Solving for the extrapolation wavenumber

Developing an expression for the extrapolation wavenumber requires isolating one of the wavenumbers in equation 25 (herein assumed to be coordinate $\xi_3$). Rearranging the results of expanding equation 25 by introducing indicies i,j=1,2,3 yields

$$\begin{eqnarray} m^{33}k_\xi_3^2 + \left(2m^{13}k_\xi_1+ 2m^{23}k_\xi_2- i\,n_3 ... ...right)- m^{11}k_\xi_1^2 - m^{22}k_\xi_2^2 - 2m^{12}k_\xi_1k_\xi_2.\end{eqnarray}$$ (26)

An expression for wavenumber $k_\xi_3$ can be obtained by completing the square

$$\begin{eqnarray} \,& m^{33} \left(k_\xi_3+ \frac{ \left(2 m^{13}k_\xi_1+ 2m^{23... ...m^{23}\,n_3}{ m^{33} }\right)- \frac{n_3^2 }{ m^{33}}. \,\nonumber\end{eqnarray}$$ (27)

Isolating wavenumber $k_\xi_3$ yields  

$$k_\xi_3= - a_1 k_\xi_1 - a_2 k_\xi_2 + i a_3 \pm \left[ a_... ... i\,k_\xi_1 + a_9 i\,k_\xi_2 - a_{10}^2 \right]^{\frac{1}{2}},$$ (28)

where a_i are non-stationary coefficients given by  

$$\mathbf{a} = \left[ \frac{g^{13}}{ g^{33}} \;\;\;\;\; \frac{... ...}\right]\;\;\;\;\; \frac{ n_3 }{ m^{33} } \right]^{\mathbf{T}}.$$ (29)

Note that the coefficients contain a mixture of m^ij and g^ij terms, and that positive definite terms, a₄, a₅, a₆ and a₁₀ in equation 28are squared, such that the familiar Cartesian split-step Fourier correction is recovered.

For general 2D situations, the coefficients in equation 29 reduce to  

$$\mathbf{a} = \left[ \frac{g^{13}}{ g^{33}} \;\;\;\;\; 0 \;\;... ...;\;\; 0 \;\;\;\;\; \frac{ n_3 }{ m^{33} } \right]^{\mathbf{T}},$$ (30)

while a strictly orthogonal 2D mesh leads to the following coefficients  

$$\mathbf{a} = \left[ 0 \;\;\;\;\; 0 \;\;\;\;\; \frac{ n_3 } {... ...;\;\; 0 \;\;\;\;\; \frac{ n_3 }{ m^{33} } \right]^{\mathbf{T}}.$$ (31)