BOUNDARY VALUE PROBLEM WITH
INTEGRAL CONDITIONS FOR A LINEAR
THIRD-ORDER EQUATION
M. DENCHE AND A. MEMOU
Received 6 March 2003 and in revised form 29 July 2003
We prove the existence and uniqueness of a strong solution for a linear
third-order equation with integral boundary conditions. The proof uses
energy inequalities and the density of the range of the generated operator.
1. Introduction
In the rectangle Ω = [0, 1] × [0, T ], we consider the equation
£u =
∂u
∂3 u ∂
+
a(x,
t)
= f(x, t),
∂x
∂t3 ∂x
(1.1a)
with the initial conditions
u(x, 0) = 0,
∂u
(x, 0) = 0,
∂t
x ∈ (0, 1),
(1.1b)
the final condition
∂2 u
(x, T ) = 0,
∂t2
x ∈ (0, 1),
(1.1c)
the Dirichlet condition
u(0, t) = 0,
∀t ∈ (0, T ),
Copyright c 2003 Hindawi Publishing Corporation
Journal of Applied Mathematics 2003:11 (2003) 553–567
2000 Mathematics Subject Classification: 35B45, 35G10
URL: http://dx.doi.org/10.1155/S1110757X03303031
(1.1d)
554
Boundary value problem with integral conditions
and the integral condition
1
u(x, t)dx = 0,
∀t ∈ (0, T ).
(1.1e)
0
In addition, we assume that the function a(x, t) is bounded with
0 < a0 ≤ a(x, t) ≤ a1 ,
(1.2)
and has bounded partial derivatives such that
ck ≤
∂k a
(x, t) ≤ ck , ∀x ∈ (0, 1), t ∈ (0, T ), k = 1, 3, with c1 ≥ 0,
∂tk
∂a
(x, t) ≤ b1 , for (x, t) ∈ Ω.
∂x
(1.3)
Various problems arising in heat conduction [4, 6, 14, 15], chemical engineering [9], underground water flow [13], thermoelasticity [21], and
plasmaphysics [19] can be reduced to the nonlocal problems with integral boundary conditions. This type of boundary value problems has
been investigated in [1, 2, 3, 5, 6, 7, 9, 14, 15, 16, 20, 23] for parabolic
equations, in [18, 22] for hyperbolic equations, and in [10, 11, 12] for
mixed-type equations. The basic tool in [4, 10, 11, 12, 16, 23] is the energy inequality method which, of course, requires appropriate multipliers and functional spaces. In this paper, we extend this method to the
study of a linear third-order partial differential equation. This type of
problems is encountered in the study of thermal conductivity [17] and
microscale heat transfer [8].
2. Preliminaries
In this paper, we prove the existence and uniqueness of a strong solution of problem (1.1). For this, we consider the solution of problem
(1.1) as a solution of the operator equation Lu = F, where L is the operator with domain of definition D(L) consisting of functions u ∈ E such
√
that 1 − x(∂k+1 u/∂tk ∂x)(x, t) ∈ L2 (Ω), k = 0, 3 and u satisfies conditions
(1.1d) and (1.1e). The operator L is considered from E to F, where E is
the Banach space of the functions u, u ∈ L2 (Ω), with the finite norm
(1 − x)2 ∂3 u 2 ∂2 u 2
2
uE =
∂t3 + ∂x2 dx dt
2
Ω
(2.1)
(1 − x)2 ∂u 2
+ |u|2 dx dt,
+
2 ∂x Ω
M. Denche and A. Memou
555
and F is the Hilbert space of the functions F = (f, 0, 0, 0), f ∈ L2 (Ω), with
the finite norm
F2F = (1 − x)2 |f|2 dx dt.
(2.2)
Ω
Then we establish an energy inequality
uE ≤ kLuF ,
∀u ∈ D(L),
(2.3)
and we show that the operator L has the closure L.
Definition 2.1. A solution of the operator equation Lu = F is called a
strong solution of problem (1.1).
Inequality (2.3) can be extended to u ∈ D(L), that is,
uE ≤ kLuF ,
∀u ∈ D L .
(2.4)
From this inequality, we obtain the uniqueness of a strong solution, if
it exists, and the equality of the sets R(L) and R(L). Thus, to prove the
existence of a strong solution of problem (1.1) for any F ∈ F, it remains
to prove that the set R(L) is dense in F.
