Theoretical Mathematics & Applications, vol.6, no.1, 2016, 1-14
ISSN: 1792-9687 (print), 1792-9709 (online)
Scienpress Ltd, 2016
Numerical solutions of integro-differential
equation with purely integral condition
by using Laplace transform method
Moussa Zakari Djibibe1 , Nadjime Pindra2 and Kokou Tcharie3
Abstract
The aims of this paper are to prove the existence, uniqueness and
continuous dependence upon the data of solution of following intogrodifferential hyperbolique equation with purely integral conditions.
Z t
∂2u
∂2u
∂u
+ αu(x, t) =
ϑ(t − s)u(x, s) ds, 0 < x < `, 0 < t ≤ T,
−a 2 −c
∂t2
∂x
∂x
0
u(x, 0) = φ(x),
0 < x < `,
∂u
(x, 0) = χ(x), 0 < x < `,
∂t
Z `
u(x, t) dx = E(t), 0 < t ≤ T,
0
Z `
xu(x, t) dx = G(t), 0 < t ≤ T.
0
The proofs are based on a priori estimates and Laplace transform method.
We present a numerical approximate solution to integro-differential equation with integral conditions. A Laplace transform method is described
1
2
3
Département de Mathématiques, Université de Lomé, PO Box 1515 Lomé, Togo.
E-mail: zakari.djibibe@gmail.com
Département de Mathématiques, Université de Lomé, Togo.
E-mail: nadjimepindra@gmail.com
Département de Mathématiques, Université de Lomé, Togo.
E-mail: tkokou09@yahoo.fr
Article Info: Received : July 31, 2015. Revised : August 30, 2015.
Published online : November 22, 2015.
2
Numerical solutions of integro-differential equation...
for the solution of considered equation. Following Laplace transform
of the original problem, an appropriate method of solving differential
equations is used to solve the resultat time-independent modified equation and solution is inverted numerically back into the time domain.
Numerical results are provided to show the accuracy of the proposed
method.
Mathematics Subject Classification: 35B45; 35L82; 44A10
Keywords: Laplace transform method; Integral conditions; Continuous dependence; Integro-differential
1
Introduction
In this paper, we deal with a class of hyperbolic integro-differential equation
with purely nonlocal conditions. The precise statement of the problem is a
follows : let ` > 0, T > 0, and Ω = {(x, t) ∈ R2 : 0 < x < `, 0 < t < T }. We
shall determine a solution u, in Ω of the differential equation
∂ 2u
∂ 2u
∂u
−a
−c +αu(x, t) =
2
2
∂t
∂x
∂x
Z
t
ϑ(t − s)u(x, s) ds,
0 < x < `,
0
0 < t ≤, T
(1)
satisfying the initial conditions
u(x, 0) = φ(x), 0 < x < `,
∂u
(x, 0) = χ(x), 0 < x < `,
∂t
(2)
(3)
and the integral conditions
Z
`
u(x, t) dx = E(t),
Z
0 < t ≤ T,
(4)
0
`
xu(x, t) dx = G(t),
0 < t ≤ T,
(5)
0
where ϑ, ξ, φ, E, G are known functions, `, T and a are pasitive constants,
and α, c are the reals.
M.Z. Djibibe, N. Pindra and K.Tcharie
3
Assumption 1.1. For all x, t) ∈ Ω, we assume that
ϑ0 ≤ ϑ(x, t) ≤ ϑ1 , ϑ0 > 0, ϑ1 . > 0
φ is continuously derivate on (0, `).
The notion of nonlocal condition has been introduced to extend the study
of the classical initial value problems and it is more precise for describing
nature phenomena than the classical condition since more information is taken
into account, thereby decreasing the negative effects incurred by a possibly
erroneous single measurement taken at the initial value. The inportance of
nonlocal conditions in many applications is discussed in [10], [16].
Mathematical modelling by evolution problems with a nonlocal constraint
of the form
Z 1
1
u(x, t) dx = ζ(t)
1−` `
is encountered in heat transmission theory, thermoelasticity, chemical engineering, underground water flow, and plasma physic.
