Lesson 5

Method of Weighted Residuals

Classical Solution Technique

The fundamental problem in calculus of variations is to obtain 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 a second-order governing equation (no first derivatives) as

Where and are prescribed values

22

V

1 dfJ(f ) f 2 f dV

2 dx

Classical solution (continued)

An equation containing first-order derivatives may not have a corresponding variational function. In some cases, a pseudovariational function can be used

where C=C(x)

22

r

V

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

2 dx dx

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

Rayleigh-Ritz MethodFEM variational approach attributed to Lord Rayleigh

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

Let

Assume a quadratic function

with boundary conditions

ni 1

i ii 1

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

21 2 3T C C x C x

1

2 3

T(0) 0 C 0

T(L) 0 C C L

R-R Method (continued)Hence,

Now integrate

Thus

23T C (x Lx)

3

dTC (2x L)

dx

22

V

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

2 dx

2 2 23 3

V

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

2

R-R Method (continued)which becomes

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

Therefore,

2 3 33 3AkC L AC QL

J6 6

3 33

3

2AkC LJ AQL0

C 6 6

2

3

Q Q(Lx x )C or T

2k 2k

Variation Methods

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

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

– (2) is twice differentiable in x1<x<x2

– (3) minimizes the integral2

1

x

x

duJ F x,u, dx

dx

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

Variational methods (continued)

For higher derivatives of u,

Hence,

2

1

x(1) (2) (n)

x

J F x,u,u ,u ,..., u dx2 n

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

F d F d F d F( 1) 0

u dx u dx u dx u

Variational methods (continued)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)

Variational methods (continued)Example:

Find the functional statement for the 2-D heat diffusion equation

Applying the Euler-Lagrange relation

x y

T Tk k Q 0

x x y y

x y

F d F d F0

u dx u dy u

Variational methods (continued)we find that

which yields the final functional form

x yx y

22yx

1 2 3

22yx

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 d

2 dx 2 dy

A Rayleigh-Ritz Example

Begin with the equation

with boundary conditions

First find the variational statement (J)

2

2

d uu x 0 0 x 1

dx

u(0) u(1) 0

R-R example (continued)The variational statement is

Assume a quadratic approximation

with boundary conditions

1 22

0

du duJ x,u, u 2xu dx

dx dx

20 1 2u(x) a a x a x

0

1 2

u(0) a 0

u(1) a a 0

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 21 1 1 10

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

11

1

J 3a 1 50 a

a 10 6 18

5u(x) x(1 x)

18

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 simple example - Conduction of heat in a rod of length L with source term Q.

2

2

L

d TK Q 0 x LdxdTK q for x 0dx

T T for x L

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 xK 2K

weak statement (continued)

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

20

A f 0 in Bu=g on

A B= , ,x x x x

weak statement (continued)Since u = cii(x,y) is an approximate function,

substitution into the above equation may not satisfy the equation. We set the equation equal to an error ()

We now introduce a set of weighting functions (test functions) Wi, and construct an inner product (Wi,) that is set to zero – this forces the error of the approximate differential equation to zero (average).

Au f

weak statement (continued)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

weak statement (continued)

Integrating by parts (Green-Gauss Theorem)

Problem: Use Galerkin’s Method to solve

i ii i

u u u udy dx f x, y dxdy 0

x y x x y y

2

2

d uf 0 0<x<L

dxu=0 x=0

du=0 x=L

dx

weak statement (continued)

The inner product is

or

The weak statement becomes

1

i i0, dx 0

2L 2

i 1 1 120

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

dx

L LL 1 1 11 0 10 0

du d dcdx fdx 0

dx dx dx

weak statement (continued)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

2f

u x 2xL2

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

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

Least Squares ExampleLet

For the previous example problem

2

i i i

, dxdy 2 dxdy 0c c c

21 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

Least Squares Example (continued)

Thus,

Some observations:(1) no integration by parts required – not a weak form(2) the Neumann boundary conditions do not appear naturally

(3) in this case i must satisfy global boundary conditions, which is difficult in most problems

L

10

1

2

2 2c f 2dx 0

fc

2f

u x 2xL2