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0 on ha},
and the functional of potential energy .C2(v) = 2lvl1 - (f,v)o,
where Iv112
= (Vv, Vv)o.
1.1. Unilateral Boundary Value Problems
7
Then the problem: find u c K2 such that C2(u) < C2(u) Vv E K2
(1.14)
is a variational formulation of problem P2.
The relation between the solution of problem P2 and the variational problem (1.14) is quite analogous to the case of problem P1. Problems (1.6) and (1.14) are called the primal variational formulations of the one-sided problems P1 and P2, respectively. Although the methods of solution for both of these primal problems differ only insignificantly, we
shall see that the difference between the corresponding dual variational formulations is essential.
1.1.11. Dual Variational Formulation. We can formulate problems P1 and P2 in such a way that the unknown is not the function u but rather its gradient, which is often an important and interesting quantity from the physical point of view. However, instead of the direct transformation of problems P1 and P2, we will derive variational formulations which are dual with respect to the primal variational formulations (1.6) and (1.14). Let us introduce the set
H(div, fl) = {q I q E [L2(fl)]", div q E L2(fl)),
where the divergence operator is understood in the sense of distributions:
Jo
q vcp dx = -
Jn
p div q dx `dip E Co (fl).
Theorem 1.3. Let q E H(div,12), w E H1/2(r). Then the formula (q - v,
w).=J(q .v v+vdivq)dx,
with v E H'(ll) and ryv = w, defines a functional q v E H-1/2(r). Proof. Almost identical with that of theorem 1.2. O We shall write s > 0 on r (or I',) for a functional s E H-1/2(r) if (s, iv) > 0 Vv E K, (K2, respectively). Let us introduce the class of admissible functions U1 = {q I q E
[L2(tl)]"+1,
q = [q, qn+1), q E H(div, fl),
q"+1 = f + div q, q v > 0 on I'},
1. Unilateral Problems for Scalar Functions
8
and the functional of the complementary energy 1 n+1 S, (q)
_
Ilgi IIo
i=1
The problem: find q° E tl1 such that
S1(4)
S, (q)
(1.15)
Vq E u1
will be called the dual variational formulation with respect to the problem (1.6).
Let us consider problem P2. Introduce the set tl2 = {g I q E H(div, f2), div q + f = 0,
q v > 0 on I'a}
and the functional of complementary energy S2(q) _
II4iII00
i=1
The problem: find q° E tl2 such that
S2(9) < S2(q)
(1.16)
Vq E U2
will be called the dual variational formulation with respect to the problem (1.14).
Theorem 1.4. Both of the primal problems (1.6) and (1.14), as well as both of the dual problems (1.15) and (1.16), has a unique solution.
Proof. Based on the following general result.
0
Theorem 1.5. Let 3 be a strictly convex, continuous functional defined on a reflexive Banach space X. Let K c X be a closed convex set and let 3 be coercive on K, i.e.,
vEK, IIvII-'oo=* 1(v)-++oo.
(1.17)
Then there is one and only one solution of the problem:
3(u) = min on K. (Proof is found in the book by FOik and Kufner (1978). See theorem 26.8.)
1.1. Unilateral Boundary Value Problems
9
The functionals Li, i = 1, 2 are continuous, strictly convex, and coercive in the spaces V1 = H1(1Z) and V2 = {v E H1(11) I7v = 0 on I'T1 respectively.
Indeed, take for example L2. It has a second differential which satisfies D2L2(u, v, v) = IvI
>
- CIIvjI,
Vv E V21 (mes ru > 0),
where C = const > 0. Hence, L2 is strictly convex and satisfies (1.17). It is not difficult to verify that the sets Ki are closed and convex in Vi. In order to prove existence and uniqueness for problem (1.15), we first transform it to an equivalent problem, formulated only for "reduced" vector functions q E H(div,11). Namely, after the substitution qn+1 = f + div q we have a new equivalent problem: find q E llo such that I (4) < I (q)
(1.18)
bq E Uo,
where
on F), n
I(q) =
2
1: Ilgllo+Ildivgllo
+(f,div')o.
i-t
If the norm in the space H(div, fl) is introduced by
/n
\ 1/2 IIgIIH(div,n) = t Ilgillo + IIdivgllo) i=1
then H(div, ll) is a Hilbert space. Uo is closed and convex in H(div, ll). To prove for example the closedness of 110, it is sufficient to realize that the mapping q - q v of the space H(div,11) into H-1/2(r) is linear and continuous. The functional I is continuous on H(div,1Z), strictly convex, and coercive. Hence, according to theorem 1.5 we obtain the existence and uniqueness of the solution to problem (1.15). We proceed analogously with problem (1.16). 112 is closed and convex in the space H(div,11), and the functional $2 is continuous and strictly convex
in H(div,1l). Since q E 112, IIgIIH(a1n)
o0
= i-1
Ilgillo + Ilfllo
S2 (q) - +oo, $2 is also coercive in 112.
-' oo
1. Unilateral Problems for Scalar Functions
10
1.1.12. Relation Between the Primal and Dual Variational Formulations. We will now show how the solutions of the dual problems are related to those of the primal problems. To this end we will use the saddlepoint method (the min-max method), the sequel of which will also enable us to derive the mixed variational formulation of the original problem P1 or P2, corresponding to the canonical variational principles of Hellinger-Reissner type (see Hlav£tek and (1981)). Let us consider problem P1 together with its variational formulations.
Theorem 1.6. If u is a solution of (1.6) and q° a solution of (1.15), then qi°
8u = a zi ,
2.
(
= 1, 2, ... n,
1.19
)
q0
n+1 - u,
S1(q°) +.L1(u) = 0.
(1.20)
Proof. For v E H1(Q) let us denote [L2(fl)]n+1,
C'(v) = [Vv, v] E
M = [L2(jj)]n+1'
=K1xM and let us introduce the Lagrangian functional )((v, N; q) =
2
II NII2 - (f, v)o + (q, ((v) - N),
where v E K1, N E M, q E M, and II
denote the norm and the
' II,
scalar product in [L2(c2)]n+1, respectively. We easily find that sup (q, e(-) - M) qEM
0
forty=C(v)
+oo
for N
e (v),
and thus
inf .£1(v) =
vEK1
inf
sup )-((v, N; q).
(v,NJE'W qEM
(P)
In the theory of duality (e.g., Ekeland and Temam (1974)), the problem P'
sup
inf
qEM (v,N)EW
)((v, N; q)
1.1. Unilateral Boundary Value Problems
11
is called the dual problem to P. The question arises when both values coincide.
Let us denote So (q) =
inf
[v,NJEW
X (v, N; q)
and let us try to explicitly calculate this functional. First of all, we have So(q)
0.
It is easily verified that the elements of the form pp belong to H+1/2(r) for every p > 0. Consequently, PC
H
-' +00,
Thus, we can formally write
i fC1(v)=Hnf
sup H+
(r
)
{Li(v)-(p,ryv)}.
Let us write X (v, µ) = C1(v) - (lp,'Yv). We shall establish the relation between the solution of P1 and the saddle point of X in H1 (n) x
H+1/2(I').
Theorem 2.1. A pair (w, A) E H1(1) X H+-1/2 (r) is a saddle point of X in H1(n) x H+1/2(I') if and only if w = u,
a = 8u/8v,
where u E K1 is a solution of Pi.
Proof. (i) Let (w, A) be a saddle point of X in H1(1) X H+ 1/2(I,): (w, A) E H1(n) X H+1/2(I') :
X(w,µ) :5X(w,A) s X(v;A)
d(v,µ) E H1(1) x H+112(r).
(2.2)
Substituting into this inequality first u = B, then p = 2A, we obtain (A, "w) = 0,
(2.3)1
and consequently -(ip, ryw) < 0
V11
E
H+-112
(r).
(2.3)2
Hence, of necessity w E K1. Taking into account (2.1), the second inequality in (2.2) and (2.3)1,2, as well as the definition of X, we conclude C1(w)
C1(v) - (µ,'Yv) < C1(v)
Vv E K1.
1.1. Unilateral Boundary Value Problems
15
This means that w E K1 is an element which minimizes Cl in K1. Such an element, as we know, is unique; hence, w = u. The functional assumes its minimum in H1(fl) at the point v = w. Therefore, (w, v)1 - (A, 7v) = (f, v)o
Vv E H'(1).
Using Green's formula (1.10) together with (1.1), we finally obtain
(aw/av, "v) - (A, ryv) = 0 Vv E H1(0),
which implies aw/av = A. (ii) Let u be a solution of P1. Then, (1.13) implies (u,au/av) E H1(0) x H+1/2(r). Further,
X (u, au/av)
='C1 (u) - (au/av, 7u) >- £1(u) - (,U, 7u)
= X(u,µ) Vp E H; 1/2(r). Similarly,
X(u,au/av) - X(v, au/ail) = C1(u) -£1(v)+ (aujav,'yv-'yu)
-
2 II u
- v 11
1
+ (u, u - v)1 + (f, v - u)o + (au/av, -1v - 7u)
_ -2Iju - v11i < 0 VV E H'(fl) by virtue of Green's formula (1.10) and also (1.1). Thus, we have verified that (u, (9 u/av) is a saddle point of X in Hl (0) X H+-1/2 (r). Remark 2.1. By using partial variations 8u X, 8µX with respect to the variables v, p, it is possible to equivalently characterize (2.2) by the following relations '(see Ekeland-Temam (1974)): (w, A) E H1(fl) x H+1/2(r), Vv E H1(0) 4=a (w, v) 1 - (A, v)
8 X (w, A; v) = 0
(f, v)o 8,4X(w, A;,u- A)
0 independent of h such that 0 > Bo. In other words, when refining the partition of fl, the triangles of the given triangulation do not reduce to segments. Thus, any regular system of triangulations is actually characterized by the single parameter h > 0. In the case of a regular system of triangulations, therefore, we shall use the simpler symbol {Th} instead of {7h,01-
Remark 3.1. When solving problems with various types of boundary conditions prescribed on certain parts of the boundary r, we subject the system of triangulations to the additional assumption of being compatible
with the partition of r. That is, each point of r at which the boundary condition changes must be a vertex of a certain triangle Ti E Th.
Let {Th}, h -+ 0+ be a regular system of triangulations of (2. The nodes of a triangulation (i.e., vertices of Ti's) lying on r will be denoted by
1.1. Unilateral Boundary Value Problems
19
a1, ... a,.,,,(h). Each Th will be associated with a finite-dimensional space Vh of piece-wise linear functions:
Vh={vIvEC(n), VTEP1(T) VTETh}, where Pk(T) (k > 0 is an integer) denotes the set of all polynomials of degree at most k with the definition domain T. Let us further define
Kh={vhEVh Ivh(ai)>0 Vi=1,2,...m(h)}. It is easily seen that Kh is a convex closed subset of Vh, Kh C K Vh > 0. To obtain an approximation of P1, we will make use of a modification of the Ritz method, with which the reader is familiar as one of the possible methods of numerical solution of variational equations. Problem (1.6) will be replaced by the problem of minimization of L'1 in Kh. That is, we look for uh E Kh such that £1(uh) : C1(vh)
VVh E Kh.
A complex natural question arises, which is what is the relation between u and uh, and more precisely, whether I (u - uh 11 1 -+ 0, h -+ 0+, or possibly, what is the rate of this convergence if expressed in powers of h. We shall study these issues in the subsequent text.
Remark 3.2. Problem (3.1) is already suitable for computer realization. It is easily seen that dim Vh = M(h), where M(h) is the number of all nodes of the given triangulation Th. Let {(Pi}t_1 be such elements of Vh that (pi(A1) = 6i; (Kronecker's symbol). Then, evidently, {,Pt}Mlh) forms a basis of Vh. Moreover, {Ai}i_ll1h
M(h) vh(A1)cpi(x)
vh(x)
VVh E Vh.
i-1
Let T : Vh --+ RM(h) be the isomorphism given by
T vh = a = (a1, a2) ... , aM(h)) E RM(h)
VVh E Vh,
where the components of the vector a are the coordinates of Vh with respect to {cpi }; ih). By means of T we can identify the set Kh with a convex closed subset KM(h) C RM(h): KM(h) = T (Kh)
{a E RM(h) I at > 0 Vi E I},
1. Unilateral Problems for Scalar Functions
20
where I c {1,. .. , M(h) } is the set of indices, which in the given numeration, correspond to the vertices a; E I'. Problem (3.1) is then equivalent to the problem of finding a` = (ai, ... , am* (h)) E KM(h) such that
y(a*) < Y(a) Va E KM(h).
(3.2)
Here,
Y (a) = Li(T -la) =
2 (a, Act) RM(h) - (3, a)RM(h),
T-1 : R`u(h) -+ Vh is the mapping inverse to T, A = ((cPi,coj)1)M(hi is the stiffness matrix, 3 = ((f,'P1)o,
..., (f, PM(h))O) is the vector resulting
by integrating the right-hand side f, and
denotes the scalar product in RM(h). The desired solution uh E Kh is then determined from the formula
uh =
T_la*
M(h) j=1
Thus, (3.1) leads to the problem of quadratic programming seen in (3.2). At this point, let us briefly mention the algorithm which facilitates the approximate solution of minimization problems of the type (3.2). The definition of the set KM(h) implies that we can write
KM(h) = Kl x K2 x ... x KM(h), where K' = [0, +oo) for Z 'E I, K' = R1 provided %'V I. It follows from this formula for the convex set KM(h) that each of the variables a1, ... , aM(h) is subjected to at most one constraint (provided i E I), and, moreover, that this constraint involves no other variable. In this sense, the variables are
separated. In order to find the minimum of the quadratic function / on the convex closed set KM(h) of the above-mentioned type, it is advantageous to adopt the following generalization of the well-known SOR method (see Glowinski, Lions, and I'Wmolieres (1976)): choose ao E KM1 arbitrarily; if a(n) E KM is already known, we successively correct its individual components in accordance with the following scheme: a(n+1/2)
+>aija1 -
_ah y
a!,+ 1)
ail
Px: ((1 -w)a(n)
'Here we write M instead of M(h).
j>i
1.1. Unilateral Boundary Value Problems
21
and then set a("+1) = (,,i"+1),
Here, PK; stands for the
projection of R' to K' and w > 0 is the relaxation parameter, whose proper choice increases the rate convergence of a(') to a* (the minimum point of in KM). In the case just considered, we have l,.
PK: (a) =
if a E [0, +oo),
a
ifa 0 : a(v,v) > aIvI2
Vv E V
(3.5')
instead of (3.5). In this case, a solution of problems P, Ph generally need not exist, and even if it does exist, it need not be unique (the reason why this is so is concretely illustrated in section 1.1.6. Nevertheless, if a solution of P, Ph exists, then it suffices to replace the symbol 11 11 in (3.10), (3.10') by the symbol I I in order to obtain a bound for the seminorm of the error uh - U. We can analogously generalize theorem 3.1 on the convergence of the approximate solutions Uh to u. Let us formulate the variant, which will be used in the subsequent text. Let V, H be two Hilbert spaces. Let V c H and let the embedding be totally continuous. Let -
IIv112=Iv12+IIvIIH
where II
- IIH
VV EV,
denotes the norm in H. If / is coercive on UhE(o,1)Kh, that
is,
hl1o Y(vh) = +00, IIvh!I -- +00,
vh E Kh,
and if (3.5'), (3.8), and (3.9) hold, then
provided u, uh exist and u is uniquely determined.
1.1.33. A Priori Bound for Problem P1. Let us now go back to the approximation of P1 in the form described in section 1.1.31. We will show how to apply the results of the previous section to this particular case.
1.1. Unilateral Boundary Value Problems
25
Theorem 3.2. Let a solution u E H2(f2) n K,u c W1,°° (r)3 satisfy au/av E L°°(r), and let the set of points from r at which u changes from u = 0 to u > 0 be finite. Then llu - uhlll < c(u)h,
h --+ 0+,
(3.11)
where c(u) is a positive constant that depends only on u.4
Proof. As Kh C K Vh E (0, 1), we can use the relation (3.10') to obtain the required bound. Let us set Vh = rhu, where rhu denotes the piecewise linear Lagrange interpolation of the function u. As
rhu(a,) = u(a,) > 0 bpi = 1,..., we have rhu E Kh. Using Green's formula and (1.1), we obtain
a(u, rhu - u)
= (u, rhu - u)1
u+u,rhu-u)o+fr (f, rhu - u)o +
C7v(rhu-u)ds
Jr av (rhu - u) ds.
Substituting into the right-hand side of (3.10') and using classical interpolation properties of the function rhu, we conclude that
llu-Uhll1
= (u-uh,u-uh)1
(uh - u, rhu - u)1 + Jr av (rhu - u)ds 2lluh - ulll + 21lrhu - ulll + fr av(rhu - u)ds 2 lluh - ull1 + ch2lu12 p +
fr
av(rhu - u)ds.
(3.12)
It remains to estimate the integral over r. We divide the boundary points into two groups:
ro={xEr:u(x)=0},
'Let r = U!"_1A,A,+1 be the boundary of a polygonal domain U. We define: u E i.e., the trace u on A; A,+1 and its first derivative in the direction of the side A,A,+1 is a bounded measurable function of one variable (the parameter of the side A,A,+1) 41n the sequel, c denotes a general positive constant, which may assume different W 1.O° (r) # UI A. A, E W 1,oo (A,A,+1) Vi =
values at different points of our exposition. If we want to explicitly express its dependence on parameters t 1, t2, .. , t we write c = c(t1, l2, .. , t,).
1. Unilateral Problems for Scalar Functions
26
r+ = (x E r : u(x) > o). Let alai+1 C ro. Then, rhu(ai) = u(ai) = rhu(ai+l) = u(ai+1) = 0, and consequently rhu - 0 on alai+1. Hence,
au
(rhu - u)ds = 0.
(3.13)
If alai+1 c r, then (1.2) implies that au/am =_ 0 on alai+1, and (3.13) again holds. Let T be the set of all alai+l whose interiors contain points from both r+, ro. Making use of the assumptions on the smoothness of u, au/av on the boundary, and of the interpolation properties of rhulr we obtain (rhu - u)ds 09V
<
0 on F}.
1. Unilateral Problems for Scalar Functions
28
A vector function qh E Uoh will be called an approximation of the dual problem (1.18), provided I(qh) 0, and BR-1 has only one nonzero element in each row. Instead of problem (4.2) we now have the problem
G(z) = 3(R-'z) = min in the set
Z = {z E RN, BR-1 > 01,
which we are able to effectively solve, for example, by the superrelaxation method with an additional projection (see section 1.1.3).
1.1.411. A Priori Error Bounds. In order to make an a priori estimate of the error q° - qh, we will use the method of so called one-sided approximations (see Mosco and Strang (1974)).
Lemma 4.1. Let Y(v) be a real-valued functional defined on a convex closed subset M of a Banach reflexive space X. Let us assume that lY has both the first and the second differential (in a Gateaux sense) and that there exist positive constants ao, c such that
aollzll2 < D2Y(u;z,z) < cllzll2 Vu E M, dz E X.
(4.3)
Let Mh C M be a convex closed subset. Denote by u and Uh the elements minimizing / in M and Mh, respectively. Let us assume that there exists wh E Mh such that 2u - Wh E M. Then 1/2
Ilu-uhll
/(u') + 1aoIluh - ull2, as
DtY(u,uh-u)>0.
1. Unilateral Problems for Scalar Functions
30
For any v E Mh we may write Y (v)
= y (u) + Dy (u, v - u) + 2 D2y (u + 91(v - u); v - u, v - u) y(uh).
Substituting V = Wh and v = 2u - wh into the condition Dy (u, v - u) > 0, we obtain Dy (u, wh - u) = 0, hence
y(uh)
0 such that
of = oh + E, < t vs c rm, which leads to a contradiction with the definition of the maximal element. If (C2) is fulfilled, then we can write t(si) - Oh (s.i) = It (Sj)
- Y h (si) I2 < h3It
ds2
22
(z) (si - z)dz,
I[o,rm = h3lt[2,r--
This yields the bound (4.10).
Lemma 4.3. Let rpm E C(r,r,,), 1 < m < m, be piecewise linear functions with vertices determined by the triangulation Th. Then there exists a function wh E [Vh]2 such that
wh.V=Pm on rm vm,
(4.11)
1/2
11WN12
Ch- 1/211
j=1
where cp is a function whose restrictions to rm coincide with <pri for all m.
Proof. Let us consider a boundary strip 11h, which consists of all triangles
T E Th such that T n r # 0 (we regard T as a closed set). Put wNbi) = 0, j = 1, 2, at all vertices b, E t - r. Then supp wh C fih and it is sufficient to determine and estimate the values wh(ai) at the vertices ai E r. 10 Let ai be a vertex of the polygonal boundary. Let us denote by cp+, ip- the limits of the function (p at the vertex ai from the right and left, respectively, and by v+, v- the corresponding unit outer normals. The values w4 at the point ai are given by
wh.v =P_' wh.v+=rp+ and
w (ai)
lp- 1) Isinail-1,
j = 1,2,
(4.12)
1.1. Unilateral Boundary Value Problems
33
where a; is the inner angle of IF at the vertex ai. 2° Let ai c I',,., be a vertex of Th but not one of r. Let Ivkl = max{Iv1I,
IV2I}.
