- Email: [email protected]
On the homotopy analysis method for the exact solutions of Helmholtz equation
Chaos, Solitons and Fractals 41 (2009) 1873–1879
Contents lists available at ScienceDirect
Chaos, Solitons and Fractals journal homepage: www.elsevier.com/locate/chaos
On the homotopy analysis method for the exact solutions of Helmholtz equation A.K. Alomari *, M.S.M. Noorani, R. Nazar School of Mathematical Sciences, Faculty of Science and Technology, Universiti Kebangsaan Malaysia, 43600 UKM Bangi, Selangor, Malaysia
a r t i c l e
i n f o
Article history: Accepted 30 July 2008
a b s t r a c t In this paper, the exact solutions of Helmholtz equation are obtained by means of the homotopy analysis method (HAM). This analytical method is employed to give approximate analytical solutions of Helmholtz equation. The auxiliary parameter h in the HAM solutions has provided a convenient way of controlling the convergence region of series solutions. It is also shown that the solutions which are obtained by the Adomian decomposition method (ADM) and variational iteration method (VIM) are special cases of the solution obtained by HAM. Ó 2008 Elsevier Ltd. All rights reserved.
1. Introduction Most of the scientific phenomena are modelled by ordinary or partial differential equations. Analytical solutions of these equations may well describe the various phenomena in science and nature, such as vibrations, solitons and propagation with a finite speed. Nowadays, there are several methods that can be used to get these analytical solutions. For example, one can use the Adomian decomposition method [1]. Also, Prof. Ji-Huan He has proposed two powerful methods, namely the variational iteration method (VIM) [2–6] and the homotopy perturbation method (HPM) [7–10] to obtain the analytical solutions for linear and nonlinear ordinary and partial differential equations. On the other hand, the homotopy analysis method (HAM) is developed and proposed by Liao [11] in 1992. Liao keeps sufficient room for experimenting with the convergence of the approximation by introducing a certain auxiliary parameter h and also, at times, one auxiliary non-zero function [12]. The method is a powerful and easy-to-use analytic method for nonlinear problems and has been successfully applied to solve many types of nonlinear problems in science and engineering by several authors [13–24] and the references therein. HAM has been effectively employed to solve several well-known partial differential equations such as the Laplace equation with Dirichlet and Neumann boundary conditions [25], the generalized Hirota–Satsuma coupled KdV equation [26] and the Benjamin–Bona–Mahony–Burgers (BBMB) equation [27]. Further, the Helmholtz equation is also one of the most important physics equations. This equation appears in diverse phenomena such as elastic waves in solids including vibrating string, bars, membranes, sounds or acoustics, electromagnetic waves, nuclear reactors, and the Lamb problem in geophysics [28–30]. Many researchers have solved this equation by different methods, namely, in 2004, El-Sayed and Kaya [31] used Adomian decomposition method (ADM) in comparison to the finite-difference method to solve this equation. Momani and Abuasad [32] in 2006 solved the same problem using variational iteration method (VIM). In this paper, for the first time we implement HAM for finding the exact/approximate solutions of Helmholtz equation, namely, we consider * Corresponding author. E-mail address: [email protected] (A.K. Alomari). 0960-0779/$ - see front matter Ó 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.chaos.2008.07.038
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r2 u þ f ðx; yÞu ¼ gðx; yÞ; with the boundary and initial conditions
uð0; yÞ ¼ w1 ðyÞ;
ux ð0; yÞ ¼ w2 ðyÞ;
uðx; 0Þ ¼ w3 ðxÞ;
ux ðx; 0Þ ¼ w4 ðxÞ;
