null-field approach for laplace problems with circular boundaries using degenerate kernels
DESCRIPTION
Null-field approach for Laplace problems with circular boundaries using degenerate kernels. 沈文成 陳正宗 時間: 13:20 ~ 14:00 地點:河工二館 307 室. BEM course May 13, 2008 (typicalBVP-L.ppt). Outlines. Motivation and literature review Mathematical formulation Expansions of fundamental solution - PowerPoint PPT PresentationTRANSCRIPT
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Null-field approach for Laplace Null-field approach for Laplace problems with circular problems with circular boundaries using degenerate boundaries using degenerate kernelskernels
沈文成 陳正宗 時間: 13:20 ~ 14:00地點:河工二館 307 室BEM course May 13, 2008 (typicalBVP-L.ppt)
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OutlinesOutlines Motivation and literature reviewMotivation and literature review Mathematical formulationMathematical formulation
Expansions of fundamental solutionExpansions of fundamental solution and boundary densityand boundary density
Adaptive observer systemAdaptive observer system Vector decomposition techniqueVector decomposition technique Linear algebraic equationLinear algebraic equation
Numerical examplesNumerical examples Degenerate scaleDegenerate scale ConclusionsConclusions
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OutlinesOutlines Motivation and literature reviewMotivation and literature review Mathematical formulationMathematical formulation
Expansions of fundamental solutionExpansions of fundamental solution and boundary densityand boundary density
Adaptive observer systemAdaptive observer system Vector decomposition techniqueVector decomposition technique Linear algebraic equationLinear algebraic equation
Numerical examplesNumerical examples Degenerate scaleDegenerate scale ConclusionsConclusions
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Motivation and literature reviewMotivation and literature review
Fictitious Fictitious BEMBEM
BEM/BEM/BIEMBIEM
Null-field Null-field approachapproach
Bump Bump contourcontour
Limit Limit processprocess
Singular and hypersiSingular and hypersingularngular RegulRegularar
Improper Improper integralintegral
CPV and CPV and HPVHPV
Ill-Ill-posedposed
FictitiFictitious ous
bounboundarydary
CollocatCollocation ion
pointpoint
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Present approachPresent approach
1. No principal 1. No principal valuevalue2. Well-2. Well-posedposed
(s, x)eK
(s, x)iK
Advantages of Advantages of degenerate kerneldegenerate kernel
(x) (s, x) (s) (s)BK dBj f=ò
DegeneratDegenerate kernele kernel
Fundamental Fundamental solutionsolutionCPV and CPV and
HPVHPVNo principal No principal
valuevalue
(x) (s)(x) (s) (s)B
db Baj f=ò 2
1 1( ), ( )x s x s
O O- -
(x) (s)a b
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Engineering problem with arbitrary Engineering problem with arbitrary geometriesgeometries
Degenerate Degenerate boundaryboundary
Circular Circular boundaryboundary
Straight Straight boundaryboundary
Elliptic Elliptic boundaryboundary
a(Fourier (Fourier series)series)
(Legendre poly(Legendre polynomial)nomial) (Chebyshev poly(Chebyshev polynomial)nomial)
(Mathieu (Mathieu function)function)
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Motivation and literature reviewMotivation and literature review
Analytical methods for solving Laplace problems with circular holesConformal Conformal mappingmapping
Bipolar Bipolar coordinatecoordinate
Special Special solutionsolution
Limited to doubly Limited to doubly connected domainconnected domain
Lebedev, Skalskaya and Uyand, 1979, “Work problem in applied mathematics”, Dover Publications
Chen and Weng, 2001, “Torsion of a circular compound bar with imperfect interface”, ASME Journal of Applied Mechanics
Honein, Honein and Hermann, 1992, “On two circular inclusions in harmonic problem”, Quarterly of Applied Mathematics
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Fourier series approximationFourier series approximation Ling (1943) - Ling (1943) - torsiontorsion of a circular tube of a circular tube Caulk et al. (1983) - Caulk et al. (1983) - steady heat conducsteady heat conductiontion with circular holes with circular holes Bird and Steele (1992) - Bird and Steele (1992) - harmonic and bharmonic and biharmoniciharmonic problems with circular holes problems with circular holes Mogilevskaya et al. (2002) - Mogilevskaya et al. (2002) - elasticityelasticity pr problems with circular boundariesoblems with circular boundaries
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Contribution and goalContribution and goal
However, they didn’t employ the However, they didn’t employ the null-field integral equationnull-field integral equation and and degenerate kernelsdegenerate kernels to fully capture to fully capture the circular boundary, although the circular boundary, although they all employed they all employed Fourier series Fourier series expansionexpansion..
To develop a To develop a systematic approachsystematic approach for solving Laplace problems with for solving Laplace problems with multiple holesmultiple holes is our goal. is our goal.
