Seoul National
University
Introduction to
Finite Element Method
Chapter 1
Introduction to Finite Element Method
Seoul National University
Aerospace Structures Laboratory
Introduction to Finite Element Method
Course description
This course is an introduction to Finite Element Method(FEM) which is
an essential technology in engineering and mathematical physics.
FEM has been generally used for structural analyses such as simulation tool
of static or dynamic behavior of elastic or non-elastic solids and structures.
This method is now used to analyze fluid flow problems,
heat transfer phenomena and electromagnetic field,
geo-physics problems and etc..
This FEM class is designed to give the students the ability to program
their own simulation code for the specific research fields
based on what they learned in the class.
Course Website; http://aeroguy.snu.ac.kr
Course material
Lecture Note based on the book by
Becker, Carey and Oden, Finite Elements - An Introduction Vol.1 1981.
Teaching Assistant
Jung Eun Lee ([email protected] 301-1357 Tel. 7389)
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Introduction to Finite Element Method
Course Introduction
Students will learn how to solve partial differential equations(PDEs) by
numerical method in this course. Especially, the elliptic boundary value
problems(BVPs) discretized by finite element technique by using the method
of weighted residual and Galerkin approximation will be mainly considered.
For thorough understanding on the Finite Element algorithm, students are
required to program their own Finite Element code of one-dimensional BVPs
according to the theories and algorithm they learned in the course. Direct
time integration methods and eigen-value system solvers for the parabolic and
the hyperbolic PDEs along with the semi-discretization are also taught briefly
in this course. In addition to these classical theories, recent technologies
including efficient algorithms and parallel computing techniques will be
introduced. For further understanding of FEM, IPSAP, which is developed as an
efficient parallel finite element program and DIAMOND which is a pre-post
GUI program for IPSAP (http://ipsap.snu.ac.kr) will be used to learn the
convergence behaviour of the numerical solutions of partial differential
equations upon mesh refinement. The homework and term project will be
given to exercise the methods they learn
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Introduction to Finite Element Method
References
- David S. Burnett, "Finite Element Analysis."
Addison Wesley, 1987.
- Zienkiewitz and Taylor, "The Finite Element Method.", McGraw Hill, 2005
- Klaus J. Bathe, "Finite Element Procedures.", Prentice Hall, 1996.
- Thomas J. R. Hughes, "The Finite Element Method: Linear Static and
Dynamic Finite Element Analysis ." , Prentice Hall, 2000
- R.D. Cook, D.S Malkus and M.E. Plesha, "Concepts and Applications
of Finite Element Analysis," 4th ed. Wiley, 2002
- IPSAP (Internet Parallel Structural Analysis Program), DIAMOND Manual.
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1.1 Sources of the Problems
1.1 Sources of the Problems
1) 1 - D Heat Flow
Energy conservation law ＋ Fourier's law
Section area A, length dx,
q( x)  heat flux, Q( x)  heat source
By the energy conservation law,
(A  ΔA)(q  Δq)  Aq  Q(x)Ax
ΔA, Δq  0 ,
dT ( x)
dx
d
dT ( x)
 (k ( x) A( x)
)  Q( x) A( x)
dx
dx
From Fourier's law,
q  k ( x)
If A(x ) is constant
 (kT ' )'  Q( x)
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d
(qA)  QA
dx
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1.1 Sources of the Problems
2) Elastic Rod
Force equilibrium ＋ Hooke's law
σ : stress,
f(x) : body force
(  Δ)(A  ΔA)  A  f ( x) Ax  0

d
( ( x) A( x))  f ( x) A( x)
dx
du ( x) ; E (x) : , young's modulus , u (x) : displacement
dx
d
du ( x)
 ( E ( x) A( x)
)  f ( x) A( x)
dx
dx
Hooke's law  ( x)  E ( x)
Or distributed load per unit length,

d
du
( EA )  F ( x)
dx
dx
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F ( x)  f ( x) A( x)
Essential BC(EBC) : u is given
Natural BC(NBC) : σ is given
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1.1 Sources of the Problems
3) Cable deflection
Balance of transverse force ＋ tension
F  fx  kw  ( F  F )  0
dF
 k ( x) w( x)  f ( x)
dx
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F : transverse component of tension , T
f(x) : distributed transverse force
w : transverse displacement
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1.1 Sources of the Problems
If we assume that Θ is angle between the cable and the horizontal axis,
then the equilibrium gives the equation,
F ( x)  T ( x) sin 
 T ( x) tan   T ( x)
dw
dx
  (T ( x) w' )' kw  f ( x)
EBC : w
is given
NBC : T or f is given
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1.2 General form of Two point Boundary value Problem
General form:

