On the 4D Nonlinear Schr\" odinger equation with combined terms

arXiv:1711.10362v1 [math.AP] 13 Nov 2017

¨ ON THE 4D NONLINEAR SCHRODINGER EQUATION WITH COMBINED TERMS UNDER THE ENERGY THRESHOLD CHANGXING MIAO, TENGFEI ZHAO, AND JIQIANG ZHENG

Abstract. In this paper, we consider the longtime dynamics of the solutions to focusing energy-critical Schr¨ odinger equation with a defocusing energysubcritical perturbation term under a ground state energy threshold in four spatial dimension. This extends the results in Miao et al. (Commun Math Phys 318(3):767-808, 2013, The dynamics of the NLS with the combined terms in five and higher dimensions. Some topics in harmonic analysis and applications, advanced lectures in mathematics, ALM34, Higher Education Press, Beijing, pp 265-298, 2015) to four dimension without radial assumption and the proof of scattering is based on the interaction Morawetz estimates developed in Dodson (Global well-posedness and scattering for the focusing, energy-critical nonlinear Schr¨ oinger problem in dimension d = 4 for initial data below a ground state threshold, arXiv:1409.1950), the main ingredients of which requires us to overcome the logarithmic failure in the double Duhamel argument in four dimensions.

1. Introduction In this paper, we consider the nonlinear Schrödinger equation with combined terms in four spatial dimension, i.e. ( 4 i∂t u + ∆u = −|u|2 u + |u| 3 u, (t, x) ∈ R × R4 (CNLS) u(0, x) = ϕ(x) ∈ H 1 (R4 ), P 2 ∂xj is the Laplace operator. where ∆ = 1≤j≤4

If u is a solution to (CNLS), Z Z 2 M (u) = |u| dx and E(u) = R4

R4

10 1 1 4 3 2 3 |∇u| − |u| + |u| dx 2 4 10

are often said to be the mass and energy of u respectively. Through a standard method, one can check that the mass and energy of a smooth solution to 2 4/3 (CNLS) conserve in time. The nonlinear term |u| u is H˙ 1 (R4 ) critical and |u| u 1/2 4 ˙ is H (R ) critical according to the standard scaling analysis. In general, the energy critical NLS is given by ( 4 i∂t u + ∆u = µ|u| d−2 u, (t, x) ∈ R × Rd (NLS) u(0, x) = ϕ(x) ∈ H˙ 1 (Rd ) 2000 Mathematics Subject Classification. Primary: 35L70, Secondary: 35Q55. Key words and phrases. Nonlinear Schr¨ odinger Equation; Longtime dynamics; Interaction Morawetz esimates Scattering; Threshold Energy. 1

2

MIAO, ZHAO, AND ZHENG

where µ = ±1 and d ≥ 3. For the local well-posedness of (NLS) in H˙ 1 (Rd ) or H 1 (Rd ), we refer to [6, 7, 8]. If µ = 1, we call the Cauchy problem (NLS) is defocusing. There are many results in the literature considering the defocusing cases. In [5], Bourgain first developed a method of reduction of energy and proved that any solution to defocusing (NLS) with radial initial data is scattering in spatial dimension three. Colliander. etc.[11] removed the radial assumption by exploiting a interaction Morawetz estimate, and we refer [21] for another proof, which is based on the long time Strichartz estimates. This result was extended in Ryckman and Visan [27] for dimension four (for another proof, we refer to [33])and Visan [32] for five and higher dimensions by using the Morawetz estimates. For the focusing case (µ = −1), Kenig and Merle [18] gave out a decomposition of the region if the energy of the solutions under a ground state threshold for radial solutions and they proved that in the dimensions d ∈ {3, 4, 5}, one of the two regions for global well-posedness and scattering and the other region for finite time blow-up. The corresponding results in the cases of five and higher dimensions were proved by Killip and Visan in [20] without radial assumption by employing the double Duhamel formula trick. Dodson [13] obtained global well-posedness and scattering results in dimension four under a ground state threshold by developing the long time Strichartz estimates. But the analog nonradial case of dimension three remains open up to now. For scattering results of the defocusing energy-subcritical NLS, we refer to [10, 12, 14, 25, 26, 30] and the reference therein. There are also a quantity of results for the Cauchy problem (NLS) with a energysubcritical nonlinearity perturbation. In [31], Tao, Visan, and Zhang proved the scattering results in H 1 (Rd ) of defocusing (NLS) with a defocusing perturbation 4 4 and d ≥ 3) or with focusing |u|p u ( d4 ≤ p < d−2 and d ≥ 3) and |u|p u ( d4 ≤ p < d−2 the small mass condition. The latter results was extended in Killip, Oh, Pocovnicu, and Visan [19], where the scattering results were obtained for the solutions to the defocusing (NLS) with a focusing term |u|2 u in three spatial dimension if their initial datum belong to a certain region given by rescaling. Miao, Xu, and Zhao [23] proved scattering and finite time blowup results for the focusing (NLS) perturbed by a H˙ 1/2 -critical defocusing term in spatial dimension three. More precisely, for every radial initial data u0 ∈ H 1 (R3 ) below a ground state threshold, the corresponding solution u is globally well-posed and scattering, if its energy functional is nonnegative or blows up at finite time, if its energy functional is negative. And in [24], they extends the result to five and higher spatial dimension without the radial assumption. We refer to [2, 3, 4] for the focusing energy critical NLS with some focusing perturbation terms. In this article, we consider the longtime dynamics behavior of (CNLS) below the energy threshold. First, we consider the variational derivation of the energy. As in [16], we denote ϕλα,β (x) = eαλ ϕ(e−βλ x), where (α, β, λ) ∈ R3 and ϕ : R4 → C. We will restrict (α, β) in a region, Ω = {(α, β) : α ≥ 0, 5α + 6β ≥ 0, (α, β) 6= (0, 0)} .

4D NLS WITH COMBINED TERMS

3

And for any functional F of H 1 (R4 ), we define the variation derivation by  d   Lα,β F (φ) = F (φλα,β ). dλ  λ=0

For each (α, β) ∈ Ω, we define

Kα,β (φ) = Lα,β E(φ) = (α+β)

Z

6 |∇φ(x)|2 − |φ(x)|4 dx+ α+ β 5

Z

10

|φ(x)| 3 dx,

whenever φ is a function in H 1 (R4 ). Consider the minimization problem mα,β = inf{E(ϕ) : ϕ ∈ H 1 (R4 ) \ {0}, Kα,β (ϕ) = 0}. Since the nonlinear term in (CNLS) is H˙ 1 -critical growth with the H˙ 1 -subcritical perturbation, we will use the modified energy later Z 1 4 1 2 |∇u| − |u| dx. Ec (u) = 4 R4 2 In this paper we will study the solutions to (CNLS) which start from the following two regions below the minimum mα,β , + Kα,β = ϕ ∈ H 1 (R4 ) : E(ϕ) < mα,β , Kα,β (ϕ) ≥ 0 (1.1) and

− Kα,β = ϕ ∈ H 1 (R4 ) : E(ϕ) < mα,β , Kα,β (ϕ) < 0 ,

(1.2)

where (α, β) ∈ Ω. Before stating the main theorem, we give the definition of scattering which will be used later. A global solution u to (CNLS) with u(0) = u0 ∈ H 1 (R4 ) is scattering, if there exist u± ∈ H 1 (R4 ) such that

lim u(t) − eit∆ u± 1 4 = 0. t→±∞

H (R )

Now we state our main theorem, which describes the dynamics of (CNLS) under the threshold mα,β .

+ Theorem 1.1. Let (α, β) ∈ Ω. For each ϕ ∈ Kα,β , we have that if u is the solution to (CNLS) with u(0, x) = ϕ, then u is global well-posed and scattering in H 1 (R4 ). − On the other hand, if ϕ is a radial function in Kα,β , then the solution u with the initial data ϕ will blow up at finite time.

Remark 1.2. By using the same argument in this paper one can also obtain the corresponding results of (CNLS) if |u|4/3 u is replaced by some more general defocusing |u|p u(1 < p < 2) terms. To prove the main theorem, we need the following property of the minimal mα,β . Proposition 1.3. For (α, β) ∈ Ω, we have 1 mα,β = Ec (W ) = k∇W k2L2 (R4 ) , 4 where W (x) is the ground state of the massless equation −∆ϕ = |ϕ|2 ϕ, given by 2 −1 . W (x) = 1 + |x|8 We remark that this proposition implies that mα,β is independent of (α, β). We + − also have the properties of Kα,β and Kα,β ,

+ − Proposition 1.4. The regions Kα,β and Kα,β do not depend on (α, β) if (α, β) ∈ Ω.

4


+ − Based on this property, we can denote K+ = Kα,β and K− = Kα,β for simplicity. We will also prove the energy trapping property which manifest another important property of the regions K+ and K− .

Proposition 1.5. Let u : I × R4 → C be the solution to (CNLS) with initial data u(0, x) = u0 (x) ∈ H 1 (R4 ). Then, we have (1) If u0 ∈ K− , then for each t ∈ I, u(t) ∈ K− and k∇u(t)kL2x > k∇W kL2x .

(1.3)

(2) If u0 ∈ K+ , then for each t ∈ I, u(t) ∈ K+ and k∇u(t)kL2x < k∇W kL2x .

(1.4)

This and the energy conservation law shows that if u is a solution to (CNLS) with the initial data u(0, x) ∈ K+ or K− , the solution flow {u(t), t ∈ I} will remain in the regions K+ or K− , where I is the maximal lifespan of u. Hence this proposition brings about many conveniences to our proof. To prove the blowup results, we use the same method as in [18, 23, 24]. Indeed, the estimation of the differentials of the localized virial identity helps us preclude the global existence of the solution which starts from K− . On the other hand, to prove the scattering results, by local well-posedness theory in Section 2, it suffices to show that the global scattering size of u ∈ K+ is bounded by certain constant. To this end, we turn to a proof of contradiction. More precisely, suppose the energy threshold E∗ is less than m, thus there exists a sequence of solutions un in K+ , with the property that E(un ) → E∗ and kun kST (R) → ∞ as n → ∞, where ST (R) is the scattering norm we will define later. By making use of the linear and nonlinear profile decomposition and the stability lemma given by [3], we can obtain a critical element uc (t, x). We will also prove a crucial compactness property of the critical element dynamics, that is, {uc (t), 0 ≤ t < ∞} is precompact in H˙ s (R4 ) after module the translation symmetry, for all s ∈ (0, 1]. Finally, in the last step, extinction of the critical element uc , we use the interaction Morawetz estimates to deduce that uc actually vanishes. This is a contradiction to the the local well-poesdness theory(which implies that the solution with small initial data is global well-posed and scattering). It is worthwhile to note that, since we consider the scattering problem in H 1 (R4 ), the mass of uc conserves with time and remains bounded, which makes this step more easier than the analog step in [13]. This is the main reason that why we do not use the longtime Strichartz estimates for the critical element uc . This paper is organized as follows: In the first part of Section 2, we give the notations in this paper and recall some basic harmonic analysis tools and the localwellposed theory of (CNLS). We prove Proposition 1.3–1.5 by introducing the variation method in the second part of Section 2. Section 3 will prove the finite time blowup for solutions in K− . In Section 4, we will construct the linear and nonlinear profile decomposition of the solutions sequence to (CNLS). In Section 5, we will finish the proof of the main theorem.


