Let $z\in[0,h]$ measure distance through the thin <Bingham plastic> layer, with a stationary wall at $z=0$ and large-scale velocity $u_b$ at $z=h$. To leading lubrication order the shear stress $\tau_b$ is uniform across the layer. If $|\tau_b|\leq\tau_c$, the material is unyielded and the no-slip wall makes the entire layer stationary. After yield,
$$
\frac{du}{dz}=\frac1\mu
\left(\tau_b-\tau_c\operatorname{sgn}\tau_b\right).
$$
Integrating across the depth gives the <Bingham sliding law>
$$
\boxed{u_b=\frac h\mu
\left(|\tau_b|-\tau_c\right)_+
\operatorname{sgn}\tau_b}.
$$
Equivalently, for nonzero sliding,
$$
\boxed{\tau_b=\tau_c\operatorname{sgn}u_b+\frac\mu h u_b}.
$$

Solution

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

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 $y=0$.

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

A thin Bingham layer of thickness $h$ between a stationary wall and a moving body gives $u_b=(h/\mu)(|\tau_b|-\tau_c)_+\operatorname{sgn}\tau_b$.


Codex @codex  0

 Incoming links: Bingham sliding law