Theory and Modern Applications

# Computation of eigenvalues of discontinuous dirac system using Hermite interpolation technique

## Abstract

We use the derivative sampling theorem (Hermite interpolations) to compute eigenvalues of a discontinuous regular Dirac systems with transmission conditions at the point of discontinuity numerically. We closely follow the analysis derived by Levitan and Sargsjan (1975) to establish the needed relations. We use recently derived estimates for the truncation and amplitude errors to compute error bounds. Numerical examples, illustrations and comparisons with the sinc methods are exhibited.

Mathematical Subject Classification 2010: 34L16; 94A20; 65L15.

## 1 Introduction

Let σ > 0 and ${\text{PW}}_{\sigma }^{2}$ be the Paley-Wiener space of all L2()-entire functions of exponential type type σ. Assume that $f\left(t\right)\in {\text{PW}}_{\sigma }^{2}\subset {\text{PW}}_{2\sigma }^{2}$. Then f(t) can be reconstructed via the sampling series

$f\left(t\right)=\sum _{n=-\infty }^{\infty }\left[f\left(\frac{n\pi }{\sigma }\right){S}_{n}^{2}\left(t\right)+{f}^{\prime }\left(\frac{n\pi }{\sigma }\right)\frac{\text{sin}\left(\sigma t-n\pi \right)}{\sigma }{S}_{n}\left(t\right)\right],$
(1)

where S n (t) is the sequences of sinc functions

${S}_{n}\left(t\right):=\left\{\begin{array}{cc}\hfill \frac{\text{sin}\left(\sigma t-n\pi \right)}{\left(\sigma t-n\pi \right)},\hfill & \hfill t\ne \frac{n\pi }{\sigma },\hfill \\ \hfill 1,\hfill & \hfill t=\frac{n\pi }{\sigma }.\hfill \end{array}\right\$
(2)

Series (1) converges absolutely and uniformly on (cf. [14]). Sometimes, series (1) is called the derivative sampling theorem. Our task is to use formula (1) to compute eigenvalues of Dirac systems numerically. This approach is a fully new technique that uses the recently obtained estimates for the truncation and amplitude errors associated with (1) (cf. [5]). Both types of errors normally appear in numerical techniques that use interpolation procedures. In the following we summarize these estimates. The truncation error associated with (1) is defined to be

${R}_{N}\left(f\right)\left(t\right):=f\left(t\right)-{f}_{N}\left(t\right),\phantom{\rule{1em}{0ex}}N\in {ℤ}^{+},\phantom{\rule{1em}{0ex}}t\in ℝ,$
(3)

where f N (t) is the truncated series

${f}_{N}\left(t\right)=\sum _{|n|\le N}\left[f\left(\frac{n\pi }{\sigma }\right){S}_{n}^{2}\left(t\right)+{f}^{\prime }\left(\frac{n\pi }{\sigma }\right)\frac{\text{sin}\left(\sigma t-n\pi \right)}{\sigma }{S}_{n}\left(t\right)\right].$
(4)

It is proved in [5] that if $f\text{(}t\text{)}\in {\text{PW}}_{\sigma }^{2}$ and f(t) is sufficiently smooth in the sense that there exists k + such that tkf(t) L2(), then, for t , |t| < Nπ/σ, we have

$\begin{array}{ll}\hfill |{R}_{N}\left(f\right)\left(t\right)|\le {T}_{N,k,\sigma }\left(t\right)& :=\frac{{\xi }_{k,\sigma }{E}_{k}|\text{sin}\sigma t{|}^{2}}{\sqrt{3}{\left(N+1\right)}^{k}}\left(\frac{1}{{\left(N\pi -\sigma t\right)}^{3/2}}+\frac{1}{{\left(N\pi +\sigma t\right)}^{3/2}}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\frac{{\xi }_{k,\sigma }\left(\sigma {E}_{k}+k\phantom{\rule{0.3em}{0ex}}{E}_{k-1}\right)|\text{sin}\sigma t{|}^{2}}{\sigma {\left(N+1\right)}^{k}}\left(\frac{1}{\sqrt{N\pi -\sigma t}}+\frac{1}{\sqrt{N\pi +\sigma t}}\right),\phantom{\rule{2em}{0ex}}\end{array}$
(5)

where the constants E k and ξk,σare given by

${E}_{k}:=\sqrt{\underset{-\infty }{\overset{\infty }{\int }}|{t}^{k}f\left(t\right){|}^{2}dt},\phantom{\rule{1em}{0ex}}{\xi }_{k,\sigma }:=\frac{{\sigma }^{k+1/2}}{{\pi }^{k+1}\sqrt{1-{4}^{-k}}}.$
(6)

The amplitude error occurs when approximate samples are used instead of the exact ones, which we can not compute. It is defined to be

(7)

where $\stackrel{̃}{f}\left(\frac{n\pi }{\sigma }\right)$ and ${\stackrel{̃}{f}}^{\prime }-\left(\frac{n\pi }{\sigma }\right)$ are approximate samples of $f\left(\frac{n\pi }{\sigma }\right)$ and ${f}^{\prime }-\left(\frac{n\pi }{\sigma }\right)$, respectively. Let us assume that the differences ${\epsilon }_{n}:=f\left(\frac{n\pi }{\sigma }\right)-\stackrel{̃}{f}\left(\frac{n\pi }{\sigma }\right),\phantom{\rule{2.77695pt}{0ex}}{\epsilon }_{n}^{\prime }:={f}^{\prime }\left(\frac{n\pi }{\sigma }\right)-{\stackrel{̃}{f}}^{\prime }-\left(\frac{n\pi }{\sigma }\right),\phantom{\rule{2.77695pt}{0ex}}n\in ℤ,$are bounded by a positive number ε, i.e. $|{\epsilon }_{n}|,|{\epsilon }_{n}^{\prime }|\phantom{\rule{2.77695pt}{0ex}}\le \epsilon .$If $f\text{(}t\text{)}\in {\text{PW}}_{\sigma }^{2}$ satisfies the natural decay conditions

$\left|{\epsilon }_{n}\right|\le \left|f\left(\frac{n\pi }{\sigma }\right)\right|,\phantom{\rule{1em}{0ex}}\phantom{\rule{0.3em}{0ex}}\left|{{\epsilon }^{\prime }}_{n}\right|\le \left|{f}^{\prime }\left(\frac{n\pi }{\sigma }\right)\right|,$
(8)
$\left|f\left(t\right)\right|\le \frac{{M}_{f}}{|t{|}^{\ell +1}},\phantom{\rule{1em}{0ex}}\phantom{\rule{0.3em}{0ex}}t\in ℝ\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\left\{0\right\},$
(9)

0 < λ ≤ 1, then for $0<\epsilon \le \text{min}\left\{\pi /\sigma ,\phantom{\rule{2.77695pt}{0ex}}\sigma /\pi ,\phantom{\rule{2.77695pt}{0ex}}1/\sqrt{e}\right\},$ we have, [5],

$||\mathcal{A}\left(\epsilon ,\phantom{\rule{2.77695pt}{0ex}}f\right)|{|}_{\infty }\le \frac{4{e}^{1/4}}{\sigma \left(\ell +1\right)}\left\{\sqrt{3}e\left(1+\sigma \right)+\left(\left(\pi /\sigma \right)A+{M}_{f}\right)\rho \left(\epsilon \right)+\left(\sigma +2+\text{log}\left(2\right)\right){M}_{f}\right\}\epsilon \text{log}\left(1/\epsilon \right),$
(10)

where

$A:=\frac{3\sigma }{\pi }\left(|f\left(0\right)|+{M}_{f}{\left(\frac{\sigma }{\pi }\right)}^{\ell }\right),\phantom{\rule{1em}{0ex}}\rho \left(\epsilon \right):=\gamma +10\text{log}\left(1/\epsilon \right),$
(11)

and $\gamma :=\underset{n\to \infty }{\text{lim}}\left[{\sum }_{k=1}^{n}\frac{1}{k}-\text{log}n\right]\cong 0.577216$is the Euler-Mascheroni constant.

The classical [6] sampling theorem of Whittaker, Kotel'nikov and Shannon (WKS) for $f\in {\text{PW}}_{\sigma }^{2}$is the series representation

$f\left(t\right)=\sum _{n=-\infty }^{\infty }f\left(\frac{n\pi }{\sigma }\right){S}_{n}\left(t\right),\phantom{\rule{1em}{0ex}}t\in ℝ,$
(12)

where the convergence is absolute and uniform on and it is uniform on compact sets of (cf. [68]). Series (12), which is of Lagrange interpolation type, has been used to compute eigenvalues of second order eigenvalue problems (see e.g. [913]). The use of (12) in numerical analysis is known as the sinc-method established by Stenger (cf. [1416]). In [11, 12], the authors applied (12) and the regularized sinc method to compute eigenvalues of Dirac systems with a derivation of the error estimates as given by [17, 18]. The regularized sinc method; a method which is based on (WKS) but applied to regularized functions. Hence avoiding any (multiple) integration and keeping the number of terms in the Cardinal series manageable. It has been demonstrated that the method is capable of delivering higher order estimates of the eigenvalues at a very low cost. The aim of this article is to investigate the possibilities of using Hermite interpolations rather than Lagrange interpolations, to compute the eigenvalues numerically. Notice that, due to Paley-Wiener's theorem [19] $f\in {\text{PW}}_{\sigma }^{2}$ if and only if there is g(·)L2(-σ, σ) such that

$f\left(t\right)=\frac{1}{\sqrt{2\pi }}\underset{-\sigma }{\overset{\sigma }{\int }}g\left(x\right){e}^{ixt}dx.$
(13)

Therefore ${f}^{\prime }\left(t\right)\in {\text{PW}}_{\sigma }^{2},$ i.e, f′(t) also has an expansion of the form (12). However, f′(t) can be also obtained by term-by-term differentiation formula of (12)

${f}^{\prime }\left(t\right)=\sum _{n=-\infty }^{\infty }f\left(\frac{n\pi }{\sigma }\right){S}_{n}^{\prime }\left(t\right),$
(14)

see [[6], p. 52] for convergence. Thus the use of Hermite interpolations will not cost any additional computational efforts since the samples $f\left(\frac{n\pi }{\sigma }\right)$ will be used to compute both f(t) and f′(t) according to (12) and (14), respectively. We would like to mention that works in direction of computing eigenvalues with the new method, Hermite interpolation technique, are few (see e.g. [5]). Also articles in computing of eigenvalues with discontinuous are few (see [2022]). However the computing of eigenvalues by Hermite interpolation technique which has discontinuity conditions, do not exist as for as we know. The next section contains some preliminary results. The method with error estimates are contained in Section three. The last section involves some illustrative examples.