Jump Conditions
d
du ( x)
du ( x)
[k ( x)
]  C ( x)
 b( x)u ( x)  f ( x) (1)
dx
dx
dx
4
   i
x  i i  1,2,3,4

du ( x1 ) 
  k ( x1 )
0

dx 

i
: x1 is material discontinuous point
but k, c, b are discontinuous
(2)

du ( x2 )  ˆ : point source
  k ( x2 )

f
dx 


du ( x3 ) 
: a point of discontinuous distributed source
  k ( x3 )

0
(ku' )' is not to be defined
dx 

 A x  Ax  Ax
BC's
du (0)
  0u (0)   0
dx
du (l )
l
  l u (l )   l
dx
0
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at x  0
(3)
at x  l
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1.3 Variational Formulation
Define the residual,
r ( x)  (ku' )'cu'bu  f
x  i i  1,2,3,4
And then Eq (1) can be written by

i
r ( x)v( x)dx  0
v( x)  {proper fn}  H
(4)
And we have
xi
rvdx


ku
'
v
|
fvdx  0
xi 1   ( ku' v ' cu ' v  buv ) dx  

i
i
i  1,2,3,4
i
4

i 1
i

l
x0  0, x4  l
0
Therefore,

l
0
l
4
0
i 1
(ku' v'cu ' v  buv)dx   fvdx   ku' v | xxii1
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(5)
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1.3 Variational Formulation
Also,
4
 ku' v |
i 1
 ku' v |   ku' v x  x
3
xi
xi 1
l
0

k (l )
l
i 1
i
{ l  (l )}v(l ) 
k (0)
0
{ 0   0u (0)}v(0)
 0  fˆv( x2 )  0
Note that the set of eqs (1) (2) (3) is equivalent to the following statement,
Find u ( x )  H such that

l
0
l

u (l )v(l )  k (0) 0 u (0)v(0)
l
0


fvdx  k (l ) l v(l )  k (0) 0 v(0)  fˆv( x2 )

0
(ku' v'cu ' v  buv)dx  k (l )

l
0
(6)
v  H
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1.3 Variational Formulation
◉ What is the characteristics of Variational "Weak" Formulation(VWF) ?
To find “ twice differentiable solution” to satisfy the second-order differential equation (1)
 to find the solutions from the square integrable “functions and their derivatives”
Therefore, the condition for the solutions becomes “weak”
• Characteristics of VWF (6)
1) Solutions which satisfy weaker conditions are acceptable.
This means that solutions can be found in wider range of function space,
that is,
l
H 1 (0,1)  {v(x)| [v(x)2  (v')2 ]dx  M, M is finite }
0
2) Jump Condition and Boundary Condition are included in one integral equation.
3) If C=0, it becomes symmetric form
4) Various boundary conditions can be easily adopted
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1.4 Galerkin Approximation
Since the dimension of function space H 1 (0,1) is infinite, the functions can be written.


j 1
i 1
u ( x)   j  j ( x), v( x)   i  i
where,  (x) is a basis function of H 1 (0,1)
i
But, we need to find a set of approximated solutions in the finite dimension to be solved by
computer. The functions should be approximated as the functions in the finite dimension,
N
N
j 1
i 1
uh ( x)   j j ( x), vh ( x)   ii
where, i (x) is basis functions of finite subspace, H h [ H 1 (0,1)]
And then, the problem (6) becomes the following statement,
Find uh  H h such that

l
0
l

u h (l )vk (l )  k (0) 0 u h (0)vh (0)
l
0
 (l )
 ( 0)
fvh dx  fˆvh ( x2 )  k (l )
vk (l )  k (0)
vh (0), vh  H h
 (l )
 ( 0)
(kuh ' vh ' cu h ' vh  bu h vh )dx  k (l )

l
0
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1.4 Galerkin Approximation
Or,
N
N
N
N
N
 N

k


'


'

c


'



b




0  j j j i i i j j j i i i j j j i i i dx
l

 k (l ) l
l

l
0
N
0 N
j  j j ( xl )i ii ( xl )  k (0)  j  j j ( x0 )i ii ( x0 )
0
N
N
l N
0 N
ˆ
f  ii dx  f  ii ( x2 )  k (l )  ii ( xl )  k (0)  ii ( x0 )
l i
0 i
i
i
N
N
N
vh    ii ( x)  H h
(8)
i 1
From the relations between
vh and  i , the expression becomes
vh  { i }  R N
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1.4 Galerkin Approximation
By arrangement of the previous equation (8), we have the equation
N
 l

l
0


k

'

