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(ℝ)-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 ℝ (cf. [1–4]). 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

$$R_N\left(f\right)\left(t\right):=f\left(t\right)-f_N\left(t\right),N\in {\mathbb{Z}}^{+},t\in ℝ,$$

(3)

where f _N (t) is the truncated series

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

(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 ∈ ℤ⁺ such that t^kf(t) ∈ L²(ℝ), then, for t ∈ ℝ, |t| < Nπ/σ, we have

$$\begin{array}{ll}\hfill \left|R_N\left(f\right)\left(t\right)\right|\le T_{N,k,\sigma }\left(t\right)& :=\frac{{\xi }_{k,\sigma }{E}_k|\text{sin}\sigma t{|}^2}{\sqrt{3}{\left(N+1\right)}^k}\left(\frac{1}{{\left(N\pi -\sigma t\right)}^{3/2}}+\frac{1}{{\left(N\pi +\sigma t\right)}^{3/2}}\right)\\ +\frac{{\xi }_{k,\sigma }\left(\sigma E_k+k{E}_{k-1}\right)|\text{sin}\sigma t{|}^2}{\sigma {\left(N+1\right)}^k}\left(\frac{1}{\sqrt{N\pi -\sigma t}}+\frac{1}{\sqrt{N\pi +\sigma t}}\right),\end{array}$$

(5)

where the constants E _k and ξ_k,σare given by

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

(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

$$\begin{array}{ll}\hfill \mathcal{A}\left(\epsilon ,f\right)\left(t\right)& =\sum _{n=-\infty }^{\infty }\left[\left\{f\left(\frac{n\pi }{\sigma }\right)-\tilde{f}\left(\frac{n\pi }{\sigma }\right)\right\}{S}_n^2\left(t\right)\right.\\ \left.+\left\{f^{\prime }\left(\frac{n\pi }{\sigma }\right)-{\tilde{f}}^{\prime }\left(\frac{n\pi }{\sigma }\right)\right\}\frac{\text{sin}\left(\sigma t-n\pi \right)}{\sigma }{S}_n\left(t\right)\right],t\in ℝ,\end{array}$$

(7)

where $\tilde{f}\left(\frac{n\pi }{\sigma }\right)$ and ${\tilde{f}}^{\prime }-\left(\frac{n\pi }{\sigma }\right)$ 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 ${\epsilon }_n:=f\left(\frac{n\pi }{\sigma }\right)-\tilde{f}\left(\frac{n\pi }{\sigma }\right),{\epsilon }_n^{\prime }:=f^{\prime }\left(\frac{n\pi }{\sigma }\right)-{\tilde{f}}^{\prime }-\left(\frac{n\pi }{\sigma }\right),n\in \mathbb{Z},$are bounded by a positive number ε, i.e. $\left|{\epsilon }_n\right|,\left|{\epsilon }_n^{\prime }\right|\le \epsilon .$If $f\text{(}t\text{)}\in {\text{PW}}_{\sigma }^2$ satisfies the natural decay conditions

$$\left|{\epsilon }_n\right|\le \left|f\left(\frac{n\pi }{\sigma }\right)\right|,\left|{{\epsilon }^{\prime }}_n\right|\le \left|f^{\prime }\left(\frac{n\pi }{\sigma }\right)\right|,$$

(8)

$$\left|f\left(t\right)\right|\le \frac{M_f}{|t{|}^{\ell +1}},t\in ℝ-\left\{0\right\},$$

(9)

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

$$\left|\right|\mathcal{A}\left(\epsilon ,f\right)|{|}_{\infty }\le \frac{4e^{1/4}}{\sigma \left(\ell +1\right)}\left\{\sqrt{3}e\left(1+\sigma \right)+\left(\left(\pi /\sigma \right)A+M_f\right)\rho \left(\epsilon \right)+\left(\sigma +2+\text{log}\left(2\right)\right)M_f\right\}\epsilon \text{log}\left(1/\epsilon \right),$$

(10)

where

$$A:=\frac{3\sigma }{\pi }\left(\left|f\left(0\right)\right|+M_f{\left(\frac{\sigma }{\pi }\right)}^{\ell }\right),\rho \left(\epsilon \right):=\gamma +10\text{log}\left(1/\epsilon \right),$$

(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

$$f\left(t\right)=\sum _{n=-\infty }^{\infty }f\left(\frac{n\pi }{\sigma }\right)S_n\left(t\right),t\in ℝ,$$

(12)

where the convergence is absolute and uniform on ℝ and it is uniform on compact sets of ℂ (cf. [6–8]). Series (12), which is of Lagrange interpolation type, has been used to compute eigenvalues of second order eigenvalue problems (see e.g. [9–13]). The use of (12) in numerical analysis is known as the sinc-method established by Stenger (cf. [14–16]). 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 [20–22]). 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

$$u_2^{\prime }\left(x\right)-r_1\left(x\right)u_1\left(x\right)=\lambda u_1\left(x\right),u_1^{\prime }\left(x\right)+r_2\left(x\right)u_2\left(x\right)=-\lambda u_2\left(x\right),x\in \left[-1,0\right)\cup \left(0,1\right],$$

(15)

$$U_1\left(\mathit{u}\right):=\text{sin}\alpha u_1\left(-1\right)+\text{cos}\alpha u_2\left(-1\right)=0,$$

(16)

$$U_2\left(\mathit{u}\right):=\text{sin}\beta u_1\left(1\right)+\text{cos}\beta u_2\left(1\right)=0,$$

(17)

and transmission conditions

$$U_3\left(\mathit{u}\right):=u_1\left(0^{-}\right)-\delta u_1\left(0^{+}\right)=0,$$

(18)

$$U_4\left(\mathit{u}\right):=u_2\left(0^{-}\right)-\delta u_2\left(0^{+}\right)=0,$$

(19)

where λ ∈ ℂ; the real valued function r₁(·) and r₂(·) are continuous in [−1, 0) and (0, 1], and have finite limits $r_1\left(0^{\pm }\right):=\underset{x\to 0^{\pm }}{\text{lim}}{r}_1\left(x\right),r_2\left(0^{\pm }\right):=\underset{x\to 0^{\pm }}{\text{lim}}{r}_2\left(x\right);\delta \in ℝ,\alpha ,\beta \in \left[0,\pi \right)$and δ ≠ 0.

