# A Singular Sturm-Liouville Problem with Limit Circle Endpoints and Eigenparameter Dependent Boundary Conditions

Authors: Jinming Cai; Zhaowen Zheng
Journal: Discrete Dynamics in Nature and Society
 (2017)

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## Abstract

In this paper, we investigate a class of discontinuous singular Sturm-Liouville problems with limit circle endpoints and eigenparameter dependent boundary conditions. Operator formulation is constructed and asymptotic formulas for eigenvalues and fundamental solutions are given. Moreover, the completeness of eigenfunctions is discussed.


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## Body

## 1. Introduction

It is well known that many topics in mathematical physics require the investigation of eigenvalues and eigenfunctions of the Sturm-Liouville problems. The theory of regular Sturm-Liouville problems is well built; since the foundation work of Weyl on limit-point/limit-circle classification [1], the singular Sturm-Liouville problems (see [2–7] for real coefficients and [8] for complex coefficients) and more general Hamiltonian systems (see [9, 10]) are widely researched. Meanwhile, a large number of researchers are interested in the discontinuous Sturm-Liouville problem with inner discontinuous points, since these problems are of wide applications in engineering and mechanics (see [11–25]). Various physics applications of this kind of problems are found, such as oscillation of linear or nonlinear equation (see [26–29]) and heat and mass transfer problems (see [30]).The regular Sturm-Liouville problems with transmission conditions containing an eigenparameter on one of the boundary conditions have received a lot of attention in research (see [18–22]). Based on these results, some researchers studied the regular Sturm-Liouville problems with eigenparameter on both of the boundary conditions (see [23–25]). In these papers, Yang and Wang in [18] considered a Sturm-Liouville problem with discontinuities at two points and eigenparameter dependent boundary condition at one endpoint; they obtained the fundamental solutions and gave the asymptotic formulas of eigenvalues and fundamental solutions. Further, they studied the discontinuous Sturm-Liouville problem with eigenparameter boundary conditions at two endpoints in [24] and extended the results of [18] to finite discontinuities case. In papers [19, 22, 23, 25], the authors obtained the estimations of eigenvalues and eigenfunctions of the discontinuous Sturm-Liouville problem with one inner point, containing an eigenparameter in the boundary condition. Şen et al. considered the Sturm-Liouville problem with two inner points containing an eigenparameter in the boundary condition and got similar result, respectively (see [20, 21]). Besides, the authors also discussed the completeness of the eigenfunctions of a regular discontinuous Sturm-Liouville problem in papers [18, 25]. All of them researched the regular Sturm-Liouville problem. However, little is known about the singular Sturm-Liouville problems with limit-circle endpoints.We will consider the following singular discontinuous Sturm-Liouville problem with two limit-circle endpoints and eigenparameter in the boundary conditions:(1)Ly≔-pxy′x′+qxyx=λyx,x∈a,ξ∪ξ,b,-∞<a<ξ<b<∞,with eigenparameter dependent conditions at the endpoints a and b:(2)L1y≔a1y,ua-a2y,va=λb1y,ua-b2y,va,(3)L2y≔c1y,ub-c2y,vb=λd1y,ub-d2y,vb;the transmission conditions at x=ξ are(4)L3y≔yξ+0-δ1yξ-0-δ2y′ξ-0=0,L4y≔y′ξ+0-δ3yξ-0-δ4y′ξ-0=0,(we assume that not all ξj(1≤j≤4) are equal to zero, since, in this case, the boundary value problem (1)–(3) has no transmission conditions), where both a and b are limit-circle points, u1(x) and v1(x) are linearly independent real-valued solutions of equation -pxy′x′+q(x)y(x)=0 on (a,ξ), and u2(x) and v2(x) are linearly independent real-valued solutions of equation -pxy′x′+q(x)y(x)=0 on (ξ,b) and satisfy [u1,v1](a)≠0, [u2,v2](b)≠0, where [u,v]=p(uv′¯-u′v¯) is the sesquilinear form; p(x)=1/p12 for x∈(a,ξ), p(x)=1/p22 for x∈(ξ,b), and λ∈C is a spectral parameter; q(x) is a real-valued continuous function on (a,ξ)∪(ξ,b) and has finite limits q(ξ±0)=limx→(ξ±0)q(x); pi,ai,bi,ci,di(i=1,2), and δi(i=1,2,3,4) are nonzero real numbers.For the convenience, we set(5)u=u1,x∈a,ξ,u2,x∈ξ,b,v=v1,x∈a,ξ,v2,x∈ξ,b.Moreover, we assume that (6)ρ=b1a1b2a2>0,γ=δ1δ3δ2δ4>0,θ=c1c2d1d2>0.Here, we research a singular Sturm-Liouville problem with two limit-circle endpoints and the parameterλ is not only in the equation but also in the boundary conditions. Based on the modified inner product, we define a new self-adjoint operator A such that the eigenvalues of such a problem are coincided with those of A. We rebuild its fundamental solutions, get the asymptotic formulas for eigenvalues and eigenfunctions, and also discuss the completeness of its eigenfunctions.At first, we introduce the following lemmas.Lemma 1 (the Lagrange identity, see [5]).
LetD(L)={y∣y∈L2[a,b],y′∈AC[a,b],Ly∈L2[a,b]}, where, for any interval I⊂R, L2(I) denotes the set of functions on I which are square integrable on I. For any y∈D(L), z∈D(L∗), one has (7)z¯Ly-yL∗z¯=ddxy,zx,where L∗ is the adjoint expression of L, which is given by L∗z=-p¯z′′+q¯z.Lemma 2 (Green’s formula, see [5]).
LetI=[a,b]; for arbitrary y∈D(L),z∈D(L∗), D(L) and D(L∗) are defined as above; one has (8)Ly,z-y,L∗z=∫abz¯Ly-yL∗z¯dx=y,zb-y,za. Since Ly is of limit-circle type at x=a, ∫ax1u¯Lydx,∫ax1yL∗u¯dx and [y,u](x1) exist ∀x∈(a,x1). So [y,u](a)=limx→a+⁡[y,u](x) exists. Similarly [y,v](a),[y,u](b) and [y,v](b) exist.Lemma 3 (see [5]).
For anyf,g,α,β∈D(L), while D(L) is defined in Lemma 1, one has (9)f,αf,βg,αg,β=f,g¯α¯,β.


