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Consider the problem

\begin{align*} (P) \begin{cases} - \Delta u &= |u|^{q-2}u \ \text{in} \ \Omega,\\ u &= 0 \ \text{on} \ \partial \Omega, \end{cases} \end{align*}

where $\Omega \subset \mathbb{R}^N$ is a bounded domain and $2 < q < +\infty$.

I'm trying to find an ordered pair of subsolution and supersolution for this problem in order to find a positive solution for the problem.

A subsolution is a function $\underline{u} \in C^2(\Omega) \cap C(\overline{\Omega})$ such that

\begin{align*} \begin{cases} - \Delta \underline{u} &\leq |\underline{u}|^{q-2}\underline{u} \ \text{in} \ \Omega,\\ \underline{u} &\leq 0 \ \text{on} \ \partial \Omega. \end{cases} \end{align*}

A supersolution is a function $\overline{u} \in C^2(\Omega) \cap C(\overline{\Omega})$ such that

\begin{align*} \begin{cases} - \Delta \overline{u} &\geq |\overline{u}|^{q-2}\overline{u} \ \text{in} \ \Omega,\\ \overline{u} &\geq 0 \ \text{on} \ \partial \Omega. \end{cases} \end{align*}

The subsolution and the supersolution are ordered if

\begin{align*} \begin{cases} \underline{u} &\leq \overline{u} \ \text{in} \ \Omega,\\ \underline{u} &\leq 0 \leq \overline{u} \ \text{on} \ \partial \Omega. \end{cases} \end{align*}

I could show that $\underline{u} := 0$ and $\overline{u} := K w$, where $K > 0$ is a constant small enough and $w$ is a positive solution for the linear problem

\begin{align*} (P_1) \begin{cases} - \Delta w &= 1 \ \text{in} \ \Omega,\\ w &= 0 \ \text{on} \ \partial \Omega, \end{cases} \end{align*}

is an ordered pair of subsolution and supersolution using comparison principle, but I couldn't show that the solution $u$, which is between $\underline{u}$ and $\overline{u}$, is positive in $\Omega$.

I would like to know if there is some way to show that the problem $(P)$ has a positive solution using the sub-supersolution method.

Thanks in advance!

George
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  • It is part of the conclusion of the method of sub and supersolutions that the solution $u$ satisfies $\underline u \leqslant u \leqslant \bar u$. Since you set $\underline u =0$, this immediately implies that $u\geqslant 0$. Is your issue with showing that $u>0$ in $\Omega$? – JackT Sep 20 '22 at 07:55
  • Also, I thought the method of sub and supersolutions requires $f' \in L^\infty(\mathbb R)$ (where the equation you care about is $-\Delta u =f(u)$ with $f:\mathbb R \to \mathbb R$) which doesn't hold here. Are you using another form of the theorem? – JackT Sep 20 '22 at 07:56
  • @JackT, yes, I want to show that $u > 0$ in $\Omega$. Concerning to the method of sub and supersolutions, the theorem that I know is this: if $- \Delta u = f(x,u)$ with Dirichlet condition, $\Omega \subset \mathbb{R}^N$ a bounded domain with smooth boundary, $f \in C^1(\overline{\Omega} \times \mathbb{R}, \mathbb{R})$ and if there is an ordered pair of sub and supersolutions for this problem, then exists a solution $u \in C^2(\overline{\Omega})$ for the problem such that $\underline{u} \leq u \leq \overline{u}$. – George Sep 21 '22 at 16:33

1 Answers1

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No it is not possible without further assumptions on either $q$ or $\Omega$. Indeed, if $\Omega$ is star-shaped and $q > \frac{2n}{n-2}$ then there are no non-trivial (by non-trivial we mean $u \not\equiv 0$) solutions. A proof of this can be found in Theorem 3 of Section 9.4.2 in Partial Differential Equations by L. Evans.

JackT
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