# Exact and Numerical Solutions of Poisson Equation

Authors: Selçuk Yıldırım
Journal: Mathematical Problems in Engineering
 (2008)

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## Abstract

Homotopy perturbation method (HPM) and boundary element method (BEM) for calculating the exact and numerical solutions of Poisson equation with appropriate boundary and initial conditions are presented. Exact solutions of electrostatic potential problems defined by Poisson equation are found using HPM given boundary and initial conditions. The same problems are also solved using the BEM. The cell integration approach is used for solving Poisson equation by BEM. The problem region containing the charge density is subdivided into triangular elements. In addition, this paper presents a numerical comparison with the HPM and BEM.


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## Body

## 1. Introduction

It is well known that there are many linear and
nonlinear partial equations in various fields of science and engineering. The
solution of these equations can be obtained by many different methods.  In recent years, the studies of the analytical
solutions for the linear or nonlinear evolution equations have captivated the
attention of many authors. The numerical and seminumerical/analytic solution of
linear or nonlinear, ordinary differential equation or partial differential
equation has been extensively studied in the resent years. There are several
methods have been developed and used in different problems [1–3]. The homotopy
perturbation method is relatively new and useful for obtaining both analytical
and numerical approximations of linear or nonlinear differential equations [4–7].
This method yields a very rapid convergence of the solution series. The
applications of homotopy perturbation method among scientists received more attention
recently [8–10]. In this study, we will first concentrate on analytical
solution of Poisson equation, using frequently in electrical engineering, in
the form of Taylor
series by homotopy perturbation method [11–13].The boundary element method is a numerical
technique to solve boundary value problems represented by linear partial differential
equations [14] and has some important advantages. The main advantage of the BEM
is that it replaces the original problem with an integral equation defined on
the boundary of the solution domain. For the case of a homogeneous partial
differential equation, the BEM requires only the discretization on the boundary
of the domain [15]. If the simulation domain is free from the electric charge,
the governing equation is known as Laplace equation. The BEM computes an approximate solution for the boundary integral
formulation of Laplace’s equation by
discretizing the problem boundary into separate elements, each containing a
number of collocation nodes.The distribution of the electrostatic potential
can be determined by solving Poisson equation, if there is charge density in problem domain. In this case, the boundary integral equation obtained from Poisson
equation has a domain integral. In the BEM, several methods had been developed
for solving this integral. These methods are commonly known as cell integration
approach, dual reciprocity method (DRM) and multiple reciprocity method (MRM) [16].The electric field is related to the charge
density by the divergence relationship(1.1)E=electricfield,∇E=ρε0,ρ=chargedensity,ε0=permittivity,and the electric field is related to the electric potential by a
gradient relationship (1.2)E=−∇V.Therefore the potential is related to the
charge density by Poisson equation:(1.3)∇⋅∇V=∇2V=−ρε0,V=electricpotential.


## 2. Theory of the Numerical Methods

### 2.1. Homotopy Perturbation Method

Homotopy perturbation method has
been suggested to solve boundary value problems in [17–19]. According to this
method, a homotopy with an imbedding parameter p∈[0,1] is constructed and the imbedding parameter is
considered as a “small parameter”. Here, homotopy perturbation method is used
to solve analytic solution of Poisson equation with given
boundary conditions.To illustrate this
method, we consider the following nonlinear differential equation:(2.1)A(u)−f(r)=0,r∈Ω, with boundary condition (2.2)B(u,∂u∂n)=0,r∈Γ, where A(u) is written as follows: (2.3)A(u)=L(u)+N(u).A is a general differential operator, B is a boundary operator, f(r) is a known analytical function and Γ is the boundary of the domain Ω.
The operator A can be generally divided into two parts L and N,
where L is linear operator and N is nonlinear operator. Thus, (2.1) can be
rewritten as follows: (2.4)L(u)+N(u)−f(r)=0.By the homotopy technique
[20], we obtain a homotopy ν(r,p):Ω×[0,1]→ℜ satisfying (2.5)H(ν,p)=(1−p)[L(ν)−L(u0)]+p[A(ν)−f(r)]=0,p∈[0,1],r∈Ω,where p∈[0,1] is an embedding parameter and u0 is an initial approximation of (2.1), which
satisfies the boundary conditions. Clearly, from (2.5), we have(2.6)H⁡(ν,0)=L(ν)=L(u0)=0,H⁡(ν,1)=A(ν)=f(r)=0, the changing process of p from zero to unity is just that of ν(r,p) from u0(r) to u(r).
In topology this is called deformation and L(ν)−L(u0), A(ν)−f(r) are called homotopic.We considerν as follows: (2.7)ν=ν0+pν1+p2ν2+p3ν3+⋯=∑n=0∞pnνn.According to homotopy
perturbation method, an acceptable approximation solution of (2.4) can be
explained as a series of the power of p,(2.8)u=limp→1ν=ν0+ν1+ν2+ν3+⋯=∑n=0∞νn.Convergence  of the
series  (2.8) is given in [20, 21]. Besides, the same results have been  discussed in [22–24].


