内部具有不连续性(即具有内部点条件或转移条件)的微分算子是解决热传导问题、弦的振动问题等物理问题的有力工具。例如在研究复合材料叠层板块的热量传导问题时，由于不同板块的热传导系数不同，板块与板块的结合部分热传导系数改变，则相应问题变为具有不连续性的问题 [1] ；在研究中间有结点的弦振动问题时，需要考虑在一点处附加的质量，结点两侧会产生一定的关联关系，这种关系赋予了适当的“转移条件” [2] 。经典的微分算子问题，谱参数只出现在微分方程中，但随着力学、数学物理等领域的发展，谱参数不仅出现在微分方程中，还可能出现在边界条件和转移条件中。

关于边界条件含谱参数的对称微分算子的研究，Mukhtaro和Hao等人考虑了边界条件一端含特征参数的二阶微分算子，得到了算子的谱性质以及格林函数和预解算子等 [3] [4] 。李昆和郑召文 [5] 进一步考虑了两端边界条件均含谱参数的不连续二阶微分算子的谱性质。随后，此类问题被推广到具有两个不连续点以及更高阶的情形 [6] [7] 。

关于转移条件含谱参数的对称微分算子(边界条件可能含谱参数，也可能不含谱参数)的研究，2005年，Akdoğan [8] 考虑了边界条件和转移条件均含谱参数的二阶边值问题的特征值及其对应特征函数的渐近性。2015年，上述问题被推广到具有两个转移点的情形 [9] 。同年，郭永霞 [10] 研究了多个转移条件含谱参数的二阶Sturm-Liouville算子的谱及逆谱问题。之后，关于内部点条件含有谱参数的对称微分算子问题又涌现出了一些较新的研究成果 [11] [12] 。

另外，以上研究中微分算式系的数函数均为实值函数，当系数函数为复值函数时，由此生成的算子为非对称微分算子，其中，J-对称微分算子就是一类特殊的非对称微分算子。

关于J-对称微分算子的J-自伴扩张问题，Knowles I，Race D，Shang Z J等对J-对称微分算子J-自伴扩张域进行了解析刻画 [13] - [15] 。之后，J-对称微分算子的J-自伴扩张问题还被推广到具有转移条件的情形 [16] 。另外，关于J-对称微分算子其他方面的研究，包括格林函数及预解算子，谱的离散性等也有一些基础性的工作 [17] 。以上关于J-对称微分算子的研究中，谱参数只出现在微分方程中，于是受到边界条件和转移条件含有谱参数的对称微分算子研究成果的启发，我们要问，当边界条件和转移条件均含有谱参数时，一般的二阶复系数微分方程形式下所生成的微分算子是否具有J-自伴性？格林函数的表达式是什么？这就是本文将要考虑的问题。

本节给出C-自伴算子及J-自伴算子的相关定义和结论。

设H是可分的Hilbert空间， $\left(\cdot ,\cdot \right)$表示H上的内积。

定义1 若H上的算子C满足： $C^2=I$，则C是幂等的；若

$C\left(\alpha x+\beta y\right)=\bar{\alpha }x+\bar{\beta }y,\forall \alpha ,\beta \in ℂ,x,y\in H$,

则C是反线性的；若

$\left(x,y\right)=\left(Cy,Cx\right),\forall x,y\in H$,

则C是等距的。H上幂等的等距反线性算子称为C-算子。

定义2 设A： $D\left(A\right)\to H$是闭稠密线性算子，C是H上的C-算子，若 $\forall f,g\in D\left(A\right)$，有

$\left(CAf,g\right)=\left(CAg,f\right)$

即 $A\subset C{A}^{\ast }C$，则A是C-对称算子。若 $A=C{A}^{\ast }C$，则A是C-自伴算子。

定义3 设C是H上的C-算子，H上的双线性型 $\left[\cdot ,\cdot \right]$表示为

$\left[x,y\right]=\left(x,Cy\right),\forall x,y\in H,$

如果 $\left[x,y\right]=\left(x,Cy\right)=0$，则称x与y在 $\left[\cdot ,\cdot \right]$下是正交的或C-正交的。

定义4 设A是H上的稠定线性算子，若

$\left(Ax,Jy\right)=\left(x,JAy\right),\forall x,y\in D\left(A\right),$

则A是J-对称算子。其中J是H上取复共轭的算子，即 $\forall x\in H,Jx=\bar{x}$。

引理1 [17] A是J-对称算子的充要条件是： $JA\subset A^{\ast }J$ ( $A\subset J{A}^{\ast }J$，或 $JAJ\subset A^{\ast }$)

定义5 如果 $JA=A^{\ast }J$ ( $A=J{A}^{\ast }J$，或 $JAJ=A^{\ast }$)，则A是J-自伴算子。

注1 [17] C-对称和C-自伴有时也称为J-对称和J-自伴，但在Krein空间中C-自伴与J-自伴不同。

考虑一般形式的二阶复系数阶微分方程

$l\left(y\right):=-{\left(p\left(x\right)y^{\prime }\right)}^{\prime }+q\left(x\right)y=\lambda w\left(x\right)y,$ (1)

