# Finite-difference migration

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 9 295 - 366 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 9.29

Derive the finite-difference migration equation (9.29d).

### Background

The wave equation (2.5a) becomes, for two dimensions,

 {\begin{aligned}{\frac {1}{V^{2}}}{\frac {\partial ^{2}\psi }{\partial t^{2}}}=({\frac {\partial ^{2}\psi }{\partial x^{2}}}+{\frac {\partial ^{2}\psi }{\partial z^{2}}}).\end{aligned}} (9.29a)

We can simplify equation (9.29a) by replacing $t$ with $t^{*}=t-z/V$ (see problem 9.28). To replace derivatives in the $(x,\;z,\;t)$ coordinate system with derivatives in the $(x,\;z,\;t^{*})$ system, we follow the procedure used in problem 2.6:

{\begin{aligned}{\frac {\partial \psi }{\partial t}}={\frac {\partial \psi ^{*}}{\partial t^{*}}}{\frac {\partial t^{*}}{\partial t}}={\frac {\partial \psi ^{*}}{\partial t^{*}}};\quad {\frac {\partial ^{2}\psi }{\partial t^{2}}}={\frac {\partial ^{2}\psi ^{*}}{\partial t^{*2}}};\\{\frac {\partial \psi }{\partial x}}={\frac {\partial \psi ^{*}}{\partial x}};\quad {\frac {\partial ^{2}\psi }{\partial x^{2}}}={\frac {\partial ^{2}\psi ^{*}}{\partial x^{2}}};\\{\frac {\partial \psi }{\partial z}}={\frac {\partial \psi ^{*}}{\partial z}}+{\frac {\partial \psi ^{*}}{\partial t^{*}}}{\frac {\partial t^{*}}{\partial z}}={\frac {\partial \psi ^{*}}{\partial z}}-{\frac {1}{V}}{\frac {\partial \psi ^{*}}{\partial t^{*}}};\\{\frac {\partial ^{2}\psi }{\partial z^{2}}}={\frac {\partial ^{2}\psi ^{*}}{\partial z^{2}}}+{\frac {\partial ^{2}\psi ^{*}}{\partial z\partial t^{*}}}{\frac {\partial t^{*}}{\partial z}}-{\frac {1}{V}}{\frac {\partial ^{2}\psi ^{*}}{\partial z\partial t^{*}}}-{\frac {1}{V}}{\frac {\partial ^{2}\psi ^{*}}{\partial t^{*2}}}{\frac {\partial t^{*}}{\partial z}}\\={\frac {\partial ^{2}\psi ^{*}}{\partial z^{2}}}-{\frac {2}{V}}{\frac {\partial ^{2}\psi ^{*}}{\partial z\partial t^{*}}}+{\frac {1}{V^{2}}}{\frac {\partial ^{2}\psi ^{*}}{\partial t^{*2}}}.\end{aligned}} Substituting these expressions into equation (9.29a), we obtain the result

 {\begin{aligned}{\frac {\partial ^{2}\psi ^{*}}{\partial x^{2}}}+{\frac {\partial ^{2}\psi ^{*}}{\partial z^{2}}}-{\frac {2}{V}}{\frac {\partial ^{2}\psi ^{*}}{\partial z\partial t^{*}}}=0.\end{aligned}} (9.29b)

However, ${\frac {\partial ^{2}\psi ^{*}}{\partial z^{2}}}$ varies slowly because the coordinate system rides along with the wavefront and so we omit it. This gives the simplified wave equation,

 {\begin{aligned}{\frac {\partial ^{2}\psi ^{*}}{\partial x^{2}}}-{\frac {2}{V}}{\frac {\partial ^{2}\psi ^{*}}{\partial z\partial t^{*}}}\approx 0.\end{aligned}} (9.29c)

Using the method of finite differences discussed below, this equation can be changed to

 {\begin{aligned}\psi (x,\;z,\;t^{*})&={\frac {2\Delta z\Delta t^{*}(\Delta x)^{2}}{2(\Delta x)^{2}-V\Delta z\Delta t^{*}}}\left[{\frac {\psi ^{*}(x,z-\Delta z,t^{*})}{\Delta z\Delta t^{*}}}+{\frac {\psi ^{*}(x,z,t^{*}-\Delta t^{*})}{\Delta z\Delta t^{*}}}\right.\\&\quad -{\frac {V\psi ^{*}(x-\Delta x,z,t^{*})}{(\Delta x)^{2}}}-{\frac {\psi ^{*}(x,z-\Delta z,t^{*}-\Delta t^{*})}{\Delta z\Delta t^{*}}}\\&\left.+{\frac {V\psi ^{*}(x-2\Delta x,z,t^{*})}{2(\Delta x)^{2}}}\right].\end{aligned}} (9.29d)

Approximate solutions of differential equations can be found using the method of finite differences. If we denote the value of $f(x)$ at $x_{1}$ by the symbol $f_{1}$ , an approximate value of the derivative is

 {\begin{aligned}{\hbox{d}}f/{\hbox{d}}x\approx {\frac {f_{2}-f_{1}}{x_{2}-x_{1}}}={\frac {\Delta f}{\Delta x}},\end{aligned}} (9.29e)

the error decreasing as $\Delta x\to 0$ . The second derivative at a given point can be found by finding the difference between two first derivatives close to the given point and dividing the difference by the distance between the two points. Derivatives with respect to more than one variable can be found using the same principle.

