On the Extension Problem for Singular Accretive ...iknowles/papers/1986.jde.pdfclass of singular...

On the Extension Problem for Singular Accretive Differential Operators

Department of Pure Mathematics, Unioersity College, CardqT CFI I X L , Wa1e.r. United Kingdom

AND

Department o/ Mathematics, University of Alabama at Birmingham,

Birmingham, Alabama 35294

Received February 15, 1985; revised April 9, 1985

A linear operator T with domain 9 ( T ) in a Hilbert space 2 is said to be accretive if

for all y in g ( T ) , and maximal accretive if it is accretive and has no proper accretive extension. Operators of this general type, in which the numerical range

O ( T ) = {(Tu, u): u ~ 9 ( T ) , Ilujl = 1 ) (1.2)

is restricted to lie in a half-plane, have proved very useful in applications (see [8, Sect. 71 for a partial survey, noting that T is called dissipative if - T is accretive in the above sense; see also [7, 14, 18, 231).

In applications involving differential equations a major problem is that one is usually only given a formal differential expression and a function domain on which it gives rise to an accretive operator, and one has to prescribe a (possibly) larger domain on which the differential expression is maximal accretive. More generally, one would like to describe, explicitly, boundary conditions for all possible maximal accretive differential operators derivable from a given formal differential expression.

264 0022-0396/86 $3.00 Copyright 4- I986 by Academlc Press. Inc. All rights of reproduction tn any iorm reserved

SINGULAR ACCRETIVE DIFFERENTIAL OPERATORS

We consider here differential expressions r of the general form

,vhere I is an interval of the real axis, and the coefficients o > 0, p ,,..., p,, are real-valued and such that the minimal operator, To ,

associated with r (see Sect. 2) is accretive in the weighted Hilbert space, L;~(I), of functions f satisfying

A solution to the above extension problem is given in [ l l ] for the case that r in (1.3) is a regular differential operator in the sense of [20, Sect. 151, i.e., that the interval I is closed, and the coefficient functions n; p o i , p ,,..., p,, are Lebesgue integrable on I. Our main aim here is to consider the case when the given operator r is singular, i.e., when either I is not closed or at least one of the functions mentioned above fails to be in L1(I).

The methods used here are similar to those of [ I 1 ] in that we ultimately appeal to the abstract extension theory of Phillips ([25]; see [ I I , Sect. 23 for an outline). However, the technical difficulties are somewhat more severe in the singular case; in particuIar the proof of the crucial embedding result (Lemma 3.6) is more involved. In Section 2 we present some preliminary differential operator theory. The main results appear in Sec- tion 3, notably Theorem 3.1 1 and Corollary 3.12. The Appendix contains a detailed outline of the adjoint construction technique of Brown and Krall [4], including proofs for the singular case, which is not covered in [4].

It should be noted that we solve the extension problem for a specific class of singular differential expressions, namely those satisfying conditions ( A 2 ) and (A3) of Sections 2 and 3, (i.e., roughly, we assume that functions in the appropriate maximal domains all have finite energy integrals). In particular, all of the formal operators r above are of limit-point type at the singular end-point(s) of I. It is not known what happens if these assumptions are dropped. Thus, the extension problem is open even for the simplest operators, ry = - y" + q(x) y, of limit-circle type.

Once again we acknowledge with gratitude the assistance given by Professor R. C. Brown in adapting the techniques in [4] to the present situation.

266 EVANS AND KNOWLES

Throughout the paper we consider formal differential expressions of the form (1.3) with coefficient functions o > 0 , po # 0 a.e., and

assumed to be real and Lebesgue integrable in any closed subinterval of I. The expression 7 is said to be regular on I if I is finite and closed, and the functions (2 .1 ) are integrable on I ; otherwise 7 is said to be singular. In addition, the left-hand end-point, a, of I is regular if a > - co, I is closed at a, and the functions (2.1 ) are integable on every interval [a, Dl c I, and singular otherwise, with a similar definition for the right-hand endpoint, b, ofI . In lieu of further smoothness requirements on the coefficients, and following [20 , Sect. 15.21 we define the formal quasi-derivatives (up to order 2 n ) of a function y to be the functions y[O] = y, y['],. . . , yC2"] given by

where y(" is the usual ith derivative. The expression 7 is then given by

If p , , _ , has i continuous derivatives, then ~ ( y ) is also given by (1 .3 ) . Let L f , ( I ) denote the Hilbert space of all complex-valued functions f on I

satisfying (1.4) . For w = 1 , we denote L,',(I) by L 2 ( I ) . Set

9 = { y E L i ( I ) : yCk1 exists and is absolutely continuous on compact subintervals of I, 0 < k 6 2r2 - 1, and 7.v E L,Z , ( I ) ) . (2.4)

