Finite difference migration - Stanford...

Finite difference migration

Monday, December 3, 12

Parabolic equation

kz =�

v

�1� v2k2

x

�2P


Parabolic equation

kz =�

v

�1� v2k2

x

�2P

ikz =�

�z

Substitute definition of a derivative and multiply by p


Parabolic equation

�P

�z=

i�

v(z)

�1� v(z)2k2

x

�2P

kz =�

v

�1� v2k2

x

�2P

ikz =�

�z

Substitute definition of a derivative and multiply by p


Parabolic equation

�P

�z=

i�

v(z)

�1� v(z)2k2

x

�2P


Parabolic equation

�P

�z=

i�

v(z)

�1� v(z)2k2

x

�2P

We can’t construct a finite difference

stencil of a square root


Parabolic equation

�P

�z=

i�

v(z)

�1� v(z)2k2

x

�2P

�P

�z=

i�

v(z)

�1� v(z)2k2

x

2�2+

v(z)4k4x

8�4+ ...

�P

Replace with the power series for

square root


Parabolic equation

�P

�z=

i�

v(z)

�1� v(z)2k2

x

�2P

�P

�z=

i�

v(z)

�1� v(z)2k2

x

2�2+

v(z)4k4x

8�4+ ...

�P

Truncate after the second term


Parabolic equation

�P

�z=

i�

v(z)

�1� v(z)2k2

x

�2P

�P

�z=

i�

v(z)

�1� v(z)2k2

x

2�2+

v(z)4k4x

8�4+ ...

�P

�P

�z=

iw

v(x, z)P +

v(x, z)�iw2

�2P

�x2


Parabolic equation

�P

�z=

i�

v(z)

�1� v(z)2k2

x

�2P

�P

�z=

i�

v(z)

�1� v(z)2k2

x

2�2+

v(z)4k4x

8�4+ ...

�P

�P

�z=

iw

v(x, z)P +

v(x, z)�iw2

�2P

�x2

Error in approximation increases with

propagation angle


2-D Heat equationq -temperaturet - time - conductivityx,y space�

@T

@t

=

✓�

@

2

@x

2+ �

@

2

@y

2

◆T


Splittingq -temperaturet - time - conductivityx,y space�

@T

@t

=

✓�

@

2

@x

2+ �

@

2

@y

2

◆T

@T

@t

= 2�@

2T

@x

2

@T

@t= 2�

@2T

@y2

Iterate {

}


SPLITTING

INITIAL SOLVE IN X SOLVE IN Y


Parabolic equation@P

@z

=i!

v(x, z)P +

v(x, z)

�i!2

@

2P

@x

2



@z

=i!

v(x, z)P +

v(x, z)

�i!2

@

2P

@x

2

@P

@z

=i!

v(z)P + i!

✓1

v(x, z)� 1

v(z)

◆P +

v(x, z)

�i!2

@

2P

@x

2



@z

=i!

v(x, z)P +

v(x, z)

�i!2

@

2P

@x

2

@P

@z

=i!

v(z)P + i!

✓1

v(x, z)� 1

v(z)

◆P +

v(x, z)

�i!2

@

2P

@x

2

Shift



@z

=i!

v(x, z)P +

v(x, z)

�i!2

@

2P

@x

2

@P

@z

=i!

v(z)P + i!

✓1

v(x, z)� 1

v(z)

◆P +

v(x, z)

�i!2

@

2P

@x

2

Shift Thin lens



@z

=i!

v(x, z)P +

v(x, z)

�i!2

@

2P

@x

2

@P

@z

=i!

v(z)P + i!

✓1

v(x, z)� 1

v(z)

◆P +

v(x, z)

�i!2

@

2P

@x

2

Shift Thin lens Diffraction


Explicit vs implicitdq

dt= 2rq

qt+1 � qt = 2r�tqt+1 + qt

2



dt= 2rq

q(t+�t)� q(t)

�t= 2rq


2



dt= 2rq

q(t+�t)� q(t)

�t= 2rq

0

1

2

3


2



dt= 2rq

q(t+�t)� q(t)

�t= 2rq

0

1

2

3


2



dt= 2rq

q(t+�t)� q(t)

�t= 2rq

0

1

2

3


2



dt= 2rq

q(t+�t)� q(t)

