Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 196, pp. 1–13.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
EXISTENCE OF SOLUTIONS TO FRACTIONAL-ORDER
IMPULSIVE HYPERBOLIC PARTIAL DIFFERENTIAL
INCLUSIONS
SAÏD ABBAS, MOUFFAK BENCHOHRA
Abstract. In this article we use the upper and lower solution method combined with a fixed point theorem for condensing multivalued maps, due to
Martelli, to study the existence of solutions to impulsive partial hyperbolic
differential inclusions at fixed instants of impulse.
1. Introduction
The theory of differential equations and inclusions of fractional order play a
very important role in describing some real world problems. For example some
problems in physics, mechanics, viscoelasticity, electrochemistry, control, porous
media, electromagnetic, etc. (see [16, 31]). Recently, numerous research papers
and monographs have appeared devoted to fractional differential equations, for
example see the monographs of Abbas et al [7], Kilbas et al [22], Lakshmikantham
et al [24], and Malinowska and Torres [28], and the papers of Abbas and Benchohra
[2, 5], Abbas et al [1, 6], Belarbi et al [8], Benchohra and Ntouyas [10], Kilbas et al
[20], Kilbas and Marzan [21], Semenchuk [32],Vityuk and Golushkov [34], and the
references therein.
The method of upper and lower solutions has been successfully applied to study
the existence of solutions for fractional order ordinary and partial partial differential
equations and inclusions. See the monographs by Benchohra et al [9], Heikkila and
Lakshmikantham [15], Ladde et al [26], the papers of Abbas and Benchohra [3, 4],
Benchohra and Ntouyas [10] and the references therein.
This article deals with the existence of solutions to impulsive fractional order
initial value problems (IVP for short), for the system
(c Dθrk u)(x, y) ∈ F (x, y, u(x, y)),
u(x+
k , y)
=
u(x−
k , y)
+
if (x, y) ∈ Jk ; k = 0, . . . , m;
Ik (u(x−
k , y)),
if y ∈ [0, b], k = 1, . . . , m;
2000 Mathematics Subject Classification. 26A33, 34A60.
Key words and phrases. Impulsive hyperbolic differential inclusions; fractional order;
upper solution; lower solution; left-sided mixed Riemann-Liouville integral;
Caputo fractional-order derivative; fixed point.
c
2014
Texas State University - San Marcos.
Submitted February 3, 2014. Published September 18, 2014.
1
(1.1)
(1.2)
2
S. ABBAS, M. BENCHOHRA
EJDE-2014/196


