一、线性齐次时变状态方程的解
\boldsymbol{\dot{x}}=A_{(t)}\boldsymbol{x}_{(t)}\quad \Rightarrow\quad \boldsymbol{x}_{(t)}=\varPhi_{(t,t_{0})}\boldsymbol{x}_{_{(t_0)}}
\end{align}$$
只有当 $A_{(t)}$ 和 $\int _{t_{0}}^{t} A(\tau)\, d\tau$ 满足乘法可交换条件，时变系统的自由解才可以写为封闭的幂级数形式：
$$\begin{align}
\varPhi(t,t_{0})= \exp \left[\int _{t_{0}}^{t} A(\tau) \mathrm{d}\tau \right] \boldsymbol{x}_{(t_0)}
\end{align}$$
如果不满足乘法可交换条件，只能用级数近似：
$$\begin{align}
\varPhi(t,t_{0})=I+ \int _{t_{0}}^{t}A(\tau_{0}) \, d\tau_{0}+ \int _{t_{0}}^{t} \int _{t_{0}}^{\tau_{0}} A(\tau_{1}) \, d\tau_{1}d \tau_{0}+ \int _{t_{0}}^{t} \int _{t_{0}}^{\tau_{0}} \int _{t_{0}}^{\tau_{1}} A(\tau_{2}) \, d\tau_{2}d \tau_{1}d \tau_{0}+\cdots
\end{align}$$
#### 时变状态转移矩阵的基本性质
$$\begin{gathered}
\varPhi(t,t)=I \\
\varPhi(t_{2},t_{1})\varPhi(t_{1},t_{0})=\varPhi (t_{2},t_{0}) \\
\varPhi(t,t_{0})=\varPhi^{-1}(t_{0},t) \\
\dot{\varPhi}(t,t_{0})= A_{(t)}\varPhi(t,t_{0})
\end{gathered}$$
### 二、线性非齐次时变状态方程的解
$$\begin{align}
\boldsymbol{\dot{x}}_{(t)}=A_{(t)}\boldsymbol{x}_{(t)}+B_{(t)}\boldsymbol{u}_{(t)}\quad \Rightarrow\quad
\boldsymbol{x}_{(t)}=\varPhi(t,t_{0})\boldsymbol{x}_{(t_0)}+\int _{t_{0}}^{t} \varPhi(t,\tau)B(\tau)\boldsymbol{u}(\tau)\, d\tau
\end{align}$$
$$\begin{align}
& \boldsymbol{\dot{x}}_{(t)} =A\boldsymbol{x}_{(t)}+B\boldsymbol{u}_{(t)} \\
\; {\color{red}\Rightarrow} \; & e^{ -At } \boldsymbol{\dot{x}_{(t)}}= e^{ -At } A\boldsymbol{x}_{(t)}+e^{ -At }B\boldsymbol{u}_{(t)} \\
\; {\color{red}\Rightarrow} \; & \dfrac{\mathrm{d} (\; e^{ -At }\boldsymbol{x}_{(t)} \;)}{\mathrm{d} t} =e^{ -At }B\boldsymbol{u}_{(t)} \\
\end{align}$$
两边积分得到：
$$\begin{align}
& \int _{t_{0}}^{t} \dfrac{\mathrm{d} (e^{ -A\tau }\boldsymbol{x}_{(\tau)})}{\mathrm{d} t}\, d\tau = \int _{t_{0}}^{t} e^{ -A\tau }B\boldsymbol{u}_{(\tau)} \, d\tau \\
\; {\color{red}\Rightarrow} \; & \left. e^{ -A\tau }\boldsymbol{x}_{(\tau)}\right\rvert ^{t}_{t_{0}} =\int _{t_{0}}^{t} e^{ -A\tau }B\boldsymbol{u}_{(\tau)} \, d\tau
\\
\; {\color{red}\Rightarrow} \; & e^{ -At }\boldsymbol{x}_{(t)}=e^{ -At_{0} }\boldsymbol{x}_{t_{0}}+\int _{t_{0}}^{t} e^{ -At }B\boldsymbol{u}_{(\tau)} \, d\tau \\
\; {\color{red}\Rightarrow} \; & \boldsymbol{x}_{(t)}=e^{ A(t-t_{0}) }\boldsymbol{x}_{(t_{0} )}+e^{ At } \int _{t_{0}}^{t} e^{ -A\tau }B\boldsymbol{u}_{(t)} \, d\tau
\end{align}$$
