# The Modified Parseval Equality of Sturm-Liouville Problems with Transmission Conditions

Authors: Mudan Bai; Jiong Sun; Siqin Yao
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We consider the Sturm-Liouville (S-L) problems with  very general transmission conditions on a finite interval. Firstly, we obtain the sufficient and necessary condition forλ being an eigenvalue of the S-L problems
by constructing  the fundamental solutions of the problems and prove that the eigenvalues of the S-L problems  are bounded below and are countably infinite. Furthermore, the asymptotic formulas of the eigenvalues and eigenfunctions of the S-L problems are obtained. Finally,
we derive the eigenfunction expansion for Green's function  of the S-L problems with transmission conditions and establish the modified Parseval equality in the associated Hilbert space.


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## Body

## 1. Introduction

The Sturm-Liouville (S-L) theory, as an active area of research in pure and applied mathematics, plays an important role in solving many problems in mathematical physics and is concerned in many publications [1–7]. It is well known that for the classical S-L problems, the solutions or the derivatives of the solutions are continuous on the interval, but these conditions cannot be satisfied in many practical physical problems. So, a class of S-L operators with “discontinuity,” that is, the S-L problems with transmission conditions at an interior point, are concerned by many mathematical and physical researchers [8–10]. Such conditions are known by various names including transmission conditions [11, 12], interface conditions [13–15], jump conditions [16], and discontinuous conditions [17, 18].In this paper, we consider the following Sturm-Liouville equation:(1)ly:=-y′′+q(x)y=λy,x∈I=[-1,0)∪(0,1],
with boundary conditions:
(2)l1y:=α1y(-1)+α2y′(-1)=0,(3)l2y:=β1y(1)+β2y′(1)=0,
and transmission conditions:
(4)l3y:=y(0+)-γ1y(0-)-γ2y′(0-)=0,l4y:=y′(0+)-γ3y(0-)-γ4y′(0-)=0,
where λ is a complex eigenparameter and q∈L(I,ℝ); notice that the potential function q(x) guarantee y(0±) and y′(0±) in (4) makes sense (see Theorem 1); all coefficients of the boundary and transmission conditions are real numbers. Throughout this paper, we assume that α12+α22≠0,  β12+β22≠0, and
(5)ρ=|γ1γ2γ3γ4|>0.
We derive the eigenfunction expansion for Green’s function of the S-L problem (1)–(4) and establish the modified Parseval equality of the S-L problem with very general transmission conditions at one inner point 0 of the finite interval [-1,1].The organization of this paper is as follows. After the Introduction, we construct the basic solutions of S-L equation (1) with transmission conditions (4) and obtain the sufficient and necessary condition for λ being an eigenvalue of the S-L problem in Section 2. In Section 3, the asymptotic formulas for eigenvalues and eigenfunctions of the S-L problem are obtained by using the asymptotic expressions of the solutions, and we prove that the eigenvalues of the S-L problem are bounded below and are countably infinite. Section 4 contains the eigenfunction expansion for Green’s function of the S-L problem (1)–(4) and we establish the modified Parseval equality in the associated Hilbert space by using the eigenfunction expansion for Green’s function.


## 2. The Basic Solutions and Eigenvalues

We construct the basic solutions of S-L equation (1) with transmission conditions (4) and obtain the sufficient and necessary condition for λ being an eigenvalue of the S-L problem in this section. At first, we prove the existence of finite limits for all solution y of (1) and its derivative at both sides of 0 point in the following theorem.Theorem 1.
Assume the coefficient functionq(x) in S-L equation (1) is real-valued and Lebesgue integrable on [-1,0) and (0,1], that is, q∈L(I,ℝ); then for the solution y of (1), the limits
(6)y(0±)=limx→0±y(x),y′(0±)=limx→0±y′(x)
exist and are finite.In order to prove Theorem1, we need the following lemma.Lemma 2 (see [19]).
LetJ=(a,b),-∞≤a<b≤∞. Let Mn,m(L(a,b),ℂ) denote the set of n×m matrices with entries from L((a,b),ℂ)(n,m∈ℕ).(1)
Suppose thatP,F satisfy
(7)P∈Mn(L(a,c),ℂ);F∈Mn,m(L(a,c),ℂ)for somec∈(a,b), where Mn(L(a,c),ℂ)=Mn,n(L(a,c),ℂ). For some u∈J and C∈Mn,m(ℂ), let Y be the solution of
(8)Y′=PY+F,Y(u)=C.ThenY(a)=limt→a+Y(t) exists and is finite.(2)
Suppose that(9)P∈Mn(L(c,b),ℂ);F∈Mn,m(L(c,b),ℂ)for somec∈(a,b). For some u∈J and C∈Mn,m(ℂ), let Y be the solution of (8) on J. Then, Y(b)=limt→b-Y(t) exists and is finite.Proof of Theorem1.
Leta=0,b=1, and let Y=(yy′),P=(01q-λ0),andF=(00). Then (1) is equivalent to the equation Y′=PY+F on (0,1]. From Lemma 2, we know that Y(0+)=limt→0+Y(t) exists and is finite. This implies that
(10)y(0+)=limx→0+y(x),y′(0+)=limx→0+y′(x)
exist and are finite.