 u(x, 0) = ϕ(x), x ∈ [0, a],
u(0, y) = ψ(y), y ∈ [0, b],


ϕ(0) = ψ(0),
(1.3)
where J0 = [0, x1 ] × [0, b], Jk := (xk , xk+1 ] × [0, b], k = 1, . . . , m, θk = (xk , 0),
k = 0, . . . , m, a, b > 0, θ = (0, 0), c Dθr is the fractional caputo derivative of order
r = (r1 , r2 ) ∈ (0, 1] × (0, 1], 0 = x0 < x1 < · · · < xm < xm+1 = a, F : J × Rn →
P(Rn ) is a compact valued multivalued map, P(Rn ) is the family of all subsets of
Rn , Ik : Rn → Rn , k = 1, . . . , m are given functions, ϕ : [0, a] → Rn , ψ : [0, b] → Rn
−
are given absolutely continuous functions. Here u(x+
k , y) and u(xk , y) denote the
right and left limits of u(x, y) at x = xk , respectively.
In this article, we provide sufficient conditions for the existence of solutions for
the problem (1.1)-(1.3). Our approach is based on the existence of upper and
lower solutions and on a fixed point theorem for condensing multivalued maps, due
to Martelli [29]. The present results extend those considered with integer order
derivative [9, 11, 18, 19, 25, 30] and those with fractional derivative and without
impulses [21].
2. Preliminaries
In this section, we introduce notation and preliminary facts which are used
throughout this paper. By C(J) we denote the Banach space of all continuous
functions from J to Rn with the norm
kwk∞ = sup kw(x, y)k,
(x,y)∈J
where k · k denotes a suitable norm on Rn . As usual, by AC(J) we denote the
space of absolutely continuous functions from J into Rn and L1 (J) is the space of
Lebesgue-integrable functions w : J → Rn with the norm
Z aZ b
kwk1 =
kw(x, y)k dy dx.
0
0
Definition 2.1 ([34]). Let r = (r1 , r2 ) ∈ (0, ∞) × (0, ∞), θ = (0, 0) and u ∈ L1 (J).
The left-sided mixed Riemann-Liouville integral of order r of u is defined as
Z xZ y
1
(x − s)r1 −1 (y − t)r2 −1 u(s, t) dt ds.
(Iθr u)(x, y) =
Γ(r1 )Γ(r2 ) 0 0
In particular,
(Iθθ u)(x, y) = u(x, y), (Iθσ u)(x, y) =
Z
x
Z
y
u(s, t) dt ds; for almost all (x, y) ∈ J,
0
0
where σ = (1, 1) For instance, Iθr u exists for all r1 , r2 ∈ (0, ∞), when u ∈ L1 (J).
Note also that when u ∈ C(J), then (Iθr u) ∈ C(J), moreover
(Iθr u)(x, 0) = (Iθr u)(0, y) = 0;
x ∈ [0, a], y ∈ [0, b].
Example 2.2. Let λ, ω ∈ (−1, 0) ∪ (0, ∞) and r = (r1 , r2 ) ∈ (0, ∞) × (0, ∞), then
Iθr xλ y ω =
Γ(1 + λ)Γ(1 + ω)
xλ+r1 y ω+r2 ,
Γ(1 + λ + r1 )Γ(1 + ω + r2 )
for almost all (x, y) ∈ J.
2
By 1 − r we mean (1 − r1 , 1 − r2 ) ∈ [0, 1) × [0, 1). Denote by Dxy
:=
mixed second order partial derivative.
∂2
∂x∂y ,
the
EJDE-2014/196
EXISTENCE OF SOLUTIONS
3
Definition 2.3 ([34]). Let r ∈ (0, 1] × (0, 1] and u ∈ L1 (J). The Caputo fractionalorder derivative of order r of u is defined by the expression
c
2
Dθr u(x, y) = (Iθ1−r Dxy
u)(x, y)
1
=
Γ(1 − r1 )Γ(1 − r2 )
Z
x
Z
0
y
0
2
Dst
u(s, t)
dt ds.
r
(x − s) 1 (y − t)r2
The case σ = (1, 1) is included and we have
2
(Dθσ u)(x, y) = (c Dθσ u)(x, y) = (Dxy
u)(x, y),
for almost all (x, y) ∈ J.
Example 2.4. Let λ, ω ∈ (−1, 0) ∪ (0, ∞) and r = (r1 , r2 ) ∈ (0, 1] × (0, 1], then
Dθr xλ y ω =
Γ(1 + λ)Γ(1 + ω)
xλ−r1 y ω−r2 ,
Γ(1 + λ − r1 )Γ(1 + ω − r2 )
for almost all (x, y) ∈ J.
Let a1 ∈ [0, a], z + = (a1 , 0) ∈ J, Jz = [a1 , a] × [0, b], r1 , r2 > 0 and r = (r1 , r2 ).
For u ∈ L1 (Jz , Rn ), the expression
Z xZ y
1
r
(x − s)r1 −1 (y − t)r2 −1 u(s, t) dt ds,
(Iz+ u)(x, y) =
Γ(r1 )Γ(r2 ) a+
0
1
is called the left-sided mixed Riemann-Liouville integral of order r of u.
2
Definition 2.5 ([34]). For u ∈ L1 (Jz , Rn ) where Dxy
u is Lebesque integrable on
[xk , xk+1 ] × [0, b], k = 0, . . . , m, the Caputo fractional-order derivative of order r
2
of u is defined by the expression (c Dzr+ f )(x, y) = (Iz1−r
+ Dxy f )(x, y). The RiemannLiouville fractional-order derivative of order r of u is defined by (Dzr+ f )(x, y) =
2 1−r
Iz+ f )(x, y).
(Dxy
We need also some properties of set-valued Maps. Let (X, k · k) be a Banach
space. Denote P(X) = {Y ∈ X : Y 6= ∅}, Pcl (X) = {Y ∈ P(X) : Y closed},
Pb (X) = {Y ∈ P(X) : Y bounded}, Pcp (X) = {Y ∈ P(X) : Y compact} and
Pcp,cv (X) = {Y ∈ P(X) : Y compact and convex}.
Definition 2.6. A multivalued map T : X → P(X) is convex (closed) valued if
T (x) is convex (closed) for all x ∈ X. T is bounded on bounded sets if T (B) =
∪x∈B T (x) is bounded in X for all B ∈ Pb (X) (i.e. supx∈B supy∈T (x) kyk < ∞). T
