# Discontinuous Sturm-Liouville Problems and Associated Sampling Theories

Authors: M. M. Tharwat
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

This paper investigates the sampling analysis associated with discontinuous Sturm-Liouville problems with eigenvalue parameters in two boundary conditions and with transmission conditions at the point of discontinuity. We closely follow the analysis derived by Fulton (1977) to establish the needed relations for the derivations of the sampling theorems including the construction of Green's function as well as the eigenfunction expansion theorem. We derive sampling representations for transforms whose kernels are either solutions or Green's functions. In the special case, when our problem is continuous, the obtained results coincide with the corresponding results in the work of Annaby and Tharwat (2006).


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## Body

## 1. Introduction

The recovery of entire functions from a discrete sequence of points is an important problem from mathematical and practical points of view. For instance, in signal processing it is needed to reconstruct (recover) a signal (function) from its values at a sequence of samples. If this aim is achieved, then an analog (continuous) signal can be transformed into a digital (discrete) one and then it can be recovered by the receiver. If the signal is band limited, the sampling process can be done via the celebrated Whittaker, Shannon, and Kotel’nikov (WKS) sampling theorem [1–3]. By a band-limited signal with band width σ, σ>0, that is, the signal contains no frequencies higher than σ/2π cycles per second (cps), we mean a function in the Paley-Wiener space PWσ2 of entire functions of exponential type at most σ which are L2(ℝ)-functions when restricted to ℝ. This space is characterized by the following relation which is due to Paley and Wiener [4, 5]:(1.1)f(t)∈PWσ2⟺f(t)=12π∫-σσeixtg(x)dx,forsomefunctiong(⋅)∈L2(-σ,σ).
Now WKS [6, 7] sampling theorem states the following.Theorem 1.1 (WKS).
Iff(t)∈PWσ2, then it is completely determined from its values at the points tk=kπ/σ, k∈ℤ, by means of the formula
(1.2)f(t)=∑k=-∞∞f(tk)sinc⁡σ(t-tk),t∈C,
where
(1.3)sinc⁡t={sin⁡⁡tt,t≠0,1,t=0.
The sampling series (1.2) is absolutely and uniformly convergent on compact subsets of ℂ, uniformly convergent on ℝ and converges in the norm of L2(ℝ), see [6, 8, 9].The WKS sampling theorem has been generalized in many different ways. Here we are interested in two extensions. The first is concerned with replacing the equidistant sampling points by more general ones, which is important from practical and theoretical point of view. The following theorem which is known in some literature as Paley-Wiener theorem, [5] gives a sampling theorem with a more general class of sampling points. Although the theorem in its final form may be attributed to Levinson [10] and Kadec [11], it could be named after Paley and Wiener who first derive the theorem in a more restrictive form, see [6, 7] for more details.Theorem 1.2 (Paley and Wiener).
Let{tk}, k∈ℤ be a sequence of real numbers satisfying
(1.4)D:=sup⁡k∈Z⁡|tk-kπσ|<π4σ,
and let G(t) be the entire function defined by the canonical product
(1.5)G(t):=(t-t0)∏k=1∞(1-ttk)(1-tt-k).
Then, for any f∈PWσ2(1.6)f(t)=∑k∈Zf(tk)G(t)G′(tk)(t-tk),t∈C.
The series (1.6) converges uniformly on compact subsets of ℂ.The WKS sampling theorem is a special case of this theorem because if we choosetk=kπ/σ=-t-k, then (1.7)G(t)=t∏k=1∞(1-ttk)(1+ttk)=t∏k=1∞(1-(tσ/π)2k2)=sin⁡⁡tσσ,G′(tk)=(-1)k.
Expansion (1.6) is of Lagrange-type interpolation.The second extension of WKS sampling theorem is the theorem of Kramer [12]. In this theorem sampling representations were given for integral transforms whose kernels are more general than exp(ixt).Theorem 1.3 (Kramer).
Let I be a finite closed interval,K(·,t):I×ℂ→ℂ a function continuous in t such that K(·,t)∈L2(I) for all t∈ℂ, and let {tk}k∈ℤ be a sequence of real numbers such that {K(·,tk)}k∈ℤ is a complete orthogonal set in L2(I). Suppose that
(1.8)f(t)=∫IK(x,t)g(x)dx,g(⋅)∈L2(I).
Then
(1.9)f(t)=∑k∈Zf(tk)∫IK(x,t)K(x,tk)¯dx∥K(⋅,tk)∥L2(I)2.
Series (1.9) converges uniformly wherever ∥K(·,t)∥L2(I) as a function of t is bounded.Again Kramer’s theorem is a generalization of WKS theorem. If we takeK(x,t)=eitx,I=[-σ,σ],tk=kπ/σ, then (1.9) will be (1.2).The relationship between both extensions of WKS sampling theorem has been investigated extensively. Starting from a function theory approach, cf. [13], it is proved in [14] that if K(x,t), x∈I, t∈ℂ satisfies some analyticity conditions, then Kramer’s sampling formula (1.9) turns out to be a Lagrange interpolation one, see also [15–17]. In another direction, it is shown that Kramer’s expansion (1.9) could be written as a Lagrange-type interpolation formula if K(·,t) and tk are extracted from ordinary differential operators, see the survey [18] and the references cited therein. The present work is a continuation of the second direction mentioned above. We prove that integral transforms associated with second-order eigenvalue problems with an eigenparameter appearing in the boundary conditions and also with an internal point of discontinuity can also be reconstructed in a sampling form of Lagrange interpolation type. We would like to mention that works in direction of sampling associated with eigenproblems with an eigenparameter in the boundary conditions are few, see, for example, [19, 20]. Also papers in sampling with discontinuous eigenproblems are few, see [21–23]. However sampling theories associated with eigenproblems, which contain eigenparameter in the boundary conditions and have at the same time discontinuity conditions, do not exist as for as we know. Our investigation will be the first in that direction, introducing a good example. To achieve our aim we will briefly study the spectral analysis of the problem. Then we derive two sampling theorems using solutions and Green’s function, respectively.


## 2. The Eigenvalue Problem

In this section we define our boundary value problem and state some of its properties. Consider the boundary value problem(2.1)l(y):=-r(x)u′′(x)+q(x)u(x)=λu(x),x∈[-1,0)∪(0,1],
with boundary conditions(2.2)U1(u):=(α1′λ-α1)u(-1)-(α2′λ-α2)u′(-1)=0,(2.3)U2(u):=(β1′λ+β1)u(1)-(β2′λ+β2)u′(1)=0,
and transmission conditions(2.4)U3(u):=γ1u(-0)-δ1u(+0)=0,U4(u):=γ2u′(-0)-δ2u′(+0)=0,
where λ is a complex spectral parameter; r(x)=r12 for x∈[-1,0), r(x)=r22 for x∈(0,1]; r1>0 and r2>0 are given real number; q(x) is a given real-valued function, which is continuous in [-1,0) and (0,1] and has a finite limit q(±0)=limx→±0q(x); γi,δi, αi,βi,αi′,βi′ (i=1,2) are real numbers; γi≠0, δi≠0 (i=1,2); ρ and γ are given by(2.5)ρ:=det⁡⁡(α1′α1α2′α2)>0,γ:=det⁡⁡(β1′β1β2′β2)>0.
