MATH 212 Advanced Calculus 2 for Electrical Engineering
Advanced Calculus 2 for Nanotechnology Engineering
NE 217
Finite-Element Methods
in One Dimension
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 One Dimension
Outline
This topic discusses an introduction to finite-element methods
– Background
– Justification
– Uniform test functions
• On equally spaced points
• On unequally spaced points
2
Finite-element Methods in One Dimension
Outcomes Based Learning Objectives
By the end of this laboratory, you will:
– The concept of a test function
– How we can approximate Laplace’s equation on unequally spaced
points
3
Finite-element Methods in One Dimension
Background
The most significant problem with finite differences:
– Seldom does nature line up on a grid
In 1d, finite differences form equally spaced points on a line:
We need to be able to move the points around so that:
– The points will match the geometry of the actual shape, and
– We can add more points in areas of interest
4
Finite-element Methods in One Dimension
Justification
In higher dimensions, examples from mechanical engineering
quickly spring to mind:
– The crumple zones of a car
User:MrMambo
5
Finite-element Methods in One Dimension
Justification
However, we may need this flexibility even in one dimension
– Consider heat diffusion along a rod containing three regions
• Copper
• A transition from copper to aluminium
• Aluminium
The points do not even line up with the transitions in the materials
6
Finite-element Methods in One Dimension
Justification
We could add more points:
Now we are solving a system of 41 equations and unknowns…
7
Finite-element Methods in One Dimension
Justification
What we really want:
– More points close to and in the transition
Such a system is much simpler than the previous idea…
8
Finite-element Methods in One Dimension
Test Functions
We will look at a solution that produces the same result as finite
differences; however, we will be able to generalize it
– The generalization can be extended to higher dimensions, too
9
Finite-element Methods in One Dimension
Test Functions
Starting with the equally spaced intervals, define a point a piecewise
constant test function for each interval:
1 x1  x  x3
0 otherwise
2  x   
10
Finite-element Methods in One Dimension
Test Functions
Starting with the equally spaced intervals, define a point a piecewise
constant test function for each interval:
1 x2  x  x4
0 otherwise
3  x   
11
Finite-element Methods in One Dimension
The Unknown Solution
We know the solution passes through unknown points with the
constraints:
u(a) = ua = u1
u(b) = ub = u2
12
Finite-element Methods in One Dimension
Piecewise Linear Approximations
We will approximation the solution through piecewise linear
functions:
u1  x   u1
x  x2
x  x1
 u2
x1  x2
x2  x1
13
Finite-element Methods in One Dimension
Piecewise Linear Approximations
We will approximation the solution through piecewise linear
functions:
u1  x   u1
x  x2
x  x1
 u2
x1  x2
x2  x1
1
At x1: u1  x1   u1
0
x1  x2
x x
 u2 1 1  u1
x1  x2
x2  x1
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Finite-element Methods in One Dimension
Piecewise Linear Approximations
We will approximation the solution through piecewise linear
functions:
u1  x   u1
x  x2
x  x1
 u2
x1  x2
x2  x1
0
At x2: u1  x2   u1
1
x2  x2
x x
 u2 2 1  u2
x1  x2
x2  x1
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Finite-element Methods in One Dimension
Piecewise Linear Approximations
We can write a similar piecewise-linear on each interval
u2  x   u 2
x  x3
x  x2
 u3
x2  x3
x3  x2
16
Finite-element Methods in One Dimension
The Target Equation
Next, we have Laplace’s equation in one dimension:
d2
u  x  0
2
dx
Define:
def
d2
V  x  2 u  x
dx
we have the equation
V  x  0
17
Finite-element Methods in One Dimension
The Integral
If V  x   0 , it follows that
  x V  x   0
for any test function (x) and therefore
b
   x V  x  dx  0
a
d2
In this case, however, we defined V  x  
u  x  and thus
2
dx
def
 d2

a   x   dx2 u  x   dx  0
b
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Finite-element Methods in One Dimension
The Integral
Consider the first test function 1(x) :
3
 d2

 d2

a  1  x   dx 2 u  x   dx  x  dx 2 u  x   dx  0
1
b
x
19
Finite-element Methods in One Dimension
The Integral
If, however, we approximate u(x) on that interval by the piecewise
constant function
x  x3
x  x2
x  x1
x  x2
u1  x   u1
 u2
u2  x   u 2
 u3
x1  x2
x2  x1
x2  x3
x3  x2
d2
d2
u x  2 u2  x   0
we already have that
2 1 
dx
dx
We get no additional information!
20
Finite-element Methods in One Dimension
The System of Linear Equations
Thus, we approximate the function u(x) by
x  x1
 x  x2
u

u
2
 1 x x
x2  x1
 1 2
x  x2
 x  x3
 u3
u2
x3  x2
u  x   u  x    x2  x3


x  xn
x  xn 1

u

u
n
 n 1 x  x
xn  xn 1
n 1
n

def
where we define uk  x   uk
x1  x  x2
x2  x  x3
xn 1  x  xn
x  xk 1
x  xk
 uk 1
on xk  x  xk 1
xk  xk 1
xk 1  xk
21
Finite-element Methods in One Dimension
Integration by Parts
However, take
 d2


