iii solution of pde’s using variational principles introduction euler-lagrange equations method of...
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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
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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
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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
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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
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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 (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
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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
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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
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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
dxdy 2uf u u J[u]D
2y
2x
y)f(x, y)u(x,2
D
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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
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• 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
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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
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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
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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
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• 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
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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
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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 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
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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
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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
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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
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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
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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
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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
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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)
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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
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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
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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
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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
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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
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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