3. An energy inequality and its applications
Theorem 3.1. For any function u ∈ D(L), there exists the a priori estimate
uE ≤ kLuF ,
(3.1)
where
2 17 exp(ct) 5 + 4 b1 / c3 − 3cc2 + 3c2 c1 − c3 a1 − b12 + 1
,
k =
min 1, a20 , c3 − 3cc2 + 3c2 c1 − c3 a1 − b12
2
(3.2)
with the constant c satisfying
sup
(x,t)∈Ω
1 ∂a
a ∂t
≤ c < inf
(x,t)∈Ω
1 ∂a
+1 ,
a ∂t
2
c3 − 3cc2 + 3c2 c1 − c3 a1 − b1 > 0,
c2 − 2cc1 + c2 a1 − c1 + ca1 < 0.
(3.3)
556
Boundary value problem with integral conditions
Proof. Let
Mu = (1 − x)2
∂3 u
∂3 u
+
2(1
−
x)J
,
x
∂t3
∂t3
(3.4)
where
Jx u =
x
u(ζ, t)dζ.
(3.5)
0
We consider the quadratic form
Φ(u, u) = Re
Ω
exp(−ct)£uMu dx dt,
(3.6)
with the constant c satisfying (3.3), obtained by multiplying (1.1a) by
exp(−ct)Mu, integrating over Ω, and taking the real part. Substituting
the expression of Mu in (3.6), we obtain
Re
Ω
exp(−ct)£uMu dx dt
= Re
Ω
3 2
2 ∂ u exp(−ct)(1 − x) 3 dx dt
∂t
∂3 u ∂3 u
+ 2 Re exp(−ct)(1 − x) 3 Jx 3 dx dt
∂t
∂t
Ω
∂u
∂
a(x, t)
Mu dx dt.
+ Re exp(−ct)
∂x
∂x
Ω
(3.7)
Integrating the last two terms on the right-hand side by parts with respect to x in (3.7) and using the Dirichlet condition (1.1d), we obtain
1
2 Re
0
1
∂3 u 2
∂3 u ∂3 u
J
dx
=
exp(−ct)
J
x
x ∂t3 dx,
∂t3 ∂t3
0
∂u
∂
a
Mu dx dt
Re exp(−ct)
∂x
∂x
Ω
∂u ∂4 u
= − Re exp(−ct)(1 − x)2 a
dx dt
∂x ∂t3 ∂x
Ω
∂a
∂3 u
− 2 Re exp(−ct) uJx 3 dx dt
∂x
∂t
Ω
3
∂u
− 2 Re exp(−ct)au 3 dx dt.
∂t
Ω
(1 − x) exp(−ct)
(3.8)
(3.9)
M. Denche and A. Memou
557
Integrating each term by parts in (3.9) with respect to t and using the
initial and final conditions (1.1b) and (1.1c), we get
∂u
∂
a
Mu dx dt
Re exp(−ct)
∂x
∂x
Ω
∂a
∂3 u
= −2 Re exp(−ct) uJx 3 dx dt
∂x
∂t
Ω
3
∂2 a
∂a
2 ∂a
3
−c a
− 3c 2 + 3c
+ exp(−ct)
∂t
∂t3
∂t
Ω
(1 − x)2 ∂u 2
2
+ |u| dx dt
×
2 ∂x (1 − x)2 ∂2 u 2 ∂u 2
∂a
− ca
+
− 3 exp(−ct)
dx dt
∂t
2 ∂t∂x ∂t Ω
1
(1 − x)2 ∂2 u 2 ∂u 2
+ exp(−ct)a
∂t∂x + ∂t dx
2
0
T =t
1
2
∂a
∂a
(1 − x)2 ∂u 2
2
2
+
c
− exp(−ct)
−
2c
a
+
|u|
dx
2
∂t
2
∂x
∂t
0
1
t=T
∂u ∂2 u ∂u
∂a
− ca (1 − x)2
+ 2u
+ Re exp(−ct)
dx.
∂t
∂t∂x ∂x
∂t T =t
0
(3.10)
Substituting (3.8) and (3.10) in (3.7) and using conditions (1.2), (1.3),
and (3.3), we obtain
3 2
∂ u
exp(−ct)(1 − x)2 3 dx dt
∂t
Ω
+ exp(−ct) c3 − 3cc2 + 3c2 c1 − c3 a1 − b12
Ω
(1 − x)2 ∂u 2
2
×
+ |u| dx dt
2 ∂x ≤ Re exp(−ct)£uM u dx dt.