In some literature conditions (2.4) are called compatability conditions, see, for example, [24]. To formulate a theoretic approach to problem (2.1)–(2.4) we define the Hilbert space H:=L2(-1,1)⊕ℂ2 with an inner product(2.6)〈f(⋅),g(⋅)〉H:=1r12∫-10f(x)g¯(x)dx+1r22∫01f(x)g¯(x)dx+1ρf1g1¯+1γf2g2¯,
where
(2.7)f(x)=(f(x)f1f2),g(x)=(g(x)g1g2)∈H,f(·),g(·)∈L2(-1,1) and fi,gi∈ℂ, i=1,2. For convenience we put(2.8)(R-1(u)R1(u)R-1′(u)R1′(u)):=(α1u(-1)-α2u′(-1)β1u(1)-β2u′(1)α1′u(-1)-α2′u′(-1)β1′u(1)-β2′u′(1)).For functionf(x), which is defined on [-1,0)∪(0,1] and has finite limit f(±0):=limx→±0f(x), by f(1)(x) and f(2)(x) we denote the functions
(2.9)f(1)(x)={f(x),x∈[-1,0),f(-0),x=0,f(2)(x)={f(x),x∈(0,1],f(+0),x=0,
which are defined on I1:=[-1,0] and I2:=[0,1], respectively.In the following we will define the minimal closed operator inH associated with the differential expression ℓ, cf. [25, 26].Let𝒟(A)⊆H be the set of all(2.10)f(x)=(f(x)R-1′(f)R1′(f))∈H
such that f(i)(·), f(i)′(·) are absolutely continuous in Ii, i=1,2,ℓ(f)∈L2(-1,0)⊕L2(0,1) and U3(f)=U4(f)=0. Define the operator A:𝒟(A)→H by(2.11)A(f(x)R-1′(f)R1′(f))=(l(f)R-1(f)-R1(f)),(f(x)R-1′(f)R1′(f))∈D(A).
The eigenvalues and the eigenfunctions of the problem (2.1)–(2.4) are defined as the eigenvalues and the first components of the corresponding eigenelements of the operator A, respectively.Theorem 2.1.
Letγ1γ2=δ1δ2. Then, the operator A is symmetric.Proof.
Forf(·),g(·)∈𝒟(A)(2.12)〈Af(⋅),g(⋅)〉H=1r12∫-10l(f(x))g¯(x)dx+1r22∫01l(f(x))g¯(x)dx+1ρR-1(f)R-1′(g¯)-1γR1(f)R1′(g¯).
By two partial integration we obtain
(2.13)〈Af(⋅),g(⋅)〉H=〈f(⋅),Ag(⋅)〉H+W(f,g¯;-0)-W(f,g¯;-1)+W(f,g¯;1)-W(f,g¯;+0)+1ρ(R-1(f)R-1′(g¯)-R-1′(f)R-1(g¯))+1γ(R1′(f)R1(g¯)-R1(f)R1′(g¯)),
where, as usual, by W(f,g;x) we denote the Wronskian of the functions f and g(2.14)W(f,g;x):=f(x)g′(x)-f′(x)g(x).
Since f(x) and g¯(x) are satisfied the boundary condition (2.2)-(2.3) and transmission conditions (2.4) we get
(2.15)R-1(f)R-1′(g¯)-R-1′(f)R-1(g¯)=ρW(f,g¯;-1),R1′(f)R1(g¯)-R1(f)R1′(g¯)=-γW(f,g¯;1),γ1γ2W(f,g¯;-0)=δ1δ2W(f,g¯;+0).
Finally substituting (2.15) in (2.13) then we have
(2.16)〈Af(⋅),g(⋅)〉H=〈f(⋅),Ag(⋅)〉H,f(⋅),g(⋅)∈D(A),
thus, the operator A is Hermitian. The symmetry of A arises from the well-known fact that 𝒟(A) is dense in H see, for example, [24].Corollary 2.2.
All eigenvalues of the problem (2.1)–(2.4) are real.We can now assume that all eigenfunctions of the problem (2.1)–(2.4) are real valued.Corollary 2.3.
Letλ1 and λ2 be two different eigenvalues of the problem (2.1)–(2.4). Then the corresponding eigenfunctions u1 and u2 of this problem are orthogonal in the sense of
(2.17)1r12∫-10u1(x)u2(x)dx+1r22∫01u1(x)u2(x)dx+1ρR-1′(u1)R-1′(u2)+1γR1′(u1)R1′(u2)=0.Proof.
Formula (2.17) follows immediately from the orthogonality of corresponding eigenelements
(2.18)u1(x)=(u1(x)R-1′(u1)R1′(u1)),u2(x)=(u2(x)R-1′(u2)R1′(u2))
in the Hilbert space H.Now, we will construct a special fundamental system of solutions of the equation (2.1) for λ not being an eigenvalue. Let us consider the next initial value problem:(2.19)-r12u′′(x)+q(x)u(x)=λu(x),x∈(-1,0),(2.20)u(-1)=λα2′-α2,u′(-1)=λα1′-α1.
By virtue of Theorem  1.5 in [27] this problem has a unique solution u=φ1(x)=φ1λ(x), which is an entire function of λ∈ℂ for each fixed x∈[-1,0]. Similarly, employing the same method as in proof of Theorem  1.5 in [27], we see that the problem(2.21)-r22u′′(x)+q(x)u(x)=λu(x),x∈(0,1),(2.22)u(1)=λβ2′+β2,u′(1)=λβ1′+β1,
has a unique solution u=χ2(x)=χ2λ(x) which is an entire function of parameter λ for each fixed x∈[0,1].Now the functionsφ2λ(x) and χ1λ(x) are defined in terms of φ1λ(x) and χ2λ(x) as follows: the initial-value problem,(2.23)-r22u′′(x)+q(x)u(x)=λu(x),x∈(0,1),(2.24)u(0)=γ1δ1φ1λ(0),u′(0)=γ2δ2φ1λ′(0),
which contains the entire functions of eigenparameter λ (in the right-hand side), has unique solution u=φ2λ(x) for each λ∈ℂ.Similarly, the following problem also has a unique solutionu=χ1(x)=χ1λ(x):(2.25)-r12u′′(x)+q(x)u(x)=λu(x),x∈(-1,0),(2.26)u(0)=δ1γ1χ2λ(0),u′(0)=δ2γ2χ2λ′(0).Since the WronskiansW(φiλ,χiλ;x) are independent on variable x∈Ii (i=1,2) and φiλ(x) and χiλ(x) are the entire functions of the parameter λ for each x∈Ii (i=1,2), then the functions(2.27)ωi(λ):=W(φiλ,χiλ;x),x∈Ii,i=1,2,
are the entire functions of parameter λ.Lemma 2.4.
If the conditionγ1γ2=δ1δ2 is satisfied, then the equality ω1(λ)=ω2(λ) holds for each λ∈ℂ.Proof.
Taking into account (2.24) and (2.26), a short calculation gives γ1γ2W(φ1λ,χ1λ;0)=δ1δ2W(φ2λ,χ2λ;0), so ω1(λ)=ω2(λ) for each λ∈ℂ.Corollary 2.5.
The zeros of the functionsω1(λ) and ω2(λ) coincide.Let us construct two basic solutions of (2.1) as(2.28)φλ(x)={φ1λ(x),x∈[-1,0),φ2λ(x),x∈(0,1],χλ(x)={χ1λ(x),x∈[-1,0),χ2λ(x),x∈(0,1].
By virtue of (2.24) and (2.26) these solutions satisfy both transmission conditions (2.4).Now we may introduce to the consideration the characteristic functionω(λ) as(2.29)ω(λ):=ω1(λ)=ω2(λ).Theorem 2.6.
The eigenvalues of the problem (2.1)–(2.4) are coincided zeros of the function ω(λ).Proof.
Letω(λ0)=0. Then W(φ1λ0,χ1λ0;x)=0, and so the functions φ1λ0(x) and χ1λ0(x) are linearly dependent, that is,
(2.30)χ1λ0(x)=kφ1λ0(x),x∈[-1,0],for somek≠0.
