一类边界条件中含谱参数的Dirac算子的谱性质

Spectral Properties of a Class of Dirac Operators with Eigenparameter in the Boundary Conditions

摘要: 本文考虑了一类内部具有两个不连续点且边界条件依赖谱参数的Dirac算子的谱性质。首先通过引入适当的Hilbert空间并在其上定义新的自伴算子，使得所考虑问题的特征值与该算子的特征值一致。然后通过构造基本解得到了特征值的一些性质。最后给出了问题的Green函数和预解算子。

Abstract: In this paper, we consider the spectral properties of a class of Dirac operators with two internal discontinuities and spectral parameter-dependent boundary conditions. First, the eigenvalues of the problem under consideration are made to coincide with the eigenvalues of the operator by introducing a suitable Hilbert space and defining a new self-adjoint operator on it. Then some properties of the eigenvalues are obtained by constructing the basic solution. Finally, Green’s function and the resolvent operator of the problem are given.

1. 引言

Dirac算子又称为AKNS算子，在描述量子力学中的能量和原子的内力问题中至关重要。Dirac算子的谱性质是算子谱理论的重要组成部分，诸多学者对此进行了一系列研究，并取得了一些优秀研究成果[1]-[4]。近些年，越来越多的学者对内部具有不连续点的Dirac算子的谱性质进行了研究[5]-[11]。如：2013年，Tharwat M. M.等人在文献[5]中研究了内部具有一个不连续点的Dirac系统，首先讨论了边值问题的特征值和特征函数的部分性质，然后得到了格林矩阵和扩张定理。2018年，袁兆年在其硕士论文[6]中考虑了内部具有一个不连续点且两个边界条件均含谱参数的Dirac算子的谱问题，证明了算子的自伴性，讨论了其特征值和特征函数的性质，给出了格林函数和预解算子。

本文拟在上述研究的基础上，将[6]中的一个不连续点推广为两个不连续点，考虑一类内部具有两个不连续点且边界条件依赖谱参数的Dirac算子的谱性质，即考虑如下由(1)~(7)构成的Dirac系统

$\tau y\left(x\right):=J{y}^{\prime }\left(x\right)+Q\left(x\right)y\left(x\right)=\lambda y\left(x\right),x\in I,$ (1)

其中

$J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right),Q\left(x\right)=\left(\begin{array}{cc}p\left(x\right)& q\left(x\right)\\ q\left(x\right)& -p\left(x\right)\end{array}\right),y\left(x\right)=\left(\begin{array}{c}{y}_1\left(x\right)\\ y_2\left(x\right)\end{array}\right),$

$p\left(x\right),q\left(x\right)$ 为定义在$I=\left[0,{\xi }_1\right)\cup \left({\xi }_1,{\xi }_2\right)\cup \left({\xi }_2,\pi \right]$ 上的实值连续函数，$\lambda$ 是谱参数，

满足边界条件

$L_{1}y:={\alpha }_1{y}_1\left(0\right)-{\alpha }_2{y}_2\left(0\right)-\lambda \left({{\alpha }^{\prime }}_1{y}_1\left(0\right)-{{\alpha }^{\prime }}_2{y}_2\left(0\right)\right)=0,$ (2)

$L_{2}y:={\beta }_1{y}_1\left(\pi \right)-{\beta }_2{y}_2\left(\pi \right)+\lambda \left({{\beta }^{\prime }}_1{y}_1\left(\pi \right)-{{\beta }^{\prime }}_2{y}_2\left(\pi \right)\right)=0,$ (3)

在间断点${\xi }_1,{\xi }_2$ 处满足转移条件

$L_{3}y:={\gamma }_1{y}_1\left({\xi }_1-0\right)+{\gamma }_2{y}_2\left({\xi }_1-0\right)+{\delta }_1{y}_1\left({\xi }_1+0\right)+{\delta }_2{y}_2\left({\xi }_1+0\right)=0,$ (4)

$L_{4}y:={{\gamma }^{\prime }}_1{y}_1\left({\xi }_1-0\right)+{{\gamma }^{\prime }}_2{y}_2\left({\xi }_1-0\right)+{{\delta }^{\prime }}_1{y}_1\left({\xi }_1+0\right)+{{\delta }^{\prime }}_2{y}_2\left({\xi }_1+0\right)=0,$ (5)

$L_{5}y:={\gamma }_3{y}_1\left({\xi }_2-0\right)+{\gamma }_4{y}_2\left({\xi }_2-0\right)+{\delta }_3{y}_1\left({\xi }_2+0\right)+{\delta }_4{y}_2\left({\xi }_2+0\right)=0,$ (6)

$L_{6}y:={{\gamma }^{\prime }}_3{y}_1\left({\xi }_2-0\right)+{{\gamma }^{\prime }}_4{y}_2\left({\xi }_2-0\right)+{{\delta }^{\prime }}_3{y}_1\left({\xi }_2+0\right)+{{\delta }^{\prime }}_4{y}_2\left({\xi }_2+0\right)=0,$ (7)

其中${\alpha }_i,{{\alpha }^{\prime }}_i,{\beta }_i,{{\beta }^{\prime }}_i,{\gamma }_j,{{\gamma }^{\prime }}_j,{\delta }_j,{{\delta }^{\prime }}_j\left(i=1,2,j=\overline{1,4}\right)$均为实数，且假设边界条件中的参数满足

${\rho }_1=\left|\begin{array}{ll}{\alpha }_1\hfill & {{\alpha }^{\prime }}_1\hfill \\ {\alpha }_2\hfill & {{\alpha }^{\prime }}_2\hfill \end{array}\right|>0,{\rho }_2=\left|\begin{array}{ll}{\beta }_1\hfill & {{\beta }^{\prime }}_1\hfill \\ {\beta }_2\hfill & {{\beta }^{\prime }}_2\hfill \end{array}\right|>0,$

转移条件中的参数满足

$d_1=\left|\begin{array}{ll}{\gamma }_1\hfill & {{\gamma }^{\prime }}_1\hfill \\ {\gamma }_2\hfill & {{\gamma }^{\prime }}_2\hfill \end{array}\right|>0,d_2=\left|\begin{array}{ll}{\delta }_1\hfill & {{\delta }^{\prime }}_1\hfill \\ {\delta }_2\hfill & {{\delta }^{\prime }}_2\hfill \end{array}\right|>0,d_3=\left|\begin{array}{ll}{\gamma }_3\hfill & {{\gamma }^{\prime }}_3\hfill \\ {\gamma }_4\hfill & {{\gamma }^{\prime }}_4\hfill \end{array}\right|>0,d_4=\left|\begin{array}{ll}{\delta }_3\hfill & {{\delta }^{\prime }}_3\hfill \\ {\delta }_4\hfill & {{\delta }^{\prime }}_4\hfill \end{array}\right|>0.$

若对某个$\lambda \in ℂ$ ，使得$y\left(x,\lambda \right)=\left(\begin{array}{c}{y}_1\left(x,\lambda \right)\\ y_2\left(x,\lambda \right)\end{array}\right)$ 是方程(1)的非平凡解，且满足边界条件(2)，(3)和转移条件(4)~(7)，则称$\lambda$ 是问题(1)~(7)的一个特征值，相应的非平凡解$y\left(x,\lambda \right)$ 称为对应于特征值$\lambda$ 的一个特征函数。

2. 算子公式

若

${\mathscr{H}}_1:=\left\{f\left(x\right)=\left(\begin{array}{c}{f}_1\left(x\right)\\ f_2\left(x\right)\end{array}\right):f_1\left(x\right),f_2\left(x\right)\in {ℒ}^2\left(I\right)\right\},$

对于任意的$f\left(x\right)={\left(f_1\left(x\right),f_2\left(x\right)\right)}^{\text{T}}$ ，$g\left(x\right)={\left(g_1\left(x\right),g_2\left(x\right)\right)}^{\text{T}}\in {\mathscr{H}}_1$ ，在${\mathscr{H}}_1$ 内定义内积为

${\langle f\left(x\right),g\left(x\right)\rangle }_1={\displaystyle {\int }_0^{{\xi }_1}{f}^{\text{T}}\left(x\right)\overline{g}\left(x\right)\text{d}x}+\frac{d_2}{d_1}{\displaystyle {\int }_{{\xi }_1}^{{\xi }_2}{f}^{\text{T}}}\left(x\right)\overline{g}\left(x\right)\text{d}x+\frac{d_2{d}_4}{d_1{d}_3}{\displaystyle {\int }_{{\xi }_{{}_2}}^{\pi }{f}^{\text{T}}}\left(x\right)\overline{g}\left(x\right)\text{d}x,$

易知${\mathscr{H}}_1$ 是完备的。

定义

$\mathscr{H}={\mathscr{H}}_1\oplus ℂ\oplus ℂ,$

对于任意的$F={\left(f\left(x\right),\widetilde{f_1},\widetilde{f_2}\right)}^{\text{T}}$，$G={\left(g\left(x\right),\widetilde{g_1},\widetilde{g_2}\right)}^{\text{T}}\in \mathscr{H}$，在$\mathscr{H}$ 内定义内积为

