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Finite Element Method

A presentation byProf. S.V. Kulkarni

Electrical Engineering Department

IIT Bombay

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1 October 2009

Main steps of the FEM method

Discretization into "elements"

FE formulation: Approximation of PDE over elements

Assembly of element equations

Solution of global linear system

Prof. SV Kulkarni, EE Dept,Indian Institute of Technology Bombay

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1 October 2009

Classification of approximation Methods

Minimization of functional -

`variational'/`Raleigh-Ritz'

Weighted residual methods

1. Point collocation2. Sub-domain collocation

3. Least squares

4. Galerkin

Prof. SV Kulkarni, EE Dept,Indian Institute of Technology Bombay

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1 October 2009

Variational methodsReferences: M. N. O. Sadiku, Numerical techniques in electromagnetics, 2nd Ed, CRC press, 2001. J. Jin, The finite element method in electromagnetics, John Wiley and Sons, 1993.

Find a function extremizing the functional subjected toboundary conditions.

dxyyxFyI

b

a= ),,()(

Aay =)(

Functional:

Boundary conditions: Bby =)(and

x is independent variable

y is dependent variable

Prof. SV Kulkarni, EE Dept,Indian Institute of Technology Bombay

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1 October 2009

Variational methodsyy +

)(xhy =

a b

y

x

dxdy

y

Variational symbol

y for 0=x

Differential symbol d

dy for finite dx

As FFFyyy ++ ,

yy

Fy

y

FF

+

=

Analogous differential is

ydy

Fdy

y

Fdx

x

FdF

+

+

=

Prof. SV Kulkarni, EE Dept,Indian Institute of Technology Bombay

8/6/2019 Fem Theory Final

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1 October 2009

Variational methods To have minimum value of functional

0=I)()( yIhyII +=

y hy +

I

y

[ ] [ ]( , , ) ( , , )b b

a a

F x y h y h dx F x y y dx = + +

[ ]dxyyxFhyyxhFIb

a

yy + ),,(),,( Expanding first term using Taylor series

Integrating by parts leads to

( ) ( ) ( )

b b

b

a

a a

F F d F I h x dx h x h x dx

y y dx y

+

Prof. SV Kulkarni, EE Dept,Indian Institute of Technology Bombay

8/6/2019 Fem Theory Final

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1 October 2009

Variational methodsb

a

b

a

xhy

Fdxxh

y

F

dx

d

y

FI

+

)()(

b

a

dxxhy

F

dx

d

y

FI )(

As 0)()( == bhah

In order to I vanish, the integrand must vanish

0=

y

F

dx

d

y

F

Prof. SV Kulkarni, EE Dept,Indian Institute of Technology Bombay

8/6/2019 Fem Theory Final

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1 October 2009

Variational methods Necessary condition for functional to have extremum is to

satisfy Euler Lagrange equation by a given function.

0=

y

F

dx

d

y

F

This is for one independent and one dependent variable.

Prof. SV Kulkarni, EE Dept,Indian Institute of Technology Bombay

8/6/2019 Fem Theory Final

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1 October 2009

Variational methodsFunctional Euler Lagrange equation

dxyyxFyI

b

a= ),,()(

0=

y

F

dx

d

y

F

dsuuuyxFuIs

yx= ),,,,()( 0=

yx u

F

yu

F

xu

F

dsvvuuvuyxFuIs

yxyx= ),,,,,,,()( 0=

yx u

F

yu

F

xu

F

0=

yx v

F

yv

F

xv

F

dxyyyyxFyI

b

a

n

= ),...,,,,()( ( ) ( ) ( ) 0)1(....22

=++ nyn

nn

yyy Fdx

dF

dx

dF

dx

dF

Prof. SV Kulkarni, EE Dept,Indian Institute of Technology Bombay

8/6/2019 Fem Theory Final

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1 October 2009

Variational methods Consider functional ( ) )1(),(

2

1)(

22

+= dxdyyxfI

s

yx

Euler Lagrange equation is

)2(0 =

yx

Fy

Fx

F

From (1) and (2), ),(2 yxf=

Solving Poissons equation is equivalent to extremizing thefunctional

( ) ),(21 22 yxfF yx +=

Prof. SV Kulkarni, EE Dept,Indian Institute of Technology Bombay

8/6/2019 Fem Theory Final

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1 October 2009

Variational methodsEquation PDE Functional

[ ]dvIv

=

2||

2

1)( 0

2= Laplace

Poisson f= 2 [ ]dvfIv

= 2||21

)(2

Homogeneouswave

Inhomogeneouswave

Diffusion

0

22=+

k

01

2

2= tt

u

fk =+ 22

02

=+ tk

[ ]dvkI v=

222

||2

1

)(

dtdvu

It v

t

+=

2

2

2 1||2

1)(

[ ]dvkIt v

t = 2||21

)( 2

[ ]dvfkI v = 2||21

)(222

Prof. SV Kulkarni, EE Dept,Indian Institute of Technology Bombay

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1 October 2009

Construction of functional from PDEs

Earlier, we have proved that Eulers equation produces the

governing differential equation corresponding to a given

functional or variational principle.

Now, let us see procedure of constructing a functional

formulation for a given differential equation.

Suppose, we have one dependent variable u and twoindependent variables x and y. We are interested in finding

the functional associated with the Poissons equation.2

2

2

( , )

0

0

u f x y

or u f

u f u

=

=

= Prof. SV Kulkarni, EE Dept,

Indian Institute of Technology Bombay

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1 October 2009

(multiplied both sides by variation of the independent variable).

Integrate over the domain of problem (Iis a functional),

Integrating by parts the equation

2 0u f udxdy I = = 2

I u u dxdy f udxdy =

2 2

2 2

u u I udxdy f udxdy

x y

= +

( ) ( )u uu u u uu dx dy u dy dx f udxdy

x x x y y y

=

Prof. SV Kulkarni, EE Dept,Indian Institute of Technology Bombay

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1 October 2009

Now

( ) ( )u uu u u uu dx dy u dy dx f udxdy x x x y y y

= ( ) ( )u uu u u u

f u dx dy udy u dx

x x y y x y

= +

2

F F is a function of onlyu usayx x

=

Prof. SV Kulkarni, EE Dept,Indian Institute of Technology Bombay

y

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1 October 2009

The last two terms vanish if we assume either the

homogeneous Dirichlet or Neumann condition at theboundaries.

( )2

FF 2 2

uu u u uy

y x x x x x

= = =

( )

21

2

uu u

x x x

=

22

22

u u u u I fu dxdy udy udx x y x y

= +

Prof. SV Kulkarni, EE Dept,Indian Institute of Technology Bombay

is taken inside, since it isindependent of xby definition,

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Hence

And

22

2u u I fu dxdyx y

= +

221

22

u u I fu dxdy

x y

= +

Prof. SV Kulkarni, EE Dept,Indian Institute of Technology Bombay

8/6/2019 Fem Theory Final

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1 October 2009

Example Consider a second order differential equation,

with boundary conditions

The exact solution of above equation is

2

21, 0 1

d ux x

dx= +