is called upper semi-continuous (u.s.c.) on X if for each x0 ∈ X, the set T (x0 ) is
a nonempty closed subset of X, and if for each open set N of X containing T (x0 ),
there exists an open neighborhood N0 of x0 such that T (N0 ) ⊆ N . T is lower
semi-continuous (l.s.c.) if the set {x ∈ X : T (x) ∩ A 6= ∅} is open for any open
subset A ⊆ X. T is said to be completely continuous if T (B) is relatively compact
for every B ∈ Pb (X). T has a fixed point if there is x ∈ X such that x ∈ T (x).
The fixed point set of the multivalued operator T will be denoted by F ixT . A
multivalued map G : X → Pcl (Rn ) is said to be measurable if for every v ∈ Rn , the
function x 7→ d(v, G(x)) = inf{kv − zk : z ∈ G(x)} is measurable.
Lemma 2.7. [17] Let G be a completely continuous multivalued map with nonempty
compact values, then G is u.s.c. if and only if G has a closed graph (i.e. un → u,
wn → w, wn ∈ G(un ) imply w ∈ G(u)).
Definition 2.8. A multivalued map F : J ×Rn → P(Rn ) is said to be Carathéodory
if
(i) (x, y) 7→ F (x, y, u) is measurable for each u ∈ Rn ;
4
S. ABBAS, M. BENCHOHRA
EJDE-2014/196
(ii) u 7→ F (x, y, u) is upper semicontinuous for almost all (x, y) ∈ J.
F is said to be L1 -Carathéodory if (i), (ii) and the following condition holds;
(iii) for each c > 0, there exists σc ∈ L1 (J, R+ ) such that
kF (x, y, u)kP = sup{kf k : f ∈ F (x, y, u)}
≤ σc (x, y)
or all kuk ≤ c and for a.e. (x, y) ∈ J.
For each u ∈ C(J), define the set of selections of F by
SF,u = {w ∈ L1 (J) : w(x, y) ∈ F (x, y, u(x, y)) a.e. (x, y) ∈ J}.
Let (X, d) be a metric space induced from the normed space (X, k · k). Consider
Hd : P(X) × P(X) → R+ ∪ {∞} given by
Hd (A, B) = max{sup d(a, B), sup d(A, b)},
a∈A
b∈B
where d(A, b) = inf a∈A d(a, b), d(a, B) = inf b∈B d(a, b). Then (Pb,cl (X), Hd ) is a
metric space and (Pcl (X), Hd ) is a generalized metric space (see [23]). For more
details on multi-valued maps we refer the reader to the books of Deimling [12],
Gorniewicz [13], Graef et al [14], Hu and Papageorgiou [17] and Tolstonogov [33].
Lemma 2.9 ([27]). Let X be a Banach space. Let F : J × X → Pcp,cv (X) be an
L1 -Carathéodory multivalued map and let Λ be a linear continuous mapping from
L1 (J, X) to C(J, X), then the operator
Λ ◦ SF : C(J, X) → Pcp,cv (C(J, X)),
u 7→ (Λ ◦ SF )(u) := Λ(SF,u )
is a closed graph operator in C(J, X) × C(J, X).
Lemma 2.10 ([29]). (Martelli) Let X be a Banach space and N : X → Pcl,cv (X)
be an u. s. c. and condensing map. If the set Ω := {u ∈ X : λN (u) =
N (u) f or some λ > 1} is bounded, then N has a fixed point.
3. Main Result
To define the solutions of problems (1.1)-(1.3), we shall consider the Banach
space
P C = u : J → Rn : u ∈ C(Jk ); k = 0, . . . , m, and there exist u(x−
k , y)
+
−
and u(xk , y); y ∈ [0, b], k = 1, . . . , m, with u(xk , y) = u(xk , y) ,
with the norm
kukP C = sup ku(x, y)k.
(x,y)∈J
Definition 3.1. A function u ∈ P C ∩ ∪m
k=0 AC(Jk ) whose r-derivative exists on
Jk is said to be a solution of (1.1)-(1.3) if there exists a function f ∈ L1 (J) with
f (x, y) ∈ F (x, y, u(x, y)) such that u satisfies (c Dθrk u)(x, y) = f (x, y) on Jk , k =
0, . . . m and conditions (1.2), (1.3) are satisfied.
Let z, z̄ ∈ C(J) be such that
z(x, y) = (z1 (x, y), z2 (x, y), . . . , zn (x, y)),
(x, y) ∈ J,
z̄(x, y) = (z̄1 (x, y), z̄2 (x, y), . . . , z̄n (x, y)),
(x, y) ∈ J.
and
EJDE-2014/196
EXISTENCE OF SOLUTIONS
5
The notation z ≤ z̄ means that
zi (x, y) ≤ z̄i (x, y)
for i = 1, . . . , n.
Definition 3.2. A function z ∈ P C ∩ ∪m
k=0 AC(Jk ) is said to be a lower solution of
(1.1)-(1.3) if there exists a function f ∈ L1 (J) with f (x, y) ∈ F (x, y, u(x, y)) such
that z satisfies
(c Dθrk z)(x, y) ≤ f (x, y, z(x, y)),
z(x+
k , y)
≤
z(x−
k , y)
+
Ik (z(x−
k , y)),
on Jk ;
if y ∈ [0, b], k = 1, . . . , m;
z(x, 0) ≤ ϕ(x), x ∈ [0, a];
z(0, y) ≤ ψ(y),
y ∈ [0, b];
z(0, 0) ≤ ϕ(0).
The function z is said to be an upper solution of (1.1)-(1.3) if the reversed inequalities hold.
Let h ∈ C(Jk ), k = 1, . . . , m and set
µ(x, y) := ϕ(x) + ψ(y) − ϕ(0),
(x, y) ∈ J.
For the existence of solutions for problem (1.1)-(1.3), we need the following lemma.
Lemma 3.3 ([4]). Let r1 , r2 ∈ (0, 1] and let h : J → Rn be continuous. A function
u is a solution of the fractional integral equation

