# Energy Scattering for Schrödinger Equation with Exponential Nonlinearity in Two Dimensions

Authors: Shuxia Wang
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

When the spatial dimensionsn=2, the initial data u0∈H1, and the
Hamiltonian H(u0)≤1, we prove that the scattering operator is well defined in the whole energy space H1(ℝ2)
for nonlinear Schrödinger equation
with exponential nonlinearity (eλ|u|2-1)u, where 0<λ<4π.


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## Body

## 1. Introduction

We consider the Cauchy problem for the following nonlinear Schrödinger equation:(1)iut+∆u=f(u),(2)f(u):=(eλ|u|2-1)u,
in two spatial dimensions with initial data u0∈H1 and 0<λ<4π. Solutions of the above problem satisfy the conservation of mass and Hamiltonian:
(3)M(u;t):=∫ℝ2|u|2dx=M(u0),H(u;t):=∫ℝ2(|∇u|2+F(u))dx=H(u0),
where
(4)F(u)=1λ(eλ|u|2-λ|u|2-1).Nakamura and Ozawa [1] showed the existence and uniqueness of the scattering operator of (1) with (2). Then, Wang [2] proved the smoothness of this scattering operator. However, both of these results are based on the assumption of small initial data u0. In this paper, we remove this assumption and show that for arbitrary initial data u0∈H1(ℝ2) and H(u0)≤1, the scattering operator is always well defined.Wang et al. [3] proved the energy scattering theory of (1) with f(u)=(eλ|u|2-1-λ|u|2-(λ2/2)|u|4)u, where λ∈ℝ and the spatial dimension n=1. Ibrahim et al. [4] showed the existence and asymptotic completeness of the wave operators for (1) with f(u)=(eλ|u|2-1-λ|u|2)u when the spatial dimensions n=2, λ=4π, and H(u0)≤1. Under the same assumptions as [4], Colliander et al. [5] proved the global well-posedness of (1) with (2).Theorem 1.
Assume thatu0∈H1(ℝ2), H(u0)≤1, and λ=4π. Then problem (1) with (2) has a unique global solution u in the class C(ℝ,H1(ℝ2)).Remark 2.
In fact, by the proof in [5], the global well-posedness of (1) with (2) is also true for 0<λ≤4π.In this paper, we further study the scattering of this problem. Note thatf(u)=(eλ|u|2-1)u=∑k=1∞‍(λk/k!)|u|2ku. Nakanishi [6] proved the existence of the scattering operators in the whole energy space H1(ℝ2) for (1) with f(u)=|u|pu when p>2. Then, Killip et al. [7] and Dodson [8] proved the existence of the scattering operators in L2(ℝ2) for (1) with f(u)=|u|2u. Inspired by these two works, we use the concentration compactness method, which was introduced by Kenig and Merle in [9], to prove the existence of the scattering operators for (1) with (2).For convenience, we write (1) and (2) together; that is,
(5)iut+∆u=f(u):=(eλ|u|2-1)u,u(0,x)=u0,
where u0∈H1(ℝ2) and 0<λ<4π. Our main result is as follows.Theorem 3.
Assume that the initial datau0∈H1(ℝ2), H(u0)≤1, and 0<λ<4π. Let u be a global solution of (5). Then
(6)∥u∥Lt,x4(ℝ×ℝ2)<∞.In Section2, Lemma 9 will show us that Theorem 3 implies the following scattering result.Theorem 4.
Assume that the initial datau0∈H1(ℝ2), H(u0)≤1, and 0<λ<4π. Then the solution of (5) is scattering in the energy space H1(ℝ2).We will prove Theorem3 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 H1 sequence which will be used to prove concentration compactness.Now, we introduce some notations:(7)G(u):=u-f(u)-F(u)=eλ|u|2|u|2-1λ(eλ|u|2-1)G(u)=∑k=1∞kλk|u|2k+2(k+1)!,E=E(u;t):=M(u;t)+H(u;t).We define(8)∥u∥Hqs(ℝ2):=∥(I-∆)s/2u∥Lq(ℝ2),∥u∥H˙qs(ℝ2):=∥(-∆)s/2u∥Lq(ℝ2).For Banach spaceX=Hqs(ℝ2), H˙qs(ℝ2), or Lq(ℝ2), we denote
(9)∥u∥Lp(ℝ;X):=(∫ℝ∥u(t)∥Xpdt)1/p.Whenq=r, LtqLxr is abbreviated to  Lt,xq. When q or r is infinity or when the domain ℝ×ℝ2 is replaced by I×ℝ2, we make the usual modifications. Specially, we denote
(10)S(u):=∥u∥Lt,x4(ℝ×ℝ2)4.Fort0∈ℝ, we split S(u)=S≤t0(u)+S≥t0(u), where
(11)S≤t0(u):=∫-∞t0∫ℝ2|u(t,x)|4dxdt,S≥t0(u):=∫t0+∞∫ℝ2|u(t,x)|4dxdt.For any two Banach spacesX and Y, ∥·∥X∩Y:=max{∥·∥X,∥·∥Y}. C denotes positive constant. If C depends upon some parameters, such as λ, we will indicate this with C(λ).Remark 5.
Note that0<λ<4π in Theorem 3; we only need to prove the result for 0<λ<4(1-4ε)2π, ε∈(0,1/8). Hence, we always suppose that 0<λ<4(1-4ε)2π in the context.
Moreover, we always suppose that the initial datau0 of (5) satisfies u0∈H1(ℝ2) and H(u0)≤1.


## 2. Nonlinear Estimates

In order to estimate (2), we need the following Trudinger-type inequality.Lemma 6 (see [10]).
Letλ∈[0,4π). Then for all u∈H1(ℝ2) satisfying ∥∇u∥L2(ℝ2)≤1, one has
(12)∫ℝ2(eλ|u|2-1)dx≤C(λ)∥u∥L2(ℝ2)2.Note that for for allα≥1,
(13)(eλ|u|2-1)α≤eλα|u|2-1.By Lemma6 and Hölder inequality, for λ∈(0,4π) and for all β≥0, we have
(14)∫ℝ2(eλ|u|2-1)|u|βdx≤∥eλ|u|2-1∥Lx1/(1-ε)(ℝ2)∥u∥Lxβ/ε(ℝ2)β≤∥u∥L2(ℝ2)2∥u∥H1(ℝ2)β≤C(λ,β)∥u∥L2(ℝ2)2,
and thus
(15)∫ℝ2(eλ|u|2-λ|u|2-1)dx≤λ∫ℝ2(eλ|u|2-1)|u|2dx≤C(λ)∥u∥L2(ℝ2)2.Lemma 7 (Strichartz estimates).
Fors=0 or 1,
(16)2≤r,p<∞,1γ(p)+1p=12,1p+1p′=1,
(the pairs (γ(p),p) were called admissible pairs) we have
(17)∥ei(t-t0)∆u(t0)∥Lγ(p)(ℝ;Hps)≤C∥u(t0)∥Hs(ℝ2),∥∫t0tei(t-τ)∆f(u(τ))dτ∥Lγ(p)(ℝ;Hps)≤C∥f(u)∥Lγ(r)′(ℝ;Hr′s).Lemma 8 (see [3, Proposition 2.3]).
