A Sturm-Liouville Problem with a Discontinuous Coefficient and Containing an Eigenparameter in the Boundary Condition

Abstract

We study a Sturm-Liouville operator with eigenparameter-dependent boundary conditions and transmission conditions at two interior points. We give an operator-theoretic formulation, construct fundamental solutions, investigate some properties of the eigenvalues and corresponding eigenfunctions of the discontinuous Sturm-Liouville problem and then obtain asymptotic formulas for the eigenvalues and eigenfunctions and find Green function of the discontinuous Sturm-Liouville problem.

1. Introduction

In recent years, more and more researchers are interested in the discontinuous Sturm-Liouville problem for its application in physics (see [1–16]). Such problems are connected with discontinuous material properties, such as heat and mass transfer, varied assortment of physical transfer problems, vibrating string problems when the string loaded additionally with point masses, and diffraction problems. Moreover, there has been a growing interest in Sturm-Liouville problems with eigenparameter-dependent boundary conditions; that is, the eigenparameter appears not only in the differential equations but also in the boundary conditions of the problems (see [1–16] and corresponding bibliography).

In this paper, following [8] we consider the boundary value problem for the differential equation 𝜏𝑢∶=−𝑢^′′+𝑞(𝑥)𝑢=𝜆𝜔(𝑥)𝑢, (1) for 𝑥∈[−1,ℎ₁)∪(ℎ₁,ℎ₂)∪(ℎ₂,1] (i.e., 𝑥 belongs to [−1,1], but the two inner points are 𝑥=ℎ₁ and 𝑥=ℎ₂), where 𝑞(𝑥) is a real-valued function, continuous in [−1,ℎ₁), (ℎ₁,ℎ₂), and (ℎ₂,1] with the finite limits 𝑞(±ℎ₁)=lim_{𝑥→±ℎ1}, 𝑞(±ℎ₂)=lim_{𝑥→±ℎ2}; 𝜔(𝑥) is a discontinuous weight function such that 𝜔(𝑥)=𝜔²₁ for 𝑥∈[−1,ℎ₁), 𝜔(𝑥)=𝜔²₂ for 𝑥∈(ℎ₁,ℎ₂), and 𝜔(𝑥)=𝜔²₃ for 𝑥∈(ℎ₂,1], 𝜔>0 together with the standard boundary condition at 𝑥=−1𝐿₁𝑢∶=𝑢^′(−1)=0, (2) the spectral parameter-dependent boundary condition at 𝑥=1𝐿₂𝑢∶=𝜆𝛽^′₁𝑢(1)−𝛽^′₂𝑢^′(1)+𝛽₁𝑢(1)−𝛽₂𝑢^′(1)=0, (3) and the four transmission conditions at the points of discontinuity 𝑥=ℎ₁ and 𝑥=ℎ₂𝐿₃𝑢∶=𝛾₁𝑢ℎ₁−0−𝛿₁𝑢ℎ₁𝐿+0=0,₄𝑢∶=𝛾₂𝑢^′ℎ₁−0−𝛿₂𝑢^′ℎ₁𝐿+0=0,₅𝑢∶=𝛾₃𝑢ℎ₂−0−𝛿₃𝑢ℎ₂𝐿+0=0,₆𝑢∶=𝛾₄𝑢^′ℎ₂−0−𝛿₄𝑢^′ℎ₂+0=0, (4)(5)(6)(7) in the Hilbert space 𝐿₂(−1,ℎ₁)⊕𝐿₂(ℎ₁,ℎ₂)⊕𝐿₂(ℎ₂,1), where 𝜆∈ℂ is a complex spectral parameter; and all coefficients of the boundary and transmission conditions are real constants. We assume that 𝛽^′₂≠0 and |𝛽₁|+|𝛽₂|≠0. Moreover, we will assume that 𝜌∶=𝛽^′₁𝛽₂−𝛽₁𝛽^′₂>0. Some special cases of this problem arise after application of the method of separation of variables to the diverse assortment of physical problems, heat and mass transfer problems (e.g., see [11]), vibrating string problems when the string was loaded additionally with point masses (e.g., see [11]), and a thermal conduction problem for a thin laminated plate.

2. Operator-Theoretic Formulation of the Problem

In this section, we introduce a special inner product in the Hilbert space (𝐿₂(−1,ℎ₁)⊕𝐿₂(ℎ₁,ℎ₂)⊕𝐿₂(ℎ₂,1))⊕ℂ and define a linear operator 𝐴 in it so that the problem (1)–(5) can be interpreted as the eigenvalue problem for 𝐴. To this end, we define a new Hilbert space inner product on 𝐻∶=(𝐿₂(−1,ℎ₁)⊕𝐿₂(ℎ₁,ℎ₂)⊕𝐿₂(ℎ₂,1))⊕ℂ by ⟨𝐹,𝐺⟩_𝐻=𝜔²₁^ℎ1₋₁𝑓(𝑥)𝑔(𝑥)𝑑𝑥+𝜔²₂𝛿₁𝛿₂𝛾₁𝛾₂^ℎ2_ℎ1𝑓(𝑥)𝑔(𝑥)𝑑𝑥+𝜔²₃𝛿₁𝛿₂𝛿₃𝛿₄𝛾₁𝛾₂𝛾₃𝛾₄¹_ℎ2𝑓(𝑥)𝑔(𝑥)+𝛿𝑑𝑥₁𝛿₂𝛿₃𝛿₄𝜌𝛾₁𝛾₂𝛾₃𝛾₄𝑓₁𝑔₁, (8) for 𝐹=_𝑓^𝑓(𝑥)₁ and 𝐺=_𝑔^𝑔(𝑥)₁∈𝐻. For convenience, we will use the notations 𝑅₁(𝑢)∶=𝛽₁𝑢(1)−𝛽₂𝑢^′(1),𝑅^′₁(𝑢)∶=𝛽^′₁𝑢(1)−𝛽^′₂𝑢^′(1). (9) In this Hilbert space, we construct the operator 𝐴∶𝐻→𝐻 with domain 𝐷(𝐴)=𝑓(𝑥)𝑓𝐹=₁∣𝑓(𝑥),𝑓^′(𝑥)areabsolutelycontinuousin1,ℎ₁∪ℎ₁,ℎ₂∪ℎ₂𝑓ℎ,1andhavefinitelimits₁±0,𝑓ℎ₂±0,𝑓^′ℎ₁±0,𝑓^′ℎ₂±0,𝜏𝑓∈𝐿₂−1,ℎ₁⊕𝐿₂ℎ₁,ℎ₂⊕𝐿₂ℎ₂,𝐿,1₁𝑓=𝐿₃𝑓=𝐿₄𝑓=𝐿₅𝑓=𝐿₆𝑓𝑓=0,₁=𝑅^′₁𝑓 (10) which acts by the rule 1𝐴𝐹=𝜔(𝑥)−𝑓^′′+𝑞(𝑥)𝑓−𝑅₁𝑓𝑓(𝑥)𝑅with𝐹=_′1𝑓∈𝐷(𝐴). (11) Thus we can pose the boundary value transmission problem (1)–(7) in 𝐻 as 𝑢(𝑥)𝑅𝐴𝑈=𝜆𝑈,𝑈∶=_′1(𝑢)∈𝐷(𝐴). (12) It is readily verified that the eigenvalues of 𝐴 coincide with those of the problem (1)–(7).

