Seoul National
University
Introduction to
Finite Element Method
Chapter 4
Boundary Value Problem in 2D
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4.1 Introduction
 The principle ingredients of the finite element method for
approximate solutions of problems from the 1-D setting are :
1.The formulation of the problem in a variational framework in which the
appropriate space H of admissible functions is identified.
2.The construction of a finite element mesh and piecewise-polynomial basis
functions defined on the mesh, which generate a finite-dimensional
subspace of H.
3. The construction of an approximation of the variational boundary value
problem on a finite element subspace Hh of H. This entails the calculation
of element matrices and the generation of a sparse system of linear
algebraic equations in the values of the approximate solution of nodal
points in the mesh.
4. The solution of the algebraic system.
5. The examination of the characteristics of the approximate solution and, if
possible, an estimation of the inherent approximation error.
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4.1 Introduction
These steps in the previous page form the basis of most finite element methods for
not only one-dimensional problems, but, more importantly, for boundary value
problems in two and three dimensions.
Characteristic differences between one and two dimensional problems is listed in
the following table
1-D BVP
- ODE , spatial variable x
(independent variable)
- Boundary values at both ends
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2-D BVP
-PDE, independent variables (x, y)
- Boundary conditions along the
curvature boundary of twodimensional space
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4.2 Description of the problem
․ Domain (open)
    
where

 : Closure of 
 : Open domain + Bounded ( differential equation )
: Boundary
The independent variables are
i
 x  x( s ) 

 y  y( s) 
is a curvature by 
u  u ( x, y ) , and differential operator is


j
x
y
Directional Derivative of u in n direction is n  inx  jn y
therefore,
u  n  nx
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u
u u
 ny

x
y n
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4.2 Description of the problem
(a) Representation of the vector-valued function    ( x, y ) ;
(b) Resolution of the flux  (s ) at the point s on
tangential flux components n(s) to 
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
into normal flux 
n
( s)
and
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4.2 Description of the problem
(a) Distribution of normal flux on
boundary  of subregion 
(b) Magnitude of normal flux on
boundary of square region
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4.2 Description of the problem
Define 2D BVP corresponding to 1D BVP;
Find
u ( x, y ) ,
( x, y)    R 2 which satisfies
   [k ( x, y)u]  b( x, y)u( x, y)  f ( x, y)
( x, y)  i
i  1,2
and the Jump Condition

u 

k
(
x
,
y
)
 0,


n 

s    1   2
and the Boundary Condition
 ( s)
u
  ( s )u ( s )   ( s )
n
ⅰ)
 ( s)  0
Essential B.C
ⅱ)
 (s)  0
Neumann B.C
ⅲ)
  0,   0
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Mixed B.C
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4.2 Description of the problem
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4.3 Variational BVP
Residual is defined as follows:
r ( x, y )    (ku )  bu  f
and then,
    (ku)  bu  f v( x, y)dxdy  0,
i
v( x, y )
Gauss - Green theorem
 uvdxdy   uvnds   uvdxdy

d
therefore,
    (ku ) vdxdy   ku  vdxdy   vku  nds
i
cf)
i


 i
uv.i   uvni ds   u.i v
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 i
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4.3 Variational BVP
and then,

i
from
(ku  v  buv)dxdy   k
 i
u
vds   fvdxdy, i  1,2

n
1  2    


2
(ku  v  buv)dxdy    k
from   1  2  
i 1
 i
u
vds   fvdxdy

n
 k

 u 
u
vds    k
vds


n
 n 
1

  k  (  u ) vds  0,



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v
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4.3 Variational BVP
Therefore 2D BVP can be expressed as follows:
Find


u V such that
(ku  v  buv)dxdy  

k

k
uvds  


vds   fvdxdy,

v  V
In Summary, transformation of the differential equation,
   (ku )  bu  f x  



u


 u   s   
n


to weak form can be stated by
“Find a function u ( x, y )  H such that


(ku  v  buv)dxdy  
k



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k

vds   fvdxdy,


uvds
v  H  H 1 () "
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4.3 Variational BVP
When u , v  L2 () and u , v  L2 (),
the previous variation al equation is integrable with the smooth functions b, c.
And this means that u , v  H 1 ()
where

H ()  v | 
v  v dxdy  M . M finite
and note that
H 1  L2
L2 ()  v |  v dxdy  M . M finite
2

