Transcript Document

Boundary Collocation Methods:
Review and Application to
Composite Media
P. A. Ramachandran
Washington University
St. Louis, MO
Lecture Presented at UNLV
4-25-03
OUTLINE OF LECTURE

Introduction: Boundary Collocation

MFS: Diffusion-Reaction (D-R) Problem

Trefftz method: Laplace Equation

Trefftz: D-R Problem

MFS for Composite Regions

Effective Media Properties.

Concluding Remarks
ST. LOUIS ARCH
The Gate Way of the West
Boundary Collocation
CONSIDER
Lu=0
where L is a linear differential operator.
Find or choose a set of basis function Gi which satisfy PDE
SOLUTION IS EXPRESSED AS:
N
u   bi Gi
i 1
where bi are expansion coefficients.
Fit the boundary conditions at selected points called the
collocation points to find bi
PROTOTYPE EQUATIONS

Laplace Equation
 2u  0
 Diffusion-Reaction Problems
 2u  2u  0

 2  Helmholtz equation
Poisson equation and Non-linear and time
dependent problems
 2u  f  f  x , u , u x 
 These can be reduced to linear operator type by
use of particular solutions; u = v + w;
 MESH FREE METHODS
APPLICATION AREAS

LAPLACE EQUATON
 HEAT TRANSFER
 POTENTIAL FLOW
 ELECTRIC POTENTIAL;
 ELECTROCHEMICAL SYSTEMS
 DIFFUSION WITH REACTION (Porous catalyst)
 ACOUSTICS: (Helmholtz Equation)
 STOKES FLOW: VECTOR POISSON EQUATION
 LINEAR ELASTICITY: VECTOR EQUATION
 MAGNETIC FIELD: VECTOR EQUATION.
MFS for Diffusion-Reaction Problem
•CONSIDER
 u  u  0
2
2
on 
•SOLUTION IS EXPRESSED AS:
n
u   bi Gi

i 1
Gi  K 0 (ri )
Gi is a solution to PDE;
Source point
Collocation point
Also known as fundamental solution
or free space Greens function
ri is the distance from source point i to any field point.
Gradient at the boundary
N
 G 
p   bi 
   bi Qi
 n i i 1
i 1
N
where Qi   G   1 K1 (ri )x  xi n x  y  yi n y 
 n i
2r
Constants bi, determined using boundary collocation to satisfy
boundary condition. Choose a set of ‘M’ Collocation Points are
located on the boundary. Then for each point, j, we have:
n
u j   bi G ji or
i 1
Dirichlet
or
n
p j   bi Q ji
i 1
Neumann
Equations can be assembled in a matrix form

(COE )(b )  ( RHS )
(COE) is Gji or Qji;
(RHS) is either uj or pj
Size of COE matrix is M (collocation Points) x N( source points).
M=N
Direct Collocation
M>N
Least Square Fitting
MATLAB: b = COE \ RHS’
or b = pinv (COE) *RHS
Estimate of Solution Accuracy (chi-square error)
MD
MN
i 1
i 1
 2   (u  u~) 2   ( p  ~p ) 2
Test Problem 1
 2u
 2u
2



