Analysis of the new homotopy perturbation method for linear and nonlinear problems

  • Ali Demir1Email author,

    Affiliated with

    • Sertaç Erman1,

      Affiliated with

      • Berrak Özgür1 and

        Affiliated with

        • Esra Korkmaz2

          Affiliated with

          Boundary Value Problems20132013:61

          DOI: 10.1186/1687-2770-2013-61

          Received: 14 December 2012

          Accepted: 6 March 2013

          Published: 26 March 2013

          Abstract

          In this article, a new homotopy technique is presented for the mathematical analysis of finding the solution of a first-order inhomogeneous partial differential equation (PDE) u x ( x , y ) + a ( x , y ) u y ( x , y ) + b ( x , y ) g ( u ) = f ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq1_HTML.gif. The homotopy perturbation method (HPM) and the decomposition of a source function are used together to develop this new technique. The homotopy constructed in this technique is based on the decomposition of a source function. Various decompositions of source functions lead to various homotopies. Using the fact that the decomposition of a source function affects the convergence of a solution leads us to development of a new method for the decomposition of a source function to accelerate the convergence of a solution. The purpose of this study is to show that constructing the proper homotopy by decomposing f ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq2_HTML.gif in a correct way determines the solution with less computational work than using the existing approach while supplying quantitatively reliable results. Moreover, this method can be generalized to all inhomogeneous PDE problems.

          1 Introduction

          Many problems in the fields of physics, engineering and biology are modeled by linear and nonlinear PDEs. In recent years, the homotopy perturbation method (HPM) has been employed to solve these PDEs [16]. HPM has been used extensively to solve nonlinear boundary and initial value problems. Therefore HPM is of great interest to many researchers and scientists. HPM, first presented by Ji Huan He [79], is a powerful mathematical tool to investigate a wide variety of problems arising in different fields. It is obtained by successfully coupling homotopy theory in topology with perturbation theory. In HPM, a complicated problem under study is continuously deformed into a simple problem which is easy to solve to obtain an analytic or approximate solution [10].

          In this work, a new homotopy perturbation technique is proposed to find the solution u ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq3_HTML.gif based on the decomposition of a right-hand side function f ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq2_HTML.gif which leads to the construction of new homotopies. It is shown that decomposing f ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq2_HTML.gif in a correct way, requiring less computational work than the existing approach, helps us to determine the solution. The results reveal that the proposed method is very effective and simple.

          The paper is organized as follows. In Section 2, an analysis of the new homotopy perturbation method is given. The construction of a new homotopy based on the decomposition of a source function for a first-order inhomogeneous PDE is covered in Section 3. Some examples are given to illustrate the construction of a new homotopy in Section 4. Finally, some concluding remarks are given in Section 5.

          2 New homotopy perturbation method

          To illustrate the basic ideas of the homotopy analysis method, we consider the following nonlinear differential equation:
          A ( u ) f ( r ) = 0 , r Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ1_HTML.gif
          (1)
          with the boundary conditions
          B ( u , u n ) = 0 , r Γ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ2_HTML.gif
          (2)

          where A is a general differential operator, B is a boundary operator, f ( r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq4_HTML.gif is a known analytical function, and Γ is the boundary of domain Ω.

          The operator A in (1) can be rewritten as a sum of L and N, where L and N are linear and nonlinear parts of A, respectively, as follows:
          L ( u ) + N ( u ) f ( r ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equa_HTML.gif
          By the homotopy technique, we construct the homotopy
          H ( v , p ) = ( 1 p ) ( L ( v ) L ( u 0 ) ) + p ( A ( v ) f ( r ) ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ3_HTML.gif
          (3)
          which is equivalent to
          H ( v , p ) = L ( v ) L ( u 0 ) + p L ( u 0 ) + p ( N ( v ) f ( r ) ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ4_HTML.gif
          (4)
          where
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equb_HTML.gif

          p is an embedding parameter, u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq5_HTML.gif is an initial approximation of (1), which satisfies the boundary conditions. As p changes from zero to unity, v ( r , p ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq6_HTML.gif changes from u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq5_HTML.gif to u ( r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq7_HTML.gif. In this technique, the convergence of a solution depends on the choice of u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq5_HTML.gif, that is, we can have different approximate solutions for different u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq5_HTML.gif.

