next up previous print clean
Next: Examples Up: Finite-difference solutions to the Previous: Finite-difference solutions to the

$15^\circ$ wave-equation in a 3-D Riemannian space

A simple way of deriving the $15^\circ$ equation is by a second order Taylor series expansion of the extrapolation wavenumber $k_\zeta$ relative to $k_\xi$ and $k_\eta$:
   \begin{eqnarray}
k_\zeta\left(k_\xi,k_\eta\right)\approx
 k_\zeta\left(k_\xi=0,k...
 ...\partial^2 k_\zeta}{\partial k_\eta^2} \right\vert _0 k_\eta^2 \;.\end{eqnarray}
(21)
If we plug equation (20) into equation (21), we obtain an equivalent form for the $15^\circ$ equation in a semi-orthogonal 3-D Riemannian space:
   \begin{eqnarray}
k_\zeta\approx i \frac{\c_{\zeta}}{2\c_{\zeta\zeta}} + \; k_o
+...
 ...k_o^2} -\frac{\c_{\xi\eta}}{\c_{\zeta\zeta}}\right]k_\xi k_\eta\;.\end{eqnarray}
(22)

For the particular case of Cartesian coordinates ($\c_{\xi}=\c_{\eta}=\c_{\zeta}=0, \c_{\xi\xi}=\c_{\eta\eta}=\c_{\zeta\zeta}=1, \c_{\xi\eta}=0, \; k_o=\omega\ss$),
\begin{displaymath}
k_\zeta\approx \omega\ss - \frac{1}{2\omega\ss} \left(k_\xi^2 + k_\eta^2 \right)\;,\end{displaymath} (23)
which is the usual form of the $15^\circ$ equation.


next up previous print clean
Next: Examples Up: Finite-difference solutions to the Previous: Finite-difference solutions to the
Stanford Exploration Project
10/14/2003