1 Null-field approach for Laplace problems with circular boundaries using degenerate kernels 研究生：沈文成指導教授：陳正宗教授時間： 10:30 ~ 12:00 地點：河工二館

1

Null-field approach for Laplace proNull-field approach for Laplace problems with circular boundaries usinblems with circular boundaries using degenerate kernelsg degenerate kernels

研究生：沈文成指導教授：陳正宗教授時間： 10:30 ~ 12:00地點：河工二館 307 室

碩士論文口試

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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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4

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

5

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 solutionsolution

CPV and CPV and HPVHPV

No 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

6

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)

7

Motivation and literature reviewMotivation and literature review

Analytical methods for solving Laplace problems with circ

ular 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

8

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 conduc

tiontion with circular holes with circular holes Bird and Steele (1992) - Bird and Steele (1992) - harmonic and harmonic and

biharmonicbiharmonic problems with circular hol problems with circular holeses

Mogilevskaya et al. (2002) - Mogilevskaya et al. (2002) - elasticityelasticity pr problems with circular boundariesoblems with circular boundaries

9

Contribution and goalContribution and goal

However, they didn’t employ the However, they didn’t employ the nnull-field integral equationull-field integral equation and and degedegenerate kernelsnerate kernels to fully capture the ci to fully capture the circular boundary, although they all ercular boundary, although they all employed mployed Fourier series expansionFourier series expansion..

To develop a To develop a systematic approachsystematic approach f for solving Laplace problems with or solving Laplace problems with mmultiple holesultiple holes is our goal. is our goal.

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OutlinesOutlines





11

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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13

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

UR

U R m Rm

rq r f q f r

q r f r q f rr

¥

=

¥

=

ìïï = - - ³ïïïï=íïï = - - >ïïïïî

å

å

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

å

å

14

Separable form of fundamental Separable form of fundamental solution (1D)solution (1D)

-10 10 20

2

4

6

8

10

Us,x

2

1

2

1

(x) (s), s x

(s, x)

(s) (x), x s

i ii

i ii

a b

U

a b

=

=

ìïï ³ïïïï=íïï >ïïïïî

å

å

1(s x), s x

1 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,x

s

Separable Separable propertyproperty

continuocontinuousus

discontidiscontinuousnuous

1, s x2(s, x)1, x s

2

T

ìïï >ïïï=íï -ï >ïïïî

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

UR

U R m Rm

rq r f q f r

q r f r q f rr

¥

=

¥

=


å

å

x ( , )r f=

16

Boundary density discretizationBoundary density discretization

Fourier Fourier seriesseries

Ex . constant Ex . constant elementelement

Present Present methodmethod

Conventional Conventional BEMBEM

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Adaptive observer systemAdaptive observer system

( , )r f

collocation collocation pointpoint

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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 ULr

pz x z x

r r f¶ ¶

= - + - +¶ ¶

(s, x) 1 (s, x)(s, x) cos( ) cos( )

2

T TM r

pz 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)T

M 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

uM u dB L t dB D

uM u dB L t dB D

r r

ff

p

p

¶= - Î

¶¶

= - Î¶

ò ò

ò ò

n

t

nt

t

n

True normal True normal directiondirection

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22

{ }

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

t

t

t t

t

M

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 U

U U UU

U U U

L

L

M M O M

L

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 11 1 1 1 1

0 1 12 2 2 2 2

0 1 13 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 U

U U U U U

U U U U U

U U U U U

ff ff f

ff ff f

ff ff f

ff ff

é ù=ê úë ûU

L

L

L

M M M O M M

L 20 1 1

2 1 2 1 2 1 2 1 2 1

( )

( ) ( ) ( ) ( ) ( )M

c c s Mc Mspk M pk M pk M pk M pk MU U U U U

f

ff 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

24

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

25

Comparisons of conventional BEM Comparisons of conventional BEM and present methodand present method