5

2. Basic estimates and variational method 2.1. Basic Tools. First we will present some harmonic analysis tools which will be used later. We define the Fourier transform on R4 by Z 1 b f (ξ) := 4π2 e−ix·ξ f (x) dx. R4

Based on this, for each s ∈ R, we define the differentiation operator φ(∇) by \ (ξ) = ϕ(iξ)fˆ(ξ). Hence we can define the homogeneous Sobolev norms by ϕ(∇)f kukH˙ s (R4 ) = k|∇|s ukL2 (R4 ) and the inhomogeneous Sobolev norms by kuk2H s (R4 ) = kuk2H˙ s (R4 ) + kuk2L2(R4 ) , for s ∈ R. We will also use the following two lemmas, which deal with the fractional derivatives. Lemma 2.1 (Fractional product rule, [9]). Let s ∈ (0, 1] and f, g ∈ S(R4 ), then we have k|∇|s (f g)kLp (R4 ) . k|∇|s f kLq1 (R4 ) kgkLr1 (R4 ) + k|f kLq2 (R4 ) k∇|s gkLr2 (R4 ) , where

1 p

=

1 q1

+

1 1 r1 , p

=

1 q2

+

1 r2

(2.1)

and 1 ≤ p ≤ q1 , q2 , r1 , r2 < ∞.

Lemma 2.2 (Fratianal chain rule, [9]). Let G ∈ C 1 (C) and s ∈ (0, 1], then for any Schwartz function u we have k|∇|s G(u)kLp (R4 ) . kG′ (u)kLq (R4 ) k|∇|s ukLr (R4 ) , where

1 p

=

1 q

+

1 r

(2.2)

and 1 ≤ p ≤ q, r < ∞.

Now, we are ready to introduce the Strichartz estimates and give the local wellposedness of (CNLS). First, we say a pair of exponents (q, r) is Schrödinger H˙ s admissible in dimension four if 2 1 1 −s =4 − q 2 r and 2 ≤ q, r ≤ ∞. And we denote the dual exponent q ′ to q ∈ (1, ∞) by

1 q′

+ 1q = 1.

Lemma 2.3 (Strichartz estimates, [15, 17, 28]). Let (q, r) and (e q , re) be Schr¨ odinger ′ ′ L2 -admissible pairs in dimension four. If ϕ ∈ L2 (R4 ) and f ∈ Lqte Lrxe (R × R4 ), then we have keit∆ ϕ(x)kLqt Lrx (R×R4 ) .kϕkL2 (R4 ) .

Z t

ei(t−τ )∆ f (τ, x)dτ q .kf kLqe′ Lre′ (R×R4 ) .

r 4 0

Lt Lx (R×R )

t

(2.3) (2.4)

x

If I ⊆ R is a interval, we define some time-spatial Strichartz spaces by 4 4 2 4 2 S(I) =L∞ t (I; L (R )) ∩ Lt (I; L (R )),

12

W1 (I) =L6t (I; L6 (R4 )),

V1 (I) = L6t (I; L 5 (R4 )),

W2 (I) =L4t (I; L4 (R4 )),

V2 (I) = L4t (I; L 3 (R4 )),

8

ST (I) =W1 (I) ∩ W2 (I), V0 (I) = L3t (I; L3 (R4 )). By the definition of the Schrödinger admissible pairs, one can find that W1 is H˙ 1 1 admissible, W2 is H˙ 2 -admissible and V0 , V1 , V2 is L2 -admissible. On the other hand, standard arguments show that if a solution u to (CNLS) is global, with

6


kukST (R) < +∞, then it scatters. In view of this, we define kukST (I) as the scattering size of u on time interval I if u is a solution to Cauchy problem (CNLS). For the sake of later use, by the H¨ older inequality and Lemma 2.2, we give some nonlinear estimates:

s

s

|∇| (|u|2 u) 3/2

|∇| u , for s = 0, 1 (2.5) kuk2 4 ≤ Lt,x (I×R )

and

s

|∇| (|u| 34 u) 3/2 ≤ |∇|s u V L (I×R4 )

4

0 (I)

t,x

W1 (I)

V0 (I)

3 kukW , 2 (I)

for s = 0, 1.

(2.6)

As a consequence of the Strichartz estimates and the nonlinear estimates, one can obtain the local theory of (CNLS). Theorem 2.4 (Local well-posedness, [3]). Let u0 ∈ H 1 (R4 ) and I be a time interval with 0 ∈ I. We have: (1) There exists δ = δ(ku0 kH 1 (R4 ) ) > 0 such that: If

h∇ieit∆ u0 ≤ δ, (2.7) V2 (I)

then there exists a solution u ∈ C(I; H 1 (R4 )) to the Cauchy problem (CNLS) with the following properties: u(0) =u0 ,

(2.8)

kh∇iukS(I) ≤ku0 kH 1 ,

(2.9)

kh∇iukV2 (I) ≤2kh∇ieit∆ u0 kV2 (I) . 1

(2.10)

4

Furthermore, assume that u ∈ C(Imax ; H (R )) is a solution to (CNLS), where Imax is the maximal lifespan of u. Note that Imax must be open by (1). (2) The mass and energy conservation laws hold true, ∀ t, t0 ∈ Imax , M (u(t)) = M (u(t0 )),

(2.11)

E(u(t)) = E(u(t0 )). (3) Let Tmax = sup Imax . If Tmax < +∞, then

(2.12)

kukST ([T,Tmax )) = ∞ for any T ∈ Imax .

(2.13)

A similar result holds when Tmin = inf Imax > −∞. (4) If kukST (Imax ) < ∞, then Imax = R and there exist φ± ∈ H 1 (R4 ) such that lim ku(t) − eit∆ φ± kH 1 (R4 ) = 0.

t→±∞

(2.14)

Similar to [3, Proposition 5.6], we have the Perturbation theory. Proposition 2.5 (Perturbation theory,[3]). Let I be an interval, u ∈ C(I; H 1 (R4 )) be a solution to (CNLS) and u ˜ be a function in C(I; H 1 (R4 )). Let A > 0 and t1 ∈ I such that kukL∞ (I;H 1 ) + ku(t1 ) − u ˜(t1 )kH 1 + k˜ ukST (I) ≤ A. (2.15) Then there exists δ > 0 depending on A such that if

4

˜ 3 u + |˜ u |2 u ˜ − |˜ u| 3 u ≤δ, (2.16)

h∇i (i∂t + ∆)˜ L 2 (I×R4 )

˜(t1 )] ≤δ. (2.17)

h∇iei(t−t1 )∆ [u(t1 ) − u V2 (I)

Then we have kh∇iukS(I) < ∞.


7

2.2. Variational methods. First we recall the energy of the solution u to (CNLS) Z Z Z 1 3 10 1 |∇u(t, x)|2 dx − |u(t, x)|4 dx + |u(t, x)| 3 dx, E(u)(t) = (2.18) 2 4 10 and the modified energy Z Z 1 1 Ec (u)(t) = |∇u(t, x)|2 dx − |u(t, x)|4 dx. (2.19) 2 4

Recall in the introduction, for any functional F of H 1 , we define its variation differential by  d  λ  F (φ ) = F [(α − βx · ∇)φ](x) , Lα,β F (φ) = α,β dλ  λ=0

where φλα,β (x) = eαλ φ(e−βλ x), if (α, β, λ) ∈ R3 . Thus we have Z Z 10 6 |∇φ(x)|2 − |φ(x)|4 dx+ α+ β Kα,β (φ) := Lα,β E(φ) = (α+β) |φ(x)| 3 dx. 5 By the definition of the region of (α, β), Ω = {(α, β) : α ≥ 0, 5α + 6β ≥ 0, (α, β) 6= (0, 0)} ,

(2.20)

it is easy to check that α + β > 0 if (α, β) ∈ Ω. Proposition 2.6. For any (α, β) ∈ Ω, if {ϕn }n≥1 is a sequence in H 1 (R4 ) with lim kϕn kH˙ 1 = 0,

n→∞

then, we have Kα,β (ϕn ) > 0

(2.21)

for sufficient large n. Proof. From the Sobolev inequalities kϕn kL4 (R4 ) . kϕn kH˙ 1 (R4 ) , we have kϕn k4L4 (R4 ) = o(kϕn k2H˙ 1 (R4 ) ) as n → ∞, which together with (2.20) implies (2.21). We define µ by n o 10 µ = max 2(α + β), α + 4β = 3

(

2(α + β) if 2α ≤ −3β, 10 3 α + 4β if 2α ≥ −3β.

By a direct calculation, we have Lemma 2.7.

and

 10  α+β kϕk4 4 − 2α+3β kϕk 3 10 if 2α ≤ −3β, L 2 5 L 3 µ − Lα,β E(ϕ) =  2 α + β kϕk2˙ 1 + 1 α kϕk4 4 if 2α ≥ −3β. H L 3 6

Lα,β (µ − Lα,β )E(ϕ)  10 2(α + β)2 kϕk4 4 − 2 (2α + 3β)(5α + 6β)kϕk 3 10 if 2α ≤ −3β, L 15 3 L =  2 α + β (2α + 2β) kϕk2˙ 1 + 2 α(α + β) kϕk4 4 if 2α ≥ −3β. H L 3 3

(2.22)

(2.23)

8


This lemma impies that Hα,β (ϕ) > 0 and Lα,β Hα,β (ϕ) > 0 for ϕ ∈ H 1 (R4 )\{0}, where Hα,β is given by Lα,β E(ϕ) Hα,β (ϕ) = 1 − µ  (2.24)  1 kϕk4 − 2α+3β kϕk 10 3 if 2α ≤ −3β, 10 L4 10(α+β) L 3 = 42α+3β 2 α  kϕk ˙ 1 + kϕk4 4 if 2α ≥ −3β. H

10α+12β

20α+24β

L

As in the introduction, we define mα,β = inf E(ϕ) : ϕ ∈ H 1 (R4 )\{0}, Kα,β (ϕ) = 0 .

Lemma 2.8. For (α, β) ∈ Ω, we have mα,β = inf Hα,β (ϕ) : ϕ ∈ H 1 (R4 )\{0}, Kα,β (ϕ) ≤ 0 = inf Hα,β (ϕ) : ϕ ∈ H 1 (R4 )\{0}, Kα,β (ϕ) < 0 . Proof. Let m′α,β = inf Hα,β (ϕ) : ϕ ∈ H 1 (R4 )\{0}, Kα,β (ϕ) ≤ 0 and m′′α,β = inf Hα,β (ϕ) : ϕ ∈ H 1 (R4 )\{0}, Kα,β (ϕ) < 0 .

(2.25)

If Kα,β (ϕ) = 0, then we have Hα,β (ϕ) = E(ϕ). Hence we have mα,β ≥ m′α,β .

(2.26)

On the other hand, if Kα,β (ϕ) < 0, from Proposition 2.6, there exists λ0 < 0 such that 0 Kα,β (ϕλα,β ) = 0. By Lemma 2.7, we have Lα,β Hα,β (φ) > 0 for any φ ∈ H 1 , which implies 0 Hα,β (ϕλα,β ) < Hα,β (ϕ).

This implies mα,β ≤ m′α,β , which together with (2.26) implies mα,β = m′α,β . For the second equality in (2.25), one can easily find that m′α,β ≤ m′′α,β .

(2.27)

For any ϕ ∈ H 1 (R4 )\{0} such that Kα,β (ϕ) ≤ 0, from Lemma 2.7, we have Lα,β Kα,β (ϕ) = µKα,β (ϕ) − Hα,β (ϕ) < 0. This implies for λ > 0, Kα,β (ϕλα,β ) < 0. And by definition of Hα,β , we have as λ → 0, Hα,β (ϕλα,β ) → Hα,β (ϕ), which implies m′α,β ≥ m′′α,β . 10 Next we will use the (H˙ 1 -invariant) scaling argument to remove the L 3 -term (the lower regularity quantity than H˙ 1 ) in Kα,β , that is, to replace the constrained c condition Kα,β (ϕ) < 0 by Kα,β (ϕ) < 0, where Z Z c Kα,β (ϕ) = Lα,β Ec (ϕ) = (α + β) |∇ϕ(x)|2 dx − (α + β) |ϕ(x)|4 dx.