Consequently, χλ0(x) satisfied the boundary condition (2.3), so the function χλ0(x) is an eigenfunction of the problem (2.1)–(2.4) corresponding to the eigenvalue λ0.
Now letu0(x) be any eigenfunction corresponding to the eigenvalue λ0, but ω(λ0)≠0. Then the functions φiλ0(x),χiλ0(x) are linearly independent on Ii,i=1,2. Thus, u0(x) may be represented as in the form
(2.31)u0(x)={c1φ1λ0(x)+c2χ1λ0(x),x∈[-1,0),c3φ2λ0(x)+c4χ2λ0(x),x∈(0,1],
where at least one of the constants ci,  i=1,2,3,4, is not zero.
Consider the equations(2.32)Ui(u0(x))=0,i=1,2,3,4
as the homogenous system of linear equations of the variables ci,  i=1,2,3,4, and taking into account (2.24) and (2.26), it follows that the determinant of this system is
(2.33)|0-ω1(λ0)0000ω2(λ0)0γ1φ1λ0(0)γ1χ1λ0(0)-δ1φ2λ0(0)-δ1χ2λ0(0)γ2φ1λ0′(0)γ2χ1λ0′(0)-δ2φ2λ0′(0)-δ2χ2λ0′(0)|=δ1δ2ω1(λ0)ω22(λ0)≠0.
Thus, the system (2.32) has only trivial solution ci=0,  i=1,2,3,4, and so we get contradiction which completes the proof.Lemma 2.7.
Ifλ=λ0 is an eigenvalue, then φλ0(x) and χλ0(x) are linearly dependent.Proof.
Sinceλ0 is an eigenvalue, then from Theorem 2.6 we have W(φiλ0,χiλi;x)=ω(λ0)=0,   i=1,2. Therefore
(2.34)χiλ0(x)=kiφiλ0(x),i=1,2,
for some ki≠0, i=1,2. Now, we must show that k1=k2. Suppose if possible that k1≠k2. Taking into account the definitions of solution φiλ0(x) and φiλ0(x), i=1,2, from the equalities (2.34) we get
(2.35)U3(χλ0)=γ1χλ0(-0)-δ1χλ0(+0)=γ1χ1λ0(0)-δ1χ2λ0(0)=γ1k1φ1λ0(0)-δ1k2φ2λ0(0)=δ1k1φ2λ0(0)-δ1k2φ2λ0(0)=δ1(k1-k2)φ2λ0(0).
Since U3(χλ0)=0, δ1≠0, and k1-k2≠0 it follows that
(2.36)φ2λ0(0)=0.
By the same procedure from the equality U4(χλ0)=0 we can derive that
(2.37)φ2λ0′(0)=0.
From the fact that φ2λ0(x) is a solution of (2.1) on [0,1] and satisfied the initial conditions (2.36) and (2.37) it follows that φ2λ0(x)=0 identically on [0,1], because of the well-known existence and uniqueness theorem for the initial value problems of the ordinary linear differential equations.
By using (2.24), (2.36), and (2.37) we may also find
(2.38)φ1λ0(0)=φ1λ0′(0)=0.
For latter discussion for φ2λ0(x), it follows that φ1λ0(x)=0 identically on [-1,0]. Therefore φλ0(x)=0 identically on [-1,0)∪(0,1]. But this is contradicted with (2.20), which completes the proof.Corollary 2.8.
Ifλ=λ0 is an eigenvalue, then both φλ0(x) and χλ0(x) are eigenfunctions corresponding to this eigenvalue.Lemma 2.9.
If the conditionγ1γ2=δ1δ2 is satisfied, then all eigenvalues λn are simple zeros of ω(λ).Proof.
Since(2.39)1r12∫-10l(φλ(x))φλ0(x)dx+1r22∫01l(φλ(x))φλ0(x)dx=1r12∫-10φλ(x)l(φλ0(x))dx+1r22∫01φλ(x)l(φλ0(x))dx+W(φλ,φλ0;1)-W(φλ,φλ0;-1),
then
(2.40)(λ-λ0)[1r12∫-10φλ(x)φλ0(x)dx+1r22∫01φλ(x)φλ0(x)dx]=W(φλ,φλ0;1)-(λ-λ0)ρ,
for any λ. Since
(2.41)χλ0(x)=k0φλ0(x),x∈[-1,0)∪(0,1],
for some k0≠0, then
(2.42)W(φλ,φλn;1)=1knW(φλ,χλn;1)=1kn(λnR1′(φλ)+R1(φλ))=1kn[ω(λ)-(λ-λn)R1′(φλ)]=(λ-λn)1kn[ω(λ)λ-λn-R1′(φλ)].
Substituting (2.42) in (2.40) and letting λ→λn we get
(2.43)1r12∫-10(φλn(x))2dx+1r22∫01(φλ0(x))2dx=1kn[ω′(λn)-R1′(φλn)]-ρ.
Now putting
(2.44)R1′(φλn)=1knR1′(χλn)=γkn
in (2.43) it yields ω′(λn)≠0, which completes the proof.Ifλn, n=0,1,2,… denote the zeros of ω(λ), then the three-component vectors(2.45)Φn(x)≔(φλn(x)R-1′(φλn)R1′(φλn))
are the corresponding eigenvectors of the operator A satisfying the orthogonality relation(2.46)〈Φn(⋅),Φm(⋅)〉H=0forn≠m.
Here {φλn(·)}n=0∞ will be the sequence of eigenfunctions of (2.1)–(2.4) corresponding to the eigenvalues {λn}n=0∞. We denote by Ψn(·) the normalized eigenvectors(2.47)Ψn(x)≔Φn(x)‖Φn(⋅)‖H=(ψn(x)R-1′(ψn)R1′(ψn)).
Because of simplicity of the eigenvalues, we find nonzeros constants kn such that(2.48)χλn(x):=knφλn(x),x∈[-1,0)∪(0,1],n=0,1,….
To study the completeness of the eigenvectors of A, and hence the completeness of the eigenfunctions of (2.1)–(2.4), we construct the resolvent of A as well as Green’s function of problem (2.1)–(2.4). We assume without any loss of generality that λ=0 is not an eigenvalue of A. Otherwise, from discreteness of eigenvalues, we can find a real number c such that c≠λn for all n and replace the eigenparameter λ by λ-c. Now let λ∈ℂ not be an eigenvalue of A and consider the inhomogeneous problem(2.49)(λI-A)u(x)=f(x),forf(x)=(f(x)f1f2)∈H,u(x)=(u(x)R-1′(u)R1′(u))∈D(A),
and I is the identity operator. Since(2.50)(λI-A)u(x)=λ(u(x)R-1′(u)R1′(u))-(l(u(x))R-1(u)-R1(u))=(f(x)f1f2),
then we have(2.51)(λ-l)u(x)=f(x),x∈[-1,0)∪(0,1],(2.52)f1=λR-1′(u)-R-1(u),f2=λR1′(u)+R1(u).
Now, we can represent the general solution of (2.51) in the following form:(2.53)u(x,λ)={A1φ1λ(x)+B1χ1λ(x),x∈[-1,0),A2φ2λ(x)+B2χ2λ(x),x∈(0,1].
Applying the method of variation of the constants to (2.53), thus, the functions A1(x,λ), B1(x,λ) and A2(x,λ), B2(x,λ) satisfy the linear system of equations(2.54)A1′(x,λ)φ1λ(x)+B1′(x,λ)χ1λ(x)=0,A1′(x,λ)φ1λ′(x)+B1′(x,λ)χ1λ′(x)=f(x)r12,x∈[-1,0),A2′(x,λ)φ2λ(x)+B2′(x,λ)χ2λ(x)=0,A2′(x,λ)φ2λ′(x)+B2′(x,λ)χ2λ′(x)=f(x)r22,x∈(0,1].
