线性时变状态方程的解

\begin{array}{r} \dot{x} = A_{(t)} x_{(t)} \Rightarrow x_{(t)} = Φ_{(t, t_{0})} x_{_{(t_{0})}} \end{array}

只有当 $A_{(t)}$ 和 $\int_{t_{0}}^{t} A (τ) d τ$ 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{array}{r} Φ (t, t_{0}) = \exp [\int_{t_{0}}^{t} A (τ) d τ] x_{(t_{0})} \end{array}

如果不满足乘法可交换条件，只能用级数近似：

\begin{array}{r} Φ (t, t_{0}) = I + \int_{t_{0}}^{t} A (τ_{0}) d τ_{0} + \int_{t_{0}}^{t} \int_{t_{0}}^{τ_{0}} A (τ_{1}) d τ_{1} d τ_{0} + \int_{t_{0}}^{t} \int_{t_{0}}^{τ_{0}} \int_{t_{0}}^{τ_{1}} A (τ_{2}) d τ_{2} d τ_{1} d τ_{0} + \dots \end{array}

\begin{matrix} Φ (t, t) = I \\ Φ (t_{2}, t_{1}) Φ (t_{1}, t_{0}) = Φ (t_{2}, t_{0}) \\ Φ (t, t_{0}) = Φ^{- 1} (t_{0}, t) \\ \dot{Φ} (t, t_{0}) = A_{(t)} Φ (t, t_{0}) \end{matrix}

\begin{array}{r} {\dot{x}}_{(t)} = A_{(t)} x_{(t)} + B_{(t)} u_{(t)} \Rightarrow x_{(t)} = Φ (t, t_{0}) x_{(t_{0})} + \int_{t_{0}}^{t} Φ (t, τ) B (τ) u (τ) d τ \end{array}

\begin{aligned} {\dot{x}}_{(t)} = A x_{(t)} + B u_{(t)} \\ \Rightarrow & e^{- A t} {\dot{x}}_{(t)} = e^{- A t} A x_{(t)} + e^{- A t} B u_{(t)} \\ \Rightarrow & \frac{d (e^{- A t} x_{(t)})}{d t} = e^{- A t} B u_{(t)} \end{aligned}

两边积分得到：

\begin{aligned} \int_{t_{0}}^{t} \frac{d (e^{- A τ} x_{(τ)})}{d t} d τ = \int_{t_{0}}^{t} e^{- A τ} B u_{(τ)} d τ \\ \Rightarrow & {e^{- A τ} x_{(τ)} |}_{t_{0}}^{t} = \int_{t_{0}}^{t} e^{- A τ} B u_{(τ)} d τ \\ \Rightarrow & e^{- A t} x_{(t)} = e^{- A t_{0}} x_{t_{0}} + \int_{t_{0}}^{t} e^{- A t} B u_{(τ)} d τ \\ \Rightarrow & x_{(t)} = e^{A (t - t_{0})} x_{(t_{0})} + e^{A t} \int_{t_{0}}^{t} e^{- A τ} B u_{(t)} d τ \end{aligned}