## 2. Operator Formulation

In this section, we introduce the Hilbert spaceH=L2(a,ξ)⊕L2(ξ,b)⊕C2; the inner product on H is defined by (10)F,G=γp12∫aξfxgx¯dx+p22∫ξbfxgx¯dx+γρf1g1¯+1θf2g2¯,for F=(f,f1,f2),G=(g,g1,g2)∈H,f,g∈L2((a,ξ)∪(ξ,b)). For the convenience, we use the following notations: H1=L2(a,ξ)⊕L2(ξ,b), (11)Tay=b1y,ua-b2y,va,Ta′y=a1y,ua-a2y,va,Tby=d1y,ub-d2y,vb,Tb′y=c1y,ub-c2y,vb. Besides, we introduce the operator A in the Hilbert space H as follows: (12)DA=F=fx,f1,f2∈H∣fx,f′xareabsolutelycontinuousona,ξ∪ξ,bwithfinitelimitsfξ±0,f′ξ±0andsatisfyLf∈L2a,ξ∪ξ,b,L3f=L4f=0,f1=Taf,f2=Tbf,which acts by the rule(13)AF=Lf,Ta′f,Tb′fwith F=(f,Ta(f),Tb(f))∈D(A). Now we can rewrite problem (1)–(4) in the operator form AF=λF, for F=(f,Ta(f),Tb(f))∈D(A). Obviously, we have the following lemmas.Lemma 4.
The eigenvalues and eigenfunctions of problem (1)–(4) are corresponding to the eigenvalues and the first component of the corresponding eigenfunctions of operator A, respectively.Lemma 5.
The domainD(A) is dense in H.Proof.
LetF=(f(x),f1,f2)∈H,F⊥D(A) and let C0∞ be a functional set which has compact support and can be differential infinitely such that ϑ1(x)∈C0∞(a,ξ),ϑ2(x)∈C0∞(ξ,b). Since C0∞⊕0⊕0⊂D(A), for ∀G=(g(x),0,0)∈C0∞⊕0⊕0 is orthogonal to F, namely, (14)F,G=γp12∫aξfxgx¯dx+p22∫ξbfxgx¯dx=0, so f(x)=0. For all V=(ν(x),ν1,0)∈D(A),〈F,V〉=(γ/ρ)f1ν1¯=0, f1=0, since ν1 is arbitrary. Similarly, one gets f2=0. Above all, one has F=(0,0,0) which proves the assertion.Definition 6.
LetW(f,g;x)=fxg′(x)-f′(x)g(x) denote the Wronskian of functions f(x) and g(x). One has (15)Wf,g;a=p12f,g¯a,Wf,g;b=p22f,g¯b.Theorem 7.
OperatorA is symmetric.Proof.
For eachF,G∈D(A), by the definition of inner product and operator A, we can get(16)AF,G-F,AG=γWf,g¯;ξ-0-Wf,g¯;ξ+0+Wf,g¯;b-γWf,g¯;a+γρTa′fTag¯-TafTa′g¯+1θTb′fTbg¯-TbfTb′g¯;then (4) implies that(17)Wf,g¯;ξ+0=γWf,g¯;ξ-0;using boundary condition (2), we get(18)γρTa′fTag¯-TafTa′g¯=γWf,g¯;a,and, by boundary condition (3), we have(19)1θTb′fTbg¯-TbfTb′g¯=Wf,g¯;b.Substituting (17)–(19) into (16) yields 〈AF,G〉-〈F,AG〉=0. This completes the proof.Moreover, we have the following conclusion.Theorem 8.
