ISSN 1749-3889 (print), 1749-3897 (online)
International Journal of Nonlinear Science
Vol.16(2013) No.2,pp.152-163
On the Existence Uniqueness and Solution of the Nonlinear Volterra Partial
Integro-Differential Equations
A. Tari ∗
Department of Mathematics, Shahed University, P. O. Box 18151-159, Tehran, Iran
(Received 13 August 2012 , accepted 31 May 2013)
Abstract: In this paper, we study a form of the nonlinear Volterra partial integro-differential equations (NVPIDEs). First, we prove a theorem and some corollaries about existence and uniqueness of solution of the
NVPIDEs. Next, we develop the two-dimensional reduced differential transform method (RDTM) for solving
a class of the NVPIDEs. For this purpose, based on some preliminary results of the reduced differential transform, we prove some theorems to develop RDTM for the aforementioned equations. The paper concludes
with some numerical examples to demonstrate the accuracy of the presented method.
Keywords: nonlinear volterra partial integro-differential equations; differential transform; reduced differential transform.
1
Introduction
The Volterra partial integro-differential equations enjoy many applications in Physics, Mechanics and other applied sciences [1], however, to solve these equations, not many numerical methods of high accuracy exist (see, for example [2-3]).
Here, we investigate such types of equations and develop RDTM for solving them.
The concept of differential transform was first introduced by Zhou [4] for solving linear and nonlinear initial value problems in electrical analysis. Recently, a reduced form of the differential transform was introduced and developed for some
equations. Some instances follow. In [5], this method was used in solving partial differential equations, while in [6], it was
used to solve the fractional partial differential equations. In [7], the RDTM was developed for solving nonlinear PDEs
with proportional delay. In [8], the RDTM was applied for solving gas dynamics equation. In [9], it was applied to solve
the linear and nonlinear wave equations and in [10] to solve generalized KDV equation. And quite recently, the authors
of [11] developed the RDTM for Wu-Zhang equation and the authors of [12] for fractional Benney-Lin equation.
As mentioned above, the subject of this paper is applying RDTM for solving the linear and nonlinear Volterra partial
integro-differential equations. For this purpose, we consider the Volterra partial integro-differential equations of the form
∫ t∫ x
Du(x, t) −
K(x, t, y, z, u(y, z))dydz = f (x, t), x ∈ [0, b], t ∈ [0, d]
(1)
0
0
with some supplementary conditions, where Du(x, t) denotes the differential part of equation and the functions K and f
are assumed to be continuous on their domains. Note should be made that throughout the rest of this paper, b and d are
taken to be finite.
2
Existence and uniqueness of solution
In this section, we investigate the existence and uniqueness of solution of the equation (1). To this end, we first consider
the equations of the form
∫ t∫ x
u(x, t) −
K(x, t, y, z, u(y, z))dydz = f (x, t), x ∈ [0, b], t ∈ [0, d]
(2)
0
∗ Corresponding
author.
0
E-mail address: [email protected], [email protected]
c
Copyright⃝World
Academic Press, World Academic Union
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A. Tari : On the Existence, Uniqueness and Solution of the Nonlinear · · ·
153
and we extend the theorem 4.1 from [13], to two-dimensional case in the following theorem.
Theorem 2.1 Let the functions f and K be continuous for all x, y ∈ [0, b], z, t ∈ [0, d] and −∞ < u < ∞. Let also the
kernel K satisfies the Lipschitz condition, i.e.
|K(x, t, y, z, u1 ) − K(x, t, y, z, u2 )| ≤ L|u1 − u2 |
(3)
where L is independent of x, t, y, z, u1 and u2 .
Then the equation (2) has a unique continuous solution.
Proof. We define
u0 (x, t) = f (x, t)
∫ t∫
x
un (x, t) = f (x, t) +
K(x, t, y, z, un−1 (y, z))dydz, n = 1, 2, ...
0
(4)
0
then we have
∫ t∫
x
un (x, t) − un−1 (x, t) =
0
[
]
K(x, t, y, z, un−1 (y, z)) − K(x, t, y, z, un−2 (y, z)) dydz.
