2013/06/13 page 1 複變數邊界積分方程推導 national taiwan ocean university msvlab...

2013/06/13

複變數邊界積分方程推導National Taiwan Ocean University

MSVLABDepartment of Harbor and River Engineering

Date: June, 13, 2013

Student: Jia-Wei Lee (李家瑋 )Advisor: Jeng-Tzong Chen (陳正宗 )

2013/06/13

Outline

1. Revisit the boundary integral equation in R2

2. Green’s third identity in C3. Derivation of boundary integral

equation in C4. RVBIE vs. CVBIE

2013/06/13

Green’s third identity FdA F ndS

Divergence theorem in R2

F is a vector function

let F v u ( )

( ) .....(2)

v u dA v u ndS

uv u v u dA v dSn

let F u v ( )

( ) .....(1)

u v dA u v ndS

vu v u v dA u dSn

(1) (2) ( ) v uu v v u dA udS v dSn n

Green’s third identity

u and v are scalar functions

2013/06/13

x

Boundary integral equation in R2

2D Laplace problem

( ) 0,x xu

( ) v uu v v u dA udS v dSn n

Auxiliary system

Unknown field u

( , )x sv U( , ) ( )x s x sU

1( , ) ln2

x s x sU

( , ) ( )( ) ( ) ( ) ( , ) ( )x x

x s xs x x x s xU uu u dS U dSn n

Fundamental solution

2013/06/13

BIE in real number 2R R

( ) ( , ) ( ) ( ) ( , ) ( ) ( ),x s x s s s x s s xu T u dS U t dS

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

2x s x s s s x s s xu C PV T u dS U t dS

20 ( , ) ( ) ( ) ( , ) ( ) ( ), \s x s s s x s s x RT u dS U t dS

1( , ) ln2

s x s xU


Singular boundary integral equation

2D Laplace equation

x

( , )( , )s

s xs x UTn

2013/06/13

Outline




2013/06/13

Complex analysis

2CD iz x y

2( )

2

z zxz x iyz x iy i z zy

2CD iz x y

Cauchy-Riemann operator

( ) ( ) ( )w z u z iv z

Complex number

Complex function

2 2

2 2C C C CD D D Dx y

2D Laplace operator

Cauchy-Riemann operator (Conjugate form)

2013/06/13

Holomorphic function & Harmonic function

u vx yu vy x

Harmonic functionHolomorphic function (Analytic function)

1 ( )( ) 02C

w zD w zz

Cauchy-Riemann equation

1 ( )( )2C

w zD w zz

Exist( ) 0C CD D w z

( ) 0C CD D w z

( ) 0w z

2D Laplace equation

( ) 0 ( ) 0C C CD w z D D w z

( ) 0CD w z ( ) 0( ) 0 ( ) 0

C C

C C

D D w zD w z or D w z

2013/06/13

Gauss theorem (Green theorem) in C

( ) ( ) ( ) ( ) ......(1)

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

( ) ( ) ( ) ( ) ......(3)

( ) ( ) ( )

u z v z dxdy u z dy v z dxx y

v z u z dxdy v z dy u z dxx y

u z v z dxdy u z dy v z dxx y

v z u z dxdy v zx y

( ) ......(4)dy u z dx

Q P dxdy Pdx Qdyx y

Green theorem in R2

2013/06/13

Gauss theorem (Green theorem) in C

(1) (2)

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( )2 ( ) ( )

( ) 1 ( )2

i

u z v z v z u zi dxdy u z iv z dy u z iv z idxx y x y

f z dxdy f z i dx idyz

f z dxdy f z dzz i

( ) ( ) ( ) ( ) ......(1)u z v z dxdy u z dy v z dxx y

( ) ( ) ( ) ( ) ......(2)v z u z dxdy v z dy u z dxx y

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Gauss theorem (Green theorem) in C( ) ( ) ( ) ( ) ......(3)u z v z dxdy u z dy v z dxx y

( ) ( ) ( ) ( ) ......(4)v z u z dxdy v z dy u z dxx y

(3) (4)

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( )2 ( ) ( )

( ) 1 ( )2

i

u z v z v z u zi dxdy u z iv z dy u z iv z idxx y x y

f z dxdy f z i dx idyz

f z dxdy f z dzz i

( ) ( ) ( )f z u z iv z

Conjugate form

2013/06/13

Green’s third identity in C( )( ) ( , ) w zlet f z U z sz

2( , ) ( ) ( ) 1 ( )( , ) ( , ) ......(5)2

U z s w z w z w zU z s dxdy U z s dzz z z z i z

( , )( ) ( )U z slet f z w zz

2 ( , ) ( , ) ( ) 1 ( , )( , ) ( ) ......(6)2

U z s U z s w z U z sw z z dxdy w z dzz z z z i z

( ) 1 ( )2

f z dxdy f z dzz i

( ) 1 ( )2

f z dxdy f z dzz i

2 2

(6) (5)

