South China University of Technology
Potential and Field
Xiaobao Yang
Department of Physics
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ODE and PDE
(1) initial-value problems, which involve timedependent equations with given initial conditions;
(2) boundary-value problems, which involve differential
equations with specified boundary conditions;
(3) eigenvalue problems, which involve solutions for
selected parameters (eigenvalues) in the equations.
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Invisible Cloak by Metamaterials
LIGHT
SOURCE
LIGHT
RAYS
OBJECT
METAMATERIAL
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Undetectable vs Invisible
隐形飞机
吸收掉而不是反射掉来自雷达的能量
谐振型 相消干涉和衰减结合
宽频带型 碳-耗能塑料材料
负折射
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Maxwell
Laplace
Poisson
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Maxwell Equations
Name
Differential form
Integral form
Gauss’s law:
Gauss’s law for magnetism:
Maxwell-Faraday equation
(Faraday’s law of induction):
Ampere’s circuital law
(with Maxwell's correction):
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Iterative method
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Laplace’s Equation
In a region containing no charges, the electronic potentials obeys:
 V
2
 V 
2
V
x
 V

2
x
2

x
 V
2

2
y
2
2

V ( i , j , k )  V ( i  1, j , k )
x
1
x
[
 V ( i  12 )
x

 V
z
, or
 V ( i  12 )
x
2
 0
V
x
]

V ( i  1, j , k )  V ( i  1, j , k )
2x
V ( i  1, j , k )  V ( i  1, j , k )  2(V ( i , j , k )
(x)
2
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Laplace’s Equation
 V
2
x

2
 V
2
y

2
 V
2
z
2

x  y  z
V ( i  1, j , k )  V ( i  1, j , k )  2V ( i , j , k )
(x)
2
V ( i , j  1, k )  V ( i , j  1, k )  2V ( i , j , k )
(y )
2
V ( i , j , k  1)  V ( i , j , k  1)  2V ( i , j , k )
 V
(z )
x
V (i , j , k ) 
1
2
2
2
 V
2

y
2
 V
2

z
2
 0
[V ( i  1, j , k )  V ( i  1, j , k )  V ( i , j  1, k )  V ( i , j  1, k )
6
 V ( i , j , k  1)  V ( i , j , k  1)]
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Laplace’s Equation
V (i , j , k ) 
1
[V ( i  1, j , k )  V ( i  1, j , k )  V ( i , j  1, k )  V ( i , j  1, k )
6
 V ( i , j , k  1)  V ( i , j , k  1)]
V0(i,j,k) V1(i,j,k)V2(i,j,k)…
Relaxation algorithm with boundary condition.
Why it works?
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Laplace’s Equation
Take it as a diffusion equation for a new function V :
V ( x , y , z , t )
t
 V
2
 D(
x
2
 V
2

y
2
 V
2

z
2
)
The iteration in the relaxation algorithm is considered as an
evolution of time for
V ( x, y, z, t   )
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Application of Laplace’s Equation
Example of application:
V=-1
y
+1
-1
V=+1
+1
x
-1
Solid lines: metallic wall,
Dashed lines: insulating wall
Z: infinite (2-D problem)
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Application of Laplace’s Equation
 V
2
x
 V
2

2
x
2

V ( i  1, j )  V ( i  1, j )  2V ( i , j )
(x)
V ( i , j  1)  V ( i , j  1)  2V ( i , j )
(x)
V (i , j ) 
1
2
x  y
2
 V
2
x
2
 V
2

y
2
 0
[V ( i  1, j )  V ( i  1, j )  V ( i , j  1)  V ( i , j  1)
4
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Application of Laplace’s Equation
Example of application:
y
V=-1
V=+1
x
Refer to code
“prismV.m”
Schematic cross-section of a hollow metallic
prism with a solid, metallic inner conductor.
Z direction is assumed infinite.
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Application of Laplace’s Equation
Example of application:
y
V=+1
V=-1
V=0
x
Schematic of two capacitor plates held at V=1
(left plate) and V=-1 (right plate) The square
boundary surronding the plates is held at V=0.
Plates and box are assumed to be infinite
along Z direction.
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Convergence: Jacobi vs Gauss-Seidel
L x L grid problem
►Jacobi method:
V n ew ( i , j ) 
1
4
[V o ld ( i  1, j )  V o ld ( i  1, j )  V o ld ( i , j  1)  V o ld ( i , j  1)]
Convergence scaled by L4.
One factor of L2 from the increase of required number of iterations,
while another factor of L2 arises from loops due to increased grid points.
►Gauss-Seidel method:
V new ( i , j ) 
1
4
[V old ( i  1, j )  V new ( i  1, j )  V old ( i , j  1)  V new ( i , j  1)
In the code it can be simplified as:
V (i , j ) 
1
[V ( i  1, j )  V ( i  1, j )  V ( i , j  1)  V ( i , j  1)
4
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Speed up convergence: simultaneous over-relaxation (SOR)
L x L grid problem
 V ( i , j )  V ( i , j )  V ( i , j ),
*
Where V* is the new value obtained by Gauss-Seidel method.
To speed up convergence,:
V new ( i , j )    V ( i , j )  V old ( i , j )
Generally, if α≥2, the method does not converge.
If α<1  “under-relaxation”;
If 1<α<2  “over-relaxation”;
It is tested that the best choice for α is:
Refer to code “plateV_SOR.c”
 
