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A CERTAIN CLASS OF PERTURBED SYMMETRIC OPERATORS

A. HEBBECHE

Received 3 February 2004 and in revised form 20 May 2004

The generalized resolvents for a certain class of perturbed symmetric operators with equal and finite deficiency indices are investigated. Using the Weinstein-Aronszajn formula, we give a classification of the spectrum.

1. Introduction

The present paper is concerned with the study of spectral properties for a certain class of linear symmetric operator T, defined in the Hilbert spaceH of the formT =A+ B, whereAis a closed linear symmetric operator, with nondensely defined domain in general,D(A)H, andBis a finite-rank operator of the form

B f = n k=1

akf,ykyk, (1.1)

where y1,y2,. . .,yn is a linearly independant system inH,a1,a2,. . .,anR. We remark that the operatorT can be considered as a perturbation of the operatorAby the finite- rank operatorB.

The case whenA is a first-order or second-order differential operator in the spaces L2(0, 2π),L2(0,) or in the Hilbert space of vector-valued functions, andB is a one- dimensional perturbation (n=1), has been studied by many authors (see, e.g., [9,20, 24]).

In particular, certain integrodifferential equations of the above type occur in quantum mechanical scattering theory [8].

In this paper, the generalized resolvents of perturbed symmetric operatorTwith equal and finite deficiency indices are investigated. Using the Weinstein-Aronszajn formula (see, e.g., [18]), we give a classification of the spectrum. Finally, the obtained results are applied to the study of two classes of first-order and second-order differential operators.

We note that the spectral theory of perturbed symmetric and selfadjoint operators have been investigated using various methods by many authors [3,4,5,6,11,12,13,14, 15,16,17,21,22].

Copyright©2005 Hindawi Publishing Corporation Journal of Applied Mathematics 2005:1 (2005) 81–92 DOI:10.1155/JAM.2005.81

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2. Preliminaries

LetA be a closed symmetric operator with nondensely defined domain in a separable Hilbert spaceHwith equal deficiency indices (m,m), andm <. We denote byρ(A) the resolvent set of the operatorA, the resolvent operatorRλ(A) ofAis defined asRλ(A)= (AλI)1. The complement ofρ(A) in the complex plane is called the spectrum ofA and denoted byσ(A). There is a decomposition of the spectrumσ(A) into three disjoint subsets, at least one of which is not empty [1,2,10]:

σ(A)=Pσ(A)(A)PCσ(A), (2.1) (A) is called the point spectrum,(A) the continuous spectrum, andPCσ(A) the point-continuous spectrum. We denote the essential spectrum of the operatorAbyσe(A)

=(A)PCσ(A).

For arbitraryλC, we denotePλ=Nλ(D(A)Nλ), whereNλ=HΘ(AλI)D(A) is the deficiency subspace of the operatorA[1,2].

It is known [23] thatPλ= {0} if and only ifD(A)=H, and if D(A)=H, then the subset

Gλ=

[ϕ,ψ]Nλ×Nλ:ϕψD(A) (2.2) is a graph of the isometric operatorXλwith domainPλand values inPλ.

We denote bythe set of linear operatorsFdefined fromNitoNi, such that F 1. For each analytic operator-valued functionF(λ) inC+, withC+= {λ: Imλ >0}, and values in, we introduce the setΩF() consisting of elementshNisuch that

λ→∞lim∈C+ε

|λ|

hF(λ)h<, (2.3) whereCε+= {λC+:ε <argλ < πε}, 0< ε < π/2.

It is known [27] thatΩF() is a vector space and for eachhF(),

λ→∞lim∈C+ε

F(λ)h=F0()h (2.4)

exists in the sense of the strong topology, andF0() is an isometric operator.

According to the theory of ˇStraus [28], the generalized resolvents ofAare given by the formula

Rλ(A)=Rλ=

AF(λ)λI1, Rλ=Rλ, λC+, (2.5) whereAF(λ)is an extension ofAwhich is determined by the functionF(λ), whose values are operators from the deficiency subspaceNi to the deficiency subspaceNisuch that

F(λ)1 andF(λ) satisfy the condition

F0(=Xiψ, forψ=0 only, (2.6)

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thenAF(λ) is a restriction onH of a selfadjoint operator defined in a certain extended Hilbert space and is called quasiselfadjoint extension of the operatorA[28] defined on D(AF(λ))=D(A) + (F(λ)I)Niby

AF(λ)f+F(λ)ϕϕ=A f+iF(λ)ϕ+iϕ, f D(A),ϕNi. (2.7) For selfadjoint extensions with exit in the space in which acts the considered operators, see, for example, [12,21] and the references therein.