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OutlinesOutlines Motivation and literature reviewMotivation and literature review Mathematical formulationMathematical formulation
Expansions of fundamental solutionExpansions of fundamental solution and boundary densityand boundary density
Adaptive observer systemAdaptive observer system Vector decomposition techniqueVector decomposition technique Linear algebraic equationLinear algebraic equation
Numerical examplesNumerical examples Degenerate scaleDegenerate scale ConclusionsConclusions
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Boundary integral equation and null-Boundary integral equation and null-field integral equationfield integral equation
2 (x) (s, x) (s) (s) (s, x) (s) (s), xB B
u T u dB U t dB Dp = - Îò ò0 (s, x) (s) (s) (s, x) (s) (s), x c
B BT u dB U t dB D= - Îò ò
s
s
(s, x) ln x s ln
(s, x)(s, x)
(s)(s)
U r
UT
ut
= - =¶= ¶
¶= ¶
n
n
x
D
xcD
x
D xcD
Interior Interior casecase
Exterior Exterior casecase
Null-field integral Null-field integral equationequation
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OutlinesOutlines Motivation and literature reviewMotivation and literature review Mathematical formulationMathematical formulation
Expansions of fundamental solutionExpansions of fundamental solution and boundary densityand boundary density
Adaptive observer systemAdaptive observer system Vector decomposition techniqueVector decomposition technique Linear algebraic equationLinear algebraic equation
Numerical examplesNumerical examples Degenerate scaleDegenerate scale ConclusionsConclusions
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Expansions of fundamental solution Expansions of fundamental solution and boundary densityand boundary density
Degenerate kernel - fundamental Degenerate kernel - fundamental solutionsolution
Fourier series expansions - boundary Fourier series expansions - boundary densitydensity
1
1
1( , ; , ) ln ( ) cos ( ),(s, x)
1( , ; , ) ln ( ) cos ( ),
i m
m
e m
m
U R R m Rm R
URU R m R
m
rq r f q f r
q r f r q f rr
¥
=¥
=
ìïï = - - ³ïïïï=íïï = - - >ïïïïî
åå
01
01
(s) ( cos sin ), s
(s) ( cos sin ), s
M
n nn
M
n nn
u a a n b n B
t p p n q n B
q q
q q
=
=
= + + Î
= + + Î
åå
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Separable form of fundamental Separable form of fundamental solution (1D)solution (1D)
-10 10 20
2
4
6
8
10
Us,x2
1
2
1
(x) (s), s x(s, x)
(s) (x), x s
i ii
i ii
a bU
a b
=
=
ìïï ³ïïïï=íïï >ïïïïî
åå
1 (s x), s x1 2(s, x)
12 (x s), x s2
U r
ìïï - ³ïïï= =íïï - >ïïïî
-10 10 20
-0.4
-0.2
0.2
0.4
Ts,xs
Separable Separable propertyproperty
continuocontinuousus
discontidiscontinuousnuous
1 , s x2(s, x)1, x s2
T
ìïï >ïïï=íï -ï >ïïïî
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15-20 -15 -10 -5 0 5 10 15 20-20
-15
-10
-5
0
5
10
15
20
Separable form of fundamental Separable form of fundamental solution (2D)solution (2D)
-20 -15 -10 -5 0 5 10 15 20-20
-15
-10
-5
0
5
10
15
20
Ro
s ( , )R q=
x ( , )r f=
iU
eU
r
1
1
1( , ; , ) ln ( ) cos ( ),(s, x)
1( , ; , ) ln ( ) cos ( ),
i m
m
e m
m
U R R m Rm R
URU R m R
m
rq r f q f r
q r f r q f rr
¥
=¥
=
ìïï = - - ³ïïïï=íïï = - - >ïïïïî
åå
x ( , )r f=
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Boundary density discretizationBoundary density discretizationFourier Fourier seriesseries
Ex . constant Ex . constant elementelement
Present Present methodmethod
Conventional Conventional BEMBEM
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OutlinesOutlines Motivation and literature reviewMotivation and literature review Mathematical formulationMathematical formulation
Expansions of fundamental solutionExpansions of fundamental solution and boundary densityand boundary density
Adaptive observer systemAdaptive observer system Vector decomposition techniqueVector decomposition technique Linear algebraic equationLinear algebraic equation
Numerical examplesNumerical examples Degenerate scaleDegenerate scale ConclusionsConclusions
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Adaptive observer systemAdaptive observer system
( , )r f
collocation collocation pointpoint
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OutlinesOutlines Motivation and literature reviewMotivation and literature review Mathematical formulationMathematical formulation
Expansions of fundamental solutionExpansions of fundamental solution and boundary densityand boundary density
Adaptive observer systemAdaptive observer system Vector decomposition techniqueVector decomposition technique Linear algebraic equationLinear algebraic equation
Numerical examplesNumerical examples Degenerate scaleDegenerate scale ConclusionsConclusions
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Vector decomposition technique for Vector decomposition technique for potential gradientpotential gradient
zx
z x-
(s, x) 1 (s, x)(s, x) cos( ) cos( )2
U ULrpz x z x
r r f¶ ¶= - + - +¶ ¶
(s, x) 1 (s, x)(s, x) cos( ) cos( )2
T TM rpz x z x
r r f¶ ¶= - + - +¶ ¶
Special case Special case (concentric case) :(concentric case) :
z x=(s, x)(s, x) ULr r
¶= ¶(s, x)(s, x) TM r r
¶= ¶
Non-Non-concentric concentric
case:case:
(x)2 (s, x) (s) (s) (s, x) (s) (s), x
(x)2 (s, x) (s) (s) (s, x) (s) (s), x
B B
B B
u M u dB L t dB D