2
and excises.
1  / L
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Potential and Fields Near Electric Charges
When charges are concluded, the electronic potentials obeys Poisson EQ:
 V
2
 V 
2
x
2
 V
2

y
2
 V
2

z
2


+q
0
x  y  z
V (i , j , k ) 
1
[V ( i  1, j , k )  V ( i  1, j , k )  V ( i , j  1, k )  V ( i , j  1, k )
6
 V ( i , j , k  1)  V ( i , j , k  1)] 
 ( i , j , k )(  x )
2
6 0
All the relaxation methods are applicable here.
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Potential and Fields Near Electric Charges
Set V=0 at the boundary by assuming the box is large enough.
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Negative refraction
n 
 22
11
假如说“隐身衣”是想让别人看不到你，
那么负折射透镜就是让你看到别人看不到的东西。
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Finite-difference time-domain method
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Integration
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Thomas Simpson
is best remembered for his work on
interpolation and numerical methods
of integration
Gauss, Karl Friedrich
German mathematician who is
sometimes called the
"prince of mathematics."
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Simpson integration
将 这 些 区 域 的 面 积 加 起 来 ， 从 1 ~ n  1个 区 域 ， 得 到 ：
1
S = [ f ( x1 )  4 f ( x 2 )  2 f ( x 3 )  4 f ( x 4 )  ...  f ( x n )] x
3
总结系数规律是：
首尾2项的系数是1/3
其它的偶数项的系数是4/3
奇数项的系数是2/3
x1
x2
x3
x4
x5
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抛物线法积分
Numerical Integration (Simpson method)
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Integration: Number of Nodes vs Precision
Generally, the numerical integration could be
expressed as following with n+1 integration nodes:

b
a

n
f ( x) dx   Ak f ( xk )
k 0
Is it possible to have a precision with an
higher order than n in terms of
polynomial?
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Basis of the Gaussian Quadrature Rule
Revisit of Trapezoid method:
b

f ( x ) dx  c1 f ( a )  c 2 f ( b )

ba
2
a
f (a ) 
ba
f (b )
2
The two-point Gauss Quadrature Rule is an extension of the
Trapezoidal Rule approximation where the arguments of the
function are not predetermined as a and b but as unknowns
x1 and x2. In the two-point Gauss Quadrature Rule, the integral
is approximated as
b
I 

f ( x ) dx  c1 f ( x1 )  c 2 f ( x 2 )
a
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Basis of the Gaussian Quadrature Rule
The four unknowns x1, x2, c1 and c2 are found by assuming that the
formula gives exact results for integrating a general third order
polynomial,
f ( x )  a 0  a1 x  a 2 x  a 3 x .
2
Hence
b

a
b
f ( x ) dx 
 a
 a1 x  a 2 x  a 3 x
2
0
3
3
 dx
a
b

x
x
x 
  a 0 x  a1
 a2
 a3

2
3
4

a
2
3
4
 b2  a2 
 b3  a3 
 b4  a4 
 a 0  b  a   a1 
  a2 
  a3 

2
3
4






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Basis of the Gaussian Quadrature Rule
b

It follows that
f ( x ) dx  c1  a 0  a1 x1  a 2 x1  a 3 x1
2
  c 2  a 0  a1 x 2  a 2 x 2  a 3 x 2
3
2
3

a
Equating Equations the two previous two expressions yield
 b2  a2 
 b3  a3 
 b4  a4 
a 0  b  a   a1 
  a2 
  a3 

2
3
4






 c1  a 0  a1 x1  a 2 x1  a 3 x1
2
 a 0  c1  c 2   a 1  c 1 x 1  c 2
  c a  a x  a x  a x 
x   a c x  c x   a c x  c
3
2
2
0
1
2
2
2
2
2
1 1
3
2
3
2
2
2
2
3
3
1 1
x
2 2
3

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Basis of the Gaussian Quadrature Rule
Since the constants a0, a1, a2, a3
are arbitrary, we have:
b  a  c1  c 2
b a
2
2
b a
3
b a
4
ba
x1  
 2
 c1 x1  c 2 x 2
2
2
c1 
4
 c1 x1  c 2 x 2
3
1  ba




