Consider the Fucik eigenvalue problem

\left\{ \begin{aligned} -\Delta u &= \alpha (u^+)^{p-1} - \beta (u^-)^{p-1}, &&x \in \Omega,\\ u&=0, &&x \in \partial\Omega, \end{aligned} \right.

where $u = u^+ - u^-$, and $u^\pm := \max{\pm u, 0}$.

In [1] it is proved that the first nontrivial curve of the Fucik spectrum can be described as a set of points $(s + c(s), c(s))$, where $s \in \mathbb{R}$ and $c(s)$ defined by

$c(s) = \inf_{\gamma \in \Gamma} \max_{u \in \gamma[-1,1]} \left(\int_{\Omega}|\nabla u|^p \, dx - s \int_{\Omega}|u^+|^p \, dx \right).$

Here

$\Gamma := \{\gamma \in C([-1,1], S):~ \gamma(-1) = -\varphi_1,~ \gamma(1) = \varphi_1 \},$

where $S :=\{w \in W_0^{1,p}:~ \|w\|_{L^p}=1\}$ and $\varphi_1$ is the first eigenfunction.

There is another characterization of the first nontrivial curve of the Fucik spectrum. Namely, consider

$\alpha^*(\beta) := \inf\left\{ \frac{\int_{\Omega}|\nabla u^-|^p \, dx}{\int_{\Omega}|u^-|^p \, dx}:~ u \in W_0^{1,p},~ u^\pm \not\equiv 0,~ \frac{\int_{\Omega}|\nabla u^+|^p \, dx}{\int_{\Omega}|u^+|^p \, dx} = \beta \right\}.$

Note that the admissible set for this minimization problem is nonempty for all $\beta > \lambda_1(p)$. This definition is, in essence, the same as of Theorem 1.2 in [2] for the linear case $p=2$ (see also [3]), and it was pointed out in that works that for $p>1$ this definition is also ok. Let us prove this fact explicitly.

Proposition. The set of points $(\alpha^*(\beta), \beta)$ is the first nontrivial curve of the Fucik spectrum.

Proof. The main idea is to switch between the parametrizations: $c(s)$ parametrized by diagonals, while $\alpha^*(\beta)$ is parametrized by horizontal lines. Note that $c(s)$ is strictly decreasing (see Propositions 4.1 in [1]), i.e., $c(s) > c(s’)$ whenever $s < s’$; moreover, $c(s) \to \lambda_1(p)$ as $s \to +\infty$, see Proposition 4.4 in [1]. Thus, for each $\beta > \lambda_1(p)$ there exists unique $s \in \mathbb{R}$ such that $\beta = c(s)$. (See figure below). Notice that the $c(s)$ is constructed in [1] only for $s \geq 0$ and then the constructed part is reflected with respect to the bisector $\alpha = \beta$. However, it doesn’t cause troubles.

Let us show now that $\alpha^*(c(s)) = s + c(s)$ for any $c(s) = \beta > \lambda_1(p)$. Note first that the eigenfunction which corresponds to $(\alpha, \beta) = (s+c(s), c(s))$ is always an admissible point for $\alpha^*(c(s))$, and hence $\alpha^*(c(s)) \leq s + c(s)$. Suppose, by contradiction, that $\alpha^*(c(s)) < s + c(s)$ for some $s$. Then, by definition of $\alpha^*(c(s))$, there have to exist a function $u \in W_0^{1,p}$ such that

$\alpha^*(c(s)) \leq \frac{\int_{\Omega}|\nabla u^-|^p \, dx}{\int_{\Omega}|u^-|^p \, dx} < s+ c(s) \quad \text{and} \quad \frac{\int_{\Omega}|\nabla u^+|^p \, dx}{\int_{\Omega}|u^+|^p \, dx} = \beta = c(s).$

Due to the continuity and monotonicity of $c(s)$ (see Proposition 4.1 in [1]), there exists $s_0$ such that

$\frac{\int_{\Omega}|\nabla u^-|^p \, dx}{\int_{\Omega}|u^-|^p \, dx} = s_0 + c(s) < s_0 + c(s_0) \quad \text{and} \quad \frac{\int_{\Omega}|\nabla u^+|^p \, dx}{\int_{\Omega}|u^+|^p \, dx} = \beta < c(s_0),$

or, equivalently,

$\int_{\Omega}|\nabla u^-|^p \, dx < (s_0 + c(s_0))\int_{\Omega}|u^-|^p \, dx \quad \text{and} \quad \int_{\Omega}|\nabla u^+|^p \, dx < c(s_0) \int_{\Omega}|u^+|^p \, dx,$

which is, in fact, the main contradictory assumption in the proof of Theorem 3.1 in [1] (see also the proof of Lemma 5.3, (5.10) in [1]). Thus, proceeding exactly as in the proof of Theorem 3.1 in [1], we obtain a contradiction to the definition of $c(s_0)$.

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