The comparison of optimal homotopy asymptotic method and homotopy perturbation method to solve Fisher equation

Transcription

1 Computational Methods for Differential Equations Vol 4, No, 206, pp The comparison of optimal homotopy asymptotic method and homotopy perturbation method to solve Fisher equation Zainab Ayati Department of Engineering sciences, Faculty of Technology and Engineering East of Guilan, University of Guilan PC4489-Rudsar-Vajargah,Iran Sima Ahmady Department of Mathematics, Payame Noor University, POBox , Tehran, Iran Abstract In recent years, numerous approaches have been applied for finding the solutions of functional equations One of them is the optimal homotopy asymptotic method In current paper, this method has been applied for obtaining the approximate solution of Fisher equation The reliability of the method will be shown by solving some examples of various kinds and comparing the obtained outcomes with the results of homotopy Perturbation method Keywords Optimal Homotopy Asymptotic method, Homotopy Perturbation method, Fisher equation 200 Mathematics Subject Classification 34K28, 35A25, 4Axx Introduction In the past two decades, partial differential equations have been the subject of many studies, owing to their importance in the modeling of many phenomena in the areas sciences [3-7] Fisher s equation occurs in chemical kinetics and population dynamics which include problems such as neutron population in a nuclear reaction, nonlinear evolution of a population in a one-dimensional habitat, logistic population growth models, flame propagation, neurophysiology, autocatalytic chemical reactions, and branching Brownian motion processes [-2] Wazwaz et al used Adomian Decomposition method (ADM) for the exact solutions of Fisher s equation and a nonlinear diffusion equation of the Fisher s type [3] Matinfar et al used Homotopy perturbation method (HPM), variational iterative method (VIM) and modified VIM for Fisher s equation, Generalized Fisher s equation and nonlinear diffusion equation of the Fisher s type [4-7] Received: 7 July 206 ; Accepted: 9 November 206 Corresponding author 43

2 44 Z AYATI AND S AHMADY The objective of this paper is to show the effectiveness of optimal homotopy asymptotic method (OHAM) for the solution of Fisher s equation In present work, we are dealing with the approximate solution of the Fisher equation as follows u t = u xx + α u( α u), () where α is known constant 2 Basic idea of OHAM To illustrate the basic concept of optimal homotopy asymptotic method [8,9], consider the following nonlinear differential equation A(u(x)) + g(x) = 0, x Ω, (2) with boundary conditions B(u, u/ n) = 0, x Γ, where A is a general differential operator, B is a boundary operator, g(x) is a known analytic function, and Γ is the boundary of the domain Ω Generally speaking the operator A can be divided into two parts L and N, where L is a linear, while N is a nonlinear operator Therefore, Eq(2) can be rewritten as follows L(u) + N(u) + g(x) = 0 (22) For applying optimal homotopy asymptotic method we construct a homotopy h(v(x, p), p) : R [0, ] R which satisfies ( p) [L(v(x, p)) + g(x)] = H(p) [L(v(x, p)) + g(x) + N(v(x, p))], (23) where x Rand p [0, ] is an embedding parameter, H(p) is a nonzero auxiliary function for p 0, H(0) = 0 and v(x, p) is an unknown function Obviously, when p = 0 and p = it holds that v(x, 0) = u 0 (x) and v(x, ) = u(x) respectively Thus, as p varies from 0 to, the solution v(x, p) approaches from u 0 (x) to u(x) where u 0 (x) is obtained from Eq4 For p = 0, we have ( L(u 0 (x)) + g(x) = 0, B u 0, du ) 0 = 0 (24) dx Next, we choose auxiliary function H(p) in the form H(p) = pc + p 2 c 2 +, where c, c 2, constants to be determined H(p) can be expressed in many forms as reported by V Marinca et al [8-] To get an approximate solution, we expand v(x, p, c i ) in Taylor s series about p in the following manner, v(x, p, c i ) = u 0 (x) + u k (x, c, c 2,, c k )p k (25) k=

