众所周知, 三阶微分方程起源于应用数学和物理学的许多不同领域中, 如带有固定或变化横截面的屈曲梁的挠度、三层梁、地球引力吹积的涨潮等. 因此, 对于三阶微分算子边值问题的研究越来越受到了人们广泛的关注, 从正则的到具有转移条件的, 再到一个边界条件中含有谱参数的情形, 都有一些相应的研究结论. 如: 2019年, 牛天在其硕士论文^[1]中得到了三阶正则微分算子自伴边界条件标准型的完全描述, 同时, 利用亏指数理论将正则情况推广到奇异情况, 最后研究了实耦合型自伴边界条件下三阶正则微分算子特征值关于问题的依赖性. 同年, Uǧurlu在文献[2]中考虑了三阶正则微分表达式分别同分离, 实耦合和复耦合边界条件所产生的边值问题, 验证了问题的自伴性以及问题的特征值对条件中给定参数的依赖性. Uǧurlu在文献[3]中考虑的是具有转移条件的情形, 证明了该类算子特征值是实的, 同时给出特征值关于给定参数的微分表达式. 在上述文献的基础上, 我们研究了几类三阶微分算子, 其中包括一个边界条件中含有谱参数的情形, 具有转移条件且一个边界条件中含有谱参数的情形.

本文拟讨论一类具有转移条件且两个边界条件带有谱参数的三阶微分算子. 类似于文献[4-6]的方法, 根据给定问题构造新空间和新算子, 使得新空间上新算子的特征值和原问题的特征值是一致的; 然后证明新算子定义域的稠密性, 新算子的对称性, 进而证明该算子的自伴性, 从而推出特征值是实的. 最后研究了所考虑问题的特征值的性质, 得到问题的谱只有点谱.

本文考虑三阶微分方程

(1.1) $ \begin{equation} ly = \frac{1}{w} (-{\rm i} y^{(3)}+q y) = \lambda y, x \in J, \end{equation} $

其中$ J = [a, c)\cup (c, b] $, $ -\infty<a<c<b<\infty $, 其中$ q $和$ w $是实值连续函数, $ w(x) $几乎处处大于零, $ \lambda \in {\Bbb C} $是谱参数. 对应的边界条件为

(1.2) $ \begin{eqnarray} && l_{1} y = \left(\alpha_{1} \lambda+\beta_{1}\right) y(a)-\left(\alpha_{2} \lambda+\beta_{2}\right) y^{[2]}(a) = 0, \end{eqnarray} $

(1.3) $ \begin{eqnarray} &&l_{2} y = ({\rm i}+\tan \gamma) y^{[1]}(a)-(1+{\rm i} \tan \gamma) y^{[1]}(b) = 0, \end{eqnarray} $

(1.4) $ \begin{eqnarray} && l_{3} y = \left(\alpha_{3} \lambda+\beta_{3}\right) y(b)-\left(\alpha_{4} \lambda+\beta_{4}\right) y^{[2]}(b) = 0, \end{eqnarray} $

其中$ \alpha_{i}, \beta_{i}, \gamma $均属于$ {{\Bbb R}}\ (i = 1, 2) $.

转移条件为

(1.5) $ \begin{eqnarray} &&l_{4}y = y(c+)-\alpha_{11} y(c-)+\alpha_{12} y^{[2]}(c-) = 0, \end{eqnarray} $

(1.6) $ \begin{eqnarray} && l_{5}y = y^{[1]}(c+)-y^{[1]}(c-) = 0, \end{eqnarray} $

(1.7) $ \begin{eqnarray} &&l_{6}y = y^{[2]}(c+)-\beta_{11} y(c-)+\beta_{12} y^{[2]}(c-) = 0, \end{eqnarray} $

其中拟导数定义为

(1.8) $ \begin{equation} y^{[0]} = y, y^{[1]} = -\frac{1+{\rm i}}{\sqrt{2}} y^{\prime}, y^{[2]} = {\rm i}y^{\prime \prime}-{\rm i}qy, \end{equation} $

$ y^{[0]}, y^{[1]}, y^{[2]}\in AC[J] $且$ ly\in L_{w}^{2}(J) $.

假设

$ \rho_{1} = \alpha_{2} \beta_{1}-\alpha_{1} \beta_{2}>0, \qquad \rho_{2} = \alpha_{3} \beta_{4}-\alpha_{4} \beta_{3}>0, \qquad \theta = \alpha_{12} \beta_{11}-\alpha_{11} \beta_{12}>0, $

为了更方便的考虑边值问题(1.1)–(1.7), 在$ H_{1} = L_{w}^{2} $上定义内积如下

(1.9) $ \begin{equation} [f(x), g(x)]_{1} = \int_{a}^{c} f_{1}(x) \bar{g}_{1}(x) w_{1}(x){\rm d}x+\int_{c}^{b} f_{2}(x) \bar{g}_{2}(x) w_{2}(x){\rm d}x, \end{equation} $

其中$ f(x) = \left\{\begin{array}{l} f_{1}(x), x \in[a, c) \\ f_{2}(x), x \in(c, b] \end{array}\right., $$ g(x) = \left\{\begin{array}{l} g_{1}(x), x \in[a, c) \\ g_{2}(x), x \in(c, b] \end{array}\right., $$ w(x) = \left\{\begin{array}{l} w_{1}(x), x \in[a, c) \\ w_{2}(x), x \in(c, b] \end{array}\right., $易证$ (H_{1}, [\cdot, \cdot]_{1}) $是Hilbert空间.

