一类具有复权函数和两个不连续点的Dirac算子的逆问题

The inverse spectral problems of a class of Dirac operators with complex weight function and two discontinuities

摘要

本文考虑了一类具有转移条件和复值权函数的Dirac算子的逆问题，采用了谱映射的方法得到了具有两个不连续点的Dirac算子的势函数在整个区间能被两组谱或Weyl函数唯一确定。

Abstract

In this paper, the inverse problem of a class of Dirac operators with jump conditions and complex-valued weight functions is studied. The spectral mapping method is used to obtain that the potential function of Dirac operators with two discontinuities can be uniquely determined by two groups of spectra or Weyl-type functions on the whole interval.

关键词

Dirac算子 / 势函数 / Weyl函数

Key words

Dirac operator / the potential funcrtions / Weyl-type functions

近年来，许多学者对微分算子的逆问题进行了大量研究并已经取得了许多重要的研究成果^[1-6]。因为微分算子的逆问题相对于正问题来说更加具有实用价值，所以越来越多的学者对微分算子的逆问题产生了兴趣。2000年Yurko研究了Sturm-Liouville算子中具有转移条件的边值问题^[7-8]，并且后续在文献中提出了谱映射的思想，成功运用谱映射的思想证明了Sturm-Liouville算子逆问题的唯一性^[9]。后来Amirov把谱映射的方法推广到了Dirac算子中，这种方法将逆问题简化为主方程，使Dirac算子逆问题的唯一性问题对应于Banach空间中线性方程组解存在的唯一性问题，就可以证明Dirac算子中势函数的唯一性，并且可以得到Dirac算子可解的充要条件^[10]。2018年Yurko^[11]和2021年Zhang等^[12]分别研究了Sturm-Liouville算子和Dirac算子具有复值权函数和转移条件的逆问题，证明了势函数在整个区间能被Weyl函数或两组谱唯一确定；本文将Dirac算子具有复值权函数和转移条件的逆问题推广到具有两个不连续点和复值权函数的Dirac算子的逆问题中，类似地，运用谱映射的方法得到势函数在整个区间能被Weyl函数或两组谱唯一确定。

1 预备知识

在x\in \left[\mathrm{0}\right.,\left.b_{\mathrm{1}}\right)\bigcup \left(b_{\mathrm{1}},b_{\mathrm{2}}\right)\bigcup \left(b_{\mathrm{2}}\right.,\left.T\right]区间考虑如下Dirac方程

B{y}^{\text{'}}\left(x\right)+Q\left(x\right)y\left(x\right)=\lambda r\left(x\right)y\left(x\right)(1)

具有分离型边界条件\left(\mathrm{0}\le \alpha ,\beta \le \pi \right)

U\left(y\right)\begin{array}{c}:\end{array}=y_{\mathrm{1}}\left(\mathrm{0}\right)\mathrm{c}\mathrm{o}\mathrm{s}\alpha +y_{\mathrm{2}}\left(\mathrm{0}\right)\mathrm{s}\mathrm{i}\mathrm{n}\alpha =\mathrm{0}(2)

V\left(y\right)\begin{array}{c}:\end{array}=y_{\mathrm{1}}\left(T\right)\mathrm{c}\mathrm{o}\mathrm{s}\beta +y_{\mathrm{2}}\left(T\right)\mathrm{s}\mathrm{i}\mathrm{n}\beta =\mathrm{0}(3)

在点b_{\mathrm{1}},b_{\mathrm{2}}处有转移条件

\begin{array}{c}{y}_{\mathrm{1}}\left(b_{\mathrm{1}}+\mathrm{0}\right)=d_{\mathrm{1}}{y}_{\mathrm{1}}\left(b_{\mathrm{1}}-\mathrm{0}\right),\\ y_{\mathrm{2}}\left(b_{\mathrm{1}}+\mathrm{0}\right)=d_{\mathrm{2}}{y}_{\mathrm{2}}\left(b_{\mathrm{1}}-\mathrm{0}\right)+d_{\mathrm{3}}{y}_{\mathrm{1}}\left(b_{\mathrm{1}}-\mathrm{0}\right)\end{array}(4)

\begin{array}{c}{y}_{\mathrm{1}}\left(b_{\mathrm{2}}+\mathrm{0}\right)=d_{\mathrm{2}}{y}_{\mathrm{1}}\left(b_{\mathrm{2}}-\mathrm{0}\right),\\ y_{\mathrm{2}}\left(b_{\mathrm{2}}+\mathrm{0}\right)=d_{\mathrm{3}}{y}_{\mathrm{2}}\left(b_{\mathrm{2}}-\mathrm{0}\right)+d_{\mathrm{4}}{y}_{\mathrm{1}}\left(b_{\mathrm{2}}-\mathrm{0}\right)\end{array}(5)

其中

B=\left(\begin{array}{cc}\mathrm{0}& \mathrm{1}\\ -\mathrm{1}& \mathrm{0}\end{array}\right),Q\left(x\right)=\left(\begin{array}{cc}p\left(x\right)& q\left(x\right)\\ q\left(x\right)& -p\left(x\right)\end{array}\right)，

y\left(x\right)=\left(\begin{array}{c}{y}_{\mathrm{1}}\left(x\right)\\ y_{\mathrm{2}}\left(x\right)\end{array}\right),r\left(x\right)=\left\{\begin{array}{cc}{a}_{\mathrm{1}},& x\in \left.\left[\mathrm{0},b_{\mathrm{1}}\right.\right)\\ a_{\mathrm{2}},& x\in \left(b_{\mathrm{1}},b_{\mathrm{2}}\right)\\ a_{\mathrm{3}},& x\in \left.\left(b_{\mathrm{2}},T\right.\right]\end{array}\right.(6)

\lambda是谱参数，p\left(x\right),q\left(x\right),r\left(x\right)是复值函数，p\left(x\right),q\left(x\right)在区间\left[\mathrm{0},b_{\mathrm{1}}\right],\left[b_{\mathrm{1}},b_{\mathrm{2}}\right],\left[b_{\mathrm{2}},T\right]上绝对连续，可能有如下情况发生：

p\left(b_i+\mathrm{0}\right)\ne p\left(b_i-\mathrm{0}\right),q\left(b_i+\mathrm{0}\right)\ne q\left(b_i-\mathrm{0}\right),\left(i=\mathrm{1,2}\right)且a_k,d_j\left(k=\overline{\mathrm{1,3}};j=\overline{\mathrm{1,4}}\right)是复数，假设a_{\mathrm{1}}{a}_{\mathrm{2}}{a}_{\mathrm{3}},d_{\mathrm{1}}{d}_{\mathrm{2}},d_{\mathrm{1}}{d_{\mathrm{2}}}^{\mathrm{2}}{d}_{\mathrm{3}},\left(d_{\mathrm{2}}+d_{\mathrm{3}}-i{d}_{\mathrm{4}}\right)\left(d_{\mathrm{1}}+d_{\mathrm{2}}-i{d}_{\mathrm{3}}\right),\left(d_{\mathrm{2}}+d_{\mathrm{3}}-i{d}_{\mathrm{4}}\right)\left(-d_{\mathrm{1}}+d_{\mathrm{2}}+i{d}_{\mathrm{3}}\right),\left(-d_{\mathrm{2}}+d_{\mathrm{3}}+i{d}_{\mathrm{4}}\right)\left(-d_{\mathrm{1}}+d_{\mathrm{2}}-i{d}_{\mathrm{3}}\right),\left(-d_{\mathrm{2}}+d_{\mathrm{3}}+i{d}_{\mathrm{4}}\right)\left(d_{\mathrm{1}}+d_{\mathrm{2}}+i{d}_{\mathrm{3}}\right),\left(-d_{\mathrm{2}}+d_{\mathrm{3}}-i{d}_{\mathrm{4}}\right)\left(d_{\mathrm{1}}+d_{\mathrm{2}}-i{d}_{\mathrm{3}}\right),\left(-d_{\mathrm{2}}+d_{\mathrm{3}}-i{d}_{\mathrm{4}}\right)\left(-d_{\mathrm{1}}+d_{\mathrm{2}}+i{d}_{\mathrm{3}}\right),\left(d_{\mathrm{2}}+d_{\mathrm{3}}+i{d}_{\mathrm{4}}\right)\left(-d_{\mathrm{1}}+d_{\mathrm{2}}-i{d}_{\mathrm{3}}\right),\left(d_{\mathrm{2}}+d_{\mathrm{3}}+i{d}_{\mathrm{4}}\right)\left(d_{\mathrm{1}}+d_{\mathrm{2}}+i{d}_{\mathrm{3}}\right)是非零的,并用L=L\left(Q\left(x\right),r\left(x\right),\alpha ,\beta \right)表示式(1)~(6)。当边界条件为y_{\mathrm{1}}\left(\mathrm{0}\right)\mathrm{s}\mathrm{i}\mathrm{n}\alpha -y_{\mathrm{2}}\left(\mathrm{0}\right)\mathrm{c}\mathrm{o}\mathrm{s}\alpha =V\left(y\right)=\mathrm{0}时，用L_{\mathrm{0}}表示。

