0
1 0 1
, , ( , ),..., ( , ), ,
0
, , ,..., ,
, 0 , , ,..., , ,
m m n
n n
n
t x u x t u x t u x t
x x a
f u x t u x t
t a t a
t x a a a m n N
 ∂ ∂  
 ∂ ∂  
   =
 
 
 
∂ ∂
     
 ∂   ∂   
 
≥ ∈
( , ,1 2 , ) u x x xn 1 2 1 1 2
0
( , , , )
( , , , , ) 1
!
n
k n
n k
n
x
u x x x
U x x x k
k x
−
=
 
∂ 
=  
 ∂ 
 
 
ISSN 1818-4952
© IDOSI Publications, 2014
DOI: 10.5829/idosi.wasj.2014.30.03.1841
Application of Projected Differential Transform Method on Nonlinear Partial Differential Equations with Proportional Delay in One Variable
Tarig M. Elzaki
Department of Mathematics, Faculty of Sciences and Arts-Alkamil, King Abdulaziz University, Jeddah-Saudi Arabia
Submitted: Dec 11, 2013; Accepted: Feb 5, 2014; Published:Feb 9, 2014
Abstract:In this paper we solve nonlinear partial differential equations with proportional delay in the variable t using the projected differential transform method. The purpose of the method is obtained analytical or approximate solutions of some nonlinear partial differential equations with proportional delay. This method is more efficient and easy to handle such partial differential equations with proportional delay in comparison to other methods. The result showed the efficiency, accuracy and validation of the projected differential transform method.
Keywords: Projected Differential Transform Method Nonlinear Partial Differential Equations with Proportional Delay Exact Solutions.
INTRODUCTION In this work we will present the projected differential Several numerical and analytical techniques including solve some nonlinear partial differential equations with the Adomian's decomposition method [1], differential proportional delay in the variable t.
transform method [2-9], homotopy perturbation method In this paper we consider the following nonlinear [10-13] and projected differential transform method [14,15], partial differential equations with proportional delay in the have been developed for solving nonlinear delay variablet.
differential equations
Delay partial differential equations arise from various applications, like biology, medicine, control theory,
climate models and many others. Their independent (1)
variables are time t and one or more dimensional variable x , which usually represents position in space but may also represent relative DNA content, size of cells, or their maturation level, or other values. The solutions
(dependent variables) of delay partial differential Projected Differential Transform Methods: In this equations may represent temperature, voltage, or section we introduce the projected differential transform concentrations or densities of various particles, for method [14, 15] which is modified method of the example cells, bacteria, chemicals, animals and so on. differential transform method.
Delay partial differential equations depend on their
solutions at retarded variables or/and they depend on Definition:The basic definition of projected differential smooth averages of the solutions over some retarded transform method of function is defined as intervals. Although delay partial differential equations are
also called partial functional differential equations, the class of functional equations is wider than the class of delay equations.
transform method and some theorems with their proofs to
(2)
( , ,1 2 , ) u x x xn ( , ,1 2 , 1, ) U x x xn− k
( , ,1 2 , 1, ) U x x xn− k
1 2 1 2 1 0
( , , , )n ( , , ., n , )( )k
k o
u x x x ∞U x x x − k x x
=
=
∑
− 
( ) ( ) ( ) ( )
(
1 21 2 1) (
1 21 2 11 2) (
1 2 1)
1 , ,..., , ,..., , ,...,
, ,..., , , ,..., , , ,..., ,
n n n
n n n
If z x x x u x x x v x x x
Then z x x x k u x x− x k v x x− x k−
= ±
= ±
( ) ( ) ( )
(
1 21 2 1)
1 2(
1 2 1)
2 , ,..., , ,...,
, ,..., , , ,..., ,
n n
n n
Ifz x x x cu x x x
Then z x x x − k cu x x x − k
=
=
( ) ( ) ( )
( ) ( ) ( )
1 2 1 2
1 2 1 1 2 1
, ,..., 3 , ,...,
, ,..., , 1 , ,..., , 1
n n
n
n n
du x x x Ifz x x x
dx
Then z x x x − k k u x x x − k
=
= + +
( ) ( ) ( )
( ) ( ) ( )
1 2 1 2
1 2 1 1 2 1
, ,..., 4 , ,...,
, ,..., , ! , ,..., ,
!
n n
n n
n
n n
d u x x x Ifz x x x
dx
Then z x x x k k n u x x x k n
− k −
=
= + +
( ) ( ) ( ) ( )
( )
( ) ( )
1 2 1 2 1 2
1 2 1
1 2 1 1 2 1
0
5 , ,..., , ,..., , ,...,
, ,..., ,
, ,..., , , ,..., ,
n n n
n k
n n
m
If z x x x u x x x v x x x
Then z x x x k
u x x x m v x x x k m
−
− −
=
=
=
∑
−( ) ( ) ( )
( ) ( )
( )
( )
( )
1 3 2
1 2 2 1
1 2 1 1 2
2 1 2 1 2
1 2 1
1 1 2 1 1
0 0 0 0
2 1 2 1 2 1
1 1
6 , ,..., , ,..., , ,..., ... , ,..., , ,..., ,
... , ,..., , , ,..., ,
... ,
n
n n
n n
n n n
n
k k k
k
n
k k k k
n n
If z x x x u x x x
u x x x u x x x Then
z x x x k
u x x x k
u x x x k k
u x
−
− −
−
−
= = = =
−
−
=
=
−
×
∑ ∑ ∑ ∑
( )
(
1 2 2 1 1 1)
1 2,..., , , ,..., ,
n n n
n n n
x x k k
u x x x k k
− − −
− −
−
−
( , ), ( , ) Z x k U x k
( , ), ( , ) z x t u x t a N∈
( , ) ,t z x t u x
a
 