Theorem 1. The operator 𝐴 is symmetric.

Proof. Let 𝐹=_𝑅^𝑓(𝑥)_′1(𝑓) and 𝐺=_𝑅^𝑔(𝑥)_′1(𝑔) be arbitrary elements of 𝐷(𝐴). Twice integrating by parts, we find ⟨𝐴𝐹,𝐺⟩_𝐻−⟨𝐹,𝐴𝐺⟩_𝐻=𝑊𝑓,𝑔;ℎ₁−0−𝑊𝑓,+𝛿𝑔;−1₁𝛿₂𝛾₁𝛾₂𝑊𝑓,𝑔;ℎ₂−0−𝑊𝑓,𝑔;ℎ₁+𝛿+0₁𝛿₂𝛿₃𝛿₄𝛾₁𝛾₂𝛾₃𝛾₄𝑊𝑓,𝑔;1−𝑊𝑓,𝑔;ℎ₂+𝛿+0₁𝛿₂𝛿₃𝛿₄𝜌𝛾₁𝛾₂𝛾₃𝛾₄𝑅^′₁𝑓𝑅₁𝑔−𝑅₁𝑓𝑅^′₁𝑔, (13) where, as usual, 𝑊(𝑓,𝑔;𝑥) denotes the Wronskian of 𝑓 and 𝑔; that is, 𝑊𝑓,𝑔;𝑥∶=𝑓(𝑥)𝑔^′(𝑥)−𝑓^′(𝑥)𝑔(𝑥). (14) Since 𝐹,𝐺∈𝐷(𝐴), the first components of these elements, that is, 𝑓 and 𝑔, satisfy the boundary condition (2). From this fact, we easily see that 𝑊𝑓,𝑔;−1=0. (15) Further, as 𝑓 and 𝑔 also satisfy both transmission conditions, we obtain 𝑊𝑓,𝑔;ℎ₁=𝛿−0¹𝛿₂𝛾₁𝛾₂𝑊𝑓,𝑔;ℎ₁,𝑊+0𝑓,𝑔;ℎ₂=𝛿−0¹𝛿₂𝛿₃𝛿₄𝛾₁𝛾₂𝛾₃𝛾₄𝑊𝑓,𝑔;ℎ₂+0. (16) Moreover, the direct calculations give 𝑅^′₁𝑓𝑅₁𝑔−𝑅₁𝑓𝑅^′₁𝑔=−𝜌𝑊𝑓,𝑔;1. (17) Now, inserting (15)–(17) in (13), we have ⟨𝐴𝐹,𝐺⟩_𝐻=⟨𝐹,𝐴𝐺⟩_𝐻(𝐹,𝐺∈𝐷(𝐴)), (18) and so 𝐴 is symmetric.

Recalling that the eigenvalues of (1)–(7) coincide with the eigenvalues of 𝐴, we have the next corollary.

Corollary 2. All eigenvalues of (1)–(7) are real.

Since all eigenvalues are real, it is enough to study only the real-valued eigenfunctions. Therefore, we can now assume that all eigenfunctions of (1)–(7) are real-valued.

