A new homotopy analysis method for finding the exact solution of systems of partial differential equations

Selçuk J. Appl. Math. Selçuk Journal of Vol. 13. No. 1. pp. 41-56, 2012 Applied Mathematics

A New Homotopy Analysis Method for Finding the Exact Solution of Systems of Partial Differential Equations

M. Matinfar1, M. Saeidy2, B. Gharahsuflu3

1,2_{Department of Mathematics and Computer Science, University of Mazandaran,}

Babolsar, Iran

e-mail: 1m .m atinfar@ um z.ac.ir,2m .saidy@ um z.ac.ir

3_{Department of Mathematics, Qaemshahr Branch, Islamic Azad University, Qaemshahr,}

Iran

e-mail: b.gharahsuflu@ hotm ail.com

Received Date: September 5, 2011 Accepted Date: March 19, 2012

Abstract. In this paper, the application of a new homotopy analysis method presented for obtaining solutions of systems of non-linear partial differential equations. Theoretical considerations are discussed. To explain the capability and reliability of the new method some examples are provided. The results show that the new technique is very effective and convenient and comparison of the obtained solutions of this new method with those of applying homotopy analysis method have high accuracy.

Key words: New homotopy analysis method; System of partial differential equations.

2000 Mathematics Subject Classification: 65N99. 1. Introduction

In the last two decades with the rapid development of nonlinear science, there has appeared ever increasing interest of scientists and engineers in the analytical techniques for nonlinear problems. The purpose of this paper is to use new ho-motopy analysis methd that briefly is called NHAM, to a system of differential equations that arise in many areas of mathematics, engineering and physical sciences and illustrate the advantages and simplicity of NHAM as compared to HAM. These equations are often too complicated to be solved exactly and even if an exact solution is obtained, the required calculations may be too compli-cated. Most scientific problems and phenomena occur nonlinearly. Except a

limited number of these problems, most of them do not have precise analytical solution, thus we have to use various approximate analytical methods. Very recently, many powerful methods have been presented, such as the Adomian decomposition method (ADM) [8, 18], the variational iteration method (VIM) [9, 10, 15], the homotopy perturbation method (HPM) [5, 7, 12, 16, 17], the dif-ferential transform method [3, 6], the homotopy analysis method (HAM) [1, 13, 14] and the others [11]. At first, we explain the new modification of HAM that is called NHAM for finding the exact solution of systems of partial differential equations is presented. The applicability of the new method is verified by the numerical results that obtained of implement method for three examples that are shown in section 3, and also comparisons between this method and HAM are illustrated in this section. The conclusions appear in section 4.

2. Basic Idea of NHAM

To illustrate of basic idea of NHAM, We consider the general form of a system of PDEs can be considered as the following [4]:

(1) ∂ui

∂t + Ni(x1, x2, · · · , xn−1, t, u1, · · · , un) = gi(x1, x2, · · · , xn−1, t) , with the following initial conditions:

ui(x1, x2, ..., xn−1, t0) = fi(x1, x2, ..., xn−1) , i = 1, · · · , n,

where N1, · · · , Nn are non-linear operators, which usually depend on the un-known functions ui and their derivatives, xj for j = 1, ..., n − 1 denotes inde-pendent variable and g1, g2, · · · , gn are inhomogeneous terms. For simplicity, we ignore all boundary or initial conditions, which can be treated in the sim-ilar way. For solving Eq.(1) by means of NHAM we constructs the so-called zero-order deformation equation

(2) (1−q) ∙ ∂φ_i ∂t − ui,0(r, t) ¸ −q ~iHi ∙ ∂φ_i ∂t + Ni(r, t, φ1, ..., φn) − gi(r, t) ¸ = 0, where r = (x1, ..., xn−1), q ∈ [0, 1] is the embedding parameter, ~i6= 0 is a non-zero auxiliary parameter, Hi(r, t) 6= 0 is an auxiliary function, ui,0(x1, ..., xn−1, t) is an initial guess of ui(r, t), φi(r, t; q) is a unknown function, respectively. It is important, that one has great freedom to choose auxiliary things in NHAM. Obviously, when q = 0 and q = 1, it holds

φ_i(r, t; 0) = Z t

t0

ui,0(r, t1)dt1, φ_i(r, t; 1) = ui(r, t).