We denote by the set of analytic operator functionsF(λ) in C+ with values in satisfying the condition (2.6).

Remark 2.1. To each selfadjoint extension of the operatorAcorresponds a certain con- stant operator functionF(λ)=V, whereVis an isometric operator defined fromNiover Nisatisfying the conditionV ψ=Xiψforψ=0 only, and reciprocally.

We denote byA˚a selfadjoint extension ofAand we introduce the operator

U˚λλ0=A˚λ0IA˚λI1, Imλ >0. (2.8) We note that (see [19])

U˚λλ0Nλ0=Nλ, (Imλ)Imλ0

=0. (2.9)

We denote by

ϕ(1)i(2)i ,. . .,ϕ(m)i (2.10) a basis ofNi. From (2.9),ϕ(k)λ =U˚λiϕ(k)i ,k=1, 2,. . .,mform a basis forNλ. In particular, the vectors

ϕ(k)i =˚ (k)i , k=1, 2,. . .,m, (2.11) whereU˚=U˚iiis the Cayley transform [1,2] ofA, form an orthogonal basis of˚ Ni.

To get a convenient formula of the generalized resolvents ofA, we will need the fol- lowing notation:

Φλµ=µ)ϕ(k)λ ,ϕ(s)µ mk,s=1, C(λ)=Φλi1Φλ(i), (2.12) whereE is the identity matrix of orderm, Ω(λ) is an analytic matrix function in C+ corresponding, in the bases (2.10) and (2.11), to the operator functionF(λ)∈ ℵandϕλ=(1)λ ,. . .,ϕ(m)λ )t, (fλ)t=((f(1)λ ),. . ., (f(m)λ )),tdenotes the transpose, and (ϕλ,g) is defined analogously.

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In what follows, we denote byΦthe set of matricesΩ(λ),λC+, associated in the bases (2.10) and (2.11) to the operator functionsF(λ)∈ ℵ.

According to the notation used in [7], the generalized resolvents ofAare given by Rλ(A)f =Rλf =R˚λf +fλtEΩ(λ)C(λ)Ω(λ)E1Φλi1ϕλ,

Rλ=Rλ, λC+, (2.13)

whereR˚λis the resolvent ofA˚andΩ(λ)Φ.

Remark 2.2. The formula (2.13) defines a resolvent of a selfadjoint extension ofAif and only ifΩ(λ) is a unitary constant matrix.

3. Resolvent and spectrum of a symmetric perturbed operator

LetT=A+Bbe defined onD(T)=D(A), whereAis a linear closed symmetric operator inHandBis a finite-rank operator.

Lemma3.1. Forλρ(A)ρ(T), the resolventRλ(T)of the operatorTis given by

Rλ(T)=Rλ(A)Rλ(A)I+BRλ(A)1BRλ(A). (3.1) Proof. Forλρ(A)ρ(T), the operator

Rλ(A)I+BRλ(A)1=Rλ(T) (3.2) exists and is bounded. Then, we get

(TλI)Rλ(A)Rλ(A)I+BRλ(A)1BRλ(A)

=(AλI+B)Rλ(A)Rλ(A)I+BRλ(A)1BRλ(A)

=I+BRλ(A)

I+BRλ(A)I+BRλ(A)1BRλ(A)=I

(3.3)

as required.

Remark 3.2. If BRλ(A) <1, then from (3.1), we obtain Rλ(T)=Rλ(A)I+BRλ(A)1=Rλ(A)

k=0

(1)kBRλ(A)k. (3.4) Now, the aim is to give a convenient expression of (I+BRλ(A))1in a more specific case.

So, we study in detail the case whenBis a finite-rank operator. Then, B f =

n k=1

akf,ykyk, f H, (3.5) wherea1,a2,. . .,anR;{y1,y2,. . .,yn}is a linearly independent system inH. If we put

I+BRλ(A)1BRλ(A)f =y, (3.6)

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we have

y=BRλ(A)fBRλ(A)y, (3.7)

then,yImB, so that

y= n k=1

ckyk. (3.8)

From (3.7) and (3.8), we get n k=1

ckyk=BRλ(A)f n k=1

ckBRλ(A)yk, (3.9)

with

ck+ak n j=1

cjRλ(A)yj,yk=akRλ(A)f,yk. (3.10)

The determinant∆(λ) of the system (3.10) is given by

∆(λ)=detδk j+akRλ(A)yj,yknk,j=1, (3.11) whereδk j is the Kronecker symbol. If we suppose that∆(λ)=0, the solution of (3.10) is given by

ck=ck(λ;f)=

f,∆k(λ)

∆(λ) , k=1, 2,. . .,n, (3.12) where ∆k(λ) is the determinant obtained from ∆(λ) by replacing the kth column by [ajRλ(A)yj]nj=1. So, from (3.1), we have

Rλ(T)f =Rλ(A)f n k=1

f,∆k(λ)

∆(λ) Rλ(A)yk. (3.13)

This completes the proof of the following theorem.

