Fundamental Journal of Mathematics and Applications

Öz

For a bounded and smooth enough domain $\Omega$ in $\mathbb{R}^{n}$, with $n\geq2,$ we consider the problem $-\Delta u=au^{-\beta}+\lambda h\left( .,u\right) $ in $\Omega,$ $u=0$ on $\partial\Omega,$ $u>0$ in $\Omega,$ where $\lambda>0,$ $0<\beta<3,$ $a\in L^{\infty}\left( \Omega\right) ,$ $ess\,inf\,\,(a)>0,$ and with $h=h\left( x,s\right) \in C\left( \overline{\Omega}\times\left[ 0,\infty\right) \right) $ positive on $\Omega\times\left( 0,\infty\right) $ and such that, for any $x\in\Omega,$ $h\left( x,.\right) $ is strictly convex on $\left( 0,\infty\right) $, nondecreasing, belongs to $C^{2}\left( 0,\infty\right) ,$ and satisfies, for some $p\in\left( 1,\frac{n+2}{n-2}\right) ,$ that $\lim_{s\rightarrow\infty }\frac{h_{s}\left( x,s\right) }{s^{p}}=0$ and $\lim_{s\rightarrow\infty }\frac{h\left( x,s\right) }{s^{p}}=k\left( x\right) ,$ in both limits uniformly respect to $x\in\overline{\Omega}$, and with $k\in C\left( \overline{\Omega}\right)$ such that $\min_{\overline{\Omega}}k>0.$ Under these assumptions it is known the existence of $\Sigma > 0 $ such that for $ \lambda =0 $ and $ \lambda = \Sigma $ the above problem has exactly a weak solution, whereas for $ \lambda \in \left( 0, \Sigma \right) $ it has at least two weak solutions, and no weak solutions exist if $ \lambda > \Sigma $. For such a $ \Sigma $ we prove that for $ \lambda \in \left( 0, \Sigma \right) $ the considered problem has it has exactly two weak solutions.