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Global well-posedness and scattering for nonlinear Schrödinger equations with combined nonlinearities in the radial case
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ScienceDirect J. Differential Equations 261 (2016) 2881–2934 www.elsevier.com/locate/jde
Global well-posedness and scattering for nonlinear Schrödinger equations with combined nonlinearities in the radial case Xing Cheng a , Changxing Miao b , Lifeng Zhao c,∗ a College of Science, Hohai University, Nanjing 210098, Jiangsu, China b Institute of Applied Physics and Computational Mathematics, P. O. Box 8009, Beijing 100088, China c Wu Wen-Tsun Key Laboratory of Mathematics and School of Mathematical Sciences, University of Science
and Technology of China, Hefei 230026, Anhui, China Received 20 February 2016 Available online 1 June 2016
Abstract We consider the Cauchy problem for the nonlinear Schrödinger equation with combined nonlinearities, one of which is defocusing mass-critical and the other is focusing energy-critical or energy-subcritical. The threshold is given by means of variational argument. We establish the profile decomposition in H 1 (Rd ) and then utilize the concentration–compactness method to show the global wellposedness and scattering versus blowup in H 1 (Rd ) below the threshold for radial data when d ≤ 4. © 2016 Elsevier Inc. All rights reserved.
MSC: 35L70; 35Q55 Keywords: Nonlinear Schrödinger equation; Combined nonlinearities; Global wellposedness; Scattering; Blowup
Contents 1.
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2882
* Corresponding author.
E-mail addresses: [email protected] (X. Cheng), [email protected] (C. Miao), [email protected] (L. Zhao). http://dx.doi.org/10.1016/j.jde.2016.04.031 0022-0396/© 2016 Elsevier Inc. All rights reserved.
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2. Variational estimates . . . . . . . . . . . . . 3. Wellposedness and perturbation theory 4. Linear profile decomposition . . . . . . . 5. Extraction of a critical element . . . . . . 6. Extinction of the critical element . . . . 7. Blow-up . . . . . . . . . . . . . . . . . . . . . Acknowledgments . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . .
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1. Introduction We will consider the Cauchy problem for the nonlinear Schrödinger equation 4 i∂t u + u = |u| d u − |u|p−1 u, u(0) = u0 ∈ H 1 (Rd ), where u : R × Rd → C is a complex-valued function, 1 + 1 + d4 < p < ∞, d = 1, 2. The equation (1.1) has the following mass and energy: M(u) = |u|2 dx,
E(u) =
4 d
Rd 2 1 2 |∇u|
−
p+1 1 p+1 |u|
+
d 2(d+2) |u|
4 d−2 ,
(1.1) d = 3, 4 and
(1.2) 2(d+2) d dx.
(1.3)
Rd
The equation (1.1) is a special case of the general nonlinear Schrödinger equation with combined nonlinearities i∂t u + u = μ1 |u|p1 −1 u + μ2 |u|p2 −1 u, (1.4) u(0) = u0 ∈ H 1 (Rd ), 4 where 1 < p1 , p2 ≤ 1 + d−2 , for d ≥ 3, 1 < p1 , p2 < ∞, for d = 1, 2, μ1 , μ2 ∈ {±1}. This equation arises in the study of the Hartree approximation and quasi-classically within the frame of the secondary quantization in the boson gas with many body δ-function interaction. It can also be used to describe the effect of saturation of nonlinear refractive index. At the same time, the equation of nuclear hydrodynamics with effective Skyrme’s forces reduces quasi-classically to (1.4). For more physical background, we refer the reader to [5,31] and the references therein. The local theory for (1.4) in the energy space follows from the standard method of T. Cazenave, F.B. Weissler [10], see also [9]. Proceeded by [37], T. Tao, M. Visan and X. Zhang considered various cases in [33]. They proved global wellposedness and scattering of the solution 4 to the equation (1.4) for finite energy data when μ1 = μ2 = +1 and 1 + d4 ≤ p1 < p2 ≤ 1 + d−2 , 4 4 d ≥ 3. The case when p1 = 1 + d and p2 = 1 + d−2 is the most difficult. In [33], the low frequencies of the solution are well approximated by the L2 critical problem and the high frequencies
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are well approximated by the energy-critical problem. The medium frequencies are controlled by the Morawetz estimates. In [27], C. Miao, G. Xu and L. Zhao considered the case where μ1 = +1, μ2 = −1 and 4 , d = 3. The threshold was given by variational method due to 1 + d4 < p1 < p2 = 1 + d−2 the energy trapping property as in [16]. They established the linear profile decomposition in H 1 (Rd ) in the spirit of [16]. By using this new profile decomposition, they reduced the scattering problem to the extinction of the critical element. The critical element can then be excluded by using the virial identity. They showed the dichotomy of global wellposedness and scattering versus blow-up phenomenon below the threshold for radial solutions. The radial assumption was removed in dimensions five and higher in [28]. 4 If μ1 = μ2 = −1 and 1 + d4 < p1 < p2 = 1 + d−2 , d ≥ 5, the Cauchy problem was considered in [1,2]. After giving existence of the ground state based on the idea in [7] and [16], they showed a sufficient and necessary condition for the scattering in the spirit of [18] by using the profile decomposition in H˙ 1 (Rd ) and the global wellposedness and scattering result in [22]. In this paper, we aim to look for the suitable threshold to study the global well-posedness and scattering versus blowup of (1.1). Before stating the main theorem, we introduce some notad tions. For ϕ ∈ H 1 (Rd ), we denote the scaling (Tλ ϕ)(x) = λ 2 ϕ(λx). For any ω > 0, we have the Lyapunov functional Sω (ϕ) = E(ϕ) + 12 ωM(ϕ).
(1.5)
We also denote the scaling derivative of Sω (ϕ) by K(ϕ), d d K(ϕ) = L Sω (ϕ) = dλ S (T ϕ) = dλ E(Tλ ϕ) λ=1 ω λ λ=1 2(d+2) p+1 d + d+2 |ϕ| d dx. = |∇ϕ|2 − d(p−1) 2(p+1) |ϕ|
(1.6)
Rd
Let mω = inf{Sω (ϕ) : ϕ ∈ H 1 (Rd ) \ {0}, K(ϕ) = 0},
(1.7)
then we have 4 Proposition 1.1. For 1 + d4 < p < 1 + d−2 , d ≥ 3, and 1 + d4 < p < ∞, d = 1, 2, ω > 0, we have mω > 0. Moreover, mω = Sω (Q), where Q ∈ H 1 (Rd ) is the ground state of 4
−ωQ + Q + |Q|p−1 Q − |Q| d Q = 0. Proposition 1.2. For p = 1 +
4 d−2 ,
d ≥ 3, we have for all ω > 0, mω = E 0 (W ),
where
(1.8)
(1.9)
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E (W ) = 0
2 1 2 |∇W |
−
2d d−2 d−2 2d |W |
dx,
Rd
and W ∈ H˙ 1 (Rd ) is unique positive radial solution of 4
−W = |W | d−2 W.
(1.10)
Remark 1. 4 (1) When p = 1 + d−2 , we note mω = E 0 (W ), which indicates that mω is independent of ω, so we can define m := mω = E 0 (W ) in this case. Let
Aω,+ = {ϕ ∈ H 1 (Rd ) : Sω (ϕ) < mω , K(ϕ) ≥ 0}, Aω,− = {ϕ ∈ H 1 (Rd ) : Sω (ϕ) < mω , K(ϕ) < 0}, A+ = {ϕ ∈ H 1 (Rd ) : E(ϕ) < m, K(ϕ) ≥ 0}, A− = {ϕ ∈ H 1 (Rd ) : E(ϕ) < m, K(ϕ) < 0}. (2) Aω,± and A± are non-empty by using the same argument in [27]. In fact, we note ϕ = 0 belongs to both Aω,+ and A+ , so Aω,+ and A+ are nonempty. On the other hand, we can d easily verify that ϕ(x) = − 2 Q( x ) belongs to Aω,− , when is sufficiently small, similarly, d
by using some truncation of ϕ = − 2 W ( x ) to make sure ϕ ∈ H 1 (Rd ), we can show A− is nonempty. (3) We note for any u0 ∈ A+ , we have E(u0 ) < m, we can take ω > 0 small enough such that E(u0 ) + 12 ωM(u0 ) < m, thus u0 ∈ Aω,+ . Similarly, for any u0 ∈ A− , we can still take ω > 0 4 , small enough such that u0 ∈ Aω,− . Thus, our result below in the situation p = 1 + d−2 4 4 d = 3, 4 can be proved as in the case 1 + d < p < 1 + d−2 , d ∈ {3, 4}. So we will give 4 the proof of the following theorem in a unified form regardless of 1 + d4 < p < 1 + d−2 , 4 4 d ∈ {3, 4}, 1 + d < p < ∞, d ∈ {1, 2} or p = 1 + d−2 , d = 3, 4. Now, we state our result. Theorem 1.3. 4 (1) For 1 + d4 < p < 1 + d−2 , d ∈ {3, 4}, 1 + d4 < p < ∞, d ∈ {1, 2}, and u0 is radial, we have for any ω > 0, (i) if u0 ∈ Aω,+ , the solution u to (1.1) exists globally and scatters in H 1 (Rd ); 4 (ii) if u0 ∈ Aω,− , for d ≥ 2, p ≤ min(5, 1 + d−2 ), the solution u blows up in finite time. 4 (2) For p = 1 + d−2 , d ∈ {3, 4}, and u0 is radial, we have (i) if u0 ∈ A+ , the solution u to (1.1) exists globally and scatters in H 1 (Rd ); (ii) if u0 ∈ A− , the solution u blows up in finite time.
For the nonlinear Schrödinger equation with combined nonlinearities, we focus on the different roles played by the two nonlinearities. Generally speaking, the main barrier of the local
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theory is the higher order term while the lower order term is dominant in the global behavior. We prove our main theorem by the compactness–contradiction method initiated by C.E. Kenig and F. Merle [18]. In the argument, the linear profile decomposition plays an important role. In previous works such as [27,28], the authors considered the equation
4
i∂t u + u = |u|p−1 u − |u| d−2 u, u(0) = u0 ∈ H 1 (Rd ),
(1.11)
4 2 where 1 + d4 < p < 1 + d−2 , d ≥ 3. Since the lower order term is H˙ s (Rd ) (0 < s = d2 − p−1 < 1) critical, the solution of (1.11) is expected to behave like that of the defocusing equation i∂t u + u = |u|p−1 u. The critical space of this defocusing equation is H˙ s (0 < s < 1), so it ∇
1 d is reasonable to apply the H˙ s -profile decomposition to |∇| s u for u ∈ H (R ). Since the symmetry group in H˙ s is of the same type as in H˙ 1 case, we may equivalently apply the H˙ 1 -profile 1 d 2 d decomposition to ∇
|∇| u for u ∈ H (R ). However, the lower term in (1.1) is L (R )-critical. The symmetry group in L2 (Rd ) is different from that in H˙ 1 (Rd ) due to the Galilean symmetry. Therefore, it is natural that the L2 -profile decomposition is applied to ∇ u for u ∈ H 1 (Rd ) in this paper. We explore the profile decomposition in use to get better estimate (4.14) for the remainder. This estimate provides spacetime control for both the remainder wnk and its derivative |∇|wnk . This makes remarkable difference with the profile decomposition for H 1 (Rd ) data obtained both in [8] and [27]. In fact, R. Carles and S. Keraani [8] only provided the control over the remainder, while the authors in [27] only provided the control over the derivative of the remainder. Although we make more delicate analysis due to the stronger control in the perturbation theorem, we get stronger compactness for the critical element in H 1 (Rd ), which is stronger than the compactness in H˙ s (0 < s ≤ 1) obtained in [27,28]. We expect our result will be extended to higher dimensions (d ≥ 5) since all the arguments make sense except the long-time perturbation. However, the exotic Strichartz estimates in [14,36] seem useless to establish the long-time perturbation in our case because the mass-critical term in our equation cannot be controlled properly in the Sobolev spaces. The radial assumption is expected to be removed in a forthcoming paper. The rest of the paper is organized as follows. After introducing some notations and preliminaries, we give the threshold in Section 2. Moreover, we show the energy-trapping properties for the set Aω,± in this section. The local wellposedness and perturbation theory are stated in Section 3. In Section 4, we derive the linear profile decomposition for data in H 1 (Rd ). Then we argue by contradiction. We reduce to the existence of a critical element in Section 5 and show the extinction of such a critical element in Section 6. To make the results complete, we show the existence of blowup solutions in Section 7.
Notation and preliminaries We will use the notation X Y whenever there exists some positive constant C so that X ≤ CY . Similarly, we will use X ∼ Y if X Y X. We define the Fourier transform on Rd to be fˆ(ξ ) =
1 (2π)d Rd
e−ixξ f (x) dx,
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s f (ξ ) = |ξ |s fˆ(ξ ). We and for s ∈ R, the fractional differential operators |∇|s is defined by |∇| s s s ˆ also define ∇ by ∇ f (ξ ) = (1 + |ξ |) f (ξ ). We define the homogeneous Sobolev norms
f H˙ s (Rd ) = |∇|s f L2 (Rd ) , and inhomogeneous Sobolev norms f H s (Rd ) = ∇ s f L2 (Rd ) . We define the radial Lebesgue and Sobolev space as follows: for 1 < q < ∞, s ∈ R, we define q
Lrad (Rd ) = {f ∈ Lq (Rd ) : f is radial function}, s Hrad (Rd ) = {f ∈ H s (Rd ) : f is radial function}, s (Rd ) = {f ∈ H˙ s (Rd ) : f is radial function}. H˙ rad
We use the notation on (1) to denote a quantity which tends to 0, as n → ∞. q For I ⊂ R, we use Lt Lrx (I × Rd ) to denote the spacetime norm u Lqt Lr (I ×Rd ) =
x
I q
|u(t, x)| dx r
qr
q1 dt
.
Rd
q
When q = r, we abbreviate Lt Lrx as Lt,x . We also recall Duhamel’s formula t u(t) = e
i(t−t0 )
u(t0 ) − i
ei(t−s) (iut + u)(s)ds.
(1.12)
t0
We say that a pair of exponents (q, r) is L2 -admissible if (q, r, d) = (2, ∞, 2), d ≥ 1.
2 q
+
d r
=
d 2
and 2 ≤ q, r ≤ ∞,
Lemma 1.4 (Strichartz estimate, [17]). Let I be a compact time interval and let u : I × Rd → C be a solution to the forced Schrödinger equation i∂t u + u = G for some function G, then we have u Lq Lr (I ×Rd ) u(t0 ) L2x (Rd ) + G t
x
q˜
Lt Lrx˜ (I ×Rd )
for any t0 ∈ I and any L2 -admissible exponents (q, r), (q, ˜ r˜ ). If I × Rd is a spacetime slab, we define the Strichartz norm S 0 (I ) by u S 0 (I ) = sup u Lq Lr (I ×Rd ) , t
x
(1.13)
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where the sup is taken over all L2 -admissible pairs (q, r). When d = 2, we need to modify the norm a little, where the sup is taken over all L2 -admissible pairs with q ≥ 2 + , for > 0 arbitrary small. We also define S 1 (I ) norm by u S 1 (I ) = ∇ u S 0 (I ) .
(1.14)
2. Variational estimates In this section, we prove Proposition 1.1 and 1.2. We show the existence of the ground state together with the energy-trapping property for Aω,± , which will be used to show the scattering and blow-up. Let K(ϕ) = KQ (ϕ) + KN (ϕ), where Q K (ϕ) = |∇ϕ|2 dx, KN (ϕ) = − d(p−1) 2(p+1)
Rd
|ϕ|p+1 dx +
d d+2
|ϕ|
2(d+2) d
dx.
We have the following basic fact about K(Tλ ϕ): Lemma 2.1 (Dynamic behavior of K under the scaling). For any ϕ ∈ H 1 (Rd ) \ {0}, there exists a unique λ0 (ϕ) > 0 that ⎧ ⎪ ⎨> 0, K(Tλ ϕ) = 0, ⎪ ⎩ < 0,
0 < λ < λ0 (ϕ), λ = λ0 (ϕ), λ > λ0 (ϕ).
(2.1)
Proof. An easy computation gives 1 K(Tλ ϕ) = λ2
|∇ϕ|2 +
2(d+2) d d d+2 |ϕ|
−
d(p−1) d2 (p−1)−2 |ϕ|p+1 dx, 2(p+1) λ
which implies K(Tλ ϕ) > 0 for λ > 0 sufficiently small. By d dλ
1 K(Tλ ϕ) λ2
= − d(p−1) 2(p+1)
d
d (p−1)−3 2 2 (p − 1) − 2 λ
|ϕ|p+1 dx < 0,
we see λ12 K(Tλ ϕ) is monotone decreasing with respect to λ > 0. Since 2(d+2) d p+1 2 2 d d K(Tλ ϕ) = λ ∇ϕ L2 + d+2 ϕ 2(d+2) − λ 2 (p−1) d(p−1) 2(p+1) ϕ Lp+1 L
d
→ −∞, as λ → ∞, there exists a unique λ0 > 0 such that K(Tλ0 ϕ) = 0 and (2.1) follows.
2
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Now, we show the positivity of K near 0 in the energy space. Lemma 2.2. For any bounded sequence ϕn ∈ H 1 (Rd ) \ {0} with KQ (ϕn ) → 0, as n → ∞, then for n large enough, we have K(ϕn ) > 0. Proof. By assumption, ∇ϕn L2 → 0, as n → ∞. Due to the interpolation and Sobolev inequalities, we have p+1− d2 (p−1)
p+1
ϕn Lp+1 ϕn L2
d
Since d2 (p − 1) > 2, for n large enough, we see p+1 K(ϕn ) = |∇ϕn |2 − d(p−1) + 2(p+1) |ϕn | Rd
≥
|∇ϕn | dx − o 2
Rd
.
2(d+2) d d d+2 |ϕn |
|∇ϕn |2 dx
dx
Rd
|∇ϕn |2 dx > 0.
∼
(p−1)
∇ϕn L2 2
2
Rd
Lemma 2.3. For any ϕ ∈ H 1 (Rd ), we have (2 − L)Sω (ϕ) = ω ϕ 2L2 + L(2 − L)Sω (ϕ) =
p+1 d(p−1)−4 2(p+1) ϕ Lp+1 ,
p+1 d(p−1)(d(p−1)−4) ϕ Lp+1 . 4(p+1)
Proof. Direct computation shows that L ∇ϕ 2L2 = 2 ∇ϕ 2L2 , p+1
p+1
L ϕ Lp+1 = d2 (p − 1) ϕ Lp+1 , L ϕ
2(d+2) d 2(d+2) L d
= 2 ϕ
2(d+2) d 2(d+2) L d
,
hence we have (2 − L)Sω (ϕ) = 2Sω (ϕ) − K(ϕ) = ω ϕ 2L2 +
d(p−1)−4 2(p+1)
|ϕ|p+1 dx, Rd
L(2 − L)Sω (ϕ) = ωL ϕ 2L2 +
p+1 d(p−1)−4 2(p+1) L ϕ Lp+1
=
p+1 d(p−1)−4 2(p+1) L ϕ Lp+1
=
p+1 (d(p−1)−4)d(p−1) ϕ Lp+1 . 4(p+1)
2
(2.2) (2.3)
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Due to the lack of positivity of Sω (ϕ), we introduce a non-negative functional Hω (ϕ) = 1 − L2 Sω (ϕ) = Sω (ϕ) − 12 K(ϕ) = ω2 ϕ 2L2 +
(2.4)
p+1 d(p−1)−4 4(p+1) ϕ Lp+1 ,
then for any ϕ ∈ H 1 (Rd ) \ {0}, we have Hω (ϕ) ≥ 0, LHω (ϕ) ≥ 0. Proposition 2.4. mω = inf{Hω (ϕ) : ϕ ∈ H 1 \ {0}, K(ϕ) ≤ 0} = inf{Hω (ϕ) : ϕ ∈ H 1 \ {0}, K(ϕ) < 0}.
(2.5)
Proof. Since Sω (ϕ) = Hω (ϕ) when K(ϕ) = 0, mω ≥ inf{Hω (ϕ) : ϕ ∈ H 1 \ {0}, K(ϕ) ≤ 0}. Claim: mω ≤ inf{Hω (ϕ) : ϕ ∈ H 1 \ {0}, K(ϕ) < 0}. In fact, ∀ ϕ ∈ H 1 \ {0} with K(ϕ) < 0, by Lemma 2.1, there exists 0 < λ0 < 1 that K(Tλ0 ϕ) = 0, then the fact LHω ≥ 0 implies Sω (Tλ0 ϕ) = Hω (Tλ0 ϕ) ≤ Hω (ϕ). It suffices to show inf{Hω (ϕ) : ϕ ∈ H 1 \ {0}, K(ϕ) ≤ 0} ≥ inf{Hω (ϕ) : ϕ ∈ H 1 \ {0}, K(ϕ) < 0}.
(2.6)
In fact, for any ϕ ∈ H 1 \ {0} with K(ϕ) ≤ 0, by (2.3), we know that L K(ϕ) = 2K(ϕ) −
p+1 d(p−1)(d(p−1)−4) ϕ Lp+1 4(p+1)
< 0,
(2.7)
then for any λ > 1, we have K(Tλ ϕ) < 0, and as λ → 1, Hω (Tλ ϕ) = ω2 Tλ ϕ 2L2 + = ω2 ϕ 2L2 +
p+1 d(p−1)−4 d2 (p−1) ϕ Lp+1 4(p+1) λ
→ ω2 ϕ 2L2 + This shows (2.6) and completes the proof.
p+1 d(p−1)−4 4(p+1) Tλ ϕ Lp+1
p+1 d(p−1)−4 4(p+1) ϕ Lp+1
= Hω (ϕ).
2
4 We now give the value of mω for 1 + d4 < p < 1 + d−2 , d ≥ 3, and 1 + d4 < p < ∞, d = 1, 2, namely prove Proposition 1.1.
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Proof of Proposition 1.1. Let ϕn ∈ H 1 (Rd ) be a minimizing sequence for (2.5), namely K(ϕn ) ≤ 0, ϕn = 0, Hω (ϕn ) mω , as n → ∞. Let ϕn∗ be the Schwartz symmetrization of ϕn , i.e. the radial decreasing rearrangement. Since the symmetrization preserves the nonlinear parts and does not increase the H˙ 1 part, we have ϕn∗ = 0, K(ϕn∗ ) ≤ K(ϕn ) ≤ 0 and Hω (ϕn∗ ) = Hω (ϕn ) → mω , as n → ∞. Then by Lemma 2.1 and (2.7), there exists 0 < λn ≤ 1 such that ψn = Tλn ϕn∗ satisfies ψn = 0, K(ψn ) = 0, Sω (ψn ) = Hω (ψn ) → mω , as n → ∞. Moreover, direct computation gives 2 ω 2 ψn L2
+ 12 ∇ψn 2L2 ≤ Sω (ψn ) +
4 d(p−1)−4 Hω (ψn ),
which implies the boundedness of ψn in H 1 (Rd ). Then, ψn converges weakly to some ψ in H 1 (Rd ), up to a subsequence. Since ψn is radial, 2d it also converges strongly in Lq (Rd ) for 2 < q < d−2 , d ≥ 3 and 2 < q < ∞, d = 1, 2. Thus, K(ψ) ≤ lim inf K(ψn ) = 0, Hω (ψ) ≤ lim inf Hω (ψn ) = mω . Moreover, ψ = 0. In fact, if ψ = 0, n→∞
n→∞
then K(ψn ) = 0 implies K Q (ψn ) = −K N (ψn ) → 0, as n → ∞, and by Lemma 2.2, we have K(ψn ) > 0 for n large, a contradiction. Since K(ψ) ≤ 0 and ψ = 0, we have Hω (ψ) ≥ mω , so we have Hω (ψ) = mω and K(ψ) ≤ 0. By scaling, we may replace ψ by its rescaling, so that K(ψ) = 0, Sω (ψ) = Hω (ψ) ≤ mω and ψ = 0. Then ψ is a minimizer and mω = Hω (ψ) > 0. By variational theory, there is a Lagrange multiplier η ∈ R such that Sω (ψ) = ηK (ψ). We note 0 = K(ψ) = LSω (ψ) = Sω (ψ)Lψ = ηK (ψ)Lψ = η · L2 Sω (ψ). By (2.3) and LSω (ψ) = 0, we have L2 Sω (ψ) = 2LSω (ψ) −
p+1 d(p−1)(d(p−1)−4) ψ Lp+1 4(p+1) p+1
= − d(p−1)(d(p−1)−4) ψ Lp+1 < 0, 4(p+1) 4
therefore η = 0 and ψ is a solution to −ωQ + Q + |Q|p−1 Q − |Q| d Q = 0. The minimality of Sω (Q) among the solutions is clear from (1.7), since every solution Q in H 1 (Rd ) of (1.8) satisfies K(Q) = Sω (Q)LQ = 0. 2
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4 We now turn to find the ground state in the case p = 1 + d−2 , d ≥ 3, that is to prove Proposition 1.2. Let 2d K0 (ϕ) = |∇ϕ|2 − |ϕ| d−2 dx, (2.8) Rd 2d
H0 (ϕ) = d1 ϕ d−22d .
(2.9)
L d−2
We will show Lemma 2.5. For p = 1 +
4 d−2 ,
d ≥ 3,
mω = inf{H 0 (ϕ) : ϕ ∈ H 1 \ {0}, K0 (ϕ) < 0}
(2.10)
= inf{H 0 (ϕ) : ϕ ∈ H 1 \ {0}, K0 (ϕ) ≤ 0}. Proof. Since K0 (ϕ) ≤ K(ϕ), H0 (ϕ) ≤ Hω (ϕ), it follows that
mω = inf{Hω (ϕ) : ϕ ∈ H 1 \ {0}, K(ϕ) < 0} ≥ inf{H0 (ϕ) : ϕ ∈ H 1 \ {0}, K0 (ϕ) < 0}. Hence, in order to show the first equality, it suffices to show inf{Hω (ϕ) : ϕ ∈ H 1 \ {0}, K(ϕ) < 0} ≤ inf{H0 (ϕ) : ϕ ∈ H 1 \ {0}, K0 (ϕ) < 0}. For any ϕ ∈ H 1 \ {0} with K0 (ϕ) < 0, taking Tλ ϕ(x) = λ K(Tλ ϕ) =
2d
|∇ϕ|2 − |ϕ| d−2 +
d−2 2
(2.11)
ϕ(λx), we have as λ → ∞,
2(d+2) − d4 d |ϕ| d d+2 λ
dx → K0 (ϕ),
Rd 2d
Hω (Tλ ϕ) = ω2 λ−2 ϕ 2L2 + d1 ϕ d−22d → H0 (ϕ). L d−2
This gives (2.11) and completes the proof of the first equality. For the second equality, it suffices to show inf{H0 (ϕ) : ϕ ∈ H 1 \ {0}, K0 (ϕ) < 0} ≤ inf{H0 (ϕ) : ϕ ∈ H 1 \ {0}, K0 (ϕ) ≤ 0}. For any ϕ ∈ H 1 \ {0} with K0 (ϕ) ≤ 0 and L K0 (ϕ) =
2|∇ϕ|2 −
2d 2d d−2 d−2 |ϕ|
dx = 2K0 (ϕ) −
which implies K0 (Tλ ϕ) < 0, for λ > 1. We also have
2d
d−2 4 d−2 ϕ 2d L d−2
< 0,
(2.12)
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934 2d
2d
H0 (Tλ ϕ) = d1 λ d−2 ϕ d−22d → H0 (ϕ), as λ → 1, L d−2
so we obtain (2.12) and complete the proof.
2
Proof of Proposition 1.2. By Lemma 2.5, we have 2d 2d d−2 d−2 1 2 1 mω = inf d ϕ 2d : ϕ ∈ H \ {0}, ∇ϕ L2 ≤ ϕ 2d L d−2
≥ inf
≥ inf
2d
2 1 d ∇ϕ L2
≥ inf
L d−2
L d−2
⎧ ⎪ ⎪ ⎨
⎛
1 ∇ϕ 2L2 ⎪d
⎪ ⎩
⎧ ⎨ 1
⎩d
: ϕ ∈ H 1 \ {0}, ∇ϕ 2L2 ≤ ϕ d−22d ⎜ ⎝
∇ϕ L2 ϕ 2d
⎞ d−2 2
∇ϕ 2 2
L 2d d−2 ϕ 2d L d−2
⎟ ⎠
: ϕ ∈ H 1 \ {0}
d : ϕ ∈ H˙ 1 \ {0}
L d−2
⎫ ⎬ ⎭
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
= d1 (Cd∗ )−d ,
where Cd∗ is the sharp Sobolev constant in Rd , that is ϕ
2d
L d−2 (Rd )
≤ Cd∗ ∇ϕ L2 , ∀ ϕ ∈ H˙ 1 (Rd ),
(2.13)
with the equality is attained by W (see [3,32]) and d1 (Cd∗ )−d = E 0 (W ). On the other hand, by the density of H 1 (Rd ) in H˙ 1 (Rd ), we can find ϕn ∈ H 1 \ {0}, such that ϕn → W in H˙ 1 (Rd ), as n → ∞. Then we have H0 (ϕn ) → H0 (W ) = E 0 (W ), as n → ∞, by Lemma 2.5, H0 (ϕn ) ≥ mω for n large enough. So we have mω ≤ E 0 (W ). Thus, we obtain (1.9). 2 Next we show the energy-trapping properties of Aω,± . Lemma 2.6. For 1 + d4 < p ≤ 1 + ϕ ∈ H 1 (Rd ) with K(ϕ) ≥ 0, d(p−1)−4 d(p−1)
2 1 2 |∇ϕ|
+
d ≥ 3, and 1 +
4 d−2 ,
2(d+2) d d 2(d+2) |ϕ|
4 d
< p < ∞, d = 1, 2, we have for any
dx ≤ E(ϕ) ≤
2 1 2 |∇ϕ|
+
2(d+2) d d 2(d+2) |ϕ|
Proof. On the one hand, E(ϕ) =
−
p+1 1 p+1 |ϕ|
2 1 2 |∇ϕ|
+
2(d+2) d d 2(d+2) |ϕ|
≤ On the other hand, since
+
2(d+2) d d 2(d+2) |ϕ|
2 1 2 |∇ϕ|
dx.
dx
dx.