Many methods were used to investigate the existence and uniqueness of the
solution of mixed problems which combine classical and integral conditions. J.
R. Cannon [5] used the potentiel method, combining a Dirichlet and an intégral
condition for a parabolic equation. L.A. Mouravey and V. Philinovoski [9] used
the maximum principle, combining a Neumann and an integral condition for
heat equation. M.Z. Djibibe and K. Tcharie [13], Ionkin [6] and L. Bougoffa
[4] used the Fourier method for same purpose. Recently, mixed problems with
integral conditions for generalization of equation (1) have been treated using
the energy-integral method. See M.Z. Djibibe and K. Tcharie [11] , M.Z.
Djibibe and K. Tcharie [12], M.Z. Djibibe el al. [14], [15], N.I. Yurchuk [16],
[17], M. Mesloub, A. Bouziani and N. Kechkar [8]. Differently to these works,
in the present paper we combine a priori estimate and Fourier’s method to
prove existence and uniqueness solution of the problem (1)- (5).
The results obtained in this paper generalize the results of [1], and constitute a new contribution to this emerging field of research.
It can be a part in the contribution of the development of a priori estimates
an Laplace methods for solving such problems.
The questions related to these problems are so miscellaneous that the elaboration of a general theory is still premature. Therefore, the investigationof
these problems requires at every time a separate study.
4
Numerical solutions of integro-differential equation...
The remainder of the paper is organized as follows. After this introduction,
in section 2, we present some preliminaries and basic lemmas. Then in Section
3, we establish a priori estimate. Finally, in section 4, we prove existence
solution.
2
Preliminaries
We transform the problem (1)-(5) with nonhomogeneous boundary conditions (4) and (5) into a problem with homogeneous boundary conditions. For
this, we introduce a new unknow function defined by u(x, y) = v(x, t)+w(x, t),
where
2` − 4x
4x − 2`
4x − 2`
w(x, t) =
E(t) +
G(t) =
(G(t) − E(t)).
4
4
`
`
`4
Then, problem becomes :
∂2v
∂2v
∂v
−
a
−c
+ αv(x, t) = f (x, t),
2
2
∂t
∂x
∂x
v(x, 0) = ϕ(x), 0 < x < `,
∂v
(x, 0) = ψ(x), 0 < x < `,
∂t
Z `
v(x, t) dx = 0, 0 < t ≤ T,
0
Z `
xv(x, t) dx = 0, 0 < t ≤ T,
0 < x < `,
0 < t ≤ T,
(6)
(7)
(8)
(9)
(10)
0
where
4x − 2`
(G(0) − E(0)),
`4
4x − 2` 0
(G (0) − E 0 (0)),
ψ(x) = χ(x) −
`4
ϕ(x) = φ(x) −
Z
t
f (x, t) =
ϑ(t − s)u(x, s) ds −
0
2(2x − `) 00
(G (t) − E 00 (t))
4
`
2(2αx − α` − 2c
(G(t) − E(t)).
`4
Instead of searching for the function u, we search for the function v. So the
solution of problem (1), (2), (3), (4) and (5) will be given by u(x, t) = v(x, t) +
w(x, t).
−
5
M.Z. Djibibe, N. Pindra and K.Tcharie
Lemma 2.1 (Gronwall). Let x(t) ≥ 0, h(t), y(t) the integrables fonctions on [a; b]. If
Z t
y(t) ≤ h(t) +
x(τ )y(τ ) dτ, ∀t ∈ [a; b]
a
then
Z
Z
t
y(t) ≤ h(t) +
t
h(τ )x(τ ) exp(
a
x(s) ds) dτ,
∀t ∈ [a ; b].
τ
In particular, if x(t) ≡ c is invariable fonction and h(t) is an increasing fonction, then
y(t) ≤ h(t)ec(t−a) , ∀t ∈ [a ; b].
3
Main Results
3.1
Uniqueness and Continuous Dependence of the solution
Theorem 3.1. If v(x, t) is a solution of problem (1), (2), (3), (4) and (5),
then we have a priori estimates
¶
µZ `
Z `µ µ ¶
Z `
∂u
2
2
2
2
sup
Jx
ϕ (x) dx +
Jx2 ψ(x) dx
+ Jx u dx + u dx ≤A
∂t
0≤t≤T 0
0
0
¶
Z `
+
Jx2 ϕ(x) dx .
