iii solution of pde’s using variational principles introduction euler-lagrange equations method of...

III Solution of pde’s using variational principles

• Introduction• Euler-Lagrange equations• Method of Ritz for minimising functionals• Weighted residual methods• The Finite Element Method

4.1 Introduction

IntroductionVariational principles

• Variational principles are familiar in mechanics •the ‘best’ approximate wave function for the ground state of a quantum system is the one with the minimum energy•The path between two endpoints (t1, t2) in configuration space taken by a particle is the one for which the action is minimised

•Energy or Action is a function of a function or functions •Wave function or particle positions and velocities•A function of a function is called a functional

•A functional is minimal if its functional derivative is zero•This condition can be expressed as a partial differential equation

IntroductionHamilton’s principal of least action

2

t

1t

N11N11 dt (t)).q..., (t),

.q(t),

.q(t),q(t),...,q(t),L(q Action

L = T – V is the Lagrangian (t)q1

1t

2t

(t)q2

(t)q1

The path actually taken is the one for which infinitesimal variations in the path result in no change in the action

IntroductionHamilton’s principal of least action

• The condition that a particular function is the one that minimises the value of a functional can be expressed as a partial differential equation

• We are therefore presented with an alternative method for solving partial differential equations besides directly seeking an analytical or numerical solution

• We can solve the partial differential equation by finding the function which minimises a functional

• Lagrange’s equations arise from the condition that the action be minimal 0

qL -

qL

dtd

ii

4.2 Euler-Lagrange Equations

 • Let J[y(x)] be the functional • Denote the function that minimises J[y] and satisfies boundary conditions specified in the problem by

• Let (x) be an arbitrary function which is zero at the boundaries in the problem so that + (x) is an arbitrary function that satisfies the boundary conditions

• is a number which will tend to zero

b

adx )y'y,F(x, J[y]

dx

dy y'

y

y

Euler-Lagrange Equations

Functionals

b

adx )''y,yF(x, (x)]yJ[

a b

A

(x)

y(x)

B (x)y

(x) (x)y

x

Functional

Boundary conditionsy(a) = Ay(b) = B

Function )J(

0dx )'y,yF(x,

dx )''y,yF(x,d

d

d

)dJ(

b

a

b

a

Euler-Lagrange Equations

Functionals

0y'

F

dx

d

y

F if 0

d

dJ

dx y'

F

dx

d

y

F

dx 'y'

F

y

F

d

dJ

'y'

F

y

F 0

y'

y'

F

y

y

F

x

x

F

F

b

a

b

a

•

• y is the solution to a pde as well as being the function which minimises F[x,y,y’]

• We can therefore solve a pde by finding the function which minimises the corresponding functional

y y

• Electrostatic potential u(x,y) inside region D SF p 362

• Charges with density f(x,y) inside the square

• Boundary condition zero potential on boundary

•Potential energy functional

•Euler-Lagrange equation

4.3 Method of Ritz for minimising functionals

dxdy 2uf u u J[u]D

2y

2x

y)f(x, y)u(x,2

D

Method of Ritz for minimising functionalsElectrostatic potential problem

etc.

y)(x, xy y)(x,

y)(x,y y)(x,

y)(x, x y)(x,

y)(x,y y)(x,

y)(x, x y)(x,

y)-x)(1- xy(1 y)(x,

16

12

5

12

4

13

12

1

Basis set which satisfies boundary conditions

00.2

0.40.6

0.8

1 0

0.2

0.4

0.6

0.8

1

0

0.02

0.04

0.06

00.2

0.40.6

0.8

1

00.2

0.40.6

0.81 0

0.2

0.4

0.6

0.8

1

0

0.01

0.02

0.03

00.2

0.40.6

0.81

1

2

• Series expansion of solution

•Substitute into functional

•Differentiate wrt cj

Method of Ritz for minimising functionalsElectrostatic potential problem

y)(x, c y)u(x,N

1 iii

dydx cf 2 y

c x

c )J(cD

N

1 iii

2N

1 i

ii

2N

1 i

iii

dydx f cyy

xx

2 c

JD j

N

1 ii

jiji

j

Method of Ritz for minimising functionalsElectrostatic potential problem• Functional minimised when

• Linear equations to be solved for ci

Aij.cj = bi

 where

dydx yy

xx

AD

jijiij

dydx y)(x,y)f(x,- bD ii

0 c

J

j

4.4 Weighted residual methods

• For some pde’s no corresponding functional can be found

• Define a residual (solution error) and minimise this

• Let L be a differential operator containing spatial derivatives D is the region of interest bounded by surface S

• An IBVP is specified by

BC S x t)(x,f t)u(x,

IC D x(x) u(x,0)

PDE 0 t D x u Lu

s

t

Weighted residual methodsTrial solution and residuals

•Define pde and IC residuals

n

1 iiisI

tTTE

(x)u (0)c - (x,0)u - (x) (x)R

t))(x,(u - t)(x,Lu t)(x,R

•Trial solution

n

1 iiisT (x)u (t)c t)(x,u t)(x,u

S x 0 (x)u

t)(x,f t)(x,u

i

ss

• RE and RI are zero if uT(x,t) is an exact solution

ui(x) are basis functions

• The weighted residual method generates and approximate solution in which RE and RI are minimised

• Additional basis set (set of weighting functions) wi(x)