u M u dB L t dB D
r r
ff
p
p
¶ = - ζ¶ = - ζ
ò òò ò
n
t
nt
t
n
True normal True normal directiondirection
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OutlinesOutlines Motivation and literature reviewMotivation and literature review Mathematical formulationMathematical formulation
Expansions of fundamental solutionExpansions of fundamental solution and boundary densityand boundary density
Adaptive observer systemAdaptive observer system Vector decomposition techniqueVector decomposition technique Linear algebraic equationLinear algebraic equation
Numerical examplesNumerical examples Degenerate scaleDegenerate scale ConclusionsConclusions
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{ }
0
1
2
N
ì üï ïï ïï ïï ïï ïï ïï ïï ï=í ýï ïï ïï ïï ïï ïï ïï ïï ïî þ
tt
t t
tM
Linear algebraic equationLinear algebraic equation
[ ]{ } [ ]{ }U t T u=
[ ]00 01 0
10 11 1
0 1
N
N
N N NN
é ùê úê úê ú= ê úê úê úê úë û
U U UU U U
U
U U U
LL
M M O ML
whwhereere
Column vector of Column vector of Fourier coefficientsFourier coefficients(Nth routing circle)(Nth routing circle)
0B1B
Index of Index of collocation collocation
circlecircle
Index of Index of routing circle routing circle
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Explicit form of each submatrix [Explicit form of each submatrix [UUpkpk] an] and vector {d vector {ttkk}}0 1 1
1 1 1 1 10 1 1
2 2 2 2 20 1 1
3 3 3 3 3
0 1 12 2 2 2
( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
c c s Mc Mspk pk pk pk pkc c s Mc Mspk pk pk pk pkc c s Mc Mspk pk pk pk pk
pk
c c s Mc Mspk M pk M pk M pk M pk
U U U U UU U U U UU U U U U
U U U U U
ff ff fff ff fff ff f
ff ff
é ù=ê úë ûU
LLL
M M M O M ML 2
0 1 12 1 2 1 2 1 2 1 2 1
( )( ) ( ) ( ) ( ) ( )
Mc c s Mc Mspk M pk M pk M pk M pk MU U U U U
fff ff f+ + + + +
é ùê úê úê úê úê úê úê úê úê úê úê úê úë ûL
{ } { }0 1 1
Tk k k k kk M Mp p q p q=t L
1f
2f
3f
2Mf
2 1Mf +
Fourier Fourier coefficientscoefficients
Truncated Truncated terms of terms of Fourier seriesFourier series
Number of Number of collocation pointscollocation points
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Flowchart of present methodFlowchart of present method
0 [ (s, x) (s) (s, x) (s)] (s)BT u U t dB= -ò
Potential Potential of domain of domain
pointpointAnalytiAnalyticalcal
NumeriNumericalcal
Adaptive Adaptive observer observer systemsystem
DegeneratDegenerate kernele kernel
Fourier Fourier seriesseries
Linear algebraic Linear algebraic equation equation
Collocation point and Collocation point and matching B.C.matching B.C.
Fourier Fourier coefficientscoefficients
Vector Vector decompodecompo
sitionsition
Potential Potential gradientgradient
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Comparisons of conventional BEM Comparisons of conventional BEM and present methodand present method
BoundaryBoundarydensitydensity
discretizatdiscretizationionAuxiliaryAuxiliarysystemsystem FormulatFormulationion ObserObserverver
systesystemm
SingularSingularityity
ConventiConventionalonalBEMBEM
Constant,Constant,Linear,Linear,
QurdraturQurdrature…e…FundameFundamentalntalsolutionsolution
BoundarBoundaryy
integralintegralequationequation
FixedFixedobserobserverver
systesystemm
CPV, RPCPV, RPVVand HPVand HPV
PresentPresentmethodmethod
FourierFourierseriesseries
expansioexpansionn
DegeneraDegeneratetekernelkernel
Null-Null-fieldfield
integralintegralequationequation
AdaptiAdaptiveve
obserobserverver
systesystemm
NoNoprincipprincip
alalvaluevalue
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OutlinesOutlines Motivation and literature reviewMotivation and literature review Mathematical formulationMathematical formulation
Expansions of fundamental solutionExpansions of fundamental solution and boundary densityand boundary density
Adaptive observer systemAdaptive observer system Vector decomposition techniqueVector decomposition technique Linear algebraic equationLinear algebraic equation
Numerical examplesNumerical examples Degenerate scaleDegenerate scale ConclusionsConclusions
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Numerical examplesNumerical examples Steady state heat conduction problemsSteady state heat conduction problems Electrostatic potential of wiresElectrostatic potential of wires Flow of an ideal fluid pass cylindersFlow of an ideal fluid pass cylinders A circular bar under torqueA circular bar under torque An infinite medium under antiplane sheAn infinite medium under antiplane shearar Half-plane problemsHalf-plane problems
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Numerical examplesNumerical examples Steady state heat conduction problemsSteady state heat conduction problems Electrostatic potential of wiresElectrostatic potential of wires Flow of an ideal fluid pass cylindersFlow of an ideal fluid pass cylinders A circular bar under torqueA circular bar under torque An infinite medium under antiplane sheAn infinite medium under antiplane shearar Half-plane problemsHalf-plane problems
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Steady state heat conduction Steady state heat conduction problemsproblems
Case Case 11
Case Case 22
1u=
0u=
1 2.5a =2 1.0a =
1u=
1u=
0u=
0 2.0R =
a
a