OperatorA is self-adjoint.Proof.
A is self-adjoint if and only if, for each F=(f(x),Ta(f),Tb(f))∈D(A), 〈AF,W〉=〈F,U〉 for some U∈H implying that W∈D(A) and AW=U, where W=(w(x),h,r),U=(μ(x),k,s). Concretely, we should prove that the following properties hold:(i)
w(x),w′(x) are absolutely continuous on (a,ξ)∪(ξ,b),Lw∈L2((a,ξ)∪(ξ,b)).(ii)
h=b1[w,u](a)-b2[w,v](a), r=d1[w,u](b)-d2[w,v](b).(iii)
Liw=0,i=3,4.(iv)
μ(x)=Lw.(v)
k=a1[w,u](a)-a2[w,v](a); s=c1[w,u](b)-c2[w,v](b).
For an arbitrary pointF∈C0∞⊕0⊕0⊂D(A), we have (20)γp12∫aξLfxwx¯dx+p22∫ξbLfxwx¯dx=γp12∫aξfxμx¯dx+p22∫ξbfxμx¯dx;that is, 〈Lf,w〉1=〈f,μ〉1. According to the classical Sturm-Liouville theory, (i) and (iv) hold. So 〈AF,W〉=〈F,U〉,∀F∈D(A).(21)Lf,w1+γρTa′fh¯+1θTb′fr¯=f,Lw1+γρTafk¯+1θTbfs¯.Besides, we have(22)Lf,w1=f,Lw1+γWf,w¯;ξ-0-Wf,w¯;ξ+0+Wf,w¯;b-γWf,w¯;a.Substituting (22) into (21), we get(23)γρTafk¯-Ta′fh¯+1θTbfs¯-Tb′fr¯=γWf,w¯;ξ-0-Wf,w¯;ξ+0+Wf,w¯;b-γWf,w¯;a.By Naimark’s Patching Lemma [4], there exists F∈D(A) such that(24)f,ub=f,vb=fξ±0=f′ξ±0=0,f,ua=b2,f,va=b1.Substituting (24) into (23), we have h=b1[w,u](a)-b2[w,v](a). Further, there exists F∈D(A) such that(25)f,ua=f,va=fξ±0=f′ξ±0=0,f,ub=d2,f,vb=d1.Analogously, we can get r=d1[w,u](b)-d2[w,v](b). So (ii) holds. Similarly, one proves (v). Next, let F∈D(A) such that(26)f,ua=f,va=f,ub=f,vb=fξ+0=0,fξ-0=-δ2,f′ξ-0=δ1,f′ξ+0=γ.Then, by (23), we have L3w=0. Similarly, we can get L4w=0. So A is a self-adjoint operator.From the properties of self-adjoint operators, we have the following corollaries.Corollary 9.
All eigenvalues of the singular Sturm-Liouville problem (1)–(4) are real.Corollary 10.
Letλ1 and λ2 be two different eigenvalues of the singular Sturm-Liouville problem (1)–(4); then the corresponding eigenfunctions f(x) and g(x) are orthogonal in the sense of (27)γp12∫aξfxgx¯dx+p22∫ξbfxgx¯dx+γρf1g1¯+1θf2g2¯=0.