Thus, as q increases from 0 to 1, the solution φ_i(r, t; q) varies from Z t

t0

to the solution ui(r, t). Expanding φi(r, t; q) in Taylor series with respect to q, we have (3) φ_i(r, t; q) = ui,0(r, t) + +∞ X m=1 ui,m(r, t)qm, where (4) ui,m(r, t) = 1 m! ∂m_φ i(r, t; q) ∂qm |q=0.

If the initial guess, the auxiliary parameter ~i, and the auxiliary function are so properly chosen, the series (3) converges at q = 1, then we have

(5) ui(r, t) = ui,0(r, t) + +∞ X m=1

ui,m(r, t),

which must be one of solutions of original nonlinear equation, as proved by [2]. As ~i= −1 and Hi(x1, x2, ..., xn−1, t) = 1, Eq.(2) becomes

(6) (1 − q)[∂φi(r, t; q)_∂t − ui,0(r, t)] + q [ ∂φi

∂t + Ni(r, t, φ1, ..., φn) − gi(r, t)] = 0, which is used mostly in the new homotopy perturbation method [12], where as the solution obtained directly, without using Taylor series [5, 12]. According to the definition, the governing equation can be deduced from the zero-order deformation Eq.(2). Define the vector

−

→_{ui,n(r, t) = {ui,0(r, t), ui,1(r, t), . . . , ui,n(r, t)}.}

Differentiating Eq.(2) m times with respect to the embedding parameter q and then setting q = 0 and finally dividing them by m!, we have the so-called mth-order deformation equation

(7) ∂

∂t[ui,m(r, t)−χmui,m−1(r, t)] = ~iHi(r, t)Ri,m(−→u1,m−1(r, t), . . . , −→un,m−1(r, t)), where Ri,m(−→u1,m−1, . . . , −→un,m−1) = 1 (m − 1)! ∂m−1h∂φi ∂t + Ni(r, t, φ1, ..., φn) − gi i ∂qm−1 |q=0, and (8) χ_m= ½ 0, _{m ≤ 1,} 1, m > 1.

Therefore, the solution of Eq.(1), can be readily obtained by ui(r, t) =

+∞ X j=0

In practice, all terms of series ui(r, t) = P+j=0∞ui,j(r, t) cannot be determined and so we use an approximation of the solution by the following truncated series

μ_i,m(r, t) = m_X−1 j=0 ui,j(r, t), where ui(r, t) = lim m→+∞μi,m(r, t).

It should be emphasized that ui,m(r, t) for m > 1 is governed by the linear Eq.(7) under the linear boundary conditions that come from original problem, which can be easily solved by symbolic computation software such as Matlab. Because we have a great freedom to choose ui,0(r, t) in NHAM, we suppose that the initial approximation of the solution of Eq.(1) is in the following form: (9) ui,0(r, t) = fi(r) + +∞ X j=0 ai,j(r) Z t t0 Pj(t1)dt1,

where ai,j(r), are unknown coefficients and P0(t), P1(t), P2(t), ... are specific functions. With substituting Eq.(9) into mth-order deformation Eq.(7) we ob-tain the other components as follows

(10)

ui,1(r, t) = ~i Rt

t0[Hi(r, t1)Ri,1(−

→_{u1,0(r, t1), . . . , −}→_{un,0(r, t1))]dt1,} ui,2(r, t) = ui,1+ ~i R_tt₀[Hi(r, t1)Ri,2(−→u1,1(r, t1), . . . , −→un,1(r, t1))]dt1,

.. . ui,k(r, t) = ui,k−1+ ~i Rt t0[Hi(r, t1)Ri,k(− →_u1,k −1(r, t1), . . . , −→un,k−1(r, t1))]dt1, .. .

Now if we solve these equations in such a way that ui,1(r, t) = 0, then Eqs.(10) yield

ui,1(r, t) = ui,2(r, t) = · · · = 0.

Therefore the exact solution may be obtained as the following ui(r, t) = ui,0(r, t) = fi(r) + +∞ X j=0 ai,j(r) Z t t0 Pj(t)dt1.

It is worth mentioning that if gi(x1, x2, ..., xn−1, t) and ui,0(x1, x2, ..., xn−1, t), are analytic around t = t0, then their Taylor series can be defined as

ui,0(r, t) = +∞ X j=0 ai,j(r) (t − t0)n, gi(r, t) = +∞ X j=0 bi,j(r) (t − t0)n,

which can be used in Eqs.(10), where ai,j(r) are unknown coefficients which must be computed, and bi,j(r) are known ones. To show the capability of the method, NHAM has been applied to some examples in the next section. If Eq.(1) admits unique solution, then this method will produce the unique solution and if Eq.(1) does not possess unique solution, the NHAM will give a solution among many other (possible) solutions.

