# Spectral Analysis ofq-Sturm-Liouville Problem with the Spectral Parameter in the Boundary Condition

Authors: Aytekin Eryılmaz
Journal: Journal of Function Spaces and Applications
 (2012)

---

## Abstract

This paper is concerned withq-Sturm-Liouville  boundary value problem in the Hilbert space with a spectral parameter in the boundary condition. We construct a self-adjoint dilation of the maximal dissipative q-difference operator and its incoming and outcoming spectral representations, which make it possible to determine the scattering matrix of the dilation. We prove theorems on the completeness of the system of eigenvalues and eigenvectors of operator generated by boundary value problem.


---

## Body

## 1.  Introduction

Spectral analysis of Sturm-Liouville and Schrödinger differential equations with a spectral parameter in the boundary conditions has been analyzed intensively (see [1–16]). Then spectral analysis of discrete equations became an interesting subject in this field. So there is a substantial literature on this subject (see [10, 17–19]).There has recently been great interest in quantum calculus and many works have been devoted to some problems ofq-difference equation. In particular, we refer the reader to consult the reference [20] for some definitions and theorems on q-derivative, q-integration, q-exponential function, q-trigonometric function, q-Taylor formula, q-Beta and Gamma functions, Euler-Maclaurin formula, anf so forth. In [21], Adıvar and Bohner investigated the eigenvalues and the spectral singularities of non-selfa-djoint q-difference equations of second order with spectral singularities. In [12], Huseynov and Bairamov examined the properties of eigenvalues and eigenvectors of a quadratic pencil of q-difference equations. In [22], Agarwal examined spectral analysis of self-adjoint equations. In [23], Shi and Wu presented several classes of explicit self-adjoint Sturm-Liouville difference operators with either a non-Hermitian leading coefficient function, or a non-Hermitian potential function, or a nondefinite weight function, or a non-self-adjoint boundary condition. In [24], Annaby and Mansour studied a q-analogue of Sturm-Liouville eigenvalue problems and formulated a self-adjoint q-difference operator in a Hilbert space. They also discussed properties of the eigenvalues and the eigenfunctions.In this paper, we considerq-Sturm-Liouville Problem and define an adequate Hilbert space. Our main target of the present paper is to study q-Sturm-Liouville boundary value problem in case of dissipation at the right endpoint of (0,a) and with the spectral parameter at zero. The maximal dissipative q-Sturm-Liouville operator is constructed using [25, 26] and Lax-Phillips scattering theory in [27]. Then we constructed a functional model of dissipative operator by means of the incoming and outcoming spectral representations and defined its characteristic function in terms of the solutions of the corresponding q-Sturm-Liouville equation. By combining the results of Nagy-Foiaş and Lax-Phillips, characteristic function is expressed with scattering matrix and the dilation of dissipative operator is set up. Finally, we give theorems on completeness of the system of eigenvectors and associated vectors of the dissipative q-difference operator.Letq be a positive number with 0<q<1,A⊂ℝ,  and a∈ℂ.  A q-difference equation is an equation that contains q-derivatives of a function defined on A. Let y(x) be  a complex-valued function on x∈A. The q-difference operator Dq is defined by
(1.1)Dqy(x)=y(qx)-y(x)μ(x),∀x∈A,
where  μ(x)=(q-1)x. The q-derivative at zero is defined by
(1.2)Dqy(0)=limn→∞y(qnx)-y(0)qnx,x∈A,
if the limit exists and does not depend on  x. A right inverse to Dq, the Jackson q-integration, is given by
(1.3)∫0xf(t)dqt=x(1-q)∑n=0∞qnf(qnx),x∈A,
provided that the series converges, and
(1.4)∫abf(t)dqt=∫0bf(t)dqt-∫0af(t)dqt,a,b∈A.LetLq2(0,a) be the space of all complex-valued functions defined on  [0,a]  such that
(1.5)∥f∥:=(∫0a|f(x)|dqx)1/2<∞.
The space Lq2(0,a) is a separable Hilbert space with the inner product
(1.6)(f,g):=∫0af(x)g(x)¯dqx,f,g∈Lq2(0,a).We will consider the basic Sturm-Liouville equation(1.7)l(y):=-1qDq-1Dqy(x)+v(x)y(x),0≤x≤a<+∞,
where v(x) is defined on [0,a] and continuous at zero. The q-Wronskian of y1(x),y2(x) is defined to be
(1.8)Wq(y1,y2)(x):=y1(x)Dqy2(x)-y2(x)Dqy1(x),x∈[0,a].
