes equations from first principles. JAUNZIEMis, W_ (1967) Coritmi um Mechanics_ New York: Macmillan Publishing

Company, Inc. 604 pp. A graduate teat that rakes a "modern" rigorous approach to the subject_ Emphasizes elasticity.

Bihla'cr$raphr

573

SEDüv, L. I. (1971) A Course in ConrinuNm Mechanics (translation from Russian edited by J. R. M. Raduk). Groningen, The Netherlands: Wolters-Noordhoff Publishing, four volumes of 242, 309, 340, and 305 pp. "In recent years, the need for the introduction of a course iii continuum mechanics at more advanced levels at universities and technical high schools has become clear; it is needed as a general base for the theoretical development of thermodynamics, electromagnetism, hydrodynamics, gas dynamics, elasticity. plasticity, creep and of many other groups of phenomena of physics and mechanics, The generality and, at the same time, inseparability of these branches, which at a first glance may appear to be autonomous subdivisions of mechanics and physics, bring about the need to consider them] as a single unit." SOMMERFELD, A. ( 1950)_ "Mechanics of Defarmable Bodies." Vol_ 11 of Lectures urn Theoretical Physic.. New York: Academic Press, Inc. 396 pp, (Paperback edition: 1964 "My aim is to give the reader a vivid picture of the vast and varied material that comes within the scope of theory when a reasonably elevated vantage point is chosen. " Unlike the other references on continuum mechanics, this requires no knowledge of tensors. TRtJESO]ELL, C_, and W. Not.'. (1965) ' . The Non-linear Field Theories of Mechanics," Encyclopedia of Phy. irx, S. F kigge, ed., Vol. 3/3, New York : Springer-Verlag New

York, inc. 602 pp. " The maximum mathematical generality consistent with concrete, definile physical interpretation is sought." TRIJESUI:iwdrnnrayrtelic Slability. New York: Oxford University Press, Inc. 654 pp. A compendium of results on the linear stability theory of stratified fluids, of rotating fluids, and of many related flows_ CatJRANT, R., and K. 0_ FRIEDRICHS_ ( 1948) Supersonic Fiait and Shock]eru!s New York: Academic Press, Inc. 464 pp.

The classical authority on the theory of compressible flow_ "The book has been written by mathematicians seeking to understand in a rational

Bibliography

574

way a fascinating field of physical reality, and willing to accept compromise with empirical approach_" G O.n5rr_irr. S. (ed.). (1938) Modern Develapments is Fluid Dynamics_ New York: Oxford University Press, Inc. 702 pp. The first four chapters (187 pages) still form an exccllenl introduction to viscous flow. HAPrEI., J., and H_ BRENNER. (1%5) Low Reynolds Number Hydrodynamics. PrenticeHall, Englewood Cliffs. N.J. 553 pp. "The subject matter is largely confined to a development of the macroscopic properties of fluid-particle systems from first principles." KINSMAN, B. (1965) Wind Waves. Englewood Cliffs. N.J.: Prentice-Hall, Inc. 676 pp. "This is a very personal book. It is riot she ;rush about waves. It is an effort to communicate to you what I have so far managed to understand about waves." Full of interesting theory, and with magnificent pictures lo remind the reader that water waves are wet_ LAMtt, li. 0932) Hydrodynamics, 6th ed_ New York: Cambridge University Press_ 738 pp. Also available from Dover Publications, Inc., New York. An encyclopedic collection of classical results. LANlu ta, L. D.. and F.. H. 1.1FSr-UI1rt. ( 1959) F7uid Mechanics. Elmsford, N.Y. Pergarnun Press, Inc_ 536 pp. Translated from the Russian by J. R Sykes and W. }I_ Reid. "The nature of the book is largely determined by the fact that it describes fluid mechanics as a branch of theoretical physics." LEVICH, V. G (1962) Physicochemical Hydrodynamics. Prentice-Ball, Englewood Cliffs, N.J. 700 pp. "Hy physicochemical hydrodynamics we understand the aggregate of problems dealing with the effect of fluid flow on chemical or physicochemical transformations as well as the effect of physicochemical factors on fluid flow" (V. Levich)_ "Higher mathematics is employed throughout the text, neither as an end in itself nor as an elegant dress for the subject matter, but as a potent implement for sharpening physical understanding" (L. Scriven, Editor's foreword) Mu.NE-THOMPSON. L. M. (1968) Theoretical Hydrodynamics, 5th ed. London: Macmillan & Company Ltd_ 743 pp_ A text stressing inviscid fluids and complex variable met hods_ New York ; PRA NOTL, L., and O. G. TIETJENS. (1934) Applied ff,i dro- andAero-mechanics, McGraw-Hill hook Company_ 270 pp. Also available from Dover Publications, inc., New York, 1957, as Fundamentals of Hydro and Aero-Mechanics_ Brief and inexpensive, with considerable physical insight_ RosENtiEen, I.., ed. (1963) Laminar Boundary Layers. New York: Oxford University Press. Inc. 687 pp. " An account of the development. structure, and stability of laminar boundary layers in incompressible fluids together with a description of the associated experimental techniques. SEEd Ett, R. J., and G. TEMPLE (eds.). (1965) Research frontiers to Raid Dynamics. New. York: Iriterscience Publishers, a division of John Wiley & Sons, Inc. 788 pp_ A collection of useful survey articles. SERRIN, J. B. (1959) . 'Mathematical Principles of Classical Fluid Mechanics," in Esuyciopedia of Physics, S_ FlOgge, ed., Vol. 8/1, pp_ 125-263. New York: SpringerVeriag New York, [Tic_ -