Let H be the Hilbert space

$$\mathit{H}:=\left\{\mathit{u}\left(x\right)=\left(\begin{array}{c}\hfill u_1\left(x\right)\hfill \\ \hfill u_2\left(x\right)\hfill \end{array}\right),u_1\left(x\right),u_2\left(x\right)\in L^2\left(-1,0\right)\oplus L^2\left(0,1\right)\right\}.$$

(20)

The inner product of H is defined by

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

(21)

where ⊤ denotes the matrix transpose,

$$\mathit{u}\left(x\right)=\left(\begin{array}{c}\hfill u_1\left(x\right)\hfill \\ \hfill u_2\left(x\right)\hfill \end{array}\right),\mathit{v}\left(x\right)=\left(\begin{array}{c}\hfill v_1\left(x\right)\hfill \\ \hfill v_2\left(x\right)\hfill \end{array}\right)\in \mathit{H},u_i\left(\cdot \right),v_i\left(\cdot \right)\in L^2\left(-1,1\right),i=1,2.$$

Equation (15) can be written as

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

(22)

where

$$A=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 0\hfill \end{array}\right),P\left(x\right)=\left(\begin{array}{cc}\hfill r_1\left(x\right)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill r_2\left(x\right)\hfill \end{array}\right),\mathit{u}\left(x\right)=\left(\begin{array}{c}\hfill u_1\left(x\right)\hfill \\ \hfill u_2\left(x\right)\hfill \end{array}\right).$$

(23)

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

$$u_{\left(1\right)}\left(x\right)=\left\{\begin{array}{cc}\hfill u\left(x\right),\hfill & \hfill x\in \left[-1,0\right);\hfill \\ \hfill u\left(0^{-}\right),\hfill & \hfill x=0\hfill \end{array}\right.,u_{\left(2\right)}\left(x\right)=\left\{\begin{array}{cc}\hfill u\left(x\right),\hfill & \hfill x\in \left(0,1\right]\hfill \\ \hfill u\left(0^{+}\right),\hfill & \hfill x=0;\hfill \end{array}\right.$$

(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],

$$\frac{d}{dx}\{u_1(x){\overline{u}}_2(x)-{\overline{u}}_1(x)u_2(x)\}=({\overline{\lambda }}_0-{\text{λ}}_{\text{0}})\left\{\right|u_1(x){|}^2+|u_2(x){|}^2\}.$$

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

$$\begin{array}{ll}\hfill \left({\bar{\lambda }}_0-{\lambda }_0\right)\left[\underset{-1}{\overset{0}{\int }}\left(|u_1\left(x\right){|}^2+|u_2\left(x\right){|}^2\right)dx\right]& =u_1\left(0^{-}\right){ū}_2\left(0^{-}\right)-{ū}_1\left(0^{-}\right)u_2\left(0^{-}\right)\\ -\left[u_1\left(-1\right){ū}_2\left(-1\right)-{ū}_1\left(-1\right)u_2\left(-1\right)\right],\end{array}$$

(25)

$$\begin{array}{ll}\hfill \left({\bar{\lambda }}_0-{\lambda }_0\right)\left[\underset{0}{\overset{1}{\int }}\left(|u_1\left(x\right){|}^2+|u_2\left(x\right){|}^2\right)dx\right]& =u_1\left(1\right){ū}_2\left(1\right)-{ū}_1\left(1\right)u_2\left(1\right)\\ -\left[u_1\left(0^{+}\right){ū}_2\left(0^{+}\right)-{ū}_1\left(0^{+}\right)u_2\left(0^{+}\right)\right].\end{array}$$

(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 ${\lambda }_{\text{0}}\ne {\bar{\lambda }}_{\text{0}},$ 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

$$\underset{-1}{\overset{0}{\int }}{\mathit{u}}^{\top }\left(x,{\lambda }_1\right)\mathit{v}\left(x,{\lambda }_2\right)dx+{\delta }^2\underset{0}{\overset{1}{\int }}{\mathit{u}}^{\top }\left(x,{\lambda }_1\right)\mathit{v}\left(x,{\lambda }_2\right)dx=0.$$

(28)

Proof. By (15) we obtain

$$\frac{d}{dx}\left\{u_1\left(x,{\lambda }_1\right)v_2\left(x,{\lambda }_2\right)-u_2\left(x,{\lambda }_2\right)v_1\left(x,{\lambda }_1\right)\right\}=\left({\lambda }_2-{\lambda }_1\right)\left\{u_1\left(x,{\lambda }_1\right)v_1\left(x,{\lambda }_2\right)+u_2\left(x,{\lambda }_1\right)v_2\left(x,{\lambda }_2\right)\right\}.$$

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 λ₁≠λ₂.

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:

$$u_2^{{}^{\prime }}\left(x\right)-r_1\left(x\right)u_1\left(x\right)=\lambda u_1\left(x\right),u_1^{{}^{\prime }}\left(x\right)+r_2\left(x\right)u_2\left(x\right)=-\lambda u_2\left(x\right),x\in \left(-1,0\right),$$

(29)

$$u_1\left(-1\right)=\text{cos}\alpha ,u_2\left(-1\right)=-\text{sin}\alpha .$$

(30)

By virtue of Theorem 1.1 in [23] this problem has a unique solution $\mathit{u}=\left(\begin{array}{c}\hfill {\varphi }_{11}\left(x,\lambda \right)\hfill \\ \hfill {\varphi }_{21}\left(x,\lambda \right)\hfill \end{array}\right),$which is an entire function of λ ∈ ℂ 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

$$u_2^{\prime }\left(x\right)-r_1\left(x\right)u_1\left(x\right)=\lambda u_1\left(x\right),u_1^{\prime }\left(x\right)+r_2\left(x\right)u_2\left(x\right)=-\lambda u_2\left(x\right),x\in \left(0,1\right),$$

(31)

$$u_1\left(1\right)=\text{cos}\beta ,u_2\left(1\right)=-\text{sin}\beta .$$

(32)

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

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

$$u_2^{\prime }\left(x\right)-r_1\left(x\right)u_1\left(x\right)=\lambda u_1\left(x\right),u_1^{\prime }\left(x\right)+r_2\left(x\right)u_2\left(x\right)=-\lambda u_2\left(x\right),x\in \left(0,1\right),$$

(33)

$$u_1\left(0\right)=\frac{1}{\delta }{\varphi }_{11}\left(0,\lambda \right),u_2\left(0\right)=\frac{1}{\delta }{\varphi }_{21}\left(0,\lambda \right),$$

(34)

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

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

$$u_2^{\prime }\left(x\right)-r_1\left(x\right)u_1\left(x\right)=\lambda u_1\left(x\right),u_1^{\prime }\left(x\right)+r_2\left(x\right)u_2\left(x\right)=-\lambda u_2\left(x\right),x\in \left(-1,0\right),$$

(35)

$$u_1\left(0\right)=\delta {\chi }_{12}\left(0,\lambda \right),u_2\left(0\right)=\delta {\chi }_{22}\left(0,\lambda \right).$$

(36)