令$ H = H_{1} \oplus {\Bbb C}_{\rho_{1}} \oplus {\Bbb C}_{\rho_{2}} $, 对于任意的$ F = (f(x), h, r)^{T}, G = (g(x), k, s)^{T}, h, r, k, s $均为复数, 在$ H $内定义内积如下

(1.10) $ \begin{equation} [F, G] = [f(x), g(x)]_{1}+\frac{1}{\rho_{1}} h \bar{k}+\frac{1}{\rho_{2}} r \bar{s}, \end{equation} $

显然$ H $是一个Hilbert空间. 定义算子$ A $如下

$ \begin{eqnarray*} A\left(\begin{array}{c} f \\ \alpha_{1} f(a)-\alpha_{2} f^{[2]}(a)\\ \alpha_{3} f(b)-\alpha_{4} f^{[2]}(b) \end{array}\right) = \left(\begin{array}{c} w^{-1}\left\{-{\rm i}{f}^{(3)}+q f\right\} \\ -\beta_{1} f(a)+\beta_{2} f^{[2]}(a)\\ -\beta_{3} f(b)+\beta_{4} f^{[2]}(b) \end{array}\right), \end{eqnarray*} $

$ D(A) = \left\{(f(x), h, r)^{T} \in H \Bigg| \begin{array}{l} f_{1}(x), f_{1}^{[1]}(x), f_{1}^{[2]}(x) \in A C[a, c); f_{2}(x), f_{2}^{[1]}(x), \\ f_{2}^{[2]}(x) \in A C(c, b], l(f)\in H_{1}, l_{i}(f) = 0(i = 2, 4, 5, 6), \\ h = \alpha_{1} f(a)-\alpha_{2} f^{[2]}(a), r = \alpha_{3} f(b)-\alpha_{4} f^{[2]}(b). \end{array}\right\}, $

显然有如下结论.

定理1.1  问题$ (1.1) $–$ (1.7) $的特征值与算子$ A $的特征值一致, 且特征函数是算子$ A $的相应的特征函数的第一个分量.

根据文献[3], 有拉格朗日恒等式

(1.11) $ \begin{equation} [Ly, z]_{1}-[y, Lz]_{1} = [y \bar{z}]|_{a}^{c-}+[y \bar{z}]|_{c+}^{b}, \end{equation} $

并有$ [y \bar{z}]|_{t_{1}}^{t_{2}} = [y \bar{z}]\left(t_{2}\right)-[y \bar{z}]\left(t_{1}\right) $, 其中

(1.12) $ \begin{equation} [y z] = yz^{[2]}-y^{[2]}z+{\rm i}{y}^{[1]}z^{[1]}, \end{equation} $

此处$ y^{[r]} $为$ y $的$ r(r = 0, 1, 2) $阶拟导数, 见定义(1.8).

定理2.1  算子$ A $的定义域$ D(A) $在$ H $中稠密.

证  假设$ D(A) $在$ H $中按内积(1.9)–(1.10)是不稠密的, 那么一定存在一个非零元素$ F \in H $, 对于$ \forall G \in D(A) $, 有$ [F, G] = 0 $. 其中$ F = (f(x), h, r)^{T}, G = (g(x), k, s)^{T} $.

令$ C_{0}^{\infty} $表示如下函数集合

$ \varphi(x) = \left\{\begin{array}{l} \varphi_{1}(x), x \in[a, c), \\ \varphi_{2}(x), x \in(c, b], \end{array}\right. $

其中$ \varphi_{1}(x) \in C_{0}^{\infty}[a, c), \varphi_{2}(x)\in C_{0}^{\infty}(c, b] $, 显然$ C_{0}^{\infty}\oplus \{0\}\oplus \{0\}\subseteq D(A) $.

设$ U = (u(x), 0, 0)^{T}\in C_{0}^{\infty}\oplus \{0\}\oplus \{0\} $, 则$ F\perp U $. 根据内积定义有下列式子成立

$ [F, U] = \int_{a}^{c}f_{1}(x)\bar{u}_{1}(x)w_{1}(x){\rm d}x+\int_{c}^{b}f_{2}(x)\bar{u}_{2}(x)w_{2}(x){\rm d}x = 0, $

又知$ \bar{C}_{0}^{\infty} = H_{1} $, 故$ f_1(x) = 0, f_2(x) = 0 $, 即$ f(x) = 0 $.

因$ G = (g(x), k, s)^{T}\in D(A) $, 有

$ \begin{eqnarray*} [F, G] = [f(x), g(x)]_{1}+\frac{1}{\rho_{1}}h(\bar{k})+\frac{1}{\rho_{2}}r(\bar{s}) = \frac{1}{\rho_{1}}h(\bar{k})+\frac{1}{\rho_{2}} r(\bar{s}) = 0, \end{eqnarray*} $

由于$ k, s $是任意选取的, 故$ h = 0, r = 0 $. 于是$ F = (f(x), h, r)^{T} = (0, 0, 0)^{T} $, 这与最初的假设矛盾, 综上$ D(A) $在$ H $中是稠密的.

