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I am interested in finding references regarding estimates of the form $$ \| D^2 u\|_{L^2(\Omega)} \leq C(\|f\|_{L^2(\Omega)}+\|g\|_{S} )$$ where $\|D^2 u\|_{L^2(\Omega)}^2 = \sum\limits_{i,j \in \{1,2\}} \int_{\Omega}(\partial_{x_i x_j}u)^2$ and $$ \begin{cases} -\Delta u = f & \text{ in }\Omega \\ u = 0 & \text{ on } \Gamma_D \\ \frac{\partial u}{\partial n} = g & \text{ on } \Gamma_N \end{cases} $$ and $\Omega\subset \Bbb{R}^2$ is polygonal (bounded, convex) such that the solution $u$ belongs to $H^2(\Omega)$ (in particular $g$ is regular enough). The space $S$ for $g$ is $H^{1/2}(\Gamma_N)$ (but I am interested in other options if available). The boundaries $\Gamma_D, \Gamma_N$ are both non-void and make a partition for $\partial\Omega$.

I was able to find in the book of Grisvard (Elliptic problems in nonsmooth domains) a similar result when $g\equiv 0$. However, I could not find this precise estimate for the non-homogeneous Neumann boundary conditions.

Can you point me to a reference that contains proofs of such estimates? In particular, I am interested in finding precisely the constant $C$ (if possible).

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  • $\begingroup$ In Grisvard's book, mixed boundary conditions is covered, as he considers a general boundary condition operator.. $\endgroup$
    – username
    Commented Mar 26, 2022 at 16:10
  • $\begingroup$ Since your domain is convex, it is worth trying to do the calculation by hand, using Morawetz multipliers (testing against $\chi \cdot Du$ where $\chi$ is an extension of the normal vector into the interior) to check if you can't get simply $C=1$. $\endgroup$
    – username
    Commented Mar 26, 2022 at 16:14
  • $\begingroup$ @username: Thank you for your suggestions. Indeed, in Grisvard's book, there is a section on a priori estimates, where such kind of estimates is presented. However, it is in a very general setting, and in the case of the Laplacian, I thought there may be a more direct reference where the computations are more straightforward. I will try your second suggestion. $\endgroup$ Commented Mar 26, 2022 at 23:19
  • $\begingroup$ @username: it turns out it is possible to obtain an estimate by integrating by parts by hand. It is weird that I couldn't find any explicit references regarding this. $\endgroup$ Commented Apr 9, 2022 at 20:07

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