I am trying to get some intuition regarding Sobolev spaces, specifically in regards to minimizing a functional in those. I have some open domain $\Omega\subset \mathbb{R}^d$ with a sufficiently regular boundary. I have a closed subset $K\subset \Omega$ also with a sufficiently regular boundary. Then the problem reads $$\min_{u \in\mathcal{U}}\int_{\Omega\setminus K} \nabla u(x) \cdot \nabla u(x)\,dx, \,\text{such that } u(x)=f(x), \, x\in K.$$
Here $\nabla u$ can be interpreted using weak derivatives. As I understand it, one option for $\mathcal{U}$ here would be $H^1(\Omega\setminus K)$. Another I have seen, considers the Beppo-Levi or homogeneous Sobolev space $\dot{H}^1(\Omega\setminus K)$ (they do consider higher derivatives, but for simplicity I have chosen $k=1$ here). As far as I can tell the two differ only in their norms, where the latter has a semi-norm that does not depend on $\|f\|_2$. I understand this as $\dot{H}^1(\Omega\setminus K)$ having slightly different equivalence classes, where for instance it cannot differentiate between functions up to a constant (I assume that higher order versions wouldn't be able to differentiate between functions up to some multivariate polynomial of degree $k-1$).
Is this a correct interpretation? What would be the advantage of considering the latter homogeneous Sobolev spaces? I am asking mainly in the context of the fact that Beppo-Levi spaces are considered by Duchon and Meinguet for polyharmonic scattered data interpolation. Note that their considerations are for $\Omega=\mathbb{R}^d$. Could I not consider non-homogeneous ones for the same problem? My understanding is that they consider Beppo-Levi spaces because the variational problem above uses exactly the norm of those. However, wouldn't the minimizer be unique also in $H^1(\Omega\setminus K)$ due to the boundary conditions on $\partial K$ (I assume that the BCs on $\partial\Omega$ are homogeneous Neumann)? E.g. in the classical setting I would have gotten a Laplace equation with Dirichlet conditions on $\partial K$.