LECTURES
ON ELLIPTIC BOUNDARY VALUE PROBLEMS SHMUEL AGMON
AMS CHELSEA PUBLISHING American Mathematical Socien
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LECTURES
ON ELLIPTIC BOUNDARY VALUE PROBLEMS SHMUEL AGMON
AMS CHELSEA PUBLISHING American Mathematical Socien
I'ruvidcncc, Rhode Island
LECTURES ON ELLIPTIC BOUNDARY VALUE PROBLEMS
Lectures on ELLIPTIC BOUNDARY VALUE PROBLEMS
by
SHMUEL AGMON Professor of Mathematics The Hebrew University of Jerusalem
Prepared for Publication by B. Frank Jones, Jr. with the assistance of George W. Batten, Jr.
D. VAN NOSTRAND COMPANY, INC. PRINCETON, NEW JERSEY
TORONTO
NEW YORK
LONDON
D. VAN NOSTRAND COMPANY, INC. 120 Alexander St., Princeton, New Jersey (Principal office) 24 West 40 Street, New York 18, New York D. VAN NOSTRAND COMPANY, LTD.
358, Kensington High Street, London, W.14, England D. VAN NOSTRAND COMPANY (Canada), IIrD.
25 Hollinger Road, Toronto 16, Canada COPYRIGHT © 1965, BY
D. VAN NOSTRAND COMPANY, INC. Published simultaneously in Canada by D. VAN NOSTRAND COMPANY (Canada), LTD.
No reproduction in any form of this book, in whole or in part (except for brief quotation in critical articles or reviews), may be made without written authorization from the publishers.
PRINTED IN THE UNITED STATES OF AMERICA
PREFACE This book reproduces with few corrections notes of lectures given at the Summer Institute for Advanced Graduate Students held at the William Rice University from July 1, 1963, to August 24, 1963. The Summer Institute was sponsored by the National Science Foundation and was directed by Professor Jim Douglas, Jr., of Rice University. The subject matter of these lectures is elliptic boundary valued problems. In recent years considerable advances have been made in developing a general theory for such problems. It is the purpose of these lectures to present some selected topics of this theory. We consider elliptic problems only in the framework of the L 2 theory. This approach is particularly simple and elegant. The hard core of the theory is certain fundamental L2 differential inequalities. The discussion of most topics, with the exception of that of eigenvalue problems, follows more or less along wellknown lines. The treatment of eigenvalue problems is perhaps less standard and differs in some important details from that given in the literature. This approach yields a very general form of the theorem on the asymptotic distribution of eigenvalues of elliptic operators. Only a few references are given throughout the text. The literature on elliptic differential equations is very extensive. A comprehensive bibliography on elliptic and other differential problems is to be found in the book by J. L. Lions, Equations diffe4entielles opeiationelles, SpringerVerlag, 1961. These lectures were prepared for publication by Professor B. Frank Jones, Jr., with the assistance of Dr. George W.Batten, Jr. I am greatly indebted to them both. Professor Jones also took upon himself the trouble of inserting explanatory and complementary material in several places. I am particularly grateful to him. I would also like to thank Professor Jim Douglas for his active interest in the publication of these lectures. Shmuel Agmon
Jerusalem
CONTENTS Section
0 Notations and Conventions 1 1 Calculus of L 2 DerivativesLocal Properties 1 2 Calculus of L 2 DerivativesGlobal Properties 11 3 Some Inequalities 17
4 Elliptic Operators 45 5 Local Existence Theory 47 6 Local Regularity of Solutions of Elliptic Systems 51 7 Garding's Inequality 71 8 Global Existence 90
9 Global Regularity of Solutions of Strongly Elliptic Equations 103 10 Coerciveness 134 11 Coerciveness Results of Aronszajn and Smith 151 12 Some Results on Linear Transformations on a Hilbert Space 175 13 Spectral Theory of Abstract Operators 208 14 Eigenvalue Problems for Elliptic Equations; the SelfAdjoint Case 230 15 NonSelfAdjoint Eigenvalue Problems 261 16 Completeness of the Eigenfunctions 278
0. Notations and Conventions
The following notations and conventions will be used. En will denote real nspace. For any points x = (xl,...,xn) and y = (yl,...,yn) CEn, i
IXI = (X12 +...+Xn2)VZ, x
y = X1y1 +...+Xnyn.
For an index or exponent a = (a1,...,an), whose components are ,Integers, jai = a1 +...+ an; from the context it will be clear whether norm is intended. Also, a! = this norm or for N = 0611...,an)r
(a//
N!(a_.S)!
For any x
a,!.....an!
_
a!
S1!...$ !(alY1)1...(anOn)!
En' a= (al, ... Pan),
x==x...x = 1
«
n
In particular, this will beused with the differentiation operator: 1
n
0X1«1
Also a S
3xn«n
a 1 5 s 1'... Pan "
On'
These conventions greatly simplify many expressions. For example, Taylor's series for a function f(x) has the form Q D°`f (0) xa,
We will also find it convenient to have a special set inclusion: CC. We will writ 01 CCSI2 if and only if Q1 and 02 are open, SZl is Compact and Ell C 02. For any function u, the notation supp (u) will be used to denote the
support of v; i.e., the closure of the set (x:u(x) # 01.
1. Calculus of L2 DerivativesLocal Properties In this section SZ is an open set in ndimensional Euclidean space E. For any nonnegative integer in, Ca' (SI) is the class of m times 1
2
Elliptic Boundary Value Problems
continuous differentiable (complexvalued) functions on SZ and C°°(SZ) = fl Cm(1l). Also Co (Sl) is the subset of C°°(&1) consisting mD of functions having compact support contained in U. The functions in C`a (1) will be called test functions for Q. Now some norms and seminorms will be defined. Definition 1.1. For u C Cm(fl) ID"ul 2dx]1/' .
[f
I Jul lm,jj
SZ I"l_<m
If
IIvIIm,SZ are both finite, then there exists
(u, v)m,c = f
D"u D°`vdx.
Sl H <m
Likewise, for u C Cm(Sl), Iulm,SI
[fn
I
ID"ul2dx]'/x
Q lalm
This latter quantity is of course only a seminorm: lulm,Sl = 0 u = 0. The notation (u, v)m = (u, v)m.fl. l lul lm = l Iul lm,I lulm lulm,Sl may also be used if Sl is fixed during a discussion. Definition 1.2. Cm%l) is the subset of Cm(Sl) consisting of functions u with I Jul Im,ci < °°.
Definition 1.3. Hm(Sl) is the completion of Cm"U1) with respect to the norm II llm,j2
Since C`°*(S1) is obviously an inner product space with respect to the inner product (u, v)m.c, it follows that Hm(Sl) is itself a Hilbert space. According to the general concept of completion for a metric space, the members of H ( 0 ) can be assumed to be equivalence classes of Cauchy sequences of elements of Cm"(0). However, in the present case another characterization of Hm(S1) is possible. Indeed, if {uk{ is a Cauchy sequence in Cm*U1), then for any fixed a, Iai < m, it follows from the inequality
.
If
l Duk _ D"uJf
2dx]1/1 _
0 chosen so that S (x, 2rx) C Oi. Since F is compact and covered by IS (x, rx): x E F1, it is covered by a finite number of these spheres, k = 1,...,m. say by IS (xk, rx k Let Ci=
U XkE7
S (Xk, rx ) i
k
Then Ci has the desired properties. Now let (Ci* } be a collection of compact sets satisfying Ci Cint Ci* C Ci*,C 0i. Let iii* be the function which equals 1 on Ci* and vanishes elsewhere. Choose c > 0 less than dist (Ci, a Ci*)
and dist (C,*, a U. Let Ji = JEIi*. Then lii E Co (0,) and Vii (x) = 1 for x E Ci. Finally, let 61 = V/ 1
6i =(1'A,)' (102)...(1Oi )Oi, i> 1. Then ei E Co (0,) and
8
Elliptic Boundary Value Problems
e= i v
ton U C,JF.Q.E.D. 1= i
Terminology: The collection of functions 6, is called a partition of unity subordinate to the open covering 10,1 of F. Definition 1.8. u E Wm°c(Q) [H°c(SZ)] if any x E I has an open neighborhood 0 C S2 such that u E Wm(0) [Ifm (0)].
Note that Hb0c(Sl) C W10c (H). For m = 0: H0l 0c(1l) = W2 c(1l) = L2 c(S2)
THEOREM 1.10. If u E W. (11) [Hm (12)], then for any S21 C C S1
Jrn u in W. (Q 1) [Hm (S21)] as c . 0. Proof. Clearly Jeu is well defined in S21 for e > 0 sufficiently small and it is enough to show that D«J( u D°`u in L2(121) for 0 _', jal 5 in. This however follows immediately from Theorem 1.8 and Theorem 1.7 since for a sufficiently small D° Je u (x) = JED"u (x)
D°`u
in .L.2 (111). Q. E. D. THEOREM 1.11. If u E Wm loc (S) [H u E Wm (S21) [Hm (H 1)] for any S21 C C Q.
loc (il)], then
Proof. Suppose u E Wm loc (1k). Since 521 is compact, it can be covered by a finite number if open subsets 01,...10k of f2, with
u E W. (0,), i = 1,...,k. Thus the restriction of u to 0; admits weak derivatives D°`u E L2 (0,) , jai S m, which we denote by ui. From the uniqueness of weak derivatives it follows further that ui = u!' almost everywhere in 0, n Oi, 1S i, j S k. Thus after correction on nullsets we .obtain functions u°` E L 2 ( U 0) such that u°` = the 1=1
weak derivative D°`u in Wm(01), lal < m and i = 1,...,k. Let now (61} be a partition of unity subordinate to 1011; k
161 1 on S21. For any q
Co (Q l), let 0, = 0 6,, i = 1,...,k;
then (, is a test function on 0, and k cb =
1=i
1 on 521.
Calculus of L2 derivativesLocal properties
sec. 1
9
Since u E W.(01), and since the support of 01 is contained in 01 f1521,
f
u
a.
(1)1«I
g1dx =
f
uDa(k 1dx
Q1
Q1
for jai < m, i = 1,...,k. After summing on i, we have
f
u a . qSdx =
U1
(1)1'1 f u . DaOdx, a1
la I < m.
Thus, v E W. (521).
Suppose u E Hm°°(12) C W1 °(12). Then by the first part of the proof, u E Wm(122) with 121 CC 122 CC 52. Then by the corollary to Theorem
1.10, Jfu  u in Wm(1 1). Since J1u E C°°(121), this implies u E Hm(&11). Q.E.D. Corollary. If 121 ccci, then the functions in W10c(12), restricted to 121, are in H,,(1 1); i.e., u E Wn°°(52) implies u E H,.(121) Proof. If u C Wb°c(12), then JEu . u in Wm(Q1) as c  0 and JEu C C°°
by Theorem 1.5. Q.E.D. THEOREM 1.12. Hm °(12) = Wm0 c(1)
Proof. This follows immediately from the above corollary. Q.E.D. THEOREM 1.13. (Leibnitz's rule). If u C W .. (Q) [H.(0)], and if v C C'(12) has bounded derivatives of all orders < m, then uv C Wm(12) [Ha(lt)] and
(1
Da(uv)
=2
()D13uDa_13v.
Proof. If u C Hm(12), then there is a sequence Iuk{ CCm"(Q) such
that Dauk . D u in L2(12) for 0 < lal < m. Since v has bounded derivatives in 12, Da(ukv ) _ f3 0. Let 1, = Oi fl a SZ. Then u E Wm(En  T). For let 0 be any test function on En  r. Then K = supp (q) fl supp (u) C n. Let CC Co W) be 1 in K. Then CO is a test function on SZ and we have
sec. 2
Calculus of L2 derivativesGlobal properties
13
fr uD"cdx = Kf uD"(tc)dx
En
= f uD°(tc)dx _ (1)H«H f Dau tOdx
0
(1)1"1 f D"u cdx K
_ (1) I "I
j
D«u cdx,
En F
where in the last integral we define D°`u as zero outside U. Hence
u' Wm(En  '). Let rr = F  ry' where y 1 is the vector associated with 01 by the segment property and 0 < r < min (1, S1y1l 1). Then rr c 01, for if z E I'r then dist (z, O,):< dist (z, r) G Iry1l < S. Also, I'r n ?Z = 0,
asi
for if z E rT fl fl, then z c f1 fl 01 and z + ry 1 f F, so that z + ry' C S1, contradicting the segment property. Since I'r is compact, dist (I'r , 0)
14
Elliptic Boundary Value Problems
Let u'= u(x + ry'). Then U' C Wm (En l'd.); since SZ C :En  F,we have, a fortiori, u' C W m (fi). Since (D'u') (x) = (D°`u)(x + ry'), D'u' D°`u on L2(SZ) by a familiar theorem on Lebesgue integration. Hence, u' u in Wm(c1) as r + 0. Thus it is sufficient to approximate u' by functions uk E Co En). To obtain such a sequence we simply take the mollified sequence uk = J1/k u, k = 1, 2,.... Clearly uk E Co (En) (Theorem 1. 5). Since dist (hr, 0) > 0 it follows from Theorem 1:8 that (D«uk)(X) = ( ,/kD«u)(X) for x E 9, Jai < m
for all k > [dist (I,T , 9)1 1. From this and from Theorem 1.7 we have: D°`uk + D°`u' in L 2(c) for jai < in. This shows that uk is the desired approximating sequence of u'. Q.E.D.
As a corollary we have THEOREM 2.2. If 0 has the segment property, then Hm(c) = Wm(SZ) This theorem is important for obtaining properties of Hm(f2), for frequently it is easier to obtain them for W. (Q). As an example, we prove THEOREM 2.3. If SZ has the segment property, if u E Hi(SZ), and if D°`u E Hk(11) for Jai = j, then u E Hi+k(SZ).
Proof. It is sufficient to prove the theorem for H replaced by W. This
is trivial. Q.E.D. Likewise, using the fact that H'O '(SZ) = Wm °(SZ) for any 0 (Theorem 1.12), we have
THEOREM 2.4. For any open set 52, if u E Hi°°(c), and if D"u E
H'"(9) for lal = j, then u E 111"(0). We will now extend the idea of generalized derivatives to differential
operators. Let A(D)
l'
l <m
be a linear differential operator with constant coefficients. Let a,,(1)1'i D'.
AD) = I
jl
m
sec. 2
Calculus of L2 derivativesGlobal properties
15
If u E Cm (9), then integration by parts shows that f uAqS dx = f Auk dx for any di E Co (SZ). More generally we make the following definition.
Definition 2.2. Let u be a locally integrable function on ci If there exists a locally integrable function f on Q such that
f uA0 dx = f fc dx sz
fZ
for all 95 E Co (c), then we will say that Au = f exists weakly and that u is a weak solution of Au = f. Note that the weak existence of Au does not imply the existence of any weak derivatives of u. However, weak existence does imply strong existence in the sense of the following theorem. THEOREM 2.4. If u, f E L2 loc (a), and if u is a weak solution of A. u = f., i = 11 ...,m, on &I, then there is a sequence I uk I CC°° (En) such that for any 91 CC 0, uk + u and Ai uk fi in L2 (a2) as k + .
Proof. Extend u to En, defining u = 0 outside Q. Let uk = Jl/k u. As in Theorem 1.8, we see that
A. J1Ik u (x) = Jl/k A, u (x) if dist (x, 3 fl) > 1/k. It follows from Theorem 1.7 that A A. uk = )_as k + o. Q.E.D. Let A°` (i;) = D A (6). We will call A" = A°` (D) a subordinate of A.
Ai Jl ,k u = J1 /k `9i u + Ai u = fi in L
(fi
ll
We have the following analog of Leibnitz's rule.
