# Stokes' paradox

In the science of fluid flow, Stokes' paradox is the phenomenon that there can be no creeping flow of a fluid around a disk in two dimensions; or, equivalently, the fact there is no non-trivial steady-state solution for the Stokes equations around an infinitely long cylinder. This is opposed to the 3-dimensional case, where Stokes' method provides a solution to the problem of flow around a sphere.
Stokes' paradox was resolved by Carl Wilhelm Oseen in 1910, by introducing the Oseen equations which improve upon the Stokes equations – by adding convective acceleration.

## Derivation

The velocity vector $\mathbf {u}$ of the fluid may be written in terms of the stream function $\psi$ as
    $\mathbf {u} =\left({\frac {\partial \psi }{\partial y}},-{\frac {\partial \psi }{\partial x}}\right).$
The stream function in a Stokes flow problem, $\psi$ satisfies the biharmonic equation. By regarding the $(x,y)$-plane as the complex plane, the problem may be dealt with using methods of complex analysis. In this approach, $\psi$ is either the real or imaginary part of
    ${\bar {z}}f(z)+g(z)$.
Here $z=x+iy$, where $i$ is the imaginary unit, ${\bar {z}}=x-iy$, and $f(z),g(z)$ are holomorphic functions outside of the disk. We will take the real part without loss of generality. Now the function $u$, defined by $u=u_{x}+iu_{y}$ is introduced. $u$ can be written as $u=-2i{\frac {\partial \psi }{\partial {\bar {z}}}}$, or ${\frac {1}{2}}iu={\frac {\partial \psi }{\partial {\bar {z}}}}$ (using the Wirtinger derivatives). This is calculated to be equal to
    ${\frac {1}{2}}iu=f(z)+z{\bar {f\prime }}(z)+{\bar {g\prime }}(z).$
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:
    $\lim _{z\to \infty }u=1,$
    $u=0,$
whenever $|z|=1$, and by representing the functions $f,g$ as Laurent series:
    $f(z)=\sum _{n=-\infty }^{\infty }f_{n}z^{n},\quad g(z)=\sum _{n=-\infty }^{\infty }g_{n}z^{n},$
the first condition implies $f_{n}=0,g_{n}=0$ for all $n\geq 2$.
Using the polar form of $z$ results in $z^{n}=r^{n}e^{in\theta },{\bar {z}}^{n}=r^{n}e^{-in\theta }$. After deriving the series form of u, substituting this into it along with $r=1$, and changing some indices, the second boundary condition translates to
    $\sum _{n=-\infty }^{\infty }e^{in\theta }\left(f_{n}+(2-n){\bar {f}}_{2-n}+(1-n){\bar {g}}_{1-n}\right)=0.$
Since the complex trigonometric functions $e^{in\theta }$ compose a linearly independent set, it follows that all coefficients in the series are zero. Examining these conditions for every $n$ after taking into account the condition at infinity shows that $f$ and $g$ are necessarily of the form
    $f(z)=az+b,\quad g(z)=-bz+c,$
where $a$ is an imaginary number (opposite to its own complex conjugate), and $b$ and $c$ are complex numbers. Substituting this into $u$ gives the result that $u=0$ globally, compelling both $u_{x}$ and $u_{y}$ 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.

## Resolution

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 $r$.
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.

## Unsteady-state flow around a circular cylinder

On the contrary to Stokes' paradox, there exists the unsteady-state solution of the same problem which models a fluid flow moving around a circular cylinder with Reynolds number being small. This solution can be given by explicit formula in terms of vorticity of the flow's vector field.

### Formula of the Stokes Flow around a circular cylinder

The vorticity of Stokes' flow is given by the following relation: $w_{k}(t,r)=W_{|k|,|k|-1}^{-1}\left[e^{-\lambda ^{2}t}W_{|k|,|k|-1}[w_{k}(0,\cdot )](\lambda )\right](t,r).$
Here $w_{k}(t,r)$ - are the Fourier coefficients of the vorticity's expansion by polar angle which are defined on $(r_{0},\infty )$, $r_{0}$ - radius of the cylinder, $W_{|k|,|k|-1}$, $W_{|k|,|k|-1}^{-1}$ are the direct and inverse special Weber's transforms, and initial function for vorticity $w_{k}(0,r)$ satisfies no-slip boundary condition.
Special Weber's transform has a non-trivial kernel, but from the no-slip condition follows orthogonality of the vorticity flow to the kernel.

