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Solving partial differential equations (PDEs)

Hans Fangohr

Engineering and the EnvironmentUniversity of Southampton

United Kingdomfangohr@soton.ac.uk

May 3, 2012

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Outline I

1 Introduction: what are PDEs?

2 Computing derivatives using finite differences

3 Diffusion equation

4 Recipe to solve 1d diffusion equation

5 Boundary conditions, numerics, performance

6 Finite elements

7 Summary

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This lecture

tries to compress several years of material into 45 minutes

has lecture notes and code available for download athttp://www.soton.ac.uk/fangohr/teaching/comp6024

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http://www.soton.ac.uk/~fangohr/teaching/comp6024http://www.soton.ac.uk/~fangohr/teaching/comp6024http://www.soton.ac.uk/~fangohr/teaching/comp6024http://www.soton.ac.uk/~fangohr/teaching/comp6024http://find/

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What are partial differential equations (PDEs)

Ordinary Differential Equations (ODEs)one independent variable, for example t in

d2x

dt2 =

k

mx

often the indepent variable t is the timesolution is function x(t)

important for dynamical systems, population growth,control, moving particles

Partial Differential Equations (ODEs)multiple independent variables, for example t, x and y in

u

t =D2ux2 +

2u

y2

solution is function u(t,x,y)important for fluid dynamics, chemistry,electromagnetism, . . . , generally problems with spatial

resolution4 / 4 7

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2d Diffusion equation

ut

=D 2ux2

+ 2

uy2

u(t,x,y) is the concentration [mol/m3]

t is the time [s]

x is the x-coordinate [m]

y is the y-coordinate [m]

D is the diffusion coefficient [m2/s]

Also known as Ficks second law. The heat equation has thesame structure (and urepresents the temperature).Example:http://www.youtube.com/watch?v=WC6Kj5ySWkQ

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http://www.youtube.com/watch?v=WC6Kj5ySWkQhttp://www.youtube.com/watch?v=WC6Kj5ySWkQhttp://find/

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Examples of PDEs

Cahn Hilliard Equation(phase separation)

Fluid dynamics (including ocean and atmospheric models,plasma physics, gas turbine and aircraft modelling)Structural mechanics and vibrations, superconductivity,

micromagnetics, . . .6 / 4 7

http://en.wikipedia.org/wiki/Cahn%E2%80%93Hilliard_equationhttp://en.wikipedia.org/wiki/Cahn%E2%80%93Hilliard_equationhttp://find/

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Computing derivatives

using finite differences

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O

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Overview

Motivation:

We need derivatives of functions for example foroptimisation and root finding algorithmsNot always is the function analytically known (but weare usually able to compute the function numerically)The material presented here forms the basis of thefinite-difference technique that is commonly used tosolve ordinary and partial differential equations.

The following slides show

the forward difference technique

the backward difference technique and thecentral difference technique to approximate thederivative of a function.We also derive the accuracy of each of these methods.

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Th 1 d

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The 1st derivative

(Possible) Definition of the derivative (or differentialoperator d

dx)

f(x) df

dx(x) = lim

h0

f(x+h) f(x)

h

Use differenceoperator to approximate differential

operator

f(x) =df

dx(x) = lim

h0

f(x+h) f(x)

h

f(x+h) f(x)

h

can now compute an approximation off(x) simply byevaluating f (twice).

This is called the forward differencebecause we use f(x)and f(x+h).

Important questions: How accurate isthisapproximation?9 / 4 7

A f h f d diff

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Accuracy of the forward difference

Formal derivation using the Taylor series off around x

f(x+h) = n=0

hn f(n)

(x)n!

= f(x) +hf(x) +h2f(x)

2! +h3

f(x)3!

+. . .

Rearranging for f(x)

hf(x) =f(x+h) f(x) h2f(x)

2! h3

f(x)3!

. . .

f(x) =

1

h f(x+h) f(x) h2 f

(x)

2!

h

3 f(x)

3!

. . .=

f(x+h) f(x)

h

h2f(x)2! h

3 f(x)3!

h . . .

= f(x+h) f(x)

h

hf(x)

2! h2

f(x)

3! . . .