'

c


'

b


dx

k
(
l
)

(
l
)

(
l
)

k
(
0
)

(
0
)

(
0
)


i 
j 
i
j
i j
i j
i
j
i
j
0


i 1
j

1
l
0



N
(9)
 l


 (l )
  fi dx  fˆi ( x2 )  k (l )
i (l )  k (0) 0 i (0)  0, { j}  R N
l
0
0

Or, we can write
N

i  j Kij  Fi   0

i 1
 j 1

N
i , the eq. (9) becomes,
From the condition of
N
 K
j 1
j
{i }  R
ij
 Fi  0
i  1,2,..., N
Finally, we obtain the N- dimensional simultaneous equation (10).
N
K 
j 1
ij
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j
 Fi
i  1,2,..., N
(10)
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1.4 Galerkin Approximation
where,
l
l

i (l ) j (l )  k (0) 0 i (0) j (0)
l
0

 (l )
fi dx  fˆi ( x2 )  k (l )
i (l )  k (0) 0 i (0)
l
0
K ij   (ki ' j 'ci j 'bi j )dx  k (l )
0
Fi  
l
0
If we use the matrix form in direct notation, we write
Kα  F
Solutions could be obtained by solving the equation (10)
α  K 1F
And, finally, the approximated solution for the set of equation (1) can be
expressed in the form of the linear combination with the  as
N
u ( x)   j j ( x)
h
j 1
We note that matrix K is symmetric if
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c( x)  0
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1.5 Symmetrization of the ODE
If c( x)  0 , the matrix of the eq. (10) becomes nonsymmetric.
This nonsymmetry requires more memory space and the computation time in
computation. It will be desirable if we can make the equation symmetric.
We start from the given DE,
 ku' '(k 'c)u 'bu  f
To change the equation into the symmetric form, we try to find some function g (x )
which satisfies the relation,
g ( x)[ ku' '(k 'c)u ' ]  [ g ( x)ku' ]'
and then we have the symmetrized equation (11) through manipulation,
( gk )'  (k'  c) g  kg'  gc  0
dg
c


 g  k dx 
 [ g ( x)ku' ]' gbu  gf

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c
g  exp (   dx)
k
(11)
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1.6 Trial-Solution Method
1.6.1 General procedure and methods for the approximated solutions
u ( x)  uh ( x) : Trial Solution
This is generally expressed as Linear Combination(LC)
of known functions
• General procedure of Trial Solution Methods
1. Construct the trial solution, u h (x) , in terms of basis functions in the approximated space
2. Determine the method to obtain optimal solution u h (x)
3. Predict the accuracy of u h (x)
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1.6 Trial-Solution Method
1. uh ( x)   0 ( x) 
(
N
 
j 1
j
j
( x)
 0 ( x) need be kept for boundary condition)
2. Method to obtain the Best possible solution
a. Method of weighted residual(MWR)
(1) Collocation method -- Collocation FEM
(2) Subdomain method
-- Subdomain FEM
(3) Least-square method -- LS FEM
(4) Galerkin method
-- Galerkin FEM
b. Ritz variation method(RVM) : For the Min/max problem
3. If we set e( x)  u ( x)  uh ( x)
as the point-wise error
1

2
2
the global error norm can be defined by e 1  [(e)  (e' ) ]dx
2
0
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1.6.2 Example
For the generalized 1-D PDE in eqn (1), let us have an example problem with
the coefficients values of k ( x)  x, c( x)  0, b( x)  0, f ( x)  
2
, 1 x  2
2
x
Then, the equation is

.
d  du 
2
x


, 1  x  2 without JCs
dx  dx 
x2
 0  0,  0  1,  0  2
u (1)  2

BC : 
du 
1
1
1

x


u
'




1
,


0
,





0
0
0

dx
2
4
4


x 2

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1.6.2 Example
At first, we can consider polynomial forms as approximated solution,
uh ( x)  d1  d 2 x  d 3 x 2    d N x N 1
If we take N=4, then the trial solution will be
uh ( x)  d1  d 2 x  d 3 x 2  d 4 x 3
With the BC, we have
 uh (1)  d1  d 2  d 3  d 4  2


