Electronic Journal of Differential Equations, Vol. 2010(2010), No. 91, pp. 1–17.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
IMPULSIVE DISCONTINUOUS HYPERBOLIC PARTIAL
DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER ON
BANACH ALGEBRAS
SAÏD ABBAS, RAVI P. AGARWAL, MOUFFAK BENCHOHRA
Abstract. This article studies the existence of solutions and extremal solutions to partial hyperbolic differential equations of fractional order with impulses in Banach algebras under Lipschitz and Carathéodory conditions and
certain monotonicity conditions.
1. Introduction
This article studies the existence of solutions to fractional order initial-value
problems (IV P for short), for the system
u(x, y)
c r
D0
= g(x, y, u(x, y)), (x, y) ∈ J, x 6= xk , k = 1, . . . , m, (1.1)
f (x, y, u(x, y))
−
−
u(x+
k , y) = u(xk , y) + Ik (u(xk , y)),
u(x, 0) = ϕ(x),
u(0, y) = ψ(y),
y ∈ [0, b]; k = 1, . . . , m,
(1.2)
x ∈ [0, a], y ∈ [0, b],
(1.3)
where J = [0, a] × [0, b], a, b > 0, c D0r is the Caputo’s fractional derivative of
order r = (r1 , r2 ) ∈ (0, 1] × (0, 1], 0 = x0 < x1 < · · · < xm < xm+1 = a, f :
J × Rn → Rn \{0Rn }, g : J × Rn → Rn and Ik : Rn → Rn , k = 1, . . . , m are given
functions satisfying suitable conditions and ϕ : [0, a] → Rn , ψ : [0, b] → Rn are
given absolutely continuous functions with ϕ(0) = ψ(0).
There has been a significant development in ordinary and partial fractional differential equations in recent years; see the monographs of Kilbas [20], Lakshmikantham et al. [22], Podlubny [24], Samko [25], the papers by Abbas and Benchohra
[1, 2, 3], Agarwal et al. [5], Belarbi et al. [7], Benchohra et al. [8, 9, 10], Diethelm [16], Vityuk and Golushkov [27] and the references therein. We can find
numerous applications of differential equations of fractional order in viscoelasticity,
electrochemistry, control, porous media, electromagnetic, etc. (see [18, 23]).
The theory of impulsive differential equations have become important in some
mathematical models of real processes and phenomena studied in physics, chemical
technology, population dynamics, biotechnology and economics. There has been a
2000 Mathematics Subject Classification. 26A33.
Key words and phrases. Impulsive differential equations; fractional order; upper solution;
lower solution; extremal solutions; left-sided mixed Riemann-Liouville integral;
Caputo fractional-order derivative; fixed point; Banach algebra.
c
2010
Texas State University - San Marcos.
Submitted May 6, 2010. Published July 7, 2010.
1
2
S. ABBAS, R. P. AGARWAL, M. BENCHOHRA
EJDE-2010/91
significant development in impulse theory in recent years, especially in the area of
impulsive differential equations and inclusions with fixed moments; see the monographs of Benchohra et al. [11], Lakshmikantham et al. [21], and Samoilenko and
Perestyuk [26], and the references therein.
In this article, we prove the existence of extremal solutions under discontinuous
nonlinearity under certain Lipschitz and monotonicity conditions. These results
extend to the Banach algebra setting those considered with integer order derivative
[12, 19] and those with fractional derivative [4]. Also, we extend some results
considered on Banach algebras with integer order derivative and without impulses
[6, 15]. Finally, an example illustrating the abstract results is presented in the last
Section.
This paper initiates the study of fractional hyperbolic differential equations with
impulses on Banach algebras.
2. Preliminaries
In this section, we introduce notation, definitions, and preliminary facts which
are used throughout this paper. Let L∞ (J, Rn ) be the Banach space of measurable
functions u : J → Rn which are bounded, equipped with the norm
kukL∞ = inf{c > 0 : ku(x, y)k ≤ c, a.e. (x, y) ∈ J},
where k · k denotes a suitable complete norm on Rn . By L1 (J, Rn ) we denote the
space of Lebesgue-integrable functions u : J → Rn with the norm
Z aZ b
kuk1 =
ku(x, y)k dy dx.
0
0
AC(J, Rn ) is the space of absolutely continuous valued functions on J. Denote by
∂2
2
the mixed second order partial derivative. In all what follows set
Dxy
:= ∂x∂y
Jk := (xk , xk+1 ] × [0, b],
k = 0, 1, . . . , m.
Consider the space
P C(J, Rn ) = u : J → Rn : u ∈ C(Jk , Rn ); k = 1, . . . , m, and there exist
+
−
u(x−
k , y), u(xk , y); k = 1, . . . , m, with u(xk , y) = u(xk , y) .
This set is a Banach space with the norm
kukP C = sup ku(x, y)k.
(x,y)∈J
Define a multiplication “ · ” by
(u · v)(x, y) = u(x, y)v(x, y)
for (x, y) ∈ J.
Then P C(J, Rn ) is a Banach algebra with the above norm and multiplication.
Let a1 ∈ [0, a], z + = (a+
1 , 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
(Iz+ u)(x, y) =
(x − s)r1 −1 (y − t)r2 −1 u(s, t) dt ds,
Γ(r1 )Γ(r2 ) a+
0
1
where Γ(.) is the Euler gamma function, is called the left-sided mixed RiemannLiouville integral of order r.
EJDE-2010/91
FRACTIONAL-ORDER IMPULSIVE PDES
3
Definition 2.1 ([27]). For u ∈ L1 (Jz , Rn ), the Caputo fractional-order derivative
2
of order r is defined by the expression (c Dzr+ u)(x, y) = (Iz1−r
+ Dxy u)(x, y).
Let X be a Banach algebra with norm k · k. An operator T : X → X is called
compact if T (S) is a compact subset of X for any S ⊂ X. Similarly T : X → X is
called totally bounded if T maps a bounded subset of X into the relatively compact