where w1 ðyÞ; w2 ðyÞ; w3 ðxÞ, and w4 ðxÞ are given functions. By the freedom of choosing the auxiliary parameter h , we found that HAM gives the exact/approximate solution of the Helmholtz equation in a more accurate and effective manner. 2. Basic idea of HAM Consider the following differential equation:
N½uðx; yÞ ¼ 0; where N is a nonlinear operator, x and y denote the independent variables and uðx; yÞ is an unknown function. For simplicity, we ignore all boundary or initial conditions, which can be treated in a similar way. By means of the generalization of the traditional homotopy method, Liao [11] constructs the so-called zeroth-order deformation equation
ð1 qÞL½/ðx; y; qÞ u0 ðx; yÞ ¼ qhHðx; yÞN½/ðx; y; qÞ;
ð1Þ
where q 2 ½0; 1 is the embedding parameter, h–0 is an auxiliary parameter, L is an auxiliary linear operator, /ðx; y; qÞ is an unknown function, u0 ðx; yÞ is an initial guess of uðx; yÞ and Hðx; yÞ denotes a non-zero auxiliary function. Obviously, when q ¼ 0 and q ¼ 1, both
/ðx; y; 0Þ ¼ u0 ðx; yÞ;
/ðx; y; 1Þ ¼ uðx; yÞ;
hold, respectively. Thus as q increases from 0 to 1, the solution /ðx; y; qÞ varies from the initial guess u0 ðx; yÞ to the solution uðx; yÞ. Expanding /ðx; y; qÞ in Taylor series with respect to q, one has
/ðx; y; qÞ ¼ u0 ðx; yÞ þ
þ1 X
um ðx; yÞqm ;
ð2Þ
m¼1
where
um ðx; yÞ ¼
1 om /ðx; y; qÞ : m! oqm q¼0
If the auxiliary linear operator, the initial guess and the auxiliary parameter h are properly chosen, the series (2) converges at q ¼ 1, and thus one has
uðx; yÞ ¼ u0 ðx; yÞ þ
þ1 X
um ðx; yÞ:
ð3Þ
m¼1
Define the vectors
~ un ¼ fu0 ðx; yÞ; u1 ðx; yÞ; . . . ; un ðx; yÞg: Differentiating the zeroth-order deformation Eq. (1) m-times with respect to q and then dividing them by m! and finally setting q ¼ 0, we get the following mth-order deformation equation
L½um ðx; yÞ vm um1 ðx; yÞ ¼ Hðx; yÞhRm ð~ um1 Þ;
ð4Þ
1 om1 N½/ðx; y; qÞ Rm ð~ um1 Þ ¼ ðm 1Þ! oqm1
ð5Þ
where
; q¼0
and
vm ¼
0; m 6 1; 1; m > 1:
It should be emphasized that um ðx; yÞ for m P 1 is governed by the linear Eq. (4) with linear boundary conditions that come from the original problem, which can be solved by the symbolic computation softwares such as MAPLE and MATHEMATICA.
A.K. Alomari et al. / Chaos, Solitons and Fractals 41 (2009) 1873–1879
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3. Applications To illustrate the effectiveness of HAM, we shall consider two test examples. Comparison with the ADM and VIM shall also be made. In these examples, we choose the solution expressed by the form
uðx; yÞ ¼ f1 ðx; yÞ þ f2 ðx; yÞ
1 X
an xn :
ð6Þ
n¼0
From the style of this solution and the form of Helmholtz equation, the linear operator of these two examples is taken as
L½/ðx; y; qÞ ¼
o2 /ðx; y; qÞ ; ox2
ð7Þ
with the property
L½c1 ðyÞx þ c2 ðyÞ ¼ 0; where ci ðyÞði ¼ 1; 2Þ are integral constants. We take the inverse linear operator of Eq. (4) as
um ðx; yÞ ¼ vm um1 ðx; yÞ þ L1 ½Hðx; yÞhRm ð~ um1 Þ;
ð8Þ
and for simplification we assume that Hðx; yÞ ¼ 1. According to (7), the solution in Eq. (8) becomes
um ðx; yÞ ¼ vm um1 ðx; yÞ þ h
Z
x
0
Z
x
Rm ð~ um1 ðf; yÞÞdfdf þ c1 ðyÞx þ c2 ðyÞ;
ð9Þ
0
where ci ðyÞði ¼ 1; 2Þ are integral constants determined by the initial conditions given in the examples. Example 1. Consider the following case of Helmholtz equation:
o2 uðx; yÞ o2 uðx; yÞ þ uðx; yÞ ¼ 0; ox2 oy2
ð10Þ
subject to the initial conditions
uð0; yÞ ¼ y;
ux ð0; yÞ ¼ y þ coshðyÞ:
ð11Þ
To solve this example by HAM, we determine the solution expressed in Eq. (6), and we take
f1 ðx; yÞ ¼ x coshðyÞ;
f 2 ðx; yÞ ¼ y;
where the nonlinear operator is
Nu ¼
o2 /ðx; y; qÞ o2 /ðx; y; qÞ þ /ðx; y; qÞ: ox2 oy2
ð12Þ
By Eq. (5), we have
Rð~ um1 ðx; yÞÞ ¼
o2 um1 ðx; yÞ o2 um1 ðx; yÞ þ um1 ðx; yÞ: oy2 ox2
From the style of the solution (6) and the initial conditions (11), the initial guess is
u0 ðx; yÞ ¼ yð1 þ xÞ þ x coshðyÞ: By applying Eq. (9) and using Rð~ um1 ðx; yÞÞ in Eq. (13), along with the initial conditions
uð0; yÞ ¼ 0;
ux ð0; yÞ ¼ 0;
we have
u1 ðx; yÞ ¼ y
2 hx hx3 ; þ 2! 3!
