FEM: Introduction and Weighted Residual Methods

Weighted Residual Methods
Mohammad Tawfik
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Mohammad Tawfik

Mohammad Tawfik
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Objectives
• In this section we will be introduced to the
general classification of approximate
methods
• Special attention will be paid for the
weighted residual method
• Derivation of a system of linear equations
to approximate the solution of an ODE will
be presented using different techniques

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Classification of Approximate
Solutions of D.E.’s
• Discrete Coordinate Method
– Finite difference Methods
– Stepwise integration methods
• Euler method
• Runge-Kutta methods
• Etc…
• Distributed Coordinate Method

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Distributed Coordinate Methods
• Weighted Residual Methods
– Interior Residual
• Collocation
• Galrekin
• Finite Element
– Boundary Residual
• Boundary Element Method
• Stationary Functional Methods
– Reyligh-Ritz methods
– Finite Element method

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Basic Concepts
• A linear differential equation may be written in the form:
    xgxfL 
• Where L(.) is a linear differential operator.
• An approximate solution maybe of the form:
   

n
i
ii xaxf
1


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Basic Concepts
• Applying the differential operator on the approximate
solution, you get:
        
     0
1
1












xgxLa
xgxaLxgxfL
n
i
ii
n
i
ii


      xRxgxLa
n
i
ii 1

Residue

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Handling the Residue
• The weighted residual methods are all
based on minimizing the value of the
residue.
• Since the residue can not be zero over the
whole domain, different techniques were
introduced.

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General Weighted Residual
Method

Mohammad Tawfik
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Objective of WRM
• As any other numerical method, the
objective is to obtain of algebraic
equations, that, when solved, produce a
result with an acceptable accuracy.
• If we are seeking the values of ai that
would reduce the Residue (R(x)) allover
the domain, we may integrate the residue
over the domain and evaluate it!

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Evaluating the Residue
      xRxgxLa
n
i
ii 1

            xRxgxLaxLaxLa nn   ...2211
n unknown variables
       0
1






   Domain
n
i
ii
Domain
dxxgxLadxxR 
One equation!!!

Mohammad Tawfik
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Using Weighting Functions
• If you can select n different weighting
functions, you will produce n equations!
• You will end up with n equations in n
variables.
           0
1






   Domain
n
i
iij
Domain
j dxxgxLaxwdxxRxw 

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Collocation Method
• The idea behind the collocation method is
similar to that behind the buttons of your
shirt!
• Assume a solution, then force the residue
to be zero at the collocation points

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Collocation Method
  0jxR
 
     0
1



j
n
i
jii
j
xFxLa
xR


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Example Problem

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The bar tensile problem
 
0/
00
'
02
2





dxdulx
ux
sBC
xF
x
u
EA

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Bar application
  02
2



xF
x
u
EA
   

n
i
ii xaxu
1

     xRxF
dx
xd
aEA
n
i
i
i 1
2
2

Applying the collocation method
    0
1
2
2

j
n
i
ji
i xF
dx
xd
aEA


Mohammad Tawfik
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In Matrix Form
 
 
 








































nnnnnn
n
n
xF
xF
xF
a
a
a
kkk
kkk
kkk

2
1
2
1
21
22212
12111
...
...
...
Solve the above system for the “generalized
coordinates” ai to get the solution for u(x)
 
jxx
i
ij
dx
xd
EAk

 2
2


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Notes on the trial functions
• They should be at least twice
differentiable!
• They should satisfy all boundary
conditions!
• Those are called the “Admissibility
Conditions”.

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Using Admissible Functions
• For a constant forcing function, F(x)=f
• The strain at the free end of the bar should
be zero (slope of displacement is zero).
We may use:
  






l
x
Sinx
2



Mohammad Tawfik
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Using the function into the DE:
• Since we only have one term in the series,
we will select one collocation point!
• The midpoint is a reasonable choice!
 













l
x
Sin
l
EA
dx
xd
EA
22
2
2
2

   faSin
l
EA 




















 1
2
42


Mohammad Tawfik
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Solving:
• Then, the approximate
solution for this problem is:
• Which gives the maximum
displacement to be:
• And maximum strain to be:
    EA
fl
EA
fl
SinlEA
f
a
2
2
2
21 57.0
24
42


