Wavelet based Green's function approach

to 2D PDEs

Kevin Amaratunga

�

and John R. Williams

y

October, 1992

�

Graduate Student, Intelligent Engineering Systems Laboratory, Massachusetts

Institute of Technology, Cambridge, MA 02139, USA.

y

Associate Professor, Intelligent Engineering Systems Laboratory, Massachusetts

Institute of Technology, Cambridge, MA 02139, USA.

Abstract

In this paper we describe how wavelets may be used to solve

partial di�erential equations. These problems are currently solved

by techniques such as �nite di�erences, �nite elements and multi-

grid. The wavelet method, however, o�ers several advantages over

traditional methods. Wavelets have the ability to represent func-

tions at di�erent levels of resolution, thereby providing a logical

means of developing a hierarchy of solutions. Furthermore, com-

pactly supported wavelets (such as those due to Daubechies [1]) are

localized in space, which means that the solution can be re�ned in

regions of high gradient, e.g. stress concentrations, without having

to regenerate the mesh for the entire problem.

In order to demonstrate the wavelet technique, we consider Pois-

son's equation in two dimensions. By comparison with a simple

�nite di�erence solution to this problem with periodic boundary

conditions we show how a wavelet technique may be e�ciently de-

veloped. Dirichlet boundary conditions are then imposed, using the

capacitance matrix method decribed by Proskurowski and Widlund

[2] and others. The convergence of the wavelet solutions are exam-

ined and they are found to compare extremely favourably to the

�nite di�erence solutions. Preliminary investigations also indicate

that the wavelet technique is a strong contender to the �nite element

method.

1 Introduction

Wavelets are a family of orthonormal functions which are characterized

by the translation and dilation of a single function (x). This family of

functions, denoted by

m;k

(x) and given by

m;k

(x) = 2

m

2

(2

m

x� k); m;k � Z

is a basis for the space of square integrable functions L

2

(R) i.e.

f(x) =

P

m

P

k

d

m;k

m;k

(x) � L

2

(R)

Wavelets are derived from scaling functions i.e. functions which sat-

isfy the recursion

�(x) =

P

k

a

k

�(2x� k)

in which a �nite number of the �lter coe�cients a

k

are nonzero. Any L

2

(R)

function f(x) may be approximated at resolution m by

P

m

(f)(x) =

P

k

c

m;k

�

m;k

(x) k � Z

where, using Daubechies notation [1], P

m

f represents the projection of the

function f onto the space of scaling functions at resolution m.

�

m;k

(x) = 2

m

2

�(2

m

x� k); k � Z

is a scaling function basis for the scale m approximation of L

2

(R). The set

of approximations P

m

(f)(x) constitutes a multiresolution representation of

the function f(x) [3].

In two dimensions, the space of square integrable functions is L

2

(R

2

)

and any function f(x; y) which lies in this space may be expressed in terms

of the orthonormal basis

i;k

(x)

j;l

(y); i; k; j; l � Z

This is simply the tensor product of the one dimensional bases in the two

coordinate directions, x and y. f(x; y) may be represented at resolution m

by

P

m

(f)(x; y) =

P

k

P

l

c

m;k;l

�

m;k

(x)�

m;l

(y) k; l � Z

For a greater insight into the properties of wavelets and their con-

struction, the reader is referred to [4].

This work details how a hierarchy of wavelet solutions to two dimen-

sional partial di�erential equations may be developed using scaling func-

tion bases. In order to demonstrate the wavelet technique, we consider

Poisson's equation in two dimensions i.e.

u

;xx

+ u

;yy

= f

where u = u(x; y), f = f(x; y).

2 Finite di�erence solution of the periodic

problem

Consider the problem

u

;xx

+ u

;yy

= f (1)

where u and f are periodic functions in x and y. Let d

x

be the period in

the x direction and d

y

be the period in the y direction. Then

u(0; y) = u(d

x

; y)

f(0; y) = f(d

x

; y)

u(x; 0) = u(x; d

y

)

f(x; 0) = f(x; d

y

)

Suppose now that we have an n

x

x n

y

mesh discretization of the rectangular

region [0; d

x

], [0; d

y

], so that

u

j;k

= u(jh; kh)

f

j;k

= f(jh; kh)

where

j = 0; 1; 2; ::: n

x

� 1

k = 0; 1; 2; ::: n

y

� 1

and

h =

d

x

n

x

=

d

y

n

y

The �nite di�erence approximation to u

;xx

is then

(u

xx

)

j;k

=

1

h

(

u

j+1;k

�u

j;k

h

+

u

j;k

�u

j�1;k

h

) =

u

j+1;k

�2u

j;k

+u

j�1;k

h

2

Similarly,

(u

yy

)

j;k

=

u

j;k+1

�2u

j;k

+u

j;k�1

h

2

and so the discrete form of equation (1) is

u

j+1;k

� 2u

j;k

+ u

j�1;k

+ u

j;k+1

� 2u

j;k

+ u

j;k�1

= h

2

f

j;k

j = 0; 1; 2; ::: n

x

� 1

k = 0; 1; 2; ::: n

y

� 1

or

u

j+1;k

+ u

j;k+1

� 4u

j;k

+ u

j�1;k

+ u

j;k�1

= h

2

f

j;k

Noting that u

�1;k

= u

n

x

�1;k

, u

n

x

;k

= u

0;k

, u

j;�1

= u

j;n

y

�1

and u

j;n

y

= u

j;0

,

this system of equations can be written in matrix form as

2

6

6

6

6

6

6

6

6

6

6

6

4

T I 0 0 ::: 0 I

I T I 0 ::: 0 0

0 I T I ::: 0 0

0 0 I T ::: 0 0

::: ::: ::: ::: ::: ::: :::

0 0 0 0 ::: T I

I 0 0 0 ::: I T

3

7

7

7

7

7

7

7

7

7

7

7

5

2

6

6

6

6

6

6

6

6

6

6

6

4

U

0

U

1

U

2

U

3

:::