(3.11)
Ω
Again, substituting the expression of Mu in (3.11) and using elementary
inequality, we get
558
Boundary value problem with integral conditions
(1 − x)2 ∂3 u 2
exp(−ct)
dx dt
2 ∂t3 Ω
+ exp(−ct) c3 − 3cc2 + 3c2 c1 − c3 a1 − b12
Ω
(1 − x)2 ∂u 2
+ |u|2 dx dt
×
2 ∂x ≤ 17 exp(−ct)(1 − x)2 |f|2 dx dt.
(3.12)
Ω
By virtue of (1.1a), we have
2 2
∂ u (1 − x)2
dx dt
a0 2 2
∂x
Ω
3 2
2
2
2 ∂ u ≤ (1 − x) |f| dx dt + 2(1 − x) 3 dx dt
∂t
Ω
Ω
2
(1 − x)2 ∂u 2
+ 4 b12
∂x + |u| dx dt.
2
Ω
(3.13)
This last inequality combined with (3.12) yields
(1 − x)2 ∂3 u 2
dx dt
2 ∂t3 Ω
(1 − x)2 ∂u 2
2 3
2
2
+ |u| dx dt
+
c3 − 3cc2 + 3c c1 − c a1 − b1
2 ∂x Ω
(1 − x)2 ∂2 u 2
+ a20
dx dt
2 ∂x2 Ω
4b12
≤ 17 exp(cT ) 5 + +1
c3 − 3cc2 + 3c2 c1 − c3 a1 − b12
× (1 − x)2 |f|2 dx dt.
Ω
(3.14)
Thus, this inequality implies
(1 − x)2 ∂3 u 2 ∂2 u 2
(1 − x)2 ∂u 2
+ |u|2 dx dt
∂t3 + ∂x2 dx dt +
2
2 ∂x Ω
Ω
≤ k2 (1 − x)2 |f|2 dx dt,
Ω
(3.15)
M. Denche and A. Memou
559
where
17 exp(cT ) 5 + 4b12 / c3 − 3cc2 + 3c2 c1 − c3 a1 − b12 + 1
k =
.
min 1, a20 , c3 − 3cc2 + 3c2 c1 − c3 a1 − b12
2
(3.16)
Then,
uE ≤ kLuF ,
∀u ∈ D(L).
(3.17)
Thus, we obtain the desired inequality.
Lemma 3.2. The operator L from E to F admits a closure.
Proof. Suppose that (un ) ∈ D(L) is a sequence such that
un −→ 0
Lun −→ F
in E,
(3.18)
in F.
We need to show that F = 0. We introduce the operator
£0v =
∂3 v
−(1 − x)2 3
∂t
∂ ∂
2
a(x, t)
(1 − x) v ,
+
∂x
∂x
(3.19)
with domain D(£ 0 ) consisting of functions v ∈ W22,3 (Ω) satisfying
∂v = 0,
∂t t=0
v|t=0 = 0,
∂2 v = 0,
∂t2 t=0
v|x=0 = 0,
∂v = 0.
∂x x=0
(3.20)
We note that D(£ 0 ) is dense in the Hilbert space obtained by completing
L2 (Ω) with respect to the norm
Ω
(1 − x)2 |v|2 dx dt = v2 .
(3.21)
Since
(1 − x) fv dx dt = lim
2
Ω
n→+∞ Ω
(1 − x)2 £un v dx dt
= lim
n→+∞ Ω
(3.22)
un £ 0 v dx dt = 0,
for any function v ∈ D(£ 0 ), it follows that f = 0.
560
Boundary value problem with integral conditions
Theorem 3.1 is valid for a strong solution, then we have the inequality
uE ≤ kLuF ,
∀u ∈ D L .
(3.23)
Hence we obtain the following corollary.
Corollary 3.3. A strong solution of problem (1.1) is unique if it exists, and
depends continuously on F.
Corollary 3.4. The range R(L) of the operator L is closed in F, and R(L) =
R(L).
4. Solvability of problem (1.1)
To prove the solvability of problem (1.1), it is sufficient to show that R(L)
is dense in F. The proof is based on the following lemma.
