MATH 212 Advanced Calculus 2 for Electrical Engineering
Advanced Calculus 2 for Nanotechnology Engineering
NE 217
1-D Finite-Element Methods
with Poisson’s Equation
Douglas Wilhelm Harder
Department of Electrical and Computer Engineering
University of Waterloo
Waterloo, Ontario, Canada
Copyright © 2011 by Douglas Wilhelm Harder. All rights reserved.
1-D Finite-element Methods with Poisson’s Equation
Outline
This topic discusses an introduction to finite-element methods
– Review of Poisson’s equation
– Defining a new kernel V(x)
– Approximate solutions using uniform test functions
2
1-D Finite-element Methods with Poisson’s Equation
Outcomes Based Learning Objectives
By the end of this laboratory, you will:
– Understand how to approximate the heat-conduction/diffusion and wave
equations in two and three dimensions
– You will understand the differences between insulated and Dirichlet
boundary conditions
3
1-D Finite-element Methods with Poisson’s Equation
The Target Equation
Recall the first of Maxwell’s equations (Gauss’s equation):
E 

0
If we are attempting to solve for the underlying potential function,
under the assumption that it is a conservative field, we have
 2u 

0
This is in the form of Poisson’s equation
4
1-D Finite-element Methods with Poisson’s Equation
The Target Equation
In one dimension, this simplifies to:
  x
d2
u  x 
2
dx
0
Define:
  x
d2
V  x  2 u  x 
dx
0
def
and thus we are solving for
V  x  0
5
1-D Finite-element Methods with Poisson’s Equation
The Integral
If V  x   0 , it follows that
f  x V  x   0
for any test function f(x) and therefore
b
 f  x V  x  dx  0
a
Substituting the alterative definition of V(x) into this equation, we get
  x 
 d2
f
x
u
x

a    dx2    0  dx  0
b
6
1-D Finite-element Methods with Poisson’s Equation
The Integral
Consider the first test function f1(x) :
x3
b
  x 
 d2
 d2

a f 1  x   dx 2 u  x   dx  x  dx 2 u  x    0  dx  0
1
7
1-D Finite-element Methods with Poisson’s Equation
Integration by Parts
Again, take
  x 
 d2
f
x
u
x

a    dx2    0  dx  0
but before we apply integration by parts, expand the integral:
b
b
  x
 d2

f
x
u
x
dx

f
x
a    dx2    a    0 dx  0
b
Everything in the second integral is known: bring it to the right:
b
  x
 d2

a f  x   dx2 u  x   dx  a f  x   0 dx
b
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1-D Finite-element Methods with Poisson’s Equation
Integration by Parts
The left-hand integral is no different from before,
b
b
  x
 d2

f
x
u
x
dx

f
x
a    dx2    a    0 dx
and performing integration by parts, we have
  x

d

d
 d

f
x
u
x

f
x
u
x
dx

f
x
dx











 




0
 dx
 a a  dx
 dx


a
b
b
b
  x
d
d
d
 d

f  b   u  x   f  a   u  x     f  x   u  x   dx   f  x 
dx
dx
dx
dx
dx





0
a
a
x b
x a
b
b
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1-D Finite-element Methods with Poisson’s Equation
Integration by Parts
First, substituting in the first test function:
b
b
  x
 d2

f
x
u
x
dx

f
x
a 2    d 2 x    a 2    0 dx
  x  x1     x  x3 
which yields
x3
x3
1
1
1 x
  dx
d
d


d
 d

f
x
u
x

f
x
u
x

f
x
u
x
dx

f
x
 
 
 
2   
2 
 2  3
 2  1



dx
dx
dx
dx
0




x1 
x1
x  x3
x  x1
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1-D Finite-element Methods with Poisson’s Equation
The System of Linear Equations
Again, recall that we approximated the solution by unknown
piecewise linear functions:
x  x1
 x  x2
u

u
2
 1 x x
x2  x1
 1 2
x  x2
 x  x3
 u3
u2
x3  x2
u  x    x2  x3


x  xn
x  xn 1

u

u
n
 n 1 x  x
xn  xn 1
n 1
n

x1  x  x2
x2  x  x3
xn 1  x  xn
x  xk 1
x  xk
 uk 1
where we define uk  x   uk
on xk  x  xk 1
xk  xk 1
xk 1  xk
def
11
1-D Finite-element Methods with Poisson’s Equation
Unequally Spaced Points
Recall that we approximated the left-hand integral by substituting the
piecewise linear functions to get:
b
  x

 d2
a fk  x   d 2 x u  x   dx  a fk  x   0 dx
b
k 1
  x

 d2

dx
x
u
x  d 2 x    x  0 dx
k 1
k 1
xk 1
x
uk  uk 1 k 1   x 
uk 1  uk
dx
 
2
2
xk  xk 1 xk 1  0
xk 1  xk
x



1
1
1

2

u
2

 k 1
x

x
k 1 
 xk  xk 1 xk 1  xk
 k


1
2

u

 k 1
 xk 1  xk

k 1
  x

dx
 uk 1  
0

xk 1
x
12
1-D Finite-element Methods with Poisson’s Equation
Integration by Parts
That is, our linear equations are:





