Am26242246

P. R. Mistry, V. H. Pradhan / International Journal of Engineering Research and Applications
(IJERA) ISSN: 2248-9622 www.ijera.com
Vol. 2, Issue 6, November- December 2012, pp.242-246
Analytic Solution Of Burger’s Equations By Variational Iteration
Method
P. R. Mistry* and V. H. Pradhan**
*
(Department of Applied Mathematics & Humanities, S.V.N.I.T., Surat-395007, India)
**
(Departments of Applied Mathematics & Humanities, S.V.N.I.T., Surat-395007, India)

Abstract
By means of variational iteration method the 2 VARIATIONAL ITERATION METHOD:
solutions of (1+1), (1+2) and (1+3) dimensional Consider the following differential equation:
Burger equations are exactly obtained. In this Lu  Nu  h ( x, t ) (2.1)
paper, He's variational iteration method is where L is a linear operator, N a nonlinear
introduced to overcome the difficulty arising in
calculating Adomian polynomials. operator, and an h ( x, t ) is the source
inhomogeneous term. The VIM was proposed by
Key words: Burger’s equation, Nonlinear time “He”, where a correctional functional for equation
dependent partial differential equations, Variational (1.4.1) can be written as
iteration method and Lagrange multiplier. un 1 (t )  un (t ) 
t
, n  0 (2.2)
1 Introduction
We often come with non linear partial
  ( Lu ( )  Nu ( )  h ( )) d
0
n
 n

differential equations obtained through
Where  is a general Lagrange multiplier
mathematical models of scientific phenomena.
which can be identified optimally via the variational
There are some methods to obtain approximate theory. The subscript n indicates the n th
solution of this kind of equation. Some of them are approximation and 
u n is considered as a restricted
numerical methods, homotopy analysis, Exp-
variation, i.e.  un  0 . It is clear that the

function method, and linearization of the equation
successive approximations un , n  0 can be
[1, 6, 7]. In 1999, the variational iteration method
established by determining  , so we first
was developed by mathematician “He”. This method
determine the Lagrange multiplier  that will be
is used for solving linear and non linear differential identified optimally via integration by parts. The
equations. The method introduces a reliable and successive approximation un ( x, t ), n  0 of the
efficient process for a wide variety of scientific and solution u( x, t ) will be readily obtained using the
engineering applications [1, 2, 4, 5,]. It is based on Lagrange multiplier obtained and by using any
Lagrange multiplier and it has the merits of selective function u 0 . The initial values u( x,0)
simplicity and easy execution. This method avoids and ut ( x,0) are usually used for selecting the
linearization of the problem. zeroth approximation u 0 . With  determined, then
In this paper exact solution of (1+1), (1+2) several approximation un ( x, t ), n  0 , follow
and (1+3) dimensional Burger equation [3] has been immediately. Consequently the exact solution may
obtained by variational iteration method. The be obtained by using
mentioned problem has been solved by u  lim un (2.3)
n
N.Taghizadeh etc. [3] by Homotopy Perturbation
Method and Reduced Differential Transformation 3. APPLICATION OF VIM FOR
BURGER’S EQUATION:
Method here in the present paper we have solved the 3.1 (1+1)-Dimensional Burgers equation
same problem by variational iteration mehod.. u  u    2u 
 u     2   0 (3.1.1)
t  x   x 

242 | P a g e

u4  {{{{x(1  t  t 2 2  t 3 3  t 4 4
I.C.: u  x,0  x (3.1.2)
13t 5 5 2t 6 6 29t 7 7 71t 8 8
   
Now following the variational iteration 15 3 63 252
method given in the above section we get the 86t 
9 9
22t 
10 10
5t 
11 11
t12 12
following functional    
567 315 189 126
un 1 ( x, t )  un ( x, t )  t 
13 13
t 
14 14
t 
15 15
   )}}}} (3.1.9)
 un   2u   567 3969 59535
 u 
t
(3.1.3)

0  t
   un n
 x


   2n   d
 x  
u5  {{{{{x(1  t  t 2 2  t 3 3  t 4 4
43t 6 6 13t 7 7 943t 8 8
t 5 5    
Stationary conditions can be obtained as follows: 45 15 1260
3497t 9 9 27523t10 10 1477t11 11
 '    0  
5670 56700 4050
(3.1.4)
1     17779t  12 12
13141t  13 13
1019t14 14
 t   
68040 73710 8820
The Lagrange multiplier can therefore be 63283t  15 15
43363t  16 16
1080013t17 17
simply identified as   1 , and substituting this   
893025 1058400 48580560
value of Lagrange multiplier into the functional
(3.1.3) gives the following iteration equation. 2588t  18 18
162179t  19 19
16511t 20 20
  
229635 30541455 7144200
un 1 ( x, t )  un ( x, t )  207509t  21 21
557t 22 22
2447t 23 23
   
t
 un  u    2 un   (3.1.5) 225042300 1666980 22504230
  t    un xn
0 
  2
  x
  d
 16927t 24 24 5309t 25 25 t 26 26
 