3. Numerical Example

In order to illustrate the effectiveness of the method discussed above, some example of systems of non-linear partial equations are presented.

Example 3.1. Consider the following system of three-dimensional partial dif-ferential equations: (11) ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ ∂y1 ∂t − y2 ∂y1 ∂x − ∂y1 ∂y ∂y2 ∂t = 1 − x + y + t, ∂y2 ∂t − y1 ∂y2 ∂x − ∂y2 ∂y ∂y1 ∂t = 1 − x − y − t, the initial conditions are given by

y1(x, y, 0) = x + y − 1, y2(x, y, 0) = x − y + 1, the exact solutions are

y1(x, y, t) = x + y + t − 1, y2(x, y, t) = x − y − t + 1. First we apply the HAM approach and then the NHAM approach. HAM approach:

Solving the system (11) by the HAM with ~i= −1, Hi(x, y, t) = 1, for i = 1, 2, and y1,0(x, y, t) = x + y − 1 and y2,0(x, y, t) = x − y + 1, we obtain

⎧ ⎨ ⎩ y1,1(x, y, t) = 2t +1₂t2, y2,1(x, y, t) = −1₂t2, y2,1(x, y, t) = −1₂t2, ⎧ ⎨ ⎩ y1,2(x, y, t) = −1₆t3−1₂t2, y2,2(x, y, t) =1₂t2+1₆t3− 2t, y2,2(x, y, t) =1₂t2+1₆t3− 2t, ⎧ ⎨ ⎩ y1,3(x, y, t) =1₃t3+₂₄1t4−1₂t2− 2t, y2,3(x, y, t) = −₂₄1t4+1₂t2, y2,3(x, y, t) = −₂₄1t4+1₂t2, .. .

Therefore, the approximate solution of Example (3.1), can be readily obtained by ⎧ ⎨ ⎩ y1(x, y, t) =P+_m=0∞ y1,m(x, y, t) = −1 + x + y + 2t −₅₀₄₀1 t7+₄₀1t5−1₂t3+ · · · , y2(x, y, t) =P+m=0∞ y2,m(x, y, t) = 1 + x − y + 2t +50401 t7− 1 40t5+ 1 2t3+ · · · . NHAM approach:

To solve the system (11) by the NHAM, for simplicity we define the system of nonlinear operators as ⎧ ⎪ ⎨ ⎪ ⎩ N1[φ1(x, y, t; p), φ2(x, y, t; p)] = ∂φ1 ∂t − φ2 ∂φ1 ∂x − ∂φ1 ∂y ∂φ2 ∂t − 1 + x − y − t, N2[φ1(x, y, t; p), φ2(x, y, t; p)] = ∂φ2 ∂t − φ1 ∂φ2 ∂x − ∂φ2 ∂y ∂φ1 ∂t − 1 + x + y + t. Using above definition, we construct the system zeroth-order deformation equa-tions (1−p) ∙_∂φ i(x, y, t; p) ∂t − yi,0(x, y, t) ¸ = p ~iHi(x, y, t) Ni[φ1, φ2], (i = 1, 2). For p = 0 and p = 1, we can write

φ_i(x, y, t; 0) = Z t

0

yi,0(x, y, t1)dt1, φi(x, y, t; 1) = yi(x, y, t), (i = 1, 2). Thus, we obtain the system of mth-order deformation equations as follows (12)

∂

∂t[yi,m(x, y, t) − χmyi,m−1(x, y, t)] = ~i Hi(x, y, t) Ri,m(−→y1,m−1, −→y2,m−1), where ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ R1,m= ∂y1,m−1_∂t −Pmk=0−1 h

y2,k∂y1,m−1−k_∂x +∂y_∂y1,k∂y2,m−1−k_∂t i

+ (1 − χm)(−1 + x − y − t), R2,m= ∂y2,m−1_∂t −Pmk=0−1

h

y1,k∂y2,m−1−k_∂x +∂y_∂y2,k∂y1,m−1−k_∂t i

+ (1 − χm)(−1 + x + y + t). Now, the solution of system (12), for (m ≥ 1) is yi,m(x, y, t) = χmyi,m−1(x, y, t)+ ~i

