Let us consider the system of differential equation $\dot x=Ax+h(t),$ where A is a constant matrix of dimension $n\geq2$ and $h$ is continuous on $[0,\infty).$ All eigenvalues of $A$ have a negative real part. It is known, that $h(t)\to 0$ for $t\to\infty$ implies that all solutions $x(t)$ of $\dot x=Ax+h(t)$ converge to $0$ for $t\to\infty.$ Is the converse implication true too? (All solutions $x(t)\to 0$ implies $h(t)\to 0$).
Thank you.
On
More precisely, the equivalence holds if for some vector norm $\left\Vert\cdot\right\Vert,$ \begin{equation} \sup\limits_{0\leq u\leq1}\left\Vert\int\limits_{t}^{t+u}h(s)ds\right\Vert\to 0 \end{equation} as $t\to\infty.$
For the details and (long) proof see Theorem B in:
Strauss, A.; Yorke, J. A., Perturbing uniform asymptotically stable nonlinear systems, J. Differ. Equations 6, 452-483 (1969). ZBL0182.12103.
No. Consider $h$ which is $0$ everywhere except near $n^2$ for $n\in\mathbb N$ where it has bumps of height $1$ and width $1/n$: $$ h(t) = \sum_{n\in\mathbb N} 1_{[n^2-1/n,n^2+1/n]}(t). $$ Consider now for instance $x'(t)=-x(t)+h(t)$. All trajectories become eventually positive. Then it's clear that the effect of $h$ perturbs $x$ by at most $2/n$ when $t$ passes the time $n$: $$ x(n+1/n)-x(n-1/n) \leq \int_{n^2-1/n}^{n^2+1/n} h(t)\,dt=2/n. $$ From this you should be able to conclude that $x$ goes to $0$ nonetheless, because in the interval $[n^2,(n+1)^2]$ of length $\sim n$ it has plenty of time to become exponentially close to $0$.