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Steady state heat conduction Steady state heat conduction problemsproblems
Case Case 33
Case Case 44
0un
¶ =¶
1u=
0u=
0un
¶ =¶
0 2.0R =
a
a
a
1u=
0u=
0un
¶ =¶
1u=
0 2.0R =
a
a
a
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Case 1: Isothermal lineCase 1: Isothermal line
Exact Exact solutionsolution(Carrier and (Carrier and Pearson)Pearson)
BEM-BEPO2DBEM-BEPO2D(N=21)(N=21)
FEM-ABAQUSFEM-ABAQUS(1854 (1854 elements)elements)
Present Present methodmethod(M=10)(M=10)-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
-1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
-1 .5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5-2 .5
-2
-1 .5
-1
-0 .5
0
0.5
1
1.5
2
2.5
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0 90 180 270 360
Degr ee ( )
0
1
2
3
Relat
ive e
rror
of f
lux o
n th
e sm
all c
ircle
(%) B E M -B E P O2 D (N = 2 1 )
P r es ent met hod (M = 1 0 )Tr efft z met hod (N T= 2 1 )M FS (N M = 2 1 ) (a1 '= 3 .0 , a2 '= 0 .7 )
Relative error of flux on the small Relative error of flux on the small circlecircle
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Convergence test - Parseval’s sum for Convergence test - Parseval’s sum for Fourier coefficientsFourier coefficients
0 4 8 12 16 20
Ter ms of Four ier s er ies (M )
1 0
1 1
1 2
1 3
1 4
1 5
Pars
eval'
s su
m
0 4 8 12 16 20
Ter ms of Four ier s er ies (M )
2
2.4
2.8
3.2
3.6
Pars
eval'
s su
m
22 2 2 2
00 1
( ) 2 ( )M
n nn
f d a a bp
q q p p=
+ +åò B&Parseval’s Parseval’s sumsum
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Case 2: Isothermal lineCase 2: Isothermal line
Caulk’s data (1983)Caulk’s data (1983)IMA Journal of Applied MatheIMA Journal of Applied Mathematicsmatics Present Present method method (M=10)(M=10)
FEM-ABAQUSFEM-ABAQUS(6502 (6502 elements)elements)
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
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Case 3: Isothermal lineCase 3: Isothermal line
FEM-ABAQUSFEM-ABAQUS(8050 (8050 elements)elements)
Present Present method method (M=10)(M=10)
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Caulk’s data (1983)Caulk’s data (1983)IMA Journal of Applied MatheIMA Journal of Applied Mathematicsmatics
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Case 4: Isothermal lineCase 4: Isothermal line
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
FEM-ABAQUSFEM-ABAQUS(8050 (8050 elements)elements)
Present Present method method (M=10)(M=10)
Caulk’s data (1983)Caulk’s data (1983)IMA Journal of Applied MatheIMA Journal of Applied Mathematicsmatics
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37
Numerical examplesNumerical examples Steady state heat conduction problemsSteady state heat conduction problems Electrostatic potential of wiresElectrostatic potential of wires Flow of an ideal fluid pass cylindersFlow of an ideal fluid pass cylinders A circular bar under torqueA circular bar under torque An infinite medium under antiplane sheAn infinite medium under antiplane shearar Half-plane problemsHalf-plane problems
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38
Electrostatic potential of wiresElectrostatic potential of wires
Hexagonal Hexagonal electrostatic electrostatic
potentialpotential
Two parallel cylinders Two parallel cylinders held positive and held positive and
negative potentialsnegative potentials
1u=- 1u=
2l
aa1u=
1u=-1u=
1u=-
1u= 1u=-
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39
Contour plot of potentialContour plot of potential
Exact solution (LebeExact solution (Lebedev et al.)dev et al.) Present Present method method (M=10)(M=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 10-10
-8
-6
-4
-2
0
2
4
6
8
10
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40
Contour plot of potentialContour plot of potential
-10 -8 -6 -4 -2 0 2 4 6 8 10-10
-8
-6
-4
-2
0
2
4
6
8
10
Onishi’s data Onishi’s data (1991)(1991) Present Present method method (M=10)(M=10)
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41
Numerical examplesNumerical examples Steady state heat conduction problemsSteady state heat conduction problems Electrostatic potential of wiresElectrostatic potential of wires Flow of an ideal fluid pass cylindersFlow of an ideal fluid pass cylinders A circular bar under torqueA circular bar under torque An infinite medium under antiplane sheAn infinite medium under antiplane shearar Half-plane problemsHalf-plane problems
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42
Flow of an ideal fluid pass two Flow of an ideal fluid pass two parallel cylindersparallel cylinders
is the velocity of flow far is the velocity of flow far from the cylindersfrom the cylinders is the incident angleis the incident anglev¥
g
v¥
g
2l
a a
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43
Velocity field in different incident Velocity field in different incident angleangle
-14 -12 -10 -8 -6 -4 -2 0 2 4-10
-8
-6
-4
-2
0
2
4
6
8
10
-14 -12 -10 -8 -6 -4 -2 0 2 4-10
-8
-6
-4
-2
0
2
4
6
8
10
Present Present method method (M=10)(M=10)
180g= o
Present Present method method (M=10)(M=10)
135g= o
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44