We define the maximal operator for .r in L t ( I ) , T, , to be

The minimal operator for 7 in L ; ( I ) , To , is defined to be the closure of the restriction of T, to the functions in 23 that vanish outside a finite interval [a, p ] interior to I. It is known [20, Sect. 171 that

T,* = T I . (2 .6 )


observe also that the Lagrange formula is [20, p. 501,

the corresponding Green's formula

where

For y, z E 9, we have the differentiated Dirichlet formula

I7

- U T ( ~ ) + 1 p,, -. JJ~)z( ' ) = / D(y , z ) ( x ) } ' . r = O

(2.9)

For a later use we also need a version of the well-known Sobolev inequality:

LEMMA 2.1. (c.f. [ I , Theorem 3.91). Let u and u'"' be in L 2 ( J ) for some real intervai J . Then for any x in J and any E > 0 there is a constant K = K(E) such that for any k nith 0 < k < n - 1,

where l l . l l J denotes !he usuai norm on L2(J ) .

We assume throughout that

( A l ) To is an accretive operator on L:,(I), and

(A2) the Dirichlet integral for T,

exists (possibly as an improper integral) for y, z in 9, and forms a positive definite inner product on the maximal domain 9; this inner product is


denoted by (., .), and the corresponding norm by / I . / / , . Assume further that for some c > 0 ,

Observe that ( A l ) is a consequence of ( A 2 ) . For if u E 9 vanishes outside an interval [a, PI interior to I, (Tau, u ) = llul/L > 0. This continues to hold for U E 9 ( T o ) and hence To 3 0. In fact, To is actually symmetric here. Observe that ( A 2 ) is clearly satisfied if, for example, p, 3 0 , 0 < i < n - 1 , and p , 3 EW on I, for some positive constant E.

In this case, in addition to ( A l ) and ( A 2 ) , we also assume

( A 3 ) At each singular end-point c of I, limy + D ( y , z ) ( x ) = 0 , for all y, z in 9 = 9 ( T , ) .

As an immediate consequence, we have from (2.9) that the integral in (A2) exists when I = [a , h) , with a a regular endpoint, and also

for all J, z in 53. ( A 3 ) is satisfied if, for example, p, 3 E > 0 and p , 3 0, 1 6 r 6 n - 1. For an account of the relationships between conditions of type ( A 2 ) and ( A 3 ) we refer to 1 1 3 , 161. As the coefficient functions (2.1) are real, it follows from ( A 3 ) and 1123 that the deficiency indices of To are (n, n ) if r is singular at only one endpoint of I, and To has deficiency indices (0 ,O) if r is singular at both endpoints of I. In the latter case, as To is a nonnegative self-adjoint operator, and hence maximal accretive by ( A l ) , the extension problem becomes trivial. Thus we may assume without loss that I= [a , b), and r is regular at a and singular at h. In this case we have from [20, pp. 71, 781 that

Set H = ~ t , [ a , b ) x Lfo[a , b ) and P = G ( T , ) . Under the indefinite inner product

where 17 = { u , , u 2 ) and G = j L ' ~ , z l Z ) are in H, and (., . ) is the usual inner product on L;>[a, h ) , H is a Krein space. As in [ l l ] , for 17 = { u , ~ , u ) we have

SINGULAR ACCRETIVE DIFFERENTIAL OPERATORS 269

and thus P is a closed positive subspace of H with a trivial null space. From (2.6) and [ I 1, (2 .16)] we have that P'= G ( - T I ) , and from [ l l , (2.13)] P 1 = M + O M w h e r e M + = P ' n H + . T h u s i i = j u , - T , u ) E M + if and only if - T I u = u. AS the deficiency indices of T o are ( n , n ) it follows

that

consequently there exist n real functions M , , , it1,, ..., it,,, in 9 orthogonal in ~ i [ a , b ) , and such that

~ h u s M + , the linear span of { w , , ..., rrl,,}, is a positive definite subspace of H.

