02 finite differences per gadin

8/12/2019 02 Finite Differences Per Gadin

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Finite Differences

Finite Difference Approximations

Simple geophysical partial differential equationsFinite differences - definitionsFinite-difference approximations to pdes

Exercises Acoustic wave equation in 2DSeismometer equationsDiffusion-reaction equation

Finite differences and Taylor ExpansionStability -> The Courant CriterionNumerical dispersion

1Computational Seismology

dx x f dx x f f x )()( + +


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Finite Differences

)( 222

22

zyx

tspcp

++=

+= The acousticwave equation- seismology

- acoustics- oceanography- meteorology

Diffusion, advection,Reaction- geodynamics- oceanography- meteorology- geochemistry- sedimentology

- geophysical fluid dynamics

P pressure

c acoustic wave speed

s sources

P pressure


s sources

p RC C C k C t += v

C tracer concentrationk diffusivity

v flow velocity

R

reactivity

p sources

C tracer concentrationk diffusivity

v flow velocity

R reactivity

p sources

Partial Differential Equations in Geophysics


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Finite Differences

Finite differences

Finite volumes

- time-dependent PDEs- seismic wave propagation- geophysical fluid dynamics- Maxwells equations- Ground penetrating radar -> robust, simple concept, easy to

parallelize, regular grids, explicit method

Finite elements- static and time-dependent PDEs- seismic wave propagation- geophysical fluid dynamics- all problems-> implicit approach, matrix inversion, well founded,

irregular grids, more complex algorithms,engineering problems

- time-dependent PDEs- seismic wave propagation- mainly fluid dynamics-> robust, simple concept, irregular grids, explicit

method

Numerical methods: properties


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Finite Differences

Particle-basedmethods

Pseudo spectralmethods

- lattice gas methods- molecular dynamics- granular problems- fluid flow- earthquake simulations-> very heterogeneous problems, nonlinear problems

Boundary elementmethods

- problems with boundaries (rupture)

- based on analytical solutions- only discretization of planes-> good for problems with special boundary conditions

(rupture, cracks, etc)

- orthogonal basis functions, special case of FD- spectral accuracy of space derivatives- wave propagation, ground penetrating radar -> regular grids, explicit method, problems with

strongly heterogeneous media

Other numerical methods


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Finite Differences

What is a finite difference?

Common definitions of the derivative of f(x):

dx

x f dx x f

f dx x)()(

lim0+=

dx

dx x f x f f

dx x

)()(lim

0

=

dxdx x f dx x f

f dx

x 2)()(

lim0

+=

These are all correct definitions in the limit dx->0.

But we want dx to remain FINITE


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Finite Differences

What is a finite difference?

The equivalent approximations of the derivatives are:

dx x f dx x f f x )()( + +

dxdx x f x f f x

)()(

dxdx x f dx x f

f x 2)()( +

forward difference

backward difference

centered difference


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The big question :

How good are the FD approximations?

This leads us to Taylor series....



Taylor Series

Taylor series are expansions of a function f(x) for somefinite distance dx to f(x+dx)

What happens, if we use this expression for

dx x f dx x f

f x)()( +

+

?

...)(!4

)(!3

)(!2

)(dx)()( ''''4

'''3

''2

' ++= x f dx x f dx x f dx x f x f dx x f



Taylor Series

... that leads to :

The error of the first derivative using the forward

formulation is of order dx .

Is this the case for other formulations of the derivative?Lets check!

)()(

...)(!3

)(!2

)(dx1)()(

'

'''3

''2

'

dxO x f

x f dx

x f dx

x f dxdx

x f dx x f

+=

+++=+



... with the centered formulation we get:

The error of the first derivative using the centered

approximation is of order dx2.

This is an important results: it DOES matter which formulationwe use. The centered scheme is more accurate!

Taylor Series

)()(

...)(!3

)(dx1)2/()2/(

2'

'''3

'

dxO x f

x f dx x f dxdx

dx x f dx x f

+=

++=+



x j 1 x j x j + 1 x j + 2 x j + 3

f x j( )

dx h

desired x location

What is the (approximate) value of the function or its (first,second ..) derivative at the desired location ?

How can we calculate the weights for the neighboring points?

x

f(x)

Alternative Derivation



Lets try Taylors Expansion

f x( )

dx

x

f(x)

dx x f x f dx x f )(')()( ++ (1)(2)

we are looking for something like

f x w f xi j

i

index j j L

( ) ( )

( ),( ) ( )

=1

Alternative Derivation

dx x f x f dx x f )(')()(



deriving the second-order scheme

a f a f a f d x+ + '

b f b f b f d x ' + + + + a f b f a b f a b f d x( ) ( ) '

the solution to this equation for a and b leads toa system of equations which can be cast in matrix form

dxba

ba

/1

0

=

=+

0

1

=

=+

ba

baInterpolation Derivative

2nd order weights



Taylor Operators

... in matrix form ...