Let L0 denote the closure of the minimal operator generated by (1.7) and by D0 its domain. Besides, we denote by D the set of all functions y(x) from Lq2(0,a) such that y(x) and Dqy(x) are continuous in [0,a) and l(y)∈Lq2(0,a);D is the domain of the maximal operator L. Furthermore, L=L0* [2, 4, 13]. Suppose that the operator L0 has defect index (2,2).For everyy,z∈D we have q-Lagrange’s identity [24]
(1.9)(Ly,z)-(y,Lz)=[y,z¯](a)-[y,z¯](0),
where [y,z¯]:=y(x)Dq-1z(x)¯-Dq-1y(x)z(x)¯.


## 2. Construction of the Dissipative Operator

Consider boundary value problem governed by(2.1)(ly)=λy,y∈D,
subject to the boundary conditions
(2.2)y(a)-hDq-1y(a)=0,Imh>0,(2.3)α1y(0)-α2Dq-1y(0)=λ(α1′y(0)-α2′Dq-1y(0)),
where λ is spectral parameter and α1,α2,α1′,α2′∈ℝ and α is defined by
(2.4)α:=|α1′α1α2′α2|=α1′α2-α1α2′>0.
For convenience we assume
(2.5)R0(y):=α1y(0)-α2Dq-1y(0),R0′(y):=α1′y(0)-α2′Dq-1y(0),N1a(y):=y(a),N2a(y):=Dq-1y(a),N10(y):=y(0),N20(y):=Dq-1y(0)∞,Ra(y):=N2a(y)-hN1a(y).Lemma 2.1.
For arbitraryy,z∈D,  let one suppose that R0(z¯)=R0(z)¯,R0′(z¯)=R0′(z)¯, then one has the following.Proof.
(2.6)
[
y
,
z
]
0
=
1
α
[
R
0
(
y
)
R
0
′
(
z
)
¯
-
R
0
′
(
y
)
R
0
(
z
)
¯
]
,
(2.7)
1
α
[
R
0
(
y
)
R
0
′
(
z
)
¯
-
R
0
′
(
y
)
R
0
(
z
)
¯
]
=
1
α
[
(
α
1
y
(
0
)
-
α
2
D
q
-
1
y
(
0
)
)
(
α
1
′
z
(
0
)
-
α
2
′
D
q
-
1
z
(
0
)
¯
)
-
(
α
1
′
y
(
0
)
-
α
2
′
D
q
-
1
y
(
0
)
)
(
α
1
z
(
0
)
-
α
2
D
q
-
1
z
(
0
)
¯
)
]
=
1
α
[
(
α
1
′
α
2
-
α
1
α
2
′
)
(
y
(
0
)
¯
D
q
-
1
z
(
0
)
-
D
q
-
1
y
(
0
)
z
(
0
)
¯
)
]
=
[
y
,
z
]
0
.
Let  θ1,  θ2 denote the solutions of (2.1) satisfying the conditions
(2.8)N10(θ2)=α2-α2′λ,N20(θ2)=α1-α1′λ,N1a(θ1)=h,N2a(θ1)=1.
Then from (2.3) we have
(2.9)Δ(λ)=[θ1,θ2]x=-[θ2,θ1]x=-[θ2,θ1]0=-1α[R0(θ1)R0′(θ2)¯-R0′(θ1)R0(θ2)¯]=R0(θ2)-λR0′(θ2),Δ(λ)=[θ1,θ2]x=-[θ2,θ1]x=-[θ2,θ1]a=-(y(a)Dq-1z(a)¯-z(a)Dq-1y(a)¯)=-(y(a)-hDq-1y(a)¯)=-(N2a(θ1)-hN1a(θ1)).
We let
(2.10)G(x,ξ,λ)=-1Δ(λ){θ2(ξ,λ)θ1(x,λ),x<ξθ1(x,λ)θ2(ξ,λ),ξ<x}.
It can be shown that G(x,ξ,λ) satisfies (2.1) and boundary conditions (2.2)–(2.3). G(x,ξ,λ) is a Green function of the boundary value problem (2.1)–(2.3). Thus, we obtain that  the G(x,ξ,λ) is a Hilbert-Schmidt kernel and the solution of the boundary value problem can be expressed by
(2.11)y(x,λ)=∫0aG(x,ξ,λ)y(ξ,λ)dξ=Rλy.
Thus Rλ is a Hilbert Schmidt operator on space Lq2(0,a). The spectrum of the boundary value problem coincides with the roots of the equation Δ(λ)=0. Since Δ is analytic and not identical to zero, it means that the function Δ has at most a countable number of isolated zeros with finite multiplicity and possible limit points at infinity.Suppose thatf(1)∈L2[0,a),  f(2)∈ℂ, then we denote linear space H=Lq2(0,a)⊕ℂ with two component of elements of f^=(f(1)f(2)). If α>0 and f^=(f(1)f(2)),  g^=(g(1)g(2))∈H, then the formula (2.12)(f^,g^)=∫0af(1)(x)g¯(1)dqx+1αf(2)g¯(2)
defines an inner product in Hilbert space H. Let us define operator of Ah:H→Hwith equalities suitable for boundary value problem
(2.13)D(Ah)={f^=(f(1)f(2))∈H:f(1)∈D,Ra(f(1))=0,f(2)=R0′(f(1))},Ahf^=l~(f^):=(l(f(1))R0(f(1))).
