2
$\begingroup$

We consider the one dimensional cubic nonlinear Shr\"odinger equation (NLS): $$i\partial_{t}\phi (x,t) +\Delta \phi (x,t)= \pm |\phi (x,t)|^{2} \phi(x,t), \ (x, t\in \mathbb R),$$ $$\phi (x,0) = \phi_{0}(x)\in H^{s}(\mathbb R);$$ where $H^{s}(\mathbb R)$ is usual Sobolev space.

In 1978 Giniberg-Velo have shown that the above NLS is globally wellposed: that is, for the initial data in $\phi_{0}\in H^{1}(\mathbb R)$, the NLS has a uniqe solution in $C(\mathbb R, H^{1}(\mathbb R)).$

My Question: Let $\phi_{0}\in H^{s}(\mathbb R), (0<s<1).$ Then what can we say about the local and global well posedness of the above NLS ? (If it is well know just the proper refeence will be o.k for me)

Thanks,

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

For the NLS in 1D, quintic is $L^2$ critical. So you are quite comfortably in the subcritical regime. Indeed, you have the result of

Tsutsumi, Yoshio; "$L^2$-solutions for nonlinear Schrödinger equations and nonlinear groups"

which implies for $H^s$ (and hence $L^2$) initial data, you have global existence of solutions in the space $C(\mathbb{R}; L^2(\mathbb{R}))$ intersected with some Strichartz space.

For local solutions in $H^s$, in the sense of having solutions in the space $C((T_{min}, T_{max}); H^s(\mathbb{R}))$, you can look at

Kato, Tosio; "On nonlinear Schrödinger equations. II. $H^s$-solutions and unconditional well-posedness"

(Remark, the nonlinearities considered in Kato's paper allow more general forms than that of Tsutsumi's.)

The main problem here is that we have no direct conservation law on the level of $H^s$, and so global well-posedness cannot follow, as in the $L^2$ case, from local well-posedness. On the other hand, the Strichartz type estimates used in Kato's work is sufficient to allow a direct proof of global well posedness provided that the initial data in $H^s$ is sufficiently small.

(Remark: for general results concerning NLS, a good reference book is Cazenave's Semilinear Schroedinger equations,)

However, global wellposedness for the cubic NLS is true in 1 spatial dimension. Here we can use the fact that we are looking at an algebraic nonlinearity: that the right hand side is $\pm \phi\bar{\phi} \phi$. In particular, for this equation we can prove persistence of regularity. For this equation the statement is

Proposition (Prop. 3.11, p132, Tao, Nonlinear dispersive equations)
Let $u\in C((t_1,t_2); H^s(\mathbb{R}))$ be a (local) solution for $s\geq 0$, with $t_1 < 0 < t_2$. Then there exists a constant $C$ depending on $s$ such that $$ \|u\|_{L^\infty((t_1,t_2); H^s(\mathbb{R}))} \leq \|u(0)\|_{H^s} \exp (C \|u\|^3_{L^2((t_1,t_2); L^\infty(\mathbb{R}))}) $$ provided that the right hand side of the estimate is finite.

It turns out that for the 1D Schrodinger equation you can control $L^4_t L^\infty_x$ using Strichartz estimates from $L^2$ of initial data, and so restricting to finite intervals you also pick up $L^2_t L^\infty_x$ which then allows you to use the $L^2$ global well-posedness result above to obtain global wellposedness in any $H^s$ for $s > 0$; however, we lose control over the $H^s$ norm over time. If you run the argument using the above proposition, you get that $\|u(t)\|_{H^s(\mathbb{R})}$ is allowed to grow exponentially in $t$. For more details see Section 3.4 in Tao, op. cit.

Improve this answer
$\endgroup$
9
  • $\begingroup$ @WW; Thanks a lot; If I understood correctly: you are trying to say: Using the above Proposition(3.11) one can conclude the following: For the given initial data $\phi_{0}\in H^{s}(\mathbb R) (s>0),$ the cubic NLS has a unique global solution in $C(\mathbb R, H^{s}(\mathbb R)).$ This fact I have been trying to figure out this from Section 3.4 from the above mention Tao's book; but I am unable to see the clear proof. Kindly would you tell me bit more? (Thanks, and sorry to disturb u again) Please also let me know if am missing something. $\endgroup$
    – Inquisitive
    Commented Mar 22, 2015 at 15:42
  • $\begingroup$ What part do you not understand? The proposition above immediately implies that the solution which we guarantee to exist in $C(\mathbb{R}, L^2)$ is in fact in $C(\mathbb{R}, H^s)$. Uniqueness follows trivially since we already have the result on local wellposedness from Kato, $\endgroup$
    – Willie Wong
    Commented Mar 23, 2015 at 8:24
  • 1
    $\begingroup$ @Inquisitive: you don't use Kato for local existence. Persistence of regularity allows you to say that for initial data in $H^s$ and for any interval (finite or infinite) including the initial time over which the right hand side of the inequality is finite, the solution is in $H^s$ on that interval. The Strichartz estimates allow you to show that for any finite interval the RHS is finite. Therefore you can exhaust $\mathbb{R}$ with a sequence of increasing finite intervals and obtain the desired result. As I mentioned, doing so you lose control of the $H^s$ norm as $t\to \pm\infty$. $\endgroup$
    – Willie Wong
    Commented Mar 23, 2015 at 12:39
  • 1
    $\begingroup$ The argument above gives that "there exists (uniquely) a solution that is continuous as a function from $\mathbb{R}$ to $H^s$." So I am guessing that you are asking whether the $L^\infty_t H^s_x$ norm is bounded. For initial data below $H^1$ I think this is an open problem as of 10 years ago. But I don't know if it has been resolved since. $\endgroup$
    – Willie Wong
    Commented Mar 30, 2015 at 7:56
  • 1
    $\begingroup$ @Inquisitive: as an exercise, can you prove that there exists a unique continuous solution to the ODE $y' = y$ for all times? Note that the solution is not bounded. If you can't, you should sit down and first figure that out yourself. If you can, then the proof in the NLS case is entirely analogous. $\endgroup$
    – Willie Wong
    Commented Mar 30, 2015 at 10:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .