Bulletin of the Brazilian Mathematical Society, New Series ( IF 0.7 ) Pub Date : 2023-10-28 , DOI: 10.1007/s00574-023-00371-7 V. Ambrosio , H. Bueno , A. H. S. Medeiros , G. A. Pereira
In this paper, we deal with the convergence of the fractional relativistic Schrodinger operator
$$\begin{aligned}(-\Delta + m^2)^s\quad \text {as}\ s\rightarrow 1^-.\end{aligned}$$Intuitively, this operator converges to \((-\Delta +m^2)\) but the proof of this result is not so simple and it is based on a careful analysis of the involved modified Bessel function \(K_\nu \). The convergence of the operator makes natural the following question: do the solutions of the problem
$$\begin{aligned} (-\Delta + m^2)^s w=f(w) \, \, \text{ in } \mathbb {R}^N \end{aligned}$$converge, in a suitable sense, to a solution u of the problem
$$\begin{aligned} (-\Delta + m^2)v=f(v)\,\, \text{ in } \mathbb {R}^N \ \text{ as } s\rightarrow 1^-? \end{aligned}$$We will show that the answer is affirmative under reasonable hypotheses. The proof of the convergence of solutions is obtained by combining an immersion result for Bessel potential spaces with the uniform convergence on compact sets. To the best of our knowledge, the analogous result for the fractional Laplacian operator \((-\Delta )^{s}\) has not been established yet.
中文翻译：
分数相对论薛定谔算子的收敛性
在本文中，我们处理分数相对论薛定谔算子的收敛性
$$\begin{对齐}(-\Delta + m^2)^s\quad \text {as}\ s\rightarrow 1^-.\end{对齐}$$直观上，这个算子收敛于\((-\Delta +m^2)\)但这个结果的证明并不那么简单，它是基于对所涉及的修正贝塞尔函数\(K_\nu \)的仔细分析。算子的收敛自然会产生以下问题：求问题的解
$$\begin{对齐} (-\Delta + m^2)^sw=f(w) \, \, \text{ in } \mathbb {R}^N \end{对齐}$$在适当的意义上收敛到问题的解u
$$\begin{对齐} (-\Delta + m^2)v=f(v)\,\, \text{ in } \mathbb {R}^N \ \text{ as } s\rightarrow 1^- ？\end{对齐}$$我们将证明在合理的假设下答案是肯定的。通过将贝塞尔势空间的浸入结果与紧集上的一致收敛相结合，获得了解收敛性的证明。据我们所知，分数拉普拉斯算子\((-\Delta )^{s}\)的类似结果尚未建立。
京公网安备 11010802027423号