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A f h f d diff (2)

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Accuracy of the forward difference (2)

f(x) = f(x+h) f(x)

h hf(x)

2! h2

f(x)

3! . . . Eforw(h)

f(x) = f(x+h) f(x)

h +Eforw(h)

Therefore, the error term Eforw(h) is

Eforw(

h) =

hf(x)

2! h2

f(x)

3! . . .

Can also be expressed as

f(x) = f(x+h) f(x)

h + O(h)

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Th 1 t d i ti i th b k d diff

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The 1st derivative using the backward difference

Another definition of the derivative (or differential

operator ddx)

df

dx(x) = lim

h0

f(x) f(x h)

h

Use difference operator to approximate differentialoperator

df

dx(x) = lim

h0

f(x) f(x h)

h

f(x) f(x h)

h

This is called the backward differencebecause we usef(x) and f(x h).

How accurate is the backward difference?

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A f th b k d diff

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Accuracy of the backward difference

Formal derivation using the Taylor Series off around x

f(x h) = f(x) hf(x) +h2 f(x)2!

h3 f(x)3!

+. . .

Rearranging for f(x)

hf(x) =f(x) f(x h) +h2 f(x)2!

h3 f(x)3!

. . .

f(x) = 1

h

f(x) f(x h) +h2

f(x)2!

h3f(x)

3! . . .

= f(x) f(x h)

h +

h2f(x)2! h

3 f(x)3!

h . . .

= f(x) f(x h)

h +h

f(x)2!

h2f(x)

3! . . .

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A f th b k d diff (2)

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Accuracy of the backward difference (2)

f(x) = f(x) f(x h)

h +hf(x)

2! h2f(x)

3! . . . Eback(h)

f(x) = f(x) f(x h)

h +Eback(h) (1)

Therefore, the error term Eback(h) is

Eback(h) =hf(x)

2!

h2f(x)

3!

. . .

Can also be expressed as

f(x) = f(x) f(x h)

h + O(h)

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Combining backward and forward differences (1)

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Combining backward and forward differences (1)

The approximations are

forward:

f(x) = f(x+h) f(x)

h +Eforw(h) (2)

backward

f(x) = f

(x

) f

(x h

)h +Eback(h) (3)

Eforw(h) = hf(x)

2! h2

f(x)3!

h3f(x)

4! h4

f(x)5!

. . .

Eback(h) = hf(x)

2! h2

f(x)3!

+h3f(x)

4! h4

f(x)5!

+. . .

Add equations (2) and (3) together, then the error cancels

partly! 15/47

Combining backward and forward differences (2)

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Combining backward and forward differences (2)

Add these lines together

f(x) =

f(x+h) f(x)

h +Eforw(h)

f(x) = f(x) f(x h)

h +Eback(h)

2f(x) = f(x+h) f(x h)

h +Eforw(h) +Eback(h)

Adding the error terms:

Eforw(h) +Eback(h) = 2h2 f

(x)3!

2h4f(x)

5! . . .

The combined (central) difference operator is

f(x) = f(x+h) f(x h)

2h +Ecent(h)

with

Ecent(h) = h2f(x)

3! h4f(x)

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Central difference

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Central difference

Can be derived (as on previous slides) by adding forwardand backward difference

Can also be interpreted geometrically by defining thedifferential operator as

df

dx(x) = lim

h

0

f(x+h) f(x h)

2h

and taking the finite difference form

df

dx(x)

f(x+h) f(x h)

2h

Error of the central difference is only O

(h2

), i.e. betterthan forward or backward difference

It is generally the case that symmetric differences

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Example (1)

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Example (1)

Using forward difference to estimate the derivative off(x) = exp(x)

f(x) fforw = f(x+h) f(x)

h

=exp(x+h) exp(x)

hNumerical example:

h= 0.1, x= 1

f(1) fforw(1.0) = exp(1.1)exp(1)

0.1 = 2.8588

Exact answers is f(1.0) = exp(1) = 2.71828

(Central diff: fcent(1.0) = exp(1+0.1)exp(10.1)

0.2 = 2.72281)

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Example (2)

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Example (2)

Comparison: forward difference, central difference and exact

derivative off(x) = exp(x)

01

2 34 5

x

0

2 0

4 0

6 0

8 0

1 0 0

1 2 0

1 4 0

d

f

/

d

x

(

x

)

