Write $\tau_d=\rho gH\sin\theta$. For unidirectional velocity $\mathbf u=(u(y,z),0,0)$, the nontrivial force balances are
$$
\frac{\partial\tau_{xy}}{\partial y}
+\frac{\partial\tau_{xz}}{\partial z}
+\rho g\sin\theta=0,
\qquad
\frac{\partial p}{\partial y}=0,
\qquad
\frac{\partial p}{\partial z}=-\rho g\cos\theta.
$$
For the <power-law fluid>,
$$
\tau_{xy}=K|\dot\gamma|^{n-1}u_y,
\qquad
\tau_{xz}=K|\dot\gamma|^{n-1}u_z,
\qquad
|\dot\gamma|=(u_y^2+u_z^2)^{1/2}.
$$
At the free surface $z=H$, $p=p_{\rm atm}$ and $\tau_{xz}=0$. At the bed $z=0$, continuity of tangential velocity and traction gives the local sliding condition
$$
u(y,0)=\frac h\mu[\tau_b(y)-\tau_c(y)],
\qquad
\tau_b(y)=\tau_{xz}(y,0),
$$
because the stated inequalities keep the downhill bed yielded. The velocity and horizontal shear traction are continuous across $y=0$, and the flow tends to uniform states as $y\to\pm\infty$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 352 / 1 / b / i / Solution

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A power-law fluid has shear stress $\tau=K|\dot\gamma|^{n-1}\dot\gamma$. It is shear thinning for $n<1$, Newtonian for $n=1$, and shear thickening for $n>1$.

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