RxRy
1
µ(x, y) + Γ(r1 )Γ(r
(x − s)r1 −1 (y − t)r2 −1 h(s, t) dt ds;


0 0
2)


 if (x, y) ∈ [0, x1 ] × [0, b],




Pk

µ(x, y) + i=1 (Ii (u(x−
, y)) − Ii (u(x−
i , 0)))
u(x, y) =
Pk R xi iR y
1
r1 −1

(x
−
s)
(y − t)r2 −1 h(s, t) dt ds
+

i
i=1 xi−1 0
Γ(r1 )Γ(r2 )


R
R
x
y

1

(x − s)r1 −1 (y − t)r2 −1 h(s, t) dt ds;
+ Γ(r1 )Γ(r


xk 0
2)


if (x, y) ∈ (xk , xk+1 ] × [0, b], k = 1, . . . , m,
if and only if u is a solution of the fractional IVP
c
u(x+
k , y)
=
Dr u(x, y) = h(x, y),
u(x−
k , y)
+
Ik (u(x−
k , y)),
(x, y) ∈ Jk ,
y ∈ [0, b], k = 1, . . . , m.
To study problem (1.1)-(1.3), we first list the following hypotheses:
(H1) F : J × Rn → Pcp,cv (Rn ) is L1 -Carathéodory;
(H2) There exist v and w ∈ P C ∩ AC(Jk ), k = 0, . . . , m, lower and upper
solutions for the problem (1.1)-(1.3) such that v(x, y) ≤ w(x, y) for each
(x, y) ∈ J;
(H3) For each y ∈ [0, b], we have
v(x+
k , y) ≤
min
−
u∈[v(x−
k ,y),w(xk ,y)]
Ik (u) ≤
max
−
u∈[v(x−
k ,y),w(xk ,y)]
Ik (u) ≤ w(x+
k , y),
with k = 1, . . . , m.
Theorem 3.4. Assume that hypotheses (H1)-(H3) hold. Then problem (1.1)-(1.3)
has at least one solution u such that
v(x, y) ≤ u(x, y) ≤ w(x, y),
for all (x, y) ∈ J.
6
S. ABBAS, M. BENCHOHRA
EJDE-2014/196
Proof. We transform problem (1.1)-(1.3) into a fixed point problem. Consider the
modified problem
(c Dθrk u)(x, y) ∈ F (x, y, g(u(x, y))),
if (x, y) ∈ Jk , k = 0, . . . , m;
−
−
−
u(x+
k , y) = u(xk , y) + Ik (g(xk , y, u(xk , y))),
u(x, 0) = ϕ(x),
if y ∈ [0, b], k = 1, . . . , m;
x ∈ [0, a], u(0, y) = ψ(y) ; y ∈ [0, b], ϕ(0) = ψ(0),
where g : P C → P C be the truncation


v(x, y),
(gu)(x, y) = u(x, y),


w(x, y),
(3.1)
(3.2)
(3.3)
operator defined by
u(x, y) < v(x, y),
v(x, y) ≤ u(x, y) ≤ w(x, y),
w(x, y) < u(x, y).
A solution to (3.1)-(3.3) is a fixed point of the operator N : P C → P(P C) defined
by


h ∈ P C : h(x, y) = µ(x, y)



−
−
−
−
+ P
0<xk <x (Ik (g(xk , Ry, u(xRk , y))) − Ik (g(xk , 0, u(xk , 0))))
P
N (u) =
x
y
k
1
r1 −1
+ Γ(r1 )Γ(r
(y − t)r2 −1 f (s, t) dt ds

0<xk <x xk−1 0 (xk − s)
2)