And let

c Hα,β (ϕ)

L α,β Ec (ϕ) = = 1− µ ¯

(

1 4 4 kϕkL4 2α+3β 10α+12β

if 2α ≤ −3β, 2 α kϕkH˙ 1 + 20α+24β kϕk4L4

if

2α ≥ −3β.


Lemma 2.9. For (α, β) ∈ Ω, we have mα,β = inf Hα,β (ϕ) : ϕ ∈ H 1 (R4 )\{0}, = inf Hα,β (ϕ) : ϕ ∈ H 1 (R4 )\{0}, c = inf Hα,β (ϕ) : ϕ ∈ H 1 (R4 )\{0}, c = inf Hα,β (ϕ) : ϕ ∈ H 1 (R4 )\{0}, c = inf Hα,β (ϕ) : ϕ ∈ H 1 (R4 )\{0}, (1)

(2)

9

c Kα,β (ϕ) ≤ 0 c Kα,β (ϕ) < 0 c Kα,β (ϕ) < 0 c Kα,β (ϕ) c Kα,β (ϕ)

≤0

(2.28)

=0 .

Proof. First, we denote mα,β and mα,β by (1) c mα,β = inf Hα,β (ϕ) : ϕ ∈ H 1 (R4 )\{0}, Kα,β (ϕ) ≤ 0

and

(2) c mα,β = inf Hα,β (ϕ) : ϕ ∈ H 1 (R4 )\{0}, Kα,β (ϕ) < 0 .

c From the definition of Hα,β and the fact that Kα,β (ϕ) ≥ Kα,β (ϕ), we have (2)

mα,β ≤ m′′α,β = mα,β .

(2.29)

(2)

c To prove mα,β ≥ m′′α,β , for any ϕ ∈ H 1 (R4 )\{0} with Kα,β (ϕ) < 0, we have 10 6 2 c (ϕ), Kα,β (ϕλ1,−1 ) = (α + β) k∇ϕk2L2 − kϕk4L4 + e− 3 λ α + β kϕk 3 10 → Kα,β L 3 5 (2)

as λ → +∞. And for λ > 0, Hα,β (ϕλ1,−1 ) ≤ Hα,β (ϕ), which implies mα,β ≥ m′′α,β . On the other hand, it is trivial that (1)

(2)

mα,β ≤ mα,β .

(2.30)

(1)

(2)

c (ϕ) ≤ 0, Next, we need to prove mα,β ≤ mα,β . Let ϕ ∈ H 1 (R4 )\{0} such that Kα,β then from 2 4 c Lα,β Kα,β (ϕ) = 2(α + β)2 k∇ϕkL2 − 2 kϕkL4 < 0, c we have Kα,β (ϕλα,β ) < 0, for λ > 0. As above we have Hα,β (ϕλα,β ) → Hα,β (ϕ), as (2)

(1)

λ → 0. This implies mα,β ≤ mα,β . Second, we denote c c (ϕ) : ϕ ∈ H 1 (R4 )\{0}, Kα,β (ϕ) = 0 . mcα,β = inf Hα,β One can easily find that c c inf Hα,β (ϕ) : ϕ ∈ H 1 (R4 )\{0}, Kα,β (ϕ) ≤ 0 c c ≤ inf Hα,β (ϕ) : ϕ ∈ H 1 (R4 )\{0}, Kα,β (ϕ) < 0 .

(2.31)

c c Let ϕ ∈ H 1 (R4 )\{0} such that Kα,β (ϕ) ≤ 0. For λ > 0, we have Kα,β (ϕλα,β ) < 0, and c c (ϕ), as λ → 0. Hα,β (ϕλα,β ) → Hα,β Hence, we have c c inf Hα,β (ϕ) : ϕ ∈ H 1 (R4 )\{0}, Kα,β (ϕ) ≤ 0 c (2.32) c ≥ inf Hα,β (ϕ) : ϕ ∈ H 1 (R4 )\{0}, Kα,β (ϕ) < 0 .

It is trivial that

c c inf Hα,β (ϕ) : ϕ ∈ H 1 (R4 )\{0}, Kα,β (ϕ) ≤ 0 ≤ mcα,β .

(2.33)

10


c Let ϕ ∈ H 1 (R4 )\{0} such that Kα,β (ϕ) < 0. Then there exist λ0 < 0 such that c 0 Kα,β (ϕλα,β ) = 0.

(2.34)

c c 0 From Lα,β Hα,β (φ) > 0 for any φ ∈ H 1 (R4 )\{0}, we have Hα,β (ϕλα,β ) ≤ Hα,β (ϕ), which implies

c c inf Hα,β (ϕ) : ϕ ∈ H 1 (R4 )\{0}, Kα,β (ϕ) ≤ 0 ≥ mcα,β .

(2.35)

Combine (2.31), (2.32), (2.33) (2.34) (2.35) together, we can obtain the last two inequalities in (2.28). Finally, we just need to show mcα,β = mα,β . It is trivial that mcα,β ≤ mα,β . By c the definition of Hα,β and Hα,β , we just need to show mcα,β ≥ mα,β in the case of 2α < −3β. c c Let φ ∈ H 1 (R4 )\{0} such that Kα,β (ϕ) < 0. Then for any λ ∈ R, Kα,β (ϕλ1,−1 ) = c Kα,β (ϕ), and c Hα,β (ϕλ1,−1 ) = Hα,β (ϕ) −

10 2α + 3β − 2 λ c e 3 kϕk 3 10 → Hα,β (ϕ) as λ → +∞. L 3 10(α + β)

This implies mcα,β ≥ mα,β . Therefore, we complete the proof of Lemma 2.9.

c We remark that by Lemma 2.9 and the definition of mcα,β and Hα,β , we have

o n 2 4 mα,β = mcα,β = inf Ec (ϕ) : ϕ ∈ H 1 (R4 )\{0}, k∇ϕkL2 = kϕkL4 .

(2.36)

which implies that mα,β is independent of (α, β) if (α, β) ∈ Ω. Hence we can denote m by m = mα,β for any (α, β) ∈ Ω. Now, we can make use of the sharp Sobolev constant in [1, 29] to compute the minimization m, which also shows Proposition 1.3. Lemma 2.10. For the minimization mα,β , we have m = Ec (W ) Proof. From Lemma 2.9 and (2.36), we have o n 2 4 m = inf Ec (ϕ) : ϕ ∈ H 1 (R4 )\{0}, k∇ϕkL2 = kϕkL4 1 2 2 4 1 4 k∇ϕkL2 : ϕ ∈ H (R )\{0}, k∇ϕkL2 = kϕkL4 = inf 4 1 2 2 4 1 4 ≥ inf k∇ϕkL2 : ϕ ∈ H (R )\{0}, k∇ϕkL2 ≤ kϕkL4 , 4

(2.37)


11

where the equality holds if and only if the minimization is attained by some ϕ with 2 4 k∇ϕkL2 = kϕkL4 . Next, we have 1 2 2 4 1 4 inf k∇ϕkL2 : ϕ ∈ H (R )\{0}, k∇ϕkL2 ≤ kϕkL4 4 1 2 2 4 1 4 ˙ k∇ϕkL2 : ϕ ∈ H (R )\{0}, k∇ϕkL2 ≤ kϕkL4 = inf 4 ) ( 4 (2.38) 1 k∇ϕkL2 2 4 = inf : ϕ ∈ H˙ 1 (R4 )\{0}, k∇ϕkL2 ≤ kϕkL4 4 kϕkL4 1 = kW k4L4 4 =Ec (W ),

where we used the fact that H 1 is dense in H˙ 1 and the sharp Sobolev inequality 2 4 kϕkL4 (R4 ) ≤ kW k−1 L4 (R4 ) k∇ϕkL (R ) .

Hence we can define + Kα,β

and

+ Kα,β

− Kα,β

and by = ϕ ∈ H 1 (R4 ) : E(ϕ) < m, Kα,β (ϕ) ≥ 0

− Kα,β = ϕ ∈ H 1 (R4 ) : E(ϕ) < m, Kα,β (ϕ) < 0 . + Kα,β

(2.39) (2.40)

− Kα,β

Now we are going to prove that the regions and are independent of (α, β). The proof is mainly based on the minimal property of mα,β and the relation between energy E and its variational derivatives Kα,β , which is similar to Lemma 2.9 in [16]. ± Lemma 2.11. If (α, β) ∈ Ω, then Kα,β are independent of (α, β).

Proof. From the definition of Ω in (2.20), we have α + β > 0, if (α, β) ∈ Ω. First, + − Kα,β ∪ Kα,β is independent of (α, β), since + − Kα,β ∪ Kα,β = ϕ ∈ H 1 (R4 ) : E(ϕ) < m . (2.41)

+ + For ϕ ∈ Kα,β , by the definition of m, if Kα,β (ϕ) = 0, then ϕ = 0. Hence if ϕ ∈ Kα,β and ϕ 6= 0, then we have Kα,β (ϕ) > 0. From this we have the scaling variation of + ϕ ∈ Kα,β , ϕλα,β such that the following properties:

(1) By the definition of m, Kα,β (ϕλα,β ) > 0 is preserved, provided E(ϕλα,β ) < m; (2) E(ϕλα,β ) does not increase, as λ ≤ 0 decreases, if Lα,β E(ϕλα,β ) = Kα,β (ϕλα,β ) > 0. + − ′ ′ If there exists ϕ ∈ Kα,β ∩ Kα ′ ,β ′ , where (α, β), (α , β ) ∈ Ω. By the definition of m and the fact that m > 0, we have ϕ 6= 0 and Kα,β (ϕ) > 0. Then, from the above two facts, we have E(ϕλα,β ) < m, for λ ≤ 0. And by α+β > 0, we have ϕλα,β → 0 in H˙ 1 , as λ → −∞. But by Proposition 2.6, there exists 0 0 λ0 < 0 such that Kα′ ,β ′ (ϕλα,β ) = 0, which together with E(ϕλα,β ) < m contradicts + − the minimization of m. Hence Kα,β ∩ Kα′ ,β ′ = ∅ for any (α, β), (α′ , β ′ ) ∈ Ω, which and (2.41) imply the claim.

12


After the computation of the minimization m, we next give some variational estimates. Lemma 2.12. For any ϕ ∈ H 1 (R4 ) with K2,−1 (ϕ) ≥ 0, we have Z Z 1 1 1 10 3 10 2 3 ≤ E(ϕ) ≤ |∇ϕ| + |ϕ| |∇ϕ|2 + |ϕ| 3 . 10 10 R4 2 R4 4

(2.42)

Proof. On one hand, the right hand side of (2.42) is trivial. On the other hand, by the definition of E and Kα,β , we have Z 1 1 1 10 E(ϕ) = |∇ϕ|2 + |ϕ| 3 + K2,−1 (ϕ), 10 4 R4 4

which implies the left hand side of (2.42).

At the last of this section, we give the uniform bounds on the scaling derivative functional Kα,β (ϕ) with the energy E(ϕ) below the threshold m, which plays an important role for the blow-up and scattering analysis. Lemma 2.13. For any ϕ ∈ H 1 with E(ϕ) < m. (1) If K2,−1 (ϕ) < 0, then K2,−1 (ϕ) ≤ − and

8 m − E(ϕ) , 3

(2.43)

k∇ϕkL2x > k∇W kL2x . (2) If K2,−1 (ϕ) ≥ 0, then 8 K2,−1 (ϕ) ≥ min (m − E(ϕ)), 3 and

10 1

∇ϕ 2 2 + 1 ϕ 3 10 L L 3 5 100

(2.44)

k∇ϕkL2x < k∇W kL2x .

