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

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 for 𝜆 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.

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: 𝑙𝑦∶=−𝑦^′′+𝑞(𝑥)𝑦=𝜆𝑦,𝑥∈𝐼=[−1,0)∪(0,1], (1) with boundary conditions: 𝑙₁𝑦∶=𝛼₁𝑦(−1)+𝛼₂𝑦^′()𝑙−1=0,₂𝑦∶=𝛽₁𝑦(1)+𝛽₂𝑦^′(1)=0, (2)(3) and transmission conditions: 𝑙₃𝑦∶=𝑦(0+)−𝛾₁𝑦(0−)−𝛾₂𝑦^′()𝑙0−=0,₄𝑦∶=𝑦^′(0+)−𝛾₃𝑦(0−)−𝛾₄𝑦^′(0−)=0, (4) 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 𝛼²₁+𝛼²₂≠0,  𝛽²₁+𝛽²₂≠0, and 𝛾𝜌=¹𝛾₂𝛾₃𝛾₄>0. (5) 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 function 𝑞(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 𝑦(0±)=lim_𝑥→0±𝑦(𝑥),𝑦^′(0±)=lim_𝑥→0±𝑦^′(𝑥) (6) exist and are finite.

In order to prove Theorem 1, we need the following lemma.

Lemma 2 (see [19]). Let 𝐽=(𝑎,𝑏),−∞≤𝑎<𝑏≤∞. Let 𝑀_𝑛,𝑚(𝐿(𝑎,𝑏),ℂ) denote the set of 𝑛×𝑚 matrices with entries from 𝐿((𝑎,𝑏),ℂ)(𝑛,𝑚∈ℕ).(1) Suppose that 𝑃,𝐹 satisfy 𝑃∈𝑀_𝑛(𝐿(𝑎,𝑐),ℂ);𝐹∈𝑀_𝑛,𝑚(𝐿(𝑎,𝑐),ℂ) (7) for some 𝑐∈(𝑎,𝑏), where 𝑀_𝑛(𝐿(𝑎,𝑐),ℂ)=𝑀_𝑛,𝑛(𝐿(𝑎,𝑐),ℂ). For some 𝑢∈𝐽 and 𝐶∈𝑀_𝑛,𝑚(ℂ), let 𝑌 be the solution of 𝑌^′=𝑃𝑌+𝐹,𝑌(𝑢)=𝐶. (8) Then 𝑌(𝑎)=lim_𝑡→𝑎+𝑌(𝑡) exists and is finite. (2)Suppose that 𝑃∈𝑀_𝑛(𝐿(𝑐,𝑏),ℂ);𝐹∈𝑀_𝑛,𝑚(𝐿(𝑐,𝑏),ℂ) (9) for some 𝑐∈(𝑎,𝑏). For some 𝑢∈𝐽 and 𝐶∈𝑀_𝑛,𝑚(ℂ), let 𝑌 be the solution of (8) on 𝐽. Then, 𝑌(𝑏)=lim_{𝑡→𝑏−}𝑌(𝑡) exists and is finite.

Proof of Theorem 1. Let 𝑎=0,𝑏=1, and let 𝑌=^𝑦_𝑦^′,𝑃=⁰¹_𝑞−𝜆0,and𝐹=⁰₀. Then (1) is equivalent to the equation 𝑌^′=𝑃𝑌+𝐹 on (0,1]. From Lemma 2, we know that 𝑌(0+)=lim_𝑡→0+𝑌(𝑡) exists and is finite. This implies that 𝑦(0+)=lim_𝑥→0+𝑦(𝑥),𝑦^′(0+)=lim_𝑥→0+𝑦^′(𝑥) (10) exist and are finite.
Let 𝑎=−1,𝑏=0. Then, (1) is equivalent to the equation 𝑌^′=𝑃𝑌+𝐹 on [−1,0). Since 𝑌(0−)=lim_𝑡→0−𝑌(𝑡) exists and is finite by Lemma 2, the limits 𝑦(0−)=lim_𝑥→0−𝑦(𝑥),𝑦^′(0−)=lim_𝑥→0−𝑦^′(𝑥) (11) exist and are finite.

Let us define a new inner product in 𝐿²(𝐼) as follows, which is associated with transmission conditions (4) and useful to investigate the S-L problem (1)–(4): 𝑓,𝑔=𝜌⁰₋₁𝑓₁𝑔₁𝑑𝑥+¹₀𝑓₂𝑔₂𝑑𝑥,for𝑓,𝑔∈𝐿²(𝐼), (12) 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 solutions 𝜙(𝑥,𝜆) and 𝜒(𝑥,𝜆) of S-L equation (1) by the following procedure. At first, we consider the initial-value problem−𝑦^′′(𝑥)[),𝑦()+𝑞𝑦=𝜆𝑦,𝑥∈−1,0−1=𝛼₂,𝑦^′(−1)=−𝛼₁. (13) 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 −𝑦^′′(𝑥)(],𝑦()+𝑞𝑦=𝜆𝑦,𝑥∈0,10+=𝛾₁𝜙₁(0−,𝜆)+𝛾₂𝜙^′₁(),𝑦0−,𝜆_′(0+)=𝛾₃𝜙₁(0−,𝜆)+𝛾₄𝜙^′₁(0−,𝜆) (14) 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 𝜙()=𝜙𝑥,𝜆¹()[),𝜙𝑥,𝜆,𝑥∈−1,0₂(𝑥,𝜆),𝑥∈(0,1]. (15) 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 −𝑦^′′(𝑥)(],𝑦(1)+𝑞𝑦=𝜆𝑦,𝑥∈0,1=𝛽₂,𝑦^′(1)=−𝛽₁ (16) has a solution 𝜒₂(𝑥,𝜆) which is an entire function of 𝜆 for each fixed 𝑥∈(0,1]. Similarly, the initial-value problem −𝑦^′′(𝑥)[),𝑦()=1+𝑞𝑦=𝜆𝑦,𝑥∈−1,00−𝜌𝛾₄𝜒₂(0+,𝜆)−𝛾₂𝜒^′₂(),𝑦0+,𝜆_′()10−=−𝜌𝛾₃𝜒₂(0+,𝜆)−𝛾₁𝜒^′₂(0+,𝜆) (17) also has a unique solution 𝜒₁(𝑥,𝜆), which is an entire function of 𝜆 for each fixed 𝑥∈[−1,0). Let 𝜒()=𝜒𝑥,𝜆¹()[),𝜒𝑥,𝜆,𝑥∈−1,0₂(𝑥,𝜆),𝑥∈(0,1]. (18) 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 Wronskians 𝜔_𝑗(𝜆)∶=𝑊(𝜙_𝑗(𝑥,𝜆),𝜒_𝑗(𝑥,𝜆))(𝑗=1,2) are independent of the variable 𝑥.