Numerical examplesNumerical examples Steady state heat conduction problemsSteady state heat conduction problems Electrostatic potential of wiresElectrostatic potential of wires Flow of an ideal fluid pass cylindersFlow of an ideal fluid pass cylinders A circular bar under torqueA circular bar under torque An infinite medium under antiplane sheAn infinite medium under antiplane shearar Half-plane problemsHalf-plane problems
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45
Torsion bar with circular holes Torsion bar with circular holes removedremoved
The warping The warping functionfunction
Boundary conditionBoundary condition
wherewhere
2 ( ) 0,x x DjÑ = Î
j
sin cosk k k kx ynj q q¶ = -¶ kB
2 2cos , sini ii ix b y b
N Np p= =
2 kNp
a
a
ab q
R
oonn
TorqTorqueue
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46
Axial displacement with two circular Axial displacement with two circular holesholes
Present Present method method (M=10)(M=10)
Caulk’s data (1983)Caulk’s data (1983)ASME Journal of Applied MechASME Journal of Applied Mechanicsanics-2
-1 .5
-1
-0 .5
0
0.5
1
1.5
2
-2-1.5-1-0 .500.511.52
Dashed line: exact Dashed line: exact solutionsolutionSolid line: first-order Solid line: first-order solutionsolution
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47
Axial displacement with three Axial displacement with three circular holescircular holes
Present Present method method (M=10)(M=10)
-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
Caulk’s data (1983)Caulk’s data (1983)ASME Journal of Applied MechASME Journal of Applied Mechanicsanics
Dashed line: exact Dashed line: exact solutionsolutionSolid line: first-order Solid line: first-order solutionsolution
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48
Axial displacement with four circular Axial displacement with four circular holesholes
Present Present method method (M=10)(M=10)
-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
Caulk’s data (1983)Caulk’s data (1983)ASME Journal of Applied MechASME Journal of Applied Mechanicsanics
Dashed line: exact Dashed line: exact solutionsolutionSolid line: first-order Solid line: first-order solutionsolution
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49
Numerical examplesNumerical examples Steady state heat conduction problemsSteady state heat conduction problems Electrostatic potential of wiresElectrostatic potential of wires Flow of an ideal fluid pass cylindersFlow of an ideal fluid pass cylinders A circular bar under torqueA circular bar under torque An infinite medium under antiplane sheAn infinite medium under antiplane shearar Half-plane problemsHalf-plane problems
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50
Infinite medium under antiplane shearInfinite medium under antiplane shearThe displacementThe displacement
Boundary conditionBoundary condition
Total displacementTotal displacement
t
m
sw2 ( ) 0,sw x x DÑ = Î
( ) sinsw xn
t qm
¶ =¶
sw w w¥= +
oonn
kB
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51
Shear stress Shear stress σz around the hole of radiu around the hole of radius as a11 (x axis) (x axis)
0 1 2 3 4 5 6 (in r adians )
- 2
0
2
4
6
8
z
/ (
aro
un
d h
ole
wit
h r
ad
ius
a1)
d/a1 = 0 .0 1d/a1 = 0 .1d/a1 = 2 .0s ingle hole
Present Present method method (M=20)(M=20)
Honein’s data (1Honein’s data (1992)992)Quarterly of Applied MathQuarterly of Applied Mathematicsematics
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52
Shear stress Shear stress σz around the hole of radiu around the hole of radius as a11 (y axis) (y axis)
0 1 2 3 4 5 6 (in r adians )
- 2
0
2
4
6
8
z
/ (
aro
un
d h
ole
wit
h r
ad
ius
a1)
d/a1= 0 .0 1d/a1= 0 .1d/a1= 2 .0
Present Present method method (M=20)(M=20)
Honein’s data (1Honein’s data (1992)992)Quarterly of Applied MathQuarterly of Applied Mathematicsematics
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53
Shear stress Shear stress σz around the hole of radiu around the hole of radius as a11 (45 degrees) (45 degrees)
0 1 2 3 4 5 6 (in r adians )
- 2
0
2
4
6
8
1 0
z
/ (
aro
un
d h
ole
wit
h r
ad
ius
a1)
d/a1= 0 .0 1d/a1= 0 .1d/a1= 2 .0
Present Present method method (M=20)(M=20)
Honein’s data (1Honein’s data (1992)992)Quarterly of Applied MathQuarterly of Applied Mathematicsematics
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54
Shear stress Shear stress σz around the hole of radiu around the hole of radius as a11 (Touching) (Touching)
0 1 2 3 4 5 6 (in r adians )
- 2
0
2
4
6
8
1 0
z
/ (
aro
un
d h
ole
wit
h r
ad
ius
a1)
M = 1 0M = 2 0M = 3 0M = 4 0
Present Present methodmethod
discontidiscontinuousnuous
discontidiscontinuousnuous
1a 2a
Honein’s data (1Honein’s data (1992)992)Quarterly of Applied MathQuarterly of Applied MathematicsematicsGibb’s Gibb’s
phenomenophenomenonn
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55
Two equivalent approachesTwo equivalent approaches
0 sinw R q=
0R
d
2a
1a
sinwn
q¶ =¶
0R
d
2a
1a
Displacement Displacement approachapproach
Stress Stress approachapproach
Present Present methodmethod
Bird and Steele Bird and Steele (1992)(1992)ASME Journal of Applied ASME Journal of Applied MechanicsMechanics
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56
Shear stress Shear stress σz around the hole of radiu around the hole of radius as a11
0 90 180 270 360
-2
0
2
4
R 0= 7 .5R 0= 1 5 .0R 0= 3 0 .0
0 90 180 270 360