Remind that a linear operator Ah with domain D(Ah) in Hilbert space H is called dissipative if Im(Ahf,f)≥0 for all f∈D(Ah) and maximal dissipative if it does not have a proper extension.Definition 2.2.
If the system of vectors ofy0,y1,y2,…,yn corresponding to the eigenvalue λ0  is
(2.14)l(y0)=λ0y0,R0(y0)-λR0′(y0)=0,Ra(y0)=0,l(ys)-λ0ys-ys-1=0,R0(ys)-λR0′(ys)-R0′(ys-1)=0,Ra(ys)=0,s=1,2,…,n,
then the system of vectors of y0,y1,y2,…,yn corresponding to the eigenvalue λ0 is called a chain of eigenvectors and associated vectors of boundary value problem (2.2)–(2.12).Since the operatorAh is dissipative in H and from Definition 2.2, we have the following.Lemma 2.3.
The eigenvalue of boundary value problem (2.1)–(2.3) coincides with the eigenvalue of dissipative Ah operator. Additionally each  chain of eigenvectors and associated vectors y0,y1,y2,…,yn corresponding to the eigenvalue λ0 corresponds to the chain  eigenvectors and associated vectors y^0,y^1,y^2,…,y^n corresponding to the same eigenvalue λ0 of dissipative Ah operator. In this case, the equality
(2.15)y^k=(ykR0′(yk)),k=0,1,2,…,n
holds.Proof.
y
^
0
∈
D
(
A
h
) and Ahy^0=λ0y^0, then the equality l(y0)=λ0y0,  R0(y0)-λR0′(y0)=0,R1(y0)=R2(y0)=0 takes place; that is, y0 is an eigenfunction of the problem. Conversely, if conditions (2.14) are realized, then (y0R0′(y0))=y^0∈D(Ah) and Ahy^0=λ0y^0,y^0 is an eigenvector of the operator Ah. If y^0,y^1,y^2,…,y^n are a chain of the eigenvectors and associated vectors of the operator Ah corresponding to the eigenvalue λ0, then by implementing the conditions y^k∈D(Ah)(k=0,1,2,…,n) and equality Ahy^0=λ0y^0,  Ahy^s=λ0y^s+y^s-1,  s=1,2,…,n,we get the equality (2.15), where y0,y1,y2,…,yn are the first components of the vectors y^0,y^1,y^2,…,y^n. On the contrary, on the basis of the elements y0,y1,y2,…,yn corresponding to (2.1)–(2.3), one can construct the vectors y^k=(ykR0′(yk)) for which y^k∈D(Ah)(k=0,1,2,…,n) and Ahy^0=λ0y^0,  Ahy^s=λ0y^s+y^s-1,s=1,2,…,n.Theorem 2.4.
The operatorAh is maximal dissipative in the space H.Proof.
Lety^∈D(Ah). From (2.6), we have
(2.16)(Ahy^,y^)-(y^,Ahy^)=[y1,y1]a-[y1,y1¯]0+1α[R0(y1)R0′(y1)¯-R0′(y1)R0(y1)¯]=[y1,y1]a=2Imh(Dq-1y1(a))2.
It follows from that Im(Ahy^,y^)=Imh(Dq-1y1(a))2≥0,  Ah is a dissipative operator in H. Let us prove that Ah is maximal dissipative operator in the space H. It is sufficient to check that
(2.17)(Ah-λI)D(Ah)=H,Imλ<0.
To prove (2.17), let  F∈H,  Imλ<0  and put
(2.18)Γ=((G~x,F¯)R0′[(G~x,F¯)]),
where
(2.19)G~x=(G(x,ξ,λ)R0′[G(x,ξ,λ)])=(G(x,ξ,λ)-1Δ(λ)θ1(x,λ)α),G(x,ξ,λ)=-1Δ(λ){θ2(ξ,λ)θ1(x,λ),x<ξθ1(x,λ)θ2(ξ,λ),ξ<x}.
The function x→(G(x,ξ,λ),F1¯) satisfies the equation l(y)-λy=F1(0≤x<∞) and the boundary conditions (2.1)–(2.3). Moreover, for all F∈H and for Imλ<0, we arrive at Γ∈D(Ah). For each F∈H and for Imλ<0, we have (Ah-λI)Γ=F. Consequently, in the case of  Imλ<0, the result is (Ah-λI)D(Ah)=H. Hence, Theorem 2.4 is proved.