As 2vk > 1, we have IvkI > 11V2. Let us choose
wk (ai) = vk'jo(ai),
while the remaining component wPh (ai) vanishes: wh (ai) = 0, (p 54 k). Then, evidently Iw (ai)I < / Ico(a;)I, (4.13) 1, 2. As a consequence of (4.12), (4.13) we obtain the estimate ma2llwi IIc(n) < CIIrIIr,oo.
(4.14)
For an (a, /3)-regular system of triangulations, the so-called inverse inequality holds (e.g., Ciarlet (1978)): IIw, II1,n < Ch-1IIw, II°,n.
(4.15)
On the other hand, we obtain from (4.14) IIwjhjIo,n
= IIw2 Ilo,n,, 5 (mes f1h)1/2II'w3 IIc(n) < Ch1/2IIVIIr,oo
(4.16)
as mes flh < Ch. Combining (4.15) and (4.16), we arrive at the bound (4.11).
Proof of Theorem 4.1. Let Oh be the function defined in lemma 4.2. Let us set So=qi.v-Oh,
and construct the function wh E [Vh]2 according to lemma 4.3. Then, the function th = qI wh E [Vh]2 satisfies the conditions (4.7), (4.8). Indeed, on each side Fm we have
-
th'v=q1 v-Vm=Oh') and (4.9) yields (4.7). Besides,
II (4 -q°)jII1 0 on ra}. We will now work with vector solenoidal functions (that is, vectors with zero
divergence). To this end, we shall use linear finite elements on triangles, which were introduced by Veubeke and Hogge (1972). Let us recall the construction of these spaces. On each triangle T from the triangulation Th, we define a vector function .M(T) _ {q I q E [P1(T)]2, div q = 0).
Further, we introduce the space of solenoidal finite elements
Nh={gIglTE.M(T) VTETh,
=0 VxETnT'}. The last condition means that the "flow" q - v is continuous when passing through the common side of any two adjacent triangles. We easily verify that .Nh is a linear finite-dimensional set and Nh c
H (div, n), since for each qh E Mh we have div qh = 0 in the sense of distributions. Let us define a linear continuous mapping liT E C([H1(T)]2, [P1(T)]2)
1. Unilateral Problems for Scalar Functions
38
by the conditions
f[(q.v)s -(HTqv)s1]vds=O Vv EP1(S,) on each side Si of the triangle T. Further, let R(fl) = {q E [H'(fl)2 I div q = 0). If we define a mapping rh on )Z(11) so that
rhglT = HTq VT E Th, then it can be proved (see Haslinger and HlavaZ_Iek (1976)) that (4.25)
rh E £()Z (0), ,Nh),
Iiq - rhgllo,n < Ch2lgl2,n
Vq E [H2(tl)]2-
(4.26)
In addition, let us assume that there is such a function G that G E R(f1),
G . v = -) - v on r,,.
(4.27)
Denote -a - v = g and construct a function gh E L2(I'a,) such that on each side s C I'a, s E T h , the restriction g h I s coincides with the L2 (S)_ projection of the function g onto the subspace Pl (S). Thus, gh is piecewise linear and can have jumps at the nodes of the triangulation. Define an approximation of the set 1120,
u20={pENh
IP-V>ghonl'a}.
(Since gh > g does not generally hold on l'a, the set U20 need not be a subset of U20.)
By substituting the particular solution we can transform the dual problem (1.16) to an equivalent problem:
find p° E U20 such that
J(P°) 5 J(p)
Vp E U20,
(4.28)
where 2
J(p) = 2IIPII0 + (A,P)o
A vector function ph E 1120 will be called an approximation of the dual problem (4.28), if (4.29) J(Ph) J(P) Vp E 1120.
1.1. Unilateral Boundary Value Problems
39
Figure 1 As U 20 is convex and closed in [L2 (11)12, problem (4.29) has a unique solu-
tion.
Algorithm for the Solution of Problem (4.29). For the nodal parameters in .M (Te) let us take the limit values of the "flow" p - v at the vertices, and let us denote them (see Figure 1) by pe
1'
e
e
Te E Th.
The following identity holds in each triangle Te E Th: 11(Q2 + Q3) + l204 + Nb) + l3(Q1+ M6) = 0.
In the triangle Te let us denote PITS = pe and we = [Pl (a1), P2 (a1), pe (a2), p2(a2), pe(a3), p2 (a3)IT.
Then,
we = C9,
(4.30)
1. Unilateral Problems for Scalar Functions
40
where the (6 x 6)-matrix Ce is regular, because its inverse matrix is
1
=
0
0
0
0
0
v211
0
0
L23)
LM
L(1)
0
0
0
0
L( 2 )
L(2 )
0
0
0
0
v1
112
0
0
0
L(3)
1/23)
1
Ce
0
0
1/13)
0 L11}
0 (2)
0 2
where v(k) stands for the unit outer normal to the side Sk = akak+1 (k = 1, 2, 3, a4 = al).
If gh
on ra in the form (4.32)
? 9h(aJ)
f
at all vertices of all sides S C ra. Thus, we have
pEu20bQEB={QERN 16 which fulfills all conditions of the forms (4.30), (4.31), and (4.32)}, and the problem (4.29) is transformed to an equivalent problem: 2 QT A,6
- bT Q =min in B.
(4.33)
In this way, we have obtained a quadratic programming problem, which can be solved, for example, by Uzawa's method (see Cea (1971), Chapter 4, Section 5.1), the method of feasible directions (Zoutendijk (1960), (1966)), etc. Another algorithm arises from already choosing the base functions in Nh in such a way that they a priori fulfill the conditions of the forms (4.30) and (4.31).
1.1.421. A Priori Error Bound. In order to obtain a bound for the error p° - ph, we again apply the method of one-sided approximations. Let us first define the sets
C = {gEH(div,1) Idivq=0, Ch= C (1 Nh = {q E Xh I q v > 0
onra}, on
ra}.
Under the assumption (4.27), we have
p°-G`UEC.
(4.34)
According to (4.25), we have rhG E Nh and, moreover, (rhG) . v = 9h
on ra,
because on each side S c ra, S E Th, (rhG) v is, by the definition of the mapping rh equal to the L2(S)-projection onto P1(S) of the function G v = g. Thus, we obtain ph - r,LG = Uh E Ch,
(4.35)
1. Unilateral Problems for Scalar Functions
42
and this implies that
pEu20 t=ip-rhG=VhEChLemma 4.4. Let there exist Wh E Ch such that 2U - Wh E C. Then, IIPO-Ph1Io
(4.36)
s IIU - WhIlo+IIG - rhGIIO.
Proof. Set p = C + WH. Then, p E U20 and 2p° - p = 2(G + U) - (G + Wh) = G + (2U - Wh) E
1120.
The solution p0 satisfies the inequalities
DJ(p°,P-P )>0, DJ (p0,2p°-p-P) =DJ(p°,PO-P)?0, where
DJ(p, q) _ (p, q)o + (A, q)o. Thus, we have
0 = DJ(p°, p - p°) _
(p o, Wh
- U)o + (A, Wh - U)o.
(4.37).
Second, set p = G + Uh E 2120. Since p - p° = Uh - U, we obtain
0 < DJ(p°, p - p°) = (P°, Uh - U)o + (A, Uh - U)0.
(4.38)
Finally, setting p = rhG + Wh E 1120, we/ can write
0 0 by (4.24). We easily verify that (4.41) holds. Moreover,
II'h - thllp an,_ra = i
mes(anj-Ira) < CII th - tbhllo,an nr,
G C11 0h - thllo,ra.
Now (4.42) is derived by combining (4.46) and (4.47).
(4.47)
1.1. Unilateral Boundary Value Problems
45
Lemma 4.6. Let a piecewise linear function ip on r be given, whose nodes coincide with the vertices of an (ca, (j)-regular system of triangulations {Th} (ep is generally discontinuous at its nodes), such that w ds = 0,
j = 1, ... , ?.
(4.48)
Then there exists a vector function wh E JVh such that on I', IIU'hIIo,n < Ch-1/'JJS0J!o,r.
(4.49)
Proof. Let us again consider a boundary layer nh C f2, which is formed by all (closed) triangles T E Th such that T n r 0 0. Evidently Oh = U 1Zh) j=1
where flh is adjacent to the polygon 8Uj. We shall construct wh e Mh by means of suitably chosen parameters of the flow /3 (see algorithm for the solution of the problem (4.29)) in such a way that supp wh C 11h Consider a layer 12h. On (90j we choose the parameters of the flow equal to the corresponding values of the function cp, as we let them vanish on 812h-8flj. On the sides which connect vertices of 8flj with those of
8flh-8flj, we set Rk = 0 at the inner vertices on 812h-8f2j) while the parameters Pi at the external vertices on 8flj remain free to be suitably chosen later. Each of the sides li, i = 1,. .. , n is associated with one unknown parameter Qi (see Figure 2). 1' Let us first assume that each Ti E 11h has at most one side lying on 8flj. The conditions of the form (4.30), (4.31) generate a system of n equations !i B = b,
where
A = -li, Ai,i+1 = 1i+1, Ant = 11
i = 1, 2,..., n, i = 1, ... , n -
while all the other elements of the matrix A are zeros. Further, bi = -lm6P,n+VM+)
(4.50)
1. Unilateral Problems for Scalar Functions
46
Figure 2
or bi = 0 provided Ti n 8113 reduces to a single vertex. The assumption (4.48) implies n
Ebi=0, i=1
hence, we can omit the last equation of the system (4.50). If we put ,61 = 0, then the system has the solution
i-1 lei = di 1 E bp,
i = 2, 3, ... , n.
(4.51)
P=1
2° Let a triangle Tk E 81h have two sides lq and 1q+1 on 8113 (Figure 3). Then, we obtain the following equation for Tk:
(i3k +Yk)lk = -lq(coq +SPq) - lq+1(cq+l +'P[) Then we can set /3 = 0, for instance, and calculate Nk. The remaining system of equations again has the form (4.50), being of the shape of the system corresponding to the "truncated" triangulation Rh-Tk, with Qk and Qk playing the roles of given external parameters of the flow. 3° Equation (4.51), together with the (a, #)-regularity of the system, Th implies S.
{mil=21
1
cp ds
2hml l
i = 1,...,n;
'901
ISol ds
0 : Ib(v, µ) I < M1 IIvII IpI
d(v, p) E V x L,
(5.2)
and let us define a functional X : V x L -+ R by X (v, µ) = Y (v) + b(v, µ) - [g, µ],
(5.3)
where g E L' is fixed.
Finally, let K C V, A C L be nonempty, convex, closed subsets. In the following, we will assume that A is either (CC) a convex cone with its
vertex at BL s (the zero element of L) and K = V, or (BC), a bounded subset of L. 6This means that if p E A, then pµ E A for every p > 0. In what follows OX denotes the zero element of a linear space X.
1. Unilateral Problems for Scalar Functions
52
Let (u, A) be a saddle point of X in K x A: )1(u,µ) < X(u, A) < )1(v, A)
V(v, p) E K x A,
(P)
or, equivalently (see Ekeland and Temam (1974))
(u, A) E K x A such that
a(u,v-u)+b(v-u,A)> (f,v-u) VvEK, b(u,,u - A)
[g, a -)A]
(P')
bµ E A.
We easily verify that the first component of u is a solution of a certain minimization problem. Indeed, we have
Lemma 5.1. Set j(v) = supA {b(v,,u) - [g, p]}. Then,
y(u) +j(u) = Kin{y(v) +j(v)}.
(P)
Proof. The first inequality in P yields j (u) = sup{b(u,.u) - [g, p]} < b(u, A) - [9, A]. A
On the other hand, the converse inequality is evident and hence, j (u) = b(u, A) - [g, A].
This, together with the second inequality in (F), immediately implies the assertion.
Definition 5.1. Problem P will be called the mixed variational formulation of the minimization problem P. As concerns the very existence, or even the uniqueness of (u, A) satisfying F, it is possible to make use of the well-known results of convex analysis (e.g., Cea (1971), Ekeland and Temam (1974), etc.). We recall here those results, which will be referred to frequently in the following. As a consequence of the V-ellipticity of the form a, the functional y (v) + j(v) is strictly convex in V; hence, the first component of the saddle point (if it exists), is uniquely determined. Let us now discuss the conditions which would guarantee the uniqueness of the second component of the saddle point. Let uz additionally assume that K is a convex closed cone with its vertex at B1, ,iA us define
K* = {v E K I - v E K}.
1.1. Unilateral Boundary Value Problems
53
Lemma 5.2. Let
b(v,p)=0 dvEK*
u=9L.
(5.4)
The second component of the saddle point also is uniquely determined.
Proof. Let (u, A1), (u, A2) be saddle points of N in K x A. As K is a cone with its vertex at Bv, we can first choose v = 0v, and then v = 2u, thus eventually obtaining
a(u,v)+b(v,A,)> (f,v) VvEK, i=1,2. Restricting ourselves to the trial functions v E K*, the previous inequalities reduce to equations as a consequence of the linearity of K*: a(u, v) + b(v, A;) = (f, v) Vv E K*,
i=1,2.
Subtracting one from the other, we find b(v, Al - A2) = 0 Vv E K*,
and (5.4) yields Al = A2.
Next we will introduce conditions guaranteeing the existence of a solution (u, A).
Lemma 5.3. Let (CC) hold and let there exists a constant fi > 0 such that
supb >QIi I V
VpEL.
(5.5)
hull
Then there exists exactly one solution (u, A) of problem P.
Proof. The uniqueness follows from the fact that (5.5) implies (5.4), from K* = V, and finally, from the V-ellipticity of the form a. Its existence is proved in Brezzi, Hager, and Raviart (1979). If we assume that A is bounded, the situation is much simpler. Lemma 5.4. Let (BC) hold. Then there exists a solution (u, A) for problem A.
Proof. The assertion is a consequence of a more general result (see proposition 2.2, Chapter 6, Ekeland and Temam (1974)). Remark 5.3. In concrete cases, the proof of existence of a solution usually proceeds in such a way that we first "guess" the solution, and only then verify that it really satisfies P or P'.
1. Unilateral Problems for Scalar Functions
54
Remark 5.4. Until now, we have assumed that the form a is symmetric. If a is a general, continuous, V-elliptic form (not necessarily symmetric), then we start from P'. We are looking for (u, A) E K x A that satisfies the inequalities from P. Analogously to lemma 5.1, it is possible to verify that u E K satisfies the inequality
a(u,v-u)+j(v)-3(u)> (f,v-u) Vv EK. In this case, P will be called the mixed formulation of P'. Thus, it is evident that P is more general than P, as it does not require the symmetry of a. It is possible to formulate conditions analogous to those mentioned before, which guarantee uniqueness and existence of the solution (u, A)
for problem P. However, throughout this book we will only encounter problems with a symmetric form a.
1.1.52. Approximation of the Mixed Variational Formulation and Error Bounds. Let {Vh} and {LH}, h, H E (0,1) be systems of finite-dimensional subspaces of V and L, respectively. Let Kh and Am be nonempty convex closed subsets of Vh and LH, respectively. It need not generally hold that Kh C K, Am C A. Each pair (h, H) E (0, 1) will be associated with the set Kh x AH. Further, we assume Am either to be (CCH)
a cone with its vertex at BL, and Kh = Vh Vh, H E (0, 1), or (BCH) a uniformly bounded subset of L, that is, there exists a positive constant c > 0 such that µH I < C
d/lH E Am VH E (0, 1).
By an approximation of problem P we mean to find a saddle point (uh,AH) of the Lagrangian X on Kh x AH: X(uh,/AH) : X(uh,AH) < X(vh,AH)
V(Vh,AH) E Kh X Am
or equivalently
find (uh, AH) E Kh x Am such that a(uh,, vh - uh) + b(vh - uh, AH) > (f, vh - uh) b(uh, AH - AH) :5I9, pH - AH]
\Ivh E Kh,
(PPH)
d/ZH E AH.
Denote
7H(Vh) = sup{b(vh,/AH) - [9,PH1} AR
As in the continuous case, we can verify that the first component uh E Kh is a solution of the minimization problem
7(uh)+5H(uh) :5Y(vh)+iH(vh) dvh E Kh.
(5.6)
1.1. Unilateral Boundary Value Problems
55
The proof is left to the reader as an easy exercise. The question of existence or uniqueness of the solution of problems phH in a finite dimension is simpler than it was with the continuous case, as is evident from the following lemma.
Lemma 5.5. Let (BCH) hold. Then there exists a solution of problem phH Moreover, its first component is uniquely determined.
Proof. The existence of a solution is a consequence of a more general assertion (see the proof of lemma 5.4); the uniqueness of the first component is a consequence of the V-ellipticityof a.
Lemma 5.6. Let (CCH) hold and let KhH = {Vh E Vh 11H(uh) < +001-
If KhH 7/ 0 (that is, KhH is a set with nonempty interior), then there exists a solution of PhH with a uniquely determined first component.
Proof. We easily see that KhH = {vh E Vh I b(Uh,PH) 5 [9,PH]
V/SH E AH}.
Indeed, if Vh E KhH, then jH(vh) = 0; otherwise, it would be jH(vh) = +oo (see section 1.1.2). By virtue of (5.6), we have
uh E KhH, y(uh) < Y(vh)
VVh E KhH.
Now the existence of a solution (uh, AH) is a consequence of the fact that Kh°H is nonempty (e.g., Cea (1971)). The uniqueness of the first component follows from the V-ellipticity of the form a.
Remark 5.5. Let
Kh= {vhEKh I - VhEKh}. If
b(Uh, PH) = 0
VVh E Kh = PH = BL,
then even the second component of the saddle point is uniquely determined. The proof completely coincides with that for the continuous case. Now we will study the mutual relation between (uh, AH) and (u, A). To this end, we first establish a bound for the error Ilu - uhII.
Lemma 5.7. Let {u, A} and {uh, AH} be solutions of P and phH, respectively. Then IIu-uhll2
0 is a given parameter, and (D : V i-+ L is the mapping defined by
(µ, (v)) = b(v, µ) - [g, µ] V(v, µ) E V x L, where ( , ) denotes the scalar product in L. Since b : V X L ---+ R is continuous, we have I((v) - -O(w) I < M1 11v - wII
Vv,wEV.
Thus, all the assumptions of section 1, Chap. 7 of Ekeland and Temam (1974) are fulfilled, and hence we have:
Theorem 5.5. There exist numbers p1, p2, 0 < P1 < p2, such that for p" E (P1, p2) the algorithm defined by the formulas (5.30), (5.31) is convergent in the following sense:
un -+u, n-+oo in V. Remark 5.10. This theorem guarantees only the convergence to the first component u. The convergence of An to A will be dealt with in Chapter 2, in connection with the Signorini problem with friction. In order to acquire a better comprehension of this and preceding sec-
tions, let us go back to the approximation of the problem (2.5). Let Vh, LH, AH be defined in the same way as in remark 5.1. Let (uh, AH)
be a saddle point of N in Vh x AH. In this case K = V, Kh = Vh, AH C A VH E (0, 1) and it is again possible to verify the validity of each of the conditions (5.13)-(5.17). Hence, uh u, h -+ 0+ in H1(fl), AH -+ A, H -+ 0+ in L2 (t). Moreover, uh is uniquely determined. As we know,
1.1. Unilateral Boundary Value Problems
61
a sufficient condition for the uniqueness of AH is (see remark 5.5 and the definition of X):
Jr VhµH ds = 0 dvh E Kh = Vh
µH = 8y.
(5.32)
It is evident that this condition need not be generally fulfilled. Let h = H (that is the partition TH of the boundary r is generated by the boundary nodes of the triangulation Th of the domain f2), and let us assume that the nodes a1, ... , an(h) form an equidistant partition of r. Then, (5.32) is equivalent to the system of linear algebraic equations
Al+/12=0 /12+/13=0
11 + Fim. = 0,
Iii = AH la;ai+1 .
If m is an even number, then this system has a nontrivial solution. On the other hand, we easily find out when (5.32) is fulfilled. Let M(H) be the number of segments of the partition TH, and m(h) the number of boundary nodes of Th. Then, (5.32) represents a homogeneous system of m(h) equations for M(H) unknowns. Hence, it is sufficient that this system be "overdetermined" and therefore have solely the trivial solution. To get an approximation of the saddle point (uh, AH) of the Lagrangian X in Vh x AH, we use Uzawa's algorithm. In this particular case we set V = Vh, L = LH, A = Vh, B = AH, X = X. Now, we can write (5.30) and (5.31) in the following explicit form: choose A E AH arbitrarily; use it to calculate uh E Vh, then AH, uh, etc. If we know AH E AH, then we find uh E Vh such that
(vuh, Vvh)0 = (f, Vh) 0 - Jr A I Vhd$ VVh E Vh; Then,
AH 1 = ll(A - Pnuh), P. > 0, n = 1,2,..., A,,
where
[J(9)lb,b,+i =
A
1
ifiri(q)>1
iri(9)
if iri(9) E l-1, ll.
