World Applied Sciences Journal 18 (12): 1839-1846, 2012 ISSN 1818-4952
© IDOSI Publications, 2012
DOI: 10.5829/idosi.wasj.2012.18.12.2700
Corresponding Author: V.G. Gupta, Department of Mathematics, University of Rajasthan, Jaipur-302055, India
Applications of Homotopy Analysis Transform Method
for Solving Various Nonlinear Equations
V.G. Gupta and Sumit Gupta
Department of Mathematics, University of Rajasthan, Jaipur-302055, India Jagan Nath Gupta Institute of Engineering and Technology, Sitapura, Jaipur, India
Abstract: In this paper, we apply Homotopy Analysis Transform Method (HATM) for solving various nonlinear equations. This method is the combined for of the homotopy analysis method and Laplace transform method. HATM is applied without any discretization or restrictive assumption and avoids round-off errors which may lead the solution in closed form. The results reveal that the HATM is very effective, convenient and quite accurate to system of nonlinear equations. Also it is shown that the Adomain decomposition method and homotopy perturbation method and Variational iteration method are special case of homotopy analysis transform method = −1
Key words: Homotopy analysis method • Laplace transform method • coupled equations • approximate
solution• exact solution INTRODUCTION
Analytical methods have made a comeback in research methodology after taking a backseat to the numerical techniques for the latter half of the preceding century. The advantage of analytical methods are manifolds, the main being that they give a much better insight than the numbers crunched by a computer using a purely numerical algorithm. Most new nonlinear equations do-not have a precise analytic solution; so numerical methods have largely been used to handle these equation. In recent years, many authors have paid attention to studying the solutions of nonlinear partial differential equations by various methods. Among these are Adomain Decomposition method [1-4], the tanh-method [5], the sine-cosine tanh-method [6, 7], the differential transform method [8, 9], the Variational iteration method [10-17] and the Laplace decomposition method [18-22]. One of the well known perturbation technique is Homotopy Perturbation Method (HPM), first proposed by Ji Huan He by combining the standard homotopy and classical perturbation technique for solving various linear, nonlinear initial and boundary values problems [23-33] and has been modified later by some scientists to obtain more accurate results, rapid convergence and reduce the amount of computation [34-37]. Also homotopy perturbation method is combined with Laplace transform method [38], Sumudu transform [39] and Variational iteration method [40] to produce highly effective techniques for nonlinear problems.
Recently, Liao [41] proposed a powerful analytical method, namely the Homotopy Analysis Method (HAM), for solving various linear and nonlinear partial differential equations. Different from perturbation techniques, the homotopy analysis method does not depend upon any small or large parameters. The HAM was successfully applied to solve many nonlinear problems [42-61]. During the last two decades; M.Khan [56] proposed a technique by combining the homotopy analysis method and Laplace transform method, first time in literature, namely Homotopy Analysis Transform Method (HATM) for solving various nonlinear problems. The main advantage of this problem is that we can accelerate the convergence rate, minimize iterative times, accordingly save the computation time and evaluate the efficiency. Several examples are given to access the reliability of HATM.
HOMOTOPY ANALYSIS METHOD Consider the following differential equation
N[u( )] 0τ = (1)
where N is a nonlinear operator, τ is the independent variable and u(τ) is an unknown function. For simplicity, we ignore all boundary or initial conditions, which can be treated in the similar way. By means of generalizing the traditional homotopy method, constructs the zero-order deformation equation, given by Liao.
