Fractional Order Riemann-Liouville Integral Equations with Multiple Time Delays

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1 Applied Mathematics E-Notes, 12(212), c ISSN Available free at mirror sites of Fractional Order Riemann-Liouville Integral Equations with Multiple Time Delays Saïd Abbas y, Mou ak Benchohra z Received 12 November 211 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, viscoelasticity, 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 Kilbas et al. [14], Miller and Ross [18], Samko et al. [21], the papers of Abbas and Benchohra [1, 2], Abbas et al. [3], Belarbi et al. [4], Benchohra et al. [5, 6, 7], Diethelm [8], Kilbas and Marzan [15], Mainardi [16], Podlubny et al [2], 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) b i (t)u(t i ) 2 I F (t; u(t)) if t 2 [; T ]; where i < t 2 [; T ]; b i : [; T ]! R; i = 1; : : : ; n are continuous functions, and F : [; 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) = g i (x; y)u(x i ; y i )I r f(x; y; u(x; y)) if (x; y) 2 J := [; a][; b]; (1) Mathematics Subject Classi cations: 26A33 y Laboratoire de Mathématiques, Université de Saïda, B.P. 138, 2, Saïda, Algérie z Laboratoire de Mathématiques, Université de Sidi Bel-Abbès, B.P. 89, 22, Sidi Bel-Abbès, Algérie 79

2 8 Fractional Integral Equations with Multiple Delays u(x; y) = (x; y); if (x; y) 2 ~ J := [ ; a] [ ; b]n(; a] (; b]; (2) where a; b > ; = (; ); i ; i ; i = 1; : : : ; m; = max ;:::;m f i g; = max ;:::;m f i g; I r is the left-sided mixed Riemann-Liouville integral of order r = (r 1 ; r 2 ) 2 (; 1) (; 1); f : J R n! R n ; g i : J! R; i = 1; : : : ; m are given continuous functions, and : J ~! R n is a given continuous function such that and (x; ) = (; y) = g i (x; )(x i ; i ); x 2 [; a]; g i (; y)( i ; y i ); y 2 [; 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 from J into R n with the norm kwk 1 = sup kw(x; y)k; (x;y)2j where k:k denotes a suitable complete norm on R n : Also, C := C([ ; a] [ ; b]) is a Banach space endowed with the norm kwk C = sup kw(x; y)k: (x;y)2[ ;a][ ;b] As usual, by L 1 (J) we denote the space of Lebesgue-integrable functions w : J! R n with the norm kwk L 1 = Z a Z b kw(x; y)kdydx: DEFINITION 1 ([23]). Let r = (r 1 ; r 2 ) 2 (; 1)(; 1); = (; ) and u 2 L 1 (J): The left-sided mixed Riemann-Liouville integral of order r of u is de ned by (I r u)(x; y) = 1 (r 1 ) (r 2 ) (x s) r1 1 (y t) r2 1 u(s; t)dtds: In particular, (I u)(x; y) = u(x; y); (I u)(x; y) = u(s; t)dtds for almost all (x; y) 2 J;

3 S. Abbas and M. Benchohra 81 where = (1; 1): For instance, I ru exists for all r 1; r 2 2 (; 1); when u 2 L 1 (J): Note also that when u 2 C(J); then (I r u) 2 C(J); moreover (I r u)(x; ) = (I r u)(; y) = ; x 2 [; a]; y 2 [; b]: EXAMPLE 1. Let ;! 2 ( 1; 1) and r = (r 1 ; r 2 ) 2 (; 1) (; 1); then I r x y! = (1 ) (1!) (1 r 1 ) (1! r 2 ) xr1 y!r2 for almost all (x; y) 2 J: 3 Existence of Solutions Let us start by de ning what we mean by a solution of the problem (1)-(2). DEFINITION 2. A function u 2 C is said to be a solution of (1)-(2) if u satis es equation (1) on J and condition (2) on ~ J: Set B = max ;:::;m ( sup jg i (x; y)j (x;y)2j ) : THEOREM 1. Assume (H 1 ) There exists a positive function h 2 C(J) such that kf(x; y; u)k h(x; y); for all (x; y) 2 J and u 2 R n : If mb < 1, then problem (1)-(2) has at least one solution u on [ ; a] [ ; b]: PROOF. Transform problem (1)-(2) into a xed point problem. Consider the operator N : C! C de ned by, N(u)(x; y) = ( (x; y); (x; y) 2 ~ J; P m g i(x; y)u(x i ; y i ) I r f(x; y; u(x; y)); (x; y) 2 J: (3) The problem of nding the solutions of problem (1)-(2) is reduced to nding the solutions of the operator equation N(u) = u: Let R R 1 mb where R = a r1 b r2 h (1 r 1 ) (1 r 2 ) ; and h = khk 1 ; and consider the set B R = fu 2 C : kuk C Rg:

4 82 Fractional Integral Equations with Multiple Delays It is clear that B R is a closed bounded and convex subset of C: For every u 2 B R and (x; y) 2 J we obtain by (H 1 ) that kn(u)(x; y)k jg i (x; y)j ku(x i ; y i )k 1 (r 1 ) (r 2 ) mbkuk C 1 (r 1 ) (r 2 ) (x s) r1 1 (y t) r2 1 kf(s; t; u(s; t))kdtds mbkuk C h a r1 b r2 (1 r 1 ) (1 r 2 ) mbr (1 mb)r = R: (x s) r1 1 (y t) r2 1 h(s; t)dtds On the other hand, for every u 2 B R and (x; y) 2 ~ J; we obtain So we obtain that kn(u)(x; y)k = k(x; y)k R: kn(u)k C R: That is, N(B R ) B R : Since f is bounded on B R ; thus N(B R ) is equicontinuous and the Schauder xed point theorem shows that N has at least one xed point u 2 B R which is solution of (1)-(2). For the uniqueness we prove the following Theorem THEOREM 2. Assume that following hypothesis holds: (H 2 ) There exists a positive function l 2 C(J) such that for each (x; y) 2 J and u; v 2 R n : kf(x; y; u) f(x; y; v)k l(x; y)ku vk; If mb (1 r 1 ) (1 r 2 ) a r1 b r2 l (1 r 1 ) (1 r 2 ) < 1; (4) where l = klk 1, then problem (1)-(2) has a unique solution on [ ; a] [ ; b]: PROOF. Consider the operator N de ned in (3). Then by (H 2 ); for every u; v 2 C

5 S. Abbas and M. Benchohra 83 and (x; y) 2 J we have Thus kn(u)(x; y) N(v)(x; y)k kn(u) jg i (x; y)j ku(x i ; y i ) v(x i ; y)k 1 (x s) r1 1 (y t) r2 1 (r 1 ) (r 2 ) kf(s; t; u(s; t)) f(s; t; v(s; t))kdtds mbku vk 1 Z 1 x Z y (x s) r1 1 (y t) r2 1 (r 1 ) (r 2 ) l(s; t)ku vk C dtds mbku vk 1 l a r1 b r2 = mb (1 r 1 ) (1 r 2 ) ku vk C l a r1 b r2 ku vk C : (1 r 1 ) (1 r 2 ) N(v)k C mb (1 r 1) (1 r 2 ) a r1 b r2 l ku (1 r 1 ) (1 r 2 ) Hence by (4), we have that N 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 ([1]). (Nonlinear alternative of Leray-Schauder type) By U we denote the closure of U and the boundary of U respectively. Let X be a Banach space and C a nonempty convex subset of X: Let U a nonempty open subset of C with 2 U and T : U! C continuous and compact operator. Then either (a) T has xed points, or (b) there exist u and 2 (; 1) with u = 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! [; 1) be a real function and!(:; :) be a nonnegative, locally integrable function on J: If there are constants c > and < r 1 ; r 2 < 1 such that (x; y)!(x; y) c then there exists a constant = (r 1 ; r 2 ) such that (x; y)!(x; y) c (s; t) dtds; (x s) r1 r2 (y t)!(s; t) dtds; (x s) r1 r2 (y t) for every (x; y) 2 J: Now, we present an existence result for the problem (1)-(2) based on the Nonlinear alternative of Leray-Schauder type. THEOREM 4. Assume vk C