其中 $x\in I=\left[a,c\right)\cup \left(c,b\right]$， $p^{-1}\left(x\right)$， $q\left(x\right)$是 $\left[a,c\right),\left(c,b\right]$上的连续复值函数，且 $\underset{x\to c\pm }{\lim }q\left((x)\right)=q\left(c\pm \right)$存在有限极限， $w\in L^1\left(I,ℝ\right),w>0$， $\lambda \in ℂ$是谱参数。

含有谱参数的耦合型边界条件为

$AY\left(a\right)+BY\left(b\right)=0,$ (2)

含有谱参数的转移条件为

$CY\left(c-\right)+DY\left(c+\right)=0,$ (3)

其中

$A=\left(\begin{array}{cc}{\alpha }_1-\lambda {\gamma }_1& {\alpha }_2\\ {\beta }_1& 0\end{array}\right),B=\left(\begin{array}{cc}{\alpha }_3& 0\\ {\beta }_2-\lambda {\gamma }_2& {\beta }_3\end{array}\right),C=\left(\begin{array}{cc}{\theta }_1-\lambda {\delta }_1& {\theta }_2\\ {\sigma }_1& 0\end{array}\right),D=\left(\begin{array}{cc}{\theta }_3& 0\\ {\sigma }_2-\lambda {\delta }_2& {\sigma }_3\end{array}\right)$,

$Y(\cdot )={\left(y(\cdot ),y^{\left[1\right]}(\cdot )\right)}^{\text{T}},y^{\left[1\right]}=p{y}^{\prime },{\alpha }_i,{\beta }_i,{\theta }_i,{\sigma }_i,{\gamma }_j,{\delta }_j\in ℝ,i=1,2,3,j=1,2$,

且假定

$-{\alpha }_2{\beta }_1=-{\theta }_2{\sigma }_1=\rho >0,{\alpha }_3{\beta }_3={\theta }_3{\sigma }_3=\eta >0$. (4)

定义6 令微分算式 $l\left(y\right):=-{\left(p\left(x\right)y^{\prime }\right)}^{\prime }+q\left(x\right)y$，由微分算式 $l\left(y\right)$生成的最大算子 $T_M$的定义为

$T_M\left(y\right)=l\left(y\right),y\in D_M,D_M=\left\{y\in L_w^2\left(I\right)|y,y^{\left[1\right]}\in AC\left(I\right),l\left(y\right)\in L_w^2\left(I\right)\right\},$

其中 $D_M$称为 $l\left(y\right)$的最大算子域。

引理2 [18] 对于任意的复数组 ${\xi }_1,{\xi }_2,{\xi }_3,{\xi }_4,{\tau }_1,{\tau }_2,{\tau }_3,{\tau }_4$，一定存在一个函数 $f\in D_M$，使得

$f\left(a\right)={\xi }_1,f\left(b\right)={\xi }_2,f\left(c+\right)={\xi }_3,f\left(c-\right)={\xi }_4,f^{\left[1\right]}\left(a\right)={\tau }_1,f^{\left[1\right]}\left(b\right)={\tau }_2,f^{\left[1\right]}\left(c+\right)={\tau }_3,f^{\left[1\right]}\left(c-\right)={\tau }_4$.

在空间 $L_w^2\left(I\right)$中定义内积

${\langle f,g\rangle }_w=\rho {\displaystyle {\int }_a^{c}f\bar{g}w\text{d}x}+\eta {\displaystyle {\int }_c^{b}f\bar{g}w\text{d}x},\forall f,g\in L_w^2\left(I\right)$,

易知 $H_1\left(L_w^2\left(I\right),\langle \cdot ,\cdot \rangle \right)$是Hilbert空间。在直和空间 $ℍ=H_1\oplus {ℂ}^4$中定义内积

$\langle F,G\rangle =f,g_w+m_1{f}_1\overline{g_1}+m_2{f}_2\overline{g_2}+m_3{f}_3\overline{g_3}+m_4{f}_4\overline{g_4}$,

$\forall F={\left(f,f_1,f_2,f_3,f_4\right)}^{\text{T}},G={\left(g,g_1,g_2,g_3,g_4\right)}^{\text{T}}\in \mathscr{H}$,

则 $\mathscr{H}$为Hilbert空间。其中

$f\left(x\right),g\left(x\right)\in H_1,f_i,g_i\in ℂ,i=1,2,3,4,$

$\begin{array}{l}{f}_1={\gamma }_{1}f\left(a\right),f_2={\gamma }_{2}f\left(b\right),f_3={\delta }_{1}f\left(c-\right),f_4={\delta }_{2}f\left(c+\right),\\ g_1={\gamma }_{1}g\left(a\right),g_2={\gamma }_{2}g\left(b\right),g_3={\delta }_{1}g\left(c-\right),g_4={\delta }_{2}g\left(c+\right),\end{array}$