## 2 The eigenvalue problem

In this section we closely follow the analysis derived by [23] to establish the needed relations (see also [24]). We consider the Dirac system

${u}_{2}^{\prime }\left(x\right)-{r}_{1}\left(x\right){u}_{1}\left(x\right)=\lambda {u}_{1}\left(x\right),\phantom{\rule{1em}{0ex}}{u}_{1}^{\prime }\left(x\right)+{r}_{2}\left(x\right){u}_{2}\left(x\right)=-\lambda {u}_{2}\left(x\right),\phantom{\rule{1em}{0ex}}x\in \left[-1,0\right)\cup \left(0,1\right],$
(15)
${U}_{1}\left(\mathbit{u}\right):=\text{sin}\alpha {u}_{1}\left(-1\right)+\text{cos}\alpha {u}_{2}\left(-1\right)=0,$
(16)
${U}_{2}\left(\mathbit{u}\right):=\text{sin}\beta {u}_{1}\left(1\right)+\text{cos}\beta {u}_{2}\left(1\right)=0,$
(17)

and transmission conditions

${U}_{3}\left(\mathbit{u}\right):={u}_{1}\left({0}^{-}\right)-\delta {u}_{1}\left({0}^{+}\right)=0,$
(18)
${U}_{4}\left(\mathbit{u}\right):={u}_{2}\left({0}^{-}\right)-\delta {u}_{2}\left({0}^{+}\right)=0,$
(19)

where λ ; the real valued function r1(·) and r2(·) are continuous in [−1, 0) and (0, 1], and have finite limits ${r}_{1}\left({0}^{±}\right):=\underset{x\to {0}^{±}}{\text{lim}}{r}_{1}\left(x\right),{r}_{2}\left({0}^{±}\right):=\underset{x\to {0}^{±}}{\text{lim}}{r}_{2}\left(x\right);\delta \in ℝ,\alpha ,\beta \in \left[0,\phantom{\rule{2.77695pt}{0ex}}\pi \right)$and δ ≠ 0.

Let H be the Hilbert space

$\mathbit{H}\phantom{\rule{2.77695pt}{0ex}}:=\left\{\mathbit{u}\left(x\right)=\left(\begin{array}{c}\hfill {u}_{1}\left(x\right)\hfill \\ \hfill {u}_{2}\left(x\right)\hfill \end{array}\right),\phantom{\rule{2.77695pt}{0ex}}{u}_{1}\left(x\right),\phantom{\rule{2.77695pt}{0ex}}{u}_{2}\left(x\right)\in {L}^{2}\left(-1,0\right)\oplus {L}^{2}\left(0,1\right)\right\}.$
(20)

The inner product of H is defined by

${〈\mathbit{u}\left(\cdot \right),\mathbit{v}\left(\cdot \right)〉}_{\mathbit{H}}:=\underset{-1}{\overset{0}{\int }}{\mathbit{u}}^{\top }\left(x\right)\stackrel{̄}{\mathbit{v}}\left(x\right)dx+{\delta }^{2}\underset{0}{\overset{1}{\int }}{\mathbit{u}}^{\top }\left(x\right)\stackrel{̄}{\mathbit{v}}\left(x\right)dx,$
(21)

where denotes the matrix transpose,

$\mathbit{u}\left(x\right)=\left(\begin{array}{c}\hfill {u}_{1}\left(x\right)\hfill \\ \hfill {u}_{2}\left(x\right)\hfill \end{array}\right),\phantom{\rule{2.77695pt}{0ex}}\mathbit{v}\left(x\right)=\left(\begin{array}{c}\hfill {v}_{1}\left(x\right)\hfill \\ \hfill {v}_{2}\left(x\right)\hfill \end{array}\right)\in \mathbit{H},\phantom{\rule{1em}{0ex}}{u}_{i}\left(\cdot \right),\phantom{\rule{2.77695pt}{0ex}}{v}_{i}\left(\cdot \right)\in {L}^{2}\left(-1,\phantom{\rule{2.77695pt}{0ex}}1\right),\phantom{\rule{1em}{0ex}}i=1,\phantom{\rule{2.77695pt}{0ex}}2.$

Equation (15) can be written as

$\ell \left(\mathbit{u}\right):=A{\mathbit{u}}^{\prime }\left(x\right)-P\left(x\right)\mathbit{u}\left(x\right)=\lambda \mathbit{u}\left(x\right),$
(22)

where

$A=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 0\hfill \end{array}\right),\phantom{\rule{1em}{0ex}}P\left(x\right)=\left(\begin{array}{cc}\hfill {r}_{1}\left(x\right)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {r}_{2}\left(x\right)\hfill \end{array}\right),\phantom{\rule{1em}{0ex}}\mathbit{u}\left(x\right)=\left(\begin{array}{c}\hfill {u}_{1}\left(x\right)\hfill \\ \hfill {u}_{2}\left(x\right)\hfill \end{array}\right).$
(23)

For functions u(x), which defined on [−1, 0) (0, 1] and has finite limit $u\left(±0\right):=\underset{x\to ±0}{\text{lim}}u\left(x\right),$ by u(1)(x) and u(2)(x) we denote the functions

${u}_{\left(1\right)}\left(x\right)=\left\{\begin{array}{cc}\hfill u\left(x\right),\hfill & \hfill x\in \left[-1,0\right);\hfill \\ \hfill u\left({0}^{-}\right),\hfill & \hfill x=0\hfill \end{array}\right\,\phantom{\rule{0.3em}{0ex}}{u}_{\left(2\right)}\left(x\right)=\left\{\begin{array}{cc}\hfill u\left(x\right),\hfill & \hfill x\in \left(0,1\right]\hfill \\ \hfill u\left({0}^{+}\right),\hfill & \hfill x=0;\hfill \end{array}\right\$
(24)

which are defined on Γ1 := [−1, 0] and Γ2 := 0[1] respectively.

In the following lemma, we will prove that the eigenvalues of the problem (15)-(19) are real.

Lemma 2.1 The eigenvalues of the problem (15)-(19) are real.

Proof. Assume the contrary that λ0 is a nonreal eigenvalue of problem (15)-(19). Let $\left(\begin{array}{c}\hfill {u}_{1}\left(x\right)\hfill \\ \hfill {u}_{2}\left(x\right)\hfill \end{array}\right)$ be a corresponding (non-trivial) eigenfunction. By (15), we have, for x [−1, 0) ( 0, 1],

$\frac{d}{dx}\left\{{u}_{1}\left(x\right){\overline{u}}_{2}\left(x\right)-{\overline{u}}_{1}\left(x\right){u}_{2}\left(x\right)\right\}=\left({\overline{\lambda }}_{0}-{\text{λ}}_{\text{0}}\right)\left\{|{u}_{1}\left(x\right){|}^{2}+|{u}_{2}\left(x\right){|}^{2}\right\}.$

Integrating the above equation through [−1, 0) and (0, 1], we obtain

$\begin{array}{ll}\hfill \left({\stackrel{̄}{\lambda }}_{0}-{\lambda }_{0}\right)\left[\underset{-1}{\overset{0}{\int }}\left(|{u}_{1}\left(x\right){|}^{2}+|{u}_{2}\left(x\right){|}^{2}\right)dx\right]& ={u}_{1}\left({0}^{-}\right){ū}_{2}\left({0}^{-}\right)-{ū}_{1}\left({0}^{-}\right){u}_{2}\left({0}^{-}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}-\left[{u}_{1}\left(-1\right){ū}_{2}\left(-1\right)-{ū}_{1}\left(-1\right){u}_{2}\left(-1\right)\right],\phantom{\rule{2em}{0ex}}\end{array}$
(25)
$\begin{array}{ll}\hfill \left({\stackrel{̄}{\lambda }}_{0}-{\lambda }_{0}\right)\left[\underset{0}{\overset{1}{\int }}\left(|{u}_{1}\left(x\right){|}^{2}+|{u}_{2}\left(x\right){|}^{2}\right)dx\right]& ={u}_{1}\left(1\right){ū}_{2}\left(1\right)-{ū}_{1}\left(1\right){u}_{2}\left(1\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}-\left[{u}_{1}\left({0}^{+}\right){ū}_{2}\left({0}^{+}\right)-{ū}_{1}\left({0}^{+}\right){u}_{2}\left({0}^{+}\right)\right].\phantom{\rule{2em}{0ex}}\end{array}$
(26)

Then from (16), (17) and transmission conditions, we have respectively

${u}_{1}\left(-1\right){ū}_{2}\left(-1\right)-{ū}_{1}\left(-1\right){u}_{2}\left(-1\right)=0,$
${u}_{1}\left(1\right){ū}_{2}\left(1\right)-{ū}_{1}\left(1\right){u}_{2}\left(1\right)=0$

and

${u}_{1}\left({0}^{-}\right){ū}_{2}\left({0}^{-}\right)-{ū}_{1}\left({0}^{-}\right){u}_{2}\left({0}^{-}\right)={\delta }^{2}\left[{u}_{1}\left({0}^{+}\right){ū}_{2}\left({0}^{+}\right)-{ū}_{1}\left({0}^{+}\right){u}_{2}\left({0}^{+}\right)\right].$

Since ${\lambda }_{\text{0}}\ne {\stackrel{̄}{\lambda }}_{\text{0}},$ it follows from the last three equations and (25), (26) that

$\underset{-1}{\overset{0}{\int }}\left(|{u}_{1}\left(x\right){|}^{2}+|{u}_{2}\left(x\right){|}^{2}\right)dx+{\delta }^{2}\underset{0}{\overset{1}{\int }}\left(|{u}_{1}\left(x\right){|}^{2}+|{u}_{2}\left(x\right){|}^{2}\right)dx=0.$
(27)

Then u i (x) = 0, i =1, 2 and this is contradiction. Consequently, λ0 must be real.