U

n

y

�2

U

n

y

�1

3

7

7

7

7

7

7

7

7

7

7

7

5

= h

2

2

6

6

6

6

6

6

6

6

6

6

6

4

f

0

f

1

f

2

f

3

:::

f

n

y

�2

f

n

y

�1

3

7

7

7

7

7

7

7

7

7

7

7

5

(2)

where I is the identity matrix,

T =

2

6

6

6

6

6

6

6

6

6

6

6

4

�4 1 0 0 ::: 0 1

1 �4 1 0 ::: 0 0

0 1 �4 1 ::: 0 0

0 0 1 �4 ::: 0 0

::: ::: ::: ::: ::: ::: :::

0 0 0 0 ::: �4 1

1 0 0 0 ::: 1 �4

3

7

7

7

7

7

7

7

7

7

7

7

5

(n

x

;n

x

)

U

k

=

2

6

6

6

6

6

6

6

6

6

6

6

4

u

0;k

u

1;k

u

2;k

u

3;k

:::

u

n

x

�2;k

u

n

x

�1;k

3

7

7

7

7

7

7

7

7

7

7

7

5

f

k

=

2

6

6

6

6

6

6

6

6

6

6

6

4

f

0;k

f

1;k

f

2;k

f

3;k

:::

f

n

x

�2;k

f

n

x

�1;k

3

7

7

7

7

7

7

7

7

7

7

7

5

Now, the matrix T is a circulant matrix and the matrix on the left

hand side of equation (2) is a block circulant matrix i.e. a circulant matrix

whose elements are matrices in themselves. Thus equation (2) may be

rewritten as

U

k�1

+ TU

k

+ U

k+1

= h

2

f

k

; k = 0; 1; 2; ::: n

y

� 1 (3)

This form of the equations may be e�ciently solved using the Fast

Fourier Transform (FFT). Let F

n

be the nxn Fourier matrix [5] i.e.

F

n

=

1

n

2

6

6

6

6

6

6

6

6

6

6

6

4

1 1 1 1 ::: 1 1

1 ! !

2

!

3

::: !

n�2

!

n�1

1 !

2

!

4

!

6

::: !

2(n�2)

!

2(n�1)

1 !

3

!

6

!

9

::: !

3(n�2)

!

3(n�1)

::: ::: ::: ::: ::: ::: :::

1 !

n�2

!

2(n�2)

!

3(n�2)

::: !

(n�2)

2

!

(n�2)(n�1)

1 !

n�1

!

2(n�1)

!

3(n�1)

::: !

(n�2)(n�1)

!

(n�1)

2

3

7

7

7

7

7

7

7

7

7

7

7

5

(n;n)

Where ! = e

2�i

n

and i =

p

�1. Then, the discrete Fourier transform of an

n-dimensional vector a is

a = F

n

�1

a = n

�

F

n

a

Furthermore, if A is an n x n circulant matrix, then F

n

�1

AF

n

is a diagonal

matrix containing the eigenvalues of A. These eigenvalues may also be

obtained by taking the discrete Fourier transform of the �rst column of A.

Multiplying equation (3) by F

n

x

�1

, therefore, we get

V

k�1

+ �V

k

+ V

k+1

= h

2

c

k

; k = 0; 1; 2; ::: n

y

� 1 (4)

where

� = F

n

x

�1

TF

n

x

=

2

6

6

6

6

6

6

6

6

6

6

6

4

�

0

0 0 0 ::: 0 0

0 �

1

0 0 ::: 0 0

0 0 �

2

0 ::: 0 0

0 0 0 �

3

::: 0 0

::: ::: ::: ::: ::: ::: :::

0 0 0 0 ::: �

n

x

�2

0

0 0 0 0 ::: 0 �

n

x

�1

3

7

7

7

7

7

7

7

7

7

7

7

5

(n

x

;n

x

)

V

k

=

2

6

6

6

6

6

6

6

6

6

6

6

4

v

0;k

v

1;k

v

2;k

v

3;k

:::

v

n

x

�2;k

v

n

x

�1;k

3

7

7

7

7

7

7

7

7

7

7

7

5

=

^

U

k

c

k

=

2

6

6

6

6

6

6

6

6

6

6

6

4

c

0;k

c

1;k

c

2;k

c

3;k

:::

c

n

x

�2;k

c

n

x

�1;k

3

7

7

7

7

7

7

7

7

7

7

7

5

=

^

f

k

Since � is diagonal, each set of equations in (4) is a decoupled system in

itself i.e. the equations reduce to

v

j;k�1

+ �

j

v

j;k

+ v

j;k+1

= h

2

c

j;k

j = 0; 1; 2; ::: n

x

� 1

k = 0; 1; 2; ::: n

y

� 1

or

2

6

6

6

6

6

6

6

6

6

6

6

4

�

j

1 0 0 ::: 0 1

1 �

j

1 0 ::: 0 0

0 1 �

j

1 ::: 0 0

0 0 1 �

j

::: 0 0

::: ::: ::: ::: ::: ::: :::

0 0 0 0 ::: �

j

1

1 0 0 0 ::: 1 �

j

3

7

7

7

7

7

7

7

7

7

7

7

5

2

6

6

6

6

6

6

6

6

6

6

6

4

v

j;0

v

j;1

v

j;2

v

j;3

:::

v

j;n

y

�2

v

j;n

y

�1

3

7

7

7

7

7

7

7

7

7

7

7

5

= h

2

2

6

6

6

6

6

6

6

6

6

6

6

4

c

j;0

c

j;1

c

j;2

c

j;3

:::

c

j;n

y

�2

c

j;n

y

�1

3

7

7

7

7

7

7

7

7

7

7

7

5

(5)

j = 0; 1; 2; ::: n

x

� 1

Once again, a circulant matrix is encountered. In fact the left hand side of

equation (5) represents a discrete convolution. It follows from the proper-

ties of the Fourier matrix that

(F

n

y

�1

2

6

6

6

6

6

6

6

6

6

6

6

4

�

j

1

0

0

:::