$\langle F,G\rangle ={\langle f\left(x\right),g\left(x\right)\rangle }_1+\frac{1}{{\rho }_1}\widetilde{f_1}\overline{\widetilde{g_1}}+\frac{d_2{d}_4}{d_1{d}_3}\frac{1}{{\rho }_2}\widetilde{f_2}\overline{\widetilde{{\text{g}}_2}},$

可知$\mathscr{H}$ 是一个完备的内积空间。

为了方便描述，我们引入以下符号

$\begin{array}{l}{M}_0\left(y\right)={\alpha }_1{y}_1\left(0\right)-{\alpha }_2{y}_2\left(0\right),{M^{\prime }}_0\left(y\right)={{\alpha }^{\prime }}_1{y}_1\left(0\right)-{{\alpha }^{\prime }}_2{y}_2\left(0\right),\\ M_{\pi }\left(y\right)={\beta }_1{y}_1\left(\pi \right)-{\beta }_2{y}_2\left(\pi \right),{M^{\prime }}_{\pi }\left(y\right)={{\beta }^{\prime }}_1{y}_1\left(\pi \right)-{{\beta }^{\prime }}_2{y}_2\left(\pi \right),\end{array}$

由边界条件可知：$M_0\left(y\right)=\lambda {M^{\prime }}_0\left(y\right)$ ，$M_{\pi }\left(y\right)=-\lambda {M^{\prime }}_{\pi }\left(y\right)$ 。

在Hilbert空间$\mathscr{H}$ 中定义算子$\mathcal{T}$ ，其定义域为

$\mathcal{D}\left(\mathcal{T}\right)=\left\{F=\left(\begin{array}{c}f\left(x\right)\\ \widetilde{f_1}\\ \widetilde{f_2}\end{array}\right)\in \mathscr{H}|\begin{array}{c}{f}_1\left(x\right),f_2\left(x\right)\in \mathcal{A}\mathcal{C}\left(\left[0,{\xi }_1\right)\cup \left({\xi }_1,{\xi }_2\right)\cup \left({\xi }_2,\pi \right]\right),\\ f_{\text{1}}\left({\xi }_{\text{1}}\pm \text{0}\right),f_{\text{2}}\left({\xi }_{\text{1}}\pm \text{0}\right),f_{\text{1}}\left({\xi }_{\text{2}}\pm \text{0}\right),f_{\text{2}}\left({\xi }_{\text{2}}\pm \text{0}\right)存在,\\ 并且满足\tau f\in {\mathscr{H}}_1,L_{3}f=L_{4}f=L_{5}f=L_{6}f=0,\widetilde{f_1}={M^{\prime }}_0\left(f\right),\widetilde{f_2}={M^{\prime }}_{\pi }\left(f\right)\end{array}\right\},$

并满足

$\mathcal{T}\left(\begin{array}{c}f\\ \widetilde{f_1}\\ \widetilde{f_2}\end{array}\right)=\left(\begin{array}{c}\tau f\\ M_0\left(f\right)\\ -M_{\pi }\left(f\right)\end{array}\right).$

因此，我们所考虑的问题(1)~(7)就可改写为如下算子形式$\mathcal{T}F=\lambda F$ 。

此时，问题(1)~(7)的特征值与算子$\mathcal{T}$ 的特征值一致，且特征函数为算子$\mathcal{T}$ 相应特征函数的第一个分量。

定理2.1 算子$\mathcal{T}$ 的定义域$\mathcal{D}\left(\mathcal{T}\right)$ 在$\mathscr{H}$ 中稠密，即$\overline{\mathcal{D}\left(\mathcal{T}\right)}=\mathscr{H}$ 。

证明 设$F={\left(f\left(x\right),\widetilde{f_1},\widetilde{f_2}\right)}^{\text{T}}\in \mathscr{H}$ (其中$f\left(x\right)={\left(f_1\left(x\right),f_2\left(x\right)\right)}^{\text{T}}$ )且$F\perp \mathcal{D}\left(\mathcal{T}\right)$ ，则只需要证明 $F={\left(0,0,0\right)}^{\text{T}}$ 。

令$C_0^{\infty }$ 表示如下函数集合

$\phi \left(x\right)=\left(\begin{array}{l}{\phi }_1\left(x\right)\hfill \\ {\phi }_2\left(x\right)\hfill \end{array}\right),$

其中${\phi }_1\left(x\right)\subset C_0^{\infty }\left(I\right)$ ，${\phi }_2\left(x\right)\subset C_0^{\infty }\left(I\right)$ 。因为$C_0^{\infty }\left(I\right)$ 在${ℒ}^2$ 中稠密，即$\overline{C_0^{\infty }}={\mathscr{H}}_1$ ，于是有$C_0^{\infty }\oplus 0\oplus 0\subset \mathcal{D}\left(\mathcal{T}\right)$ ，故对于任意的$G={\left(g\left(x\right),\widetilde{g_1},\widetilde{g_2}\right)}^{\text{T}}\in C_0^{\infty }\oplus 0\oplus 0$ (其中$g\left(x\right)={\left(g_1\left(x\right),g_2\left(x\right)\right)}^{\text{T}}$ )，根据内积的定义有

$\langle F,G\rangle ={\displaystyle {\int }_0^{{\xi }_1}{f}^T}\left(x\right)\overline{g}\left(x\right)\text{d}x+\frac{d_2}{d_1}{\displaystyle {\int }_{{\xi }_1}^{{\xi }_2}{f}^T}\left(x\right)\overline{g}\left(x\right)\text{d}x+\frac{d_2{d}_4}{d_1{d}_3}{\displaystyle {\int }_{{\xi }_2}^{\pi }{f}^T}\left(x\right)\overline{g}\left(x\right)\text{d}x+0+0=0,$

又因为$\overline{C_0^{\infty }}={\mathscr{H}}_1$ ，所以$f\left(x\right)=0$ ，从而$F={\left(0,\widetilde{f_1},\widetilde{f_2}\right)}^T$ 。因而对于任意的$V={\left(v\left(x\right),\widetilde{v_1},\widetilde{v_2}\right)}^{\text{T}}\in \mathcal{D}\left(\mathcal{T}\right)$，根据内积的定义有

$\langle F,V\rangle =\frac{1}{{\rho }_1}\widetilde{f_1}\overline{\widetilde{v_1}}+\frac{d_2{d}_4}{d_1{d}_3}\frac{1}{{\rho }_2}\widetilde{f_2}\overline{\widetilde{v_2}}=0,$

又因为${\rho }_1,{\rho }_2,d_1,d_2,d_3,d_4$ 均大于零且$\widetilde{v_1},\widetilde{v_2}$是任取的，所以$\widetilde{f_1}=0,\widetilde{f_2}=0$。因此$F={\left(0,0,0\right)}^{\text{T}}$ ，故定理得证。

定理2.2 算子$\mathcal{T}$ 是自伴的。

证明 令$F={\left(f\left(x\right),\widetilde{f_1},\widetilde{f_2}\right)}^{\text{T}}$，$G={\left(g\left(x\right),\widetilde{g_1},\widetilde{g_2}\right)}^{\text{T}}\in \mathcal{D}\left(\mathcal{T}\right)$，定义$f$ 和$g$ 的朗斯基行列式为

$W\left(f,g;x\right)=f_1\left(x\right)g_2\left(x\right)-f_2\left(x\right)g_1\left(x\right),$

根据内积和算子$\mathcal{T}$ 的定义，我们有

$\begin{array}{c}\langle \mathcal{T}F,G\rangle -\langle F,\mathcal{T}G\rangle =-\left[W\left(f,\overline{g};{\xi }_1-0\right)-W\left(f,\overline{g};0\right)\right]-\frac{d_2}{d_1}\left[W\left(f,\overline{g};{\xi }_2-0\right)-W\left(f,\overline{g};{\xi }_1+0\right)\right]\\ -\frac{d_2{d}_4}{d_1{d}_3}\left[W\left(f,\overline{g};\pi \right)-W\left(f,\overline{g};{\xi }_2+0\right)\right]+\frac{1}{{\rho }_1}\left[M_0\left(f\right){M^{\prime }}_0\left(\overline{g}\right)-{M^{\prime }}_0\left(f\right)M_0\left(\overline{g}\right)\right]\\ +\frac{d_2{d}_4}{d_1{d}_3}\frac{1}{{\rho }_2}\left[-M_{\pi }\left(f\right){M^{\prime }}_{\pi }\left(\overline{g}\right)+{M^{\prime }}_{\pi }\left(f\right)M_{\pi }\left(\overline{g}\right)\right],\end{array}$(8)

由转移条件(4)~(7)，可得

$W\left(f,\overline{g};{\xi }_1-0\right)=\frac{d_2}{d_1}W\left(f,\overline{g};{\xi }_1+0\right),$ (9)

