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TRANSCRIPT
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.
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