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Suppose that $f:\Bbb R^2\to\Bbb R$ is a continuous non-linearity and consider the following semi-linear elliptic PDE given by: $$-\Delta u=f(x,u),\;\;x\in\Omega\subset\Bbb R^n,\tag{1}\label{1}$$ To avoid the mention of critical Sobolev exponents and to narrow down the scope of the answer, let us assume that $n=2$ and $\Omega$ is bounded. Assume that $u\in H^1_0(\Omega)$ is a weak solution of \eqref{1}.


Q1. Under what assumptions on $f$ can we show that (i) $u\in H^2(\Omega)$, (ii) recover a classical solution of \eqref{1}?


Remark 1. There are numerous conditions on $f$ that guarantee the existence of $u$. For $n=2$, see this question. For $n\geq 3$, see the the book Semi-linear elliptic equations for beginners by Badiale and Serra.

Remark 2. In Evan's Partial Differential Equtions book, Chpater 6 exercise 4, if $f(x,u)=g(x)+h(u)$, where $g\in L^2(\mathbb{R}^n)$, $h(0)=0$, $h$ is smooth, and $h'\geq 0$, then a solution $u\in H^1(\Bbb R^n)$ to \eqref{1} is in fact in $H^2(\Bbb R^n)$.

Remark 3. From this paper, where $\Omega$ is the unit ball in $\Bbb R^n$ with $n\geq 2$, and $f$ is Lipschitz with $\partial_uf\geq -M$ for some $M\geq 0$, I quote the following sentence in the introduction:

We remark that solutions to \eqref{1} are already in $W^{2,p}(\Omega)\cap C^{1,\alpha}(\Omega)$ for $p<\infty$ and $\alpha<1$ (since $f$ is bounded).

I think I can workout a proof as to why $u\in H^2(\Omega)$, but not $u\in C^{1,\alpha}(\Omega)$ for $\alpha<1$, which brings us to the second question.


Question 2. How does one obtain $u\in C^{1,\alpha}(\Omega)$?


Remark 4. In the answer, assume any needed regularity on $\Omega$ to obtain regularity of $u$.

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    $\begingroup$ If you know that the right hand side $f(x,u)$ is in $L^2$, then $u \in H^2$. Then boundedness of $f$ suffices, as well as $|f(x,u)| \leq a+b|u|$. A bootstrap argument (under similar hypotheses) yields $u \in W^{2,p}$ and then $u \in C^{1,\alpha}$ by Sobolev embedding. $\endgroup$ Commented Sep 22, 2021 at 7:35

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There is a "standard" bootstrap argument which can be used to show regularity for semilinear equations. I sketch it here under the assumption that $|f(x, u)| \leq C(1 + |u|^p)$ for some $0 < p < \frac{2n}{n-2}$ in $n \geq 2$ (I know the question was posed in $n = 2$ but it is useful to see the role played by the critical exponent).

Assume we know that $u$ is in $L^q$ for some $p < q < \infty$, and that $\Omega$ is of class $C^{1,1}$. Then $\Delta u \in L^{q/p}$, and so by the Calderon-Zygmund theorem $u \in W^{2, q/p}$. The Calderon-Zygmund theorem is not always stated in this form (i.e. on domains), but this version may be found in Gilbarg-Trudinger (Section 9.6). Applying the Sobolev embedding gives that $u \in L^{q'}$, where $q' = \frac{nq}{np - 2q}$ (or $\infty$ if $2q > np$).

Notice that $q' > q$ if and only if $q > \frac{n}{2}(p - 1)$, and moreover $q'/q$ is an increasing function of $q$. This means that as long as we start with $u \in L^q$ for $q > \frac{n}{2}(p - 1)$, we may apply this procedure repeatedly to get that $u \in L^{q_*}$ for any $q_* < \infty$, with the number of steps needed always finite and depending only on $q, q_*$. For a sufficiently large value of $q_*$, $u \in W^{2, q_*}$ implies $\nabla u \in C^{0, \alpha}$ for $ \alpha \in (0, 1)$, from Sobolev embeddings.

The point, then, is to check that $u \in L^q$ for $q > \frac{n}{2}(p - 1)$. This is not guaranteed by the equation in general, but it does help that we know that $u \in W^{1, 2}$. This embeds into $L^{\frac{2n}{n-2}}$, so if $n = 2$ and $p$ is anything or $n > 2$ and $p < \frac{n + 2}{n - 2}$ we succeed.

This kind of argument is "sharp" in the sense that there are examples of solutions to semilinear equations which are not bounded. The most standard is $u(x) = |x|^{-\frac{2}{p - 1}}$ being a distributional solution of $-\Delta u = au^p$ for some $a \in \mathbb{R}$: this $u$ is not in $L^q$ for $ q \geq \frac{n}{2}(p - 1)$, and it is not in $W^{1, 2}$ unless $p > \frac{n + 2}{n - 2}$ (and $n \geq 3$). It is not sharp in other ways: it is not sharp at any of the endpoint cases, and the dependence on $x$ in the assumptions on $f(x, u)$ here is not sharp. Therefore if the question is, in $n = 2$ what are the optimal assumptions on $f$ to ensure regularity of $W^{1, 2}$ solutions, this does not give a complete answer.

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  • $\begingroup$ Thank you for this well written answer! However I do have one concern: many authors in the reference I have linked put various assumptions on $f'$ to guarantee a solution of \eqref{1}. I noticed that here, we did not use information about $f'$. Does this mean that regularity is not affected by $f'$? $\endgroup$
    – UserA
    Commented Oct 1, 2021 at 5:09
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    $\begingroup$ Roughly speaking, you always expect $u$ to be two derivatives better than $f$. For $C^{1, \alpha}$ regularity of a bounded $u$, then, boundedness (or less) of $f$ is sufficient, and the argument above shows that this is still valid without a priori knowledge of $u$ being bounded. For higher regularity ($C^{2, \alpha}$, etc.) you will need more information about the smoothness of $f$. $\endgroup$
    – user378654
    Commented Oct 2, 2021 at 4:07
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    $\begingroup$ Assumptions on $f'$ in existence theory are usually of two types: either they are really there to guarantee correct asymptotic behavior of $f$ near infinity (or possibly near $0$) in a strong way, or they are there because the approach towards existence relies on differentiating the equation in some sense (e.g. arguments based on implicit function theorems). $\endgroup$
    – user378654
    Commented Oct 2, 2021 at 4:15
  • $\begingroup$ And might you have any idea why one these assumptions is that $f'$ doesn't hit an eigenvalue of $\Delta$' (or lies strictly between two consecutive eigenvalues, etc...)? And does this assumption in particular have anything to do with regularity? $\endgroup$
    – UserA
    Commented Oct 2, 2021 at 7:09

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