3. Asymptotic Formulas for Eigenvalues and Fundamental Solutions

Let us define fundamental solutions 𝜙()=𝜙𝑥,𝜆¹(𝑥,𝜆),𝑥∈−1,ℎ₁,𝜙₂(𝑥,𝜆),𝑥∈ℎ₁,ℎ₂,𝜙₃(𝑥,𝜆),𝑥∈ℎ₂,𝜒()=𝜒,1𝑥,𝜆¹(𝑥,𝜆),𝑥∈−1,ℎ₁,𝜒₂(𝑥,𝜆),𝑥∈ℎ₁,ℎ₂,𝜒₃(𝑥,𝜆),𝑥∈ℎ₂,1, (19) of (1) by the following procedure. We first consider the next initial value problem: −𝑢^′′+𝑞(𝑥)𝑢=𝜆𝜔²₁𝑢,𝑥∈−1,ℎ₁,𝑢()𝑢−1=1,_′(−1)=0. (20)(21)(22) By virtue of [1, Theorem 1.5], the problem (20)–(22) has a unique solution 𝑢=𝜙₁(𝑥,𝜆) which is an entire function of 𝜆∈ℂ for each fixed 𝑥∈[−1,ℎ₁]. Similarly, −𝑢^′′+𝑞(𝑥)𝑢=𝜆𝜔²₂𝑢,𝑥∈ℎ₁,ℎ₂,𝑢ℎ₁=𝛾₁𝛿₁𝜙₁ℎ₁,𝑢,𝜆_′ℎ₁=𝛾₂𝛿₂𝜙^′₁ℎ₁,𝜆, (23)(24)(25) have a unique solution 𝑢=𝜙₂(𝑥,𝜆) which is an entire function of 𝜆∈ℂ for each fixed 𝑥∈[ℎ₁,ℎ₂]. Continuing in this manner, −𝑢^′′+𝑞(𝑥)𝑢=𝜆𝜔²₃𝑢,𝑥∈ℎ₂,𝑢ℎ,1₂=𝛾₃𝛿₃𝜙₂ℎ₂,𝑢,𝜆_′ℎ₂=𝛾₄𝛿₄𝜙^′₂ℎ₂,𝜆, (26)(27)(28) have a unique solution 𝑢=𝜙₃(𝑥,𝜆) which is an entire function of 𝜆∈ℂ for each fixed 𝑥∈[ℎ₂,1]. Slightly modifying the method of [1, Theorem 1.5], we can prove the initial value problem −𝑢^′′+𝑞(𝑥)𝑢=𝜆𝜔²₃𝑢,𝑥∈ℎ₂,𝑢(1),1=𝛽^′₂𝜆+𝛽₂,𝑢^′(1)=𝛽^′₁𝜆+𝛽₁, (29)(30)(31) have a unique solution 𝑢=𝜒₃(𝑥,𝜆) which is an entire function of spectral parameter 𝜆∈ℂ for each fixed 𝑥∈[ℎ₂,1]. Similarly, −𝑢^′′+𝑞(𝑥)𝑢=𝜆𝜔²₂𝑢,𝑥∈ℎ₁,ℎ₂,𝑢ℎ₂=𝛿₃𝛾₃𝜒₃ℎ₂,𝑢,𝜆_′ℎ₂=𝛿₄𝛾₄𝜒^′₃ℎ₂,𝜆, (32)(33)(34) have a unique solution 𝑢=𝜒₂(𝑥,𝜆) which is an entire function of 𝜆∈ℂ for each fixed 𝑥∈[ℎ₁,ℎ₂]. Continuing in this manner, −𝑢^′′+𝑞(𝑥)𝑢=𝜆𝜔²₃𝑢,𝑥∈−1,ℎ₁,𝑢ℎ₁=𝛿₁𝛾₁𝜒₂ℎ₁,𝑢,𝜆_′ℎ₁=𝛿₂𝛾₂𝜒^′₂ℎ₁,𝜆, (35)(36)(37) have a unique solution 𝑢=𝜒₁(𝑥,𝜆) which is an entire function of 𝜆∈ℂ for each fixed 𝑥∈[−1,ℎ₁].

By virtue of (21) and (22), the solution 𝜙(𝑥,𝜆) satisfies the first boundary condition (2). Moreover, by (24), (25), (27), and (28), 𝜙(𝑥,𝜆) satisfies also transmission conditions (4)–(7). Similarly, by (30), (31), (33), (34), (36), and (37) the other solution 𝜒(𝑥,𝜆) satisfies the second boundary condition (3) and transmission conditions (4)–(7). It is well known from the theory of ordinary differential equations that each of the Wronskians Δ₁(𝜆)=𝑊(𝜙₁(𝑥,𝜆),𝜒₁(𝑥,𝜆)), Δ₂(𝜆)=𝑊(𝜙₂(𝑥,𝜆),𝜒₂(𝑥,𝜆)), and Δ₃(𝜆)=𝑊(𝜙₃(𝑥,𝜆),𝜒₃(𝑥,𝜆)) is independent of 𝑥 in [−1,ℎ₁], [ℎ₁,ℎ₂], and [ℎ₂,1], respectively.

Lemma 3. The equality Δ₁(𝜆)=(𝛿₁𝛿₂/𝛾₁𝛾₂)Δ₂(𝜆)=(𝛿₁𝛿₂𝛿₃𝛿₄/𝛾₁𝛾₂𝛾₃𝛾₄)Δ₃(𝜆) holds for each 𝜆∈ℂ.

Proof. Since the above Wronskians are independent of 𝑥, using (27), (28), (30), (31), (33), (34), (36), and (37), we find Δ₁(𝜆)=𝜙₁ℎ₁,𝜆𝜒^′₁ℎ₁,𝜆−𝜙^′₁ℎ₁,𝜆𝜒₁ℎ₁=𝛿,𝜆₁𝛾₁𝜙₂ℎ₁𝛿,𝜆²𝛾₂𝜒^′₂ℎ₁−𝛿,𝜆₂𝛾₂𝜙^′₂ℎ₁𝛿,𝜆¹𝛾₁𝜒₂ℎ₁=𝛿,𝜆₁𝛿₂𝛾₁𝛾₂Δ₂(𝜆)=𝛿₁𝛿₃𝛾₁𝛾₃𝜙₃ℎ₂𝛿,𝜆²𝛿₄𝛾₂𝛾₄𝜒^′₃ℎ₂−𝛿,𝜆₂𝛿₄𝛾₂𝛾₄𝜙^′₃ℎ₂𝛿,𝜆¹𝛿₃𝛾₁𝛾₃𝜒₃ℎ₂=𝛿,𝜆₁𝛿₂𝛿₃𝛿₄𝛾₁𝛾₂𝛾₃𝛾₄Δ₃(𝜆). (38)

Corollary 4. The zeros of Δ₁(𝜆), Δ₂(𝜆), and Δ₃(𝜆) coincide.

In view of Lemma 3, we denote Δ₁(𝜆), (𝛿₁𝛿₂/𝛾₁𝛾₂)Δ₂(𝜆), and (𝛿₁𝛿₂𝛿₃𝛿₄/𝛾₁𝛾₂𝛾₃𝛾₄)Δ₃(𝜆) by Δ(𝜆). Recalling the definitions of 𝜙_𝑖(𝑥,𝜆) and 𝜒_𝑖(𝑥,𝜆), we infer the next corollary.

Corollary 5. The function Δ(𝜆) is an entire function.

Theorem 6. The eigenvalues of (1)–(7) coincide with the zeros of Δ(𝜆).

