## The Self-Adjointness and Dependence of Eigenvalues of Fourth-Order Differential Operator with Eigenparameters in the Boundary Conditions

Yan Wenwen, Xu Meizhen,

College of Sciences, Inner Mongolia University of Technology, Hohhot 010051

 基金资助: 国家自然科学基金.  11561051内蒙古自然科学基金.  2021MS01020

 Fund supported: the NSFC.  11561051the NSF of Inner Mongolia.  2021MS01020

Abstract

In this paper we consider the self-adjointness and the dependence of eigenvalues of a class of discontinuous fourth-order differential operator with eigenparameters in the boundary conditions of one endpoint. By constructing a linear operator T associated with problem in a suitable Hilbert space, the study of the above problem is transformed into the research of the operator in this space, and the self-adjointness of this operator T is proved. In addition, on the basis of the self-adjointness of the operator T, we show that the eigenvalues are not only continuously but also smoothly dependent on the parameters of the problem, and give the corresponding differential expressions. In particular, giving the Fréchet derivative of the eigenvalue with respect to the eigenparameter-dependent boundary condition coefficient matrix, and the first-order derivatives of the eigenvalue with respect to the left and right sides of the inner discontinuity point c.

Keywords： Fourth-order differential operator ; Transmission condition ; Self-adjointness ; Dependence of eigenvalue ; Fréchet derivative

Yan Wenwen, Xu Meizhen. The Self-Adjointness and Dependence of Eigenvalues of Fourth-Order Differential Operator with Eigenparameters in the Boundary Conditions. Acta Mathematica Scientia[J], 2022, 42(3): 671-693 doi:

## 2 预备知识

$$$l(y): = (p_{2}(x)y'')''-(p_{1}(x)y')'+q(x)y = \lambda w(x)y, x\in J' = (a', c)\cup(c, b').$$$

$\begin{eqnarray} & &l_{1}y: = (\alpha_{1}\lambda+\beta_{1})y(a)-(\alpha_{2}\lambda+\beta_{2})((p_{2}y'')'-p_{1}y')(a) = 0, \end{eqnarray}$

$\begin{eqnarray} & &l_{2}y: = (\alpha_{3}\lambda+\beta_{3})y'(a)-(\alpha_{4}\lambda+\beta_{4})(p_{2}y'')(a) = 0, \end{eqnarray}$

$\begin{eqnarray} &&l_{3}y: = \cos\gamma y(b)-\sin\gamma ((p_{2}y'')'-p_{1}y')(b) = 0, \end{eqnarray}$

$\begin{eqnarray} &&l_{4}y: = \cos\gamma y'(b)-\sin\gamma (p_{2}y'')(b) = 0 \end{eqnarray}$

$$$CY(c-)+DY(c+) = 0,$$$

$$$p_{2}, p_{1}, q, w: J'\rightarrow {\mathbb R}, \frac{1}{p_{2}}, p_{1}, q, w\, \in L^{1}_{loc}(J'), w>0{\quad} a.e. {\quad} J', Y = \left(\begin{array}{c}y\\y'\\p_{2}y''\\(p_{2}y'')'-p_{1}y'\end{array}\right),$$$

$$$\alpha_{i}, \beta_{i}\in {\mathbb R}, i = 1, 2, 3, 4. \theta_{1} = \alpha_{1}\beta_{2}-\beta_{1}\alpha_{2}>0, \theta_{2} = \alpha_{4}\beta_{3}-\beta_{4}\alpha_{3}>0, \gamma\in (0, \pi].$$$

$Y(c\pm) = \lim\limits_{x\rightarrow c\pm}Y(x). C = (c_{ij}), D = (d_{ij}) $$4\times 4 实矩阵, det C = \eta^{2}, det D = \xi^{2},$$ \eta>0, \xi>0$, 且满足

$\begin{eqnarray} \xi CJ_{0}C^{T} = \eta DJ_{0}D^{T}, \, J_{0} = \left(\begin{array}{cccc}0&\ 0\ &0&\ 1\\0&0&-1&\ 0\\ 0&1&0&\ 0\\-1&0&0&\ 0\end{array}\right), \end{eqnarray}$

$C^{T}$表示矩阵$C$的转置, $J_{0}^{T} = J_{0}^{-1} = -J_{0}, J_{0}\cdot J_{0} = -E$, 其中$E$为四阶单位矩阵.

$\begin{eqnarray} M(f) = \beta_{1}f(a)-\beta_{2}((p_{2}f'')'-p_{1}f')(a), M'(f) = \alpha_{1}f(a)-\alpha_{2}((p_{2}f'')'-p_{1}f')(a), \end{eqnarray}$

$\begin{eqnarray} N(f) = \beta_{3}f'(a)-\beta_{4}(p_{2}f'')(a), N'(f) = \alpha_{3}f'(a)-\alpha_{4}(p_{2}f'')(a), \end{eqnarray}$

${\cal F} = (f, -M'(f), N'(f))^{T},$

$\begin{eqnarray} T{\cal F} = (\frac{l(f)}{w}, M(f), -N(f))^{T} = (\lambda f, -\lambda M'(f), \lambda N'(f))^{T} = \lambda {\cal F}. \end{eqnarray}$