Lemma 4. The equality 𝜔₂(𝜆)=𝜌𝜔₁(𝜆) holds for each 𝜆∈ℂ.

Proof. Since the Wronskians 𝜔_𝑗(𝜆)∶=𝑊(𝜙_𝑗(𝑥,𝜆),𝜒_𝑗(𝑥,𝜆))(𝑗=1,2) are independent of the variable 𝑥, then by (14) and (17), 𝜔₂(𝜆)=𝜔₂(𝜆)_𝑥=0+=𝜙₂(0+,𝜆)𝜒₂()𝜙0+,𝜆_′2(0+,𝜆)𝜒^′₂()=𝛾0+,𝜆₁𝜙₁(0−,𝜆)+𝛾₂𝜙^′₁(0−,𝜆)𝛾₁𝜒₁(0−,𝜆)+𝛾₂𝜒^′₁()𝛾0−,𝜆₃𝜙₁(0−,𝜆)+𝛾₄𝜙^′₁(0−,𝜆)𝛾₃𝜒₁(0−,𝜆)+𝛾₄𝜒^′₁(0−,𝜆)=𝜌𝜔₁(𝜆). (19)

Let 𝜔(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. Let 𝜆₀ 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(𝜙_𝑗(𝑥,𝜆₀),𝜒_𝑗(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 𝑦𝑥,𝜆₀=𝑐₁𝜙₁𝑥,𝜆₀+𝑐₂𝜒₁𝑥,𝜆₀[),𝑑,𝑥∈−1,0₁𝜙₂𝑥,𝜆₀+𝑑₂𝜒₂𝑥,𝜆₀,𝑥∈(0,1], (20) 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: 𝑑₂𝜒₂0+,𝜆₀−𝑐₁𝛾₁𝜙₁0−,𝜆₀+𝛾₂𝜙^′₁0−,𝜆₀𝑑=0,₂𝜒^′₂0+,𝜆₀−𝑐₁𝛾₃𝜙₁0−,𝜆₀+𝛾₄𝜙^′₁0−,𝜆₀=0. (21) 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, let 𝜔(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, 𝜒₁𝑥,𝜆₀=𝑘₁𝜙₁𝑥,𝜆₀,𝑥∈[−1,0), (22) for some 𝑘₁≠0. Then 𝛼₁𝜒−1,𝜆₀+𝛼₂𝜒^′−1,𝜆₀=𝛼₁𝜒₁−1,𝜆₀+𝛼₂𝜒^′₁−1,𝜆₀=𝑘₁𝛼₁𝜙₁−1,𝜆₀+𝛼₂𝜙^′₁−1,𝜆₀=0. (23) 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. Let 𝜆=𝑢+𝑖𝑣. We differentiate the equation 𝑙𝜒(𝑥,𝜆)=𝜆𝜒(𝑥,𝜆) with respect to 𝜆 and have 𝑙𝜒_𝜆(𝑥,𝜆)=𝜆𝜒_𝜆(𝑥,𝜆)+𝜒(𝑥,𝜆). Then 𝑙𝜒_𝜆,𝜙−𝜒_𝜆,𝑙𝜙=⟨𝜆𝜒_𝜆+𝜒,𝜙⟩−⟨𝜒_𝜆,𝜆𝜙⟩=⟨𝜒,𝜙⟩+2𝑖𝑣⟨𝜒_𝜆,𝜙⟩. (24) By integration by parts and some calculations ⟨𝑙𝜒_𝜆,𝜙⟩−⟨𝜒_𝜆,𝑙𝜙⟩=𝜌⁰₋₁−𝜒_𝜆^′′(𝑥,𝜆)+𝑞(𝑥)𝜒_𝜆(𝑥,𝜆)+𝜙(𝑥,𝜆)𝑑𝑥₁₀−𝜒_𝜆^′′(𝑥,𝜆)+𝑞(𝑥)𝜒_𝜆(𝑥,𝜆)𝜙(𝑥,𝜆)𝑑𝑥−𝜌⁰₋₁𝜒_𝜆(𝑥,𝜆)−𝜙^′′()(𝑥)𝜙(−𝑥,𝜆+𝑞𝑥,𝜆)𝑑𝑥₁₀𝜒_𝜆(𝑥,𝜆)−𝜙^′′()(𝑥)𝜙()𝜒𝑥,𝜆+𝑞𝑥,𝜆𝑑𝑥=𝜌_1𝜆(𝑥,𝜆)𝜙^′₁(𝑥,𝜆)−𝜒^′_1𝜆(𝑥,𝜆)𝜙₁(𝑥,𝜆)⁰⁻₋₁+𝜒_2𝜆(𝑥,𝜆)𝜙^′₂(𝑥,𝜆)−𝜒^′_2𝜆(𝑥,𝜆)𝜙₂(𝑥,𝜆)¹₀₊=𝜌𝛼₂𝜒^′_𝜆(−1,𝜆)+𝛼₁𝜒_𝜆(−1,𝜆) (25) from (13), (14), (16), and (17) where 𝜒_𝑗𝜆(𝑥,𝜆),𝜒^′_𝑗𝜆(𝑥,𝜆) are the derivatives of 𝜒_𝑗(𝑥,𝜆),𝜒^′_𝑗(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 𝜙^′₁(−1,𝜆) given in (13), 𝜔^′(𝜆)_𝑥=−1=𝑑𝜔₁(𝜆)𝑑𝜆_𝑥=−1=𝑑𝜙𝑑𝜆¹(−1,𝜆)𝜒^′₁(−1,𝜆)−𝜙₁(−1,𝜆)𝜒^′₁(−1,𝜆)=𝛼₂𝜒^′_1𝜆(−1,𝜆)+𝛼₁𝜒_1𝜆(−1,𝜆). (26) So, by (24), (25), and (26), 𝜔^′(𝜆)=1𝜌𝜒,𝜙+2𝑖𝑣𝜒_𝜆,𝜙. (27) 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, 𝜒_𝑗(𝑥,𝜆₀)=𝑐_𝑗𝜙_𝑗(𝑥,𝜆₀) for some 𝑐_𝑗≠0(𝑗=1,2). And from (4), 𝜒₂0+,𝜆₀=𝑐₁𝛾₁𝜙₁0−,𝜆₀+𝛾₂𝜙^′₁0−,𝜆₀=𝑐₁𝜙₂0+,𝜆₀,𝜒^′₂0+,𝜆₀=𝑐₁𝛾₃𝜙₁0−,𝜆₀+𝛾₄𝜙^′₁0−,𝜆₀=𝑐₁𝜙^′₂0+,𝜆₀, (28) we get 𝑐₁=𝑐₂≠0. Thus, (27) becomes 𝜔^′𝜆₀=1𝜌𝑐⟨𝜒,𝜙⟩=¹𝜌⟨𝜙,𝜙⟩≠0. (29) 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. Let 𝜆=𝑠², 𝑠=𝜎+𝑖𝑡. If 𝛼₂≠0, then one has the following asymptotic expressions: 𝑑^𝑘𝑑𝑥^𝑘𝜙₁(𝑥,𝜆)=𝛼₂𝑑^𝑘𝑑𝑥^𝑘cos𝑠(𝑥+1)+𝑂|𝑠|^𝑘−1𝑒^{|𝑡|(𝑥+1)},𝑘=0,1, (30) as |𝜆|→∞. If 𝛼₂=0, then 𝑑^𝑘𝑑𝑥^𝑘𝜙₁()𝛼𝑥,𝜆=−¹𝑠𝑑^𝑘𝑑𝑥^𝑘sin𝑠(𝑥+1)+𝑂|𝑠|^𝑘−2𝑒^{|𝑡|(𝑥+1)},𝑘=0,1, (31) as |𝜆|→∞.