�t= 2rq

0

1

2

3


2

(1� r�t)qt+1 + (1 + r�t)qt = 0


1-D Heat equation: Difference approximation

q -temperaturet - time - conductivityx- space

⇥q

⇥t= �

⇥2q

⇥x2�

qt+1 � qt

�tqx+1 � 2qx + qx�1

�x2


Discrete space at initial t

x

t0


Calculate right sidex

⇥q

⇥t= �

⇥2q

⇥x2

1 1-2

Approximates derivative at location

t0


Combine two filters

x

t

t0

t1

⇥q

⇥t= �

⇥2q

⇥x2


Combine two filters

x

t

t0

t1

⇥q

⇥t= �

⇥2q

⇥x2

Known values

Unknown values


x

t

t0

t1

⇥q

⇥t= �

⇥2q

⇥x2

qxt+1 � qx

t = ⇥qx+1t � 2qx

t + qx�1t

�x2

qxt+1 � qx

t � �(qx+1t � 2qx

t + qx�1t ) = 0

Rewriting equations


Heat flowreal q(12),d(12)nx=12; a = 8.; write(6,'(/"a =",f5.2)') a; alpha = .5*ado ix= 1,6 { q(ix) = 0.} # Initial temperature stepdo ix= 7,12 { q(ix) = 1.}do it= 1,4 { write(6,'(20f6.2)') (q(ix),ix=1,nx) d(1) = 0.; d(nx) = 0. do ix= 2, nx-1 d(ix) = q(ix) + alpha*(q(ix-1)-2.*q(ix)+q(ix+1)) call rtris( nx, alpha, -alpha, (1.+2.*alpha), -alpha, alpha, d, q) }call exit(0); end


Different versions of alpha

alpha = 0.33 0.00 0.00 0.00 0.00 0.00 0.00 1.00 1.00 1.00 1.00 1.00 1.00 0.00 0.00 0.00 0.00 0.00 0.33 0.67 1.00 1.00 1.00 1.00 1.00 0.00 0.00 0.00 0.00 0.11 0.33 0.67 0.89 1.00 1.00 1.00 1.00 0.00 0.00 0.00 0.04 0.15 0.37 0.63 0.85 0.96 1.00 1.00 1.00 0.00 0.00 0.01 0.06 0.19 0.38 0.62 0.81 0.94 0.99 1.00 1.00 0.00 0.00 0.02 0.09 0.21 0.40 0.60 0.79 0.91 0.98 1.00 1.00

alpha = 0.67 0.00 0.00 0.00 0.00 0.00 0.00 1.00 1.00 1.00 1.00 1.00 1.00 0.00 0.00 0.00 0.00 0.00 0.67 0.33 1.00 1.00 1.00 1.00 1.00 0.00 0.00 0.00 0.00 0.44 0.00 1.00 0.56 1.00 1.00 1.00 1.00 0.00 0.00 0.00 0.30 -0.15 0.96 0.04 1.15 0.70 1.00 1.00 1.00 0.00 0.00 0.20 -0.20 0.89 -0.39 1.39 0.11 1.20 0.80 1.00 1.00 0.13 0.13 -0.20 0.79 -0.69 1.65 -0.65 1.69 0.21 1.20 0.87 0.87


Different versions of alpha

alpha = 0.33 0.00 0.00 0.00 0.00 0.00 0.00 1.00 1.00 1.00 1.00 1.00 1.00 0.00 0.00 0.00 0.00 0.00 0.33 0.67 1.00 1.00 1.00 1.00 1.00 0.00 0.00 0.00 0.00 0.11 0.33 0.67 0.89 1.00 1.00 1.00 1.00 0.00 0.00 0.00 0.04 0.15 0.37 0.63 0.85 0.96 1.00 1.00 1.00 0.00 0.00 0.01 0.06 0.19 0.38 0.62 0.81 0.94 0.99 1.00 1.00 0.00 0.00 0.02 0.09 0.21 0.40 0.60 0.79 0.91 0.98 1.00 1.00

alpha = 0.67 0.00 0.00 0.00 0.00 0.00 0.00 1.00 1.00 1.00 1.00 1.00 1.00 0.00 0.00 0.00 0.00 0.00 0.67 0.33 1.00 1.00 1.00 1.00 1.00 0.00 0.00 0.00 0.00 0.44 0.00 1.00 0.56 1.00 1.00 1.00 1.00 0.00 0.00 0.00 0.30 -0.15 0.96 0.04 1.15 0.70 1.00 1.00 1.00 0.00 0.00 0.20 -0.20 0.89 -0.39 1.39 0.11 1.20 0.80 1.00 1.00 0.13 0.13 -0.20 0.79 -0.69 1.65 -0.65 1.69 0.21 1.20 0.87 0.87

Solution is unstable for small sampling in x compared to sigma. Problems when heat travels

more than one grid cell per time step.