Proof. Let Δ(𝜆₀)=0. Then, 𝑊(𝜙₁(𝑥,𝜆₀),𝜒₁(𝑥,𝜆₀))=0 for all 𝑥∈[−1,ℎ₁]. Consequently, the functions 𝜙₁(𝑥,𝜆₀) and 𝜒₁(𝑥,𝜆₀) are linearly dependent; that is, 𝜒₁(𝑥,𝜆₀)=𝑘𝜙₁(𝑥,𝜆₀), 𝑥∈[−1,ℎ₁], for some 𝑘≠0. By (21) and (22), from this equality, we have 𝜒^′−1,𝜆₀=𝑘𝜙^′₁−1,𝜆₀=0, (39) and so 𝜒(𝑥,𝜆₀) satisfies the first boundary condition (2). Recalling that the solution 𝜒(𝑥,𝜆₀) also satisfies the other boundary condition (3) and transmission conditions (4)–(7), we conclude that 𝜒(𝑥,𝜆₀) is an eigenfunction of (1)–(7); that is, 𝜆₀ is an eigenvalue. Thus, each zero of Δ(𝜆) is an eigenvalue. Now, let 𝜆₀ be an eigenvalue, and let 𝑢₀(𝑥) be an eigenfunction with this eigenvalue. Suppose that Δ(𝜆₀)≠0, whence 𝑊(𝜙₁(𝑥,𝜆₀),𝜒₁(𝑥,𝜆₀))≠0, 𝑊(𝜙₂(𝑥,𝜆₀),𝜒₂(𝑥,𝜆₀))≠0, and 𝑊(𝜙₃(𝑥,𝜆₀),𝜒₃(𝑥,𝜆₀))≠0. From this, by virtue of the well-known properties of Wronskians, it follows that each of the pairs 𝜙₁(𝑥,𝜆₀), 𝜒₁(𝑥,𝜆₀); 𝜙₂(𝑥,𝜆₀), 𝜒₂(𝑥,𝜆₀) and 𝜙₃(𝑥,𝜆₀), 𝜒₃(𝑥,𝜆₀) is linearly independent. Therefore, the solution 𝑢₀(𝑥) of (1) may be represented as 𝑢₀(𝑥)=𝑐₁𝜙₁𝑥,𝜆₀+𝑐₂𝜒₁𝑥,𝜆₀,𝑥∈−1,ℎ₁,𝑐₃𝜙₂𝑥,𝜆₀+𝑐₄𝜒₂𝑥,𝜆₀,𝑥∈ℎ₁,ℎ₂,𝑐₅𝜙₃𝑥,𝜆₀+𝑐₆𝜒₃𝑥,𝜆₀,𝑥∈ℎ₂,1, (40) where at least one of the coefficients 𝑐_𝑖(𝑖=1,6) is not zero. Considering the true equalities 𝐿_𝜐𝑢₀(𝑥)=0,𝜐=1,6 (41) as the homogenous system of linear equations in the variables 𝑐_𝑖(𝑖=1,6) and taking (24), (25), (27), (28), (33), (34), (36), and (37) into account, we see that the determinant of this system is equal to −((𝛿₁𝛿₂𝛿₃𝛿₄)²/𝛾₁𝛾₂𝛾₃𝛾₄)Δ⁴(𝜆₀), and so it does not vanish by assumption. Consequently, the system (41) has the only trivial solution 𝑐_𝑖=0(𝑖=1,6). We thus get at a contradiction, which completes the proof.

Theorem 7. Let 𝜆=𝜇² and Im𝜇=𝑡. Then, the following asymptotic equalities hold as |𝜆|→∞: 𝜙₁^(𝑘)()=𝑑𝑥,𝜆^𝑘𝑑𝑥^𝑘cos𝜇𝜔₁(1𝑥+1)+𝑂𝜇^1−𝑘exp|𝑡|𝜔₁(),𝜙𝑥+1_2(𝑘)()=𝛾𝑥,𝜆¹𝛿₁𝑑^𝑘𝑑𝑥^𝑘cos𝜇𝜔₂𝑥+𝜔₁ℎ₁+𝜔₁1+𝑂𝜇^1−𝑘exp|𝑡|𝜔₂𝑥+𝜔₁ℎ₁+𝜔₁,𝜙₃^(𝑘)()=𝛾𝑥,𝜆¹𝛾₃𝛿₁𝛿₃𝑑^𝑘𝑑𝑥^𝑘cos𝜇𝜔₃𝑥+𝜔₂ℎ₂+𝜔₁1+𝑂𝜇^1−𝑘exp|𝑡|𝜔₃𝑥+𝜔₂ℎ₂+𝜔₁, (42)(43)(44) for 𝑘=0 and 𝑘=1. Moreover, each of these asymptotic equalities holds uniformly for 𝑥.