$W\left(f,\overline{g};{\xi }_2-0\right)=\frac{d_4}{d_3}W\left(f,\overline{g};{\xi }_2+0\right),$ (10)

并且容易得到

$M_0\left(f\right){M^{\prime }}_0\left(\overline{g}\right)-{M^{\prime }}_0\left(f\right)M_0\left(\overline{g}\right)=-{\rho }_{1}W\left(f,\overline{g};0\right),$ (11)

$-M_{\pi }\left(f\right){M^{\prime }}_{\pi }\left(\overline{g}\right)+{M^{\prime }}_{\pi }\left(f\right)M_{\pi }\left(\overline{g}\right)={\rho }_{2}W\left(f,\overline{g};\pi \right).$ (12)

将(9)~(12)代入(8)即可得$\langle \mathcal{T}F,G\rangle =\langle F,\mathcal{T}G\rangle$ ，故算子$\mathcal{T}$ 是对称的。

下面只需证明：对于任意的$F={\left(f\left(x\right),{M^{\prime }}_0\left(f\right),{M^{\prime }}_{\pi }\left(f\right)\right)}^{\text{T}}\in \mathcal{D}\left(\mathcal{T}\right)$ ，若$\langle \mathcal{T}F,W\rangle =\langle F,U\rangle$ 成立，则$W\in D\left(\mathcal{T}\right)$ 且$\mathcal{T}W=U$ ，其中$W={\left(w\left(x\right),\widetilde{w_1},\widetilde{w_2}\right)}^{\text{T}}$，$U={\left(u\left(x\right),\widetilde{u_1},\widetilde{u_2}\right)}^{\text{T}}$，$w\left(x\right)={\left(w_1\left(x\right),w_2\left(x\right)\right)}^{\text{T}}$ ， $u\left(x\right)={\left(u_1\left(x\right),u_2\left(x\right)\right)}^{\text{T}}$ 。

要证如上结论成立，即要证如下七条成立：(i) $w_1\left(x\right),w_2\left(x\right)\in \mathcal{A}\mathcal{C}\left(\left[0,{\xi }_1\right)\cup \left({\xi }_1,{\xi }_2\right)\cup \left({\xi }_2,\pi \right]\right)$ ，且$\tau w\in {\mathscr{H}}_1$ ；(ii) $\widetilde{w_1}={{\alpha }^{\prime }}_1{w}_1\left(0\right)-{{\alpha }^{\prime }}_2{w}_2\left(0\right)$ ；(iii) $\widetilde{w_2}={{\beta }^{\prime }}_1{w}_1\left(\pi \right)-{{\beta }^{\prime }}_2{w}_2\left(\pi \right)$ ；(iv) $L_{3}w=L_{4}w=L_{5}w=L_{6}w=0$ ；(v) $u\left(x\right)=\tau w\left(x\right)$ ；(vi) $\widetilde{u_1}={\alpha }_1{w}_1\left(0\right)-{\alpha }_2{w}_2\left(0\right)$ ；(vii) $\widetilde{u_2}=-\left({\beta }_1{w}_1\left(\pi \right)-{\beta }_2{w}_2\left(\pi \right)\right)$ 。

对于任意的$F={\left(f\left(x\right),0,0\right)}^{\text{T}}\in {\mathcal{C}}_0^{\infty }\oplus \left\{0\right\}\oplus \left\{0\right\}\subset \mathcal{D}\left(\mathcal{T}\right)$ ，有$\langle \mathcal{T}F,W\rangle =\langle F,U\rangle$ ，根据内积的定义，可得${\langle \tau f\left(x\right),w\left(x\right)\rangle }_1={\langle f\left(x\right),u\left(x\right)\rangle }_1$ ，显然(i)成立。

因为$\mathcal{T}$ 是对称算子，所以有${\langle \tau f\left(x\right),w\left(x\right)\rangle }_1={\langle f\left(x\right),\tau w\left(x\right)\rangle }_1$ ，由(i)的证明过程又可知 ${\langle \tau f\left(x\right),w\left(x\right)\rangle }_1={\langle f\left(x\right),u\left(x\right)\rangle }_1$ ，故$\tau w\left(x\right)=u\left(x\right)$ ，即(v)成立。

通过(v)可知，对于任意的$F\in \mathcal{D}\left(\mathcal{T}\right)$ ，$\langle \mathcal{T}F,W\rangle =\langle F,U\rangle$ 可表示为

$\begin{array}{l}{\langle \tau f\left(x\right),w\left(x\right)\rangle }_1-{\langle f\left(x\right),\tau w\left(x\right)\rangle }_1\\ =\frac{1}{{\rho }_1}\left[{M^{\prime }}_0\left(f\right)\overline{\widetilde{u_1}}-M_0\left(f\right)\overline{\widetilde{w_1}}\right]+\frac{d_2{d}_4}{d_1{d}_3}\frac{1}{{\rho }_2}\left[{M^{\prime }}_{\pi }\left(f\right)\overline{\widetilde{u_2}}+M_{\pi }\left(f\right)\overline{\widetilde{w_2}}\right],\end{array}$ (13)

又由于

$\begin{array}{l}{\langle \tau f\left(x\right),w\left(x\right)\rangle }_1-{\langle f\left(x\right),\tau w\left(x\right)\rangle }_1\\ =-\left[W\left(f,\overline{w};{\xi }_1-0\right)-W\left(f,\overline{w};0\right)\right]-\frac{d_2}{d_1}\left[W\left(f,\overline{w};{\xi }_2-0\right)-W\left(f,\overline{w};{\xi }_1+0\right)\right]\\ -\frac{d_2{d}_4}{d_1{d}_3}\left[W\left(f,\overline{w};\pi \right)-W\left(f,\overline{w};{\xi }_2+0\right)\right],\end{array}$ (14)

故结合(13)和(14)可得

$\begin{array}{l}\frac{1}{{\rho }_1}\left[{M^{\prime }}_0\left(f\right)\overline{\widetilde{u_1}}-M_0\left(f\right)\overline{\widetilde{w_1}}\right]+\frac{d_2{d}_4}{d_1{d}_3}\frac{1}{{\rho }_2}\left[{M^{\prime }}_{\pi }\left(f\right)\overline{\widetilde{u_2}}+M_{\pi }\left(f\right)\overline{\widetilde{w_2}}\right]\\ =-\left[W\left(f,\overline{w};{\xi }_1-0\right)-W\left(f,\overline{w};0\right)\right]-\frac{d_2}{d_1}\left[W\left(f,\overline{w};{\xi }_2-0\right)-W\left(f,\overline{w};{\xi }_1+0\right)\right]\\ -\frac{d_2{d}_4}{d_1{d}_3}\left[W\left(f,\overline{w};\pi \right)-W\left(f,\overline{w};{\xi }_2+0\right)\right],\end{array}$(15)

再由Naimark patching lemma可得，存在$F\in \mathcal{D}\left(\mathcal{T}\right)$ ，使得

$f_1\left(\pi \right)=f_2\left(\pi \right)=f_1\left({\xi }_1\pm 0\right)=f_2\left({\xi }_1\pm 0\right)=f_1\left({\xi }_2\pm 0\right)=f_2\left({\xi }_2\pm 0\right)=0,f_1\left(0\right)={{\alpha }^{\prime }}_2,f_2\left(0\right)={{\alpha }^{\prime }}_1,$

此时

${M^{\prime }}_0\left(f\right)={M^{\prime }}_{\pi }\left(f\right)=M_{\pi }\left(f\right)=0,M_0\left(f\right)={\alpha }_1{{\alpha }^{\prime }}_2-{\alpha }_2{{\alpha }^{\prime }}_1,$

$W\left(f,\overline{w};{\xi }_1-0\right)=W\left(f,\overline{w};{\xi }_1+0\right)=W\left(f,\overline{w};{\xi }_2-0\right)=W\left(f,\overline{w};{\xi }_2+0\right)=W\left(f,\overline{w};\pi \right)=0,$

将上述式子代入(15)，可得

$-\frac{1}{{\rho }_1}\left({\alpha }_1{{\alpha }^{\prime }}_2-{\alpha }_2{{\alpha }^{\prime }}_1\right)\overline{\widetilde{w_1}}=f_1\left(0\right){\overline{w}}_2\left(0\right)-f_2\left(0\right){\overline{w}}_1\left(0\right),$

进一步有$\widetilde{w_1}={{\alpha }^{\prime }}_1{w}_1\left(0\right)-{{\alpha }^{\prime }}_2{w}_2\left(0\right)$ ，即(ii)得证。运用同样的方法，我们可以证明$\widetilde{u_1}={\alpha }_1{w}_1\left(0\right)-{\alpha }_2{w}_2\left(0\right)$ 成立，即(vi)得证。类似的，运用Naimark patching lemma可得(iii)，(iv)，(vii)成立。因此$\mathcal{T}$ 是自伴算子。