A p p r o x i m a t i o n s o f d f / d x f o r f ( x ) = e x p ( x )

f o r w a r d h = 1

c e n t r a l h = 1

f o r w a r d h = 0 . 0 0 0 1

e x a c t

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Summary

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Summary

Can approximate derivatives offnumerically using onlyfunction evaluations off

size of step hvery important

central differences has smallest error term

name formula error

forward f(x) = f(x+h)f(x)h

O(h)

backward f(x) = f(x)f(xh)h

O(h)

central f(x) = f(x+h)

f(x

h)

2h O(h2)

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Appendix: source to compute figure on page 19 I

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Appendix: source to compute figure on page19I

EPS=1 # v er y l ar ge E PS to p ro vo ke i na c cu ra cy

def f o r w a r d d i f f ( f , x , h = E P S ) :# df / dx = ( f (x +h )- f( x) )/ h + O (h )

return ( f (x + h) -f ( x) )/ h

def b a c k w a r d d i f f ( f , x , h = E P S ) :

# df / dx = ( f (x )- f(x -h ) )/ h + O (h )

return ( f (x ) -f (x - h) )/ h

def c e n t r a l d i f f ( f , x , h = E P S ) :

# df / dx = ( f( x+ h) - f (x -h ))/ h + O (h ^2)

return ( f ( x +h ) - f ( x -h ) ) /( 2* h )

if _ _n a me _ _ = = " _ _ m a in _ _ " :

# c r ea te e xa mp le p lo t

import pylab

import n um py as np

a =0 # l ef t a nd

b =5 # r ig ht l im it s f or x

N=11 # s t e p s

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Appendix: source to compute figure on page 19 II

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Appendix: source to compute figure on page19II

def f ( x ) :

" "" O u r t es t f un ti on w it h

c o n ve n ie n t p r op e rt y t ha t

df / dx = f "" " return n p . e x p ( x )

x s = n p . l i n s p a c e ( a , b , N )

f or wa rd = []

f o rw a rd _ sm a ll _ h = [ ]

c en tr al = []

for x in xs :

f o rw a rd . a p pe n d ( f o r wa r d di f f ( f , x ) )

c e nt r al . a p pe n d ( c e n tr a l di f f ( f , x ) )

f o r w a r d _ s m a l l _ h . a p p e n d (

f o r w a r d d i f f ( f , x , h = 1 e - 4 ) )

p y l a b . f i g u r e ( f i g s i z e = ( 6 , 4 ) )p y l a b . a x i s ( [ a , b , 0 , n p . e x p ( b ) ] )

p y l ab . p l ot ( x s , f o rw a rd , ^ , l a b e l = f o r w ar d h = % g % E P S )

p y l ab . p l ot ( x s , c e nt r al , x , l a b e l = c e n t ra l h = % g % E P S )

pylab.plot(xs,forward_small_h ,o,

l a be l = f o r wa r d h = % g % 1 e - 4)

x s fi n e = n p . l i ns p ac e ( a , b , N * 10 0 )

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Appendix: source to compute figure on page 19 III

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Appendix: source to compute figure on page19III

p y l a b . p l o t ( x s f i n e , f ( x s f i n e ) , - , l a b e l = e x a c t )

p y l a b . g r i d ( )

p y la b . l e g en d ( l o c = u pp e r l ef t )

p y l a b . x l a b e l ( " x " )

p y l a b . y l a b e l ( " d f / d x ( x ) " )p y la b . t i t le ( " A p p r o x im a t io n s o f d f / dx f or f ( x ) = e xp ( x ) " )

p y l a b . p l o t ( )

pylab. savefig( central -and- forward - difference .pdf )

p y l a b . s h o w ( )

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Note: Eulers (integration) method derivation

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Note: Euler s (integration) method derivation

using finite difference operator

Use forward difference operator to approximatedifferential operator

dy

dx(x) = lim

h0

y(x+h) y(x)

h

y(x+h) y(x)

h

Change differential to difference operator in dydx

=f(x, y)

f(x, y) = dy

dx

y(x+h) y(x)

h

hf(x, y) y(x+h) y(x)= yi+1 = yi+hf(xi, yi)

Eulers method (for ODEs) can be derived from theforward difference operator.

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Note: Newtons (root finding) method

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Note: Newton s (root finding) method

derivation from Taylor series

We are looking for a root, i.e.

we are looking for a x so thatf(x) = 0.

