某耦合边界条件下四阶常微分算子特征值的秩

The Rank of Eigenvalue of a Four-Order Ordinary Differential Operator under Coupled Boundary Condition

摘要: 本文讨论了一类具有耦合边界条件的四阶常微分算子，利用该算子相应算式的拉格朗日等式，得到问题特征值的秩与某整函数的零点重数之间的联系。

Abstract: In this paper, a class of four-order ordinary differential operators with coupled boundary conditions is discussed. By using the Lagrange equation of the operator's corresponding formula, the relation between the rank of the eigenvalue of the problem and the zero point weight of an entire function is obtained.

1. 问题的提出

在微分算子的领域中，许多学者对特征值问题进行了研究，如文献 [1] - [7] ，这些文献的研究方法均为将特征值问题的讨论转化为一个整函数零点的讨论，本文研究了一类具有耦合边界条件的四阶微分算子，得到问题特征值的秩与某整函数的零点重数之间的联系。

记 $L\equiv -\frac{{\text{d}}^4}{\text{d}{x}^4}+q\left(x\right)$ ， $x\in \left(0,1\right)$ ， $q\left(x\right)\in C\left[0,1\right]$ 。考虑以下耦合边界条件下的特征值问题：

$Ly=-y^{\left(4\right)}+q\left(x\right)y=\lambda y$ (1)

$y\left(0\right)=y^{\prime }\left(1\right),y^{\prime }\left(0\right)=y^{″}\left(1\right),y^{″}\left(0\right)=y^{\left(3\right)}\left(1\right),y^{\left(3\right)}\left(0\right)=y\left(1\right)$ (2)

设 $Q_i\left(x,\lambda \right)$ $\left(i=1,2,3,4\right)$ 是四个满足(1)式的解，它们分别在 $x=0$ 处满足

$\left(\begin{array}{c}{Q}_1\left(0,\lambda \right)\\ {Q^{\prime }}_1\left(0,\lambda \right)\\ {Q^{″}}_1\left(0,\lambda \right)\\ Q_1^{\left(3\right)}\left(0,\lambda \right)\end{array}\right)=\left(\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}{Q}_2\left(0,\lambda \right)\\ {Q^{\prime }}_2\left(0,\lambda \right)\\ {Q^{″}}_2\left(0,\lambda \right)\\ Q_2^{\left(3\right)}\left(0,\lambda \right)\end{array}\right)=\left(\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right)$ ,

$\left(\begin{array}{c}{Q}_3\left(0,\lambda \right)\\ {Q^{\prime }}_3\left(0,\lambda \right)\\ {Q^{″}}_3\left(0,\lambda \right)\\ Q_3^{\left(3\right)}\left(0,\lambda \right)\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right),\left(\begin{array}{c}{Q}_4\left(0,\lambda \right)\\ {Q^{\prime }}_4\left(0,\lambda \right)\\ {Q^{″}}_4\left(0,\lambda \right)\\ Q_4^{\left(3\right)}\left(0,\lambda \right)\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\\ 1\end{array}\right)$ ,

显然，Wronsky行列式 $W\left[Q_1\left(x\right),Q_2\left(x\right),Q_3\left(x\right),Q_4\left(x\right)\right]=1$ ， $Q_1\left(x\right),Q_2\left(x\right),Q_3\left(x\right),Q_4\left(x\right)$ 是(1)的四个线性无关的解，且可以构成(1)的基础解系。

设 $y\left(x\right)={\displaystyle \underset{i=1}{\overset{4}{\sum }}{c}_i}{Q}_i\left(x,{\lambda }_0\right)$ 是问题(1) (2)的特征函数，其中 ${\lambda }_0$ 是特征值，则 $y\left(x\right)$ 满足(2)式，即：

$\{\begin{array}{c}{c}_1=c_1{Q^{\prime }}_1\left(1,{\lambda }_0\right)+c_2{Q^{\prime }}_2\left(1,{\lambda }_0\right)+c_3{Q^{\prime }}_3\left(1,{\lambda }_0\right)+c_4{Q^{\prime }}_4\left(1,{\lambda }_0\right)\\ c_2=c_1{Q^{″}}_1\left(1,{\lambda }_0\right)+c_2{Q^{″}}_2\left(1,{\lambda }_0\right)+c_3{Q^{″}}_3\left(1,{\lambda }_0\right)+c_4{Q^{″}}_4\left(1,{\lambda }_0\right)\\ c_3=c_1{Q^{‴}}_1\left(1,{\lambda }_0\right)+c_2{Q^{‴}}_2\left(1,{\lambda }_0\right)+c_3{Q^{‴}}_3\left(1,{\lambda }_0\right)+c_4{Q^{‴}}_4\left(1,{\lambda }_0\right)\\ c_4=c_1{Q}_1\left(1,{\lambda }_0\right)+c_2{Q}_2\left(1,{\lambda }_0\right)+c_3{Q}_3\left(1,{\lambda }_0\right)+c_4{Q}_4\left(1,{\lambda }_0\right)\end{array}$ (3)