3 CMDE Vol 4, No, 206, pp Substituting Eq (25) into Eq (23) and equating the coefficient of like powers of p, the following linear equations will be obtained Zeroth order problem is given by Eq (24) and the first order problem is given by following equation ( L(u (x)) + g(x) = c N 0 (u 0 (x)), B u, du ) = 0 dx The general governing equations for u k (x) are given by: L (u k (x)) L (u k (x)) = c k N 0 (u 0 (x)) c i [L (u k i (x)) + N k i (u 0 (x), u (x),, u k (x))], k + i= ( ) du k k = 2, 3, B u k = 0 dx Where N m (u 0 (x), u (x),, u m (x)) is the coefficient of p m in the expansion of N(v(x, p)) about the embedding parameter p N(v(x, p, c i )) = N 0 (u 0 (x)) + N m (u 0, u, u 2,, u m )p m m= It has been observed that the convergence of the series (25) depends upon the auxiliary constants c, c 2,, if it is convergent at p =, one has v(x, c i ) = u 0 (x) + u m (x, c, c 2,, c k ) k= The result of the m th order approximations are given m u(x, c, c 2,, c m ) = u 0 (x) + u i (x, c, c 2,, c i ) (26) Substituting Eq (26) into Eq (22), leads to the following equation i= R(x, c, c 2,, c m ) = L( u(x, c, c 2,, c m )) + g(x) + N( u(x, c, c 2,, c m )) (27) If R = 0 then u will be the exact solution Generally, it does not happen, especially in nonlinear problems There are several methods to find the optimal c i such as Galerkin method, Ritz method, collocation method and least square method In present paper, least square method has been applied to achieve the goal Therefore, the following functional equation will be constructed J(c, c 2,, c m ) = b And for minimizing it, we have a R 2 (x, c, c 2,, c m )dx (28) J c = J c 2 = = J c m = 0 (29)

4 46 Z AYATI AND S AHMADY After determining these constants, the approximate solution (of order m) will be obtained 3 Solution of the Fisher equation by HPM and OHPM 3 Solving by HPM Homotopy perturbation method has been well-addressed in [2] introduce the method and apply it directly Consider Eq () with the following initial condition u(x, 0) = f(x) So we skip to According to the homotopy perturbation method, we construct the following homotopy or H(v, p) = ( p)(v t (u 0 ) t ) + p(v t v xx αv + α 2 v 2 ) = 0, v t (u 0 ) t + p((u 0 ) t v xx αv + α 2 v 2 ) = 0 (3) To solve Eq (3) by homotopy perturbation method, let s consider the solution v as the summation of a series, v = v i p i (32) i=0 i=0 Substituting (32) into (3) leads to v i t pi u 0 = p t u 0 t + i=0 2 v i x 2 pi + α By equating the terms with the identical powers in p, we derive ( v i p i α 2 i=0 i=0 ) 2 v i p i p 0 : (v 0 ) t (u 0 ) t = 0, p : (v ) t + (u 0 ) t (v 0 ) xx αv 0 + α 2 (v 0 ) 2 = 0, p 2 : (v 2 ) t (v ) xx αv + 2 α 2 v 0 v = 0, p 3 : (v 3 ) t (v 2 ) xx αv 2 + α 2 ( 2v 0 v 2 + (v ) 2) = 0, For the sake of simplicity, let s take v 0 = u 0 = f(x), so we have v = t 0 ( (u 0) t + (v 0 ) xx + αv 0 α 2 (v 0 ) 2 )dt, v 2 = t 0 ((v ) xx + αv 2 α 2 v 0 v )dt, (33) (34)

5 CMDE Vol 4, No, 206, pp Setting p=, results in an approximation to the solution of Eq() So u = lim p v = v 0 + v + v Solving by OHAM: According to Eq (23), we construct the following homotopy or ( p)(v t ) = (c p + c 2 p 2 + c 3 p 3 )(v t v xx αv + α 2 v 2 ), v t = pv t + (c p + c 2 p 2 + c 3 p 3 )(v t v xx αv + α 2 v 2 ) (35) Let s consider the solution v as the summation of a series (25), and by equating the terms with the identical powers in p, the following equations will be obtained p 0 : (v 0 ) t = 0, p : (v ) t = (v 0 ) t + c (v 0 ) t c (v 0 ) xx c α v 0 + c α 2 (v 0 ) 2, p 2 : (v 2 ) t = (v ) t + c (v ) t c (v ) xx c αv + 2c α 2 v 0 v + c 2 (v 0 ) t c 2 (v 0 ) xx c 2 αv 0 + c 2 α 2 (v 0 ) 2, p 3 : (v 3 ) t = (v 2 ) t + c (v 2 ) t c (v 2 ) xx c αv 2 + c α 2 ( (v ) 2 + 2v 0 v 2 ) + c2 (v ) t c 2 (v ) xx c 2 αv + 2c 2 α 2 v 0 v + c 3 (v 0 ) t c 3 (v 0 ) xx c 3 αv 0 + c 3 α 2 (v 0 ) 2, let s take v 0 = u 0 = f(x), so we have v = t 0 ((v 0) t + c (v 0 ) t c (v 0 ) xx c αv 0 + c α 2 (v 0 ) 2 )dt, v 2 = t 0 ((v ) t + c (v ) t c (v ) xx c αv + 2c α 2 v 0 v + c 2 (v 0 ) t c 2 (v 0 ) xx c 2 αv 0 + c 2 α 2 (v 0 ) 2 )dt, So, by considering an approximation with (m + ) terms as u(x, c, c 2,, c m ) = u 0 (x) + m i= u i(x, c, c 2,, c i ), and by using eq s (27)-(29) we compute c i s 4 Example To illustrate the method some examples are provided In each example, we defined values of α in prior to applying the method

