MATH 212 NE 217 Douglas Wilhelm Harder Department of Electrical and Computer Engineering University of Waterloo Waterloo, Ontario, Canada Copyright © 2011

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

Finite-Element Methodsin Two Dimensions

Finite-element Methods in Two Dimensions

2

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


3

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


4

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:

– Now use a corollary of Stokes’ theorem:

R R R

uv u v v u

u v uv v u

u v dxdy uv dxdy v u dxdy

R R R

u v dxdy uvd v u dxdy

s

Ref: http://en.wikipedia.org/wiki/Vector_calculus_identities


5

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:

– Now use a corollary of Stokes’ theorem:

R R R

u u u

u u u

u dxdy u dxdy u dxdy

v v v

v v v

v v v

R R R

u v dxdy u d v u dxdy

v s


6

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:

The ultimate goal is to do this in 2 and 3 dimensions– In order to achieve this, we need one further idea

2

0

,,

x yu x y


7

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


8

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


9


We have seen how using finite differences, we:– Create a uniform grid

– Assign points to either the boundary or the interior of the region


10


Unfortunately, uniform grids make a poor approximation of most real-world situations


11


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


12


In one dimension, the regions between points were intervals– Similarly, we must divide the region into a number of sub-regions


13


Our goal will be to find approximations of the solution on the interior of the region R


14


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


15


From wikipedia: a tessellation of a 2D magnetostatic configuration

User: Zureks


16

Defining Test Functions

In one dimension, for each interiorpoint, xk, we used the two adjacentintervals upon which we definedthe test functions– The support of the test function

6(x) is the interval [x5, x7]

6(x)

7(x)


17


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


18


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


19


Similarly, we will label each triangles with Rk, in this case:

Rk,1 Rk,2 Rk,3 Rk,4 Rk,5 Rk,6


20


On each region Rk, we will define a piecewise planar test function

– The test function is k(x, y)

– It is defined to be zero outside Rk

– It is also useful to have it zero on the boundary ∂Rk


21


The formula for each plane k, ℓ(x, y) defined on Rk, ℓ is a plane of the form x + y + – To find it, we use linear algebra


22

Finding Interpolating Planes

Given three points in the plane

(x1, y1), (x2, y2), (x3, y3)

if we want to find the interpolating plane x + y + that passes through these points z1, z2, and z3, respectively, we must solve the system of linear equations

1 1 1

2 2 2

3 3 3

x y z

x y z

x y z


23


Rewritten using matrices and vectors, we must solve:

1 1 1

2 2 2

3 3 3

1

1

1

x y z

x y z

x y z


24


Example, given the points in the plane

(1.3, 5.4), (2.9, 7.0) and (6.5, 4.9)


25



(1.3, 5.4), (2.9, 7.0) and (6.5, 4.9)

we want to find the polynomial x + y + that passes through the points 2.8, 1.6 and 3.9, respectively


26



(1.3, 5.4), (2.9, 7.0) and (6.5, 4.9)

we want to find the polynomial x + y + that passes through the points 2.8, 1.6 and 3.9, respectively

We must therefore solve

>> [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

1.3 5.4 1 2.8

2.9 7.0 1 1.6

6.5 4.9 1 3.9


27


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


28

Using the Test Function

Now, going back to the problem:– We have the test function k(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

, , 0k

k

R

x y V x y dxdy


29


Again, let’s consider Poisson’s equation:

We defined

and thus our integral is

2

0

,, , 0

k

k

R

x yx y u x y dxdy

2

0

,,

x yu x y

2

0

,, ,

def x yV x y u x y


30


Expanding the integral, we get

where the right-hand side is reasonably easy to calculate

The left-hand side, however, requires furthercalculus

2

0

,, , ,

k k

k k

R R

x yx y u x y dxdy x y dxdy


31


Using the 2-dimensional equivalent of integration-by-parts, we get:

2, , , , , ,k k k

k k k

R R R

x y u x y dxdy x y u x y d x y u x y dxdy

s


32


Using the 2-dimensional equivalent of the fundamental theorem of calculus:

However, recall that we specifically chose a test function that is zero on the boundary ∂Rk

– Therefore, the first term disappears!

