Page 1
1
Study on the Green’s functions for Laplace problems with circular and
spherical boundaries by using the image
method Reporter: H.C. Shieh
Adviser: Dr. J.T. ChenDepartment of Harbor and River Engineering,
National Taiwan Ocean UniversityJuly 25, 2009
碩士學位論文口試報告
Page 2
2
FrameMotivation and literature review
Two-dimensional Green’s functionGreen’s
function
Conclusions
MFS (Image method)
Trefftz method
BVP without sources
Page 3
3
Numerical methods
Numerical methods
Boundary Element MethodFinite Element Method Meshless Method
Page 4
4
Method of fundamental solutions
MN
jjj xsUcxu
1
),()(
is the fundamental solution),( xsU
Interior case Exterior case
This method was proposed by Kupradze in 1964.
Page 5
5
Optimal source location
Conventional MFS Alves & Antunes
GoodNot Good
?
Page 6
6
The simplest image method
Neumann boundary Neumann boundary conditioncondition DirichletDirichlet boundary conditio boundary conditionn
Mirror
Page 7
7
Conventional method to determine the image location
R
R’
O
a rr’
aOR
ORa
ORa
ORa
PR
RP
'''
2
''
OR a aOR
a OR OR
P
AB
aa
O R’RO
PPLord Kelvin(1824~1907) (1949, 相似三角形 )
Greenberg (1971, 取巧法 )
Page 8
8
Image location (Chen and Wu, 2006)
a
s 's2'
'ss
R
a
aR
R
a
R
1
1ln cos ( )
s
m
sm
ma
RR
m
1
1ln cos ( )
m
m
a mm
R
a
a s2
''s
s
aR
R
a
R
R
a
1
1ln cos ( )
ms
m
a mm
R
a
1
1ln cos ( )
m
m
a
RR m
m
Rigid body term
's
u=0
u=0
Page 9
9
2-D Degenerate kernal
1
1
),(cos)(1
ln
,)(cos)(1
lnln
m
m
m
m
RmR
m
RmRm
Rr
s( , )R q
R
r
rx( , )r f
x( , )r f
o
iU
eU
References:
W. C. Chen, A study of free terms and rigid body modes in the dual BEM, NTOU Master Thesis, 2001.
C. S. Wu,Degenerate scale analysis for membrane and plate problems using the meshless method and boundary element method, NTOU Master Thesis, 2004
rsxU ln),(
Page 10
10
Addition theorem & degenerate kernel
Addition theorem Subtraction theorem
sxsx eee
sxsxsx sinsincoscos)cos( sin( ) sin cos cos sinx s x s x s
/x s x se e e cos( ) cos cos sin sinx s x s x s sin( ) sin cos cos sinx s x s x s
Degenerate kernel for Laplace problem
1-D
2-D
RmR
m
RmRm
Rr
m
m
m
m
,)(cos)(1
ln
,)(cos)(1
lnln
1
1
sxifxs
sxifsxr
,
,
sx
Page 11
11
3-D degenerate kernel
11 0
11 0
1 ( )!cos ( ) (cos ) (cos ) ,
( )!1
1 ( )!cos ( ) (cos ) (cos ) ,
( )!
nni m m
m n n nn m
nne m m
m n n nn m
n mU m P P R
R n m R
r n m RU m P P R
n m
1, 02 , 1,2,...,m
mm
s ( , , )( , , )
xs R
x
exterior
x
interior
Page 12
12
Outline
Motivation and literature reviewDerivation of 2-D Green’s function
by using the image methodTrefftz method and MFS
Image method (special MFS)Trefftz method
Equivalence of solutions derived by Trefftz method and MFS
Boundary value problem without sourceConclusions
Page 13
13
Eccentric annulus
b
a
d
s
sR
Governing equation:
2 ( , ) ( ),G x s x s x
Dirichlet boundary condition:
1 2( , ) 0,G x s x B B= Î U
Case 1
B1
B2
Page 14
14
Eccentric problem
b
a
d
sGoverning equation:
2 ( , ) ( ),G x s x s x
Dirichlet boundary condition:
1 2( , ) 0,G x s x B B= Î U
Case 2
B1
B2
Page 15
15
Half plane with circular hole problem
( , )sR
Governing equation:
2 ( , ) ( ),G x s x s x
Dirichlet boundary condition: s
1 2( , ) 0,G x s x B B= Î U
u1=0
u2=0
Case 3
B1
B2
Page 16
16
Bipolar coordinates
221 lnln rr
1r 2r 1r
2r
221 lnln rr
x
121 lnln rr1r
2r1cs 2cs
Page 17
17
Bipolar coordinates
Eccentric annulus A half plane with a hole An infinite plane with double holes
focus
Page 18
18
Annular (EABE, 2009) to eccentric case
s s1s2s4s3
s6s5
…. ….