## 3.  Self-Adjoint Dilation of Dissipative Operator

We first construct the self-adjoint dilation of the operatorAh. Let us add the “incoming” and “outgoing” subspaces D-=L2(-∞,0] and D+=L2[0,∞) to H=Lq2(0,a)⊕ℂ. The orthogonal sum H=D-⊕H⊕D+  is called main Hilbert space of the dilation. In the space ℋ we consider the operator ℒh on the set D(ℒh), its elements consisting of vectors w=〈φ-,y,φ+〉, generated by the expression
(3.1)ℒ〈φ-,y^,φ+〉=〈idφ-dξ,l~(y^),idφ+dξ〉.
satisfying the conditions: φ-∈W21(-∞,0],  φ+∈W21[0,∞),  y^∈H,  y^=(y1(x)y2),  y1∈D,  y2=R0(y1), and
(3.2)y(a)-hDq-1y(a)=βφ-(0),y(a)-h¯Dq-1y(a)=βφ+(0),
where W21(·,·) are Sobolev spaces and β2:=2Imh,  β>0. Then we have the following.Theorem 3.1.
The operatorℒh is self-adjoint in ℋ and it is a self-adjoint dilation of the operator Ah.Proof.
We first prove thatℒh is symmetric in ℋ. Namely (ℒhf,g)ℋ-(f,ℒhg)ℋ=0. Let f,g∈D(ℒh),  f=〈φ-,y^,φ+〉 and g=〈ψ-,z^,ψ+〉. Then we have
(3.3)(ℒhf,g)ℋ-(f,ℒhg)ℋ=(ℒ〈φ-,y^,φ+〉,〈ψ-,z^,ψ+〉)-(〈φ-,y^,φ+〉,ℒ〈ψ-,z^,ψ+〉),=[y1,z1¯]a-[y1,z1¯]0+1α[R0(y1)R0′(z1)¯-R0′(y1)R0(z1)¯]+iψ-(0)φ¯-(0)-iφ+(0)ψ¯+(0),(ℒhf,g)ℋ-(f,ℒhg)ℋ=[y1,z1¯]a+iψ-(0)φ¯-(0)-iφ+(0)ψ¯+(0).
On the other hand,
(3.4)iψ-(0)φ¯-(0)-iφ+(0)ψ¯+(0)=iβ2(y(a)-hDq-1y(a))(z(a)-hDq-1z(a))¯-iβ2(y(a)-h¯Dq-1y(a))(z(a)-h¯Dq-1z(a))¯,=iβ2[(h-h¯)y(a)Dq-1z(a)¯-Dq-1y(a)z(a)¯].
By (3.3), we have
(3.5)iψ-(0)φ¯-(0)-iφ+(0)ψ¯+(0)=-[y1,z1¯]a.
From equalities (3.3) and (3.5), we have (ℒhf,g)ℋ-(f,ℒhg)ℋ=0. Thus, ℒh is a symmetric operator. To prove that ℒh is self-adjoint, we need to show that ℒh⊆ℒh*. We consider the bilinear form (ℒhf,g)ℋ on elements g=〈ψ-,z^,ψ+〉∈D(ℒh*), where f=〈φ-,y^,φ+〉∈D(ℒh), φ∓∈W21(ℝ∓),φ∓(0)=0. Integrating by parts, we get ℒh*g=〈i(dψ-/dξ),z^*,i(dψ+/dξ)〉, where ψ∓∈W21(ℝ∓), z^*∈H. Similarly, if f=〈0,y^,0〉∈D(ℒh), then integrating by parts in (ℒhf,g)ℋ, we obtain
(3.6)ℒh*g=ℒ*〈ψ-,z^,ψ+〉=〈idψ-dξ,l~(z^),idψ+dξ〉,z1∈D,z2=R0′(z1).
Consequently, we have (ℒhf,g)ℋ=(f,ℒhg)ℋ, for each f∈D(ℒh) by (3.6), where the operator ℒ is defined by (3.1). Therefore, the sum of the integrated terms in the bilinear form (ℒhf,g)ℋ must be equal to zero:
(3.7)[y1,z1¯]a-[y1,z1¯]0+1α[R0(y1)R0′(z1)¯-R0′(y1)R0(z1)¯]+iφ-(0)′ψ¯-(0)-iφ′+(0)ψ¯+(0)=0.
Then by (2.6), we get
(3.8)[y1,z1¯]a+iφ-(0)′ψ¯-(0)-iφ′+(0)ψ¯+(0)=0.
From the boundary conditions for ℒh, we have
(3.9)y(a)=βφ-(0)-h1iβ(φ-(0)-φ+(0)),Dq-1y(a)=iβ(φ-(0)-φ+(0)).