0
(1 q)L[ ( ;q) u ( )] q H( ) N[ ( ;q)]− φ τ − τ = τ φ τ (2) where q∈[0,1] is an embedding parameter,≠0is a non zero auxiliary parameter, H(τ)≠0 is a non zero auxiliary function, L is an auxiliary linear operator, u0(τ) is an initial guess of u(τ) and φ(τ;q) is an unknown function. It is important that one has great freedom to choose auxiliary things in HAM. Obviously, when q = 0 and q = 1, it holds
0
( ;0) u ( ) and ( ;1) u( )
φ τ = τ φ τ = τ (3)
Thus, as q increases from 0 to 1, the solution φ(τ;q) varies from the initial guess u0(τ) to the solution u(τ). Expanding φ(τ;q) by Taylor series with respect to q, we get m 0 m m 1 ( ;q) u ( ) ∞ u ( )q = φ τ = τ +
∑
τ (4) where m m m 1 ( ;q) u ( ) q 0 m! q ∂ φ τ τ = = ∂ (5)If the auxiliary linear operator, the initial guess, the auxiliary parameterand the auxiliary function are so properly chosen, the series (5) converges at q = 1, then we have 0 m m 1 u(r,t) u (r,t) ∞ u (r,t) = = +
∑
(6)which must be one of the solutions of the original nonlinear equation, as proven by Liao [21]. As = −1 and H(τ) = 1, equation (3) becomes
0
(1 q)L[ ( ;q) u ( )] qN[ ( ;q)] 0− φ τ − τ + φ τ = (7) The governing equation can be deduced from the zero-order deformation equation (3). Define the vector
{
}
n 0 1 n
u = u ( ) , u ( ) , u ( )τ τ τ
(8)
Differentiating equation (3) m-times with respect to the embedding parameter q, then setting q = 0 and finally dividing them by m!, we obtain the mth-order deformation equation.
[
m m m 1]
m m 1 L u ( )τ − χ u ( )− τ =H( )R (u )τ − (9) where m 1 m m 1 1 N[ ( ;q)]m 1 R (u ) q 0 m 1! q − − − ∂ φ τ = = − ∂ (10) and m 0, m 1 1, m 1 ≤ χ = > (11)it should be emphasized that um(τ) for m≥1 is governed by the linear boundary conditions that come from original problem, which can be easily solved by symbolic computation software such as Maple, Mathematica and Matlab. If equation (2) admits unique solution, then this method will produce the unique solution. If equation (2) does not posses a unique solution, the HAM will give a solution among many other possible solutions.
HOMOTOPY ANALYSIS TRANSFORM METHOD (HATM)
Consider equation N[u(x)] = g(x), where N represents a general nonlinear ordinary or partial differential equation including both linear and nonlinear terms. The linear terms are decompose into L+R, where L is the highest order linear operator and R is the remaining of the linear operator. Thus, the equation can be written as [56]
Lu R u Nu g(x)+ + = (12) where Nu, indicates the nonlinear terms. By applying Laplace transform on both sides of Eq. (13), we get
L[Lu R u Nu g(x)]+ + = (13) Using the differentiation property of Laplace transform, we get
n
n k 1 (n k) k 1
s L[u] s u− − (0) L[Ru] L[Nu] L[g(x)]
= −
∑
+ + = (14) On simplifying n k 1 (n k) n n k 1 1 1L[u] s u (0) L[Ru] L[Nu] 0
s s
− − =
−
∑
+ + = (15)We define the nonlinear operator n k 1 ( n k ) n k 1 1 N[ (x,t;q)] L (x,t;q) s (x,t;q)(0) s − − = φ = φ −
∑
φ n 1 L[ (x,t;q)] L[R (x,t;q)] s + φ + φ (16)where φ(x,t;q) is a real functions of x, t and q. We construct a homotopy as
0
(1 q)L[ (x,t;q) u (x,t)] q H(x,t)N[u(x,t)]− φ − = (17) where q∈[0,1] is an embedding parameter, ≠0 is a non zero auxiliary parameter, H(x,τ) ≠ 0 is a non zero auxiliary function, L is an auxiliary linear operator, u0(x,τ) is an initial guess of u(τ0 and φ(x,t;q) is an unknown function. It is important that one has great freedom to choose auxiliary things in HAM. Obviously, when q = 0 and q = 1, it holds
0
(x,t;0) u(x,t)and (x,t;1) u(x,t)
φ = φ = (18)
respectively. Thus, as q increases from 0 to 1, the solution φ(t;q) varies from the initial guess u0(τ) to the solution u(τ). Expanding φ(t;q) by Taylor series with respect to q, we get m 0 m m 1 (x,t;q) u (x,t) ∞ u (x,t)q = φ = +
∑
(19) where m m m 1 (x,t;q) u (x,t) q 0 m ! q ∂ φ = = ∂ (20)If the auxiliary linear operator, the initial guess, the auxiliary parameter and the auxiliary function are so properly chosen, the series (19) converges at q = 1, then we have 0 m m 1 u(r,t) u (r,t) ∞u (r,t) = = +
∑
(21)which must be one of the solutions of the original nonlinear equation. The governing equation can be deduced from the zero-order deformation equation (17). Define the vector
{
}
n 0 1 n
u = u (x,t),u (x,t), u (x,t)
(22)
Differentiating equation (17) m-times with respect to the embedding parameter q, then setting q = 0 and finally dividing them by m!, we obtain the mth-order deformation equation.
