= Solution
The shear stress decreases monotonically from the fibre to zero at the free surface, so its maximum is
$$
\tau_{rz}(a)=\frac{\Delta p}{2L}\frac{b^2-a^2}{a}.
$$
Flow occurs precisely when this exceeds the yield stress:
$$
\boxed{\Delta p>\frac{2La\sigma_y}{b^2-a^2}.}
$$
When this condition holds, there is one <yield surface> $r=r_y$ determined by
$$
\frac G2\left(\frac{b^2}{r_y}-r_y\right)=\sigma_y.
$$
The region $a\leq r<r_y$ is yielded and the outer region $r_y\leq r\leq b$ 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 $u(a)=0$ gives
$$
u(r)=\frac1\eta\left[
\frac G2\left\{b^2\log\frac ra-\frac{r^2-a^2}{2}\right\}
-\sigma_y(r-a)
\right],
\qquad a\leq r\leq r_y.
$$
The unyielded layer translates without shearing at the plug speed
$$
\boxed{u_p=u(r_y)=\frac1\eta\left[
\frac G2\left\{b^2\log\frac{r_y}{a}-\frac{r_y^2-a^2}{2}\right\}
-\sigma_y(r_y-a)
\right].}
$$
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