Proof. The asymptotic formulas for 𝜙₁(𝑥,𝜆) follow from the similar formulas of Lemma 1.7 in [20].

The asymptotic formulas for 𝜙₂(𝑥,𝜆) are as follows.

Lemma 8. Let 𝜆=𝑠²,  𝑠=𝜎+𝑖𝑡; then 𝜙₂(𝑥,𝜆) has the following asymptotic formulas as |𝜆|→∞:(1) if 𝛼₂≠0,𝛾₂≠0, then 𝑑^𝑘𝑑𝑥^𝑘𝜙₂(𝑥,𝜆)=−𝛼₂𝛾₂𝑑𝑠sin𝑠^𝑘𝑑𝑥^𝑘cos𝑠𝑥+𝑂𝑒^|𝑡|𝑥, (32)(2) if 𝛼₂≠0,𝛾₂=0, then 𝑑^𝑘𝑑𝑥^𝑘𝜙₂(𝑥,𝜆)=𝛼₂𝛾₁𝑑cos𝑠^𝑘𝑑𝑥^𝑘cos𝑠𝑥−𝛾₄𝑑sin𝑠^𝑘𝑑𝑥^𝑘|𝑠|sin𝑠𝑥+𝑂⁻¹𝑒^|𝑡|𝑥, (33)(3) if 𝛼₂=0,𝛾₂≠0, then 𝑑^𝑘𝑑𝑥^𝑘𝜙₂(𝑥,𝜆)=−𝛼₁𝛾₂𝑑cos𝑠^𝑘𝑑𝑥^𝑘cos𝑠𝑥+𝑂|𝑠|⁻¹𝑒^|𝑡|𝑥, (34)(4) if 𝛼₂=0,𝛾₂=0, then 𝑑^𝑘𝑑𝑥^𝑘𝜙₂()𝛼𝑥,𝜆=−¹𝑠𝛾₁𝑑sin𝑠^𝑘𝑑𝑥^𝑘cos𝑠𝑥+𝛾₄𝑑cos𝑠^𝑘𝑑𝑥^𝑘|𝑠|sin𝑠𝑥+𝑂⁻²𝑒^|𝑡|𝑥 (35)for 𝑘=0,1.

Proof. Let 𝜆=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: 𝜙₂(𝑥,𝜆)=𝛾₁𝜙₁(0−,𝜆)+𝛾₂𝜙^′₁()+10−,𝜆cos𝑠𝑥𝑠𝛾₃𝜙₁(0−,𝜆)+𝛾₄𝜙^′₁()+10−,𝜆sin𝑠𝑥𝑠^𝑥₀sin[𝑠(𝑥−𝑧)]𝑞(𝑧)𝜙₂(𝑧,𝜆)𝑑𝑧. (36) Let 𝛼₂≠0. Substituting (30), for 𝑘=0, in (36), we have 𝜙₂(𝑥,𝜆)=−𝛼₂𝛾₂𝑠sin𝑠cos𝑠𝑥+𝛼₂𝛾₁cos𝑠cos𝑠𝑥−𝛾₄+1sin𝑠sin𝑠𝑥𝑠𝛼₂𝛾₃+1cos𝑠sin𝑠𝑥𝑠^𝑥₀sin[𝑠(𝑥−𝑧)]𝑞(𝑧)𝜙₂()1𝑧,𝜆𝑑𝑧+𝑂𝑠𝑒^{|𝑡|(𝑥+1)}. (37) 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: 𝐹(𝑥,𝜆)=−𝛼₂𝛾₂𝑒^{−|𝑡|𝑥}+1sin𝑠cos𝑠𝑥|𝑠|𝛼₂𝑒^{−|𝑡|𝑥}𝛾₁cos𝑠cos𝑠𝑥−𝛾₄+1sin𝑠sin𝑠𝑥|𝑠|²𝛼₂𝛾₃𝑒^{−|𝑡|𝑥}+1cos𝑠sin𝑠𝑥|𝑠|^𝑥₀sin[𝑠(𝑥−𝑧)]𝑒^{−|𝑡|(𝑥−𝑧)}𝑞(𝑧)𝐹()1𝑧,𝜆𝑑𝑧+𝑂|𝑠|²𝑒^|𝑡|. (38) Let 𝜇(𝜆)=max_{−1≤𝑥≤1}|𝐹(𝑥,𝜆)|. Then 𝜇(𝜆)≤𝛼₂𝛾₂+𝑀₀|𝑠| (39) 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).