-2
0
2
4
R 0 = 7 .5R 0 = 1 5 .0R 0 = 3 0 .0
Present Present method method (M=20)(M=20)
Present Present method method (M=20)(M=20)
Steele’s data Steele’s data (1992)(1992)
Stress Stress approachapproach
Displacement Displacement approachapproachHonein’s data Honein’s data (1992)(1992)5.35.3
48485.35.34949
4.64.64747
5.35.34545
13.1313.13%%
0.020.02%%
AnalytiAnalyticalcal
0.060.06%%
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57
Convergence of stress σzat =45 degrees versus R0
0 30 60 90 120 150
R adius R 0
0
2
4
6
8
z
at
=45
deg
rees
Equivalent dis placement appr oachEquivalent s t r es s appr oach
0 sinw R q=
sinwn
q¶ =¶
0t=
0t=
0R
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58
Three circular holes with centers on Three circular holes with centers on the x axisthe x axis
0 1 2 3 4 5 6 (in r adians )
- 2
0
2
4
6
8
z
/ (
arou
nd h
ole w
ith ra
dius
a 1)
d/a1= 2 .0d/a1= 0 .1d/a1= 0 .0 1
1a2a3a
y
x
t
m
dd
![Page 59: Null-field approach for Laplace problems with circular boundaries using degenerate kernels](https://reader034.vdocuments.net/reader034/viewer/2022042901/5681376e550346895d9f07e3/html5/thumbnails/59.jpg)
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Three circular holes with centers on Three circular holes with centers on the y axisthe y axis
0 1 2 3 4 5 6 (in r adians )
- 2
- 1
0
1
2
z
/ (a
roun
d ho
le wi
th ra
dius
a 1)
d/a1= 2 .0d/a1= 0 .1d/a1= 0 .0 1
x
y
1a
2a
3a
t
m
d
d
![Page 60: Null-field approach for Laplace problems with circular boundaries using degenerate kernels](https://reader034.vdocuments.net/reader034/viewer/2022042901/5681376e550346895d9f07e3/html5/thumbnails/60.jpg)
60
Three circular holes with centers on Three circular holes with centers on the line making 45 degreesthe line making 45 degrees
0 1 2 3 4 5 6 (in r adians )
- 2
0
2
4
6
8
1 0
z
/ (a
roun
d ho
le wi
th ra
dius
a 1)
d/a1= 2 .0d/a1= 0 .1d/a1= 0 .0 1
1a
2a
3a
x
y
t
m d
d
![Page 61: Null-field approach for Laplace problems with circular boundaries using degenerate kernels](https://reader034.vdocuments.net/reader034/viewer/2022042901/5681376e550346895d9f07e3/html5/thumbnails/61.jpg)
61
Numerical examplesNumerical examples Steady state heat conduction problemsSteady state heat conduction problems Electrostatic potential of wiresElectrostatic potential of wires Flow of an ideal fluid pass cylindersFlow of an ideal fluid pass cylinders A circular bar under torqueA circular bar under torque An infinite medium under antiplane sheAn infinite medium under antiplane shearar Half-plane problemsHalf-plane problems
![Page 62: Null-field approach for Laplace problems with circular boundaries using degenerate kernels](https://reader034.vdocuments.net/reader034/viewer/2022042901/5681376e550346895d9f07e3/html5/thumbnails/62.jpg)
62
Half-plane problemsHalf-plane problems
Dirichlet boundary cDirichlet boundary conditionondition(Lebedev et al.)(Lebedev et al.)Mixed-type boundary coMixed-type boundary conditionndition(Lebedev et al.)(Lebedev et al.)
0u=
1u=
1B
2B
0u=
1un
¶ =¶
1B
2B
h ha a
![Page 63: Null-field approach for Laplace problems with circular boundaries using degenerate kernels](https://reader034.vdocuments.net/reader034/viewer/2022042901/5681376e550346895d9f07e3/html5/thumbnails/63.jpg)
63
Dirichlet problemDirichlet problem
Exact solution (LebeExact solution (Lebedev et al.)dev et al.) Present Present method method (M=10)(M=10)
IsothermIsothermal lineal line
- 1 0 - 8 - 6 - 4 - 2 0 2 4 6 8 1 0- 1 0
- 8
- 6
- 4
- 2
0
- 1 0 - 8 - 6 - 4 - 2 0 2 4 6 8 1 0- 1 0
- 8
- 6
- 4
- 2
0
![Page 64: Null-field approach for Laplace problems with circular boundaries using degenerate kernels](https://reader034.vdocuments.net/reader034/viewer/2022042901/5681376e550346895d9f07e3/html5/thumbnails/64.jpg)
64
Mixed-type problemMixed-type problem
Exact solution (LebeExact solution (Lebedev et al.)dev et al.) Present Present method method (M=10)(M=10)
IsothermIsothermal lineal line
- 1 0 - 8 - 6 - 4 - 2 0 2 4 6 8 1 0- 1 0
- 8
- 6
- 4
- 2
0
- 1 0 - 8 - 6 - 4 - 2 0 2 4 6 8 1 0- 1 0
- 8
- 6
- 4
- 2
0
![Page 65: Null-field approach for Laplace problems with circular boundaries using degenerate kernels](https://reader034.vdocuments.net/reader034/viewer/2022042901/5681376e550346895d9f07e3/html5/thumbnails/65.jpg)
65
OutlinesOutlines Motivation and literature reviewMotivation and literature review Mathematical formulationMathematical formulation
Expansions of fundamental solutionExpansions of fundamental solution and boundary densityand boundary density
Adaptive observer systemAdaptive observer system Vector decomposition techniqueVector decomposition technique Linear algebraic equationLinear algebraic equation
Numerical examplesNumerical examples Degenerate scaleDegenerate scale ConclusionsConclusions
![Page 66: Null-field approach for Laplace problems with circular boundaries using degenerate kernels](https://reader034.vdocuments.net/reader034/viewer/2022042901/5681376e550346895d9f07e3/html5/thumbnails/66.jpg)
66
Numerical instability in BEMNumerical instability in BEM2r
1ra Annular Annular
casecaseInterior Interior
casecaseMax Max errorerror
DegeneratDegenerate scalee scale
u specified= u specified=
International Journal International Journal forforNumerical Methods in Numerical Methods in EngineeringEngineering