Z t 0 [H

in other words ⎧ ⎨ ⎩ y1,1(x, y, t) = ~1 R₀t[H1(x, y, t1)R1,1(−→y1,0, −→y2,0)] dt1, y2,1(x, y, t) = ~2 R₀t[H2(x, y, t1)R2,1(−→y1,0, −→y2,0)] dt1, ⎧ ⎨ ⎩ y1,2(x, y, t) = y1,1(x, y, t) + ~1 R₀t[H1(x, y, t1)R1,2(−→y1,1, −→y2,1)] dt1, y2,2(x, y, t) = y2,1(x, y, t) + ~2 R₀t[H2(x, y, t1)R2,2(−→y1,1, −→y2,1)] dt1, .. . ⎧ ⎨ ⎩ y1,j+1(x, y, t) = y1,j(x, y, t) + ~1 R t 0[H1(x, y, t1)R1,j+1(−→y1,j, −→y2,j)] dt1, y2,j+1(x, y, t) = y2,j(x, y, t) + ~2 R₀t[H2(x, y, t1)R2,j+1(−→y1,j, −→y2,j)] dt1, .. .

For solving the above systems by means of NHAM, considering that H1(x, y, t1) = H2(x, y, t1) = 1, ~1= ~2= −1, with initial approximations as follows

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ y1,0(x, y, t) = x + y − 1 +R₀thP+j=0∞aj(x, y)P1,j(t) i , y2,0(x, y, t) = x − y + 1 +R t 0 hP+∞ j=0bj(x, y)P2,j(t) i ,

such that Pi,j(t) = tj, for i = 1, 2. Solving the above equations for y1,1(x, y, t) and y2,1(x, y, t) with considering the above information leads to the result

y1,1(x, y, t) = [−a0(x, y) + b0(x, y) + 2] t + [− 1 2a1(x, y) + 1 2b0(x, y) +1 2(x − y + 1)a0x(x, y) + 1 2b1(x, y) + 1 2a0y(x, y)b0(x, y) + 1 2]t 2 + [−1 3a2(x, y) + 1 6b1(x, y) + 1 6(x − y + 1)a1x(x, y) + 1 3a0x(x, y) × b0(x, y) + 1 3b2(x, y) + 1 3a0y(x, y)b1(x, y) + 1 6a1y(x, y)b0(x, y)]t 3 + [−1 4a3(x, y) + 1 12b2(x, y) + 1 12(x − y + 1)a2x(x, y) + 1 8a0x(x, y) × b1(x, y) + 1 8a1x(x, y)b0(x, y) + 1 4b3(x, y) + 1 4a0y(x, y)b2(x, y) +1 8a1y(x, y)b1(x, y) + 1 12a2y(x, y)b0(x, y)]t 4 + · · · ,

y2,1(x, y, t) = (−b0(x, y) − a0(x, y))t + [− 1 2b1(x, y) + 1 2a0(x, y) +1 2(x + y − 1)b0x(x, y) − 1 2a1(x, y) + 1 2a0(x, y)b0y(x, y) − 1 2]t 2 + [−1 3b2(x, y) + 1 6a1(x, y) + 1 6(x + y − 1)b1x(x, y) + 1 3a0(x, y) × b0x(x, y) − 1 3a2(x, y) + 1 6a0(x, y)b1y(x, y) + 1 3a1(x, y)b0y(x, y)]t 3 + [−1₄b3(x, y) + 1 12a2(x, y) + 1 12(x + y − 1)b2x(x, y) + 1 8a1(x, y) × b0x(x, y) + 1 8a0(x, y)b1x(x, y) − 1 4a2(x, y) + 1 4a2(x, y)b0y(x, y) +1 8a1(x, y)b1y(x, y) + 1 12a0(x, y)b2y(x, y)]t 4 + · · · .

By the vanishing of y1,1(x, y, t) and y2,1(x, y, t) the coefficients aj(x, y) and bj(x, y) for j = 1, 2, 3, ... are determined as

a0(x, y) = 1, a1(x, y) = a2(x, y) = a3(x, y) = · · · = 0, b0(x, y) = −1, b1(x, y) = b2(x, y) = b3(x, y) = · · · = 0, therefore we obtain the solution of Eq.(11) as

y1(x, y, t) = y1,0(x, y, t) = x + y − 1 + a0(x, y)t + 1 2a1(x, y)t 2₊1 3a2(x, y)t 3 +1 4a3(x, y)t 4 + · · · = x + y + t − 1, y2(x, y, t) = y2,0(x, y, t) = x − y + 1 + b0(x, y)t + 1 2b1(x, y)t 2₊1 3b2(x, y)t 3 +1 4b3(x, y)t 4 + · · · = x − y − t + 1, which is an exact solution for Eq.(11).

