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. $h=0$.

  • $\begingroup$ What is the integral with respect to? $\endgroup$ – Sepide Feb 13 at 17:37
  • $\begingroup$ The variable $t$. $\endgroup$ – Chrystomath Feb 13 at 19:33

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