6 84 Fractional Integral Equations with Multiple Delays (H 3 ) There exist positive functions p; q 2 C(J) such that kf(x; y; u)k p(x; y) q(x; y)kuk; for all (x; y) 2 J and u 2 R n : If mb < 1; then problem (1)-(2) has at least one solution on [ ; a] [ ; b]: PROOF. Consider the operator N 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 set U C with u 6= N(u); for 2 (; 1) and u Let u 2 C and u = N(u) for some < < 1: Thus for each (x; y) 2 J; we have u(x; y) = g i (x; y)u(x i ; y i ) I r f(x; y; u(x; y)): This implies by (H 3 ) that, for each (x; y) 2 J; we have ku(x; y)k mbku(x; y)k q (r 1 ) (r 2 ) p a r1 b r2 (1 r 1 ) (1 r 2 ) (x s) r1 1 (y t) r2 1 u(s; t)dtds; where p = kpk 1 and q = kqk 1 : Thus, for each (x; y) 2 J; we get where and ku(x; y)k p a r1 b r2 (1 mb) (1 r 1 ) (1 r 2 ) q (1 mb) (r 1 ) (r 2 ) w c w := (x s) r1 1 (y t) r2 1 u(s; t)dtds (x s) r1 1 (y t) r2 1 u(s; t)dtds; p a r1 b r2 (1 mb) (1 r 1 ) (1 r 2 ) c := q (1 mb) (r 1 ) (r 2 ) : From Lemma 1, there exists := (r 1 ; r 2 ) > such that, for each (x; y) 2 J; we get kuk 1 w 1 c (x s) r1 1 (y t) r2 1 dtds b w 1 car1 r2 := M: r 1 r f 2 Set M := maxfkk; f Mg and U = fu 2 C : kuk C < M 1g:

7 S. Abbas and M. Benchohra 85 By our choice of U; there is no u such that u = N(u); for 2 (; 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) = x3 y 8 u(x 3 4 ; y 3) x4 y 2 12 u(x 2; y 1 2 ) 1 4 u(x 1; y 3 2 ) I r f(x; y; u); if (x; y) 2 J := [; 1] [; 1]; (5) u(x; y) = ; if (x; y) 2 ~ J := [ 2; 1] [ 3; 1]n(; 1] (; 1]; (6) where m = 3; r = ( 1 2 ; 1 5 ) and Set f(x; y; u) = e xy 1 1 juj : g 1 (x; y) = x3 y 8 ; g 2(x; y) = x4 y 2 12 ; g 3(x; y) = 1 4 : We have B = 1 4 and jf(x; y; u)j e xy ; for all (x; y) 2 J and u 2 R: Then condition (H 1 ) is satis ed and mb = 3 4 < 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) = x3 y 8 u(x 1; y 1 2 ) x4 y 2 12 u(x 2 5 ; y 3 4 ) 1 u(x 3; y 2) 8 I r f(x; y; u); if (x; y) 2 J := [; 1] [; 1]; (7) u(x; y) = (x; y); if (x; y) 2 ~ J := [ 3; 1] [ 2; 1]n(; 1] (; 1]; (8) where m = 3; r = ( 1 2 ; 1 xy juj 5 ); f(x; y; u) = 2 1juj and : J ~! R is continuous with (x; ) = 1 8 (x 3; 2); (; y) = 1 ( 3; y 2); x; y 2 [; 1]: (9) 8 Notice that condition (9) is satis ed by : Set g 1 (x; y) = x3 y 8 ; g 2(x; y) = x4 y 2 12 ; g 3(x; y) = 1 8 : We have B = 1 8 : It is clear that f satis es (H 2) with l = 1 1. 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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