具有周期加紧支势的一维薛定谔算子
1D Schrödinger operator with periodic plus compactly supported potentials
摘要：
我们考虑一维薛定谔算子$Hy=-y''+(p+q)y$，其周期势$p$加上实直线上的紧支势$q$。$H$的谱由一个绝对连续的部分加上每个谱隙$\g_n\ne \es, n\geq 0$中有限个简单本征值组成，其中$\g_0$是无界的间隙。我们证明了以下结果：1）确定了大半径圆盘中共振态的分布; 2）确定了共振态的禁带; 3）确定了本征值和反束缚态的渐近性; 4）如果$q_0=\int_\R qdx=0$，则粗略地说，在每个非简并带隙$\g_n$中，如果$n$足够大，则存在两个本征值和零个反束缚态或零个本征值和两个反束缚态5）如果$H$在连续谱中有无穷多个能隙，那么对于任何序列$\s=(\s)_1^\iy, \s_n\in \{0,2\}$，存在一个紧支势$q$，使得$H$在每个能隙$\g_n$中有$\s_n$束缚态和$2-\s_n$反束缚态，对于足够大的$n$. 6)对于任何$q$（具有$q_0=0$），$\s=(\s_n)_{1}^\iy$，其中$\s_n\in \{0,2\}$和对于任何序列$\d=(\d_n)_1^\iy\in \ell^2, \d_n>0$，存在潜在的$p\in L^2(0,1)$，使得对于足够大的$n$，每个间隙长度$|\g_n|=\d_n, n\ge 1$和$H$在每个间隙$\g_n\ne \es$中具有精确的$\s_n$本征值和$2-\s_n$反结合态。
英文摘要：
We consider the 1D Schrödinger operator $Hy=-y''+(p+q)y$ with a periodic potential $p$ plus compactly supported potential $q$ on the real line. The spectrum of $H$ consists of an absolutely continuous part plus a finite number of simple eigenvalues in each spectral gap $\g_n\ne \es, n\geq 0$, where $\g_0$ is unbounded gap. We prove the following results: 1) we determine the distribution of resonances in the disk with large radius, 2) a forbidden domain for the resonances is specified, 3) the asymptotics of eigenvalues and antibound states are determined, 4) if $q_0=\int_\R qdx=0$, then roughly speaking in each nondegenerate gap $\g_n$ for $n$ large enough there are two eigenvalues and zero antibound state or zero eigenvalues and two antibound states, 5) if $H$ has infinitely many gaps in the continuous spectrum, then for any sequence $\s=(\s)_1^\iy, \s_n\in \{0,2\}$, there exists a compactly supported potential $q$ such that $H$ has $\s_n$ bound states and $2-\s_n$ antibound states in each gap $\g_n$ for $n$ large enough. 6) For any $q$ (with $q_0=0$), $\s=(\s_n)_{1}^\iy$, where $\s_n\in \{0,2\}$ and for any sequence $\d=(\d_n)_1^\iy\in \ell^2, \d_n>0$ there exists a potential $p\in L^2(0,1)$ such that each gap length $|\g_n|=\d_n, n\ge 1$ and $H$ has exactly $\s_n$ eigenvalues and $2-\s_n$ antibound states in each gap $\g_n\ne \es$ for $n$ large enough.
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