Abstract

Abstract Let Ω ⊂ R n \Omega \subset {{\bf{R}}}^{n} be a smooth bounded domain. In this article, we prove a result of which the following is a by-product: Let q ∈ ] 0 , 1 [ q\in ]0,1{[} , α ∈ L ∞ ( Ω ) \alpha \in {L}^{\infty }\left(\Omega ) , with α > 0 \alpha \gt 0 , and k ∈ N k\in {\bf{N}} . Then, the problem − tan ∫ Ω ∣ ∇ u ( x ) ∣ 2 d x Δ u = α ( x ) u q in Ω u > 0 in Ω u = 0 on ∂ Ω ( k − 1 ) π < ∫ Ω ∣ ∇ u ( x ) ∣ 2 d x < ( k − 1 ) π + π 2 \left\{\begin{array}{ll}-\tan \left(\mathop{\displaystyle \int }\limits_{\Omega }| \nabla u\left(x){| }^{2}{\rm{d}}x\right)\Delta u=\alpha \left(x){u}^{q}\hspace{1.0em}& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}\Omega \\ u\gt 0\hspace{1.0em}& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}\Omega \\ u=0\hspace{1.0em}& \hspace{0.1em}\text{on}\hspace{0.1em}\hspace{0.33em}\partial \Omega \\ \left(k-1)\pi \lt \mathop{\displaystyle \int }\limits_{\Omega }| \nabla u\left(x){| }^{2}{\rm{d}}x\lt \left(k-1)\pi +\frac{\pi }{2}\hspace{1.0em}\end{array}\right. has a unique weak solution u ˜ \tilde{u} , which is the unique global minimum in H 0 1 ( Ω ) {H}_{0}^{1}\left(\Omega ) of the functional u → 1 2 tan ∫ Ω ∣ ∇ u ˜ ( x ) ∣ 2 d x ∫ Ω ∣ ∇ u ( x ) ∣ 2 d x − 1 q + 1 ∫ Ω α ( x ) ∣ u + ( x ) ∣ q + 1 d x , u\to \frac{1}{2}\tan \left(\mathop{\int }\limits_{\Omega }| \nabla \tilde{u}\left(x){| }^{2}{\rm{d}}x\right)\mathop{\int }\limits_{\Omega }| \nabla u\left(x){| }^{2}{\rm{d}}x-\frac{1}{q+1}\mathop{\int }\limits_{\Omega }\alpha \left(x)| {u}^{+}\left(x){| }^{q+1}{\rm{d}}x, where u + = max { 0 , u } {u}^{+}=\max \left\{0,u\right\} .

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