## 3. Asymptotic Approximation of Fundamental Solutions

In this section, we construct the fundamental solutions of problem (1)–(4) and get the asymptotic approximation for fundamental solutions.Lemma 11 (see [5]).
Let the real-valued functionq(x) be continuous on I=(a,ξ)∪(ξ,b), and let f(λ),g(λ) be given entire functions. Then, for any λ∈C, the equation (28)Ly≔-pxy′x′+qxyx=λyx,x∈a,ξ∪ξ,bhas unique solution y=y(x,λ) satisfying the initial conditions (29)ya,λ=fλ,y′a,λ=gλ.For each fixed x∈(a,ξ)∪(ξ,b),y(x,λ) is an entire function of λ.Here, we define fundamental solutionsφ(x,λ) and χ(x,λ) of (1) by the following procedure: (30)φx,λ=φ1x,λ,x∈a,ξ,φ2x,λ,x∈ξ,b,χx,λ=χ1x,λ,x∈a,ξ,χ2x,λ,x∈ξ,b.Letφ1(x,λ) be the solution of (1) on the interval (a,ξ), which satisfies the initial conditions(31)φ1a,λ≔y,ua,λ=a2-λb2,φ1′a,λ≔y,va,λ=a1-λb1;by virtue of Lemma 11, we can define the solution φ2(x,λ) of (1) on (ξ,b) by the initial conditions(32)φ2ξ+0φ2′ξ+0=δ1φ1ξ-0+δ2φ1′ξ-0δ3φ1ξ-0+δ4φ1′ξ-0.Analogously, we define the solutions χ2(x,λ) and χ1(x,λ) of (1) by the initial conditions(33)χ2b,λ≔y,ub,λ=c2-λd2,χ2′b,λ≔y,vb,λ=c1-λd1,χ1ξ-0χ1′ξ-0=δ4χ2ξ+0-δ2χ2′ξ+0γδ1χ2′ξ+0-δ3χ2ξ+0γ.Now, we consider Wronskian(34)Wiλ≔Wφi,χi;x=φix,λχi′x,λ-φi′x,λχix,λi=1,2. By the dependence of solutions of initial value problems on the parameter, one has that Wiλ(i=1,2) are entire functions of λ and are independent of x.Lemma 12.
For everyλ∈C, W2(λ)=γW1(λ).Proof.
By the definition ofWi(λ), we have (35)W1λ=φ1ξ-0,λχ1′ξ-0,λ-φ1′ξ-0,λχ1ξ-0,λ,W2λ=φ2ξ+0,λχ2′ξ+0,λ-φ2′ξ+0,λχ2ξ+0,λ; using the transmission conditions (4), simple computation gives(36)Wφ2,χ2;ξ+0=γWφ1,χ1;ξ-0. Thus, for each λ∈C, we have W2(λ)=γW1(λ). This completes the proof.Besides, we setW(λ)≔W1(λ)=(1/γ)w2(λ).Theorem 13.
The eigenvalues of problem (1)–(4) coincide with the zeros of the function W(λ).Proof.
Letν0(x,λ0) be any eigenfunction corresponding to eigenvalue λ0; then the function ν0(x,λ0) may be represented in the form (37)ν0x,λ0=m1φ1x,λ0+m2χ1x,λ0,x∈a,ξ,m3φ2x,λ0+m4χ2x,λ0,x∈ξ,b,where at least one of the constants mi(i=1,2,3,4) is not zero. We should show that W(λ0)=0. Suppose to the contrary that there exists λ0∈R such that W(λ0)=W1(λ0)=(1/γ)W2(λ0)≠0. Since eigenfunction ν0(x,λ0) satisfies both boundary and transmission conditions (2)–(4), we have Liν0(x,λ0)=0(i=1,2,3,4), while the determinant of coefficient matrix is not zero, so we obtain