When horizontal shear dominates, $u=U(y)+O(\epsilon)$ and
$$
\tau_{xy}=K|U'|^{n-1}U'+O(\epsilon).
$$
Integrating downhill momentum through $0<z<H$ gives
$$
\tau_b=\tau_d+HK\frac d{dy}
\left(|U'|^{n-1}U'\right).
$$
The <Bingham sliding law> then yields the ODE
$$
\boxed{HK\frac d{dy}\left(|U'|^{n-1}U'\right)
-\frac\mu hU+\tau_d-\tau_c(y)=0}.
$$
Define
$$
U_-:=\frac h\mu(\tau_d-\tau_1),
\qquad
U_+:=\frac h\mu(\tau_d-\tau_2).
$$
Then
$$
\boxed{U(-\infty)=U_->U_+=U(+\infty)}.
$$
The speed decreases monotonically across the material transition, with a smooth margin layer centered at $y=0$.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-352/1/b/ii/solution