Applied Mathematics E-Notes, 12(2012), 79-87 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
Fractional Order Riemann-Liouville Integral
Equations with Multiple Time Delays
Saïd Abbasy, Mou¤ak Benchohraz
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, 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 [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)
m
X
bi (t)u(t
i)
i=1
2 I F (t; u(t)) if t 2 [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) =
m
X
i=1
gi (x; y)u(x
i; y
i )+I
r
f (x; y; u(x; y)) if (x; y) 2 J := [0; a] [0; b]; (1)
Mathematics Subject Classi…cations: 26A33
de Mathématiques, Université de Saïda, B.P. 138, 20000, Saïda, Algérie
z Laboratoire de Mathématiques, Université de Sidi Bel-Abbès, B.P. 89, 22000, Sidi Bel-Abbès,
Algérie
y Laboratoire
79
80
Fractional Integral Equations with Multiple Delays
u(x; y) = (x; y); if (x; y) 2 J~ := [
; a]
[
; b]n(0; a]
(0; b];
(2)
where a; b > 0;
= (0; 0); i ; i
0; i = 1; : : : ; m;
= maxi=1;:::;m f i g;
=
maxi=1;:::;m f i g; I r is the left-sided mixed Riemann-Liouville integral of order r =
(r1 ; r2 ) 2 (0; 1) (0; 1); f : J Rn ! Rn ; gi : J ! R; i = 1; : : : ; m are given
continuous functions, and : J~ ! Rn is a given continuous function such that
(x; 0) =
m
X
gi (x; 0) (x
m
X
gi (0; y) (
i;
i );
x 2 [0; a];
i );
y 2 [0; b]:
i=1
and
(0; y) =
i; y
i=1
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 Rn with the norm
kwk1 = sup kw(x; y)k;
(x;y)2J
where k:k denotes a suitable complete norm on Rn : Also, C := C([
Banach space endowed with the norm
kwkC =
sup
(x;y)2[
;a] [
;b]
; a]
[
; b]) is a
kw(x; y)k:
As usual, by L1 (J) we denote the space of Lebesgue-integrable functions w : J ! Rn
with the norm
Z aZ b
kwkL1 =
kw(x; y)kdydx:
0
0
DEFINITION 1 ([23]). Let r = (r1 ; r2 ) 2 (0; 1) (0; 1); = (0; 0) and u 2 L1 (J):
The left-sided mixed Riemann-Liouville integral of order r of u is de…ned by
Z xZ y
1
(x s)r1 1 (y t)r2 1 u(s; t)dtds:
(I r u)(x; y) =
(r1 ) (r2 ) 0 0
In particular,
(I u)(x; y) = u(x; y); (I u)(x; y) =
Z
0
x
Z
0
y
u(s; t)dtds for almost all (x; y) 2 J;
S. Abbas and M. Benchohra
81
where = (1; 1):
For instance, I r u exists for all r1 ; r2 2 (0; 1); when u 2 L1 (J): Note also that when
u 2 C(J); then (I r u) 2 C(J); moreover
(I r u)(x; 0) = (I r u)(0; y) = 0; x 2 [0; a]; y 2 [0; b]:
EXAMPLE 1. Let ; ! 2 ( 1; 1) and r = (r1 ; r2 ) 2 (0; 1)
I r x y! =
3
(1 + ) (1 + !)
x
(1 + + r1 ) (1 + ! + r2 )
+r1 !+r2
y
(0; 1); then
for almost all (x; y) 2 J:
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
i=1;:::;m
(
)
sup jgi (x; y)j :
(x;y)2J
THEOREM 1. Assume
(H1 ) 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 Rn :
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,
(
~
(x; y);
(x; y) 2 J;
N (u)(x; y) = Pm
(3)
r
i; y
i ) + I f (x; y; u(x; y)); (x; y) 2 J:
i=1 gi (x; y)u(x
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 1 RmB where
R =
ar1 br2 h
;
(1 + r1 ) (1 + r2 )
and h = khk1 ; and consider the set
BR = fu 2 C : kukC
Rg:
82
Fractional Integral Equations with Multiple Delays
It is clear that BR is a closed bounded and convex subset of C: For every u 2 BR and
(x; y) 2 J we obtain by (H1 ) that
m
X
kN (u)(x; y)k
i=1
+
jgi (x; y)j ku(x
1
(r1 ) (r2 )
Z
x
Z
i; y
y
(x
0
0
i )k
s)r1 1 (y t)r2 1 kf (s; t; u(s; t))kdtds
Z xZ y
(x s)r1 1 (y t)r2 1 h(s; t)dtds
1
(r1 ) (r2 ) 0 0
ar1 br2
mBkukC + h
(1 + r1 ) (1 + r2 )
mBR + (1 mB)R = R:
mBkukC +
~ we obtain
On the other hand, for every u 2 BR and (x; y) 2 J;
kN (u)(x; y)k = k (x; y)k
R:
So we obtain that
kN (u)kC
R:
That is, N (BR ) BR : Since f is bounded on BR ; thus N (BR ) is equicontinuous and
the Schauder …xed point theorem shows that N has at least one …xed point u 2 BR
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 function l 2 C(J) such that
kf (x; y; u)
f (x; y; v)k