令\phi \left(x,\lambda \right)和\omega \left(x,\lambda \right)是方程(1)满足如下初始条件和转移条件(4)和(5)的解

\begin{array}{l}\phi \left(\mathrm{0},\lambda \right)=\left(\begin{array}{c}{\phi }_{\mathrm{1}}\left(\mathrm{0},\lambda \right)\\ {\phi }_{\mathrm{2}}\left(\mathrm{0},\lambda \right)\end{array}\right)=\left(\begin{array}{c}\mathrm{s}\mathrm{i}\mathrm{n}\alpha \\ -\mathrm{c}\mathrm{o}\mathrm{s}\alpha \end{array}\right),\\ \omega \left(\mathrm{0},\lambda \right)=\left(\begin{array}{c}{\omega }_{\mathrm{1}}\left(\mathrm{0},\lambda \right)\\ {\omega }_{\mathrm{2}}\left(\mathrm{0},\lambda \right)\end{array}\right)=\left(\begin{array}{c}\mathrm{c}\mathrm{o}\mathrm{s}\alpha \\ \mathrm{s}\mathrm{i}\mathrm{n}\alpha \end{array}\right)\mathrm{。}\end{array}

记式(1)~(6)的特征函数是\mathrm{\Delta }\left(\lambda \right)=V\left(\phi \right)，那么这个函数是关于\lambda的一阶整函数，并且它的零点{\left\{{\lambda }_n\right\}}_{n\in \mathbb{Z}}与L的特征值一致。

令\varphi \left(x,\lambda \right)=\left(\begin{array}{c}{\varphi }_{\mathrm{1}}\left(x,\lambda \right)\\ {\varphi }_{\mathrm{2}}\left(x,\lambda \right)\end{array}\right)是方程(1)满足式(4)、式(5)和边界条件U(\varphi )=\mathrm{1},V\left(\varphi \right)=\mathrm{0}的解，则有：

\varphi \left(x,\lambda \right)=\omega \left(x,\lambda \right)+M\left(\lambda \right)\phi \left(x,\lambda \right)(7)

记M\left(\lambda \right)是L=L\left(Q\left(x\right),r\left(x\right),\alpha ,\beta \right)的Weyl函数，代入式(3)得到

V\left(\varphi \right)=V\left(\omega \right)+M\left(\lambda \right)V\left(\phi \right)=\mathrm{0}(8)

接下来由式(8)可得

M\left(\lambda \right)=-\frac{V\left(\omega \right)}{V\left(\phi \right)}=-\frac{{\mathrm{\Delta }}_{\mathrm{0}}\left(\lambda \right)}{\mathrm{\Delta }\left(\lambda \right)}(9)

其中，{\mathrm{\Delta }}_{\mathrm{0}}\left(\lambda \right)=V\left(\omega \right)是L_{\mathrm{0}}的特征函数，且{\mathrm{\Delta }}_{\mathrm{0}}\left(\lambda \right)是关于\lambda的一阶整函数，它的零点{\left\{{\lambda }_{n\mathrm{0}}\right\}}_{n\in \mathbb{Z}}是L_{\mathrm{0}}的特征值。

记

\Psi \left(x,\lambda \right)=\left(\begin{array}{cc}{\phi }_{\mathrm{1}}\left(x,\lambda \right)& {\varphi }_{\mathrm{1}}\left(x,\lambda \right)\\ {\phi }_{\mathrm{2}}\left(x,\lambda \right)& {\varphi }_{\mathrm{2}}\left(x,\lambda \right)\end{array}\right)。

由上式和式(4)、(5)可得：

(10)

为了解决如下逆问题，首先令a_k,d_j\left(k=\overline{\mathrm{1},\mathrm{3}}\right.；\left.j=\overline{\mathrm{1},\mathrm{4}}\right)和\alpha是已知的。

逆问题1　已知Weyl函数M\left(\lambda \right)，确定Q\left(x\right)和\beta。

逆问题2　已知两组谱{\left\{{\lambda }_n\right\}}_{n\in \mathbb{Z}},{\left\{{\lambda }_{n\mathrm{0}}\right\}}_{n\in \mathbb{Z}}和\beta，确定Q\left(x\right)。

令\tilde{L}是与L形式相同但系数不同的边值问题，如果v表示与L有关的量，那么，\tilde{v}也同样能表示与\tilde{L}有关的类似的量，则有b=\tilde{b}，T=\tilde{T}，a_k={\tilde{a}}_k\left(k=\overline{\mathrm{1},\mathrm{3}}\right)，d_j={\tilde{d}}_j\left(j=\overline{\mathrm{1},\mathrm{4}}\right)，并且\alpha =\tilde{\alpha }。

2 重要引理

首先引入一组符号：{\mathrm{\Omega }}_k^{\pm }:=\left\{\lambda \in \mathbb{C}:\pm \mathrm{I}\mathrm{m}\left(\lambda a_k\right)>\mathrm{0}\right\}，A_k\left(\lambda ,x\right)=\left\{\begin{array}{c}{\mathrm{\Lambda }}_k\left(-\lambda x\right),\\ {\mathrm{\Lambda }}_k\left(\lambda x\right),\end{array}\right.\begin{array}{c}\lambda \in {\mathrm{\Omega }}_k^{+}\\ \lambda \in {\mathrm{\Omega }}_k^{-}\end{array}，{\mathrm{\Lambda }}_k\left(\lambda \right)=\mathrm{e}\mathrm{x}\mathrm{p}\left(i\lambda a_k\right)。

令c_{\mathrm{0}}=\mathrm{0},c_{\mathrm{1}}=b_{\mathrm{1}},c_{\mathrm{2}}=b_{\mathrm{2}},c_{\mathrm{3}}=T。

类似于文献[12]的引理3.1可以得到方程(1)的Birkhoff型解。

引理1 对于k=\mathrm{1},\mathrm{2},\mathrm{3},x\in \left[c_{k-\mathrm{1}},c_k\right],\lambda \in {\mathrm{\Omega }}_k^{\pm }，存在方程(1)的一组基本解\left\{e_k\left(x,\lambda \right),E_k\left(x,\lambda \right)\right\}。

e_k\left(x,\lambda \right)=\left(\begin{array}{c}{e}_{k\mathrm{1}}\left(x,\lambda \right)\\ e_{k\mathrm{2}}\left(x,\lambda \right)\end{array}\right)=\left(\begin{array}{c}i{e}^{-i\lambda a_k\left(x-c_{k-\mathrm{1}}\right)}\left[\mathrm{1}\right]\\ -e^{-i\lambda a_k\left(x-c_{k-\mathrm{1}}\right)}\left[\mathrm{1}\right]\end{array}\right)(11)

E_k\left(x,\lambda \right)=\left(\begin{array}{c}{E}_{k\mathrm{1}}\left(x,\lambda \right)\\ E_{k\mathrm{2}}\left(x,\lambda \right)\end{array}\right)=\left(\begin{array}{c}-i{e}^{i\lambda a_k\left(x-c_{k-\mathrm{1}}\right)}\left[\mathrm{1}\right]\\ -e^{i\lambda a_k\left(x-c_{k-\mathrm{1}}\right)}\left[\mathrm{1}\right]\end{array}\right)(12)

