Wavelets trong Electromagnetics và mô hình thiết bị P4 potx

CHAPTER FOUR
Wavelets in Boundary
Integral Equations
Numerical treatment of integral equations can be found in classic books [1, 2]. In
this chapter the integral equations obtained from field analysis of electromagnetic
wave scattering, radiating, and guiding problems are solved by the wavelet expansion
method [3–7]. The integral equations are converted into a system of linear algebraic
equations. The subsectional bases, namely the pulses or piecewise sinusoidal (PWS)
modes, are replaced by a set of orthogonal wavelets. In the numerical example we
demonstrate that while the PWS basis yields a full matrix, the wavelet expansion
results in a nearly diagonal or nearly block-diagonal matrix; both approaches re-
sult in very close answers. However, as the geometry of the problem becomes more
complicated, and consequently the resulting matrix size increases greatly, the ad-
vantages of having a nearly diagonal matrix over a full matrix will become more
profound.
4.1 WAVELETS IN ELECTROMAGNETICS
Galerkin’s method is a zero residual method if the basis functions are orthogonal and
complete, and thus Galerkin’s method with orthogonal basis functions is generally
more accurate and rapidly convergent. Two types of orthogonal basis functions are
frequently utilized for electromagnetic field computation. Mode expansion method
(or mode-matching method) has often been applied to solve problems due to various
discontinuities in waveguides, finlines, and microstrip lines. Generally, this technique
is useful when the geometry of the structure can be identified as consisting of two
or more regions, which each belongings to a separable coordinate system. The basic
idea in the mode expansion procedure is to expand the unknown fields in the indi-
vidual regions in terms of their respective normal modes. In fact the mode expansion
method is identical to Galerkin’s method which uses the normal mode functions as
100
Wavelets in Electromagnetics and Device Modeling. George W. Pan
Copyright
¶ 2003 John Wiley & Sons, Inc.

ISBN: 0-471-41901-X
WAVELETS IN ELECTROMAGNETICS 101
the basis functions. Quite often the normal modes are made of the classical orthogo-
nal series systems such as trigonometric, Legendre, Bessel, Hermite, and Chebyshev.
Owing to the orthogonality of the normal modes, a sparse system of linear algebraic
equations is expected to be generated by the mode expansion method. For general
cases of arbitrary geometries and material distributions, however, the mode functions
are often too difficult to be constructed.
The second class of orthogonal basis functions consists of a group of subsectional
bases, each of which is defined only in a given subsection of the solution domain.
An advantage of the subsectional bases is the localization property, that is, each of
the expansion coefficients affects the approximation of the unknown function only
over a subdomain of the region of interest. Thus, often not only does this class of
computations simplify the computation, but it also leads easily to convergent solu-
tions. In the subsectional basis systems, generally, only partial orthogonality can be
attained; only the pair of bases whose supporting regions do not overlap are orthog-
onal. Moreover the higher the continuity order of the constructed bases is rendered,
the larger the required supporting region. Hence there exists a trade-off between the
orthogonality and continuity for the subsectional basis systems.
Even if complete orthogonal bases with higher-order continuity are hard to build,
the subsectional bases with certain continuity order can be constructed widely (e.g.
by using polynomial interpolation functions). The finite element method, which has
been universally applied in engineering, is a subsectional basis method. So is the
boundary element method. Because of the kind of orthogonality, or, say, localization
that exists in subsectional basis systems, the differential operator equations may yield
sparse systems of linear algebraic equations by using subsectional bases. However,
it is also noted that the subsectional basis systems do not necessarily convert the
integral operator equations into sparse systems of linear algebraic equations.
Orthogonal wavelets have several properties that are fascinating for electromag-
netic field computations. First, wavelets are sets of orthonormal bases of L

2
(R).
They are problem-independent orthogonal bases and thus are suitable for numeri-
cal computations for general cases. Second, the trade-off between orthogonality and
continuity is well balanced in orthogonal wavelet systems because now the orthog-
onality always holds, whether the supporting regions are overlapping or not. One
can build an orthogonal wavelet system with any order of regularity, expecting larger
supporting regions as higher orders of regularity are selected. Third, in addition to
the advantages of the traditional orthogonal basis systems, orthogonal wavelets have
a cancellation property such that they are much more certain to yield sparse systems
of linear algebraic equations.
Furthermore orthogonal wavelets have localization properties in both the spatial
and spectral domains. Therefore the decorrelation of the expansion coefficients oc-
curs both in the space and Fourier domains. Nevertheless, according to the theory
of multigrid processing, one can improve convergence by operating on both fine
and coarse grids to reduce both the “high-frequency” and “low-frequency” compo-
nent errors between the approximate and exact solutions. In contrast, the traditional
way of operating only on fine grids reduces only the “high-frequency” component.
The expansion with subsectional bases actually is equivalent to the expansion on
102 WAVELETS IN BOUNDARY INTEGRAL EQUATIONS
the finest scale only (in fact, the pulse function is equivalent to the scalet of Haar’s
bases). On the contrary, the multiresolution analysis implemented by wavelet ex-
pansion provides a multigrid method. Finally, the pyramid scheme employed in the
wavelet analysis provides fast algorithms.
4.2 LINEAR OPERATORS
Functional spaces and linear operators were presented systematically and rigorously
in Chapter 1. In this section we will only quote the minimum prerequisite knowledge
for the method of moment applications.
I
NNER PRODUCT  f, g. An inner product  f, g on a complex linear space is a

complex-valued scalar satisfying
 f, g=
g, f 
α f +βg, h=
α f, h+βg, h
 f, f =f 
2

