3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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. $h=0$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.