Flexure of the lithosphere

Remarks

The equation is the Kirchhoff–Love plate equation in one dimension, or the Euler–Bernoulli beam equation with an elastic foundation, the buoyancy of the mantle acting as a restoring force. Let $a = \big( 4D/((\rho_m - \rho_i) g) \big)^{1/4}$ be the flexural parameter. Example: take a line load $V_0$ per unit length at $x = 0$. The solution is even in $x$, and for $x \geq 0$ it is $$w(x) = w_0 \, e^{-x/a} \left( \cos \frac{x}{a} + \sin \frac{x}{a} \right), \qquad w_0 = \frac{V_0 a^3}{8D}.$$ Check: for $x > 0$ it solves the equation with $q = 0$, and $w'(0) = 0$. The jump of $D w'''$ across $x = 0$ is $8 D w_0 / a^3 = V_0$. The deflection changes sign at $x = 3\pi a/4$, and the bulge is highest at $x = \pi a$, with height $w_0 e^{-\pi} \approx 0.04 \, w_0$. Non-example: with $D = 0$ the equation becomes $w = q/((\rho_m - \rho_i) g)$, which is local isostasy, with no moat and no bulge.