由定理2.2易得如下两推论成立。

推论2.1 问题(1)~(7)的特征值均是实的。

推论2.2 问题(1)~(7)的对应于不同特征值${\lambda }_1,{\lambda }_2$ 的特征函数$y\left(x,{\lambda }_1\right)=\left(\begin{array}{c}{y}_1\left(x,{\lambda }_1\right)\\ y_2\left(x,{\lambda }_1\right)\end{array}\right)$ 和 $z\left(x,{\lambda }_2\right)=\left(\begin{array}{c}{z}_1\left(x,{\lambda }_2\right)\\ z_2\left(x,{\lambda }_2\right)\end{array}\right)$ 在如下意义下是正交的，即

$$ $\begin{array}{l}{\displaystyle {\int }_0^{{\xi }_1}\left[y_1\left(x,{\lambda }_1\right){\overline{z}}_1\left(x,{\lambda }_2\right)+y_2\left(x,{\lambda }_1\right){\overline{z}}_2\left(x,{\lambda }_2\right)\right]\text{d}x}+\frac{d_2}{d_1}{\displaystyle {\int }_{{\xi }_1}^{{\xi }_2}\left[y_1\left(x,{\lambda }_1\right){\overline{z}}_1\left(x,{\lambda }_2\right)+y_2\left(x,{\lambda }_1\right){\overline{z}}_2\left(x,{\lambda }_2\right)\right]\text{d}x}\\ +\frac{d_2{d}_4}{d_1{d}_3}{\displaystyle {\int }_{{\xi }_2}^{\pi }\left[y_1\left(x,{\lambda }_1\right){\overline{z}}_1\left(x,{\lambda }_2\right)+y_2\left(x,{\lambda }_1\right){\overline{z}}_2\left(x,{\lambda }_2\right)\right]\text{d}x}+\frac{1}{{\rho }_1}\widetilde{y_1}\overline{\widetilde{z_1}}+\frac{d_2{d}_4}{d_1{d}_3}\frac{1}{{\rho }_2}\widetilde{y_2}\overline{\widetilde{z_2}}=0.\end{array}$

3. 基本解及特征值的性质

我们构造(1)的两个基本解

$\Lambda \left(x,\lambda \right)=\{\begin{array}{ll}\left(\begin{array}{l}{\varphi }_1\left(x,\lambda \right)\hfill \\ {\psi }_1\left(x,\lambda \right)\hfill \end{array}\right),\hfill & x\in \left[0,{\xi }_1\right),\hfill \\ \left(\begin{array}{l}{\varphi }_2\left(x,\lambda \right)\hfill \\ {\psi }_2\left(x,\lambda \right)\hfill \end{array}\right),\hfill & x\in \left({\xi }_1,{\xi }_2\right),\hfill \\ \left(\begin{array}{l}{\varphi }_3\left(x,\lambda \right)\hfill \\ {\psi }_3\left(x,\lambda \right)\hfill \end{array}\right),\hfill & x\in \left({\xi }_2,\pi \right].\hfill \end{array}\Gamma \left(x,\lambda \right)=\{\begin{array}{ll}\left(\begin{array}{l}{\theta }_1\left(x,\lambda \right)\hfill \\ {\chi }_1\left(x,\lambda \right)\hfill \end{array}\right),\hfill & x\in \left[0,{\xi }_1\right),\hfill \\ \left(\begin{array}{l}{\theta }_2\left(x,\lambda \right)\hfill \\ {\chi }_2\left(x,\lambda \right)\hfill \end{array}\right),\hfill & x\in \left({\xi }_1,{\xi }_2\right),\hfill \\ \left(\begin{array}{l}{\theta }_3\left(x,\lambda \right)\hfill \\ {\chi }_3\left(x,\lambda \right)\hfill \end{array}\right),\hfill & x\in \left({\xi }_2,\pi \right].\hfill \end{array}$

其中$\left(\begin{array}{c}{\varphi }_1\left(x,\lambda \right)\\ {\psi }_1\left(x,\lambda \right)\end{array}\right)$ 为(1)在$\left[0,{\xi }_1\right)$ 上满足初始条件$\left(\begin{array}{c}{\varphi }_1\left(0,\lambda \right)\\ {\psi }_1\left(0,\lambda \right)\end{array}\right)=\left(\begin{array}{c}{\alpha }_2-\lambda {{\alpha }^{\prime }}_2\\ {\alpha }_1-\lambda {{\alpha }^{\prime }}_1\end{array}\right)$ 的解；$\left(\begin{array}{c}{\varphi }_2\left(x,\lambda \right)\\ {\psi }_2\left(x,\lambda \right)\end{array}\right)$ 为(1)在$\left({\xi }_1,{\xi }_2\right)$ 上满足初始条件$\left(\begin{array}{c}{\varphi }_2\left({\xi }_1+0,\lambda \right)\\ {\psi }_2\left({\xi }_1+0,\lambda \right)\end{array}\right)=BC\left(\begin{array}{c}{\varphi }_1\left({\xi }_1-0,\lambda \right)\\ {\psi }_1\left({\xi }_1-0,\lambda \right)\end{array}\right)$ 的解，这里$B={\left(\begin{array}{cc}{\delta }_1& {\delta }_2\\ {{\delta }^{\prime }}_1& {{\delta }^{\prime }}_2\end{array}\right)}^{-1}$ ，$C=\left(\begin{array}{cc}-{\gamma }_1& -{\gamma }_2\\ -{{\gamma }^{\prime }}_1& -{{\gamma }^{\prime }}_2\end{array}\right)$ ；$\left(\begin{array}{c}{\varphi }_3\left(x,\lambda \right)\\ {\psi }_3\left(x,\lambda \right)\end{array}\right)$ 为(1)在$\left({\xi }_2,\pi \right]$ 上满足初始条件$\left(\begin{array}{c}{\varphi }_3\left({\xi }_2+0,\lambda \right)\\ {\psi }_3\left({\xi }_2+0,\lambda \right)\end{array}\right)=DE\left(\begin{array}{c}{\varphi }_2\left({\xi }_2-0,\lambda \right)\\ {\psi }_2\left({\xi }_2-0,\lambda \right)\end{array}\right)$ 的解，这里$D={\left(\begin{array}{cc}{\delta }_3& {\delta }_4\\ {{\delta }^{\prime }}_3& {{\delta }^{\prime }}_4\end{array}\right)}^{-1}$ ， $E=\left(\begin{array}{cc}-{\gamma }_3& -{\gamma }_4\\ -{{\gamma }^{\prime }}_3& -{{\gamma }^{\prime }}_4\end{array}\right)$ 。

类似的，$\left(\begin{array}{c}{\theta }_3\left(x,\lambda \right)\\ {\chi }_3\left(x,\lambda \right)\end{array}\right)$ 为(1)在$\left({\xi }_2,\pi \right]$ 上满足初始条件$\left(\begin{array}{c}{\theta }_3\left(\pi ,\lambda \right)\\ {\chi }_3\left(\pi ,\lambda \right)\end{array}\right)=\left(\begin{array}{c}{\beta }_2+\lambda {{\beta }^{\prime }}_2\\ {\beta }_1+\lambda {{\beta }^{\prime }}_1\end{array}\right)$ 的解；$\left(\begin{array}{c}{\theta }_2\left(x,\lambda \right)\\ {\chi }_2\left(x,\lambda \right)\end{array}\right)$ 为(1)在$\left({\xi }_1,{\xi }_2\right)$ 上满足初始条件$\left(\begin{array}{c}{\theta }_2\left({\xi }_2-0,\lambda \right)\\ {\chi }_2\left({\xi }_2-0,\lambda \right)\end{array}\right)=E^{-1}{D}^{-1}\left(\begin{array}{c}{\theta }_3\left({\xi }_2+0,\lambda \right)\\ {\chi }_3\left({\xi }_2+0,\lambda \right)\end{array}\right)$ 的解；$\left(\begin{array}{c}{\theta }_1\left(x,\lambda \right)\\ {\chi }_1\left(x,\lambda \right)\end{array}\right)$ 为(1)在$\left[0,{\xi }_1\right)$ 上满足初始条件$\left(\begin{array}{c}{\theta }_1\left({\xi }_1-0,\lambda \right)\\ {\chi }_1\left({\xi }_1-0,\lambda \right)\end{array}\right)=C^{-1}{B}^{-1}\left(\begin{array}{c}{\theta }_2\left({\xi }_1+0,\lambda \right)\\ {\chi }_2\left({\xi }_1+0,\lambda \right)\end{array}\right)$ 的解。

下面考虑朗斯基行列式$W_i\left(\lambda \right)={\varphi }_i{\chi }_i-{\theta }_i{\psi }_i,i=\overline{1,3}$。易知对于任意的$x\in \left[0,{\xi }_1\right),\left({\xi }_1,{\xi }_2\right),\left({\xi }_2,\pi \right]$ ，$W_1\left(\lambda \right)$ ，$W_2\left(\lambda \right)$ 和$W_3\left(\lambda \right)$ 均是关于$\lambda$ 的整函数。

