finite element method. fem approaches there are two main methods that are applied to solving...

Finite Element Method

FEM Approaches

There are two main methods that are applied to solving problems using the FEM method.

Variational methods that use the classical Rayleigh-Ritz technique

Galerkin’s methods which is one technique in a family of methods referred to as method of weighted residuals.

Variational Approach

In solving problems arising in physics and engineering it is often possible to replace the problem of integrating a differential equation by the equivalent problem of seeking a function that gives a minimum value of some integral. Problems of this type are called variational problems.

The methods that allow us to reduce the problem of integrating a differential equation to the equivalent variational problem are usually called variational methods.


What is a functional?

dxyyxFyIb

a

),,()(

subjected to the boundary conditions

BbyAay )(,)(

functional

Goal: Find a function F(x,y,y’) for which the functional I(y) has an extremum (usually a minimum)


Some problems:

dxyyxFyIb

a

),,()( NOT:

Question: Are there situations in which the function F(x,y,y’) for which the functional I(y) in minimized is ALSO a solution to the PDE and BCs??

.....0),( 22 ort

UUoryxfU


Question: Are there situations in which the function F(x,y,y’) for which the functional I(y) in minimized is ALSO a solution to the PDE and BCs??

Answer: Yes, all PDEs typically found in physics and engineering have functionals or variational equations whose solution is equivalent to solving the PDE directly.


Deriving those equations can be done using the calculus of variations (beyond the scope of this class).

Answer: Yes, all PDEs typically found in physics and engineering have functionals or variational equations whose solution is equivalent to solving the PDE directly.


Name of equations PDE Variational principle

Homogeneous

wave equation with sources

Homogeneous

wave equation without sources

Diffusion equation

Homogenous Poisson’s equation

Homogenous Laplace’s equation

gk 22

022 k

02

t

k

g2

02

dvgkIv 2

2

1)( 222

dvkIv 222

2

1)(

dvdtt

kIt v

22

2

1)(

dvgIv 2

2

1)(

2

dvIv 2

2

1)(


The finite element method (FEM) has its origin in the field of structural analysis. However, since then the method has been employed in nearly all areas of computational physics and engineering.

The FEM method, while more difficult to program than either the finite difference (FD) or method of moments (MOM), is a more powerful and versatile numerical technique for handling problems involving complex geometries and inhomogeneous media.

Basic concept

Although the behaviour may be complex when viewed over a largeregion, a simple approximation may suffice over a small subregion.

The region is divided up into finite elements.

(usually, triangles or squares,but can be more complicated)

Regardless of the shape the field is approximated by a different expression over each element, maintaining continuity at adjoiningelements.

Solution Strategy: Variational Approach

The equations to be solved are usually stated not in terms of field the variables but in terms of an integral-type functional such as energy.

The functional is chosen such that the field solution makes the functional stationary

The total functional is the sum of the integral over each element


The finite element method (FEM) involves basically four steps:

(1) Discretize the solution region into a finite number of subregions or elements

(2) Derive the governing equations for each element based on either a variational approach or Galerkin’s method

(3) Assemble all the elements together in the solution space.

(4) Solve the resulting system of equations


CREATING THE MESH


Finite Element Method in 2D

CREATING THE MESH


• divide the geometry into elements(in 2D triangular elements are common)• each element has a number of nodesattached.

In this figure there are 8 elements (E1-E8)and 9 nodes (N1-N9)


The mesh is often described using two tables(element table and a node table)

Element TableElement# Node1 Node2 Node31 4 7 82 4 8 53 8 9 5......8 2 6 3

Node TableNode# x y1 0 1 2 0.5 1 3 1 1 ......9 1 0

2D FEM: Right Triangular Single Element

The simplest element is the right triangle (0,1,2)

x

y

1

2

h

h

0

The potential can becalculated as

,x y a bx cy

a,b,c unknown

! Smart to specify theinterpolation at the vertices!

Then: 1 0 2 00,x y x y

h h

Other elements

Square element: consider as two triangles. This is ok, but doesnot lead to a smooth interpolation over the square. Better to usea higher order interpolation that uses four nodal values

,x y a bx cy dxy This scheme leads to:

0 12

84 5

36 7 8

01

1

8 ii

Element expansion

In general we wish to approximate the unknown function, (x,y),Inside an element by an interpolation of its values at the vertices.

element shape functions or basis functions. In FEM theseare usually interpolation functions

unknown anywherewithin the element

Provided the valuesat the vertices are known

N

iii yxyx

1

),(),(

Number of vertices

Generalized Triangular

Develop the Governing Equations for a Single Element

x

y

Ve(x,y)

1

2

3

Ve1

(x1, y1)

Ve2

(x2, y2)

Ve3

(x3, y3)

cybxayxVe ),(

Assume the unknown variable (V), in eachelement can be found as a linear interpolationof its value at the three nodes.

