Weighted Residual
Methods
1
Formulation of FEM Model
Formulation of FEM Model

Direct Method
Variational Method
Weighted Residuals
• Several approaches can be used to transform the physical
formulation of a problem to its finite element discrete analogue.
• If the physical formulation of the problem is described as a
differential equation, then the most popular solution method is
the Method of Weighted Residuals.
• If the physical problem can be formulated as the minimization
of a functional, then the Variational Formulation is usually used.
2
Formulation of FEM Model
Finite element method is used to solve physical problems
Solid Mechanics
Fluid Mechanics
Heat Transfer
Electrostatics
Electromagnetism
….
Physical problems are governed by differential equations which satisfy
Boundary conditions
Initial conditions
One variable: Ordinary differential equation (ODE)
Multiple independent variables: Partial differential equation (PDE)
3
Physical problems
Axially loaded elastic bar
A(x) = cross section at x
b(x) = body force distribution
y
(force per unit length)
x E(x) = Young’s modulus
u(x) = displacement of the bar at x
x
x=0
x=L
Differential equation governing the response of the bar
d 
du 
AE

  b  0;
dx 
dx 
0 xL
Second order differential equations
Requires 2 boundary conditions for solution
4
Physical problems
Axially loaded elastic bar
y
x
x
x=L
x=0
Boundary conditions (examples)
u0
u 1
at x  0 Dirichlet/ displacement bc
at x  L
u  0 at x  0
du
Neumann/ force bc
EA
 F at x  L
dx
Differential equation + Boundary conditions = Strong form of the
“boundary value problem”
5
Physical problems
Flexible string
y
x=0
x=L
x
x
S
S = tensile force in string
p(x) = lateral force distribution
(force per unit length)
w(x) = lateral deflection of the
string in the y-direction
S
p(x)
Differential equation governing the response of the bar
d 2u
S 2  p  0;
dx
0 xL
Second order differential equations
Requires 2 boundary conditions for solution
6
Physical problems
Heat conduction in a fin
y
x
x
x=0
x =L
A(x) = cross section at x
Q(x) = heat input per unit length per
unit time [J/sm]
k(x) = thermal conductivity [J/oC ms]
T(x) = temperature of the fin at x
Q(x)
Differential equation governing the response of the fin
d 
dT 
Ak

  Q  0;
dx 
dx 
0 xL
Second order differential equations
Requires 2 boundary conditions for solution
7
Physical problems
Heat conduction in a fin
y
x
x
x=0
x=L
Q(x)
Boundary conditions (examples)
T  0 at x  0
dT
k
 h at x  L
dx
Dirichlet/ displacement bc
Neumann/ force bc
8
Physical problems
Fluid flow through a porous medium (e.g., flow of water through a dam)
y
x
x
x=0
x=L
A(x) = cross section at x
Q(x) = fluid input per unit volume
per unit time
k(x) = permeability constant
j(x) = fluid head
Q(x)
Boundary conditions (examples)
Differential equation
d  dj 
k
  Q  0;
dx  dx 
0 xL
j  0 at x  0
dj
k
Second order differential equations
Requires 2 boundary conditions for solution
9
dx
Known head
 h at x  L Known velocity
Physical problems
10
Physical problems
11
Formulation of FEM Model
Observe:
1. All the cases we considered lead to very similar differential
equations and boundary conditions.
2. In 1D it is easy to analytically solve these equations
3. Not so in 2 and 3D especially when the geometry of the domain is
complex: need to solve approximately
4. We’ll learn how to solve these equations in 1D. The approximation
techniques easily translate to 2 and 3D, no matter how complex the
geometry
12
Formulation of FEM Model
A generic problem in 1D
d 2u
 x  0;
2
dx
u  0 at x  0
u  1 at x  1
0  x 1
Analytical solution
1
7
u ( x)   x 3  x
6
6
Assume that we do not know this solution.
13
Formulation of FEM Model
A generic problem in 1D
A general algorithm for approximate solution:
Guess u( x)  a0j o ( x)  a1j1 ( x)  a2j 2 ( x)  ...
where jo(x), j1(x),… are “known” functions and ao, a1, etc are constants
chosen such that the approximate solution
Satisfies the differential equation
Satisfies the boundary conditions
i.e.,
d 2jo ( x)
d 2j1 ( x)
d 2j 2 ( x)
a0
 a1
 a2
 ...  x  0;
2
2
2
dx
dx
dx
a0jo (0)  a1j1 (0)  a2j 2 (0)  ...  0
0  x 1
a0jo (1)  a1j1 (1)  a2j 2 (1)  ...  1
Solve for unknowns ao, a1, etc and plug them back into
This is your approximate
u ( x)  a0j o ( x)  a1j1 ( x)  a2j 2 ( x)  ...
solution to the strong form
14
Formulation of FEM Model
Solution of Continuous Systems – Fundamental Concepts
Exact solutions
limited to simple geometries and boundary & loading conditions
Approximate Solutions
Reduce the continuous-system mathematical model to a discrete idealization
Variational
Weighted Residual Methods
 Rayleigh Ritz Method
 Galerkin
 Least Square
 Collocation
 Subdomain
15
Weighted Residual Methods
Weighted Residual Formulations
Consider a general representation of a governing equation on a region V
Lu  0
L is a differential operator
eg. For Axial element
d 
du 
EA