( , ) ( ) 1 ( , ) 1 ( )( ) ( , ) ( ) ( , )2 2

U z s w z U z s w zw z U z s dxdy w z dz U z s dzz z z z i z i z

2013/06/13

Outline




2013/06/13

x

Boundary integral equation in C

2D Laplace problem

2 ( )4 0,w z zz z

2 2( , ) ( ) 1 ( , ) 1 ( )( ) ( , ) ( ) ( , )2 2

U z s w z U z s w zw z U z s dxdy w z dz U z s dzz z z z i z i z

Auxiliary system

Two unknown fields w u iv

( , )U z s

2 ( , )4 4 ( )U z s z sz z

2( , ) ln ln ( )( )U z s z s z s z s Fundamental solution

1 ( , ) 1 ( )( ) ( ) ( , )2 2

U z s w zw s w z dz U z s dzi z i z

2013/06/13

BIE in complex number

221 1 ( )( ) ( ) ,

2ln

l2

n w sw z w s dss z

s z ds zisi s

2

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

lnln

2C PV w sw z w s ds ds z

i is

s zs

zs

2

21 1 ( )0 ( ) , \2 2

lnln

s zs z w sw s ds ds z C

i i ss

C C


2D Laplace equation

z

2013/06/13

Outline




2013/06/13

RVBIE vs. CVBIE RVBIE

( ) ( , ) ( ) ( ) ( , ) ( ) ( ),

ln1 1 ( )( ) ( ) ( ) ln ( ),2 2s s

x s x s s s x s s x

s x sx s s s x s x

u T u dS U t dS

uu u dS dSn n

CVBIE

22

1 ( , ) 1 ( )( ) ( ) ( , ) ,2 2

ln1 1 ( )( ) ( ) ln ,2 2

U s z w sw z w s ds U s z ds zi s i s

s z w sw z w s ds s z ds zi s i s

1( , ) ln2

s x s xU

2( , ) lnU z s z s

1 ln ( )( ) ( ) ln ,2

1 ln ( )( ) ( ) ln ,2

s ss s

s ss s

r u su z u s dt r dt zn n

r v sv z v s dt r dt zn n

12

,12

s ss s

s s

s s

is n tds dn idt

ds dn idti

s n t

2013/06/13

The endThanks for your kind attentions

http://msvlab.hre.ntou.edu.tw/Welcome to visit the web site of MSVLAB/NTOU

2013/06/13

BIE in real number 2R C

( ) ( , ) ( ) ( ) ( , ) ( ) ( ),x s x s s s x s s xu T u dS U t dS

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

2x s x s s s x s s xu C PV T u dS U t dS

20 ( , ) ( ) ( ) ( , ) ( ) ( ), \s x s s s x s s x RT u dS U t dS

(1)0 ( )

( , )4

s x i H krU



2D Helmholtz equation

x

( , )( , )s

s xs x UTn

2( ) ( ) 0xk u

2013/06/13

Conventional CVBIE vs. Present CVBIE

Conventional CVBIE

2

1 ( )( ) ,2

ln1( ) ( ) ,2

w sw z ds zi s z

s zw z w s ds z

i s

Present CVBIE

2

22

1 ( ) 1 ( )( ) ln ,2 2

ln1 1 ( )( ) ( ) ln ,2 2

w s w sw z ds s z ds zi s z i s

s z w sw z w s ds s z ds zi s i s

2ln1 ln[( )( )]s z s z s zs z s s

2013/06/13

Gauss theorem (Divergence theorem) inR

FdA F ndS

Green theorem in R

let F U u

F is a vector function

2

( )

( )

U u dA U u ndS

uU u U u dA U dSn

let F u U 2

( )

( )

u U dA u U ndS

uU u U u dA U dSn

2013/06/13

Q & A• CVBIE的優缺點。• 有沒有實際的問題，一定要用新導到的 CVBIE求解，假若使用傳統的 CVBIE是無法求解的。• 如何用 CVBIE求解外域問題。• 若 u和 v的邊界條件類型不一樣，還能保留 C

VBIE的優點嗎 ?• 引入退化核，對於一些特別的幾何• 外形，是否也可以得到半解析解。• 可否將 CVBIE推廣至 Helmholtz problem。

2013/06/13

Q & A(克氏分析 )• Clifford BIE的優缺點。• 簡單的例題應用。• 實際程式的運算。• 廣義 stokes’ theorem 如何退回去一般在向量微積分所學到的 Stokes’ theorem。• 可否將 Clifford BIE推廣至 Helmholtz pro

blem。• 若引入退化核，對於一些特別的幾何外形，是否也可以得到半解析解。

2013/06/13

Singular BIE from Borel-Pompeiu formula

(

( ) 1 ( )2

( ) ( ), ( , ) ( , )) 1 ( )

2( ) 1 ( )

2

f s

w z dxdy

d d f s d

w z dzz i

w z f s x y

f s d d f s di

s

s

s i

s

( )( )