FIND a, b and c

c

b

a

yx

yx

yx

V

V

V

e

e

e

33

22

11

3

2

1

1

1

1


Develop the Governing Equations for a Single Triangular Element

x

y

Ve(x,y)

1

2

3

Ve1

(x1, y1)

Ve2

(x2, y2)

Ve3

(x3, y3)

c

b

a

yx

yx

yx

V

V

V

e

e

e

33

22

11

3

2

1

1

1

1

Solve for the unknowns (a, b and c) and thensubstituting that result into

cybxayxVe ),(


x

y

Ve(x,y)

1

2

3

Ve1

(x1, y1)

Ve2

(x2, y2)

Ve3

(x3, y3)

3

1

),(),(i

eiie VyxyxV

where

12131312

122112213

311331132

233223321

2

1

2

1),(

2

1),(

2

1),(

yyxxyyxxA

yxxxyyyxyxA

yx

yxxxyyyxyxA

yx

yxxxyyyxyxA

yx

Shape or basis functions(only function of geometry)


x

y

Ve(x,y)

1

2

3

Ve1

(x1, y1)

Ve2

(x2, y2)

Ve3

(x3, y3)

12131312

122112213

311331132

233223321

2

1

2

1),(

2

1),(

2

1),(

yyxxyyxxA

yxxxyyyxyxA

yx

yxxxyyyxyxA

yx

yxxxyyyxyxA

yx

Properties of the shape functions

1),(

,0

,1),(

3

1

i

i

i

yx

ji

jiyx

2D Example:Laplace’s Equation

When the quantity being sought (here the electrostatic potential V)

Recall: The true potential is known to minimize the electrostaticfield energy.

dvVVIv 2

2

1)(

0),(2 yxV

Create the Mesh

Step #1: Discretize the surface/volume to be solved into small finite elements. Label each element and its associated nodes

1

2

3

4

1 2 3

Basic concept: Review

Step #2: Approximate the complex solution within the whole regionin terms of a sum of solutions found within each element

1

2

3

41 2 3

eN

ee yxVyxV

1

),(),(

FEM Laplace’s Equation

Approximate the solution within each element in terms ofeach value at the corresponding nodes. (i.e. interpolation)

(x1,y1)

(x2,y2)

(x3,y3) Element

1 2 3 3 2 2 3 3 2

2 3 1 1 3 3 1 1 3

3 1 2 2 3 1 2 2 1

1,

21

,21

,2

x y x y x y y y x x x yA



3

1

),(),(i

eiie VyxyxV


x

y

Ve(x,y)

1

2

3

Ve1

(x1, y1)

Ve2

(x2, y2)

Ve3

(x3, y3)

To derive an equation for each element we plug in our linear approximation into an functional expression.

dvyxVdvVyxVIv i

iei

v ieiie

23

1

23

1

),(2

1),(

2

1)(

3

1

),(),(i

eiie VyxyxV

dvVVIv

ee 2

2

1)(

eji j

j

v

ieie VdsyxyxVVI

3

1

3

1

),(),(2

1)(


x

y

Ve(x,y)

1

2

3

Ve1

(x1, y1)

Ve2

(x2, y2)

Ve3

(x3, y3)

To derive an equation for each element we plug in our linear approximation into an functional expression.

dvyxVdvVyxVIv i

iei

v ieiie

23

1

23

1

),(2

1),(

2

1)(

3

1

),(),(i

eiie VyxyxV

dvVVIv

ee 2

2

1)(

eji j

j

v

ieie VdsyxyxVVI

3

1

3

1

),(),(2

1)(

1 2 3 3 2 2 3 3 2

2 3 1 1 3 3 1 1 3

3 1 2 2 3 1 2 2 1

1,

21

,21

,2





x

y

Ve(x,y)

1

2

3

Ve1

(x1, y1)

Ve2

(x2, y2)

Ve3

(x3, y3)

Example: Laplace’s Equation

dsyxyxVVI j

v

ii j

eie ),(),(2

1)(

3

1

3

1

dxdyyxyxC j

v

ieij ),(),(

Let

eet

ee VCVVI2

1)(

where

3

2

1

e

e

e

e

V

V

V

V and

eee

eee

eee

e

CCC

CCC

CCC

C

333231

232221

131211

element coefficient matrix


Example: Laplace’s Equation

dxdyyxyxC j

v

ieij ),(),(

eet

ee VCVVI2

1)(

23322

1QPQPA

1 2 3 3 2 2 3 3 2

2 3 1 1 3 3 1 1 3

3 1 2 2 3 1 2 2 1

1,

21

,21

,2




jijieij QQPP

AC

4

1

x

y

Ve(x,y)

