1-d finite-element methods with poisson’s equation

MATH 212NE 217

Douglas Wilhelm Harder

Department of Electrical and Computer Engineering

University of Waterloo

Waterloo, Ontario, Canada

Copyright © 2011 by Douglas Wilhelm Harder. All rights reserved.

Advanced Calculus 2 for Electrical EngineeringAdvanced Calculus 2 for Nanotechnology Engineering

1-D Finite-Element Methodswith Poisson’s Equation

1-D Finite-element Methods with Poisson’s Equation

2

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


3

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


4

The Target Equation

Recall the first of Maxwell’s equations (Gauss’s equation):

If we are attempting to solve for the underlying potential function, under the assumption that it is a conservative field, we have

This is in the form of Poisson’s equation

0

E

2

0

u


5

The Target Equation

In one dimension, this simplifies to:

Define:

and thus we are solving for

2

20

xdu x

dx

2

20

def xdV x u x

dx

0V x


6

The Integral

If , it follows that

for any test function (x) and therefore

Substituting the alterative definition of V(x) into this equation, we get

0x V x

0V x

0b

a

x V x dx

2

20

0b

a

xdx u x dxdx


7

The Integral

Consider the first test function 1(x) :

3

1

2 2

2 210

0xb

a x

xd dx u x dx u x dxdx dx


8

Integration by Parts

Again, take

but before we apply integration by parts, expand the integral:

Everything in the second integral is known: bring it to the right:

2

20

0b

a

xdx u x dxdx

2

20

0b b

a a

xdx u x dx x dxdx

2

20

b b

a a

xdx u x dx x dxdx


9


The left-hand integral is no different from before,

and performing integration by parts, we have

0

0

b b b

a aa

b b

a ax b x a

xd d dx u x x u x dx x dxdx dx dx

xd d d db u x a u x x u x dx x dx

dx dx dx dx

2

20

b b

a a

xdx u x dx x dxdx


10


First, substituting in the first test function:

which yields

3 3

1 13 1

2 3 2 1 2 20

x x

x xx x x x

xd d d dx u x x u x x u x dx x dxdx dx dx dx

2

2 220

b b

a a

xdx u x dx x dxd x

1 1 1

1 3x x x x


11

The System of Linear Equations

Again, recall that we approximated the solution by unknown piecewise linear functions:

where we define on

2 11 2 1 2

1 2 2 1

3 22 3 2 3

2 3 3 2

11 1

1 1

n nn n n n

n n n n

x x x xu u x x xx x x x

x x x xu u x x xx x x xu x

x x x xu u x x x

x x x x

11

1 1

defk k

k k kk k k k

x x x xu x u u

x x x x

1k kx x x


12

Unequally Spaced Points

Recall that we approximated the left-hand integral by substituting the piecewise linear functions to get:

1 1

1 1

1

1

1

2

20

2

20

1 1

1 1 0

1 1 11 1 1 1 0

2 2

1 1 1 12 2 2

k k

k k

k

k

k

b b

k k

a a

x x

x x

x

k k k k

k k k k x

k k kk k k k k k k k x

xdx u x dx x dxd x

xdu x dx dx

d x

xu u u udx

x x x x

xu u u dx

x x x x x x x x

1kx


13


That is, our linear equations are:

This time, let’s take an actual example

1

1

1 1 11 1 1 1 0

1 1 1 1 1

2

k

k

x

k k kk k k k k k k k x

xu u u dx

x x x x x x x x


14


Consider the equation

with the boundary conditionsu(0) = 0

u(1) = 0

2

2 42

sind

u x u x xdx


15


The solution to the boundary value problem

u(0) = 0

u(1) = 0

is the exact function

2

2 42

sind

u x u x xdx

2 42

1 13 1 4 5cos cos

16u x x x x x


16


We know in one dimension if the right-hand side is close to zero, the solution is a straight line

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];

2

2 42

sind

u x u x xdx


17


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;


18

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 = zeros ( n, 1 );

for k = 1:n b(k) = 0.5*int( rho, x(k), x(k + 2) ); end b(1) = b(1) - idx(1)*uab(1); b(end) = b(end) - idx(end)*uab(end); v = [uab(1); M \ b; uab(2)];end

int( rho, a, b ) approximates b

a

x dx


19

The Test Functions

The right-hand function of

is also straight-forward:

function [ u ] = rho( x ) u = sin( pi*x ).^4;end

2

2 42

sind

u x u x xdx


20

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' );


21

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


22

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' );


23


Understanding that the right-hand side has a greater influence in the centre,

appropriately changing the sample points yielded a significantly better approximation

2

2 42

sind

u x u x xdx


24

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


25

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


26

References

[1] Glyn James, Advanced Modern Engineering Mathematics, 4th Ed., Prentice Hall, 2011, §§9.2-3.


27

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

[email protected]

1-d finite-element methods with poisson’s equation

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