Weighted Residual Methods - mashayekhi.iut.ac.ir

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Weighted Residual

Methods

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Formulation of FEM Model Direct Method

Variational Method

Weighted Residuals

Formulation of FEM Model

• Several approaches can be used to transform the physical

formulation of a problem to its finite element discrete analogue.

• If the physical formulation of the problem is described as a

differential equation, then the most popular solution method is

the Method of Weighted Residuals.

• If the physical problem can be formulated as the minimization

of a functional, then the Variational Formulation is usually used.

Finite element method is used to solve physical problems

Solid Mechanics

Fluid Mechanics

Heat Transfer

Electrostatics

Electromagnetism

….

Physical problems are governed by differential equations which satisfy

Boundary conditions

Initial conditions

One variable: Ordinary differential equation (ODE)

Multiple independent variables: Partial differential equation (PDE)

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Axially loaded elastic bar

x

y

x=0 x=L

A(x) = cross section at x

b(x) = body force distribution

(force per unit length)

E(x) = Young’s modulus

u(x) = displacement of the bar at x x

Differential equation governing the response of the bar

Lxbdx

duAE

dx

d

0;0

Second order differential equations

Requires 2 boundary conditions for solution

4

Physical problems

5

x

y

x=0 x=L

x

Boundary conditions (examples)

Lxatu

xatu

1

00 Dirichlet/ displacement bc

LxatFdx

duEA

xatu

00

Neumann/ force bc

Differential equation + Boundary conditions = Strong form of the

“boundary value problem”

Physical problems Axially loaded elastic bar

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Flexible string

x

y

x=0 x=L x

S S p(x)

S = tensile force in string

p(x) = lateral force distribution

(force per unit length)

w(x) = lateral deflection of the

string in the y-direction

Differential equation governing the response of the bar

Lxpdx

udS 0;0

2

2



Physical problems

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Heat conduction in a fin

Differential equation governing the response of the fin




Q(x) = heat input per unit length per

unit time [J/sm]

k(x) = thermal conductivity [J/oC ms]

T(x) = temperature of the fin at x

x

y

x=0 x=L

x

Q(x)

LxQdx

dTAk

dx

d

0;0

Physical problems

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Dirichlet/ displacement bc

Lxathdx

dTk

xatT

00

Neumann/ force bc

x

y

x=0 x=L

x

Q(x)

Physical problems Heat conduction in a fin

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Fluid flow through a porous medium (e.g., flow of water through a dam)

Differential equation




Q(x) = fluid input per unit volume

per unit time

k(x) = permeability constant

j(x) = fluid head

x

y

x=0 x=L

x

Q(x)

LxQdx

dk

dx

d

0;0

j

Physical problems


Known head

Lxathdx

dk

xat

j

j 00

Known velocity

10

Physical problems

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Physical problems

12

Observe:

1. All the cases we considered lead to very similar differential

equations and boundary conditions.

2. In 1D it is easy to analytically solve these equations

3. Not so in 2 and 3D especially when the geometry of the domain is

complex: need to solve approximately

4. We’ll learn how to solve these equations in 1D. The approximation

techniques easily translate to 2 and 3D, no matter how complex the

geometry


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Finite Element Method

Integral Formulation

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Some Mathematical Concepts

Simply connected domain: If any two points of the domain can be

Joint by a line lying entirely within the domain

Class of a domain: A function of several variables is said to be of

Class in a domain if all its partial derivatives up to and

including the mth order exist and are continuous in

)(mC

0C F is continuous (i.e. exist but may not be

continuous.)

yfxf / , /

Boundary Value Problems: A differential equation (DE) is said to be

a BVP if the dependent variable and possibly its derivatives are

required to take specified values on the boundary.