We have an initial guess x0 which we refine in subsequent iterations:

xi+1=xi hi where hi = f(xi)

f(xi). (4)

.

This equation can be derived from the Taylor series off around x.Suppose we guess the root to be at x and x+h is the actuallocation of the root (so h is unknown and f(x+h) = 0):

f(x

+h

) = f

(x

) +hf

(x

) +. . .

0 = f(x) +hf(x) +. . .

= 0 f(x) +hf(x)

h f(x)

f(x). (5)

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The diffusion equation

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Diffusion equation

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q

The 2d operator

2

x2

+

2

y2

is called the Laplaceoperator, so that we can also write

u

t =D

2

x2+

2

y2

= Du

The diffusion equation (with constant diffusioncoefficient D) reads u

t =Duwhere the Laplace

operator depends on the number dof spatial dimensions

d= 1: = 2

x2

d= 2: = 2x2

+ 2y2

d= 3: = 2

x2+

2

y2+

2

z2

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1d Diffusion equation ut = D ux2

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q t x2

In one spatial dimension, the diffusion equation reads

ut

=D 2u

x2

This is the equation we will use as an example.

Lets assume an initial concentrationu(x, t0) =

12

exp

(xxmean)2

2

with xmean= 0and

width = 0.5.

64

2 0 24

6

x

0 . 0

0 . 1

0 . 2

0 . 3

0 . 4

0 . 5

0 . 6

0 . 7

0 . 8

u

(

x

,

t

_

0

)

( x , t

0

)

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1d Diffusion eqn ut = D ux2

, time integration I

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q t x2 , g

Let us assume that we have some way of computingD 2u

x2 at time t0 and lets call this g(x, t0), i.e.

g(x, t0) D2u(x, t0)

x2

We like to solve

u(x, t)

t =g(x, t0)

to compute u(x, t1) at some later time t1.

Use finite difference time integration scheme:

Introduce a time step size hso that t1 =t0+h.

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1d Diffusion eqn ut = D ux2

, time integration II

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q t x2 , g

Change differential operator to forward difference operator

g(x, t0) = u(x, t)t

= limh0

u(x, t0+h) u(x, t0)h

(6)

u(x, t0+h) u(x, t0)

h (7)

Rearrange to find u(x, t1) u(x, t0+h) gives

u(x, t1) u(x, t0) +hg(x, t0)

We can generalise this using ti=t0+ih to read

u(x, ti+1) u(x, ti) +hg(x, ti) (8)

If we can find g(x, ti), we can compute u(x, ti+1)

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1d Diffusion eqn ut = D ux2 , spatial part I

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q t x p p

u

t =D

2u

x2 =g(x, t)

Need to compute g(x, t) =D 2u(x,t)x2

for a givenu(x, t).

Can ignore the time dependence here, and obtain

g(x) =D2u(x)

x2 .

Recall that

2

ux2 = x ux

and we that know how to compute ux

using centraldifferences.

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Second order derivatives from finite differences I

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Recall central difference equation for first order derivative

df

dx

(x) f(x+h) f(x h)

2hwill be more convenient to replace hby 1

2h:

df

dx(x)

f(x+ 12

h) f(x 12

h)

h

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Second order derivatives from finite differences II

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Apply the central difference equation twice to obtain d2f

dx2:

d2f

dx2(x) =

d

dx

df

dx(x)

d

dx

f(x+ 1

2h) f(x 1

2h)

h

= 1

h

d

dxf

x+

1

2h

d

dxf

x

1

2h

1

h f(x+h) f(x)

h

f(x) f(x h)

h =

f(x+h) 2f(x) +f(x h)

h2 (9)

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Recipe to solve ut = D ux2

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t x

1 Discretise solution u(x, t) into discrete values2 uij u(xj, ti)where

xj x0+jx andti t0+it.

3 Start with time iteration i= 04 Need to know configuration u(x, ti).

5 Then compute g(x, ti) =D2ux2

using finite differences(9).