"

Bibliography

575

"Our intent ... is to present in a mathematically correct way, in concise form, and with more than passing attention to the foundations, the principles of classical fluid mechanics." STOKER, J_ J. (1957) Winer Waves. New York: Interscience Publishers_ 567 pp_ 'l -he classical theoretical book on the subject. WFtrrt;iM. G. R_ (1974) Linear and Nonlinear Waves. New York: John Wiley & Sons. Inc_ 636 pp. "This book is an account of the underlying mathematical theory with emphasis on the unifying ideas and the main points that illuminate the behavior of waves The emphasis is on the conceptually more difficult nonlinear theory_" Most, but by no means all. applications are taken from fluid mechanics. 'I -he two principal journals are Journal of Fluid Medu nres and Physics of F7ut s. Beginning in 1969, Annual Reviews of Fluid Mechanics, a collection of authoritative surnmarks of research progress, has been published by Annual Reviews, Inc., Palo Alto, Calif_

Elasticity K. F. (1975) Wave Marion in Elastic Solids_ Oxford: Clarendon Press, 649 pp. "The purpose of this book is to present, in one place and in a fairly comprehensive manner, air intermediate-level coverage of nearly all of the major topics of elastic wave propagation in solids . - - - Throughout the book, emphasis has been placed on showing results ... from both theoretical and experimental studies." GREEN, A. E., and W. ZERNA_ (1954) Theoretical Elasticity. New York Oxford University Press, inc. 442 pp. This book is mainly concerned with three aspects of elasticity theory which have GRAFF,

„

atrcednio tyars.fnielcdomats,plexvrib methods for two dimensional problems ... and shell theory-" LOVE, A. E. H. (1944) A Treatise ran the Alatlrerrmtiral Theory of Elusnriry, 4th ed. New York: Dover Publications. Inc. 643 pp. "ft is hoped . - . to present a fair picture of the subject in its various aspects, as a mathematical theory, having important relations to general physics. and valuable applications to engineering." Although not easy to read, this is the classical reference work PRESCOTT, J. (1924) Applied Elasticity. Essex, England: Longman Group Ltd. 666 pp. Reissued by Dover Publications, Inc„ New York, 1946_ "In writing this book 1 have tried to sec the subject from the point of view of the engineer rather [than] from that of the mathematician." SOKOLVLKOF , L S. (1956) Mathematical Theory of Elasticity. New York : McGraw-Hill Book Company. 476 pp. Perhaps the best single reference. "This book represents an attempt to present several aspects of the theory of elasticity from a unified point of view and to indicate, along with the familiar methods of solution of the field equations of elasticity, some newer general methods of solution of the two-dimensional problemsTIMOSHENitO, S_ P., and I_ N_ Goonim. (1970) Theory of Elasticity, 3rd ed. New York: McGraw-Hill Book Company. 567 pp_

576

Bibliography

"The primary intention [is] to provide for engineers. in as simple a form as the subject allows. the essential fundamental knowledge of the theory of elasticity together with a compilation of solutions of special problems that are important in engineering practice and design." A leading journal is the Journal of Applied Mechanics, More than 2000 pages of up-to-date information on Solid mechanics can be found in Encyclopedia of Physics, S. Flagge, ed., Vols. 6a/I-3; .Mechanics of Solids f-111, C. Truesdell, ed. Berlin' Springer-Veriag, 1471

Variational Methods AKHIEZak, N. I. (l%2) Hue Calculus of Variations. New York: Blaisdell Publishing Company. 247 pp. Survey of main problems and methods. Buss, G. A_ (1944 Lectures on the Calculus of Variations. Chicago: University of Chicago Press. 292 pp. A more advanced work with emphasis on such matters as higher dimensionality and variable end points. See also Bliss's treatise Calculus of - Vtwiaiion.s. Mathematical Association of America. 1944, COURANT. R. (1957) Calcidus of Variations_ New York University Lecture Notes, revised and amended by J. Moser. 280 pp. (typewritten). These excellent notes are available in many libraries. COURANT, R.. and D. HILBERT. (1953) Methods of Mathematical Physics, Vol. i. New York Interscience Publishers, a division of John Wiley & Sons, Inc. 577 pp. Chapter 1V packs a great deal of material into l 10 pages. Seventy pages of applications to eigenvalue problems comprise Chapter VI. FORSYTH, A. It (1960) Calculus of Variations. New York: Dover Publications, inc. Paperback edition of 1927 treatise. 656 pp. "Though it does not purport to he a history, the gradual growth of the successive tests has governed the arrangement." GELFANU, I. M_. and S. V. Foulthl- (1963) Calculus of Variritions, revised and translated by R. A. Silverman. Englewood Cliffs, NJ.: Prentice-Hall, Inc. 232 pp. A modern introduction with considerable material on applications_ HEST , M. R. (196114 L'ult UJus uj` Variartorts and Optimal Control dory. New York : John Wiley & Sons. Inc. 405 pp. A unified rigorous approach to a large class of problems in calculus of variations and control theory_ LANCZOB, C_ (1949) The Variational Principles of Mechanics_ University of Toronto Press, Toronto. 307 pp. The variational principles of mechanics are firmly rooted in the soil of that great century of liberalism which starts with Descartes and ends with the French Revolution and which has witnessed the lives of Leibniz, Spinoza, Goe the, and Johann Sebastian Bach. It is the only period of cosmic thinking in the entire history of Europe since the time of the Greeks. 11- the author has succeeded in conveying an inkling of that cosmic spirit, his effort will be amply rewarded" (from the Preface).

Bibliography

577

MIIC}rt_LN, S. G. (1964) Variational Methods in Marhesnurical Physics. Elmsford, N_Y_ Pergamon Press, Inc. 5R2 pp. Advanced and comprehensive treatment centering on direct me thods. PARS, L. A. (1962) An Introduction so she Calculus of Variations. New York : John Wiley & Sons, Inc. 350 pp. "My aim has been to provide a clear and rigorous exposition of the fundamental theory, without losing sight of the practical applications •- in geometry and dynamics and physics." WEINSTOCK, R. (1952) Calculus of VariotiO ns. New York: McGraw-Hill Book Company. 326 pp. An elementary introduction with an extensive treatment of applications. WoormOUSE, R. (1964) A History of the Calculus of Variarinns in she Eighteenth Ceniw y New York: Chelsea Publishing Company, Inc. 154 pp_ Originally published in 1810; contains very interesting accounts of t he pioneering work of the Bernoullis, Taylor, Euler, etc. FR1F1]MAp.i, k3. 0956) Principles and Techniques of Applivd Murhernrirics. New York: John Wiley & Sons, Inc. 315 pp. A studyofabstract linear spaces, operators defined on such spaces, and their use in systematizing the methods and techniques for solving problems in applied mathematics_ .