THEOREM 2.5. Let A be a linear differential operator of order m with constant coefficients. Suppose that u E L2 1oc (fl) and that Au and A°` u exist weakly for Jul S m. If C E Cm (SZ), then A (Cu)
exists weakly and A (Cu) _
E loI < m
D« A°` a!
u.
Elliptic Boundary Value Problems
16
Proof. First assume that u E C°° (&1). Then by Leibnitz's rule we have a
A (bu) =
I
15
1
a« m
« \
N) DO
DNS RS« a«
IRI<m
H«
Da_
A13
113 < m a!
D«
'
u
OID«u
m
u,
since
!
Ale(= 1«I
a
DO em
a
a!
<m
I
S 1, then it holds for j = k  1. Under the hypothesis that (3.8) holds for j = k < m we have by (3.9) that
IUIk  1 S Y (E
Iul2 + E  (k  1)
uI
)
< Y fey (Em I JU12 + C k IUIO) +
y2 (Em
,(k 1)
Iul2
(k  1) lull J
+(k  1) IUIO)
and the theorem follows by induction. Q.E.D. As a corollary we have THEOREM 3.4. If SZ is bounded and has the restricted cone property, if c > 0, and if u E H. (Il) for some m 2, then there exists a number C. = CE (cl, m), which depends only on Q, m, and c, such that
I"Im,a+CEIu10.0. Ilullm1,ci
For convenience, we now collect some facts about Fourier series. We will use the notation Q = {x: Ixkl < 1/21. The cube Q will serve as a fundamental domain for various classes of periodic functions.
Elliptic Boundary Value Problems
26
befinition 3.1. CO rC' ] is the class of functions u (x) defined on En which are periodic of period 1 in each variable and which are m times continuously differentiable [infinitely differenti
able]. Also, H# is the completion of C' (or C') with respect to the norm II IIm,
.
We may obviously identify the elements of him with functions de
fined on Q, and, with this identification HM CH. (Q). Let u E C'. Then u has a Fourier series expansion: (3.10)
u(x)=Ibee2 e 6
We use the notation Ito stand for summation over all e_ (&. 6 where l;1,...,6n are integers. This series and the derived series
(3.11)
D°` u (x) = (27ri)I be
e27r'e . X `
6
are uniformly convergent, and, by Parseval 's identity (3.12)
=(2'r)2I"L Ib6I2 62"
ID°` uI o. p
We will use the notation
(3.13)
IIuIIm = [
(161 Euclidean norm.)
Ibo.....oI2 +6
Ibel2
Iel2m
]/2.
Some Inequalities
sec. .3
27
THEOREM 3.5. There is a constant Cm > 1 such that
cm 1 Ilul lm s IluI Im,
Q
< Cm I Iul lm
holds for all u E C' Proof. Consider the ratio lei  2m
(27)2 111 62, for 6 an

III<m
ntuple of integers, e # 0. This ratio is bounded above for all such e. Therefore, since by (3.12)
llullm. Q =
I«l
IM ID' ul2. Q = I
Ibe12
1m (27)21le2« l«I
the result follows immediately. Q.E.D. is the Consequently, the completion of C# in the norm Ilm, Q same as its competion in the norm ll Ilm. Therefore, we may identify H' with a class of formal Fourier series be e27ix ' 6 suppose such that the expression (3.13) is finite. Now is such that the expression that the formal series I be e27 X
is finite, and let uk (x)
(3.13)
I6Isk
b
e271x
Then
Dec
Uk is a Cauchy sequence in L2 (Q) for I a I < in, so that, by the RieszFischer theorem, D" uk converges in L2 (Q), a l < M. Thus,
nk E C° , uk  u in L2 (Q), and D°` Uk converges, al < m. Hence, uk u in Hm (Q), so that u E H# . Therefore, we may identify Hm with the collection of all formal Fourier series u = Y b, e27Tix
6
such that
#
I lul I m < no.
Moreover, if u E H# has the Fourier series m
28
Elliptic Boundary Value Problems
u = E bee 27Tix 6 then for lal S in, D" u has the Fourier series
D" u = y bt (27Ti e27Tix S S
the series converging in L2 (Q). THEOREM 3.6. If {u 1I is a bounded sequence in Hm with the formal Fourier series u i (x) =
' 2b 6
. J
x
e21116
1, and if for fixed 6, 1be0converges to be as
if m
 ; (Q) as j . .
{u1} converges in Hm
Proof. Let
bt
Pu_ r
T
S" u = r
'
lei
t
lei n/2,
let f be the largest integer less than m n/2 (f = m [n/2]  1). Then u can be modified on a set of measure zero so that u E Cf (11). Moreover, for any r 1, Ial S f, xE 0,
ID'a(x)l
n/2,
1 Ot rk  1 fk (r) dr12 = I
Ot rk
ft r2k 0
/Z (n + 1) [7./2 (n  1) fk (r)] drI 2
n
t rn  1 Ifk (r)I2 dr
dr 0
t2k  n
=2k 1 n Thus
Ot
rn  1
If (r)12 dr. k
Elliptic Boundary Value Problems
34
t2k  n {k 1 Iu (0)I2 < (k + 1) 11=0 t21 if (t)I2 + k2
2kn
1
ft
rn
0
Ifk (r)I2 dr}.
Now integrate this inequality over Gh for some h < h0. The volume
of Gh is (vhn, where m is a constant. Since 0 < t < h, we have (3.14)
k
whn Iu (0)I 2 < (k + 1)
+


1
1= 0
Gh
f
k2
h21 f
2kn Gh
If1(t)I2 dx
t2k  n rh rn  1 Ifk (r)I2 drdx 0
Let y be a generic constant depending only on w, in, and h0. Since f1 is a linear combination of derivatives of u, the first term on the right in (3.14) is bounded by k  1
y
h21 Iu12
1=0
h
To estimate the second term, we introduce polar coordinates. Let a be the ordinary surface measure on the surface of the unit sphere, and let A be that part of the surface of the unit sphere subtended by the cone Gh . We then have the formula for integration using the 0
spherical measure or (3.15)
f gdx = f f Gh
Ao
h
gtn
dtda
.
Using (3.15), the second term on the right side of (3.14) becomes
35
Some inequalities
sec. 3
t2kn r it rn1if k(T)2drdx 0
1 Ifk (r)I2 dr] to  ldtdu
= f f [t2k n
1 dt) (f rh rn  1 Ifk (r)I2 drdu
t 2k
0
=h
2k
Ifk (T)I2 dX
f
G
2k
it
h2 k
2
0.
Q
for
Proof. Let h be fixed and let r be a fixed integer larger than 2h. Now apply Theorem 3.13 to the case Q = NQ = (x: lxkl < N/21 and
Sl'=(Nr) Q, where N>r. For xEdS2and y1
',
Elliptic Boundary Value Problems
44
Ix _ yJ 2 r/2 > h, so by Theorem 3.13 i
Ilbh ul Im1, (N1) QS Ilullm', NQ*
By periodicity this inequality may be written (N  r)n 1181h
m1. Q
Nn IIuII2
Q.
Now simply divide by N' and let N oo to obtain the result. Q.E.D. THEOREM 3.15. Assume that 0 has the segment property. If u E H. (a), and if there is a number C such that for every Q' CC 12, Ibh ulJm, SZ' < C for all h sufficiently small, then u E H., +1 (ci) and
Proof. First we prove the theorem with m = 0. By the weak compactness theorem there is a sequence lhk} of real numbers with
0 and functions u. E L2 (c') such that 5' u ui in L2 (i') as
hk
k
i = 1,...,n. Clearly IIuj I0,
k
ui 0 dx = lim
C.
For any 0 E Co (c'),
f 8h u/ dx
k+oo Q
k
=lim f uSih k'oo
k
,
dx
uD, 0 dx.
As this relation holds for any c' CC ci, of is uniquely determined in
ci and is equal to the weak derivative D. u there. Hence uEW1
ion
(c)=H1 ioc (a)
Now proceed by induction. Suppose the theorem is true if m is replaced by m  1 . As above, there is a sequence lhk } such that u
$r,
DuinL2(SZ'),i=1,...,n. By Theorem 3.12 D.uEHm(ci'),
k
i = 1,...,n. Hence u E Wm+1 (SZ') by Theorem 2.4. That IID1 uIIm,
SZ'
5 C is immediate. Therefore u E Wm+1 (c) and Theorem 2.2 implies u EHm+1 (c). Q.E.D.
sec. 4
Elliptic Operators
45
THEOREM 3.16. Let Er = lx: IxI < r, x,, > 01, r > 0. If u E L2 (YER), and if there is a number C such that for every R' < R and all h sufficiently small 1I5h
SC for i = 1,...,n  1,
uHIo,
R'
then the weak derivatives D. u, i = 1,...,n  1, exist as functions in L2 (IR) and 11I), ul (o, ER S C,
i = 1,...,n  1,
Proof. The proof is the same as that of Theorem 3.15. Q.E.D. 4. Elliptic Operators Let A (x, D) be a linear differential operator of order P, that is A (x, D) =
I SP
as (x) D«
where the coefficients a« (x) are complex valued functions defined
in some open set Sl in E. (It is assumed that not all coefficients a with Ial = P vanish indentically.) We associate with A (x, D) the (1,...,n) of degree P: homogeneous polynomial in
A'(x,6)=I
a. (x)
.
a_P
The corresponding differential operator A' (x, D) is called the principal part of A (x, D). We now introduce the notion of an elliptic operator.
Definition 4.1. A = A (x, D) is said to be elliptic at a point Y° if and only if for any real e # 0, A' (x°, 6) # 0. A is uniformly elliptic in a domain Sl if and only if there is a constant C such that C1I6IPSIA'(x,6) 1 ,,' cIelP
46
Elliptic Boundary Value Problems
for all real 6 A 0 and all x E Sl. A is strongly elliptic if and only if there is a function C (x) such that
R(C(x)A'(x,.))>0 for all real 6 / 0. Note that uniform ellipticity and strong ellipticity are distinct properties of differential operators. In discussing strongly elliptic operators we will assume that a normalization has been made so that C (x) can be taken to be identically constant. Note that strongly elliptic operators are necessarily of even order. The first result concerns the order of an elliptic operator. THEOREM 4.1. Let A be an elliptic operator of order Pat x0. If 1° the coefficients of A' are real, or if
2°n>3, then f is even.
and let P (.',n) = A' (x°, ).
Proof. Let
P (i',
n) = b0
p
+ bl ( e )
Then
P +...+bg
where b. (6') is a homogeneous polynomial in 6' of degree i, i = 1,...,n. Since A' (x°, en) = P (0, 1) = bo and A is elliptic at x°, bo # 0. Clearly, in case 10 it is sufficient to assume that bo > 0. Let ' be fixed, 6' / 0. If P is odd, P (6', 6n) has the same sign as 6n for sufficiently large I6nL Since P (', en) is real and depends continuously on 4n, there is a real value of 6. for which P (4',n) = 0. This contradicts the ellipticity of A. Hence f is even. In case 2°, for ,' 0 let N+ (6') [N' (,')] be the number of complex zeros 6. of P (e, ) = 0 having positive [negative] imaginary part.
Since A is elliptic, there are no real zeros, so that N+(')+ N(C') = P. We will show that N+(4') = N(4'); from this it follows that P is even.
Note that P ( ',  n) _ ( 1)P P (i', 6.). Thus, any zero of P (l;', 6) is also a zero of P ( 6',  6); hence N+ (f') = N ( '). Now we use Rouche's theorem to show that for '0 zeros N+ (.') in the upper half plane is constant for
0, the number of
' near
0'.
Let
sec. 5
Local Existence Theory
47
I' be a contour in the upper half plane containing all of the zeros of P (eo, en) which lie in the upper half plane. Then P (eo, en) does
not vanish on r. Since P (f', ) is a continuous function of for e' sufficiently near eo IP (e'a,
P (6', n)( < IP (6o,
on I'.
Hence, by Rouche's theorem, P (f', en) and P (60, 6n) have the same number of zeros within F. Applying the same technique to the lower close to 60. Therehalf plane, it follows that N+ (60') = N+ (c') for fore, N+ (6') is a continuous function of 6' for 0. Since n Z 3, 16' E E,, _1: 6' / 01 is a connected subset of En_1; N+ (6') is a continuous:, integervalued function on a connected set, and therefore must be constant. Therefore, N+ (6') = N+ ( 61) = N (c'); it follows that £=N + (6')+N  (C') = 2N+ (6') must be even. Q.E.D. Two classical examples of elliptic operators are the Laplacian A = D 2+...+D 2 and, for n = 2, the CauchyRiemann operator 1/s(D.  iD2).
5. Local Existence Theory
In this section it will be extablished that the elliptic equation Au = f always has weak solutions in small neighborhoods, if f E L2 and mild assumptions are made on the coefficients in the differential operator A. First we treat the case which A has constant coefficients and coincides with its principal part. We shall also momentarily restrict our considerations to periodic f. Recall that Q is the cube (x: (xk( < 1/21. THEOREM 5.1. Let A (D)
I
a« D°` be an elliptic operator
HI=P
or order Phaving constant coefficients. Then for every f E L2(Q) such that f fdx = 0 there is a unique u satisfying Q
uEHp,
Elliptic Boundary Value Problems
48
Au = f,
f udx=0. Q
Denote this uniquely determined u as T' f. Then there is a constant N depending only on A such that IIT* flip Q S N IIfHI O, Q Moreover, if f has the Fourier series expansion f (x)
(5.1)
cr e27'1
.
S
then e
u (x) = (277i)`f
(5.2)
e2rrirr
A
where E' stands for summation over all 6 # 0,1,..., n integers. e Proof. Note first that if f E L2 (Q) has the Fourier series expansion Ice then c(a,,a) = 0 if and only if f fdx = 0. Now supQ
and u = Y' be e2nix't*. Then Au = .' be A (2rrie)
pose u E H
(2rri) fl' A (1;) be 6
e277ix'&.
Therefore, if f is given by (5.1)
As A is elliptic, A ()
and Au=
0, so that b, = ce/A (e). Therefore, u is uniquely determined by the relation (5.2). Obviously, the function given by (5.2) is indeed a solution of Au = f. Finally, as IA (e)I 2 c 1 161 p IIuI IP = (20f [1'
Ic612
5 (2n)t c
_
(2n)fc ilfll
IA (, )I21C12p]''
Ic612]'/2
0 for
Local Existence Theory
sec. 5 Since I
I
I
I
M
and I
I
I
I
49
m. Q are equivalent for all m, the estimate
on T'r f is proved. Q.E.D. It will also be necessary to solve the equation Au = f for functions f which do not have mean value zero. Since in this section we wish only to state that at least one solution u exists, it will be sufficient here to construct a single particular solution of the equation Au = 1. This can be accomplished very simply in the present case of A having constant coefficients and A = A'. For since A is elliptic, the coefficient in A of DP=a 0. If u0 (x) =
then a Now if f E L2 (Q), let (5.3)
µ (f)
XP
XP/A,
cannot vanish: if a = ay, 0'...'0),
P
I
then Auo = 1.
fdx. Q
Then the function f  µ (f) has mean value zero, and u' = T" (f is a solution of Au' = f  it (f). Now define u = Tf, where
(f)
Tf = T' (f  lc (f) ) + µ (f) u0.
(5.4)
Then Au = [f  µ (f)] + 1< (f) = f, and we have the estimate
(5.5)
IITfIIg 0 s N IIf it (f)JI., , + Its (f)( IluolIP.
