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Let $g=\dot{\gamma }$ denote the constant rate in the simple shear flowThenWhen $\xi =1$, the structure equation contains the Jaumann derivative. In a steady homogeneous flow it becomesSolving its component equations gives

The $b_2$ term has no $xy$ component because ${\alpha }_{xx}+{\alpha }_{yy}=0$. Thusand the shear viscosity isIt exhibits shear thinning whenever $b_1{c}_1{c}_2>0$: it decreases from ${\eta }_0+b_1{c}_2/c_1$ at zero shear rate to the solvent plateau ${\eta }_0$ at large shear rate. If the product vanishes, the viscosity is constant.

For the diagonal stresses,The two normal-stress differences are therefore

In either half-plane, let $U_{\infty }$ denote its far-field value. Multiplying the ODE by $U^{\prime }$ and integrating from the far field givesSince the profile decreases,Continuity of $U^{\prime }$ at $y=0$ gives

For $n=1$, let $\ell =A^{-1}=\sqrt{hHK/\mu }$ and $\Delta U=U_{-}-U_{+}$. ThenThe Newtonian disturbance therefore has exponential tails.

For $n<1$ and $y>0$, put $W=U-U_{+}$. Integration givesThe corresponding expression on $y<0$ follows by reflection about $(0,U_0)$. A shear thinning power-law material has algebraic rather than exponential margin tails. Measurements of how rapidly an ice-stream speed approaches its far-field values could therefore distinguish an approximately Newtonian rheology from power-law shear thinning and estimate $n$.

The trace isSubstituting it into $F=(L^2-3)/(L^2-T)$ givesFor $\beta \ll 1$, $F=1+\beta +O({\beta }^2)$. For $\beta \gg 1$, $F\sim {\beta }^{1/3}$. The effective shear viscosity is thereforeThe 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 ${\eta }_s+{\eta }_p$.