Let1<r<p<∞ be fixed indices. Then for any q∈[p,∞),
(18)∥u∥Lq(ℝ2)≤C(p,r)q1/r′+p/rq∥u∥Lp(ℝ2)p/q∥u∥H˙r2/r(ℝ2)1-p/q.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.Lemma 9 (Theorem3 implies Theorem 4).
Letu be a global solution of (5), H(u)≤1, and ∥u∥Lt,x4(ℝ×ℝ2)<∞. Then, for all admissible pairs, we have
(19)∥u∥Lγ(p)(ℝ;Hp1)<∞.
Moreover, there existu±∈H1 such that
(20)limt→±∞∥u-eit∆u±∥H1(ℝ2)=0.Proof.
DefiningX(I)=L2/(1-2ε)(I;H1/ε1), Y(I)=L4(I;H41), by Strichartz estimates, (14) and (15),
(21)∥u∥X(I)∩Y(I)≤C∥u(S)∥H1(ℝ2)+C∥λ|u|2u∥L4/3(I;H4/31)+C∥(eλ|u|2-λ|u|2-1)u∥L2/(1+2ε)(I;H1/(1-ε)1)≤C(E)+C∥u∥Lt,x4(I×ℝ2)2∥u∥Y(I)+C∥u∥Lt,x4(I×ℝ2)8ε∥u∥X(I)×∥(eλ|u|2-λ|u|2-1)u-8ε∥Lt∞Lx1/(1-4ε)(I×ℝ2)+C∥u∥Lt,x4(I×ℝ2)8ε∥u∥X(I)×∥(eλ|u|2-1)|u|2-8ε∥Lt∞Lx1/(1-4ε)(I×ℝ2)≤C(E)+C(E)(∥u∥Lt,x4(I×ℝ2)2+∥u∥Lt,x4(I×ℝ2)8ε)∥u∥X(I)∩Y(I).
Using the same way as in Bourgain [12], one can split ℝ into finitely many pairwise disjoint intervals:
(22)ℝ=⋃j=1JIj,∥u∥Lt,x4(Ij×ℝ2)≤η,C(E)(η2+η8ε)≤12.
By (21),
(23)∥u∥L2/(1-2ε)(Ij;H1/ε1)∩L4(Ij;H41)≤C(E).
Since∥u∥L∞(Ij;H1)≤C(E) and ε∈(0,1/8) can be chosen small arbitrarily, by interpolation,
(24)∥u∥Lγ(p)(Ij;Hp1)≤C(E),
for all admissible pairs and j=1,2,…,J. The desired result (19) follows.
Let(25)u±=u0-i∫0±∞e-iτ∆f(u(τ))dτ.
By (19) and (21),
(26)∥u±∥H1(ℝ2)≲1+∥u∥Y(0,±∞)3+∥u∥X(0,±∞)∩Y(0,±∞)1+8ε≲1.
Thus,u± were well defined and belong to H1. Since
(27)∥u-eit∆u±∥H1(ℝ2)=∥∫t±∞ei(t-τ)∆f(u(τ))dτ∥H1(ℝ2)∥u-eit∆u±∥H1(ℝ2)≲∥u∥Y(t,±∞)3+∥u∥X(t,±∞)∩Y(t,±∞)1+8ε,∥u∥X(0,±∞)∩Y(0,±∞)≲1,
we must have
(28)limt→±∞∥u-eit∆u±∥H1(ℝ2)≲limt→±∞(∥u∥Y(t,±∞)3+∥u∥X(t,±∞)∩Y(t,±∞)1+8ε)=0.
(20) was proved.


## 3. Stability

Lemma 10 (stability).
For anyA>0 and σ>0, there exists δ>0 with the following property: suppose that u:[0,∞)×ℝ2→ℂ satisfies ∥∇u∥Lx2(ℝ2)≤1 for all t∈[0,∞), ∥u∥Lt,x4([0,∞)×ℝ2)≤A and approximately solves (5) in the sense that
(29)∥∫0tei(t-τ)∆(iut+∆u-f(u))(τ)dτ∥Lt,x4∩Lt2/(1-2ε)Lx1/ε([0,∞)×ℝ2)≤δ.
Then for any initial datav(0)∈H1(ℝ2) satisfying H(v(0))≤1 and ∥u(0)-v(0)∥Lx2(ℝ2)≤δ, there is a unique global solution v to (5) satisfying ∥u-v∥Lt,x4([0,∞)×ℝ2)≤σ.Proof.
Denotev=u+w, then
(30)iwt+∆w=(f(u+w)-f(u))-(iut+∆u-f(u))
and ∥w(0)∥L2(ℝ2)≤δ. Let X=Lt∞Lx2∩Lt,x4∩Lt2/(1-2ε)Lx1/ε. By the similar estimates as (21), we have
(31)∥w∥X≤δ+∥∫0tei(t-τ)∆(f(u+w)-f(u))(τ)dτ∥X+δ≤2δ+C∥|u+w|2(u+w)-|u|2u∥Lt,x4/3+C∥(eλ|u+w|2-λ|u+w|2-1)(u+w)m-(eλ|u|2-λ|u|2-1)u∥Lt2/(1+2ε)Lx1/(1-ε)≤2δ+C∥|u+w|2|w|+|u|2|w|∥Lt,x4/3+C∥(eλ|u+w|2-λ|u+w|2-1)|w|m+(eλ|u|2-λ|u|2-1)|w|∥Lt2/(1+2ε)Lx1/(1-ε)+C∥(eλ|u+w|2-1)|u+w|2|w|m+(eλ|u|2-1)|u|2|w|∥Lt2/(1+2ε)Lx1/(1-ε)≤2δ+C(∥u∥Lt,x42+∥u+w∥Lt,x42)∥w∥Lt,x4+C(∥u∥Lt,x48ε+∥u+w∥Lt,x48ε)∥w∥Lt2/(1-2ε)Lx1/ε≤2δ+C(∥u∥Lt,x42+∥u∥Lt,x48ε)∥w∥X+∥w∥X1+8ε+∥w∥X3.
Then we subdivide the time interval[0,∞) into finite subintervals Ij=[tj,tj+1], j=1,…,J, J=J(A) such that
(32)C(∥u∥Lt,x4(Ij×ℝ2)2+∥u∥Lt,x4(Ij×ℝ2)8ε)<14
for each j. Let δ=δ(A,J) be small such that
(33)C82Jδ≪1,82Jδ≪σ.
Then by (31) on I1×ℝ2, we have ∥w∥X(I1×ℝ2)≤8δ and
(34)∥w(t2)∥Lx2(ℝ2)≤∥ei(t2-t1)∆w(t1)∥Lx2(ℝ2)+∥∫t1t2ei(t2-τ)∆(f(u+w)-f(u))(τ)dτ∥Lx2(ℝ2)+δ≤2δ+δ+C(∥u∥Lt,x4(I1×ℝ2)2+∥u∥Lt,x4(I1×ℝ2)8ε)∥w∥X(I1×ℝ2)+∥w∥X(I1×ℝ2)1+8ε+∥w∥X(I1×ℝ2)3≤2δ+4·8δ≤82δ.