R
R

x
y
1
+
(x − s)r1 −1 (y − t)r2 −1 f (s, t) dt ds,
Γ(r1 )Γ(r2 )
xk
0
where
1
f ∈ S̃F,g(u)
1
= f ∈ SF,g(u)
: f (x, y) ≥ f1 (x, y) on A1 and f (x, y) ≤ f2 (x, y) on A2 ,
A1 = {(x, y) ∈ J : u(x, y) < v(x, y) ≤ w(x, y)},
A2 = {(x, y) ∈ J : u(x, y) ≤ w(x, y) < u(x, y)},
1
SF,g(u)
= {f ∈ L1 (J) : f (x, y) ∈ F (x, y, g(u(x, y))), for (x, y) ∈ J}.
Remark 3.5. (A) For each u ∈ P C, the set S̃F,g(u) is nonempty. In fact, (H1)
implies there exists f3 ∈ SF,g(u) , so we set
f = f1 χA1 + f2 χA2 + f3 χA3 ,
where χAi is the characteristic function of Ai ; i = 1, 2, 3 and
A3 = {(x, y) ∈ J : v(x, y) ≤ u(x, y) ≤ w(x, y)}.
Then, by decomposability, f ∈ S̃F,g(u) .
(B) By the definition of g it is clear that F (., ., g(u)(., .)) is an L1 -Carathéodory
multi-valued map with compact convex values and there exists φ ∈ C(J, R+ ) such
that
kF (x, y, g(u(x, y)))kP ≤ φ(x, y); for each (x, y) ∈ Jand u ∈ Rn .
Set
φ∗ := sup φ(x, y).
(x,y)∈J
(C) By the definition of g and from (H3) we have
+
u(x+
k , y) ≤ Ik (g(xk , y, u(xk , y))) ≤ w(xk , y); y ∈ [0, b]; k = 1, . . . , m.
EJDE-2014/196
EXISTENCE OF SOLUTIONS
7
From Lemma 3.3 and the fact that g(u) = u for all v ≤ u ≤ w, the problem of
finding the solutions of the IVP (1.1)-(1.3) is reduced to finding the solutions of
the operator equation N (u) = u. We shall show that N is a completely continuous
multivalued map, u.s.c. with convex closed values. The proof will be given in
several steps.
Step 1: N (u) is convex for each u ∈ P C. If h1 , h2 belong to N (u), then there exist
1
f1 , f2 ∈ S̃F,g(u)
such that for each (x, y) ∈ J we have
X
−
−
−
(Ik (g(x−
(hi u)(x, y) = µ(x, y) +
k , y, u(xk , y))) − Ik (g(xk , 0, u(xk , 0))))
0<xk <x
X Z xk Z y
1
(xk − s)r1 −1 (y − t)r2 −1 fi (s, t) dt ds
+
Γ(r1 )Γ(r2 ) 0<x <x xk−1 0
k
Z xZ y
1
(x − s)r1 −1 (y − t)r2 −1 fi (s, t) dt ds.
+
Γ(r1 )Γ(r2 ) xk 0
Let 0 ≤ ξ ≤ 1. Then, for each (x, y) ∈ J, we have
(ξh1 + (1 − ξ)h2 )(x, y)
X
X
−
−
= µ(x, y) +
(Ik (g(x−
(Ik (g(x−
k , y, u(xk , y)))) −
k , 0, u(xk , 0))))
0<xk <x
+
0<xk <x
X Z
1
Γ(r1 )Γ(r2 ) 0<x
k <x
xk
Z
xk−1
y
(xk − s)r1 −1 (y − t)r2 −1
0
× [ξf1 (s, t) + (1 − ξ)f2 (s, t)] dt ds
Z xZ y
1
(x − s)r1 −1 (y − t)r2 −1 [ξf1 (s, t) + (1 − ξ)f2 (s, t)] dt ds.
+
Γ(r1 )Γ(r2 ) xk 0
1
Since S̃F,g(u)
is convex (because F has convex values), we have
ξh1 + (1 − ξ)h2 ∈ G(u).
Step 2: N sends bounded sets of P C into bounded sets. We can prove that N (P C)
is bounded. It is sufficient to show that there exists a positive constant ` such that
for each h ∈ N (u), u ∈ P C one has khk∞ ≤ `. If h ∈ N (u), then there exists
1
f ∈ S̃F,g(u)
such that for each (x, y) ∈ J we have
X
−
−
−
(hu)(x, y) = µ(x, y) +
(Ik (g(x−