subset of X. Finally T : X → X is called completely continuous operator if it
is continuous and totally bounded operator on X. It is clear that every compact
operator is totally bounded, but the converse may not be true.
Definition 2.2. A function γ : J × Rn → Rn is called Carathéodory’s if
(i) the function (x, y) → γ(x, y, u) is measurable for each u ∈ Rn ,
(ii) the function u → γ(x, y, u) is continuous for almost each (x, y) ∈ J.
A non-empty closed set K in a Banach algebra X is called a cone if
(i) K + K ⊆ K,
(ii) λK ⊆ K for λ ∈ R, λ ≥ 0 and
(iii) {−K} ∩ K = 0, where 0 is the zero element of X.
The cone K is called to be positive if
(iv) K ◦ K ⊆ K, where “◦” is a multiplication composition in X.
We introduce an order relation, ≤, in X as follows. Let u, v ∈ X. Then u ≤ v
if and only if v − u ∈ K. A cone K is called to be normal if the norm k · k is
monotone increasing on K. It is known that if the cone K is normal in X, then
every order-bounded set in X is norm-bounded.
Lemma 2.3 ([14]). Let K be a positive cone in a real Banach algebra X and let
u1 , u2 , v1 , v2 ∈ K be such that u1 ≤ v1 and u2 ≤ v2 . Then u1 u2 ≤ v1 v2 .
For any v, w ∈ X, v ≤ w, the order interval [v, w] is a set in X given by
[v, w] = {u ∈ X : v ≤ u ≤ w}.
The nonlinear alternative of Schaefer type proved by Dhage [13] is embodied in
the following theorem.
Theorem 2.4. Let X be a Banach algebra and let A, B : X → X be two operators
satisfying
(a) A is Lipschitz with a Lipschitz constant α,
(b) B is compact and continuous, and
(c) αM < 1, where M = kB(X)k := sup{kBuk : u ∈ X}.
Then either
(i) the equation λ[Au Bu] = u has a solution for 0 < λ < 1, or
(ii) the set E = {u ∈ X : λ[Au Bu] = u, 0 < λ < 1} is unbounded.
We use the following fixed point theorems by Dhage [14] for proving the existence
of extremal solutions for our problem under certain monotonicity conditions.
Theorem 2.5. Let K be a cone in a Banach algebra X and let v, w ∈ X. Suppose
that A, B : [v, w] → K are two operators such that
(a) A is completely continuous,
(b) B is totally bounded,
(c) Au1 Bu2 ∈ [v, w] for all u1 , u2 ∈ [v, w], and
4
S. ABBAS, R. P. AGARWAL, M. BENCHOHRA
EJDE-2010/91
(d) A and B are nondecreasing.
Further if the cone K is positive and normal, then the operator equation Au Bu = u
has a least and a greatest positive solution in [v, w].
Theorem 2.6. Let K be a cone in a Banach algebra X and let v, w ∈ X. Suppose
that A, B : [v, w] → K are two operators such that
(a)
(b)
(c)
(d)
A is Lipschitz with a Lipschitz constant α,
B is totally bounded,
Au1 Bu2 ∈ [v, w] for all u1 , u2 ∈ [v, w], and
A and B are nondecreasing.
Further if the cone K is positive and normal, then the operator equation Au Bu = u
has least and a greatest positive solution in [v, w], whenever αM < 1, where M =
kB([v, w])k := sup{kBuk : u ∈ [v, w]}.
Remark 2.7. Note that hypothesis (c) of Theorems 2.5 and 2.6 holds if the operators A and B are positive monotone increasing and there exist elements v and w
in X such that v ≤ Av Bv and Aw Bw ≤ w.
3. Auxiliary Results
Let us start by defining what we mean by a solution of problem (1.1)-(1.3). Set
J 0 := J\{(x1 , y), . . . , (xm , y), y ∈ [0, b]}.
Definition 3.1. A function u ∈ P C(J, Rn ) whose r-derivative exists on J 0 is said
to be a solution of (1.1)-(1.3) if
u(x,y)
f (x,y,u(x,y)) is absolutely
u(x,y)
c r
D0 f (x,y,u(x,y))
= g(x, y, u(x, y))
(i) the function (x, y) 7→
continuous, and
(ii) u satisfies
(1.3) are satisfied.
on J 0 and conditions (1.2),
Let f ∈ C([xk , xk+1 ] × [0, b], Rn \{0Rn }), g ∈ L1 ([xk , xk+1 ] × [0, b], Rn ), zk =
(xk , 0), and
µ0,k (x, y) =
u(x+
u(x, 0) u(x+
k , 0)
k , y)
+
−
,
f (x, 0) f (x+
f (x+
k , y)
k , 0)
k = 0, . . . , m.
For the existence of solutions for the problem (1.1)-(1.3), we need the following
lemma.
Lemma 3.2. A function u ∈ AC([xk , xk+1 ] × [0, b], Rn ), k = 0, . . . , m is a solution
of the differential equation
u
c r
(x, y) = g(x, y), (x, y) ∈ [xk , xk+1 ] × [0, b],
(3.1)
Dzk
f
if and only if u(x, y) satisfies
u(x, y) = f (x, y) µ0,k (x, y) + (Izrk g)(x, y) ,
(x, y) ∈ [xk , xk+1 ] × [0, b].
(3.2)
Proof. Let u(x, y) be a solution of (3.1). Then, taking into account the definition
of the derivative c Dzrk , we have
2 u
Iz1−r
)(x, y) = g(x, y).
+ (Dxy
f
k
EJDE-2010/91
Hence, we obtain
then
Since
FRACTIONAL-ORDER IMPULSIVE PDES
5
2 u
Izr+ (Iz1−r
Dxy
)(x, y) = (Izr+ g)(x, y),
k
k
k
f
2 u
Iz1+ (Dxy
)(x, y) = (Izr+ g)(x, y).
k
k
f
u(x+
u(x, y) u(x, 0) u(x+
2 u
k , y)
k , 0)
Iz1+ (Dxy
)(x, y) =
−
−
+
,
+
k
f
f (x, y) f (x, 0) f (xk , y) f (x+
k , 0)
we have
u(x, y) = f (x, y) µ0,k (x, y) + (Izrk g)(x, y) .
Now let u(x, y) satisfies (3.2). It is clear that u(x, y) satisfies (3.1).
Corollary 3.3. The function u ∈ AC([xk , xk+1 ] × [0, b], Rn ), k = 0, . . . , m is a
solution of the differential equation (1.1) if and only if u satisfies the equation
u(x, y) = f (x, y, u(x, y)) µk (x, y)
Z xZ y
(3.3)
1
(x − s)r1 −1 (y − t)r2 −1 g(s, t, u(s, t)) dt ds ,
+
Γ(r1 )Γ(r2 ) 0 0
for (x, y) ∈ [xk , xk+1 ] × [0, b], where
µk (x, y) =
u(x+
u(x+
u(x, 0)
k , 0)
k , y)
+
−
,
+
+
+
f (x, 0, u(x, 0)) f (xk , y, u(xk , y)) f (xk , 0, u(x+
k , 0))
k = 0, . . . , m.
Let µ0 := µ0,0 .
Lemma 3.4. Let 0 < r1 , r2 ≤ 1 and let f : J → Rn \{0Rn }, g : J → Rn be
continuous. A function u is a solution of the fractional integral equation

