Journal of Applied Mathematics and Physics
Vol.05 No.09(2017), Article ID:79617,6 pages
10.4236/jamp.2017.59162
Non-Scattering of the Solution of the Nonlinear Schrödinger Equation on the Torus
Xing Cheng, Qiang Li
College of Science, Hohai University, Nanjing, China
Received: September 8, 2017; Accepted: October 13, 2017; Published: October 16, 2017
ABSTRACT
In this article, we will show non-scattering of the solution of the nonlinear Schrodinger equation on the torus. The result extends the result of Colliander, J., Keel, M., Staffilani, G., Takaoka, H. and Tao, T. for the cubic nonlinear Schrödinger equation on 2-dimensional torus.
Keywords:
Schrödinger Equation, Torus, Scattering
1. Introduction
In this article, we will consider the nonlinear Schrödinger equation on d-dimensional torus:
(1)
where . This kind of nonlinear Schrödinger equations has been studied intensively in the last two decades. See [1] [2] [3] [4].
Many mathematicians believe this equation does not have nontrivial solutions which scatter, i.e. which approach a solution to the linear equation at time . Colliander, J., Keel, M., Staffilani, G., Takaoka, H. and Tao, T. [5] consider the cubic nonlinear Schrödinger equation on two dimensional torus, and prove the solution cannot scatter to free solution in .
As in [5], the explicit solution
(2)
where cannot converge to a free solution due to
the presence of the phase rotation which is caused by the nonlinearity. We will show this is the only solution that scatters modulo phase rotation in in the sense that there exists and function such that
is of the form (2), which then reveals that no solution of the nonlinear Schrödinger equation on torus can scatter to free solution.
2. Main Theorem
In this section, we will present the main theorem in this article. We will show the only solution that scatters modulo phase rotation is of the form (2).
Theorem 1 (No non-trivial solution scatters modulo phase rotation). Let be an solution to (1) which scatters modulo phase in , then u is of the form (2) for some .
To prove this theorem, we first need some lemmas.
Lemma 2 (Pre-compactness). For ,
is pre-compact in .
Proof. It is equivalent to show is pre-compact in after taking Fourier transforms. By monotone convergence, , there exists such that . We can conclude is covered by finitely many balls of radius , and the claim follows.
□
Lemma 3 (Diamagnetic inequality) If , then and
.
Proof. It suffices to verify this when is smooth. For any , we have
,
and hence
. Taking distributional limits as , we obtain the claim.
□
Lemma 4 ( has no step functions). Let be such that takes at most two value. Then u is constant.
Proof. We may assume that u takes 0 and 1 only, thus . On the one hand, differentiating this we obtain , thus . On the other hand, since , , and thus , therefore is constant.
□
Proof of Theorem 1. Let
be as above. We may assume
has non-zero mass. From Lemma 2, we see that
is precompact in
. Thus we can find a sequence
such that
in
, as
. Applying Lemma 2 and passing to a subsequence, we can also assume
in
, as
Since u has non-zero mass, we see also has non-zeromass. Let be the time-translated solution , thus in . Let be solution to NLS with initial data . By the local well-posedness theory in we conclude that converge uniformly in to on every compact time interval [-T,T].
On the other hand, by hypothesis, we have
Since , we conclude that
By taking limits, we conclude that
for some .
In particular, since has non-zero mass, we have
From (1) and Sobolev, we see that is continuously differentiable in , and so from the above identity we see that α is continuously differentiable in time. Now we apply to both sides of (3). Using the NLS equation, we conclude that
And thus by (1.3), we have whenever
.
Thus we see that , takes at most two values. By Lemma 3 and Lemma 4, we conclude that is constant in , by (3), we see that the same is true for . By mass conservation we conclude that is also constant in time.
Since has non-zero mass, we can thus write
where is some function and . Since was in , we see that is in also for every t. Differentiating the identity , We see that is imaginary almost everywhere for j = 1, 2, thus is an imaginary multiple of for almost every (t, x). This implies the imaginary vector field is curl-free and thus by Hodge theory, we may write for some which is locally in uniformly in t. This implies that , thus by adjusting by a constant independent of space. We may assume , thus .
Applying to both sides, we conclude that
in the sense of distributions.
Since A is non-zero, we conclude that
Taking imaginary parts, we conclude that and in particular at time t = 0, is a harmonic function from to . On the other hand, from the identity , we know that is periodic, so has at most linear growth. Thus must in fact be linear. Descending back to , we conclude that
Thus, we have
Since
but is a multiple of by a phase, thus . Applying phase rotation, we may assume , thus , and we have
.
From mass and energy conservation, we conclude
and
.
On the other hand, from Hölder’s inequality, we have
then
Thus, we must have
.
Thus, u is constant in space, and thus it is of the form
□
Applying (1), we see that
then
so
,
so , which is exactly the form of (2).
Cite this paper
Cheng, X. and Li, Q. (2017) Non-Scattering of the Solution of the Nonlinear Schrödinger Equation on the Torus. Journal of Applied Mathematics and Physics, 5, 1917-1922. https://doi.org/10.4236/jamp.2017.59162
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