Finite Differences

Finite Difference Approximations

Simple geophysical partial differential equationsFinite differences - definitionsFinite-difference approximations to pde‘s

ExercisesAcoustic wave equation in 2DSeismometer equationsDiffusion-reaction equation

Finite differences and Taylor ExpansionStability -> The Courant CriterionNumerical dispersion

1Computational Seismology

dxxfdxxffx)()( −+

≈∂ +

Finite Differences

)( 222

22

zyx

ts pcp

∂+∂+∂=Δ

+Δ=∂ The acoustic wave 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

c acoustic wave speed

s sources

pRCCCkCt +−∇•−Δ= v∂

C tracer concentration

k diffusivity

v flow velocity

R reactivity

p sources

C tracer concentration

k diffusivity

v flow velocity

R reactivity

p sources

Partial Differential Equations in Geophysics

Finite Differences

Finite differences

Finite volumes

- time-dependent PDEs- seismic wave propagation- geophysical fluid dynamics- Maxwell’s 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

Finite Differences

Particle-based methods

Pseudospectralmethods

- 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

Finite Differences

What is a finite difference?

Common definitions of the derivative of f(x):

dxxfdxxff

dxx)()(lim

0

−+=∂

→

dxdxxfxff

dxx)()(lim

0

−−=∂

→

dxdxxfdxxff

dxx 2)()(lim

0

−−+=∂

→

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

But we want dx to remain FINITE

Finite Differences

What is a finite difference?

The equivalent approximations of the derivatives are:

dxxfdxxffx)()( −+

≈∂ +

dxdxxfxffx

)()( −−≈∂ −

dxdxxfdxxffx 2

)()( −−+≈∂

forward difference

backward difference

centered difference

Finite Differences

The big question:

How good are the FD approximations?

This leads us to Taylor series....

Finite Differences

Taylor Series

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

What happens, if we use this expression for

dxxfdxxffx)()( −+

≈∂ + ?

...)(!4

)(!3

)(!2

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

'''3

''2

' ±+±+±=± xfdxxfdxxfdxxfxfdxxf

Finite Differences

Taylor Series

... that leads to :

The error of the first derivative using the forwardformulation is of order dx.

Is this the case for other formulations of the derivative?Let’s check!

)()(

...)(!3

)(!2

)(dx1)()(

'

'''3

''2

'

dxOxf

xfdxxfdxxfdxdx

xfdxxf

+=

⎥⎦

⎤⎢⎣

⎡+++=

−+

Finite Differences

... 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

'

dxOxf

xfdxxfdxdx

dxxfdxxf

+=

⎥⎦

⎤⎢⎣

⎡++=

−−+

Finite Differences

xj − 1 xj xj + 1 xj + 2 xj + 3

f xj( )

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

Finite Differences

Lets’ try Taylor’s Expansion

f x( )

dx

x

f(x)

dxxfxfdxxf )(')()( +≈+ (1)(2)

we are looking for something like

f x w f xiji

index jj L

( ) ( )( )

,( ) ( )≈ ∑

=1

Alternative Derivation

dxxfxfdxxf )(')()( −≈−

Finite Differences

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 to a system of equations which can be cast in matrix form

dxbaba

/10

=−=+

01

=−=+

baba

Interpolation Derivative

2nd order weights

Finite Differences

Taylor Operators

... in matrix form ...

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛− 0

111

11ba

Interpolation Derivative

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

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛− dxb

a/10

1111

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

01

1111 1

ba

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

dxba

/10

1111 1

Finite Differences

... and the result ...

Interpolation Derivative

Can we generalise this idea to longer operators?

