LAUSR

AbstractWe consider the following elliptic problem: {−div(|∇u|p−2∇u|y|ap)=|u|q−2u|y|bq+f(x)in Ω,u=0on ∂Ω,$$ \textstyle\begin{cases} -\operatorname{div} ( \frac{ \vert \nabla u \vert ^{p-2} \nabla u}{ \vert y \vert ^{ap}} ) = \frac { \vert u \vert ^{q-2} u}{ \vert y \vert ^{bq}} + f(x) & \mbox{in } \Omega,\\ u = 0 & \mbox{on } \partial\Omega, \end{cases} $$ in an unbounded cylindrical domain Ω:={(y,z)∈Rm+1×RN−m−1;01$p>1$ and A,B∈R+$A,B\in\mathbb{R}_{+}$. Let pN,m∗:=p(N−m)N−m−p$p^{*}_{N,m}:=\frac {p(N-m)}{N-m-p}$. We show that pN,m∗$p^{*}_{N,m}$ is the true critical exponent for this problem. The starting point for a variational approach to this problem is the known Maz’ja’s inequality (Sobolev Spaces, 1980) which guarantees, for the q previously defined, that the energy functional associated with this problem is well defined. This inequality generalizes the inequalities of Sobolev (p=2,a=0 and b=0)$(p=2, a=0 \mbox{ and } b=0)$ and Hardy (p=2,a=0 and b=1)$(p=2, a=0 \mbox{ and } b=1)$. Under certain conditions on the parameters a and b, using the principle of symmetric criticality and variational methods, we prove that the problem has at least one solution in the case f≡0$f\equiv0$ and at least two solutions in the case f≢0$f \not\equiv0$, if p