          Let us decompose the source function as f ( r ) = f 1 ( r ) + f 2 ( r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq8_HTML.gif. If we take
          L ( u 0 ) = f 1 ( r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equc_HTML.gif
          in (3), we obtain the following homotopy:
          H ( v , p ) = ( 1 p ) ( L ( v ) f 1 ( r ) ) + p ( A ( v ) f ( r ) ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ5_HTML.gif
          (5)
          which is equivalent to
          H ( v , p ) = L ( v ) f 1 ( r ) + p ( N ( v ) f 2 ( r ) ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ6_HTML.gif
          (6)
          Obviously, from (6) we have
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equd_HTML.gif

          As the embedding parameter p changes from zero to unity, v ( r , p ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq6_HTML.gif changes from L 1 ( f 1 ( r ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq9_HTML.gif to u ( r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq7_HTML.gif.

          According to He’s homotopy perturbation method, we can first use the embedding parameter p as a small parameter and assume that the solution of (6) can be written as a power series in p as follows:
          v = v 0 + p v 1 + p 2 v 2 + p 3 v 3 + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ7_HTML.gif
          (7)
          Setting p = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq10_HTML.gif, we get the approximate solution of (1)
          u = lim p 1 v = v 0 + v 1 + v 2 + v 3 + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Eque_HTML.gif

          In the next section, it is shown that the decomposition of the right-hand side function has a great effect on the amount of calculation and the speed of convergence of the solution.

          3 Construction of a new homotopy based on the decomposition of a source function

          Let us consider the following boundary value problem with the following inhomogeneous PDE:
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ8_HTML.gif
          (8)
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ9_HTML.gif
          (9)
          where a, b, g and f are continuous functions in some region of the plane and g ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq11_HTML.gif. By solving this boundary value problem by the homotopy perturbation method, we obtain an approximate or exact solution u ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq3_HTML.gif. Before proceeding further, let us introduce the integral operator S defined in the following form:
          S ( ) = 0 x ( ) d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ10_HTML.gif
          (10)
          Then the derivative of operator S with respect to y is defined as
          S y ( ) = y 0 x ( ) d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equf_HTML.gif
          Rewriting f ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq2_HTML.gif as a sum of two functions f ( x , y ) = f 1 ( x , y ) + f 2 ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq12_HTML.gif and then constructing a homotopy in equation (6), we have
          H ( v , p ) = ( 1 p ) ( v x f 1 ) + p ( v x + a ( x , y ) v y + b ( x , y ) g ( v ) f ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ11_HTML.gif
          (11)
          which is equivalent to
          H ( v , p ) = v x ( x , y ) f 1 ( x , y ) + p ( a ( x , y ) v y ( x , y ) + b ( x , y ) g ( v ( x , y ) ) f 2 ( x , y ) ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ12_HTML.gif
          (12)
          Substituting (7) into (12), and equating the coefficient of the terms by the same power in p, we have
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ13_HTML.gif
          (13)
          If there exists the relationship
          a ( x , y ) [ S y ( f 1 ( x , y ) ) + h y ( y ) ] + b ( x , y ) g ( S ( f 1 ( x , y ) ) + h ( y ) ) = f 2 ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ14_HTML.gif
          (14)
          between f 1 ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq13_HTML.gif and f 2 ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq14_HTML.gif, then we have from (13)
          ( v 1 ) x = 0 , v 1 ( 0 , y ) = 0 v 1 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equg_HTML.gif
          and
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equh_HTML.gif
          Hence the approximate or exact solution of problem (8)-(9) is obtained as
          u ( x , y ) = v 0 ( x , y ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equi_HTML.gif
          and the source function is obtained from (14) in the following form:
          f ( x , y ) = f 1 + a ( x , y ) [ S y ( f 1 ) + h y ( y ) ] + b ( x , y ) g ( S ( f 1 ) + h ( y ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ15_HTML.gif
          (15)

          Therefore equation (15) is a necessary condition to accelerate the convergence. If equation (15) has a solution, then we obtain the approximate or exact solution of problem (8)-(9) in two steps.