0

1

3

7

7

7

7

7

7

7

7

7

7

7

5

) : (F

n

y

�1

2

6

6

6

6

6

6

6

6

6

6

6

4

v

j;0

v

j;1

v

j;2

v

j;3

:::

v

j;n

y

�2

v

j;n

y

�1

3

7

7

7

7

7

7

7

7

7

7

7

5

) = h

2

F

n

y

�1

2

6

6

6

6

6

6

6

6

6

6

6

4

c

j;0

c

j;1

c

j;2

c

j;3

:::

c

j;n

y

�2

c

j;n

y

�1

3

7

7

7

7

7

7

7

7

7

7

7

5

where the . denotes component by component multiplication of the paren-

thesized vectors. Transposing this system of equations and expanding for

j = 0; 1; 2; ::: n

x

� 1,

(LF

n

y

�1

) : (V F

n

y

�1

) = h

2

cF

n

y

�1

(6)

where

L =

2

6

6

6

6

6

6

6

6

6

6

6

4

�

0

1 0 0 ::: 0 1

�

1

1 0 0 ::: 0 1

�

2

1 0 0 ::: 0 1

�

3

1 0 0 ::: 0 1

::: ::: ::: ::: ::: ::: :::

�

n

x

�2

1 0 0 ::: 0 1

�

n

x

�1

1 0 0 ::: 0 1

3

7

7

7

7

7

7

7

7

7

7

7

5

(n

x

;n

y

)

V =

2

6

6

6

6

6

6

6

6

6

6

6

4

v

0;0

v

0;1

v

0;2

v

0;3

::: v

0;n

y

�2

v

0;n

y

�1

v

1;0

v

1;1

v

1;2

v

1;3

::: v

1;n

y

�2

v

1;n

y

�1

v

2;0

v

2;1

v

2;2

v

2;3

::: v

2;n

y

�2

v

2;n

y

�1

v

3;0

v

3;1

v

3;2

v

3;3

::: v

3;n

y

�2

v

3;n

y

�1

::: ::: ::: ::: ::: ::: :::

v

n

x

�2;0

v

n

x

�2;1

v

n

x

�2;2

v

n

x

�2;3

::: v

n

x

�2;n

y

�2

v

n

x

�2;n

y

�1

v

n

x

�1;0

v

n

x

�1;1

v

n

x

�1;2

v

n

x

�1;3

::: v

n

x

�1;n

y

�2

v

n

x

�1;n

y

�1

3

7

7

7

7

7

7

7

7

7

7

7

5

and

c =

2

6

6

6

6

6

6

6

6

6

6

6

4

c

0;0

c

0;1

c

0;2

c

0;3

::: c

0;n

y

�2

c

0;n

y

�1

c

1;0

c

1;1

c

1;2

c

1;3

::: c

1;n

y

�2

c

1;n

y

�1

c

2;0

c

2;1

c

2;2

c

2;3

::: c

2;n

y

�2

c

2;n

y

�1

c

3;0

c

3;1

c

3;2

c

3;3

::: c

3;n

y

�2

c

3;n

y

�1

::: ::: ::: ::: ::: ::: :::

c

n

x

�2;0

c

n

x

�2;1

c

n

x

�2;2

c

n

x

�2;3

::: c

n

x

�2;n

y

�2

c

n

x

�2;n

y

�1

c

n

x

�1;0

c

n

x

�1;1

c

n

x

�1;2

c

n

x

�1;3

::: c

n

x

�1;n

y

�2

c

n

x

�1;n

y

�1

3

7

7

7

7

7

7

7

7

7

7

7

5

From the properties of the Fourier matrix, the �rst column of A is

the inverse Fourier transform of the �rst column of L. Hence,

F

n

x

L =

2

6

6

6

6

6

6

6

6

6

6

6

4

�4 1 0 0 ::: 0 1

1 0 0 0 ::: 0 0

0 0 0 0 ::: 0 0

0 0 0 0 ::: 0 0

::: ::: ::: ::: ::: ::: :::

0 0 0 0 ::: 0 0

1 0 0 0 ::: 0 0

3

7

7

7

7

7

7

7

7

7

7

7

5

(n

x

;n

y

)

= K (say)

K is referred to as the convolution kernel. Thus

L = F

n

x

�1

K

Also, from the de�nitions of V

k

and c

k

,

V = F

n

x

�1

U

and

c = F

n

x

�1

f

where

U =

2

6

6

6

6

6

6

6

6

6

6

6

4

u

0;0

u

0;1

u

0;2

u

0;3

::: u

0;n

y

�2

u

0;n

y

�1

u

1;0

u

1;1

u

1;2

u

1;3

::: u

1;n

y

�2

u

1;n

y

�1

u

2;0

u

2;1

u

2;2

u

2;3

::: u

2;n

y

�2

u

2;n

y

�1

u

3;0

u

3;1

u

3;2

u

3;3

::: u

3;n

y

�2

u

3;n

y

�1

::: ::: ::: ::: ::: ::: :::

u

n

x

�2;0

u

n

x

�2;1

u

n

x

�2;2

u

n

x

�2;3

::: u

n

x

�2;n

y

�2

u

n

x

�2;n

y

�1

u

n

x

�1;0

u

n

x

�1;1

u

n

x

�1;2

u

n

x

�1;3

::: u

n

x

�1;n

y

�2

u

n

x

�1;n

y

�1

3

7

7

7

7

7

7

7

7

7

7

7

5

and

f =

2

6

6

6

6

6

6

6

6

6

6

6

4

f

0;0

f

0;1

f

0;2

f

0;3

::: f

0;n

y

�2

f

0;n

y

�1

f

1;0

f

1;1

f

1;2

f

1;3

::: f

1;n

y

�2

f

1;n

y

�1

f

2;0

f

2;1

f

2;2

f

2;3

::: f

2;n

y

�2

f

2;n

y

�1

f

3;0

f

3;1

f

3;2

f

3;3

::: f

3;n

y

�2

f

3;n

y

�1

::: ::: ::: ::: ::: ::: :::

f

n

x

�2;0

f

n

x

�2;1

f

n

x

�2;2

f

n

x

�2;3

::: f

n

x

�2;n

y

�2

f

n

x

�2;n

y

�1

f

n

x

�1;0

f

n

x

�1;1

f

n

x

�1;2

f

n

x

�1;3

::: f

n

x

�1;n

y

�2

f

n

x

�1;n

y

�1

3

7

7

7

7

7

7

7

7

7

7

7

5

Equation (6) may then be written as

^

~

K :

^

~

U = h

2

^

~

f

where

^

~

A = F

n

x

�1

AF

n

y

�1

denotes the two-dimensional FFT of an n

x

x n

y

matrix A. Then the

solution to equation (1) is obtained by taking the inverse two-dimensional

FFT of

^

~

U = h

2

^

~

f =

^

~

K

where / denotes component by component division.