Theorem3.3. Let λρ(A)such that∆(λ)=0. Then,λρ(T)and the resolvent of the operatorTis given by (3.13).

Remark 3.4. From (3.13), we note that the resolventRλ(T) is a perturbation ofRλ(A) by a finite-rank operator.

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Remark 3.5. For the particular casen=1 anda1=1, the formula (3.13) was established in [9].

Remark 3.6. Ifλρ(A) such that∆(λ)=0, thenλis an eigenvalue of the operatorT.

Proof. We can show that there exists an element ψ=

n k=1

αkyk (3.14)

such thatRλ(A)ψis an eigenvector of the operatorT, corresponding to the eigenvalueλ.

Consequently, we have ak

n j=1

αjRλ(A)yj,yk+αk=0, k=1,n. (3.15)

Since the determinant of this system∆(λ)=0, it admits a nontrivial solution, which gives

the desired result.

Theorem3.7. Letµbe a fixed complex number. Then, the following holds.

(a)Ifµρ(A)and∆(µ)=0, thenµρ(T).

(b)Ifµρ(A)and∆(µ)=0, thenµ(T)and the multiplicity ofµas an eigenvalue ofTis equal to the order of the zero of∆(λ)atµ.

(c)Ifµ(A)andµof multiplicityk >0and ifµis a pole of∆(λ)of multiplicity p (kp), then

(1)fork > p, it holds thatµ(T)of multiplicity(kp), (2)fork=p, it holds thatµρ(T).

(d)Ifµ(A)is neither a zero, nor a pole of∆(λ), thenµPσ(T).

(e)Ifµ(A)of multiplicityk andµis a root of the function∆(λ)of order p, then µPσ(T)of order(k+p).

(f)The essential spectraσe(A)andσe(T), respectively of the operatorsAandT, coincide.

Proof. It is sufficient to evaluate the function

C(λ)=detI+BRλ(A). (3.16)

To this end, letyImB. Then, BRλ(A)y=

n k=1

ak

y,Rλ(A)yk

yk, (3.17)

it is clear thatC(λ)=∆(λ), and the function∆(λ) is meromorphic inρ(A)Pσ(A). From the formula of Weinstein and Aronszajn [18], we have

ϑ(λ;T)=ϑ(λ;A) +ϑ(λ;∆), (3.18)

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where

ϑ(λ;A)=

0 ifλρ(A),

k ifλPσ(A) and of multiplicityk, + otherwise,

ϑ(λ;∆)=

k ifλis a zero of∆(λ) of orderk,

k ifλis a pole of∆(λ) of orderk, 0 for otherλΩ,

(3.19)

which gives the desired result.

4. Generalized resolvents

Now, we suppose thatAis a symmetric operator with deficiency indices (m,m),m <. Lemma4.1. LetλCsuch thatImλ >0andϕλ(A)Nλ(A). Then, the elementϕλ(T), defined by the formula

ϕλ(T)=D(λ)ϕλ(A)=ϕλ(A) n k=1

ϕλ(A),˚gk(λ)

˚∆(λ) RλA˚yk, (4.1) is an element of the deficiency subspaceNλ(T), where

D(λ)=IRλA˚I+BRλA˚1B=IRλT˚B, g˚k(λ)=A˚λI∆˚k(λ), (4.2)

˚∆(λ)and˚∆k(λ)are defined similarly as∆(λ)andk(λ)in the formula (3.13) by putting the operatorA˚instead of the operatorA.

Proof. Since the operatorsA˚andT˚=A˚+Bare selfadjoint and λis nonreal, then λ ρ(A)˚ ρ(T). In addition, from˚ Theorem 3.3 we have˚∆(λ)=0. Furthermore, for each

f D(A)=D(T), we have

T˚λIf,D(λ)ϕλ(A)=

D(λ)T˚λIfλ(A)

=

IBRλT˚T˚λIfλ(A)

=A˚λIfλ(A)

=0,

(4.3)

and the equality

ϕλ(T)=ϕλ(A) n k=1

ϕλ(A),g˚k(λ)

˚∆(λ) RλA˚yk (4.4)

results from (3.13).