When 𝜆 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. Let 𝜆=𝑠²,  𝑠=𝜎+𝑖𝑡; then 𝜔(𝜆) has the following asymptotic formulas for large enough 𝜆.(1) If 𝛼₂≠0,𝛾₂≠0,𝛽₂≠0, then 𝜔(𝜆)1=−𝜌𝛼₂𝛽₂𝛾₂𝑠²sin²𝑠+𝑂|𝑠|𝑒^|𝑡|. (40)(2) If 𝛼₂≠0,𝛾₂≠0,𝛽₂=0, then 𝜔(𝜆)=1𝜌𝛼₂𝛽₁𝛾₂𝑠sin𝑠cos𝑠+𝑂𝑒^|𝑡|. (41)(3) If 𝛼₂≠0,𝛾₂=0,𝛽₂≠0, then 𝜔(𝜆)=1𝜌𝛼₂𝛽₂𝛾₁+𝛾₄𝑠sin𝑠cos𝑠+𝑂𝑒^|𝑡|. (42)(4) If 𝛼₂≠0,𝛾₂=0,𝛽₂=0, then 𝜔(𝜆)=1𝜌𝛼₂𝛽₁𝛾₄sin²𝑠−𝛾₁cos²𝑠+𝑂|𝑠|⁻¹𝑒^|𝑡|. (43)(5) If 𝛼₂=0,𝛾₂≠0,𝛽₂≠0, then 𝜔(𝜆)1=−𝜌𝛼₁𝛽₂𝛾₂𝑠sin𝑠cos𝑠+𝑂𝑒^|𝑡|. (44)(6) If 𝛼₂=0,𝛾₂≠0,𝛽₂=0, then 𝜔(𝜆)=1𝜌𝛼₁𝛽₁𝛾₂cos²𝑠+𝑂|𝑠|⁻¹𝑒^|𝑡|. (45)(7) If 𝛼₂=0,𝛾₂=0,𝛽₂≠0, then 𝜔(𝜆)=1𝜌𝛼₁𝛽₂𝛾₁sin²𝑠−𝛾₄cos²𝑠+𝑂|𝑠|⁻¹𝑒^|𝑡|. (46)(8) If 𝛼₂=0,𝛾₂=0,𝛽₂=0, then 𝜔(𝜆)=1𝛼𝜌𝑠¹𝛽₁𝛾₁+𝛾₄sin𝑠cos𝑠+𝑂|𝑠|⁻²𝑒^|𝑡|. (47)

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 Theorem 5, 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 𝜔(𝜆)=𝜔^∗(𝜆)+𝛿(𝜆), (48) where 𝜔^∗(𝜆)=−(1/𝜌)𝛼₂𝛽₂𝛾₂𝑠²sin²𝑠,𝛿(𝜆)=𝑂(|𝑠|𝑒^|𝑡|) for the case 𝛼₂≠0,𝛾₂≠0,𝛽₂≠0 from Lemma 9.
Let 𝜆=𝑠²,  𝑠=𝜎+𝑖𝑡 and let Γ^′_𝑛=𝜆=𝑠²=(𝜎+𝑖𝑡)²∣|𝜎|=1𝑛+2,Γ𝜋,0≤𝑡≤𝑛_𝑛′′=𝜆=𝑠²=(𝜎+𝑖𝑡)²∣|𝜎|≤1𝑛+2𝜋,𝑡=𝑛, (49) and let Γ_𝑛=Γ^′_𝑛∪Γ_𝑛^′′. Then Γ_𝑛 is a closed curve on the plane of 𝜆. Next, we prove that |𝜔^∗(𝜆)|>|𝛿(𝜆)| on Γ_𝑛. When |𝜆| is big enough, on Γ^′_𝑛||=1sin𝑠2𝑒^𝑖𝑠−𝑒^−𝑖𝑠=12𝑒^{𝑖(𝑛+(1/2))𝜋}𝑒^−𝑡−𝑒^{−𝑖(𝑛+(1/2))𝜋}𝑒^𝑡=12𝑖𝑒^−𝑡+𝑖𝑒^𝑡=12𝑒^𝑡1+𝑒^−2t>𝑒^|𝑡|4. (50) Similarly, when |𝜆| is big enough, on Γ_𝑛^′′||=1sin𝑠2𝑒^𝑖𝜎𝜋𝑒^−𝑛−𝑒^{−𝑖𝜎𝜋}𝑒^𝑛=12𝑒^𝑛1−𝑒^{2𝑖𝜎𝜋}𝑒^−2𝑛>𝑒^𝑛4. (51) We obtain 𝜔^∗(𝜆)_Γ𝑛=−1𝜌𝛼₂𝛽₂𝛾₂𝑠²sin²𝑠_Γ𝑛>𝑛𝑒4𝜌^|𝑡|𝛼₂𝛽₂𝛾₂𝑠sin𝑠. (52) So, 𝜔^∗(𝜆)_Γ𝑛>|𝛿(𝜆)|_Γ𝑛. (53) By Rouche’s Theorem in [21], 𝜔(𝜆) and 𝜔^∗(𝜆) have the same zeros interior of Γ_𝑛=Γ^′_𝑛∪Γ_𝑛^′′ as follows: 0²,𝜋²,(2𝜋)²,…,(𝑛𝜋)². (54)
Letting 𝑠=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. Let 𝜆₁,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 −∞<𝜆₁≤𝜆₂≤⋯𝜆_𝑛⋯⟶∞. (55) And {𝜑_𝑛;𝑛∈ℕ} is complete in 𝐻 and 𝜑_𝑛,𝜑_𝑚=1,𝑛=𝑚,0,𝑛≠𝑚. (56)