• Find ci which minimise residuals according to

• RE and RI then become functions of the expansion coefficients ci

Weighted residual methodsWeighting functions

0 (x)dx(x)Rw

0 t)dx(x,(x)Rw

D Ii

D Ei

Weighted residual methodsWeighting functions

• Bubnov-Galerkin methodwi(x) = ui(x) i.e. basis functions themselves

• Least squares method

i

Ei c

R2(x)w

0 c

)(cJ

0 c

)(cJ

0 (x)dxR)(cJ

0 t)dx(x,R)(cJ

i

iI

i

iE

D2IiI

D2EiE

Positive definite functionals u(x) real

Conditions for minima

4.5 The Finite Element Method

• Variational methods that use basis functions that extend over the entire region of interest are

•not readily adaptable from one problem to another•not suited for problems with complex boundary shapes

• Finite element method employs a simple, adaptable basis set

The finite element methodComputational fluid dynamics websites

• Gallery of Fluid Dynamics• Introduction to CFD• CFD resources online• CFD at Glasgow University

Vortex Shedding around a Square CylinderCentre for Marine Vessel Development and ResearchDepartment of Mechanical EngineeringDalhousie University, Nova Scotia

Computational fluid dynamics (CFD) websitesVortex shedding illustrations by CFDnet

The finite element method Mesh generation

Local coordinate axes andnode numbers

Global coordinate axes

1

23

Finer mesh elements in regions where the solution varies rapidly

Meshes may be regular or irregular polygons

Definition of local and global coordinate axes and node numberings

The finite element method Example: bar under stress

• Define mesh• Define local and global node numbering• Make local/global node mapping• Compute contributions to functional from each element• Assemble matrix and solve resulting equations

1F

2Fi

T1i

T

The finite element method Example: bar under stress

• Variational principle

• W = virtual work done on system by external forces (F) and load (T)

• U = elastic strain energy of bar

• W = U or (U – W) = = 0

dxx

xdx

du

2

AEx

x

T(x)u(x)dxuFuFΠ2

1

22

1

1122

The finite element method Example: bar under stress

dxx

xdx

)d(u

2

AEx

x

)dxT(x)(u

)(uF)(uF)Π(u

2

1

22

1

222111

dxx

x

dx

d

dx

duAE

x

x

dx TFFd

dΠ 2

1

2

1

2211

• Eliminate d/dx using integration by parts

The finite element method Example: bar under stress

dxx

x

dx

duAE

dx

d

dx

duAE

dx x

x

dx

duAE

dx

dxxdx

duAE dx

x

x

dx

d

dx

duAE

2

1

12

2

1

2

1

2

1

|

0 T(x)dx

duAE

dx

d

0 xdx

duAEF 0

xdx

duAEF

22

11 ||

Differential equation being solved

Boundary conditions

The finite element method Example: bar under stress

• Introduce a finite element basis to solve the minimisation problem [u(x)] = 0

• Assume linear displacement function  

u(X) = + X

ui(X) = + Xi

uj(X) = + Xj•Solve for coefficients

ij

ijji1 X - X

Xu - Xu

ij

ij2 X - X

u - u X is the local

displacement variable

u(X)

i jX

The finite element method Example: bar under stress

•Substitute to obtain finite elements

u(X) = u1 + u2

ij

j1 X - X

X - X N

ij

i2 X - X

X - X N

• u1 and u2 are coefficients of the basis functions N1 and N2

N1

N2

u(X) = [N1 N2] (u)

The finite element method Example: bar under stress

• Potential energy functional Grandin pp91ff

dx T(x)u(x) dxdx

du

2

AE uF - uF- [u(x)]

2x

1x

22x

1x

2211

1-

1 u u

X - X

1

u

u 1 1-

X - X

1

X - X

u - u

dx

duji

ijj

i

ijij

ij

j

ij

i2

ij

2

ij

ij2

u

u

11-

1-1u u

X - X

1

X - X

u - u

dx

du

The finite element method Example: bar under stress

matrix stiffnessElement 11-

1-1

X - X

1

2

EA [k]

q [k]. . q 2

1

u

u

11-

1-1]u u [

X - X

1

2

EA

dX u

u

11-

1-1]u u [

X - X

1

2

EA U

ij

T

j

iji

ij

jX

iXj

iji2

ij

• Strain energy dxdx

du

2

AE energy train

22x

1x

s

per element

The finite element method Example: bar under stress

j

ijiNF u

u ]F [F - Venergy potential force Node

• Node force potential energy

dx T(x)u(x) Venergy potential load dDistribute2

x

1x

T

• Distributed load potential energy

dX u

u

X - X

X - X

X - X

X - XT(X)- V

j

i

ij

i

ij

jjX

iXT

The finite element method Example: bar under stress

• Energy functional for one element

0 u i

j

ijX

iX21

j

iji

j

iji u

u . ]N [N T(X) dX

u

u . ]F F [

u

u .k . ]u u [

2

1

• Equilibrium condition for all i

0

1 . ]N [N T(X) dX

0

1 . ]F F [

0

1 .k . ]u [u

2

1 u

u .k . 0] 1 [

2

1

u

jX

iX

21

jijij

i

i

The finite element method Example: bar under stress

• Equilibrium condition for one element

2

1jX

iXj

i

j

i

N

N T(X) dX

F

F

u

u .k

• Assemble matrix for global displacement vector

TFu .k

The finite element method Example: bar under stress

element labelsn N

N )T(X dX

...

0

0

F

...

u

u

u

...100

1210

0121

0011

K

2n

1njX

iXnnn

1

3

3

1

TF

u

• Solve resulting linear equations for u