hello...............this is lesson 6

7/29/2019 hello...............this is lesson 6

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Lesson 5

Method of Weighted Residuals


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Classical Solution Technique

The fundamental problem in calculus of variations is toobtain a function f(x) such that small variations in the

function f(x) will not change the original function

The variational function can be written in general form for asecond-order governing equation (no first derivatives) as

Where ,, and are prescribed values

2

2

V

1 dfJ(f ) f 2 f dV

2 dx

= +


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Classical solution (continued)

An equation containing first-order derivatives may

not have a corresponding variational function. In

some cases, a pseudovariational function can beused

where C=C(x)

2

2

r

V

1 dC dCJ(C) D Cu K C 2mC dV

2 dx dx

= + +


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Classical solution (continued)Example:

Consider a rod of length L. The equation defining heat transfer in the rod is

with boundary conditions

Integrating twice, one obtains

Applying boundary conditions, the final result is

2

2

d T Q

dx k=

T(0) T(L) 0= =

2

1 2

QxT C x C

2k= + +

2Q(Lx x )T

2k

=


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Rayleigh-Ritz Method

FEM variational approach attributed to Lord Rayleigh

(1842-1919) & Walter Ritz (1878-1909)

Let

Assume a quadratic function


ni 1

i i

i 1

T(x) C x (where C are the unknowns)

=

=

2

1 2 3T C C x C x= + +

1

2 3

T(0) 0 C 0

T(L) 0 C C L

= == =


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R-R Method (continued)

Hence,

Now integrate

Thus

2

3T C (x Lx)=

3

dT

C (2x L)dx =

2

2

V

1 dfJ f 2 f dV (let f T, k, 0, Q)

2 dx

= +

2 2 2

3 3

V

1J kC (2x L) 2QC (x Lx) Adx

2 =


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R-R Method (continued)

which becomes

To find the value of C3 that makes J a minimum,

Therefore,

2 3 3

3 3AkC L AC QLJ6 6

= +

3 3

3

3

2AkC LJ AQL0

C 6 6

= + =

2

3

Q Q(Lx x )C or T

2k 2k

= =


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Variation Methods

Given a function u(x), the following constraints must

be met

(1) satisfies the constraints u(x1)=u1 andu(x2)=u2

(2) is twice differentiable in x1


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Variational methods (continued)

Then it can be shown that u(x) is also the solution of

the Euler-Lagrange equation

where

(1)

d F F0

dx u u

=

i(i )

i

d uu

dx=


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For higher derivatives of u,

Hence,( )

2

1

x

(1) (2) (n)

xJ F x, u, u , u ,..., u dx=

2 n

n(1) 2 (2) n (n)

F d F d F d F( 1) 0

u dx u dx u dx u + = L


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In 2-D, the constraints are(1) satisfies the constraint u = u0 on

(2) is twice differentiable in domain A(x,y)

(3) minimizes the functional

and

u uJ F x, y,u, , d

x y

=

F F F0

u x ( u / x) y ( u / y)

=


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Example:

Find the functional statement for the 2-D heat

diffusion equation

Applying the Euler-Lagrange relation

x yT Tk k Q 0

x x y y + + =

x y

F d F d F0

u dx u dy u =


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we find that

which yields the final functional form

x y

x y

22

yx1 2 3

22

yx

F T F T Fk , k , Q

u x u y u

kk dT dTF C , F C , F QT C

2 dx 2 dy

kk dT dTJ(T) QT d2 dx 2 dy

= + = + = +

= +


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A Rayleigh-Ritz Example

Begin with the equation


First find the variational statement (J)

2

2

d uu x 0 0 x 1

dx+ + =

u(0) u(1) 0= =


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R-R example (continued)

The variational statement is

Assume a quadratic approximation


12

2

0

du duJ x, u, u 2xu dx

dx dx

=

2

0 1 2u(x) a a x a x= + +

0

1 2

u(0) a 0

u(1) a a 0

= == + =


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R-R example (continued)Thus,

Now,

To be a minimum,

Finally

1 1 1

1

u(x) a [x(1 x)] a (x)

dua (1 2x)

dx

=

=

1 2 2 2 2 2 2

1 1 1 10

J(a ) a (1 2x) a x (1 x) 2a x (1 x) dx =

1

11

J 3a 1 5

0 aa 10 6 18

= = =

5u(x) x(1 x)

18=


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The Weak Statement

Method of Weighted Residuals - one does not

need a strong mathematical background to use

FEM. However, one must be able to integrate.

To illustrate the MWR, let us begin with a simpleexample - Conduction of heat in a rod of length L

with source term Q.

2

2

L

d TK Q 0 x Ldx

dTK q for x 0

dxT T for x L

= < < = =

= =


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weak statement (continued)

Integrating,

or

This analytical solution serves as a useful benchmark

for verifying the numerical approach.

( ) ( ) ( )L y

L x 0

q 1T x T L x Q z dz dyK K = + +

( ) ( )

( )

2 2L

q QT x T L x L x

K 2K

= + +


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There are basically two ways to numerically solve

this equation using the FEM: Rayleigh Ritz

Method and the Galerkin Method ( which

produces a weak statement)

Consider

T2 0

2

0

A f 0 in Bu=g on

A B= , ,x x x x

=

+ +

L L


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Hence,

Example:

The inner product becomes

( )i i j jd A c f d 0

+ =

( )2 2

2 2

u uf x, y

x y

=

( ) ( )2 2

i i i 2 2

u u, dxdy f x, y dxdy 0

x y

= = + + =


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Integrating by parts (Green-Gauss Theorem)

Problem: Use Galerkins Method to solve

( )i ii iu u u u

dy dx f x, y dxdy 0x y x x y y

+ + + =

2

2

d uf 0 0


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The inner product is

or

The weak statement becomes

( )1

i i0

, dx 0 = =

( )2

L2

i 1 1 120

d uf dx 0 where u=c c x 2xL

dx

+ = =

L LL 1 1 1

1 0 10 0

du d dcdx fdx 0

dx dx dx

+ =


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Since

we obtain

This is the same as the Rayleigh-Ritz Method

but no variational principle is required.

du0 at x=L

dx=

( )

1

2

fc

2

fu x 2xL

2

=

=


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Weighting Function Choices

Galerkin

Least Squares

Method of Moments

( )i i

i

W

, 0

=

=

( )i

, which is the square of the errorc

( )i

i

x , 0 i=0,1,2,

W any set of linearly independent functions

=

=

L


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Weighting Function Choices (continued)

Collocation

Sub domain

( )

( )( )i

i i

i x

W x x

x x , 0

=

=

i

1 for x in subinterval iW

0 for x outside i

=


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Least Squares Example

Let

For the previous example problem

( ) 2

i i i

, dxdy 2 dxdy 0c c c

=

( )

( )

2

1 1 1

1

2

12

u c c x 2xL

uc 2x 2L

x

u2c

x

= =

=

=

( )i i

1

, 2 dx 0c c

2c f

= =

= +


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hello...............this is lesson 6

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