0

1

0 , )0(,10 )( gdx

duxduxf

dx

dua

dx

d

x

Example:

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Initial Value Problem: An IVP is one in which the dependent

variable and possibly its derivatives are specified initially at t = 0

0

0

002

2

, )0(,0 vdt

duuuttfau

dt

ud

t

Example:

Initial and Boundary Value Problem:

)()0,(), (), (),0(

0 10 ) ,(

00

1

0

0

xuxutgx

uatdtu

ttandxfortxft

u

x

ua

x

x

Eigenvalue Problem: the problem of determining value l of such that

0, 0)0(

10

1

xdx

duu

xudx

dua

dx

dl

ionEigenfunctu

Eigenvalue

l

Example:

Example:

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Integration-by-Part Formula:

)()()()()( avawbvbwdxdx

dwvdx

dx

dvw

dx

dvwv

dx

dwwv

dx

db

a

b

a

Next

First

)()()()(2

2

adx

duawb

dx

dubwdx

dx

dw

dx

dudx

dx

udw

b

a

b

a

Similarly

)()()()()()()()(3

3

3

3

2

2

2

2

2

2

2

2

4

4

adx

wdavb

dx

wdbvb

dx

dvb

dx

wda

dx

dva

dx

wddx

dx

vd

dx

wddx

dx

wdv

b

a

b

a

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Gradient Theorem

FdsndxdyFdxdyFgrad ˆ

But jninnjy

Fi

x

FF yx

ˆ,

Thus

Fdsjnindxdyj

y

Fi

x

Fyx

or

dsFndxdyy

F

dsFndxdyx

F

y

x

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Divergence Theorem

GdsndxdyGdxdydivG .ˆ .

dsGnGndxdy

y

G

x

Gyyxx

yx )(

Using gradient and divergence theorem, the following relations can

Be derived! (Exercise)

wGdsnGdxdywdxdywG ˆ) ( ) ( and

wds

n

GdxdyGwdxdywG )) .(( ) ( 2

(*)

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The components of equation (*) are:

wGdsnGdxdyy

wdxdyw

y

G

wGdsnGdxdyx

wdxdyw

x

G

y

x

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An integral in the form of

b

adx

duuxuudxuuxFuI '), (, )',,()(

where integrand F(x,u,u’) is a given function of arguments x, u, u’

is called a functional (a function of function) .

A functional is said to be linear if and only if:

)()()( vIuIvuI scalarsare ,

A functional B(u,v) is said to be bilinear if it is linear in each of its

arguments

),(),(),( 2121 vuBvuBvuuB Linearity in the first argument

),(),(),( 2121 vuBvuBvvuB Linearity in the second argument

Functionals

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Functionals

A bilinear form B(u,v) is symmetric in its arguments if

),(),( uvBvuB

Example of linear functional is

L

MLdx

dvvfdxvI

0

0)()(

Example of bilinear functional is

L

dxdx

dw

dx

dvawvB

0

),(

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25


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27


28


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Consider the function )',,( uuxFF for fixed value of x, F only

depends on

The change v in u, where is constant and v is a function, is

called variation of u and denoted by:

vu Variational Symbol

''

uu

Fu

u

FF

In analogy with the total differential of a function

Note that

''du

u

Fdu

u

Fdx

x

FdF

The Variational Symbol

',uu

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1

1

11

2

2

2112

2

1

211221

2121

)()(

)(

)(

FFnF

F

FFFF

F

F

FFFFFF

FFFF

nn

Also

Furthermore

b

a

b

a

dxxudxxu

dx

duuv

dx

dvv

dx

du

dx

d

)()(

)('')()(

The Variational Symbol

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Weak Formulation of BVP

Weighted – integral and weak formulation

00 , )0(

0) ()(

Qdx

duauu

Lxxqdx

duxa

dx

d

Lx

Consider the following DE

Transverse deflection of a cable

Axial deformation of a bar

Heat transfer

Flow through pipes

Flow through porous media

Electrostatics

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There are 3 steps in the development of a weak form, if exists,

of any DE.

STEP 1:

Move all expression in DE to one side, multiply by w (weight

function) and integral over the domain.

L

dxqdx

dua

dx

dw

0

0)(

Weighted-integral or weighted-residual

N

j

jjN cUu1

0N linearly independent equation for w and

obtain N equation for Ncc ,,1

(+)

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STEP 2

1-The integral (+) allows to obtain N independent equations

2- The approximation function, , should be differentiable as many

times as called for the original DE.

3- The approximation function should satisfy the BCs.

4- If the differentiation is distributed between w and then the

resulting integral form has weaker continuity conditions.

Such a weighted-integral statement is called weak form.

The weak form formulation has two main characteristics:

-requires weaker continuity on the dependent variable and often

results in a symmetric set of algebraic equations.

