Further thoughts on the energy estimate

Question

I put my question about the energy estimate two days ago. And finally I can get

$$\frac{d}{dt}\|x\|^2=2\|x\|\frac{d}{dt}\|x\|=\frac{d}{dt}\langle x,x \rangle=2Re\langle x, Ax\rangle$$

If I have the estimate $Re\langle x,Ax\rangle \leq 2\alpha\|x\|^2$ for some $\alpha\in{\bf R}$, then I will get

$$\|x(t)\|\frac{d}{dt}\|x(t)\| \leq 2\alpha\|x(t)\|^2 \qquad\text{for all}\quad t\in {\bf R} \qquad(*)$$

Here are my questions:

1. Can one deduce from (*) that $\frac{d}{dt}\|x(t)\|\leq 2\alpha\|x(t)\|$ without worrying about that $x(t)=0$ may happen for some $t$?
2. If I can get $\frac{d}{dt}\|x(t)\|\leq 2\alpha\|x(t)\|$, can I use the method of solving the linear system $$\frac{d}{dt}\|x(t)\|= 2\alpha\|x(t)\|$$ to get $$\|x(t)\| \leq e^{2\alpha t}\|x(0)\|?$$

My thoughts on the second question:

Turn the inequalities $\frac{d}{dt}\|x(t)\|\leq 2\alpha\|x(t)\|$ into the non-homogeneous equation $\frac{d}{dt}\|x(t)\|= 2\alpha\|x(t)\|-\epsilon(t)$, then hopefully one can get the estimate.

Answer 1

It is much easier to just write

$$ \frac{d}{dt} \|x\|^2 \leq 4\alpha \| x(t) \|^2 $$

which then implies by Gronwall's inequality

$$ \|x(t)\|^2 \leq e^{4\alpha t} \|x(0)\|^2 $$

Now you can take square root.