6 48 Z AYATI AND S AHMADY Example If we take α = and u(x, 0) = f(x) = λ, Eq () turns into u t = u xx + u u 2, where λis constant The exact solution is given by u(x, t) = λe t λ + λe t Solution via HPM According to HP M the following homotopy can be constructed H(v, p) = ( p)(v t (u 0 ) t ) + p(v t v xx v + v 2 ) = 0, or v t (u 0 ) t + p((u 0 ) t v xx v + v 2 ) = 0 v t (u 0 ) t + p((u 0 ) t v xx v + v 2 ) = 0 Substitute v = i=0 v ip i in above Eqation and equating the coefficients of the terms with the identical powers of p, leads to p 0 : (v 0 ) t (u 0 ) t = 0, p : (v ) t + (u 0 ) t (v 0 ) xx v 0 + (v 0 ) 2 = 0, p 2 : (v 2 ) t (v ) xx v + 2v 0 v = 0, p 3 : (v 3 ) t (v 2 ) xx v 2 + 2v 0 v 2 + (v ) 2 = 0, Assume v 0 = λ, then v = λ(λ )t, v 2 = 3 2 λ2 t λt2 + λ 3 t 2, v 3 = 7 6 λ2 t λt3 + 2λ 3 t 3 λ 4 t 3, Consider approximation for four terms v v 0 + v + v 2 + v 3 solution of equation will be obtained as the following form v(x, t) = λ λ 2 t + λt 3 2 λ2 t λt2 + λ 3 t λ2 t λt3 + 2λ 3 t 3 λ 4 t 3 Solution via OHAM According to the OHAM, by applying Eq (23), we derive ( p)(v t ) = (c p + c 2 p 2 + c 3 p 3 )(v t v xx v + v 2 )

7 CMDE Vol 4, No, 206, pp By substituting Eq (25) into above equation, and equating the coefficients of the terms with the identical powers of p, leads to (v 0 ) t = 0, (v ) t = (v 0 ) t + c (v 0 ) t c (v 0 ) xx c v 0 + c (v 0 ) 2, (v 2 ) t = (v ) t + c (v ) t c (v ) xx c v + 2c v 0 v + c 2 (v 0 ) t c 2 (v 0 ) xx c 2 v 0 + c 2 (v 0 ) 2, (v 3 ) t = (v 2 ) t + c (v 2 ) t c (v 2 ) xx c v 2 + c ( (v ) 2 + 2v 0 v 2 ) + c2 (v ) t c 2 (v ) xx c 2 v + 2c 2 v 0 v + c 3 (v 0 ) t c 3 (v 0 ) xx c 3 v 0 + c 3 (v 0 ) 2, Assume v 0 = λ, then v = c λ(λ )t, v 2 = c λ 2 t c λt + c 2 λ 2 t c 2 λt 3 2 c2 λ 2 t c2 λt 2 + c 2 λ 3 t 2 c 2 λt + c 2 λ 2 t, v 3 = c λt 2c 2 λt + c 2 λt 2 2c c 2 λt + 2c c 2 λ 2 t + c c 2 λt 2 3c c 2 λ 2 t 2 +2c c 2 λ 3 t 2 c 2 λt + c 2 λ 2 t + c λ 2 t + 2c 2 λ 2 t 3c 2 λ 2 t 2 + 2c 2 λ 3 t 2 + c 3 λ 4 t 3 c 3 λt + c 3 λ 2 t + c 3 λ 2 t c 3 λt 3c 3 λ 2 t 2 + c 3 λt 2 + 2c 3 λ 3 t c3 λ 2 t 3 6 c3 λt 3 2c 3 λ 3 t 3, Therefore, the four terms approximation using OHAM for solution will be obtained as follows v = 3c λt 3c 2 λt c2 λt 2 2c c 2 λt + 2c c 2 λ 2 t + c c 2 λt 2 3c c 2 λ 2 t 2 +2c c 2 λ 3 t 2 2c 2 λt + 2c 2 λ 2 t + 3c λ 2 t + 3c 2 λ 2 t 9 2 c2 λ 2 t 2 + 3c 2 λ 3 t 2 + λ +c 3 λ 4 t 3 c 3 λt + c 3 λ 2 t + c 3 λ 2 t c 3 λt 3c 3 λ 2 t 2 + c 3 λt 2 + 2c 3 λ 3 t c3 λ 2 t 3 6 c3 λt 3 2c 3 λ 3 t 3 The values of c i s are obtained by least square method c = 0/ , c 2 = 0/ , c 3 = 0/ Table Comparison of OHAM and HPM for λ = /5 t Exact HPM OHAM Error(HPM) Error(OHAM) 0/ / / / / / / /2 / / / / / /3 / / / / / /4 / / / / / /5 / / / / / It should be noted that by increasing the amount oft, the errors of OHAM have been less than the error of HPM

