Title: Pointwise estimates for the fundamental solutions of higher order Schrödinger equations in odd dimensions II: high dimensional case Show Chinese title

Title: 高维情形下奇数维更高阶薛定谔方程基本解的逐点估计 II：高维情况

Abstract: In this paper, for any odd $n$ and any integer $m\geq1$ with $n>4m$, we study the fundamental solution of the higher order Schr\"{o}dinger equation \begin{equation*} \mathrm{i}\partial_tu(x,t)=((-\Delta)^m+V(x))u(x,t),\quad t\in \mathbb{R},\,\,x\in \mathbb{R}^n, \end{equation*} where $V$ is a real-valued $C^{\frac{n+1}{2}-2m}$ potential with certain decay. Let $P_{ac}(H)$ denote the projection onto the absolutely continuous spectrum space of $H=(-\Delta)^m+V$, and assume that $H$ has no positive embedded eigenvalue. Our main result says that $e^{-\mathrm{i}tH}P_{ac}(H)$ has integral kernel $K(t,x,y)$ satisfying \begin{equation*} |K(t, x,y)|\le C(1+|t|)^{-(\frac{n}{2m}-\sigma)}(1+|t|^{-\frac{n}{2 m}})\left(1+|t|^{-\frac{1}{2 m}}|x-y|\right)^{-\frac{n(m-1)}{2 m-1}},\quad t\neq0,\,x,y\in\mathbb{R}^n, \end{equation*} where $\sigma=2$ if $0$ is an eigenvalue of $H$, and $\sigma=0$ otherwise. A similar result for smoothing operators $H^\frac{\alpha}{2m}e^{-\mathrm{i}tH}P_{ac}(H)$ is also given. The regularity condition $V\in C^{\frac{n+1}{2}-2m}$ is optimal in the second order case, and it also seems optimal when $m>1$. Show Chinese abstract

Abstract: 在本文中，对于任何奇数$n$和任何整数$m\geq1$且$n>4m$，我们研究高阶薛定谔方程的基本解 \begin{equation*} \mathrm{i}\partial_tu(x,t)=((-\Delta)^m+V(x))u(x,t),\quad t\in \mathbb{R},\,\,x\in \mathbb{R}^n, \end{equation*}其中$V$是一个具有特定衰减的实值$C^{\frac{n+1}{2}-2m}$势。 设$P_{ac}(H)$表示到$H=(-\Delta)^m+V$绝对连续谱空间的投影，并假设$H$没有正的嵌入本征值。 我们的主要结果表明，$e^{-\mathrm{i}tH}P_{ac}(H)$有积分核$K(t,x,y)$，满足\begin{equation*} |K(t, x,y)|\le C(1+|t|)^{-(\frac{n}{2m}-\sigma)}(1+|t|^{-\frac{n}{2 m}})\left(1+|t|^{-\frac{1}{2 m}}|x-y|\right)^{-\frac{n(m-1)}{2 m-1}},\quad t\neq0,\,x,y\in\mathbb{R}^n, \end{equation*}，其中如果$0$是$H$的特征值，则$\sigma=2$，否则$\sigma=0$。 对于平滑算子 $H^\frac{\alpha}{2m}e^{-\mathrm{i}tH}P_{ac}(H)$ 的类似结果也给出了。 正则性条件 $V\in C^{\frac{n+1}{2}-2m}$ 在二阶情况下是最优的，当 $m>1$ 时似乎也是最优的。