k , y, u(xk , y))) − Ik (g(xk , 0, u(xk , 0))))
0<xk <x
X Z xk Z y
1
(xk − s)r1 −1 (y − t)r2 −1 f (s, t) dt ds
+
Γ(r1 )Γ(r2 ) 0<x <x xk−1 0
k
Z xZ y
1
+
(x − s)r1 −1 (y − t)r2 −1 f (s, t) dt ds.
Γ(r1 )Γ(r2 ) xk 0
Then, for each (x, y) ∈ J we get
k(hu)(x, y)k = kµ(x, y)k + 2
m
X
k=1
+
max (kv(x+
k , y)k, kw(xk , y)k)
y∈[0,b]
8
S. ABBAS, M. BENCHOHRA
EJDE-2014/196
X Z xk Z y
φ∗
(xk − s)r1 −1 (y − t)r2 −1 dt ds
Γ(r1 )Γ(r2 ) 0<x <x xk−1 0
k
Z xZ y
φ∗
(x − s)r1 −1 (y − t)r2 −1 dt ds.
+
Γ(r1 )Γ(r2 ) xk 0
+
Thus,
kuk∞ ≤ kµk∞ + 2
m
X
k=1
+
max (kv(x+
k , y)k, kw(xk , y)k) +
y∈[0,b]
2ar1 br2 φ∗
:= `.
Γ(r1 + 1)Γ(r2 + 1)
Step 3: N sends bounded sets of P C into equicontinuous sets. Let (τ1 , y1 ),
(τ2 , y2 ) ∈ J, τ1 < τ2 , y1 < y2 and Bρ = {u ∈ P C : kuk∞ ≤ ρ} be a bonded
1
set of P C. For each u ∈ Bρ and h ∈ N (u), there exists f ∈ S̃F,g(u)
such that for
each (x, y) ∈ J we have
k(hu)(τ2 , y2 ) − h(u)(τ1 , y1 )k
m
X
≤ kµ(τ1 , y1 ) − µ(τ2 , y2 )k +
−
(kIk (g(x−
k , y1 , u(xk , y1 )))
k=1
−
− Ik (g(x−
k , y2 , u(xk , y2 )))k)
m Z xk Z y1
X
1
+
(xk − s)r1 −1 [(y2 − t)r2 −1 − (y1 − t)r2 −1 ]
Γ(r1 )Γ(r2 )
xk−1 0
k=1
× kf (s, t)k dt ds
m Z xk Z y2
X
1
+
(xk − s)r1 −1 (y2 − t)r2 −1 kf (s, t)k dt ds
Γ(r1 )Γ(r2 )
x
y
1
k−1
k=1
Z τ1 Z y1
1
[(τ2 − s)r1 −1 (y2 − t)r2 −1 − (τ1 − s)r1 −1 (y1 − t)r2 −1 ]
+
Γ(r1 )Γ(r2 ) 0
0
× kf (s, t)k dt ds
Z τ2 Z y2
1
+
(τ2 − s)r1 −1 (y2 − t)r2 −1 kf (s, t)k dt ds
Γ(r1 )Γ(r2 ) τ1 y1
Z τ1 Z y2
1
+
(τ2 − s)r1 −1 (y2 − t)r2 −1 kf (s, t)k dt ds
Γ(r1 )Γ(r2 ) 0
y
Z τ2 Z 1y1
1
+
(τ2 − s)r1 −1 (y2 − t)r2 −1 kf (s, t)k dt ds
Γ(r1 )Γ(r2 ) τ1 0
≤ kµ(τ1 , y1 ) − µ(τ2 , y2 )k
+
m
X
−
−
−
(kIk (g(x−
k , y1 , u(xk , y1 ))) − Ik (g(xk , y2 , u(xk , y2 )))k)
k=1
m
+
X
φ∗
Γ(r1 )Γ(r2 )
∗
+
φ
Γ(r1 )Γ(r2 )
Z
xk
k=1 xk−1
m Z xk
X
k=1
xk−1
Z
y1
(xk − s)r1 −1 [(y2 − t)r2 −1 − (y1 − t)r2 −1 ] dt ds
0
Z
y2
y1
(xk − s)r1 −1 (y2 − t)r2 −1 dt ds
EJDE-2014/196
EXISTENCE OF SOLUTIONS
9
Z τ1 Z y1
φ∗
[(τ2 − s)r1 −1 (y2 − t)r2 −1 − (τ1 − s)r1 −1 (y1 − t)r2 −1 ] dt ds
Γ(r1 )Γ(r2 ) 0
0
Z τ2 Z y2
φ∗
+
(τ2 − s)r1 −1 (y2 − t)r2 −1 dt ds
Γ(r1 )Γ(r2 ) τ1 y1
Z τ1 Z y2
φ∗
(τ2 − s)r1 −1 (y2 − t)r2 −1 dt ds
+
Γ(r1 )Γ(r2 ) 0
y1
Z τ2 Z y1
φ∗
+
(τ2 − s)r1 −1 (y2 − t)r2 −1 dt ds.
Γ(r1 )Γ(r2 ) τ1 0
+
As τ1 → τ2 and y1 → y2 , the right-hand side of the above inequality tends to zero.
As a consequence of Steps 1 to 3 together with the Arzelá-Ascoli theorem, we can
conclude that N is completely continuous and therefore a condensing multivalued
map.
Step 4: N has a closed graph. Let un → u∗ , hn ∈ N (un ) and hn → h∗ . We need
1
to show that h∗ ∈ N (u∗ ). hn ∈ N (un ) means that there exists fn ∈ S̃F,g(u
such
n)
that, for each (x, y) ∈ J, we have
X
−
−
−
