具有磁场和快速衰减电位的分数阶Schrödinger方程的半经典态

Xiaoming An, Xian Yang
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摘要

本文处理以下分数磁性Schrödinger方程\begin{document}$ \varepsilon^{2s}(-\Delta)^s_{A/\varepsilon} u +V(x)u = |u|^{p-2}u, \ x\in{\mathbb R}^N, $\end{document}是一个参数,\begin{document}$ s\in(0,1) $\end{document}, \begin{document}$ N\geq3 $\end{document}, \begin{document}$ 2+2s/(N-2s),\begin{document}$ A\in C^{0,\alpha}({\mathbb R}^N,{\mathbb R}^N) $\end{document} with \begin{document}$ \alpha\in(0,1) $\end{document}是一个磁场,\ begin{document}$ V:{\mathbb R}^N\到{\mathbb R} $\end{document}是一个非负的连续电位。通过变分方法和惩罚思想,我们证明了问题有一组解集中在\begin{document}$ V $\end{document}的局部极小值处,即\begin{document}$ \varepsilon\到0 $\end{document}。\begin{document}$ V $\end{document}的衰减率没有限制。特别地,可以紧凑地支持\begin{document}$ V $\end{document}。\begin{document}$ A $\end{document}的出现和\begin{document}$ (-\Delta)^s $\end{document}的非局部化使得证明比[7]中的证明更加困难,[7]考虑了\begin{document}$ A\equiv 0 $\end{document}的情况。
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Semi-classical states for fractional Schrödinger equations with magnetic fields and fast decaying potentials

This paper deals with the following fractional magnetic Schrödinger equations

where \begin{document}$ \varepsilon>0 $\end{document} is a parameter, \begin{document}$ s\in(0,1) $\end{document}, \begin{document}$ N\geq3 $\end{document}, \begin{document}$ 2+2s/(N-2s), \begin{document}$ A\in C^{0,\alpha}({\mathbb R}^N,{\mathbb R}^N) $\end{document} with \begin{document}$ \alpha\in(0,1] $\end{document} is a magnetic field, \begin{document}$ V:{\mathbb R}^N\to{\mathbb R} $\end{document} is a nonnegative continuous potential. By variational methods and penalized idea, we show that the problem has a family of solutions concentrating at a local minimum of \begin{document}$ V $\end{document} as \begin{document}$ \varepsilon\to 0 $\end{document}. There is no restriction on the decay rates of \begin{document}$ V $\end{document}. Especially, \begin{document}$ V $\end{document} can be compactly supported. The appearance of \begin{document}$ A $\end{document} and the nonlocal of \begin{document}$ (-\Delta)^s $\end{document} makes the proof more difficult than that in [7], which considered the case \begin{document}$ A\equiv 0 $\end{document}.

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