(11)
0
∂u
Proof. Appying Jx to (1), Multiplying with Jx
and integrating the re∂t
sults obtained over Ωτ = (0, `) × (0, τ ). Observe that
Z `Z τ µ 2 ¶ µ ¶
Z `Z τ µ 2 ¶ µ ¶
∂ v
∂v
∂v
∂ v
Jx
Jx
Jx
dt dx − a
Jx
dt dx
2
2
∂t
∂t
∂x
∂t
0
0
0
0
Z `Z
µ
τ
−c
Jx
0
0
∂v
∂x
¶
µ
∂v
Jx
∂t
Z `Z τ
=
0
0
¶
¶
∂v
dt dx + α
Jx v(x, t)Jx
dt dx
∂t
0
0
µZ t
¶ µ ¶
∂v
Jx
ϑ(t − s)u(x, s) ds Jx
dt dx, (12)
∂t
0
Z `Z
τ
µ
6
Numerical solutions of integro-differential equation...
Z
x
where Jx (u(x, t) =
u(ξ, t) dξ. Successive intrgration by parts of integrals
0
on the left-hand of (12) are straight-foward but somewhat tedious. We give
only their results
µ
¶
Z
Z
Z
Z
1 ` 2
1 ` 2 ∂u(x, τ )
α ` 2
α ` 2
dx −
J
J ψ(x) dx +
J u dx −
J ϕ(x) dx
2 0 x
∂t
2 0 x
2 0 x
2 0 x
Z
Z
Z
a ` 2
a ` 2
∂u
+
u dx −
ϕ (x) dx = a
u(`, t) dxdt
2 0
2
∂t
Ωτ
µ ¶ 0
¶ µ ¶
Z µ
Z
∂u
∂u
∂u
dxdt −
a (0, t) + u(0, t) Jx
dxdt
+c
uJx
∂t
∂x
∂t
Ωτ
Ωτ
¶ µ ¶
Z ` Z τ µZ t
∂v
+
Jx
ϑ(t − s)u(x, s) ds Jx
dt dx.
(13)
∂t
0
0
0
By the Cauchy inequality,
¶ µ ¶
Z
Z ` Z τ µZ t
∂v
T ϑ0
Jx2 u dx dt
Jx
ϑ(t − s)u(x, s) ds Jx
dt dx ≤
∂t
2
ΩT
0
0
0
Z TZ ` µ ¶
∂u
ϑ0 T
+
Jx2
dx dτ, (14)
2 0 0
∂t
µ ¶
Z
Z
Z
∂u
1
1
∂u
2
2
u dxdt +
Jx
dxdt, (15)
u(`, t) dxdt ≤
∂t
2 ΩT
2 ΩT
∂t
Ωτ
µ ¶
µ ¶
Z
Z
Z
∂u
∂u
1
1
2
2
uJx
u dxdt +
Jx
dxdt ≤
dxdt. (16)
∂t
2 ΩT
2 ΩT
∂t
Ωτ
Substuting (16) into (15), yields
µ
¶
Z
Z
Z
Z
1 ` 2 ∂u(x, τ )
α ` 2
a ` 2
a ` 2
J
dx +
J u dx +
u dx ≤
ϕ (x) dx
2 0 x
∂t
2 0 x
2 0
2 0
Z
Z
Z
1 ` 2
α ` 2
|a| + T ϑ0 + |c|
+
Jx ψ(x) dxAutof orm +
Jx ϕ(x) dx +
u2 dxdt
2 0
2 0
2
ΩT
µ ¶
Z
Z
ϑ0 T + |c| + 1
∂u
T ϑ0
+
Jx2
Jx2 u dxdt.
