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: $$w_k(t,r) = W^{-1}_{|k|,|k|-1} \left [ e^{-\lambda^2 t} W_{|k|,|k|-1} [w_k(0,\cdot)](\lambda) \right ](t,r).$$

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, $$W_{|k|,|k|-1}$$, $$W^{-1}_{|k|,|k|-1}$$ 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.

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 $$W_{k,k- 1}[f](\lambda) = \int_{r_0}^\infty \frac{J_{k}(\lambda s)Y_{k- 1}(\lambda r_0) - Y_{k}(\lambda s)J_{k- 1}(\lambda r_0)} {\sqrt{J_{k- 1}^2(\lambda r_0) + Y_{k- 1}^2(\lambda r_0)}}  f(s) s ds,$$ 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 $$W^{-1}_{k,k- 1}[\hat f](r) = \int_{0}^\infty \frac {J_{k}(\lambda r)Y_{k- 1}(\lambda r_0) - Y_{k}(\lambda s)J_{k- 1}(\lambda r_0)}{\sqrt{J_{k- 1}^2(\lambda r_0) + Y_{k- 1}^2(\lambda r_0)}} \hat f (\lambda) \lambda d\lambda.$$

Due to non-triviality of the kernel, the inversion identity $$ f(r) = W^{-1}_{k,k-1}\left [W_{k,k-1} [f] \right ](r) $$ 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 $$rdr$$: $$ \int_{r_0}^\infty \frac 1{r^k} f(r) r dr = 0,~k>1. $$

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 $$d_k=\delta_{\vert k \vert,1} (v_{\infty,y} + i k v_{\infty,x}), $$ $$\delta_{\vert k \vert,1}$$ 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 $$\Delta_k w_k(t,r) = \frac 1r \frac {\partial}{\partial r}\left(r \frac {\partial}{\partial r}w_k(t,r)\right) - \frac{k^2}{r^2} w_k(t,r).$$

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 $$ w\equiv 0$$.