! 2 2 h x4 h x5 x3 x2 2 2 þ ðh þ h Þ ðh þ h Þ ; 4! 5! 6 2 ! 3 6 3 7 h x h x u3 ðx; yÞ ¼ y þ Mðx; hÞ ; 6! 7!
u2 ðx; yÞ ¼ y
.. . where
Mðx; hÞ ¼
x5 2 x4 2 x3 x2 3 3 2 3 2 3 ðh þ h Þ þ ð h þ h Þ ðh þ 2h þ h Þ ðh þ 2h þ h Þ: 60 12 6 2
ð13Þ
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By the same manner, we have
un ðx; yÞ ¼ y
ðhÞn x2n ðhÞn x2nþ1 þ þ Z n ðx; hÞ ; ð2nÞ! ð2n þ 1Þ!
ð14Þ
where Z n ðx; hÞ is a polynomial function of degree less than 2n. When h ¼ 1, the previous few terms will be as follows:
1 2 1 3 yx þ yx ; 2! 3! 1 1 u2 ðx; yÞ ¼ yx4 þ yx5 ; 4! 5! 1 1 u3 ðx; yÞ ¼ yx6 þ yx7 ; 6! 7! u1 ðx; yÞ ¼
which are the same as those obtained by ADM and VIM [31,32] (i.e. ADM and VIM are special cases of HAM) and by Eq. (3), the solution is
x2 x3 x4 x5 uðx; yÞ ¼ x coshðyÞ þ y 1 þ x þ þ þ þ þ : 2! 3! 4! 5! By Taylor series, the solution uðx; yÞ in a closed form is
uðx; yÞ ’ x coshðyÞ þ yex ; which is the exact solution of this example when jxj < 1. In Section 4 of this paper, we will also see the numerical solution at different values of h. Example 2. Consider the following second Helmholtz equation:
o2 uðx; yÞ o2 uðx; yÞ þ þ 5uðx; yÞ ¼ 0; ox2 oy2
ð15Þ
subject to the initial conditions
uð0; yÞ ¼ y;
ux ð0; yÞ ¼ 3 sinhð2yÞ:
ð16Þ
To solve this example by HAM, we determine the solution expressed in Eq. (6), and we take
f1 ðx; yÞ ¼ 0;
f 2 ðx; yÞ ¼ sinhð2yÞ:
where the nonlinear operator is
Nu ¼
o2 /ðx; y; qÞ o2 /ðx; y; qÞ þ þ 5/ðx; y; qÞ: ox2 oy2
ð17Þ
According to Eq. (5) we have
Rð~ um1 ðx; yÞÞ ¼
o2 um1 ðx; yÞ o2 um1 ðx; yÞ þ þ 5um1 ðx; yÞ: oy2 ox2
From the style of the solution (6) and the initial conditions (16), the initial guess is
u0 ðx; yÞ ¼ 3x sinhð2yÞ: By applying Eq. (9) and using Rð~ um1 ðx; yÞÞ in Eq. (18), along with the initial conditions
uð0; yÞ ¼ 0;
ux ð0; yÞ ¼ 0;
we have
! ð3xÞ3 h ; u1 ðx; yÞ ¼ sinhð2yÞ 3! ! 2 ð3xÞ5 h 9 3 2 u2 ðx; yÞ ¼ sinhð2yÞ þ x ðh þ h Þ ; 2 5! u3 ðx; yÞ ¼ sinhð2yÞ
! 3 ð3xÞ7 h 648 5 3 9 2 3 2 x ðh þ h Þ þ x3 ðh þ h þ 2h Þ ; þ 140 2 7!
.. . and by the same manner, we have
ð18Þ
A.K. Alomari et al. / Chaos, Solitons and Fractals 41 (2009) 1873–1879
un ðx; yÞ ¼ sinhð2yÞ ð hÞ n
! ð3xÞ2nþ1 þ g n ðx; hÞ ; ð2n þ 1Þ!
1877
ð19Þ
where g n ðx; h Þ is a polynomial function of degree less than ð2n þ 1Þ. When h ¼ 1, the previous few terms will be as follows:
ð3xÞ3 sinhð2yÞ; 3! 5 ð3xÞ u2 ðx; yÞ ¼ sinhð2yÞ; 5! 7 ð3xÞ sinhð2yÞ; u3 ðx; yÞ ¼ 7! u1 ðx; yÞ ¼
ð20Þ
which are the same as those obtained by ADM and VIM [31,32] (i.e. ADM and VIM are special cases of HAM). Therefore, by Eq. (3) the general solution is in the form
! ð3xÞ3 ð3xÞ5 ð3xÞ7 uðx; yÞ ¼ sinhð2yÞ 3x þ þ ; 3! 5! 7!
ð21Þ
and thus, by Taylor series, the exact solution of uðx; yÞ in a closed form is
uðx; yÞ ’ sinhð2yÞ sinð3xÞ;
ð22Þ
which is the exact solution of this example when jxj < 1=3, and in the next section, we will also see the numerical solution at different values of h .