(2.14)
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
K(ϕ) =
|∇ϕ|2 −
d(p−1) p+1 2(p+1) |ϕ|
+
2(d+2) d d d+2 |ϕ|
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dx ≥ 0,
we have
|ϕ|
1 p+1
p+1
dx ≤
2 d(p−1)
|∇ϕ|2 +
2(d+2) d d d+2 |ϕ|
dx.
So E(ϕ) =
−
p+1 1 p+1 |ϕ|
2 1 2 |∇ϕ|
+
2(d+2) d d 2(d+2) |ϕ|
≥ =
1 2
−
2 d(p−1)
|∇ϕ|2 +
=
d(p−1)−4 d(p−1)
+
2(d+2) d d 2(d+2) |ϕ|
2 1 2 |∇ϕ|
2 1 2 |∇ϕ|
+
dx 2 |∇ϕ|2 + dx − d(p−1)
d 2(d+2)
−
2 (d+2)(p−1)
2(d+2) d d 2(d+2) |ϕ|
|ϕ|
2(d+2) d d d+2 |ϕ| 2(d+2) d
dx
dx
dx. 2
4 Proposition 2.7 (Energy-trapping for Aω,− ). For 1 + d4 < p ≤ 1 + d−2 , d ≥ 3, 1 + d4 < p < ∞, d = 1, 2, and u0 ∈ Aω,− . Let u be the solution of (1.1), and Imax be the lifespan of u, then
K(u(t)) < − mω − Sω (u(t)) , ∀ t ∈ Imax .
(2.15)
Proof. We first claim that K(u(t)) < 0, for t ∈ Imax . Indeed, since u0 ∈ Aω,− , we have by the mass and energy conservation that Sω (u(t)) < mω , for t ∈ Imax . If K(u(t0 )) ≥ 0 for some t0 ∈ Imax , then there is t1 ∈ Imax , such that K(u(t1 )) = 0. So we have Sω (u(t1 )) ≥ mω , which contradicts Sω (u(t)) < mω , for all t ∈ Imax , so we have K(u(t)) < 0 for t ∈ Imax . Next, we turn to (2.15). By the above claim, for any t ∈ Imax , there exists 0 < λ(t) < 1 such that K(Tλ(t) u(t)) = 0, which together with the definition of mω shows that Sω (Tλ(t) u(t)) ≥ mω . By Lemma 2.3, L2 Sω (u(t)) = 2L Sω (u(t)) − = 2K(u(t)) −
p+1 d(p−1)(d(p−1)−4) u(t) Lp+1 4(p+1)
p+1 d(p−1)(d(p−1)−4) u(t) Lp+1 4(p+1)
< 0,
we have d Sω (u(t)) > Sω (Tλ(t) u(t)) + (1 − λ(t)) dλ S (T u(t)) λ=1 ω λ = Sω (Tλ(t) u(t)) + (1 − λ(t))K(u(t)) > mω + K(u(t)), so K(u(t)) < −(mω − Sω (u(t))).
2
Before discussing the energy-trapping for Aω,+ , we first show ∀ ω > 0, Aω,+ is bounded in H 1 (Rd ).
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
Lemma 2.8. Let ω > 0 and u ∈ Aω,+ , then we have u 2L2 ≤
2mω ω ,
∇u 2L2 mω ,
and hence u 2H 1 mω +
mω ω .
(2.16)
Proof. By u ∈ Aω,+ , we have Hω (u) ≤ Sω (u) ≤ mω , which implies 2mω ω .
u 2L2 ≤
The boundedness of ∇u L2 follows from 1 mω ≥ Sω (u) ≥ E(u) = 12 |∇u|2 − p+1 |u|p+1 + 2 1 2 |∇u|
≥ ≥
−
1 2
+
2 d(p−1)
2(d+2) d d 2(d+2) |u|
dx −
2(d+2) d d 2(d+2) |u|
2 d(p−1)
dx
∇u 2L2 +
d d+2 u
2(d+2) d 2(d+2) L d
∇u 2L2 ,
where the first inequality is given by K(u) ≥ 0.
2
Proposition 2.9 (Energy-trapping for Aω,+ ). For u0 ∈ Aω,+ , let u be a solution of (1.1), with Imax the lifespan, we have δ > 0 depending on d, p and ω such that for t ∈ Imax , K(u(t)) ≥ min
d(p−1)−4 d(p−1)
∇u(t) 2L2
+
d d+2 u(t)
2(d+2) d 2(d+2) L d
, δ mω − Sω (u(t)) .
(2.17)
Proof. Direct computation shows that Sω (Tλ u(t)) = ω2 Tλ u(t) 2L2 + E(Tλ u(t)) 2 ω = 2 u(t) L2 + 12 λ2 |∇u(t)|2 + and d2 S (T u(t)) = dλ2 ω λ
Since
|∇u(t)|2 +
2(d+2) d d d+2 |u(t)|
K(Tλ u(t)) =
2(d+2) 2 d d 2(d+2) λ |u(t)|
λ2 |∇u(t)|2 −
−
−
d 1 2 (p−1) |u(t)|p+1 dx p+1 λ
d(p−1)(d(p−1)−2) d2 (p−1)−2 λ |u(t)|p+1 dx. 4(p+1)
d(p−1) d2 (p−1) |u(t)|p+1 2(p+1) λ
+
2(d+2) 2 d d d+2 λ |u(t)|
dx,
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
2895
we have d2 S (T u(t)) = − λ12 K(Tλ u(t)) + λ22 dλ2 ω λ
K(Tλ u(t)) +
(4−d(p−1))d(p−1) 8(p+1)
|Tλ u(t)|
p+1
dx . (2.18)
Case I. K(u(t)) −
(d(p−1)−4)d(p−1) 8(p+1)
|u(t)|p+1 dx ≥ 0, for t ∈ Imax .
(2.19)
In this case, K(u(t)) =
|∇u|2 −
d(p−1) p+1 2(p+1) |u|
|∇u|2 +
2(d+2) d d d+2 |u|
≥
+
2(d+2) d d d+2 |u|
dx −
dx
4 d(p−1)−4 K(u(t)),
thus K(u(t)) ≥
d(p−1)−4 d(p−1)
|∇u(t)|2 +
2(d+2) d d d+2 |u(t)|
dx.
Case II. K(u(t)) −
(d(p−1)−4)d(p−1) 8(p+1)
|u(t)|p+1 dx < 0, for t ∈ Imax .
(2.20)
In this case, ∇u(t) 2L2 <
d 2 (p−1)2 8(p+1)
|u(t)|p+1 dx −
d d+2
|u(t)|
2(d+2) d
dx,
which implies u(t) = 0, ∀ t ∈ Imax , and p+1 d 2 (p−1)2 8(p+1) u(t) Lp+1
≥ ∇u(t) 2L2 .
(2.21)
Since u0 ∈ Aω,+ and u(t) = 0, ∀ t ∈ Imax , together with the definition of mω , we have K(u(t)) > 0 by similar argument as in the claim in Proposition 2.7. Therefore, by Lemma 2.1, there is λ(t) > 1 such that K(Tλ(t) u(t)) = 0
(2.22)
K(Tλ u(t)) > 0 for 1 ≤ λ < λ(t).
(2.23)
and
By (2.22), we have
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934 d p+1 d(p−1) 2 (p−1)−2 u(t) 2(p+1) λ(t) Lp+1
∇u(t) 2L2 −
+
d d+2 u(t)
2(d+2) d 2(d+2) L d
= 0,
so by Lemma 2.8 together with the interpolation and Sobolev inequalities, we have λ(t)
d 2 (p−1)−2
p+1 u(t) Lp+1
=
2(p+1) d(p−1)
≤
2(p+1) d(p−1)
≤
2(p+1) d(p−1)
∇u(t) 2L2 +
d d+2 u(t)
∇u(t) 2L2 1 + Cd
2(d+2) d 2(d+2) L d
4 d
+ Cd u(t) L2 ∇u(t) 2L2
2mω ω
2 d
∇u(t) 2L2 ,
where Cd is some constant depending on d related to the Sobolev inequality. Thus, we see by (2.21), 2(p+1) d(p−1)
1 + Cd
2mω ω
2 d
d
p+1
λ(t)2 ∇u(t) 2L2 ≥ λ(t) 2 (p−1) u(t) Lp+1 ≥
d 8(p+1) λ(t) 2 (p−1) ∇u(t) 2L2 , d 2 (p−1)2
which yields λ(t) ≤
d 4 (p − 1)
1 + Cd
2mω ω
1 2 d (p−1)−2 d
2
(2.24)
.
Because d dλ
1 K(Tλ u) λ2
d dλ
1 λ2
= − d(p−1) 2(p+1)
d
d (p−1)−3 2 2 (p − 1) − 2 λ
|u|p+1 dx ≤ 0,
|Tλ u|p+1 dx = d2 (p − 1) − 2 |u|p+1 dx ≥ 0,
we have d dλ
1 λ2
K(Tλ u) +
(4−d(p−1))d(p−1) 8(p+1)
|Tλ u|p+1 dx
≤ 0.
(2.25)
Collecting (2.20) and (2.25), we have 1 λ2
K(Tλ u(t)) +
(4−d(p−1))d(p−1) 8(p+1)
|Tλ u(t)|
Hence, (2.18), (2.23) and (2.26) shows for 1 ≤ λ ≤ λ(t),
p+1
dx < 0, for λ ≥ 1.
(2.26)
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934 d2 S (T u(t)) dλ2 ω λ
=−
1 K(Tλ u(t)) + λ22 λ2
<−
1 K(Tλ u(t)) ≤ 0. λ2
2897
K(Tλ u(t)) +
(4−d(p−1))d(p−1) 8(p+1)
|Tλ u(t)|
p+1
dx (2.27)
Combining (2.24) and (2.27), we obtain ⎞ ⎛ 1 2 d (p−1)−2 2 ⎝ d (p − 1) 1 + Cd 2mω d − 1⎠ K(u(t)) 4 ω d ≥ (λ(t) − 1) dλ S (T u(t)) λ=1 ω λ ≥ Sω (Tλ(t) u(t)) − Sω (u(t)) ≥ mω − Sω (u(t)). Thus, there is δ > 0 depending on d, p and ω that K(u(t)) ≥ min
d(p−1)−4 d(p−1)
∇u(t) 2L2
+
d d+2 u(t)
2(d+2) d 2(d+2) L d
, δ mω − Sω (u(t)) .
2
3. Wellposedness and perturbation theory In this section, we present the local wellposedness theory and the perturbation theory for (1.1). We start by recording the wellposedness theory. For the proof we refer to [9,10,23]. Proposition 3.1. (i) (Local existence) Let φ ∈ H 1 (Rd ), I be an interval, t0 ∈ I and A > 0. Assume that φ H 1 ≤ A, and there exists δ > 0 depending on A that ∇ ei(t−t0 ) φ
2(d+2)
Lt,x d
(I ×Rd )
≤ δ,
then there exists a unique solution u ∈ C(I, H 1 (Rd )) to (1.1) such that u(t0 ) =φ, u S 1 (I ) φ H 1 , ∇ u 2(d+2) ≤2 ∇ ei(t−t0 ) φ Lt,x d
(I ×Rd )
2(d+2)
Lt,x d
(I ×Rd )
.
As a consequence, we have the small data global existence: if φ H 1 is sufficiently small, then u is a global solution with u S 1 (R) φ H 1 .
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
(ii) (Unconditional uniqueness) Suppose u1 , u2 ∈ C(I, H 1 (Rd )) are two solutions of (1.1) with u1 (t0 ) = u2 (t0 ) for some t0 ∈ I , then u1 = u2 . Let u ∈ C((Tmin , Tmax ), H 1 (Rd )) be the maximal-lifespan solution to (1.1), then we have (iii) (Conservation laws) For any t, t0 ∈ (Tmin , Tmax ), M(u(t)) = M(u(t0 )), E(u(t)) = E(u(t0 )), Sω (u(t)) = Sω (u(t0 )), for any ω > 0, P(u(t)) = ∇u(t, x)u(t, x) dx = P(u(t0 )). Rd
(iv) (Blow-up criterion) If Tmax < ∞, then u
(d+2)(p−1) 2 ((T , Tmax )×Rd )
2(d+2)
Lt,x d
∩Lt,x
= ∞, ∀ Tmin < T < Tmax .
A similar result holds, if Tmin > −∞. (v) (Scattering) If u
(d+2)(p−1) 2 ((Tmin ,Tmax )×Rd )
2(d+2)
Lt,x d
∩Lt,x
< ∞,
(3.1)
then Tmax = ∞, Tmin = −∞, and there exist u± ∈ H 1 (Rd ) such that lim u(t) − eit u+ 1 = lim u(t) − eit u− 1 = 0. t→∞
t→−∞
H
H
(3.2)
In the following, we will give the long-time perturbation theory when d ≤ 4. Proposition 3.2 (Long-time perturbation). Let I be a compact time interval and let w be an approximate solution to (1.1) on I × Rd in the sense that 4
i∂t w + w = |w| d w − |w|p−1 w + e for some function e. Assume that
w
w L∞ 1 d ≤ A1 , t Hx (I ×R )
(3.3)
≤B
(3.4)
2(d+2)
Lt,x d
(d+2)(p−1) 2 (I ×Rd )
∩Lt,x
for some A1 , B > 0. Let t0 ∈ I and u(t0 ) close to w(t0 ) in the sense that u(t0 ) − w(t0 ) Hx1 (Rd ) ≤ A2 for some A2 > 0.
(3.5)
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
Assume also the smallness conditions ∇ ei(t−t0 ) (u(t0 ) − w(t0 ))
2(d+2)
Lt,x d
∇ e
(I ×Rd )
2899
≤ δ,
(3.6)
≤δ
(3.7)
2(d+2)
Lt,xd+4 (I ×Rd )
for some 0 < δ ≤ δ1 , where δ1 = δ1 (A1 , A2 , B) is a small constant. Then there exists a solution u to (1.1) on I × Rd with the specified initial data u(t0 ) at time t = t0 that satisfies u − w
≤C(A1 , A2 , B)δ α ,
(3.8)
u − w S 1 (I ) ≤C(A1 , A2 , B)A2 ,
(3.9)
(d+2)(p−1) 2 (I ×Rd )
2(d+2)
Lt,x d
∩Lt,x
u S 1 (I ) ≤C(A1 , A2 , B), where 0 < α <
(3.10)
4 d(p−1) .
To show the long-time perturbation theory, we will first give the following short-time perturbation theory. Lemma 3.3 (Short-time perturbation). Let I be a compact time interval and let w be an approximate solution to (1.1) on I × Rd in the sense that 4
i∂t w + w = |w| d w − |w|p−1 w + e for some function e. Suppose we also have the energy bound w L∞ 1 d ≤ A1 t Hx (I ×R )
(3.11)
for some constant A1 > 0. Let t0 ∈ I and let u(t0 ) ∈ H 1 (Rd ) be close to w(t0 ) in the sense that u(t0 ) − w(t0 ) Hx1 ≤ A2
(3.12)
for some A2 > 0. Moreover, assume the smallness conditions ∇ w
2(d+2)
Lt,x d
(I ×Rd )
+ w
(d+2)(p−1) 2 (I ×Rd )
≤ δ0 ,
(3.13)
≤ δ,
(3.14)
≤δ
(3.15)
Lt,x
∇ ei(t−t0 ) (u(t0 ) − w(t0 ))
2(d+2)
Lt,x d
∇ e
(I ×Rd )
2(d+2)
Lt,xd+4 (I ×Rd )
for some 0 < δ ≤ δ0 , where δ0 = δ0 (A1 , A2 ) > 0 is a small constant.
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
Then, there exists a solution u ∈ S 1 (I ) to (1.1) on I × Rd with the specified initial data u(t0 ) at time t = t0 that satisfies u − w
δα ,
(3.16)
u − w S 1 (I ) A2 + δ α ,
(3.17)
u S 1 (I ) A1 + A2 ,
(3.18)
(d+2)(p−1) 2(d+2) Lt,x d ∩Lt,x 2 (I ×Rd )
∇ ((i∂t + )(u − w) + e) where 0 < α <
δ , α
2(d+2)
Lt,xd+4 (I ×Rd )
(3.19)
4 d(p−1) .
Proof. By the wellposedness theory, it suffices to prove (3.16)–(3.19) as a priori estimate, that is, we assume that the solution u already exists and belongs to S 1 (I ). By time symmetry, we may assume t0 = inf I . Let v = u − w, then v satisfies 4
4
i∂t v + v = |w + v| d (w + v) − |w + v|p−1 (w + v) − |w| d w + |w|p−1 w − e, and v(t0 ) = u(t0 ) − w(t0 ). For T ∈ I , define S(T ) = ∇ ((i∂t + )v + e)
.
2(d+2)
Lt,xd+4 ([t0 ,T ]×Rd )
We will now work entirely on the slab [t0 , T ] × Rd . ∇ v
2(d+2)
Lt,x d
+ v
≤ ∇ ei(t−t0 ) v(t0 )
(d+2)(p−1) 2
Lt,x
2(d+2) Lt,x d
+ ∇ ((i∂t + )v + e) S(T ) + δ α , (0 < α < where we use the fact i(t−t0 ) v(t0 ) e
+ ei(t−t0 ) v(t0 )
(d+2)(p−1) 2
Lt,x
+ ∇ e
2(d+2)
Lt,xd+4
2(d+2)
Lt,xd+4
4 d(p−1)
< 1)
(3.20)
(d+2)(p−1) 2
Lt,x
∇ ei(t−t0 ) v(t0 )
2d(d+2)(p−1) (d+2)(p−1) d(d+2)(p−1)−8 2 Lx
Lt
4 1− 4 4 1− 4 d(p−1) d(p−1) ∇ ei(t−t0 ) v(t0 ) 2(d+2) ∇ ei(t−t0 ) v(t0 ) ∞ 2 δ d(p−1) A2 d(p−1) ≤ δ α . Lt,x d
Lt Lx
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
2901
On the other hand, since 4
4
4
|∇((i∂t + )v + e)| |∇w||v| d + |∇v||w + v| d + |∇w||v||w| d −1 + |∇w||v|p−1 + |∇v||w + v|p−1 + |∇w||v||w|p−2 and 4
4
|(i∂t + )v + e| |w||v| d + |w + v| d |v| + |w||v|p−1 + |w + v|p−1 |v|, we have 4
S(T ) ∇ w
2(d+2) Lt,x d
4
(v, w) d 2(d+2) + ∇ v
2(d+2) Lt,x d
Lt,x d
+ ∇ w
2(d+2) Lt,x d
p−1
(v, w)
(d+2)(p−1) Lt,x 2
(v, w) d 2(d+2) Lt,x d
+ ∇ v
2(d+2) Lt,x d
(v, w)
p−1 (d+2)(p−1) 2
Lt,x
4 4 p−1 δ0 (S(T ) + δ α ) d + δ0d + (S(T ) + δ α )p−1 + δ0 4
p−1
+ (δ0d + δ0
4
)(S(T ) + δ α ) + (S(T ) + δ α )1+ d + (S(T ) + δ α )p .
By the continuity argument, we can take δ0 = δ0 (A1 , A2 ) sufficiently small, then S(T ) δ α , ∀ T ∈ I,
(3.21)
which implies (3.19). We also have u − w
(d+2)(p−1) 2
2(d+2)
Lt,x d
∩Lt,x
ei(t−t0 ) v(t0 )
(d+2)(p−1) 2(d+2) Lt,x d ∩Lt,x 2
+ ∇ ((i∂t + )v + e)
2(d+2) Lt,xd+4
+ ∇ e
2(d+2) Lt,xd+4
δα ,
which is (3.16). To obtain (3.17), we see u − w S 1 (I ) u(t0 ) − w(t0 ) H 1 + ∇ ((i∂t + )v + e)
2(d+2)
Lt,xd+4
+ ∇ e
2(d+2)
Lt,xd+4
A2 + S(t) + δ A2 + δ α . We now show (3.18). Using Strichartz estimate, (3.11) and (3.13), we get 4 w S 1 (I ) w(t0 ) H 1 + ∇ (|w| d w − |w|p−1 w) 2(d+2) + ∇ e Lt,xd+4
4
A1 + w d 2(d+2) ∇ w Lt,x d 4
p−1
A1 + (δ0d + δ0
2(d+2) Lt,x d
) w S 1 (I ) + δ.
p−1
+ w
(d+2)(p−1) 2
Lt,x
2(d+2)
Lt,xd+4 (I ×Rd )
∇ w
2(d+2)
Lt,x d
+δ
2902
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
By the continuity argument, we have w S 1 (I ) A1 , provided δ0 is sufficiently small depending on A1 . This together with (3.20), (3.21) and u = v +w yields ∇ u
≤ ∇ v
2(d+2)
Lt,x d
+ ∇ w
2(d+2)
Lt,x d
2(d+2)
Lt,x d
A1 .
While, we have by (3.16), (3.20) and (3.21) u
2(d+2)
Lt,x d
(d+2)(p−1) 2
∩Lt,x
≤ v
(d+2)(p−1) 2
2(d+2)
∩Lt,x
Lt,x d
+ w
2(d+2)
Lt,x d
δα
(d+2)(p−1) 2
∩Lt,x
Combining these with Strichartz estimate, we obtain 4 u S 1 (I ) u(t0 ) H 1 + ∇ (|u| d u − |u|p−1 u) 4
A1 + A2 + u d 2(d+2) ∇ u Lt,x d
2(d+2) Lt,x d
2(d+2)
Lt,xd+4
+ u
p−1 (d+2)(p−1) 2
∇ u
Lt,x
2(d+2)
Lt,x d
4
A1 + A2 + δ d α A1 + δ (p−1)α A1 A1 + A2 , which proves (3.18), provided δ0 is sufficiently small depending on A1 and A2 .
2
We now show the long-time perturbation theory. Proof of Proposition 3.2. We will derive Proposition 3.2 from Lemma 3.3 by an iterative procedure. First, we will assume without loss of generality that t0 = inf I . Let δ0 = δ0 (A1 , 2A2 ) be as in Lemma 3.3. The first step is to establish an S 1 bound on w. In order to do so, we subdivide I into N0 ∼ d 2(d+2) 1+ M subintervals Ik such that 0 w
(d+2)(p−1) 2 (I
2(d+2)
Lt,x d
∩Lt,x
k ×R
d)
On each subinterval Ik , we have 4 d w − |w|p−1 w) w S 1 (Ik ) w L∞ + ∇ (|w| 1 t Hx 4
A1 + w d 2(d+2) ∇ w Lt,x 4
d
p−1
A1 + (δ0d + δ0
2(d+2) Lt,x d
2(d+2)
Lt,xd+4
+ ∇ e
p−1
+ w
) w S 1 (Ik ) + δ.
∼ δ0 .
(d+2)(p−1) 2
Lt,x
2(d+2)
Lt,xd+4 (Ik ×Rd )
∇ w
2(d+2)
Lt,x d
+δ
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
2903
By the continuity argument, we have w S 1 (Ik ) A2 , provided δ0 is sufficiently small depending on A1 . Summing these bounds over all the intervals Ik , we obtain w S 1 (I ) ≤ C(A1 , B, δ0 ), which implies ∇ w
2(d+2)
(I ×Rd )
Lt,x d
+ w
(d+2)(p−1) 2 (I ×Rd )
≤ C(A1 , B, δ0 ).
Lt,x
This allows us to subdivide I into N1 = C(B, δ0 ) subintervals Jk = [tk , tk+1 ] such that ∇ w
2(d+2)
(Jk ×Rd )
Lt,x d
+ w
(d+2)(p−1) 2 (J
Lt,x
k ×R
d)
≤ δ0 .
Choosing δ1 = δ1 (N1 , A1 , A2 ) sufficiently small, we apply Lemma 3.3 to obtain for each k, and all 0 < δ < δ1 , u − w
2(d+2)
Lt,x d
(d+2)(p−1) 2 (J
∩Lt,x
k ×R
d)
≤ C(k)δ α ,
u − w S 1 (Jk ) ≤ C(k)(A2 + δ α ), u S 1 (Jk ) ≤ C(k)(A1 + A2 ), ∇ ((i∂t + )(u − w) + e)
2(d+2)
Lt,xd+4 (Jk ×Rd )
≤ C(k)δ α ,
provided we can show (3.5), (3.6) hold with t0 replaced by tk . We verify this using an inductive argument. We have u(tk+1 ) − w(tk+1 ) H 1 u(t0 ) − w(t0 ) H 1 + ∇ e
2(d+2)
Lt,xd+4 ([t0 ,tk+1 ]×Rd )
+ ∇ ((i∂t + )(u − w) + e)
2(d+2)
Lt,xd+4 ([t0 ,tk+1 ]×Rd )
k
A2 + δ +
C(j )δ α , j =0
and
∇ ei(t−tk+1 ) (u(tk+1 ) − w(tk+1 )) ∇ ei(t−t0 ) (u(t0 ) − w(t0 ))
2(d+2)
Lt,x d
+ ∇ ((i∂t + )(u − w) + e) k
δ+
C(j )δ α . j =0
Here, C(j ) depends only on j , A1 , A2 , δ0 .
2(d+2)
Lt,x d
(I ×Rd )
2(d+2)
Lt,xd+4
+ ∇ e
2(d+2)
Lt,xd+4
2904
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
Choosing δ1 sufficiently small depending on N1 , A1 , A2 , we can continue the inductive argument. This concludes the proof of Proposition 3.2. 2 4. Linear profile decomposition The linear profile decomposition was first established by H. Bahouri and P. Gérard [4] for the energy critical wave equation in H˙ 1 (Rd ). Later, S. Keraani [19] proved the linear profile decomposition for the energy critical Schrödinger equation in H˙ 1 (Rd ). At almost the same time, F. Merle and L. Vega [26] gave the linear profile decomposition for the mass critical Schrödinger equation in L2 (Rd ), then R. Carles and S. Keraani [8] established this in L2 (R). In [20], S. Keraani use the L2 profile decomposition to describe the minimal mass blowup solution. Later, P. Bégout and A. Vargas [6] extend these results to L2 (Rd ), d ≥ 3. In this section, we will 1 (Rd ) by using the linear profile decomposition in give the linear profile decomposition in Hrad 2 d Lrad (R ). We first review the linear profile decomposition in L2rad (Rd ) in the following. Lemma 4.1 (Profile decomposition in L2rad (Rd ), [6,8,26,33]). Let {φn }n≥1 be a bounded sequence in L2rad (Rd ). Then up to passing to a subsequence of {φn }n≥1 , there exists a sequence of j j functions φ j ∈ L2rad (Rd ) and (hn , tn )n≥1 ⊂ (0, ∞) × R, with hm n j hn
+
j
hn hm n
+
j
|tn −tnm | j
(hn )2
→ ∞, as n → ∞, for j = m,
(4.1)
where j
j
j
j
tn j (hn )2
j
hn → h∞ ∈ {0, 1, ∞}, hn = 1 if h∞ = 1, τn = −
j
j
→ τ∞ ∈ [−∞, ∞], as n → ∞,
(4.2) (4.3)
such that ∀ k ≥ 1, there exists rnk ∈ L2rad (Rd ), k j
φn (x) =
Tn φ j (x) + rnk (x),
(4.4)
j =1 j
here Tn is defined by Tn φ(x) = e−itn j
j
1 j d
φ
(hn ) 2
x j hn
. The remainder rnk satisfies
lim sup eit rnk Lq Lr (R×Rd ) → 0, as k → ∞, n→∞
t
x
(4.5)
where (q, r) is L2 -admissible, and 2 < q < ∞ when d ≥ 2, 4 < q < ∞ when d = 1. We also have as n → ∞, it j j it m m j → 0, Tn (φ j ), Tnm (φ m ) L2 → 0, for j = m, (4.6) e Tn (φ )e Tn (φ ) d+2 Lt,xd (R×Rd )
and ∀ 1 ≤ j ≤ k, Tn (φ j ), rnk L2 → 0, (Tn )−1 rnk 0 in L2 (Rd ). j
j
(4.7)
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
2905
As a consequence, we have the mass decoupling property: k
φn 2L2 −
j =1
φ j 2L2 − rnk 2L2 → 0, as n → ∞, ∀ k ≥ 1.