1
u
'
(
2
)

d

4
d

12
d


2
3
4
 h
4
Elimination of  3 ,  4 with above two equation gives the equation,
1
uh ( x)  2  ( x  1)  d1 ( x  1)( x  3)  d 2 ( x  1)( x 2  x  11)
4
 0 ( x)  11 ( x)   22 ( x)
Now we have two undetermined coefficients, 1 ,  2 .
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1.6.3 Approximate solutions by MWR
The residual is
R( x)  xu( x)' '
2
x2
MWR (Method of Weighted Residual) states


R( x)v( x)dx  0 v  H
where, v (x )
is weight function, or Test function
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(1) The Collocation Method
Select number of node points ( xi ) same as the unknown number of  j
By using the definition of dirac delta,


R( j , x) ( x  xi )dx  0 , namely, weight function v( x)   ( x  xi ) i  1,2,, N
 R( j , x1 )  0

N
 R( j , x2 )  0
 
  K ij j  Fi

j 1

 R( j , x N )  0

4
5
x1  , x2 
3
3
4
11
1  4 2 
3
8
8
97
1  13 2 
3
100
Here we have N=2 and select
 1  2.0993,  2  0.3560
1
4
du
1 1
 h ( x)   x h   ( x  2)  4.1986 x( x  2)  1.068 x( x  2)( x  2)
dx 2 4
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And then the solutions are uh ( x)  2  ( x  1)  2.0993( x  1)( x  3)  0.3560( x  1)( x 2  x  11)
(1) The Collocation Method
Accuracy Check
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(2) Sub-Domain Method
We assume that he average of residual in each sub-domain be zero
1
R( x, d j )dx  0
i  1,2, , N


x
i
xi
This is called the sub-domain method. These domains could be overlapped.
v( x) 
1
xi 1 xi xi
2
1
H ( x  xi 1 )  H ( x  xi )
xi
xi  xi  xi 1
For the previous example with x1  (1, 1.5), x2  (1.5, 2)
 1.5  1
2
     4( x  1)d1  3(3x 2  4)d 2  2 dx  0
x 
1  4
 2 1
2
     4( x  1)d1  3(3x 2  4)d 2  2 dx  0
x 
 1.5  4
9
19 3
63
11
and then, 1
d1  d 2 
,
d1  d 2 
 d1  2.5417, d 2  0.4259
2
8
24 2
8
24
1
 uh ( x)  2  ( x  1)  2.5417( x  1)( x  3)  0.4259( x  1)( x 2  x  11)
4
1 1
 h ( x)   ( x  2)  5.0834 x( x  2)  1.2777 x( x  2)( x  2)
2 4
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(2) Sub-Domain Method
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(3) The Least-Square Method
Determine  j , to minimize the integral value of square of residual
 2 2
in whole domain;
R ( x,  j )dx  0, i  1,2, , N
 i 1
By differentiation,
2
R( x,  )

1
R( x,  j )
Weight function becomes,
j
 i
v( x) 
In the case of the previous example,
R( x,  j )
1
then, two equations are
 4( x  1)
R( x,  j )
 i
R( x,  j )

 3(3x 2  4)
2
 1
2


4
(
x

1
)


3
(
3
x
 4) 2 

1
1  4
2
1
2


4
(
x

1
)


3
(
3
x
 4) 2 

1
1  4
2
7
 16
 1  27 2  8 ln 2 
2
3

711
33
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2 
 271 
5
4

By integral,
dx  0
2
4( x  1)dx  0
2 
x 
2
3(3x 2  4)dx  0
2 
x 
1  2.3155,
 2  0.3816
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(3) The Least-Square Method
Therefore,
1
uh ( x)  2  ( x  1)  2.3155( x  1)( x  3)  0.3816( x  1)( x 2  x  11)
4
1 1
 h ( x)   ( x  2)  4.6310 x( x  2)  1.1448 x( x  2)( x  2)
2 4
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(4) Galerkin Method
Select the weight function from the same basis with Trial function
N
This means that , v( x)    ii ( x)
i 1
and then,
N
  R ( x, 
i 1
j
)  ii dx  0
  R( x,  j )i ( x)dx  0
 i
i  1,2, , N
Applying to the previous example, we have
1
2
2
(


4
(
x

1
)


3
(
3
x

4
)


)( x  1)( x  3)dx  0
1
2
2
 4
x
1
2
2
2
(


4
(
x

1
)


3
(
3
x

4
)