## 3 算子$T$的自共轭性

$f(x) $$H_{1} 中正交于 \widetilde{C^{\infty}_{0}},$$ f(x)$为零. 设${\cal G} = (g(x), k_{1}, 0)^{T}\in \widetilde{C^{\infty}_{0}}\oplus\{0\} (\widetilde{C^{\infty}_{0}}\oplus\{0\}\subset\widetilde{C^{\infty}_{0}}\oplus\{0\}\oplus\{0\}), $$\langle {\cal F}, {\cal G}\rangle = \langle f, g\rangle _{1}+\eta\rho_{1}h_{1}\overline{k}_{1} = 0, 由于 k_{1} = -M'(g) 是任意选取的, 故 h_{1} = 0. 又设 {\cal G} = (g(x), 0, k_{2})^{T}\in D(T),$$ \langle {\cal F}, {\cal G}\rangle = \langle f, g\rangle_{1}+\eta\rho_{2}h_{2}\overline{k}_{2} = 0,$由于$k_{2} = N'(g)$是任意选取的, 故$h_{2} = 0.$于是${\cal F} = (0, 0, 0)^{T}.$所以, 与$D(T)$正交的只有零元素, 从而证得$D(T) $$H 中是稠密的. 定理3.1 线性算子 T 是定义在 H 上的自共轭算子. 对任意的 {\cal F}, {\cal G}\in D(T), 由分部积分法可得 其中 \begin{eqnarray} [f, g](x)& = &f(x)((p_{2}\overline{g}'')'-p_{1}\overline{g}')(x)-f'(x)(p_{2}\overline{g}'')(x){}\\ &&+(p_{2}f'')(x)\overline{g}'(x)-((p_{2}f'')'-p_{1}f')(x)\overline{g}(x) {} \\ & = &-G^{\ast}(x)J_{0}F(x), \end{eqnarray} G^{\ast} 表示向量 G 的共轭转置, G = (g, g', p_{2}g'', (p_{2}g'')'-p_{1}g')^{T}. l_{3}f = 0, l_{3}\overline{g} = 0, 可知 f(b)((p_{2}\overline{g}'')'-p_{1}\overline{g}')(b)-((p_{2}f'')'-p_{1}f') (b)\overline{g}(b) = 0. 再由 l_{4}f = 0, l_{4}\overline{g} = 0, 可知 f'(b)(p_{2}\overline{g}'')(b) -(p_{2}f'')(b)\overline{g}'(b) = 0. 因此, [f, g](b) = 0. 由(2.10)式和(2.11) 式分别可以得到如下的(3.2) 式和(3.3) 式 $$\rho_{1}(M'(f)M(\overline{g})-M(f)M'(\overline{g})) = -f(a)((p_{2}\overline{g}'')'-p_{1}\overline{g}')(a) +((p_{2}f'')'-p_{1}f')(a)\overline{g}(a),$$ $$\rho_{2}(N'(f)N(\overline{g})-N(f)N'(\overline{g})) = f'(a)(p_{2}\overline{g}'')(a)-(p_{2}f'')(a)\overline{g}'(a).$$ 再由(3.2)和(3.3) 式得 由转移条件(2.6) 知 \begin{eqnarray} F(c-) = -C^{-1}D F(c+), {\quad} G^{\ast}(c-) = -G^{\ast}(c+)D^{\ast}(C^{\ast})^{-1}, \end{eqnarray} 将(3.4) 式代入(3.1) 式并结合(2.9) 式可得 \begin{eqnarray} \eta[f, g](c-) = -\eta G^{\ast}(c+)D^{\ast}(C^{\ast})^{-1}J_{0}C^{-1}DF(c+) = \xi[f, g](c+), \end{eqnarray} 因此, \langle T{\cal F}, {\cal G}\rangle = \langle {\cal F}, T{\cal G}\rangle, 故算子 T 是对称的. 接下来根据自伴算子的定义证明算子 T 是自伴的. 下面只需证明: 若对任意的 {\cal F} = (f(x), -M'(f), N'(f))^{T}\in D(T),$$ \langle T{\cal F}, {\cal Z}\rangle = \langle {\cal F}, {\cal U}\rangle$成立, 则${\cal Z}\in D(T) $$T{\cal Z} = {\cal U}. 其中, {\cal Z} = (z(x), m_{1}, m_{2})^{T}, {\cal U} = (u(x), n_{1}, n_{2})^{T}, ( \rm{i} ) z(x), z'(x), (p_{2}z'')(x), ((p_{2}z'')'-p_{1}z')(x)\in AC_{loc}(J), \frac{l(z)}{w}\in H_{1}; ( \rm{ii} ) m_{1} = -M'(z) = -[\alpha_{1}z(a)-\alpha_{2}((p_{2}z'')'-p_{1}z')(a)], m_{2} = N'(z) = \alpha_{3}z'(a)-\alpha_{4}(p_{2}z'')(a); ( \rm{iii} ) l_{i}z = 0, i = 3, 4; ( \rm{iv} ) u(x) = \frac{l(z)(x)}{w}; ( \rm{v} ) CZ(c-)+DZ(c+) = 0; ( \rm{vi} ) n_{1} = M(z) = \beta_{1}z(a)-\beta_{2}((p_{2}z'')'-p_{1}z')(a), n_{2} = -N(z) = -(\beta_{3}z'(a)-\beta_{4}(p_{2}z'')(a)). 对任意的 {\cal F}\in \widetilde{C^{\infty}_{0}}\oplus\{0\}\oplus\{0\}\subset D(T),$$ \langle T{\cal F}, {\cal Z}\rangle = \langle {\cal F}, {\cal U}\rangle$