### 2.2. Boundary Element Method

Consider the Poisson equation(2.9)∇2u=b0,where b0 is a known function (for the electrostatic
problems, according to Gauss law is b0=−ρ/ε0).We can develop the boundary element method for
the solution of∇2u=b0 in a two-dimensional domain Ω.
We must first form an integral equation from the Poisson equation by using a
weighted integral equation and then use the Green-Gauss theorem: (2.10)∫Ω(∇2u−b0)w0dΩ=∫Γ∂u∂nw0dΓ−∫Ω∇u⋅∇w0dΩ.To derive the starting equation for the
boundary element method, we use the Green-Gauss theorem again on the second
integral. This gives(2.11)∫Ωu(∇2w0)dΩ−∫Ωbw0dΩ=∫Γu∂w0∂ndΓ−∫Γw0∂u∂ndΓ, and
thus, the boundary integral equations are obtained for a domain Ω with boundary Γ,
where u potential, ∂u/∂n,
is derivative with respect to normal of u and w0 is the known fundamental solution to Laplace’s equation applied at point   ξ(w0=−(1/2π)ℓnr). r=(ξ−x)2+(η−y)2 (singular at the point (ξ,η)∈Ω).Then,
using the property of the Dirac delta from (2.11), (2.12)∫Ωu(∇2w0)dΩ=−∫Ωuδ(ξ−x,η−y)dΩ=−u(ξ,η),(ξ,η)∈Ω,that
is, the domain integral has  been replaced by a point value [25].Thus, from Poisson equation the boundary
integral equation is obtained on the boundary:(2.13)c(ξ)u(ξ)+∫Γu∂w0∂ndΓ+∫Ωb0w0dΩ=∫Γw0∂u∂ndΓ,  where (2.14)c(ξ)={1inΩ,12onΓ.The  boundary integral
equation for the internal points is(2.15)u(ξ)=∫Γw0∂u∂ndΓ−∫Γu∂w0∂ndΓ−∫Ωb0w0dΩ.


#### 2.2.1. Cell Integration Approach

One of solution of domain integral
in the BEM is cell integration approach which is the problem region subdivided
to triangular elements as done in the finite element method (Figure1). Domain
integral is solved with respect to relationship between all cells and each
boundary node by Gauss quadrature method.Figure 1
Subdivided regions.The domain integral in
(2.15) for each boundary point i can be written as (2.16)di=∫Ωb0w0dΩ=∑e=1M[∑k=1Rωk(b0w0)k]Ωe,where the integral approximated by a
summation over different cells. In  (2.16), M is the total number of cells
describing the domain Ω, ωk is the Gauss integration weights and Ωe is the area of cell e. Besides, the function (b0w0) needs to be evaluated at integration point’s k on each cell by 1 to R, see [26].In this study, a Matlab
program has been developed to solve the Poisson equation with BEM by using cell
integration approach. This program calculates the potentials in the problem
domain.