2, , , , , ,k k k

k k k

R R R

x y u x y dxdy x y u x y d x y u x y dxdy

s

0


33


Therefore, we have the easier integral:

and therefore we have

2, , , ,k k

k k

R R

x y u x y dxdy x y u x y dxdy

0

,, , ,

k k

k k

R R



34


We can calculate the right-hand side: both are known

The left-hand side has the unknown function u(x, y)…

0

,, , ,

k k

k k

R R



35


We can calculate the right-hand side: both are known

First, rewrite the left-hand side as integrals over each sub-region

0

,, , ,

k k

k k

R R


,

,1

0

, ,

,,

k

k

k

n

k

R

k

R

x y u x y dxdy

x yx y dxdy

Let nk be the number of triangles touching xk

– in this case, 6


36

Finding the Test Function Gradient

The test function defined on each sub-region is planar, that is

thus we can easily calculate:

,

,

,

,

,,

k

k

k

x yx

x yx y

y

, ,k x y x y

,

,1

0

, ,

,,

k

k

k

n

k

R

k

R

x y u x y dxdy

x yx y dxdy


37


To find the gradient of this plane, we must find the vector

,

,1

0

, ,

,,

k

k

k

n

k

R

k

R

x y u x y dxdy

x yx y dxdy

1 1

,1 2 2

,2 3 3

: 1 1

: 1 0

: 1 0

k

k

k

x y

x y

x y

x

x

x


38


Solving for is as follows:

,

6

,1

0

, ,

,,

k

k

k

R

k

R

x y u x y dxdy

x yx y dxdy

1

2

3 2 3

1 1 1 1

2 2 2 2

3 3 3 3

1 1

det 0 1

0 1

1 1

det 1 det 1

1 1

y

y

y y ya

x y x y

x y x y

x y x y


39


Solving for is as follows:

,

6

,1

0

, ,

,,

k

k

k

R

k

R

x y u x y dxdy

x yx y dxdy

1

2

3 3 2

1 1 1 1

2 2 2 2

3 3 3 3

1 1

det 0 1

0 1

1 1

det 1 det 1

1 1

x

x

x x xx y x y

x y x y

x y x y


40

Approximating the Solution Gradient

The next step is to approximate the unknown

,

6

,1

0

, ,

,,

k

k

k

R

k

R

x y u x y dxdy

x yx y dxdy

,u x y


41


The actual unknown function u(x, y) an unknown and likely non-linear function– You don’t get paid $100 000/a to find a linear function…


42


Recall from 1-D finite element methods:– We approximated the solution with an unknown piecewise

linear function


43


We are attempting to approximate u(x, y) at a number of interior points where the solution is defined on the border


44


We can do the same as we did in one dimension:– Approximate the solution by a piecewise planer approximation


45


As before, we can find the interpolating polynomial:

,

6

,1

0

, ,

,,

k

k

k

R

k

R

x y u x y dxdy

x yx y dxdy

1 1

,1 2 2 ,1

,2 3 3 ,2

: 1

: 1

: 1

k k

k k

k k

x y a u

x y b u

x y c u

x

x

x


46


To find the gradient of this triangle, we must find the vector

,

6

,1

0

, ,

,,

k

k

k

R

k

R

x y u x y dxdy

x yx y dxdy

1 1

,1 2 2 ,1

,2 3 3 ,2

: 1

: 1

: 1

k k

k k

k k

x y a u

x y b u

x y c u

x

x

x

a

b


47


Solving for a is as follows:

,

6

,1

0

, ,

,,

k

k

k

R

k

R

x y u x y dxdy

x yx y dxdy

1

,1 2

,2 3 2 1 ,2 ,1 3 3 1 ,1 2 ,2

1 1 1 1

2 2 2 2

3 3 3 3

1

det 1

1

1 1

det 1 det 1

1 1

k

k

k k k k k k k

u y

u y

u y u y y u u y u y y u y ua

x y x y

x y x y

x y x y


48


Solving for b is as follows:

,

6

,1

0

, ,

,,

k

k

k

R

k

R

x y u x y dxdy

x yx y dxdy

1

2 ,1

3 ,2 1 ,1 3 2 ,2 1 ,2 2 ,1 3

1 1 1 1

2 2 2 2

3 3 3 3

1

det 1

1

1 1

det 1 det 1

1 1

k

k

k k k k k k k

x u

x u

x u x u x u x u x u u x u xb

x y x y

x y x y

x y x y


49

Approximating the Double Integral

Thus, we can find both gradients:

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

,1 ,1

, , , ,k k

k

R R

ax y u x y dxdy dxdy

b

,1

1kR

a b dxdy


50


Finally, we note that the right-hand integral

is just the area of the region Rk,1, call it Ak,1:

,1 ,1

, , , 1k k

k

R R

x y u x y dxdy a b dxdy

1 1

,1 2 2

3 3

11

det 12

1k

x y

A x y

x y


51


Thus, for each interior point xk, we must calculate:

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

, ,

, ,1 1 0

,, , ,

k k

k k

n n

k k

R R



52


If the ratio is a constant, the right-hand integral can be

simplified further:

where Ak is the area of the region Rk

,

,1 0

, 1,

3

k

k

n

k k

R

x yx y dxdy A

0

,x y


53

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


54

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

2 ,u 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


55

Example 1

Consider the following region and we are trying to approximate Laplace’s equation ( = 0) with the following region and boundary values:


56

Example 1

We can define the system as follows:pts1 = [ 0 -1 -1 % 1 -1 0 1 % 2 0 0 -Inf % 3 1 0 1 % 4 -1 1 0 % 5 0 1 -Inf % 6 0 2 -4]'; % 7obj1 = fe_system( 0, pts1 );


57

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] )


58

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] );


59

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 -1.0000 0 0 -1.0000 0 0 0 1.0000 1.0000 2.0000 -1.0000 1.0000 -0.1053 1.0000 0 -1.4211 -4.0000

plot3( u1(1,:), u1(2,:), u1(3,:), 'ko' )

The plot was augmented in CorelDRAW!


60

Example 2

Consider the following region and we are trying to approximate Poisson’s equation with = 4, the following region and boundary values:


61

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]'; % 13obj2 = fe_system( 4, pts2 );


62

Example 2

We then define the various tessellations:obj2 = fe_update( obj2, 4, [1 2 7 9 6] );obj2 = fe_update( obj2, 6, [1 4 9 12 11 8 3] );obj2 = fe_update( obj2, 7, [2 5 10 9 4] );obj2 = fe_update( obj2, 9, [4 7 10 13 12 6] );obj2 = fe_update( obj2, 12, [6 9 13 11] );


63

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 0 1.0000 1.0000 1.0000 2.0000 2.0000 3.0000 2.9111 10.0000 8.0000 5.5663 10.0000

plot3( u2(1,:), u2(2,:), u2(3,:), 'o' )


64

Example 2

Here we see the boundary points and the actual solution (light blue) and the approximations (in red)


65

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…


66

What’s Next?

The next step is to go to three dimensions– Divide the region into tetrahedral regions

– 3-dimensional tessellations

Tessellation User: Tom Ruen

Robert Webb's Great Stella softwarehttp://www.software3d.com/Stella.html


67

References

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


68

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]

Documents

MATH 212 NE 217 Douglas Wilhelm Harder Department of Electrical and Computer Engineering University of Waterloo Waterloo, Ontario, Canada Copyright © 2011