= Solution
Use a <time-harmonic wave> with time dependence $e^{-i\omega t}$ and let the same symbols denote phasors. The <wave amplitude> vector obeys
$$
\frac{d}{dx}\begin{pmatrix}\phi_+\\\phi_-\end{pmatrix}=\begin{pmatrix}i\omega/\beta&g\\g&-i\omega/\beta\end{pmatrix}\begin{pmatrix}\phi_+\\\phi_-\end{pmatrix}.
$$
At fixed material contrast, the stated low-frequency condition permits neglecting the diagonal <frequency> term to zeroth order. All remaining coefficient <matrices> are scalar multiples of the same exchange <matrix>, so they commute. With
$$
L=\int_{x_a}^{x_b}g\,dx=\frac12\log\frac{Z_b}{Z_a},\qquad C=\cosh L,\quad H=\sinh L,
$$
the transfer relation is
$$
\begin{pmatrix}\phi_+(x_b)\\\phi_-(x_b)\end{pmatrix}=\begin{pmatrix}C&H\\H&C\end{pmatrix}\begin{pmatrix}\phi_+(x_a)\\\phi_-(x_a)\end{pmatrix}.
$$
Solve the second row for the outgoing left <wave amplitude> and substitute in the first. Since $C^2-H^2=1$, this gives the <zero-frequency scattering through an elastic layer> coefficients
$$
\boxed{R_{aa}=\frac{Z_a-Z_b}{Z_a+Z_b},\quad R_{bb}=\frac{Z_b-Z_a}{Z_a+Z_b},\quad T_{ab}=T_{ba}=\frac{2\sqrt{Z_aZ_b}}{Z_a+Z_b}.}
$$
Thus the zero-frequency limit depends only on the endpoint <seismic impedances>. Monotonicity supports the given scale comparison but is not needed to integrate the zero-frequency equation. If $Z_a=Z_b$ in a monotone profile, <seismic impedance> is constant: there is no directional coupling, with only <wave phase> accumulation at nonzero <frequency>.
Back to article page