(5)
0
We also introduce the function ϕn (x, t) as
ϕ0 (x, t) = u0 (x, t)
ϕn (x, t) = un (x, t) − un−1 (x, t), n = 1, 2, ...
so
un (x, t) =
n
∑
(6)
ϕi (x, t)
i=0
and
∫ t∫
x
|ϕn (x, t)| ≤ L
|ϕn−1 (y, z)|dydz
0
0
which implies
|ϕn (x, t)| ≤
M (Lxt)n
, ∀x ∈ [0, b], t ∈ [0, d]
(n!)2
where
M=
max
0≤x≤b, 0≤t≤d
|f (x, t)|
so the sequence {ϕn (x, t)} is convergence by Weierstrass M-test.
Therefore
∞
∑
u(x, t) =
ϕn (x, t)
(7)
n=0
exists and is a continuous function.
Now we prove that the continuous function u(x, t) in (7) satisfies the equation (2). To this end, we set
u(x, t) = un (x, t) + ∆n (x, t).
Then from (4) we have
∫ t∫
x
u(x, t) − ∆n (x, t) = f (x, t) +
K(x, t, y, z, u(y, z) − ∆n−1 (y, z))dydz
0
0
or equivalently
∫ t∫
u(x, t) − f (x, t) −
x
K(x, t, y, z, u(y, z))dydz = ∆n (x, t)+
0
0
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International Journal of Nonlinear Science, Vol.16(2013), No.2, pp. 152-163
∫ t∫
0
x
[
]
K(x, t, y, z, u(y, z) − ∆n−1 (y, z)) − K(x, t, y, z, u(y, z)) dydz
0
and by using the Lipschitz condition (3) we obtain
∫ t∫ x
u(x, t) − f (x, t) −
≤ |∆n (x, t)| + Lbd∥∆n−1 ∥
K(x,
t,
y,
z,
u(y,
z))dydz
0
(9)
0
where
∥∆n−1 ∥ =
max
0≤x≤b, 0≤t≤d
|∆n−1 (x, t)|.
But from (8) we have
lim |∆n (x, t)| = 0
n→∞
therefore when n → ∞, (9) yields
∫ t∫
x
u(x, t) −
K(x, t, y, z, u(y, z))dydz = f (x, t)
0
0
and therefore u(x, t) is a solution of the equation (2).
Finally, we prove uniqueness. To this end, let u
e(x, t) be another continuous solution of (2), then we have
∫ t∫ x[
]
u(x, t) − u
e(x, t) =
K(x, t, y, z, u(y, z)) − K(x, t, y, z, u
e(y, z)) dydz
0
0
and using (3) implies
∫ t∫
x
|u(x, t) − u
e(x, t)| ≤ L
|u(y, z) − u
e(y, z)|dydz.
0
(10)
0
But |u(y, z) − u
e(y, z)| is bounded, so we set
B=
max
0≤x≤b, 0≤t≤d
|u(x, t) − u
e(x, t)|
then from (10) we obtain
|u(x, t) − u
e(x, t)| ≤ BLxt
and using this relation repeatedly yields
|u(x, t) − u
e(x, t)| ≤ B
(Lxt)n
(Lbd)n
≤
B
(n!)2
(n!)2
(11)
and passing to the limit n → ∞ in (11) yields u(x, t) = u
e(x, t).
As the results of the above theorem, we consider some cases of the Volterra partial integro-differential equations and
investigate the existence and uniqueness of solution of them.
Corollary 2.1 If the hypothesis of theorem 2.1 hold, then the equation
∫ t∫ x
∂u(x, t)
+ u(x, t) −
K(x, t, y, z, u(y, z))dydz = f (x, t), x ∈ [0, b], t ∈ [0, d]
∂t
0
0
(12)
with initial condition u(x, 0) = α1 (x), where α1 (x) is a continuous function, has a unique continuous solution.
