International Journal of Pure and Applied Mathematics
Volume 78 No. 3 2012, 299-308
ISSN: 1311-8080 (printed version)
url: http://www.ijpam.eu
AP
ijpam.eu
APPLICATION OF DIFFERENTIAL TRANSFORMATION
METHOD FOR SOLVING A FOURTH-ORDER
PARABOLIC PARTIAL DIFFERENTIAL EQUATIONS
B. Soltanalizadeh
Department of Mathematics
Sarab Branch
Islamic Azad University
Sarab, IRAN
Abstract: In this research, the Differential Transformation method (DTM)
has been utilized to solve a fourth-order parabolic partial differential equations.
This method can be used to obtain the exact solutions of this equation. In the
end, some numerical tests are presented to demonstrate the effectiveness and
efficiency of the proposed method.
AMS Subject Classification: 35K25, 65M70
Key Words: differential transformation method, spectral method, parabolic
equation
1. Introduction
In this paper the Differential transformation method has been utilized for solving the following forth order partial differential equations expectrul
∂4u
∂2u
+
α
= f (x, t)
∂t2
∂x4
(1)
where t is time, x is a spatial coordinate, α is a positive constant and f (x, t)
Received:
April 22, 2011
c 2012 Academic Publications, Ltd.
url: www.acadpubl.eu
300
B. Soltanalizadeh
are continuous functions.
A family of Bspline methods has been considered for the numerical solution
of the problem (1) by H. Caglar and N. Caglar[23]. To solve these types of
problems various methods are proposed by different authors [24, 25, 26, 29, 3,
4, 6, 10, 11, 12].
The concept of the DTM was first proposed by Zhou [1], who solved linear
and nonlinear problems in electrical circuit problems. Chen and Ho [2] developed this method for partial differential equations and Ayaz [13] applied it
to the system of differential equations. During recent years many authors has
been used this method for solving various types of equations. For example,
this method has been used for differential algebraic equations[19], partial differential equations[16, 17, 18], fractional differential equations[21, 22], Volterra
integral equations[30] and Difference equations[20]. In [5, 7, 8, 9], this method
has been utilized for Telegraph, Kuramoto-Sivashinsky, RLW and Kawahara
equations. S. Shahmorad et al. developed DTM to fractional-order integrodifferential equations with nonlocal boundary conditions[28] and class of twodimensional Volterra integral equations[27]. Abazari et. al. applied this method
for Burgers[14], Schrödinger equations[15].
2. The Definitions and Operations of DT
2.1. The One-Dimensional Differential Transform
The basic definitions and operations of one-dimensional DT are introduced in
[1, 2] as follows:
Definition 2.1. If u(t) is analytic in the time domain T then
dk u(t)
= ϕ(t, k), ∀t ∈ T.
(2)
dtk
for t = ti , ϕ(t, k) = ϕ(ti , k), where k belong to the non-negative integer,
denoted as the K domain. Therefore, equation (2) can be rewritten as
h dk u(t) i
Ui (k) = ϕ(ti , k) =
, ∀t ∈ T.
(3)
dtk t=ti
where Ui (k) is called the spectrum of u(t) at t = ti in the K domain.
Definition 2.2. If ut is analytic, then it can be shown as
∞
X
(t − ti )k
U (k).
u(t) =
k!
k=0
(4)
APPLICATION OF DIFFERENTIAL TRANSFORMATION...
301
Equation (4) is known as the inverse transformation of U(k). If U(k) is
defined as
h dk q(t)u(t) i
U (k) = M (k)
, k = 0, 1, 2, ....
(5)
dtk
t=ti
then the function u(t) can be described as
∞
u(t) =
1 X (t − ti )k U (k)
.
q(t)
k!
M (k)
(6)
k=0
where M (k) 6= 0, q(t) 6= 0. M (k) is called the weighting factor and q(t) is
regarded as a kernel corresponding to u(t). If M (k) = 1 and q(t) = 1 then Eqs.