Image point+
-
Source point
sxsxGm ln2
1),(
4 3 4 2 4 1 41
lim ln ln ln lnN
i i i iN
i
x s x s x s x s remainder term
0),(1
BxsxG 0),(
2
BxsxG
Page 19
19
Series of images
-4 -2 0 2
x
-1
0
1
y
Gra p h 1
Im ag e p o in t
The final images
sc1 sc2
4 3 4 2 4 1 41
1 1 2 2
1( , ) {ln lim ln ln ln ln
2( ) ln ( ) ln ( )}
N
i i i iN
i
c c
G x s x s x s x s x s x s
c N x s c N x s e N
Page 20
20
Numerical approach to determine c1(N), c2(N) and e(N)
( ) 0-6-2.696 10 e N
1( )c N -0.8625
2 ( )c N -0.1375
0 2 4 6 8 10
N
-1
-0 .8
-0.6
-0.4
-0.2
0
V alu e
C o effic ien tc 1(N )
c 2(N )
e (N )
Coefficients sc1 sc2
Page 21
21
Contour plot of eccentric annulus problem
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-1 -0.5 0 0.5 1 1.5 2 2.5 3-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Image method Analytical solution
(bipolar coordinates )
Dirichlet boundary for the eccentric case 1
Page 22
22
Analytical derivation of locationfor the two frozen points
d
a
b
2cR
2cs 1cs
1cRc2
1 1 1
2
11
2
11
1
2
1ln ( ) cosln ( ,)c c c
cc
c
cc
m
c
mRa m
m as x s a R
R a aR
a R R
ff¥
=- -å® - = >
ß
= Þ =
2 21
2
2 2 2
1ln ( ) cos ( )ln ,m
mc c
c
c c c
aR mx a R
ms
Rs ff
¥
=- -- <å® =
cRR cc 212
4 2 2 4 2 2 2 2 42 2 2
2
a a b b a d dc
b d
d
- + +=
-Þ
-
Page 23
23
Eccentric case
4 3 4 2 4 1
1 1 2 2
41
1( , ) ln lim ln ln ln ln
( )
2
l ( )n l (n )
N
i i i i
c
i
c
NG x s x s x s x s x s x
c N s c N s e
s
x Nx
1 2 and c cs s focuses
Image sources
True source
Page 24
24
Contour plot of eccentric annulus
- 1 - 0 . 8 - 0 . 6 - 0 . 4 - 0 . 2 0 0 . 2 0 . 4 0 . 6 0 . 8 1- 1
- 0 . 8
- 0 . 6
- 0 . 4
- 0 . 2
0
0 . 2
0 . 4
0 . 6
0 . 8
1
Image method
- 1 - 0 . 8 - 0 . 6 - 0 . 4 - 0 . 2 0 0 . 2 0 . 4 0 . 6 0 . 8 1- 1
- 0 . 8
- 0 . 6
- 0 . 4
- 0 . 2
0
0 . 2
0 . 4
0 . 6
0 . 8
1
Null-field BIE approach (addition theorem and superposition technique)
Page 25
25
A half plane with a circular hole
4 3 4 2 4 1
1 1 2 2
41
1( , ) ln lim ln ln ln ln
( )
2
l ( )n l (n )
N
i i i i
c
i
c
NG x s x s x s x s x s x
c N s c N s e
s
x Nx
image source
true source
1 2 and c cs s focuses
Page 26
26
Contour plot of half plane problem
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 40
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
Image method Null-field BIE approach (addition theorem and superposition technique)
Page 27
27
Linking of MFS and image method
s
Ä
1s
Ä
2s
e
3s
e
4s
e
5s
e
6s
Ä
7s
ÄK
8s
ÄK
MFS (special case) Conventional MFS
1
1( ) ln ( , )
2
N
j jj
u x x s w U x s
4 3 4 21
1( ) {ln
2
lim ln ln
( ) ( ) ln }
N
i iNi
u x x s
x s x s
c N d N
s
Page 28
28
Image method versus MFS
N N
Ku p
K
N
large
3 3
Ku p
K
optimal
All the strength need to be determined.