(4.8)
Proof. We only need to show (4.2), (4.5) and (4.7). Other statements in the theorem are stated in the profile decomposition in L2rad (Rd ) proved in [6,8,26,33]. Without loss of generality, we assume that the sequence is up to a subsequence in the following. j j j To show (4.2), we only need to prove that we may take h∞ and hn to be 1 when h∞ ∈ (0, ∞). j j In fact, if hn → h∞ ∈ (0, ∞), as n → ∞, we have j −itn j · 1 e φ (x) j j d =e
j
−itn
hn
(hn ) 2 1
d
j
φ
j
·
j
h∞
(h∞ ) 2
j −itn (x) + e
1 j d
φ
j
(hn ) 2
· j hn
(x) − e
j
−itn
1 j
d
φ
(h∞ ) 2
j
·
j
h∞
(x) .
Note that −it j j j · −itn j 1 1 · e n φ φ (x) − e (x) j j 2 j d j d h h n ∞ (hn ) 2 (h∞ ) 2 L d j 2 j = φ j (x) − hjn φ j hjn x → 0, as n → ∞. h∞ h∞ 2 L
j
So we can put e−itn
1 j d (hn ) 2
φj
· j hn
j
(x) − e−itn
by the Strichartz estimate. We now shift φ j (x) by
1 j d (h∞ ) 2
1
j d (h∞ ) 2
φj
φj
·
j h∞
x j h∞
(x) into the remainder term, j j j , and (hn , tn ) by (1, tn ). It is
easy to see that (4.1)–(4.3) are not affected. Thus, we conclude (4.2). For the profile decomposition in [6,8,26,33], the remainder rnk satisfies lim sup eit rnk
2(d+2)
Lt,x d
n→∞
(R×Rd )
→ 0, as k → ∞.
By using interpolation, the Strichartz estimate and (4.8), we obtain (4.5). To show (4.7), we see in [6,8,26,33], we already have (Tn )−1 rn 0 in L2 (Rd ), as n → ∞. j
j
Since rnk (x) = rn (x) −
k ! m=j +1
j
(4.9)
Tnm (φ m )(x), when 1 ≤ j < k, so by (4.1) and (4.9),
(Tn )−1 rnk = (Tn )−1 rn − j
j
j
k
(Tn )−1 Tnm φ m 0 in L2 (Rd ), as n → ∞. j
m=j +1
2
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
1 (Rd ). We can now show the linear profile decomposition in Hrad 1 (Rd ). Then up to passing to a subTheorem 4.2. Let {ϕn }n≥1 be a bounded sequence in Hrad 1 (Rd ) and (hj , t j ) sequence of {ϕn }n≥1 , there exists a sequence of functions ϕ j ∈ Hrad n n n≥1 ⊂ (0, ∞) × R, with when n → ∞, j
j
tn j (hn )2
j
j
j hn hm n
hm n j hn
τn = −
j
→ τ∞ ∈ [−∞, ∞],
(4.10)
hn → h∞ ∈ {0, 1, ∞}, +
+
j |tn −tnm | j (hn )2
(4.11)
→ ∞, ∀ j = m,
(4.12)
1 (Rd ), such that ∀ k ∈ N, there exists wnk ∈ Hrad k
∇ −1 Gn (eit ∇ ϕ j ) + eit wnk j
eit φn = j =1 k
=
j j −1 Gn (hn ) ∇ −1 ∇ eit ϕ j + eit wnk
(4.13)
j =1 k
j
∇ −1 hn ∇ ei(t−tn ) gn ϕ j + eit wnk , j
=
j
j =1
where j (gn ϕ)(x) =
1
x
ϕ , j j d hn (hn ) 2 j 1 t − tn x j , j . (Gn v)(t, x) = j d v j (hn )2 hn (hn ) 2 The remainder wnk satisfies lim sup ∇ eit wnk Lqt Lr (R×Rd ) → 0, as k → ∞, x
n→∞
(4.14)
where (q, r) is L2 -admissible, and 2 < q < ∞ when d ≥ 2, 4 < q < ∞ when d = 1. Moreover, we have the following decoupling properties: For any k ∈ N, let σn = (hn )−1 ∇ −1 ∇ , j
then
j
(4.15)
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
s 2 |∇| φn 2 − L
k j =1
2 s j j iτnj j 2 ϕ 2 − |∇|s wnk L2 → 0, s = 0, 1, |∇| gn σn e L
j j j E gn σn eiτn ϕ j − E(wnk ) → 0,
k
E(φn ) −
2907
(4.16)
(4.17)
j =1
j j j Sω gn σn eiτn ϕ j − Sω (wnk ) → 0,
k
Sω (φn ) −
(4.18)
j =1
j j j K gn σn eiτn ϕ j − K(wnk ) → 0,
k
K(φn ) −
(4.19)
j =1
j j j Hω gn σn eiτn ϕ j − Hω (wnk ) → 0, as n → ∞.
k
Hω (φn ) −
(4.20)
j =1
Proof. We divide the proof into three steps. Step 1. Applying Lemma 4.1 to { ∇ φn }, we have k
∇ φn (x) =
e
j
−itn
1 j
d
(hn ) 2
j =1
∇ ϕ
j
· j
hn
(x) + ∇ wnk (x),
(4.21)
where j
j
tn j (hn )2
j
j
τn = −
j
→ τ∞ ∈ [−∞, ∞], j
j
hn → h∞ ∈ {0, 1, ∞}, hn = 1 if h∞ = 1, j
hn hm n
+
hm n j hn
+
j
|tn −tnm | j
(hn )2
(4.22)
→ ∞, for j = m, as n → ∞,
lim sup ∇ eit wnk Lq Lr (R×Rd ) → 0, as k → ∞, t
n→∞
x
where (q, r) is L2 -admissible and 2 < q < ∞ when d ≥ 2, 4 < q < ∞ when d = 1. Moreover, we have k
∀ k ≥ 1,
∇ φn 2L2 −
j =1
2 ∇ ϕ j 2 − ∇ wnk 2L2 → 0, as n → ∞. L
By (4.21), we have k
e
it
φn (x) =
e j =1
it
j
−1 −itn
∇
e
1 j
d
(hn ) 2
· ∇ ϕ x + eit wnk (x) j hn j
2908
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934 k
∇ −1 Gn (eit ∇ ϕ j )(x) + eit wnk (x) j
= j =1 k
Gn ( (hn )−1 ∇ −1 ∇ eit ϕ j ) + eit wnk (x) j
=
j
j =1 k
j
∇ −1 hn ∇ ei(t−tn ) gn ϕ j + eit wnk (x), j
=
j
j =1 j
j
j
where Gn (eit ϕ)(x) = eit Tn ϕ(x) and Tn is defined in Lemma 4.1. So we have obtain (4.10), (4.11), (4.12), (4.13), (4.14). We note by (4.7), for 1 ≤ m ≤ j , (Tnm )−1 ∇ wn 0 in L2rad (Rd ), as n → ∞. j
(4.23)
Step 2. We now turn to (4.16). By (4.13), we have j
tn j j −i j gn σn e (hn )2 ϕ j (x) + wnk (x),
k
ϕn (x) = j =1
so we only need to show the orthogonality: "
tn tnm # − m 2 j j − (hj )2 j m m (h μ ∇ gn σn e n ϕ , μ ∇ gn σn e n ) ϕ m j
→ 0, ∀ j = m,
L2
"
# tn j j − j μ ∇ gn σn e (hn )2 ϕ j , μ ∇ wnk j
→ 0, for 1 ≤ j ≤ k, as n → ∞,
L2
where μ =
|∇| 1 ∇ , ∇ .
"
This follows from
tn tnm # − m 2 j j − (hj )2 j m m (h μ ∇ gn σn e n ϕ , μ ∇ gn σn e n ) ϕ m j
L2
" $ % j j = gn μ ∇j ∇j σn e hn
" j gn μ
= "
= e
∇ j hn
hn
∇ e
tnm i m (hn )2
" m,j
= Sn μ
j t − jn (hn )2
j t − jn (hn )2
j (gnm )−1 gn e
∇ j hn
ϕ
j
ϕ j , gnm μ
, gnm μ h∇m n
j t − jn (hn )2
∇ ϕ j , μ
∇ hm n
μ
∇ j hn
∇ ϕ m
∇ hm n
$
∇ hm n
%
σnm e
tnm − m (hn )2
# ϕm L2
# tnm − m 2 m (h ) n ϕ ∇ e
L2
∇ ϕ , μ
# L2
j
∇ hm n
# ∇ ϕ
→ 0, as n → ∞,
m L2
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
2909
where m,j Sn ϕ(x) =
hm n j hn
j
d
2
e
−i
tn −tnm 2 (hm n)
e
j j ihn (ξn −ξnm )x
j
x−
ϕ
j
j
xn −xnm +2tn (ξn −ξnm ) hm n
j
hn hm n
→ 0, as n → ∞, ∀ ϕ ∈ L2 , by (4.12), and "
"
# tn j j − (hj )2 j k μ ∇ gn σn e n ϕ , μ ∇ wn j
L2
j gn μ
= "
$
∇ j hn
# tn % j − (hj )2 j k σn e n ϕ , μ ∇ wn j
∇ j hn
# j = Tn μ ∇j ∇ ϕ j , μ ∇ wnk
L2
hn
"
L2
# 2 j = Tn μ ∇j ∇ ϕ j , ∇ wnk hn
L2
" # 2 j −1 j k ∇ = μ j ∇ ϕ , (Tn ) ∇ wn hn
L2
→ 0, as n → ∞, ∀ 1 ≤ j ≤ k, by (4.23). Step 3. Since we already have (4.16), to obtain (4.17), (4.18), (4.19), (4.20), it suffices to show k p+1
ϕn Lp+1 −
j =1
j j iτnj j p+1 k p+1 ϕ p+1 − wn p+1 → 0, as n → ∞, for 1 + gn σn e L
L
4 d
≤p≤1+
4 d−2 .
(4.24) 1 ∈ R, then the refined Fatou Lemma (see [22,25]) shows that Suppose that τ∞ 1 1 iτn1 1 p+1 1 p+1 p+1 lim ϕn Lp+1 − gn σn e ϕ p+1 − wn p+1 n→∞
L
L
p+1 p+1 p+1 d 1 = lim (h1n ) 2 (1−p) (gn1 )−1 ϕn p+1 − σn1 eiτn ϕ 1 p+1 − (gn1 )−1 wn1 p+1 n→∞ L L L p+1 1 d(1−p) 1 iτ 1 1 p+1 1 d(1−p) 1 iτn1 1 −1 1 −1 2(1+p) 2(1+p) n σn e (σn ) (Tn ) ϕn − (hn ) σn e ϕ = lim (hn ) p+1 n→∞ p+1 L
L
1 d(1−p) 1 iτ 1 1 −1 1 −1 1 p+1 2(1+p) σ e n (σ ) − (Tn ) wn n n p+1 . (hn ) L
We see 1 −1 1 −1 (σn ) (Tn ) ϕn
Lp+1
and
∇ (σn1 )−1 (Tn1 )−1 ϕn
L2
ϕn H 1
2910
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
∇
d(1−p)
(h1n ) 2(1+p) σn1 =
(h1n )
(1 + | h∇1 |) n
∇
= (h1n ) =
d(p−1) 2(p+1)
d(p−1) 2(p+1)
d(p−1) 1− d(p−1) (1 + | h∇1 |) 2(p+1) (1 + | h∇1 |) 2(p+1) n n
∇
(h1n + |∇|)
d(p−1) 2(p+1)
1− d(p−1) 2(p+1)
(1 + | h∇1 |)
,
n
then 1 d(1−p) 1 iτ 1 1 −1 1 −1 (h ) 2(1+p) σ e n (σ ) (T ) ϕn n n n n p+1 L 1 ∇
eiτn (σn1 )−1 (Tn1 )−1 ϕn p+1 |∇| d(p−1) 2(p+1) L 1 −1 1 −1 iτn1 1 −1 1 −1 ∇ e (σn ) (Tn ) ϕn 2 (σn ) (Tn ) ϕn
H1
L
.
So we have 1 1 iτn1 1 p+1 1 p+1 p+1 lim ϕn Lp+1 − gn σn e ϕ p+1 − wn p+1 = 0. n→∞ L L 1 = ±∞, then by the dispersive estimate, we obtain Next, suppose that τ∞ 1 1 iτn1 1 p+1 1 p+1 p+1 lim ϕn Lp+1 − gn σn e ϕ p+1 − wn p+1 n→∞
L
L
(4.25)
p+1 p+1 d(1−p) 1 1 ≤ lim (h1n ) 2 σn1 eiτn (σn1 )−1 (Tn1 )−1 ϕn p+1 − σn1 eiτn (σn1 )−1 (Tn1 )−1 wn1 p+1 n→∞ L L p+1 d(1−p) 1 + lim (h1n ) 2 σn1 eiτn ϕ 1 p+1 n→∞ L p+1 d(1−p) 1 iτn1 1 −1 1 −1 1 p+1 σ 1 eiτn1 (σ 1 )−1 (T 1 )−1 ϕn σ lim (h1n ) 2 − e (σ ) (T ) w n n n n n n dx n n→∞
+ lim (h1n ) n→∞
d(1−p) 2
1 iτn1 1 p+1 ϕ p+1 σn e L
1 iτn1 1 −1 1 −1 p 1 lim (hn ) (σn ) (Tn ) ϕn σ n e n→∞ p 1 1 + σn1 eiτn (σn1 )−1 (Tn1 )−1 wn1 σn1 eiτn ϕ 1 dx p+1 d(1−p) 1 + lim (h1n ) 2 σn1 eiτn ϕ 1 p+1 n→∞ L 1 d(1−p) 1 iτ 1 1 −1 1 −1 p 2(1+p) n ≤ lim (hn ) σn e (σn ) (Tn ) ϕn d(1−p) 2
n→∞
Lp+1
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
1 d(1−p) 1 iτ 1 1 −1 1 −1 1 p 2(1+p) σ e n (σ ) + ) (T ) w (h n n n n n
2911
Lp+1
1 d(1−p) 1 iτ 1 1 1 d(1−p) 1 iτ 1 1 p+1 (h ) 2(1+p) σ e n ϕ 2(1+p) σ e n ϕ · ) + lim (h n n n p+1 n→∞ n p+1 L L p p 1 1 ∇
iτn 1 −1 ≤ lim d(p−1) eiτn (σn1 )−1 (Tn1 )−1 ϕn + ∇
(σn ) (Tn1 )−1 wn1 d(p−1) e n→∞ p+1 |∇| 2(p+1) |∇| 2(p+1) Lp+1 L p+1 iτn1 1 iτn1 1 · ∇
ϕ + lim ∇
ϕ d(p−1) e d(p−1) e |∇| 2(p+1) p+1 n→∞ |∇| 2(p+1) p+1 L L p p 1 −1 1 −1 1 1 −1 1 −1 ∇
iτn1 1 lim (σn ) (Tn ) ϕn 1 + (σn ) (Tn ) wn 1 d(p−1) e ϕ n→∞ |∇| 2(p+1) p+1 H H L p+1 iτn1 1 + lim ∇
ϕ = 0. d(p−1) e n→∞ p+1 |∇| 2(p+1) L
Thus, we have proved p+1 p+1 1 p+1 ϕn Lp+1 − gn1 σn1 eiτn ϕ 1 p+1 − wn1 p+1 → 0, as n → ∞. L
L
(4.26)
Similarly, we can show p+1 p+1 p+1 2 1 wn p+1 − gn2 σn2 eiτn ϕ 2 p+1 − wn2 p+1 → 0, as n → ∞, L
L
L
(4.27)
which together with (4.26) shows (4.24) with k = 2. Moreover, repeating this procedure, we obtain (4.24) for any k ≥ 1. 2 5. Extraction of a critical element In this section, we show the existence of the critical element in the radial case by using the profile decomposition and the long-time perturbation theory. By Proposition 3.1(v), it suffices for Theorem 1.3 to show that any solution u to (1.1) with u0 ∈ Aω,+ satisfies u
2(d+2)
Lt,x d
(d+2)(p−1) 2 (I
∩Lt,x
max ×R
d)
< ∞,
where Imax denotes the maximal interval where u exists. To this end, for m > 0, let & u is a radial solution to (1.1) ' ω (m) = sup u K(Imax ) : with u0 ∈ Aω,+ and Sω (u) ≤ m with u K(I ) := u
2(d+2)
Lt,x d
(d+2)(p−1) 2 (I ×Rd )
∩Lt,x
and define
(5.1)
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
m∗ω = sup{m > 0 : ω (m) < ∞}.
(5.2)
If u0 ∈ Aω,+ with Sω (u0 ) ≤ m sufficiently small, then Lemma 2.8 shows u H 1 1. Hence, Proposition 3.1(i) gives the finiteness of ω (m), which implies m∗ω > 0. Now our aim is to show m∗ω ≥ mω . Suppose by contradiction that m∗ω < mω , we will show the existence of the critical element. In fact, by the definition of m∗ω , we can take a sequence {un } of solutions (up to time translations) to (1.1) that and Sω (un ) → m∗ω , as n → ∞,
un (t) ∈ Aω,+ , for t ∈ In ,
lim un K([0,sup In )) = lim un K((inf In ,0]) = ∞,
n→∞
n→∞
(5.3) (5.4)
where In denotes the maximal interval of un including 0. By Lemma 2.8, sup un 2L∞ H 1 (I t
n
x
n ×R
d)
mω +
mω . ω
(5.5)
Applying Theorem 4.2 to {un (0)} and obtain some subsequence of {un (0)} (still denoted by the 1 (Rd ) and (hj , t j ) same symbol), then there exists ϕ j ∈ Hrad n n n≥1 of sequences in (0, ∞) × R, with j
j
tn j (hn )2
j
j
τn = −
j
→ τ∞ ∈ [−∞, ∞],
hn → h∞ ∈ {0, 1, ∞}, j
hn hm n
+
hm n j hn
+
j
|tn −tnm | j
(hn )2
→ ∞, as n → ∞, ∀ j = m,
(5.6)
1 (Rd ), such that, ∀ k ∈ N, there exists wnk ∈ Hrad k
e
it
un (0) =
j j Gn (hn )−1 ∇ −1 ∇ eit ϕ j + eit wnk
j =1 k
j
∇ −1 hn ∇ ei(t−tn ) gn ϕ j + eit wnk .
(5.7)
lim sup ∇ eit wnk Lq Lr (R×Rd ) → 0, as k → ∞,
(5.8)
j
=
j
j =1
The remainder wnk satisfies
n→∞
t
x
where (q, r) is L2 -admissible, 2 < q < ∞ when d ≥ 2 and 4 < q < ∞ when d = 1. Moreover, j j for any k ∈ N, s = 0, 1, let σn = (hn )−1 ∇ −1 ∇ , we have
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
s |∇| un (0)2 2 − L
k j =1
2913
s j j iτnj j 2 ϕ 2 − |∇|s wnk 2L2 → 0, |∇| gn σn e
(5.9)
L
j j j E gn σn eiτn ϕ j − E(wnk ) → 0,
k
E(un (0)) −
(5.10)
j =1
j j j Sω gn σn eiτn ϕ j − Sω (wnk ) → 0,
k
Sω (un (0)) −
(5.11)
j =1
j j j K gn σn eiτn ϕ j − K(wnk ) → 0,
k
K(un (0)) −
(5.12)
j =1
j j j Hω gn σn eiτn ϕ j − Hω (wnk ) → 0, as n → ∞.
k
Hω (un (0)) −
(5.13)
j =1
Using Strichartz estimate, (5.9) and (5.5), we get sup lim sup ∇ eit wnk S 0 (Rd ) sup lim sup wnk H 1 < ∞.
k∈N
n→∞
k∈N
(5.14)
n→∞
j
j
Next, we construct the nonlinear profile. Let Un be the solution to (1.1) with Un (0) = j j j j j j j ∇ −1 hn ∇ e−itn gn ϕ j , so that eit Un (0) = Gn (σn eit ϕ j ). Thus, Un satisfies j j j 4 j j j i∂t Un + Un = |Un | d Un − |Un |p−1 Un , j
Un (0) = ∇ −1 hn ∇ e−itn gn ϕ j j
j
j
Undoing the transformations Gn and then apply (σn )−1 , we have j
j
4
(i∂t + )vn (t) = (σn )−1 (|σn vn | d σn vn − (hn ) j vn (0) = ϕ j j
j
j j
j j
j
d+4 dp 2 − 2
j j
j j
|σn vn |p−1 σn vn ),
(5.15)
where vn = (σn )−1 (Gn )−1 Un . j
j
j
j
j
We now take the formal limit of vn as n → ∞, and denote the limit is uj , then we see uj satisfies the following equation: j
(1) When h∞ = 0, ⎧ ⎨(i∂t + )uj = 0,
4 ∇ j d−2 ∇ j j ⎩(i∂t + ) ∇
|∇| u = − |∇| u |∇| u ,
if p < 1 +
4 d−2 ,
d = 3, 4, or p < ∞, d = 1, 2,
if p = 1 +
4 d−2 ,
d = 3, 4, (5.16)
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934 j
(2) When h∞ = 1, 4
(i∂t + )uj = |uj | d uj − |uj |p−1 uj ,
(5.17)
j
(3) When h∞ = ∞, 4
(i∂t + ) ∇ uj = | ∇ uj | d ∇ uj ,
(5.18)
with uj (0) = ϕ j . j We then define the nonlinear profile to be uj . The unique existence of uj around t = τ∞ is j j known in all cases, including h∞ = ∞ and τ∞ = ±∞ (the latter corresponding to the existence of the wave operators), by using the standard iteration with the Strichartz estimate. Let I j = j j (Tmin , Tmax ) be the maximal interval where the nonlinear profile uj exists, we see from the construction of the nonlinear profile that uj ∈ C(I j , H 1 (Rd )) and j j j u (τn ) − eiτn ϕ j
H1
→ 0, as n → ∞.
(5.19)
j
(5.20)
Let j
j
un = Gn σn uj , j
j
j
j
j
j
j
j
then, the lifespan of un is In = ((hn )2 Tmin + tn , (hn )2 Tmax + tn ). j Now, ∀ j ∈ N, we define the space ST∞ by ⎧ 2(d+2) ⎪ ⎪ Lt,xd−2 , ⎪ ⎪ ⎨ (d+2)(p−1) 2(d+2) j 2 ST∞ = L ∩ Lt,xd , t,x ⎪ ⎪ ⎪ ⎪ ⎩ 2(d+2) Lt,xd ,
j
if h∞ = 0 with p is energy-critical exponents, j
j
if h∞ = 0 with p is energy-subcritical exponents, or h∞ = 1, j
if h∞ = ∞.
For the linear profile decomposition (5.7), we can give the corresponding nonlinear profile decomposition k j
u
un (t),
(5.21)
j =1 it w k is a good approximation for u provided that each nonlinear profile we will show u
Lemma 5.1 (Classification of nonlinear profiles). Let uj be a non-trivial nonlinear profile. Then ∀t ∈ Ij,
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
⎧ Aω,+ , ⎪ ⎪ ⎪ 2 d ⎨ L (R ), j δ∞ uj (t) ∈ ⎪ H 1 (Rd ), ⎪ ⎪ ⎩ 0 A+ ,
2915
j
h∞ = 1, j if h∞ = ∞, j if h∞ = 0 with p is energy-subcritical exponents, j if h∞ = 0 with p is energy-critical exponents,
(5.22)
where ⎧ j j ⎪ if h∞ = 0 with p is energy-subcritical exponents or h∞ = 1, ⎨I, j j δ∞ = ∇ , if h∞ = ∞, ⎪ ⎩ ∇
j |∇| , if h∞ = 0 with p is energy-critical exponents, and A0+ = {ϕ ∈ H˙ 1 (Rd ) : E 0 (ϕ) < W, K0 (ϕ) < K0 (W )}. By using the variational analysis, we can easily show A0+ is just the scattering set of the energycritical focusing nonlinear Schrödinger equation in [18]. Proof. Since uj is non-trivial, we see from (5.19) that the corresponding linear profile ϕ j is also non-trivial. We shall show m + m∗ j ω j j ω Hω gn σn eiτn ϕ j < , (5.23) 2 j j j (5.24) K gn σn eiτn ϕ j > 0, m + m∗ j ω j j ω , Sω gn σn eiτn ϕ j < 2
(5.25)
for any j and sufficient large n. It follows from (5.3), K(un (0)) > 0 and (5.13) that m∗ω + on (1) = Sω (un (0)) ≥ Hω (un (0)) =
k
j j j Hω gn σn eiτn ϕ j + Hω (wnk ) + on (1). (5.26)
j =1
Hence, (5.26) together with m∗ω < mω and the positivity of Hω show (5.23). Moreover, (5.23) with (2.5) show (5.24). By (2.14), (5.24), we have j j j E gn σn eiτn ϕ j ≥ 0, for n large enough, j ≥ 1, and so j j j Sω gn σn eiτn ϕ j ≥ 0, for n large enough, j ≥ 1. this together with (5.3) and (5.11) gives (5.25).
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
We shall show (5.22). j Case I. h∞ = 1. It follows from (5.19) that j j j j j u (τn ) − gn σn eiτn ϕ j
H1
→ 0, as n → ∞.
(5.27)
This together with (5.23) gives us that mω + m∗ω j lim Hω uj (τn ) ≤ , n→∞ 2 which together with (2.5) shows j K uj (τn ) > 0, for n large enough. Moreover, (5.25) together with (5.27) yields that j Sω uj (τn ) < mω , for n large enough. Hence, j uj τn ) ∈ Aω,+ for n large enough. Then, we have (5.22) in this case. j j Case II. h∞ = ∞. It is easily seen that δ∞ uj (t) = ∇ uj (t) ∈ L2 (Rd ), ∀ t ∈ I j . j Case III. h∞ = 0 with p is energy-subcritical exponents. It is easily seen that uj ∈ H 1 (Rd ). j Case IV. h∞ = 0 with p is energy-critical exponents. Similar to Case I, we can obtain (5.22) in this case. 2 When j large enough, we have the following basic fact about uj : j
j
Lemma 5.2. There exists j0 ∈ N such that Tmin = −∞, Tmax = ∞ for j > j0 and uj 2S 1 (R) j >j0
ϕ j 2H 1 < ∞.
(5.28)
j >j0
Proof. By (5.5) and (5.9), we have ∞ j =1
which shows
∞ ! j =1
2 j ∇ (e−itn ϕ j ) 2 < ∞, L
ϕ j 2H 1 < ∞, and therefore ϕ j H 1 → 0, as j → ∞. By the small data
global wellposedness and scattering theory together with (5.19), we obtain when j large enough, uj S 1 (R) ϕ j H 1 and so we have the desired result. 2
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
2917
Lemma 5.3. In the nonlinear profile decomposition (5.21), if j
δ∞ uj ST j ((T j ∞
j
j d min ,Tmax )×R )
< ∞ for 1 ≤ j ≤ k,
(5.29)
j
then, we have Tmin = −∞, Tmax = ∞, and j
un
(d+2)(p−1) Lt,x 2 (R×Rd )
j + ∇ un
j
2 d L∞ t Lx (R×R )
+ ∇ un
2(d+2)
Lt,x d
(R×Rd )
∇ uj S 0 (R) 1, (5.30)
for 1 ≤ j ≤ k, and there exists B > 0 such that lim sup n→∞
+ ∇ u
(d+2)(p−1) Lt,x 2 (R×Rd )
+ ∇ u
2(d+2) d 2 L∞ (R×Rd ) t Lx ∩Lt,x
j
2 d L∞ t Lx (R×R )
≤ B.
(5.31)
j
Proof. If h∞ = ∞, then δ∞ uj is a solution to the mass-critical defocusing nonlinear Schrödinger j j equation. Since I j coincides with the maximal interval where δ∞ uj exists, we have Tmin = −∞, j Tmax = ∞. j Similarly, when h∞ = 0 with p is energy-subcritical exponents, uj is just the solution of j j j the free Schrödinger equation, we still have Tmin = −∞, Tmax = ∞. When h∞ = 0 with j j p is energy-critical exponents, δ∞ u is a solution to the energy-critical defocusing nonlinear Schrödinger equation. By using a easy variational analysis, we can easily see A0+ is just the scatj tering set of the energy-critical defocusing nonlinear Schrödinger equation. So δ∞uj is globally j j exists with Tmin = −∞, Tmax = ∞. j j When h∞ = 1, we have δ∞ uj is a solution to (1.1) and therefore, Proposition 3.1(v) together j j with (5.29) shows Tmin = −∞, Tmax = ∞. Using the Strichartz estimate and (5.29), we get ∇ uj S 0 (R) 1, for 1 ≤ j ≤ k, so it suffices to show j
un
(d+2)(p−1) Lt,x 2 (R×Rd )
j + ∇ un
j
2 d L∞ t Lx (R×R )
+ ∇ un
2(d+2)
Lt,x d
(R×Rd )
uj S 1 (R) .