Example 3.2. Consider the following system of two-dimensional partial differ-ential equations: ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ ∂y1 ∂x − y2 ∂y1 ∂t + y1 ∂y2 ∂t = −1 + e x_{sin t,} ∂y2 ∂x + ∂y1 ∂t ∂y2 ∂x + ∂y2 ∂t ∂y1 ∂x = −1 − e −x_{cos t,} the boundary conditions are given by

y1(0, t) = sin t, y2(0, t) = cos t, the exact solutions are

First we apply the HAM approach and then the NHAM approach. HAM approach:

Solving the system (13) by the HAM with ~i = −1, Hi(x, t) = 1, for i = 1, 2, and y1,0(x, t) = sin t and y2,0(x, t) = cos t, we obtain

⎧ ⎨ ⎩ y1,1(x, t) = exsin t − sin t, y2,1(x, t) = −x + e−xcos t − cos t, ⎧ ⎨ ⎩ y1,2(x, t) = ex− e−x− 2x −1₂x2cos t,

y2,2(x, t) = −e−xcos2t + exsin2t + 2 cos2t + x cos t − 1, ..

.

Therefore, the approximate solution of Example (3.2), can be readily obtained by ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ y1(x, t) =P+m=0∞ y1,m(x, t) = exsin t + ex− e−x− 2x −1₂x2cos t + · · · , y2(x, t) =P+m=0∞ y2,m(x, t) = −e−xcos2t + exsin2t + e−xcos t + 2 cos2t

+ x cos t − x − 1 + · · · .

NHAM approach:

To solve the system (13) by the NHAM, for simplicity we define the system of nonlinear operators as ⎧ ⎨ ⎩ N1[φ1(x, t; p), φ2(x, t; p)] = ∂φ1 ∂x − φ2 ∂φ1 ∂t + φ1 ∂φ2 ∂t + 1 − exsin t, N2[φ1(x, t; p), φ2(x, t; p)] = ∂φ2 ∂x + ∂φ1 ∂t ∂φ2 ∂x + ∂φ2 ∂t ∂φ1 ∂x + 1 + e−xcos t. Thus, we obtain the system of mth-order deformation equations as follows (14) ∂

∂x[yi,m(x, t) − χmyi,m−1(x, t)] = ~i Hi(x, t) Ri,m(−→y1,m−1, −→y2,m−1), where ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ R1,m(−→y1,m−1, −→y2,m−1) =∂y1,m−1∂x + Pm−1 k=0 h y1,k∂y2,m−1−k_∂t − y2,k∂y1,m−1−k_∂t i + (1 − χm)(1 − exsin t), R2,m(−→y1,m−1, −→y2,m−1) =∂y2,m−1∂x + Pm−1 k=0 h_∂y 1,k ∂t ∂y2,m−1−k ∂x + ∂y2,k ∂t ∂y1,m−1−k ∂x i + (1 − χm)(1 + e−xcos t).

Now, the solution of system (14), for (m ≥ 1) is yi,m(x, t) = χmyi,m−1(x, t) + ~i Z x 0 [H i(x1, t)Ri,m(−→y1,m−1, −→y2,m−1)] dx1, in other words ⎧ ⎨ ⎩ y1,1(x, t) = ~1 R x 0 [H1(x1, t)R1,1(−→y1,0, −→y2,0)] dx1, y2,1(x, t) = ~2 R₀x[H2(x1, t)R2,1(−→y1,0, −→y2,0)] dx1, ⎧ ⎨ ⎩ y1,2(x, t) = y1,1(x, t) + ~1 R₀x[H1(x1, t)R1,2(−→y1,1, −→y2,1)] dx1, y2,2(x, t) = y2,1(x, t) + ~2 Rx 0 [H2(x1, t)R2,2(−→y1,1, −→y2,1)] dx1, .. .⎧ ⎨ ⎩ y1,j+1(x, t) = y1,j(x, t) + ~1 R₀x[H1(x1, t)R1,j+1(−→y1,j, −→y2,j)] dx1, y2,j+1(x, t) = y2,j(x, t) + ~2 R₀x[H2(x1, t)R2,j+1(−→y1,j, −→y2,j)] dx1, .. .