$m_1=-\frac{\rho }{{\alpha }_2{\gamma }_1},m_2=\frac{\eta }{{\beta }_3{\gamma }_2},m_3=\frac{\rho }{{\theta }_2{\delta }_1},m_4=-\frac{\eta }{{\sigma }_3{\delta }_2},$

在Hilbert空间 $\mathscr{H}$上定义算子如下

$TF=\left(\begin{array}{c}\frac{l\left(f\right)}{w}\\ {\alpha }_{1}f\left(a\right)+{\alpha }_2{f}^{\left[1\right]}\left(a\right)+{\alpha }_{3}f(b)\\ {\beta }_{1}f\left(a\right)+{\beta }_{2}f\left(b\right)+{\beta }_3{f}^{\left[1\right]}\left(b\right)\\ {\theta }_{1}f\left(c-\right)+{\theta }_2{f}^{\left[1\right]}\left(c-\right)+{\theta }_{3}f\left(c+\right)\\ {\sigma }_{1}f\left(c-\right)+{\sigma }_{2}f\left(c+\right)+{\sigma }_3{f}^{\left[1\right]}\left(c+\right)\end{array}\right)=\left(\begin{array}{c}\lambda f\\ \lambda f_1\\ \lambda f_2\\ \lambda f_3\\ \lambda f_4\end{array}\right)=\lambda F$,

$\forall F={\left(f,f_1,f_2,f_3,f_4\right)}^{\text{T}}\in \mathscr{H}$,

其中，算子T的定义域为

$\begin{array}{l}D\left(T\right)=\{{\left(f\left(x\right),f_1,f_2,f_3,f_4\right)}^{\text{T}}\in \mathscr{H}|f,f^{[1]}\in A{C}_{loc}\left(I\right),l\left(f\right)\in H_1,AF\left(a\right)+BF\left(b\right)=0,\\ {}_{}{}^{{}_{}{}^{}}CF\left(c-\right)+DF\left(c+\right)=0,f_1={\gamma }_{1}f\left(a\right),f_2={\gamma }_{2}f\left(b\right),f_3={\delta }_{1}f\left(c-\right),f_4={\delta }_{2}f\left(c+\right)\}.\end{array}$

为简便，令

$\begin{array}{l}{M}_1\left(f\right)={\alpha }_{1}f\left(a\right)+{\alpha }_2{f}^{\left[1\right]}\left(a\right)+{\alpha }_{3}f\left(b\right),M_2\left(f\right)={\beta }_{1}f\left(a\right)+{\beta }_{2}f\left(b\right)+{\beta }_3{f}^{\left[1\right]}\left(b\right),\\ N_1\left(f\right)={\theta }_{1}f\left(c-\right)+{\theta }_2{f}^{\left[1\right]}\left(c-\right)+{\theta }_{3}f\left(c+\right),N_2\left(f\right)={\sigma }_{1}f\left(c-\right)+{\sigma }_{2}f\left(c+\right)+{\sigma }_3{f}^{\left[1\right]}\left(c+\right),\end{array}$ (5)

${M^{\prime }}_1\left(f\right)={\gamma }_{1}f\left(a\right),{M^{\prime }}_2\left(f\right)={\gamma }_{2}f\left(b\right),{N^{\prime }}_1\left(f\right)={\delta }_{1}f\left(c-\right),{N^{\prime }}_2\left(f\right)={\delta }_{2}f\left(c+\right),$ (6)

即

$F={\left(f,{M^{\prime }}_1\left(f\right),{M^{\prime }}_2\left(f\right),{N^{\prime }}_1\left(f\right),{N^{\prime }}_2\left(f\right)\right)}^{\text{T}}$,

于是，问题(1)-(3)可转化为算子问题 $TF=\lambda F$。

引理3 $D\left(T\right)$在 $\mathscr{H}$中稠密。

证明 设 $F={\left(f\left(x\right),f_1,f_2,f_3,f_4\right)}^{\text{T}}\in \mathscr{H}$，且 $F\perp D\left(T\right)$。令 $\widetilde{C_0^{\infty }}$表示下列函数的集合

$\varphi \left(x\right)=\{\begin{array}{l}{\varphi }_1\left(x\right),x\in \left[a,c\right),\\ {\varphi }_2\left(x\right),x\in \left(c,b\right],\end{array}$