Q s, and a« is measurable and locally bounded for ja( S f  s. Now weak solutions of differential equations with nonconstant coefficients are defined in the same fashion as weak solutions of equations with constant coefficients; cf. Definition 2.2. Thus, suppose that A (x, D) is ssmooth in fl, s2 Q, and suppose u is a CQ solution of Au = fin 0. Then if 0 E Co (Sl), integration by parts shows that (6.1)
(Au, 95)0, S2 = (u, A* 0)0.
S2
where (6.2)
A* d' 
IQ}SQ
(^ 1)«I D« (a o)
The operator defined by (6.2) is called the formal adjoint of A. Since a« E CI«I, upon performing the indicated differentiations in
(6.2) and using Leibnitz's rule, it is seen that A* is differential operator of order Q with continuous coefficients. In fact, the coefficient of DR q in (6.2) is precisely
Elliptic Boundary Value Problems
52 (6.3)
I
ISe 13 S a
(
1)1°`1 (13) D`13
a« ;
Thereas a« E CI'l e+9, the expression (6.3) belongs to fore, if A is ssmooth, s 2 P, then also A* is ssmooth, and an easy 6131e+s.
computation shows that A*" = (A*)* = A.
Also, (6.3) shows that the coefficient of DS 0 in A* 0 for 1,31 = e ; consequently, the principal part of A* is ( 1)eA' . is just ( 1)g
Thus, A is elliptic if and only if A* is elliptic. Now we make a definition of weak solution based upon (6.1). For convenience we will frequently write ( , ) in place of ( , ) 0. fl. Definition 6.2. Let u and f be locally integrable in Q and let A (x, D) be esmooth and of order e. Then u is a weak solution in f
of Au = f if for all 0 E Co (il) (6.4)
(f, 0) = (u, A* 0).
Now suppose u is a weak solution in SZ of Au = f, and also that f E L 2 (Sl). Then by the CauchySchwarz inequality
I(u, A*O) = I(f, c )I < Ilfllo. sZ He HO, Q
Thus, there is a constant C such that (6.5)
1(u,
A* O)1 S C 110110, 0, all O E Co ([i).
In proving the first .regularity theorems of this section, we shall not even need to assume that u is a weak solution of Au = f, i.e., that (6.4) is satisfied, but only that (6.5) is satisfied. Since for sufficiently smooth coefficients, A = A**, and since A* and A are simultaneously elliptic, in dealing with inequalities like (6.5) we shall replace A by At This is more convenient notation and it gives more general results. We shall also derive results using a generalization of (6.5). To motivate the considerations suppose that u E II ' 0c (SZ), that 0 S j 5 Q, and that A is jsmooth. Then for cb E Co (Q), D°`_R (u. as D°` q) _ (a7 u, D°` c) = (D'3 (i u), ¢), where DS is any
sec. 6
Local regularity of solutions of Elliptic Systems
53
derivative such that $ S a and I13I = IaI  +1, and then we obtain the estimate for 0 E Co (S2) and Ial > 0  j ((u, a« D« qS)I S const I
ID«R
0I I o, ci
S const I cI PJ. St,
where the constant depends only on u, a«, and supp (0). For at S P j we have by the CauchySchwarz inequality that I(u, a« D« (b)I S const IIcl IgJ, 12.
Summing over all a, at S P, we obtain (6.6)
I(u, Ac)I S const IItIIPJ, Q, 0 E Co (ci)
'Note that (6.5) is just the case j = P. It will be shown below that if (6.6) holds and A is elliptic and jsmooth, then u E HJ 1oc (i); cf. Theorem 6.3.
The key to all the following regularity results is now given in a lemma. We have to work hard to establish the lemma; having the lemma, all the other regularity results given in this section are proved with great ease. a« (x) D« be an elliptic LEMMA 6.1. Let A (x, D) I«I=P
operator of order Q in the cube Q and let as E CO and satisfy a Lipschitz condition
Ia« (x)a« (y)l SK Ix  YI.
Let E. be the largest constant such that E0 DIPS IA (0, 01, 6 real. Then there exists a positive number w0 depending on E0, n, and f, such that if Ia« (x)  a« (0)I S w0, then the following assertion is valid.
Elliptic Boundary Value Problems
54
IfuEL2(Q)andifforallvEC' (6.7)
R(u, Av)j S C IIVIIP1.
then u E H 1 (Q) and IIUIII.Q s Y (C + IIUIIo,Q),
where y depends only on Eo, n, P, and K.
Proof. We first prove the result under the assumption that u has mean value zero, i.e., that f udx = 0. Then we extend u to a periodic Q
function on En, so that we have u E Ho. Let i be fixed and consider the function Sh u, which we write Sh u for the time being (cf. Definition 3.3). Note that 8h u E Ho and 5h u has mean value zero. Let vh E HE be the unique solution of A (0, D) vh = Sh u having mean value zero; the existence of vh is guaranteed by Theorem 5.1, where it is also shown that (6.8)
IIVhIIP,Q S N Iish UIIO,Q,
where N depends only on E0, n, and P.
Since HP is the completion of C# with respect to the norm I P. Q, it follows that (6.7) holds not only for v E C#, but also for v E H. I
Now S_h vh E ffP, since vh c He, and therefore we may in (6.7) insert for v the function S_h vh: We obtain I(u, A5_h vh)I .S C
Itsh
VhIIQ'.1.Q.
By Theorem 3.14 and by (6.8) (6.9)
I(u, AS_h vh)I S C IIVhH f.Q
s CN Ilsh UII0.Q
sec. 6
Local regularity of solutions of Elliptic Systems
55
Now since D« 6_h Vh = 6_h D« vh. (6.10)
(u, A6_h vh) =
(u, a« D« (6_h vh) )
I
P
«I p
(u, a« 6_h D« vh).
Now we need a formula analogous to Leibnitz's rule for derivatives: for two functions f (x) and g (x) 6_h (fg) = f (x) 6_h 9 + 9 (x  he') 6h f. Applying this formula in the case f = a« and g = D« vh, (6.10) becomes (u, A6_h vh) = I«dIP (u, 6h (a© D« vh) )
(6.11)
E (u, D« vh (x  he;) « Q
6_h a«)
= i IEP (6h u, a© D« vh) _
(8h 
EP (u, D« vh (x  hei) . 6h a«)
u, A (x, D) vh) E P (u, D« vh (x  he')
6_h a«);
the second equality follows from the periodicity of a«, vh, and u. Next, since A (0, D) vh = 6h u, (6.11) implies (6.12)
(u, AS_h vh) + (6h u, 6h u) = (6h u, [A (0, D)  A (x, D)] vh) II
(u, D« vh (x  he`) 6_h a«).
Elliptic Boundary Value Problems
56
Now let
la, (x)  a« (0)\.
su
Let p = p (n, Q) be the number of derivatives D" of order lal = P. Then the CauchySchwarz inequality applied to (6.12) implies I(u, A6_h Vh)I + IISh ullO,Q
IISh till O,Q (OP HVhIIP Q
+P IIuHIO,Q K IIVhIIP,Q
(We have used 15h a«l S K, a consequence of the Lipschitz condition
assumed on a,..) Combining this inequality and (6.9), (6.8), we obtain (6.13)
IISh ullo,Q S CN l'Sh ullo.Q + PN IISh ullo,Q + pKN hulI O.Q 1lsh ulI O,Q
We choose (jo = 1/2pN. Then Cu 5 cuo, so that after dividing both sides of (6.13) by 115h ullo.Q. IISh uH IO.Q S CN +(1/2)I ISh ul l0,Q + PKN Hullo, Q'
thus, IISh ullO,Q 5 2CN + 2pKN llullo,Q.
Then, if y'1 = max (2N, 2pKN), Ish ul I o.Q S y1 (C + l lul I o,Q), i = 1,...,n.
This result holds for all positive h; as u E HO (Q), Theorem 3.15 implies that u E H (Q) and 1
Hull 1,Q S y, (C +
llu%,Q).
Local regularity of solutions of Elliptic Systems
sec. 6
57
This completes the proof in the case that u has mean value zero. We now eliminate this assumption. This is relatively simple. In fact, wo can be left unchanged. Let u E L2 (Q) and let µ (u) = f udx. We now show that u' = u  p (u), which has mean value zero, Q
satisfies all the requirements of the lemma. Now
(u', Av) = (u, Av)  p (u)
(6.14)
Av x. Q
The quantity f Avdx is a sume of terms of the form f as D" vdx for Q
Q
jai = F. For each such a there is an index i such that D" = D1 DO,
I=e1. Thus,forvEC# f a« D°` vdx = f a« D1 DO vdx Q
Q
f a. 5n DO vdx
= lim
n+o
Q
 f 6' n a«
= lim
h+o
DO vdx,
Q
since as and v are periodic. Thus, the Lipschitz continuity of as 1aiplies
I f as D" vdxl < K I Q
IDO vI dx
Q
S K [f IDa vl2 dx]1
Q
s K I Ivlle,.Q Hence, If Avdxl s PK I ICI I f,.Q
58
Elliptic Boundary Value Problems
Using this inequality in (6.14), together with the inequality IP (U)1 S Ilul l o,Q, we find
(u', Av) I s I(u, Av) I + PK
IIUI1 o,Q I lvl lp1.Q
Then applying (6.7), it follows that for all v E Cm I(u', Av)I s [C + pK IIUI IO,Q]
I lvl lp1, Q.
Since u' has mean value zero and since we have already proved the lemma for such u', it follows that u' E H1 (Q) and I lu'lI1,Q s yl (C + pK I Iullo,Q + Ilullo,Q).
Since (lull 1,Q = Ilu' + k (U)lI1,Q s Ilu'111,Q + Ilullo,Q, it follows that
Ou'II1,Q + IP (U)I 111111.4
IJulI1,Q s y (C + IJulIo.Q),
where y = max [yl,
(pK + 1) + 1]. Q.E.D. (Whew) We can easily, generalize this result to obtain REGULARITY THEOREMS (Theorems 6.26.7). as (x) D°` be uniformly THEOREM 6.2. Let A(x, D) = y1
«Ie
elliptic in J2, and let E be the largest constant such that
EI6I?IA(x,61,xE1t,6real. Let a« be Lipschitz continuous for I«I = f:
a. (x)a«(y)I SK Ixyl; and let a« be bounded and measurable for IaI
I?:
a. (x) I S M.
Let u EL2 (Q) and suppose that for all qS E Co (9)
sec. 6 (6.15)
Local regularity of solutions of Elliptic Systems I(u, AqS)l ,IC
59
ie1.St
I
Then u E H 1 10 ° (12), and for every [1' CC 9 (6.16)
IIulI1,Q'sy (C+ IICII".Q),
'where y depends only on E, n, P, K, M, the diameter of SZ', and the
distance from 9 to f. proof. We make several reductions to simpler cases. First, we may assume A coincides with its principal part A'. For if B = A A', then
I(u, A' q)I s l(u, AO)I + I(u, BO)I
+ IIUIIoM C1 I*le1,Q,
S C I IcS
Where C1 depends only on n, f, and M. Thus, an estimate like (6.15) holds with A replaced by A' and C replaced by C + C 1 I u H 10. . "therefore, if (6.16) were proved in the case in which A has no lower order terms, we would have I
IIuII1,cz'Sy(C+C1 IIullo,,+II u II",Q). And the result would hold also in the general case. Also, since the result is of a local nature, it is sufficient to obtain the estimate (6.16) with 9' replaced by a neighborhood 0 of a fixed X°, as long as 0 depends only on the quantities E, n, f, K and M. And then by a coordinate transformation it is obviously sufficient to
beat the case x0=0and[=Q. Let w o be the number whose existence is guaranteed by Lemma 6.1 $nd let S be a fixed number so small that la. (x)  a« (0)1 s wo for 191 S S, and 0 < S < 1/4; S depends only on K and w0 Let be a fixed real function in C (Q) such that C (x) = 1 for 1x1 S 8/2,
o
C(x)=o for I x l 2 S, 0 C Then for any v CC,, Cv E Co (Q), and upon applying (6.15) with _
0 and any 0 E Co (En) (7.1)
IiI1 E
S YO
(fm1 I0Im,E
n +
Er
I0IO.E n
72
Elliptic Boundary Value Problems
Proof. We shall use the notation W for the Fourier transform of e'X
f 0 (x) E
dx.
n
Integration by parts shows that for 0 E Co (En)
(20nl' 1En
Dk 0
_ (2rr)n/Z
(x) e'X 6
A
(x) i k e i.6
A
Dk
0 En
iekh (0 Therefore, if 4 E Co (En) and laI = j < in,
D O(0_(i6)`0(0Then Parseval's identity gives lD' 012 ,E
_ f 621 A1 (0
2 de
En
n
=
/11 C2a 10
CI 2=E 1 1
(
12 de
+
621 io (G)I2 d6 12>E 1
=E
i01 2 de + Emi En
2>E
I
161
By another application of Parseval's identity, Da 0I2
O.En
sEi
t
2 IO.E
+ YE
m'
whence the lemma follows. Q.E.D.
1
2
l
2m I 2 d
sec. 7
Garding's Inequality
73
Corollary 1. There is a constant yo = yo(n, m) such that for
0<ES1andfor(b ECp(En) E1rn
2 Yll,n1,En S YO (E I0Im,En +
O,En).
Lemma 7.1 is really a convexity theorem since the range on f is not bounded above. Indeed, we have Corollary 2, as follows: Corollary 2. There is a constant yl = y1 (n, m) such that for all t}3 E Co (En) (7.2)
01
I1I,,En s Y1
I0Im/,En
O,En
j
Proof. Let E
AI2/1
O,En
I0I2/1
m,En
Then the two terms on the right side of (7.1) are equal and (7.2) Vesults. Q.E.D. In fact, Corollary 2 implies Lemma 7.1: this is trivial if j = 0 or 1 = m; otherwise, use young's inequality Iabl
< a1 lala + 01 Ibla
r a1 + 61 = 1, a, 0 > 0, taking a = I'I21/m EI(rni)/m, b = m,E n
0
2(rn1)/rn O,En
j(m1)/m
a=m/j, /3=m/(mj). We shall say that Sl has bounded width S d if and only if there is a ne I such that each line parallel to 1 intersects I in a set whose ameter is no greater than d. The following lemma is the POINCARE' INEQUALITY. LEMMA 7.3. If 1 has bounded width S d, then
S ydm'
for all 4) E Co (Q), 0 < j S m  1, where y is a constant de
nding only on m and n.
Elliptic Boundary Value Problems
74
Proof. Let I' be a line parallel to 1, and assume that x° and x° + q are points of 1' fl aQ such that 1' fl S1 is contained in the segment between x° and x° + q. By defining (k to vanish outside H, we can
E C0 (En). Let
assume
f(t)=0(x°+tIgllq) Then f (0) = 0, so that t
f (t) = f f' (r) dr. 0
By the CauchySchwarz inequality,
If (t)I2 < t ft f' (r)I2 dr < d
(7.3)
If, (r)I2 dr. 00
0
Hence
f°° If (t)I2 dt = fd If (t)I2 dt 5 d2 f°° If, (t)12 dt. 00
«.
0
Now express I(kI0.Sl as an iterated integral with one of the integrations taken in the direction of 1. From the inequality above it follows that I0I0
S d2 I0Ii.Si.
Applying this inequality to Di 0, IDi
01
O.Q S d2 IDi
012
Summing over all i, we obtain
I'I1.cIS2d2I0I2.9* Proceeding in this manner,
sec. 7
Garding's Inequality
75
I0Ii.SZ S yd ICI,+i.ii
for0<jSm1. Q.E.D. Although the Poincare'inequality, as it stands, is completely sufficient for our purposes (indeed, we shall need the inequality only in the case that Q is itself bounded), for the sake of completeness we present the following generalization: LEMMA 7.4. Assume that there are positive constants 8 and d such that for every x E S1 there is a point y for which Ix  yI 5 d and {z: z  yl < 81 fl Sl = 0. Then there is a constant c depending only on n and d/8, such that I0I;.Si i
for any
(dc)mi I0Im.ci, 0< j S m  1,
E Co U.