Only three coefficients are required to be determined.
Conventional MFS Image method
Page 29
29
Outline
Motivation and literature reviewDerivation of 2-D Green’s function
by using the image methodTrefftz method and MFS
Image method (special MFS)Trefftz method
Equivalence of solutions derived by Trefftz method and MFS
Boundary value problem without sourcesConclusions
Page 30
30
Trefftz method
The method was proposed by Trefftz in 1926.
TN
jjjcxu
1
)(
mm mm sin,cos,1 mm mm sin,cos,ln
Interior case Exterior case
is the jth T-complete functionj
Page 31
31
Trefftz method and MFS
Method Trefftz method MFS
Definition
Figure sketch
Base , (T-complete function) , r=|x-s|
G. E.
Match B. C. Determine cj Determine wj
( , ) lnU x s r
1( ) ( , )
N
j jj
u x w U x s
( )2 0u xÑ = ( )2 0u xÑ =
D
u(x)
~x
s
Du(x)
~x
r
~s
is the number of complete functions MN is the number of source points in the MFS
1( )
M
j jj
u x c
j
Page 32
32
Derivation of 3-D Green’s function by using the image method
Interior problem
Exterior problem
Page 33
33
The weighting of the image source in the
3-D problem
y
z
1a
x
y
1 a
x
z
),,(2
sR
as ),,( sRs
),,( sRs
),,(2
sR
as
Interior problem Exterior problem
1sR
a
1sR
a
Page 34
34
The image group
22 15 91
1 5 9 4 32 3
2 32 1
2 6 10 4 22
2 32 13 7 11
3 7 11 4 12 3
2
4 8 2
, ( ), ( ) ( )
, ( ), ( ) ( )
, ( ), ( ) ( )
, (
nn
nn
nn
aR a RR b b b b b b bw w w w
b R b R a b R a R aa a a a a a a a a
w w w wR bR R b b R R b R baR a R a Rb b b b b b b
w w w wbR a b R a a b R a a a aa a a a
w wb b b
3
2 112 43
), ( ) ( )nn
a a a a aw w
b b b b b b
2 2 2 2 21
1 5 4 32 2
2 2 2 2 21
2 6 4 22 2
2 2 2 2 21
3 7 4 12 2 2 2 2
2 2 2 2 21
4 8 42 2 2 2 2
, ........ ( )
, ....... ( )
, ... ( )
, ... ( )
nn
nn
nn
nn
b b b b bR R R
R R a R a
a a a a aR R R
R R b R b
b R b R b b R bR R R
a a a a a
a R a R a a R aR R R
b b b b b
Obtain image weighted
Obtain image location
s
Ä
1s
Ä
2s
e
3s
e
4s
e
5s
e
6s
Ä
7s
ÄK
8s
ÄK
Page 35
35
Interpolation functions
a
b
1),(),( sxGsxG ba
( ) ( )( , ) ( , ) ( , ) ( , ),
( ) ( )m m b m a
b a a bG x s G x s G x s G x s a b
b a b a
4 3 4 2 4 1 4
1 4 3 4 2 4 1 4
1 1( , ) lim
4
( ) 1
( ) ( )
Ni i i i
Ni i i i i
N Ns s
s s
w w w wG x s
x s x s x s x s x s
R a a b Ra a
b R b a b R b a
Page 36
36
Analytical derivation
4 3 4 2 4 1 4
1 4 3 4 2 4 1 4
1 1 ( )( , ) lim ( )
4
Ni i i i
Ni i i i i
w w w w d NG x s c N
x s x s x s x s x s
Page 37
37
Numerical solution
a
b
4 3 4 2 4 1 4
1 4 3 4 2 4 1 4
4 3 4 2 4 1 4
1 4 3 4 2 4 1 4
1 1 ( )( , ) ( ) 0
4
1 1 ( )( , ) ( ) 0
4
Ni i i i
aia a i a i a i a i
Ni i i i
bib b i b i b i b i
w w w w d NG x s c N
x s x s x s x s x s a
w w w w d NG x s c N
x s x s x s x s x s b
0
0
)(
)(1
1
11
1
1
14
4
14
14
24
24
34
34
14
4
14
14
24
24
34
34
Nd
Nc
b
a
sx
w
sx
w
sx
w
sx
w
sx
sx
w
sx
w
sx
w
sx
w
sx
iib
i
ib
i
ib
i
ib
i
b
iia
i
ia
i
ia
i
ia
i
a
Page 38
38
Numerical and analytic ways to determine c(N) and d(N)
0 2 4 6 8 10
N
0
0.02
0.04
0.06
0.08
0.1c (N ) a n d d (N )
A n a ly tica l c (N )N u m er ica l c (N )A n a ly tica l d (N )N u m er ica l d (N )
Coefficients
Page 39
39
Derivation of 3-D Green’s function by using the Trefftz Method
1G
2G
11 GG
22 GG
PART 1 PART 2
PART 1
11 0
11 0
1 1 ( )!cos ( ) (cos ) (cos ) ,
4 ( )!( , )
1 1 ( )!cos ( ) (cos ) (cos ) ,
4 ( )!