The negative subspace, M , of P' may be characterized as follows. Let i = { u , - T I ~ ~ ) ~ M p . Then, if G , = { i c , , - T , W , } = {w,, rv,) we have for l < i < n ,

0 = Q(C, s,) = ( - T I u , M I , ) + ( u , - T I M ' , )

= [ u , w , ] ( a ) + 2(u , IV,)

= D ( w , , u ) ( a ) - D ( u , w , ) (a ) + 2(u , w,). (3 .4 )

u 2 ( x ) = [u i2 , r 2 1 ( x ) , ..., u [ " ] ( x ) ] ~ ,

( x ) . . . w f n l ( x )

W , ( x ) = ( ) . . . , 1 I . . . w',5"l (X)

[ ;

w,52;1, " ( x ) . . . H';';] ( x )

W d x ) = [ 2 / 1 I ]

\L'?n . . . M : ~ ; ] ( . Y ) 1, (3.5)

Then i i = { u , - T I u } E M if and only if


Note that det W , ( a ) #O, since {w,: i = 1, ..., n } is linearly independent, and hence W ; ' ( a ) exists. Consequently, ii = { u , - T I u ) E M if and only if

Define the space X by

where Hn is defined to be the completion of 9 with respect to the inner product (., .),. We identify P' with a subspace of X via the association

where ii = ju, - T , u } E P' and li = {u , u , ( a ) , u , ( a ) } E X , the vectors u,(a) being defined by (3.5). F o r li= { u , a , , a , } and C = { v , P I , P , ) in X (with u, v E H,, and a, , p i E C " ) define a n inner product e l ( . , . ) by

Q I ( ~ = - ( a , , P 2 > - ( a , , P I > - 2(u, v),, (3.8)

where (., . ) is the usual Euclidean inner product. Then, if ii, C E P' and Zi-f iEX, C H G E X ,

Thus the correspondence (3 .7) is a one-to-one inner product preserving map of P' into X. We also have

LEMMA 3.1. X is a Pontrjagin space of index n (i.e., a IT,, space).

Proof: This is a consequence of the decomposition

X = X + O X -

where

and any l i = { ~ r , a , p } , U E H , , , a , p ~ @ " can be written as l i = l i + + & - > where

li = { u , i ( a + P ) , t ( a + P ) ) , (3.10)

il+ = ( 0 , t ( a - P ) , { ( P - a l l . .

SINGULAR ACCRETIVE DIFFERENTIAL OPERATORS 27 1

The strong norm on X (with respect to ( 3 . 9 ) ) is given for li = { u , a , P ) by

We show eventually that A. , the intrinsic completion of M _ , may be identified with the strong closure of M - in X. We need several lemmas; the proofs are very close to those given for Lemmas 3.2, 3.3, 3.4, and 3.5,

in [ I l l .

LEMMA 3.2. Denore 6)s L rhe Q,-orrhogonal complement of M + in X. Tjen L is rhe ser of all { u , a , , a , ) in X suclz tlzat

a 2 = W l l ( a ) { W 2 ( a ) a , + 2C(u, w)). (3 .12)

LEMMA 3.3. Under rhe identification (3.7) , M + is a positive dejinite suhspace of X of dimension n.

LEMMA 3.4. L is a negative defnite suhspace of X.

LEMMA 3.5. The srrong topology (generared by (3 .11)) and the intrinsic topology (generated by ( 3 . 8 ) ) are equivalenr on L, and L is intrinsically closed.

Corresponding to [ 1 1 , Lemma 3.61, we have

LEMMA 3.6. Assume rhat

for some constants p > 0 and q > 0. Then the set

is dense in H , in the norm / I . / I , , .

Proof: As in the Proof of 1 1 1 , Lemma 3.61, any u E H,, can be identified with a function having ( n - 1 ) absolutely continuous derivatives on [a, a + q ] . Also for any E > 0 there exists a constant K ( E ) > 0 such that

l u C k 1 ( a ) 1 2 d ~ ~ l ~ l / : , + K ( E ) I I U I I ~ , U E H,, O d k d n - 1 .

Define

hCu, v l = (u , u ) ~ + ( a 2 , v , ( a ) > , u, V E H,,


where

a 2 = W ; : ( a ) { W 2 ( a ) u l ( u ) + 2C(u, w)}.

By the method of [ l l , Lemma 3.61 it follows that h [ . , . ] is a closed sectorial form in Lt,[u, b ) . Also the restriction S of T I to B satisfies

( S u , v ) = h [ u , v], u E B, v E H ,

on account of (3.6) . S is therefore sectorial and since M - is negative, S is also accretive. To complete the proof we prove that - 1 lies in the resolvent set of S .

