Open Access

Galerkin method applied to telegraph integro-differential equation with a weighted integral condition

Boundary Value Problems20132013:102

DOI: 10.1186/1687-2770-2013-102

Received: 23 November 2012

Accepted: 10 April 2013

Published: 24 April 2013

Abstract

In this work, we study a telegraph integro-differential equation with a weighted integral condition. By means of the Galerkin method, we establish the existence and uniqueness of a generalized solution.

MSC:35L05, 35L20, 35L99.

Keywords

integro-differential equation integral conditions approximate solution Galerkin method

1 Introduction

In this work, we consider the following hyperbolic integro-differential equation with integral conditions:
l u = 2 u t 2 a 2 ( x , t ) 2 u x 2 + c ( x , t ) u = f ( x , t ) + 0 t α ( t s ) K ( s , u ( x , s ) ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ1_HTML.gif
(1.1)
for all ( x , t ) Q = ( 0 , 1 ) × ( 0 , T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq1_HTML.gif, subject to the initial conditions
u ( x , 0 ) = φ ( x ) , u t ( x , 0 ) = ψ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ2_HTML.gif
(1.2)
and the weighted integral conditions
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ3_HTML.gif
(1.3)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ4_HTML.gif
(1.4)

where f, φ, ψ, h, a, c, α and K are given functions.

Various problems arising in heat conduction [15], chemical engineering [6], thermoelasticity [7], and plasma physics [8] can be modeled by the nonlocal problems. Boundary value problems with integral conditions constitute a very interesting and important class of problems. These nonlocal conditions arise mostly when the data on the boundary cannot be measured directly. Recall that the presence of an integral term in boundary conditions can complicate the application of classical methods of functional analysis in the theoretical study of nonlocal problems, therefore, several methods have been proposed for overcoming the difficulties arising from nonlocal conditions; see Beilin [1], Cannon et al. [2, 8], and Dehghan et al. [3, 4, 9].

Numerical solutions are introduced to obtain approximations for the solution of partial differential equations when the analytical solutions are difficult or impossible to obtain due to complicated geometry or boundary conditions. In the area of numerical analysis, the Galerkin method is a class of methods for converting a continuous operator problem to a discrete problem. In principle, it is the equivalent of applying the method of a variation of parameters to a function space, by converting the equation to a weak formulation, hence in this approach we choose a system of linearly independent functions such that they satisfy the given homogeneous boundary condition, and they are dense in a function space containing the exact solution of the above boundary value problem.

The advantage of this approach is not only to establish the existence and uniqueness of the solution, but it is also a very effective method in the study of the approximate solution and its convergence.

In this paper, we study the hyperbolic integro-differential equation (1.1) with a Volterra operator of the form 0 t α ( t s ) K ( s , u ( x , s ) ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq2_HTML.gif in the second member, which appears in the modelling of the quasi-static flexure of a thermo-elastic rod and has been studied in [9, 10] under different boundary conditions, by means of the Rothe method. Let us mention that different methods are used to solve similar integro-differential equations, for example, in [11, 12] the authors have established the existence and uniqueness of the solution using Rothe’s method of an integro-differential equation. In [10, 13], the authors have used Rothe’s method and the techniques of [7] to prove the existence, uniqueness and continuous dependence of a strong solution to a quasi-linear integro-differential equation. In [6], the local existence and uniqueness of a classical solution of an abstract second-order integro-differential equation in a Banach space have been investigated by using the theory of an analytic semi-groups and contraction mapping theorem. In [14, 15] the authors investigated a telegraph equation with non-local integral conditions by means of the Galerkin method.

This paper is organized as follows: In the next section, we define the generalized solution and the functional spaces. In Section 3, we prove that the generalized solution if it exists is unique. The existence of the generalized solution by using the Galerkin method is established in the fourth section, and for this, we construct an approximation solution of the problem (1.1)-(1.4). We prove that we can extract a subsequence, which converges to the desired generalized solution. An application is included to illustrate that corresponding assumptions are satisfied.