方程组(3)的系数行列式为 $k\left({\lambda }_0\right)$ ，即：

$k\left({\lambda }_0\right)=\left|\begin{array}{cccc}{Q^{\prime }}_1\left(1,{\lambda }_0\right)-1& {Q^{″}}_1\left(1,{\lambda }_0\right)& Q_1^{(3)}\left(1,{\lambda }_0\right)& Q_1\left(1,{\lambda }_0\right)\\ {Q^{\prime }}_2\left(1,{\lambda }_0\right)& {Q^{″}}_2\left(1,{\lambda }_0\right)-1& Q_2^{(3)}\left(1,{\lambda }_0\right)& Q_2\left(1,{\lambda }_0\right)\\ {Q^{\prime }}_3\left(1,{\lambda }_0\right)& {Q^{″}}_3\left(1,{\lambda }_0\right)& Q_3^{(3)}\left(1,{\lambda }_0\right)-1& Q_3\left(1,{\lambda }_0\right)\\ {Q^{\prime }}_4\left(1,{\lambda }_0\right)& {Q^{″}}_4\left(1,{\lambda }_0\right)& Q_4^{(3)}\left(1,{\lambda }_0\right)& Q_4\left(1,{\lambda }_0\right)-1\end{array}\right|$ (4)

方程组(3)可以看作是未知数 $c_1,c_2,c_3,c_4$ 的方程组，由特征函数的定义知，此方程组存在非零解，故系数行列式 $k\left({\lambda }_0\right)=0$ 。类似于文献 [1] 中命题1，2的证明，我们可以得到如下两个命题。

命题1：记

显然， $k\left(\lambda \right)=\left|\Omega \left(\lambda \right)\right|$ 。 $R\left(\lambda \right)$ 是特征值的秩，则以下两结论成立：

i) ${\lambda }_0$ 为问题(1)(2)的特征值 $\iff k\left({\lambda }_0\right)=0$ 。

ii) $R\left({\lambda }_0\right)+R\left(\Omega \left({\lambda }_0\right)\right)=4$ 。

命题2：设 $H_i\left(x,\lambda \right)\left(i=1,2,3,4\right)$ 是四个满足(1)式的解，记为

$\left(\begin{array}{c}{H}_1\left(x,\lambda \right)\\ H_2\left(x,\lambda \right)\\ H_3\left(x,\lambda \right)\\ H_4\left(x,\lambda \right)\end{array}\right)=A\left(\begin{array}{c}{Q}_1\left(x,\lambda \right)\\ Q_2\left(x,\lambda \right)\\ Q_3\left(x,\lambda \right)\\ Q_4\left(x,\lambda \right)\end{array}\right),A=\left(\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ a_{21}& a{}_{22}& a_{23}& a_{24}\\ a_{31}& a_{32}& a_{33}& a_{34}\\ a_{41}& a_{42}& a_{43}& a_{44}\end{array}\right)$ . (5)

记 $T_i={\left(a_{i1},a_{i2},a_{i3},a_{i4}\right)}^{{\rm T}}$ ， $i=1,2,3,4$ ，其中 $T_i\ne 0$ 。如果 $k\left({\lambda }_0\right)=0$ 且 ${\Omega }^{{\rm T}}\left(\lambda \right)T_i=0$ ，则：

i) $H_i\left(x,{\lambda }_0\right)$ 是问题(1)(2)的特征函数，则满足

$\{\begin{array}{l}{H}_i\left(0,{\lambda }_0\right)={H^{\prime }}_i\left(1,{\lambda }_0\right)=a_{i1}\hfill \\ {H^{\prime }}_i\left(0,{\lambda }_0\right)={H^{″}}_i\left(1,{\lambda }_0\right)=a_{i2}\hfill \\ {H^{″}}_i\left(0,{\lambda }_0\right)=H_i^{\left(3\right)}\left(1,{\lambda }_0\right)=a_{i3}\hfill \\ H_i^{\left(3\right)}\left(0,{\lambda }_0\right)=H_i\left(1,{\lambda }_0\right)=a_{i4}\hfill \end{array}\left(i=1,2,3,4\right)$

ii) $H_1,H_2,H_3,H_4$ 的极大无关组个数为 $R\left(A\right)$ 。

引理1：算式 $L\equiv -\frac{{\text{d}}^4}{\text{d}{x}^4}+q\left(x\right)$ 有以下的Lagrange恒等式

${\displaystyle {\int }_a^{b}g\left(Lf\right)\text{d}x}-{\displaystyle {\int }_a^{b}f\left(Lg\right)\text{d}x}={\left[\left|\begin{array}{cc}f& g\\ f^{‴}& g^{‴}\end{array}\right|-\left|\begin{array}{cc}{f}^{\prime }& g^{\prime }\\ f^{″}& g^{″}\end{array}\right|\right]}_a^b$