(4) and (6) are equivalent. In this paper, the transformation with M (k) = 1/k!
and q(t) = 1 is applied. Thus from equation (6), we have
U (k) =
1 h dk u(t) i
,
k! dtk t=ti
k = 0, 1, 2, ....
(7)
Using the differential transform, a differential equation in the domain of interest
can be transformed to be an algebraic equation in the K domain and u(t) can
be obtained by finite-term Taylor series plus a remainder, as
u(t) =
n
n
k=0
k=0
X
1 X (t − ti )k U (k)
+ Rn+1 (t) =
(t − t0 )k U (k) + Rn+1 (t). (8)
q(t)
k!
M (k)
In order to speed up the convergence rate and improve the accuracy of calculation, the entire domain of t needs to be split into sub-domains.
2.2. The Two-Dimensional Differential Transform
Consider a function of two variables w(x, t), and suppose that it can be represented as a product of two single-variable functions, i.e., w(x, t) = f (x)g(t).
Based on the properties of one-dimensional differential transform, function
w(x, t) can be represented as
w(x, t) =
∞ X
∞
X
W (i, j)xi tj .
(9)
i=0 j=0
where W (i, j) is called the spectrum of w(x, y). Now we introduce the basic
definitions and operations of two-dimensional DT as follows.
Definition 2.3. If w(x, t) is analytic and continuously differentiable with
respect to time t in the domain of interest, then
302
B. Soltanalizadeh
W (h, k) =
i
1 h ∂ k+h
w(x,
t)
k!h! ∂xk ∂th
x=x0 ,
t=t0
,
(10)
where the spectrum function W (k, h) is the transformed function, which is also
called the T-function. Let w(x, y) as the original function while the uppercase
W (k, h) stands for the transformed function. Now we define The differential
inverse transform of W (k, h) as following:
w(x, t) =
∞
∞ X
X
W (k, h)(x − x0 )k (t − t0 )h .
(11)
k=0 h=0
Using equation (10) in equation (9), we have
w(x, t) =
∞ X
∞
i
X
1 h ∂ k+h
w(x,
t)
k!h! ∂xk ∂th
x=x0 ,
k=0 h=0
t=t0
xk t h
=
∞ X
∞
X
W (k, h)xk th . (12)
k=0 h=0
Now from the above definitions and equation (11) and (12), we can obtain some
of the fundamental mathematical operations performed by two-dimensional differential transform in Table 1.
3. Application of the DTM
In this section, we apply the DTM to the presented fourth order equation.
Consider the equation (1),
∂4u
∂2u
+
α
= f (x, t)
∂t2
∂x4
(13)
u(x, 0) = f (x), 0 ≤ x ≤ L.
(14)
with the initial conditions
and
∂u(x, 0)
= g(x), 0 ≤ x ≤ L.
(15)
∂t
Let U (k, h) as the differential transform of u(x, t). Applying Table 1., equation (1) and Definition 2.3 when x0 = t0 = 0, we get the differential transform
APPLICATION OF DIFFERENTIAL TRANSFORMATION...
Original function
Transformed function
w(x, t) = u(x, t) ± v(x, t)
W (k, h) = U (k, h) ± V (k, h)
w(x, t) = cu(x, t)
W (k, h) = cU (k, h)
∂
u(x, t)
∂x
∂ r+s
w(x, t) =
u(x, t)
∂xr ∂ts
w(x, t) =
W (k, h) = (k + 1)U (k + 1, h)
(k + r)!(h + s)!
U (k + r, h + s)
k!h!
k X
h
X
U (r, h − s)V (k − r, s)
W (k, h) =
W (k, h) =
w(x, t) = u(x, t)v(x, t)
w(x, t) =
303
∂u(x, t) ∂v(x, t)
∂x
∂x
r=0 s=0
k X
h
X
W (h, k) =
(r + 1)(k − r + 1)
r=0 s=0
×U (r + 1, h − s)V (k − r + 1, s)
Table 1: Operations of the two-dimensional differential transform
version of equation (13) as following:
(h+2)!
h! U (k, h
= f (k, h).