Using the same analysis as above, we can get∥w∥X(I2×ℝ2)≤83δ. Iterating this for I2,I3,…,IJ, we obtain ∥w∥X([0,∞)×ℝ2)≤82Jδ≤σ; the desired result was obtained.


## 4. Linear Profile Decomposition

In this section, we will give the linear profile decomposition for Schrödinger equation inH1(ℝ2). First, we give some definitions and lemmas.Definition 11 (symmetry group, [13]).
For any phaseθ∈ℝ/2πℤ, position x0∈ℝ2, frequency ξ0∈ℝ2, and scaling parameter λ>0, we define the unitary transformation gθ,ξ0,x0,λ:Lx2(ℝ2)→Lx2(ℝ2) by the formula
(35)gθ,ξ0,x0,λf(x):=1λeiθeix·ξ0f(x-x0λ).
We letG be the collection of such transformations; this is a group with identity g0,0,0,1, inverse gθ,ξ0,x0,λ-1=g-θ-x0ξ0,-λξ0,-x0/λ,1/λ, and group law
(36)gθ,ξ0,x0,λgθ′,ξ0′,x0′,λ′=gθ+θ′-x0ξ0′/λ,ξ0+ξ0′/λ,x0+λx0′,λλ′.
Ifu:I×ℝ2→ℂ is a function, we define Tgθ,ξ0,x0,λu:λ2I×ℝ2→ℂ, where λ2I:={λ2t:t∈I} by the formula
(37)(Tgθ,ξ0,x0,λu)(t,x):=1λeiθeix·ξ0e-it|ξ0|2u(tλ2,x-x0-2ξ0tλ),
or equivalently
(38)(Tgθ,ξ0,x0,λu)(t)=gθ-t|ξ0|2,ξ0,x0+2ξ0t,λ(u(tλ2)).Ifg∈G, we can easily prove that M(Tgu)=M(u) and S(Tgu)=S(u).Definition 12 (enlarged group, [13]).
For any phaseθ∈ℝ/2πℤ, position x0∈ℝ2, frequency ξ0∈ℝ2, scaling parameter λ>0, and time t0, we define the unitary transformation gθ,ξ0,x0,λ,t0:Lx2(ℝ2)→Lx2(ℝ2) by the formula
(39)gθ,ξ0,x0,λ,t0=gθ,ξ0,x0,λeit0∆,
or in other words
(40)gθ,ξ0,x0,λ,t0f(x):=1λeiθeix·ξ0(eit0∆f)(x-x0λ).
LetG′ be the collection of such transformations. We also let G′ act on global space-time function u:ℝ×ℝ2→ℂ by defining
(41)Tgθ,ξ0,x0,λ,t0u(t,x):=1λeiθeix·ξ0e-it|ξ0|2(eit0∆u)(tλ2,x-x0-2ξ0tλ),
or equivalently
(42)(Tgθ,ξ0,x0,λ,t0u)(t)=gθ-t|ξ0|2,ξ0,x0+2ξ0t,λ,t0(u(tλ2)).Lemma 13 (linear profiles forL2 sequence, [14]).
Letun be a bounded sequence in Lx2(ℝ2). Then (after passing to a subsequence if necessary) there exists a family ϕ(j), j=1,2,… of functions in Lx2(ℝ2) and group elements gn(j)∈G′ for j,n=1,2,… such that one has the decomposition
(43)un=∑j=1lgn(j)ϕ(j)+wn(l)
for all l=1,2,…; here, wn(l)∈Lx2(ℝ2) is such that its linear evolution has asymptotically vanishing scattering size:
(44)liml→∞limsupn→∞S(eit∆wn(l))=0.
Moreover, for anyj≠j′,
(45)λn(j′)λn(j)+λn(j)λn(j′)+λn(j)λn(j′)|ξn(j)-ξn(j′)|2+|xn(j)-xn(j′)|2λn(j)λn(j′)+|tn(j)(λn(j))2-tn(j′)(λn(j′))2|λn(j)λn(j′)⟶∞.
Furthermore, for anyl≥1, one has the mass decoupling property
(46)limn→∞[M(un)-∑j=1lM(ϕ(j))-M(wn(l))]=0.
For any j≤l, we have
(47)(gn(j))-1wn(l)⟶0weaklyinLx2(ℝ2).Remark 14.
If the orthogonal condition (45) holds, then (see [14])
(48)limn→∞〈gn(j)ϕ(j),gn(j′)ϕ(j′)〉L2(ℝ2)=0,j≠j′,limn→∞〈gn(j)ϕ(j),wn(l)〉L2(ℝ2)=0.
Moreover, ifv(j),v(j′)∈Lt,x4(ℝ×ℝ2), then (see [14, 15]), for any 0<θ<1,
(49)limn→∞∥|Tgn(j)v(j)|1-θ|Tgn(j′)v(j′)|θ∥Lt,x4(ℝ×ℝ2)=0.
If v(1),…,v(l)∈Lt,x4(ℝ×ℝ2), then (see [16, Lemma 5.5])
(50)limn→∞S(∑j=1lgn(j)v(j))≤∑j=1lS(v(j)).Remark 15.
As each linear profileϕ(j) in Lemma 13 is constructed in the sense that
(51)eitn(j)∆{eiξn(j)·xλn(j)un(λn(j)x)}(x+xn(j))⟶ϕ(j)
weakly in Lx2(ℝ2) (see [14]), after passing to a subsequence in n, rearrangement, translation, and refining ϕ(j) accordingly, we may assume that the parameters satisfy the following properties:(i)
tn(j)→±∞ as n→∞, or tn(j)≡0 for all n,j;(ii)
λn(j)→0 or ∞ as n→∞, or λn(j)≡1 for all n,j;(iii)
|ξn(j)|→∞ as n→∞, or ξn(j)≡ξ(j) with |ξ(j)|<∞;(iv)
whenλn(j)≡1, ξn(j)≡ξ(j) and |ξ(j)|<∞, we can let ξ(j)≡0.Our main result in this section is the following lemma.Lemma 16 (linear profiles forH1 sequence).
Letun be a bounded sequence in H1(ℝ2). Then up to a subsequence, for any J≥1, there exists a sequence ϕαJ in H1(ℝ2) and a sequence of group elements gnα=gθnα,ξnα,xnα,λnα,tnα∈G′ such that
(52)un=∑α=1JgnαϕαJ+R(n,J).
Here, for eachα, λnα and ξnα must satisfy
(53)"λnα≡1andξnα≡0,"or"λnα⟶∞."R(n,J)∈H1(ℝ2) is such that
(54)limJ→∞limsupn→∞S(eit∆R(n,J))=0.
