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21

Solving equation

Equation (21) can be written:  
 \begin{displaymath}
\partial_z = \frac{v_z}{2v} \frac {1} {1 + n^2 \partial_{xx}}.\end{displaymath} (39)
As for Equation (31), we use a Crank-Nicolson scheme for the second derivative in x. Equation (39) becomes:  
 \begin{displaymath}
\Delta_1 p_{z+1}^{x-1} + \left(1-2\Delta_1+\Delta_2 \right) ...
 ...1-2\Delta_1-\Delta_2 \right) p_{z }^{x} + 
\Delta_1 p_{z }^{x }\end{displaymath} (40)
where
\begin{eqnarray}
\Delta_1 &=& \frac{n^2}{\Delta x^2},\\ \Delta_2 &=&-\frac{v_z \Delta z}{4v}.\end{eqnarray} (41)
(42)
Equation (40) is solved the same way as (35). B We provide here the algorithm of our mixed-domain implementation of both $\Lambda^{-1}$ and $\Gamma$ corrections. The notations refer to a single spatial axis, but they can be easily extended to two axes. The $\Lambda^{-1}$correction is only applied to the source wavefield, and is only applied once, at the surface, not for each depth step. The correction is $P_{z=0}=-\frac{1}{2ik_z}Q$ . Because it generates strong wraparound that translates as noise in the final image, it is recommended to pad the data with zeros, apply the correction, then window it to a smaller size for downward continuation.

For each frequency, to downward continue one step with corrected amplitudes ($\Gamma$operator) we Fourier transform to the wavenumber domain:

In the wavenumber domain, we perform the following operations: Three separate Fourier transforms are needed to transform back to the spatial domain: In the spatial domain we perform: The $\Gamma$ correction for the receiver wavefield is computed the same way. Once both wavefields have been computed, the imaging condition (9) is applied for each depth.

In laterally varying velocity the terms in the spatial wavenumber domain that contain kx do not commute with terms that contain functions of x, namely v and vz. We took care to apply the terms in the right order inside the $\Gamma$ operator. Outside $\Gamma$,things change. Strictly mathematically speaking, $\Gamma$ does not commute with the split-step, and the right order of applying the operators remains to be studied. As in the case of split-step (for which the two terms that compose it do not commute with each other either), the differences in output may not warrant the cost of the extra Fourier transforms. C At depth z and at each point x we compute:
\begin{displaymath}
I_x = \frac{A_0}{A_x} I_0,\end{displaymath} (43)
where A0 and Ax are the areas of the unitary wavefront and a particular wavefront respectively. For spherical wavefronts,
In 2-D: $A_0=2\pi R_0 =2\pi$, $A_x=2\pi R~~$ hence $I_x=\frac{1}{R }I_0$.
In 3-D: $A_0=4\pi R_0^2=4\pi$, $A_x=4\pi R^2$ hence $I_x=\frac{1}{R^2}I_0$.
To be able to compute the perturbed amplitude Ix from the unitary amplitude I0, we need to compute R, the radius of the spherical wavefront. Two cases where the wavefronts are spherical are media of constant velocity and constant vertical velocity gradient.


 
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Next: Constant velocity media Up: REFERENCES Previous: Solving equation 20
Stanford Exploration Project
7/8/2003