h
i
RxRy
1

f (x, y) µ0 (x, y) + Γ(r1 )Γ(r
(x − s)r1 −1 (y − t)r2 −1 g(s, t) dt ds ,

)
0
0

2




if (x, y) ∈ [0, x1 ] × [0, b],



h

Pk I (u(x− ,y)) I (u(x− ,0)) 
f (x, y) µ0 (x, y) + i=1 i f (x+i,y) − i f (x+i,0)
u(x, y) =
i
i
Pk R xi R y

1
r1 −1
r2 −1

+
(x
−
s)
(y
−
t)
g(s, t) dt ds
i

i=1
Γ(r
)Γ(r
)
x
0
1
2
i−1

i

R
R

x
y
1
r
−1
r
−1

1
2

+ Γ(r1 )Γ(r2 ) xk 0 (x − s)
(y − t)
g(s, t) dt ds ,



 if (x, y) ∈ (x , x
] × [0, b], k = 1, . . . , m,
k
k+1
(3.4)
if and only if u is a solution of the fractional initial-value problem
c r u
D ( )(x, y) = g(x, y), (x, y) ∈ J 0 ,
f
+
−
u(xk , y) = u(xk , y) + Ik (u(x−
k , y)), y ∈ [0, b], k = 1, . . . , m.
(3.5)
(3.6)
Proof. Assume u satisfies (3.5)-(3.6). If (x, y) ∈ [0, x1 ] × [0, b], then
c r u
D ( )(x, y) = g(x, y).
f
Lemma 3.2 implies
Z xZ y
1
0
u(x, y) = f (x, y) µ (x, y) +
(x − s)r1 −1 (y − t)r2 −1 h(s, t) dt ds .
Γ(r1 )Γ(r2 ) 0 0
6
S. ABBAS, R. P. AGARWAL, M. BENCHOHRA
EJDE-2010/91
If (x, y) ∈ (x1 , x2 ] × [0, b], then Lemma 3.2 implies
u(x, y)
= f (x, y) µ0,1 (x, y) +
Z
1
Γ(r1 )Γ(r2 )
x
y
Z
(x − s)r1 −1 (y − t)r2 −1 g(s, t) dt ds
0
x1
ϕ(x)
u(x+
u(x+
1 , y)
1 , 0)
= f (x, y)
+
−
+
f (x, 0) f (x1 , y) f (x+
1 , 0)
Z xZ y
1
(x − s)r1 −1 (y − t)r2 −1 g(s, t) dt ds
+
Γ(r1 )Γ(r2 ) x1 0
ϕ(x)
u(x−
u(x−
I1 (u(x−
I1 (u(x−
1 , y)
1 , 0)
1 , y))
1 , 0))
= f (x, y)
+
−
+
−
+
+
+
+
f (x, 0) f (x1 , y) f (x1 , 0)
f (x1 , y)
f (x1 , 0)
Z xZ y
1
(x − s)r1 −1 (y − t)r2 −1 g(s, t) dt ds
+
Γ(r1 )Γ(r2 ) x1 0
ϕ(x)
u(x1 , y)
u(x1 , 0)
I1 (u(x−
I1 (u(x−
1 , y))
1 , 0))
= f (x, y)
+
−
+
−
+
+
+
f (x, 0) f (x1 , y) f (x1 , 0)
f (x1 , y)
f (x+
1 , 0)
Z xZ y
1
(x − s)r1 −1 (y − t)r2 −1 g(s, t) dt ds
+
Γ(r1 )Γ(r2 ) x1 0
ϕ(x)
ψ(y)
u(0, 0) I1 (u(x−
I1 (u(x−
1 , y))
1 , 0))
= f (x, y)
+
−
+
−
+
+
f (x, 0) f (0, y) f (0, 0)
f (x1 , y)
f (x1 , 0)
Z x1 Z y
1
(x1 − s)r1 −1 (y − t)r2 −1 g(s, t) dt ds
+
Γ(r1 )Γ(r2 ) 0
0
Z xZ y
1
+
(x − s)r1 −1 (y − t)r2 −1 g(s, t) dt ds
Γ(r1 )Γ(r2 ) x1 0
I1 (u(x−
I1 (u(x−
1 , 0))
1 , y))
−
= f (x, y) µ0 (x, y) +
+
+
f (x , y)
f (x1 , 0)
Z x1 Z y 1
1
+
(x1 − s)r1 −1 (y − t)r2 −1 g(s, t) dt ds
Γ(r1 )Γ(r2 ) 0
0
Z xZ y
1
(x − s)r1 −1 (y − t)r2 −1 g(s, t) dt ds .
+
Γ(r1 )Γ(r2 ) x1 0
If (x, y) ∈ (x2 , x3 ] × [0, b], then from Lemma 3.2 we obtain
u(x, y)
= f (x, y) µ0,2 (x, y) +
1
Γ(r1 )Γ(r2 )
Z
x
x2
Z
y
(x − s)r1 −1 (y − t)r2 −1 g(s, t) dt ds
0
ϕ(x)
u(x+
u(x+
2 , y)
2 , 0)
= f (x, y)
+
−
+
f (x, 0) f (x2 , y) f (x+
2 , 0)
Z xZ y
1
+
(x − s)r1 −1 (y − t)r2 −1 g(s, t) dt ds
Γ(r1 )Γ(r2 ) x2 0
ϕ(x)
u(x−
u(x−
I2 (u(x−
I2 (u(x−
2 , y)
2 , 0)
2 , y))
2 , 0))
= f (x, y)
+
−
+
−
+
+
+
f (x, 0) f (x2 , y) f (x2 , 0)
f (x2 , y)
f (x+
2 , 0)
Z xZ y
1
+
(x − s)r1 −1 (y − t)r2 −1 g(s, t) dt ds
Γ(r1 )Γ(r2 ) x2 0
EJDE-2010/91
FRACTIONAL-ORDER IMPULSIVE PDES
7
ϕ(x)
u(x2 , y)
u(x2 , 0)
I2 (u(x−
I2 (u(x−
2 , y))
2 , 0))
+
= f (x, y)
−
+
−
+
+
+
f (x, 0) f (x2 , y) f (x2 , 0)
f (x2 , y)
f (x+
2 , 0)
Z xZ y
1
(x − s)r1 −1 (y − t)r2 −1 g(s, t) dt ds
+
Γ(r1 )Γ(r2 ) x2 0
I1 (u(x−
I1 (u(x−
I2 (u(x−
I2 (u(x−
1 , y))
1 , 0))
2 , y))
2 , 0))
= f (x, y) µ0 (x, y) +
−
+
−
+
+
+
+
f (x , y)
f (x1 , 0)
f (x2 , y)
f (x2 , 0)
Z x1 Z y 1
1
(x1 − s)r1 −1 (y − t)r2 −1 g(s, t) dt ds
+
Γ(r1 )Γ(r2 ) 0
0
Z x2 Z y
1
+
(x2 − s)r1 −1 (y − t)r2 −1 g(s, t) dt ds
Γ(r1 )Γ(r2 ) x1 0
Z xZ y
1
+
(x − s)r1 −1 (y − t)r2 −1 g(s, t) dt ds .
Γ(r1 )Γ(r2 ) x2 0
If (x, y) ∈ (xk , xk+1 ] × [0, b] then again from Lemma 3.2 we get (3.4).
Conversely, assume that u satisfies the impulsive fractional integral equation
(3.4). If (x, y) ∈ [0, x1 ] × [0, b] and using the fact that c Dr is the left inverse of I r
we get
c r u
D ( )(x, y) = g(x, y), for each (x, y) ∈ [0, x1 ] × [0, b].
f
If (x, y) ∈ [xk , xk+1 ) × [0, b], k = 1, . . . , m and using the fact that c Dr C = 0, where
C is a constant, we get
c r u
D ( )(x, y) = g(x, y), for each (x, y) ∈ [xk , xk+1 ) × [0, b].
f
Also, we can easily show that
−
−
u(x+
k , y) = u(xk , y) + Ik (u(xk , y)),
y ∈ [0, b], k = 1, . . . , m.
Let µ := µ0 .
Corollary 3.5. Let 0 < r1 , r2 ≤ 1 and let f : J ×Rn → Rn \{0Rn }, g : J ×Rn → Rn
be continuous. A function u is a solution of the fractional integral equation
h