Proof. For any ntuple a = (al,...,a,) of integers, let Q. = (x:
n''(ak + 1)d, k = 1,...,nI. Then E,, = U Q,,. We shall prove something similar to the last estimate with Sl replaced by Q. fl Q. Then, by summing over all a, the lemma will follow for Q. n117akd < xk S
Assume that (k = 0 outside St. Assume that x' E Q. fl 1, and let y be the corresponding point guaranteed by the hypothesis. Since for
Elliptic Boundary Value Problems
76
x E Q., Ix  x' j < d, the triangle inequality gives Ix  yI < 2d. Let S = z: ly  zj < 81. Now integrate 1012 over Q.  S; the integral can be expressed in polar coordinates about y as I0I2 dx S f fed I012 rn1drdo, r Q«"s 28
where c is the spherical measure and E the surface of the unit sphere r = 1 . By (7.3), we have for &< r S 2d I2 dr1 , rn1
012 rn1 < 2d r2d 1 art
8
S 2d
f2d
(2d)n1
a!
1 2 dr 1
art
r2d
S (2d)" 8'
al 2 rn 1 dr 11
1
8
r1
where a0/ar1 is the derivative of 0 in the radial direction from y. Thus, fed
1012 rn1 dr < (2d)n+1 S1n r2d 8
8
2 rn1 dr. ar
Since a0/ar is a particular directional derivative of 0, it must be bounded in magnitude by the gradient of 0; that is, at
12
0 such that IB[u, u]I 2 CI I u I IM for all u E Hm(0), then the GDP has a unique solution.
Proof. We wish to find a function uo E Hm(fl) such that B[0, uo] _ (0, f)  B[0, g] for all 0 C Ca (12). Such a function is provided by the LaxMilgram theorem with F(0) = (0, f)  B[0, g]. Note that IF((k)l < II'IIoiIfiIo + KII(kIIm1Ig.[Im s (IIfIIo + KIIg1I,,,)II0IIm, which implies F is a bounded linear functional on the Hilbert space Hm(fl). Q.E.D. Before proceeding further it will be convenient to prove the following compactness theorem which is a (simplified) version of RELLICH'S THEOREM (cf. Theorem 3.8): THEOREM 8.3. If Sl is bounded and if j < m, then the identity map of Hm(fl) into Hi(!l) is compact.
Proof. We may clearly assume j = m  1. Suppose that I c H" H (Q) is a sequence such that I uk1 I m. i S c. By the definition ffm ([1), there is a sequence {Ok E E C) (S2) such that I uk 'k I I m.H < 1/k. As in the proof of Rellich's theorem, assume thatTcQ = {x:Ixkl < 1/21, and that cbk has been extended to Q by defining Tk = 0 outside Q. By Theorem 3.7 14kI has a subsequence {'k 1 which converges in H._1(Q). Since I
I
m
Elliptic Boundary Value Problems
100
lluk'  ukJI Im1,Q s I1ukt  Ok+ 110kt  rkj1Im1,11 +IlOkt 
01;
whenever a distinction between the two is necessary, we shall indicate which should be considered. We shall write G' = GR, and G" = GR,,, assuming that R' < R and R" = '/2 (R' + R). The symbol will denote a real function which is infinitely differentiable on E. and has its
sec. 9
Global Regularity
107
support in {x: IxI < R}. Note that S need not vanish on the flat part of the boundary of the hemisphere. Now we are ready to prove the fundamental regularity lemma. The
essential tool will be Garding's inequality. LEMMA 9.2. Assume 1°
that the bilinear form B[v, u] =
E (Dav
,GR
I/aaIISm
has bounded (IaaRI,9M) measurable coefficients in GR, and that, for j al = m aap satisfies a uniform Lipschitz condition:
Iaap(x)  aap(y)I S Klx  yI,
x, y E G
that the associated quadratic form is uniformly strongly elliptic: for i; real and x E Q E
aaR(x)i; a+O 2 EIeI 2m
IaI=m IR1=m
2°
3°
that u E Hm(GR) and that Cu E H.(GR) for all CE Co({x: IxI < R}) (Note that this condition is needed only if GR is a hemisphere; it is automatically satisfied if GR is a sphere.); and that there is a number C such that for all 0 E Ca(GR)
IB[c', u]I 1, this method will not work. Instead, we shall use a lemma which guarantees the existence of weak derivatives of a function known to have certain pure derivatives and known to satisfy a certain inequality: LEMMA 9.3. For any r > 0, let Lr = Ix: Ixl < r, xn > 01. Let R be a positive number and let j be a positive integer. Assume
sec. 9 1°
Global Regularity
113
that u E L2(IR), that u has weak derivatives Dfu E L2(1R),
i = 1,...,n  1; and 2°
(9.3)
that there is a constant C and an integer m.2! j such that I(Dn 0, u)I s CI 101
for all 0 C Co (YR ).
Then for every R' < R, u i H1(IR.) and n1
I IUI Ii,IR. < y(C + 1 I ID;UI O,yR + I jUI
where y = y(n, in, R, R,).
Proof. The proof is by transforming the boundary problem into an interior problem by a type of reflection, and then using the local results of Theorem 6.3. First, observe that inequality (9.3) holds for any function 0 E since JET + in Hm(IR), by Theorems 1.7 and 1.8. Moreover, (9.3) holds for any function 0 E Cm(IR) such that for some 8 > 0, O(x) = 0 for R  3 < IxI < R and DnO(x) = 0 for xn = 0, s = 0,...,m (note that it then follows that D°`96 = 0 for x" = 0, Ia(S m): to see this, let
Ox  Ee") for
x E I R8, X. _ E,
91((x) = 0
for all other
x;
in H. (4) as e . 0.
then 0,(x) E Co and 0E Now we extend u to En: let 0
(9.4)
for
IxI
R, xn > 0,
u(x', xn) = 2m+1
:
Aku(x',  k1xn)
for
xn < 0,
k1
where x' = (x1,...,xn_1); here the Ak's are constants which are chosen so that u is sufficiently differentiable (in the weak sense) for xn = 0.
Elliptic Boundary Value Problems
114
In order for the function and its tangential derivatives (that is, derivatives D" with an = 0) to match nicely, it is sufficient to assume that 2m+1
Ak = 1.
E
kl
For the normal derivatives, we must have 2m+1
k1xn)]Ix
AkDn[u(x',
Dnu(x', Xn)Ix 0 =
n
k1
n
0
2m+1
= Dnu(xl, xn)Ix 0 n
Ak( k) k1
';
i.e., 2m+1
1
(9.5)
Ak( k)_8
k1
Thus, in order to have derivatives up to order 2m  1, it is sufficient to assume that (9.5) holds for s = 0,...,2m  1. This imposes only 2m conditions on the 2m + 1 quantities Ak, k = 1,...,2m + 1. Because we will need (9.5) to hold for s =  1 later, we take JAkI to be the (unique) set of real numbers satisfying (9.5) for s =  1, 0, 1,...,2m 1. (The determinant of the coefficients in the system of equations (9.5) is a Vandermonde determinant for the quantities  k1, k = 1,...,2m + 1, and thus is not zero.) Note that the argument  klxn in (9.4) is chosen so that the value of u at any point in SR = (x: IxI < R1 depends on only the values of u at a finite number of points in FR. Thus, u E L2(SR) and I Ivl I
o,sR S YI IvI I O.R.
Moreover, if jai 2m  1, if an = 0, and if D'u exists weakly in'R, then D'u exists in SR and it is the extension (by (9.4)) of D"u on IR To see this, let E Co (SR) and consider
f D"q SR
udx = f
xn>0
D"q
udx + f
xn0
=f x>0
k
ki
f
xn0
where we have substituted  kxn for xn. Recall that an = 0.) Thus,
f S
D°`O
udx = f
R
D°00
udx,
IR
where 2m+1
00(x) = O(x) + 2
kAkO(X',  kXn).
k1
By (9.5) with s =  1, 00(x', 0) = 0; however, (bo is not necessarily a test function on IR, since its support may intersect { x: xn = 01. To circumvent this problem, let p(A) be a infinitely differentiable function on the real line with p(A) = 0 for A S 1, p(A) = 1 for A 2 2. Let E(X) = p(E 1I n I ).
Then, D,/E = 0 if i 4 n, so that
f
SED° c
. udx = f
udx.
Since .00 is a test function on 4, the existence of D°`u on 2R yields
f
ED°`c 0
udx = ( 1)11 f E/ oD"udx. ER
YR
Thus,
f D"O udx = f 4D"00 udx + f R
4
4
(1  <E)D°`q
0
udx
Elliptic Boundary Value Problems
116
= ( 1)II f Cf 0D°`u dx + f (1  QD"c0
udx.
As a tends to zero, the second term on the right tends to zero, and the first term tends to be obvious limit, so that f D°`c . udx = (  1)I°'1 f
O0D"udx
SR
1R 2m+1
f
kxn))D«u(x)dx
kAkc(x',
(c(x) + E k1
TR
_ (1) l"I [f c(x)D"u(x)dx xn>0 2m + 1
+f
AkD°Cu(x',  k1Xn)dx],
E
O(X)
x 0
D2mo , udx + n
xn>0
n
udx + f
D2mLr
xn0
Dnm[ (x`,kxn)] u(x)dx,
x >0 n
where we have substituted kxn for xn. Let, for xn > 0, q/'(x) = t(i(x', xn) 
2m+1 k,
Ak(k)12m./.(x', k
k.1
n
Then
(9.7)
(D 2nm/J, Ct)O'SR
U)0'1R,
(D
moreover, by the assumption that (9.5) holds for s = 1,...,2m  1, we have
DstO*= 0 for xn = 0, s = 0,.... 2m. n Note that also tO*= 0 for R  S S I x I 5 R, xn > 0, some S > 0. Let = Dmt(i*, then, by the remark made at the beginning of the proof, (9.3) n holds for 0: 2mw '/,i
l(Dn
u)O.y RI = I(DnO'u)O, _R S
Cl I
Y*
Thus, from (9.7), (9.8)
J(Dnm
, u)o.SRI s C I I
*112mR
YCI I'f' 112mJ,SR
R.
118
Elliptic Boundary Value Problems
for any t A ECo (SR). For i / n, we can use the fact that u has the weak derivatives Di in SR, together with the estimate (9.6), to obtain (9.9)
(Dm , u)O,SRI = I(DImJo, DIu)O,SR
< IIVII2mj.SR IIDIUIIO.SR S YI IDIUI I.,ERI I'I 12mj.SR
Now let A be the elliptic operator of order 2m given by A=
D 2m;
; I
from (9.8) and (9.9) we see that n1
I(At, U)O,SRI I< Y(C +, I 1 fID,UII O' R)
for all 0 E Co (SR). Thus, the interior regularity theorm (Theorem 6.3) can be applied: for R' < R we have u E Hj(SR) so that, in particular, u E Hj(IR,), and n1 I IUD I1.): R,
S I Iul I j,SR, < Y(C + I1
I IDIUI I O,1R + I IUI I O,YR)
Q.E.D.
One additional lemma is needed before proceeding again to the boundary regularity question. LEMMA 9.4. Let G = (x: IxI O. If u E Hm+1(G) and if Cu E Hm(G) for all CC Co (}x: IxI < R}), then 6DIu E Hm(G) for all C,; Co ({x: IxI < R}), i = 1, 2,...,n  1.
Proof. Let CECo({x: IxI < R})be fixed, and let i / n. Defining u as zero outside G, we have, for x E G, (9.10)
C(X)8'u(X) = 8,(Cu)(x)  6,8(x)
u(x + he1).
sec. 9
Global Regularity
119
Clearly, 8h(Cu) E HO(G) for sufficiently small Ihl. Also, 61C E Coo ((x: Ixl < RI) for sufficiently small Ihl, so that 8' C(x) u(x + he') E HO(G) for sufficiently small lh I, by the assumption on the behavior of u. Therefore, C& 'u E HO(G) for small Ihl.
For suitable R' < R, supp (0 fl G c G'. By the proof of Theorem 3.13, we have for small lhl Ils'(CU)11.,,G = Ilsh(SU)IIm,Gn IISUIIm+1,G
1, we observe that u E Hm +I (G"' ); hence, by Lemma 9.4' o tD,U E Hm(GO1), i = 1,...,n  1, for any CE Co({x: Ixl < R"').
sec. 9
Global Regularity
121
Furthermore, for 0E Co(G"'), B[q, D,u] =
E
(Daq,aaQD1D,u)o,G",
IRS<m
(DaO,D;(aaRDRu))o.G
mj 1). In the case of the sphere, inequality (9.13) holds for i = 1,...,n; this completes the proof in this case, since u E Hm+((G") and by (9.13) and the inductive hypothesis llul lm+j,G" S y(C + IlUIIm.G)'
In the case of the hemisphere, we still must show that these relations hold for i = n. The proof is completed by using Lemma 9.3 to show that D n u q Hm+(1(G'). For E Co(G") (9.14)
a
B[c, u] = (Dn o, IRI<m
n
me .P
DOu)o,GIf
(DaO, aapDRu)a,G".
+
mj 0 there is a constant C. such that I
l U I l2m+k,Sl s y(I IAull
st + Ell H 2m+k,g + CEI lUI I
O,si).
Choose c so small that yE 5 1/2. Then IIUIl2m+k$
s 2y(IIAUI Ik,i + CEllUl IO,1)
and (9.25) holds in this case. The extension to the case in which A' need not have constant coefficients is similar to the proof of Garding's inequality. For, consider the form
for U E C2m+k(SZ). Since the coefficients of A are in Ck('l), this is a Dirichlet form of the type we have considered. We wish to prove a version of Garding's inequality: (9.27)
B(u)
COIIUIl2m+k
A0IIUII0,
for the subclass of functions satisfying: D°`u = 0 on 3l for lal S m  1. We have succeeded in establishing this in the case that A' has constant coefficients. Exactly as in the proof of Lemma 7.9 and its corollary, we can establish (9.27) 1 ocally: that is, (9.27) holds if supp(u)has sufficiently small diameter. Of course, in Garding's inequality we assumed that u E Co (Sl). But the proof showed that we really needed only two facts concerning the class of functions u: (1) the validity of an interpolation inequality, and (2) the fact that for
Elliptic Boundary Value Problems
134
C E Co (En), Cu is in the same class of functions as u. These are both valid in our case. Thus, the interpolation inequality of Theorem 3.4 implies (9.27) holds locally. And the partition of unity argument of pp. 8486 then implies the estimate (9.27) for all U q: C2m+k(n) satisfying: D°`u = 0 on a&Z for Iai < m  1. Q.E.D.
10. Coerciveness
This chapter contains a treatment of some boundary value problems other than the Dirichlet problem. We shall need the following special form of Green's formula, which we have already mentioned (vid. equation 8.2). For simplicity, we shall assume throughout this section that the differential operators and the bounded region & are sufficiently smooth.
THEOREM 10.1. Let A be a strongly elliptic operator of order Q in Q. Then there exist linear differential operators Nk(x, D) for x E 3l, such that Nk is of order k, the boundary of & is nowhere char
acteristic for each Nk, and for all u, v E Q(f1 Cl) (A*v, u)o.SZ  (v, Au)o.,Q = iIo fu NQ1iv .
an do,
where di/c9ni is the exterior normal derivative of order j at 9&1.