2 Notation and definition

Let L 2 ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq3_HTML.gif be the usual space of Lebesgue square integrable real functions on ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq4_HTML.gif whose inner product and norm will be denoted respectively by ( , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq5_HTML.gif and https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq6_HTML.gif. Denote by H ( Q ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq7_HTML.gif the Sobolev space consisting of all functions u L 2 ( Q ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq8_HTML.gif having weak derivatives in L 2 ( Q ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq9_HTML.gif, with the norm
u H ( Q ) 2 = 0 T 0 1 [ ( x 1 u ( ξ , t ) d ξ ) 2 + ( u ( x , t ) ) 2 + ( x 1 u t ( ξ , t ) d ξ ) 2 ] d x d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equa_HTML.gif
Let us define the generalized solution of the problem (1.1)-(1.4). Suppose that u is a solution of this problem, multiply both sides of equation (1.1) by x 1 ( h ( ξ ) h ( x ) ) v t ( ξ , t ) d ξ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq10_HTML.gif, where v H T ( Q ) = { v ( x , t ) H ( Q ) , v t ( x , T ) = 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq11_HTML.gif, integrate by parts the resultant equation over the domain Q, use the conditions (1.2), (1.3), (1.4) and the fact that v t ( x , T ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq12_HTML.gif, we obtain
Q l u ( x 1 ( h ( ξ ) h ( x ) ) v t ( ξ , t ) d ξ ) d x d t = I 1 I 2 + I 3 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equb_HTML.gif
where
I 1 = Q u t t ( x , t ) ( x 1 ( h ( ξ ) h ( x ) ) v t ( ξ , t ) d ξ ) d x d t = Q h ( x ) ( x 1 u t d ξ ) ( x 1 v t t ( ξ , t ) d ξ ) d x d t , I 2 = Q a 2 ( x , t ) u x x ( x , t ) ( x 1 ( h ( ξ ) h ( x ) ) v t ( ξ , t ) d ξ ) d x d t = 2 Q ( a a x ) x u ( x 1 ( h ( ξ ) h ( x ) ) v t ( ξ , t ) d ξ ) d x d t Q ( 4 a a x h ( x ) + a 2 h ( 2 ) ( x ) ) u ( x 1 v t ( ξ , t ) d ξ ) d x d t + Q a 2 h ( x ) u v t d x d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equc_HTML.gif
and
I 3 = Q c ( x , t ) u ( x , t ) ( x 1 ( h ( ξ ) h ( x ) ) v t ( ξ , t ) d ξ ) d x d t = Q c h ( x ) ( x 1 u ( ξ , t ) d ξ ) ( x 1 v t ( ξ , t ) d ξ ) d x d t + Q c x ( x 1 u ( ξ , t ) d ξ ) ( x 1 ( h ( ξ ) h ( x ) ) v t ( ξ , t ) d ξ ) d x d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equd_HTML.gif
Calculating I 1 I 2 + I 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq13_HTML.gif, we deduce
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ5_HTML.gif
(2.1)

Definition 1 By a generalized solution of problem (1.1)-(1.4), we mean a function u H ( Q ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq14_HTML.gif satisfying for all v H T ( Q ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq15_HTML.gif the identity (2.1).

3 Uniqueness of generalized solution

For solving the problem, we make the following hypotheses:

(H1) The functions a and c are nonnegative and satisfy on Q
0 < a 0 a ( x , t ) A 0 , | a t , a x , a x t , a x x , a x x x | A 1 , 0 < c 0 c ( x , t ) C 0 , | c x , c t | C 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Eque_HTML.gif

The function α is continuous and denote l = max 0 t T | α ( t ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq16_HTML.gif.

(H2) The function h C 2 ( [ 0 , 1 ] , R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq17_HTML.gif, h https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq18_HTML.gif is nonnegative and satisfy for all x ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq19_HTML.gif
( | h ( x ) | , h ( x ) , | h ( 2 ) ( x ) | ) k 1 , 0 < k 0 h ( x ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equf_HTML.gif
(H3) The operator K ( t , u ( x , t ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq20_HTML.gif is linear with respect to u and continuous according to the both variables t and u and satisfies for all u H ( Q ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq21_HTML.gif and ( x , t ) Q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq22_HTML.gif
| K ( t , u ( x , t ) ) | | u ( x , t ) | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equg_HTML.gif

Now we shall show that the generalized solution of problem (1.1)-(1.4) if it exists is unique.