引理2：设 $H_i\left(x,{\lambda }_0\right)$ $\left(i=1,2,3,\mathrm{4\; <!--\; mn-->}\right)$ 是(5)式所给出的特征函数。 ${\lambda }_0$ 是 $k\left(\lambda \right)$ 的零点，则有：

$\Omega \left({\lambda }_0\right)B=\left(\begin{array}{cccc}-a_{12}-a_{14}& -a_{22}-a_{24}& -a_{32}-a_{34}& -a_{42}-a_{44}\\ a_{11}+a_{13}& a_{21}+a_{23}& a_{31}+a_{33}& a_{41}+a_{43}\\ -a_{12}-a_{14}& -a_{22}-a_{24}& -a_{32}-a_{34}& -a_{42}-a_{44}\\ a_{11}+a_{13}& a_{21}+a_{23}& a_{31}+a_{33}& a_{41}+a_{43}\end{array}\right)$

其中

$B=\left(\begin{array}{cccc}{a}_{12}& a_{22}& a_{32}& a_{42}\\ -a_{11}& -a_{21}& -a_{31}& -a_{41}\\ a_{14}& a_{24}& a_{34}& a_{44}\\ -a_{13}& -a_{23}& -a_{33}& -a_{43}\end{array}\right)$

证明：利用引理1有

$\begin{array}{c}0={\displaystyle {\int }_0^1{Q}_j\left(x,{\lambda }_0\right)}\left(L{H}_i\left(x,{\lambda }_0\right)\right)\text{d}x-{\displaystyle {\int }_0^1\left(L{Q}_j\left(x,{\lambda }_0\right)\right)H_i\left(x,{\lambda }_0\right)\text{d}x}\\ ={\left[\left|\begin{array}{cc}{H}_i\left(x,{\lambda }_0\right)& Q_j\left(x,{\lambda }_0\right)\\ H_i^{\left(3\right)}\left(x,{\lambda }_0\right)& Q_j^{\left(3\right)}\left(x,{\lambda }_0\right)\end{array}\right|-\left|\begin{array}{cc}{H^{\prime }}_i\left(x,{\lambda }_0\right)& {Q^{\prime }}_j\left(x,{\lambda }_0\right)\\ {H^{″}}_i\left(x{\lambda }_0\right)& {Q^{″}}_j\left(x,{\lambda }_0\right)\end{array}\right|\right]}_0^1\\ =H_i\left(1,{\lambda }_0\right)Q_j^{\left(3\right)}\left(1,{\lambda }_0\right)-H_i^{\left(3\right)}\left(1,{\lambda }_0\right)Q_j\left(1,{\lambda }_0\right)-H_i\left(0,{\lambda }_0\right)Q_j^{\left(3\right)}\left(0,{\lambda }_0\right)+H_i^{\left(3\right)}\left(0,{\lambda }_0\right)Q_j\left(0,{\lambda }_0\right)\\ -{H^{\prime }}_i\left(1,{\lambda }_0\right){Q^{″}}_j\left(1,{\lambda }_0\right)+{H^{″}}_i\left(1,{\lambda }_0\right){Q^{\prime }}_j\left(1,{\lambda }_0\right)+{H^{\prime }}_i\left(0,{\lambda }_0\right){Q^{″}}_j\left(0,{\lambda }_0\right)-{H^{″}}_i\left(0,{\lambda }_0\right){Q^{\prime }}_j\left(0,{\lambda }_0\right)\end{array}$

$\begin{array}{l}=a_{i4}{Q}_j^{(3)}\left(1,{\lambda }_0\right)-a_{i3}{Q}_j\left(1,{\lambda }_0\right)-a_{i1}{Q}_j^{\left(3\right)}\left(0,{\lambda }_0\right)+a_{i4}{Q}_j\left(0,{\lambda }_0\right)-a_{i1}{Q^{″}}_j\left(1,{\lambda }_0\right)\\ +a_{i2}{Q^{\prime }}_j\left(1,{\lambda }_0\right)+a_{i2}{Q^{″}}_j\left(0,{\lambda }_0\right)-a_{i3}{Q^{\prime }}_j\left(0,{\lambda }_0\right)\end{array}$

分别取 $i,j=1,2,3,4$ ，即得结论。

引理3：设 $H_i\left(x,{\lambda }_0\right)$ 是(5)式所给出的特征函数 $\left(i=1,2,3,4\right)$ ， ${\lambda }_0$ 是特征值。简记 $H_i\left(x,{\lambda }_0\right)=H_i$ ，记一元函数 $H_i^{(j-1)}\left(1,\lambda \right)$ 关于 $\lambda$ 求导后在处的取值为 ${\stackrel{\dot{}}{H}}_i^{\left(j-1\right)}\left(1,{\lambda }_0\right)$ ，记两个函数 $H{}_i$ ， $H_j$ 的内积为 $\langle H_i,H_j\rangle$ ，其定义为

