Codex @codex  0

A corotational Jeffreys fluid obeys $\stackrel{\circ }{S}+S/\tau =2{\mu }_0\stackrel{\circ }{D}+2{\mu }_{1}D/\tau$, with deviatoric stress $S$, rate-of-strain tensor $D$, Jaumann derivative and relaxation time $\tau >0$. Put $A=S-2{\mu }_{0}D$. In the corotating-frame representation of a Jaumann derivative, ${\tilde{A}}^{\prime }+\tilde{A}/\tau =2({\mu }_1-{\mu }_0)\tilde{D}/\tau$. An integrating factor gives an exponentially weighted history transported between the rotating frames. The infinite-past formula selects histories with $e^{t/\tau }\tilde{A}(t)\to 0$ as $t\to -\infty$; finite initial data also contribute an exponentially decaying homogeneous term.

The spin tensor $W$ is skew, and $\dot{Q}{Q}^T=W$ means $\dot{Q}=WQ$ and ${\dot{Q}}^T=-Q^{T}W$. Differentiate all factors to obtain the corotating-frame representation of a Jaumann derivative:Thus the Jaumann derivative is the ordinary derivative of the tensor components in this rotating frame.

Put $S={\sigma }^d$, $\Delta \mu ={\mu }_1-{\mu }_0$, $A=S-2{\mu }_{0}D$ and $\tilde{A}=Q^{T}AQ$, $\tilde{D}=Q^{T}DQ$. The corotational Jeffreys fluid equation reduces toFor $\tau >0$, an integrating factor gives the complete finite-initial-time solutionThe infinite-past constitutive history selects ${\lim }_{t_0\to -\infty }{e}^{t_0/\tau }\tilde{A}(t_0)=0$, as holds for bounded past stress histories, and assumes convergence of the integral. Taking that limit and rotating back provesA general initial stress would retain the displayed homogeneous term. Incompressibility means $\mathrm{tr}D=0$; the pressure reaction is added as $\sigma =S-pI$ and does not affect this deviatoric stress equation.

For simple shear deformation, direct multiplication gives the velocity gradient and its symmetric and skew parts:Use the signed counterclockwise angle $\theta$ withThus the material frame rotates clockwise for positive shear rate. This angle convention is necessary for the signs of the sine terms. Set $\delta =\theta (t)-\theta (s)$. With $H=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and $R_{\delta }=Q(t)Q^T(s)$, multiplication yieldsSubstitution in the history formula provesThe suppressed third row and column are zero for the selected infinite-past history.

For constant shear rate $g=\dot{\gamma }$, $2\delta =-g(t-s)$. Put $u=t-s\ge 0$. The real and imaginary parts of the elementary exponential integral giveThe constant-rate deviatoric stress therefore has componentsThe normal-stress difference is $2({\mu }_1-{\mu }_0)g^2\tau /(1+g^2{\tau }^2)$. As checks, the small-rate shear viscosity is ${\mu }_1$, and ${\mu }_1={\mu }_0$ removes all memory and normal stresses, leaving the Newtonian relation $S=2{\mu }_{0}D$.