1
1
1
1
u


u


 k 1 
 k 1 
x

x
x

x
x

x
k 1 
k 1
k 1
k 
 k
 k
 xk 1  xk

1 k 1   x 
dx
 uk 1  
2 xk 1  0

x
This time, let’s take an actual example
13
1-D Finite-element Methods with Poisson’s Equation
Integration by Parts
Consider the equation
d2
 u  x   2 u  x   sin 4  x 
dx
with the boundary conditions
2
u(0) = 0
u(1) = 0
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1-D Finite-element Methods with Poisson’s Equation
Integration by Parts
The solution to the boundary value problem
d2
 u  x   2 u  x   sin 4  x 
dx
2
u(0) = 0
u(1) = 0
is the exact function
u  x 
1
1

2
4
3
x
x

1

4

5cos

x

cos

x










16 
2

15
1-D Finite-element Methods with Poisson’s Equation
Integration by Parts
We know in one dimension if the right-hand side is close to zero, the
solution is a straight line
d2
2
 u  x   2 u  x   sin 4  x 
dx
Thus, choose 9 interior points with a focus on the centre:
>> x_uneq = [0 0.16 0.25 0.33 0.41 0.5 0.59 0.67 0.75 0.84 1];
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1-D Finite-element Methods with Poisson’s Equation
Integration by Parts
We will compare this approximation with the approximation found
using 9 equally spaced interior points
– The finite difference approximation
>> x_eq = 0:0.1:1;
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1-D Finite-element Methods with Poisson’s Equation
The Test Functions
The function to find the approximations is straight-forward:
function [ v ] = uniform1d( x, uab, rho )
n = length( x ) - 2;
idx = 1./diff(x);
M = diag( -(idx( 1:end - 1 ) + idx( 2: end )) ) + ...
diag( idx( 2:end - 1 ), -1 ) + ...
diag( idx( 2:end - 1 ), +1 );
b
b = zeros ( n, 1 );
int( rho, a, b ) approximates
   x dx
for k = 1:n
b(k) = 0.5*int( rho, x(k), x(k + 2) );
end
a
b(1)
= b(1)
- idx(1)*uab(1);
b(end) = b(end) - idx(end)*uab(end);
v = [uab(1); M \ b; uab(2)];
end
18
1-D Finite-element Methods with Poisson’s Equation
The Test Functions
The right-hand function of
2
d
2u  x   2 u  x   sin 4  x 
dx
is also straight-forward:
function [ u ] = rho( x )
u = sin( pi*x ).^4;
end
19
1-D Finite-element Methods with Poisson’s Equation
The Equations
Thus, we can find our two approximations:
>>
>>
>>
>>
>>
x_eq = (0:0.1:1)';
plot( x_eq, uniform1d( x_eq, [0, 0], @rho ), 'b+' );
hold on
x_uneq = [0 0.16 0.25 0.33 0.41 0.5 0.59 0.67 0.75 0.84 1]';
plot( x_uneq, uniform1d( x_uneq, [0, 0], @rho ), 'rx' );
20
1-D Finite-element Methods with Poisson’s Equation
The Equations
It is difficult to see which is the better function, therefore create a
function storing the actual solution (as found in Maple):
function u = u(x)
u = -1/16*(
(cos(pi*x).^4 - 5*cos(pi*x).^2 + 4)/pi^2 +
3*x.*(x - 1)
);
end
...
...
...
21
1-D Finite-element Methods with Poisson’s Equation
The Equations
Instead, plotting the errors:
>>
>>
>>
>>
>>
x_eq = 0:0.1:1;
plot( x_eq, uniform1d( x_eq, [0, 0], @rho ) - u(x_eq), 'b+' );
hold on
xnu = [0 0.16 0.25 0.33 0.41 0.5 0.59 0.67 0.75 0.84 1];
plot( x_uneq, uniform1d( x_uneq, [0, 0], @rho ) - u(x_uneq), 'rx' );
22
1-D Finite-element Methods with Poisson’s Equation
Integration by Parts
Understanding that the right-hand side has a greater influence in the
centre,
d2
2
 u  x   2 u  x   sin 4  x 
dx
appropriately changing the sample points yielded a significantly
better approximation
23
1-D Finite-element Methods with Poisson’s Equation
Summary
In this topic, we have generalized Laplace’s equation to Poisson’s
equation
– Used the same uniform test functions
– We looked at a problem for which there is an exact solution
• Changing the points allowed us to get better approximations
24
1-D Finite-element Methods with Poisson’s Equation
What’s Next?
The impulse function (the derivative of a step function) is difficult to
deal with…
We will next consider test functions that avoid this…
– The test functions will be tents
– This generalizes to higher dimensions
25
1-D Finite-element Methods with Poisson’s Equation
References
[1] Glyn James, Advanced Modern Engineering Mathematics, 4th Ed.,
Prentice Hall, 2011, §§9.2-3.
26
1-D Finite-element Methods with Poisson’s Equation
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
27