- The natural BCs are included in the weak form, and therefore the

approximation function is required to satisfy only the essential BCs.

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Returning to our example:

0 0)(000

LLL

dx

duwadxwq

dx

dua

dx

dwdxwq

dx

dua

dx

dw

Secondary Variable (SV):

Coefficient of weight function and its derivatives

xndx

duaQ )( Natural Boundary Conditions (NBC)

Primary Variable (PV): The dependent variable of the problem

u Essential Boundary Conditions (EBC)

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000

LL

dx

duwadxwq

dx

dua

dx

dw

000

Lx

x

x

x

L

ndx

duwan

dx

duwadxwq

dx

dua

dx

dw

0)()( 0

0

L

L

wQwQdxwqdx

dua

dx

dw

Lxn

xn

x

x

1

0 1Note that

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STEP 3:

The last step is to impose the actual BCs of the problem w has to

satisfy the homogeneous form of specified EBC.

In weak formulation w has the meaning of a virtual change in PV.

If PV is specified at a point, its variation is zero.

0)0()0( 0 wuu

0Qdx

duan

dx

dua

LxLx

x

NBC

000

Lx

x

x

x

L

ndx

duwan

dx

duwadxwq

dx

dua

dx

dw

Thus

0)( 0

0

QLwdxwq

dx

dua

dx

dwL

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Linear and Bilinear Forms

0)( 0

00

QLwwqdxdx

dx

dua

dx

dwLL

),( uwB )(wl

0)(),( wluwB

Bilinear and symmetric in w and u ),( uwB

)(wl Linear

)(),( wluwB

Therefore, problem associated with the DE can be stated as one of

finding the solution u such that

holds for any w satisfies the homogeneous form of the EBC and

continuity condition implied by the weak form

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Assume wuu *

Variational solution

Satisfy EBC

Actual solution

Satisfy EBC+NBC

Satisfy the homogeneous

Form of EBC

Looking at the definition of the variational symbol, w is the variation of the

solution, i.e. uw

Then )(),()(),( uluuBwluwB

L L

LL L

ulQLuuqdxQLuuqdxul

uuBdxdx

du

dx

duadx

dx

duadx

dx

du

dx

udauuB

0 0

00

00 0

2

)]([)()()(

),(2

1

2

1

2),(

(#)

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)(),(2

1)(

0)(0)(),(2

10)(),(

uluuBuI

uIuluuBuluuB

Substituting in (#), we have:

In general, the relation ),(2

1),( uuBuuB holds only if

),( uwB is bilinear and symmetric and )(wl is linear

If B(w,u) is not linear but symmetric the functional I(u) can

be derived but not from (##). (see Oden & Reddy, 1976, Reddy 1986)

(##)

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Equation represents the necessary condition for the

functional I (u) to have an extremum value. For solid mechanics,

I(u) represents the total potential energy functional and the

statement of the total potential energy principle.

0)( uI

Of all admissible function u, that which makes the total potential

energy I(u) a minimum also satisfies the differential equation and

natural boundary condition in (+).

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Example 1

Some Examples

Consider the following DE which arise in the study of the deflection

of a cable or heat transfer in a fin (when c = 0).

1, 0)0(

10 0)(

1

2

xdx

duau

xforxcudx

dua

dx

d

Step 1

0)(

1

0

2

dxxcu

dx

dua

dx

dw

Step 2

0

1

0

1

0

2

dx

duwadxwxcuw

dx

du

dx

dwa

0)0(

1

0)0(

1

w

NBCdx

dua

EBCu

x

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Example 1

Some Examples

1

0

2

1

0

)1()(

),(

wdxwxwl

dxcuwdx

du

dx

dwauwB

or

0)(),( wluwB

B is bilinear and symmetric and l is linear! (prove)

Thus we can compute the quadratic functional form

0)1(

1

0

2

1

0

wdxwxdxcuw

dx

du

dx

dwa

Step 3

)1(22

1)(

1

0

22

2

udxuxcudx

duauI

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Example 2

Some Examples

Consider the following fourth-order DE (elastic bending of beam)