In fact, j
un
(d+2)(p−1) Lt,x 2
d 2 j j − |∇| 2 p−1 Gn σn uj ∼
2d(d+2)(p−1) (d+2)(p−1) d(d+2)(p−1)−8 2 Lx
Lt
d j − 2 ∇
|(hn )−1 ∇| 2 p−1 uj 2d(d+2)(p−1) j −1 (d+2)(p−1) (hn ) ∇
d(d+2)(p−1)−8 2 Lt Lx
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
∇ uj
2d(d+2)(p−1) (d+2)(p−1) d(d+2)(p−1)−8 2 Lt Lx
∇ uj
S 0 (R)
,
where the Hörmander–Mihlin multiplier theorem is used in the second inequality. j ∇ un
2 d L∞ t Lx (R×R )
= ∇ uj
2 L∞ t Lx
j ∇ un
2(d+2) Lt,x d (R×Rd )
≤ uj
S 1 (R)
= ∇ uj
,
≤ uj
2(d+2) Lt,x d (R×Rd )
S 1 (R)
.
Hence (5.30) follows. We now turn to (5.31). By q q j j j un − |un |q−1 |um un ≤ Ck,q n |, 1 < q < ∞, 1≤j ≤k 1≤j ≤k 1≤j =m≤k we have (d+2)(p−1) 2 j u n 1≤j ≤k (d+2)(p−1) 2 Lt,x
(d+2)(p−1) 2 j un (d+2)(p−1)
≤ 1≤j ≤k
(R×Rd )
Lt,x
2
+ Ck
(R×Rd ) j
|un | 1≤j =m≤k
(d+2)(p−1) −1 2
|um n | dxdt.
(5.32)
R×Rd
We see by (5.30) and Lemma 5.2 that j
un 1≤j ≤k
(d+2)(p−1) 2 (d+2)(p−1) Lt,x 2 (R×Rd )
(d+2)(p−1)
1≤j ≤j0
uj S 1 (R)2
+
ϕ j 2H 1 < ∞.
(5.33)
j >j0
Next, we consider the second term on the right side of (5.32). By the Hölder inequality, (5.6) and (5.30), we have
j
|un |
(d+2)(p−1) −1 2
j
m |um n | dxdt un un
R×Rd
j un um n
(d+2)(p−1) Lt,x 4
(d+2)(p−1) 4
j
un
(d+2)(p−1) −2 2 (d+2)(p−1) Lt,x 2
→ 0, as n → ∞.
Lt,x
Plugging (5.33) and (5.34) into (5.32), we obtain that there is B1 > 0 such that lim sup u
Similarly, we have
(d+2)(p−1) 2 (R×Rd )
Lt,x
≤ B1 .
(5.34)
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
lim sup ∇ u
2(d+2)
Lt,x d
n→∞
(R×Rd )
2919
≤ B1
and lim sup ∇ u
2 d L∞ t Lx (R×R )
n→∞
≤ B1 .
2
Thus, we obtain (5.31).
Lemma 5.4 (At least one bad profile). Let j0 be as in Lemma 5.2, then there exists 1 ≤ j ≤ j0 such that j
δ∞ uj ST j ((T j ∞
j d min ,Tmax )×R )
= ∞.
Proof. We argue by contradiction. Assume j
δ∞ uj ST j ((T j
j d min ,Tmax )×R )
∞
< ∞, ∀ 1 ≤ j ≤ j0 .
Combining this with Lemma 5.2, we have j
δ∞ uj ST j ((T j ∞
j d min ,Tmax )×R )
< ∞, ∀ j ≥ 1.
(5.35)
j
This together with Lemma 5.3 implies un exists globally in time for j ≥ 1 and hence so does it w k . u 0 such that it k lim sup ∇ (u
(5.36)
n→∞
it k lim sup u
(d+2)(p−1) 2(d+2) 2 ∩Lt,x d (R×Rd )
Moreover, it follows from (5.7) with t = 0 and (5.19) that k un (0) − u
≤ j =1
Hence,
≤ B.
Lt,x
j j ∇ eiτn ϕ j − uj (τn )
L2
H1
−→ 0, as n → ∞.
(5.37)
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
k un (0) − u
H1
→ 0, as n → ∞.
(5.38)
Next, we claim that as k → ∞, it k
n→∞
2(d+2)
Lt,xd+4
→ 0,
(5.39)
4
where N (u) = |u| d u − |u|p−1 u. Before proving this, we remark that (5.39) together with the long-time perturbation theory leads to a contradiction. Indeed, by (5.36), (5.37), (5.38), we conclude as a consequence of Proposition 3.2 that un
2(d+2)
Lt,x d
< ∞,
(d+2)(p−1) 2 (R×Rd )
∩Lt,x
when n large enough, which contradicts (5.4). Hence, Lemma 5.4 holds. It remains to prove (5.39). Note that it k it k wn ) − N (u
=
Hence, it suffices for (5.39) to show that k j lim (i∂t + )un − N (u
n n→∞ j =1
2(d+2) Lt,xd+4
= 0, ∀ k ∈ N,
(5.40)
→ 0, as k → ∞.
(5.41)
and it k
n→∞
2(d+2)
Lt,xd+4
First, we show (5.40). j When h∞ = 1, we have j
j
(i∂t + )un = N (un ). j
When h∞ = 0 with p is energy-subcritical exponent, we have j
(i∂t + )un = 0. j
When h∞ = 0 with p is energy-critical exponent,
∇ j |∇| u
satisfies the energy-critical focusing j
nonlinear Schrödinger equation, these cases can be deal with similar to the case h∞ = ∞ below.
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
2921
j
When h∞ = ∞, ∇ uj satisfies the mass-critical defocusing nonlinear Schrödinger equation, < ∞ shows uj S 1 (R) < ∞. In this case, we have this together with ∇ uj 2(d+2) Lt,x d
j
(R×Rd )
j j j 4 j j j |Gn δ∞ u | d Gn δ∞ u 4 j j j j j j 1 = N (un ) + ∇
|Gn δ∞ uj | d Gn δ∞ uj − N (un ) ,
(i∂t + )un =
1 ∇
we will show 4 j j j j j 1 (|Gn δ∞ uj | d Gn δ∞ uj ) − N (un ) ∇ ∇
We note 4 j j j j j 1 (|Gn δ∞ uj | d Gn δ∞ uj ) − N (un ) ∇ ∇
j j ≤ ∇ (|un |p−1 un )
2(d+2)
Lt,xd+4
j j ≤ ∇ (|un |p−1 un )
2(d+2) Lt,xd+4
2(d+2)
Lt,xd+4
→ 0, as n → ∞.
(5.42)
2(d+2)
Lt,xd+4
j 4 j + ∇ |un | d un −
1 ∇
4 j j j j |Gn δ∞ uj | d Gn δ∞ uj
4 j 4 j j j j j + ∇ (|un | d un ) − |Gn δ∞ uj | d Gn δ∞ uj
2(d+2)
2(d+2)
Lt,xd+4
.
Lt,xd+4
We see j j ∇ (|un |p−1 un )
2(d+2)
Lt,xd+4
p−1 j j ≤ un (d+2)(p−1) ∇ un 2
2(d+2)
Lt,x d
Lt,x
2 j (− d2 + p−1 )(p−1) j j p−1 σn u (d+2)(p−1) Lt,x 2
j j −1 j Gn (hn ) ∇ σn uj
2 j (− d2 + p−1 )(p−1) j j p−1 σn u (d+2)(p−1) Lt,x 2
∇ uj
= (hn ) = (hn )
(d+2)(p−1) 2
By ∇ uj ∈ S 0 (R) ∈ Lt j σn uj
(d+2)(p−1) 2
Lt,x
2(d+2)
.
Lt,x d
2d(d+2)(p−1)
Lxd(d+2)(p−1)−8 (R × Rd ), we may assume uj ∈ C0∞ (R × Rd ), then
∇
j = j −1 u (hn ) ∇
(d+2)(p−1) 2
Lt,x
∇ uj
j
j
∇ (|un |p−1 un )
2(d+2)
Lt,xd+4
(d+2)(p−1) 2
Lt,x
so we have
We also have
2(d+2)
Lt,x d
→ 0, as n → ∞.
< ∞,
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
j 4 j |∇|(|un | d un )
2(d+2)
Lt,xd+4
→ 0, as n → ∞.
(5.43)
Indeed, we may assume uj ∈ C0∞ (R × Rd ), then j 4 j |∇|(|un | d un )
2(d+2) Lt,xd+4
4 jd j ≤ un 2(d+2) |∇|un Lt,x d
4 j d j = (hn )−1 σn uj 2(d+2) Lt,x d
j 4 j The above proof also implies |un | d |∇|un
|(hj )−1 ∇| j nj ∇ u (hn )−1 ∇
2(d+2)
Lt,x d
→ 0, as n → ∞. Therefore, we obtain, when
2(d+2)
Lt,xd+4
n → ∞, 4 j 4 j j j j j ∇ (|un | d un ) − |Gn δ∞ uj | d Gn δ∞ uj
2(d+2)
Lt,x d
2(d+2)
Lt,xd+4
4 j 4 j j 4 j j j j j = |un | d un + |∇|(|un | d un ) − |Gn δ∞ uj | d Gn δ∞ uj 4 j 4 j j j j j = |un | d ∇ un − |Gn δ∞ uj | d Gn δ∞ uj
2(d+2)
Lt,xd+4
2(d+2)
Lt,xd+4
+ on (1)
4 4 j j j j j j j j = |Gn σn uj | d ∇ Gn σn uj − |Gn δ∞ uj | d Gn δ∞ uj
2(d+2)
Lt,xd+4
+ on (1)
4 4 4 j j j j j j j j j j j j = |Gn σn uj | d − |Gn δ∞ uj | d ∇ Gn σn uj + |Gn δ∞ uj | d ∇ Gn σn uj − Gn δ∞ uj + on (1) 4 j j j j j |Gn (σn − δ∞ )uj | d ∇ Gn σn uj
4 j j j j j j + |Gn δ∞ uj | d ∇ Gn σn uj − Gn δ∞ uj
2(d+2) Lt,xd+4
4 j j j j j j j + |Gn (σn − δ∞ )uj ||Gn δ∞ uj | d −1 ∇ Gn σn uj
2(d+2)
Lt,xd+4
4 d j j j j j Gn (σn − δ∞ )uj 2(d+2) ∇ Gn σn uj Lt,x d
2(d+2)
Lt,x d
2(d+2)
Lt,x d
2(d+2) Lt,x d
4 j j j d −1 j j Gn δ∞ u 2(d+2) ∇ Gn σn uj Lt,x d
4 j j d j Gn (σn − δ∞ )uj 2(d+2) ∇ uj Lt,x d
4 j d + δ∞ uj 2(d+2) Lt,x d
+ on (1)
Lt,x d
4 j j d j j j j + Gn δ∞ uj 2(d+2) ∇ Gn σn uj − Gn δ∞ uj j j j + Gn (σn − δ∞ )uj
2(d+2)
Lt,xd+4
2(d+2)
Lt,x d
j −1 j j (hn ) ∇ σn uj − δ∞ uj
2(d+2)
Lt,x d
2(d+2)
Lt,x d
+ on (1)
2(d+2)
Lt,xd+4
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
j j + (σn − δ∞ )uj
2(d+2) Lt,x d
4 j j d −1 j −1 j δ∞ u 2(d+2) (hn ) ∇ σn uj Lt,x d
4 d j j j ∼ Gn (σn − δ∞ )uj 2(d+2) ∇ uj Lt,x d
j j + (σn − δ∞ )uj
2(d+2) Lt,x d
2(d+2)
Lt,x d
2923
+ on (1)
2(d+2)
Lt,x d
4 d ∇ uj 2(d+2) + on (1) → 0.
(5.44)
Lt,x d
Thus, k j ∇
(i∂t + )un − N (u
N (un ) − N ( un ) j =1 j =1
2(d+2)
Lt,xd+4
2(d+2) Lt,xd+4
+ on (1) → 0, as n → ∞,
where we use the fact that by (5.6) and Lemma 5.3, ⎞ ⎛ k k j j ⎠ ∇ ⎝ N (u ) − N ( u ) n n j =1 j =1
→ 0, as n → ∞,
2(d+2) Lt,xd+4
and (5.40) follows. We now turn to (5.41). By the Fundamental Theorem of Calculus, 1
Fz v + θ (u − v) dθ + (u − v)
F (u) − F (v) = (u − v) 0
1
Fz¯ v + θ (u − v) dθ,
0
we obtain it k
2(d+2)
Lt,xd+4
p−1 it k |u
2(d+2) Lt,xd+4
4 d it w k + |u
(5.45)
2(d+2)
Lt,xd+4
p d+4 d + eit wnk 2(d+2)p + eit wnk 2(d+2) Lt,xd+4
(5.46)
Lt,x d
p−1 + |u
4 it k d + |u
2(d+2)
+ |eit wnk |p−1 ∇u
4 + |eit wnk | d ∇u
2(d+2)
2(d+2) Lt,xd+4 2(d+2) Lt,xd+4
(5.47)
Lt,xd+4
Lt,xd+4
(5.48)
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
+ |eit wnk |p−1 ∇eit wnk
2(d+2) Lt,xd+4
p−2 it k
2(d+2) Lt,xd+4
4 + |eit wnk | d ∇eit wnk
(5.49)
2(d+2)
Lt,xd+4
4−d d e it w k ∇u
2(d+2)
(5.50)
.
Lt,xd+4
We now estimate the terms (5.46) and (5.49). By the Hölder inequality and (5.8), we have p d d+4 − (5.46) |∇| 2 2p eit wnk 2(d+2)p d+4
Lt
p ∇ eit wnk 2(d+2)p Lt
d+4
d+4 d + eit wnk 2(d+2)
2d(d+2)p d(d+2)p−2(d+4) Lx
2d(d+2)p d(d+2)p−2(d+4) Lx
Lt,x d
d+4 d + eit wnk 2(d+2) → 0, as n → ∞, k → ∞, Lt,x d
p−1 4 ) ( d (5.49) ≤ eit wnk (d+2)(p−1) + eit wnk 2(d+2) · ∇eit wnk 2 Lt,x 2
Lt,x
p−1 ( |∇|sp eit wnk (d+2)(p−1) Lt
2d(d+2)(p−1) d(d+2)(p−1)−8 Lx
2
2(d+2)
Lt,x d
4 ) d + eit wnk 2(d+2) · ∇eit wnk Lt,x 2
2(d+2)
Lt,x d
→ 0, as n → ∞, k → ∞. By Lemma 5.3 and (5.8), we also have for (5.48) that lim lim
k→∞ n→∞
≤ lim lim
it k p−1
k→∞ n→∞
2(d+2) Lt,xd+4
4 + |eit wnk | d ∇u
2(d+2) Lt,xd+4
4 ( ) it k p−1 d + eit wnk 2(d+2) · ∇ u
Lt,x
2(d+2)
Lt,x d
We now consider the terms of the form
2(d+2)
Lt,xd+4
, 1+
4 d
≤q ≤1+
4 d−2 ,
s = 0, 1,
which corresponds to (5.45), (5.47). By the Hölder inequality, (5.8), (5.14), (5.31), we have
2(d+2) Lt,xd+4
q−1 s it k |∇| e w u
2
2(d+2)
Lt,xd+4
s it k |∇| e wn
q−1 s it k |∇| e wn u
→ 0, as n → ∞, k → ∞. Thus, we obtain for 1 +
4 d
≤q ≤1+
4 d−2 ,
s = 0, 1,
2(d+2)
Lt,x d 2(d+2)
Lt,x d
= 0.
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
q−1 lim |u
n→∞
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→ 0, as k → ∞.
2(d+2)
Lt,xd+4
We now estimate (5.50). For q = p, 1 + d4 , we have
≤ eit wnk
(d+2)(q−1) Lt,x 2
2(d+2) Lt,x d
|∇|sq eit wnk
2(d+2)d(q−1) (d+2)(q−1) d(d+2)(q−1)−8 2 Lt Lx
as n → ∞, k → ∞. Thus, (5.41) follows.
Lt,x
2(d+2) Lt,x d
2
We can now show the main result in this section: Proposition 5.5 (Existence of the critical element). Suppose m∗ω < mω , then there exists a global 1 (Rd )) to (1.1) such that solution uc ∈ C(R, Hrad uc (t) ∈ Aω,+ and Sω (uc (t)) = m∗ω , for t ∈ R, uc
(d+2)(p−1) 2(d+2) 2 ∩Lt,x d ([0,∞)×Rd )
= uc
Lt,x
(5.51)
(d+2)(p−1) 2(d+2) 2 ∩Lt,x d ((−∞,0]×Rd )
= ∞.
(5.52)
Lt,x
Proof. By Lemma 5.2, Lemma 5.4 and reordering indices, there exists J ≤ j0 such that ⎧ j j ⎨ δ∞ u ST j ((T j
j
d ∞ min ,Tmax )×R ) j j ⎩ δ∞ u j j j ST∞ ((Tmin ,Tmax )×Rd )
= ∞,
for 1 ≤ j ≤ J,
< ∞,
for j ≥ J.
(5.53)
We see from Lemma 5.1 that for 1 ≤ j ≤ J , t ∈ I j , ⎧ Aω,+ , ⎪ ⎪ ⎪ ⎨L2 (Rd ), j δ∞ uj (t) ∈ ⎪ H 1 (Rd ), ⎪ ⎪ ⎩ 0 A+ ,
j
h∞ = 1, j if h∞ = ∞, j if h∞ = 0 with p is energy-subcritical exponents, j if h∞ = 0 with p is energy-critical exponents,
(5.54)
j
On the hand, we can see δ∞ uj is a solution to the mass-critical defocusing nonlinear Schrödinger j equation when h∞ = ∞, the scattering theorem of the mass-critical defocusing nonlinear Schrödinger equation [11–13,21,24,34,35] together with (5.54) shows that j
j
δ∞ uj ST j (I j ) = δ∞ uj ∞
j
2(d+2) Lt,x d (I j ×Rd )
j
< ∞, if h∞ = ∞. j
On the other hand, δ∞ uj is a solution to the free Schrödinger equation when h∞ = 0 with p is energy-subcritical exponents or a solution the energy-critical focusing nonlinear Schrödinger
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934 j
equation when if h∞ = 0 with p is energy-critical exponents, then by the scattering theorem in [18,22] together with (5.54) shows that j
δ∞ uj ST j (I j ) < ∞. ∞
Hence, we have j
j
h∞ = 1, δ∞ = I for 1 ≤ j ≤ J.
(5.55)
From (5.9), (5.11) and (5.13), we have J
∇ un (0) 2L2 − J
Sω (un (0)) −
j =1
2 2 ∇ ϕ j 2 − ∇ wnJ 2 → 0, L
L
j
Sω (e−itn ϕ j ) − Sω (wnJ ) → 0,
(5.56)
j =1 J
Hω (un (0)) −
j
Hω (e−itn ϕ j ) − Hω (wnJ ) → 0, as n → ∞.
(5.57)
j =1
Since K(un (0)) > 0, we have Hω (un (0)) ≤ Sω (un (0)) by (2.4). It follows from (5.57) and (5.3) that j
Hω (e−itn ϕ j ), Hω (wnJ ) ≤ Sω (un (0)) <
mω + m∗ω , 2
for 1 ≤ j ≤ J and n large enough. By (2.5), we have j
K(e−itn ϕ j ) > 0, K(wnJ ) > 0, for 1 ≤ j ≤ J and n large enough,
(5.58)
which together with Lemma 2.6 shows j
Sω (e−itn ϕ j ) ≥ 0, Sω (wnJ ) ≥ 0,
(5.59)
for 1 ≤ j ≤ J and n large enough. We shall show J = 1. Assume for a contradiction that J ≥ 2. Then, it follows from (5.3) and (5.56) that j
lim sup Sω (e−itn ϕ j ) < m∗ω , ∀ 1 ≤ j ≤ J, n→∞
which together with (5.19) shows
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
2927
Sω (uj (t)) < m∗ω , ∀ 1 ≤ j ≤ J, t ∈ (Tmin , Tmax ). j
j
Since uj is a solution to (1.1), it follows from the definition of m∗ω that uj
(d+2)(p−1) 2(d+2) j j 2 ∩Lt,x d ((Tmin ,Tmax )×Rd )
< ∞, ∀ 1 ≤ j ≤ J.
Lt,x
This contradicts (5.53). Thus, we have J = 1. Since u1 (d+2)(p−1) 2(d+2) = ∞, we have Lt,x
2
∩Lt,x d
1 ,T 1 )×Rd ) ((Tmin max
1 1 Sω (u1 (t)) ≥ m∗ω , for t ∈ (Tmin , Tmax ).
(5.60)
1 , T 1 ), S (u1 (t)) ≤ m∗ . On the other hand, by (5.19), (5.56) and (5.59), we get for any t ∈ (Tmin ω max ω Combining this with (5.60), we obtain 1 1 Sω (u1 (t)) = m∗ω , for t ∈ (Tmin , Tmax ).
(5.61)
By (5.19), we have Sω (u1 (t)) = lim Sω (e−itn ϕ 1 ), 1
n→∞
this together with (5.3), (5.56) and (5.61) shows Sω (wn1 ) → 0, as n → ∞.
(5.62)
Hence, Lemma 2.6 together with (5.58) and (5.62) shows wn1 H 1 → 0, as n → ∞.
(5.63)
un (0) − e−itn ϕ 1 H 1 → 0, as n → ∞.
(5.64)
We see from (5.7) and (5.63) that 1
1 = −∞, T 1 = ∞. Assume for a contradiction that T 1 < ∞. Let Now, we shall show Tmin max max 1 1 ) such that t T 1 and put {tn } be a sequence in (Tmin , Tmax n max 1 1 u˜ n (t) = u1 (t + tn ) with I˜n = (Tmin − tn , Tmax − tn ).
We see that {u˜ n } satisfies u˜ n (t) ∈ Aω,+ for t ∈ I˜n , and Sω (u˜ n ) = m∗ω , u˜ n
(d+2)(p−1) 2(d+2) 2 ∩Lt,x d (I˜n ×Rd )
Lt,x
= ∞.
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
Then, we can apply the above argument as deriving (5.64) to this sequence and find that there 1 (Rd ), a sequence {τ } with τ = lim τ ∈ [−∞, ∞] such that exists a non-trivial ψ ∈ Hrad n ∞ n n→∞
lim u1 (tn ) − e−iτn ψ H 1 = lim u˜ n (0) − e−iτn ψ H 1 = 0.
n→∞
n→∞
This together with the Strichartz estimate yields ∇ (eit u1 (tn ) − ei(t−τn ) ψ)
2(d+2)
Lt,x d
(R×Rd )
→ 0, as n → ∞.
(5.65)
Case 1. τ∞ = ±∞. By the dispersive estimate for the free solution, for any compact interval I , we have → 0, as n → ∞, ∇ ei(t−τn ) ψ 2(d+2) Lt,x d
(I ×Rd )
this together with (5.65) yields ∇ eit u1 (tn )
2(d+2)
Lt,x d
(I ×Rd )
→ 0, as n → ∞.
(5.66)
Case 2. τ∞ ∈ R. For any interval I with τ∞ ∈ I and |I | 1, we have by (5.65), lim ∇ eit u1 (tn ) 2(d+2) = ∇ ei(t−τ∞ ) ψ 2(d+2) 1. n→∞
(I ×Rd )
Lt,x d
Lt,x d
(I ×Rd )
(5.67)
1 , which is a Then, Proposition 3.1 together with (5.66), (5.67) implies that u1 exists beyond Tmax 1 1 contradiction. Thus, Tmax = ∞. Similarly, we have Tmin = −∞. Therefore, u1 is a global solution and it is just the desired critical element uc satisfying (5.51) and (5.52). 2
We now show the trajectory of the critical element is precompact in the energy space 1 (Rd ). Hrad Proposition 5.6 (Compactness of the critical element). Let uc be the critical element in Proposi1 (Rd ). tion 5.5, then {uc (t) : t ∈ R} is precompact in Hrad Proof. For {tn } ⊂ R, if tn → t ∗ ∈ R, as n → ∞, then we see by the continuity of uc (t) in t that uc (tn ) → uc (t ∗ ) in H 1 (Rd ), as n → ∞. If tn → ∞. Applying the above argument as deriving (5.64) to uc (t + tn ), there exist tn ∈ R 1 (Rd ) that and φ ∈ Hrad
uc (tn ) − e−itn φ → 0 in H 1 (Rd ), as n → ∞. (i) If tn → −∞, then we have
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
∇ eit uc (tn )
= ∇ eit φ
2(d+2) Lt,x d ([0,∞)×Rd )
2(d+2)
Lt,x d
([−tn ,∞)×Rd )
2929
+ on (1) → 0, as n → ∞.
Hence, we can solve (1.1) for t > tn globally by iteration with small Strichartz norm when n large enough, which contradicts uc
2(d+2)
Lt,x d
(d+2)(p−1) 2 ([0,∞)×Rd )
∩Lt,x
= ∞.
(ii) If tn → ∞, then we have ∇ eit uc (tn )
2(d+2)
Lt,x d
((−∞,0]×Rd )
= ∇ eit φ
2(d+2)
Lt,x d
((−∞,−tn ]×Rd )
+ on (1) → 0, as n → ∞.
Hence, uc can solve (1.1) for t < tn when n large enough with diminishing Strichartz norm, which contradicts uc
2(d+2)
(d+2)(p−1) 2 ((−∞,0])×Rd )
∩Lt,x
Lt,x d
= ∞.
1 (Rd ). Thus tn is bounded, which implies that tn is precompact, so is uc (tn ) in Hrad Similar argument makes sense when tn → −∞, we will omit the proof. 2
We define for R > 0, E˜R (u(t)) =
|∇u(t)|2 + |u(t)|
2(d+2) d
+ |u(t)|p+1 dx,
|x|≥R
then by the compactness of the critical element, we have Corollary 5.7. Let uc be the critical element in Proposition 5.5, then for any > 0, there exists R0 () > 0 such that E˜R0 (uc (t)) ≤ E(uc ), ∀ t ∈ R. 6. Extinction of the critical element In this section, we prove the non-existence of the critical element by deriving a contradiction from Proposition 5.6 and the virial identity in the radial case. For a bounded real-valued function φ ∈ C ∞ (Rd ), we can define the virial quantity: VR (t) =
φR (x)|u(t, x)|2 dx, where φR (x) = R 2 φ( |x| R ), ∀ R > 0.
Rd
Then, for u ∈ C(I ; H 1 (Rd )), we have
(6.1)
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
VR (t) = 2R VR (t) = 4
x φ ( |x| R ) |x| · ∇u(t, x) u(t, x) dx,
·
Rd
(6.2)
∂j ∂k φR (x)∂j u(t, x)∂k u(t, x) dx −
−
2 φR (x)|u(t, x)|2 dx
2(p−1) p+1
φR (x)|u(t, x)|p+1 dx +
4 d+2
φR (x)|u(t, x)|
2(d+2) d
dx.
(6.3)
Theorem 6.1. There does not exist the radial critical element uc of (1.1) in Proposition 5.5. Proof. Let the weight function φ in (6.1) be a smooth, radial function satisfying 0 ≤ φ ≤ 1, and φ(x) =
|x| ≤ 1, |x| ≥ 2.
|x|2 , 0,
On the one hand, by (6.2), we have ∀ R > 0, |VR (t)| R, t ∈ R.
(6.4)
On the other hand, by (6.3), we have VR (t) = 4
φR (r)|∇uc (t, x)|2 dx −
Rd
−
2 φR (x)|uc (t, x)|2 dx
2(p−1) p+1
p+1
φR (x)|uc (t, x)|
dx +
4 d+2
φR (x)|uc (t, x)|
2(d+2) d
dx
≥ 8K(uc (t)) −
|uc (t)|2 dx
C R2 R≤|x|≤2R
|∇uc (t)|2 + |uc (t)|p+1 + |uc (t)|
−C
2(d+2) d
dx.
(6.5)
R≤|x|≤2R
By Lemma 2.9 and Lemma 2.6, we have 8K(uc (t)) ≥ min
d(p−1)−4 d(p−1)
∇uc (t) 2L2
+
d d+2 uc (t)
2(d+2) d 2(d+2) L d
, δ mω − Sω (uc (t))
E(uc (t)). Thus, choosing > 0 small enough and R = R() sufficiently large, then by Corollary 5.7, we get VR (t) E(uc (t)) = E(u0 ), this together with (6.4) implies for any T > 0,
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
2931
T T · E(u0 ) VR (t) dt = |VR (T ) − VR (0)| R. 0
Taking T large enough, we obtain a contradiction unless uc ≡ 0, which is impossible due to = ∞. 2 uc 2(d+2) (d+2)(p−1) Lt,x d
∩Lt,x
2
(R×Rd )
7. Blow-up We will show the blow-up result in Theorem 1.3. Let the weight function φ in (6.1) be a smooth, radial function ([29]) satisfying φ(r) = r 2 for r ≤ 1, φ (r) ≤ 2 and φ(r) is constant for r ≥ 3. By (6.3), we have VR (t) ≤
4 +
2|∇u|2 −
d(p−1) p+1 p+1 |u|
+
2(d+2) 2d d d+2 |u|
|u(t, x)|2 dx + C
C R2 R≤|x|≤3R
= 8K(u) +
dx
|u|p+1 + |u|
2(d+2) d
dx (7.1)
R≤|x|≤3R
|u|2 dx + C
C R2 R≤|x|≤3R
|u(t)|p+1 + |u(t)|
2(d+2) d
dx.
