MATH 212 Advanced Calculus 2 for Electrical Engineering
Advanced Calculus 2 for Nanotechnology Engineering
NE 217
Finite-Element Methods
in Two Dimensions
Douglas Wilhelm Harder
Department of Electrical and Computer Engineering
University of Waterloo
Waterloo, Ontario, Canada
Copyright © 2011 by Douglas Wilhelm Harder. All rights reserved.
Finite-element Methods in Two Dimensions
Outline
This topic discusses an introduction to finite-element methods in two
dimensions
–
–
–
–
–
–
What we did in one dimension
Dividing up a 2-D region
Defining planar test functions on those regions
Converting the test functions to systems of linear equations
Solving the system
Examples
2
Finite-element Methods in Two Dimensions
Outcomes Based Learning Objectives
By the end of this laboratory, you will:
– You will understand dividing regions into triangular sub-regions
(tessellations)
– Understand the mathematical theory for performing finite-element
methods in two dimensions
3
Finite-element Methods in Two Dimensions
Integration-by-Parts in Higher Dimensions
Before we start, we need a higher-dimensional version of integration
by parts:
– Using the properties of the gradient:
   
u  v     uv   v  u 
 u  v  dxdy    uv  dxdy   v  u  dxdy
  uv   u v  v u
R
R
R
– Now use a corollary of Stokes’ theorem:
 u  v  dxdy   uvds   v  u  dxdy
R
R
R
Ref: http://en.wikipedia.org/wiki/Vector_calculus_identities
4
Finite-element Methods in Two Dimensions
Integration-by-Parts in Higher Dimensions
This variation has the product of two scalar-valued functions
replaced with a product of scalar- and vector-valued functions
– Using the properties of the gradient:
 
 
u    v      uv    u   v
 u    v  dxdy      uv  dxdy   u   vdxdy
   uv   u  v    v u
R
R
R
– Now use a corollary of Stokes’ theorem:
 u  v  dxdy   uv  ds   v  u  dxdy
R
R
R
5
Finite-element Methods in Two Dimensions
Poisson’s Equation in Two Dimensions
At this point, we have generalized the finite-difference method to
allow arbitrarily spaced points in one dimension to solve systems
such as Poisson’s equation:
  x, y 
2
 u  x, y  
0
The ultimate goal is to do this in 2 and 3 dimensions
– In order to achieve this, we need one further idea
6
Finite-element Methods in Two Dimensions
Extending Test Functions
In the previous
– Use triangles to divide up the region instead of intervals
• Tessellations
– Approximate the solution using interpolating polynomials
Intervals
Tessellation
7
Finite-element Methods in Two Dimensions
Partitioning a Region
Given a region R, a partial differential equation defines conditions on
the boundary ∂R and we attempt to find an approximation of the
solution in the region
8
Finite-element Methods in Two Dimensions
Partitioning a Region
We have seen how using finite differences, we:
– Create a uniform grid
– Assign points to either the boundary or the interior of the region
9
Finite-element Methods in Two Dimensions
Partitioning a Region
Unfortunately, uniform grids make a poor approximation of most
real-world situations
10
Finite-element Methods in Two Dimensions
Partitioning a Region
It would be much more desirable to allow the user to select the
placement of points where appropriate:
– Non-linear boundaries can be fitted
– Points can be concentrated at regions of interest
Let N be the number interior points
– in this case, 22
11
Finite-element Methods in Two Dimensions
Partitioning a Region
In one dimension, the regions between points were intervals
– Similarly, we must divide the region into a number of sub-regions
12
Finite-element Methods in Two Dimensions
Partitioning a Region
Our goal will be to find approximations of the solution on the interior
of the region R
13
Finite-element Methods in Two Dimensions
Partitioning a Region
We will divide the region into triangular elements:
– This forms a connected graph
– The collection of all triangles is a tessellation
– Tessella is the Latin word for a small piece of a mosaic
14
Finite-element Methods in Two Dimensions
Partitioning a Region
From wikipedia: a tessellation of a 2D magnetostatic configuration
User: Zureks
15
Finite-element Methods in Two Dimensions
Defining Test Functions
In one dimension, for each interior
point, xk, we used the two adjacent
intervals upon which we defined
the test functions
f6(x)
– The support of the test function
f6(x) is the interval [x5, x7]
f7(x)
16
Finite-element Methods in Two Dimensions
Defining Test Functions
In two dimensions, given the point xk, we will define a test function
on the region comprised of triangles adjacent to xk
– Call that region Rk
17
Finite-element Methods in Two Dimensions
Defining Test Functions
For ease of reference, given a point xk, we will number the adjacent
points, in this case:
xk,1 xk,2 xk,3 xk,4 xk,5 xk,6
18
Finite-element Methods in Two Dimensions
Defining Test Functions
Similarly, we will label each triangles with Rk, in this case:
Rk,1 Rk,2 Rk,3 Rk,4 Rk,5 Rk,6
19
Finite-element Methods in Two Dimensions
Defining Test Functions
On each region Rk, we will define a piecewise planar test function
– The test function is fk(x, y)
– It is defined to be zero outside Rk
– It is also useful to have it zero on the boundary ∂Rk
20
Finite-element Methods in Two Dimensions
Defining Test Functions
The formula for each plane fk, ℓ(x, y) defined on Rk, ℓ is a plane of the
form ax + by + g
– To find it, we use linear algebra
21
Finite-element Methods in Two Dimensions
Finding Interpolating Planes
Given three points in the plane
(x1, y1), (x2, y2), (x3, y3)
if we want to find the interpolating plane ax + by + g that passes
through these points z1, z2, and z3, respectively, we must solve the
system of linear equations
a x1  b y1  g  z1
a x2  b y2  g  z2
a x3  b y3  g  z3
22
Finite-element Methods in Two Dimensions
Finding Interpolating Planes
Rewritten using matrices and vectors, we must solve:
 x1