> 0iff = 0
= 0iff = 0,
where the overbar denotes the complex conjugate.
O
PERATOR L. The linear operator L and its corresponding equation are given as
Lf = g.
For instance, the Poisson equation is
− 
2
φ = ρ,
where the linear operator
L =− 
2
.
The adjoint L
a
is defined by
Lf, g=f, L
a
g.
An adjoint operator is self-adjoint if L
a

= L.
The inverse operator of L is denoted as L
−1
. For instance, the formal solution to
(4.3.1) is
f = L
−1
g.
In numerical computations we use a matrix to represent a linear operator.
METHOD OF MOMENTS (MoM) 103
4.3 METHOD OF MOMENTS (MoM)
Consider an operator equation
Lf = g, (4.3.1)
where L is a linear operator, f is the unknown function, and g is a given excitation.
We first expand the unknown function f (x ) in terms of the basis functions f
n
(x)
with unknown coefficients α
n
, namely
f =

n
α
n
f
n
.
Thus
L


n
α
n
f
n
= g.
Multiplying both sides of (4.3.1) by the weighting (testing) function w
m
and taking
the inner product ·, ·, we obtain

n
α
n
w
m
, Lf
n
=w
m
, g.
In matrix form, it appears as
[l
mn
]|α=|g, (4.3.2)
where
|α=






α
1
α
2
.
.
.
α
N





N ×1
←− unknown,
[l
mn
]=


w
1
, Lf
1
, w
1

, Lf
2
 ···
w
2
, Lf
1
, w
2
, Lf
2
 ···
···


N ×N
←− evaluated,
and
|g=



w
1
, g
w
2
, g
.
.

.



N ×1
.
Formally, equation (4.3.2) is solved to yield
|α=[l
mn
]
−1
|g.
104 WAVELETS IN BOUNDARY INTEGRAL EQUATIONS
There are two kinds of popular schemes in the method of moments:
(i) Pulse-delta scheme. Basis functions f
n
= pulse functions, and testing func-
tion
w
m
= δ(x
m
− x), Dirac delta function (point matching).
(ii) Galerkin scheme w
m
= f
m
.
The pulse-delta scheme is equivalent to the rectangular rule in the numerical integra-
tion, and the Galerkin scheme is a zero residual method.

Example Charged Conducting Plate (zero thickness).
A charged plate is de-
picted in Fig. 4.1. Find the charge distribution.
Solution
The electrostatic potential at any point (x, y, z) in space is given by
V (x , y, z) =

a
−a
dx


a
−a
dy

σ(x

, y

)
4π|r − r

|
with the unknown charge density σ(x

, y

). An integral equation of the first kind is
then formulated as

V (r) =

v
G(r, r

)σ (r

)d
3
r

,
where the Green’s function is G(r, r

) = 1/4π|r − r

|, and the potential on the
plate surface is a constant V.
0
0.2
0.4
0.6
0.8
1
0
0.2
0.4
0.6
0.8
1

0
0.5
1
1.5
2
2.5
x 10
−3
FIGURE 4.1 Charge distribution q/ on a square plate.
METHOD OF MOMENTS (MoM) 105
The capacitance of the plate can be found by
C =
q
V
=
1
V

a
−a
dx

a
−a
dyσ(x, y).
Therefore the main problem is to solve σ(x

, y

) of the integral equation by the MoM,

namely

a
−a
G(r, r

)σ (r

) ds

= V ,
where G(r, r

) is also called the integral kernel. We present this example because of
its simple Green’s function, and clear physical meaning. The numerical procedures
may be outlined in the following steps:
(i) Define the basis functions to be the pulse functions
f
n
(x

, y

) =

1,(x

, y

)on S

n
0, on all other S
m
, m = n,
that is, α
n
applies only on S
n
.
(ii) Approximate the charge by
σ(x, y) ≈
N

n=1
α
n
f
n
.
(iii) Convert the operator equation into a matrix equation. The operator equation
is
Lσ = V,
or in the explicit form

G(r, r

)


n

α
n
f
n

ds

= V .
Applying the linearity of the operator, we obtain

n
α
n

S
n
f
n
dx

dy

4π

(x − x

)
2
+ (y − y


)
2
+ (z − z

)
2
= V (x, y).
Take the inner product of the equation above with the testing function
w
m
(x, y) = δ(x
m
− x)δ(y
m
− y),
where (x
m
, y
m
) is the midpoint of the patch S
m
. The corresponding system
of equations is formed with
106 WAVELETS IN BOUNDARY INTEGRAL EQUATIONS
RHS =w
m
, V =

V (x , y)δ(x
m

− x)δ(y
m
− y) dx dy
= V (x
m
, y
m
)
and
LHS =

n
α
n

S
n
dx

dy


dx dy
δ(x
m
− x)δ(y
m
− y)
4π


(x − x

)
2
+ (y − y

)
2
=

n
α
n

S
n
dx

dy

1
4π

(x
m
− x

)
2
+ (y

m
− y

)
2
.
Thus a matrix equation converted from the operator equation is





l
11
l
12
··· l
1N
l
21
l
22
··· l
2N
.
.
.
l
N 1
l

N 2
··· l
NN










α
1
α
2
.
.
.
α
N





=






V (x
1
, y
1
)
V (x
2
, y
2
)
.
.
.
V (x
N
, y
N
)