By integrating from two sides of the relation (12) with respect to t we obtain
u(x, t) = α1 (x)+
∫ t{
∫ t∫
−u(x, t) +
0
0
}
x
K(x, t, y, z, u(y, z))dydz + f (x, t) dt, x ∈ [0, b], t ∈ [0, d]
0
so if we set
∫ t∫
x
K1 (x, t, y, z, u) = −u(x, t) +
K(x, t, y, z, u(y, z))dydz + f (x, t)
0
0
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then we have
|K1 (x, t, y, z, u1 ) − K1 (x, t, y, z, u2 )|
∫ t∫
x
≤ |u1 − u2 | +
|K(x, t, y, z, u1 ) − K(x, t, y, z, u2 )|dydz
0
0
≤ (1 + bdL)|u1 − u2 |.
Hence the kernel of the equation (13) satisfies in Lipschitz condition and therefore by theorem 2.1 it has a unique continuous solution.
Corollary 2.2 If the hypothesis of theorem 2.1 hold for the equation
∫ t∫
∂ 2 u(x, t)
+ a(x, t)u(x, t) −
∂t2
x
K(x, t, y, z, u(y, z))dydz = f (x, t), x ∈ [0, b], t ∈ [0, d]
0
(14)
0
with initial conditions
∂
u(x, 0) = α2 (x)
∂t
u(x, 0) = α1 (x),
and if a, α1 and α2 are continuous functions for x ∈ [0, b], t ∈ [0, d], then the problem has a unique continuous solution.
By integrating twice from both sides of the relation (14) with respect to t we obtain
u(x, t) = α1 (x) + α2 (x)t+
∫ t∫ t{
∫ t∫
−a(x, t)u(x, t) +
0
0
0
}
K(x, t, y, z, u(y, z))dydz + f (x, t) dtdt, x ∈ [0, b], t ∈ [0, d]
x
(15)
0
then by setting
∫ t∫
K1 (x, t, y, z, u) = −a(x, t)u(x, t) +
x
K(x, t, y, z, u(y, z))dydz + f (x, t)
0
0
we have
|K1 (x, t, y, z, u1 ) − K1 (x, t, y, z, u2 )|
∫ t∫
x
≤ |a(x, t)(u1 − u2 )| +
|K(x, t, y, z, u1 ) − K(x, t, y, z, u2 )|dydz
0
0
≤ (M + bdL)|u1 − u2 |
where
M=
max
0≤x≤b, 0≤t≤d
|a(x, t)|
so by theorem 2.1 the equation (15) has a unique continuous solution.
Corollary 2.3 Let the assumptions of theorem 2.1 satisfy for the equation
∂ 2 u(x, t)
+ a(x, t)u(x, t) −
∂x2
∫ t∫
x
K(x, t, y, z, u(y, z))dydz = f (x, t), x ∈ [0, b], t ∈ [0, d]
0
0
with conditions
u(0, t) = β1 (t),
∂
u(0, t) = β2 (t).
∂x
Furthermore assume that the functions a, β1 and β2 are continuous for x ∈ [0, b], t ∈ [0, d].
Then this problem has a unique continuous solution.
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International Journal of Nonlinear Science, Vol.16(2013), No.2, pp. 152-163
Integrating twice from both sides of (16) with respect to x implies that
u(x, t) = β1 (t) + β2 (t)x+
∫ x∫ x{
∫ t∫
−a(x, t)u(x, t) +
0
0
0
x
}
K(x, t, y, z, u(y, z))dydz + f (x, t) dxdx, x ∈ [0, b], t ∈ [0, d]
(17)
0
and let
∫ t∫
K1 (x, t, y, z, u) = −a(x, t)u(x, t) +
x
K(x, t, y, z, u(y, z))dydz + f (x, t)
0
0
we have
|K1 (x, t, y, z, u1 ) − K1 (x, t, y, z, u2 )| ≤ (M + bdL)|u1 − u2 |
where
M=
3
max
0≤x≤b, 0≤t≤d
|a(x, t)|
Two-dimensional RDT
In this section, we remind some preliminary results of the two-dimensional RDT. Then we prove some theorems to develop
the two-dimensional RDT for solving the NVPIDEs.
Definition 3.1 The two-dimensional RDT of the analytic bivariate function f (x, t) at t0 is defined by (see [5], [9-11])
[
]
1 ∂ n f (x, t)
Fn (x) =
(18)
n!