Proof. Asymptotic formulas for 𝜙₁(𝑥,𝜆) and 𝜙₂(𝑥,𝜆) are found in [9, Lemma 1.7] and [8, Theorem 3.2], respectively. But the formulas for 𝜙₃(𝑥,𝜆) need individual considerations, since this solution is defined by the initial condition with some special nonstandard form. The initial-value problem (26)–(28) can be transformed into the equivalent integral equation 𝑢(𝑥)=𝛾₃𝛿₃𝜙₂ℎ₂,𝜆cos𝜇𝜔₃𝛾𝑥+⁴𝜇𝜔₃𝛿₄𝜙^′₂ℎ₂,𝜆sin𝜇𝜔₃𝑥+𝜔₃𝜇^𝑥_ℎ2sin𝜇𝜔₃𝑥−𝑦𝑞𝑦𝑢𝑦𝑑𝑦. (45) Inserting (43) in (45), we have 𝜙₃()=𝛾𝑥,𝜆¹𝛾₃𝛿₁𝛿₃cos𝜇𝜔₃𝑥+𝜔₂ℎ₂+𝜔₁+𝜔₃𝜇^𝑥_ℎ2sin𝜇𝜔₃𝑥−𝑦𝑞𝑦𝜙₃1𝑦,𝜆𝑑𝑦+𝑂𝜇exp|𝑡|𝜔₃𝑥+𝜔₂ℎ₂+𝜔₁. (46) Multiplying this by exp(−|𝑡|(𝜔₃𝑥+𝜔₂ℎ₂+𝜔₁)) and denoting 𝐹(𝑥,𝜆)=exp(−|𝑡|(𝜔₃𝑥+𝜔₂ℎ₂+𝜔₁))𝜙₃(𝑥,𝜆), we have the next “asymptotic integral equation”: 𝐹()=𝛾𝑥,𝜆¹𝛾₃𝛿₁𝛿₃exp−|𝑡|𝜔₃𝑥+𝜔₂ℎ₂+𝜔₁×cos𝜇𝜔₃𝑥+𝜔₂ℎ₂+𝜔₁+𝜔₃𝜇^𝑥_ℎ2sin𝜇𝜔₃−|𝑡|𝜔𝑥−𝑦×exp₃𝑞𝑦𝐹1𝑥−𝑦𝑦,𝜆𝑑𝑦+𝑂𝜇. (47) Putting 𝑀(𝜆)=max_{𝑥∈[ℎ2,1]}|𝐹(𝑥,𝜆)|, from the last equation, we derive that 𝑀(𝜆)≤𝑀₀𝛾₁𝛾₃𝛿₁𝛿₃+1𝜇, (48) for some 𝑀₀>0. Consequently, 𝑀(𝜆)=𝑂(1) as |𝜆|→∞, and so 𝜙₃(𝑥,𝜆) = 𝑂(exp(|𝑡|(𝜔₃𝑥+𝜔₂ℎ₂+𝜔₁))) as |𝜆|→∞. Inserting the integral term of (46) yields (44) for 𝑘=0. The case 𝑘=1 of (44) follows at once on differentiating (43) and making the same procedure as in the case 𝑘=0.

Theorem 8. Let 𝜆=𝜇², 𝜇=𝜎+𝑖𝑡. Then, the following asymptotic formula holds for the eigenvalues of the boundary value transmission problem (1)–(7): 𝜇_𝑛=𝜋(𝑛−1)𝜔₃+𝜔₂ℎ₂+𝜔₁1+𝑂𝑛. (49)

Proof. Putting 𝑥=1 in Δ₃(𝜆)=𝜙₃(𝑥,𝜆)𝜒^′₃(𝑥,𝜆)−𝜙^′₃(𝑥,𝜆)𝜒₃(𝑥,𝜆) and inserting 𝜒₃(1,𝜆)=𝛽^′₂𝜆+𝛽₂, 𝜒^′₃(1,𝜆)=𝛽^′₁𝜆+𝛽₁, we have the following representation for Δ₃(𝜆): Δ₃(𝜆)=𝛽^′₁𝜆+𝛽₁𝜙₃(1,𝜆)−𝛽^′₂𝜆+𝛽₂𝜙^′₃(1,𝜆). (50) Putting 𝑥=1 in (44) and inserting the result in (50), we derive now that Δ₃(𝜆)=𝛿₂𝛿₄𝛾₂𝛾₄𝜔₃𝛽^′₂𝜇³sin𝜇𝜔₃+𝜔₂ℎ₂+𝜔₁+𝑂𝜇²exp2|𝑡|𝜔+𝜔₂ℎ₂+𝜔₁. (51) By applying the well-known Rouché theorem which asserts that if 𝑓(𝑧) and 𝑔(𝑧) are analytic inside and on a closed contour Γ and |𝑔(𝑧)|<|𝑓(𝑧)| on Γ, then 𝑓(𝑧) and 𝑓(𝑧)+𝑔(𝑧) have the same number of zeros inside Γ provided that the zeros are counted with multiplicity on a sufficiently large contour, it follows that Δ₃(𝜆) has the same number of zeros inside the contour as the leading term in (51). Hence, if 𝜆₀<𝜆₁<𝜆₂… are the zeros of Δ₃(𝜆) and 𝜇²_𝑛=𝜆_𝑛, we have 𝜋(𝑛−1)𝜔₃+𝜔₂ℎ₂+𝜔₁+𝛿_𝑛, (52) for sufficiently large 𝑛, where |𝛿_𝑛|<𝜋/4(𝜔₃+𝜔₂ℎ₂+𝜔₁) for sufficiently large 𝑛. By putting in (51), we have 𝛿_𝑛=𝑂(1/𝑛), and the proof is completed.

Theorem 9. The following asymptotic formula holds for the eigenfunctions 𝜙_𝜆𝑛(𝑥)=𝜙₁𝑥,𝜆_𝑛,𝑥∈−1,ℎ₁,𝜙₂𝑥,𝜆_𝑛,𝑥∈ℎ₁,ℎ₂,𝜙₃𝑥,𝜆_𝑛,𝑥∈ℎ₂,1, (53) of (1)–(7):

𝜙_𝜆𝑛(𝑥)=𝜔cos¹𝜋(𝑛−1)(𝑥+1)𝜔₂+𝜔₁1+𝑂𝑛,𝑥∈−1,ℎ₁,𝛾₁𝛿₁𝜔cos²𝑥+𝜔₁ℎ₁+𝜔₁𝜋(𝑛−1)𝜔₂+𝜔₁ℎ₁+𝜔₁1+𝑂𝑛,𝑥∈ℎ₁,ℎ₂,𝛾₁𝛾₃𝛿₁𝛿₃𝜔cos³𝑥+𝜔₂ℎ₂+𝜔₁𝜋(𝑛−1)𝜔₃+𝜔₂ℎ₂+𝜔₁1+𝑂𝑛,𝑥∈ℎ₂,1. (54)