.
In choosing pulse basis functions, the charge is assumed to be a constant over
a subarea (patch), namely
l

mn
=

S
n
dx

dy

1
4π

(x
m
− x

)
2
+ (y
m
− y

)
2
.
(iv) In handling integral equations, singularity occurs when the field point (x, y),
in this case (x
m
, y
m

), lies with in the domain of integration, S
n
. For the di-
agonal element, l
mn
(m = n), the integrand experiences a singularity which
must be treated carefully. Analytical removal, pole extraction by an asymp-
tote, and folding technique are among the popular methods for handling sin-
gularities [8, 9, 10]. In the present case the method of analytic removal is
applicable:
l
11
=

x
1
+b
x
1
−b
dx

y
1
+b
y
1
−b
dy
1

4π

(x − x
1
)
2
+ (y − y
1
)
2
=

b
−b
dx

b
−b
dy
1
4π

x
2
+ y
2
=
b
4π


1
−1
du

1
−1
dv
1
√
u
2
+ v
2
,
where

1
−1
du

dv
1
√
u
2
+ v
2
= 8

π/4

θ=0

1/ cos θ
ρ=0
ρ dρ dθ
ρ
FUNCTIONAL EXPANSION OF A GIVEN FUNCTION 107
= 8

π/4
0
d sin θ
cos
2
θ
= 8

1/
√
2
0
dτ
1 − τ
2
= 8 ·
1
2

1/
√

2
0

1
1 − τ
+
1
1 + τ

dτ
= 8ln

3 + 2
√
2 = 8ln(1 +
√
2).
The numerical solution of the charge distribution on the plate is depicted in Fig. 4.1.
4.4 FUNCTIONAL EXPANSION OF A GIVEN FUNCTION
In the previous section the MoM was briefly discussed. The MoM is a powerful
numerical algorithm, and it has been employed to solve electromagnetic problems
for a half-century. Unfortunately, the MoM matrix is full. With the help of wavelets,
one can obtain sparse impedance matrices.
We begin with the expansion of a given function in the wavelet bases. It is easier
to expand a given function in a wavelet basis than to expand an unknown function
in wavelets while solving the corresponding integral equation by the method of mo-
ments. The experience we gain here will be applied in the wavelet-based MoM.
From the multiresolution analysis (MRA), the nested subspaces can be decom-
posed as
V

m+1
= W
m
⊕ V
m
= W
m
⊕ W
m−1
+ V
m−1
= W
m−1
⊕ W
m−2
⊕ W
m−3
⊕···
and
⊕
m∈Z
W
m
= L
2
(R).
Therefore {ψ
m,n
}
m,n∈Z

is an orthonormal (o.n.) basis of L
2
(R).Forall f (x ) ∈
L
2
(R),wehave
f (x) =

m,n
 f (x), ψ
m,n
(x)ψ
m,n
(x).
In practice, we can only approximate a given physical phenomenon with finite pre-
cision. Mathematically the approximation is to project a function from the L
2
onto a
subspace V
m+1
= V
m
⊕ W
m
, namely
f (x)  A
m+1
f (x) :=

n

s
m+1
n
ϕ
m+1,n
(x),
108 WAVELETS IN BOUNDARY INTEGRAL EQUATIONS
where s
m+1
n
=f (x), ϕ
m+1,n
, A
m+1
f (x) is the approximation of f (x) at resolution
level 2
m+1
and A
m+1
is the projection operator. As m →∞,
A
m
f (x) = f (x).
Since
V
m+1
= V
m
⊕ W
m

,
it follows that
f (x)  A
m+1
f (x) = A
m
f (x) + B
m
f (x),
where
B
m
f (x) =

n
d
m
n
ψ
m,n
(x)
and
d
m
n
=f (x), ψ
m,n
(x).
Continuing the process, we obtain
A

m+1
f (x) = A
m
1
f (x) +
m

m

=m
1
B
m

f (x), (4.4.1)
where m
1
is a prescribed number, representing the lowest resolution level.
Example
Expand f (x) in terms of Daubechies wavelets N = 2, where f (x) =
1 −|x | for |x |≤1.
Solution
The function f (x) is defined on [−1, 1], but the Daubechies are with
supp {ϕ}=[0, 3] and supp {ψ}=[−1, 2]. Therefore we cannot use ϕ
0,0
nor ψ
0,0
because they are too wide. Let us choose
f (x) ∼ f
4

∈ V
4
.
By the MRA
V
4
= V
2
⊕ W
2
⊕ W
3
,
where the lowest resolution level m
1
= 2. Thus
f (x) ∼

n
 f (x), ϕ
2,n
(x)ϕ
2,n
(x)
+

p
 f (x), ψ
2, p
(x)ψ

2, p
(x) +

k
 f (x), ψ
3,k
(x)ψ
3,k
(x),
where
ϕ
j,n
(x) = 2
j/2
ϕ(2
j
x −n),
ψ
j,k
(x) = 2
j/2
ψ(2
j
x −k).
FUNCTIONAL EXPANSION OF A GIVEN FUNCTION 109
From the supports of supp {ϕ}, supp {ψ} and the scale, we have
supp {ϕ
2,0
(x)}=


