= Solution
The $xy$ component of the stationary <constitutive equation> is
$$
T_{xy}-\tau gT_{yy}+\frac{\alpha g}{2}(T_{xx}+T_{yy})=\mu g.
$$
Insert $T_{yy}=-\alpha gT_{xy}$ and $T_{xx}=(2\tau-\alpha)gT_{xy}$. The coefficient of $g^2T_{xy}$ becomes $\tau\alpha+\alpha(\tau-\alpha)=\alpha(2\tau-\alpha)$, giving
$$
\boxed{T_{xy}=\frac{\mu\dot\gamma}{1+\alpha(2\tau-\alpha)\dot\gamma^2},\qquad \eta_{\rm sh}=\frac{T_{xy}}{\dot\gamma}=\frac{\mu}{1+\alpha(2\tau-\alpha)\dot\gamma^2}.}
$$
For $C=\alpha(2\tau-\alpha)>0$, the small-rate <shear stress> is $T_{xy}=\mu\dot\gamma-\mu C\dot\gamma^3+O(\dot\gamma^5)$. At large positive rate, $T_{xy}\sim\mu/(C\dot\gamma)$ and the <shear viscosity> decreases as $\mu/(C\dot\gamma^2)$. The <shear stress> is odd in rate, reaches a positive maximum at
$$
\dot\gamma_*=C^{-1/2},\qquad T_{xy,*}=\frac{\mu}{2\sqrt C},\qquad
\frac{dT_{xy}}{d\dot\gamma}=\mu\frac{1-C\dot\gamma^2}{(1+C\dot\gamma^2)^2},
$$
and decreases thereafter. The original plot displays this homogeneous constitutive curve. Its decreasing branch can permit <shear banding>; the local model alone supplies neither an interface-selection rule nor a guaranteed stable homogeneous realization.
\Image[/past-exam-of-the-advanced-fluid-dynamics-course-of-the-university-of-oxford/2016-viscoelastic-shear-response.png]
{title=Homogeneous shear stress and the two temporal branches of the linear response}
{description=The steady curve uses q equal to the shear rate times the square root of alpha times (2 tau minus alpha). The spectral panel uses K equal to the wavenumber magnitude times the square root of tau times nu. These are distinct nondimensional variables.}
This is a <Johnson-Segalman fluid> without an added <Newtonian fluid> solvent contribution. To identify conventions explicitly, set $a=1-\alpha/\tau$ and $L=G^T=E+W$, with $W$ the <spin tensor>. The derivative appearing in the <constitutive equation> becomes
$$
D_t\mathsf T-W\mathsf T+\mathsf TW-a(E\mathsf T+\mathsf TE),
$$
the <Gordon--Schowalter derivative>. Its usual slip interval $|a|\le1$ corresponds to $0\le\alpha\le2\tau$, while a nonsingular genuinely <shear thinning> curve requires
$$
\boxed{\tau>0,\quad\mu>0,\quad0<\alpha<2\tau.}
$$
At $\alpha=0$ the model is the <Upper-convected Maxwell model>; at $\alpha=2\tau$ it is the <lower-convected derivative> counterpart, and both have constant <shear viscosity> in this test. At $\alpha=\tau$ the derivative is the <Jaumann derivative>, giving the <corotational Maxwell fluid>. Merely requiring $\alpha>0$ is insufficient: for $\alpha=3\tau$ the denominator is $1-3\tau^2\dot\gamma^2$, which vanishes at $|\dot\gamma|=1/(\sqrt3\tau)$ and then gives a negative steady <shear viscosity>. There is no solvent-induced large-rate plateau in the stated equation.
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