The trace is
$$
T=\frac3F+\frac{2a^2}{F^3}.
$$
Substituting it into $F=(L^2-3)/(L^2-T)$ gives
$$
\boxed{F^2(F-1)=\beta},
\qquad
\boxed{\beta=\frac{2\lambda^2\dot\gamma^2}{L^2}}.
$$
For $\beta\ll1$, $F=1+\beta+O(\beta^2)$. For $\beta\gg1$, $F\sim\beta^{1/3}$. The effective shear viscosity is therefore
$$
\eta_{\rm eff}=\eta_s+\frac{\eta_p}{F}
\sim
\begin{cases}
\eta_s+\eta_p,&\lambda|\dot\gamma|\ll L,\\
\eta_s+\eta_p
\left(\dfrac{L^2}{2\lambda^2\dot\gamma^2}\right)^{1/3},
&\lambda|\dot\gamma|\gg L.
\end{cases}
$$
The FENE-P curve decreases from $\eta_s+\eta_p$ toward the solvent plateau $\eta_s$, displaying <shear thinning>. Oldroyd-B has $F=1$ and remains at the constant value $\eta_s+\eta_p$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 352 / 2 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-352/2/b/ii/solution