Lemma 4.1. Suppose that a(x, t) and its derivatives ∂4 a/∂t3 ∂x and ∂2 a/∂t∂x
are bounded. Let D0 (L) = {u ∈ D(L) : u(x, 0) = 0, (∂u/∂t)(x, 0) = 0, (∂2 u/
∂t2 )(x, T ) = 0}. If, for u ∈ D0 (L) and for some functions w ∈ L2 (Ω),
Ω
(1 − x)£uw dx dt = 0,
(4.1)
then w = 0.
Proof. Equality (4.1) can be written as follows:
∂3 u
(1 − x)w 3 dx dt = −
∂t
Ω
x
∂u
∂
w
a(1 − x)
dζ dx dt.
w−
∂x
0 1−ζ
Ω ∂x
(4.2)
For a given w(x, t), we introduce the function v(x, t) such that
v(x, t) = w(x, t) −
x
0
From (4.3), we conclude that
1
0
w(ζ, t)
dζ.
1−ζ
(4.3)
v(x, t)dx = 0, and thus, we have
∂3 u
Nv dx dt = −
3
Ω ∂t
Ω
A(t)uv dx dt,
where A(t)u = (∂/∂x)(a(1 − x)(∂u/∂x)) and Nv = (1 − x)v + Jv.
(4.4)
M. Denche and A. Memou
561
Following [23], we introduce the smoothing operators
3 −1
∂
,
Jε−1 = I−
∂t3
3 −1
−1 ∗
∂
Jε
= I +
,
∂t3
(4.5)
with respect to t, which provide the solutions of the respective problems
g − g∗
∂3 gε
= g,
∂t3
∂3 g ∗
+ 3 = g,
∂t
g (0) = 0,
∂g
(0) = 0,
∂t
∂2 g
(T ) = 0,
∂t2
g∗ (0)
∂g∗
(T ) = 0,
∂t
∂2 g∗
(T ) = 0.
∂t2
= 0,
(4.6)
We also have the following properties: for any g ∈ L2 (0, T ), the functions
J−1 (g), (J−1 )∗ g ∈ W23 (0, T ). If g ∈ D(L), then J−1 (g) ∈ D(L) and we have
∗
lim J−1 g − g L2 [0,T ] = 0
lim J−1 g − g L2 [0,T ] = 0
for −→ 0,
for −→ 0.
(4.7)
Substituting the function u in (4.4) by the smoothing function uε and
using the relation
A(t)uε = Jε−1 Au − Jε−1 β (t)uε ,
(4.8)
where
β (t)uε = 3
∂A(t) ∂2 uε ∂3 A(t)
∂2 A(t) ∂uε
+
3
uε ,
+
∂t
∂t ∂2 t
∂t2
∂t3
(4.9)
we obtain
−
Ω
uN
∂3 vε ∗
dx dt =
∂3 t
Ω
A(t)uvε∗ dx dt − Ω
β (t)uε vε∗ dx dt.
(4.10)
Passing to the limit, the equality in the relation (4.10) remains true for all
functions u ∈ L2 (Ω) such that (1 − x)(∂u/∂x), (∂/∂x)((1 − x)(∂u/∂x)) ∈
L2 (Ω), and satisfying condition (1.1d).
562
Boundary value problem with integral conditions
The operator A(t) has a continuous inverse in L2 (0, 1) defined by
ζ
1
1
g(η, t)dη dζ
0 1 − ζ a(ζ, t) 0
x
1
1
dζ,
+ C(t)
0 1 − ζ a(ζ, t)
A−1 (t)g = −
x
(4.11)
where
1 ζ
dζ/a(ζ, t) 0 g(η, t)dη
0
.
C(t) =
1 dζ/a(ζ,
t)
0
(4.12)
1
Then, we have 0 A−1 (t)g dx = 0, hence the function uε = (Jε )−1 u can be
represented in the form
−1
uε = Jε A−1 (t)A(t)u.
(4.13)
Then
x
∂4 a −1
1
Bε (t)g = 3 Jε
g(η, t)dη − C(t)
a(x, t)
∂t ∂x
0
x
3
ax
∂ a −1 g
− 2
g(η, t)dη − C(t)
+ 3 Jε
a a (x, t)
∂t
0
x
2
∂ ∂ a ∂ −1 1
J
g(η,
t)dη
−
C(t)
+3
∂t ∂t2 ∂x ∂t ε a(x, t)
0
x
ax
∂a ∂ −1 g
Jε
− 2
g(η, t)dη − C(t) .