=  
( , ) ( , ) U x kk Z x k
= a ( , ) mm ,t ,
z x t u x t a
∂  
=∂  
( )! ( , )
( , )
! k m
k m U x k m z x k
k a +
+ +
=
( , ) , 1 ,
k k k
kz x t k u x t k k u x t
a a
t t a t
   
∂ = ∂   = ∂   
∂ ∂    ∂   
0 0
( , ) 1 , ! ( , )
k k
k k k k
t t
z x t u x t k U x k
t = a t a = a
∂  = ∂   =
 ∂  ∂  
   
   
0
1 ( , ) 1 ! ( , ) ( , )
( , )
! !
k
k k k
t
z x t k U x k U x k Z x k
k t = k a a
∂   
=  ∂  =  =
0 0
( , ) 1 , ( )! ( , )
k k m
k k m k m k m
t t
z x t u x t k m U x k m
t a t a a
+
+ + +
= =
∂  = ∂   = + +
 ∂  ∂  
   
   
0
1 ( , ) ( , )
!
1 ( )! ( , ) ( )! ( , )
! !
k
k t
k m k m
z x t Z x k
k t
k m U x k m k m U x k m
k a k a
=
+ +
∂ 
=  
 ∂ 
 
+ + + +
 
=  =
Such that is the original function Application of ProjectedDifferential Transform Method and is projected transform function.
And the differential inverse transform of approximating equation (1).
is defined as
(3)
The fundamental theorems of the projected differential transform are
Theorems 1:
Note that c is a constant and n is a nonnegative integer.
on Equation (1): In this section we introduce some theorems in projected differential transform method for
Theorem 2: Let a re the projected differential transform with respect tot of and
, then
(I) If , then
(ii) If then
Proof:
(I) From (2), we have
Therefore
Then by (2), we get
Such that k = 0,1,2,...., (ii) From (i), we have
Then
( , ), ( , ), ( , ) Z x k U x k V x k
( , ), ( , ), ( , ) z x t u x t v x t
( , ) ,t ,t
z x t u x v x
a b
   
=     
0
( , ) ( , )
( , ) k r k r , 0,1,2,...,
r
U x r V x k r
Z x k k
a b −
=
=
∑
− = ∞0
( , ) , ,
1 , 1 ,
k k
k k
k k k r
r k k r k r
r
t t
z x t u x v x
a b
t t
k u x t v x t
r a t a b x a
−
− −
=
 
∂ = ∂      
∂ ∂     
   
  ∂   ∂  
=
∑
   ∂    ∂   0 0
0
( , ) ! ( , )
( )! ( , ) ! ( , ) ( , )
k k
k r
t r
k
k r r k r
r
z x t k r U x r
t r a
k r V x k r k U x r V x k r
b a b
= =
− −
=
∂  =   
 ∂     
   
 
− − −
 =
 
 
∑
∑
0
( , ) ( , )
( , ) k r k r , 0,1,2,...,
r
U x r V x k r
Z x k k
a b −
=
=
∑
− = ∞( , ), ( , ), ( , ) Z x k U x k V x k
( , ), ( , ), ( , ) z x t u x t v x t , , ,
a b n m N∈
( , ) nn ,t mm ,t
z x t u x v x
a a
t x
∂   ∂  
=∂  ∂  
0
( )!( )!
( , ) ( , )
( , ) , 0,1,2,...,
!( )!
k
r n k r m r
r n k r m U x r n V x k r m
Z k x k
r k r a + b − +
=
+ − +
+ − +
= = ∞
∑
−0
( , ) , ,
1 , 1 ,
k k n m
k k n m
k r n k r m
r n r n k r m k r m
r
t t
z x t u x v x
a a
t t t t
k u x t v x t
r a t a b t a
+ − +
+ + − + − +
=
 
