Lectures on Elliptic Boundary Value Problems

&j3

Yea

D"Odx.

Thus, by (7.10) (7.13)

f

IR(c)I _
IDY0I ID8tIdx.

IYI+I8I<2m-1

Now the cubes Q. overlap in such a fashion that any fixed point contained in exactly 2" distinct cubes, except for points on $U

!-1

OQ;, a set of measure zero.

f

e-l Q.

is

Thus

IDYcI IDSct dx = 2° f IDYc6j ID30Idx; 1Z

using this relation in (7.13),

IR(O)I < K

f

E

I DYci DScIdx.

IYI+I3I<2m-1 Q

By the Cauchy-Schwarz inequality IR(O)I < F.

E

IIDYcH Io,1Z I ID8 0I l o,Si

IYI+I0I<2m-1 (7.14)

IOIIm.Q II0I Im-1.j2

Thus, (7.12) implies (7.15)

18(c¢) Z

i-1

K ICI Im, I Im-1,i .

But the diameter of supp is no greater than the diameter of Q', which, by choice, is less than p. Thus, Lemma 7.9 implies that

Elliptic Boundary Value Problems

86

yE0 15,O m.0'

Hence, (7.15) implies 00

(7.16)

JZB(O) 2 yEo ill

- KIM IL.cl

Iz'n,cl

I

The same type argument as was used to derive (7.15) from (7.12) shows that f

I"I HI

I i1' f C,2ID"0I2dx + R1(o) M

H"

f ID"cI 2dx + R1(0) mf

=

where R 1(qS)

R1

is a quadratic form satisfying an estimate similar to

(7.14): R1(O)


Combining these results with (7.16), J1

,h) >- yEOI I0I Imnfl

- KI

I(bI

'Q I IOIIm..l,ci

As in the proof of Lemma 7.8, this implies the required estimate in Theorem 7.6. Q.E.D. We shall conclude this section with a proof that Garding's inequality is very nearly equivalent to the ellipticity of the quadratic form B.

THEOREM 7.12. Let as (x) be bounded measurable functions in an open set 12 for Ial 5 m, I f3I < m, and let

sec. 7

Garding's Inequality

87

f aB(x)Dao D13O dx.

B(O) _ lalsm Ia Ism

0 be given constants.

Let CO > 0 and to 1°

If for all cb C Co (Sl)

_%B(O)2coII0IIm,SZ-Aol10ll2

(7.17)

then for almost all x C SZ and for all real 6 (7.18)

+R aap(x) a z c°

E

R

Iai_-

l-'" 2°

If for all

(7.19)

e.za

E

C Co (cl)

RW Iz °Il0llm,0-Xoll0llo,QY

then for almost all x C Q and for all real aa13(x)6a+RI ., co

(7.20)

62Q

IaI=-

If moreover 12 is connected and a a,8 (x) C C°(S2) for lal = IPI = m, then there exists a real number 0 such that for all x C 12 and for all al 6

(7.21)

ei0

aa,8 (x)t~a+QJ2 c

E IaI=IaI=m

/

1, 62a, Ial=m

"Where c is some positive constant independent of x and 6. Proof. Let e 1 and 62 be any non-zero real vectors and let 11, z C Co(&2) be such that supp (1/1) and supp ('2) are disjoint. Set O(x) =

1(W)eirl~1

+

Y

We shall apply either (7.17) or (7.19) to the function 0 and let r -. oo.

Elliptic Boundary Value Problems

88

So in computing B(O), cIIm,ii, and I(kIIoSZ, it will be necessary to keep track only on those terms containing the factor r-2m. Thus, since supp (c1i1) fl supp (cli2) = 0, Leibnitz's rule implies

f

0.(x)

rzm I

=

(,,)a+(3

J-1

.(x) dx + ...

11

IaI=I0I=m

f a aq(x)Iqjj (x)I2 dx + ...,

aI=IfI=m

where the other terms are of lower order in r that 2m. Likewise, (7.23).

I ICI m,SZ = 7- 2m

I (6I)2 f i0,(x)I2 dx + ...; II0I12,SZ = 0 + -1

Assuming (7.19), if we use this special function 0, the estimates (7.22) and (7.23) imply, after letting r w, that 2

(7.24)

1 ;.1

2 (e')a+p f aaR(x)1',(x)z dx( Sl IaI=lei=m 2

co

(6j)21 I I

-1

,(x) 2 dx.

SZ

This inequality has been derived for functions ci1, 02 E C of 12)- But

since Co(c) is dense in L2(c), (7.24) is valid under the assumptions that 01, Vj2 E L2(11), supp (c/i1) flsupp (02) _ 0. Now let x1 and x2 be distinct points of SZ, and let for j = 1, 2

Ix - x'I < p, 0,(x) = 0,

ix - x'I 2 p,

where P'. is a non-negative number. If x 1 and x 2 are in the Lebesque sets of all the functions aaf(x), IaI = /3I = in, then applying (7.24) to the special functions 0i and letting p -a 0, we obtain (7.25)

I E p2

aap(x')(Sj)a+RI

1

IaI=INI=m

sec. 7

Garding's Inequality

cy P

.89

2

F (S')2«

o

1' 1

I«l_m

To prove (7.20) let x1 be any point in the Lebesque set of all the functions aafi(x), Iai =101 = m, let p1 = 1 and p2 = 0. Then (7.25) becomes a

aN(x 1)(S

1)a+fil > c0 Ia'=m

which is just (7.20). Likewise, (7.18) can be proved by exactly the same type argument.

Finally, assume that' a a (x) is continuous for Jal = 101 = m and assume (7.19). Then (7.15) holds for all x', x2 E Sl, x1 x2; and the also implies that (7.25) holds even if x 1 = x2. -Thus, continuity of aafi if we let p 1 = 1, p2 = \/p for p >_ 0, and if we define

H(x, ) =

E aaR(x)e+R, 1a1=A=M

then (7.25) becomes (7.26)

IH(x1, 61) + pH(x2, 62))

co,(IS 1I2m +

PI62I2m),

an inequality which holds for all x1, x2 E S1, all real vectors 61, 62, and all numbers p 0. Let Z be the set of numbers of the form H(x, 6 for x E Q, 6 real. Since Sl is connected and H is continuous, Z is a connected subset of the complex plane. The set Z contains, together with any nonzero complex number z, all the numbers of the form rz, 0 < r < .. Therefore, we need only show that the angle between the line from 0 to H(xl, e1) ano the line from 0 to H(x2, 62) is less than rr - 8 for some S > 0, S independent of x1, x2, e1, 62. Obviously, we may assume I C1I = Jell = 1. Then (7.26) becomes IH(x1 61) + pH(x2. 62)I Z co(1 + P).

Since the coefficients

aa13

are bounded, we have IH(x, 61 S K for

90

Elliptic Boundary Value Problems

161 = 1. Now let H(xi, 1) = rie'ej; then Irlele1

+ pr2eie2l

> CO, (1 + p)

Let p = r1/r2; then leiel

+ e'e21

1

2 co'(ri + r2l) 2 2cp/K.

But this means that 01 - 02 cannot be too near an odd mutiple of ii. Q.E.D.

8. Global Existence

In this section we shall assume that A is a strongly elliptic operator of even order P = 2m which has been normalized so that

(- 1)mRA'(x, 6> 0

eI0

for

(cf. Definition 4.1). Consider the problem of finding a function u such that

Au=f u=yf0

au='1 (8.1)

in on

St,

an,

on

an,

an am-lu

an' 1

m-1

on

an,

where a/an indicates differentiation in the direction of the exterior normal to an. This is the Dirichlet problem for the elliptic operator Ain12. The existence theory developed in this section is for the Dirichlet problem. However, the formulation of the problem given above is fraught with difficulties. For instance, considerations involving normal differentiations at the boundary are quite complicated, and, indeed, we would like to consider domains whose boundaries may fail to have tangent planes. Therefore, we shall reformulate the Dirichlet boundary conditions in (8.1).

sec. 8

Global Existence

91

The reformulation can be motivated in the following way. Suppose that the boundary and the solution u are sufficiently smooth. Then the condition u = (Ao on all automatically prescribes all derivatives of u in the directions tangent to a11. Likewise, the remaining boundivy conditions automatically prescribe all derivatives of u on alt in Which enter at most m - 1 differentiations in the direction normal to Therefore, under sufficient smoothness conditions, the Dirichlet boundary conditions of u automatically prescribe at alt all derivatives p"u, jaI < m - 1. Thus, instead of all normal derivatives of order ess than m, we could prescribe for u all derivatives of order less than m on att. However, we obviously cannot prescribe arbitrarily on 311 all devatives of u of order less than in, since these derivatives are not independent. One way-out of this difficulty is to list a number of Compatibility conditions these derivatives must satisfy. However, an easier procedure is simply to-postulate the existence of a function (x) which is in C'_ I (1l) and whose derivatives at a1Z of order less than m are precisely the prescribed derivatives for the solution of the Dirichlet problem. Thus, the Dirichlet boundary conditions for u now 6sume the form: A.

D"u=Dg

on

a11,

aI <m-1 .

There are many ways of proving the existence of functions u satisfying systems such as (8.1). One of the more important techniques was first used by Riemann to prove the existence of a solution of the Dirichlet problem for Laplace's equation. This method makes use of the Dirichlet principle: a solution u of Laplace's equation on 1Z minimizes the Dirichlet integral,

J 11

(au )2 dx,

-1 ax

among all smooth functions u which satisfy the given boundary conditions. As a generalization, the following program could be used to prove the existence of solution of (8.1). In previous sections we have already discussed the meaning of the statement that a function u is a weak solution of an equation Au = f. In section 5 we proved the existence of weak solutions of elliptic equations, and in section 6 we established several regularity results,

92

Elliptic Boundary Value Problems

allowing us to give conditions on the smoothness of A and f which automatically insure that weak solutions are sufficiently regular to be classical smooth solutions. For boundary-value problems the procedure is similar. We shall first formulate a weak sense in which a function has certain boundary values, and shall then in sections 9 and 10 give conditions on the data which allow the conclusion that the weak solution of the boundary-value problem is indeed a classical solution. We shall now give several formulations of the weak boundary conditions. The discussion will be motivated by considering the situation in which the functions treated and the boundary are sufficiently smooth. Suppose u is a smooth solution of Au = f. Let v be a smooth function, and assume that f1 has a smooth boundary. By Green's formula we have (8.2)

am-I

(A*v, u) - (v, Au) =

J.0 fH Nam-i-iv

I

-

an

do.

This formula is non-trivial; it will be derived and discussed in section 10. Here Nk is a linear differential operator of order k. If D°`v = 0 on a(l for 0 5 jai S m - 1, then v is a function with zero Dirichlet data. For such a function, and for u satisfying (8.1), the relation (8.2) becomes (8.3)

m-1

(A*v, u) - (v, f) = E f Nam-1-;v ct do.

;'0 do

Conversely, suppose that u satisfies (8.3) for every function v having 2m continuous derivatives up to the boundary and zero Dirichlet data. Then u satisfies (8.1) in the following weak sense. If v E Co(Q), then (A*v, u) = (v, f),

so that u is a weak solution of Au = fin Q. For the boundary conditions, let g be any smooth function such that atg/anj _ . on aQ, 0 < j < m - 1. Note that (8.2) gives m-I

(A*v, g) - (v, Ag) =j.oE on f N2m-1-tv

do,

sec. 8

93

Global Existence

where now v is assumed to have zero Dirichlet data. This subtracted from (8.3) gives (8.4)

(A*v, u-g)-(v, Au-Ag)=0.

Comparing this relation with (8.2) with u replaced by u - g, it follows that (8'5)

m1 E

;- o

N2m-l-,v al(u - g) da= 0 ale

;

for all sufficiently smooth v having zero Dirichlet data. But it is reasonable to expect that the functions N2m_1_;v for 0 j S m - 1 can be essentially arbitrary for such v, so that (8.5) implies --L' That is, aJu/anj d ](u - g)/an' = 0, 0 Z 0 S j S in - 1. Thus, u satisfies (8.1). Thus, we might make the definition that u is a generalized solution of (8.1) if (8.3) holds for all smooth v having zero Dirichlet data. However, this is somewhat inconvenient because of the presence of the boundary integrals in (8.3). An alternate definition follows from (8.4). If g is any smooth function having the Dirichlet data alg/ani = 0; on M, 0 S j s m - 1, then we might define u to be a generalized solution of (8.1) if (A*v, u - g) = (v, f) - (v, Ag)

for all smooth v having zero Dirichlet data. This is evidently a satisfactory definition, since no boundary integrals appear in this formulation. Indeed, this is almost our ultimate definition. It still needs modification so that something analogous to Dirichlet's integral for Laplace's equation appears. We now proceed to the definition that will be used here. If both gS and t/ are smooth functions having zero Dirichlet data, then (8.2) shows (A"S, ,b) - (95, Ac) = 0.

This result is obtained by transferring all the differentiations on t' 1in the integral (0, AO) to 0 by integration by parts. However, we could transfer only some of the differentiations to obtain different formulas; since q and 0 both have zero Dirichlet data, and since the boundary

94

Elliptic Boundary Value Problems

integrals contain only terms of the form Da/ . D18t/i with ial + 1/3 < 2m - 1, it is seen that all the boundary integrals vanish. In particular, suppose we integrate by parts the expression (0, Ai/i) in some manner so that no derivative for either 0 or 0 of order greater than m results. Thus, we obtain some formula (8.6)

I

(0, As/i) =

f c aS(x)Dao DO O dx.

l a lam

(There is no reason to expect that there is a unique set of functions ca,6 (x) with this property; we shall give an example below to show that there are in general many distinct possibilities.)

We shall set aaa(x) - cax; the bilinear form (8.7)

!,

B[O, 0] =

(DaO, aaaD,80)

IaISm

is called a Dirichlet bilinear form corresponding to the operator A. The relation (8.6) can be expressed (8.8)

(0, AV,) = B[q, i/i],

all smooth c¢, Vi having zero Dirichlet data.

Note that if in (8.6) we transfer all the differentiations of 0 by.integrating by parts, we have formally ((k, At/i) =

Y_

(-1)H (0, Da(aapD1 li) )

Ia Ism

Thus, since even the functions 0 in Co(SZ) are dense in LZ(SZ), we have (8.9)

Ai/i =

I IaIS. INISm

(-1)I-I Da(aaSDA1i).

sec. 8

Global Existence

95

Now we motivate our final choice for definition of weak solution. Assume all the functions and ale are. sufficiently smooth, and let g be a smooth function having the Dirichlet data aig/ant = /P 0S j S m - 1.

If u is a solution of (8.1), then u - g has zero Dirichlet data. Thus, if 0 has zero Dirichlet data, (8.8) implies

(0, A(u - g)) = B[0, u - g]; or setting uo = u - g, using Au = f,

(0, f-Ag)=B[S, u}. i

Thus, we shall say (approximately) that u is a generalized solution of (8.8) if u = uo + g,`where uo is a function having zero Dirichlet data and satisfying B[¢, uo] = (0, f) - (0, Ag),

all 0 having zero Dirichlet data. Or, if we look only at the test functions 0 E C o(Jj), we require only that (8.10)

B[0, ua] = (0, f) - B[0, g], all 0 E Co UZ).

(Here we have used the relation (8.8) with i' = g; the relation holds since all derivatives of 0 are zero at a1Z.) After a few definitions, we shall make the statement (8.10) precise. First, we give some examples. (-1)Ia! Dan

as in (8.9), then the formal adjoint of A is given by (8.11)

A*=

(-1)Ial DaapaDS.

Thus, if aap = aj6a , then A = A*, a relation which is expressed by say-

Elliptic Boundary Value Problems

96

ing that A is formally self-adjoint. Note that in the case aag= aj8a, the Dirichlet form B[O, t(r] is Hermitian symmetric; that is, B[c,

]=Bci].

Moreover, if A = A*, then some Dirichlet form for A is symmetric in

the sense that aaR = asa, For, suppose A has some Dirichlet form B1[qS, 0] =

I

(DaO, ca'DRJ).

IaISm

I0Isp1

Here A is assumed to be any linear differential operator, not necessarily of the form (8.9). Then A* has the Dirichlet form (8.12)

(Dao, FP--a Dj3V

B 210, 0] _ laI<m IRI<m

as shown by a simple computation. But if A A*, then B2 is also a Dirichlet form for A. Thus, '/21B 1 + B2) is also a Dirichlet form for A, and '/21B1[qS,

] + B2[0, 0]} _

(D a!, aa1DQc], IQa

Ism

INCm

where aaf3 =

Thus, aap =

'/z (ca,8 + cc

.

, as was desired.

2. The biharmonic operator in two-dimensional space is

A2=(D 1 2+D 2 2)2. For A 2 there are the two distinct Dirichlet forms:

Global Existence

.sec. 8

97

B 1 [0, 0] = (AO, AO);

B2[0, w]=([D12-D22]0, [D12-D2210)+4 (D1D20, D1D2J) The B2 is a Dirichlet form corresponding to A2 follows from the identity (D12 + D22)2 = (D12 -D22)2 + 4(D1D2)2.

Note that from B1 and B 2 one obtains infinitely many distinct Dirichlet forms: B = tB 1 +(1- t)B2, t/and arbitrary real number. Now we give a rigorous d4inition of weak solution of (8.1). 0

Definition 8.1. H/(S2) is the completion of C '0Q) under the norm II

IIm,SI.

Clearly, Hm(SZ) can be identified with the closure in HM(0) of

C(); is a closed subspace of H (SZ). The functions u 0(' thus, H (St) m m in Hm(11) should be thought of as having zero Dirichlet data (D"u = 0

on c3, lal < m - 1) in some vague sense. We shall now even weaken the concept of differential operator. We shall assume that a bilinear form B is given: (8.13)

I

B[v, u] =

(Dav, aa, DRu)0"c,

la lSm lN/

l5m

and we assume only that the coefficients a (x) are bounded and a,8 measurable in 0, and that the leading coefficients aap(x), I a I = 1,81 = m, are uniformly continuous in Q. With such a form is associated a formal differential operator (8.14)

Au =

I

(-1)lal Da(aa,SDQu),

lalSm

in accordance with (8.9). It must be emphasized that A is just a symbol, and may not be a differential operator at all, since the coefficients aap are not assumed to be differentiable. Note that if we start with a genuine differential operator A, and if B is a Dirichlet form associated with

Elliptic Boundary Value Problems

98

A, then the formal operator associated with B is just the differential operator A itself. Finally, note that if A is a differential operator, then its principal part is A' = (-1)m

F

aap Da+R.

I/a l=m

IaHm Thus,

Iaa l =m

(

If'=m

Hence, the assumption of strong ellipticity of A is equivalent to the assumption of strong ellipticity of the quadratic form B[0, 0] (cf. Definition 7.1). We can now define THE GENERALIZED DIRICHLET PROBLEM: given g E Hm(Sl) and fE L2(Sl), find a function u E Hm(Sl) such that 1)

u - g E Hm(Sl),

2)

B[(k, u] = (c, f)

for all

0 E Co(Sl).

Alternatively, by setting u0 = u og this can be stated in the equivalent form, `find a function uo E H,n(f) such that B[0, u0] = (0, f) B[95, g]." Such a function u will be called a generalized solution of Au = f with the same Dirichlet data as g. It will be convenient to abbreviate 'generalized Direchlet problem' to GDP. The motivation for this definition of generalized solution is given in the discussion leading to (8.10). Our first task is to show that in certain cases the GDP has a solution. The regularity theory in section 6 will then show that this solution is a solution in the classical sense, if enough assumptions are made on the regularity of A, f, g, and 91i. The proof of existence makes use of the following LAX-MILGRAM theorem, which is derived and at the same time is a generalization of the Riesz-Frechet representation theorem (see L. Nirenberg, On

sec. 8

Global Existence

99

Elliptic Partial Differential Equations, Annali della scuola Normale Superior di Pisa, v. 13 (1959 pp. 134-135). THEOREM 8.1. Let B[u, v] be a bilinear form on a Hilbert space H with norm I II If there are positive constants c1 and c2 such that I

IB[u, vii < c1I ICI I

I ICI 1,

IB[u, u] 2 c2IIuII2,

for all u, v E H, and if F(x) is,,a bounded linear functional on H, then there exist unique v, w E H such that F(x) = B[x, v] = B[w, x],

all x E H.

Observe that the Dirichlet bilinear form satisfies the first condition of the Lax-Milgram theorem. Indeed, since the coefficients a are aR bounded, the Cauchy-Schwarz. inequality implies I B [v, u] I S K I I v I m I u I I

I

I

We now have THEOREM 8.2. If there is a constant c > 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 Lax-Milgram 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 Im-1,Q s I1ukt - Ok+ 110kt - rkj1Im-1,11 +Il-Okt -


ukjllm-1.si

I10k1-Oki1lm-1,1I+kj1

as k,, k.

00,

it follows that {uk I converges in Hm_iUI). Q.E.D. We shall call the form B"[u, v] = B[v, u]

the adjoint of B. The formal differential operator associated with B* is given in accordance with (8.14) by

(-1)lal Da(aSaD0v)

A -v = I CZ

If sm

In case both A and A* are differential operators, then A* is the formal adjoint of A in the usual sense; cf. (8.11). Now assume that for some positive constant co, (8.15)

1B[u, u]l 2 col luJ 12 .S1,

all

u F Hm(SZ).

By theorem 8.2, the GDP has a unique solution. In particular, if we take g = 0, corresponding to zeroo Dirichlet data, there is for each f E L2(1l) a unique element u E Hm(1I) such that (8.16)

B[qS, u] = (0, f), all 0 E Co (f.

Note that since both sides of (8.16) are continuous functions of 0 for 0 in Hm(IZ), we can even assert that (8.16) holds for all (k E Hm(ct). Now we introduce the notation u = Tf. Thus, (8.16) becomes

sec. 8 (8.17)

Global Existence

B[c, Tf] - (0, f),

101

all

0

¢i E Hm (Q).

= Tf;

Now T maps L2(1I) into Hm(Sl), so in (8.17) we may take therefore,

c0IITfI 2B[Tf, Tf]I (Tf, f)I TfII0, I

IIfIlo,S-l

I TfI Im,S1 I,I-f1 I 0.0

Hence,

0

so that T is a bounded linear transformation of L2(1)) into Hm(1l). Now (8.15) implies also that IB*[u, n]I collul 12 ,SZ Therefore,

for each h E L20 there exists a unique v = T1h E H,,(SZ) such that B*[i/i, TIM = (0, h) for all tl' E H.(!1). Or, by the definition of B*, (8.18)

B[T lh, t0i] = (h, i0i),

all

11

(S1).

Moreover, T 1 is a bounded linear transformation of L2 (1l) into H',1(Q). Now we can obviously think of T and T 1 as being linear transformations mapping L2(Sl) into L2(S1). With this indentification we have THEOREM 8.4. In L2(fl), T1 = T*. That is,

(h, Tf),= (T Ih, f), all

f, h E L2(fl).

Proof. By (8.17) with 0 = T 1h, B[T 1h, Tf] _ (T 1h, f);

by (8.18) with 0 = Tf, B[T 1h, Tf] _ (h, Tf).

Elliptic Boundary Value Problems

102

Comparing these expressions, the result is immediate. Q.E.D. The previous results were derived under the assumption that B satisfies (8.15). Now if A is a differential operator which is strongly and uniformly elliptic, then we have seen that the quadratic form B[c, 95] is uniformly strongly elliptic. Therefore, under the assumpis uniformly conis bounded and measurable, and tions that aa,8 aa,8 tinuous in fl for lal = 1,81 = m, we have at least the validity of Garding's inequality; cf. Theorem 7.6. Moreover, a consequence of Theorem 8.2 and the corollary to Lemma 7.9 is that in sufficiently small subsets of H the GDP for Au = f is always uniquely solvable. More generally, we have THEOREM 8.5. Assume Garding's inequality holds: for some constants c0 > 0 and A0 Z 0, and for all 4 E Co 0 JRB[0, -k] z c0ll0I I2 fl - A0IIkI

lo,fl

Then for any A Z a0 the GDP for Au + Au = f has a unique solution. Moreover, if SZ is bounded, then A satisfies the Fredholm alternative; that is, if N(A) is the null space of A (the set of functions u I Ho m such that Au = 0 in the generalized sence) and if N(A*) is the null space of A*, then N(A) and N(A*) are finite demensional and have the same dimension. Also, the GDP for Au = f and u having zero Dirichlet data has a solution if and only if f is orthogonal (in L2(fl) ) to N(A*).

Proof. Let B + A be the bilinear form B[v, u] + A [v, u]0"(). Then for A z A0, `.R(B + k)[0, 0] = JRB[o, 0] + AI ltll o,fl z c0I Im,fl , so that Theorem 8.2 applies to B + A. It remains to verify the Fredholm

alternative. Let TO be the operator associated with B +'k0 according to the above analysis; that is, TO is a bounded linear transformation of 0

L2(SZ) into Hm(SZ), satisfying

(8.19)

(B + A0)[0, T0f]

=

(0, )) all

H0m(fl),

0E

Hm (fZ)

(cf. 8.17). Now if u E then u is a generalized solution of Au = f precisely if B[c, u] = (0, f), all O C Hm(fl). This is true if and only if (B + A0)[0, u] = (0, f + A0u), all 0 E Hm(fl). By (8.19), this is equivalent to the assertion that u = T0(f + A0u). That is, for

sec. 9

103

Global Regularity

0

U E H. and f E L2 (8.20)

B[c, u] = (0, f),

all

0 E Hm(S2)

u - AoTou = f.

In particular, for u E Hm, u E N(A) <----> B[0, u] = 0, all E Hm(f2) u - A0Tou = 0. Thus, if N(1 - AOT0) is the null space of 1 - XoTo, considered as an operator in L2(1l), N(A) CN(1 - A0T0). Next, suppose u E NO - A0T0); then u E L2(f) and u = AoTou, and, since To maps L2(f) into H. (Q), Tou E H,,,(o1Z). But then u = AOTou E ff (S2).

By (8.20), B[0, u] = 0 for all 0 E H, (R). Thus, u E N(A). We conclude that N(A) = N(1 - A0T0). Likewise, N(A*) = N(1 - A0To*).

Next, To, considered as a mapping of L2(1l) into L2(f), is really a composition of To, the mapping of L2(Sl) into 1Y (Q), followed by the identity mapping of H,.( q) into H0U1) = L2(Q). This identity mapping is compact, by Theorem 8.3: Thus-To is a compact operator on L2(f2) and we can apply the Riesz-Schauder theory of compact operators. Thus, NO - AoT o) and NO - \0T*0) have the same finite dimension. B[c, u] = (0, f), all 0 E Also, Au = f is solvable for u u - A0T0u = f, by (8.20). By the Riesz-Schauder theory, Hm(S2) the range of 1 - A0T0 is precisely the orthogonal complement of N(1 - A0To), whichh, as we have seen is N(A*). Thus, Au = f is solvable for u E HM(S2) if and only if f is orthogonal to N(A*). Q.E.D. 9. Global Regularity of Solutions of Strongly Elliptic Equations

In section 6 we discussed local regularity results for elliptic systems. In the present section we are again confronted with the problem of regularity. This time, however, we shall obtain regularity properties up to the boundary of the domain; moreover, we shall in the process obtain some more local regularity theorems. It should be noted that we have already given results which show that the solution of a generalized Dirichlet problem for an elliptic equation is regular in Q. For suppose u E Hm(f2) satisfies B[0, u] _ (4, f) for all 0 E Co (1l), where B is a Dirichlet form corresponding to an elliptic operator A. Suppose that A has coefficients which are smooth enough to guarantee that B[c, u] = W o, u). Then we have the result that u is actually a weak solution of the equation Au = I.