$\langle H_i,H_j\rangle ={\displaystyle {\int }_0^1{H}_i}\left(x\right)H_j\left(x\right)\text{d}x$ ,

这样可以得到16个恒等式，写成矩阵形式如下：

$\begin{array}{l}\left(\begin{array}{cccc}\langle H_1,H_1\rangle & \langle H_1,H_2\rangle & \langle H_1,H_3\rangle & \langle H_1,H_4\rangle \\ \langle H_2,H_1\rangle & \langle H_2,H_2\rangle & \langle H_2,H_3\rangle & \langle H_2,H_4\rangle \\ \langle H_3,H_1\rangle & \langle H_3,H_2\rangle & \langle H_3,H_3\rangle & \langle H_3,H_4\rangle \\ \langle H_4,H_1\rangle & \langle H_4,H_2\rangle & \langle H_4,H_3\rangle & \langle H_4,H_4\rangle \end{array}\right)\\ =\left(\begin{array}{cccc}{{\stackrel{\dot{}}{H}}^{\prime }}_1& {{\stackrel{\dot{}}{H}}^{″}}_1& {\stackrel{\dot{}}{H}}_1^{\left(3\right)}& {\stackrel{\dot{}}{H}}_1\\ {{\stackrel{\dot{}}{H}}^{\prime }}_2& {\stackrel{\dot{}}{{H^{\prime }}^{\prime }}}_2& {\stackrel{\dot{}}{H}}_2^{\left(3\right)}& {\stackrel{\dot{}}{H}}_2\\ {{\stackrel{\dot{}}{H}}^{\prime }}_3& {\stackrel{\dot{}}{{H^{\prime }}^{\prime }}}_3& {\stackrel{\dot{}}{H}}_3^{\left(3\right)}& {\stackrel{\dot{}}{H}}_3\\ {{\stackrel{\dot{}}{H}}^{\prime }}_4& {\stackrel{\dot{}}{{H^{\prime }}^{\prime }}}_4& {\stackrel{\dot{}}{H}}_4^{\left(3\right)}& {\stackrel{\dot{}}{H}}_4\end{array}\right)\left(\begin{array}{cccc}-a_{12}& -a_{22}& -a_{32}& -a_{42}\\ a_{11}& a_{21}& a_{31}& a_{41}\\ -a_{14}& -a_{24}& -a_{34}& -a_{44}\\ a_{13}& a_{23}& a_{33}& a_{43}\end{array}\right)\end{array}$

证明：

$\begin{array}{c}\langle H_i,H_j\rangle =\underset{\lambda \to {\lambda }_0}{{\displaystyle \lim }}\frac{1}{\lambda -{\lambda }_0}\left(\lambda -{\lambda }_0\right){\displaystyle {\int }_0^1{H}_i}\left(x,\lambda \right)H_j\left(x,{\lambda }_0\right)\text{d}x\\ =\underset{\lambda \to {\lambda }_0}{{\displaystyle \lim }}\frac{1}{\lambda -{\lambda }_0}\left[{\displaystyle {\int }_0^1{H}_j}\left(x,{\lambda }_0\right)\left(L{H}_i\left(x,\lambda \right)\right)\text{d}x-{\displaystyle {\int }_0^1{H}_i}\left(x,\lambda \right)\left(L{H}_j\left(x,{\lambda }_0\right)\right)\text{d}x\right]\\ =\underset{\lambda \to {\lambda }_0}{{\displaystyle \lim }}\frac{1}{\lambda -{\lambda }_0}\left\{{\left|\begin{array}{cc}{H}_i\left(x,\lambda \right)& H_j\left(x,{\lambda }_0\right)\\ H_i^{\left(3\right)}\left(x,\lambda \right)& H_j^{\left(3\right)}\left(x,{\lambda }_0\right)\end{array}\right|}_0^1-{\left|\begin{array}{cc}{H^{\prime }}_i\left(x,\lambda \right)& {H^{\prime }}_j\left(x,{\lambda }_0\right)\\ {H^{″}}_i\left(x,\lambda \right)& {H^{″}}_j\left(x,{\lambda }_0\right)\end{array}\right|}_0^1\right\}\end{array}$