+
∂t ∂t
a a (x, t)
0
(4.14)
The adjoint of Bε (t) has the form
Bε∗ (t)
1 −1 ∗ ∂3 a
3 −1 ∗ ∂ ∂a ∂h
J
=
h + Jε
a ε
a
∂t ∂t ∂t
∂t3
x 1/a(η, t) dη Gε h (1),
+ Gε h (x) − 01 1/a(x, t) dx
0
(4.15)
M. Denche and A. Memou
563
where
G h (x) =
x 2
3 −1 ∗ ∂
∂ ∂h
−
J
a(ζ, t) ∂t ∂t∂ζ ∂t
0
∂a 1 −1 ∗ ∂ ∂a ∂h
J
+3
∂ζ a2 (ζ, t) ∂t ∂t ∂t
4
∂a 1 −1 ∗ ∂3 a
1 −1 ∗ ∂ a
J
Jε
h
dζ.
h
+
−
a(ζ, t)
∂ζ a2 (ζ, t) ε
∂t3 ∂ζ
∂t3
(4.16)
Consequently, equality (4.10) becomes
∂3 v∗
− uN 3ε dx dt =
∂t
Ω
Ω
A(t)uhε dx dt,
(4.17)
where hε = vε∗ − εBε∗ vε∗ .
The left-hand side of (4.17) is a continuous linear functional of u.
Hence the function hε has the derivatives (1 − x)(∂hε /∂x), (∂/∂x)((1 −
x)(∂hε /∂x)) ∈ L2 (Ω) and the following conditions are satisfied: hε |x=0 =
0, hε |x=1 = 0, and (1 − x)(∂hε /∂x)|x=1 = 0.
From the equality
∂vε∗
1 −1 ∗ ∂3 a
∂hε
= I − ε Jε
(1 − x)
(1
−
x)
∂x
a
∂x
∂t3
∂v∗
1 ∗ ∂ ∂a ∂
(1 − x) ε ,
− 3ε Jε−1
a
∂t ∂t ∂t
∂x
(4.18)
and since the operator (Jε−1 )∗ is bounded in L2 (Ω), for sufficiently small ε,
we have ε(1/a)(Jε−1 )∗ (∂3 a/∂t3 ) < 1. Hence the operator I −ε(1/a)(Jε−1 )∗
(∂3 a/∂t3 ) has a bounded inverse in L2 (Ω). We conclude that (1 − x)(∂vε∗ /
∂x) ∈ L2 (Ω).
Similarly, we conclude that (∂/∂x)((1 − x)(∂vε∗ /∂x)) exists and belongs to L2 (Ω), and the following conditions are satisfied:
vε∗ |x=0 = 0,
vε∗ |x=1 = 0,
(1 − x)
∂vε∗ = 0.
∂x x=1
(4.19)
t η T
Substituting u = 0 0 ζ exp(cτ)vε∗ (τ)dτ dζ dη in (4.4), where the constant
c satisfies (3.3), we obtain
Ω
exp(ct)vε∗ Nv dx dt
=−
Ω
A(t)uv dx dt.
(4.20)
564
Boundary value problem with integral conditions
Using the properties of smoothing operators, we have
Ω
exp(ct)vε∗ Nv dx dt
=−
Ω
A(t)uvε∗ dx dt − ε
Ω
A(t)u
∂3 vε∗
dx dt,
∂t3
(4.21)
and from
∂3 vε∗
ε Re A(t)u 3 dx dt = ε
∂t
Ω
∂u ∂ ∂3 vε∗
(1 − x)a
dx dt
∂x ∂x ∂t3
Ω
∂a ∂u ∂2 ∂vε∗
dx dt
= −ε Re (1 − x)
∂t ∂x ∂t2 ∂x
Ω
∂a ∂2 u ∂ ∂vε∗
+ ε Re (1 − x)
dx dt
∂t ∂t∂x ∂t ∂x
Ω
∂v∗ 2
+ ε a exp(−ct)(1 − x) ε dx dt
∂x
Ω
∂a ∂2 u ∂vε∗
dx dt,
+ ε Re (1 − x)
∂t ∂t∂x ∂x
Ω
(4.22)
we have
ε Re
Ω
A(t)u
∂3 vε∗
dx dt
∂t3
∂v∗ 2
ε
≥ ε a exp(+ct)(1 − x)
dx dt
∂x Ω
3 2
∂u 1 ∂a 2
exp(−ct) 2 dx dt
− ε (1 − x)
4a ∂t
∂t ∂x
Ω
∂v∗ 2
ε
− ε a exp(+ct)(1 − x)
dx dt
∂x Ω
2
∂u 1 − x ∂a 2
exp(−ct) dx dt
−ε
∂t
∂x
Ω 2
2
1 − x ∂3 vε∗ − ε exp(+ct)
dx dt
2 ∂t2 ∂x Ω
2
1 ∂2 vε∗ − ε exp(+ct) dx dt
2 ∂t∂x Ω
2 2
∂u
1 − x ∂a 2
dx dt.