  






l
x
Sin
EA
fl
xu
2
57.0
2

   5.057.0
2
 exact
EA
fl
lu
   0.19.00  exact
EA
lf
ux

Mohammad Tawfik
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The Subdomain Method
• The idea behind the
subdomain method is
to force the integral
of the residue to be
equal to zero on a
subinterval of the
domain

Mohammad Tawfik
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The Subdomain Method
  0
1

j
j
x
x
dxxR
     0
11
1
  


j
j
j
j
x
x
n
i
x
x
ii dxxgdxxLa 

Mohammad Tawfik
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Bar application
  02
2



xF
x
u
EA
   

n
i
ii xaxu
1

     xRxF
dx
xd
aEA
n
i
i
i 1
2
2

Applying the subdomain method
    



11
1
2
2 j
j
j
j
x
x
n
i
x
x
i
i dxxFdx
dx
xd
aEA


Mohammad Tawfik
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In Matrix Form
     


















 11
2
2 j
j
j
j
x
x
i
x
x
i
dxxFadx
dx
xd
EA


Mohammad Tawfik
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Using the function into the DE:
• Since we only have one term in the series,
we will select one subdomain!
 













l
x
Sin
l
EA
dx
xd
EA
22
2
2
2

 




























ll
fdxadx
l
x
Sin
l
EA
0
1
0
2
22


Mohammad Tawfik
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Solving:
• Then, the approximate
solution for this problem is:
• Which gives the maximum
displacement to be:
• And maximum strain to be:
  EA
fl
EA
fl
lEA
fl
a
22
1 637.0
2
2


  






l
x
Sin
EA
fl
xu
2
637.0
2

   5.0637.0
2
 exact
EA
fl
lu
   0.10.10  exact
EA
lf
ux
   fla
l
x
Cos
l
EA
l



























1
0
22


Mohammad Tawfik
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The Galerkin Method
• Galerkin suggested that the residue
should be multiplied by a weighting
function that is a part of the suggested
solution then the integration is performed
over the whole domain!!!
• Actually, it turned out to be a VERY
GOOD idea

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The Galerkin Method
    0Domain
j dxxxR 
         0
1
   Domain
j
n
i Domain
iji dxxgxdxxLxa 

Mohammad Tawfik
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Bar application
  02
2



xF
x
u
EA
   

n
i
ii xaxu
1

     xRxF
dx
xd
aEA
n
i
i
i 1
2
2

Applying Galerkin method
         
 Domain
j
n
i Domain
i
ji dxxFxdx
dx
xd
xaEA 


1
2
2

Mohammad Tawfik
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In Matrix Form
         

















 Domain
ji
Domain
i
j dxxFxadx
dx
xd
xEA 

 2
2

Mohammad Tawfik
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Same conditions on the functions
are applied
• They should be at least twice
differentiable!
• They should satisfy all boundary
conditions!
• Let’s use the same function as in the
collocation method:
  






l
x
Sinx
2



Mohammad Tawfik
#WikiCourses
Substituting with the approximate
solution:
         
 Domain
j
n
i Domain
i
ji dxxFxdx
dx
xd
xaEA 


1
2
2




























l
l
fdx
l
x
Sin
dx
l
x
Sin
l
x
Sina
l
EA
0
0
1
2
2
222



 ll
a
l
EA
2
22
1
2







EA
fll
EA
f
a
2
3
2
1 52.0
16



Mohammad Tawfik
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Substituting with the approximate
solution: (Int. by Parts)
         
 Domain
j
n
i Domain
i
ji dxxFxdx
dx
xd
xaEA 


1
2
2

 ll
a
l
EA
2
22
1
2







EA
fll
EA
f
a
2
3
2
1 52.0
16


   
       



Domain
ij
l
i
j
Domain
i
j
dx
dx
xd
dx
xd
dx
xd
x
dx
dx
xd
x




0
2
2
Zero!

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What did we gain?
• The functions are required to be less
differentiable
• Not all boundary conditions need to be
satisfied
• The matrix became symmetric!

Mohammad Tawfik
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Summary
• We may solve differential equations using a
series of functions with different weights.
• When those functions are used, Residue
appears in the differential equation
• The weights of the functions may be determined
to minimize the residue by different techniques
• One very important technique is the Galerkin
method.

FEM: Introduction and Weighted Residual Methods

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FEM: Introduction and Weighted Residual Methods