 x2
x
 3
y1 1  a   z1 
   
y2 1   b    z 2 
y3 1  g   z3 
23
Finite-element Methods in Two Dimensions
Finding Interpolating Planes
Example, given the points in the plane
(1.3, 5.4), (2.9, 7.0) and (6.5, 4.9)
24
Finite-element Methods in Two Dimensions
Finding Interpolating Planes
Example, given the points in the plane
(1.3, 5.4), (2.9, 7.0) and (6.5, 4.9)
we want to find the polynomial ax + by + g that passes through the
points 2.8, 1.6 and 3.9, respectively
25
Finite-element Methods in Two Dimensions
Finding Interpolating Planes
Example, given the points in the plane
(1.3, 5.4), (2.9, 7.0) and (6.5, 4.9)
we want to find the polynomial ax + by + g that passes through the
points 2.8, 1.6 and 3.9, respectively
We must therefore solve
 1.3 5.4 1 a   2.8 

   
2.9
7.0
1

 b    1.6 
 6.5 4.9 1 g   3.9 

   
>> [1.3 5.4 1; 2.9 7.0 1; 6.5 4.9 1] \ [2.8 1.6 3.9]'
ans =
0.1272
-0.8772
7.3715
26
Finite-element Methods in Two Dimensions
Finding Interpolating Planes
From
>> [1.3 5.4 1; 2.9 7.0 1; 6.5 4.9 1] \ [2.8 1.6 3.9]'
ans =
0.1272
-0.8772
7.3715
we have that the interpolating plane is
0.1272x – 0.8772y + 7.3715
27
Finite-element Methods in Two Dimensions
Using the Test Function
Now, going back to the problem:
– We have the test function fk(x, y) and we have a function V(x, y) that is
known to satisfy the formula
V(x, y) = 0
As we did in one dimension, it must therefore be true that
 f  x, y V  x, y  dxdy  0
k
Rk
28
Finite-element Methods in Two Dimensions
Using the Test Function
Again, let’s consider Poisson’s equation:
  x, y 
 u  x, y  
0
2
We defined
def
V  x, y   2u  x, y  
and thus our integral is
  x, y 
0
  x, y  
 2
R fk  x, y    u  x, y    0  dxdy  0
k
29
Finite-element Methods in Two Dimensions
Using the Test Function
Expanding the integral, we get
  x, y 
R fk  x, y   u  x, y  dxdy  R fk  x, y   0 dxdy
2
k
k
where the right-hand side is reasonably easy to calculate
The left-hand side, however, requires further
calculus
30
Finite-element Methods in Two Dimensions
Using the Test Function
Using the 2-dimensional equivalent of integration-by-parts, we get:
2
f
x
,
y

u  x, y  dxdy 


 k
Rk
 f  x, y  u  x, y   ds   f  x, y  u  x, y  dxdy
k
Rk
k
Rk
31
Finite-element Methods in Two Dimensions
Using the Test Function
Using the 2-dimensional equivalent of the fundamental theorem of
calculus:
0
 f  x, y   u  x, y  dxdy   f  x, y  u  x, y   ds   f  x, y  u  x, y  dxdy
2
k
Rk
k
Rk
k
Rk
However, recall that we specifically chose a test function that is zero
on the boundary ∂Rk
– Therefore, the first term disappears!
32
Finite-element Methods in Two Dimensions
Using the Test Function
Therefore, we have the easier integral:
2
f
x
,
y

u  x, y  dxdy    fk  x, y  u  x, y  dxdy


 k
Rk
Rk
and therefore we have
  x, y 
 fk  x, y  u  x, y  dxdy   fk  x, y 
dxdy
0
R
R
k
k
33
Finite-element Methods in Two Dimensions
Using the Test Function
We can calculate the right-hand side: both are known
  x, y 
 fk  x, y  u  x, y  dxdy   fk  x, y 
dxdy
0
R
R
k
k
The left-hand side has the unknown function u(x, y)…
34
Finite-element Methods in Two Dimensions
Using the Test Function
We can calculate the right-hand side: both are known
  x, y 
 fk  x, y  u  x, y  dxdy   fk  x, y 
dxdy
0
R
R
k
k
First, rewrite the left-hand side as integrals over each sub-region
nk
  fk ,  x, y  u  x, y  dxdy 
1 Rk ,
  x, y 
R fk  x, y   0 dxdy
k
Let nk be the number of triangles touching xk
– in this case, 6
35
Finite-element Methods in Two Dimensions
Finding the Test Function Gradient
The test function defined on each sub-region is planar, that is
fk ,  x, y   a x  b y  g
thus we can easily calculate:
 

f
x
,
y


 x k ,
 a 
 
fk ,  x, y   
  f  x, y    b 
 y k ,



nk
  fk ,  x, y  u  x, y  dxdy 
1 Rk ,
  x, y 
R fk  x, y   0 dxdy
k
36
Finite-element Methods in Two Dimensions
Finding the Test Function Gradient
a 
To find the gradient of this plane, we must find the vector  
b 
x k :  x1 y1 1  a   1 

   
x k ,1 :  x2 y2 1  b    0 
x k ,2 :  x3 y3 1  g   0 
nk
  fk ,  x, y  u  x, y  dxdy 
1 Rk ,
  x, y 
R fk  x, y   0 dxdy
k
37
Finite-element Methods in Two Dimensions
Finding the Test Function Gradient
Solving for a is as follows:
 1 y1 1


