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Let $M \simeq [0,1]^2 / \sim$ be a Moebius band, where we identify $\{0\} \times [0,1]$ and $\{1\} \times [0,1]$ by $(0,y) \sim (1,1-y)$ for all $y \in [0,1]$. Let $C = [0,1] \times \{1/2\}$ be the central circle of the Moebius band. We identify $M$ (and $C$) with their smooth embeddings in $\mathbb{R}^3$.

Consider the following mixed boundary value problem on $M \setminus C$: \begin{cases} \Delta u + \lambda u = 0 &\text{ in $\mathrm{int} \, M \setminus C$} \\ u = 0 &\text{ on $\partial M$} \\ \frac{\partial u}{\partial \nu} = 0 &\text{ on $C$} \end{cases} where $\Delta$ is the metric Laplacian on $M$, $\lambda \in \mathbb{R}$ is a constant. (In other words, $u$ is an eigenfunction of the Laplacian.) Is this is a well-defined problem? If yes, what is the correct weak formulation? If not, what could go wrong?

As far as I can tell, the issue could be that $\partial(M \setminus C) = \partial M \cup C$ is not a Lipschitz boundary in the classical sense, because the interior of $M \setminus C$ lies on both sides of $C$. However, $M \setminus C$ can be isometrically deformed into a "Moebius strip with two twists", and a boundary with two connected components, corresponding to $\partial M$ and $C$. If permissible, such a deformation could get rid of the issue, as after it $M \setminus C$ would lie on one side of $C$.

However, I still can't quite see how one would make sense of the space "$H_0^1(\mathrm{int} \, M\setminus C \cup C)$", which is the right space for the weak formulation according to the answer to this question.

Note: As I didn't get any replies on here, I asked the same question on Math.Overflow.

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    As you said, $M \setminus C$ is a "Moebius strip with two twists" and hence it is just a regular strip $S^1 \times [0,1]$ with no twist at all. The problem is that you can consider either $u \in H^1(\operatorname{int} M)$ or $u \in H^1(\operatorname{int} M \setminus C)$. Which one are you interested in? – 2017-01-15
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    The case $u \in H^1(\mathrm{int} \, M \setminus C)$. Surely considering $u \in H^1(\mathrm{int} \, M)$ would make the problem over-determined? – 2017-01-15
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    @Michał Do you have a literature reference for the problem? I reckon I could make a heuristic argument along the lines described in the question, but I'm trouble giving formal justification. – 2017-01-18

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I assume that we are interested in $u \in H^1(M \setminus C)$. Our space $M \setminus C$ is isometrically equivalent to an ordinary strip $X = S^1 \times I$ (where $S^1$ has length $2$ and $I$ has length $1/2$). The boundary consists of two circles: $\Gamma_1$ (corresponding to $C$) and $\Gamma_2$ (corresponding to $\partial M$). As in this question, take your function space $H$ to be the closure of $C_c^\infty(X \cup \Gamma_1)$ (or $C_0^1(X \cup \Gamma_1)$) wrt. the norm $H^1(X)$. Here $C_c^\infty(X \cup \Gamma_1)$ stands for functions smooth up to $\Gamma_1$ and vanishing on some neighborhood of $\Gamma_2$.

This space encodes the Dirichlet boundary data $u = 0$ on $\Gamma_2$. The Neumann condition $\frac{\partial u}{\partial \nu} = 0$ on $\Gamma_1$ has to be included in the weak formulation, which can be tested with functions from $C_c^\infty(X \cup \Gamma_1)$, hence also from $H$.

As a side comment: if one is interested in $u \in H^1(M)$ one has to require additionally "continuity across $C$", i.e. twe two one-sided traces on $C$ should agree. This translates to the property that the trace of $u \in H^1(X)$ on $\Gamma_1 \cong S^1$ is $\mathbb{Z}_2$-invariant.

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    My question was more about giving a formal justification that the two problems (on $M \setminus C$ and $X$) are equivalent, and in particular justifying why it's not an issue that $X$ is a [manifold with boundary](https://en.wikipedia.org/wiki/Manifold#Manifold_with_boundary), whereas $M \setminus C$ is not. – 2017-01-20
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    These two spaces are isometric, is this what you want to see explained? Also, I don't understand the problem with boundaries. – 2017-01-21