8 50 Z AYATI AND S AHMADY Example 2 In Eq (), set α = 6 and u(x, 0) = f(x) = (+e x ) 2 The exact solution is given by u(x, t) = ( + e x 5t ) 2, u t = u xx + 6u 36u 2 Solution via HPM According to the HPM the following homotopy can be constructed v t (u 0 ) t + p ( (u 0 ) t v xx 6v + 36v 2) = 0 Substitute v = v 0 + pv + p 2 v 2 + in above equation and equating the coefficients of the terms with the identical powers of p, leads to (v 0 ) t (u 0 ) t = 0, (v ) t + (u 0 ) t (v 0 ) xx 6v v 2 0 = 0, (v 2 ) t (v ) xx 6v + 72v 0 v = 0, (v 3 ) t (v 2 ) xx 6v v 0 v v 2 = 0, Assume v 0 = (+e x ) 2, then v = 0 ( e 2x + e x 3 ) t ( + e x ) 4, v 2 = 5t2 (0e 4x +5e 3x 26e 2x +89e x +98) (+e x ) 6, v 3 = 5 (00e 6x +25e 5x 472e 4x 3794e 3x e 2x +293e x 9548) 3, (+e x ) 8 Consider approximation for four terms v v 0 + v + v 2 + v 3 solution of equation will be obtained as the following form v = 3 (3 90t + 8e x + 50e 6x t e 5x t e 6x t + 525e 5x t 2 (+e x ) e 4x t e 5x t 290e 4x t e 3x t e 4x t 4890e 3x t e 2x t 3 60e 3x t 590e 2x t 2 390e 2x t +500e 6x t e x t e x t 2 330e x t + 8e 5x +3e 6x 97740t e 3x + 45e 4x t e 3x ) Solution via OHAM According to the OHAM, we derive ( p)(v t ) = (c p + c 2 p 2 + c 3 p 3 )(v t v xx 6v + 36v 2 )