  P( x)  0
dx 
dx 
d
d
L
EA 
dx
dx
u
Assume approximate solution
then
Lu  R
16
  P( x)
Weighted Residual Methods
Weighted Residual Formulations
Lu  R
Lu  0
Exact
Approximate
ERROR  L  u   R
Objective:
Define u so that weighted average of Error vanishes
Set Error relative to a weighting function w
 w L(u) dV  0
or
V
 w R dV  0
V
17
Weighted Residual Methods
Weighted Residual Formulations
w
 w L(u ) dV  0
V
w 1
ERROR
18
Weighted Residual Methods
Weighted Residual Formulations
w
 w L(u ) dV  0
V
ERROR
19
Appendix
20
Weighted Residual Methods
 Start with the integral form of governing equations
 Assume functional form for trial (interpolation, shape) functions
 Minimize errors (residuals) with selected weighting functions
x j
Power series

sin jx, cos jx Fourier series

w j ( x)   L j ( x)
Lagrange

Hermite
 H j ( x)
T ( x )
Chebychev
j

21
Weighted Residual Methods
 Assume certain profile (trial or shape function) between nodes
J
u ( x)   a j j ( x)
j 1
but
 L(u )  R( x)  0 ,

  w R dx   wL(u )dx  0
22
Residual
Weighted Residual
Weighted Residual Methods
 In general, we deal with the numerical integration of trial or
interpolation functions
 Trial functions:
constant, linear, quadratic, sinusoidal, Chebychev
polynomial, ….
 Weighting functions:
subdomain, collocation, least square, Galerkin, ….
 w( x, y, z) R( x, y, z)dxdydz   wL(u)dv  0
V
V
23
General Formulation
 Weighted Residual Methods (WRMs)
 Construct an approximate solution
J
u ( x, y, z )  uo ( x, y, z)   a j  j ( x, y, z)
j 1
Chosen to satisfy I.C./B.C.s if possible
 Steady problems – system of algebraic equations for trial
function j (x,y,z)
 Transient problems – system of ODEs in time
24
Weighted Residual Methods
 Consider one-dimensional diffusion equation
 L(u )  0

 L(u )  R ( x)  0
Exact solution
Approximation
 In general, R 0 with increasing J (higher-order)
 w
m
( x, y, z ) R( x, y, z )dxdydz  0, m  1, 2,
 Weak form – integral form, discontinuity allowed
(discontinuous function and/or slope)
25
,M
Weighted Residual Methods
 Weak form – integral formulation
Differential Form:


Exact Integral Form:

Discretization :
L(u )  0
 w(x, y,z)L(u )dxdydz  0
 w (x, y,z)L(u)dxdydz  0
m
 R  0, but “weighted R” = 0
 Choices of shape or interpolation functions?
 Choices of weighting functions?
26
Subdomain Method
 w (x, y,z)L(u )dxdydz  0
m

L(u )  R  0
Equivalent to finite volume method
1,
wm  
0,
in Dm
outside Dm
 w L(u)dxdydz  
m


;
Dm
R( x, y, z )dv  0
Dm : numerical element (arbitrary control volume)
Dm may be overlapped
27
Collocation Method
 w (x, y,z)L(u )dxdydz  0
m

L(u )  R  0
;
Zero residuals at selected locations (xm, ym, zm)
wm  x     x  xm 
 w  x R  x  dv     x  x  R( x )dv
m
m
 R  xm   R  xm , ym , zm 

No control on the residuals between nodes
28
Least Square Method
 w (x, y,z)L(u )dxdydz  0
m

;
L(u )  R  0
Minimize the square error

R
2
R ( x, y, z, am )dxdydz  0  
Rdxdydz  0

am V
am
V
R
wm  x  
am
Square error
R2  0
1 
 wm  x R  x  dx  2 am
29
2
R
 dx  0
R2  0
Galerkin Method
 w (x, y,z)L(u )dxdydz  0
m