m m m 1 m m 1
L u (x,t) − χ u (x,t)− =H(x,t)R (u ) − (23)
Applying inverse Laplace transform we get 1 m m m 1 m m 1 u (x,t) u L H(x,t)R (u )− − − = χ + (24) where m 1 m m 1 m 1 1 N[ (x,t;q)] R (u ) q 0 m 1! q − − − ∂ φ = = − ∂ (25) and m 0, m 1 1, m 1 ≤ χ = > (26) APPLICATIONS
Example 4.1: Let us consider the following problem
t x
u u u+ =0 (27)
with the initial conditions u(x,0)= −x
To solve equation (27) by means of the homotopy analysis method
Let us consider the following linear operator (x,t;q) L (x,t;q) t ∂φ φ = ∂ (28)
with the property that 1
L c =0 (29)
which implies that
t 1 0 L ( )− = ()dt
∫
(30)Taking Laplace transform of equation (27) both of sides subject to the initial condition, we get
x x 1 Lu(x,t) L u u 0 s s + + = (31)
We now define the nonlinear operator as
x 1 (x,t;q) N[ (x,t;q)] L (x,t;q) L (x,t;q) 0 s s x ∂φ φ = φ + + φ = ∂ (32)
and then the mth-order deformation equation is given by
m m m 1 m m 1
L u (x,t) − χ u (x,t)− =H(x,t)R (u ) − (33)
Taking inverse Laplace transform of Eq. (33), we get 1 m m m 1 m m 1 u (x,t) u L H(x,t)R (u )− − − = χ + where
(
)
(
)
m 1 m 1 i m m 1 m 1 m i i 0 x 1 u R u L u 1 L u s s x − − − − − = ∂ = + − χ + ∂ ∑
(34)let us take the initial approximation as 0
u(x,t )= −x the other components are as follows
( )(
)
1 u (x,t)= −x −t( ) ( )
{
2 2 2}
2 u (x,t)= −x −t −t+t (35)( ) (
{
)
2 3 2 2 2 2 3}
3 u (x,t)= −x −t +2 t +2 t −t 2 t− − t and so onthen the approximate solution at = −1 is given by x
u(x,t) , t 1 t 1
= <
− (36)
which is an exact solution and is same as obtained by HAM [57] and VIM [58].
Example 4.2: Let us consider the following problem
2 2
t x xxx
u ( u )+ +(u ) =0 (37) with the initial condition
u(x,0) x=
To solve equation (37) by means of the homotopy analysis method
Let us consider the following linear operator (x,t;q) L (x,t;q) t ∂φ φ = ∂ (38)
with the property that 1
L c =0 (39)
which implies that
t 1 0 L ( )− = ()dt
∫
(40)Taking Laplace transform of equation (37) both of sides subject to the initial condition, we get
( ) ( )
2 2 x xxx x 1 Lu(x,t) L u u 0 s s − + + = (41)we now define the nonlinear operator as
(
)
(
)
2 x 2 xxx (x,t;q) x 1 N (x,t;q) L (x,t;q) L 0 s s (x,t;q) φ + φ = φ − + = φ (42)and then the mth-order deformation equation is given by
m m m 1 m m 1 L u (x,t) − χ u (x,t)− =H(x,t)R (u ) − (43) where
(
)
(
)
m m 1 m 1 m 3 m 1 m 1 i m 1 i 3 i m 1 i i 0 i 0 x R u L u 1 s 1L u u u u s x x − − − − − − − − = = = − −χ ∂ ∂ + ∂ +∂ ∑
∑
(44)Let us take the initial approximation as 0
u(x,t) x=
the other components are given by 1 u(x,t) 2 x t=
(
)
2 u (x,t) 2 xt 2 t= + + 1(
)
2 3 u (x,t) 2 x t 2 t= + + 1 (45)(
)
3 4 u( x,t ) 2 xt 2 t= + + 1 and so on then the approximate solution at = −1 is given by x u(x,t) 1 2t = + (46)
which is an exact solution and is same as obtained by HAM [57] and VIM [58].