R≤|x|≤3R
Since u is radial, we have the following radial Sobolev inequalities p+1
u Lp+1 (|x|≥R) ≤ 2(d+2) d 2(d+2) L d (|x|≥R)
u
≤
p+3
C (d−1)(p−1) 2 R
C 2(d−1) R d
p−1
u L22(|x|≥R) ∇u L22(|x|≥R) , 2(d+1)
2
d u L2 (|x|≥R) ∇u Ld 2 (|x|≥R) ,
this together with (7.1), the mass conservation and Young’s inequality shows ∀ > 0, there exists R large enough that VR (t) ≤ 8K(u) + ∇u(t) 2L2 + . By K(u) < 0, energy, mass conservation and Lemma 2.7, we see K(u(t)) < − (mω − Sω (u(t))) , ∀ t ∈ Imax , thus ∇u(t) 2L2 < So we have by (7.2),
p+1 d(p−1) 2(p+1) u(t) Lp+1
− (mω − Sω (u)).
(7.2)
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
VR (t) ≤ 8K(u) + ∇u(t) 2L2 + = 16Sω (u) − 8ω u 2L2 + < 16Sω (u) − 8ω u 2L2 +
p+1 16−4d(p−1) u(t) Lp+1 p+1
16−4d(p−1) p+1
+ ∇u(t) 2L2 + p+1 + d(p−1) 2(p+1) u(t) Lp+1 − (mω − Sω (u)) + .
Here we take > 0 small enough such that 16−4d(p−1) + p+1 We also note by K(u) < 0 and Proposition 2.4, mω ≤ Hω (u(t)) = ω2 u(t) 2L2 +
d(p−1) 2(p+1)
< 0.
p+1 d(p−1)−4 4(p+1) u(t) Lp+1 ,
so 4(p+1) d(p−1)−4
p+1 mω − ω2 u(t) 2L2 ≤ u(t) Lp+1 .
Thus, we have VR (t) ≤ 16Sω (u) − 8ω u 2L2 − (mω − Sω (u)) + d(p−1) 4(p+1) 2 ω + 16−4d(p−1) + − u(t) m . ω 2 p+1 2(p+1) d(p−1)−4 2 L By Sω (u) < mω and the energy, mass conservation, we see there exists δ1 > 0 small enough such that Sω (u) ≤ (1 − δ1 )mω . Then, we get VR (t) ≤ 16(1 − δ1 )mω − 8ω u 2L2 − (mω − Sω (u)) + 4(p+1) 2 ω + − 4d(p−1)−16 + d(p−1) p+1 2(p+1) d(p−1)−4 mω − 2 u L2 ω 4(p+1) d(p−1) d(p−1) 4(p+1) 2 = − 16δ1 + δ1 − d(p−1)−4 2(p+1) mω − 2(p+1) · d(p−1)−4 · 2 u L2 + 2d(p−1) mω + . ≤ − 16δ1 + δ1 − d(p−1)−4 We can take > 0 small enough that VR (t) ≤ −4δ1 mω , which implies that u must blow up in finite time. Remark 2. The blowup is shown for p ≤ 5, which leads to the restriction of the blowup result to d ≥ 2. This is a technical restriction. See also [15,29,30] for some related discussion. Acknowledgments The authors are grateful to Ze Li and Zihua Guo for pointing out a mistake in the original version of the article, which has now been corrected. X. Cheng is grateful to K. Nakanishi for discussing on the combined nonlinear Schrödinger equation in Central China Normal University in 2015. X. Cheng has been partially supported by the NSFC grant of China (No. 11526072) and also “the Fundamental Research Funds for the Central Universities” (No. 2014B14214). C. Miao has been partially supported by the NSFC grant of China (No. 11231006, No. 11371059). C. Miao is also supported by Beijing Center of Mathematics and Information Interdisciplinary
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Science. L. Zhao has been partially supported by the NSFC grant of China (No. 10901148, No. 11371337). L. Zhao is also supported by Youth Innovation Promotion Association CAS. References [1] T. Akahori, S. Ibrahim, H. Kikuchi, H. Nawa, Existence of a ground state and blow-up problem for a nonlinear Schrödinger equation with critical growth, Differential Integral Equations 25 (3–4) (2012) 383–402, MR 2917888. [2] T. Akahori, S. Ibrahim, H. Kikuchi, H. Nawa, Existence of a ground state and scattering for a nonlinear Schrödinger equation with critical growth, Selecta Math. (N.S.) 19 (2) (2013) 545–609, MR 3090237. [3] T. Aubin, Équations différentielles non linéaires et probléme de Yamabe concernant la courbure scalaire, J. Math. Pures Appl. (9) 55 (3) (1976) 269–296, MR 0431287 (55 #4288). [4] H. Bahouri, P. Gérard, High frequency approximation of solutions to critical nonlinear wave equations, Amer. J. Math. 121 (1) (1999) 131–175, MR 1705001 (2000i:35123). [5] I.V. Barashenkov, A.D. Gocheva, V.G. Makhankov, I.V. Puzynin, Stability of the soliton-like “bubbles”, Phys. D: Nonlinear Phenom. 34 (1–2) (1989) 240–254, MR 0982390 (90b:35196). [6] P. Bégout, A. Vargas, Mass concentration phenomena for the L2 -critical nonlinear Schrödinger equation, Trans. Amer. Math. Soc. 359 (11) (2007) 5257–5282, MR 2327030 (2008g:35190). [7] H. Brézis, L. Nirenberg, Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents, Comm. Pure Appl. Math. 36 (4) (1983) 437–477, MR 0709644 (84h:35059). [8] R. Carles, S. Keraani, On the role of quadratic oscillations in nonlinear Schrödinger equations II. The L2 -critical case, Trans. Amer. Math. Soc. 359 (1) (2007) 33–62, MR 2247881 (2008a:35260). [9] T. Cazenave, Semilinear Schrödinger Equations, Courant Lecture Notes in Mathematics, vol. 10, New York University, Courant Institute of Mathematical Sciences/American Mathematical Society, New York/Providence, RI, ISBN 0-8218-3399-5, 2003, MR 2002047. [10] T. Cazenave, F.B. Weissler, The Cauchy problem for the critical nonlinear Schrödinger equation in H s , Nonlinear Anal. 14 (10) (1990) 807–836, MR 1055532 (91j:35252). [11] B. Dodson, Global well-posedness and scattering for the defocusing, L2 -critical, nonlinear Schrödinger equation when d ≥ 3, J. Amer. Math. Soc. 25 (2) (2012) 429–463, MR2869023. [12] B. Dodson, Global well-posedness and scattering for the defocusing, L2 -critical, nonlinear Schrödinger equation when d = 2, arXiv:1006.1375. [13] B. Dodson, Global well-posedness and scattering for the defocusing, L2 -critical, nonlinear Schrödinger equation when d = 1, arXiv:1010.0040. [14] D. Foschi, Inhomogeneous Strichartz estimates, J. Hyperbolic Differ. Equ. 2 (1) (2005) 1–24, MR 2134950 (2006a:35043). [15] J. Holmer, S. Roudenko, On blow-up solutions to the 3D cubic nonlinear Schrödinger equation, Appl. Math. Res. Express. AMRX 1 (2007) abm004, MR2354447 (2008i:35227). [16] S. Ibrahim, N. Masmoudi, K. Nakanishi, Scattering threshold for the focusing nonlinear Klein–Gordon equation, Anal. PDE 4 (3) (2011) 405–460, MR 2872122. [17] M. Keel, T. Tao, Endpoint Strichartz estimates, Amer. J. Math. 120 (5) (1998) 955–980, MR 1646048 (2000d:35018). [18] C.E. Kenig, F. Merle, Global well-posedness, scattering and blow-up for the energy-critical, focusing, non-linear Schrödinger equation in the radial case, Invent. Math. 166 (3) (2006) 645–675, MR 2257393 (2007g:35232). [19] S. Keraani, On the defect of compactness for the Strichartz estimates of the Schrödinger equations, J. Differential Equations 175 (2) (2001) 353–392, MR 1855973 (2002j:35281). [20] S. Keraani, On the blow up phenomenon of the critical Schrödinger equation, J. Funct. Anal. 235 (1) (2006) 171–192, MR 2216444 (2007e:35260). [21] R. Killip, T. Tao, M. Visan, The cubic nonlinear Schrödinger equation in two dimensions with radial data, J. Eur. Math. Soc. (JEMS) 11 (6) (2009) 1203–1258, MR 2557134 (2010m:35487). [22] R. Killip, M. Visan, The focusing energy-critical nonlinear Schrödinger equation in dimensions five and higher, Amer. J. Math. 132 (2) (2010) 361–424, MR 2654778 (2011e:35357). [23] R. Killip, M. Visan, Nonlinear Schrödinger Equations at Critical Regularity, Proceedings for the Clay Summer School “Evolution Equations”, Eidgenössische technische Hochschule, Zürich, 2008. [24] R. Killip, M. Visan, X. Zhang, The mass-critical nonlinear Schrödinger equation with radial data in dimensions three and higher, Anal. PDE 1 (2) (2008) 229–266, MR 2472890 (2011b:35487). [25] E.H. Lieb, M. Loss, Analysis, Graduate Studies in Mathematics, vol. 14, American Mathematical Society, Providence, RI, ISBN 0-8218-2783-9, 2001, MR 1817225 (2001i:00001).
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ScienceDirect J. Differential Equations 261 (2016) 2881–2934 www.elsevier.com/locate/jde
Global well-posedness and scattering for nonlinear Schrödinger equations with combined nonlinearities in the radial case Xing Cheng a , Changxing Miao b , Lifeng Zhao c,∗ a College of Science, Hohai University, Nanjing 210098, Jiangsu, China b Institute of Applied Physics and Computational Mathematics, P. O. Box 8009, Beijing 100088, China c Wu Wen-Tsun Key Laboratory of Mathematics and School of Mathematical Sciences, University of Science
and Technology of China, Hefei 230026, Anhui, China Received 20 February 2016 Available online 1 June 2016
Abstract We consider the Cauchy problem for the nonlinear Schrödinger equation with combined nonlinearities, one of which is defocusing mass-critical and the other is focusing energy-critical or energy-subcritical. The threshold is given by means of variational argument. We establish the profile decomposition in H 1 (Rd ) and then utilize the concentration–compactness method to show the global wellposedness and scattering versus blowup in H 1 (Rd ) below the threshold for radial data when d ≤ 4. © 2016 Elsevier Inc. All rights reserved.
MSC: 35L70; 35Q55 Keywords: Nonlinear Schrödinger equation; Combined nonlinearities; Global wellposedness; Scattering; Blowup
Contents 1.
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2882
* Corresponding author.
E-mail addresses: [email protected] (X. Cheng), [email protected] (C. Miao), [email protected] (L. Zhao). http://dx.doi.org/10.1016/j.jde.2016.04.031 0022-0396/© 2016 Elsevier Inc. All rights reserved.
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2. Variational estimates . . . . . . . . . . . . . 3. Wellposedness and perturbation theory 4. Linear profile decomposition . . . . . . . 5. Extraction of a critical element . . . . . . 6. Extinction of the critical element . . . . 7. Blow-up . . . . . . . . . . . . . . . . . . . . . Acknowledgments . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . .
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1. Introduction We will consider the Cauchy problem for the nonlinear Schrödinger equation 4 i∂t u + u = |u| d u − |u|p−1 u, u(0) = u0 ∈ H 1 (Rd ), where u : R × Rd → C is a complex-valued function, 1 + 1 + d4 < p < ∞, d = 1, 2. The equation (1.1) has the following mass and energy: M(u) = |u|2 dx,
E(u) =
4 d
Rd 2 1 2 |∇u|
−
p+1 1 p+1 |u|
+
d 2(d+2) |u|
4 d−2 ,
(1.1) d = 3, 4 and
(1.2) 2(d+2) d dx.
(1.3)
Rd
The equation (1.1) is a special case of the general nonlinear Schrödinger equation with combined nonlinearities i∂t u + u = μ1 |u|p1 −1 u + μ2 |u|p2 −1 u, (1.4) u(0) = u0 ∈ H 1 (Rd ), 4 where 1 < p1 , p2 ≤ 1 + d−2 , for d ≥ 3, 1 < p1 , p2 < ∞, for d = 1, 2, μ1 , μ2 ∈ {±1}. This equation arises in the study of the Hartree approximation and quasi-classically within the frame of the secondary quantization in the boson gas with many body δ-function interaction. It can also be used to describe the effect of saturation of nonlinear refractive index. At the same time, the equation of nuclear hydrodynamics with effective Skyrme’s forces reduces quasi-classically to (1.4). For more physical background, we refer the reader to [5,31] and the references therein. The local theory for (1.4) in the energy space follows from the standard method of T. Cazenave, F.B. Weissler [10], see also [9]. Proceeded by [37], T. Tao, M. Visan and X. Zhang considered various cases in [33]. They proved global wellposedness and scattering of the solution 4 to the equation (1.4) for finite energy data when μ1 = μ2 = +1 and 1 + d4 ≤ p1 < p2 ≤ 1 + d−2 , 4 4 d ≥ 3. The case when p1 = 1 + d and p2 = 1 + d−2 is the most difficult. In [33], the low frequencies of the solution are well approximated by the L2 critical problem and the high frequencies
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are well approximated by the energy-critical problem. The medium frequencies are controlled by the Morawetz estimates. In [27], C. Miao, G. Xu and L. Zhao considered the case where μ1 = +1, μ2 = −1 and 4 , d = 3. The threshold was given by variational method due to 1 + d4 < p1 < p2 = 1 + d−2 the energy trapping property as in [16]. They established the linear profile decomposition in H 1 (Rd ) in the spirit of [16]. By using this new profile decomposition, they reduced the scattering problem to the extinction of the critical element. The critical element can then be excluded by using the virial identity. They showed the dichotomy of global wellposedness and scattering versus blow-up phenomenon below the threshold for radial solutions. The radial assumption was removed in dimensions five and higher in [28]. 4 If μ1 = μ2 = −1 and 1 + d4 < p1 < p2 = 1 + d−2 , d ≥ 5, the Cauchy problem was considered in [1,2]. After giving existence of the ground state based on the idea in [7] and [16], they showed a sufficient and necessary condition for the scattering in the spirit of [18] by using the profile decomposition in H˙ 1 (Rd ) and the global wellposedness and scattering result in [22]. In this paper, we aim to look for the suitable threshold to study the global well-posedness and scattering versus blowup of (1.1). Before stating the main theorem, we introduce some notad tions. For ϕ ∈ H 1 (Rd ), we denote the scaling (Tλ ϕ)(x) = λ 2 ϕ(λx). For any ω > 0, we have the Lyapunov functional Sω (ϕ) = E(ϕ) + 12 ωM(ϕ).
(1.5)
We also denote the scaling derivative of Sω (ϕ) by K(ϕ), d d K(ϕ) = L Sω (ϕ) = dλ S (T ϕ) = dλ E(Tλ ϕ) λ=1 ω λ λ=1 2(d+2) p+1 d + d+2 |ϕ| d dx. = |∇ϕ|2 − d(p−1) 2(p+1) |ϕ|
(1.6)
Rd
Let mω = inf{Sω (ϕ) : ϕ ∈ H 1 (Rd ) \ {0}, K(ϕ) = 0},
(1.7)
then we have 4 Proposition 1.1. For 1 + d4 < p < 1 + d−2 , d ≥ 3, and 1 + d4 < p < ∞, d = 1, 2, ω > 0, we have mω > 0. Moreover, mω = Sω (Q), where Q ∈ H 1 (Rd ) is the ground state of 4
−ωQ + Q + |Q|p−1 Q − |Q| d Q = 0. Proposition 1.2. For p = 1 +
4 d−2 ,
d ≥ 3, we have for all ω > 0, mω = E 0 (W ),
where
(1.8)
(1.9)
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E (W ) = 0
2 1 2 |∇W |
−
2d d−2 d−2 2d |W |
dx,
Rd
and W ∈ H˙ 1 (Rd ) is unique positive radial solution of 4
−W = |W | d−2 W.
(1.10)
Remark 1. 4 (1) When p = 1 + d−2 , we note mω = E 0 (W ), which indicates that mω is independent of ω, so we can define m := mω = E 0 (W ) in this case. Let
Aω,+ = {ϕ ∈ H 1 (Rd ) : Sω (ϕ) < mω , K(ϕ) ≥ 0}, Aω,− = {ϕ ∈ H 1 (Rd ) : Sω (ϕ) < mω , K(ϕ) < 0}, A+ = {ϕ ∈ H 1 (Rd ) : E(ϕ) < m, K(ϕ) ≥ 0}, A− = {ϕ ∈ H 1 (Rd ) : E(ϕ) < m, K(ϕ) < 0}. (2) Aω,± and A± are non-empty by using the same argument in [27]. In fact, we note ϕ = 0 belongs to both Aω,+ and A+ , so Aω,+ and A+ are nonempty. On the other hand, we can d easily verify that ϕ(x) = − 2 Q( x ) belongs to Aω,− , when is sufficiently small, similarly, d
by using some truncation of ϕ = − 2 W ( x ) to make sure ϕ ∈ H 1 (Rd ), we can show A− is nonempty. (3) We note for any u0 ∈ A+ , we have E(u0 ) < m, we can take ω > 0 small enough such that E(u0 ) + 12 ωM(u0 ) < m, thus u0 ∈ Aω,+ . Similarly, for any u0 ∈ A− , we can still take ω > 0 4 , small enough such that u0 ∈ Aω,− . Thus, our result below in the situation p = 1 + d−2 4 4 d = 3, 4 can be proved as in the case 1 + d < p < 1 + d−2 , d ∈ {3, 4}. So we will give 4 the proof of the following theorem in a unified form regardless of 1 + d4 < p < 1 + d−2 , 4 4 d ∈ {3, 4}, 1 + d < p < ∞, d ∈ {1, 2} or p = 1 + d−2 , d = 3, 4. Now, we state our result. Theorem 1.3. 4 (1) For 1 + d4 < p < 1 + d−2 , d ∈ {3, 4}, 1 + d4 < p < ∞, d ∈ {1, 2}, and u0 is radial, we have for any ω > 0, (i) if u0 ∈ Aω,+ , the solution u to (1.1) exists globally and scatters in H 1 (Rd ); 4 (ii) if u0 ∈ Aω,− , for d ≥ 2, p ≤ min(5, 1 + d−2 ), the solution u blows up in finite time. 4 (2) For p = 1 + d−2 , d ∈ {3, 4}, and u0 is radial, we have (i) if u0 ∈ A+ , the solution u to (1.1) exists globally and scatters in H 1 (Rd ); (ii) if u0 ∈ A− , the solution u blows up in finite time.
For the nonlinear Schrödinger equation with combined nonlinearities, we focus on the different roles played by the two nonlinearities. Generally speaking, the main barrier of the local
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theory is the higher order term while the lower order term is dominant in the global behavior. We prove our main theorem by the compactness–contradiction method initiated by C.E. Kenig and F. Merle [18]. In the argument, the linear profile decomposition plays an important role. In previous works such as [27,28], the authors considered the equation
4
i∂t u + u = |u|p−1 u − |u| d−2 u, u(0) = u0 ∈ H 1 (Rd ),
(1.11)
4 2 where 1 + d4 < p < 1 + d−2 , d ≥ 3. Since the lower order term is H˙ s (Rd ) (0 < s = d2 − p−1 < 1) critical, the solution of (1.11) is expected to behave like that of the defocusing equation i∂t u + u = |u|p−1 u. The critical space of this defocusing equation is H˙ s (0 < s < 1), so it ∇
1 d is reasonable to apply the H˙ s -profile decomposition to |∇| s u for u ∈ H (R ). Since the symmetry group in H˙ s is of the same type as in H˙ 1 case, we may equivalently apply the H˙ 1 -profile 1 d 2 d decomposition to ∇
|∇| u for u ∈ H (R ). However, the lower term in (1.1) is L (R )-critical. The symmetry group in L2 (Rd ) is different from that in H˙ 1 (Rd ) due to the Galilean symmetry. Therefore, it is natural that the L2 -profile decomposition is applied to ∇ u for u ∈ H 1 (Rd ) in this paper. We explore the profile decomposition in use to get better estimate (4.14) for the remainder. This estimate provides spacetime control for both the remainder wnk and its derivative |∇|wnk . This makes remarkable difference with the profile decomposition for H 1 (Rd ) data obtained both in [8] and [27]. In fact, R. Carles and S. Keraani [8] only provided the control over the remainder, while the authors in [27] only provided the control over the derivative of the remainder. Although we make more delicate analysis due to the stronger control in the perturbation theorem, we get stronger compactness for the critical element in H 1 (Rd ), which is stronger than the compactness in H˙ s (0 < s ≤ 1) obtained in [27,28]. We expect our result will be extended to higher dimensions (d ≥ 5) since all the arguments make sense except the long-time perturbation. However, the exotic Strichartz estimates in [14,36] seem useless to establish the long-time perturbation in our case because the mass-critical term in our equation cannot be controlled properly in the Sobolev spaces. The radial assumption is expected to be removed in a forthcoming paper. The rest of the paper is organized as follows. After introducing some notations and preliminaries, we give the threshold in Section 2. Moreover, we show the energy-trapping properties for the set Aω,± in this section. The local wellposedness and perturbation theory are stated in Section 3. In Section 4, we derive the linear profile decomposition for data in H 1 (Rd ). Then we argue by contradiction. We reduce to the existence of a critical element in Section 5 and show the extinction of such a critical element in Section 6. To make the results complete, we show the existence of blowup solutions in Section 7.
Notation and preliminaries We will use the notation X Y whenever there exists some positive constant C so that X ≤ CY . Similarly, we will use X ∼ Y if X Y X. We define the Fourier transform on Rd to be fˆ(ξ ) =
1 (2π)d Rd
e−ixξ f (x) dx,
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s f (ξ ) = |ξ |s fˆ(ξ ). We and for s ∈ R, the fractional differential operators |∇|s is defined by |∇| s s s ˆ also define ∇ by ∇ f (ξ ) = (1 + |ξ |) f (ξ ). We define the homogeneous Sobolev norms
f H˙ s (Rd ) = |∇|s f L2 (Rd ) , and inhomogeneous Sobolev norms f H s (Rd ) = ∇ s f L2 (Rd ) . We define the radial Lebesgue and Sobolev space as follows: for 1 < q < ∞, s ∈ R, we define q
Lrad (Rd ) = {f ∈ Lq (Rd ) : f is radial function}, s Hrad (Rd ) = {f ∈ H s (Rd ) : f is radial function}, s (Rd ) = {f ∈ H˙ s (Rd ) : f is radial function}. H˙ rad
We use the notation on (1) to denote a quantity which tends to 0, as n → ∞. q For I ⊂ R, we use Lt Lrx (I × Rd ) to denote the spacetime norm u Lqt Lr (I ×Rd ) =
x
I q
|u(t, x)| dx r
qr
q1 dt
.
Rd
q
When q = r, we abbreviate Lt Lrx as Lt,x . We also recall Duhamel’s formula t u(t) = e
i(t−t0 )
u(t0 ) − i
ei(t−s) (iut + u)(s)ds.
(1.12)
t0
We say that a pair of exponents (q, r) is L2 -admissible if (q, r, d) = (2, ∞, 2), d ≥ 1.
2 q
+
d r
=
d 2
and 2 ≤ q, r ≤ ∞,
Lemma 1.4 (Strichartz estimate, [17]). Let I be a compact time interval and let u : I × Rd → C be a solution to the forced Schrödinger equation i∂t u + u = G for some function G, then we have u Lq Lr (I ×Rd ) u(t0 ) L2x (Rd ) + G t
x
q˜
Lt Lrx˜ (I ×Rd )
for any t0 ∈ I and any L2 -admissible exponents (q, r), (q, ˜ r˜ ). If I × Rd is a spacetime slab, we define the Strichartz norm S 0 (I ) by u S 0 (I ) = sup u Lq Lr (I ×Rd ) , t
x
(1.13)
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where the sup is taken over all L2 -admissible pairs (q, r). When d = 2, we need to modify the norm a little, where the sup is taken over all L2 -admissible pairs with q ≥ 2 + , for > 0 arbitrary small. We also define S 1 (I ) norm by u S 1 (I ) = ∇ u S 0 (I ) .
(1.14)
2. Variational estimates In this section, we prove Proposition 1.1 and 1.2. We show the existence of the ground state together with the energy-trapping property for Aω,± , which will be used to show the scattering and blow-up. Let K(ϕ) = KQ (ϕ) + KN (ϕ), where Q K (ϕ) = |∇ϕ|2 dx, KN (ϕ) = − d(p−1) 2(p+1)
Rd
|ϕ|p+1 dx +
d d+2
|ϕ|
2(d+2) d
dx.
We have the following basic fact about K(Tλ ϕ): Lemma 2.1 (Dynamic behavior of K under the scaling). For any ϕ ∈ H 1 (Rd ) \ {0}, there exists a unique λ0 (ϕ) > 0 that ⎧ ⎪ ⎨> 0, K(Tλ ϕ) = 0, ⎪ ⎩ < 0,
0 < λ < λ0 (ϕ), λ = λ0 (ϕ), λ > λ0 (ϕ).
(2.1)
Proof. An easy computation gives 1 K(Tλ ϕ) = λ2
|∇ϕ|2 +
2(d+2) d d d+2 |ϕ|
−
d(p−1) d2 (p−1)−2 |ϕ|p+1 dx, 2(p+1) λ
which implies K(Tλ ϕ) > 0 for λ > 0 sufficiently small. By d dλ
1 K(Tλ ϕ) λ2
= − d(p−1) 2(p+1)
d
d (p−1)−3 2 2 (p − 1) − 2 λ
|ϕ|p+1 dx < 0,
we see λ12 K(Tλ ϕ) is monotone decreasing with respect to λ > 0. Since 2(d+2) d p+1 2 2 d d K(Tλ ϕ) = λ ∇ϕ L2 + d+2 ϕ 2(d+2) − λ 2 (p−1) d(p−1) 2(p+1) ϕ Lp+1 L
d
→ −∞, as λ → ∞, there exists a unique λ0 > 0 such that K(Tλ0 ϕ) = 0 and (2.1) follows.
2
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Now, we show the positivity of K near 0 in the energy space. Lemma 2.2. For any bounded sequence ϕn ∈ H 1 (Rd ) \ {0} with KQ (ϕn ) → 0, as n → ∞, then for n large enough, we have K(ϕn ) > 0. Proof. By assumption, ∇ϕn L2 → 0, as n → ∞. Due to the interpolation and Sobolev inequalities, we have p+1− d2 (p−1)
p+1
ϕn Lp+1 ϕn L2
d
Since d2 (p − 1) > 2, for n large enough, we see p+1 K(ϕn ) = |∇ϕn |2 − d(p−1) + 2(p+1) |ϕn | Rd
≥
|∇ϕn | dx − o 2
Rd
.
2(d+2) d d d+2 |ϕn |
|∇ϕn |2 dx
dx
Rd
|∇ϕn |2 dx > 0.
∼
(p−1)
∇ϕn L2 2
2
Rd
Lemma 2.3. For any ϕ ∈ H 1 (Rd ), we have (2 − L)Sω (ϕ) = ω ϕ 2L2 + L(2 − L)Sω (ϕ) =
p+1 d(p−1)−4 2(p+1) ϕ Lp+1 ,
p+1 d(p−1)(d(p−1)−4) ϕ Lp+1 . 4(p+1)
Proof. Direct computation shows that L ∇ϕ 2L2 = 2 ∇ϕ 2L2 , p+1
p+1
L ϕ Lp+1 = d2 (p − 1) ϕ Lp+1 , L ϕ
2(d+2) d 2(d+2) L d
= 2 ϕ
2(d+2) d 2(d+2) L d
,
hence we have (2 − L)Sω (ϕ) = 2Sω (ϕ) − K(ϕ) = ω ϕ 2L2 +
d(p−1)−4 2(p+1)
|ϕ|p+1 dx, Rd
L(2 − L)Sω (ϕ) = ωL ϕ 2L2 +
p+1 d(p−1)−4 2(p+1) L ϕ Lp+1
=
p+1 d(p−1)−4 2(p+1) L ϕ Lp+1
=
p+1 (d(p−1)−4)d(p−1) ϕ Lp+1 . 4(p+1)
2
(2.2) (2.3)
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
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Due to the lack of positivity of Sω (ϕ), we introduce a non-negative functional Hω (ϕ) = 1 − L2 Sω (ϕ) = Sω (ϕ) − 12 K(ϕ) = ω2 ϕ 2L2 +
(2.4)
p+1 d(p−1)−4 4(p+1) ϕ Lp+1 ,
then for any ϕ ∈ H 1 (Rd ) \ {0}, we have Hω (ϕ) ≥ 0, LHω (ϕ) ≥ 0. Proposition 2.4. mω = inf{Hω (ϕ) : ϕ ∈ H 1 \ {0}, K(ϕ) ≤ 0} = inf{Hω (ϕ) : ϕ ∈ H 1 \ {0}, K(ϕ) < 0}.