(17)
dx dτ +
2
∂t
2
ΩT
ΩT
The right-hand side of (17) is independent of τ, hence, replacing the left-hand
side by the upper bound with respect to τ and by Gronwall Lemma, we get
¶
¶
µZ `
Z `µ µ
∂u(x, τ )
2
2
2
λT
+ Jx u dx + u dx ≤ e
sup
Jx
ϕ2 (x) dx
∂t
0≤τ ≤T 0
0
¶
Z `
Z `
+
Jx2 ψ(x) dx +
Jx2 ϕ(x) dx ,
(18)
0
0
7
M.Z. Djibibe, N. Pindra and K.Tcharie
where
µ
max
λ=
¶
1 α T ϑ0 + |c| + 1 T ϑ0 + |a| + |c|
, ,
,
2 2
2
2
µ
¶
.
1 α a
min
, ,
2 2 2
From (18), we obtain the priori estimates (11). This complete the proof of
Theoreme (3.1), with A = eλT .
Corollary 3.2. If problem (1), (2), (3), (4) and (5) has a solution, then
this solution is unique and depends continuously on (ϕ, ψ).
3.2
Existence of the Solution
Laplace transform is widely used in the area of engineering technology and
mathematical science. There are many problems whose solution may be found
in terms of the Laplace. In fact, it is an efficient method for solving many
diffential equations and partial differential equation. The mains difficult of
the method of Laplace domain into the real domain.
Hence in this section, we apply the technique of the Laplace transform to
find solutions of the problem (1)-(5).
Suppose that v(x, t) is défined and is of the exponential order for t ≥ 0,
there exists λ > 0, β > 0 and t0 > 0 such that |v(x, t)| ≤ λeβt for t ≥ t0 . Then
the Laplace transform V (x, s) includind the function v(x, t) is defined by
Z
+∞
V (x, s) =
v(x, t)e−st dt,
(19)
0
where s is know as a Laplace variable and V is a function in the Laplace
domain.
Applying the Laplace transform on both sides of (14), we obtain
Z
+∞
0
∂ 2 v −st
e dt−a
∂t2
Z
+∞
∂ 2 v −st
e dt − c
∂x2
Z
+∞
∂v −st
e dt+
∂x
0
0
Z +∞
Z +∞ Z t
−st
+α
v(x, t)e dt =
ϑ(t − s)v(x, s) ds dt. (20)
0
0
0
8
Numerical solutions of integro-differential equation...
The standard integration by parts of the terms on the left-hand of (20), leads
Z
+∞
Z
0
Z
0
+∞
+∞
0
∂ 2 v −st
e dt = −ψ(x) − sϕ(x) + s2 V (x, s),
2
∂t
d2 V (x, s)
∂ 2 v −st
e
dt
=
,
∂x2
dx2
∂v −st
dV (x, s)
e dt =
.
∂x
dx
(21)
(22)
(23)
Substuting (21), (22) and (23) into (20), we get
d2 V (x, s)
dV (x, s)
−c
+ (α + s2 )V (x, s) = f (x, s) + ψ(x) + sϕ(x), (24)
2
dx
dx
Z +∞ Z t
where f (x, s) =
ϑ(t − s)v(x, s) ds dt.
−a
0
Similarly, we have
0
Z
Z
`
V (x, s) dx = 0,
(25)
xV (x, s) dx = 0.
(26)
0
`
0
Now, we distinguish the following cas :
I Case 1 : If c2 + 4aα + 4as2 = 0.
I Case 2 : If c2 + 4aα + 4as2 > 0.
I Case 3 : If c2 + 4aα + 4as2 < 0.
In this article, we only deal Case 1 and Case 2
For Case 1, that is c2 + 4aα + 4as2 = 0, the solution general of (26) is given
by
V (x, s) = (C1 (s)x + C2 (s))e
c
x
− 2a
1
−
a
Z
x
c
(x − τ )(f (τ, s) + ψ(τ ) + sϕ(τ ))e 2a τ dτ.