定理3.1 对于每一个$\lambda \in ℂ$ ，等式

$W_3\left(\lambda \right)=\frac{d_3}{d_4}{W}_2\left(\lambda \right)=\frac{d_1{d}_3}{d_2{d}_4}{W}_1\left(\lambda \right)$

成立。

证明 根据$W_i\left(\lambda \right)$ 的定义及转移条件可得结论成立。

下面，我们记

$W\left(\lambda \right):=W_3\left(\lambda \right)=\frac{d_3}{d_4}{W}_2\left(\lambda \right)=\frac{d_1{d}_3}{d_2{d}_4}{W}_1\left(\lambda \right).$

定理3.2 ${\lambda }_0\in ℂ$ 是问题(1)~(7)的一个特征值，当且仅当${\lambda }_0$ 是$W\left(\lambda \right)$ 的零点，即$W\left({\lambda }_0\right)=0$ 。

证明(必要性) 设${\lambda }_0\in ℂ$ 是问题(1)~(7)的一个特征值，$y\left(x,{\lambda }_0\right)=\left(\begin{array}{l}{y}_1\left(x,{\lambda }_0\right)\hfill \\ y_2\left(x,{\lambda }_0\right)\hfill \end{array}\right)$ 是${\lambda }_0$ 对应的特征函数。

令

$y\left(x,{\lambda }_0\right)=\{\begin{array}{ll}{r}_1\left(\begin{array}{l}{\varphi }_1\left(x,{\lambda }_0\right)\hfill \\ {\psi }_1\left(x,{\lambda }_0\right)\hfill \end{array}\right)+r_2\left(\begin{array}{l}{\theta }_1\left(x,{\lambda }_0\right)\hfill \\ {\chi }_1\left(x,{\lambda }_0\right)\hfill \end{array}\right),\hfill & x\in \left[0,{\xi }_1\right),\hfill \\ r_3\left(\begin{array}{l}{\varphi }_2\left(x,{\lambda }_0\right)\hfill \\ {\psi }_2\left(x,{\lambda }_0\right)\hfill \end{array}\right)+r_4\left(\begin{array}{l}{\theta }_2\left(x,{\lambda }_0\right)\hfill \\ {\chi }_2\left(x,{\lambda }_0\right)\hfill \end{array}\right),\hfill & x\in \left({\xi }_1,{\xi }_2\right),\hfill \\ r_5\left(\begin{array}{l}{\varphi }_3\left(x,{\lambda }_0\right)\hfill \\ {\psi }_3\left(x,{\lambda }_0\right)\hfill \end{array}\right)+r_6\left(\begin{array}{l}{\theta }_3\left(x,{\lambda }_0\right)\hfill \\ {\chi }_3\left(x,{\lambda }_0\right)\hfill \end{array}\right),\hfill & x\in \left({\xi }_2,\pi \right],\hfill \end{array}$

则$r_i\left(i=\overline{1,6}\right)$ 至少有一个不为零，下面用反证法证明$W\left({\lambda }_0\right)=0$ 。

假设$W\left({\lambda }_0\right)\ne 0$ ，即$W_3\left({\lambda }_0\right)=\frac{d_3}{d_4}{W}_2\left({\lambda }_0\right)=\frac{d_1{d}_3}{d_2{d}_4}{W}_1\left({\lambda }_0\right)\ne 0$ ，则由$y\left(x,{\lambda }_0\right)$ 满足边界条件(2)，且 $\left(\begin{array}{c}{\varphi }_1\left(0,{\lambda }_0\right)\\ {\psi }_1\left(0,{\lambda }_0\right)\end{array}\right)=\left(\begin{array}{c}{\alpha }_2-{\lambda }_0{{\alpha }^{\prime }}_2\\ {\alpha }_1-{\lambda }_0{{\alpha }^{\prime }}_1\end{array}\right)$ 成立，可得

$r_2\left[\left({\alpha }_1-{\lambda }_0{{\alpha }^{\prime }}_1\right){\theta }_1\left(0,{\lambda }_0\right)-\left({\alpha }_2-{\lambda }_0{{\alpha }^{\prime }}_2\right){\chi }_1\left(0,{\lambda }_0\right)\right]=0,$

又因为$W_1\left({\lambda }_0\right)\ne 0$ ，所以$r_2=0$ 。同理，$y\left(x,{\lambda }_0\right)$ 满足边界条件(3)，且$\left(\begin{array}{c}{\theta }_3\left(\pi ,{\lambda }_0\right)\\ {\chi }_3\left(\pi ,{\lambda }_0\right)\end{array}\right)=\left(\begin{array}{c}{\beta }_2+{\lambda }_0{{\beta }^{\prime }}_2\\ {\beta }_1+{\lambda }_0{{\beta }^{\prime }}_1\end{array}\right)$ 成立，可得$r_5=0$ 。

又因为$y\left(x,{\lambda }_0\right)$ 满足转移条件(4)，(5)，故可得

$\left(\begin{array}{c}{r}_1{\varphi }_1\left({\xi }_1-0,{\lambda }_0\right)\\ r_1{\psi }_1\left({\xi }_1-0,{\lambda }_0\right)\end{array}\right)={\left(\begin{array}{cc}-{\gamma }_1& -{\gamma }_2\\ -{{\gamma }^{\prime }}_1& -{{\gamma }^{\prime }}_2\end{array}\right)}^{-1}\left(\begin{array}{cc}{\delta }_1& {\delta }_2\\ {{\delta }^{\prime }}_1& {{\delta }^{\prime }}_2\end{array}\right)\left(\begin{array}{cc}{\varphi }_2\left({\xi }_1+0,{\lambda }_0\right)& {\theta }_2\left({\xi }_1+0,{\lambda }_0\right)\\ {\psi }_2\left({\xi }_1+0,{\lambda }_0\right)& {\chi }_2\left({\xi }_1+0,{\lambda }_0\right)\end{array}\right)\left(\begin{array}{c}{r}_3\\ r_4\end{array}\right),$

又

$\left(\begin{array}{c}{\varphi }_1\left({\xi }_1-0,{\lambda }_0\right)\\ {\psi }_1\left({\xi }_1-0,{\lambda }_0\right)\end{array}\right)=C^{-1}{B}^{-1}\left(\begin{array}{c}{\varphi }_2\left({\xi }_1+0,{\lambda }_0\right)\\ {\psi }_2\left({\xi }_1+0,{\lambda }_0\right)\end{array}\right),$

从而得

$\left(\begin{array}{cc}{\varphi }_2\left({\xi }_1+0,{\lambda }_0\right)& {\theta }_2\left({\xi }_1+0,{\lambda }_0\right)\\ {\psi }_2\left({\xi }_1+0,{\lambda }_0\right)& {\chi }_2\left({\xi }_1+0,{\lambda }_0\right)\end{array}\right)\left(\begin{array}{c}{r}_3-r_1\\ r_4\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right),$

又因为$W_2\left({\lambda }_0\right)\ne 0$ ，所以$r_3-r_1=0$ ，$r_4=0$ 。

同理，因为$y\left(x,{\lambda }_0\right)$ 满足转移条件(6)，(7)，类似的可得$r_3=0,r_6=0$ 。

综上可得，$r_1=r_2=r_3=r_4=r_5=r_6=0$ ，这与$r_i\left(i=\overline{1,6}\right)$ 至少有一个不为零矛盾，故$W\left({\lambda }_0\right)=0$ 。

(充分性) 令$W\left({\lambda }_0\right)=0$ ，又因为$d_1,d_2,d_3,d_4$ 都不等于0，故可得$W_i\left({\lambda }_0\right)=0$ ，$i=\overline{1,3}$ ，所以存在$k_i\ne 0$ ，使得$\left(\begin{array}{c}{\varphi }_i\left(x,{\lambda }_0\right)\\ {\psi }_i\left(x,{\lambda }_0\right)\end{array}\right)=k_i\left(\begin{array}{c}{\theta }_i\left(x,{\lambda }_0\right)\\ {\chi }_i\left(x,{\lambda }_0\right)\end{array}\right)$ ，其中$i=\overline{1,3}$ 。从而可得，$\Lambda \left(x,{\lambda }_0\right)$ 满足边界条件(3)，$\Gamma \left(x,{\lambda }_0\right)$ 满足边界条件(2)。又因为$\Lambda \left(x,{\lambda }_0\right)$ 和$\Gamma \left(x,{\lambda }_0\right)$ 都满足转移条件(4)~(7)，故$\Lambda \left(x,{\lambda }_0\right)$ 和$\Gamma \left(x,{\lambda }_0\right)$ 是特征值${\lambda }_0$ 对应的特征函数。

定理3.3 问题(1)~(7)的特征值是单重的。

证明 设$y\left(x,\lambda \right)$ 是问题(1)~(7)的任一解，则有

$\tau y\left(x,\lambda \right)=J{y}^{\prime }\left(x,\lambda \right)+Q\left(x\right)y\left(x,\lambda \right)=\lambda y\left(x,\lambda \right),$

