Karthik M.tech. Project Report

7/31/2019 Karthik M.tech. Project Report

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3-D ADI Poisson Solver

A Project Report

Submitted in partial fulfilment of the

requirements for the Degree of

Master of Technology

in

Computational Science

by

V.Karthik

Supercomputer Education and Research Center

Indian Institute of Science

BANGALORE 560 012

June 7, 2011


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Acknowledgements

I would like to thank my advisor Dr.Atanu Mohanty for his valuable guidance and support

through out the project. I would like to mention that his passion for teaching, interest

towards students encourages us to perceive our studies and research with much more

interest.

I would like to thank all the people of SERC for providing this platform of study to

us.

I would like to thank my friends Govind, Sufal, Avinash Bhat, Mritunjay, Avinash

Dash for all the nice time we had together studying and having fun. Special thanks tomy friends Appala Naidu and Brahmanandam.

I would like to thank all my family members for being with me all the time and giving

continuous support and encouragment.

Finally, I would like to thank god for his blessings through out.

i


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Abstract

Alternating Direction Implicit(ADI) method is a Finite Difference Method(FDM) to solve

partial differential equations. We employ ADI method to solve Poissons equation for

dirichlet boundary conditions.The discretised equation is modified near the boundaries to

incorporate arbitrary domain shapes. The ADI Poisson solver developed, works for both

two or three dimensions. ADI method performs much better than traditional iterative

methods because of usage of efficient tri-diagonal solvers

ii


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Contents

Acknowledgements i

Abstract ii

1 Introduction 11.1 Poissons Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Applications of Poissons equation . . . . . . . . . . . . . . . . . . . . . . 21.3 Properties of Poissons equation . . . . . . . . . . . . . . . . . . . . . . . 5

2 Analytical Solution to Poissons Equation 72.1 Solution strategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Solution of Laplace equation on square domain: . . . . . . . . . . . . . . 10

2.2.1 Solution of Poisson equation on square domain: . . . . . . . . . . 132.2.2 Solution of Poisson equatin on square domain with given boundary

conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.2.3 Solution of Laplace equation on an annular ring: . . . . . . . . . . 152.2.4 Solution of Laplace equation on circular domain: . . . . . . . . . . 162.2.5 Solution of Poissons equation on circular domain: . . . . . . . . . 17

3 Solution of Poissons equation using Finite Difference Methods 203.1 Finite Difference Methods . . . . . . . . . . . . . . . . . . . . . . . . . . 203.2 Mesh/Grid Generation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.3 Discretised Poissons equation . . . . . . . . . . . . . . . . . . . . . . . . 213.4 Matrix Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.5 Iterative Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.5.1 Jacobi Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.5.2 Gauss-Siedel Method . . . . . . . . . . . . . . . . . . . . . . . . . 253.5.3 Successive-over Relaxation Method . . . . . . . . . . . . . . . . . 25

3.6 Convergence of Iterative Methods . . . . . . . . . . . . . . . . . . . . . . 25

4 Alternating Direction Implicit(ADI) Method 274.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.2 Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.3 Matrix Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

iii


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CONTENTS iv

4.4 ADI Method Implementation . . . . . . . . . . . . . . . . . . . . . . . . 31

4.5 Solution of Laplace equation on Square domain . . . . . . . . . . . . . . 324.5.1 Problem Statement . . . . . . . . . . . . . . . . . . . . . . . . . . 324.5.2 Numerical Solution . . . . . . . . . . . . . . . . . . . . . . . . . . 324.5.3 Error Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324.5.4 Choice of relaxation parameter . . . . . . . . . . . . . . . . . . . 34

4.6 Solution of Poisson Equation on square domain . . . . . . . . . . . . . . 364.6.1 Problem Statement . . . . . . . . . . . . . . . . . . . . . . . . . . 364.6.2 Governing Equation and Numerical Solution . . . . . . . . . . . . 364.6.3 Error Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

5 ADI Method on various domains 395.1 ADI Solver Implementation Steps . . . . . . . . . . . . . . . . . . . . . . 395.2 FDM formula for irregular boundary . . . . . . . . . . . . . . . . . . . . 395.3 Solution of Laplace equation on a ring domain . . . . . . . . . . . . . . . 405.4 Solution of Poisson equation on a circular disk domain . . . . . . . . . . 425.5 Solution of Laplace equation on a hyperbola domain . . . . . . . . . . . . 43

6 ADI Method on 3-D Domains 486.1 3-D Poisson Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486.2 3-D ADI method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496.3 Solution of Laplace Equation on the Cube . . . . . . . . . . . . . . . . . 49

7 Conclusions and Future Work 567.1 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 567.2 Future Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

References 57


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List of Tables

4.1 Comparision of iterations taken by various iterative methods . . . . . . . 35

4.2 Cyclic ADI parameters performance comparision . . . . . . . . . . . . . . 36

v


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List of Figures

2.1 Superposition principle . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2 Superposition principle . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.3 Solution of Laplace Equation on Square domain . . . . . . . . . . . . . . 122.4 Analytical Solution of Poisson Equation on Square domain . . . . . . . . 14

3.1 Domain Grid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4.1 Square Domain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324.2 Laplace equation Solution using ADI method on Square domain . . . . . 334.3 Error plot for ADI method in Square domain . . . . . . . . . . . . . . . . 334.4 Error plot for ADI Method in the interior of Square domain . . . . . . . 344.5 Poisson function for square plate . . . . . . . . . . . . . . . . . . . . . . 37

4.6 Numerical solution of Poisson equation on square domain . . . . . . . . . 374.7 Error plot for Poisson equation on square domain . . . . . . . . . . . . . 384.8 Error plot in the interior of the domain for Poisson equation on square

domain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

5.1 A curved boundary with mesh . . . . . . . . . . . . . . . . . . . . . . . 405.2 Mesh point O with neighbours NEWS . . . . . . . . . . . . . . . . . . . 405.3 Ring Domain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415.4 Laplace equation Solution using ADI method on ring domain . . . . . . . 425.5 Error plots for solution of Laplace equation on ring domain . . . . . . . . 435.6 Poisson equation Solution using ADI method on circular disk . . . . . . . 44

5.7 Error plots for solution of Poisson equation on circular disk . . . . . . . . 455.8 Hyperbola Domain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455.9 Laplace equation Solution using ADI method on hyperbola domain . . . 465.10 Error plots for solution of Laplace equation on hyperbola domain . . . . 47

6.1 A point in 3D domain . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496.2 Solution of Laplace Equation on Cube domain using ADI method . . . . 546.3 Error plot on Cube domain . . . . . . . . . . . . . . . . . . . . . . . . . 55

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Chapter 1

Introduction

1.1 Poissons Equation

Poissons Equation is an elliptic linear inhomogeneous partial differential equation of the

second order. It is given by

2

u = f (1.1)where 2(also denoted with ) stands for the Laplacian operator, and f is called as

load/source function. The solution of the equation u, is the unknown scalar potential

function.