;
L(u )  R  0
Weighting function = trial (interpolation) function
wm  x   m  x 
 w  x R  x  dx     x  R  x  dx
m

m
For orthogonal polynomials, the residual R is orthogonal to
every member of a complete set!
30
Numerical Accuracy









How do we determine the most accurate method?
How should the error be “weighted”?
Zero average error?
Least square error?
Least rms error?
Minimum error within selected domain?
Minimum (zero) error at selected points?
Minimax – minimize the maximum error?
Some functions have fairly uniform error distributions comparing
to the others
31
Application to an ODE
 Consider a simple ODE (Initial value problem)
 dy
  y 0 , 0 x1
x

y

e
dx


 y (0)  1
 Use global method with only one element
 Select a trial function of the form of
N
y  1 ajx j
j 1
 Automatically satisfy the auxiliary condition
 aj = constant, not a function of time
32
Application to an ODE
 Consider a cubic interpolation function with N = 3
N
y  1   a j x j  1  a1 x  a2 x 2  a3 x 3
j 1
 QUESTION: Which cubic polynomial gives the best fit to the
exact (exponential function) solution?
 Definition of best fit?
 Zero average error, least square, least rms, …?
33
Residual
 Substitute the trial function into governing equation
N
N
N

dy
j 1
j 
R  L( y ) 
 y   ja j x   1   a j x   1   a j x j 1 ( j  x)
dx
j 1
j 1
j 1


 For cubic interpolation function N = 3
R( x)  1  a1 (1  x)  a2 (2x  x 2 )  a3 (3x 2  x 3 )
 (a1  1)  (2a2  a1 ) x  (3a3  a2 ) x 2  a3 x 3  0
 The residual is a cubic polynomial  R  0
 Determine the optimal values of aj to minimize the error (under
pre-selected weighting functions)
34
Subdomain Method
 Zero average error in each subdomain
D1
x0

1
0
x1
D3
x2
Uniform spacing
x3
xm
a
a
a


R( x)dx  0  (a1  1) x  (a2  1 ) x 2  (a3  2 ) x 3  3 x 4 
xm1
2
3
4  xm1

wm Rdx  

m  1 :


m  2 :


m  3 :

D2
xm
1/ 3

0
Rdx  0

2/3

1
1/ 3
2/3
Rdx  0
Rdx  0
5
8
11
1
a1  a2 
a3 
18
81
324
3
3
20
69
1
 a1  a2 
a3 
18
81
324
3
1
26
163
1
 a1 
a2 
a3 
18
81
324
3

 Note: R(0) = 0.0156  0, R(1) = 0.0155  0
35
1.0156 
ai  0.4219 
 0.2813 
Subdomain Method
Net area under each curve = 0
y  1  1.0156 x  0.4219x 2  0.2812x 3
R  0.0156  0.1719x  0.4219x 2  0.2812x 3
Zero average error in each subdomain
36
Subdomain Method
 Nonuniform subdomains?
D1
D2
x0
x1
D1
x0
m  1 :


m  2 :

m  3 :

D3
x2
D2
x1
D3
x2

x1

x2

x3
x0
x1
x2
x3
Grid clustering in highgradient regions
x3
Rdx  0
Different coefficients
for different choices of
subdomains
Rdx  0
Rdx  0
37
Least Square Method
 Minimum square errors over the entire domain
N
R( x)  1   am (mx m1  x m )

m 1

1
0
wm Rdx  
1
0
1
   (mx
0
m 1
R
 mx m 1  x m
am
R( x)
R( x)dx  0
am
N
 x )dx   a j   mj x m  j 2  ( j  m) x m  j 1  x m  j  dx
0
m
1
j 1
 For arbitrary N (symmetric matrix)
 mj

1
1
m
aj 
1 


 1 
m  j 1 
m 1 m 1
j 1
 m  j 1
N
38
Least Square Method
 For cubic interpolation function (N=3)
R   a1  1   2a2  a1  x   3a3  a2  x 2  a3 x 3
1