Theorem 12. The following asymptotic formulas hold for eigenvalues and eigenfunctions of the S-L problem (1)–(4) for large enough 𝑛∈ℕ.(1) If 𝛼₂≠0,𝛾₂≠0,𝛽₂≠0, then 𝜆_𝑛=()1𝑛−1𝜋+𝑂𝑛,𝜙𝑥,𝜆_𝑛=𝛼₂()𝜋()1cos𝑛−1𝑥+1+𝑂𝑛[),𝑂(1)(].,𝑥∈−1,0,𝑥∈0,1 (57)(2) If 𝛼₂≠0,𝛾₂≠0,𝛽₂=0, then 𝜆_𝑛=(𝑛−1)21𝜋+𝑂𝑛,𝜙𝑥,𝜆_𝑛=𝛼₂()cos𝑛−12𝜋()1𝑥+1+𝑂𝑛[),𝛼,𝑥∈−1,0₂𝛾₂(𝑛−1)2()𝜋cos𝑛−12(1),𝑂(1)(].𝜋𝑥+𝑂)(𝑘𝑖𝑠𝑒𝑣𝑒𝑛)(𝑘𝑖𝑠𝑜𝑑𝑑,𝑥∈0,1 (58)(3) If 𝛼₂≠0,𝛾₂=0,𝛽₂≠0, then 𝜆_𝑛=(𝑛−1)21𝜋+𝑂𝑛,𝜙𝑥,𝜆_𝑛=𝛼₂()cos𝑛−12𝜋()1𝑥+1+𝑂𝑛[),𝑂1,𝑥∈−1,0𝑛,𝑥∈(0,1]. (59)(4) If 𝛼₂≠0,𝛾₂=0,𝛽₂=0, then 𝜆_𝑛=(𝑛−1)𝜋±arctan𝛾₁𝛾₄1+𝑂𝑛,𝜙𝑥,𝜆_𝑛=𝛼₂cos(𝑛−1)𝜋±arctan𝛾₁𝛾₄×()1𝑥+1+𝑂𝑛[),𝛼,𝑥∈−1,0₂𝛾₁cos(𝑛−1)𝜋±arctan𝛾₁𝛾₄×cos(𝑛−1)𝜋±arctan𝛾₁𝛾₄𝑥−𝛾₄sin(𝑛−1)𝜋±arctan𝛾₁𝛾₄×sin(𝑛−1)𝜋±arctan𝛾₁𝛾₄𝑥,𝑥∈(0,1]. (60)(5) If 𝛼₂=0,𝛾₂≠0,𝛽₂≠0, then 𝜆_𝑛=(𝑛−1)21+𝑂𝑛,𝜙𝑥,𝜆_𝑛=−2𝛼₁()𝜋()𝑛−1sin𝑛−12𝜋()1𝑥+1+𝑂𝑛²[),𝛼,𝑥∈−1,0₁𝛾₂()cos𝑛−121𝜋𝑥+𝑂𝑛)𝑂1(𝑘𝑖𝑠𝑜𝑑𝑑𝑛(𝑘𝑖𝑠𝑒𝑣𝑒𝑛),𝑥∈(0,1]. (61)(6) If 𝛼₂=0,𝛾₂≠0,𝛽₂=0, then 𝜆_𝑛=1𝑛−21𝜋+𝑂𝑛,𝜙𝑥,𝜆_𝑛=−𝛼₁((𝜋1𝑛−1/2))sin𝑛−2()1×𝜋𝑥+1+𝑂𝑛²[),𝑂1,𝑥∈−1,0𝑛,𝑥∈(0,1]. (62)(7) If 𝛼₂=0,𝛾₂=0,𝛽₂≠0, then 𝜆_𝑛=(𝑛−1)𝜋±arctan𝛾₄𝛾₁1+𝑂𝑛,𝜙𝑥,𝜆_𝑛=−𝛼₁(𝑛−1)𝜋±arctan𝛾₄/𝛾₁×sin(𝑛−1)𝜋±arctan𝛾₄𝛾₁×()1𝑥+1+𝑂𝑛²[),−𝛼,𝑥∈−1,0₁(𝑛−1)𝜋±arctan𝛾₄/𝛾₁×𝛾₁sin(𝑛−1)𝜋±arctan𝛾₄𝛾₁×cos(𝑛−1)𝜋±arctan𝛾₄𝛾₁𝑥+𝛾₄cos(𝑛−1)𝜋±arctan𝛾₄𝛾₁×sin(𝑛−1)𝜋±arctan𝛾₄𝛾₁𝑥1+𝑂𝑛²,𝑥∈(0,1]. (63)(8) If 𝛼₂=0,𝛾₂=0,𝛽₂=0, then 𝜆_𝑛=(𝑛−1)21𝜋+𝑂𝑛,𝜙𝑥,𝜆_𝑛=−2𝛼₁()𝜋()𝑛−1sin𝑛−12𝜋()1𝑥+1+𝑂𝑛²[),𝑂1,𝑥∈−1,0𝑛²,𝑥∈(0,1]. (64)(65)

Proof. By Theorem 5, 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: 𝜆_𝑛=()1𝑛−1𝜋+𝑂𝑛, (66) for the case 𝛼₂≠0,𝛾₂≠0,𝛽₂≠0 when 𝑛 is big enough. By using the asymptotic formulas of eigenvalues 𝜆_𝑛, the corresponding eigenfunctions 𝜙(𝑥,𝜆_𝑛) have the following asymptotic formulas: 𝜙𝑥,𝜆_𝑛=𝛼₂()𝜋()1cos𝑛−1𝑥+1+𝑂𝑛[),𝑂(1)(],𝑥∈−1,0,𝑥∈0,1 (67) 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(𝜑_𝑛(𝑥)𝜑_𝑛(𝜉)/𝜆_𝑛) converges absolutely and uniformly where 𝜑_𝑛(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: 𝐺()=−1𝑥,𝜉,𝜆𝜌𝜙₁(𝜉,𝜆)𝜒₁(𝑥,𝜆)𝜔₁−1,−1≤𝜉≤𝑥<0,𝜌𝜙₁(𝑥,𝜆)𝜒₁(𝜉,𝜆)𝜔₁−1,−1≤𝑥≤𝜉<0,𝜌𝜙₁(𝑥,𝜆)𝜒₂(𝜉,𝜆)𝜔₂−𝜙,−1≤𝑥<0,0<𝜉≤1,₁(𝜉,𝜆)𝜒₂(𝑥,𝜆)𝜔₁−𝜙,−1≤𝜉<0,0<𝑥≤1,₂(𝜉,𝜆)𝜒₂(𝑥,𝜆)𝜔₂−𝜙,0<𝜉≤𝑥≤1,₂(𝑥,𝜆)𝜒₂(𝜉,𝜆)𝜔₂,0<𝑥≤𝜉≤1. (68)

Let us consider the nonhomogeneous differential equation 𝑙𝑦−𝜆𝑦=𝑓(𝑥) (69) 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. Let 𝑓∈𝐻. Then the function 𝑦(𝑥,𝜆)=𝜌⁰₋₁𝐺(𝑥,𝜉,𝜆)𝑓(𝜉)𝑑𝜉+¹₀𝐺(𝑥,𝜉,𝜆)𝑓(𝜉)𝑑𝜉 (70) 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 𝑦()=−1𝑥,𝜆𝜔₁𝜒₁(𝑥,𝜆)∫^𝑥₋₁𝜙₁()𝑓(𝜉)−1𝜉,𝜆𝑑𝜉𝜔₁𝜙₁(𝑥,𝜆)∫⁰_𝑥𝜒₁()𝑓(𝜉)−1𝜉,𝜆𝑑𝜉𝜔₂𝜙₁(𝑥,𝜆)∫¹₀𝜒₂()𝑓(𝜉)[),−1𝜉,𝜆𝑑𝜉,𝑥∈−1,0𝜔₁𝜒₂(𝑥,𝜆)∫⁰₋₁𝜙₁()𝑓(𝜉)−1𝜉,𝜆𝑑𝜉𝜔₂𝜒₂(𝑥,𝜆)∫^𝑥₀𝜙₂()𝑓(𝜉)−1𝜉,𝜆𝑑𝜉𝜔₂𝜙₂(𝑥,𝜆)∫¹_𝑥𝜒₂(𝜉,𝜆)𝑓(𝜉)𝑑𝜉,𝑥∈(0,1]. (71) By differentiating twice 𝑦(𝑥,𝜆) and by the definitions of solutions 𝜙_𝑗(𝑥,𝜆) and 𝜒_𝑗(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. Let 𝜆_𝑛 be the eigenvalue of the S-L problem (1)–(4) and let 𝜑_𝑛(𝑥) be the corresponding normalized eigenfunction. Then, 𝐺(𝑥,𝜉)=−^∞_𝑛=1𝜑_𝑛(𝑥)𝜑_𝑛(𝜉)𝜆_𝑛. (72)

In order to prove Theorem 14, we need the following lemma.

Lemma 15. The S-L problem (1)–(4) is equivalent to 𝑦(𝑥,𝜆)−𝜆𝜌⁰₋₁𝐺(𝑥,𝜉)𝑦(𝜉)𝑑𝜉+¹₀𝐺(𝑥,𝜉)𝑦(𝜉)𝑑𝜉=0. (73)

Proof. By Theorem 13, we know that 𝑦(𝑥,𝜆)=𝜌⁰₋₁𝐺(𝑥,𝜉)𝑓(𝜉)𝑑𝜉+¹₀𝐺(𝑥,𝜉)𝑓(𝜉)𝑑𝜉 (74) 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 𝑦(𝑥,𝜆)=𝜌⁰₋₁𝐺(𝑥,𝜉)𝑓(𝜉)𝑑𝜉+¹₀𝐺(𝑥,𝜉)𝑓(𝜉)𝑑𝜉, (75) 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 𝑦(𝑥,𝜆)−𝜆𝜌⁰₋₁𝐺(𝑥,𝜉)𝑦(𝜉)𝑑𝜉+¹₀𝐺(𝑥,𝜉)𝑦(𝜉)𝑑𝜉=0. (76)

