Non-Scattering of the Solution of the Nonlinear Schrödinger Equation on the Torus

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.

Many mathematicians believe this equation does not have nontrivial solutions which scatter, i.e. which approach a solution to the linear equation at time t = ∞.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  1 (  ).

2𝜋𝜋ℤ
,  ∈ ℤ  cannot converge to a free solution e  ∆ u + due to the presence of the phase rotation e ||  −1  which is caused by the nonlinearity.
We will show this is the only solution that scatters modulo phase rotation in H 1 in the sense that there exists  + ∈  1 (  ) and function : ℝ → ℝ/2πℤ 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.

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).
To prove this theorem, we first need some lemmas.
Lemma 2 (Pre-compactness).For Proof.It suffices to verify this when  is smooth.For any ϵ > 0, we have

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Lemma 4 (H 1 has no step functions).Let  ∈ H 1 (  ) be such that () takes at most two value.Then u is constant.
On the other hand, by hypothesis, ∀  we have By taking limits, we conclude that In particular, since  + has non-zero mass, we have 2 � () ∆  + d.

𝕋𝕋 𝑑𝑑
From (1) and Sobolev, we see that (t) is continuously differentiable in  −1 (  ), and so from the above identity we see that α is continuously differentiable in time.Now we apply i ∂  + ∆ to both sides of (3).Using the NLS equation, we conclude that Since + has non-zero mass, we can thus write where : ℝ ×   →  1 is some function and  > 0. Since  + was in  1 , we see that () is in  1 also for every t.Differentiating the identity  � = 1, 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 ω: ℝ × ℝ  → ℝ which is locally in  1 uniformly in t.
Since A is non-zero, we conclude that Taking imaginary parts, we conclude that ∆ω = 0 and in particular at time t = 0, ω is a harmonic function from ℝ  to ℝ. 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   (0,) =  ( + ) , for some  ∈ ℤ  ,  ∈ ℝ/2ℤ .Thus, we have but  −  ∆  + is a multiple of   by a phase, thus  + () = A    .Applying phase rotation, we may assume  = 0, thus  + () = A  , and we have �() −   ()  −∆ A  �  1 (  ) → 0, as  → ∞.
From mass and energy conservation, we conclude On the other hand, from Hölder's inequality, we have so (, ) = e  (0)  ||  −1  , which is exactly the form of (2).