= 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 $z=0$. Let
$$
\eta=\widehat\eta e^{ikx+\sigma t},
\qquad
\mathbf u=\nabla\times(\psi\widehat{\mathbf y}),
$$
so $u=-\psi_z$ 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, leaving
$$
\boxed{\psi=(A+Bz)e^{-kz}e^{ikx+\sigma t}.}
$$
Thus
$$
u=(kA-B+kBz)e^{-kz}e^{ikx+\sigma t},
\qquad
w=ik(A+Bz)e^{-kz}e^{ikx+\sigma t},
$$
and the perturbation pressure obtained from the Stokes equation is
$$
\boxed{p'=2i\mu kB e^{-kz}e^{ikx+\sigma t}.}
$$
The linearized zero-<shear stress> condition at $z=0$ is
$$
\mu(u_z+w_x)=2\mu k(B-kA)e^{ikx+\sigma t}=0,
$$
so $B=kA$. The water is hydrostatic, and displacement of the density interface gives the normal-stress condition
$$
-p'+2\mu w_z=(\rho_w-\rho)g\eta.
$$
After $B=kA$, one has $w_z(0)=0$, so
$$
-2i\mu k^2A=(\rho_w-\rho)g\widehat\eta.
$$
The <kinematic boundary condition> $\eta_t=w(0)$ gives $\sigma\widehat\eta=ikA$. Eliminating $A$ yields
$$
\boxed{\sigma=-\frac{(\rho_w-\rho)g}{2\mu k}<0.}
$$
Hence a corrugation of wavelength $\lambda=2\pi/k$ decays on the timescale
$$
\boxed{\tau(\lambda)=\frac1{|\sigma|}
=\frac{4\pi\mu}{(\rho_w-\rho)g\lambda}.}
$$
Hydrostatic 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.
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