R
 1 x
   1  x  Rdx  0
m  1, w1 
0

a
1


1
R
2
 2x  x
   2x  x 2  Rdx  0
m  2, w2 
0
a2


1
R
2
3
 3x  x    3x 2  x 3  Rdx  0
m  3, w3 
0
a3

 Nonuniform weighting of residuals over the domain
39
Least Square Method
 Cubic trial function
 mj
mj 
m
aj 



j 1
 m  j 1 m  j  1  m  1
1
1
1
1

m  1 , 3 a1  4 a2  5 a3  2
1.0131 

1
8
2
2

0.4255 
m

2
,
a

a

a


a


1
2
3
i


4
15
3
3

0.2797 
1
2
33
3

m  3 , 5 a1  3 a2  35 a3  4

3
 R(0) = 0.0131  0, R(1) = 0.0151  0
40
Least Square Method
Weighted average errors = 0
Minimum sqaure error
41
Least Square Method
Weighted average error =
Net area under curve = 0
Weighted average error = 0
Weighted average error = 0
42
Galerkin Method
 Weighting function = Trial function
0
1
2
3


(
x
)

x
,
x
,
x
,
x
,
 m

m1
w
(
x
)


(
x
)

x

m
 m
, x N 1
N
R( x)  1   am (mx m1  x m )
m 1

1
0
1
wm Rdx   x
0
m 1
1
R( x)dx  0    x
m 1
0
N
1
j 1
0
dx   a j 

j
1 
1

a



 j
j

m

1
j

m
m
j 1 

N
N
S
j 1
mj
a j  dm
43

SA  D
 jx
m j 2
 x m j 1  dx
Galerkin Method
 For cubic interpolation function (N=3)
R   a1  1   2a2  a1  x   3a3  a2  x 2  a3 x 3
m  1, W  x0  1
1


1
m  2, W2  x  x

m  3, W3  x 2

1
  Rdx  0
0
1
  xRdx  0
0
1
  x 2 Rdx  0
0
 Small weighting of residuals near x = 0
 Largest weight for residuals near x = 1
44
Galerkin Method
 Cubic trial function

j
1  1
a


j 

j 1
 m  j 1 m  j  m
1
2
3
a1  a2  a3  1
2
3
4
 1.0141 
1
5
11
1
a1  a2  a3   ai  0.4225 
6
12
20
2
0.2817 
1
3
13
1
a1  a2  a3 
12
10
30
3
3

m  1,


m  2,


m  3,

2
3

 y( x)  1  1.0141x  0.4225x  0.2817 x

2
3
R
(
x
)

0
.
0141

0
.
1691x

0
.
4226
x

0
.
2817
x


 R(0) = 0.0141  0, R(1) = 0.0141  0
45
Galerkin Method
W1 = 1
Weighting functions
W2 = x
W3 = x2
Weighted residuals
xR
x2R
R
46
Galerkin Method
Order of
approximation:
Linear, Quadratic, and Cubic
Trial functions
47
Galerkin Method
48
Galerkin Method
 Alternative choice of weighting functions
R  (a1  1)  (2a2  a1 ) x  (3a3  a2 ) x 2  a3 x 3
3
 1  (1  x)a1  (2x  x 2 )a2  (3x 2  x 3 )a3  1   a j j
j 1
m  1, W  1  x
1


2
m

2
,
W

2x

x

2

m  3, W3  3x 2  x 3

1
  (1  x ) Rdx  0
0
1
  (2x  x 2 ) Rdx  0
0
1
  (3x 2  x 3 ) Rdx  0
0
 More uniform weighting functions
 Identical to the least square method
49
Collocation Method
 R = L(u) = 0 at collocation points
x0

1
0
x1
x2
R(u )  0 but R(u )  0
wm Rdx  R( xm )  (a1  1)  (2a2  a1 ) xm  (3a3  a2 ) xm2  a3 xm3
m  1 : x1  0

1

m

2
:
x


2
2

m  3 : x3  1
 a1  1
1
3
5
a1  a2  a3  1
2
4
8
 a2  2a3  1

y ( x)  1  x 
 1   1 
ai   3 / 7   0.4286 
 2 / 7  0.2857 
3 2 2 3
5
x  x  y (1)  2  e  2.71828
7
7
7
 Identical to Galerkin method if the residuals are evaluated at x =
0.1127, 0.5, 0.8873
50
Collocation Method
Zero residuals at collocation points
R(0) = R(0.5) = R(1) = 0
But y  yexact at
collocation points
y(1) = 2.7142857  e
51
Taylor-series Expansion
 Truncated Taylor-series
2
3
x
x
e  1 x  
2 ! 3!
x
 1   1 
 ai  1 / 2    0.5 
1 / 6  0.1667 
 R(0) = 0, R(1) =  1/6 =  0.1667  0
 Poor approximation at x = 1
 Power series has highly nonuniform error distribution
52
Interpolation Functions
53
Numerical Accuracy
54
Interpolation Functions
Comparison of
numerical errors
for weighted
residual methods
55