4. Comparison and discussion In this section, we present the comparison of the analytical results between HAM, VIM and ADM, particularly for the Helmholtz equation of Examples 1 and 2, as well as the discussion on the freedom of choosing h that consequently helps in achieving the most accurate and closest results to the exact solutions. The convergent regions of h are presented by
4.3 4.2 4.1
(1,1)
4 3.9 3.8 3.7 3.6 3.5 3.4 3.3 -2.5
-2
-1.5
-1
-0.5
0
Fig. 1. The h-curve of the 15th-order approximation for Example 1.
10 8
(1, 1)
6 4 2 0 -2 -2
-1.5
-1
-0.5
0
Fig. 2. The h -curve of the 10-th order approximation for Example 2.
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A.K. Alomari et al. / Chaos, Solitons and Fractals 41 (2009) 1873–1879
45000 = =
40000
=
35000
1.2 0.7 1
2)
25000
(
30000
20000 15000 10000 5000 0 0
2
4
6
8
10
Fig. 3. The solutions of Example 1 at different values of h after 15-iterations.
150
50
(
1)
100
0 -50 -100 0
0.5
1
1.5
2
2.5
3
3.5
4
Fig. 4. Comparison between HAM (at h ¼ 0:89), VIM and ADM solutions with exact solution for Example 2 after 10-iterations.
the h-curves as shown in Figs. 1 and 2, for Examples 1 and 2, respectively. For these two figures of h-curves, we fixed the values of x and y to be 1. Further, Fig. 3 shows the graphs of the HAM solutions for Example 1 with different values of h in comparison with the exact solution. It is also shown in this figure that HAM has successfully solved this example under the freedom of choosing h. The freedom to choose h in the HAM solution has provided a convenient way of controlling the convergence region of series solutions. The special cases of VIM and ADM also appear in this example. In Example 2, we mentioned that when h ¼ 1, we obtained the same solution as obtained by ADM and VIM. Thus, Fig. 4 shows that when h ¼ 1, the series solution is different from the exact solution for larger values of x, but by the freedom to choose h in HAM, we obtained the exact solution of Helmholtz equation at h ¼ 0:89. Therefore, we can say that HAM is more powerful and flexible than VIM and ADM to solve these examples of Helmholtz equation. 5. Conclusions In this paper, the HAM was used to obtain the exact solutions of the Helmholtz equation using the PC-based MAPLE package for illustrations and for generating analytical results. The comparison between the HAM, VIM and ADM was made and it was found that HAM is more effective than ADM and VIM, at least for those particular examples. Hence, it may be concluded that this method is a powerful and an efficient technique in finding the exact solutions for wide classes of problems. Furthermore, the advantage of this method is the fast convergence of the solutions by means of the auxiliary parameter h and the freedom of choosing h for HAM gives us more accuracy than VIM and ADM. It is also worth mentioning to this end that for both examples considered, we have shown that VIM and ADM are special cases of HAM. References [1] [2] [3] [4] [5] [6] [7] [8] [9]
Adomian G. Solving frontier problems of physics: the decomposition method. Boston: Kluwer; 1994. He JH. Variational iteration method for autonomous ordinary differential systems. Appl Math Comput 2000;114:115–23. He JH. Some asymptotic methods for strongly nonlinear equations. Int J Mod Phys B 2006;20:1141–99. He JH. Variational iteration method – a kind of non-linear analytical technique: some examples. Int J Non-Linear Mech 1999;34:699–708. He JH. Variational iteration method-some recent results and new interpretations. J Comput Appl Math 2007;207:3–17. He JH, Wan YQ, Guo Q. An iteration formulation for normalized diode characteristics. Int J Circ Theor Appl 2004;32:629–32. He JH. A simple perturbation approach to Blasius’ equation. Appl Math Comput 2003;140:217–22. He JH. Non-perturbative methods for strongly nonlinear problems. Dissertation. Berlin: de-Verlag im Internet GmbH; 2006. He JH. Homotopy perturbation technique. Comput Meth Appl Mech Eng 1999;178:257–62.