exp(−ct)
−ε
∂t
∂x∂t Ω 2
(4.23)
M. Denche and A. Memou
565
Integrating the first term on the right-hand side by parts in (4.21), we
obtain
Re
A(t)uvε∗ dx dt
2 2
∂u
∂a
3
dx dt
− ca (1 − x) exp(−ct)
≥−
2 Ω
∂t
∂t∂x 2 2 ∂u
∂a
1 1
dx
− ca +
(1 − x) exp(−ct) a − 2 0
∂t
∂t∂x
t=T
1
2 2
∂u 1
∂a 2 ∂a
∂a
−
(1−x) exp(−ct)
−2c +c a+ −ca dx
2 0
∂t
∂t
∂x t=T
∂t2
3
2
∂u 1
∂2 a
∂a
∂a
+
− c3 a dx dt.
(1 − x) exp(−ct)
− 3c 2 + 3c2
3
2 Ω
∂t
∂x
∂t
∂t
(4.24)
Ω
Combining (4.23) and (4.24), we get
Re
exp(ct)vε∗ Nv dx dt
∂2 u 2
3
dx dt
(1 − x) exp(−ct) c1 − ca0 ≤
2 Ω
∂t∂x ∂2 u 2 1 1
dx
−
(1 − x) exp(−ct) a0 − c1 − ca1 2 0
∂t∂x t=T
1
2 ∂u 1
2
+
(1 − x) exp(−ct) c2 − 2c1 c − c a1 − c1 + ca1 dx
2 0
∂x t=T
2
∂u 1
(1 − x) exp(−ct) c3 − 3c2 c + 3c2 c1 − c3 a1 dx dt
−
2 Ω
∂x
c12 ∂3 u 2
dx dt
(1 − x) exp(−ct)
+ε
4a0 ∂t2 ∂x Ω
c2 ∂u 2
+ (1 − x) exp(−ct) 1 dx dt
2 ∂x
Ω
3 ∗ 2
∂v 1−x
exp(ct) 2 ε dx dt
+
2
∂t ∂x
Ω
c2 ∂2 u 2
dx dt
+ (1 − x) exp(−ct) 1 2 ∂t∂x Ω
2 ∗ 2
∂ vε 1−x
+
exp(ct)
dx dt .
∂t∂x Ω 2
Ω
(4.25)
566
Boundary value problem with integral conditions
Using conditions (3.3) and inequalities (4.23) and (4.24), we obtain
Re
Ω
exp(ct)vNv dx dt ≤ 0,
as ε −→ 0.
(4.26)
Since Re Ω exp(ct)vJ x v dx dt = 0, then v = 0 a.e.
Finally, from the equality (1 − x)v + Jx v = (1 − x)w, we conclude w = 0.
Theorem 4.2. The range R(L) of L coincides with F.
Proof. Since F is Hilbert space, then R(L) = F if and only if the relation
Ω
(1 − x)2 £uf dx dt = 0,
(4.27)
for arbitrary u ∈ D0 (L) and F ∈ F, implies that f = 0.
Taking u ∈ D0 (L) in (4.27) and using Lemma 4.1, we obtain that w =
(1 − x)f = 0, then f = 0.
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M. Denche: Laboratoire Equations Différentielles, Département de Mathématiques, Faculté des Sciences, Université Mentouri, 25000 Constantine, Algeria
E-mail address: denech@wissal.dz
A. Memou: Laboratoire Equations Différentielles, Département de Mathématiques, Faculté des Sciences, Université Mentouri, 25000 Constantine, Algeria