$\begin{array}{l}=\underset{\lambda \to {\lambda }_0}{{\displaystyle \lim }}\frac{1}{\lambda -{\lambda }_0}\{\left[H_i\left(1,\lambda \right){H^{‴}}_j\left(1,{\lambda }_0\right)-H_j\left(1,{\lambda }_0\right){H^{‴}}_i\left(1,\lambda \right)-H_i\left(0,\lambda \right){H^{‴}}_j\left(0,{\lambda }_0\right)+H_j\left(0,{\lambda }_0\right){H^{‴}}_i\left(0,\lambda \right)\right]\\ +\left[-{H^{\prime }}_i\left(1,\lambda \right){H^{″}}_j\left(1,{\lambda }_0\right)+{H^{\prime }}_j\left(1,{\lambda }_0\right){H^{″}}_i\left(1,\lambda \right)+{H^{\prime }}_i\left(0,\lambda \right){H^{″}}_j\left(0,{\lambda }_0\right)-{H^{\prime }}_j\left(0,{\lambda }_0\right){H^{″}}_i\left(0,\lambda \right)\right]\}\\ =\underset{\lambda \to {\lambda }_0}{{\displaystyle \lim }}\frac{1}{\lambda -{\lambda }_0}\left[a_{j3}{H}_i\left(1,\lambda \right)-a_{j4}{H}_i^{\left(3\right)}\left(1,\lambda \right)-a_{j2}{H^{\prime }}_i\left(1,\lambda \right)+a_{j1}{H^{″}}_i\left(1,\lambda \right)+h\right]\end{array}$

其中 $h=-a_{i1}{a}_{j4}+a_{i4}{a}_{j1}+a_{i2}{a}_{j3}-a_{i3}{a}_{j2}$ ，接着利用洛必达法则，可得

$\langle H_i,H_j\rangle =a_{j3}{\stackrel{\dot{}}{H}}_i\left(1,{\lambda }_0\right)-a_{j4}{\stackrel{\dot{}}{H}}_i^{(3)}\left(1,{\lambda }_0\right)-a_{j2}{{\stackrel{\dot{}}{H}}^{\prime }}_i\left(1,{\lambda }_0\right)+a_{j1}{{\stackrel{\dot{}}{H}}^{″}}_i\left(1,{\lambda }_0\right)$

当 $i,j$ 分别取1，2，3，4时，就可以得到定理的结论。

2. 主要结果与证明

定理1：设 ${\lambda }_0$ 为问题(1) (2)的特征值，其秩记为 $R\left({\lambda }_0\right)$ ，则

1) $R\left({\lambda }_0\right)=4$ 的充要条件是 $k\left({\lambda }_0\right)=k^{\prime }\left({\lambda }_0\right)=k^{″}\left({\lambda }_0\right)=k^{\left(3\right)}\left({\lambda }_0\right)=0$ 且 $k^{\left(4\right)}\left({\lambda }_0\right)\ne 0$ 。

2) 当满足条件 $a_{i2}+a_{i4}=0$ 或 $a_{i1}+a_{i3}=0\left(i=1,2,3,4\right)$ 时， $R\left({\lambda }_0\right)=3$ 的充要条件是

3) 当满足条件 $a_{i2}+a_{i4}=0$ 和 $a_{i1}+a_{i3}=0\left(i=1,2,3,4\right)$ 时， $R\left({\lambda }_0\right)=2$ 的充要条件是

$k\left({\lambda }_0\right)=k^{\prime }\left({\lambda }_0\right)=0,k^{″}\left({\lambda }_0\right)\ne 0$

4) 当满足条件 $a_{i2}+a_{i4}=0$ 和时，的充要条件是

$k\left({\lambda }_0\right)=0,k^{\prime }\left({\lambda }_0\right)\ne 0$

证明：(1)~(4)的充分性显然成立。下面证明必要性。

1) $R\left(A\right)=4$ ， ${\lambda }_0$ 对应4个线性无关的特征函数 $H_i\left(x,{\lambda }_0\right)$ ， $i=1,2,3,4$ 。

$\left|A\right|k\left({\lambda }_0\right)=\left|A\Omega \left({\lambda }_0\right)\right|=\left|\begin{array}{cccc}{H^{\prime }}_1\left(1,{\lambda }_0\right)-a_{11}& {H^{″}}_1\left(1,{\lambda }_0\right)-a_{12}& {H^{‴}}_1\left(1,{\lambda }_0\right)-a_{13}& H_1\left(1,{\lambda }_0\right)-a_{14}\\ {H^{\prime }}_2\left(1,{\lambda }_0\right)-a_{21}& {H^{″}}_2\left(1,{\lambda }_0\right)-a_{22}& {H^{‴}}_2\left(1,{\lambda }_0\right)-a_{23}& H_2\left(1,{\lambda }_0\right)-a_{24}\\ {H^{\prime }}_3\left(1,{\lambda }_0\right)-a_{31}& {H^{″}}_3\left(1,{\lambda }_0\right)-a_{32}& {H^{‴}}_3\left(1,{\lambda }_0\right)-a_{33}& H_3\left(1,{\lambda }_0\right)-a_{34}\\ {H^{\prime }}_4\left(1,{\lambda }_0\right)-a_{41}& {H^{″}}_4\left(1,{\lambda }_0\right)-a_{42}& {H^{‴}}_4\left(1,{\lambda }_0\right)-a_{43}& H_4\left(1,{\lambda }_0\right)-a_{44}\end{array}\right|=0$