In Cartesian coordinate coordinate system, the Poisson equation is given by

2u

x2+

2u

y2+

2u

z2= f(x , y, z ) (1.2)

where x , y, z are the independent space dimensions.

When the Poisson equation is satisfied by a scalar potential in a given domain . We

can find the scalar potential inside the domain by solving the Poisson equation with the

help of boundary conditions prescribed at the boundary .

1


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Chapter 1. Introduction 2

1.2 Applications of Poissons equation

Poisson equation arises in a variety of branches mathematical and physical contexts rang-

ing through elasticity and solid mechanics, fluid mechanics, electromagnetism, potential

theory, heat conduction, geometry, probability, number theory, and many more [4, 5].

Poisson equation is mainly used to describe the equilibrium physical phenomena under

the presence of source particles or load function. It is used to describe the steady state

conditions of a given system as it involves no time variable.

Electrostatics:

In electrostatics, it is often necessary to find the electric field ( E) for a given charge

distribution in a region[3].

Directly solving for E requires solving two equations.

E = 0 E =

40

As E is a vector, it involves solving for all the vector components which is cumber-

some. This process can be simplified with help of Poissons equation.

We can find the electric potential in the region by solving Poissons equation.

2V(x,y,z) = 140

(x,y,z)

where (x,y,z) is a charge distribution in the given volume. Then electric field is given

simply by the gradient of the potential.

The advantages of second method is that it involves only scalars and requires to solve

only one partial differential equation(Poissons equation) involving voltage instead of two

involving gradient and curl.


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Derivation of Poissons equation in Electrostatics

In a given volume, Gausss law in differential form states that the divergence of the

electric field in a region is proportional to charge enclosed in the region.

E = 140

(1.3)

where stands for charge density. In the absence of changing magnetic field B, Faradays

law of induction gives

E = Bt

= 0

Since curl of electric field is zero, the electric field can be given by the gradient of the

electric potential.

E = V (1.4)

Combining the equations 1.2 and 1.2, we get the Poisson equation for electric potential

2V = 140

(1.5)

Stationary Heat Equation

Poissons equation also governs heat flow problems that are steady, that is, time inde-

pendent. The thermal equilibrium of bodies(eg. flat plates) can be described by the

Poissons equation in the presence of heat sources:

2T(x, y) = h(x, y)

where T(x,y) represents the temperature, h(x,y) represents the external heat source and

stands for the thermal conductivity.


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Steady state Wave equation

The steady state displacement of a membrane(like drum skin) follows Poissons equa-

tion. The inhomogenuous term represents an external forcing over the surface of the

membrane. The governing equation is given by

2u(x, y) = P(x, y)T

P(x, y) represents pressure, T represent Tension and u(x, y) represents the displacement

of the membrane.

Fluid Dynamics

The velocity potential of a steady incompressible, irrotational planar flow of fluid follows

Laplace equation. The gradient of the velocity potential gives the velocity.

Elasticity

Poissons equation arises in the study of torsion of prismatic bars. The stress distribution

over a non-circular cross-section of a twisted bar can be determined by finding the stress

function (x, y) which satisfies the two-dimensional Poisson equation.

2 = 2G

G is shear modulus. is the angle of twist per unit length

Gravitational Potential

To study the forces acting in a gravitational field filled with masses, Poisson equation is

also useful.The gravitational potential is given by the equation

2 = 4G


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stands for Gravitational potential, G stands for universal gravitational constant and

stands for mass density. Besides above,Poissons equation has applications in image

processing, number theory etc.

1.3 Properties of Poissons equation

Poissons equation and its homogeneous counter-part Laplace equation has several in-

teresting properties which help in finding their solutions or make their study interesting

and useful in many areas which are worth mentioning here.

1. Harmonic functions: The Solutions of Laplace equations are called as harmonic

functions. They have continuous second-order partial-derivaties.

2. Mean value property: The value of the harmonic function at any interior point is

equal to the mean value of the function on any circle in the domain with its center

at that point.

3. Maximum modulus property: Solution of Laplace equation cannot have maximum

or minimum in the interior of the domain. Consequently, the maximum and the

minimum are taken on the boundary of the domain

4. Analytic functions: The real and imaginery part of analytic functions are harmonic.

They remain harmonic under conformal mapping so that conformal mapping be-

comes a powerful tool in solving boundary value problems for Laplace equation by

mapping complex domains on to simpler domains like circle.

The complex representation of potential has several interpretations. For example,

in electrostatics, the equal real parts correspond to equi-potential lines and the

equal imaginery parts correspond to lines of electric force.In heat problems, they

are isotherms and lines of heat flow.

5. Linearity : Laplace equation is second order linear PDE. So the principle of super-

position hold good. So new solutions can be obtained by the summation of two


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existing solutions. This is very useful in obtaining solution of Laplace equation like

complex problems can be constructed by summing simple solutions.

6. Uniqueness: We can derive from above properties that Laplace equation and Pois-

sons equation has unique stable solution for Dirichlet or Neumann boundary con-

ditions for closed surfaces. So that if a solution is found to Laplace equation by any

method then any other solution is just different represention of the same solution.


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Chapter 2

Analytical Solution to PoissonsEquation

In this chapter we introduce various traditional analytical methods used to solve partial

differential equations. Finding analytical solutions helps in verifying the correctness of

solution obtained from the numerical method.

2.1 Solution strategies

Some general techniques used to solve Partial Differential Equations analytically are as

follows

1. Separation of variables: This is the most common method used to solve partial

differential equations on simple domains. The idea is to express the solution as the

product of the functions of the variables involved. For example, in two variables,

Solution u(x,y) can be written as

u(x, y) = F(x)G(y)

This separates out the partial differential equation in to two or three ordinary dif-

ferential equations which are related by a common constant(usually eigen values).

7


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Chapter 2. Analytical Solution to Poissons Equation 8

These ordinary differential equations are solved using the boundary conditions that

are already transformed in to simpler conditions involving only one of the variable.

Thus separation of variables is a very useful technique when domain boundary

is symmetric to the variables involved and boundary conditions specified involve

those variables only.

2. Conformal mapping: These are useful to reduce a complicated domain in to a sim-

pler domain like disk. Complex analysis is useful in this method. Components

of analytic functions satisfy laplace equation and they remain analytic in confor-

mal maps. So that Laplace equation is solved on the simpler domain and inverse

mapped to get the solution on the complicated domain.

3. Greens function : Greens function characterise the response of the system for a

unit impulse. In simpler terms, it gives the solution of the system for a unit load

at a particular point which when integrated over the domain with required scaling

gives the solution for the given load. Solution of Laplace equation on circular disk

can be derived in this method with the help of Poissons Integral formula and also

inhomogeneous equation like Poisson equation can be solved using this method.

But this method involves integrals which may be tough to evaluate.

4. Integral transforms Integral transforms like Fourier transforms are useful when the

domain considered is infinite domain.