Theorem 2 Assume that f L 2 ( Q ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq23_HTML.gif, φ , ψ L 2 ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq24_HTML.gif and hypotheses (H1)-(H3) hold, then the generalized solution of problem (1.1)-(1.4) if it exists is unique.

Proof Suppose that there exists two different generalized solutions u 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq25_HTML.gif and u 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq26_HTML.gif of the problem (1.1)-(1.4), then u = u 1 u 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq27_HTML.gif is a generalized solution of the problem (1.1)-(1.4) with φ = ψ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq28_HTML.gif and second member F = 0 t α ( t s ) ( K ( s , u 1 ( x , s ) ) K ( s , u 2 ( x , s ) ) ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq29_HTML.gif. We shall prove that u = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq30_HTML.gif in Q. Let v H T ( Q ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq31_HTML.gif and denote for 0 τ T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq32_HTML.gif.

Q τ = { ( x , t ) Q , 0 < x < 1 , 0 < t τ } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq33_HTML.gif. Consider the function v such that
v ( x , t ) = { u ( x , t ) , 0 t τ , 0 , τ t T . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equh_HTML.gif
Substituting v into identity (2.1), it follows
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ6_HTML.gif
(3.1)
Integrating by parts it yields
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ7_HTML.gif
(3.2)
Applying Cauchy inequality, ϵ-inequality and the hypotheses on the operator K to the last term in the right-hand side of (3.2), we get
2 | Q τ ( 0 t α ( t s ) ( K ( s , u 1 ( x , s ) ) K ( s , u 2 ( x , s ) ) ) d s ) × ( x 1 ( h ( ξ ) h ( x ) ) u t ( ξ , t ) d ξ ) d x d t | 2 ( Q τ | 0 t α ( t s ) ( K ( s , u 1 ( x , s ) ) K ( s , u 2 ( x , s ) ) ) d s | 2 d x d t ) 1 2 × ( Q τ | x 1 ( h ( ξ ) h ( x ) ) u t ( ξ , t ) d ξ | 2 d x d t ) 1 2 ( max | α ( t ) | ) 2 Q τ | 0 t ( K ( s , u 1 ( x , s ) ) K ( s , u 2 ( x , s ) ) ) d s | 2 d x d t + 4 k 1 2 Q τ ( x 1 u t ( ξ , t ) d ξ ) 2 d x d t l 2 τ Q τ | K ( s , u 1 ( x , s ) ) K ( s , u 2 ( x , s ) ) | 2 d s d x + 4 k 1 2 Q τ ( x 1 u t ( ξ , t ) d ξ ) 2 d x d t l 2 τ K ( t , u 1 ) K ( t , u 2 ) L 2 ( Q τ ) 2 + 4 k 1 2 Q τ ( x 1 u t ( ξ , t ) d ξ ) 2 d x d t l 2 τ u 1 u 2 L 2 ( Q τ ) 2 + 4 k 1 2 Q τ ( x 1 u t ( ξ , t ) d ξ ) 2 d x d t l 2 τ u L 2 ( Q τ ) 2 + 4 k 1 2 Q τ ( x 1 u t ( ξ , t ) d ξ ) 2 d x d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equi_HTML.gif
Applying similar inequalities with ε = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq34_HTML.gif, for the second, the third and the fourth terms in the right-hand side of (3.1) then using conditions (H1)-(H3), we obtain
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ8_HTML.gif
(3.3)
denote
M = k 1 max ( 6 A 0 A 1 + A 0 2 + l 2 T k 1 , 3 C 1 + 4 k 1 ) , m = k 0 min ( 1 , a 0 2 , c 0 ) , L = M m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equj_HTML.gif
then (3.3) becomes
0 1 [ ( x 1 u t ( ξ , τ ) d ξ ) 2 d x + | u ( x , τ ) | 2 + ( x 1 u ( ξ , τ ) d ξ ) 2 ] d x L Q τ [ ( x 1 u t ( ξ , t ) d ξ ) 2 + | u ( x , t ) | 2 + ( x 1 u ( x , t ) d ξ ) 2 ] d x d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ9_HTML.gif
(3.4)
Gronwall inequality implies
0 1 [ ( x 1 u t ( ξ , τ ) d ξ ) 2 d x + | u ( x , τ ) | 2 + ( x 1 u ( ξ , τ ) d ξ ) 2 ] d x 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equk_HTML.gif