因为 ${H^{\prime }}_i\left(1,{\lambda }_0\right)=a_{i1},{H^{″}}_i\left(1,{\lambda }_0\right)=a_{i2},{H^{‴}}_i\left(1,{\lambda }_0\right)=a_{i3},H_i\left(1,{\lambda }_0\right)=a_{i4}$ ，且 $\left|A\right|\ne 0$ ，故 $k\left({\lambda }_0\right)=0$ 。

又因为求导后展开的每个行列式中至少有一行未被求导，从而行列式 $\left|A\right|k^{\prime }\left({\lambda }_0\right)=\left|A\right|k^{″}\left({\lambda }_0\right)=\left|A\right|k^{‴}\left({\lambda }_0\right)=0$ 。同理可得 $k^{\prime }\left({\lambda }_0\right)=k^{″}\left({\lambda }_0\right)=k^{‴}\left({\lambda }_0\right)=0$ 。

对 $\left|A\right|k\left(\lambda \right)$ 关于 $\lambda$ 求四阶导数后，在 $\lambda ={\lambda }_0$ 处只剩一项。

$\left|A\right|k^{\left(4\right)}\left({\lambda }_0\right)=\left|\begin{array}{cccc}{{\stackrel{\dot{}}{H}}^{\prime }}_1\left(1,{\lambda }_0\right)& {{\stackrel{\dot{}}{H}}^{″}}_1\left(1,{\lambda }_0\right)& {\stackrel{\dot{}}{H}}_1^{\left(3\right)}\left(1,{\lambda }_0\right)& {\stackrel{\dot{}}{H}}_1\left(1,{\lambda }_0\right)\\ {{\stackrel{\dot{}}{H}}^{\prime }}_2\left(1,{\lambda }_0\right)& {{\stackrel{\dot{}}{H}}^{″}}_2\left(1,{\lambda }_0\right)& {\stackrel{\dot{}}{H}}_2^{\left(3\right)}\left(1,{\lambda }_0\right)& {\stackrel{\dot{}}{H}}_2\left(1,{\lambda }_0\right)\\ {{\stackrel{\dot{}}{H}}^{\prime }}_3\left(1,{\lambda }_0\right)& {{\stackrel{\dot{}}{H}}^{″}}_3\left(1,{\lambda }_0\right)& {\stackrel{\dot{}}{H}}_3^{\left(3\right)}\left(1,{\lambda }_0\right)& {\stackrel{\dot{}}{H}}_3\left(1,{\lambda }_0\right)\\ {{\stackrel{\dot{}}{H}}^{\prime }}_4\left(1,{\lambda }_0\right)& {{\stackrel{\dot{}}{H}}^{″}}_4\left(1,{\lambda }_0\right)& {\stackrel{\dot{}}{H}}_4^{\left(3\right)}\left(1,{\lambda }_0\right)& {\stackrel{\dot{}}{H}}_4\left(1,{\lambda }_0\right)\end{array}\right|$

$\left|A\right|k^{\left(4\right)}\left({\lambda }_0\right)\left|B\right|=\left|\begin{array}{cccc}\langle H_1,H_1\rangle & \langle H_1,H_2\rangle & \langle H_1,H_3\rangle & \langle H_1,H_4\rangle \\ \langle H_2,H_1\rangle & \langle H_2,H_2\rangle & \langle H_2,H_3\rangle & \langle H_2,H_4\rangle \\ \langle H_3,H_1\rangle & \langle H_3,H_2\rangle & \langle H_3,H_3\rangle & \langle H_3,H_4\rangle \\ \langle H_4,H_1\rangle & \langle H_4,H_2\rangle & \langle H_4,H_3\rangle & \langle H_4,H_4\rangle \end{array}\right|$

上式右端是Crame行列式，故由文献 [2] 知， $\left|A\right|k^{\left(4\right)}\left({\lambda }_0\right)\left|B\right|\ne 0$ ，故 $k^{\left(4\right)}\left({\lambda }_0\right)\ne 0$ 。

2) 先证在 $a_{i2}+a_{i4}=0$ 条件下的结论，另一种情况类似证明。

$R\left(A\right)=3$ ，可以生成3个线性无关的特征函数，不妨取为：

$\left(\begin{array}{c}{H}_1\\ H_2\\ H_4\end{array}\right)=\left(\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ a_{21}& a_{22}& a_{23}& a_{24}\\ a_{41}& a_{42}& a_{43}& a_{44}\end{array}\right)\left(\begin{array}{c}{Q}_1\left(1,{\lambda }_0\right)\\ Q_2\left(1,{\lambda }_0\right)\\ Q_3\left(1,{\lambda }_0\right)\\ Q_4\left(1,{\lambda }_0\right)\end{array}\right)$

