# How to show the uniqueness of the solution of the time dependent linear system in $\mathbb{R}^n$?

Consider the time dependent linear system in $$\mathbb{R}^n$$:

$$\dot{x} = Ax + b(t),$$ where $$b: \mathbb{R}^n \rightarrow \mathbb{R}^n$$ is continuous. Prove that

$$x(t) = e^{At}x_0 + \int_0^t e^{A(t-s)}b(s)d(s), \tag{1}$$ for $$t \in \mathbb{R}$$ is the unique solution satisfying the initial condition $$x(0) = x_0$$.

My try: I tried to take the derivative with respect to Leibniz rule to show that $$(1)$$ satisfies the systems of differential equation.

$$\dot{x}(t) = Ae^{At}x_0 +\frac{ \mathrm{d}}{\mathrm{d}t}\bigg(\int_0^t e^{A(t-s)}b(s)d(s)\bigg)$$

Also, I have no idea how to prove uniqueness.

If there were two solutions with the same $$x_0$$, then their difference $$h$$ would satisfy $$\dot{h}=Ah,\quad h(0)=0,$$ equivalent to $$h(t)=\int_0^tAh$$
Since this is reminiscent of Banach's theorem, let $$Ty(t):=\int_0^tAy$$. Then $$\|T(u-v)\|=\sup_t|\int_0^t A(u-v)| \le \|A\|\|u-v\|\int_0^t1\le\|A\||t|\|u-v\|$$ where $$\|u\|$$ is the sup-norm. So for $$t$$ small enough, the mapping $$T$$ is a contraction mapping and has a unique fixed point. In fact, clearly, this is zero, i.e. $$h=0$$.
• The variable $t$. – Chrystomath Feb 13 at 19:33