,

(2.45)

(2.46)

Proof. By Lemma 2.7, for any ϕ ∈ H 1 , we have 8 4 4 2 2 L22,−1 E(ϕ) = L2,−1 E(ϕ) − ∇ϕ L2 − ϕ L4 . 3 3 3 Let j(λ) = E(ϕλ2,−1 ), then we have

2

4 8 ′ 2 4 j (λ) − e2λ ∇ϕ L2 − e4λ ϕ L4 . 3 3 3 Case I: If K2,−1 (ϕ) < 0, then by j ′′ (λ) =

(2.47)

lim kϕλ2,−1 k2H˙ 1 = 0,

λ→−∞

Proposition 2.6 and the continuity of K2,−1 in λ, there exists a negative number 0 λ0 < 0 such that K2,−1 (ϕλ2,−1 ) = 0, and K2,−1 (ϕλ2,−1 ) < 0, ∀ λ ∈ (λ0 , 0).


13

0 By the definition of m, we obtain j(λ0 ) = E(ϕλ2,−1 ) ≥ m. Now by integrating (2.47) over [λ0 , 0], we have Z Z 0 8 0 ′ j (λ) dλ. j ′′ (λ) dλ ≤ 3 λ0 λ0

This yields that K2,−1 (ϕ) = j ′ (0) − j ′ (λ0 ) ≤

8 8 (j(0) − j(λ0 )) ≤ − m − E(ϕ) , 3 3

which implies (2.43). Since K2,−1 (ϕ) < 0, we have by Lemma 2.8

1

ϕ 4 4 . (2.48) L 4 where we have used the fact that K2,−1 (ϕ) < 0 in the second inequality. By the −4 fact m = 41 (C4∗ ) and the Sharp Sobolev inequality, we have

4

∇ϕ 2 ≥ (C ∗ )−4 ϕ 4 4 > (4m)2 , (2.49) 4 L L

2 which implies that ∇ϕ L2 > 4m = k∇W k2L2 . Thus, we conclude (2.44). x Case II: K2,−1 (ϕ) ≥ 0. We divide it into two subcases:

4 When 4K2,−1 (ϕ) ≥ ϕ L4 . Since Z Z 4 1 10 4 2 4 ϕ dx = − 4 K2,−1 (ϕ) + ∇ϕ + ϕ 3 dx, 3 R4 3 3 15 R4 m ≤ H2,−1 (ϕ)
0 and (2.50) holds on [0, λ0 ]. 0 By K2,−1 (ϕλ2,−1 ) = j ′ (λ0 ) = 0, we know that 0
k∇W kL2x .

(2.51)

+ + (2) If u0 ∈ Kα,β , then for each t ∈ I, u(t) ∈ Kα,β and

k∇u(t)kL2x < k∇W kL2x . Remark 2.15. By sharp Sobolev embedding: kϕk4L4 ≤ x

(2.52) k∇ϕk4L2

x

k∇W k4 2

, we know that if

Lx

k∇ϕkL2x < k∇W kL2x , then Z Z Z 1 1 1 |∇ϕ|2 dx − |ϕ|4 dx > |∇ϕ|2 > 0. Ec (ϕ) = 2 R4 4 R4 4 R4 3. Blow up of K− In this section, we will prove that the radial solution of (CNLS) in K− blows up at finite time. Let φ be a smooth, radial function satisfying |∂r2 φ(r)| ≤ 2, φ(r) = r2 for r ≤ 1, and φ(r) = 0 for r ≥ 3. For R ≥ 1, we define Z |x| 2 φR (x) = R φ( ) and VR (x) = φR (x)|u(t, x)|2 dx. R R4 Let u(t, x) be a radial solution to (CNLS), then by a direct computation, we have Z [u∂j u](t, x)∂j φR (x)dx, ∂t VR (x) = 2Im (3.1) R4

and

∂t2 VR (x)

Z

∂j ∂k φR (x)∂k u(t, x)∂j u(t, x)dx =4Re 4 Z R 4 10 2 2 4 3 ∆ φR |u| + ∆φR (|u| − |u| ) dx − 5 R4 Z Z 4 10 2 2 4 ′′ 2 3 ∆ φR |u(t, x)| + ∆φR (x)(|u| − |u| )(t, x) dx φR |∇u| dx − =4 5 R4 R4 Z 8 10 ≤4 2|∇u|2 − 2|u|4 + |u| 3 (t, x)dx 5 4 R Z Z i h C 10 + 2 |u|2 dx + C |u|4 + |u| 3 dx. R R≤|x|≤3R R≤|x|≤3R


15

By the radial Sobolev inequality, we have 1 1 c kf kL∞(|x|≥R) ≤ 3 kf kL2 2 (|x|≥R) k∇f kL2 2 (|x|≥R) . x x R2 Therefore, by mass conservation and Young’s inequality, we know that for any ǫ > 0 there exist sufficiently large R such that

2 ∂t2 V (t) ≤8K2,−1 (u(t)) + ǫ ∇u(t, x) L2 + ǫ2 10

2 16 =32E(u(t)) − (8 − ǫ)k∇u(t, x) L2 − kuk 3 10 + ǫ2 . 5 Lx3

This together with Lemma 2.14 and E(u) < (1 − δ)m for some δ > 0 implies that ∂t2 VR (t) ≤ −32δm + Cεk∇W k2L2 + Cε2 .

(3.2)

Finally, if we choose ε sufficient small, we can obtain ∂t2 VR (t) ≤ −16δ2m, which implies that u blows up in finite time. 4. Profile decomposition In this section, we give the profile decomposition of (CNLS) by the strategy in [3, 16, 22, 23, 24]. First, we give some notations for later use. x−xj For j, n ∈ N, we denote Tnj by Tnj ϕ(x) = h1j ϕ( hj n ), where ϕ ∈ H 1 (R4 ) and (xjn , hjn ) ∈ R4 × (0, 1]. We define τnj = σnj

=

tjn (hjn )2

n

n

and the multiplier

j −1 −1 (hn ) ∇ h∇i hjn

.

(4.1)

4.1. Linear profile decomposition. Now we give the linear profile decomposition for the solutions sequence to free Schrödinger equation in the inhomogeneous space H 1 (R4 ). Proposition 4.1 (Linear profile decomposition). Let vn (t, x) = eit∆ ϕn (x) be 1 4 a sequence of the free Schr¨ odinger solutions with bounded Then j H (R ) norm. up to a subsequence, there exist K ∈ {1, 2, · · · , ∞}, ϕ j∈[1,K) ⊂ H 1 (R4 ) and j j j tn , xn , hn n∈N ⊂ R × R4 × (0, 1] satisfying vn (t, x) =

k X

vnj (t, x) + eit∆ ωnk , for any k ∈ [1, K),

(4.2)

j=1

where

vnj (t, x) = h∇i

−1

j

|∇|eit∆ Tnj e−iτn ∆ h∇i|∇|−1 ϕj j

= eit∆ Tnj σnj e−iτn ∆ ϕj

(4.3)

i(t−tjn )∆

=e Tnj σnj ϕj and ωnk : k ∈ [1, K), n ≥ 1 ⊂ H 1 (R4 ). The error terms ωnk of the linear profile decomposition (4.2) have the properties:

lim lim k|∇|−1 h∇ieit∆ ωnk kLqt (R;Lr (R4 )) = 0, for any H˙ 1 -admissible pair (q, r),

k→K n→∞

(4.4)

16


and for any j ≤ k ∈ [1, K). j lim eiτn ∆ (Tnj )−1 |∇|−1 h∇iωnk = 0 weakly in H˙ 1 (R4 ).

n→∞

For any l < j ≤ K, and k ∈ [1, K), we have hl tj − tl xj − xl lim log nj + n l 2n + n l n = ∞, n→∞ (hn ) hn hn k X j k M (vn (0)) − M (ωn ) = 0, lim M (vn (0)) − n→+∞ j=1

and

k X j k E(vn (0)) − E(ωn ) = 0, lim E(vn (0)) − n→+∞ j=1

(4.5)

(4.6)

(4.7)

(4.8)

k X lim K2,−1 (vn (0)) − K2,−1 (vnj (0)) − K2,−1 (ωnk ) = 0. (4.9) n→+∞ j=1 Moreover, for each fixed j, the sequence hjn n∈N is either going to 0 or identically 1 for all n.

Note that, by k ∈ [1, K) we mean that 1 ≤ k ≤ K if K < +∞ and 1 ≤ k < ∞ if K = ∞. Proof. The proof is given by [3, Lemma 2.2] based on [22, Theroem 1.6], where the linear profile decomposition dealt with the sequences in H˙ 1 . And the property (4.5) is from [20, lemma 2.10]. We also have the following property for a free Schrödinger solution sequence if it is in K+ . Corollary 4.2. Suppose that vn (t, x) is a sequence of the free Schr¨ odinger solution with bounded H 1 (R4 ) norm satisfying that vn (0) ∈ K+ and E(vn (0)) ≤ m0 < m. Let vn (t, x) =

k X

vnj (t, x) + eit∆ ωnk

j=1

be the linear profile decomposition given by Proposition 4.1. Then for sufficiently large n and all j, k ∈ [1, K), we have vnj (0), wnk (x) ∈ K+ . Moreover for all j < K, we have 0 ≤ lim E(vnj (0)) ≤ lim E(vnj (0)) ≤ lim E(vn (0)), n→∞

n→∞

(4.10)

n→∞

where the last inequality becomes equality only if K = 1 and ωn1 (x) → 0 in H˙ 1 as n → ∞.


17

Proof. Since vn (0) ∈ K+ and E(vn (0)) ≤ m0 , by Lemma 2.14, we have lim k∇vn (0)kL2 (R4 ) < k∇W kL2 (R4 ) .

n→∞

And from the linear profile decomposition, up to a subsequence, we have lim kvnj (0)k2H 1 (R4 ) + kωnk k2H 1 (R4 ) ≤ lim kvn (0)k2H 1 (R4 ) < ∞, n→∞

n→∞

(4.11)

and kvnj (0)k2H˙ 1 (R4 ) + kωnk k2H˙ 1 (R4 ) < kW k2H˙ 1 (R4 ) for large n and any j < k ∈ [1, K). Then also by Lemma 2.14, we have vnj (0) and ωnk positive energy, which together with (4.8) implies (4.10). Thus, vnj (0), wnk ∈ K+ follows from the definition of K+ .