(2.5)
Proof. Since Sω (ϕ) = Hω (ϕ) when K(ϕ) = 0, mω ≥ inf{Hω (ϕ) : ϕ ∈ H 1 \ {0}, K(ϕ) ≤ 0}. Claim: mω ≤ inf{Hω (ϕ) : ϕ ∈ H 1 \ {0}, K(ϕ) < 0}. In fact, ∀ ϕ ∈ H 1 \ {0} with K(ϕ) < 0, by Lemma 2.1, there exists 0 < λ0 < 1 that K(Tλ0 ϕ) = 0, then the fact LHω ≥ 0 implies Sω (Tλ0 ϕ) = Hω (Tλ0 ϕ) ≤ Hω (ϕ). It suffices to show inf{Hω (ϕ) : ϕ ∈ H 1 \ {0}, K(ϕ) ≤ 0} ≥ inf{Hω (ϕ) : ϕ ∈ H 1 \ {0}, K(ϕ) < 0}.
(2.6)
In fact, for any ϕ ∈ H 1 \ {0} with K(ϕ) ≤ 0, by (2.3), we know that L K(ϕ) = 2K(ϕ) −
p+1 d(p−1)(d(p−1)−4) ϕ Lp+1 4(p+1)
< 0,
(2.7)
then for any λ > 1, we have K(Tλ ϕ) < 0, and as λ → 1, Hω (Tλ ϕ) = ω2 Tλ ϕ 2L2 + = ω2 ϕ 2L2 +
p+1 d(p−1)−4 d2 (p−1) ϕ Lp+1 4(p+1) λ
→ ω2 ϕ 2L2 + This shows (2.6) and completes the proof.
p+1 d(p−1)−4 4(p+1) Tλ ϕ Lp+1
p+1 d(p−1)−4 4(p+1) ϕ Lp+1
= Hω (ϕ).
2
4 We now give the value of mω for 1 + d4 < p < 1 + d−2 , d ≥ 3, and 1 + d4 < p < ∞, d = 1, 2, namely prove Proposition 1.1.
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Proof of Proposition 1.1. Let ϕn ∈ H 1 (Rd ) be a minimizing sequence for (2.5), namely K(ϕn ) ≤ 0, ϕn = 0, Hω (ϕn ) mω , as n → ∞. Let ϕn∗ be the Schwartz symmetrization of ϕn , i.e. the radial decreasing rearrangement. Since the symmetrization preserves the nonlinear parts and does not increase the H˙ 1 part, we have ϕn∗ = 0, K(ϕn∗ ) ≤ K(ϕn ) ≤ 0 and Hω (ϕn∗ ) = Hω (ϕn ) → mω , as n → ∞. Then by Lemma 2.1 and (2.7), there exists 0 < λn ≤ 1 such that ψn = Tλn ϕn∗ satisfies ψn = 0, K(ψn ) = 0, Sω (ψn ) = Hω (ψn ) → mω , as n → ∞. Moreover, direct computation gives 2 ω 2 ψn L2
+ 12 ∇ψn 2L2 ≤ Sω (ψn ) +
4 d(p−1)−4 Hω (ψn ),
which implies the boundedness of ψn in H 1 (Rd ). Then, ψn converges weakly to some ψ in H 1 (Rd ), up to a subsequence. Since ψn is radial, 2d it also converges strongly in Lq (Rd ) for 2 < q < d−2 , d ≥ 3 and 2 < q < ∞, d = 1, 2. Thus, K(ψ) ≤ lim inf K(ψn ) = 0, Hω (ψ) ≤ lim inf Hω (ψn ) = mω . Moreover, ψ = 0. In fact, if ψ = 0, n→∞
n→∞
then K(ψn ) = 0 implies K Q (ψn ) = −K N (ψn ) → 0, as n → ∞, and by Lemma 2.2, we have K(ψn ) > 0 for n large, a contradiction. Since K(ψ) ≤ 0 and ψ = 0, we have Hω (ψ) ≥ mω , so we have Hω (ψ) = mω and K(ψ) ≤ 0. By scaling, we may replace ψ by its rescaling, so that K(ψ) = 0, Sω (ψ) = Hω (ψ) ≤ mω and ψ = 0. Then ψ is a minimizer and mω = Hω (ψ) > 0. By variational theory, there is a Lagrange multiplier η ∈ R such that Sω (ψ) = ηK (ψ). We note 0 = K(ψ) = LSω (ψ) = Sω (ψ)Lψ = ηK (ψ)Lψ = η · L2 Sω (ψ). By (2.3) and LSω (ψ) = 0, we have L2 Sω (ψ) = 2LSω (ψ) −
p+1 d(p−1)(d(p−1)−4) ψ Lp+1 4(p+1) p+1
= − d(p−1)(d(p−1)−4) ψ Lp+1 < 0, 4(p+1) 4
therefore η = 0 and ψ is a solution to −ωQ + Q + |Q|p−1 Q − |Q| d Q = 0. The minimality of Sω (Q) among the solutions is clear from (1.7), since every solution Q in H 1 (Rd ) of (1.8) satisfies K(Q) = Sω (Q)LQ = 0. 2
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
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4 We now turn to find the ground state in the case p = 1 + d−2 , d ≥ 3, that is to prove Proposition 1.2. Let 2d K0 (ϕ) = |∇ϕ|2 − |ϕ| d−2 dx, (2.8) Rd 2d
H0 (ϕ) = d1 ϕ d−22d .
(2.9)
L d−2
We will show Lemma 2.5. For p = 1 +
4 d−2 ,
d ≥ 3,
mω = inf{H 0 (ϕ) : ϕ ∈ H 1 \ {0}, K0 (ϕ) < 0}
(2.10)
= inf{H 0 (ϕ) : ϕ ∈ H 1 \ {0}, K0 (ϕ) ≤ 0}. Proof. Since K0 (ϕ) ≤ K(ϕ), H0 (ϕ) ≤ Hω (ϕ), it follows that
mω = inf{Hω (ϕ) : ϕ ∈ H 1 \ {0}, K(ϕ) < 0} ≥ inf{H0 (ϕ) : ϕ ∈ H 1 \ {0}, K0 (ϕ) < 0}. Hence, in order to show the first equality, it suffices to show inf{Hω (ϕ) : ϕ ∈ H 1 \ {0}, K(ϕ) < 0} ≤ inf{H0 (ϕ) : ϕ ∈ H 1 \ {0}, K0 (ϕ) < 0}. For any ϕ ∈ H 1 \ {0} with K0 (ϕ) < 0, taking Tλ ϕ(x) = λ K(Tλ ϕ) =
2d
|∇ϕ|2 − |ϕ| d−2 +
d−2 2
(2.11)
ϕ(λx), we have as λ → ∞,
2(d+2) − d4 d |ϕ| d d+2 λ
dx → K0 (ϕ),
Rd 2d
Hω (Tλ ϕ) = ω2 λ−2 ϕ 2L2 + d1 ϕ d−22d → H0 (ϕ). L d−2
This gives (2.11) and completes the proof of the first equality. For the second equality, it suffices to show inf{H0 (ϕ) : ϕ ∈ H 1 \ {0}, K0 (ϕ) < 0} ≤ inf{H0 (ϕ) : ϕ ∈ H 1 \ {0}, K0 (ϕ) ≤ 0}. For any ϕ ∈ H 1 \ {0} with K0 (ϕ) ≤ 0 and L K0 (ϕ) =
2|∇ϕ|2 −
2d 2d d−2 d−2 |ϕ|
dx = 2K0 (ϕ) −
which implies K0 (Tλ ϕ) < 0, for λ > 1. We also have
2d
d−2 4 d−2 ϕ 2d L d−2
< 0,
(2.12)
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934 2d
2d
H0 (Tλ ϕ) = d1 λ d−2 ϕ d−22d → H0 (ϕ), as λ → 1, L d−2
so we obtain (2.12) and complete the proof.
2
Proof of Proposition 1.2. By Lemma 2.5, we have 2d 2d d−2 d−2 1 2 1 mω = inf d ϕ 2d : ϕ ∈ H \ {0}, ∇ϕ L2 ≤ ϕ 2d L d−2
≥ inf
≥ inf
2d
2 1 d ∇ϕ L2
≥ inf
L d−2
L d−2
⎧ ⎪ ⎪ ⎨
⎛
1 ∇ϕ 2L2 ⎪d
⎪ ⎩
⎧ ⎨ 1
⎩d
: ϕ ∈ H 1 \ {0}, ∇ϕ 2L2 ≤ ϕ d−22d ⎜ ⎝
∇ϕ L2 ϕ 2d
⎞ d−2 2
∇ϕ 2 2
L 2d d−2 ϕ 2d L d−2
⎟ ⎠
: ϕ ∈ H 1 \ {0}
d : ϕ ∈ H˙ 1 \ {0}
L d−2
⎫ ⎬ ⎭
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
= d1 (Cd∗ )−d ,
where Cd∗ is the sharp Sobolev constant in Rd , that is ϕ
2d
L d−2 (Rd )
≤ Cd∗ ∇ϕ L2 , ∀ ϕ ∈ H˙ 1 (Rd ),
(2.13)
with the equality is attained by W (see [3,32]) and d1 (Cd∗ )−d = E 0 (W ). On the other hand, by the density of H 1 (Rd ) in H˙ 1 (Rd ), we can find ϕn ∈ H 1 \ {0}, such that ϕn → W in H˙ 1 (Rd ), as n → ∞. Then we have H0 (ϕn ) → H0 (W ) = E 0 (W ), as n → ∞, by Lemma 2.5, H0 (ϕn ) ≥ mω for n large enough. So we have mω ≤ E 0 (W ). Thus, we obtain (1.9). 2 Next we show the energy-trapping properties of Aω,± . Lemma 2.6. For 1 + d4 < p ≤ 1 + ϕ ∈ H 1 (Rd ) with K(ϕ) ≥ 0, d(p−1)−4 d(p−1)
2 1 2 |∇ϕ|
+
d ≥ 3, and 1 +
4 d−2 ,
2(d+2) d d 2(d+2) |ϕ|
4 d
< p < ∞, d = 1, 2, we have for any
dx ≤ E(ϕ) ≤
2 1 2 |∇ϕ|
+
2(d+2) d d 2(d+2) |ϕ|
Proof. On the one hand, E(ϕ) =
−
p+1 1 p+1 |ϕ|
2 1 2 |∇ϕ|
+
2(d+2) d d 2(d+2) |ϕ|
≤ On the other hand, since
+
2(d+2) d d 2(d+2) |ϕ|
2 1 2 |∇ϕ|
dx.
dx
dx.
(2.14)
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
K(ϕ) =
|∇ϕ|2 −
d(p−1) p+1 2(p+1) |ϕ|
+
2(d+2) d d d+2 |ϕ|
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dx ≥ 0,
we have
|ϕ|
1 p+1
p+1
dx ≤
2 d(p−1)
|∇ϕ|2 +
2(d+2) d d d+2 |ϕ|
dx.
So E(ϕ) =
−
p+1 1 p+1 |ϕ|
2 1 2 |∇ϕ|
+
2(d+2) d d 2(d+2) |ϕ|
≥ =
1 2
−
2 d(p−1)
|∇ϕ|2 +
=
d(p−1)−4 d(p−1)
+
2(d+2) d d 2(d+2) |ϕ|
2 1 2 |∇ϕ|
2 1 2 |∇ϕ|
+
dx 2 |∇ϕ|2 + dx − d(p−1)
d 2(d+2)
−
2 (d+2)(p−1)
2(d+2) d d 2(d+2) |ϕ|
|ϕ|
2(d+2) d d d+2 |ϕ| 2(d+2) d
dx
dx
dx. 2
4 Proposition 2.7 (Energy-trapping for Aω,− ). For 1 + d4 < p ≤ 1 + d−2 , d ≥ 3, 1 + d4 < p < ∞, d = 1, 2, and u0 ∈ Aω,− . Let u be the solution of (1.1), and Imax be the lifespan of u, then
K(u(t)) < − mω − Sω (u(t)) , ∀ t ∈ Imax .
(2.15)
Proof. We first claim that K(u(t)) < 0, for t ∈ Imax . Indeed, since u0 ∈ Aω,− , we have by the mass and energy conservation that Sω (u(t)) < mω , for t ∈ Imax . If K(u(t0 )) ≥ 0 for some t0 ∈ Imax , then there is t1 ∈ Imax , such that K(u(t1 )) = 0. So we have Sω (u(t1 )) ≥ mω , which contradicts Sω (u(t)) < mω , for all t ∈ Imax , so we have K(u(t)) < 0 for t ∈ Imax . Next, we turn to (2.15). By the above claim, for any t ∈ Imax , there exists 0 < λ(t) < 1 such that K(Tλ(t) u(t)) = 0, which together with the definition of mω shows that Sω (Tλ(t) u(t)) ≥ mω . By Lemma 2.3, L2 Sω (u(t)) = 2L Sω (u(t)) − = 2K(u(t)) −
p+1 d(p−1)(d(p−1)−4) u(t) Lp+1 4(p+1)
p+1 d(p−1)(d(p−1)−4) u(t) Lp+1 4(p+1)
< 0,
we have d Sω (u(t)) > Sω (Tλ(t) u(t)) + (1 − λ(t)) dλ S (T u(t)) λ=1 ω λ = Sω (Tλ(t) u(t)) + (1 − λ(t))K(u(t)) > mω + K(u(t)), so K(u(t)) < −(mω − Sω (u(t))).
2
Before discussing the energy-trapping for Aω,+ , we first show ∀ ω > 0, Aω,+ is bounded in H 1 (Rd ).
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
Lemma 2.8. Let ω > 0 and u ∈ Aω,+ , then we have u 2L2 ≤
2mω ω ,
∇u 2L2 mω ,
and hence u 2H 1 mω +
mω ω .
(2.16)
Proof. By u ∈ Aω,+ , we have Hω (u) ≤ Sω (u) ≤ mω , which implies 2mω ω .
u 2L2 ≤
The boundedness of ∇u L2 follows from 1 mω ≥ Sω (u) ≥ E(u) = 12 |∇u|2 − p+1 |u|p+1 + 2 1 2 |∇u|
≥ ≥
−
1 2
+
2 d(p−1)
2(d+2) d d 2(d+2) |u|
dx −
2(d+2) d d 2(d+2) |u|
2 d(p−1)
dx
∇u 2L2 +
d d+2 u
2(d+2) d 2(d+2) L d
∇u 2L2 ,
where the first inequality is given by K(u) ≥ 0.
2
Proposition 2.9 (Energy-trapping for Aω,+ ). For u0 ∈ Aω,+ , let u be a solution of (1.1), with Imax the lifespan, we have δ > 0 depending on d, p and ω such that for t ∈ Imax , K(u(t)) ≥ min
d(p−1)−4 d(p−1)
∇u(t) 2L2
+
d d+2 u(t)
2(d+2) d 2(d+2) L d
, δ mω − Sω (u(t)) .
(2.17)
Proof. Direct computation shows that Sω (Tλ u(t)) = ω2 Tλ u(t) 2L2 + E(Tλ u(t)) 2 ω = 2 u(t) L2 + 12 λ2 |∇u(t)|2 + and d2 S (T u(t)) = dλ2 ω λ
Since
|∇u(t)|2 +
2(d+2) d d d+2 |u(t)|
K(Tλ u(t)) =
2(d+2) 2 d d 2(d+2) λ |u(t)|
λ2 |∇u(t)|2 −
−
−
d 1 2 (p−1) |u(t)|p+1 dx p+1 λ
d(p−1)(d(p−1)−2) d2 (p−1)−2 λ |u(t)|p+1 dx. 4(p+1)
d(p−1) d2 (p−1) |u(t)|p+1 2(p+1) λ
+
2(d+2) 2 d d d+2 λ |u(t)|
dx,
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
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we have d2 S (T u(t)) = − λ12 K(Tλ u(t)) + λ22 dλ2 ω λ
K(Tλ u(t)) +
(4−d(p−1))d(p−1) 8(p+1)
|Tλ u(t)|
p+1
dx . (2.18)
Case I. K(u(t)) −
(d(p−1)−4)d(p−1) 8(p+1)
|u(t)|p+1 dx ≥ 0, for t ∈ Imax .
(2.19)
In this case, K(u(t)) =
|∇u|2 −
d(p−1) p+1 2(p+1) |u|
|∇u|2 +
2(d+2) d d d+2 |u|
≥
+
2(d+2) d d d+2 |u|
dx −
dx
4 d(p−1)−4 K(u(t)),
thus K(u(t)) ≥
d(p−1)−4 d(p−1)
|∇u(t)|2 +
2(d+2) d d d+2 |u(t)|
dx.
Case II. K(u(t)) −
(d(p−1)−4)d(p−1) 8(p+1)
|u(t)|p+1 dx < 0, for t ∈ Imax .
(2.20)
In this case, ∇u(t) 2L2 <
d 2 (p−1)2 8(p+1)
|u(t)|p+1 dx −
d d+2
|u(t)|
2(d+2) d
dx,
which implies u(t) = 0, ∀ t ∈ Imax , and p+1 d 2 (p−1)2 8(p+1) u(t) Lp+1
≥ ∇u(t) 2L2 .
(2.21)
Since u0 ∈ Aω,+ and u(t) = 0, ∀ t ∈ Imax , together with the definition of mω , we have K(u(t)) > 0 by similar argument as in the claim in Proposition 2.7. Therefore, by Lemma 2.1, there is λ(t) > 1 such that K(Tλ(t) u(t)) = 0
(2.22)
K(Tλ u(t)) > 0 for 1 ≤ λ < λ(t).
(2.23)
and
By (2.22), we have
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934 d p+1 d(p−1) 2 (p−1)−2 u(t) 2(p+1) λ(t) Lp+1
∇u(t) 2L2 −
+
d d+2 u(t)
2(d+2) d 2(d+2) L d
= 0,
so by Lemma 2.8 together with the interpolation and Sobolev inequalities, we have λ(t)
d 2 (p−1)−2
p+1 u(t) Lp+1
=
2(p+1) d(p−1)
≤
2(p+1) d(p−1)
≤
2(p+1) d(p−1)
∇u(t) 2L2 +
d d+2 u(t)
∇u(t) 2L2 1 + Cd
2(d+2) d 2(d+2) L d
4 d
+ Cd u(t) L2 ∇u(t) 2L2
2mω ω
2 d
∇u(t) 2L2 ,
where Cd is some constant depending on d related to the Sobolev inequality. Thus, we see by (2.21), 2(p+1) d(p−1)
1 + Cd
2mω ω
2 d
d
p+1
λ(t)2 ∇u(t) 2L2 ≥ λ(t) 2 (p−1) u(t) Lp+1 ≥
d 8(p+1) λ(t) 2 (p−1) ∇u(t) 2L2 , d 2 (p−1)2
which yields λ(t) ≤
d 4 (p − 1)
1 + Cd
2mω ω
1 2 d (p−1)−2 d
2
(2.24)
.
Because d dλ
1 K(Tλ u) λ2
d dλ
1 λ2
= − d(p−1) 2(p+1)
d
d (p−1)−3 2 2 (p − 1) − 2 λ
|u|p+1 dx ≤ 0,
|Tλ u|p+1 dx = d2 (p − 1) − 2 |u|p+1 dx ≥ 0,
we have d dλ
1 λ2
K(Tλ u) +
(4−d(p−1))d(p−1) 8(p+1)
|Tλ u|p+1 dx
≤ 0.
(2.25)
Collecting (2.20) and (2.25), we have 1 λ2
K(Tλ u(t)) +
(4−d(p−1))d(p−1) 8(p+1)
|Tλ u(t)|
Hence, (2.18), (2.23) and (2.26) shows for 1 ≤ λ ≤ λ(t),
p+1
dx < 0, for λ ≥ 1.
(2.26)
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934 d2 S (T u(t)) dλ2 ω λ
=−
1 K(Tλ u(t)) + λ22 λ2
<−
1 K(Tλ u(t)) ≤ 0. λ2
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K(Tλ u(t)) +
(4−d(p−1))d(p−1) 8(p+1)
|Tλ u(t)|
p+1
dx (2.27)
Combining (2.24) and (2.27), we obtain ⎞ ⎛ 1 2 d (p−1)−2 2 ⎝ d (p − 1) 1 + Cd 2mω d − 1⎠ K(u(t)) 4 ω d ≥ (λ(t) − 1) dλ S (T u(t)) λ=1 ω λ ≥ Sω (Tλ(t) u(t)) − Sω (u(t)) ≥ mω − Sω (u(t)). Thus, there is δ > 0 depending on d, p and ω that K(u(t)) ≥ min
d(p−1)−4 d(p−1)
∇u(t) 2L2
+
d d+2 u(t)
2(d+2) d 2(d+2) L d
, δ mω − Sω (u(t)) .
2
3. Wellposedness and perturbation theory In this section, we present the local wellposedness theory and the perturbation theory for (1.1). We start by recording the wellposedness theory. For the proof we refer to [9,10,23]. Proposition 3.1. (i) (Local existence) Let φ ∈ H 1 (Rd ), I be an interval, t0 ∈ I and A > 0. Assume that φ H 1 ≤ A, and there exists δ > 0 depending on A that ∇ ei(t−t0 ) φ
2(d+2)
Lt,x d
(I ×Rd )
≤ δ,
then there exists a unique solution u ∈ C(I, H 1 (Rd )) to (1.1) such that u(t0 ) =φ, u S 1 (I ) φ H 1 , ∇ u 2(d+2) ≤2 ∇ ei(t−t0 ) φ Lt,x d
(I ×Rd )
2(d+2)
Lt,x d
(I ×Rd )
.
As a consequence, we have the small data global existence: if φ H 1 is sufficiently small, then u is a global solution with u S 1 (R) φ H 1 .
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(ii) (Unconditional uniqueness) Suppose u1 , u2 ∈ C(I, H 1 (Rd )) are two solutions of (1.1) with u1 (t0 ) = u2 (t0 ) for some t0 ∈ I , then u1 = u2 . Let u ∈ C((Tmin , Tmax ), H 1 (Rd )) be the maximal-lifespan solution to (1.1), then we have (iii) (Conservation laws) For any t, t0 ∈ (Tmin , Tmax ), M(u(t)) = M(u(t0 )), E(u(t)) = E(u(t0 )), Sω (u(t)) = Sω (u(t0 )), for any ω > 0, P(u(t)) = ∇u(t, x)u(t, x) dx = P(u(t0 )). Rd
(iv) (Blow-up criterion) If Tmax < ∞, then u
(d+2)(p−1) 2 ((T , Tmax )×Rd )
2(d+2)
Lt,x d
∩Lt,x
= ∞, ∀ Tmin < T < Tmax .
A similar result holds, if Tmin > −∞. (v) (Scattering) If u
(d+2)(p−1) 2 ((Tmin ,Tmax )×Rd )
2(d+2)
Lt,x d
∩Lt,x
< ∞,
(3.1)
then Tmax = ∞, Tmin = −∞, and there exist u± ∈ H 1 (Rd ) such that lim u(t) − eit u+ 1 = lim u(t) − eit u− 1 = 0. t→∞
t→−∞
H
H
(3.2)
In the following, we will give the long-time perturbation theory when d ≤ 4. Proposition 3.2 (Long-time perturbation). Let I be a compact time interval and let w be an approximate solution to (1.1) on I × Rd in the sense that 4
i∂t w + w = |w| d w − |w|p−1 w + e for some function e. Assume that
w
w L∞ 1 d ≤ A1 , t Hx (I ×R )
(3.3)
≤B
(3.4)
2(d+2)
Lt,x d
(d+2)(p−1) 2 (I ×Rd )
∩Lt,x
for some A1 , B > 0. Let t0 ∈ I and u(t0 ) close to w(t0 ) in the sense that u(t0 ) − w(t0 ) Hx1 (Rd ) ≤ A2 for some A2 > 0.
(3.5)
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
Assume also the smallness conditions ∇ ei(t−t0 ) (u(t0 ) − w(t0 ))
2(d+2)
Lt,x d
∇ e
(I ×Rd )
2899
≤ δ,
(3.6)
≤δ
(3.7)
2(d+2)
Lt,xd+4 (I ×Rd )
for some 0 < δ ≤ δ1 , where δ1 = δ1 (A1 , A2 , B) is a small constant. Then there exists a solution u to (1.1) on I × Rd with the specified initial data u(t0 ) at time t = t0 that satisfies u − w
≤C(A1 , A2 , B)δ α ,
(3.8)
u − w S 1 (I ) ≤C(A1 , A2 , B)A2 ,
(3.9)
(d+2)(p−1) 2 (I ×Rd )
2(d+2)
Lt,x d
∩Lt,x
u S 1 (I ) ≤C(A1 , A2 , B), where 0 < α <
(3.10)
4 d(p−1) .
To show the long-time perturbation theory, we will first give the following short-time perturbation theory. Lemma 3.3 (Short-time perturbation). Let I be a compact time interval and let w be an approximate solution to (1.1) on I × Rd in the sense that 4
i∂t w + w = |w| d w − |w|p−1 w + e for some function e. Suppose we also have the energy bound w L∞ 1 d ≤ A1 t Hx (I ×R )
(3.11)
for some constant A1 > 0. Let t0 ∈ I and let u(t0 ) ∈ H 1 (Rd ) be close to w(t0 ) in the sense that u(t0 ) − w(t0 ) Hx1 ≤ A2
(3.12)
for some A2 > 0. Moreover, assume the smallness conditions ∇ w
2(d+2)
Lt,x d
(I ×Rd )
+ w
(d+2)(p−1) 2 (I ×Rd )
≤ δ0 ,
(3.13)
≤ δ,
(3.14)
≤δ
(3.15)
Lt,x
∇ ei(t−t0 ) (u(t0 ) − w(t0 ))
2(d+2)
Lt,x d
∇ e
(I ×Rd )
2(d+2)
Lt,xd+4 (I ×Rd )
for some 0 < δ ≤ δ0 , where δ0 = δ0 (A1 , A2 ) > 0 is a small constant.
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
Then, there exists a solution u ∈ S 1 (I ) to (1.1) on I × Rd with the specified initial data u(t0 ) at time t = t0 that satisfies u − w
δα ,
(3.16)
u − w S 1 (I ) A2 + δ α ,
(3.17)
u S 1 (I ) A1 + A2 ,
(3.18)
(d+2)(p−1) 2(d+2) Lt,x d ∩Lt,x 2 (I ×Rd )
∇ ((i∂t + )(u − w) + e) where 0 < α <
δ , α
2(d+2)
Lt,xd+4 (I ×Rd )
(3.19)
4 d(p−1) .
Proof. By the wellposedness theory, it suffices to prove (3.16)–(3.19) as a priori estimate, that is, we assume that the solution u already exists and belongs to S 1 (I ). By time symmetry, we may assume t0 = inf I . Let v = u − w, then v satisfies 4
4
i∂t v + v = |w + v| d (w + v) − |w + v|p−1 (w + v) − |w| d w + |w|p−1 w − e, and v(t0 ) = u(t0 ) − w(t0 ). For T ∈ I , define S(T ) = ∇ ((i∂t + )v + e)
.
2(d+2)
Lt,xd+4 ([t0 ,T ]×Rd )
We will now work entirely on the slab [t0 , T ] × Rd . ∇ v
2(d+2)
Lt,x d
+ v
≤ ∇ ei(t−t0 ) v(t0 )
(d+2)(p−1) 2
Lt,x
2(d+2) Lt,x d
+ ∇ ((i∂t + )v + e) S(T ) + δ α , (0 < α < where we use the fact i(t−t0 ) v(t0 ) e
+ ei(t−t0 ) v(t0 )
(d+2)(p−1) 2
Lt,x
+ ∇ e
2(d+2)
Lt,xd+4
2(d+2)
Lt,xd+4
4 d(p−1)
< 1)
(3.20)
(d+2)(p−1) 2
Lt,x
∇ ei(t−t0 ) v(t0 )
2d(d+2)(p−1) (d+2)(p−1) d(d+2)(p−1)−8 2 Lx
Lt
4 1− 4 4 1− 4 d(p−1) d(p−1) ∇ ei(t−t0 ) v(t0 ) 2(d+2) ∇ ei(t−t0 ) v(t0 ) ∞ 2 δ d(p−1) A2 d(p−1) ≤ δ α . Lt,x d
Lt Lx
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
2901
On the other hand, since 4
4
4
|∇((i∂t + )v + e)| |∇w||v| d + |∇v||w + v| d + |∇w||v||w| d −1 + |∇w||v|p−1 + |∇v||w + v|p−1 + |∇w||v||w|p−2 and 4
4
|(i∂t + )v + e| |w||v| d + |w + v| d |v| + |w||v|p−1 + |w + v|p−1 |v|, we have 4
S(T ) ∇ w
2(d+2) Lt,x d
4
(v, w) d 2(d+2) + ∇ v
2(d+2) Lt,x d
Lt,x d
+ ∇ w
2(d+2) Lt,x d
p−1
(v, w)
(d+2)(p−1) Lt,x 2
(v, w) d 2(d+2) Lt,x d
+ ∇ v
2(d+2) Lt,x d
(v, w)
p−1 (d+2)(p−1) 2
Lt,x
4 4 p−1 δ0 (S(T ) + δ α ) d + δ0d + (S(T ) + δ α )p−1 + δ0 4
p−1
+ (δ0d + δ0
4
)(S(T ) + δ α ) + (S(T ) + δ α )1+ d + (S(T ) + δ α )p .