Afterwards, by (3.8) we get
(3.10)βφ-(0)-h1iβ(φ-(0)-φ+(0))z(a)¯-iβ(φ-(0)-φ+(0))Dq-1z(a)¯=iφ+(0)ψ¯+(0)-iφ-(0)ψ¯-(0).
Comparing the coefficients of  φ-(0) in  (3.10), we obtain
(3.11)iβ2-h1βz(a)+1βDq-1z(a)¯=φ-(0)
or
(3.12)z(a)-hDq-1z(a)=βψ-(0).
Similarly, comparing the coefficients of  φ+(0) in   (3.10) we get
(3.13)z(a)-h¯Dq-1z(a)=βψ+(0).
Therefore conditions (3.12) and (3.13) imply  D(ℒh*)⊆D(ℒh),  hence ℒh=ℒh*.
The self-adjoint operatorℒh generates on ℋ a unitary group Ut=exp(iℒht) (t∈ℝ+=(0,∞)). Let us denote by P:ℋ→H and P1:H→ℋ the mapping acting according to the formulae P:〈φ-,y^,φ+〉→y^ and P1:y^→〈0,y^,0〉. Let Zt:=PUtP1,t≥0, by using Ut. The family {Zt}(t≥0) of operators is a strongly continuous semigroup of completely nonunitary contraction on H. Let us denote by Bh the generator of this semigroup:Bhy^=limt→+0(it)-1(Zty^-y^). The domain of Bh consists of all the vectors for which the limit exists. The operator Bh is dissipative. The operator ℒh is called the self-adjoint dilation of Bh (see [2, 9, 18]). We show that  Bh=Ah,  hence  ℒh  is self-adjoint dilation of  Bh. To show this, it is sufficient to verify the equality
(3.14)P(ℒh-λI)-1P1y^=(Ah-λI)-1y^,y^∈H,Imh<0.
For this purpose, we set (ℒh-λI)-1P1y^=g=〈ψ-,z^,ψ+〉 which implies that (ℒh-λI)g=P1y^, and hence l~(z^)-λz^=y^,ψ-(ξ)=ψ-(0)e-iλξ and ψ+(ξ)=ψ+(0)e-iλξ. Since g∈D(ℒh), then ψ-∈L2(-∞,0), and it follows that ψ-(0)=0, and consequently z satisfies the boundary condition z(a)-hDq-1z(a)=0. Therefore, z^∈D(Ah), and since point λ with Imλ<0 cannot be an eigenvalue of dissipative operator, it follows that  ψ+(0) is obtained from the formula  ψ+(0)=β-1(z(a)-h¯Dq-1z(a)). Thus
(3.15)(ℒh-λI)-1P1y^=〈0,(Ah-λI)-1y^,β-1(z(a)-h¯Dq-1z(a))〉
for y^ and Imλ<0. On applying the mapping  P, we obtain (3.14), and(3.16)(Ah-λI)-1=P(ℒh-λI)-1P1=-iP∫0∞Ute-iλtdtP1=-i∫∞Zte-iλtdt=(Bh-λI)-1,Imλ<0,
so this clearly shows that Ah=Bh.The unitary group{Ut} has an important property which makes it possible to apply it to the Lax-Phillips [27], that is, it has orthogonal incoming and outcoming subspaces D-=〈L2(-∞,0),0,0〉 and  D+=〈0,0,L2(0,∞)〉 having the following properties:(1)
UtD-⊂D-,  t≤0  and  UtD+⊂D+,  t≥0;(2)
∩t≤0UtD-=∩t≥0UtD+={0};(3)
∪t≥0UtD-=∪t≤0UtD+=ℋ;(4)
D-⊥D+.To be able to prove property(1) for D+ (the proof for D- is similar), we set ℛλ=(ℒh-λI)-1. For all λ, with  Imλ<0 and for any  f=〈0,0,φ+〉∈D+, we have
(3.17)ℛλf=〈0,0,-ie-iλξ∫0ξeiλsφ+(s)ds〉,
as  ℛλf∈D+. Therefore, if  g⊥D+, then
(3.18)0=(ℛλf,g)ℋ=-i∫0∞e-iλt(Utf,g)ℋdt,Imλ<0
which implies that (Utf,g)ℋ=0 for all t≥0. Hence, for t≥0,  UtD+⊂D+, and property (1) has been proved.In order to prove property(2), we define the mappings P+:ℋ→L2(0,∞) and P1+:L2(0,∞)→D+ as follows: P+:〈φ-,y^,φ+〉→φ+ and P1+:φ→〈0,0,φ〉, respectively. We take into consideration that the semigroup of isometries Ut+:=P+UtP1+(t≥0) is a one-sided shift in L2(0,∞). Indeed, the generator of the semigroup of the one-sided shift Vt in L2(0,∞) is the differential operator i(d/dξ) with the boundary condition φ(0)=0. On the other hand, the generator S of the semigroup of isometries Ut+(t≥0) is the operator Sφ=P+ℒhP1+φ=P+ℒh〈0,0,φ〉=P+〈0,0,i(d/dξ)φ〉=i(d/dξ)φ, where φ∈W21(0,∞) and φ(0)=0. Since a semigroup is uniquely determined by its generator, it follows that Ut+=Vt, and hence
(3.19)⋂t≥0UtD+=〈0,0,⋂t≤0VtL2(0,∞)〉={0},
so, the proof of property (2) is completed.Definition 3.2.