5. Superposition principle Laplace equation is linear so that combination of solutionsis also a solution. We can find solution for complicated boundary condition by

splitting it in to smaller problems.

Solution of Laplace equation 2U = 0 on square domain of length L with twoboundaries x = 0 , y = L having potential one can be found by U = U1 + U2 where

2U1 = 0 with U1(x, L) = 0 and 2U2 = 0 with U2(0, y) = 1. This is described infigure (2.1).

Solution of Poisson equation for given boundary conditions can be found by adding


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Figure 2.1: Superposition principle

solution of poisson equation for homogeneous zero boundary condition and partic-

ular solution of laplace equation for given boundary conditions.This is described

in figure(2.2)

Figure 2.2: Superposition principle

Mathematically solution of 2U = f(x, y) for U = g(x, y) on boundary isgiven by U = U1 + U2 where 2U1 = f(x, y) for U1(x, y) = 0 on boundary and

2U2 = 0 with U2(x, y) = g(x, y) on .


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6. Eigen function expansions: Solution of inhomogenuous differential equations like

Poissons equation can be found by expressing the source/load function as an ex-

pansion of eigen functions of the linear operator and finding the coefficients/eigen

values. Fourier series is used in rectangular domain and Fourier-Bessel series is

used in circular domains.

2.2 Solution of Laplace equation on square domain:

Problem Statement

To solve Laplace equation

uxx + uyy = 0 (2.1)

in the interior of a square domain of length L with the Dirichlet boundary conditions.

u(0, y) = 0 for 0 < y L (2.2)

u(L, y) = 0 for 0 < y L (2.3)u(x, 0) = 0 for 0 x L (2.4)u(x, L) = V for0 < x < L (2.5)

i.e. three sides are kept at zero boundary conditions and another side(y = L) is main-

tained at a constant potential V.

Separation of Variables

By the method of separation of variables, we split the partial differential equation (2.1)

in to two ordinary differential equations and solve for the potential u in the interior of

the domain. The first step is to assume the solution as the product of functions which

depend on only one variable i.e.

u(x, y) = F(x) G(y) (2.6)


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Substituting in equation (2.1) we get,

G(y)F(x) + F(x)G(y) = 0

Dividing by U(x, y) we get,F(x)

F(x)+

G(y)

G(y)= 0

We write the above equation as

F

(x)F(x)

= G

(y)G(y)

= 2

Since the LHS in the above equation involes only x variable and RHS is only a function

of y variable, we conclude that this can happen only if 2 is constant.

This forms the two ordinary differential equations

F(x) + 2F(x) = 0 (2.7)

G(y) 2G(y) = 0 (2.8)

The solution of (2.7) is given by

F(x) = A sin(x) + B cos(x) (2.9)

Applying boundary conditions (2.2), we get B=0 in the above equation.

Applying the boundary condition (2.3), we get =nL where n = 1, 2 . . .

The solution of (2.8) is given by

F(x) = Cey + Dey

= C1 sinh(y) + D1 cosh(y)


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Using boundary condition (2.4), we get D1 = 0 in the above equation. Now using

equation (2.6), the solution of Laplace equation can be given as follows:

u(x, y) =n=1

cn sinh

ny

L

sin

nx

L

(2.10)

Fourier Series

In order to determine the solution completely, we need to determine the coefficients cn.

It can be determined by the fourth boundary condition given.Assume on y = L, the

potential is f(x). We can observe that equation (2.10) is in the form of fourier sine series.

Then the coefficient cn is given by

cn =2

L sinh(n)

L0

f(x)sin

nx

L

dx (2.11)

For the given problem f(x) = V which gives the solution,

u(x, y) =

n=1 ,n :odd

4 V

n sinh(n) sinhny

L

sinnx

L

(2.12)

0

0.5

1

0

0.5

10

0.5

1

xy

u(x,y

)

Figure 2.3: Solution of Laplace Equation on Square domain.Boundary Conditions: u(x,y) = 0 on three sides and u(x,1) = 1

No. of terms in fourier series used = 200


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2.2.1 Solution of Poisson equation on square domain:

Problem Statement

To solve Poisson equation

uxx + uyy = 140

(x, y) (2.13)

in a square domain of length L with homogeneous zero boundary conditions and constant

charge density 0 in the interior of the domain. i.e.

(x, y) = 0 for 0 < x < L, 0 < y < L (2.14)

Eigenfunctions

The solution to the Poissons equation 2u = q(x, y) for homogeneous dirichlet bound-ary conditions can be given in terms of eigenfunctions of the Laplacian operator: sin

nxL

sinmyL

i.e. Let the solution vector u be:

u =

m,n=1

Km,n sin

nxL

sin

myL

Then,

2u = uxx + uyy=

(m2 + n2)2

L2Km,n sin

nx

L

sin

my

L

=(m2 + n2)2

L2 u

So in order to find the solution of Poisson equation, we express charge density in

terms of eigen functions sinnxL

sinmyL

with the help of two dimensional Fourier

sine series

(x, y) =

m,n=1

Qn,m sin

nx

L

sin

my

L

(2.15)


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where Qn,m is given by the fourier series coefficient:

Qn,m =4

L2

L0

L0

q(x, y)sin

nx

L

sin

my

L

dxdy (2.16)

Then the solution to Poissons equation is given by,

u(x, y) =

m,n=1

Qn,mL2

(m2 + n2)2sin

nx

L

sin

my

L

(2.17)

For the given problem, fourier series coefficients of unit charge density is given by,

Qn,m =4

L2

L0

L0

040

sin

nx

L

sin

my

L

dxdy

=16

nm20

40n, m: odd otherwise zero

Thus from the equation (2.17), the solution for the given problem is

u(x, y) =

m,n=1,odd

0

40

16L2

m n (m2 + n2)4 sinnx

L

sinmy

L

(2.18)

02

46

0

2

4

60

0.5

1

1.5

2

x 1010

xy

Voltage

Figure 2.4: Analytical Solution of Poisson Equation on Square domainwith zero boundary value and f(x,y) = 9e9.

