[T050101] 设 \(A\) 为 \(n\times n\) 常数矩阵, \(\Phi(t)\) 是方程组 \(X'=AX\) 的标准基解矩阵 \((\Phi(0)=E)\), 证明 \(\Phi(t)\Phi^{-1}(t_0)=\Phi(t-t_0)\), 其中 \(t_0\) 是常数.
证 由题设可知 \(\Phi'(t)=A\Phi(t)\), 将 \(t\) 换为 \(t-t_0\), 则 \(\Phi'(t-t_0)=A\Phi(t-t_0)\), 故 \(\Phi(t-t_0)\) 是方程组 \(X'=AX\) 的解矩阵. 又当 \(t=t_0\) 时 \(\det\Phi(0)=1\), 故 \(\Phi(t-t_0)\) 是方程组 \(X'=AX\) 的基解矩阵. 于是存在 \(n\) 阶非奇异常数矩阵 \(C\), 使得 \(\Phi(t-t_0)=\Phi(t)C\), 再令 \(t=t_0\), 有 \(E=\Phi(t_0)C\), 得 \(C=\Phi^{-1}(t_0)\), 故 \(\Phi(t-t_0)=\Phi(t)\Phi^{-1}(t_0)\). #
[T050102] (Liouville公式) 设 \(X_1(t),X_2(t),\cdots,X_n(t)\) 是齐次线性方程组 \(X'(t)=A(t)X(t)\) 的任意 \(n\) 个解, 则它们的 Wronski 行列式 \(W(t)\) 满足一阶线性微分方程
\[W'(t)=\left[a_{11}(t)+a_{22}(t)+\cdots+a_{nn}(t)\right]W(t), \]因而有
\[W(t)=W(t_0)\cdot e^{\int_{t_0}^t\left[a_{11}(t)+a_{22}(t)+\cdots+a_{nn}(t)\right]\mathrm{~d}s},\quad t,t_0\in[a,b]. \]证 记 \(X_i(t)=\left(x_{i1}(t),\cdots,x_{in}(t)\right)^T\), 则
\[W(t)=\left|X_1(t),\cdots,X_n(t)\right|=\begin{vmatrix} x_{11}(t) & x_{21}(t) & \cdots & x_{n1}(t)\\ x_{12}(t) & x_{22}(t) & \cdots & x_{n2}(t)\\ \vdots&\vdots&&\vdots\\ x_{1n}(t) & x_{2n}(t) & \cdots & x_{nn}(t) \end{vmatrix} \]由行列式求导法则知
\[\begin{aligned} W'(t)=&\begin{vmatrix} x'_{11}(t) & x'_{21}(t) & \cdots & x'_{n1}(t)\\ x_{12}(t) & x_{22}(t) & \cdots & x_{n2}(t)\\ \vdots&\vdots&&\vdots\\ x_{1n}(t) & x_{2n}(t) & \cdots & x_{nn}(t) \end{vmatrix}+ \begin{vmatrix} x_{11}(t) & x_{21}(t) & \cdots & x_{n1}(t)\\ x'_{12}(t) & x'_{22}(t) & \cdots & x'_{n2}(t)\\ \vdots&\vdots&&\vdots\\ x_{1n}(t) & x_{2n}(t) & \cdots & x_{nn}(t) \end{vmatrix}\\&+\cdots + \begin{vmatrix} x_{11}(t) & x_{21}(t) & \cdots & x_{n1}(t)\\ x_{12}(t) & x_{22}(t) & \cdots & x_{n2}(t)\\ \vdots&\vdots&&\vdots\\ x'_{1n}(t) & x'_{2n}(t) & \cdots & x'_{nn}(t) \end{vmatrix}\\ =&\begin{vmatrix} \sum\limits_{i=1}^na_{1i}x_{1i}(t) & \sum\limits_{i=1}^na_{1i}x_{2i}(t) & \cdots & \sum\limits_{i=1}^na_{1i}x_{ni}(t)\\ x_{12}(t) & x_{22}(t) & \cdots & x_{n2}(t)\\ \vdots&\vdots&&\vdots\\ x_{1n}(t) & x_{2n}(t) & \cdots & x_{nn}(t) \end{vmatrix}+ \begin{vmatrix} x_{11}(t) & x_{21}(t) & \cdots & x_{n1}(t)\\ \sum\limits_{i=1}^na_{2i}x_{1i}(t) & \sum\limits_{i=1}^na_{2i}x_{1i}(t) & \cdots & \sum\limits_{i=1}^na_{2i}x_{1i}(t)\\ \vdots&\vdots&&\vdots\\ x_{1n}(t) & x_{2n}(t) & \cdots & x_{nn}(t) \end{vmatrix}\\&+\cdots + \begin{vmatrix} x_{11}(t) & x_{21}(t) & \cdots & x_{n1}(t)\\ x_{12}(t) & x_{22}(t) & \cdots & x_{n2}(t)\\ \vdots&\vdots&&\vdots\\ \sum\limits_{i=1}^na_{ni}x_{1i}(t) & \sum\limits_{i=1}^na_{ni}x_{1i}(t) & \cdots & \sum\limits_{i=1}^na_{ni}x_{1i}(t) \end{vmatrix}\\ =&\begin{vmatrix} a_{11}x_{11}(t) & a_{11}x_{21}(t) & \cdots & a_{11}x_{n1}(t)\\ x_{12}(t) & x_{22}(t) & \cdots & x_{n2}(t)\\ \vdots&\vdots&&\vdots\\ x_{1n}(t) & x_{2n}(t) & \cdots & x_{nn}(t) \end{vmatrix}+\cdots + \begin{vmatrix} x_{11}(t) & x_{21}(t) & \cdots & x_{n1}(t)\\ x_{12}(t) & x_{22}(t) & \cdots & x_{n2}(t)\\ \vdots&\vdots&&\vdots\\ a_{nn}x_{1n}(t) & a_{nn}x_{2n}(t) & \cdots & a_{nn}x_{nn}(t) \end{vmatrix}\\ \\ =&[a_{11}(t)+a_{22}(t)+\cdots+a_{nn}(t)]W(t). \end{aligned} \]求解上述微分方程得
\[W(t)=ce^{\int_{t_0}^t\left[a_{11}(t)+a_{22}(t)+\cdots+a_{nn}(t)\right]\mathrm{~d}s} \]又 \(W(t_0)=c\), 故
\[W(t)=W(t_0)\cdot e^{\int_{t_0}^t\left[a_{11}(t)+a_{22}(t)+\cdots+a_{nn}(t)\right]\mathrm{~d}s},\quad t,t_0\in[a,b].\quad\quad \# \] 标签：11,5.1,Phi,nn,vmatrix,cdots,一阶,习题,vdots From： https://www.cnblogs.com/hznudmh/p/18048595