( ) 1

1

( ) 0, \2

,( ) ( )02

C

w sLet f ss z

w s d d w s ds zs s z i

w s ws

s d ddsi s z s s z

z

Borel-Pompeiu formula

22

22 2

22 2 1 2

ln( ) 1( ) ln ,

( ) ( ) 1 1 ( )ln ln

( ) 1 ( ) ( )ln ln (

2

) ( ), ( )2

w s d d w s w ss z ds s z d d w s C Cs s z i s s s

s zw sLet f s s zs s s z

w s w s w ss z d d s z dss s s s z i s

2

2 21 ( ) 1 ( ) ( )0 ln ln2 2

w s w s w sds s z ds s z d di s z i s s s

2 ( )( ) , 0w sw s let it be a harmonic functions s

21 ( ) 1 ( )0 ln , \

2 2Cw s w sds s z ds z

i s z i s

Singular BIE for \Cz

2013/06/13

BIE in complex number

21 1 1 ( )( ) ( ) ,2 2

ln s z w sw z w s ds ds zi s z i s

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

2 2l

2nC PV w sw z w s ds ds z

i i sz

zs

s

2ln1 1 1 ( )0 ( ) , \

2 2s w sw s ds ds z C

i s z i sz

2CD iz x y

2CD iz x y

2 2

2 2 2C C C CD D D Dx y

C CSingular boundary integral equation

2D Laplace equation

z

2013/06/13

BIE in Clifford number 0,nCl

( ) ( , ) ( ) ( ) ( ) ( , ) ( ) ( ) ( ),x s x s s s s x s s s xu E n u dS U n u dS

( ) ( ) . . . ( , ) ( ) ( ) ( ) ( , ) ( ) ( ) ( ),x x s x s s s s x s s s x

in

n

u C PV E n u dS U n u dS

0,0 ( , ) ( ) ( ) ( ) ( , ) ( ) ( ) ( ), \s x s s s s x s s s x R nE n u dS U n u dS

2 , , 1,2,...,i j j i ije e e e i j n 1

xn

i ii

e x

( ) ( )x xu e u

0,0,R n

nCl


1( ), () s xs x

s x nn

E

2

1 , 121( , ) ln , 2,

21 1 1 , 3, 4, 5,...

2

s x

s x s x

s x nn

n

U n

nn

22

( )2

n

n n


2013/06/13

BIE in Clifford number nCl

( ) ( , ) ( ) ( ) ( ) ( , ) ( ) ( ) ( ),x s x s s s s x s s s xu E n u dS U n u dS

( )

( ) . . . ( , ) ( ) ( ) ( ) ( , ) ( ) ( ) ( ),x x s x s s s s x s s s x

in

n

u C PV E n u dS U n u dS

0 ( , ) ( ) ( ) ( ) ( , ) ( ) ( ) ( ), \s x s s s s x s s s x RnE n u dS U n u dS

2 , , 1,2,...,i j j i ije e e e i j n 1

xn

i ii

e x

( ) ( )x xu e u

RnnCl


1( ), () s xs x

s x nn

E

2

1 , 121( , ) ln , 2,

21 1 1 , 3, 4, 5,...

2

s x

s x s x

s x nn

n

U n

nn

22

( )2

n

n n


( ) ( , ) ( , ) ( ) ( )x s x s x x x sE E

( ) ( , ) ( , ) ( ) ( )x s x s x x x sU U

2013/06/13

( ) ( , ) ( , ) ( ) ( )x s x s x x x sE E

( ) ( , ) ( , ) ( ) ( )x s x s x x x sU U

( ) ( , ) ( , ) ( ) ( , )x s x s x x s xU U E

( ) ( ) ( , )( ) ( , ) ( )( ) ( , )( )

x x s xx s x xx s xs x

UUE

1

n

ii i

ex

1

n

ii i

ex

22

21

n

ii i

ex

( ) ( , ) ( , ) ( ) ( )x s x s x x x sE E

( ) ( , ) ( , ) ( ) ( )x s x s x x x sU U

( ) ( ) ( , )( ) ( , ) ( )( ) ( , )( )

x x s xx s x xx s xs x

UUE

1

n

ii i

ex

2

21

n

ii i

ex

0,0,R n

nCl RnnCl

2 , , 1,2,...,i j j i ije e e e i j n 2 , , 1,2,...,i j j i ije e e e i j n

( ) ( , ) ( , ) ( ) ( , )x s x s x x s xU U E

nCl0,nCl and

2013/06/13

What do I want to do ?

( ) ( ) ( )w z u z iv z

Try to extend complex valued BIE to deal with 2D Helmholtz problem

2( ) ( ) 0w z k w z 2D Helmholtz equation

21 1 1 ( )( ) ( ) ln ,2 2

sz s s s z s zs z s

ww w d di i

( ) 0w z Harmonic function

OK

Now2

2

( ) ( ) 0( ) ( ) ( )

( ) ( ) 0R R

R II I

w z k w zw z w z iw z

w z k w z

Non harmonic function Analytical function ??

1 1 ( )( ) ( ) ,2 2

? ? sz s s s zs

ww w d di i

First

2013/06/13

What do I want to do ?

Try to extend complex valued BIE to Clifford BEM

and

deal with 2D Helmholtz problem by using the Clifford BEM

Second

2013/06/13 page 1 複變數邊界積分方程推導 national taiwan ocean university msvlab...

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