Journal of Function Spaces

Journal of Function Spaces / 2015 / Article

Research Article | Open Access

Volume 2015 |Article ID 758410 | https://doi.org/10.1155/2015/758410

A. Guezane-Lakoud, N. Bendjazia, R. Khaldi, "An Approximation Method for Solving Volterra Integrodifferential Equations with a Weighted Integral Condition", Journal of Function Spaces, vol. 2015, Article ID 758410, 10 pages, 2015. https://doi.org/10.1155/2015/758410

An Approximation Method for Solving Volterra Integrodifferential Equations with a Weighted Integral Condition

Academic Editor: Richard I. Avery
Received19 Feb 2015
Revised23 Apr 2015
Accepted23 Apr 2015
Published13 May 2015

Abstract

We apply the reproducing kernel Hilbert space (RKHS) method for getting analytical and approximate solutions for second-order hyperbolic integrodifferential equations with a weighted integral condition. The analytical solution is represented in the form of series; thus, the n-terms approximate solutions are obtained. The results of the numerical examples are compared with the exact solutions to illustrate the accuracy and the effectivity of this method.

1. Introduction

In functional analysis, a reproducing kernel Hilbert space is a Hilbert space of functions in which pointwise evaluation is a continuous linear functional. The subject was originally developed by Aronszajn in 1950 (see [1]).

Reproducing kernel theory has important applications in numerical analysis, differential equations, integral equations, probability, and statistics [24]. Recently, using the RKHS method, the authors in [521] have discussed singular linear two-point boundary value problems, singular nonlinear two-point periodic boundary value problems, nonlinear system of boundary value problems, initial value problems, singular integral equations, nonlinear partial differential equations, operator equations, and fourth-order integrodifferential equations (IDEs), differential equations with integral condition.

The purpose of this paper is to extend the application of the reproducing kernel Hilbert space method to solve a second-order boundary value problem of IDEs of the following Volterra type:for all , subject to the initial conditionsand the weighted integral conditionwhere , , , , , and are given functions, such that The operator is linear with respect to and continuous according to both variables and . The function is continuous and denote .

In order to put initial conditions and integral boundary conditions of (1) into the reproducing kernel space constructed in the following sections, we have to homogenize these conditions. For this, let ; then the problem can be converted into the following form:for all , subject to the initial conditionsand the weighted integral condition

The theoretical aspects of (1) with conditions (2)–(4) have been studied by Galerkin method in [22]; however, no numerical method was presented. In fact, by this procedure, the authors have found several difficulties to find the discretization of this problem due to the presence of the integral condition. Therefore, to overcome these difficulties, in the present work, we propose the RKHS method which is a simple and effective method for obtaining the numerical solution that converges rapidly to the exact solution with a small error and with a good precision. Moreover, the partial derivatives of are also convergent to the partial of . So, from this work, we show the applicability of the RKHS method to solve this kind of problem without discretization such as Galerkin and Rothe’s method. Furthermore, the RKHS method has an advantage that it is the possibility to pick any point in the domain of integration as well as in the approximate solutions and all its partial derivatives up to order two that will be applicable.

In recent years, a growing interest has been devoted to the study of IDEs which are a combination of differential and Volterra-Fredholm integral equations. IDEs appear in various fields of science such as physics, biology, and engineering. We should mention also that integrodifferential equations are usually difficult to solve analytically.

Recently, various kinds of numerical methods have been used for efficient approximation solution. In [23], authors have presented a practical matrix method for solving nonlinear Volterra-Fredholm integrodifferential equations. In [24], authors have given a numerical solution of nonlinear Volterra-Fredholm integrodifferential equations using the spectral homotopy analysis method. In [2528], authors have obtained the representation of the exact solution for the nonlinear Volterra-Fredholm integral equations by using the reproducing kernel method.

To sum up, in this paper, we define several reproducing kernel spaces in Section 2, whereas, in Section 3, we introduce a linear operator, a complete normal orthogonal system, and some essential results. In Section 4, we provide the main results and the exact and approximate solution of (1)–(4) and we also develop an iterative method for this kind of problems. Finally, in Section 5, the results of some numerical examples are presented and are compared with the analytical solution to confirm the good accuracy of the presented method.

2. Reproducing Kernel Spaces

In this section, we define some useful reproducing kernel spaces.

Definition 1. Let Define, respectively, the inner product and the norm in byWe have the following result.

