Fractional Order Riemann-Liouville Integral Equations with Multiple Time Delays

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Fractional Order Riemann-Liouville Integral Equations with Multiple Time Delays

Saïd Abbas

^y

, Mou¤ak Benchohra

^z

Received 12 November 2011

Abstract

In the present article we investigate the existence and uniqueness of solutions for a system of integral equations of fractional order by using some …xed point theorems. Also we illustrate our results with some examples.

1 Introduction

The idea of fractional calculus and fractional order integral equations has been a subject of interest not only among mathematicians, but also among physicists and engineers.

Indeed, we can …nd numerous applications in rheology, control, porous media, vis- coelasticity, electrochemistry, electromagnetism, etc. [9, 11, 16, 17, 19]. There has been a signi…cant development in ordinary and partial fractional di¤erential equations in recent years; see the monographs of Kilbaset al. [14], Miller and Ross [18], Samko et al. [21], the papers of Abbas and Benchohra [1, 2], Abbas et al. [3], Belarbiet al.

[4], Benchohra et al. [5, 6, 7], Diethelm [8], Kilbas and Marzan [15], Mainardi [16], Podlubny et al [20], Vityuk [22], Vityuk and Golushkov [23], and Zhang [24] and the references therein.

In [13], R. W. Ibrahim and H. A. Jalab studied the existence of solutions of the following fractional integral inclusion

u(t) Xm

i=1

bi(t)u(t i)2I F(t; u(t))ift2[0; T];

where i < t 2 [0; T]; bi : [0; T] ! R; i = 1; : : : ; n are continuous functions, and F : [0; T] R! P(R)is a given multivalued map.

This paper concerned with the existence and uniqueness of solutions for the following fractional order integral equations for the system

u(x; y) = Xm

i=1

gi(x; y)u(x _i; y _i)+I^rf(x; y; u(x; y))if(x; y)2J := [0; a] [0; b]; (1)

Mathematics Sub ject Classi…cations: 26A33

yLaboratoire de Mathématiques, Université de Saïda, B.P. 138, 20000, Saïda, Algérie

zLaboratoire de Mathématiques, Université de Sidi Bel-Abbès, B.P. 89, 22000, Sidi Bel-Abbès, Algérie

79

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u(x; y) = (x; y); if(x; y)2J~:= [ ; a] [ ; b]n(0; a] (0; b]; (2) where a; b > 0; = (0;0); _i; _i 0; i = 1; : : : ; m; = maxi=1;:::;mf ig; = maxi=1;:::;mf ig; I^r is the left-sided mixed Riemann-Liouville integral of order r = (r₁; r₂) 2 (0;1) (0;1); f : J Rⁿ ! Rⁿ; g_i : J ! R; i = 1; : : : ; m are given continuous functions, and : ~J !Rⁿ is a given continuous function such that

(x;0) = Xm

i=1

gi(x;0) (x _i; _i); x2[0; a];

and

(0; y) = Xm

i=1

g_i(0; y) ( _i; y _i); y2[0; b]:

We present three results for the problem (1)-(2), the …rst one is based on Schauder’s

…xed point theorem (Theorem 1), the second one is a uniqueness of the solution by using the Banach …xed point theorem (Theorem 2) and the last one on the nonlinear alternative of Leray-Schauder type (Theorem 4).

2 Preliminaries

In this section, we introduce notations, de…nitions, and preliminary facts which are used throughout this paper. By C(J) we denote the Banach space of all continuous functions fromJ intoRⁿ with the norm

kwk1= sup

(x;y)2Jkw(x; y)k;

wherek:kdenotes a suitable complete norm onRⁿ:Also,C:=C([ ; a] [ ; b])is a Banach space endowed with the norm

kwkC= sup

(x;y)2[ ;a] [ ;b]kw(x; y)k:

As usual, by L¹(J)we denote the space of Lebesgue-integrable functionsw:J !Rⁿ with the norm

kwkL¹= Z a

0

Z b

0 kw(x; y)kdydx:

DEFINITION 1 ([23]). Letr= (r1; r2)2(0;1) (0;1); = (0;0)andu2L¹(J):

The left-sided mixed Riemann-Liouville integral of orderrofuis de…ned by (I^ru)(x; y) = 1

(r1) (r2) Z x

0

Z y 0

(x s)^r¹ ¹(y t)^r² ¹u(s; t)dtds:

In particular,

(I u)(x; y) =u(x; y); (I u)(x; y) = Z x

0

Z y 0

u(s; t)dtdsfor almost all(x; y)2J;

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where = (1;1):

For instance,I^ruexists for allr₁; r₂2(0;1);whenu2L¹(J):Note also that when u2C(J);then(I^ru)2C(J);moreover

(I^ru)(x;0) = (I^ru)(0; y) = 0; x2[0; a]; y2[0; b]:

EXAMPLE 1. Let ; !2( 1;1)and r= (r₁; r₂)2(0;1) (0;1);then I^rx y^!= (1 + ) (1 +!)