hence u ( x , τ ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq35_HTML.gif, for all x ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq36_HTML.gif and τ ( 0 , T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq37_HTML.gif, then u = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq30_HTML.gif in Q. Thus, the uniqueness is proved. □

4 Existence of generalized solution

In order to prove the existence of the generalized solution we apply Galerkin method.

Theorem 3 Assume that the assumptions of Theorem  2 hold, then the problem (1.1)-(1.4) has a unique solution u H ( Q ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq14_HTML.gif.

Proof Let { w k ( x ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq38_HTML.gif be a fundamental system in H ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq39_HTML.gif, such that
( w k , w i ) = δ k , i = { 1 , k = i , 0 , k i . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equl_HTML.gif
We have to find for each n N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq40_HTML.gif, the approximate solution of the problem (1.1)-(1.4) which has the following form:
u ( n ) = k = 1 n β k ( t ) w k ( x ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ10_HTML.gif
(4.1)
Denote
φ ( n ) ( x ) = k = 1 n φ k w k ( x ) , ψ ( n ) ( x ) = k = 1 n ψ k w k ( x ) , β k ( 0 ) = φ k , β k ( 0 ) = ψ k , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ11_HTML.gif
(4.2)
the approximate of the functions φ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq41_HTML.gif and ψ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq42_HTML.gif. Substituting the approximate solution in equation (1.1), multiplying both sides by x 1 ( h ( ξ ) h ( x ) ) w i ( ξ ) d ξ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq43_HTML.gif, then integrating according to x on ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq44_HTML.gif, we get
0 1 ( u t t ( n ) ( x , t ) a 2 ( x , t ) u x x ( n ) ( x , t ) + c ( x , t ) u ( n ) ( x , t ) ) ( x 1 ( h ( ξ ) h ( x ) ) w i ( ξ ) d ξ ) d x = 0 1 f ( x , t ) ( x 1 ( h ( ξ ) h ( x ) ) w i ( ξ ) d ξ ) d x + 0 1 0 t α ( t s ) K ( s , u ( n ) ( x , s ) ) d s ( x 1 ( h ( ξ ) h ( x ) ) w i ( ξ ) d ξ ) d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ12_HTML.gif
(4.3)
Substituting (4.1) in (4.3), we get
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ13_HTML.gif
(4.4)
Integrating by parts in L 2 ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq45_HTML.gif the left-hand side of (4.4) yields
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ14_HTML.gif
(4.5)
Denote
θ k , i = ( h ( x ) x 1 w k ( ξ ) d ξ , x 1 w i ( ξ ) d ξ ) L 2 ( 0 , 1 ) , σ k , i = ( 2 ( a a x ) x w k ( x ) , x 1 ( h ( ξ ) h ( x ) ) w i ( ξ ) d ξ ) L 2 ( 0 , 1 ) ( ( 4 a a x h + a 2 h ( 2 ) ) w k ( x ) , x 1 w i ( ξ ) d ξ ) L 2 ( 0 , 1 ) a 2 h ( w k ( x ) , w i ( x ) ) L 2 ( 0 , 1 ) ( c h x 1 w k ( ξ ) d ξ , x 1 w i ( ξ ) d ξ ) L 2 ( 0 , 1 ) + ( c x x 1 w k ( ξ ) d ξ , x 1 ( h ( ξ ) h ( x ) ) w i ( ξ ) d ξ ) L 2 ( 0 , 1 ) , F i ( t ) = 0 1 f ( x , t ) ( x 1 ( h ( ξ ) h ( x ) ) w i ( ξ ) d ξ ) d x + k = 1 n 0 1 ( 0 t α ( t s ) K ( s , β k ( s ) w k ( x ) ) d s ) ( x 1 ( h ( ξ ) h ( x ) ) w i ( ξ ) d ξ ) d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equm_HTML.gif
then (4.5) becomes
k = 1 n β k ( t ) θ k , i + β k ( t ) σ k , i = F i ( t ) , β k ( 0 ) = φ k , β k ( 0 ) = ψ k . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equn_HTML.gif