det  0 y2 1
 0 y 1
y2  y3
3

 
a
 x1 y1 1
 x1 y1 1




det  x2 y2 1 det  x2 y2 1
 x y 1
 x y 1
3
3
 3

 3

6
  fk ,  x, y  u  x, y  dxdy 
1 Rk ,
  x, y 
R fk  x, y   0 dxdy
k
38
Finite-element Methods in Two Dimensions
Finding the Test Function Gradient
Solving for b is as follows:
 x1 1 1


det  x2 0 1
 x 0 1
x3  x2
 3
 
b
 x1 y1 1
 x1 y1 1




det  x2 y2 1 det  x2 y2 1
 x y 1
 x y 1
3
3
 3

 3

6
  fk ,  x, y  u  x, y  dxdy 
1 Rk ,
  x, y 
R fk  x, y   0 dxdy
k
39
Finite-element Methods in Two Dimensions
Approximating the Solution Gradient
The next step is to approximate the unknown
u  x, y 
6
  fk ,  x, y  u  x, y  dxdy 
1 Rk ,
  x, y 
R fk  x, y   0 dxdy
k
40
Finite-element Methods in Two Dimensions
Approximating the Solution Gradient
The actual unknown function u(x, y) an unknown and likely nonlinear function
– You don’t get paid $100 000/a to find a linear function…
41
Finite-element Methods in Two Dimensions
Approximating the Solution Gradient
Recall from 1-D finite element methods:
– We approximated the solution with an unknown piecewise
linear function
42
Finite-element Methods in Two Dimensions
Approximating the Solution Gradient
We are attempting to approximate u(x, y) at a number of interior
points where the solution is defined on the border
43
Finite-element Methods in Two Dimensions
Approximating the Solution Gradient
We can do the same as we did in one dimension:
– Approximate the solution by a piecewise planer approximation
44
Finite-element Methods in Two Dimensions
Approximating the Solution Gradient
As before, we can find the interpolating polynomial:
x k :  x1

x k ,1 :  x2
x k ,2 :  x3
y1 1  a   uk 

  
y2 1  b    uk ,1 
y3 1  c   uk ,2 
6
  fk ,  x, y  u  x, y  dxdy 
1 Rk ,
  x, y 
R fk  x, y   0 dxdy
k
45
Finite-element Methods in Two Dimensions
Approximating the Solution Gradient
a
To find the gradient of this triangle, we must find the vector  
b
x k :  x1 y1 1  a   uk 


  
x k ,1 :  x2 y2 1  b    uk ,1 
x k ,2 :  x3 y3 1  c   uk ,2 
6
  fk ,  x, y  u  x, y  dxdy 
1 Rk ,
  x, y 
R fk  x, y   0 dxdy
k
46
Finite-element Methods in Two Dimensions
Approximating the Solution Gradient
Solving for a is as follows:
 uk y1 1


det  uk ,1 y2 1
u
y3 1 uk y2  y1uk ,2  uk ,1 y3  uk y3  y1uk ,1  y2uk ,2
k ,2

a

 x1 y1 1
 x1 y1 1




det  x2 y2 1
det  x2 y2 1
 x y 1
 x y 1
3
3
 3

 3

6
  fk ,  x, y  u  x, y  dxdy 
1 Rk ,
  x, y 
R fk  x, y   0 dxdy
k
47
Finite-element Methods in Two Dimensions
Approximating the Solution Gradient
Solving for b is as follows:
 x1 uk 1


det  x2 uk ,1 1
x u
1 x1uk ,1  x3uk  x2uk ,2  x1uk ,2  uk x2  uk ,1 x3
3
k ,2

b

 x1 y1 1
 x1 y1 1




det  x2 y2 1
det  x2 y2 1
 x y 1
 x y 1
3
3
 3

 3

6
  fk ,  x, y  u  x, y  dxdy 
1 Rk ,
  x, y 
R fk  x, y   0 dxdy
k
48
Finite-element Methods in Two Dimensions
Approximating the Double Integral
Thus, we can find both gradients:
 fk ,  x, y  u  x, y  dxdy 
Rk ,1
a
 a , b    b  dxdy
 
