Solution of Pdes Using Variational Principles

8/13/2019 Solution of Pdes Using Variational Principles

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III Solution of pdes using variational

principles

Introduction

Euler-Lagrange equations

Method of Ritz for minimising functionals

Weighted residual methods

The Finite Element Method

4.1 Introduction


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Introduction

Variational principles

Variational principles are familiar in mechanicsthe 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 actionis minimised

Energy or Action is a function of a function or functionsWave function or particle positions and velocities

A function of a function is called a functional

A functional is minimal if its functional der ivativeis zeroThis condition can be expressed as a partial differential equation


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Introduction

Hamiltons principal of least action

2t

1t

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

.q(t),

.q(t),q(t),...,q(t),L(qAction

L = TV 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


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Introduction

Hamiltons principal of least action

The condition thata particular function is the one thatminimises the value of a functional can be expressed as a

partial differential equation

We are therefore presented with an alternative method forsolving 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

Lagranges equations arise from the condition that the

action be minimal 0q

L-q

L

dt

d

ii


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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 h(x) be an arbitrary function which is zero at the

boundaries in the problem so that + eh(x) is an arbitrary

function that satisfies the boundary conditions

eis a number which will tend to zero

b

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

dy

y'

y

y


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Euler-Lagrange Equations

Functionals

b

a

dx)''y,yF(x,(x)]yJ[ eheheh

a b

A

(x)h

y(x)

B(x)y(x)(x)y eh

x

Functional

Boundary conditions

y(a) = Ay(b) = B

Function )J(e

e0dx)'y,yF(x,

dx)''y,yF(x,dd

d)dJ(

b

a

b

a

ehehe

ehehee

e


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Euler-Lagrange Equations

Functionals

0y'

F

dx

d

y

F

if0d

dJ

dxy'

F

dx

d

y

F

dx'y'

F

y

F

d

dJ

'y'

F

y

F0

y'

y'

F

y

y

F

x

x

F

F

b

a

b

a

e

h

hhe

hh

eeee

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

ehyy


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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

dxdy2ufuuJ[u]D

2y

2x

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

D


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Electrostatic potential problem

etc.

y)(x,xyy)(x,y)(x,yy)(x,

y)(x,xy)(x,

y)(x,yy)(x,

y)(x,xy)(x,

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

16

1

2

5

12

4

13

12

1

Basis set which satisfies boundary conditions

0

0.2

0.4

0.6

0.8

1 0

0.2

0.4

0.6

0.8

1

0

0.02

0.04

0.06

0

0.2

0.4

0.6

0.8

1

0

0.2

0.4

0.6

0.8

1 0

0.2

0.4

0.6

0.8

1

0

0.01

0.02

0.03

0

0.2

0.4

0.6

0.8

1

1

2


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Series expansion of solution

Substitute into functional

Differentiate wrt cj



y)(x,cy)u(x,N

1iii

dydxcf2y

cx

c)J(cD

N

1iii

2N

1i

ii

2N

1i

iii

dydxfcyy

xx

2c

J

D j

N

1i

i

jiji

j


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Functional minimised when

Linear equations to be solved for ci

Aij.cj= bi

where

dydxyy

xx

AD

jijiij

dydxy)(x,y)f(x,-bD

ii

0c

J

j


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4.4 Weighted residual methods

For some pdes 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

BCS xt)(x,ft)u(x,

ICD x(x)u(x,0)PDE0 tD xuLu

s

t


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Trial solution and residuals

Define pde and IC residuals

n

1iiisI

tTTE

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

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

Trial solution

n

1iiisT(x)u(t)ct)(x,ut)(x,u

Sx0(x)u

t)(x,ft)(x,u

i

ss

REand RIare zero if uT(x,t) is an exact solution

ui(x) are basis functions


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The weighted residual method generates andapproximate solution in which REand RIare minimised

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

Find ciwhich minimise residuals according to

REand RIthen become functions of the expansion

coefficients ci


Weighting functions

0(x)dx(x)Rw

0t)dx(x,(x)Rw

D Ii

D Ei


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Weighting functions

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

Least squares method

i

Ei c

R2(x)w

0

c

)(cJ

0c

)(cJ

0(x)dxR)(cJ

0t)dx(x,R)(cJ

i

iI

i

iE

D

2IiI

D2EiE

Positive definite functionals u(x) real

Conditions for minima


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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