l(x; y)ku
vk;
for each (x; y) 2 J and u; v 2 Rn :
If
mB (1 + r1 ) (1 + r2 ) + ar1 br2 l
< 1;
(1 + r1 ) (1 + r2 )
where l = klk1 , then problem (1)-(2) has a unique solution on [
(4)
; a]
[
; b]:
PROOF. Consider the operator N de…ned in (3). Then by (H2 ); for every u; v 2 C
S. Abbas and M. Benchohra
83
and (x; y) 2 J we have
kN (u)(x; y)
m
X
N (v)(x; y)k
i=1
jgi (x; y)j ku(x
i; y
i)
v(x
i ; y)k
Z xZ y
1
(x s)r1 1 (y t)r2 1
(r1 ) (r2 ) 0 0
kf (s; t; u(s; t)) f (s; t; v(s; t))kdtds
mBku vk1
Z xZ y
1
(x s)r1 1 (y t)r2 1
+
(r1 ) (r2 ) 0 0
l(s; t)ku vkC dtds
ar1 br2
mBku vk1 + l
ku vkC
(1 + r1 ) (1 + r2 )
l ar1 br2
ku vkC :
=
mB +
(1 + r1 ) (1 + r2 )
+
Thus
mB (1 + r1 ) (1 + r2 ) + ar1 br2 l
ku vkC
(1 + r1 ) (1 + r2 )
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 ([10]). (Nonlinear alternative of Leray-Schauder type) By U and @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
0 2 U and T : U ! C continuous and compact operator. Then either
kN (u)
N (v)kC
(a) T has …xed points, or
(b) there exist u 2 @U and
2 (0; 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 ! [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
Z xZ y
(s; t)
(x; y) !(x; y) + c
dtds;
r1 (y
(x
s)
t)r2
0
0
then there exists a constant = (r1 ; r2 ) such that
Z xZ y
!(s; t)
(x; y) !(x; y) + c
dtds;
s)r1 (y t)r2
0
0 (x
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
84
Fractional Integral Equations with Multiple Delays
(H3 ) There exist positive functions p; q 2 C(J) such that
p(x; y) + q(x; y)kuk; for all (x; y) 2 J and u 2 Rn :
kf (x; y; u)k
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 (0; 1) and u 2 @U: Let u 2 C and u = N (u) for some 0 < < 1: Thus for
each (x; y) 2 J; we have
u(x; y) =
m
X
gi (x; y)u(x
i; y
i)
+ I r f (x; y; u(x; y)):
i=1
This implies by (H3 ) that, for each (x; y) 2 J; we have
ku(x; y)k
p ar1 br2
mBku(x; y)k +
(1 + r1 ) (1 + r2 )
Z xZ y
q
+
(x s)r1 1 (y t)r2
(r1 ) (r2 ) 0 0
1
u(s; t)dtds;
where p = kpk1 and q = kqk1 : Thus, for each (x; y) 2 J; we get
ku(x; y)k
p ar1 br2
(1 mB) (1 + r1 ) (1 + r2 )
Z xZ y
q
+
(x s)r1 1 (y t)r2
(1 mB) (r1 ) (r2 ) 0 0
Z xZ y
w+c
(x s)r1 1 (y t)r2 1 u(s; t)dtds;
0
u(s; t)dtds
0
where
w :=
(1
p ar1 br2
mB) (1 + r1 ) (1 + r2 )
and
c :=
q
:
mB) (r1 ) (r2 )
(1
From Lemma 1, there exists
kuk1
1
:= (r1 ; r2 ) > 0 such that, for each (x; y) 2 J; we get
Z xZ y
w 1+c
(x s)r1 1 (y t)r2 1 dtds
0
0
c ar1 br2
w 1+
r1 r2
fg and
Set M := maxfk k; M
f:
:= M
U = fu 2 C : kukC < M + 1g:
S. Abbas and M. Benchohra
85
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
x3 y
3
x4 y 2
1
1
u(x
; y 3) +
u(x 2; y
) + u(x 1; y
8
4
12
2
4
+I r f (x; y; u); if (x; y) 2 J := [0; 1] [0; 1];
u(x; y) = 0; if (x; y) 2 J~ := [ 2; 1] [ 3; 1]n(0; 1] (0; 1];
u(x; y) =
3
)
2
(5)
(6)
where m = 3; r = ( 21 ; 51 ) and
f (x; y; u) = ex+y
Set
g1 (x; y) =
We have B =
1
4
1
:
1 + juj
x3 y
x4 y 2
1
; g2 (x; y) =
; g3 (x; y) = :
8
12
4
and
jf (x; y; u)j
ex+y ; for all (x; y) 2 J and u 2 R:
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
x3 y
1
x4 y 2
2
3
1
u(x 1; y
)+
u(x
;y
) + u(x 3; y 2)
8
2
12
5
4
8
+I r f (x; y; u); if (x; y) 2 J := [0; 1] [0; 1];
(7)
~
u(x; y) = (x; y); if (x; y) 2 J := [ 3; 1] [ 2; 1]n(0; 1] (0; 1];
(8)
x+y juj
1 1
where m = 3; r = ( 2 ; 5 ); f (x; y; u) = 20 1+juj and : J~ ! R is continuous with
u(x; y) =
1
1
(x 3; 2); (0; y) =
( 3; y 2); x; y 2 [0; 1]:
(9)
8
8
Notice that condition (9) is satis…ed by
0:
Set
x4 y 2
1
x3 y
g1 (x; y) =
; g2 (x; y) =
; g3 (x; y) = :
8
12
8
1
We have B = 18 : It is clear that f satis…es (H2 ) with l = 10
. 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]:
(x; 0) =
Acknowledgement. The authors are grateful to the referee for his/her remarks.
86
Fractional Integral Equations with Multiple Delays
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