mi=0(i=1,2,3,4), which is a contradiction; then W(λ0)=0. Conversely, let λ=λ0 be a zero of function W(λ); then W(λ0)=W1(λ0)=(1/γ)W2(λ0)=0; therefore, χi(x,λ0)=kφi(x,λ0)(i=1,2), for some k≠0. Since both φ2(x,λ0) and χ2(x,λ0) satisfy the boundary condition (3),(38)φx,λ0=φ1x,λ0,x∈a,ξ,φ2x,λ0,x∈ξ,bsatisfies problem (1)–(4). So function φ(x,λ0) is an eigenfunction of problem (1)–(4) corresponding to eigenvalue λ0. This completes the proof.Theorem 14.
The eigenvalues of problem (1)–(4) are analytically single.Proof.
Letλ=s+it, and we use the following notions for simplicity: φ=φ(x,λ), φ1λ=∂φ1/∂λ, and φ1λ′=∂φ1′/∂λ. We differentiate the equation Aχ=λχ with respect to λ to obtain(39)Aχλ=χ+λχλ.Using integration by parts, we get(40)Aχλ,φ1-χλ,Aφ1=γχ1λφ1′¯-χ1λ′φ1¯aξ+χ2λφ2′¯-χ2λ′φ2¯ξb.Substituting (39) and Aφ=λφ into the left side of (40), we have (41)λχλ,φ1-χλ,λφ1=χ,φ1+2itχλ,φ1. Moreover,(42)γχ1λφ1′¯-χ1λ′φ1¯aξ+χ2λφ2′¯-χ2λ′φ2¯ξb=d1φ2¯b-d2φ2′¯b-γa1-λb1χ1λa-a2-λb2χ1λ′a. By (31), we observe that(43)W′λ=φ1a,λχ1′a,λ-φ1′a,λχ1a,λ=b1χ1a-b2χ1′a+a2-λb2χ1λ′a-a1-λb1χ1λa; then (40) becomes(44)γW′λ=χ,φ1+2itχλ,φ1+γb1χ1a-b2χ1′a-d1φ¯2b-d2φ¯2′b.Next, let μ0 be an arbitrary zero of W(λ). Since W(μ0)=0, we obtain φi(x,μ0)=kχi(x,μ0)(i=1,2)(k≠0), k∈R. Noting that μ0 is real, a short calculation (44) becomes (45)γW′μ0=kγp12∫aξχ1x2dx+p22∫ξbχ2x2dx+ρk+θk.Since ρ>0, θ>0, γ>0, and k≠0, W′(μ0)≠0. Hence, the analytic multiplicity of μ0 is one, which completes the proof.Lemma 15.
Letλ=s2,s=σ+it. Then the following integral and differential equations hold for k=0 and k=1:(46)dkdxkφ1x,λ=a2-s2b2dkdxkcos⁡p1sx-a-a1-s2b1p1sdkdxksin⁡p1sx-a+p1s∫axdkdxksin⁡p1sx-τqτφ1τdτ,(47)dkdxkφ2x,λ=φ2ξ+0dkdxkcos⁡p2sx-ξ+1p2sφ2′ξ+0dkdxksin⁡p2sx-ξ+p2s∫ξxdkdxksin⁡p2sx-τqτφ2τdτ.Proof.
For the case ofk=0, consider φ1(x,λ) as the solution of the following nonhomogeneous problem:(48)-pxyx′′+λyx=qxyxφ1a,λ=a2-λb2,φ1′a,λ=a1-λb1.Using the method of constant variation, φ1(x,λ) satisfies(49)φ1x,λ=a2-s2b2cos⁡p1sx-a-a1-s2b1p1sdkdxksin⁡p1sx-a+p1s∫axsin⁡p1sx-τqτφ1τdτ. Then, differentiating it with respect to x, we have (46). The proof for (47) is similar, so we omit the details.Similarly, we have the following theorem.Lemma 16.
Letλ=s2,s=σ+it. Then the following integral and differential equations hold for k=0 and k=1:(50)dkdxkχ1x,λ=δ4χ2ξ+0-δ2χ2′ξ+0γdkdxkcos⁡p1sx-ξ+δ1χ2′ξ+0-δ3χ2ξ+0p1sγdkdxksin⁡p1sx-ξ-p1s∫xξdkdxksin⁡p1sx-τqτχ1τdτ,dkdxkχ2x,λ=c2-s2d2dkdxkcos⁡p2sx-b+c1-s2d1p2sdkdxksin⁡p2sx-b-p2s∫xbdkdxksin⁡p2sx-τqτχ2τdτ.Lemma 17.