## 2.1. Homotopy Perturbation Method

Homotopy perturbation method has
been suggested to solve boundary value problems in [17–19]. According to this
method, a homotopy with an imbedding parameter p∈[0,1] is constructed and the imbedding parameter is
considered as a “small parameter”. Here, homotopy perturbation method is used
to solve analytic solution of Poisson equation with given
boundary conditions.To illustrate this
method, we consider the following nonlinear differential equation:(2.1)A(u)−f(r)=0,r∈Ω, with boundary condition (2.2)B(u,∂u∂n)=0,r∈Γ, where A(u) is written as follows: (2.3)A(u)=L(u)+N(u).A is a general differential operator, B is a boundary operator, f(r) is a known analytical function and Γ is the boundary of the domain Ω.
The operator A can be generally divided into two parts L and N,
where L is linear operator and N is nonlinear operator. Thus, (2.1) can be
rewritten as follows: (2.4)L(u)+N(u)−f(r)=0.By the homotopy technique
[20], we obtain a homotopy ν(r,p):Ω×[0,1]→ℜ satisfying (2.5)H(ν,p)=(1−p)[L(ν)−L(u0)]+p[A(ν)−f(r)]=0,p∈[0,1],r∈Ω,where p∈[0,1] is an embedding parameter and u0 is an initial approximation of (2.1), which
satisfies the boundary conditions. Clearly, from (2.5), we have(2.6)H⁡(ν,0)=L(ν)=L(u0)=0,H⁡(ν,1)=A(ν)=f(r)=0, the changing process of p from zero to unity is just that of ν(r,p) from u0(r) to u(r).
In topology this is called deformation and L(ν)−L(u0), A(ν)−f(r) are called homotopic.We considerν as follows: (2.7)ν=ν0+pν1+p2ν2+p3ν3+⋯=∑n=0∞pnνn.According to homotopy
perturbation method, an acceptable approximation solution of (2.4) can be
explained as a series of the power of p,(2.8)u=limp→1ν=ν0+ν1+ν2+ν3+⋯=∑n=0∞νn.Convergence  of the
series  (2.8) is given in [20, 21]. Besides, the same results have been  discussed in [22–24].


## 2.2. Boundary Element Method

Consider the Poisson equation(2.9)∇2u=b0,where b0 is a known function (for the electrostatic
problems, according to Gauss law is b0=−ρ/ε0).We can develop the boundary element method for
the solution of∇2u=b0 in a two-dimensional domain Ω.
We must first form an integral equation from the Poisson equation by using a
weighted integral equation and then use the Green-Gauss theorem: (2.10)∫Ω(∇2u−b0)w0dΩ=∫Γ∂u∂nw0dΓ−∫Ω∇u⋅∇w0dΩ.To derive the starting equation for the
boundary element method, we use the Green-Gauss theorem again on the second
integral. This gives(2.11)∫Ωu(∇2w0)dΩ−∫Ωbw0dΩ=∫Γu∂w0∂ndΓ−∫Γw0∂u∂ndΓ, and
thus, the boundary integral equations are obtained for a domain Ω with boundary Γ,
where u potential, ∂u/∂n,
is derivative with respect to normal of u and w0 is the known fundamental solution to Laplace’s equation applied at point   ξ(w0=−(1/2π)ℓnr). r=(ξ−x)2+(η−y)2 (singular at the point (ξ,η)∈Ω).Then,
using the property of the Dirac delta from (2.11), (2.12)∫Ωu(∇2w0)dΩ=−∫Ωuδ(ξ−x,η−y)dΩ=−u(ξ,η),(ξ,η)∈Ω,that
is, the domain integral has  been replaced by a point value [25].Thus, from Poisson equation the boundary
integral equation is obtained on the boundary:(2.13)c(ξ)u(ξ)+∫Γu∂w0∂ndΓ+∫Ωb0w0dΩ=∫Γw0∂u∂ndΓ,  where (2.14)c(ξ)={1inΩ,12onΓ.The  boundary integral
equation for the internal points is(2.15)u(ξ)=∫Γw0∂u∂ndΓ−∫Γu∂w0∂ndΓ−∫Ωb0w0dΩ.


