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Am26242246

  1. 1. 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.242-246Analytic 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)AbstractBy 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, Hes variational iteration method is where L is a linear operator, N a nonlinearintroduced to overcome the difficulty arising incalculating Adomian polynomials. operator, and an h ( x, t ) is the source inhomogeneous term. The VIM was proposed byKey words: Burger’s equation, Nonlinear time “He”, where a correctional functional for equationdependent partial differential equations, Variational (1.4.1) can be written asiteration 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  ndifferential equations obtained through Where  is a general Lagrange multipliermathematical models of scientific phenomena. which can be identified optimally via the variationalThere are some methods to obtain approximate theory. The subscript n indicates the n thsolution of this kind of equation. Some of them are approximation and  u n is considered as a restrictednumerical 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 firstwas developed by mathematician “He”. This method determine the Lagrange multiplier  that will beis used for solving linear and non linear differential identified optimally via integration by parts. Theequations. The method introduces a reliable and successive approximation un ( x, t ), n  0 of theefficient process for a wide variety of scientific and solution u( x, t ) will be readily obtained using theengineering applications [1, 2, 4, 5,]. It is based on Lagrange multiplier obtained and by using anyLagrange 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 thelinearization 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 , followand (1+3) dimensional Burger equation [3] has been immediately. Consequently the exact solution mayobtained by variational iteration method. The be obtained by usingmentioned problem has been solved by u  lim un (2.3) nN.Taghizadeh etc. [3] by Homotopy PerturbationMethod 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 equationsame problem by variational iteration mehod.. u  u    2u   u     2   0 (3.1.1) t  x   x  242 | P a g e
  2. 2. 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 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 126un 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
  3. 3. 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.242-246un 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 40     2n  2n    x y       16t 4 y 4  416 5 5 t x 15Stationary conditions can be obtained as follows: 416 5 5 128 6 6  t y  t x    0 15 3 128 6 6 3712 7 71     (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 63value 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 567un 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 12 120     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 567this selection in (3.2.5) we obtain the following 8192 13 13 16384t14 x 14successive approximations:  t y  567 3969u1  {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}} 3u3  {{{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
  4. 4. 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.242-246u5  {{{{{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 3u  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 23.3 (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 6I.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 7get the following functional 243 7 7  t z }}} 7 245 | P a g e
  5. 5. 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.242-246u4  {{{{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 partialu  x, y, z, t   u1  u2  u3  ... (3.3.10) differential equations, The journal of Mathematics and Computer Science, 2, x yzu  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 handling3.4 (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, ComputersI.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) weget the following exact solution.u  x1 , x2 , x3 ,.............., xn , t   x1  x2  x3  x4  ............  xn  (3.4.3) 1  n t4. Conclusion In this paper, the variational iterationmethod has been successfully applied to finding thesolution of (1+1), (1+2) and (1+3) dimensionalBurger’s equations. The solution obtained by thevariational iteration method is an infinite powerseries for appropriate initial condition, which can, inturn, be expressed in a closed form, the exactsolution. The results show that the variational 246 | P a g e

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