Second-order fluid

A second-order fluid is the quadratic, weak-flow expansion of an isotropic simple-fluid constitutive equation. In steady simple shear it predicts first and second normal-stress differences $N_1={\Psi }_1{\dot{\gamma }}^2$ and $N_2={\Psi }_2{\dot{\gamma }}^2$.

For a sphere of radius $a$ rotating with angular velocity $\Omega$, the primary Stokes flow has $v_{\varphi }=K\sin \theta /r^2$, $K=\Omega a^3$. In the weak, inertialess second-order fluid correction, put $c={\psi }_1+2{\psi }_2$. The stream function in the title hasBoth meridional components vanish at $r=a$ and decay at infinity. Indeed, $E_{r\varphi }=-3K\sin \theta /(2r^3)$ gives $[\nabla \times \nabla \cdot (2c{E}^2){]}_{\varphi }=72c{K}^2\sin \theta \cos \theta /r^8$, and the curl of the momentum equation reduces to $(d^2/d{r}^2-6/r^2{)}^{2}H=72c{K}^2/(\mu r^7)$. For $c>0$ the secondary flow enters towards the equator and leaves near the poles; $c<0$ reverses it and $c=0$ removes this correction. The meridional circulation is even in the sign of sphere rotation. Inertial secondary flow is excluded from this calculation.

Write the non-Newtonian extra stress as $S=2({\psi }_1+2{\psi }_2)E^2-{\psi }_1\stackrel{\circ }{E}$, with the Jaumann derivative. For a steady Stokes flow, $\nabla \times \nabla \cdot \stackrel{\circ }{E}=0$. If ${\psi }_1+2{\psi }_2=0$, the extra force is a gradient and only alters pressure. In a planar incompressible flow, $E^2=q\mathrm{diag}(1,1,0)$, so its divergence is also a gradient; no coefficient restriction is needed. A Newtonian velocity with prescribed boundary velocities consequently solves the second-order model with an altered pressure. On the regular weak-flow branch, Uniqueness of Stokes flow makes each velocity correction vanish. This does not establish uniqueness of arbitrary large nonlinear branches or the absence of higher-order effects in a real material.