### 2.2.1. Cell Integration Approach

One of solution of domain integral
in the BEM is cell integration approach which is the problem region subdivided
to triangular elements as done in the finite element method (Figure1). Domain
integral is solved with respect to relationship between all cells and each
boundary node by Gauss quadrature method.Figure 1
Subdivided regions.The domain integral in
(2.15) for each boundary point i can be written as (2.16)di=∫Ωb0w0dΩ=∑e=1M[∑k=1Rωk(b0w0)k]Ωe,where the integral approximated by a
summation over different cells. In  (2.16), M is the total number of cells
describing the domain Ω, ωk is the Gauss integration weights and Ωe is the area of cell e. Besides, the function (b0w0) needs to be evaluated at integration point’s k on each cell by 1 to R, see [26].In this study, a Matlab
program has been developed to solve the Poisson equation with BEM by using cell
integration approach. This program calculates the potentials in the problem
domain.


## 2.2.1. Cell Integration Approach

One of solution of domain integral
in the BEM is cell integration approach which is the problem region subdivided
to triangular elements as done in the finite element method (Figure1). Domain
integral is solved with respect to relationship between all cells and each
boundary node by Gauss quadrature method.Figure 1
Subdivided regions.The domain integral in
(2.15) for each boundary point i can be written as (2.16)di=∫Ωb0w0dΩ=∑e=1M[∑k=1Rωk(b0w0)k]Ωe,where the integral approximated by a
summation over different cells. In  (2.16), M is the total number of cells
describing the domain Ω, ωk is the Gauss integration weights and Ωe is the area of cell e. Besides, the function (b0w0) needs to be evaluated at integration point’s k on each cell by 1 to R, see [26].In this study, a Matlab
program has been developed to solve the Poisson equation with BEM by using cell
integration approach. This program calculates the potentials in the problem
domain.


## 3. Implementation of Homotopy Perturbation Method to Poisson Equation

### 3.1. Case 1

First, let us investigate exact
solution in they-direction of Poisson equation (3.1)∂2u∂x2+∂2u∂y2+ρε0=0,with the initial condition(3.2)u(0,y)=ρa22ε0(1−32π3Cosh(πy/2a)Cosh(πb/2a)),and with the Dirichlet boundary
conditions (Figure 2); u=0, on x=∓1 and y=∓1 (coordinates; −1≤x≤1 and −1≤y≤1).Figure 2
The problem domain and boundary conditions.To investigate the
solution of (3.1), we can construct a homotopy as follows: (3.3)(1−p)[Y′′−y0′′]+p[Y′′+Y′′+ρε0]=0,where Y′′=∂2Y/∂y2,  Y′′=∂2Y/∂x2,
and  p∈[0,1],  with initial approximation Y0=u0=(ρa2/2ε0)(1−(32/π3)(Cosh(πy/2a)/Cosh(πb/2a))).
The solution of (3.1) can be expressed in a series in p:(3.4)Y=Y0+pY1+p2Y2+p3Y3+⋯.Then, substituting (3.4)
into (3.3), and arranging the coefficients of “p’’ powers, we have (3.5)Y0′′+pY1′′+p2Y2′′+p3Y3′′−y0′′+py0′′+pY0′′+p2Y1′′+p3Y2′′+p4Y3′′+pρε0+⋯=0,  where the Yi(x,t),  i=1,2,3,…, are
functions to be determined. We have to solve the following system which
includes four equations with four unknowns:(3.6)p0:Y0′′−y0′′=0,p1:Y1=−∬(∂2y0∂x2)dxdx−∬(∂2Y0∂y2)dxdx−∬(ρε0)dxdx,p2:Y2=−∬(∂2Y1∂y2)dxdx,p3:Y3=−∬(∂2Y2∂y2)dxdx.To found unknownsY1,Y2,Y3,…,
we must use the initial condition (3.2) for the above system, then we obtain (3.7)Y0=ρa22ε0(1−32π3Cosh(πy/2a)Cosh(πb/2a)),Y1=16ρa2π3ε0x22!(π2a)2Cosh(πy/2a)Cosh(πb/2a)−x22ρε0,Y2=16ρa2π3ε0x44!(π2a)4Cosh(πy/2a)Cosh(πb/2a),Y3=16ρa2π3ε0x66!(π2a)6Cosh(πy/2a)Cosh(πb/2a),⋮Thus, as considering (3.4)
with (3.7) and using Taylor
series, we obtain the analytical solutions as(3.8)u=ρa22ε0−16ρa2π3ε0Cosh(πy/2a)Cosh(πb/2a)[1−12!(πx2a)2+14!(πx2a)4−16!(πx2a)6+⋯]−x22ρε0.Therefore, the exact
solution ofu(x,y) in closed form is (3.9)u(x,y)=ρa22ε0(1−x2a2)−16ρa2π3ε0Cosh(πy/2a)Cosh(πb/2a)Cosπx2a.⁡