$$$\langle \frac{l(f)}{w}, z\rangle_{1} = \langle f, u\rangle_{1},$$$

$\begin{eqnarray} &&-\eta\rho_{1}(M(f)\overline{m}_{1}+M'(f)\overline{n}_{1})+\eta\rho_{2}(N(f)\overline{m}_{2} +N'(f)\overline{n}_{2}){}\\ & = &-\eta[f, z](c-)+\eta[f, z](a)-\xi[f, z](b)+\xi[f, z](c+){}\\ & = &\eta Z^{\ast}(c-)J_{0}F(c-)-\eta Z^{\ast}(a)J_{0}F(a)+\xi Z^{\ast}(b)J_{0}F(b)-\xi Z^{\ast}(c+)J_{0}F(c+). \end{eqnarray}$

$\begin{eqnarray} F(b) = F(c-) = F(c+) = 0, \end{eqnarray}$

$$$f'(c-) = (p_{2}f'')(c-) = ((p_{2}f'')'-p_{1}f')(c-) = 0, f(c-)\neq 0,$$$

$\begin{eqnarray} &&-\eta f(c-)((p_{2}\overline{z}'')'-p_{1}\overline{z}')(c-)+\xi[f(c+)((p_{2}\overline{z}'')'-p_{1}\overline{z}')(c+){}\\ &&-f'(c+)(p_{2}\overline{z}'')(c+)+(p_{2}f'')(c+)\overline{z}'(c+)-((p_{2}f'')'-p_{1}f')(c+)\overline{z}(c+)] = 0, \end{eqnarray}$

## 4 特征值和特征函数的连续性

$$$A_{\lambda}Y(a)+BY(b) = 0,$$$

$\psi_{11}, \psi_{12} $$\psi_{13}, \psi_{14} 是方程(2.1) 在区间[ a, c ) 满足如下初始条件的线性无关解 其中 E 是四阶单位矩阵, \Psi_{1j}(x) = (\psi_{1j}(x), \psi_{1j}'(x), p_{2}\psi''_{1j}(x), ((p_{2}\psi_{1j}'')'-p_{1}\psi_{1j}')(x))^{T}$$ (j = 1, 2, 3, 4).$它们的Wronskian与变量$x$无关, 且是关于特征参数$\lambda$的整函数.

$\psi_{21}, \psi_{22} $$\psi_{23}, \psi_{24} 是方程(2.1) 在区间 (c, b] 满足如下初始条件的线性无关解 其中 \Psi_{2j}(x) = (\psi_{2j}(x), \psi_{2j}'(x), p_{2}\psi''_{2j}(x), ((p_{2}\psi_{2j}'')' -p_{1}\psi_{2j}')(x))^{T},$$ (j = 1, 2, 3, 4).$且令

$\Delta(\lambda)$为判别函数.

这与文献[12, Theorem 3.1] 讨论的边界条件两端含有特征参数的证明过程类似, 因此省略.

$\widetilde{p_{1}}, \widetilde{q}, \widetilde{w}$可类似定义.

令$\Delta(\lambda) = \Delta(\omega, \lambda) = {\rm det}(A_{\lambda}+B\Phi(b, \lambda)).$由引理4.1可知, 问题的特征值正是判别函数的零点. 对$\omega\in\Omega, \lambda(\omega)$是算子$T$的特征值当且仅当$\Delta(\omega, \lambda(\omega)) = 0.$对于$\forall \omega\in\Omega, \Delta(\omega, \lambda) $$\lambda 的整函数且在 \omega 处连续(见文献[33, Theorem 2.7, 2.8]). 又由算子 T 的自伴性可知, \mu 是孤立特征值, 从而 \Delta(\omega_{0}, \mu) = 0$$ \Delta(\omega_{0}, \lambda)$关于$\lambda$不是常数, 因此, $\exists \rho>0, $$\lambda\in S_{\rho}: = \{\lambda\in {\mathbb C}: |\lambda-\mu| = \rho\}, \Delta(\omega_{0}, \lambda)\neq0. 由方程解对初值和参数的连续性定理(见文献[34, p248, (9.17.4)]), 即可得定理. 引理4.2 假设(2.7)–(2.9) 成立, t_{0}\in [a, c)\cup(c, b] \cup\{c+, c-\}, d, k, m, n\in{\mathbb C}. 初值问题 的唯一解 y = y(\cdot, t_{0}, d, k, m, n, C, D, 1/p_{2}, p_{1}, q, w) 是关于任一变量的连续函数. 即对 \forall \varepsilon>0,$$ \exists \delta>0,$

可参考文献[15, Lemma 3.2], 虽然本文讨论的边界条件含有谱参数, 但证明过程类似.