Proof. Inserting (44) in the integral term of (51), we easily see that ^𝑥_ℎ2sin𝜇𝜔₃𝑥−𝑦𝑞𝑦𝜙₃|𝑡|𝜔𝑦,𝜆𝑑𝑦=𝑂exp₃𝑥+𝜔₂ℎ₂+𝜔₁. (55)   Inserting in (42) yields 𝜙₃()=𝛾𝑥,𝜆¹𝛾₃𝛿₁𝛿₃cos𝜇𝜔₃𝑥+𝜔₂ℎ₂+𝜔₁1+𝑂𝜇exp|𝑡|𝜔₃𝑥+𝜔₂ℎ₂+𝜔₁. (56) We already know that all eigenvalues are real. Furthermore, putting 𝜆=−𝐻, 𝐻>0 in (51), we infer that 𝜔(−𝐻)→∞ as 𝐻→+∞, and so 𝜔(−𝐻)≠0 for sufficiently large 𝑅>0. Consequently, the set of eigenvalues is bounded below. Letting 𝜆_𝑛=𝜇_𝑛 in (56), we now obtain 𝜙₃𝑥,𝜆_𝑛=𝛾₁𝛾₃𝛿₁𝛿₃cos𝜇_𝑛𝜔₃𝑥+𝜔₂ℎ₂+𝜔₁1+𝑂𝜇_𝑛, (57) since 𝑡_𝑛=𝑙𝑚𝜇_𝑛 for sufficiently large 𝑛. After some calculation, we easily see that cos𝜇_𝑛𝜔₃𝑥+𝜔₂ℎ₂+𝜔₁𝜔=cos³𝑥+𝜔₂ℎ₂+𝜔₁𝜋(𝑛−1)𝜔₃+𝜔₂ℎ₂+𝜔₁1+𝑂𝑛. (58) Consequently, 𝜙₃𝑥,𝜆_𝑛=𝛾₁𝛾₃𝛿₁𝛿₃𝜔cos³𝑥+𝜔₂ℎ₂+𝜔₁𝜋(𝑛−1)𝜔₃+𝜔₂ℎ₂+𝜔₁1+𝑂𝑛. (59) In a similar method, we can deduce that 𝜙₂𝑥,𝜆_𝑛=𝛾₁𝛿₁𝜔cos²𝑥+𝜔₁ℎ₁+𝜔₁𝜋(𝑛−1)𝜔₂+𝜔₁ℎ₁+𝜔₁1+𝑂𝑛,𝜙₁𝑥,𝜆_𝑛𝜔=cos¹𝜋(𝑛−1)(𝑥+1)𝜔₂+𝜔₁1+𝑂𝑛. (60) Thus is the proof of the theorem, completed.

4. Asymptotic Formulas for Normalized Eigenfunctions

We will first obtain the norms of the eigenelements Φ_𝑛𝜙∶=^𝜆𝑛(𝑥)𝑅^′₁𝜙_𝜆𝑛. (61) It is evident that the two-component vectors Φ_𝑛𝜙∶=^𝜆𝑛(𝑥)𝑅^′₁𝜙_𝜆𝑛,𝑛=0,1,2,…, (62) are the eigenelements of 𝐴 with eigenvalue 𝜆_𝑛. For 𝑛≠𝑚, Φ_𝑛,Φ_𝑚𝐻=0,𝑛,𝑚=0,1,2,… (63) since 𝐴 is symmetric. Putting 𝜓_𝑛𝜙∶=^𝜆𝑛(𝑥)Φ_𝑛𝐻, (64) we see easily that the eigenelements Ψ_𝑛=𝜓_𝜆𝑛(𝑥)𝑅^′₁𝜓_𝜆𝑛,𝑛,𝑚=0,1,2,… (65) are orthonormal. That is, 𝐴Ψ_𝑛=𝜆_𝑛Ψ_𝑛 and ⟨Ψ_𝑛,Ψ_𝑚⟩_𝐻=𝛿_𝑛𝑚, where 𝛿_𝑛𝑚 is the Kronecker delta.

Lemma 10. The following asymptotic equality holds: 𝑅^′₁𝜙_𝜆𝑛1=𝑂𝑛. (66)

Proof. It follows that from the equalities Δ₃(𝜆_𝑛)=0 that 𝜆_𝑛𝑅^′₁𝜙_3𝜆𝑛+𝑅₁𝜙_3𝜆𝑛=0. (67) From (44), we obtain 𝑅₁𝜙_3𝜆𝑛=𝛽₁𝜙_3𝜆𝑛(1)−𝛽₂𝜙^′_3𝜆𝑛(1)=𝛽₁𝑂(1)−𝛽₂𝑂𝑠_𝑛. (68) Applying Theorem 8, we now have 𝑅₁𝜙_3𝜆𝑛=𝑂(𝑛). (69) Inserting (69) in (67) and using Theorem 8, we find 𝑅^′₁𝜙_3𝜆𝑛1=−𝜆_𝑛𝑅₁𝜙_3𝜆𝑛1=𝑂𝑛. (70) The proof is completed.

Theorem 11. Let Φ_𝑛 be defined as in (61). Then, the following asymptotic formula holds for the norms ‖Φ_𝑛‖_𝐻 of the eigenelements Φ_𝑛∶Φ_𝑛𝐻=𝜔²₁𝛿₁𝛿₃𝛾₂𝛾₄+𝜔²₂𝛿₂𝛿₃𝛾₁𝛾₄+𝜔²₃𝛿₂𝛿₄𝛾₁𝛾₃2𝛿₁𝛿₃𝛾₂𝛾₄1+𝑂𝑛. (71)