Since λ is not an eigenvalue and W(φiλ(x),χiλ(x);x)≠0, i=1,2, each of the linear systems in (2.54) has a unique solution which leads(2.55)A1(x,λ)=1r12ω(λ)∫x0f(ξ)χ1λ(ξ)dξ+A1,B1(x,λ)=1r12ω(λ)∫-1xf(ξ)φ1λ(ξ)dξ+B1,x∈[-1,0),A2(x,λ)=1r22ω(λ)∫x1f(ξ)χ2λ(ξ)dξ+A2,B2(x,λ)=1r22ω(λ)∫0xf(ξ)φ2λ(ξ)dξ+B2,x∈(0,1],
where A1,A2,B1, and B2 are arbitrary constants. Substituting (2.55) into (2.53), we obtain the solution of (2.51)(2.56)u(x,λ)={φ1λ(x)r12ω(λ)∫x0f(ξ)χ1λ(ξ)dξ+χ1λ(x)r12ω(λ)∫-1xf(ξ)φ1λ(ξ)dξ+A1φ1λ(x)+B1χ1λ(x),x∈[-1,0),φ2λ(x)r22ω(λ)∫x1f(ξ)χ2λ(ξ)dξ+χ2λ(x)r22ω(λ)∫0xf(ξ)φ2λ(ξ)dξ+A2φ2λ(x)+B2χ2λ(x),x∈(0,1].
Then from (2.52) and the transmission conditions (2.4) we get (2.57)A1=1r22ω(λ)∫01f(ξ)χ2λ(ξ)dξ+f2ω(λ),B1=-f1ω(λ),A2=f2ω(λ),B2=1r12ω(λ)∫-10f(ξ)φ1λ(ξ)dξ-f1ω(λ).
Then (2.56) can be written as(2.58)u(x,λ)=f2ω(λ)φλ(x)-f1ω(λ)χλ(x)+χλ(x)ω(λ)∫-1xf(ξ)r(ξ)φλ(ξ)dξ+φλ(x)ω(λ)∫x1f(ξ)r(ξ)χλ(ξ)dξ,x,ξ∈[-1,0)∪(0,1].
Hence, we have(2.59)u(x)=(λI-A)-1f(x)=(f2ω(λ)φλ(x)-f1ω(λ)χλ(x)+∫-11G(x,ξ,λ)f(ξ)r(ξ)dξR-1′(u)R1′(u)),
where(2.60)G(x,ξ,λ)={χλ(x)φλ(ξ)ω(λ),-1≤ξ≤x≤1,x≠0,ξ≠0,χλ(ξ)φλ(x)ω(λ),-1≤x≤ξ≤1,x≠0,ξ≠0,
is the unique Green’s function of problem (2.1)–(2.4). Obviously G(x,ξ,λ) is a meromorphic function of λ, for every (x,ξ)∈([-1,0)∪(0,1])2, which has simple poles only at the eigenvalues. Although Green’s function looks as simple as that of Sturm-Liouville problems, cf., for example, [28], it is a rather complicated because of the transmission conditions, see the example at the end of this paper.Lemma 2.10.
The operatorA is self-adjoint in H.Proof.
SinceA is a symmetric densely defined operator, then it is sufficient to show that the deficiency spaces are the null spaces and hence A=A*. Indeed, if f(x)=(f(x)f1f2)∈H and λ is a nonreal number, then taking
(2.61)u(x)=(f2ω(λ)φλ(x)-f1ω(λ)χλ(x)+∫-11G(x,ξ,λ)f(ξ)r(ξ)dξR-1′(u)R1′(u))
implies that u∈𝒟(A). Since G(x,ξ,λ) satisfies conditions (2.2)–(2.4), then (A-λI)u(x)=f(x). Now we prove that the inverse of (A-λI) exists. If Au(x)=λu(x), then
(2.62)(λ¯-λ)〈u(⋅),u(⋅)〉H=〈u(⋅),λu(⋅)〉H-〈λu(⋅),u(⋅)〉H=〈u(⋅),Au(⋅)〉H-〈Au(⋅),u(⋅)〉H=0(sinceAissymmetric).
Since λ∉ℝ, we have λ¯-λ≠0. Thus 〈u(·),u(·)〉H=0, that is, u=0. Then R(λ;A)≔(A-λI)-1, the resolvent operator of A, exists. Thus
(2.63)R(λ;A)f=(A-λI)-1f=u.
Take λ=±i. The domains of (A-iI)-1 and (A+iI)-1 are exactly H. Consequently the ranges of (A-iI) and (A+iI) are also H. Hence the deficiency spaces of A are
(2.64)N-i≔N(A*+iI)=R(A-iI)⊥=H⊥={0},Ni≔N(A*-iI)=R(A+iI)⊥=H⊥={0}.
Hence A is self-adjoint.The next theorem is an eigenfunction expansion theorem, which is similar to that established by Fulton in [29].Theorem 2.11.
(i) Foru(·)∈H(2.65)‖u(⋅)‖H2=∑n=-∞∞|〈u(⋅),Ψn(⋅)〉H|2.(ii) For u(·)∈𝒟(A)(2.66)u(x)=∑n=-∞∞〈u(⋅),Ψn(⋅)〉HΨn(x),
with the series being absolutely and uniformly convergent in the first component for on [-1,0)∪(0,1] and absolutely convergent in the second component.Proof.
The proof is similar to that in [29, pages 298-299].


## 3. Asymptotic Formulas of Eigenvalues and Eigenfunctions

Now we derive first- and second-order asymptotics of the eigenvalues and eigenfunctions similar to the classical techniques of [27, 30] and [29], see also [25, 26]. We begin by proving some lemmas.Lemma 3.1.
Letφλ(x) be the solutions of (2.1) defined in Section 2, and let λ=s2. Then the following integral equations hold for k=0 and k=1:
(3.1)dkdxkφ1λ(x)=(-α2+s2α2′)dkdxkcos⁡[s(x+1)r1]-(-α1+s2α1′)r1sdkdxksin⁡[s(x+1)r1]+1r1s∫-1xdkdxksin⁡[s(x-y)r1]q(y)φ1λ(y)dy,(3.2)dkdxkφ2λ(x)=γ1δ1φ1λ(-0)dkdxkcos⁡⁡[sxr2]+r2γ2δ2sφ1λ′(-0)dkdxksin⁡⁡[sxr2]+1r2s∫0xdkdxksin⁡⁡[s(x-y)r2]q(y)φ2λ(y)dy.Proof.
For proving it is enough substitutes2φ1λ(y)+r(y)φ1λ′′(y) and s2φ2λ(y)+r(y)φ2λ′′(y) instead of q(y)φ1λ(y) and q(y)φ2λ(y) in the integral terms of the (3.1) and (3.2), respectively, and integrate by parts twice.Lemma 3.2.
Letλ=s2, Ims=t. Then the functions φiλ(x) have the following asymptotic representations for |λ|→∞, which hold uniformly for x∈Ii (i=1,2):
(3.3)dkdxkφ1λ(x)=s2α2′dkdxkcos⁡[s(x+1)r1]+O(|s|k+1e|t|((x+1)/r1)),k=0,1,(3.4)dkdxkφ2λ(x)=s2α2′[γ1δ1cos⁡[sr1]dkdxkcos⁡[sxr2]-r2γ2δ2r1sin⁡[sr1]dkdxksin⁡[sxr2]]+O(|s|k+1e|t|((r1x+r2)/r1r2)),k=0,1,
if  α2′≠0,
(3.5)dkdxkφ1λ(x)=-sr1α1′dkdxksin⁡[s(x+1)r1]+O(|s|ke|t|((x+1)/r1)),k=0,1,(3.6)dkdxkφ2λ(x)=-sα1′[γ1r1δ1sin⁡[sr1]dkdxkcos⁡[sxr2]+r2γ2δ2cos⁡[sr1]dkdxksin⁡[sxr2]]+O(|s|ke|t|((r1x+r2)/r1r2)),k=0,1,
if  α2′=0.Proof.
