Sphere falling in viscous fluid boundary condition A sphere, radius $a$, falls at constant velocity $U_0$ through an incompressible
viscous fluid. I am told to assume that
\begin{equation}
\mathbf{u} = U_0\mathbf{e}_z + \tilde{\mathbf{u}}(\mathbf{x})
\end{equation}
where
\begin{equation}
\tilde{\mathbf{u}}(\mathbf{x})=\nabla\times\nabla\times(U_0f(r)\mathbf{e}_z).
\end{equation}
From this I have shown that
\begin{equation}
\tilde{\mathbf{u}}(\mathbf{x})=-U_0\Delta f \mathbf{e}_z
\end{equation}
and by taking the curl of the equations for Stokes flow,
\begin{equation}
\Delta^2 f= \text{ constant.}
\end{equation}
where $\Delta^2 f=\Delta(\Delta f)$. I also know that
\begin{equation}
\Delta f\rightarrow0 \text{ as } r\rightarrow\infty.
\end{equation}
Here is where I am confused
I need to show that $f(r)=Ar+\frac{B}{r}$. I tried to do so by the Ansatz $f=r^a$ and found that $a=-1,0,1,2,4$ are all suitable for $\Delta^2 f=$ constant to hold. So $f(r)=Ar+\frac{B}{r}+C+Dr^2+Er^4.$ To satisfy $\Delta f\rightarrow0 \text{ as } r\rightarrow\infty$, I set $D=E=0$.
But I cannot explain why $C=0$. Why is this true?
I also don't know how to set $A,B$
I think that $\textbf{u}=U_0\textbf{e}_z \text{ at } r=a$ for the no-slip condition, which sets $\tilde{\textbf{u}}=-U_0\Delta f\textbf{e}_z=0 \text{ at } r=a \implies A=0$.
So now I have $f(r)=\frac{B}{r}$, and somehow I have to use the fact that $\nabla\times\textbf{e}_\phi=\frac{cot\theta}{r}\textbf{e}_r-\frac{1}{r}\textbf{e}_\theta$ to find $B$. I am completely lost here.
 A: The free-stream and boundary conditions are $\mathbf{u}(\mathbf{x}) \to U_0\mathbf{e}_z = U_0\cos \theta \,\mathbf{e}_r - U_0 \sin \theta \,\mathbf{e}_\theta $ as $r \to \infty$ and $\mathbf{u}(\mathbf{x}) =\mathbf{0}$ at $r = a$. For axisymmetric flow the disturbance velocity  $\tilde{\mathbf{u}}(\mathbf{x})$ has non-zero components $\tilde{u}_r$ and $\tilde{u}_\theta$ in the radial and angular directions and the free-stream and boundary conditions reduce to
$$\tag{1} \tilde{u}_r \to 0, \,\, \tilde{u}_\theta \to 0 \,\, \text{ as } \,\, r \to \infty$$
$$\tag{2}\tilde{u}_r = -U_0\cos\theta,\,\, \tilde{u}_\theta = U_0\sin \theta \,\, \text{ at } \,\, r = a$$
This problem can be framed in terms of a streamfunction $\psi$ such that
$$\tilde{u}_r = \frac{1}{r^2 \sin \theta}\frac{\partial \psi}{\partial \theta}, \,\, \tilde{u}_\theta = \frac{-1}{r \sin \theta}\frac{\partial \psi}{\partial r}$$
Similar to your approach, the streamfunction can be shown to have the general solution
$$\psi = \left(\frac{A}{r} + Br + Cr^2 + Dr^4\right) \sin^2 \theta,$$
whence
$$\tag{3}\tilde{u}_r = \frac{2 \cos \theta}{r^2}\left(\frac{A}{r} + Br + Cr^2 + Dr^4\right), \\ \tilde{u}_\theta = \frac{- \sin \theta}{r}\left(-\frac{A}{r^2} + B + 2Cr + 4Dr^3\right)$$
Applying condition (1) we get $C= D = 0$.  Applying condition (2) we get
$$\left.\tilde{u}_r\right|_{r = a} = 2 \cos \theta\left(\frac{A}{a^3} + \frac{B}{a} \right) = - U_0 \cos \theta, \\ \left.\tilde{u}_\theta\right|_{r = a} = -\sin \theta\left(-\frac{A}{a^3} + \frac{B}{a} \right) =  U_0 \sin \theta$$
Solving, we get  $A= \frac{U_0a^3}{4}$ and $B= \frac{-3U_0a}{4}$ and
$$\tilde{u}_r = U_0 \cos\theta \left(\frac{ a^3}{2r^3} - \frac{3a}{2r}\right), \,\,\tilde{u}_\theta = -U_0 \sin\theta \left(-\frac{ a^3}{4r^3} - \frac{3a}{4r}\right)$$

Linking Approaches
Starting with the ansatz $\tilde{\mathbf{u}}=\nabla\times\nabla\times(U_0f(r)\mathbf{e}_z)$, we have
$$\nabla\times(U_0f(r)\mathbf{e}_z) = -U_0\sin \theta \frac{df}{dr} \mathbf{e}_\phi,$$
and
$$\tag{4}\tilde{\mathbf{u}} = \nabla \times \left( -U_0\sin \theta \frac{df}{dr} \mathbf{e}_\phi\right) = \frac{1}{r \sin \theta} \frac{\partial}{\partial \theta}\left(- U_0\sin^2 \theta \frac{df}{dr} \right)\mathbf{e}_r - \frac{1}{r } \frac{\partial}{\partial r}\left(- U_0r\sin \theta \frac{df}{dr} \right)\mathbf{e}_\theta \\ = \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta}\left(\underbrace{- U_0\sin^2 \theta \,r\frac{df}{dr}}_{\psi} \right)\mathbf{e}_r - \frac{1}{r \sin \theta } \frac{\partial}{\partial r}\left(\underbrace{- U_0\sin^2 \theta \,r\frac{df}{dr}}_{\psi} \right)\mathbf{e}_\theta$$
(Here you see the appearance of the streamfunction $\psi$).  Taking the curl of the Stokes equations and some additional steps it can be shown that the problem reduces to
$$\left(\frac{\partial^2}{\partial r^2} + \frac{\sin \theta}{r^2} \frac{\partial}{\partial \theta}\left( \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \right) \right)^2 \left(U_0 \sin^2 \theta \, r \frac{df}{dr} \right) =0,$$
and with further simplification to
$$\left(\frac{d^2}{d r^2} - \frac{2}{r^2} \right)^2 \left( r \frac{df}{dr} \right) =0$$
This admits the general solution
$$r \frac{df}{dr} = \frac{A}{r} + Br + Cr^2 + Dr^4$$
Substituting into (4), solving for the components $\tilde{u}_r$ and $\tilde{u}_\theta$, and applying the boundary conditions as before gives the solution for Stokes flow past a sphere.