Moreover, for anyα≠α′, one has the same orthogonal conditions as (45). For any J≥1, one has the following decoupling properties:
(55)limn→∞{∥un∥L2(ℝ2)2-∑α=1J∥ϕαJ∥L2(ℝ2)2-∥R(n,J)∥L2(ℝ2)2}=0,(56)limn→∞{∥un∥H˙1(ℝ2)2-∑α=1J∥gnαϕαJ∥H˙1(ℝ2)2-∥R(n,J)∥H˙1(ℝ2)2}=0,(57)limJ→∞limsupn→∞{H(un)-∑α=1JH(gnαϕαJ)-H(R(n,J))}=0.Proof.
Let(58)□k=ℱ-1χkℱ,χk={12k-1<|ξ|≤2k,0else.
Then, we have(59)un=∑k=-∞+∞□kun:=∑|k|≤N□kun+RN,∥un∥L2(ℝ2)2=∑|k|≤N∥□kun∥L2(ℝ2)2+∥RN∥L2(ℝ2)2,∥un∥H˙1(ℝ2)2=∑|k|≤N∥□kun∥H˙1(ℝ2)2+∥RN∥H˙1(ℝ2)2,limN→∞limsupn→∞∥RN∥L2(ℝ2)=0.
By Lemma13, after passing to a subsequence if necessary, we can obtain
(60)□kun=∑j=1lkgnk(j)ϕk(j)+wnk(lk)
with the stated properties (i)–(iv) in Remark 15 and (43)–(47). Denote
(61)Λ1,0={(k,j)∣|k|≤N,1≤j≤lk,λnk(j)≡1,ξnk(j)≡0},Λ1,∞={(k,j)∣|k|≤N,1≤j≤lk,λnk(j)≡1,|ξnk(j)|⟶∞},Λ0={(k,j)∣|k|≤N,1≤j≤lk,λnk(j)⟶0},Λ∞,0={{|ξk(j)|<2k-1}(k,j)∣|k|≤N,1≤j≤lk,λnk(j)⟶∞,ξnk(j)≡ξk(j),|ξk(j)|<2k-1},Λ∞,1={{|ξk(j)|∈[2k-1,2k]}(k,j)∣|k|≤N,1≤j≤lk,λnk(j)⟶∞,ξnk(j)≡ξk(j),|ξk(j)|∈[2k-1,2k]},Λ∞,∞={{ξnk(j)≡ξk(j),|ξk(j)|>2k}(k,j)∣|k|≤N,1≤j≤lk,λnk(j)⟶∞,|ξnk(j)|⟶∞orξnk(j)≡ξk(j),|ξk(j)|>2k}.
Step
1. We prove that
(62)un=∑(k,j)∈Λ1,0∪Λ∞,1gnk(j)ϕk(j)+R
with ϕk(j)∈H1 and for each fixed N,
(63)limn→∞{∥un∥L2(ℝ2)2-∑(k,j)∈Λ1,0∪Λ∞,1∥ϕk(j)∥L2(ℝ2)2l-∥R∥L2(ℝ2)2{∑(k,j)∈Λ1,0∪Λ∞,1}}=0,limn→∞{∥un∥H˙1(ℝ2)2-∑(k,j)∈Λ1,0∪Λ∞,1∥gnk(j)ϕk(j)∥H˙1(ℝ2)2l-∥R∥H˙1(ℝ2)2{∑(k,j)∈Λ1,0∪Λ∞,1}}=0,(64)limN→∞limlk→∞limsupn→∞S(eit∆R)=0,
where
(65)R=RN+Rw,Rw=∑|k|≤Nwnk(lk).
By (44) and limN→∞limsupn→∞∥RN∥L2(ℝ2)=0, (64) holds obviously. For (62), we prove it by induction. For every k, suppose that
(66)□kun=gnk(1)ϕk(1)+wnk(1).
Case
1. If (k,1)∈Λ1,∞∪Λ0∪Λ∞,0∪Λ∞,∞, we have ϕk(1)=0.
In fact, by (66),
(67)ϕk(1)=(gnk(1))-1□kun-(gnk(1))-1wnk(1).
Thus,(68)∥ϕk(1)∥L2(ℝ2)2=〈(gnk(1))-1□kun-(gnk(1))-1wnk(1),ϕk(1)〉L2(ℝ2)=〈un,□kgnk(1)ϕk(1)〉L2(ℝ2)-〈(gnk(1))-1wnk(1),ϕk(1)〉L2(ℝ2).
Using (47),
(69)〈(gnk(1))-1wnk(1),ϕk(1)〉L2(ℝ2)⟶0asn⟶∞.
By direct calculation,(70)□kgnk(1)ϕk(1)=ℱ-1χk(ξ)λnk(1)eiθnk(1)e-ixnk(1)(ξ+ξnk(1))e-itnk(1)(λnk(1))2|ξ+ξnk(1)|2×ϕk(1)^(λnk(1)(ξ+ξnk(1)))=1λnk(1)eiθnk(1)eix·ξnk(1)(ℱ-1χk(ξλnk(1)+ξnk(1))e-itnk(1)|ξ|2ϕk(1)^)×(x-xnk(1)λnk(1)).
Letn→∞. When (k,1)∈Λ1,∞,
(71)∥□kgnk(1)ϕk(1)∥L2(ℝ2)2≤∫2k-1≤|ξ+ξnk(1)|≤2k|ℱϕk(1)|2dξ⟶0.
When (k,1)∈Λ0,
(72)∥□kgnk(1)ϕk(1)∥L2(ℝ2)2≤∫λnk(1)2k-1≤|ξ+λnk(1)ξnk(1)|≤λnk(1)2k|ℱϕk(1)|2dξ⟶0.
When (k,1)∈Λ∞,0,
(73)∥□kgnk(1)ϕk(1)∥L2(ℝ2)2≤∫|ξ|≥λnk(1)(2k-1-|ξk(1)|)|ℱϕk(1)|2dξ⟶0.
When (k,1)∈Λ∞,∞,
(74)∥□kgnk(1)ϕk(1)∥L2(ℝ2)2≤∫|ξ|≥λnk(1)(|ξnk(1)|-2k)|ℱϕk(1)|2dξ⟶0.
By (68)–(74), ∥ϕk(1)∥L2(ℝ2)2=0 and thus ϕk(1)=0.
Case
2. If (k,1)∈Λ1,0∪Λ∞,1, we can prove
(75)∥gnk(1)ϕk(1)-□kgnk(1)ϕk(1)∥L2(ℝ2)⟶0asn⟶∞.
By absorbing the error intownk(1), we can suppose gnk(1)ϕk(1)=□kgnk(1)ϕk(1). Since □kgnk(1)ϕk(1)∈H1 for each fixed n, we must have ϕk(1)∈H1.
Now, we begin to prove (75). Let χAk(1) be the characteristic function of the set Ak(1) and PAk(1)=ℱ-1χAk(1)ℱ, and then
(76)gnk(1)(PAk(1)ϕk(1)+PAk(1)(gnk(1))-1wnk(1))=PA~k(1)(gnk(1)ϕk(1)+wnk(1))=PA~k(1)□kun,
where
(77)PA~k(1)=ℱ-1χAk(1)(λnk(1)(ξ-ξnk(1)))ℱ.
Note that(78)〈PAk(1)ϕk(1),PAk(1)(gnk(1))-1wnk(1)〉L2(ℝ2)=〈PAk(1)ϕk(1),(gnk(1))-1wnk(1)〉L2(ℝ2)⟶0.