0
(27)
M.Z. Djibibe, N. Pindra and K.Tcharie
9
Putting the intagral conditions (25) and (26) in (27), we get
µ
µ
¶¶
c ¢
c` −c`
c ¡
2 − 1 + e 2a
C1 (s) +
1 − e− 2a C2 (s)
2a
2a
Z `Z x
c
=2
(x − y)(f (τ, s) + ψ(τ ) + sϕ(τ ))e 2a τ dτ dx,
0
0
4a2 + (c2 `2 − 4ac` − 4a1 ))C1 (s) + (2a − (2a + `c))C2 (s)
Z `Z x
c
2
= 4a
x(x − τ )(f (τ, s) + ψ(τ ) + sϕ(τ ))e 2a τ dτ dx.
0
0
For Case 1, that is c2 + 4aα + 4as2 > 0, Using the method of variation of
parameter, to solve (26), we have the general solution as
!
c2 + 4a(α + s2 )
x
V (x, s) = C1 (s)exp −
2a
Ã
!
p
c + c2 + 4a(α + s2 )
2
+ C2 (s)exp −
x −p
2a
c2 + 4a(α + s2 )
Ã
!
p
Z x
2 + 4a(α + s2 )
c(x−τ )
c
×
(f (τ, s) + ψ(τ ) + sϕ(τ )) e− 2a sinh
(x − τ ) dτ,
2a
0
Ã
c−
p
(28)
where C1 et C2 are arbitrary functions of s.
Substuting (28) into (25) and (26), we have
!
p
c2 + 4a(α + s2 )
x dx
C1 (s)
exp −
2a
0
Ã
!
p
Z `
c + c2 + 4a(α + s2 )
2
+ C2 (s)
exp −
x dx = p
2a
c2 + 4a(α + s2 )
0
Ã
!
p
Z `Z x
2 + 4a(α + s2 )
c(x−τ )
c
×
(f (τ, s) + ψ(τ ) + sϕ(τ )] e− 2a sinh
(x − τ ) dτ dx,
2a
0
0
Z
`
Ã
c−
(29)
10
Numerical solutions of integro-differential equation...
!
p
c2 + 4a(α + s2 )
C1 (s)
x dx
xexp −
2a
0
Ã
!
p
Z `
c + c2 + 4a(α + s2 )
2
+ C2 (s)
xexp −
x dx = p
2a
c2 + 4a(α + s2 )
0
Ã
!
p
Z `Z x
2 + 4a(α + s2 )
c(x−τ )
c
×
x (f (τ, s) + ψ(τ ) + sϕ(τ )] e− 2a sinh
(x − τ ) dτ dx,
2a
0
0
Z
`
Ã
c−
(30)
where
Ã
! Ã
!−1 Ã
!
C1 (s)
a11 (s) a12 (s)
b1 (s)
=
×
,
C2 (s)
a21 (s) a22 (s)
b2 (s)
and
!
p
c2 + 4a(α + s2 )
a11 (s) =
exp −
x dx,
(31)
2a
0
Ã
!
p
Z `
c + c2 + 4a(α + s2 )
a12 (s) =
exp −
x dx,
(32)
2a
0
Ã
!
p
Z `
2
2
c − c + 4a(α + s )
a21 (s) =
xexp −
x dx,
(33)
2a
0
Ã
!
p
Z `
c + c2 + 4a(α + s2 )
a22 (s) =
xexp −
x dx,
(34)
2a
0
Z `Z x
c(x−τ )
a
(f (τ, s) + ψ(τ ) + sϕ(τ )] e− 2a
(35)
b1 (s) = − p
c2 + 4a(α + s2 ) 0 0
!
Ãp
c2 + 4a(α + s2 )
(x − τ ) dτ dx,
(36)
× sinh
2a
Z `Z x
c(x−τ )
a
b2 (s) = − p
x (f (τ, s) + ψ(τ ) + sϕ(τ )] e− 2a
c2 + 4a(α + s2 ) 0 0
(37)
Ãp
!
c2 + 4a(α + s2 )
× sinh
(x − τ ) dτ dx.
(38)
2a
Z
`
Ã
c−
If it is not possible to calculate the integral directly, then we calculate tem
numerically. If the Laplace inversion is possibly computed directly for (28) and
11
M.Z. Djibibe, N. Pindra and K.Tcharie
(38), we obtain our solution explicitly. Otherwise, we use the suitable approximate method, then we use the numerical inversion of the Laplace transform.