定理2.2  线性算子$ A $是对称的.

证  对于$ F, G\in D(A) $,

$ F = \left(f(x), \alpha_{1} f(a)-\alpha_{2} f^{[2]}(a), \alpha_{3} f(b)-\alpha_{4}f^{[2]}(b)\right)^{T}, $

$ G = \left(g(x), \alpha_{1} g(a)-\alpha_{2}g^{[2]}(a), \alpha_{3} g(b)-\alpha_{4} g^{[2]}(b)\right)^{T}, $

由$ A $的定义, 知

$ AF = \left(l f(x), \beta_{2} f^{[2]}(a)-\beta_{1} f(a), \beta_{4} f^{[2]}(b)-\beta_{3} f(b)\right)^{T}, $

$ AG = \left(l g(x), \beta_{2} g^{[2]}(a)-\beta_{1} g(a), \beta_{4} g^{[2]}(b)-\beta_{3} g(b)\right)^{T}, $

(2.1) $ \begin{eqnarray} [AF, G]& = &[lf(x), g(x)]_{1}+\frac{1}{\rho_{1}}\left(\beta_{2} f^{[2]}(a)-\beta_{1} f(a)\right)\left(\alpha_{1} \bar{g}(a)-\alpha_{2} \bar{g}^{[2]}(a)\right) \\ &&+\frac{1}{\rho_{2}}\left(\beta_{4}f^{[2]}(b)-\beta_{3}f(b)\right)\left(\alpha_{3} \bar{g}(b)-\alpha_{4}\bar{g}^{[2]}(b)\right), \end{eqnarray} $

(2.2) $ \begin{eqnarray} [F, AG]& = &[f(x), lg(x)]_{1}+\frac{1}{\rho_{1}}\left(\alpha_{1}f(a)-\alpha_{2}f^{[2]}(a)\right)\left(\beta_{2}\bar{g}^{[2]}(a)-\beta_{1} \bar{g}(a)\right) \\ &&+\frac{1}{\rho_{2}}\left(\alpha_{3}f(b)-\alpha_{4}f^{[2]}(b)\right)\left(\beta_{4}\bar{g}^{[2]}(b)-\beta_{3}\bar{g}(b)\right), \end{eqnarray} $

根据拉格朗日恒等式、边界条件以及转移条件, 有下式成立

$ \begin{array}{c} [AF, G]-[F, AG] = f(c-)\bar{g}^{[2]}(c-)-f^{[2]}(c-)\bar{g}(c-)-\left(f(c-)\bar{g}^{[2]}(c-)-f^{[2]}(c-)\bar{g}(c-)\right) = 0, \end{array} $

因此$ [AF, G] = [F, AG] $, 故算子$ A $是对称的.

定理2.3   线性算子$ A $是自伴的.

证  由于$ A $是对称的, 要证$ A $在$ H $中是自伴的, 只需要证明: 若对于任意的$ F = \left(f(x), \alpha_{1}f(a)-\alpha_{2}f^{[2]}(a), \alpha_{3}f(b)-\alpha_{4}f^{[2]}(b)\right)^{T}\in D(A) $, 有若: $ [AF, V] = [F, U] $成立, 则$ V\in D(A) $且$ AV = U $ (其中$ V = (v(x), m_{1}, m_{2})^{T}, U = (u(x), n_{1}, n_{2})^{T} $).

要证如上结论成立, 即要证明如下七条成立:

(1) $ v_{1}(x), v_{1}^{[1]}(x), v_{1}^{[2]}(x)\in AC[a, c), v_{2}(x), v_{2}^{[1]}(x), v_{2}^{[2]}(x)\in AC(c, b], lv(x)\in H_{1} $;

(2) $ m_{1} = \alpha_{1}v(a)-\alpha_{2}v^{[2]}(a) $;

(3) $ m_{2} = \alpha_{3}v(b)-\alpha_{4}v^{[2]}(b) $;

(4) $ l_{2}v = l_{4}v = l_{5}v = l_{6}v = 0 $;

(5) $ u(x) = lv(x) $;

(6) $ n_{1} = \beta_{2}u^{[2]}(a)-\beta_{1}u(a) $;

(7) $ n_{2} = \beta_{4}u^{[2]}(b)-\beta_{3}u(b) $;

具体证明过程如下: 对于任意的$ F = (f(x), 0, 0)\in D(A) $, 有$ [AF, V] = [F, U] $, 根据内积的定义, 这时

$ {[AF, V] = [lf(x), v(x)]_{1}, [F, U] = [f(x), u(x)]_{1}}, $

即$ [lf(x), v(x)]_{1} = [f(x), u(x)]_{1}, $由标准的Sturm-Liouville理论可知$ v(x)\in D(l) $, 故(1)成立.

因为算子$ A $的对称性, 所以有$ [lf(x), v(x)]_{1} = [f(x), lv(x)]_{1} $, 又根据$ (1) $中的证明有$ [lf(x), v(x)]_{1} = [f(x), u(x)]_{1} $, 综上所述(5)得证.