f
(x,
y,
u(x,
y))
µ(x, y)


i

RxRy


1
r1 −1
r2 −1

+
(x
−
s)
(y
−
t)
g(s,
t,
u(s,
t))
dt
ds
,

Γ(r1 )Γ(r2 ) 0 0





if (x, y) ∈ [0, x1 ] × [0, b],


h
Ii (u(x−
Ii (u(x−
u(x, y) = f (x, y, u(x, y)) µ(x, y) + Pk
i ,0))
i ,y))
−
+
i=1 f (x+ ,y,u(x+ ,y))

f (x+

i
i
i ,0,u(xi ,0))

Pk R xi R y

1
r1 −1
r2 −1

+
(x
−
s)
(y
−
t)
g(s,
t,
u(s,
t)) dt ds

i
xi−1 0
 Γ(r1 )Γ(r2 ) R i=1
i

R

x
y
1

+ Γ(r )Γ(r
(x − s)r1 −1 (y − t)r2 −1 g(s, t, u(s, t)) dt ds ,

xk 0
1
2)



if (x, y) ∈ (xk , xk+1 ] × [0, b], k = 1, . . . , m,
(3.7)
if and only if u is a solution of the fractional initial-value problem
u(x, y)
c r
D
= g(x, y, u(x, y)), (x, y) ∈ J 0 ,
(3.8)
f (x, y, u(x, y))
−
−
u(x+
k , y) = u(xk , y) + Ik (u(xk , y)), y ∈ [0, b], k = 1, . . . , m.
(3.9)
8
S. ABBAS, R. P. AGARWAL, M. BENCHOHRA
EJDE-2010/91
4. Existence of Solutions
In this section, we are concerned with the existence of solutions for the problem
(1.1)-(1.3). The following hypotheses will be used in the sequel.
(A1) The function f is continuous on J × Rn .
(A2) There exists a function α ∈ C(J, R+ ) such that
kf (x, y, u) − f (x, y, u)k ≤ α(x, y)ku − uk;
for all (x, y) ∈ J and u, u ∈ Rn .
(A3) The function g is Carathéodory, and there exists h ∈ L∞ (J, R+ ) such that
kg(x, y, u)k ≤ h(x, y);
a.e. (x, y) ∈ J, for all u ∈ Rn .
(A4) There exists a function β ∈ C(J, R+ ) such that
I (u) k
≤ β(x, y); for all (x, y) ∈ J and u ∈ Rn .
f (x, y, u)
Theorem 4.1. Assume that hypotheses (A1)–(A4) hold. If
h
2ar1 br2 khkL∞ i
kαk∞ kµk∞ + 2mkβk∞ +
< 1,
Γ(r1 + 1)Γ(r2 + 1)
(4.1)
Then the initial-value problem (1.1)-(1.3) has at least one solution on J.
Proof. Let X := P C(J, Rn ). Define two operators A and B on X by
Au(x, y) = f (x, y, u(x, y));
(x, y) ∈ J,
(4.2)
and
Bu(x, y) = µ(x, y) +
m X
i=1
1
+
Γ(r1 )Γ(r2 )
+
1
Γ(r1 )Γ(r2 )
Ii (u(x−
Ii (u(x−
i , y))
i , 0))
−
+
+
+
+
f (xi , y, u(xi , y)) f (xi , 0, u(xi , 0))
m Z xi Z y
X
(xi − s)r1 −1 (y − t)r2 −1 g(s, t, u(s, t)) dt ds
i=1
x
Z
xm
xi−1
Z
0
y
(x − s)r1 −1 (y − t)r2 −1 g(s, t, u(s, t)) dt ds;
0
(4.3)
with (x, y) ∈ J. Solving (1.1)-(1.3) is equivalent to solving (3.3), which is further
equivalent to solving the operator equation
Au(x, y) Bu(x, y) = u(x, y),
(x, y) ∈ J.
(4.4)
We show that operators A and B satisfy all the assumptions of Theorem 2.4. First
we shall show that A is a Lipschitz. Let u1 , u2 ∈ X. Then by (A2),
kAu1 (x, y) − Au2 (x, y)k = kf (x, y, u1 (x, y)) − f (x, y, u2 (x, y))k
≤ α(x, y)ku1 (x, y) − u2 (x, y)k
≤ kαk∞ ku1 − u2 kP C .
Taking the maximum over (x, y), in the above inequality yields
kAu1 − Au2 kP C ≤ kαk∞ ku1 − u2 kP C ,
and so A is a Lipschitz with a Lipschitz constant kαk∞ .
EJDE-2010/91
FRACTIONAL-ORDER IMPULSIVE PDES
9
Next, we show that B is compact operator on X. Let {un } be a sequence in X.
From (A3) and (A4) it follows that
kBun kP C ≤ kµk∞ + 2mkβk∞ +
2ar1 br2 khkL∞
.
Γ(r1 + 1)Γ(r2 + 1)
As a result {Bun : n ∈ N} is a uniformly bounded set in X.
Let (τ1 , y1 ), (τ2 , y2 ) ∈ J, τ1 < τ2 and y1 < y2 , then for each (x, y) ∈ J,
kB(un )(τ2 , y2 ) − B(un )(τ1 , y1 )k
m
X
≤ kµ(τ1 , y1 ) − µ(τ2 , y2 )k +
m
X
1
+
Γ(r1 )Γ(r2 )
k=1
Z
xk
xk−1
Z
k=1
y1
Ik (u(x−
Ik (u(x−
k , y1 ))
k , y2 ))
−
+
+
+
+
f (xk , y1 , u(xi , y1 )) f (xk , y2 , u(xi , y2 ))
(xk − s)r1 −1 [(y2 − t)r2 −1 − (y1 − t)r2 −1 ]
0