540101520 675126900 595350
As stated before, we can select Initial 2t 27 27 13t 28 28 t 29 29
condition given in the equation (3.1.2) and using   
this selection in (3.1.5) we obtain the following 6751269 315059220 236294415
successive approximations: t 30 30 t 31 31
  )}}}}}(3.1.10)
3544416225 109876902975
u1  {x  tx } (3.1.6)
1 u  x, t   u1  u2  u3  .... (3.1.11)
u2  {{x   (tx  t 2 x  t 3 x 2 )}} (3.1.7)
3 x
u  x, t   (3.1.12)
1
u3  {{{x  tx  t 2 x 2  t 3 x 3 1  t
3
3.2 (2+1)-Dimensional Burger’s equation
1 3 2
 (tx  t 2 x  t x ) 
3 u  u u 
(3.1.8)
2 4 3  u  u  
 (tx  t x  t x  t x
2 3 2
t  x y 
3 (3.2.1)
  2u  2u 
1 1 1
 t 5 x 4  t 6 x 5  t 7 x 6 )}}}  2  2 0
3 9 63  x y 
I.C.: u  x, y,0  x  y (3.2.2)

Now following the variational iteration method, we
get the following functional

243 | P a g e

un 1 ( x, t )  un ( x, t )  u4  {{{{x  y  2tx
 un  u u    2ty  4t 2 x 2 
    un n  un n  
t
 t  x y   (3.2.3) 4t 2 y 2  8t 3 x 3
     2u  2u   d
 8t 3 y 3  16t 4 x 4
0
    2n  2n  
 x y  
    16t 4 y 4 
416 5 5
t x
15
Stationary conditions can be obtained as follows: 416 5 5 128 6 6
 t y  t x
 '    0
15 3
128 6 6 3712 7 7
1    
(3.2.4)
 t y  t x
 t 3 63
3712 7 7 4544 8 8
The Lagrange multiplier can therefore be  t y  t x
simply identified as   1 , and substituting this 63 63
value of Lagrange multiplier into the functional 4544 8 8 44032 9 9
(3.2.3) gives the following iteration equation.  t y  t x
63 567
un 1 ( x, t )  un ( x, t )  44032 9 9 22528 10 10
 t y  t x
567 315
 un  u u  
    un n  un n   
22528 10 10 10240 11 11
t y  t x
 t  x y  
t
(3.2.5)
    2u  2u   d
315 189
10240 11 11 2048 12 12
0
    2n  2n    t y  t x
 x y  
   189 63

2048 12 12 8192 13 13
As stated before, we can select Initial  t y  t x
condition given in the equation (3.2.2) and using 63 567
this selection in (3.2.5) we obtain the following
8192 13 13 16384t14 x 14
successive approximations:  t y 
567 3969
u1  {x  y  2t ( x  y ) } (3.2.6) 16384t y
14 14
32768t15 x 15
 
u2  {{x  y  2tx  2ty  3969 59535
8 32768t y
15 15
4t 2 x 2  4t 2 y 2  t 3 x 3 (3.2.7)  }}}} (3.2.9)
3 59535
8
 t 3 y 3}}
3
u3  {{{x  y  2tx  2ty 
4t 2 x 2  4t 2 y 2  8t 3 x 3  8t 3 y 3
32 32 32
 t 4 x 4  t 4 y 4  t 5 x 5 (3.2.8)
3 3 3
32 64 64
 t 5 y 5  t 6 x 6  t 6 y 6
3 9 9
128 7 7 128 7 7
 t x  t y }}}
63 63

244 | P a g e

u5  {{{{{x  y  2tx  2ty un 1 ( x, t )  un ( x, t ) 
4t 2 x 2  4t 2 y 2  8t 3 x 3  un  u u u  
    un n  un n  un n  
8t y  16t x  16t y
3 3 4 4 4 4 t
t  x y z  

(3.3.3)
2752 6 6     2u  2u  2u   d
32t 5 x 5  32t 5 y 5  t x 0
   2  2  2 
n n n 
45  x y z  
  
2752 6 6 1664 7 7
 t y  t x
45 15 Stationary conditions can be obtained as follows:
1664 7 7 60352 8 8
 t y  t x  '    0
15 315 (3.3.4)
60352 8 8 895232t 9 x 9 1      t
 t y 
315 2835
The Lagrange multiplier can therefore be
895232t y 9 9
7045888t10 x 10 simply identified as   1 , and substituting this
 
2835 14175 value of Lagrange multiplier into the functional
(3.3.3) gives the following iteration equation.
7045888t y 10 10
1512448t11 x 11
 
14175 2025 un 1 ( x, t )  un ( x, t ) 
1512448t y 11 11
9102848t12 x 12  un
   u u u  
2025 8505     un n  un n  un n  
 t  x y z  
t
(3.3.5)
9102848t y 53825536t13 x 13     2u  2u  2u   d
12 12
 