No. of terms in fourier series used = 100


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2.2.2 Solution of Poisson equatin on square domain with given

boundary conditions

2.2.3 Solution of Laplace equation on an annular ring:

To solve Laplace equation

2u = 0

on a ring domain with inner radius R1 and outer radius R2 with boundary conditions.

f(r, ) =

1 r = R1

0 r = R2(2.19)

We express Laplace equation in polar coordinates so that boundary becomes symmetric

to the variables involved. Laplace equation in polar coordinates is given by:

1

r

r

r

ru

+

1

r22

2u = 0

General solution

Using separation of variables in polar coordinates, let us say,

u(r, ) = R(r) ()

gives us two ordinary differential equations,

r2R + rR R = 0 (2.20) + = 0 (2.21)

Solution of equation(2.21) is given in terms of trigonmetric functions because it needs

to be periodic ( + 2) = (). Equation (2.20) is in the form of second order Euler

ordinary differential equation. The solutions can be given in terms of eigen functions as

follows:


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n = 0, () = 1, R(r) = 1 and R(r) = ln r

n > 0, () = sin(n), cos(n) R(r) = rn and R(r) = rn

Thus general solution of Laplace Equation on ring is:

u(r, ) = a0 +n

anrn cos(n) +n

anrn cos(n)

+ b0 ln r +n

bnrn sin(n) +n

bnrn sin(n) (2.22)

Since the given boundary conditions does not depend on the parameter we can say

solution is in the form

u(r, ) = c1 + c2 ln r

Solving with the boundary conditions (2.19), we get

u(r, ) =ln R2 ln rln R2 ln R1 (2.23)

2.2.4 Solution of Laplace equation on circular domain:

Problem Statement

To solve Laplace equation2u

r2+

1

r

u

r+

1

r22u

2= 0 (2.24)

on a circular domain of radius R with boundary conditions u(R, ) = f(). For a circular

disk, the terms containing ln r , rn in equation (2.22) have singularities as r 0 sothat solution on disk is given by,

u(r, ) = a0 +n=1

(an rn cos n + bn r

n sin n) (2.25)

The coefficients can be determined by using the boundary conditions and finding fourier

coefficients.

Solution on circular disk can also be given in terms of Greens function which is given


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by Poissons kernel, P(x, s). The influence due to a small element s on the boundary at

a point x in the interior of the boundary is given by

P(x, s) =R2 x2x s2

where the denominator measures squared distance from x to s and numerator roughly

measures distance from x to boundary. In polar coordinates, s can be given by single

angular coordinate [, ) If x has coordinates (x, y), then

P(x, s) =R2 (x2 + y2)

(x Rcos())2 + (y Rsin())2

Converting to polar coordinates x is represented by (r, ) and dividing by R2 we get,

P(x, s) =1 rR

21 2rR

cos( ) +rR

2

So the solution on circular disk is given by Poissons Integral formula:

u(r, ) =1

2

f()1 rR

21 2rR

cos( ) +rR

2 d (2.26)

The solution can be viewed as a sort of convolution of boundary function with Poissons

kernel in circular domain.

2.2.5 Solution of Poissons equation on circular domain:

Problem Statement

To solve Poisson equation 2u = q(r, ) i.e.

2u

r2+

1

r

u

r+

1

r22u

2= q(r, ) (2.27)


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on a circular domain of radius R with homogeneous zero boundary conditions and unit

source function inside the domain i.e. q(r, ) = 1 for r < R

Bessel Equation derivation

Let f(r, ) be an eigen function of the Laplacian operator then the eigen vector equation

is given by

f = 2f (2.28)

for some eigen value 2.

Applying separation of variables, f(r) = R(r)() in the polar coordinates form we get,

R(r)() + 1rR(r)() + 1

r2R(r)() = 2R(r)()

Dividing by R(r)()R(r)R(r) +

R(r)rR(r) +

()

r2()= 2

Multiplying with r2 and rearranging the terms we get,

r2R(r)R(r) + r

R(r)R(r) +

2r2 = ()

()

To satisfy the periodic boundary condtions on , both LHS and RHS should be equal to

a negative constant, so that equations become,

()m2

() = 0 (2.29)

r2R(r) + rR(r) + (2r2 m2)R(r) = 0 (2.30)

Equation (2.30) is known as Bessels equation. The solution of (2.30) is given in terms

of Bessels function of first kind Jn(r) and Bessel functions of second kind Yn(r) . The

solution of (2.29) is given by cos(n) and sin(n).

Yn(x) is unbounded at the centre, so that Yn(x) is not a solution for the circular

disk. In order to satisfy homogeneous zero boundary conditions we need Jn(R) = 0. So


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we consider a sequence of roots of Bessel function = knm which are irrational. Then

solution to (2.28) is given by Fourier-Bessel expansion

f(r, ) =n=0

m=1

AnmJn

knmr

R

cos(n) +

n=1

m=1

BnmJn

knmr

R

sin(n)

where Anm and Bnm are real-valued coefficients.

To find the solution of Poissons equation given in equation (2.27), we find Fourier-

Bessel expansion of q(r, ).

q(r, ) =n=0

m=1

AnmJn

knmr

R

cos(n) +

n=1

m=1

BnmJn

knmr

R

sin(n)

The coefficients are given by the formula

Anm =2

R2J2n+1(knm)

R0

q(r, )Jn

knmr

R

cos(n) rdrd

Bnm =

2

R2J2n+1(knm)R0 q(r, )Jnknmr

R

sin(n) rdrd

Then the solution of Poissons equation is given by,

u(r, ) = n=0

m=1

R2Anmk2nm

Jn

knmr

R

cos(n)

n=1

m=1

R2Bnmk2nm

Jn

knmr

R

sin(n)

(2.31)

Consider a unit radius circular disk with uniform charge density then its Fourier-Bessel

expansion in terms of J0(kmrR

)is given by

q(r, ) = 1 =n=0

2

k0mJ1(k0m)J0

k0mr

R

Then the solution of Poissons equation is given by

u(r, ) =n=0

2R2

k30mJ1(k0m)J0

k0mr

R

(2.32)

where radius R = 1


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Chapter 3

Solution of Poissons equation usingFinite Difference Methods

It is not always possible to solve differential equation analytically. For example, it

is not possible to get analytical solution to Poissons Equation for L-shaped domain.

Even when analytical solutions exist it may be tough to calculate i.e. it may involveevaluating complicated integrals. In such situations, numerical methods are employed to

get approximate solutions to some desired level of accuracy.

Various numerical methods exist for solving differential equations like finite element

methods, finite difference methods, boundary element method etc. Here we employ finite

difference methods to solve Poisson equation.

3.1 Finite Difference Methods

The Finite Difference Methods(FDM) involve approximating the equation involving

derivates with difference equation. For example,df

dxis taken as

f(x + h) f(x)h

20


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Chapter 3. Solution of Poissons equation using Finite Difference Methods21

3.2 Mesh/Grid Generation

To apply the difference equation we first divide the solution region of potential function

u(x, y) in to rectangles or meshes. For a 2D mesh, consider the distance between two

adjacent mesh points as x = hx, y = hy. Then coordinates of any point P with index

(i, j) on 2D mesh is given by

x = i hx y = j hy

where hx =xmax xmin

no of divisions + 1hy =

ymax yminno of divisions + 1

.

Here xmax, xmin, ymax, ymin stand for boundary extremes.

Figure 3.1: Domain Grid

3.3 Discretised Poissons equation

Now the discretised Poissons equation at point P(x,y) for the potential u(x,y) is given

by centralised difference formula is given by,

ui+1,j 2ui,j + ui1,jh2x

+ui,j+1 2ui,j + ui,j1

h2y

= f(x, y) (3.1)


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where ui,j = u(xi, yj) and f(x, y) is the load function. With the above difference formula,

the accuracy of the calculated solution depends on the number of divisions of the mesh

and the error is O(h2).