0, , 0)0(

)0(

0 0)()(

2

2

02

2

2

2

2

2

LxLxdx

wdb

dx

dM

dx

wdb

dx

dww

Lxforxfdx

wdb

dx

d

Step 1

0)(0

2

2

2

2

dxf

dx

wdb

dx

dv

L

Step 2

0

0

2

2

0

2

2

LL

dx

wdb

dx

dvdxvf

dx

wdb

dx

d

dx

dv

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Example 2

Some Examples

0

0

2

2

2

2

0

2

2

2

2

LL

dx

wdb

dx

dv

dx

wdb

dx

dvdxvf

dx

wd

dx

vdb

) (

) (

2

2

2

2

momentBendingMdx

wdb

forceShearVdx

wdb

dx

d

( )( )

( )( )

2

2

x L

2

02

x L

dw 0w 0 0

dx

dv 0v 0 0

dx

d d wb 0

dx dx

d wb M

dx

B.C

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Example 2

Some Examples

L 2 2

02 2

x L0

d v d w dvb vf dx M 0

dx dx dx

Step 3

or )(),( vlwvB

0

0

0

2

2

2

2

)(

),(

Mdx

dvvfdxvl

dxdx

wd

dx

vdbwvB

Lx

L

L

where

The functional I (w) can be written as:

0

0

2

2

2

2)( M

dx

dwdxwf

dx

wdbwI

Lx

L

Symmetric&Bilinear Linear

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Example 3

Some Examples

Consider a 2D heat transfer problem

02

2

2

2

inqy

T

x

Tk

)(0 xTT

)(

TT

x

Tk

)(ˆ yqx

Tk

x

T

n

T

0ˆ q

Convection

B C

D A

a

b

x

Y

Steady heat conduction in a two-dimensional domain

Insulated

q0 : uniform heat generation

k : conductivity of the isotropic material

T : temperature

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Example 3

Some Examples

Step 1

002

2

2

2

dxdyqy

T

x

Tkw

Step 2

0 x y

w T w T T Tk wq dxdy wk n n ds 0

x x y y x y

nyx qn

Tkn

y

Tn

x

Tk

T=Primary variable

=Secondary variable (heat flux) nq

0) 1,0 (

0)() 0,1 (

)( ) 1,0 (

)(ˆ ) 0,1 (

4

3

02

1

n

TnnDAon

TTn

TknnCDon

xTnnBCon

yqnnABon

yx

yx

yx

yx

(*)

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Example 3

Some Examples

Step 3

b b

n

yx

dyTyaTyawdyyqyw

dswdsTTwdsn

Tkdswq

dsn

Twkdsn

y

Tn

x

Twk

0 0

),(),()(ˆ),0(

)0()(0

4321

Substituting in (*) we have

0),(),()(ˆ),0(0 0

0

dyTyaTyawdyyqywdxdywqy

T

y

w

x

T

x

wk

b b

)(),( wlTwB

w should satisfy

the EBC

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Example 3

Some Examples

b

dyyaTyawdxdyy

T

y

w

x

T

x

wkTwB

0

),(),(),(

dyTyawdyyqywdxdywqwl

b b

0 0

0 ),()(ˆ),0()(

The quadratic functional is given by:

b b

dyTyaTyaTdyyqyTdxdyTqdxdyy

T

x

TkTI

0 0

2

0

22

),(2),(2

1)(ˆ),0(

2)(

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Conclusions

1- The weak form of a DE is the same as the statement of the total

potential energy.

2- Outside solid mechanics I(u) may not have meaning of energy but

it is still a use mathematical tools.

3- Every DE admits a weighted-integral statement, or a weak form

exists for every DE of order two or higher.

4- Not every DE admits a functional formulation. For a DE to have

a functional formulation, its bilinear form should be symmetric in its

argument.

5- Variational or FE methods do not require a functional, a weak

form of the equation is sufficient.

6- If a DE has a functional, the weak form is obtained by taking its

first variation.


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References

1- An Introduction to the Finite Element Method, by: J. N. Reddy, 3rd

ed., McGraw-Hill Education (2005). (chapter 2)

2- Energy Principles and Variational Methods in Applied Mechanics, by:

J. N. Reddy, 2nd ed., John Wiley (2002). (chapter 7)

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Weighted Residual Methods - mashayekhi.iut.ac.ir