Consider the following Stokes equation (to be understood in the classic form) , where $B$ is a ball of center $0$ and radius $a$ :
\begin{aligned} - \mu \Delta u + \nabla p= 0 \text{ in } \mathbb{R}^3 \setminus B\\ div \, u=0 \text{ in } \mathbb{R}^3 \setminus B\\ u = v \text{ on } \partial B\\ u(x) \underset{|x| \rightarrow +\infty}{\longrightarrow} 0 \end{aligned}
where $v(x)= \lambda U(x-x_f) \bar{d}$ with
$U(x)$ is the Oseen tensor defined by $U(x):= \frac{1}{8 \pi \mu} (\frac{I_3}{|x|} + \frac{x \otimes x}{|x|^3})$, and $\mu$ is the constant fluid viscosity.
$\bar{d}$ is a given directionnal vector of norm one and $x_f = a(1+\beta)\bar{d}$ is a given fixed position outside $B$, with $\beta<< 1$.
I would like to derive explicit formulas for $u$ and eventually $p$. Notes that $v$ is a solution to the following point-force problem : \begin{aligned} - \mu \Delta v + \nabla p_v= \lambda \delta(x-x_f) \bar{d} \text{ in } \mathbb{R}^3 \\ div \, v=0 \text{ in } \mathbb{R}^3\\ v(x) \underset{|x| \rightarrow +\infty}{\longrightarrow} 0. \end{aligned}
Classic litterature such as
[Microhydrodynamics: principles and selected applications,Kim, Sangtae and Karrila, Seppo J,2013]
and
[A physical introduction to suspension dynamics, Guazzelli, Elisabeth and Morris, Jeffrey F,2011]
solves the desired problem with $v$ being a simple translation ($v=U$ constant), a rotation ($v=w \wedge x$) or a rate-of-strain ($v = E \cdot x$).
Any help or remarks are welcomed !
Edit
Using the kelvin's transformation in the sphere as suggested in the comment, we pose
$$x^*= \frac{a^2}{|x|^2}x \text{ and } u^*(x^*)= \frac{|x|}{a^2}u(x).$$
The main property of this transformation is that it conserves harmonicity as
$$\Delta u^\star (x^\star) = \frac{a^4}{|x^*|^5} \Delta u(x).$$ However, the divergence-free condition doesn't have a good interaction with the Kelvin transform since
$$div \, u^* (x^*)= \frac{|x|}{a^2} div \, u(x) + \frac{2 x \cdot u (x)}{a^2|x|}.$$
Therefore, the Kelvin transformation of my Stokes problem leads to find $u^\star$ such that
\begin{aligned} - \mu \Delta u^\star(x^\star) + \nabla p^\star(x^\star)= 0 \text{ in } B\\ div \, u^\star (x^\star)= \frac{u(x) \cdot x}{a^2 |x|} \text{ in } B\\ u^\star(x^\star) = v(x) \text{ on } \partial B\\ u^\star(x^\star) \underset{|x^\star| \rightarrow 0}{\longrightarrow} 0 \end{aligned} and this problem is not computable easily since it is not anymore a Stokes problem. Therefore I'm not sure Kelvin transformation is the suitable method to this problem.
Edit 2
Denoting $w:= u -v$, we get that $w$ is a solution to
\begin{aligned} - \mu \Delta w + \nabla p_w= \lambda \delta(x-x_f) \bar{d} \text{ in } \mathbb{R}^3 \setminus B\\ div \, w =0 \text{ in } \mathbb{R}^3 \setminus B\\ w = 0 \text{ on } \partial B\\ w(x) \underset{|x| \rightarrow +\infty}{\longrightarrow} 0 \end{aligned}
Adapting the method of images of a grounded sphere near a point charge (which i believe is finally linked to the Kelvin transform), I find an explicit solution for $w$ :
$$w(x) = \frac{\lambda}{8 \pi \mu} \left[ \frac{I_d}{r_1(x)} + \frac{(x-A) \otimes (x-A) }{ r_1(x)^3} - \frac{(1+\beta)^{-1} I_d}{r_2(x)} - (1+\beta)^{-1} \frac{(x-B) \otimes (x-B) }{ r_2(x)^3} \right] \bar{d}$$ with $A=((1+\beta)a,0,0)^t$, $B=(a (1+\beta)^{-1},0,0)^t$, $$r_1(x)=\sqrt{|x|^2 +(1+\beta)^2 a^2 - 2 a (1+\beta)|x| \sin \theta \cos \alpha}$$ and $$r_2(x)=\sqrt{|x|^2 + (1+\beta)^{-2} a^2 - 2 a |x| (1+\beta)^{-1} \sin \theta \cos \alpha}.$$
(Note that both $\theta$ and $\alpha$ depends on $x$ as $\theta= \arccos(x_3/|x|)$ and $\alpha=\arcsin(x_2/|x|)$.)
I'm having trouble proving that $w$ is indeed a solution because I didn't compute the pressure $p_w$, however we can easily see that $w$ verifies the condition at the sphere boundary. If we suppose $w$ is indeed a solution, then I will simply have
$$u|_{\mathbb{R}^3 \setminus B} = (w + v)|_{\mathbb{R}^3 \setminus B}$$ as the solution to my problem.