3.4 Matrix Formulation

A system of equations are obtained when the difference formula is applied at each in-

terior mesh point. For 1 i, j no of divisions we get N2 number of equations,(N =no of divisions). These equations can be put in matrix form as follows:

AU = B,

Vector of unknowns : U = [u11, u21, ..., uN1, u12, ...uN2, ..., uNN]T,

Right hand side vector : B = h2 [f11, ..., fN1, ...,f1N, ..., fNN]T


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The N2 N2 matrix A is given by

A =

4 1 11 . . . . . . . . .

. . .. . . 1 . . .1 4 1

1 4 1 . . .. . . 1 . . . . . . . . .

. . . . . . . . . 1 . . .1 1 4 . . .

. . .. . . 1

. . .. . .

. . .

. . . . . . . . .

. . .. . . 1

1 4 1. . . 1 . . . . . .

. . .. . .

. . . 11 1 4

(3.2)

The solution of Poissons equation can thus be found by solving above equations.

3.5 Iterative Methods

To solve the above system of equations usual elimination methods cannot be applied.

Especially when the number of divisions become more, the amount of storage required

to store intermediate matrices produced during the computation by elimination methods

become very large.

Iterative methods are employed to solve system of equations. This is because they

take advantage of the sparsity structure of the A matrix. They require storage of just

the same as A matrix (5N2 in our problem) and can be terminated when we get the


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solution with sufficient amount of accuracy.

In an iterative method, we begin with an initial vector X(0), and generate a sequence

of vectors

X(0) X(1) X(2)

which converge towards the desired solution X. Each iteration requires very less amount

of work comparable to that of multiplication of A with a vector which is very less when

matrix A is sparse. So we can carry out a large number of iterations.

3.5.1 Jacobi Method

This is the simplest of the iterative methods. A system of n equations where ith

equation(1 i n) is given byn

j=1

aij xj = bi

can be solved by transforming each equation in to a suitable format. The ith equation

is rewritten as

xi =

bi n

j=1j=n

aij xjaii

(3.3)

Jacobi method involves finding next iteration vector X(I+1) from current vector X(I)

by substituting x(I+1) in LHS and x(I) in RHS of equation 3.3. So each iteration of the

Jacobi method is given by n explicit steps. The ith step involves calculating x(I)i using

the formula:

x(I+1)i =

bi n

j=1j=naij x(I)j

aii(3.4)


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3.5.2 Gauss-Siedel Method

Gauss Siedel Method uses new values as soon as they are computed. It involves using

(i-1) new values of X in RHS in the calculation of xi.

x(I+1)i =

bi i1j=1

aij x(I+1)j n

j=i+1

aij x(I)jaii

(3.5)

3.5.3 Successive-over Relaxation Method

In this method, in order to improve convergence, we extrapolate the solution vector

obtained from Gauss-Siedel method. Extrapolation is done in form of weighted average

of the previous iteration solution and new solution from Gauss-Siedel method with the

help of relaxation parameter .

x(I+1)j = (1 )x(I)j + x (3.6)

where x is obtained using gauss-siedel method as in 3.5.2.

3.6 Convergence of Iterative Methods

In order to study the convergence of iterative methods, iterative method can be formu-

lated in terms of matrix notation. For each iterative method, we have a new matrix M

which expresses iteration step for solution of matrix equation AX = B as

MX(I+1) = (MA)X(I) + B X(I+1) = (IM1A)X(I) + B


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The convergence of iterative method depends on the spectral radius () of the (IM1A)matrix.

For an iterative method to converge, the spectral radius () should be less thanone.

The smaller the spectral radius, faster the convergence of iterative method is.

Given below are the results from [8],

For consistently ordered matrices, Spectral radius of Jacobi method (J) andGauss-Siedel method (H) are related by

(H) = (J)2

For SOR method, the optimal parameter b is given by ,

b = 21 +

1 (J)2

(H(b) =

(J)

1 +

1 (J)2

2

Therefore Gauss-Siedel method converges twice as fast as Jacobi method and SORmethod converges N times as fast as Jacobi method.

ADI method, an iterative method to solve Poisson equation is introduced in next

chapter and is shown to be superior than above classical iterative methods.


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Chapter 4

Alternating Direction Implicit(ADI)Method

Alternating Direction Implicit(ADI) method is an iterative method to solve system of

equations arising from the finite difference discretizations of parabolic and elliptic partial

differential equations. It was first introduced in mid-1950s by Peaceman and Rachford.

4.1 Motivation

When dealing with multi-dimensional domains, inversion of coefficient matrix A (3.2)

requires lot of effort. But if we do it in steps by solving for each dimension once we get

a tridiagonal system to solve which can be carried out in O(n) time. So, as the iteration

steps varies in direction every time, the name is given as alternation direction method.Also since the system of equations needs to solved implicitly, this is an implicit method.

4.2 Procedure

The idea of the ADI method is to execute each iteration in two or three steps( equal to

the number of dimensions). The finite difference formula (6.2) is written in such a way

that derivates involving one spatial direction are in left hand side of the equation and

27


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Chapter 4. Alternating Direction Implicit(ADI) Method 28

remaining terms are carried to the right hand side.

ui+1,j 2ui,j + ui1,jh2x

= f(x, y) ui,j1 2ui,j + ui,j+1h2y

(4.1)

ui,j1 2ui,j + ui,j+1h2y

= f(x, y) ui1,j 2ui,j + ui+1,jh2y

(4.2)

Using above equations, each iteration is described in two steps. U(I+1

2) is obtained

by using 4.1 and then U(I+1) is obtained from 4.2

u(I+12 )

i+1,j 2u(I+ 12 )i,j + u(I+ 12 )i1,jh2x

= f(x, y) u(I)i,j1 2u(I)i,j + u(I)i,j+1

h2y(4.3)

u(I+1)i,j1 2u(I+1)i,j + u(I+1)i,j+1

h2y= f(x, y) u

(I+ 12)

i1,j 2u(I+1

2)

i,j + u(I+ 1

2)

i+1,j

h2y(4.4)

So in each iteration, the discrete equations are solved first in one spatial direction and

then in the other direction. Also the system of equations should be solved implicitly

leading to the terminology, alternating direction implicit method.

4.3 Matrix Formulation

The difference equations given in 4.4 when applied to each point on the domain gives

rise to system of equations which can be given in matrix equations.

Hu = B

V u (4.5)

V u = B Hu (4.6)

where

H: Horizontal dimension matrix,contains coefficients for unknowns in x direction

V : Vertical dimension matrix, contains for coefficients for unknowns in y-direction.

B : RHS vector In mathematical terms,

w =2ui,j

ui1,j

ui+1,j

h2x if w = Hu


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w =2ui,j ui,j1 ui,j+1

h2

y

if w = V u

In simple words, H and V arise as the central difference approximations of the corre-

sponding terms in the Poissons equation.

