Migration in (z, t)-space

Migration and data synthesis may be envisioned in ( z' , t' )-space on the following table, which contains the upcoming wave U:  

$$\begin{tabular} {c\vert c\vert c\vert c\vert c\vert c\vert} ... ... & 0 & 0 \\ & & & & & \\ $t'$\space & & & & & \\ \end{tabular}$$ (48)

In this table the observed upcoming wave at the earth's surface z' = 0 is denoted by u_t. The migrated section, denoted by c_t, is depicted along the diagonal because the imaging condition of exploding reflectors at time t = 0 is represented in retarded space as

$$\begin{eqnarray} z' \ \ \ &=&\ \ \ z \\ t' \ \ \ &=&\ \ \ t \ +\ z/v \ \ \ \ \ \ \ (+\ \rm{for\ up} ) \\ 0 \eq t \ \ \ &=&\ \ \ t' \ -\ z' /v\end{eqnarray}$$ (49) (50) (51)

The best-focused migration need not fall on the 45$^\circ$ line as depicted in (48); it might be on any line or curve as determined by the earth velocity. This curve forms the basis for velocity determination. You couldn't determine velocity this way in the frequency-domain.

The equation for upcoming waves U in retarded coordinates (t' , x' , z' ) is  

$${ \partial^2 U \over \partial z' \, \partial t' } \eq -\ {v \over 2 }\ { \partial^2 U \ \over \partial { x' } ^2 }$$ (52)

Next, Fourier transform the x-axis. This assumes that v is a constant function of x and that the x-dependence of U is the sinusoidal function $\exp ( i k_x x )$.Thus,  

$$0 \eq \left( {v \over 2 }\ { k_x^2 } \ -\ { \partial^2 \over \partial z' \, \partial t' } \right) \ U$$ (53)

Now this partial-differential equation will be discretized with respect to t' and z'. Matrix notation will be used, but the notation does not refer to matrix algebra. Instead the matrices refer to differencing stars that may be placed on the $(t' ,\, z' )$-plane of (48). Let ${{\rm *}}$ denote convolution in $(z,\,t)$-space. A succession of derivatives is really a convolution, so the concept of $(\partial / \partial z) \,( \partial / \partial t) \ =$$\partial^2 / \partial z \partial t$ is expressed by  

$$[ \ -1 \ \ \ +1 \ ] \ \ \ { {\rm *} } \ \ \ \left[ \matrix {... ...eft[ \begin{array} {rr} 1 & -1 \\ -1 & 1 \\ \end{array}\right]$$ (54)

Thus, the differenced form of (53) is  

$$0 \eq \left\{ {v \over 2 }\ { \Delta z' \, \Delta t' \over 4... ... & -1 \\ -1 & 1 \\ \end{array}\right] \ \right\} \ {{\rm *}}\ U$$ (55)

The 1/4 enters in because the average of U is taken over four places on the mesh.

The sum of the two operators always has $\vert b\vert \ \ge \ \vert s\vert$ in the form  

$$0 \eq \ \ \left[ \matrix { \matrix { \ s \cr \ b } \matrix { b\ \cr s\ } } \right] \ \ \ {{\rm *}}\ \ U$$ (56)

Now the differencing star in (56) will be used to fill the table (48) with values for U.

Given the three values of U in the boxes, a missing one, M, may be determined by either of the implied two operations  

$$\begin{tabular} {ccc} \begin{tabular} {\vert c\vert c\vert}... ...umn{1}{c\vert}{ } & \\ \cline{2-2} \end{tabular}\end{tabular}$$ (57)

It turns out that because $\vert b\vert \ \ge \ \vert s\vert$, the implied filling operations by  

$$\begin{tabular} {ccc} \begin{tabular} {\vert c\vert c\vert}... ...lumn{1}{\vert c}{ } \\ \cline{1-1} \end{tabular}\end{tabular}$$ (58)

are unstable. It is obvious that there would be a zero-divide problem if s were equal to 0, and it is not difficult to do the stability analysis that shows that (58) causes exponential growth of small disturbances.

It is a worthwhile exercise to make the zero-dip assumption ( k_x = 0 ) and use the numerical values in the operator of (56) to fill in the elements of the table (48). It will be found that the values of u_t move laterally in z across the table with no change, predicting, as the table should, that c_t = u_t. Slow change in z suggests that we have oversampled the z-axis. In practice, effort is saved by sampling the z-axis with fewer points than are used to sample the t-axis.