For solving the above systems by means of NHAM, considering that H1(x1, t) = H2(x1, t) = 1, ~1= ~2= −1, with initial approximations as follows

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ y1,0(x, t) = sin t +R₀xhP_j=0+∞aj(t)P1,j(x1) i , y2,0(x, t) = cos t +R₀xhPj=0+∞bj(t)P2,j(x1) i ,

such that Pi,j(x) = xj, for i = 1, 2. Solving the above equations for y1,1(x, t) and y2,1(x, t) with considering the above information leads to the result

y1,1(x, t) = [−a0(t) + sin t] x + [− 1 2a1(t) + a0₀(t) cos t 2 + b0(t) cos t 2 − b0₀(t) sin t 2 +a0(t) sin t 2 + sin t 2 ]x 2 + [−1₃a2(t) + a0₁(t) cos t 6 + b1(t) cos t 6 +b0(t)a 0 0(t) 3 − b0₁(t) sin t 6 + a1(t) sin t 6 − b0₀(t)a0(t) 3 + sin t 6 ]x 3 + [−1 4a3(t) + a0₂(t) cos t 12 + b2(t) cos t 12 + b1(t)a 0 0(t) 8 + b0(t)a 0 1(t) 8 −b 0 2(t) sin t 12 + a2(t) sin t 12 − b0₀(t)a1(t) 8 − b0₁(t)a0(t) 8 + sin t 24 ]x 4 + · · · ,

y2,1(x, t) = [−b0(t) − b0(t) cos t + a0(t) sin t − cos t − 1]x + [− 1 2b1(t) −b1(t) cos t₂ −b0(t)a 0 0(t) 2 + a1(t) sin t 2 − b0₀(t)a0(t) 2 + cos t 2 ]x 2 + [−1₃b2(t) − b2(t) cos t 3 − b0(t)a 0 1(t) 6 − b1(t)a 0 0(t) 3 + a2(t) sin t 3 −b 0 1(t)a0(t) 6 − b0₀(t)a1(t) 3 − cos t 6 ]x 3 + [−1₄b3(t) − b3(t) cos t 4 −b0(t)a 0 2(t) 12 − b1(t)a 0 1(t) 8 − b2(t)a 0 0(t) 4 + a3(t) sin t 4 − b0₀(t)a2(t) 4 −b 0 1(t)a1(t) 8 − b0₂(t)a0(t) 12 + cos t 24 ]x 4 + · · · .

By the vanishing of y1,1(x, t) and y2,1(x, t) the coefficients aj(t) and bj(t) for j = 1, 2, 3, ... are determined as

½

a0(t) = sin t, a1(t) = sin t, a2(t) = 1₂sin t, a3(t) = 1₆sin t, a4(t) =₂₄1 sin t, · · ·

½

b0(t) = − cos t, b1(t) = cos t, b2(t) = −1₂cos t, b3(t) = 1₆cos t, b4(t) = −₂₄1 cos t, · · ·

Therefore we obtain the solution of Eq.(13) as y1(x, t) = y1,0(x, t) = sin t + a0(t)x + 1 2a1(t)x 2₊1 3a2(t)x 3 + · · · = exsin t, y2(x, t) = y2,0(x, t) = cos t + b0(t)x + 1 2b1(t)x 2₊1 3b2(t)x 3 + · · · = e−xcos t, which is an exact solution for Eq.(13).

Example 3.3. Consider the following non-linear system of inhomogeneous partial differential equations:

(15) ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ∂y1 ∂t − ∂y3 ∂x ∂y2 ∂t − 1 2 ∂y3 ∂t ∂2y1 ∂x2 = −4xt, ∂y2 ∂t − ∂y3 ∂t ∂2_y 1 ∂x2 = 6t, ∂y3 ∂t − ∂2_y 1 ∂x2 − ∂y2 ∂x ∂y3 ∂t = 4xt − 2t − 2, the initial conditions are given by

y1(x, 0) = x2+ 1, y2(x, 0)x2− 1, y3(x, 0) = x2− 1, the exact solutions are

First we apply the HAM approach and then the NHAM approach. HAM approach:

Solving the system (15) by the HAM with ~i = −1, Hi(x, t) = 1, for i = 1, 2, 3, and y1,0(x, t) = x2+ 1, y2,0(x, t) = x2− 1, and y3,0(x, t) = x2− 1, we obtain

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ y1,1(x, t) = −2xt2, y2,1(x, t) = 3t2, y3,1(x, t) = 2xt2− t2, ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ y1,2(x, t) = 1₂[16x − 2]t2, y2,2(x, t) = 12[8x − 4]t2, y3,2(x, t) = x[4x − 2]t2, ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ y1,3(x, t) = 3t4+ [x(8x − 4) + x(4x − 2)]t2, y2,3(x, t) = 2x[4x − 2]t2, y3,2(x, t) = 2x2[4x − 2]t2, .. .