          However, it is not always possible to decompose the source function f ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq2_HTML.gif in such a way that the functions f 1 ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq13_HTML.gif and f 2 ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq14_HTML.gif have the relationship in (14). If we have such a case, then we are looking for an arbitrary P ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq15_HTML.gif such that the functions f 1 ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq13_HTML.gif and f 2 ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq14_HTML.gif have the following relationship:
          a ( x , y ) [ S y ( f 1 ) + h y ( y ) ] + b ( x , y ) g ( S ( f 1 ) + h ( y ) ) = f 2 + P ( x , y ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ16_HTML.gif
          (16)

          In this case, we get the approximate or exact solution of the problem in more than two steps. Moreover, we can get the solution in the form of series.

          4 Analysis of the new homotopy perturbation method for linear problems

          In this section we illustrate the theory of the new homotopy perturbation method given in Sections 2 and 3 for some special f ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq2_HTML.gif in the following linear problem:
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ17_HTML.gif
          (17)
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ18_HTML.gif
          (18)

          where a and b are constants.

          4.1 Case 1: the source function f ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq2_HTML.gif is a polynomial

          We consider that h ( y ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq16_HTML.gif and f ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq2_HTML.gif is an n th-order polynomial in problem (17)-(18). Hence f ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq2_HTML.gif can be written in the following form:
          f ( x , y ) = k = 0 n α + β = k A α β x α y β , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ19_HTML.gif
          (19)

          where A α β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq17_HTML.gif are constants and α, β are natural numbers.

          If the polynomial f ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq2_HTML.gif is decomposed as
          f ( x , y ) = f 1 ( x , y ) + a S y ( f 1 ( x , y ) ) + b S ( f 1 ( x , y ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ20_HTML.gif
          (20)
          from (15), we can get the solution of the problem in two steps. We take f 1 ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq13_HTML.gif as an n th-order polynomial given as
          f 1 ( x , y ) = k = 0 n α + β = k c α β x α y β , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ21_HTML.gif
          (21)
          where c α β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq18_HTML.gif are constants. Then S ( f 1 ( x , y ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq19_HTML.gif becomes
          S ( f 1 ( x , y ) ) = k = 0 n α + β = k c α β α + 1 x α + 1 y β , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ22_HTML.gif
          (22)
          and S y ( f 1 ( x , y ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq20_HTML.gif becomes
          S y ( f 1 ( x , y ) ) = k = 0 n α + β = k β > 1 β c α β α + 1 x α + 1 y β 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ23_HTML.gif
          (23)
          In order to determine the unknown coefficients c α β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq18_HTML.gif in terms of A α β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq17_HTML.gif, we substitute (21), (22), and (23) into (20)
          f ( x , y ) = k = 0 n α + β = k c α β x α y β + k = 0 n α + β = k β > 1 β a c α β α + 1 x α + 1 y β 1 + k = 0 n α + β = k b c α β α + 1 x α + 1 y β , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ24_HTML.gif
          (24)
          and then using (19), we get the following equations:
          A 00 = c 00 , A 10 = c 10 + b c 00 , A 01 = c 01 , A 20 = c 20 + b 2 c 10 , A 02 = c 02 , A 11 = c 11 + 2 a c 02 + b c 01 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equj_HTML.gif
          Then we obtain
          c 00 = A 00 , c 10 = A 10 b A 00 , c 01 = A 01 , c 20 = A 20 b 2 ( A 10 b A 00 ) , c 02 = A 02 , c 11 = A 11 2 a A 02 b A 01 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equk_HTML.gif
          As a result, constructing the following homotopy:
          H ( v , p ) = ( 1 p ) ( v x f 1 ( x , y ) ) + p ( v x + a v y + b v f ( x , y ) ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equl_HTML.gif

          leads us to reaching the solution in two steps.

          4.2 Case 2: the source function f ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq2_HTML.gif is a sum of two functions r ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq21_HTML.gif and t ( y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq22_HTML.gif