Lemma 2.2 Let λ1 and λ2 be two different eigenvalues of the problem (15)-(19). Then the corresponding eigenfunctions u(x, λ1) and v(x, λ2) of this problem satisfy the following equality

$\underset{-1}{\overset{0}{\int }}{\mathbit{u}}^{\top }\left(x,\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{1}\right)\mathbit{v}\left(x,\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{2}\right)\phantom{\rule{2.77695pt}{0ex}}dx+{\delta }^{2}\underset{0}{\overset{1}{\int }}{\mathbit{u}}^{\top }\left(x,\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{1}\right)\mathbit{v}\left(x,\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{2}\right)\phantom{\rule{2.77695pt}{0ex}}dx=0.$
(28)

Proof. By (15) we obtain

$\frac{d}{dx}\left\{{u}_{1}\left(x,\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{1}\right){v}_{2}\left(x,\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{2}\right)-{u}_{2}\left(x,\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{2}\right){v}_{1}\left(x,\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{1}\right)\right\}=\left({\lambda }_{2}-{\lambda }_{1}\right)\left\{{u}_{1}\left(x,\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{1}\right){v}_{1}\left(x,\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{2}\right)+{u}_{2}\left(x,\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{1}\right){v}_{2}\left(x,\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{2}\right)\right\}.$

Integrating the above equation through [−1, 0) and (0, 1], and taking into account u(x, λ1) and v(x, λ2) satisfy (16)-(19), we obtain (28), where λ1λ2.

Now, we shall construct a special fundamental system of solutions of the Equation (15) for λ not being an eigenvalue. Let us consider the next initial value problem:

${u}_{2}^{{\phantom{\rule{0.1em}{0ex}}}^{\prime }}\left(x\right)-{r}_{1}\left(x\right){u}_{1}\left(x\right)=\lambda {u}_{1}\left(x\right),\phantom{\rule{1em}{0ex}}{u}_{1}^{{\phantom{\rule{0.1em}{0ex}}}^{\prime }}\left(x\right)+{r}_{2}\left(x\right){u}_{2}\left(x\right)=-\lambda {u}_{2}\left(x\right),\phantom{\rule{1em}{0ex}}x\in \left(-1,0\right),$
(29)
${u}_{1}\left(-1\right)=\text{cos}\alpha ,\phantom{\rule{1em}{0ex}}{u}_{2}\left(-1\right)=-\text{sin}\alpha .$
(30)

By virtue of Theorem 1.1 in [23] this problem has a unique solution $\mathbit{u}=\left(\begin{array}{c}\hfill {\varphi }_{11}\left(x,\lambda \right)\hfill \\ \hfill {\varphi }_{21}\left(x,\lambda \right)\hfill \end{array}\right),$which is an entire function of λ for each fixed x [−1, 0]. Similarly, employing the same method as in proof of Theorem 1.1 in [23], we see that the problem

${u}_{2}^{\prime }\left(x\right)-{r}_{1}\left(x\right){u}_{1}\left(x\right)=\lambda {u}_{1}\left(x\right),\phantom{\rule{1em}{0ex}}{u}_{1}^{\prime }\left(x\right)+{r}_{2}\left(x\right){u}_{2}\left(x\right)=-\lambda {u}_{2}\left(x\right),\phantom{\rule{1em}{0ex}}x\in \left(0,1\right),$
(31)
${u}_{1}\left(1\right)=\text{cos}\beta ,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{0.3em}{0ex}}{u}_{2}\left(1\right)=-\text{sin}\beta .$
(32)

has a unique solution $\mathbit{u}=\left(\begin{array}{c}\hfill {\chi }_{12}\left(x,\lambda \right)\hfill \\ \hfill {\chi }_{22}\left(x,\lambda \right)\hfill \end{array}\right)$ which is an entire function of parameter λ for each fixed x [0.1].

Now the functions φi 2(x, λ) and χi 1(x, λ) are defined in terms of φi 1(x, λ) and χi 2(x, λ), i =1, 2, respectively, as follows: The initial-value problem,

${u}_{2}^{\prime }\left(x\right)-{r}_{1}\left(x\right){u}_{1}\left(x\right)=\lambda {u}_{1}\left(x\right),\phantom{\rule{1em}{0ex}}{u}_{1}^{\prime }\left(x\right)+{r}_{2}\left(x\right){u}_{2}\left(x\right)=-\lambda {u}_{2}\left(x\right),\phantom{\rule{1em}{0ex}}x\in \left(0,\phantom{\rule{2.77695pt}{0ex}}1\right),$
(33)
${u}_{1}\left(0\right)=\frac{1}{\delta }{\varphi }_{11}\left(0,\phantom{\rule{2.77695pt}{0ex}}\lambda \right),\phantom{\rule{2.77695pt}{0ex}}{u}_{2}\left(0\right)=\frac{1}{\delta }{\varphi }_{21}\left(0,\phantom{\rule{2.77695pt}{0ex}}\lambda \right),$
(34)

has unique solution $\mathbit{u}=\left(\begin{array}{c}\hfill {\varphi }_{12}\left(x,\lambda \right)\hfill \\ \hfill {\varphi }_{22}\left(x,\lambda \right)\hfill \end{array}\right)$for each λ .

Similarly, the following problem also has a unique solution $\mathbit{u}=\left(\begin{array}{c}\hfill {\chi }_{11}\left(x,\lambda \right)\hfill \\ \hfill {\chi }_{21}\left(x,\lambda \right)\hfill \end{array}\right):$

${u}_{2}^{\prime }\left(x\right)-{r}_{1}\left(x\right){u}_{1}\left(x\right)=\lambda {u}_{1}\left(x\right),\phantom{\rule{1em}{0ex}}{u}_{1}^{\prime }\left(x\right)+{r}_{2}\left(x\right){u}_{2}\left(x\right)=-\lambda {u}_{2}\left(x\right),\phantom{\rule{1em}{0ex}}x\in \left(-1,\phantom{\rule{2.77695pt}{0ex}}0\right),$
(35)
${u}_{1}\left(0\right)=\delta {\chi }_{12}\left(0,\phantom{\rule{2.77695pt}{0ex}}\lambda \right),\phantom{\rule{1em}{0ex}}{u}_{2}\left(0\right)=\delta {\chi }_{22}\left(0,\phantom{\rule{2.77695pt}{0ex}}\lambda \right).$
(36)

Let us construct two basic solutions of the equation (15) as

$\mathbit{\varphi }\left(\cdot ,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)=\left(\begin{array}{c}\hfill {\varphi }_{1}\left(\cdot ,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)\hfill \\ \hfill {\varphi }_{2}\left(\cdot ,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)\hfill \end{array}\right),\phantom{\rule{1em}{0ex}}\mathbit{\chi }\left(\cdot ,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)=\left(\begin{array}{c}\hfill {\chi }_{1}\left(\cdot ,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)\hfill \\ \hfill {\chi }_{2}\left(\cdot ,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)\hfill \end{array}\right),$

where

${\varphi }_{1}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)=\left\{\begin{array}{cc}\hfill {\varphi }_{11}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right),\hfill & \hfill x\in \left[-1,0\right)\hfill \\ \hfill {\varphi }_{12}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right),\hfill & \hfill x\in \left(0,1\right]\hfill \end{array}\right\,\phantom{\rule{1em}{0ex}}{\varphi }_{2}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)=\left\{\begin{array}{cc}\hfill {\varphi }_{21}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right),\hfill & \hfill x\in \left[-1,0\right)\hfill \\ \hfill {\varphi }_{22}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right),\hfill & \hfill x\in \left(0,1\right]\hfill \end{array},\right\$
(37)
${\chi }_{1}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)=\left\{\begin{array}{cc}\hfill {\chi }_{11}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right),\hfill & \hfill x\in \left[-1,0\right)\hfill \\ \hfill {\chi }_{12}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right),\hfill & \hfill x\in \left(0,1\right]\hfill \end{array},\right\\phantom{\rule{1em}{0ex}}{\chi }_{2}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)=\left\{\begin{array}{cc}\hfill {\chi }_{21}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right),\hfill & \hfill x\in \left[-1,0\right)\hfill \\ \hfill {\chi }_{22}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right),\hfill & \hfill x\in \left(0,1\right]\hfill \end{array},\right\$
(38)

By virtue of Equations (34) and (36) these solutions satisfy both transmission conditions (18) and (19). These functions are entire in λ for all x [−1, 0) (0, 1].