)(
x

1
)(
x
 x  11)dx  0
1
2
2
 4
x
41
29
 5





 8 ln 2

1
2
5
6
 3
 41
81
211
1  2.1378,  2  0.3477





 24 ln 2

1
2
2
16
 5
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(4) Galerkin Method
Solutions :
1
uh ( x)  2  ( x  1)  2.1378( x  1)( x  3)  0.3477( x  1)( x 2  x  11)
4
1 1
 h ( x)   ( x  2)  4.2756 x( x  2)  1.0431x( x  2)( x  2)
2 4
<check>
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1.6.4 Ritz Variational Method
(1) Calculus of Variation
We define a functional,
1 b
J [v( x)]   [k ( x)v' ( x) 2  bv 2  2 f ( x)v( x)]dx
2 a
Now we introduce a class of minimization problems
as the following minimization problem,
“Find u  H 01 (a, b)
J (u )  J (v)
such that
v  H 01 (a, b)
“
Let us use the variational method to obtain minimizer u (x)
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(1) Calculus of Variation
Substitute the following function for v in J (v)
 ( x)  u ( x)  v( x)  u ( x)  u ( x)
J ( )  J (u )  J (u; v)   2 2 J (v)
b
where,
J (u; v)   (ku' v'buv  fv)dx : first vari ation
a
1 b
2
2
(
kv
'

bv
)dx
: second variation

a
2
J (u; v) could be obtained by J (u; v)   J (u  v)

 0
If J (u; v)  0 , the extremum value could be found
 2 J (v ) 

The positiveness of the second variation means that u (x) minimize functional.
To have the non-negative value of
J (u  v)  J (u)  J (u; v)   2 2 J (v)
and the arbitrariness of  forces the first variation being zero,
J (u; v)  0
b
b
a
a
v   (ku' v  buv)dx   fvdx
v  H 01 (a, b)
and then we obtain the equation,  (ku' )'bu  f ,  BC after the Integrated by Parts.
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(2) Ritz approximation
For the previous minimization problem, we first select the proper trial-function and
then substitute this to the functional to obtain unknowns minimizing the functional value.
This method is called the Ritz approximation.
N
Let us select
u h ( x)   j j ( x)
j 1
And we substitute this into the functional,
J (uh )  J h ( j )
Minimization with respect to  j means
J h
 0 i  1,2,  , N
 i
Applying this method to the previous example, the functional for minimization becomes
2
2
2  1  du 
1 
du


J (u )    x   2 u  dx  ( x )u 
1 2
dx  1

  dx  x 
The minimization of the approximated functional gives
41
29
5
 1   2  8 ln 2 
5
6
3
 41
81
211
 1   2  24 ln 2 
2
16
 5
Note that the equation is same as the one from Galerkin Method
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Review
Galerkin
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◎ Formally Self Adjoint
With the definition of Inner Product (u , v) L2 ( a ,b )  ab u ( x)v( x)dx,
Let us visit the differential equation,  (ku' )'bu  f
We define an operator A such that
and then, ( Au, v) L2  a Auvdx
 d d


k

b
 dx dx
u  f


 Au  f
b

b
a
b  du dv
 d d



k

b
uvdx

k

buv

 dx dx

dx
a  dx dx




b

d  dv 
   u  k   ubv dx
a
 dx  dx 

b 
d dv

  u  k
 bv dx  (u , Av) L2
a
 dx dx

We call this A as the “formally Self Adjoint operator”
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1.7 Finite Element basis function
(1) Basic principles of Finite Element Basis Function
a. Simple expression for easy numerical integration
b. Smoothness requirement of the basis function such the integral in the variational
h
1
formulation should be meaningful, H  H
h
c. At the ith node xi , u ( xi )   i should satisfy
This orthonormal condition in the finite dimensional space means
N
u ( xi )    j j ( xi )   i
h
j 1
1
  j ( xi )   ij (Kronecker delta )  
0
i j
i j
 Basis functions which are easily generalized and formulated are
necessary.
 Finite element mesh: divide the given domain with finite elements.
Node: boundary nodal point constructing element.
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1.7 Finite Element basis function
(2) Simple example of Finite Element basis function
Piecewise linear function
In the case of
u (a)  0  u (b)
it is called as Hat function
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( x  xi 1 ) / hi , xi 1  x  xi

