Let us prove that the system $x′ = Ax + b(x)$ where $\|b(x)\| \le C \|x\|^2$ is locally exponentially stable around $0$ if A is diagonal and all elements on the diagonal are real and negative.
My attempt:
The solution of the system is given by:
$$x(t)= e^{(t-t_0)A}x_0 + \int_{t_0}^{t} e^{(t-s)A}b(s) \,ds$$
Thus:
$$\|x(t)\| \le e^{(t-t_0)\|A\|}\|x_0\| + \int_{t_0}^{t} e^{t\|A\|}\|b(s)\| \,ds$$
$$ \le e^{(t-t_0)\|A\|}\|x_0\| + \int_{t_0}^{t} e^{t\|A\|}C\|x(s)\|^2 \,ds $$
Besides, if we consider the system $y'=Ay$ with the same $A$, we already know that the system is exponentially stable around $0$. I am looking for an expression of the form:
$$ \|x(t)\| \le k\|x_0\| e^{-\lambda (t-t_0)} $$
for all $\|x_0\| < c$ with $c, k$ and $\lambda$ positive constants. How can I continue?
Edit:
Since matrix $A$ is Hurwitz:
$$ \|e^{(t-t_0)A} x_0\|\le k\|x_0\| e^{-\lambda (t-t_0)} $$
Hence:
$$\|x(t)\| \le k e^{-\lambda (t-t_0)} \|x_0\| + \int_{t_0}^{t} ke^{-\lambda (t-s)} \|x_0\| C \|x(s)\|^2 \,ds$$
$$ e^{\lambda t} \|x(t)\| \le k e^{\lambda t_0} \|x_0\| + \int_{t_0}^{t} k \|x_0\| C e^{\lambda s} \|x(s)\|^2 \,ds$$
We consider $\|x_0\| \le \epsilon$. I was thinking of using Grönwall's inequality but the exponent 2 does not permit this.