x
u
x
a    dx2    dx  0
and performing integration by parts, we have
b
b

d

d
 d


x
u
x


x
u
x









 

 dx  0

 dx
 a a  dx
 dx


b
d
d
d
 d

  b   u  x     a   u  x       x   u  x   dx  0
dx
 dx
 dx
 dx

a
x b
x a
b
22
Finite-element Methods in One Dimension
Integration by Parts
First, substituting in the first test function:
b
 d2


x
u
x
a 2    d 2 x    dx  0
  x  x1     x  x3 
which yields
x3
1
1
d
d


d
 d


x
u
x


x
u
x


x
u
x
 
 
   dx  0
2   
 2  3
 2  1


dx
dx
dx


 dx

x1 
x  x3
x  x1
23
Finite-element Methods in One Dimension
Integration by Parts
Now, given
x
3
d
d
d

u
x

u
x


x

x


x

x
u
x











1
3 


 dx  0

 dx
 dx
 dx

x1
x  x3
x  x1
substitute our approximation:
x3
d
d
d

u
x

u
x


x

x


x

x
u
x











1
3 


 dx  0

 dx
 dx
 dx

x1
x  x3
x  x1
Remember that on the interval of interest,
 x  x2
u1
 u2

 u1  x  x1  x  x2  x1  x2
u  x  

u
x
x

x

x
 2   2
3
u x  x3  u
3
 2 x2  x3
x  x1
x2  x1
x1  x  x2
x  x2
x3  x2
x2  x  x3
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Finite-element Methods in One Dimension
Integration by Parts
Differentiating and substituting in the two approximations yields
u3  u2
u2  u1
2
2
0
x3  x2
x2  x1
but as the denominators are equal (equal width intervals), this
simplifies to
u3  2u2  u1  0
This is the exact same linear equation we got from Laplace’s
equation using finite differences
25
Finite-element Methods in One Dimension
The System of Linear Equations
If we were to repeat this at each interval, we would have:
 2

 1








1
2
1
1
2
1
1
2
1
1
2
1
1
2
1
  u2   ua 
u   0 

 3  
  u4   0 

  
u

0

 5  
  u6   0 

  
1   u7   0 
 

2 
 u8   ub 
9k
k 1
This has the regular solution uk 
ua 
ub
8
8
which is a straight line…
26
Finite-element Methods in One Dimension
Unequally Spaced Points
What this tells us, however, is that we have a method that can allow
us to use arbitrary sized intervals:
27
Finite-element Methods in One Dimension
The Test Functions
The most obvious generalization is to use similar test functions:
2  x 
3  x 
4  x 
5  x 
28
Finite-element Methods in One Dimension
The Equations
The most obvious generalization is to use similar test functions:
u3  u2 u2  u1

0
x3  x2 x2  x1
u4  u3 u3  u2

0
x4  x3 x3  x2
u5  u4 u4  u3

0
x5  x4 x4  x3
u6  u5 u5  u4

0
x6  x5 x5  x4
29
Finite-element Methods in One Dimension
The Equations
Assuming the width of the intervals are either h or 2h, we get:
u3  u2 u2  u1

0
x3  x2 x2  x1
u4  u3 u3  u2

0
x4  x3 x3  x2
u5  u4 u4  u3

0
x5  x4 x4  x3
u6  u5 u5  u4

0
x6  x5 x5  x4
u3  u2
h
u4  u3
h
u5  u4
2h
u6  u5
2h
u2  u1
2h
u u
 3 2
h
u u
 4 3
h
u u
 5 4
2h

0
0
0
0
30
Finite-element Methods in One Dimension
The Equations
Multiply by h:
u3  u2
h
u4  u3
h
u5  u4
2h
u6  u5
2h
u2  u1
2h
u u
 3 2
h
u u
 4 3
h
u u
 5 4
2h

0
u3  u2  12 u2  12 u1  0
0
0
u4  u3  u3  u2  0
1
2
1
2
u5  12 u4  u4  u3  0
u6  12 u5  12 u5  12 u4  0
0
31
Finite-element Methods in One Dimension
The Equations
This gives us the system of linear equations
 1.5 1
  u2   0.5u1 

u   0 
1

2
1


 3   

1 1.5 0.5   u4   0 


  
u

0.5
u
0.5
1

 5  
6
32
Finite-element Methods in One Dimension
The Equations
Solving this yields points on a straight line
 3u1  u6 


4
 u2   5u  3u 
1
6
 

8
 u3   
 u4   u1  u6 

  

2
 u5  
 u  3u 
6
 1


4

1
4
3
8
1
2
3
4
33
Finite-element Methods in One Dimension
Summary
We have looked an alternate approach to approximating solutions to
partial differential equations
–
–
–
–
Used test functions
Used integration by parts
The result gave us a system of linear equations
The solution was an approximation
• In this case, with Laplace’s equation in one dimension, it was exact
34
Finite-element Methods in One Dimension
What’s Next?
We will next consider
– Replacing Laplace’s equation with Poisson’s equation
– Use non-uniform test functions
• Tent functions
35
Finite-element Methods in One Dimension
References
[1] Glyn James, Advanced Modern Engineering Mathematics, 4th Ed.,
Prentice Hall, 2011, §§9.2-3.
36
Finite-element Methods in One Dimension
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
37