对于任意的$ F\in D(A) $, 方程$ [AF, V] = [F, U] $, 有下式成立

$ \begin{eqnarray*} &&[lf(x), v(x)]_{1}+\frac{1}{\rho_{1}}\left(\beta_{2}f^{[2]}(a)-\beta_{1}f(a)\right)\bar{m}_{1}+\frac{1}{\rho_{2}}\left(\beta_{4} f^{[2]}(b)-\beta_{3}f(b)\right)\bar{m}_{2}, \\ & = &[f(x), u(x)]_{1}+\frac{1}{\rho_{1}}\left(\alpha_{1}f(a)-\alpha_{2}f^{[2]}(a)\right)\bar{n}_{1}+\frac{1}{\rho_{2}}\left(\alpha_{3} f(b)-\alpha_{4}f^{[2]}(b)\right)\bar{n}_{2}, \end{eqnarray*} $

再由$ (5) $可知

(2.3) $ \begin{eqnarray} [lf(x), v(x)]_{1}-[f(x), lv(x)]_{1} & = &\frac{1}{\rho_{1}}[\left(\alpha_{1}f(a)-\alpha_{2}f^{[2]}(a)\right)\bar{n}_{1}-\left(\beta_{2} f^{[2]}(a)-\beta_{1}f(a)\right)\bar{m}_{1}] \\ &&+\frac{1}{\rho_{2}}[\left(\alpha_{3}f(b)-\alpha_{4}f^{[2]}(b)\right)\bar{n}_{2}-\left(\beta_{4}f^{[2]}(b)-\beta_{3}f(b)\right)\bar{m}_{2}], {\qquad} \end{eqnarray} $

又由算子$ A $的对称性, 有

(2.4) $ \begin{eqnarray} [lf(x), v(x)]_{1}-[f(x), lv(x)]_{1} = f^{[2]}(a)\bar{v}(a)-f(a)\bar{v}^{[2]}(a)+f(b)\bar{v}^{[2]}(b) -f^{[2]}(b)\bar{v}(b), \end{eqnarray} $

由$ (2.3) = (2.4), $根据Naimark Patching Lemma, 存在这样的$ F = (f(x), \alpha_{1}f(a)-\alpha_{2}f^{[2]}(a), \alpha_{3}f(b)-\alpha_{4}f^{[2]}(b))^{T}\in D(A) $, 使得

$ f(a) = f^{[1]}(a) = f^{[2]}(a) = f(c-) = f^{[1]}(c-) = f^{[2]}(c-) = f(c+) = f^{[1]}(c+) = f^{[2]}(c+) = 0, $

有下式成立

$ f(b)\bar{v}^{[2]}(b)-f^{[2]}(b)\bar{v}(b) = \frac{1}{\rho_{2}}[\left(\alpha_{3}f(b)-\alpha_{4}f^{[2]}(b)\right)\bar{n}_{2}-\left(\beta_{4}f^{[2]}(b)-\beta_{3}f(b)\right)\bar{m}_{2}], $

即有$ n_{2} = \beta_{4}v^{[2]}(b)-\beta_{3}v(b), m_{2} = \alpha_{3}v(b)-\alpha_{4}v^{[2]}(b) $, (3)(7) 成立, 利用同样的方法我们可得(2)和(6).

现在只需证明$ (4) $成立. 由(1.11)和(1.12)式可知

(2.5) $ \begin{eqnarray} &&[lf(x), v(x)]_{1}-[f(x), lv(x)]_{1} \\ & = &f(b)\bar{v}^{[2]}(b) -f^{[2]}(b)\bar{v}(b)+{\rm i} f^{[1]}(b)\bar{v}^{[1]}(b)+f^{[2]}(a)\bar{v}(a)-f(a)\bar{v}^{[2]}(a)\\ &&-{\rm i} f^{[1]}(a)\bar{v}^{[1]}(a)+f(c-)\bar{v}^{[2]}(c-) -f^{[2]}(c-)\bar{v}(c-)+{\rm i} f^{[1]}(c-)\bar{v}^{[1]}(c-)\\ &&+f^{[2]}(c+)\bar{v}(c+)-f(c+)\bar{v}^{[2]}(c+)-{\rm i} f^{[1]}(c+)\bar{v}^{[1]}(c+), \end{eqnarray} $

结合(2.4)和(2.5)式可得

(2.6) $ \begin{eqnarray} &&{\rm i} f^{[1]}(b)\bar{v}^{[1]}(b)-{\rm i} f^{[1]}(a)\bar{v}^{[1]}(a) +f(c-)\bar{v}^{[2]}(c-) -f^{[2]}(c-)\bar{v}(c-)\\ && +{\rm i} f^{[1]}(c-)\bar{v}^{[1]}(c-)+f^{[2]}(c+)\bar{v}(c+) -f(c+)\bar{v}^{[2]}(c+)-{\rm i} f^{[1]}(c+)\bar{v}^{[1]}(c+) = 0, \end{eqnarray} $

根据Naimark Patching Lemma, 存在这样的$ F = (f(x), \alpha_{1}f(a)-\alpha_{2}f^{[2]}(a), \alpha_{3}f(b)-\alpha_{4}f^{[2]}(b))^{T} $$ \in D(A) $, 使得$ f(c-) = f^{[1]}(c-) = f^{[2]}(c-) = f(c+) = f^{[1]}(c+) = f^{[2]}(c+) = 0, f^{[1]}(a) = ({\rm i}+\tan\gamma), $$ f^{[1]}(b) = 1+{\rm i}\tan\gamma $代入(2.6)式中, 可得$ l_{2}(v) = 0 $, 同理可证, $ l_{4}(v) = l_{5}(v) = l_{6}(v) = 0 $, 故线性算子$ A $在$ H $中是自伴的.