-1
ifiri(q) 0, its solution v fulfills v E H1+E(fl) and Ilvlil+e,n
0 independent of h, H > 0 such that Jr vhµHds > 4IIpHII-1/2,r VAH E LH,
up IIUhI
provided that h/H is "suitably" small. In the above quoted paper, the rate of convergence of (uh, AH) to (u, A) is analyzed in detail for problem P2. We will come back to these problems once more in the next chapter when studying contact problems with friction.
1.1.6
Semicoercive Problems
In this section, we will study a problem analogous to P2 except for the additional assumption r,, = 0. Thus, let us consider the following onesided boundary value problem:
-L1u=f infl, 7For the definition of the Sobolev spaces with a fractional derivative we refer the reader to Nedas (1967) and Aubin (1972).
1.1. Unilateral Boundary Value Problems
63
u > 0, au/av > 0, ua/u8v = 0 on r.
(6.1)
Set
K= {vEH1(fl) (v>0 a.e. on r}, Y(v) = 2IvIi - (f,v)o,
f E L' (0).
A function u E K will be called a variational solution of problem (6.1) if y (u) < y (v)
Vv E K.
(6.2)
If a variational solution is sufficiently smooth, then it fulfills the point relations (6.1). The proof proceeds similarly to that of problem P1 in section 1.1.1.
Let us now deal with the problem of existence and uniqueness of (6.2). Since r,, = 0, Friedrich's inequality does not hold, and hence the functional
1, is not coercive in H'(fl). On the other hand, as follows from theorem 1.5, section 1.1.11, the coercivity of 1, only on K is already sufficient for the existence of a solution. We shall now formulate conditions which guarantee this property. First of all, let us show that a solution of the problem (6.2) cannot exist for arbitrary right-hand sides f E L2(f2). Indeed, we have
Lemma 6.1. If there exists a solution of problem (6.2), then (f, 1)0 < 0.
(6.3)
Proof. (6.2) is equivalent to
(Vu, Vv - Vu)o ? (f, v - u)o Vv E K. Substituting here the function u + 1 for v, where obviously u + 1 E K, we immediately conclude (6.3). Thus, a necessary condition for the existence of a solution is that the mean value of f in 0 be nonpositive. If the mean value of f is even negative, then we have
Theorem 6.1. Let (f, 1)o < 0.
Then there is one and only one solution of (6.2).
Proof. Uniqueness. Let u1, u2 be solutions of (6.2), that is,
(Vui, Vv - Vui) > (f, v - ui)o Vv E K.
1. Unilateral Problems for Scalar Functions
64
Substituting first the function u2i and then ul for v, after subtracting the resulting inequalities we obtain lug -
ull2,
0. Let us assume Ch > 0. Then we can find ch > 0, such that uh - ch > 0 on r, and hence uh_- Ch E Kh. By virtue of (6.4) we have .7 (uh - Eh) =.(uh) + Ch(!, 1) 0 < Y(uh),
which contradicts the assumption that uh E Kh is a minimizing element for 1f in Kh.
This lemma implies that there exists at least one node ai E r (we do not know it explicitly) such that uh(ai) = 0. This fact is the basis for a method of numerical solution. Let us arbitrarily choose a tangent point ai c F and define Vh
{Vh E Vh I vh(ai) = 0},
Kh={vhEVh I vh>0 onF}. Now we replace problem (6.9) by
find ah E Kh
Y(u'h) G Y(Vh)
dvh E Kh.
(6.9')
We easily verify that the stiffness matrix A formed from the base elements of Vh already is positive definite and results from A through deletion of the corresponding row and column of A. Thus, to solve problem (6.9') we can again, for example, use the superrelaxation method with additional projection. Nevertheless, it is then necessary to verify whether ai was correctly chosen; that is, whether uh(ai) = 0 really holds. A critierion for this verification may be, for instance, the calculation of auh/8v in a neighborhood of ai. if ash/8v < 0, then it is necessary to choose another point, at which we fix the solution. However, the method of conjugate gradients (see PsenRnii and Danilin (1975)) is much serviceable. This method makes it possible to find the minimum of the quadratic form given by a positive semidefinite matrix A with constraints of the form
Vi, a)R" = bi, (7i, o') R'
0 holds a.e. on F, hence (6.33) holds as well. 2° Now let us consider an arbitrary trapezoidal domain F,,,,, 1 < m < M. If F,,, is a parallelogram, we reduce it to a single line segment G`,,,, fl G',,,+i by dilating G,,,, and G,,,+,, and we set
P(') = 0 in 3(G;;, U G'L), p(2) = 0 in 3(G,,+i U G;,+i)
in (6.30), (6.31), (6.32), where p0) are components of the vector p in the local basis of G,,,,+i. Thus, we obtain the continuity of the flow p v on the segment G,,,, n
Thus, let us consider a general trapezoid F,,,. Let us introduce a new coordinate system (yi, y2) by the mapping xi = y1 y2 X2 =y2
x = Ty -
(6.34)
where the origin of the local Cartesian system (x1i x2) is at the point of intersection of the lines A$ and OD (see Figure 6). Then the trapezoid
F,,,= x E R 2 1 a
0 on the whole boundary r; hence p e Uo n [C°° (f))]2. 3° We still have to prove (6.26). There exists a sequence pn E [Co (Sl)]2
such that Ilpn p°llo,n -+ 0 for n -- oo. Let Eopi be the extension of pi (i = 1, 2) by zero outside ft; define (pn ) (y) = Eopi (ky). Then the following estimates are valid: Ilp
- (pn)illo.n
ECl(pi,)'
1. Unilateral Problems for Scalar Functions
82
IIPt - (P"); Ilo,n < IIPn - p IIo,n + IIEP°Ilo,ko. o. (For the detailed proofs we refer the reader to the paper Hlava&k (1980b).) If we use these estimates and the identity Ep° = p from lemma 6.5, we infer lip, - P IIo < IIPE - (P")E Ilo + II(P")E -P'110 + lip" - P'110
< 3IIP' - P°IIo + eC2(p") + IIPIIk(6.44) For a given rl > 0 we can find p" and E (depending on p") such that each of the three terms on the right-hand side of (6.44) will be less than a rl. Finally, we choose )i (depending on pE) sufficiently small and such that IIRxp - PEIIo
supSo(q) = sup[-Sg(q)] = -inf S9(q) = M
for
-S9(q°).
(7.6)
U1
If we set q = 4 = [Vu, u], then again as in the proof of theorem 1.6, we find 4 E 111. Indeed, according to theorem 1.1, inequality (1.8) holds for all
V E Kg. Substituting v - u = ±p, cp E CO '(0), we find that Du = u - f, hence div 4 = 4n+1 - f. Insert u = G + w, v = C + w in (1.8), where w E K1, w E K1. Then
(u,w-w)i > (f,w-w)o dwEK1
(7.7)
1.1. Unilateral Boundary Value Problems
85
and setting w = 0, w = 2w we obtain (u, w) i = (f, w)o,
(u, w)i > (f, w)o Vw E Ki.
(7.9)
Hence for all w c K1, (4
' v,7w) _ (Vu, Vw)o + (w, Au)o = (Du, pw)o + (w, u)o (f,w)o = (u,w)i - (f,W)o > 0.
Consequently, 4 E Ui. Further, using (7.8) we obtain
au,7u - 9) = (ou, pw)o + (w, Du)o (Du, Qw)o + (w, u - f )o = (u, w)i - (f, w)o = 0, hence
(4 ' v, 9) =
au,7u) = (pu, pu)o + (u, Lu)o
(vu, pu)o + (u, u - f)o = IIuIIi - (f, u)o. Substituting q = 4 into Sg, we obtain
-S9(4) = -2II4II2 + (49)
- 1 IIuIIi + IIuII, - (f,u)o = Li(u).
(7.10)
By virtue of (7.6), this implies that the functional (-Sg) assumes its maximum at the point 4. The uniqueness of problem (7.2) implies 4 = q°. Relation (7.5) follows from this and (7.10). 0
1.1.71. Approximation of the Primal Problem. Let us now consider problem Pg in a planar polygonal domain fl C R2, whose vertices will be
denoted by Ai, ... , A,,. Let {Th}, h -- 0+ be a regular system of triangulations of 0. Each Th will again be associated with a finite-dimensional space
Vh = {Vh E C(ll) IvhIT e P1(T) VT E Th}, and a convex closed set
Kgh = {vh E VhIVh(ai) ? g(aj),
i = 1,..., m(h)},
1. Unilateral Problems for Scalar Functions
86
where ai i = 1.... , rn(h) are the nodes of the triangulation Th, which lie on r. The approximation of problem Pg is defined in the usual way: find Uh E Kgh such that £1(uh) :5 C1(vh)
VVh E Kgh.
(7.11)
As C1 fulfills all the assumptions of theorem 1.5 on Kgh, for every h > 0 there exists a unique solution Uh of problem (7.11). Let us study the relation between u, uh. The situation is now more complicated due to the fact that Kgh are generally not subsets of Kg. First, we will deal with the rate of convergence of uh to u, provided the solution u is sufficiently smooth.
Theorem 7.2. Let the solution u fulfill u E H2(11) n K, u c W1,00(r) and let the set of points from r at which u = g changes to u > g be finite. Moreover, let g E H2(Ai, Ai+1), i = 1,... , n. Then IIu-uhl1 g(ai) Vi = 1,...,m(h), which follows from the definition of Kgh, we infer that uh > Rhg, where Rhg is the piecewise linear Lagrangian interpolation of the function g on r. We simul-
taneously have 8u/av > 0 a.e. on r and consequently
f
_
av (9 - uh)ds < fr_ TV (9 < Ch 2
Rh9)ds
0 on r} and define: by an approximation qh E Uoh of the dual problem we mean a solution of the problem I©(qh) G Ig(q)
Vq E Uoh.
(7.14)
1. Unilateral Problems for Scalar Functions
88
This problem has a unique solution, since u0h is convex and closed in H(div, f1), Ig is quadratic and strictly convex. The last term of the functional Ig reduces to the integral v, g) = - J 4 ' v gds
r
V E [Vh]2.
(7.15)
As concerns the error I' - qh, we can apply the method from section 1.1.41, only adding the term (7.15) to the functional I. Hence, we obtain
Theorem 7.4. Let us assume q E [H2(fl)12 and q0 v E H2(r,,,) on each of the polygonal boundary.
side Then
2
1I4; - qh112 + Ildiv(go - gh)IIo)1/2 = 0(h) i=1
holds for any (a,#)-regular system of triangulations.
Remark 7.1. If qh is a solution of problem (7.14), then
ah={gh,g2,f+divgh}EU1 is an approximation of the original dual problem (7.2). By virtue of theorems 7.1 and 7.4, we obtain, under the assumptions listed above, 2
au
q;h
= 0(h), Ildiv qh + f - ullo = 0(h). 0
:.1
1.1.73. A Posteriori Error Bounds and a Two-Sided Energy Estimate. Let us assume that we have evaluated approximations iih E K9 and q"H E Uotj of the primal and the dual problem, respectively. (Notice that if gh > g is not valid on r, then Kgh is not a subset of Kg, and hence the above defined approximation by the finite element cannot be used!) Then, error bounds for both primal and dual approximations can be established.
Theorem 7.5. Let uh E Kg and qH E UOH. Then H
IIilh - uIIl
f, (-Au-f)(u-,p)=0, u > V a.e. in 0,
u=0
r.
(1.2)
The domain 11 can be divided into two subsets 12o, n+, where
11o={xE0 u(x)=4p(x)},
n+={xEn u(x)>cp(x)}. If x E fl+, then (1.2) implies that -Au(x) = f (x). Notice that the partition of 0 into flo, t2+ is not a priori known. Therefore, this partition is one of the unknowns of the problem considered. Definition 1.1. The problem (1.1) will be called the primal variational formulation of the problem with an inner obstacle. Its classical (pointwise) formulation is given by (1.2). For the same reasons as in section 1.1.11, we will formulate problem (1.2) in terms of the gradient components. In what follows we assume q7 E Ho (1l). Let
Qt = {q E [L2(n)]2 I (q, Vv)o > (f, v)o Vv E Ko}, where
Ko={vEHo(1l) I v>0 a.e. on fl). Remark 1.1. We easily find that
qEQ! bdivq+f _ (f, v*)o
Vv* E Ko,
that is, VuEQf. Let us now prove S(Vu) _ (q-Vu,Vsp)o VgEQf.
1. Unilateral Problems for Scalar Functions
92
By virtue of (1.7) we obtain
(vu, q - Vu)o - (q - Vu, ow)o = (q, 0(u -'p))o - (vu, 0(u -'p))o (q, pu*)o - (Vu, pu*)o = (q, Vu*)o - (f, u*)o ? 0, since u* E Ko and q E Qf . As q* is uniquely determined, the identity q* = pu obviously holds. Definition 1.2. Problem (1.3) will be called the dual variational formulation of the problem with an inner obstacle.
Remark 1.2. Let us introduce still another equivalent formulation of problem (1.3), which will be useful for an approximation of the original problem. Let q E [L2(cl)]2 be such that
divq= -f in 0. Then evidently
Qf = +Qo, where
Qo = {q E [L2(1)]2 I (q, Vv)o > 0 Vv E Ko} is the set of vector functions whose divergence (in the sense of distributions) is nonpositive. Denote
S(q)=S(q+q),
gEQo.
Let q° E Qo be the element for which S assumes its minimum in Qo (such an element exists and is unique, as S again is a quadratic, strictly convex, and continuous functional in Qo ):
S(q°) 5 S(q) dq E Qo .
(1.8)
Then q + q° minimizes S in Qf , that is, q* = q + qo.
Remark 1.3. Let a solution u of problem (1.2) be smooth enough to guarantee Du = div(pu) E L2(0). Then instead of Qf we can consider the set Qf (div, f2) defined by Q f (div, fE) = {q E H(div, 0) 1(div q + f, v)o < 0 Vv E K°}.
The definition of the operator of divergence implies the following equivalent version of the functional S:
(q) = S (q) =
2
]IgIIO + (div q, ,p)o
Vq E Q f (div, fl).
1.2. Problems with Inner Obstacles
93
As an exercise, the reader can prove the following assertion: let the solution u of (1.1) satisfy Au E L2(1l). Then q* = Vu E H(div, tt) minimizes S in Q1 (div, fl), that is, (q*)
(q)
`dq E Q1 (div,fl).
The proof is analogous to that of theorem 1.1.
1.2.2
The Mixed Variational Formulation
Problem (1.3) represents a minimization problem with a constraint. This constraint can be removed by introducing the Lagrange multiplier in a suitable way. Let us now describe this procedure. Let us define the Lagrangian X in [L2(fl)j2 x H'(0) by 1(q, v) =
2
[JgJJ' - (q, v')o + (f, v)o - (q, Vv)o, (2.1)
(q, v) E [L2 (11)12 X H01 (f2).
Theorem 2.1. There exists exactly one saddle point (q*, v*) of the Lagrangian )( in [L2(IZ)[2 x Ko, and
v* = u - V,
(2.2)
q* = Du,
where u is a solution of the problem (1.1). Proof. (i) First we will show that every saddle point of )( in [L2(f1)12 X Ko satisfies (2.2). Let (q*, v*) be a saddle point of)( in the set just mentioned. Then necessarily (q* ,
q)o - (vSo, q)o - (vv*, q)o = 0 Vq E [L2(12)]2,
(f, vo - v*)o - (q*, p(v - v*))o < 0 by E Ko.
(2.3) (2.4)
The last inequality implies
q* = 7(p + v*);
(2.5)
hence, by substituting in (2.4) we obtain
v*), 0(v - v*))o ? (f, v - v*)o by E Ko. This inequality can be written in the form
(o('+v*), o(v+P)-o(v*+ca))o ? (f,
dv E Ko. (2.6)
1. Unilateral Problems for Scalar Functions
94
The function v + yo E K for every v E K0. Now (2.6) is equivalent to the fact that the function v* + V is a solution of (1.1). The rest follows from (2.5).
(ii) Let u E K be a solution of (1.1). Then we can write u
sp + u*,
u* E Ko,
and consequently
VU =pip+Vu' Evaluating the scalar product of both sides of this identity with q E [L2 (fl)]2,
we obtain (2.3) with q* = vu, v* = u - cp. On the other hand, u as a solution of (1.1) satisfies (1.1'), which by virtue of (i) can be equivalently written in the form (Du, Vv - Vu*)o >- (f, v - u*)o Vv E Ko,
which is exactly (2.4) with q* = vu, v* = u* = u - cp. The pair (Vu, u ,p) E [L2(i1)12 X Ko is thus a saddle point of )( on the above mentioned set.
Definition 2.1. The problem of finding a saddle point of )( given by (2.1) on the set [L2(fl)]2 x Ko is called the mixed variational formulation of the problem with an inner obstacle. Remark 2.1. If Au E L2 (0), then we can replace the Lagrangian l by )i, which is in the set H(div, 12) x L+(f2) defined by (q, v)
2114112
+ (div q,,p)o + (f, v)o + (div q, v)o.
(2.7)
Then we look for a saddle point (q*, v*) of the Lagrangian N in the set H(div,f2) x L2 (0). Again, (2.2) can be shown to be valid.
1.2.3
Solution of the Primal Problem by the Finite Element Method
In what follows we shall assume that fZ C R2 is a bounded polygonal domain.
Let { Th }, h -i 0+ be a regular system of triangulation. We associate each Th with a finite-dimensional space Vh C Ho (fl), where Vh = {vh E C(f2) IvhIT E P1(T) VT E Th, vh = 0
on r).
We denote the inner nodes of the triangulation Th successively by al, a2, ... a,n(h), and define Kh = {vh E Vh I vh(ai) > cp(ai) Vi = 1, 2,..., m(h)}.
1.2. Problems with Inner Obstacles
95
Kh is a convex closed subset of Vh for every h E (0, 1). Nevertheless, Kh C K does not generally hold. By an approximation of the primal formulation of the problem with an inner obstacle, we mean 1/a function uh E Kh which satisfies y (uh) < Y(vh)
dvh E Kh.
Our aim is to establish the order of the error flu - uh111. We have
Theorem 3.1. Let cp E H2(fl), u c H2(1l) fl K. Then (lu - uh111 < c(u, f,V)h,
h
0+.
Proof. We apply (3.10) of section 1.1.32 to our particular case: lu - uhl1
- (f, u - Vh)0 + (f, uh - v)0 + (V(uh - U), V(Vh - u))o
+(Vu, p(v - uh))o + (vu, V(vh - u))o Vv E K, VVh E Kh. (3.1) The Green formula together with the inclusion Kh, K C Ho (0) Vh > 0 implies
(Qu, v(v - uh) )o = (- A u, v - uh)o dv E K, (Du, V(vh - u))0 = (- A U, Vh - u)0 dvh E Kh. By substituting into (3.1) we obtain
lu - uhll _< (- Au- f,v-uh)0+(-Au-f,vh- u)0 (3.1') + (V(uh - u), 0(vh - u))o Vv E K, dvh E Kh. In (3.1') we set Vh = rhu, where rhu is the piecewise linear Lagrangian
interpolation of the function u. This choice is justified by rhu E Kh. The following inequalities now follow from the well-known interpolation properties of Vh:
I(- o u - f, rhu - u)ol < 11- Au - fllollrhu - ullo < c(u, f)h2,
(3.2)
1(0(uh - u), 17 (rhu - u))01 C 2luh - ul1 + 2lrhu - ul1 21uh
- ul2 + c(u)h2.
(3.3)
The estimate of the first term on the right-hand side of (3.1') is a little more complicated. The function used to establish a bound for this term is defined by v = sup{rp, uh}. It can be shown that v E K, and therefore this function can be used to estimate (3.1'). Let f2-
Ell
P(X)
1. Unilateral Problems for Scalar Functions
96
fl+ = {x E fl I Uh(X) > P(x)}.
The definition of Kh implies that uh(a;) > c (a;) for i = 1,..., m(h), and hence also uh > rhu in 11. This, together with the definition of v and the inequality -Au - f > 0 a.e. in fl, implies that we can write
(- L U- f, V - uh)0 = (-A U- f, 'p - uh)O,CT(- A u - f, 'P - rh'P)O,r)-
0, such that IIVO - wno II.
2E.
TThis assumption is not necessary; ro E C(0) is sufficient.
1.2. Problems with Inner Obstacles
97
As w,, E Co (f1), we can construct its piecewise linear Lagrangian interpolation rhwryo . Evidently, rhWno E Koh and for h > 0 sufficiently small, Ilwno - rhWno II1
Ze.
This together with (3.5) and the triangle inequality lIvo - rhwn.o 111 0 in 0 for all h > 0, we also have v - y> > 0 in 0. Now the assertion on the convergence of uh to u follows from theorem 3.1, section 1.1.32.