### 3.2. Case 2

Let us investigate exact solution in
thex-direction of Poisson equation (3.10)∂2u∂x2+∂2u∂y2+ρε0=0,with the initial condition (3.11)u(x,0)=ρb22ε0(1−32π3Cosh(πx/2b)Cosh(πa/2b)),and with the Dirichlet boundary
conditions. To investigate the solution of (3.10), we can construct a homotopy
as follows: (3.12)(1−p)[Y′′−y0′′]+p[Y′′+Y′′+ρε0]=0.After that, substituting
(3.4) into (3.12), and arranging the coefficients of “p’’  powers, we have to solve the following
system including four equations with four unknowns: (3.13)p0:Y0′′−y0′′=0,p1:Y1=−∬(∂2y0∂x2)dydy−∬(∂2Y0∂y2)dydy−∬(ρε0)dydy,p2:Y2=−∬(∂2Y1∂x2)dydy,p3:Y3=−∬(∂2Y2∂x2)dydy.As found unknownsY1,Y2,Y3,…,
we have exact solution of (3.10): (3.14)u(x,y)=ρb22ε0−16ρb2π3ε0Cosh(πx/2b)Cosh(πa/2b)[1−12!(πy2b)2+14!(πy2b)4−16!(πy2b)6+⋯]−y22ρε0.Therefore, the exact
solution ofu(x,y) in closed form
is (3.15)u(x,y)=ρb22ε0(1−y2b2)−16ρb2π3ε0Cosh(πx/2b)Cosh(πa/2b)Cosπy2b.The equipotential lines
obtained using exact solution and numerical results have been shown in Figures3–5 (for ρ/ε0=1). These results then are compared in Tables 1 and  2  (for ρ/ε0=1 and ρ/ε0=50).Table 1
The comparison of potentials
(Volts) for   ρ/ε0=1 at y=0(a=1,b=1).
xHPMHPMBEM|HPM(y)-BEM||HPM(x)-BEM|(y-direction)(x-direction)1.00.0−0.016024550.00.00.016024550.90.0628285040.0522438130.0631779290.0003494250.0109341160.80.1164491770.1094415020.1163565080.0000926690.0069150060.70.1616346830.1569827140.1612746020.0003600810.0042918880.60.1991191520.1960428980.1987401170.0003790350.0026972190.50.2295801090.2275878070.2294200390.0001600700.0018322320.40.2536217870.2523973830.2536042460.0000136240.0012068630.30.2717602480.2710850370.2718099010.0000496530.0007248640.20.2844106800.2841128170.2847971090.0003864290.0006842920.10.2918771700.2918028330.2921173950.0002402250.0003145620.00.2943452180.2943452180.2945696730.0002244550.000224455Table 2
The comparison of
potentials (Volts) for ρ/ε0=50 at x=0(a=1,b=1).
yHPMHPMBEM|HPM(y)-BEM||HPM(x)-BEM|(y-direction)(x-direction)1.0−0.801227540.00.00.801227540.00.92.6121906813.1414252173.1587224600.5465317790.0172972430.85.4720751015.8224588825.8151003540.3430252530.0073585280.77.8491357438.0817341568.0835583160.2344225730.0018241600.69.8021449249.9559576309.9301000410.1279551170.0258575890.511.3793903811.4790054811.465986980.086596600.013018500.412.6198691712.6810893512.676980490.057111320.004108860.313.5542518713.5880124113.602517710.048265840.014505300.214.2056408914.2205340114.225550530.019909640.005016520.114.5901416814.5938585214.625672290.035530610.031813770.014.7172609414.7172609414.728483670.011222730.01122273Figure 3
Equipotential lines forρ/ε0=1  (using HPM, x-direction).Figure 4
Equipotential lines forρ/ε0=1   (using HPM, y-direction).Figure 5
Equipotential lines forρ/ε0=1 (using BEM).Tables1 and 2 compare the exact HPM and approximate
BEM of the Poission equation for ρ/ε0=1 and ρ/ε0=50,
respectively. Tables 1 and 2 show that the differences between HPM and BEM for
both directions x and y. The differences clearly show that the
results of the approximate BEM introduced in this study are acceptable.