Leta=-1,b=0. Then, (1) is equivalent to the equation Y′=PY+F on [-1,0). Since Y(0-)=limt→0-Y(t) exists and is finite by Lemma 2, the limits
(11)y(0-)=limx→0-y(x),y′(0-)=limx→0-y′(x)
exist and are finite.Let us define a new inner product inL2(I) as follows, which is associated with transmission conditions (4) and useful to investigate the S-L problem (1)–(4):
(12)〈f,g〉=ρ∫-10‍f1g¯1dx+∫01‍f2g¯2dx,forf,g∈L2(I),
where f1=f(x)|[-1,0),f2=f(x)|(0,1]. It is easy to verify that (L2(I),〈·,·〉) is a Hilbert space. For simplicity, it is denoted by H. The norm induced by the inner product is denoted by ∥·∥H. We consider the S-L problem (1)–(4) in the associated Hilbert space H.Theorem 3 (see [14]).
The S-L problem (1)–(4) is self-adjoint.We construct two basic solutionsϕ(x,λ) and χ(x,λ) of S-L equation (1) by the following procedure. At first, we consider the initial-value problem(13)-y′′+q(x)y=λy,x∈[-1,0),y(-1)=α2,y′(-1)=-α1.
By virtue of Theorem 1.5 in [20], the problem (13) has a unique solution ϕ1(x,λ) for each λ∈ℂ, which is an entire function of λ for each fixed x∈[-1,0).Similarly, the initial-value problem(14)-y′′+q(x)y=λy,x∈(0,1],y(0+)=γ1ϕ1(0-,λ)+γ2ϕ1′(0-,λ),y′(0+)=γ3ϕ1(0-,λ)+γ4ϕ1′(0-,λ)
has a solution ϕ2(x,λ) for each λ∈ℂ. Moreover, ϕ2(x,λ) is an entire function of λ for each fixed x∈(0,1]. We define a function ϕ(x,λ) on x∈I=[-1,0)∪(0,1] by
(15)ϕ(x,λ)={ϕ1(x,λ),x∈[-1,0),ϕ2(x,λ),x∈(0,1].
Obviously, ϕ(x,λ) satisfies S-L equation (1), the boundary condition (2), and both transmission conditions (4); that is, ϕ(x,λ) is a solution of S-L equation (1).As same as above, the initial-value problem(16)-y′′+q(x)y=λy,x∈(0,1],y(1)=β2,y′(1)=-β1
has a solution χ2(x,λ) which is an entire function of λ for each fixed x∈(0,1]. Similarly, the initial-value problem
(17)-y′′+q(x)y=λy,x∈[-1,0),y(0-)=1ρ(γ4χ2(0+,λ)-γ2χ2′(0+,λ)),y′(0-)=-1ρ(γ3χ2(0+,λ)-γ1χ2′(0+,λ))
also has a unique solution χ1(x,λ), which is an entire function of λ for each fixed x∈[-1,0). Let
(18)χ(x,λ)={χ1(x,λ),x∈[-1,0),χ2(x,λ),x∈(0,1].
By (16) and (17), χ(x,λ) satisfies S-L equation (1) and conditions (3) and (4); that is, χ(x,λ) is another solution of S-L equation (1).It is well known that, from the ordinary linear differential equation theory, the Wronskiansωj(λ):=W(ϕj(x,λ),χj(x,λ))(j=1,2) are independent of the variable x.Lemma 4.
The equalityω2(λ)=ρω1(λ) holds for each λ∈ℂ.Proof.
Since the Wronskiansωj(λ):=W(ϕj(x,λ),χj(x,λ))(j=1,2) are independent of the variable x, then by (14) and (17),
(19)ω2(λ)=ω2(λ)|x=0+=|ϕ2(0+,λ)χ2(0+,λ)ϕ2′(0+,λ)χ2′(0+,λ)|=|γ1ϕ1(0-,λ)+γ2ϕ1′(0-,λ)γ1χ1(0-,λ)+γ2χ1′(0-,λ)γ3ϕ1(0-,λ)+γ4ϕ1′(0-,λ)γ3χ1(0-,λ)+γ4χ1′(0-,λ)|=ρω1(λ).Letω(λ)=ω1(λ). Then ω(λ)=(1/ρ)ω2(λ) and is an entire function of λ.Theorem 5.
The eigenvalues of the S-L problem (1)–(4) coincide with the zeros of the function ω(λ).Proof.