The H matrix for the the domain described in figure (3.1) is given by:

H =1

h2x

2 1 1

1

. . .. . .

. . .

. . .. . . 1 . . .1 2 1

1 2 1 . . .. . . 1 . . . . . . . . .

. . .. . .

. . . 1 . . .1 1 2 . . .

. ..

. .. 1. . . . . . . . .

. . . . . . . . .

. . . . . . 11 2 1

. . . 1 . . . . . .. . .

. . .. . . 1

1 1 2

(4.7)

The diagonal entries of H matrix correspond to the current point( xi, yj) for which the

equation is being generated. H matrix row contains a subdiagonal entry for the preced-

ing point(x = xi1) and a superdiagonal entry for the next point in x-direction. If the

adjacent point is a boundary point, then correspondingly sub-diagonal or super-diagonal

entry may be zero and the rhs vector is updated with the value at the boundary.


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The V matrix for the the domain described in figure (3.1) is given by:

V =1

h2y

2 1. . . . . .

. . .. . .

2 11 2 . . .

. . .. . .

. . .

. . . . . . . . .

1 2 . . .. . .

. . . 1. . . . . . . . .

. . .. . .

. . .

. . .. . . 1

1 2. . . . . .

. . . . . .

1 2

(4.8)

Each row of V matrix corresponds to an equation for a point(xi, yj) in the interior of the

domain in vertical direction. One entry in lower triangular part is non-zero corresponding

to the preceding point (y = yj1) in the vertical dimension and an entry in upper

triangular part is non-zero corresponding to the next point (y = yj+1)in the vertical

dimension.

This matrix formulation can also be viewed as the coefficient matrix A described in

(3.2) being split in to parts in the ADI method depending on the number of dimensions

of the domain. For two dimensions,

A = H+ V


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The B vector is obtained by evaluation the Poisson load/source function and accumulat-

ing the boundary values if the adjacent points to the given point are boundary points.

4.4 ADI Method Implementation

Relaxation parameter r is introduced in the equations 4.6 similar to Gauss-Siedel method

to enhance the convergence. Now the steps in an iteration of the ADI method can be

summarised as follows:

(H+ rI)u(n+1

2) = (rI V)u(n) + B (4.9)

(V + rI)u(n+1) = (rI H)u(n+ 12 ) + B (4.10)

Each iteration step involves solving a matrix equation which is costly if we employ

Gaussian elimination (O(n3) where n = N2). But we can observer that H matrix is

tri-diagonal i.e. it contains non-zero entries only in diagonal, sub-diagonal and super-

diagonal elements. So to solve system of equations (4.9), we can employ tri-diagonal

solve which involves only O(n) work.

Algorithm Tri-diagonal-solve can be slightly modified to solve for equations (4.10)

where elimination of non-zero lower-triangular entry below this diagonal element is done

and corresponding row is modified instead of next row. We can observe that this algo-

rithm also involves only O(n) work.

So the implementation of an iteration requires only O(n) to solve the matrix equations

given in the steps of ADI method. But it is still costlier than classical elimination

methods. This may seem to be a disadvantage. But the advantage of ADI method

springs from the fact that it converges at much faster rate by clever choice of relaxation

parameter r.


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4.5 Solution of Laplace equation on Square domain

4.5.1 Problem Statement

To find the steady state temperature distribution on a unit square plate(length L = 1)

with boundary conditions with boundary conditions unit temperature on top edge(y =

L, 0 < x < L) and zero on the remaining edges.

Figure 4.1: Square Domain

4.5.2 Numerical Solution

The problem involves finding the solution of Laplace equation 2u = 0 on the squaredomain. Solving the problem using ADI method involves generating H and V matrices

as given in equation (4.7) and (4.8) and RHS vector B contains non-zero entries only for

index j = N rows of the domain given in (3.1). Iterations steps described in (4.9) and

(4.10) are applied.

4.5.3 Error Analysis

The relative error plot given in figure(4.3) is obtained by taking ratio of difference between

analytical solution (fig. 2.3) and numerical solution (fig. 4.2) with analytical solution.


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0

0.5

1

0

0.5

10

0.5

1

xy

Temperature

Figure 4.2: Temperature distribution on unit square plate.Solution of Laplace Equation using ADI Method.

Side Length =1 No. of Divisions = 99

The relative error plot indicates two peaks at the corners of the square which are due

to discontinuity in the function value on the boundary( sudden jump from 0 to 1).

Analytically, error can be explained with the help of Gibbs Phenomenon [1] ( overshoot

of Fourier series approximation near the discontinuity).

0

0.5

1

0

0.5

10

0.01

0.02

0.03

xy

RelativeError

Figure 4.3: Relative error plot between ADI solution and analytical solutionfor Laplace equation on Square domain. No. of Divisions = 99

Generally two different methods [11] are used to to reduce or localise the error arising

due to the discontinuity in the boundary shape/ boundary function


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Table 4.2: Cyclic ADI parameters performance comparision

Number of r=1 m=4 m=8Divisions Iterations taken Iterations taken Big ADI steps Iterations taken Big ADI steps

10 36 17 4 25 3100 2981 33 8 33 4500 55304 49 12 41 5

1000 57 14 49 62000 69 17 49 63000 77 19 57 7

r[0] : 3.365050 r[1] : 0.001197 r[2] : 0.116132 r[3] : 0.034673

r[4] : 1.270268 r[5] : 0.003170 r[6] : 0.388272 r[7] : 0.010371

The comparision of iterations taken with constant and varying relaxation parameters is

given in table 4.2.

4.6 Solution of Poisson Equation on square domain

4.6.1 Problem Statement

To find the electro static potential inside a square domain of length(L = 5) with homo-

geneous zero boundary conditions with unit charge density distributed inside the domain

4.6.2 Governing Equation and Numerical SolutionThe electro static potential(Voltage) in a closed region with charge density contained

inside satisfies Poisson equation

2V = 140

(4.11)

where V stands for Voltage or electro static potential. The value of the constant, permi-

tivity of free space is 140

is 9109 For the given problem, charge density = 1Coulomb.

The plot of solution obtained by ADI method is given in 4.6. The potential is in the


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Figure 4.5: Poisson function for square plate

shape of a dome as the edges of the plate are maintained at ground potential.

02

46

0

2

4

60

0.5

1

1.5

2

x 1010

xy

Voltage

Figure 4.6: Numerical Solution of Poisson equation on square domain of length 5m.No. of Divisions = 100

4.6.3 Error Analysis

Figure (4.7) shows relative error plot containing four peaks due to the discontinuity in

the shape of the boundary at the corners of the square domain. The reduced error plot

in the interior of the region is shown in (4.8).


