The Finite Difference Method for
the Helmholtz Equation with
Applications to Cloaking
Li Zhang
Introduction
•
In the past few years, scientists have made
great progress in the field of cloaking.
Cloaking involves making an object invisible or
undetectable to electromagnetic waves.
•
In this paper, we will use the numerical
solutions to visualize the effect that the
cloaking constructions have on a traveling
waves.
2
The Helmholtz Equation
• The wave equation
2w
2

c
w
2
t
(1)
models the propagation of a wave travelling
through a given medium at a constant speed
c.
2
2
 2  2
x
y
3
• If we assume the solution w is separable,
then we can write
w( x, y, t )  u( x, y)v(t )
Substituting this into (1) gives
 2v
u 2  c 2 vu
t
which can be rewritten as
1  v u

2
2
c v t
u
2
( 2)
4
• If we assume both sides are equal to the
constant  s 2 , then solving (1) reduces to
solving the two equations
  2v 2 2
 2 c s v 0
 t
 u  s 2u  0
(3)
( 4)
•
• Equation (3) has solutions of the form
v(t )  c1 cos(t )  c2 sin(t )
• where   cs
5
Approximating Derivatives
• To apply the finite difference method, we will
need to estimate the derivatives of a function.
• Then from Taylor’s formula,
h 2 ''
h3 '''
f ( x  h)  f ( x)  hf ( x) 
f ( x) 
f ( x)  
2!
3!
(5)
h 2 ''
h3 '''
f ( x  h)  f ( x)  hf ( x) 
f ( x) 
f ( x)  
2!
3!
( 6)
'
'
6
• Subtracting them and so we can approximate
the derivative of f using the formula
f ( x  h)  f ( x  h)
f ( x) 
2h
'
(7 )
• Furthermore , adding (5) and (6) gives the
approximate
f ( x  h)  2 f ( x )  f ( x  h)
f ' ' ( x) 
h2
(8)
7
In the similarly way, we can approximate the
partial derivatives of f as follows
u
u ( x  h, y )  u ( x  h, y )
( x, y ) 
x
2h
u
u ( x, y  k )  u ( x, y  k )
( x, y ) 
y
2k
 2u
u ( x  h, y )  2u ( x, y )  u ( x  h, y )
( x, y ) 
x 2
h2
 2u
u( x, y  k )  2u( x, y)  u( x, y  k )
(
x
,
y
)

y 2
k2
(9 )
(10)
(11)
(12)
8
The Finite Difference Method
• Let R  a, b c, d  be a rectangle in R 2 , and
consider the boundary value problem
2


u

s
u0
•
in R
(13)

u ( x, y )  f ( x, y ) on R
•
• Then apply the finite difference method to
solve this problem.
9
k2
r 2
h
Letting
,this we can get the following
five-point formula,
ru ( x  h, y)  ru ( x  h, y)  u( x, y  k )
 u( x, y  k )  (s k  2r  2)u( x, y)  0
2
2
10
Consider the boundary value problem
•
in R

 u  u  0x
u( x, y)  sin( )
•
on R

6

• Using a 70  70 lattice resulted in the solution
pictured in Figure (1)
11
• Similarly, the solution to
 u  u  0
•
in

u( x, y)   1
•
on

x2  y 2

• is pictured in Figure (2)
R
R
12
Transformations
• A divergence form operator acting on function
u  C 2 ( R2 )
• is a differential operator L of the form
Lu  div( Au)
r x
r y
F ( x, y )  ((1  ) , (1  ) )
2 r
2 r
• with r  x 2  y 2
13
Then for a differentiable function, define the matrix
•
 F1
 x ( x, y )
DF  
 F2 ( x, y )
 x
A( x, y ) 
F1