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛2/12/1

ba

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛1

12

1dxb

a

Let us start by extending the Taylor expansion beyond f(x±dx):

Interpolation and difference weights

Finite Differences

'''!3

)2(''!2)2(')2()2(

32

fdxfdxfdxfdxxf −+−≈−*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

fdxfdxfdxfdxxf −+−≈−

'''!3)(''

!2)(')()(

32

fdxfdxfdxfdxxf +++≈+

'''!3

)2(''!2)2(')2()2(

32

fdxfdxfdxfdxxf +++≈+

Higher order operators

Finite Differences

≈+++ +++−−− dfcfbfaf

++++ )( dcbaf

+++−− )22(' dcbadxf

++++ )222

2(''2 dcbafdx

)68

61

61

68('''3 dcbafdx ++−−

... we can now ask for the coefficients a,b,c,d, so that theleft-hand-side yields either f,f’,f’’,f’’’ ...

Higher order operators

Finite Differences

1=+++ dcba

022 =++−− dcba

0222

2 =+++ dcba

068

61

61

68

=++−− dcba

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

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

Linear system

Finite Differences

High-order interpolation

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−

−−

0001

6/86/16/16/822/12/1221121111

dcba

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

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−

−

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

6/13/23/26/1

dcba

... Interpolation ...

Finite Differences

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−

−−

00/10

6/86/16/16/822/12/1221121111

dx

dcba

... with the result ...

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−

−=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

6/13/43/4

6/1

21dx

dcba

... first derivative ...

First derivative

Finite Differences

Our first FD algorithm (ac1d.m) !

)( 222

22

zyx

ts pcp

∂+∂+∂=Δ

+Δ=∂ P pressure

c acoustic wave speed

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 finite

Difference method.

Solution:Solution:

[ ]2

2

22

)()(2

)()(2)()(

sdtdttptp

dxxpxpdxxpdxdtcdttp

+−−+

−+−+=+

Finite Differences

Problems: Stability

[ ]2

2

22

)()(2

)()(2)()(

sdtdttptp

dxxpxpdxxpdx

dtcdttp

+−−+

−+−+=+

1≈≤ εdxdtc

Stability:

Careful analysis using harmonic functions shows that a stable numerical calculation is subject to special conditions (conditional stability). This holds for many numerical problems.(Derivation on the board).

Stability:

Careful analysis using harmonic functions shows that a stable numerical calculation is subject to special conditions (conditional stability). This holds for many numerical problems.(Derivation on the board).

Finite Differences

Problems: Dispersion

[ ]2

2

22

)()(2

)()(2)()(

sdtdttptp

dxxpxpdxxpdx

dtcdttp

+−−+

−+−+=+

Dispersion:

The numerical approximation has artificial dispersion,

in other words, the wave speed becomes frequency dependent (Derivation in the board). You have to find a frequency bandwidth where this effect is small.

The solution is to use a sufficient number of grid points per wavelength.

Dispersion:

The numerical approximation has artificial dispersion,in other words, the wave speed becomes frequency dependent (Derivation in the board). You have to find a frequency bandwidth where this effect is small.The solution is to use a sufficient number of grid points per wavelength.

True velocity

Finite Differences

Our first FD code!

[ ]2

2

22

)()(2

)()(2)()(

sdtdttptp

dxxpxpdxxpdx

dtcdttp

+−−+

−+−+=+

% Time stepping

for i=1:nt,

% FD

disp(sprintf(' Time step : %i',i));

for j=2:nx-1d2p(j)=(p(j+1)-2*p(j)+p(j-1))/dx^2; % space derivative

endpnew=2*p-pold+d2p*dt^2; % time extrapolationpnew(nx/2)=pnew(nx/2)+src(i)*dt^2; % add source termpold=p; % time levelsp=pnew;p(1)=0; % set boundaries pressure freep(nx)=0;

% Display plot(x,p,'b-')title(' FD ')drawnow

end

Finite Differences

Snapshot Example

0 1000 2000 3000 4000 50000

500

1000

1500

2000

2500

3000

Distance (km)

Tim

e (s

) Velocity 5 km/s

Finite Differences

Seismogram Dispersion

Finite Differences

Finite Differences - Summary

Conceptually the most simple of the numerical methods and can be learned quite quicklyDepending on the physical problem FD methods are conditionally stable (relation between time and space increment)FD methods have difficulties concerning the accurate implementation 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