则右端 $3\times 4$ 阶矩阵至少存在一个三阶子式非零，不妨取为：

$\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{14}\\ a_{21}& a_{22}& a_{24}\\ a_{41}& a_{42}& a_{44}\end{array}\right|\ne 0$

此行列式展开的六项中至少有一项非零，为讨论方便取为 $a_{11}{a}_{22}{a}_{44}\ne 0$ 。

$k\left(\lambda \right)=\frac{1}{a_{11}{a}_{22}{a}_{44}}\left|\begin{array}{cccc}{a}_{11}\left[{Q^{\prime }}_1\left(1,\lambda \right)-1\right]& a_{11}{Q^{″}}_1\left(1,\lambda \right)& a_{11}{Q}_1^{\left(3\right)}\left(1,\lambda \right)& a_{11}{Q}_1\left(1,\lambda \right)\\ a_{22}{Q^{\prime }}_2\left(1,\lambda \right)& a_{22}\left[{Q^{″}}_2\left(1,\lambda \right)-1\right]& a_{22}{Q}_2^{\left(3\right)}\left(1,\lambda \right)& a_{22}{Q}_2\left(1,\lambda \right)\\ {Q^{\prime }}_3\left(1,\lambda \right)& {Q^{″}}_3\left(1,\lambda \right)& Q_3^{\left(3\right)}\left(1,\lambda \right)-1& Q_3\left(1,\lambda \right)\\ a_{44}{Q^{\prime }}_4\left(1,\lambda \right)& a_{44}{Q^{″}}_4\left(1,\lambda \right)& a_{44}{Q}_4^{\left(3\right)}\left(1,\lambda \right)& a_{44}\left(Q_4\left(1,\lambda \right)-1\right)\end{array}\right|$

经过初等变换得

$k\left(\lambda \right)=\frac{1}{a_{11}{a}_{22}{a}_{44}}\left|\begin{array}{cccc}{H^{\prime }}_1\left(1,\lambda \right)-a_{11}& {H^{″}}_1\left(1,\lambda \right)-a_{12}& H_1^{\left(3\right)}\left(1,\lambda \right)-a_{13}& H_1\left(1,\lambda \right)-a_{14}\\ {H^{\prime }}_2\left(1,\lambda \right)-a_{21}& {H^{″}}_2\left(1,\lambda \right)-a_{22}& H_2^{\left(3\right)}\left(1,\lambda \right)-a_{23}& H_2\left(1,\lambda \right)-a_{24}\\ {Q^{\prime }}_3\left(1,\lambda \right)& {Q^{″}}_3\left(1,\lambda \right)& Q_3^{\left(3\right)}\left(1,\lambda \right)-1& Q_3\left(1,\lambda \right)\\ {H^{\prime }}_4\left(1,\lambda \right)-a_{41}& {H^{″}}_4\left(1,\lambda \right)-a_{42}& H_4^{\left(3\right)}\left(1,\lambda \right)-a_{43}& H_4\left(1,\lambda \right)-a_{44}\end{array}\right|$

显然 $k\left({\lambda }_0\right)=0$ ， $k^{\prime }\left({\lambda }_0\right)=0$ ， $k^{″}\left({\lambda }_0\right)$ 。

$k^{(3)}\left({\lambda }_0\right)=\frac{1}{a_{11}{a}_{22}{a}_{44}}\left|\begin{array}{cccc}{\stackrel{\dot{}}{H^{\prime }}}_1\left(1,{\lambda }_0\right)& {\stackrel{\dot{}}{H^{″}}}_1\left(1,{\lambda }_0\right)& {\stackrel{\dot{}}{H}}_1^{\left(3\right)}\left(1,{\lambda }_0\right)& {\stackrel{\dot{}}{H}}_1\left(1,{\lambda }_0\right)\\ {\stackrel{\dot{}}{H^{\prime }}}_2\left(1,{\lambda }_0\right)& {\stackrel{\dot{}}{H^{″}}}_2\left(1,{\lambda }_0\right)& {\stackrel{\dot{}}{H}}_2^{\left(3\right)}\left(1,{\lambda }_0\right)& {\stackrel{\dot{}}{H}}_2\left(1,{\lambda }_0\right)\\ {Q^{\prime }}_3\left(1,{\lambda }_0\right)& {Q^{″}}_3\left(1,{\lambda }_0\right)& Q_3^{\left(3\right)}\left(1,{\lambda }_0\right)-1& Q_3\left(1,{\lambda }_0\right)\\ {{\stackrel{\dot{}}{H}}^{\prime }}_4\left(1,{\lambda }_0\right)& {{\stackrel{\dot{}}{H}}^{″}}_4\left(1,{\lambda }_0\right)& {\stackrel{\dot{}}{H}}_4^{\left(3\right)}\left(1,{\lambda }_0\right)& {\stackrel{\dot{}}{H}}_4\left(1,{\lambda }_0\right)\end{array}\right|$