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

$$\mathit{\varphi }\left(\cdot ,\lambda \right)=\left(\begin{array}{c}\hfill {\varphi }_1\left(\cdot ,\lambda \right)\hfill \\ \hfill {\varphi }_2\left(\cdot ,\lambda \right)\hfill \end{array}\right),\mathit{\chi }\left(\cdot ,\lambda \right)=\left(\begin{array}{c}\hfill {\chi }_1\left(\cdot ,\lambda \right)\hfill \\ \hfill {\chi }_2\left(\cdot ,\lambda \right)\hfill \end{array}\right),$$

where

$${\varphi }_1\left(x,\lambda \right)=\left\{\begin{array}{cc}\hfill {\varphi }_{11}\left(x,\lambda \right),\hfill & \hfill x\in \left[-1,0\right)\hfill \\ \hfill {\varphi }_{12}\left(x,\lambda \right),\hfill & \hfill x\in \left(0,1\right]\hfill \end{array}\right.,{\varphi }_2\left(x,\lambda \right)=\left\{\begin{array}{cc}\hfill {\varphi }_{21}\left(x,\lambda \right),\hfill & \hfill x\in \left[-1,0\right)\hfill \\ \hfill {\varphi }_{22}\left(x,\lambda \right),\hfill & \hfill x\in \left(0,1\right]\hfill \end{array},\right.$$

(37)

$${\chi }_1\left(x,\lambda \right)=\left\{\begin{array}{cc}\hfill {\chi }_{11}\left(x,\lambda \right),\hfill & \hfill x\in \left[-1,0\right)\hfill \\ \hfill {\chi }_{12}\left(x,\lambda \right),\hfill & \hfill x\in \left(0,1\right]\hfill \end{array},\right.{\chi }_2\left(x,\lambda \right)=\left\{\begin{array}{cc}\hfill {\chi }_{21}\left(x,\lambda \right),\hfill & \hfill x\in \left[-1,0\right)\hfill \\ \hfill {\chi }_{22}\left(x,\lambda \right),\hfill & \hfill x\in \left(0,1\right]\hfill \end{array},\right.$$

(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.,

$$W\left(\mathit{\varphi },\mathit{\chi }\right)\left(\cdot ,\lambda \right):=\left|\begin{array}{cc}\hfill {\varphi }_1\left(\cdot ,\lambda \right)\hfill & \hfill {\varphi }_2\left(\cdot ,\lambda \right)\hfill \\ \hfill {\chi }_1\left(\cdot ,\lambda \right)\hfill & \hfill {\chi }_2\left(\cdot ,\lambda \right)\hfill \end{array}\right|.$$

It is obvious that the Wronskian

$${\Omega }_i\left(\lambda \right):=W\left(\mathit{\varphi },\mathit{\chi }\right)\left(x,\lambda \right)={\varphi }_{1i}\left(x,\lambda \right){\chi }_{2i}\left(x,\lambda \right)-{\varphi }_{2i}\left(x,\lambda \right){\chi }_{1i}\left(x,\lambda \right),x\in {\Gamma }_i,i=1,2$$

(39)

are independent of x ∈ Γ_i 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 ∈ ℂ.

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 ∈ ℂ, λ ≠ μ,

$$\frac{d}{dx}\left\{{\varphi }_1\left(x,\lambda \right){\varphi }_2\left(x,\mu \right)-{\varphi }_1\left(x,\mu \right){\varphi }_2\left(x,\lambda \right)\right\}=\left(\mu -\lambda \right)\left\{{\varphi }_1\left(x,\lambda \right){\varphi }_1\left(x,\mu \right)+{\varphi }_2\left(x,\lambda \right){\varphi }_2\left(x,\mu \right)\right\}.$$

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

$$\begin{array}{ll}\hfill {\varphi }_{12}\left(1,\lambda \right){\varphi }_{22}\left(1,\mu \right)& -{\varphi }_{12}\left(1,\mu \right){\varphi }_{22}\left(1,\lambda \right)=\\ \left(\mu -\lambda \right)\left(\underset{-1}{\overset{0}{\int }}\left({\varphi }_{11}\left(x,\lambda \right){\varphi }_{11}\left(x,\mu \right)+{\varphi }_{21}\left(x,\lambda \right){\varphi }_{21}\left(x,\mu \right)\right)dx\right.\\ \left.+{\delta }^2\underset{0}{\overset{1}{\int }}\left({\varphi }_{12}\left(x,\lambda \right){\varphi }_{12}\left(x,\mu \right)+{\varphi }_{22}\left(x,\lambda \right){\varphi }_{22}\left(x,\mu \right)\right)dx\right).\end{array}$$

(41)

Dividing both sides of (41) by (λ − µ) and by letting µ → λ, we arrive to the relation

$$\begin{array}{l}{\varphi }_{\text{22}}\left(\text{1},\text{ λ}\right)\frac{\partial {\varphi }_{\text{12}}\left(\text{1},\text{ λ}\right)}{\partial \text{λ}}-{\varphi }_{\text{12}}\left(\text{1},\text{ λ}\right)\frac{\partial {\varphi }_{\text{22}}\left(\text{1},\text{ λ}\right)}{\partial \text{λ}}=-({\displaystyle \underset{-1}{\overset{0}{\int }}\left(|{\varphi }_{\text{11}}\left(x,\lambda \right)\left|{}^{\text{2}}+\right|{\varphi }_{\text{21}}\left(x,\text{ λ}\right){|}^2\right)dx}\\ +{\delta }^2{\displaystyle \underset{0}{\overset{1}{\int }}\left(|{\varphi }_{\text{12}}\left(x,\text{λ}\right)\left|{}^{\text{2}}+\right|{\varphi }_{\text{22}}\left(x,\text{λ}\right){|}^2\right)dx}).\end{array}$$

(42)

We show that equation

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

(43)

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

$$\text{sin}\beta {\varphi }_{12}\left(1,{\lambda }^{*}\right)+\text{cos}\beta {\varphi }_{22}\left(1,{\lambda }^{*}\right)=0,$$

(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

$${\varphi }_{22}\left(1,{\lambda }^{*}\right)\frac{\partial {\varphi }_{12}\left(1,{\lambda }^{*}\right)}{\partial \lambda }-{\varphi }_{12}\left(1,{\lambda }^{*}\right)\frac{\partial {\varphi }_{22}\left(1,{\lambda }^{*}\right)}{\partial \lambda }$$

(46)

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

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

(47)

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

Here ${\left\{\varphi \left(\cdot ,{\lambda }_n\right)\right\}}_{n=-\infty }^{\infty }$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

$$\text{sin}\alpha {\chi }_{11}\left(-1,\lambda \right)+\text{cos}\alpha {\chi }_{21}\left(-1,\lambda \right)=\Omega \left(\lambda \right).$$