× g(s, t, u(s, t)) dt ds
m Z xk Z y2
X
1
(xk − s)r1 −1 (y2 − t)r2 −1 kg(s, t, u(s, t))k dt ds
+
Γ(r1 )Γ(r2 )
k=1 xk−1 y1
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
× g(s, t, u(s, t)) dt ds
Z τ2 Z y2
1
+
(τ2 − s)r1 −1 (y2 − t)r2 −1 kg(s, t, u(s, t))k dt ds
Γ(r1 )Γ(r2 ) τ1 y1
Z τ1 Z y2
1
+
(τ2 − s)r1 −1 (y2 − t)r2 −1 kg(s, t, u(s, t))k dt ds
Γ(r1 )Γ(r2 ) 0
y1
Z τ2 Z y1
1
+
(τ2 − s)r1 −1 (y2 − t)r2 −1 kg(s, t, u(s, t))k dt ds
Γ(r1 )Γ(r2 ) τ1 0
m −
X
Ik (u(x−
Ik (u(xk , y1 ))
k , y2 ))
≤ kµ(τ1 , y1 ) − µ(τ2 , y2 )k +
−
+
+
+
+
f (xk , y1 , u(xi , y1 )) f (xk , y2 , u(xi , y2 ))
k=1
Z y1
m Z
khkL∞ X xk
+
(xk − s)r1 −1 [(y2 − t)r2 −1 − (y1 − t)r2 −1 ] dt ds
Γ(r1 )Γ(r2 )
x
0
k−1
k=1
m Z xk Z y2
X
khkL∞
+
(xk − s)r1 −1 (y2 − t)r2 −1 dt ds
Γ(r1 )Γ(r2 )
k=1 xk−1 y1
Z τ1 Z y1
∞
khkL
+
[(τ2 − s)r1 −1 (y2 − t)r2 −1 − (τ1 − s)r1 −1 (y1 − t)r2 −1 ] dt ds
Γ(r1 )Γ(r2 ) 0
Z τ2 Z0 y2
khkL∞
+
(τ2 − s)r1 −1 (y2 − t)r2 −1 dt ds
Γ(r1 )Γ(r2 ) τ1 y1
Z τ1 Z y2
khkL∞
+
(τ2 − s)r1 −1 (y2 − t)r2 −1 dt ds
Γ(r1 )Γ(r2 ) 0
y1
Z τ2 Z y1
khkL∞
+
(τ2 − s)r1 −1 (y2 − t)r2 −1 dt ds
Γ(r1 )Γ(r2 ) τ1 0
10
S. ABBAS, R. P. AGARWAL, M. BENCHOHRA
m X
≤ kµ(τ1 , y1 ) − µ(τ2 , y2 )k +
m
khkL∞ X
+
Γ(r1 )Γ(r2 )
+
khkL∞
Γ(r1 )Γ(r2 )
Z
xk
xk−1
k=1
m Z xk
X
k=1
xk−1
Z
k=1
y1
EJDE-2010/91
Ik (u(x−
Ik (u(x−
k , y1 ))
k , y2 ))
−
+
+
+
+
f (xk , y1 , u(xi , y1 )) f (xk , y2 , u(xi , y2 ))
(xk − s)r1 −1 [(y2 − t)r2 −1 − (y1 − t)r2 −1 ] dt ds
0
Z
y2
(xk − s)r1 −1 (y2 − t)r2 −1 dt ds
y1
khkL∞
+
[2y r2 (τ2 − τ1 )r1 + 2τ2r1 (y2 − y1 )r2
Γ(r1 + 1)Γ(r2 + 1) 2
+ τ1r1 y1r2 − τ2r1 y2r2 − 2(τ2 − τ1 )r1 (y2 − y1 )r2 ].
As τ1 → τ2 and y1 → y2 , the right-hand side of the above inequality tends to zero.
From this we conclude that {Bun : n ∈ N} is an equicontinuous set in X. Hence
B : X → X is compact by Arzelà-Ascoli theorem. Moreover,
M = kB(X)k ≤ kµk∞ + 2mkβk∞ +
2ar1 br2 khkL∞
,
Γ(r1 + 1)Γ(r2 + 1)
and so,
αM ≤ kαk∞ kµk∞ + 2mkβk∞ +
2ar1 br2 khkL∞ < 1,
Γ(r1 + 1)Γ(r2 + 1)
by assumption (4.1). To finish, it remain to show that either the conclusion (i) or
the conclusion (ii) of Theorem 2.4 holds. We now will show that the conclusion (ii)
is not possible. Let u ∈ X be any solution to (1.1)-(1.3), then for any λ ∈ (0, 1) we
have
u(x, y)
h
X = λf (x, y, u(x, y)) µ(x, y) +
0<xk <x
Ik (u(x−
Ik (u(x−
k , y))
k , 0))
−
+
+
f (x+
f (x+
k , y, u(xi , y))
k , 0, u(xi , 0))
X Z xk Z y
1
+
(xk − s)r1 −1 (y − t)r2 −1 g(s, t, u(s, t)) dt ds
Γ(r1 )Γ(r2 ) 0<x <x xk−1 0
k
Z xZ y
i
1
(x − s)r1 −1 (y − t)r2 −1 g(s, t, u(s, t)) dt ds .
+
Γ(r1 )Γ(r2 ) xk 0
for (x, y) ∈ J. Therefore,
2ar1 br2 khkL∞ ku(x, y)k ≤ kf (x, y, u(x, y))k kµ(x, y)k + 2mkβk∞ +
Γ(r1 + 1)Γ(r2 + 1)
≤ kf (x, y, u(x, y)) − f (x, y, 0)k + kf (x, y, 0)k
2ar1 br2 khkL∞ × kµ(x, y)k + 2mkβk∞ +
Γ(r1 + 1)Γ(r2 + 1)
2ar1 br2 khkL∞ ≤ kαk∞ ku(x, y)k + f ∗ kµ(x, y)k + 2mkβk∞ +
Γ(r1 + 1)Γ(r2 + 1)
r1 r2
2a b khkL∞ ≤ [kαk∞ kukP C + f ∗ ] kµk∞ + 2mkβk∞ +
,
Γ(r1 + 1)Γ(r2 + 1)
EJDE-2010/91
FRACTIONAL-ORDER IMPULSIVE PDES
11
where f ∗ = sup{kf (x, y, 0)k : (x, y) ∈ J}, and consequently
2ar1 br2 khkL∞ f ∗ kµk∞ + 2mkβk∞ + Γ(r
1 +1)Γ(r2 +1)
kukP C ≤
:= M.
2ar1 br2 khkL∞ 1 − kαk∞ kµk∞ + 2mkβk∞ + Γ(r
1 +1)Γ(r2 +1)
Thus the conclusion (ii) of Theorem 2.4 does not hold. Therefore (1.1)-(1.3) has a
solution on J.
5. Existence of extremal solutions
