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Mixed domain -- Fourier finite-differences

The pseudo-screen solution to equation (13) derives from a fourth-order expansion of the square-root around $(a_0,b_0)$ and $(a,b)$:
$\displaystyle k_\tau$ $\textstyle \approx$ $\displaystyle \omega a \left [1+\frac{1}{2} \left ( \frac{b k_\gamma }{a \omega...
...ht )^2+ \frac{1}{8} \left ( \frac{b k_\gamma }{a \omega } \right )^4\right ]\;,$ (47)
$\displaystyle {k_\tau }_0$ $\textstyle \approx$ $\displaystyle \omega a_0 \left [1+\frac{1}{2} \left ( \frac{b_0k_\gamma }{a_0 \...
...)^2+ \frac{1}{8} \left ( \frac{b_0k_\gamma }{a_0 \omega } \right )^4\right ]\;.$ (48)

If we subtract equations (A-16) and (A-17), we obtain the following expression for the wavenumber along the extrapolation direction $k_\tau $:
$\displaystyle k_\tau \approx {k_\tau }_0+ \omega \left (a-a_0\right )$ $\textstyle +$ $\displaystyle \frac{1}{2}\omega \left [a \left (\frac{b }{a }\right )^2- a_0 \l...
...frac{b_0}{a_0}\right )^2\right ] \left ( \frac{ k_\gamma }{ \omega } \right )^2$  
  $\textstyle +$ $\displaystyle \frac{1}{8}\omega \left [a \left (\frac{b }{a }\right )^4- a_0 \l...
...c{b_0}{a_0}\right )^4\right ] \left ( \frac{ k_\gamma }{ \omega } \right )^4\;.$ (49)

We can make the notations
$\displaystyle \delta_1$ $\textstyle =$ $\displaystyle a \left (\frac{b }{a }\right )^2- a_0 \left (\frac{b_0}{a_0}\right )^2\;,$ (50)
$\displaystyle \delta_2$ $\textstyle =$ $\displaystyle a \left (\frac{b }{a }\right )^4- a_0 \left (\frac{b_0}{a_0}\right )^4\;,$ (51)

therefore equation (A-18) can be written as
\begin{displaymath}
k_\tau = {k_\tau }_0+ \omega \left (a-a_0\right )
+ \frac{1}...
...ga \delta_2 \left ( \frac{ k_\gamma }{ \omega } \right )^4\;.
\end{displaymath} (52)

Using the approximation

\begin{displaymath}
\frac{1}{2}\delta_1 u^2 + \frac{1}{8}\delta_2 u^4 \approx
\...
...{1}{2}\delta_1^2 u^2}
{\delta_1-\frac{1}{4}\delta_2 u^2} \;,
\end{displaymath} (53)

we can write
\begin{displaymath}
k_\tau = {k_\tau }_0+ \omega \left (a-a_0\right )
+\omega \f...
...}\delta_2 \left ( \frac{ k_\gamma }{ \omega } \right )^2} \;.
\end{displaymath} (54)

If we make the notations

$\displaystyle \nu$ $\textstyle =$ $\displaystyle \frac{1}{2}\delta_1^2 \;,$ (55)
$\displaystyle \mu$ $\textstyle =$ $\displaystyle \delta_1 \;,$ (56)
$\displaystyle \rho$ $\textstyle =$ $\displaystyle \frac{1}{4}\delta_2 \;,$ (57)

we obtain the mixed-domain Fourier finite-differences solution to the one-way wave equation in Riemannian coordinates:
\begin{displaymath}
k_\tau \approx {k_\tau }_0+ \omega \left (a-a_0\right )+
\...
...{\mu-\rho \left ( \frac{ k_\gamma }{ \omega } \right )^2} \;.
\end{displaymath} (58)


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Next: Bibliography Up: Sava and Fomel: Riemannian Previous: Mixed domain

2007-10-07