Elsevier

Applied Mathematics and Computation

Volume 227, 15 January 2014, Pages 755-761
Applied Mathematics and Computation

Fractional order integral equations of two independent variables

https://doi.org/10.1016/j.amc.2013.10.086Get rights and content

Abstract

In this paper, we present some results concerning the existence, the uniqueness and the attractivity of solutions for some functional integral equations of Riemann–Liouville fractional order, by using some fixed point theorems.

Introduction

Fractional integral equations have recently been applied in various areas of engineering, science, finance, applied mathematics, and bio-engineering and others. However, many researchers remain unaware of this field. There has been a significant development in ordinary and partial fractional differential and integral equations in recent years; see the monographs of Baleanu et al. [4], Kilbas et al. [13], Miller and Ross [15], Lakshmikantham et al. [14], Podlubny [20], Samko et al. [22]. Recently some interesting results on the attractivity of the solutions of some classes of integral equations have been obtained by Abbas et al. [1], [2], Banaś et al. [5], [6], [7], Darwish et al. [8], Dhage [9], [10], [11], Pachpatte [18], [19] and the references therein.

In [17], Mureşan proved some results concerning the existence, uniqueness, data dependence and comparison theorems, by applying some results from Picard and weakly Picard operators’ theory [21], for the following functional integral equation of the formx(t)=α+fx,0g(t)x(s)ds,x(h(t));t[0,T],where T>0,f:[0,T]×Rn×RnRn, g,h:[0,T][0,T]. In this paper we improve the above results for the following partial integral equation of Riemann–Liouville fractional order of the formu(x,y)=μ(x,y)+f(x,y,Iθru(x,y),u(x,y));(x,y)J[0,a]×[0,b],where θ=(0,0),r=(r1,r2),r1,r2(0,),μ:JRn,f:J×Rn×RnRn are given continuous functions, Iθr is the left-sided mixed Riemann–Liouville integral of order r.

Next, we prove some results concerning the existence and the attractivity of solutions for the following partial Riemann–Liouville fractional order integral equation of the formu(x,y)=μ(x,y)+f(x,y,Iθru(x,y),u(x,y));(x,y)JR+×[0,b],where b>0,R+=[0,) and μ:JRn,f:J×Rn×RnRn are given continuous functions.

Our investigations are conducted in Banach spaces with an application of Banach’s contraction principle and Schauder’s fixed point theorem for the existence and uniqueness of solutions of Eq. (2). We use the Schauder fixed point theorem for the existence of solutions of Eq. (3), and we prove that all solutions are globally asymptotically stable. Also, we present some examples illustrating the applicability of the imposed conditions.

Section snippets

Preliminaries

In this section, we introduce notations, definitions, and preliminary facts which are used throughout this paper. By C(J) we denote the Banach space of all continuous functions from J into Rn with the normw=sup(x,y)Jw(x,y),where . denotes a suitable complete norm on Rn.

Let E be the space of functions wC(J), which fulfill the following condition:M0:w(x,y)Meλ(x+y),for(x,y)J,where λ is a positive constant. In the space E we define the normwE=sup(x,y)J{w(x,y)e-λ(x+y)}.According

Main Results

Let us start by defining what we mean by a solution of Eq. (2).

Definition 3.1

A function uE is said to be a solution of (2) if u satisfies Eq. (2) on J.

Now, we shall prove the following theorem concerning the existence and uniqueness of a solution of Eq. (2).

Setf(x,y)=f(x,y,0,0).

Theorem 3.2

Assume that the following hypotheses hold

  • (H1)

    The functions μ and f are in E,

  • (H2)

    There exist constants L1,L2>0 such thatf(x,x,u,v)-f(x,y,u,v)L2u-u+L1v-vfor each (x,y)J and u,v,u,vRn.

IfL1+L2ar1br2Γ(1+r1)Γ(1+r2)1,then Eq. (2)

Examples

As applications and to illustrate our results, we present the following examples.

Example 4.1

Consider the following integral equation of fractional orderu(x,y)=(x+y2)eλ(x+y)+2+|u(x,y)|+|Iθru(x,y)|ex+y+10(1+|u(x,y)|)(1+|Iθru(x,y)|);(x,y)[0,1]×[0,1].Set μ(x,y)=(x+y2)eλ(x+y) andf(x,y,u,v)=e-(x+y+10)1+|u|+e-(x+y+10)1+|v|;(x,y)[0,1]×[0,1].We have f(x,y)=2e-(x+y+10), then μ,fE and condition (H1) is satisfied.

For each u,vR and (x,y)[0,1]×[0,1], we have|f(x,y,u(x,y))-f(x,y,v(x,y))|1e10|u-v|.Hence condition

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