104

Elliptic Boundary Value Problems

Thus, the local regularity results of section 6, especially Theorem 6.3, give conditions on A and f which allow the conclusion that u is as regular in St as we please. We shall not dwell on this point, however, since the regularity theorems of this section imply results both in the interior and at the boundary of S2.0 The method consists of using Garding's inequality to show that if

a solution u already has a certain regularity, say u E N.(Q), then u actually has more regularity, say u CH. +,(Q). As in section 6, the principal lemmas (Lemmas 9.2 and 9.3) require a considerable amount of work; the more precise regularity theorems then follow with relative ease. We will consider only the Dirichlet problem with zero Dirichlet data. As a first result we have LEMMA 9.1. Let x° E an and assume that the coordinates can be rotated so that some neighborhood U of x°has the representation

U = ly: f(y')-ho
jy' I < ro

.

Let u E H 1(1l) and assume that u is defined and continuous in u fl S2. Then u(x°) = 0.

sec. 9

Global Regularity

105

Proof. Let Ok E Co O. For fixed y' such that l y' I < r0, and for

0
Ok(y', Ay') - h) = f

DAdyn,

f(y )

since Ok vanishes on as By the Cauchy-Schwarz inequality, 0k(3'', f(y') - h)I2 < h

f 1(y ');

)-h

f(

IDn(kkI2 dyn'

so that, on integrating With respect to y', for r < r°,

f

I0k(y', f(y') - h)I2dy' S h f

Gh

ly'I
where Gn = (y: f(y') - h
IDA12dy,

Iy'I
h1

f f)d I<'

10k(y', f(Y')

- h)I2 dy'dh

°

h1

f

h Gh f

0

IDnck 12dy

1/2h1 f 0h

I DnOkl2dy, 1

where 0 < h 1

< Let Ok

U in H l (S2); we see that

n1

h11

f f °

I

Iu(y',

f(y')-h)I2dy'dh

Y. I
-<

'/2h1 f G

1/2 h 1 I u

IDnul2dy h1

.

Elliptic Boundary Value Problems

106

Since u is continuous, on letting h 1

f

0 we obtain

I u(y', f(y')) 12dy' = 0,

ly'1<<

so that u(y', f(y')) = 0 for

l y' I


From this lemma it follows that if u E H m(Q), if x° E t91 satisfies the condition of the lemma, if D"u is continuous in a neighborhood of x° in f , and if Ial < m - 1, then D"u vanishes at x°. Consequently, if we succeed in proving the continuity of D°`u up to a portion of aQ, then it follows automatically that D"u vanishes on that portion of afl,

IaI<m-1. As in section 6, it will not be necessary to assume that u satisfies certain identities in order to establish regularity results. Instead, we shall assume only that certain inequalities hold. If, for example, u satisfies B[O, u] = (0, f)

for all 0 E C (1l), then

o IB[c, u]l
where C is some number depending on n, in, fI, u. Even this inequality is more than is needed. Indeed, we shall assume only that IB[O, u]I < CII9I Im-j.Q °

for some j: I < j< m; and we shall show that if u E Hm(Q), this inequality implies that u E Hm+j(Q). This fact will first be proved for a neighborhood of a part of the boundary which is flat. Since the principal lemma can be stated for either a sphere or a hemisphere, corresponding to interior or boundary estimates, respectively, the following notation will be convenient. By G = GR, we shall denote either the sphere {x: Ixl < RI or the hemisphere lx: Ixl < R, x. > 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
where either

108

Elliptic Boundary Value Problems

i = 1,...,n, if GR is a sphere, or i = 1,...,n - 1, if GR is a hemisphere.

(1) (2)

Proof. An argument, familiar by now, shows that without loss of generality we may assume that aaR = 0 for Ial < m; for if hypothesis is replaced by 3° of the lemma holds, then it certainly holds if aaQ zero for j al < m, since from 3° and the inequality, (Dao, aapDRu)o,G +

18[o, u]l = I

alI m

1N1Sm

IRI<m

l

>

I

I

Ia1 m

(Dao, aai3Dsu)o,G

1

(Dac, aaQDRu) - C1I1011.-1,GIIuIIm,GI

Y-

lal=m IRI<-m

it follows that, for any 0 E Co (G)

I

(Dao, aapDau)1 s (C + C11IUI1m,G)II011m-1,G,

Iai=m

I13ICm

where C 1 depends on m, n, and M. The remainder of the proof consists of an estimation of difference

quotients. Let C be a function as described above, such that C = 1 on G' and = 0 outside G". Throughout the proof the function will be fixed. For 0 < h < R - R", let vh = Sh(Cu); if G is a sphere [hemisphere], i can take on any of the values 1,...,n [1,...,n - 11. By 2° Cu E Hm(GR). Since i n in the case of the hemisphere, it is not difficult to see that vh = &h(Cu) E HO (GR). Moreover, since we may assume C __ 0 for Ix1 2 R" (and xn >'n 0 in the case of the hemisphere), we may consider u to be defined (- 0) for such x, and then we obtain a version of Leibnitz's rule: (a)DSC

.

Da-j9u,

Sa valid for x E Gr anyr > 0. Here Da-,8u(x) can be defined to be anything (say zero) outside of GR, and the formula remains valid.

sec. 9

Global Regularity

109

in GR. There-

By this convention we also have DI3vh = E Ca(G) fore, for any (9.1)

B[ O, vh] _

(DQ'O, IQI=m

13 Sm

Let k be a generic constant depending onlyJ on m, n, M, K, E, R, and R' (k will also depend on , but is fixed for fixed R and R°). In (9.1) we shall use Leibnitz's rule; we then obtain terms of the form $hD,8-y( Dyu. The proof of Theorem 3413 implies that for Iyl S m - 1

IDP-yC Dyu I1,GR+h

lIsh(DP_Y - Dyu)O GR

(9.2)

= II DR-'C

Sk UIIm,G' Thus, (9.1) implies

I (DaO, aa)3h(CDRu))I + kl

B[,k, vh]I S I

01 Im,G I IUI Im,G.

IQ I=m IN//

ICm

Now the first term on the right can be estimated as follows: I

I

(Dao, aQ,8 8h(CDI8u) )I

I I

=m

INICm

E /QI=m 1

(Dac,

5ha°10

(x + he')D)8u(x + he')I

110

Elliptic Boundary Value Problems (DacS,

I

.

where we have made use of the Lipschitz condition on aap. Thus, IB[O, VhII S I

(Dad, 8,(aaR$DRu))I +kII(kIIm,GIIUIIm,G

Y-

IQI=m 13 Ism

=I

-Y- (C8-hDQ(k, aapD)9u)I +kll01Im,GIIuIIm,G. lal=m 1131-m

We have used the vanishing of C outside of G" to transfer 8h from the

right side of (,) to the left. Using Leibnitz's rule and the proof of Theorem 3.13 again, we see that IB[cS, vh]I S I

Y-

(Da(Cs-h" aapDRu)l + kII0IIm,GIIuIIm,G

QI-m

INISm

= IB[Cs_hcS, u]I + kI I(kI Im,G I IuI Im,G.

The function CS-ho is a test function on G; here we use again the fact that for the case of a hemisphere, i n. Therefore, by hypothesis 3° of the lemma,

IB[Cs-6(k' u]I s CI0-hcl Im-1,G Thus

IB[O, vh1l s CIIC8-hOIlm_1,G + kIIoI Im,GIIuI Im,G = k(C + IIuI Im,G)Il I Im,G.

Again we have used a modification of Theorem 3.13; cf. (9.2).

Global Regularity

sec. 9

111

Since this inequality holds for every E Ca(G), by continuity it holds 0 for all 0 E Hm(G). In particular, it holds for 0 = vh. Thus, B[vh, Vh]I -< kIIVh(Im.G(C + IIUIIm,G)' BY

Garding's inequality, Theorem 7.6, IVhI

21n, G

S y0-lE'1[.%B[Vh, Vh]+XoIIVhI1O,G] 01E-1[kIIvhIIm.G(C + IIOIIm,G) * aoll%'hl 2O,G]

< kIIVhIIm.G(C + IIUIIm,G + IIVhIkO,G) IIUIIm,G +

S

'where we have utilized (9.2) with 8 = y = 0 and m replaced by 1. Dividing by I IVh I Im,G'

lsh(SU)L,G

[IVh[Im,G S k(C +

IIUIIm,G)'

By Theorems 3.15 and 3.16, this implies that D4u E Hm(G') and that D1CuI Im,G' s k(C + IIuIIm,G)'

= 1 on G', the conclusion of the lemma follows. Q.E.D. Since this lemma can be applied to any sphere contained in a domain f, it implies interior regularity. More than this, it provides regularity properties up to a flat boundary. This is true even for the normal detivative D,,u, although the proof above certainly does not show this. For m = 1, the situation is quite simple. If u satisfies the equation Since

B195, u]

= (0, f)'m for all 0 E C(il), then for any sphere S CSl we

have IB[O, u].1 -< IIfIlo,SII0II1-1,S for all 0 E Ca(S), and the lemma can be applied in S. It follows that u E H2 `(SZ) and satisfies Au = f weakly in Sl, if A is the elliptic operator with Dirichlet form B. If GR C Q is a hemisphere with the flat part on its boundary contained in (A2, and if the coordinate system is rotated so that the flat part of 49GR lies in the plane x,, = 0, then it follows from the lemma that, for R' < R, D.u E H1(GR,), i = 1,...,n - 1, whence D.Dju E Ho(GR.),

Elliptic Boundary Value Problems

112

i = 1,...,n - 1, j = 1,...,n. Since B is uniformly strongly elliptic, the coefficient ann is bounded away from zero; indeed, Rann 2 E. Hence, assuming A is really a differential operator, the equation Au = f can be written n-1

D2nu

-

ann(f -

n

n

1 ai.D7D,u i_l f-1

b,D,u - cu). 1-1

Since all of the terms on the right are in Ho(GR,), it follows that Dnu E Ho(GR,) and that u E H2(GR,). Moreover, by the estimates obtained in Lemma 9.2 for the derivatives D.D.u, i / n, IDnuHI 1.GR,

Y(C + II°II1,G)-

Thus, IIu1I2.G, < Y(C +

IIUIII,G)

Even in case 6K1 is not flat, but is sufficiently smooth, the same result holds. This will be seen later; cf. Theorem 9.7. Note further that if n < 3, then 2 > n/2, so that Sobolev's inequality implies that, after modification on a set of measure zero, u E C°(S2). Therefore, Lemma 9.1 implies that u - 0 on c311. Thus, for second-order elliptic equations Au = f, with only the assumption that f E L2(Q) and weak assumptions on the smoothness of A, it follov:s that any solution u and vanishes on tAl. Moreover, the regularity in H 1(S2) is in C results of section 6 allow us easily to give smoothness conditions on A and f which show that u is even in C2(Q) and is thus a genuine solution of the entire Dirichlet problem. We shall later in this section obtain regularity results that allow the same conclusions for any n and any in.

If m > 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 n-1

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 R-8, 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', - k-1xn)

for

xn < 0,

k-1

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

k-l

For the normal derivatives, we must have 2m+1

-k-1xn)]Ix

AkDn[u(x',

Dnu(x', Xn)Ix -0 =

n

k-1

n

-0

2m+1

= Dnu(xl, xn)Ix -0 n

Ak(- k) k-1

';

i.e., 2m+1

1

(9.5)

Ak(- k)_8

k-1

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 - k-1, k = 1,...,2m + 1, and thus is not zero.) Note that the argument - k-lxn 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

xn<0

D°`q

udx

sec. 9

Global Regularity

115

2m+1

=f

D°`O udx + I

x >0

=f x>0

k

k-i

f

xn<0

D°`O(x) . u(x', - k-lxn)dx

2m+1

udx + 2 k-i

D°`f

kAk f

D°`O(x', - kxn) u(x)dx,

xn>0

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).

k-1

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)1-1 f E/ oD"udx. ER

Y-R

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)I-I 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 k-1

T-R

_ (-1) l"I [f c(x)D"u(x)dx xn>0 2m + 1

+f

AkD°Cu(x', - k-1Xn)dx],

E

O(X)

x <0

k"1

n

where we have substituted - k-lxn for xn. This proves that the extension of D"u is, indeed, the a-th derivative of the extension of u in SR. Moreover, from (9.4) it follows that for a = jet I ID1ulI0,SR S YI IDiul IO,j,R, 1 =

1,...,n - 1.

Let 1/r E Co (SR ). Then (D2m1&, u) O,SR n

f x

=f

=f

xn>0

D2mo , udx + n

xn>0

n

udx + f

D2mLr

xn<0

n

2m+1

E k-1

Af

k x <0

D`m (x). u(x', n

n

adx +

I,

k-I

- k-lx n)dx

n

2m+1

D2mi/r

D2m0 udx n

Akf k

xn>0

(Dn2m,/,) (x', -kxn)

u x dx

sec. 9

117

Global Regularity

2m+1 2 q, Dnm

=f

, udx +

I

Akkl-2mf

k-i

xn >0

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)1-2m./.(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' 112m-J,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(DIm-Jo, DIu)O,SR

< IIVII2m-j.SR IIDIUIIO.SR S YI IDIUI I.,ERI I'I 12m-j.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 n-1

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 n-1 I IUD I1.): R,

S I Iul I j,SR, < Y(C + I-1

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 <

by Leibnitz's rule, Then (9.10) implies (9.11)

I ISshU1Im,Gs y1IlUI Im+1,G 0

for small Ihl, where y1 is independent of h. Since Hm(G) is a Hilbert space, any closed sphere {v: Ilvllm S A} in HO(G) is weakly sequentially compact. By (9.11), CShu is uniformly bounded in H(G), and so a subsequence C8 u converges weakly in HO (G). In particular, for k

any

E C' (G)

lim f CSh u . cdx = lim - f u61 h (C )dx

kG

k

k-ioo

G

k

f uD G

= fG D,u -

Ccidx.

Hence, C5;, u converges weakly in HO (G) to D,u, and so M,u E HO (G Q.E.D.

k

Now we can extend Lemma 9.2. We shall show that u and all of its derivatives of order S j are smooth up to the flat part of the boundary of G. The theorem is stated precisely in Lemma 9.6. We will use the notation 00 for the class of functions which have bounded, continuous derivatives of all orders < k on fZ.

Elliptic Boundary Value Problems

120

Definition 9.1. A Dirichlet bilinear form B[k, 1/]

=

E (Dao, aap DRqf)o.SZ lalsm

Ialsm

is right j-smooth in 0 if the coefficients aas are bounded and measur-

ableinQandaap C°`l+' m(Q)forlal+j-m>0. LEMMA 9.5. Assume that j < m and 1°

that B is uniformly strongly elliptic and right j-smooth in G;

2°

that u E H,,, (G), that Cu E H(G) for all and

3°

that there is a positive number C such that

E Co ({x:

xl < Rj);

JB[c, UIl S CIIOIIm-j,G

for all 0 E Co (G).

Then, for any R° < R, u E Hm+i(G') and there is a constant y = y(m, n, B, R, R') such that IIUIlm+j,G' S y(C + IIUIIm,G)

Proof. The proof is by induction on j. Observe that the lemma is trivial for j = 0. Thus, it suffices to establish an induction step. As in the proof of Lemma 9.2, we assume, without loss of generality, that aap = 0 for I aI + j - m -< 0. Assume that 1 S j S m and that the lemma holds if j is replaced by j - 1; we will show that it holds for j. Let R1O = 1/2(R + R10), and G= GR,,,. Since IB[

,

U]l : CIl IIm-j,G S CIIi1Im-(j-1).G,

the inductive hypothesis gives u E Hm+j-1(G"'). For j = 1, D,u E Hm+j-1(G"), i = 1,...,n - 1, by Lemma 9.2. To establish a similar result for j > 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

m-j
m-j<jaj<m IPIsm

- E (DaD,b, aap DRu)o m-j
Gm

NISm

I

(Dac, D1aaR. DRu)Q.G"1

m-jj/
(9.12)

B[DIou]

-

(D ac, D,a E aQ m-j
DRu)o,G'".

Consider any term in the sum on the right; integrate by parts lal m + j - 1 times (which is possible, since, by the inductive hypothesis, u E Hm+j-1(G"')), transfering lal - m + j - 1 differentiations from to the other side of (, )o.GI,,. By the Cauchy-Schwarz inequality, I(Dao, D,aap , DI u)o,GI S

YII(kIIm-j+1.Gllullm+j-1,G,

note that we have used the fact that the derivatives of D,aaR up to order at - m + j - 1 are bounded. since B is right j-smooth. Thus, by hypothesis 3° and (9.12), JB[O, D,u]l < CIID,0IIm-j, G" + YI I0llm-i+1,G"' Ilullm+j-1,G°,

Elliptic Boundary Value Problems

122

y(C + I

Im-(+1,Gm.

Here y is a generic constant depending only on in, n, B, R, R'. Therefore, D,u satisfies the conditions of the lemma with j replaced by j - 1, and G replaced by G"'. By the inductive hypothesis it follows that Diu E Hm+j-1(G") and that (9.13)

IIDiUIlm+j-1,G" S y(C+ lIUllm+j-1,G"' + IIDiu!Im,G"') < y(C +

IIUIIm+j-1,(,".)

(note that m + 1 S m + j - 1 since j > 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".

+

m-j
For any term in the sum on the right, there is an i Dac

=

(9.15)

n such that

DeiDQ-e 10. Thus

(DaO, aapD(3u)0,G,i = -(Da-e`0, D,(aapDRu))0,Gn

= - (Da-e 0, DiaQR . D13u)o.G"

sec. 9

Global Regularity

123 (Da_et 0,

-

aa13 D 16D ju)OIGI).

Estimate the right side by transferring lal - m + j - 1 differentiations Da-eI0 to the other sides of (,) and applying the Cauchy-Schwarz inequality. Since B is right j-smooth, the derivatives of aaR which ,occur will be bounded. For the first term we have from

I(Da eio,

Daa,8

A)O,GII S

YIIOI Im-j.G^ IIUI

for the second, by (9.13) we have I (Da

eil, aapDOD

fu)O,G" I S YI IOI Im-j.G

" YI IOIIm-j,G"(C + I IUllm+j-1,G).

Thus, (9.15) implies I(Dao, aa,8 DRU)O,G" I S YII0IIm-j.G'(C + I!UIlm+j-1,Gm)

Therefore, from (9.14), I(Dn 0,

I

DRU)a. Gn I S IB[O, u]l + YI 101 Im

a

1RIsm

(C + I IUI Im+j-1,Gm).

Let

V-

a IF'Ism

DRu.

me°,R

Then, by hypothesis 3°, I(Dnc. v)O.G,,I < CI I0IIm-j.Gn + YII0

Im_JsGo(C +

S Y(C+ I'IUllm+j-1,Gm)Il0llm-j.G"

.

IUIIm+j-1.G m)

Elliptic Boundary Value Problems

124

From the inductive hypothesis,

Hull."_1,G"' s y(C+

IIUIImoG)

Therefore, I(Dn o, V)o,Gn s y(C + I IUI Im,G)I

I0IIm-j,G',

so that, in G", v satisfies the conditions of Lemma 9.3. (Note that Dtvc L2(G") for i n, since we have shown D,u E Hm+j_i(G"), inn.) Hence, Lemma 9.3 implies v E HI(G') and n-1

iiVII;,G' s y(C + IIUIIm,G + Y1 IIDI°IIo.G" + IIVIIO,G") s y(c + I I U I Im,G)'

(We have used (9.13) in this estimate.) Since, by the ellipticity of B, a 0, it follows that mer',men

Dmu = (a

men,men

I amen.

)-1(v -

n

INIsm 14m en

0 N

E H1 (G'),

and that IIDnuilj,G' s y(C +

IIUIIm,G)

Q.E.D.

From Lemma 9.5 we can easily get global regularity theorems for strongly elliptic equations. As a first result we have THEOREM 9.6 Assume 1°

that for some j, 1S j S m, B[c, cli] is a right j-smooth, uniformly strongly elliptic Dirichlet bilinear form in G;

2°

that u E H. (G), that for every 0

H(G); 311

that f E L2(G) and that

Co ((x:

I x I < R}), u E

sec. 9

Global Regularity

125

B[0, u] = (0, f)0,G for every 0 E Co(G). Then u E Hm+,(G') and 1IUI

Im+;,G' < Y( If1Io,G +

IIUIIm,G)'

Proof. Hypothesis 3° of Lemma 9.5 is immediate, since IB[0, u]I =1(0, f)0,G s If110,GII IIO,G s

1110,00I1m_j,G

Therefore, the lemma applies and the theorem is proved. Q.E.D. As an extension, we have THEOREM 9.7. If hypotheses 1°, 2°, and 3° of Theorem 9.6 hold, and if for some V2 0 4°

f E Hk(G)

5°

B is right (m + k)-smooth,

then u E H 2.+k (GI) and IUI12m+k,G' S Y(11fhIk,G +

IIUIIr,G)'

Proof. The proof is by induction on k. By Theorem 9.6 the conclusion holds for k = 0. Suppose that k 1 and that the theorem is true if k is replaced by k - 1. Then u E Hem+k-1(G"), and (9.16)

IIU112m+k-1,G" S

Y(Ilfllk-1,G.+

IIUHr,G)

< Y(I IfI Ik,G + llUllm,G)'

Thus, for any i = 1,...,n - 1, and for 0 E Co(G") B[DI95, ul =

E (D,Da0, aQSA)0,G" lalsm R lsm

Elliptic Boundary Value Problems

126

(D2

,

D,(aQRDRu))o0o.,

IaI<m

Rlsm

E (Dao, D,aQa . DRu)o,o I 0 kSm

E

(Dao, aafDOD;u)o.Go"

IQI<m

Let a' = D .aQR; in each term in the first sum on the right we transfer all the differentiations from 0 to the other side of (, ), obtaining B[D44, u] = -

(-1)IQI(0, Da(ai DRu))ooG" - B[0, D;u]

_ - (0, A,u)o,G" - B[0, D,u], where

A. =

E (-1)IQIDaaQRDO IaIsm

101<-

Note that the smoothness of B implies aQR i C;QI+k(G) C C JaI+1(G), since k > 1. Thus, A, is a differential operator of order at most 2m, having bounded coefficients. Now B[D,c, u] = (D,q, f)o,G" = - (0, D,f)o'Gto; thus,

B[q, D,u] = ((k, D,f)o,G" - (0, A,u)o,G" = ((k, f,)O,Ggo,

sec. 9

Global Regularity

127

where ft = Dif - Aiu. Clearly, fi i Hk_1(G"). Moreover, IIfiIIk-1,G° s

IIDIf!Ik-1,Gu +

Y(I IfIIk,G" + I

IIAiuHIk-1,Gn

IUII2m+k-1,G");

by (9.16),

Ilfillk-1,G" < Y(Ilfllk; + According to Lemma 9.4, 4Du E Hm(G"), so t hat, by the inductive hypothesis applied to Diu, Diu E H2m+k-1(G and (9.17)

IIDiu

12m+k-1,G' -Y01 fillk-1,G" + IIDiuIIm,G")

< Y(lfillk-1,G + Hull2m,Gn) Y(IIfII

k,G + IIuIIm,G),

where (9.16) has been used again. Thus, we have obtained the required estimates on D to for i n. To treat the normal derivative, let A be the elliptic operator A=

E (-1)IQIDaaaRDO. Ia!.IfI-<m

The operator A can be written as a genuine differential operator with bounded coefficients, since aQR I C Jal+k(G) c C Ial(G). By (8.8) and hypothesis 3°, (0, Au) f), all E Co(G), since we know already that u E H2m(G"). Thus, Au = f, and so, by Leibnitz's rule, (9.18)

(-1)'°a

Dnmu = f +

me°,me°

baDau, IQI<2m

where the coefficients ba belong to C* (G), and b

2me

n = 0. Now

is bounded away from zero, since the Dirichlet form B is men me n uniformly strongly elliptic on G. Since, by the results above, every a

Elliptic Boundary Value Problems

128

term on the right in (9.18) is in Hk(G') and satisfies bounds as given by (9.17), it follows that Dnmu E Hk(G') and (9.19)

IIDnmuIIk,G'
Since D2mu C Hk(G') and D,u E Hem{k_i(G') for i n, it follows that D"u E W k(G') for I a I = 2m. Therefore, Theorem 2.2 and Theorem 2.3 imply that u C Hem+k(GI). The estimates we have derived, (9.16) and (9.19), then show that IIVII2m+k.G' S Y(IIfIIk,G + IIVIIm,G)' Q.E.D.

In order to extend the above results to arbitrary domains with smooth boundaries, we make

Definition 9.2. For k 2 1, an open set SZ is said to be of class Ck if 1°

about every x°E c9 there is an open neighborhood U such that for some i, u fl t9 has the representation xi = 9W )p X'= Xl,...,xl_1, Xi+1,...,Xn) C U 'P

where U' is the projection of U on the hyperplane xi = 0, and g E Ck(U'); and if 2°

U fl n is contained in the half cylinder{x: xi > g(x'), x' E U' }.

Observe that such a region u fl 9 can be mapped, by a mapping of class Ck, onto an open set with a flat part of its boundary correspond-

ing to u fl do: simply let (9.20)

Y; = xt - g(x');

then u no is mapped into ly: y, > 01. In fact, we can assume that U fl a is mapped onto a hemisphere: about x° there is an open neighborhood which is carried into a sphere centered at the image of x° having radius small enough so that the sphere is contained in the image of U; since cA2 is compact, a finite number of these neighborhoods cover it.

sec. 9

Global Regularity

129

From these remarks, it is also seen that if (1 is bounded and of class Ck with k 2 1, then S2 has the restricted cone property. For, the Jacobian of the transformation (9.20) is one. THEOREM 9.8. Assume 1°

that S2 (bounded or not) is of class C2m, that Ixl < R},

= S2 nix:

i

2°

that B[95, tti] is a Dirichlet,bilinear form of order m satisfying Garding's inequajity-in f2,

3°

that u E I0 (S2),

4°

that f E L2(S2), and that

(9.21)

B[c, u] = (0, f)0,12

for all

Co (12).

If for some j, 1S j S m, B is right j-smooth in fl, then for every

R,

u E Hm+j(SlR) and

(9.22)

IIUIlm+,.[

R

s Y(IIflIa.- + IIulla.u).

If for some k Z 0, j2 is of class C2-+k, f E Hk(9), and B is right (m 4 k)-smooth, then u E H2m+k(1ZR) and (9.23)

I lul I2m+k.0R S Y(I fl Ik.S2 + Ilul 0.9)'

Remark. If S2 is bounded, then the theorem holds with S'R replaced

by Q. This statement is also true if only 912 is bounded or if Sl is ultimately cylindrical (i.e., it for sufficiently large R, n nix: Ixl. Ri consists of a finite number of truncated cylinders).

Proof. By the definition of the class C2m, every point in ) has a neighborhood which can be mapped into either a sphere or a hemisphere by a mapping of class C2m[C2m+k if I E C2m+k]. Only a finite number of such neighborhoods are needed to cover S2R. Theorem 9.6 and 9.7 can be applied in each of the mapped neighborhoods; since the mappings perserve the desired properties of u, we obtain

Elliptic Boundary Value Problems

130

(e.g., for the second part of the theorem) for each of the neighborhoods U that u C Hem+k(U) and IIUII2m+k,U S Y(IIfl Ik,c +

IIUIIm,fl)

Since a finite number of neighborhoods U cover SZR' we obtain u E W10c

(cZ) and

(9.24)

IIUI I2m+k,0R

2m+k

Y(I Ifl Ik,SZ + IIUIlm,si).

Thus, indeed u E H 2m+k (C1 R ). To obtain (9.23) we need only apply 0 we may Garding's inequality (Theorem 7.6). For, since u E Hm(Sl), H" in (9.21) take = u to obtain B[u, ul = (u, f) (,,11.

