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Journal of Function Spaces and Applications
Volume 2013 (2013), Article ID 968603, 13 pages
Energy Scattering for Schrödinger Equation with Exponential Nonlinearity in Two Dimensions
School of Mathematical Sciences, Peking University, Beijing 100871, China
Received 9 January 2013; Accepted 24 February 2013
Academic Editor: Baoxiang Wang
Copyright © 2013 Shuxia Wang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
When the spatial dimensions , the initial data , and the Hamiltonian , we prove that the scattering operator is well defined in the whole energy space for nonlinear Schrödinger equation with exponential nonlinearity , where .
We consider the Cauchy problem for the following nonlinear Schrödinger equation: in two spatial dimensions with initial data and . Solutions of the above problem satisfy the conservation of mass and Hamiltonian: where
Nakamura and Ozawa  showed the existence and uniqueness of the scattering operator of (1) with (2). Then, Wang  proved the smoothness of this scattering operator. However, both of these results are based on the assumption of small initial data . In this paper, we remove this assumption and show that for arbitrary initial data and , the scattering operator is always well defined.
Wang et al.  proved the energy scattering theory of (1) with , where and the spatial dimension . Ibrahim et al.  showed the existence and asymptotic completeness of the wave operators for (1) with when the spatial dimensions , , and . Under the same assumptions as , Colliander et al.  proved the global well-posedness of (1) with (2).
In this paper, we further study the scattering of this problem. Note that . Nakanishi  proved the existence of the scattering operators in the whole energy space for (1) with when . Then, Killip et al.  and Dodson  proved the existence of the scattering operators in for (1) with . Inspired by these two works, we use the concentration compactness method, which was introduced by Kenig and Merle in , to prove the existence of the scattering operators for (1) with (2).
Theorem 3. Assume that the initial data , , and . Let be a global solution of (5). Then
Theorem 4. Assume that the initial data , , and . Then the solution of (5) is scattering in the energy space .
We will prove Theorem 3 by contradiction in Section 5. In Section 2, we give some nonlinear estimates. In Section 3, we prove the stability of solutions. In Section 4, we give a new profile decomposition for sequence which will be used to prove concentration compactness.
Now, we introduce some notations:
For Banach space , , or , we denote
When , is abbreviated to . When or is infinity or when the domain is replaced by , we make the usual modifications. Specially, we denote
For , we split , where
For any two Banach spaces and , . denotes positive constant. If depends upon some parameters, such as , we will indicate this with .
2. Nonlinear Estimates
In order to estimate (2), we need the following Trudinger-type inequality.
Lemma 6 (see ). Let . Then for all satisfying , one has
Note that for for all ,
By Lemma 6 and Hölder inequality, for and for all , we have and thus
Lemma 7 (Strichartz estimates). For or , (the pairs were called admissible pairs) we have
Lemma 8 (see [3, Proposition 2.3]). Let be fixed indices. Then for any ,
As shown in [6, 11], to obtain the scattering result, it suffices to show that any finite energy solution has a finite global space-time norm. In fact, if Theorem 3 is true, we have the following theorem.
Proof. Defining , , by Strichartz estimates, (14) and (15),
Using the same way as in Bourgain , one can split into finitely many pairwise disjoint intervals:
Since and can be chosen small arbitrarily, by interpolation, for all admissible pairs and . The desired result (19) follows.
By (19) and (21),
Thus, were well defined and belong to . Since we must have (20) was proved.
Lemma 10 (stability). For any and , there exists with the following property: suppose that satisfies for all , and approximately solves (5) in the sense that
Then for any initial data satisfying and , there is a unique global solution to (5) satisfying .
Proof. Denote , then
and . Let . By the similar estimates as (21), we have
Then we subdivide the time interval into finite subintervals , , such that for each . Let be small such that
Then by (31) on , we have and
Using the same analysis as above, we can get . Iterating this for , we obtain ; the desired result was obtained.
4. Linear Profile Decomposition
In this section, we will give the linear profile decomposition for Schrödinger equation in . First, we give some definitions and lemmas.
Definition 11 (symmetry group, ). For any phase , position , frequency , and scaling parameter , we define the unitary transformation by the formula
We let be the collection of such transformations; this is a group with identity , inverse , and group law
If is a function, we define , where by the formula or equivalently
If , we can easily prove that and .
Definition 12 (enlarged group, ). For any phase , position , frequency , scaling parameter , and time , we define the unitary transformation by the formula
or in other words
Let be the collection of such transformations. We also let act on global space-time function by defining or equivalently
Lemma 13 (linear profiles for sequence, ). Let be a bounded sequence in . Then (after passing to a subsequence if necessary) there exists a family , of functions in and group elements for such that one has the decomposition
for all ; here, is such that its linear evolution has asymptotically vanishing scattering size:
Moreover, for any ,
Furthermore, for any , one has the mass decoupling property For any , we have
Remark 15. As each linear profile in Lemma 13 is constructed in the sense that weakly in (see ), after passing to a subsequence in , rearrangement, translation, and refining accordingly, we may assume that the parameters satisfy the following properties:(i) as , or for all ;(ii) or as , or for all ;(iii) as , or with ;(iv)when , and , we can let .
Our main result in this section is the following lemma.
Lemma 16 (linear profiles for sequence). Let be a bounded sequence in . Then up to a subsequence, for any , there exists a sequence in and a sequence of group elements such that
Here, for each , and must satisfy is such that
Moreover, for any , one has the same orthogonal conditions as (45). For any , one has the following decoupling properties:
Then, we have
By Lemma 13, after passing to a subsequence if necessary, we can obtain with the stated properties (i)–(iv) in Remark 15 and (43)–(47). Denote
Step 1. We prove that with and for each fixed , where
By (44) and , (64) holds obviously. For (62), we prove it by induction. For every , suppose that
Case 1. If , we have .
In fact, by (66),
By direct calculation,
Let . When , When , When , When ,
By (68)–(74), and thus .
Case 2. If , we can prove
By absorbing the error into , we can suppose . Since for each fixed , we must have .
Now, we begin to prove (75). Let be the characteristic function of the set and , and then where
Note that We have
When , we have . Choosing , then by (79), , the desired result follows.
When and , we have
When and , we denote and . The line (when , we use the line instead) separates the frequency space into two half-planes. We let to be the half-plane which contains the point , and then
By (79), we have . Note that (75) holds.
When and , let be the half-plane which does NOT contain the point ; we can prove (75) similarly as above.
By the proof above, we get and . Denote and suppose
Repeating the proof above, we can get , , and ; by induction, we obtain (62).
By the orthogonal condition (45), following the proof in , we can obtain that for fix and for all , (63) were proved.
Step 2. For arbitrary , we define if the orthogonal condition (45) is NOT true for any subsequence; that is,
By the definition above, if , we have
Note that By Remark 15, we can put these two profiles together as one profile. Then, by denoting / as , we can obtain the sequence , ; and (52)–(56) were proved.
Specially, since for each , we have for fixed and , and hence for any fixed and .
Step 3. We prove (57) now. By (56), we only need to prove that for all , ,
As and for , By (54), we have
We separate the set into two subsets:
Hence, in order to prove (90), one only needs to prove
If and , for a function , we have
By approximating by in and sending , we have . Note that ; we obtain for all .
If and , we have orthogonal condition for any . Thus,