Let f E Lf , [a , b ) have compact support, but otherwise be arbitrary. We try to solve (t + 1 ) y = f so that y E 9 ( S ) and y = y( f ) is bounded on L i r a , b ) . Choose w,, n + 1 6 i 6 2n, to be linearly independent solutions of (7 + 1 ) y = 0 satisfying

Each of the functions w,, n + 1 d i d 2n, as well as all linear combinations of them are not in 9. To see this, let v be a nontrivial linear combination of these functions that lies in .9. Then v L k l ( a ) = 0 , Od k < n - 1, and by (3.1)

thus, v = 0 , a contradiction. The general solution of (7 + 1 ) .v = f is given by (c.f. [20, Eq. (241,

P. 591)

where v, is given by the Wronskian quotient,

Asf'has compact support, and no linear combination of the functions \ I * / ; ?

n + 1 d k <2n, can lie in 9, it follows that for these values of k, a k = - [I: il,! cof: Thus (3.16) becomes

!


First, consider the operator M defined on functi0ns.f with compact support in 9 by

7,r

( M f ( Y ) = ( Y ) ' o f - ( Y ) jh f (3.19) k = I k = N + l ,

Observe that (r + 1 ) ( M f ) =.J; and that

I,, h

( M f ) '"(a) = - x M ! [ ~ ~ ( U ) j wu1,f; (3.20) 1 = 1 r + I O

for 0 6 k < 2n - 1 . Thus for n + 1 < r < 2n, and setting h, = 1; ov, .L

= 0, from (3.15).

Let Q be the restriction of r to the domain

It follows from standard theory [20, Theorem 4', p. 791 that Q is a self- adjoint extension of To. As the latter is positive, the essential spectrum of Q lies in [0, a). Furthermore, if y is a solution of ( r + 1 ) y = 0 satisfying [ y , w ,] (a) = 0, n + 1 < i 6 2n, then y is a linear combination of w,, n + 1 6 i62n. To see this, let y = xP=, crkwk. Then

i.e., Aa = o, where a = ( a , ,..., and A is the n x n matrix with the entry [w , , w,](a) in row j and column k. From (3.15) one can show that A = W4(a)(W1(a))' where W,(a) is defined in (3.5). We know already that det W,(a ) # 0. As { w , ,..., w,,} are linearly independent solutions of ( T + 1) y = 0, their Wronskian, W, is nonsingular. As the columns of W involving w,, n + 1 < j < 2n, are linearly independent, it follows from (3.15) again that det W,(a) # 0. Consequently A is nonsingular, and a = 0, as required. Now, as w,, n + 1 6 k < 2n, and all their linear combinations, lie


outside 9, it follows that - 1 cannot be an eigenvalue for Q. Thus - 1 must lie in the resolvent set for Q, and hence ( Q + 1 ) ' exists as a bounded linear operator defined on all of Lt,[a, 6 ) .

Returning to (3.19), it is clear from (3.21) that for f with compact support, M f and ( Q + 1 )-'f are solutions of ( z + 1 ) y = f in 9 satisfying the same boundary conditions, and thus M f = ( Q + 1 ) I f :

We now choose a , , ..., a , so that the function y defined in (3.18) lies in 9 ( S ) . From (3.4), y E 9 ( S ) if and only if for 1 6 id n,

On substituting (3.18) into this identity we obtain

Now from (3.1 ) for 1 6 k, i 6 n,

[ w k , w , ] ( a ) = ( z w , , w k ) - ( w , , Z C L ' ~ ) = 0.

Similarly, as z ( M f ) = f - M f ,

Consequently, setting a = ( a , ,..., a,)T,

( G a ) , = - 2(MJ w , ) + ( w i , j), i = 1 ,..., n, (3.23)

where G is the Gram matrix for { w I , ..., w,,}. AS G is nonsingular, (3.23) determines a uniquely. Also, when f E Ltu[a, 6 ) does not have compact support, one can replace M by ( Q + 1 ) ' in (3.23) and (3.18) so that

Thus, from (3.23), j7= y( f ) is a bounded function off on L,;>[a, 6 ) . This completes the proof.

LEMMA 3.7. Assuming (3.13), the strong conzpletion of M - in X is given by

M = { { u , u , ( a ) , a , ) : U E H , , , a z ~ C 1 ' , a n d

a r = W ; ' ( a ) [ W . ( a ) u , ( a ) + 2C(u, w ) ] )

Proof: Using Lemma 3.6, the proof is very similar to the proof of Lemma 3.71. 1


LEMMA 3.8. Assurning (3.1 3 ) , the howzdarj? space H as a subspace qf'x is given

F ? = M + O M = j ju , u l ( a ) , a ) : u ~ H , , a n d a ~ C 1 ' j .

pro05 It is clear from Lemmas 3.5 and 3.7 that M = M. Thus we only need show that each i;= ( v , ' v , ( a ) , a ) , where U E H , , and ~ E C " are arbitrary, may be written in the form

for some constants a , , I d i d n, and some ti = ju, u l ( a ) , a , ) E M. We choose u E H,, so that

where the numbers a, are to be determined so that li E M , i.e., so that for l , < i < n ,

