Consider the time-dependent linear system in $\mathbb{R}^n$ $$\dot{x} = A x + b(t)$$ where $b: \mathbb{R} \rightarrow \mathbb{R}^n$ is continuous. Suppose $$\lim_{t \rightarrow \infty} b(t) = 0$$ and that all eigenvalues of $A$ have negative real parts. Prove that regardless of the initial condition, the solution $x(t)$ approaches $0$ as $t \rightarrow \infty$.
My try:
We know the unique solution satisfying the initial condition $x(0) = x_0$ for $t \in \mathbb{R}$ is
$$
x(t) = e^{At}x_0 + \int_0^t e^{A(t-s)}b(s)d(s)
$$
Also, we know that when all eigenvalues have negative real parts, $\|e^{At}x_0\| \leq Me^{-c}\|x_0\|$ for all $t$ where $\|\cdot\|=\|\cdot\|_2$ and some positive $c$ and $M$. Therefore taking the norm from both sides and using triangle inequality we have:
$$
\|x(t)\| = \|e^{At}x_0 + \int_0^t e^{A(t-s)}b(s)d(s)\|
$$
$$
\|x(t)\| \leq \|e^{At}x_0\| + \|\int_0^t e^{A(t-s)}b(s)d(s)\|
$$
$$
\|x(t)\| \leq Me^{-c}\|x_0\| + \|\int_0^t e^{A(t-s)}b(s)d(s)\|
$$
$$
\|x(t)\| \leq Me^{-c}\|x_0\| + \int_0^t \|e^{A(t-s)}\|\|b(s)\|d(s)
$$
I do not know how to handle the integral part.
Best Answer
As pointed out, we have \begin{align} \|e^{tA} f\|_2 \leq& \|e^{tA}\|_\text{op}\|f\|_2 = \sqrt{\lambda_\max(e^{tA^*} e^{tA})}\|f\|_2\leq e^{-ct}\|f\|_2 \end{align} for some $c>0$.
Then it follows \begin{align} \|x(t)\| \leq& \left(e^{-ct}\|x_0\|+ \int^t_0 e^{-c(t-s)}\|b(s)\|\ ds \right)\\ =&\ \left(e^{-ct}\|x_0\|+ \int^{t_0}_0 e^{-c(t-s)}\|b(s)\|\ ds+ \int^t_{t_0} e^{-c(t-s)}\|b(s)\|\ ds \right). \end{align} Since $b(t)$ is continuous and $b(t) \rightarrow 0$ as $t\rightarrow \infty$, then $\|b(t)\|\leq M$ for all $t$ and there exists $t_0$ such that for all $t>t_0$ we have that $\|b(t)\| <\varepsilon$. Hence it follows \begin{align} \|x(t)\| \leq& \left(e^{-ct}\|x_0\| + M\int^{t_0}_0 e^{-c(t-s)}\ ds + \varepsilon\int^t_{t_0} e^{-c(t-s)}\ ds \right)\\ \leq&\ \left(e^{-ct}\|x_0\|+ \frac{M}{c}(e^{-c(t-t_0)}-e^{-ct}) + \frac{\varepsilon}{c} \right) \end{align} which means \begin{align} \limsup_{t\rightarrow \infty} \|x(t)\| \leq \frac{\varepsilon}{c} \end{align} for any $\varepsilon>0$. Thus $\lim_{t\rightarrow \infty} \|x(t)\| = \limsup_{t\rightarrow \infty} \|x(t)\| = 0$.