Let $\Omega\subset { \bf R}^n$ be a smooth bounded domain. In this paper, we prove a result of which the following is a by-product: Let $q\in ]0,1[$, $\alpha\in L^{ \infty}(\Omega)$, with $\alpha>0$, and $k\in { \bf N}$. Then, the problem $$\cases { -\tan\left(\int_{ \Omega}|\nabla u(x)|^2dx\right)\Delta u= \alpha(x)u^q & in $\Omega$\cr & \cr u>0 & in $\Omega$\cr & \cr u=0 & on $\partial \Omega$ \cr & \cr (k-1)\pi<\int_{ \Omega}|\nabla u(x)|^2dx<(k-1)\pi+{ { \pi}\over { 2}} \cr}$$ has a unique weak solution $\tilde u$ which is the unique global minimum in $H^1_0(\Omega)$ of the functional $$u\to { { 1}\over { 2}}\tan\left (\int_{ \Omega}|\nabla\tilde u(x)|^2dx\right)\int_{ \Omega}|\nabla u(x)|^2dx-{ { 1}\over { q+1}}\int_{ \Omega}\alpha(x)|u^+(x)|^{ q+1}dx\ ,$$ where $u^+=\max\{ 0,u\}$.
Categories: math.AP