Theorem 2. The space is a complete reproducing kernel space; that is, for each fixed , there exists simply and such that for any . The reproducing kernel is given bywhere

Proof. Through several integrations by parts of (10), we obtain Since , we have ThenNote that the property of the reproducing kernel ; then is the solution of the following generalized differential equation:with the boundary conditionsAs , it follows that Let us find the coefficients , , and .
Since , we haveSince , it follows thatThrough (17)–(20), the unknown coefficients of (11) can be obtained.

Definition 3. Define the space The inner product and the norm in are defined, respectively, by The space is a complete reproducing kernel space, and its reproducing kernel is given by

Definition 4. DefineThe inner product and the norm in are defined, respectively, by is a reproducing kernel space and its reproducing kernel function is such that, for any ,

Definition 5. Define The inner product and the norm in are defined, respectively, by

Definition 6. Define The inner product and the norm in are defined, respectively, by is a complete reproducing kernel space; thus, its reproducing kernel function is (see [11]) where

3. A Bounded Linear Operator on

Define differential operator : bytherefore, (1) can be converted into an equivalent form as follows:where ,  , and

It is easy to show that is a bounded linear operator from to . Now, we choose a countable subset in and we define

Lemma 7. Assume that is dense in ; then, is a complete system in , and

Applying Gram-Schmidt process, we obtain an orthogonal basis of , such thatwhere are orthogonal coefficients.

4. The Exact and Approximate Solution

Lemma 8. If is dense in , then the exact solution of (35) is

Proof. The exact solution can be expanded in the Fourier series in terms of normal orthogonal basis in :

4.1. Iterative Method

We construct an iterative method to obtain the approximate solutions of (1) in the reproducing kernel space . Let

Then

Define the initial function and the -term approximation to bywhere the coefficients are given as

Theorem 9. If is dense in , then where is given by (44).

Theorem 10. Suppose that and ; then

Proof. From the preceding theorem, we have soFrom and (49), we get

Proposition 11. Suppose that as ; then

Now, the approximate solution can be obtained by taking finitely many terms in the series representation of and

5. Numerical Examples

In order to calculate the approximate solution , we take

Example 1. Consider the following hyperbolic integrodifferential equation: for all , subject to the initial conditions and the weighted integral conditionwhereThe exact solution is given by . In order to homogenize the initial conditions, we put ; then the problem can be converted into the following form: where

The numerical results are presented in Tables 1, 2, 3, and 4 and Figures 1 and 2.


)

(0.00, 0.01)−0.0000333333−0.0000333637−0.0000333333
(0.10, 0.05)−0.000820406−0.00082375−0.000820406
(0.15, 0.10)−0.00321499−0.003237990.000022998−0.00321499
(0.35, 0.20)−0.0105706−0.01067730.000106649−0.01057060.0000198708
(0.49, 0.31)−0.0183567−0.01852770.000170909−0.01838850.0000317253
(0.57, 0.45)−0.0273509−0.02756920.000218296−0.02735090.0000385202
(0.69, 0.52)−000800865−0.008047270.0000386212−0.00800874
(0.73, 0.71)0.00592180.005981970.00006017280.005949410.0000276115
(0.81, 0.73)0.05572780.05606040.0003326160.05580070.0000728933
(0.99, 0.99)0.360580.3622130.001633330.3607560.000176542



(0.00, 0.01)0.000000.00000.00000−0.00666667−0.00667043
(0.10, 0.05)0.0002629270.000263362−0.0328162−0.03281620.000065
(0.15, 0.10)0.001618340.0016219−0.0642998−0.06442480.000124
(0.35, 0.20)0.01676270.01679860.0000359083−0.105706−0.1059010.000194
(0.49, 0.31)0.06076560.06089140.000125813−0.118431−0.1186150.00018
(0.57, 0.45)0.1555740.1558660.000291519−0.12156−0.1216890.00012
(0.69, 0.52)0.2687010.2691160.000415711−0.0308025−0.03078290.000019
(0.73, 0.71)0.5419480.542520.000572150.01668110.01672370.000042
(0.81, 0.73)0.665010.6655380.0005276740.1526790.1527250.000045
(0.99, 0.99)1.657581.6590.001425690.7284440.7279130.00053



