题海 · 解答 · p.543
\[
=\left[\boldsymbol{f}^{\mathrm{T}}(\boldsymbol{x})\left(\frac{\partial \boldsymbol{f}}{\partial \boldsymbol{x}}\right)^{\mathrm{T}}+\left(\frac{\partial \boldsymbol{f}}{\partial \boldsymbol{x}}\right) \boldsymbol{f}(\boldsymbol{x})\right]^{\mathrm{T}} \boldsymbol{f}(\boldsymbol{x})=2\left[\left(\frac{\partial \boldsymbol{f}}{\partial \boldsymbol{x}}\right) \boldsymbol{f}(\boldsymbol{x})\right]^{\mathrm{T}} \boldsymbol{f}(\boldsymbol{x})
\]
其中\(\left(\dfrac{\partial \boldsymbol{f}}{\partial \boldsymbol{x}}\right) \boldsymbol{f}(\boldsymbol{x})=\boldsymbol{f}^{\mathrm{T}}(\boldsymbol{x})\left(\dfrac{\partial \boldsymbol{f}}{\partial \boldsymbol{x}}\right)^{\mathrm{T}}\)。由于
\[
\boldsymbol{f}^{\mathrm{T}}(\boldsymbol{x})\left(\frac{\partial \boldsymbol{f}}{\partial \boldsymbol{x}}\right) \boldsymbol{f}(\boldsymbol{x})+\left[\left(\frac{\partial \boldsymbol{f}}{\partial \boldsymbol{x}}\right) \boldsymbol{f}(\boldsymbol{x})\right]^{\mathrm{T}} \boldsymbol{f}(\boldsymbol{x})=\boldsymbol{f}^{\mathrm{T}}(\boldsymbol{x})\left(\frac{\partial \boldsymbol{f}}{\partial \boldsymbol{x}}\right) \boldsymbol{f}(\boldsymbol{x})+\boldsymbol{f}^{\mathrm{T}}(\boldsymbol{x})\left(\frac{\partial \boldsymbol{f}}{\partial \boldsymbol{x}}\right)^{\mathrm{T}} \boldsymbol{f}(\boldsymbol{x})
\]
\[
=\boldsymbol{f}^{\mathrm{T}}(\boldsymbol{x})\left[\left(\frac{\partial \boldsymbol{f}}{\partial \boldsymbol{x}}\right)+\left(\frac{\partial \boldsymbol{f}}{\partial \boldsymbol{x}}\right)^{\mathrm{T}}\right] \boldsymbol{f}(\boldsymbol{x})
\]
\[
=\boldsymbol{f}^{\mathrm{T}}(\boldsymbol{x})\left[\boldsymbol{F}(\boldsymbol{x})+\boldsymbol{F}^{\mathrm{T}}(\boldsymbol{x})\right] \boldsymbol{f}(\boldsymbol{x})
\]
故当\(\boldsymbol{F}(\boldsymbol{x})+\boldsymbol{F}^{\mathrm{T}}(\boldsymbol{x})\)负定时,有\(\dfrac{\mathrm{d} V}{\mathrm{d} t}<0\)。表明系统的原点平衡状态\(\boldsymbol{x}_e=0\)为大范围渐近稳定。
考研参考题
9-64
设系统状态方程为
\[
\begin{bmatrix}\dot{x}_1\\ \dot{x}_2\end{bmatrix}=\begin{bmatrix}0 & 1\\ -a_1 & -a_2\end{bmatrix}=\begin{bmatrix}x_1\\ x_2\end{bmatrix}
\]
试用李雅普诺夫第二法证明:当\(a_1>0,a_2>0\)时,系统的原点是大范围一致渐近稳定的。
证明 因系统矩阵
\[
\boldsymbol{A}=\begin{bmatrix}0 & 1\\ -a_1 & -a_2\end{bmatrix}
\]
当\(a_1>0,a_2>0\)时,\(\boldsymbol{A}\)为常值非奇异阵,原点\(x_e=0\)为唯一平衡状态。
\(x_e=0\)大范围一致渐近稳定的充分必要条件是:对于给定的正定实对称矩阵\(\boldsymbol{Q}\),存在唯一的正定实对称矩阵\(\boldsymbol{P}\),使
\[
\boldsymbol{A}^{\mathrm{T}} \boldsymbol{P}+\boldsymbol{P} \boldsymbol{A}=-\boldsymbol{Q}
\]
其中,李雅普诺夫函数\(V(\boldsymbol{x})=\boldsymbol{x}^{\mathrm{T}} \boldsymbol{P} \boldsymbol{x}\)。
令\(\boldsymbol{Q}=\boldsymbol{I}\),并设\(\boldsymbol{P}=\begin{bmatrix}p_{11} & p_{12}\\ p_{12} & p_{12}\end{bmatrix}\),因
\[
\begin{bmatrix}0 & -a_1\\ 1 & -a_2\end{bmatrix}\begin{bmatrix}p_{11} & p_{12}\\ p_{12} & p_{22}\end{bmatrix}+\begin{bmatrix}p_{11} & p_{12}\\ p_{12} & p_{22}\end{bmatrix}\begin{bmatrix}0 & 1\\ -a_1 & -a_2\end{bmatrix}=-\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}
\]
于是有
\[
-2a_1 p_{12}=-1
\]
\[
p_{11}-a_1 p_{22}-a_2 p_{12}=0
\]
\[
2(p_{12}-a_2 p_{22})=-1
\]
解得
\[
\boldsymbol{P}=\begin{bmatrix}\dfrac{a_1^2+a_2^2+a_1}{2a_1a_2} & \dfrac{1}{2a_1}\\[2ex] \dfrac{1}{2a_1} & \dfrac{a_1+1}{2a_1a_2}\end{bmatrix}
\]
因为当\(a_1>0,a_2>0\)时
\[
p_{11}=\frac{a_1^2+a_2^2+a_1}{2a_1a_2}>0
\]
\[
\det \boldsymbol{P}=p_{11}p_{22}-p_{12}^2=\frac{(a_1+1)(a_1^2+a_2^2+a_1)}{4a_1^2a_2^2}-\frac{1}{4a_1^2}
\]
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