若$\lambda(\omega_{0})$是单重特征值, 则$\Delta'(\lambda(\omega_{0}))\neq0$. 因为$\Delta(\lambda) $$\lambda 的整函数, 由定理4.1可知结论成立. 定义4.1 设 u 满足边值问题(2.1)–(2.6), u_{1} = -M'(u), u_{2} = N'(u), 且有 $$\eta\int_{a}^{c}u\overline{u}w{\rm d}x+\xi\int_{c}^{b}u\overline{u}w{\rm d}x+\eta\rho_{1}u_{1}\overline{u}_{1}+\eta\rho_{2}u_{2}\overline{u}_{2} = 1$$ 成立, 则称 (u, u_{1}, u_{2})^{T} 为正规化特征向量. 定理4.2 假设记号同定理4.1. 设特征值 \lambda(\omega) (\omega\in \Omega)$$ \Omega $$\omega_{0} 的某个邻域 M 内所有 \omega$$ l\ (l = 1, 2, 3, 4)$重特征值, 令$(u_{k}(x, \omega_{0}), u_{k1}(\omega_{0}), u_{k2}(\omega_{0}))^{T}\in H, $$k = 1, 2, \cdots, l 是算子 T 对应于 l 重特征值 \lambda(\omega_{0}) (\omega_{0}\in \Omega) 的线性无关的正规化特征向量. 则存在 l 个对应于特征值 \lambda(\omega) 的线性无关的正规化特征向量 (u_{k}(x, \omega), u_{k1}(\omega), u_{k2}(\omega))^{T}\in H, k = 1, 2, \cdots, l, 使得在 \Omega 中当 \omega\rightarrow \omega_{0} 时有 \begin{eqnarray} &&u_{k}(x, \omega)\rightarrow u_{k}(x, \omega_{0}), u_{k}'(x, \omega)\rightarrow u_{k}'(x, \omega_{0}), (p_{2}u_{k}'')(x, \omega)\rightarrow (p_{2}u_{k}'')(x, \omega_{0}), {}\\ &&((p_{2}u_{k}'')'-p_{1}u_{k}')(x, \omega)\rightarrow ((p_{2}u_{k}'')'-p_{1}u_{k}')(x, \omega_{0}), \\ &&u_{k1}(\omega)\rightarrow u_{k1}(\omega_{0}), u_{k2}(\omega)\rightarrow u_{k2}(\omega_{0}){} \end{eqnarray} 在区间 J 上一致成立. 设 \lambda(\omega_{0}) 是算子 T 的单重特征值, (y(x, \omega_{0}), y_{1}(\omega_{0}), y_{2}(\omega_{0}))^{T}\in H 是其对应的特征向量, 且满足 由引理4.3可知, 存在一个 \omega_{0} 的邻域 M 使得对任意的 \omega\in M, \lambda(\omega) 是单重的. 根据定理4.1知, 当 \omega\rightarrow \omega_{0}$$ \lambda(\omega)\rightarrow \lambda(\omega_{0})$成立. 当$\omega\rightarrow \omega_{0}$时, 边界条件矩阵满足$(A_{\lambda}, B)_{4\times8}(\omega)\rightarrow (A_{\lambda}, B)_{4\times8}(\omega_{0}).$由文献[4, Theorem 3.2] 可知, 当$\omega\rightarrow \omega_{0}$时, 存在特征值$\lambda(\omega)$对应的特征向量$(y(x, \omega), y_{1}(\omega), y_{2}(\omega))^{T}\in H,$使其第一个分量$y(x, \omega)$在区间$J$上满足

$\begin{eqnarray} &&\| y(x, \omega)\|^{2} = \eta\int_{a}^{c}| y(x, \omega)|^{2}w{\rm d}x+\xi\int_{c}^{b}| y(x, \omega)|^{2}w{\rm d}x = 1, {}\\ &&y(x, \omega)\rightarrow y(x, \omega_{0}), y'(x, \omega)\rightarrow y'(x, \omega_{0}), (p_{2}y'')(x, \omega)\rightarrow (p_{2}y'')(x, \omega_{0}), \\ &&((p_{2}y'')'-p_{1}y')(x, \omega)\rightarrow ((p_{2}y'')'-p_{1}y')(x, \omega_{0}).{} \end{eqnarray}$

$$$y_{1}(\omega)\rightarrow y_{1}(\omega_{0}), y_{2}(\omega)\rightarrow y_{2}(\omega_{0}).$$$

$\begin{eqnarray} &&(u(x, \omega_{0}), u_{1}(\omega_{0}), u_{2}(\omega_{0}))^{T} = \frac{(y(x, \omega_{0}), y_{1}(\omega_{0}), y_{2}(\omega_{0}))^{T}}{\|(y(x, \omega_{0}), y_{1}(\omega_{0}), y_{2}(\omega_{0}))^{T}\|}, \end{eqnarray}$

$\begin{eqnarray} &&u'(x, \omega_{0}) = \frac{y'(x, \omega_{0})}{\|(y(x, \omega_{0}), y_{1}(\omega_{0}), y_{2}(\omega_{0}))^{T}\|}, \end{eqnarray}$

$\begin{eqnarray} &&(p_{2}u'')(x, \omega_{0}) = \frac{(p_{2}y'')(x, \omega_{0})}{\|(y(x, \omega_{0}), y_{1}(\omega_{0}), y_{2}(\omega_{0}))^{T}\|}, \end{eqnarray}$