1. Unilateral Problems for Scalar Functions
98
1.2.4
Solution of the Dual Problem by the Finite Element Method
1.2.41. Approximation of the Dual Formulation of the Problem with an Inner Obstacle. In this section we will study the approximation of the dual formulation of the problem with an inner obstacle. To this end we will again use the Ritz method. Let QOh C Q0 , h E (0, 1), be "finite-dimensional approximations" of the convex set Qo and set Qjh = q + QOh]
where q" E [L2(f1)]2 is a particular solution of the equation
div q = - f in 0. Evidently, Qfh C Qf for every h E (0, 1). By an approximation of the dual formulation of the problem with an inner obstacle, we mean the problem to find qh E Qfh such that S(qh) S S(qh)
Vqh E Qfh.
(4.1)
We have
Theorem 4.1. For every h > 0 there exists exactly one solution of problem (1.4), and llq* - ghllo < {(q* - qh, oco)o + (q* - qh, q* - qh)0 + (q*, qh - q*)o} (4.2) holds for any qh E Q fh.
Proof. The existence and the uniqueness of the solution of (4.1) is a consequence of the fact that S is convex and coercive on a convex closed subset Qfh. The inequality (4.2) is just the transcription of (3.10') from section 1.1.32 for our particular case and notation. We also use the inclusion
QfhCQf VhE(0,1).
O
Remark 4.1. Let the reader not be confused by the fact that for estimating the error JJq* - gh11o we use the relations (3.10) or, as the case may
be (3.10') from section 1.1.32, which were formally used to estimate the errors of the approximations of the primal variational formulations. What is substantial is that the dual variational formulation of our problem has the same character as the problem which was formulated in a general setting in section 1.1.32, that is, the problem of minimization of a quadratic functional on a convex closed subset of a Hilbert space.
1.2. Problems with Inner Obstacles
1.2.42.
99
Construction of the Sets QJh and Their Approximate
Properties. To be able to solve problem (4.1), we have to specify the choice of the sets Qoh. For this purpose, we will use the finite element method. The construction will be similar to that used in section 1.1.42 for the approximation of the dual formulation of problem P2. Let fl be a bounded polygonal domain, {Th}, h -+ 0+ a regular system of triangulations of fl. Let us define Qoh = {q IgIT E (P1(T))2 VT E Th, (q - v)T + (q - v)T' = 0
`dxETf1T', divq 0} and c is a positive constant independent of h.
Proof. To prove (4.9) we shall use the relation (4.7). Put qh = rhq°, where rh has the same meaning as in the previous section. This choice is justified by rhq° E Qoh. Then (4.5) implies (q* - qh, q° - rhq°)oI
1IIq* - ghII + 1IIq° - rhgollo
= 2 IIq* - qh 0 + 0(h2').
(4.10)
Now let us write
(div(4 - rh4 ), u - V)o = E (div(q° - rh4 ), u - p)o,T. TETk
If T C fZ+t for some t = 1, ... , p, then
div q* = div(ou) _ - f in T by virtue of (1.2)2. Hence,
div q° = -div q - f = f - f = 0 in T. Taking into account the definition of rhq° and particularly (4.25), section 1.1.42, we obtain div(rhq°) IT = 0 in T, and consequently (div(q° - rhq°),u - 'p)o,T = 0. (4.11)
If T E Th is such that T C tZ- it1 f)+t, then u = cp a.e. in T and (4.11) again holds. Let I be the system of all T E Th which fulfill T- iZ+t # 0 but T C iZ+t.
Set flt h = 0 n 0+,. Then, in virtue of the facts proven above, we can write (div(q° - rhq°, u - co)o,o I
>
on rK,
(1.10)
u''=u",n" i r n'=-n'. n
Let us briefly indicate how to derive condition (1.10), which essentially represents the condition of nonpenetrating of the bodies into each other: Let us assume that before the deformation both bodies n' and n" had contact along the whole arc rK (see Figure 9). Let us identify the axis xl
2.1. Formulation of Contact Problems
113
Figure 9
with the normal n" and the axis x2 with the tangent t" at a certain point 0 E rK. During the deformation process the points 0' E 811' and 0" E 811" are generally displaced in a different way, nonetheless, always to that the body 11" cannot penetrate the body 11'. This condition yields U'1'(0, 0) 0 and u', + u,.' + +' < 0 on rK. There exists w E V such that wn w, = 0 on rK, hence v = u + w E K. Condition (1.28) together with the inequality Tn 0 on rK implies
0
0.
Denoting ul - u2 = z, we have A(z, z) _< 0. Now the condition (1.4) implies
that ei (z) = 0 Vi, j, hence z E R n V. If R nV = {0}, then z = 0 and the solution is unique.
If z # 0, denote u2 = u, ul = u + z. Then A(u, z) = A(z, z) = 0,
L(z) = 0, L(u) = L(u + z) which contradicts assumption (2.2). Hence, again z = 0. C(u) _ .C(u + z)
Example 2.1. Let ro consist of line segments parallel to the xl-axis (see Figure 8). Then
V nR= {zI z' = (0,0), z" = (a,0), aERl}. Assume that ni > 0 almost everywhere on rK and that there is x E rK with ni(x) > 0. Then,
KnR={yjy'=(0,0), y"=(a,0), a 0. Then (see Hlav£Zek, NeZas (1970), Part I, remark 4)
H= V E) Rv = {v E V I p(v) = 0}.
(2.11)
Proof of Theorem 2.3. 1° Every v E H satisfies Korn's inequality (see Hlav£Zek, NeZas (1970), Part I, remark 3.4) c1IIIvIIji 1colvl2 - L(v) >
CIIIVIII2 - IILII
111'1111,
which implies the coerciveness of £ on the subspace H. 2° Since L is also quadratic, convex and the set k is convex and closed, there exists a solution u of problem (2.8). Let u2, ul be two solutions of problem (2.8). Similarly to the proof .
of theorem 2.1, we derive z = ul - u2 E RV. Since z E H, we have z c Ry n H = {0}, hence, the solution is unique. 3° By virtue of (2.7) we obtain
£(v)_£(v+y) byERv.
(2.14)
Moreover, for the orthogonal projection PH : V --+ H we have
PH(K)=KnH.
(2.15)
Indeed, let v E K. By virtue of (2.3) and (2.6) we obtain PHV = v - PR,, v,
R* = Ry,
(PHV)' + (PHV )' = vn + vn < 0 on rK, hence PHV E K n H. The converse inclusion
K n H = PH(K n H) C PH(K) is trivial. Now let u be a weak solution of problem Pl. Using (2.14), we can write £(PHV) = C(PHV + PR., V) _ £(v)
Vv E V.
Further, PHU E K n H by (2.15),
£(PHU) = £(u) 0. Then
yERv - {0} = L(y) =a1V1"¢0, hence the condition (2.19) is fulfilled. When the situation corresponds to that shown in Figure 12, then it is also easy to verify (2.21).
Proof of Theorem 2.4. 1° Let us first consider the case K n R = {0}. We shall use the following Abstract Theorem 1 (see 2.2):
(1975), theorem
Let Jul be a seminorm in a Hilbert space H with a norm llull. Let us define a subspace
R={uEHIJul =0}.
2. One-Sided Contact of Elastic Bodies
128
Figure 12
ro
ru Figure 13
2.2. Existence and Uniqueness of Solution
129
Assume dine R < oo and C1llull cllul12 Vu E H.
(2.23)
We apply this abstract theorem to our case with H = V, R = V n R = Rv, defining lv) according to (2.13) and Q(u) = 2
f
1(uI + u`n')+12ds. K
In order to verify (2.22), we use an inequality of Korn's type (see Hlava ek and NeUs (1970)) and the decomposition
V=QeRv. Thus, we obtain for all u E V the inequality IIIulIIi =1IlPQullli + IIIPRv uIIIi
0 (see Figure 15) and
Vi' =
Jcv'
Fi dx + J PP'ds > 0.
rr
(2.28)
Then C is coercive on KE and there exists a unique solution of problem P2.
2. One-Sided Contact of Elastic Bodies
132
Figure 15
Proof. Let us set
0
Po(v) =
(v4 - v')dn, Ja
Vp = {v G v Ipo(v) = 0}.
Then
RnVp={o}.
(2.29)
Indeed,
RnVpCRv={z=(z',z")Iz'=0, z"=(c,0), cER'}. The identity po(v) = 0 yields /b
(b
0 = I zf dry = c J cos(e, xl)dr7 = c = 0. JJa
a
By means of (2.29) we can prove an inequality of Korn's type (see Hlav£6ek, N0as (1970)): Ivl > Cl11vll11
Vv E Vp.
Let v E V. Define y E Rv by the relations
y'=0, yi with
d=
=Po(v)d-1,
/b
Ja
y2" =0,
cos(e, xl)d,i.
(2.30)
2.2. Existence and Uniqueness of Solution
133
It is easily verified that the difference Pv = v - y satisfies b
P0 (PV) = P0 (V) - Po(y) = P0 (V) - f Po(v)d-1 cos(e, x1)dri = 0, a
hence, Pv E V. With the help of (2.30) we can write 'C (V) =
1 A(Pv, Pv) - L(Pv) - L(y) ? CIIIIPvIiii - C2IIIPvIII1 - y"Vi'. (2.31)
When IIIvIII1 -+ oo then at least one of the norms IIIPvIIIl, IIIy!II1 tends to infinity. Moreover, V E KE = p0 (v) C
f
b
e dry < +oo,
(2.32)
a
IIIyIIiI = Iyil `f dx J (1"
1/2
=
IPo(v)Id-1(mestl")112.
(2.33)
10 Let IIIylil1 -- oo. Then/(2.32) together with (2.33) imply -p0(v) -
+oo, and hence -yi -+ oo. As C1IIIPvIIIi - C2IIIPvIIi1 > C3 > -00,
we conclude from (2.31) and (2.28) that .C(v) -- +oo. 2° Let IIIPvIIIl -' +oo. Then (2.32), (2.28) yield £1(PV) = CIIIIPv1II1- C21IIPVIII1 - +00,
£2(y) = -y"V1' = -Po(v)d-1Vi" > -d-1V1'
f
b
edrl > -oo.
a
By virtue of (2.31) we have
£(v) > L1(Pv) + C2 (Y) -++00. Thus, we have proved that .C is coercive on K. Since KE is a closed convex subset of V and the functional C is convex and continuous on V, a solution of problem P2 exists. Uniqueness is a consequence of condition (2.28). Indeed, we first prove-
as in the proof of theorem 2.1-that the two solutions u1 and u2 differ from each other by an element z E Rv with L(z) = 0. On the other hand, however, L(z) = cV1", c E R1. Condition (2.28) implies c = 0, that is,
z=0.
2. One-Sided Contact of Elastic Bodies
134
2.3
Solution of Primal Problems by the Finite Element Method
In this section we will study approximations of contact problems by the finite element method. We will describe a construction of finite-dimensional approximations of the set of admissible displacements, which will be used for defining approximate solutions. This will be done first for the contact problems with a bounded zone of contact, the contact zone being given first by a piecewise linear curve, then by a smooth one. Subsequently, we will
deal in the same way with problems with an increasing zone of contact. Further, we will discuss the mutual relation of the approximate and the exact solution. We will also find the rate of convergence, provided the exact solution is sufficiently smooth.
2.3.1
Approximation of the Problem with a Bounded Zone of Contact
Let us consider problem P1, using the symbols V, K, L, A, and L in the same sense as in section 2.1.3.
I. First, let us assume that f2', f2" C R2 are bounded domains with polygonal boundaries 8f2', 8f2". In this case, we can write rK in the form m
rK = U rK,i, i=1
where rK,; is a closed line segment with an initial point Ai and an endpoint Ai+1. Let Th, Th" denote triangulations of polygonal domains 0' and it". Here we observe the current rules, which were formulated in section 1.1.31. Naturally, we assume that both Th' and Th' are consistent with the respective decompositions of the boundaries 812' and aft". Moreover, the nodes lying on rK belong to both the triangulations. The pair {Th, Th'} defines a decomposition of the set f2 = 0' U 0". More frequently, we will use a simpler notation, namely, Th = {Th, Th'}. Th is said to be regular if both Th, Th' are regular. We associate every triangulation Th with a finite dimensional space Vh, given by Vh = {Vh E [C(jn')]2 X [C( ")]2 n VIvhIT E IP1(T)12 VT E Th}.
(3.1)
Let a'' , j = 1, ... , rn; be the vertices of Th lying on rK,i (ai = A;, a;,, Ait 1), i = 1, ... , m, and let ni be the unit vector of the outer normal of ,
the side rK,i with respect to W. Let us define
Kh={vhEVhIn''(vh-vh)(a1) 0 by the relation wa (x) = w(x + Ae').
For N < CA with C > 0 we have the identity RXwa = 0 on r(2), and IIRXwa - wII I
we conclude from (3.27) that Hu2 - u' 11, - 0,
A -. 0+,
X < CA.
Group 3. Let j < k and let By contain a point P E 1'K n r,. We introduce a new Cartesian coordinate system with its origin at P, whose x1-axis coincides with rK (see Figure 19). In that case
un+uR=-u2+u2oKh = K, the closure being taken with respect to the norm of N1(f2) (see remark 3.9, section 1.1.32).
In section 2.2 we have formulated sufficient conditions guaranteeing the coerciveness of 1/ on K. In some special cases, these results guarantee the coerciveness of / on Uh>oKh. 3The constant c in the estimates IIuhII1/2,rh < cIIuhII1,r.LJ1E,,
IIuaIII/2,rh < cllu'h'Ijl,n"uu;v;, generally depends on h. Nevertheless, it can be shown that for h > 0 sufficiently small, c can be estimated from above independently of h.
2. One-Sided Contact of Elastic Bodies
154
For example, let rK contain a segment I, and let us define KI = {v E V l v'' - v' < 0 on I}.
Then, evidently the convex sets Kh defined by means of (3.4) satisfy the inclusion Kh C KI for all h E (0, 1). Since Uh>oKh C KI, the coerciveness of iJ on this union follows for example from the coerciveness of 1I on Ki. The coerciveness of l,/ on Uh>oKh can be studied even in more complicated cases (see Haslinger (1979), where these problems are studied for some semicoercive cases of the Signorini problem). We do not intend to discuss the problem of density of Uh>oKh in K here. Nonetheless, let us mention that the corresponding density result can be obtained by modifying the proof of lemma 3.2.
Remark 3.6. If we want to establish the convergence of the approximate solutions uh to a nonregular solution u, then with regard to the fact that Kh ¢ K we have to verify the implication
vh E Kh, vh-, v, h - 0+ (weakly) in V = v E K. For the Signorini problem, this was accomplished in Haslinger (1979). (See also lemma 3.7 of the next section.)
2.3.32. Increasing Zone of Contact. Let us consider problem P2 and its approximation P2h, defined in section 2.3.2. We will study the rate of convergence of uh to u, provided u is sufficiently smooth. Then we will prove the convergence of uh to a nonregular solution u. To this end, we will make use of the results of section 2.1.2 and 2.2.2.
Theorem 3.4. Let P2, P2h have solutions u, uh, respectively. Let u E X2 (0) n KE, of E wl,-(rK), u' E w1'O°(r, ), f'f" E C2([a,bl). Further,
let us assume that the number of points in [a, bl at which the inequality
u't - u't < e changes into the identity u't - u' = e (e = f" - f') is finite. Then,
Iu - uhl < c(u, f', f")h for an arbitrary regular system {Th}, h - 0+. Proof. We proceed analogously to the proof of theorem 3.3. We start with relation (3.8). After integrating by parts we can write it in the form
Colu - uhl2 < A(uh - u, Vh - u) + (T'e(u), vhf -
+ (Te (u), vhf - uE)o,rx + (TT (u), v4 - uh f)o,rx+ (TfM"(u, vf'
Vu EK,h, VEKE,
- uhf)o rx
2.3. Solution of Primal Problems
155
where all symbols have the same meaning as in section 2.1.2. Let the function vh be chosen to be the corresponding P-interpolation of the exact solution, constructed by using the isoparametric technique (see Zlamal (1973)):
Vh = ut,
where
UIIT=*(uIToFT)oFTi VTETh. Here T = FT (T), and fr is the operator of the linear Lagrangian interpolation of T (see section 2.3.2). The definition of the function uI implies that uI E KEh. The well-known approximative properties of uI (see Zlamal (1973)) imply
IA(uh - u, uI - u)I G
2luh - u12 + c[[ui
u]]1
2luh - u12 + ch2Illu[[[2,n.
(3.40)
Let us write (TE(u), ule - uE)o,r' + (T'(u), uie - uF)o,r,,
f Te(u){(u1F - ute) - (uf b
=
a
uf)}dn,
where
TT(u) 'If TE'(u)(eos a")-1 = -Te'(u)(cos ai)-1. Denote
Wh(rl) = u' e(f"(rl), rl) - ute(f'(n), rl),
(rl) = u' (f"(rl), rl) - u' (f'(rl),rl), rl E [a,b). The functions u'I(f'(q), q), u' (f"(rl), rl) are piecewise linear in the variable r7 on [a, b] with vertices at points C3. Since £ E R2 is a constant vector, Wh is a piecewise linear function on [a, b] as well. Let us divide the interval [a, b] into two disjoint subsets r°, r-, where
r°={r7E[a,b]lUl -u' =E},
r-={, E[a,b][uE-u' <E}. if [C;, C;+1] C r°, then the function Wh is a piecewise linear Lagrangian interpolation of a on [C;, C;+1], and
f
c,+1
c,+,
c,
Te(u)(Wh(rl) - U(rl))drl = f
c,
Te(u)(Wh - E)dn
2. One-Sided Contact of Elastic Bodies
156
0 for all v E K, then A) = X2 (0, A) = 0.
inf X2 (v, vEK
Denoting
KF p = {A E S (X2 (v, A) > 0 Vv E K},
(4.13)
we thus obtain provided A E K+ p,
0
inf X2 (v, A) =
(4.14)
-oo provided A V KF. p.
vEK
Lemma 4.5. Let there exist a solution u of the primal problem. Then KF p is a nonempty closed and convex subset of S.
Proof. We will show that the stress tensor r(u) E KF p. Indeed, we have
rij(u)eij(v - u)dx > L(v - u) Vv E K. Substituting here v = u + w, where w is an arbitrary element of K, we obtain
X2(w, r(u)) = /n rjj(u)ejj(w)dx - L(w) > 0 t/w E K. The fact that KF p is both closed and convex is evident. The relations (4.10), (4.12), and (4.14) imply that f W
-S(A) provided A E KF p,
XU u 'v ' A) --
-oo
where
S(A) =
0
2
f
n
provided A ¢ KF p,
aijklAijAkldx.
Hence, we have
inf
sup [-S(A)]
sug IN v)Ew
aEKF.r
zEKfi.r
S (A).
If there exists a saddle point of X on V x S, then
-
inf
AE Kp
p
S(A) _ -S(r(u)) = .C(u),
(4.15)
where u is the solution of the primal problem. Indeed, this follows from lemmas 4.2, 4.1, and 4.4.
2. One-Sided Contact of Elastic Bodies
168
On the other hand, lemma 4.3 guarantees the existence of a saddle point provided there exists a solution of the primal problem. In section 2.2.1 we dealt with the problem of existence and uniqueness of a solution of the primal problem. According to theorem 2.2, sufficient conditions for the existence of a solution are
L(y) J Fiv;dx + n
n'uan"
I
[T(A)v+
P;v;ds.
JJJrr
Substituting vM = ±joi E Co (0M), M = ',", we find (4.18). Thus, we obtain
Ti(A)v;ds> J P;v;ds `dvEK. rr Choosing v; = ±Oi such that the traces of 0 have a support in rr, we
f
n 'uaa1
obtain (4.19).
Now take v' = 0 and v" such that vn = 0, vt' = f+o i on r°i where the support of o is in r°. Hence, (4.20) follows. It remains to analyze the inequality IrK
[Tn(A')v'n + Tt(A')vt + Tn(A")v' +Tt(A")vt']ds > 0 \/v E K. (4.23)
Let v E V be such that vt = vt = 0 and v'', = -v' = ±(p on rK, E Co (rK). Thus, we find that
fr,)T
- Tn(A")]pds = 0.
Hence,
Tn(A') = T,(A")
on rK.
2. One-Sided Contact of Elastic Bodies
170
If we choose v E V such that vn = vn = 0, vt = 0, v' = fop on 1'K, then (4.23) implies that Tt(A') = 0 on rK. The identity Tt(A") = 0 can be analogously derived. Finally, we have
f
v") ds > 0 Vv E K,
Jr K
hence, T (A) < 0 by virtue of the inequality vn + vn < 0 that holds on I'K. 2° Let us multiply (4.18) by a function v,, where v E K, and integrate by parts over ft. TThen,
0=-
fo
.1,
av'dx+
axj
fo
a;;njv;ds
F; v;dx+
- fn .1,je,j(v)dx+ L(v) + J
fooluen"
r
Tn(A)(v' + vn)ds.
The definition of K, together with (4.22), implies that the last integral is nonnegative, hence .1 E KF K.