Remark 4.2. We note that ifϕλ(A)=0, thenϕλ(T)=0.

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Proof. If we suppose the contrary, we obtainRλ(T)Bϕ˚ λ(A)=ϕλ(A), which gives˚ λ(A)= λϕλ(A). This leads to a contradiction, since a selfadjoint operator can not have nonreal

eigenvalues.

Remark 4.3. IfD(A) is dense inH, thenϕλ(A) andϕλ(T) are, respectively, eigenfunctions of the operatorsAandT, corresponding to the eigenvaluesλ.

Let ϕ(k)i (T)=D(i)ϕ(k)λ (A), k=1, 2,. . .,m, defined by the formula (4.1). If ϕ(1)i (A), ϕ(2)i (A),. . .(m)i (A) is a basis of the deficiency subspaceNi(A) of the operatorA, then ϕ(1)i (T),ϕ(2)i (T),. . .,ϕ(m)i (T) is a basis of the deficiency subspaceNi(T) of the operatorT.

Putting

U˚λλ0(T)˚ =T˚λ0IRλT˚, ϕ(k)λ (T)=U˚λiT˚ϕ(k)i (T), k=1, 2,. . .,m, ϕλ(T)=

ϕ(1)λ (T),. . .,ϕ(m)λ (T)t, Φλµ(T)=µ)ϕ(k)λ (T),ϕ(j)µ (T)mk,j=1, (4.5) C(λ)=Φλi1(T)Φλ(i)(T) denotes the characteristic matrix of the operatorT, andω(λ) the corresponding matrix of orderm×m, in the basesϕ(1)i (T),ϕ(2)i (T),. . .(m)i (T) and ϕ(1)i(T),ϕ(2)i(T),. . .,ϕ(m)i (T).

Theorem4.4. The set of all generalized resolvents of the operatorTis given by

Rλ(T)f =RλT˚f+f,ϕλ(T)tEω(λ)C(λ)ω(λ)E1Φλi1(T)ϕλ(T), f H, (4.6) where

RλT˚f =RλA˚f n k=1

f,∆˚k(λ)

˚∆(λ) RλA˚yk. (4.7) Proof. The proof results fromLemma 4.1and formula (2.13).

We denote, respectively, byAωandTωthe quasiselfadjoint extensions of operatorsA andTcorresponding to the operator functionF(λ), defined by the matrixω(λ).

Remark 4.5. To selfadjoint extensions of these operators correspond the constant unitary matricesω=i j].

Theorem4.6. Suppose thaty1,y2,. . .,ynImA,µis an eigenvalue of the quasiselfadjoint extensionAωof the operatorA,µ(Aω). Ifµρ(A)˚ and˚∆(µ)=0,thenµis an eigen- value of the operatorTω=Aω+Band the corresponding eigenfunctionϕµ(Tω)is given by

ϕµTω=D(µ)ϕµAω=ϕµAω n k=1

ϕµAωgk(µ)

˚∆(µ) RµA˚yk, (4.8) whereϕµ(Aω)is the eigenfunction of the operatorAω, corresponding to the eigenvalueµ.

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Proof. Sincey1,y2,. . .,ynImA, thenµ(A)ImA. We also have ϕµ

Tω

=D(µ)ϕµ

Aω

=ϕµ

Aω

Rµ

T˚µ(A)=ϕµ

Aω

ψµ, (4.9) where

ψµ=RµT˚µ(A)D(A). (4.10) Then,

Tωϕµ Tω

=Tω ϕµ

Aω

ψµ

=

Aω+BϕµAωTωRµT˚µ(A)

=µϕµAω+µAωµAω+µRµT˚µ(A)

=µϕµTω.

(4.11)

5. Applications

5.1. Perturbed first-order differential operator. Consider inL2(0, 2π) the operatorT= A+B, whereAis defined byAy=iywith domainD(A)=H01(0, 2π) andBis given by

(B y)(x)= n k=1

ak

y,yk

yk(x), (5.1)

wherey1,y2,. . .,ynL2(0, 2π) andakR, for allk=1,n. From [1,2], the operatorAis regular symmetric of deficiency indices (1, 1) and each selfadjoint extension ofAhas a discrete spectrum.

