= Solution
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> gives
$$
\boxed{\sigma'_{ij}
=2\left(\frac{\tau_0}{\dot\gamma}+\mu_0\right)D_{ij}.}
$$
The <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 if
$$
S:E\leq\tau_0(2E:E)^{1/2}\qquad
\text{for every symmetric trace-free }E.
$$
The <Cauchy-Schwarz inequality> shows that this is equivalent to $\|S\|_F\leq\sqrt2\tau_0$; necessity follows by choosing $E$ proportional to $S$. Hence the no-strain-rate restriction is
$$
\boxed{\left(\frac12\sigma'_{ij}\sigma'_{ij}\right)^{1/2}\leq\tau_0.}
$$
The 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 $D_{12}=D_{21}=u'/2$, so $\dot\gamma=|u'|$ and the yielded <shear stress> law is
$$
\sigma_{12}=\tau_0\operatorname{sign}u'+\mu_0u'.
$$
<Force balance> gives $G+\sigma_{12,y}=0$. The prescribed odd, continuous shear <stress> fixes its integration constant to zero, proving
$$
\boxed{\sigma_{12}(y)=-Gy.}
$$
For $\mu_0>0$ the inverse shear law is
$$
u'(y)=
\begin{cases}
0,&|Gy|\leq\tau_0,\\
(-Gy+\tau_0\operatorname{sign}y)/\mu_0,&|Gy|>\tau_0.
\end{cases}
$$
The original PDF also calls $u$ odd. That assumption is inconsistent with a nonzero flow between identical fixed plates: the inverse shear law makes $u'$ odd, whereas the derivative of an odd $u$ is even. Both can hold only if $u'=0$, and the no-slip wall values then give $u=0$. Thus \b[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 $Gh\leq\tau_0$, 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 $|y|\leq y_0$. In the upper yielded layer $u'=(-Gy+\tau_0)/\mu_0$, and integration from $u(h)=0$ gives
$$
u(y)=\frac G{2\mu_0}\left[(h-y_0)^2-(y-y_0)^2\right]
\qquad(y_0\leq y\leq h).
$$
In 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 is
$$
\boxed{\text{Nonzero flow requires }G>\tau_0/h,\qquad
u(y)=\frac G{2\mu_0}\left[(h-y_0)^2-(|y|-y_0)_+^2\right].}
$$
The 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>.
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