### Derivation
#### Special Weber's transform

Special Weber's transform is an important tool in solving problems of the hydrodynamics. It is defined for $k\in \mathbb {R}$ as $W_{k,k-1}[f](\lambda )=\int _{r_{0}}^{\infty }{\frac {J_{k}(\lambda s)Y_{k-1}(\lambda r_{0})-Y_{k}(\lambda s)J_{k-1}(\lambda r_{0})}{\sqrt {J_{k-1}^{2}(\lambda r_{0})+Y_{k-1}^{2}(\lambda r_{0})}}}f(s)sds,$ where $J_{k}$, $Y_{k}$ are the Bessel functions of the first and second kind respectively. For $k>1$ it has a non-trivial kernel which consists of the functions $C/r^{k}\in \ker(W_{k,k-1})$.
The inverse transform is given by the formula $W_{k,k-1}^{-1}[{\hat {f}}](r)=\int _{0}^{\infty }{\frac {J_{k}(\lambda r)Y_{k-1}(\lambda r_{0})-Y_{k}(\lambda s)J_{k-1}(\lambda r_{0})}{\sqrt {J_{k-1}^{2}(\lambda r_{0})+Y_{k-1}^{2}(\lambda r_{0})}}}{\hat {f}}(\lambda )\lambda d\lambda .$
Due to non-triviality of the kernel, the inversion identity $f(r)=W_{k,k-1}^{-1}\left[W_{k,k-1}[f]\right](r)$ is valid if $k\leq 1$. Also it is valid in the case of $k>1$ but only for functions, which are orthogonal to the kernel of $W_{k,k-1}$ in $L_{2}(r_{0},\infty )$ with infinitesimal element $rdr$: $\int _{r_{0}}^{\infty }{\frac {1}{r^{k}}}f(r)rdr=0,~k>1.$

#### No-slip condition and Biot–Savart law

In exterior of the disc of radius $r_{0}$ $B_{r_{0}}=\{\mathbf {x} \in \mathbb {R} ^{2},~\vert \mathbf {x} \vert >r_{0}\}$ the Biot-Savart law $\mathbf {v} (\mathbf {x} )={\frac {1}{2\pi }}\int _{B_{r_{0}}}{\frac {(\mathbf {x} -\mathbf {y} )^{\perp }}{\vert \mathbf {x} -\mathbf {y} \vert ^{2}}}w(\mathbf {y} )\operatorname {d\mathbf {y} } +\mathbf {v} _{\infty },$ restores the velocity field $\mathbf {v} (\mathbf {x} )$ which is induced by the vorticity $w(\mathbf {x} )$ with zero-circularity and given constant velocity $\mathbf {v} _{\infty }$ at infinity.
No-slip condition for $\mathbf {x} \in S_{r_{0}}=\{\mathbf {x} \in \mathbb {R} ^{2},~\vert \mathbf {x} \vert =r_{0}\}$ ${\frac {1}{2\pi }}\int _{B_{r_{0}}}{\frac {(\mathbf {x} -\mathbf {y} )^{\perp }}{\vert \mathbf {x} -\mathbf {y} \vert ^{2}}}w(\mathbf {y} )\operatorname {d\mathbf {y} } +\mathbf {v} _{\infty }=0$ leads to the relations for $k\in \mathbf {Z}$: $\int _{r_{0}}^{\infty }r^{-\vert k\vert +1}w_{k}(r)dr=d_{k},$ where $d_{k}=\delta _{\vert k\vert ,1}(v_{\infty ,y}+ikv_{\infty ,x}),$ $\delta _{\vert k\vert ,1}$ is the Kronecker delta, $v_{\infty ,x}$, $v_{\infty ,y}$ are the cartesian coordinates of $\mathbf {v} _{\infty }$.
In particular, from the no-slip condition follows orthogonality the vorticity to the kernel of the Weber's transform $W_{k,k-1}$: $\int _{r_{0}}^{\infty }r^{-\vert k\vert +1}w_{k}(r)dr=0~for~|k|>1.$

#### Vorticity flow and its boundary condition

Vorticity $w(t,\mathbf {x} )$ for Stokes flow satisfies to the vorticity equation ${\frac {\partial w(t,\mathbf {x} )}{\partial t}}-\Delta w=0,$ or in terms of the Fourier coefficients in the expansion by polar angle ${\frac {\partial w_{k}(t,r)}{\partial t}}-\Delta w_{k}=0,$ where $\Delta _{k}w_{k}(t,r)={\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial }{\partial r}}w_{k}(t,r)\right)-{\frac {k^{2}}{r^{2}}}w_{k}(t,r).$
From no-slip condition follows ${\frac {d}{dt}}\int _{r_{0}}^{\infty }r^{-\vert k\vert +1}w_{k}(t,r)dr=0.$
Finally, integrating by parts, we obtain the Robin boundary condition for the vorticity: $\int _{r_{0}}^{\infty }s^{-|k|+1}\Delta _{k}w_{k}(t,r)dr=-r_{0}^{-|k|}\left(r_{0}{\frac {\partial w_{k}(t,r)}{\partial r}}{\Big |}_{r=r_{0}}+|k|w_{k}(t,r_{0})\right)=0.$ Then the solution of the boundary-value problem can be expressed via Weber's integral above.

### Remark

Formula for vorticity can give another explanation of the Stokes' Paradox. The functions ${\frac {C}{r^{k}}}\in ker(\Delta _{k}),~k>1$ belong to the kernel of $\Delta _{k}$ and generate the stationary solutions of the vorticity equation with Robin-type boundary condition. From the arguments above any Stokes' vorticity flow with no-slip boundary condition must be orthogonal to the obtained stationary solutions. That is only possible for $w\equiv 0$.