By the continuity argument, we can take δ0 = δ0 (A1 , A2 ) sufficiently small, then S(T ) δ α , ∀ T ∈ I,
(3.21)
which implies (3.19). We also have u − w
(d+2)(p−1) 2
2(d+2)
Lt,x d
∩Lt,x
ei(t−t0 ) v(t0 )
(d+2)(p−1) 2(d+2) Lt,x d ∩Lt,x 2
+ ∇ ((i∂t + )v + e)
2(d+2) Lt,xd+4
+ ∇ e
2(d+2) Lt,xd+4
δα ,
which is (3.16). To obtain (3.17), we see u − w S 1 (I ) u(t0 ) − w(t0 ) H 1 + ∇ ((i∂t + )v + e)
2(d+2)
Lt,xd+4
+ ∇ e
2(d+2)
Lt,xd+4
A2 + S(t) + δ A2 + δ α . We now show (3.18). Using Strichartz estimate, (3.11) and (3.13), we get 4 w S 1 (I ) w(t0 ) H 1 + ∇ (|w| d w − |w|p−1 w) 2(d+2) + ∇ e Lt,xd+4
4
A1 + w d 2(d+2) ∇ w Lt,x d 4
p−1
A1 + (δ0d + δ0
2(d+2) Lt,x d
) w S 1 (I ) + δ.
p−1
+ w
(d+2)(p−1) 2
Lt,x
2(d+2)
Lt,xd+4 (I ×Rd )
∇ w
2(d+2)
Lt,x d
+δ
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
By the continuity argument, we have w S 1 (I ) A1 , provided δ0 is sufficiently small depending on A1 . This together with (3.20), (3.21) and u = v +w yields ∇ u
≤ ∇ v
2(d+2)
Lt,x d
+ ∇ w
2(d+2)
Lt,x d
2(d+2)
Lt,x d
A1 .
While, we have by (3.16), (3.20) and (3.21) u
2(d+2)
Lt,x d
(d+2)(p−1) 2
∩Lt,x
≤ v
(d+2)(p−1) 2
2(d+2)
∩Lt,x
Lt,x d
+ w
2(d+2)
Lt,x d
δα
(d+2)(p−1) 2
∩Lt,x
Combining these with Strichartz estimate, we obtain 4 u S 1 (I ) u(t0 ) H 1 + ∇ (|u| d u − |u|p−1 u) 4
A1 + A2 + u d 2(d+2) ∇ u Lt,x d
2(d+2) Lt,x d
2(d+2)
Lt,xd+4
+ u
p−1 (d+2)(p−1) 2
∇ u
Lt,x
2(d+2)
Lt,x d
4
A1 + A2 + δ d α A1 + δ (p−1)α A1 A1 + A2 , which proves (3.18), provided δ0 is sufficiently small depending on A1 and A2 .
2
We now show the long-time perturbation theory. Proof of Proposition 3.2. We will derive Proposition 3.2 from Lemma 3.3 by an iterative procedure. First, we will assume without loss of generality that t0 = inf I . Let δ0 = δ0 (A1 , 2A2 ) be as in Lemma 3.3. The first step is to establish an S 1 bound on w. In order to do so, we subdivide I into N0 ∼ d 2(d+2) 1+ M subintervals Ik such that 0 w
(d+2)(p−1) 2 (I
2(d+2)
Lt,x d
∩Lt,x
k ×R
d)
On each subinterval Ik , we have 4 d w − |w|p−1 w) w S 1 (Ik ) w L∞ + ∇ (|w| 1 t Hx 4
A1 + w d 2(d+2) ∇ w Lt,x 4
d
p−1
A1 + (δ0d + δ0
2(d+2) Lt,x d
2(d+2)
Lt,xd+4
+ ∇ e
p−1
+ w
) w S 1 (Ik ) + δ.
∼ δ0 .
(d+2)(p−1) 2
Lt,x
2(d+2)
Lt,xd+4 (Ik ×Rd )
∇ w
2(d+2)
Lt,x d
+δ
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
2903
By the continuity argument, we have w S 1 (Ik ) A2 , provided δ0 is sufficiently small depending on A1 . Summing these bounds over all the intervals Ik , we obtain w S 1 (I ) ≤ C(A1 , B, δ0 ), which implies ∇ w
2(d+2)
(I ×Rd )
Lt,x d
+ w
(d+2)(p−1) 2 (I ×Rd )
≤ C(A1 , B, δ0 ).
Lt,x
This allows us to subdivide I into N1 = C(B, δ0 ) subintervals Jk = [tk , tk+1 ] such that ∇ w
2(d+2)
(Jk ×Rd )
Lt,x d
+ w
(d+2)(p−1) 2 (J
Lt,x
k ×R
d)
≤ δ0 .
Choosing δ1 = δ1 (N1 , A1 , A2 ) sufficiently small, we apply Lemma 3.3 to obtain for each k, and all 0 < δ < δ1 , u − w
2(d+2)
Lt,x d
(d+2)(p−1) 2 (J
∩Lt,x
k ×R
d)
≤ C(k)δ α ,
u − w S 1 (Jk ) ≤ C(k)(A2 + δ α ), u S 1 (Jk ) ≤ C(k)(A1 + A2 ), ∇ ((i∂t + )(u − w) + e)
2(d+2)
Lt,xd+4 (Jk ×Rd )
≤ C(k)δ α ,
provided we can show (3.5), (3.6) hold with t0 replaced by tk . We verify this using an inductive argument. We have u(tk+1 ) − w(tk+1 ) H 1 u(t0 ) − w(t0 ) H 1 + ∇ e
2(d+2)
Lt,xd+4 ([t0 ,tk+1 ]×Rd )
+ ∇ ((i∂t + )(u − w) + e)
2(d+2)
Lt,xd+4 ([t0 ,tk+1 ]×Rd )
k
A2 + δ +
C(j )δ α , j =0
and
∇ ei(t−tk+1 ) (u(tk+1 ) − w(tk+1 )) ∇ ei(t−t0 ) (u(t0 ) − w(t0 ))
2(d+2)
Lt,x d
+ ∇ ((i∂t + )(u − w) + e) k
δ+
C(j )δ α . j =0
Here, C(j ) depends only on j , A1 , A2 , δ0 .
2(d+2)
Lt,x d
(I ×Rd )
2(d+2)
Lt,xd+4
+ ∇ e
2(d+2)
Lt,xd+4
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
Choosing δ1 sufficiently small depending on N1 , A1 , A2 , we can continue the inductive argument. This concludes the proof of Proposition 3.2. 2 4. Linear profile decomposition The linear profile decomposition was first established by H. Bahouri and P. Gérard [4] for the energy critical wave equation in H˙ 1 (Rd ). Later, S. Keraani [19] proved the linear profile decomposition for the energy critical Schrödinger equation in H˙ 1 (Rd ). At almost the same time, F. Merle and L. Vega [26] gave the linear profile decomposition for the mass critical Schrödinger equation in L2 (Rd ), then R. Carles and S. Keraani [8] established this in L2 (R). In [20], S. Keraani use the L2 profile decomposition to describe the minimal mass blowup solution. Later, P. Bégout and A. Vargas [6] extend these results to L2 (Rd ), d ≥ 3. In this section, we will 1 (Rd ) by using the linear profile decomposition in give the linear profile decomposition in Hrad 2 d Lrad (R ). We first review the linear profile decomposition in L2rad (Rd ) in the following. Lemma 4.1 (Profile decomposition in L2rad (Rd ), [6,8,26,33]). Let {φn }n≥1 be a bounded sequence in L2rad (Rd ). Then up to passing to a subsequence of {φn }n≥1 , there exists a sequence of j j functions φ j ∈ L2rad (Rd ) and (hn , tn )n≥1 ⊂ (0, ∞) × R, with hm n j hn
+
j
hn hm n
+
j
|tn −tnm | j
(hn )2
→ ∞, as n → ∞, for j = m,
(4.1)
where j
j
j
j
tn j (hn )2
j
hn → h∞ ∈ {0, 1, ∞}, hn = 1 if h∞ = 1, τn = −
j
j
→ τ∞ ∈ [−∞, ∞], as n → ∞,
(4.2) (4.3)
such that ∀ k ≥ 1, there exists rnk ∈ L2rad (Rd ), k j
φn (x) =
Tn φ j (x) + rnk (x),
(4.4)
j =1 j
here Tn is defined by Tn φ(x) = e−itn j
j
1 j d
φ
(hn ) 2
x j hn
. The remainder rnk satisfies
lim sup eit rnk Lq Lr (R×Rd ) → 0, as k → ∞, n→∞
t
x
(4.5)
where (q, r) is L2 -admissible, and 2 < q < ∞ when d ≥ 2, 4 < q < ∞ when d = 1. We also have as n → ∞, it j j it m m j → 0, Tn (φ j ), Tnm (φ m ) L2 → 0, for j = m, (4.6) e Tn (φ )e Tn (φ ) d+2 Lt,xd (R×Rd )
and ∀ 1 ≤ j ≤ k, Tn (φ j ), rnk L2 → 0, (Tn )−1 rnk 0 in L2 (Rd ). j
j
(4.7)
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
2905
As a consequence, we have the mass decoupling property: k
φn 2L2 −
j =1
φ j 2L2 − rnk 2L2 → 0, as n → ∞, ∀ k ≥ 1.
(4.8)
Proof. We only need to show (4.2), (4.5) and (4.7). Other statements in the theorem are stated in the profile decomposition in L2rad (Rd ) proved in [6,8,26,33]. Without loss of generality, we assume that the sequence is up to a subsequence in the following. j j j To show (4.2), we only need to prove that we may take h∞ and hn to be 1 when h∞ ∈ (0, ∞). j j In fact, if hn → h∞ ∈ (0, ∞), as n → ∞, we have j −itn j · 1 e φ (x) j j d =e
j
−itn
hn
(hn ) 2 1
d
j
φ
j
·
j
h∞
(h∞ ) 2
j −itn (x) + e
1 j d
φ
j
(hn ) 2
· j hn
(x) − e
j
−itn
1 j
d
φ
(h∞ ) 2
j
·
j
h∞
(x) .
Note that −it j j j · −itn j 1 1 · e n φ φ (x) − e (x) j j 2 j d j d h h n ∞ (hn ) 2 (h∞ ) 2 L d j 2 j = φ j (x) − hjn φ j hjn x → 0, as n → ∞. h∞ h∞ 2 L
j
So we can put e−itn
1 j d (hn ) 2
φj
· j hn
j
(x) − e−itn
by the Strichartz estimate. We now shift φ j (x) by
1 j d (h∞ ) 2
1
j d (h∞ ) 2
φj
φj
·
j h∞
x j h∞
(x) into the remainder term, j j j , and (hn , tn ) by (1, tn ). It is
easy to see that (4.1)–(4.3) are not affected. Thus, we conclude (4.2). For the profile decomposition in [6,8,26,33], the remainder rnk satisfies lim sup eit rnk
2(d+2)
Lt,x d
n→∞
(R×Rd )
→ 0, as k → ∞.
By using interpolation, the Strichartz estimate and (4.8), we obtain (4.5). To show (4.7), we see in [6,8,26,33], we already have (Tn )−1 rn 0 in L2 (Rd ), as n → ∞. j
j
Since rnk (x) = rn (x) −
k ! m=j +1
j
(4.9)
Tnm (φ m )(x), when 1 ≤ j < k, so by (4.1) and (4.9),
(Tn )−1 rnk = (Tn )−1 rn − j
j
j
k
(Tn )−1 Tnm φ m 0 in L2 (Rd ), as n → ∞. j
m=j +1
2
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
1 (Rd ). We can now show the linear profile decomposition in Hrad 1 (Rd ). Then up to passing to a subTheorem 4.2. Let {ϕn }n≥1 be a bounded sequence in Hrad 1 (Rd ) and (hj , t j ) sequence of {ϕn }n≥1 , there exists a sequence of functions ϕ j ∈ Hrad n n n≥1 ⊂ (0, ∞) × R, with when n → ∞, j
j
tn j (hn )2
j
j
j hn hm n
hm n j hn
τn = −
j
→ τ∞ ∈ [−∞, ∞],
(4.10)
hn → h∞ ∈ {0, 1, ∞}, +
+
j |tn −tnm | j (hn )2
(4.11)
→ ∞, ∀ j = m,
(4.12)
1 (Rd ), such that ∀ k ∈ N, there exists wnk ∈ Hrad k
∇ −1 Gn (eit ∇ ϕ j ) + eit wnk j
eit φn = j =1 k
=
j j −1 Gn (hn ) ∇ −1 ∇ eit ϕ j + eit wnk
(4.13)
j =1 k
j
∇ −1 hn ∇ ei(t−tn ) gn ϕ j + eit wnk , j
=
j
j =1
where j (gn ϕ)(x) =
1
x
ϕ , j j d hn (hn ) 2 j 1 t − tn x j , j . (Gn v)(t, x) = j d v j (hn )2 hn (hn ) 2 The remainder wnk satisfies lim sup ∇ eit wnk Lqt Lr (R×Rd ) → 0, as k → ∞, x
n→∞
(4.14)
where (q, r) is L2 -admissible, and 2 < q < ∞ when d ≥ 2, 4 < q < ∞ when d = 1. Moreover, we have the following decoupling properties: For any k ∈ N, let σn = (hn )−1 ∇ −1 ∇ , j
then
j
(4.15)
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
s 2 |∇| φn 2 − L
k j =1
2 s j j iτnj j 2 ϕ 2 − |∇|s wnk L2 → 0, s = 0, 1, |∇| gn σn e L
j j j E gn σn eiτn ϕ j − E(wnk ) → 0,
k
E(φn ) −
2907
(4.16)
(4.17)
j =1
j j j Sω gn σn eiτn ϕ j − Sω (wnk ) → 0,
k
Sω (φn ) −
(4.18)
j =1
j j j K gn σn eiτn ϕ j − K(wnk ) → 0,
k
K(φn ) −
(4.19)
j =1
j j j Hω gn σn eiτn ϕ j − Hω (wnk ) → 0, as n → ∞.
k
Hω (φn ) −
(4.20)
j =1
Proof. We divide the proof into three steps. Step 1. Applying Lemma 4.1 to { ∇ φn }, we have k
∇ φn (x) =
e
j
−itn
1 j
d
(hn ) 2
j =1
∇ ϕ
j
· j
hn
(x) + ∇ wnk (x),
(4.21)
where j
j
tn j (hn )2
j
j
τn = −
j
→ τ∞ ∈ [−∞, ∞], j
j
hn → h∞ ∈ {0, 1, ∞}, hn = 1 if h∞ = 1, j
hn hm n
+
hm n j hn
+
j
|tn −tnm | j
(hn )2
(4.22)
→ ∞, for j = m, as n → ∞,
lim sup ∇ eit wnk Lq Lr (R×Rd ) → 0, as k → ∞, t
n→∞
x
where (q, r) is L2 -admissible and 2 < q < ∞ when d ≥ 2, 4 < q < ∞ when d = 1. Moreover, we have k
∀ k ≥ 1,
∇ φn 2L2 −
j =1
2 ∇ ϕ j 2 − ∇ wnk 2L2 → 0, as n → ∞. L
By (4.21), we have k
e
it
φn (x) =
e j =1
it
j
−1 −itn
∇
e
1 j
d
(hn ) 2
· ∇ ϕ x + eit wnk (x) j hn j
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934 k
∇ −1 Gn (eit ∇ ϕ j )(x) + eit wnk (x) j
= j =1 k
Gn ( (hn )−1 ∇ −1 ∇ eit ϕ j ) + eit wnk (x) j
=
j
j =1 k
j
∇ −1 hn ∇ ei(t−tn ) gn ϕ j + eit wnk (x), j
=
j
j =1 j
j
j
where Gn (eit ϕ)(x) = eit Tn ϕ(x) and Tn is defined in Lemma 4.1. So we have obtain (4.10), (4.11), (4.12), (4.13), (4.14). We note by (4.7), for 1 ≤ m ≤ j , (Tnm )−1 ∇ wn 0 in L2rad (Rd ), as n → ∞. j
(4.23)
Step 2. We now turn to (4.16). By (4.13), we have j
tn j j −i j gn σn e (hn )2 ϕ j (x) + wnk (x),
k
ϕn (x) = j =1
so we only need to show the orthogonality: "
tn tnm # − m 2 j j − (hj )2 j m m (h μ ∇ gn σn e n ϕ , μ ∇ gn σn e n ) ϕ m j
→ 0, ∀ j = m,
L2
"
# tn j j − j μ ∇ gn σn e (hn )2 ϕ j , μ ∇ wnk j
→ 0, for 1 ≤ j ≤ k, as n → ∞,
L2
where μ =
|∇| 1 ∇ , ∇ .
"
This follows from
tn tnm # − m 2 j j − (hj )2 j m m (h μ ∇ gn σn e n ϕ , μ ∇ gn σn e n ) ϕ m j
L2
" $ % j j = gn μ ∇j ∇j σn e hn
" j gn μ
= "
= e
∇ j hn
hn
∇ e
tnm i m (hn )2
" m,j
= Sn μ
j t − jn (hn )2
j t − jn (hn )2
j (gnm )−1 gn e
∇ j hn
ϕ
j
ϕ j , gnm μ
, gnm μ h∇m n
j t − jn (hn )2
∇ ϕ j , μ
∇ hm n
μ
∇ j hn
∇ ϕ m
∇ hm n
$
∇ hm n
%
σnm e
tnm − m (hn )2
# ϕm L2
# tnm − m 2 m (h ) n ϕ ∇ e
L2
∇ ϕ , μ
# L2
j
∇ hm n
# ∇ ϕ
→ 0, as n → ∞,
m L2
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
2909
where m,j Sn ϕ(x) =
hm n j hn
j
d
2
e
−i
tn −tnm 2 (hm n)
e
j j ihn (ξn −ξnm )x
j
x−
ϕ
j
j
xn −xnm +2tn (ξn −ξnm ) hm n
j
hn hm n
→ 0, as n → ∞, ∀ ϕ ∈ L2 , by (4.12), and "
"
# tn j j − (hj )2 j k μ ∇ gn σn e n ϕ , μ ∇ wn j
L2
j gn μ
= "
$
∇ j hn
# tn % j − (hj )2 j k σn e n ϕ , μ ∇ wn j
∇ j hn
# j = Tn μ ∇j ∇ ϕ j , μ ∇ wnk
L2
hn
"
L2
# 2 j = Tn μ ∇j ∇ ϕ j , ∇ wnk hn
L2
" # 2 j −1 j k ∇ = μ j ∇ ϕ , (Tn ) ∇ wn hn
L2
→ 0, as n → ∞, ∀ 1 ≤ j ≤ k, by (4.23). Step 3. Since we already have (4.16), to obtain (4.17), (4.18), (4.19), (4.20), it suffices to show k p+1
ϕn Lp+1 −
j =1
j j iτnj j p+1 k p+1 ϕ p+1 − wn p+1 → 0, as n → ∞, for 1 + gn σn e L
L
4 d
≤p≤1+
4 d−2 .
(4.24) 1 ∈ R, then the refined Fatou Lemma (see [22,25]) shows that Suppose that τ∞ 1 1 iτn1 1 p+1 1 p+1 p+1 lim ϕn Lp+1 − gn σn e ϕ p+1 − wn p+1 n→∞
L
L
p+1 p+1 p+1 d 1 = lim (h1n ) 2 (1−p) (gn1 )−1 ϕn p+1 − σn1 eiτn ϕ 1 p+1 − (gn1 )−1 wn1 p+1 n→∞ L L L p+1 1 d(1−p) 1 iτ 1 1 p+1 1 d(1−p) 1 iτn1 1 −1 1 −1 2(1+p) 2(1+p) n σn e (σn ) (Tn ) ϕn − (hn ) σn e ϕ = lim (hn ) p+1 n→∞ p+1 L
L
1 d(1−p) 1 iτ 1 1 −1 1 −1 1 p+1 2(1+p) σ e n (σ ) − (Tn ) wn n n p+1 . (hn ) L
We see 1 −1 1 −1 (σn ) (Tn ) ϕn
Lp+1
and
∇ (σn1 )−1 (Tn1 )−1 ϕn
L2
ϕn H 1
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
∇
d(1−p)
(h1n ) 2(1+p) σn1 =
(h1n )
(1 + | h∇1 |) n
∇
= (h1n ) =
d(p−1) 2(p+1)
d(p−1) 2(p+1)
d(p−1) 1− d(p−1) (1 + | h∇1 |) 2(p+1) (1 + | h∇1 |) 2(p+1) n n
∇
(h1n + |∇|)
d(p−1) 2(p+1)
1− d(p−1) 2(p+1)
(1 + | h∇1 |)
,
n
then 1 d(1−p) 1 iτ 1 1 −1 1 −1 (h ) 2(1+p) σ e n (σ ) (T ) ϕn n n n n p+1 L 1 ∇
eiτn (σn1 )−1 (Tn1 )−1 ϕn p+1 |∇| d(p−1) 2(p+1) L 1 −1 1 −1 iτn1 1 −1 1 −1 ∇ e (σn ) (Tn ) ϕn 2 (σn ) (Tn ) ϕn
H1
L
.
So we have 1 1 iτn1 1 p+1 1 p+1 p+1 lim ϕn Lp+1 − gn σn e ϕ p+1 − wn p+1 = 0. n→∞ L L 1 = ±∞, then by the dispersive estimate, we obtain Next, suppose that τ∞ 1 1 iτn1 1 p+1 1 p+1 p+1 lim ϕn Lp+1 − gn σn e ϕ p+1 − wn p+1 n→∞
L
L
(4.25)
p+1 p+1 d(1−p) 1 1 ≤ lim (h1n ) 2 σn1 eiτn (σn1 )−1 (Tn1 )−1 ϕn p+1 − σn1 eiτn (σn1 )−1 (Tn1 )−1 wn1 p+1 n→∞ L L p+1 d(1−p) 1 + lim (h1n ) 2 σn1 eiτn ϕ 1 p+1 n→∞ L p+1 d(1−p) 1 iτn1 1 −1 1 −1 1 p+1 σ 1 eiτn1 (σ 1 )−1 (T 1 )−1 ϕn σ lim (h1n ) 2 − e (σ ) (T ) w n n n n n n dx n n→∞
+ lim (h1n ) n→∞
d(1−p) 2
1 iτn1 1 p+1 ϕ p+1 σn e L
1 iτn1 1 −1 1 −1 p 1 lim (hn ) (σn ) (Tn ) ϕn σ n e n→∞ p 1 1 + σn1 eiτn (σn1 )−1 (Tn1 )−1 wn1 σn1 eiτn ϕ 1 dx p+1 d(1−p) 1 + lim (h1n ) 2 σn1 eiτn ϕ 1 p+1 n→∞ L 1 d(1−p) 1 iτ 1 1 −1 1 −1 p 2(1+p) n ≤ lim (hn ) σn e (σn ) (Tn ) ϕn d(1−p) 2
n→∞
Lp+1
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
1 d(1−p) 1 iτ 1 1 −1 1 −1 1 p 2(1+p) σ e n (σ ) + ) (T ) w (h n n n n n
2911
Lp+1
1 d(1−p) 1 iτ 1 1 1 d(1−p) 1 iτ 1 1 p+1 (h ) 2(1+p) σ e n ϕ 2(1+p) σ e n ϕ · ) + lim (h n n n p+1 n→∞ n p+1 L L p p 1 1 ∇
iτn 1 −1 ≤ lim d(p−1) eiτn (σn1 )−1 (Tn1 )−1 ϕn + ∇
(σn ) (Tn1 )−1 wn1 d(p−1) e n→∞ p+1 |∇| 2(p+1) |∇| 2(p+1) Lp+1 L p+1 iτn1 1 iτn1 1 · ∇
ϕ + lim ∇
ϕ d(p−1) e d(p−1) e |∇| 2(p+1) p+1 n→∞ |∇| 2(p+1) p+1 L L p p 1 −1 1 −1 1 1 −1 1 −1 ∇
iτn1 1 lim (σn ) (Tn ) ϕn 1 + (σn ) (Tn ) wn 1 d(p−1) e ϕ n→∞ |∇| 2(p+1) p+1 H H L p+1 iτn1 1 + lim ∇
ϕ = 0. d(p−1) e n→∞ p+1 |∇| 2(p+1) L
Thus, we have proved p+1 p+1 1 p+1 ϕn Lp+1 − gn1 σn1 eiτn ϕ 1 p+1 − wn1 p+1 → 0, as n → ∞. L
L
(4.26)
Similarly, we can show p+1 p+1 p+1 2 1 wn p+1 − gn2 σn2 eiτn ϕ 2 p+1 − wn2 p+1 → 0, as n → ∞, L
L
L
(4.27)
which together with (4.26) shows (4.24) with k = 2. Moreover, repeating this procedure, we obtain (4.24) for any k ≥ 1. 2 5. Extraction of a critical element In this section, we show the existence of the critical element in the radial case by using the profile decomposition and the long-time perturbation theory. By Proposition 3.1(v), it suffices for Theorem 1.3 to show that any solution u to (1.1) with u0 ∈ Aω,+ satisfies u
2(d+2)
Lt,x d
(d+2)(p−1) 2 (I
∩Lt,x
max ×R
d)
< ∞,
where Imax denotes the maximal interval where u exists. To this end, for m > 0, let & u is a radial solution to (1.1) ' ω (m) = sup u K(Imax ) : with u0 ∈ Aω,+ and Sω (u) ≤ m with u K(I ) := u
2(d+2)
Lt,x d
(d+2)(p−1) 2 (I ×Rd )
∩Lt,x
and define
(5.1)
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
m∗ω = sup{m > 0 : ω (m) < ∞}.
(5.2)
If u0 ∈ Aω,+ with Sω (u0 ) ≤ m sufficiently small, then Lemma 2.8 shows u H 1 1. Hence, Proposition 3.1(i) gives the finiteness of ω (m), which implies m∗ω > 0. Now our aim is to show m∗ω ≥ mω . Suppose by contradiction that m∗ω < mω , we will show the existence of the critical element. In fact, by the definition of m∗ω , we can take a sequence {un } of solutions (up to time translations) to (1.1) that and Sω (un ) → m∗ω , as n → ∞,
un (t) ∈ Aω,+ , for t ∈ In ,
lim un K([0,sup In )) = lim un K((inf In ,0]) = ∞,
n→∞
n→∞
(5.3) (5.4)
where In denotes the maximal interval of un including 0. By Lemma 2.8, sup un 2L∞ H 1 (I t
n
x
n ×R
d)
mω +
mω . ω
(5.5)
Applying Theorem 4.2 to {un (0)} and obtain some subsequence of {un (0)} (still denoted by the 1 (Rd ) and (hj , t j ) same symbol), then there exists ϕ j ∈ Hrad n n n≥1 of sequences in (0, ∞) × R, with j
j
tn j (hn )2
j
j
τn = −
j
→ τ∞ ∈ [−∞, ∞],
hn → h∞ ∈ {0, 1, ∞}, j
hn hm n
+
hm n j hn
+
j
|tn −tnm | j
(hn )2
→ ∞, as n → ∞, ∀ j = m,
(5.6)
1 (Rd ), such that, ∀ k ∈ N, there exists wnk ∈ Hrad k
e
it
un (0) =
j j Gn (hn )−1 ∇ −1 ∇ eit ϕ j + eit wnk
j =1 k
j
∇ −1 hn ∇ ei(t−tn ) gn ϕ j + eit wnk .
(5.7)
lim sup ∇ eit wnk Lq Lr (R×Rd ) → 0, as k → ∞,
(5.8)
j
=
j
j =1
The remainder wnk satisfies
n→∞
t
x
where (q, r) is L2 -admissible, 2 < q < ∞ when d ≥ 2 and 4 < q < ∞ when d = 1. Moreover, j j for any k ∈ N, s = 0, 1, let σn = (hn )−1 ∇ −1 ∇ , we have
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
s |∇| un (0)2 2 − L
k j =1
2913
s j j iτnj j 2 ϕ 2 − |∇|s wnk 2L2 → 0, |∇| gn σn e
(5.9)
L
j j j E gn σn eiτn ϕ j − E(wnk ) → 0,
k
E(un (0)) −
(5.10)
j =1
j j j Sω gn σn eiτn ϕ j − Sω (wnk ) → 0,
k
Sω (un (0)) −
(5.11)
j =1
j j j K gn σn eiτn ϕ j − K(wnk ) → 0,
k
K(un (0)) −
(5.12)
j =1
j j j Hω gn σn eiτn ϕ j − Hω (wnk ) → 0, as n → ∞.
k
Hω (un (0)) −
(5.13)
j =1
Using Strichartz estimate, (5.9) and (5.5), we get sup lim sup ∇ eit wnk S 0 (Rd ) sup lim sup wnk H 1 < ∞.
k∈N
n→∞
k∈N
(5.14)
n→∞
j
j
Next, we construct the nonlinear profile. Let Un be the solution to (1.1) with Un (0) = j j j j j j j ∇ −1 hn ∇ e−itn gn ϕ j , so that eit Un (0) = Gn (σn eit ϕ j ). Thus, Un satisfies j j j 4 j j j i∂t Un + Un = |Un | d Un − |Un |p−1 Un , j
Un (0) = ∇ −1 hn ∇ e−itn gn ϕ j j
j
j
Undoing the transformations Gn and then apply (σn )−1 , we have j
j
4
(i∂t + )vn (t) = (σn )−1 (|σn vn | d σn vn − (hn ) j vn (0) = ϕ j j
j
j j
j j
j
d+4 dp 2 − 2
j j
j j
|σn vn |p−1 σn vn ),
(5.15)
where vn = (σn )−1 (Gn )−1 Un . j
j
j
j
j
We now take the formal limit of vn as n → ∞, and denote the limit is uj , then we see uj satisfies the following equation: j
(1) When h∞ = 0, ⎧ ⎨(i∂t + )uj = 0,
4 ∇ j d−2 ∇ j j ⎩(i∂t + ) ∇
|∇| u = − |∇| u |∇| u ,
if p < 1 +
4 d−2 ,
d = 3, 4, or p < ∞, d = 1, 2,
if p = 1 +
4 d−2 ,
d = 3, 4, (5.16)
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934 j
(2) When h∞ = 1, 4
(i∂t + )uj = |uj | d uj − |uj |p−1 uj ,
(5.17)
j
(3) When h∞ = ∞, 4
(i∂t + ) ∇ uj = | ∇ uj | d ∇ uj ,
(5.18)
with uj (0) = ϕ j . j We then define the nonlinear profile to be uj . The unique existence of uj around t = τ∞ is j j known in all cases, including h∞ = ∞ and τ∞ = ±∞ (the latter corresponding to the existence of the wave operators), by using the standard iteration with the Strichartz estimate. Let I j = j j (Tmin , Tmax ) be the maximal interval where the nonlinear profile uj exists, we see from the construction of the nonlinear profile that uj ∈ C(I j , H 1 (Rd )) and j j j u (τn ) − eiτn ϕ j
H1
→ 0, as n → ∞.