(1 + +r1) (1 +!+r2)x ^+r¹y^!+r² for almost all(x; y)2J:

3 Existence of Solutions

Let us start by de…ning what we mean by a solution of the problem (1)-(2).

DEFINITION 2. A functionu2C is said to be a solution of (1)-(2) if usatis…es equation (1) onJ and condition (2) onJ :~

Set

B= max

i=1;:::;m

( sup

(x;y)2Jjgi(x; y)j )

:

THEOREM 1. Assume

(H₁) There exists a positive functionh2C(J)such that

kf(x; y; u)k h(x; y); for all(x; y)2J andu2Rⁿ:

IfmB <1, then problem (1)-(2) has at least one solutionuon[ ; a] [ ; b]:

PROOF. Transform problem (1)-(2) into a …xed point problem. Consider the op- eratorN :C!Cde…ned by,

N(u)(x; y) =

( (x; y); (x; y)2J ;~

Pm

i=1gi(x; y)u(x _i; y _i) +I^rf(x; y; u(x; y)); (x; y)2J: (3) The problem of …nding the solutions of problem (1)-(2) is reduced to …nding the solutions of the operator equationN(u) =u:LetR ₁^R_mB where

R = a^r¹b^r²h (1 +r₁) (1 +r₂); andh =khk1;and consider the set

BR=fu2C:kuk^C Rg:

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It is clear that B_R is a closed bounded and convex subset ofC:For everyu2B_R and (x; y)2J we obtain by(H1)that

kN(u)(x; y)k

Xm

i=1

jgi(x; y)j ku(x _i; y _i)k

+ 1

(r₁) (r₂) Z x

0

Z y 0

(x s)^r¹ ¹(y t)^r² ¹kf(s; t; u(s; t))kdtds mBkuk^C+ 1

(r₁) (r₂) Z x

0

Z y 0

(x s)^r¹ ¹(y t)^r² ¹h(s; t)dtds mBkuk^C+h a^r¹b^r²

(1 +r1) (1 +r2) mBR+ (1 mB)R=R:

On the other hand, for everyu2B_R and(x; y)2J ;~ we obtain kN(u)(x; y)k=k (x; y)k R:

So we obtain that

kN(u)kC R:

That is,N(B_R) B_R:Sincef is bounded on B_R;thusN(B_R)is equicontinuous and the Schauder …xed point theorem shows that N has at least one …xed point u 2BR

which is solution of (1)-(2).

For the uniqueness we prove the following Theorem THEOREM 2. Assume that following hypothesis holds:

(H2) There exists a positive functionl2C(J)such that

kf(x; y; u) f(x; y; v)k l(x; y)ku vk; for each(x; y)2J andu; v2Rⁿ:

If

mB (1 +r₁) (1 +r₂) +a^r¹b^r²l

(1 +r1) (1 +r2) <1; (4)

where l =klk1, then problem (1)-(2) has a unique solution on[ ; a] [ ; b]:

PROOF. Consider the operatorN de…ned in (3). Then by(H2);for everyu; v2C

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and(x; y)2J we have kN(u)(x; y) N(v)(x; y)k

Xm

i=1

jgi(x; y)j ku(x _i; y _i) v(x _i; y)k

+ 1

(r₁) (r₂) Z x

0

Z y 0

(x s)^r¹ ¹(y t)^r² ¹ kf(s; t; u(s; t)) f(s; t; v(s; t))kdtds mBku vk1

+ 1

(r1) (r2) Z x

0

Z y 0

(x s)^r¹ ¹(y t)^r² ¹ l(s; t)ku vk^Cdtds

mBku vk1+l a^r¹b^r²

(1 +r₁) (1 +r₂)ku vk^C

= mB+ l a^r¹b^r²

(1 +r1) (1 +r2) ku vkC: Thus

kN(u) N(v)k^C mB (1 +r1) (1 +r2) +a^r¹b^r²l

(1 +r1) (1 +r2) ku vk^C

Hence by (4), we have thatN is a contraction mapping. Then in view of Banach …xed point Theorem, N has a unique …xed point which is solution of problem (1)-(2).

THEOREM 3 ([10]). (Nonlinear alternative of Leray-Schauder type) ByU and@U we denote the closure of U and the boundary of U respectively. Let X be a Banach space andCa nonempty convex subset ofX:LetU a nonempty open subset ofCwith 02U andT :U !C continuous and compact operator. Then either

(a) T has …xed points, or

(b) there existu2@U and 2(0;1)withu= T(u):

In the sequel we use the following version of Gronwall’s Lemma for two independent variables and singular kernel.