=

0

1

11

11

b

a

Interpolation Derivative

... so that the solution for the weights is ...

=

dxb

a

/1

0

11

11

=

0

1

11

11 1

b

a

=

dxb

a

/1

0

11

11 1


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'''!3

)2(''

!2)2(

')2()2(32

f dx

f dx

f dx f dx x f +*a |

*b |

*c |

*d |

... again we are looking for the coefficients a,b,c,d with whichthe function values at x(2)dx have to be multiplied in order

to obtain the interpolated value or the first (or second) derivative!

... Let us add up all these equations like in the previous case ...

'''!3)(''

!2)(')()(

32 f dx f dx f dx f dx x f +

'''

!3

)(''

!2

)(')()(

32

f dx

f dx

f dx f dx x f ++++

'''!3

)2(''

!2)2(

')2()2(32

f dx

f dx

f dx f dx x f ++++

Higher order operators



+++ +++ df cf bf af

++++ )( d cba f

+++ )22(' d cbadxf

++++ )222

2(''2 d cb

a f dx

)68

61

61

68

('''3

d cba f dx ++

... we can now ask for the coefficients a,b,c,d, so that the

left-hand-side yields either f,f,f,f ...

Higher order operators



1=+++ d cba

022 =++ d cba

0222

2 =+++ d cba

06

8

6

1

6

1

6

8 =++ d cba

... if you want the interpolated value ...

... you need to solve the matrix system ...

Linear system



High-order interpolation

=

0

00

1

6/86/16/16/8

22/12/122112

1111

d

cb

a

... with the result after inverting the matrix on the lhs ...

=

6/1

3/2

3/2

6/1

d

c

b

a

... Interpolation ...



=

0

0/1

0

6/86/16/16/8

22/12/122112

1111

dx

d

cb

a

... with the result ...

=

6/1

3/4

3/4

6/1

21dx

d

c

b

a

... first derivative ...

First derivative



Our first FD algorithm (ac1d.m) !

)( 222

22

zyx

tspcp

++=

+= P pressure


s sources

P pressurec

acoustic wave speed

s sources

Problem:

Solve the 1D acoustic wave equation using the finite

Difference method.

Problem:

Solve the 1D acoustic wave equation using the finiteDifference method.

Solution:Solution:

[ ]

2

2

22

)()(2

)()(2)()(

sdt dt t pt p

dx x p x pdx x pdx

dt cdt t p

++

++=+


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Problems: Dispersion

[ ]2

2

22

)()(2

)()(2)()(

sdt dt t pt p

dx x p x pdx x pdx

dt cdt t p

++

++=+

Dispersion:

The numericalapproximation has

artificial dispersion, in other words, the wavespeed becomes frequencydependent (Derivation inthe board).You have to find afrequency bandwidthwhere this effect is small.

The solution is to use a

sufficient number of gridpoints per wavelength.

Dispersion:

The numericalapproximation hasartificial dispersion,in other words, the wavespeed becomes frequencydependent (Derivation inthe board).

You have to find afrequency bandwidthwhere this effect is small.The solution is to use asufficient number of gridpoints per wavelength.

True velocity



Our first FD code!

[ ]2

2

22

)()(2

)()(2)()(

sdt dt t pt p

dx x p x pdx x pdx

dt cdt t p

++

++=+

% Ti me st eppi ng

f o r i =1: nt ,

% FD

di s p( s pr i nt f ( ' Ti me st ep : %i ' , i ) ) ;

f o r j =2: nx- 1d2p( j ) =( p( j +1) - 2*p( j ) +p( j - 1) ) / dx 2; % space der i vat i ve

endpnew=2*p- pol d+d2p*dt 2; % t i me ext r apol at i onpnew( nx/ 2) =pnew( nx/ 2) +sr c( i ) *dt 2; % add sour ce t er mpol d=p; % t i me l evel s

p=pnew;p( 1) =0; % set boundar i es pr essur e f r eep( nx) =0;

% Di spl aypl ot ( x, p, ' b- ' )t i t l e( ' FD ' )dr awnow

end



Snapshot Example

0 1000 2000 3000 4000 50000

500

1000

1500

2000

2500

3000

Distance (km)

T i m

e ( s )

Velocity 5 km/s


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Finite Differences - Summary

Conceptually the most simple of the numerical methods and

can be learned quite quicklyDepending on the physical problem FD methods areconditionally stable (relation between time and spaceincrement)

FD methods have difficulties concerning the accurateimplementation of boundary conditions (e.g. free surfaces,absorbing boundaries)FD methods are usually explicit and therefore very easy to

implement and efficient on parallel computersFD methods work best on regular, rectangular grids

02 finite differences per gadin

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