研究生：吳清森指導教授：陳正宗　教授　　　　　陳義麟　博士

1 力學聲響振動研究室 (MSVLAB)

Degenerate scale analysis for membrane and plate problems using the meshless method and

boundary element method

研究生：吳清森指導教授：陳正宗　教授　　　　　陳義麟　博士

國立台灣海洋大學河海工程學系結構組

碩士班論文口試

日期 : 2004/06/16 09:00-10:20


Frame of the thesis

Chapter 1 Introduction

Free term and Jump term

Chapter 5 Free terms for plate problem

(Biharmonic problem)

Degenerate kernel

Chapter 2 Green’s function and Poisson integral formula

(Laplace problem)

Chapte3 BIEM and BEM for degenerate scale problem

(Laplace and biharmonic problem)

Chapter 4 Meshless method for degenerate scale problem

(Laplace and biharmonic problem)

Chapter 6 Conclusions and further research


Literature review (Engineering background)

Engineers Year Problem Journal Treatment

He et al.

(China)

1996 2-D potential and plane elasticity problem

Computer & Structures

CNME

Necessary and sufficient boundary integral formulation (NSBIE)

Zhou et al.

(China)

1999 2-D elasticity problem

CNME Boundary contour method

(BCM)

He, W. J.

(China)

2000 BIE---thin plate Computer & Structures

Equivalent BIE

Chen et al.

(Taiwan)

2002 Plane elasticity

(Dual BIEM)

IJNME Hypersingular formulation

Chen et al.

(Taiwan)

2002 2-D Laplace equation

(Degenerate kernel)

EABE Combined Helmholtz exterior integral equation formulation (CHEEF)

Chen et al.

(Taiwan)

2003 2-D Laplace and Navier problem

IJNME Addition of rigid body term

Hypersingular formulation

CHEEF concept


Literature review (Mathematical background)

Mathematician Year Problem Journal Treatments and logarithm capacity

Soren Christiansen

(Denmark)

1982 Detect non-unique solution

Applicable Analysis

Scaling method

Restriction method

Constanda

(U. K.)

1995 Non-unique solution in plane elasticity problem

Quart. Appl. Math.

First kind integral equations

Martin et al.

(France)

1996 Invertibility of single layer potential operator

Integr. Equat.

Oper. Th.

Logarithm capacity

Soren Christiansen

(Denmark)

1998 Investigation of direct BIE for biharmonic

problem

JCAM Logarithm capacity

Soren Christiansen

(Denmark)

2001 Detecting non-uniqueness of solution

through SVD

JCAM Logarithm capacity

2

3

2

110 ,,,

eeee

4

3

2

110 ,,,

eeee

2

110 ,,

eee


Motivation

(1) BIEM, BEM

(2) MFS, Trefftz Method

Methods Techniques

(1) Degenerate kernel

(2) Circulants

Membrane

(Laplace equation)

Plate

(biharmonic equation)

Statics

Degenerate scale problem

1-D case (Euler beam)


S

x

rx (field point): variable

s (source point): fixed

Degenerate kernel

jjj

E

jjj

I

sxxBsAsxU

sxsBxAsxU

sxU

),()(),(

),()(),(

),(

x

sO1

R

iUeU

A

),(

),(

Rs

x

O2

x

s

R

iUeU

B

O1

O2


Alternative derivations for the Poisson integral formula


G. E.: xxu ,0)(2

B. C. :

)(fu

a

Derivation of the Poisson integral formula

Traditional method

R 'R

Image source

Null-field integral equation method

Reciprocal radii method

Poisson integral formula

Image concept

Methods

Free of image concept

Searching the image point

Degenerate kernel

2

0 22

22

)()cos(22

1),( df

aa

au


Null-field integral equation in conjunction with degenerate kernels

xsdBstxsUsdBsuxsTxuB

IF

B

IF ,)()(),()()(),()(2

a

Bx ),( ),( Rs

Bx

BcE

FB

EF xsdBstxsUsdBsuxsT ),()(),()()(),(0

1

0 ))sin()cos(()()(n

nn nbnaafsu Boundary densities: .))sin()cos(()(

10

n

nn nqnppst

Degenerate kernel

Unknown coefficients

dfma

dnbnaama

u

m

m

nnn

m

m

2

0 1

2

01

01

)()](cos[)(212

1

))sin()cos(()](cos[)(212

1),(

c

unknown

specified

Fundamental solution

Green’s identity

,ln),( rsxU F


Degenerate scale for plate analysis using the BIEM and BEM


Engineering problem governed by biharmonic equation

1. Plane elasticity:

2. Slow viscous flow (Stokes’ Flow):

3. Solid mechanics (Plate problem):

functionstressAiry:,04

functionstream:,04

ntdisplacemelateraluu :,04


Problem statement

uniform pressure

a

B

w=constant

0)(,0)( xxu

: flexure rigidity

: uniform distributed load

: domain of interest

)(xu

D

)(xw

Governing equation: xD

xwxu ,

)()(4

Boundary condition: Bxxxu ,0)(,0)(

Splitting method

Governing equation: xxu ,0)(*4

Boundary condition: Bxxxxuxu ),()(),()(****

: deflection of the circular plate


Boundary integral equations for plate

),()}(),()(),()(),()(),({)(8 sdBsuxsVsxsMsmxssvxsUxuB

),()}(),()(),()(),()(),({)(8 sdBsuxsVsxsMsmxssvxsUxB

),()}(),()(),()(),()(),({)(8 sdBsuxsVsxsMsmxssvxsUxm mmB

mm

),()}(),()(),()(),()(),({)(8 sdBsuxsVsxsMsmxssvxsUxv vvB

vv

)(, xK

)(, xmK

)(, xvK

(2) Slope

(3) Normal moment

(4) Effective shear force

(1) Displacement )(, sK

)(, smK )(, svK


Operators

nK

)(

)(

2

2

2 )()1()()(

nK

m

))(

()1()(

)(22

tntnK

v

Slope

Normal moment

Effective shear force

t

n


Kernel functions

)(8),(4 sxsxUx Fundamental solution:

)ln(),( 2 rrsxU

Kernel functions:)),((),( , xsUKxs s

)),((),( , xsUKxsM sm

)),((),( , xsUKxsV sv


)(, sK

)(, smK )(, svK


Degenerate kernels for biharmonic operator

RmRmm

mRmm

RRRRRxsU

RmR

mmm

R

mmRRxsU

m m

m

m m

mI

m m

m

m m

mE

xsU,)](cos[

)1(

1)](cos[

)1(

1)cos()ln21(ln)ln1(),(

,)](cos[)1(

1)](cos[

)1(

1)cos()ln21(ln)ln1(),(

2 21

222

2 21

222

),(

1 2 2

11

2 11 1

22

)],(cos[1

1)](cos[

)1(

2)cos()ln21()ln1(2),(

,)](cos[)1(

2)](cos[

1

1)cos()ln23()ln21(),(

),( m m m

m

m

mI

m m

m

m m

mE

RmRm

mRmm

mRRRxsU

RmR

mm

mm

R

mR

RxsU

xsU

RmRmm

mm

RmRRR

Rxs

RmR

mm

R

mm

mRxs

m m

m

m m

mI

m m m

m

m

mE

xs,)](cos[

)1(

2)](cos[

1

1)cos()ln23()ln21(),(

,)](cos[1

1)](cos[

)1(

2)cos()ln21()ln1(2),(

2 11 1

22

1 2 2

11),(

RmRm

mm

Rm

mR

Rxs

RmR

m

mm

R

m

mRxs

m m

m

m m

mI

m m

m

m m

mE

xs,)]([cos

1

2)]([cos

1

2)cos()ln23(

2),(

,)](cos[1

2)]([cos

1

2)cos()ln23(

2),(

2 1

1

1 1

1

2 1

1

1 1

1),(


Mathematical analysis --- Discrete model

c

cc

f

f

m

vSM

2

1][

NN

c

U

USM

44

][

For the clamped circular plate (u and are specified):

)(sv)(sm

a}]{[}]{[}{ 1 mvUf c

}]{[}]{[}{ 2 mvUf c

,u formulation:


Circulant

NNN

NNN

NN

N

zzzzz

zzzz

zzzz

zzzz

U

22012321

3201222

221012

12210

][

12,...,2,1,0,),,,()],,,([)

2

1(

)2

1(

NmaaaUdaaaUz

m

mmm

NNe Ni

,1,..,2,1,0,2

2

NNezzN

m

N

mi

m

N

m

mm

U ),1(,...,2,1,0,12

0

2

212

0

][

daaaU

aaaUm m

N

mN

U

2

0

12

0

][

)]0,,,()[cos(

)]0,,,([)cos(lim

a1

23

45

2N-12N-22N-3

2N


Eigenvalues of the four matrices

NNlalll

laa

laaU

l

),1(,...,3,2,)1)(1(

2

1),ln22

3(

0),ln21(2

3

3

3

][

NNlalll

l

laa

laa

l

),1(,...,3,2,]1

1

)1(

2[

1),ln22

5(

0),ln1(4

2

2

2

][

NNlall

l

l

laa

laaU

l

),1(,...,3,2,])1(

2

1

1[

1),ln22

5(

0),ln1(4

2

2

2

][

NNlal

l

l

l

laa

la

l

),1(,...,3,2,]1

2

1

2[

1),ln22

3(

0,4][

U

U

kernel kernel

kernelkernel


Determinant

1

11

11

0

0

0

0

][

U

U

U

UcSM

N

N

UU

U

UcSM)1(

][][][][ )(det]det[

NNllll

alaa

laaa

),1(,...,3,2,)1)(1(

21)),ln(1(4

0],))(ln()ln(1[8

2

42

42

242

0)ln(1 aDegenerate scale

a)(sv

)(sm


,u

mu, 0lnln)ln1)(1(2)ln21)(1( 2 aaaa

vu, 0)ln2ln24()ln1( aaa

m,

v,

vm,

Degenerate scales for the clamped case Degenerate scales for the simply-supported case

0ln1 a

02ln)ln1()ln1( aaa

0ln2)ln23( aa42C

6 options

Formulation Equation of the degenerate scale in the BEM

Never zero


),()}(),()(),()(),()(),({)(8 sdBsuxsVsxsMsmxssvxsUxB

),()}(),()(),()(),()(),({)(8 sdBsuxsVsxsMsmxssvxsUxm mm

Bmm

),()}(),()(),()(),()(),({)(8 sdBsuxsVsxsMsmxssvxsUxv vvB

vv


Degenerate scale

0 0.2 0.4 0.6 0.8 1

R adius a

1E-005

0.0001

0.001

0.01

0.1

1

10

D e t

0 .3 6 8a

1ea

0 0.2 0.4 0.6 0.8 1

R ad iu s a

1E-013

1E-012

1E-011

1E-010

1E-009

1E-008

1E-007

1E-006

1E-005

0.0001

0.001

0.01

0.1

1

10

100

D et

0.071

0.368

a

u formulation vu formulation 0)ln2ln24()ln1( aaa 0ln1 a3.0


Degenerate scales for the free case

,u

mu,

vu,

m,

v,

vm,

Formulation Equation of the degenerate scale in the BEM

Never zero except three rigid body modes







Relationship between the Laplace problem and biharmonic problem

(a) translation:

),( u

(b) rotation:

cos4

1),(

au

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

sin4

1),(

au

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

constant


Nontrivial modes in FEM and BEM

FEM BEM

Rigid body mode Spurious mode

(Hour-glass mode)

(zero-energy mode)

Rigid body modes Spurious mode

(Null-field)

Physically realizable Mathematical

realizable Physically realizable Mathematical

realizable

Q4 or Q8Q4 or Q8

0u


Number of degenerate scales (Laplace problem)

Laplace problem:

0 0.25 0.5 0.75 1a

-1-0.500.5

-0.3-0.2

-0.10

0.1

F,a0 0.25 0.5 0.75 1

a

-1-0.500.5

0 0.25 0.5 0.75 1a

-1-0.5

00.5

0

20

40

F,a0 0.25 0.5 0.75 1

a

-1-0.5

00.5

UT formulation:

LM formulation:

0)ln( a

No degenerate scale


Number of degenerate scales (biharmonic problem)

0 0.25 0.5 0.75 1a

-1-0.5

00.5

-1

0

1

F,a0 0.25 0.5 0.75 1

a

-1-0.5

00.5

0 0.25 0.5 0.75 1a

-1-0.5

00.5

-1

0

1

2

F,a0 0.25 0.5 0.75 1

a

-1-0.5

00.5

0.2 0.4 0.6 0.8 1 1.2 1.4-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.2 0.4 0.6 0.8 1 1.2 1.4-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0 0.25 0.5 0.75 1a

-1-0.500.5

-6

-4

-2

0

F,a0 0.25 0.5 0.75 1

a

-1-0.500.5

0.2 0.4 0.6 0.8 1 1.2 1.4-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

u

mu

vu

formulation

formulation

formulation

0 0.2 0 .4 0 .6 0 .8 1

R ad ius a

1E -005

0.0001

0.001

0.01

0.1

1

10

D e t

0 .3 6 8

0 0.2 0.4 0.6 0.8 1

R ad iu s a

1E-013

1E-012

1E-011

1E-010

1E-009

1E-008

1E-007

1E-006

1E-005

0.0001

0.001

0.01

0.1

1

10

100

D et

0.071

0.368


0 0.25 0.5 0.75 1a

-1-0.5

00.5

-3

-2

-1

0

F,a0 0.25 0.5 0.75 1

a

-1-0.5

00.5

0.2 0.4 0.6 0.8 1 1.2 1.4-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

Number of degenerate scales (biharmonic problem)

0 0.25 0.5 0.75 1a

-1-0.5

00.5

-2

0

2

4

F,a0 0.25 0.5 0.75 1

a

-1-0.5

00.5

0 0.25 0.5 0.75 1

a

-1-0.50

0.5

0

5

10

F,a0 0.25 0.5 0.75 1

a

0.2 0.4 0.6 0.8 1 1.2 1.4-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.2 0.4 0.6 0.8 1 1.2 1.4-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

m

v

vm

formulation

formulation

formulationNo degenerate

scale occurs


Illustrative example (JFM, Mill 1977)

a

02 u0

n

u

1n

u1

0 We adopt the null-field integral equation in conjunction with degenerate kernel to derive the analytic solution.

- 1 - 0 . 8 - 0 . 6 - 0 . 4 - 0 . 2 0 0 . 2 0 . 4 0 . 6 0 . 8 1

B I E M ( N = 2 0 )

- 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

Exact so lution

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

2 1 1 011 1 1( , ) (1 )[ ( tan ) ( tan )]

2 1 2 1 2

r ru r r tan tan

r r

- 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

Exact solution : M=20 M=50

0u

2sin]

)(8[)22(

8

1)(

11

222

m

Rm

R

RRxu

M

mm

m


On the equivalence of the Trefftz method and MFS for Laplace and biharmonic equations


Trefftz method and MFS

Method Trefftz method MFS

Definition

Base uj(x) (T-complete function) , r=|x-s|

G. E. L u(x)=0, L U(x,s)=0, (singularity at s)

Match B. C. Determine cj Determine wj

TN

jjj xucxu

1

)()(

MN

jjj sxUwxu

1

),()(

Dx

is the number of complete functions TN

MN is the number of source points in the MFS

Dx

s

Du(x)

~x

r

~s

D

u(x)

~x

)(),( rsxU


Statement for Laplace problem

Two-dimensional Laplace problem with a circular domain:

Dxxu ,0)(2G.E. : B.C. : Bxuxu ,)(

B

D

a aD

B

Interior : Exterior :

N

n

n

N

n

n nbnaaau

11

0 )sin()cos(),( Analytical solution:


By matching the boundary condition at a

,00 aa

Tnn

n Nna

aa ,...2,1,

Tnn

n Nna

bb ,...2,1

TT N

n

nn

N

n

nn

I nabnaaaau

11

0 )sin()cos(),(

TT N

n

nn

N

n

nn

E nabnaaaaau

11

0 )sin()cos(ln),(

Interior problem:

Exterior problem: ,

ln

100 a

aa

Tnn

n Nnaaa ,...2,1,

Tnn

n Nnbab ,...2,1

Derivation of unknown coefficients(Trefftz method)

Field solution:Interior :

Exterior :

T-complete set functions :

)sin(),cos(,1 nn nn

)sin(),cos(,ln nn nn Interior:

Exterior:


Degenerate kernel :

RnR

nRU

RnRn

RRUrRU

n

ne

n

ni

1

1

)),(cos()(1

)ln(),;,(

)),(cos()(1

)ln(),;,(ln),,,(

MN

jj Rca

1

0 )ln(

M

N

jj

njn

nNnn

Rnc

a M

,...2,1,)cos()1

(1

1

M

N

jj

njn

nNnn

Rnc

b M

,...2,1,)sin()1

(1

1

20],))(cos()(1

)[ln(),(

1 1

MN

j n

jn

jI n

R

a

nRcau

20],))(cos()(1

)[ln(),(

1 1

MN

j n

jn

jE n

a

R

nacau

Interior problem:

Exterior problem:

MN

j

jcaa

1

0)ln(

1

M

N

j

jn

jnn NnnR

ncaa

M

,...2,1,)cos()(1

1

M

N

j

jn

jnn NnnR

ncba

M

,...2,1,)sin()(1

1

Field solution: Interior :

Exterior :

Derivation of unknown coefficients(MFS)


,2

2

1

1

0

N

N

b

a

b

a

b

a

a

w

12

2

5

4

3

2

1

N

N

c

c

c

c

c

c

c

c

cKw Trefftz MFS

Relationship between the two methods

)sin()1

(1

)sin()1

(1

)sin()1

(1

)sin()1

(1

)cos()1

(1

)cos()1

(1

)cos()1

(1

)cos()1

(1

)sin(1

)sin(1

)sin(1

)sin(1

)cos(1

)cos(1

)cos(1

)cos(1

lnlnlnln

12321

12321

12321

12321

NNNNN

NNNNN

N

N

I

NRN

NRN

NRN

NRN

NRN

NRN

NRN

NRN

RRRR

RRRR

RRRR

K

)sin()(1

)sin()(1

)sin()(1

)sin()(1

)cos()(1

)cos()(1

)cos()(1

)cos()(1

)sin()sin()sin()sin(

)cos()cos()cos()cos(

1111

12321

12321

12321

12321

NNNNN

NNNNN

N

N

E

NRN

NRN

NRN

NRN

NRN

NRN

NRN

NRN

RRRR

RRRR

K

Interior:

Exterior:

By setting 12 TN

Trefftz method

MFS

= MN = 12 N


)12()12(1221

1221

1221

1221

1221

1221

)sin()sin()sin(

)cos()cos()cos(

)2sin()2sin()2sin(

)2cos()2cos()2cos(

)sin()sin()sin(

)cos()cos()cos(

111

][

NNN

N

N

N

N

N

NNN

NNN

T

)12()12(2

12000

02

1200

02

120

0012

][][

NN

T

N

N

NN

TT

Matrix TTK R

numbernaturalNN

TN

N

,02

)12(]det[

2

1


Matrix RT

N

aR ill-posed problem


0ln R

)12()12(

2

2

)1

(1

0000

0)1

(1

000

00

)1

(2

1

)1

(2

1

01

00

00001

0

00000)ln(

][

NN

N

N

IR

RN

RN

R

R

R

R

R

T

)12()12(

2

2

)(1

0000

0)(1

000

00

)(2

1

)(2

1000

00000

000001

][

NN

N

N

ER

RN

RN

R

R

R

R

T

)1( R

N

aR ill-posed problem

Degenerate scale problem?

)ln(

)ln(

a

a


Circulants

12212

222

1210][

NNNNN CaCaCaIaU

where

00001

10000

00100

00010

12

NC

dkUmkmU

N

mNk )cos()0,(

1)cos()0,(lim

2

0

12

0

2

1

1

212 1)(

ln2)2(]det[

N

NI

R

a

R

RaNUInterior

:

Degenerate scale problem (R=1)

a

R=1fail

2

1

1

212 1)(

ln2)2(]det[

N

NE

a

R

a

aRNUExterior

:

Nonunique problem (a=1)

a=1R

fail


Numerical Examples

0)(:.. xuEG

DB

a

x

y)3cos()( xu

)3cos(),( 3 rru

)3cos(1

ln),(3

r

rcru

Simply-connected problem Multiply-connected problem

0u 1u

5.2ln

ln),(

u

5.22 a

11 a

0u 1u

X

Y

}cos816

cos8116{

2ln2

1),(

2

2

u

11 a

5.22 a

D

Ba

)3cos()( xu


Numerical Example 1

Trefftz method for the simply-connected problem

Interior problem Exterior problem

Exact solution Numerical solution Exact solution Numerical solution

5 Points: B.C. aliasing

base deficiency

9 Points:

a=1 5 Points:

B.C. aliasing

Failure ( )9 Points:

Failure ( )

a=2 5 Points:

B.C.

aliasing

9 Points:

-1 .0 -0 .8 -0 .6 -0 .4 -0 .2 0 .0 0 .2 0 .4 0 .6 0 .8 1 .0

E xact so lu tion

-1 .0

-0 .8

-0 .6

-0 .4

-0 .2

0 .0

0 .2

0 .4

0 .6

0 .8

1 .0

-1 .0 -0 .8 -0 .6 -0 .4 -0 .2 0 .0 0 .2 0 .4 0 .6 0 .8 1 .0

T refftz m eth od (5 sets)