4.2. Nonlinear Profile decomposition. Now we are ready to give the construction of the nonlinear profile decomposition. The strategy here is the same as the 3D case in [23] and higher dimensional case in [3, 24]. Let vn (t, x) and un (t, x) be solutions to the free Schrödinger equation and the (CNLS) respectively. And assume that vn (t, x) and un (t, x) have the same initial datum ϕn (x). By Proposition 4.1, we have the linear profile decomposition vn (t, x) =

k X

vnj (t, x) + eit∆ ωnk

j=1

=

k X

(4.12) e

i(t−tjn )∆

Tnj σnj ϕj

+e

it∆

ωnk .

j=1

j j Let ujn (t, x) be the solution to (CNLS) with the initial data un (0, x) = vn (0, x). If j t−t we define Unj (t, x) by ujn (t, x) = Tnj σnj Unj (hj )n2 , then Unj satisfies n ( 2 (i∂t + ∆)Unj = −(σnj )−1 f1 (σnj Unj ) + (hjn ) 3 (σnj )−1 f2 (σnj Unj ) (4.13) j Unj (−τnj ) = e−iτn ∆ ϕj 10

with f1 (u) = |u|2 u and f2 (u) = |u| 3 u. j Next, let {U∞ }j≥1 be the solutions of the limit equation of above equation. More precisely, for each j, we denote the limit of τnj and σnj by j τ∞ = lim τnj ∈ [−∞, +∞], n→∞

and j σ∞

=

(

1, h∇i |∇| ,

hjn ≡ 1; hjn → 0 as n → ∞.

j Then U∞ satisfies

when hjn ≡ 1, or

j j j (i∂t + ∆)U∞ = −f1 (U∞ ) + f2 (U∞ ) j j j −iτ∞ ∆ j U∞ (−τ∞ ) = e ϕ ,

j (i∂t + ∆)U∞ j j U∞ (−τ∞ )

j −1 j j = −(σ∞ ) f1 (σ∞ U∞ ) j −iτ∞ ∆ j =e ϕ ,

(4.14)

(4.15)

18


j when hjn → 0 as n → ∞. When τ∞ = ±∞, we consider (4.14) and (4.15) as the j final value problems, which means that U∞ (t, x) satisfies

j

j

lim U∞ (−τnj ) − e−iτn ∆ ϕj = 0. (4.16) 1 4 n→∞

H (R )

j Based on this, we define the nonlinear profile by uj(n) (t, x) = Tnj σnj U∞ which satisfies

( when hjn ≡ 1, or (

(i∂t + ∆)uj(n) j uj(n) tjn − τ∞

= −f1 (uj(n) ) + f2 (uj(n) ) j

= e−iτ∞ ∆ ϕj ,

(i∂t + ∆)uj(n) j uj(n) tjn − τ∞ (hjn )2

|∇| = − h∇i f1 j

h∇i j |∇| u(n) j

= e−iτ∞ ∆ ϕ ,

t−tjn (hjn )2

,

(4.17)

(4.18)

when hjn → 0 as n → ∞. Then we define the nonlinear profile decomposition of un (t, x) by u ˜kn :=

k X

uj(n) + eit∆ ωnk .

(4.19)

j=1

Next we will show that each nonlinear profile such that uj(n) ∈ K+ , if un (0) ∈ K+ for each n. Lemma 4.3. Suppose un (t, x) is a solution sequence of the (CNLS) with the initial datum un (0), which is bounded in H 1 (R4 ) with un (0) ∈ K+ . Then we have the nonlinear decomposition given by u˜kn =

k X

uj(n) + eit∆ ωnk .

j=1

j Let I j be the maximal lifespan of U∞ , Then for t ∈ I j , we have that + K , hjn ≡ 1; j j σ∞ U∞ (t) ∈ Kc+ , hjn → 0 as n → ∞,

(4.20)

where Kc+ is defined by Kc+ := ϕ ∈ H˙ 1 (R4 ) : Ec (ϕ) < Ec (W ), k∇ϕk2L2x ≥ kϕk4L4x . j

Proof. By Corollary 4.2, we have that vnj (0) = e−itn ∆ Tnj σnj ϕj ∈ K+ . Then from the definition of Tnj and the fact that |hjn | ≤ 1, we have by Lemma 2.14 j

j

ke−iτn ∆ σnj ϕj kH˙ 1 (R4 ) = ke−itn ∆ Tnj σnj ϕj kH˙ 1 (R4 ) < kW kH˙ 1 (R4 ) . j j j If hjn ≡ 1, then σnj = σ∞ ≡ 1. Hence from (4.16), we have σ∞ U∞ (−τnj ) ∈ K+ , + j j j for large n. By Lemma 2.14, we have σ∞ U∞ (t) ∈ K for t ∈ I . On the other hand, if hjn → 0, then we have

j −iτnj ∆ j j j j kσ∞ U∞ (−τ∞ )kH˙ 1 (R4 ) = lim σ∞ e ϕ < kW kH˙ 1 (R4 ) . (4.21) n→∞

˙ 1 (R4 ) H


19

j j j This implies Ec (σ∞ U∞ (−τ∞ )) > 0 by Remark 2.15. And by the Sobolev embedding 1 4 4 d H˙ (R ) ֒→ L (R ), we have j

j j j Ec (σ∞ U∞ (−τ∞ )) = lim Ec (σnj e−iτn ∆ ϕj ) = lim Ec (vnj (0)) < Ec (W ). n→∞

j

Hence, if t ∈ I , we have

n→∞

j j σ∞ U∞ (t)

∈

Kc+ ,

which ends the proof.

j Lemma 4.4. Let U∞ be constructed by above discussion and I j be the maximal j lifespan of U∞ . Then there exists j0 large enough such that the following property: X 2 X

j 2

ϕj 1 4 < ∞.

h∇iU∞ (4.22) ≤ H (R ) S(R) j≥j0

j≥j0

Proof. For any δ > 0, by the linear decomposition, there exist two constants j0 and n0 large enough with the property that: If n ≥ n0 , then we have

X X X 2

−iτnj ∆ j 2

j −iτnj ∆ j j 2

ϕj 1 4 = ϕ 1 4 = e σ ϕ

e

T

1 4 < δ. n n H (R ) H (R )

j≥j0

H (R )

j≥j0

j≥j0

If hjn ≡ 1, by small data theory of (CNLS), we have

j 2

h∇iU j 2

∞ S(R) . ϕ H 1 (R4 ) .

If hjn → 0, by the Strichartz estimates and the Mikhlin multiplier theorem, we have

j j j

h∇iU∞ . ϕj H 1 (R4 ) + |∇|f1 (σ∞ U∞ ) L2 (R;L4/3 (R4 )) S(R) x t

j

j j 2 j

h∇iU∞ 6 . ϕ 1 4 + σ∞ U∞ 12/5 4 H (R )

W1 (R)

Lt (R;Lx

(R ))

j 2 j 1/3 j 2/3

h∇iU∞

h∇iU∞ . ϕj H 1 (R4 ) + h∇iU∞ 2 4 4 2 4 S(R) Lt (R;Lx (R )) L∞ t (R;Lx (R ))

j

3 j . ϕ H 1 (R4 ) + h∇iU∞ . (4.23) S(R)

j Hence, by the continuity argument, for sufficiently small δ > 0, we have h∇iU∞ . S(R)

j

ϕ 1 4 . Combining this two cases together, we have H (R )

X 2 X

j 2

ϕj 1 4 .

h∇iU∞ . H (R ) S(R) j≥j0

j≥j0

This ends the proof of this lemma.

jRecall

ST (I) = W1 (I) ∩ W2 (I) in Section 2.1. Now we define the scattering size

σ U j j of σ j U j for each j by ∞ ∞ ST (I) ∞ ∞ W1 (I) ∩ W2 (I), for hjn ≡ 1; ST j (I) = W1 (I), hj∞ = 0. Lemma 4.5. Let k0 ∈ N such that for any 1 ≤ j ≤ k0 , j j kσ∞ U∞ kST j (I j ) < ∞.

(4.24)

j

Then we have I = ∞ and

j + uj(n)

∇u(n) V0 (R)

ST (R)

j . kh∇iU∞ kS(R) . 1, for 1 ≤ j ≤ k0 .

(4.25)

20


And there exists B > 0 such that: for any given k ∈ N and 1 ≤ k ≤ k0 , there exists Nk ∈ N s.t  

k

k

X j

  X (4.26) u(n) + ∇uj(n) sup   ≤ B.

n≥Nk

j=1

j=1 ST (R)

V0 (R)

Furthermore, if (4.24) holds for any j ∈ N, then, for any p ∈ {12/7, 2} and j ≥ 1, we have

lim lim uj(n) |∇|s eit∆ ωnk p × 4 = 0, and s ∈ {0, 1}. (4.27) k→∞ n→∞

Lt,x (R R )

Proof. First, I j = R follows from the local well-posedness theory of (4.14) and (4.15) and the scattering size condition of (4.24). Next, for the first inequality in (4.25), by employing the Mikhlin multiplier theorem, we have

j j

j j j = σnj U∞ . σ∞ U∞ W1 (R) . h∇iU∞ , (4.28)

u(n) W1 (R) S(R) W1 (R)

j

u(n)

W2 (R)

1/2

≤ uj(n)

W1 (R)

j 1/2

u(n)

V0 (R)

(4.29)

j 1/2 j 1/2 . h∇iU∞ (hjn )1/2 σnj U∞ S(R) V0 (R)

j . h∇iU∞ , S(R)

j

∇u(n)

and

V0 (R)

j j

j j = ∇σnj U∞ . ∇σ∞ U∞ S(R) . h∇iU∞ . V0 (R) S(R)

Hence to finish (4.25), it suffices to show

h∇iU j ∞ S(R) . 1.

(4.30)

(4.31)

If hjn → 0, by using the Strichartz estimates, (4.16), and (4.24), we have

j j −1 j j

h∇iU∞ . ϕj H 1 (R4 ) + h∇i(σ∞ ) f1 (σ∞ U∞ (4.32) ) L2 (R;L4/3 (R4 )) S(R) x t

j

j j 2

j . ϕ H 1 (R4 ) + σ∞ U∞ W1 (R) h∇iU∞ 12/5 L6t (R;Lx (R4 ))

j

j 2/3 j 1/3

h∇iU∞ . ϕ H 1 (R4 ) + h∇iU∞ . 2 4 L∞ L2t (R;L4x (R4 )) t (R;Lx (R ))

In this case, similar to Lemma 4.4, ϕj H 1 (R4 ) . 1 and by Lemma 4.3, we have

j

h∇iU∞ . 1. Thus, by the weighted H¨ odelr inequality, we have ∞ 2 4 Lt (R;Lx (R ))

j j

h∇iσ∞ U∞ . 1. S(R)

j On the other hand, if hjn ≡ 1, then σ∞ = 1. Similar to the former case, by the condition (4.24), we have

j j j

h∇iU∞ . 1 + h∇i f1 (U∞ ) + f2 (U∞ ) L2 (R;L4/3 (R4 )) S(R) x t (4.33) . 1.

By using the standard continuity argument again, we have (4.31) in this case, which implies (4.25).


21

Second, to prove (4.26), we use X q q X X j q−1 j ′ X j j .k,q − u u(n) , 1 < q < ∞. u(n) u (n) (n) 1≤j≤k 1≤j≤k 1≤j′ ≤k 1≤j≤k j ′ 6=j

to obtain

X

6

uj(n)

1≤j≤k

W1 (R)

≤

X

j 6

u(n)

W1 (R)

1≤j≤k

+ Ck

X

1≤j≤k

1≤j ′ ≤k j ′ 6=j

Without loss of generality, we assume that k ≥ j0 in

X X X

j 6

j 6 + .

u(n)

u(n) 1≤j≤k

W1 (R)

W1 (R)

1≤j≤j0

X Z Z

R4

R

j 5 j ′ u(n) u(n) dxdt.

(4.34) Lemma 4.4. Hence, we have

j 6

u(n)

j0 0, which guarantees that for any 1 ≤ k ≤ k0 , ∃Nk′′ ∈ N such that

k

k X

X j j

≤ B1 . (4.38) u + sup ∇u sup

(n)

(n)

n≥Nk j=1 n≥Nk′′ j=1

V0 (R)

W2 (R)

Hence, by taking B = max{B0 , B1 } and Nk = max{Nk′ , Nk′′ }, we have (4.26). Finally, we turn to the proof of (4.27). For the case s = 0, using the H¨ older inequality and the Mikhlin multiplier theorem, we have as n → ∞

j it∆ k

u(n) e ωn 12/7 2 Lt,x ∩Lt,x

h i

it∆ k

j

eit∆ ωnk

e ωn . u(n) + W (R) W (R) 1 2 V0 (R) (4.39) h

j j j

it∆ k 1/2 it∆ k 1/2 i it∆ k . hn σn U∞ L3 e ωn W1 (R) + e ωn W1 (R) e ωn V0 (R) t,x

1/2 i

1/2

1/2 h

j . h∇iU∞ S(R) eit∆ ωnk W1 (R) eit∆ ωnk W1 (R) + ωnk H 1 (R4 ) → 0, where we used the fact that (4.4) and ωnk ∈ H 1 (R4 ) in Proposition 4.1.