2(u5 !+',I = - D(M' , , u ) ( a ) + ( a , , w , , , ( a ) ) (3.26)

by (3.4). Note that from (3.24)

where

a , = W l l ( a ) { W , ( a ) u l ( a ) + 2C(u, w ) ) . (3.28)

On solving (3.27) for a , and substituting this and (3.25) into (3.26) we obtain, for 1 6 i 6 n,


Thus, if a = ( a , ,..., a,)=, and G is the Gram matrix for {w, ,..., w , ) , we have

(Ga l , = 2(v , 1vi) + D ( w , , v ) ( a ) - ( a , w, . , (a ) ) . (3.29)

As G is invertible, (3.29) defines the vector a uniquely. With this value for a, and u, a , given by (3.25) and (3.28), l i e M, as required. I

We now prove

THEOREM 3.9. Under the assumption (3.13), an operator T is a maximal accretive extension of To if and only if its adjoint, T*, is a restriction of T , to a domain of the form

for some functions 4") E H,, and vectors p'" E Cn, 1 < i < n, satisfying

Proof: As H is a Z7,, space, all maximal positive subspaces of H have dimension n, and are generated by a set { q ? ( ' ) = {q$(" ,@\l ) (a) ,p '" ) : 1 d i d n ) , where the 4"' and p'" satisfy (3.31). Each maximal negative subspace, fi, in H is the Q,-orthogonal complement in H of such a subspace. By [ l l , Theorem 2.23,

N = { { u , - T , u ) : C = { u , u , ( a ) , u 2 ( a ) } E H

satisfies Q , (zi, q?"') = 0, 1 < i < n )

is the graph of the adjoint of a maximal accretive extension of T o , and all such extensions are found in this way.

Remark. Examples of these extensions may be constructed by a similar method to that used in the remark following [ l l , Theorem 3.91.


Once again (cf. [ l 1, Theorem 3 .10] ) , the realization of H given in Lemma 3.8 is not unique. The next theorem shows that any maximal

subspace of X gives rise to a maximal accretive extension of T,.

THEOREM 3.10. Under the assuniptio~i (3.13), an operator T is a iiinsiniol extension of T o ( f and only i f iis arijoinr, T* , is o restriction of' T I to

a dor71ain of rlze .form

for somefuncrions 4"' E H,, and oecrors p"', q (" E @ I 1 , 1 ,< i < n, sarisll.ing rlze conditions

Prooj: From Theorem 3.9, it is only necessary to show that each choice of {$'"= { d " ' , p"', I ' ~ ' } : 1 < i d n } satisfying (3.33) defines a maximal accretive restriction of T , .

For 4 = ( 4 , p, q ) in X we define a mapping h from X to H by

h ( 4 ) = $ = { d + 0, P , I + 2Q,(a)}, (3.34)

where Q satisfies

As before, in the proof of [ l l , Theorem 3.101, the conditions (3.35) define Q uniquely, as there are precisely n linearly independent solutions of t (Q) = 0 in 9 ( T , ) in this case. Observe that, by construction, for l j =

{u , u l ( a ) , u 2 ( a ) } in P',

Also, the mapping h is linear (from the uniqueness of d ) , and

for any 4 in X. The latter identity follows from the identities


valid for any 4 in H , and 8 in 9 ( T l ) with T , 0 = 0 ; these may be derived from (3.1) .

Consider now the maximal positive subspace B in X generated by {d"): 1 < i< n ) . It follows from (3.37) and the linearity of h that h ( B ) c H is a positive subspace. Furthermore, by (3.37) again, h is one-to-one when restricted to a positive subspace of X. Consequently h ( B ) has dimension n and is therefore a maximal positive subspace of H. The remainder of the proof now follows that of 1 1 1 , Theorem 3.101. 1

We are now in a position to describe the maximal accretive extensions of T o explicitly, via Theorem A.8 of the Appendix. For a given y E Cn and 4 =

(4 ( " , ..., 4( ' i ) )T , ~ ( ' I E H,, the partial adjoints for r in this case are defined by

THEOREM 3.1 1 . An operator T is a maximal accretive extension of T o i f and only if it has the form

9 ( T ) = { z E Lf,[a, b ) : r: ,- is absolutely continuous,

for some y E cr7, ~ ( ' 1 , q ( i ) E cn, 4( ' ) E H,,, 1 < i < n, satisfying yT4 = 2kT4, and (3.33).

