Non-Scattering of the Solution of the Nonlinear Schrödinger Equation on the Torus ()
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).