Letλ=s2,s=σ+it. Then, for k=0,1, φ(x,λ) have the following estimations.
Case 1. If b2≠0, then(51)dkdxkφ1x,λ=-s2b2dkdxkcos⁡p1sx-a+Osk+1etp1x-a,dkdxkφ2x,λ=p1δ2s3b2sin⁡p1sξ-adkdxkcos⁡p2sx-a+Osk+2etp1ξ-a+p2x-a.
Case 2. If b2=0, then(52)dkdxkφ1x,λ=-b1sp1dkdxksin⁡p1sx-a+Osketp1x-a,dkdxkφ2x,λ=-δ2b1s2cos⁡p1sξ-adkdxkcos⁡p2sx-ξ+Osk+1etp1ξ-a+p2x-ξ.
Each of these asymptotic equalities holds uniformly forx as |λ|→∞.Proof.
The proof of formulas forφ1(x,λ) is identical to those of Titchmarsh’s proof for φ(x,λ) (see [6]), so we only give the proof of formulas for φ2(x,λ), namely, equality (51); the other equalities are similar.
Fork=0, by the estimations of φ1(x,λ) and φ1′(x,λ), we have(53)φ1ξ-0,λ=-s2b2cos⁡p1sξ-a+Osetp1ξ-a,φ1′ξ-0,λ=p1s3b2sin⁡p1sξ-a+Os2etp1ξ-a; then, substituting (53) into (47) and observing (32), we have(54)φ2x,λ=p1δ2s3b2sin⁡p1sξ-acos⁡p2sx-a+Os2etp1ξ-a+p2x-ξ; differentiating (54) with respect to x, we have (51).Lemma 18.
Letλ=s2,s=σ+it. Then, for k=0,1, χ(x,λ) have the following estimations.
Case 1. If d2≠0, then(55)dkdxkχ1x,λ=-p2d2δ2s3γsin⁡p2sξ-bdkdxkcos⁡p1sx-ξ+Osk+2etp1x-ξ+p2ξ-b,dkdxkχ2x,λ=-d2s2dkdxkcos⁡p2sx-b+Osk+1etp2x-b.
Case 2. If d2=0, then(56)dkdxkχ1x,λ=-d1δ2s2γcos⁡p2sξ-bdkdxkcos⁡p1sx-ξ+Osk+1etp1x-ξ+p2ξ-b,dkdxkχ2x,λ=-d1sp2dkdxksin⁡p2sx-b+Osketp2x-b. Each of these asymptotic equalities holds uniformly for x as |λ|→∞.Theorem 19.
Letλ=s2, s=σ+it. Then the function W2(λ) has the following asymptotic representations.
Case 1. If b2≠0 and d2≠0, then(57)W2λ=δ2p1p2b2d2s6sin⁡p1sξ-asin⁡p2sb-ξ+Os5etp1ξ-a+p2b-ξ.
Case 2. If b2≠0 and d2=0, then(58)W2λ=-δ2p1b2d1s5sin⁡p1sξ-acos⁡p2sb-ξ+Os4etp1ξ-a+p2b-ξ.
Case 3. If b2=0 and d2≠0, then(59)W2λ=-δ2p2b1d2s5cos⁡p1sξ-asin⁡p2sb-ξ+Os4etp1ξ-a+p2b-ξ.
Case 4. If b2=0 and d2=0, then(60)W2λ=δ2b1d1s4cos⁡p1sξ-acos⁡p2sb-ξ+Os3etp1ξ-a+p2b-ξ.Proof.
By the definition ofW2(λ), we have(61)W2λ=φ2b,λχ2′b,λ-φ2′b,λχ2b,λ=c1-λd1φ2b,λ-c2-λd2φ2′b,λ. According to the equalities of φ2(x,λ) and φ2′(x,λ) in Lemma 17, we can obtain the formulas of W2(λ) in this theorem.Corollary 20.
The eigenvalues of the boundary value problem (1)–(4) are bounded below.Proof.
Puttings=it(t>0) in Theorem 19, we can obtain W2(λ)=W2(-t2)→∞ (t→∞). Hence, W2(-t2)≠0 for sufficiently negative λ and sufficiently large λ. This completes the proof.