其中\left[\mathrm{1}\right]=\mathrm{1}+\mathrm{{\rm O}}\left({\lambda }^{-\mathrm{1}}\right)，\lambda \to \mathrm{\infty }，当\lambda \in {\mathrm{\Omega }}_k^{\pm }，\left|\lambda \right|>{\lambda }^{\mathrm{*}}时，向量函数e_k(x,\lambda )和E_k\left(x,\lambda \right)在x属于\left[c_{k-\mathrm{1}},c_k\right]上连续，且是解析的。

接下来类似于文献[12]把\lambda复平面划分为一些扇形：

D_j\begin{array}{c}:\end{array}=\left\{\lambda :\mathrm{a}\mathrm{r}\mathrm{g}\lambda \in \left({\theta }_j,{\theta }_{j+\mathrm{1}}\right)\right\},j=\overline{\mathrm{1},\mathrm{3}s},

{\theta }_{\mathrm{3}s+\mathrm{1}}\begin{array}{c}:\end{array}={\theta }_{\mathrm{1}},

使得\mathrm{R}\mathrm{e}\left(i\lambda a_k\right),k=\mathrm{1},\mathrm{2},\mathrm{3},在每一部分D_j内保持符号，显然在\lambda复平面上的划分依赖于a_k，由此决定了s，且\mathrm{1}\le s\le \mathrm{3}，对充分小的\varsigma >\mathrm{0}，则有

S_{j,\varsigma }\begin{array}{c}:\end{array}=\left\{\lambda :\mathrm{a}\mathrm{r}\mathrm{g}\lambda \in \left({\theta }_j+\varsigma ,{\theta }_{j+\mathrm{1}}-\varsigma \right)\right\}S_{\varsigma }={\displaystyle \underset{j=\mathrm{1}}{\overset{\mathrm{3}s}{\cup }}}{S}_{j,\varsigma }。

在x\in \left[c_{k-\mathrm{1}},c_k\right]中，当\lambda \in S_{\varsigma }时，式(11)和(12)的余项估计\mathrm{{\rm O}}\left({\lambda }^{-\mathrm{1}}\right)是一致的。

引理2 当\lambda \in S_{\varsigma }，\left|\lambda \right|>{\lambda }^{\mathrm{*}},j=\mathrm{1},\mathrm{2}，并且x在区间\left[\mathrm{0},b_{\mathrm{1}}\right],\left[b_{\mathrm{1}},b_{\mathrm{2}}\right]或\left[b_{\mathrm{2}},T\right]上时，下列估计成立。

\left|{\phi }_j\left(x,\lambda \right)\right|\le C\left|A_{\mathrm{1}}\left(\lambda ,x\right)\right|,x\in \left[\mathrm{0},b_{\mathrm{1}}\right](13)

\left|{\phi }_j\left(x,\lambda \right)\right|\le C\left|A_{\mathrm{1}}\left(\lambda ,b_{\mathrm{1}}\right)A_{\mathrm{2}}\left(\lambda ,x-b_{\mathrm{1}}\right)\right|,x\in \left[b_{\mathrm{1}},b_{\mathrm{2}}\right](14)

\begin{array}{c}\left|{\phi }_j\left(x,\lambda \right)\right|\le C\left|A_{\mathrm{1}}\left(\lambda ,b_{\mathrm{1}}\right)A_{\mathrm{2}}\left(\lambda ,b_{\mathrm{2}}-b_{\mathrm{1}}\right)A_{\mathrm{3}}\left(\lambda ,x-b_{\mathrm{2}}\right)\right|,\\ x\in \left[b_{\mathrm{2}},T\right]\end{array}(15)

\left|{\phi }_j\left(x,\lambda \right)-{\tilde{\phi }}_j\left(x,\lambda \right)\right|\le \frac{C}{\left|\lambda \right|}\left|A_{\mathrm{1}}\left(\lambda ,x\right)\right|,x\in \left[\mathrm{0},b_{\mathrm{1}}\right](16)

\begin{array}{c}\left|{\phi }_j\left(x,\lambda \right)-{\tilde{\phi }}_j\left(x,\lambda \right)\right|\le \frac{C}{\left|\lambda \right|}\left|A_{\mathrm{1}}\left(\lambda ,b_{\mathrm{1}}\right)A_{\mathrm{2}}\left(\lambda ,x-b_{\mathrm{1}}\right)\right|,\\ x\in \left[b_{\mathrm{1}},b_{\mathrm{2}}\right]\end{array}(17)

\begin{array}{c}\left|{\phi }_j\left(x,\lambda \right)-{\tilde{\phi }}_j\left(x,\lambda \right)\right|\le \\ \frac{C}{\left|\lambda \right|}\left|A_{\mathrm{1}}\left(\lambda ,b_{\mathrm{1}}\right)A_{\mathrm{2}}\left(\lambda ,b_{\mathrm{2}}-b_{\mathrm{1}}\right)A_{\mathrm{3}}\left(\lambda ,x-b_{\mathrm{2}}\right)\right|,\\ x\in \left[b_{\mathrm{2}},T\right]\end{array}(18)

在区间\left[\mathrm{0},b_{\mathrm{1}}\right]上，有\phi \left(b_{\mathrm{1}},\lambda \right)=\phi \left(b_{\mathrm{1}}-\mathrm{0},\lambda \right)，在区间\left[b_{\mathrm{1}},b_{\mathrm{2}}\right]上，有\phi \left(b_{\mathrm{1}},\lambda \right)=\phi \left(b_{\mathrm{1}}+\mathrm{0},\lambda \right),\phi \left(b_{\mathrm{2}},\lambda \right)=\phi \left(b_{\mathrm{2}}-\mathrm{0},\lambda \right)，在区间\left[b_{\mathrm{2}},T\right]上，有\phi \left(b_{\mathrm{2}},\lambda \right)=\phi \left(b_{\mathrm{2}}+\mathrm{0},\lambda \right)。

证明 因为方程(1)有基础解系\left\{e_k\left(x,\lambda \right),E_k\left(x,\lambda \right)\right\}，因此，

\begin{array}{c}\phi \left(x,\lambda \right)={\alpha }_k\left(\lambda \right)e_k\left(x,\lambda \right)+{\beta }_k\left(\lambda \right)E_k\left(x,\lambda \right),\\ k=\mathrm{1},\mathrm{2},\mathrm{3},x\in \left[c_{k-\mathrm{1}},c_k\right]\end{array}(19)

其中{\alpha }_k\left(\lambda \right)和{\beta }_k\left(\lambda \right)不依赖于x，根据\phi \left(x,\lambda \right)满足的初始条件和式(11)、(12)，并且当\lambda \in {\mathrm{\Omega }}_{\mathrm{1}}^{\pm },\left|\lambda \right|>{\lambda }^{\mathrm{*}}，\left|\lambda \right|\to \mathrm{\infty }时，得{\alpha }_{\mathrm{1}}\left(\lambda \right),{\beta }_{\mathrm{1}}\left(\lambda \right)有

\mathrm{2}{\alpha }_{\mathrm{1}}\left(\lambda \right)=e^{-i\alpha }[\mathrm{1}],\mathrm{2}{\beta }_{\mathrm{1}}\left(\lambda \right)=e^{i\alpha }[\mathrm{1}](20)

当\lambda \in {\mathrm{\Omega }}_{\mathrm{2}}^{\pm },\left|\lambda \right|>{\lambda }^{\mathrm{*}}，\left|\lambda \right|\to \mathrm{\infty }且S_{\varsigma }中余项估计一致时，运用式(4)、(5)、(11)和(12)得{\alpha }_{\mathrm{2}}\left(\lambda \right),{\beta }_{\mathrm{2}}\left(\lambda \right)：

\begin{array}{l}\mathrm{4}{\alpha }_{\mathrm{2}}\left(\lambda \right)=\left(d_{\mathrm{1}}+d_{\mathrm{2}}-i{d}_{\mathrm{3}}\right)e^{-i\left(\alpha +\lambda a_{\mathrm{1}}{b}_{\mathrm{1}}\right)}\left[\mathrm{1}\right]\\ +\left(-d_{\mathrm{1}}+d_{\mathrm{2}}+i{d}_{\mathrm{3}}\right)e^{i\left(\alpha +\lambda a_{\mathrm{1}}{b}_{\mathrm{1}}\right)}\left[\mathrm{1}\right]\end{array}\begin{array}{l}\mathrm{4}{\beta }_{\mathrm{2}}\left(\lambda \right)=\left(-d_{\mathrm{1}}+d_{\mathrm{2}}-i{d}_{\mathrm{3}}\right)e^{-i\left(\alpha +\lambda a_{\mathrm{1}}{b}_{\mathrm{1}}\right)}\left[\mathrm{1}\right]\\ +\left(d_{\mathrm{1}}+d_{\mathrm{2}}+i{d}_{\mathrm{3}}\right)e^{i\left(\alpha +\lambda a_{\mathrm{1}}{b}_{\mathrm{1}}\right)}\left[\mathrm{1}\right]\end{array}(21)