The linear operatorA with domain D(A) acting in the Hilbert space H is called completely non-self-adjoint (or  simple)if there is no invariant subspace M⊆D(A)(M≠{0}) of the operator A on which the restriction A to M is self-adjoint.To prove property(3) of the incoming and outcoming subspaces, let us prove following lemma.Lemma 3.3.
The operatorAh is completely non-self-adjoint (simple).Proof.
LetH′⊂H be a nontrivial subspace in which Ah induces a self-adjoint operator Ah′ with domain  D(Ah′)=H′∩D(Ah).  If   f^∈D(Ah′),  then  f^∈D(Ah'*) and
(3.20)ddt∥eiAh′tf^∥H2=ddt(eiAh′tf^,eiAh′tf^)H=i(Ah′eiAh′tf^,eiAh′tf^)-i(eiAh′tf^,Ah′eiAh′tf^)
and taking  g^=eiAh′tf^,  we have
(3.21)0=i(Ah′g^,g^)H-i(g^,Ah′g^)H=i[g1,g1¯]a-i[g1,g1¯]0+iα[R0(g1)R0′(g1)¯-R0′(y1)R0(g1)¯]=-2Imh(Dq-1y1(a))2=-β2(Dq-1y1(a))2.
Since  f^∈D(Ah′),  Ah′ holds condition above. Moreover, eigenvectors of the operator Ah′ should also hold this condition. Therefore, for the eigenvectors  y^(λ) of the operator Ah acting in H′ and the eigenvectors of the operator Ah′, we have Dq-1y1(a)=0. From the boundary conditions, we get y1(a)=0 and y^(x,λ)=0. Consequently, by the theorem on expansion in the eigenvectors of the self-adjoint operator Ah′, we obtain H′={0}. Hence the operator Ah is simple. The proof is completed.Let us defineH-=∪t≥0UtD-¯,  H+=∪t≤0UtD+¯.Lemma 3.4.
The equalityH-+H+=ℋ holds.Proof.
Considering property(1) of the subspace D+, it is easy to show that the subspace ℋ′=ℋ⊝(H-+H+) is invariant relative to the group {Ut} and has the form ℋ′=〈0,H′,0〉, where H′ is a subspace in H. Therefore, if the subspace ℋ′ (and hence also H′)  was nontrivial, then the unitary group {Ut} restricted to this subspace would be a unitary part of the group {Ut′}, and hence the restriction Bh′ of Bh to H′ would be a self-adjoint operator in H′. Since the operator Bh is simple, it follows that  H′={0}. The lemma is proved.Assume thatφ(λ) and ψ(λ) are solutions of l(y)=λy satisfying the conditions
(3.22)φ1(0,λ)=0,φ2(0,λ)=1,ψ1(0,λ)=1,ψ2(0,λ)=0.θ(x,λ)=φ(x,λ)+ma(λ)ψ(x,λ)∈Lq2(0,a),Imλ>0.
The Titchmarsh-Weyl function ma(λ) is a meromorphic function on the complex plane ℂ with a countable number of poles on the real axis. Further, it is possible to show that the function ma(λ) possesses the following properties: Imma(λ)≥0 for all Imλ>0, and ma(λ)¯=ma(λ¯)  for all  λ∈ℂ,  except the real poles ma(λ). We set
(3.23)Sh(λ):=ma(λ)-hma(λ)-h¯,(3.24)Uλ-(x,ξ,ζ)=〈e-iλξ,(ma(λ)-h)-1αθ(x,λ),Sh¯(λ)e-iλζ〉.
We note that the vectors Uλ-(x,ξ,ζ) for real λ do not belong to the space ℋ. However, Uλ-(x,ξ,ζ) satisfies the equation ℒU=λU and the corresponding boundary conditions for the operator ℒH. By means of vector Uλ-(x,ξ,ζ), we define the transformation F-:f→f~-(λ) by
(3.25)(F-f)(λ)≔f~-(λ)≔12π(f,Uλ-)ℋ,
on the vectors f=〈φ-,y^,φ+〉 in which φ-(ξ),  φ+(ζ),  y(x) are smooth, compactly supported functions.Lemma 3.5.