Let W (φ, χ)(·, λ) denote the Wronskian of φ(·, λ) and χ(·, λ) defined in [[25], p. 194], i.e.,

$W\left(\mathbit{\varphi },\phantom{\rule{2.77695pt}{0ex}}\mathbit{\chi }\right)\left(\cdot ,\phantom{\rule{2.77695pt}{0ex}}\lambda \right):=\left|\begin{array}{cc}\hfill {\varphi }_{1}\left(\cdot ,\lambda \right)\hfill & \hfill {\varphi }_{2}\left(\cdot ,\lambda \right)\hfill \\ \hfill {\chi }_{1}\left(\cdot ,\lambda \right)\hfill & \hfill {\chi }_{2}\left(\cdot ,\lambda \right)\hfill \end{array}\right|.$

It is obvious that the Wronskian

${\Omega }_{i}\left(\lambda \right):=W\left(\mathbit{\varphi },\phantom{\rule{2.77695pt}{0ex}}\mathbit{\chi }\right)\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)={\varphi }_{1i}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right){\chi }_{2i}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)-{\varphi }_{2i}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right){\chi }_{1i}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right),\phantom{\rule{1em}{0ex}}x\in {\Gamma }_{i},\phantom{\rule{1em}{0ex}}i=1,\phantom{\rule{2.77695pt}{0ex}}2$
(39)

are independent of x Γi and are entire functions. Taking into account (34) and (36), a short calculation gives

${\Omega }_{1}\left(\lambda \right)={\delta }^{2}{\Omega }_{2}\left(\lambda \right),$

for each λ .

Corollary 2.3 The zeros of the functions Ω1(λ) and Ω2(λ) coincide.

Then, we may introduce to the consideration the characteristic function Ω(λ) as

$\Omega \left(\lambda \right):={\Omega }_{1}\left(\lambda \right)={\delta }^{2}{\Omega }_{2}\left(\lambda \right).$
(40)

In the following lemma, we show that all eigenvalues of the problem (15)-(19) are simple.

Lemma 2.4 All eigenvalues of problem (15)-(19) are just zeros of the function Ω(λ). Moreover, every zero of Ω(λ) has multiplicity one.

Proof. Since the functions φ1(x, λ) and φ2(x, λ) satisfy the boundary condition (16) and both transmission conditions (18) and (19), to find the eigenvalues of the (15)-(19) we have to insert the functions φ1(x, λ) and φ2(x, λ) in the boundary condition (17) and find the roots of this equation.

By (15) we obtain for λ, µ , λμ,

$\frac{d}{dx}\left\{{\varphi }_{1}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right){\varphi }_{2}\left(x,\phantom{\rule{2.77695pt}{0ex}}\mu \right)-{\varphi }_{1}\left(x,\phantom{\rule{2.77695pt}{0ex}}\mu \right){\varphi }_{2}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)\right\}=\left(\mu -\lambda \right)\left\{{\varphi }_{1}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right){\varphi }_{1}\left(x,\phantom{\rule{2.77695pt}{0ex}}\mu \right)+{\varphi }_{2}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right){\varphi }_{2}\left(x,\phantom{\rule{2.77695pt}{0ex}}\mu \right)\right\}.$

Integrating the above equation through [−1, 0) and (0, 1], and taking into account the initial conditions (30), (34) and (36), we obtain

(41)

Dividing both sides of (41) by (λµ) and by letting µλ, we arrive to the relation

(42)

We show that equation

$\Omega \left(\lambda \right)=-{\delta }^{2}\left(\text{sin}\beta {\varphi }_{12}\left(1,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)+\text{cos}\beta {\varphi }_{22}\left(1,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)\right)=0$
(43)

has only simple roots. Assume the converse, i.e., Equation (43) has a double root λ, say. Then the following two equations hold

$\text{sin}\beta {\varphi }_{12}\left(1,\phantom{\rule{2.77695pt}{0ex}}{\lambda }^{*}\right)+\text{cos}\beta {\varphi }_{22}\left(1,\phantom{\rule{2.77695pt}{0ex}}{\lambda }^{*}\right)=0,$
(44)
$\text{sin}\beta \frac{\partial {\varphi }_{12}\left(1,{\lambda }^{*}\right)}{\partial \lambda }+\text{cos}\beta \frac{\partial {\varphi }_{22}\left(1,{\lambda }^{*}\right)}{\partial \lambda }=0.$
(45)

The Equations (44) and (45) imply that

${\varphi }_{22}\left(1,\phantom{\rule{2.77695pt}{0ex}}{\lambda }^{*}\right)\frac{\partial {\varphi }_{12}\left(1,{\lambda }^{*}\right)}{\partial \lambda }-{\varphi }_{12}\left(1,\phantom{\rule{2.77695pt}{0ex}}{\lambda }^{*}\right)\frac{\partial {\varphi }_{22}\left(1,{\lambda }^{*}\right)}{\partial \lambda }$
(46)

Combining (46) and (42), with λ = λ*, we obtain

$\underset{-1}{\overset{0}{\int }}\left(|{\varphi }_{11}\left(x,\phantom{\rule{2.77695pt}{0ex}},\phantom{\rule{2.77695pt}{0ex}}{\lambda }^{*}\right){|}^{2}+|{\varphi }_{21}\left(x,\phantom{\rule{2.77695pt}{0ex}},\phantom{\rule{2.77695pt}{0ex}}{\lambda }^{*}\right){|}^{2}\right)\phantom{\rule{2.77695pt}{0ex}}dx+{\delta }^{2}\underset{0}{\overset{1}{\int }}\left(|{\varphi }_{12}\left(x,\phantom{\rule{2.77695pt}{0ex}},\phantom{\rule{2.77695pt}{0ex}}{\lambda }^{*}\right){|}^{2}+|{\varphi }_{22}\left(x,\phantom{\rule{2.77695pt}{0ex}},\phantom{\rule{2.77695pt}{0ex}}{\lambda }^{*}\right){|}^{2}\right)\phantom{\rule{2.77695pt}{0ex}}dx=0.$
(47)

It follows that φ1(x, λ*)=φ2(x, λ*)=0, which is impossible. This proves the lemma.

Here ${\left\{\varphi \left(\cdot ,\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{n}\right)\right\}}_{n=-\infty }^{\infty }$will be a sequence of eigen-vector-functions of (15)-(19) corresponding to the eigenvalues ${\left\{{\lambda }_{n}\right\}}_{n=-\infty }^{\infty }.$Since χ(·, λ) satisfies (17)-(19), then the eigenvalues are also determined via

$\text{sin}\alpha {\chi }_{11}\left(-1,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)+\text{cos}\alpha {\chi }_{21}\left(-1,\lambda \right)=\Omega \left(\lambda \right).$
(48)

Therefore ${\left\{\mathbit{\chi }\left(\cdot ,\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{n}\right)\right\}}_{n=-\infty }^{\infty }$ is another set of eigen-vector-functions which is related by ${\left\{\mathbit{\varphi }\left(\cdot ,\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{n}\right)\right\}}_{n=-\infty }^{\infty }$ with

$\mathbit{\chi }\left(x,\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{n}\right)={c}_{n}\varphi \left(x,\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{n}\right),\phantom{\rule{1em}{0ex}}x\in \left[-1,0\right)\cup \left(0,1\right],\phantom{\rule{1em}{0ex}}n\in ℤ,$
(49)

Where c n ≠ 0 are non-zero constants, since all eigenvalues are simple. Since the eigenvalues are all real, we can take the eigen-vector-functions to be real valued.

Since φ(·, λ) satisfies (16), then the eigenvalues of the problem (15)-(19) are the zeros of the function

$\Omega \left(\lambda \right)=-{\delta }^{2}\left(\text{sin}\beta {\varphi }_{12}\left(1,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)+\text{cos}\beta {\varphi }_{22}\left(1,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)\right).$
(50)

Notice that both φ(·, λ) and Ω(λ) are entire functions of λ. Now we shall transform Equations (15), (30), (34) and (37) into the integral equations (see [25]),

${\varphi }_{11}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)=\text{cos}\left(\lambda \left(x+1\right)-\alpha \right)-{\mathcal{S}}_{-1,1}{\varphi }_{11}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)-{\stackrel{̃}{\mathcal{S}}}_{-1,2}{\varphi }_{21}-\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right),$
(51)
${\varphi }_{21}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)=\text{sin}\left(\lambda \left(x+1\right)-\alpha \right)+{\stackrel{̃}{\mathcal{S}}}_{-1,1}{\varphi }_{11}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)-{\mathcal{S}}_{-1,2}{\varphi }_{21}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right),$
(52)
${\varphi }_{12}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)=\frac{1}{\delta }{\varphi }_{11}\left({0}^{-},\phantom{\rule{2.77695pt}{0ex}}\lambda \right)\text{cos}\left(\lambda x\right)-\frac{1}{\delta }{\varphi }_{21}\left({0}^{-},\phantom{\rule{2.77695pt}{0ex}}\lambda \right)\text{sin}\left(\lambda x\right)-{\mathcal{S}}_{0,1}{\varphi }_{12}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)-{\stackrel{̃}{\mathcal{S}}}_{0,2}{\varphi }_{22}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right),$
(53)
${\varphi }_{22}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)=\frac{1}{\delta }{\varphi }_{11}\left({0}^{-},\phantom{\rule{2.77695pt}{0ex}}\lambda \right)\text{sin}\left(\lambda x\right)+\frac{1}{\delta }{\varphi }_{21}\left({0}^{-},\phantom{\rule{2.77695pt}{0ex}}\lambda \right)\text{cos}{\left(\lambda x\right)}^{-}+{\stackrel{̃}{\mathcal{S}}}_{0,1}{\varphi }_{12}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)-{\mathcal{S}}_{0,2}{\varphi }_{22}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right),$
(54)

where ${\mathcal{S}}_{-1,i},\phantom{\rule{2.77695pt}{0ex}}{\stackrel{̃}{\mathcal{S}}}_{-1,i},\phantom{\rule{2.77695pt}{0ex}}{\mathcal{S}}_{0,i}$ and ${\stackrel{̃}{\mathcal{S}}}_{0,i},\phantom{\rule{2.77695pt}{0ex}}i=1,2,$ are the Volterra integral operators defined by

