Mixed-domain solutions

Mixed-domain solutions to the one-way wave equation usually consist of terms computed in the Fourier domain for a reference of the extrapolation medium, followed by a finite-differences correction applied in the space-domain. For equation (9), a generic mixed-domain solution has the form:  

$$k_\tau\approx {k_\tau}_0+ \omega\left (a-a_0\right )+ \ome... ... )^2} {\mu-\rho\left (\frac{ k_\gamma}{ \omega}\right )^2} \;,$$ (13)

where a₀ and b₀ are reference values for the medium characterized by the parameters a and b, and the coefficients $\mu$, $\nu$ and $\rho$ take different forms according to the type of approximation. As for usual Cartesian coordinates, ${k_\tau}_0$ is applied in the Fourier domain, and the other two terms are applied in the space domain. If we limit the space-domain correction to the thin lens term, $\omega\left (a-a_0\right )$, we obtain the equivalent of split-step Fourier (SSF) method Stoffa et al. (1990) in Riemannian coordinates.

Appendix A details the derivations for two types of expansions: pseudo-screen Huang et al. (1999), and Fourier finite-differences Biondi (2002); Ristow and Ruhl (1994).

• Pseudo-screen:

  The coefficients for the pseudo-screen solution to equation (13) are  

  $$\left\{ \begin{array} {l} \nu = a_0\left [c_1\left (\frac{a}... ...\ \rho=3b\left (\frac{b_0}{a_0}\right )^2\;,\end{array}\right.$$ (14)

  where a₀ and b₀ are reference values for the medium characterized by parameters a and b. In the special case of Cartesian coordinates, a=s and b=1, equation (13) with coefficients equation (14) takes the familiar form

  $$k_\tau\approx {k_\tau}_0+ \omega \left [1+ \frac{ \frac{ c_1... ...k_\gamma}{ \omega}\right )^2} \right ]\left (s-s_0 \right )\;,$$ (15)

  where the coefficients c₁ and b take different values for different orders of the finite-differences term: c₁₅=(c₁,b)=(0.50,0.00) and c₄₅=(c₁,b)=(0.50,0.25). When (c₁,b)=(0.00,0.00) we obtain the usual split-step Fourier equation.

• Fourier finite-differences:

  The coefficients for the Fourier finite-differences solution to equation (13) are  

  $$\left\{ \begin{array} {l} \nu = \frac{1}{2}\delta_1^2 \;, \\... ...delta_1 \;, \\ \rho= \frac{1}{4}\delta_2 \;,\end{array}\right.$$ (16)

  where

  $$\left\{ \begin{array} {l} \delta_1 = a\left (\frac{b }{a }\r... ... )^4- a_0\left (\frac{b_0}{a_0}\right )^4\;.\end{array}\right.$$ (17)

  a₀ and b₀ are reference values for the medium characterized by the parameters a and b. In the special case of Cartesian coordinates, a=s and b=1, equation (13) with coefficients equation (16) takes the familiar form:

  $$k_\tau\approx {k_\tau}_0+ \omega \left [1+ \frac{ \frac{c_1}... ...k_\gamma}{ \omega}\right )^2} \right ]\left (s-s_0 \right )\;,$$ (18)

  where the coefficients c₁ and b take different values for different orders of the finite-differences term: c₁=0.5, b=0.0 for $15^\circ$, or c₁=0.5, b=0.25 for $45^\circ$.When c₁=b=0.0 we obtain the usual split-step Fourier equation.