u 0
2
2
x
y
 =10;
u=1
u=1
u=1
u=1
Geometry and Boundary Condition
Dirichlet condition
Source point
Collocation point
Results for Test Problem 1
• 40 Collocation Points; 40 source points
located on a circle; Circle radius 1 to 10.
• Center concentration = 2.5294
• Average Flux = 872.6
• Chi-Square error 2e-06 to 2e-14.
• Condition number O(1.E+15)
• Convergence test was done with M = 160
and 20 to 160 source points.
Gradients along the bottom for test problem #1;
M = 160; source radius = 3.0; N = 160
10
Gradient
8
6
4
2
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
distance from the corner
0.8
0.9
1
Variations on the Theme:
Method of Complete Solutions
• Regular Solutions: Chan and Tanaka
Gi  I 0 ( ri )
• Example: Circular domain with dirichlet
• Completeness of the basis functions?
• Method of complete solutions
Gi  AK0 (ri )  BI 0 ( ri )
• Annular region: Both functions are needed.
Comparison of K0 and I0 basis functions
for test problem #1.
12
KO radius = 3.0; M =
160
8
I0 radius = 3.
6
I0; radius = 0.8
4
2
19
17
15
13
11
9
7
5
3
0
1
Gradients
10
Collocation points along the bottom
Test Problem 2
du
0
dn
du
0
dn
u =1
u =1
Results for Test Problem # 2
Gradient vs x-location at the bottom boundary for test
problem #2. Number of collocation points = 160
12
Gradient
10
8
Ns - 160; Radius = 3
6
Ns = 60
4
Ns = 20
2
0
0
0.5
1
distance from the left corner
1.5
Least square error as a function of source ponts for
test problem #2; Collocation points = 160
chi-squared error
0.02
0.015
0.01
0.005
0
-0.005
0
20
40
60
80
100
Number of source points
120
140
160
SOURCE POINT LOCATION FOR MIXED
BOUNDARY CONDITIONS
Singularity at the points where the boundary conditions
change. Special functions may be needed to capture
singularity.
HEURISTIC RULE: Points close to the Neumann Boundary
.
MFS for Laplace equation
Basis Functions
1, ln ri,
i = 1 to N
ri = distance from the source point, i.
N+1
Coefficients to be found by either direct collocation
or by the least square method
Laplace Equation: Example Test Problem (Golberg and Chen)
!
Least Square error (n=39)
Chi squared error
Test Problem: Oval of Cassini
3.00E-02
2.50E-02
2.00E-02
1.50E-02
1.00E-02
5.00E-03
0.00E+00
-5.00E-03 0
2
4
6
8
Source radius
10
12
Jewel Box
Botanical Garden
St.Louis
Missouri
Forest Park
Method of Regular Solutions: Trefftz Method
For Laplace Equation on a simple domain the following basis
functions hold.
Obtained by Separation of Variables.
b0 = 1
r
b1 = r cos 
b2 = r sin 
b3 = r2 cos 2
b4 = r2 sin 2
etc.
Solution for u is then
NB
u   bi Gi
i 1
NB = Number of basis function

r,  polar coordinate from
some interior point,
preferably the centroid of
the area
Trefftz Method: Illustration
Test Problem: Oval of Cassini (Golberg and Chen)
Boundary Condition: u = exp(x)cos(y)
Least Square fitting of B.C. was used .
Treftz Series converges rapidly with the following coefficients:
b0 = 1
b1 = 0.5
b3 = 0.1667
b5 = 0.0417
b6 = 0.0083
b7 = 0.0014
b9 = 0.0002
Coefficients are Fourier’s series of the analytic continuation of the
boundary conditions on an unit circle.
Advantage: Only a few terms are needed . No need to locate sources.
No source points needed
Trefftz Functions: Derivations
1  u
1  2u
(r ) 
0
2
2
r r r
r 
Separation of Variables in Polar Coordinates
u  R(r )T ( )
2 n 
T2 d 1
2 d T
d 2R
dR
2
r
r
 n2 R  0
dr
dr 2
Trignometric
Euler-Cauchy
T-Treftz Method: Multiple Domains
Complete Set of Basis Functions are therefore as follows;
1, ln r
rn cos n
rn sin n
n = 1,2,….etc.
1
1
sin n
cos n
n
n
r
r
n = 1,2,….etc.
Center (origin)
T-Treftz Method: Diffusion-Reaction Problem
1  u
1  2u
2
(r )  2