Since the proof of the formulae forφ1λ(x) is identical to Titchmarshs proof of similar results for φλ(x) (see [27, Lemma  1.7 page 9-10]), we may formulate them without proving them here. Therefore we will prove only the formulas for φ2λ(x). Let α2≠0. Then according to (3.3)
(3.7)φ1λ(-0)=s2α2′cos⁡[sr1]+O(|s|e|t|/r1),φ1λ′(-0)=-s3α2′r1sin⁡[sr1]+O(|s|2e|t|/r1).
Substituting (3.7) into (3.2) (for k=0), we get
(3.8)φ2λ(x)=γ1s2α2′δ1cos⁡[sr1]cos⁡[sxr2]-r2s2α2′γ2δ2r1sin⁡[sr1]sin⁡[sxr2]+1r2s∫0xsin⁡[s(x-y)r2]q(y)φ2λ(y)dy+O(|s|e|t|((r1x+r2)/r1r2)).
Multiplying (3.8) by |s|-2e-|t|((r1x+r2)/r1r2) and denoting
(3.9)Fλ(x)=|s|-2e-|t|((r1x+r2)/r1r2)φ2λ(x)
we get
(3.10)Fλ(x)=|s|-2e-|t|((r1x+r2)/r1r2){s2α2′γ1δ1cos⁡[sr1]cos⁡[sxr2]-r2s2α2′γ2δ2r1sin⁡[sr1]sin⁡[sxr2]}+1r2s∫0xsin⁡[s(x-y)r2]q(y)e-|t|((x-y)/r2)Fλ(y)dy+O(|s|-1).
Denoting M(λ)≔max0≤x≤1|Fλ(x)| from the last formula, it follows that
(3.11)M(λ)≤|γ1||α2′||δ1|+r2|α2′||γ2||δ2|r1+M(λ)r2|s|∫01|q(y)|dy+M0|s|
for some M0>0. From this, it follows that M(λ)=𝒪(1) as λ→∞, so
(3.12)φ2λ(x)=O(|s|2e|t|((r1x+r2)/r1r2)).
Substituting (3.12) into the integral on the right of (3.8) yields (3.4) for k=0. The case k=1 of (3.4) follows by applying the same procedure as in the case k=0. The case α2′=0 is proved analogically.Lemma 3.3.
Letλ=s2, Ims=t. Then the characteristic function ω(λ) has the following asymptotic representations.
Case 1.
Ifβ2′≠0 and α2′≠0, then
(3.13)ω(λ)=α2′β2′s5[γ1r2δ1cos⁡[sr1]sin⁡[sr2]+γ2δ2r1sin⁡[sr1]cos⁡[sr2]]+O(|s|4e|t|((r1+r2)/r1r2)).
Case 2.
Ifβ2′≠0 and α2′=0, then
(3.14)ω(λ)=α1′β2′s4[-γ1r1r2δ1sin⁡[sr1]sin⁡[sr2]+γ2δ2cos⁡[sr1]cos⁡[sr2]]+O(|s|3e|t|((r1+r2)/r1r2)).
Case 3.
Ifβ2′=0 and α2′≠0, then
(3.15)ω(λ)=β1′α2′s4[γ1δ1cos⁡[sr1]cos⁡[sr2]-r2γ2δ2r1sin⁡[sr1]sin⁡[sr2]]+O(|s|3e|t|((r1+r2)/r1r2)).
Case 4.
Ifβ2′=0 and α2′=0, then
(3.16)ω(λ)=-β1′α1′s3[γ1r1δ1sin⁡[sr1]cos⁡[sr2]+r2γ2δ2cos⁡[sr1]sin⁡[sr2]]+O(|s|2e|t|((r1+r2)/r1r2)).Proof.
The proof is immediate by substituting (3.4) and (3.6) into the representation
(3.17)ω(λ)=(λβ1′+β1)φ2λ(1)-(λβ2′+β2)φ2λ′(1).Corollary 3.4.
The eigenvalues of the problem (2.1)–(2.4) are bounded below.Proof.
Puttings=it (t>0) in the above formulae, it follows that ω(-t2)→∞ as t→∞. Hence, ω(λ)≠0 for λ negative and sufficiently large.Now we can obtain the asymptotic approximation formula for the eigenvalues of the considered problem (2.1)–(2.4). Since the eigenvalues coincide with the zeros of the entire function ω(λ), it follows that they have no finite limit. Moreover, we know from Corollaries 2.2 and 3.4 that all eigenvalues are real and bounded below. Therefore, we may renumber them as λ0≤λ1≤λ2,…, listed according to their multiplicity.Theorem 3.5.
The eigenvaluesλn=sn2, n=0,1,2,…, of the problem (2.1)–(2.4) have the following asymptotic representation for n→∞, with γ1δ2r1-γ2δ1r2=0.
Case 1.
Ifβ2′≠0 and α2′≠0, then
(3.18)sn=r1r2r1+r2(n-1)π+O(n-1).
Case 2.
Ifβ2′≠0 and α2′=0, then
(3.19)sn=r1r2r1+r2(n-12)π+O(n-1).
Case 3.
Ifβ2′=0 and α2′≠0, then
(3.20)sn=r1r2r1+r2(n-12)π+O(n-1).
Case 4.
Ifβ2′=0 and α2′=0, then
(3.21)sn=r1r2r1+r2nπ+O(n-1).Proof.
We will only consider the first case. From (3.13) we have
(3.22)ω(λ)=α2′β2′γ2s5δ2r1sin⁡⁡[r1+r2r1r2s]+O(|s|4e|t|((r1+r2)/r1r2)).
We will apply the well-known Rouche theorem, which asserts that if f(λ) and g(λ) are analytic inside and on a closed contour C and |g(λ)|<|f(λ)| on C, then f(λ) and f(λ)+g(λ) have the same number of zeros inside C, provided that each zero is counted according to its multiplicity. It follows that ω(λ) has the same number of zeros inside the contour as the leading term in (3.22). If λ0≤λ1≤λ2,…, are the zeros of ω(λ) and λn=sn2, we have
(3.23)sn=r1r2r1+r2(n-1)π+δn,
for sufficiently large n, where |δn|<π/4, for sufficiently large n. By putting in (3.22) we have δn=𝒪(n-1), so the proof is completed for Case 1. The proof for the other cases is similar.Then from (3.3)–(3.6) (for k=0) and the above theorem, the asymptotic behavior of the eigenfunctions(3.24)φλn(x)={φ1λn(x),x∈[-1,0),φ2λn(x),x∈(0,1],
of (2.1)–(2.4) is given by, γ1δ2r1-γ2δ1r2=0,(3.25)φλn(x)={{α2′cos⁡⁡(r2(n-1)πr1+r2(x+1))+O(n-1),x∈[-1,0),γ1α2′δ1cos⁡⁡((n-1)πr1+r2(r1x+r2))+O(n-1),x∈(0,1],ifβ2′≠0,α2′≠0,{-r1α1′sin⁡⁡(r2(n-1/2)πr1+r2(x+1))+O(n-1),x∈[-1,0),-γ1r1α1′δ1sin⁡⁡((n-1/2)πr1+r2(r1x+r2))+O(n-1),x∈(0,1],ifβ2′≠0,α2′=0,{α2′cos⁡⁡(r2(n-1/2)πr1+r2(x+1))+O(n-1),x∈[-1,0),γ1α2′δ1cos⁡⁡((n-1/2)πr1+r2(r1x+r2))+O(n-1),x∈(0,1],ifβ2′=0,α2′≠0,{-r1α1′sin⁡⁡(r2nπr1+r2(x+1))+O(n-1),x∈[-1,0),-γ1r1α1′δ1sin⁡⁡(nπr1+r2(r1x+r2))+O(n-1),x∈(0,1],ifβ2′=0,α2′=0.