STAKGOLD, 1. (1967) Boundary Vrdue Problems of Mathematical Physics. New York: Macmillan Publishing Company, Inc. Two volumes of 340 and 40g pp_ Eigenfunction expansions and Green's functions are the principal tools developed for the solution to linear boundary value problems.

F., and B, SZ. -NpcJY. (1955) Functional Analysis. New York Frederick Ungar Publishing Company, Inc. 467 pp. A classic introduction to the study of linear operators_ Applied mathematicians may find this book more readable than many successors. for it seems to use the organizing force thane an end in itself. abstrcionme

Hi nts a n d Answer s SUCTION 1_ I

L Hint: itetord the given displacement as et imposed of three mutually perpendicular displacements along the axes and project the whole figure_ 2. Hinz' Take scalar products with Orr and ERN, respectively.

— I o U -I D D

7.L =

L}

U 1

£' il _ felr1. e4}1• /e t► 1`

S. Hint:

SEC 11oN

6. R, 1 = (cos 6)6 o f (

I — cos

triM ► r; e, j

(sin

1.2

14:, ►r n r,-

S EC'l'1ON 1.4

4. (si)

(} 2

U

l

0

I

I

U t)

(bp

(a, 6, I I

SFCrI{7N 2.1 22. The results requested here are derived in Section U. 24. (a) ab r(eb i ha) + 1(ab bal. IN Hint: Choose the unit vectors as coordinate vectors. -

SECTION

2.2

4. (a) C'orrespi tiding to the eigenvalues 1. 2i, and —21, the required eigensectors are multiples of (1, 0, U), (0, 2i. 1), and (0, —2i, 2), respectively. 5 (a) Eigenvalues are 0, t, — t. Corresponding eigcnvectors are proportional to e'" £121 + ie131 „ £I l I — 1£131 SEt'11oN 23

2, Hint: Use Theorem 1.2.9 with A, = f, i . = mol l+ + grad ¢ A v. 5. (g) curl r 1 6. [bl Hint: [^ A u] p = 4 t pik(^u^1^ xr ^t+;1[^xrj. 9. (a) DO - t NjPr = $- W,fev,lDxjl.- .uw.,1• (e) Hrnf: In your calculations the term $za,AA r shoulid appear. Modify i t by means o f the chain rule. ^

SECTION

9. Hint :

3.1

Employ an argument used in the method of separation of variables. 579

Nr,trs

5 80

and Answers

SECTION 12

Hint: Try tr=u(z,r),u =w= O. 14. (c) Hint : Take advantage of symmetry; find the force on the side along x = 1 and 11.

triple it. SECTION 3.3

3. (a) The product of Lr(x, O)and the integral gives the decrease, caused by the slowing of flow in the boundary layer, of the volume flux along the boundary. To first approximation, the same decrease would be caused by an outward displacement of the inviscid flow by a distance 6 1 .

(b) 15 2(x) _ fo ❑u(x.(x,Y)0) 1 ^

y) u (x' dy U(x, 0)

SECTION 3.4 -Ar3 10. (C) T is proportional la t exp ( —r 1 /41a). (d) Hint: Suppose that the heat were initially concentrated at r = re . SECTION 4-L

6. flint: Use (32b). 12. (b) Hint: If a particular displacement field can be found SO that (36) is satisfied, (36) is still satisfied if the field is augmented by an arbitrary rigid body rotation. 13. flint: Contemplate (39). Also see the end of Section 4,314, (b) Hint: Show that each of the two terrils vanishes separately. 15. (b) Hint: QLL.1 —

Q1,3 — Q13.3

x1 t