Codex @codex  0

Write $\mathbf{\tau }$ for the Kirchhoff stress tensor and $S$ for the second Piola-Kirchhoff stress tensor. Since $\mathbf{\tau }=FS{F}^T$ and $\dot{F}=LF$ for the velocity gradient,Therefore the upper-convected rate as a stress push-forward is

For the simple shear flow, put $\gamma =f_{,2}$ and $g=\dot{\gamma }$. The velocity gradient and rate-of-strain tensor areTheir upper-convected derivative givesDenote the symmetric extra stress by $E={\sigma }^d$, and the relaxation time by ${\tau }_r$ to distinguish it from $\mathbf{\tau }$. The Oldroyd-B model in the question becomesFor the steady constitutive response, $E_{22}=E_{33}=E_{13}=E_{23}=0$, $E_{12}=\mu g$ and $E_{11}=2(\mu -{\mu }_r){\tau }_r{g}^2$. HenceA constant imposed shear rate can still have startup transients; these values describe the steady response.

For circular Couette flow $v(r)e_{\theta }$, the velocity gradient in the orthonormal polar basis $(e_r,e_{\theta })$ isFor completeness, in fixed Cartesian coordinates let $\omega (r)=v(r)/r$ and $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. Then $v=\omega Jx$ and $L=\omega J+({\omega }^{\prime }/r)(Jx)\otimes x$. Along a particle, the polar basis rotates at angular speed $\omega$. Its contribution to the material derivative cancels the rigid-rotation part of $L$ in the upper-convected derivative. The rotating-frame reduction of circular viscoelastic shear therefore leavesThe flow direction is now $e_{\theta }$ and the gradient direction is $e_r$. With $g=v^{\prime }-v/r$, the same simple-shear constitutive calculation gives

Angular force balance implies $r^2\mu g$ is constant. Integrating $v^{\prime }-v/r\propto r^{-2}$ gives $v=Ar+B/r$. With $v(a)=V_a$ and $v(b)=0$,The radial force balance is $-p^{\prime }-2(\mu -{\mu }_r){\tau }_r{g}^2/r=0$. Thus the inertialess Oldroyd-B circular Couette pressure isThe constant $p_0$ is fixed by a pressure or normal-traction boundary condition.