# Derivative and integral operators

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 9.31a

The operator ${\displaystyle f_{t}=[-1,\;+1]}$ is called the “derivative operator”; explain why.

### Solution

In terms of finite differences (see problem 9.29), the derivative of a function is approximately equal to the change in values between two adjacent points divided by the distance between the two points. Convolving ${\displaystyle f_{t}}$ with ${\displaystyle g_{t}}$ gives values of the finite-difference derivative of ${\displaystyle g_{t}}$ for different values of ${\displaystyle t}$. To show this, we find the ${\displaystyle (n+1)^{th}}$ term of ${\displaystyle h_{t}=f_{t}*g_{t}}$. From equation (9.2b), we have

{\displaystyle {\begin{aligned}h_{n}=\mathop {\sum } \limits _{k=0}^{}f_{k}g_{n-k}=[-g_{0},\;(-g_{1}+g_{0}),\;.\;.\;.\;(-g_{n}+g_{n-1})].\end{aligned}}}

Figure 9.30a.  Seismic section. (i) Unmigrated; (ii) migrated.
Figure 9.30b.  Interpreted migrated seismic section.

Each term (except the first) equals ${\displaystyle -\Delta g}$ at ${\displaystyle =n\Delta }$. Taking ${\displaystyle \Delta =1}$, we see that ${\displaystyle h_{n}}$ is minus the finite difference derivative at ${\displaystyle t=n\Delta }$.

## Problem 9.31b

What is the integral operator?

### Solution

The integral, ${\displaystyle \int _{g_{0}}^{g_{n}}\ g(t)\,\mathrm {d} t}$, for continuous functions becomes a summation for digital functions, the equivalent of the above integral being ${\displaystyle \Delta \mathop {\sum } \limits {_{g_{0}}^{g_{n}}}g_{t}}$. If we take the operator ${\displaystyle f_{t}=[1,1,1,\ldots ..,1]}$ where both ${\displaystyle f_{t}}$ and ${\displaystyle g_{t}}$ have ${\displaystyle (n+1)}$ terms, we obtain the following for the terms of ${\displaystyle f_{t}*g_{t}}$ (omitting ${\displaystyle \Delta }$):

{\displaystyle {\begin{aligned}&[(g_{0}),(g_{0}+g_{1}),...,(g_{0}+g_{1}+\cdots +g_{r}),(g_{0}+g_{1}+\cdots +g_{n}),\\&\quad (g_{1}+g_{2}+\cdots +g_{n}),(g_{2}+g_{3}+\cdots +g_{n}),...,(g_{n-1}+g_{n}),(g_{n}).\end{aligned}}}

The terms are equivalent to the integrals between 0 and ${\displaystyle r}$ for ${\displaystyle 0\leq r\leq n}$. After the value ${\displaystyle r=n}$ is reached, the upper limit remains ${\displaystyle n}$ while the lower limit is successively 1, 2, . . . , ${\displaystyle n}$. Thus we conclude that the operator [1, 1, 1, . . . , 1] is an integral operator. By taking ${\displaystyle f_{t}}$ with fewer terms than ${\displaystyle g_{t}}$, we can get the equivalent of an integral over selected parts of ${\displaystyle g_{t}}$.