22


For the case s = 1, from the fact that

(

hjn j

j j

)h∇iU → 0, as n → ∞, if hjn → 0

(

j ∞

σ U − σ j U j hn +|∇| S(R) n ∞ ∞ ∞ W1 (R) . 0, hjn ≡ 1,

we have that for any ε > 0, we can find v j ∈ Cc∞ (R × R4 ) and Nj ∈ N such that

j j

σn U∞ − v j ≤ ε for n > Nj . W (R) 1

Hence, by the H¨ older inequality and the Strichartz estimates we have

j

u(n) |∇|eit∆ ωnk 2 Lt,x (R×R4 )

j j i ∆ ωnk t(hjn )2 + tjn , xhjn + xjn L2 (R×R4 ) (hjn )2 . σn U∞ |∇|e t,x

j . σnj U∞ − v j W1 (R) ωnk H 1 (R4 )

+ v j |∇|ei ∆ ωnk t(hjn )2 + tjn , xhjn + xjn L2 (R×R4 ) (hjn )2 .

(4.40)

t,x

j

4

Assume supp v ⊂ {(t, x) ∈ R × R : |t| ≤ Tj , |x| ≤ Rj } and let ω ˜ nj,k (x) = j eiτn ∆ ωnk (xhjn + xjn ), then by using Lemma 2.5 in [20] and (4.4) in Proposition 4.1, we have

h i j 2 j

(hjn )2

|∇|ei(t(hn ) +tn )∆ ωnk (xhjn + xjn ) 2 Lt,x (|t|≤Tj ,|x|≤Rj )

=hjn |∇|eit∆ ω ˜ nj,k L2 (|t|≤Tj ,|x|≤Rj ) t,x (4.41)

j,k 2/3

j 1/9 1/18 it∆ j,k 1/3

∇˜ ω 2 4 .h (Tj ) (Rj ) e ω ˜ n

n

W1 (R)

n

Lx (R )

2/3

1/3 .(Tj )1/9 (Rj )1/18 eit∆ ωnk W1 (R) ∇ωnk L2 (R4 ) → 0, x

as k, n → ∞. On the other hand, for p = 12/7, we define 0, hjn → 0; Vj = j U∞ , hjn ≡ 1.

Then we have

j 1/2 j j

(hn ) σn U∞ − V j

W2 (R)

j . |∇|1/2 (hjn )1/2 σnj U∞ − |∇|1/2 V j

V2

→ 0,

as n → ∞. From this we can find v j ∈ Cc∞ (R × R4 ) with the

property that for any j η > 0, there exists Nj ∈ N, such that (hjn )1/2 σnj U∞ − v j W2 (R) < η, if n > Nj . By the H¨ older inequality and the Strichartz estimates, we have

j

u(n) |∇|eit∆ ωnk 12/7 Lt,x

j 2 j

j 1/2 j j

. (hn ) σn U∞ · |∇|ei(t(hn ) +hn )∆ ωnk (xhjn + xjn ) 12/7 (hjn )2 Lt,x

j 2 j

.η |∇|ei(t(hn ) +hn )∆ ωnk (xhjn + xjn ) 3 (hjn )2 (4.42) Lt ,x

h i j 2 j

(hjn )2 + v j |∇|eit(hn ) +tn ∆ ωnk (xhjn + xjn ) 12/7 Lt,x (R×R4 )

h i

j 2 j

.η |∇|ωnk L2 (R4 ) + v j |∇|eit(hn ) +tn ∆ ωnk (xhjn + xjn ) 2 (hjn )2 . 4 Lt,x (R×R )


23

Similar to (4.41), we have the last term of (4.42) tends to 0, as k, n → ∞, which together with (4.40), (4.41) implies (4.27). This ends the proof of Lemma 4.5. j be as above. Assume that Proposition 4.6. Let ujn and U∞ j j kσ∞ U∞ kST j (I j ) < ∞,

holds for any j ≥ 1.

(4.43)

Then we have kun kST (R) < ∞,

for sufficient large n.

(4.44)

Proof. To prove (4.44), we will use the perturbation theory (Proposition 2.5). More precisely, we need to prove that the nonlinear profile u ˜kn constructed above is actually a sequence of approximation solutions. j First, by Lemma 4.5 and our assumption of U∞ in (4.43), we have that uj(n) is a global solution and u ˜kn is a function of t defined on the whole R. Next, from the inequalities (4.26) in Lemma 4.5 and H 1 (R4 ) bounded condition of ωnk (x) in Corollary 4.2, we have that there exists a constant B > 0 satisfying the property: there exists N1,k > 0, such that

k sup u ñ ST (R) ≤ B. (4.45) n≥N1,k

In fact, the linear profile decomposition implies k

X

j

j

un (0) − u ˜kn (0) H 1 (R4 ) ≤

vn (0) − u(n) (0) j=1

H 1 (R4 )

k

X

−τnj ∆ j

j ϕ − U∞ (−τnj ) .

e

H 1 (R4 )

j=1

(4.46) .

From this and (4.16), we have that for any k ∈ N, there exists N2,k > 0 such that

sup un (0) − u ˜kn (0) H 1 (R4 ) ≤ 1. (4.47) n≥N2,k

Similarly, by the Strichartz estimates and (4.16), for any k ∈ N, we can find N3,k ∈ N such that

(4.48) sup h∇ieit∆ un (0) − u˜kn (0) V (R) ≤ δ, 2

n≥N3,k

where the δ > 0 is from the perturbation theory and depends on B and the uniform bound of kun (0)kH 1 ((R4 )) . Hence, by Proposition 2.5, (4.44) holds if we could show that there exist k0 , N0 ∈ N such that if k ≥ k0 and n ≥ N0 ,

h∇i (i∂t + ∆)˜ ukn + f1 (˜ ukn ) − f2 (˜ ukn ) 3 ≤ δ. (4.49) 2 (R×R4 ) Lt,x

To do this, if we define f (u) = f1 (u) − f2 (u), then we have   k X (i∂t + ∆)˜ ukn + f (˜ ukn ) =f (˜ ukn ) − f  uj(n)  j=1

+

k X j=1

  k X (i∂t + ∆)uj(n) + f  uj(n)  . j=1

(4.50)

24


Hence, we just need to show that

  

k X

j  k f u  u lim lim h∇i − f ˜ n

(n) k→∞ n→∞

j=1

= 0,

(4.51)

3 2 (R×R4 ) Lt,x

and

  

k k X X

j j 

  lim lim h∇i (i∂t + ∆)u(n) + f u(n)

n→∞ k→∞

j=1 j=1

= 0.

(4.52)

3 2 (R×R4 ) Lt,x

First, we will prove (4.51). Let u(x) and w(x) be two functions defined on R4 , then if 1 < r ≤ 2, we have |u + w|r (u + w) − |u|r u . (|u|r + |w|r )|w|, (4.53) and

|∇ (|u + w|r (u + w) − |u|r u)| . |u|r |∇w| + (|∇u| + |∇w|)|w|r .

(4.54)

By these two inequalities and the H¨ older inequality, to prove (4.51), we just need to estimate

h

i

it∆ k 43

|e ωn | + |eit∆ ωnk |2 |eit∆ ωnk | 23 Lt,x (R×R4 )

h i (4.55) 4

+ |eit∆ ωnk | 3 + |eit∆ ωnk |2 |∇eit∆ ωnk | 32 , Lt,x (R×R4 )

h k i X

it∆ k 4

|e ωn | 3 + |eit∆ ωnk |2 ∇uj(n)

j=1

,

(4.56)

3 2 (R×R4 ) Lt,x

and





k k X

X

4 j j 2  it∆ k

| 3 + | u | u | |e ω | n (n) (n)

j=1

23 j=1 Lt,x (R×R4 )





k k X

X

4 j j 2 it∆ k | 3 + | u | + . u | |∇e ω | n (n) (n)

j=1

23 j=1 4

(4.57)

Lt,x (R×R )

Now deal with (4.55). Using the H¨ older inequalities, we have

it∆ k 2/3 it∆ k 5/3

it∆ k 4/3 it∆ k

e ωn

e ωn , ≤

|e ωn | |e ωn | 32 V0 (R) W1 (R) 4 Lt,x (R×R )

it∆ k 2 it∆ k

|e ωn | |e ωn |

3

2 (R×R4 ) Lt,x

2 ≤ eit∆ ωnk W1 (R) eit∆ ωnk V0 (R) ,

it∆ k 4/3

|e ωn | |∇eit∆ ωnk |

2/3

2/3

≤ eit∆ ωnk W (R) eit∆ ωnk V (R) ∇eit∆ ωnk V 1

it∆ k 2

|e ωn | |∇eit∆ ωnk |

3 2 (R×R4 ) Lt,x

(4.59)

3

2 (R×R4 ) Lt,x

and

(4.58)

0

0

(4.60) , (R)

2

≤ eit∆ ωnk W1 (R) ∇eit∆ ωnk V0 (R) .

(4.61)


25

Then, by the Strichartz estimates and combining the above four inequalities with the fact that kωnk kH 1 (R4 ) ≤ C, we have i h

2

2/3 (4.62) (4.55) . eit∆ ωnk W1 (R) + eit∆ ωnk W1 (R) .

Hence, by the property (4.4) of ωnk , we obtain that (4.55) tends to 0, as n, k → ∞. From the same method of (4.55) and the fact

X

k j

. B, ∇u(n) sup n≥Nk j=1

V0 (R)

in (4.26) from Lemma 4.5. we can deal with (4.56). To deal with (4.57), we consider the terms having the form of

X

X

k j 4/3 s it∆ k

k j 2 s it∆ k

u(n) · |∇| e ωn u(n) · |∇| e ωn +

23

j=1

j=1 4

,

3 2 (R×R4 ) Lt,x

Lt,x (R×R )

(4.63)

where s ∈ {0, 1}. By the H¨ older inequality and (4.26) in Lemma 4.5, we have k

X

uj(n) |4/3 · |∇|s eit∆ ωnk

|

k

1/3

X

uj(n) . j=1

W2 (R)

k

X

uj(n) |∇|s eit∆ ωnk · 12/7

Lt,x (R×R4 )

k

X

uj(n) |2 · |∇|s eit∆ ωnk

| j=1

k

X

uj(n) . j=1

W1 (R)

(4.64)

,

3

2 (R×R4 ) Lt,x

k

X

uj(n) |∇|s eit∆ ωnk ·

L2t,x (R×R4 )

j=1

k

X

uj(n) |∇|s eit∆ ωnk .

L2t,x (R×R4 )

j=1

12/7

Lt,x (R×R4 )

j=1

k

X

uj(n) |∇|s eit∆ ωnk . j=1

and

3

2 (R×R4 ) Lt,x

j=1

(4.65)

.