∂tn
t=t0
and its inverse transform is defined as
f (x, t) =
∞
∑
Fn (x)(t − t0 )n .
(19)
n=0
Note that the relations (18) and (19) imply
f (x, t) =
[
]
∞
∑
1 ∂ n f (x, t)
(t − t0 )n .
n
n!
∂t
t=t0
n=0
(20)
In the following theorem, we summarize some fundamental properties of the two dimensional reduced differential transform (see [9-11]).
Theorem 3.1 Let Fn (x), Un (x) and Vn (x) be the two-dimensional reduced differential transforms of the functions
f (x, t), u(x, t) and v(x, t) at t0 = 0 respectively, then we have
a. If f (x, t) = u(x, t) ± v(x, t), then
Fn (x) = Un (x) ± Vn (x).
b. If f (x, t) = au(x, t), then
Fn (x) = aUn (x).
c. If f (x, t) = u(x, t)v(x, t), then
Fn (x) =
n
∑
Un (x)Vn−k (x).
k=0
d. If f (x, t) = xk tl , then
Fn (x) = xk δn,l .
e. If f (x, t) = sin(ax + bt), then
Fn (x) =
bn
nπ
sin(ax +
).
n!
2
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157
f. If f (x, t) = cos(ax + bt), then
Fn (x) =
bn
nπ
cos(ax +
).
n!
2
g. If f (x, t) = eax+bt , then
Fn (x) =
bn ax
e .
n!
For applying the reduced differential transform to the differential part of the equation (1), we state the following theorem(See [9-10]).
Theorem 3.2 Let Fn (x), Un (x) and Vn (x) be the two-dimensional reduced differential transforms of the functions
f (x, t), u(x, t) and v(x, t) at t0 = 0 respectively, then we have
a. If f (x, t) =
∂ r u(x,t)
∂xr ,
r ∈ N , then
Fn (x) =
b. If f (x, t) =
∂ s u(x,t)
,
∂ts
dr
Un (x).
dxr
s ∈ N , then
Fn (x) = (n + 1)(n + 2) · · · (n + s)Un+s (x).
c. If f (x, t) =
∂ r+s u(x,t)
∂ r x∂ s t ,
r, s ∈ N , then
Fn (x) = (n + 1)(n + 2) · · · (n + s)
dr
Un+s (x).
dxr
Now for applying the reduced differential transform to the integral part of (1), we give the following theorem.
Theorem 3.3 Assume that Un (x), Vn (x), Hn (x) and Gn (x) be the differential transforms of the functions u(x, t), v(x, t),
h(x, t) and g(x, t) at t0 = 0 respectively, then we have
a. If g(x, t) =
∫t∫x
0
0
u(y, z)dydz, then
x ∈ [0, b]
∫
1 x
Gn (x) =
Un−1 (y)dy,
n 0
G0 (x) = 0,
b. If g(x, t) =
∫t∫x
0
0
n = 1, 2, . . . .
(21)
u(y, z)v(y, z)dydz, then
G0 (x) = 0,
x ∈ [0, b]
n−1 ∫
1∑ x
Gn (x) =
Uk (y)Vn−k−1 (y)dy,
n
0
c. If g(x, t) = h(x, t)
∫t∫x
0
0
n = 1, 2, . . . .
(22)
k=0
u(y, z)dydz, then
G0 (x) = 0,
Gn (x) =
x ∈ [0, b]
n−1
∑
k=0
1
Hk (x)
n−k
∫
x
Un−k−1 (y)dy,
n = 1, 2, . . . .
0
Proof. It is obvious that G0 (x) = 0 in all three parts.
∑∞
a. Since Un (x) is the reduced differential transform of u(x, t) at t0 = 0, we have u(x, t) = n=0 Un (x)tn thus
)
∫
∫ t ∫ x (∑
∞
∞
∑
1 n+1 x
n
Un (y)dy
t
g(x, t) =
Un (y)z
dydz =
n+1
0
0
0
n=0
n=0
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International Journal of Nonlinear Science, Vol.16(2013), No.2, pp. 152-163
therefore
Gn+1 (x) =
or
Gn (x) =
1
n+1
1
n
∫
∫
x
Un (y)dy,
n = 0, 1, . . .
