Galerkin method

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Computational Engineering
Galerkin Method
Yijian Zhan
Ning Ma

Galerkin Method
Engineering problems: differential
equations with boundary conditions.
Generally denoted as: D(U)=0; B(U)=0
Our task: to find the function U which
satisfies the given differential equations
and boundary conditions.
Reality: difficult, even impossible to solve
the problem analytically

Galerkin Method
In practical cases we often apply
approximation.
One of the approximation methods:
GalerkinMethod, invented by Russian
mathematician Boris Grigoryevich Galerkin.

Galerkin Method
Related knowledge
Inner product of functions
Basis of a vector space of functions

Galerkin Method
Inner product
Inner product of two functions in a certain
domain:
shows the inner
product of f(x) and g(x) on the interval [ a,
b ].
*One important property: orthogonality
If , f and g are orthogonal to each
other;
**If for arbitrary w(x), =0, f(x) 0, ( ) ( )
b
a
f g f x g x dx  ,0fg  ,wf 

Galerkin Method
Basis of a space
V: a function space
Basis of V: a set of linear independent
functions
Any function could be uniquely
written as the linear combination of the
basis:0
{ ( )}
ii
Sx


 ()f x V 0
( ) ( )
jj
j
f x c x




Galerkin Method
Weighted residual methods
A weighted residual method uses a finite
number of functions .
The differential equation of the problem is
D(U)=0 on the boundary B(U), for example:
on B[U]=[a,b].
where “L” is a differential operator and “f”
is a given function. We have to solve the
D.E. to obtain U.0
{ ( )}
n
ii
x
 ( ) ( ( )) ( ) 0D U L U x f x  

Galerkin method
Weighted residual
Step 1.
Introduce a “trial solution” of U:
to replace U(x)
: finite number of basis functions
: unknown coefficients
* Residual is defined as:0
1
( ) ( ) ( )
n
jj
j
U u x x c x

    j
c ()
j
x ( ) [ ( )] [ ( )] ( )R x D u x L u x f x  

Galerkin Method
Weighted residual
Step 2.
Choose “arbitrary” “weight functions” w(x),
let:
With the concepts of “inner product” and
“orthogonality”, we have:
The inner product of the weight function
and the residual is zero, which means that
the trial function partially satisfies the
problem.
So, our goal: to construct such u(x), ( ) , ( ) ( ){ [ ( )]} 0
b
a
w R x w D u w x D u x dx    

Galerkin Method
Weighted residual
Step 3.
Galerkinweighted residual method:
choose weight function w from the basis
functions , then
These are a set of n-order linear
equations. Solve it, obtain all of the
coefficients .j
 0
1
, [ ( )] ( ){ [ ( ) ( )]} 0
n
b
j j j j
a
j
w R D u dx x D x c x dx   

      j
c

Galerkin Method
Weighted residual
Step 4.
The “trial solution”
is the approximation solution we want.0
1
( ) ( ) ( )
n
jj
j
u x x c x



Galerkin Method Example
Solve the differential equation:
with the boundary condition:( ( )) ''( ) ( ) 2 (1 ) 0D y x y x y x x x     (0) 0, (1) 0yy

Galerkin Method Example
Step 1.
Choose trial function:
We make n=3, and0
1
( ) ( ) ( )
n
ii
i
y x x c x

 0
1
22
2
33
3
0,
( 1),
( 1)
( 1)
xx
xx
xx









Galerkin Method Example
Step 2.
The “weight functions” are the same as
the basis functions
Step 3.
Substitute the trial function y(x) intoi
 0
1
, [ ( )] ( ){ [ ( ) ( )]} 0
n
b
j j j j
a
j
w R D u dx x D x c x dx   

     

Galerkin Method Example
Step 4.
i=1,2,3; we have three equations with
three unknown coefficients 1 2 3
,,c c c 312
312
312
4351
0
15 10 84 315
615 111
0
70 84 630 13860
734 611
0
315 315 13860 60060
ccc
ccc
ccc
    
   
    

Galerkin Method Example
Step 5.
Solve this linear equation set, get:
Obtain the approximation solution1
2
3
1370
0.18521
7397
50688
0.185203
273689
132
0.00626989
21053
c
c
c
   

    3
1
( ) ( )
ii
i
y x c x



Galerkin Method Example
GalerkinGalerkinsolutionsolution Analytic solutionAnalytic solution

References
1. O. C. Zienkiewicz, R. L. Taylor, Finite
Element Method, Vol 1, The Basis, 2000
2. Galerkinmethod, Wikipedia:
http://en.wikipedia.org/wiki/Galerkin_method#cite_note-BrennerScott-1
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