Also, we may apply Garding's inequality to u, obtaining COI IUI I, S 1B[u, ul + A0l

s I(u,

IUIIO,-

f)0.01 +'lplIullo.SZ

IUIIO,Sl1IfI Io,Q + AOIIUI12,n

s (1/2+A0)IIUIII,SZ+1/21IfIIo,SZ Thus,

IIuI Im.

-

s co-'1(1 + A0)II IuI I O,SZ +

c012 4I lfl l o,fl.

Substituting this in (9.24) yields (9.23). Likewise, (9.22) is proved. Q.E.D.

As a corollary we have

sec. 9

Global Regularity

131

THEOREM 9.9. Assume 1°

that SZ is of class C°° and is bounded,

2°

that A is a uniformly strongly elliptic operator whose coefficients are in the class C°°(SZ),

3°

that f E C°°((), and

4°

that u E Hm (SZ) is asolutigh of the GDP for Au = f with

infinitely differentiable-irichlet data. Then u can be modified on a set-of measure zero so that u E C°°(SZ).

Proof. We are given f, g E C°G((), and uo = u - g satisfies B[c, uo1 = (0, f) - B[di, g] for all (k I Co(&1), where B is some Dirichlet form corresponding to

A. This is from the definition of GDP. Since the coefficients of A are smooth, we have B[q, g] = (0, Ag), so that for all 0 E Co (c ) B[/, u0] = ((k, f - Ag).

Now the second part of Theorem 9.8 applies, and k can be any positive integer. Thus, uo E Hem+k(SZ) for every positive k. By Theorem 3.9, uo is equivalent to a function in Cj([) for every positive j (note that SZ has the restricted cone property). Since u = uo + g, the result follows. Q.E.D. The results of this section can be used as a priori bounds. We shall first need LEMMA 9.10. If ( is bounded and in class C', if u E Cm (TI) and if D"u = 0 for I a I S m- 1, then u E Hm (0).

Proof. First we prove the theorem locally. Let x° E Q; x° has a neighborhood U such that St fl u can be mapped onto the hemisphere IR ='x: Ixl < R, xn > 01 by a mapping of class Cm(SZ fl U). Therefore, we can consider the restriction of u ton n U as a function in Cm(F,R). Assume that supp(u) C (x: I x I < R'}, R' < R, and let

Elliptic Boundary Value Problems

132

u(x - Een) uE (W)

for

X - Een E

R,

_

otherwise.

0

Fore < R - R', uE Co(YR) CHm(ER). Clearly, uE - u in H.(LR) as o e - 0. Hence u E 9.(Y. R). To prove the result globally, use a special partition of unity as in the proof of Garding's inequality, Q.E.D. THEOREM 9.11. If aaD«

A=

.

l

Z2m

is a uniformly strongly elliptic operator on Sl, if for k Z 0, Sl is bounded and in class C2m+k, if a, E Ck(St), and if u E C2a'+k(5) satisfies

Du =0 onr9lfor alSm-1, then (9.25)

I1 u1 1 2m+k,St

Y(I IAuI

lk.si + Il

o,&.

Proof. First, assume that A = A' has constant coefficients. By Lemma 9.10, u E H.O. Moreover, f = Au E Hk(Sl). Since, for cb E

o(St),

B[¢, u] = (0, Au)o,SZ = (0, f)(,,2,

where B is any Dirichlet bilinear form for A, it follows from Theorem 9.8 that (9.26)

IIUII2m+k, l S.y(llAullk,H + 1lullo.S1),

so that the theorem holds in this case. Now assume that A' has constant coefficients. Ia1<2m-1

so that

Then

133

Global Regularity

,sec. 9

IA'ul Ik.cl s I IAul Ik,cl + Yl IUI l2m+k-11 k > 0.

Thus, by (9.26) applied to A' instead of A, we have k,Sl + I IUI l 2 .+k-1.Q + I IUI I O,SI).

I

By the interpolation inequality, Theorem 3.4, for any c > 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. 84-86 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(f-1 Cl) (A*v, u)o.SZ - (v, Au)o.,Q = iIo fu NQ-1-iv .

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

an-l.)

p

v, Dnnu)

v, D n

-

a -1 . udx', D « vDn p

f Ix'I
where x' = (x1,... , xn_ 1) and dx' = A, ... dyn- 1. Therefore n-1u)

(-1)«1+'...+«n-1+1(D°+env,

- (v, Dau)

Dn

On applying this argument an - 1 more times, we obtain (-1)I«I(D«v, u) - (v,

DM U)

« -1

f

(-1)I«' I+k

k-0

D«'+ke"V

.

Ix'I
D«° n

-k-1

udx'.

Since A is e-smooth, an i CI«I(SZ), and the identity above applies to

a «v. Thus, ((-1)I«ID«(a«v), u) - (v, a«u) « -1 nc

=

(-1)Ia

G

k-0

I

n .-. o+ D«ke(av)

f

+k

«n^ k.. 1

.D

udx'.

n

Ix' I
On summing over a, we obtain (A* v, u) - (v, Au) « -1 n

Ia

k-0

(-1)

«pl+k

f

1-0

Ix'
I«IJ

-k-1

e (a av) Dnn

udx'

Ix'I
f

f-1

-

«°+k" D

n

D«-(J+1)e (a«v) Dnudx'.

Elliptic Boundary Value Problems

136

Let v)

Ng_1_jv =

on j+1

1

I«I Then the surface xn = 0 is non characteristic for NQ_1_j, since Ng_1_j contains the term (10.1)

(_1)Q-1

where a(o,...,o,e) cannot vanish since A is elliptic. Also,

f-1 (10.2)

(A*v, u) - (v, Au) = 1

f

Nf-1-jv . (-Dn)judx',

Ix'I
so that the theorem is proved in the case of a hemisphere and v vanishing for Ixi Z R'. In (10.2) the normal derivative -Dn can be replaced by any non-tangential derivative, whose direction is a sufficiently smooth function of x'. Such an oblique derivative D. can be represented as a linear combination of D 1,...,Dn. Thus for some dj X 0, djD1 = (-Dn)' + I«Isj

d«,ID«'

where dj and d.,j are smooth functions of x'. Since v vanishes for 1x' j > R', an integration by parts gives

f d1Nf_1_jv adx' Ix'L
f

Nf-1_jv . (-Dn)judx'

Ix+J
f

+

IaISj I x'I
Ng_1_jv d,,,jD°`udx'

sec. 10

Coerciveness

(10.3)

=

f

137

NQ-1-1v . (-Dn)1udx'

I z I
f

+ 1-1-<j « len

(-l)1«ID«'(d,

1NQ-1_1v) . (-Dn)«nudx'

I x'I
where a' = (al... ian- I Y 0). Note that an < j. Hence, the term with the oblique derivative of o;der j can be represented as a term with the normal derivative of order j plus terms in which the normal derivative is of order less than j. Moreover, the operator

I«'I<1-«

n

associated with (-D n) ' u is of order at most P-1-a., and does not contain the pure normal derivative of order f - 1 - an. The relation (10.3) can be written (10.4)

(NP-1-1v, (-Dn)1u)' =

(d/NP-1-1v, D8u)'

1-1

«

-«I (Niv, (-Dn) nu)', n

where

(0, &)' =

f

O(x')O(x')dx'.

.'I
Applying (10.4) with j = Q-1, we obtain Q-1

-0

(NP-1-1v, (-Dn)'u)'

(dQ_1Nov,

-« n

-0

DP-1u

NQ-1,P-J-« v, n

(-Dn)

nu)'

Elliptic Boundary Value Problems

138

E-2

+ I (NE-1-,v, (-D,,)Ju)' j-0 E-2

_ (Nav, DE-1u)' + jYo (NEt 1-jv,

(10.5)

where No = dE-1No, N _ 1 _ 1 =NE-1-j-NE_1,E-1-j,

0SjSE-2.

Since Np_1,E-1-j does not contain the derivative De-1-j, but NE-1_j does contain this derivative, the surface xn = 0 is non characteristic for NP _ 1_1, 0 S j 5 E - 2. Since dp_ 1 X 0, this surface is non charac-

teristic for N. Now a relation similar to (10.4) holds with NE_1_j replaced by 1-j, 0 < j < E - 2. Therefore, the argument just given shows that

NQ

a

2 2 (NE_1-1v, (-Dn)ju) - (Nlv, Da2 U) + E'3 2 (NE_1_ v, (-Dn)ju) , j-o 1

j-o

where the surface xn = 0 is non characteristic for NQ Therefore, (10.5) becomes F'

j-0

(NE-1-,v, (-D,,)'v)' = (N1v, DE-1u)1 + (N lv, DE-2u), E-3 + Y_

j-0 (NE-1-jv,

An induction argument then shows that Y1

E

E1 (NE_1_jv, (-Dn)'u)' =

j-0

j-0

(NQ_1_Iv'

Du)',

0 S j S E - 2.

Coerciveness

sec. 10

139

where the operator NQ_i_, is of order j and the surface x,, = 0 is noncharacteristic for this operator. Therefore, (10.2) becomes f- i 0 Dsudx', f NQ-1_,v (10.6) (A*v, u)--.(v, Au) _ Ix'I
where NQ_1_ = NQ , Now we continue with the proof of the theorem.

If v vanishes for ixl 2 R', then the theorem holds trivially for the sphere Ix: Ixl < R1, since the boundary terms vanish. Now consider an arbitrary bounded region Sl E Cf. Let 1011 be a finite open covering of a Q, of the type described in the remarks following Definition 9.2; i.e., there is a smooth mapping carrying 01 fl 1) onto a hemisphere. Let ISJI be a finite collection of spheres such that S S. C Sl and SZ - U 01 C U Si. Let 1011 and I S' l be sets sat-

isfying cc S.,

01 cc 0i,

3 CU0!,

S'

Let IQ and 1C'I be collections of functions in C°°(E,,) such that 0, supp(Q CO', supp(C') C Sj, and Y-C1 + LC' = 1 on [Z. Let v1 = C1v, v' = C'v. By the remarks above,

1 Z 0,

(10.7)

(A*v', u) - (v', Au) = 0.

The boundary regions are more difficult to handle. Consider a particular boundary neighborhood 0, and let r1 be the mapping which carries 0, f Il onto a hemisphere H. The mapping r1 carries the operators A and A* into smooth linear differential operators A, and A, respectively, on the hemisphere. The operator Al is the formal adjoint of A: if u, v E Co (01 fl 11), then r1 carries (Au, v)0 0 _

[(u, A*v)0

0a

Pa 1

1] into (A1u, v)O.H[(u, A,v)o,H], since the Jacobian of

the transformation is I. If Al is the formal adjoint of A,, then we must have (v, A,v)o,H = (A,u, v)o,H = (u, Aiv)0.H

140

Elliptic Boundary Value Problems

for all u, v E Co (N); hence, A = Ai . Thus, by the first part of the proof, for Ai, Ai, we have an identity of the form (10.6) with the oblique derivative D. chosen in such a direction that its image under r;-1 is a normal derivative on an.. Thus, after mapping back into 0, n SI, we obtain (A*v,, u)o PQ

-

(v,, Au)0

PQ

e -1

j 0

falZno, Ne-1-jtv1 .

aiu do, anj

where Nk,, is a linear differential operator of order k, and the surface all n 0, is non characteristic for Nk, C Recalling (10.7), we sum over all i to obtain (A*v, u) - (v, Au) e-1 i

=j

0

I fnu ast 0

Ne-1_1,,v,

fall

atlj

aju da j

da,

since v, vanishes outside 0,. Therefore, we must show that

1,Ne-l-,,,v, = Ne-1-jv,

1 -j, for where Ne_1_i is a linear differential operator of order which all is non characteristic. By Leibnitz's rule, we have

ae-l-iv

j +...,

where rl, does not vanish and... represents terms in the tangential derivatives and lower order derivatives of v. Observe that the sign if A is suitably normalized (cf. (10.1)). Thus, of -'Rrl, is

sec. 10

Coerciveness

141

1-Iv +..., anP-1-j

v,

where Vq, cannot vanish since the numbers 9171, all have the same sign and 2,C, = 1,-C, -, 0. Thus, the result is proved. Q.E.D. As a corqllary to the proof, we have THEOREM 10.2. Let (Dau, aapDRv)

B[u, v] = aSm

0,12

IRlsm

be a bilinear form which is uniformly elliptic in SZ (&1 bounded while aap are sufficiently smooth functions defined in TI). Then there are linear differential operators Nk(x, D), x E asp of order k, k = 0,...,m such that for u, v E C2m(SZ), m-1

31u B[u, v] = (u, Av)o,n + j-0 Jdo anj N

m

- 1-j vdv,

where

(-1)IaIDaaafD13,

A= lal<m 1,8 1s-

and the surface auu is non characteristic for Nk.

Definition 10.1. Let V be a linear space,

H.41

(j2) C V CH.(12).

A bilinear form

B[u, v] _

2 (Dau, as,8D13v) o,SZ laIsm 191:<-

is coercive over V if and only if there are constants co and ao such that c0>0and (10.8)

IRB[u, ul>- cal

aollullo,D

1,

Elliptic Boundary Value Problems

142

for all u E V. B[u, v] is strongly coercive over V if (10.8) holds with

A0=0. From Theorem 7.12, we see that a coercive bilinear form must be strongly elliptic. Indeed, it would suffice to assume (10.8) with `RB[u, u] replaced by IB[u, u]l. Then Theorem 7.12 shows we could multiply B by a complex number to achieve strong ellipticity for B. Assume that B[u, v] is strongly coercive (a strongly coercive quadratic form can be obtained from one that is just coercive by adding Iko(u, v)opc). Then, if f E L2(S2), by the Lax-Milgram theorem there is a unique u E V such that (10.9)

B[v, u] = (v, f)O,Q

for all v E V. For this we must assume V is closed in Hm(cI). Thus, we have solved the problem of finding for each f E L2(1) a function u i V such that (10.9) is satisfied for all v E V. In case O V = H ,.(Q), this is just the generalized Dirichlet problem already O discussed. In case Hm(I) is a proper subspace of V, then we have no longer solved the GDP, but some other type of problem, which we can consider as an abstract boundary-value problem. The question now arises concerning the regularity of the solution u. The interior regularity is no different from the theory we presented in section 9, or, alternatively, the interior regularity theory of section 6. Thus, in case f and B are sufficiently regular, then we can give precise statements concerning the regularity of u. Then, under sufficient assumptions, u E H'° (f ), and since (10.9) is satisfied a fortiori for all v E Co (i2), u is a weak solution of Au = f, where (-1)IaIDaaapDO.

A= laI<m

Iflsm As usual, the question of boundary regularity is more complicated. We can reduce the considerations to the case of a hemisphere {x: Ixl < R, x,, > 01, by making sufficient smoothness assumptions on Q. Then we need to consider functions C E Co (1x: Ixl < R}), as in the results of section 9. If one examines the proofs in section 9, it is evident that the same arguments will work in our case if we make

143

Coerciveness

sec. 10

two assumptions of the nature of V. First, if v E V, then v E V. Second, if v E V, then sufficiently small translations of £v tangent to

the surface x. = 0 also:give elements of V. Thus, if both these assumptions hold, then one can establish the boundary regularity of U.

As an ei ample, consider the case in which V = Hm(1Z). Both the above conditions are obviously satisfied, so that u is highly regular in fZ in case f, B, and SZ are sufficiently regular. Let us then assume that everything is sufficiently regular for the following manipulations. We shall show that u is actually a solution of a certain boundary-value problem. For, according to Theorem 10.2, M-1

B[v, u] = (v, AU)p.Q +

aiV

fau an i

N2m-1-jUdv

for all v E C2m(1l). But (10.9) implies B[v, u] = (v, Au)0 , so we obtain m-1

J-u

ajV

Nzm-l-Judo = O.

fc31Z

Since the functions aiv/ani are essentially arbitrary, this implies that on o1Z (10.10)

N2m_1_JU = 0,

j = 0,...,m - 1.

Recall that N2m_1_J is order 2m - 1 - j, and N2m_1_Ju contains the a2m-1-iu/an2m-1-i with a non-vanishing coefficient. Thus, term the boundary conditions for u are specified by m differential operators on a1Z, having orders m, m + 1,...,2m - I. Note that in this case the boundary conditions actually depend on the choice of the Dirichlet form B. These are so-called natural boundary conditions. The idea can be generalized. Let Bo,...,Bk-1 be sufficiently smooth linear differential operators on a1Z, where k 5 m and the order of B. is less than m. Assume that the orders of B.,...Bk_1 are distinct, and that ale is non characteristic for these operators. In this case V shall be the closure in Hm(1Z) of the set of functions 0 ci Cm(fZ), such that B,O = 0 on ai, i = 0,...,k - 1. If k < m, let Bk,...,Bm_1 be differen-

Elliptic Boundary Value Problems

144

tial operators on act, such that (A2 is non characteristic for these operators, and if 0 < j < m - 1, then, for some i, BI has order j. Then, by the proof of Theorem 10.2, we obtain a Green's formula m-1

B[v, u] = (v, Au)o.,Q + I f Biv M2m-l-iudv, i-o act where M2m-1-i is an operator for which act is non characteristic. If u E V and (10.9) holds for all v E V, then we obtain m-1

Io fad Biv M2m-l-iuda= O. v E C-(Sl), B,v = 0 on act, i = 0,...,k - 1. Thus, since Bkv,.... Bm_1v can be chosen essentially arbitrarily, we obtain (10.11)

M2m-1-iu = 0

on

afl, j = k,...,m - 1.

Thus, in addition to the given boundary conditions, Bou=...=Bk-1u = 0 on acl, the function u also satisfies the natural boundary conditions (10.11). Note that if k = m, then no natural boundary conditions a appear.

It is therefore seen that the critical fact needed for treatment of general boundary-value problems is the coerciveness of a quadratic form. For an example of the. usefulness of this criterion, let us consider the case of second-order problems (m = 1), and a strongly elliptic operator (10.12)

n

A= -

1,i-1

n

a11D1D

I a1D,

+ 1-1

+ a,

where ali is real, and for real e, n

(10.13)

E 1,1-1

By the procedure given in section 8 we can associate with A a

Dirichlet bilinear form,

sec. 10

Coerciveness

(10.14)

B[v, u] =

145

n Y.

n

(Div, b1u)

(Div,

n

+

Y. (v, ciDiu) + (v, du), i-1

in a non-unique fashion. The form B comes from integrating by parts so that (10.15)

B[v, u] = (v, Au)

is satisfied in case v, u E Co (St). Now one possible selection for B has the principal part (10.16)

n

f a,,DiuD,udx, fZ

so that the condition (10.13) implies the ellipticity of this particular form, and, hence, by (10.15) the ellipticity of B. Note that Garding's inequality is trivial for this second-order case, since the integrand in (10.16) is already bounded below by n

Eo

i-1

IDiuI2

so that the inequality

I

i.1-1

f aiJDiuD1udx

Eolul i.fl

SZ

is immediate. In higher order problems it is no longer the case that the integrand in B[u, u] is positive. Since the lower order terms in (10.14) can be easily estimated in terms of u1 i S with a small coefficient, and l u l o,Q, it follows that B is coercive over H1(Sl) if we assume only (10.13). As we have remarked, the boundary conditions are connected with

the choice of B. In the second-order case, the computations are fairly simple. Let us write A in the form

Elliptic Boundary Value Problems

146

A=-

n

n

n

i-i

i-i

aiiD1Di+ E a.D,+ F. 0Di+a,

where, of course, by comparison with (10.12) we must have

aii + aii = aii + aii,

ai+0i=ai. Then, if ni is the ith component of the exterior normal to M, Green's formula gives (v, a1Diu)

(D1(aiv), u) + f

vaiunido,

r7Sl

- (v, aiiDiDiu) = (D.(aiiv), Diu) - fdo vai,Diunido, assuming aii to be real. Thus, summing these expressions gives

(v, Au) = B[v, u] + f

[- Ii,i aiin1Diu + !iainiulvdo,

for a certain bilinear form B. Thus, the associated boundary condition for u is (10.17)

- 1i.iaiiniDiu + Y-tainiu = 0.

In case aii = aii = aii, the expression Y-i.iaiiniDiu is called the conormal derivative of u'corresponding to the operator A; if A = A (the Laplacian), the conormal derivative becomes Iin1Diu = au/an, the normal derivative. Usually, of course, the boundary condition is given in advance. Thus, it is important to be able to pick the form B in such a manner as to obtain the given boundary condition. That is, aii and ai must be chosen in such a fashion that (10.17) coincides with the given boundary condition. We shall now demonstrate how this can be accomplished in the case that A = A. Suppose then thi t the given boundary condition is

sec. 10

Coerciveness

an +

147

gau + au = 0,

where r = (rl,...,rn) is, a given smoothly varying unit tangent vector. Since the problem. .of choosing the quantities a, is trivial, we concentrate orii the choice of ati. Now the given boundary condition can be written

an+.igr3Diu+au=0. Since A = A, the coefficients aii must satisfy (10.18)

alt = 1,

ati+aii=0, iT j. we shall show that we may indeed choose

a,i = ail + (nirl - nirt)g. Here Sii is the Kronecker 45-function while the functions n,, r, and g which a-priori are defined only on the boundary are supposed to possess smooth extensions into Sl. Using these extensions, ati are smooth functions in &1 which obviously satisfy (10.18). Moreover, the boundary condition (10.17) becomes

0 = l'i.jsii

+

(niri

- nirt)g]niDiu - Etatntu

= Y,iniDiu + g!ir1DiuI ni - gYiniDiuYinirt - Y,a,n;u

an- +g

au+0-Iiatntu,

since on %TI En; = 1 and Fniri = 0. Therefore, this boundary condition is equivalent to the given one, if the quantities ai are chosen correctly.

Elliptic Boundary Value Problems

148

We can also treat mixed boundary-value problems using this procedure. For instance, suppose we wish to obtain a Dirichlet form giving the boundary conditions

(9f +

au

O on a 1 Sl,

u = O on a29,

where a,9 and a2sl are disjoint; and their union is 9 1. Then we take for V the closure in H ,(Q) of the class of functions in C'(5) which vanish vanish on a2sl. And then the Dirichlet form must be chosen in such a manner that the boundary condition on a,sl emerges as a natural boundary condition. It should be noted that our regularity theorems hold except in neighborhoods of points in a1Sl n a2sl. Our theory fails there because at such points tangential translations of functions in V do not remain in V. And, indeed, singularities actually occur at such points. To conclude this section, we shall briefly mention some a priori estimates. Let A be a sufficiently smooch, uniformly strongly elliptic operator on Sl, let SZ E C2, and suppose that u E H2(Sl) n C1(Sl)

satisfies

X

+au=0 on oSZ,

where a/ae is a smoothly varying directional derivative that is not tangent to asl. From the regularity theorems, we obtain the a priori estimate I Jul I

2,S1

S C(l lAul I o,Sl +

I Jul I

o,Si).

We can obtain such results for strongly elliptic second order problems which are not real, such as problems for the operator D + D2 + i e10D 2, where 0 is a constant, 101 < rr.

Coerciveness

sec. 10

149

n

B'[v, ul _

],)- 1 (Dtv, atjDju)o,SZ

corresponding to the principal part A' of a strongly elliptic operator, n

A' 1,1' 1

Now n

2

'B'[u, u]

7. 1[(Diu, aliDiu)o.0 + (a,1Diu, D,u)o.SZ]

n

_

1,1-1

(Diu, a*jDiu)o.Q,

where a*., = 1/2(a}j + a. For real

'

r

n 1

n

r

a1J Slej

r 1a1Je,

rj > 0.

Hence, for some positive constatn co, n

calel

a1jete/ 1.1- I

It follows that RB'[u, u] = f

n i.i-1

D1ua 1iDiudx

r.

Co

I-1 2

2

ID lul o,Q

= Co11111,9-

Elliptic Boundary Value Problems

150

Therefore, B' [u, u] is coercive over H 1(1). Hence, we obtain the regularity results and the a priori estimates. As an application, we present the following trick which will be useful later. Let A be a uniformly strongly elliptic second order operator over SZ, normalized in the usual fashion, such that A has real coeffi-

cients. For x = (x1,...,xd, let (x, r) be a vector in Ea+1 For some constant 0,101 < Tr, let

L=A-e'0DT on the cylindrical domain F = SZ x E 1. Then L is uniformly strongly

elliptic on F, as is easily verified. Suppose that u E H2(F) fl C1(F) satisfies

d + a(x)u = 0 on aF, 0 -f

where f is a smoothly varying vector parallel to the I plane, but not tangent to c%1. Then we have the a priori estimate Jul I2,r s C(I ILul l o,F + IJul

We note that if B[v, u] is a Dirichlet bilinear form corresponding to A, then

B,[v, u] = f

B[v(x, t), u(x, t)] dt + eie(Dty, Dtu)o,r

is a Dirichlet form corresponding to L. Finally, we present a brief remark concerning the biharmonic operator in two variables. As mentioned on p. 96, there are the following two distinct Dirichlet forms corresponding to A 2. The first, B 1 (AO, At/i), comes from the Green's formula (v, A 2u) = (Av, Au) + f

(v

AU

an

and gives the associated boundary conditions

)do,

sec. 11

Coerciveness results of Aronszajn and Smith

151

Au = 0,

an

Du = 0.

This is not regular boundary-value problem; the form B 1 is not coerciv "over H2(n) since the space of harmonic functions in SZ has infinit'emension. However, the second, V1 = ([D2 - D2]0, [Di

- D2] 2]1p) +4(DID 0, DID

leads to a regular boundary problem. This form is coercive over H2(&2), since the polynomials 62 - 62 and 26162 have no common complex zero, except 6= 0. Cf. Theorem 11.9. More generally, it can be shown that if one considers the Dirichlet form tB1 + (1 - t)B2, t a real parameter, then this form is coercive over H2(1) if and only if - 3 < t < 1. For a proof see S. Agmon, Proceedings of the International Symposium on Linear Spaces, Pergamon Press, Jerusalem 1961, pp. 1-13. 11. Coerciveness Results of Aronszajn and Smith

In this section we shall give some very precise conditions for a special type of quadratic form to be coercive over Hm(SZ). Before obtaining these results, we need two auxiliaries: some results of Calderon and Zygmund, and the Sobolev representation formula. Definition 11.1. A Calderon-Zygmund kernel is a measurable function on E,, - 101 satisfying the conditions: 1°

k is (positively) homogeneous of degree -n; that is, k(ax) = a-"k(x),

2°

a > 0,

x 4 0;

and

for lal = 1 k(a) is a bounded measurable function on the unit sphere Y., and k has mean value zero; that is, f k(a)da = 0.

I

Elliptic Boundary Value Problems

152

In general, a function h(x) is said to be (positively) homogeneous of degree r if h(ax) = a'h(x) for a > 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 Calderon-Zygmund 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 Calderdn-Zygmund 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 non-negative 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)rn-1 dadr = - f f r° eo

0

0

h(Q) p

r n- 11

(r)ar n-1darlr,

which simplifies to f

(11.1)

ear) dr f (Dih)(a)da = - f

w

Since f r-lp(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. 479-481.

c. 11

Coerciveness results of Aronszajn and Smith

153

LEMMA 11.2. Let k be a Calderon-Zygmund 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 Calderon-Zygmund 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 Calderon-Zygmund 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

1-ndy.

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

R-ioo

ff

h(x - y)f(y)D;O(x)dydx.


By Green's formula, we have for fixed y (11.7) - f

h(x - y)O(x) R

h(x - y)D1O(x)dx = - f

E
Y'

dax

x-Y I -R -f

h(x - Y)c(x)Y

" dax

Ix-yI-c

+ f

D;h(x - y)q(x)dx.