8505 36855 0
   2  2  2 
n n n 
 x y z  
53825536t y 13 13
4173824t14 x 14   

36855 2205 As stated before, we can select Initial condition
4173824t y 14 14
2073657344t15 x 15 given in the equation (3.3.2) and using this selection
  in (3.3.5) we obtain the following successive
2205 893025 approximations:
2073657344t y 15 15

893025 u1  {x  y  z  3t ( x  y  z ) } (3.3.6)
88807424t16 x 16 u2  {{x  y  z  3tx  3ty
  .......}}}}} (3.2.10)
33075  3tz  9t 2 x 2  9t 2 y 2
(3.3.7)
 9t 2 z 2  9t 3 x 3
u  x, t   u1  u2  u3  .... (3.2.11)
x y  9t 3 y 3  9t 3 z 3}}
u  x, y , t   (3.2.12) u3  {{{x  y  z  3tx  3ty
1  2 t
3tz  9t 2 x 2  9t 2 y 2
3.3 (3+1)-Dimensional Burger’s equation
u  u u u  9t 2 z 2  27t 3 x 3  27t 3 y 3
 u  u  u 
t  x y z  27t 3 z 3  54t 4 x 4  54t 4 y 4
(3.3.1)
  2u  2u  2u  54t 4 z 4  81t 5 x 5  81t 5 y 5 (3.3.8)
  2  2  2   0
 x y z  81t z  81t x  81t y
5 5 6 6 6 6

I.C.: u  x, y, z,0  x  yz (3.3.2) 81t 6 z 6 
243 7 7 243 7 7
t x  t y
Now following the variational iteration method, we 7 7
get the following functional 243 7 7
 t z }}}
7

245 | P a g e

u4  {{{{x  y  z  3tx  3ty iteration method is a powerful mathematical tool to
solving Burger’s equations. In our work, we use the
 3tz  9t 2 x 2  9t 2 y 2  9t 2 z 2 Mathematica Package to calculate the series
obtained from the variational iteration method.
 27t 3 x 3  27t 3 y 3  27t 3 z 3
 81t 4 x 4  81t 4 y 4  81t 4 z 4 References
1. Sadighi A and Ganji D, Exact solution of
1053 5 5 1053 5 5
 t x  t y nonlinear diffusion equations by variational
5 5 iteration method, Commuters and
Mathematics with Applications, 54, (2007),
1053 5 5
 t z  486t 6 x 6 pp. 1112-1121.
5 2. He.J.H. Variational iteration method -- a
1053 5 5 kind of non-linear analytical technique:
 t z  486t 6 x 6 some examples, Internat. J. Nonlinear
5 Mech. 34, (1999).,pp. 699–708
486t 6 y 6  486t 6 z 6 3. N.Taghizadeh, M.Akbari and
A.Ghelichzadeh, Exact Solution of Burgers
7047 7 7 7047 7 7
 t x  t y Equations by Homotopy Perturbation
7 7 Method and Reduced Differential
Transformation Method, Australian Journal
7047 7 7 51759 8 8
 t z  t x of Basic and Applied Sciences, 5(5):
7 28 (2011),pp. 580-589,
 ...........}}}} (3.3.9) 4. Salehpoor E. and Jafari H., Variational
iteration method: A tools for solving partial
u  x, y, z, t   u1  u2  u3  ... (3.3.10) differential equations, The journal of
Mathematics and Computer Science, 2,
x yz
u  x, y , z , t   (3.3.11) (2011)pp. 388-393.
1  3 t 5. Wazwaz A. M.., The variational iteration
method: A Powerful scheme for handling
3.4 (n+1)-Dimensional Burger’s equation linear and nonlinear diffusion equations.
Computer and Mathematics with
u  u u u u  Applications 54, (2007) pp.933-939.
 u u u  ......  u  6. M.A. Abdou and A.A. Soliman,
t  x1 x2 x3 xn  Variational iteration method for solving
Burger’s and coupled Burger’s equations
  2u  2u  2u  2u  Journal of Computational and Applied
   2  2  2  ....  0 Mathematics 181 (2005), pp. 245 – 251.
 x1 x2 x3 xn 2 
7. Junfeng Lu, Variational iteration method
(3.4.1) for solving a nonlinear system of second-
u  x1 , x2 , x3 ,.........., xn , 0  order boundary value problems, Computers
I.C.: (3.4.2) and Mathematics with Applications 54
 x1  x2  ......  xn (2007), pp.1133–1138.
Similarly, we apply VIM on the equation (3.4.1) we
get the following exact solution.
u  x1 , x2 , x3 ,.............., xn , t  
 x1  x2  x3  x4  ............  xn  (3.4.3)

1  n t

4. Conclusion
In this paper, the variational iteration
method has been successfully applied to finding the
solution of (1+1), (1+2) and (1+3) dimensional
Burger’s equations. The solution obtained by the
variational iteration method is an infinite power
series for appropriate initial condition, which can, in
turn, be expressed in a closed form, the exact
solution. The results show that the variational

246 | P a g e

Am26242246

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