Engineering Engineering AnalysisAnalysiswith Boundary with Boundary Elements Elements
Matrix Matrix singularsingular
ErroErrorr
SinguSingularlarvaluevalue
![Page 67: Null-field approach for Laplace problems with circular boundaries using degenerate kernels](https://reader034.vdocuments.net/reader034/viewer/2022042901/5681376e550346895d9f07e3/html5/thumbnails/67.jpg)
67
2 21 1 1 1 1 1 2 2 2 1 2 1
2 2
2 21 1 1 2 1 2 2 2 2 2 2 2
2 2
2 21 1 1 2 1 1 2 1 2 2 2 2 1 2
2 2
2 ln cos sin 2 ln ( )cos ( )sin
2 ln cos sin 2 ln ( )cos ( )sin
2 ln cos sin 2 ln ( )cos ( )sinM M M
a aa a a a a a a
a aa a a a a a a
a aa a a a a a a
p p f p f p r p f p fr r
p p f p f p r p f p fr r
p p f p f p r p f pr r+ + +
L L
L L
M M M O M M M O
L 2 1
1 11 1 1 1 1 1 2 2 2 1 2 1
1 1
1 11 1 1 2 1 2 2 2 2 2 2 2
1 1
1 11 1 1 2 1 1 2 1 2 2 2 2 1
1 1
2 ln ( )cos ( )sin 2 ln cos sin
2 ln ( )cos ( )sin 2 ln cos sin
2 ln ( )cos ( )sin 2 ln cos
M
M M M
a a a a a a a aa a
a a a a a a a aa a
a a a a a a aa a
f
r rp p f p f p p f p f
r rp p f p f p p f p f
r rp p f p f p p f
+
+ + +
L
L L
L L
M M M O M M M O
L
1,0
1,1
1,1
1,
1,
2,0
2,1
2,1
2,
2,2 2 1sin
M
M
M
MM
ppq
pqppq
pqap f +
é ùê úì üê úï ïï ïê úï ïï ïê úï ïê úï ïï ïê úï ïï ïê úï ïê úï ïï ïê úï ïï ïê úï ïê úï ïï ïê úï ïï ïê úï ïï ïí ýê úï ïï ïê úï ïï ïê úï ïï ïê úï ïê úï ïï ïê úï ïï ïê úï ïê úï ïïê úïïê úïê úïïê úïïî þê úïïê úê úë ûê ú
M
M
L
[ ]
1,0
1,1
1,1
1,
1,
2,0
2,1
2,1
2,
2,
M
M
M
M
aab
abaab
ab
ì üï ïï ïï ïï ïï ïï ïï ïï ïï ïï ïï ïï ïï ïï ïï ïï ïï ïï ïï ïï ïï ï= í ýï ïï ïï ïï ïï ïï ïï ïï ïï ïï ïï ïï ïï ïï ï ïï ï ïï ï ïï ï ïï ï ïï ï ïï ï ïï ï ïî þï ï ïï ï ï
T
M
M
Degenerate scale in the multiply Degenerate scale in the multiply connected problemconnected problem
a1 =1.0, influence matrix [U] is singular
1a
![Page 68: Null-field approach for Laplace problems with circular boundaries using degenerate kernels](https://reader034.vdocuments.net/reader034/viewer/2022042901/5681376e550346895d9f07e3/html5/thumbnails/68.jpg)
68
Treatments of degenerate scale Treatments of degenerate scale problemproblem
Method of adding a Method of adding a rigid body termrigid body term
CHEEF conceptCHEEF concept
[ ]{ } [ ]{ }=U t T u
(s, x) (s, x)mU U c= +
12 a cp+ 1 12 (ln )a a cpé ù+ê úê úë ûL
M O1 12 lna apé ùê úê úë û
LM O
SinguSingularlar
[ ]{ } [ ]{ }=U t T u
SinguSingularlar
Auxiliary Auxiliary constraint constraint { } { }=w t v u
[ ]{ } [ ]{ }=U t T u[ ] { } [ ] { }é ù é ùê ú ê ú=ê ú ê úê ú ê úë û ë ûU T
t uw vNonsingulaNonsingula
rr
1a
CHEEF CHEEF pointpoint
Promote Promote rankrank
![Page 69: Null-field approach for Laplace problems with circular boundaries using degenerate kernels](https://reader034.vdocuments.net/reader034/viewer/2022042901/5681376e550346895d9f07e3/html5/thumbnails/69.jpg)
69
0.5 1 1.5 2 2.5 3
R adius a1
0
0.1
0.2
0.3
0.4
0.5
1
P r es ent met hodA dding a CH EEF poing (5 .0 ,5 .0 )A dding a r igid body t er m (c= 1 .0 )
The minimum singular value versus The minimum singular value versus radius aradius a11
DegeneratDegenerate scalee scale
1a
Numerical Numerical failurefailure
![Page 70: Null-field approach for Laplace problems with circular boundaries using degenerate kernels](https://reader034.vdocuments.net/reader034/viewer/2022042901/5681376e550346895d9f07e3/html5/thumbnails/70.jpg)
70
OutlinesOutlines Motivation and literature reviewMotivation and literature review Mathematical formulationMathematical formulation
Expansions of fundamental solutionExpansions of fundamental solution and boundary densityand boundary density
Adaptive observer systemAdaptive observer system Vector decomposition techniqueVector decomposition technique Linear algebraic equationLinear algebraic equation
Numerical examplesNumerical examples Degenerate scaleDegenerate scale ConclusionsConclusions
![Page 71: Null-field approach for Laplace problems with circular boundaries using degenerate kernels](https://reader034.vdocuments.net/reader034/viewer/2022042901/5681376e550346895d9f07e3/html5/thumbnails/71.jpg)
71
ConclusionsConclusions A systematic approach using A systematic approach using degenerate kdegenerate kernelsernels, , Fourier seriesFourier series and and null-field integral null-field integral equationequation has been successfully proposed t has been successfully proposed to solve Laplace problems with circular bouo solve Laplace problems with circular boundaries.ndaries. Numerical results Numerical results agree wellagree well with available with available exact solutions, Caulk’s data, Onishi’s dexact solutions, Caulk’s data, Onishi’s data and FEM (ABAQUS) for ata and FEM (ABAQUS) for only few terms oonly few terms of Fourier seriesf Fourier series..
![Page 72: Null-field approach for Laplace problems with circular boundaries using degenerate kernels](https://reader034.vdocuments.net/reader034/viewer/2022042901/5681376e550346895d9f07e3/html5/thumbnails/72.jpg)
72
ConclusionsConclusions Method of adding a rigid body termMethod of adding a rigid body term and and CHEEF CHEEF
approachapproach have been successfully adopted to have been successfully adopted to overcome the overcome the degenerate scale for multiply degenerate scale for multiply connected problemconnected problem..