All these asymptotic formulae hold uniformly for x.


## 4. The Sampling Theorem

In this section we derive two sampling theorems associated with problem (2.1)–(2.4). For convenience we may assume that the eigenvectors of A are real valued.Theorem 4.1.
Consider the boundary value problem (2.1)–(2.4), and let
(4.1)φλ(x)={φ1λ(x),x∈[-1,0),φ2λ(x),x∈(0,1],
be the solution defined above. Let g(·)∈L2(-1,1) and
(4.2)F(λ)=1r12∫-10g(x)φ1λ(x)dx+1r22∫01g(x)φ2λ(x)dx.
Then F(λ) is an entire function of exponential type 2 that can be reconstructed from its values at the points {λn}n=0∞ via the sampling formula
(4.3)F(λ)=∑n=0∞F(λn)ω(λ)(λ-λn)ω′(λn).
The series (4.3) converges absolutely on ℂ and uniformly on compact subset of ℂ. Here ω(λ) is the entire function defined in (2.29).Proof.
Relation (4.2) can be rewritten as an inner product of H as follows
(4.4)F(λ)=〈g(⋅),Φλ(⋅)〉H=1r12∫-10g(x)φ1λ(x)dx+1r22∫01g(x)φ2λ(x)dx,
where
(4.5)g(x)=(g(x)00),Φλ(x)=(φλ(x)R-1′(ϕλ)R1′(ϕλ))∈H.
Both g(·) and Φλ(·) can be expanded in terms of the orthogonal basis on eigenfunctions, that is,
(4.6)g(x)=∑n=0∞ĝ(n)Φn(x)‖Φn(⋅)‖H2,Φλ(x)=∑n=0∞Φ̂λ(n)Φn(x)‖Φn(⋅)‖H2,
where ĝ(n) and Φ̂λ(n) are the fourier coefficients
(4.7)ĝ(n)=〈g(⋅),Φn(⋅)〉H=1r12∫-10g(x)φ1λn(x)dx+1r22∫01g(x)φ2λn(x)dx=F(λn).
Applying Parseval’s identity to (4.4) and using (4.7), we obtain
(4.8)F(λ)=∑n=0∞F(λn)〈Φn(⋅),Φλ(⋅)〉H‖Φn(⋅)‖H2.
Now we calculate Φ̂λ(n)=〈Φn(·),Φλ(·)〉H and ∥Φn(·)∥H. Let λ∈ℂ not be an eigenvalue and n∈ℕ. To prove (4.3) we need to show that
(4.9)〈Φn(⋅),Φλ(⋅)〉H‖Φn(⋅)‖H2=ω(λ)(λ-λn)ω′(λ),n=0,1,2,….
By the definition of the inner product of H, we have
(4.10)〈Φλ(⋅),Φn(⋅)〉H=1r12∫-10φ1λ(x)φ1λn(x)dx+1r22∫01φ2λ(x)φ2λn(x)dx+1ρR-1′(φλ)R-1′(φλn)+1γR1′(φλ)R1′(φλn).
Since
(4.11)1r12∫-10l(φ1λ(x))φ1λn(x)dx+1r22∫01l(φ2λ(x))φ2λn(x)dx=1r12∫-10φ1λ(x)l(φ1λn(x))dx+1r22∫01φ2λ(x)l(φ2λn(x))dx+W(φ1λ,φ1λn;-0)-W(φ1λ,φ1λn;-1)-W(φ2λ,φ2λn;+0)+W(φ2λ,φ2λn;1),
then, from (2.20) and (2.24), (4.11) becomes
(4.12)(λ-λn)[1r12∫-10φ1λ(x)φ1λn(x)dx+1r22∫01φ2λ(x)φ2λn(x)dx]=W(φ2λ,φ2λn;1)-(λ-λn)ρ.
Thus
(4.13)1r12∫-10φ1λ(x)φ1λn(x)dx+1r22∫01φ2λ(x)φ2λn(x)dx=W(φ2λ,φ2λn;1)λ-λn-ρ.
From (2.48), (2.22), and (2.8), the Wronskian of φ2λn and φ2λ at x=1 will be
(4.14)W(φ2λ,φ2λn;1)=φ2λ(1)φ2λn′(1)-φ2λ′(1)φ2λn(1)=kn-1[χ2λn′(1)φ2λ(1)-χ2λn(1)φ2λ′(1)]=kn-1[(β1′λn+β1)φ2λ(1)-(β2′λn+β2)φ2λ′(1)]=kn-1[ω(λ)+(λn-λ)R1′(φλ)].
Relations (2.48) and R1′(χλn)=γ and the linearity of the boundary conditions yield
(4.15)1γR1′(φλ)R1′(φλn)=kn-1γR1′(φλ)R1′(χλn)=kn-1R1′(φλ).
Substituting from (4.13), (4.14), (4.15), and R-1′(φλ)=R-1′(φλn)=-ρ into (4.10), we get
(4.16)〈Φλ(⋅),Φn(⋅)〉H=kn-1ω(λ)λ-λn.
Letting λ→λn in (4.16) and since the zeros of ω(λ) are simple, we have
(4.17)〈Φn(⋅),Φn(⋅)〉H=‖Φn(⋅)‖H2=kn-1ω′(λn).
Therefore from (4.16) and (4.17) we establish (4.9). Since λ and n are arbitrary, then (4.3) is proved with a pointwise convergence on ℂ, since the case λ=λn is trivial.
Now we investigate the convergence of (4.3). First we prove that it is absolutely convergent on ℂ. Using Cauchy-Schwarz’ inequality for λ∈ℂ,
(4.18)∑n=0∞|F(λn)ω(λ)(λ-λn)ω′(λn)|≤(∑n=0∞|〈g(⋅),Φn(⋅)〉H|2‖Φn(⋅)‖H2)1/2(∑n=0∞|〈Φn(⋅),Φλ(⋅)〉H|2‖Φn(⋅)‖H2)1/2.
Since g(·), Φλ(·)∈H, then both series in the right-hand side of (4.18) converge. Thus series (4.3) converges absolutely on ℂ. For uniform convergence let M⊂ℂ be compact. Let λ∈M and N>0. Define σN(λ) to be
(4.19)σN(λ)≔|F(λ)-∑n=0NF(λn)ω(λ)(λ-λn)ω′(λn)|.
Using the same method developed above
(4.20)σN(λ)≤(∑n=N+1∞|〈g(⋅),Φn(⋅)〉H|2‖Φn(⋅)‖H2)1/2(∑n=N+1∞|〈Φn(⋅),Φλ(⋅)〉H|2‖Φn(⋅)‖H2)1/2.
Therefore
(4.21)σN(λ)≤‖Φλ(⋅)‖H(∑n=N+1∞|〈g(⋅),Φn(⋅)〉H|2‖Φn(⋅)‖H2)1/2.
Since [-1,1]×M is compact, then, cf., for example, [31, page 225], we can find a positive constant CM such that
(4.22)‖Φλ(⋅)‖H≤CM,∀λ∈M.