$\begin{eqnarray} &&((p_{2}u'')'-p_{1}u')(x, \omega_{0}) = \frac{((p_{2}y'')'-p_{1}y')(x, \omega_{0})}{\|(y(x, \omega_{0}), y_{1}(\omega_{0}), y_{2}(\omega_{0}))^{T}\|}. \end{eqnarray}$

$\begin{eqnarray} &&\| y_{k}(x, \omega)\|^{2} = \eta\int_{a}^{c}| y_{k}(x, \omega)|^{2}w{\rm d}x+\xi\int_{c}^{b}| y_{k}(x, \omega)|^{2}w{\rm d}x = 1, {}\\ &&y_{k}(x, \omega)\rightarrow y_{k}(x, \omega_{0}), y_{k}'(x, \omega)\rightarrow y_{k}'(x, \omega_{0}), (p_{2}y_{k}'')(x, \omega)\rightarrow (p_{2}y_{k}'')(x, \omega_{0}), \\ &&((p_{2}y_{k}'')'-p_{1}y_{k}')(x, \omega)\rightarrow ((p_{2}y_{k}'')'-p_{1}y_{k}')(x, \omega_{0}).{} \end{eqnarray}$

## 5 特征值的可微性

$\begin{eqnarray} &&[\lambda(\gamma+\varepsilon)-\lambda(\gamma)][\eta\int_{a}^{c}u\overline{v}w{\rm d}x+\xi\int_{c}^{b}u\overline{v}w{\rm d}x]{}\\ & = &\eta\bigg[\int_{a}^{c}[(p_{2}\overline{v}'')''-(p_{1}\overline{v}')']u{\rm d}x-\int_{a}^{c}[(p_{2}u'')''-(p_{1}u')']\overline{v}{\rm d}x\bigg]{}\\ &&+\xi\bigg[\int_{c}^{b}[(p_{2}\overline{v}'')''-(p_{1}\overline{v}')']u{\rm d}x-\int_{c}^{b}[(p_{2}u'')''-(p_{1}u')']\overline{v}{\rm d}x\bigg]{}\\ & = &\eta[u, v](c-)-\eta[u, v](a)+\xi[u, v](b)-\xi[u, v](c+). \end{eqnarray}$

$\begin{eqnarray} & &[\lambda(\gamma+\varepsilon)-\lambda(\gamma)]\eta\rho_{1}u_{1}\overline{v}_{1}{}\\ & = &\eta\rho_{1}u_{1}(\beta_{1}\overline{v}(a)-\beta_{2}((p_{2}\overline{v}'')'-p_{1}\overline{v}')(a)) -\eta\rho_{1}(\beta_{1}u(a)-\beta_{2}((p_{2}u'')'-p_{1}u')(a))\overline{v}_{1}{}\\ & = &\eta[u(a)((p_{2}\overline{v}'')'-p_{1}\overline{v}')(a)-((p_{2}u'')'-p_{1}u')(a)\overline{v}(a)]. \end{eqnarray}$

$\begin{eqnarray} &&[\lambda(\gamma+\varepsilon)-\lambda(\gamma)]\eta\rho_{2}u_{2}\overline{v}_{2}{}\\ & = &\eta\rho_{2}u_{2}(-\beta_{3}\overline{v}'(a)+\beta_{4}(p_{2}\overline{v}'')(a)) -\eta\rho_{2}(-\beta_{3}u'(a)+\beta_{4}(p_{2}u'')(a))\overline{v}_{2}{}\\ & = &-\eta[u'(a)(p_{2}\overline{v}'')(a)-(p_{2}u'')(a)\overline{v}'(a)]. \end{eqnarray}$

$\begin{eqnarray} &&[\lambda(A+L)-\lambda(A)][\eta\int_{a}^{c}u\overline{v}w{\rm d}x+\xi\int_{c}^{b}u\overline{v}w{\rm d}x] = -\eta[u, v](a){}\\ & = &-\eta(u, -u', p_{2}u'', -((p_{2}u'')'-p_{1}u'))(a)E \left(\begin{array}{ccccc} (p_{2}\overline{v}'')'-p_{1}\overline{v}'\\ p_{2}\overline{v}''\\ \overline{v}'\\ \overline{v}\end{array}\right)(a). \end{eqnarray}$

$A+L = \left(\begin{array}{ccccc} \widetilde{\alpha_{1}}&0&0&\ \widetilde{\beta_{1}}\\ 0&\ \widetilde{\alpha_{3}}\ &\widetilde{\beta_{3}}&\ 0\\ 0&\widetilde{\alpha_{4}}&\widetilde{\beta_{4}}&\ 0\\ \widetilde{\alpha_{2}}&0&0&\ \widetilde{\beta_{2}}\end{array}\right)$, 然后根据边界条件(2.2) 得