0,
3
4

supp {ψ
2,0
(x)}=

−
1
4
,
1
2

supp {ψ
3,0
(x)}=

−
1
8
,
1
4

.
It can be easily verified:
(1) For j = 2,ϕ
2,−6

(x) is the leftmost scalet that intercepts −1and
supp {ϕ
2,−6
(x)}=

−
6
4
, −
3
4

.
Note that ϕ
2,−5
(x) also intersects −1, with supp {ϕ
2,−5
(x)}=[−5/4, −2/4].
However, it is only next to the leftmost scalet. We find n =−6 by substituting
n into
2
2
x −(−n) = 0 (left edge),
2
2
x −(−n) = 3 (right edge).
The integer n is selected such that x solved from the right edge equation is just
on the right of −1. When n =−7 is used in the left and right edge equations,
the resultant interval does not intersect −1. In the same way, the rightmost
scalet that intercepts 1 is found as ϕ

2,3
(x) and supp {ϕ
2,3
(x)}=[3/4, 6/4].
(2) For j = 3, the leftmost ψ
3,n
(x) that intercepts −1isn =−9. In fact
supp {ψ
3,−9
(x)}=[−10/8, −7/8], as solved from
2
3
x −(−9) =−1 ⇒ x =−
10
8
,
2
3
x −(−9) = 2 ⇒ x =−
7
8
.
In the same manner, the rightmost ψ
3,n
that intercepts 1 is n = 8.
(3) For j = 2, ψ
2,−5
(x) is the leftmost basis that intercepts −1and
supp {ψ
2,−5

(x)}=

−
6
4
, −
3
4

.
The rightmost basis that intercepts 1 is ψ
2,4
and supp {ψ
2,4
(x)}=[3/4, 6/4].
In conclusion,
f (x) ≈
3

n=−6
 f,ϕ
2,n
ϕ
2,n
(x) +
4

m=−5
 f,ψ
2,m

ψ
2,m
(x) +
8

k=−9
 f,ψ
3,k
ψ
3,k
(x).
110 WAVELETS IN BOUNDARY INTEGRAL EQUATIONS
approach by Daubechies Wavelet
1.5
1.0
0.5
0.0
−
0.5
−1.0 −
0.5 0.0 0.5
1
.0
f(x)
f(x)=1−|x|,
order = 2, lowest = 2, highest = 4
FIGURE 4.2 Expansion of f (x) in Daubechies wavelets of N = 2.
Figure 4.2 depicts the function f (x) and its wavelet expansion in V
4
. It can be seen

that the two agree well.
4.5 OPERATOR EXPANSION: NONSTANDARD FORM
We solve integral equations and expand a 1D function f (x

) in terms of wavelet basis
functions by (4.4.1). We substitute this expansion into the integral equation, and then
test using Galerkin’s procedure for the unprimed variable x. The corresponding ma-
trix represents an approximation of the operator. In most cases only linear operators
are discussed. An integral operator T is given as
(Tf)(x) =

K (x, y) f (y) dy, (4.5.1)
where K (x, y) is the integral kernel. For instance, the integral equation

G(x, x

)σ (x

) dx

= V (x)
is that for a 1D problem σ(x

) with a 2D kernel K (x, y) = G(x, x

).
OPERATOR EXPANSION: NONSTANDARD FORM 111
4.5.1 Operator Expansion in Haar Wavelets
Expanding the kernel into a two-dimensional Haar series, we have
K (x, y) =


I,I

α
II

ψ
I
(x)ψ
I

(y) +

I,I

β
II

ψ
I
(x)ϕ
I

(y)
+

I,I

γ
II


ϕ
I
(x)ψ
I

(y), (4.5.2)
where
α
II

=

K (x, y)ψ
I
(x)ψ
I

(y) dx dy,
β
II

=

K (x, y)ψ
I
(x)ϕ
I

(y) dx dy, (4.5.3)

γ
II

=

K (x, y)ϕ
I
(x)ψ
I

(y) dx dy.
The previous expansion may be classified as two categories: standard form and non-
standard form.
C
ASE 1. STANDARD FORM
I = I
−j,k
, I
−j,k
=[2
j
k, 2
j
(k + 1)],
I

= I
−j

,k


, I
−j

,k

=[2
j

k, 2
j

(k

+ 1)].
Note that the combination of II

in (4.5.2) and (4.5.3) experiences all possible levels,
with j = j

and j = j

.
C
ASE 2. NONSTANDARD FORM
I = I
−j,k
, I
−j,k
=[2

j
k, 2
j
(k + 1)],
I

= I
−j,k

(instead of I
−j

,k

).
In contrast to the standard form, only II

with equal levels appear in (4.5.2) and
(4.5.3).
In this section only the nonstandard form is studied. To simplify the notation, we
use
α
j
k,k

= α
I
j,k,
I
j,k


,
β
j
k,k

= β
I
j,k,
I
j,k

,
γ
j
k,k

= γ
I
j,k,
I
j,k

.
112 WAVELETS IN BOUNDARY INTEGRAL EQUATIONS
Equation (4.5.2) is referred to as the nonstandard form. Substituting (4.5.2) into
(4.5.1), we obtain
T ( f )(x) =

I

ψ
I
(x)

I

α
II


d
I

  
ψ
I

(y) f (y) dy
+

I
ψ
I
(x)

I

β
II



s
I

  
ϕ
I

(y) f (y) dy
+

I
ϕ
I
(x)

