【习题】5.1 一阶线性微分方程的基本概念

合集 - 常微分方程(14)

1.【习题】2.1 分离变量法2024-01-222.【习题】2.2 变量替换法2024-01-223.【习题】2.3 积分因子法2024-01-234.【习题】2.4 参数法2024-01-235.【习题】3.1 Picard存在唯一性定理2024-01-256.【习题】3.2 不动点定理和解的存在性2024-02-097.【习题】3.3 解的延拓2024-02-258.【习题】3.4 解对初值和参数的连续性和可微性2024-02-259.【习题】4.2 高阶微分方程的一般理论2024-02-2710.【习题】4.3 常系数齐次线性方程的待定指数函数法2024-02-2911.【习题】4.4 常系数非齐次线性方程的待定系数法2024-03-01

12.【习题】5.1 一阶线性微分方程的基本概念2024-03-02

13.【习题】5.2 一阶常系数线性微分方程组2024-03-0214.常微分方程选题2024-03-22

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[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 $\Phi(t-t_0)=\Phi(t)\Phi^{-1}(t_0)$. #

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[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

$$W(t)=W(t_0)\cdot e^{\int_{t_0}^t\left[a_{11}(t)+a_{22}(t)+\cdots+a_{nn}(t)\right]\mathrm{~d}s},\quad t,t_0\in[a,b].\quad\quad \#$$