Proof. First, consider the case that & is the hemisphere #x: xj < R, xn > 01 and v vanishes for (xi 2 R', where R' < R. For a derivaD,D'ef with i n, an integration by parts gives tive D°` =  (D;v,
D` u)  (v, D«u) = 0, t
since the boundary terms vanish. Indeed, after a1+...+an integrations, we obtain (1)11(D«'v, Dnnu)
partial
 (v, D«u) = 0,
where a' = a  anen. Further partial integrations contribute boundary terms:
Coerciveness
sec. 10 (D« I
«
135
(Da +e n
anl.)
p
v, Dnnu)
v, D n

a 1 . udx', D « vDn p
f Ix'I 0 and x / 0. Note that if h is differentiable, then Dih is homogeneous of degree r  1. The following result exhibits a wide class of CalderonZygmund kernels. LEMMA 11.1. Let h E C1(En  101) and let h be homogeneous of degree 1  n. Then for i = 1,...,n, D,h is a CalderdnZygmund kernel.
Proof. We have remarked that D,h is homogeneous of degree n. Thus, we need only exhibit that Dih has mean value zero. Let p(t) be a nonnegative function of a real variable such that p E C(E1), supp (p) C [1, 2], and fOQ dt = 1. Now an integration by parts shows that
f
Dih(x)
p(lxi)dx =
En
 fE
h(x)p'(IxI) i
t
dx,
n
where p'(t) = dp/dt. Next we introduce polar coordinates x = ra, r = Ix I, in each of these integrals to obtain
f oo f (D,h) (o) p(r)rn1 dadr =  f f r° eo
0
0
h(Q) p
r n 11
(r)ar n1darlr,
which simplifies to f
(11.1)
ear) dr f (Dih)(a)da =  f
w
Since f rlp(r)dr = 1 and since f 0
p'(r)dr f h(a)oida.
p'(r)dr = 0, (11.1) becomes 0
fI (Dih)(a)da = 0. Q.E.D. We shall need also the following basic lemma of Calderon and Zygmund,
which we state without proof. For an elementary proof see A. Zygmund, on singular integrals, Rendiconti di Matematica 16 (1957), pp. 479481.
c. 11
Coerciveness results of Aronszajn and Smith
153
LEMMA 11.2. Let k be a CalderonZygmund kernel, let 0 < E < R, and let (11.2)
k(x), c < IxI < R,
0, Ixj S E, xI
R.
Then if kE,R is the Fourier transform of k. R1 there exists a constant c, independent of c and R, such that for all e C En IkE,R(
I S C.
Moreover, for all e C En there exists k(O = lira &+0
kE,R('
R,m
Remark. The notation k is used here since this function behaves somewhat like a classical Fourier transform, even though k itself does not have a Fourier transform in the usual sense. In the distriA bution sense, k is actually the Fourier transform of k, though we shall not need this fact. As a corollary of this lemma we have LEMMA 11.3. Let k be a CalderonZygmund kernel and let f C L2(En). Let kE,R*f be the convolution kE R(x  Y)f(y)dy.
(kE,R *f)(x) ' f
E
'
n
Then, kE,R*f converges in L2(En) as c , 0, R , oc. If k*f denotes the limit in L2(E), then the following estimate holds: (11.3)
S(2rr)n/2c1'fj I o,E
Jjk*fHHo,E n
n
,
where c is the constant appearing in Lemma 11.2. Proof. Using a familiar fact about the Fourier transform of a convolution,
Elliptic Boundary Value Problems
154
hn (2u)n/2kE,Rf.
kE R*f =
(11.4)
Thus, Parseval's relation shows IIkE,R*f  kE°.R'*fII0,E
= (2i)nlI[kER  k(i,RI]fII
E
n
(277)n f En
IkE °
R(x)
n
kc',r'(x)I2 h
If(x)I2dx.
Using the previous lemma, the integrand in the last expression is bounded by the integrable function 4c2I?(x)12, and the integrand tends to zero a.e. as c, c'  0, R, R'  o. By Lebesgue's dominated convergence theorem, the integral tends to zero, so that by the completeness of L2(E), kfR*f converges to a function k*f in L2(E). By (11.4) and Lemma 11.2,
R(
Ik
* f ( x ) 12 < (21T)nc
12.
If we integrate this expression over En and use Parseval's relation, (11.3) follows. Q.E.D. Now we state a result we shall frequently use in the remainder of this section. THEOREM 11.4. Let h E C1(En  {01), let h be homogeneous of degree 1  n, and let k' be the CalderonZygmund kernel given by k' = D .h. Let (11.5)
C, = f h(a)o1da.
I
Let f be a bounded measurable function having compact support in En. Then h*f E HOC (En), and h*f has weak derivatives given by the formula
D.(h*f) = k'*f + c,f.
sec. 11
Coerciveness results of Aronszajn and Smith
155
Moreover, for a certain constant c depending only on h, I
ID,(h*f) I
IO,En S
cI IfI IO,En.
Remark.! Although the last estimate shows that really D,(h*f) E L2(En), still we must write h*f E H10c(En), since h*f itself may not be in L2(En). Proof,. We have
fEn
Ih(x  Y)I If(y)Idy s max Ih(a)I sup If(y)I f Ial
Y
1
n
unn cr, Ix  YI
1ndy.
Since f is bounded and has compact support, it follows that h*f exists as an absolutely convergent integral, and is uniformly bounded. Let 0 E Co(En). Then (11.6)
f
(h*f)(x)D,O(x)dx = lim 
En
Rioo
ff
h(x  y)f(y)D;O(x)dydx.
0 for any v. Since the set of functions u, of the above form is not of finite dimension, when A takes on all complex values, Lemma 11.7 implies that B cannot be coercive over H.W). Q.E.D. Ngw_a partial converse of the second part of this theorem will be given. The only additional assumption is a requirement on the domain. The proof is taken from K. T. Smith, Inequalities for formally positive integrodifferential forms, Bull. Amer. Math. Soc. 67(1961), so that B[LA,
368370.
THEOREM 11.9. Let be homogeneous polynomials of degree m, having constant coefficients, and let have no common nonzero complex zero. Let (1 be a bounded open set having the restricted cone property. Then the quadratic form in (11.13) is coercive over Hm(U); that is, there exists a constant C such that for all u E Hm(cl) N
(11.14)
Iuf l m,II
S C[kY1
I
1 Pk(D)uH o.SZ +
Iuj j o,St].
Proof. According to Hilbert's Nullstellensatz, if P(t,`) is any polynomial which vanishes at all common complex zeros of P of P1,...,PN, in the sense that there exist polynomials A1( ),...,AN(e), such that Pp = N 7
k1
AkPk. Since the only common zero of the polynomials P1,.... PN
is zero, the requirement on P is simply that P(0) = 0. Notice that if the above formula holds for Pp, then a similar expression is obviously valid for PP', any p' > p. For a proof of the Nullstellensatz, cf. Van der Waerden, Modern Algebra, II, Ungar Pub. Co., pp. 56. Now we apply this result to the polynomial P(6 = e1. We then obtain for some sufficiently large p and for some polynomials Akf k )
162
Elliptic Boundary Value Problems N
E Ak1 1(6Pk(e, j = 1,...,N. k'1
(11.15)
Therefore, if m' 2 pn and if l al = m', then al Z p for some j, so that (11.15) implies that for some polynomials N
Y A'
(11.16)
k`1
lal = m'.
We can certainly assume m' 2 m. (This is actually implied by (11. 16), though we need not use this fact.) Now if Aak(6) is the part of the polynomial A«k(e) containing terms of degree precisely m'  m, then the homogeneity of Pk(0 implies (11.17)
r=
N
E A«k(M(e), lal = m'.
k`1
If we interpret (11,17) as a formula for derivatives, it follows that (11.18)
N
D" =
k l
A«k(D)Pk(D), lal = m',
where A.k is a homogeneous polynomial of degree m'  m. Obviously, it is sufficient in proving (11.14) to assume u E C°°(En), by the approximation theorem, Theorem 2.1. Let 10,1 be a finite open covering of Si and let C, be the cones guaranteed by assumption that Sl has the restricted cone property. If h is the minimum height of the cones C', we may refine the covering (01#, if necessary, to insure that the diameter of 0, is less than h. We have for any x c Sl fl 0, that
x+C,C[1. Now suppose first that u has its support in 0,. Since the diameter of 0, is less than the height of C,, it follows that if x E 0i the cone x f C1 has the spherical part of its boundary outside 0,, and therefore outside the support of u. Thus, we may apply the Sobolev representation formula (Theorem 11.5) to obtain for each fixed x c Sl fl 0, (11.19)
u(x) =
Y
11,
fci c«(y)D°`u(x + y)dy,
sec. 11
Coerciveness results of Aronszajn and Smith
163
where 0«(y) E C°°(En  {01), 0.(y) is homogeneous of degree m'  n, and supp [O.(y/lyl)] CAi, where At is the portion of the unit sphere subtended by C1. Now utilizing the formula (11.18) for D", the representation (11.19) becomes N
(11.20) ;i u(x) =
E
Y_
k1
fC
c«(y)A«k(D)Pk(D)u(x + y)dy. 1
We wishto ntegrate by parts in (11.20), moving all the differentiations in A.k(D) so that they apply to 0«. Note that there are no boundary contributions, since u(x + y) vanishes for y near the spherical part of
Xi and 0a(y) vanishes for y near the lateral part of aC1. Thus, since A«k(D) has constant coefficients and is homogeneous of degree m'  in, (11.20) becomes
(11.21)
N
I
u(x) _
k I
lalm'
f A ,k(D)O«(y) . Pk(D)u(x + y)dy
Let (_1)m ' m Aa (D )Oa (y). k
Ok(y)
I«I m'
Then Ok is homogeneous of degree m' n  (m'  m)  in  n; in particular Ok is integrable over C., so the integration by parts leading to (11.21) is fully justified, and we have N
(11.22)
u(x) = E f
k1 C.
Ok(y)Pk(D)u(x + y)dy, x E Sl fl 0,.
i
Also tAkE C°°(En  101) and supp [fk(y/lyl)] CA,. We now need to apply Theorem 11.4. To do so, it is necessary that the formula (11.22) be modified to contain integrals over all of En,
not just over C,. To accomplish this, let
Elliptic Boundary Value Problems
164
Pk(D)u(y), k(y) _
y E SZ,
o, y' fl.
Also let C1 be the infinite cone which subtends the protion A, of E. Since supp (u) C 0,, u(x + y) vanishes for y E C,'  C,, and the integration in (11.22) may be taken over all of C,' with no change in the value of the integrals. Next, the integration may be performed over En, since z/ik(y) vanishes outside C. Thus, (11.22) becomes N
U(x)
k'1 E n
t/!k(y)Wk(x + y)dy, x c a n o,,
Note that we must restrict x to be in SZ n 0, for the validity of this formula. Now replace x + y by y in this formula to obtain N
(11.23)
u(x) = Y f k'1
E
ck(x  y)wk(y)dy, x c 0, n Si,
where t/!k(y) = t" k( y). Since t/!k E CO0(En  (01) and 1/!k is homoge
neous of degree m  n, we may differentiate formally the relation (11.23) a total of m  1 times to obtain for 191 = m  1 (11.24)
D"U(x) =
kN f
c? k(x  y)W k(y)dy, x c 0, n fi, n
/
where t/iok = D' Vik is homogeneous of degree m  n  (m  1) n+ and is in C°°(En  101). This formula is valid because &/ip k(y) is integrable near y =0. By Theorem 11.4 it now follows that each of the functions wk(x) = f z/isk(x  y)wk(y)dy has weak derivatives of first n
order which satisfy
IID'((ASk Wk)O'En S CI
IWkIlO,En
Rec. 11
Coerciveness results of Aronszajn and Smith
165
Thus, (11.24) implies N
I
I D,
Dsu 0.0 S ki
I
I DI(T a 'k)I I 0,E"
N
cllwkII0,E n
k1
N
=c
I IPk(D)uI IO'n.
Since this holds for all 0, 101 = m, and all i = 1,...,N, we obtain N
IUIm,flScl kIl IIPk(D)ullo.cz By the interpolation result of Theorem 3.4, we have N (11.255)
C2[
IIUI Im.SZ
I IPk(D)ul I O,SZ + Ilul I 0.0],
which is just (11.14) for the case in which supp (u) C 01. Thus, we have established (11.14) locally. The procedure for obtaining the global result is precisely the same as for obtaining Garding's inequality from the local result in Lemma 7.9; cf. pp. 8183. Therefore, (11.14) is established. Q.E.D. We now extend this result to show that the existence of certain weak derivatives of u implies the existence of all weak derivatives ,up to a certain order.
THEOREM 11.10. Let P1(6,...,PN(6 be homogeneous polynomials of degree m, having constant coefficients, and having no common nonzero complex zero. Let 1 be a bounded open set having the restricted cone property. Let u EL2(0) and let Pk(D)u exist weakly and belong to L2(9), k = 1,...,N. Then u E Hm(SZ), and N
I IUI Im.SZ S C[
k
I
I
I IPk(D)ul 1,,,n + I Jul l o.&.
Elliptic Boundary Value Problems
166
Proof. Obviously, once we have shown that u E Hm(Sl), the previous theorem implies the estimate on I IuI Im.St Thus, we need only show that u C Hm(SI). It is easily seen that u E Hm `(SZ). For, extend u to vanish outside 0, and let uE = JEu, where J. is the mollifier introduced in Definition 1.7. Then uE E Co (En) and, by Theorem 2.4, uE . u in L2 '"(0), Pk(D)u( Pk(D)u in L2 `(Q), k = 1,...,N. If S is a sphere such that S CC 0, then (11.14) implies N Lk£1 I IPk(D)uEl I O,S + I IUEI I O,S].
Cr
IIuEI
Im,S S
Therefore, it follows that u E . u in L2 (S) and I IuEI Im.sis bounded as c  0. By Theorem 3.12 it follows that u E HM(S). Since S is any sphere whose closure is contained in 2, it follows that u E Hm `(S1). Now let 10',1 be a finite open covering of (Sl and let {C,} be the associated cones such that x + C . C 0 for x c 1 fl 0',. By approximating each open set 0; by finite unions of spheres contained in 0i, we may actually assume that there is a finite open covering 0.} of an such that each set Ot is a sphere whose diameter is less than the height of C1. Let 12, be the union of all the cones x + C; for x c f2 fl O. Obviously, Sly itself has the restricted cone property. By decreasing the size of the sets Of, if necessary, we can also assume that for any fixed unit vector e in the cone C,, the translated set 1i,E = Ee + [1, satisfies [2, CC SZ, if c is sufficiently small. Let uE(x) = u(x + E ) for x E SZ.. Since x + E6 E Q,..E for x q [,, and since we have shown already that u C H10 °([2) C Hm(Slf,f), it follow; : that uE C Hm(Sl). Moreover, we have the estimate NN
I IUEI Im.sl. S C[ G II Pk(D)uEI I O.Sl. + I IuEI I O.Sl J, r
k'1
which follows from (11.14) since Sl; has the restricted cone property. But then it follows, since uE is just a translation of u, that N 11
UEI 1 m,[, .S C1 k11 1 IPk(D)uIIO,0 + I Jul I 0,I2].
sec. 11
Coerciveness results of Aronszajn and Smith
167
Therefore, when c . 0, uE  u in L 2 (& .) and I I uEl l m,g is bounded. Applying Theorem 3.12 again it follows that u E Hm(Qi) CHm(cI n 0). Since this holds for each i, it follows that D"u E L2(&) for I a I S m. ) whose D"u belong As mentioned op page 10, any function in W10C m to L2(&1), laf S m, is in Wm(cI). By Theorem 2.2, it follows that uE Q.E.D. Thisresult will prove extremely useful in section 13. In that appli
catjon, it will prove essential that no assumption is necessary on the smoothness of 4. This emphasizes the importance of Smith's proof, since Aronszajn's original proof required a smooth boundary. Specifically, we shall need the foll owing corollary of the theorem. Corollary. Let 12 be a bounded open set having the restricted cone property and let u E L2(2). Suppose that Dku exists weakly and belongs to L2(&1), k = 1,...,n. Then u E Hm(f1) and for some constant C depending only on [1 and m n
I IUl Im.s s
I IDk ul I o.SZ + I Ivll o.01
For the case of variable coefficients we have THEOREM 11.11. Let [1 be a bounded open set having the restricted cone property. Let P1(x, D),...,PN,(x, D) be differential operators of order m whose coefficients are bounded in il and for which the coefficients of the principal parts Pk (x, D) are continuous in El. Assume that 1°
for each x°E Cl the polynomials Pk(x°, D) possess no common nonzero real zero; and
2°
the polynomials Pk(x°, D) posses no comfor each x° E mon nonzero complex zero.