定理2.4   问题$ (1.1) $–$ (1.7) $的所有特征值及相应的特征函数均是实的.

证  由于原问题的特征值与算子$ A $的特征值是一致的, 且自共轭算子$ A $的特征值是实的, 故该推论成立.

注2.1   问题(1.1)–(1.7)的所有特征值是实的, 如果$ \lambda_{1}, \lambda_{2} $是两个不同的特征值, 那么问题所对应的特征函数$ f(x), g(x) $在如下定义的内积的意义下是正交的

$ \begin{eqnarray*} &&\int_{a}^{c}f_{1}(x)\bar{g}_{1}(x)w_{1}(x){\rm d}x+\frac{1}{\rho_{1}}\left(\alpha_{1} f(a)-\alpha_{2}f^{[2]}(a)\right)\left(\alpha_{1}\bar{g}(a)-\alpha_{2}\bar{g}^{[2]}(a)\right)\\ &&+\int_{c}^{b}f_{2}(x)\bar{g}_{2}(x)w_{2}(x){\rm d}x +\frac{1}{\rho_{2}}\left(\alpha_{3}f(b)-\alpha_{4}f^{[2]}(b)\right)\left(\alpha_{3}\bar{g}(b)-\alpha_{4}\bar{g}^{[2]}(b)\right) = 0. \end{eqnarray*} $

引理3.1  令实值函数$ q(x) $在$ [a, b] $上是连续的, $ f_{i}(\lambda)(i = 1, 2, 3) $是给定的整函数, 对于$ \forall\lambda \in {\Bbb C} $, 方程$ -{\rm i} u^{(3)}(x)+q(x) u(x) = \lambda w(x) u(x), x \in[a, c) \cup(c, b] $, 满足初始条件$ u(a) = f_{1}(\lambda), u^{[1]}(a) = f_{2}(\lambda), u^{[2]}(a) = f_{3}(\lambda) $ (或$ u(b) = f_{1}(\lambda), u^{[1]}(b) = f_{2}(\lambda), u^{[2]}(b) = f_{3}(\lambda)) $的解$ u = u(x, \lambda) $存在且唯一, 并对于每一个固定的$ x\in[a, b], u(x, \lambda) $是$ \lambda $的整函数.

证  在常微分方程中由解的存在唯一性, 我们可得到该结论.

设$ \varphi_{1}(x, \lambda), \chi_{1}(x, \lambda), \psi_{1}(x, \lambda) $分别是方程

$ -{\rm i}{y}^{(3)}+q y = \lambda wy, x \in[a, c), $

满足初始条件

$ \begin{eqnarray*} &&\varphi_{1}(a, \lambda) = 1, \varphi_{1}^{[1]}(a, \lambda) = 0, \varphi_{1}^{[2]}(a, \lambda) = 0\nonumber, \\ &&\chi_{1}(a, \lambda) = 0, \chi_{1}^{[1]}(a, \lambda) = 1, \chi_{1}^{[2]}(a, \lambda) = 0, \\ &&\psi_{1}(a, \lambda) = 0, \psi_{1}^{[1]}(a, \lambda) = 0, \psi_{1}^{[2]}(a, \lambda) = 1 \end{eqnarray*} $

的线性无关解.

它的朗斯基行列式独立于变量$ x, $记为$ \omega_{1}(\lambda), $则

$ \begin{eqnarray*} \omega_{1}(\lambda) = \left|\begin{array}{ccc} \varphi_{1}(a, \lambda) & \chi_{1}(a, \lambda) & \psi_{1}(a, \lambda) \\ \varphi_{1}^{[1]}(a, \lambda) &{ } \chi_{1}^{[1]}(a, \lambda){ } & \psi_{1}^{[1]}(a, \lambda) \\ \varphi_{1}^{[2]}(a, \lambda) & \chi_{1}^{[2]}(a, \lambda) & \psi_{1}^{[2]}(a, \lambda) \end{array}\right| = 1. \end{eqnarray*} $

设$ \varphi_{2}(x, \lambda), \chi_{2}(x, \lambda), \psi_{2}(x, \lambda) $是方程

$ -{\rm i}{y}^{(3)}+qy = \lambda wy, x \in(c, b], $

满足条件(1.5)–(1.7)的解, 函数$ \varphi_{2}(x, \lambda) $, $ \chi_{2}(x, \lambda) $, $ \psi_{2}(x, \lambda) $的朗斯基行列式独立于变量$ x $, 记为$ \omega_{2}(\lambda) $, 则

$ \begin{eqnarray*} \omega_{2}(\lambda) = \left|\begin{array}{ccc} \varphi_{2}(c+, \lambda) & \chi_{2}(c+, \lambda) & \psi_{2}(c+, \lambda) \\ \varphi_{2}^{[1]}(c+, \lambda) & { } \chi_{2}^{[1]}(c+, \lambda) { }& \psi_{2}^{[1]}(c+, \lambda) \\ \varphi_{2}^{[2]}(c+, \lambda) & \chi_{2}^{[2]}(c+, \lambda) & \psi_{2}^{[2]}(c+, \lambda) \end{array}\right| = \theta\omega_{1}(\lambda), \end{eqnarray*} $