  ( xi 1  x) / hi 1 , xi  x  xi 1
0
, x  xi 1 or x  xi 1

1 / hi , xi 1  x  xi

   1 / hi 1 , xi  x  xi 1
0
, x  xi 1 or x  xi 1

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1.7 Finite Element basis function
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1.7 Finite Element basis function
(3) Finite Element Interpolation
Lagrange Finite Element
a. Mapping function at ( xi , xi 1) to (-1,1),
x  ( xi , xi 1 )

b.
2 x  ( xi  xi 1 )
xi 1  xi
  (1,1)
: master element
i ( x)   i ( )
 i ( j )   ij , kth Lagrange polynomial
 i ( ) 
(  1 )(   2 )  (   i 1 )(   i 1 )  (   k 1 )
( i  1 )  ( i   i 1 )( i   i 1 )  ( i   k 1 )
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1.7 Finite Element basis function
c. first,
1  1, 2  1
  2
1
 (1   )
(1   2 ) 2
  1
1
 2 ( ) 
 (  1)
( 2  1 ) 2
 1 ( ) 
second,
1  1,  2  0, 3  1
1
2
 2 ( )  1   2
 1 ( )   (  1)
1
2
 3 ( )   (  1)
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1.7 Finite Element basis function
1
1
3
3
1
1
4
1
 2
 9
 1 ( )  (  )(  )(  1) / ( )(  )( 2)  (1   )( 2  )
3
3
3
9
 3
 16
third, 1  1,  2   ,  3  ,  4  1
2
4  27 2
1
 2
(  1)(  )
3
3  16
3
 3
1
2  27
1
 4 2
 3 ( )  (  1)(  )(  1) / ( )( )(  )  (1   2 )(  )
3
3  16
3
 3 3
1
3
 2 ( )  (  1)(  )(  1) / ( )(  )(  ) 
1
3
1  4 2  9
1
(  1)( 2  )
3  3 3  16
9
 4 ( )  (  1)(  )(  ) / 2( )( ) 

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1.7 Finite Element basis function
(4) Characteristics of Lagrange FE Basis function
a. Slopes of solution will be discontinuous at the boundary node point
between elements since they are Lagrange functions in H 1
b. All  j are 1, which means that the summation of all shape functions becomes 1.
N
N
j 1
j 1
uh ( )   j j ( )  uh ( )   j ( )  1
c. Summation of all differentiated basis functions becomes zero.
N

j 1
j
' ( )  uh ' ( )  0
The characteristics b. and c. could be used to verify shape functions when you
do the Finite Element programming.
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1.8 Galerkin Finite Approximation
We now know that the system of differential equation (1) can be changed
by MWR and Galerkin approximation into a simultaneous equation,
N
K
j 1
ij
  j  Fi
(10)
where, K ij  [ki ' j ' ci j 'bi j ]dx  k (l )  l i (l ) j (l )  k (0)  0 i (0) j (0)

l
l
0
0


Fi   fi dx  fˆi ( x2 )  k (l ) l i (l )  k (0) 0 i (0)
l
0
l
0
And then, the next step may be how to implement the concept of finite elements
to the equation (10).
Specially, convenient and efficient construction and operation of the K matrix
and F vector in the equation are very important for this method.
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1.8 Galerkin Finite Approximation
Therefore, we will consider the following items step by step in the following lectures
1. How to construct K ij and Fi ?
․ Take advantage of the Finite Element Concept
․ Efficient numerical scheme to obtain the values for the matrix and vector
․ Element by Element construction of the matrix and vector
- Assembly of element matrix and vector from FE concept
2. How to solve the simultaneous equations efficiently
- efficient solver to take advantage of the sparseness of FE matrix
Band Solver
----Solving the band structure of FE matrix
Skyline Solver
----Utilization of sparseness of FE matrix
Frontal Solver
----Best utilization of FE concept
Iterative method
----In the case of Large DOF problems
(For example, Pre-Conditioned Conjugate Gradient Method)
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1.8 Galerkin Finite Approximation
 Calculation process:
(1) Finite element in the range of (0, l ) , divide by  e
Node at x  x2 to satisfy jump[|  |] x2  fˆ
Take one element ; e  (s1 , s2 )
Weak form at Element  e can be written by

s2
s1
s2
(ku' v'cu ' v  buv)dx   fvdx   ( s1 )v( s1 )   ( s2 )v( s2 )
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s1
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1.8 Galerkin Finite Approximation
By Galerkin approximation in element  e

s2
s1
(kuhe ' vhe 'cuhe ' vhe  buhe vhe )dx
s2
  fvhe dx   ( s1 )vhe ( s1 )   ( s2 )vhe ( s2 ) vhe  H 1
s1
Set
Ne
 e
e e
u h ( x)   u j j ( x)