对上式两端关于$\lambda$ 求导，可得

$\tau y_{\lambda }\left(x,\lambda \right)=J{y^{\prime }}_{\lambda }\left(x,\lambda \right)+Q\left(x\right)y_{\lambda }\left(x,\lambda \right)=y\left(x,\lambda \right)+\lambda y_{\lambda }\left(x,\lambda \right),$

其中，$y_{\lambda }\left(x,\lambda \right)$ 是$y\left(x,\lambda \right)$ 对$\lambda$ 求偏导。

由推论2.1可知，问题(1)~(7)的特征值均是实的，则对于任意的两个解$y\left(x,\lambda \right)$ 和$z\left(x,\lambda \right)$ ，有

${\langle \tau y_{\lambda },z\rangle }_1-{\langle y_{\lambda },\tau z\rangle }_1={\langle y+\lambda y_{\lambda },z\rangle }_1-{\langle y_{\lambda },\lambda z\rangle }_1={\langle y,z\rangle }_1.$ (16)

通过分部积分，并进一步整理得

${\langle \tau y_{\lambda },z\rangle }_1-{\langle y_{\lambda },\tau z\rangle }_1={{\alpha }^{\prime }}_1{\overline{\theta }}_1\left(0,\lambda \right)-{{\alpha }^{\prime }}_2{\overline{\chi }}_1\left(0,\lambda \right)+\frac{d_2{d}_4}{d_1{d}_3}\left[\left({\beta }_2+\lambda {{\beta }^{\prime }}_2\right){\psi }_{3\lambda }\left(\pi ,\lambda \right)-\left({\beta }_1+\lambda {{\beta }^{\prime }}_1\right){\varphi }_{3\lambda }\left(\pi ,\lambda \right)\right].$ (17)

由$W\left(\lambda \right)$ 的定义和(16)，(17)，可得

$\begin{array}{c}{W^{\prime }\left(\lambda \right)|}_{x=\pi }=\frac{\text{d}{W}_3\left(\lambda \right)}{\text{d}\lambda }{|}_{x=\pi }=\frac{\text{d}\left[{\varphi }_3\left(\pi ,\lambda \right){\chi }_3\left(\pi ,\lambda \right)-{\theta }_3\left(\pi ,\lambda \right){\psi }_3\left(\pi ,\lambda \right)\right]}{\text{d}\lambda }\\ \text{=}\frac{d_1{d}_3}{d_2{d}_4}\left[-{\langle y,z\rangle }_1+{{\alpha }^{\prime }}_1{\overline{\theta }}_1\left(0,\lambda \right)-{{\alpha }^{\prime }}_2{\overline{\chi }}_1\left(0,\lambda \right)\right]-{{\beta }^{\prime }}_2{\psi }_3\left(\pi ,\lambda \right)+{{\beta }^{\prime }}_1{\varphi }_3\left(\pi ,\lambda \right).\end{array}$(18)

设${\lambda }_0$ 是问题(1)~(7)的特征值，则有$W\left({\lambda }_0\right)=0$ 。因此存在非零常数$c_j$ ，使得 $\left(\begin{array}{c}{\theta }_j\left(x,{\lambda }_0\right)\\ {\chi }_j\left(x,{\lambda }_0\right)\end{array}\right)=c_j\left(\begin{array}{c}{\varphi }_j\left(x,{\lambda }_0\right)\\ {\psi }_j\left(x,{\lambda }_0\right)\end{array}\right),j=\overline{1,3}$，由转移条件可得$c_1=c_2=c_3\ne 0$ ，且有$\Gamma \left(x,{\lambda }_0\right)=c_1\Lambda \left(x,{\lambda }_0\right)$ 。

因为${\alpha }_1,{{\alpha }^{\prime }}_1,{\alpha }_2,{{\alpha }^{\prime }}_2,{\beta }_1,{{\beta }^{\prime }}_1,{\beta }_2,{{\beta }^{\prime }}_2$ 都是实的，$d_1,d_2,d_3,d_4,{\rho }_1,{\rho }_2$ 均大于零，方程(18)变为

$W^{\prime }\left({\lambda }_0\right)=-\frac{d_1{d}_3}{d_2{d}_4}\left\{{\overline{c}}_1\left({{\displaystyle {\int }_0^{{\xi }_1}\left|y\left(x,{\lambda }_0\right)\right|}}^2\text{d}x+\frac{d_2}{d_1}{\displaystyle {\int }_{{\xi }_1}^{{\xi }_2}{\left|y\left(x,{\lambda }_0\right)\right|}^2\text{d}x}+\frac{d_2{d}_4}{d_1{d}_3}{\displaystyle {\int }_{{\xi }_2}^{\pi }{\left|y\left(x,{\lambda }_0\right)\right|}^2\text{d}x}\right)+c_1{\rho }_1+\frac{1}{c_1}\frac{d_2{d}_4}{d_1{d}_3}{\rho }_2\right\}\ne 0,$

因此，问题(1)~(7)的特征值${\lambda }_0$ 是单重的。

4. 格林函数和预解算子

这部分我们将考虑问题(1)~(7)的Green函数和预解算子。

设$\lambda$ 不是$\mathcal{T}$ 的特征值。显然算子方程

$\left(\mathcal{T}-\lambda ℐ\right)Y=F,F={\left(f\left(x\right),\widetilde{f_1},\widetilde{f_2}\right)}^{\text{T}}\in \mathscr{H},f\left(x\right)={\left(f_1\left(x\right),f_2\left(x\right)\right)}^{\text{T}},$

等价于如下Dirac系统

$J{y}^{\prime }\left(x\right)=-Q\left(x\right)y\left(x\right)+\lambda y\left(x\right)+f\left(x\right),x\in I,$ (19)

和边界条件

${\alpha }_1{y}_1\left(0\right)-{\alpha }_2{y}_2\left(0\right)-\lambda \left({{\alpha }^{\prime }}_1{y}_1\left(0\right)-{{\alpha }^{\prime }}_2{y}_2\left(0\right)\right)=\widetilde{f_1},$ (20)

${\beta }_1{y}_1\left(\pi \right)-{\beta }_2{y}_2\left(\pi \right)+\lambda \left({{\beta }^{\prime }}_1{y}_1\left(\pi \right)-{{\beta }^{\prime }}_2{y}_2\left(\pi \right)\right)=-\widetilde{f_2},$ (21)

以及转移条件(4)~(7)构成的问题。下面我们运用常数变易法来进行求解。

引理4.1 [6]若$\Phi \left(x\right)$ 为线性齐次方程组

$Y^{\prime }=A\left(x\right)Y$

的一个基本解矩阵，则线性非齐次方程组

$Y^{\prime }=A\left(x\right)Y+F\left(x\right)$

的通解为

$Y\left(x\right)=\Phi \left(x\right)C+\Phi \left(x\right){\displaystyle {\int }_n^x{\Phi }^{-1}}\left(t\right)F\left(t\right)\text{d}t,$ (22)

这里$C\in {ℝ}^n$ 为任意的常变量，$A\left(x\right)$ 和$F\left(x\right)$ 是定义在含$n$ 的某一区间上的连续函数。

上述(19)等价于如下方程组

$\left(\begin{array}{c}{y^{\prime }}_1\left(x\right)\\ {y^{\prime }}_2\left(x\right)\end{array}\right)=\left(\begin{array}{cc}q\left(x\right)& -p\left(x\right)-\lambda \\ -p\left(x\right)+\lambda & -q\left(x\right)\end{array}\right)\left(\begin{array}{c}{y}_1\left(x\right)\\ y_2\left(x\right)\end{array}\right)+\left(\begin{array}{c}-f_2\left(x\right)\\ f_1\left(x\right)\end{array}\right).$

由于$\lambda$ 不是$\mathcal{T}$ 的特征值，故$\Lambda \left(x,\lambda \right)$ 和$\Gamma \left(x,\lambda \right)$ 在$x\in \left[0,{\xi }_1\right)\cup \left({\xi }_1,{\xi }_2\right)\cup \left({\xi }_2,\pi \right]$ 上构成的朗斯基行列式$W\left(\lambda \right)\ne 0$ ，所以构成了(1)的一个基本解矩阵。当$x\in \left[0,{\xi }_1\right)$ 时，对应的基本解矩阵为$\left(\begin{array}{cc}{\varphi }_1\left(x,\lambda \right)& {\theta }_1\left(x,\lambda \right)\\ {\psi }_1\left(x,\lambda \right)& {\chi }_1\left(x,\lambda \right)\end{array}\right)$ ，其逆矩阵为$\left(\begin{array}{cc}\frac{{\chi }_1\left(x,\lambda \right)}{W_1\left(\lambda \right)}& -\frac{{\theta }_1\left(x,\lambda \right)}{W_1\left(\lambda \right)}\\ -\frac{{\psi }_1\left(x,\lambda \right)}{W_1\left(\lambda \right)}& \frac{{\varphi }_1\left(x,\lambda \right)}{W_1\left(\lambda \right)}\end{array}\right)$ ，在(22)中取$n=0$ ，此时(19)的解为