The transformationF- isometrically maps H- onto L2(ℝ). For all vectors f,g∈H- the Parseval equality and the inversion formulae hold:
(3.26)(f,g)ℋ=(f~-,g~-)L2=∫-∞∞f~-(λ)g~-(λ)-dλ,f=12π∫-∞∞f~-(λ)Uλ-dλ,
where  f~-(λ)=(F-f)(λ) and g~-(λ)=(F-g)(λ).Proof.
Forf,g∈D-,  f=〈φ-,0,0〉,  g=〈ψ+,0,0〉, with Paley-Wiener theorem, we have
(3.27)f~-(λ)=12π(f,Uλ-)ℋ=12π∫-∞0φ-(ξ)e-iλξdξ∈H-2,
and by using usual Parseval equality for Fourier integrals
(3.28)(f,g)ℋ=∫-∞∞φ-(ξ)ψ-(ξ)¯dξ=∫-∞∞f~-(λ)g~-(λ)¯dλ=(F-f,F-g)L2.
Here, H±2 denote the Hardy classes in L2(ℝ) consisting of the functions analytically extendible to the upper and lower half-planes, respectively.
We now extend to the Parseval equality to the whole ofH-. We consider in H- the dense set of H-′ of the vectors obtained as follows from the smooth, compactly supported functions in  D-:f∈H-′  if  f=UTf0,  f0=〈φ-,0,0〉,  φ-∈C0∞(-∞,0), where T=Tf  is a nonnegative number depending on  f.  If f,g∈H-′, then for  T>Tf  and T>Tg we have U-Tf,U-Tg∈D-; moreover, the first components of these vectors belong to C0∞(-∞,0). Therefore, since the operators Ut(t∈ℝ) are unitary, by the equality
(3.29)F-Utf=(Utf,Uλ-)ℋ=eiλt(f,Uλ-)ℋ=eiλtF-f,
we have
(3.30)(f,g)ℋ=(U-Tf,U-Tg)ℋ=(F-U-Tf,F-U-Tg)L2(eiλTF-f,eiλTF-g)L2=(f~,g~)L2.
By taking the closure (3.30), we obtain the Parseval equality for the space  H-. The inversion formula is obtained from the Parseval equality if all integrals in it are considered as limits in the of integrals over finite intervals. Finally F-H-=∪t≥0F-UtD-¯=∪t≥0eiλtH-2¯=L2(ℝ), that is, F- maps H- onto the whole of L2(ℝ). The lemma is proved.We set(3.31)Uλ+(x,ξ,ζ)=〈Sh(λ)e-iλξ,(ma(λ)-h¯)-1αθ(x,λ),e-iλζ〉.
We note that the vectors Uλ+(x,ξ,ζ) for real λ do not belong to the space ℋ. However, Uλ+(x,ξ,ζ) satisfies the equation ℒU=λU and the corresponding boundary conditions for the operator ℒH. With the help of vector Uλ+(x,ξ,ζ), we define the transformation F+:f→f~+(λ) by (F+f)(λ):=f~+(λ):=(1/2π)(f,Uλ+)ℋ on the vectors f=〈φ-,y^,φ+〉 in which φ-(ξ),  φ+(ζ) and y(x) are smooth, compactly supported functions.Lemma 3.6.
The transformationF+ isometrically maps H+ onto L2(ℝ). For all vectors  f,g∈H+  the Parseval equality and the inversion formula hold:
(3.32)(f,g)ℋ=(f~+,g~+)L2=∫-∞∞f~+(λ)g~+(λ)¯dλ,f=12π∫-∞∞f~+(λ)Uλ+dλ,
where  f~+(λ)=(F+f)(λ)  and   g~+(λ)=(F+g)(λ).Proof.
The proof is analogous to Lemma3.5.It is obvious that the matrix-valued functionSh(λ) is meromorphic in ℂ and all poles are in the lower half-plane. From (3.23), |Sh(λ)|≤1 for Imλ>0; and Sh(λ) is the unitary matrix for all λ∈ℝ. Therefore, it explicitly follows from the formulae for the vectors Uλ- and Uλ+ that
(3.33)Uλ+=Sh(λ)Uλ-.
It follows from Lemmas 3.5 and 3.6 that H-=H+. Together with Lemma 3.4, this shows that H-=H+=ℋ; therefore, property (3) above has been proved for the incoming and outcoming subspaces. Finally property (4) is clear.Thus, the transformationF- isometrically maps H- onto L2(ℝ) with the subspace D- mapped onto H-2 and the operators Ut are transformed into the operators of multiplication by eiλt. This means that F- is the incoming spectral representation for the group {Ut}. Similarly, F+ is the outgoing spectral representation for the group {Ut}. It follows from (3.33) that the passage from the F- representation of an element f∈ℋ to its F+ representation is accomplished as f~+(λ)=Sh-1(λ)f~-(λ). Consequently, according to [27] we have proved the following.Theorem 3.7.