nnm m
m n n snn ms s
Fnn
m msm n n sn
n m
n mm P P R
R n m RG x s
Rn mm P P R
n m
Page 40
40
Boundary value problem
1( , )
TN
T j jj
G x s c
11 GG
22 GG
Interior:
)(cos)sin(),(cos)cos(,1 m
n
nm
n
n PmPm
Exterior:
)(cos)sin(),(cos)cos(,1 )1()1(
m
n
nm
n
n PmPm
( 1)0000
1 0
( 1)
( , ) [ (cos )cos( ) (cos )cos( )
(cos )sin( ) (cos )sin( )]
nn m n m
T nm n nm nn m
n m n m
nm n nm n
BG x s A A P m B P m
C P m D P m
00
00
4( )
4
s
s
s
s
R a
R b aAB a b R
R b a
2 1 2 1
1 2 1 2 1
2 1 2 1 2 1
1 2 1 2 1
( )!(cos )cos( )
4 ( )!
( )!(cos )cos( )
4 ( )!
n nmm s
nn n nsnm
n n nnm s mm
nn n ns
R an mP m
n m R b aA
B a b Rn mP m
n m R b a
2 1 2 1
1 2 1 2 1
2 1 2 1 2 1
1 2 1 2 1
( )!(cos )sin( )
4 ( )!
( )!(cos )sin( )
4 ( )!
n nmm s
nn n n
nm
n n nnm s mm
nn n ns
R an mP m
n m R b aC
D a b Rn mP m
n m R b a
PART 2
Page 41
41
PART 1 + PART 2 :
1G2G11 GG
22 GG
( , ) ( , ) ( , )F TG x s G x s G x s
11 0
11 0
1 1 ( )!cos ( ) (cos ) (cos ) ,
4 ( )!( )
1 1 ( )!cos ( ) (cos ) (cos ) ,
4 ( )!
nnm m
m n n snn ms s
Fnn
m msm n n sn
n m
n mm P P R
R n m RG x
Rn mm P P R
n m
( 1)0000
1 0
( 1)
( , ) [ (cos )cos( ) (cos )cos( )
(cos )sin( ) (cos )sin( )]
nn m n m
T nm n nm nn m
n m n m
nm n nm n
BG x s A A P m B P m
C P m D P m
( 1)0000
1 0
( 1)
1( , ) [ (cos )cos( ) (cos )cos( )
4
(cos )sin( ) (cos )sin( )],
nn m n m
nm n nm nn m
n m n m
nm n nm n
BG x s A A P m B P m
x s
C P m D P m
Page 42
42
Results
-10 -8 -6 -4 -2 0 2 4 6 8 10-10
-8
-6
-4
-2
0
2
4
6
8
10
-10 -8 -6 -4 -2 0 2 4 6 8 10-10
-8
-6
-4
-2
0
2
4
6
8
10
Trefftz method (x-y plane) Image method (x-y plane)
Page 43
43
Outline
Motivation and literature reviewDerivation of 2-D Green’s function
by using the image methodTrefftz method and MFS
Image method (special MFS)Trefftz method
Equivalence of solutions derived by Trefftz method and MFS
Boundary value problem without sourcesConclusions
Page 44
44
Trefftz solution
( 1)0000
1 0
( 1)
1( , ) [ (cos )cos( ) (cos )cos( )
4
(cos )sin( ) (cos )sin( )],
nn m n m
nm n nm nn m
n m n m
nm n nm n
BG x s A A P m B P m
x s
C P m D P m
2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1
1 1 2 1 2 11 0
( )1 1( , ) + +
4 ( ) ( )
( )! + cos[ ( )] (cos )
4 ( )! ( )
s s
s s
n n n n n n n nnmm s s
nn n n nn m s
R a a b RG x s
x s R b a R b a
R a a b a Rn mm P
n m R b a
Without loss of generality
Page 45
45
Mathematical equivalence the Trefftz method
and MFS Trefftz method series expand
2 1 2 1 2 1 2 1
2 1 2 1 1 2 1 2 1 1 1 2 1 2 1 1 2 1 2 1(cos )
( ) ( ) ( )
n n n n n n n nm
nn n n n n n n n n n n n
R a a b a RP
b a R b a R b a b a