I

γ
II


d
I

  
ψ
I

(y) f (y) dy,

where I and I

always have the same length of [2
j
k, 2
j
(k +1)],andI = I
−j,k
, I

=
I
−j,k

are understood.
Define a projection operator
P
−j
f =

k
 f,ϕ
I
−j,k
ϕ
I
−j,k
, j = 0, 1, ,n.
The integral operator T is then approximated by T
0

, according to the projection P
0
of prespecified precision
Tf ∼ T
0
f = P
0
(T (P
0
f )) or T ∼ T
0
= P
0
TP
0
,
where the first P
0
represents testing and the second P
0
is for expansion. The non-
standard decomposition yields
P
0
TP
0
={(P
0
− P
−1

)T (
Q
−1
  
P
0
− P
−1
) + (P
0
− P
−1
)TP
−1
+ P
−1
T (P
0
− P
−1
)}+P
−1
TP
−1
={Q
−1
TQ
−1
+ Q
−1

TP
−1
+ P
−1
TQ
−1
}+P
−1
TP
−1
,
where the last term on the right-hand side is similar to the left-hand side, but is one
level down. The first equal mark in the previous operator equation can be verified by
a
2
= (a − b)
2
+ (a − b)b + b(a − b) + b
2
with a ↔ P
0
, b ↔ P
−1
, and no commutation is allowed.
Repeating this process, on the −1, −2, ,levels, we have
P
0
TP
0
=

n

j=1
(P
−j+1
TP
−j+1
− P
−j
TP
−j
) + P
−n
TP
−n
=
n

j=1
{P
−j+1
− P
−j
)T (
Q
−j
  
P
−j+1
− P

−j
) + (P
−j+1
− P
−j
)TP
−j
OPERATOR EXPANSION: NONSTANDARD FORM 113
+ P
−j
T (P
−j+1
− P
−j
)}+P
−n
TP
−n
···
=
n

j=1
[Q
−j
TQ
−j
+ Q
−j
TP

−j
+ P
−j
TQ
−j
]+P
−n
TP
−n
. (4.5.4)
Equation (4.5.4) is for decomposing T
0
into a summation of contributions from dif-
ferent levels of wavelets and scales referred to as the telescopic series. The formulas,
derived in this subsection based on Haar, apply to general wavelet systems.
4.5.2 Operator Expansion in General Wavelet Systems
We now readily to expand operators in general wavelet systems, either compactly or
infinitely supported. Let T be a linear operator
T : L
2
(R) → L
2
(R).
The projection operator
P
j
: L
2
(R) → V
j

provides that
(P
j
f )(x) =

k
 f,ϕ
j,k
ϕ
j,k
(x).
Expanding T in telescopic series, we obtain
T =

j∈Z
Q
j
TQ
j
+ Q
j
TP
j
+ P
j
TQ
j
,
where
Q

j
= P
j+1
− P
j
.
Let T be an integral operator
(Tf)(x ) =

K (x, x

) f (x

) dx

T ∼ T
j
= P
j
TP
j
,
then the kernel K (x, x

) can be expanded in a nonstandard form
K (x, x

) =
m
h−1


m=m
l

n,k

{α
m
n,k

ψ
m,n
(x)ψ
m,k

(x

) + β
m
n,k

ψ
m,n
(x)ϕ
m,k

(x

)
+ γ

m
n,k

ϕ
m,n
(x)ψ
m,k

(x

)}+

n,k

s
m
l
n,k

ϕ
m
l
,n
(x)ϕ
m
l
,k

(x


), (4.5.5)
114 WAVELETS IN BOUNDARY INTEGRAL EQUATIONS
where
α
m
n,k

=K

(x, x

), ψ
m,n
(x)ψ
m,k

(x

),
β
m
n,k

=K

(x, x

), ψ
m,n
(x)ϕ

m,k

(x

),
γ
m
n,k

=K

(x, x

), ϕ
m,n
(x)ψ
m,k

(x

),
s
m
l
n,k

=K

(x, x


), ϕ
m
l,n
(x)ϕ
m
l,k

(x

). (4.5.6)
This leads to a fast wavelet transform algorithm, which will be discussed later in
Section 4.8. For further in depth information, readers are referred to [3].
4.5.3 Numerical Example
A plane wave is impinging on a conducting screen with two slots as shown in Fig. 4.3,
where the two slots are at [0.1λ, 1.1λ] and [−0.1λ, −1.1λ]. Find the magnetic cur-
rent M = E ׈n on the slots [5].
Formulation
(1) Using the equivalence principle, we can close the slots with magnetic current
M =−ˆn × E.
(2) Applying image theory, we can put the image magnetic current, and then
remove the conducting plane
Region a
(µ
0
,ε
0
)
Region b
(µ
0

,ε
0
)
Double Slot L
Conducting Screen L
−
x
z
z=0
H
y
inc
FIGURE 4.3 Diffraction of two apertures.
OPERATOR EXPANSION: NONSTANDARD FORM 115
0.0 0.2 0.4 0.6 0.8 1.0 1.2
Slot width
0.0
5.0
10.0
15.0
20.0
Magnetic current magnitude
Wavelet solution
MoM solution
FIGURE 4.4 Magnitude of magnetic current.
M = M
+
+ M
−
= 2M

y
ˆy.
Equivalently, there is a source consisting of two magnetic current sheets in
free space.
(3) In the wave equations
∇×∇×E − k
2
E =−jωµJ −∇×M,
∇×∇×H − k
2
H =−jωM +∇×J,
duality theorem has been applied. We choose the second equation above for
the formulation, namely
∇×∇×H − k
2
H =−jωM.
Using a vector identity, we have
∇(∇·H) −∇
2
H − k
2
H =−jωM.
Because of the zero divergence of the magnetic field, we obtain
(∇
2
+ k
2
)H
s
= jωM