The The stress concentrationstress concentration due to due to different different orientationsorientations was discussed by using present was discussed by using present method.method.
Engineering problemsEngineering problems with with circular boundariescircular boundaries which satisfy the which satisfy the Laplace equationLaplace equation can be solved can be solved by using the proposed approach in a by using the proposed approach in a more more efficient and accurate mannerefficient and accurate manner..
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The endThe end
Thanks for your kind attentions.Thanks for your kind attentions.Your comments will be highly apprYour comments will be highly appreciated.eciated.
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Further researchFurther research Expansion to general boundary, e.g. elExpansion to general boundary, e.g. elliptic, straight, degenerate.liptic, straight, degenerate. Antiplane problem with rigid inclusioAntiplane problem with rigid inclusionn Expansion to three-dimensional problExpansion to three-dimensional problemem Bi-center expansion techniqueBi-center expansion technique
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Derivation of degenerate kernelDerivation of degenerate kernel Graf’s addition theoremGraf’s addition theorem Complex variableComplex variable
s xs ( , ) , x ( , )R z zq r f= = = =
x sln x s ln z z- = - Real Real partpart
x x xs x s s s
1s s s
1ln( ) ln[( )(1 )] ln( ) ln(1 ) ln( ) ( )mm
z z zz z z z zz z m z
¥
=- = - = + - = - å
( )x
1 1 1 1s
1 1 1 1( ) ( ) ( ) [ ] ( ) cos ( )i
m m m i m mi
m m m m
z e e mm z m Re m R m R
ff q
qr r r q f
¥ ¥ ¥ ¥-
= = = == = = -å å å å
Real Real partpart
IfIf s xz z-
1
1
1( , ; , ) ln ( ) cos ( ),(s, x)
1( , ; , ) ln ( ) cos ( ),
i m
m
e m
m
U R R m Rm R
URU R m R
m
rq r f q f r
q r f r q f rr
¥
=¥
=
ìïï = - - ³ïïïï=íïï = - - >ïïïïî
åå
ln R
2
2 3
1
1ln(1 ) (1 )11 1( )2 31 m
m
x dx x x dxx
x x x
xm
¥
=
- =- =- + + +-=- + + +
=-
ò ò
å
L
L
0k ®
Bessel’s Bessel’s functionfunction
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Non-unique solutionsNon-unique solutions
Non-unique Non-unique solutionssolutions
Rigid body Rigid body solutionsolutionfor Neumann for Neumann problemsproblems
Critical size of thCritical size of theedomain in plane domain in plane BVPs BVPs Hypersingular formulatiHypersingular formulationonfor multiply connected for multiply connected problemsproblems
u specifiedn
¶ =¶ u specified=1a= 1a=
MathematicalMathematically andly andphysically physically realizablerealizable
Mathematically Mathematically realizablerealizable
Mathematically Mathematically realizablerealizable
[ ]{ } [ ]{ }=U t T u[ ]{ } [ ]{ }[ ]{ } [ ]{ }
==
U t T u
L t M u
[ ]{ } [ ]{ }[ ]{ } [ ]{ }
==
U t T u
L t M uDegeneratDegenerate scalee scale
Non-Non-uniquenesuniquenesss
2 0uÑ =2 0uÑ =
2 0uÑ =
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Non-unique solutions in direct BEMNon-unique solutions in direct BEMDomain of Domain of interestinterest
SingularSingularformulationformulation
HypersinHypersingulargularformulatiformulationon
SimplySimplyconnecconnec
tedteddomaidomai
nn
InteriInterioror
casecase
a=1.0a=1.0 NANA
ExterExteriorior
casecase
a=1.0a=1.0 a is a is arbitrararbitrar
yy
MultiplMultiplyy
connecconnectedted
domaidomainn
AnnulAnnularar
casecase
a=1.0a=1.0 a is a is arbitrararbitrar
yyEccenEccentrictriccasecase
a=1.0a=1.0 a is a is arbitrararbitrar
yy
a
a
a
a
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Solutions of half-plane problemSolutions of half-plane problem
1u=-
1u= 1u=
Half-plane Half-plane problemproblem
Infinite Infinite problemproblem
Image Image conceptconcept
Anti-symmetry Anti-symmetry propertyproperty
(s; x, x ) ln x s ln x sU ¢ ¢= - - -
1s(s; x, x ) 0BU ΢ =0u=
1u=
1B
2B 2B
1B
1B
x¢
s
x rr¢
2B
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FormulationFormulation
2 (x) (s; x, x ) (s) (s) (s; x, x ) (s) (s), xB B
u T u dB U t dB Dp ¢ ¢= - Îò ò0 (s; x, x ) (s) (s) (s;x, x ) (s) (s), x c
B BT u dB U t dB D¢ ¢= - Îò ò
1
1
1
1
1( , ; , , , ) ln ( ) cos ( )
1ln ( ) cos ( ),(s; x, x )
1( , ; , , , ) ln ( ) cos ( )
1ln ( ) cos ( ),
i m
m
m
m
e m
m
m
m
U R R mm R
R m Rm
URU R m
mR m R
m
rq r f r f q f
r q f r rr
q r f r f r q fr
r q f r rr
¥
=¥
=¥
=¥
=
ìïï ¢ ¢= - -ïïïïïïï ¢ ¢ ¢- + - > ³ïï ¢ïï¢=ïíïï ¢ ¢= - -ïïïïïïï ¢ ¢ ¢- + - > >ïï ¢ïïîï
åååå