Example 4.3: Let us consider the following coupled system of equation
( )
t xx x x u u− −2 uu + u v =0 (47)( )
t xx x x v v− −2 vv + u v =0 with the initial conditionsu(x,0) sinx, v(x,0) sinx= =
To solve equation (47) by means of the homotopy analysis method
Let us consider the following linear operator (x,t;q) L (x,t;q) t ∂φ φ = ∂ (48)
with the property that 1
L c =0 (49)
which implies that
t 1 0 L ( )− = ()dt
∫
(50)Taking Laplace transform of equation (47) both of sides subject to the initial condition, we get
( )
xx x x sinx Lu(x,t) L u 2u u u v 0 s − + − − + = and( )
xx x x sinx L v(x,t) L v 2vv u v 0 s − + − − + = (51)we now define the nonlinear operator as 2 1 1 1 1 2 1 sinx (x,t;q) (x,t;q) N (x,t;q) L (x,t;q) L L 2 (x,t;q) s x x ∂ φ ∂φ φ = φ − + − φ ∂ ∂
(
1(x,t;q) (x,t;q)2)
x φ φ (52) and 2 2 2 2 2 2 2 sinx (x,t;q) (x,t;q) N (x,t;q) L (x,t;q) L L 2 (x,t;q) s x x ∂ φ ∂φ φ = φ − + − φ ∂ ∂ (
φ1(x,t;q) (x,t;q)φ2)
x (53)then the mth-order deformation equation is given by
m m m 1 m m 1
L u (x,t) − χ u (x,t)− =H(x,t)R (u ) − (54)
and
m m m 1 m m 1
L v (x,t) − χ v (x,t)− =H(x,t)R (v ) − (55)
Taking inverse Laplace transform of Eq. (54) and Eq. (55), we get 1 m m m 1 m m 1 u (x,t) u L H(x,t)R (u )− − − = χ + (56) 1 m m m 1 m m 1 v (x,t) v L H(x,t)R (v )− − − = χ + (57) where
(
)
(
)
2 m 1 m 1 m 1 i m 1(
i m 1 i)
m m 1 m 1 m 2 i i 0 i 0 u v sinx u u R u L u (x,t) 1 L 2 u s x x x − − − − − − − − − = = ∂ ∂ ∂ = − − χ + ∂ − ∂ − ∂ ∑
∑
(
)
(
)
2 m 1 m 1 m 1 i m 1(
i m 1 i)
m m 1 m 1 m 2 i i 0 i 0 u v sinx u u R u L u (x,t) 1 L 2 u s x x x − − − − − − − − − = = ∂ ∂ ∂ = − − χ + ∂ − ∂ − ∂ ∑
∑
(58) and(
)
(
)
2 m 1 m 1 m 1 i m 1(
i m 1 i)
m m 1 m 1 m 2 i i 0 i 0 u v sinx v v R v L v (x,t) 1 L 2 v s x x x − − − − − − − − − = = ∂ ∂ ∂ = − − χ + ∂ − ∂ − ∂ ∑
∑
(59)Let us take the initial approximation
0
u(x,t) sinx=
and
0
v (x,t) sinx= (60)
1 u(x,t)=tsinx 1 v(x,t)=tsinx
(
)
2 2 2 2 t u( x,t) sinx t 2! = + + (
)
2 2 2 2 t v (x,t) sinx t 2! = + + (61)(
) (
)
3 3 2 3 3 2 2 3 t u (x,t) sinx 2 t t 3! = + + + + + (
) (
)
3 3 2 3 3 2 2 3 t v( x,t) sinx 2 t t 3! = + + + + + and so on then the approximate solution at = −1 is given by t u(x,t) e sinx= − (62) and t v(x,t) e sinx= −
which is an exact solution and is same as obtained by HAM [57] and VIM [58].
CONCLUSIONS
In this paper, the homotopy analysis transform method (HATM) is successfully applied to solve many nonlinear problems. It is apparently seen that HATM is very powerful and efficient technique in finding analytical solutions for wider class of problems. They also do not require large computer memory and discretization of variable x. The results show that HATM is powerful mathematical tool for solving nonlinear equations. Also it has been shown that VIM is the special case of HATM and HAM.
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