j 1

N
v e ( x)  e v e e ( x)

i
i
 h
i 1
where N e is number of nodes in  e
e
and  i is basis function in  e
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1.8 Galerkin Finite Approximation
and then, we have equation in this element
Kijeu ej  fi e   (s1 ) ie (s1 )   (s2 ) ie (s2 )
where,
 K e  s2 (k e ' e 'c e ' e  b e e )dx
i
j
i
j
i
j
 ij s1

s2
 f i e   f ie dx
s1

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1.8 Galerkin Finite Approximation
(2) Assemble
Let us look at three element in neighbor.
#
#
and then,
e  1 element : K ije 1u ej 1  f i e 1   e 1 ( s1 )1i   e 1 ( s2 ) N ei
e
element : K ijeu ej  f i e   e ( s1 )1i   e ( s2 ) N ei
e  1 element : K ije 1u ej 1  f i e 1   e 1 ( s1 )1i   e 1 ( s2 ) N ei
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1.8 Galerkin Finite Approximation
Since the flux should be in equilibrium, we have
 e (s1 )   e1 (s2 ),  e1 (s1 )   e ( s2 )
and

l
0

1

2
 
N
Therefore Global Stiffness Matrix and Global Force Vector can be obtained
NN
K ij   K ije
e 1
NN
Fi   f i e  ( jump )   1 ( s1 ) i1   NN ( s2 ) iN
e 1
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1.8 Galerkin Finite Approximation
(3) Prescription Boundary and jump condition
du
Prescription of the boundary conditions with    k
dx
will produce,
  k (  u ) / 


  k  k   j j ( x )

 j
0
0 N
  ( s1 )  k
k
 j j (0)

0
 0 j 1
1

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NN


( s2 )   k l  k l
l
l
N
   (l )
j 1
j
j
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1.8 Galerkin Finite Approximation
Jump condition;
N
N
i 1
i
fˆ vh ( x2 )  fˆ   ii ( x2 )  fˆ   i nx
2
Finally, we have the Global matrix and the Global force vector
NN
K ij   K ije  k
e 1
l

 iN jN  k 0  i1 j1
l
0


Fi   f i e  fˆ inx 2  k l  iN  k 0  i1
NN
e 1
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l
0
Aerospace Structures Laboratory
1.8 Galerkin Finite Approximation
Three kinds of Boundary conditions
(a) Dirichle BC
 0   l  0  u (0) 
0

, u (l )  l
0
l
ⅰ) Eliminate unknown u1 and u N , then solve the remain (N-2) dof problem
ⅱ) Apply
 0   l  10 30
This approach is an application of the Exterior penalty method.
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1.8 Galerkin Finite Approximation
(b) Neumann BC
The slope of the solution ( or flux ) will be prescribed
in both side of boundary points
u ' (0) 
0

, u ' (l )  l ,  0   l  0
0
l
(c) Mixed BC
It is given in the form of   Au  B ,
This form is same as the general boundary condition,
au' u  
in the set of equation in (1).
Then an appropriate values of α, β, γ will prescribe the boundary
condition
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1.9 Error Estimation, Accuracy, Convergence
(a) various error norm
e : pointwise error
e
e
e

E
 ke' be dx
1
1/ 2
2
0
 e' e dx
  e dx
H1
1

2
2
or
: energy norm, H 1 norm
1/ 2
0
1
0
2
2
1/ 2
0
: root mean square norm, H 0 norm,
L2 norm
e