当\lambda \in {\mathrm{\Omega }}_{\mathrm{3}}^{\pm },\left|\lambda \right|>{\lambda }^{\mathrm{*}}，\left|\lambda \right|\to \mathrm{\infty }时，由{\alpha }_{\mathrm{2}}\left(\lambda \right),{\beta }_{\mathrm{2}}\left(\lambda \right)和转移条件可得{\alpha }_{\mathrm{3}}\left(\lambda \right),{\beta }_{\mathrm{3}}\left(\lambda \right)：

\begin{array}{l}\mathrm{8}{\alpha }_{\mathrm{3}}\left(\lambda \right)=\left(d_{\mathrm{2}}+d_{\mathrm{3}}-i{d}_{\mathrm{4}}\right)\left(d_{\mathrm{1}}+d_{\mathrm{2}}-i{d}_{\mathrm{3}}\right)e^{-i\left(\alpha +\lambda \left(a_{\mathrm{1}}{b}_{\mathrm{1}}+a_{\mathrm{2}}\left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right)\right)}\left[\mathrm{1}\right]\\ +\left(d_{\mathrm{2}}+d_{\mathrm{3}}-i{d}_{\mathrm{4}}\right)\left(-d_{\mathrm{1}}+d_{\mathrm{2}}+i{d}_{\mathrm{3}}\right)e^{i\left(\alpha +\lambda \left(a_{\mathrm{1}}{b}_{\mathrm{1}}-a_{\mathrm{2}}\left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right)\right)}\left[\mathrm{1}\right]\\ +\left(-d_{\mathrm{2}}+d_{\mathrm{3}}+i{d}_{\mathrm{4}}\right)\left(-d_{\mathrm{1}}+d_{\mathrm{2}}-i{d}_{\mathrm{3}}\right)e^{-i\left(\alpha +\lambda \left(a_{\mathrm{1}}{b}_{\mathrm{1}}-a_{\mathrm{2}}\left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right)\right)}\left[\mathrm{1}\right]\\ +\left(-d_{\mathrm{2}}+d_{\mathrm{3}}+i{d}_{\mathrm{4}}\right)\left(d_{\mathrm{1}}+d_{\mathrm{2}}+i{d}_{\mathrm{3}}\right)e^{i\left(\alpha +\lambda \left(a_{\mathrm{1}}{b}_{\mathrm{1}}+a_{\mathrm{2}}\left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right)\right)}\left[\mathrm{1}\right]\end{array}(22)

\begin{array}{l}\mathrm{8}{\beta }_{\mathrm{3}}\left(\lambda \right)=\left(-d_{\mathrm{2}}+d_{\mathrm{3}}-i{d}_{\mathrm{4}}\right)\left(d_{\mathrm{1}}+d_{\mathrm{2}}-i{d}_{\mathrm{3}}\right)e^{-i\left(\alpha +\lambda \left(a_{\mathrm{1}}{b}_{\mathrm{1}}+a_{\mathrm{2}}\left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right)\right)}\left[\mathrm{1}\right]\\ +\left(-d_{\mathrm{2}}+d_{\mathrm{3}}-i{d}_{\mathrm{4}}\right)\left(-d_{\mathrm{1}}+d_{\mathrm{2}}+i{d}_{\mathrm{3}}\right)e^{i\left(\alpha +\lambda \left(a_{\mathrm{1}}{b}_{\mathrm{1}}-a_{\mathrm{2}}\left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right)\right)}\left[\mathrm{1}\right]\\ +\left(d_{\mathrm{2}}+d_{\mathrm{3}}+i{d}_{\mathrm{4}}\right)\left(-d_{\mathrm{1}}+d_{\mathrm{2}}-i{d}_{\mathrm{3}}\right)e^{-i\left(\alpha +\lambda \left(a_{\mathrm{1}}{b}_{\mathrm{1}}-a_{\mathrm{2}}\left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right)\right)}\left[\mathrm{1}\right]\\ +\left(d_{\mathrm{2}}+d_{\mathrm{3}}+i{d}_{\mathrm{4}}\right)\left(d_{\mathrm{1}}+d_{\mathrm{2}}+i{d}_{\mathrm{3}}\right)e^{i\left(\alpha +\lambda \left(a_{\mathrm{1}}{b}_{\mathrm{1}}+a_{\mathrm{2}}\left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right)\right)}\left[\mathrm{1}\right]\end{array}(23)

根据所求出的{\alpha }_k\left(\lambda \right),{\beta }_k\left(\lambda \right)渐进式，代入式(19)，可知引理2的估计式(13)~(15)成立，并且式(11)、(12)、(20)~(23)不依赖于Q\left(x\right),\beta，则估计式(16)~(18)成立。

引理3 当\lambda \in S_{\varsigma }，\left|\lambda \right|>{\lambda }^{\mathrm{*}},j=\mathrm{1},\mathrm{2}，其中x在区间\left[\mathrm{0},b_{\mathrm{1}}\right],\left[b_{\mathrm{1}},b_{\mathrm{2}}\right]或\left[b_{\mathrm{2}},T\right]上时，下列估计成立。

\left|{\varphi }_j\left(x,\lambda \right)\right|\le C\left|{A_{\mathrm{1}}}^{-\mathrm{1}}\left(\lambda ,x\right)\right|,x\in \left[\mathrm{0},b_{\mathrm{1}}\right](24)

\left|{\varphi }_j\left(x,\lambda \right)\right|\le C\left|{A_{\mathrm{1}}}^{-\mathrm{1}}\left(\lambda ,b_{\mathrm{1}}\right){A_{\mathrm{2}}}^{-\mathrm{1}}\left(\lambda ,x-b_{\mathrm{1}}\right)\right|,x\in \left[b_{\mathrm{1}},b_{\mathrm{2}}\right](25)

\begin{array}{c}\left|{\varphi }_j\left(x,\lambda \right)\right|\le \\ C\left|{A_{\mathrm{1}}}^{-\mathrm{1}}\left(\lambda ,b_{\mathrm{1}}\right){A_{\mathrm{2}}}^{-\mathrm{1}}\left(\lambda ,b_{\mathrm{2}}-b_{\mathrm{1}}\right){A_{\mathrm{3}}}^{-\mathrm{1}}\left(\lambda ,x-b_{\mathrm{2}}\right)\right|,\\ x\in \left[b_{\mathrm{2}},T\right]\end{array}(26)

\left|{\varphi }_j\left(x,\lambda \right)-{\tilde{\varphi }}_j\left(x,\lambda \right)\right|\le \frac{C}{\left|\lambda \right|}\left|{A_{\mathrm{1}}}^{-\mathrm{1}}\left(\lambda ,x\right)\right|,x\in \left[\mathrm{0},b_{\mathrm{1}}\right](27)

\begin{array}{c}\left|{\varphi }_j\left(x,\lambda \right)-{\tilde{\varphi }}_j\left(x,\lambda \right)\right|\le \\ \frac{C}{\left|\lambda \right|}\left|{A_{\mathrm{1}}}^{-\mathrm{1}}\left(\lambda ,b_{\mathrm{1}}\right){A_{\mathrm{2}}}^{-\mathrm{1}}\left(\lambda ,x-b_{\mathrm{1}}\right)\right|,x\in \left[b_{\mathrm{1}},b_{\mathrm{2}}\right]\end{array}(28)

\begin{array}{c}\left|{\varphi }_j\left(x,\lambda \right)-{\tilde{\varphi }}_j\left(x,\lambda \right)\right|\le \\ \frac{C}{\left|\lambda \right|}\left|{A_{\mathrm{1}}}^{-\mathrm{1}}\left(\lambda ,b_{\mathrm{1}}\right){A_{\mathrm{2}}}^{-\mathrm{1}}\left(\lambda ,b_{\mathrm{2}}-b_{\mathrm{1}}\right){A_{\mathrm{3}}}^{-\mathrm{1}}\left(\lambda ,x-b_{\mathrm{2}}\right)\right|\\ x\in \left[b_{\mathrm{2}},T\right]\end{array}(29)