### Solution

We use points spaced at intervals $\Delta x$ , $\Delta z$ , $\Delta t^{*}$ to evaluate the two derivatives in equation (9.29c):

{\begin{aligned}{\frac {\partial \psi ^{*}}{\partial x}}&=[\psi ^{*}\;(x,\;z,\;t^{*})-\psi ^{*}(x-\Delta x,\;z,\;t^{*})]/\Delta x;\\{\frac {\partial ^{2}\psi ^{*}}{\partial x^{2}}}&=\{[\psi ^{*}(x,\;z,\;t^{*})-\psi ^{*}(x-\Delta x,\;z,\;t^{*})]-[\psi ^{*}(x-\Delta x,\;z,\;t^{*})\\&\quad -\psi ^{*}(x-2\Delta x,\;z,\;t^{*})]\}/(\Delta x)^{2}\\&=[\psi ^{*}(x,\;z,\;t^{*})-2\psi ^{*}(x-\Delta x,\;z,\;t^{*})+\psi ^{*}(x-2\Delta x,\;z,\;t^{*})]/(\Delta x)^{2};\\{\frac {\partial \psi ^{*}}{\partial z}}&=[\psi ^{*}(x,\;z,\;t^{*})-\psi ^{*}(z,\;z-\Delta z,\;t^{*})]/\Delta z;\\{\frac {\partial ^{2}\psi ^{*}}{\partial z\partial t^{*}}}&=\{\psi ^{*}(x,\;z,\;t^{*})-[\psi ^{*}(x,\;z-\Delta z,\;t^{*})-\psi ^{*}(x,\;z,\;t^{*}-\Delta t^{*})\\&\quad -\psi ^{*}(x,\;z-\Delta z,\;t^{*}-\Delta t^{*})]\}/(\Delta z\Delta t^{*}).\\\end{aligned}} Writing equation (9.29c) in the form

{\begin{aligned}{\frac {\partial ^{2}\psi ^{*}}{\partial x^{2}}}={\frac {2\partial ^{2}\psi ^{*}}{V\partial z\partial t^{*}}},\end{aligned}} we substitute the above values of the two derivatives and obtain

{\begin{aligned}&[\psi ^{*}(x,\;z,\;t^{*})-2\psi ^{*}(x-\Delta x,\;z,\;t^{*})+\psi ^{*}(x-2\Delta x,\;z,\;t^{*})]/(\Delta x)^{2}\\&\quad =(2/V)\{[\psi ^{*}(x,\;z,\;t^{*})-\psi ^{*}(x,\;z-\Delta z,\;t^{*})]-[\psi ^{*}(x,\;z,\;t^{*}-\Delta t^{*})\\&\qquad -\psi ^{*}(x,\;z-\Delta z,\;t^{*}-\Delta t^{*})]\}/(\Delta z\Delta t^{*}).\end{aligned}} Rearranging, we have

{\begin{aligned}&\psi ^{*}(x,\;z,\;t^{*})\left({\frac {1}{(\Delta x)^{2}}}-{\frac {2}{V\Delta z\Delta t^{*}}}\right)\\&\quad =[2\psi ^{*}\;(x-\Delta x,\;z,\;t^{*})-\psi ^{*}(x-2\Delta x,\;z,\;t^{*})]/(\Delta x)^{2}\\&\qquad +{\frac {2}{V}}[-\psi ^{*}(x,\;z-\Delta z,\;t^{*})-\psi ^{*}(x,\;z,\;t^{*}-\Delta t^{*})\\&\qquad +\psi ^{*}(x,\;z-\Delta z,\;t^{*}-\Delta t^{*})]/(\Delta z\Delta t^{*}).\end{aligned}} Solving for $\psi ^{*}(x,\;z,\;t^{*})$ gives

{\begin{aligned}\psi ^{*}(x,\;z,\;t^{*})&=\left({\frac {2(\Delta x)^{2}V\Delta z\Delta t^{*}}{V\Delta z\Delta t^{*}-2(\Delta x)^{2}}}\right)\\&\quad \times \{[2\psi ^{*}\;(x-\Delta x,\;z,\;t^{*})-\psi ^{*}(x-2\Delta x,\;z,\;t^{*})]\\&\quad +{\frac {2}{V}}[-\psi ^{*}(x,\;z-\Delta z,\;t^{*})-\psi ^{*}(x,\;z,\;t^{*}-\Delta t)\\&\quad +\psi ^{*}(x,\;z-\Delta z,\;t^{*}-\Delta t^{*})]/(\Delta z\Delta t^{*})\}.\end{aligned}} Multiplying the first bracket by $-1/V$ and the second one by $-V/2$ , we get

{\begin{aligned}\psi ^{*}(x,\;z,\;t^{*})&=\left[{\frac {2(\Delta x)^{2}\Delta z\Delta t^{*}}{(\Delta x)^{2}-V\Delta z\Delta t^{*}}}\right]\left[{\frac {\psi ^{*}(x,z-\Delta z,t^{*})}{\Delta z\Delta t^{*}}}+{\frac {\psi ^{*}(x,z,t^{*}-\Delta t^{*})}{\Delta z\Delta t^{*}}}\quad -{\frac {\psi ^{*}(x,z-\Delta z,t^{*}-\Delta t^{*})}{\Delta z\Delta t^{*}}}-{\frac {V\psi ^{*}(x-\Delta x,z,t^{*})}{(\Delta x)^{2}}}\quad +{\frac {V\psi ^{*}(x-2\Delta x,z,t^{*})}{2(\Delta x)^{2}}}\right],\end{aligned}} which is equation (9.29d).