(48)

Therefore ${\left\{\mathit{\chi }\left(\cdot ,{\lambda }_n\right)\right\}}_{n=-\infty }^{\infty }$ is another set of eigen-vector-functions which is related by ${\left\{\mathit{\varphi }\left(\cdot ,{\lambda }_n\right)\right\}}_{n=-\infty }^{\infty }$ with

$$\mathit{\chi }\left(x,{\lambda }_n\right)=c_n\varphi \left(x,{\lambda }_n\right),x\in \left[-1,0\right)\cup \left(0,1\right],n\in \mathbb{Z},$$

(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

$$\Omega \left(\lambda \right)=-{\delta }^2\left(\text{sin}\beta {\varphi }_{12}\left(1,\lambda \right)+\text{cos}\beta {\varphi }_{22}\left(1,\lambda \right)\right).$$

(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]),

$${\varphi }_{11}\left(x,\lambda \right)=\text{cos}\left(\lambda \left(x+1\right)-\alpha \right)-{\mathcal{S}}_{-1,1}{\varphi }_{11}\left(x,\lambda \right)-{\tilde{\mathcal{S}}}_{-1,2}{\varphi }_{21}-\left(x,\lambda \right),$$

(51)

$${\varphi }_{21}\left(x,\lambda \right)=\text{sin}\left(\lambda \left(x+1\right)-\alpha \right)+{\tilde{\mathcal{S}}}_{-1,1}{\varphi }_{11}\left(x,\lambda \right)-{\mathcal{S}}_{-1,2}{\varphi }_{21}\left(x,\lambda \right),$$

(52)

$${\varphi }_{12}\left(x,\lambda \right)=\frac{1}{\delta }{\varphi }_{11}\left(0^{-},\lambda \right)\text{cos}\left(\lambda x\right)-\frac{1}{\delta }{\varphi }_{21}\left(0^{-},\lambda \right)\text{sin}\left(\lambda x\right)-{\mathcal{S}}_{0,1}{\varphi }_{12}\left(x,\lambda \right)-{\tilde{\mathcal{S}}}_{0,2}{\varphi }_{22}\left(x,\lambda \right),$$

(53)

$${\varphi }_{22}\left(x,\lambda \right)=\frac{1}{\delta }{\varphi }_{11}\left(0^{-},\lambda \right)\text{sin}\left(\lambda x\right)+\frac{1}{\delta }{\varphi }_{21}\left(0^{-},\lambda \right)\text{cos}{\left(\lambda x\right)}^{-}+{\tilde{\mathcal{S}}}_{0,1}{\varphi }_{12}\left(x,\lambda \right)-{\mathcal{S}}_{0,2}{\varphi }_{22}\left(x,\lambda \right),$$

(54)

where ${\mathcal{S}}_{-1,i},{\tilde{\mathcal{S}}}_{-1,i},{\mathcal{S}}_{0,i}$ and ${\tilde{\mathcal{S}}}_{0,i},i=1,2,$ are the Volterra integral operators defined by

$$\begin{array}{ll}\hfill {\mathcal{S}}_{-1,i}\phi \left(x,\lambda \right)& :=\underset{-1}{\overset{x}{\int }}\text{sin}\lambda \left(x-t\right)r_i\left(t\right)\phi \left(t,\lambda \right)dt,{\tilde{\mathcal{S}}}_{-1,i}\phi \left(x,\lambda \right):=\underset{-1}{\overset{x}{\int }}\text{cos}\lambda \left(x-t\right)r_i\left(t\right)\phi \left(t,\lambda \right)dt,\\ \hfill {\mathcal{S}}_{0,i}\phi \left(x,\lambda \right)& :=\underset{0}{\overset{x}{\int }}\text{sin}\lambda \left(x-t\right)r_i\left(t\right)\phi \left(t,\lambda \right)dt,{\tilde{\mathcal{S}}}_{0,i}\phi \left(x,\lambda \right):=\underset{0}{\overset{x}{\int }}\text{cos}\lambda \left(x-t\right)r_i\left(t\right)\phi \left(t,\lambda \right)dt.\end{array}$$

For convenience, we define the constants

$$\begin{array}{l}{c}_1:=\underset{-1}{\overset{0}{\int }}\left[\left|r_1\left(t\right)\right|+\left|r_2\left(t\right)\right|\right]dt,c_2:=c_1\text{exp}\left(c_1\right),\\ c_3:=\underset{0}{\overset{1}{\int }}\left[\left|r_1\left(t\right)\right|+\left|r_2\left(t\right)\right|\right]dt,c_4:=c_2+\frac{2}{\left|\delta \right|}\left(1+c_2\right).\end{array}$$

(55)

Define h_−1,i(·, λ) and h_0,i(·, λ), i = 1, 2, to be

$$\left.\begin{array}{c}\hfill h_{-1,1}\left(x,\lambda \right):={\mathcal{S}}_{-1,1}{\varphi }_{11}\left(x,\lambda \right)+{\tilde{\mathcal{S}}}_{-1,2}{\varphi }_{21}\left(x,\lambda \right),\hfill \\ \hfill h_{-1,2}\left(x,\lambda \right):={\tilde{\mathcal{S}}}_{-1,1}{\varphi }_{11}\left(x,\lambda \right)+{\mathcal{S}}_{-1,2}{\varphi }_{21}\left(x,\lambda \right),\hfill \end{array}\right\}$$

(56)

$$\left.\begin{array}{c}{h}_{0,1}\left(x,\lambda \right):={\mathcal{S}}_{0,1}{\varphi }_{12}\left(x,\lambda \right)+{\tilde{\mathcal{S}}}_{0,2}{\varphi }_{22}\left(x,\lambda \right),\\ h_{0,2}\left(x,\lambda \right):={\tilde{\mathcal{S}}}_{0,1}{\varphi }_{12}\left(x,\lambda \right)-{\mathcal{S}}_{0,2}{\varphi }_{22}\left(x,\lambda \right).\end{array}\right\}$$

(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

$$\left|h_{-1,1}\left(x,\lambda \right)\right|,\left|h_{-1,2}\left(x,\lambda \right)\right|\le 2c_2{e}^{\left|\Im \lambda \right|\left(x+1\right)},\lambda \in ℂ.$$

(58)