BoundaryBoundarydensitydensity

discretizatdiscretizationion

AuxiliaryAuxiliarysystemsystem

FormulatFormulationion

ObserObserverver

systesystemm

SingularSingularityity

ConventiConventionalonal

BEMBEM

Constant,Constant,Linear,Linear,

QurdraturQurdrature…e…

FundameFundamentalntal

solutionsolution

BoundarBoundaryy

integralintegral

equationequation

FixedFixed

obserobserverver

systesystemm

CPV, RPCPV, RPVV

and HPVand HPV

PresentPresentmethodmethod

FourierFourier

seriesseries

expansioexpansionn

DegeneraDegeneratete

kernelkernel

Null-Null-fieldfield

integralintegral

equationequation

AdaptiAdaptiveve

obserobserverver

systesystemm

NoNo

principprincipalal

valuevalue

26

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27

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 she

arar Half-plane problemsHalf-plane problems

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29

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

30

Steady state heat conduction Steady state heat conduction problemsproblems

Case Case 33

Case Case 44

0u

n

¶=

¶

1u=

0u=

0u

n

¶=

¶

0 2.0R =

a

a

a

1u=

0u=

0u

n

¶=

¶

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

32

0 90 180 270 360

Degr ee ( )

0

1

2

3

Rel

ativ

e er

ror

of f

lux

on t

he

sm

all

circ

le (

%)

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

33

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

Par

sev

al's

sum

0 4 8 12 16 20

Ter ms of Four ier s er ies (M )

2

2.4

2.8

3.2

3.6

Par

sev

al's

sum

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

34


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)


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

35



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


36


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


elements)elements)Present Present method method (M=10)(M=10)


37




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=-

39

Contour plot of potentialContour plot of potential

Exact solution (LebeExact solution (Lebedev et al.)dev et al.)


-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

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)


41




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 angle

v¥

g

v¥

g

2l

a a

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


180g= o


135g= o

44




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 yn

jq q

¶= -

¶ kB

2 2cos , sini i

i ix b y b

N N

p p= =

2 k

N

p

a

a

ab q

R

oonn

TorqTorqueue

46

Axial displacement with two circular Axial displacement with two circular holesholes


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 solutionsolution

Solid line: first-order Solid line: first-order solutionsolution

47

Axial displacement with three Axial displacement with three circular holescircular holes


-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




48

Axial displacement with four circular Axial displacement with four circular holesholes


-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




49




50

Infinite medium under antiplane shearInfinite medium under antiplane shear

The displacementThe displacement

Boundary conditionBoundary condition

Total displacementTotal displacement

t

m

sw2 ( ) 0,sw x x DÑ = Î

( )sin

sw x

n

tq

m¶

=¶

sw w w¥= +

oonn

kB

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 1

d/a1 = 0 .1

d/a1 = 2 .0

s ingle hole


Honein’s data (1Honein’s data (1992)992)Quarterly of Applied MathQuarterly of Applied Mathematicsematics

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

adiu

s a

1)

d/a1= 0 .0 1

d/a1= 0 .1

d/a1= 2 .0



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

und

hole

wit

h ra

dius

a1

)

d/a1= 0 .0 1

d/a1= 0 .1

d/a1= 2 .0



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




1a 2a


Gibb’s Gibb’s phenomenophenomeno

nn

55

Two equivalent approachesTwo equivalent approaches

0 sinw R q=

0R

d

2a

1a

sinw

nq

¶=

¶

0R

d

2a

1a

Displacement Displacement approachapproach

Stress Stress approachapproach


Bird and Steele Bird and Steele (1992)(1992)

ASME Journal of Applied ASME Journal of Applied MechanicsMechanics

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 .5

R 0= 1 5 .0

R 0= 3 0 .0

0 90 180 270 360

-2

0

2

4

R 0= 7 .5

R 0= 1 5 .0

R 0= 3 0 .0



Steele’s data Steele’s data (1992)(1992)

Stress Stress approachapproach

Displacement Displacement approachapproachHonein’s data Honein’s data

(1992)(1992)5.35.34848

5.35.34949

4.64.64747

5.35.34545

13.1313.13%%

0.020.02%%

AnalytiAnalyticalcal

0.060.06%%

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 oach

Equivalent s t r es s appr oach

0 sinw R q=

sinw

nq

¶=

¶

0t=

0t=

0R

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

/

(aro

und

hol

e w

ith

radi

us

a1

)

d/a1= 2 .0

d/a1= 0 .1

d/a1= 0 .0 1

1a2a3a

y

x

t

m

dd

59

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

/

(aro

un

d h

ole

wit

h r

ad

ius

a1)

d/a1= 2 .0

d/a1= 0 .1

d/a1= 0 .0 1

x

y

1a

2a

3a

t

m

d

d

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

/

(aro

und

hole

wit

h ra

dius

a1)

