I have a question about well-posedness of the following problem. First, I'll demonstrate the problem on an example, but in general, I'd say it's a grossly over-defined BVP: $\nabla^2 \varphi = 0$ outside of a bounded region of space (i.e. outside of a ball, cylinder, ... potato, ...) with $\vec{\nabla} \varphi = \vec{F}$ in the limit of being far away from the origin (far away from that bounded region of space) and $\vec{n} \times \vec{\nabla} \varphi = \vec{0}$ when at the boundary of an object, with $\vec{n}$ being the normal vector of the surface of the object (almost looks like a Neumann problem, over-defined by one dimension).
Here's an example when "object" = ball. Let's have a perfectly conductive ball of radius $R$ sitting at the origin. We turn on the electric field, which, far away from the ball, is of the form $\vec{E} = E_0 \hat{z}$ (uniform field for simplicity pointing in the $z$-direction). How is the electric field deformed near the sphere?
Now I'm familiar with the solution and it is indeed possible to write down both electric field and potential outside explicitly $$ \vec{E} (r,\theta) = E_0 \left( 1 + \frac{2 R^3}{r^3} \right) \cos \theta \, \hat{r} - E_0 \left( 1 - \frac{R^3}{r^3} \right) \sin \theta \, \hat{\theta} = E_0 \left[ \left( 1 - \frac{R^3}{r^3} \right) \hat{z} + \frac{3 r^3 z}{r^4} \hat{r} \right] $$ where hat vectors are unit vectors in the corresponding direction.
The way to derive this is to exploit that $\vec{\nabla} \times \vec{E} = 0$ (Faraday), therefore $\vec{E} = - \vec{\nabla} \varphi$ and since there's no charges, from Gauss' law we have $\nabla^2 \varphi = 0$ in the region outside of the sphere. Then we write down the most general function that satisfies $\nabla^2 \varphi = 0$ in spherical coordinates with azimuthal symmetry $$ \varphi (r, \theta) = \sum_{\ell = 0}^\infty \left( \alpha_\ell r^\ell + \frac{\beta_\ell}{r^{\ell+1}} \right) P_\ell (\cos \theta) $$ where $P_\ell$ are Legendre polynomials. Considering that far away from the sphere the field has the form $\vec{E} = E_0 \hat{z} = E_0 \left( \cos \theta \, \hat{r} - \sin \theta \, \hat{\theta} \right)$, we right away pick $\ell = 1$ and $\alpha_1 = - E_0$. Then by imposing $\hat{\theta} \cdot \vec{E} = 0$ at the boundary of the sphere, we restrict the solution further by $B_1 = E_0 R^3$.
Now, my question is: this is obviously neither a Dirichlet nor Neumann problem. Firstly, we are somehow able to impose the full gradient of $\varphi$ at $r \to \infty$ (which is the "outer" boundary of our region). This almost seems like a Neumann problem, but with Neumann problem we can only prescribe $\vec{n} \cdot \vec{\nabla} \varphi$. Secondly, at the boundary of the sphere we again prescribe a very weird boundary condition: $\vec{t}_{1,2} \cdot \vec{\nabla} \varphi = 0$ where $\vec{t}_{1,2}$ is the basis of tangential space ot the sphere. This can also be written as $\hat{r} \times \vec{\nabla} \varphi = \vec{0}$ where $\hat{r}$ serves as the normal vector at the surface of the sphere. Therefore, we are somehow able to prescribe more than we should be able to, yet everything in the end works out smoothly and we get a unique solution (there's a question mark here, because uniqueness wasn't proved - not that I'm aware of any other function that would satisfy these conditions).
What kind of problem is this? It cannot be either Dirichlet or Neumann problem, but do mathematicians recognize this kind of BVP and give it a name? Can we take an object of arbitrary shape and pose the same problem for it? How would you prove existence and uniqueness of the solution (does it always exist and is it unique)? Just to be clear, we're solving Laplace equation $\nabla^2 \varphi = 0$ outside of a bounded conductive object with $\varphi \to - E_0 z$ when far away from said object and $\vec{n} \times \vec{\nabla} \varphi = \vec{0}$ at the surface of the object, where $\vec{n}$ is the normal vector of the surface. If this problem isn't well defined, i.e. there are objects for which the solution does not exist or exists but is not unique, I would very much like to see a counterexample (so I can nag someone with a CNC to make me such object and do an experiment).
Can we generalize the kind of field we submerge the object into? Can we say the background field is of the form $\vec{E} = E_x (x,y,z) \hat{x} + E_y (x,y,z) \hat{y} + E_z (x,y,z) \hat{z}$ (of course, it must be irrotational, so that we're still in the realm of electrostatics)? I'm asking this because the boundary condition of the original problem, $\vec{E} = E_0 \hat{z}$ seems almost like the sophomore's dream: the field far away is already in the shape that is one of the Legendre polynomials, i.e. it seems like this problem was naturally set up to be solved this way, but can we modify the background field so it's not so easy?