Let $U \subset R^n$ be a regular bounded domain having the topology of a ball. Then, the boundary value problem for $\omega\in \Omega^2(U)$, $$ d\omega = 0 \qquad \delta\omega = \sigma \qquad \mathfrak{t}(\omega) = 0 $$ is solvable provided that $\delta\sigma=0$; here $\mathfrak{t}(\omega)$ is the tangential component of $\omega$ on $\partial U$, $$ \mathfrak{t}(\omega)(X,Y) = \omega(P_\parallel(X),P_\parallel(Y)). $$ A solution exists also if the Dirichlet boundary condition is replaced by the Neumann condition $\mathfrak{n}(\omega) = 0$.

This statement remains correct if this system is for a vector-valued 2-form $\omega\in\Omega^2(U;TU)$ (solve the boundary value problem for each component $\omega^i$). Note that in this case $d$ and $\delta$ have to be replaced with the covariant exterior differential and co-differential $d^\nabla$ and $\delta^\nabla$, but the connection here is flat.

A vector-value 2-form $\omega$ can be decomposed in a collar neighborhood of $\partial U$ into a normal and a tangential component. Let $r$ be the distance from the boundary in a collar neighborhood, then we can define $$ \omega^\perp(X,Y) = (\omega(X,Y),\partial_r)\partial_r, $$ and $\omega^\parallel = \omega - \omega^\perp$.

The question is the following: is the boundary-value problem with mixed boundary conditions, $$ d^\nabla\omega = 0 \qquad \delta^\nabla\omega = \sigma \qquad \mathfrak{t}(\omega^\parallel) = 0 \qquad \mathfrak{n}(\omega^\perp) = 0 $$ solvable?

In fact, what I really need is that $\omega$ satisfies in addition $d \operatorname{tr}\omega =0$. These mixed boundary conditions guaranteed that this is indeed the case. The problem itself arises from the theory of elasticity, attempting to derive a potential theory (like Airy and Beltrami potentials) in arbitrary dimension