其中， ${\varphi }_1\left(x\right)\in C_0^{\infty }\left[a,c\right),{\varphi }_2\left(x\right)\in C_0^{\infty }\left(c,b\right]$。于是， $\widetilde{C_0^{\infty }}\oplus \left\{0\right\}\oplus \left\{0\right\}\oplus \left\{0\right\}\oplus \left\{0\right\}\subset D\left(T\right),\left(0\in ℂ\right)$。设 $U=\left(u\left(x\right),0,0,0,0\right)\in \widetilde{C_0^{\infty }}\oplus \left\{0\right\}\oplus \left\{0\right\}\oplus \left\{0\right\}\oplus \left\{0\right\}$。则 $F\perp U$。由 $\langle F,U\rangle ={\langle f,u\rangle }_1=0$知 $f\left(x\right)$在 $H_1$中正交于 $\widetilde{C_0^{\infty }}$，故 $f\left(x\right)$为零。设 $G\left(x\right)=\left(g\left(x\right),g_1,0,0,0\right)\in \widetilde{C_0^{\infty }}\oplus \left\{0\right\}\oplus \left\{0\right\}\oplus \left\{0\right\}$，则 $\langle F,G\rangle ={\langle f,g\rangle }_w+m_1{f}_1\overline{g_1}=0$，由于 $g_1$是任意选取的，故 $f_1=0$。设 $G\left(x\right)=\left(g\left(x\right),0,g_2,0,0\right)\in \widetilde{C_0^{\infty }}\oplus \left\{0\right\}\oplus \left\{0\right\}\oplus \left\{0\right\}$，则 $\langle F,G\rangle ={\langle f,g\rangle }_w+m_2{f}_2\overline{g_2}=0$，由于 $g_2$是任意选取的，故 $f_2=0$。同理可以证明 $f_3=0,f_4=0$。于是 $F={\left(0,0,0,0,0\right)}^{\text{T}}$。所以与 $D\left(T\right)$正交的只有零元素，从而证得 $D\left(T\right)$在 $\mathscr{H}$中稠密。

定理1 算子T是定义在 $\mathscr{H}$中的J-自伴算子。

证明 对 $\forall F,G\in D\left(T\right)$，由分部积分得

$\begin{array}{c}\langle TF,JG\rangle ={\langle \frac{l\left(f\right)}{w},Jg\rangle }_w+m_1{M}_1\left(f\right)\overline{J{M^{\prime }}_1\left(g\right)}+m_2{M}_2\left(f\right)\overline{J{M^{\prime }}_2\left(g\right)}+m_3{N}_1\left(f\right)\overline{J{N^{\prime }}_1\left(g\right)}+m_4{N}_2\left(f\right)\overline{J{N^{\prime }}_2\left(g\right)}\\ =\langle F,JTG\rangle +\rho {\left[f,g\right]}_a^{c-}+\eta {\left[f,g\right]}_{c+}^b+m_1\left[M_1\left(f\right){M^{\prime }}_1\left(g\right)-{M^{\prime }}_1\left(f\right)M_1\left(g\right)\right]\\ +m_2\left[M_2\left(f\right){M^{\prime }}_2\left(g\right)-{M^{\prime }}_2\left(f\right)M_2\left(g\right)\right]+m_3\left[N_1\left(f\right){N^{\prime }}_1\left(g\right)-{N^{\prime }}_1\left(f\right)N_1\left(g\right)\right]\\ +m_4\left[N_2\left(f\right){N^{\prime }}_2\left(g\right)-{N^{\prime }}_2\left(f\right)N_2\left(g\right)\right],\end{array}$

其中

$\left[f,g\right]\left(x\right)=f\left(x\right)g^{\left[1\right]}\left(x\right)-f^{\left[1\right]}\left(x\right)g\left(x\right),$ (7)

根据(4)，(5)，(6)式，经推导计算可得

$m_1\left[M_1\left(f\right){M^{\prime }}_1\left(g\right)-{M^{\prime }}_1\left(f\right)M_1\left(g\right)\right]+m_2\left[M_2\left(f\right){M^{\prime }}_2\left(g\right)-{M^{\prime }}_2\left(f\right)M_2\left(g\right)\right]=\rho \left[f,g\right]\left(a\right)-\eta \left[f,g\right]\left(b\right),$

$m_3\left[N_1\left(f\right){N^{\prime }}_1\left(g\right)-{N^{\prime }}_1\left(f\right)N_1\left(g\right)\right]+m_4\left[N_2\left(f\right){N^{\prime }}_2\left(g\right)-{N^{\prime }}_2\left(f\right)N_2\left(g\right)\right]=\eta \left[f,g\right]\left(c+\right)-\rho \left[f,g\right]\left(c-\right),$

因此

$\langle TF,JG\rangle =\langle F,JTG\rangle$,

故算子T是J-对称的。接下来利用J-自伴算子的定义来证明算子T的J-自伴性。

下面只需证明：若对于任意的 $F={\left(f,{M^{\prime }}_1\left(f\right),{M^{\prime }}_2\left(f\right),{N^{\prime }}_1\left(f\right),{N^{\prime }}_2\left(f\right)\right)}^{\text{T}}\in D\left(T\right)$，有 $\langle TF,JG\rangle =\langle F,Z\rangle$成立，则 $G\in D\left(T\right)$且 $JTG=Z$，其中 $G={\left(g,g_1,g_2,g_3,g_4\right)}^{\text{T}}$， $Z={\left(z,z_1,z_2,z_3,z_4\right)}^{\text{T}}$，即

(i) $g\left(x\right),g^{\left[1\right]}\left(x\right)\in A{C}_{loc}\left(I\right),\frac{l\left(g\right)}{w}\in H_1$;

(ii) $g_1={\gamma }_{1}g\left(a\right),g_2={\gamma }_{2}g\left(b\right),g_3={\delta }_{1}g\left(c-\right),g_4={\delta }_{2}g\left(c+\right)$;

