Let $A$ be a constant $n\times n$-matrix and $u(t) \in\mathbb{R}^n$ be a continuous function defined on $\mathbb{R}$. Assume that the real part of any eigenvalue of A is negative. Show that solutions of $$\frac{d}{dt}x = Ax + u(t) $$ are asymptotically stable.
What I know :
Asymptotically stable : A solution $\psi(t)$ of is asymptotically stable $\iff$ $\psi(t)$ is stable and for any $t_0\in[a,+1)$,there exists $\zeta >0$ such that : $$\left\|\xi-\psi\left(\mathrm{t}_{0}\right)\right\|<\zeta \Longrightarrow \lim _{t \rightarrow \infty}\left\|\varphi\left(t ; t_{0}, \xi\right)-\psi({t})\right\| = 0$$
Unique solution of initial value problem : $x(t_0) = ξ$, is $$φ(t;t_0, ξ) = X(t − t_0)ξ +\int_{t_o}^t X(t − s) u(s) ds$$ where $ X(t) = \exp(tA)$. There exist $\alpha>0$ and $M_0>0$ such that $ \|X(t)\| \leq M_0e^{-t\alpha}$
Properties of matrix : $\|xA\| \leq \|x\| \|A\|$ and $\|A + B\|\leq \|A\|+\|B\|$
Properties solution stable : Stability of a solution $\psi(t) \Leftrightarrow$ Stability of the null solution $y(t) \equiv 0$ (but I don't know if it's true for the asymptotically stability.
What I did :
$\|\phi(t;t_0,\xi)-\psi(t)\| = \|X(t − t_0)\xi +\int_{t_o}^t X(t − s) u(s) ds-\psi(t)||$
$\|φ(t;t_0,\xi)-\psi(t)\| \leq \| X(t − t_0)\xi-\psi(t)\| +\|\int_{t_o}^t X(t − s) u(s) ds\|$
$\|\phi(t;t_0,\xi)-\psi(t)\| \leq \|M_0e^{-t\alpha}\xi - \psi(t)\| +\|\int_{t_o}^t M_0e^{-(t-s)\alpha}u(s) ds\|$
But I don't know how to finish and conclude, does anybody have an idea ?
Let $\varphi(t)$ and $\psi(t)$ be two different solutions of the system. Consider the difference $$ y(t)=\varphi(t)-\psi(t). $$ The derivative is $$ \frac{dy}{dt}=\frac{d\varphi}{dt}-\frac{d\psi}{dt}= (A\varphi+u(t))-(A\psi+u(t))=A(\varphi(t)-\psi(t))=Ay. $$ This implies $$ y(t)=e^{A(t-t_0)}(\varphi(t_0)-\psi(t_0)) $$ and $$ \|\varphi(t)-\psi(t)\|=\|y(t)\|<Me^{-\gamma t}\| \varphi(t_0)-\psi(t_0) \|, $$ where $M$ and $\gamma$ are some positive constants. This implies not only asymptotic, but also exponential stability of all solutions of the system.