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

$$\begin{array}{ll}\hfill \left|h_{-1,1}\left(x,\lambda \right)\right|& \le \left|{\mathcal{S}}_{-1,1}\text{cos}\left(\lambda \left(x+1\right)-\alpha \right)\right|+\left|{\tilde{\mathcal{S}}}_{-1,2}\text{sin}\left(\lambda \left(x-+1\right)-\alpha \right)\right|+\left|{\mathcal{S}}_{-1,1}{h}_{-1,1}\left(x,\lambda \right)\right|\\ +\left|{\tilde{\mathcal{S}}}_{-1,2}{h}_{-1,2}\left(x,\lambda \right)\right|\\ \le e^{\left|\Im \lambda \right|\left(x+1\right)}\underset{-1}{\overset{x}{\int }}\left[\left|r_1\left(t\right)\right|\left|h_{-1,1}\left(t,\lambda \right)\right|+\left|r_2\left(t\right)\right|\left|h_{-1,2}\left(t,\lambda \right)\right|\right]e^{-\left|\Im \lambda \right|\left(t+1\right)}dt\\ +2e^{\left|\Im \lambda \right|\left(x+1\right)}\underset{-1}{\overset{x}{\int }}\left[\left|r_1\left(t\right)\right|+\left|r_2\left(t\right)\right|\right]dt\\ \le 2c_1{e}^{\left|\Im \lambda \right|\left(x+1\right)}+e^{\left|\Im \lambda \right|\left(x+1\right)}\underset{-1}{\overset{x}{\int }}\left[\left|r_1\left(t\right)\right|\left|h_{-1,1}\left(t,\lambda \right)\right|+\left|r_2\left(t\right)\right|\left|h_{-1,2}\left(t,\lambda \right)\right|\right]e^{-\left|\Im \lambda \right|\left(t+1\right)}dt.\end{array}$$

The above inequality can be reduced to

$$e^{-\left|\Im \lambda \right|\left(x+1\right)}\left|h_{-1,1}\left(x,\lambda \right)\right|\le 2c_1+\underset{-1}{\overset{x}{\int }}\left[\left|r_1\left(t\right)\right|\left|h_{-1,1}\left(t,\lambda \right)\right|+\left|r_2\left(t\right)\right|\left|h_{-1,2}\left(t,\lambda \right)\right|\right]e^{-\left|\Im \lambda \right|\left(t+1\right)}dt.$$

(59)

Similarly, we can prove that

$$e^{-\left|\Im \lambda \right|\left(x+1\right)}\left|h_{-1,2}\left(x,\lambda \right)\right|\le 2c_1+\underset{-1}{\overset{x}{\int }}\left[\left|r_1\left(t\right)\right|\left|h_{-1,1}\left(t,\lambda \right)\right|+\left|r_2\left(t\right)\right|\left|h_{-1,2}\left(t,\lambda \right)\right|\right]e^{-\left|\Im \lambda \right|\left(t+1\right)}dt.$$

(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 h_0,1(·, λ) and h_0,2(·, λ).

Lemma 2.6 The functions h_0,1(x, λ) and h_0,2(x, λ) are entire in λ for any fixed x ∈ (0, 1] and satisfy the growth condition

$$\left|h_{0,1}\left(x,\lambda \right)\right|,\left|h_{0,2}\left(x,\lambda \right)\right|\le 2c_3{c}_4{e}^{\left|\Im \lambda \right|\left(x+1\right)},\lambda \in ℂ.$$

(61)

Proof. Since $h_{0,1}\left(x,\lambda \right)={\mathcal{S}}_{0,1}{\varphi }_{11}\left(x,\lambda \right)+{\tilde{\mathcal{S}}}_{0,2}{\varphi }_{21}\left(x,\lambda \right),$ then from (53) and (54) we obtain

$$\begin{array}{ll}\hfill h_{0,1}\left(x,\lambda \right)& =\frac{1}{\delta }{\varphi }_{11}\left(0^{-},\lambda \right){\mathcal{S}}_{0,1}\text{cos}\left(\lambda x\right)-\frac{1}{\delta }{\varphi }_{21}\left(0^{-},\lambda \right){\tilde{\mathcal{S}}}_{0,1}\text{sin}\left(\lambda x\right)-{\mathcal{S}}_{0,1}{h}_{-1,2}\left(x,\lambda \right)\\ +\frac{1}{\delta }{\varphi }_{11}\left(0^{-},\lambda \right){\tilde{\mathcal{S}}}_{0,2}\text{sin}\left({\lambda }_X\right)+\frac{1}{\delta }{\varphi }_{21}\left(0^{-},\lambda \right){\tilde{\mathcal{S}}}_{0,2}\text{cos}\left({\lambda }_X\right)+{\tilde{\mathcal{S}}}_{0,2}{h}_{-1,2}\left(x,\lambda \right).\end{array}$$

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

$$\begin{array}{ll}\hfill h_{0,1}\left(x,\lambda \right)& \le \frac{1}{\left|\delta \right|}\left|{\varphi }_{11}\left(0^{-},\lambda \right)\right|\left|{\mathcal{S}}_{0,1}\text{cos}\left(\lambda x\right)\right|+\frac{1}{\left|\delta \right|}\left|{\varphi }_{21}\left(0^{-},\lambda \right)\right|\left|{\mathcal{S}}_{0,1}\text{sin}\left(\lambda x\right)\right|\\ +\left|{\mathcal{S}}_{0,1}{h}_{-1,2}\left(x,\lambda \right)\right|+\frac{1}{\left|\delta \right|}\left|{\varphi }_{11}\left(0^{-},\lambda \right)\right|\left|{\tilde{\mathcal{S}}}_{0,2}\text{sin}\left(\lambda x\right)\right|\\ +\frac{1}{\left|\delta \right|}\left|{\varphi }_{21}\left(0^{-},\lambda \right)\right|\left|{\tilde{\mathcal{S}}}_{0,2}\text{cos}\left(\lambda x\right)\right|-+\left|{\tilde{\mathcal{S}}}_{0,2}{h}_{-1,2}\left(x,\lambda \right)\right|\\ \le 2\left(c_2+\frac{2}{\left|\delta \right|}\left(1+c_2\right)\right)c_3{e}^{\left|\Im \lambda \right|\left(x+1\right)}=2c_3{c}_4{e}^{\left|\Im \lambda \right|\left(x+1\right)}.\end{array}$$

Similarly, we can prove that

$$h_{0,2}\left(x,\lambda \right)\le 2c_3{c}_4{e}^{\left|\Im \lambda \right|\left(x+1\right)}.$$

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

$$\mathcal{U}\left(\lambda \right):={\delta }^2\text{sin}\beta h_{0,1}\left(1,\lambda \right)-{\delta }^2\text{cos}\beta h_{0,2}\left(1,\lambda \right)-\delta \text{sin}\left(\lambda -\beta \right)h_{-1,1}\left(0^{-},\lambda \right)-\delta \text{cos}\left(\lambda +\beta \right)h_{-1,2}\left(0^{-},\lambda \right),$$

(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

$$\left|\mathcal{U}\left(\lambda \right)\right|\le M{e}^{2\left|\Im \lambda \right|},$$