The case n = 1 here is particularly simple. We consider the operator

assuming that l / p o , p , are locally integrable on [a , b ) , p r l l '= a, PO 3 p 0 > 0 on [ a , a + q ] , where q > 0 , and p , > p > 0 on [a , b ) . These assumptions ensure that conditions (3.13), ( A l ) , ( A 2 ) , and ( A 3 ) are satisfied (see [ l o , Corollary 1, p. 881). Then we have

COROLLARY 3.12. Under the above assumptions on r , T is a ma..ci17?al


ocn.etiue e.rtension of To if and only if T is the Dirichlet extension of T o , or it has the .form

r E L 2 [ a , b): r and po

are locally ah solute!,^ continuous orz [a , b)

T: ,- E L 2 [ a , b), and

for some a , b E @, ci # 0, and 4 in H , satisfying

In this case T* is the restrictiorz of T , with domain

Proof: This follows directly from Theorem 3.1 1. As 6 = (4 , a, 8 ) must be a one-dimensional subspace of H, x @ x @, it follows from (3.39) that ci

or b must be nonzero. Consequently one can eliminate ( from the parametric boundary conditions T,+ z(a) = (a , T : z(a) = - ( b t o obtain the (Dirichlet) boundary condition z(a) = 0 when ci (and hence 4) is zero, or, if cr # 0, the boundary condition ciz: z(a) + BT,+ z(a) = 0. I

Remarks. (1) Observe that in Theorem 3.11 we must have the $("# 6 in X in order to maintain the correct dimensions.

(2) In Corollary 3.12, on setting 4 = 0, ci = 0, b # 0 in (3.39) we get the Dirichlet extension of T o , a positive (hence maximal accretive) self- adjoint operator. Similarly, the Neumann extension is given by 4 = 0, b = 0, a # 0.

APPENDIX

We consider here in detail the method, due to Brown and Krall [4], by which one obtains the maximal accretive extensions of T o , given the


specific form of their adjoints. The treatment follows that of [4,5], except for the modifications needed to handle quasi-differential expressions.

First, we consider the theory for the regular case in which the expression t is defined on a finite closed interval [a , b ] of the real axis. Let S be the restriction of T , defined by the conditions

where p'", v'", q'j), p'" denote fixed vectors in C", 1 6 i6 2n, and the d'", 1 6 i 6 2n, denote given functions in H,. Assume also that 9 ( S ) is dense in L i [ a , b ] . The latter assumption simplifies the presentation somewhat, and is sufficient for our present needs, as the operators S will be closed and maximal accretive, and therefore densely defined by [25, Theorem 2.31. Our objective here is to determine the specific form of the operator S*.

Let go = .9(To) n .9(S), where . 9 (To) is given by [20, p. 621,

Using (2.9), for y in go , the conditions ( A . l ) become

for 1 6 i6 2n and r = ( l / o ) y12n1. Let { u k : 1 6 k 6 2 n } be a fixed fundamen- tal set of solutions for the equation t u = 0. Then we have

LEMMA A. 1 . Assume that the uk are real, and let r E L f, [a , b ] . Then

lj" and only if r = (1101) y12"] for son~e J) in 9 , .

Proof. If r = (1101) yC2"1 for some y in go, then it is clear from (A.2) that O,r = 0, 1 d i 6 2n. Also, by (2 .7) and since t u , = 0, 1 < i < 2n,

Conversely, if r satisfies (A .3 ) , then

By standard theory (see, e.g., [20, p.59]), ( l / w ) !I["" = r. Moreover, as ( u , , r ) = 0 , 1 d i d 2n, r E I . (T,) ' = & ( T o ) ; consequently y E 9 ( T o ) . ina all^,


ii ,r = 0, 1 6 i 6 2n, and E 9 ( T o ) , it follows that 1, E 9 ( S ) and the proof is complete. 1

Remark. Lemma A.l may also be stated as

here 9 denotes the subspace spanned by ( u , , #"I: 1 6 i 6 2nj and S o denotes the restriction of S to 9 , .

LEMMA A.2. [ I s , p. 71. Suppose E., 45, ,..., $,, are linear functionals defined on a linear space. If ker 2 3 n;=, ker $,, then ,i is a linear combination of the ,functionals $,.

Let y E C2" and # " ) E H,,, 1 6 i 6 2n, be given. The formal partial adjoint expressions for r are defined by

LEMMA A.3. Let z E 9 ( S * ) . Then there exists a vector y E C2" such that r: z, 0 6 k 6 2n - 1, are absolutely continuous on [a , b ] and S*z =

( 1 1 ~ ) 7 2 , ~ .