We have
!
c2 + 4a(α + s2 )
a11 (s) =
x dx
exp −
2a
0
Ã
!
p
n
c − c2 + 4a(α + s2 )
1X
=
ki exp −
(xi + 1) ,
2 i=1
2a
Z
`
Ã
c−
p
!
p
c2 + 4a(α + s2 )
a12 (s) =
exp −
x dx
2a
0
Ã
!
p
n
c + c2 + 4a(α + s2 )
1X
=
ki exp −
(xi + 1) ,
2 i=1
2a
Z
`
Ã
!
p
c2 + 4a(α + s2 )
a21 (s) =
xexp −
x dx
2a
0
Ã
!
p
n
c − c2 + 4a(α + s2 )
1X
=
ki (xi + 1)exp −
(xi + 1) ,
4 i=1
2a
Z
`
Ã
c+
(39)
!
c2 + 4a(α + s2 )
a22 (s) =
xexp −
x dx
2a
0
Ã
!
p
n
c + c2 + 4a(α + s2 )
1X
=
ki (xi + 1)exp −
(xi + 1) ,
4 i=1
2a
Z
`
Ã
c−
c+
p
(40)
b1 (s) = − p
a
Z `Z
x
(f (τ, s) + ψ(τ ) + sϕ(τ )] e−
c2 + 4a(α + s2 ) 0 0
!
Ãp
c2 + 4a(α + s2 )
(x − τ ) dτ dx
× sinh
2a
c(x−τ )
2a
12
Numerical solutions of integro-differential equation...
¶
µ
¶
· µ
1
1
=− p
ki f
(xi + 1), s + ψ
(xi + 1
2
2
4 c2 + 4a(α + s2 ) i=1
µ
¶
µ
·µ
¶
1
c
1
+ 1 − ( (xi + 1) exp −
1 − (xi + 1) xj
2
2a
2
Ãp
µ
¶¸¶ X
n
c2 + 4a(α + s2 )
1
sinh
+ 1 + (xi + 1)
2
2a
j=1
· ·µ
¶
µ
¶¸
¸¶
1
1
1
1
×
1 − (xi + 1) xj + 1 + (xi + 1) − (xi + 1)
2
2
2
2
n
X
a
a
b2 (s) = − p
Z `Z
x
x (f (τ, s) + ψ(τ ) + sϕ(τ )] e−
c(x−τ )
2a
c2 + 4a(α + s2 ) 0 0
Ãp
!
c2 + 4a(α + s2 )
× sinh
(x − τ ) dτ dx
2a
· µ
¶
µ
¶
1
1
ki f
b2 (s) = − p
(xi + 1), s + ψ
(xi + 1
2
2
4 c2 + 4a(α + s2 ) i=1
µ
¶¸ µ
¶ µ ·µ
¶
1
1
1
1
+sϕ
(xi + 1
1 − ( (xi + 1)
1 − (xi + 1) xj
2
2
2
2
¶¸¶
µ
·µ
¶
µ
c
1
1
exp −
1 − (xi + 1) xj
+ 1 + (xi + 1)
2
2a
2
Ãp
µ
¶¸¶ X
n
c2 + 4a(α + s2 )
1
+ 1 + (xi + 1)
sinh
2
2a
j=1
· ·µ
¶
µ
¶¸
¸¶
1
1
1
1
×
1 − (xi + 1) xj + 1 + (xi + 1) − (xi + 1) .
2
2
2
2
a
n
X
where xi and ki are the abscissa and weights, defined as
xi : ith
zero of Pn (x),
ki =
2
(1 −
x2i )(P 0 (x))2
Their tabulated values can be found in [7] for different values of N.
Using Stehfest’s algorithm, the time domain solution is approximated as
µ
¶
min(k,n)
2n
X
ln 2 X
k ln 2
in (2i)!(−1)k+n
v(x, t) '
V x,
.
t k=1 2
t
3 (n − i)!i!(i − 1)!(k − i)!(2i − k)!
k+1
5
i=4
2
(41)
M.Z. Djibibe, N. Pindra and K.Tcharie
13
References
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