Letλ0 be an eigenvalue of the S-L problem (1)–(4) and let y(x,λ0) be any corresponding eigenfunction. Let us assume that ω(λ0)≠0. Then, W(ϕj(x,λ0),χj(x,λ0))≠0(j=1,2). Consequently, each pair of functions ϕ1(x,λ0),χ1(x,λ0) and ϕ2(x,λ0),χ2(x,λ0) is linearly independent. Therefore, the solution y(x,λ0) of (1) may be represented in the form
(20)y(x,λ0)={c1ϕ1(x,λ0)+c2χ1(x,λ0),x∈[-1,0),d1ϕ2(x,λ0)+d2χ2(x,λ0),x∈(0,1],
where at least one of the constants c1,c2,d1,d2 is not zero. Since y(x,λ0) satisfies boundary conditions (2) and (3), we obtain c2=0,  d1=0. And substituting (20) in condition (4), we get c1 and d2 satisfying the following equations:
(21)d2χ2(0+,λ0)-c1(γ1ϕ1(0-,λ0)+γ2ϕ1′(0-,λ0))=0,d2χ2′(0+,λ0)-c1(γ3ϕ1(0-,λ0)+γ4ϕ1′(0-,λ0))=0.
From (14), the determinant of the matrix of coefficients in the last equations is equal to ω2(λ0) which is not zero by the assumption. Hence, c1=0 and d2=0. This is a contradiction. Thus, each eigenvalue of the S-L problem (1)–(4) is a zero of the function ω(λ).
Conversely, letω(λ0)=0. Then, W(ϕ1(x,λ0),χ1(x,λ0))=0 for all x∈[-1,0). Consequently, the functions ϕ1(x,λ0) and χ1(x,λ0) are linearly dependent. That is,
(22)χ1(x,λ0)=k1ϕ1(x,λ0),x∈[-1,0),
for some k1≠0. Then
(23)α1χ(-1,λ0)+α2χ′(-1,λ0)=α1χ1(-1,λ0)+α2χ1′(-1,λ0)=k1(α1ϕ1(-1,λ0)+α2ϕ1′(-1,λ0))=0.
So, χ(x,λ0) satisfies condition (2). And the solution χ(x,λ0) satisfies condition (3) and both transmission condition (4) from its construction. Thus χ(x,λ0) would be an eigenfunction of the S-L problem (1)–(4) for the eigenvalue λ0.Lemma 6.
The eigenvalues of the S-L problem (1)–(4) are simple.Proof.
Letλ=u+iv. We differentiate the equation lχ(x,λ)=λχ(x,λ) with respect to λ and have lχλ(x,λ)=λχλ(x,λ)+χ(x,λ). Then
(24)〈lχλ,ϕ〉-〈χλ,lϕ〉=〈λχλ+χ,ϕ〉-〈χλ,λϕ〉=〈χ,ϕ〉+2iv〈χλ,ϕ〉.
By integration by parts and some calculations
(25)〈lχλ,ϕ〉-〈χλ,lϕ〉=ρ∫-10‍(-χλ′′(x,λ)+q(x)χλ(x,λ))ϕ(x,λ)¯dx+∫01‍(-χλ′′(x,λ)+q(x)χλ(x,λ))ϕ(x,λ)¯dx-ρ∫-10‍χλ(x,λ)(-ϕ′′(x,λ)+q(x)ϕ(x,λ))¯dx-∫01‍χλ(x,λ)(-ϕ′′(x,λ)+q(x)ϕ(x,λ))¯dx=ρ[(χ1λ(x,λ)ϕ1′(x,λ)¯-χ1λ′(x,λ)ϕ1(x,λ)¯)|-10-pppppl+(χ2λ(x,λ)ϕ2′(x,λ)¯-χ2λ′(x,λ)ϕ2(x,λ)¯)|0+1]=ρ(α2χλ′(-1,λ)+α1χλ(-1,λ))
from (13), (14), (16), and (17) where χjλ(x,λ),χjλ′(x,λ) are the derivatives of χj(x,λ),χj′(x,λ) with respect to λ(j=1,2). Since ω(λ) is independent of the variable x and the value of ϕ1(-1,λ) and ϕ1′(-1,λ) given in (13),
(26)ω′(λ)|x=-1=dω1(λ)dλ|x=-1=ddλ(ϕ1(-1,λ)χ1′(-1,λ)pppppl-ϕ1(-1,λ)χ1′(-1,λ))=α2χ1λ′(-1,λ)+α1χ1λ(-1,λ).
So, by (24), (25), and (26),
(27)ω′(λ)=1ρ(〈χ,ϕ〉+2iv〈χλ,ϕ〉).
Let λ0 be an eigenvalue of the S-L problem (1)–(4). Then λ0 is real. Since ω(λ0)=0 for the eigenvalue of the S-L problem (1)–(4) from Theorem 5, χj(x,λ0)=cjϕj(x,λ0) for some cj≠0(j=1,2). And from (4),
(28)χ2(0+,λ0)=c1(γ1ϕ1(0-,λ0)+γ2ϕ1′(0-,λ0))=c1ϕ2(0+,λ0),χ2′(0+,λ0)=c1(γ3ϕ1(0-,λ0)+γ4ϕ1′(0-,λ0))=c1ϕ2′(0+,λ0),
we get c1=c2≠0. Thus, (27) becomes
(29)ω′(λ0)=1ρ〈χ,ϕ〉=c1ρ〈ϕ,ϕ〉≠0.
Hence λ0 is simple.


