Let $\varepsilon > 0$ be fixed, and let $\Omega \subset \mathbb{R}^d$ be a bounded domain with smooth boundary. Consider the following (Neumann) elliptic problem \begin{cases} \varepsilon \Delta u - u + f = 0 \quad \text{in} \ \Omega\\ \partial_n u = 0 \quad \text{on} \ \partial \Omega \end{cases} where $f \in C^{0, 1}(\overline{\Omega})$ is non-negative on $\overline{\Omega}$ and not identically $0$, and $\partial_n$ denotes the normal derivative (taken with respect to the outward normal vector). I managed to prove, using a Lax-Milgram argument, that the problem has a unique solution $u \in \{u \in H^1(\Omega) \colon -\Delta u \in L^2(\Omega)\}$. How may I show that the solution $u$ is positive? Furthermore, how may we deduce (using some sort of elliptic regularity) that the solution is $C^2(\overline{\Omega})$? On a side note, may we say that $u \in \{u \in H^1(\Omega) \colon -\Delta u \in L^2(\Omega)\}$ implies $u \in H^2(\Omega)$?

  • $\begingroup$ The Schauder theory will get you the regularity. But the solution need not be positive unless more is known about $f$ (is $f$ positive?). $\endgroup$
    – Jeff
    Nov 19, 2017 at 3:36
  • $\begingroup$ @Jeff You're right, I have edited my question. $\endgroup$
    – bgsk
    Nov 19, 2017 at 8:17
  • $\begingroup$ If it is $C^2$ then it is positive by maximum principle $\endgroup$
    – user99914
    Nov 19, 2017 at 8:27
  • $\begingroup$ @JohnMa Is it possible to prove the positivity without using the fact that it is $C^2$? I'm merely asking this question because in the paper where I found this problem, the authors say that existence and positivity are classical, and they only mention the elliptic regularity afterwards. $\endgroup$
    – bgsk
    Nov 19, 2017 at 8:30
  • $\begingroup$ @Jeff I don't want to come off as annoying, but could you be more accurate on "Schauder theory"? I haven't really studied elliptic equations all that much, and I can't seem to find the precise result in Gilbert and Trudinger's book. $\endgroup$
    – bgsk
    Nov 19, 2017 at 10:32

1 Answer 1


The weak solution $u\in H^1(\Omega)$ of your PDE is the unique minimizer of

$$I(u) = \int_\Omega \frac{1}{2}\varepsilon |\nabla u|^2 + \frac{1}{2} u^2 - uf \, dx.$$

If $f$ is nonnegative, then $I(u_+)\leq I(u)$ for all $u\in H^1(\Omega)$ where $u_+=\max\{u,0\}$. This shows that $u\geq 0$ almost everywhere.

I don't have a copy of Gilbarg-Trudinger with me. Look in the Schauder estimates section for estimates that control the $C^{2,\alpha}$ norm of $u$ by the $C^{0,\alpha}$ norm of $f$. This will prove $u\in C^{2,\alpha}$.


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