Then there exists a constant C such that for all u E H (Cl) N
(11.26)
IluI Im.C < C[kIl IlPkuIIomC + IIul10,C1.
Elliptic Boundary Value Problems
168
Proof. First we will prove the theorem for a neighborhood of the boundary. Let x° E dG. Then the polynomials Pk(x°, ), k = 1,...,N, have no common nonzero complex zeros. Let 0 be a small neighborhood of xa since SZ has the restricted cone property, we may assume that u = 0 n fZ has this property. By Theorem 11.9 there is a constant C1 such that for u E HM(U) N
IIuIIm.U S C1(k
IIPk(x°, D)uII0,U + Ilullo,U).
Now
Pk(x, D)u =
E Pk,a(x)D°`u °C Ism
= Pk(x°, D)u +
[Pk., (x)  Pk,«(x°)]D"u I°CI'm
P,,,(x)D°`u.
+
I° I<m1
Hence, for some constant C2, IIPk(x° D)ullo.U 5 IIPk(x, D)uII0,U + C2[sup
lPk,a(x)  Pk,, (x')I
XC U
I°GIm
+ lull,,,1,U]. Thus,
N IlulIm,U
S C1 kEl IIPk(x, D)ullo.U
'lull
m,v
sec. 11
Coerciveness results of Aronszajn and Smith + NC1C2 sup
169
lPk,a(X)  Pk,a(xo)l I lul lm U
x1 U
lalm
1S kSN
+NC1C2
Ilullm1 +
Clllul lo,U
$y 'the interpolation inequality, there is for c > 0 a constant CE, such that
NC1C21lullm,,U
s Ellullm,U +
CEIIuHIO,U
Furthermore, since, for lal = m, pk,a(x) is continuous on rl, U can be taken small enough that NC1C2 Sup .Cu
lP" c(X)  Pk,a(X )l < E.
1«1m
1SkSN Thus,
N IJul. lm,u s C1
k
l IIPk(x, D)ulIO,U + 2EIIulI.,U + (C1 + CE)1IulIO,U
Choose c < 1/4. Then we have N
I lullm,U S
IIPk(x, D)ullo,u + llul lo,ul,
where C = 2(C1 + CE). Now let °1,...,OV be a finite covering of 3S1 by
such sets 0; thus, for u E H. (n n 0,), N
(11.27)
Ilullm.uno1 s V
Let 00= U01. 11
C[ Y1 IIPk(x,
k
+ llullo,&n0 l D)ullo,nn01
Elliptic Boundary Value Problems
170
Let U CC SZ' CC S1, where U is chosen such that 1  00 CU. Let N
I IPk(x D)uI10 ri.
B(U) =
Then B(u) is a uniformly strongly elliptic quadratic form on U. For, since Pi(x, 6),...,PN(x, 6 possess no common nonzero real zero for x E SZ, N
Pk(x, 612
x E W,
0,
161=1
Hence, for x E U and 161 = 1 this quantity is bounded away from zero.
It follows that there is a constant E0 such that for real band x E U, N
kEl IPk(x, )I2 z E0I612m.
Furthermore, the coefficients of B are uniformly continuous on U. By Garding's inequality there are constants yo > 0 and Ao 10 such that for u E H°(U) RB(u)> yo IJulIm,U
AoI ul
2o,U'
in other words,
N IlulIm,U

y0 1 k
1
1
Pk(x, D)ulIO,U +Aoyo 11IulIo,U
By the CauchySchwarz inequality, it follows that there is a constant C such that IIuII..U
 Cl
N
IIPk(x, D)uII0,U + IIuIIO,u].
As above, the interpolation theorem yields
sec. 11
Coerciveness results of Aronszajn and Smith
171
N IIUIIm,U
s C[kYIIIPk(x, D)uIIO.U + IIuIIo,u].
On combining this inequality with (11.27), it is seen that we have actually proved a local version of (11.26). As in the proof of Garding's inequality, the global version now follows from a familiar argument utilizing a partition of unity. Q.E.D. The original result of Aronszajn can be expressed as follows. Sup'pose,SZ is of class C' and consider the condition 3°
for each x° E r3Sl let n be the unit normal vector to dQ at x°. For each real tangent vector 6 at x° the polynomials in 7 given by Pk(x°, 6+ rn) possess no common nonzero complex zero.
The theorem of Aronszajn states that if Cl is of class Cm and Cl is bounded, then the conditions 1° and 3° are necessary and sufficient for the validity of (11.26). We shall not give the proof of this result.
Remark. The estimates guaranteed by the preceding theorems can be extended to more general classes of norms: for instance, LP norms and norms involving the Holder continuity of derivatives of u. We conclude this section with an application of Sobolev's representation formula and the CalderonZygmund theorem already mentioned. The theorem will not be needed in the following but is of con
siderable interest in itself. THEOREM 11.12. (CALDERON'S EXTENSION THEOREM). Let Cl be a bounded domain having the restricted cone property. Then there exists a bounded linear transformation & of Hm(T) into Hm(E,,) such that for every u E Hm(C) the restriction of iSu to Cl coincides with u.
Proof. We prove the theorem first for u c Cm(Cl) fl Hm(T).
Let }0,}" 1 be a finite open covering of r3Cl with associated cones {C, } such that x + C, c Q for x c Cl f1 0,.. As in the proof of Theorem
3.10 we add to this covering an open set 00 cc Q such that { 0i }v o is an open covering of Cl; let Co be a sufficiently small cone such that x + Co CU for x E 00. We can arrange this covering such that
Elliptic Boundary Value Problems
172
the diameter of 01 is less than the height of C1. Next let 0; cc 0, be
is still a covering of I. Let (C1 be a partition of unity subordinate to (01}, such that 1 = 1 on I. Then u = such that }Q1,1l o V
io
1"0
ui, where u, = uC1 is in C°`() and supp(u,) c 01. To extend u to
En it is clearly sufficient to extend each u1. By the Sobolev representation formula we have for x i SI fl O1 (11.28)
I f c",(y)Du,(x + y)dy,
u,(x) = I
i
m
where 0", E C°O(En  10}), 0", is homogeneous of degree m  n, and supp [0",(y/I yI)] (:A,, the portion of I subtended by C1. As in the proof of Theorem 11.9, we shall arrange things so the
integration in (11.28) is taken over all of En. First, let (11.29)
D"u1(y),
y E a,
w",(Y) =
yE9.
0,
Then (11.28) can be written (11.30)
u1(x) =
I f
",(y)w",(x + y)dy, x E S2 O.
I"I m E n Now let >1i, E Co (En) satisfy i&,(x) = 1 for x E 011, t0i1(x) = 0 for x
01.
(11.31)
Since u1(x) = 0 for x Q 07, (11. 30) implies Y
u,(x) = 0,(x)
I"Im
fE
95«1(x  Y)w",(Y)dy,
y). Now let
Here we have set ¢«1(y) (11.32)
v,(x)
,(x)
If
1"1m En Then we set
x E Il.
n
c«,(x  Y)w",(Y)dY,
x E En.
sec. 11
Coerciveness results of Aronszajn and Smith v
v(x) = Y v1(x),
(11.33)
173
x E En.
10
By the very construction, v(x) = u(x) on SZ. Now as in the proof of Theorem 11.9, and by Leibnitz's rule,
(11'34)
Dav1(x) _
(a)DRYV 1(x) Y 0, let g = (A 1 
Then the two relations imply
(A1T)g=0,
 1f.
T)k
1
Elliptic Boundary Value Problems
194
(A2 
T)k2g
= 0.
The first follows from the definition of,,g and the second by multiplying the second equality in (12.21) by (AL1  T) 1 . But now it is seen that Tg = A1g, so that l(A2  T)k2g = (A2  A1)k2g 0. Since Al A2,
g=O,or,(A1T)
k
fk 0. Ifk11=0,then f=0;ifk11>0,
then we obtain (A1  T) 1 f = 0. Proceeding in this manner, it follows that f = 0, and the lemma is proved for v = 2. Now suppose the lemma to be valid for v  1, v , 3; we then establish the result for v. We have (AIT)kIfi=0,
(12.22)
11.
But since f E R(P), it follows that f = P2f, and we have from (12.28) 1 < IIP(Pf  Qf)I I < I IPI I 1(P  Q)fI I < IIPI I IIP  Q11. This contradicts the hypothesis. Therefore, R(P) has dimension not
greater than that of R(Q). Q.E.D. As a corollary we have LEMMA 12.16. Let P be a nonzero projection on a Hilbert space, and let IPkI be a sequence of projections such that lim I IPk  PI I = 0. k+oo
Then for all sufficiently large k, the ranges of Pk have the same dimension as the range of P. Proof. For sufficiently large k we have I IPk I I < 2.1 IPI I and also and the ;l1Pk  P I I < 1/211PI I  1 . Therefore also I IPk PI I < I IPkH result follows immediately from the lemma. Q.E.D. THEOREM 12.17. Let T be a transformation having finite doubleform on a Hilbert space X, and let {A( be the sequence of character4stic values of T, each repeated a number of times equal to its multiliplicity. Then for all A E pm(T), the modified resolvent TA has finite double norm, and (12.29)
Tr(TTA) = A12, (X 1

) + tr(T 2) 1
t
2
AI
Elliptic Boundary Value Problems
198
Proof. First note that since TA = T(1  AT1., Theorem 12.10 implies that Tx has finite doublenorm. Let be an orthonormal basis in X, and let t'j Let TN be the transformation defined by (12.18). As shown in the proof of Theorem
12.11, IIITTNIII0 Now TN is essentially a transformation on a Hilbert space of dimension N, and so TN has characteristic values AJ,N, of which there are at most N, counted according to their multiplicity. Moreover, using a triangular representation of the matrix of TN, the formula (12.29) for TN is obvious; thus, (12.30)
Atr(TN(TN)A)
(A
1
j,N
_A
1
j,N
Note that in this finitedimensional case,tr(T2 _ Since the trace function is linear, we have
A2
tr(TTA)  tr(TN(TN)A) = tr[(T  TN)TX] + tr[TN(TA  (TN)A)]
Applying (12.19), we obtain (12.31)
Itr(TTA)  tr(TN(TN)A)I s IIIT  TNIII IIITAIII +I
I TN I
I
(TN)XI
I I
1
1
.
Now
T,A  (TN)A = T(1  AT)'  TN(1  ATN)r =(TTN)(1AT)'+TN[(1AT)'(1ATN)1]
=(TTN)(1Al)7'+TN(1_AT)i[1AT N(1AT)]'
(1ATN)1 = (T  TN)(1 
AT)1 + ATN(1

AT)'(T
 TN)(1  ATN)
sec. 12
Some results on Linear Transformations
199
Therefore, Theorem 12.10 implies IIS[11(1AT)1II+IAI IIITNIII 11(lXT)111
I TA  (TN)A
,
I I(I  ATN)1I I ) I I T  TNI I I I
Therefore, by (12.3), III T,  (TN)AI I I  0 uniformly for A bounded and
a fixed positive distance away from characteristic values of T and TN. Hence, (12.31) implies that tr(TN(TN)A)  tr(TTA), and (12.30) implies (12.32)
Atr(TTA) = lim It N4m
(1 Aj,N  A
1
),
Aj,N
the limit being uniform for k bounded and a fixed positive distance from {At{ and
tA..NI.
Now some results of operational calculus are needed. See, for instance, Taylor, Introduction to Functional Analysis, Wiley, 1958, pp. 287307. Let A. be fixed. Since the characteristic values of T are isolated, a circle of sufficiently small radius 2E with center X1 contains only the eigenvalue X1 of T. The result of operational calculus we shall need is that the operator (12.33) 21ri
f
(µ T)1dµ
is a continuous projection onto the generalized eigenspace M(AJ).
Likewise, the operator (12.34)
iff
21ri
(µ  TN)1dµ
IµA, 1IE
is a continuous projection onto the direct sum of the generalized eigenspaces of TN corresponding to eigenvalues of TN within a distance of c to X7'. Now since (µ  T)1 exists for Iµ  k7 1I = E, also, for all sufficiently large N, (µ TN)1 exists for Iµ  A7 11 = E. Moreover, (µ  TN)1 converges uniformly to (µ  T)1 on the circle
Elliptic Boundary Value Problems
200
Ip  A 1I = E. All this follows since I ITN  TI I  0.
But this implies that the projection (12.34) converges uniformly to the projection (12.33). By the corollary to Lemma 12.15, it follows that the total multiplicity of all eigenvalues of TN within E of X7 1 is the same as the multipli(City m of A 1. Moreover, by taking c smaller and smaller, it follows that among the eigenvalues of TN there are precisely m (counted according to multiplicity) which converge to AJ 1. Therefore, the characteristic values Ia/.N I can be arranged in such a manner that A/.N A/, for all A1. In case T has only a finite number of characteristic values, the remaining characteristic values of TN must tend to infinity. For, if they had a finite limit point, that limit point would necessarily be a characteristic val ue of T. This follows since if (p  T)1 exists, then also (A  T)1 exists for IA  pl < 2E (some c > 0), whence also (A  TN)' exists for IA  pl < E, all N sufficiently large. Now let (12.35)
Atr(TTA).
F(A)
Note that since 1
(12.36)
A
1
and since
1/
I2
< IIITIIIZ
(cf. Theorem 12.14), it follows that the series in the definition of F(A) converges uniformly for A in a compact subset of p .. (T). By (12.32) and the previous discussion, we have for any integer n F(A) =
I /Zn X1 '\  1)  lim N*o (
1
/
1Zn
(
1
1.N 
X
1
I.N
),
Some results on Linear Transformations
sec. 12
201
the convergences being uniform in any compact subset of pm(T). Therefore, F(k) is an entire function of A. We shall estimate F(A) for I Al < r, taking n so large that IA/ 12 2r, 2r for j 2 n. Thus, I AI, N
I
by (12,36)
S/2n l
+ lim sup I
A/  AI 1A/I
N 01M
A
IA1
j,N Al IAI,NI
S jZnLkL + lim sup I/>IAI 1% 2Ixj12
Nroo
< 21A1(IIITIII2 + N
1 1A.,N12
IIITNII12)
= 41IITI!I2IA!,
by Theorem 12.14. Therefore, F(A) is an entire function satisfying IFU)I S const IAI for all X. Hence, F is linear: F(A) = aA + b, some constants a, b. Letting A , 0 in (12_35), it follows that b = 0. Thus, dividing (12.35) by A gives A)A, tr(TTA) = a
Letting A e 0 in this relation, we obtain
1  tr(T 2) =a. A2 I
Q.E.D.