证明 由基础解系\left\{e_k\left(x,\lambda \right),E_k\left(x,\lambda \right)\right\}组成的解\varphi \left(x,\lambda \right)如下：

\begin{array}{c}\varphi \left(x,\lambda \right)={\vartheta }_k\left(\lambda \right)e_k\left(x,\lambda \right)+{\sigma }_k\left(\lambda \right)E_k\left(x,\lambda \right),\\ k=\mathrm{1},\mathrm{2},\mathrm{3},x\in \left[c_{k-\mathrm{1}},c_k\right]\end{array}(30)

令{\vartheta }_k\left(\lambda \right)和{\sigma }_k\left(\lambda \right)不依赖于x，已知\varphi \left(x,\lambda \right)满足式(4)、(5)、(30)和边界条件U(\varphi )=\mathrm{1},V\left(\varphi \right)=\mathrm{0}由此得到{\vartheta }_k\left(\lambda \right)和{\sigma }_k\left(\lambda \right)线性表示：

\begin{array}{c}{\vartheta }_{\mathrm{1}}\left(\lambda \right)U\left(e_{\mathrm{1}}\left(x,\lambda \right)\right)+{\sigma }_{\mathrm{1}}\left(\lambda \right)U\left(E_{\mathrm{1}}\left(x,\lambda \right)\right)=\mathrm{1}\\ {\vartheta }_{\mathrm{3}}\left(\lambda \right)V\left(e_{\mathrm{3}}\left(x,\lambda \right)\right)+{\sigma }_{\mathrm{3}}\left(\lambda \right)V\left(E_{\mathrm{3}}\left(x,\lambda \right)\right)=\mathrm{0}\\ {\vartheta }_{\mathrm{2}}\left(\lambda \right)e_{\mathrm{21}}\left(b_{\mathrm{1}},\lambda \right)+{\sigma }_{\mathrm{2}}\left(\lambda \right)E_{\mathrm{21}}\left(b_{\mathrm{1}},\lambda \right)\\ -d_{\mathrm{1}}\left({\vartheta }_{\mathrm{1}}\left(\lambda \right)e_{\mathrm{11}}\left(b_{\mathrm{1}},\lambda \right)+{\sigma }_{\mathrm{1}}\left(\lambda \right)E_{\mathrm{11}}\left(b_{\mathrm{1}},\lambda \right)\right)=\mathrm{0}\\ {\vartheta }_{\mathrm{2}}\left(\lambda \right)e_{\mathrm{22}}\left(b_{\mathrm{1}},\lambda \right)+{\sigma }_{\mathrm{2}}\left(\lambda \right)E_{\mathrm{22}}\left(b_{\mathrm{1}},\lambda \right)\\ -d_{\mathrm{2}}\left({\vartheta }_{\mathrm{1}}\left(\lambda \right)e_{\mathrm{12}}\left(b_{\mathrm{1}},\lambda \right)+{\sigma }_{\mathrm{1}}\left(\lambda \right)E_{\mathrm{12}}\left(b_{\mathrm{1}},\lambda \right)\right)\\ -d_{\mathrm{3}}\left({\vartheta }_{\mathrm{1}}\left(\lambda \right)e_{\mathrm{11}}\left(b_{\mathrm{1}},\lambda \right)+{\sigma }_{\mathrm{1}}\left(\lambda \right)E_{\mathrm{11}}\left(b_{\mathrm{1}},\lambda \right)\right)=\mathrm{0}\\ {\vartheta }_{\mathrm{3}}\left(\lambda \right)e_{\mathrm{31}}\left(b_{\mathrm{2}},\lambda \right)+{\sigma }_{\mathrm{3}}\left(\lambda \right)E_{\mathrm{31}}\left(b_{\mathrm{2}},\lambda \right)\\ -d_{\mathrm{2}}\left({\vartheta }_{\mathrm{2}}\left(\lambda \right)e_{\mathrm{21}}\left(b_{\mathrm{2}},\lambda \right)+{\sigma }_{\mathrm{2}}\left(\lambda \right)E_{\mathrm{21}}\left(b_{\mathrm{2}},\lambda \right)\right)=\mathrm{0}\\ {\vartheta }_{\mathrm{3}}\left(\lambda \right)e_{\mathrm{32}}\left(b_{\mathrm{2}},\lambda \right)+{\sigma }_{\mathrm{3}}\left(\lambda \right)E_{\mathrm{32}}\left(b_{\mathrm{2}},\lambda \right)\\ -d_{\mathrm{3}}\left({\vartheta }_{\mathrm{2}}\left(\lambda \right)e_{\mathrm{22}}\left(b_{\mathrm{2}},\lambda \right)+{\sigma }_{\mathrm{2}}\left(\lambda \right)E_{\mathrm{22}}\left(b_{\mathrm{2}},\lambda \right)\right)\\ -d_{\mathrm{4}}\left({\vartheta }_{\mathrm{2}}\left(\lambda \right)e_{\mathrm{21}}\left(b_{\mathrm{2}},\lambda \right)+{\sigma }_{\mathrm{2}}\left(\lambda \right)E_{\mathrm{21}}\left(b_{\mathrm{2}},\lambda \right)\right)=\mathrm{0}\end{array}

用\kappa \left(\lambda \right)表示这个行列式，根据式(11)、(12)，且在S_{\varsigma }中有\left|\lambda \right|\to \mathrm{\infty }，可得：

\begin{array}{l}\kappa \left(\lambda \right)=\left(d_{\mathrm{2}}+d_{\mathrm{3}}-i{d}_{\mathrm{4}}\right)\left(d_{\mathrm{1}}+d_{\mathrm{2}}-i{d}_{\mathrm{3}}\right){\mathrm{\Lambda }}_{\mathrm{1}}\left(-\lambda b_{\mathrm{1}}\right)\\ {\mathrm{\Lambda }}_{\mathrm{2}}\left(-\lambda \left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right){\mathrm{\Lambda }}_{\mathrm{3}}\left(-\lambda \left(T-b_{\mathrm{2}}\right)\right)\left(-e^{i\left(\beta -\alpha \right)}\right)\left[\mathrm{1}\right]\\ +\left(d_{\mathrm{2}}+d_{\mathrm{3}}-i{d}_{\mathrm{4}}\right)\left(-d_{\mathrm{1}}+d_{\mathrm{2}}+i{d}_{\mathrm{3}}\right){\mathrm{\Lambda }}_{\mathrm{1}}\left(\lambda b_{\mathrm{1}}\right)\\ {\mathrm{\Lambda }}_{\mathrm{2}}\left(-\lambda \left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right){\mathrm{\Lambda }}_{\mathrm{3}}\left(-\lambda \left(T-b_{\mathrm{2}}\right)\right)\left(-e^{i\left(\beta +\alpha \right)}\right)\left[\mathrm{1}\right]\\ +\left(-d_{\mathrm{2}}+d_{\mathrm{3}}+i{d}_{\mathrm{4}}\right)\left(-d_{\mathrm{1}}+d_{\mathrm{2}}-i{d}_{\mathrm{3}}\right){\mathrm{\Lambda }}_{\mathrm{1}}\left(-\lambda b_{\mathrm{1}}\right)\\ {\mathrm{\Lambda }}_{\mathrm{2}}\left(\lambda \left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right){\mathrm{\Lambda }}_{\mathrm{3}}\left(-\lambda \left(T-b_{\mathrm{2}}\right)\right)\left(-e^{i\left(\beta -\alpha \right)}\right)\left[\mathrm{1}\right]\\ +\left(-d_{\mathrm{2}}+d_{\mathrm{3}}+i{d}_{\mathrm{4}}\right)\left(d_{\mathrm{1}}+d_{\mathrm{2}}+i{d}_{\mathrm{3}}\right){\mathrm{\Lambda }}_{\mathrm{1}}\left(\lambda b_{\mathrm{1}}\right)\\ {\mathrm{\Lambda }}_{\mathrm{2}}\left(\lambda \left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right){\mathrm{\Lambda }}_{\mathrm{3}}\left(-\lambda \left(T-b_{\mathrm{2}}\right)\right)\left(-e^{i\left(\beta +\alpha \right)}\right)\left[\mathrm{1}\right]\\ +\left(-d_{\mathrm{2}}+d_{\mathrm{3}}-i{d}_{\mathrm{4}}\right)\left(d_{\mathrm{1}}+d_{\mathrm{2}}-i{d}_{\mathrm{3}}\right){\mathrm{\Lambda }}_{\mathrm{1}}\left(-\lambda b_{\mathrm{1}}\right)\end{array}

