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Let $M,N$ be smooth, connected, compact, oriented, two-dimensional Riemannian manifolds, with $C^k$ boundaries.

Let $f:M \to N$ be a Lipschitz continuous weakly harmonic map**, and assume that $f(\partial M) \subseteq \partial N$,$f|_{M^\circ}(M^\circ) \subseteq N^\circ$. ($M^\circ,N^\circ$ are the interiors).

Then by known regularity results, $f|_{M^\circ}:M^\circ \to N^\circ$ is smooth. Now suppose that $f|_{M^\circ}:\partial M \to \partial N$ is $C^k$.

Is $f:M \to N \in C^k$ up to the boundary?

**I use the definition of weak harmonicity which is via the second fundamental form of an isometric embedding $N \to \mathbb{R}^D$.

I don't know the answer even when the boundaries are smooth.

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    $\begingroup$ I am bit confused: if $\partial M \neq \emptyset$ and $\varphi: M \to N$ is a Lipschitz continuous map, then there are cases where there is an energy-minimising map $u$ in the homotopy class of $\varphi$ with $u \mid \partial M = \varphi \mid \partial M$. (For example, if I am not mistaken, Lemaire's 'Boundary value problems for harmonic and minimal maps of surfaces into manifolds' shows this is possible for $\varphi \in C^0 \cap W^{1,2}(M,N)$ if $\pi_2(N) = 0$.) If $\varphi$ is chosen so as not to be $C^1$ on the boundary, then $u$ would not be $C^1$ up the boundary either, no? $\endgroup$
    – Leo Moos
    Commented Oct 20, 2020 at 17:12
  • $\begingroup$ OK, it seems you are right. So I guess a more reasonable version of the question would be to assume that the restriction of our weakly harmonic map to the boundary is $C^1$, and ask whether $f$ is $C^1$ on all $M$. $\endgroup$
    – Asaf Shachar
    Commented Oct 20, 2020 at 19:44
  • $\begingroup$ I have now modified the question to assume that the restriction of the map to the boundary is regular. $\endgroup$
    – Asaf Shachar
    Commented Feb 28, 2021 at 16:07

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