In the particular case of Poisson's equation, the convolution kernel

is singular. As a consequence of this singularity, the �rst element of

^

~

K is

zero.

(

^

~

K)

0;0

= 0

To avoid division by zero, (

^

~

K)

0;0

may be set to arbitrary nonzero value. If

the function f is chosen such that its mean over the period is zero in both

the x and the y directions, then the value assigned to (

^

~

K)

0;0

is immaterial.

3 Wavelet-Galerkin solution of the periodic

problem

The wavelet-Galerkin method [6, 7, 8, 9, 10] entails representing the solu-

tion u and the right hand side f as expansions of scaling functions at a

particular scale m. For the purposes of the current work, it will su�ce to

say that the scaling function � is de�ned by a dilation equation of the form

�(x) =

P

1

k=�1

a

k

�(2x� k)

and that the values of the scaling function may be calculated using this

recursion. Compactly supported scaling functions, such as those belonging

to the Daubechies family of wavelets [1], have a �nite number of nonzero

�lter coe�cients a

k

. The number of nonzero �lter coe�cients is denoted

by N . Figure 1 depicts the Daubechies 6 coe�cient scaling function. For a

more detailed description of scaling functions, their construction and their

properties, reference is made to [11], [12], [13] or most introductory works

on wavelets.

The wavelet-Galerkin solution of the periodic problem is slightly more

complicated than the �nite di�erence solution, since the solution procedure

consists of solving a set of simultaneous equations in wavelet space and not

in physical space. This means that we have to transform the right hand

side function into wavelet space, solve the set of simultaneous equations

to get the solution in wavelet space, and then transform the solution from

wavelet space back into physical space. Consider the same problem as

before

u

;xx

+ u

;yy

= f (7)

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

x

phi(

x)

Figure 1: Daubechies D6 scaling function

The wavelet-Galerkin approximation to the solution u(x; y) at scale

m is

u(x; y) =

X

k

X

l

~c

k;l

2

m

2

�(2

m

x� k)2

m

2

�(2

m

y � l) k; l � Z (8)

~c

k;l

are the wavelet coe�cients of u i.e. they de�ne the solution in wavelet

space. The transformation from wavelet space to physical space (or vice

versa) can be easily accomplished using the FFT if the wavelet expansion

is expressed as a discrete convolution. To do this make the substitutions

X = 2

m

x

Y = 2

m

y

Then

U(X;Y ) = u(x; y) =

X

k

X

l

c

k;l

�(X � k)�(Y � l) c

k;l

= 2

m

~c

k;l

(9)

Now u(x; y) is periodic in x and y with periods d

x

and d

y

, so that U(X;Y ) is

periodic inX and Y with periods 2

m

d

x

and 2

m

d

y

. Assuming that d

x

; d

y

� Z,

so that n

x

= 2

m

d

x

� Z and n

y

= 2

m

d

y

� Z then c

k;l

is periodic in k and l

with periods n

x

and n

y

.

Now consider an n

x

x n

y

mesh discretization of U(X;Y ), obtained by

letting X and Y take integer values only. This gives the values of u(x; y)

at all the dyadic points (x; y) = (2

�m

X; 2

�m

Y ) i.e. the discretization of

u(x; y) depends on the scale we have chosen (or vice versa). Thus

U

i;j

= U(i �X; j �Y ) = U(i; j)

i = 0; 1; 2; ::: n

x

� 1

j = 0; 1; 2; ::: n

y

� 1

Equation (9) may then be written as

U

i;j

=

P

k

P

l

c

k;l

�

i�k

�

j�l

=

P

k

P

l

c

i�k;j�l

�

k

�

l

where �

k

= �(k). In matrix form this becomes

U = �

n

x

c �

n

y

T

(10)

in which

U =

2

6

6

6

6

6

6

6

6

6

6

6

4

U

0;0

U

0;1

U

0;2

U

0;3

::: U

0;n

y

�2

U

0;n

y

�1

U

1;0

U

1;1

U

1;2

U

1;3

::: U

1;n

y

�2

U

1;n

y

�1

U

2;0

U

2;1

U

2;2

U

2;3

::: U

2;n

y

�2

U

2;n

y

�1

U

3;0

U

3;1

U

3;2

U

3;3

::: U

3;n

y

�2

U

3;n

y

�1

::: ::: ::: ::: ::: ::: :::

U

n

x

�2;0

U

n

x

�2;1

U

n

x

�2;2

U

n

x

�2;3

::: U

n

x

�2;n

y

�2

U

n

x

�2;n

y

�1

U

n

x

�1;0

U

n

x

�1;1

U

n

x

�1;2

U

n

x

�1;3

::: U

n

x

�1;n

y

�2

U

n

x

�1;n

y

�1

3

7

7

7

7

7

7

7

7

7

7

7

5

c =

2

6

6

6

6

6

6

6

6

6

6

6

4

c

0;0

c

0;1

c

0;2

c

0;3

::: c

0;n

y

�2

c

0;n

y

�1

c

1;0

c

1;1

c

1;2

c

1;3

::: c

1;n

y

�2

c

1;n

y

�1

c

2;0

c

2;1

c

2;2

c

2;3

::: c

2;n

y

�2

c

2;n

y

�1

c

3;0

c

3;1

c

3;2

c

3;3

::: c

3;n

y

�2

c

3;n

y

�1

::: ::: ::: ::: ::: ::: :::

c

n

x

�2;0

c

n

x

�2;1

c

n

x

�2;2

c

n

x

�2;3

::: c

n

x

�2;n

y

�2

c

n

x

�2;n

y

�1

c

n

x

�1;0

c

n

x

�1;1

c

n

x

�1;2

c

n

x

�1;3

::: c

n

x

�1;n

y

�2

c

n

x

�1;n

y

�1

3

7

7

7

7

7

7

7

7

7

7

7

5

and

�

n

=

2

6

6

6

6

6

6

6

6

6

6

6

6

6

4

0 0 0 ::: �

N�2

::: �

2

�

1

�

1

0 0 ::: 0 ::: �

3

�

2

�

2

�

1

0 ::: 0 ::: �

4

�

3

::: ::: ::: ::: ::: ::: ::: :::