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02

46

0

2

4

60

2

4

6

8

x 103

xy

RelativeError

Figure 4.7: Error plot for Poisson equation on Square domain. No. of Divisions = 100

02

46

0

2

4

60

1

2

3

4

x 104

xy

RelativeError

Figure 4.8: Error plot in the interior of the domain for Poisson equation on Squaredomain. No. of Divisions = 100


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Chapter 5

ADI Method on various domains

ADI method was introduced in the previous chapter and was shown to be able to solve

Laplace and Poisson equation with good accuracy on square domain. In this chapter,

we extend ADI method to various domains in two dimensions.

5.1 ADI Solver Implementation Steps

The ADI solver takes the following conditions as inputs. Find Boundary Extremes: This

is useful for the generation of mesh. The bounding box of the domain is determined.

The

5.2 FDM formula for irregular boundary

Finite Difference approximation formula given in equation(6.2) holds when the boundary

is aligned with the grid points. If the boundary is curved and doesnt intersect at mesh

point as given in the figure (5.1) then the discretisation should take in to account that

adjacent point is not at a distance h but a fraction of it, say nh.

A generalised discretisation formula taking into account the fractional distances as

39


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Chapter 5. ADI Method on various domains 40

Figure 5.1: A curved boundary with mesh

Figure 5.2: Mesh point O with neighbours NEWS

shown in figure (5.2) is given in [4].It is as follows:

2uo = 2h2

uR

r(r + l)+

uNn(n + s)

+uL

l(l + r)+

uSs(s + n)

lr + nslrns

uo

(5.1)

5.3 Solution of Laplace equation on a ring domain

Problem statement

To find the electrostatic potential in an annular ring domain(figure 5.3) whose outer

radius(R2 = 1) boundary is maintained at ground voltage and inner radius (R1 = 0.25)


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Figure 5.3: Ring Domain

boundary at voltage 1V

u(r, ) = 1, r = R1

= 0, r = R2

Numerical Solution

The solution involves solving laplace equation 2u = 0 in the interior of ring domain.In cartesian coordinates, ring has curved boundary so that discretisation as described in

equation (5.1) is applied at points near to the boundary. The solution obtained is given

in the figure5.4). The voltage increases from interior boundary to exterior boundary in

a logarithmic fashion.

Error Analysis

The analytical solution for the given problem is obtained from equation(2.23). The error

plot and relative error plot for the ring domain is given in figure (5.5).


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1

0

1

1

0

10

0.5

1

xy

Voltage

Figure 5.4: Voltage inside ring domain.Solution of Laplace Equation using ADI Method.

Inner Radius = 0.25 Outer Radius =1 No. of Divisions = 100

5.4 Solution of Poisson equation on a circular disk

domain

Problem statement

To find the solution of Poissons equation 2u = 1 on a circular disk of radius R = 1with homogeneous zero boundary conditions.

Numerical Solution

The problem can be viewed as uniform charge density inside circular disk for electro

static potential or uniform pressure on a membrane. The potential will be maximum atthe centre of the disk. The solution plot is given in figure (5.6)

Error Analysis

The analytical solution for the given problem can be obtained by equation (2.32). Then

the error plot is obtained as given in figure(5.7)


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1

0

1

1

0

11

0.5

0

0.5

1

x 104

xy

Error

(a) Actual Error

1

0

1

1

0

10

1

2

3

4

x 104

xy

RelativeError

(b) Relative Error

Figure 5.5: Error plots for solution of Laplace equation on ring domain

5.5 Solution of Laplace equation on a hyperbola do-

main

Problem statement

To solve Laplace equation uxx + uyy = 0 in the interior of region bounded by a pair of

hyperbolas x2 y2 = a2 and y2 x2 = b2 with boundary conditions

U(x, y) = V when x2 y2 = a2

= 0 when y2 x2 = b2

Analytical Solution

We can see that the solution is symmetric along the x, y axes so that the solution may

not contain any x, y terms. So we can expect a polynomial solution as:

U(x, y) = c0 + c1x2 + c2y

2


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1 0.5 0 0.5 1101

0

0.05

0.1

0.15

0.2

0.25

xy

u(x,y

)

Figure 5.6: Solution of Poisson equation on circular disk.Radius = 1 Source function f(x,y) = 1 No. of Divisions = 100

To satisfy Laplace equation, the necessary condition is c1 = c2So the general solution is:

U(x, y) = c0 + c1(x2 y2)

By applying boundary conditions, we get

c0 + c1(a2) = V

c0 + c1(b2) = 0

Solving we get, c0 = V

b2

a2+b2

c1 = V

1a2+b2

Therefore solution is:

U(x, y) = V

b2

a2 + b2

+

1

a2 + b2

x2 y2)

Numerical Solution

The domain is not closed so it cannot be solved directly with ADI solver. We consider

x=-3a, 3a and y= -3b, 3b as boundary extremes and consider them as boundaries. Very

few points fall on this line so that the solution is accurate enough. The solution in the

interior gives a saddle shape surface i.e. it can not have maximum or minimum in the


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1

0

1

1

0

14

2

0

2

4

x 106

xy

Error

(a) Actual Error

1

0

1

1

0

12

1

0

1

2

x 106

xy

Re

lativeError

(b) Relative Error in interior

Figure 5.7: Error plots for solution of Poisson equation on circular disk

Figure 5.8: Hyperbola Domain


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interior of the surface. Iso-lines are hyperbolas increasing in magnitude in xdirection

and decreasing in ydirection. The solution plot is given in the figure(5.9).

10

0

10

10

0

100

0.5

1

xy

Voltage

Figure 5.9: Solution of Laplace equation on hyperbola domain.a = 2 b = 2 V = 1 No. of Divisions = 100

x [3a, 3a] = [6, 6] y [3b, 3b] = [6, 6]

Error Analysis

The error plot indicates four peaks at the corners where the boundary is discontin-

uous(in the modified version). However the error is very minimum in the interior

region(figure:5.10).


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Chapter 6

ADI Method on 3-D Domains

In this chapter, ADI method formulation is explained for 3-Dimensional Domains.

6.1 3-D Poisson Equation

3D Poisson Equation in cartesian coordinate system is given by:

uxx + uyy + uzz = f(x , y, z ) (6.1)

where f(x,y,z) stands for load function at point (x,y,z) and u(x,y,z) stands for poten-

tial that needs to be determined. Similar to the finite difference descretisation described

in section(3.2) the discretised equation at point, as given in figure(6.1) described by index

(i,j,k) with coordinates xi = i

hx, yj = j

hy, zk = k

hz is given by

ui+1,j,k 2ui,j,k + ui1,j,kh2x

+ui,j+1,k 2ui,j,k + ui,j1,k

h2y

+ui,j,k+1 2ui,j,k + ui,j,k1

h2z= f(x,y,z) (6.2)

48


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Chapter 6. ADI Method on 3-D Domains 49

Figure 6.1: A point in 3D domain

6.2 3-D ADI method

For three dimensional domains, Each iteration of ADI method involves three steps to get

the solution.