## 4. Asymptotic Formulas for Eigenvalues and Eigenfunctions

In this section, we can get the asymptotic formulas for the eigenvalues and eigenfunctions of the singular Sturm-Liouville problem (1)–(4). Since the eigenvalues coincide with the zeros of the entire function W(λ), it follows that they have no finite limits.Theorem 21.
The eigenvaluesλn=sn2 (n=0,1,2,…) of problem (1)–(4) have the following asymptotic representations as n→∞.
Case 1. If b2≠0 and d2≠0, then(62)sn′=n-1πp1ξ-a+O1n,sn′′=n-1πp2b-ξ+O1n.
Case 2. If b2≠0 and d2=0, then(63)sn′=n-1πp1ξ-a+O1n,sn′′=n-1/2πp2b-ξ+O1n.
Case 3. If b2=0 and d2≠0, then(64)sn′=n-1/2πp1ξ-a+O1n,sn′′=n-1πp2b-ξ+O1n.
Case 4. If b2=0 and d2=0, then(65)sn′=n-1/2πp1ξ-a+O1n,sn′′=n-1/2πp2b-ξ+O1n.Proof.
By applying the well-known Rouche theorem, we can obtain these conclusions (see [3] Theorem 2.3).According to Theorem21 and Lemmas 17 and 18, we can obtain the following asymptotic representations of the eigenfunctions φ(x,λn) and χ(x,λn).Theorem 22.
The eigenfunctionsφ(x,λn) and χ(x,λn)(n=0,1,2,…) of problem (1)–(4) have the following asymptotic representations as n→∞.
Case 1. If b2≠0 and d2≠0, then(66)φx,λn′=-b2n-1πp1ξ-a2cos⁡n-1πξ-ax-a+On,x∈a,ξ,On2,x∈ξ,b,φx,λn′′=-b2n-1πp2b-ξ2cos⁡p1n-1πp2b-ξx-a+On,x∈a,ξ,p1b2δ2n-1πp2b-ξ3sin⁡p1n-1πp2b-ξξ-acos⁡n-1πb-ξx-a+On2,x∈ξ,b,χx,λn′=-p2d2δ2γn-1πp1ξ-a3sin⁡p2n-1πp1ξ-acos⁡n-1πξ-ax-ξ+On2,x∈a,ξ,-d2n-1πp1ξ-a2cos⁡p2n-1πp1ξ-ax-b+On,x∈ξ,b,χx,λn′′=On2,x∈a,ξ,-d2n-1πp2b-ξ2cos⁡n-1πb-ξx-b+On,x∈ξ,b.
Case 2. If b2≠0 and d2=0, then(67)φx,λn′=-b2n-1πp1ξ-a2cos⁡n-1πξ-ax-a+On,x∈a,ξ,On2,x∈ξ,b,φx,λn′′=-b2n-1/2πp2b-ξ2cos⁡p1n-1/2πp2b-ξx-a+On,x∈a,ξ,p1b2δ2n-1/2πp2b-ξ3sin⁡p1n-1/2πp2b-ξξ-acos⁡n-1/2πb-ξx-a+On2,x∈ξ,b,χx,λn′=d1δ2γn-1πp1ξ-a2cos⁡p2n-1πp1ξ-acos⁡n-1πξ-ax-ξ+On,x∈a,ξ,-d1p2n-1πp1ξ-asin⁡p2n-1πp1ξ-ax-b+O1,x∈ξ,b,χx,λn′′=On,x∈a,ξ,-d1p2n-1/2πp2b-ξsin⁡n-1/2πb-ξx-b+O1,x∈ξ,b.
Case 3. If b2=0 and d2≠0, then(68)φx,λn′=-b1n-1/2πp1ξ-asin⁡n-1/2πξ-ax-a+O1,x∈a,ξ,On,x∈ξ,b,φx,λn′′=-b1n-1πp2b-ξsin⁡p1n-1πp2b-ξx-a+O1,x∈a,ξ,-b1δ2n-1πp2b-ξ2cos⁡p1n-1πp2b-ξξ-acos⁡n-1πb-ξx-ξ+On,x∈ξ,b,χx,λn′=-p2d2δ2γn-1/2πp1ξ-a3sin⁡p2n-1/2πp1ξ-acos⁡n-1/2πξ-ax-ξ+On2,x∈a,ξ,-d2n-1/2πp1ξ-a2cos⁡p2n-1/2πp1ξ-ax-b+On,x∈ξ,b,χx,λn′′=On2,x∈a,ξ,-d2n-1πp2b-ξ2cos⁡n-1πb-ξx-b+On,x∈ξ,b.
Case 4. If b2=0 and d2=0, then(69)φx,λn′=-b1n-1/2πp1ξ-asin⁡n-1/2πξ-ax-a+O1,x∈a,ξ,On,x∈ξ,b,φx,λn′′=-b1n-1/2πp2b-ξsin⁡p1n-1/2πp2b-ξx-a+O1,x∈a,ξ,-b1δ2n-1/2πp2b-ξ2cos⁡p1n-1/2πp2b-ξξ-acos⁡n-1/2πb-ξx-ξ+On,x∈ξ,b,χx,λn′=d1δ2γn-1/2πp1ξ-a2cos⁡p2n-1/2πp1ξ-acos⁡n-1/2πξ-ax-ξ+On,x∈a,ξ,-d1p2n-1/2πp1ξ-asin⁡p2n-1/2πp1ξ-ax-b+O1,x∈ξ,b,χx,λn′′=On,x∈a,ξ,-d1p2n-1/2πp2b-ξsin⁡n-1/2πb-ξx-b+O1,x∈ξ,b.