Consequently, we obtain a Cauchy system of second-order integro-differential equations with smooth coefficients, so it has one and only one solution that for every n there exists a unique sequence u ( n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq46_HTML.gif that satisfies (4.3). □

Lemma 4 The sequence ( u ( n ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq47_HTML.gif is bounded.

Proof Multiplying (4.3) by β i ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq48_HTML.gif then summing with respect to i from 1 to n it yields
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ15_HTML.gif
(4.6)
Integrating (4.6) over t from 0 to τ we obtain
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ16_HTML.gif
(4.7)
Thanks to Cauchy inequality, ϵ-inequality, the hypotheses on the operator K to the last term in the right-hand side of (4.7), we get
2 | Q τ 0 t a ( t s ) K ( s , u ( n ) ( x , s ) ) d s ( x 1 ( h ( ξ ) h ( x ) ) u t ( n ) ( ξ , t ) d ξ ) d x d t | l 2 τ K ( t , u ( n ) ) L 2 ( Q τ ) 2 + 4 k 1 2 Q τ ( x 1 u t ( ξ , t ) d ξ ) 2 d x d t l 2 τ u ( n ) L 2 ( Q τ ) 2 + 4 k 1 2 Q τ ( x 1 u t ( n ) ( ξ , t ) d ξ ) 2 d x d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equo_HTML.gif
Using similar inequalities for the second, the third and the fourth terms in the right-hand side of (4.7), then regrouping the same terms yields
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ17_HTML.gif
(4.8)
Let L ˜ = M ˜ / m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq49_HTML.gif, where
M ˜ = k 1 max ( 3 C 1 + 4 A 0 A 1 + 4 A 1 2 + 2 + 4 k 1 , 10 A 0 A 1 + A 0 2 + 4 A 1 2 + l 2 T k 1 , C 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equp_HTML.gif
then (4.8) becomes
0 1 [ ( x 1 u ( n ) ( ξ , τ ) d ξ ) 2 + u ( n ) ( x , τ ) 2 + ( x 1 u t ( n ) ( ξ , τ ) d ξ ) 2 ] d x L ˜ [ 0 1 [ ( x 1 φ ( n ) ( ξ ) d ξ ) 2 + ( φ ( n ) ( x ) ) 2 + ( x 1 ψ ( n ) ( ξ ) d ξ ) 2 ] d x + f 2 + Q τ [ ( x 1 u ( n ) ( ξ , t ) d ξ ) 2 + | u ( n ) ( x , t ) | 2 + ( x 1 u t ( n ) ( ξ , t ) d ξ ) 2 ] d x d t ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ18_HTML.gif
(4.9)
Now, we apply Gronwall lemma to get
0 1 [ ( x 1 u t ( ξ , τ ) d ξ ) 2 + | u ( x , τ ) | 2 + ( x 1 u ( ξ , τ ) d ξ ) 2 ] d x e τ L ˜ ( 0 1 [ ( x 1 φ ( n ) ( ξ ) d ξ ) 2 + ( φ ( n ) ( x ) ) 2 + ( x 1 ψ ( n ) ( ξ ) d ξ ) 2 ] d x + f 2 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ19_HTML.gif
(4.10)
Integrating (4.10) according to τ on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq50_HTML.gif yields
u ( n ) H ( Q ) 2 T e T L ˜ ( φ ( n ) H ( 0 , 1 ) 2 + ψ ( n ) 2 + f L 2 ( Q ) 2 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ20_HTML.gif
(4.11)

Thus inequality (4.11) implies the boundedness of the sequence u ( n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq46_HTML.gif. □

Remark 5 We have proved that the sequence { u ( n ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq51_HTML.gif is bounded, so we can extract a subsequence, which we denote by { u ( n k ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq52_HTML.gif that is weakly convergent. Now we prove that its limit is the desired solution of the problem (1.1)-(1.4).

