Consider the Fucik eigenvalue problem
\[\left\{ \begin{aligned} \Delta u &= \alpha (u^+)^{p1}  \beta (u^)^{p1}, &&x \in \Omega,\\ u&=0, &&x \in \partial\Omega, \end{aligned} \right.\]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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Last modified: 14May21
Bibliography

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