= Solution
Let $h$ be the water-layer thickness and $z=0$ the horizontal bedrock. In the <hydrostatic approximation>, with constant ice-interface pressure,
$$
p(x,z,t)=p_0+\rho_wg[h(x,t)-z],\qquad p_x=\rho_wg h_x.
$$
The prescribed two-wall stress balance gives
$$
\frac{2\mu u}{\delta_v}=-\rho_wg h h_x,\qquad
\boxed{u=-D h h_x,\quad q=hu=-D h^2h_x,\quad D=\frac{\rho_wg\delta_v}{2\mu}>0.}
$$
In this <subglacial current with constant viscous wall layers>, the uniform-pressure ice roof is treated as a movable confining boundary; the closure neglects inertia in comparison with pressure and wall stress. The bulk may be turbulent, while the specified wall <viscous boundary layer> supplies the linear stress relation. A different turbulent drag law would give a different transport coefficient and similarity exponent.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-65-subglacial-current.png]
{title=Subglacial water layer with constant interface pressure, hydrostatic driving and two viscous wall layers}
{height=480}
Without melting, the depth-integrated <continuity equation> gives $h_t+q_x=0$, hence
$$
\boxed{h_t=D\partial_x(h^2h_x).}
$$
To write this in the PDF's displayed plus-sign form, its constant must be
$$
\boxed{\gamma=-D=-\frac{\rho_wg\delta_v}{2\mu}.}
$$
The signed value is essential. If the printed $\gamma$ were interpreted as a positive diffusivity, its equation would drive water uphill and be backward parabolic. For example, linearizing around a uniform depth $H$ would give a Fourier perturbation growth rate $+\gamma H^2k^2$, whereas the derived physical equation damps it at $-DH^2k^2$. The plus-sign form is consistent only with the negative $\gamma$ above.
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