2.4.1
Approximation of the Dual Problem
To obtain an approximation of the dual problem, it is necessary to construct finite-dimensional approximations of the set KF p of admissible stress fields.
To this end we first find a particular solution A of the nonhomogeneous equations (4.18), (4.19), and then we will write A = A +,r for A E KF p where r is a "self-equilibriated" stress field. Since the resultant of the system of forces F,, P, (which act on the body 0") is generally nonzero, we must introduce reactions (normal loads) Tn(A)
on r K, which naturally act on both bodies 0' and ft". Lemma 4.7. If A E S fulfills conditions (4.18), (4.19), then
frx T(")ds = - V0, F,dx + fr P,ds + L T,(")ds] ,
i = 1, 2. (4.24)
P roof. Immediately follows from (4.18) and (4.19):
r
- fo"
F,dx=
c7'Kdx= a" fo axj
an
M'n'ds 3 S3
jr Pi ds+Jrp T,(1")ds+f T1(")ds, K
2.4. Dual Variational Formulation
171
and this yields (4.24).
With regard to lemma 4.7, we can select the simplest distribution of reaction pressure forces T,,(.1) on rK.
Example 4.1. Let ro consist of line segments parallel to the xl-axis, while
rK is such a segment that ni > 0 on rK. We can choose a E KF p such that (4.25) Tn (A') = T. (All) = g on rK, where
g=-
[fo,
F1 dx +
f
P1 ds]
J dx2 = const.
(4.26)
K
r
Indeed, using the identities T1(A) = Tt (a) = 0 on ro and dx2 = n' ds, 0 on rK, we find that the choice (4.25), (4.26) satisfies condition (4.24). Further, we know that (4.16)1 is a necessary condition for the existence of solution, which in our case means (see example 2.1): Tt(A)
Vi'=
f
Fldx+
ci"
f
r T,
Plds>0. -
(4.27)
Hence g < 0, and condition (4.22) is also fulfilled.
Example 4.2. Let ro and rK be the same as in the previous example and let n2 > 0. We can choose T,, (A') = T (A") = 0 on rK, and
Tt(a'!) _ -V"/
f
dx1 = const.
on rK.
(4.28)
rK
Then evidently a §t KF p, unless V1" = 0 holds.
Example 4.3. Let ro = 0 and let rK be a segment parallel to the x1-axis. We can choose A such that
T. (X") = -T2 (A") = V2'/ IrK ds = g on K,
(4.29)
where
V2' = f F2 dx + f P2ds. r i" Since V2" < 0 is a necessary condition for the existence of solution (according to (4.16)1), we have Tn(A) < 0 on rK, and hence (4.22) holds as well.
Remark 4.3. It is not necessary for .1 to belong to the set KF p. Nonetheless, from the practical point of view it is suitable that A satisfy T; (A) =
2. One-Sided Contact of Elastic Bodies
172
const on rK. Namely, this is the case when we can construct internal approximations of the set KF P . This offers some advantages. An algorithm for the construction of A will be presented in section 2.4.13. In the following, let us consider the situation from example 4.1 and show
in detail the approximation of the set KFP. Let A satisfy condition
L We easily find that
AEKFP
A
rEUo,
with
uo= {rESI f r3e,,(v)dx>_-g J (vn+v;;)ds Vv EK}. l Jrx )))
Lemma 4.8.
1°
Let r E Uo be sufficiently smooth. Then r fulfills the
homogeneous equations (4.18), (4.19), (4.20), (4.21) and
Tn(r') = Tn(r") < -g on rK.
(4.31)
2° Let O be divided into a finite number of closed subdomains Kr,
Krf1Ka=0 for r#s r
(K denotes the interior of the set Kr). Let r E S fulfill the homogeneous 0 equations (4.18) in each subdomain Kr, the homogeneous boundary conditions (4.19), (4.20), (4.21), and the inequalities (4.31). Moreover, let the stress tensor T(r) be continuous when crossing an arbitrary common boundary of two adjacent subdomains, that is,
T(r)IK, +T(r)IK, = 0 on Kr n K8,
Vr 54 s.
Then, r E llo.
Proof. Analogous to that of lemma 4.6. Remark 4.4. If condition (4.27) is fulfilled, then Uo is nonempty, since it contains the zero element. Besides, UO is a convex and closed subset of S. 11
2.4. Dual Variational Formulation
173
Substituting a = a + r into definition (4.17) of the dual problem, we obtain an equivalent dual problem: find r° E Uo such that
J(r°) < J(r) dr E Uo,
(4.32)
where
J(r) = 2 J aijk,Tij (Tk1 +
2Akl)dx.
n
2.4.11. Equlibrium Model of Finite Elements. Since the admissible stress fields in problem (4.31) are required to satisfy the homogeneous equations of equilibrium, it is necessary to construct finite-dimensional subspaces of tensor fields with the same property. For this purpose, we can use the equilibrium model of finite elements, proposed by Watwood and Hartz (1968).
This model consists of triangular block elements, which are formed by joining the vertices of a general triangle with its center of gravity. In each subtriangle we define three linear functions-the components of the selfequilibriated stress field. The stress vector is continuous when crossing any boundary between two subtriangles. In a triangle K let us define the space of self-equilibriated linear stress fields 7-11 = 191 + 1992 x2 + 193x2,
722 = 194 + 195xp1 + P6 X2,
712 = 721 = 197 - 196x1 - 192x2 11
where /j E R7 is an arbitrary vector. Immediately we see that arij1axj = 0 in K for every r E M(K), i = 1, 2.
Further, let us consider a block element K = Us_1Ki according to Figure 22 and define
N(K) _ {r = (r', r2, r3) Jr' = rlKi E M(Ki), T(7-') +T(ri-1) = 0 on Oai, i = 1,2,3}. (The last condition expresses the continuity of the stress vectors on the sides 0ai.)
Theorem 4.1. Given an arbitrary exterior load T of the triangle K, such that (i) it is linear along each side of the triangle K, (ii) it satisfies the conditions of the total equilibrium, then there exists a unique stress field r E N(K) such that
T(r) = T on 8K. Proof. See
(1979), theorem 2.1.
2. One-Sided Contact of Elastic Bodies
174
a,
Figure 22
Let G be a bounded polygonal domain, h E (0, 1] a real parameter, and Th a triangulation of the domain G. Let us define
h = max diam K, KETh
Nh(G)={rES(G)IrIKEN(K) VKETh, T(r)]K+T(r)IK' =0 VKnK'}, E(G) = {r E [H'(G)]4 n S(G) l8ri;/8x; = 0 in G, i=1,2}. Theorem 4.2. There exists a linear continuous mapping rh: E(G) -+ Nh(G) such that for every r E E(G) n [H2(G)]4 the estimate h r - rhrII[L2(G)]4 < Ch2hI rhI [H2(G)14
(4.33)
holds, where C is independent of h and r, provided the system of triangulations {Th} is regular.
Proof. See Hlav£Lek (1979), theorem 2.5. Estimate (4.33) also follows from the results of Johnson and Mercier (1978).
Remark 4.5. The mapping rh from theorem 4.2 is defined locally on every block element K E Th in the following way: the stress vector Tk (r) on each side of S; C 8K is projected in L2 (S3) to the subspace Pl (Sj) of linear functions. These projections of T uniquely determine the stress field rhrIK E N(K), as follows from theorem 4.1.
2.4. Dual Variational Formulation
175
Remark 4.6. Every stress field r E Nh(G) satisfies the homogeneous equations of equilibrium 8r;, /8xj = 0, i = 1, 2, in the domain G in the sense of distributions.
2.4.12. Applications of the Equilibrium Model. Let us assume that both 11' and Cl" are bounded polygonal domains. Define the approximations
of the set uo by Uoh = ho n Nh(fl), where
Nh (D) = {(r ,
I rM
E Nh(f1M), M= 1,111We say that rh E uoh is an approximation of the solution of the dual problem, if
J(rh) < J(r) br E Itoh.
(4.34)
Lemma 4.9. If Vi" > 0 (see (4.27)), then there exists a unique solution of problem (4.34).
Proof. The set Nh(12) evidently is a linear finite-dimensional subset of S, hence it is closed and convex. Using remark 4.4, we deduce that Uh is also closed, convex, and nonempty. As the functional J is differentiable and strictly convex, this easily yields the existence and uniqueness of rh. An algorithm for finding rh will be presented in the next section. Here we will deal with an estimate of the error 11A - ah110,o = 1170 - rh110,n,
with X _ .1 + r°, )h = X + rh, 11 11°,o being the norm in [L2(0)]4. The main result is included in the following theorem.
Theorem 4.3. Let ro consist of line segments parallel to the xl-axis and let rK be a segment such that ni > 0 on rK. Let us assume that r°IOM E [H2(flM]4, M = ',", and Tn(r°) E H2(rK). Let the system of triangulations {Th} be (a, /I)-regular and satisfy the following conditions:
between 1'K and ro in the domain Ii" and between rK and ru in fl', the triangulation Th is inscribed into smooth "vaulted strips" with bounded curvature and a slope 181 < 8 < 2 (A independent of h), which are perpendicular to 1'o and 1'K (see Figure 23). Then the estimate 11x0 - rhIlo,n < C(r°)h3/2,
(4.35)
with C independent of h holds.
Proof. Based on the idea of unilateral approximations like the proof of theorem 6.6 in section 1.1.62. A detailed proof may be found in Haslinger and Hlava ek (1981 b).
2. One-Sided Contact of Elastic Bodies
176
Figure 23
Remark 4.7. Let us briefly consider the situation from example 3, that is,
let r0 = 0 and rK be a segment parallel to the xl-axis. Let us construct a particular stress field A (which satisfies condition (4.30)), define Uo (the equivalent dual problem (4.32)), Uoh, and the approximations of the solution of (4.34). If V2" < 0, then uo contains the zero element and there is a unique approximation rh. Then an analog of theorem 4.3 holds, by only replacing the condition concerning the vaulted strips by the following condition: The triangulation Th in fl" includes a fixed rectangle AUBC (see Figure 24) independent of h, with AU = I'K, which is divided into rectangular elements. The triangulation Ti, in fl' fulfills the same conditions as in theorem 4.3.
2.4.13. Algorithm for Approximations of the Dual Problem. Let us first present a survey of the results by Watwood and Hartz (1968), which can be immediately used in the algorithm for the solution of problem (4.34), that is, for finding the stress field rh E Uoh The behavior of the stress field can be expressed on each subtriangle K1 in the following way:
7=
Tl l 722 T12
2.4. Dual Variational Formulation
177
A
U
rK Figure 24
E
VA0 0
AEo
0
0
0
xl
0
x2
vfA-
0
x1
0
x2
0
v/A
0
0 0
-x2 -xl
S = MS,
(4.36)
where S E R7 is the vector of coefficients, E is the Young module, Eo a dimensionless quantity equal to the reference module, and A the area of the triangle K1. Let the origin of the Cartesian coordinates (x1i X2) coincide with the
center of gravity of K;. Let us consider a material, homogeneous and isotropic in K;, with the Poisson constant v. Then mjklT,njrkldx = ST fS, 1K;
where the matrix f has the form a A, ,OA,
aA
tE
0 0
0
''A,
0
0
a61i
symmetry
0
0 0 0
0 0 0
0 0 0
fi61,
a61 +''62, (Q +'')612, a62+'761,
fE
a612, 9512,
a612,
fKj
962,
t being the thickness of the element K1, Si = A-1
;
1,2,
a62,
2. One-Sided Contact of Elastic Bodies
178
xix2dxldx2, 1K; a, ,B, 7 constants, which in the case of a plane stress assume the values 812 = A-1
a=1, /j=-a,
ry=2(1+a),
while in the case of a plane strain,
8°-o(1+Q), 7=2(1+a).
a=1-Q2,
Similarly, we could find
l
Ki
am,klrm?.lkidx
\
(fr,
aT B1M dx I S = Bo S,
with AT = (.111, A22, )'12), M the matrix from formula (4.36), and B-1 the (3 x 3)-matrix of the inverse Hooke's Law (e = B-1r). Each component of the stress vector on the side a;a;+i of the triangle
K can be expressed in terms of the exterior parameters S* E R4 and a continuous parameter p E [-1, 1] as follows: T1(P) = Sl + S2 P,
T2 (P) = S3 + S4 P.
For instance, let us consider the side a2a3 (see Figure 22). Then, at the point a2 we have p = -1, and at a3 we have p = 1 and S
*
1
2lafA
(a)
e s, a
where la is the length of the side a2a3i while C is the following (4 x 7)matrix: 2V(A) (Y, - Y3), (a)
C=
0
0
2
(A)(X2-X3),
0
o 2
0
0 0
0
(A)(X2-X2),
-2
X22 - Xy,
-(X2- X2)2,
0
0
-2(X2Y2-X3Ys), -2(X2-X3)(Y2-Y3),
-(Xi-Xj),
-(YY -Y, ),
(X2-X3)2,
(Y2-Y3)2,
y,2 -Y3 ,
2 (X2 Y2 - X3 Y3),
o
-2(X2-X3)(Y2-Y3),
0
(Y2-Y3)2,
.
J
(b)
By a cyclic permutation of indices we can find the matrices and C. Let us denote the exterior parameters on side b by S5, Ss, S*, S8, and on side c by S9, S10, Sf1, and S12. Then, the total vector S* E R12 satisfies S* = CS,
(4.37)
2.4. Dual Variational Formulation
179
(a)
(b)
(c)
where the (12 x 7)-matrix C consists of the matrices C, C, C:
C
dal
C=
1
1
2/
The conditions of continuity of the stress vector on the common sides have the form
S; +S* =0,
(4.38)
where the indices i, j correspond to the same basis function, but to adjacent triangles. From the definition of uoh = Uo fl Nh (Il), it is easily seen that r C uoh if and only if all the constraints of the form (4.38) hold,
S; = 0 on each side aiai+l c r,
(4.39)
S; tl + Sj+2t2 = 0 on ro u rK (4.40) (where tk are the components of the tangent vector and (4.40) holds independently of TM E Nh(I1M), M = ',"), and
[Sa n1 + Si+2n2 - (Si+ini + Si+3nz)]n, < -g on rK, [S.1 n1 + Sj+2n2 + (Sj+1n1 + SJ+3n2)Io, < -g on rK,
(4.41)
conditions of the form (4.38) hold on the common sides of any two triangles belonging to rK.
Instead of working with all components of the vector S, it is recommended to reduce these parameters by eliminating the interior degrees of freedom in each triangular block. Let us write the conditions of continuity on the segments Oai (see Figure 22) in the form AuS = 0,
(4.42)
where A,, is a (12 x 21)-matrix and S a (21 x 1)-matrix. There exists a regular (21 x 21)-matrix Q such that
A,Q = [I
01,
(4.43)
where I is the unit matrix. Naturally, the matrix Q is not uniquely determined. It even suffices to replace I in (4.43) by an arbitrary regular (12 x 12)-matrix. Besides, in the following we need only the last nine columns of the matrix Q (which form the matrix Q1). The identity (4.43)
2. One-Sided Contact of Elastic Bodies
180
can be obtained, for example, by the Gauss elimination of the matrix Au. Let us carry out the transformation u l
S = QS = [Qo
Q11
l
S1
(4.44) J
cutting Q between its 12th and 13th column and S in a corresponding fashion. After substituting in (4.42), we obtain
S"=0, and the transformation (4.44) reduces to (4.45)
S = Q1S1,
with Q1 a (21 x 9)-matrix and S1 a (9 x 1)-matrix. The parameters Sk, k = 1, ... , 9, are the degrees of freedom of the triangular block element. In the following, we will write Zk = Sk , that is, S1 = Z. Now the functional J(r) of the equivalent dual problem can be written in terms of Z:
J(r) 3
/
KEr,, i.1 \
3
KErh i=1 fK,
amjklrmj(rkl +2Aki)dx
(STFS+BS)
\\
2STfS+boS`= //
\KErh
(ZTQTFQiZ+BoQ17i f = ZTAZ+bTZ=y(Z);
/
KErt,
where the vectors S and Z successively correspond to a subtriangle, to a triangular block, or to the whole triangulation. The (N x N)-matrix A is now positive definite.
Substituting (4.37) and (4.45) into the conditions of the form (4.38)(4.41), we obtain constraints
DZ = 0,
(4.46)
EZ < -g,
(4.47)
where D, E are matrices of the types (r1 x N), (r2 x N), respectively, and g is a vector whose all components are equal to g. Let us define a set
B = {Z E RN ] Z fulfills (4.46) and (4.47)).
2.4. Dual Variational Formulation
181
Thus, we arrive at the problem to find v E B such that 1,(Q) < 1,(Z)
dZ E B.
(4.48)
This problem can be solved, for example, by applying Uzawa's algorithm
(see Cea (1971), Chapter 4, section 5.1). Denotelr = ri + r2 and
B=[E
I
G=[90
J,
the matrices being of the types (r x N), (r x 1), respectively. Define the set of Lagrangian multipliers
A={yERrI yj>0 for j=r1+1,...,r}. We choose y° E A and calculate z° E RN from the system
Az° = -b - BTyo If we have y', z', then the values y"+1, z"+1 are determined from the conditions
yn+1 = PA [yn + p(Bz" + G)),
Az"+1 = -b - BT n+1 Y
where PA denotes the projection to the set A (that is, (PAt); = ti for j= (PA t)3 = max{0, tj } for j = ri + 1, ... , r), and p E R is a sufficiently small parameter.
It can be proven that z" - or in RN for n -' oo, where a is the solution of (4.48) (which is unique according to lemma 4.9), provided the matrix B has the full rank, that is, r.
In conclusion, let us suggest the construction of a particular stress field A. Let us again consider the situation in example 4.1 and choose
T"(A') = T"(.X") = g on I'K (see (4.25), (4.26)). In accordance with the interpretation of the set KF P and (4.30) (cf. lemmas 4.6 and 4.8), we can proceed as follows:
Choose P) E S, which fulfills conditions (4.18) in 0 = f2' U fl" (by direct integration with respect to xi or x2). Assume that F is a constant and P a piecewise linear vector field. Then, are linear polynomials and Ti (P)) is linear on each side of the polygonal boundary 812' U 8fl". We put a = a(1) + A(2) and seek for A(2) in the space Nh(f)), imposing the following boundary conditions: T(A(2)) = P
- T(a(i))
on r,,
(4.49)
2. One-Sided Contact of Elastic Bodies
182
Tt(),(2) ) _ -Tt(A 11)
on ro, Tt(),(2)) = Tt(A(2)") _ -Tt(A(1)) on rK, Tn(A(2)")
=g-
Tn(A(1))
on I'K.
(4.50) (4.51) (4.52)
Since the right-hand sides in (4.49)-(4.52) are piecewise linear functions, there exists A(2) E Nh(fl) which satisfies these conditions. We can construct
it using the procedure introduced above. We use the parameters Z and formulae (4.36), (4.45) and (4.37), and write conditions of continuity of form
(4.38), as well as the boundary conditions (4.49)-(4.52) in terms of Z via (4.37) and (4.45). The undetermined reactions T,, (A(2)) on I'o and T1()(2))
on ru can be chosen in such a way that the resulting system of linear equations is solvable. The choice of these reactions is in accordance with the conditions of total equilibrium of the bodies fY and fl", respectively.
2.5
Contact Problems with Friction
In the preceding part of the book we studied the contact problem of two elastic bodies without friction, when the tangent component of the stress vector on the contact zone is Tt = 0. It is clear that the assumption of zero friction between 0', 0" does not fully comport with the real situation and consequently, it is desirable to include the influence of friction in our considerations. For the sake of simplicity, we will study the contact between an elastic body it and a perfectly rigid foundation. Extension to the contact of two elastic bodies is possible (see Jarugek (1982)). The finite nonzero friction will be expressed heuristically, by means of Coulomb's Law: on the
contact surface rK of the elastic body with the perfectly rigid foundation we assume
un < 0,
Tn < 0,
unTn = 0
Tt=T - nTn, ITtJ 0. In this case, (5.2) is replaced by ITtf 0, P E L2(rp). Let a closed convex set of admissible displacements K be given by
K={vEH'(fl)1 v=u° on r,,,
onrK}.
(5.4)
A function u E K is called a weak solution of the Signorini problem with friction for given g,,, if Vv E K : a(u, v - u) + f Fgn(Ivtl - jutJ)ds
rx
fFi(v1-u1)dx+jPi(v---u1)ds(5.5) r where
Ut = U - nun, a(u, v) = / Ci jkmfij(U)Ekm(v)dx, n
Cijkm are bounded measurable functions in fl that fulfill the conditions of symmetry (1.3) and of positive definiteness (1.4) in the domain fl. Similar to section 2.1.3, we will prove formal equivalence of the classical and weak formulations of our problem. Let us show in more detail how the friction conditions (5.3) will be derived. To do so, we will assume that we have already proved the validity of the equilibrium conditions
arij+Fi=0 axj
in 0,
i=1,2,
the boundary conditions
rjnj=Pi onrp, i=1,2, and the unilateral boundary conditions (5.1). Integrating by parts in (5.5) and employing all the above conditions, we obtain J L.