$\begin{array}{l}y\left(x,\lambda \right)=\left(\begin{array}{c}{c}_1{\varphi }_1\left(x,\lambda \right)+c_2{\theta }_1\left(x,\lambda \right)\\ c_1{\psi }_1\left(x,\lambda \right)+c_2{\chi }_1\left(x,\lambda \right)\end{array}\right)+\frac{1}{W_1\left(\lambda \right)}\\ \times \left(\begin{array}{c}{\displaystyle {\int }_0^x\left\{\left[{\theta }_1\left(x,\lambda \right){\varphi }_1\left(t,\lambda \right)-{\varphi }_1\left(x,\lambda \right){\theta }_1\left(t,\lambda \right)\right]f_1\left(t\right)+\left[{\theta }_1\left(x,\lambda \right){\psi }_1\left(t,\lambda \right)-{\varphi }_1\left(x,\lambda \right){\chi }_1\left(t,\lambda \right)\right]f_2\left(t\right)\right\}}\text{d}t\\ {\displaystyle {\int }_0^x\left\{\left[{\chi }_1\left(x,\lambda \right){\varphi }_1\left(t,\lambda \right)-{\psi }_1\left(x,\lambda \right){\theta }_1\left(t,\lambda \right)\right]f_1\left(t\right)+\left[{\chi }_1\left(x,\lambda \right){\psi }_1\left(t,\lambda \right)-{\psi }_1\left(x,\lambda \right){\chi }_1\left(t,\lambda \right)\right]f_2\left(t\right)\right\}}\text{d}t\end{array}\right),\end{array}$ (23)

将上式代入$x=0$ 点的边界条件(20)，得$c_2=-\frac{\widetilde{f_1}}{W_1\left(\lambda \right)}$ 。

当$x\in \left({\xi }_1,{\xi }_2\right)$ 时，在(22)中取$n={\xi }_1$ ，此时(19)的解为

$\begin{array}{l}y\left(x,\lambda \right)=\left(\begin{array}{c}{c}_3{\varphi }_2\left(x,\lambda \right)+c_4{\theta }_2\left(x,\lambda \right)\\ c_3{\psi }_2\left(x,\lambda \right)+c_4{\chi }_2\left(x,\lambda \right)\end{array}\right)+\frac{1}{W_2\left(\lambda \right)}\\ \times \left(\begin{array}{c}{\displaystyle {\int }_{{\xi }_1}^x\left\{\left[{\theta }_2\left(x,\lambda \right){\varphi }_2\left(t,\lambda \right)-{\varphi }_2\left(x,\lambda \right){\theta }_2\left(t,\lambda \right)\right]f_1\left(t\right)+\left[{\theta }_2\left(x,\lambda \right){\psi }_2\left(t,\lambda \right)-{\varphi }_2\left(x,\lambda \right){\chi }_2\left(t,\lambda \right)\right]f_2\left(t\right)\right\}}\text{d}t\\ {\displaystyle {\int }_{{\xi }_1}^x\left\{\left[{\chi }_2\left(x,\lambda \right){\varphi }_2\left(t,\lambda \right)-{\psi }_2\left(x,\lambda \right){\theta }_2\left(t,\lambda \right)\right]f_1\left(t\right)+\left[{\chi }_2\left(x,\lambda \right){\psi }_2\left(t,\lambda \right)-{\psi }_2\left(x,\lambda \right){\chi }_2\left(t,\lambda \right)\right]f_2\left(t\right)\right\}}\text{d}t\end{array}\right).\end{array}$ (24)

当$x\in \left({\xi }_2,\pi \right]$ 时，在(22)中取$n=\pi$ ，此时(19)的解为

$\begin{array}{l}y\left(x,\lambda \right)=\left(\begin{array}{c}{c}_5{\varphi }_3\left(x,\lambda \right)+c_6{\theta }_3\left(x,\lambda \right)\\ c_5{\psi }_3\left(x,\lambda \right)+c_6{\chi }_3\left(x,\lambda \right)\end{array}\right)-\frac{1}{W_3\left(\lambda \right)}\\ \times \left(\begin{array}{c}{\displaystyle {\int }_x^{\pi }\left\{\left[{\theta }_3\left(x,\lambda \right){\varphi }_3\left(t,\lambda \right)-{\varphi }_3\left(x,\lambda \right){\theta }_3\left(t,\lambda \right)\right]f_1\left(t\right)+\left[{\theta }_3\left(x,\lambda \right){\psi }_3\left(t,\lambda \right)-{\varphi }_3\left(x,\lambda \right){\chi }_3\left(t,\lambda \right)\right]f_2\left(t\right)\right\}}\text{d}t\\ {\displaystyle {\int }_x^{\pi }\left\{\left[{\chi }_3\left(x,\lambda \right){\varphi }_3\left(t,\lambda \right)-{\psi }_3\left(x,\lambda \right){\theta }_3\left(t,\lambda \right)\right]f_1\left(t\right)+\left[{\chi }_3\left(x,\lambda \right){\psi }_3\left(t,\lambda \right)-{\psi }_3\left(x,\lambda \right){\chi }_3\left(t,\lambda \right)\right]f_2\left(t\right)\right\}}\text{d}t\end{array}\right),\end{array}$ (25)

将上式代入$x=\pi$ 点的边界条件(21)，得$c_5=-\frac{\widetilde{f_2}}{W_3\left(\lambda \right)}$ 。

接下来考虑转移条件(4)~(7)。将$c_2=-\frac{\widetilde{f_1}}{W_1\left(\lambda \right)}$ 和$c_5=-\frac{\widetilde{f_2}}{W_3\left(\lambda \right)}$ 分别代入(23)和(25)。由转移条件(4)，(5)可知$y\left({\xi }_1+0,\lambda \right)=BCy\left({\xi }_1-0,\lambda \right)$ ，进一步有

$\begin{array}{c}\left(\begin{array}{c}{c}_3{\varphi }_2\left({\xi }_1+0,\lambda \right)+c_4{\theta }_2\left({\xi }_1+0,\lambda \right)\\ c_3{\psi }_2\left({\xi }_1+0,\lambda \right)+c_4{\chi }_2\left({\xi }_1+0,\lambda \right)\end{array}\right)=\left(\begin{array}{c}{c}_1{\varphi }_2\left({\xi }_1+0,\lambda \right)-\frac{\widetilde{f_1}}{W_1\left(\lambda \right)}{\theta }_2\left({\xi }_1+0,\lambda \right)\\ c_1{\psi }_2\left({\xi }_1+0,\lambda \right)-\frac{\widetilde{f_1}}{W_1\left(\lambda \right)}{\chi }_2\left({\xi }_1+0,\lambda \right)\end{array}\right)\\ +\frac{1}{W_1\left(\lambda \right)}\left(\begin{array}{c}{\displaystyle {\int }_0^{{\xi }_1}\left[{\theta }_2\left({\xi }_1+0,\lambda \right){\varphi }_1\left(t,\lambda \right)-{\varphi }_2\left({\xi }_1+0,\lambda \right){\theta }_1\left(t,\lambda \right)\right]f_1\left(t\right)\text{d}t}\\ {\displaystyle {\int }_0^{{\xi }_1}\left[{\chi }_2\left({\xi }_1+0,\lambda \right){\varphi }_1\left(t,\lambda \right)-{\psi }_2\left({\xi }_1+0,\lambda \right){\theta }_1\left(t,\lambda \right)\right]f_1\left(t\right)\text{d}t}\end{array}\right)\\ +\frac{1}{W_1\left(\lambda \right)}\left(\begin{array}{c}{\displaystyle {\int }_0^{{\xi }_1}\left[{\theta }_2\left({\xi }_1+0,\lambda \right){\psi }_1\left(t,\lambda \right)-{\varphi }_2\left({\xi }_1+0,\lambda \right){\chi }_1\left(t,\lambda \right)\right]f_2\left(t\right)\text{d}t}\\ {\displaystyle {\int }_0^{{\xi }_1}\left[{\chi }_2\left({\xi }_1+0,\lambda \right){\psi }_1\left(t,\lambda \right)-{\psi }_2\left({\xi }_1+0,\lambda \right){\chi }_1\left(t,\lambda \right)\right]f_2\left(t\right)\text{d}t}\end{array}\right)\end{array}$ (26)

由转移条件(6)，(7)可知$y\left({\xi }_2+0,\lambda \right)=DEy\left({\xi }_2-0,\lambda \right)$ ，进一步有

(27)