Image method series expand
1212
12
12
12
24
222
121
9
91
5
51
1
1
1
nn
nn
n
n
n
nn
n
nnn
n
nn
n
n
n
n
n
n
ab
R
ba
bR
b
Ra
aR
b
b
R
R
b
Rw
Rw
Rw
s s1s2s4 s3 s5 s9s7
)(1121211
1212
12
12
11
12
12
42
1
2
110
1016
612
2
nnnn
nn
n
n
nn
n
nnn
n
nn
n
n
n
n
n
n
n
abR
ba
baR
a
Rb
a
bR
a
R
a
R
aRw
Rw
Rw
s s1 s3s2s4s6s8s10
s s1s2s4 s3 s5 s9s7
)(1)()(
12121
12
12
12
112
12
144
44
2
2
122
22
1
7
71
3
3
nnn
nn
n
n
nn
nn
nn
nn
nn
nn
n
n
n
n
abR
a
baRb
a
Rb
a
a
b
Rb
a
a
b
Rw
Rw
s s1 s3s2s4s6s8s10
)(1)()(
12121
12
12
12
112
12
14
4
2
2
12
2
1
881
44
nnn
nn
n
n
nn
nn
nn
nn
nn
nn
n
n
n
n
ab
Ra
ba
bRa
b
Ra
b
a
b
Ra
b
aRw
Rw
Page 46
46
Equivalence of solutions derived by Trefftz method and image method (special MFS)
Trefftz method MFS (image method)
1, cos( ) (cos ),
sin( ) (cos )
, , 0,1,2,3, , ,
1,2, , ,
n m
n
n m
n
m P
m P
m
n
r f q
r f q
= ¥
= ¥
K L
L
1,
j
j Nx s
-Î
-
Equivalence
addition theorem
linkage
3-D
True source
Page 47
47
Outline
Motivation and literature reviewDerivation of 2-D Green’s function
by using the image methodTrefftz method and MFS
Image method (special MFS)Trefftz method
Equivalence of solutions derived by Trefftz method and MFS
Boundary value problem without sourcesConclusions
Page 48
48
An infinite plane with two circular holes (anti-symmetric BC)
B2
y
x
B1
u=V=V1=-1 u=V=V2=1
a=1.0 d=10
2 ( ) 0,u x x D
d
a a2c
Page 49
49
Animation - An infinite plane with two
circular holes
u=-1 u=1
s1 s2s3s4sc1 sc2
)lnln()(lim)( 21 ooNrrNqxu
)(ln)(ln)(
)lnlnln(ln
2211
14142434
NesxNcsxNc
sxsxsxsx
cc
N
iiiii
)(Nq )(Nq
Page 50
50
Numerical approach to determine q(N), c1(N), c2(N) and e(N)
0 4 8N
-0 .6
-0 .4
-0 .2
0
0.2
0.4
0.6
c 1(N )
c 2(N )
e(N )q (N )q(N)=e(N)=0
436.0sinh
)(1
1
ac
VNc
436.0sinh
)(1
2
ac
VNc
Coefficients
Page 51
51
Contour plot of an infinite plane with two circular holes (antisymmetric case)
Image solution bipolar coordinates
-10-8
-6-4
-20
24
68
-14
-12
-10 -8 -6 -4 -2 0 2 4 -10-8
-6-4
-20
24
68
-14
-12
-10 -8 -6 -4 -2 0 2 4
null-field BIEM
-10-8
-6-4
-20
24
68
-14
-12
-10 -8 -6 -4 -2 0 2 4
Page 52
52
An infinite space with two cavities (anti-
symmetric BC)
B2
y
z
x
B1
u=V1=-1 u=V2=1 a=1.0 d=5.0
2 ( ) 0,u x x D
2
2
1
1
14
4
14
14
24
24
34
34
21
)()(
)()11
()(lim)(
c
a
c
a
N
ii
i
i
i
i
i
i
i
oo
a
N
sx
Nc
sx
Nc
sx
w
sx
w
sx
w
sx
w
rrNqxu
Page 53
53
Numerical approach to determine q(N), c1(N) and c2(N)
0 4 8N
-0 .4
0
0.4
0.8
1.2
V a lu e
c 1(N )
c 2(N )
q (N )
1)(lim
NqN
0)(lim,0)(lim 21
NcNcNN
Coefficients
Page 54
54
Contour plot of an infinite space with two spherical cavities
Bispherical coordinates Image method Null-field BIE