Tt(vt - ut)ds + f 3gn(1vtj- jutj)ds > 0 Vv E K.
(5.6)
K
Let vEKhave the form v=u±'i,where i/in=0on rK,10=0on ru. Then
fTt(±tbt)ds+jIgn(lut±btI_IUtl)dsO
V O, 0,, = 0 on FK,
2. One-Sided Contact of Elastic Bodies
184 and hence
±
IrK
Tods < fr 3gl1'tlds, K
that is,
frK
Tt itds
=
Irx
Hence, the first inequality in (5.3) easily follows. Since ut < lutl on rK, we obtain from the results just proven that
Ttut + Now let ,b be such that into (5.6) we obtain
0
on rK.
(5.7)
0, of = -ut on rK. Substituting v = u + tP
- J Ttutds - fr 3 g,,1ut1ds > 0. rK
K
Hence and from (5.7), we conclude
Tt ut + Fg l ut i = 0 on rK, which is an equivalent expression of the remaining conditions in (5.3). In the sequel, we will assume that the relation between the stress tensor and the strain tensor is described by Hooke's Law for homogeneous isotropic bodies. In that case, we have T;j = )t&,ekk + 2pe,j,
where a, p > 0 are Lame's constants. Furthermore, we will assume that rp = 0, rK is a sufficiently differentiable part of aft, and similarly, that T'2: 0 is a sufficiently smooth function with a compact support in rK. By H11'(80) let us denote the space of traces of functions from H' (t1) (the meaning of this notation will be seen from the following, see also NeLas (1967), and Fueik, John, and Kufner (1977)). Further, by H- 112(an) let us denote the space of functionals over H1/2(8f1). Let H1/2(rK) C H1/2(afl)
be the space of such v's that vanish on r,,, and denote by H-1/2(rK) the dual to H112(rK) We will say that gn E H-1/2(rK) is < 0 if the duality fulfills (g,,, v) < 0 for all v > 0, v E H1/2(rK). Defining (Fg,,, v) ` (g,,, Fv) for v E H112(8f1), we see that definition (5.5) can be extended to g E H-1/2(rk), gn < 0, in the form
uEK, VvEK, a(u,v-u)-(Fgn,1vtI - lutI)>J F;(v,-ui)dx. (5.8) n
2.5. Contact Problems with Friction
185
Let us recall that w E H1(11) implies JwJ E H' (fl), and 11 IwI 111,0
(Iwlll,a.
(5.9)
From definition (5.8) it is seen that the function u satisfies in fl (in the sense of distributions) the system of Lame's equations (A + i)Dui + M
8 a xi
(div u) = -Fi;
(5.10)
therefore, it is reasonable to define T,,(u) for a solution u of problem (5.8) (on the basis of Green's theorem) by (T. (u), vn)
4 a(u, v)
- J n fiv;dx,
(5.11)
for v = 0 on r,,, vt = 0 on rK. Hence, Tn(u E H-1/2(I'K)). Thus we define: u E K is a solution of the Signorini problem with friction, if (5.8) holds and gn = T,, (u). Consequently, if we define a mapping t : then our task is to find a fixed point of this mapping. Since compactness of the mapping cannot be expected and the authors have not succeeded in finding any kind of monotonicity for it, none of the classical fixed point theorems or methods from the theory of monotone operators can be applied. This is why the theory to be explained is a little more complicated. In this book we will show that the mapping just mentioned is a weakly continuous mapping from L2(rK) into itself. Then the existence of a fixed point follows from the so-called "weak Schauder theorem," which, for the reader's convenience, we will prove for a special case of the Hilbert space, namely, the separable one. Naturally, it is necessary to find a closed convex set which the mapping maps into itself. This will be achieved by the smallness of the friction coefficient. We will also introduce (without proof) estimates of the friction coefficient which comport with our theory. They appear to suit practical requirements. We will present all the main ideas, as well as methods for the proofs. For the sake of brevity we will not repeat analogous proofs, leaving them to the kind reader. For detailed proofs we refer him or her to the paper by Jarugek, and Haslinger (1980); however, spaces H-112+a(rK), 0 < a < 1/4 are considered there instead of L2(rK) (for the definition, see below).
Schauder Theorem (Weak Version). Let H be a separable Hilbert space.
Let A be a mapping from K c H into K, where K is a closed,
bounded, convex set. Let A be a weakly continuous operator, that is, let u, u (weak convergence) imply Aun Au. Then there is a fixed point
of the operator A in K, that is, u E K such that Au = u.
2. One-Sided Contact of Elastic Bodies
186
Proof. Let yi E K, i = 1, 2, ... , be such points that co{yl, y2, ...} = K, be an orthonormal basis where co stands for the convex hull. Let in H and Pn the projector to the subspace Hn of linear combinations of XI, x2i ... , xn. Let Ek = 1/k. There exist points Yi, ya,... , ym(k) such that for x E K, max
aminm(k) II PkA(x) - Pky;II
ai(x).
Sk(x)
i=1
i=1
Finally, let Kk = co{yi, y2, ... , ym(k) }. Evidently, Sk (Kk) C Kk. Since the operator A is weakly continuous, the operator Sk is continuous from Kk into Kk, hence by Brouwer's theorem? there is a fixed point zk E Kk, Skzk = zk. Now we choose a subsequence zk1 z E K. However, for
xEKwe have m(k)
m(k)
IIPkA(x) - PkSS(x)II =
ai(x)(PkA(x) - Pkyi)/
ai(x) t, I(w, A(zkt) - Skt(zkt))I = I (w, Pk,A(zkt) - Pk,Skt(zkt))1 k > 0, the dual space to Hk (Rl ); if f E H-k(R'), we will write the duality in the form (f, w). Let us consider the strip P and let rK = {x E R2; z2 = 0}, r,, = {z E
R2
11.
Further, let u° E Hl (P), let u° = 0 on rK, F E L2(P), g E H-1/2(Rl),
3 E D (Rl ), I > 0. We will write g,,, < 0 if (g,,, w) < 0 for w > 0, w E H1/2(Rl). Thus, consider gn < 0 and let K = {v E Hl(P);v < 0 on rK, v = u° on r,,}. A function u E Hl(P) solves the Signorini contact problem in the strip P with a prescribed normal force, if
uEK, VvEK, f Fi(vi - ui)dx 4 (F, v - u)o,p. (5.30)
a(u, v - u) - (Ign, Ivll - Full)
P
Theorem 5.1. There exists a unique solution of problem (5.30), and the estimate c[Iluolll,p
IIulli,P
cllvIIi p
(5.33)
implies the coerciveness of the functional on K: lim
jjvjj-+oo,vEK
T (v) = oo.
(5.34)
However, the functional T (v) is convex, and thus also weakly lower- semi-
continuous (for every c E Rii, the set of v E K with 7(v) < c is convex, closed, and hence weakly closed, which yields the implication vn - v limn_,. T (vn) > T (v) ). This makes the application of the fundamental theorem 1.5 from Chapter 1 possible. The uniqueness of solution: let wl, w2 be two solutions. Then a(w2, wl - w2) > (F, wl - w2)o p + (39n, Iwi I - Iwi l),
a(wl, w2
- w') ? (F, w2
- wl)o p + (3g,,, Iwi l - Iwi l),
(5.35)
2. One-Sided Contact of Elastic Bodies
190
wl, w2 - wl) < 0, consequently w1 = w2 by (5.33). Now set v = u° in (5.30). We obtain
which implies a(w2
a(u - uo, u - uo) : (Fgn, lull) - (F, u° - u)o,p - a(uo, u - uo)
< -(F, u° - u)o,p - a(u°, u - u°),
(5.36)
which together with Korn's inequality yields Ilu - uolll,p 0, and let At = ti, t > 0. Then the solution u of problem (5.30) with friction coefficient Ft satisfies II'II1,1/2,P < C1tIIgnIIp,R1 +
BThis assumption can be weakened.
e2(Iiu°II1,1/2,P + IIFII1,P).
(5.59)
2.5. Contact Problems with Friction
195
Proof. Set v = u_h - u° h + u° in (5.30). This yields
a(u, U-h - Uh + u° - u) - t(3gn, lull-h - lull) > (F, u-h - u_ h + u° - u)O,p.
(5.60)
By a translation of coordinates we obtain from (5.30)
a(u-h, v-h - u-h) - t(.3 h(gn)-h, IUiI-h - lull-h) (F_h, v_h - u_h)O,P.
(5.61)
Substituting here v = uh - uh + 0°, we obtain from (5.60), (5.61)
a(u_h - U, u-h - u) < a(u-h - U, u- h - u°) + t(,3 h(gn)-h
3gn, lull-h - lull) - (F - F_h, u_h - ua h - (u - u°))°,P.
(5.62)
First of all, notice that
(3-h(gn)-h - 3gn, lull-h - lull) = ((gn)-h - 9nr 3(lu1l-h - lull)) + ((g.)-h, 3h - fl(Iu11-h - lull)).
(5.63)
Now w E Hl /2(R') satisfies IiTwli1/2,R'
(5.64)
< ciiwlll/2,R=,
II(3 h - 3)wlil/2,R' < clhlllwlll/2
(5.65)
while for w E H'(P) we have I (F - F-h, u-h - U + u° - U'-h) 1 (3, y -
x)Rm
by E KE.
(5.84)
Let us write y = (y1, Y2)', jr = (fl, f2)', yl, fl E R", y2, f2 E Rk. Analogously, we divide the matrix C into blocks
C-_ (
C11 C'21
C12 C22
where C11, C22 are square matrices of orders n, k, respectively, while C12, C21 are rectangular of types n x k and k x n, respectively. Evidently C'12 = C211'
Now let us choose y in (5.84) so that y1 = xl + z1, y2 = x2i zl E R" arbitrary. Then y E KE. After multiplication we obtain zi (Cllxi + C12x2) = zi f1 dz1 E R", or
C11x1 + C12x2 = fl.
(5.85)
Now let us choose y E KE in (5.84) so that yl = xl, y2 = z2i Bz2 < 0 arbitrary. After multiplication we obtain (z2 - x2)T (C21x1 + C22x2)
(Z2 - x2) f2.
(5.86)
2. One-Sided Contact of Elastic Bodies
204
From (5.85) we can express xi in the form
xi = C1'(fl - C12x2), which after substituting into (5.86) yields the following relation for x2: (Z2 - x2)T Cx2 > (z2 - x2)T f,
(5.87)
where f = f2 - C21Cll1 f1, C = C22 - C'.21C111C12. Now (5.86) implies that x2 is the minimum of the quadratic function i,/(x2) = (z2,Cx2)R` (12, x2)Rk on a convex closed set KE, where
-
KE={x2ERkIBx2 0 and Tn(u) E L2(rK). Moreover, let there exist a constant Q > 0 independent of h, H and such that sup ((µ1H, Vhn)o,r,c + (1A2H, yht)o,r,r ) > $IIµH II-1i2,rx IIvhIII
IIuhII 1
J
holds for arbitrary µH = (PH1,pH2) E LH x LH. Then
IIu - uhII H'(())2 = o(H9), H -- 0+,
(S)
2. One-Sided Contact of Elastic Bodies
206
11A - AHII-1/2,rK = o(H4),
H - 0+,
where q" = min(q; 1/4).
Naturally, the crucial problem is that of when condition (S) is fulfilled. This problem is also solved in the above mentioned paper.
Uzawa's method can again be used for the realization of the mixed variational formulation. While the partial dualization, consisting merely of removing the nondifferentiable term 3(v), meant transforming (5.78) into a quadratic programming problem, now the total dualization yields Kh = Vh and (5.78) is equivalent to the problem of solving a system of linear algebraic equations, in which some components of the right-hand side are
being corrected. The reader can easily see that a partial elimination of free components of the displacement vector uh again leads to forming an effective algorithm of solution.
Remark 5.6. Let us show another useful variational formulation of problem (5.70), which involves only the quantities defined on FK. Consider a triplet of functions (F, µl, gµ2), where F E ]L2(f0)]2, µl E A1, P2 E A2 with A1i A2 defined in the preceding remark. This triplet defines a generalized force F E V. The corresponding field of displacements w E V is given by the relation (5.88) w = G(F) = 6(p1,p2), where G : V -p V is the Green operator of our problem. The reciprocal variational formulation of the Signorini problem with given friction is the problem of finding (Al, A2) E Al X A2 such that
SA2) (Al, 5 S(µ1,µ2)
d(µ1,µ2) E Al X A2,
(5.89)
where
S (µl,µ2) =
1
1
2
(µ1, G' (µ1, µ2) . n) +
2
(µ29, G(µ1, µ2) - t).
It can be shown that between (5.70) and (5.89) the following relation holds:
Al = Tn(u),
gA2 = Tt(u).
(u solves (5.70).) In this form the reciprocal variational formulation is hardly realizable, since the explicit form of G is known only for several particular cases. For this reason it is necessary to use an approximation of G. For instance, the inverse matrix to the stiffness matrix of our problem may serve as such an approximation. The application of this formulation
2.5. Contact Problems with Friction
207
to the solution of contact problems without friction is thoroughly dealt with in the work of Oden and Kukuchi (1979). It is of interest to note here that the Signorini problem with friction governed by Coulomb's Law, if expressed in terms of the recriprocal variational formulation, leads to the solution of the so-called quasivariational inequality, with the convex set depending on the solution itself. Indeed, let K;g be the convex, closed subset of IH-1/2(rK)]2 defined by K3g = {14 = (Al, 142), Al < 0,
]u21 < F9 a (A2, vt)
+ (Fg,lvtl) >0 VvEV}, gEH+1/2(rK). Let (A1(9), A2 (9)) E K,3g be the minimum of the quadratic functional S (Al, ,42) =
2
(Al, G(141, 02) . n) +
2
(µ2, G(141, µ2)
t)
on K3,. Again we can show that )1(g) =
-X2(g) = Tt(u), where u solves the Signorini problem with given friction 3g E H+112(rK). Define a mapping : H+112(rK) -. H+1/2(rK) by
(3g) _ -3x1(9). If g = -a1(g), then the solution of the Signorini problem with given friction equals to -3.X1 is the solution of the Signorini problem with friction governed by Coulomb's Law (5.2). The detailed analysis of the approximation based on the reciprocal variational formulation of the Signorini problem with given friction is given in the paper by Haslinger and Panagiotopoulos (1982).
2.5.42. Alternating Iterations. The contact problem for two elastic bodies with friction can be solved in yet another way, which was suggested in a paper by Panagiotopoulos (1975) and recommended in a research report by Frederikson, Rydholm, and Sjobolm (1977). Each step of the algorithm consists of two partial problems. 1st step:
1.1. Unilateral contact with a given shear force. Tt = F(o) (Choose, on rK. 1.2. Friction with a given normal force T = F(,1) on rK. We compute Tt = Ft(1) on rK. In the i-th step (i > 2) we solve the following partial problems: i.l. Unilateral contact with a given shear force Tt = Ft('-'). We compute T,i = F('). e.g., F(°) = 0.) We compute T,, =
2. One-Sided Contact of Elastic Bodies
208
i.2. Friction with a given normal force TI, = F('). We compute Tt =
F(') As stopping test of the iteration process we may choose, for instance, a sufficiently small change of the normal contact forces IIF(')
- Fn'-')II < c,.IIF(')II
and the friction forces
Ilk') - F('-1)II < etIIF(')II Thus, the algorithm alternates partial problems of two different types. In the following sections we will study both of them in more detail, considering the case of bounded contact zone.
2.5.421. Unilateral Contact with a Given Shear Force. As before we assume that
an' n an" = rK, an, = r u r, u rK, ant, = rp u r7' u rK, and the following inequalities hold on rK:
un + un < 0,
(5.90)
Tn=T;' 0
(5.114)
be fulfilled, which guarantees the existence and uniqueness of solution of the primal problem. First, we establish the condition of total equilibrium of forces and reactions on the body tZ":
- Jn
Fldx =
rrK
1
P1ds+J
Ti (a)ds =
1rr
Plds+ J (TT (a)ni+Ftti)ds. rX
This implies
fK
Tn (A)dx2 = -Vi t-
(5.115)
Thus, let T, (a)
T, ('X)
_ `Vi
(fdX2
X
J
-1 = go.
(5.116)
The c onstant go is negative due to condition (5.114). Thus, the tensor field a fulfills the condition (A,e(v)) = L(v) + Lt(v) + frK go(v, + vn)ds
dv E V.
(5.117)
Remark 5.10. If, for example, F1 = F° = const, F2 = 0, and P and Ft are piecewise linear, then A can be constructed as follows: A = AF + AO, where
F X11 = -Fi0x1,
F F ,12 = X22 = 0 in Cl,
2. One-Sided Contact of Elastic Bodies
212
and A' E Nh (fl) fulfills the modified boundary conditions
T(A0) = P - T(AF) on r,, TL(A°) _ .1°2 = 0 on ro,
Tt(A°) = Tt'(A°) = Ft -Tt(AF) on rK. By (5.117) we easily derive that
AEKtA-rEUo, where
U. = { r E S
>-go
ll
rx
(v',s+vn)ds VvEK}. 1
We proceed further in the same way as in section 2.4.1. Analogs to lemmas 4.8, 4.9 are valid, too, if we write go instead of g and V, instead of Vi". If we use the results of section 2.4.11, we also establish the error estimates from section 2.4.12 and the algorithm from section 2.4.13.
2.5.422. Realizability of the Algorithm of Alternating Iterations. Now we will consider a model of a simpler Signorini problem in order to study some questions connected with the possibility of realization of the algorithm of alternating iterations in the semicoercive case, that is, the situation when there exist nontrivial admissible displacements of the rigid body. In order to grasp the core of the problem, let us consider only one elastic
body of a trapezoidal shape, loaded by uniformly distributed horizontal forces and resting on a perfectly rigid foundation rK, where the friction occurs (see Figure 25). So we have the case that fl" _- fl, W is a perfectly rigid body, that is, we have the so-called Signorini problem with friction. Let us first collect the achieved results on existence and then complete the analysis of uniqueness of solution in the case of the partial problem of friction with given normal force.
A. Unilateral contact with given shear force. We assume that nl > 0, n2 < 0 holds on rK; further,
an = r, u ro u rK and a shear force Ft E L2(rK) is given. We denote the potential energy (cf. (5.94), (5.95)) by
('a(v) =
2
A(v, v) - L(v) - Lt (v),
2.5. Contact Problems with Friction
213
Figure 25 where L(v) _ f P1vlds
P1 = const > 0,
r
Lt(v) = f Ftvtds. x
In our case we have
V = {v E [H'(12)I2 v - v2 = 0 on I'o},
onrK}, and the primal problem reads:
find u E K such that ,C. (u) < Z. (v)
Vv E K.
(5.118)
Theorem 5.13. A solution u of problem (5.118) exists only if
L(y)+Lt(y) 0.
rX
Indeed, in our case we have
KnR={(yi,y2)Iyi=a0 a.e.
2.5. Contact Problems with Friction
215
and a normal force Fn E L°°(I'K) be given. On rK we have Tn = Fn, IT,,I < g,
where g = 3IF,aI,
ITtI. ut=0, ITtI = g =Ttut < 0. We denote the potential energy by
Lb(v) = 1A(v, v) - L(v) - L,,, (v) + j(v), with L
(v) =rKf Fvds,
1(v) =
gvt ds. IrK
The primal problem reads: find
u E V such that
Lb(u) 0, hence,
u1-u2=yERf1 V, Tt(u1) = Tt(u2) = T. The condition of total equilibrium yields
L1(y) + fr T tytds = 0.
(5.122)
Tt = g a. e. on rK.
(5.123)
x
Let us assume
Then we obtain by (5.122)
-L1(y) = fr gytds = 1C
fric
gl yt Ids = ±j(y),
and condition (5.121) yields y = 0. Similarly, the assumption
Tt = -g a.e. on rK
(5.124)
2.5. Contact Problems with Fiction
217
leads to the conclusion y = 0. Now let us assume
-g < Tt < g on a set E C rK, Denoting u2
meas E > 0.
(5.125)
u, ul = u + y, we have
ut = 0,
ut + yt = 0 on E,
at, = 0 on E, which yields y = 0. The last case to be dealt with is that with
hence, yt
Tt = g on ri;
meas rl > 0,
Tt = -g on r2; meas r2 > 0,
(5.126)
where rl u r2 = rK (up to a set of zero measure). Then evidently
ut < 0,
ut + yt < 0 on rl,
ut>0, ut+yt>0
on r2.
Let us assume that yt = at, > 0 (the case yt < 0 is solved analogously). Hence,
ut < -yt < 0 on r1, ut > 0 on r2.
(5.127)
However, we have ut E H1/2(rK), which contradicts (5.127). Hence, again
y=0.
C. Realizability of the algorithm. Let us denote problem (5.18) for
Ft = F('-1), ti = 1, 2, ... , as problem (ia T), and problem (5.120) for Fn = F(,'}, i = 1,2,..., as problem (ib T). Recall that the algorithm defined at the beginning of section 2.5.22 consists of the successive solution of the problems
(1a T), (lb T), where F(°) is chosen, and F(') =
(2a T), (2b T),..., where u is the solution of the
problem (ia T), while Fdl) = Tt (u), where u is the solution of the problem (jb T).
