 Solution

Because the wavelength is much smaller than the shelf thickness, the ice-shelf corrugation relaxation may be treated as Stokes flow in the half-space $z>0$, with the mean ice-ocean boundary at $z=0$. Letso $u=-{\psi }_z$ and $w={\psi }_x$. Taking the curl of the Stokes equation gives the Biharmonic stream function for planar Stokes flow equation ${\nabla }^4\psi =0$. Decay as $z\to \infty$ removes the growing modes, leavingThusand the perturbation pressure obtained from the Stokes equation is

The linearized zero-shear stress condition at $z=0$ isso $B=kA$. The water is hydrostatic, and displacement of the density interface gives the normal-stress conditionAfter $B=kA$, one has $w_z(0)=0$, soThe kinematic boundary condition ${\eta }_t=w(0)$ gives $\sigma \hat{\eta }=ikA$. Eliminating $A$ yieldsHence a corrugation of wavelength $\lambda =2\pi /k$ decays on the timescaleHydrostatic buoyancy supplies the restoring stress, while viscous deformation over depth $O(k^{-1})$ supplies the resistance. Shorter wavelengths deform a shallower but more strongly sheared layer and therefore have a longer decay time in this gravity-only model. As ice is advected away from the grounding line, long corrugations should disappear first, leaving progressively shorter-wavelength structure farther downstream.