1

2

3

Ve1

(x1, y1)

Ve2

(x2, y2)

Ve3

(x3, y3)

213

132

321

yyP

yyP

yyP

123

212

231

xxQ

xxQ

xxQ

Solution for triangular elements


x

y

Ve(x,y)

1

2

3

Ve1

(x1, y1)

Ve2

(x2, y2)

Ve3

(x3, y3)dxdyyxyxC j

v

ieij ),(),(

Main property of the coefficient matrix:

eji

eij CC


Assembling of All Elements Together

Solve for the unknowns (a, b and c) and thensubstituting that result into

N

ee yxIyxI

1

),(),(

VCVyxIyxI tN

ee 2

1),(),(

1

where

NV

V

V

V

V

....3

2

1

and

NNN

N

e

CC

C

CC

CCCC

C

.........

...............

............

.........

...

1

31

2221

1131211

Global coefficient matrix


Assembling of All Elements Together

NNN

N

e

CC

C

CC

CCCC

C

.........

...............

............

.........

...

1

31

2221

1131211

Coefficients are only non-zero for elements that touch each other

1

2 4

3

5

E1

1-4-2(1-2-3) E2

1-3-4(1-2-3)

E3

3-5-4(1-2-3)

322

323

321

332

333

233

122

331

232

123

231

121

312

313

223

311

122

221

132

133

131

213

112

212

113

211

111

00

0

00

0

CCC

CCCCCCCCC

CCCCCC

CCC

CCCCCC

C


Setting up and solving the resulting equations

The solution for the node values is the ones that result in the functional obtaining its minimum value!

VCVyxIyxI tN

ee 2

1),(),(

1

NkV

yxI

V

yxI

V

yxI

V

yxI

k

N

,....,2,10),(

0),(

......),(),(

21



NkV

yxI

k

,....,2,10),(

Example for five nodes:

5

4

3

2

1

5554535251

4544434241

3534333231

2524232221

1514131211

54321

V

V

V

V

V

CCCCC

CCCCC

CCCCC

CCCCC

CCCCC

VVVVVI

0

2

515414313212

1551441331221111

CVCVCVCV

CVCVCVCVCVV

I



NkV

yxI

k

,....,2,10),(

Example for five nodes:

0

2

515414313212

1551441331221111

CVCVCVCV

CVCVCVCVCVV

I

In general:

NkCVN

iiki ,....3,2,10

1

Iterative solution method

At node k in a mesh with n nodes, we have the solution

e.g. see previous slide

We note that Cki = 0 if node k is not directly connected to node i,only nodes that are directly connected to node k contribute to Vk.

We apply this iteratively to all free nodes (not at fixed boundary value) where the potential is unknown. Initially we assign either0, random or some average value to each free node and then iterate on the above equation until convergence is met.

n

kiikii

kkk CV

CV

,1

1


Example: Poisson’s Equation

dvVyxgyxVyxVIv i

eiii

eiii

eiie

3

1

3

1

23

1

),(),(2),(2

1)(

3

1

),(),(i

eiie VyxyxV

ejj

v

ii j

eiejj

v

ii j

eie gdsyxyxVVdsyxyxVVI

),(),(),(),(2

1)(

3

1

3

1

3

1

3

1

),(2 yxgV

dvgVVVIv 2

2

1)(

2

3

1

),(),(i

eii gyxyxg

Equations for single element:


Example: Poisson’s Equation

ejj

v

ii j

eiejj

v

ii j

eie gdsyxyxVVdsyxyxVVI

),(),(),(),(2

1)(

3

1

3

1

3

1

3

1

),(2 yxgV dvgVVVI

v 2

2

1)(

2


eet

eeet

ee gTVVCVVI 2

1)(

where

3

2

1

e

e

e

e

V

V

V

V and

dsyxyxC j

v

ieij ),(),(

dsyxyxT j

v

ie

ij ),(),(


Example: Poisson’s Equation ),(2 yxgV

Equations for single element: eet

eeet

ee gTVVCVVI 2

1)(

dsyxyxC j

v

ieij ),(),( dsyxyxT j

v

ie

ij ),(),(

23322

1QPQPA

jijieij QQPP

AC

4

1

213

132

321

yyP

yyP

yyP

123

212

231

xxQ

xxQ

xxQ


23322

1QPQPA

jiA

jiAT e

ij 6/

12/

213

132

321

yyP

yyP

yyP

123

212

231

xxQ

xxQ

xxQ




dvgVVVIv 2

2

1)(

2

Put all the elements together:

gTVVCVVIVI ttN

ee

2

1)()(

1

Construct matrix and then solve:

Nk

V

gTVVCV

V

VI

k

tt

k

,.....3,2,1021

)(



Example:

Nk

V

gTVVCV

V

VI

k

tt

k

,.....3,2,102

1)(

02

1

5

4

3

2

1

5554535251

4544434241

3534333231

2524232221

1514131211

54321

5

4

3

2

1

5554535251

4544434241

3534333231

2524232221

1514131211

543211

g

g

g

g

g

TTTTT

TTTTT

TTTTT

TTTTT

TTTTT

VVVVV

V

V

V

V

V

CCCCC

CCCCC

CCCCC

CCCCC

CCCCC

VVVVVV

5

11

11

5

21

111

11

iii

iii gT

CCV

CVResult in:

In General:

N

iiki

kk

N

kiikii

kkk gT

CCV

CV

1)(1

11


Example: Homogenous Wave Equation

dvyxgyxyxkyxIv i

eiii

eiii

eiii

eiie

3

1

3

1

23

1

2

23

1

),(),(2),(),(2

1)(

3

1

),(),(i

eiie yxyx

ejj

v

ii j

eiejj

v

ii j

ei

ejj

v

ii j

eie

gdsyxyxdsyxyxk

dsyxyxI

),(),(),(),(2

),(),(2

1)(

3

1

3

1

3

1

3

1

2

3

1

3

1

),(),(),( 22 yxgyxkyx

3

1

),(),(i

eii gyxyxg

Equations for single element: dvgkIv 2

2

1)( 222


Example: Wave Equation


eet

eeet

eeet

ee gTTk

CI 22

1)(

2

where

3

2

1

e

e

e

e and

dsyxyxC j

v

ieij ),(),(

dsyxyxT j

v

ie

ij ),(),(

),(),(),( 22 yxgyxkyx

ejj

v

ii j

eiejj

v

ii j

ei

ejj

v

ii j

eie

gdsyxyxdsyxyxk

dsyxyxI

),(),(),(),(2

),(),(2

1)(

3

1

3

1

3

1

3

1

2

3

1

3

1


Example: Wave Equationwith no sources


eet

eeet

ee Tk

CI 22

1)(

2

where

3

2

1

e

e

e

e and

dsyxyxC j

v

ieij ),(),(

dsyxyxT j

v

ie

ij ),(),(

0),(),( 22 yxkyx

ejj

v

ii j

eiejj

v

ii j

eie dsyxyxk

dsyxyxI

),(),(2

),(),(2

1)(

3

1

3

1

23

1

3

1


Example: Wave Equation


Tk

CII ttN

ee 22

1)()(

2

1

0),(),( 22 yxkyx

x

y

z

Finite Element Method: Waveguide

TM Modes

22tz kk

zjkz

zeyxzyxE ),(),,(

TE Modes

zjkz

zeyxzyxH ),(),,(

where 0),(),( 22 yxkyx t

and


TM Modes


Tk

CII tttN

ee 22

1)()(

2

1

0),(),( 22 yxkyx t

Band Matrix Solution: Break the matrices up into sub-matrices that distinguishesfree-nodes from those that are known or prescribed nodes (i.e. boundary values):

p

f

pppf

fpffpf

t

p

f

pppf

fpffpf TT

TTkCC

CCI

22

1 2

zjkz

zeyxzyxE ),(),,(


TM Modes

0),(),( 22 yxkyx t

Minimize the functional:

p

f

pppf

fpffpf

t

p

f

pppf

fpffpf TT

TTkCC

CCI

22

1 2

zjkz

zeyxzyxE ),(),,(

0

0

2

p

ffpfft

p

ffpff

f

TTkCC

I


TM Modes

0),(),( 22 yxkyx t

zjkz

zeyxzyxE ),(),,(

For TM modes: 0 psince tangential E field must vanish on PEC boundary

02

p

f

fpfftp

f

fpff TTkCC

02 ffftff TkC


TM Modes 0),(),( 22 yxkyx tzjk

zzeyxzyxE ),(),,(

02 ffftff TkC

Pre-multiply both sides by1

ffT

0

0

0

21

121

f

ftffff

ffffftffff

IA

IkCT

TTkCT

Where:

2

1

t

ffff

k

CTA


Project #2: Write a FEM program that calculates the mode shapes and cutoff frequencies of the TM modes for the air filled ridge waveguideshown below.

h

w wd

t

1

1

r

r

PEC Walls

finite element method. fem approaches there are two main methods that are applied to solving...

Documents

variational problems

variational methods

variational equations

equivalent variational

variational approachin

variational approachquestion

variational approachwhat

fem method