$\begin{eqnarray} &&[\lambda(A+L)-\lambda(A)]\eta\rho_{1}u_{1}\overline{v}_{1}{}\\ & = &\eta\rho_{1}u_{1}(\widetilde{\beta_{1}}\overline{v}(a)-\widetilde{\beta_{2}}((p_{2}\overline{v}'')'-p_{1}\overline{v}')(a)) -\eta\rho_{1}(\beta_{1}u(a)-\beta_{2}((p_{2}u'')'-p_{1}u')(a))\overline{v}_{1}{}\\ & = &\eta\rho_{1}(u, -u', p_{2}u'', -((p_{2}u'')'-p_{1}u'))(a) \left(\begin{array}{ccccc} -\alpha_{1}\\0\\0\\ -\alpha_{2}\end{array}\right) (-\widetilde{\beta_{2}}, 0, 0, \widetilde{\beta_{1}}) \left(\begin{array}{ccccc}(p_{2}\overline{v}'')'-p_{1}\overline{v}'\\p_{2}\overline{v}''{\nonumber}\\ \overline{v}'\\ \overline{v}\end{array}\right)(a){}\\ &&+\eta\rho_{1}(u, -u', p_{2}u'', -((p_{2}u'')'-p_{1}u'))(a) \left(\begin{array}{ccccc}-\beta_{1}\\0\\0\\ -\beta_{2} \end{array}\right)(\widetilde{\alpha_{2}}, 0, 0, -\widetilde{\alpha_{1}}) \left(\begin{array}{ccccc} (p_{2}\overline{v}'')'-p_{1}\overline{v}'\\p_{2}\overline{v}''\\ \overline{v}'{\nonumber}\\ \overline{v}\end{array}\right)(a){}\\ & = &\eta\rho_{1}(u, -u', p_{2}u'', -((p_{2}u'')'-p_{1}u'))(a){}\\ &&\cdot\left(\begin{array}{ccccc} \alpha_{1}\widetilde{\beta_{2}}-\beta_{1}\widetilde{\alpha_{2}}&\ 0\ &0&\ -\alpha_{1}\widetilde{\beta_{1}}+\beta_{1}\widetilde{\alpha_{1}}\\ 0&0&0&\ 0\\ 0&0&0&\ 0\\ \alpha_{2}\widetilde{\beta_{2}}-\beta_{2}\widetilde{\alpha_{2}}&0&0& \ -\alpha_{2}\widetilde{\beta_{1}}+\beta_{2}\widetilde{\alpha_{1}}\end{array}\right) \left(\begin{array}{ccccc} (p_{2}\overline{v}'')'-p_{1}\overline{v}'\\p_{2}\overline{v}''\\ \overline{v}'\\ \overline{v}\end{array}\right)(a). \end{eqnarray}$

$\begin{eqnarray} &&[\lambda(A+L)-\lambda(A)]\eta\rho_{2}u_{2}\overline{v}_{2}{}\\ & = &\eta\rho_{2}u_{2}(-\widetilde{\beta_{3}}\overline{v}'(a)+\widetilde{\beta_{4}}(p_{2}\overline{v}'')(a)) -\eta\rho_{2}(-\beta_{3}u'(a)+\beta_{4}(p_{2}u'')(a))\overline{v}_{2}{}\\ & = &\eta\rho_{2}(u, -u', p_{2}u'', -((p_{2}u'')'-p_{1}u'))(a)\left(\begin{array}{ccccc}0\\ -\alpha_{3}\\ -\alpha_{4}\\0\end{array}\right)(0, \widetilde{\beta_{4}}, -\widetilde{\beta_{3}}, 0)\left(\begin{array}{ccccc}(p_{2}\overline{v}'')'-p_{1}\overline{v}'\\p_{2}\overline{v}''\\ \overline{v}'\\ \overline{v}\end{array}\right)(a){}\\ &&+\eta\rho_{2}(u, -u', p_{2}u'', -((p_{2}u'')'-p_{1}u'))(a)\left(\begin{array}{ccccc}0\\ -\beta_{3}\\ -\beta_{4}\\0\end{array}\right)(0, -\widetilde{\alpha_{4}}, \widetilde{\alpha_{3}}, 0)\left(\begin{array}{ccccc}(p_{2}\overline{v}'')'-p_{1}\overline{v}'\\p_{2}\overline{v}''\\ \overline{v}'\\ \overline{v}\end{array}\right)(a){}\\ & = &\eta\rho_{2}(u, -u', p_{2}u'', -((p_{2}u'')'-p_{1}u'))(a){}\\ &&\cdot\left(\begin{array}{ccccc}0&0&0&0\\ 0&\ -\alpha_{3}\widetilde{\beta_{4}}+\beta_{3}\widetilde{\alpha_{4}}\ & \alpha_{3}\widetilde{\beta_{3}}-\beta_{3}\widetilde{\alpha_{3}}&\ 0\\ 0&-\alpha_{4}\widetilde{\beta_{4}} +\beta_{4}\widetilde{\alpha_{4}}& \alpha_{4}\widetilde{\beta_{3}}-\beta_{4}\widetilde{\alpha_{3}}&\ 0\\ 0&0&0&\ 0\end{array}\right) \left(\begin{array}{ccccc}(p_{2}\overline{v}'')'-p_{1}\overline{v}'\\p_{2}\overline{v}''\\ \overline{v}'\\ \overline{v}\end{array}\right)(a). \end{eqnarray}$

$L\rightarrow0$, 由定义4.1和定理4.2可得到(5.4) 式成立.

