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Put $S={\sigma }^d-2{\mu }_{r}D$ in the Oldroyd-A model. It obeys ${\mathcal{D}}_{l}S+S/\tau =2(\mu -{\mu }_r)D/\tau$. Pulling back as $F^{T}SF$ converts this to an ordinary linear relaxation equation. With decayed remote-past initial memory, its solution isAt a finite initial time, include the homogeneous transported term $e^{-(t-t_0)/\tau }F(t{)}^{-T}F(t_0{)}^{T}S(t_0)F(t_0)F(t{)}^{-1}$. The integral-only law is not equivalent to all arbitrary initial-value solutions without a memory condition.

Differentiate the pulled-back Kirchhoff stress tensor $T=F^T\tau F$, using the velocity gradient convention $L=\dot{F}{F}^{-1}$:Thus the lower-convected stress pull-back givesFor an explicit objective time derivative check, superpose $x^{\ast }=Q(t)x+c(t)$, with $Q^{T}Q=I$ and $\Omega =\dot{Q}{Q}^T$ skew. Then ${\tau }^{\ast }=Q\tau Q^T$ and $L^{\ast }=QL{Q}^T+\Omega$, whereasThe additional terms in $(L^{\ast }{)}^T{\tau }^{\ast }+{\tau }^{\ast }{L}^{\ast }$ are $-\Omega {\tau }^{\ast }+{\tau }^{\ast }\Omega$. They cancel the extra material-derivative terms, proving $\boxed{{\mathcal{D}}_l{\tau }^{\ast }=Q({\mathcal{D}}_l\tau )Q^T}$. The translation has no effect on a stress or velocity gradient.

For the incompressible Oldroyd-A model, avoid using the same symbol for a stress tensor and its relaxation time by referring to $\tau >0$ below only as the scalar relaxation time. Define $S={\sigma }^d-2{\mu }_{r}D$. The constitutive equation becomesBy the pull-back identity, $A=F^{T}SF$ satisfiesAn integrating factor givesPush forward by $F(t{)}^{-T}$ and $F(t{)}^{-1}$. For a convergent, fading remote-past history with the transported homogeneous term tending to zero as $t_0\to -\infty$, this proves the fading-memory representation of the Oldroyd-A model:For arbitrary finite-time initial data retain $e^{-(t-t_0)/\tau }F(t{)}^{-T}A(t_0)F(t{)}^{-1}$. Thus the printed integral-only expression incorporates an initial-memory condition; it is not an unconditional replacement for all initial-value solutions.

For time-dependent simple shear,Put $\Delta \gamma =\gamma (s)-\gamma (t)$. The relative deformation is $F(s)F(t{)}^{-1}=\left(\begin{array}{cc}1& \Delta \gamma \\ 0& 1\end{array}\right)$, so the entire transported integrand simplifies toConsequently the explicit nonpressure stresses areSymmetry gives ${\sigma }_{21}^d={\sigma }_{12}^d$, and the unshown third-direction nonpressure components vanish. The superscript $d$ here labels the paper's nonpressure constitutive stress; it need not have zero trace. Subtracting $pI$ supplies the total Cauchy stress tensor.

For constant shear rate $s_0$, use $\gamma (s)-\gamma (t)=-s_0(t-s)$ and the integrals ${\int }_0^{\infty }{e}^{-a/\tau }da=\tau$ and ${\int }_0^{\infty }a{e}^{-a/\tau }da={\tau }^2$. This gives the steady simple shear of an Oldroyd-A fluid resultand thereforeThe incompressibility pressure cancels from this normal-stress difference. With the usual $\mu \ge {\mu }_r\ge 0$, it is nonnegative, while the second normal-stress difference equals its negative.

Write the velocity gradient aswhere $\mathbf{\Omega }$ is the spin tensor. Expanding the upper-convected derivative in the structure equation givesFor $\xi =0$, the first four terms reproduce the upper-convected derivative. Settingtherefore givesWith polymeric stress ${\mathbf{\tau }}_p=b_1\mathbf{\alpha }$, this is the Oldroyd-B model. For $c_1>0$, its relaxation time and polymer viscosity areIndeed the total stress obeys

For $\xi =2$, the coefficient of
$\dot{\mathbf{\gamma }}\mathbf{\alpha }+\mathbf{\alpha }\dot{\mathbf{\gamma }}$
is $+1/2$, so the objective derivative becomes the lower-convected derivative. Hencerecovers the Oldroyd-A model, again with $c_1>0$ for a finite positive relaxation time. Parameter choices such as $b_1{c}_2=0$ give the degenerate Newtonian limit.