= Solution
Write $q=\dot\gamma\tau$ and $f=1+\alpha\operatorname{tr}A$. With $L_{ij}=\partial_jv_i$, the steady homogeneous <upper-convected derivative> is $-LA-AL^T$. Its diagonal and shear equations give
$$
f a_{22}=f a_{33}=0,\qquad f a_{12}=q,\qquad f a_{11}=2q a_{12}.
$$
On the physical branch $f>0$, the other off-diagonal components vanish too. Thus $a_{22}=a_{33}=0$, $a_{11}=2a_{12}^2$ and
$$
\boxed{q=(1+2\alpha a_{12}^2)a_{12}.}
$$
For $\alpha\geq0$ the right side has derivative $1+6\alpha a_{12}^2>0$, so the shear stress determines a unique shear rate. The <shear viscosity> is $G_0\tau/f$ and decreases with shear magnitude when $\alpha>0$. This is the shear part of <shear and pipe flow of an affine linear PTT fluid>.
For fully developed pipe flow take $\mathbf v=w(r)\widehat{\mathbf z}$ and the signed axial <pressure gradient> $dp/dz=\Delta p$. Regularity at the axis and the supplied momentum balance determine the shear stress without using the constitutive equation:
$$
\frac{d}{dr}(r\sigma_{rz})=\Delta p r,\qquad
\sigma_{rz}=\frac{\Delta p r}{2},\qquad a_{rz}=\frac{\Delta p r}{2G_0}.
$$
Locally the constitutive equations have the same shear form, with axial flow replacing the Cartesian flow direction and $\dot\gamma=w'(r)$ carrying its sign. Hence
$$
w'(r)=\frac1\tau(a_{rz}+2\alpha a_{rz}^3)
=\frac{\Delta p r}{2G_0\tau}+\frac{\alpha\Delta p^3 r^3}{4G_0^3\tau}.
$$
Integrating and imposing <no-slip boundary conditions> at $r=R$ gives
$$
\boxed{w(r)=-\frac{\Delta p}{4G_0\tau}(R^2-r^2)-\frac{\alpha\Delta p^3}{16G_0^3\tau}(R^4-r^4).}
$$
A negative pressure gradient produces positive axial flow. At $\alpha=0$ this is the ordinary <Hagen-Poiseuille flow>; positive $\alpha$ increases the flux at a fixed pressure drop through <shear thinning>. The nonzero axial normal stress varies radially, but its axial derivative vanishes in fully developed flow, so it does not change the axial momentum balance used here.
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