 max e( x)
0 x 1
: Maximum norm, L norm
infinity norm
All the definition of error norm should satisfy the following convergence property,
  0  e  0 that is, un  u (convergenc e)
Seoul National University
Aerospace Structures Laboratory
1.9 Error Estimation, Accuracy, Convergence
(b) a- priori error estimate
The general error estimation is possible for the simple equation (1)
Let say,
k ( x)  1
c( x)  0
b( x )  1 ,
homogeneou s BC, No jump
then, the variational form is
or
l
l
0
0
 (u' v'uv)dx   ( f v)dx
①
(u, v) H 1  ( f , v) H 0 v  H10
②
Seoul National University
Aerospace Structures Laboratory
1.9 Error Estimation, Accuracy, Convergence
From the variational statement in finite dimension,
(uh , vh ) H 1  ( f ( x), vh ) H 0 v  Hh  H10
③
The ② - ③ will result
(u  uh , vh ) H 1  0
vh  H h
then,
(e, v h )1  0
v h  H h
④
h
This means that the error and v are orthogonal each other.
(w. r. t  H norm
)
1
Seoul National University
Aerospace Structures Laboratory
1.9 Error Estimation, Accuracy, Convergence
From the Cauchy - Schwartz Inequality,
(u, v)V  C u V u V
⑤
The equation ② can be written as
l
 (v' w'vw)dx  (v, w)
0
H1
C v
H1
w H1
l
l
0
0
 C (  (v'2 v 2 )dx)  (  ( w'2  w2 )dx)
⑥
Therefore, we have an inequality equation,
e
2
H1
 (e, e) H 1  (e, u  uh ) H 1
 (e, u  vh  vh  uh ) H 1
 (e, u  vh ) H 1  (e, vh  uh ) E  C e
Seoul National University
H
1
u  vh
H1
⑦
Aerospace Structures Laboratory
1.9 Error Estimation, Accuracy, Convergence
If we assume wh  H 0h be a special FE weight function interpolating u (x)
Let us have a linear interpolation function such that
wB  wA
(  x A )  wA
h
and then,
wA  w( x A ), wB  w( xB )
wh ( ) 
⑧
x A    xB
Three term Taylor series expansion of u (x) at x A
is
1
u ( )  u ( x A )  u ' ( x A )(  x A )  u ' ' ( )(  x A ) 2
2
where the last term in ⑨ is the remainder, and xA    xB
Seoul National University
⑨
Aerospace Structures Laboratory
1.9 Error Estimation, Accuracy, Convergence
Since u( x A )  wh ( x A ) , the equation will be
u ( )  wh ( )  u ' ( x A )(  x A ) 
wB  wA
1
(  x A )  u ' ' ( )(  x A ) 2
h
2
w  wA 
1

2
  u' ( xA )  B
(


x
)

u
'
'
(

)(


x
)

A
A
h
2


⑩
w  wA 
1

2
  u' ( xA )  B
(  x A )  u ' ' ( )(  x A )
h
2


(From the Triangular Inequality )
Seoul National University
Aerospace Structures Laboratory
1.9 Error Estimation, Accuracy, Convergence
2
If e(x ) has bounded second derivative, i.e., u( x)  C then e(x) can be written as
e( x)  e( )  e' ( )( x   ) 
e' ' ( )
( x   )2
2!
⑪
If we choose the point  as the extreme point of e(x )
and note that, from the interpolation property,
e( xA )  0  e( xB )
Since e' ( )  0 and
e( )  
e' ' ( )
( x   )2
2
and
e' ' ( )  h 
e( ) 
 
2 2
Seoul National University
2
⑫
Aerospace Structures Laboratory
1.9 Error Estimation, Accuracy, Convergence
Then,
e' ' ( )  h 2 
 
max e( x)  e( ) 
x
2  2 

u ' ' ( ) 1 2
h  C1h 2
2 4
⑬
since the interpolation function in this example is linear polynomial.
Let say,
E ( x)  e' ( x)  u' ( x)  wh ' ( x)
By the same procedure as the previous process,
E ( x)  E ( )  E ' ( )( x   )
⑭
 max E(x)  C2 h
⑮
x
Seoul National University
Aerospace Structures Laboratory
1.9 Error Estimation, Accuracy, Convergence
After some arrangements, we obtain
2
u  wh
2
H1
2
 C3[max u ' ( x)  wh ' ( x)  max u ( x)  wh ( x) ]
0 x4
0 xl
 C3[C12 h 4  C22 h 2 ]
 C h2
⑯
whe n h is small value
In summary from the previous results, the final error estimates are
e
e
e
2
L

1
H
 C * u  vh
2
L
 C * u  wh
2
L
 Ch 2
 max e( x)  Ch2
 C * u  vh
Seoul National University
1
H
 C * u  wh
1
H
 Ch
Aerospace Structures Laboratory
1.9 Error Estimation, Accuracy, Convergence
In general form, it could written by
e
H
m
 Ch  u r ,
u  min( k  1  m, r  m)
where k is complete polynomial order of the interpolat ion function
and u  H r (), v h  H h  H m ()
When the convergenc e study is carried out, the logarithmi c expression is helpful.
Let E  e
H
m
 Ch  u r , and then the inequality equation w ill be in the form,
ln E  ln C    ln h
where the norm of u is included in C .
The equation can be drawn in logarithmi c axis of E and h.
Then the convergenc e ratio  will be the slope of the line curve.
Seoul National University
Aerospace Structures Laboratory