Past exam of the theoretical mechanics course of the University of Oxford / 2007 / A / 4

Two-dimensional incompressibility permits $u={\psi }_y$, $v=-{\psi }_x$. Taking the curl of the Stokes momentum equation eliminates pressure and gives ${\nabla }^2(v_x-u_y)=0$. Since $v_x-u_y=-{\nabla }^2\psi$, the stream function obeys $\boxed{{\nabla }^4\psi =0}$.

Scale radius by $a$, velocity by $U$ and stream function by $Ua$. In polar coordinates $u_r=r^{-1}{\psi }_{\theta }$, $u_{\theta }=-{\psi }_r$. No penetration makes $\psi$ constant on the cylinder; choose that constant zero. No slip also requires ${\psi }_r=0$ at $r=1$. The uniform stream has $\psi =r\sin \theta$, specifying the far-field condition.

The boundary forcing has angular mode $\sin \theta$, so its corresponding mode must have $\psi =f(r)\sin \theta$. Define $L_1=d^2/d{r}^2+r^{-1}d/dr-r^{-2}$. Then ${\nabla }^4\psi =(L_1^{2}f)\sin \theta$, while $L_1{r}^s=(s^2-1)r^{s-2}$. Thus the indicial equation of $L_1^2$ is $(s^2-1)[(s-2{)}^2-1]=(s-3)(s-1{)}^2(s+1)=0$, givingThe repeated exponent one supplies the logarithmic term. The condition $f^{\prime }\to 1$ first forces $A=B=0$ and $C=1$. Then $f(1)=0$ forces $D=-1$, but $f^{\prime }(1)=C-D=2\ne 0$. No additional Fourier mode can repair the missing boundary condition in this angular mode. Therefore the unbounded-domain Stokes problem has no solution with both no slip and a nonzero uniform far stream.

This is Stokes' paradox. It identifies a nonuniform approximation, rather than the absence of physical flow. For any small but nonzero Reynolds number, convective inertia becomes important sufficiently far away. An inner Stokes region must be matched to an outer Oseen or Navier-Stokes region that retains that inertia. Alternatively, finite external boundaries introduce a different well-posed physical problem; they cannot be sent to infinity while keeping the naive Stokes solution unchanged.