= Solution
The local motion is simple shear with flow direction $z$, gradient direction $r$, and vorticity direction $\theta$. Writing $q=du/dr$, a <second-order fluid> has first and second normal-stress differences
$$
N_1=\tau_{zz}-\tau_{rr}=\Psi_1q^2,
\qquad
N_2=\tau_{rr}-\tau_{\theta\theta}=\Psi_2q^2.
$$
Up to an isotropic contribution absorbed into pressure, a convenient representation is
$$
\tau_{zz}=(\Psi_1+\Psi_2)q^2,
\qquad \tau_{rr}=\Psi_2q^2,
\qquad \tau_{\theta\theta}=0.
$$
Thus both the flow-direction and gradient-direction normal stresses can be nonzero, while the radial momentum equation becomes
$$
\boxed{\frac{dp}{dr}
=\frac1r\frac{d(r\tau_{rr})}{dr}-\frac{\tau_{\theta\theta}}r
=\frac{\Psi_2}{r}\frac{d}{dr}(rq^2).}
$$
The first normal-stress coefficient affects $\tau_{zz}$ but cancels from radial balance.
In the yielded layer,
$$
q=\frac1\eta\left[
\frac G2\left(\frac{b^2}{r}-r\right)-\sigma_y
\right]>0.
$$
Moreover,
$$
\frac{d}{dr}(rq^2)
=\frac q\eta\left[-\frac G2\left(\frac{b^2}{r}+3r\right)-\sigma_y\right]<0.
$$
Consequently $dp/dr$ has the sign opposite to $\Psi_2$ and is independent of $\Psi_1$. For the common case $\Psi_2<0$, pressure increases radially outward toward the free surface.
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