hn (x, y) = µ(x, y) +
(Ik (g(x−
k , y, un (xk , y))) − Ik (g(xk , 0, un (xk , 0))))
0<xk <x
X Z xk Z y
1
+
(xk − s)r1 −1 (y − t)r2 −1 fn (s, t) dt ds
Γ(r1 )Γ(r2 ) 0<x <x xk−1 0
k
Z xZ y
1
+
(x − s)r1 −1 (y − t)r2 −1 fn (s, t) dt ds.
Γ(r1 )Γ(r2 ) xk 0
1
We must show that there exists f∗ ∈ S̃F,g(u
such that, for each (x, y) ∈ J,
∗)
X
−
−
−
h∗ (x, y) = µ(x, y) +
(Ik (g(x−
k , y, u∗ (xk , y))) − Ik (g(xk , 0, u∗ (xk , 0))))
0<xk <x
X Z xk Z y
1
+
(xk − s)r1 −1 (y − t)r2 −1 f∗ (s, t) dt ds
Γ(r1 )Γ(r2 ) 0<x <x xk−1 0
k
Z xZ y
1
(x − s)r1 −1 (y − t)r2 −1 f∗ (s, t) dt ds.
+
Γ(r1 )Γ(r2 ) xk 0
Now, we consider the linear continuous operator Λ : L1 (J) → C(J) defined by
f 7→ Λ(f )(x, y),
X
−
−
−
(Λf )(x, y) =
(Ik (g(x−
k , y, u(xk , y))) − Ik (g(xk , 0, u(xk , 0))))
0<xk <x
X Z xk Z y
1
(xk − s)r1 −1 (y − t)r2 −1 f (s, t) dt ds
Γ(r1 )Γ(r2 ) 0<x <x xk−1 0
k
Z xZ y
1
+
(x − s)r1 −1 (y − t)r2 −1 f (s, t) dt ds.
Γ(r1 )Γ(r2 ) xk 0
+
From Lemma 2.9, it follows that Λ ◦ S̃F1 is a closed graph operator. Clearly we have
h
i
X
−
−
−
(Ik (g(x−
hn (x, y) − µ(x, y) −
k , y, un (xk , y))) − Ik (g(xk , 0, un (xk , 0))))
0<xk <x
10
S. ABBAS, M. BENCHOHRA
EJDE-2014/196
h
i
X
−
−
−
− h∗ (x, y) − µ(x, y) −
(Ik (g(x−
k , y, u∗ (xk , y))) − Ik (g(xk , 0, u∗ (xk , 0)))) ≤
1
Γ(r1 )Γ(r2 ) x
+
0<xk <x
xk Z y
Z
X
xk−1
1 <xk <x
1
Γ(r1 )Γ(r2 )
Z
x
xk
Z
(xk − s)r1 −1 (y − t)r2 −1 kfn (s, t) − f∗ (s, t)k dt ds
0
y
(x − s)r1 −1 (y − t)r2 −1 kfn (s, t) − f∗ (s, t)k dt ds → 0,
0
as n → ∞. Moreover, from the definition of Λ, we have
h
i
X
−
−
−
hn (x, y) − µ(x, y) −
(Ik (g(x−
k , y, un (xk , y))) − Ik (g(xk , 0, un (xk , 0))))
0<xk <x
1
).
∈ Λ(S̃F,g(u
n)
1
Since un → u∗ , it follows from Lemma 2.9 that, for some f∗ ∈ Λ(S̃F,g(u
), we have
∗)
X
−
−
−
h∗ (x, y)µ (x, y) −
(Ik (g(x−
k , y, u∗ (xk , y))) − Ik (g(xk , 0, u∗ (xk , 0))))
0<xk <x
X Z xk Z y
1
=
(xk − s)r1 −1 (y − t)r2 −1 f∗ (s, t) dt ds
Γ(r1 )Γ(r2 ) 0<x <x xk−1 0
k
Z xZ y
1
(x − s)r1 −1 (y − t)r2 −1 f∗ (s, t) dt ds, (x, y) ∈ J.
+
Γ(r1 )Γ(r2 ) xk 0
From Lemma 2.7, we can conclude that N is u.s.c.
Step 5: The set Ω = {u ∈ P C : λu = N (u) for some λ > 1} in bounded. Let
1
u ∈ Ω. Then, there exists f ∈ Λ(S̃F,g(u)
), such that
X
−
−
−
λu(x, y) = µ(x, y) +
(Ik (g(x−
k , y, u(xk , y))) − Ik (g(xk , 0, u(xk , 0))))
0<xk <x
X Z xk Z y
1
(xk − s)r1 −1 (y − t)r2 −1 f (s, t) dt ds
Γ(r1 )Γ(r2 ) 0<x <x xk−1 0
k
Z xZ y
1
+
(x − s)r1 −1 (y − t)r2 −1 f (s, t) dt ds.
Γ(r1 )Γ(r2 ) xk 0
+
As in Step 2, this implies that for each (x, y) ∈ J, we have
kuk∞ ≤ kµk∞ + 2
m
X
k=1
+
max (kv(x+
k , y)k, kw(xk , y)k) +
y∈[0,b]
2ar1 br2 φ∗
= `.
Γ(r1 + 1)Γ(r2 + 1)
This shows that Ω is bounded. As a consequence of Lemma 2.10, we deduce that