Proof: Observe first that by (2 .9) and assumption (A.2) , the positive self-adjoint restriction T , of T , with Dirichlet boundary conditions is invertible. Thus 9 ( T , ) = B ( T , ) = L,2,[a, b ] . Let y ~ $ 3 ~ and choose p in 9 so that rp = S*z. Then for z in 9 ( S * )

i.e., ( z - p, r y ) = 0. Set ,i( f ) = (f, z - p) . Then by ( A S ) , the kernel of contains the set { r y :

y E g o } , which by Lemma A.l is exactly the intersection of the kernels of the functionals

$ [ ( r ) = (r , #("), i = 1 ,..., 2n,

$2,1+ i ( " ) = ( r , uO, i = 1 , ..., 2n.


Consequently, by Lemma A.2, there exist constants c , , ..., c2, and a vector y = (yI , y2 ,..., y2n)T E C2" such that

Consequently, z - yT+ E 9, and

= r ( p ) (as zu, = 0)

= S*z. I (A.6)

LEMMA A.4. (Green's formula). Let y E 9, let z: z , 0 < k < 2n - 1, be absolutely continuous on [a , b ] and ( 1 1 ~ ) zz , z he in Lt,[a, b ] . Then

[ ( 2 7 . ~ - I'G) a= [D((z-yT+), y ) D(.Y, i ) ] b + (y, y T + ) , (A.7)

Proof.


LEMMA A.5. If' r E D ( S * ) then z sati.yfies the parar7~etric endpolrlt houtl- (b,:,' conditions

Z,?

T ( a ) - , n < k , < 2 n - 1, i = I

2,,

T ; r ( b ) = C i p t i I > O < k < i ? - 1, i = 1

where 5 = (51 ,..., is any vector equivalent to y in the sense that kT$ =

iyT$. If {d"): 1 < i < 2n} is linearly independent then 5 = ty.

Proof: By Lemma (A.3) there exists a vector y such that T: r, 0 < k < 2n - 1, are absolutely continuous on [a, b], and ( l / w ) z,',z E

Lt,[a, b]. Hence by Green's formula (Lemma A.4) for fixed z , the functional 1.: 9 -t C defined by

has the property that its kernel contains 9 ( S ) . By Lemma A.2, there exists a vector 5 E C2" such that A(y) = Cf" ,, U,(y ) . Hence

Consequently,


As the right-hand side of the last equation depends only on the values of y at a and b, it follows that each side is identically zero for all y E 9. Clearly yT$= 25'4. Furthermore from [20, Lemma 2, p. 631, one can choose y ~ 9 , such that y,(a), y,(b), y,(a), y,(b) take arbitrary prescribed values. Thus (A.8) follows. I

We now have

THEOREM A.6. The adjoint, S*, of the restriction S of T, defined by the boundary conditions (A.l) is defined by

B = { z ~ L i [ a , b ] : there exists y ~ C ~ " s u c h that ~ $ 2 , O < k < 2 n - 1 , defined by (A.4) are absolutely continuous on [a, b], ( l l o ) r & z E L:] [a, b] and z satisfies the boundary conditions ('4.811,

S*z = ( l l o ) r&z, z E B = 9 ( S * ) .

Proof: By Lemmas A.3 and A.5, it is clear that 9 ( S * ) c B, and S*z= r&z for all z in 9 ( S * ) . Conversely, if z E B, it follows from Green's identity (Lemma A.4) that for all y E 9 (S)

Consequently, z E 9 ( S * ) and the proof is complete. I Finally, we consider the singular case when I= [a, b ) is half-open, and r

is regular at a and singular a t 6. Following the method outlined in [5] we consider a restriction S of the maximal operator T , defined by

9 ( S ) = { ~ € g : U , y = O , i = l , ..., n) , (A.10)

where

and 4(') E H,,, p('), q") E C", 1 < i < n. Assume that B ( S ) is dense in L?"[a, 6). As usual we denote the minimal operator for r on [a, b ) by TO (see (3.2)). Observe that as the graph of S is an orthogonal complement, S is a closed operator. By analogy with (A.4), and for y E C" (depending on z), we define the partial adjoint expressions for r by