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He JH. Application of homotopy perturbation method to nonlinear wave equations. Chaos, Solitons & Fractals 2005;26:695–700. Liao SJ. The proposed homotopy analysis technique for the solution of nonlinear problems. PhD thesis. Shanghai Jiao Tong University; 1992. Sajid M, Hayat T. The application of homotopy analysis method to thin film flows of a third order fluid. Chaos, Solitons & Fractals 2008;38:506–15. Liao SJ. On the homotopy analysis method for nonlinear problems. Appl Math Comput 2004;147:499–513. Liao SJ. Beyond perturbation: introduction to homotopy analysis method. Boca Raton: Chapman and Hall/CRC Press; 2003. Liao SJ. A new branch of solutions of boundary-layer flows over an impermeable stretched plate. Int J Heat Mass Transfer 2005;48:2529–39. Ayub M, Rasheed A, Hayat T. Exact flow of a third grade fluid past a porous plate using homotopy analysis method. Int J Eng Sci 2003;41:2091–103. Wu W, Liao SJ. Solving solitary waves with discontinuity by means of the homotopy analysis method. Chaos, Solitons & Fractals 2005;26:177–85. Hayat T, Sajid M. On analytic solution for thin film flow of a fourth grade fluid down a vertical cylinder. Phys Lett A 2007;361:316–22. Abbasbandy S. Solitary wave solutions to the modified form of Camassa Holm equation by means of the homotopy analysis method. Chaos, Solitons & Fractals 2009;39:428–35. Liao SJ. An analytic approximation of the drag coefficient for the viscous flow past a sphere. Int J Non-Linear Mech 2002;37:1–18. Allan F. Construction of analytic solution to chaotic dynamical systems using the homotopy analysis method. Chaos, Solitons & Fractals 2009;39:1744–52. Wang C, Wu Y, Wu W. Solving the nonlinear periodic wave problems with the homotopy analysis method. Wave Motion 2004;41:329–37. Wu Y, Wang C, Liao SJ. Solving the one-loop soliton solution of the Vakhnenko equation by means of the homotopy analysis method. Chaos, Solitons & Fractals 2005;23:1733–40. Sajid M, Hayat T, Asghar S. On the analytic solution of the steady flow of a fourth grade fluid. Phys Lett A 2006;355:18–26. Inc M. On exact solution of Laplace equation with Dirichlet and Neumann boundary conditions by the homotopy analysis method. Phys Lett A 2007;365:412–5. Abbasbandy S. The application of homotopy analysis method to solve a generalized Hirota–Satsuma coupled KdV equation. Phys Lett A 2007;361:478–83. Fakhari A, Domairry G, Ebrahimpour R. Approximate explicit solutions of nonlinear BBMB equations by homotopy analysis method and comparison with the exact solution. Phys Lett A 2007;368:64–8. Burden RL, Faires JD. Numerical analysis. Boston: PWS Publishing Company; 1993. Gerald CF, Wheatley PO. Applied numerical analysis. California: Addison Wesley; 1994. Debnath L. Nonlinear partial differential equations for scientists and engineers. Boston: Birkhauser; 2004. El-Sayed SM, Kaya D. Comparing numerical methods for Helmholtz equation model problem. Appl Math Comput 2004;150:763–73. Momani S, Abuasad S. Application of He’s variational iteration method to Helmholtz equation. Chaos, Solitons & Fractals 2006;27:1119–23.
Contents lists available at ScienceDirect
Chaos, Solitons and Fractals journal homepage: www.elsevier.com/locate/chaos
On the homotopy analysis method for the exact solutions of Helmholtz equation A.K. Alomari *, M.S.M. Noorani, R. Nazar School of Mathematical Sciences, Faculty of Science and Technology, Universiti Kebangsaan Malaysia, 43600 UKM Bangi, Selangor, Malaysia
a r t i c l e
i n f o
Article history: Accepted 30 July 2008
a b s t r a c t In this paper, the exact solutions of Helmholtz equation are obtained by means of the homotopy analysis method (HAM). This analytical method is employed to give approximate analytical solutions of Helmholtz equation. The auxiliary parameter h in the HAM solutions has provided a convenient way of controlling the convergence region of series solutions. It is also shown that the solutions which are obtained by the Adomian decomposition method (ADM) and variational iteration method (VIM) are special cases of the solution obtained by HAM. Ó 2008 Elsevier Ltd. All rights reserved.