$\begin{array}{ll}\hfill {\mathcal{S}}_{-1,i}\phi \left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)& :=\underset{-1}{\overset{x}{\int }}\text{sin}\phantom{\rule{2.77695pt}{0ex}}\lambda \left(x-t\right){r}_{i}\left(t\right)\phi \left(t,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)dt,\phantom{\rule{1em}{0ex}}{\stackrel{̃}{\mathcal{S}}}_{-1,i}\phi \left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right):=\underset{-1}{\overset{x}{\int }}\text{cos}\lambda \left(x-t\right){r}_{i}\left(t\right)\phi \left(t,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)\phantom{\rule{2.77695pt}{0ex}}dt,\phantom{\rule{0.3em}{0ex}}\phantom{\rule{2em}{0ex}}\\ \hfill {\mathcal{S}}_{0,i}\phi \left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)& :=\underset{0}{\overset{x}{\int }}\text{sin}\phantom{\rule{2.77695pt}{0ex}}\lambda \left(x-t\right){r}_{i}\left(t\right)\phi \left(t,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)dt,\phantom{\rule{1em}{0ex}}{\stackrel{̃}{\mathcal{S}}}_{0,i}\phi \left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right):=\underset{0}{\overset{x}{\int }}\text{cos}\lambda \left(x-t\right){r}_{i}\left(t\right)\phi \left(t,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)\phantom{\rule{2.77695pt}{0ex}}dt.\phantom{\rule{2em}{0ex}}\end{array}$

For convenience, we define the constants

$\begin{array}{l}{c}_{1}:=\underset{-1}{\overset{0}{\int }}\left[|{r}_{1}\left(t\right)|+|{r}_{2}\left(t\right)|\right]\phantom{\rule{2.77695pt}{0ex}}dt,\phantom{\rule{1em}{0ex}}{c}_{2}:={c}_{1}\text{exp}\left({c}_{1}\right),\phantom{\rule{2em}{0ex}}\\ {c}_{3}:=\underset{0}{\overset{1}{\int }}\left[|{r}_{1}\left(t\right)|+|{r}_{2}\left(t\right)|\right]\phantom{\rule{2.77695pt}{0ex}}dt,\phantom{\rule{1em}{0ex}}{c}_{4}:={c}_{2}+\frac{2}{|\delta |}\left(1+{c}_{2}\right).\phantom{\rule{2em}{0ex}}\end{array}$
(55)

Define h−1,i(·, λ) and h0,i(·, λ), i = 1, 2, to be

(56)
(57)

Lemma 2.5 The functions h−1,1(x, λ) and h−1,2(x, λ) are entire in λ for any fixed x [−1, 0) and satisfy the growth condition

$|{h}_{-1,1}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|,\phantom{\rule{2.77695pt}{0ex}}|{h}_{-1,2}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|\le 2{c}_{2}{e}^{|\Im \lambda |\left(x+1\right)},\phantom{\rule{1em}{0ex}}\lambda \in ℂ.$
(58)

Proof. Since ${h}_{-1,1}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)={\mathcal{S}}_{-1,1}{\varphi }_{11}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)+{\stackrel{̃}{\mathcal{S}}}_{-1,2}{\varphi }_{21}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right),$ then from (51) and (52) we obtain ${h}_{-1,1}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)={\mathcal{S}}_{-1,1}\text{cos}\left(\lambda \left(x+1\right)-\alpha \right)+{\stackrel{̃}{\mathcal{S}}}_{-1,2}\text{sin}\left(\lambda \left(x+1\right)-\alpha \right)-{\mathcal{S}}_{-1,1}{h}_{-1,1}-\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)+{\stackrel{̃}{\mathcal{S}}}_{-1,2}{h}_{-1,2}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)$ Using the inequalities $|\text{sin}z|\phantom{\rule{2.77695pt}{0ex}}\le {e}^{|\Im z|}$ and leads for λ to

$\begin{array}{ll}\hfill |{h}_{-1,1}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|& \le \phantom{\rule{2.77695pt}{0ex}}|{\mathcal{S}}_{-1,1}\text{cos}\left(\lambda \left(x+1\right)-\alpha \right)|+|{\stackrel{̃}{\mathcal{S}}}_{-1,2}\text{sin}\left(\lambda \left(x-+1\right)-\alpha \right)|+|{\mathcal{S}}_{-1,1}{h}_{-1,1}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+|{\stackrel{̃}{\mathcal{S}}}_{-1,2}{h}_{-1,2}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|\phantom{\rule{2em}{0ex}}\\ \le {e}^{|\Im \lambda |\left(x+1\right)}\underset{-1}{\overset{x}{\int }}\left[|{r}_{1}\left(t\right)||{h}_{-1,1}\left(t,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|+|{r}_{2}\left(t\right)||{h}_{-1,2}\left(t,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|\right]\phantom{\rule{2.77695pt}{0ex}}{e}^{-|\Im \lambda |\left(t+1\right)}\phantom{\rule{2.77695pt}{0ex}}dt\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+2{e}^{|\Im \lambda |\left(x+1\right)}\underset{-1}{\overset{x}{\int }}\left[|{r}_{1}\left(t\right)|+|{r}_{2}\left(t\right)|\right]\phantom{\rule{2.77695pt}{0ex}}dt\phantom{\rule{2em}{0ex}}\\ \le 2{c}_{1}{e}^{|\Im \lambda |\left(x+1\right)}+{e}^{|\Im \lambda |\left(x+1\right)}\underset{-1}{\overset{x}{\int }}\left[|{r}_{1}\left(t\right)||{h}_{-1,1}\left(t,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|+|{r}_{2}\left(t\right)||{h}_{-1,2}\left(t,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|\right]\phantom{\rule{2.77695pt}{0ex}}{e}^{-|\Im \lambda |\left(t+1\right)}\phantom{\rule{2.77695pt}{0ex}}dt.\phantom{\rule{2em}{0ex}}\end{array}$

The above inequality can be reduced to

${e}^{-|\Im \lambda |\left(x+1\right)}|{h}_{-1,1}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|\phantom{\rule{2.77695pt}{0ex}}\le 2{c}_{1}+\underset{-1}{\overset{x}{\int }}\left[|{r}_{1}\left(t\right)||{h}_{-1,1}\left(t,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|+|{r}_{2}\left(t\right)||{h}_{-1,2}\left(t,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|\right]\phantom{\rule{2.77695pt}{0ex}}{e}^{-|\Im \lambda |\left(t+1\right)}\phantom{\rule{2.77695pt}{0ex}}dt.$
(59)

Similarly, we can prove that

${e}^{-|\Im \lambda |\left(x+1\right)}|{h}_{-1,2}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|\phantom{\rule{2.77695pt}{0ex}}\le 2{c}_{1}+\underset{-1}{\overset{x}{\int }}\left[|{r}_{1}\left(t\right)||{h}_{-1,1}\left(t,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|+|{r}_{2}\left(t\right)||{h}_{-1,2}\left(t,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|\right]\phantom{\rule{2.77695pt}{0ex}}{e}^{-|\Im \lambda |\left(t+1\right)}\phantom{\rule{2.77695pt}{0ex}}dt.$
(60)

Then from (58) and (59) and and Lemma 3.1 of [[25], pp. 204], we obtain (58).

In a similar manner, we will prove the following lemma for h0,1(·, λ) and h0,2(·, λ).

Lemma 2.6 The functions h0,1(x, λ) and h0,2(x, λ) are entire in λ for any fixed x (0, 1] and satisfy the growth condition

$|{h}_{0,1}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|,|{h}_{0,2}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|\phantom{\rule{2.77695pt}{0ex}}\le 2{c}_{3}{c}_{4}{e}^{|\Im \lambda |\left(x+1\right)},\phantom{\rule{1em}{0ex}}\lambda \in ℂ.$
(61)

Proof. Since ${h}_{0,1}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)={\mathcal{S}}_{0,1}{\varphi }_{11}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)+{\stackrel{̃}{\mathcal{S}}}_{0,2}{\varphi }_{21}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right),$ then from (53) and (54) we obtain

$\begin{array}{ll}\hfill {h}_{0,1}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)& =\frac{1}{\delta }{\varphi }_{11}\left({0}^{-},\phantom{\rule{2.77695pt}{0ex}}\lambda \right){\mathcal{S}}_{0,1}\text{cos}\left(\lambda x\right)-\frac{1}{\delta }{\varphi }_{21}\left({0}^{-},\phantom{\rule{2.77695pt}{0ex}}\lambda \right){\stackrel{̃}{\mathcal{S}}}_{0,1}\text{sin}\left(\lambda x\right)-{\mathcal{S}}_{0,1}{h}_{-1,2}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\frac{1}{\delta }{\varphi }_{11}\left({0}^{-},\phantom{\rule{2.77695pt}{0ex}}\lambda \right){\stackrel{̃}{\mathcal{S}}}_{0,2}\text{sin}\left({\lambda }_{X}\right)+\frac{1}{\delta }{\varphi }_{21}\left({0}^{-},\phantom{\rule{2.77695pt}{0ex}}\lambda \right){\stackrel{̃}{\mathcal{S}}}_{0,2}\text{cos}\left({\lambda }_{X}\right)+{\stackrel{̃}{\mathcal{S}}}_{0,2}{h}_{-1,2}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right).\phantom{\rule{2em}{0ex}}\end{array}$

