Translating cylinder in viscous fluid

[MichielM]

Homework Statement

I have a cylinder of radius R translating (in radial direction) with velocity U in a viscous fluid at Re<<1. Find the force exerted on the cylinder per unit length

Homework Equations

Re<<1 so the Navier-Stokes equation simplifies to the Stokes equation:
\nabla P=\mu \nabla^2 v
The cylinder makes use of polar coordinates useful so the stream function for this problem is defined as:
\frac{1}{r}\frac{\partial \Psi}{\partial \theta}=u_r
-\frac{\partial \Psi}{\partial r}=u_{\theta}

The Attempt at a Solution

Given the problem statement and the choice for polar coordinates, the boundary conditions become:
u_r(R)=U \cos\theta
u_{\theta}(R)=-U \sin\theta
u_r(\infty)=u_{\theta}(\infty)=0

With help of the continuity equation the Stokes equation can be turned into:
\nabla^4 \Psi=0
And taking \Psi=\sin \theta f(r) the equation further 'simplifies' to:
\left[\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}-\frac{1}{r^2}\right]^2f(r)=0

A general solution to this equation is f(r)=A r^3+Br \ln r + C r+\frac{D}{r}

Using the definition of the stream function it follows that:
u_r=\frac{f(r)}{r}\cos \theta and u_{\theta}=-\sin\theta \frac{\partial f(r)}{\partial r}.

The boundary conditions for f(r) then become:
f(R)=U R
\frac{\partial f(R)}{\partial r}=U
\frac{f(\infty)}{\infty}\cos \theta=0
-\sin \theta \frac{\partial f(\infty)}{\partial r}=0

The two infinity conditions yield A=B=C=0 and for the other two I find:

\frac{D}{R}=U R and \frac{-D}{R^2}=U

And that is where I get stuck, because I will not be able to satisfy these conditions, so what did I do wrong?

 

[MichielM]

mmm, seems like I found the famous Stokes' paradox. For other people reading this and wondering how to solve it: take a look at Oseen's approximation method