= j
ω
2
µ
ωµ
M
= j
k
η
M(r

),
116 WAVELETS IN BOUNDARY INTEGRAL EQUATIONS
where η =

µ

is the intrinsic impedance. Thus the formal solution of the
scattered field
H
s
=

G · j
k
η
M(r

)dr


,
where the dyadic Green’s function,
G, satisfies
∇×∇×
G + µ
∂
2
∂t
2
G =−I δ(r − r

).
In the frequency domain
G =

I +
∇∇
k
2

G
0
with
G
0
=−
1
4π|r − r

|

e
jk|r−r

|
.
For the scalar case we have
H
s
y
=
jk
η

V
(2M
y
)G
0
dv

.
For 2D structures the 2D free-space Green’s function is
G
0
=−
1
4 j
H
(2)
0

(k|
␳
−
␳

|).
It follows that
H
s
y
(x) =
−k
2η

H
(2)
0

−k

(x − x

)
2
+ (z − z

)
2

M

y
(x

) dx

=
−k
2η

H
(2)
0
(k|x − x

|)M
y
(x

) dx

.
Applying the boundary condition
H
in
y
(x) + H
s
y
(x)|
z=0

= 0,
we end with

H
(2)
0
(k|x − x

|)M
y
(x

) dx

=
η
2k
H
in
(x).
OPERATOR EXPANSION: NONSTANDARD FORM 117
Using x = λu, normalized by wavelength, we arrive at

H
(2)
0
(kλ|u − u

|)M
y

(u

)λ du

=
η
2k
H
in
(u)
or

H
(2)
0
(2π|u − u

|)M
y
(u

) du

=
η
π
H
in
(u).
For convenience, we use x instead of u


L
H
(2)
0
(2π| x − x

|)M
y
(x

) dx

=
η
π
H
in
y
(x), (4.5.7)
where x has been normalized by wavelength λ,and
L =[−1.1, −0.1]

[0.1, 1.1].
Edge Treatment.
In boundary value problems edges must be properly handled, or
else the solution can have a nonphysical meaning. If the edges are not treated, the
solution is oscillatory in nature, and this behavior is inaccurate.
Consider the LHS of (4.5.7)
LHS =



b
a
+

d
c

(H
(2)
0
M
y
(x

) dx

)
=

d
c
[H
(2)
0
(2π| x − x

|) + H
(2)

0
(2π| x + x

|)]M
y
(x

) dx

,
where a =−1.1, b =−0.1, c = 0.1, and d = 1.1.
First, let us take a close look at the integral

d
c
. Near the right edge of the slot,
some of the basis functions, say ϕ
j,n
, may not be completely supported in the interval
(c, d). There seem to be two choices for us.
(1) Chop off ϕ
j,n
(x) for the portion x > d, (denoted as ϕ
c
j,n
(x)). However, this
basis is incomplete. Therefore we have destroyed the orthogonality, and

ϕ
c

j,n
(x)ϕ
i,m
(x) dx = 0.
(2) Remove the incomplete basis functions from the expansion. However, the
interval will no longer be covered completely by the basis functions.
It turns out that neither of the previous ideas work. To solve the integral equations
on bounded intervals using wavelets as basis functions, the treatment of edges must
be carried out with caution. There are several techniques, and we list the most com-
monly used ones below:
118 WAVELETS IN BOUNDARY INTEGRAL EQUATIONS
•
Coordinate transformation.
•
Periodic wavelets.
•
Intervallic wavelets.
•
Weighted wavelets.
Here we apply coordinate transformation to
I =

b
a
f (x

)G(x, x

) dx


. (4.5.8)
Using the transform
x =
b −a
π
tan
−1
t +
b +a
2
,
we will map x :[a, b] to t : (−∞, +∞).
The Jacobian is
dx
dt
=
b − a
π
1
1 + t
2
.
As a result
I =
b −a
π

∞
−∞
f (x(t


))G(t, t

)
1
1 + t
2
dt

.
Since the wavelets are defined on the real line R as
⊕ W
m
= L
2
(R),
m∈Z
no edge exists in the transformed domain. Thus, in the transform domain t, we can
allocate the scalets and wavelets as much as we like since the interval is not bounded.
Physically, those basis functions of large t in magnitude are compressed in the orig-
inal physical coordinates. The rapidly varying M
y
near the two edges in physical
space has been stretched horizontally in the transform domain. Hence the expansion
approximates the function better. Figure 4.5 shows the basis functions in the original
coordinate system. It can be seen clearly that more basis functions are placed near
the two edges of the slot so that the singular behavior of the fields there is modeled
more precisely.
In the case where the incident wave is not normal to the screen, there will be
two integrals in the integral equation. The same transform can be applied to both of

them. How far should we put the wavelet bases in the t-axis? One could set up a stop
criterion in terms of the relative error of the consequent solutions with a different
number of wavelet bases.
OPERATOR EXPANSION: NONSTANDARD FORM 119
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1
0.4
0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
φ
0,0
→
x
Scaling functions of level J=0 in the x domain
FIGURE 4.5 Compressed Daubechies scalets as basis functions.
Matrix Equations.
We will discuss the problem in the physical space, although it
applies to the transform domain as well. Let us assume that the unknown function
f (x) ∈ L
2
(R) is projected to the highest resolution subspace as f
m
h
∈ V

m
h
:
f (x)