6 Then compute u(x, ti+1) based on g(x, ti) using (8)7 increase ito i+ 1, then go back to5.

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A sample solution ut = D ux2 ,I

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x

import n um py as np

import m a t pl o t li b . p y p l ot a s p lt

import m a t p lo t l ib . a n i ma t io n a s a n i ma t io n

a , b = - 5 , 5 # size of box

N = 51 # n um be r of s u bd i vi s io n s

x = n p . l i n s p a c e ( a , b , N ) # p o si t i on s o f s u bd i v is i o ns

h = x [ 1 ] - x [ 0 ] # d i s c r et i s at i o n s t ep s iz e i n x - d i r e c ti o n

def t o t a l ( u ) :" "" C o m pu te s t ot al n um be r of m ol es i n u . "" "

return ( ( b - a ) / f l o a t ( N ) * n p . s u m ( u ) )

def g a u s s d i s t r ( m e a n , s i g m a , x ) :

" "" R e tu rn g au ss d i st r ib u ti o n f or g iv en n um py a rr ay x " ""

return 1 . / ( s i g m a * n p . s q r t ( 2 * n p . p i ) ) * n p . e x p (

- 0 . 5 * ( x - m e a n ) * * 2 / s i g m a * * 2 )

# s t a r ti n g c o n fi g u ra t i on f or u ( x , t0 )

u = g a u ss d is t r ( m e an = 0 . , s i gm a = 0 .5 , x = x )

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A sample solution ut = D ux2 ,II

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def c om pu te _g ( u , D , h ):

" "" g i ve n a u (x , t ) in a rr ay , c om pu te g (x , t )= D * d ^2 u / dx ^ 2

u si ng c en tr al d if f er e nc es w it h s pa ci ng h ,

a nd r et ur n g (x , t ). " "" d 2 u_ d x2 = n p . z er o s ( u . sh ap e , n p . f lo a t )

for i in r a n g e ( 1 , l e n ( u ) - 1 ) :

d 2u _d x2 [ i ] = ( u [ i +1 ] - 2* u [ i ]+ u [ i -1 ]) / h * *2

# s p ec ia l c as es a t b ou nd ar y : a ss um e N eu ma n b ou nd ar y

# c o nd it io ns , i . e . no c ha ng e o f u o ve r b ou nd ar y

# so t ha t u [ 0] - u [ - 1] =0 a nd t hu s u [ - 1] = u [0 ]

i =0d 2u _d x2 [ i ] = ( u [i + 1] - 2 * u[ i ]+ u [ i ]) / h ** 2

# s am e at o th er e nd so t ha t u [N -1] - u [ N ]=0

# a nd t hu s u [ N ]= u [ N -1 ]

i = l e n ( u ) - 1

d 2u _d x2 [ i ] = ( u [i ] - 2* u [ i ]+ u [ i - 1] )/ h * *2

return D * d 2 u _ d x 2

def a d va n ce _ ti m e ( u , g , dt ) :

"" " Gi ve n t he a rr ay u , t he r at e of c ha ng e a rr ay g ,

an d a t im es te p dt , c om pu te th e s ol ut io n fo r u

a ft er t , u si ng s im pl e E ul er m et ho d . " ""

u = u + d t * g

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A sample solution ut = D ux2 ,III

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return u

# s ho w e xa mp le , q ui ck a nd d ir tl y , l ot s o f g lo ba l v ar i ab le s

dt = 0.01 # s te p s iz e or t imes t e p sb e f o re u p d at i n g gr a p h = 2 0 # p l ot ti ng is s lo w

D = 1. # D i f f us i on c o e ff i c ie n t

s te ps do ne = 0 # k ee p t ra ck of i te r at io ns

def d o _ s t e p s ( j , n s t e p s = s t e p s b e f o r e u p d a t i n g g r a p h ) :

" "" F u n ct io n c al le d by F u nc A ni m at io n c la ss . C om pu te s

n s te p s i te r at i on s , i . e . c a rr i es f o rw a rd s o lu t io n f ro m u ( x , t _i ) t o u ( x , t_ { i + n s t ep s } ) .