Let A = [c, d l be an arbitrary finite interval in [a, h), and define S, as


the restriction of Ti , , (the maximal operator for r on A ) with boundary conditions

where (., . )D.A and (. , .), denote the inner products on H,,(A) and L:o(A) Further, let So,, be the restriction of S , defined by the boun-

dary conditions

j ,[ '](c) = l . [ ' l (d ) = 0, 0 6 i 6 2n - 1,

(i.e., g(S0.d) = 9( TO,,) n 9 ( S A ) ) ; also define S,; on L:o(A) by

9(S, f ) = z E Lf,(A): r,+ 2 , 0 6 i6 2n - 1, is absolutely i 1

continuous on A, and - r2,z E L?>(A) O

By Lemma A.4, suitably modified, for y E 9 ( T i , , ) and z t 9 ( S ; ),

( T I , A Y > Z ) A - ( Y , S , ~ Z ) A = [ Y , z ] i ( d ) - [ Y , z ] ~ ( c ) + ( y , Y ~ $ ) D , A , (A.13)

where

[ Y , Z ] I ( ~ ) = D ( ( Z - Y ~ $ ) , y ) ( l ) - D ( y , ? ) ( [ I . (A.14)

Let S + be defined in L i [ a , h ) by

9 ( S + ) = ( z t L i la , b) : r,+ z, 0 6 i6 2n - 1, is locally absolutely

1 continuous on [a, h) , and - r2,z t Lk[a, b )

0

Also, let So be the restriction of S to the domain 9 ( S , ) = 9 ( T 0 ) n 9 ( S ) , where 9 ( T o ) is defined in (3.2). I t follows from (A.13) that for y t 9 and z ~ 9 ( S + ) , [ y , z l l ( h ) = lim,+, [ y , z ] , ( t ) exists and


LEMMA A.7. Assume that ( A 2 ) holds, and that T is of limit-point type at b. Then

where 9 denotes the subspace spanned by the set {u , } u {$"': 1 < i < n } , where { u , ) is a basis for &"(TI ) , the null-space of the maximal operator T , .

Proof: Observe first that as To has an invertible extension (cf. the proof of Lemma A.3), ,@(To) is closed; as T,* = T I , it follows that

% ( T o ) = N ( T , ) L . (A . 16)

Let Z E %(So) . Then z = ~y for some y in B ( S o ) . Consequently, for u, in - { " ( T I )

Also, as To =z, where is the restriction of T to functions in 9 with compact support contained in (a, b ) , one can choose { y,} c 9 ( T b ) so that y, -+ y and ~ y , -+ ~y in L i [ a , b). In addition, it follows from (A.2) and (2.9) that y, + y in the norm \ ] . / I ,. Thus for 1 < i < n we have

= lim ( ~ y , , , 4 ( j 1 ) n - 3:

This means that z E 9'. Conversely, if z E 9', then certainly z E y / t - ( ~ l ) L )

and hence z E ,@(To) from (A.16). Thus, z = ~ y , where E 9 ( T o ) . Finally, from (A.17), it follows that ( y , 4 ( " ) , = 0, 1 < i < n, and hence that ?. E 2?(SO), as required. 1

THEOREM A.8. I f To has deficiency indices (n, n ) (i.e., T is of linqit-point t?pe at b ) and (A.2) holds, then S* is [he resfricfion of St 10 funcfions saf isfying


,$,here 5 = (51 ,..., 5,,lT is any vector equivalent to y in the sense that ST+ = fyT4. If {d( j ' : 1 d i d n } is linearly independent, then 5 = fy.

proof: Set B = { Z E 9 ( S f ): z satisfies (A .18) ) . Observe that, by Lern- maA.7, and the method of Lemma A.3, we have S* c S + . Observe also that for ,V E 9 ( S ) , z E g ( S + ) ,

by [20, p. 781 as y and z - yT4 lie in 9 . For A = [a, d l , let S,,, denote the of S A satisfying y C ' ] ( d ) = 0, 0 < i < 2n - 1, and ( y , y T ~ ) , = 0.

Then S,,, is a 2n-dimensional extension of So,,. Let v , , v,, ..., v,,, be a basis for 9(S2, , ) /9(S0, , ) . If these functions are defined to be zero in [c, b ) , then as dim 9 ( S ) / 9 ( S o ) = 2n, we have

9 ( S ) = 9 ( S o ) i [ { v , : 1 d i d 2 n } ] . (A.20)

Now, for Y E 9 ( S O ) we have from Lemma A.7 and (2.9) that

and thus, from (A.19) and (A.20) that for Y E 9 ( S ) and Z E 9 ( S + ) ,

Consequently, from (A.15), and a similar argument to that used in the proof of Lemma A.5 we have 9 ( S * ) c B. Conversely, let z E B. For y € 9 ( S 0 ) it is clear from (A.15) and (A.21) that

( S y , z ) = ( y , S + z ) . (A.22)

As (A.22) also holds for y = vi for any i, 1 6 i 6 2n, it follows from (A.20) and (A.22) that Z E 9 ( S * ) , as required. 1

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