Consider the periodic strip $\Omega=\mathbb{T}\times[0,1]$ where $\mathbb{T}$ is the 1D torus with period 1. We consider the mixed Dirichlet/Neumann problem $$-\Delta u=f$$ with boundary conditions $$u(x,0)=0,\,\partial_y u(x,1)=0$$ for all $x\in \mathbb{T}$. I'd like to say that the Laplacian associated with these boundary conditions is invertible, with compact inverse with elliptic regularity estimates of the form $$\|u\|_{H^2}\le C\|f\|_{L^2}.$$ Here's what I know: i) such elliptic estimates are classical for Dirichlet and Neumann boundary conditions, separately. ii) in the case of arbitrary smooth domain $\Omega$ with homogeneous Dirichlet boundary conditions on a boundary portion $\Gamma\subset\partial\Omega$ and homogeneous Neumann boundary conditions on the remainder $\partial\Omega\setminus\Gamma$, the solution $u$ maybe be singular (say, near the boundary where there's a transition from Dirichlet to Neumann boundary conditions) so that, in particular, $u\notin H^2$.

However, for the problem that I've stated, we need not worry about a nonsmooth transition from Dirichlet to Neumann boundary conditions (the corresponding boundary portions are strictly disjoint), so I suspect that elliptic regularity maybe holds. Are there any references for such a problem?

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    $\begingroup$ You can consider the resolvent of the Laplacian with Dirichlet b.c at y=0,1 and periodic boundary conditions in x, and that with Neumann b.c. at y=0,1 and glue together using cut-off functions to get solvability and a-priori estimates for a large spectral parameter λ. Then you go back to λ=0 since the operator has a compact resolvent and is injective. I admit that this is a little vague...but I can give a reference for the method, if needed. $\endgroup$ Mar 18, 2021 at 9:29

1 Answer 1


There are probably different (and more general) ways of doing that, but in your specific case I think you can prove everything with rather elementary arguments. (Maybe that's more or less what Giorgio Metafune had in mind).

Start by "extending" $f$ into a function $\tilde f$ in the following way:

  • $\tilde f(x,y) = f(x,y)$ for $y\in(0,1)$,
  • $\tilde f(x,-y) = -\tilde f(x,y)$ for $y\in(0,1)$,
  • $\tilde f(x,1+y) = \tilde f(x,1-y)$ for $y\in(0,2)$,
  • $\tilde f$ is 4-periodic in $y$.

This would be the natural way of periodically extending the solution $u$ in a smooth way, according to the boundary conditions you have.

Next, consider

\begin{equation*} -\Delta \tilde u = \tilde f \end{equation*}

on $\mathbb{T} \times [-1,3]$ with periodic boundary conditions in $y$, or if you prefer in $\mathbb{T} \times \tilde{\mathbb{T}}$, where $\tilde{\mathbb{T}}$ is the 1D torus of length 4. You can easily study this new problem (for instance using Fourier series since everything is now periodic), and get the estimates you want.

It only remains to show that $\tilde u$ restricted to $\mathbb{T} \times [0,1]$ satisfies the original equation (trivial), with the original the boundary conditions, which is the case because $\tilde u$ has the same symmetries that we imposed on $\tilde f$ (this is again rather straightforward to see at the level of Fourier series, but there are probably other ways of showing it).

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    $\begingroup$ Easier than what I proposed. I guess the last reflection should be even around $y=1$, since you want Neumann. $\endgroup$ Mar 18, 2021 at 11:47
  • $\begingroup$ Indeed, thanks for the correction! $\endgroup$
    – Maxime
    Mar 18, 2021 at 12:40

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