Rk ,1
If you look back, all values are either known
– The positions of the points xk, xk,1 and xk,2:
x1, x2, x3, y1, y2 , y3
or unknown constants uk, uk,1 and uk,2 :
– Take them out of the integral
 a a  b b   1dxdy
Rk ,1
49
Finite-element Methods in Two Dimensions
Approximating the Double Integral
Finally, we note that the right-hand integral
 f  x, y  u  x, y  dxdy  a a  b b   1dxdy
k,
Rk ,1
Rk ,1
is just the area of the region Rk,1, call it Ak,1:
 x1
1

Ak ,1  det  x2
2
x
 3
y1 1

y2 1 
y3 1
50
Finite-element Methods in Two Dimensions
Approximating the Double Integral
Thus, for each interior point xk, we must calculate:
nk
nk
1 Rk ,
1 Rk ,
  fk ,  x, y  u  x, y  dxdy    fk ,  x, y 
  x, y 
dxdy
0
For each xk, there are nk adjacent triangles Rk,ℓ, and each of these is
associated given three points, call them
x1 = (x1, y1), x2 = (x2, y2), x3 = (x3, y3)
and the unknowns associated with, call them u1, u2 and u3:
– Each interior point ends up with nk linear equations that are summed
together forming a single linear equation
51
Finite-element Methods in Two Dimensions
Approximating the Double Integral
  x, y 
  is a constant, the right-hand integral can be
If the ratio
0
simplified further:
  x, y 
1
f
x
,
y
dxdy