Hence, to prove (4.51), we just need show k

X

uj(n) |∇|s eit∆ ωnk lim lim

k→∞ n→∞

4 Lp t,x (R×R )

j=1

= 0,

where p ∈ {12/7, 2} and s ∈ {0, 1}. By Lemma 4.4, for any ε > 0, there exists J(ε) > 0 such that

X X 1

j 2 j 2 kh∇iU∞ kS(R) < ε2 . ≤

u(n) 4 ST (R) J(ε)≤j≤k

j≥J(ε)

(4.66)

(4.67)

26


Using the same method of proving (4.26), by the almost decoupling condition (4.6), we have can find Nk,ε > 0 such that: if n > Nk,ε , then 1/2 

2

X X

j  + 1 ε < ε, (4.68) ≤ uj(n)

u(n)

2 ST (R) ST (R) J(ε)≤j≤k J(ε)≤j≤k

where in the last step, we used the boundedness of ϕj H 1 . Then, for p ∈ {12/7, 2} and s ∈ {0, 1}, by the H¨ older inequality and the Strichartz estimates, we have

X

X

s it∆ k

j j s it∆ k

|∇| e ω

u ≤ u |∇| e ω n V0 (R) . ε. n (n) (n)

p

J(ε)≤j≤k J(ε)≤j≤k 4 ST (R)

Lt,x (R×R )

(4.69) On the other hand, for p ∈ {12/7, 2} and s ∈ {0, 1}, by (4.27) in Lemma 4.5, we have

X

j s it∆ k = 0, (4.70) u |∇| e ω lim lim n (n)

k→∞ n→∞

p 1≤j≤J(ε) 4 Lt,x (R×R )

which together with (4.69) implies (4.66). This finishes the proof of (4.51).

Proof of (4.52). Recall that uj(n) satisfy that (4.17) and (4.18), then we have

  

k k X X

s X j  j

  (4.71) u(n) f u(n) − f (4.52) .

|∇|

3/2 s=0,1 j=1 j=1 4 Lt,x (R×R ) i h X

s X j j j −1 (4.72) (σ∞ ) f1 σ∞ u(n) − f1 uj(n) 3/2 +

|∇| 4 s=0,1

+

Lt,x (R×R )

1≤j≤k j hn →0

X

s X

f2 uj(n)

|∇|

s=0,1

For (4.71), using a elementary X X s=0,1

plus

X

s=0,1

3/2

Lt,x (R×R4 )

1≤j≤k j hn →0

1≤j,j ′ ≤k j6=j ′

X

1≤j,j ′ ≤k j6=j ′

.

(4.73)

inequality, (4.71) can be estimated by

j 2 ′

u |∇|s uj (n) 3/2

(n) 4

(4.74)

Lt,x (R×R )

j 4/3 ′

u |∇|s uj (n)

(n)

.

(4.75)

3/2

Lt,x (R×R4 )

For (4.74), we use density. By

and

min(1 − hj , hj )

j j

n n j j j

σn U∞ − σ∞ U∞ W1 (R) . →0 h∇iU

∞

hjn + |∇| S(R)

j j j j

∇ σn U∞ − σ∞ U∞ V

0 (R)

min(1 − hj , hj )

n n j → 0, h∇iU . ∞

hjn + |∇| S(R)

(4.76)

(4.77)


27

we have for any ε > 0, there exists U j , V j ∈ Cc∞ (R × R4 ) such that for sufficiently large n,

! !

t − tjn t − tjn

j j j

j j j j j j j + ∇ Tn σn U∞ − Tn V < ε,

Tn σn U∞ − Tn U

(hjn )2 W (R) (hjn )2 V (R) 1 0 (4.78) by density. Hence

! ! ′ X X t − tjn

j j 2 t − tjn s j j′ · |∇| Tn V . (4.74) . ε +

|Tn U | j 2 j′ 2

(hn ) (hn ) L3/2 (R×R4 ) s=0,1 1≤j,j′ ≤k t,x

j6=j ′

(4.79) → ∞, we have

! ! ′

t − tjn

j j 2 t − tjn

j′ j′ · |∇|Tn V (4.80)

|Tn U | j 2 j′ 2

(hn ) (hn ) L3/2 (R×R4 ) t,x

!

i h j′ j j′ 2 j′ j′ j ′ t(h ) + t − t xh

n n n n + xn − xn j 2 j −2 j 2 j′ .(hn ) (hn ) |U | (t, x) · |∇|V ,

3/2 (hjn )2 hjn Lt,x (R×R4 )

2

′ ′

.(hjn )2 (hjn )−2 U j L∞ (R×R4 ) |∇|V j 3/2 → 0, 4 If

hjn

′ hjn

Lt,x (R×R )

t,x

as n →j ∞. h If jn′ → 0, we have hn

! ! ′

t − tjn

j j 2 t − tjn j′ j′ (4.81) · |∇|Tn V

|Tn U | ′

(hjn )2 (hjn )2 L3/2 (R×R4 ) t,x

!

i h j j j′ j 2 j j′ ′ ′ xh + x − x t(h ) + t − t

n n n n n , n .(hjn )2 (hjn )−2 |U j |2 (t, x) · |∇|V j

j′ 2 j′

3/2

(hn ) hn

Lt,x (R×R4 )

′

.(hjn )2 (hjn )−2 → 0,

as n → ∞. If there exists Cj such that

hjn

′ hjn

′

+

hjn hjn

< Cj for any n, then by the almost

decoupling condition (4.6), we have ′ tj − tj xj ′ − xj n n n n → ∞, as n → ∞, j 2 + (hn ) hjn

where by symmetry we assume that j < j ′ . Then for sufficiently large n, we have

! ! ′

t − tjn

j j 2 t − tjn

j′ j′ · |∇|Tn V (4.82)

|Tn U |

′

(hjn )2 (hjn )2 L3/2 (R×R4 ) t,x

! ′

i h j′ 2 j′ j j′ ′ t(hn ) + tn − tn xhn + xjn − xjn

j 2 j −2 j 2 j′ .(hn ) (hn ) |U | (t, x) · |∇|V ,

3/2

(hjn )2 hjn

Lt,x (R×R4 )

=0,

28


where we used the fact that U j , V j ∈ Cc∞ (R × R4 ). On the other hand, we need to consider

X

j 2 j′ .

|u(n) | u(n) 3/2 4

(4.83)

Lt,x (R×R )

j,j ′

j j

If hjn ≡ 1, then we have σnj U∞ = U∞ ≤ C. If hjn → 0, then we have S(R) S(R)

j j = (hjn )2 σnj U∞ that uj(n) . hjn h∇iU∞ → 0. Hence by the similar V0 S(R) V0 (R)

argument as before, we have for any ε > 0, there exists U j , V j ∈ Cc (R × R) such that

j

+ uj(n) − Tnj V j 0, by the Strichartz estimates, we can find δ > 0, such that

it∆

e φ + ∇eit∆ φ L3 ([t −t′ −δ,t −t′ +δ]) < ε. W ([t −t′ −δ,t −t′ +δ]) 1

n

n

n

This and (5.6) imply that

it∆

e uc (tn ) W

n

t,x

1 ([−δ,δ])

n

n

n

+ ∇eit∆ uc (tn ) L3

t,x ([−δ,δ])

n

< ε,

for sufficiently large n (up to a subsequence). Then by the Strichartz estimates and the H¨ older inequality, we have kuc (t)kW1 ([tn −δ,tn +δ]) + k∇uc (t)kL3t,x ([tn −δ,tn +δ]) .ε + k∇ [f1 (uc (t)) + f2 (uc (t))]kL3/2 ([tn −δ,tn +δ]) t,x

.ε +

kuc (t)k2W1 ([tn −δ,tn +δ]) 4

(5.7)

k∇uc (t)kV0 ([tn −δ,tn +δ]) 1/3

2/3

3 k∇uc (t)kV0 ([tn −δ,tn +δ]) k∇uc (t)kL2 + kuc (t)kW 1 ([tn −δ,tn +δ])

t,x ([tn −δ,tn +δ])

.

By the standard continuous method and the fact that uc ∈ K+ , we have that kuc (t)kW1 ([tn −δ,tn +δ]) < +∞. On the other hand, by Sobolev embedding relation H˙ 1 (R4 ) ֒→ L4 (R4 ), we have that kuc (t)kW2 ([tn −δ,tn +δ]) < +∞. Hence kuc kST ([tn −δ,tn +δ]) < +∞, which contradicts with the fact that uc is not a global solution.


31

By the local well-posedness theory, if we take ε small enough, the solution uc (t) exists on [tn − δ, tn + δ], which contradicts with the choice of tn . Hence we have uc is a global solution. 5.2. Compactness of the critical element. In this subsection, for the critical elements as obtained above, we give out their compactness property and an important corollary. Proposition 5.2. Let uc be a forward critical element, i.e. kuc kST ([0,∞)) = ∞.

(5.8)

Then there exists x(t) : (0, ∞) → R4 such that the set {uc (t, x − x(t)) : 0 < t < ∞} is precompact in H˙ s for any s ∈ (0, 1]. Proof. By the observation that the mass of uc conserves of time, we just need to prove the precompacness property in the space H˙ 1 (R4 ). To do this, for any sequence {tn } ⊂ [0, ∞), we need find a sequence {xn } ∈ R4 such that the set {uc (tn , x + xn )} is precompact in H˙ 1 (R4 ). If there exists a subsequence {tnk } of {tn } converges, the claim follows from the time continuousness of uc . If tn converges to ∞, by profile decomposition and the discussion in Lemma 5.1, there exist a sequence (xn , t′n ) ∈ R4 × [0, ∞) and a function ϕ ∈ H 1 such that

′

→ 0, as n → ∞.

uc (tn ) − e−itn ∆ ϕ(x − xn ) ˙ 1 (R4 ) H

t′n

Hence we will be done if converges to a constant. Thus we just need to preclude the two cases below: (1) If t′n → −∞, by the Strichartz estimates, we have

it∆

it∆

1/2 it∆

e uc (tn )

∇e uc (tn ) 3 + + e u (t )

|∇|

3 c n ST ([0,∞)) Lt,x ([0,∞)) Lt,x ([0,∞))

′

. h∇ieit∆ ϕ S([−t′ ,∞)) + uc (tn ) − e−itn ∆ ϕ(x − xn ) . ˙ 1/2 (R4 ) ˙ 1 ∩H H

n

−it′n ∆

Notice that the mass of uc and e ϕ(x − xn ) is bounded, which together with interpolation implies H˙ 1/2 (R4 ) norm of their difference also tends to 0. Hence for any ε > 0, by Strichartz estimates and the fractional chain rule, we have

kuc kL6 ([tn ,∞)) + kuc kL4 ([tn ,∞)) + k∇uc kL3 ([tn ,∞)) + |∇|1/2 uc 3 t,x t,x t,x Lt,x ([tn ,∞))

2 .ε + kuc kL6 ([tn ,∞)) k∇uc kL3 ([tn ,∞)) + |∇|1/2 uc 3 t,x t,x Lt,x ([tn ,∞))

1/2 4/3 + kuc kL4 ([tn ,∞)) k∇uc kL3 ([tn ,∞)) + |∇| uc 3 , t,x

t,x

Lt,x ([tn ,∞))

by taking n sufficiently large. By the standard continuous method, we can find that kuc kST ([tn ,∞)) < +∞, which contradicts with (5.8).

