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

{c}

The Oldroyd-A model is the lower-convected counterpart of the <Oldroyd-B model>: its polymeric stress evolves using a <lower-convected derivative>.

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 Incoming links: Oldroyd-A model