Analytical Solutions of Boundary Values Problem of 2D and 3D Poisson and Biharmonic Equations by Homotopy Decomposition Method

and Applied Analysis 3 3. Solutions of the Main Problems Problem 1. Consider the following equation ∂ 2 u ∂x + ∂ 2 u ∂y = sin (πx) sin (πy) ; u (x, y) = 0 along the boundaries, 0 ≤ x, y ≤ 1; u x (0, y) = − sin (yπ) 2π . (11) The exact solution of the previous equation is given as u (x, y) = sin (xπ) sin (πy) −2π . (12) In the view of the homotopy decomposition method, (11) can be first transformed to u (x, y) = u (0, y) − sin (πy) 2π x + ∫ x 0 (x − τ) [sin (πτ) sin (πy) − u yy (τ, y)] , u (x, y, p) = ∞ ∑ n=0 p n u n (x, y) . (13) Following the decomposition techniques, we obtain the following equation ∞ ∑ n=0 p n u n (x, y) = T (x, y) + p∫ x 0 (x − τ) [ sin (πτ) sin (πy)


Introduction
The numerical solution of Poisson equations and biharmonic equations is an important problem in numerical analysis.A vast arrangement of investigating effort has been published on the development of numerical solution of Poisson equations and biharmonic equations.The finite difference schemes of second and fourth order for the solution of Poisson's equation in polar coordinates have been derived by Mittal and Gahlaut [1].A numerical method to interpolate the source terms of Poisson's equation by using B-spline approximation has been devised by Perrey-Debain and ter Morsche [2].Sutmann and Steffen [3] proposed compact approximation schemes for the Laplace operator of fourth and sixth order; the schemes are based on a Padé approximation of the Taylor expansion for the discretized Laplace operator.Ge [4] used fourth-order compact difference discretization scheme with unequal mesh sizes in different coordinate directions to solve a 3D Poisson equation on a cubic domain.Gumerov and Duraiswami [5] developed a complete translation theory for the biharmonic equation in three dimensions.Khattar et al. [6] derived a fourth-order finite difference approximation based on arithmetic average discretization for the solution of three-dimensional nonlinear biharmonic partial differential equations on a 19-point compact stencil using coupled approach.Altas et al. [7] used multigrid and preconditioned Krylov iterative methods to solve three-dimensional nonlinear biharmonic partial differential equations.Jeon [8] derived scalar boundary integral equation formulas for both interior and exterior biharmonic equations with the Dirichlet boundary data.A spectral collocation method for numerically solving two-dimensional biharmonic boundaryvalue problems has been reported in [9].An indirect radialbasis-function collocation method for numerically solving biharmonic boundary-value problems has been reported in [10].A high-order boundary integral equation method for the solution of biharmonic equations has been presented in [11].A Galerkin boundary node method for solving biharmonic problems was developed in [12].An integral collocation approach based on Chebyshev polynomials for numerically solving biharmonic equations for the case of irregularly shaped domains has been developed by Mai-Duy et al. [13].A numerical method, based on neural-network-based functions, for solving partial differential equations has been in [14].Mai-Duy and Tanner [15] presented a collocation method based on a Cartesian grid and a 1D integrated radial basis function scheme for numerically solving partial differential equations in rectangular domains and Haar wavelet presented in [16].The aim of this paper is to solve these problems via the homotopy decomposition method.

Method
In this study we follow the method of [17][18][19][20].In order to illustrate the basic idea of this method we consider a general nonlinear nonhomogeneous partial differential equation with initial conditions of the following form subject to the initial conditions where  is a known function,  is the general nonlinear differential operator, and  represents a linear differential operator.The method's first step here is to apply the inverse operator   /  of on both sides (1) to obtain ( (, )) +  ( (, )) +  (, )  ⋅ ⋅ ⋅ . ( The multi-integrals in (3) can be transformed to So that (3) can be reformulated as Using the homotopy scheme the solution of the previous integral equation is given in a series form as and the nonlinear term can be decomposed as where  ∈ (0, 1] is an embedding parameter.H  () is He's polynomials [21] that can be generated by The homotopy decomposition method is obtained by the graceful coupling of decomposition method with He's polynomials and is given by with Comparing the terms of the same power of  gives the solutions of various orders.The initial guess of the approximation is (, ).Some further related results can be seen in [22][23][24][25].
Proof.The number of computations including product, addition, subtraction, and division are as follows.
In step 2  0 : 0 because it is obtained directly from the initial conditions  1 : 3 . . .

Solutions of the Main Problems
Problem 1.Consider the following equation The exact solution of the previous equation is given as In the view of the homotopy decomposition method, ( 11) can be first transformed to Following the decomposition techniques, we obtain the following equation Comparing the terms of the same power of  leads to 1 (, ) = 0 along the boundaries, The following solutions are obtained:  ( Therefore in general for any  > 8 we have This is the exact solution of the problem.Figures 1 and 2 show the comparison of the exact solution and the approximated one for  = 4.The approximate solution and the exact solution are compared in Figures 1 and 2, respectively.
The numerical errors for  = 4 are evaluated in Table 1.
Following the discussion presented earlier we obtain the following set of integral equations: The following solutions are obtained:  3  3! + () 5  5! − () 7  7! + () 9  9! − () 3  11! ]sin() sin() Therefore, for any  ≥ 6, the partial sum is given as Thus And this is the exact solution to the problem.One can evaluate error committed by choosing the  first terms in the series solutions, in the same manner as in Table 1.The accuracy of the results is estimated by error function Problem 3. Let us consider the following biharmonic equation for which the exact solution is The aim of this part is to compare the numerical results obtained via HDM and the method used in [26].Applying the steps involved in the HDM, we arrive at the following: In the same manner, one can obtain the remaining term by using the following recursive formula: In this paper we consider only the first six terms of the series solution as follows: To access the accuracy of the method used in paper, we compare in Table 2 the numerical results of the above equation, the solution obtained in [26] with the exact solution.
Figures 3 and 4 are the graphical representation of the previous solution.We have plotted the solution for (31) in Figure 3 and showed absolute value of the solution in Figure 4.
Theorem 2. Let  be a nonzero natural number and let (x, y) ∈ [0, 1] × [0, 1]; then two dimensional biharmonic equation of form with (, ) = 0 along the boundaries has an exact solution as follows Proof.Use the step of the homotopy decomposition method.

Conclusion
In this paper the recent homotopy decomposition [18][19][20][21] is used to solve the 2D and 3D Poisson equations and biharmonic equations.The method is chosen because it does not require the linearization or assumptions of weak nonlinearity, the solutions are generated in the form of general solution, and it is more realistic compared to the method of simplifying the physical problems.The method does not require any corrected function any Lagrange multiplier and it avoids repeated terms in the series solutions compared to the existing decomposition method including the variational iteration and the Adomian decomposition method.
The approximated solutions obtained converge to the exact solution as  tends to infinity.The numerical values are presented in Table 1 shows that the method is very efficient and accurate.

5 2
In the same manner one can obtain the rest of the components.But for eight terms were computed and the asymptotic solution is given by  =8 (, ) sin () sin () .

Table 1 :
Evaluation of numerical errors for  = 4.