【习题】5.1 一阶线性微分方程的基本概念
[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
\[W(t)=W(t_0)\cdot e^{\int_{t_0}^t\left[a_{11}(t)+a_{22}(t)+\cdots+a_{nn}(t)\right]\mathrm{~d}s},\quad t,t_0\in[a,b].\quad\quad \#
\]