求解(26)和(27)，可得$c_1,c_3,c_4,c_6$ 。

将上述求得的$c_1,c_2,c_3,c_4,c_5,c_6$ 代入解的表达式，我们即可得到(19)满足边界条件(20)，(21)以及转

移条件(4)~(7)的解$y\left(x,\lambda \right)$ ，将其写成分量$y\left(x,\lambda \right)=\left(\begin{array}{c}{y}_1\left(x,\lambda \right)\\ y_2\left(x,\lambda \right)\end{array}\right)$ 的形式，如下所示

若令$G\left(x,t,\lambda \right)=\left(\begin{array}{cc}{G}_{11}\left(x,t,\lambda \right)& G_{12}\left(x,t,\lambda \right)\\ G_{21}\left(x,t,\lambda \right)& G_{22}\left(x,t,\lambda \right)\end{array}\right)$ ，其中

$G_{11}\left(x,t,\lambda \right)=\{\begin{array}{c}\frac{1}{W_1\left(\lambda \right)}{\theta }_i\left(x,\lambda \right){\varphi }_j\left(t,\lambda \right),0\le t\le x\le \pi ,x\ne {\xi }_1,{\xi }_2,t\ne {\xi }_1,{\xi }_2,\\ \frac{1}{W_1\left(\lambda \right)}{\varphi }_i\left(x,\lambda \right){\theta }_j\left(t,\lambda \right),0\le x\le t\le \pi ,x\ne {\xi }_1,{\xi }_2,t\ne {\xi }_1,{\xi }_2;\end{array}$

$G_{12}\left(x,t,\lambda \right)=\{\begin{array}{c}\frac{1}{W_1\left(\lambda \right)}{\chi }_i\left(x,\lambda \right){\varphi }_j\left(t,\lambda \right),0\le t\le x\le \pi ,x\ne {\xi }_1,{\xi }_2,t\ne {\xi }_1,{\xi }_2,\\ \frac{1}{W_1\left(\lambda \right)}{\psi }_i\left(x,\lambda \right){\theta }_j\left(t,\lambda \right),0\le x\le t\le \pi ,x\ne {\xi }_1,{\xi }_2,t\ne {\xi }_1,{\xi }_2;\end{array}$

$G_{21}\left(x,t,\lambda \right)=\{\begin{array}{c}\frac{1}{W_1\left(\lambda \right)}{\theta }_i\left(x,\lambda \right){\psi }_j\left(t,\lambda \right),0\le t\le x\le \pi ,x\ne {\xi }_1,{\xi }_2,t\ne {\xi }_1,{\xi }_2,\\ \frac{1}{W_1\left(\lambda \right)}{\varphi }_i\left(x,\lambda \right){\chi }_j\left(t,\lambda \right),0\le x\le t\le \pi ,x\ne {\xi }_1,{\xi }_2,t\ne {\xi }_1,{\xi }_2;\end{array}$

$G_{22}\left(x,t,\lambda \right)=\{\begin{array}{c}\frac{1}{W_1\left(\lambda \right)}{\chi }_i\left(x,\lambda \right){\psi }_j\left(t,\lambda \right),0\le t\le x\le \pi ,x\ne {\xi }_1,{\xi }_2,t\ne {\xi }_1,{\xi }_2,\\ \frac{1}{W_1\left(\lambda \right)}{\psi }_i\left(x,\lambda \right){\chi }_j\left(t,\lambda \right),0\le x\le t\le \pi ,x\ne {\xi }_1,{\xi }_2,t\ne {\xi }_1,{\xi }_2;\end{array}$

$i=\overline{1,3},j=\overline{1,3}.$

又因为

$W_3\left(\lambda \right)=\frac{d_3}{d_4}{W}_2\left(\lambda \right)=\frac{d_1{d}_3}{d_2{d}_4}{W}_1\left(\lambda \right),$

则可得

$\begin{array}{c}y(x,\lambda )={\displaystyle {\int }_0^{{\xi }_1}{G}^{\text{T}}}\left(x,t,\lambda \right)f\left(t\right)\text{d}t+\frac{d_2}{d_1}{\displaystyle {\int }_{{\xi }_1}^{{\xi }_2}{G}^{\text{T}}}\left(x,t,\lambda \right)f\left(t\right)\text{d}t+\frac{d_2{d}_4}{d_1{d}_3}{\displaystyle {\int }_{{\xi }_2}^{\pi }{G}^{\text{T}}}\left(x,t,\lambda \right)f\left(t\right)\text{d}t\\ -\frac{\widetilde{f_1}}{W_1\left(\lambda \right)}\Gamma \left(x,\lambda \right)-\frac{d_2{d}_4}{d_1{d}_3}\frac{\widetilde{f_2}}{W_1\left(\lambda \right)}\Lambda \left(x,\lambda \right).\end{array}$(28)

另一方面，把$t$ 看作是积分变量，我们有

$\begin{array}{c}{M^{\prime }}_0\left(G\left(x,\cdot ,\lambda \right)\right)={{\alpha }^{\prime }}_1{G}_1\left(x,0,\lambda \right)-{{\alpha }^{\prime }}_2{G}_2\left(x,0,\lambda \right)\\ =\frac{1}{W_1\left(\lambda \right)}\Gamma \left(x,\lambda \right)\left[{{\alpha }^{\prime }}_1{\varphi }_1\left(0,\lambda \right)-{{\alpha }^{\prime }}_2{\psi }_1\left(0,\lambda \right)\right]\\ =\frac{1}{W_1\left(\lambda \right)}\Gamma \left(x,\lambda \right)\left[{{\alpha }^{\prime }}_1{\alpha }_2-{\alpha }_1{{\alpha }^{\prime }}_2\right]\\ =-\frac{{\rho }_1}{W_1\left(\lambda \right)}\Gamma \left(x,\lambda \right),\end{array}$ (29)

类似可得

${M^{\prime }}_{\pi }\left(G\left(x,\cdot ,\lambda \right)\right)=-\frac{{\rho }_2}{W_1\left(\lambda \right)}\Lambda \left(x,\lambda \right).$ (30)

将(29)，(30)代入(28)，等式(28)又可表示为

$\begin{array}{c}y\left(x,\lambda \right)={\displaystyle {\int }_0^{{\xi }_1}{G}^{\text{T}}}\left(x,t,\lambda \right)f\left(t\right)\text{d}t+\frac{d_2}{d_1}{\displaystyle {\int }_{{\xi }_1}^{{\xi }_2}{G}^{\text{T}}}\left(x,t,\lambda \right)f\left(t\right)\text{d}t+\frac{d_2{d}_4}{d_1{d}_3}{\displaystyle {\int }_{{\xi }_2}^{\pi }{G}^{\text{T}}}\left(x,t,\lambda \right)f\left(t\right)\text{d}t\\ +\frac{1}{{\rho }_1}{M^{\prime }}_0\left(G\left(x,\cdot ,\lambda \right)\right)\widetilde{f_1}+\frac{d_2{d}_4}{d_1{d}_3}\frac{1}{{\rho }_2}{M^{\prime }}_{\pi }\left(G\left(x,\cdot ,\lambda \right)\right)\widetilde{f_2}.\end{array}$(31)

现在定义

${\tilde{G}}_{x,\lambda }=\left(\begin{array}{c}G\left(x,\cdot ,\lambda \right)\\ {M^{\prime }}_0\left(G\left(x,\cdot ,\lambda \right)\right)\\ {M^{\prime }}_{\pi }\left(G\left(x,\cdot ,\lambda \right)\right)\end{array}\right),F=\left(\begin{array}{c}f\left(x\right)\\ \widetilde{f_1}\\ \widetilde{f_2}\end{array}\right),\overline{F}=\left(\begin{array}{c}\overline{f\left(x\right)}\\ \overline{\widetilde{f_1}}\\ \overline{\widetilde{f_2}}\end{array}\right),$

此时，(31)可表示为

$y\left(x,\lambda \right)=\langle {\tilde{G}}_{x,\lambda },\overline{F}\rangle .$

故预解算子$ℛ\left(\lambda ,\mathcal{T}\right)={\left(\mathcal{T}-\lambda ℐ\right)}^{-1}$ 可表示为如下形式

$ℛ\left(\lambda ,\mathcal{T}\right)F=\left(\begin{array}{c}\langle {\tilde{G}}_{x,\lambda },\overline{F}\rangle \\ {M^{\prime }}_0\left(\langle {\tilde{G}}_{x,\lambda },\overline{F}\rangle \right)\\ {M^{\prime }}_{\pi }\left(\langle {\tilde{G}}_{x,\lambda },\overline{F}\rangle \right)\end{array}\right).$

5. 总结与展望

在文献[6]的基础上，本文研究了一类内部具有两个不连续点且边界条件均含谱参数的Dirac算子。分析了该算子的特征值的性质，给出了问题的格林函数和预解算子。本文将内部不连续点的个数从一个推广到两个，丰富了微分算子的谱理论。今后也可以进一步研究内部不连续点多于两个的情形。

基金项目

国家自然科学基金资助项目(12261066)，内蒙古自然科学基金(2017MS(LH)0103, 2023LHMS01015)。

NOTES

^*通讯作者。