Dips greater than 90 degrees

Migration of dips greater than $90^\circ$ requires careful handling of evanescent energy. As this is being written, most migration-by-depth-extrapolation programs ignore or set to zero the energy that turns evanescent. The proper thing to do with energy becoming evanescent at depth z is to save it for a second pass upward . The upward pass begins from the bottom of the section with a zero downgoing wave. As the downgoing wave is extrapolated upward, the saved evanescent energy is reintroduced. As usual, the images are withdrawn from the wave at time t = 0.

To illustrate the concept, a program will be sketched that makes two images, first the usual image of the top side of the reflector, and second the image of the under side. The images may be viewed separately or summed.

The program makes the simplifying restriction on the velocity that $dv/dz \ge 0$.Because of this assumption, evanescent energy can be stored ``in place'' and ignored until the return pass. It is worth noting that the second pass is cheaper than the first pass because the region in which evanescence never occurred, $\vert k\vert < \vert \omega \vert / v( \tau_{\rm max} )$, need not be processed.

# first pass of conventional phase-shift migration. 
$P(\omega, k_x)\ =\$FT[u(t,x)] 
For $\ \tau =\triangle\tau$, $\ 2\triangle\tau$,                                 ..., $\tau_{\rm max}$ {
		 For all kx {
		 		 Uimage$(k_x, \tau)\ =\ 0$.		 		 For all $\omega \gt \mid k\mid\ u(\tau)$ {
		 		 		 $C\ =\$ exp$(-i\omega\triangle\tau$                     $\sqrt{1-v(\tau)^2k_x^2/\omega^2}$ 
		 		 		 $P(\omega, k_x)\ =\ P(\omega, k_x) \ \ast\ C$ 
		 		 		 Uimage$(k_x,\tau)\ =\$Uimage$(k_x,\tau)\ +\ P(\omega,k_x)$		 		 		 }
		 		 }
		 uimage$(x, \tau)\ =\$FT[Uimage$(k_x, \tau)]$ 
		 }  
 

# Second pass for underside image. 
For $\ \tau =\tau_{\rm max}$, $\ \tau_{\rm max} - \triangle\tau$,$\ \tau_{\rm max} - 2\triangle\tau$,  ..., 0 {
		 For all kx {
		 		 Dimage$(k_x, \tau)\ =\ 0$.		 		 For all $\omega = \mid k\mid\$              $v(\tau)\$ to $\omega = \mid k\mid\$              $v(\tau_{\rm max}$ {
		 		 		 # The wave changes direction but so                        does $\triangle\tau$ 
		 		 		 $C\ =\ \exp(-i\omega\triangle\tau$                     $\sqrt{1-v(\tau)^2k_x^2/\omega^2}$ 
		 		 		 $P(\omega, k_x)\ =\ P(\omega, k_x) \ \ast\ C$ 
		 		 		 Dimage$(k_x,\tau)\ =\$Dimage$(k_x,\tau)\ +\ P(\omega,k_x)$		 		 		 }
		 		 }
		 dimage$(x, \tau)\ =\$FT[Dimage$(k_x, \tau)]$ 
		 }