因此, $ \varphi_{2}(x, \lambda), \chi_{2}(x, \lambda), \psi_{2}(x, \lambda) $在$ (c, b] $上是线性无关的. 现在区间$ J = [a, c)\cup(c, b] $上定义函数

$ \begin{eqnarray*} \varphi(x, \lambda) = \left\{\begin{array}{l} \varphi_{1}(x, \lambda), x\in[a, c), \\ \varphi_{2}(x, \lambda), x\in(c, b], \end{array}\right.\\ \chi(x, \lambda) = \left\{\begin{array}{l} \chi_{1}(x, \lambda), x\in[a, c), \\ \chi_{2}(x, \lambda), x\in(c, b], \end{array}\right.\\ \psi(x, \lambda) = \left\{\begin{array}{l} \psi_{1}(x, \lambda), x\in[a, c), \\ \psi_{2}(x, \lambda), x\in(c, b], \end{array}\right.\\ \end{eqnarray*} $

则$ \varphi(x, \lambda), \chi(x, \lambda), \psi(x, \lambda) $是方程$ ly = \lambda y $在$ J = [a, c)\cup(c, b] $上满足转移条件的线性无关的解.

引理3.2^[7]   设$ u(x, \lambda) = \left\{\begin{array}{l} u_{1}(x, \lambda), x\in[a, c), \\ u_{2}(x, \lambda), x\in(c, b] \end{array}\right. $是方程$ ly = \lambda y $的任意一个解, 则它可以表示为

$ u(x, \lambda) = \left\{\begin{array}{l} c_{1}\varphi_{1}(x, \lambda)+c_{2}\chi_{1}(x, \lambda)+c_{3}\psi_{1}(x, \lambda), x\in[a, c), \\ c_{4}\varphi_{2}(x, \lambda)+c_{5}\chi_{2}(x, \lambda)+c_{6}\psi_{2}(x, \lambda), x\in(c, b], \end{array}\right. $

其中$ c_{i}\in{\Bbb C}(i = 1, 2, \cdots6) $. 若$ u(x, \lambda) $满足转移条件, 则$ c_{1} = c_{4}, c_{2} = c_{5}, c_{3} = c_{6} $.

下面令

$ \Phi_{1}(x, \lambda) = \left(\begin{array}{ccc} \varphi_{1}(x, \lambda) & \chi_{1}(x, \lambda) & \psi_{1}(x, \lambda) \\ \varphi_{1}^{[1]}(x, \lambda) & { }\chi_{1}^{[1]}(x, \lambda) { }& \psi_{1}^{[1]}(x, \lambda) \\ \varphi_{1}^{[2]}(x, \lambda) & \chi_{1}^{[2]}(x, \lambda) & \psi_{1}^{[2]}(x, \lambda) \end{array}\right), x\in[a, c), $

$ \Phi_{2}(x, \lambda) = \left(\begin{array}{ccc} \varphi_{2}(x, \lambda) & \chi_{2}(x, \lambda) & \psi_{2}(x, \lambda) \\ \varphi_{2}^{[1]}(x, \lambda) &{ } \chi_{1}^{[1]}(x, \lambda){ } & \psi_{2}^{[1]}(x, \lambda) \\ \varphi_{2}^{[2]}(x, \lambda) & \chi_{2}^{[2]}(x, \lambda) & \psi_{2}^{[2]}(x, \lambda) \end{array}\right), x\in(c, b], $

其中$ \Phi_{1}(x, \lambda) $和$ \Phi_{2}(x, \lambda) $在$ c $点处的值分别由左右极限定义, 设

$ \Phi(x, \lambda) = \left\{\begin{array}{l} \Phi_{1}(x, \lambda), x\in[a, c) , \\ \Phi_{2}(x, \lambda), x\in(c, b], \end{array}\right. $

其中

$ \Phi_{1}(c, \lambda) = \Phi(c-, \lambda), \Phi_{2}(c, \lambda) = \Phi(c+, \lambda), $

对于任意的$ x\in [a, c)\cup(c, b], \Phi(x, \lambda) $是关于$ \lambda $的整函数.

现将边值问题的边界条件写为矩阵形式, 则

$ A_{\lambda}Y(a)+B_{\lambda}Y(b) = 0, Y(x) = \left(y(x), y^{[1]}(x), y^{[2]}(x)\right)^{T}, $

其中

$ A_{\lambda} = \left(\begin{array}{ccc} \left(\alpha_{1}\lambda+\beta_{1}\right) & 0 & -\left(\alpha_{2}\lambda+\beta_{2}\right) \\ 0 & {\rm i}+\tan\theta & 0 \\ 0 & 0 & 0 \end{array}\right), $

$ B_{\lambda} = \left(\begin{array}{ccc} 0 & 0 & 0 \\ 0 & -(1+{\rm i}\tan\theta) & 0 \\ (\alpha_{3}\lambda+\beta_{3})& 0 & -(\alpha_{4}\lambda+\beta_{4}) \end{array}\right). $