2 2
6
( ) ( ) 0
0 (10 )
k u x
k
-8 -6 -4 -2 0 2 4 6 8-8
-6
-4
-2
0
2
4
6
8
-8-6
-4-2
02
46
8-8 -6 -4 -2 0 2 4 6 8
x-y planey
z
x
-8-6
-4-2
02
46
8-8 -6 -4 -2 0 2 4 6 8
Page 56
56
Outline
Motivation and literature reviewDerivation of 2-D Green’s function
by using the image methodTrefftz method and MFS
Image method (special MFS)Trefftz method
Equivalence of solutions derived by Trefftz method and MFS
Boundary value problem without sourcesConclusions
Page 57
57
Optimal location of MFS
• Depends on loading (image location)
• Depends on geometry (frozen image point)
Page 58
58
Final images to bipolar (bispherical) focus
2-D 3-D
Bipolar coordinates Bispherical coordinates
Page 59
59
Equivalence of Trefftz method and MFS
3-D
Trefftz method MFS (image method)
Page 60
60
Image solution for BVP without sources
x
x
y
y
Page 61
61
Thanks for your kind attentionsYou can get more information from our website
http://msvlab.hre.ntou.edu.tw/
The end
Page 62
62
A half plane with a circular hole
2007, Ke J. N. 2009, Image method
a
b
Page 63
63
An infinite plane with two circular holes subject to Neumann boundary
ab
d
s
01 t02 t
Page 64
64
Extra terms of complementary solutions
x
y
01 tt 2 0t t
( , )S R
O1 O2
1 2
4 3 4 2 4 1 4
1 1 2 2 1 2
1( , ) {ln [ln ln ln ln ]
2( )ln ( )ln ( )ln ( )ln
N
i i i ii
c c O O
G x s s x s x s x s x s x
c N s x c N s x d N s x d N s x
Two complementary solutions
f1 f2
Source point
Frozen point
Page 65
65
The method provide of JW. Lee
d1 d2
Frozen point
M
jjMO
M
jjMO xxdxxd
122
111 lnlimlnlnlimln
complementary solutions
Page 66
66
The total potential
1 1 1 2
4 3 4 2 4 1 4
1 2 1 2
1( ) {ln [ln ln ln ln ]
2( )ln ( )ln ( )ln ( )ln
N
i i i ii
f f O O
u x s x s x s x s x s x
f N s x f N s x d N s x d N s x
1 1
1
2
4 3 4 2 4 1 4
1 2
11
21
1( ) {ln [ln ln ln ln ]
2( )ln ( )ln
( ) ln lim ln
( ) ln lim ln
N
i i i ii
f fM
O jM
jM
O jM
j
u x s x s x s x s x s x
f N s x f N s x
d N x x
d N x x
Where the N=M
Page 67
67
Results
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
Image method)
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
Null-field BIE approach (addition theorem and superposition technique)
Page 68
68
Conclusions
The analytical solutions derived by the Trefftz method and MFS were proved to be mathematically equivalent for Green’s functions of the concentric sphere.
In the concentric sphere case, we can find final two frozen image points (one at origin and one at infinity). Their singularity strength can be determined numerically and analytically in a consistent manner.
