[proofplan] We remove the homogeneous dynamics by multiplying by $e^{-tA}$. The transformed inhomogeneous term is an indefinite [Lebesgue integral](/page/Lebesgue%20Integral) of an $L^1$ function, hence is absolutely continuous and differentiates to its integrand almost everywhere. Multiplying back by $e^{tA}$ gives the stated formula and verifies the differential equation. Uniqueness follows by applying the same integrating-factor transformation to the difference of two trajectories. [/proofplan]

[step:Define the candidate trajectory by an integrating factor]Choose a Borel measurable representative of the equivalence class $u \in L^1([0,T];\mathbb{R}^m)$, and denote this representative again by \begin{align*} u: [0,T] \to \mathbb{R}^m. \end{align*} For a [linear map](/page/Linear%20Map) $L: \mathbb{R}^m \to \mathbb{R}^n$, let \begin{align*} \|L\|_{\mathrm{op}}=\sup\{|Lz|: z \in \mathbb{R}^m, |z|\le 1\} \end{align*} denote its Euclidean operator norm. Define the measurable map \begin{align*} f: [0,T] \to \mathbb{R}^n,\qquad f(\tau)=e^{-\tau A}Bu(\tau). \end{align*} Since $\tau \mapsto e^{-\tau A}B$ is continuous on the compact interval $[0,T]$, there is a finite constant \begin{align*} M=\sup_{\tau \in [0,T]}\|e^{-\tau A}B\|_{\mathrm{op}}<\infty. \end{align*} Thus $|f(\tau)| \le M|u(\tau)|$ for $\mathcal{L}^1$-a.e. $\tau \in [0,T]$, so $f \in L^1([0,T];\mathbb{R}^n)$. Define the indefinite integral \begin{align*} F: [0,T] \to \mathbb{R}^n,\qquad F(t)=\int_0^t f(\tau)\,d\mathcal{L}^1(\tau). \end{align*} By the absolute [continuity theorem](/theorems/1145) for indefinite Lebesgue integrals applied to the function $f \in L^1([0,T];\mathbb{R}^n)$, $F$ is absolutely continuous and \begin{align*} \dot{F}(t)=f(t)=e^{-tA}Bu(t) \end{align*} for $\mathcal{L}^1$-a.e. $t \in [0,T]$. Now define \begin{align*} x: [0,T] \to \mathbb{R}^n,\qquad x(t)=e^{tA}(x_0+F(t)). \end{align*} The map $t \mapsto e^{tA}$ is continuously differentiable, and $F$ is absolutely continuous, so $x$ is absolutely continuous.[/step]

[guided]The point of the integrating factor is to convert the forced equation into an equation whose right-hand side is directly integrable. Since $u \in L^1([0,T];\mathbb{R}^m)$ is an equivalence class of [measurable functions](/page/Measurable%20Functions) modulo equality almost everywhere, we first choose a Borel measurable representative and denote it again by \begin{align*} u: [0,T] \to \mathbb{R}^m. \end{align*} Changing this representative on a set of $\mathcal{L}^1$-measure zero does not change any of the Lebesgue integrals below. For a linear map $L: \mathbb{R}^m \to \mathbb{R}^n$, define its Euclidean operator norm by \begin{align*} \|L\|_{\mathrm{op}}=\sup\{|Lz|: z \in \mathbb{R}^m, |z|\le 1\}. \end{align*} We now define \begin{align*} f: [0,T] \to \mathbb{R}^n,\qquad f(\tau)=e^{-\tau A}Bu(\tau). \end{align*} This is the forcing term after removing the homogeneous flow. We must check that this expression is integrable before using it inside an indefinite integral. The map $\tau \mapsto e^{-\tau A}B$ is continuous because the matrix exponential is smooth in $\tau$, and $[0,T]$ is compact. Therefore \begin{align*} M=\sup_{\tau \in [0,T]}\|e^{-\tau A}B\|_{\mathrm{op}} \end{align*} is finite. For $\mathcal{L}^1$-a.e. $\tau \in [0,T]$, \begin{align*} |f(\tau)|=|e^{-\tau A}Bu(\tau)| \le \|e^{-\tau A}B\|_{\mathrm{op}}|u(\tau)| \le M|u(\tau)|. \end{align*} Since $u \in L^1([0,T];\mathbb{R}^m)$, this domination proves $f \in L^1([0,T];\mathbb{R}^n)$. We may therefore define \begin{align*} F: [0,T] \to \mathbb{R}^n,\qquad F(t)=\int_0^t f(\tau)\,d\mathcal{L}^1(\tau). \end{align*} The absolute continuity theorem for indefinite Lebesgue integrals applies because $f \in L^1([0,T];\mathbb{R}^n)$, and it gives that $F$ is absolutely continuous with derivative \begin{align*} \dot{F}(t)=f(t)=e^{-tA}Bu(t) \end{align*} for $\mathcal{L}^1$-a.e. $t \in [0,T]$. Finally define \begin{align*} x: [0,T] \to \mathbb{R}^n,\qquad x(t)=e^{tA}(x_0+F(t)). \end{align*} This is exactly the result of solving for the transformed variable and then multiplying back by the homogeneous flow. Because $t \mapsto e^{tA}$ is continuously differentiable and $F$ is absolutely continuous, their product is absolutely continuous.[/guided]