= M
y
(x) =
m
h−1

m=m
l

n
M
ψ
m,n
ψ
m,n
(x) +

n
M
ϕ
m
l
,n
ϕ
m

l
,n
(x)
=

n
(M
ψ
0,n
ψ
0,n
(x) + M
ϕ
0,n
ϕ
0,n
(x)) +

n
M
ψ
1,n
ψ
1,n
(x)
+

n
M
ψ

2,n
ψ
2,n
(x) +

n
M
ψ
3,n
ψ
3,n
(x), (4.5.9)
where M
ψ
m,n
, M
ϕ
m
l
,n
are unknown, and we use m
l
= 0, m
h
= 4 to be specific.
The Green’s function, according to Eq. (4.5.5), is
G(x, x

) =
m

h
−1

m=m
l

n,k

{α
m
n,k

ψ
m,n
(x)ψ
m,k

(x

)
+ β
m
n,k

ψ
m,n
(x)ϕ
m,k

(x


) + γ
m
n,k

ϕ
m,n
(x)ψ
m,k

(x

)}
+

n,k

s
m
l
n,k

ϕ
m
l
,n
(x)ϕ
m
l
,k


(x

). (4.5.10)
Substituting (4.5.9) and (4.5.10) into (4.5.8) and making the inner product with the
testing functions ϕ
m
l
,p
(x), ψ
m
l
,p
(x) and ψ
m, p
(x) according to the Galerkin proce-
dure, we obtain a set of system equations. To be more specific yet not too tedious,
120 WAVELETS IN BOUNDARY INTEGRAL EQUATIONS
we assume that m
l
= 0andm
h
= 2. For the case of

dxϕ
0, p
(x) testing, we arrive
at

dxϕ

m
l
,p
(x)

G(x, x

) f (x

) dx

=

dx dx

ϕ
0, p
(x)

n,k

{[α
0
n,k

ψ
0,n
(x)ψ
0,k


(x

)
+ β
0
n,k

ψ
0,n
(x)ϕ
0,k

(x

) + γ
0
n,k

ϕ
0,n
(x)ψ
0,k

(x

) + s
0
n,k

ϕ

0,n
(x)ϕ
0,k

(x

)
+ α
1
n,k

ψ
1,n
(x)ψ
1,k

(x

) + β
1
n,k

ψ
1,n
(x)ϕ
1,k

(x

) + γ

1
n,k

ϕ
1,n
(x)ψ
1,k

(x

)]}

q

[M
ϕ
0,n
ϕ
0,q

(x

) + M
ψ
0,n
ψ
0,q

(x


) + M
ψ
1,n
ψ
1,q

(x

)]
=

n,k

,q

α
0
n,k

[M
ϕ
0,n

dx

ψ
0,k

(x


)ϕ
0,q

(x

)

dxψ
0,n
(x)ϕ
0, p
(x)
+ M
ψ
0,n

dx

ψ
0,k

(x

)ψ
0,q

(x

)


dxψ
0,n
(x)ϕ
0, p
(x)
+ M
ψ
1,n

dx

ψ
0,k

(x

)ψ
1,q

(x

)

dxψ
0,n
(x)ϕ
0, p
(x)]+β
0
n,k


{
···,
where α, β, γ, s are pre-evaluated according to (4.5.6). The inner product of the
right-hand side with testing function ϕ
0, p
(x) will result in a complex number in
general. Thus we arrive at set of algebraic equations.
4.6 PERIODIC WAVELETS
4.6.1 Construction of Periodic Wavelets
Consider a periodic function with period 1, namely
f (x + 1) = f (x) ⇔ f (x − 1) = f (x).
Then, the wavelet coefficients on a given scale j
 f,ψ
j,k
=f,ψ
j,k+2
j
.
Show.
RHS =

f (x)2
j/2
ψ(2
j
x − k −2
j
) dx
=


f (x)ψ[2
j
(x −1) −k]2
j/2
dx
=

f (u + 1)ψ(2
j
u − k)2
j/2
du
PERIODIC WAVELETS 121
=

f (u)ψ(2
j
u − k)2
j/2
du
=f,ψ
j,k

= LHS.
Thus a periodic MRA on [0, 1] can be constructed by periodizing the basis functions
as
ϕ
p
j,k

=

∈Z
ϕ
j,k
(x +) for 0 ≤ k < 2
j
, j > 0,
ψ
p
j,k
=

∈Z
ψ
j,k
(x +) for 0 ≤ k < 2
j
, j > 0,
where superscript p stands for periodic.
The subspace V
p
j
has a dimension of 2
j
,and
V
p
j
= span{ϕ

p
j,k
, k ∈ Z }.
We do not consider the cases j < 0, meaning the stretched wavelets. For j ≤ 0, it
can be shown that
ϕ
p
j,k
(x) = 2
j/2


ϕ(2
j
x −k + 2
j
) = 2
−j/2
, constant, (4.6.1)
ψ
p
j,k
(x) = 2
j/2


ψ(2
j
x −k + 2
j

) = 0.
We prove (4.6.1) in two steps, using mathematical induction although other proofs
are also possible.
Proof. First, we show that (4.6.1) is held for j = 0, that is,

