 Solution

Basal sliding allows ice to pass over bed obstacles by enhanced deformation and regelation: pressure-induced melting on the upstream side and refreezing downstream transfer ice around small obstacles. Cavitation opens water-filled gaps on the downstream side. Their size and the remaining ice-bed contact depend on effective pressure in subglacial drainage $N=p_{\mathrm{ice}}-p_{\mathrm{water}}$ as well as basal shear stress $\tau$. At fixed $\tau$, reducing $N$ generally enlarges cavities and reduces resistance, giving a sliding law $u(\tau ,N)$ with $u_N<0$ in this regime.

A linked cavity drainage system connects these cavities through passages along the bed. Under the prescribed decreasing relation $N(Q)$, the chain rule gives $\boxed{u_Q=u_N{N}^{\prime }(Q)>0}$. This conclusion uses that hydraulic relation; it need not apply to a different drainage regime.

With $t=\epsilon \tau$, the ice continuity equation becomes $H_{\tau }+\epsilon (uH{)}_x=\epsilon s^{\prime }(x)$. For bounded spatial gradients over $\tau =O(1)$, $H$ is independent of the fast seasonal time to leading order. Meanwhile $S^{\prime }(Q)=\epsilon \kappa$ gives $S_t=\kappa Q_{\tau }$, soLet $J(\tau )={\int }_0^{\tau }M(r)dr$. Direct differentiation proves thatsatisfies the equation and $Q(0,\tau )=0$. This is the boundary-generated solution, or a periodic-forcing solution with the forcing continued into its past. An independently prescribed initial profile would also need matching in the region whose characteristic curves have not yet reached the upstream boundary.

If $M>0$, $J$ is invertible and rearranging gives $x=[\tau -J^{-1}(J(\tau )-\kappa Q)]/\kappa$. For $M=1-\delta \cos \tau$ with $|\delta |<1$,Substitution, followed by the sine-difference identity, givesFor fixed $Q$ and nonzero first-order amplitude, the extrema occur at $\tau =\kappa Q/2+n\pi +O(\delta )$. Since $x=Q+O(\delta )$, the seasonal phase lag in linked cavity drainage gives the stated phase relation $\boxed{\tau \approx \kappa x/2(\mathrm{mod}\pi )}$. Maxima and minima alternate; when $\sin (\kappa Q/2)=0$, the first-order approximation is flat and does not locate individual extrema.