E
Suppose that f(y) = 0 for IyI 2 A, and suppose q(x) = 0 for IxI > B. Then if IyI < A and R > A + B, it follows that for Ix - yI = R we have IxI > R - A > B, so that c(x) = 0. Thus, if y E supp (f) and R > A + B, then the integral in (11.7) taken over the sphere Ix - yI = R vanishes. Also, if in the integral over the sphere Ix - yI = c we set x = y + ea, then we have

Elliptic Boundary Value Problems

156

- f

Y)o(x)yi-x,dax

h(x -

h (Q) ):(n - I 0

= f

E

+ EQ)O Endo 1

+ ca)da

+ f h(a)a,[0(y + ca) - 0(y)]da, E

by (11.5). The last integral tends to zero with c uniformly for y E En,

since D,0 is bounded. Thus, from (11.7) we have for R sufficiently large and for y E supp (f)

-f

h(x - Y)D,O(x)dx = f

k,',R(x - Y)o(x)dx + c,o( ) + JE(y),

En

E
where lim JE(y) = 0, uniformly for all y. Here kE R is defined as in H0 (11.2). Substituting this is (11.6),

-f

En

(h*f)(x)D,c(x)dx = lim f f(y)dyf kER(x - y)O(x)dx + c, f f(y)o(y)dy E-,0 E n R-?oo

lima f

= lim f E-4 0

R-,oo

En

n

kE,R(x - Y)f(Y)dy + c,f

cb(x)dx f n

R-,oo

E

n

*f)(x)o(x)dx + c, f

E n

f(Y)O(Y)dy n

En

f(x)o(x)dx.

By Lemma 11.3, kE,R*f converges in L2(En) to k'*f. Thus, we have

-f

(h*f)(x)D,O(x)dx = f [(k'*f)(x) + c,f(x)]o(x)dx. n

Q.E.D.

n

We now pass to the proof of the SOBOLEV REPRESENTATION FORMULA:

sec. 11

Coerciveness results of Aronszajn and Smith

157

THEOREM 11.5. Let C be an open cone of height h and opening 0 with vertex at the origin. That is, for some unit vector 60, the axis of the cone, C = 16: 161 < h, 6. 6° > I6l cos 01. Let I be the surface of ,the unit sphere, and let A be the portion of E subtended by C. Let 0'E C00(Y.), supp (0) C A, and suppose

,

f q(o)do = (-i)m (m - 1)!

E

where do is the surface element on E. Let u E Cm(C ), where u = 0 for Ixl Z h - S, some S > 0. Then

f m! 0(Y) yf D°`u(y)dy

u(0)

c a!

lyj jyI°

Proof. Let o be any unit vector such that o 6°> cos 0, and consider the function of a single real variable, u(to), 0 S t S h. Expand this function in its Taylor series about the point t = h. Then, since u(ta) vanishes for t . h - S, u(0) _m-1)1)1 fh tm-I dmm u(ta)dt. ( Multiply this relation by 0(01) and integrate over A to obtain (11.8)

u(O) = f D(a)da fhtm

' dtm u(ta)dt.

Now

(11.9)

dtm

u(to) _

m!

0-(D"u)(ta).

This formula can be proved by induction on m. For, if it is valid for a given value of m and we differentiate once more, we obtain (11.10)

dm +i

dtm;'

u(to) _ I«I -m

m! a« Y or (D"" a! i

i-i

u)(ta),

Elliptic Boundary Value Problems

158

so that, for any j3, 1 P I = m + 1, the coefficient in (11.10) of o8DRu is

n

m!

M!,

i-1 ( - e')!

=(m+1)!. Thus, (11.9) is established by induction, as it is obvious for the case m = 0.

Substituting (11.9) into (11.8), tm-i m!

fh

u(0) _

a!

I«I-m A

t(a)a°`(D"u)(ta)dtda.

Setting y = ta* u(0) =

fh

I«II

)"(D"U)(y)dtda.

IyI ) Iylm-1(

a!

IyI

A Since the formula for volume integration in polar coordinates is dy = tn-ldtda, we obtain u(0) _

I f ! 0(y)

I-I-m C a!

Y«IYI -°D"u(Y)dy.

IYI

Q.E.D.

The preliminaries having been completed, we shall now discuss some important coerciveness results due originally to Aronszajn; we shall give proofs of K. T. Smith. First, two lemmas are needed. LEMMA 11.6. Let SZ be a bounded domain having the segment

property. Let B[u, u] be a coercive quadratic form over a subspace

sec. 11

Coerciveness results of Aronszajn and Smith

159

L of H(SZ). If for all u E L u]I s

(11.11)

M-1,0,

CIIUII2

then L is /finite-dimensional.

Proof-) Since B is coercive over L, (11.1) and Definition 10.1 imply (1,1. 12)

colIUI Im S (C + ko)I

IUIIm-1

for

u E L.

Although L CHm(SZ), we shall consider L as a subspace of Hm_10; let L be the closure of L in the space Hm_1(u1). It follows from (11.12) that L CHmU) and that (11.12) holds for u E L. Let {ukI C L, IIukIIm_1 S 1. By (11.12), the norms IIukIIm are bounded. Thus, Rellich's theorem, Theorem 3.8, implies that a subsequence of {ukI converges in Hm_1(i2). Therefore, the unit ball in L is compact. But if the unit ball in a normed linear space is compact, the space must have finite dimension. Thus L is finite-dimensional. Q.E.D. LEMMA 11.7. Let SZ be a bounded domain having the segment

property. Let B[u, u] be a coercive quadratic form over a subspace V of Hm(SZ), and suppose RB[u, u] 0 for u E V. Then the set

L={u: uEV, RB[u, u]=01 is linear and of finite dimension.

Proof. By Lemma 11.6. we need only show that L is linear. For

u, vEV, let B 1[u, v] = 1/2B[u, v] + 1/2B[v, u].

Then B 1 [u, u[ = 91B [u, u] 0 and B 1 [u, v] = B 1 [v, u]. Thus B 1 is a Hermitian symmetric, non-negative definite, bilinear form on V, and the Cauchy-Schwarz inequality must be valid for B 1: IB1[u, VII
RB[v, v]1'/2.

Elliptic Boundary Value Problems

160

Thus, if u, v E L, then B [u, v] = 0, and 1

B1 [u + v, u+v]=B1[u, u]+2%B1[u, v]+B1[v, v] = 0,

or, JRB[u + v, u + v] = 0, so that u + v E L. Obviously, L is closed under scalar multiplication. Thus, L is linear. Q.E.D. We now turn to the results of Aronszajn and Smith. The first result is almost trivial, using Garding's inequality and the previous lemma. THEOREM 11.8. Let P1(e),...,PN(6 be homogeneous polynomials of degree m, having constant coefficients. For u, v E Hm(S2) let (11.13)

B[u, v] _

f Pk(D)uPk(D)vdx, J2

where Cl is a bounded domain. 1°

A necessary and sufficient condition that B be coercive 0

over Hm(SZ) is that there is no common non-zero real zero of 220

If Cl has the segment property, a necessary condition that B be coercive over Hm(9) is that there is no common nonzero complex zero of n

Proof. Note that the non-vanishing of Y

for all non-

zero . is equivalent to the ellipticity of the form B[u, u]. Thus, in 1° the sufficiency is an immediate result of Garding's inequality, and the necessity follows from the converse of Garding's inequality, Theorem 7.12. Thus, 1° is merely a restatement of previously derived results. To prove 2°, suppose z = (z1,...,zn) is a common non-zero complex u E Hm(O) since SZ is bounded. zero of P1,...,PN. Let uA(x) = Now

Pk(D)uA = Pk(Az)eAZ-R

sec. 11

Coerciveness results of Aronszajn and Smith

161

_ AmPk(z)exz-x = 0,

0. Obviously, RB[v, v].> 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 integro-differential forms, Bull. Amer. Math. Soc. 67(1961), so that B[LA,

368-370.

THEOREM 11.9. Let be homogeneous polynomials of degree m, having constant coefficients, and let have no common non-zero 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-

k-1

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. 5-6. 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-

1-1--,

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_

k-1

fC

c«(y)A«k(D)Pk(D)u(x + y)dy. 1

We wish-to 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

lal-m'

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

k-1 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 k-i

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 non-zero 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. This-result 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 non-zero real zero; and

2°

the polynomials Pk(x°, D) posses no comfor each x° E mon non-zero 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 non-zero 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<m-1

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°GI-m

+ 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

lal-m

1S kSN

+NC1C2

Ilullm-1 +

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«1-m

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. 1-1

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 non-zero 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 Cauchy-Schwarz 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 non-zero 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 Calderon-Zygmund 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"I-m

fE

95«1(x - Y)w",(Y)dy,

y). Now let

Here we have set ¢«1(y) (11.32)

v,(x)

,(x)

If

1"1-m 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.

1-0

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)DR-YV 1(x) Y-
. fDYta,(x - y)

WQ1(Y)dy,

Y

if

1. Here DYE«1 is homogeneous of degree m - n - IYI We now need, to estimate the L2 norm of In this computation K will stand for a generic constant depending only on SI, the functions 95a1 and Vi,, and m. Suppose f2 C U 0' C{x; IxI < at. Then the L2 0

norm of any term on the right in (11.34) is no greater than IDS-Y `1(x)I2dxjf DY0«1(Y) . Wa1(x - Y)dYI2]'/2

K[f

n

I

(11.35)

IYIm-nIYIIwa,(x - Y)I dyj2]%,

S K[fo,dxl f I

En

where we have again replaced x - y by y in the inner integral, and have used the homogeneity of DY95«1. By the Minkowski inequality, (11.35) is no greater than lyI2(m-n-IYI)Iw«1(X-

KfE dy[f n

IYIm-n-IYldy[f01

=Kf n

(11.36)

y)I2dx]14

7

= Kf IYIm-n-IYIdy[f E n

Iw«1(x - Y)I2dx]'/

I

Eo!i

ID«u,(X - Y)I2dx

For x E 01, and x - y E 9, we have xj < a and Ix - yI < a, so that IyI < 2a. Thus, since IyI S m - 1, (11.36) is no greater than

Elliptic Boundary Value Problems

174

K(2a)m-lyl-IJ,.I<2Blyl-n+ Idyluilm.i

S Kluilm.f2.

Therefore, summing over y in (11.34), we have (11.37)

IViIIm-I,Q S KIUlm,Q'

Now we need to estimate D1D13vi for j = 1,...,n and 1131 = m - 1. In computing DiDgvi from (11.34), terms arise which can be estimated

exactly as above, and we also have a term of the form (11.38)

Vji(x)D1 f D13Oai(x - y) . wai(y)dy.

Since D13cat is homogeneous of degree 1 - n, the inequality of Calderdn and Zygmund (Theorem 11.4) implies that the £2 norm of (11.38) is no greater than K11w.ill o,E S KIUilm.si. Thus we have n

IVtlm,O S Klut,lm,,Q

Combining this with (11.37) yields (11.39)

S KlUilm.S2'

llvil

Now (11.33) implies that V

(11.40)

11v1

m,E

S K '-O lull..(

Next, Leibnitz's rule implies v S KIIUI lm.ci.

Therefore, (11.40) implies IIVI lm,-" S Kl Jul lm,,Q n

Therefore, the correspondence u -. v is a linear transformation of cm (n) fl H. (cI)into H. (En) with the properties

sec. 12 I

l

Some Results on Linear Transformations

175

'I Im,En < Kl lul lm,U.

v=uonU. Since:"this transformation is a bounded transformation from a dense subset of HmUI) into Hm(En), it may be extended in a unique fashion to'Hm(Tl). Moreover, it follows that the extended mapping still satisfies the requirement v = u almost everywhere on Q. Q.E.D.

Remarks. The same argument shows that we can use LP norms (1 < p < oo) everywhere instead of L2 norms. The only additional tool is the Calderon-Zygmund estimate for LP instead of L2. We could also obtain estimates for extensions of u preserving Holder continuity of derivatives of u. Here a slight change appears, since the extension of D°`u, outside c can no longer be w,. (cf. (11.29)), but must be an extension perserving the Holder continuity of D°`u..

12. Some Results on Linear Transformations on a Hilbert Space

This section is devoted to preliminary material needed in the discussion of eigenvalue problems in section 13. It is divided into three parts. Part 1. ELEMENTARY SPECTRAL THEORY

Definition 12.1. Let A be a linear transformation defined on a (not necessarily closed) subspace D(A) of a complex Hilbert space X; D(A) is the domain of A. Let R(A) be the range of A; that is, R(A) = lAf: f E X j. The resolvent set of A, denoted p(A), is the set of all complex numbers A such that A - A is one-to-one, R(A - A) = X, and the inverse (A - A)-' is a bounded linear transformation on X. Here the notation A stands for the transformation mapping f into Al. all f E X. The spectrum of A, denoted a(A), is the set of all complex numbers not in p(A). THEOREM 12.1. The resolvent set of A is open, and (A - A)-' is an analytic function on p(A). If A is bounded, then p(A) D lA:

IAI > IJAllj.

Elliptic Boundary Value Problems

176

Proof. If ko E p(A), then

A-A=(Ao-A)+(A-ao) (12.1)

_

[1

- (Ao - A)(Ao - A)-11(,k0 - A).

The operator 1 - O LO - AA O - A)-1 is. invertible if

I I(A

0 - A)(1k o - A)-111 < 1,

or, if

- A)-111-1-

I'A0 - AI < I I(A0

Moreover, we have the Neumann expansion [1

- (A0 -

A)(Ao

DO

-

A)-11-1 =

Y.

k 0

(A0 - k)k(X0 -

A)-k

where (A0 - A)'k = [(A0 - A)-11k. Therefore, (12.1) implies that for IA0 - AI < II(A0 - A)-111-1, A E p(A), and the continuity of (A0 - A)-1 implies that (12.2)

(A - A)-' =

(A o

-

,q)-k-1

k-0

Thus, (A - A)-1 is expressed in terms of a power series about any point A0 E p(A); that is to say, (k - A)-1 is analytic. Finally, if A is bounded and Al I> IIAII, then

A-A=A(1-A 'A). and I IA-1AI I < 1. Thus, the above argument shows that k E p(A). Q.E.D.

Definition 12.2. The function (A - A)-1, defined on p(A), is the resolvent of A. Note that (12.2) implies II(A -

A)-'I I S

kY0

Iao - kIkII()Lo -

A)-1IIk+1.

sec. 12

Some results on Linear Transformations

177

Summing this geometric series, we obtain a very useful estimate (12.3)

1 l(, k-

I Vo

A)-'Is

A) -'I I :

Iko - AI <

1- l'ko - Xl I I'Ao-A)-1ll I I(A0 -

A)-ill-1

Frequently in dealing with integral equations, we shall consider -relations-of the form

f=g+,kAf, in which it is desired to solve for fin terms of g. In case 1 - AA is invertible, i.e., A-1 E p(A), the solution will be given by f =(1 - XA)-1g (the case A = 0 is obviously without interest). It will prove to be ,kA)-1 itself, very fruitful, however, not to analyze the operator (1 but, instead, A(1 - AA)-1. One reason for this is that the range of A(1 - kA)- 1 is the same as the range of A, and we shall frequently have some very precise information about the range of A. Note that 1 = (1 - AA)(1 -,kA)-' _ (1 -,kA)-1

- AA(1 - \A)-1,

so that (12.4)

(1 -

AA)-1

= 1 + AA(1 - kA)-

This notion is in the following definition. Definition 12.3. The modified resolvent set pm(A) of A is the set of all non-zero complex numbers A such that 1 - AA is one-to-one, R(1 - kA) = X, and (1 - \A)-1 is a bounded linear transformation on X. If A E p. (A), then the transformation AX = A(1 - kA)-1 is the modified resolvent of A. Note that (12.4) can be expressed in the form (12.5)

(1 - AA)-1 = 1 + kAA.

Elliptic Boundary Value Problems

178

Also note that (12.6)

A ipm(A)f=->A-1Ep(A).

It should be remarked that if the linear transformation is closed, then the condition in Definition 12.1 that (A - A)-1 be bounded is superfluous. For, if A - A is one-to-one and onto, then (A - A)-1 is closed (since A is closed) and defined everywhere on X. Then the closed graph theorem implies that (A - A)-1 is bounded. Likewise, if A is closed, then the condition in Definition 12.3 that (1 - AA)-1 be bounded is superfluous. It should also be remarked that if D(A) = X, and if A E p(A), then A and (A - A)-1 commute. For, in any case, (A - A)(A - A)-1 = 1 and (A - A)-1(A - A) is the identity on D(A). Thus, if D(A) = X, then (A - A)'1(A - A) = 1. Likewise, if D(A) = X and A E pm(A), then A and AA commute.

We now state the analog of Theorem 12.1 for modified resolvents.

THEOREM 12.2. The modified resolvent set of A is open, and AA is an analytic function on pm(A). If A0 E pm(A), then for IA - A01 < IIAA

0

II-1,A/ 0,AEpm(A)and 00

(12.7)

AA = E (A k0

1, 0

in the sense of convergence in the operator norm. Also, IIAA II

(12.8)

I IAAI I S 1 _

JAX

IA - X 0 I IAA 0 1

I,

IA - A0I <

If A is bounded, pm(A) a {A: 0 < Al < AII`'(-

Proof. We have

1-AA=1-A0A+(A0-A)A [1 - (A - Ao)A(1 - A0A)-1](1 - AQA)

I

I-'

sec. 12

Some results on Linear Transformations

= [1 - (A Thus, if JA -X01 <

AO)AAu ](1

179

- '\0A).

1, we as before, [ 1 - ' have, ,

;'(1 - AA)-' _ (1- AQA)-'

k10

(A - Ao)kA

.

0

Multiply-i-ng by A, we obtain (12.7). The other assertions are then immediate. Q.E.D. In the situation of interest to us, A will be a differential operator, and there will exist some number A in the resolvent set of A. For example, if we are interested in the generalized Dirichlet problem with

zero Dirichlet data, X = L2(fl) and A is defined on H.O. By Theorem 8.5, if Garding's inequality holds, then - A E p(A) for all sufficiently large real A. Also, if ( is bounded, then (A + A)-' is compact. If we consider the differential operator A + A instead of A, then this differential operator is itself invertible and has a compact inverse. The abstract case of this situation is a special case of what we shall consider. A simple result shall now be given. THEOREM 12.3. Let 0 E p(A) and let T = A-'. Then for X / 0, A E pm(T) if and only if A E p(A); if A E pm(T), then (12.9)

TA = (A - A)-'.

Proof. We have (A - A)T = AT - AT

=AT - 1. The assertions about pm(T) and p(A) are now easily checked, and then (12.9) is immediate. Q.E.D. Definition 12.4. A complex number A is an eigenvalue of A if there exists a non-zero vector f such that Af = Af. A complex number A is a characteristic value of A if there exists a non-zero vector g such that AAg = g.

180

Elliptic Boundary Value Problems

Note that every characteristic value of A is non-zero and that k is a characteristic value of A if and only if k-1 is an eigenvalue of A. Definition 12.5. Let k be an eigenvalue of A. A non-zero vector f is a generalized eigenvector of A corresponding to k if for some positive integer k, (k - A)kf = 0. The set of all generalized eignenvectors of A corresponding to k, together with the origin in X, forms a subspace of X, whose dimension is the multiplicity of the eigenvalue X. Likewise, if k is a characteristic value of A, then a non-zero vector f is a generalized characteristic vector of A corresponding to k if for some positive integer k, (1 - kA)kf = 0.

If we use the notation N(A) for the null space of a transformation A, then f is a generalized eigenvector of T corresponding to k if and only if f 0 and f is in the set 00

M(k) = U N((k - T)k). k-1

Thus, the multiplicity of k is the dimension of this set. Note that this set is a subspace of X, since N((k - T)k) CN((k - T)k+1), k We shall call M(k) the generalized eigenspace of the operator T

1.

corresponding to X.

THEOREM 12.4. Let T be defined on X, and let ko E pm(T). If 1, then k is an eigenvalue of T if and = 1, then k E p(T). If kko only if 11 is an eigenvalue of TX . Furthermore, the multiplicity AX 0

If

0

0

of k as an eigevalue of T is the same as the multiplicity of k1-kk 0

as an eigenvalue of TX. Indeed, the set of all generalized eigen0

vectors of T corresponding to k is the same as the set of all generalized eigenvectors of Tk corresponding to k 0

1--kko

Proof. The first assertion follows from (12.6). Suppose kko Then by definition of TX (Definition 12.3), 0

1.

Some results on Linear Transformations

sec. 12 Tk 0

-

k

=

[T

-

1-AA0

k (1 - k0T)](1 - k0T)-1 1-kko XX

= [(1 +

)T -

°

1-AA o

=

181

1

k

](1 - k0T)-1

1-kko

(T - k)(1 - k0T)-1.

1-kk0

Therefore, since T -,k and (1 - k0T)-1 commute,

(TX - 1

(12.10)

11

° )k =

(1-

(T - k)k(1 - k0T)-k

1

0

The relation (12.10) for k = 1 implies that T - k is one-to-one if and only if TX - k is one-to-one. Thus, k is an eigenvalue of 0

T <-->

k

1-kk0

1-XX 0

is an eigenvalue for T1k. Moreover, (12.10) implies that o

)kf=0 Q.E.D.

(T - k)kf = 0 <--> (Tko-1 A 0

Before giving the next definition, some notation will be convenient. For 0 real and a > 0, let C(O, a) = {rei0: r > al.

Definition 12.6. Let A be a linear transformation defined on a subspace D(A) of a Hilbert space X. A vector e'0 in the complex plane is a direction of minimal growth of the resolvent of A if for some positive a, -(0, a) C p(A) and I I(k - A)-'I I = 0(IAI--1) for k E ;(O, a),

Al ILikewise, a vector e'0 is a direction of minimal growth of the modified resolvent of A if for some positive a, _(O, a) C p.(A) and IIAAII = 0(IAI-') for k E E(O, a),

I

Al

In general, we cannot expect the resolvent, or modified resolvent, to decay faster along any ray than 0(IAI-1). This is shown by the following theorem.

Elliptic Boundary Value Problems

182

THEOREM 12.5. Suppose that for some positive constant a and for some sequence of complex numbers k,, in p(A) such that IAnI the following estimate holds: I(An -

A)-'I I

S alAn 1-1-

Then a >- 1. Correspondingly, if for some sequence of complex numbers in pM(A) such that IAn I the estimate

"

IIAA n II s

aIAnP-1

holds, then either A = 0 or a 2 1. Proof. By (12.3) it follows that A E p(A) if IA - An < a-1IAn I, and we have an estimate alAnI -1

II(A-A)-111 <

I-1

1-IA-AnIa IAn

a-1IAnI-IA-AnI

Now assume a < 1. Then a = (1 + 2e)-1 for some positive c. Also all of the circles {A: IA - AnI < (1 + e) IAnI I are contained in p(A), and for A in such a circle (12.11) implies (12.12)

I I(A - A)-1

<

1

(1 + 2c)IAnI-(1 + E)IAnI

_

1

EIAnI

Thus, since such circles cover the complex plane, (12.12) implies (A - A)-1 exists and is uniformly bounded in the complex plane. As (A - A)'1 is also analytic on p(A) = complex plane (by Theorem 12.1), Liouville's theorem implies (A - A)-1 is constant. But then A - A is constant, an absurdity. For the proof of the second part, the same procedure as given above shows that if a < 1, then AA is constant. But since I IAA -, 0, this n

I

I

Some results on Linear Transformations

sec. 12

183

implies AA = 0 for all A. That is, A(1 - AA)-' = 0 for all A. Multiply-

ing this relation by 1 - AA implies A = 0. Q.E.D. THEOREM 12.6. The set of numbers 0 such that e'0 is a direction of minimal gtowth of the resolvent [modified resolvent] of A is open.

Proof. Suppose that

A(AA)-III < IAI

for A i E(O, a). By (12.3), µ E p(A) if Jµ - Al < K ' CAI and A E -(0, a), and we have an estimate II(µ -

A)-'

KIAI-1-Iµ-AIKIAI-.

I<

Thus, if A i _(O, a) and if Iµ - Al < V2K-'IAI, then I I(µ _ A)-1II <

2KIAI-1

Since

IµI < IAI + µ -XI <(l+ 1 )IAI,

2K

we have

II(µ_A)-III<2K+1 Ii

On estimate which holds for all µ such that Iµ - AI < %2K-1IAI for some

E (O, a). An easy computation then shows that 1/j is a direction of A minimal growth of the resolvent of A if IL - 01 < sin-' 1/2K). similar argument gives the result for the directions of minimal growth of the modified resolvent. Q.E.D.

Elliptic Boundary Value Problems

184

We now give some examples.

THEOREM 12.7. Let A be a self-adjoint operator on the Hilbert space X. Then any non-real direction ere is a direction of minimal growth of both the resolvent and the modified resolvent of A. Furthermore, for arg A = 0 (12.13)

[I(A - A)-'II s Icsc 01 !XI-1,

(12.14)

IIAAI I S Icsc 01 IAI-1.

Proof. Since A is self-adjoint, (Af, f) is real for f C D(A). Thus, if A = IAIe'°, then

9[((A - A)f, f)] = Al Isine Jf, f).

Therefore, the Cauchy-Schwarz inequality implies IIfII2 < IAI 1I csc BI I.I(A-A)fII IIfII;

dividing by (12.15)

I f 11,

I IfI I s Icsc 01 IAI-11Ll(A - A)I IflI

This inequality shows that N(A - A) = {0) for all non-real A. Thus, for non-real A, N((A - A)*) = N(A - A) _ {0(. But the orthogonal com-

plement of the range of A - A is precisely NO - A). Therefore, the range of A - A is dense in X. Since (12.15) implies that the range of A - A is closed (here we use the fact that a self-adjoint transformation is closed), it follows that A - A is one-to-one and onto; and another application of (12.15) implies I I(A - A)-11I S Icsc BI IAI-'.

To obtain the result for the modified resolvent of A, note that for non-real A and f E D(A) 91((1 - AA)f, Af)] = IAl sin 9I IAf1l2.

Therefore, applying the Cauchy-Schwartz inequality, dividing by I If I I,

Some results on Linear Transformations

sec. 12 I IAf l l

185

s Icsc 01 I A-11 I I(1- AA)fl 1.

Putting in the last inequality f = (1 - AA)-1g, g an arbitrary element in X, (12.14) follows. Q.E.D. We also have the following result. THEOREM 12.8. Let A be a linear transformation with domain D(A) C X, such that for all f q D(A) `R(Af,,f) z Aollfl12,

for some real number A0. Suppose that some number Al is in the resolvent set of A, and that JRA1 < A0. Then the half-plane {A: JRA < Ao) is contained in the resolvent set of A, and every direction ei0

with -

< 0 < 2 is a direction of minimal growth of the resolvent of

A.

Proof. We have for f E D(A) 5t((A

- A)f, f)

`RA 1.1 f 112

JqAl

(fRA

lfl

- fR(Af, f)

2 - Ao11f112

- Ao)IIfl12.

Therefore, if `RA < A0, the Cauchy-Schwarz inequality implies

- ((A - A)f, f)

11112 = A

o-511 A

1


o-

A II(A - A)f l

l

I If 112

Or,

(12.16)

llflI

Ao

1RA 11(A-A)fhI.

Elliptic Boundary Value Problems

186

Since Al

I I(A1

p(A), (12.16) implies that for f = (A1 - A)-'g

- A)-1gl I S

Ao

-1RA1 IIgII.