          Let us take f ( x , y ) = r ( x ) + t ( y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq23_HTML.gif in problem (17)-(18). Since f ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq2_HTML.gif is a continuous function, r ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq21_HTML.gif and t ( y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq22_HTML.gif are also continuous functions. Based on (15), f ( x , y ) = r ( x ) + t ( y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq23_HTML.gif is decomposed as follows:
          f ( x , y ) = r 1 ( x ) + t 1 ( y ) + a S y ( r 1 ( x ) + t 1 ( y ) ) + a h y ( y ) + b S ( r 1 ( x ) + t 1 ( y ) ) + b h ( y ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ25_HTML.gif
          (25)
          Then we can get the solution of the problem in two steps. Hence the function f ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq2_HTML.gif can be rewritten in the following form:
          f ( x , y ) = r 1 ( x ) + t 1 ( y ) + a t 1 y ( y ) x + a h y ( y ) + b 0 x r 1 ( x ) d x + b t 1 ( y ) x + b h ( y ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ26_HTML.gif
          (26)
          Since f ( x , y ) = r ( x ) + t ( y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq23_HTML.gif, we have
          a t 1 y ( y ) x + b t 1 ( y ) x = 0 for  x 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ27_HTML.gif
          (27)
          The solutions of the above ordinary differential equation are
          t 1 ( y ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ28_HTML.gif
          (28)
          and
          t 1 ( y ) = e b a y . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ29_HTML.gif
          (29)
          On the other hand, we have
          t ( y ) = a h y ( y ) + b h ( y ) + t 1 ( y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ30_HTML.gif
          (30)
          and
          r ( x ) = b 0 x r 1 ( x ) d x + r 1 ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ31_HTML.gif
          (31)
          If either (28) or (29) satisfy equation (30), and equation (31) has a solution, then it is possible to decompose the function f ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq2_HTML.gif to accelerate the convergence of the solution. As a result, by constructing the following homotopy:
          H ( v , p ) = ( 1 p ) ( v x r 1 ( x ) t 1 ( y ) ) + p ( v x + a v y + b v r ( x ) t ( y ) ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equm_HTML.gif

          we reach the solution in two steps.

          5 Numerical examples

          5.1 Example 1

          Consider the inhomogeneous linear boundary value problem (BVP) with constant coefficients
          u x u y + u = e y + e x , u ( 0 , y ) = e y . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ32_HTML.gif
          (32)
          By constructing the following homotopy:
          H ( v , p ) = ( 1 p ) ( v x r 1 ( x ) t 1 ( y ) ) + p ( v x v y + v e x e y ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ33_HTML.gif
          (33)
          which is equivalent to
          H ( v , p ) = v x r 1 ( x ) t 1 ( y ) + p ( v y + v + r 1 ( x ) + t 1 ( y ) e x e y ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ34_HTML.gif
          (34)
          we get the functions r 1 ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq24_HTML.gif, t 1 ( y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq25_HTML.gif as it is explained in Section 4.2. It is obvious that t 1 ( y ) = e y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq26_HTML.gif satisfies equation (30). Moreover, from equation (31), r 1 ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq24_HTML.gif can be found as follows:
          r 1 ( x ) = e x + e x 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ35_HTML.gif
          (35)
          If we put the function r 1 ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq24_HTML.gif and t 1 ( y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq25_HTML.gif in the homotopy (33) or (34), we get
          H ( v , p ) = v x e x + e x 2 e y + p ( v y + v + e x e x 2 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ36_HTML.gif
          (36)
          and we get the following:
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equn_HTML.gif
          Hence, the solution u ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq3_HTML.gif becomes
          u ( x , y ) = v 0 = e x e x 2 + e y x + e y , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ37_HTML.gif
          (37)

          which is the exact solution of the problem with minimum amount of calculation.

          5.2 Example 2

          Consider the following inhomogeneous linear BVP with variable coefficients:
          u x + y u y u = 2 y 2 + 2 x y 2 , u ( 0 , y ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ38_HTML.gif
          (38)
          In this problem, the right-hand side function f ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq2_HTML.gif is a third-order polynomial. From (15), we have
          f ( x , y ) = f 1 ( x , y ) + y S y ( f 1 ( x , y ) ) S ( f 1 ( x , y ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ39_HTML.gif
          (39)
          Based on the source function f ( x , y ) = 2 y 2 + 2 x y 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq27_HTML.gif, if we take f 1 ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq13_HTML.gif as
          f 1 ( x , y ) = a 1 x 3 + a 2 x 2 y + a 3 x y 2 + a 4 y 3 + a 5 x 2 + a 6 x y + a 7 y 2 + a 8 x + a 9 y + a 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ40_HTML.gif
          (40)
          and substitute (40) into (39), we obtain
          a 7 = 2 , a 1 = a 2 = a 3 = a 4 = a 5 = a 6 = a 8 = a 9 = a 0 = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equo_HTML.gif
          Based on these results, we have f 1 ( x , y ) = 2 y 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq28_HTML.gif, f 2 ( x , y ) = 2 x y 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq29_HTML.gif and the homotopy is constructed in the following form:
          H ( v , p ) = v x 2 y 2 + p ( y v y v 2 x y 2 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ41_HTML.gif
          (41)
          and we get the following:
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equp_HTML.gif
          Hence, we obtain the exact solution in the following form:
          u ( x , y ) = v 0 = 2 x y 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ42_HTML.gif
          (42)

          with short-length calculation.