Dispersion


Numerical differentiation

A Taylor series or LaGrange interpolation of points can be used to find the derivatives. The Taylor series

expansion is defined as:


First Derivative at a

i-2 i-1 i i+1 i+2



i-2 i-1 i i+1 i+2



i-2 i-1 i i+1 i+2



i-2 i-1 i i+1 i+2



i-2 i-1 i i+1 i+2



i-2 i-1 i i+1 i+2



Use the Taylor series expansion to represent three points about single location:



Assume that the data points are equally spaced and the equations can be written as:



Assume that the data points are equally spaced and the equations can be written as:


Forward differentiation

For a forward first derivative, subtract eqn[2] from eqn[1]:







Rearrange the equation:













As the Δx gets smaller the error will get smaller







The error is defined as:






Backward differentiation


Backward differentiationSubtract eqn[3] from eqn[2]:










Central differentiation


Central differentiationSubtract eqn[3] from eqn[1]:










Differential Error

Notice that the errors of the forward and backward 1st derivative of the equations have an error of the order of

O(Δx) and the central differentiation has an error of order O(Δx2). The central difference has an better

accuracy and lower error that the others. This can be improved by using more terms to model the first

derivative.


Numerical If you want to improve the accuracy and

decrease the error you will need to eliminate terms :







=

0




=

0

=

0




=

0

=

00


Higher Order Errors in The terms become :





The terms become A=-3, B= 4 and C=-1



The terms become A=-3, B= 4 and C=-1


Higher order 1st derivative

i-2 i-1 i i+1 i+2



i-2 i-1 i i+1 i+2



i-2 i-1 i i+1 i+2



i-2 i-1 i i+1 i+2



i-2 i-1 i i+1 i+2

Parabolic curve



i-2 i-1 i i+1 i+2

Parabolic curve

Forward difference



i-2 i-1 i i+1 i+2

Parabolic curve

Forward difference



i-2 i-1 i i+1 i+2

Parabolic curve

Backward difference

Forward difference


Higher Order Derivatives

To find higher derivatives, use the Taylor series expansions of term and eliminate the terms from the

sum of equations. To improve the error in the problem add additional terms.


2nd Derivative of the It will require three terms to get a central 2nd derivative of

discrete set of data.







=

0




=

0

=

0




=

0

=

0

=

#


2nd order Central The terms become :





The terms become A=1,B=-2 and C=1. Therefore



The terms become A=1,B=-2 and C=1. Therefore


Effect of stencil size

0

0.25

0.5

0.75

1

2 4 6 8 10 12 14 16 18 20

Volu

me

size

Oder of the approximation

Domain size


Where to take the space derivative?

qx,t+1 � qx,t = �(qx+1,t � 2qx,t + qx�1,t)

Known

Unknown

x

t0t1t2



qx,t+1 � qx,t = �(qx+1,t � 2qx,t + qx�1,t)

Known

Unknown

x

t0t1t2

Center x derivative

qx,t+1 � qx,t =�

2(qx+1,t � 2qx,t + qx�1,t + qx+1,t+1 � 2qx,t+1 + qx�1,t+1)



x

t0t1t2

qx,t+1 � qx,t =�




x

t0t1t2

qx,t+1 � qx,t =�


qx,t+1 ��

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t



x

t0t1t2

qx,t+1 � qx,t =�


qx,t+1 ��

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t

dataMonday, December 3, 12


x

t0t1t2

qx,t+1 � qx,t =�


qx,t+1 ��

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t



x

t0t1t2

qx,t+1 � qx,t =�


qx,t+1 ��

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t

a =�

2



x

t0t1t2

qx,t+1 � qx,t =�


qx,t+1 ��

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t

a =�

2



x

t0t1t2

qx,t+1 � qx,t =�


qx,t+1 ��

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t

�a

a =�

2



x

t0t1t2

qx,t+1 � qx,t =�


qx,t+1 ��

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t

�a

a =�

2

1 + 2a



x

t0t1t2

qx,t+1 � qx,t =�


qx,t+1 ��

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t

�a

a =�

2

�a1 + 2a



x

t0t1t2

qx,t+1 ��

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t

�a

a =�

2

�a1 + 2a

�a �a1 + 2a



x

t0t1t2

qx,t+1 ��

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t

�a

a =�

2

�a1 + 2a

�a �a1 + 2a

�a 1 + 2a �a



x

t0t1t2

qx,t+1 ��

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t

�a

a =�

2

�a1 + 2a

�a �a1 + 2a

�a 1 + 2a �a

�a 1 + 2a �a



Boundary condition

x

t0t1t2

qx,t+1 ��

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t

�a

a =�

2

�a1 + 2a

�a �a1 + 2a

�a 1 + 2a �a

�a 1 + 2a �a

?�a

�a?