Lemma 6 The limit of the subsequence { u ( n k ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq53_HTML.gif is the solution of the problem (1.1)-(1.4).

Proof We shall prove that the limit of the subsequence { u ( n k ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq54_HTML.gif satisfies the identity (2.1). Let θ k ( t ) C 2 ( 0 , T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq55_HTML.gif, such that θ k ( T ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq56_HTML.gif, let us prove that identity (2.1) holds for any functions v ( x , t ) = k = 1 n θ k ( t ) w k ( x ) H T ( Q ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq57_HTML.gif. Since the set S n = { v ( x , t ) = k = 1 n θ k ( t ) w k ( x ) , θ k ( t ) C 2 ( 0 , T ) , θ k ( T ) = 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq58_HTML.gif is such that n = 1 S n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq59_HTML.gif is dense in H T ( Q ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq60_HTML.gif, it suffices to prove (2.1) for v S n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq61_HTML.gif. Multiplying (4.3) by θ k ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq62_HTML.gif, summing according to k from 1 to n, then integrating over t from 0 to T, we obtain
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ21_HTML.gif
(4.12)
Denote by u the weak limit of the subsequence { u ( n k ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq52_HTML.gif when k tends to infinity. Hence,
| Q 0 t α ( t s ) K ( s , u ( n k ) ( x , s ) ) d s ( x 1 ( h ( ξ ) h ( x ) ) v t ( ξ , t ) d ξ ) d x d t Q 0 t α ( t s ) K ( s , u ( x , s ) ) d s ( x 1 ( h ( ξ ) h ( x ) ) v t ( ξ , t ) d ξ ) d x d t | = | Q ( 0 t α ( t s ) ( K ( s , u ( n k ) ( x , s ) ) K ( s , u ( x , s ) ) ) d s ) × ( x 1 ( h ( ξ ) h ( x ) ) v t ( ξ , t ) d ξ ) d x d t | 2 k l T ( Q ( x 1 | v t | d ξ ) 2 d x d t ) 1 / 2 × ( Q | K ( s , u ( n k ) ( x , s ) ) K ( s , u ( x , s ) ) | 2 d s d x ) 1 / 2 = 2 k l T ( Q ( x 1 | v t | d ξ ) 2 d x d t ) 1 / 2 K ( t , u ( n k ) ) K ( t , u ) L 2 ( Q ) 2 k l T ( Q ( x 1 | v t | d ξ ) 2 d x d t ) 1 / 2 u ( n k ) u L 2 ( Q ) 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equq_HTML.gif

Finally, by passing to the limit in (4.12), we get that the limit u satisfies (2.1). □

Example 7 Consider the following boundary value problem for hyperbolic integro-differential equation for 0 < x < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq63_HTML.gif, 0 < t T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq64_HTML.gif:
2 u t 2 ( x , t ) 1 8 2 u x 2 ( x , t ) + 1 2 π 2 u ( x , t ) = 2 π 2 cos ( 2 π x ) e π t ( t + 1 π [ e π t 1 ] ) sin ( 2 π x ) 2 π 2 + 0 t ( t s ) ( 0 x u ( r , s ) d r ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ22_HTML.gif
(4.13)
subject to the initial conditions
u ( x , 0 ) = cos ( 2 π x ) , u t ( x , 0 ) = π cos ( 2 π x ) , 0 < x < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ23_HTML.gif
(4.14)
and the weighted integral condition
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ24_HTML.gif
(4.15)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_Equ25_HTML.gif
(4.16)

where h ( x ) = 1 cos ( π x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq65_HTML.gif. It is easy to prove that assumptions (H1)-(H3) are satisfied, then from Theorems 2 and 3, and we deduce that the problem (4.13)-(4.16) has a unique generalized solution in the sense of Definition 1. Moreover, the function u ( x , t ) = e π t cos ( 2 π x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-102/MediaObjects/13661_2012_Article_356_IEq66_HTML.gif is the solution of this problem.