2
3

ba
x2  
 2
 c1 x1  c 2 x 2
3
3
4
2
Thus,
3
c2 
 1  b  a


2
 3 
ba
2
ba
2
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Basis of the Gaussian Quadrature Rule
Hence Two-Point Gaussian Quadrature Rule
b
 f ( x ) dx 
a
c1 f  x1   c 2 f  x 2  =
ba
2
ba 1  ba ba ba 1  ba
f
f






2
2
2
2
2
3

 3




It gives exact results for any third order polynomial.
That is, the Gaussian Quadrature gives a precision of 2n-1.
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Standard Gaussian Quadrature
In handbooks, coefficients and
arguments given for n-point
Gauss Quadrature Rule are
given for integrals
( x  m t  c; m 
ba
2
1

1
;c 
a b
)
2
n
g ( x ) dx 

i 1
ci g ( xi )
b

a
1
f ( x)dx 

1
b a ba
ba
f 
x
dx

2
2
2


[a,b][-1,1]
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Arguments and Weighing Factors
for n-point Gauss Quadrature Formulas
1
Table 1: Weighting factors c and function
arguments x used in Gauss Quadrature
Formulas.
n
 g ( x ) dx   c g ( x )
i
1
i
i 1
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Arguments and Weighing Factors
for n-point Gauss Quadrature Formulas
Table 1 (cont.) : Weighting factors c and function
arguments x used in Gauss Quadrature Formulas.
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Example for Gauss Quadrature
a) Use two-point Gauss Quadrature Rule to approximate the
distance covered by a rocket from t=8 to t=30 as given by
30

140000



x    2000 ln 
 9.8 t  dt

 140000  2100 t 

8 
b) Find the true error, E
t
for part (a).
c) Also, find the absolute relative true error,
a
for part (a).
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Solution
First, change the limits of integration from [8,30] to [-1,1], by
previous relations as follows
30

f ( t ) dt 
8
30  8
2
1

1
30  8 
 30  8
f 
x
 dx
2 
 2
1
 11  f  11 x  19  dx
1
Next, get weighting factors and function argument values from
Table 1, for the two point rule,
c1 
1 . 000000000
x1   0.577350269
c2 
1 . 000000000
x 2  0 . 577350269
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Solution (cont.)
Now we can use the Gauss Quadrature formula
1
11  f  11 x  19  dx  11c1 f  11 x1  19   11c 2 f 11 x 2  19 
1
 11 f  11(  0.5773503)  19   11 f 11(0.5773503)  19 
 11 f (12.64915)  11 f (25.35085)
 11(296.8317 )  11(708.4811)
 11058.44 m
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Solution (cont)
b) The true error, E , is
t
E t  T rue V alue  A pproxim ate V alue
 11061.34  11058.44
 2.9000 m
c) The absolute relative true error,  t , is (Exact value =
11061.34m)
t 
11061.34  11058.44
 100%
11061.34
 0.0262%
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A comparison between Gauss Quadrature and Newton-Cotes methods
I 

1
0
sin x
Set x=(t+1)/2, we get:
dx
x
I 

s in ( t 2 1 )
1
1
t 1
dt
2-point Gaussian qudrature:
sin
I 
1
(  0.5773503  1)
2
 0.5773503  1
sin

1
(0.5773503  1)
2
 0.9460411
0.5773503  1
3-point Gaussian qudrature:
sin
I  0.5555556 
1
(0.7745907  1)
2
0.7745907  1
s in
 0 .5 5 5 5 5 5 6 
sin
 0.8888889 
1
2
0 1
1
( 0 .7 7 4 5 9 0 7  1)
2
 = 0 .9 4 6 0 8 3 1
0 .7 7 4 5 9 0 7  1
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A comparison between Gauss Quadrature
and Newton-Cotes methods

The exact value should be 0.9460831...

用梯形公式，对积分区间[0，1]二分了11次用
2049个函数值，才可得到7位准确数字。

用Simpson公式对区间二分3次，用了9个函数
值，得到同样的结果。

用Gauss公式仅用了3个函数值，就得到结果。
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A summary on numerical integration

梯形求积公式和抛物线求积公式是低精度的方法
，但对于光滑性较差的函数有时比用高精度方法
能得到更好的效果。


Gauss型求积，它的节点是不规则的，所以当
节点增加时，前面的计算的函数值不能被后面利
用。计算过程比较麻烦，但精度高，特别是对计
算无穷区间上的积分和广义积分，则是其他方法
所不能比的。
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Homework




5.7, 5.8, 5.11
Sending to 17273799@qq.com
when ready
For lecture notes, refer to
http://www.compphys.cn/~xby
ang/
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