Boundary condition

x

t0t1t2

qx,t+1 ��

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t

�a

a =�

2

�a1 + 2a

�a �a1 + 2a

�a 1 + 2a �a

�a 1 + 2a �a

�a

�a


Boundary condition

x

t0t1t2

qx,t+1 ��

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t

�a

a =�

2

�a1 + 2a

�a �a1 + 2a

�a 1 + 2a �a

�a 1 + 2a �a

�a

�aeleft

eright


Boundary conditionqx,t+1 �

�

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t



�

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t

Rewrite in a general form

dj = + +qjaj bjqj+1 cj qj�1



�

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t

Write a general relation between

any two qj

dj = + +qj

qj�1= ej�1 fj�1+

aj bjqj+1 cj qj�1

qj



�

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t

Write a general relation between

any two qj

dj = + +qj

= ejqj+1 + fjqj�1= ej�1 fj�1+

aj bjqj+1 cj qj�1

qjqj



�

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t

+ +qj

= ejqj+1 + fj

ej�1 fj�1+aj bjqj+1 cj

qj

qj( )

Substitute into our definition of dj

dj =



�

2(qx+1,t+1 � 2qx,t+1 + qx�1,t+1) =

�

2(qx+1,t � 2qx,t + qx�1,t) + qx,t

+qj

= ej qj+1+ fj

ej�1

fj�1ajbj

qj+1cj

qj

Rearrange

dj=

bjcj ej�1

cj��


Rtris


Implicit solver


€

dPdz

=iω

v(x,z)P +

v−iω2

δ 2Pδx 2

Solve in two stages by splitting:

€

1: dPdz

=v

−iω2δ 2Pδx 2

€

2 : dPdz

=iωvP


Diffrac'on solu'on

• Use the Crank-‐Nicolson method…

€

1: dPdz

=v

−iω2δ 2Pδx 2

€

pz+1x − pz

x

Δz=

v−iω2

pzx+1 − 2pz

x + pzx−1

2Δx 2+pz+1x+1 − 2pz+1

x + pz+1x−1

2Δx 2⎛

⎝ ⎜

⎞

⎠ ⎟

€

Set α =vΔz

−iω4Δx 2

€

pz+1x − pz

x = α pzx+1 − 2pz

x + pzx−1 + pz+1

x+1 − 2pz+1x + pz+1

x−1( )


Diffrac'on solu'on

• Same form as heat equa'on–Can solve with tridiagonal solver!€

−αpz+1x+1 + (1+ 2α)pz+1

x −αpz+1x−1 = αpz

x+1 + (1− 2α)pzx +αpz

x−1


Time shiA solu'on

• Easily solved analy'cally:

€

2 : dPdz

=iωvP

€

P(z + Δz) = P(z)eiΔzωv


52


52

Wavelength to guarantee that we repeat exactly nlam

'mes


52

Frequency sampling given wavelength


53

Convert frequency sampling to 'me


53

Figure out the phase given distance away

from focus


54

Matrix coefficients


54

Data


55

Diffrac'on term


55

ShiA term


55

How the wavefield evolves at this depth (note the Fourier transform)


56


56

Note the 90 phase shiA


56

Zero slope

boundary


57


57


57

Reflec'on off

boundary


57

Reflec'on off

boundary


Things to do

• Slide on boundary condi'ons

58


Things to do

• 9.5–Note phases moving up (Circle vs Parabola)–Pulse coming back from the edges–Smoothing in space would solve prob lem

• 9.6/9.7–Note the difference in boundary condi'ons–Talk about why (how off is the boundary)

• 9.8–Velocity gradient–Not decaying in amplitude and increasing amp diagonal59


• Migra'on program–Talk about how similar to downward con'nua'on

• 9.9–Not the smearing/inaccuracies

• Argument throwing away is we only want solu'on• Go through Muir

60


• Figure 9.11–Note small improvement to 9.5

• Figure 9.12–Look up and show 1/6 trip

• Figure 9.13–Shows rays–Energy moving leA

61


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