+ 2) + α (k+4)!
k! U (k + 4, h)
(16)
By the first initial condition we get
∞
X
U (k, 0)xk =
∞
X
f k (0)
k=0
k=0
k!
xk .
(17)
which implies
f k (0)
, k = 0, 1, 2, ...,
k!
and from the second initial condition and Table.1. we have
U (k, 0) =
∞
X
k
U (k, 1)x =
∞
X
g k (0)
k=0
k=0
k!
xk .
(18)
(19)
which implies
U (k, 1) =
gk (0)
,
k!
k = 0, 1, 2, ....
(20)
304
B. Soltanalizadeh
Therefore, the value of U (k, 0) and U (k, 1), for k = 0, 1, 2, ..., can be obtained
from Eqs. (18,20). Then by using equation (16), we find the remainder values
of U as following:
U (k, h + 2) =
k!
1
f (k, h) −
U (k + 4, h) ,
(h + 1)(h + 2)
(k + 4)!
k = 0, 1, ..., h = 0, 1, .... (21)
Example 3.1. Consider the following fourth order equation
∂4u
∂2u
+
2
= 3ex+t
∂t2
∂x4
with the initial conditions
u(x, 0) = ex , 0 ≤ x ≤ L.
and
∂u(x, 0)
= ex , 0 ≤ x ≤ L.
∂t
From the first initial condition and equation (18), we have
U (i, 0) =
1
, i = 0, 1, 2...
i!
and from the second initial condition and equation (20), we get
U (i, 1) =
1
, i = 0, 1, 2...
i!
Now if we utilize the equation (21), then the remainder value of U (k, h) can be
computed as follows:
U (0, 2) =
1
1
1
1
, U (1, 2) = , U (2, 2) =
, U (3, 2) =
, ...
2!
2!
2!2!
3!2!
U (0, 3) =
1
1
1
1
, U (1, 3) = , U (2, 3) =
, U (3, 3) =
, ...
3!
3!
2!3!
3!3!
U (0, 4) =
1
1
1
1
, U (1, 4) = , U (2, 4) =
, U (3, 4) =
, ...
4!
4!
2!4!
3!4!
and
and
APPLICATION OF DIFFERENTIAL TRANSFORMATION...
305
..
..
By continuing this process, we obtain
1
1 2
t + ... + tn + x + xt
2!
n!
1 2
1 n
1
1
1 2 2
1 n n
+ xt + ... + xt + + x2 + x2 t +
x t + ... +
x t + ...,
2!
n!
2!
2!
2!2!
n!n!
which is the Taylor expansion of the
u(x, t) ≃ 1 + t +
u(x, t) = ex+t
which is the exact solution of the Example 3.1.
Example 3.2. Consider the following fourth order equation
∂4u
∂2u
− 2 4 = −sin(x)et
2
∂t
∂x
with the initial conditions
u(x, 0) = sin(x), 0 ≤ x ≤ L.
and
∂u(x, 0)
= sin(x), 0 ≤ x ≤ L.
∂t
By applying equation (18) to the first initial condition we have
(
0,
for i is even,
U (i, 0) =
(−1)i−1/2
,
for i is odd,
i!
and for the second initial condition, we have from e q. (20)
(
0,
for i is even,
U (i, 1) =
(−1)i−1/2
,
for i is odd,
i!
and from the equation (21), the remainder value of U can be found ass follows:
(
0,
for i is even,
U (i, 2) =
(22)
(−1)i−1/2
,
for i is odd,
i!2!
By continuing this process, we obtain
u(x, t) = sin(x)et
which is the exact solution of the Example 3.2.
306
B. Soltanalizadeh
4. Conclusions
In this paper, we solved a fourth order equation with initial conditions by
the Differential transformation method. We find the analytical solutions by
applying this method and obtaining a simple iterative process. Consequently,
it is seen that this method can be an alternative way for the solution of partial
differential equations that have no analytic solutions.
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