(65)

where

$$M:=4\left|\delta \right|\left(c_2+\left|\delta \right|c_3{c}_4\right).$$

(66)

Proof. From (63), we have

$$\begin{array}{ll}\hfill \left|\mathcal{U}\left(\lambda \right)\right|& \le \left|\delta {|}^2\right|\text{sin}\beta \left|\right|h_{0,1}\left(1,\lambda \right)|+|\delta {|}^2\left|\text{cos}\beta \right|\left|h_{0,2}\left(1,\lambda \right)\right|\\ +\left|\delta \right|\left|\text{sin}\left(\lambda -\beta \right)\right|\left|h_{-1,1}\left(0^{-},\lambda \right)\right|+\left|\delta \right|\left|\text{cos}\left(\lambda +\beta \right)\right|\left|h_{-1,2}\left(0^{-},\lambda \right)\right|.\end{array}$$

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

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

$${\mathcal{G}}_{\theta ,m}\left(\lambda \right):={\left(\frac{\text{sin}\theta \lambda }{\theta \lambda }\right)}^m\mathcal{U}\left(\lambda \right),\lambda \in ℂ.$$

(67)

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

$$\left|{\mathcal{G}}_{\theta ,m}\left(\lambda \right)\right|\le \frac{c_0^{m}M}{{\left(1+\theta \left|\lambda \right|\right)}^m}{e}^{\left|\Im \lambda \right|\left(2+m\theta \right)},\lambda \in ℂ.$$

(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 $\left|\frac{\text{sin}z}{z}\right|\le \frac{c_0}{1+\left|z\right|}{e}^{\left|\Im z\right|},$where c₀ ≃ 1.72, cf. [26], and (65), we obtain

$$\begin{array}{ll}\hfill \left|{\lambda }^{m-1}{\mathcal{G}}_{\theta ,m}\left(\lambda \right)\right|& \le {\left(\frac{c_0}{1+\theta \left|\lambda \right|}\right)}^m{e}^{\left|\Im z\right|m\theta }\cdot M|\lambda {|}^{m-1}{e}^{2\left|\Im \lambda \right|}\\ =\frac{c_0^{m}M|\lambda {|}^{m-1}}{{\left(1+\theta \left|\lambda \right|\right)}^m}{e}^{\left|\Im \lambda \right|\left(2+m\theta \right)},\lambda \in ℂ.\end{array}$$

(70)

Therefore if λ ∈ ℝ we have

$$\left|{\lambda }^{m-1}{\mathcal{G}}_{\theta ,m}\left(\lambda \right)\right|\le \frac{c_0^{m}M|\lambda {|}^{m-1}}{{\left(1+\theta \left|\lambda \right|\right)}^m},\lambda \in ℝ,$$

(71)

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

$$\underset{-\infty }{\overset{\infty }{\int }}|{\lambda }^{m-1}{\mathcal{G}}_{\theta ,m}\left(\lambda \right){|}^{2}d\lambda \le c_0^{2m}{M}^2\underset{-\infty }{\overset{\infty }{\int }}\frac{|\lambda {|}^{2m-2}}{{\left(1+\theta \left|\lambda \right|\right)}^{2m}}d\lambda =2c_0^{2m}{M}^2{\xi }_0^2.$$

(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+mθ. 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

$${\mathcal{G}}_{\theta ,m}\left(\lambda \right)=\sum _{n=-\infty }^{\infty }\left[{\mathcal{G}}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right)S_n^2\left(\lambda \right)+{\mathcal{G}}_{\theta ,m}^{{}^{\prime }}\left(\frac{n\pi }{\sigma }\right)\frac{\text{sin}\left(\sigma \lambda -n\pi \right)}{\sigma }{S}_n\left(\lambda \right)\right].$$

(73)

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

$${\mathcal{G}}_{\theta ,m,N}\left(\lambda \right):=\sum _{n=-N}^N\left[{\mathcal{G}}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right)S_n^2\left(\lambda \right)+{\mathcal{G}}_{\theta ,m}^{{}^{\prime }}\left(\frac{n\pi }{\sigma }\right)\frac{\text{sin}\left(\sigma \lambda -n\pi \right)}{\sigma }{S}_n\left(\lambda \right)\right].$$

(74)

Since all eigenvalues are real, then from now on we restrict ourselves to λ ∈ ℝ. 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

$$|{\mathcal{G}}_{\theta ,m}\left(\lambda \right)-{\mathcal{G}}_{\theta ,m,N}\left(\lambda \right)|\le T_{N,m-1,\sigma }\left(\lambda \right),$$

(75)

where

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

(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 ${\left\{{\tilde{\mathcal{G}}}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right)\right\}}_{n=-N}^N$ and ${\left\{{\tilde{\mathcal{G}}}_{\theta ,m}^{{}^{\prime }}\left(\frac{n\pi }{\sigma }\right)\right\}}_{n=-N}^N$ 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 ${\tilde{\mathcal{G}}}_{\theta ,m,N}\left(\lambda \right),$ which approximates ${{\mathcal{G}}_{\theta }}_{,m,N}\left(\lambda \right)$

$${\tilde{\mathcal{G}}}_{\theta ,m,N}\left(\lambda \right):=\sum _{n=-N}^N\left[{\tilde{\mathcal{G}}}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right)S_n^2\left(\lambda \right)+{{\tilde{\mathcal{G}}}^{\prime }}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right)\frac{\text{sin}\left(\sigma \lambda -n\pi \right)}{\sigma }{S}_n\left(\lambda \right)\right],N>m.$$

(77)

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

$$\left|{\mathcal{G}}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right)-{\tilde{\mathcal{G}}}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right)|<\epsilon ,|{{\mathcal{G}}^{\prime }}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right)-{{\tilde{\mathcal{G}}}^{\prime }}_{\theta ,m}\left(\frac{n\pi }{\sigma }\right)\right|<\epsilon ,$$

(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 $0<\epsilon \le \text{min}\left\{\pi /\sigma ,\sigma /\pi ,1/\sqrt{e}\right\}$ we have

$$|{\mathcal{G}}_{\theta ,m,N}\left(\lambda \right)-{\tilde{\mathcal{G}}}_{\theta ,m,N}\left(\lambda \right)|\le \mathcal{A}\left(\epsilon \right),\lambda \in ℝ,$$

(79)

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

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

(80)