## 3. Asymptotic Formulas for Eigenvalues and Eigenfunctions

In this section, the asymptotic formulas for eigenvalues and eigenfunctions of the S-L problem (1)–(4) are obtained by using the asymptotic expressions of the solutions. At first, we calculate the asymptotic expressions of the solutions.Lemma 7.
Letλ=s2, s=σ+it. If α2≠0, then one has the following asymptotic expressions:
(30)dkdxkϕ1(x,λ)=α2dkdxkcoss(x+1)+O(|s|k-1e|t|(x+1)),pppppppppppppppppppppppppppppppk=0,1,
as |λ|→∞. If α2=0, then
(31)dkdxkϕ1(x,λ)=-α1sdkdxksins(x+1)+O(|s|k-2e|t|(x+1)),pppppppppppppppppppppppppppppppppk=0,1,
as |λ|→∞.Proof.
The asymptotic formulas forϕ1(x,λ) follow from the similar formulas of Lemma 1.7 in [20].The asymptotic formulas forϕ2(x,λ) are as follows.Lemma 8.
Letλ=s2,  s=σ+it; then ϕ2(x,λ) has the following asymptotic formulas as |λ|→∞:(1)
ifα2≠0,γ2≠0, then
(32)dkdxkϕ2(x,λ)=-α2γ2ssinsdkdxkcossx+O(e|t|x),(2)
ifα2≠0,γ2=0, then
(33)dkdxkϕ2(x,λ)=α2(γ1cossdkdxkcossx-γ4sinsdkdxksinsx)+O(|s|-1e|t|x),(3)
ifα2=0,γ2≠0, then
(34)dkdxkϕ2(x,λ)=-α1γ2cossdkdxkcossx+O(|s|-1e|t|x),(4)
ifα2=0,γ2=0, then
(35)dkdxkϕ2(x,λ)=-α1s(γ1sinsdkdxkcossx+γ4cossdkdxksinsx)+O(|s|-2e|t|x)for k=0,1.Proof.
Letλ=s2,  s=σ+it.  ϕ2(x,λ) satisfies (14). By the method of variation of constants, we have ϕ2(x,λ) satisfying the following integral equation:
(36)ϕ2(x,λ)=(γ1ϕ1(0-,λ)+γ2ϕ1′(0-,λ))cossx+1s(γ3ϕ1(0-,λ)+γ4ϕ1′(0-,λ))sinsx+1s∫0x‍sin[s(x-z)]q(z)ϕ2(z,λ)dz.
Let α2≠0. Substituting (30), for k=0, in (36), we have
(37)ϕ2(x,λ)=-α2γ2ssinscossx+α2(γ1cosscossx-γ4sinssinsx)+1sα2γ3cosssinsx+1s∫0x‍sin[s(x-z)]q(z)ϕ2(z,λ)dz+O(1se|t|(x+1)).
We consider the case γ2≠0. Let ϕ2(x,λ)=|s|e|t|xF(x,λ). Multiplying (37) by |s|-1e-|t|x, we have the following:
(38)F(x,λ)=-α2γ2e-|t|xsinscossx+1|s|α2e-|t|x(γ1cosscossx-γ4sinssinsx)+1|s|2α2γ3e-|t|xcosssinsx+1|s|∫0x‍sin[s(x-z)]e-|t|(x-z)q(z)F(z,λ)dz+O(1|s|2e|t|).
Let μ(λ)=max-1≤x≤1|F(x,λ)|. Then
(39)μ(λ)≤|α2γ2|+M0|s|
for some M0>0. Consequently, μ(λ)=O(1) as |λ|→∞. So, ϕ2(x,λ)=|s|e|t|xF(x,λ)=O(|s|e|t|x) as |λ|→∞. Substituting the asymptotic equality in (37) gives (32) for k=0. And the case for k=1 is obtained by differentiating (37) and by some similar calculations.
Similarly, we can prove (33), (34), and (35).Whenλ is big enough, the asymptotic formulas for ϕ2(x,λ) are obtained in Lemma 8. Substituting the asymptotic formulas of ϕ2(x,λ) and the value of χ2(x,λ) at x=1, which is given in (16), into ω(λ)=(1/ρ)ω2(λ), we can establish the following lemma.Lemma 9.
Letλ=s2,  s=σ+it; then ω(λ) has the following asymptotic formulas for large enough λ.(1)
Ifα2≠0,γ2≠0,β2≠0, then
(40)ω(λ)=-1ρα2β2γ2s2sin2s+O(|s|e|t|).(2)
Ifα2≠0,γ2≠0,β2=0, then
(41)ω(λ)=1ρα2β1γ2ssinscoss+O(e|t|).(3)
Ifα2≠0,γ2=0,β2≠0, then
(42)ω(λ)=1ρα2β2(γ1+γ4)ssinscoss+O(e|t|).(4)
Ifα2≠0,γ2=0,β2=0, then
(43)ω(λ)=1ρα2β1(γ4sin2s-γ1cos2s)+O(|s|-1e|t|).(5)
Ifα2=0,γ2≠0,β2≠0, then
(44)ω(λ)=-1ρα1β2γ2ssinscoss+O(e|t|).(6)
Ifα2=0,γ2≠0,β2=0, then
(45)ω(λ)=1ρα1β1γ2cos2s+O(|s|-1e|t|).(7)
Ifα2=0,γ2=0,β2≠0, then
(46)ω(λ)=1ρα1β2(γ1sin2s-γ4cos2s)+O(|s|-1e|t|).(8)
Ifα2=0,γ2=0,β2=0, then
(47)ω(λ)=1ρsα1β1(γ1+γ4)sinscoss+O(|s|-2e|t|).Theorem 10.