-1 .0

-0 .8

-0 .6

-0 .4

-0 .2

0 .0

0 .2

0 .4

0 .6

0 .8

1 .0

-1 .0 -0 .8 -0 .6 -0 .4 -0 .2 0 .0 0 .2 0 .4 0 .6 0 .8 1 .0

T refftz m eth od (9 sets)

-1 .0

-0 .8

-0 .6

-0 .4

-0 .2

0 .0

0 .2

0 .4

0 .6

0 .8

1 .0

-5 .0 -4 .0 -3 .0 -2 .0 -1 .0 0 .0 1 .0 2 .0 3 .0 4 .0 5 .0

T refftz so lu tion (9 se ts)

-5 .0

-4 .0

-3 .0

-2 .0

-1 .0

0 .0

1 .0

2 .0

3 .0

4 .0

5 .0

-5 .0 -4 .0 -3 .0 -2 .0 -1 .0 0 .0 1 .0 2 .0 3 .0 4 .0 5 .0

T refftz so lu tio n (5 se ts )

-5 .0

-4 .0

-3 .0

-2 .0

-1 .0

0 .0

1 .0

2 .0

3 .0

4 .0

5 .0

ln

ln- 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5

- 5

- 4

- 3

- 2

- 1

0

1

2

3

4

5

- 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5- 5

- 4

- 3

- 2

- 1

0

1

2

3

4

5


MFS for the simply-connected problem

Interior problem Exterior problem


5 Points:

B.C.

aliasing

9 Points:

20 Points:

a=1: 5 Points:

B.C. aliasing

Failure ( ln a)

9 Points:

Failure ( ln a)

a=2: 5 Points:

B.C.

Aliasing

9 Points:

55 Points:

Numerical Example 2

-1 .0 -0 .8 -0 .6 -0 .4 -0 .2 0 .0 0 .2 0 .4 0 .6 0 .8 1 .0

E xact so lu tion

-1 .0

-0 .8

-0 .6

-0 .4

-0 .2

0 .0

0 .2

0 .4

0 .6

0 .8

1 .0

-1 .0 -0 .8 -0 .6 -0 .4 -0 .2 0 .0 0 .2 0 .4 0 .6 0 .8 1 .0

M F S (5 sets)

-1 .0

-0 .8

-0 .6

-0 .4

-0 .2

0 .0

0 .2

0 .4

0 .6

0 .8

1 .0

-1 .0 -0 .8 -0 .6 -0 .4 -0 .2 0 .0 0 .2 0 .4 0 .6 0 .8 1 .0

M F S (9 se ts)

-1 .0

-0 .8

-0 .6

-0 .4

-0 .2

0 .0

0 .2

0 .4

0 .6

0 .8

1 .0

-1 .0 -0 .8 -0 .6 -0 .4 -0 .2 0 .0 0 .2 0 .4 0 .6 0 .8 1 .0

M F S (55 sets)

-1 .0

-0 .8

-0 .6

-0 .4

-0 .2

0 .0

0 .2

0 .4

0 .6

0 .8

1 .0

-5 .0 -4 .0 -3 .0 -2 .0 -1 .0 0 .0 1 .0 2 .0 3 .0 4 .0 5 .0

E x a c t so lu tio n (r= 1 )

-5 .0

-4 .0

-3 .0

-2 .0

-1 .0

0 .0

1 .0

2 .0

3 .0

4 .0

5 .0

-5 .0 -4 .0 -3 .0 -2 .0 -1 .0 0 .0 1 .0 2 .0 3 .0 4 .0 5 .0

E x a c t so lu tio n (r = 2 )

-5 .0

-4 .0

-3 .0

-2 .0

-1 .0

0 .0

1 .0

2 .0

3 .0

4 .0

5 .0

-5 .0 -4 .0 -3 .0 -2 .0 -1 .0 0 .0 1 .0 2 .0 3 .0 4 .0 5 .0

M F S (5 se ts )

-5 .0

-4 .0

-3 .0

-2 .0

-1 .0

0 .0

1 .0

2 .0

3 .0

4 .0

5 .0

-5 .0 -4 .0 -3 .0 -2 .0 -1 .0 0 .0 1 .0 2 .0 3 .0 4 .0 5 .0

M F S (9 se ts )

-5 .0

-4 .0

-3 .0

-2 .0

-1 .0

0 .0

1 .0

2 .0

3 .0

4 .0

5 .0

-5 .0 -4 .0 -3 .0 -2 .0 -1 .0 0 .0 1 .0 2 .0 3 .0 4 .0 5 .0

M F S (6 5 se ts )

-5 .0

-4 .0

-3 .0

-2 .0

-1 .0

0 .0

1 .0

2 .0

3 .0

4 .0

5 .0


Numerical Example 3

Trefftz method for the multiply-connected problem

Concentric circle Eccentric circle


26 Points 26 Points

6 Points

14 Points

26 Points

-1 .5 -1 .0 -0 .5 0 .0 0 .5 1 .0 1 .5 2 .0 2 .5 3 .0 3 .5

E x a ct so lu tio n

-2 .5

-2 .0

-1 .5

-1 .0

-0 .5

0 .0

0 .5

1 .0

1 .5

2 .0

2 .5

-1 .5 -1 .0 -0 .5 0 .0 0 .5 1 .0 1 .5 2 .0 2 .5 3 .0 3 .5

T refftz m eth o d (tt= 1 )

