Let $f:\mathbb{R}^n\to \mathbb{R}^n$ be a continuously differentiable function and $\mathbf{A}$ be a constant matrix. Suppose that $\| e^{At}\|\leq 2^{-9t}$ and $\|f(x)\|\leq 5\|x\|$ for all $x\in\mathbb{R}^n$.
Consider the differential equation $$x'=Ax+f(x),\ \ x(0)=\xi.$$ Show that if $\phi(t)=\begin{bmatrix}\phi_1(t)\\\cdots\\ \phi_n(t) \end{bmatrix}$ is a solution of the above differential equation then $\lim\limits_{t\to\infty}\phi_i(t)=0$, for all $i\in \{1,\cdots,n\}$.
How to deal with this kind of problem? Usually, $f$ is considered to be a fucntion of $t$, but here, it's a function of $x$, which makes it difficult.
I tried and I think it works. First, the ODE $$x'=Ax+g(t,x),\ \ x(\tau)=\xi$$ can be written as $$x=e^{(t-\tau)A}\xi+\int_\tau^te^{(t-s)A}g(s,x(s))ds.$$ In the above ODE, $g(t,x)=f(x).$ Now, taking the norm, we get that\begin{eqnarray*} \|x\|&\leq& \|e^{(t-\tau)A}\xi\|+\int_\tau^t\|e^{(t-s)A}f(x(s))\|ds\\ &\leq&\|e^{(t-\tau)A}\|\|\xi\|+\int_\tau^t\|e^{(t-s)A}\|\|f(x(s))\|ds\\ &\leq & 2^{-9(t-\tau)}\|\xi\|+5\int_\tau^t2^{-9(t-s)}\|x\|.ds \end{eqnarray*} therefore, we get that $$\|x\|2^{9(t-\tau)}\leq \|\xi\|+5\int_\tau^t2^{9(s-\tau)}\|x\|ds.$$ Use Gronwall’s inequality to get $$\|x\|2^{9(t-\tau)}\leq \|\xi\|e^{5(t-\tau)}\Rightarrow \|x\|\leq \|\xi\|2^{-4t},$$ where I used $\tau=0$. I think the result is clear, now. Please, check if I made some error.