Here

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

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

$$\begin{array}{ll}\hfill \mathcal{K}\left({\lambda }^{*}\right)+{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\tilde{\mathcal{G}}}_{\theta ,m,N}\left({\lambda }^{*}\right)& ={\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\tilde{\mathcal{G}}}_{\theta ,m,N}\left({\lambda }^{*}\right)-{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\mathcal{G}}_{\theta ,m}\left({\lambda }^{*}\right)\\ =\left[{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\tilde{\mathcal{G}}}_{\theta ,m,N}\left({\lambda }^{*}\right)-{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\mathcal{G}}_{\theta ,m,N}\left({\lambda }^{*}\right)\right]\\ +\left[{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\mathcal{G}}_{\theta ,m,N}\left({\lambda }^{*}\right)-{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\mathcal{G}}_{\theta ,m}\left({\lambda }^{*}\right)\right]\end{array}$$

and so

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

Since $\mathcal{K}\left({\lambda }^{*}\right)+{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\tilde{\mathcal{G}}}_{\theta ,m,N}\left({\lambda }^{*}\right)$ 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

$$\begin{array}{ll}\hfill -{\left|\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right|}^{-m}& \left(T_{N,m-1,\sigma }\left({\lambda }^{*}\right)+\mathcal{A}\left(\epsilon \right)\right)\le \mathcal{K}\left({\lambda }^{*}\right)\\ +{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}{\tilde{\mathcal{G}}}_{\theta ,m,N}\left({\lambda }^{*}\right)\le {\left|\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right|}^{-m}\left(T_{N,m-1,\sigma }\left({\lambda }^{*}\right)+\mathcal{A}\left(\epsilon \right)\right).\end{array}$$

(81)

Its solution is an interval containing λ*, and over which the graph $\mathcal{K}\left({\lambda }^{*}\right)+{\left(\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right)}^{-m}\times {\tilde{\mathcal{G}}}_{\theta ,m,N}\left({\lambda }^{*}\right)$ 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

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

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

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

in a neighborhood of λ*. Hence the graph of $\mathcal{K}\left(\lambda \right)+{\left(\frac{\text{sin}\theta \lambda }{\theta \lambda }\right)}^{-m}{\tilde{\mathcal{G}}}_{\theta ,m,N}\left(\lambda \right)$ 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

$$I_{\epsilon ,N}:=\left[a_{-}\left({\lambda }^{*},N,\epsilon \right),a_{+}\left({\lambda }^{*},N,\epsilon \right)\right]$$

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 ∈ ℤ⁺ and sufficiently small ε such that λ*∈I _ε,N for N > N₀. Moreover

$$\left[a_{-}\left({\lambda }^{*},N,\epsilon \right),a_{+}\left({\lambda }^{*},N,\epsilon \right)\right]\to \left\{{\lambda }^{*}\right\}\text{as }N\to \infty \text{and }\epsilon \to 0.$$

(84)

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

$$\mathcal{K}\left(\lambda \right)+{\left(\frac{\text{sin}\theta \lambda }{\theta \lambda }\right)}^{-m}{\tilde{\mathcal{G}}}_{\theta ,m,N_0}\left(\lambda \right)=\pm {\left|\frac{\text{sin}\theta \lambda }{\theta \lambda }\right|}^{-m}\left(T_{N_0,m-1,\sigma }\left(\lambda \right)+\mathcal{A}\left(\epsilon \right)\right)$$

has two distinct solutions which we denote by a _(λ*,N₀,ε) ≤ a₊(λ*,N₀,ε). The decay of T_N,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 ${\tilde{\mathcal{G}}}_{\theta ,m,N}\left(\lambda \right)\to {\mathcal{G}}_{\theta ,m}\left(\lambda \right)$ as N → ∞ and as ε → 0. Hence by taking the limit we obtain

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

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

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

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

$$\left|\Omega \left(\lambda \right)-{\tilde{\Omega }}_N\left(\lambda \right)\right|\le {\left|\frac{\text{sin}\theta \lambda }{\theta \lambda }\right|}^{-m}\left(T_{N,m-1,\sigma }\left(\lambda \right)+\mathcal{A}\left(\epsilon \right)\right),\left|\lambda \right|<\frac{N\pi }{\sigma }$$

(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 ${\tilde{\Omega }}_N\left({\lambda }_N\right)=0.$ From (85) we have $\left|{\tilde{\Omega }}_N\left({\lambda }^{*}\right)\right|\le {\left|\frac{\text{sin}\theta {\lambda }^{*}}{\theta {\lambda }^{*}}\right|}^{-m}\left(T_{N,m-1,\sigma }\left({\lambda }^{*}\right)+\mathcal{A}\left(\epsilon \right)\right).$ Now we estimate the error |λ*-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

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

(86)

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

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

(87)

Using the mean value theorem yields that for some ζ∈J _ε,N :=[min(λ*, λ _N ), max(λ*, λ _N )],

$$\left|\left({\lambda }^{*}-{\lambda }_N\right){\Omega }^{\prime }\left(\zeta \right)\right|\le {\left|\frac{\text{sin}\theta {\lambda }_N}{\theta {\lambda }_N}\right|}^{-m}\left(T_{N,m-1,\sigma }\left({\lambda }_N\right)+\mathcal{A}\left(\epsilon \right)\right),\zeta \in J_{\epsilon ,N}\subset I_{\epsilon ,N}.$$

(88)

Since the eigenvalues are simple, then for sufficiently large $N\underset{\zeta \in I_{\epsilon ,N}}{\text{inf}}\left|{\Omega }^{\prime }\left(\zeta \right)\right|>0$ and we get (86).

4 Numerical examples

This section includes two detailed worked examples illustrating the above technique. By E _S and E _H we mean the absolute errors associated with the results of the classical sinc method and our new method (Hermite interpolations) respectively. All examples are computed in [22] with the classical sinc method. We indicate in these examples the effect of the amplitude error in the method by determining enclosure intervals for different values of ε. We also indicate the effect of the parameters m and θ by several choices. Every example is accompanied with six figures illustrating $\Omega \left(\lambda \right),\tilde{\Omega }\left(\lambda \right)$ and the enclosure curves dominating the zeros. Recall that a_±(λ) are defined by

$$a\pm \left(\lambda \right)={\Omega }_N\left(\lambda \right)\pm {\left|\frac{\text{sin}\theta \lambda }{\theta \lambda }\right|}^{-m}\left(T_{N,m-1,\sigma }\left(\lambda \right)+\mathcal{A}\left(\epsilon \right)\right),\left|\lambda \right|<\frac{N_{\pi }}{\sigma }.$$

(89)

Recall also that the enclosure interval I _ε,N :=[a_-,a₊] is determined by solving

$$a_{\pm }\left(\lambda \right)=0,\left|\lambda \right|<\frac{N\pi }{\sigma }.$$

(90)

Example 1 Consider the system

$$u_2^{\prime }\left(x\right)-r\left(x\right)u_1\left(x\right)=\lambda u_1\left(x\right),u_1^{\prime }\left(x\right)+r\left(x\right)u_2\left(x\right)=-\lambda u_2\left(x\right),x\in \left[-1,0\right)\cup \left(0,1\right],$$