The eigenvalues of the S-L problem (1)–(4) are bounded below and countably infinite; the unique cluster point is infinity.Proof.
From Theorem5, the eigenvalues of the S-L problem (1)–(4) are zeros of the function ω(λ) which is an entire function of λ. And the asymptotic formulas of the function ω(λ) are obtained in Lemma 9. Let
(48)ω(λ)=ω*(λ)+δ(λ),
where ω*(λ)=-(1/ρ)α2β2γ2s2sin2s,δ(λ)=O(|s|e|t|) for the case α2≠0,γ2≠0,β2≠0 from Lemma 9.
Letλ=s2,  s=σ+it and let
(49)Γn′={λ=s2=(σ+it)2∣|σ|=(n+12)π,0≤t≤n},Γn′′={λ=s2=(σ+it)2∣|σ|≤(n+12)π,t=n},
and let Γn=Γn′∪Γn′′. Then Γn is a closed curve on the plane of λ. Next, we prove that |ω*(λ)|>|δ(λ)| on Γn. When |λ| is big enough, on Γn′(50)|sins|=12|eis-e-is|=12|ei(n+(1/2))πe-t-e-i(n+(1/2))πet|=12|ie-t+iet|=12et|1+e-2t|>e|t|4.
Similarly, when |λ| is big enough, on Γn′′(51)|sins|=12|eiσπe-n-e-iσπen|=12en|1-e2iσπe-2n|>en4.
We obtain
(52)|ω*(λ)|Γn=|-1ρα2β2γ2s2sin2s|Γn>n4ρe|t||α2β2γ2ssins|.
So,
(53)|ω*(λ)|Γn>|δ(λ)|Γn.
By Rouche’s Theorem in [21], ω(λ) and ω*(λ) have the same zeros interior of Γn=Γn′∪Γn′′ as follows:
(54)02,π2,(2π)2,…,(nπ)2.
Lettings=it, we can prove that |ω*(λ)|Γn>|δ(λ)|Γn. Hence, ω(λ)=ω(-t2)≠0 as |t|→∞. Thus, ω(λ) only has finite negative zeros. The proof of other cases in Lemma 9 is similar to the proof of the case α2≠0,γ2≠0,β2≠0.Theorem 11.
Letλ1,λ2,λ3,…, be the collection of all eigenvalues of the S-L problem (1)–(4) and let φ1(x),φ2(x),… be the corresponding normalized eigenfunctions. Then
(55)-∞<λ1≤λ2≤⋯λn⋯⟶∞.
And {φn;n∈ℕ} is complete in H and
(56)〈φn,φm〉={1,n=m,0,n≠m.Theorem 12.
The following asymptotic formulas hold for eigenvalues and eigenfunctions of the S-L problem (1)–(4) for large enough n∈ℕ.(1)
Ifα2≠0,γ2≠0,β2≠0, then
(57)λn=(n-1)π+O(1n),ϕ(x,λn)={α2cos(n-1)π(x+1)+O(1n),x∈[-1,0),O(1),x∈(0,1].(2)
Ifα2≠0,γ2≠0,β2=0, then
(58)λn=(n-1)2π+O(1n),ϕ(x,λn)={α2cos(n-1)2π(x+1)+O(1n),x∈[-1,0),α2γ2(n-1)2πcos(n-1)2πx+O(1)(kiseven),O(1)(kisodd),x∈(0,1].(3)
Ifα2≠0,γ2=0,β2≠0, then
(59)λn=(n-1)2π+O(1n),ϕ(x,λn)={α2cos(n-1)2π(x+1)+O(1n),x∈[-1,0),O(1n),x∈(0,1].(4)
Ifα2≠0,γ2=0,β2=0, then
(60)λn=(n-1)π±arctanγ1γ4+O(1n),ϕ(x,λn)={α2cos[(n-1)π±arctanγ1γ4]×(x+1)+O(1n),ppppppx∈[-1,0),α2(γ1cos[(n-1)π±arctanγ1γ4]×cos[(n-1)π±arctanγ1γ4]x-γ4sin[(n-1)π±arctanγ1γ4]×sin[γ1γ4(n-1)π±arctanγ1γ4]x),ppppppppppppppppppppppx∈(0,1].(5)
Ifα2=0,γ2≠0,β2≠0, then
(61)λn=(n-1)2+O(1n),ϕ(x,λn)={-2α1(n-1)πsin(n-1)2π(x+1)+O(1n2),x∈[-1,0),α1γ2cos(n-1)2πx+O(1n)(kisodd)O(1n)(kiseven),x∈(0,1].(6)
Ifα2=0,γ2≠0,β2=0, then
(62)λn=(n-12)π+O(1n),ϕ(x,λn)={-α1(n-(1/2))πsin(n-12)×π(x+1)+O(1n2),x∈[-1,0),O(1n),x∈(0,1].(7)
Ifα2=0,γ2=0,β2≠0, then
(63)λn=(n-1)π±arctanγ4γ1+O(1n),ϕ(x,λn)={-α1(n-1)π±arctanγ4/γ1×sin[(n-1)π±arctanγ4γ1]×(x+1)+O(1n2),x∈[-1,0),-α1(n-1)π±arctanγ4/γ1×(γ1sin[(n-1)π±arctanγ4γ1]pppppl×cos[(n-1)π±arctanγ4γ1]xpppppl+γ4cos[(n-1)π±arctanγ4γ1]pppppl×sin[γ4γ1(n-1)π±arctanγ4γ1]x)+O(1n2),ppppppppplx∈(0,1].(8)
Ifα2=0,γ2=0,β2=0, then
(64)λn=(n-1)2π+O(1n),(65)ϕ(x,λn)={-2α1(n-1)πsin(n-1)2π(x+1)+O(1n2),x∈[-1,0),O(1n2),x∈(0,1].Proof.