(iii) $z_1=J\left[{\alpha }_{1}g\left(a\right)+{\alpha }_2{g}^{\left[1\right]}\left(a\right)+{\alpha }_{3}g\left(b\right)\right]$, $z_2=J\left[{\beta }_{1}g\left(a\right)+{\beta }_{2}g\left(b\right)+{\beta }_3{g}^{\left[1\right]}\left(b\right)\right]$,

$z_3=J\left[{\theta }_{1}g\left(c-\right)+{\theta }_2{g}^{\left[1\right]}\left(c-\right)+{\theta }_{3}g\left(c+\right)\right]$, $z_4=J\left[{\sigma }_{1}g\left(c-\right)+{\sigma }_{2}g\left(c+\right)+{\sigma }_3{g}^{\left[1\right]}\left(c+\right)\right]$;

(iv) $z\left(x\right)=-{\left(p{g}^{\prime }\right)}^{\prime }+qg$。

对 $\forall F\in C_0^{\infty }\oplus {\left\{0\right\}}^4\subset D\left(T\right)$，由 $\langle TF,JG\rangle =\langle F,Z\rangle$可得

$\rho {\displaystyle {\int }_a^{c}l\left(f\right)gw\text{d}x}+\eta {\displaystyle {\int }_c^{b}l\left(f\right)gw\text{d}x}=\rho {\displaystyle {\int }_a^{c}f\bar{z}w\text{d}x}+\eta {\displaystyle {\int }_c^{b}f\bar{z}w\text{d}x}$,

即 ${\langle \frac{l\left(f\right)}{w},Jg\rangle }_w={\langle f,z\rangle }_w$，由标准Sturm-Liouville理论，(i)和(iv)成立。

由 $\langle TF,JG\rangle =\langle F,Z\rangle$及(iv)可知

$\begin{array}{l}{\langle \frac{l\left(f\right)}{w},g\rangle }_w+m_1{M}_1\left(f\right)\overline{J{g}_1}+m_2{M}_2\left(f\right)\overline{J{g}_2}+m_3{N}_1\left(f\right)\overline{J{g}_3}+m_4{N}_2\left(f\right)\overline{J{g}_4}\\ ={\langle f,\frac{Jl\left(g\right)}{w}\rangle }_w+m_1{M^{\prime }}_1\left(f\right)\overline{z_1}+m_2{M^{\prime }}_2\left(f\right)\overline{z_2}+m_3{N^{\prime }}_1\left(f\right)\overline{z_3}+m_4{N^{\prime }}_2\left(f\right)\overline{z_4},\end{array}$

所以

$\begin{array}{c}{\langle \frac{l\left(f\right)}{w},Jg\rangle }_w={\langle f,\frac{Jl\left(g\right)}{w}\rangle }_w+m_1\left[{M^{\prime }}_1\left(f\right)\overline{z_1}-M_1\left(f\right)g_1\right]+m_2\left[{M^{\prime }}_2\left(f\right)\overline{z_2}-M_2\left(f\right)g_2\right]\\ +m_3\left[{N^{\prime }}_1\left(f\right)\overline{z_3}-N_1\left(f\right)g_3\right]+m_4\left[{N^{\prime }}_2\left(f\right)\overline{z_4}-N_2\left(f\right)g_4\right].\end{array}$

另一方面，

${\langle \frac{l\left(f\right)}{w},Jg\rangle }_w={\langle f,\frac{Jl\left(g\right)}{w}\rangle }_w+\rho \left[f,g\right]\left(c-\right)-\rho \left[f,g\right]\left(c-\right)+\eta \left[f,g\right]\left(b\right)-\eta \left[f,g\right]\left(c+\right),$

因此，结合(7)式有

$\begin{array}{l}{m}_1\left[{M^{\prime }}_1\left(f\right)\overline{z_1}-M_1\left(f\right)g_1\right]+m_2\left[{M^{\prime }}_2\left(f\right)\overline{z_2}-M_2\left(f\right)g_2\right]\\ +m_3\left[{N^{\prime }}_1\left(f\right)\overline{z_3}-N_1\left(f\right)g_3\right]+m_4\left[{N^{\prime }}_2\left(f\right)\overline{z_4}-N_2\left(f\right)g_4\right]\\ =\rho \left[f,g\right]\left(c-\right)-\rho \left[f,g\right]\left(a\right)+\eta \left[f,g\right]\left(b\right)-\eta \left[f,g\right]\left(c+\right)\\ =\rho {\mathcal{G}}^{\text{T}}\left(c-\right)ℰℱ\left(c-\right)-\rho {\mathcal{G}}^{\text{T}}\left(a\right)ℰℱ\left(a\right)+\eta {\mathcal{G}}^{\text{T}}\left(b\right)ℰℱ\left(b\right)-\eta {\mathcal{G}}^{\text{T}}\left(c+\right)ℰℱ\left(c+\right).\end{array}$ (8)