( x, y ) 
y

F2
( x, y ) 

y
1
det DF ( x, y )
where
F ( x, y)  ( F1 ( x, y), F2 ( x, y))
DF ( x, y)DF ( x, y)
T
( x , y )  F 1 ( x , y )
14
A Generalization of the Helmholtz
Equation
• In this section, we are interested in a general
Helmholtz equation of the form
2
Lu  s u  0
• Given a 2  2 matrix A, we wish to apply the
finite difference method to solve a boundary
value problem
 Lu  s 2u  0
•
in R

u ( x, y )  f ( x, y )
•
on R
15
• where R  a, b c, d  as before, and
• For simplicity, let  u  ux and  u  uy
• If we assume
2
1
 a11
A
a21
Lu  div( Au)
a12 
a22 
• we can write
 a11 a12   1u   a111u  a12  2u 
Au  





a
a

u
a

u

a

u
22 2 
 21 22   2   21 1
16
• Then we can express Lu as
2
2
Lu  div( Au )    i (aij  j u )
(14)
i 1 j 1
• In the same manner as before from the Taylor
series, we can show that
1
1 ( 2u )( x, y ) 
( 2u ( x  h, y )   2u ( x  h, y ))
2h

1 u ( x  h, y  k )  u ( x  h, y  k ) u ( x  h, y  k )  u ( x  h, y  k )
(

)
2h
2k
2k

u ( x  h, y  k )  u ( x  h, y  k )  u ( x  h, y  k )  u ( x  h, y  k )
4hk
(15)
17
• Utilizing (9)-(12)along with(15) in (14) allows
us to approximate the equation Lu  s 2u  0
• with the nine-point formula
(
(
a12  a21
a
b
a  a21
)u ( x  h, y  k )  ( 222  2 )u ( x, y  k )  ( 12
)u ( x  h, y  k ) 
4hk
k
2k
4hk
(
a11 b1
2a11 2a22
a11 b1
2

)
u
(
x

h
,
y
)

(
s


)
u
(
x
,
y
)

(
 )u ( x  h, y ) 
2
2
2
2
h
2h
h
k
h
2h
a12  a21
a
b
a  a21
)u ( x  h, y  k )  ( 222  2 )u ( x, y  k )  ( 12
)u ( x  h, y  k )  0
4hk
k
2k
4hk
• with
2
b j    i aij
i 1
18
Applications to Cloaking
• We use a specific example to solve the
boundary value problem.
• Consider the transformation
r x
r y
F ( x, y )  ((1  ) , (1  ) )
2 r
2 r
• where
r  x2  y2
19
• Using the transformation as discussed in last
section, we can calculate to conclude
 2  r x2
 3

r
DF ( x, y )   2r
  xy
r3




2  r y2 
 3
2r
r 
 r
(2r  1) x 2


3
r

1
(
r

1
)
r
A( x, y )  
  (2r  1) xy
3

(
r

1
)
r


xy
r3
(2r  1) xy 

(r  1)r 3 
r
(2r  1) y 3 

r  1 (r  1)r 3 

20
• For example, let R be the square  3,3  3,3
• Consider  Lu  u  0
u( x, y)  sin(x )

6

• Using the MATLAB program , the numerical
solution is shown in Figure (3)
21
• For the equation
 Lu  u  0

u( x, y)   1
2
2

x

y

• The numerical solutions are Figure (4).
22
• Compare Figure (1) with Figure (3)

 u  u  0x
u( x, y)  sin( )

6


 Lu  u  0x
u( x, y)  sin( )

6

23
• Figure (2) and Figure (4)
 u  u  0

u( x, y)   1
2
2

x

y

 Lu  u  0

u( x, y)   1
2
2

x

y

24
Indistinguishable waves
•
Two waves are indistinguishable, if their
boundary values and normal derivatives are
identical.
30  30 40  40 50  50 60  60 70  70 80  80
Figure(1)and Figure(3) 0.377
0.214
0.116
0.062
0.034
0.018
Figure(2)and Figure(4) 0.161
0.044
0.018
0.009
0.005
0.003
25
• We set up the initial value

 2v
v 0

2
t

v (0)  1, v (0)  0

t
26
27
Thank you!
28