Declarations

Acknowledgements

The authors would like to thank the referees for their valuable suggestions.

Authors’ Affiliations

(1)
Faculty of Sciences, Laboratory of Advanced Materials, Badji Mokhtar-Annaba University
(2)
Faculty of Sciences, Laboratory LASEA, Badji Mokhtar-Annaba University

References

  1. Beilin SA: Existence of solutions for one dimensional wave equations with nonlocal conditions. Electron. J. Differ. Equ. 2001, 76: 1-8.MathSciNet
  2. Cannon JR: The solution of the heat equation subject to the specification of energy. Q. Appl. Math. 1963, 21: 155-160.MathSciNet
  3. Dehghan M: On the solution of an initial-boundary value problem that combines Neumann and integral condition for the wave equation. Numer. Methods Partial Differ. Equ. 2005, 21: 24-40. 10.1002/num.20019View Article
  4. Dehghan M: A finite difference method for a non-local boundary value problem for two dimensional heat equation. Appl. Math. Comput. 2000, 112: 133-142. 10.1016/S0096-3003(99)00055-7MathSciNetView Article
  5. Ionkin NI: Solutions of boundary value problem in heat conduction theory with nonlocal boundary conditions. Differ. Uravn. (Minsk) 1977, 13: 294-304.MathSciNet
  6. Dubey RS: Existence of the unique solution to abstract second order semilinear integrodifferential equations. Nonlinear Dyn. Syst. Theory 2010, 10(4):375-386.MathSciNet
  7. Crandall MG, Souganidis P: Convergence of difference approximations of quasilinear evolution equations. Nonlinear Anal. 1986, 10: 425-445. 10.1016/0362-546X(86)90049-0MathSciNetView Article
  8. Cannon JR, Lin Y, Wang S: An implicit finite difference scheme for the diffusion equation subject to mass specification. Int. J. Eng. Sci. 1990, 28: 573-578. 10.1016/0020-7225(90)90086-XMathSciNetView Article
  9. Dehghan M: Second order schemes for a boundary value problem with Neumann boundary conditions. Appl. Math. Comput. 2002, 138: 173-184. 10.1016/S0377-0427(00)00452-0View Article
  10. Bahuguna D: Quasilinear integrodifferential equations in Banach spaces. Nonlinear Anal. 1995, 24: 175-183. 10.1016/0362-546X(94)E0049-MMathSciNetView Article
  11. Kumar K, Kumar R, Shukla RK: Nonlocal parabolic integro-differential equations with delay. Int. J. Appl. Math. Res. 2012, 1(4):549-564.View Article
  12. Guezane-Lakoud A, Chaoui A: Rothe method applied to semilinear hyperbolic integro-differential equation with integral conditions. Int. J. Open Probl. Comput. Sci. Math. 2011, 4: 1-14.MathSciNet
  13. Dabas J, Bahuguna D: An integro-differential equation with an integral boundary condition. Math. Comput. Model. 2009, 50: 123-131. 10.1016/j.mcm.2009.03.002MathSciNetView Article
  14. Guezane-Lakoud A, Dabas J, Bahuguna D: Existence and uniqueness of generalized solutions to a telegraph equation with an integral boundary condition via Galerkin method. Int. J. Math. Math. Sc 2011., 2011: Article ID 451492
  15. Guezane-Lakoud A, Bendjazia N: Galerkin method for solving a telegraph equation with a weighted integral condition. Int. J. Open Probl. Complex Anal. 2012, 5: 41-53.

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© Guezane-Lakoud et al.; licensee Springer. 2013

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