Proof. By (54), we have ^ℎ1₋₁𝜙_𝜆𝑛(𝑥)²1𝑑𝑥=21+𝑂𝑛_ℎ2ℎ1,𝜙_𝜆𝑛(𝑥)²𝛾𝑑𝑥=²¹2𝛿²₁1+𝑂𝑛_1ℎ2,𝜙_𝜆𝑛(𝑥)²𝛾𝑑𝑥=²¹2𝛿²₁𝛾²₃𝛿²₃1+𝑂𝑛. (72)(73)(74) Using (66), (72), (73), and (74), we obtain Φ_𝑛²_𝐻=𝜔²₁^ℎ1₋₁𝜙_𝜆𝑛(𝑥)²𝑑𝑥+𝜔²₂𝛿₁𝛿₂𝛾₁𝛾₂^ℎ2_ℎ1𝜙_𝜆𝑛(𝑥)²𝑑𝑥+𝜔²₃𝛿₁𝛿₂𝛿₃𝛿₄𝛾₁𝛾₂𝛾₃𝛾₄¹_ℎ2𝜙_𝜆𝑛(𝑥)²+𝛿𝑑𝑥₁𝛿₂𝛿₃𝛿₄𝜌𝛾₁𝛾₂𝛾₃𝛾₄𝑅^′₁𝜙_𝜆𝑛²=𝜔²₁121+𝑂𝑛+𝜔²₂𝛿₁𝛿₂𝛾₁𝛾₂𝛾²₁2𝛿²₁1+𝑂𝑛+𝜔²₃𝛿₁𝛿₂𝛿₃𝛿₄𝛾₁𝛾₂𝛾₃𝛾₄𝛾²₁2𝛿²₁𝛾²₃𝛿²₃1+𝑂𝑛+𝛿₁𝛿₂𝛿₃𝛿₄𝜌𝛾₁𝛾₂𝛾₃𝛾₄𝑂1𝑛²=𝜔²₁𝛿₁𝛿₃𝛾₂𝛾₄+𝜔²₂𝛿₂𝛿₃𝛾₁𝛾₄+𝜔²₃𝛿₂𝛿₄𝛾₁𝛾₃2𝛿₁𝛿₃𝛾₂𝛾₄1+𝑂𝑛. (75) Consequently, Φ_𝑛𝐻=𝜔²₁𝛿₁𝛿₃𝛾₂𝛾₄+𝜔²₂𝛿₂𝛿₃𝛾₁𝛾₄+𝜔²₃𝛿₂𝛿₄𝛾₁𝛾₃2𝛿₁𝛿₃𝛾₂𝛾₄1+𝑂𝑛, (76) which proves (71).

Theorem 12. The first components of the normalized eigenelements (65) have the following asymptotic representation as 𝑛→∞:

𝜓_𝑛(𝑥)=^{2𝛿1𝛿3𝛾2𝛾4𝜔21𝛿1𝛿3𝛾2𝛾4+𝜔22𝛿2𝛿3𝛾1𝛾4+𝜔23𝛿2𝛿4𝛾1𝛾3𝜔cos1𝜋()𝑛−1)(𝑥+1𝜔1+𝜔21+𝑂𝑛,𝑥∈−1,ℎ1,𝛾1}_𝛿1^{2𝛿1𝛿3𝛾2𝛾4}_{𝜔21𝛿1𝛿3𝛾2𝛾4+𝜔22𝛿2𝛿3𝛾1𝛾4+𝜔23𝛿2𝛿4𝛾1𝛾3}^{𝜋(𝜔cos𝑛−1)2𝑥+𝜔1ℎ1+𝜔1}_{𝜔2+𝜔1ℎ1+𝜔1}^1+𝑂_𝑛^ℎ,𝑥∈₁^,ℎ₂^,_{𝛾1𝛿1𝛾3𝛿32𝛿1𝛿3𝛾2𝛾4𝜔21𝛿1𝛿3𝛾2𝛾4+𝜔22𝛿2𝛿3𝛾1𝛾4+𝜔23𝛿2𝛿4𝛾1𝛾3𝜋(𝜔cos𝑛−1)3𝑥+𝜔2ℎ2+𝜔1𝜔3+𝜔2ℎ2+𝜔11+𝑂𝑛ℎ,𝑥∈2.,1} (77)

Proof. From (71), it follows that 1Φ_𝑛𝐻=2𝛿₁𝛿₃𝛾₂𝛾₄𝜔²₁𝛿₁𝛿₃𝛾₂𝛾₄+𝜔²₂𝛿₂𝛿₃𝛾₁𝛾₄+𝜔²₃𝛿₂𝛿₄𝛾₁𝛾₃1+𝑂𝑛. (78) Putting (54) and (78) in (64), we find the needed asymptotic formula (77).