-2 .5

-2 .0

-1 .5

-1 .0

-0 .5

0 .0

0 .5

1 .0

1 .5

2 .0

2 .5

-1 .5 -1 .0 -0 .5 0 .0 0 .5 1 .0 1 .5 2 .0 2 .5 3 .0 3 .5


-2 .5

-2 .0

-1 .5

-1 .0

-0 .5

0 .0

0 .5

1 .0

1 .5

2 .0

2 .5

-1 .5 -1 .0 -0 .5 0 .0 0 .5 1 .0 1 .5 2 .0 2 .5 3 .0 3 .5


-2 .5

-2 .0

-1 .5

-1 .0

-0 .5

0 .0

0 .5

1 .0

1 .5

2 .0

2 .5

-1 .5 -1 .0 -0 .5 0 .0 0 .5 1 .0 1 .5 2 .0 2 .5 3 .0 3 .5


-2 .5

-2 .0

-1 .5

-1 .0

-0 .5

0 .0

0 .5

1 .0

1 .5

2 .0

2 .5

-1 .5 -1 .0 -0 .5 0 .0 0 .5 1 .0 1 .5 2 .0 2 .5 3 .0 3 .5


-2 .5

-2 .0

-1 .5

-1 .0

-0 .5

0 .0

0 .5

1 .0

1 .5

2 .0

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

5.2ln

ln),(

u


Numerical Example 4

MFS for the multiply-connected problem

Concentric circle Eccentric circle


Inner circle: 20 points outer circle: 60points

Inner circle: a1=0.9 outer circle :a2=2.6



Inner: 20points; outer: 60points; inner a1=0.9

outer a2=2.6 outer a2 =3.0

outer a2 =4.0 outer a2 =10.0

Inner: 20points; outer: 60points; outer a22.6

inner a1=0.5 inner a1 =0.3- 3 - 2 - 1 0 1 2 3

- 3

- 2

- 1

0

1

2

3

- 3 - 2 - 1 0 1 2 3

- 3

- 2

- 1

0

1

2

3

5.2ln

ln),(

u

- 3 - 2 - 1 0 1 2 3

- 3

- 2

- 1

0

1

2

3

}cos816

cos8116{

2ln2

1),(

2

2

u

- 3 - 2 - 1 0 1 2 3

- 3

- 2

- 1

0

1

2

3

- 3 - 2 - 1 0 1 2 3

- 3

- 2

- 1

0

1

2

3

- 3 - 2 - 1 0 1 2 3

- 3

- 2

- 1

0

1

2

3

- 3 - 2 - 1 0 1 2 3

- 3

- 2

- 1

0

1

2

3

- 3 - 2 - 1 0 1 2 3

- 3

- 2

- 1

0

1

2

3

- 3 - 2 - 1 0 1 2 3

- 3

- 2

- 1

0

1

2

3


Trefftz method and MFS for biharmonic equation

)(sin)(cos)()(sin)(cos),(1

2

1

220

110

* mdmccmbmaauTTTT N

m

mm

N

m

mm

N

m

mm

N

m

mm

Analytical solution:

TN

j

jj xugxu1

* )()(Field solution :

T-complete functions: )sin(),cos(,),sin(),cos(,1 222 mmmm mmmm

TN

j

jjx

xhn

xux

1

* )()(

)(

Trefftz method:

MFS:

,),()(1

MN

jjj sxUvxu

MM N

jjj

N

j x

jj

x

sxvn

sxUvx

n

xu

11

),(),(

)()(

Field solution :


Relationship between the Trefftz method and MFS

Coefficients of the

Trefftz method Coefficients of the

MFS Mapping matrix

[K]

1)24(24

14

4

3

2

1

)24()24(21

21

21

21

22

21

2

22

21

2

21

21

222

1)24(

1

1

0

1

1

0

)sin(1

)1(

1)sin(

1

)1(

1)sin(

1

)1(

1

)cos(1

)1(

1)cos(

1

)1(

1)cos(

1

)1(

1

)sin(2

1)sin(

2

1)sin(

2

1

)cos(2

1)cos(

2

1)cos(

2

1ln1ln1ln1

)sin()1

()1(

1)sin()

1(

)1(

1)sin()

1(

)1(

1

)cos()1

()1(

1)cos()

1(

)1(

1)cos()

1(

)1(

1

)sin()ln1()sin()ln1()sin()ln1(

)cos()ln1()cos()ln1()cos()ln1(

lnlnln

NN

N

NNNNNN

NNNN

N

N

NNNN

NNNN

N

N

NN

N

N

N

v

v

v

v

v

v

NRNN

NRNN

NRNN

NRNN

NRNN

NRNN

RRR

RRR

RRR

NRNN

NRNN

NRNN

NRNN

NRNN

NRNN

RRRRRR

RRRRRR

RRRRRR

d

c

d

c

c

b

a

b

a

a

M

M

M

M

M

M

M

M


Decomposition of the K matrix

][][][ TTK R

)24()24(241421

241421

241421

241421

241421

241421

241421

241421

sinsinsinsin

coscoscoscos

sinsinsinsin

coscoscoscos

1111

sinsinsinsin

coscoscoscos

sinsinsinsin

coscoscoscos

1111

][

NNNN

NN

NN

NN

NN

NN

NN

NN

NNNN

NNNN

NNNN

NNNN

T

)24()24(

2

2

2

)1(

1)1(

1

2

12

1ln1

)1(

1)1(

1

)ln21(

)ln21(

ln

][

NNN

N

N

N

R

NNR

NNR

R

R

RNNR

NNR

RR

RR

RR

T

1

2

3

12 N

4

0)12(2]det[ 12 NNT


)24()24(

2

2

2

)1(

1)1(

1

2

12

1ln1

)1(

1)1(

1

)ln21(

)ln21(

ln

][

NNN

N

N

N

R

NNR

NNR

R

R

RNNR

NNR

RR

RR

RR

T

Diagonal matrix TR

Existence of the degenerate scales

12

10 ,,

eeeRNonuniqueness

(in numerical aspect)

Rln

Rln21

Rln1 Degenerate scale problem

0

0 1)24()24()24(1)24( }{][}{ NjNNNj vKg O. K.!