(H+ rI)z(i+1/3) = (rI V N)z(i) + B(V + rI)z(i+2/3) = (rI

H

N)z(i+1/3) + B

(N + rI) z(i+1) = (rI H V)z(i+2/3) + B (6.3)

H: Horizontal dimension Matrix

V: Vertical dimension Matrix

N: Normal dimension Matrix

Thus the coefficent matrix A is split as A = H+ V + N

6.3 Solution of Laplace Equation on the Cube

Problem statement

To solve 3D Laplace equation

uxx + uyy + uzz = 0


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on the cube domain with side length L with boundary conditions

U(x , y, z ) = V z = L, 0 < x < L, 0 < y < L

= 0 onthe other five faces.

i.e. forx, y, z [0, L]U(0, y , z ) = 0

U(L , y, z ) = 0

U(x, 0, z) = 0

U(x,L,z) = 0

U(x,y, 0) = 0

Analytical Solution:

Analytical solution is obtained by the method of separation of variables i.e. solution isexpressed as product of three functions each of single variable.

U(x,y,z) = F(x)G(y)H(z)

Substituting in Laplace equation:

G(y)H(z)F(x) + F(x)H(z)G(y) + F(x)G(y)H(z) = 0

Dividing the equation by U(x,y,z) we get,

F(x)

F(x)+

G(y)

G(y)+

H(z)

H(z)= 0

We write the above equation as:

F(x)

F(x) = G(y)

G(y) H(z)

H(z) = 2


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Since the first term involves only x variable and RHS doesnt have any x variable terms

we conclude that 2 is a constant. Similarly we define a constant 2 as

G(x)

G(x)= F

(y)

F(y) H

(z)

H(z)= 2

Let 2 = 2 + 2 then,

H(z)

H(z)= F

(y)

F(y) G

(z)

G(z)= 2

General solution to the above equation is given by

H(z) = A sinh(z) + B cosh(z)

Applying the boundary condition U(x,y, 0) = 0, we get H(0) = 0 so that B = 0 which

gives,

H(z) = A sinh(z)

Solution of F(x) + 2F(x) = 0 is given by,

F(x) = Csin(x) + D cos(x)

Applying the boundary condition U(0,y,z) = 0 we get, D = 0

Applying the boundary condition U(L,y,z) = 0 we get, = nL

, n = 1,2,...

which gives

F(x) = Csin

nL

x

Similarly, we get

G(y) = Esin

m

Ly

So the solution to the Laplace equation can be written as


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U(x,y,z) =

n,m=1

An,m sin

nL

x

sin

mL

y

sinh (z)

where =

2 + 2 =

(n2 + m2)

LTo determine the coefficient An,m we use the remaining boundary condition U(x , y, L) =

f(x, y) = V

f(x, y) =

n,m=1An,m sinh (L) sin

n

Lx

sin

m

Ly

We see that [An,m sinh (L)] are the coefficients of 2D Fourier sine series of f(x, y)

Multiply both sides of the above equation with sinjL

x

sinlL

y

and integrate over

x and y. Using orthogonality of sine functions,

L0

dx sin

n

Lx

sin

j

Lx

=

L/2 if j = n

0 j = n

An,m =1

sinh(L)

4

L2

Lx=0

Ly=0

f(x, y) sin

n

Lx

sin

m

Ly

dxdy

=1

sinh(L)

4

L2VLx=0

sin

n

Lx

dxLy=0

sin

m

Ly

dy

=1

sinh(L)

4

L2V

cos

n

LxL

0

cos

m

LyL

0

=4 V [1 cos(n)][1 cos(m)]

sinh(L) L2

Therefore solution to Laplace equation on cube with above boundary conditions is given

by,

U(x , y, z ) =

n,m=1

4 V [1 (1)n] [1 (1)m]L2 sinh(L)

sin

n

Lx

sin

m

Ly

sinh(z)


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Since even terms become zero we get,

U(x,y,z) =

n,m=1

16V

L2 sinh(L)sin

n

Lx

sin

m

Ly

sinh(z) n, m odd

Numerical Solution

Numerical Solution obtained using ADI method is given in figure (6.2). The slices at

x = 0.1, 0.25, 0.5, 0.75, 0.9 are given. The solution increases exponentially from 0 to 1

along z axis.

Error Analysis

Error plot is given in figure(??). The error is more near the top four edges i.e. intersection

z = 1 plane with x=0,x=1,y=0,y=1 planes due to discontinuity of boundary function at

that edges. The error is very minimum of the order of 104 in the interior region. Thus

ADI method works well for 3-D Domains also.


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Figure 6.2: Solution of Laplace Equation on Cube domain using ADI method.

No. of Divisions =100


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Figure 6.3: Error plot for Laplace equation on Cube domain

No. of Divisions =100


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Chapter 7

Conclusions and Future Work

7.1 Conclusion

Poisson Solver based on Alternating Direction Implicit(ADI) method is developed. The

solver takes as input the geometry of the domain, boundary conditions and source func-

tion produces the potential distribution inside the domain. The convergence of ADImethod is observed to be much better than convergence of other classical iterative meth-

ods. The working of ADI poisson solver on various geometries like circle,ring, hyperbola

are demonstrated. The ADI solver is extended to work for three dimensions also.

7.2 Future Work

A Graphical User Interface (GUI) needs to be developed which can simplify theinput of geometry,description of boundary conditions and source function.

The optimal parameters for different geometries needs to be determined.

The working of the method should be checked on much more complicated domains.

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References

[1] Arfken, G. B., Weber, H.J., Mathematical Methods for Physicists, Elsevier, Noida,

(2005), Sixth Edition.

[2] Foley, J.D., et al., Computer Graphics: Principles and Practice, Addison Wesley,

Reading, (1990), Second Edition.

[3] Griffiths, D. J., Introduction to Electrodynamics, Prentice Hall, New Jersey, (1999).

[4] Kreyszig, E., Advanced Engineering Mathematics, John Wiley & Sons (1993), Seventh

Edition.

[5] Olver, P.J., Introduction to Partial Differential Equations, available at, http://www.

math.umn.edu/~olver/pdn.html.

[6] Peaceman, D. W., Rachford, H. H., Jr., The Numerical Solution of Parabolic and

Elliptic Differential Equations, Journal of the Society for Industrial and Applied

Mathematics 3 (1955), 2841.

[7] Pivato, M., Linear Partial Differential Equations and Fourier Theory, Cambridge

University Press, (2009).

[8] Stoer, J., Bulirsch, R., Introduction to Numerical Analysis, Springer Verlag, New

York, (1993), Second Edition.

[9] Saad, Y., Iterative Methods for Sparse Linear Systems, SIAM, Philadelphia, (1993),

Second Edition.

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REFERENCES 58

[10] Varga, R.S., Matrix Iterative Analysis, Springer-Verlag, Heidelberg, (2000), Second

Revised and Expanded Edition.

[11] Zenger, C., Gietl, H., Improved Difference Schemes for the Dirichlet Problem of

Poissons Equation in the Neighbourhood of Corners Numerische Mathematik, 30,

(1978), 315332.

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