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The finite element method

Computational fluid dynamics websites

Gallery of Fluid DynamicsIntroduction to CFD

CFD resources online

CFD at Glasgow University

Vortex Shedding around a Square Cylinder

Centre for Marine Vessel Development and Research

Department of Mechanical Engineering

Dalhousie University, Nova Scotia

Computational fluid dynamics (CFD) websites

Vortex shedding illustrations by CFDnet
http://www.eng.vt.edu/fluids/msc/gallery/gall.htmhttp://www.cham.co.uk/website/new/cfdintro.htmhttp://www.cfd-online.com/Resources/misc.htmlhttp://www.aero.gla.ac.uk/Research/CFD/education/course/CALF/index/nindx.htmlhttp://www.cfd-online.com/Resources/misc.htmlhttp://cfdnet.com/nwt/http://cfdnet.com/nwt/http://www.cfd-online.com/Resources/misc.htmlhttp://www.aero.gla.ac.uk/Research/CFD/education/course/CALF/index/nindx.htmlhttp://www.cfd-online.com/Resources/misc.htmlhttp://www.cham.co.uk/website/new/cfdintro.htmhttp://www.eng.vt.edu/fluids/msc/gallery/gall.htm


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Mesh generation

Local coordinate axes and

node numbers

Global coordinate axes

1

23

Finer mesh elements inregions where the

solution varies rapidly

Meshes may be regularor irregular polygons

Definition of local and global

coordinate axes and node numberings


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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

2F

iT

1iT


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Variational principle

W = virtual work done on system by external forces (F)

and load (T)

U = elastic strain energy of bar

W =U or (UW) =P= 0

dx

x

xdx

du

2

AEx

x

T(x)u(x)dxuFuF2

1

22

1

1122


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dx

x

x

dx

)d(u

2

AEx

x

)dxT(x)(u

)(uF)(uF)(u2

1

22

1

222111

eheh

eheheh

dx

x

x

dx

d

dx

duAE

x

x

dxTFF

d

d2

1

2

1

2211 hhhh

e

Eliminate dh/dx using integration by parts


22/30



dxx

x

dxduAE

dxd

dxduAE

dx

x

x

dx

duAE

dx

dxx

dx

duAEdx

x

x

dx

d

dx

duAE

2

1

12

2

1

21

2

1

|

hhh

hhh

0T(x)

dx

duAE

dx

d

0xdxduAEF0xdx

duAEF2

21

1 ||

Differential equation being solved

Boundaryconditions


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

Assume linear displacement function

u(X) = a1+ a2X

ui(X) = a1+ a2Xi

uj(X) = a1+ a2Xj

Solve for coefficients a

ij

ijji1

X-X

Xu-Xua

ij

ij2

X-X

u-ua X is the local

displacement variable

u(X)

i jX


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Substitute to obtain finite elements

u(X) =N1u1+N2u2

ij

j1

X-XX-XN

ij

i2

X-XX-XN

u1and u2are coefficients of the

basis functions N1and N2

N1

N2

u(X) = [N1N2] (u)

Th fi it l t th d


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Potential energy functional Grandin pp91ff

dxT(x)u(x)dxdx

du

2

AEuF-uF-[u(x)]

2x

1x

22x

1x

2211

P

1-

1uu

X-X

1

u

u11-

X-X

1

X-X

u-u

dx

duji

ijj

i

ijij

ij

j

i

ji2

ij

2

ij

ij2

u

u

11-

1-1uu

X-X

1

X-X

u-u

dx

du

Th fi it l t th d


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matrixstiffnessElement11-

1-1

X-X

1

2

EA[k]

q[k]..q2

1

u

u

11-

1-1]uu[

X-X

1

2

EA

dX

u

u

11-

1-1]uu[

X-X

1

2

EAU

ij

T

j

iji

ij

jX

iXj

iji2

ij

Strain energy dxdx

du

2

AEenergytrain

22x

1x

s

per element

Th fi it l t th d


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j

ijiNF u

u]F[F-VenergypotentialforceNode

Node force potential energy

dxT(x)u(x)VenergypotentialloaddDistribute2x

1x

T

Distributed load potential energy

dXu

u

X-X

X-X

X-X

X-XT(X)-V

j

i

ij

i

ij

jjX

iX

T

Th fi i l h d


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Energy functional for one element

0ui

P

P

j

ijX

iX

21j

iji

j

iji u

u.]N[NT(X)dX

u

u.]FF[

u

u.k.]uu[

2

1

Equilibrium condition for all i

P

0

1.]N[NT(X)dX

0

1.]FF[

0

1.k.]u[u

2

1

u

u.k.0]1[

2

1

u

jX

iX

21

jijij

i

i


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Equilibrium condition for one element

2

1jX

iXj

i

j

i

N

NT(X)dX

F

F

u

u.k

Assemble matrix for global displacement vector

TFu .k

Th fi it l t th d


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elementlabelsnN

N)T(XdX

...

0

0

F

...

u

u

u

...100

1210

0121

0011

K

2n

1njX

i

Xnnn

1

3

3

1

TF

u

Solve resulting linear equations for u

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Solution of Pdes Using Variational Principles