�

N�2

�

N�3

�

N�4

::: ::: ::: 0 0

0 �

N�2

�

N�3

::: ::: ::: 0 0

::: ::: ::: ::: ::: ::: ::: :::

0 0 0 ::: �

N�3

::: �

1

0

3

7

7

7

7

7

7

7

7

7

7

7

7

7

5

(n;n)

Taking the two dimensional FFT of equation (10),

^

~

U = F

n

x

�1

�

n

x

c �

n

y

T

F

n

y

�1

or

^

~

U = (F

n

x

�1

�

n

x

F

n

x

) (F

n

x

�1

c F

n

y

�1

) (F

n

y

�

n

y

T

F

n

y

�1

)

But �

n

(n = n

x

; n

y

) are circulant matrices and so

F

n

x

�1

�

n

x

F

n

x

= M

n

x

F

n

y

�

n

y

T

F

n

y

�1

= F

n

y

�1

�

n

y

F

n

y

= M

n

y

where M

n

(n = n

x

; n

y

) are of the form

M

n

=

2

6

6

6

6

6

6

6

6

6

6

6

4

�

0

0 0 0 ::: 0 0

0 �

1

0 0 ::: 0 0

0 0 �

2

0 ::: 0 0

0 0 0 �

3

::: 0 0

::: ::: ::: ::: ::: ::: :::

0 0 0 0 ::: �

n�2

0

0 0 0 0 ::: 0 �

n�1

3

7

7

7

7

7

7

7

7

7

7

7

5

(n;n)

Thus

^

~

U = M

n

x

^

~c M

n

y

(11)

Equation (11) may be rewritten as

^

~

U = B

x

: B

y

:

^

~c (12)

where

B

x

=

2

6

6

6

6

6

6

6

6

6

6

6

4

�

0

�

0

�

0

�

0

::: �

0

�

0

�

1

�

1

�

1

�

1

::: �

1

�

1

�

2

�

2

�

2

�

2

::: �

2

�

2

�

3

�

3

�

3

�

3

::: �

3

�

3

::: ::: ::: ::: ::: ::: :::

�

n

x

�2

�

n

x

�2

�

n

x

�2

�

n

x

�2

::: �

n

x

�2

�

n

x

�2

�

n

x

�1

�

n

x

�1

�

n

x

�1

�

n

x

�1

::: �

n

x

�1

�

n

x

�1

3

7

7

7

7

7

7

7

7

7

7

7

5

(n

x

;n

y

)

B

y

=

2

6

6

6

6

6

6

6

6

6

6

6

4

�

0

�

1

�

2

�

3

::: �

n

y

�2

�

n

y

�1

�

0

�

1

�

2

�

3

::: �

n

y

�2

�

n

y

�1

�

0

�

1

�

2

�

3

::: �

n

y

�2

�

n

y

�1

�

0

�

1

�

2

�

3

::: �

n

y

�2

�

n

y

�1

::: ::: ::: ::: ::: ::: :::

�

0

�

1

�

2

�

3

::: �

n

y

�2

�

n

y

�1

�

0

�

1

�

2

�

3

::: �

n

y

�2

�

n

y

�1

3

7

7

7

7

7

7

7

7

7

7

7

5

(n

x

;n

y

)

But from the properties of the Fourier matrix

B

x

=

^

~

K

�

x

and

B

y

=

^

~

K

�

y

where

K

�

x

=

2

6

6

6

6

6

6

6

6

6

6

6

6

6

4

0 0 0 ::: 0 ::: 0 0

�

1

0 0 ::: 0 ::: 0 0

�

2

0 0 ::: 0 ::: 0 0

::: ::: ::: ::: ::: ::: ::: :::

�

N�2

0 0 ::: ::: ::: 0 0

0 0 0 ::: ::: ::: 0 0

::: ::: ::: ::: ::: ::: ::: :::

0 0 0 ::: 0 ::: 0 0

3

7

7

7

7

7

7

7

7

7

7

7

7

7

5

(n

x

;n

y

)

and

K

�

y

=

2

6

6

6

6

6

6

6

6

6

6

6

6

6

4

0 �

1

�

2

::: �

N�2

::: 0 0

0 0 0 ::: 0 ::: 0 0

0 0 0 ::: 0 ::: 0 0

::: ::: ::: ::: ::: ::: ::: :::

0 0 0 ::: ::: ::: 0 0

0 0 0 ::: ::: ::: 0 0

::: ::: ::: ::: ::: ::: ::: :::

0 0 0 ::: 0 ::: 0 0

3

7

7

7

7

7

7

7

7

7

7

7

7

7

5

(n

x

;n

y

)

Hence equation (12) becomes

^

~

U =

^

~

K

�

x

:

^

~

K

�

y

:

^

~c (13)

giving a simple relationship between U and its wavelet coe�cients c.

Similar relationships to equations (9) and (13) exist for the right

hand side function, f i.e.

F (X;Y ) = f(x; y) =

X

k

X

l

g

k;l

�(X � k)�(Y � l) (14)

and

^

~

F =

^

~

K

�

x

:

^

~

K

�

y

:

^

~g (15)

where g is the matrix of wavelet coe�cients of F (X;Y ).

The di�erential equation in wavelet space may now be formulated by

substituting equations (9) and (14) into equation (7):

@

2

@x

2

P

k

P

l

c

k;l

�(X � k)�(Y � l) +

@

2

@y

2

P

k

P

l

c

k;l

�(X � k)�(Y � l) =

P

k

P

l

g

k;l

�(X � k)�(Y � l)

i.e.

