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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.

1 Answers 1

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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.

  • 2
    Let me add the following general principle: when working over a Hilbert space, it is better to work with the squared norm, rather than just the norm itself. Why? the norm operator is not "differentiable" near 0. Think $\mathbb{R}^n$ and $x\mapsto |x|^2$ versus $x\mapsto |x|$.2011-04-13