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Mixed domain

One solution of (10) is:
         \begin{eqnarray}

P'_z &=& P_z e^{- \Gamma \Delta z} \\ 
P_{z+\Delta z} &=& P'_z e^{\mp \Lambda \Delta z}.\end{eqnarray} (23)
(24)
We implemented (24) with the split-step method Stoffa et al. (1990). To implement (23) we take a Taylor expansion of the $\Gamma$ term:
\begin{displaymath}
\Gamma = \frac{v_z}{2v} \frac{1}{1-\left(\frac{vk_x}{\omega}...
 ...\omega}\right)^2 +
 \left(\frac{vk_x}{\omega}\right)^4 \right].\end{displaymath} (25)
Equation (23) becomes:  
 \begin{displaymath}
P'_z = P_z 
e^{-\frac{v_z\Delta z }{2v }} 
e^{-\frac{v_z\Del...
 ...^2}{2 \omega^2}}
e^{-\frac{v_z\Delta z v^3 k_x^4}{2 \omega^4}}.\end{displaymath} (26)
The last exponentials cannot be computed in a single step because the velocity terms and the horizontal wavenumber do not belong to the same domain. One solution consists of expanding the exponentials in a Taylor series and keeping only the terms of order smaller or equal to 4 in kx:
\begin{displaymath}
e^{-\frac{v_z\Delta z v k_x^2}{2 \omega^2}}e^{-\frac{v_z\Del...
 ...frac{v_z\Delta z}{4}-v\right)\left(\frac{k_x}{\omega}\right)^4.\end{displaymath} (27)
The $\Gamma$ correction is:
\begin{displaymath}
P'_z = P_z e^{-\frac{v_z\Delta z }{2v }}
 \left[1-\frac{v_z\...
 ...\Delta z}{4}-v\right)\left(\frac{k_x}{\omega}\right)^4 \right].\end{displaymath} (28)
We give an algorithm of the mixed-domain implementation in Appendix B.
next up previous print clean
Next: Results Up: Implementing the operator Previous: Finite differences
Stanford Exploration Project
7/8/2003