By Theorem5, we know that the eigenvalues of the S-L problem (1)–(4) coincide with the zeros of the function ω(λ). From Lemma 9 and Theorem 10, the entire function ω(λ) has a sequence of zeros:
(66)λn=(n-1)π+O(1n),
for the case α2≠0,γ2≠0,β2≠0 when n is big enough. By using the asymptotic formulas of eigenvalues λn, the corresponding eigenfunctions ϕ(x,λn) have the following asymptotic formulas:
(67)ϕ(x,λn)={α2cos(n-1)π(x+1)+O(1n),x∈[-1,0),O(1),x∈(0,1]
for the case α2≠0,γ2≠0,β2≠0. Similarly, we can obtain the other cases.By the asymptotic formulas for the eigenvalues and eigenfunctions of the S-L problem (1)–(4) in the above theorem, we know that the series ∑n=1∞‍(φn(x)φn(ξ)¯/λn) converges absolutely and uniformly where φn(x) is the normalized eigenfunction.Green’s function of the classical S-L problem is studied in [22]. Here, we obtain Green’s function of the S-L problem (1)–(4) as follows:
(68)G(x,ξ,λ)={-1ρϕ1(ξ,λ)χ1(x,λ)ω1,-1≤ξ≤x<0,-1ρϕ1(x,λ)χ1(ξ,λ)ω1,-1≤x≤ξ<0,-1ρϕ1(x,λ)χ2(ξ,λ)ω2,-1≤x<0,0<ξ≤1,-ϕ1(ξ,λ)χ2(x,λ)ω1,-1≤ξ<0,0<x≤1,-ϕ2(ξ,λ)χ2(x,λ)ω2,0<ξ≤x≤1,-ϕ2(x,λ)χ2(ξ,λ)ω2,0<x≤ξ≤1.Let us consider the nonhomogeneous differential equation(69)ly-λy=f(x)
together with boundary and transmission conditions (2)–(4) where f∈H. Green’s functions provide a powerful method for solving the linear nonhomogeneous equations.Theorem 13.
Letf∈H. Then the function
(70)y(x,λ)=ρ∫-10‍G(x,ξ,λ)f(ξ)dξ+∫01‍G(x,ξ,λ)f(ξ)dξ
satisfies (69) and both boundary and transmission conditions (2)–(4), where G(x,ξ,λ) is the Green function defined in (68).Proof.
Putting (68) in (70), we have
(71)y(x,λ)={-1ω1χ1(x,λ)∫-1x‍ϕ1(ξ,λ)f(ξ)dξ-1ω1ϕ1(x,λ)∫x0‍χ1(ξ,λ)f(ξ)dξ-1ω2ϕ1(x,λ)∫01‍χ2(ξ,λ)f(ξ)dξ,x∈[-1,0),-1ω1χ2(x,λ)∫-10‍ϕ1(ξ,λ)f(ξ)dξ-1ω2χ2(x,λ)∫0x‍ϕ2(ξ,λ)f(ξ)dξ-1ω2ϕ2(x,λ)∫x1‍χ2(ξ,λ)f(ξ)dξ,x∈(0,1].
By differentiating twice y(x,λ) and by the definitions of solutions ϕj(x,λ) and χj(x,λ)(j=1,2), we obtain y′′(x,λ)=(q(x)-λ)y(x,λ)-f(x). So, the function y(x,λ) defined in (70) is the solution of (69). And by calculations, we may prove that (71) satisfies both boundary and transmission conditions (2)–(4).


## 4. Eigenfunction Expansion for Green’s Function and the Modified Parseval Equality

In this section, we derive the eigenfunction expansion for Green’s function of the S-L problem (1)–(4) and establish the modified Parseval equality in the associated Hilbert space H. Without loss of generality, we assume that λ=0 is not an eigenvalue. Let G(x,ξ)=G(x,ξ,0).Theorem 14.