5. Green's Function

Let 𝐴∶𝐻→𝐻 be defined by (10) and (11), and let 𝜆 not be an eigenvalue of 𝐴. For finding the resolvent operator 𝑅(𝜆,𝐴)=(𝜆𝐼−𝐴)⁻¹, consider the operator equation (𝜆𝐼−𝐴)𝑈=𝐹 (79) for 𝐹=_𝑓^𝑓(𝑥)₁∈𝐻. This operator equation is equivalent to the inhomogeneous differential equation 𝑢^′′+𝜆𝜔(𝑥)−𝑞(𝑥)𝑢=𝑓(𝑥)for𝑥∈−1,ℎ₁∪ℎ₁,ℎ₂∪ℎ₂,1 (80) subject to the inhomogeneous boundary conditions 𝑢^′()𝜆𝛽−1=0,_′1𝑢(1)−𝛽^′₂𝑢^′(1)+𝛽₁𝑢(1)−𝛽₂𝑢^′(1)=𝑓₁ (81) and the homogeneous transmission conditions 𝛾₁𝑢ℎ₁−0−𝛿₁𝑢ℎ₁𝛾+0=0,₂𝑢^′ℎ₁−0−𝛿₂𝑢^′ℎ₁𝛾+0=0,₃𝑢ℎ₂−0−𝛿₃𝑢ℎ₂𝛾+0=0,₄𝑢^′ℎ₂−0−𝛿₄𝑢^′ℎ₂+0=0. (82) By applying the same techniques as in [8], we can prove that the problem (80)–(82) has a unique solution 𝑢(𝑥,𝜆) which can be represented as𝑢()=𝜒𝑥,𝜆¹(𝑥,𝜆)Δ₁(𝜆)𝜔²₁^𝑥₋₁𝜙₁𝑓𝑦+𝜙𝑦,𝜆𝑑𝑦₁(𝑥,𝜆)Δ₁(𝜆)𝜔²₁^ℎ1_𝑥𝜒₁𝑓𝑦𝛿𝑦,𝜆𝑑𝑦+¹𝛿₂𝛾₁𝛾₂𝜔²₂^ℎ2_ℎ1𝜒₂𝑓𝑦+𝛿𝑦,𝜆𝑑𝑦₁𝛿₂𝛿₃𝛿₄𝛾₁𝛾₂𝛾₃𝛾₄𝜔²₃¹_ℎ2𝜒₃𝑓𝑦𝛿𝑦,𝜆𝑑𝑦+¹𝛿₂𝛿₃𝛿₄𝛾₁𝛾₂𝛾₃𝛾₄𝑓₁,𝑥∈−1,ℎ₁,𝜒₂(𝑥,𝜆)Δ₂(𝜆)𝛾₁𝛾₂𝛿₁𝛿₂𝜔²₁^ℎ1₋₁𝜙₁𝑦,𝜆𝑓𝑦𝑑𝑦+𝜔²₂^𝑥_ℎ1𝜙₂𝑓𝑦×𝜙𝑦,𝜆𝑑𝑦₂(𝑥,𝜆)Δ₂(𝜆)𝜔²₂^ℎ2_𝑥𝜒₂𝑓𝑦+𝛿𝑦,𝜆𝑑𝑦₃𝛿₄𝛾₃𝛾₄𝜔²₃¹_ℎ2𝜒₃𝑓𝑦𝛿𝑦,𝜆𝑑𝑦+³𝛿₄𝛾₃𝛾₄𝑓₁,𝑥∈ℎ₁,ℎ₂,𝜒₃(𝑥,𝜆)Δ₃(𝜆)𝛾₁𝛾₂𝛾₃𝛾₄𝛿₁𝛿₂𝛿₃𝛿₄𝜔²₁^ℎ1₋₁𝜙₁𝑓𝑦+𝛾𝑦,𝜆𝑑𝑦₃𝛾₄𝛿₃𝛿₄𝜔²₂^ℎ2_ℎ1𝜙₂𝑦,𝜆𝑓𝑦𝑑𝑦+𝜔²₃^𝑥_ℎ2𝜙₃𝑓𝑦+𝜙𝑦,𝜆𝑑𝑦₃(𝑥,𝜆)Δ₃(𝜆)𝜔²₃¹_𝑥𝜒₃𝑦,𝜆𝑓𝑦𝑑𝑦+𝑓₁,𝑥∈ℎ₂,1. (83)Putting 𝐺=𝜒()𝜙𝑥,𝑦,𝜆𝑥,𝜆𝑦,𝜆𝜔(𝜆)𝜙()𝜒for−1≤𝑦≤𝑥≤1,𝑥,𝜆𝑦,𝜆𝜔(𝜆)for−1≤𝑥≤𝑦≤1, (84) where 𝑥≠0 and 𝑦≠0, we reduce (73) to 𝑢(𝑥,𝜆)=𝜔²₁^ℎ1₋₁𝐺𝑥,𝑦,𝜆𝑓𝑦𝑑𝑦+𝜔²₂𝛿₁𝛿₂𝛾₁𝛾₂^ℎ2_ℎ1𝐺𝑥,𝑦,𝜆𝑓𝑦𝑑𝑦+𝜔²₃𝛿₁𝛿₂𝛿₃𝛿₄𝛾₁𝛾₂𝛾₃𝛾₄¹_ℎ2𝐺𝑥,𝑦,𝜆𝑓𝑦𝑑𝑦+𝑓₁𝜙(𝑥,𝜆)𝜔(𝜆). (85) On the other hand, by applying 𝑅^′₁ to Green's function with respect to the variable 𝑦 and recalling that 𝜒(𝑥,𝜆) satisfies the initial conditions (30) and (31), we have 𝑅^′₁(𝐺(𝑥,⋅,𝜆))=𝛽^′₁𝐺(𝑥,1,𝜆)−𝛽^′₂𝜕𝐺(𝑥,1,𝜆)=𝛾𝜕𝑦₁𝛾₂𝛾₃𝛾₄𝛿₁𝛿₂𝛿₃𝛿₄𝜙(𝑥,𝜆)𝜔(𝜆)𝛽^′₁𝜒(1,𝜆)−𝛽^′₂𝜒^′()=𝛾1,𝜆₁𝛾₂𝛾₃𝛾₄𝛿₁𝛿₂𝛿₃𝛿₄𝜙(𝑥,𝜆)𝜔(𝜆)×𝛽^′₁𝛽^′₂𝜆+𝛽₂−𝛽^′₂𝛽^′₁𝜆+𝛽₁=𝜌𝛾₁𝛾₂𝛾₃𝛾₄𝛿₁𝛿₂𝛿₃𝛿₄𝜙(𝑥,𝜆)𝜔(𝜆). (86)   Inserting in (85) yields 𝑢(𝑥,𝜆)=𝜔²₁^ℎ1₋₁𝐺𝑥,𝑦,𝜆𝑓𝑦𝑑𝑦+𝜔²₂𝛿₁𝛿₂𝛾₁𝛾₂^ℎ2_ℎ1𝐺𝑥,𝑦,𝜆𝑓𝑦𝑑𝑦+𝜔²₃𝛿₁𝛿₂𝛿₃𝛿₄𝛾₁𝛾₂𝛾₃𝛾₄¹_ℎ2𝐺𝑓𝑦+𝑓𝑥,𝑦,𝜆𝑑𝑦₁𝛿₁𝛿₂𝛿₃𝛿₄𝜌𝛾₁𝛾₂𝛾₃𝛾₄𝑅^′₁(𝐺(𝑥,⋅,𝜆)). (87) We now introduce 𝐺_𝑥,𝜆=𝐺()𝑅𝑥,⋅,𝜆_′1(𝐺(𝑥,⋅,𝜆)), (88) which we will call Green’s element of (80)–(82). The last formula (87) then takes the form 𝑢(𝑥,𝜆)=𝐺_𝑥,𝜆,𝐹, (89) where by 𝐹 we mean 𝐹=^𝑓(𝑥)_𝑓1.