1. Introduction Most of the scientific phenomena are modelled by ordinary or partial differential equations. Analytical solutions of these equations may well describe the various phenomena in science and nature, such as vibrations, solitons and propagation with a finite speed. Nowadays, there are several methods that can be used to get these analytical solutions. For example, one can use the Adomian decomposition method [1]. Also, Prof. Ji-Huan He has proposed two powerful methods, namely the variational iteration method (VIM) [2–6] and the homotopy perturbation method (HPM) [7–10] to obtain the analytical solutions for linear and nonlinear ordinary and partial differential equations. On the other hand, the homotopy analysis method (HAM) is developed and proposed by Liao [11] in 1992. Liao keeps sufficient room for experimenting with the convergence of the approximation by introducing a certain auxiliary parameter h and also, at times, one auxiliary non-zero function [12]. The method is a powerful and easy-to-use analytic method for nonlinear problems and has been successfully applied to solve many types of nonlinear problems in science and engineering by several authors [13–24] and the references therein. HAM has been effectively employed to solve several well-known partial differential equations such as the Laplace equation with Dirichlet and Neumann boundary conditions [25], the generalized Hirota–Satsuma coupled KdV equation [26] and the Benjamin–Bona–Mahony–Burgers (BBMB) equation [27]. Further, the Helmholtz equation is also one of the most important physics equations. This equation appears in diverse phenomena such as elastic waves in solids including vibrating string, bars, membranes, sounds or acoustics, electromagnetic waves, nuclear reactors, and the Lamb problem in geophysics [28–30]. Many researchers have solved this equation by different methods, namely, in 2004, El-Sayed and Kaya [31] used Adomian decomposition method (ADM) in comparison to the finite-difference method to solve this equation. Momani and Abuasad [32] in 2006 solved the same problem using variational iteration method (VIM). In this paper, for the first time we implement HAM for finding the exact/approximate solutions of Helmholtz equation, namely, we consider * Corresponding author. E-mail address: [email protected] (A.K. Alomari). 0960-0779/$ - see front matter Ó 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.chaos.2008.07.038
1874
A.K. Alomari et al. / Chaos, Solitons and Fractals 41 (2009) 1873–1879
r2 u þ f ðx; yÞu ¼ gðx; yÞ; with the boundary and initial conditions
uð0; yÞ ¼ w1 ðyÞ;
ux ð0; yÞ ¼ w2 ðyÞ;
uðx; 0Þ ¼ w3 ðxÞ;
ux ðx; 0Þ ¼ w4 ðxÞ;
where w1 ðyÞ; w2 ðyÞ; w3 ðxÞ, and w4 ðxÞ are given functions. By the freedom of choosing the auxiliary parameter h , we found that HAM gives the exact/approximate solution of the Helmholtz equation in a more accurate and effective manner. 2. Basic idea of HAM Consider the following differential equation:
N½uðx; yÞ ¼ 0; where N is a nonlinear operator, x and y denote the independent variables and uðx; yÞ is an unknown function. For simplicity, we ignore all boundary or initial conditions, which can be treated in a similar way. By means of the generalization of the traditional homotopy method, Liao [11] constructs the so-called zeroth-order deformation equation
ð1 qÞL½/ðx; y; qÞ u0 ðx; yÞ ¼ qhHðx; yÞN½/ðx; y; qÞ;
ð1Þ
where q 2 ½0; 1 is the embedding parameter, h–0 is an auxiliary parameter, L is an auxiliary linear operator, /ðx; y; qÞ is an unknown function, u0 ðx; yÞ is an initial guess of uðx; yÞ and Hðx; yÞ denotes a non-zero auxiliary function. Obviously, when q ¼ 0 and q ¼ 1, both
/ðx; y; 0Þ ¼ u0 ðx; yÞ;
/ðx; y; 1Þ ¼ uðx; yÞ;
hold, respectively. Thus as q increases from 0 to 1, the solution /ðx; y; qÞ varies from the initial guess u0 ðx; yÞ to the solution uðx; yÞ. Expanding /ðx; y; qÞ in Taylor series with respect to q, one has
/ðx; y; qÞ ¼ u0 ðx; yÞ þ
þ1 X
um ðx; yÞqm ;
ð2Þ
m¼1
where
um ðx; yÞ ¼
1 om /ðx; y; qÞ : m! oqm q¼0
If the auxiliary linear operator, the initial guess and the auxiliary parameter h are properly chosen, the series (2) converges at q ¼ 1, and thus one has
uðx; yÞ ¼ u0 ðx; yÞ þ
þ1 X
um ðx; yÞ:
ð3Þ
m¼1
Define the vectors
~ un ¼ fu0 ðx; yÞ; u1 ðx; yÞ; . . . ; un ðx; yÞg: Differentiating the zeroth-order deformation Eq. (1) m-times with respect to q and then dividing them by m! and finally setting q ¼ 0, we get the following mth-order deformation equation
L½um ðx; yÞ vm um1 ðx; yÞ ¼ Hðx; yÞhRm ð~ um1 Þ;
ð4Þ
1 om1 N½/ðx; y; qÞ Rm ð~ um1 Þ ¼ ðm 1Þ! oqm1
ð5Þ
where
; q¼0
and
vm ¼
0; m 6 1; 1; m > 1:
It should be emphasized that um ðx; yÞ for m P 1 is governed by the linear Eq. (4) with linear boundary conditions that come from the original problem, which can be solved by the symbolic computation softwares such as MAPLE and MATHEMATICA.
