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Write $q=\dot{\gamma }\tau$ and $f=1+\alpha \mathrm{tr}A$. With $L_{ij}={\partial }_j{v}_i$, the steady homogeneous upper-convected derivative is $-LA-A{L}^T$. Its diagonal and shear equations giveOn the physical branch $f>0$, the other off-diagonal components vanish too. Thus $a_{22}=a_{33}=0$, $a_{11}=2a_{12}^2$ andFor $\alpha \ge 0$ the right side has derivative $1+6\alpha a_{12}^2>0$, so the shear stress determines a unique shear rate. The shear viscosity is $G_0\tau /f$ and decreases with shear magnitude when $\alpha >0$. This is the shear part of shear and pipe flow of an affine linear PTT fluid.

For fully developed pipe flow take $\mathbf{v}=w(r)\hat{\mathbf{z}}$ and the signed axial pressure gradient $dp/dz=\Delta p$. Regularity at the axis and the supplied momentum balance determine the shear stress without using the constitutive equation:Locally the constitutive equations have the same shear form, with axial flow replacing the Cartesian flow direction and $\dot{\gamma }=w^{\prime }(r)$ carrying its sign. HenceIntegrating and imposing no-slip boundary conditions at $r=R$ givesA negative pressure gradient produces positive axial flow. At $\alpha =0$ this is the ordinary Hagen-Poiseuille flow; positive $\alpha$ increases the flux at a fixed pressure drop through shear thinning. The nonzero axial normal stress varies radially, but its axial derivative vanishes in fully developed flow, so it does not change the axial momentum balance used here.

Let the flow direction be $x$, the gradient direction $y$, and the vorticity direction $z$, with simple shear flow $u=(sy,0,0)$ and signed shear rate $s=\dot{\gamma }$. Use the gradient convention established in part (a):At steady homogeneous shear the ordinary time derivative and advection vanish. Put $X={\sigma }_{xx}$, $Y={\sigma }_{yy}$ and $S={\sigma }_{xy}$. The corotational Maxwell fluid law then givesThe $zz$ equation gives ${\sigma }_{zz}=0$, and the unforced $xz,yz$ equations force those components to zero. Solving the three displayed equations gives the steady shear of a corotational Maxwell fluid:The deviatoric stress is indeed traceless. Its isotropic pressure contribution is irrelevant to either shear viscosity or a normal-stress difference.

Consequently the steady shear viscosity isFor $s>0$, $d{\eta }_{\mathrm{sh}}/ds=-2\mu {\lambda }^{2}s/(1+{\lambda }^2{s}^2{)}^2<0$ when $\mu ,\lambda >0$; it decreases with $|s|$ and tends to zero at large rate. This proves shear thinning directly, without confusing a decreasing viscosity with a necessarily decreasing shear stress.

With the standard definitions $N_1={\sigma }_{xx}-{\sigma }_{yy}$ and $N_2={\sigma }_{yy}-{\sigma }_{zz}$,At zero shear both vanish; at small shear they are quadratic, consistently with part (b). The signs and the fact $|N_2|<N_1$ agree qualitatively with many polymeric-fluid experiments. However the fixed ratio $|N_2|/N_1=1/2$ often overestimates the measured second difference and cannot capture variation of that ratio with material and shear rate. For example, measurements on polyethylene show a smaller, rate-dependent relative second normal-stress difference. Thus the qualitative normal-stress signs are reasonable, but the predicted relative magnitude is not generally quantitatively satisfactory; these signs are not a universal statement about every non-Newtonian fluid.

At order ${\alpha }^0$, the preceding equations giveAt order ${\alpha }^1$, the quadratic term is evaluated on ${\mathbf{\tau }}^0$. Solving first the $yy$ equation, then $xy$, then $xx$, givesThus the apparent shear viscosity isso positive $\alpha$ produces shear thinning. The expansion requires $\alpha \ll 1$ and $\alpha (\lambda g{)}^2\ll 1$; it cannot describe arbitrarily high shear rates even when $\alpha$ is numerically small.

Using the normal-stress difference definitionswe findIn the low-rate limit, $-2{\Psi }_2/{\Psi }_1=\alpha +O({\alpha }^2)$. A cone-and-plate or parallel-plate rheometer can measure shear stress and normal thrust over a low-rate range; combining ${\Psi }_1$ and ${\Psi }_2$ then estimates $\alpha$ independently of the viscosity scale.

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 ${\eta }_s+{\eta }_p$.

In steady simple shear flow, the tensor equation gives $a_{22}=a_{33}=0$, $a_{11}=2a_{12}^2$ and the equation in the title. A circular pipe with axial pressure gradient $\Delta p$ has $G_0{a}_{rz}=\Delta pr/2$. Thus $w^{\prime }=\Delta pr/(2G_0\tau )+\alpha \Delta p^3{r}^3/(4G_0^3\tau )$, and no-slip boundary conditions givePositive $\alpha$ produces shear thinning; the sign of the pressure gradient determines the direction of the flow.

For simple shear flow $u=(\dot{\gamma }y,0,0)$, put $s=\dot{\gamma }$ and $\ell =\lambda s$. The steady deviatoric stress isThis follows by setting ${\omega }_{xy}=-s$, ${\omega }_{yx}=s$ in the corotational Maxwell fluid equation: ${\sigma }_{xx}=\lambda s{\sigma }_{xy}$, ${\sigma }_{yy}=-\lambda s{\sigma }_{xy}$, and ${\sigma }_{xy}+\lambda s({\sigma }_{xx}-{\sigma }_{yy})/2=\mu s$. Hence the shear viscosity is $\mu /(1+{\lambda }^2{s}^2)$, which exhibits shear thinning, while $N_1=2\mu \lambda s^2/(1+{\lambda }^2{s}^2)$ and $N_2=-N_1/2$. These differences are independent of the isotropic pressure.