It is found that final image points terminate at the two focuses of the bipolar (bispherical) coordinates for all the cases
Page 69
69
Numerical examples 1: Eccentric annulus
- 1 - 0 . 8 - 0 . 6 - 0 . 4 - 0 . 2 0 0 . 2 0 . 4 0 . 6 0 . 8 1- 1
- 0 . 8
- 0 . 6
- 0 . 4
- 0 . 2
0
0 . 2
0 . 4
0 . 6
0 . 8
1
Image method (50+2 points)
u1=0u2=0 ( , )ss R
a
b
Page 70
70
Numerical examples 3: An infinite plane with double holes
ab
d
x
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
Image method (20+4+10 point)
t1=0t2=0
Page 71
71
Animation- an infinite plane with double holes
s
1s2s
3s4s
The final images terminate at the focus
1ds 2ds
Multipole expansion
4 3 4 2 4 1 4
11 1 2
1 1 2
2
12
1
2
( ) ( )
1( , ) {ln [ln ln ln ln ]
2
ln lim ln ln
( )ln (
m
n
l
l
li n
)N
i i i ii
M
f f
d d
M
j jM Mj j
d
G x s x s x s x s x s x s
x x s x x s
f N x s f N
s N
x s
N d s
1fs2fs
true source
1 2 and c cs s focus
image source
2dsand1ds
Multipoles
t1=0 t2=0
Page 72
72
Equivalence of solutions derived by Trefftz method and MFS
Trefftz solution
( 1)0000
1 0
( 1)
1( , ) [ (cos )cos( ) (cos )cos( )
4
(cos )sin( ) (cos )sin( )],
nn m n m
nm n nm nn m
n m n m
nm n nm n
BG x s A A P m B P m
x s
C P m D P m
00
00
1( )4
s
s
s
s
R a
R b aAB a b R
R b a
Image solution
4 3 4 2 4 1 4
1 4 3 4 2 4 1 4
1 1 ( )( , ) lim ( )
4
Ni i i i
Ni i i i i
w w w w d NG x s c N
x s x s x s x s x s
)(
)(
)(
)(
)(
abR
Rba
b
a
abR
aR
b
a
Nd
Nc
s
s
N
s
s
N
The same
Page 73
73
The simplest MFS
1-D Rod
2211 ),(),()( PsxUPsxUxu 1P
2P
1s 2s
1s2s
1P2P
1s2s
2P1P
where U(x,s) is the fundamental solution.
0 l
Page 74
74
Present method- MFS (Image method)
……
4 3 4 2 4 1 4
1 4 3 4 2 4 1 4
1 1( , ) lim
4
Ni i i i
mN
i i i i i
w w w wG x s remainder term
x s x s x s x s x s
Page 75
75
An infinite space with two cavities (symmetric BC)
B2
y
z
x
B1
u=V1=1 u=V2=1
q q Obtain image weighting4 4
4 34 4
4 54 2
4 5
4 24 1
4 2
4 34
4 3
ii
i
ii
i
ii
i
ii
i
aww
d Raw
wRaw
wd R
aww
R
--
-
--
-
--
-
-
-
=-
=
=-
=
1 2
4 3 4 2 4 1 4 1 2
11 2 4 3 4 2 4 1 4
( ) ( )1 1( ) lim ( )
s sNs i i i i
Nio o i i i i c c
w w w w c N c Nu x q N
r r x s x s x s x s x s x s
Obtain image location2
4 34 4
2
4 24 5
2
4 14 2
2
44 3
ii
ii
ii
ii
aR d
d Ra
RR
aR d
d Ra
RR
--
--
--
-
= --
=
= --
=
1qs2qs
33sqw44sqw
d
a a
Page 76
76
The strength of two frozen points and q(N)
The strength of c1(N), c2(N) and q(N)
0.1)( Nq
1( ) -1.691750044745917E-014c N
2 ( ) -1.720415903023510E-014c N
Page 77
77
Contour plot of an infinite space with two spherical cavities (symmetric case)
3-D Bipolar coordinates
Bispherical coordinates
Image method Null-field BIE
2 2( ) ( ) 0
0
k u x
k
x-y planey
z
x
-8 -6 -4 -2 0 2 4 6 8-8
-6
-4
-2
0
2
4
6
8-8
-6-4
-20
24
68
-8 -6 -4 -2 0 2 4 6 8
-8 -6 -4 -2 0 2 4 6 8-8
-6
-4
-2
0
2
4
6
8
Page 78
78
Illustrative examples – An eccentric annulus
1a2 5 .b
1d
y
x1c 2c
u=V1=0 u=V2=1
B1
B2
c c
r1
r2
Page 79
79
Numerical approach to determine q(N), c1(N), c2(N) and e(N)
0 4 8N
-1
0
1
2
V alu e
c 1(N )
c 2(N )
e(N )q (N )q(N)=0 (exact)
e(N)=2 (exact)
c1(N)=1.44 (exact)
c2(N)=-1.44 (exact)
Page 80
80
Contour plot of eccentric annulus
-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
Image solution bipolar coordinates null-field BIEM
Page 81
81
Outline
Motivation and literature reviewDerivation of 2-D Green’s function
by using the image methodTrefftz method and MFS
Image method (special MFS)Trefftz method
Equivalence of solutions derived by Trefftz method and MFS
Boundary value problem without sourceConclusions
Page 82
82
Optimal source location
Conventional MFS Alves CJS & Antunes PRS
Not good Good
Page 83
83
MFS-Image location and weighting-interior (Chen and Wu, 2006)
b
),,( x
1
1R
b
),,( Rs ),,('
2
R
bs
y
x
z
11 0
1 1 ( )!cos ( ) (cos ) (cos )
( )!