∂ = ∂ ∂   ∂  
∂ ∂ ∂  ∂  
   
  ∂   ∂  
=
∑
   ∂    ∂   0 0
0
( , ) ( )! ( , )
( )! ( , )
!( )!( )! ( , ) ( , )
!( )!
k k
k r n
t r
k r m k
r n k r m r
z x t k r n U x r n
t r a
k r m V x k r m b
k r n k r m U x r n V x k r m r k r a b
= +
=
− +
+ − +
=
∂  =   + + 
 ∂    
   
 
− + − +
 
 
 
+ − + + − +
= −
∑
∑
0
( )!( )!
( , ) ( , )
( , ) , 0,1,2,...,
!( )!
k
r n k r m r
r n k r m U x r n V x k r m
Z k x k
r k r a + b − +
=
+ − +
+ − +
= = ∞
∑
−2 , . , ( , ) , , 0, ( ,0)
2 2
t t
u x u x xu x t x t u x x
t
  ∂   = ≥ =
  ∂  
   
0 1
( 1) ( , ) ( , 1)
2 ( , )
2 .2
k
r k r r
k r U x r U x k r− + xU x k
=
− + − + =
∑
( ,0) U x =x
( ,1) , ( ,2) , ( ,3) ,
2 6
( ,4) ,..., ( , )
24 !
x x
U x x U x U x
x x
U x U x n
n
= = =
= =
0
( , )
! k t
k
u x t x t xe k
∞
=
=
∑
=Theorem 3: Let are the projected Therefore differential transform with respect to t of
respectively and a, b, N, then
If , then
Proof: From theorem (2) and Leibnitz, we get
Therefore
And from equation (2), we get
Theorem 4: Let are the projected differential transform with respect to t of
respectively and , then
If , then
Proof:From theorem (7) and Leibnitz, we get
And from equation (2), we get
Applications:In this section we illustrate some examples to explain the presented method. We choose examples have exact solutions.
Example 1:In the first example, consider the following nonlinear first-order partial differential equation with proportional delay in the variable t.
(4)
Using the projected differential transform method with respect to t of Eq. (4), yields
(5)
The projected differential transform method with respect tot of initial condition is , then by this equation and Eq. (5), we have
Substituting in Eq.(3), to get the exact solution of Eq.
(4), in the form
Example 2: Consider the following nonlinear second-order partial differential equation with proportional delay in the variablet.
2
( , ) 42 , . , ( , ) 0, , 0
2 2
u x t u x t u x t u x t x t
x t
t
∂ ∂ − ∂∂   ∂∂  + = ≥
( ,0) ( ,0) , u x
u x x x
t
= ∂ =
∂
0 1
( 2)( 1) ( , 2) ( , )
( , )( 1) ( , 1)
4 0
2 .2
k
r k r r
k k U x k U x k
U x r k r U x k r
x − +
=
+ + + +
∂ − + − +
−
∑
∂ =( ,0) ( ,0) , u x
u x x x
t
= ∂ =
∂
( ,0) ( ,1) U x =U x =x
( ,2) , ( ,3) , ( ,4) ,..., ( , )
2 6 24 !
x x x x
U x U x U x U x n
= = = =n
0
( , )
! k t
k
u x t x t xe k
∞
=
=
∑
=2 2
2 2
( , )
16 , . , cos 0 , , 0
2 2
t t u x t
u x u x x x t
t t x
∂   ∂   + ∂ = ≥
∂   ∂   ∂
( ,0)
( ,0) 0, u x cos
u x x
t
= ∂ =
∂
( 1)( 2)( 1)
k r+ k r− + k r− +
( ,0) ( ,0) 0, u x cos
u x x
t
= ∂ =
∂
( ,0) 0, ( ,1) cos
U x = U x = x
cos cos
( ,2) 0 , ( ,3) , ( ,4) 0, ( ,5) ,...
3! 5!
x x
U x = U x = U x = U x =
( , ) sinh cos u x t = t x
(6) And the projected differential transform with respect
With the initial conditions
(7)
Taking the projected differential transform method with respect to t of Eq. (6), we have
(8)
And the projected differential transform with respect tot of initial conditions will be
(9) Where U(x,k) is the projected differential transform method with respect to t of U(x,t)
Using Eqs. (8) and (9), we have
Substituting in Eq.(3), to get the exact solution of Eq.
(6), in the form
Example 3: Consider the following nonlinear second-order partial differential equation with proportional delay in the variablet.
(10)
With the conditions
(11) Taking the projected differential transform with respect to t of Eq. (10), we have
tot of initial condition will be (13) Using Eqs. (12) and (13), we have
Substituting in Eq.(3), to get the exact solution of Eq.
(10), in the form
CONCLUSION
In this paper we applied the projected differential transformation technique to solve nonlinear partial differential equation with proportional delay in the variablet. The obtained results show that this method is powerful and meaningful for solving partial differential equations with proportional delays. In fact, PDTM is very efficient methods to find the numerical and analytic solutions of partial differential-difference equations, delay partial differential equations as well as integral equations.
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