Therefore, the approximate solution of Example (3.3), can be readily obtained by ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ y1(x, t) =P+m=0∞ y1,m(x, t) ≈ x2− t2+ 1, y2(x, t) =P+_m=0∞ y2,m(x, t) ≈ x2+ t2− 1, y3(x, t) =P+m=0∞ y3,m(x, t) ≈ x2− t2− 1. NHAM approach:

To solve the system (15) by the NHAM, for simplicity we define the system of nonlinear operators as ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ N1[φ1(x, t; p), φ2(x, t; p), φ3(x, t; p)] = ∂φ1 ∂t − ∂φ3 ∂x ∂φ2 ∂t − 1 2 ∂φ3 ∂t ∂2φ1 ∂x2 + 4xt, N2[φ1(x, t; p), φ2(x, t; p), φ3(x, t; p)] = ∂φ2 ∂t − ∂φ3 ∂t ∂2φ1 ∂x2 − 6t, N3[φ1(x, t; p), φ2(x, t; p), φ3(x, t; p)] = ∂φ3 ∂t − ∂2φ1 ∂x2 − ∂φ2 ∂x ∂φ3 ∂t − 4xt + 2t + 2.

Thus, we obtain the system of mth-order deformation equations as follows (16)

∂

∂t[yi,m(x, t) − χmyi,m−1(x, t)] = ~i Hi(x, t) Ri,m(−→y1,m−1, −→y2,m−1, −→y3,m−1), where ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ R1,m(−→y1,m−1, −→y2,m−1, −→y3,m−1) = ∂y1,m−1∂t − Pm−1 k=0[ ∂y3,k ∂x ∂y2,m−1−k ∂t +1₂∂y3,k ∂t ∂2_y 1,m−1−k ∂x2 ] + (1 − χm)(4xt), R2,m(−→y1,m−1, −→y2,m−1, −→y3,m−1) = ∂y2,m−1∂t − Pm−1 k=0 h_∂y 3,k ∂t ∂2y1,m−1−k ∂x2 i + (1 − χm)(−6t), R3,m(−→y1,m−1, −→y2,m−1, −→y3,m−1) = ∂y3,m−1∂t − ∂2_y 1,m−1 ∂x2 − Pm−1 k=0 h_∂y 2,k ∂x ∂y3,m−1−k ∂t i + (1 − χm)(−4xt + 2t + 2).

Now, the solution of system (16), for (m ≥ 1) is yi,m(x, t) = χmyi,m−1(x, t)+~i Z t 0 [H i(x, t1)Ri,m(−→y1,m−1, −→y2,m−1, −→y3,m−1)] dt1, in other words ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ y1,1(x, t) = ~1 Rt 0[H1(x, t1)R1,1(−→y1,0, −→y2,0, −→y3,0)] dt1, y2,1(x, t) = ~2 Rt 0[H2(x, t1)R2,1(−→y1,0, −→y2,0, −→y3,0)] dt1, y3,1(x, t) = ~3 R t 0[H3(x, t1)R3,1(−→y1,0, −→y2,0, −→y3,0)] dt1, ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ y1,2(x, t) = y1,1(x, t) + ~1 R₀t[H1(x, t1)R1,2(−→y1,1, −→y2,1, −→y3,1)] dt1, y2,2(x, t) = y2,1(x, t) + ~2 R₀t[H2(x, t1)R2,2(−→y1,1, −→y2,1, −→y3,1)] dt1, y3,2(x, t) = y3,1(x, t) + ~3 R₀t[H3(x, t1)R3,2(−→y1,1, −→y2,1, −→y3,1)] dt1, .. . ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ y1,j+1(x, t) = y1,j(x, t) + ~1 Rt 0[H1(x, t1)R1,j+1(−→y1,j, −→y2,j, −→y3,j)] dt1, y2,j+1(x, t) = y2,j(x, t) + ~2 R t 0[H2(x, t1)R2,j+1(−→y1,j, −→y2,j, −→y3,j)] dt1, y3,j+1(x, t) = y3,j(x, t) + ~3 R₀t[H3(x, t1)R3,j+1(−→y1,j, −→y2,j, −→y3,j)] dt1, .. .