Theorem 5.19. Let nl > 0, n2 < 0. Let 3 = const > 0. fo =
rK
F(°)tlds < 0,
(5.128)
2. One-Sided Contact of Elastic Bodies
218
P1 ds + fo { 1+
rK
ti, ) > 0.
(5.129)
Then the conditions (5.119), (5.121), which guarantee the existence and uniqueness of solution, are fulfilled for all approximation problems (ia T), (ib T), respectively, i = 1, 2, ... , and the identity
frx
Ft')tlds = fo
holds for all i.
Proof. Problem (ia T) involves the total equilibrium condition
fr P1ds +
+ F,')nl)ds
0,
JK
while problem (ib T) involves condition
J Pl ds + J (Ft')nl + F')ti)ds = 0. r
K
(Eb)
This immediately implies
f: =
IrK
Ftids = j
F1tlds = ... = fo
rK
Hence, for problem (ia T) we have
Vl =
P1ds + fi-1 = J Pids + fo. rr
r
However, by (5.128) and ti = -n2 > 0, we evidently have fo
11+!-I) irt,
0, that is, condition (5.119) holds. Consider now problem (ib T) and verify condition (5.121). On the one hand, we obtain from (Eb),
JrrPlds + Jr
F'nlds = - fo;
(5.130)
K
on the other hand we can write after substituting from (5.130): fr K
gtlds = f T F(') Itjds = -3t1 K
FN`)ds
2.5. Contact Problems with Friction
It,
fo + J
219 Plds)
.
(5.131)
nl
From assumption (5.129) we find
fo+
r.
nl Pids> -fo It, I -
hence,
nll (fo + 1 Plds) > -fo = Vol, which is condition (5.121) with respect to equations (5.131), (5.130).
Remark 5.11. Theorem 5.19 expresses the fact that the mean value of the shear force coincides for all the iterations. Thus, the iteration solution depend on the initial choice of Ft(o) . Hence, it also follows that there is no single limit of all the iteration solutions common for all initial choices of Ft(o)
Chapter 3
Problems of the Theory of Plasticity In this chapter we will deal with variational inequalities of evolution that result from some problems of plasticity. Let us note that we will consider processes that depend on the history of loading, that is, irreversible ones. Elasto-plasticity, which has come into fashion lately, is evidently a mere special case of nonlinear elasticity with generally nonlinear Hooke's Law
ai.i c9eii -8AIn addition to nonlinearity of the inequalities (which is of geometrical character as regards contact problems), we have here a physical nonlinearity (0.1), which naturally requires further linearization if approximate methods
are to be used. For details we refer the reader to the books by Washizu (1968), Duvaut and Lions (1972), NeZas and HlaviZek (1981), and Ka6anov (1974). This theory is justly criticized for its insensitiveness to the history of loading as well as for the inadequacy of the nonlinear relation (0.1) between the Cauchy stress tensor and the tensor of small strains. In this chapter we will consider the so-called flow theory of plasticity.
It is rate independent, which limits its validity to shorter time periods, so that it cannot describe such phenomena as creep or fading memory. It is interesting that the method of penalization considered (which will be used to prove the existence of solution) by itself has some features of an approximate method. It approximates the given problem by problems for an elasto-inelastic material with internal state variables (for details, see the book by Naas and Hlav£&k (1981), which includes further references
222
3. Problems of the Theory of Plasticity
concerning this approach). These models are of independent physical significance, and they are sensitive to creep and fading memory. When studying flow theory we will follow the above quoted book by NeLas and Hlava&k, naturally introducing the problem in a somewhat more general form (which was dealt with as a mathematical model in the fundamental works by Quoc Son Nguyen (1973), Halphen and Quoc Son Nguyen (1975), and theoretically by the above mentioned method in the paper by Ne?!as and Travni6ek (1978), which was preceded by the papers by Ne6as and Travnicek (1980), Tr'evni6ek (1976) ). The same problems were dealt with by a method based on evolution equations with maximally monotone operators in a number of papers by Groger, (see Groger (1979), (1978a), (1978b), (1977), (1980)). Groger and Nefas (1979), and Groger, Necas, Tr'avni6ek (1979). A partial theoretical solution is also given in the book by Duvaut and Lions (1972), and in C. Johnson (1976a). Groger's research is continued by Hiihnlich (1979).
To introduce the reader to the problem, let us first consider some simple yet typical examples. First of all, let us realize that we are going to follow the loading process together with the corresponding course of solution, which, in other words, means that the quantities observed will be functions of both time and space. Since a physical nonlinearity is involved (passing to the tensor of finite strains is still an unmanaged affair), we will sketch its character in the one-dimensional case of the relation between the stress
and strain (that is, by the graph of the stress-strain relation at a fixed point of the one-dimensional bar in question, e = e(t), a = Q(t) and the graph (e(t), v(t)) belongs to the plotted line). Moreover, we assume that the deformation e is the sum of the elastic and plastic deformations, that is, e = e + p, and that the relation between e, v is linear; for the sake of simplicity, let a = e. The tensor c is assumeed to satisfy E = dy , the
compatibility condition, where u is the displacement vector, while the stress tensor is assumed to satisfy dz = 0, the condition of equilibrium.
Thus, the linear elasticity is represented by the reversible process in Figure 26 and simply means p = 0. The elasto-pe rfec tly-plas tic case is shown in Figure 27. The solution in the latter case thus fulfills the condition Io(t)a < vo. Let us now assume that by increasing (or decreasing) the stress we would obtain the graph in Figure 28 and, moreover, when reaching the point ±Qo and then decreasing (increasing) the stress again, we should stay on the line o = E. In other words, this means that in the domain to _< ao no increment of the plastic deformation occurs. Let us assume that the loading reaches the stress o , and at this point let the stress fall. If we now go back along
the line'AB, then there is no change of the plastic deformation, and the
3. Problems of the Theory of Plasticity
223
Figure 26
Q*
C
Figure 27
224
3. Problems of the Theory of Plasticity
Figure 28
3. Problems of the Theory of Plasticity
Figure 29
225
3. Problems of the Theory of Plasticity
226
domain of elasticity has grown to the set Jul < Ql. Thus, we obtain the so-called isotropic strain-hardening. Let us again consider the growth process of the stress along the curve
in Figure 29. Let again no growth of the plastic deformation occur in the domain Jul < ao. Let us again reach the point A and go back along the line AB. This means that now no change of the plastic deformation
occurs in the domain Iv - (al - oo)l < ao. Consequently, the center of the convex set Jul _< Qo has shifted to the point vl - co. Thus, we obtain the so-called kinematic hardening. Naturally, the reader can imagine a combination of both hardenings; then we speak about an elasto-plastic material with hardening. The reader certainly correctly understands (we have actually mentioned the fact at the beginning) that the theoretical treatment of these problems has been successfully completed only quite recently, though the problems were formulated as early as in the 1950s (for example by Koiter (1960) and Hodge (1959)). Besides, the elasto-perfectly plastic case is much idealized and the results obtained are less complete than in the case with hardening. This is connected with the simple fact that the graph in Figure 28 does not determine e from the values of o. As concerns the case with hardening, we will solve it in a little more general setting, following the paper by News and (1978). We will also formulate in detail the special cases of both the isotropic and kinematic hardening by introducing the yield surface.
3.1
Prandtl-Reuss Model of Plastic Flow
We will first give the classical formulation of the problem without insisting on precision. Thus, in addition to the domain considered fl, let a time interval [0, T] be given; it will be seen from the formulation that we can
introduce another parameter t' E 10, rol such that t' = t'(t) and dt'/dt > 0 in 10, T]. Let F(t) be the vector function of the body forces and g(t) the vector function of the given stress vector on the boundary. We will deal with the traction boundary value problem; the reader will easily formulate and solve the other boundary value problems. Naturally, for every time moment t E (0, T] we assume the conditions of total equilibrium:
j F(t)dx + f(x x F(t))dx +
g(t)dS = 0,
(1.1)
f(x x g(t))dS = 0.
(1.2)
J o
3.1. Prandtl-Reuss Model of Plastic Flow
227
The stress tensor r = r(t) fulfills the condition of equilibrium
arij(t) +F;(t) = 0 Vt e [O, T] , i = 1 , 2 , 3. axj
(1.3)
As concerns the strain tensor e, we assume that it can be written as the sum of two symmetric tensors, e = e +P,
(1.4)
where e is the elastic and p the plastic part. The compatibility of deformations is expressed by the fact that there is a displacement vector u such that
_ E'
1
aui
auj
2
ax -
+ ax-i
(1.5)
Further, let the function of plasticity f (a) be given, which is assumed to fulfill f (0) = 0, f (a) > 0 for a # 0, and to be convex in a. Of course, we assume that f (a) is invariant when replacing aij by aji. Let ao > 0 and assume that the solution fulfills f (a) < ao For a and e, let us assume the linear Hooke's Law eij = Aijklakl,
(1.7)
with Ai jkl being functions only of x E il. Now let us assume that the increment of p can be nonvanishing only for a with f (a) = ao, and that the condition of normality holds:
pija ;
with A>0.
(1.8)
The solution a is supposed to fulfill
aijvj = gi on all.
(1.9)
Let us assume that a is the solution of our problem, and that r is another tensor which satisfies (1.3), (1.6), and (1.9). Then we can formally prove
that Jn
Aijklbkl(rij - aij)dx > 0.
(1.10)
3. Problems of the Theory of Plasticity
228
Indeed, the compatibility condition implies 0
f ijj(rsj - a;j)dx,
(1.11)
n
but then (1.8) yields
p;j(r j - a;j) > 0,
(1.12)
which together with (1.11) and (1.7) gives (1.10).
For simplicity we will assume in this section that the loading process started with zero values F(0) = 0 and g(0) = 0. Evidently, this leads to the initial condition c(O) = 0. In order to be able to define the weak solution, let us first introduce some auxiliary spaces. If as usual S E [L 2(0)19, or c S = a;j = aji, let Co ([0, TI, S) be the space of continuously differentiable functions vanishing for t = 0 and with values in S. Similarly we define Co([0,T], [L2(tl)]3), Co ([0, T], [L2 (an)]3).
In Ca ([0, T], S) let us introduce the inner product (1.13)
f T (T(t), Q(t))dt
(with (r, a) = fo r;ja;jdx), and hence the norm T
lIrII =
J
1/2
(i (t), T(t))dt
,
(1.14)
and let us find the completion of Co in this norm. Thus, we obtain the space Ho ([0, T], S). In the same way, we introduce Ho QO, TI, IL 2( 11)13) and Ho ([0, T], [L2(an)]3). In the same way as in the space of numerical functions, we immediately conclude 11r(tl) - r(t2)II s :5 Itl - t211/ZI1rI1Ho
(1.15)
which implies that the functions r from Ho are continuous in 10, T] (with values in S); similarly for [L2(1)]3, [L2(a[1)]3.
Definition 1.1. Let F E Co ([O, T], [L2(f2)]3), g E Co ([0, T], [L2(afl)]3) and let conditions (1.1), (1.2) be fulfilled for all t E [0, T]. We say that r E S fulfills (1.3) and (1.9) if r T{je{j(v)dx = n
Ja n
/
g;v;dS+ J F;v1dx n
(1.16)
3.1. Prandtl-Reuss Model of Plastic Flow
229
for all v E [W1"2(12)]3. A function v E Ho([0,T],S) is a weak solution of an elasto-inelastic body with a perfectly plastic domain, if (1.16) is fulfilled for all t E [0, T], if
f (v(t)) < ao almost everywhere in ii
(1.17)
for all t E [0, TI, and if for every r E Ho ([0, T], S) satisfying both (1.16) and (1.17) and for every t E [0, T] the inequality t
f dtI Aijklakt(rij -Qi7)dx> 0 n
(1.17')
holds. In the following, we will assume Aijkl E L°O(1l),
Aijkl = Ajikl = Aklij,
(1.18)
ynij = t1ji,
(1.19)
Aijkirlij?7kl ? c[rl[2, c > 0, f E C2(R9) and
of as
(here
82f
CC 0,
fo
[o1 (t) - o2(t), a1(t) - c2(t)]
3. Problems of the Theory of Plasticity
230
Theorem 1.2. Let the assumptions from definition 1.1 and (1.18)-(1.20) be fulfilled. Let there exist a° E Co ([0, T], S) satisfying (1.16) and f (a°(t)) + ryo°(t) < ao for t E [0, T], where I > 0. Then there exists precisely one weak solution of the first boundary value problem of the elasto-inelastic body with a perfectly plastic domain. Before proceeding to the proof of theorem 1.2, we will explain the idea leading to the introduction of the penalization functional. As we have already mentioned in the introduction, this idea leads to abstract differential equations which describe the regularized plasticity or, in other words, the plasticity with a yield surface, which is not infinitely thin." This model has a physical meaning by itself and is not rate independent. Naturally, we will go back to the formal considerations. Let us assume that if the process reaches the situation f(a(t)) = ao, there is an increase
of both the plastic deformation and the level of plasticity f(a(t)), hence generally f(a(t)) > ao, but of course f(a(t)) does not go too far from the surface f (a) = ao. This results in replacing equation (1.8) by the equation
p= E[f ao]+af, E>0.
(1.22)
For purely mathematical reasons, let us replace (1.8) by a more suitable equation of the form p
If e
- ao]+ as (1 + ([f (a) - a0]+)2) -1/2.
(1.23)
Now let us seek ac, pc in the way described at the beginning of this section; that is, by satisfying (1.1)-(1.5), (1.7), (1.9), and (1.23). Our mathematical
optimism makes us believe that in a certain sense there exist limits for e --' 0+. Since pc(t) = 0 for f (ac(t)) < ao, the same identity holds for the limit. Since }i' fulfills the condition of normality, the same holds for the limit. However, now the pc's are in a sense bounded. Since a f laa is by assumption a bounded tensor as well, we have 1
[f(ac(t)) - a0[+
< C < 00,
E [1 + ([f (a) - ao]+)2]1/2 -
hence, for e --p 0 we have f(a(t)) < ao. It turns out that our optimism has not deceived us. In the paper by NeLas and Travnflek (1980) a slightly simpler proof of existence of solution is given, which coincides with our course of proof provided the matrix A in relation (1.7) is an identity matrix.
Proof of Theorem 1.2. Let us introduce the penalization functional g(a) =
f[{([f(a) - ao]+)2 + 1}1/2 - 1]dx.
(1.24)
3.1. Prandtl-Reuss Model of Plastic Flow
231
We find that g(a) is Gateaux differentiable; that is,
- aol+aflaaij r; dx dda g(Q + ar)1a=o = Dg(a r) = fo [1[f(a) + (PO- ao]+)211/2 ,
3
(1.24')
Hence, we obtain monotonicity of Dg(a, ):
Dg(a,a - r) - Dg(r,a - r) > 0,
(1.25)
and
JDg(a,r)) < c[[a[[s[[r[[s. Let e > 0 and let us look for a', p` E Cp ([0, TI, S) such that
(1.26)
e' = Ac' (componentwise: es = A{jk(ak,),
(1.27)
= 1 If (a) -aol+af(a)laa W A'(v)af(a)
(1.28)
is the compatible strain tensor,
(1.29)
PE
8a
E [1 + ([f (a) - ao]+)211/2
EE = eE + pE
aE
(1.30)
fulfills (1.16).
Problem (1.27)-(1.30) will be transformed to the initial problem for an abstract differential equation. To this end, let So C S be the orthogonal complement to the space
E _ {E E S; E,; =
2
(a'Uixj+ aU?)
,
u E [WI,2(n)]3),
Korn's inequality (see Chapter 2.2) implies that E is closed. Hence, the condition of compatibility of the tensor a is equivalent to condition (E, w) = 0 Vw E So.
(1.31)
Now (1.16) allows So to be interpreted as the subspace of those tensors from S which satisfy the condition of equilibrium (1.3) with 3 = 0 and conditions (1.9) with g - 0. Further, let us denote by P the orthogonal projector of S to So. Conditions (1.27), (1.29), (1.31) yield P(A&E +jE) = 0,
(1.32)
PA&E = -P [.A(a)L].
(1.33)
hence, (1.28) implies
3. Problems of the Theory of Plasticity
232
PA&E = a', PA I B. Now let us look for aE = a° + &E, &E E So, and put However, the operator B has an inverse in So, since for r" E So we have
(BT,7) _ (PAT,f _ (AT,T) > CIITIis
(1.34)
Thus, we may apply the Lax-Milgram Theorem' (see Rektorys (1974)). Hence, we can transcribe (1.33) in the form
aE = -BQ° - P [Ai0 + B-'aE)8a(a°
+ B-'a) 1 .
(1.35)
Thus, our problem is transformed to that of solving the equation (1.35) in So with the initial condition aE(0) = 0. From (1.20) we find that the right-hand side of (1.35) is Lipschitzian in the variable aE, hence, there is a unique solution of (1.35) and by substituting into (1.27), (1.28), and (1.29) we determine the tensors eE, pE, eE Now let r E C([0, TI, So). From (1.27), (1.28), (1.24'), (1.31) we obtain -1
f [&E, fI dt + E
r
Dg(uE, r")dt = 0.
(1.36)
0
Put r" = ry(&E - 6,0) + aE - a° We have ft
Dg(aE,7(E - &0) + aE - a°)dt = 7g(aE(t))
0
t
+
f[DVa-(7°+a°))-Dg(a°+a°,aE-(7a°+a°))]dt; (1.37)
we have used the condition f (a° (t) +-y&° (t)) < ao, which implies Dg(-y&° + ao, aE - (7QO + &o)) = 0. Hence, the monotonicity condition (1.25) yields
fD(i(&- v0) + aE - a°)dt > ryg(a(t)). From (1.36) and (1.38) we obtain 1g(a(t)) -< c
j
0
E
Let e,, - 0 be chosen so that a",.
+
(1.38)
[a°,a°1)dt.
(1.39)
a (weakly) in Ho ([0, T], S). Since
(aE° (t), T) = J t (oE", r)dt 0
If B = B', then the well-known Riess Theorem for the inner product (Be,T) in So applies; the Lax-Milgram theorem is its immediate generalization to nonsymmetric operators.
3.1. Prandtl-Reuss Model of Plastic Flow
233
for all r E S, we have o "(t) - c(t) for all t E [0,T]. Now (1.39) yields
g(c`'(t)) < cle",
(1.40)
and since the functional g(a) is weakly lower semicontinuous by virtue of condition (1.25), we conclude
g(o(t)) < liminf g(aE"(t)) = 0.
(1.41)
n-+cc
Hence, f(a(t)) < ao. Now let r E Ho ([0, T], S) fulfill (1.16) and (1.17). For f = aE" - r we obtain from (1.36): 1
rt
0=
J0
t
[Dg(cr, aE" - r)
[cE-, aE^ - -r]dt + E
Dg(r,a- r)]dt > However, then
0
j
AQ° - r]dt.
/t
t
0:5 lim sup J [vE", r - or'- ]dt
2
[o, r]dt 0
0
-
(1.42)
lnrn nf[ae"(t), aE" (t)]
,
a'P 13 = aij - 36ijQkk,
fulfills the conditions of theorem 1.2. For f (Q) = [aD aD$3 ] 1/2 we additionally
have to prove the possibility of the limiting process a - 0.
3.1.2
Solution by Finite Elements
We shall show how to solve variational evolutionary inequalities of the type (1.10). However, we will restrict ourselves to the case when the body oc-
cupies a polyhedral domain 0 C R", n = 2, 3, and to the displacement boundary value problem (an = P,n). We will follow C. Johnson (1976b), who also proved existence and uniqueness of the solution of this problem. Let Ro stand for the space of symmetric matrices of the type (nx n) (the stress tensors). Let us assume that we are given the function of plasticity
3. Problems of the Theory of Plasticity
234
f : R, -i It, which is convex and continuous in It,, and a constant ao > 0. Denote
B={rERo, f(r) 0 VT E K(t).
(1.44)
Let us recall that this definition differs from definition 1.1 in that (1.17') results only by integrating (1.44) with respect to time. Inequality (1.44) immediately corresponds to (1.10). Zero displacement is given on the bound-
ary aft. C. Johnson proved existence and uniqueness of the weak solution of problem (1.44) (see C. Johnson (1976a)), such that or E L°°(I,S) and Q E L2 (I, S), under the following assumption:
there is X E K(t), where [ry(t)I = tmal [7(t)1,
(1.45)
and positive constants C, b such that ]9(x)] < C for almost all x E 11 and ± (1 + 6)X E P. With the aim of defining approximate solutions of problem (1.44), we introduce finite-dimensional internal approximations of the set E(t),
Eh(t) c E(t) Vt c I, 0 < h < ho.