(0.00, 0.01)0.00010000.000100171−0.666667−0.6673910.000724
(0.10, 0.05)0.002762930.00277046−0.656325−0.6582530.001928
(0.15, 0.10)0.01161830.01163740.0000190−0.642998−0.642960.0000379
(0.35, 0.20)0.05676270.05684280.0000801−0.528532−0.5283620.000169
(0.49, 0.31)0.1568660.1572010.0003349−0.382034−0.3820730.000038
(0.57, 0.45)0.3580740.358230.0001562−0.270133−0.2703180.0001851
(0.69, 0.52)0.5391010.5393540.0002531−0.0592356−0.00591330.0001026
(0.73, 0.71)1.046051.045970.00007330.02349450.02322410.00027045
(0.81, 0.73)1.197910161.197980.000069670.2091490.2082890.00085993
(0.99, 0.99)2.637682.640760.0030760.7358020.7358020.00426859



(0.00, 0.01)0.000000.00000.00000
(0.10, 0.05)0.01051710.01054020.0000231397
(0.15, 0.10)0.03236680.03243790.0000710837
(0.35, 0.20)0.1676270.1679910.000364067
(0.49, 0.31)0.3920360.3928360.000799951
(0.57, 0.45)0.691440.6925950.00115446
(0.69, 0.52)1.033461.034560.00109998
(0.73, 0.71)1.526611.526940.000326263
(0.81, 0.73)1.821951.821990.000041932
(0.99, 0.99)3.348643.349240.000598899

Example 2. Consider the following hyperbolic integrodifferential equation: where The exact solution is given by .

The numerical results are presented in Tables 5, 6, 7, and 8 and Figures 3 and 4.



(0.00, 0.01)−0.000050000−0.00005000380.0000762474
(0.10, 0.05)−0.001225000−0.001225590.000482156
(0.15, 0.10)−0.004775−0.00477850.000732779
(0.35, 0.20)−0.01510000−0.01510990.000656861
(0.49, 0.31)−0.0249764−0.02499180.00001538560.000616005
(0.57, 0.45)−0.0354578−0.03548350.00002577610.000726954
(0.69, 0.52)−0.00646256−0.006452570.00154632
(0.73, 0.71)0.01658490.01666690.00008198280.00494322
(0.81, 0.73)0.08318570.0833870.0002013470.00242045
(0.99, 0.99)0.4705460.4717070.001160790.00246689



(0.00, 0.01)0.00000.00000.000−0.010−0.0100011
(0.10, 0.05)0.00050.000499228−0.049−0.0490310.00003
(0.15, 0.10)0.0030.00300832−0.0955−0.09559130.00009
(0.35, 0.20)0.0280.02802820.000028−0.151−0.1511220.00012
(0.49, 0.31)0.0941780.09426950.00009−0.161138−0.1612590.00012
(0.57, 0.45)0.230850.2311070.00025−0.15759−0.1577790.00018
(0.69, 0.52)0.3731520.373590.00043−0.024856−0.02477790.00007
(0.73, 0.71)0.7359860.7372180.00120.0467180.04711390.00039
(0.81, 0.73)0.8632980.8646990.00140.2279060.2288050.00089
(0.99, 0.99)1.94061.944270.00360.9505980.9543430.0037



(0.00, 0.01)0.00020.000196032−1.0000−1.000220.0002
(0.10, 0.05)0.0050.005042960.00004−0.98−0.9810280.00102
(0.15, 0.10)0.020.02017710.0001−0.955−0.9564440.0014
(0.35, 0.20)0.080.07993290.00006−0.7550.00070.0001
(0.49, 0.31)0.19220.1919320.00026−0.5198−0.5208590.0010
(0.57, 0.45)0.4050.4049390.00006−0.3502−0.3511630.0009
(0.69, 0.52)0.54080.5412060.0004−0.0478−0.04734310.0004
(0.73, 0.71)1.00821.009360.00110.06580.06708970.0012
(0.81, 0.73)1.06581.065190.00060.31220.3149280.0027
(0.99, 0.99)1.94061.944270.00360.9505980.9543430.0037



(0.00, 0.01)0.0000.0000.000
(0.10, 0.05)0.020.01997860.00002
(0.15, 0.10)0.060.06020840.0002
(0.35, 0.20)0.280.2803960.00039
(0.49, 0.31)0.60760.6084020.0008
(0.57, 0.45)1.0261.027990.0019
(0.69, 0.52)1.43521.437880.002
(0.73, 0.71)2.07322.078780.005
(0.81, 0.73)2.36522.371120.005
(0.99, 0.99)3.92043.931840.011