LEMMA 1 ([12]). Let :J ![0;1)be a real function and!(:; :)be a nonnegative, locally integrable function on J: If there are constants c >0 and 0 < r₁; r₂ <1 such that

(x; y) !(x; y) +c Z x

0

Z y 0

(s; t)

(x s)^r¹(y t)^r²dtds;

then there exists a constant = (r1; r2)such that (x; y) !(x; y) + c

Z x 0

Z y 0

!(s; t)

(x s)^r¹(y t)^r²dtds;

for every (x; y)2J:

Now, we present an existence result for the problem (1)-(2) based on the Nonlinear alternative of Leray-Schauder type.

THEOREM 4. Assume

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(H₃) There exist positive functionsp; q2C(J)such that

kf(x; y; u)k p(x; y) +q(x; y)kuk; for all(x; y)2J andu2Rⁿ: IfmB <1;then problem (1)-(2) has at least one solution on[ ; a] [ ; b]:

PROOF. Consider the operatorN de…ned in (3). We shall show that the operator N is completely continuous. By the continuity of f and the Arzela-Ascoli Theorem, we can easily obtain that N is completely continuous.

A priori bounds. We shall show there exists an open setU C withu6= N(u);

for 2 (0;1) and u2@U: Letu 2C and u= N(u) for some 0 < <1:Thus for each(x; y)2J;we have

u(x; y) = Xm

i=1

gi(x; y)u(x _i; y _i) + I^rf(x; y; u(x; y)):

This implies by(H3)that, for each(x; y)2J;we have ku(x; y)k mBku(x; y)k+ p a^r¹b^r²

(1 +r₁) (1 +r₂)

+ q

(r₁) (r₂) Z x

0

Z y 0

(x s)^r¹ ¹(y t)^r² ¹u(s; t)dtds;

where p =kpk1 andq =kqk1:Thus, for each(x; y)2J; we get ku(x; y)k p a^r¹b^r²

(1 mB) (1 +r1) (1 +r2)

+ q

(1 mB) (r1) (r2) Z x

0

Z y 0

(x s)^r¹ ¹(y t)^r² ¹u(s; t)dtds w+c

Z x 0

Z y 0

(x s)^r¹ ¹(y t)^r² ¹u(s; t)dtds;

where

w:= p a^r¹b^r²

(1 mB) (1 +r1) (1 +r2) and

c:= q

(1 mB) (r1) (r2):

From Lemma 1, there exists := (r₁; r₂)>0such that, for each(x; y)2J;we get kuk1 w 1 +c

Z x 0

Z y 0

(x s)^r¹ ¹(y t)^r² ¹dtds w 1 + c a^r¹b^r²

r₁r₂ :=M :f SetM := maxfk k;Mfgand

U =fu2C:kuk^C< M + 1g:

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By our choice of U; there is no u 2 @U such that u = N(u); for 2 (0;1): As a consequence of Theorem 3, we deduce that N has a …xed point u in U which is a solution to problem (1)-(2).

4 Examples

We provide two examples.

EXAMPLE 1. As an application of our results we consider the following system of fractional integral equations of the form

u(x; y) =x³y

8 u(x 3

4; y 3) + x⁴y²

12 u(x 2; y 1 2) +1

4u(x 1; y 3 2) +I^rf(x; y; u); if(x; y)2J := [0;1] [0;1]; (5) u(x; y) = 0; if(x; y)2J~:= [ 2;1] [ 3;1]n(0;1] (0;1]; (6) where m= 3; r= (¹₂;¹₅)and

f(x; y; u) =e^x+y 1 1 +juj: Set

g₁(x; y) =x³y

8 ; g₂(x; y) =x⁴y²

12 ; g₃(x; y) = 1 4: We have B=¹₄ and

jf(x; y; u)j e^x+y; for all(x; y)2J andu2R:

Then condition (H₁) is satis…ed and mB = ³₄ < 1: In view of Theorem 1, problem (5)-(6) has a solution de…ned on[ 2;1] [ 3;1]:

EXAMPLE 2. Consider the fractional integral equation u(x; y) =x³y

8 u(x 1; y 1

2) +x⁴y²

12 u(x 2 5; y 3

4) +1

8u(x 3; y 2) +I^rf(x; y; u); if(x; y)2J := [0;1] [0;1]; (7) u(x; y) = (x; y); if(x; y)2J~:= [ 3;1] [ 2;1]n(0;1] (0;1]; (8) where m= 3; r= (¹₂;¹₅); f(x; y; u) =^x+y₂₀ ₁₊^j^u^j

juj and : ~J!Ris continuous with (x;0) = 1

8 (x 3; 2); (0; y) =1

8 ( 3; y 2); x; y2[0;1]: (9) Notice that condition (9) is satis…ed by 0:

Set

g₁(x; y) =x³y

8 ; g₂(x; y) =x⁴y²

12 ; g₃(x; y) = 1 8:

We have B = ¹₈: It is clear thatf satis…es(H2)withl =₁₀¹. A simple computation shows that condition (4) is satis…ed. Hence by Theorem 2, problem (7)-(8) has a unique solution de…ned on[ 3;1] [ 2;1]:

Acknowledgement. The authors are grateful to the referee for his/her remarks.

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