经过等值变换得

$k^{\left(3\right)}\left({\lambda }_0\right)=\frac{1}{a_{11}^2{a}_{22}^2{a}_{44}^2}\left|\begin{array}{cccc}\langle H_1,H_1\rangle & \langle H_1,H_2\rangle & \langle H_1,H_4\rangle & {\stackrel{\dot{}}{H}}_1\left(1,{\lambda }_0\right)\\ \langle H_2,H_1\rangle & \langle H_2,H_2\rangle & \langle H_2,H_4\rangle & {\stackrel{\dot{}}{H}}_2\left(1,{\lambda }_0\right)\\ -a_{12}-a_{14}& -a_{22}-a_{24}& -a_{42}-a_{44}& Q_3\left(1,{\lambda }_0\right)\\ \langle H_4,H_1\rangle & \langle H_4,H_2\rangle & \langle H_4,H_4\rangle & {\stackrel{\dot{}}{H}}_4\left(1,{\lambda }_0\right)\end{array}\right|$

当 $a_{i2}+a_{i4}=0\left(i=1,2,4\right)$ 时，

$k^{\left(3\right)}\left({\lambda }_0\right)=-\frac{Q_3\left(1,{\lambda }_0\right)}{a_{11}^2{a}_{22}^2{a}_{33}^2}\left|\begin{array}{ccc}\langle H_1,H_1\rangle & \langle H_1,H_2\rangle & \langle H_1,H_4\rangle \\ \langle H_2,H_1\rangle & \langle H_2,H_2\rangle & \langle H_2,H_4\rangle \\ \langle H_4,H_1\rangle & \langle H_4,H_2\rangle & \langle H_4,H_4\rangle \end{array}\right|$

下面说明 $Q_3\left(1,{\lambda }_0\right)\ne 0$ ，由以上假设知：

$W\left[H_1,H_2,H_4,Q_3\right]=\left|\begin{array}{cccc}{a}_{11}& a_{21}& a_{41}& 0\\ a_{12}& a{}_{22}& a_{42}& 0\\ a_{13}& a_{23}& a_{43}& 1\\ a_{14}& a_{24}& a_{44}& 0\end{array}\right|=-\left|\begin{array}{ccc}{a}_{11}& a_{21}& a_{41}\\ a_{12}& a_{22}& a_{42}\\ a_{14}& a_{24}& a_{44}\end{array}\right|\ne 0$

因此 $H_1,H_2,H_4,Q_3$ 是线性无关的。

另一方面，由引理2以及 $a_{i2}+a_{i4}=0\left(i=1,2,3,4\right)$ 知，

$\left(\begin{array}{cccc}{a}_{12}& -a_{11}& a_{14}& -a_{13}\\ a_{22}& -a_{21}& a_{24}& -a_{23}\\ a_{32}& -a_{31}& a_{34}& -a_{33}\\ a_{42}& -a_{41}& a_{44}& -a_{43}\end{array}\right)\left(\begin{array}{c}{Q^{\prime }}_3\left(1,{\lambda }_0\right)\\ {Q^{″}}_3\left(1,{\lambda }_0\right)\\ Q_3^{\left(3\right)}\left(1,{\lambda }_0\right)-1\\ Q_3\left(1,{\lambda }_0\right)\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right)$

假设 $Q_3\left(1,{\lambda }_0\right)=0$ ，由上式可得到

$\left(\begin{array}{ccc}{a}_{12}& -a_{11}& a_{14}\\ a_{22}& -a_{21}& a_{24}\\ a_{42}& -a_{41}& a_{44}\end{array}\right)\left(\begin{array}{c}{Q^{\prime }}_3\left(1,{\lambda }_0\right)\\ {Q^{″}}_3\left(1,{\lambda }_0\right)\\ Q_3^{\left(3\right)}\left(1,{\lambda }_0\right)-1\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)$

因为系数行列式不等于零，故有 ${Q^{\prime }}_3\left(1,{\lambda }_0\right)={Q^{″}}_3\left(1,{\lambda }_0\right)=0$ ， $Q_3^{\left(3\right)}\left(1,{\lambda }_0\right)=1$ ，即 $Q_3\left(1,{\lambda }_0\right)$ 也是特征函数，这与已知矛盾，故 $k^{\left(3\right)}\left({\lambda }_0\right)\ne 0$ 。

(3) (4)仿照 $R\left({\lambda }_0\right)=3$ 的方法，进行证明。

基金项目

国家自然科学基金(11661059)，内蒙古自然科学基金(2017MS(LH)0103)。