6
$\begingroup$

I'm reading the paper "Global existence and scattering for rough solutions of a nonlinear Schrödinger equation on $\mathbb{R}^3$ by Colliander, Keel, Staffilani, Takaoka and Tao. They study the following NLS $$i\partial_t\phi(x,t)+\Delta\phi(x,t)=\vert\phi(x,t)\vert^2\phi(x,t)$$ with initial datum $\phi_0\in H^s(\mathbb{R}^3)$. At the beginning of the paper they state that conservation laws and the local-in-time theory immediately yield global-in-time well-posedness for $s\geq 1$. I understand the case $s=1$ just using the conservation of the energy. But now, let's consider for instance the case $s=2$. In particular an $H^2$-solution is an $H^1$-solution, but how to control that the solution doesn't blow-up in the $H^2$-norm?

$\endgroup$
  • 5
    $\begingroup$ Could you change your title to something more descriptive, As it is now, your title is useless, and would describe a non-trivial amount of the questions here. Imagine a world where every fifth question here is titled "Help in understanding a statement in a paper" $\endgroup$ – Paul Plummer May 27 '15 at 15:56
  • 3
    $\begingroup$ I took a stab at a better title. Feel free to change it if there is something else you prefer. I also corrected the spelling of "Schrödinger". $\endgroup$ – Nate Eldredge May 27 '15 at 16:35
1
$\begingroup$

What you need is "persistence of regularity".

As you said, data in $H^1$ lead to global solutions that remain bounded in $H^1$ by conservation of mass and energy.

If you take data in $H^2$, say, you just need to check that the solution does not blow up its $H^2$-norm in finite time. (Of course, it may be the case that $\lim_{t\to\infty} \|u(t)\|_{H^2} = \infty$.)

To do this, you can run a bootstrap-type argument using Strichartz estimates. For example, on an interval $[0,T]$ you could use the $L_t^2 L_x^{6/5}$ endpoint and Sobolev embedding to estimate

\begin{align*} \|u\|_{L_t^\infty H_x^2} \lesssim \|u_0\|_{H^2} + T^{1/2}\|u\|_{L_t^\infty H_x^1}^2 \|u\|_{L_t^\infty H_x^2}. \end{align*}

Choosing $T$ sufficiently small depending on $\|u\|_{L_t^\infty H_x^1}$ (which is bounded purely in terms of the mass and energy), you can deduce that $$ \|u\|_{L_t^\infty H_x^2([0,T]\times\mathbb{R}^3)} \lesssim 2\|u_0\|_{H^2}. $$ In particular, $$ \|u(T)\|_{H^2} \lesssim 2 \|u_0\|_{H^2}. $$

Now you can run the same argument on $[T,2T]$. In particular, you see that the $H^2$-norm at most doubles on intervals of length $T$. Thus $u$ remains in $H^2$ throughout its lifespan (although the $H^2$-norm may blow up in "infinite time").

Modifications of this type of argument also give continuity in $H^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.