We have
(79)limn→∞(∥PAk(1)(gnk(1))-1wnk(1)∥L2(ℝ2)2∥PA~k(1)□kun∥L2(ℝ2)2-∥PAk(1)ϕk(1)∥L2(ℝ2)2-∥PAk(1)(gnk(1))-1wnk(1)∥L2(ℝ2)2)=0.
When(k,1)∈Λ1,0, we have PAk(1)=PA~k(1). Choosing Ak(1)={ξ||ξ|≤2k-1or|ξ|>2k}, then by (79), PAk(1)ϕk(1)=0, the desired result follows.
When(k,1)∈Λ∞,1 and |ξk(1)|∈(2k-1,2k), we have
(80)∥gnk(1)ϕk(1)-□kgnk(1)ϕk(1)∥L2(ℝ2)2≤∫|ξ+λnk(1)ξk(1)|≤λnk(1)2k-1|ℱϕk(1)|2dξ+∫|ξ+λnk(1)ξk(1)|≥λnk(1)2k|ℱϕk(1)|2dξ≤∫|ξ|≥λnk(1)(|ξk(1)|-2k-1)|ℱϕk(1)|2dξ+∫|ξ|≥λnk(1)(2k-|ξk(1)|)|ℱϕk(1)|2dξ⟶0asn⟶∞.
When(k,1)∈Λ∞,1 and |ξk(1)|=2k, we denote ξ=(ξ1,ξ2) and ξk(1)=(ξk1(1),ξk2(1)). The line ξ2=-(ξk1(1)/ξk2(1))ξ1 (when ξk2(j)=0, we use the line ξ1=0 instead) separates the frequency space ℝ2 into two half-planes. We let Ak(1) to be the half-plane which contains the point ξk(1), and then
(81)PA~k(1)□kun=ℱ-1χAk(1)(λnk(1)(ξ-ξk(1)))χkℱun=0.
By (79), we have PAk(1)ϕk(1)=0. Note that
(82)limn→∞∥gnk(1)(I-PAk(1))ϕk(1)-□kgnk(1)ϕk(j)∥L2(ℝ2)2=limn→∞∥[(1-χAk(1))(·)-χk(·λnk(1)+ξk(1))]ℱϕk(1)∥L2(ℝ2)2≤limn→∞∫ξ∈ℝ2∖(Ak(1)∪{2k-1≤|ξ/λnk(1)+ξk(1)|≤2k})|ℱϕk(1)|2dξ=0.
(75) holds.
When(k,1)∈Λ∞,1 and |ξk(1)|=2k-1, let Ak(1) be the half-plane which does NOT contain the point ξk(1); we can prove (75) similarly as above.
By the proof above, we get(83)□kun=gnk(1)ϕk(1)+wnk(1),gnk(1)ϕk(1)=□kgnk(1)ϕk(1),ϕk(1)∈H1
and (k,1)∈Λ1,0∪Λ∞,1. Denote un(1)=un-gnk(1)ϕk(1) and suppose
(84)□kun(1)=gnk(2)ϕk(2)+wnk(2).
Repeating the proof above, we can getgnk(2)ϕk(2)=□kgnk(2)ϕk(2), ϕk(2)∈H1, and (k,2)∈Λ1,0∪Λ∞,1…; by induction, we obtain (62).
By the orthogonal condition (45), following the proof in [14], we can obtain that for fix k and for all j≠j′,
(85)limn→∞〈gnk(j)ϕk(j),gnk(j′)ϕk(j′)〉H˙1(ℝ2)=limn→∞〈□kgnk(j)ϕk(j),□kgnk(j′)ϕk(j′)〉H˙1(ℝ2)=0,limn→∞〈gnk(j)ϕk(j),wnk(lk)〉H˙1(ℝ2)=limn→∞〈□kgnk(j)ϕk(j),□kwnk(lk)〉H˙1(ℝ2)=0,
(63) were proved.
Step
2. For arbitrary (k1,j1),(k2,j2)∈Λ1,0∪Λ∞,1, we define (k1,j1)~(k2,j2) if the orthogonal condition (45) is NOT true for any subsequence; that is,
(86)limsupn→∞({|xnk2(j2)-xnk1(j1)|2λnk2(j2)λnk1(j1)+|tnk2(j2)(λnk2(j2))2-tnk1(j1)(λnk1(j1))2|λnk2(j2)λnk1(j1)}λnk1(j1)λnk2(j2)+λnk2(j2)λnk1(j1)+λnk2(j2)λnk1(j1)|ξnk2(j2)-ξnk1(j1)|2m+|xnk2(j2)-xnk1(j1)|2λnk2(j2)λnk1(j1)+|tnk2(j2)(λnk2(j2))2-tnk1(j1)(λnk1(j1))2|λnk2(j2)λnk1(j1))<∞.
By the definition above, if(k1,j1)~(k2,j2), we have
(87)λnk1(j1)~λnk2(j2),λnk1(j1)ξnk1(j1)~λnk2(j2)ξnk2(j2),xnk1(j1)λnk1(j1)~xnk2(j2)λnk2(j2),tnk1(j1)~tnk2(j2).
Note that(88)gθ,ξ0,x0,λ,t0f(x):=1λeiθei(x/λ)·λξ0(eit0∆f)(xλ-x0λ).
By Remark 15, we can put these two profiles together as one profile. Then, by denoting (Λ1,0∪Λ∞,1)/~ as {1,2,…,J}, we can obtain the sequence ϕαJ, α=1,2,…,J; and (52)–(56) were proved.
Specially, sinceϕk(j)∈H1 for each (k,j)∈Λ1,0∪Λ∞,1, we have
(89)∑(k,j)∈Λ1,0∪Λ∞,1ϕk(j)∈H1
for fixed N and lk, and hence ϕαJ∈H1(ℝ2) for any fixed J and 1≤α≤J.
Step
3. We prove (57) now. By (56), we only need to prove that for all m∈ℕ, m≥2,
(90)limJ→∞limsupn→∞{∥un∥L2m(ℝ2)2m-∑α=1J∥gnαϕαJ∥L2m(ℝ2)2mkkl-∥R(n,J)∥L2m(ℝ2)2m{∑α=1J∥gnαϕαJ∥L2m(ℝ2)2m}}=0.
As(91)∥R(n,J)∥L2m(ℝ2)≲∥R(n,J)∥L4(ℝ2)1/(m-1)∥R(n,J)∥H1(ℝ2)(m-2)/(m-1)
and for 1/4<θ<1/2,
(92)∥R(n,J)∥L4(ℝ2)≤∥eit∆R(n,J)∥Lt∞Lx4(ℝ×ℝ2)≲∥eit∆R(n,J)∥H4θLx4(ℝ×ℝ2)=∥eit∆R(n,J)∥Lt,x4(ℝ×ℝ2)1-2θ∥eit∆R(n,J)∥H41/2Lx4(ℝ×ℝ2)2θ=∥eit∆R(n,J)∥Lt,x4(ℝ×ℝ2)1-2θ∥eit∆R(n,J)∥Lt4H41(ℝ×ℝ2)2θ≲∥eit∆R(n,J)∥Lt,x4(ℝ×ℝ2)1-2θ.