"""

global u , s t e p s d o n e

for i in r a n g e ( n s t e p s ) :

g = com pute_g ( u , D , h )

u = a dv an ce _t im e ( u , g , dt )

s te ps do ne += 1

t im e_ pa ss ed = s te ps do ne * dt

print ( " s t e ps d o ne = %5 d , t im e = % 8 gs , t o ta l ( u ) = %8 g " %

(stepsdone ,time_passed ,total(u )))

l . s e t _ y d a t a ( u ) # upd at e data in plot

f i g 1 . c a n v a s . d r a w ( ) # r ed ra w th e c an va s

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A sample solution ut = D ux2 ,IV

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return l ,

f ig 1 = p lt . f i gu re ( ) # s e t up a n im a ti o n

l , = p lt . p l o t ( x ,u , b - o ) # p lo t in it ia l u( x ,t )

# t he n c om pu te s ol ut io n a nd a ni ma tel i ne _ an i = a n im a t io n . F u n c A n im a t io n ( f i g1 ,

d o _s t ep s , r a n ge ( 1 0 0 0 0) )

p l t . s h o w ( )

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Boundary conditions I

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For ordinary differential equations (ODEs), we need toknow theinitial value(s)to be able to compute a solution.

For partial differential equations (PDEs), we need toknow the initial values and extra information about the

behaviour of the solution u(x, t)at the boundary of thespatial domain (i.e. at x= aand x=b in this example).

Commonly used boundary conditions are

Dirichlet boundary conditions: fix u(a) =c to some

constant.Would correspond here to some mechanism that keepsthe concentration u at position x=a constant.

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Boundary conditions II

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Neuman boundary conditions: fix the change ofu acrossthe boundary, i.e.

u

x(a) =c.

For positive/negative c this corresponds to an imposedconcentration gradient.

For c= 0, this corresponds to conservation of the atomsin the solution: as the gradient across the boundarycannot change, no atoms can move out of the box.

(Used in our program on slide 35)

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Numerical issues

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The time integration scheme we use isexplicitbecause wehave an explicit equation that tells us how to computeu(x, ti+1) based on u(x, ti) (equation (8) on slide 30)

An implicit scheme would compute u(x, ti+1) based onu(x, ti) and on u(x, ti+1).

The implicit scheme is more complicated as it requiressolving an additional equation system just to findu(x, ti+1) but allows larger step sizes t for the time.

The explicit integration scheme becomes quickly unstableift is too large. t depends on the chose spatialdiscretisation x.

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Performance issues

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Our sample code is (nearly) as slow as possibleinterpreted languageexplicit for loopsenforced small step size from explicit scheme

Solutions:Refactor for-loops into matrix operations and use(compiled) matrix library (numpy for small systems, usescipy.sparse for larger systems)Use library function to carry out time integration (will

use implicit method if required), for examplescipy.integrate.odeint.

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Finite Elements

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Another widely spread way of solving PDEs is using so-calledfinite elements.

Mathematically, the solution u(x) for a problem like2ux2

=f(x) is written as

u(x) =N

i=1uii(x) (10)

where each ui is a number (a coefficient), and each i(x)a known function of space.The i are called basisorshape functions.Each i is normally chosen to be zero for nearly all x, and

to be non-zero close to a particular node in the finiteelement mesh.By substitution (10) into the PDE, a matrix system canbe obtained, which if solved provides the coefficients

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Finite Elements vs Finite differences

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Finite differencesare mathematically much simpler andfor simple geometries (such as cuboids) easier toprogram

Finite elementshave greater flexibility in the shape of the domain,the specification and implementation of boundaryconditions is easierbut the basic mathematics and code is morecomplicated.

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Practical observation on time integration

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Usually, we solve thespatialpart of a PDE using somediscretisation scheme such as finite differences and finiteelements).

This results in a set of coupled ordinary differential

equations (where time is the independent variable).Can think of this as one ODE for every cube from ourdiscretisation.

This temporalpart is then solved using time integration

schemes for (systems of) ordinary differential equations.

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Summary

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Partial differential equations important in many contexts

If no analytical solution known, use numerics.

Discretise the problem through

finite differences (replace differential with differenceoperator, corresponds to chopping space and time in

little cuboids)finite elements (project solution on localised basisfunctions, often used with tetrahedral meshes)related methods (finite volumes, meshless methods).

Finite elements and finite difference calculations arestandard tools in many areas of engineering, physics,chemistry, but increasingly in other fields.

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changeset: 53:a22b7f13329etag: tipuser: Hans Fangohr [MBP13] date: Fri Dec 16 10:57:15 2011 +0000

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date: Fri Dec 16 10:57:15 2011 +0000

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