          5.3 Example 3

          Let us find the solution of following BVP:
          u x + u y x u = y e x y + x , u ( 0 , y ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ43_HTML.gif
          (43)
          As in the previous examples, using (15) we get
          y e x y + x = f 1 + S y ( f 1 ) x S ( f 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ44_HTML.gif
          (44)
          Based on the source function f ( x , y ) = y e x y + x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq30_HTML.gif, we conclude that f 1 ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq13_HTML.gif is in the following form:
          f 1 ( x , y ) = a 1 e x y + a 2 x e x y + a 3 x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ45_HTML.gif
          (45)
          Substituting (45) into (44), we obtain
          a 1 = a 2 = 0 , a 3 = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equq_HTML.gif
          Now, these results let us take f 1 ( x , y ) = y e x y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq31_HTML.gif, f 2 ( x , y ) = x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq32_HTML.gif and construct the homotopy in the following form:
          H ( v , p ) = v x y e x y + p ( v y x v x ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ46_HTML.gif
          (46)
          which leads to the following equations:
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equr_HTML.gif
          Hence, the exact solution becomes
          u ( x , y ) = v 0 = e x y 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ47_HTML.gif
          (47)

          which is obtained by minimum amount of calculations.

          5.4 Example 4

          Let us find the solution of the following first-order non-linear BVP:
          u x + u y u 2 = e 2 x + 3 e x 1 , u ( 0 , y ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ48_HTML.gif
          (48)
          Using (15) we get
          e 2 x + 3 e x 1 = f 1 + S y ( f 1 ) [ S ( f 1 ) ] 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ49_HTML.gif
          (49)
          Based on the source function f ( x , y ) = e 2 x + 3 e x 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq33_HTML.gif, we find f 1 ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq13_HTML.gif is in the following form:
          f 1 ( x , y ) = a 1 e x + a 2 e 2 x + a 3 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ50_HTML.gif
          (50)
          Substituting (50) into (49), we obtain
          a 1 = 1 , a 2 = a 3 = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equs_HTML.gif
          Now we can take f 1 ( x , y ) = e x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq34_HTML.gif, f 2 ( x , y ) = e 2 x + 2 e x 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq35_HTML.gif and construct the homotopy in the following form:
          H ( v , p ) = v x e x + p ( v y v 2 + e 2 x 2 e x + 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ51_HTML.gif
          (51)
          which leads to the following equations:
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equt_HTML.gif
          Hence, we obtain the exact solution
          u ( x , y ) = v 0 = e x 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_Equ52_HTML.gif
          (52)

          with short-length calculation.

          6 Conclusion

          In this paper, we employ a new homotopy perturbation method to obtain the solution of a first-order inhomogeneous PDE. In this method, each decomposition of the source function f ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq2_HTML.gif leads to a new homotopy. However, we develop a method to obtain the proper decomposition of f ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq2_HTML.gif which lets us obtain the solution with minimum computation and accelerate the convergence of the solution. This study shows that the decomposition of the source function has a great effect on the amount of computations and the acceleration of the convergence of the solution. Comparing to the standard one, decomposing the source function f ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-61/MediaObjects/13661_2012_Article_311_IEq2_HTML.gif properly is a simple and very effective tool for calculating the exact or approximate solutions with less computational work.

          Declarations

          Acknowledgements

          Dedicated to Professor Hari M Srivastava.

          The research was supported by parts by the Scientific and Technical Research Council of Turkey (TUBITAK).

          Authors’ Affiliations

          (1)
          Department of Mathematics, Kocaeli University Umuttepe
          (2)
          Ardahan University

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          Copyright

          © Demir et al.; licensee Springer 2013

          This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.