Letλn be the eigenvalue of the S-L problem (1)–(4) and let φn(x) be the corresponding normalized eigenfunction. Then,
(72)G(x,ξ)=-∑n=1∞‍φn(x)φn(ξ)¯λn.In order to prove Theorem14, we need the following lemma.Lemma 15.
The S-L problem (1)–(4) is equivalent to
(73)y(x,λ)-λ(ρ∫-10‍G(x,ξ)y(ξ)dξ+∫01‍G(x,ξ)y(ξ)dξ)=0.Proof.
By Theorem13, we know that
(74)y(x,λ)=ρ∫-10‍G(x,ξ)f(ξ)dξ+∫01‍G(x,ξ)f(ξ)dξ
satisfies -y′′(x)+q(x)y(x)=f(x) and both boundary and transmission conditions (2)–(4). Equation (69) can be written in the form -y′′(x)+q(x)y(x)=f~(x) where f~(x)=f(x)+λy. Similar to (74), the new form of (69) has a solution
(75)y(x,λ)=ρ∫-10‍G(x,ξ)f~(ξ)dξ+∫01‍G(x,ξ)f~(ξ)dξ,
which satisfies both boundary and transmission conditions (2)–(4). If f(x)≡0, then the corresponding homogeneous case is the S-L problem (1)–(4). Consequently, the S-L problem (1)–(4) is equivalent to
(76)y(x,λ)-λ(ρ∫-10‍G(x,ξ)y(ξ)dξ+∫01‍G(x,ξ)y(ξ)dξ)=0.Proof of Theorem14.
Letλn be the eigenvalue of the S-L problem (1)–(4) and let φn(x) be the corresponding normalized eigenfunction as in Theorem 11. Let P(x,ξ)=G(x,ξ)+∑n=1∞‍(φn(x)φn(ξ)¯/λn). Then P(x,ξ) is continuous and symmetric. We assume that P(x,ξ)≠0. Then, by the Fredholm integral equation, there are a number λ~ and a function y~(x)≠0 in H such that
(77)y~(x)=λ~(ρ∫-10‍P(x,ξ)y~(ξ)dξ+∫01‍P(x,ξ)y~(ξ)dξ).
By Lemma 15,
(78)φn(x)-λn(ρ∫-10‍G(x,ξ)φn(ξ)dξpppppppppl+∫01‍G(x,ξ)φn(ξ)dξ)=0.
Putting G(x,ξ)=P(x,ξ)-∑n=1∞‍(φn(x)φn(ξ)¯/λn) in (78), we obtain
(79)ρ∫-10‍P(x,ξ)φn(ξ)dξ+∫01‍P(x,ξ)φn(ξ)dξ=0.
In the following, we prove that 〈y~,φn〉=0 and y~ is an eigenfunction of the S-L problem (1)–(4). By (77) and (79)
(80)〈y~,φn〉=ρ∫-10‍y~(x)φn(x)¯dx+∫01‍y~(x)φn(x)¯dx=λ~ρ∫-10‍(ρ∫-10P(x,ξ)y~(ξ)dξppppppp+∫01P(x,ξ)y~(ξ)dξ)φn(x)¯dx+λ~∫01‍(ρ∫-10P(x,ξ)y~(ξ)dξppppppl+∫01P(x,ξ)y~(ξ)dξ)φn(x)¯dx=λ~ρ∫-10‍(ρ∫-10P(x,ξ)φn(x)¯dxppppppl+∫01P(x,ξ)φn(x)¯dx)y~(ξ)dξ+λ~∫01‍(ρ∫-10P(x,ξ)φn(x)¯dxppppppl+∫01P(x,ξ)φn(x)¯dx)y~(ξ)dξ=0.
And we have
(81)y~(x)-λ~(ρ∫-10‍G(x,ξ)y~(ξ)dξ+∫01‍G(x,ξ)y~(ξ)dξ)=y~(x)-λ~(ρ∫-10‍(P(x,ξ)-∑n=1∞‍φn(x)φn(ξ)¯λn)y~(ξ)dξppppppppppppl+∫01‍(P(x,ξ)-∑n=1∞‍φn(x)φn(ξ)¯λn)y~(ξ)dξ)=y~(x)-λ~(∑n=1∞‍(ρ∫-10‍P(x,ξ)y~(ξ)dξ+∫01‍P(x,ξ)y~(ξ)dξ)9999999999999-∑n=1∞‍φn(x)λn〈y~,φn〉)=y~(x)-λ~(ρ∫-10‍P(x,ξ)y~(ξ)dξpppppppppppp+∫01‍P(x,ξ)y~(ξ)dξ)=0
by (77). This implies that y~ is the eigenfunction of the S-L problem (1)–(4) by Lemma 15. So, by 〈y~,φn〉=0 and the completeness of the eigenfunctions in Theorem 11, we know that y~=0. Consequently, P(x,ξ)=0. The proof is completed.Lemma 16.