## 3.1. Case 1

First, let us investigate exact
solution in they-direction of Poisson equation (3.1)∂2u∂x2+∂2u∂y2+ρε0=0,with the initial condition(3.2)u(0,y)=ρa22ε0(1−32π3Cosh(πy/2a)Cosh(πb/2a)),and with the Dirichlet boundary
conditions (Figure 2); u=0, on x=∓1 and y=∓1 (coordinates; −1≤x≤1 and −1≤y≤1).Figure 2
The problem domain and boundary conditions.To investigate the
solution of (3.1), we can construct a homotopy as follows: (3.3)(1−p)[Y′′−y0′′]+p[Y′′+Y′′+ρε0]=0,where Y′′=∂2Y/∂y2,  Y′′=∂2Y/∂x2,
and  p∈[0,1],  with initial approximation Y0=u0=(ρa2/2ε0)(1−(32/π3)(Cosh(πy/2a)/Cosh(πb/2a))).
The solution of (3.1) can be expressed in a series in p:(3.4)Y=Y0+pY1+p2Y2+p3Y3+⋯.Then, substituting (3.4)
into (3.3), and arranging the coefficients of “p’’ powers, we have (3.5)Y0′′+pY1′′+p2Y2′′+p3Y3′′−y0′′+py0′′+pY0′′+p2Y1′′+p3Y2′′+p4Y3′′+pρε0+⋯=0,  where the Yi(x,t),  i=1,2,3,…, are
functions to be determined. We have to solve the following system which
includes four equations with four unknowns:(3.6)p0:Y0′′−y0′′=0,p1:Y1=−∬(∂2y0∂x2)dxdx−∬(∂2Y0∂y2)dxdx−∬(ρε0)dxdx,p2:Y2=−∬(∂2Y1∂y2)dxdx,p3:Y3=−∬(∂2Y2∂y2)dxdx.To found unknownsY1,Y2,Y3,…,
we must use the initial condition (3.2) for the above system, then we obtain (3.7)Y0=ρa22ε0(1−32π3Cosh(πy/2a)Cosh(πb/2a)),Y1=16ρa2π3ε0x22!(π2a)2Cosh(πy/2a)Cosh(πb/2a)−x22ρε0,Y2=16ρa2π3ε0x44!(π2a)4Cosh(πy/2a)Cosh(πb/2a),Y3=16ρa2π3ε0x66!(π2a)6Cosh(πy/2a)Cosh(πb/2a),⋮Thus, as considering (3.4)
with (3.7) and using Taylor
series, we obtain the analytical solutions as(3.8)u=ρa22ε0−16ρa2π3ε0Cosh(πy/2a)Cosh(πb/2a)[1−12!(πx2a)2+14!(πx2a)4−16!(πx2a)6+⋯]−x22ρε0.Therefore, the exact
solution ofu(x,y) in closed form is (3.9)u(x,y)=ρa22ε0(1−x2a2)−16ρa2π3ε0Cosh(πy/2a)Cosh(πb/2a)Cosπx2a.⁡