9 CMDE Vol 4, No, 206, pp By substituting Eq(25) into above equation, and equating the coefficients of the terms with the identical powers of p, leads to (v 0 ) t = 0, (v ) t = (v 0 ) t + c (v 0 ) t c (v 0 ) xx 6c v c (v 0 ) 2, (v 2 ) t = (v ) t + c (v ) t c (v ) xx 6c v + 72c v 0 v +c 2 (v 0 ) t c 2 (v 0 ) xx 6c 2 v c 2 (v 0 ) 2, (v 3 ) t = (v 2 ) t + c (v 2 ) t c (v 2 ) xx 6c v c v 0 v 2 +36c (v ) 2 + c 2 (v ) t c 2 (v ) xx 6c 2 v +72c 2 v 0 v c 3 (v 0 ) t c 3 (v 0 ) xx 6c 3 v c 3 (v 0 ) 2, By considering v 0 = (+e x ) 2, the following results will be obtained v = 0c(e2x +e x 3)t (+e x ) 4, v 2 = (+e x ) 6 (5t(0e 4x c 2 t 2e 4x c 2 + 5e 3x tc 2 2e 4x c 2e 4x c 2 6e 3x c 2 26e 2x tc 2 6e 3x c 6e 3x c 2 89e x tc 2 + 0e x c tc 2 + 0e x c + 0e x c 2 + 6c 2 + 6c + 6c 2 )), v 3 = 5 3 (t( 88tc 2 (+e x ) e 5x c + 30e 5x c e 5x c 3 2e 3x c e 4x c 3 78e 2x c 78e 2x c 2 78e 2x c 3 + 6e 6x c 3 + 2e 6x c e 5x c 3 + 6e 6x c + 6e 6x c 2 +6e 6x c e 5x c e 4x c 3 24e 3x c c 2 56e 2x c c e 5x c c e 3x tc e 2x t 2 c e 4x t 2 c c e 2x tc 3 472e 4x t 2 c 3 + 2e 6x c c 2 20e 5x tc 2 +56e 4x tc e 3x t 2 c 3 60e 6x tc e 5x t 2 c 3 60e 6x tc 2 20e 5x tc 3 +00e 6x t 2 c 3 88tc 3 66e x c 3 66e x c 3 36c c e x t 2 c 3 842e x tc 3 32e x c c t 2 c e 4x tc c e 3x tc c e 2x tc c 2 60e 6x tc c 2 20e 5x tc c 2 842e x tc c 2 8c 3 88tc c 8c 3 +56e 4x tc e 3x tc 2 36c 2 8c e 4x c + 42e 4x c 2 24e 3x c 2 2e 3x c 2e 3x c e 4x c 2 32e x c 2 66e x c 66e x c 2 56e 2x c e 2x tc 2 842e x c 2 t 8c )), Therefore, the four terms, approximation using OHAM for solution will be obtained as follows

10 52 Z AYATI AND S AHMADY v = 3 ( 3 270c 2 (+e x ) t 8e x 390e 2x tc e 4x tc 3 60e 3x tc 3 70e 2x tc 780e 2x tc e 6x tc e 2x t 3 c e 5x tc + 300e 5x tc 2 +50e 5x tc e 4x t 3 c 3 575e 5x t 2 c e 3x t 3 c e 6x tc + 60e 6x tc e 6x t 3 c e 5x t 3 c 3 450e 6x t 2 c 2 330e x tc t 2 c c e x t 3 c 3 300e 6x t 2 c c 2 050e 5x t 2 c c e 4x t 2 c c e 3x t 2 c c e 2x t 2 c c 2 920e x t 2 c c t 3 c 3 60e 3x tc e 2x t 2 c 3 390e 2x tc e 4x t 2 c e 5x tc e 4x tc e 3x t 2 c e 6x tc 3 050e 5x t 2 c e 6x tc 2 +50e 5x tc 3 300e 6x t 2 c 3 8e 5x 3e 6x 90tc 3 920e x t 2 c 3 330e x tc t 2 c e 4x tc c 2 20e x tc c 2 780e x tc c e x tc c e x tc c 2 660e x tc c 2 80tc c 2 90tc 3 890t 2 c 2 270tc +3870e 4x t 2 c e 3x t 2 c e 4x tc + 420e 4x tc e 2x t 2 c 2 80e 3x tc 20e 3x tc 385e x t 2 c 2 990e x tc 660e x tc 2 70e 2x tc e 4x tc 2 80e 3x tc 2 60e 3x 45e 4x 80tc 2 990e x tc 2 45e 2x ) We use least squares method to obtain the unknown convergent constants c, c 2 and c 3 So c = 0/ , c 2 = 0/ , c 3 = 2/ Table 2 Comparison of OHAM and HPM for x = t Exact HPM OHAM Error(HPM) Error(OHAM) 0/00 0/ / / / /02 0/ / / / / /04 0/ / / / / /06 0/ / / / / /08 0/ / / / / /0 0/ / / / / As it clear table like the previous example, by increasing the amount of t, the error of OHAM is less than the error of HPM 5 Conclusion In this paper, the Fisher equation has been solved by HPM and OHAM The results obtained by OHAM are very consistent in comparison to HPM It is found that OHAM compared with HPM is a reliable efficient and powerful method for solving nonlinear partial differential equations, especially for the Fisher equation Therefore, we believe that the OHAM is an expectable technique for solving linear and nonlinear Fisher equation References [] P Brazhnik, and J Tyson, on traveling wave solutions of Fisher s equation in two spatial dimensions, SIAM J Appl Math, 60 (999), [2] W Malfliet, Solitary wave solutions of nonlinear wave equations, J Phys, 60 (992),

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