Then from (51) and (52) and Lemma 2.5, we get

$\begin{array}{ll}\hfill {h}_{0,1}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)& \le \frac{1}{|\delta |}|{\varphi }_{11}\left({0}^{-},\phantom{\rule{2.77695pt}{0ex}}\lambda \right)||{\mathcal{S}}_{0,1}\text{cos}\left(\lambda x\right)|+\frac{1}{|\delta |}|{\varphi }_{21}\left({0}^{-},\phantom{\rule{2.77695pt}{0ex}}\lambda \right)||{\mathcal{S}}_{0,1\phantom{\rule{2.77695pt}{0ex}}}\text{sin}\left(\lambda x\right)|\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+|{\mathcal{S}}_{0,1}{h}_{-1,2}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|+\frac{1}{|\delta |}|{\varphi }_{11}\left({0}^{-},\phantom{\rule{2.77695pt}{0ex}}\lambda \right)||{\stackrel{̃}{\mathcal{S}}}_{0,2\phantom{\rule{2.77695pt}{0ex}}}\text{sin}\left(\lambda x\right)|\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\frac{1}{|\delta |}|{\varphi }_{21}\left({0}^{-},\phantom{\rule{2.77695pt}{0ex}}\lambda \right)||{\stackrel{̃}{\mathcal{S}}}_{0,2}\text{cos}\left(\lambda x\right)|-+|{\stackrel{̃}{\mathcal{S}}}_{0,2}{h}_{-1,2}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|\phantom{\rule{2em}{0ex}}\\ \le 2\left({c}_{2}+\frac{2}{|\delta |}\left(1+{c}_{2}\right)\right){c}_{3}{e}^{|\Im \lambda |\left(x+1\right)}=2{c}_{3}{c}_{4}{e}^{|\Im \lambda |\left(x+1\right)}.\phantom{\rule{2em}{0ex}}\end{array}$

Similarly, we can prove that

${h}_{0,2}\left(x,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)\le 2{c}_{3}{c}_{4}{e}^{|\Im \lambda |\left(x+1\right)}.$

## 3 The numerical scheme

In this section we derive the method of computing eigenvalues of problem (15)-(19) numerically. The basic idea of the scheme is to split Ω(λ) into two parts a known part $\mathcal{K}\left(\lambda \right)$ and an unknown one $\mathcal{U}\left(\lambda \right)$. Then we prove that $\mathcal{U}\left(\lambda \right)$ has an expansion of the form (1). We then approximate $\mathcal{U}\left(\lambda \right)$ in two stages. First by truncating the sampling expansion (4) and then by approximating the samples, using standard methods of solving ordinary differential equations. This produces both a truncation error and an amplitude error. We apply forms (4) and (7) to derive an estimate of the error of the technique. We first split Ω(λ) into two parts:

$\Omega \left(\lambda \right):=\mathcal{K}\left(\lambda \right)+\mathcal{U}\left(\lambda \right),$
(62)

where $\mathcal{U}\left(\lambda \right)$ is the unknown part involving integral operators

$\mathcal{U}\left(\lambda \right):={\delta }^{2}\text{sin}\beta {h}_{0,1}\left(1,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)-{\delta }^{2}\text{cos}\beta {h}_{0,2}\left(1,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)-\delta \text{sin}\left(\lambda -\beta \right){h}_{-1,1}\left({0}^{-},\phantom{\rule{2.77695pt}{0ex}}\lambda \right)-\delta \text{cos}\left(\lambda +\beta \right){h}_{-1,2}\left({0}^{-},\phantom{\rule{2.77695pt}{0ex}}\lambda \right),$
(63)

and $\mathcal{K}\left(\lambda \right)$ is the known part

$\mathcal{K}\left(\lambda \right):=-\delta \text{sin}\left(2\lambda -\alpha +\beta \right).$
(64)

Then, from Lemmas 2.5 and 2.6, we have the following result.

Lemma 3.1 The function $\mathcal{U}\left(\lambda \right)$ is entire in λ and the following estimate holds

$|\mathcal{U}\left(\lambda \right)|\phantom{\rule{2.77695pt}{0ex}}\le M{e}^{2|\Im \lambda |},$
(65)

where

$M:=4|\delta |\left({c}_{2}+|\delta |{c}_{3}{c}_{4}\right).$
(66)

Proof. From (63), we have

$\begin{array}{ll}\hfill |\mathcal{U}\left(\lambda \right)|& \le \phantom{\rule{2.77695pt}{0ex}}|\delta {|}^{2}|\text{sin}\beta ||{h}_{0,1}\left(1,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|+|\delta {|}^{2}|\text{cos}\beta ||{h}_{0,2}\left(1,\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+|\delta ||\text{sin}\left(\lambda -\beta \right)||{h}_{-1,1}\left({0}^{-},\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|+|\delta ||\text{cos}\left(\lambda +\beta \right)||{h}_{-1,2}\left({0}^{-},\phantom{\rule{2.77695pt}{0ex}}\lambda \right)|.\phantom{\rule{2em}{0ex}}\end{array}$

Using the inequalities $|\text{sin}\lambda |\le {e}^{|\Im \lambda |}$ and $|\text{cos}\lambda |\le {e}^{|\Im \lambda |}$ for λ , and Lemmas 2.5 and 2.6 imply (65).

Let θ (0, 1) and m +, m> 1 be fixed. Let ${{\mathcal{G}}_{\theta }}_{,m}\left(\lambda \right)$ be the function

${\mathcal{G}}_{\theta ,m}\left(\lambda \right):={\left(\frac{\text{sin}\theta \lambda }{\theta \lambda }\right)}^{m}\mathcal{U}\left(\lambda \right),\phantom{\rule{1em}{0ex}}\lambda \in ℂ.$
(67)

Lemma 3.2 ${{\mathcal{G}}_{\theta }}_{,m}\left(\lambda \right)$ is an entire function of λ which satisfies the estimate

$|{\mathcal{G}}_{\theta ,m}\left(\lambda \right)|\phantom{\rule{2.77695pt}{0ex}}\le \frac{{c}_{0}^{m}M}{{\left(1+\theta |\lambda |\right)}^{m}}{e}^{|\Im \lambda |\left(2+m\theta \right)},\phantom{\rule{1em}{0ex}}\lambda \in ℂ.$
(68)

Moreover, ${\lambda }^{m-1}{\mathcal{G}}_{\theta ,m}\left(\lambda \right)\in {L}^{2}\left(ℝ\right)$ and

${E}_{m-1}\left({\mathcal{G}}_{\theta ,m}\right)=\sqrt{{\int }_{-\infty }^{\infty }|{\lambda }^{m-1}{\mathcal{G}}_{\theta m}\left(\lambda \right){|}^{2}d\lambda }\le \sqrt{2}{c}_{0}^{m}M{\xi }_{0},$
(69)

where

${\xi }_{0}:=\sqrt{\frac{1}{{\theta }^{2m-1}\left(2m-1\right)}}.$

Proof. Since $\mathcal{U}\left(\lambda \right)$ is entire, then also ${{\mathcal{G}}_{\theta }}_{,m}\left(\lambda \right)$ is entire in λ. Combining the estimates $\left|\frac{\text{sin}z}{z}\right|\le \frac{{c}_{0}}{1+|z|}{e}^{|\Im z|},$where c0 1.72, cf. [26], and (65), we obtain

$\begin{array}{ll}\hfill |{\lambda }^{m-1}{\mathcal{G}}_{\theta ,m}\left(\lambda \right)|& \le {\left(\frac{{c}_{0}}{1+\theta |\lambda |}\right)}^{m}{e}^{|\Im z|m\theta }\cdot M|\lambda {|}^{m-1}{e}^{2|\Im \lambda |}\phantom{\rule{2em}{0ex}}\\ =\frac{{c}_{0}^{m}M|\lambda {|}^{m-1}}{{\left(1+\theta |\lambda |\right)}^{m}}{e}^{|\Im \lambda |\left(2+m\theta \right)},\phantom{\rule{1em}{0ex}}\lambda \in ℂ.\phantom{\rule{2em}{0ex}}\end{array}$
(70)

Therefore if λ we have

$|{\lambda }^{m-1}{\mathcal{G}}_{\theta ,m}\left(\lambda \right)|\phantom{\rule{2.77695pt}{0ex}}\le \frac{{c}_{0}^{m}M|\lambda {|}^{m-1}}{{\left(1+\theta |\lambda |\right)}^{m}},\phantom{\rule{1em}{0ex}}\lambda \in ℝ,$
(71)

i.e. ${\lambda }^{m-1}{\mathcal{G}}_{\theta ,m}\left(\lambda \right)\in {L}^{2}\left(ℝ\right).$ Moreover

$\underset{-\infty }{\overset{\infty }{\int }}|{\lambda }^{m-1}{\mathcal{G}}_{\theta ,m}\left(\lambda \right){|}^{2}d\lambda \le {c}_{0}^{2m}{M}^{2}\underset{-\infty }{\overset{\infty }{\int }}\frac{|\lambda {|}^{2m-2}}{{\left(1+\theta |\lambda |\right)}^{2m}}d\lambda =2{c}_{0}^{2m}{M}^{2}{\xi }_{0}^{2}.$
(72)