We equip the space P C(J, Rn ) with the order relation ≤ with the help of the
cone defined by
K = {u ∈ P C(J, Rn ) : u(x, y) ≥ 0, ∀(x, y) ∈ J}.
Thus u ≤ ū if and only if u(x, y) ≤ ū(x, y) for each (x, y) ∈ J.
It is well-known that the cone K is positive and normal in P C(J, Rn ) ([17]). If
u, ū ∈ C(J, Rn ) and u ≤ ū, we put
[u, u] = {u ∈ P C(J, Rn ) : u ≤ u ≤ ū}.
Definition 5.1. A function γ : J × Rn → Rn is called Chandrabhan if
(i) the function (x, y) → γ(x, y, u) is measurable for each u ∈ Rn ,
(ii) the function u → γ(x, y, u) is nondecreasing for almost each (x, y) ∈ J.
Definition 5.2. A function u(·, ·) ∈ P C(J, Rn ) is said to be a lower solution of
(1.1)-(1.3) if
h
i
u(x, y)
c r
D0
≤ g(x, y, u(x, y)), (x, y) ∈ J, x 6= xk , k = 1, . . . , m,
f (x, y, u(x, y))
−
−
u(x+
k , y) ≤ u(xk , y) + Ik (u(xk , y)), y ∈ [0, b]; k = 1, . . . , m,
u(x, 0) ≤ ϕ(x),
u(0, y) ≤ ψ(y),
(x, y) ∈ J.
Similarly a function ū(·, ·) ∈ P C(J, Rn ) is said to be an upper solution of (1.1)-(1.3)
if
h
i
ū(x, y)
c r
D0
≥ g(x, y, ū(x, y)), (x, y) ∈ J, x 6= xk , k = 1, . . . , m,
f (x, y, ū(x, y))
−
−
ū(x+
k , y) ≥ ū(xk , y) + Ik (ū(xk , y)),
ū(x, 0) ≥ ϕ(x),
y ∈ [0, b]; k = 1, . . . , m,
ū(0, y) ≥ ψ(y),
(x, y) ∈ J.
Definition 5.3. A solution uM of the problem (1.1)-(1.3) is said to be maximal if
for any other solution u to the problem (1.1)-(1.3) one has u(x, y) ≤ uM (x, y), for
all (x, y) ∈ J. Again a solution um of the problem (1.1)-(1.3) is said to be minimal
if um (x, y) ≤ u(x, y), for all (x, y) ∈ J where u is any solution of the problem
(1.1)-(1.3) on J.
The following hypotheses will be used in the sequel.
(H1) f : J × Rn+ → Rn+ \ {0}, g : J × Rn+ → Rn+ , ψ(y) ≥ 0 on [0, b] and
ϕ(0)
ϕ(x)
≥
f (x, 0, ϕ(x))
f (0, 0, ϕ(0))
for all x ∈ [0, a].
(H2) The functions f and g are Chandrabhan.
12
S. ABBAS, R. P. AGARWAL, M. BENCHOHRA
EJDE-2010/91
(H3) There exists a function h̃ ∈ L∞ (J, R+ ) such that
kg(x, y, u)k ≤ h̃(x, y),
a.e. (x, y) ∈ J, for all u ∈ Rn .
(H4) There exists a function β̃ ∈ C(J, R+ ) such that
Ik (u) ≤ β̃(x, y), for all (x, y) ∈ J, for all u ∈ Rn .
f (x, y, u)
(H5) The problem (1.1)-(1.3) has a lower solution u and an upper solution u with
u ≤ u.
Theorem 5.4. Assume that hypotheses (A2), (H1)–(H5) hold. If
kαk∞ kµk∞ + 2mkβ̃k +
2ar1 br2 kh̃kL∞ < 1,
Γ(r1 + 1)Γ(r2 + 1)
then (1.1)-(1.3) has a minimal and a maximal positive solution on J.
Proof. Let X = P C(J, Rn ) and consider a closed interval [u, u] in X which is well
defined in view of hypothesis (H5). Define two operators A, B : [u, u] → X by (4.2)
and (4.3), respectively. Clearly A and B define the operators A, B : [u, u] → K.
Now solving (1.1)-(1.3) is equivalent to solving (3.3), which is further equivalent
to solving the operator equation
Au(x, y) Bu(x, y) = u(x, y),
(x, y) ∈ J.
(5.1)
We show that operators A and B satisfy all the assumptions of Theorem 2.6. As in
Theorem 4.1 we can prove that A is Lipschitz with a Lipschitz constant kαk∞ and
B is completely continuous operator on [u, u]. Now hypothesis (H2) implies that
A and B are nondecreasing on [u, u]. To see this, let u1 , u2 ∈ [u, u] be such that
u1 ≤ u2 . Then by (H2), we obtain
Au1 (x, y) = f (x, y, u1 (x, y)) ≤ f (x, y, u2 (x, y)) = Au2 (x, y),
∀(x, y) ∈ J,
and
Bu1 (x, y)
Ik (u1 (x−
Ik (u1 (x−
k , 0))
k , y))
−
+
+
+