(2) If t′n → +∞, then we have

it∆

e uc (tn ) + ∇eit∆ uc (tn ) L3 ([0,∞)) + |∇|1/2 eit∆ uc (tn ) 3 ST ((−∞,0)) t,x Lt,x ([0,∞))

′

. h∇ieit∆ ϕ S((−∞,−t′ )) + uc (tn ) − e−itn ∆ ϕ(x − xn ) → 0. n

˙ 1/2 (R4 ) ˙ 1 ∩H H

32


By a similar way as in (1), we have kuc kST (−∞,tn ) → 0, which implies uc = 0 and contradicts with (5.8). By Proposition 5.2 and the Arzela-Ascoli theorem, we have Corollary 5.3. Let uc (t, x) be a critical element, then there exists x(t) : (0, ∞) → R4 such that: for any η > 0, there exists C(η) < ∞ such that Z Z Z |ξ|2 |ˆ u(t, ξ)|2 dξ < η |ξ|2 |ˆ u(t, ξ)|2 dξ + |∇u(t, x)|2 dx + |x−x(t)|>C(η)

|ξ|>C(η)

1 |ξ|< C(η)

5.3. Extinction of the critical element. In this part, we will finish the proof of the main theorem. We will use the interaction Morawetz estimate to exclude the critical element in Proposition 5.2. Theorem 5.4. Suppose u is a critical element in Proposition 5.2 satisfying that u ∈ K+ = {u(t) ∈ H 1 : E(u) < m, Kα,β (u) ≥ 0, (α, β) ∈ Ω}. Then we have u ≡ 0. Proof. Let ψ ∈ C0∞ (R) be a radial function such that ψ(x) = 1 when |x| ≤ 1 and ψ(x) = 0 when |x| > 2. Next, let Z φ(x − y) = ψ 2 (x − s)ψ 2 (y − s)ds. (5.9)

Hence φ(x) is supported on |x| ≤ 4. For 1 ≤ R ≤ R0 define the interaction Morawetz potential Z |x − y| (x − y)j Im[u∂j u](t, x)dxdy (5.10) MR (t) = |u(t, y)|2 φ R

and let

Z

1 MR (t)dR. (5.11) R 1≤R≤R0 By the H¨ older inequality and the Sobolev embedding theorem, we have M (t) =

sup |MR (t)| . R4 .

(5.12)

t∈R

By a direct calculation, we have d MR (t) dtZ |x − y| h 4 10 i |∇u|2 − |u|4 + |u| 3 (t, x)dxdy =2 |u|2 (t, y)φ R 5 Z |x − y| Im[u∂j u](t, x)dxdy − 2 Im[u∂j u](t, y)φ R Z |x − y| (x − y) (x − y) j k + 2 |u|2 (t, y)φ′ Re[(∂j u∂k u)](t, x)dxdy R |x − y|R Z |x − y| |x − y| h 1 2 10 i + |u|2 (t, y)φ′ |∇u|2 − |u|4 + |u| 3 (t, x)dxdy R R 2 5 Z |x − y| (x − y) (x − y) j k − 2 Im[u∂k ](t, y)φ′ Im[u∂j u](t, x)dxdy R |x − y|R

(5.13) (5.14) (5.15) (5.16) (5.17)


−

1 2

Z

h |x − y| |x − y| i |x − y| ) + φ′ |u|2 (t, y)∆ 4φ( |u|2 (t, x)dxdy. R R R

33

(5.18)

1 4 By Proposition 5.2 and conservation of mass, we have u(t, x) ∈ L∞ t H (R ). Let 1 I = (0, K) ⊆ R and 1 ≤ R0 ≤ K 5 . First we consider (5.15), (5.16) and (5.17). From the definition of φ, we have Z Z 1 ′ |x − y| |x − y| 1 ′ |x − y| (x − y)j (x − y)k dR . 1, φ dR . φ R |x − y|R R R 1≤R≤R0 R 1≤R≤R0 R

and is supported on |x − y| . R0 . Hence we have Z Z 1 [(5.15) + (5.16) + (5.17)]dRdt . K. R I 1≤R≤R0

(5.19)

Next we consider (5.18). Notice that Z i 1 h |x − y| ′ |x − y| |x − y| + φ dR ∆ 4φ R R R 1≤R≤R0 R Z 6 15 |x − y| |x − y| 1 ′′ |x − y| ′ |x − y| = φ′′′ dR + φ + φ R R3 R R2 R R|x − y| 1≤R≤R0 R 1 1 . + . (5.20) 1 + |x − y|2 |x − y| + |x − y|2 This together with the Hardy-Littlewood-Sobolev inequality, the Sobolev theorem 1 and u ∈ L∞ t Hx implies that Z Z 1 (5.18) dRdt R I 1≤R≤R0 Z Z 1 |u|2 (t, y)dxdydt . |u|2 (t, x) 1 + 2 |x − y| (5.21) I 4 .K kuk4L∞ 2 + kuk 8 t Lx 3 L∞ t Lx

.K.

Now we will estimate (5.13) and (5.14). By abuse of notations, we will write ψ(x) = ψ(|x|), for x ∈ R4 . Notice that for each s and t, there exists a ξ(s, t) such that Z

x

− s Im[eix·ξ(s,t) u∇(eix·ξ(s,t) u)](t, x)dx R Z Z x x − s Im[u∇u](t, x)dx + ξ(s, t) ψ 2 − s |u|2 (t, x)dx = ψ2 R R =0. ψ2

(5.22)

Moreover, for any s, t and ξ(s, t), the quantity Z y h x i − s ψ2 − s |∇u(t, x)|2 |u(t, y)|2 − Im[u∇u](t, x)Im[u∇u](t, y) dxdy ψ2 R R

is invariant under the Galilean transformation u(t, x) 7→ eix·ξ(s,t) u(t, x). Therefore, for any given s, t we can choose ξ(s, t) to eliminate the momentum squared terms.

34


Then, from integration by parts, we have Z x − s |∇(eix·ξ(s,t) u(t, x))|2 − |u(t, x)|4 dx (5.23) ψ2 R Z Z x x 2 = ∇ ψ − s eix·ξ(s,t) u(t, x) dx − ψ 2 − s |u(t, x)|4 dx (5.24) R R Z x x − s ∆x ψ − s dx. (5.25) + |u(t, x)|2 ψ R R ¯ Since u ∈ K+ , we have kuk ∞ ˙ 1 ≤ (1 − δ)kW k ˙ 1 by Lemma 2.14. Therefore, by Lt H

H

the sharp Sobolev inequalities, we have kukL4(R4 ) ≤

k∇ukL2 (R4 ) 1 ¯ ¯ ≤ (1 − δ)kW kH˙ 1 ≤ (1 − δ)kW kL4 (R4 ) . (5.26) kW kL4 (R4 ) kW kL4 (R4 )

Let v = eix·ξ(s,t) u, then by H¨ older inequalities we have 2 ¯ ¯ 2 |u|4 kL1 (5.24) ≥k∇(ψv)k2L2 − (1 + δ)kψ |u|4 kL1 + δkψ ¯ 2 |u|4 kL1 ¯ − δ) ¯ 2 k∇(ψv)k2 2 + δkψ ≥k∇(ψv)k2 2 − (1 + δ)(1 L

L

(5.27)

¯ 2 |u|4 kL1 . ≥δkψ For (5.25), noting that Z y x − y 1 x x 1 ψ , (5.28) + −s ∆x ψ −s ·ψ 2 −s ds . ψ R R R 4R R2 R|x − y| we have Z Z Z 1 (5.25)|u|2 (t, y)dydsdRdt (5.29) R I 1≤R≤R0 Z Z Z 1 1 2 |x − y| 1 |u|2 (t, y) dR dx dy dt + |u| (t, x) ψ . 4R R2 R|x − y| I 1≤R≤R0 R Z Z 1 2 . |u| (t, x) 1 + |u|2 (t, y)dxdydt |x − y|2 I .K. 1

1

For |x − y| ≤ R02 and R ≥ R02 , we have Z y x − s ψ2 − s ds & 1. ψ2 R R Thus, we have Z Z I

&

1 R02

Z Z I

1 R02

& ln R0

≤R≤R0

1 R

≤R≤R0

1 R

Z Z I

& ln R0

Z

|u|2 (t, y)ψ 2

1 |x−y|≤R02

1 |x−y|≤R02

Z "Z I

Z

Z

Z

(5.30)

y − s ψ2 − s |u|4 (t, x)dsdxdydRdt R R

x

|u|2 (t, y)|u|4 (t, x)dxdydRdt (5.31)

|u|2 (t, y)|u|4 (t, x)dxdydt 4

1

|x−x(t)|≤R04

|u| (t, x)dx ·

Z

2

1

|y−x(t)|≤R04

#

|u| (t, y)dy dt.


This together with (5.12),(5.19) and (5.21) implies Z 1 4 sup |MR ((t))|dR R0 & R t 1≤R≤R Z Z 0 d 1 & MR (t)dtdR R dt I 1≤R≤R0 Z Z "Z & ln R0

1

I

|u|4 (t, x)dx ·

|x−x(t)|≤R04

35

(5.32) 1

#

|u|2 (t, y)dy dt − K.

|y−x(t)|≤R04

1

Since 1 ≤ R0 ≤ K 5 , we have Z Z "Z 4 |u| (t, x)dx · ln R0 1 I

|x−x(t)|≤R04

2

1

|y−x(t)|≤R04

#

|u| (t, y)dy dt . K.

(5.33)

Furthermore, (5.33) implies that there exists a sequence tn ∈ R such that either Z |u|4 (tn , x)dx → 0, R0,n → ∞, (5.34) 1 4 |x−x(tn )|≤R0,n

or

Z

1 4 |y−x(tn )|≤R0,n

|u|2 (tn , y)dy → 0, R0,n → ∞.

(5.35)

By Corollary 5.3 and the Sobolev inequalities, (5.34) and (5.35) implies u ≡ 0. Acknowledgments:The authors thank the referee and the associated editor for their invaluable comments and suggestions which helped improve the paper greatly. This work was supported in part by the National Natural Science Foundation of China under grants 11671047. References [1] T. Aubin, Probl´ emes isop´ erim´ etriques et espaces de Sobolev. Journal of Differential Geometry, 11.4(1976), 573-598. [2] T. Akahori, S. Ibrahim, H. Kikuchi and H. Nawa, Existence of a ground state and blow-up problem for a nonlinear Schr¨ odinger equation with critical growth. Differential & Integral Equations 25.3/4(2012), 383-402. [3] T. Akahori, S. Ibrahim, H. Kikuchi and H. Nawa. Existence of a ground state and scattering for a nonlinear Schr¨ odinger equation with critical growth. Selecta Mathematica New 19.2(2013), 545-609. [4] T. Akahori, S. Ibrahim, H. Kikuchi and H. Nawa, Global dynamics above the ground state energy for the combined power-type nonlinear Schr¨ odinger equations with energy-critical growth at low frequencies. arXiv: 1510.08034v1. [5] J. Bourgain, Global wellposedness of defocusing critical nonlinear Schr¨ odinger equation in the radial case. Journal of the American Mathematical Society. 12.1(1999), 145-171. [6] T. Cazenave, Semilinear Schr¨ odinger equations. Courant Lecture Notes in Mathematics, 10, NYU, CIMS, AMS 2003: 635-637. [7] T. Cazenave and F. B. Weissler, The Cauchy problem for the nonlinear Schr¨ odinger equation in H 1 . Manuscripta Mathematica, 61.4(1988), 477-494. [8] T. Cazenave and F. B. Weissler, The Cauchy problem for the critical nonlinear Schr¨ odinger equation in H s . Nonlinear Analysis, TMA, 14.10(1990), 807-836. [9] M. Christ and M.I. Weistein, Dispersion of small amplitude solutions of the generalized Korteweg-de Vries equation. Journal of Functional Analysis, 100.1(1991), 87-109. [10] J. Colliander, M. Keel, G. Staffilani, H. Takaoka and T. Tao, Global existence and scattering for rough solutions of a nonlinear Schr¨ odinger equation on R3 . Communications on Pure & Applied Mathematics, 57.8(2004), 987-1014.

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Changxing Miao: Institute of Applied Physics and Computational Mathematics, P. O. Box 8009, Beijing, China, 100088, E-mail address: miao [email protected] Tengfei Zhao: The Graduate School of China Academy of Engineering Physics, P. O. Box 2101, Beijing, China, 100088, E-mail address: zhao [email protected] Jiqiang Zheng e Nice Sophia-Antipolis, Universit´ 06108 Nice Cedex 02, France E-mail address: [email protected], [email protected]

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