Theorem5.1. The generalized resolventRλ(Tθ)ofT, corresponding to the functionω(λ)= θ(λ), is an integral operator with kernel

K(x,t)=

1[x,2π](x) + 1 θ(λ)e2πλi+ 1

eiλ(tx)+ n k=1

θk(λ,x)φk(λ,t), (5.2) where1[x,2π](x)is the characteristic function of the interval[x, 2π],

φk(λ,t)=

θk(λ)(t), θk(λ,x)=

RλAθyk(x)

θ(λ) , (5.3)

whereRλ(Aθ), associated to the functionθ(λ), is given by Rλ

Aθ y(x)=

x

0 y(t)eiλ(tx)dt 1 θ(λ)e2πti+ 1

0 y(t)eλi(tx)dt (5.4) with

θ(λ)= δkj+ak

Rλ

Aθ

yj,yk

, (5.5)

andθkis the determinant obtained fromθ(λ)replacing thekth column by[akRλ(Aθ)yk]n1.

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Proof. The proof results from [26] andTheorem 3.3.

Corollary 5.2. Let Tθ be a selfadjoint extension of T corresponding to the functionθ,

|θ| =1.

(1)The spectrum of Tθ is simple if and only if the roots ofθ(λ) are simple and for k=0,±1,±2,. . .,∆θ(1/2 +kϕ0/2π)=0, where{1/2 +kϕ0/2π}is the spectrum ofAθ, andϕ0=argθ.

(2)σ(Tθ)=(Tθ)=E1E2, whereE1is the set of points ofσ(Aθ)= {1/2 +kϕ0/2π, k=0,±1,±2,. . .}in whichθ(λ)is analytic,E2is the set of roots ofθ(λ).

Proof. The proof results from (5.4),Theorem 3.7, andLemma 4.1.

5.2. Perturbed second-order differential operator. Consider in L2(0,) the operator T=A+B, whereAis defined by

Ay= −y+x2y (5.6)

with domainD(A) consisting of all variablesywhich satisfy (i) yL2(0,),

(ii) yis absolutely continuous on all compact subintervals of [0,[, (iii)AyL2(0,),

(IV)y(0)=y()=limx→∞y(x)=0,y(0)=y()=0, andBis given by

(B y)(x)= n k=1

ak

y,yk

yk(x), (5.7)

wherey1,y2,. . .,ynL2(0, 2π) andakIR, for allk=1,n.

From [1,2], the operatorAis symmetric of deficiency indices (1, 1). Letu1,u2be two solutions of (5.6), satisfying the initial conditions

u1(0,λ)=1, u1(x,λ)x=0=0,

u2(0,λ)=0, u2(x,λ)x=0= −1. (5.8) There exists a functionm(λ) [29] analytic inC\Rsuch that

ψ(x,λ)=u2(x,λ) +m(λ)u1(x,λ)L2(0,). (5.9) Theorem5.3. The generalized resolventsRλ(Tθ)of the operatorTare defined by

RλTθy=RλAθy n k=1

y,∆θk(λ)

θ(λ) RλAθyk, Imλ >0, (5.10)

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where

Rλ

Aθ

y=ψ(x,λ) x

0 y(s)u1(s,λ)ds+u1(x,λ)

x y(s)ψ(s,λ)ds

ψ(x,λ) θ(λ) +m(λ)

0 y(s)ψ(s,λ)ds,

(5.11)

θ(λ)=detσjk+ak

Rλ

Aθ

yj,yk

, λC+, (5.12)

withθ(λ) an arbitrary function analytic inC+ and such that Imθ(λ)0 or θ(λ)is an infinite constant.

Proof. First, we show that forλC+,∆θ(λ)=0 (then,∆θ=0). We know (see [1,2]) that for each quasiselfadjoint extension of a symmetric operator,C+is contained in the set of regular points of this operator. Then, ifλC+, we haveλρ(Aθ) andλρ(Tθ). If we suppose thatλC+and∆θ(λ)=0, fromTheorem 3.7, we obtainλ(Tθ), which is a contradiction. The formula (5.11) results from [25]. UsingTheorem 3.3, we end the

proof.

Corollary5.4. LetTθ be a selfadjoint extension associated toθIR, letλ12,. . .be the roots ofθ(λ)inρ(Aθ)and letz1,z2,. . .be the poles ofθ(λ). Then,

Tθ=

Aθ\

zi1

λj1. (5.13)

Proof. The proof results from (b) and (c) ofTheorem 3.7.

Acknowledgment

The author is grateful to the editor and the anonymous referees for their valuable com- ments and helpful suggestions which have much improved the presentation of the paper.

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A. Hebbeche: Laboratoire Equations Differentielles, Departement de Mathematiques, Facult´e des Sciences, Universit´e Mentouri, 25000 Constantine, Algeria

E-mail address:[email protected]

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