Part 3. HILBERTSCHMIDT KERNELS
We shall now give two results identifying the class of operators of finite doublenorm in case the Hilbert space is L2(fl), for some measurable subset SZ of En .
Elliptic Boundary Value Problems
202
THEOREM 12.18. LetK(x, y) be a measurable function on SZ x 0,
square integrable on Il x 0i.e., K E L2(1l xIl). If f E L2(SZ), then
f K(x, y)f(y)# converges absolutely for almost all x in Il, and represents an element Tf in L2(1l). The transformation T is a linear transformation having finite doublenorm on L2W), and (12.37)
IK(x, Y)I2dxdy.
I IITII I2 = SZxSZ
Proof. Fubini's theorem shows that, for almost all x, the function K(x, y) is measurable for y E Sl, and
fIl
IK(x, Y)I2dy
N1 or j>N1 00
n/2 and let u E Hm(0). Then u can be modified on a set of measure zero so that u E C0(SZ). Moreover, there exists a constant y, depending only on SZ and in, such that for all u(x)I s
(13.1)
y(lul01(n/2m)lu1m/2m + lulo).
Proof. By Theorem 3.9 it is enough to prove the estimate. According to Theorem 3.9, (13.2)
u(x)I < ylr(m1/in)(Iulm + rmIul o),
r 2 1.
If Iulm S ulo, then, taking r = 1, we obtain from (13.2) yl(IuIO(n/2m)Iuin/2m + Iulo\
lu(x)I
If
IuI
m > lulo, then take rm = ulmlulo1, so that (13.2) implies
lu(x)I s y1(IulmIul
1
1+(n/2m)(Juim + Jul.)
2y1lul0_(n/1m)lulm/2m =
2y1(lul1(n12m)lulm/2m + lulo).
Q.E.D.
Remark. An easy argument based on Young's inequality shows also that (13.1) implies (13.2). The argument is the same as was given on page 73 to show that the inequality (7.2) implies (7.1). As an immediate corollary we have LEMMA 13.2. Let m > n/2 and let u E Hm(SZ). Then there exists a constant yS, depending only on SI and m, such that, after modification of u on a set of measure zero,
Elliptic Boundary Value Problems
210
(13.x)
IU(X)I
lull
Y.Ilul
10(n/2m)P
X E ri .
Definition 13.1. The smallest constant ys such that (13.3) is valid for all u E Hm(SZ) is the Sobolev constant of Q.
Likewise, we can prove a similar result concerning interpolation inequalities. This result will not actually be used until the next section. LEMMA 13.3. There exists a constant y, depending only on SZ and in, such that for u E Hm(S1) and 0 S j S in,
I°ly[Mo(;/m)Iulj m+Iulo] 2ylul0
IUIIm/m.
Proof. The proof follows from Theorem 3.3 by choosing the parameter a appropriately; the procedure is just the same as given in the proof of Lemma 13.1. Q.E.D. Now we turn to the consideration of linear transformations which will later play the role of inverses to differential operators. LEMMA 13.4. Let T be a bounded linear transformation on L2(1l), such that the range of T is contained in Hm(IZ), some m > 1. Then T is a bounded linear transformation of L2(S1) into Hm(SZ).
Proof. By the closed graph theorem, the result will follow if we show that T is closed, considered as a mapping of L2(SZ) into Hm(Q). Suppose then that JCL 2(Sl), Uk u in L2(Sl), and Tuk v in Hm(SZ). Then, since T is continuous on L2(&1), Tuk ) Tu in L2(1l). But a fortiori Tuk  v in L2(1l). Therefore Tu = v. Q.E.D. Definition 13.2. Let T be a bounded linear transformation on L2(S2), such that R(T) dHm(Sz). For 0 S k S m let llTllk =
l Tl k
kL2(e).II1IIo.Q1
SUP
2(n)' l Ill l o.Q' I
I Tf l
I k.I1
Tf I k,
sec. 13
Spectral Theory of Abstract Operators
211
By Lemma 13.4 it follows that these norms all are finite; of course, is not a norm, but a seminorm, if k > 0.
TI k
THEOREM 13.5. Let T be a bounded linear transformation on L2(1), such that R(T) CHm(SZ), where m > n/2. Then T has finite doublenorm, and the following inequalities hold: (13.4)
IIITIII
(13.5)
IIITIII
n/2. Let T be the unique linear transformation of L2(S2) into V satisfying B[v, Tf] _ (v, f)O,SZ
(13.11)
for all f E L2(Sl), v V. Then the half plane {A: RA < 01 is contained in pm(T), and the negative real axis is a direction of minimal growth of the modified resolvent of T. If the characteristic values ( Al of T are arranged in order of increasing modulus, then c0 IA,I >
(16ySISZI)m/n
'
m/n
Proof. The existence of T is a consequence of the LaxMilgram theorem (Theorem 8.1). If in (13.11) we substitute v = Tf, we obtain (13.12)
R(Tf, f) = RB[Tf, Tf]
z c0IITIIIm,0 0.
Since T is bounded as a mapping of L2(cl) into L2(0), it follows that for sufficiently large At, A is in the resolvent set of T. Therefore, Theorem 12.8 shows that each A with negative real part is in the resolvent set of T. Therefore, (1  AT)1 exists for all A such that RA < 0, since 1  AT =  T), and ERA1 < 0. To obtain the estimate, note that if u = TAf = T(1  AT)1f, then, since T and (1  AT)1 commute, (1  AT)u = Tf, so that u = T(Au + f). Therefore, (13.11) implies that for all v E VA(A1
B[v, u] = (v, Au + f)7.
216
Elliptic Boundary Value Problems
i.e., B[v, TAf] = (v, ATAf + f),
`Rx < 0.
Now, if we set v = TA(, then B[TAf, TAf] = AIITAfII o + (TXf, f).
Therefore, coI
I TAf 112 s RB [T,f, TAf]
_ RAIITAfHIo + `JZ(TAf, f) < .1RAI(TAfIl0 2
+ ITAfIIOIIfII0
Dividing by I I T,Af I I o, we obtain
IITXII0< c
1
,Rx 1/2(2n  n) _ n/2, or, that m  [n/2]  1 > n/2, or, that m > n/2 + [n/2] + 1. But
this is precisely the assumption made on m. Therefore, Theorem 3.10 applies. Thus, K(x, x) is uniquely determined for x 0, and we have an estimate corresponding to (3.19). Since do, the Lebesgue measure on the diagonal of E. x En, in our case is just 2 1/"dx, (3.19) (with a = 0) becomes in our case
Elliptic Boundary Value Problems
224 [
IIK(x, x)I 2dx
0 and if the assumptions of the theorem hold for the open set all and an operator To on L2(atZ), and if K(8)
Elliptic Boundary Value Problems
226
is the kernel corresponding to T(a), then
an/2[f
IK(a)(x, x)I2dx]'/,
C1. Q.E.D.
,
sec. 14
Eigenvalue Problems; The SelfAdjoint Case
239
Part 2. EIGENVALUE PROBLEMS FOR ELLIPTIC EQUATIONS
Now we are ready for the fundamental result of this section. Because of the generality of the situations to which the theorem applies, the theorem requires quite a long statement. As in section 13, it will be assumed throughout the remainder of section 14 that SZ is a finite union of disjoint open sets, each of which has the restricted cone property. THEOREM 14.4.
Hypothesis. T is a bounded linear transfor
mation on L 2(SZ) such that both the range of T and the range of T* are contained in Hr,,(SZ), where m' > n if n is odd, m' > n + 1 if n is even. The direction ei0 is a direction of minimal growth of the modified resolvent of T. There exists an open set go contained in Q and an elliptic operator P(x, D) of order m' in go of the form P(x, D)
_
a,(x)D°`
where a. E C °(SZ °). 10
For x E go, for all real ' , and for all complex A that argA=0, P(x,
2°
For any x° I go and any positive c, there exists a neighborhood U of x°, U CSZo, and a constant CE, such that if a,,(x°)D',
Po(D) _ IAI°m'
then for all f E L2(SZ) (14.16)
IIPoTf  fIIo,U s IIfJJo,SZ + IIP*T*f  fIIo,U S EIIfHIo.SZ + CEIIT*fllm'1,SZ
where Po* is the formal adjoint of Po.
0 such
Elliptic Boundary Value Problems
240
Conclhsion. For A E pm(T), TA is an integral operator with a HilbertSchmidt kernel KA E H (S2 x fl), where e = m'  [n/2]  1. The kernel KA has a trace KA(x, x) E L2 12) on the diagonal of 0 x Q. For A E F_(O, 0), Ix! > °°,
[f IKA(x, x) I `dx]y = 0(IAI
f
K,(x, x)dx = cIAI cn/m')1 + o(IAI (n1m')1),
Q0
where c is a constant depending only on P, 0, and f2o. Indeed, if fle(x) _
(2n)n f
EnP(x, i  er e
and if 0S is the, subset of 00 consisting of points whose distance from 1 o is greater than S, and if Q,5 CS1 CCS2o, then
c = lim f p (x)dx. S,o Og8 Remark: Some care was taken in the definition of c, since the following proof does not show that pe(x) is absolutely convergent over 110. In reality, however, the function pe(x) is bounded in SZo. This follows from a much stronger version of Theorem 14.4 (compare remark at the end of Theorem 13.9) which yields pointwise asymptotic results. Thus, under the conditions of the theorem, one can show that KA(x, y) is continuous and bounded in Il x 11 for A E pm(T). Moreover, if x E f o, y E (la, x y, then as IAI , A E (9, 0), KA(x, x) =
pe(x)IAI(n/m') 1 + o(IAI(n/m')'1),
while KA(x, y) =
o(IAI(n1m')1).
sec. 14
Eigenvalue Problems; The SelfAdjoint Case
241
Proof. We first show that Hypothesis 2° implies that we also have the following. For any positive e, there exists a neighborhood U' of x°, U CfZJ and a constant CE, such that for all f E L2(fZ) and for all IAl sufficiently large, arg A = 0, (14.16)'
II(Po ,k)TXf 
f
I I
O,U
s e l l f l l o,fZ + CEIITAfl lm1,n
I I(Po  A)(TA)*f  fI I o,U s Eilfll o,fl + CEI I(TA)*fI la,l,sz
Indeed, observe that (Po  A)TAf  f = POTAf  (1  AT)'f = POT(1
 kT)'f  (1  AT)lf.
Given e > 0 let U be the neighborhood of x° for which (14.16) holds. Then, applying this inequality to (1  AT)1f, using the above relation, we obtain II(P0  A)TAf fl Io,U s EI I(1 AT)If llo,Sl + CEIITAfIIm
s EII1 AT)'Il0IIfl lo,f1 + CellTAfl lm1,fZ
Using (12.5) we have for IAI sufficiently large, arg A = 0, 11(1AT)1110 s 1 + lAl IITAII0S 1 + K
where K is some constant. This combined with the preceding inequality yield the first part of (14.16)' (one replaces e by c' = e(1 + K)1 choosing U' as the neighborhood which corresponds to e' in Hypothesis 2). The proof of the second inequality in (14. 16)' is similar. Let now flg be fixed and let 0' CC flg. Let a A. Then there exists a collection of congruent nonoverlapping cubes Q...... QN, such that N
if&= U Q,, then fZ' r1
and the estimates (14.16)' are valid
in Q., the cube concentric to Q' having twice the side length of Q,. Let x1 be the center of Q,, and let b be the length of the side of Q,. Let arg A = 0. By the results of Part 1, there exists a fundamental solution FA(x) for the elliptic operator with constant coefficients, P(x', D)  A. For arg A = 0, lsufficiently large, the modified reAl
Elliptic Boundary Value Problems
242
solvent TA exists. Suppose f E L 2(Q'); extend f to be zero outside Q'. Define an operator Sj on L2(Qi) by the formula Skf = TAI  F1*f. E Now R(TA) cR(T) CHm,(12), and, as we have seen in Part 1, HI(E) . Thus, if we restrict our attention to Q,, R(SA) C Hm , (Q;). By Lemma 13.4 and by the estimate (14.10), it follows that S,k is a bounded linear transformation from L2(Q') into The assumed relation between m' and n shows that the hypothesis of Theorem 13.10 is satisfied, so it follov;s that the HilbertSchmidt kernel KA(x, y) of TX exists, and that KA(x, x) exists as an L2 function in 12. The same remarks apply to the kernel Gk(x, y) of the operator by Theorem 13.9. It should be noted in this case that the adjoint of Sk is given by the formula
SA*f
= T, f  f
 x)f(y)dy,
Q. .
and FA*(x)  Fk(_ x) is the fundamental solution for P*(x', D)  X. Thus, R(SA*) CHm,(Q!), and Theorem 13.9 can be applied. Therefore, corresponding to the operator definition of S,A is the equation for the kernels: (14.17)
GA(x, y) = KA(x, y)  FA(x y) ,
Gj*(x, y)=K,(x, y)FA*(xy), where G * and K, are the HilbertSchmidt kernels for the operators (V and (Tx)*, respectively. Let y be the constant of (13.13) which applies to a fixed cube of side 8a. Then for b < 8o, we have by the corollary to Theorem 13.9 (14.18)
(b/8o)"" [f
JGA(x, x)l2dx]ll
4 y[(b/80)"`ISAIm'm'
+ S\
Imm')ISIOJS101.
sec. 14
Eigenvalue Problems; The SelfAdjoint Case
243
We now need to estimate the seminorms of SA appearing in this ex
pression. First we shall obtain some rough estimates. By the triangle inequality and the definition of S;A, for f E L2(QP we have (14.19)
SAfIk S ITAflk + IFA*fIk
Iare A = 0, we have by hypothesis For large Al, KIAI1.
I ITAI IO
n'/m', the coefficients of A are in C(kl)m''(SZ), and D(Qt) CHkm'(fZ). Then the spectrum of Q is discrete, and the eigenvalues of Cl have finite multiplicity. Let lAi } be the sequence of eigenvalues of d counted according to multiplicity. For k > C. let N+(A) be the number of nonnegative eigenvalues AI < A, and let N_(k) be the number of negative eigenvalues Al  A. Then Nt(A) = c±An/m' + o(A"/m') as k where
ct = (2n) " f o)*(x)dx, and
Wt(x)=Pe. 0 0, we have (14.35)
1.
1
Ct(n/km')1 + 0(t(n/km')1).
=
Ak  it 1
Using the formulas derived in Theorem 14. 4, (14.36)
c = fp(x)dx,
where (14.37)
d6
P(x) =(277)n f
En [A'(x, 1Sl]k ^ 1
The problem is now to use this information to find expressions for N+(A) and N_(A). Let q = n/m'; 0 < q < k. Then, if c1 and c2 are the real and imaginary parts of c, it follows from (14.35) that (14.38)
2.
A2kA+ t2 t
A2k + t2
= c it (q/ k)1 + 0(t(q/k)1) 1
= C2 t(q/k)1 +
A,t(q/k)1
Divide the latter relation by t, and replace t2 by t, to obtain (14.39)
E
1
= c2t(q/2k)1 + 0(t(q/2k)1).
Let N(A) = N+(A) + N(A), the number of A, such that IAJ! S A. Then
sec. 14
Eigenvalue Problems; The SelfAdjoint Case
.
1
=
r
255
f°° dN(A1/2k) o AA2k+t +t
so that (14.39) becomes r°° dN(Al/2k)= 
A  +2
0
C
t(q/2k)1
o(t(q/2k)^1).
+
Applying the Tauberian theorem of Hardy and Littlewood (Theorem 14.5), it follows that (14.40) N(A1/2k) = C2 sin(rrq/2k) Aq/2k+ o(Aq/2k). rrq/2k
or, replacing A by A2k' N(A) = 2kc2 sin(irq/2k) Aq + o(,Aq).