0
x
Un−1 (y)dy,
n = 1, 2, . . .
0
which yields
d
Gn (x) = Un−1 (x), n = 1, 2, . . . .
dx
b. By differentiating from g(x, t) with respect to t and x, we obtain
n
∂ 2 g(x, t)
= u(x, t)v(x, t)
∂t∂x
and by using part c of the theorems 3.2 and 3.1, we get
∑
d
Gn+1 (x) =
Uk (x)Vn−k (x),
dx
n
(n + 1)
n = 0, 1, . . . .
k=0
Replacing n + 1 by n gives
result.
∫ t ∫the
x
c. First we set f (x, t) = 0 0 u(y, z)dydz then by part a, we have
∫
1 x
Fn (x) =
Un−1 (y)dy, n = 1, 2, . . . .
n 0
Since
g(x, t) = h(x, t)f (x, t)
we have
Gn (x) =
n
∑
Hk (x)Fn−k (x) =
k=0
n−1
∑
k=0
1
Hk (x)
n−k
∫
x
Un−k−1 (y)dy
0
since Fn−k (x) = 0 for k = n.
Theorem 3.4 Let Un (x), Vn (x) and Gn (x) be the reduced differential transforms of the functions u(x, t), v(x, t) and
g(x, t) at t0 = 0 respectively, and v(x, t) ̸= 0 for x ∈ [0, b], t ∈ [0, d]. Then we have
a. If g(x, t) =
∫t∫x
0
u(y,z)
dydz,
0 v(y,z)
then
G0 (x) = 0,
Un (x) =
b. If g(x, t) =
1
v(x,t)
∫t∫x
0
0
x ∈ [0, b]
n
∑
(k + 1)
k=0
n = 0, 1, . . . .
(24)
n = 1, 2, . . . .
(25)
u(y, z)dydz, then
G0 (x) = 0,
n
∑
k=0
x ∈ [0, b]
1
Gk (x)Vn−k (x) =
n
Proof. a. We have
by setting ge(x, t) =
d
Gk+1 (x)Vn−k (x),
dx
∫
x
Un−1 (y)dy,
0
∂ 2 g(x, t)
u(x, t)
=
∂t∂x
v(x, t)
∂ 2 g(x,t)
∂t∂x ,
it can be written
ge(x, t)v(x, t) = u(x, t)
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159
and by using part c of the theorem 3.1, we obtain
n
∑
e k (x)Vn−k (x) = Un (x)
G
k=0
e
where G(x)
is the reduced differential transform of the function ge(x, t). Therefore
n
∑
(k + 1)
k=0
d
Gk+1 (x)Vn−k (x) = Un (x).
dx
b. We have
∫ t∫
g(x, t)v(x, t) =
x
u(y, z)dydz
0
0
and by using part a of the theorem 3.3
n
∑
Gk (x)Vn−k (x) =
k=0
4
1
n
∫
x
Un−1 (y)dy
0
Method and examples
In this section, we describe the RDTM and illustrate it by some numerical examples. Here, we consider the equations of
the form
∫ t∫ x
Du(x, t) −
K(x, t, y, z)uq (y, z)dydz = f (x, t), x ∈ [0, b], t ∈ [0, d]
(26)
0
0
with some supplementary conditions, where q is a positive integer. Obviously, in case q = 1, the equation (26) is reduced
to a linear equation.
We also assume that K has the following degenerate form
p
∑
K(x, t, y, z) =
vi (x, t)wi (y, z).
i=0
By substituting this K(x, t, y, z) into (26), we obtain
Du(x, t) −
p
∑
i=1
∫ t∫
x
wi (y, z)uq (y, z)dydz = f (x, t)
vi (x, t)
0
(27)
0
which is solvable by using RDTM.