其中， $ℱ={\left(f,f^{\left[1\right]}\right)}^{\text{T}}$, $\mathcal{G}=\left(g,g^{\left[1\right]}\right)$, $ℰ=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$。

根据引理2，结合J-自伴算子的定义可知，存在函数 $ℱ\in D\left(T\right)$，使得

$ℱ\left(c+\right)=ℱ\left(c-\right)=ℱ\left(b\right)=0,f\left(a\right)=0,f^{\prime }\left(a\right)=-\frac{{\gamma }_1}{\rho }$,

此时

${M^{\prime }}_1\left(f\right)={M^{\prime }}_2\left(f\right)={N^{\prime }}_1\left(f\right)={N^{\prime }}_2\left(f\right)=M_2\left(f\right)=N_1\left(f\right)=N_2\left(f\right)=0,M\left(f\right)=-\frac{{\alpha }_2{\gamma }_1}{\rho }$,

于是结合(8)式可知 $g_1={\gamma }_{1}g\left(a\right)$。同理，根据引理2，令

$ℱ\left(c+\right)=ℱ\left(c-\right)=ℱ\left(a\right)=0,f\left(b\right)=0,f^{\left[1\right]}\left(b\right)=-\frac{{\gamma }_2}{\eta },$ (9)

$ℱ\left(c+\right)=ℱ\left(a\right)=ℱ\left(b\right)=0,f\left(c-\right)=0,f^{\left[1\right]}\left(c-\right)=-\frac{{\delta }_1}{\rho },$ (10)

$ℱ\left(c+\right)=ℱ\left(a\right)=ℱ\left(b\right)=0,f\left(c-\right)=0,f^{\left[1\right]}\left(c-\right)=-\frac{{\delta }_1}{\rho },$ (11)

将(9)~(11)式分别代入(8)式，即可求得

$g_2={\gamma }_{2}g\left(b\right),g_3={\delta }_{1}g\left(c-\right),g_4={\delta }_{2}g\left(c+\right),$

所以(ii)成立。

利用与(ii)的证明类似的方法，可以证明(iii)成立。事实上，根据引理2，选取

$ℱ\left(c+\right)=ℱ\left(c-\right)=0,f\left(b\right)=0,f\left(a\right)=1,f^{\left[1\right]}\left(a\right)=-\frac{{\alpha }_1}{{\alpha }_2},f^{\left[1\right]}\left(b\right)=-\frac{{\beta }_1}{{\beta }_3},$ (12)

$ℱ\left(c+\right)=ℱ\left(c-\right)=0,f\left(a\right)=0,f\left(b\right)=1,f^{\left[1\right]}\left(a\right)=-\frac{{\alpha }_3}{{\alpha }_2},f^{\left[1\right]}\left(b\right)=-\frac{{\beta }_2}{{\beta }_3},$ (13)

$ℱ\left(a\right)=ℱ\left(b\right)=0,f\left(c+\right)=0,f\left(c-\right)=1,f^{\left[1\right]}\left(c-\right)=-\frac{{\theta }_1}{{\theta }_2},f^{\left[1\right]}\left(c+\right)=-\frac{{\sigma }_1}{{\sigma }_3},$ (14)

$ℱ\left(a\right)=ℱ\left(b\right)=0,f\left(c-\right)=0,f\left(c+\right)=1,f^{\left[1\right]}\left(c-\right)=-\frac{{\theta }_3}{{\theta }_2},f^{\left[1\right]}\left(c+\right)=-\frac{{\sigma }_2}{{\sigma }_3},$ (15)

将(12)~(15)式分别代入(8)式，可以得到

$\overline{z_1}={\alpha }_{1}g\left(a\right)+{\alpha }_2{g}^{\left[1\right]}\left(a\right)+{\alpha }_{3}g\left(b\right),\overline{z_2}={\beta }_{1}g\left(a\right)+{\beta }_{2}g\left(b\right)+{\beta }_3{g}^{\left[1\right]}\left(b\right)$,

$\overline{z_3}={\theta }_{1}g\left(c-\right)+{\theta }_2{g}^{\left[1\right]}\left(c-\right)+{\theta }_{3}g\left(c+\right),\overline{z_4}={\sigma }_{1}g\left(c-\right)+{\sigma }_{2}g\left(c+\right)+{\sigma }_3{g}^{\left[1\right]}\left(c+\right)$,

所以(iii)成立。

综上所述，算子T是J-自伴算子。

根据定义3及Hilbert空间 $\mathscr{H}$中内积的定义可得：

推论1若 ${\lambda }_1$和 ${\lambda }_2$是算子T两个不同的特征值，则相应的特征函数 $f\left(x\right)$与 $g\left(x\right)$在如下内积意义下是C-正交的：

$\rho {\displaystyle {\int }_a^{c}fgw\text{d}x}+\eta {\displaystyle {\int }_c^{b}fgw\text{d}x}+m_1{M^{\prime }}_1\left(f\right){M^{\prime }}_1\left(g\right)+m_2{M^{\prime }}_2\left(f\right){M^{\prime }}_2\left(g\right)+m_3{N^{\prime }}_1\left(f\right){N^{\prime }}_1\left(g\right)+m_4{N^{\prime }}_2\left(f\right){N^{\prime }}_2\left(g\right)=0$.

