= Solution
Take upward deflection as positive, use $k>0$, and write $\Delta\rho=\rho_m-\rho_c>0$. A positive added load depresses the plate, so $w$ is negative beneath it. Because bending moves the original surface while the load adds thickness, the final topography is \b[$e=s+w$]. For the cosine mode the horizontal biharmonic operator multiplies $w$ by $k^4$, giving
$$
(Dk^4+g\rho_m)w=-g\rho_cs.
$$
Eliminate $s=e-w$ to obtain
$$
\boxed{e=-\frac{Dk^4+g\Delta\rho}{g\rho_c}\,w.}
$$
The original loading equation contains $\rho_m$, whereas the relation after eliminating the added thickness contains $\Delta\rho$. These must not be interchanged prematurely.
At first order in the relief, represent a displaced <density> interface by a sinusoidal mass sheet. A sheet of surface-density <amplitude> $\Sigma$ at depth $z$ produces a downward gravity <amplitude> $2\pi G\Sigma e^{-kz}$ at the datum. This follows by solving the Fourier-transformed gravitational <Poisson equation> above and below the sheet: the decaying potential has <amplitude> $-2\pi G\Sigma e^{-k|z-z_0|}/k$, with the normal derivative jump fixed by the sheet mass.
The top interface contributes $\rho_ce$; the displaced <crust>–<mantle> interface contributes $\Delta\rho\,w$ at depth $t$. Thus
$$
\Delta g=2\pi G[\rho_ce+\Delta\rho\,w e^{-kt}].
$$
The second term is negative beneath a positive load: depression replaces denser <mantle> material by lighter <crust>. Substituting the deflection-to-topography ratio yields
$$
\boxed{\Delta g=Z(k)e,\qquad
Z(k)=2\pi G\rho_c\left[1-
\frac{e^{-kt}}{1+Dk^4/(g\Delta\rho)}\right].}
$$
This <flexural gravity admittance> combines mechanical compensation and attenuation of the buried interface's gravitational signal. It assumes small relief, a flat uniform thin plate, uniform <density> layers and the stated surface loading, without additional internal <density> anomalies. Here $t$ is the depth of the <crust>–<mantle> interface. It is not automatically the <effective elastic thickness> $T_e$; the latter enters through $D=ET_e^3/[12(1-\nu^2)]$.
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