A.K. Alomari et al. / Chaos, Solitons and Fractals 41 (2009) 1873–1879
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3. Applications To illustrate the effectiveness of HAM, we shall consider two test examples. Comparison with the ADM and VIM shall also be made. In these examples, we choose the solution expressed by the form
uðx; yÞ ¼ f1 ðx; yÞ þ f2 ðx; yÞ
1 X
an xn :
ð6Þ
n¼0
From the style of this solution and the form of Helmholtz equation, the linear operator of these two examples is taken as
L½/ðx; y; qÞ ¼
o2 /ðx; y; qÞ ; ox2
ð7Þ
with the property
L½c1 ðyÞx þ c2 ðyÞ ¼ 0; where ci ðyÞði ¼ 1; 2Þ are integral constants. We take the inverse linear operator of Eq. (4) as
um ðx; yÞ ¼ vm um1 ðx; yÞ þ L1 ½Hðx; yÞhRm ð~ um1 Þ;
ð8Þ
and for simplification we assume that Hðx; yÞ ¼ 1. According to (7), the solution in Eq. (8) becomes
um ðx; yÞ ¼ vm um1 ðx; yÞ þ h
Z
x
0
Z
x
Rm ð~ um1 ðf; yÞÞdfdf þ c1 ðyÞx þ c2 ðyÞ;
ð9Þ
0
where ci ðyÞði ¼ 1; 2Þ are integral constants determined by the initial conditions given in the examples. Example 1. Consider the following case of Helmholtz equation:
o2 uðx; yÞ o2 uðx; yÞ þ uðx; yÞ ¼ 0; ox2 oy2
ð10Þ
subject to the initial conditions
uð0; yÞ ¼ y;
ux ð0; yÞ ¼ y þ coshðyÞ:
ð11Þ
To solve this example by HAM, we determine the solution expressed in Eq. (6), and we take
f1 ðx; yÞ ¼ x coshðyÞ;
f 2 ðx; yÞ ¼ y;
where the nonlinear operator is
Nu ¼
o2 /ðx; y; qÞ o2 /ðx; y; qÞ þ /ðx; y; qÞ: ox2 oy2
ð12Þ
By Eq. (5), we have
Rð~ um1 ðx; yÞÞ ¼
o2 um1 ðx; yÞ o2 um1 ðx; yÞ þ um1 ðx; yÞ: oy2 ox2
From the style of the solution (6) and the initial conditions (11), the initial guess is
u0 ðx; yÞ ¼ yð1 þ xÞ þ x coshðyÞ: By applying Eq. (9) and using Rð~ um1 ðx; yÞÞ in Eq. (13), along with the initial conditions
uð0; yÞ ¼ 0;
ux ð0; yÞ ¼ 0;
we have
u1 ðx; yÞ ¼ y
2 hx hx3 ; þ 2! 3!
! 2 2 h x4 h x5 x3 x2 2 2 þ ðh þ h Þ ðh þ h Þ ; 4! 5! 6 2 ! 3 6 3 7 h x h x u3 ðx; yÞ ¼ y þ Mðx; hÞ ; 6! 7!
u2 ðx; yÞ ¼ y
.. . where
Mðx; hÞ ¼
x5 2 x4 2 x3 x2 3 3 2 3 2 3 ðh þ h Þ þ ð h þ h Þ ðh þ 2h þ h Þ ðh þ 2h þ h Þ: 60 12 6 2
ð13Þ
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By the same manner, we have
un ðx; yÞ ¼ y
ðhÞn x2n ðhÞn x2nþ1 þ þ Z n ðx; hÞ ; ð2nÞ! ð2n þ 1Þ!
ð14Þ
where Z n ðx; hÞ is a polynomial function of degree less than 2n. When h ¼ 1, the previous few terms will be as follows:
1 2 1 3 yx þ yx ; 2! 3! 1 1 u2 ðx; yÞ ¼ yx4 þ yx5 ; 4! 5! 1 1 u3 ðx; yÞ ¼ yx6 þ yx7 ; 6! 7! u1 ðx; yÞ ¼
which are the same as those obtained by ADM and VIM [31,32] (i.e. ADM and VIM are special cases of HAM) and by Eq. (3), the solution is
x2 x3 x4 x5 uðx; yÞ ¼ x coshðyÞ þ y 1 þ x þ þ þ þ þ : 2! 3! 4! 5! By Taylor series, the solution uðx; yÞ in a closed form is
uðx; yÞ ’ x coshðyÞ þ yex ; which is the exact solution of this example when jxj < 1. In Section 4 of this paper, we will also see the numerical solution at different values of h. Example 2. Consider the following second Helmholtz equation:
o2 uðx; yÞ o2 uðx; yÞ þ þ 5uðx; yÞ ¼ 0; ox2 oy2
ð15Þ
subject to the initial conditions
uð0; yÞ ¼ y;
ux ð0; yÞ ¼ 3 sinhð2yÞ:
ð16Þ
To solve this example by HAM, we determine the solution expressed in Eq. (6), and we take
f1 ðx; yÞ ¼ 0;
f 2 ðx; yÞ ¼ sinhð2yÞ:
where the nonlinear operator is
Nu ¼
o2 /ðx; y; qÞ o2 /ðx; y; qÞ þ þ 5/ðx; y; qÞ: ox2 oy2
ð17Þ
According to Eq. (5) we have
Rð~ um1 ðx; yÞÞ ¼
o2 um1 ðx; yÞ o2 um1 ðx; yÞ þ þ 5um1 ðx; yÞ: oy2 ox2
From the style of the solution (6) and the initial conditions (16), the initial guess is
u0 ðx; yÞ ¼ 3x sinhð2yÞ: By applying Eq. (9) and using Rð~ um1 ðx; yÞÞ in Eq. (18), along with the initial conditions
uð0; yÞ ¼ 0;
ux ð0; yÞ ¼ 0;
we have
! ð3xÞ3 h ; u1 ðx; yÞ ¼ sinhð2yÞ 3! ! 2 ð3xÞ5 h 9 3 2 u2 ðx; yÞ ¼ sinhð2yÞ þ x ðh þ h Þ ; 2 5! u3 ðx; yÞ ¼ sinhð2yÞ
! 3 ð3xÞ7 h 648 5 3 9 2 3 2 x ðh þ h Þ þ x3 ðh þ h þ 2h Þ ; þ 140 2 7!