u0
2
r r r r 
Separation of Variables: u = R(r)T()
-Solutions: Trigonometric Functions
R-Solutions: Modified Bessel Functions
I0(r),
K0(r)
In(r)cos(n),
Kn(r)sin(n)
n = 1,2,3,….etc
If r = 0 belongs to the domain, the Kn functions are excluded
Diffusion-Reaction Problem
T-Method Results for theTest Problem #1:
Square geometry; Dirichlet on all sides.
Fitted Coefficients
0 I0(r)
2.5294
In(r)cos(n), = 0; n = 1, 2,3
In(r)cos(n), 6.099 for n = 4
n=8
-2.95
n = 12
-140.5
n = 16
-2.47e+05
n = 20
613.44
In(r) sin (n) = 0;
NO Kn
Chi Square error
400
300
200
100
0
-100 0
5
10
15
Number of Functions Used
20
25
T-Method Results for theTest Problem #1:
Square geometry; Dirichlet on all sides.
Concentration
100
80
60
40
20
0
0
0.1
0.2
0.3
0.4
y-Cooridnate
Concentration profiles along the y axis.
The solution matches well with MFS.
0.5
MFS for composite
Regions: Problem
Statement
k  T 0
I
2 I
k  T 0
II
2 II
Shaded Region has different
conductivity
Motivation: Transport properties
of a composite Material
Transport Properties of Composite Media
Simple Models
Series Model
1
1
1  1
 I 
k eff
k
k II
Parallel Model
keff  1k  (1  1 )k
Variational Methods gives tighter bounds
I
II
Representation of Boundary Conditions
A Simplified Two Region Case
nII
E
nI
C
 I  E  C
E = External Boundary with prescribed boundary condition
C = Common Boundary for Region II
Also II for this case (C = II)
Matching Boundary Conditions on C
T I  T II
I
II
dT
dT
II
kI

k
dn I
dn II
on C
Representation by MFS
Region I
Region II
Source point
Collocation point
NS 1
Solution
NS 2
T  a0   ai Gi
T  b0   bi Gi
dT I NS 1
  ai Qi
dn
i 1
dT II NS 2
  bi Qi
dn
i 1
I
i 1
II
i 1
Matrix Representation of Solution
Region I
I ~I 
T G a
C
dT I ~ I 
Q a
I
dn
E
I  
  E  C
Collocation points
Region II
 II ~ II 
T G b
dT II ~ II 
Q b
II
dn
Compact Matrix Representation
Region I

I
NOTE: TC  T
IE


T
G
 E
    IC a
G 
T
 C 

due to matching condition
Matching Conditions can be expressed as

IC 
II
G (a )  G (b )
IC 
or
G (a )  G II (b )  0
and
  k II
(Q )( a )   I
k
IC
 IIC 
Q (b )  0

Equation can be compacted as

NOTE: TE
 IE

G
0    TE 
 
 IE
II  a 
 G      0 
G
II
b
k
0
IE
IIC  
 Q
Q
 

kI


Dirichlet boundary, Modify for Neumann points
Test Problem
Circular Inclusion in a Square geometry of unit length
A
T  100
B
I
dT
0
dn
II
C
T 0
D
Flux along AB = Flux along CD
keff = Average Flux on AB/100
Porosity of Region II = ¶R2, R  0.5
Porosity of Region I = 1 - ¶R2
Effective conductivity
Effective thermal conductivity; k2/k1 = 5.
5
4.5
4
3.5
3
2.5
2
1.5
1
Effective thermal
conductivity
Lower bound
Upper bound
0
0.2
0.4
0.6
Porosity of region I
0.8
1
Temperature Profiles along the y-axis
120
Temperature
100
80
k2/k1 = 0.5
60
Series2
40
Series3
20
0
-0.6
-0.4
-0.2
0
y-coordinate
0.2
0.4
0.6
Summary and Conclusions
• Trefftz method and Fundamental solution method
provide comparable results; Trefftz method is
more general (no source points needed).
• Method of complete solutions needs to be
investigated; Initial results are promising; Optimal
placement of source points? Statistical methods?
• MFS useful and easy to implement for composite
regions. Need to explore this for other problems.