Problem
$\dot x = Ax + h(t)$, where $h(t)\in \mathbb{R}^{n}$ is continuous and bounded over $\mathbb{R}$ and $A\in M_{n\times n}$ a constant matrix with all its eigenvalues having negative real part.
I have to prove that there's only one bounded solution over $(-\infty, \infty)$ and in case that $|h(t)| \to 0$ when $t \to \infty$, then all solutions tends to $0$.
My thoughts
Since A is constant with all its eigenvalues having negative real part then we can ensure that $ | \Pi_{A}(t,s) | \leq Ce^{-\alpha(t-s)}$, i.e all solutions for the homogeneous system $\dot x=Ax$ are bounded. Now, since $\dot x= Ax+h(t)$ is an inhomogeneous system, we can find the solutions by using Duhamel's formula, $x(t)=| \Pi_{A}(t,s) |x_{0} + \int | \Pi_{A}(t,r)|h(r)dr$ and after this im lost, I mean, I could bound it by $|x(t)| \leq Ce^{-\alpha(t-s)}x_{0}+\int Ce^{-\alpha(t-r)}C_{1}dr$ but I can't see where to be able to prove uniqueness of a bounded solutions. Im pretty sure that this must be some kind of theorem but I couldn't be able to find it. So, any help would be really appreciated.
Thanks so much for your help.
There being at most one bounded solution is more or less obvious: the nontrivial homogeneous solutions are unbounded at $-\infty$ so the difference of any two distinct solutions to the full equation is unbounded at $-\infty$, hence at least one of the functions in the difference was unbounded at $-\infty$.
The question of existence is a bit more tricky. Since $e^{At}$ grows really fast going backward in time, you need $x(t_0) + \int_{t_0}^t e^{-As} h(s) ds$ to decay really fast going backward in time. Thankfully that's possible: you can set $x(t_0)=\int_{-\infty}^{t_0} e^{-As} h(s) ds$ which will be a convergent integral under these assumptions.
The second part needs you to show that $\int_0^t e^{-As} h(s) ds$ doesn't grow so fast at $+\infty$ that $e^{At}$ can't send it to zero.