Thus,

I

I (A -A )

-11 1 I<

A o

1IRA

i

But (12.3) implies that A E p(A) if IA1 - AI < I 1 ( A 1 - A ) 1 ; therefore, A E p(A) if IA1 - Al < Ao - RA1. We can then extend this process (see figure), to show that A E p(A) if RA < Ao. And then (12.16) 11

1

yields the estimate I I(A -

A)-1

1

II S AO

J`ZA < Ao. !RA

From this estimate the assertion of the theorem is obvious.

sec. 12

Some results on Linear Transformations

187

Part 2. OPERATORS OF FINITE DOUBLE-NORM ON AN ABSTRACT HILBERT SPACE

THEOREM 12.9. Let T be a bounded linear transformation on a gilbert space X. If{dj1, 02,...1 and {i/ii, t/l2,...1 are orthonormal bases in X, then IIT0,II2 =

IITq,I12 00

_

i-1

IIT*Oil12

00

i,L

(To,,

;)I2;

1

it is to be understood that these sums may be either finite or infinite.

Proof. By Parseval's relation, 00

I T o,

I 12

=;y1 I(Toi,

, ) I 22.

Summing over i gives ,)I2

+y1 1y.1 uTo+'

m

o0

= E E I(0+, i-1 j-1

T*t/ij)I2

_j-1E1-1I _ ;_1 IIT*O1II2,

I,(T*O.,

0,)I2

Elliptic Boundary Value Problems

188

where we have again used Parseval's identity. Applying this argument to T* instead of T gives 00

00

i

1

IIT*O,II2 = 1 IIT**

jII2

00

Therefore all the assertions of the theorem follow.

Q.E.D.

Definition 12.7. A bounded linear transformation T on a Hilbert space has finite double-norm if the sums in Theorem 12.9 are finite. In this case, for any orthonormal sequence 10t, 02,... } the doublenorm of T is given by 00

IIITIII = [t1

I To,

1121'%

.

The next theorem gives some elementary properties. THEOREM 12.10. Let T1 and T2 each have finite double-norm, and let A be a bounded linear transformation. Then Ti, T 1 + T2, AT,, and T A each have finite double-norm, and IIIT11I1= 111TiIII; IIIaT1II I = lal 11IT1II1 for any complex number a; IIIT1 + T2111 s 11IT1111 + 111T2111:

1IIATIJII < I.IAII IIIT1111; IIITIA111 < IIAII 111T1111; IIT111 _<

IIIT1111.

Some results on Linear Transformations

sec. 12

189

Proof. The first relation follows from Theorem 12.9, the second is trivial, and the third follows from Minkowki's inequality. The fourth follows easily:

I IIAT11,112 _< I I IAI12IIT,O,112 = IIAI12IIITIII2. ,-1

,-1

And the fifth follows from the computation IIIT1AIII = III(T1A)*I11 = IIIA*T*11I1

IIA*II IIITiIII = IIAII IIIT1111

Finally, suppose f E X. Then 00

f = I (f, 1)q 1 ,-1

00

Tf = ,- 1

(f, 0,)TO,.

The Cauchy-Schwarz inequality shows that 00

I.1 Tf

I(f, ,)I

H

s[

,-1

I(f,

= IIfII 111TI11.

I211 [

,-

IITo,II2]I 1

Q.E.D.

Elliptic Boundary Value Problems

190

THEOREM 12.11. A transformation having finite double-norm is compact.

be an orthonormal basis for X. Then there Proof. Let exist numbers t't(i, j = 1, 2,...) such that 00

Tit =

t''Il.

1-1

Moreover, Parseval 's relation shows that 00

It''I2.

IITO,I12 =

Therefore, summing on i, we obtain (12.17)

I t'JI 2.

111 TI 112 =

If N is a positive integer, let _< N,

t'iN -

O,i>N or j>N.

Then let TN be the unique linear transformation satisfying 00

T N ¢1 _

f

E1

tN'tO f .

Thus,

N

(12.18)

TA =

i

t'tcl, 1 <jSN, 1

0, j > N.

Then it is obvious that TN has finite double-norm and

sec. 12

Some results on Linear Transformations

1HIT-T12 NII Therefore,

I

I

N

0o =

191

I It1J121.J-1

It1Ji2

I T - TN I I , 0, so that Theorem 12.10 implies I T - TN I I - 0. I

I

Since the range of TN is finite-dimensional, it is obvious that TN is compact. But then the fact that I T - TN I I - 0 implies T is compact. I

Q.E.D.

THEOREM 12.12. Let T and S be transformations each having finite double-norm, and let (01, 02,...} be an orthonormal basis in X. Then the series 00

2 (T95J, c1)(So i,J-1

, 95J)

is absolutely convergent and is independent of the particular orthonormal basis {01}. If the sum of the series is designated tr(TS), then (12.19)

Itr(TS)I < IIITIII IIIsIII

Proof. The absolute convergence of the series and the estimate (12.19) are immediate consequences of the Cauchy-Schwarz inequality, the definition of double-norm, and Theorem 12.9. Note that TSO, = T I (S(k1, 01)(kJ J-1 00

=

J -i

(so,, OJ)Tg1

= 2 1 (So,, OJ)(ToJ, Ok)0k J-1 k-1

Thus,

(TSO,, 01) = 1 2 (So,, 0,)(Tc., (kk)(0k, 95d J-1 k-1

Elliptic Boundary Value Problems

192

=

;

(S9,, 95;)(T95;, q,). 1

Summing on i, 00

(12.20)

(TS/,, 1,) = I (S95,, 95;)(T95;, 1) -1 i,i-1

Now let }t(i,} be another orthonormal basis. Then 00

00

(Tci, c1)(Sc1, /;) = 7jI (Sc;, ci)(TO1, 95;) 1

00

2 (S T95,,

j.1 00

= E (To,, S*c1) 1-i

--

= E 1

1

(To1, 4;)(S*c;,;),

;

1

by Parseval's relation. Thus, by the absolute convergence of the double series and Parseval's relation again, 00

00

E (T , 1)(Sc ,, c.) = 2 1 (S>/i , ,)(T* , c 1,/-1 ' ;-1 i-i 00

(S

j, T *&;)

00

= E ;- i

(TSB;, i;)

)

sec. 12

193

Some results on Linear Transformations 00

=

I (S&,,1)(T1,,)

1.1-1

,by (12.20). Q.E.D. Notation. The quantity tr(TS) is called the trace of the operator TS. Note that (12.20) can be written 00

(TSc1, 01).

tr(TS) = I- 1

Since any transformation T having finite double-norm is compact, the Riesz-Schauder theory of compact operators applies to T. In particular, a non-zero complex number A is either in the resolvent set of T or is an eigenvalue of T; the non-zero eigenvalues of T have finite multiplicity; and the only limit point of the set of eigenvalues of T is the complex number zero. The next result of immediate interest is Theorem 12.14. But first a lemma is needed.

LEMMA 12.13. Let T be a linear transformation on a linear space X, and let M(A) be the generalized eigenspace of T corresponding to the eigenvalue A. The subspaces M(A) are independent that is, if

A1,...,Av are distinct, if f E M(A,), i =1,...,v, and if I fi = 0, then

f.=0, 1SiS Proof. The proof is by induction on v. The theorem is trivial for v = 1, but this fact does not help in the proof. We first establish the lemma for v = 2. Thus, since fl + f2 = 0, we may simply assume that for some f E X, (12.21)

(A1 - T)klf = 0, k

(A2-T) 2f=0. If k 1 = 0, then it is trivial that f = 0. If k 1 > 0, let g = (A 1 -

Then the two relations imply

(A1-T)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,(A1-T)

k

fk 0. Ifk1-1=0,then f=0;ifk1-1>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 (AI-T)kIfi=0,

(12.22)

1
V

(12.23)

1 f', = 0.

I-1

Multiplying (12.23) by (X1, - T)k1', setting g, = (A1, - T)k"fi, and tak-

ing (12.23) into account, we obtain k

(Ai-T) Ig;=O, 1SiSv-1; v-1

1 gi=0.

1- 1

k

By the induction hypothesis, gi = 0; that is, (A" - T) "fi = 0, 1 S i S v - 1. But since (A1, - T)k"fi =(Ai - T)kifi = 0, the result for v = 2 implies fI = 0, 1 S i S v - 1. And then (12.23) implies also f" = 0. Q.E.D. THEOREM 12.14. Let T be a linear transformation on a separable Hilbert space X, and let T have finite double-norm. Then each non-zero eigenvalue of T has finite multiplicity and the eigenvalues of T can have only 0 as a limit point. If 1Ai } is the set of non-zero eigenvalues of T arranged in some order, where each eigenvalue is repeated the number of times equal to its multiplicity, then (12.24)

1jIA1I2

111T1112.

sec. 12

Some results on Linear Transformations

195

Proof. As we have already remarked, the non-zero eigenvalues have finite multiplicity and have 0 as the only possible limit point. Arrange the non-zero eigenvalues {A/} such that the repetitions of a given eigenvalue appear in succession. Suppose that A is a non-zero eigenvalue, and that A/ +1,...,A. +m are the members of {A/} equal to 0

0

A. Then M(A), the generalized eigenspace of T corresponding to A, has dimension m. Obviously, A - T maps M(A) into itself, so we can consider the transformation A which is the restriction of A - T to 'M (A); A is then a linear transformation on M(A). Obviously, the only eigenvalue of A is 0. A simple theorem on matrices shows that there is a basis {f/ o+ f/ +m } for M(A) such that the matrix of A with

respect to this basis is triangular. That is, (12.25)

Af.=

E 1-10+1

af, j /

o

o

0

+m,

for certain constants ai1. Moreover, a" = 0, since ail must be an eigenvalue of A. Since Afl = (A - T)fl, (12.25) implies (since also A=A/) (12.26)

Tfl= 1fl-

1-1

E

a1Jf,, jo+1SjSjo+m.

1-10+1

If this is done for each eigenvalue, then the independence of the subspaces M(A) (cf. Lemma 12.13) implies that there is a linearly independent set If 1, f2,... } such that

(12.27)

Tfl

1-1

=

A/fl + E ailf p i- 1

for certain constants ail. By the Gram-Schmidt orthogonalization process, there is an orthonormal sequence {g1, g2,...} such that

Elliptic Boundary Value Problems

196

J

gJ =

2 b'Jfi, bJJ ; 0.

i-i J

f = Ei- 1b'Jg ip 1

J

(bJJ)-1.

bJJ = 1

Therefore, (12.27) implies iI1

TgJ = E b'J[Uifi +

i-I

a1 fk]

k- 1

J-1

= bJJAJf1 + E a2Jfi i*1

J1

= bJJAJ[b1JgJ + E bl'gi] +

i-1

J11

i-1

a2J

k-1 biigk

J-1

_ AJgJ + E a3Jgi, i-1

for certain constants a3J. The orthonormality of the sequence (g1, g2,... 1 implies J_ I

IITgJ112 =

IAJ12

+

i

Ia3JI2 1

Ix J12

Summing over all j, we obtain IIITIII2 2 yJIITgJI12 z yJIXJ12.

Q.E.D.

To finish the main results of this section, an important formula is needed, which will be given in Theorem 12.17. But first a lemma is needed. Recall that a projection on a linear space is a linear transformation P satisfying the equation P2 = P.

sec. 12

Some results on Linear Transformations

197

LEMMA 12.15. Let P and Q be continuous non-zero projections on a Hilbert space X, such that IIP - QI I < IIPI1-1 Then the dimension of R(P) is not greater than the dimension of R(Q).

Proof. Suppose that R(P) has larger dimension than R(Q). Now R(Q) has dimension at least that of R(PQ), so it follows that R(PQ) is a proper subspace of R(P). Since R(P) and R(PQ) are both closed subspaces, there is a unit vector fin R(P) which is orthogonal to R(PQ). In particular, f is orthogonal to PQf, so that (12.28)

1If-PQf1I2=1+IIPQf12>1.

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 non-zero 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) = A-12, (X -1

-

) + tr(T 2) 1

t

2

AI

Elliptic Boundary Value Problems

198

Proof. First note that since TA = T(1 - AT-1., Theorem 12.10 implies that Tx has finite double-norm. 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, IIIT-TNIII-0 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 finite-dimensional case,tr(T2 _ Since the trace function is linear, we have

A-2

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 =(T-TN)(1-AT)-'+TN[(1-AT)-'-(1-ATN)-1]

=(T-TN)(1-Al)7'+TN(1_AT)-i[1-AT N-(1-AT)]'

(1-ATN)-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(1-AT)-1II+IAI IIITNIII 11(l-XT)-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 N-4m

(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 X-1 contains only the eigenvalue X-1 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

Iµ-A, 1I-E

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 jZn--LkL + lim sup I/>-IAI 1% 2Ixj12

N-roo

< 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. HILBERT-SCHMIDT KERNELS

We shall now give two results identifying the class of operators of finite double-norm 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 double-norm 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 <

Thereof re, f K(x, y)f(y)dy exists as an absolutely convergent inte(1

gral for such x, as a result of the Cauchy-Schwarz inequality; furthermore,

I(Tf)(x)I2 = I f K(x, y)f(y)dYI2

s

IK(x, Y)I2dyIIfIl2 SZ

Integrating over 0, Tf 112 <

f2xS2

K(x, y)I2dydxllfll2

Thus, T is a bounded linear transformation on L2(S2). To derive (12.37), let to,, S621... I be any orthonormal basis in L2(1l). Then for almost all x (To,)(x) = f0 K(x, Y)c,(y)dy.

Some results on Linear Transformations

sec. 12

203

Squaring and summing on i, (To

(12.38)

,)(x)1

2

=

If

K(x, Y)o,(Y) dy(2. Z

As remarked above, for fixed x the function K(x, y) is in L2(11); and

f K(x, y)o,(y)dy is just the inner product of this function with c,(y). Therefore, Parseval's relation implies that (12.38) may be written for almost all x 00

I

I(To;)(x)I2

i-1

= f IK(x, Y)I2dy

Integrating with respect to x, 00

To, I

IK(x, Y) 12dydx. 112 = SZxSZ

Q.E.D.

Definition 12.8. An operator of the type given in Theorem 12.17 is said to be an integral operator with Hilbert-Schmidt kernel K(x, y). We now give a converse to Theorem 12.18.

THEOREM 12.19. Let T be a transformation having finite doublenorm on the Hilbert space L2(1l). Then T is an integral operator with a Hilbert-Schmidt kernel.

Proof. Let 1011021 ... I be a fixed orthonormal basis in L2(1l). Let

tj' = (To,, qj), and let N

KN(x, Y) _

t"q;(x)gj(Y)

Then for N 1
f

cxcZ

IKN(x, y) - KN(x, 1 Y)I2dxdy 2

Elliptic Boundary Value Problems

204

t" 1(x)gj(y)l2dxdy

=

f

(y)dxdy

I

ltijl2

i, jSN 2

i>N1 or j>N1 00

<

00

i'N1+1 j- 1

00

It'll 2 +

00

j-N1+1 i'1

lt'jl 2.

00

Since

2 < oo, it follows that as N1, N2 - oo,

lt'ij 2 = l HHTJ

I S2XS2

l KN (x, y) - KN (x, y)l 2dXdy _ 0. 2

1

As L2(S2 X 0) is complete, there exists K(x, y) E L2(11 x 0) such that

f

SZx

l KN(x, y) - K(x, y)I2dxdy - 0.

Let T 1 be the integral operator associated with Hilbert -Schmidt kernel K; we then must show T = T 1. To see this, note that K(x, y)ci(y)cj(x)dydx

(T 19';, (kj) = S2xS2

= lim N-,00

= lim N-+Do

= t i'

f KN,(x, y)ci(y)cj(x)dydx S2XS2

,.

t' j' I ci,(x)/j(x)dx 1

SZ

f

sec. 12

Some results on Linear Transformations

205

_ (T k,, kt). Thus, T 19S1 = Tcj for all i. Since (01, 02,... } is a basis, T 1 = T. Q.E.D. To complete this section, we give two results showing how certain operations with integral operators can be carried out by performing corresponding operations on the kernels.

THEOREM 12.20. Let T be an integral operator on L2(Q) having Hilbert-Schmidt kernel K(x, y). Then the adjoint T* is an integral operator on L2(fZ) having the Hilbert-Schmidt kernel K*(x, y) _ K(y, x). Proof. If f, g E L2(Sl), then

(Tf, g) = f (Tf)(x)g(x)dx

=f

K(x, y)f(y)g(x)dxdy

Bust

=f

K(y, x)f(x)g(y)dydx

QX9 f f(x)[fH K(y, x)g(y)dy]dx

= (f, T *g).

Since this holds for all f E L2(fl), we have almost all x E SZ

(T*g)(x) = f K(y, x)g(y)dy. Q.E.D.

THEOREM 12.21. Let T. be an integral operator on L2(Sl) having Hilbert-Schmidt kernel K,(x, y), i = 1, 2. Then the integral (12.39)

K(x, y) = f K,(x, z)K2(z, y)dz

Elliptic Boundary Value Problems

206

is absolutely,/ convergent for almost all (x, y) E Sl x Sl, and T1T2 is an integral /operator whose Hilbert-Schmidt kernel is K(x, y). Moreover, K(xi"x) as defined by (12.39) exists for almost all x E Sl, and

(12.40),\ tr(T 1 T 2) = f K(x, x)dx, the integral being absolutely convergent.

Proof. We have seen that for almost all x, K1(x, z) is in L2(Sl) as a function of z; likewise, for almost all y, K2(z, y) is in L2(Sl) as a function of z. Thus, (12.39) is convergent for almost all (x, y) E Sl x Q. Also, we see that K(x, x) exists for almost all x I U. We next establish the measurability of K(x, y). Let f, g E L2(SZ). As we have seen K2(z, y)f(y) is an integrable function of y for almost all z. Moreover, f IK2(z, y)f(y)ldy is then in 12

L2(1l) as a function of z. Therefore, it follows that for almost all x, IK1(x, z)I f K2(z, y)f(y)ldy is integrable, and that Q

fQIK1(x, z)jdz IK2(z, Y)f(Y)Idy

is in L2(0). But then g(x) times this function is integrable. Since the function K1(x, z)K2(z, y)f(y)g(x) is measurable in 12 x12 x 1, it is integrable over fZ x Sl x Q. But then Fubini's theorem shows that its integral with respect to z, which is K(x, y)f(y)g(x), is measurable in fI x Q. Taking f and g to be characteristic functions of measurable sets, it then follows that K(x, y) is a measurable function on Sl X Q.

By the Cauchy-Schwarz inequality, IK(x, y)1 2 < f IK1(x, z) 2dz Sl

f 1K2(z, Y)j 2dz. Q

Integrating this over SZ x12, it is seen that K E L2(12 x fl). It remains to identify K with the operator T 1 T 2 and to prove (12.40). Now for f, g E L 2(0)'

sec. 12

Some results on Linear Transformations

207

(TIT 2f, g) = f (TIT 2f)(x)g(x)dx

= f [f K ,(x, z)(T2f)(z)dz]g(x)dx

= f [f K,(x, z){ f K2(z, Y)f(y)dy{dz]g(x)dx = f {[f K1(x, z)K2(z, Y)dz]f(y)dY[g(x)dx,

since the integrand is integrable over Il x Sl x 0. Thus, (T 1 T 2f, g) = f [K(x, y)f(y)dy]g(x)dx.

Since this holds for all g, we must have all almost all x (T 1 T2f)(x) = f K(x, y)f(y)dy.

Therefore, T 1 T2 has kernel K. Finally, we must prove (12.40). First, we have by the CauchySchwarz inequality

fIK(x, x)Idx S, f0 IK1(x, y)I IK2(y, x)Idydx

S [f

IKS(x, Y)I2dydx]1/2[f

= IIIT1III

IK2(Y,

x)2dydx]t

IIIT2111 < -.

Thus, K(x, x) is integrable over Q. Moreover, by (12.20), if 101, 02,...E is an orthonormal basis in L2(SZ), 00

tr(T 1T2) = (T 201, t1)(T 1O1, 0;) ,,j- 1 00

(12.41)

_1f 1.1-1 Ilxcl

K2(x, y)O1(y)c1(x)dxdy

Elliptic Boundary Value Problems

208

'f

K1(y, x)cb1(x)ch,(y)dxdy.

QX0

Since the functions 0j(y)cp1(x), i, j = 1, 2,..., form a complete orthonormal sequence in L2(1l X SZ), (12.41) becomes, after applying Parseval's relation, tr(T 1T2) = f

K2(x, y)K1(y, x)dxdy

OXG = f K(y, y)dy.

Q.E.D.

13. Spectral Theory of Abstract Operators

Before giving results for elliptic equations, we can give several abstract theorems in which no mention need be made of differential equations. In these abstract results and in the results of section 14, essential use will be made of Sobolev's inequality (Theorem 3.9) and Rellich's theorem (Theorem 3.8). In order that both these theorems be valid, it is sufficient to assume that t1 is a bounded open set having both the segment property and the ordinary cone property. Actually, as the proofs of these theorems show, it is enough to assume SI is a finite union of disjoint open sets of such type. This is obvious in the case of Sobolev's theorem, since a finite union of open sets which have the ordinary cone property still has the ordinary cone property. In the case of Rellich's theorem, note that the only use of the segment property for &1 was in the application of the second inequality of Theorem 3.2. But it is obvious from the proof of Theorem 3.2 that the theorem is valid for the case that Sl is a finite disjoint union of bounded open sets having the segment property. We shall be somewhat more restrictive than this, since we shall also need the restricted cone property in Theorem 13.9. We require everywhere in this section and the section following that SI be a finite disjoint union of bounded open sets, each of which has the restricted cone property. We shall use this assumption throughout these sections without mentioning it explicity.

sec. 13

Spectral Theory of Abstract Operators

209

Of the various constants appearing in estimates in this section, we wish to isolate the one appearing in the statement of Sobolev's inequality. First a lemma is needed, showing that Theorem 3.9 is really a sort of convexity result. LEMMA 13.1. Let m > 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-(m-1/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.Q-1

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 semi-norm, 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 double-norm, and the following inequalities hold: (13.4)

IIITIII

(13.5)

IIITIII


TI01-(n/2m));

ySIa1l/2IITIIm12mIITII0]-(n12m

Here 191 is the Lebesgue measure of SI, ys is the Sobolev constant of (1, and y is the constant appearing in (13.1). Proof. Let {951, 021...I be an orthonormal basis in L2U ), and let ui = T951, j = 1, 2..... By Lemma 13.1 it may be assumed that ui E C°(SZ). Let al,...aN be any complex numbers, and let f = I aA' j-1 Then, the continuity of the function ui and of u = Tf implies that N

u(x) = L1 a iu.(x)

all

x E SZ.

1-

By Lemma 13.1 we have for all x E SZ Iu(x)I S y(ITfI1--(n/2m)ITflm/2m + ITflo) S

y(ITIol-(n/2m)ITIm/2m+ ITI0)

IIfIIo.n.

That is, N (13.6)

i-1

aiui(x)I
N

121Y2.

1-1

This inequality holds for all x I Q and for all complex numbers al,...,aN. If for fixed x we take ai = ui(x), then we obtain from (13.6)

212

Elliptic Boundary Value Problems

,

N

I1 1

-1

u,(x)I

2 12

- y(ITIo1-(n/2m) ITImn/2m + ITI,)

Since this holds for all x E I, we may square and integrate over 9 to obtain N

J-1

Y2IQI(ITI0-(n /2m)ITIm/2m + ITI0)2.

IIT.O,IIO,Q s

Since this holds for all N, we have 00

-1

(n/2m)ITIm/2m + I

IIT0,1102 Sts Y21QI(ITI1 0

TI o)2.

Thus, (13.4) is proved. The proof of (13.5) is made in the same manner, using (13.3) in place of (13.1). Q.E.D. As an application of this result we have THEOREM 13.6. Let T be a bounded linear transformation on L2(1 ), such that R(T)CHm(1), where m > n/2. Let there exist at least one direction of minimal growth of the modified resolvent of T. Let {At} be the characteristic values of T, repeated according to multiplicity. Let N(t) be the number of characteristic values of modulus less than or equal to t: N(t) = E 1. Then N(t) = 0(tn/m) as t IA,Ist

and for all A E pm(T) (13.7)

tr(TTA) = Et

1 (At - A)At

Indeed, if I ITAI IOS CIAI-1 for A on the ray _(B, a), then N(t) < 4ysIUI I ITI I",/m(1 + C)2tn/m, Proof. We have by (12.5)

(1-AT)-1= 1 +ATA,

Ac

(O, a)..

t Z a.

Spectral Theory of Abstract Operators

;sec. 13

213

Therefore, l

l ( 1 - AT)-'l l 0

1 + Al II ITAI I O s 1 + C,

AE

(0, a).

Also, (13.8)

I

I TAI Im = I I T(1

- AT)-1I Im s I I TI lm l l(1 -XT)-1110

S lITllm(1 + C).

Therefore, Theorem 13.5 implies TA has finite double-norm, and we have an estimate (13.9)

IIITAIII S

ysl1I'/'llTAIIn,/2milTAllo-(n/2m)

C)n/2mC1-(n/2m)IAI-1+(n/2m)

S

yslQll/'llTllm/2m(l +

By Theorem 12.4, the characteristic values of TA are precisely

A

1

and the multiplicity of

1

is the same as the

multiplicity of the characteristic -value A. of T. Therefore, applying Theorem 12.14 to TA, we obtain the estimate valid for all A E pm(T) (13.10)

E

1 IAA-AI2

<

lTAlll2.

In case lAfl S t and lAl t, then IAA - Al S 2t. Therefore, if in (13.10) We sum only over those j for which IAII 5 t, we obtain for IAI = t, A E -(0, a),

4t2 j IAIIsr

< IIITAIII2

-yslll

Tlln/m(1 + M

C)n/mC2-(n/m)t-2+(n/m)

taking (13.9) into account. Multiplying by 4t2, we obtain

214

Elliptic Boundary Value Problems + N(t) S 4y2ISZI ITIIn/m(1 M

C)n/mC2-(n/m)tn/m.

Finally, we must prove (13.7). By Theorem 12.17 we have

+c.

1

for all I E pm(T) and for a certain constant c. For A i (O, a), we obtain from (12.19) and (13.9) I tr(T TA) 15 I I TI I I I

I

I TA I

ICI-1+(n/2m).

S const

Thus, tr(TTA) -, 0 as Al C

as kj - -. Therefore, c must vanish. Q.E.D. As a result of this estimate on N(t), we can sometimes obtain lower bounds for individual characteristic values, as we now show.

Corollary. In addition to the hypothesis of Theorem 13. 6, suppose also that TA exists for A on the ray $(O, 0), and that for such 't the estimate II TAIIo s CIAI-' is satisfied. If the characteristic values 1'kiI of T are arranged according to increasing modulus, lA11 s 'A2I ..., then for j = 1, 2,..., IA,I z [4ySl l (1 + C)2]

m/nIITII-1jn,/n.

Proof. This is an immediate consequence of the theorem, since j < N(IAJI) Q.E.D. In a sense, this is better than an asymptotic estimate, since the estimate for At is valid for all j. However, it gives only a lower bound for IAtl. Under more assumptions we shall obtain asymptotic formulas for N(t) in the next section. Let us now give an example. THEOREM 13.7. Let B[v, u] be a bilinear form over a closed subspace V of Hm(l), such that for certain positive constants K and c0 and for all u, v I V

Spectral Theory of Abstract Operators

sec. 13

IB[v, u]I

215

KIIvIIm.SZIIuIIm.U'

fRB[u, u] z collulim.Q.

Let m > 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 Lax-Milgram 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 ERA-1 < 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(A-1

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<0.

Therefore, if A is a negative real number, IITxIIo <

1c o

S ICI

Thus, it is seen that the hypothesis of the corollary to Theorem 13.6 Thus, is satisfied with C = 1. Since it follows from (13.12) that I IT I J,,,

cot, the estimate for Ia;I is a consequence of the estimate in the corollary. Q.E.D. In case the operator T in Theorem 13.5 is self-adjoint, the result can be improved to obtain the "correct" order of growth of ki. We have Theorem 13.7. bis. Suppose that the conditions of Theorem 13.7 hold and that in addition the bilinear form B[u, v] is Hermitian symmetric so that the operator T defined by (13.11) is a self-adjoint compact operator. If the characteristic values (at j of T (all of which are positive) are arranged in an increasing order, then

sec. 13 ai

Spectral Theory of Abstract Operators Co

(16y 1II)2m/n

j 2m/n,

217

j = 1, 2,....