2
1

2


Since u, v  L () at boundary  with smooth prescribed boundary functions,
the boundary integration terms in the Variational equation is integrable.
2
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4.4 Finite Element Interpolation
An example of finite element
dicretization of a domain with the
inside hole by a mesh consisting of
triangular and quadrilateral elements.
If the curved boundary exists as in the
figure, there will always be some
discretization error, since the finite
element mesh constructed as the
collection of triangular or
quadrilateral elements, will not
perfectly coincide with the given
domain. However, as the mesh is
refined, the mesh cam approximate
the domain with increasing accuracy.
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4.4 Finite Element Interpolation
4.3.1 Galerkin Approximation
Let us approximate the variational form,


(ku  v  buv)dxdy  
k


k
uvds  


vds   fvdxdy,

v  V
such that
N
uh    j j ( x, y )
j 1
N
vh   ii ( x, y )
i 1
N
uh |    j j ( s )
j 1
N
vh |    ii ( s )
i 1
We writes,


(kuh  vh  buhvh )dxdy  
k



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k

vh ds   fvh dxdy,


uhvh ds
vh  Vh  H 1 ()
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4.4 Finite Element Interpolation
Then we have,
N
N
N
N
 (k         b      )dxdy

j 1

k


j
j
i 1
N
i
i
j
j 1
j
i 1
i i
N
      ds
j 1


j
k

j
i 1
i i
N
N
   ds   f    dxdy
i 1
i i

i 1
i i
Or,
 N
 N  N

k
 i    j  k j  i  b ji dxdy     i    j 
i j ds 


 
i 1
 j 1
 i 1  j 1

N
N
k
  i 
i ds    i  fi dxdy
N
i 1


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i 1

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4.4 Finite Element Interpolation
By arrangement of the previous equation, we have the equation
 N
 N 
k
k







k






b


dxdy



ds


f

dxdy


ds



i j 
j
i
j i
i j
i


  i   i

 
 

i 1
 j 1
 i 1
N
We can write
N



K

F

i  j ij
i  0
i 1
 i 1

N
{i } R
From the condition of i
N
K 
i 1
where,
ij
j
 Fi
i  1,2,3..., N
K ij   k j  i  b ji dxdy  


Fi   fi dxdy  

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
k

k

i j ds
i ds
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4.4 Finite Element Interpolation
4.3.2 Finite element shape function
The finite element discretization of a given domain by a mesh consisting of
triangular or quadrilateral elements (there will be some discretization error)
(1) Triangular element
(a) C 0 - linear triangle
u( x, y)  a1  a2 x  a3 y
the highest order of the integrated function in K ij is
max{[ o(k )  (zero - order )  (zero - order )], [o(b)  (1st order )  (1st order )]}
if k and b are constant, the integrated function becomes second order
the order of boundary integration can be obtained as the same method
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4.4 Finite Element Interpolation
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4.4 Finite Element Interpolation
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4.4 Finite Element Interpolation
(b) High order triangle element
6 node (quadratic)
u ( x, y ) |6  a1  a2 x  a3 y  a4 x 2  a5 xy  a6 y 2
10 node (cubic)
u ( x, y ) |10  u ( x, y ) |6  a7 x 3  a8 x 2 y  a9 xy2  a10 y 3
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4.4 Finite Element Interpolation
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4.4 Finite Element Interpolation
(2) Rectangular element
(a) Tensor product ( C 0 )
4 node (linear)
u( x, y) |4  a1  a2 x  a3 y  a4 xy
9 node (quadratic)
u ( x, y ) |9 
a1
 a2 x  a3 y
 a4 x 2  a5 xy  a6 y 2
 a7 x 2 y  a8 xy2
 a9 x 2 y 2
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4.4 Finite Element Interpolation
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4.4 Finite Element Interpolation
※ Pascal triangle (3rd Prism)
1
x y
x 2 xy y 2
x 3 x 2 y xy2 y 3
x 4 x 3 y x 2 y 2 xy3 y 4
x 5 x 4 y x 3 y 2 x 2 y 3 xy4 y 5
16 node (Cubic)
→ inefficient
※ reduce the inside nodes for efficient calculation
→ serendipity element
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4.4 Finite Element Interpolation
(b) serendipity element
8 node:
2 2
exclude x y among 9 node base
u ( x, y ) |8  u ( x, y ) |10 a9 x 2 y 2
12 node:
u( x, y) |12  u( x, y) |16 a13  a14  a15  a16
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4.4 Finite Element Interpolation
(3) Interpolation error
Suppose that a smooth function g is given and assume that we wish to interpolate g by
a finite element representation gh which contains complete polynomials of degree k.
then the interpolation error satisfies,
g  gh
 e
 max e g ( x, y )  g h ( x, y )  Ch k 1
( x , y )
similarly,
g g h