\begin{array}{l}{\mathrm{\Lambda }}_{\mathrm{2}}\left(-\lambda \left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right){\mathrm{\Lambda }}_{\mathrm{3}}\left(\lambda \left(T-b_{\mathrm{2}}\right)\right)\left({e^{-}}^{i\left(\beta +\alpha \right)}\right)\left[\mathrm{1}\right]\\ +\left(-d_{\mathrm{2}}+d_{\mathrm{3}}-i{d}_{\mathrm{4}}\right)\left(-d_{\mathrm{1}}+d_{\mathrm{2}}+i{d}_{\mathrm{3}}\right){\mathrm{\Lambda }}_{\mathrm{1}}\left(\lambda b_{\mathrm{1}}\right)\\ {\mathrm{\Lambda }}_{\mathrm{2}}\left(-\lambda \left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right){\mathrm{\Lambda }}_{\mathrm{3}}\left(\lambda \left(T-b_{\mathrm{2}}\right)\right)\left(e^{-i\left(\beta -\alpha \right)}\right)\left[\mathrm{1}\right]\\ +\left(d_{\mathrm{2}}+d_{\mathrm{3}}+i{d}_{\mathrm{4}}\right)\left(-d_{\mathrm{1}}+d_{\mathrm{2}}-i{d}_{\mathrm{3}}\right){\mathrm{\Lambda }}_{\mathrm{1}}\left(-\lambda b_{\mathrm{1}}\right)\\ {\mathrm{\Lambda }}_{\mathrm{2}}\left(\lambda \left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right){\mathrm{\Lambda }}_{\mathrm{3}}\left(\lambda \left(T-b_{\mathrm{2}}\right)\right)\left(e^{-i\left(\beta +\alpha \right)}\right)\left[\mathrm{1}\right]\\ +\left(d_{\mathrm{2}}+d_{\mathrm{3}}+i{d}_{\mathrm{4}}\right)\left(d_{\mathrm{1}}+d_{\mathrm{2}}+i{d}_{\mathrm{3}}\right){\mathrm{\Lambda }}_{\mathrm{1}}\left(\lambda b_{\mathrm{1}}\right)\\ {\mathrm{\Lambda }}_{\mathrm{2}}\left(\lambda \left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right){\mathrm{\Lambda }}_{\mathrm{3}}\left(\lambda \left(T-b_{\mathrm{2}}\right)\right)\left(e^{-i\left(\beta -\alpha \right)}\right)\left[\mathrm{1}\right]\end{array}(31)

接下来运用Cramer法则能得到系数{\vartheta }_k\left(\lambda \right),{\sigma }_k\left(\lambda \right)的渐进式，代入式(30)与式(11)、(12)和(31)可得以上引理3的估计式(24)~(26)，由上文可知式(11)、(12)不依赖于Q\left(x\right)\mathrm{和}\beta，且{\vartheta }_k\left(\lambda \right)和{\sigma }_k\left(\lambda \right)不依赖于x可得估计式(27)~(29)。

引理4 （Liouville定理）有界整函数必为常数。

引理5 （Phragmen-Lindelöfs定理）设f\left(z\right)\le \rho，并且f\left(z\right)是一个有限阶的整函数，其中\alpha \le \frac{\mathrm{1}}{\rho }，令B为角\pi \alpha在z平面上的扇形，如果在B的边界上f\left(z\right)\le C，则在B内部也有f\left(z\right)\le C，且常数C相同。

3 结论及证明

定理1 当M\left(\lambda \right)=\tilde{M}\left(\lambda \right)，则有Q\left(x\right)=\tilde{Q}\left(x\right)且\beta =\tilde{\beta }，其中x\in \left.\left[\mathrm{0},b_{\mathrm{1}}\right.\right)\bigcup \left(b_{\mathrm{1}},b_{\mathrm{2}}\right)\bigcup \left.\left(b_{\mathrm{2}},T\right.\right]。

证明 定义谱映射矩阵P\left(x,\lambda \right)={\left[P_{ij\left(x,\lambda \right)}\right]}_{i.j=\mathrm{1},\mathrm{2}}，有以下表示：

P\left(x,\lambda \right)\tilde{\Psi }\left(x,\lambda \right)=\Psi \left(x,\lambda \right)(32)

由式(30)计算可得：

\left\{\begin{array}{c}\begin{array}{l}{P}_{\mathrm{11}}\left(x,\lambda \right)={\left(\mathrm{d}\mathrm{e}\mathrm{t}\tilde{\Psi }\left(x,\lambda \right)\right)}^{-\mathrm{1}}\left({\phi }_{\mathrm{1}}\left(x,\lambda \right){\tilde{\varphi }}_{\mathrm{2}}\left(x,\lambda \right)\right.\\ -\left.{\tilde{\phi }}_{\mathrm{2}}\left(x,\lambda \right){\varphi }_{\mathrm{1}}\left(x,\lambda \right)\right)\end{array}\\ \begin{array}{l}{P}_{\mathrm{12}}\left(x,\lambda \right)={\left(\mathrm{d}\mathrm{e}\mathrm{t}\tilde{\Psi }\left(x,\lambda \right)\right)}^{-\mathrm{1}}\left(-{\phi }_{\mathrm{1}}\left(x,\lambda \right){\tilde{\varphi }}_{\mathrm{1}}\left(x,\lambda \right)\right.\\ \left.+{\tilde{\phi }}_{\mathrm{1}}\left(x,\lambda \right){\varphi }_{\mathrm{1}}\left(x,\lambda \right)\right)\end{array}\end{array}\right.(33)

且以下方程成立：

\begin{array}{c}\left({\phi }_{\mathrm{1}}\left(x,\lambda \right)-{\tilde{\phi }}_{\mathrm{1}}\left(x,\lambda \right)\right){\tilde{\varphi }}_{\mathrm{2}}\left(x,\lambda \right)\\ -\left({\varphi }_{\mathrm{1}}\left(x,\lambda \right)-{\tilde{\varphi }}_{\mathrm{1}}\left(x,\lambda \right)\right){\tilde{\phi }}_{\mathrm{2}}\left(x,\lambda \right)\\ =\left({\phi }_{\mathrm{1}}\left(x,\lambda \right){\tilde{\varphi }}_{\mathrm{2}}\left(x,\lambda \right)-{\tilde{\phi }}_{\mathrm{2}}\left(x,\lambda \right){\varphi }_{\mathrm{1}}\left(x,\lambda \right)\right)\\ -\left({\tilde{\phi }}_{\mathrm{1}}\left(x,\lambda \right){\tilde{\varphi }}_{\mathrm{2}}\left(x,\lambda \right)-{\tilde{\phi }}_{\mathrm{2}}\left(x,\lambda \right){\tilde{\varphi }}_{\mathrm{1}}\left(x,\lambda \right)\right)\\ =P_{\mathrm{11}}\left(x,\lambda \right)\mathrm{d}\mathrm{e}\mathrm{t}\tilde{\Psi }\left(x,\lambda \right)-\mathrm{d}\mathrm{e}\mathrm{t}\tilde{\Psi }\left(x,\lambda \right)\end{array}

