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20

Solving equation

We define $n=\frac{v}{\omega}$.Equation 20 can be written:
\begin{displaymath}
\partial_z = \pm i
\frac
{ \alpha n \partial_{xx}}
{1+\beta n^2 \partial_{xx}},\end{displaymath} (30)
or  
 \begin{displaymath}
\partial_z +
 \beta n^2 \partial_{xxz} \mp
i \alpha n \partial_{xx } = 0.\end{displaymath} (31)
We apply the finite difference method:
\begin{eqnarray}
\partial_{z }p &=&\frac{ 1}{\Delta z} (p_{z+1}^{x }-p_{z}^{x })...
 ...}-p_{z}^{x })
 +\frac{ 1}{ \Delta x^2}(p_{z+1}^{x+1}-p_{z}^{x+1}).\end{eqnarray} (32)
(33)
(34)
After some little rearrangements, equation 31 can be written:  
 \begin{displaymath}
\Delta p_{z+1}^{x-1}+\left(1-2\Delta \right)p_{z+1}^{x}+\Del...
 ... }^{x-1}+\left(1-2\Delta^*\right)p_{z }^{x}+\Delta^*p_{z }^{x}.\end{displaymath} (35)
where
   \begin{eqnarray}
\Delta &=& \frac{\beta n^2 }{ \Delta x^2} \mp
 ~i\frac{\alpha n \Delta z}{2\Delta x^2}\end{eqnarray} (36)
For downward continuing the receiver (upgoing) wavefield, we choose the positive sign in the relation above; for downward continuation of the source (downgoing) wavefield we take the sign in $\Delta$ to be negative. To solve the $15^\circ$ equation, $(\alpha,\beta) = (0.5,0)$. For the $45^\circ$ equation, $(\alpha,\beta) = (0.5,0.25)$, and for the $65^\circ$ equation, $(\alpha,\beta) = (0.478242060,0.376369527)$Lee and Suh (1985). In our implementation, we have used the $65^\circ$equation with the ``$\frac{1}{6}$ trick'' Claerbout (1985) for improving the accuracy of the second derivative. This changes (31) and (36) into
\begin{eqnarray}
\partial_z + \left(\beta n^2 + \frac{\Delta x^2}{s}\right)
\par...
 ...} + \frac{1}{s} \right) \mp i\frac{\alpha n \Delta z}{2\Delta x^2}\end{eqnarray} (37)
(38)
where we took s = 8.13 Fomel and Claerbout (1997). Equation (35) with the new $\Delta$ gives a tridiagonal system for each frequency and depth.
next up previous print clean
Next: Solving equation 21 Up: REFERENCES Previous: REFERENCES
Stanford Exploration Project
7/8/2003