+
f (xk , y, u(xi , y)) f (xk , 0, u(xi , 0))
0<xk <x
X Z xk Z y
1
(xk − s)r1 −1 (y − t)r2 −1 g(s, t, u1 (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 g(s, t, u1 (s, t)) dt ds
Γ(r1 )Γ(r2 ) xk 0
X Ik (u2 (x− , y))
Ik (u2 (x−
k
k , 0))
−
≤ µ(x, y) +
+
+
+
+
f (xk , y, u(xi , y)) f (xk , 0, u(xi , 0))
0<xk <x
Z xk Z y
X
1
+
(xk − s)r1 −1 (y − t)r2 −1 g(s, t, u2 (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 g(s, t, u2 (s, t)) dt ds
Γ(r1 )Γ(r2 ) xk 0
= Bu2 (x, y), ∀(x, y) ∈ J.
= µ(x, y) +
X EJDE-2010/91
FRACTIONAL-ORDER IMPULSIVE PDES
13
So A and B are nondecreasing operators on [u, u]. Again hypothesis (H5) implies
u(x, y)
X = [f (x, y, u(x, y))] µ(x, y) +
0<xk <x
Ik (u(x−
Ik (u(x−
k , y))
k , 0))
−
+
+
+
+
f (xk , y, u(xi , y)) f (xk , 0, u(xi , 0))
X Z xk Z y
1
(xk − s)r1 −1 (y − t)r2 −1 g(s, t, u(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 g(s, t, u(s, t)) dt ds
+
Γ(r1 )Γ(r2 ) xk 0
X Ik (u(x− , y))
Ik (u(x−
k
k , 0))
≤ [f (x, y, u(x, y))] µ(x, y) +
−
+
+
f (x+
f (x+
k , y, u(xi , y))
k , 0, u(xi , 0))
0<xk <x
Z
Z
xk
y
X
1
(xk − s)r1 −1 (y − t)r2 −1 g(s, t, u(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 g(s, t, u(s, t)) dt ds
Γ(r1 )Γ(r2 ) xk 0
X Ik (u(x− , y))
Ik (u(x−
k
k , 0))
−
≤ [f (x, y, u(x, y))] µ(x, y) +
+
+
f (x+
f (x+
k , y, u(xi , y))
k , 0, u(xi , 0))
0<xk <x
X Z xk Z y
1
+
(xk − s)r1 −1 (y − t)r2 −1 g(s, t, u(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 g(s, t, u(s, t)) dt ds
Γ(r1 )Γ(r2 ) xk 0
≤ u(x, y),
for all (x, y) ∈ J and u ∈ [u, u]. As a result
u(x, y) ≤ Au(x, y)Bu(x, y) ≤ u(x, y),
∀(x, y) ∈ J and u ∈ [u, u].
Hence Au Bu ∈ [u, u], for all u ∈ [u, u].
Notice for any u ∈ [u, u],
M = kB([u, u])k
Ik (u(x−
Ik (u(x−
k , 0))
k , y))
−
+
+
+
+
f (xk , y, u(xi , y)) f (xk , 0, u(xi , 0))
0<xk <x
X Z xk Z y
(xk − s)r1 −1 (y − t)r2 −1 g(s, t, u(s, t)) dt ds
X ≤ kµ(x, y)k + 1
Γ(r1 )Γ(r2 ) 0<x <x xk−1 0
k
Z xZ y
1
+
(x − s)r1 −1 (y − t)r2 −1 g(s, t, u(s, t)) dt ds
Γ(r1 )Γ(r2 ) xk 0
+
≤ kµk∞ + 2mkβ̃k +
2ar1 br2 kh̃kL∞
.
Γ(r1 + 1)Γ(r2 + 1)
and so,
αM ≤ kαk∞ kµk∞ + 2mkβ̃k +
2ar1 br2 kh̃kL∞ < 1.
Γ(r1 + 1)Γ(r2 + 1)
14
S. ABBAS, R. P. AGARWAL, M. BENCHOHRA
EJDE-2010/91
Thus the operators A and B satisfy all the conditions of Theorem 2.6 and so the
operator equation (4.3) has a least and a greatest solution in [u, u]. This further
implies that the problem (1.1)-(1.3) has a minimal and a maximal positive solution
on J.
Theorem 5.5. Assume that hypotheses (A1), (H1)–(H5) hold. Then (1.1)-(1.3)
has a minimal and a maximal positive solution on J.
Proof. Let X = P C(J, Rn ). Consider the order interval [u, u] in X and define two
operators A and B on [u, u] by (4.2) and (4.3) respectively. Then the problem
(1.1)-(1.3) is transformed into an operator equation Au(x, y) Bu(x, y) = u(x, y),
(x, y) ∈ J in a Banach algebra X. Notice that(H1) implies A, B : [u, u] → K. Since
the cone K in X is normal, [u, u] is a norm bounded set in X.
Next we show that A is completely continuous on [u, u]. Now the cone K in X is
normal, so the order interval [u, u] is norm-bounded. Hence there exists a constant
r > 0 such that kuk ≤ r for all u ∈ [u, u]. As f is continuous on compact set
J × [−r, r], it attains its maximum, say M . Therefore, for any subset S of [u, u] we
have
kA(S)k = sup{kAuk : u ∈ S}