## 3.2. Case 2

Let us investigate exact solution in
thex-direction of Poisson equation (3.10)∂2u∂x2+∂2u∂y2+ρε0=0,with the initial condition (3.11)u(x,0)=ρb22ε0(1−32π3Cosh(πx/2b)Cosh(πa/2b)),and with the Dirichlet boundary
conditions. To investigate the solution of (3.10), we can construct a homotopy
as follows: (3.12)(1−p)[Y′′−y0′′]+p[Y′′+Y′′+ρε0]=0.After that, substituting
(3.4) into (3.12), and arranging the coefficients of “p’’  powers, we have to solve the following
system including four equations with four unknowns: (3.13)p0:Y0′′−y0′′=0,p1:Y1=−∬(∂2y0∂x2)dydy−∬(∂2Y0∂y2)dydy−∬(ρε0)dydy,p2:Y2=−∬(∂2Y1∂x2)dydy,p3:Y3=−∬(∂2Y2∂x2)dydy.As found unknownsY1,Y2,Y3,…,
we have exact solution of (3.10): (3.14)u(x,y)=ρb22ε0−16ρb2π3ε0Cosh(πx/2b)Cosh(πa/2b)[1−12!(πy2b)2+14!(πy2b)4−16!(πy2b)6+⋯]−y22ρε0.Therefore, the exact
solution ofu(x,y) in closed form
is (3.15)u(x,y)=ρb22ε0(1−y2b2)−16ρb2π3ε0Cosh(πx/2b)Cosh(πa/2b)Cosπy2b.The equipotential lines
obtained using exact solution and numerical results have been shown in Figures3–5 (for ρ/ε0=1). These results then are compared in Tables 1 and  2  (for ρ/ε0=1 and ρ/ε0=50).Table 1
The comparison of potentials
(Volts) for   ρ/ε0=1 at y=0(a=1,b=1).
xHPMHPMBEM|HPM(y)-BEM||HPM(x)-BEM|(y-direction)(x-direction)1.00.0−0.016024550.00.00.016024550.90.0628285040.0522438130.0631779290.0003494250.0109341160.80.1164491770.1094415020.1163565080.0000926690.0069150060.70.1616346830.1569827140.1612746020.0003600810.0042918880.60.1991191520.1960428980.1987401170.0003790350.0026972190.50.2295801090.2275878070.2294200390.0001600700.0018322320.40.2536217870.2523973830.2536042460.0000136240.0012068630.30.2717602480.2710850370.2718099010.0000496530.0007248640.20.2844106800.2841128170.2847971090.0003864290.0006842920.10.2918771700.2918028330.2921173950.0002402250.0003145620.00.2943452180.2943452180.2945696730.0002244550.000224455Table 2
The comparison of
potentials (Volts) for ρ/ε0=50 at x=0(a=1,b=1).
yHPMHPMBEM|HPM(y)-BEM||HPM(x)-BEM|(y-direction)(x-direction)1.0−0.801227540.00.00.801227540.00.92.6121906813.1414252173.1587224600.5465317790.0172972430.85.4720751015.8224588825.8151003540.3430252530.0073585280.77.8491357438.0817341568.0835583160.2344225730.0018241600.69.8021449249.9559576309.9301000410.1279551170.0258575890.511.3793903811.4790054811.465986980.086596600.013018500.412.6198691712.6810893512.676980490.057111320.004108860.313.5542518713.5880124113.602517710.048265840.014505300.214.2056408914.2205340114.225550530.019909640.005016520.114.5901416814.5938585214.625672290.035530610.031813770.014.7172609414.7172609414.728483670.011222730.01122273Figure 3
Equipotential lines forρ/ε0=1  (using HPM, x-direction).Figure 4
Equipotential lines forρ/ε0=1   (using HPM, y-direction).Figure 5
Equipotential lines forρ/ε0=1 (using BEM).Tables1 and 2 compare the exact HPM and approximate
BEM of the Poission equation for ρ/ε0=1 and ρ/ε0=50,
respectively. Tables 1 and 2 show that the differences between HPM and BEM for
both directions x and y. The differences clearly show that the
results of the approximate BEM introduced in this study are acceptable.


## 4. Conclusions

In this paper,  we proposed homotopy
perturbation method to find exact solution in thex-and y-directions of Poisson equation with
appropriate boundary and initial conditions. The numerical results of this
electrostatic potential problem have been calculated at the same boundary
conditions by BEM. These results are compared with those of HPM in Tables 1 and
2. The obtained numerical results by using
BEM are in agreement with the exact solutions obtained by HPM. This adjustment
is clearly seen in Figures 3, 4, and 5. It is shown that these methods are acceptable and
very efficient for solving electrostatic field problems with charge density.



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*Source: 578723-2008-04-14.xml*