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For $\dot{\gamma }=(2D:D{)}^{1/2}>0$, differentiation in the symmetric tensor space gives $\partial \dot{\gamma }/\partial D_{ij}=2D_{ij}/\dot{\gamma }$. Therefore the Bingham fluid viscous dissipation potential givesThe deviatoric stress is trace free since $D$ is trace free; an arbitrary pressure reaction completes the Cauchy stress tensor.

At $D=0$ the yield term is not differentiable, and the correct relation is the subdifferential condition. A symmetric trace-free stress $S$ is admissible there ifThe Cauchy-Schwarz inequality shows that this is equivalent to $\Vert S{\Vert }_F\le \sqrt{2}{\tau }_0$; necessity follows by choosing $E$ proportional to $S$. Hence the no-strain-rate restriction isThe quadratic term has zero derivative at zero and does not change this restriction. This is the unyielded part of the Bingham plastic law.

For flow along the plates, write $y=x_2$, $v=(u(y),0,0)$ and take $G=-p_{,1}>0$ as the pressure-drop magnitude. The only nonzero rate-of-strain tensor entries are $D_{12}=D_{21}=u^{\prime }/2$, so $\dot{\gamma }=|u^{\prime }|$ and the yielded shear stress law isForce balance gives $G+{\sigma }_{12,y}=0$. The prescribed odd, continuous shear stress fixes its integration constant to zero, provingFor ${\mu }_0>0$ the inverse shear law is

The original PDF also calls $u$ odd. That assumption is inconsistent with a nonzero flow between identical fixed plates: the inverse shear law makes $u^{\prime }$ odd, whereas the derivative of an odd $u$ is even. Both can hold only if $u^{\prime }=0$, and the no-slip wall values then give $u=0$. Thus with the literal odd-velocity requirement, there is no admissible flowing solution when $Gh>{\tau }_0$. The standard pressure-driven velocity is even; replacing that erroneous parity yields the intended profile below.

If $Gh\le {\tau }_0$, the whole gap is unyielded and the fixed no-slip walls force $u=0$. Otherwise let $y_0={\tau }_0/G<h$. The plug flow of a yield-stress fluid occupies $|y|\le y_0$. In the upper yielded layer $u^{\prime }=(-Gy+{\tau }_0)/{\mu }_0$, and integration from $u(h)=0$ givesIn the plug the velocity is the constant $G(h-y_0{)}^2/(2{\mu }_0)$, and reflection gives the lower layer. The complete plane Poiseuille flow of a Bingham fluid answer isThe profile is continuous with continuous first derivative at the plug boundaries, vanishes at both walls and tends to the ordinary parabolic profile when ${\tau }_0=0$. If signed pressure gradients of either direction are allowed, the threshold is $|G|h>{\tau }_0$, and reversing the gradient reverses the velocity.

The shear stress decreases monotonically from the fibre to zero at the free surface, so its maximum isFlow occurs precisely when this exceeds the yield stress:

When this condition holds, there is one yield surface $r=r_y$ determined byThe region $a\le r<r_y$ is yielded and the outer region $r_y\le r\le b$ is an unyielded plug flow of a yield-stress fluid. In the yielded region, $du/dr=({\tau }_{rz}-{\sigma }_y)/\eta$. The no-slip condition $u(a)=0$ givesThe unyielded layer translates without shearing at the plug speed

Let $w(x)$ be the downward velocity between the vertical plates. A steady vertical momentum balance givesSymmetry requires ${\tau }_{xz}(0)=0$, so the stress magnitude isThe wall stress first reaches the yield stress when $\rho g{l}_c={\tau }_y$. The onset measurement therefore gives

For $l>l_c$, define $\Delta =l-l_c$. The central region $|x|\le l_c$ is a plug flow of a yield-stress fluid, while the layers $l_c<|x|\le l$ are yielded. The Herschel–Bulkley fluid law gives, for $x\ge l_c$,Integrating from the no-slip wall $w(l)=0$ givesin the yielded layer, and the plug moves atThe volumetric flow rate per unit span is consequently

Near onset, the plug term dominates andThe observed exponent two therefore corresponds to $n=1$, a linear post-yield constitutive law at small shear rate. Far above onset, the fully yielded contribution scales asThe observed exponent four corresponds to $n=1/2$, a shear-thinning square-root law at high shear rate.

A plausible stress curve is therefore odd in $\dot{\gamma }$, starts at the two yield-stress values $\tau =\pm {\tau }_y$, and has finite linear slope just after yield:It then crosses smoothly to the concave high-rate behaviorwith the negative-rate branch fixed by odd symmetry. This combines the two power laws inferred independently from the measured flux limits.

For fixed plates $y=\pm h$, no slip, viscosity ${\mu }_0>0$, yield stress ${\tau }_0\ge 0$ and $G=-p_{,x}>0$, force balance gives shear stress $-Gy$. The central plug flow of a yield-stress fluid has half-width $y_0={\tau }_0/G$. If $y_0<h$, the velocity isIt is even in $y$; its derivative is $(-Gy+{\tau }_0\mathrm{sign}y)/{\mu }_0$ in the yielded layers and zero in the plug. If $Gh\le {\tau }_0$, there is no flow. Odd velocity is incompatible with this pressure-driven no-slip solution at nonzero flow.