(91)

$$u_{\text{1}}\text{({1) + }{u}_{\text{2}}\text{({1) = 0,}{u}_{\text{1}}\text{(1) + }{u}_{\text{2}}\text{(1) = 0,}$$

(92)

$$u_1\left(0^{-}\right)-2u_1\left(0^{+}\right)=0,u_2\left(0^{-}\right)-2u_2\left(0^{+}\right)=0.$$

(93)

Here

$$r_1\left(x\right)=r_2\left(x\right)=r\left(x\right)=\left\{\begin{array}{cc}\hfill x^2,\hfill & \hfill x\in \left[-1,0\right)\hfill \\ \hfill x,\hfill & \hfill \left(0,1\right],\hfill \end{array}\right.$$

(94)

$\alpha =\beta =\frac{\pi }{4}$ and δ = 2. Direct calculations give

$$\mathcal{K}\left(\lambda \right)=-2\text{sin}\left[2\lambda \right]$$

(95)

and

$$\Omega \left(\lambda \right)=-2\text{sin}\left[\frac{5}{6}+2\lambda \right],$$

(96)

therefore the eigenvalues are${\lambda }_k=\frac{6k\pi -5}{12},k\in \mathbb{Z}.$ The following four tables indicate the application of our technique to this problem and the effect of ε, θ and m (Tables 1, 2, 3 and 4).

Table 1 N = 20, m = 6, and θ = 1/7

Table 2 N = 20, m = 10 and θ = 1/5

Table 3 The approximation λ_k,Nand the exact solution λ _k are all inside the interval [a₋, a₊] for different values of ε, where N = 20, m = 6, θ = 1/7

Table 4 For N = 20, m = 10 and θ = 1/5, λ_k,Nand λ _k are all inside the interval [a₋, a₊] for different values of ε

In the following, the Figures 1 and 2 illustrate the comparison between Ω(λ) and${\tilde{\Omega }}_N\left(\lambda \right)$for different values of m and θ.

Figure 1

Ω( λ ), ${\tilde{\Omega }}_N\left(\lambda \right)$ with N = 20, m = 6 and θ = 1/7.

Figure 2

Ω( λ ), ${\tilde{\Omega }}_N\left(\lambda \right)$ with N = 20, m = 10 and θ = 1/5.

Figures 3 and 4, for N = 20, m = 6 and θ = 1/7, illustrate the enclosure intervals for ε = 10⁻⁸ and ε = 10⁻¹² respectively.

Figure 3

a ₊ , Ω( λ ), a _- with N = 20, m = 6, θ = 1/7 and ε = 10⁻⁸.

Figure 4

a ₊ , Ω( λ ), a ₋ with N = 20, m = 6, θ = 1/7 and ε = 10⁻¹².

Also, Figures 5 and 6 illustrate the enclosure intervals for ε = 10⁻⁸ and ε = 10⁻¹² respectively, but for m = 10, θ =1/5.

Figure 5

a ₊ , Ω( λ ), a ₋ with N = 20, m = 10, θ = 1/5 and ε = 10⁻⁸.

Figure 6

a ₊ , Ω( λ ), a ₋ with N = 20, m = 10, θ =1/5 and ε = 10⁻¹².

Example 2 In this example we consider the system

$$u_2^{\prime }\left(x\right)-r\left(x\right)u_1\left(x\right)=\lambda u_1\left(x\right),u_1^{\prime }\left(x\right)+r\left(x\right)u_2\left(x\right)=-\lambda u_2\left(x\right),x\in \left[-1,0\right)\cup \left(0,1\right],$$

(97)

$$\sqrt{3}{u}_1\left(-1\right)+u_2\left(-1\right)=0,u_1\left(1\right)+\sqrt{3}{u}_2\left(1\right)=0,$$

(98)

$$u_1\left(0^{-}\right)-3u_1\left(0^{+}\right)=0,u_2\left(0^{-}\right)-3u_2\left(0^{+}\right)=0,$$

(99)

where

$$r_1\left(x\right)=r_2\left(x\right)=r\left(x\right)=\left\{\begin{array}{cc}\hfill x,\hfill & \hfill x\in \left[-1,0\right)\hfill \\ \hfill x^2+1,\hfill & \hfill \left(0,1\right],\hfill \end{array}\right.$$

(100)

$\alpha =\frac{\pi }{3},\beta =\frac{\pi }{6}and\delta =3.$ Direct calculations give

$$\mathcal{K}\left(\lambda \right)=3\text{sin}\left[\frac{\pi }{6}-2\lambda \right]$$

(101)

and

$$\Omega \left(\lambda \right)=\frac{3}{2}\left(\text{cos}\left[\frac{5}{6}+2\lambda \right]-\sqrt{3}\text{sin}\left[\frac{5}{6}+2\lambda \right]\right),$$

(102)

therefore the eigenvalues are${\lambda }_k=\frac{\left(6k+1\right)\pi -5}{12},k\in \mathbb{Z}.$ As in the previous example, Figures 7, 8, 9, 10, 11 and 12 illustrate the results of Tables 5, 6, 7 and 8. Also, Figures 5 and 6 illustrate the enclosure intervals for ε = 10⁻⁸ and ε = 10⁻¹² respectively, but for m = 9, θ = 2/11.

Figure 7

Ω( λ ), ${\tilde{\Omega }}_N\left(\lambda \right)$ with N = 20, m = 5 and θ = 2/15.

Figure 8

Ω( λ ), ${\tilde{\Omega }}_N\left(\lambda \right)$ with N = 20, m = 9 and θ = 2/11.

Figure 9

a ₊ , Ω( λ ), a ₋ with N = 20, m = 5, θ = 2/15 and ε = 10⁻⁸.

Figure 10

a ₊ , Ω( λ ), a ₋ with N = 20, m = 5, θ = 2/15 and ε = 10⁻¹².

Figure 11

a ₊ , Ω( λ ), a ₋ with N = 20, m = 9, θ = 2/11 and ε = 10⁻⁸.

Figure 12

a ₊ , Ω( λ ), a ₋ with N = 20, m = 9, θ = 2/11 and ε = 10⁻¹².

Table 5 N = 20, m = 5, θ = 2/15

Table 6 N = 20, m =9, θ =2/11

Table 7 For N = 20, m =5 and θ = 2/15, the approximation λ_k,Nand the exact solution λ _k are all inside the interval [a₋, a₊] for different values of ε

Table 8 With N = 20, m = 9 and θ = 2/11, λ_k,Nand λ _k are all inside the interval [a₋, a₊] for different values of ε