LetC~0∞ be the set of all functions defined by
(82)ψ(x)={ψ1(x)x∈[-1,0),ψ2(x)x∈(0,1],
where ψ1(x)∈C0∞[-1,0),  ψ2(x)∈C0∞(0,1]. Then C~0∞ is dense in H.Proof.
Letf be any function in L2(I) with f(x)=f1(x),x∈[-1,0), and f(x)=f2(x),x∈(0,1]. Since C0∞[-1,0) is dense in L2[-1,0) as in [23], for f1∈L2[-1,0), there exists a function g1∈C0∞[-1,0) such that
(83)ρ∫-10‍|f1(x)-g1(x)|2dx<ϵ2.
Similarly, for f2∈L2(0,1], there exists a function g2∈C0∞(0,1] such that
(84)∫01‍|f2(x)-g2(x)|2dx<ϵ2.
Then, for any f∈H and ϵ>0, there exists g∈C~0∞ with g(x)={g1(x)x∈[-1,0),g2(x)x∈(0,1] such that
(85)∥f-g∥H2=ρ∫-10‍|f1(x)-g1(x)|2dx+∫01‍|f2(x)-g2(x)|2dx<ϵ.
Thus, C~0∞ is dense in H.In the following theorem, we prove the modified Parseval equality in the associated Hilbert spaceH.Theorem 17.
Letf∈H. Then, the modified Parseval equality holds; that is,
(86)∥f∥H2=∑n=1∞‍cn2(f),
where ∥f∥H2=〈f,f〉 and
(87)cn(f)=ρ∫-10‍f(x)φn(x)¯dx+∫01‍f(x)φn(x)¯dx.Proof.
LetC~0∞ be as in Lemma 16. At first, we prove that (86) holds for f∈C~0∞. Denote g(x)=-f′′(x)+q(x)f. Then by Theorems 13 and 14,
(88)f(x)=ρ∫-10‍G(x,ξ)g(ξ)dξ+∫01‍G(x,ξ)g(ξ)dξ=-∑n=1∞‍1λnφn(x)×(ρ∫-10‍φn(ξ)¯g(ξ)dξ+∫01‍φn(ξ)¯g(ξ)dξ).
Multiplying by φm(x)¯ and integrating the new equation, we have
(89)ρ∫-10‍φm(x)¯f(x)dx+∫01‍φm(x)¯f(x)dx=-1λm(ρ∫-10‍φm(ξ)¯g(ξ)dξ+∫01‍φm(ξ)¯g(ξ)dξ).
Then
(90)f(x)=∑n=1∞‍cn(f)φn(x),
where cn(f)=〈f,φn〉=ρ∫-10‍f(x)φn(x)¯dx+∫01‍f(x)φn(x)¯dx. Thus, for f∈C~0∞,
(91)∥f∥H2=∑n=1∞‍cn2(f).
Next, we prove that (86) holds for f∈H. For any f∈H, there exists a sequence {fk}k∈ℕ⊂C~0∞ converging to f in H since C~0∞ is dense in H. Firstly, we prove that ∑n=1∞‍cn2(f)<∞ and limk→∞∑n=1∞‍cn2(fk)=∑n=1∞‍cn2(f).
By the Cauchy-Schwartz inequality,|cn(fk)-cn(f)|=|〈fk-f,φn〉|≤∥fk-f∥H. Hence, limk→∞cn(fk)=cn(f). Since ∑n=1∞‍(cn(fk)-cn(fm))2=∑n=1∞‍cn2(fk-fm)=∥fk-fm∥H2 for fk-fm∈C~0∞,
(92)∑n=1N‍(cn(fk)-cn(fm))2≤∥fk-fm∥H2.
Let k→∞, then inequality (92) becomes ∑n=1N‍(cn(f)-cn(fm))2≤∥f-fm∥H2. Letting N→∞, we have
(93)∑n=1∞‍(cn(f)-cn(fm))2≤∥f-fm∥H2.
Then by the Minkowski inequality
(94)∑n=1∞‍cn2(f)=∑n=1∞‍(cn(f)-cn(fm)+cn(fm))2≤((∑n=1∞‍(cn(f)-cn(fm))2)1/2pppp+(∑n=1∞‍cn2(fm))1/2)2<∞
and by Ho¨lder’s inequality,
(95)|∑n=1∞‍cn2(f)-∑n=1∞‍cn2(fk)|=|∑n=1∞‍(cn(f)-cn(fk))(cn(f)+cn(fk))|≤(∑n=1∞‍(cn(f)-cn(fk))2)1/2(∑n=1∞‍(cn(f)+cn(fk))2)1/2⟶0,ask⟶∞.
This means that limk→∞∑n=1∞‍cn2(fk)=∑n=1∞‍cn2(f).
Sincefk→f in H as k→∞,  limk→∞∥fk∥H=∥f∥H. We obtain
(96)∥f∥H2=limk→∞∥fk∥H2=limk→∞∑n=1∞‍cn2(fk)=∑n=1∞‍cn2(f).



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