P

k

P

l

2

2m

c

k;l

�

00

(X � k)�(Y � l) +

P

k

P

l

2

2m

c

k;l

�(X � k)�

00

(Y � l) =

P

k

P

l

g

k;l

�(X � k)�(Y � l)

Taking the inner product of both sides, �rst with �(X � p) and then with

�(Y � q) (p; q � Z) gives

P

k

P

l

2

2m

c

k;l

R

�

00

(X � k)�(X � p)dX

R

�(Y � l)�(Y � q)dY +

P

k

P

l

2

2m

c

k;l

R

�(X � k)�(X � p)dX

R

�

00

(Y � l)�(Y � q)dY =

P

k

P

l

g

k;l

R

�(X � k)�(X � p)dX

R

�(Y � l)�(Y � q)dY

Since the translates of the scaling function are mutually orthogonal, this

simpli�es to

P

k

c

k;q

R

�

00

(X�k)�(X�p)dX +

P

l

c

p;l

R

�

00

(Y �l)�(Y �q)dY =

1

2

2m

g

p;q

or

P

k

c

k;q

p�k

+

P

l

c

p;l

q�l

=

1

2

2m

g

p;q

where

j�k

=

R

�

00

(y � k)�(y � j)dy

are the connection coe�cients described by Latto et al [7]. Remembering

that c

j;k

and g

j;k

have period n

x

and n

y

in the x and y directions, the

matrix form of the di�erential equation becomes

R

n

x

c + c R

n

y

T

=

1

2

2m

g (16)

where

R

n

=

2

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

4

0

�1

:::

2�N

:::

N�2

:::

1

1

0

:::

3�N

::: 0 :::

2

::: ::: ::: ::: ::: ::: ::: :::

N�2

N�3

:::

0

::: ::: ::: 0

0

N�2

:::

1

::: ::: ::: 0

::: ::: ::: ::: ::: ::: ::: :::

0 0 ::: ::: :::

�1

:::

2�N

2�N

0 ::: ::: :::

0

:::

3�N

::: ::: ::: ::: ::: ::: ::: :::

�1

�2

::: 0 :::

N�3

:::

0

3

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

5

(n;n)

The two dimensional FFT of equation (16) is

F

n

x

�1

R

n

x

c F

n

y

�1

+ F

n

x

�1

c R

n

y

T

F

n

y

�1

=

1

2

2m

F

n

x

�1

g F

n

y

�1

or

�

n

x

^

~c +

^

~c �

n

y

=

1

2

2m

^

~g (17)

where

F

n

x

�1

R

n

x

F

n

x

= �

n

x

F

n

y

R

n

y

T

F

n

y

�1

= F

n

y

�1

R

n

y

F

n

y

= �

n

y

and

�

n

=

2

6

6

6

6

6

6

6

6

6

6

6

4

�

0

0 0 0 ::: 0 0

0 �

1

0 0 ::: 0 0

0 0 �

2

0 ::: 0 0

0 0 0 �

3

::: 0 0

::: ::: ::: ::: ::: ::: :::

0 0 0 0 ::: �

n�2

0

0 0 0 0 ::: 0 �

n�1

3

7

7

7

7

7

7

7

7

7

7

7

5

(n;n)

Equation (17) may be rewritten as

(H

x

+ H

y

) :

^

~c =

1

2

2m

^

~g (18)

where

H

x

=

2

6

6

6

6

6

6

6

6

6

6

6

4

�

0

�

0

�

0

�

0

::: �

0

�

0

�

1

�

1

�

1

�

1

::: �

1

�

1

�

2

�

2

�

2

�

2

::: �

2

�

2

�

3

�

3

�

3

�

3

::: �

3

�

3

::: ::: ::: ::: ::: ::: :::

�

n

x

�2

�

n

x

�2

�

n

x

�2

�

n

x

�2

::: �

n

x

�2

�

n

x

�2

�

n

x

�1

�

n

x

�1

�

n

x

�1

�

n

x

�1

::: �

n

x

�1

�

n

x

�1

3

7

7

7

7

7

7

7

7

7

7

7

5

(n

x

;n

y

)

H

y

=

2

6

6

6

6

6

6

6

6

6

6

6

4

�

0

�

1

�

2

�

3

::: �

n

y

�2

�

n

y

�1

�

0

�

1

�

2

�

3

::: �

n

y

�2

�

n

y

�1

�

0

�

1

�

2

�

3

::: �

n

y

�2

�

n

y

�1

�

0

�

1

�

2

�

3

::: �

n

y

�2

�

n

y

�1

::: ::: ::: ::: ::: ::: :::

�

0

�

1

�

2

�

3

::: �

n

y

�2

�

n

y

�1

�

0

�

1

�

2

�

3

::: �

n

y

�2

�

n

y

�1

3

7

7

7

7

7

7

7

7

7

7

7

5

(n

x

;n

y

)

From the properties of the Fourier matrix, however,

H

x

=

^

~

K

x

and

H

y

=

^

~

K

y

where

K

x

=

2

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

4

0

0 ::: 0 ::: 0 ::: 0

1

0 ::: 0 ::: 0 ::: 0

::: ::: ::: ::: ::: ::: ::: :::

N�2

0 ::: 0 ::: ::: ::: 0

0 0 ::: 0 ::: ::: ::: 0

::: ::: ::: ::: ::: ::: ::: :::

0 0 ::: ::: ::: 0 ::: 0

2�N

0 ::: ::: ::: 0 ::: 0

::: ::: ::: ::: ::: ::: ::: :::

�1

0 ::: 0 ::: 0 ::: 0

3

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

5

(n

x

;n

y

)

and

K

y

=

2

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

4

0

1

:::

N�2

:::

2�N

:::

�1

0 0 ::: 0 ::: 0 ::: 0

::: ::: ::: ::: ::: ::: ::: :::

0 0 ::: 0 ::: ::: ::: 0

0 0 ::: 0 ::: ::: ::: 0

::: ::: ::: ::: ::: ::: ::: :::

0 0 ::: ::: ::: 0 ::: 0

0 0 ::: ::: ::: 0 ::: 0

::: ::: ::: ::: ::: ::: ::: :::

0 0 ::: 0 ::: 0 ::: 0

3

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

5

(n

x

;n

y

)

Let

K

= K

x

+ K

y

This is the convolution kernel of the left hand side of equation (16). Thus

equation (18) becomes

^

~

K

:

^

~c =

1

2

2m

^

~g (19)

Combining equations (13), (15) and (19)

^

~

U = (

^

~

K

�

x

:

^

~

K

�

y

) : ((

1

2

2m

^

~

F = (

^

~

K

�

x

:

^

~

K

�

y

)) =

^

~

K

)

or, simply

^

~

U =

1

2

2m

^

~

F =

^

~

K

Taking the inverse two dimensional Fourier transform gives the solution U .