A direct and important result of Lemma 51 is that ${{\mathcal{G}}_{\theta }}_{,m}\left(\lambda \right)$ belongs to the Paley-Wienerz space ${\text{PW}}_{\sigma }^{2}$ with σ = 1+. Since ${\mathcal{G}}_{\theta ,m}\left(\lambda \right)\in {\text{PW}}_{\sigma }^{2}\subset {\text{PW}}_{2\sigma }^{2},$ then we can reconstruct the functions ${{\mathcal{G}}_{\theta }}_{,m}\left(\lambda \right)$ via the following sampling formula

${\mathcal{G}}_{\theta ,m}\left(\lambda \right)=\sum _{n=-\infty }^{\infty }\left[{\mathcal{G}}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right){S}_{n}^{2}\left(\lambda \right)+{\mathcal{G}}_{\theta ,m}^{{\phantom{\rule{0.1em}{0ex}}}^{\prime }}\left(\frac{n\pi }{\sigma }\right)\frac{\text{sin}\left(\sigma \lambda -n\pi \right)}{\sigma }{S}_{n}\left(\lambda \right)\right].$
(73)

Let N +, N > m and approximate ${{\mathcal{G}}_{\theta }}_{,m}\left(\lambda \right)$ by its truncated series ${{\mathcal{G}}_{\theta }}_{,m,N}\left(\lambda \right)$, where

${\mathcal{G}}_{\theta ,m,N}\left(\lambda \right):=\sum _{n=-N}^{N}\left[{\mathcal{G}}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right){S}_{n}^{2}\left(\lambda \right)+{\mathcal{G}}_{\theta ,m}^{{\phantom{\rule{0.1em}{0ex}}}^{\prime }}\left(\frac{n\pi }{\sigma }\right)\frac{\text{sin}\left(\sigma \lambda -n\pi \right)}{\sigma }{S}_{n}\left(\lambda \right)\right].$
(74)

Since all eigenvalues are real, then from now on we restrict ourselves to λ . Since ${\lambda }^{m-1}{\mathcal{G}}_{\theta ,m}\left(\lambda \right)\in {L}^{2}\left(ℝ\right),$ the truncation error, cf. (5), is given for

$|{\mathcal{G}}_{\theta ,m}\left(\lambda \right)-{\mathcal{G}}_{\theta ,m,N}\left(\lambda \right)|\phantom{\rule{2.77695pt}{0ex}}\le {T}_{N,m-1,\sigma }\left(\lambda \right),$
(75)

where

$\begin{array}{ll}\hfill {T}_{N,m-1,\sigma }\left(\lambda \right)& :=\frac{{\xi }_{m-1,\sigma }{E}_{m-1}|\text{sin}\sigma \lambda {|}^{2}}{\sqrt{3}{\left(N+1\right)}^{m-1}}\left(\frac{1}{{\left(N\pi -\sigma \lambda \right)}^{3/2}}+\frac{1}{{\left(N\pi +\sigma \lambda \right)}^{3/2}}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\frac{{\xi }_{m-1,\sigma }\left(\sigma {E}_{m-1}+\left(m-1\right){E}_{m-2}\right)|\text{sin}\sigma \lambda {|}^{2}}{\sigma {\left(N+1\right)}^{m-1}}\left(\frac{1}{\sqrt{N\pi -\sigma \lambda }}+\frac{1}{\sqrt{N\pi +\sigma \lambda }}\right).\phantom{\rule{2em}{0ex}}\end{array}$
(76)

The samples ${\left\{{\mathcal{G}}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right)\right\}}_{n=-N}^{N}$ and ${\left\{{{\mathcal{G}}^{\prime }}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right)\right\}}_{n=-N}^{N},$ in general, are not known explicitly. So we approximate them by solving numerically 2N + 1 initial value problems at the nodes ${\left\{\frac{n\pi }{\sigma }\right\}}_{n=-N}^{N}.$ Let ${\left\{{\stackrel{̃}{\mathcal{G}}}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right)\right\}}_{n=-N}^{N}$ and ${\left\{{\stackrel{̃}{\mathcal{G}}}_{\theta ,m}^{{\phantom{\rule{0.1em}{0ex}}}^{\prime }}\left(\frac{n\pi }{\sigma }\right)\right\}}_{n=-N}^{N}$ be the approximations of the samples of ${\left\{{\mathcal{G}}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right)\right\}}_{n=-N}^{N}$ and ${\left\{{{\mathcal{G}}^{\prime }}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right)\right\}}_{n=-N}^{N},$ respectively. Now we define ${\stackrel{̃}{\mathcal{G}}}_{\theta ,m,N}\left(\lambda \right),$ which approximates ${{\mathcal{G}}_{\theta }}_{,m,N}\left(\lambda \right)$

${\stackrel{̃}{\mathcal{G}}}_{\theta ,m,N}\left(\lambda \right):=\sum _{n=-N}^{N}\left[{\stackrel{̃}{\mathcal{G}}}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right){S}_{n}^{2}\left(\lambda \right)+{{\stackrel{̃}{\mathcal{G}}}^{\prime }}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right)\frac{\text{sin}\left(\sigma \lambda -n\pi \right)}{\sigma }{S}_{n}\left(\lambda \right)\right],\phantom{\rule{1em}{0ex}}N>m.$
(77)

Using standard methods for solving initial problems, we may assume that for |n| < N,

$\left|{\mathcal{G}}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right)-{\stackrel{̃}{\mathcal{G}}}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right)|\phantom{\rule{2.77695pt}{0ex}}<\epsilon ,\phantom{\rule{1em}{0ex}}|{{\mathcal{G}}^{\prime }}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right)-{{\stackrel{̃}{\mathcal{G}}}^{\prime }}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right)\right|<\epsilon ,$
(78)

for a sufficiently small ε. From Lemma 3.2 we can see that ${{\mathcal{G}}_{\theta }}_{,m}\left(\lambda \right)$ satisfies the condition (9) when m > 1 and therefore whenever $0<\epsilon \le \text{min}\left\{\pi /\sigma ,\phantom{\rule{2.77695pt}{0ex}}\sigma /\pi ,\phantom{\rule{2.77695pt}{0ex}}1/\sqrt{e}\right\}$ we have

$|{\mathcal{G}}_{\theta ,m,N}\left(\lambda \right)-{\stackrel{̃}{\mathcal{G}}}_{\theta ,m,N}\left(\lambda \right)|\le \mathcal{A}\left(\epsilon \right),\phantom{\rule{1em}{0ex}}\lambda \in ℝ,$
(79)

where there is a positive constant ${M}_{{\mathcal{G}}_{\theta ,m}}$ for which, cf. (10),

$\mathcal{A}\left(\epsilon \right):=\frac{2{e}^{1/4}}{\sigma }\left\{\sqrt{3}e\left(1+\sigma \right)+\left(\frac{\pi }{\sigma }A+{M}_{{\mathcal{G}}_{\theta ,m}}\right)\rho \left(\epsilon \right)+\left(\sigma +2+\text{log}\left(2\right)\right){M}_{{\mathcal{G}}_{\theta ,m}}\right\}\epsilon \text{log}\left(1/\epsilon \right).$
(80)

Here

$A:=\frac{3\sigma }{\pi }\left(|{\mathcal{G}}_{\theta ,m}\left(0\right)|+\frac{\sigma }{\pi }{M}_{{\mathcal{G}}_{\theta ,m}}\right),\phantom{\rule{1em}{0ex}}\rho \left(\epsilon \right):=\gamma +10\text{log}\left(1/\epsilon \right),\phantom{\rule{1em}{0ex}}\ell =1.$

In the following we use the technique of [27] to determine enclosure intervals for the eigenvalues. Let λ be an eigenvalue, that is

$\Omega \left({\lambda }^{*}\right)=\mathcal{K}\left({\lambda }^{*}\right)+{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\mathcal{G}}_{\theta ,m}\left({\lambda }^{*}\right)=0.$

Then it follows that

$\begin{array}{ll}\hfill \mathcal{K}\left({\lambda }^{*}\right)+{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\stackrel{̃}{\mathcal{G}}}_{\theta ,m,N}\left({\lambda }^{*}\right)& ={\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\stackrel{̃}{\mathcal{G}}}_{\theta ,m,N}\left({\lambda }^{*}\right)-{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\mathcal{G}}_{\theta ,m}\left({\lambda }^{*}\right)\phantom{\rule{2em}{0ex}}\\ =\left[{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\stackrel{̃}{\mathcal{G}}}_{\theta ,m,N}\left({\lambda }^{*}\right)-{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\mathcal{G}}_{\theta ,m,N}\left({\lambda }^{*}\right)\right]\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\left[{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\mathcal{G}}_{\theta ,m,N}\left({\lambda }^{*}\right)-{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\mathcal{G}}_{\theta ,m}\left({\lambda }^{*}\right)\right]\phantom{\rule{2em}{0ex}}\end{array}$

and so

$\left|\mathcal{K}\left({\lambda }^{*}\right)+{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\stackrel{̃}{\mathcal{G}}}_{\theta ,m,N}\left({\lambda }^{*}\right)\right|\le {\left|\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right|}^{-m}\left({T}_{N,m-1,\sigma }\left({\lambda }^{*}\right)+\mathcal{A}\left(\epsilon \right)\right).$

