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Let $\Sigma$ be a compact surface and denote by $\nu$ the unit conormal for $\partial \Sigma$. Let

$$E = \left\{ \phi \in C^{2,\alpha}(\Sigma) : \int_{\Sigma} \phi \, \mathrm{d}A = 0 \right\} $$
and $$F = \left\{ \phi \in C^{0,\alpha}(\Sigma) : \int_{\Sigma} \phi \, \mathrm{d}A = 0 \right\},$$

for fixed $0 < \alpha < 1$.

I would like to know why, given $(f,h) \in F \times C^{1,\alpha}(\partial \Sigma)$, the problem

$$\begin{cases} -\Delta_{\Sigma} u = f + \frac{1}{\vert \Sigma \vert} \int_{\partial \Sigma} h \, \mathrm{d}L, \quad \text{on } \Sigma \\ \frac{\partial u}{\partial \nu} = -h, \quad \text{on } \partial \Sigma \end{cases}$$

has a solution in $E$. This came up for me in this paper, page 8. I am sorry if the question is silly, but I am not very good at PDE's.

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    $\begingroup$ This came up in which paper? $\endgroup$
    – Jochen Wengenroth
    Commented Jul 23, 2019 at 10:50
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    $\begingroup$ Is $\Sigma$ a free-boundary, or is it not? In the latter case you are just solving a Neumann problem on a domain; the fact that $\Sigma$ is a surface in $M^3$ is not that relevant. (Also, probably not a good idea to use $g$ both as the ambient metric and the boundary Neumann data.) If $\Sigma$ is a free-boundary (as you stated in the question), then things could get complicated. $\endgroup$
    – Willie Wong
    Commented Jul 23, 2019 at 12:40
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    $\begingroup$ crossposted - math.stackexchange.com/questions/3298263/… $\endgroup$
    – Calvin Khor
    Commented Feb 9, 2021 at 4:30

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