Positive solutions for Schrödinger-Poisson-Slater system with coercive potential

Yongsheng Jiang, Zhengping Wang, Huan-Song Zhou

DOI: http://dx.doi.org/10.12775/TMNA.2020.041

Abstract


We study the following Schrödinger-Poisson-Slater type system: \begin{equation}\label{eq:0.1} \begin{cases} -\Delta u + V(x) u + \lambda\phi (x) u =|u|^{p-1}u,& x\in \mathbb{R}^3, \\ -\Delta\phi = u^2, \quad \lim\limits_{|x|\to +\infty}\phi (x)=0, \end{cases} \end{equation} where $\lambda> 0$ is a parameter, $p\in (1,2)$, $V\in C(\mathbb{R}^N,\mathbb{R}_0^+)$. If $V$ is a constant, or $p \in (2,5)$ with $V\in L^{\infty}(\mathbb{R}^3)$, the above system has been considered in many papers. In this paper, we are interested in the case: $p\in (1,2)$ and $V$ being a coercive potential, i.e., $\lim\limits_{|x|\to +\infty} V(x)=\infty$. We prove that there exists $\lambda_0> 0$ such that system (\ref{eq:0.1}) has at least two positive solutions $u^0_\lambda$ and $u^1_\lambda$ for any $\lambda \in (0, \lambda_0)$. Moreover, $u^0_\lambda$ is a ground state (i.e., the least energy solution) which must blow up as $\lambda \rightarrow 0$. Particularly, when $p\in ({11}/{7},2)$ and $\lambda> 0$ is small enough, we show that the ground state of (\ref{eq:0.1}) must be non-radially symmetric even if $V(x)=V(|x|)$, such as $V(x)=|x|^2$.

Keywords


Schrödinger-Poisson-Slater system; coercive potential; ground state

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