首先，定义微分方程(1)的基本解为 $\phi \left(x,\lambda \right)=\{\begin{array}{l}{\phi }_1\left(x,\lambda \right),x\in \left[a,c\right)\\ {\phi }_2\left(x,\lambda \right),x\in \left(c,b\right]\end{array}$和 $\chi \left(x,\lambda \right)=\{\begin{array}{l}{\chi }_1\left(x,\lambda \right),x\in \left[a,c\right)\\ {\chi }_2\left(x,\lambda \right),x\in \left(c,b\right]\end{array}$。

设 ${\phi }_1\left(x,\lambda \right),{\chi }_1\left(x,\lambda \right)$是方程(1)在区间 $\left[a,c\right)$上满足如下初始条件的线性无关解

${\phi }_1\left(a,\lambda \right)=1,{\phi }_1^{\left[1\right]}\left(a,\lambda \right)=0,{\chi }_1\left(a,\lambda \right)=0,{\chi }_1^{\left[1\right]}\left(a,\lambda \right)=1,$

其中 ${\phi }^{\left[1\right]}=p{\phi }^{\prime }$，它们的Wronski行列式独立于变量x，记为 ${\omega }_1\left(\lambda \right)$，且

${\omega }_1\left(\lambda \right)={{\omega }_1\left(\lambda \right)|}_{x=a}=\left|\begin{array}{cc}{\phi }_1\left(a,\lambda \right)& {\phi }_1^{\left[1\right]}\left(a,\lambda \right)\\ {\chi }_1\left(a,\lambda \right)& {\chi }_1^{\left[1\right]}\left(a,\lambda \right)\end{array}\right|=1$.

设 ${\phi }_2\left(x,\lambda \right),{\chi }_2\left(x,\lambda \right)$是方程(1)在区间 $\left(c,b\right]$上满足如下初始条件的解

${\phi }_2\left(c+,\lambda \right)=-\frac{1}{{\theta }_3}\left[\left({\theta }_1-\lambda {\delta }_1\right){\phi }_1\left(c-,\lambda \right)+{\theta }_2{\phi }_1^{\left[1\right]}\left(c-,\lambda \right)\right],$ (16)

${\phi }_2^{\left[1\right]}\left(c+,\lambda \right)=\frac{1}{\eta }\left\{\left[\left({\sigma }_2-\lambda {\delta }_2\right)\left({\theta }_1-\lambda {\delta }_1\right)-{\theta }_3{\sigma }_1\right]{\phi }_1\left(c-,\lambda \right)+\left({\sigma }_2-\lambda {\delta }_2\right){\theta }_2{\phi }_1^{\left[1\right]}\left(c-,\lambda \right)\right\},$ (17)

${\chi }_2\left(c+,\lambda \right)=-\frac{1}{{\theta }_3}\left[\left({\theta }_1-\lambda {\delta }_1\right){\chi }_1\left(c-,\lambda \right)+{\theta }_2{\chi }_1^{\left[1\right]}\left(c-,\lambda \right)\right],$ (18)

${\chi }_2^{\left[1\right]}\left(c+,\lambda \right)=\frac{1}{\eta }\left\{\left[\left({\sigma }_2-\lambda {\delta }_2\right)\left({\theta }_1-\lambda {\delta }_1\right)-{\theta }_3{\sigma }_1\right]{\chi }_1\left(c-,\lambda \right)+\left({\sigma }_2-\lambda {\delta }_2\right){\theta }_2{\chi }_1^{\left[1\right]}\left(c-,\lambda \right)\right\}.$ (19)

函数 ${\phi }_2\left(x,\lambda \right),{\chi }_2\left(x,\lambda \right)$的Wronski行列式独立于变量x，记为 ${\omega }_2\left(\lambda \right)$。

引理4对任意的 $\lambda \in ℂ$，有 $\eta {\omega }_2\left(\lambda \right)=\rho {\omega }_1\left(\lambda \right)$成立。

证明 因为Wronski行列式独立于变量x，所以有

${\omega }_2\left(\lambda \right)={{\omega }_2\left(\lambda \right)|}_{x=c}=\left|\begin{array}{cc}{\phi }_2\left(c+,\lambda \right)& {\phi }_2^{\left[1\right]}\left(c+,\lambda \right)\\ {\chi }_2\left(c+,\lambda \right)& {\chi }_2^{\left[1\right]}\left(c+,\lambda \right)\end{array}\right|,$