Then
(4.23)σN(λ)≤CM(∑n=N+1∞|〈g(⋅),Φn(⋅)〉H|2‖Φn(⋅)‖H2)1/2
uniformly on M. In view of Parseval’s equality,
(4.24)(∑n=N+1∞|〈g(⋅),Φn(⋅)〉H|2‖Φn(⋅)‖H2)1/2⟶0asN⟶∞.
Thus σN(λ)→0 uniformly on M. Hence (4.3) converges uniformly on M. Thus F(λ) is analytic on compact subsets of ℂ and hence it is entire. From the relation
(4.25)|F(λ)|≤1r12∫-10|g(x)||φ1λ(x)|dx+1r22∫01|g(x)||φ2λ(x)|dx
and the fact that φ1λ(x) and φ2λ(x) are entire function of exponential type 2, we conclude that F(λ) is also of exponential type 2.Remark 4.2.
To see that expansion (4.3) is a Lagrange-type interpolation, we may replace ω(λ) by the canonical product
(4.26)ω̃(λ)={∏n=0∞(1-λλn),ifnoneoftheeigenvaluesiszero;λ∏n=1∞(1-λλn),ifoneoftheeigenvalues,sayλ0=0.
From Hadamard’s factorization theorem, see [4], ω(λ)=h(λ)ω̃(λ), where h(λ) is an entire function with no zeros. Thus,
(4.27)ω(λ)ω′(λn)=h(λ)ω̃(λ)h(λn)ω̃′(λn)
and (4.2), (4.3) remain valid for the function F(λ)/h(λ). Hence
(4.28)F(λ)=∑n=0∞F(λn)h(λ)ω̃(λ)h(λn)ω̃′(λn)(λ-λn).
We may redefine (4.2) by taking kernel φλ(·)/h(λ)=φ̃λ(·) to get
(4.29)F̃(λ)=F(λ)h(λ)=∑n=0∞F̃(λn)ω̃(λ)(λ-λn)ω̃′(λn).The next theorem is devoted to give interpolation sampling expansions associated with problem (2.1)–(2.4) for integral transforms whose kernels defined in terms of Green’s function. There are many results concerning the use of Green’s function in sampling theory, cf., for example, [22, 32–34]. As we see in (2.60), Green’s function G(x,ξ,λ) of problem (2.1)–(2.4) has simple poles at {λn}n=0∞. Define the function G(x,λ) to be G(x,λ)≔ω(λ)G(x,ξ0,λ), where ξ0∈[-1,0)∪(0,1] is a fixed point and ω(λ) is the function defined in (2.29) or it is the canonical product (4.26).Theorem 4.3.
Letg(·)∈L2(-1,1) and F(λ) the integral transform
(4.30)F(λ)=1r12∫-10G(x,λ)g¯(x)dx+1r22∫01G(x,λ)g¯(x)dx.
Then F(λ) is an entire function of exponential type 2 which admits the sampling representation
(4.31)F(λ)=∑n=0∞F(λn)ω(λ)(λ-λn)ω′(λn).
Series (4.31) converges absolutely on ℂ and uniformly on compact subsets of ℂ.Proof.
The integral transform (4.30) can be written as
(4.32)F(λ)=〈G(⋅,λ),g(⋅)〉H,(4.33)g(x)=(g(x)00),G(x,λ)=(G(x,λ)R-1′(G(x,λ))R1′(G(x,λ)))∈H.
Applying Parseval’s identity to (4.32) with respect to {Φn(·)}n=0∞, we obtain
(4.34)F(λ)=∑n=0∞〈G(⋅,λ),Φn(⋅)〉H〈g(⋅),Φn(⋅)〉¯H‖Φn(⋅)‖H2.
Let λ≠λn. Since each Φn(·) is an eigenvector of A, then
(4.35)(λI-A)Φn(x)=(λ-λn)Φn(x).
Thus
(4.36)(λI-A)-1Φn(x)=1λ-λnΦn(x).
From (2.59) and (4.36) we obtain
(4.37)R1′(φλn)ω(λ)φλ(ξ0)-R-1′(φλn)ω(λ)χλ(ξ0)+1r12∫-10G(x,ξ0,λ)φ1λn(x)dx+1r22∫01G(x,ξ0,λ)φ2λn(x)dx=1λ-λnφλn(ξ0).
Using R-1′(φλn)=-ρ, (2.48), and R1′(χλn)=γ, (4.37) becomes
(4.38)γkn-1ω(λ)φλ(ξ0)+ρω(λ)χλ(ξ0)+1r12∫-10G(x,ξ0,λ)φ1λn(x)dx+1r22∫01G(x,ξ0,λ)φ2λn(x)dx=1λ-λnφλn(ξ0).
Hence (4.38) can be rewritten as
(4.39)γkn-1φλ(ξ0)+ρχλ(ξ0)+1r12∫-10G(x,λ)φ1λn(x)dx+1r22∫01G(x,λ)φ2λn(x)dx=ω(λ)λ-λnφλn(ξ0).
From the definition of G(·,λ), we have
(4.40)〈G(⋅,λ),Φn(⋅)〉H=1r12∫-10G(x,λ)φ1λn(x)dx+1r22∫01G(x,λ)φ2λn(x)dx+1ρR-1′(G(x,λ))R-1′(φλn)+1γR1′(G(x,λ))R1′(φλn).
From formula (2.60), we get
(4.41)R-1′(G(x,λ))=χλ(ξ0)R-1′(φλ),R1′(G(x,λ))=φλ(ξ0)R1′(χλ).
Combining (4.41), R-1′(φλ)=R-1′(φλn)=-ρ, R1′(χλ)=R1′(χλn)=γ, and (2.48) together with (4.40) yields
(4.42)〈G(⋅,λ),Φn(⋅)〉H=1r12∫-10G(x,λ)φ1λn(x)dx+1r22∫01G(x,λ)φ2λn(x)dx+γkn-1φλ(ξ0)+ρχλ(ξ0).
Substituting from (4.39) and (4.42) gives
(4.43)〈G(⋅,λ),Φn(⋅)〉H=ω(λ)λ-λnφλn(ξ0).
As an element of H, G(·,λ) has the eigenvectors expansion
(4.44)G(x,λ)=∑i=0∞〈G(⋅,λ),Φi(⋅)〉HΦi(x)‖Φi(⋅)‖H2=∑i=0∞ω(λ)(λ-λi)φλi(ξ0)Φi(x)‖Φi(⋅)‖H2.
Taking the limit when λ→λn in (4.32), we get
(4.45)F(λn)=lim⁡λ→λn〈G(⋅,λ),g(⋅)〉H.
The interchange of the limit and summation processes is justified by the uniform convergence of the eigenvector expansion of G(x,λ) on [-1,1] on compact subsets of ℂ, cf., (2.60), (3.3)–(3.6), and (3.18)–(3.21). Making use of (4.44), we may rewrite (4.45) as
(4.46)F(λn)=lim⁡λ→λn∑i=0∞ω(λ)(λ-λi)φλi(ξ0)〈Φi(⋅),g(⋅)〉H‖Φi(⋅)‖H2=ω′(λn)φλn(ξ0)〈Φn(⋅),g(⋅)〉H‖Φn(⋅)‖H2.
The interchange of the limit and summation is justified by the asymptotic behavior of Φi(x) and ω(λ). If φλn(ξ0)≠0, then (4.46) gives
(4.47)〈g(⋅),Φn(⋅)〉¯H‖Φn(⋅)‖H2=F(λn)ω′(λn)φλn(ξ0).