Proof. From (13.12) we have (Tf, f) .-z 0 for all f so that T is a positive compact self-adjoint operator. We now use a well-known result that a positive self-adjoint operator admits a positive selfadjoint square root. More precisely, since T is compact there exists a (unique) self-adjoint positive and compact operator S such that S2 = T. Also the sequence of characteristic values of S is given by the sequence Jk1/z1 (positive square root). We observe that since Tf 0 for f 0 (an immediate consequence of (13.11)) also Sf 0 for f 0. From this it follows that the range of S is dense in L2(Sl). Using now (13.11) with v = Tf we have: B[Tf, Tf] = (Tf, f) = (S2f, f) = I ISfI I o.

Putting Sf = g, we then get c0IISgI Im = coI ITfI12 s B[Tf, Tf] = IIgII2

Since by the previous remark elements g = Sf are dense in L2(1) it follows from the last inequality that the range of S is contained in H.(Q) and that IISI Im s co1.

Consider now the modified resolvent S, along the positive imaginary axis. By (12.14) we have: I ISXI I o 5 IXJ- 1. Thus it is seen that the hypothesis of the corollary to Theorem 13.6 (for S,) is satisfied with C = 1. Using this and the estimate for I ISI Im, recalling that ()t"21 is the sequence of characteristic values of S, we obtain our theorem as a consequence of the estimate in the corollary. In Theorem 13.5 it is reasonable to expect that if m is much bigger than n/2, then more can be expected of the transformation T. This is in fact the case, as will now be shown. First, a lemma is needed.

LEMMA B.S. The open set SZ x [ is a finite union of disjoint open sets, each of which has the restricted cone property.

Elliptic Boundary Value Problems

218

Proof. Since f is a finite union of disjoint open sets having the restricted cone property, it is obviously sufficient to verify that the Cartesian product of two cones contains a cone. By shrinking the cones, we may assume that the two cones in question have the same height and same opening. Thus, let 60 and 'lo be unit vectors in E , let 0 < 0 < 77/2, let h > 0, and let C1=16: 6.6o-, 161 cos 0, C2 = ti7: rl

no 2 Iq1 cos 0, 1'11 S hl.

Vectors in En x En will be denoted (6, q). Choose the angle 01 such that cos 201 = cos 20, 0 < 01 < 7z/2, let vo be the unit vector vo = 2 /1(e0, no), and let C be the cone C = ((St, rl): (6, rl)

- vo 2 I(S' rl)I cos 01, I(6, rl)I S h}.

We shall show that C'CC1 So let (6, r1) E C; then (S

If

61 + 71

-

110)2-1/2

X

C2 and the proof will then be complete.

Z (ISI2 + 1_iI2)1/2 Cos 01.

0, then, using the Cauchy-Schwarz inequality, e .e

1+

I61?

161

J Cos 01

171 1 2

_

1771.

161 %

If a is any real number greater than one, then (1+t2)112 a _ t 2

0
Applying this with a = V/2 cos 01 and t = Inl/1 1 we obtain

I so that 6

60.>0Cos20 1_1= cos 201=Cos0, 60,2

1 e1 cos 0. This relation obviously holds also if

sec. 13

Spectral Theory of Abstract Operators

219

= 0. Moreover, I(e, 71)I < h obviously implies lei < h. Therefore,

(, 1) E C > e c C 1. Likewise, (e, rl) E C => rJ E C2, so that C c C1 X C2. Q.E.D. THEOREM 13.9. Let T be a bounded linear transformation on L2(SZ), such that R(T) CH. 0, where m > n if n is odd, m > n + 1 if n is even (equivalently, m > 2[n/2] + 1). Suppose also that R(T*) CH 0. Let P = m - [n/2] - 1. Then T is an integral operator with Hilbert-Schmidt kernel K(x, y). The function K belongs to HQ(SZ x SZ). Moreover, the trace of K on the diagonal of (Z x 1 exists in the sense of Theorem 3.10, and (13.13)

[fQ IK(x, x)

12dx]'/,

_< y[(I T I n,/m + T*

m/m)I TI 01-(n/m) + I TI 0],

where y is a constant depending only on 11 and m.

Remark: We state here without proof that under the conditions of the theorem the conclusion (13.13) could be replaced by a much stronger result. Namely, it could be shown that K(x, y) is actually continuous and bounded in (Z x (Z and that (13.13)'

IK(x, y)j < y[(ITIm/m + IT*Iln/m) ITIo-(n/m) + ITI01.

where y is a constant depending only on (Z and m. Also for the continuity of K and for the last estimate to hold it is enough to assume that (Z possesses the ordinary cone property and that m > n (also for n even). On the other hand from the proof of the theorem that follows one can deduce easily (using Sobolev's inequality and a dimensional argument) that K is continuous and satisfies (13.13)' if P > n. Proof. It follows from Theorem 13.5 that T has finite double-norm. From Theorem 12.19 it is then seen that T is an integral operator with a Hilbert-Schmidt kernel K(x, y). It remains to verify that K E HEW x (Z) and that the estimate (13.13) holds. Let IT*Im1/mITI-1/m r = max 1ITIM'ImITI-1/n' 0 0

Then we have

11.

Elliptic Boundary Value Problems

220

(13.14)

ITIm
(note: ITIO = IT*10)

1 < r.

By Theorem 12.20, T* has the Hilbert-Schmidt kernel K*(x, y) _ K(y, X).

Now consider the operator T" which maps f L2 (SZ) into D°`Tf. If lal < f, then since Tf E Hm(SZ), it follows that D"Tf is in Hm_I«I(SZ), and m - lal _-, m - P = [n/2] + 1,> n/2. Therefore, T°` is a bounded linear transformation of L2(") into Hm_e(SZ), since T is bounded as a transformation of L2(SZ) into Hm(SZ); cf. Lemma 13.4. Thus, we may apply Theorem 13.5 to T", if Iat S f. It follows then that T°` is an integral operator with a Hilbert-Schmidt kernel K°`(x, y). We could certainly estimate I I T"III by using (13.4) and (13.5), but we shall find it better to repeat the proof of Theorem 13.5 in the present situation. Suppose then that 101, 02,... is an orthonormal basis in L2(S1), and let u. = Toy. I

N

Let f = I aA, u = Tf = F, aju., where al,...,aN are complex i-I

i-1

numbers. By Sobolev's inequality it follows that, after modification on a set of measure zero, uj (i: CNI) and u E C4SZ). Therefore, if a is a fixed index with I a I < F. we have for all x E SZ N

u(x) = F1 ajuj(x), i(13.15)

Du(x) =

N F,

J.1

By Sobolev's inequality we also have the estimate for all x ID«u(x)I <

since r

u=Tf

1.

I SZ,

yr (m-/n--I« )(IuIm.Q + rmlul O.SZ),

But by Definition 13.2 this inequality implies since

sec. .

13

Spectral Theory of Abstract Operators

ID«u(x)I <

yr-(m-Yn-I"I)(ITim

221

+ rmITlo)jIfIL.

By definition of f and by (13.15), this may be written

I I afD«u.(x)I j_l

N

rmlTI0)(

Yr

-i

Iaj12)/2

This inequality is valid for all constants al,...,aN, and for all x

I Q.

Choosing aj = D°`uj(x), we obtain N

[ F,

j_l

ID"uj(x)I2]'/

yr-(m-'/n-l"I)(ITIm +rmITlo).

After squaring and integrating over SZ, and using the relation D°`uj = D-Toj = T°`0j, this becomes

I N

j-l

lIT°`Ojllo s

y2IQIr-2(m-1I,n-I«I)(ITIm

+rmITIo)2.

Therefore, as this holds for all positive integers N, we find IIIT°`III s

yIcI%r-(m-v2n-IaH)(ITIm +rmITlo).

By (13.14) this implies (13.16)

111D°`TIII = 11IT°`ll s

Since T" = D°`T, it should be expected that the kernel K°`(x, y) of T" is given by the formula K°`(x, y) = DxK(x, y). This is indeed true, if this derivative is interpreted as a weak derivative in L2(Q X as we now show. To see this, take first for a test function in SZ x SZ any function O(x, y) = a(x)b(y), where a E Co (SZ), b E Co (1). Then

f

K«(x, y)O(x, y)dxdy = f a(x)dx f K"(x, y)b(y)dy.

SZ),

Elliptic Boundary Values Problems

222

Since K°` is the kernel corresponding to T"

D"T, it follows that

for almost all x E SZ

f K«(x, y)b(y)dy

=

(D"Tb)(x).

St

Therefore,

f

K°`¢idxdy = f aD"(Tb)dx

QXQ

St

_

(-1)1"lf D°`a . Tbdx. ci

But T has the kernel K(x, y), so this relation can be expressed

xQK°`cdxdy

D"a(x)dx f K(x, y)b(y)dy

f

K(x, y)D'(a(x)b(y))dxdy

QXQ

f

KD ,6dxdy.

QXQ

Therefore, it follows immediately that (13.17)

f

K°`Odxdy = (-1)I"I f

cxc

KDXOdxdy

cxc

for any test function 0 of the form (13.18)

N O(x, Y) = E a,(x)bt(y),

where a., b1 E Co (SZ). One can then show that (13.17) holds for all test functions on i x Q. In order not to impair the continuity of the proof, this will be shown as the final stage of the proof. Thus, it follows that K(x, y) has all the weak derivatives DXK(x, y), laI < P. We also wish to show that K(x, y) has all the weak derivatives DYK(x, y), j al < P. It is precisely for this point that we need the assumptions of the theorem on the adjoint T* of T. For,

sec. 13

Spectral Theory of Abstract Operators

223

we can obviously carry out all the preceding lines of the argument with T replaced everywhere by T*. Since T* has the kernel K*(x, y) _ K(y, x), the results already obtained show that K(y, x) has all the weak derivatives D'K(y, x), lal < P, and we have an estimate analogous to (13.16), (13.19)

IIID-T*IHH

Here, the kernel of D°`T * is precisely DX K(y, x). But this means that K(x, y) has weak derivatives D;K(x, y), IaI < P. Moreover, (13.16) and (13.19), when combined with the formula (12.37) for the double-norm of an integral operator, show that for at < P JIDXKJJa,SZ.Q

2yJQJ1412n+I-IJTJ0,

llDYKII0,i1 i1 s 2yIcI'/'r1/'n+IatjTIo

Therefore, a fortiori all pure derivatives of K(x, y) of order S P exist weakly and satisfy such estimates. But then since by Lemma 13. 8, SZ x St is a disjoint union of open sets having the restricted cone property, and since the corollary to Theorem 11.10 obviously holds in such open sets, we obtain the result that K HP(Q x il), and that (13.20)

IKIP,SZXSI S

ylr'/,n+pITIo,

p S P.

Now we are in a position to apply the trace theorem (Theorem 3.10). In our case the basic open set is St x St, of dimension 2n, and we wish to consider the trace on the intersection of this set with the manifold H = I (x, x): x I En }, of dimension n. Thus, since K E HP(SZ x il), in order to apply the theorem we need that P > 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<

y2r'(P-''/(2n-n))(IKIP.j2xSl

+

/IKI o,Slx&


,

by (13.20). Therefore, (13.21)

[f IK(x, x)l2dx]'12 < 2y3rnITI0,

To finish the proof we simply need to use the definition of r. If r = 1, then (13.13) is immediate. If r > 1, then r is equal to one of the numbers 0 m 0 ' thus, M rn <

ITIm/mITlon/m + IT*Im/mlTj-n/m.

Applying this to (13.21), it follows that (13.13) is valid also in this case. Finally, we have to fulfil the promise made earlier to establish (13.17) for every (k E Co(11 x Sl). Suppose that (k is such a function. Then there exists an open set &1' cc Sl such that supp ((k) C 0' x [1'. Let C E Co (Q) be chosen such that C = 1 on W. By rescaling, if necessary, we can obviously assume that Sl x (1 is contained in a cube Q = ((x, y): Ixkl < %, lykl < 1/2, k = 1,...,nl. Suppose O(x, y) to be extended to Q by setting c = 0 in Q-supp (0). Extend 0 by periodicity to all of En x En. Then 0 is periodic and in C°°(E,, x En). Therefore, the Fourier expansion of 0 converges uniformly to 0: c(X, y) = where

rr , rl ae, r

is taken over all e, 77 with integral components. Also, y) = le 7(27ri6)°`

Now consider the function (kN(X, y) = P(X),Y.() I eI+l 1IsN

ate S7

Spectral Theory of Abstract Operators

sec. 13

I,

a6

225

r!

I61+lJi-N

By choice of , (x)e27'X'6and C(y)e27iy''1 are each in C'(0). Thus, cN(x, y) has the form (13.18), and it follows that (13.17) holds with 0 replaced by "IN: (13.22)

f

K"cb dxdy = (-1)H f

QXQ

KDxct dxdy.

S2x[1

Now it follows from Leibnitz's rule that as N -,

we have the fol-

lowing limits in L2(1lx1I): ON - y(X)S(y)c(x, y),

D'6,

-

DX(C(x)C(y)ct(x, y)).

Therefore, (13.22) becomes

f SZxtl

KD-(C(xV(y)(k)dxdy; K«C(xX(y)c6dxdy = (-1)H f [lxIl

or,

(13.23)

f supp(o)

K°`C(xX(y)(kdxdy

= (-1)IHH f

supp((k) dxdy.

If (x, y) E supp ((k) C 0' x ti', then on a neighborhood of (x, y), 4(x) = 1 and C(y) = 1. Therefore, (13.23) is the same formula as (13.17). Q.E.D. We shall later need an expression relating the dependence of the constant y in (13. 13) on the dimensions of fl. More precisely, we have the

Corollary. Under the assumptions of Theorem 13.9, let y be the constant in (13. 13). If a > 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]'/, <

au

y[an(IT(a)Im/m

+ I T*

n/m

(a)Im

/

I T (a)

11-(n/m) 0

T(a) 1 O].

Proof. For a function u(y) on SZ let x E au;

u' (x) = u(a lx), thus,

u' (ax) = u(x),

x CU.

Clearly, u E Hk(Sl) <--> u' C Hk(aQ); and

(13.24)

fft U'Ik,ac =

a/n-kluIk,Q

Now we shall define an operator on L2(ci) which corresponds to T(a). Precisely, for f C L2(SZ) let Tf be given by the expression

(Tf)' = T(a)f'. Since R(T(a )) C H. (ac), it follo. s that Tf E H. (SZ). Furthermore, it

can be easily checked that (T *f)' = T *a)f'.

Now if k = 0 or m, and if f C L2(Sl), we have from (13.24) ITfI k.ci =

ak-'/nl (Tf)'I

IflO.Q

a- '/nI f, I

= ak

kftac

O,ac

(a)fiI

k,aQ

If'Io,af

sec. 13

Spectral Theory of Abstract Operators

227

Therefore, ITlk=akIT(a)Ik. Likewise, IT*Ik=akIT(a)Ik. Substia tuting these relations in (13.13), it follows that if K(x, y) is the kernel of T, (13.25) if IK(x, x)I2dxl'4 < y[an(IT(a)Im/m + T(a)Im/m)IT (e) I0 1-n/m + IT (a) o Q

.

Finally, suppose K(a)(x, y) is the kernel of T(,,). Then for f E L2(SZ)

andy=a' (Tf)'(x) = (Tf)(a-lx) K(a-lx,

=f

Of(Ode

Q

=f

K(a-l

,

6f'(a6de

= I K(a-lx, a

1y)f°(y)a-ndy

Al

_ (T(a)f')(x) = f QK(a)(x, y)f'(y)dy. a

Therefore, K(a)(x, y) = a-nK(a-lx, a-1y). Thus, f a 2nIK(a-1x, a-1x) 2dx I U IK(a)(x, x) 2dx = au = a-nf IK(Y, Y)I2dy.

Compare this with (13.25), and the result follows. Q.E.D. THEOREM 13.10. In addition to the hypothesis of Theorem 13.9, assume that there exists a direction e'0 of minimal growth of the modified resolvent of T. Let {A1I be the sequence of characteristic values of T, repeated according to multiplicity. Then for any A which is not a characteristic value of T, the modified resolvent TA

Elliptic Boundary Value Problems

228

is an integral operator having Hilbert-Schmidt kernel K,A(x, y), and K,k E Hp(Q x W. Also K,\ has a trace on the diagonal of St x SZ, and (13.26)

[f IKA(x, x)I2dx]'1 = O(JAI - 1+(n/m))

for Al I., arg A = 0. Also, for A not a characteristic value of T, (13.27)

fK,,(x, x)dx = Et

1

.

1

Proof. If A E pm(T), then TA = T(1 - kT)-1 satisfies the same conditions as T. For, obviously R(T,\) CR(T) CHm(Sl). Since T and (1 - kT)-1 commute,R((TA)*) = R(T*(1 - AT)-1') CR(T*) CHm(fl). Therefore, by Theorem 13.9, TA has a Hilbert-Schmidt kernel KXE HF(Q x Il), and we have the estimate (13.28)

[f0IK,A(x, x)I2dx]'' S y[(ITXI m m +

1-(n/m)+

By (13.8), Now on a ray ?(9,a) we have an estimate ITAIo S ITIIm(1 + C) for such A. Also, (TX)*= T*(1 kT)-1*, so IIT,AIIm that for A on '(9, a) CIAI-1.

I I(TA)*I Im S I I T * I Im I ( 1 - AT)-1*I I I

p

IIT*IIm1I(1 - AT)-1110 IIT*I1m(1 + C),

since (1 [

-AT)-1

= 1 + ATA. Thus, (13.28) implies for A E =(9, a)

x)I2dx]'4 < y[(1 +

c)n/m(IITlIn/m + IIT*I

Im/m)

, C1-("I')

IaI-11(n/m) +

S const

IAI-1+(n/m).

clal-11

Tal'],

sec. 13

Spectral Theory of Abstract Operators

229

Thus, it remains to prove the formula (13.27). For this, note first that the corollary to Theorem 13.6 implies that const jm in for sufficiently large j, since at most a finite number IX, I of characteristic values have modulus no greater than a. Since in >n, this implies that the series !,IA, 1 converges. By Theorem 13.6, we then have for A E pm(T) tr('ATTA) = I 1

a (ai - 'A)'k.

_I

_

1 1

'ki - A (

1

)

'AI

-EI11

(13.29) 1

1

Now (1 -'AT)TA = (1 - AT)T(1 -'AT)-'

- (1 - kT)(1 - kT)-'T = T,

so that (13.30)

ATTA = Tx - T.

By Theorem 13.9, there exists f K(x, x)dx. Since Theorem 12.21 implies that the trace of ATTA is the integral of its kernel over the diagonal of 11 x Sl, (13.30) shows that tr('\TTA) = f [K,A(x, x) - K(x, x)]dx. Q

Therefore, (13. 29) implies that for all A E pm(T) f KA(x, x)dx = EI

1

+ c,

Elliptic Boundary Value Problems

230

where c is independent of A. Finally, we show that c = 0. First, the estimate (13.26) shows that f KA(x, x)dx 0 as

l

Al

By arg A = 0. Next, we show the same result for Ii (A1 Theorem 12.6 there are no characteristic values of T in an entire angle about _(0, a); thus, if AJI >a, then larg A. - 01 > 8 for some positive number 8. Now, if z = re'0 and w = se'o are the polar forms of two complex numbers, and if 8 < 10 - 951 5 1/277, then Iz - wl2 = r2 -

2rscos(0- )+s2=r2sin2(0-0)+[rcos(0-0)-s]2> r2sin2 or, z-wI

lzl sin 8. If 1/s7r< I0-0I <7r, then Iz-w12>-r2+s2>

r2, or Iz - wl z I z l . Thus in any case Iz - wl IzI sin 8 if 8 < I0-4l < 77. Applying this inequality to A. and A, we have (13.31)

Isin 8.

IAi - Al 2 max (IA1I, Al)

Therefore, if A. is outside this angle for j 2 jo, then for k C $(0, a)

I

-c

Al

It

+ csc 8

lAl).

omax(IA1

t2

.I,

From this it follows easily that I J(AJ - A)-1 0 as Al I-> ., A i $(0, a), since we know that 1tjAtl-1 converges. Therefore c = 0. Q.E.D.

14. Eigenvalue Problems for Elliptic Equations; The Self Adjoint Case Part 1. PRELIMINARY RESULTS ON FUNDAMENTAL SOLUTIONS

The results of the preceding section allow considerable sharpening in the case of operators T associated with elliptic differential operators. Before considering such cases, we need a preliminary discussion of the concept of fundamental solution. Before beginning this discussion, we need the following characterization of Hg(E,,). LEMMA 14.1. Let u E L2(En) and let ube the Fourier transform of u. Then u C Hf(En) if and only if the integral

sec. 14

Eigenvalue Problems; The Self-Adjoint Case

231

fE (1 + l i)2QIu()I2d6

(14.1)

n

converges. Moreover, (14.2)

IIUII2F,En

=I

IaILe

S2-Iti(I2dS.

n

L,\

Proof. If u I HF(En), then D°`a(4) = lation implies

that Parseval's re-

IID°`ullo,En = f 62«1u(0I2d6 n

From this relation, (14.2) follows upon addition over all a, lal S f. Conversely, suppose (14.1) is satisfied; notice then that °`u() is square integrable over En if lal < F. Therefore, (io°`U(6 is the Fourier transform of a certain function u°` in L2(En). Then Parseval's relation implies that if E Co (En),

f

uD"c6dx = f E

E

n

n

=f

n

_

(-1)I"I f u"(x)o(x)dx. E

n

Thus, u has the weak derivative Du = u°` in L2(En), lal S F. Therefore, u E WF(En) = HF(En). Q.E.D. Let A(D) be a homogeneous elliptic operator of order F having con-

stant coefficients; thus,

Elliptic Boundary Value Problems

232

A(D) _

a,,D°`,

and for some positive Eo, 2 EOIele

JA(iel =

for real e. Assume that e > n and that A is a complex number such that the equation A(ie) = A has no solution for real a Our aim is to solve the equation (14.3)

(A(D) - A)u = f

in case f E L2(En). We shall indeed find a function u E He(En) such that (14.3) is satisfied in all of En. The technique is similar to that employed in Theorem 5.1, in which f and u were periodic. There we could use techniques of Fourier series. Since now f is notperiodic, we shall have to apply Fourier transform techniques. Suppose first that u E He(En) is a solution. Since the Fourier transform of D°`u is

(i6mu( ),

it follows upon taking the Fourier transform of both sides of (14.3) that f()). Therefore,

(14.4)

U(6 =

Since

A(i6 -

Af(

A 4 0 for real e, it follows from the ellipticity of A that

for all real e (14.5)

1

< A1

const

(1 + I )e .

sec. 14

Eigenvalue Problems; The Self-Adjoint Case

233

Therefore, since f > n it follows that (A(i6 - A)'1 is in L2(En) and even in L 1(En ). Therefore, (A(ic) - A)-1 is the Fourier transform of a function in L2(En), say (2rr)n12Fk(x). By the Fourier inversion formula, (14.6)

FA(x) _ (2rr) n f n

de.

A(i

A)-1 is integrable, it follows not only that FA(x) is in L2(En), but also that FA(x) is a continuous function on En. Moreover, FA(x) decreases rapidly as Ixl oo. For, since Since

(-ixrFA(e) = D it follows from the Fourier inversion formula that for all a (14.7)

e1z D'(A(i) - W1de

(-ix)'FA(x) = (27r)"n f n

The justification of this formula comes from the fact that D (A(i) A)'1 is square integrabel for every a. Since D (A(ie) - A)-1 is integrable, the relation (14,7) implies that for all positive N and for some positive constant CN, FA(x)I S

CN

,

xEEN.

1 + IxIN

By the definition of FA, (14.4) can be written (14.8)

u = (2,)n12F'jf.

Now the rapid decrease of FA implies that the convolution FA*f has the Fourier transform

FXf = (2rr)n' 2FXf. Therefore, (14.8) implies that u and FA*f have the same Fourier trans-

Elliptic Boundary Value Problems

234

form, and must therefore coincide. That is u(x) = E FA(x - y)f(y)dy.

(14.9)

n

What we have shown is that if u E HP(E,,) and if u satisfies (14,3), then u is given by the formula (14.9). Conversely, suppose that f E L2(En) and that u is defined by (14.9). By retracing the above argument in the opposite direction, it follows that u E L2(En) and, that its Fourier transform satisfies (14.4). Therefore, the estimate (14.5) implies that (1 + I I)PIU(6I S const Ik% so that

f

const f

(1 +

n

n

Thus, Lemma 14.1 implies that u E HP(En). Then it follows that (A(D ) -

= (A(iO - A)v =

f,

so that u is a solution of (14.3). The formula (14.2) also implies that (14.10)

I I U I IP,En < const I IfI IO,E

The function FA(x) will be called the fundamental solution for the operator A(D) - A.

Now we shall investigate the formula (14.9) for the solution of (A(D) - A)u = fin the case in which f E L2(1l) for some bounded open

set Q. Regarding f as being extended to E. by setting f = 0 outside Sl, then we see from the previous results that the function (14.11)

u(x) _ f F, (x - y)f(y)dy,

x E 11,

is in HP(Sl) and satisfies (A(D) - A)u = fin Q.

The relation (14.11) represents an integral transformation of L2([l) into Hp(Sl), having a Hilbert-Schmidt kernel FA(x - y), since S1 is

sec. 14

Eigenvalue Problems; The Self-Adjoint Case

235

bounded. Since FA(x) is continuous, the kernel FA(x -- y) certainly has a trace on the diagonal of 0 x Sl, and this function is obviously the constant FA(O), since x = y on the diagonal. Recall that we have assumed that A(il;) - A 0 for real 6. If k has the polar form A = rfe'e r = IAI

1/e

> 0,

this condition is obviously equivalent to the condition that A(i) e'0 for real here we use again the homogeneity of A. If we set t~ = ry in the formula (14.6) with x = 0, then (14.12)

FA(0) = (27r)-n f -

1

En rfA(iy) (27t)-nrn-Q

( E

- rfe;B

rndy

(A(iy) - e'6)-1dY

_ (2r)-nIAI (n/e)-1 f

En

e'6)-1dy.

(A(iy) -

This formula will be needed in some later computations. Note that, taking the integral over S1 of the trace of FA(x - y), we obtain

f Fk(0)dx = ICI kj Sl

(n/Q)-1(2n)-n f

En A(iY)

1

dy

- e;0

We shall obtain analogous expressions in Part 2 of this section. Two more preliminary results are needed before giving the main results of this section. Note first that the condition, A(il;) / k for 6 real, actually implies that the values A(il;) for real 6 omit an entire angle in the complex plane. Thus, for some 8 > 0 it follows that for real 6, A(ie i if larg µ - arg kI < S. LEMMA 14.2. Let A(D) be a homogeneous elliptic operator of order f having constant coefficients, and let Eo be a positive constant such that IA(6I > E0I6Ie,

6 real.

Elliptic Boundary Value Problems

236

Let 8 be a positive number such that

A(i6

er0

for 10- 001 < 28and f real. Let 12' CC fZ C En. Then there exist positive constants Co = Co(E0, 8, f, n) and C = C(E0, 8, f, n, Il, (1') such that for all u E Hf(Sl), for 0 S k5 P, and for Iarg A - 001 < S Ial(k/f)-1[CII(A -,k)uIIO,fi

IUIk,St'

+

CIlullf-1,Q].