.. . and by the same manner, we have
ð18Þ
A.K. Alomari et al. / Chaos, Solitons and Fractals 41 (2009) 1873–1879
un ðx; yÞ ¼ sinhð2yÞ ð hÞ n
! ð3xÞ2nþ1 þ g n ðx; hÞ ; ð2n þ 1Þ!
1877
ð19Þ
where g n ðx; h Þ is a polynomial function of degree less than ð2n þ 1Þ. When h ¼ 1, the previous few terms will be as follows:
ð3xÞ3 sinhð2yÞ; 3! 5 ð3xÞ u2 ðx; yÞ ¼ sinhð2yÞ; 5! 7 ð3xÞ sinhð2yÞ; u3 ðx; yÞ ¼ 7! u1 ðx; yÞ ¼
ð20Þ
which are the same as those obtained by ADM and VIM [31,32] (i.e. ADM and VIM are special cases of HAM). Therefore, by Eq. (3) the general solution is in the form
! ð3xÞ3 ð3xÞ5 ð3xÞ7 uðx; yÞ ¼ sinhð2yÞ 3x þ þ ; 3! 5! 7!
ð21Þ
and thus, by Taylor series, the exact solution of uðx; yÞ in a closed form is
uðx; yÞ ’ sinhð2yÞ sinð3xÞ;
ð22Þ
which is the exact solution of this example when jxj < 1=3, and in the next section, we will also see the numerical solution at different values of h .
4. Comparison and discussion In this section, we present the comparison of the analytical results between HAM, VIM and ADM, particularly for the Helmholtz equation of Examples 1 and 2, as well as the discussion on the freedom of choosing h that consequently helps in achieving the most accurate and closest results to the exact solutions. The convergent regions of h are presented by
4.3 4.2 4.1
(1,1)
4 3.9 3.8 3.7 3.6 3.5 3.4 3.3 -2.5
-2
-1.5
-1
-0.5
0
Fig. 1. The h-curve of the 15th-order approximation for Example 1.
10 8
(1, 1)
6 4 2 0 -2 -2
-1.5
-1
-0.5
0
Fig. 2. The h -curve of the 10-th order approximation for Example 2.
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45000 = =
40000
=
35000
1.2 0.7 1
2)
25000
(
30000
20000 15000 10000 5000 0 0
2
4
6
8
10
Fig. 3. The solutions of Example 1 at different values of h after 15-iterations.
150
50
(
1)
100
0 -50 -100 0
0.5
1
1.5
2
2.5
3
3.5
4
Fig. 4. Comparison between HAM (at h ¼ 0:89), VIM and ADM solutions with exact solution for Example 2 after 10-iterations.
the h-curves as shown in Figs. 1 and 2, for Examples 1 and 2, respectively. For these two figures of h-curves, we fixed the values of x and y to be 1. Further, Fig. 3 shows the graphs of the HAM solutions for Example 1 with different values of h in comparison with the exact solution. It is also shown in this figure that HAM has successfully solved this example under the freedom of choosing h. The freedom to choose h in the HAM solution has provided a convenient way of controlling the convergence region of series solutions. The special cases of VIM and ADM also appear in this example. In Example 2, we mentioned that when h ¼ 1, we obtained the same solution as obtained by ADM and VIM. Thus, Fig. 4 shows that when h ¼ 1, the series solution is different from the exact solution for larger values of x, but by the freedom to choose h in HAM, we obtained the exact solution of Helmholtz equation at h ¼ 0:89. Therefore, we can say that HAM is more powerful and flexible than VIM and ADM to solve these examples of Helmholtz equation. 5. Conclusions In this paper, the HAM was used to obtain the exact solutions of the Helmholtz equation using the PC-based MAPLE package for illustrations and for generating analytical results. The comparison between the HAM, VIM and ADM was made and it was found that HAM is more effective than ADM and VIM, at least for those particular examples. Hence, it may be concluded that this method is a powerful and an efficient technique in finding the exact solutions for wide classes of problems. Furthermore, the advantage of this method is the fast convergence of the solutions by means of the auxiliary parameter h and the freedom of choosing h for HAM gives us more accuracy than VIM and ADM. It is also worth mentioning to this end that for both examples considered, we have shown that VIM and ADM are special cases of HAM. References [1] [2] [3] [4] [5] [6] [7] [8] [9]
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