Special size:

: position of the source points

The occurrence of the degenerate scales using the MFS

a

0e

1e2

1

e

Mathematics: rank-deficiency problem

(nonuniqueness problem)

Numerical failure



On the complete set of the Trefftz method and the MFS using the degenerate kernel

T-complete functions of the Trefftz method:

)sin(),cos(,),sin(),cos(,1 222 mmmm mmmm

Degenerate kernel of the MFS:

))(cos()1(

1))(cos(

)1(

)cos(2

1)cos()ln21(ln1)ln1(),(

22

2

2

322*

mRmm

mRmm

RRRRRRu

mm

m

mm

m

m=0m=0m=1m=1

m=2, 3….. m=2, 3…..


Free terms for the biharmonic equation using the dual boundary integral equation


History of free terms in the dual BEM

2-D and 3-D Laplace problem

2-D and 3-D elasticity problem

W. C. Chen thesis

2-D biharmonic problem

(1) Bump-contour method

(2) Taylor series expansion

Free terms

Dual boundary integral equation

Improper integrals


Bump-contour method

For a smooth boundary:

B

xy

B+ B-

B’ B’

B’

D

Singular point

4 ( ) 0u x

)sin,cos( s

)0,0(x

)(su

)0,1(

),1,0(

)cos,sin(

),sin,(cos

x

x

s

s

t

n

t

n

)(8),(4 sxsxUx sxrrrsxU ),ln(),( 2

Explicit forms for the sixteen kernel functions


Taylor expansion for boundary density functions

)(]sinsincossincoscos[!3

1

]sincossincos[!2

1]sincos[)()(

4333

2

32

212

1

32

221

33

31

3

2

21

22

22

22

21

2

21

Os

u

ss

u

ss

u

s

u

ss

u

s

u

s

u

s

u

s

uxusu

sn

sus

)(

)(

2

22 )(

)1()()(s

sn

sususm

ssss

s

tn

su

tn

susv

)()1(

)()(

22

Boundary

Domain

)0,0(x

B)sin,cos( s

vector component:

2,1, isxy iii


)(xu

)(2

xt

)(

2)(

2xuxt

Free terms of dual BIE for Laplace problem 2-D problem:

0

3-D problem:)(2 xu

)(3

2xt

)(2

)(3

4xuxt

BxsdBsuxsMVPHsdBstxsLVPCxt

BxsdBsuxsTVPCsdBstxsUVPRxu

BB

BB

,)()(),(...)()(),(...)(

,)()(),(...)()(),(...)(

)(])(),()(),([)(2 sdBstxsUsuxsTxuB

)(])(),()(),([)(2 sdBstxsLsuxsMxtB

)(])(),()(),([)(4 sdBstxsUsuxsTxuB

)(])(),()(),([)(4 sdBstxsLsuxsMxtB

0

BxsdBsuxsMVPHsdBstxsLVPCxt

BxsdBsuxsTVPCsdBstxsUVPRxu

BB

BB

,)()(),(...)()(),(...)(2

,)()(),(...)()(),(...)(2

Half

Half Singular point

B

partVPC ...

partVPC ...

partVPC ...

partVPC ...