Proof. By (13.31) we have for real a and for Iarg A - 001 < S (14.13)

IA(i6 - Al 2 sins max (IA(i))I, lAl)..

Suppose first that u E Hf(En), and let f = (A(D) - A)u. Then f E L2(En), and it follows from Parseval's relation and (14.4) that (14.14)

lID"uI I o,E = I ID uI I O,E n l

I

o,E

n

n

Now (14.13) implies for lal = k

(_ A(i) -.A

S

ry Ilk

IA(i) - Al k/flA(i) - At 1-(k/f) C k

S

sins JA(i

)IklCIAI1-(k/f)

< csc8 EOk/fIAI (k/f) Thus, (14.14) implies (14.15)

I ID°`ul I

Eok/PHI o,E n < csc8

II o,E Ti

n

sec. 14

Eigenvalue Problems; The Self-Adjoint Case

237

= cscS Eok/FIAT (k/e)-l I I(A - X)uI I o,E n

Now suppose only that u E 1fg(1). Then let C Co (SZ) satisfy S 1 on fZ and C=_ 1 on S2'. Let v = 4u. Then v E HH(En), and (14.15) may be applied to the function v. Note that by Leibnitz's rule (A - a)v = A(4u) _ A4u

=4Au+Bu-A4u, where Bu involves derivatives of 4, and derivatives of u of order less than F. Thus, by (14.15) with jai = k and with u replaced by Cu, I ID"(4u)I I o,E s CSC8 Eok/elal (k/e)-'[I n

I

O'E n

+.I IBuII O,E

n

s cscS Eok/eIAI (k/e)-1[I I(A - A)uil o,SZ +C1I Jul If-,.I]. The lemma then follows from the relation ID"ul I O.Q. =1 ID«(4u)I I o,SZ' S I ID"(4u)I I o,c.

Q.E.D.

We can give a simple condition that A'(x, i6 omit a certain complex number when x varies over SZ and a takes on all real values. LEMMA 14.3. Let A(x, D) be a differential operator of order e in SZ, having continuous leading coefficients and bounded, measur-

able lower order coefficients. Assume that for a certain 0 and positive numbers C, a, ICI I o,S2 S

for all real.

I

I I(A - X)OI I o,SZ,

E Co (SZ), A C

(0, a). Then A'(x, ie) - ei0

Proof. Let Vi E Co (SZ) and let

0 for all x C Q,

Elliptic Boundary Value Problems

238

O(x) =

t > 0.

For A = tPe'B, Leibnitz's rule implies that as t (A - A)O = (A'(x, D)e;te'x = (A'(x,

-

O(x) +

itt) - A)e't 'XO(x) +

04-1)

t4(A'(x, i) -

=

O(tP-1)

0(te-1).

Thus,

110h ,Q = f 10(x)12dx

<

c2 II(A - A)OLI o.Q I

Al C22

[t 2f f IA'(x, 1S)

t2Q

- e'BI21Ty(x)I2dx + O(t2f-1)]

Q

C2 f I A'(x,

e'01 211/0(x)I 2dx + 0(t-1).

Q

Letting t - o, we obtain

f [C2IA'(x, r:) -

e1012

- 1] 1O(x)I 2dx Z 0.

Q

Co W), it follows that it must also hold hold for all cii E L2(SZ). Therefore, since A'(x, is a continuous function of x,

Since this holds for all 0

C21A'(x,

eiO12

- 1 2 0,

so that for all x E Sl and all real IA'(x, iO - e'01 > C-1. Q.E.D.

,

sec. 14

Eigenvalue Problems; The Self-Adjoint 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 Self-Adjoint 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 lm-1,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 lm-1,fZ

Using (12.5) we have for IAI sufficiently large, arg A = 0, 11(1-AT)-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 non-overlapping cubes Q...... QN, such that N

if&= U Q,, then fZ' r-1

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 Hilbert-Schmidt 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*(x-y), where G * and K, are the Hilbert-Schmidt 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 Self-Adjoint Case

243

We now need to estimate the semi-norms 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, KIAI-1.

I ITAI IO <

Therefore, by (13.8) it follows that for such A I

I TAI Im, S I ITI Imi(1 + K).

Combining this with (14.10) implies IISjf m'.Q. < [ ITI Im,(1 + K) + const] IIfIIo,Q.

Thus, ISAIIm. is bounded independently of X. Therefore, Lemma 13.3 implies %1k < const (14.20)

const

m'

I S>
D-(k/m' )I I S.I I

mo

ISAI1-(k/m') 0

By Lemma 14.2, we obtain for f E L2(Q''), IFA fI o.Q, < IAI-"LwlIfII O,En + const


IAI-1

Elliptic Boundary Value Problems

244

5 const IAI-1.

Thus, by (14.20) we have for 0 < k 5 m' - 1 SIIk S ConSt

IAI(k/m'rl

Therefore, for sufficiently large JAI it follows that (14.21)

IISI

m'-1 S coast

IAI-1/m'

Now we shall obtain some sharp estimates, using the inequalities (14.16)'. We have for f E L2(Q1) I I(P(x', D) - A)TAf - fI Io,Q!

= II(P(x', D) - A)(Tkf - K,vf)II0.Q = I I(P(x', D) - A)S1fII0,Q

EIifIIO,Qi + C' TV IIm

[E+ CEITaIO/-'IIT4IIIf I10.Q, [E +

CEIAI-1/m']

I If II O,Q'.,

where we have used again Lemma 13.3; we are now using CE as a Iarg k = generic constant. It follows that for sufficiently large Al, (14.22)

II(P(x', D) - A)SAfIIo.Qt

0,

2EIIfIIO,Qj.

Now we shall again apply the estimate in Lemma 14.2, taking fZ = Qf, SZ' = Q', u = 4f. We obtain, using (14.22), SAfIk,Q. S IkI(k/m')-1[ 2EIIfIIO.Q, + CEIISAfIIm'-1,Qj

sec. 14 , Eigenvalue Problems; The Sefl-Adjoint Case

245

By the rough estimate (14.21), this implies Ill,

S$fIk,Q! S

Q;].

Therefore, we have for I AI sufficiently large, arg A = 0, SAIk < 3wEIAI(k/m')-1.

The same argument shows also that for such A, 3CJElAI(k/m1)-1.

(Sk)*Ik S

The proof is now almost completed, except for some easy, but lengthy, computations. Substituting the estimates just obtained into (14.18), we find, using (14.17),

(b/8 ,)n/2 If JKA(x, x) - FA(0)I2dx1

1/

,

Q1

y[(b/So)n3a)c2IAI(n/m')-1

+

3&,EIAI-1J;

or,

If IKA(x, x) -Fk(0)I2dx]Y Q1

.

< 6ywE(b/3a)"/2IAI

1+

3yw((b/8o)-n/2IAI-1.

Therefore, we obtain for sufficiently large JAI, after squaring, f QIIKA(x, x)

- FA(0)I2dx <

Next, summing over all the cubes Q,, we obtain since b" = All, (14.23)

E, f ,IKA(x, x) Q

- F,{(0)I2dx < (12y,))2E28 n`J-U QiI IAI(2n/m')-2 ,

1

Elliptic Boundary Value Problems

246

CE21AI(2n/m')-2

We shall use C as a generic constant for the remainder of the proof. Now we utilize the formula (14.12) for F1(0): Fk(O) = IAI(n/m')-1(2,)-n =

(P(x, i6) -

fE

n

(n/m' )-1pe(xi)

Thus, (14.23) becomes, after multiplying by (14.24)

ei"1-1de

.1i f ,I

IAI1-(n/m')Kk(X, x)

IAI2-(2n/m'),

- pe(xi)I2dx < CE2.

Qi

Now let p0(x) be the step function on ci1 , equal to p0(xi) on Q' and

equal to zero on U. - UQ1 = Slo - . Then (14.24) can be written

(6=UQ;)

[fI IAI l-(n/m')KA(x, x) - P0(x)I2dx]'' < C. Is

By the Cauchy-Schwarz inequality, if JAI 1-(n/m' )KA(x, x)dx - f &

0(x)dxI

is

fI

ICI

A(x, x) - p0(x)Idx

1-(n,/m')KA(X, x)

- pe(X)I2dXi'/2

(14.25) < CI1l0I'/2E

Applying Theorem 13.10, it follows that KA E HQ(fl x Sl), and for IAI large, arg A = 0, JAI

1-(n/m') [f IKA(X, X)I2dxl'/' < C.

247

Eigenvalue Problems; The Self-Adjoint Case

sec. 14

Therefore, the Cauchy-Schwarz inequality implies ICI

t-(n/m') f

IKA(x, x)Idx <

CI11°-EI'4.

Q0 &

Therefore, the triangle inequality applied to (14.25) shows that lim sup I

IAI,.

KA(x, x)dx

IAI

SZ o

arg A=0

- f pq(x)dx(< 0100

/2E + Cjco _

S

+ I fIspe(x)dx - f-Spe(x)dxI

+n.fK Ipe(x)Idx.

Now the integral f pe(x)dx exists, and fpe(x)dx is an approximating By choosing the cubes Q

Riemann sum for the integral

smaller and smaller, the above inequality becomes

f

lim sup arg

K,(x, x)dx - f po(x)dxl s

11o

A0

CI110

I'4E

+ CI11° - Sl'I'/2

11'

+

,Ipe(x)Idx.

%_0

Since IV was any set satisfying 11' CC l1 , and since c was any positive number, the last inequality implies lim sup I

IA

1-(n/m')

IAI

arg a.B

Since

IAI 1-(n/m')

SZ

KA(x, x)dx - f pe(x)dxj S C190- 11 °

I'/2

S

KA(x, x)dx is independent of S, and f pe(x)dx dZ 0

8

Elliptic Boundary Value Problems

248

is independent of A, it must be the case that the two limits

f

lim

KX(x, x)dx,

Ho arg A__0

and

lim

f

p (x)dx, 8

6,0 0

exist and are equal. Q.E.D. Before giving the proof of our main application of Theorem 14.4, we need a Tauberian theorem of Hardy and Littlewood, which we state without proof. For a proof see Karamata, Neuer Beweis and Verallgemeinerung der Tauberschen Satze, welch die Laplacesche and Stieltjessche Transformation betreffen, journal fvr die reine and angewandte Mathematik 164 (1931), pp. 27-39. THEOREM 14.5. Let u(A) be a non decreasing function for A > 0, let 0 < a < 1, let P be a non-negative number, and suppose that as t t

oc

f°° du(A) = Pta-1 + o(ta-1). o

k+t

Then as A (14.26)

o(A) = p sin rra Aa + o(Aa). rra

The converse ("Abelian theorem") of this result is valid, even if we omit the assumption of monotonicity of o(il). Because its proof is elementary, we include it here. Thus, we assume the asymptotic behavior (14.26) of the function o{,l). For every positive number b, we obtain by integration by parts the formula dd,k)

(14.27)

f- l+t fob l+t o

dd,k)

_

or(b)

00

o(il)

b+t + f (,l+t)2

Another integration by parts shows that

del.

sec. 14

f

249

Eigenvalue Problems; The.Sel f-Adjoint Case

e

+afmAa-1da.

dA= ba

b+t

A+t

b

Setting A = tµ in the last integral, we obtain (14.28)

Aa

fb (A+t)2

dA=

ba

+ate-i f°° A!7-! dµ.

b/tµ+l

b+t

Now a familiar formula concerning beta functions is (14.29)

f

0o

0

Aa- 1

µ+1

dµ =

77

sinrra

Therefore, (14.28) becomes (14.30)

Aa

fm b

(A + t) 2

dA= ba b+ t

to-1-ata-1 f

rra

+

sin rra

b/tL dµ.

o

a-1

µ+ 1

Multiplying this relation by p sin rra and combining with (14.27) rra yields

f- do(9) _ Pte-1

A+t

o

= f°°

1

b

(A+t)2 1

b+t

[a(A) - P

sin rra Aa]dA Ira

[o(b) - P sin rra ba] + fb do(A) 0A+t rra

- Psin na to-1 fb/t na

0

a-1

+1

,

dµ.

Given c > 0, there exists a positive number b such that for A z b la(A) - P sin rra Aal < EAa. rra

Thus, we obtain I fm o

du(k) _ pta-1I < E f°°

A+t

b

Aa

(A+t)2

+ Ip sin rra rra

Therefore, using (14.30) again,

I

dA + E

b- + f b do(A)

b+t

to-i fb/t µa-1 0

µ+1

A+t

o

dµ.

Elliptic Boundary Value Problems

250

1-e

lim sup t

I f°° do(A) o

t-,00

+t

- Pta-1I S E lim sup t1-a ram

Aa

fm b

dA

(A + t) 2

17a

=E

sin rra

Since c is arbitrary, the converse of Theorem 14.5 is proved. One further comment should be included, The Tauberian theorem is usually stated in the case that P > 0. But, having the result for P > 0, it is trivial to derive the result for the case P = 0. For suppose o(A) is non-decreasing, and f°O da(A) = o(ta-1). 0

Jl+t

Then a(A) + Aa is non-decreasing, and the converse proposition just proved shows that f°° d[o(,k) +A'] o

A+t

_

rra

sin na

to-1 +

o(ta-1).

Therefore, the Tauberian theorem shows directly that o(A) + Aa = Aa + o(Aa), or,

a(A) = o(Aa).

Therefore, the result of Theorem 14. 5 is also demonstrated in the case P = 0. The spectrum of a linear transformation on a Hilbert space is discrete if every complex number in the spectrum is an eigenvalue. THEOREM 14.6. Let A(x, D) be an elliptic operator of order m' in SZ, having continuous leading coefficients and bounded, measurable lower order coefficients. Let A be symmetric over Co (Q) in the sense that for all 0, % C Co(1l), (Aq, 0) 0'Q = (0, AO)0.H. (This holds, in particular, if the coefficients of A are sufficiently smooth, and A coincides with its formal adjoint.) Suppose that there exists an unbounded self-adjoint transformation @ on LZ(SZ), such that Co (S1) cD(Cl) CHm,(SZ) and du = Au for u E DO). Let n' = n if

sec. 14

Eigenvalue Problems; The Self-Adjoint Case

251

n is odd, n' = n + 1 if n is even. In case m' < n', suppose that there exists an odd positive integer k such that k > n'/m', the coefficients of A are in C(k-l)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 non-negative 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
on L2(1l), a well-known fact about self-adjoint transformations.

Proof. By Theorem 12.7, (A - Cl)-' exists for all non-real A, and all non-real directions e'0 are directions of minimal growth of the resolvent of (3. Moreover, since D(d) CH (5l), it follows that (A - Qr1 maps L2(1) into Hm,(Q). Therefore, (A - R)-1 is compact, by Rellich's theorem. Therefore, the Riesz-Schauder theory of compact operators implies that the spectrum of (A - d)-1 consists only of eigenvalues of finite multiplicity, whose only possible limit point is the number 0, which is also in the spectrum of (A - (3)-1. By Theorem 12.4, this implies that the spectrum of Cl is itself discrete, and the eigenvalues of Q have finite multiplicity. Therefore, there exists a real number A0 such that A0 E P(Q). By considering d - Ao instead of Q, we may now assume that T = Q-1 exists as a bounded linear transformation on L2(12). By Theorem 12.3, (A - Q)-1 = - TA, so it follows that all non-real directions are direc-

Elliptic Boundary Value Problems

252

tions of minimal growth of the modified resolvent of T. This also follows immediately, as noted above, just from the fact that T is selfadjoint. Note that R(T) = D(Q) CHm,(Q). Let us now examine carefully the case in which m' < n'. Then, if k is the integer guaranteed by the hypothesis, Tk = ((J k)_1, so that R(Tk) = D((Ik)CHk.,,(St). By the smoothness assumptions on the coefficients of A, the operator Ak exists as a differential operator, and the coefficients of Ak are continuous and bounded in Q. Moreover, the principal part of Ak is just

(A') \x, 6 = (A'(x,

))k.

Therefore, A' is an elliptic operator of order km'. Our aim is to apply Theorem 14.4 to the operator T k. We shall set k = 1 in case m' > n'. Since in any case R(Tk) CHkm,(1l) and km' > n', and since Tk is selfadjoint, it remains to verify the hypotheses 1°, 2° of that theorem, as we now do.

Since Tk = ((Jk)-1, it follows that for any f E L2(Q), (14.31)

AkTkf = f.

Now let 00 = 0, and let the operator P of Theorem 14.4 be the principal part of Ak. Thus, Ak = P+ B, where B is an operator of order less than km' having bounded coefficients. By (14.31), 1! PTkf _ fI I O'Q

(14.32)

IIBTkfI I o.S1

5 CIITkfHIkm,_l,SZ.

Now let x° E St, let U be a small neighborhood of x°, such that the coefficients of P vary by less thane in U. If PO is the operator P with coefficients evaluated and frozen at x°, then (14.32) implies IIPOTkf _ fllo.u s II(P0

-

P)TkfIIo,U+CIITkfIlkm_l,U

sec. 14

Eigenvalue Problems; The Self-Adjoint Case

s

y(IITkfIlkm'.f +

s

(14.33)

253

CIITkfIIkm'-1.l

YEIIT'IIkm'IIf IIO,Q +

CIITkfllkm'-1.I'

where y depends only on m and n. But this is precisely the first estimate of (14.16), with m' replaced by km', the order of P. Likewise, the estimate for the adjoints of the operators can be derived. Next, if e'0 is not real, then ei0 is a direction of minimal growth of the resolvent of the self adjoint operator (1, as we have remarked. Therefore, for all u E D(() and for arg A = 0,

I lul o.SZ S c

nt

I I(a - A)UI I O.Q-

JAI

In particular, since D(@) D Co (S1), we have for all I Ilk

O'n

S'onst

Co (ul)

- Aw I

I I

JAI

Therefore, Lemma 14.3 implies A'(x, it) - e'O 0. This also implies that (A'(x, i,,))k - e'0 0 for real 61 so that condition 1° of Theorem 14.4 is verified. Note that in our case it even follows that A'(x, itt)

is real for all real 6 Thus , we can apply Theorem 14.4 in the case of the operator Tk. In particular, we wish to apply the asymptotic formula for the kernel AKA of the operator (Tk)we have from Theorem 14.4 (14.34)

(n/km')-1 + O(IAI(./ km')-1).

x)dx = cIAI

where c is given in accordance with the formulas given in the statement of the theorem. Now since the characteristic values of T are just the eigenvalues {AJ} of d, the characteristic values of Tk, counted according to multiplicity, are }Ai }, By Theorem 13.10, we then have f KA(x, x) dx = Sl

1 A

A

Elliptic Boundary Value Problems

254

Comparing this with the asymptotic formula above, we have for arg A = 0 fixed, a 10 not real, =

1

CIAI(n/km')-1 + o(IAI (n/km')-1)

as IAA

oo,

IAk -A In particular, let us take for simplicity 9 = 2. Thus, letting A = it, t > 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 Self-Adjoint 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.<0

dN +(A1/2k)-f

V

A+t2

=f o

A

Ak

=E

0

dN(Al/2k)_2 f°°

A + t2

0

-

A+t2 dX

dN_(A1/2k)

dN _ (A1/2k)

A + t2

C1t(q/k)-1 + o(t(q/k)'1),

by (14.38). Let A

N(A)

= f JidN(µl/2k). 0

Then, by an integration by parts and the asymptotic formula (14.40),

Elliptic Boundary Value Problems

2%

V/,N(Al/2k)_1/2

f u-'/N(µ1/2k)dµ 0

= 2kc2 sin(irq/2k) [A2k+ 2

q

_ 1/2

rrq

o

fµ22d(µ4/2k)+OUs+2)

=2kc2 sin(rrq/2k) irq

0 q

=C

sin(7rq/2k) 2

2k dµ) +O(A2k+ 2)

2 I

A

+ o(Azk+2).

+

IT

2k

2

Thus, the converse of the Tauberian theorem implies VA

f00

dnA1/2k)_ f°° dN(A)

/

A + t2

0

\ + t2

0

= C sin(9rq/2k) t2( k 2

- i+

O(t2(2'gk-i

+2)

sin

tan(97q/2k)t(q/k)-1 + n(t(q/k)-1).

= C2

Therefore, (14.42) implies

2f0

.VA dN_(A1/2k)=[C2

X+t2

tan (97q/2k) - C, ]t(q/k)-1 + 0(t(q/k)-1).

In order to apply the Tauberian theorem, let (14.43)

N_(A)

=f

A ''UdN_(µl/21).

0

The previous asymptotic formula then shows that, after replacing t2 by t, foo

1

00

A+t

dN (;k) = 1/2LC2 tan(lT4/2k)

c11t

2) + o(t 23 i).

sec. 14

Eigenvalue Problems; The Self-Adjoint Case

257

Therefore, Theorem 14.5 implies q

1 / 2 [ c 2 tan(rrq/2k) - c 1 1

sin (5k-

"Zk

+-2rr

2

A2 k+ 2 + O(A 2k

q+ 1\2k

1

= 1/2[c2 sin (7rq/2k) - c 1 cos (7rq/2k)]

2)

2 2

1

7T

(A-2) + O(A

But from (14.43) it then follows that

i

+2).

A

N_(A11k)

= const + f µ-'/'dN (µ) 0

= 4[c2 sin (7rq/2k) - c1 cos (7rq/2k)]

1

"(2k +2 A

q

f µ-'/diL2T+2 + O(Aq/2k) 0

_ [c2 sin (7rq/2k) - c1 cos (7rq/2k)] k Agl2k + 0(,\q/2k). rrq

Replacing A by k2k, this relation, becomes (14.44)

N_(A) = [c2 sin (7rq/2k) - c1 cos (7rq/2k)] k 'k 4 +o(kq). rrq

Since N = N+ + N_, it follows from (14.44) and (14.41) that (14.45)

N+(J1) =

[c2 sin (7rq/2k) + c1 cos (7rq/2k)] k kq + o(Aq). rrq

Since q = n/m', the theorem will be proved if we now compute c1 + ic2 = c. This we proceed to do. As we have remarked, A'(x, i.e) is real for real 6 Let x be a fixed point of Sl and let p(.) = (A'(x, 1 ))k. Then (14.37) can be written in terms of its real and imaginary parts as

Elliptic Boundary Value Problems

258

Cde

(2n)np(x) = f En p(S) - r

de, i f

P(t)

=f

En p(e) 2 + 1

1

de.

En p(S)2 + 1

For 0 < t < 00 let V(t) be the region le: jp(6I < t}, and let v(t) be the measure of V(t): v(t)= IV(t)l. Then the homogeneity of A' shows that V(t) = t l /k-'V(1), so that V(t) = to/kn''V(1) = t4/k V(1).

Therefore, (2n)n9p(x) = f "O

4 v(1) f m

t(4/k)-1

k

t2+1

o

k = 9

(14.46)

dv(t)

1

t2+ 1

0

v(1)

1

ss4+21)-1

f

ds

71

v(1)

2k

dt

sin 2k

by (14.29) (we have substituted s = t2). It will be convenient to treat two cases separately. Case 1. m' is odd. In this case p(1`) is odd, so that Rp(x) = 0. Also v(1) _ I1e.

=2j1$: = 211e:

IP(ft < 111

0
= 2e (x) = 2a)-(x)

Thus, (14.46) implies

sec. 14

Eigenvalue Problems; The Self-Adjoint Case

259

p(x) = ijp(x) = i(277)-n2t)t(X) rrq

1

=

2k sin (7rq/2k)

Therefore, it follows from (14.36) that

ci + ic2 = i(2n)

n

q

k sin (nq/2k)

rrq

1

k sin (7,q/2k) Therefore, ci = 0 and sin (rrq/2k)c2 = (nq/k)ct. Hence, (14.44) and (14.45) imply Nt(k) = ctA4 + 0(1\4),

so that the theorem is proved in case m' is odd.

Case 2. m' is even. In this case p(6) = (A'(x, IO)k is even, and the ellipticity of A implies p(O does not change sign. Let z be the common sign of p(O and A'(x, i6). Then (2rr)-nRp(x)

P(6) de =f Enp(W + 1

=z

f' 0

dv(t)

t

t2+ 1

= z 4 v(1) f°° t4/k dt k

0

t2+1

= z q v(1)2k sin 2k + 2 =Z 4 v(1) 2k

as in (14.46). Also, v(1)

= 116 Vol <

111

77

cos (rrq/2k) '

Elliptic Boundary Value Problems

260

_

{

0
=l1:0
also,

w_Z(x) =fife: - 1 < zA'(x, ie) < 0j = 0. Thus, (14.36) implies

f

rr

c1 + is z = (277) "[z q

2k cos (,q/2k) Q'-

+i q

(x)dx

n

2k sin (77q/2k)

z q

77

_c

2k cos (7rq/2k)

2

+1 q

f

)

(x)dx]

rr

2k sin (7rq/2k)

c. Z

Hence, cos (rrq/2k)c1 = z(77q/2k)c2 and sin (rrq/2k)c2 = (7rq/2k)c2, so that (14.44) and (14. 45) imply N_(A) = (1/2 - 2Z)c2A° + o(AQ), N+(A) = (1/2 + 1/2z)CZAa + o(AQ).

As c_Z=0, we find for z = + N_(A) = c_A9 + o(Aq), N+()L) = c+Aq + o(Aq),

and likewise for z = -. Q.E.D. Some comments are in order concerning the theorem just proved. In case n 2, A'(x, it) has constant sign for each fixed x. The proof of this is like that of Theorem 4.1, but simpler. If A'(x, i 61) > 0 and

sec. 15

Non-Self,Adjoint Eigenvalue Problems

261

A'(x, i62) < 0, then there is a curve connecting e1 and 62 which does not pass through e= 0. Since A'(x, in is continuous, it must happen that A'(x, in = 0 for some 6 on the curve, a contradiction of the ellip-

ticity of A. Likewise, in case n = 1 and m' is even, A'(x, in has constant sign for each fixed x. (Note that our theory certainly is valid in the case n = 1, that is for ordinary differential equations.) Therefore, in either of these cases it follows that if Q is connected, either c+(x) or a)-(x) vanishes identically. Therefore, either the positive or negative eigenvalues predominate in number. This is not the case for the exceptional situation of an ordinary differential equation of odd order. For example, the eigenvalues of the problem 1 du = Au,

i dx u(0) = u(7r),

are all the even integers. This brings up another question. Even in the cases n 2 or m' even, it can actually happen that there are infinitely many eigenvalues of each sign. For second-order problems this indeed cannot happen, but there are examples involving the biharmonic operator in which there are infinitely many eigenvalues of each sign. 15. Non-Self -Adjoint Eigenvalue Problems

We still shall assume SZ to be a finite union of disjoint bounded open sets, each of which has the restricted cone property.

a,(x)D' be an elliptic

THEOREM 15.1. Let A(x, D) = IalC2m

operator or order 2m in ci, having real leading coefficients; let A be C1_1 .(Q), normalized so that A'(x, in > 0 for x E Ii, 6 real, 0. Let as E and let A* be the formal adjoint of A. Let Q be a closed, densely defined operator with domain in L2(12), such that 10

D((I) U D((f*) CH2m(ci);

Elliptic Boundary Value Problems

262

2°

for u i D(d), du = Au; for u E D(d*), d * u = A*u;

3°

all directions except the positive real axis are directions of minimal growth of the resolvent of d;

4°

in case 2m < n + 1, there exists an integer k such that k7 (n + 1)/2m, the coefficients of A are C(k-1)2m*(SZ), and D(dk) Ll D(((f*)kkH2mk(St).

Then the spectrum ofd is discrete and the eigenvalues ofd have finite multiplicity. There are only a finite number of eigenvalues outside the angle Iarg Al < E, for any positive E. If {Ai } is the sequence of eigenvalues of d, counted according to multiplicity, and if for A> 0, N(A) is the number of A. such that RAE < A, then N(A) = C0An/2m + o(An/2m) ' where

co =

(277)-n f

A'(x, ie) < 1}Idx.