 Ak
1  k ,   
3
0
R
nk
k,
where Ak is the area of the region Rk
52
Finite-element Methods in Two Dimensions
Finding the points
This will produce N linear equations and unknowns
– We can solve this system of linear equations for the unknown values
u1, u2, u3, …, uN
53
Finite-element Methods in Two Dimensions
Code For 2-D Finite Elements
A set of functions are available on the web site that set a finite
element system and solve for the interior values
– It assumes the right-hand side of Poisson’s equation is constant
2u  x, y   
fe_system( rho, pts )
Create a data structure for the above equation by
passing in a set of points (interior and boundary) with
values for the boundary points and -Inf for the
interior points
fe_update( obj, pt, alist )
Indicate the tessellation around an interior point by
passing a list of neighbouring points
fe_solve( obj )
Having called fe_update for each interior point, now
solve the system of equations returning the initial
argument pts with all values -Inf replaced with
values
fe_coeffs( x1, x2, x3 )
A helper function
54
Finite-element Methods in Two Dimensions
Example 1
Consider the following region and we are trying to approximate
Laplace’s equation ( = 0) with the following region and boundary
values:
55
Finite-element Methods in Two Dimensions
Example 1
We can define the system as follows:
pts1 = [ 0 -1 -1
-1 0 1
0 0 -Inf
1 0 1
-1 1 0
0 1 -Inf
0 2 -4]';
obj1 = fe_system( 0,
% 1
% 2
% 3
% 4
% 5
% 6
% 7
pts1 );
56
Finite-element Methods in Two Dimensions
Example 1
We define the tessellation around the point 3 by listing the adjacent
points in order (counter-clockwise):
obj1 = fe_update( obj1, 3, [4 6 2 1] );
Note, Matlab does not have proper objects, thus we
simulate object-oriented programming by replacing
obj1.fe_update( 3, [4, 6, 2, 1] )
with
obj1 = fe_update( obj1, 3, [4, 6, 2, 1] )
57
Finite-element Methods in Two Dimensions
Example 1
Similarly, we define the tessellation around the point 6 by listing the
adjacent points in order (counter-clockwise):
obj1 = fe_update( obj1, 6, [4 7 5 2 3] );
58
Finite-element Methods in Two Dimensions
Example 1
Having defined the system, we solve the system and plot the result:
u1 = fe_solve( obj1 )
u1 =
0
-1.0000 0
-1.0000 0
0
-1.0000 1.0000 -0.1053
1.0000 -1.0000 0
0
0
1.0000 1.0000 2.0000
1.0000 0
-1.4211 -4.0000
plot3( u1(1,:), u1(2,:), u1(3,:), 'ko' )
The plot was augmented in CorelDRAW!
59
Finite-element Methods in Two Dimensions
Example 2
Consider the following region and we are trying to approximate
Poisson’s equation with  = 4, the following region and boundary
values:
60
Finite-element Methods in Two Dimensions
Example 2
We can define the system as follows:
pts2 = [-2 -2 8
