Existence of positive solutions for a class of p(x)-Laplacian problems

  • G.A. Afrouzi
  • H. Ghorbani

Abstract

We consider the system \vspace{.3cm}
\begin{eqnarray*}
   \left\{\begin{array}{ll}
  -\Delta_{p(x)} u=\lambda F(x, u, v) &~in~\Omega\\
   -\Delta_{p(x)} v=\lambda G(x, u, v) & ~in~\Omega\\
     u=v= 0 & on~ \partial \Omega \hspace{2 cm }
    \end{array}\right.
    \end{eqnarray*}
 where $\Omega \subset R^N$ is a bounded domain with $C^2$ boundary $\partial
 \Omega, p(x) \in C^1(\bar{\Omega})$ is a function,
  $-\Delta_{p(x)} u = -div(|\nabla u|^{p(x)-2}\nabla
 u)$~ is called $p(x)$-Laplacian and $F(x, u, v)
 = [g(x)a(u) + f(v)],~ G(x, u, v) = [g(x)b(v) + h(u)],~ \lambda  > 0$
  is a parameter.\vspace{.2cm}

\noindent Here $g
 : [0, +\infty) \rightarrow (0, +\infty)$ is a continuous function
  and $f,~ h,~ a,~ b$ are $C^1$ nondecreasing
 functions satisfying $a(0) \geq 0,~ b(0) \geq 0$,
 \vspace{.2cm}

\noindent \hspace{1cm} $\lim_{u \rightarrow +\infty}
\frac{a(u)}{u^{P^- - 1}} = 0,~~~~ \lim_{u \rightarrow +\infty}
\frac{b(u)}{u^{P^- - 1}} = 0$,\vspace{.2cm}

\noindent \hspace{1cm}$\lim_{u \rightarrow +\infty}f(u) =
+\infty,~~~~ \lim_{u \rightarrow +\infty}h(u) = +\infty$
and\vspace{.2cm}

\noindent \hspace{1cm}
 $\lim_{u \rightarrow
+\infty}\frac{f(M(h(u))^\frac{1}{p^--1})}{u^{p^--1}} = 0,~\forall
M > 0$ \vspace{.2cm}

 \noindent We discuss the existence of positive solution via
 sub-super-solutions. Note that,we do not assume any symmetric
 domain , and we do not assume any sign conditions on $f(0),~
h(0)$.

Published
2012-11-25
Section
Articles