定理3.1   复数$ \lambda $是问题$ (1.1) $–$ (1.7) $的特征值, 当且仅当$ \lambda $满足

$ {\rm det}\left(A_{\lambda}+B_{\lambda}\Phi(b, \lambda)\right) = 0. $

证  必要性.  $ \ $设$ \lambda $是问题(1.1)–(1.7) 的特征值, $ y(x) $是相应的特征函数, 根据引理3.2知: 存在不全为零的$ c_{1}, c_{2}, c_{3} $, 使得

$ y(x, \lambda) = \left\{\begin{array}{l} c_{1}\varphi_{1}(x, \lambda)+c_{2}\chi_{1}(x, \lambda)+c_{3}\psi_{1}(x, \lambda), x\in[a, c), \\ c_{1}\varphi_{2}(x, \lambda)+c_{2}\chi_{2}(x, \lambda)+c_{3}\psi_{2}(x, \lambda), x\in(c, b]. \end{array}\right. $

将$ y(x, \lambda) $带入边界条件$ A_{\lambda}Y(a)+B_{\lambda}Y(b) = 0 $, 可得

$ A_{\lambda}\left(\begin{array}{c}c_{1}\\c_{2}\\c_{3}\end{array}\right)+B_{\lambda}\left(\begin{array}{c}c_{1}\varphi_{2}(b, \lambda)+c_{2}\chi_{2}(b, \lambda)+c_{3}\psi_{2}(b, \lambda)\\ c_{1}\varphi_{2}^{[1]}(b, \lambda)+c_{2}\chi_{2}^{[1]}(b, \lambda)+c_{3} \psi_{2}^{[1]}(b, \lambda)\\ c_{1}\varphi_{2}^{[2]}(b, \lambda)+c_{2}\chi_{2}^{[2]}(b, \lambda)+c_{3}\psi_{2}^{[2]}(b, \lambda)\end{array}\right) = 0, $

$ \left[A_{\lambda}+B_{\lambda}\left(\begin{array}{ccc} \varphi_{2}(b, \lambda) & \chi_{2}(b, \lambda) & \psi_{2}(b, \lambda) \\ \varphi_{2}^{[1]}(b, \lambda) &{ } \chi_{2}^{[1]}(b, \lambda) { }& \psi_{2}^{[1]}(b, \lambda) \\ \varphi_{2}^{[2]}(b, \lambda) & \chi_{2}^{[2]}(b, \lambda) & \psi_{2}^{[2]}(b, \lambda) \end{array}\right)\right]\left(\begin{array}{l} c_{1} \\ c_{2} \\ c_{3} \end{array}\right) = 0, $

由于$ c_{1}, c_{2}, c_{3} $不全为零. 因此$ {\rm det}\left(A_{\lambda}+B_{\lambda} \Phi(b, \lambda)\right) = 0 $.

充分性.  若$ {\rm det}\left(A_{\lambda}+B_{\lambda} \Phi(b, \lambda)\right) = 0 $, 则关于$ c_{1}, c_{2}, c_{3} $的齐次方程组

$ \left(A_{\lambda}+B_{\lambda} \Phi(b, \lambda)\right)\left(\begin{array}{c} c_{1} \\ c_{2} \\ c_{3} \end{array}\right) = 0 $

有非零解$ \left(c_{1}^{\prime}, c_{2}^{\prime}, c_{2}^{\prime}\right)^{T} $. 令

$ y(x, \lambda) = \left\{\begin{array}{l} c_{1}^{\prime} \varphi_{1}(x, \lambda)+c_{2}^{\prime} \chi_{1}(x, \lambda)+c_{3}^{\prime} \psi_{1}(x, \lambda), x \in[a, c) , \\ c_{1}^{\prime} \varphi_{2}(x, \lambda)+c_{2}^{\prime} \chi_{2}(x, \lambda)+c_{3}^{\prime} \psi_{2}(x, \lambda), x \in(c, b], \end{array}\right. $

则上式是方程$ ly = \lambda y $满足条件(1.2)–(1.7) 的非零解, 故$ \lambda $是问题(1.1)–(1.7) 的特征值.

定理3.2   问题$ (1.1) $–$ (1.7) $最多有可数个实的特征值, 并且没有有限值的聚点.

证  由定理3.1知(1.1)–(1.7) 的特征值是函数det $ (A_{\lambda}+B_{\lambda}\Phi(b, \lambda)) = 0 $的零点. 已知函数$ \Phi (b, \lambda) $是关于$ \lambda $的整函数, 又由注2.1知问题(1.1)–(1.7)的特征值为实的, 即任何复数$ \mu $不是问题(1.1)–(1.7) 的特征值, 说明整函数det $ (A_{\lambda}+B_{\lambda}\Phi (b, \lambda)) $不恒为零, 故根据整函数的零点分布定理, 我们得到问题(1.1)–(1.7)最多有可数个实的特征值, 并且没有有限值的聚点.

定理3.3   算子$ A $仅有点谱, 即

$ \sigma(A) = \sigma_{P}(A). $

证  只需证明, 若$ \lambda $不是算子$ A $的特征值, 则$ \lambda\in\rho(A) $. 由于算子$ A $是自伴算子, 故仅考虑$ \lambda $为实数的情况.