4 Incorporation of boundary conditions

4.1 The capacitance matrix method

Boundary conditions may be incorporated using the capacitance matrix

method (Proskurowski and Widlund [2], Qian and Weiss [10] and others).

Suppose that it is required to solve the problem

u

;xx

+ u

;yy

= f in S

with the Dirichlet boundary conditions

u = u

�

(x; y)

on the boundary �.

Again, suppose that u(x; y) and f(x; y) are periodic with periods d

x

and d

y

in the x and y directions, so that the region S lies in the periodic

cell [0; d

x

], [0; d

y

]. f can be made periodic by making it zero outside S.

If necessary, the function f may be extended smoothly outside S so as to

make it periodic. Let the solution to the di�erential equation with periodic

boundary conditions be v(x; y). The solution u to the di�erential equation

with Dirichlet boundary conditions may be obtained by adding in another

function w(x; y) such that

u = v + w (20)

Since v

;xx

+ v

;yy

= f in S, w must satisfy

w

;xx

+ w

;yy

= 0 in S

However, on or outside the boundary �, r

2

w = w

;xx

+ w

;yy

may

take such values as to make u satisfy the given boundary conditions. The

desired e�ect may be achieved by placing sources (or delta functions) along

a closed curve �

1

which encompasses the region S. Thus w is given by the

solution to

Figure 2: Periodic Green's function for the Laplace operator

w

;xx

+ w

;yy

= X

1

in [0; d

x

], [0; d

y

]

where

X

1

� X

1

(x; y) =

Z

�

1

X

0

(p; q) �(x� p; y � q) d�

1

; (p; q) � �

1

and �(x; y) is the delta function at (0; 0).

Let G(x; y) be the Green's function of the di�erential equation i.e.

G

;xx

+G

;yy

= �(x; y) in [0; d

x

], [0; d

y

]

The periodic Green's function may be easily computed using the periodic

solvers developed in sections 2 and 3 (see Figure 2). Thus the solution to

the boundary source problem is given by the convolution

w = G(x; y)�X

1

(x; y) =

Z

�

1

X

0

(p; q) G(x�p; y�q) d�

1

; (p; q) � �

1

(21)

In order to discretize equation (21), consider a �nite number of points

(p

j

; q

j

) j = 1; 2; 3; ::: n

�

on the curve �

1

, where n

�

is the number of mesh points lying on the

boundary �. Thus

w(x; y) =

P

j

X

0

(p

j

; q

j

) G(x� p

j

; y � q

j

)

or, simply

w(x; y) =

X

j

X

j

G(x� p

j

; y � q

j

) (22)

Now, the values of w(x; y) on the boundary � are known from the boundary

conditions and the solution to the periodic problem. Let

(x

i

; y

i

) i = 1; 2; 3; ::: n

�

be the set of mesh points lying on �. Then, from equation (20)

w

i

� w(x

i

; y

i

) = u

�

(x

i

; y

i

)� v(x

i

; y

i

) i = 1; 2; 3; ::: n

�

Equation (22) may now be written for each point (x

i

; y

i

) on the boundary:

w

i

=

X

j

X

j

G(x

i

� p

j

; y

i

� q

j

) i = 1; 2; 3; ::: n

�

(23)

In matrix form, this becomes

2

6

6

6

6

6

6

6

6

6

6

6

4

w

1

w

2

w

3

:::

:::

w

n

�

�1

w

n

�

3

7

7

7

7

7

7

7

7

7

7

7

5

=

2

6

6

6

6

6

6

6

6

6

6

6

4

G

11

G

12

G

13

::: G

1 n

�

G

21

G

22

G

23

::: G

2 n

�

G

31

G

32

G

33

::: G

3 n

�

::: ::: ::: ::: :::

::: ::: ::: ::: :::

G

n

�

�1 1

G

n

�

�1 2

G

n

�

�1 3

::: G

n

�

�1 n

�

G

n

�

1

G

n

�

2

G

n

�

3

::: G

n

�

n

�

3

7

7

7

7

7

7

7

7

7

7

7

5

2

6

6

6

6

6

6

6

6

6

6

6

4

X

1

X

2

X

3

:::

:::

X

n

�

�1

X

n

�

3

7

7

7

7

7

7

7

7

7

7

7

5

where

G

ij

= G(x

i

� p

j

; y

i

� q

j

)

Solving this set of equations yields the values of X

j

, which may then be

substituted into equation (22) to obtain w. The solution u to given di�er-

ential equation is then obtained from equation (20).

The choice of the curve �

1

is a matter of critical importance. Qian

and Weiss [10] have shown that �

1

should be o�set from the boundary �

by at least N mesh points (where N is the support of the scaling function)

in order to control the residual error arising from the �nite support of

the delta function in wavelet space. The support of the delta function in

wavelet space may be shown to be the same as that of the scaling function

by taking the inner product of the expansion

�(x; y) =

P

k

P

l

g

k;l

�(X � k)�(Y � l)

with each of the basis functions �(X � i)�(Y � j). This gives

g

i;j

= 2

2m

�(�i)�(�j)

from which the support of g

i;j

is N in both coordinate directions.

5 Convergence and computation time

The following problem was analyzed on a DEC 5000 workstation using the

�nite di�erence method as well as the wavelet method with the Daubechies

d10 wavelet:

u

;xx

+ u

;yy

= �

751

144

�

2

sin(

�

6

x) sin(

7�

4

x) sin(

3�

4

y) sin(

5�

4

y) +

7

12

�

2

cos(

�

6

x) cos(

7�

4

x) sin(

3�

4

y)sin(

5�

4

y) +

15

8

�

2

sin(

�

6

x) sin(

7�

4

x) cos(

3�

4

y) cos(

5�

4

y)

with

u(x; 1) = �

1

2

sin(

�

6

x) sin(

7�

4

x)

u(1; y) = �

1

2

p

2

sin(

3�

4

y) sin(

5�

4

y)

u(x; 2) = � sin(

�

6

x) sin(

7�

4

x)

u(2; y) = �

p

3

2

sin(

3�

4

y) sin(

5�

4

y)

The exact solution to this problem is:

u = sin(

�

6

x) sin(

7�

4

x) sin(

3�

4

y) sin(

5�

4

y)

This solution is depicted in Figure 3.