3. 令特征值$\mu = \lambda(C), $$\nu = \lambda(C+H) 对应的正规化特征向量分别为 (u, u_{1}, u_{2})^{T},$$ (v, v_{1}, v_{2})^{T},$其中$u = u(x, C), v = u(x, C+H),$我们有

$\begin{eqnarray} &&[\lambda(C+H)-\lambda(C)][\eta\int_{a}^{c}u\overline{v}w{\rm d}x+\xi\int_{c}^{b}u\overline{v}w{\rm d}x+\eta\rho_{1}u_{1}\overline{v}_{1}+\eta\rho_{2}u_{2}\overline{v}_{2}]{}\\ & = &\eta[u, v](c-)-\xi[u, v](c+){}\\ & = &\eta((p_{2}\overline{v}'')'-p_{1}\overline{v}', -p_{2}\overline{v}'', \overline{v}', -\overline{v})(c-)U(c-){}\\ &&-\xi((p_{2}\overline{v}'')'-p_{1}\overline{v}', -p_{2}\overline{v}'', \overline{v}', -\overline{v})(c+)U(c+), \end{eqnarray}$

$\xi(C+H)J_{0}(C+H)^{T} = \eta DJ_{0}D^{T}$可得$\frac{\xi}{\eta}D^{-1}(C+H)J_{0}(C+H)^{T} = J_{0}D^{T},$因此

$\begin{eqnarray} &&((p_{2}\overline{v}'')'-p_{1}\overline{v}', -p_{2}\overline{v}'', \overline{v}', -\overline{v})(c-){}\\ & = &(\overline{v}, \overline{v}', p_{2}\overline{v}'', (p_{2}\overline{v}'')'-p_{1}\overline{v}')(c-)J_{0}^{-1}{}\\ & = &-(\overline{v}, \overline{v}', p_{2}\overline{v}'', (p_{2}\overline{v}'')'-p_{1}\overline{v}')(c+)D^{\ast}[(C+H)^{\ast}]^{-1}J_{0}^{-1}{}\\ & = &-((p_{2}\overline{v}'')'-p_{1}\overline{v}', -p_{2}\overline{v}'', \overline{v}', -\overline{v})(c+)J_{0}D^{T}[(C+H)^{T}]^{-1}J_{0}^{-1}{}\\ & = &-\frac{\xi}{\eta}V^{\ast}(c+)J_{0}^{-1}D^{-1}(C+H), \end{eqnarray}$

$H\rightarrow0$, 由Fr$\acute{\rm{e}}$chet导数的定义得(5.5) 式成立. 对(5.6) 式的证明类似, 因此省略.

4. 取$h$充分小, 令特征值$\mu = \lambda(\frac{1}{p_{2}}), $$\nu = \lambda(\frac{1}{p_{2h}}) 对应的正规化特征向量分别为 (u, u_{1}, u_{2})^{T},$$ (v, v_{1}, v_{2})^{T},$其中$u = u(x, \frac{1}{p_{2}}), $$v = u(x, \frac{1}{p_{2h}}),$$ \frac{1}{p_{2h}} = \frac{1}{p_{2}}+h, $$h\in L^{1}(a, b),$$ p_{2}-p_{2h} = p_{2}p_{2h} h.$

$h\rightarrow0$, 由定义4.1和定理4.2知, (5.7) 式成立.

5. 取$h$充分小, 令特征值$\mu = \lambda(p_{1}), $$\nu = \lambda(p_{1}+h) 对应的正规化特征向量分别为 (u, u_{1}, u_{2})^{T},$$ (v, v_{1}, v_{2})^{T},$其中$u = u(x, p_{1}), v = u(x, p_{1}+h),$

$h\rightarrow0$, 由定义4.1和定理4.2知, (5.8) 式成立. (5.9)和(5.10) 式证明与此类似, 省略.

1. 固定$\omega$中除$c_{1}$之外的所有变量, 令$\lambda = \lambda(c_{1})$为特征值, 则$\lambda$是可微的且有

$\begin{eqnarray} \lambda'(c_{1}) = \eta(U')^{\ast}(c_{1}, c_{1})J_{0}U(c_{1}, c_{1}){\quad} a.e.\ c_{1}\in[a, c). \end{eqnarray}$

因为(5.22)和(5.23) 式证明类似, 我们只证明(5.22) 式. 让$|h|$充分小($h<0$), 令特征值$\mu = \lambda(c_{1}), \nu = \lambda(c_{1}+h)$对应的正规化特征向量分别为$(u, u_{1}, u_{2})^{T}, (v, v_{1}, v_{2})^{T},$其中$u = u(x, c_{1}), v = u(x, c_{1}+h),$我们有

$\begin{eqnarray} &&[\lambda(c_{1}+h)-\lambda(c_{1})][\eta\int_{a}^{c}u\overline{v}w{\rm d}x+\xi\int_{c}^{b}u\overline{v}w{\rm d}x+\eta\rho_{1}u_{1}\overline{v}_{1}+\eta\rho_{2}u_{2}\overline{v}_{2}]{}\\ & = &\eta[u, v](c-)-\xi[u, v](c+){}\\ & = &\eta\big\{u(c_{1}, c_{1})[((p_{2}\overline{u}'')'-p_{1}\overline{u}')(c_{1}, c_{1}+h)-((p_{2}\overline{u}'')'-p_{1}\overline{u}')(c_{1}+h, c_{1}+h)]{}\\ &&-u'(c_{1}, c_{1})[(p_{2}\overline{u}'')(c_{1}, c_{1}+h)-(p_{2}\overline{u}'')(c_{1}+h, c_{1}+h)]{}\\ &&+(p_{2}u'')(c_{1}, c_{1})[\overline{u}'(c_{1}, c_{1}+h)-\overline{u}'(c_{1}+h, c_{1}+h)]{}\\ &&-((p_{2}u'')'-p_{1}u')(c_{1}, c_{1})[\overline{u}(c_{1}, c_{1}+h)-\overline{u}(c_{1}+h, c_{1}+h)]\big\}, \end{eqnarray}$

