Incompressible Fluid Flow around sphere (Stokes) Stokes solved this 1851. I have a question regarding the derivation.
Following Batchelor the equations to be solved are
\begin{align}
\nabla \left( \frac{p - p_0}{\mu} \right) = \nabla^2 \vec{u} = -\nabla \times \vec{\omega} \\
\nabla \cdot \vec{u} = 0
\end{align}
with boundary conditions $\vec{u} = \vec{U}$ at the surface of the sphere where U is the velocity of the sphere in z-direction.
Also $\vec{u} \rightarrow 0, p-p_0 \rightarrow 0 \quad {\rm as} \quad r \rightarrow \infty$. The center of the sphere is instantaneous at the origin of the co-ordinate system. $\vec{\omega}=\nabla \times \vec{u}$ is the vorticity and $\mu$ the viscosity.
Since $p-p_0$ solves the Laplace equation it can be written as a series in solid harmonics
\begin{align}
\frac{p-p_0}{\mu} = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} c_{lm} I_{l}^m \quad {\rm .}
\end{align}
The m-sum is absent because of the azimutal symmetry.
He argues that $p-p_0 \propto \vec{U} \cdot \vec{r} \propto \cos \theta$ and therefore the only term $\neq 0$ is $l=1$.
I cannot see why $p-p_0 \propto \cos \theta$ should be fulfilled though.
 A: Stokes equation is linear in its dependent variables. Further its boundary condition is linearly dependent on translation velocity $\mathbf{U}$. This means that if $\{\mathbf{u},p\}$ is the solution of Stokes equation for translation velocity $\mathbf{U}$, then if the translation velocity is scaled to $\alpha\mathbf{U}$, the new solution solution corresponding to this new boundary condition is simply $\{\alpha\mathbf{u},\alpha p\}$. This suggests that both $\mathbf{u}$ and $p$ must be linear functions of $\mathbf{U}$.
In 3-dimensions, a solution of the Laplace equation is $1/r$, i.e. $\nabla^2(1/r)=0$, in which $r$ is the radial distance to a point in the flow field from the sphere center. Further solutions to the Laplace equation may be generated by taking successive gradients of $1/r$. Of course each new solution generated in this way is of higher tensorial rank than the previous one. The solutions are: $$\frac{1}{r},\nabla\frac{1}{r},\nabla\nabla\frac{1}{r},\ldots $$ 
Now the pressure field, which is a scalar field, must be linear in $\mathbf{U}$ while involving one (or more) of the solutions listed above. The only viable combination is $$p=\lambda \mathbf{U}\cdot\nabla\frac{1}{r}=-\lambda\frac{\mathbf{U}\cdot\mathbf{r}}{r^3}$$ in which $\lambda$ is a constant. This is how $p$ (or $p-p_0$, if your datum is not zero) is found to be proportional to $\cos\theta$, in which $\theta$ is the angle between the direction of motion and the position vector of a point in the flow field.
Reference: Guazzelli & Morris, A Physical Introduction to Suspension Dynamics, Cambridge University Press.
