Write the <velocity gradient> as
$$
\mathbf L=\nabla\mathbf u
=\frac12\dot{\boldsymbol\gamma}+\boldsymbol\Omega,
$$
where $\boldsymbol\Omega$ is the <spin tensor>. Expanding the <upper-convected derivative> in the structure equation gives
$$
\frac{D\boldsymbol\alpha}{Dt}
-\boldsymbol\Omega\boldsymbol\alpha
+\boldsymbol\alpha\boldsymbol\Omega
+\frac{\xi-1}{2}
\left(\dot{\boldsymbol\gamma}\boldsymbol\alpha+
\boldsymbol\alpha\dot{\boldsymbol\gamma}\right)
+c_1\boldsymbol\alpha
=c_2\dot{\boldsymbol\gamma}.
$$
For $\xi=0$, the first four terms reproduce the upper-convected derivative. Setting
$$
\boxed{\xi=0,\qquad b_2=0}
$$
therefore gives
$$
\boldsymbol\tau=\eta_0\dot{\boldsymbol\gamma}+b_1\boldsymbol\alpha,
\qquad
\boldsymbol\alpha^{\triangledown}+c_1\boldsymbol\alpha
=c_2\dot{\boldsymbol\gamma}.
$$
With polymeric stress $\boldsymbol\tau_p=b_1\boldsymbol\alpha$, this is the <Oldroyd-B model>. For $c_1>0$, its relaxation time and polymer viscosity are
$$
\lambda=\frac1{c_1},
\qquad
\eta_p=\frac{b_1c_2}{c_1}.
$$
Indeed the total stress obeys
$$
\boxed{
\boldsymbol\tau+\lambda\boldsymbol\tau^{\triangledown}
=(\eta_0+\eta_p)\dot{\boldsymbol\gamma}
+\lambda\eta_0\dot{\boldsymbol\gamma}^{\triangledown}}.
$$

For $\xi=2$, the coefficient of
$\dot{\boldsymbol\gamma}\boldsymbol\alpha+
\boldsymbol\alpha\dot{\boldsymbol\gamma}$
is $+1/2$, so the objective derivative becomes the <lower-convected derivative>. Hence
$$
\boxed{\xi=2,\qquad b_2=0}
$$
recovers the <Oldroyd-A model>, again with $c_1>0$ for a finite positive relaxation time. Parameter choices such as $b_1c_2=0$ give the degenerate Newtonian limit.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 352 / 2 / a / Solution

A plain <material derivative> of the <conformation tensor> is not an <objective time derivative>: an observer undergoing a time-dependent rigid rotation would infer a different constitutive response, and even rigid-body rotation could appear to change polymer deformation. The <upper-convected derivative> subtracts deformation and rotation carried by the velocity gradient and is frame indifferent.

The <Oldroyd-B model> is often inadequate because its Hookean dumbbells are infinitely extensible. It predicts constant shear viscosity rather than shear thinning, zero second <normal-stress difference>, and an unbounded extensional viscosity at a finite extension rate. Real polymer chains have finite extensibility and commonly exhibit shear thinning, bounded extensional stress, multiple relaxation times, and nonlinear solvent or concentration effects.

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

{wiki}

The upper-convected derivative of a contravariant tensor is $\overset{\triangledown}{\mathbf C}=D\mathbf C/Dt-(\nabla\mathbf u)\mathbf C-\mathbf C(\nabla\mathbf u)^T$.


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