Proof of Theorem 14. Let 𝜆_𝑛 be the eigenvalue of the S-L problem (1)–(4) and let 𝜑_𝑛(𝑥) be the corresponding normalized eigenfunction as in Theorem 11. Let 𝑃(𝑥,𝜉)=𝐺(𝑥,𝜉)+∑^∞_𝑛=1(𝜑_𝑛(𝑥)𝜑_𝑛(𝜉)/𝜆_𝑛). 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 𝑦(𝑥)=𝜆𝜌⁰₋₁𝑃(𝑥,𝜉)𝑦(𝜉)𝑑𝜉+¹₀𝑃(𝑥,𝜉)𝑦(𝜉)𝑑𝜉. (77) By Lemma 15, 𝜑_𝑛(𝑥)−𝜆_𝑛𝜌⁰₋₁𝐺(𝑥,𝜉)𝜑_𝑛(𝜉)+𝑑𝜉₁₀𝐺(𝑥,𝜉)𝜑_𝑛(𝜉)𝑑𝜉=0. (78) Putting 𝐺(𝑥,𝜉)=𝑃(𝑥,𝜉)−∑^∞_𝑛=1(𝜑_𝑛(𝑥)𝜑_𝑛(𝜉)/𝜆_𝑛) in (78), we obtain 𝜌⁰₋₁𝑃(𝑥,𝜉)𝜑_𝑛(𝜉)𝑑𝜉+¹₀𝑃(𝑥,𝜉)𝜑_𝑛(𝜉)𝑑𝜉=0. (79) In the following, we prove that ⟨𝑦,𝜑_𝑛⟩=0 and 𝑦 is an eigenfunction of the S-L problem (1)–(4). By (77) and (79) 𝑦,𝜑_𝑛=𝜌⁰₋₁𝑦(𝑥)𝜑_𝑛(𝑥)𝑑𝑥+¹₀𝑦(𝑥)𝜑_𝑛=(𝑥)𝑑𝑥𝜆𝜌⁰₋₁𝜌⁰₋₁𝑃(𝑥,𝜉)𝑦(𝜉)𝑑𝜉+¹₀𝑃(𝑥,𝜉)𝑦(𝜉)𝑑𝜉𝜑_𝑛+𝜆(𝑥)𝑑𝑥₁₀𝜌⁰₋₁𝑃(𝑥,𝜉)𝑦(𝜉)𝑑𝜉+¹₀𝑃(𝑥,𝜉)𝑦(𝜉)𝑑𝜉𝜑_𝑛=(𝑥)𝑑𝑥𝜆𝜌⁰₋₁𝜌⁰₋₁𝑃(𝑥,𝜉)𝜑_𝑛(𝑥)𝑑𝑥+¹₀𝑃(𝑥,𝜉)𝜑_𝑛(𝑥)𝑦(𝜉)+𝜆𝑑𝑥𝑑𝜉₁₀𝜌⁰₋₁𝑃(𝑥,𝜉)𝜑_𝑛+(𝑥)𝑑𝑥₁₀𝑃(𝑥,𝜉)𝜑_𝑛(𝑥)𝑑𝑥𝑦(𝜉)𝑑𝜉=0. (80) And we have 𝑦(𝑥)−𝜆𝜌⁰₋₁𝐺(𝑥,𝜉)𝑦(𝜉)𝑑𝜉+¹₀𝐺()𝑦(𝜉)=𝑦(𝑥)−𝜆𝜌𝑥,𝜉𝑑𝜉₀₋₁𝑃(𝑥,𝜉)−^∞_𝑛=1𝜑_𝑛(𝑥)𝜑_𝑛(𝜉)𝜆_𝑛𝑦(𝜉)+𝑑𝜉₁₀𝑃(𝑥,𝜉)−^∞_𝑛=1𝜑_𝑛(𝑥)𝜑_𝑛(𝜉)𝜆_𝑛𝑦(𝜉)=𝑦(𝑥)−𝜆𝑑𝜉𝜌⁰₋₁𝑃(𝑥,𝜉)𝑦(𝜉)𝑑𝜉+¹₀𝑃()𝑦(𝜉)−𝑥,𝜉𝑑𝜉_∞𝑛=1𝜑_𝑛(𝑥)𝜆_𝑛𝑦,𝜑_𝑛=𝑦(𝑥)−𝜆𝜌⁰₋₁𝑃()𝑦(𝜉)+𝑥,𝜉𝑑𝜉₁₀𝑃(𝑥,𝜉)𝑦(𝜉)𝑑𝜉=0 (81) by (77). This implies that 𝑦 is the eigenfunction of the S-L problem (1)–(4) by Lemma 15. So, by ⟨𝑦,𝜑_𝑛⟩=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. Let 𝐶^∞₀ be the set of all functions defined by 𝜓(𝑥)=𝜓₁(𝑥)[),𝜓𝑥∈−1,0₂(𝑥)𝑥∈(0,1], (82) where 𝜓₁(𝑥)∈𝐶^∞₀[−1,0),  𝜓₂(𝑥)∈𝐶^∞₀(0,1]. Then 𝐶^∞₀ is dense in 𝐻.

Proof. Let 𝑓 be any function in 𝐿²(𝐼) with 𝑓(𝑥)=𝑓₁(𝑥),𝑥∈[−1,0), and 𝑓(𝑥)=𝑓₂(𝑥),𝑥∈(0,1]. Since 𝐶^∞₀[−1,0) is dense in 𝐿²[−1,0) as in [23], for 𝑓₁∈𝐿²[−1,0), there exists a function 𝑔₁∈𝐶^∞₀[−1,0) such that 𝜌⁰₋₁𝑓₁(𝑥)−𝑔₁(𝑥)²𝜖𝑑𝑥<2. (83) Similarly, for 𝑓₂∈𝐿²(0,1], there exists a function 𝑔₂∈𝐶^∞₀(0,1] such that ¹₀𝑓₂(𝑥)−𝑔₂(𝑥)²𝜖𝑑𝑥<2. (84) Then, for any 𝑓∈𝐻 and 𝜖>0, there exists 𝑔∈𝐶^∞₀ with 𝑔(𝑥)=^𝑔1_𝑔^{(𝑥)𝑥∈[−1,0),}_{2(𝑥)𝑥∈(0,1]} such that 𝑓−𝑔²_𝐻=𝜌∫⁰₋₁𝑓₁(𝑥)−𝑔₁(𝑥)²𝑑𝑥+∫¹₀𝑓₂(𝑥)−𝑔₂(𝑥)²𝑑𝑥<𝜖. (85) Thus, 𝐶^∞₀ is dense in 𝐻.

In the following theorem, we prove the modified Parseval equality in the associated Hilbert space 𝐻.