Since $\mathcal{K}\left({\lambda }^{*}\right)+{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\stackrel{̃}{\mathcal{G}}}_{\theta ,m,N}\left({\lambda }^{*}\right)$ is given and, ${\left|\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right|}^{-m}\left({T}_{N,m-1,\sigma }\left({\lambda }^{*}\right)+\mathcal{A}\left(\epsilon \right)\right)$ has computable upper bound, we can define an enclosure for λ, by solving the following system of inequalities

$\begin{array}{ll}\hfill -{\left|\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right|}^{-m}& \left({T}_{N,m-1,\sigma }\left({\lambda }^{*}\right)+\mathcal{A}\left(\epsilon \right)\right)\le \mathcal{K}\left({\lambda }^{*}\right)\phantom{\rule{2em}{0ex}}\\ +{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\stackrel{̃}{\mathcal{G}}}_{\theta ,m,N}\left({\lambda }^{*}\right)\le {\left|\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right|}^{-m}\left({T}_{N,m-1,\sigma }\left({\lambda }^{*}\right)+\mathcal{A}\left(\epsilon \right)\right).\phantom{\rule{2em}{0ex}}\end{array}$
(81)

Its solution is an interval containing λ*, and over which the graph $\mathcal{K}\left({\lambda }^{*}\right)+{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}×\phantom{\rule{2.77695pt}{0ex}}{\stackrel{̃}{\mathcal{G}}}_{\theta ,m,N}\left({\lambda }^{*}\right)$ is squeezed between the graphs

$-{\left|\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right|}^{-m}\left({T}_{N,m-1,\sigma }\left({\lambda }^{*}\right)+\mathcal{A}\left(\epsilon \right)\right)$
(82)

and

${\left|\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right|}^{-m}\left({T}_{N,m-1,\sigma }\left({\lambda }^{*}\right)+\mathcal{A}\left(\epsilon \right)\right).$
(83)

Using the fact that

${\stackrel{̃}{\mathcal{G}}}_{\theta ,m,N}\left(\lambda \right)\to {\mathcal{G}}_{\theta ,m}\left(\lambda \right)$

uniformly over any compact set, and since λ* is a simple root, we obtain for large N and sufficiently small ε

$\frac{\partial }{\partial \lambda }\left(\mathcal{K}\left(\lambda \right)+{\left(\frac{\text{sin}\theta \lambda }{\theta \lambda }\right)}^{-m}{\stackrel{̃}{\mathcal{G}}}_{\theta ,m,N}\left(\lambda \right)\right)\ne 0$

in a neighborhood of λ*. Hence the graph of $\mathcal{K}\left(\lambda \right)+{\left(\frac{\text{sin}\theta \lambda }{\theta \lambda }\right)}^{-m}{\stackrel{̃}{\mathcal{G}}}_{\theta ,m,N}\left(\lambda \right)$ intersects the graphs $-{\left|\frac{\text{sin}\theta \lambda }{\theta \lambda }\right|}^{-m}\left({T}_{N,m-1,\sigma }\left(\lambda \right)+\mathcal{A}\left(\epsilon \right)\right)$ and ${\left|\frac{\text{sin}\theta \lambda }{\theta \lambda }\right|}^{-m}\left({T}_{N,m-1,\sigma }\left(\lambda \right)+\mathcal{A}\left(\epsilon \right)\right)$ at two points with abscissa a _(λ*, N, ε) ≤ a + (λ*, N, ε) the interval

${I}_{\epsilon ,N}:=\left[{a}_{-}\left({\lambda }^{*},\phantom{\rule{2.77695pt}{0ex}}N,\phantom{\rule{2.77695pt}{0ex}}\epsilon \right),\phantom{\rule{2.77695pt}{0ex}}{a}_{+}\left({\lambda }^{*},\phantom{\rule{2.77695pt}{0ex}}N,\phantom{\rule{2.77695pt}{0ex}}\epsilon \right)\right]$

and in particular λ*I ε,N . Summarizing the above discussion, we arrive at the following lemma which is similar to that of [27] for Sturm-Liouville problems.

Lemma 3.3 For any eigenvalue λ*, we can find N0 + and sufficiently small ε such that λ*I ε,N for N > N0. Moreover

(84)

Proof. Since all eigenvalues of (15)-(19) are simple, then for large N and sufficiently small ε we have $\frac{\partial }{\partial \lambda }\left(\mathcal{K}\left(\lambda \right)+{\left(\frac{\text{sin}\theta \lambda }{\theta \lambda }\right)}^{-m}{\stackrel{̃}{\mathcal{G}}}_{\theta ,m,N}\left(\lambda \right)\right)>0,$ in a neighborhood of λ*. Choose N0 such that

$\mathcal{K}\left(\lambda \right)+{\left(\frac{\text{sin}\theta \lambda }{\theta \lambda }\right)}^{-m}{\stackrel{̃}{\mathcal{G}}}_{\theta ,m,{N}_{0}}\left(\lambda \right)=±{\left|\frac{\text{sin}\theta \lambda }{\theta \lambda }\right|}^{-m}\left({T}_{{N}_{0},m-1,\sigma }\left(\lambda \right)+\mathcal{A}\left(\epsilon \right)\right)$

has two distinct solutions which we denote by a _(λ*,N0,ε) ≤ a+(λ*,N0,ε). The decay of TN,m-1,σ(λ)→0 as N → ∞ and $\mathcal{A}\left(\epsilon \right)\to 0$ as ε → 0 will ensure the existence of the solutions a _(λ*,N,ε) and a+(λ*,N,ε) as N → ∞ and ε → 0. For the second point we recall that ${\stackrel{̃}{\mathcal{G}}}_{\theta ,m,N}\left(\lambda \right)\to {\mathcal{G}}_{\theta ,m}\left(\lambda \right)$ as N → ∞ and as ε → 0. Hence by taking the limit we obtain

$\mathcal{K}\left({a}_{+}\left({\lambda }^{*},\phantom{\rule{2.77695pt}{0ex}}\infty ,\phantom{\rule{2.77695pt}{0ex}}0\right)\right)+{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\mathcal{G}}_{\theta ,m}\left({a}_{+}\left({\lambda }^{*},\phantom{\rule{2.77695pt}{0ex}}\infty ,\phantom{\rule{2.77695pt}{0ex}}0\right)\right)=0,$
$\mathcal{K}\left({a}_{-}\left({\lambda }^{*},\phantom{\rule{2.77695pt}{0ex}}\infty ,\phantom{\rule{2.77695pt}{0ex}}0\right)\right)+{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\mathcal{G}}_{\theta ,m}\left({a}_{-}\left({\lambda }^{*},\phantom{\rule{2.77695pt}{0ex}}\infty ,\phantom{\rule{2.77695pt}{0ex}}0\right)\right)=0,$

that is Ω(a+)=Ω(a-)=0. This leads us to conclude that a+ = a- = λ*, since λ* is a simple root.

Let ${\stackrel{̃}{\Omega }}_{N}\left(\lambda \right):=\mathcal{K}\left(\lambda \right)+{\left(\frac{\text{sin}\theta \lambda }{\theta \lambda }\right)}^{-m}{\stackrel{̃}{\mathcal{G}}}_{\theta ,m,N}\left(\lambda \right).$ Then (75) and (79) imply

$\left|\Omega \left(\lambda \right)-{\stackrel{̃}{\Omega }}_{N}\left(\lambda \right)\right|\le {\left|\frac{\text{sin}\theta \lambda }{\theta \lambda }\right|}^{-m}\left({T}_{N,m-1,\sigma }\left(\lambda \right)+\mathcal{A}\left(\epsilon \right)\right),\phantom{\rule{1em}{0ex}}|\lambda |\phantom{\rule{2.77695pt}{0ex}}<\frac{N\pi }{\sigma }$
(85)

and θ is chosen sufficiently small for which |θλ| < π. Let λ* be an eigenvalue and λ N be its approximation. Thus Ω(λ*) = 0 and ${\stackrel{̃}{\Omega }}_{N}\left({\lambda }_{N}\right)=0.$ From (85) we have $|{\stackrel{̃}{\Omega }}_{N}\left({\lambda }^{*}\right)|\phantom{\rule{2.77695pt}{0ex}}\le {\left|\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right|}^{-m}\left({T}_{N,m-1,\sigma }\left({\lambda }^{*}\right)+\mathcal{A}\left(\epsilon \right)\right).$ Now we estimate the error |λ* N |, for the eigenvalue λ*.

Lemma 3.4 Let λ* be an eigenvalue of (15)-(19). For sufficient large N we have the following estimate

$|{\lambda }^{*}-{\lambda }_{N}|<{\left|\frac{\text{sin}\theta {\lambda }_{N}}{\theta {\lambda }_{N}}\right|}^{-m}\frac{{T}_{N,m-1,\sigma },\left({\lambda }_{N}\right)+\mathcal{A}\left(\epsilon \right)}{\underset{\zeta \in {I}_{\epsilon ,N}}{\text{inf}}|{\Omega }^{\prime }\left(\zeta \right)|}.$
(86)

Proof. Since $\Omega \left({\lambda }_{N}\right)-{\stackrel{̃}{\Omega }}_{N}\left({\lambda }_{N}\right)=\Omega \left({\lambda }_{N}\right)-\Omega \left({\lambda }^{*}\right),$ then from (85) and after replacing λ by λ N we obtain

$|\Omega \left({\lambda }_{N}\right)-\Omega \left({\lambda }^{*}\right)|\le {\left|\frac{\text{sin}\theta {\lambda }_{N}}{\theta {\lambda }_{N}}\right|}^{-m}\left({T}_{N,m-1,\sigma }\left({\lambda }_{N}\right)+\mathcal{A}\left(\epsilon \right)\right)$