n
o
= sup
sup kf (x, y, u(x, y))k : u ∈ S
(x,y)∈J
≤ sup
n
o
sup kf (x, y, u)k : u ∈ [−r, r] ≤ M.
(x,y)∈J
This shows that A(S) is a uniformly bounded subset of X.
We note that the function f (x, y, u) is uniformly continuous on J ×[−r, r]. Therefore, for any (τ1 , y1 ), (τ2 , y2 ) ∈ J we have
kf (τ1 , y1 , u) − f (τ2 , y2 , u)k → 0
as (τ1 , y1 ) → (τ2 , y2 ),
for all u ∈ [−r, r]. Similarly for any u1 , u2 ∈ [−r, r]
kf (x, y, u1 ) − f (x, y, u2 )k → 0
as u1 → u2 ,
for all (x, y) ∈ J. Hence for any (τ1 , y1 ), (τ2 , y2 ) ∈ J and for any u ∈ S one has
kAu(τ1 , y1 ) − Au(τ2 , y2 )k = kf (τ1 , y1 , u(τ1 , y1 )) − f (τ2 , y2 , u(τ2 , y2 ))k
≤ kf (τ1 , y1 , u(τ1 , y1 )) − f (τ2 , y2 , u(τ1 , y1 )k
+ kf (τ2 , y2 , u(τ1 , y1 )) − f (τ2 , y2 , u(τ2 , y2 ))k
→0
as (τ1 , y1 ) → (τ2 , y2 ).
This shows that A(S) is an equicontinuous set in K. Now an application of ArzelàAscoli theorem yields that A is a completely continuous operator on [u, u].
Next it can be shown as in the proof of Theorem 5.4 that B is a compact operator
on [u, u]. Now an application of Theorem 2.5 yields that the problem (1.1)-(1.3)
has a minimal and maximal positive solution on J.
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6. An Example
As an application of our results we consider the following partial hyperbolic
functional differential equations of the form
u(x, y)
c r
D0
= g(x, y, u(x, y)), (x, y) ∈ [0, 1] × [0, 1],
(6.1)
f (x, y, u(x, y))
1+ 1− 1 − u
,y = u
, y + I1 u
, y , y ∈ [0, 1],
(6.2)
2
2
2
u(x, 0) = ϕ(x), x ∈ [0, 1], u(0, y) = ψ(y), y ∈ [0, 1],
(6.3)
where f, g : [0, 1] × [0, 1] × R → R and I1 : R → R are defined by
1
f (x, y, u) = x+y+10
,
e
(1 + |u|)
1
,
g(x, y, u) = x+y+8
e
(1 + u2 )
(8 + e−10 )2
I1 (u) =
.
512e10 (1 + |u|)2
The functions ϕ, ψ : [0, 1] → R are defined by
( 2
x −10
e ; if x ∈ [0, 21 ],
ϕ(x) = 22 −10
x e ; if x ∈ ( 12 , 1],
and
ψ(y) = ye−10 , for all y ∈ [0, 1].
We show that the functions ϕ, ψ, f, g and I1 satisfy all the hypotheses of Theorem
1
4.1. Clearly, the function f satisfies (A1) and (A2) with α(x, y) = ex+y+10
and
kαk∞ = 1/e10 .
Also, the function g satisfies (A3) with h(x, y) =
1
ex+y+8
and
8
khkL∞ = 1/e .
x+y
and kβk∞ =
Finally, condition (A4) holds with β(x, y) = 81e
512
computation gives µ(x, y) < 4e. Condition (4.1) holds. Indeed
h
2ar1 br2 khkL∞ i
kαk∞ kµk∞ + 2mkβk∞ +
Γ(r1 + 1)Γ(r2 + 1)
h
i
2
1
81e
2
< 10 4e +
+ 8
< 1,
e
256
e Γ(r1 + 1)Γ(r2 + 1)
81e2
512 .
A simple
for each (r1 , r2 ) ∈ (0, 1] × (0, 1]. Hence by Theorem 4.1, problem (6.1)-(6.3) has a
solution defined on [0, 1] × [0, 1].
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Saı̈d Abbas
Laboratoire de Mathématiques, Université de Saı̈da, B.P. 138, 20000, Saı̈da, Algérie
E-mail address: abbas said dz@yahoo.fr
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FRACTIONAL-ORDER IMPULSIVE PDES
17
Ravi P. Agarwal
Department of Mathematical Sciences, Florida Institute of Technology, Melboune,
Florida, 32901-6975, USA
KFUPM Chair Professor, Mathematics and Statistics Department, King Fahd University of Petroleum and Minerals, Dhahran 31261, Saudi Arabia
E-mail address: agarwal@fit.edu
Mouffak Benchohra
Laboratoire de Mathématiques, Université de Sidi Bel-Abbès, B.P. 89, 22000, Sidi BelAbbès, Algérie
E-mail address: benchohra@univ-sba.dz