Dual boundary integral equations

BxsdBsuxsVPFsdBsxsMPFsdBsmxsPFsdBsvxsUPFxuBBBB

,)()(),(..)()(),(..)()(),(..)()(),(..)(8

BxsdBsuxsVPFsdBsxsMPFsdBsmxsPFsdBsvxsUPFxBBBB

,)()(),(..)()(),(..)()(),(..)()(),(..)(8

BxsdBsuxsVPFsdBsxsMPFsdBsmxsPFsdBsvxsUPFxmB

mB

mB

mB

m ,)()(),(..)()(),(..)()(),(..)()(),(..)(8

BxsdBsuxsVPFsdBsxsMPFsdBsmxsPFsdBsvxsUPFxvB

vB

vB

vB

v ,)()(),(..)()(),(..)()(),(..)()(),(..)(8

F.P. denotes the finite part

for a smooth boundary2

1

Sixteen improper integrals

Density functions are expanded by the Taylor series


Singular behavior of the sixteen kernels

),( xsU ),( xs ),( xsM ),( xsV

),( xsU v

),( xsU m

),( xsU

),( xsv

),( xsm

),( xs

),( xsM v

),( xsM m

),( xsM

),( xsVv

),( xsVm

),( xsV

ln2O lnO

lnO

lnO

lnO

lnO

1

O

1

O

1

O

1

O

2

1

O

2

1

O

2

1

O

3

1

O

3

1

O

4

1

O


),( xsU ),( xs ),( xsM ),( xsV

),( xsU v

),( xsU m

),( xsU

),( xsv

),( xsm

),( xs

),( xsM v

),( xsM m

),( xsM

),( xsVv

),( xsVm

),( xsV

0 0 0

0 0

)(4 xu

)()1( x

)()3(4)()3(

xux

)()1(2

xm

)()1( xm

)(

)3

51()1(4

)(

x

xm

)(0)(

3

)3()1(8

)()3(2

2xu

x

xm

)()3)(1(2

xv

)()76(

3

4

)(2

2 xm

xv

2

)(16)5(3

3

)1(

)(

3

)7()1(8

)()2)(1(

x

xm

xv

3

2

2

)()1(8

)()3(16)5(3

3

)1(

)()1)(1(4)()3)(1(

2

xu

x

xmxv

Free terms due to the bump integral for the biharmonic equation


Dual boundary integral equations for the biharmonic problem

After deriving the sixteen improper integrals, we have

BxsdBsuxsVPFsdBsxsMVPRsdBsmxsVPRsdBsvxsUVPRxuBBBB

,)()(),(..)()(),(...)()(),(...)()(),(...)(4

BxsdBsuxsVPFsdBsxsMPFsdBsmxsPFsdBsvxsUPFxBBBB

,)()(),(..)()(),(..)()(),(..)()(),(..)(4

BxsdBsuxsVPFsdBsxsMPFsdBsmxsVPRsdBsvxsUVPRxmB

mB

mB

mB

m ,)()(),(..)()(),(..)()(),(...)()(),(...)(4

BxsdBsuxsVPFsdBsxsMPFsdBsmxsPFsdBsvxsUPFxvB

vB

vB

vB

v ,)()(),(..)()(),(..)()(),(..)()(),(..)(4

48 for a smooth boundary


Conclusions


1. New methods to derive the Poisson integral formula by using the degenerate kernels and the null-field integral equations.

2. The occurring mechanism of degenerate scales depends on the formulation instead of the boundary conditions.

3. It is interesting to find that the T-complete set in the Trefftz method is imbedded in the degenerate kernels of MFS.

4. We adopt the bump-contour method to derive the free terms surrounding the singularity. For a smooth boundary, the sum of free terms are half.

Conclusions


Thanks for your kind attention


Image method

xxdBxtsxUxdBxusxTsuB

FB

F ,)()(),()()(),()(2

known unknown

Image method

BxxdBxussxTsuB

G ),()(),;()(2

B

G xdBxtssxU 0)()(),;( 0|),;( BxG ssxU

B

),( Rs )','(' Rs

),( x

0|),;( BxG ssxUClosed-form Green’s function ?)',;( ssxU G


Closed-form Green’s function (Interior problem)

,)],(cos[)(1

ln),(

,)],(cos[)(1

ln),(ln)ln(),(

1

1

m

mEF

m

mIF

FRm

R

msxU

RmRm

RsxUsxrsxU

RmR

msx

m

m

,)](cos[)(

1lnln

1

RmRm

Rsxm

m

,)](cos[)(1

lnln1

R

a

RR

R

R 22

),( x

),( Rs

)','(' Rs

.lnln||ln||ln

lnln||ln||ln

lnln||ln||ln),;(

'

2'

'

Rasxsx

R

aasxsx

RasxsxssxU G

a

Image point


Series-form Green’s function (degenerate kernels)

.0,)](cos[])()[(1

)ln(

lnln||ln||ln),;(

12

'

Rma

R

Rma

R

RasxsxssxU

m

mm

G

.,)](cos[)]()[(1

)ln(

lnln||ln||ln),;(

12

'

aRma

RR

ma

RasxsxssxU

m

mm

G

a

a

),( x

),( x

),( Rs

),( Rs

)','(' Rs

)','(' Rs


Closed-form and series-form Green’s functions for interior and exterior problems

Closed-form Series-form

0,25.1R,8.0R 1,a - 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

s

- 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

s

050,M,25.1R0.8,R 1,a

- 2 - 1 . 5 - 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

s

- 2 - 1 . 5 - 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

s

0,8.0R,25.1R 1,a 050,M,8.0R,25.1R 1,a

Interior problem:

Exterior problem:


x

GG n

ssxUssxT

),;(),;(

BxxdBxussxTsuB

G ),()(),;()(2


Cosine theorem

,20,0,)()cos(2

)(

2

1

)(),;,;,(2

1),(

2

0 22

22

2

0

2

aRdf

aRRa

Ra

dafR

aRTRu G


20,0,)())(cos(212

1),(

2

01

aRdfma

RRu

mm

m

Series-form:


Degenerate scale

Rigid body mode

Solve u

u is a null field u is solved to be the rigid body solution

Discriminant

Laplace problems

Biharmonic problems

Dirichlet NeumannFree

Clamped

Simply-supported

Mathematically realizable

Physically realizable

0KDet

Flowchart of the nontrivial modes


By setting 12 TN

12

10 )ln(

N

jj Rca

12

1

12,...2,1),cos()1

(1N

jj

njn Nnn

Rnca

12,...2,1,)sin()1

(112

1

Nnn

Rncb

N

jj

njn

12

1

0

N

j

jca

12

1

12,...2,1),cos()(1

N

j

jn

jn NnnRn

ca

12,...2,1,)sin()(1

12

1

NnnRn

cb

N

j

jn

jn

Interior problem: Exterior problem:

Trefftz method

MFS

= MN = 12 N

Connection between the Trefftz method and MFS

研究生：吳清森 指導教授：陳正宗 教授 陳義麟 博士

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研究生：吳清森指導教授：陳正宗　教授　　　　　陳義麟　博士