3.1. Prandtl-Reuss Model of Plastic Flow
235
The sets Eh(t) can be constructed in R2 as sums of a particular solution X(t) of the equations of equilibrium and the subsets Eh = Nh(R), which were introduced in section 2.4.11. Then the parameter h is the maximal side of the used triangulation Th. Here we deal with block triangular finite elements and a piecewise linear stress field. The closed convex set B can be approximated as well, for instance, by the sets
Bh = {r E S I f,h(T) < ao, i = 1,...,Mh}, {{
with fi,h are certain linear functions such that Bh C B. Then
Ph={rESI T(x)EBh a.e. in R} as an approximation of the set P. If we now define
Kh(t)
Eh(t) n Ph,
then Kh c K(t) for all t c I. We introduce the discretization of the time interval I. Let N be a positive integer, k = TIN, mk, m = 0,1,...,N, I,,, = [t.-1, t.1,
T' = T(tm), arm = (r, -
rm-1)/k
Instead of the variational inequality (1.44) we shall solve the discrete problem: find o E Kh (t,,,), m = 1, ... , N, such that [aahk, T - ahk] > 0 Yr E Kh (tm), m = 1, ... , N, ahk = 0.
(1.46)
Let us assume that the following analog of assumption (1.45) holds: There exists Xh E Kh (t) and positive constants C, 6 independent of h and such that I Xh (x) I < C for almost all
xEf2and ±(1+6)XhEPh.
(1.47)
Problem (1.46) is uniquely solvable. This follows from the fact that for all m, ohk is the element which minimizes the strictly convex quadratic functional 1
[r, r] - [ahk, r]
on the closed convex set Kh(t,,,). Thus, on each time level t,,, we have to solve the quadratic programming problem.
3.1.21. A Priori Error Estimates. Let us assume that we have obtained the exact solution ohk of problem (1.46) and let us estimate the error am -
3. Problems of the Theory of Plasticity
236
ahk, where a'° = a(t,,,,) is the exact solution of the original problem (1.44).
To this end it is useful to assume that the partition of the interval I is ordered (possibly nonuniformly) in such a way that -y(t) is a monotonic function in each subinterval I,,,. However, since the proof of the a priori error estimate changes only inessentially, we will, in the sequel, consider an equidistant partition for the sake of simplicity. First, for q = (q1) ... , qN ), q"` E S, let us define
Ilglil.(S) = E kllgmllS m= 1
Lemma I.I. If (1.47) is valid, then there exist positive constants C and ko such that Ilaahkllz2(s) < C
(1.48)
for h < ho, k < ko.
Proof. Analogous to that of lemma 2 in the paper by C. Johnson (1976a) (see also Hlav6Zek (1980) or 3.2.21). Here we omit the details. O Let us define e(h, k)
rnf 11o, - r1112(s),
where
K={r=(T1,...,TN)ITmEKh(tm), m=1,...,N}. The quantity e(h, k) is actually determined by the approximation properties of the sets Eh(t), Ph and by the regularity of the solution a, provided we have any information at all about the latter.
Theorem I.S. Let assumptions (1.45), (1.47) be fulfilled and let a, ahk be solutions of problems (1.44), (1.46), respectively. Then for k sufficiently small and h < ho we have
max IIam - ahklls < C(e1/2(h, k) + k1/2). M
(1.49)
Proof. First of all, we have [Q, r - a] > 0 Yr E K(t), a.e. t E I,
(1.50)
[aahk, r - ahk) > 0 Vr E Kh (t,,), m = 1, ... , N.
(1.51)
We extend ahk to the whole interval I as follows:
ahk(t) = A(t)ahk 1 + (1 - a(t))ahk,
3.1. Prandtl-Reuss Model of Plastic Flow
237
where
fi(t) =
7(t) - _1(tm) , t E I,,,, provided 7(t,,,) 'Y(tm-1) - %(tm)
A(t) _ (t,,, - t)/k,
54
t E I,, provided ry(tm) _ l(tm-i)
Then we easily check that vhk (t) E E(t) for all t E I, taking into account the monotonicity of the function ry in each subinterval Im. The last property also implies that 0 < A(t) < 1. Since P is convex, we also have vhk(t) E P for all t E I. Summarizing, we conclude that vhk (t) E K(t) for all t E I. Now, putting r = vhk in (1.50), we obtain
[a, ahk - a] ? 0 a.e. in I. Integrating over I,,, we arrive at the inequality
-
[aam, am - am]
m + o (t) - Ohk(t)]dt.
(1.52)
Let us consider rh E K such that Ila-ThII12(S) s 2E(h,k)
and substitute r = rh in (1.51). Thus, we obtain (1.53)
laahk, rh - ahkl > 0.
By virtue of (1.52), (1.53) we come to the following inequality for the error e = a - vhk: [aem, em] C 1aahk, Th - am] + I rm I ,
where r,,, is the right-hand side of (1.52). Multiplying this inequality by k and summing over m yields: N CI)aahk 11l3(s) IITh
Max IlemIIS
-
vlt13(s) + 2k E IrmI.
(1.54)
m=1
For rm we may write 1/2
Irml
kJ
II&(t)IIS
(
(kaoz Is + k112 (f
Ilaltsds)
m
< Ck112llaahklls
I
dt
3. Problems of the Theory of Plasticity
238
< CJ
IIaIIsdt + Ckllaahkll2
Let us substitute into (1.54). Thus, we obtain the estimate max ile"`IIS < 2Clla0rhk III2(s)e(h, k)
+ Ck
l
116,
> kII aahk II S +
(t) II
sdt
< C1(e(h, k) + k),
by simply taking into account the inequality
j IId(t)Ilsdt < oo and applying lemma 1. This completes the proof of theorem 1.3.
Remark 1.1. If n = 2 and the solution a is sufficiently smooth, we can prove that e(h, k) < Ch2 with C independent of h, k by using the approximations Eh(t) = X(t) e Nh(i2), Bh = B. Here X(t) is a particular solution of the equations of equilibrium (1.3) and Nh (f2) are the spaces of piecewise linear finite elements for the equilibrium model (see section 2.4.11). (In the case n = 3 we can use composite tetrahedrons and piecewise linear stress fields-see KlI ek (1981).) The proof is a special case of that of theorem 2.4 in the next section, and hence is omitted.
3.2
Plastic Flow with Isotropic or Kinematic Hardening
We start with formal considerations as in the preceding section. Let us again assume that p = 0 provided f (a) < ao. Now we do not exclude the hardening, that is, the level of the yield surface in the course of the process increases to f (a) = a > ao, when p again fulfills the normality condition, but after another fall of stress to the level f (o) < a we have p = 0. Let
a(t) = max{ao, max f (a(t'))} o 0. If f * (a) = H(f (a)), where
H(7) =
J 7 h1/2(co)dcP,
then with a* for f*(cr) and ao = H(ao)
.
*
h(f (a)) a-« = as a . Putting 3(a, a)
f (a) - a we have p
a3.
a3. a = _ aa a, T(o,a) O.
(2.6)
Put ry = B'1/a{4 and 3(a, ry) `V f (a - B'/try) - ao. Again we have
3(a,7) 0.
(2.7)
We can also consider the combination of both the isotropic and kinematic hardenings: in the case we additionally set a = max { ao, max f (a(t') - B1/27(t')) } ,
l
o 0 and A = 0 provided 3(a,ry, a) < 0. Further, the hardening has been characterized by
a3 a3
a3 a3 +-->c>0. a7, j ary,j as as -
(2.12)
We could suggest further generalizations following the ideas of NeUs, Travnftek (1978), based on the paper by Halphen and Quoc Son Nguyen (1975). However, in this book we will keep the level of generality reached above.
Let us notice another excellent feature of the equations (2.11): the direction of the vector (P, -j, -a) coincides with that of the outward normal to the yield surface 3(a, ry, a) 0. Consequently, if 3(0, ry"", a) < 0, we obtain
Pij (Qij - oij) - 7,j (7ij - 1',j) - a(a - a) < 0,
(2.13)
which after integration over 0 yields
(P,d - a) - ('Y,7-ry)-(a,&-a) 0.
(2.15)
3.2. Plastic Flow
3.2.1
241
Existence and Uniqueness of Solution of the Plastic Flow Problem with Hardening
First of all, we call the reader's attention to the fact that the method described in this section actually coincides with that presented in section 3.1, and hence also the Prandtl-Reuss model of plastic flow can be solved in the general setting. However, the hardening condition, represented by inequality (2.12), makes it moreover possible to reversely determine A from (2.11), thus going back to the classical formulation, and also to find the tensor of plastic deformation p (that of elastic deformation e is given by the relation (1.7)), hence c = e+p. Consequently, in this case we determine the strain tensor as well. Moreover, in the case of isotropic hardening we obtain the relation
a=max{ao, max f(a(t'))}. l o_ 0 0
0
(2.22)
0
must be fulfilled for all t E [0, TI.
We will assume that I is twice continuously differentiable with respect to its arguments, that it is convex and all its second derivatives are bounded. Further, we will asume that if r E C([O, T], S), -y E C([O, T], S), a E C([0,T],L2(f2)), then there exist 6 E C([0,T],S), 6 E C([O,T],L2(f2)) such that 3(r(t),7(t) + 6(t), a(t) + /9(t)) < 0 for all t E [0, T].
Theorem 2.1. Let us consider the problem of an elasto-inclastic body with interval state variables according to definition 2.1. Let the function 3(a, ry, a) satisfy the above formulated assumptions. Then there exists a unique solution of the problem.
Proof. The uniqueness if proved as easily as in theorem 1.1. The existence will be proved again by the penalization method introduced in section 3.1. Let us seek continuously differentiable functions with values in S (or in L2(t2), respectively): a', 7E, a', eE, pE, fulfilling for all t E [0, T] the following conditions: aE fulfills (1.16), (2.23) eE = Ac r'.
(2.24)
Let us further introduce the penalization functional 9(a,'Y, a) =
Jn
({[(3(a,'Y, a))+12 + 1)1/2 - 1)dx,
(2.25)
and let us use the symbols Dog, Dig, Dag to denote its partial derivatives. Let us also assume that PE = e D.,9,
p'(0) = po E S,
(2.26)
7E _ - eD79,
'y(0) = 'Yo,
(2.27)
'if _ -1 D.9,
a&(0) = ao,
(2.28)
0 (0) = a0,
(2.29)
the tensor c = e + p is compatible.
(2.30)
3.2. Plastic Flow
243
Let P be the projector from the proof of theorem 1.2. Then we have
P(Aof +pf) = 0.
(2.31)
Now let co E C1([O,T],S) be a tensor such that a°(t) satisfies (1.16), a°(0) = ao, and let us look for a`(t) = a°(t)+of(t), a6(t) E So for all t E [0, T]. As in the proof of theorem 1.2, put
PAaf = a'.
(2.32)
Hence, (2.26)-(2.30) imply the conditions
of = -PAv° - Day, where Dog = D ,g(a° +
(2.33)
of (0) = 0,
(PA)-1a6,a),
7f = - e D, g,
7f (0) = 70,
(2.34)
of = -1 D,, g,
af(0) = ao.
(2.35)
The assumptions on the function 3(a, ry, a) imply that Dog, D7 g, Da g are Lipschitzian, and hence there exists a unique solution in the space C1([O, T], S) x Cl ([O, T], S) x C1([0, T], L2(f1)). Now put r = o + a'' -
O° - a°, 'y = if + ,yf - 6, a = of + of - P with 6 E C([0, T], S), $ E C([0, T], L2 (fl)) and 3(Q° (t) + a° (t), 6 (t), $(t)) < 0 for all t E 10, T]. Thus
we can determine p E C1([O,T],S) from (2.26). Now (2.27)-(2.28) and
/'f(a,cx)dt
(2.31) imply
f[br1dt+ J0(if, ry)dt + t
+
fI(Dg, r) + (D7g,
(2.36)
) + (D«g, a)]dt = 0.
Again we have
f +
f
t[(Dag, r) + (D7g,'y) + (D,,g, a)]dt g(af (t), -Yf (t), af(t)) - g(ao, 7o, o o)
t [
(Dog, o- Q° - °) + (D,r g, g(a'(t),
- 6) + (Dag,
- Q)]dt
-Y' (t), a' W) + fot [ (DOg(a', -y', a'), of - a° - o°
3. Problems of the Theory of Plasticity
244
-yE - 6) + (D.g(cr , 7E, a'), a'' -,0)
+ (D7 g(a', 'YE,
aE - U° - a°) - (D.rg(a0, 6, P), -y' - 6)
(D, g(dO + co, 6,
- (Da 9(a° + 0, 6, ), aE
- $)]dt > 9(o (t), YE(t), aE(t))
(2.37)
Here we have used the conditions g(010,70, ao) = 0 and F(&° +a°, 6,6) < 0, which yield Dg (e° + co, 6, f) = 0. Hence, we obtain from (2.36) and (2.37)
f
+
t
0
+
e
g(a' (t), 'YE (t), a' (t)) < C1.
(2.38)
Let en - 0, En > 0 be a sequence chosen so that c'- o- in H'((0, T), S), a in H1((0, T), L2 (fl)); we again have aE" (t) - a(t), 7E" (t) - ry (t), aE" (t) - a(t) for all t c [0, T].
7E* - - in H1((0, T), S), aE" Now (2.38) implies
g(aE"(t),'YE"(t),aE"(t))
Glen,
(2.38)
hence again,
g(a(t), 7(t), a(t)) < 0 #, F(a(t), ry(t), a(t)) < 0.
(2.39)
Let r, 6, $ be a triplet of elements from definition 2.1. Then similar to (1.42), (2.36) yields t
t
0 > fo
aE"
t
- r]dt + f (7E",'YE" - 6)dt + f (aE", aE" - Q)dt, (2.40) 0
0
and the proof can be completed as previously. Theorem 2.2. Let the assumptions of theorem 2.1 be fulfilled together with condition (2.12). Let a, -y, a be the solution of the problem. Put A = A(t, x) = 0 provided 3(a(t, x), ry(t, x), a(t, x)) < 0
(for almost all (t, x) from (0, T) x il),
a=
aF 8a a
OF 8F provided 8F OF 9. 09.7 +aa as
(2.41)
3.2. Plastic Flow
245
I (a (t, x), 7(t, x), a(t, x)) = 0
(2.42)
(for almost all (t, x) from (0, T) x f2) and solve the equation
P =A
-,
p(O) = po
(2.43)
in H'((O,T),S). Then
-a 8 F
wry,
'Y(O)=7o
(2.44)
a(0) = ao
(2.45)
(in the sense of H'((O,T),S)),
a = -a
aF ,
(in the sense of H1((O,T),L2(c l)), e(t) + p(t) is a compatible tensor.
(2.46)
o', ryE" -j ry, aE' - a, pE" - p Moreover, if En --p 0, En > 0, then or'(by the proof of theorem (2.1); that is, they converge weakly in the spaces H1((0, T), S), H1((O, T), S), H1((0, T), L2(fl)), H1((O, T), S), respectively. Denote An =
F(aE", ryE", C&) + [1 + (F(cr ",'YE", aE")+)21-1/2.
(2.47
En
Then An
) in L2((O,T),L2(l)) (= L2((0,T) x f2)), and hence, A > 0.
Further, o
-+ a in C([O,T[,S), ryE^ - ry in C((O,T[,S), aE" - a in
C([0, T], S).
Proof. Let us consider the sequences aE", ryE", aE^, pE". By virtue of the uniqueness of solution we evidently have (in the respective spaces) aE" - a, ryE" ry, a',, a. From (2.26)-(2.30) we obtain /'t
f 1
t
(aE^a - aE"[dt+
0
fo
(1!E",ry -7E")dt
t
+
1(aE",a - aE")dt > 0,
(2.48)
0
hence,
IIaln(t) - a(t)III +
IIryE"(t)
- _y(t)III
+ IIaE^(t) - a(t)IIL3(0) 0, 1 a f I < C Vi, j, Vo E R0 - Q.
aaij
3. Problems of the Theory of Plasticity
248
As an example of a function satisfying the above assumptions let us mention the von Mises yield function (for n = 3)
f(a) = (.D0D)1/a' where oD = a - 36;jvkk is the stress deviator. Let
ao=ruuro, runro=0, where each of the sets ru and ro is either empty or an open set in afl. Let us assume that we are given the (reference) vector of the body forces
F° E [C((l)]" and of the surface loads g° E [L2(ro)]". if ru = 0, let the condition of the total equilibrium be fulfilled, that is, (1.1) and (1.2) for n = 3 respectively, and
fo
(x1F2
x2F°)dx+
j(xig- x2g)ds = 0
(instead of (1.2)) for n = 2. Let the actual body forces and surface loads be
F(t, x) = ry(t)F°(x) in I x f1, g(t, x) = ry(t)g° (x)
on I x ro,
with ry : I - R a nonnegative function from C2(I), such that
3t1 > 0, y(t) = 0 bt E [0,t1]. Again, we introduce the set of statically admissible stress fields:
E(t) = E(F(t), g(t)) = {o E S ] fo o c1 (v)dx
=
Jn
F;(t)vidx+ j g;(t)v;ds VvEV}, ff r,
where
V= (V E [W 1"2((l)]" I v= 0 on ru}. Finally, let us recall the definition (see (2.3))
3(r, a) = f (r) - a. We denote
H = S x L2(f2),
(3.3)
3.2. Plastic Flow
249
{(r, a) E R, x R I F(r, a):5 0},
B
P= {(r, a) E H I (r(x), a(x)) E B a.e. in n}, K(t) _ (E(t) x L2(11)) n P, t E I.
(3.4)
Let the coefficients of the generalized Hooke's Law satisfy conditions (1.18), (1.19). Further, let a positive constant ao be given. For couples (a, a) we introduce a new symbol &, for example,
&= (a, a),
T=(T,I'),
and we define the inner products with the corresponding norms:
(&,T) =a,,r;,+afi, (a, 0. =
Izl=(T,T)1I2,
f(6,)dx, ll&= (d, a)o 2'
{&, f} = [a, r] + (a, R)o,n,
III&111= {&, &}i/2
Similarly to definition 2.1, a weak solution will be a couple & _ (a, a) E
Ho (I, S) x Hl (I, L2(0)) such that a(0) = ao, &(t) E K(t), and d {
k) - &(t) } > 0 W = (r, a) E K(t) T
,
(3.5)
holds almost everywhere in I.
Let us point out that our definitions corresponds to a(0) = 0 and to the zero displacement on r,,. The time-dependent variational inequality (2.22) from definition 2.1 results by integrating (3.5) with respect to the time variable provided -y =_ 0 (that is, without the kinematic hardening), and it is equivalent to (3.5). The existence and uniqueness of solution of (3.5) was studied for instance by C. Johnson (1978) for an = ru, and the case an = r, is dealt with in the book by and Hlav£Lek (1981). With the aim of approximately solving problem (3.5), we first introduce finite-dimensional (inner) approximations of the set E(t):
Eh(t) =X(t)+Eh,
0 0 such that
T(e(t)+p)50 VpERoxR, lpls61
3. Problems of the Theory of Plasticity
252
Hence, e(t) + p E B, which yields (3.10). Let us prove lemma 2.1 by the penalty method. Let 7r be the projection
operator to a closed convex set B in the space R, X R. Let us introduce the penalty functional
Jµ(r) =
2µIIf-7rf1I2,
µ>0, fEH.
(3.12)
Define new approximations ahkµEEh(tm)XVh,
m=0,1,...,N
by the identities (omitting for brevity the indices hkµ in the following: {eam, f} + (J,A(Um), f)o = 0
Yr E Eh X Vh,
(3.13)
0
Notice that the Gateaux derivative of J. is J"'(8)
7r&). 14
Problem (3.13) has a unique solution for every m, since b' minimizes the coercive, strictly convex and continuous functional 1111&1112
2
on the set
+ kJ,. (6) - {ofn-1, o}
x Vh, which is closed and convex in H.
Lemma Z.S. Let condition (3.7) be fulfilled. Then there are positive constants C, ko such that for k < ko, 0 < h < ho and µ > 0 we have estimates (i)
maxl<m 0.
(3.15)
Thus, we can write M
M
k{apm,pm} < - > M=1
M= 1,...,N.
(3.16)
m=1
On the other hand, we have M
M 1
k{apm, pm}
m=1
(IIIP
mIII 2_ IIIPm-1 III 2 +IIIP m-
P,-11112)
M=1
(3.17)
ZIIIPMIII2,
and hence by virtue of (3.9) we can estimate
M
M kI {ar, pm}I < Ck
N
1
IIIP'"III (1+ IIIPt1112) m=1
m=1
M=1
M < C + Ck >2 IIIPmIII2.
(3.18)
M=1
We will use the discrete analogue of the Gronwall Lemma (see Babugka, Prager, Vitasek (1966), Chapter 3, lemma 3.3): Let
M-1
`o(M) < Ik(M) + E X(r)co(r), M = 1, ... , m < N, X(r) > 0 Yr. r=0
Then
rp(m) < di(m) + r, X(r)0(r) 11 (1 + X(s)). r=0
a=r-}1
The estimates (3.16), (3.17), and (3.18) yield M-1 IIIPMI112 < C + C > K
1prIII2,
M=1,...,N.
r=0
Setting