=\mathrm{d}\mathrm{e}\mathrm{t}\tilde{\Psi }\left(x,\lambda \right)\left(P_{\mathrm{11}}\left(x,\lambda \right)-\mathrm{1}\right)(34)

因此，

\begin{array}{c}{P}_{\mathrm{11}}\left(x,\lambda \right)-\mathrm{1}={\left(\mathrm{d}\mathrm{e}\mathrm{t}\tilde{\mathrm{\Psi }}\left(x,\lambda \right)\right)}^{-\mathrm{1}}\\ \left(\left({\phi }_{\mathrm{1}}\left(x,\lambda \right)-{\tilde{\phi }}_{\mathrm{1}}\left(x,\lambda \right)\right){\tilde{\varphi }}_{\mathrm{2}}\left(x,\lambda \right)\right.\\ -\left.\left({\varphi }_{\mathrm{1}}\left(x,\lambda \right)-{\tilde{\varphi }}_{\mathrm{1}}\left(x,\lambda \right)\right){\tilde{\phi }}_{\mathrm{2}}\left(x,\lambda \right)\right)\end{array}(35)

同理，则有：

\begin{array}{c}{P}_{\mathrm{12}}\left(x,\lambda \right)={\left(\mathrm{d}\mathrm{e}\mathrm{t}\tilde{\Psi }\left(x,\lambda \right)\right)}^{-\mathrm{1}}\\ \left(-\left({\phi }_{\mathrm{1}}\left(x,\lambda \right)-{\tilde{\phi }}_{\mathrm{1}}\left(x,\lambda \right)\right){\tilde{\varphi }}_{\mathrm{1}}\left(x,\lambda \right)\right.\\ +\left.\left({\varphi }_{\mathrm{1}}\left(x,\lambda \right)-{\tilde{\varphi }}_{\mathrm{1}}\left(x,\lambda \right)\right){\tilde{\phi }}_{\mathrm{1}}\left(x,\lambda \right)\right)\end{array}(36)

令L和\tilde{L}满足定理条件，即M\left(\lambda \right)=\tilde{M}\left(\lambda \right)。

根据式(7)和式(33)可得：

\begin{array}{c}{P}_{\mathrm{11}}\left(x,\lambda \right)=\\ {\left(\mathrm{d}\mathrm{e}\mathrm{t}\tilde{\Psi }\left(x,\lambda \right)\right)}^{-\mathrm{1}}\left({\phi }_{\mathrm{1}}\left(x,\lambda \right){\tilde{S}}_{\mathrm{2}}\left(x,\lambda \right)-{\tilde{\phi }}_{\mathrm{2}}\left(x,\lambda \right)S_{\mathrm{1}}\left(x,\lambda \right)\right.\\ +\left.\left(\tilde{M}\left(\lambda \right)-M\left(\lambda \right)\right){\phi }_{\mathrm{1}}\left(x,\lambda \right){\tilde{\phi }}_{\mathrm{2}}\left(x,\lambda \right)\right),\end{array}\begin{array}{c}{P}_{\mathrm{12}}\left(x,\lambda \right)=\\ {\left(\mathrm{d}\mathrm{e}\mathrm{t}\tilde{\Psi }\left(x,\lambda \right)\right)}^{-\mathrm{1}}\left(-{\phi }_{\mathrm{1}}\left(x,\lambda \right){\tilde{S}}_{\mathrm{1}}\left(x,\lambda \right)+{\tilde{\phi }}_{\mathrm{1}}\left(x,\lambda \right)S_{\mathrm{1}}\left(x,\lambda \right)\right.\\ +\left.\left(M\left(\lambda \right)-\tilde{M}\left(\lambda \right)\right){\phi }_{\mathrm{1}}\left(x,\lambda \right){\tilde{\phi }}_{\mathrm{1}}\left(x,\lambda \right)\right)\mathrm{。}\end{array}

由已知x\in \left[\mathrm{0},T\right]，且P_{\mathrm{11}}\left(x,\lambda \right),P_{\mathrm{12}}\left(x,\lambda \right)是关于\lambda的整函数，当\lambda \in S_{\varsigma },\left|\lambda \right|>{\lambda }^{\mathrm{*}}时，通过引理2和引理3的估计式和表达式(35)，(36)，则下列估计成立：

\left|P_{\mathrm{11}}\left(x,\lambda \right)-\mathrm{1}\right|\le \frac{C}{\left|\lambda \right|},\left|P_{\mathrm{12}}\left(x,\lambda \right)\right|\le \frac{C}{\left|\lambda \right|}。

同理

\left|P_{\mathrm{22}}\left(x,\lambda \right)-\mathrm{1}\right|\le \frac{C}{\left|\lambda \right|},\left|P_{\mathrm{21}}\left(x,\lambda \right)\right|\le \frac{C}{\left|\lambda \right|}，

又根据引理4和引理5可得：

P_{\mathrm{11}}\left(x,\lambda \right)\equiv P_{\mathrm{22}}\left(x,\lambda \right)\equiv \mathrm{1},P_{\mathrm{12}}\left(x,\lambda \right)\equiv P_{\mathrm{21}}\left(x,\lambda \right)\equiv \mathrm{0}。

把式(32)代入计算可得：

{\phi }_k\left(x,\lambda \right)={\tilde{\phi }}_k\left(x,\lambda \right),{\varphi }_k\left(x,\lambda \right)={\tilde{\varphi }}_k\left(x,\lambda \right),k=\mathrm{1},\mathrm{2}(37)

接下来证明\beta =\tilde{\beta }。由式(2)可知：

{\phi }_{\mathrm{1}}\left(T,\lambda \right)\mathrm{c}\mathrm{o}\mathrm{s}\beta +{\phi }_{\mathrm{2}}\left(T,\lambda \right)\mathrm{s}\mathrm{i}\mathrm{n}\beta =\mathrm{0}，

{\phi }_{\mathrm{1}}\left(T,\lambda \right)\mathrm{c}\mathrm{o}\mathrm{s}\tilde{\beta }+{\phi }_{\mathrm{2}}\left(T,\lambda \right)\mathrm{s}\mathrm{i}\mathrm{n}\tilde{\beta }=\mathrm{0}，

且\phi \left(T,\lambda \right)是一个非零解，那么有\left|\begin{array}{cc}\mathrm{c}\mathrm{o}\mathrm{s}\beta & \mathrm{s}\mathrm{i}\mathrm{n}\beta \\ \mathrm{c}\mathrm{o}\mathrm{s}\tilde{\beta }& \mathrm{s}\mathrm{i}\mathrm{n}\tilde{\beta }\end{array}\right|=\mathrm{s}\mathrm{i}\mathrm{n}\left(\tilde{\beta }-\beta \right)=\mathrm{0}，则有\beta =\tilde{\beta }。

最后证明在式(1)中，Q\left(x\right)=\tilde{Q}\left(x\right)，其中x\in \left.\left[\mathrm{0},b_{\mathrm{1}}\right.\right)\bigcup \left(b_{\mathrm{1}},b_{\mathrm{2}}\right)\bigcup \left.\left(b_{\mathrm{2}},T\right.\right]，有

B{\phi }^{\text{'}}\left(x,\lambda \right)+Q\left(x\right)\phi \left(x,\lambda \right)=\lambda r\left(x\right)\phi \left(x,\lambda \right)(38)

B{\phi }^{\text{'}}\left(x,\lambda \right)+\tilde{Q}\left(x\right)\phi \left(x,\lambda \right)=\lambda r\left(x\right)\phi \left(x,\lambda \right)(39)

式(38)减去式(39)得\left(Q\left(x\right)-\tilde{Q}\left(x\right)\right)\phi \left(x,\lambda \right)=\mathrm{0}，同样对于\varphi \left(x,\lambda \right)，\left(Q\left(x\right)-\tilde{Q}\left(x\right)\right)\varphi \left(x,\lambda \right)=\mathrm{0}，也成立，则有：

\begin{array}{c}\left(Q\left(x\right)-\tilde{Q}\left(x\right)\right)\left(\begin{array}{cc}{\phi }_{\mathrm{1}}\left(x,\lambda \right)& {\varphi }_{\mathrm{1}}\left(x,\lambda \right)\\ {\phi }_{\mathrm{2}}\left(x,\lambda \right)& {\varphi }_{\mathrm{2}}\left(x,\lambda \right)\end{array}\right)\\ =\left(Q\left(x\right)-\tilde{Q}\left(x\right)\right)\Psi \left(x,\lambda \right)=\mathrm{0}\mathrm{。}\end{array}