Q

Proof. The proof is quite similar to that of Theorem 14.6 except that the condition of self adjointness is replaced by other restrictions. Let A0 E p(d). Then R((A0 - d)-f}= D(d) CH2m0, so that Rellich's theorem implies that (A0 -

d)-1 is

a compact transformation of L2(St)

into L2(0). Therefore, it follows that (A0 - (f)-1 has a discrete spectrum, with eigenvalues of finite multiplicity. Hence, d has the same properties. Note that this property of the spectrum of d requires only that the resolvent set ofd be non empty. By considering the operator d - Ao instead of d itself, it is seen that we may assume that 0 E p(d), and we shall thus consider the compact operator T = d'1. The assertion on the location of the eigenvalues follows immediately from Theorem 12.2, since, for any c > 0, there exists a number A. such that the set {A: JAI > AE, Iarg Al > E} is contained in

the resolvent set of d. Note that the conditions on m, n, and k are no different than those made in the statement of Theorem 14.6, concerning m', n', and k,

sec. 15

Non-Self-Adjoint Eigenvalue Problems

263

since now m' = 2m is even. As before, if 2m n + 1, set k = 1. We need to verify the hypothesis of Theorem 14.4, applied to the operator T k. Note that our assumptions 1° and 4° imply R(T k) _ D((J k) CH2mk(Sl), and R((Tk)*) = R((T*)k) = R(((f *)-k) = D((1 ) k) C

H2mk(Sl), and 2mk > 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

<

I

1-1 o

(Ak 4

t)(tL + t)

Elliptic Boundary Value Problems

264

k i

yE

i>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

Non-Self-Adjoint 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/2-k,(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 self-adjoint case. What is important, however, is that Theorem 15.1 applies to perturbations of self-adjoint 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 Self-Adjoint Eigenvalue Problems

First, suppose the transformation do of Theorem 15.2 is self-adjoint. Theorem 12.7 implies that for non-real A, Iiu[1

19,1-111(A

O,s ,

_ do)UIIo,si, u E D(d0),

or, equivalently, (15.4)

I I(X - do)-1110 s 19XI-1.

Therefore, for u E D(do) it follows from (15.3) that I

i ul l m',S S C[I I doul 10,S-t + I I UI I o,Sl]

< C[I I(A

- (1o)ulI o,S-t + 1A1

I lUllo.c + I

U1 10,01

1)19,1-111(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 self-adjoint 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,

A-Q=A-Qo-J`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 - 0-193 is an invertible mapping on Hm,_1(1l), and its inverse has norm less than 2. Therefore, since

A-a=(J1-ao)[1-(A-Qo)-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

Non-Self-Adjoint 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(A-a)-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-(b-11I0 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 non-real 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

Non--Self -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 non-zero aa(x)ap(x)S eR > 0, la -m' IQ.m or,

aa(x)eal 2 > 0.

lal-m' 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 self-adjoint and B is of lower order, we obtained estimates for A - Q just having an estimate (15.3) involving do" Of course, we also *1-I, 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 <

I

I

I I(A - Q)ul l0. O.Q.

Moreover, if the resolvent set ofd is not empty, then e'0 is a direction of minimal growth of the resolvent of Q.

Proof. Let Cbe a function in C°°(E1) satisfying IC(t)l S 1 for all t, C(t) - 1, Itl S'A, and (t) - 0, ItI -, 1. First, let u E M, and let v(x, t) = j(t)eirtu(x), where r is a positive parameter. According to the hypothesis, (15.13) is satisfied by the function v. Thus, for r r - we have by Leibnitz's rule Ik(t)elttu(x)l2m,r

S c[I

1u -

(-1)mei0uDe

+I

o,r,

I o,r]

sec. 15

Non-Self-Adjoint Eigenvalue Problems

273

r2me1O e1rtullo,r

(15.14)

+ YIIUI 12m- 1,Q +

yr2m-1IIUIIo

for a certain constant y. Now also

I f

IXI+s-2m

IT

y D0Dt(elrtu(x))I2dxdt

f1f

E

1a1+a_2m/2

r2a f DXu(x)I2dx

L

=

1-1+a-2m

2m

I

r2aIuI2

2m-a,SZ

a-0

Thus, if r Z 1, i`J'elrtu(x) 22m, S

2 ramII u I = r4mllul

2

O,St

I

2m

+ rz a

l

u2 l

I2m-a,Q

O.1l + r2IIU l2m-1,Sl

Therefore, (15. 14) implies CII Au _r2mr1Oullo,r

[r4mllullo.St+r21JUJI2m-,SZ)/ +

y-\/2[r4m-2I

luI I01S1 + I IUI

I

o.SZ

+ yv2r-1[r4mI JU11 2.Q +

r is sufficiently large, [r4mI

IUII 0,j2 + r2I IUI I2m-1, ]1/2 -< 2CI IAu - r2me1Oul 1 o c1.

,

Elliptic Boundary Value Problems

274

Hence, Iul t o,St 5 2Cr

Zm I IAu _ r2me1BuH I

O.Q.

Thus, if A = r2me'B, this becomes

Ilullo.o: 2CIXI-1II(A-a)ullo.i,

IX

large,

which is the required estimate. In case u is an arbitrary element of D(Q), let {uk} CM be chosen such that uk -. u in H2m(0). Then QUk = Auk Au = (fu in L2(1), so that the inequality for u follows from the proved inequality for uk. Finally, suppose that there exists a complex number'. in p(Q). Then (A0 _ (I)-1 is a mapping of L2(1) into D((I) CH2m(1), and so is compact by Rellich's theorem. Thus, the spectrum of (A0 - CO-' is discrete. Hence, the same holds for the spectrum of Q. For arg A = 0, JAl large, it follows from the estimate obtained above that A - ( is one-to-one, so that A is not an eigenvalue of Q. Since the spectrum of Q is discrete, A c p(Q). This estimate then implies II(X -(1)1 110

s2CXI-1.

Thus, e'0 is a direction of minimal growth of the resolvent of Q. Q.E.D. We now shall give some indications of how the abstract theorems on eigenvalues can be applied to specific problems. First, we discuss the Dirichlet problem. For convenience, we shall assume that the operator has. infinitely differentiable coefficients in Sl, and that 9 is of class C°°. Of course, these restrictions are far more than is needed, and we could easily keep track of the regularity actually needed. But for simplicity, let us assume the infinite regularity. Assuming zero Dirichlet data, the Dirichlet problem can be stated:

JA= I

in

St,

(15.15)

=0 on d,jal <m-1. Here A(x, D) shall be a strongly elliptic operator of order 2m in Sl.

sec. 15

Non-Self`Adjoint Eigenvalue Problems

275

By Garding's inequality (Theorem 7.6) and Theorem 8.2, it follows that the GDP for A + A has a unique solution, for certain complex numbers A. If A is suitably normalized, A can be any sufficiently large real number. We shall replace A by A + A, and thus assume 0 (15.15) has a unique solution u E H1(11) for every fin L2(Sf). Let u = Tf. By Theorem 9.8 it follows that u is actually in Thus, R(T) CH2m(f1). Since the same results hold for the formal adjoint A* of A, and since the solution operator associated with A* is just the adjoint of T*, it follows that R(T*) C :H 2m (Q). For the operator

d we can take T-1. Then D(@) = H2m(I) n H'(0), and du = Au for u E D(@). For, clearly D(@) = R(T) CH 2m (SZ) f1 Hm(fl). Conversely,

suppose u E H2m(fl) fl Hm(f2). Let f = Au. Then f E L2(f0), and, trivially, u = Tf. Thus, H2m(i) fl H.(Q) C R(T), and du = QTf = f = Au. Likewise, D((I*) = H2m(Q) f1Hm(fl).

We also need to check the behavior of powers of T. Again, we shall appeal to the regularity theory of section 9, and prove that R(T') C H2mk(T ). For, suppose this has been proved for k, and let u = Then u = TTkf, so that u E H2m(u) fl H,,(Sl) and Au = Tkf. Again, Theorem 9.8 implies that u E Hem+2mk(l), since Tkf E H2mk(Q), Now there are three cases we shall discuss. Tk+1f.

In case A is formally self-adjoint, then T = T*, so that d is selfadjoipt. Theorem 14.6 then applies to our case immediately. 1.

2.,; Suppose next only that A'(x, ic) is real. By passing to some 'Dirichlet form for A, it follows that A can be written in the form

I

A(x, D) =

DaaaR(x)DR;

I

IRII<m

cf. (8.9). Now let aap(x) = 1/z[aap(x) + apa(x)], and

A0(x, D)

=

II1I<m Daa' Do. a

276

Elliptic Boundary Value Problems

Then AO is formally self-adjoint (cf. (8.11)), since a0 a= 'Pal and and Ao have the same principal part. For, Leibnitz ' s rule implies Ao-

E

A

a o Da+13

a,3

//

llNl-m

I

1/4[aa,8 + aoa]Da+(3

ll-m

aap Da+O lal-m 1131-m

A'.

Thus, A = AO + B, where AO is formally self-adjoint, Ao(x, ie) is real, and B is an operator of order less than 2m. Moreover, the strong ellipticity of A implies by means of Garding's inequality that, with suitable normalization, (A0u, u)o,j2

_'kollullo,SZ

>--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 = T-1 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 self-adjoint, we obtain immediately asymptotic expressions for the umber of eigenvalues less than t. But in any case, this problem is

almost self-adjoint, 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 complex-valued function, and let JF(A)1< ClIXI-1

on two rays making an angle n/a at the origin. Let r1 < r2 <...,rk and let max

00,

IF(k)I S C2e k,

IAl.Hk

where 6 < a. Then F(A)I S C1IkI-1 everywhere between the two lines. THEOREM 16.2. Let F(A) be an entire complex-valued function of finite order p. Then for any positive c there exists a sequence such that r1 < r2 <...,rk O+E

min (F(k)I >e (Al

k

rk

The second of these theorems is a direct statement of a theorem on p. 273 in Titchmarsh, The Theory of Functions, 2nd ed., Oxford University Press (1939); the first is a Phragmen-Leindelof theorem, and can be found in a slightly different form on p. 177 of the same book.

Now we give an estimate due to Carleman. LEMMA 16.3. Let T be an operator of finite double-norm on a

Hilbert space, and let c be a positive number. Thendhere exists a sequence r1 < r2<...,rk 00, such that the modified resolvent TA exists on the circles IX: JAI = rk}, and for JA( = rk,

Completeness of the Eigenfunctions

Sec. 16

279

2+E

IIITAIII <erk

Proof. By a modification of the classical Fredholm theory of integral equations, we have a formula, 0,k

00

where 8(A) =

y.

m'0

8m,km is a complex entire function, and AX

Y_ A .,k- is an operator-valued entire function, the convergence of m'0 the latter series being in the double-norm sense. Moreover, the following estimates are valid:

80=1, I8m1

(e/m)m/21II T111--p

= T, 111A

III <

e/.(e/m)m/2IIITIIIm+1.

Also, TX exists precisely when 8(A) 0. For a derivation of these formulas, see Zaanen, Linear Analysis, 1st ed. (1953), AmsterdamGroningen, pp. 261-274. From these estimates we obtain 00

I8(U)i S 1 +

M.1

(e/m)m/2IIITIII-IXIm.

Now let µ=1eIIIT 11 2IAI 2, and split this sum into the odd and even terms, to obtain 00

k

8(A)1 < 1 + I µ k'1 kk

k+/2

M

+k'0I (k i' 12)k+/ 1`

,

280

Elliptic Boundary Value Problems

-

+µ'/ 1+(2µ)'i+ 1I k1 kk

Rk

11

-

k-1 kk k

1+(210'/1+(1+µ'I/2) F 1` k- 1 k!

= 1 + (2µ)'i

- 1 - f"'/1 + (1 + µ.'/2)e It

< (1 + (2µ)')eµ.

Therefore, IAI)e/2.11IT11121XI2

IS(A)I < (1 + e/2IIITIII

Likewise, one sees that M

III-AXIII s IIITIII +

m

Il e'%(e/m)m/211ITIIIm+ulAIm 00

S e'/2 11

TIII[1 +

I (e/m)m'21IITIHImIAIm]

M-1

ITI1121Al2.

S (e'/2 IIITI I +

elIITIII2IAI)e'/ell

Hence, both S(A) and Ax are entire functions of order not exceeding 2. Applying Theorem 16.2 to S(A), we obtain rl < r2 <...,rk -+ oo, such that 2+'/E

min

IS(,t)I > e1rk

IXI - ,k

Therefore, for XI = rk we have 2rkerk+'/E

IIITAIII s

< e

+elIITIII2rk)eeIIITII1

2+C rk

sec. 16

Completeness of the Eigenfunctions

281

for rk sufficiently large. Q.E.D. If T is a linear transformation on a Hilbert space, let sp(T) be the closed subspace spanned by the generalized eigenvectors of T, and let sp'(T) be the closed subspace spanned by the generalized eigenvectors of T which correspond to non-zero eigenvalues. In case Tf = Af and A 0, then it is clear that f E R(T). Thus, we have the trivial relation sp'(T) CR(T). Therefore, if R(T) is not dense in the Hilbert space, it cannot be expected that sp'(T) is dense. The most that could ever be expected is that sp' = R(T). We shall now give an abstract theorem containing a sufficient condition for this to occur. This theorem is found in Dunford and Schwarz, Linear Operators, II, Interscience, 1963. THEOREM 16.4. Let T be an operator of finite double-norm on a Hilbert space X, such that there exist N directions of minimal growth of the modified resolvent of T, where the angle between any two adjacent rays is less than 1/2e. (Thus, N 5.) Then sp'(T) = R(T).

Proof. The Riesz-Schauder theory of compact operators has as a consequence that the analytic function (A - T)-1 has poles at the nonzero eigenvalues of T. Thus, if Ao is a non-zero eigenvalue of T, the Laurent expansion of (A - T)-1 about ko has the form

(16.(A-

T)-'

M

k

=v-1I

ko)-v+ E Av(A- A0)v, v-o

where all the operators Bv, Av, commute with T. Therefore, 1 = (A - T)-TA - Ao) + (Ao - T)] k

00

A0)-1'+ 1 + F A1(,\

! Bv(A -

vo

v-1

k

+

v1

Bv(A0

- T)(A -

Ao)-v

- ko)v+ 1 00

+

v

E o

AV(AO

- T)(A -

Aa )v

k-i =Bk('o-T)(A-,Ao)-k+vI

[B,(A0-T)+By+1](A-Ao) v+B1+ A0(A0 -7

Elliptic Boundary Value Problems

282

00

+

v-1

[Av(A0 - T) + Av-, I(A - Ao)v'

From this it follows that

Bk(AO-T)=0, BV+1

Bv(Ao T)

v = 1,...,k - 1.

Therefore, (Ao - T)k+ 1-1BV = 0,

(16.2)

v = 1,...,k.

Now suppose f q7 X. By (16.1), k

(A -

T)- ,If

=

v-,

(k - Ao)-'Bvf

Ao)vAvf. +1 vo U -

The relation (16.2) implies that (A0 - T)Bkf = 0,

(Ao - T)2Bk-,f = 0,

(Ao -

Therefore;, the non-zero vectors among Bkf, Bk_f,.,.,B1f are all generalized eigenvectors of T corresponding to the eigenvalue AO. Next, suppose AO is a characteristic value of T. Since X0 -1 is then an eigenvalue of T, for A near Ao we obtain from (16.1) TA = T(1 - AT)-1 =

A-'T(A-, - T)-,

00

k

(16.3)

283

Completeness of the Eigenfunctions

sec. 16

=

T[V-1 I

B"-

Ao1)-v + 2

A"A-1(A

1

V-1

-

0

Now A-1(A..1 - X-1)-I/A = A-1 )-V 0

AvAv-1(_1)v(A _ Ao)-V

=

= (-1)'Ao[Ao + (A

V-1

_ (-1)"Ao µ2u

-

(v

Ao)]V-1(A - AO)-v

1)Ao{A -

Ao)v-1-µ(A - Ao)-v

If

V^1

µo

(-1)"(" µ 1) Ao+v(A -

µ-1

µ

A0)-µ-1

10,

Therefore, (16.3) may be written

TA = I y (-1)"(v µ-

1)Ao+v-1TBv(A

v-1 µ-1

- Ao)-µ+ T1(A),

where T, (A) is analytic at A0. Rearranging this series, we obtain k

k

TA = I (A - AorpP 2 (-1)"(" P. 1

v-µ

µ

1)Ao+v-1TBv)

+ T1(A)

k

(16.4)

_ Y, Cµ(A -- AD)-µ + T 1(A) . µ-1

Here, as above, it holds that for any f E X, Cµf is either zero or a generalized eigenvector of T corresponding to the eigenvalue A0 1. This follows from the corresponding fact for B,f and the formula for C µ.

Elliptic Boundary Value Problems

284

Now we shall prove that sp'(T) = R(T). Since we already know sp'(T) CR(T), it is enough to show that R(T) Csp'(T). This is equivalent to showing that the orthogonal complement of R(T) contains

that of sp'(T). Thus, assume that g is orthogonal to sp'(T) and that f E X. The theorem will be proved once we show that (Tf, g) = 0. Consider the function F(A) = (TAf, g).

This function is analytic on pm(T), and, as we have seen, has only poles at the characteristic values of T. Let A0 be any characteristic value. By (16.4), ilk follows that for A near A0, k

F(A) = E (Cpf, )(A - A0)-It + (T 1(A)f, g). µ'1 As we have shown, Cµf is either zero or is an element of sp'(T). Since g is orthogonal to sp'(T), in any case we have (C1), g) - 0. Therefore, F(A) has a removable singularity at Ao . Since this holds for any ko not in pm(T), F(A) is entire. On any of the N rays of minimal growth of the modified resolvent, of T, IF(A)I s IITAhI IIfII 11911 = 0(IAI-')

Now let c be a positive number such that the angle between any two such adjacent rays is less than n/(2 + E). By Lemma 16.3 there exist numbers rk, r1 < r2 <...,rk -, w, such that for Al I= rk 2+E

I(TAIl S IIITAIII

(erk

But now the Phragmen-Lindelof theorem, Theorem 16.1, implies that IF(A)I < CIAI 1 between two such rays. Therefore, since such angular regions fill the complex plane, IF(A)I < CIIAI-1 for all X. Therefore, F(A) is bounded in the plane, and must be constant, by Liouville's theorem. Since F(A) -, 0 as (Al -, oo, F(A) = 0; i.e., (TAI, 9) = 0. Letting A -, 0, we obtain TA -, T, so (Tf, g) = 0. Q.E.D.

Completeness of the Eig,enfunctions

sec. 16

285

For applications of this result to elliptic problems, the following theorem is useful. THEOREM 16.5. Let d be a closed, densely defined operator in the space L 2(2), such that D(d) CHm,42). Suppose that all but a finite number of directions are directions of minimal growth of the resolvent of d. In case rn' -<<'/2n, suppose there exists an integer k > n/2m' such that D(dk) CHkmI(J2). Then sp(d) = L2(f2)-

maps L2(l) into D(d) C Hm,(S2), and is therefore compact, by Rellich's theorem. By considering d - A0 instead of d, it follows that it is sufficient to suppose 0 c p(d). Let T = d'. Clearly, sp'(T) = sp(d). In case in' > '/2n, then Theorem 13.5 implies T has finite double-norm. Since TA = - (A - d)-', the directions of minimal growth of (A - d)-1 are also directions of minimal growth of TA. Therefore, Theorem 16.4 implies sp'(T) = R(T). But R(T) = D((f), so R(T) = L2(cl). Thus, sp(d) sp'(T) = L2([i), and the result follows in this case.

Proof. Let A0 C p(d). Then (d -

A0)-1

Now suppose m' <'/2n, but k > n/2m'. Then. R(Tk) = D(CQk) CHkm'(f ), so that again Theorem 13.5 implies T k has finite double-norm. Also, R(T') is dense in L2(f2), since R(T) is dense and T is continuous. That is, R(Tk) = L2(f ). Let k be the k roots of unity, and let z be any complex number such that I zj < 1. Then k

o0

k

a .(1 -

z

z)_1 = z F.

j-0

i-1

i-1

0o

CJtzj

a)

k

( r W;+1)zj+1

j0 i-1

1

Now k

E a)'+ I

t- 1

_ 0,, otherwise.

ko0

k,

Therefore,

Z L W,.(1 -

i'1

Y. kzvk = kzk(1 - zk)-1.

V-1

Elliptic Boundary Value Problems

286

k

Multiplying this relation by 1 - zk = Il (1 ;-1

and dividing by z.

k

k

(16.5)

II(1-

(u

z)=kzk-1

;y=

Although (16.5) has been derived under the assumption 0 < Izl < 1, it must actually hold for all complex z, since both sides are just polynomials in z. Therefore, if we replace z by the operator zT, a corresponding relation is obtained: k

k

U (1 - (j zT) = I a ;1 f-1

kzk-iTk-1.

;r1

Since also k

11(1-(o,zT)=1-zkTk, -1

it follows that zk Jpm(Tk) => (),z E pm(T) for all j. Assuming that zk E pm(Tk), multiply (16.6) by k

(1 - zkTk)-1 =

II1

;-

(1 - (o

zT)-1

to obtain

k

I-1

(0,(1 - (o;zT) 1 =

kzk-iTk-i(1 _zkTk)-1.

Finally, multiply both sides of this by T to obtain k (16.7)

1-1

(o.TAI z = ()

kzk-1(T k) zk

1

This relation holds for all complex z such that zk E pm(Tk).

sec. 16

Completeness of the Eigenfunctions

287

Let e'0 be a complex number all of whose kth roots are on rays of minimal growth of the modified resolvent of T. The assumptions of the theorem imply that all but a finite set of the numbers e'0 possess this property. Indeed, we didn't need to assume so much, but only enough to insure that such numbers e'B determine at least five lines such that any two adjacent lines make an angle less than 1/171. Letting z = t 1 Ike101k in (16.7), we obtain k

(Tk)tei6 = k-1t (1/k)-1e1O(1-k)/k

IT0)

1/keie/k.

Therefore, if X = te'B, II(Tk)xII s

k-1IAI(1/k)- 1

k

I-1 IIT

t I

1/keie/kII tl/keiO/kl-1

C

i- 1

I

CIXI(1/k)-1IXI-1/k <

CI>
Therefore, e'0 is a direction of minimal growth of the modified resolvent of T k. Applying Theorem 16.4 to T k, it is seen that sp' (T k) = R(T k) _ L2(SZ).

Now we shall show that sp'(Tk) Csp'(T). From this it will follow that sp(Q) = sp'(T) = L2(l), and the proof will be completed. First, if a is any positive integer, the rational function U (z - (o.)-°` can be I-1

decomposed into its partial fraction expansion. Thus, for certain complex numbers ci1, we have k

U (z - Ql/) i-1

k

i-1 V-1

ci1,(z - COi)-v

Elliptic Boundary Value Problems

288

k

Multiplying this relation by 11 (z - mwe have an identity for all, J.1 complex z, k

k

i-1 v-1

civ(z - (ji)a v U (z - (J)a =

1.

i-1

Since the left side is a polynomial in z, we obtain a corresponding identity if we replace z by the operator z-1T. And then, multiplying the resulting expression by z°`k, we obtain k

k

civzv(T - (oiz)"-v 11 (T - coiz)°` = zk, i-1 v-1 j-1

(16.8)

1

jri

a relation which holds for all complex z. Now suppose f is a generalized eigenvector of Tk corresponding to a non-zero eigenvalue X. Thus, for some positive integer a, (Tk - A)°`f = 0. Let z be any complex number such that zk = X. Then k

II (T-(o .z). Tk-k=Tk-zk= j-1 i

Thus,

k

II

i-1

(T-(oz)°`f=0. j

This implies that for any i, i = 1,...,k, and any v, v = 1,...,a,

(T - (ez)v[(T -

(oiz)°`-v

k 11

j1

(T - Wjz)"f] = 0,

jri

so that the vector

(T -

(oiz)«--v

k

11 (T - mjz)" f

J.1 j,i

is either zero or is a generalized eigenvector of T corresponding to

sec. 16

Completeness of the Eigenfunctions

289

cu,z. Hence, this vector belongs to sp'(T). But then (16.8) implies z"kf E sp'(T), and, since z 0, f E sp'(T). Thus, sp'(T') Csp'(T). Q.E.D.

This result may be applied to the Dirichlet problem for elliptic operators of order 2m. For example, if the principal part A' satisfies either the condition that A'(x, i6) is real, or the condition that larg A'(x, iOI < a, where a < 1/4rr and 2m > 1/2n, then the generalized eigenvectors span L2(Sl).

INDEX

global regularity, 103, 124 Green's formula, 92, 134, 144 Hermitian, 96 Hilbert Nullstellensatz, 161 Hilbert space, 2, 175 Hilbert-Schmidt kernel, 201, 204, 219 hypoelliptic, 66 inner product, 2 integral operator, 203 interior regularity, 111 interpolation theorem (inequality), 23, 35, 133 Laplace equation, 91, 93

a priori estimates, 67, 148 Aronszajn's theorem, 171 biharmonic operator, 96, 150, 261 boundary value problems, 134 Calderon Zygmund kernel, 151 Calderon's extension theorem, 171 Cauchy Riemann operator, 47, 71

characteristic, 134

characteristic value, 179, 214 classical solution, 92 coercive quadratic form, 142, 159 coerciveness results, 158

compact operator, 103, 190 compact support, 2

Lax-Milgram theorem, 98, 215

completeness, 278

Leibnitz rule, 9, 10 Lewy's example, 67 linear functional, 99 local existence, 47 local regularity, 51

cone property (ordinary and restricted), 11 conormal derivative, 136 convolution, 153

difference operator, 42 direction of minimal growth, 181, 212 Dirichlet (bilinear) form, 94, 120 Dirichlet boundary conditions, 90 Dirichlet data, 92, 93 Dirichlet integral, 91 Dirichlet problem, 90 double norm, 188 eigenfunctions, 278 eigenvalue, 179 elliptic operator, 45, 46 ellipticity constant, 78 formal adjoint, 51, 95 formally self-adjoint, 96 Fourier series, 25 Fourier transform, 72 Fredholm alternative, 102 fundamental solution, 230, 234 0 Garding inequality, 71, 78, 84, 86 generalized characteristic vector, 180 generalized Dirichlet problem (GDP), 98

mixed boundary value problems, 148

modified resolvent, 177 mollifier, 5 multiplicity, 180

natural boundary conditions, 143 norm, 2

normal derivatives, 91 non-self-adjoint problems, 261 null space, 102 oblique derivatives, 136, 277 ordinary diff erential equations, 261 overdetermined elliptic system, 63, 65, 66

Parseval's identity, 72, 79 partition of unity, 7, 8

perturbation, 266 Phragmen-Lindelof theorem, 278 Poincarees inequality, 73, 75 principal part, 45, 47 quadratic form, 78

Rellich theorem, 30, 99, 208 regularity theorems, 58 regularity up to the boundary, 103

generalized eigenvector, 180, 282 generalized solution, 93, 98

global existence, 90 290

Index strong derivative, 3, 4 strongly coercive form, 142 strongly elliptic operator, 46 support, 1 segment property, 11 Tauberian, 248 test function, 2 self-adjoint, 230, 250 trace of a function, 37, 38 semi-norm, 2 trace of an operator, 193 Sobolev constant, 210, 211 Sobolev's inequality, 32, 40, 208 Sobolev's representation formula, 151 uniformly elliptic, 45 162 weak compactness, 41 spectral theory, 175 weak derivative, 4 spectrum, 175 weak limit, 41 s-smooth, 51, 52 resolvent, 177 resolvent set, 175 Riesz-Schauder theory, 103, 281

291