By using the theorems 3.1, 3.2, 3.3 and 3.4 in equation (27), we obtain a recursive relation for Un (x) (reduced differential
transform of u(x, t)) which is solved by programming in MAPLE environment to obtain Un (x), n = 0, 1, ..., N . At last,
we use the relation (19) for uN (x, t)(approximate of u(x, t)) in truncated form
uN (x, t) =
N
∑
Un (x)(t − t0 )n
n=0
to obtain uN (x, t).
Note should be made that, we also need some starting values of Un (x) for solving the recursive relation that can be obtained from integro-differential equation and supplementary conditions.
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International Journal of Nonlinear Science, Vol.16(2013), No.2, pp. 152-163
Example 4.1 Consider the integro-differential equation
∂ 2 u(x, t)
− u(x, t) −
∂t2
∫ t∫
x
xey−z u(y, z)dydz =
0
0
1
xt(1 − e2x ), x, tϵ[0, 1]
2
(28)
with supplementary conditions
u(x, 0) = ex
∂u
(x, 0) = ex
∂t
which has the exact solution u(x, t) = ex+t .
By applying the RDTM on the equation (28), we obtain
[
]
∫
n−1
1
1 ∑ (−1)k x y
1
2x
Un+2 (x) =
Un (x) + x
e Un−k−1 (y)dy + xδn,1 (1 − e )
(n + 1)(n + 2)
n
k!
2
0
(29)
k=0
for n = 1, 2, . . . , N − 2.
Also from supplementary conditions we have
U0 (x) = ex , U1 (x) = ex
and by setting t = 0 in the equation (28) we obtain
U2 (x) =
1 x
e
2
finally the equation (29) leads to
1 x
e , n = 3, ..., N.
n!
Note should be added that, the obtained solution is the same as the first N + 1 terms of the Poisson series of the exact
solution. Table 1 shows the absolute errors at some points.
Un (x) =
Table 1: Errors of example 4.1
(x,t)
(0.25, 0.25)
(0.50, 0.75)
(0.50, 1.00)
(0.75, 0.75)
(0.75, 1.00)
(1.00, 1.00)
Error(N=10)
0.7832e-14
0.1860e-8
0.4503e-7
0.2388e-8
0.5782e-7
0.7424e-7
Error(N=12)
0.3000e-17
0.6645e-11
0.2850e-9
0.8532e-11
0.3660e-9
0.4700e-9
Error(N=14)
0.2000e-18
0.1767e-13
0.1344e-11
0.2270e-13
0.1726e-11
0.2217e-11
Example 4.2 We consider a nonlinear equation with variable coefficients of the form
∂ 2 u(x, t)
+ t2 u(x, t) −
∂t2
∫ t∫
x
ye3z u3 (y, z)dydz = (t3 + t − 2)ex−t
0
0
−
1
1
(3x − 1)t4 e3x − t4 , x, tϵ[0, 1]
36
36
with supplementary conditions
u(x, 0) = 0
∂u
(x, 0) = ex
∂t
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A. Tari : On the Existence, Uniqueness and Solution of the Nonlinear · · ·
161
and the exact solution u(x, t) = tex−t .
By the same way of the example 1. we obtain
[ n−1 n−k−1 n−k−r−1
∑ 3k ∫ x
1∑ ∑
1
Un+2 (x) =
yUr (y)Ul (y)Un−k−r−l−1 (y)dy
(n + 1)(n + 2) n
k! 0
r=0
k=0
−
n
∑
l=0
δk,2 Un−k (x) + ex
k=0
−
n
∑
(δk,3 + δk,1 − 2δk,0 )
k=0
(−1)n−k
(n − k)!
]
1
1
(3x − 1)e3x δn,4 − δn,4 , n = 1, 2, . . . .
36
36
(31)
And from supplementary conditions we have
U0 (x) = 0,
U1 (x) = ex .
Setting t = 0 in the both sides of (30) yields
U2 (x) = −ex
And similar to the previous examples the obtained solution is the same as the first N + 1 terms of the Poisson series of the
exact solution. Table 2 shows the absolute errors at some points.