$\begin{eqnarray} &&((p_{2}\overline{u}'')'-p_{1}\overline{u}')(c_{1}, c_{1}+h)-((p_{2}\overline{u}'')'-p_{1}\overline{u}')(c_{1}+h, c_{1}+h){}\\ & = &-\int_{c_{1}}^{c_{1}+h}[(p_{2}\overline{u}'')''-(p_{1}\overline{u}')'](s, c_{1}+h){\rm d}s{}\\ & = &-\int_{c_{1}}^{c_{1}+h}[\lambda(c_{1}+h)w(s)\overline{u}(s, c_{1}+h)-q(s)\overline{u}(s, c_{1}+h)]{\rm d}s{}\\ & = &-\lambda(c_{1}+h)\int_{c_{1}}^{c_{1}+h}w(s)\overline{u}(s, c_{1}){\rm d}s+\lambda(c_{1}+h)\int_{c_{1}}^{c_{1}+h}w(s)[\overline{u}(s, c_{1})-\overline{u}(s, c_{1}+h)]{\rm d}s{}\\ &&+\int_{c_{1}}^{c_{1}+h}q(s)\overline{u}(s, c_{1}){\rm d}s-\int_{c_{1}}^{c_{1}+h}q(s)[\overline{u}(s, c_{1}) -\overline{u}(s, c_{1}+h)]{\rm d}s \end{eqnarray}$

$\begin{eqnarray} &&(p_{2}\overline{u}'')(c_{1}, c_{1}+h)-(p_{2}\overline{u}'')(c_{1}+h, c_{1}+h){}\\ & = &-\int_{c_{1}}^{c_{1}+h}(p_{2}\overline{u}'')'(s, c_{1}+h){\rm d}s{}\\ & = &-\int_{c_{1}}^{c_{1}+h}(p_{2}\overline{u}'')'(s, c_{1}){\rm d}s+ \int_{c_{1}}^{c_{1}+h}\{[((p_{2}\overline{u}'')'-p_{1}\overline{u}')(s, c_{1}){}\\ &&-((p_{2}\overline{u}'')'-p_{1}\overline{u}')(s, c_{1}+h)]-p_{1}[\overline{u}'(s, c_{1}+h)-\overline{u}'(s, c_{1})]\}{\rm d}s, \end{eqnarray}$

让$h$充分小, 令特征值$\mu = \lambda(b), \nu = \lambda(b+h)$对应的正规化特征向量分别为$(u, u_{1}, u_{2})^{T}, (v, v_{1}, v_{2})^{T},$其中$u = u(x, b), v = u(x, b+h)$. 由边界条件(2.4)–(2.5) 和(3.5) 式得到

$\begin{eqnarray} &&[\lambda(b+h)-\lambda(b)][\eta\int_{a}^{c}u\overline{v}w{\rm d}x+\xi\int_{c}^{b}u\overline{v}w{\rm d}x+\eta\rho_{1}u_{1}\overline{v}_{1}+\eta\rho_{2}u_{2}\overline{v}_{2}]{}\\ & = &-\eta[u, v](a)+\xi[u, v](b)+\eta[u, v](a){}\\ & = &\xi[(p_{2}u'')(b, b)\overline{u}'(b, b+h)-((p_{2}u'')'-p_{1}u')(b, b)\overline{u}(b, b+h)], \end{eqnarray}$

让$h$充分小, 令特征值$\mu = \lambda(b), \nu = \lambda(b+h)$对应的正规化特征向量分别为$(u, u_{1}, u_{2})^{T}, (v, v_{1}, v_{2})^{T},$其中$u = u(x, b), v = u(x, b+h)$. 由边界条件(2.4)–(2.5) 和(3.5) 式得到

$\begin{eqnarray} &&[\lambda(b+h)-\lambda(b)][\eta\int_{a}^{c}u\overline{v}w{\rm d}x+\xi\int_{c}^{b}u\overline{v}w{\rm d}x+\eta\rho_{1}u_{1}\overline{v}_{1}+\eta\rho_{2}u_{2}\overline{v}_{2}]{}\\ & = &-\eta[u, v](a)+\xi[u, v](b)+\eta[u, v](a){}\\ & = &\xi[u(b, b)((p_{2}\overline{u}'')'-p_{1}\overline{u}')(b, b+h)-u'(b, b)(p_{2}\overline{u}'')(b, b+h)], \end{eqnarray}$

2. 固定$\omega$中除$b$之外的所有变量, 令特征值$\lambda = \lambda(b)$, 则$\lambda$是可微的且有

$\begin{eqnarray} \lambda'(b) = \xi(U')^{\ast}(b, b)J_{0}U(b, b) {\quad} a.e.\ (c, b'). \end{eqnarray}$

显然(5.32) 式是(5.27) 式和(5.29) 式的结合, 且(5.31) 式的证明同(5.32) 式类似, 因此证明过程省略.

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