Figure 4 shows the decay of the maximumresidual error with increas-

ing sample size n. The �gure clearly indicates the high rate of convergence

that is obtained with the wavelet method.

Figure 5 indicates the variation of computation time, in seconds, with

increasing sample size. The wavelet solution takes slightly longer than the

�nite di�erence solution owing to the need to transform the sample from

physical space into wavelet space and back again. This overhead becomes

less signi�cant as the sample size increases. From these results it can be

seen that the wavelet solution compares extremely favourably with the

�nite di�erence solution.

Figure 3: Solution to the test problem

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

3 3.5 4 4.5 5 5.5 6

log2(n)

log1

0(er

ror)

fd - d10 --

Figure 4: Decay in error of d10 wavelet and �nite di�erence solutions with

increasing sample size

-0.5

0

0.5

1

1.5

2

2.5

3 3.5 4 4.5 5 5.5 6

log2(n)

log1

0(tim

e)

fd - d10 --

Figure 5: Variation of computation times with increasing sample size

6 Applications of the wavelet method

The wavelet solution method for partial di�erential equations has obvious

practical applications in engineering, such as in the static and dynamic

analysis of structures and the solution of the heat equation. In engineering

problems, we often require a quick rough estimate of the solution at the

preliminary stage, which may later be re�ned as the design or investigation

progresses. Wavelets have the capability of providing a multilevel descrip-

tion of the solution (see S. Mallat's work on the application of wavelets to

multiresolution signal decomposition [3]). The multiresolution property of

wavelets, along with their orthogonality and localization properties, means

that we may obtain an initial coarse description of the solution with little

computational e�ort and then successively re�ne the solution in regions of

interest with a minimum of extra e�ort. The problem of successive re�ne-

ment is one of the main drawbacks of the �nite element method. In fact,

wavelets are applicable to any problem that can be solved using �nite ele-

ments. Preliminary research indicates that wavelets are a strong contender

to �nite elements, however, further research is still required in this area.

7 Conclusions

The wavelet method has been shown to be a powerful numerical tool for

the fast and accurate solution of partial di�erential equations. The proce-

dure described here shows that the solution to the di�erential equation is

related to the equation's right hand side by a sequence of discrete convo-

lutions which can be rapidly performed using the Fast Fourier Transform.

Although the FFT implies that the solution is periodic, we may incorporate

non periodic boundary conditions using the periodic Green's function. So-

lutions obtained using the wavelet method have been compared with those

obtained using the �nite di�erence method and the wavelet solutions have

been found to converge much faster than the �nite di�erence solutions (see

also Qian and Weiss [10]). Although the wavelet solution requires slightly

more computational e�ort than the �nite di�erence solution, the gains in

accuracy, particularly with higher order wavelets, far outweighs the in-

crease in cost. Furthermore, wavelets have the capability of representing

solutions at di�erent levels of resolution, which makes them particularly

useful for developing hierarchical solutions to engineering problems.

8 Acknowledgement

This work was funded by a grant from NTT DATA Communications Sys-

tems Corporation, Kajima Corporation and Shimizu Corporation to the In-

telligent Engineering Systems Laboratory, Massachusetts Institute of Tech-

nology.

References

[1] I. Daubechies, `Orthonormal bases of compactly supported wavelets',

Comm. Pure and Appl. Math., 41, 909-996 (1988).

[2] W. Proskurowski and O. Widlund, `On the numerical solution of

Helmholtz's equation by the capacitance matrixmethod',Math. Comp.,

30, 135, 433-468 (1976).

[3] S. G. Mallat, `A theory for multiresolution signal decomposition: the

wavelet representation', Comm. Pure and Appl. Math., 41, 7, 674-693

(1988).

[4] J. R. Williams and K. Amaratunga, `Introduction to wavelets in en-

gineering', IESL Technical Report No. 92-07, Intelligent Engineering

Systems Laboratory, M.I.T., October 1992.

[5] G. Strang, Introduction to Applied Mathematics, Wellesley-Cambridge

Press, Wellesley, MA, 1986.

[6] R. Glowinski, W. Lawton, M. Ravachol and E. Tenenbaum, `Wavelet

solution of linear and nonlinear elliptic, parabolic and hyperpolic prob-

lems in one space dimension', proc. 9th International Conference on Nu-

merical Methods in Applied Sciences and Engineering, SIAM, Philadel-

phia (1990).

[7] A. Latto, H. Resniko� and E. Tenenbaum, `The evaluation of connec-

tion coe�cients of compactly supported wavelets', to appear in proc.

French - USA workshop on Wavelets and Turbulence, Princeton Univ.,

June 1991, Springer-Verlag, 1992.

[8] A. Latto and E. Tenenbaum, `Les ondelettes a support compact et la

solution numerique de l'equation de Burgers', C. R. Acad. Sci. Paris,

311, 903-909 (1990).

[9] J. Weiss, `Wavelets and the study of two dimensional turbulence', proc.

French - USA workshop on Wavelets and Turbulence, Princeton Univ.,

June 1991, Y. Maday, Ed. Springer-Verlag, NY.

[10] S. Qian and J. Weiss, `Wavelets and the numerical solution of partial

di�erential equations', Aware Technical Report AD920318 and submit-

ted for publication to J. Comp. Phys., March 1992.

[11] K. Amaratunga, J. Williams and S. Yokoyama, `Wavelet based hier-

archical solutions of partial di�erential equations', proc. Complas III,

Third International Conference on Computational Plasticity, Funda-

mentals and Applications, Barcelona, Spain, April 1992.

[12] K. Amaratunga, J. R. Williams, S. Qian and J. Weiss, `Wavelet-

Galerkin solutions for one dimensional partial di�erential equations',

IESL Technical Report No. 92-05, Intelligent Engineering Systems Lab-

oratory, M.I.T., September 1992.

[13] G. Strang, `Wavelets and dilation equations: a brief introduction',

SIAM Review, 31, 4, 614-627 (1989).