(14.41)
rrq
We now must separate the parts N,f(A) and N_(A) out of their sum
N(A). This is just the reason we had to assume k is odd: the sign of Ak is the same as that of Al in this case. Now Ak
i
A2k+t2
1
A.>0
=
f ./ J
0
=
(14.42)
+
Ask+t2
A. n if n is odd, 2mk > n + 1 if n is even. Also, our assumption 3°, together with the relation (T k)A =
 (A  (fk)1,
implies that except for the positive real axis, all directions are directions of minimal growth of the modified resolvent of Tk. We take Slo = Sl, and for the elliptic operator P(x, D) we take that operator whose characteristic polynomial is (A'(x, ic))k. Our assumption that A'(x, is positive shows that 1°of Theorem 14.4 is verified trivially. Finally, 20 of Theorem 14.4 is checked in exactly the same manner as in the proof of Theorem 14.6. Therefore, we may now apply the conclusion of Theorem 14.4 to our case. Combining the estimate of Theorem 14.4 for the trace of the operator (Tk)A with Theorem 13.10, we obtain just as in the proof of Theorem 14.6 the asymptotic relation for fixed arg k,
I
.
1
l,kk 
= cIAt (n/k2m)1 + o(IAI (n/k2m)1)
for Al J oc, arg k L 0. Now let Q_=n/2m and let A =  t, t > 0, to ob
tain for t (15.1)
1
1
k+t
=
Ct(9/k)1 + 0(t(9/k)1).
1
Now let Al = µ1 + iv1 be the Cartesian representation of the complex number A1. The result that the eigenvalues are eventually located in any angle about the real axis implies that µ1  oc and v11IC1 , 0. Therefore, it follows that Ak = µk(1 + iv1/1L1)k = µk + and so if v1/µ1I < e for j > j0, then 1
1>10 Ak + t

k_,kkl 1
1>10 {1k + t
t o (A + t)(11 + t)
ycI
1
.
yak+t J
From this it is a consequence of (15.1) that as t > oc 1
= Ct(q/k)1 + 0(t(q/k)1).
By definition of N(A), this may be written
f' El+t dN(pl/k) =
t(q/k)1 + o(t(q/k)1).
1
0
Now we may apply the Tauberian theorem, Theorem 14.5. The re
sult is that N('L1/k) = C sin (77q/k)
,1q/k + O(ftq/k)
77q/k
Replacing µ 1 /k by A, this becomes (15.2)
'N(A) = ck sin (77q/k)Aq + o(aq), A  . 77q
Thus, all that is left is the computation of c. According to Theorem 14.4, with 6 = 77,
c = f p(x)dx, B
where p(x) _
(277)n f
dC
E,, (A'(x, IS))k + 1
If we let p(') = (A'(x, i6)k, and let
sec. 15
NonSelfAdjoint Eigenvalue Problems
265
v(t) = ate: P(, < t1l,
then we obtain, as in the proof of Theorem 14.6, p(x) =
_
de En P(6 + 1
(2ir)n f
(2)n f"
1
t+ 1
dv(t).
Since
V(t) = to/2k,(1) = tQ/kV(1),
this becomes by (14.29) 00
p(x) = (2n)"
_
(2,7)n
k
v(1) f
t(4/k1)1
F+
T4
k sin (irq/k)
dt
v(1).
Therefore,
ck
sin (irq/k) _ (2n)" f v(1)dx
lTq
_ (2ir)n f 116. A'(x, i6 < 111dx.
When this expression is compared with (15.2), the theorem is proved. Q.E.D.
Remark. To obtain results similar to those of Theorem 14.6, we need only assume in Theorem 15.1 that all directions except the positive real axis and negative real axis are directions of minimal growth of the resolvent of Q. Then we get asymptotic expressions for the number of eigenvalues Al such that 0 < Rki < A (or such that  A < JRal < 0).
Elliptic Boundary, Value Problems
266
The question now arises concerning the hypothesis 3° of Theorem 15.1. Note that Lemma 14.3 implies that a direction e'0 is of minimal growth of the resolvent of d only if A'(x, i6) omits the value e'0. In case d is self adjoint, then the remark made above allows the result of the theorem. Of course, this is not important, since we already had obtained Theorem 14.6 for the selfadjoint case. What is important, however, is that Theorem 15.1 applies to perturbations of selfadjoint operators. This concept shall now be discussed. First, a preliminary result is needed.
THEOREM 15.2. Let do be a closed, densely defined linear operator in the space L2(1), such that D(Q0) CHm,(cz). Then there exists a positive number C such that for all u E D(do) (15.3)
I
IUl
l
m',as
C(II(I oUIIo,Si + IIUIIo,c).
Proof. Let G CL2(1) x L2(St) be the graph of Q9; that is, G = {(u, Qou): u E D(Qo)l.
Since do is closed, G is a closed subspace of L2U1) x L2(1). Consider the mapping of G into H.,(1) given by (u, Qou)  u. This is a closed linear transformation of the Hilbert space G into the Hilbert
space H.,(1). For if (u1, Qouj )
(u, Qou) in L2(1) X L2(St),
uj 4 v in H., (Q),
then certainly of v in L2(1), so that v = u. By the closed graph theorem it follows that the mapping (u, aou) a u is a continuous map
ping of G into H.,(1). But this is precisely the statementof the theorem. Q.E.D. This theorem is of considerable interest in other areas than the one at hand, and we shall have more to say about it later. For the time being, however, we wish to apply this result to perturbation of selfadjoint operators.
sec. 15
267
Non SelfAdjoint Eigenvalue Problems
First, suppose the transformation do of Theorem 15.2 is selfadjoint. Theorem 12.7 implies that for nonreal A, Iiu[1
19,1111(A
O,s ,
_ do)UIIo,si, u E D(d0),
or, equivalently, (15.4)
I I(X  do)1110 s 19XI1.
Therefore, for u E D(do) it follows from (15.3) that I
i ul l m',S S C[I I doul 10,St + I I UI I o,Sl]
< C[I I(A
 (1o)ulI o,St + 1A1
I lUllo.c + I
U1 10,01
1)19,1111(A
< C[1 1(A  ('0)UII0,Q + (JAI +
_ do)UI Io,St]
 C[1 + iig_i 1 I I(A  d0)UI I O;fl Therefore,
By the convexity result of Lemma 13.3, this estimate together with (15.4) implies (1 +
(15.5)
II(X  do)11Im'r
IAI)1(1/m')
C1 19
1
with a different constant C 1. Now suppose that d = do + 53, where do is a selfadjoint transformation in L 2 (9), with D(d0) C HM, (St), and 5`3 is an operator of lower order. Precisely, assume D(53) D D(do) and that 5`3 satisfies an estimate
Elliptic Boundary Value Problems
268
(15.6)
IBullo,SZ S C2jlujlm'1,SZ
This inequality allows us to extend J3, if necessary, so that D(`f) Hm1_1(SZ). For A not real,
AQ=AQoJ`3, and, multiplying by (A  (10) 1 gives (15.7)
(A  Qo)1(A  (f) = 1  (A 
(1o)1ni;
this equation must be interpreted as being valid on D(Q0) = D(Q). Now B maps Hm,_1(11) into L2(& ), and (A  Qo)1 maps L2(1l) into D(Q0) CHm,_1(0). Therefore, (a (IOr1B is a mapping of Hm,_1(1l) into Hm_1(1l); the estimates (15.6) and (15.5) imply that the norm of (A = (10; '.3, considered as a mapping on Hm,_1(1l), is bounded by
(15.8)
c1c2
(1 +
IAI)1(1/m')
191\1
If this quantity is less than, say, 1/2, then 1  (A  0193 is an invertible mapping on Hm,_1(1l), and its inverse has norm less than 2. Therefore, since
Aa=(J1ao)[1(AQo)1B] on D(1), then there exists (A  Q)I = [1  (A 
0)
Moreover, the estimate of 2 for the norm of [1  (A  (10) 'BI', as an operator on Hm,_1(cl), shows (15.9)
11(,k
 Q)1I Im'1 S 211(A  ao)1IIm'1
Multiplying (15.7) by (A  (I)1 yields the relation
269
NonSelfAdjoint Eigenvalue Problems
sec. 15
(A(1)1_(A_(10)1+(A_(10)1B(A(1)1.
Hence, by (15.6) and (15.9) I(A  ar11l 0 S I I(A  ao)1I l 0[1 + I Ii(A  a)1I I0] < 11('k  (10) '11 0[l + C2I I(A S I (A  (10)1l 10[1 + 2C2 I(A
(1)'1Im'11
 a0) 11Im.1].
Since the expression (15.8) is less than 1/2, (15.5) implies 2C211(A 
(10)lI1.'1 S 1,
so that we obtain II(Aa)1II0S2IIA (1o)1GIo
That is, if (15. 1,16)
(AI > 2C1C2(1 +
then A E p(a) and we have the estimate (15.11)
II(A(b11I0 S2/I4AI.
One can easily see that there exists a constant K such that if (15.12)
IJAI 2 K max (I`RAI 1(1/m'), 1),
then (15.10) is satisfied. A pictorial description of the region described by (15.12) is given in the figure; we have shown on the first quadrant.
270
Elliptic Boundary Value Problems
1
Obviously, our results imply that any nonreal direction is a direction of minimal growth of the resolvent of Q. We have actually done much better, becuase our estimate (15.11) holds in a region of the A plane which includes all rays (O, a) with e'° not real, and a sufficiently large, depending on B. And since the region (15.12) is in p(@), it follows that all the sufficiently large eigenvalues of d lie in the g parabolic" type region. This is of course a much more restricted behavior of the eigenvalues than the conclusion of Theorem 15.1 can give.
Remark. If also do satisfies an estimate (@0u, u)0.
It A0 luIlo,S1,
u E D(ap),
then the argument given above allows one to show that also the negative real axis is a direction of minimal growth of the resolvent of a. Cf. also Theorem 12.8. We now wish to examine the abstract Theorem 15.2 and consider its consequences. THEOREM 15.3. Let A(x, D) be a differential operator of order m' on SZ having continuous leading coefficients and measurable, locally bounded lower order coefficients. Let a be a closed, densely defined linear operator in the space L2(SZ), such that Ca(SZ) CD(a) C H.,(Sl), and du = Au for u E D(a). Then A is elliptic.
Proof. By Theorem 15.2, there exists a positive C such that
sec. 15
NonSelf Adjoin( Eigenvalue Problems
IIullm.,c 5 C[IIAullO,c + Ilulla,n],
271
u E co(l).
Thus, the quadratic form I lAul l o, = f
I
I
aa(x)D'u(x)l'dx
0 lal<m
lalsm'
I3tm
f
a a(x)a13(x)Dau(x) . Dflu(x)dx
satisfies Garding's inequality. Therefore, the necessity of Garding's inequality, Theorem 7.12, implies that for real nonzero aa(x)ap(x)S eR > 0, la m' IQ.m or,
aa(x)eal 2 > 0.
lalm' 0. Q.E.D. That is,, A'(x, A consequence of Theorem 15.3 is that if A(x, D) is not elliptic, and we wish to close the operator defined by applying A to functions in (S2), then the extended operator Q will have a domain which includes .functions not in Hm i(z). For u E D(Q), the function Au still exists weakly, in the sense of Definition 2.2, but now Au can no longer be computed by taking the individual strong derivatives of u and forming the appropriate linear combination of them. In the discussion above of operators of the form d = Qo + fB, where QQ is selfadjoint and B is of lower order, we obtained estimates for A  Q just having an estimate (15.3) involving do" Of course, we also *1I, valid for arbitrary had to use the estimate I1(A  Qp)ll1 o S self adjoint do. We shall now show a general procedure for obtaining estimates containing a parameter A, using only estimates not involving parameters. These latter estimates will be for an operator in a space of dimension one higher.
Elliptic Boundary Value Problems
272
THEOREM 15.4. Let A(x, D) be an elliptic operator of order 2m in S2, having leading coefficients in C°(Sl) and lower order coefficients bounded and measurable in Q. Let d be a closed, densely defined operator in the space L2(S2) such that D(Q) is the closure in H2m(S2) of a linear subspace M of C2m(SZ), and Qu = Au for u E D(d). Suppose that for some real 0, and for all real e and all x E St, A'(x, io e'0.
Let D' = D" and D = a/at, and set L(x, D., Dt) = A, (x, D.)  (1)me iOD2 m, where Al is an operator having the same principal part as A, and having bounded measurable coefficients in Q. Then L is elliptic in
r=SZxE1. Suppose that for all v to and v(x, t)  0 for It1 (15.13)
C2m(r) such that v(x, to) E M for each fixed 1,
1 I'II Zm,rr s C(I ILvI I o,r + IVIIo,r),
for some positive C. Then for u E D(Q), arg k = 0, and IAI sufficiently large,
I
Iul
I o,St
A01 ul l o,ci.
Therefore, we can apply all the results dealing with operators of the form do + 93, and obtain, in particular, that all directions except the.,.,, positive real axis are directions of minimal growth of the resolvent of Q. Compare the remark preceding Theorem 15.3. Also, we have °parabolic" condensation of eigenvalues along the positive real axis, as discussed above. 3. Finally, consider the case in which we assume only that A is strongly elliptic. By suitable normalization it may be assumed that IA'(x, it): x r, S2, e realI fills out precisely the region Ia: larg Al < Bo }, where 00 < 1/277. In this case, all directions not in this sector are directions of minimal growth of the resolvent of Q. To prove this, we shall apply Theorem 15.4. Let 0 be any number satisfying 0 < 0 < 2n  Bo, and let
sec. 15
Non Self Adjoint Eigenvalue Problems
L(x, 6, r) = A'(x,
277
 (l)mei0r2m
Then
(l)mL(x,
,
r) = A'(x, i6) 
eiOr2m.
It is clear that for such 0, the values (1)mL(x, 6, r) lie in a cone properly contained in an open half space of the complex plane, so that L(x, DX, DI) is actually strongly elliptic. Therefore, Garding's inequality holds for L in a cylindrical domain fI x E 1, so that (15.13) is verified. Moreover, since d = T1 and T is compact, it follows that the spectrum of Q is discrete; in particular, the resolvent set of d is not empty. Thus, Theorem 15.4 applies to our case. We conclude this section with a brief discussion of real secondorder problems. We shall assume that n
A
ai,DiD+ .aiD.+a
and that A'(x, 6) 2 c01612 for real 6. In the case of the Neumann problem, the boundary condition has the form au/av = 0, in which au/av is the conormal derivative corresponding to A. In case A is formally selfadjoint, we obtain immediately asymptotic expressions for the umber of eigenvalues less than t. But in any case, this problem is
almost selfadjoint, as in 2. above. Therefore, all but a finite number of eigenvalues lie in a parabolic region {A: JRA 0'Aj S C15AI'/2I. This is a result first obtained by Carleman. In the case of an oblique derivative problem, we can show again that all directions except the positive real axis are directions of minimal growth. For this we consider the operator L = A + t and apply Theorem 15.4. In this case the estimate (15.13) follows from remarks made at the end of section 10. It should be noted that it is not known in this case whether the eigenvalues condense in a parabolic fashion along the positive real axis. e'0D2
Elliptic Boundary Value Problems
278
16. Completeness of the Eigenfunctions
In this section we shall state conditions under which the generalized eigenvectors of an eigenvalue problem span the Hilbert space, or, at least, a certain subspace of the Hilbert space. First we shall state some theorems on the growth of analytic functions. THEOREM 16.1. Let F(k) be an entire complexvalued function, and let JF(A)1< ClIXI1
on two rays making an angle n/a at the origin. Let r1 < r2