Stokes' paradox

The velocity vector ${\displaystyle u}$ of the fluid may be written in terms of the stream function ${\displaystyle \psi }$ as

${\displaystyle \mathbf {u} ={\begin{pmatrix}{\frac {\partial \psi }{\partial y}}&-{\frac {\partial \psi }{\partial x}}.\end{pmatrix}}}$

As the stream function in a Stokes flow problem, ${\displaystyle \psi }$ satisfies the biharmonic equation.[3] Since the plane may be regarded to as the complex plane, the problem may be dealt with using methods of complex analysis. In this approach, ${\displaystyle \psi }$ is either the real or imaginary part of

${\displaystyle {\bar {z}}f(z)+g(z)}$.[4]

Here ${\displaystyle z=x+iy}$, where ${\displaystyle i}$ is the imaginary unit, ${\displaystyle {\bar {z}}=x-iy}$, and ${\displaystyle f(z),g(z)}$ are holomorphic functions outside of the disk. We will take the real part without loss of generality. Now the function ${\displaystyle u}$, defined by ${\displaystyle u=u_{x}+iu_{y}}$ is introduced. ${\displaystyle u}$ can be written as ${\displaystyle u=-2i{\frac {\partial \psi }{\partial {\bar {z}}}}}$, or ${\displaystyle {\frac {1}{2}}iu={\frac {\partial \psi }{\partial {\bar {z}}}}}$ (using the Wirtinger derivatives). This is calculated to be equal to

${\displaystyle {\frac {1}{2}}iu=f(z)+z{\bar {f\prime }}(z)+{\bar {g\prime }}(z).}$

Without loss of generality, the disk may be assumed to be the unit disk, consisting of all complex numbers z of absolute value smaller or equal to 1.

The boundary conditions are:

${\displaystyle \lim _{z\to \infty }u=1,}$

${\displaystyle u=0,}$

whenever ${\displaystyle |z|=1}$,[1][5] and by representing the functions ${\displaystyle f,g}$ as Laurent series:[6]

${\displaystyle f(z)=\sum _{n=-\infty }^{\infty }f_{n}z^{n},\quad g(z)=\sum _{n=-\infty }^{\infty }g_{n}z^{n},}$

the first condition implies ${\displaystyle f_{n}=0,g_{n}=0}$ for all ${\displaystyle n\geq 2}$.

Using the polar form of ${\displaystyle z}$ results in ${\displaystyle z^{n}=r^{n}e^{in\theta },{\bar {z}}^{n}=r^{n}e^{-in\theta }}$. After deriving the series form of u, substituting this into it along with ${\displaystyle r=1}$, and changing some indices, the second boundary condition translates to

${\displaystyle \sum _{n=-\infty }^{\infty }e^{in\theta }\left(f_{n}+(2-n){\bar {f}}_{2-n}+(1-n){\bar {g}}_{1-n}\right)=0.}$

Since the complex trigonometric functions ${\displaystyle e^{in\theta }}$ compose a linearly independent set, it follows that all coefficients in the series are zero. Examining these conditions for every ${\displaystyle n}$ after taking into account the condition at infinity shows that ${\displaystyle f}$ and ${\displaystyle g}$ are necessarily of the form

${\displaystyle f(z)=az+b,\quad g(z)=-bz+c,}$

where ${\displaystyle a}$ is an imaginary number (opposite to its own complex conjugate), and ${\displaystyle b}$ and ${\displaystyle c}$ are complex numbers. Substituting this into ${\displaystyle u}$ gives the result that ${\displaystyle u=0}$ globally, compelling both ${\displaystyle u_{x}}$ and ${\displaystyle u_{y}}$ to be zero. Therefore, there can be no motion – the only solution is that the cylinder is at rest relative to all points of the fluid.

The paradox is caused by the limited validity of Stokes' approximation, as explained in Oseen's criticism: the validity of Stokes' equations relies on Reynolds number being small, and this condition cannot hold for arbitrarily large distances ${\displaystyle r}$.[7][2]

A correct solution for a cylinder was derived using Oseen's equations, and the same equations lead to an improved approximation of the drag force on a sphere.[8][9]

• Oseen's approximation
• Stokes' law