Combining (4.43), (4.47), and (4.34) we get (4.31) under the assumption that φλn(ξ0)≠0 for all n. If φλn(ξ0)=0, for some n, the same expansion holds with F(λn)=0. The convergence properties as well as the analytic and growth properties can be established as in Theorem 4.1.Now, we give an example exhibiting the obtained results.Example 4.4.
The boundary value problem,(4.48)-y′′(x)+q(x)y(x)=λy(x),x∈[-1,0)∪(0,1],y′(-1)=-λy(-1),y′(1)=λy(1),2y(-0)-y(+0)=0,y′(-0)-2y′(+0)=0,
is special case of the problem (2.1)–(2.4) when α1=α2′=β1=β2′=0, β2=β1′=α2=α1′=r1=r2=1, γ1=δ2=2,γ2=δ1=1, and
(4.49)q(x)={-1,x∈[-1,0),-2,x∈(0,1].
Then ρ=γ=1>0. The solutions φλ(·) and χλ(·) are
(4.50)φλ(x)={φ1λ(x)=(2ζ1λ2-ζ2λ2)sin⁡⁡ζ1λ(x+1)ζ1λ-cos⁡ζ1λ(x+1),x∈[-1,0),φ2λ(x)=2((2ζ1λ2-ζ2λ2)sin⁡⁡ζ1λζ1λ-cos⁡⁡ζ1λ)cos⁡⁡ζ2λx+(2ζ1λ2-ζ2λ2)cos⁡⁡ζ1λ+ζ1λsin⁡⁡ζ1λ2ζ2λsin⁡⁡ζ2λx,x∈(0,1],χλ(x)={χ1λ(x)=12(cos⁡⁡ζ2λ-2ζ1λ2-ζ2λ2ζ2λsin⁡⁡ζ2λ)cos⁡⁡ζ1λx+2(ζ2λζ1λsin⁡⁡ζ2λ+2ζ1λ2-ζ2λ2ζ1λcos⁡⁡ζ2λ)sin⁡⁡ζ1λx,x∈[-1,0),χ2λ(x)=cos⁡ζ2λ(x-1)+2ζ1λ2-ζ2λ2ζ2λsin⁡ζ2λ(x-1),x∈(0,1],
where ζ1λ:=λ+1 and ζ2λ:=λ+2.Here the characteristic function is
(4.51)ω(λ)=12ζ1λζ2λ{(2ζ1λ2-ζ2λ2)2ζ1λcos⁡⁡ζ1λ(-5(2ζ1λ2-ζ2λ2)ζ2λcos⁡⁡ζ2λ+(ζ2λ4-8ζ1λ2-4)sin⁡⁡ζ2λ)+sin⁡⁡ζ1λ([-ζ1λ2+4(2ζ1λ2-ζ2λ2)2ζ2λ]cos⁡⁡ζ2λ+(2ζ1λ2-ζ2λ2)(4+5ζ2λ1)sin⁡⁡ζ2λ)}.
By Theorem 4.1, the transform
(4.52)F(λ)=∫-10g(x)[(2ζ1λ2-ζ2λ2)sin⁡⁡ζ1λ(x+1)ζ1λ-cos⁡ζ1λ(x+1)]dx+∫01g(x)[2((2ζ1λ2-ζ2λ2)sin⁡⁡ζ1λζ1λ-cos⁡ζ1λ)cos⁡ζ2λx+(2ζ1λ2-ζ2λ2)cos⁡⁡ζ1λ+ζ1λsin⁡⁡ζ1λ2ζ2λsin⁡⁡ζ2λx]dx
has the following expansion:
(4.53)F(λ)=∑n=0∞F(λn)ω(λ)((2ζ1λ2-ζ2λ2)-(2ζ1λn2-ζ2λn2))ω′(λn),(4.54)ω′(λn)≔14ζ1λn3ζ2λn3[sin⁡⁡ζ1λn(2ζ2λn(-1+24λ+34λ2+11λ11)cos⁡⁡ζ2λn+(54+104λ+59λ2-5λ4)sin⁡⁡ζ2λn)+ζ1λncos⁡⁡ζ1λn(ζ2λn(-30-45λ-6λ2+5λ3)cos⁡⁡ζ2λn+(-8+24λ+41λ2+13λ3)sin⁡⁡ζ2λn)].
The Green’s function has the following form:
(4.55)G(x,ξ,λ)=12ζ1λζ2λ{(2ζ1λ2-ζ2λ2)2ζ1λcos⁡ζ1λ(-5(2ζ1λ2-ζ2λ2)ζ2λcos⁡ζ2λ+(ζ2λ4-8ζ1λ2-4)sin⁡ζ2λ)+sin⁡⁡ζ1λ([-ζ1λ2+4(2ζ1λ2-ζ2λ2)2ζ2λ]cos⁡⁡ζ2λ+(2ζ1λ2-ζ2λ2)(4+5ζ2λ1)sin⁡⁡ζ2λ)}×{((2ζ1λ2-ζ2λ2)sin⁡⁡ζ1λ(x+1)ζ1λ-cos⁡⁡ζ1λ(x+1))×[12(cos⁡ζ2λ-2ζ1λ2-ζ2λ2ζ2λsin⁡ζ2λ)cos⁡ζ1λξ+2(ζ2λζ1λsin⁡ζ2λ+2ζ1λ2-ζ2λ2ζ1λcos⁡ζ2λ)sin⁡ζ1λξ],-1≤x≤ξ<0,((2ζ1λ2-ζ2λ2)sin⁡ζ1λ(ξ+1)ζ1λ-cos⁡ζ1λ(ξ+1))×[12(cos⁡ζ2λ-2ζ1λ2-ζ2λ2ζ2λsin⁡ζ2λ)cos⁡ζ1λx+2(ζ2λζ1λsin⁡ζ2λ+2ζ1λ2-ζ2λ2ζ1λcos⁡ζ2λ)sin⁡ζ1λx],-1≤ξ≤x<0,((2ζ1λ2-ζ2λ2)sin⁡ζ1λ(ξ+1)ζ1λ-cos⁡ζ1λ(ξ+1))[cos⁡ζ2λ(x-1)+2ζ1λ2-ζ2λ2ζ2λsin⁡ζ2λ(x-1)],-1≤ξ<0,0<x≤1,((2ζ1λ2-ζ2λ2)sin⁡ζ1λ(x+1)ζ1λ-cos⁡ζ1λ(x+1))[cos⁡ζ2λ(ξ-1)+2ζ1λ2-ζ2λ2ζ2λsin⁡ζ2λ(ξ-1)],-1≤x<,0<ξ≤1,[cos⁡ζ2λ(x-1)+2ζ1λ2-ζ2λ2ζ2λsin⁡ζ2λ(x-1)]×[2((2ζ1λ2-ζ2λ2)sin⁡ζ1λζ1λ-cos⁡ζ1λ)cos⁡ζ2λξ+(2ζ1λ2-ζ2λ2)cos⁡ζ1λ+ζ1λsin⁡ζ1λ2ζ2λsin⁡ζ2λξ],0<ξ≤x≤1,[cos⁡ζ2λ(ξ-1)+2ζ1λ2-ζ2λ2ζ2λsin⁡ζ2λ(ξ-1)]×[2((2ζ1λ2-ζ2λ2)sin⁡ζ1λζ1λ-cos⁡ζ1λ)cos⁡ζ2λx+(2ζ1λ2-ζ2λ2)cos⁡ζ1λ+ζ1λsin⁡ζ1λ2ζ2λsin⁡ζ2λx],0<x≤ξ≤1.
Taking ξ∈[-1,0)∪(0,1], the transform
(4.56)F(λ)=∫-10G(x,λ)g¯(x)dx+4∫01G(x,λ)g¯(x)dx.
has a sampling representation of the type.



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*Source: 610232-2011-10-31.xml*