d/a1= 2 .0

d/a1= 0 .1

d/a1= 0 .0 1

1a

2a

3a

x

y

t

m d

d

61




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=

1u

n

¶=

¶

1B

2B

h ha a

63

Dirichlet problemDirichlet problem



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

64

Mixed-type problemMixed-type problem



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

65

OutlinesOutlines





66

Numerical instability in BEMNumerical instability in BEM

2r

1r

a Annular Annular casecase

Interior Interior casecase

Max Max errorerror

DegeneratDegenerate scalee scale

u specified= u specified=

International Journal International Journal forfor

Numerical Methods in Numerical Methods in EngineeringEngineering

Engineering Engineering AnalysisAnalysis

with Boundary with Boundary Elements Elements

Matrix Matrix singularsingular

ErroErrorr

SinguSingularlar

valuevalue

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

p

p

q

p

q

p

p

q

p

qap 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

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[ ]

1,0

1,1

1,1

1,

1,

2,0

2,1

2,1

2,

2,

M

M

M

M

a

a

b

a

b

a

a

b

a

b

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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

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 û

L

M O1 12 lna apé ù

ê úê úë û

L

M 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 û

U Tt u

w vNonsingulaNonsingula

rr

1a

CHEEF CHEEF pointpoint

Promote Promote rankrank

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 hod

A dding a CH EE F 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

70

OutlinesOutlines





71

ConclusionsConclusions

A systematic approach using A systematic approach using degenerate degenerate kernelskernels, , Fourier seriesFourier series and and null-field integnull-field integral equationral equation has been successfully propo has been successfully proposed to solve Laplace problems with circulsed to solve Laplace problems with circular boundaries.ar boundaries.

Numerical results Numerical results agree wellagree well with availabl with available exact solutions, Caulk’s data, Onishi’e exact solutions, Caulk’s data, Onishi’s data and FEM (ABAQUS) for s data and FEM (ABAQUS) for only few teronly few terms of Fourier seriesms of Fourier series..

72

ConclusionsConclusions

Method of adding a rigid body termMethod of adding a rigid body term and and CHEEF aCHEEF approachpproach have been successfully adopted to over have been successfully adopted to overcome the come the degenerate scale for multiply connectdegenerate scale for multiply connected problemed problem..

The The stress concentrationstress concentration due to due to different orientdifferent orientationsations was discussed by using present method. was discussed by using present method.

Engineering problemsEngineering problems with with circular boundariescircular boundaries which satisfy the which satisfy the Laplace equationLaplace equation can be solve can be solved by using the proposed approach in a d by using the proposed approach in a more effimore efficient and accurate mannercient and accurate manner..

73

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.

74

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

75

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( ) ( )m

m

z z zz z z z z

z z m z

¥

=

- = - = + - = - å

( )x

1 1 1 1s

1 1 1 1( ) ( ) ( ) [ ] ( ) cos ( )

im m m i m m

im m m m

z ee m

m z m Re m R m R

ff q

q

r r rq 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

UR

U R m Rm

rq r f q f r

q r f r q f rr

¥

=

¥

=


å

å

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

76

Non-unique solutionsNon-unique solutions

Non-unique Non-unique solutionssolutions

Rigid body Rigid body solutionsolution

for 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

uspecified

n

¶=

¶u specified=1a= 1a=

MathematicalMathematically andly and

physically 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 u

DegeneratDegenerate scalee scaleNon-Non-

uniquenesuniquenesss

2 0uÑ =2 0uÑ =

2 0uÑ =

77

Non-unique solutions in direct BEMNon-unique solutions in direct BEM

Domain of Domain of interestinterest

SingularSingular

formulationformulationHypersinHypersin

gulargularformulatiformulati

onon

SimplySimply

connecconnectedted

domaidomainn

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


yy

EccenEccentrictric

casecase


yy

a

a

a

a

78

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 Î¢ =0u=

1u=

1B

2B 2B

1B

1B

x¢

s

xr

r¢

2B

79

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

Rm R

mU

RU R m

m

Rm 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

å

å

å

å

Documents

1 Null-field approach for Laplace problems with circular boundaries using degenerate kernels 研究生：沈文成 指導教授：陳正宗 教授 時間： 10:30 ~ 12:00 地點：河工二館

1 Null-field approach for Laplace problems with circular boundaries using degenerate kernels 研究生：沈文成指導教授：陳正宗教授時間： 10:30 ~ 12:00 地點：河工二館