N has a fixed point which is a solution of (3.1)-(3.3) on J.
Step 6: The solution u of (3.1)-(3.3) satisfies
v(x, y) ≤ u(x, y) ≤ w(x, y),
for all (x, y) ∈ J.
Let u be the above solution to (3.1)-(3.3). We prove that
u(x, y) ≤ w(x, y)
for all (x, y) ∈ J.
−
Assume that u − w attains a positive maximum on [x+
k , xk+1 ] × [0, b] at (xk , y) ∈
+
−
[xk , xk+1 ] × [0, b], for some k = 0, . . . , m; that is,
−
(u − w)(xk , y) = max{u(x, y) − w(x, y) : (x, y) ∈ [x+
k , xk+1 ] × [0, b]} > 0,
EJDE-2014/196
EXISTENCE OF SOLUTIONS
11
for some k = 0, . . . , m. We distinguish the following cases.
−
+
−
∗
∗
Case 1. If (xk , y) ∈ (x+
k , xk+1 ) × [0, b] there exists (xk , y ) ∈ (xk , xk+1 ) × [0, b]
such that
[u(x, y ∗ ) − w(x, y ∗ )] + [u(x∗k , y) − w(x∗k , y)] − [u(x∗k , y ∗ ) − w(x∗k , y ∗ )]
(3.4)
≤ 0, for all (x, y) ∈ ([x∗k , xk ] × {y ∗ }) ∪ ({x∗k } × [y ∗ , b]),
and
u(x, y) − w(x, y) > 0,
By the definition of g, one has
c
An
for all (x, y) ∈ (x∗k , xk ] × (y ∗ , b].
Dθr u(x, y) ∈ F (x, y, w(x, y)),
for all (x, y) ∈ [x∗k , xk ] × [y ∗ , b].
integration of (3.6), on [x∗k , x] × [y ∗ , y] for each (x, y) ∈
u(x, y) + u(x∗k , y ∗ ) − u(x, y ∗ ) − u(x∗k , y)
Z xZ y
1
r1 −1
r2 −1
=
Γ(r1 )Γ(r2 )
(x − s)
x∗
k
(3.5)
(y − t)
[x∗k , xk ]
(3.6)
∗
× [y , b], yields
f (s, t) dt ds,
(3.7)
y∗
where f (x, y) ∈ F (x, y, w(x, y)). From (3.7) and using the fact that w is an upper
solution to (1.1)-(1.3) we get
u(x, y) + u(x∗k , y ∗ ) − u(x, y ∗ ) − u(x∗k , y) ≤ w(x, y) + w(x∗k , y ∗ ) − w(x, y ∗ ) − w(x∗k , y),
which gives
u(x, y) − w(x, y)
≤ [u(x, y ∗ ) − w(x, y ∗ )] + [u(x∗k , y) − w(x∗k , y)] − [u(x∗k , y ∗ ) − w(x∗k , y ∗ )].
(3.8)
Thus from (3.4), (3.5) and (3.8) we obtain the contradiction
0 < [u(x, y) − w(x, y)] ≤ [u(x, y ∗ ) − w(x, y ∗ )] + [u(x∗k , y) − w(x∗k , y)]
− [u(x∗k , y ∗ ) − w(x∗k , y ∗ )] ≤ 0,
for all (x, y) ∈ [x∗k , xk ] × [y ∗ , b].
Case 2. If xk = x+
k , k = 1, . . . , m, then
−
−
+
w(x+
k , y) < Ik (g(xk , u(xk , y))) ≤ w(xk , y),
which is a contradiction. Thus
u(x, y) ≤ w(x, y),
for all (x, y) ∈ J.
Analogously, we can prove that
u(x, y) ≥ v(x, y),
for all (x, y) ∈ J.
This shows that problem (3.1)-(3.3) has a solution u satisfying v ≤ u ≤ w which is
solution of (1.1)-(1.3).
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Saı̈d Abbas
Laboratory of Mathematics, University of Saı̈da, PO Box 138, 20000 Saı̈da, Algeria
E-mail address: abbasmsaid@yahoo.fr
Mouffak Benchohra
Laboratory of Mathematics, University of Sidi Bel-Abbès, PO Box 89, 22000, Sidi BelAbbès, Algeria
Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box
80203, Jeddah 21589, Saudi Arabia
E-mail address: benchohra@univ-sba.dz