∞
=−∞
ϕ(x +) = 1.
LHS =
−∞

=∞
ϕ(x −)
=



n
h
n
√
2ϕ(2x − 2 − n)
=



m
√
2h
m−2

ϕ(2x −m)
=

m



√
2h
m−2

ϕ(2x − m), (4.6.2)
wherewehaveusedm = 2 + n ⇒ n = m − 2.
Since we had in Chapter 3,
ˆ
h(π ) = 0,
122 WAVELETS IN BOUNDARY INTEGRAL EQUATIONS
it follows that
0 =

h
n
√
2
e
−i
ω
2
n
|

ω=2π
=
1
√
2

(−1)
n
h
n
.
Therefore

n
h
2n
=

n
h
2n+1
.
From
ˆ
h(0) = 1, we obtain

n
h
n
=

√
2. As a result

n
h
2n
=

n
h
2n+1
=
1
√
2
.
Hence we have from (4.6.2),

l
ϕ(x −l) =

m
ϕ(2x −m) =

n
ϕ(4x −n) =···=const.
Since

∞
−∞

ϕ(x) dx = 1, the constant is necessarily equal to 1.
Second, we assume that
2
j/2


ϕ(2
j
x − k +2
j
) = 2
−j/2
, j < 0,
andwewishtoshowthat
2
( j−1)/2


ϕ(2
j−1
x − k +2
j−1
) = 2
−( j−1)/2
. (4.6.3)
Applying the dilation equation to (4.6.3), we have
LHS = 2
( j−1)/2




n
√
2h
n
ϕ(2
j
x − 2k +2
j
 − n)
= 2
j/2

n


h
n
ϕ(2
j
x + 2
j
 − 2k −n).
Let 2k + n = p,thenn = p − 2k. Thus
LHS =

p
h
p−2k



2
j/2
ϕ(2
j
x − p +2
j
)
=

p
h
p−2k
2
−j/2
.
The equality is achieved by the induction assumption. Hence
LHS = 2
−j/2

p
h
p−2k
= 2
−j/2
√
2
= 2
−( j−1)/2
.

PERIODIC WAVELETS 123
In summary
(1) For large j, the wavelets are greatly compressed within [0, 1]. Hence
ϕ
p
j,k
(x) = ϕ
j,k
(x).
(2) In contrast, for small enough j,ϕ
j,k
(x) is chopped into pieces of length 1,
which are shifted onto [0, 1] and added up, yielding the periodic wavelets.
The constructed periodic wavelets of Coifman, Daubechies, and Franklin are de-
picted in Figs. 4.6, 4.7, and 4.8.
4.6.2 Properties of Periodic Wavelets
It can be verified that ψ
p
j,k
,ϕ
p
j,k
form an orthonormal basis system possessing the
same MRA properties as the regular wavelets do, for example,
ψ
p
j,k
,ϕ
p
j,k

=

1
0
ψ
p
j,k
(x)ϕ
p
j,k
(x) dx = 0
ψ
p
j,k
,ψ
p
j,k

=δ
k,k

V
p
0
⊂ V
p
1
⊂ V
p
2

⊂···
W
p
j

V
p
j
= V
p
j+1
.
V
p
j
has 2
j
basis functions {ϕ
p
j,k
;k = 0, 1, ,2
j
− 1},andL
2
([0, 1]) contains the
following basis functions
(1) ϕ
p
0,0
(x) = 1

(2) ψ
0,0
(x)
(3) ψ
p
1,0
(x)
(4) ψ
p
1,1
(x)
(5) ψ
p
2,0
(x)
(6) ψ
p
2,1
(x)
(7) ψ
p
2,2
(x)
(8) ψ
p
2,3
(x)
Let us show that ψ
p
j,k

,ψ
p
j,k

=δ
k,k

.
Proof.
ψ
p
j,k
,ψ
p
j,k

=

1
0
dxψ
p
j,k
(x )ψ
p
j,k

(x )
=


1
0
dx


2
j/2
ψ(2
j
x + 2
j
 − k)



2
j/2
ψ(2
j
x + 2
j


− k

)
124 WAVELETS IN BOUNDARY INTEGRAL EQUATIONS
x
p
0,0

Magnitude
4.00
2.00
0.00
−2.00
0.00 0.20 0.40 0.60 0.80 1.00
(a)
x
p
0,0
Magnitude
4.00
2.00
0.00
−2.00
0.00 0.20 0.40 0.60 0.80 1.00
(b)
x
p
2,3
Magnitude
4.00
2.00
0.00
−2.00
0.00 0.20 0.40 0.60 0.80 1.00
(h)
x
p
2,2

Magnitude
4.00
2.00
0.00
−2.00
0.00 0.20 0.40 0.60 0.80 1.00
(g)
x
p
2,1
Magnitude
4.00
2.00
0.00
−2.00
0.00 0.20 0.40 0.60 0.80 1.00
(f)
x
p
2,0
Magnitude
4.00
2.00
0.00
−2.00
0.00 0.20 0.40 0.60 0.80 1.00
(e)
x
p
1,1

Magnitude
4.00
2.00
0.00
−2.00
0.00 0.20 0.40 0.60 0.80 1.00
(d)
x
p
1,0
Magnitude
4.00
2.00
0.00
−2.00
0.00 0.20 0.40 0.60 0.80 1.00
(c)
FIGURE 4.6 Periodic Coifman wavelets.