The functionSh-1(λ) is the scattering matrix of the group {Ut} (of  the  self-adjoint  operator  LH).LetS(λ) be an arbitrary nonconstant inner function on the upper half-plane (the analytic function S(λ) on the upper half-plane ℂ+ is called inner function on ℂ+ if |Sh(λ)|≤1 for all λ∈ℂ+ and |Sh(λ)|=1 for almost all λ∈ℝ). Define K=H+2⊝SH+2. Then K≠{0} is a subspace of the Hilbert space H+2. We consider the semigroup of operators Zt(t≥0) acting in K according to the formula Ztφ=P[eiλtφ],φ=φ(λ)∈K, where P is the orthogonal projection from H+2 onto K. The generator of the semigroup {Zt} is denoted by
(3.34)Tφ=limt→+0(it)-1(Ztφ-φ),
in which T is a maximal dissipative operator acting in K and with the domain  D(T) consisting of all functions φ∈K, such that the limit exists. The operator T is called a model dissipative operator (we remark that this model dissipative operator, which is associated with the names of Lax-Phillips [27], is a special case of a more general model dissipative operator constructed by Nagy and Foiaş [26]). The basic assertion is that S(λ) is the characteristic functionof the operator T.LetK=〈0,H,0〉, so that ℋ=D-⊕K⨁D+. It follows from the explicit form of the unitary transformation  F- under the mapping  F-(3.35)ℋ→L2(ℝ),f→f~-(λ)=(F-f)(λ),D-→H-2,D+→ShH+2,K→H+2⊝SGH+2,Utf→(F-UtF--1f~-)(λ)=eiλtf~-(λ).
The formulae (3.35) show that operator  Ah  is a unitarily equivalent to the model dissipative operator with the characteristic function  Sh(λ).  Since the characteristic functions of unitary equivalent dissipative operator coincide (see [26]), we have thus proved the following theorem.Theorem 3.8.
The characteristic function of the maximal dissipative operatorAh coincides with the function  Sh(λ)  defined in (3.23).Using characteristic function, the spectral properties of the maximal dissipative operatorAh can be investigated. The characteristic function of the maximal dissipative operator Ah is known to lead to information of completeness about the spectral properties of this operator. For instance, the absence of a singular factor s(λ) of the characteristic function Sh(λ) in the factorization detSh(λ)=s(λ)B(λ), where B(λ) is a Blaschke product, ensures completeness of the system of eigenvectors and associated vectors of the operator Ah in the space Lq2(0,a) (see [25]).Theorem 3.9.
For all the values ofh with Imh>0, except possibly for a single value h=h0, the characteristic function Sh(λ) of the maximal dissipative operator Ah is a Blaschke product. The spectrum of Ah is purely discrete and belongs to the open upper half-plane. The operator Ah has  a countable number of isolated eigenvalues with finite multiplicity and limit points at infinity. The system of all eigenvectors and associated vectors of the operator Ah is complete in the space H.Proof.
From (3.23),  it is clear that Sh(λ) is an inner function in the upper half-plane, and it is meromorphic in the whole complex  λ-plane. Therefore, it can be factored in the form
(3.36)Sh(λ)=eiλcBh(λ),c=c(h)≥0,
where Bh(λ) is a Blaschke product. It follows from  (3.36) that
(3.37)|Sh(λ)|=|eiλc||Bh(λ)|≤e-b(h)Imλ,Imλ≥0.
Further, for ma(λ) in terms of  Sh(λ), we find from (3.23) that
(3.38)ma(λ)=h-h¯Sh(λ)Sh(λ)-1.
If c(h)>0 for a given value h(Imh>0), then (3.37) implies that limt→+∞Sh(it)=0, and then (3.24) gives us that limt→+∞ma(it)=-G. Since ma(λ) does not depend on h, this implies that c(h) can be nonzero at not more than a single point h=h0 (and further h0=-limt→+∞ma(it)). The theorem is proved.
Due to Theorem2.4, since the eigenvalues of the boundary value problem (2.1)–(2.3) and eigenvalues of the operator Ah coincide, including their multiplicity and, furthermore, for the eigenfunctions and associated functions the boundary problems (2.1)–(2.3), then theorem is interpreted as follows.Corollary 3.10.
The spectrum of the boundary value problem (2.1)–(2.3) is purely discrete and belongs to the open upper half-plane. For all the values of h with  Imλ>0, except possible for a single value h=h0, the boundary value problem (2.1)–(2.3) (h≠h0) has a countable number of isolated eigenvalues with finite multiplicity and limit points and infinity. The system of the eigenfunctions and associated functions of this problem (h≠h0) is complete in the space Lq2(0,a).



---

*Source: 736437-2012-09-27.xml*