% 1
1 -2 5
% 2
-3 -1 10
% 3
0 -1 -Inf % 4
3 -1 10
% 5
-1 0 -Inf % 6
2 0 -Inf % 7
-3 1 10
% 8
1 1 –Inf % 9
3 1 10
% 10
-2 2 8
% 11
0 2 -Inf % 12
1 3 10]'; % 13
obj2 = fe_system( 4, pts2 );
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Finite-element Methods in Two Dimensions
Example 2
We then define the various tessellations:
obj2
obj2
obj2
obj2
obj2
=
=
=
=
=
fe_update(
fe_update(
fe_update(
fe_update(
fe_update(
obj2, 4,
obj2, 6,
obj2, 7,
obj2, 9,
obj2, 12,
[1
[1
[2
[4
[6
2 7 9 6]
);
4 9 12 11 8 3] );
5 10 9 4]
);
7 10 13 12 6]
);
9 13 11]
);
62
Finite-element Methods in Two Dimensions
Example 2
Finally, we solve the system:
u2 = fe_solve( obj2 )
u =
Columns 1 through 8
-2.0000 1.0000 -3.0000 0
3.0000 -1.0000 2.0000 -3.0000
-2.0000 -2.0000 -1.0000 -1.0000 -1.0000 0
0
1.0000
8.0000 5.0000 10.0000 1.8256 10.0000 1.2883 4.6679 10.0000
Columns 9 through 13
1.0000 3.0000 -2.0000
1.0000 1.0000 2.0000
2.9111 10.0000 8.0000
0
1.0000
2.0000 3.0000
5.5663 10.0000
plot3( u2(1,:), u2(2,:), u2(3,:), 'o' )
63
Finite-element Methods in Two Dimensions
Example 2
Here we see the boundary points and the actual solution (light blue)
and the approximations (in red)
64
Finite-element Methods in Two Dimensions
Summary
This topic discusses an introduction to finite-element methods in two
dimensions
– We generalized the test functions in one dimension
– Created a tessellation of the region R
– Defined planar test functions
• Found the gradient of the planes making the test functions
– Approximated the solution with piecewise planar sections
– Created a linear equation
• This is done for each interior point
– The resulting system of linear equations may be solved
– Two examples…
65
Finite-element Methods in Two Dimensions
What’s Next?
The next step is to go to three dimensions
– Divide the region into tetrahedral regions
– 3-dimensional tessellations
Robert Webb's Great Stella software
http://www.software3d.com/Stella.html
Tessellation
User: Tom Ruen
66
Finite-element Methods in Two Dimensions
References
[1] Glyn James, Advanced Modern Engineering Mathematics, 4th Ed.,
Prentice Hall, 2011, §§9.2-3.
67
Finite-element Methods in Two Dimensions
Usage Notes
•
•
These slides are made publicly available on the web for anyone to use
If you choose to use them, or a part thereof, for a course at another
institution, I ask only three things:
– that you inform me that you are using the slides,
– that you acknowledge my work, and
– that you alert me of any mistakes which I made or changes which you make, and
allow me the option of incorporating such changes (with an acknowledgment) in
my set of slides
Sincerely,
Douglas Wilhelm Harder, MMath
dwharder@alumni.uwaterloo.ca
68