Theorem 17. Let f∈𝐻. Then, the modified Parseval equality holds; that is, 𝑓²_𝐻=^∞_𝑛=1𝑐²_𝑛𝑓, (86) where ‖𝑓‖²_𝐻=⟨𝑓,𝑓⟩ and 𝑐_𝑛𝑓=𝜌⁰₋₁𝑓(𝑥)𝜑_𝑛(𝑥)𝑑𝑥+¹₀𝑓(𝑥)𝜑_𝑛(𝑥)𝑑𝑥. (87)

Proof. Let 𝐶^∞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, 𝑓(𝑥)=𝜌⁰₋₁𝐺(𝑥,𝜉)𝑔(𝜉)𝑑𝜉+¹₀𝐺(𝑥,𝜉)𝑔(𝜉)𝑑𝜉=−^∞_𝑛=11𝜆_𝑛𝜑_𝑛(𝑥)×𝜌⁰₋₁𝜑_𝑛(𝜉)𝑔(𝜉)𝑑𝜉+¹₀𝜑_𝑛(𝜉)𝑔(𝜉)𝑑𝜉. (88) Multiplying by 𝜑_𝑚(𝑥) and integrating the new equation, we have 𝜌⁰₋₁𝜑_𝑚(𝑥)𝑓(𝑥)𝑑𝑥+¹₀𝜑_𝑚(𝑥)1(𝑥)𝑓𝑑𝑥=−𝜆_𝑚𝜌∫⁰₋₁𝜑_𝑚(𝜉)𝑔(𝜉)𝑑𝜉+∫¹₀𝜑_𝑚(𝜉)𝑔(𝜉)𝑑𝜉. (89) Then 𝑓(𝑥)=^∞_𝑛=1𝑐_𝑛𝑓𝜑_𝑛(𝑥), (90) where 𝑐_𝑛(𝑓)=⟨𝑓,𝜑_𝑛⟩=𝜌∫⁰₋₁𝑓(𝑥)𝜑_𝑛(𝑥)𝑑𝑥+∫¹₀𝑓(𝑥)𝜑_𝑛(𝑥)𝑑𝑥. Thus, for 𝑓∈𝐶^∞₀, 𝑓²_𝐻=^∞_𝑛=1𝑐²_𝑛𝑓. (91)
Next, we prove that (86) holds for 𝑓∈𝐻. For any 𝑓∈𝐻, there exists a sequence {𝑓_𝑘}_𝑘∈ℕ⊂𝐶^∞₀ converging to 𝑓 in 𝐻 since 𝐶^∞₀ is dense in 𝐻. Firstly, we prove that ∑^∞_𝑛=1𝑐²_𝑛(𝑓)<∞ and lim_𝑘→∞∑^∞_𝑛=1𝑐²_𝑛(𝑓_𝑘)=∑^∞_𝑛=1𝑐²_𝑛(𝑓).
By the Cauchy-Schwartz inequality, |𝑐_𝑛(𝑓_𝑘)−𝑐_𝑛(𝑓)|=|⟨𝑓_𝑘−𝑓,𝜑_𝑛⟩|≤‖𝑓_𝑘−𝑓‖_𝐻. Hence, lim_𝑘→∞𝑐_𝑛(𝑓_𝑘)=𝑐_𝑛(𝑓). Since ∑^∞_𝑛=1(𝑐_𝑛(𝑓_𝑘)−𝑐_𝑛(𝑓_𝑚))²=∑^∞_𝑛=1𝑐²_𝑛(𝑓_𝑘−𝑓_𝑚)=‖𝑓_𝑘−𝑓_𝑚‖²_𝐻 for 𝑓_𝑘−𝑓_𝑚∈𝐶^∞₀, ^𝑁_𝑛=1𝑐_𝑛𝑓_𝑘−𝑐_𝑛𝑓_𝑚²≤𝑓_𝑘−𝑓_𝑚²_𝐻. (92) Let 𝑘→∞, then inequality (92) becomes ∑^𝑁_𝑛=1(𝑐_𝑛(𝑓)−𝑐_𝑛(𝑓_𝑚))²≤‖𝑓−𝑓_𝑚‖²_𝐻. Letting 𝑁→∞, we have ^∞_𝑛=1𝑐_𝑛𝑓−𝑐_𝑛𝑓_𝑚²≤𝑓−𝑓_𝑚²_𝐻. (93) Then by the Minkowski inequality ^∞_𝑛=1𝑐²_𝑛𝑓=^∞_𝑛=1𝑐_𝑛𝑓−𝑐_𝑛𝑓_𝑚+𝑐_𝑛𝑓_𝑚²≤^∞_𝑛=1𝑐_𝑛𝑓−𝑐_𝑛𝑓_𝑚^21/2+^∞_𝑛=1𝑐²_𝑛𝑓_𝑚^1/22<∞ (94) and by Hölder’s inequality, ^∞_𝑛=1𝑐²_𝑛𝑓−^∞_𝑛=1𝑐²_𝑛𝑓_𝑘=^∞_𝑛=1𝑐_𝑛𝑓−𝑐_𝑛𝑓_𝑘𝑐_𝑛𝑓+𝑐_𝑛𝑓_𝑘≤^∞_𝑛=1𝑐_𝑛𝑓−𝑐_𝑛𝑓_𝑘^21/2∞_𝑛=1𝑐_𝑛𝑓+𝑐_𝑛𝑓_𝑘^21/2⟶0,as𝑘⟶∞. (95) This means that lim_𝑘→∞∑^∞_𝑛=1𝑐²_𝑛(𝑓_𝑘)=∑^∞_𝑛=1𝑐²_𝑛(𝑓).
Since 𝑓_𝑘→𝑓 in 𝐻 as 𝑘→∞,  lim_𝑘→∞‖𝑓_𝑘‖_𝐻=‖𝑓‖_𝐻. We obtain 𝑓²_𝐻=lim_𝑘→∞𝑓_𝑘²_𝐻=lim^∞_{𝑘→∞𝑛=1}𝑐²_𝑛𝑓_𝑘=^∞_𝑛=1𝑐²_𝑛𝑓. (96)

Acknowledgments

The authors thank the reviewers for their careful reading of the paper and for the constructive comments and suggestions which have led to the improvement of the presentation of the paper. This work was supported by the National Nature Science Foundation of China (11161030) and the Specialized Research Fund for the Doctoral Program of Higher Education of China (20121501120003).