/ / Article

Research Article | Open Access

Volume 2012 |Article ID 925134 | https://doi.org/10.1155/2012/925134

A. H. Bhrawy, M. M. Tharwat, A. Al-Fhaid, "Numerical Algorithms for Computing Eigenvalues of Discontinuous Dirac System Using Sinc-Gaussian Method", Abstract and Applied Analysis, vol. 2012, Article ID 925134, 13 pages, 2012. https://doi.org/10.1155/2012/925134

# Numerical Algorithms for Computing Eigenvalues of Discontinuous Dirac System Using Sinc-Gaussian Method

Accepted26 May 2012
Published30 Jul 2012

#### Abstract

The eigenvalues of a discontinuous regular Dirac systems with transmission conditions at the point of discontinuity are computed using the sinc-Gaussian method. The error analysis of this method for solving discontinuous regular Dirac system is discussed. It shows that the error decays exponentially in terms of the number of involved samples. Therefore, the accuracy of the new method is higher than the classical sinc-method. Numerical results indicating the high accuracy and effectiveness of these algorithms are presented. Comparisons with the classical sinc-method are given.

#### 1. Introduction

The mathematical modeling of many practical problems in mechanics and other areas of mathematical physics requires solutions of boundary value problems (see, e.g., ). Boundary value problems with discontinuity conditions arise in different branches of mathematics, mechanics, radio, electronics, geophysics, and other fields of natural science and technology. For example, discontinuous conditions inside an interval are connected with discontinuous or nonsmooth properties of media (see  and references there in).

Sampling theory is one of the most important mathematical tools used in communication engineering since it enables engineers to reconstruct signals from some of their sampled data. A fundamental result in information theory is the Whittaker-Kotel'nikov-Shannon (WKS) sampling theorem . It states that any , where is the space of all entire functions of exponential type which lie in when restricted to , can be reconstructed from its sampled values by the formula where Series (1.1) converges absolutely and uniformly on compact subsets of , and uniform on , cf. . Expansion (1.1) is used in several approximation problems which are known as sinc methods, see, for example, . In particular the sinc-method is used to approximate eigenvalues of boundary value problems, see, for examples, . The sinc-method has a slow rate of decay at infinity, which is as slow as . There are several attempts to improve the rate of decay. One of the interesting ways is to multiply the sinc-function in (1.1) by a kernel function, see, for example, . Let and . Assume that such that , then for we have the expansion , The speed of convergence of the series in (1.3) is determined by the decay of . But the decay of an entire function of exponential type cannot be as fast as as , for some positive , . In , Qian has introduced the following regularized sampling formula. For , and , where , which is called the Gaussian function, , and denotes the integer part of , see also [30, 31]. Qian also derived the following error bound. If , and , then In , Schmeisser and Stenger extended the operator (1.4) to the complex domain . For , and , they defined the operator , where and . Note that the summation limits in (1.6) depend on the real part of . Schmeisser and Stenger, , proved that if is an entire function of exponential type , then for , , , , we have where

The amplitude error arises when the exact values of (1.6) are replaced by the approximations . We assume that are close to , that is, there is , sufficiently small such that Let , and be fixed numbers. The authors in  proved that if (1.9) is held, then for , we have where

We are concerned with the computation of eigenvalues the Dirac system with boundary conditions and transmission conditions where ; the real valued function and are continuous in and , and have finite limits , ; , and .

In this paper we will use the sinc-Gaussian sampling formula (1.6) to compute eigenvalues of the Dirac system (1.12)–(1.16). As expected, the new method reduced the error bounds remarkably, see examples at the end of this paper. Special attention is given to the comparison of the numerical results obtained by the new method with those found by classical sinc-method. We would like to mention that works in direction of computing eigenvalues with the sinc-Gaussian, are few, see for example, . Also papers in computing of eigenvalues with discontinuous are few, see [10, 32]. However, the computing of eigenvalues by sinc-Gaussian technique which has discontinuity conditions, do not exist as for as we know.

The paper is organized as follows: Section 2 contains some preliminary results and the approximated values of the eigenvalues of the Dirac system with discontinuous. The method with error estimates are contained in Section 3. The last section involves some illustrative examples for showing the high accuracy of the proposed technique.