将(16)~(19)代入上式，经计算化简得

$\begin{array}{c}{\omega }_2\left(\lambda \right)=-\frac{1}{{\theta }_3\eta }\{\left|\begin{array}{cc}{\theta }_2{\phi }_1^{\left[1\right]}\left(c-,\lambda \right)& 0\\ \left({\sigma }_2-\lambda {\delta }_2\right){\theta }_2{\phi }_1^{\left[1\right]}\left(c-,\lambda \right)& -{\theta }_3{\sigma }_1{\chi }_1\left(c-,\lambda \right)\end{array}\right|\\ +\left|\begin{array}{cc}0& {\theta }_2{\chi }_1^{\left[1\right]}\left(c-,\lambda \right)\\ -{\theta }_3{\sigma }_1{\phi }_1\left(c-,\lambda \right)& \left({\sigma }_2-\lambda {\delta }_2\right){\theta }_2{\chi }_1^{\left[1\right]}\left(c-,\lambda \right)\end{array}\right|\}\\ =\frac{\rho }{\eta }{\omega }_1\left(\lambda \right).\end{array}$

引理5设

$y\left(x,\lambda \right)=\{\begin{array}{l}{y}_1\left(x,\lambda \right),x\in \left[a,c\right),\\ y_2\left(x,\lambda \right),x\in \left(c,b\right],\end{array}$

是微分方程(1)的任意一个解，则此解可以表示为

$y\left(x,\lambda \right)=\{\begin{array}{l}{d}_1{\phi }_1\left(x,\lambda \right)+d_2{\chi }_1\left(x,\lambda \right),x\in \left[a,c\right),\\ d_3{\phi }_2\left(x,\lambda \right)+d_4{\chi }_2\left(x,\lambda \right),x\in \left(c,b\right],\end{array}$

若 $y\left(x,\lambda \right)$满足转移条件(3)，则 $d_1=d_3,d_2=d_4$。

证明 将 $y\left(x,\lambda \right)$的表达式代入转移条件(3)，有

$\left(\begin{array}{cc}{\theta }_1-\lambda {\delta }_1& {\theta }_2\\ {\sigma }_1& 0\end{array}\right)\left(\begin{array}{c}{d}_1{\phi }_1\left(c-,\lambda \right)+d_2{\chi }_1\left(c-,\lambda \right)\\ d_1{\phi }_1^{\left[1\right]}\left(c-,\lambda \right)+d_2{\chi }_1^{\left[1\right]}\left(c-,\lambda \right)\end{array}\right)+\left(\begin{array}{cc}{\theta }_3& 0\\ {\sigma }_2-\lambda {\delta }_2& {\sigma }_3\end{array}\right)\left(\begin{array}{c}{d}_3{\phi }_2\left(c+,\lambda \right)+d_4{\chi }_2\left(c+,\lambda \right)\\ d_3{\phi }_2^{\left[1\right]}\left(c+,\lambda \right)+d_4{\chi }_2^{\left[1\right]}\left(c+,\lambda \right)\end{array}\right)=0,$ (20)

将(16)~(19)代入(20)式，得到关于 $d_1-d_3,d_2-d_4$的方程组

$\{\begin{array}{l}\left[\left({\theta }_1-\lambda {\delta }_1\right){\phi }_1\left(c-,\lambda \right)+{\theta }_2{\phi }_1^{\left[1\right]}\left(c-,\lambda \right)\right]\left(d_1-d_3\right)+\left[\left({\theta }_1-\lambda {\delta }_1\right){\chi }_1\left(c-,\lambda \right)+{\theta }_2{\chi }_1^{\left[1\right]}\left(c-,\lambda \right)\right]\left(d_2-d_4\right)=0,\\ {\sigma }_1{\phi }_1\left(c-,\lambda \right)\left(d_1-d_3\right)+{\sigma }_1{\chi }_1\left(c-,\lambda \right)\left(d_2-d_4\right)=0.\end{array}$ (21)

方程组(21)的系数行列式为

$\left|\begin{array}{cc}\left({\theta }_1-\lambda {\delta }_1\right){\phi }_1\left(c-,\lambda \right)+{\theta }_2{\phi }_1^{\left[1\right]}\left(c-,\lambda \right)& \left({\theta }_1-\lambda {\delta }_1\right){\chi }_1\left(c-,\lambda \right)+{\theta }_2{\chi }_1^{\left[1\right]}\left(c-,\lambda \right)\\ {\sigma }_1{\phi }_1\left(c-,\lambda \right)& {\sigma }_1{\chi }_1\left(c-,\lambda \right)\end{array}\right|=\rho {\omega }_1\left(\lambda \right)\ne 0,$

从而可得 $d_1=d_3,d_2=d_4$。

设 ${\text{Φ}}_1\left(x,\lambda \right)=\left(\begin{array}{cc}{\phi }_1\left(x,\lambda \right)& {\chi }_1\left(x,\lambda \right)\\ {\phi }_1^{\left[1\right]}\left(x,\lambda \right)& {\chi }_1^{\left[1\right]}\left(x,\lambda \right)\end{array}\right)$， ${\text{Φ}}_2\left(x,\lambda \right)=\left(\begin{array}{cc}{\phi }_2\left(x,\lambda \right)& {\chi }_2\left(x,\lambda \right)\\ {\phi }_2^{\left[1\right]}\left(x,\lambda \right)& {\chi }_2^{\left[1\right]}\left(x,\lambda \right)\end{array}\right)$，则有