Table 2: Errors of example 4.2
(x,t)
Error(N=10)
Error(N=12)
Error(N=14)
(0.25, 0.25)
(0.50, 0.50)
(0.75, 0.50)
(0.75, 0.75)
(0.75, 1.00)
(1.00, 1.00)
0.8248e-13
0.2122e-9
0.2724e-9
0.2306e-7
0.5345e-6
0.6862e-6
0.3921e-16
0.4046e-12
0.5195e-12
0.9925e-10
0.4102e-8
0.5268e-8
0.4000e-19
0.5585e-15
0.7172e-15
0.3090e-12
0.2276e-10
0.2922e-10
Example 4.3 We consider in this example a linear integro-differential equation as
∫ t∫ x
∂ 2 u(x, t) ∂ 2 u(x, t)
1
+
+
3u(x,
t)
−
u(y, z)dydz = −2sintcos(x + t)
2
2
∂x
∂t
0
0 cosz
−sinx − sint + sin(x + t), x, tϵ[0, 1].
with supplementary conditions
u(x, 0) = sinx
∂u
(x, 0) = cosx
∂t
and the exact solution u(x, t) = costsin(x + t).
Similar to the previous examples, we obtain
(n + 1)(n + 2)(n + 3)
[ ( 2
n−1
∑
d
d d Uk+1
Un+3 (x) = Un (x) −
(k + 1)
+ 3Uk+1 (x)
dx
dx
dx2
k=0
+(k + 2)(k + 3)Uk+3 (x) + 2
k+1
∑
r=0
rπ
(k + 1 − r)π
1
sin( )sin(x +
) + sinxδk+1,0
r!(k + 1 − r)!
2
2
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162
International Journal of Nonlinear Science, Vol.16(2013), No.2, pp. 152-163
+
1
(k + 1)π
(k + 1)π
(sin(
) − sin(x +
))
(k + 1)!
2
2
)] [
]
1
(n − k)π
cos(
)
(n − k)!
2
(33)
And from the given conditions, we have
U0 (x) = sinx, U1 (x) = cosx
and by setting t = 0 in the equation (32) we have
U2 (x) = −sinx.
Finally we obtain Un (x) for n = 3, ..., N from the recurrence relation (33).
2
U3 (x) = − cos(x)
3
U6 (x) = −
U4 (x) =
2
sin(x)
45
2
cos(x)
2835
U12 (x) =
4
sin(x)
467775
U10 (x) = −
U5 (x) =
4
cos(x)
315
U8 (x) =
2
sin(x)
14175
U11 (x) = −
U7 (x) = −
U9 (x) =
1
sin(x)
3
U13 (x) =
4
cos(x)
6081075
2
cos(x)
15
1
sin(x)
315
4
cos(x)
155925
U14 (x) = −
4
sin(x)
42567525
And thus, Table 3 shows the absolute errors for this example.
Table 3: Errors of example 4.3
5
(x,t)
Error(N=10)
Error(N=12)
Error(N=14)
(0.25, 0.25)
(0.50, 0.50)
(0.50, 0.75)
(0.75, 0.75)
(1.00, 0.75)
(1.00, 1.00)
0.5854e-11
0.1042e-7
0.8731e-6
0.6903e-6
0.4645e-6
0.9992e-5
0.9399e-14
0.6739e-10
0.1277e-7
0.1018e-7
0.6958e-8
0.2710e-6
0.1120e-16
0.3229e-12
0.1382e-9
0.1109e-9
0.7661e-10
0.5370e-8
Conclusion
In this study, we applied the reduced form of the two-dimensional differential transform method for solving the linear and
nonlinear Volterra partial integro-differential equations. For illustration purposes, we solved some examples. The results
of the examples showed that this method enjoys high accuracy. Moreover, this method can be applied to many similar
equations without linearization, discretization and perturbation. Also, since this is a simple method, it can be used by
researchers in applied sciences and engineering.
To close the discussion, it seems to me that RDTM can be developed for solving Fredholm partial integro-differential
equations, as well as systems of Volterra and Fredholm partial integro-differential equations.
Acknowledgment
The author would like to thank Dr. Sedaghat Shahmorad from University of Tabriz for his careful reading of the preliminary forms of this paper and providing me with useful comments which greatly improved the quality of this paper.
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