For solving the above systems by means of NHAM, considering that H1(x, t1) = H2(x, t1) = H3(x, t1) = 1, ~1 = ~2 = ~3 = −1, with initial approximations as

follows _⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ y1,0(x, t) = x2+ 1 + Rt 0 hP+∞ j=0aj(x)P1,j(t1) i , y2,0(x, t) = x2− 1 +R₀thP_j=0+∞bj(x)P2,j(t1) i , y3,0(x, t) = x2− 1 +R₀thPj=0+∞cj(x)P3,j(t1) i ,

such that Pi,j(x) = tj, for i = 1, 2, 3. Solving the above equations for y1,1(x, t), y2,1(x, t) and y3,1(x, t) with considering the above information leads to the result

y1,1(x, t) = [−a0(x) + 2xb0(x) + c0(x)]t + [− 1 2a1(x) + xb1(x) + 1 2b0(x)c 0 0(x) +1 2c1(x) + 1 4a 00 0(x)c0(x) − 2x]t2+ [− 1 3a2(x) + 2 3xb2(x) +1 3b1(x)c 0 0(x) + 1 6b0(x)c 0 1(x) + 1 3c2(x) + 1 12a 00 1(x)c0(x) +1 6a 00 0(x)c1(x)]t3+ [− 1 4a3(x) + 1 2xb3(x) + 1 4b2(x)c 0 0(x) +1 8b1(x)c 0 1(x) + 1 12b0(x)c 0 2(x) + 1 4c3(x) + 1 24a 00 2(x)c0(x) + 1 16a 00 1(x)c1(x) + 1 8a 00 0(x)c2(x)]t4+ · · · , y2,1(x, t) = [−b0(x) + 2c0(x)]t + [− 1 2b1(x) + c1(x) + 1 2c0(x)a 00 0(x) + 3]t2 + [−1₃b2(x) + 2 3c2(x) + 1 6c0(x)a 00 1(x) + 1 3c1(x)a 00 0(x)]t3+ [− 1 4b3(x) +1 2c3(x) + 1 12c0(x)a 00 2(x) + 1 8c1(x)a 00 1(x) + 1 4c2(x)a 00 0(x)]t4+ · · · , y3,1(x, t) = [−c0(x) + 2xc0(x)]t + [− 1 2c1(x) + 1 2a 00 0(x) + xc1(x) + b 0 0(x)c0(x) + 2x − 1]t2+ [−1₃c3(x) + 1 6a 00 1(x) + 2 3xc2(x) + 1 3b 0 0(x)c1(x) +1 6b 0 1(x)c0(x)]t3+ [− 1 4c3(x) + 1 12a 00 2(x) + 1 2xc3(x) + 1 4b 0 0(x)c2(x) +1 8b 0 1(x)c1(x) + 1 12b 0 2(x)c0(x)]t4+ · · · .

and cj(t) for j = 1, 2, 3, ... are determined as ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ a0(x) = 0, a1(x) = −2, a2(x) = a3(x) = a4(x) = · · · = 0, b0(x) = 0, b1(x) = 2, b2(x) = b3(x) = b4(x) = · · · = 0, c0(x) = 0, c1(x) = −2, c2(x) = c3(x) = c4(x) = · · · = 0, Therefore we obtain the solution of Eq.(15) as

y1(x, t) = y1,0(x, t) = x2+ 1 + a0(x)t + 1 2a1(x)t 2₊1 3a2(x)t 3 + · · · = x2− t2+ 1, y2(x, t) = y2,0(x, t) = x2− 1 + b0(x)t + 1 2b1(x)t 2₊1 3b2(x)t 3 + · · · = x2+ t2_{− 1,} y3(x, t) = y3,0(x, t) = x2− 1 + c0(x)t + 1 2c1(x)t 2₊1 3c2(x)t 3 + · · · = x2− t2− 1, which is an exact solution for Eq.(15).

4. Conclusion

In this article, a new modification of HAM, called NHAM, has been introduced for solving systems of non-linear partial differential equations. This method has been applied to three examples successfully, and exact solutions of the equations are achieved, where traditional HAM leads to an approximate solution. The examples in this paper are further confirmation of the flexibility and potential of the NHAM for solving complicated linear and non-linear initial and boundary value problems in science and engineering. The computations associated with the examples were performed using Maple. This new method easily can be employed to solve other functional equations.

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