13 weightedresidual

1
ENME 602 Spring 2007
Dr. Eng. Mohammad Tawfik
Numerical Solution of
Boundary Value Problems
Weighted Residual Methods
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

2
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
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

3
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
ψ
Basic Concepts
• Applying the differential operator on the approximate
solution, you get:
( )( ) ( ) ( ) ( )
( )( ) ( ) 0
1
1
≠−=
−⎟
⎠
⎞
⎜
⎝
⎛
=−
∑
∑
=
=
xgxLa
xgxaLxgxfL
n
i
ii
n
i
ii
ψ
ψ
( )( ) ( ) ( )xRxgxLa
n
i
ii =−∑=1
ψ
Residue

4
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.
General Weighted Residual
Method

5
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!
Evaluating the Residue
( )( ) ( ) ( )xRxgxLa
n
i
ii =−∑=1
ψ
( )( ) ( )( ) ( )( ) ( ) ( )xRxgxLaxLaxLa nn =−+++ ψψψ ...2211
n unknown variables
( ) ( )( ) ( ) 0
1
=⎟
⎠
⎞
⎜
⎝
⎛
−= ∫ ∑∫ =Domain
n
i
ii
Domain
dxxgxLadxxR ψ
One equation!!!

6
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 ψ
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

7
Collocation Method
( ) 0=jxR
( )
( )( ) ( ) 0
1
=−
=
∑=
j
n
i
jii
j
xFxLa
xR
ψ
Example Problem

8
The bar tensile problem
( )
0/
00
'
02
2
=⇒=
=⇒=
=+
∂
∂
dxdulx
ux
sBC
xF
x
u
EA
Bar application
( ) 02
2
=+
∂
∂
xF
x
u
EA
( ) ( )∑=
=
n
i
ii xaxu
1
ψ
( ) ( ) ( )xRxF
dx
xd
aEA
n
i
i
i =+∑=1
2
2
ψ
Applying the collocation method
( ) ( ) 0
1
2
2
=+∑=
j
n
i
ji
i xF
dx
xd
aEA
ψ

9
In Matrix Form
( )
( )
( )⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
−=
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
nnnnnn
n
n
xF
xF
xF
a
a
a
kkk
kkk
kkk
MMMOMM
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
ψ
Notes on the trial functions
• They should be at least twice
differentiable!
• They should satisfy all boundary
conditions!
• Those are called the “Admissibility
Conditions”.

10
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
π
ψ
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
ππψ
{ } { }faSin
l
EA −=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
− 1
2
42
ππ

11
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
≈==
πππ
( ) ⎟
⎠
⎞
⎜
⎝
⎛
≈
l
x
Sin
EA
fl
xu
2
57.0
2
π
( ) ( )5.057.0
2
=≈ exact
EA
fl
lu
( ) ( )0.19.00 =≈ exact
EA
lf
ux
Homework #11
• Solve the beam bending
problem, for beam
displacement, for a
simply supported beam
with a load placed at the
center of the beam using
– Any weighting function
– Collocation Method
• Use three term Sin series
that satisfies all BC’s
• Write a program that
produces the results for
n-term solution.
)(4
4
xF
dx
wd
=
0
)()0(
0)()0(
2
2
2
2
==
==
dx
lwd
dx
wd
lww

12
Exact Solution
12/1
10
3
15
7
412
2/10
60
13
12
)(
23
3
<<+−+−=
<<+=
x
xxx
x
xx
xw

13 weightedresidual

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