下面讨论方程$ (A-\lambda)U = F $, 其中$ U = (u(x), \alpha_{1}u(a)-\alpha_{2}u^{[2]}(a), \alpha_{3}u(b)-\alpha_{4}u^{[2]}(b))\in D(A), F = (f(x), h, r)^{T}\in H $, 由于$ A $的定义, 将问题分成初值问题

(3.1) $ \begin{equation} \left\{\begin{array}{l} (l-\lambda) u(x) = f(x), \\ u(c+) = \alpha_{11} u(c-)-\alpha_{12} u^{[2]}(c-), \\ u^{[1]}(c+) = u^{[1]}(c-), \\ u^{[2]}(c+) = \beta_{11}u(c-)- \beta_{12}u^{[2]}(c-) \end{array}\right. \end{equation} $

及方程组

(3.2) $ \begin{equation} \left\{\begin{array}{l} h = -\left(\beta_{1} u(a)-\beta_{2} u^{[2]}(a)\right)-\lambda\left(\alpha_{1} u(a)-\alpha_{2} u^{[2]}(a)\right) = 0, \\ r = -\left(\beta_{3} u(b)-\beta_{4} u^{[2]}(b)\right)-\lambda\left(\alpha_{3} u(b)-\alpha_{4} u^{[2]}(b)\right) = 0, \\ ({\rm i}+\tan \gamma) u^{[1]}(a)-(1+{\rm i} \tan \gamma) u^{[1]}(b) = 0 \end{array}\right. \end{equation} $

两部分.

由引理$ 3.2 $以及常微分方程通解的结构知问题$ (3.1) $的通解形式如下

(3.3) $ \begin{equation} u(x, \lambda) = \left\{\begin{array}{l} c_{1} \varphi_{1}(x, \lambda)+c_{2} \chi_{1}(x, \lambda)+c_{3} \psi_{1}(x, \lambda)+k_{1}(x, \lambda), x \in[a, c), \\ c_{1} \varphi_{2}(x, \lambda)+c_{2} \chi_{2}(x, \lambda)+c_{3} \psi_{2}(x, \lambda)+k_{2}(x, \lambda), x \in(c, b], \end{array}\right. \end{equation} $

其中$ k(x, \lambda) = \left\{\begin{array}{l} k_{1}(x, \lambda), x \in[a, c), \\ k_{2}(x, \lambda), x \in(c, b] \end{array}\right. $为问题(3.1)的一个特解. $ (c_{1}, c_{2}, c_{3})\in {\Bbb C}, \varphi_{1}, \varphi_{2}, $$ \chi_{1}, \chi_{2}, $$ \psi_{1}, \psi_{2} $是上面所定义的. 将通解$ (3.3) $代入方程组$ (3.2) $中, 整理得如下三式

(3.4) $ \begin{eqnarray} &&c_{1}\left(\beta_{1}+\lambda \alpha_{1}\right)-c_{3}\left(\beta_{2}+\lambda \alpha_{2}\right) = k_{1}^{[2]}(a)\left(\beta_{2}+\lambda\alpha_{2}\right)+k_{1}(a)\left(\beta_{1}+\lambda\alpha_{1}\right), \end{eqnarray} $

(3.5) $ \begin{eqnarray} &&c_{1}[(\beta_{3}+\lambda\alpha_{3})\varphi_{2}(b)-(\beta_{4}+\lambda\alpha_{4})\varphi_{2}^{[2]}(b)]+ c_{2}[(\beta_{3}+\lambda\alpha_{3})\chi_{2}(b)-(\beta_{4}+\lambda\alpha_{4})\chi_{2}^{[2]}(b)] \\ &&+c_{3}[(\beta_{3}+\lambda\alpha_{3})\psi_{2}(b) -(\beta_{4}+\lambda\alpha_{4})\psi_{2}^{[2]}(b)] = k_{2}^{[2]}(b)(\beta_{4}+\lambda\alpha_{4})- k_{2}(b)(\beta_{3}+\lambda\alpha_{3}), \end{eqnarray} $

(3.6) $ \begin{eqnarray} && \varphi_{2}^{[1]}(b)c_{1}-(\frac{{\rm i}+\tan \gamma}{1+{\rm i} \tan \gamma}-\chi_{2}^{[1]}(b))c_{2}+\psi_{2}^{[1]}(b)c_{3} = -k_{2}^{[1]}(b)+\frac{{\rm i}+\tan \gamma}{1+{\rm i} \tan \gamma}k_{1}^{[1]}(a), \end{eqnarray} $

联立(3.4)–(3.6)式, 可得关于未知数$ c_{1}, c_{2}, c_{3} $的三元一次方程组的系数行列式恰好为

$ \frac{1}{1+{\rm i} \tan \gamma}\det(A_{\lambda}+B_{\lambda}\Phi(b, \lambda)), $

因为$ \lambda $不是算子$ A $的特征值, 由定理3.1知: det$ ( A_{\lambda}+B_{\lambda}\Phi (b, \lambda))\neq 0 $, 因此$ c_{1} $, $ c_{2} $, $ c_{3} $是唯一确定的, 从而$ u $也是唯一确定的.

由以上论述证明了$ (A-\lambda I)^{-1} $定义在全空间$ H $上, 又由$ A $在$ H $上是自共轭算子及由闭图像定理知$ (A-\lambda I)^{-1} $是有界的, 说明$ \lambda\in \rho(A) $.