 C1h k
x x
g g h

 C2 h k
y y
therefore,
g  g h l  C3hk
2
g 1,
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 2 g 2 g
 g 


x
y

2

g
2
 dxdy  g 
x

2
g

y
2
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4.5 Example
For example, consider polygonal domain with k=1 and b=0;
  2u( x, y)  f ( x, y)
in 
with B.C.

  0,


  0,


  0,
  0, u  0 on 41
  0,
u
 0 on 12 , 25 , 67 , 47
n
u
 u   on 56
n
unknown : u1 , u2 ,, u6 , u7
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4.5 Example
The element 1 makes a triangle by connecting node 1, 2 and 3
k (s) (s) ~ ~
e
K ij   i j dxdy  
  ds


 (s) i j
k ~ ~
e
Fi   fi dxdy   i j ds


e
e
 k11 k12 k13 


K ij(1)   i j dx   k21 k22 k23 
(1)
k k k 
 31 32 33 
 F1 
 
Fi (1)   fi dxdy   F2 
(1)
F 
 3
K ij( 6 )
 k33 k35 k36 


  k53 k55 k56 
k k k 
 63 65 66 
i  1,2,3

 j  1,2,3
Fi ( 6 )
 F3 
 
  F5 
F 
 6
6
K   K ij( e )
G
ij
e1
Half band = 5
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4.5 Example
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4.5 Example
The assembly of the element matrices and vectors results,
We impose the essential boundary conditions, u =u =0 , then we have
1
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4
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4.6 Boundary Element Method
4.6.1 Laplace Equation
For the equation(4.1), if we set
  1, b  0  f , k  1
Then we have the Laplace equation
 2u  0 x    R 2
B.C. u  u  
n

By applying MWR, the following equation can be written.


2 d  0
 { admissible
function }
From Green-Gauss Theorem,
     d     nds  0


Applying Green-Gauss Theorem on more time, we obtain the equation


 2 d     n ds     nds  0
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


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4.6 Boundary Element Method
Assume that  is the fundamental solution of
The function
 ( x,  )
satisfies


 2   ( x,  )
p( x) ( x,  )d  p( )
The function  is called as free space Green function, fundamental function,
Elementary solution.
1
The solutions in three dimension can be written  ( x,  ) 
4r ( x,  )
In two dimensional case, we have the solution,
1
 ( x,  ) 
ln r , r ( x,  )  ( x1   1 ) 2  ( x2   2 ) 2
2
And then,
  i
 
1 1
1
1

j
   ( x1   1 )i  ( x2   2 ) j    [( x1   1 )i  ( x2   2 ) j ]
x
y
2  r
r
 2r
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4.6 Boundary Element Method
We obtain the integral equation as
 ( )   (n   n )ds

However, in the boundary region, the following relations are satisfied
  ( x) ( x,  )d  c ( )

1
i) Smooth-Boundary : c 
2
1
ii) 90° corner : c 
4
Finally, the governing equation of boundary element method can be written by
 
 
C ( )      
ds
 n
n 

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4.6 Boundary Element Method
General procedure for the BEM
i) Meshing of the boundary domain  with nodes and elements
ii) Interpolation on the mesh by the FE polynomial functions
iii) Obtain solutions in the boundary region by solving simultaneous equations
iii) From the values obtained at the boundary region,
we find the solutions in the inside region
 ( x) 
1  
 




ds,



C  n
n 
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4.6 Boundary Element Method
4.6.2 General case
 2    f ( x)
More generally, in the case of
L  f (x) , we apply the MWR method,
 (L  f )d  0
  0

If L* is the adjoint operator of L , the equation will be


L*d   ( B.C. terms) ds   d

f
Assume that  is Green solution of

L*   ( x   ) ,
then the final equation will be
C ( )   fd   ( B.C. terms) ds

Seoul National University

Aerospace Structures Laboratory
4.6 Boundary Element Method
Examples
i) L    (k )  b
ii) BEM formulation of
L   2 2 ?
iii) 3D elasticity equation, etc..
Seoul National University
Aerospace Structures Laboratory