By (54), we have
(93)limJ→∞limsupn→∞∥R(n,J)∥L2m(ℝ2)2m=0.
We separate the set1≤α≤J into two subsets:
(94)Λ1={α∣1≤α≤J,λnα≡1,ξnα≡0},Λ∞={α∣1≤α≤J,λnα⟶∞}.
Whenα∈Λ∞,
(95)limn→∞∥gnαϕαJ∥L2m(ℝ2)=limn→∞(λnα)-1+1/m∥eitnα∆ϕαJ∥L2m(ℝ2)≤limn→∞(λnα)-1+1/m∥ϕαJ∥H1(ℝ2)=0.
Hence, in order to prove (90), one only needs to prove
(96)limn→∞{∥∑α∈Λ1gnαϕαJ∥L2m(ℝ2)2m-∑α∈Λ1∥gnαϕαJ∥L2m(ℝ2)2m}=0.
Ifα∈Λ1 and tnα→∞, for a function ϕ~αJ∈H˙1/2∩L4/3, we have
(97)∥gnαϕαJ∥L4(ℝ2)≤∥gnαϕαJ-gnαϕ~αJ∥L4(ℝ2)+∥gnαϕ~αJ∥L4(ℝ2)≤∥ϕαJ-ϕ~αJ∥H˙1/2(ℝ2)+|tnα|-1/2∥ϕ~αJ∥L4/3(ℝ2).
By approximatingϕαJ by ϕ~αJ∈Cc∞ in H˙1/2 and sending n→∞, we have ∥gnαϕαJ∥L4(ℝ2)→0. Note that gnαϕαJ∈H1; we obtain ∥gnαϕαJ∥L2m(ℝ2)→0 for all m≥2.
Ifα∈Λ1 and tnα≡0, we have orthogonal condition |xnα-xnα′|→∞ for any α≠α′. Thus,
(98)limn→∞{∥∑α∈Λ1,tnα≡0gnαϕαJ∥L2m(ℝ2)2m-∑α∈Λ1,tnα≡0∥gnαϕαJ∥L2m(ℝ2)2m}=limn→∞{∥∑α∈Λ1,tnα≡0eiθnαϕαJ(·-xnα)∥L2m(ℝ2)2mm-∑α∈Λ1,tnα≡0∥ϕαJ(·-xnα)∥L2m(ℝ2)2m{∥∑α∈Λ1,tnα≡0eiθnαϕαJ(·-xnα)∥L2m(ℝ2)2m}}=0.
(96) holds and then (57) was proved.


## 5. The Proof of Theorem3

Letu be a solution of (5), H(u0)≤1; by Strichartz estimate and (21),
(99)∥u∥Lt,x4∩Lt2/(1-2ε)Lx1/ε(ℝ×ℝ2)≤C∥u0∥L2(ℝ2)+C∥u∥Lt,x4(ℝ×ℝ2)3+C(E)∥u∥Lt2/(1-2ε)Lx1/ε(ℝ×ℝ2)∥u∥Lt,x4(ℝ×ℝ2)8ε≤C∥u0∥L2(ℝ2)+C∥u∥Lt,x4∩Lt2/(1-2ε)Lx1/ε(ℝ×ℝ2)3+C(E)∥u∥Lt,x4∩Lt2/(1-2ε)Lx1/ε(ℝ×ℝ2)1+8ε.When∥u0∥L2(ℝ2)≪1, by standard continuity argument, we have
(100)∥u∥Lt,x4(ℝ×ℝ2)≤C∥u0∥L2(ℝ2)<∞.Hence, ifM(u)≪1, then ∥u∥Lt,x4(ℝ×ℝ2)<∞. In particular, we have scattering in both directions.For any massm≥0, we define
(101)A(m):=sup{S(u):uistheglobalsolutionof(5),M(u)≤m,H(u)≤1}.ThenA:[0,+∞)→[0,+∞] is a monotone increasing function of m. As A is left-continuous and finite for small m, there must exist a unique critical mass m0∈(0+∞] such that A(m) is finite for all m<m0 but infinite for all m≥m0.To prove Theorem3, one only needs to prove that the critical mass m0 is infinite. We will prove that by contradiction.Proposition 17.
Suppose that the critical massm0 is finite. Let un:ℝ×ℝ2→ℂ for n=1,2,… be a sequence of solutions and let tn∈ℝ be a sequence of times such that H(un)≤1, limsupn→∞M(un)=m0, and
(102)limn→∞S≥tn(un)=limn→∞S≤tn(un)=+∞.
Then there exists a sequence ofxn=xn(tn)∈ℝ2such that un(tn,x+xn) has a subsequence which converges strongly into Lx2(ℝ2). Especially, the Hamiltonian of the limiting function is not greater than 1.Proof.
We can taketn=0 for all n by translating un in time. Thus,
(103)limn→∞S≥0(un)=limn→∞S≤0(un)=+∞.
By Lemma16, up to a subsequence if necessary, we have
(104)un(0)=∑α∈Λ1∪Λ∞gnαϕαJ+R(n,J),
where Λ1 and Λ∞ were defined by (94). Suppose that
(105)gnα=hnαeitnα∆,
where tnα∈ℝ and hnα∈G. By (55),
(106)∑α∈Λ1∪Λ∞M(ϕαJ)≤limsupn→∞M(un(0))≤m0.
Hence,(107)supα∈Λ1∪Λ∞M(ϕαJ)≤m0.
Suppose(108)supα∈Λ1∪Λ∞M(ϕαJ)≤m0-σ.
For some σ>0, we will prove that this leads to a contradiction: defining the nonlinear profile vαJ:ℝ×ℝ2→ℂ as follows.(i)
Whenα∈Λ1,
(1)
iftnα≡0, we define vαJ to be the global solution of (5) with initial data vαJ(0)=ϕαJ;(2)
iftnα→+∞, we define vαJ to be the global solution of (5) which scatters to eit∆ϕαJ in H1 when t→+∞;(3)
iftnα→-∞, we define vαJ to be the global solution of (5) which scatters to eit∆ϕαJ in H1 when t→-∞.(ii)
Whenα∈Λ∞,
(1)
iftnα≡0, we define vαJ to be the global solution of iut+∆u=|u|2u with initial data vαJ(0)=ϕαJ;(2)
iftnα→+∞, we define vαJ to be the global solution of iut+∆u=|u|2u which scatters to eit∆ϕαJ in H1 when t→+∞;(3)
iftnα→-∞, we define vαJ to be the global solution of iut+∆u=|u|2u which scatters to eit∆ϕαJ in H1 when t→-∞.
For eachα∈Λ1, vαJ is well defined by the definition of A(m), (100), and (108). For each α∈Λ∞, vαJ is well defined by the scattering of cubic Schrödinger equation in L2(ℝ2) (see [7, 8]) and the same analysis as in Lemma 9.
Now, we define(109)u~n=∑α∈Λ1∪Λ∞Thnα[vαJ(·+tnα)]+eit∆R(n,J)
for n,J=1,2,…, and then we have the following two lemmas.
Lemma 18(110)limn→∞M(u~n(0)-un(0))=0,limJ→∞limsupn→∞∥u~n∥H˙1(ℝ2)2≤1.
Proof. Since
(111)limn→∞M(u~n(0)-un(0))≤∑α∈Λ1∪Λ∞limn→∞M(Thnα[vαJ(·+tnα)](0)-gnαϕαJ)=∑α∈Λ1∪Λ∞limn→∞M(hnα[vαJ(tnα)]-hnαeitnα∆ϕαJ)=∑α∈Λ1∪Λ∞limn→∞M(vαJ(tnα)-eitnα∆ϕαJ)=0,
we get the first equality.
Note thatThnα maps the solutions of iut+∆u=|u|2u from one to another; using (57) and energy conservation, we have
(112)limJ→∞limsupn→∞∥u~n∥H˙1(ℝ2)2≤limJ→∞limsupn→∞{∑α=1J∥Thnα[vαJ(·+tnα)]∥H˙1(ℝ2)2mona+∥R(n,J)∥H˙1(ℝ2)2{∑α=1J∥Thnα[vαJ(·+tnα)]∥H˙1(ℝ2)2}}≤limJ→∞limsupn→∞{∑α=1JH(gnαϕαJ)+H(R(n,J))}≤limJ→∞limsupn→∞H(un)≤1.
Lemma 19.If(113)limJ→∞limsupn→∞∥u~n∥Lt,x4(ℝ×ℝ2)<∞,∥vαJ∥Lt,x4(ℝ×ℝ2)<∞(∀α),then(114)limJ→∞limsupn→∞∥∫0tei(t-τ)∆((i∂t+∆)u~n-f(u~n))(τ)dτ∥X=0,whereX=Lt,x4∩Lt2/(1-2ε)Lx1/ε(ℝ×ℝ2).
Proof. Denote
(115)vnαJ=Thnα[vαJ(·+tnα)].
By the definition ofu~n, we have
(116)u~n=∑α∈Λ1∪Λ∞vnαJ+eit∆R(n,J),(i∂t+∆)u~n=∑α∈Λ1f(vnαJ)+∑α∈Λ∞|vnαJ|2vnαJ.
Thus, by triangle inequality, it suffices to show that(117)limJ→∞limsupn→∞∥∫0tei(t-τ)∆(f(u~n-eit∆R(n,J))-f(u~n){f(u~n-eit∆R(n,J))})(τ)dτ{∫0t}∥X=0,(118)limn→∞∥∫0tei(t-τ)∆(f(∑α∈Λ1∪Λ∞vnαJ)-∑α∈Λ1∪Λ∞f(vnαJ))(τ)dτ∥X=0,(119)limn→∞∥∫0tei(t-τ)∆∑α∈Λ∞(f(vnαJ)-|vnαJ|2vnαJ)(τ)dτ∥X=0.
By Lemma 18, letJ and n be sufficiently large; we have
(120)∥u~n∥H˙1(ℝ2)2≤11-4ε,∥u~n-eit∆R(n,J)∥H˙1(ℝ2)2≤11-4ε.
Since we have supposed0<λ<4(1-4ε)2π, Lemma 6 can also be used here. Then by the same estimates as (31),
(121)∥∫0tei(t-τ)∆(f(u~n-eit∆R(n,J))-f(u~n))(τ)dτ∥X≲(∥u~n∥Lt,x4(ℝ×ℝ2)2+∥eit∆R(n,J)∥Lt,x4(ℝ×ℝ2)2)×∥eit∆R(n,J)∥Lt,x4(ℝ×ℝ2)+(∥u~n∥Lt,x4(ℝ×ℝ2)8ε+∥eit∆R(n,J)∥Lt,x4(ℝ×ℝ2)8ε)×∥eit∆R(n,J)∥Lt2/(1-2ε)Lx1/ε(ℝ×ℝ2).
By (54),
(122)limJ→∞limsupn→∞∥eit∆R(n,J)∥Lt,x4(ℝ×ℝ2)=0.
As(123)∥eit∆R(n,J)∥Lt2/(1-2ε)Lx1/ε(ℝ×ℝ2)≲∥eit∆R(n,J)∥Lt,x4(ℝ×ℝ2)2ε/(1-2ε)∥eit∆R(n,J)∥Lt2/(1-ε)Lx2/ε(ℝ×ℝ2)(1-4ε)/(1-2ε),∥eit∆R(n,J)∥Lt2/(1-ε)Lx2/ε(ℝ×ℝ2)≲∥R(n,J)∥Lx2(ℝ2)<∞,
we have
(124)limJ→∞limsupn→∞∥eit∆R(n,J)∥Lt2/(1-2ε)Lx1/ε(ℝ×ℝ2)=0.
BylimJ→∞limsupn→∞∥u~n∥Lt,x4(ℝ×ℝ2)<∞, (117) was obtained.
Using Strichartz estimate,(125)∥∫0tei(t-τ)∆(f(vnαJ)-|vnαJ|2vnαJ)(τ)dτ∥X≤∥∑m=2∞|vnαJ|2mvnαJm!∥Lt,x4/3(ℝ×ℝ2)≤∑m=2∞1m!∥vnαJ∥Lt,x4(ℝ×ℝ2)∥vnαJ∥Lt,x4m(ℝ×ℝ2)2m≤∑m=2∞1m!(λnα)-2(m-1)∥vαJ∥Lt,x4(ℝ×ℝ2)∥vαJ∥Lt,x4m(ℝ×ℝ2)2m.
By Lemma8,
(126)∥vαJ∥Lt,x4m(ℝ×ℝ2)≤Cm1/2+1/2m∥vαJ∥Lt,x4(ℝ×ℝ2)1/m∥vαJ∥L∞(ℝ,H˙1)1-1/m.
Note that∥vαJ∥Lt,x4(ℝ×ℝ2)<∞. Equation (119) was obtained.
To prove Lemma 19, it is only left to prove (118). Note that
(127)|f(∑α=1Jzα)-∑α=1Jf(zα)|≲∑α≠α′|zα||eλ|zα′|2-1|.
By (57), we have
(128)∥∫0tei(t-τ)∆(f(∑α∈Λ1∪Λ∞vnαJ)-∑α∈Λ1∪Λ∞f(vnαJ))(τ)dτ∥X≲∑α≠α′(∥vnαJ(eλ|vnα′J|2-λ|vnα′J|2-1)∥Lt2/(1+2ε)Lx1/(1-ε)(ℝ×ℝ2)∥vnαJ|vnα′J|2∥Lt,x4/3(ℝ×ℝ2)m+∥vnαJ(eλ|vnα′J|2-λ|vnα′J|2-1)∥Lt2/(1+2ε)Lx1/(1-ε)(ℝ×ℝ2))≲∑α≠α′∥|vnαJ|1/2|vnα′J|1/2∥Lt,x4(ℝ×ℝ2)2∥vnα′J∥Lt,x4(ℝ×ℝ2)+∑α≠α′∥|vnαJ|ε/(1+ε)|vnα′J|1/(1+ε)∥Lt,x4(ℝ×ℝ2)4ε(1+ε)/(1-2ε)×∥vnαJ∥Lt2(1-2ε-4ε2)/(1-2ε-6ε2)Lx(1-2ε-4ε2)/ε2(ℝ×ℝ2)(1-2ε-4ε2)/(1-2ε)×∥|vnα′J|-4ε/(1-2ε)m×(eλ|vnα′J|2-λ|vnα′J|2-1)∥Lt∞Lx(1-2ε)/(1-4ε)(ℝ×ℝ2).
By (49), we immediately obtain (118).
If(129)supα∈Λ1∪Λ∞M(ϕα)≤m0-σ
for some σ>0, by the definition of A(m) and (100), we have
(130)A(m)≤Bm∀m∈[0,m0-σ],
where B=B(σ)∈(0,+∞). Then vαJ satisfies
(131)M(vαJ)=M(ϕαJ)≤m0-σ,(132)S(vαJ)≤A(M(ϕαJ))≤BM(ϕαJ).
By (50), (106), and (132), we have
(133)limJ→∞limsupn→∞S(u~n)≤Bm0.
Using Lemmas 18 and 19, we have(134)M(u~n(0)-un(0))≤δ,S(u~n)≤2Bm0,∥∫0tei(t-τ)∆((i∂t+∆)u~n-f(u~n))(τ)dτ∥Lt,x4∩Lt2/(1-2ε)Lx1/ε(ℝ×ℝ2)≤δ,
for δ>0 sufficiently small, J=J(δ), and n=n(J,δ) sufficiently large. By Lemma 10, we obtain that S(un)≤3Bm0 which contradicts (103). Thus, (129) fails for all σ>0, and then
(135)supα∈Λ1∪Λ∞M(ϕα)=m0.
Comparing this with (106), we have
(136)un(0)=hneitn∆ϕ+Rn
with tn converging to ±∞ or tn≡0, hn∈G, M(ϕ)=m0, and
(137)M(Rn)⟶0,S(eit∆Rn)⟶0asn⟶∞.
Specially, the parametersλn, ξn of hn must satisfy
(138)"λn≡1andξn≡0,"or"λn⟶∞."
Ifλn→∞, similar to the former case, we can define the approximate solution
(139)u~n=Thn[v(·+tn)]+eit∆Rn.
By the scattering of cubic Schrödinger equation, we haveS(v)<∞ and limn→∞S(u~n)<∞. By Lemmas 18, 19, and 10, for n sufficiently large, we obtain S(un)<∞ which contradicts (103).
Ifλn≡1, ξn≡0, and tn→+∞, by Strichartz estimate and monotone convergence, we have
(140)limn→∞S≥0(eit∆eitn∆ϕ)=0.
Thus,(141)limn→∞S≥0(eit∆hneitn∆ϕ)=limn→∞S≥0(eit∆eitn∆ϕ)=0.
Sincelimn→∞S(eit∆Rn)=0, we can see from (136) that
(142)limn→∞S≥0(eit∆un(0))=0.
Hence,(143)limn→∞∥∫0tei(t-τ)∆f(eit∆un(0))(τ)dτ∥Lt,x4∩Lt2/(1-2ε)Lx1/ε([0,∞)×ℝ2)=0.
By Lemma10 (with eit∆un(0) as the approximate solution and un(0) as the initial data), we have
(144)limn→∞S≥0(un)=0,
which contradicts one of the estimates in (103).
Ifλn≡1, ξn≡0, and tn→-∞, the argument is the same and we can obtain a contradiction by using the other half of (103).
Now, the only case left isλn≡1, ξn≡0, and tn≡0. In this case, we have
(145)M(un(0)-hnϕ)=M(Rn)⟶0asn⟶∞.
Thus,(hn)-1un(0)=eiθnun(0,x+xn) converges to ϕ in Lx2(ℝ2). Since H(ϕ)=H(gnϕ)≤1, after passing to a subsequence if necessary and rotating ϕ, the desired result follows.Let{un} be the sequence given in Proposition 17 and satisfy M(un)≤m0 and suppose that un(0,x+xn) converges to u0 strongly in Lx2(ℝ2); then M(u0)≤m0 and H(u0)≤1. Let u be the global solution with initial data u(0)=u0; by Lemma 10, we must have
(146)S≥0(u)=S≤0(u)=+∞.By the definition ofm0, M(u0)≥m0 and hence M(u0)=m0.Sinceu is locally in Lt,x4, for all tn∈ℝ, we have
(147)S≥tn(u)=S≤tn(u)=+∞.Using Proposition17 for {u(tn)}, we have that u(tn,x+x(tn)) converges into Lx2(ℝ2). By Ascoli-Arzela Theorem, we have the following.Proposition 18.
Suppose that the critical massm0 is finite. Then there exists a global solution u with mass m0, and for every η>0 there exists 0<C(η)<∞ such that
(148)∫|x-x(t)|≥C(η)|u(t,x)|2dx+∫|ξ-ξ(t)|≥C(η)|u^(t,ξ)|2dξ≤η
for all t∈ℝ, where the functions x,ξ:ℝ→ℝ2.Proposition 19.
The solution described in Proposition18 does not exist.Once we proved Proposition19, we can say that m0=∞ and thus Theorem 3 is true. In order to prove Proposition 19, we need the following two lemmas.Lemma 20 (see [6, Lemma 5.2]).
Let u be a global solution of (5). Then one has
(149)∬ℝ×ℝ2〈t〉2G(u)〈t〉3+|x|3dxdt≤C(E),
where 〈t〉=1+|t|2.Lemma 21 (see [6, Lemma 6.2]).
Let u be a global solution of (5). Let B be a compact subset of ℝ2. Then for any R>0 and T>0, one has
(150)∫B(R)|u(T,x)|2dx≥∫B|u(0,x)|2dx-C(E)TR,
where B(R):={x∈ℝ2∣∃y∈Bs.t.|x-y|<R}.Proof of Proposition19.
By Lemma21, choosing η sufficiently small,
(151)∫|x-x(0)|≤C(η)+R|t||u(t,x)|2dx≥∫|x-x(0)|≤C(η)|u(0,x)|2dx-C(E)R≥m0-η-C(E)R.
By Proposition18,
(152)∫|x-x(t)|≤C(η)|u(t,x)|2dx≥m0-η.
For a fixed large numberR, we must have |x(t)-x(0)|≤2C(η)+R|t|. By Lemma 20 and Hölder inequality,
(153)∞>∬ℝ×ℝ2〈t〉2|u|4〈t〉3+|x|3dxdt≳∫ℝ∫|x-x(t)|≤C(η)〈t〉2|u|4〈t〉3+|x|3dxdt≳∫ℝ∫|x-x(t)|≤C(η)|u|4〈t〉+1dxdt≳∫ℝ∫|x-x(t)|≤C(η)|u|2〈t〉+1dxdt≳(m0-η)∫ℝ1〈t〉+1dxdt=∞.
This is a contradiction. Proposition19 was obtained.



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*Source: 968603-2013-04-09.xml*