nnm m
m n n nn m
n m Rm P P
x s b n m b
1
1 0
' 1 1 ( )!cos ( ) (cos ) (cos )
' ( )! ( ')
nnm m
m n n nn m
R n m bm P P
b x s b n m R
R
bR
R
b
b
Rn
n
n
n 21
1'
)'(
The weighting of the image point
R
b
bR
b
b
R
1' 2
u=0
Page 84
84
MFS-Image location and weighting-exterior (Chen and Wu, 2006)
),,( x
1
),,( Rs
),,('2
R
as
1R
a
Ra
RPPm
mn
mn
RsxR
a
R
aPPm
mn
mn
Rsx
n
nm
nn
n
m
m
n
n
nm
nn
n
m
m
n
)'()(cos)(cos)(cos
)!(
)!(1
'
1
)(cos)(cos)(cos)!(
)!(11
1 0
11 0
aR
aR
a
R
R
a
Ra
R
R
an
n
n
n
n
n
n
n 2
1'
)'()'(
The weighting of the image point
1R
a
u=0
Page 85
85
a
a
Chen and Wu-image method (2006)
1
1
1ln ( ) cos ( ),
( , ) 1ln ( ) cos ( ),
mss
m
ms s
m s
Rm R
mU x sR m R
m R
s
2
''s
s
aR
R
a
R
R
a
s's
2''
ss
R
a
aR
R
a
R
1
1ln cos ( )
ms
m
a mm
R
a
1
1ln cos ( )
m
m
a
RR m
m
1
1ln cos ( )
s
m
sm
ma
RR
m
1
1ln cos ( )
m
m
a mm
R
a
's
Page 86
86
Analytical derivation of location for the two frozen points
b
ax
y
1cs
2cs
1cR
2 12c cR b R 1 1 1
2
11
2
11
1
2
1ln ( ) cosln ( ,)c c c
cc
c
cc
m
c
mRa m
m as x s a R
R a aR
a R R
ff¥
=- -å® - = >
ß
= Þ =
2 21
2
2 2 2
1ln ( ) cos ( )ln ,m
mc c
c
c c c
aR mx a R
ms
Rs ff
¥
=- -- <å® =
2
2
2
6 1 0
3 2 2
c
c
c c
c
aR
b R
R R
R
(0.171 & 5.828)
a=1, b=3
Page 87
87
1 2 3 4 5 6 7 8 9 10N
-0 .8
-0 .6
-0 .4
-0 .2
0
0.2
c(N
), d
(N)
and
e(N
)c (N )d (N )e (N )
Numerical approach to determine c1(N), c2(N) and e(N)
( ) 0-157.9713 10 e N
1( )c N -0.26448
2 ( )c N -0.73551
)(1 Nc)(2 Nc
)(Ne
Page 88
88
Analytical derivation of locationfor the two frozen points
d
ab
1cs
2cs
1cR
2cR
2 2
1 22 2
,c cc c
a bR R d
R R d
2 2 2
2 222
( )0c
c
a b d RR a
d
2c
4 2 2 4 2 2 2 2 42 2 2
2
a a b b a d b d dc
d
- + - - +=
cRR cc 212