Mixed domain -- Fourier finite-differences

The pseudo-screen solution to equation (9) derives from a fourth-order expansion of the square-root around (a₀,b₀) and (a,b):

$$\begin{eqnarray} k_\tau&\approx& \omega a \left [1+\frac{1}{2}\left (\frac{b k_\... ...ac{1}{8}\left (\frac{b_0k_\gamma}{a_0 \omega}\right )^4\right ]\;.\end{eqnarray}$$ (30)

If we subtract equations (30), we obtain:

$$\begin{eqnarray} k_\tau\approx {k_\tau}_0+ \omega\left (a-a_0\right ) &+&\frac{... ..._0}\right )^4\right ]\left (\frac{ k_\gamma}{ \omega}\right )^4\;.\end{eqnarray}$$ (31)

We can make the notations

$$\begin{eqnarray} \delta_1 &=& a\left (\frac{b }{a }\right )^2- a_0\left (\frac{b... ...t (\frac{b }{a }\right )^4- a_0\left (\frac{b_0}{a_0}\right )^4\;,\end{eqnarray}$$ (32) (33)

therefore equation (31) becomes

$$k_\tau= {k_\tau}_0+ \omega\left (a-a_0\right ) + \frac{1}{2}... ...{8}\omega\delta_2 \left (\frac{ k_\gamma}{ \omega}\right )^4\;.$$ (34)

With the approximation

$$\frac{1}{2}\delta_1 u^2 + \frac{1}{8}\delta_2 u^4 \approx \... ...ac{1}{2}\delta_1^2 u^2} {\delta_1-\frac{1}{4}\delta_2 u^2} \;,$$ (35)

we can write

$$k_\tau= {k_\tau}_0+ \omega\left (a-a_0\right ) +\omega\frac{... ...c{1}{4}\delta_2 \left (\frac{ k_\gamma}{ \omega}\right )^2} \;.$$ (36)

If we make the notations

$$\left\{ \begin{array} {l} \nu = \frac{1}{2}\delta_1^2 \;, \\... ...delta_1 \;, \\ \rho= \frac{1}{4}\delta_2 \;,\end{array}\right.$$ (37)

we obtain the mixed-domain Fourier finite-differences solution to the one-way wave equation in Riemannian coordinates:

$$k_\tau\approx {k_\tau}_0+ \omega\left (a-a_0\right )+ \ome... ... )^2} {\mu-\rho\left (\frac{ k_\gamma}{ \omega}\right )^2} \;.$$ (38)