由式(10)可知\Psi \left(x,\lambda \right)可逆，则\Psi \left(x,\lambda \right)不为零，则必有Q\left(x\right)=\tilde{Q}\left(x\right)，证毕。

定理2　如果全部n\in \mathbb{Z}且\beta =\tilde{\beta }，有{\lambda }_n={\tilde{\lambda }}_n，{\lambda }_{n\mathrm{0}}={\tilde{\lambda }}_{n\mathrm{0}}，则Q\left(x\right)=\tilde{Q}\left(x\right)，其中x\in \left.\left[\mathrm{0},b_{\mathrm{1}}\right.\right)\bigcup \left(b_{\mathrm{1}},b_{\mathrm{2}}\right)\bigcup\left.\left(b_{\mathrm{2}},T\right.\right]。

证明 由上文可知\mathrm{\Delta }\left(\lambda \right)=V\left(\phi \right)，在S_{\varsigma }中，\left|\lambda \right|\to\mathrm{\infty }时，由引理1和引理2的式(19)、(22)、(23)可得下列渐进公式：

\begin{array}{l}\mathrm{8}\mathrm{\Delta }\left(\lambda \right)=\left(d_{\mathrm{2}}+d_{\mathrm{3}}-i{d}_{\mathrm{4}}\right)\left(d_{\mathrm{1}}+d_{\mathrm{2}}-i{d}_{\mathrm{3}}\right){\mathrm{\Lambda }}_{\mathrm{1}}\left(-\lambda b_{\mathrm{1}}\right)\\ {\mathrm{\Lambda }}_{\mathrm{2}}\left(-\lambda \left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right){\mathrm{\Lambda }}_{\mathrm{3}}\left(-\lambda \left(T-b_{\mathrm{2}}\right)\right)\left(i{e}^{i\left(\beta -\alpha \right)}\right)\left[\mathrm{1}\right]\\ +\left(d_{\mathrm{2}}+d_{\mathrm{3}}-i{d}_{\mathrm{4}}\right)\left(-d_{\mathrm{1}}+d_{\mathrm{2}}+i{d}_{\mathrm{3}}\right){\mathrm{\Lambda }}_{\mathrm{1}}\left(\lambda b_{\mathrm{1}}\right)\\ {\mathrm{\Lambda }}_{\mathrm{2}}\left(-\lambda \left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right){\mathrm{\Lambda }}_{\mathrm{3}}\left(-\lambda \left(T-b_{\mathrm{2}}\right)\right)\left(i{e}^{i\left(\beta +\alpha \right)}\right)\left[\mathrm{1}\right]\\ +\left(-d_{\mathrm{2}}+d_{\mathrm{3}}+i{d}_{\mathrm{4}}\right)\left(-d_{\mathrm{1}}+d_{\mathrm{2}}-i{d}_{\mathrm{3}}\right){\mathrm{\Lambda }}_{\mathrm{1}}\left(-\lambda b_{\mathrm{1}}\right)\\ {\mathrm{\Lambda }}_{\mathrm{2}}\left(\lambda \left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right){\mathrm{\Lambda }}_{\mathrm{3}}\left(-\lambda \left(T-b_{\mathrm{2}}\right)\right)\left(i{e}^{i\left(\beta -\alpha \right)}\right)\left[\mathrm{1}\right]\\ +\left(-d_{\mathrm{2}}+d_{\mathrm{3}}+i{d}_{\mathrm{4}}\right)\left(d_{\mathrm{1}}+d_{\mathrm{2}}+i{d}_{\mathrm{3}}\right){\mathrm{\Lambda }}_{\mathrm{1}}\left(\lambda b_{\mathrm{1}}\right)\\ {\mathrm{\Lambda }}_{\mathrm{2}}\left(\lambda \left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right){\mathrm{\Lambda }}_{\mathrm{3}}\left(-\lambda \left(T-b_{\mathrm{2}}\right)\right)\left(i{e}^{i\left(\beta +\alpha \right)}\right)\left[\mathrm{1}\right]\\ +\left(-d_{\mathrm{2}}+d_{\mathrm{3}}-i{d}_{\mathrm{4}}\right)\left(d_{\mathrm{1}}+d_{\mathrm{2}}-i{d}_{\mathrm{3}}\right){\mathrm{\Lambda }}_{\mathrm{1}}\left(-\lambda b_{\mathrm{1}}\right)\\ {\mathrm{\Lambda }}_{\mathrm{2}}\left(-\lambda \left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right){\mathrm{\Lambda }}_{\mathrm{3}}\left(\lambda \left(T-b_{\mathrm{2}}\right)\right)\left(-i{e}^{-i\left(\beta +\alpha \right)}\right)\left[\mathrm{1}\right]\\ +\left(-d_{\mathrm{2}}+d_{\mathrm{3}}-i{d}_{\mathrm{4}}\right)\left(-d_{\mathrm{1}}+d_{\mathrm{2}}+i{d}_{\mathrm{3}}\right){\mathrm{\Lambda }}_{\mathrm{1}}\left(\lambda b_{\mathrm{1}}\right)\\ {\mathrm{\Lambda }}_{\mathrm{2}}\left(-\lambda \left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right){\mathrm{\Lambda }}_{\mathrm{3}}\left(\lambda \left(T-b_{\mathrm{2}}\right)\right)\left(-i{e}^{-i\left(\beta -\alpha \right)}\right)\left[\mathrm{1}\right]\\ +\left(d_{\mathrm{2}}+d_{\mathrm{3}}+i{d}_{\mathrm{4}}\right)\left(-d_{\mathrm{1}}+d_{\mathrm{2}}-i{d}_{\mathrm{3}}\right){\mathrm{\Lambda }}_{\mathrm{1}}\left(-\lambda b_{\mathrm{1}}\right)\\ {\mathrm{\Lambda }}_{\mathrm{2}}\left(\lambda \left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right){\mathrm{\Lambda }}_{\mathrm{3}}\left(\lambda \left(T-b_{\mathrm{2}}\right)\right)\left(-i{e}^{-i\left(\beta +\alpha \right)}\right)\left[\mathrm{1}\right]\\ +\left(d_{\mathrm{2}}+d_{\mathrm{3}}+i{d}_{\mathrm{4}}\right)\left(d_{\mathrm{1}}+d_{\mathrm{2}}+i{d}_{\mathrm{3}}\right){\mathrm{\Lambda }}_{\mathrm{1}}\left(\lambda b_{\mathrm{1}}\right)\\ {\mathrm{\Lambda }}_{\mathrm{2}}\left(\lambda \left(b_{\mathrm{2}}-b_{\mathrm{1}}\right)\right){\mathrm{\Lambda }}_{\mathrm{3}}\left(\lambda \left(T-b_{\mathrm{2}}\right)\right)\left(-i{e}^{-i\left(\beta -\alpha \right)}\right)\left[\mathrm{1}\right]\end{array}(40)

当n\in \mathbb{Z}，\beta =\tilde{\beta }时,有{\lambda }_n={\tilde{\lambda }}_n,{\lambda }_{n\mathrm{0}}={\tilde{\lambda }}_{n\mathrm{0}}。根据Hadamard Factorization定理和式(40)，可以得到\mathrm{\Delta }\left(\lambda \right)=\tilde{\mathrm{\Delta }}\left(\lambda \right)，同样能得到{\mathrm{\Delta }}_{\mathrm{0}}\left(\lambda \right)={\tilde{\mathrm{\Delta }}}_{\mathrm{0}}\left(\lambda \right)。那么，由式(9)可知，M\left(\lambda \right)=\tilde{M}\left(\lambda \right)，即满足定理1则可推出定理2成立，证毕。

参考文献

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基金资助

内蒙古自治区自然科学基金项目（2017MS(LH)0103

2021MS01020