## 5. Completeness of Eigenfunctions

In this section, we get the property of spectrum for the operatorA and discuss the completeness of the eigenfunctions of problem (1)–(4).Theorem 23.
The operatorA has only point spectrum; that is, σ(A)=σp(A).Proof.
We only need to prove that ifγ is not an eigenvalue of A, then γ∈ρ(A). Here we investigate the equation (A-γ)Z=F∈H, where γ∈R, Z=(z(x),Ta(z),Tb(z)), and F=(Lf,Ta(f),Tb(f)). Consider the initial value problem(70)-pxz′x′+qxzx-γzx=fx,x∈a,ξ∪ξ,bzξ+0-δ1zξ-0-δ2z′ξ-0=0,z′ξ+0-δ3zξ-0-δ4z′ξ-0=0.Let (71)ψx=ψ1x,x∈a,ξ,ψ2x,x∈ξ,bbe the solution of the equation -pxz′x′+q(x)z(x)-γz(x)=0 satisfying the transmission conditions (4). Let(72)ϕx=ϕ1x,x∈a,ξ,ϕ2x,x∈ξ,bbe the solution of the equation -pxz′x′+q(x)z(x)-γz(x)=f(x) satisfying (73)ϕξ+0-δ1ϕξ-0-δ2ϕ′ξ-0=0,ϕ′ξ+0-δ3ϕξ-0-δ4ϕ′ξ-0=0. Then (70) has the general solution(74)zx=nψ1x+ϕ1x,x∈a,ξ,nψ2x+ϕ2x,x∈ξ,b,where n∈C.
Asγ is not an eigenvalue of problem (1)–(4), we have(75)γb1z,ua-b2z,va-a1z,ua-a2z,va≠0,γd1z,ub-d2z,vb-c1z,cb-c2z,vb≠0.The second and third components of (A-γ)Z=F∈H mean that(76)a1z,ua-a2z,va-γb1z,ua-b2z,va=Taf,c1z,ub-c2z,vb-γd1z,ub-d2z,vb=Tbf.Substituting (74) into (76), we can get that n is uniquely solvable. So z(x) is uniquely determined. Observing that (A-γI)-1 is defined on all of H, we get that (A-γI)-1 is bounded by Theorem 8 and the closed graph theorem. Thus, γ∈ρ(A). Hence, σ(A)=σp(A).Lemma 24.
The eigenvalues of problem (1)–(4) are countably infinite and can cluster only at ∞.Lemma 25.
The operatorA has compact resolvent; that is, for each δ∈R/σp(A), (A-δI)-1 is compact on H (see [14] Theorem 6.3.3).By the above lemmas and the spectral theorem for compact operator, we obtain the following theorem.Theorem 26.
The eigenfunctions of problem (1)–(4), expanded to become eigenfunctions of A, are complete in H; that is, let {Φn=(φn(x),Ta(φn),Tb(φn));n∈N} be a maximum set of orthogonal eigenfunctions of A, where {φn(x);n∈N} are eigenfunctions of problem (1)–(4). Then, for all F∈H, F=∑n=1∞〈F,Φn〉Φn.Remark 27.
In this paper, the spectral properties of singular Sturm-Liouville problems with one inner discontinuous point are considered; if there are two or even multi-inner discontinuous pointsξ1,ξ2,…,ξn in the interval (a,b), we can obtain similar results by defining a more complicated Hilbert space.



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