[step:Rewrite the candidate in the stated convolution form] For every $t \in [0,T]$, the definition of $F$ gives \begin{align*} x(t)=e^{tA}x_0+e^{tA}\int_0^t e^{-\tau A}Bu(\tau)\,d\mathcal{L}^1(\tau). \end{align*} Since $e^{tA}$ is independent of the integration variable $\tau$, linearity of the Lebesgue integral gives \begin{align*} x(t)=e^{tA}x_0+\int_0^t e^{tA}e^{-\tau A}Bu(\tau)\,d\mathcal{L}^1(\tau). \end{align*} Because $A$ commutes with itself, the matrix exponentials satisfy $e^{tA}e^{-\tau A}=e^{(t-\tau)A}$. Hence \begin{align*} x(t)=e^{tA}x_0+\int_0^t e^{(t-\tau)A}Bu(\tau)\,d\mathcal{L}^1(\tau). \end{align*} [/step]

[step:Verify the initial condition and differential equation] At $t=0$, the integral over $[0,0]$ is zero, so \begin{align*} x(0)=e^{0A}x_0=x_0. \end{align*} For $\mathcal{L}^1$-a.e. $t \in [0,T]$ at which $\dot{F}(t)=e^{-tA}Bu(t)$, the product rule for absolutely continuous functions gives \begin{align*} \dot{x}(t)=Ae^{tA}(x_0+F(t))+e^{tA}\dot{F}(t). \end{align*} Substituting the derivative of $F$ yields \begin{align*} \dot{x}(t)=Ae^{tA}(x_0+F(t))+e^{tA}e^{-tA}Bu(t). \end{align*} Let $I_n \in \mathbb{R}^{n \times n}$ denote the identity matrix. Since $x(t)=e^{tA}(x_0+F(t))$ and $e^{tA}e^{-tA}=I_n$, this becomes \begin{align*} \dot{x}(t)=Ax(t)+Bu(t). \end{align*} Thus the candidate is an absolutely continuous solution of the initial value problem. [/step]

[step:Prove uniqueness by differentiating the transformed difference] Let $y: [0,T] \to \mathbb{R}^n$ be another absolutely continuous map satisfying $y(0)=x_0$ and \begin{align*} \dot{y}(t)=Ay(t)+Bu(t) \end{align*} for $\mathcal{L}^1$-a.e. $t \in [0,T]$. Define \begin{align*} w: [0,T] \to \mathbb{R}^n,\qquad w(t)=e^{-tA}(y(t)-x(t)). \end{align*} The map $w$ is absolutely continuous. For $\mathcal{L}^1$-a.e. $t \in [0,T]$ where both $x$ and $y$ satisfy the differential equation, the product rule gives \begin{align*} \dot{w}(t)=-Ae^{-tA}(y(t)-x(t))+e^{-tA}(\dot{y}(t)-\dot{x}(t)). \end{align*} Since \begin{align*} \dot{y}(t)-\dot{x}(t)=A(y(t)-x(t)), \end{align*} we obtain \begin{align*} \dot{w}(t)=-Ae^{-tA}(y(t)-x(t))+e^{-tA}A(y(t)-x(t)). \end{align*} The matrices $A$ and $e^{-tA}$ commute, so the two terms cancel and \begin{align*} \dot{w}(t)=0 \end{align*} for $\mathcal{L}^1$-a.e. $t \in [0,T]$. An absolutely [continuous function](/page/Continuous%20Function) with derivative zero almost everywhere is constant, hence $w(t)=w(0)$ for every $t \in [0,T]$. Since \begin{align*} w(0)=e^{0A}(y(0)-x(0))=0, \end{align*} we have $w(t)=0$ for every $t \in [0,T]$. Multiplying by $e^{tA}$ gives $y(t)=x(t)$ for every $t \in [0,T]$. Therefore the trajectory given by the formula is unique. [/step]