BEM October 07th 2015

Name Phung Duc Thuan
Student ID P46047158
Content Derive the boundary Integral Equation
HOMEWORK
Problem
Let U be a function in the region D with the boundary B. U satisfies the following
equation:

Or

ΔU x, y = f x, y
ΔU r⃗ = f r⃗

U is an unknown function need to be determined.
Green’s function G associated with Laplace’s operator has the form as follow:
ΔG r⃗, ⃗⃗⃗⃗
r =

In the polar coordinates, we have:
ΔG r⃗, r⃗⃗⃗⃗ =




r ∂r

r

∂G
∂r

⇒r


∂G
∂r

∂G
∂r

r⃗ − ⃗⃗⃗⃗
r

= 0 where r = |r⃗ − ⃗⃗⃗⃗|
r
=C

=

r

⇒ G = C lnr + D

Choose C=1 and D = 0. The Green’s function becomes:
Green’s Second Identity
Where

G(r⃗, r⃗⃗⃗⃗) = ln r

∫ (U

∂V
∂U

− V ) dS = ∫
∂n
∂n

UΔV − VΔU dV

B is a piecewise smooth contour enclosing domain D
�⃗⃗ is a normal unit vector

U, V are functions which must have continuous second order partial derivatives

Boundary Integral Equation for two domain problems

Let U be an unknown function satisfying the equation below:
ΔU r⃗ = f r⃗

G be Green’s function associated with Laplace’s operator:
ΔG r⃗, ⃗⃗⃗⃗
r =


r⃗ − ⃗⃗⃗⃗
r

Assume that U exists on the domain D with the boundary B and r⃗⃗⃗⃗ ∈ D

Green’s function is infinite at r⃗ = ⃗⃗⃗⃗
r so we enclose r⃗⃗⃗⃗ by a very small circle D�.

In the region D-D Green’s second identity becomes




+��
+��

(G

(G


∫ (�

∂G
∂U
− U ) dS = ∫
GΔU − UΔG dV
∂n
∂n
−��
∂U
∂G
− U ) dS = ∫
Gf dV
∂n
∂n
−��

∂U
∂G
∂U

∂G
Gf dV
− U ) dS + ∫ (G
− U ) dS = ∫
∂n
∂n
∂n
∂n
−��
��

Set � → 0, we obtain:
lim[∫



�→




∂n

−U

∂G
∂n

dS + ∫�



G



∂n

−U

∂G

∂n

We have three terms which need to be evaluate:
lim ∫
�→

−��

lim ∫� �
�→



�→



lim ∫� �

Gf dV − lim ∫


Gf dV = ∫

��
��

��

�→



−��

dS] = lim ∫
�→

f ln dV

=∫


Gf dV − lim ∫ f ln � dθ

=∫

Gf dV − lim � ��� ∫ fdθ

�→

�� = lim ∫�
�→



�→



��
��


�� = lim ∫� �
��
= lim ∫
�→





�→

��� �� = lim ∫



��

−��

�→

� ��

��� �� = lim ∫
�→

−��� = −2�� ⃗⃗⃗⃗


Plug back to the Green’s second identity, we obtain:

��


���� �� = 0



����
−��

���

Gf dV

∫ (�

∂U
∂G
− U ) dS + 2πU = ∫
∂n
∂n

2�� = ∫ (−�

Gf dV

∂G
∂U
+ U ) dS + ∫
Gf dV
∂n
∂n
−��