#### 2. The Approximated Eigenvalues of Dirac System

In this section we derive approximate values of the eigenvalues of problem (1.12)–(1.16). Recall that problem (1.12)–(1.16) has denumerable set of real and simple eigenvalues, compare with [33, 34]. Let be the solution of (1.12) satisfying the following initial conditions: Since satisfies (1.13), then the eigenvalues of the problem (1.12)–(1.16) are the zeros of the function Notice that both and are entire functions of and satisfies the system of integral equations, see [32, 35], where , , and , , are the Volterra integral operators defined by

For convenience, we define the constants Define and , , to be

Lemma 2.1. The functions and are entire in for any fixed and satisfy the growth condition

Proof. Since , then from (2.4) and (2.5) we obtain Using the inequalities and for , leads for to The above inequality can be reduced to Similarly, we can prove that Then from (2.12) and (2.15) and Lemma 3.1 of [34, pp. 204], we obtain (2.12).

In a similar manner, we will prove the following lemma for and .

Lemma 2.2. The functions and are entire in for any fixed and satisfy the growth condition

Proof. Since , then from (2.6) and (2.7) we obtain Then from (2.4) and (2.5) and Lemma 2.1, we get Similarly, we can prove that

#### 3. The Method and Error Analysis

In this section we derive the method of computing the eigenvalues of problem (1.12)–(1.16) numerically. We aim to approximate and hence its zeros, that is, the eigenvalues. The idea is to split into two parts, one is known and the other is unknown, but is an entire function of exponential type. Then we approximate the unknown part using (1.6) to get the approximate and then compute the approximate zeros. Now, let us split into where is the unknown part involving integral operators and is the known part Then, from Lemmas 2.1 and 2.2, we have the following result.

Lemma 3.1. The function is entire in and the following estimate holds: where

Proof. From (3.2), we have Using the inequalities and for , and Lemmas 2.1 and 2.2 imply (3.4).

Then is an entire function of exponential type . In the following we let since all eigenvalues are real. Now we approximate the function using the operator (1.6) where and and then we obtain where The samples , cannot be computed explicitly in the general case. We approximate these samples numerically by solving the initial-value problems defined by (1.12) and (2.2) to obtain the approximate values , , that is, . Accordingly we have the explicit expansion Therefore we get, compare with (1.10), Now let . From (3.7) and (3.10) we obtain Let be an eigenvalue and be its desired approximation, that is, and . From (3.11) we have . Now we define an enclosure interval for . Define the curves The curves , trap the curve of for suitably large . Hence the closure interval is determined by solving , gives an interval . Next we estimate the error for the eigenvalue .

Theorem 3.2. Let be an eigenvalue of (1.12)–(1.16) and be its approximation. Then, for , one has the following estimate: where the interval is defined above.

Proof. Replacing by in (3.11) we obtain where we have used . Using the mean value theorem yields that for some , Since is simple and is sufficiently large, then and we get (3.13).

#### 4. Numerical Examples and Comparisons

This section includes two examples illustrating the sinc-Gaussian method. In the following examples, we consider being the kth root of . Also, it is observed that the approximation and the exact solution are all inside the interval . We indicate in these two examples the effect of the amplitude error in the proposed method by determining enclosure intervals for different values of . All examples are computed in  by using the classical sinc method. We see that the sinc-Gaussian method gives remarkably better results.

Example 4.1. Consider the system subject to Here and .
Direct calculations give therefore the eigenvalues are, .
Let and denote the absolute errors associated with the results of classical sinc method and sinc-Gaussian method, respectively. In Table 1, we give comparison between the absolute error of sinc-Gaussian and the classical sinc-method.

 Sinc Exact Sinc-Gaussian 1.15412966012822 1.15412966012823 2.72492598692312 2.72492598692315

In Table 2, we observe that the approximation and the exact solution are all inside the interval for different values of .

 Exact [ , ], [ , ], [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]
, .

Example 4.2. In this example we consider the system subject to where , and .
Direct calculations give therefore the eigenvalues are , .
Table 3 gives comparison between the absolute error of sinc-Gaussian and the classical sinc-method. Moreover, in Table 4 the approximation and the exact solution are all inside the interval for different values of .

 Sinc Exact Sinc-Gaussian −3.29646018123751 −3.29645993245731 −3.29645993245735 −1.7256638185270 −1.7256636056624 −1.7256636056621 −0.154867300702813 −0.154867278867517 −0.154867278867127 1.41592898901685 1.41592904792737 1.41592904792752 2.98672509121710 2.9867253747222 2.98672537472225
 Exact [ , ], [ , ], −3.29645993245731 −3.29645993245735 −1.7256636056624 −1.7256636056621 −0.154867278867517 −0.154867278867127 1.41592904792737 <