(5.19)
j
(5.20)
Let j
j
un = Gn σn uj , j
j
j
j
j
j
j
j
then, the lifespan of un is In = ((hn )2 Tmin + tn , (hn )2 Tmax + tn ). j Now, ∀ j ∈ N, we define the space ST∞ by ⎧ 2(d+2) ⎪ ⎪ Lt,xd−2 , ⎪ ⎪ ⎨ (d+2)(p−1) 2(d+2) j 2 ST∞ = L ∩ Lt,xd , t,x ⎪ ⎪ ⎪ ⎪ ⎩ 2(d+2) Lt,xd ,
j
if h∞ = 0 with p is energy-critical exponents, j
j
if h∞ = 0 with p is energy-subcritical exponents, or h∞ = 1, j
if h∞ = ∞.
For the linear profile decomposition (5.7), we can give the corresponding nonlinear profile decomposition k j
u
un (t),
(5.21)
j =1 it w k is a good approximation for u provided that each nonlinear profile we will show u
Lemma 5.1 (Classification of nonlinear profiles). Let uj be a non-trivial nonlinear profile. Then ∀t ∈ Ij,
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
⎧ Aω,+ , ⎪ ⎪ ⎪ 2 d ⎨ L (R ), j δ∞ uj (t) ∈ ⎪ H 1 (Rd ), ⎪ ⎪ ⎩ 0 A+ ,
2915
j
h∞ = 1, j if h∞ = ∞, j if h∞ = 0 with p is energy-subcritical exponents, j if h∞ = 0 with p is energy-critical exponents,
(5.22)
where ⎧ j j ⎪ if h∞ = 0 with p is energy-subcritical exponents or h∞ = 1, ⎨I, j j δ∞ = ∇ , if h∞ = ∞, ⎪ ⎩ ∇
j |∇| , if h∞ = 0 with p is energy-critical exponents, and A0+ = {ϕ ∈ H˙ 1 (Rd ) : E 0 (ϕ) < W, K0 (ϕ) < K0 (W )}. By using the variational analysis, we can easily show A0+ is just the scattering set of the energycritical focusing nonlinear Schrödinger equation in [18]. Proof. Since uj is non-trivial, we see from (5.19) that the corresponding linear profile ϕ j is also non-trivial. We shall show m + m∗ j ω j j ω Hω gn σn eiτn ϕ j < , (5.23) 2 j j j (5.24) K gn σn eiτn ϕ j > 0, m + m∗ j ω j j ω , Sω gn σn eiτn ϕ j < 2
(5.25)
for any j and sufficient large n. It follows from (5.3), K(un (0)) > 0 and (5.13) that m∗ω + on (1) = Sω (un (0)) ≥ Hω (un (0)) =
k
j j j Hω gn σn eiτn ϕ j + Hω (wnk ) + on (1). (5.26)
j =1
Hence, (5.26) together with m∗ω < mω and the positivity of Hω show (5.23). Moreover, (5.23) with (2.5) show (5.24). By (2.14), (5.24), we have j j j E gn σn eiτn ϕ j ≥ 0, for n large enough, j ≥ 1, and so j j j Sω gn σn eiτn ϕ j ≥ 0, for n large enough, j ≥ 1. this together with (5.3) and (5.11) gives (5.25).
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
We shall show (5.22). j Case I. h∞ = 1. It follows from (5.19) that j j j j j u (τn ) − gn σn eiτn ϕ j
H1
→ 0, as n → ∞.
(5.27)
This together with (5.23) gives us that mω + m∗ω j lim Hω uj (τn ) ≤ , n→∞ 2 which together with (2.5) shows j K uj (τn ) > 0, for n large enough. Moreover, (5.25) together with (5.27) yields that j Sω uj (τn ) < mω , for n large enough. Hence, j uj τn ) ∈ Aω,+ for n large enough. Then, we have (5.22) in this case. j j Case II. h∞ = ∞. It is easily seen that δ∞ uj (t) = ∇ uj (t) ∈ L2 (Rd ), ∀ t ∈ I j . j Case III. h∞ = 0 with p is energy-subcritical exponents. It is easily seen that uj ∈ H 1 (Rd ). j Case IV. h∞ = 0 with p is energy-critical exponents. Similar to Case I, we can obtain (5.22) in this case. 2 When j large enough, we have the following basic fact about uj : j
j
Lemma 5.2. There exists j0 ∈ N such that Tmin = −∞, Tmax = ∞ for j > j0 and uj 2S 1 (R) j >j0
ϕ j 2H 1 < ∞.
(5.28)
j >j0
Proof. By (5.5) and (5.9), we have ∞ j =1
which shows
∞ ! j =1
2 j ∇ (e−itn ϕ j ) 2 < ∞, L
ϕ j 2H 1 < ∞, and therefore ϕ j H 1 → 0, as j → ∞. By the small data
global wellposedness and scattering theory together with (5.19), we obtain when j large enough, uj S 1 (R) ϕ j H 1 and so we have the desired result. 2
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
2917
Lemma 5.3. In the nonlinear profile decomposition (5.21), if j
δ∞ uj ST j ((T j ∞
j
j d min ,Tmax )×R )
< ∞ for 1 ≤ j ≤ k,
(5.29)
j
then, we have Tmin = −∞, Tmax = ∞, and j
un
(d+2)(p−1) Lt,x 2 (R×Rd )
j + ∇ un
j
2 d L∞ t Lx (R×R )
+ ∇ un
2(d+2)
Lt,x d
(R×Rd )
∇ uj S 0 (R) 1, (5.30)
for 1 ≤ j ≤ k, and there exists B > 0 such that lim sup n→∞
+ ∇ u
(d+2)(p−1) Lt,x 2 (R×Rd )
+ ∇ u
2(d+2) d 2 L∞ (R×Rd ) t Lx ∩Lt,x
j
2 d L∞ t Lx (R×R )
≤ B.
(5.31)
j
Proof. If h∞ = ∞, then δ∞ uj is a solution to the mass-critical defocusing nonlinear Schrödinger j j equation. Since I j coincides with the maximal interval where δ∞ uj exists, we have Tmin = −∞, j Tmax = ∞. j Similarly, when h∞ = 0 with p is energy-subcritical exponents, uj is just the solution of j j j the free Schrödinger equation, we still have Tmin = −∞, Tmax = ∞. When h∞ = 0 with j j p is energy-critical exponents, δ∞ u is a solution to the energy-critical defocusing nonlinear Schrödinger equation. By using a easy variational analysis, we can easily see A0+ is just the scatj tering set of the energy-critical defocusing nonlinear Schrödinger equation. So δ∞uj is globally j j exists with Tmin = −∞, Tmax = ∞. j j When h∞ = 1, we have δ∞ uj is a solution to (1.1) and therefore, Proposition 3.1(v) together j j with (5.29) shows Tmin = −∞, Tmax = ∞. Using the Strichartz estimate and (5.29), we get ∇ uj S 0 (R) 1, for 1 ≤ j ≤ k, so it suffices to show j
un
(d+2)(p−1) Lt,x 2 (R×Rd )
j + ∇ un
j
2 d L∞ t Lx (R×R )
+ ∇ un
2(d+2)
Lt,x d
(R×Rd )
uj S 1 (R) .
In fact, j
un
(d+2)(p−1) Lt,x 2
d 2 j j − |∇| 2 p−1 Gn σn uj ∼
2d(d+2)(p−1) (d+2)(p−1) d(d+2)(p−1)−8 2 Lx
Lt
d j − 2 ∇
|(hn )−1 ∇| 2 p−1 uj 2d(d+2)(p−1) j −1 (d+2)(p−1) (hn ) ∇
d(d+2)(p−1)−8 2 Lt Lx
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
∇ uj
2d(d+2)(p−1) (d+2)(p−1) d(d+2)(p−1)−8 2 Lt Lx
∇ uj
S 0 (R)
,
where the Hörmander–Mihlin multiplier theorem is used in the second inequality. j ∇ un
2 d L∞ t Lx (R×R )
= ∇ uj
2 L∞ t Lx
j ∇ un
2(d+2) Lt,x d (R×Rd )
≤ uj
S 1 (R)
= ∇ uj
,
≤ uj
2(d+2) Lt,x d (R×Rd )
S 1 (R)
.
Hence (5.30) follows. We now turn to (5.31). By q q j j j un − |un |q−1 |um un ≤ Ck,q n |, 1 < q < ∞, 1≤j ≤k 1≤j ≤k 1≤j =m≤k we have (d+2)(p−1) 2 j u n 1≤j ≤k (d+2)(p−1) 2 Lt,x
(d+2)(p−1) 2 j un (d+2)(p−1)
≤ 1≤j ≤k
(R×Rd )
Lt,x
2
+ Ck
(R×Rd ) j
|un | 1≤j =m≤k
(d+2)(p−1) −1 2
|um n | dxdt.
(5.32)
R×Rd
We see by (5.30) and Lemma 5.2 that j
un 1≤j ≤k
(d+2)(p−1) 2 (d+2)(p−1) Lt,x 2 (R×Rd )
(d+2)(p−1)
1≤j ≤j0
uj S 1 (R)2
+
ϕ j 2H 1 < ∞.
(5.33)
j >j0
Next, we consider the second term on the right side of (5.32). By the Hölder inequality, (5.6) and (5.30), we have
j
|un |
(d+2)(p−1) −1 2
j
m |um n | dxdt un un
R×Rd
j un um n
(d+2)(p−1) Lt,x 4
(d+2)(p−1) 4
j
un
(d+2)(p−1) −2 2 (d+2)(p−1) Lt,x 2
→ 0, as n → ∞.
Lt,x
Plugging (5.33) and (5.34) into (5.32), we obtain that there is B1 > 0 such that lim sup u
Similarly, we have
(d+2)(p−1) 2 (R×Rd )
Lt,x
≤ B1 .
(5.34)
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
lim sup ∇ u
2(d+2)
Lt,x d
n→∞
(R×Rd )
2919
≤ B1
and lim sup ∇ u
2 d L∞ t Lx (R×R )
n→∞
≤ B1 .
2
Thus, we obtain (5.31).
Lemma 5.4 (At least one bad profile). Let j0 be as in Lemma 5.2, then there exists 1 ≤ j ≤ j0 such that j
δ∞ uj ST j ((T j ∞
j d min ,Tmax )×R )
= ∞.
Proof. We argue by contradiction. Assume j
δ∞ uj ST j ((T j
j d min ,Tmax )×R )
∞
< ∞, ∀ 1 ≤ j ≤ j0 .
Combining this with Lemma 5.2, we have j
δ∞ uj ST j ((T j ∞
j d min ,Tmax )×R )
< ∞, ∀ j ≥ 1.
(5.35)
j
This together with Lemma 5.3 implies un exists globally in time for j ≥ 1 and hence so does it w k . u
(5.36)
n→∞
it k lim sup u
(d+2)(p−1) 2(d+2) 2 ∩Lt,x d (R×Rd )
Moreover, it follows from (5.7) with t = 0 and (5.19) that k un (0) − u
≤ j =1
Hence,
≤ B.
Lt,x
j j ∇ eiτn ϕ j − uj (τn )
L2
H1
−→ 0, as n → ∞.
(5.37)
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
k un (0) − u
H1
→ 0, as n → ∞.
(5.38)
Next, we claim that as k → ∞, it k
n→∞
2(d+2)
Lt,xd+4
→ 0,
(5.39)
4
where N (u) = |u| d u − |u|p−1 u. Before proving this, we remark that (5.39) together with the long-time perturbation theory leads to a contradiction. Indeed, by (5.36), (5.37), (5.38), we conclude as a consequence of Proposition 3.2 that un
2(d+2)
Lt,x d
< ∞,
(d+2)(p−1) 2 (R×Rd )
∩Lt,x
when n large enough, which contradicts (5.4). Hence, Lemma 5.4 holds. It remains to prove (5.39). Note that it k it k wn ) − N (u
=
Hence, it suffices for (5.39) to show that k j lim (i∂t + )un − N (u
n n→∞ j =1
2(d+2) Lt,xd+4
= 0, ∀ k ∈ N,
(5.40)
→ 0, as k → ∞.
(5.41)
and it k
n→∞
2(d+2)
Lt,xd+4
First, we show (5.40). j When h∞ = 1, we have j
j
(i∂t + )un = N (un ). j
When h∞ = 0 with p is energy-subcritical exponent, we have j
(i∂t + )un = 0. j
When h∞ = 0 with p is energy-critical exponent,
∇ j |∇| u
satisfies the energy-critical focusing j
nonlinear Schrödinger equation, these cases can be deal with similar to the case h∞ = ∞ below.
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
2921
j
When h∞ = ∞, ∇ uj satisfies the mass-critical defocusing nonlinear Schrödinger equation, < ∞ shows uj S 1 (R) < ∞. In this case, we have this together with ∇ uj 2(d+2) Lt,x d
j
(R×Rd )
j j j 4 j j j |Gn δ∞ u | d Gn δ∞ u 4 j j j j j j 1 = N (un ) + ∇
|Gn δ∞ uj | d Gn δ∞ uj − N (un ) ,
(i∂t + )un =
1 ∇
we will show 4 j j j j j 1 (|Gn δ∞ uj | d Gn δ∞ uj ) − N (un ) ∇ ∇
We note 4 j j j j j 1 (|Gn δ∞ uj | d Gn δ∞ uj ) − N (un ) ∇ ∇
j j ≤ ∇ (|un |p−1 un )
2(d+2)
Lt,xd+4
j j ≤ ∇ (|un |p−1 un )
2(d+2) Lt,xd+4
2(d+2)
Lt,xd+4
→ 0, as n → ∞.
(5.42)
2(d+2)
Lt,xd+4
j 4 j + ∇ |un | d un −
1 ∇
4 j j j j |Gn δ∞ uj | d Gn δ∞ uj
4 j 4 j j j j j + ∇ (|un | d un ) − |Gn δ∞ uj | d Gn δ∞ uj
2(d+2)
2(d+2)
Lt,xd+4
.
Lt,xd+4
We see j j ∇ (|un |p−1 un )
2(d+2)
Lt,xd+4
p−1 j j ≤ un (d+2)(p−1) ∇ un 2
2(d+2)
Lt,x d
Lt,x
2 j (− d2 + p−1 )(p−1) j j p−1 σn u (d+2)(p−1) Lt,x 2
j j −1 j Gn (hn ) ∇ σn uj
2 j (− d2 + p−1 )(p−1) j j p−1 σn u (d+2)(p−1) Lt,x 2
∇ uj
= (hn ) = (hn )
(d+2)(p−1) 2
By ∇ uj ∈ S 0 (R) ∈ Lt j σn uj
(d+2)(p−1) 2
Lt,x
2(d+2)
.
Lt,x d
2d(d+2)(p−1)
Lxd(d+2)(p−1)−8 (R × Rd ), we may assume uj ∈ C0∞ (R × Rd ), then
∇
j = j −1 u (hn ) ∇
(d+2)(p−1) 2
Lt,x
∇ uj
j
j
∇ (|un |p−1 un )
2(d+2)
Lt,xd+4
(d+2)(p−1) 2
Lt,x
so we have
We also have
2(d+2)
Lt,x d
→ 0, as n → ∞.
< ∞,
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
j 4 j |∇|(|un | d un )
2(d+2)
Lt,xd+4
→ 0, as n → ∞.
(5.43)
Indeed, we may assume uj ∈ C0∞ (R × Rd ), then j 4 j |∇|(|un | d un )
2(d+2) Lt,xd+4
4 jd j ≤ un 2(d+2) |∇|un Lt,x d
4 j d j = (hn )−1 σn uj 2(d+2) Lt,x d
j 4 j The above proof also implies |un | d |∇|un
|(hj )−1 ∇| j nj ∇ u (hn )−1 ∇
2(d+2)
Lt,x d
→ 0, as n → ∞. Therefore, we obtain, when
2(d+2)
Lt,xd+4
n → ∞, 4 j 4 j j j j j ∇ (|un | d un ) − |Gn δ∞ uj | d Gn δ∞ uj
2(d+2)
Lt,x d
2(d+2)
Lt,xd+4
4 j 4 j j 4 j j j j j = |un | d un + |∇|(|un | d un ) − |Gn δ∞ uj | d Gn δ∞ uj 4 j 4 j j j j j = |un | d ∇ un − |Gn δ∞ uj | d Gn δ∞ uj
2(d+2)
Lt,xd+4
2(d+2)
Lt,xd+4
+ on (1)
4 4 j j j j j j j j = |Gn σn uj | d ∇ Gn σn uj − |Gn δ∞ uj | d Gn δ∞ uj
2(d+2)
Lt,xd+4
+ on (1)
4 4 4 j j j j j j j j j j j j = |Gn σn uj | d − |Gn δ∞ uj | d ∇ Gn σn uj + |Gn δ∞ uj | d ∇ Gn σn uj − Gn δ∞ uj + on (1) 4 j j j j j |Gn (σn − δ∞ )uj | d ∇ Gn σn uj
4 j j j j j j + |Gn δ∞ uj | d ∇ Gn σn uj − Gn δ∞ uj
2(d+2) Lt,xd+4
4 j j j j j j j + |Gn (σn − δ∞ )uj ||Gn δ∞ uj | d −1 ∇ Gn σn uj
2(d+2)
Lt,xd+4
4 d j j j j j Gn (σn − δ∞ )uj 2(d+2) ∇ Gn σn uj Lt,x d
2(d+2)
Lt,x d
2(d+2)
Lt,x d
2(d+2) Lt,x d
4 j j j d −1 j j Gn δ∞ u 2(d+2) ∇ Gn σn uj Lt,x d
4 j j d j Gn (σn − δ∞ )uj 2(d+2) ∇ uj Lt,x d
4 j d + δ∞ uj 2(d+2) Lt,x d
+ on (1)
Lt,x d
4 j j d j j j j + Gn δ∞ uj 2(d+2) ∇ Gn σn uj − Gn δ∞ uj j j j + Gn (σn − δ∞ )uj
2(d+2)
Lt,xd+4
2(d+2)
Lt,x d
j −1 j j (hn ) ∇ σn uj − δ∞ uj
2(d+2)
Lt,x d
2(d+2)
Lt,x d
+ on (1)
2(d+2)
Lt,xd+4
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
j j + (σn − δ∞ )uj
2(d+2) Lt,x d
4 j j d −1 j −1 j δ∞ u 2(d+2) (hn ) ∇ σn uj Lt,x d
4 d j j j ∼ Gn (σn − δ∞ )uj 2(d+2) ∇ uj Lt,x d
j j + (σn − δ∞ )uj
2(d+2) Lt,x d
2(d+2)
Lt,x d
2923
+ on (1)
2(d+2)
Lt,x d
4 d ∇ uj 2(d+2) + on (1) → 0.
(5.44)
Lt,x d
Thus, k j ∇
(i∂t + )un − N (u
N (un ) − N ( un ) j =1 j =1
2(d+2)
Lt,xd+4
2(d+2) Lt,xd+4
+ on (1) → 0, as n → ∞,
where we use the fact that by (5.6) and Lemma 5.3, ⎞ ⎛ k k j j ⎠ ∇ ⎝ N (u ) − N ( u ) n n j =1 j =1
→ 0, as n → ∞,
2(d+2) Lt,xd+4
and (5.40) follows. We now turn to (5.41). By the Fundamental Theorem of Calculus, 1
Fz v + θ (u − v) dθ + (u − v)
F (u) − F (v) = (u − v) 0
1
Fz¯ v + θ (u − v) dθ,
0
we obtain it k
2(d+2)
Lt,xd+4
p−1 it k |u
2(d+2) Lt,xd+4
4 d it w k + |u
(5.45)
2(d+2)
Lt,xd+4
p d+4 d + eit wnk 2(d+2)p + eit wnk 2(d+2) Lt,xd+4
(5.46)
Lt,x d
p−1 + |u
4 it k d + |u
2(d+2)
+ |eit wnk |p−1 ∇u
4 + |eit wnk | d ∇u
2(d+2)
2(d+2) Lt,xd+4 2(d+2) Lt,xd+4
(5.47)
Lt,xd+4
Lt,xd+4
(5.48)
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
+ |eit wnk |p−1 ∇eit wnk
2(d+2) Lt,xd+4
p−2 it k
2(d+2) Lt,xd+4
4 + |eit wnk | d ∇eit wnk
(5.49)
2(d+2)
Lt,xd+4
4−d d e it w k ∇u
2(d+2)
(5.50)
.
Lt,xd+4
We now estimate the terms (5.46) and (5.49). By the Hölder inequality and (5.8), we have p d d+4 − (5.46) |∇| 2 2p eit wnk 2(d+2)p d+4
Lt
p ∇ eit wnk 2(d+2)p Lt
d+4
d+4 d + eit wnk 2(d+2)
2d(d+2)p d(d+2)p−2(d+4) Lx
2d(d+2)p d(d+2)p−2(d+4) Lx
Lt,x d
d+4 d + eit wnk 2(d+2) → 0, as n → ∞, k → ∞, Lt,x d
p−1 4 ) ( d (5.49) ≤ eit wnk (d+2)(p−1) + eit wnk 2(d+2) · ∇eit wnk 2 Lt,x 2
Lt,x
p−1 ( |∇|sp eit wnk (d+2)(p−1) Lt
2d(d+2)(p−1) d(d+2)(p−1)−8 Lx
2
2(d+2)
Lt,x d
4 ) d + eit wnk 2(d+2) · ∇eit wnk Lt,x 2
2(d+2)
Lt,x d
→ 0, as n → ∞, k → ∞. By Lemma 5.3 and (5.8), we also have for (5.48) that lim lim
k→∞ n→∞
≤ lim lim
it k p−1
k→∞ n→∞
2(d+2) Lt,xd+4
4 + |eit wnk | d ∇u
2(d+2) Lt,xd+4
4 ( ) it k p−1 d + eit wnk 2(d+2) · ∇ u
Lt,x
2(d+2)
Lt,x d
We now consider the terms of the form
2(d+2)
Lt,xd+4
, 1+
4 d
≤q ≤1+
4 d−2 ,
s = 0, 1,
which corresponds to (5.45), (5.47). By the Hölder inequality, (5.8), (5.14), (5.31), we have
2(d+2) Lt,xd+4
q−1 s it k |∇| e w u
2
2(d+2)
Lt,xd+4
s it k |∇| e wn
q−1 s it k |∇| e wn u
→ 0, as n → ∞, k → ∞. Thus, we obtain for 1 +
4 d
≤q ≤1+
4 d−2 ,
s = 0, 1,
2(d+2)
Lt,x d 2(d+2)
Lt,x d
= 0.
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
q−1 lim |u
n→∞
2925
→ 0, as k → ∞.
2(d+2)
Lt,xd+4
We now estimate (5.50). For q = p, 1 + d4 , we have
≤ eit wnk
(d+2)(q−1) Lt,x 2
2(d+2) Lt,x d
|∇|sq eit wnk
2(d+2)d(q−1) (d+2)(q−1) d(d+2)(q−1)−8 2 Lt Lx
as n → ∞, k → ∞. Thus, (5.41) follows.
Lt,x
2(d+2) Lt,x d
2
We can now show the main result in this section: Proposition 5.5 (Existence of the critical element). Suppose m∗ω < mω , then there exists a global 1 (Rd )) to (1.1) such that solution uc ∈ C(R, Hrad uc (t) ∈ Aω,+ and Sω (uc (t)) = m∗ω , for t ∈ R, uc
(d+2)(p−1) 2(d+2) 2 ∩Lt,x d ([0,∞)×Rd )
= uc
Lt,x
(5.51)
(d+2)(p−1) 2(d+2) 2 ∩Lt,x d ((−∞,0]×Rd )
= ∞.
(5.52)
Lt,x
Proof. By Lemma 5.2, Lemma 5.4 and reordering indices, there exists J ≤ j0 such that ⎧ j j ⎨ δ∞ u ST j ((T j
j
d ∞ min ,Tmax )×R ) j j ⎩ δ∞ u j j j ST∞ ((Tmin ,Tmax )×Rd )
= ∞,
for 1 ≤ j ≤ J,
< ∞,
for j ≥ J.
(5.53)
We see from Lemma 5.1 that for 1 ≤ j ≤ J , t ∈ I j , ⎧ Aω,+ , ⎪ ⎪ ⎪ ⎨L2 (Rd ), j δ∞ uj (t) ∈ ⎪ H 1 (Rd ), ⎪ ⎪ ⎩ 0 A+ ,
j
h∞ = 1, j if h∞ = ∞, j if h∞ = 0 with p is energy-subcritical exponents, j if h∞ = 0 with p is energy-critical exponents,
(5.54)
j
On the hand, we can see δ∞ uj is a solution to the mass-critical defocusing nonlinear Schrödinger j equation when h∞ = ∞, the scattering theorem of the mass-critical defocusing nonlinear Schrödinger equation [11–13,21,24,34,35] together with (5.54) shows that j
j
δ∞ uj ST j (I j ) = δ∞ uj ∞
j
2(d+2) Lt,x d (I j ×Rd )
j
< ∞, if h∞ = ∞. j
On the other hand, δ∞ uj is a solution to the free Schrödinger equation when h∞ = 0 with p is energy-subcritical exponents or a solution the energy-critical focusing nonlinear Schrödinger
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934 j
equation when if h∞ = 0 with p is energy-critical exponents, then by the scattering theorem in [18,22] together with (5.54) shows that j
δ∞ uj ST j (I j ) < ∞. ∞
Hence, we have j
j
h∞ = 1, δ∞ = I for 1 ≤ j ≤ J.
(5.55)
From (5.9), (5.11) and (5.13), we have J
∇ un (0) 2L2 − J
Sω (un (0)) −
j =1
2 2 ∇ ϕ j 2 − ∇ wnJ 2 → 0, L
L
j
Sω (e−itn ϕ j ) − Sω (wnJ ) → 0,
(5.56)
j =1 J
Hω (un (0)) −
j
Hω (e−itn ϕ j ) − Hω (wnJ ) → 0, as n → ∞.
(5.57)
j =1
Since K(un (0)) > 0, we have Hω (un (0)) ≤ Sω (un (0)) by (2.4). It follows from (5.57) and (5.3) that j
Hω (e−itn ϕ j ), Hω (wnJ ) ≤ Sω (un (0)) <
mω + m∗ω , 2
for 1 ≤ j ≤ J and n large enough. By (2.5), we have j
K(e−itn ϕ j ) > 0, K(wnJ ) > 0, for 1 ≤ j ≤ J and n large enough,
(5.58)
which together with Lemma 2.6 shows j
Sω (e−itn ϕ j ) ≥ 0, Sω (wnJ ) ≥ 0,
(5.59)
for 1 ≤ j ≤ J and n large enough. We shall show J = 1. Assume for a contradiction that J ≥ 2. Then, it follows from (5.3) and (5.56) that j
lim sup Sω (e−itn ϕ j ) < m∗ω , ∀ 1 ≤ j ≤ J, n→∞
which together with (5.19) shows
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
2927
Sω (uj (t)) < m∗ω , ∀ 1 ≤ j ≤ J, t ∈ (Tmin , Tmax ). j
j
Since uj is a solution to (1.1), it follows from the definition of m∗ω that uj
(d+2)(p−1) 2(d+2) j j 2 ∩Lt,x d ((Tmin ,Tmax )×Rd )
< ∞, ∀ 1 ≤ j ≤ J.
Lt,x
This contradicts (5.53). Thus, we have J = 1. Since u1 (d+2)(p−1) 2(d+2) = ∞, we have Lt,x
2
∩Lt,x d
1 ,T 1 )×Rd ) ((Tmin max
1 1 Sω (u1 (t)) ≥ m∗ω , for t ∈ (Tmin , Tmax ).
(5.60)
1 , T 1 ), S (u1 (t)) ≤ m∗ . On the other hand, by (5.19), (5.56) and (5.59), we get for any t ∈ (Tmin ω max ω Combining this with (5.60), we obtain 1 1 Sω (u1 (t)) = m∗ω , for t ∈ (Tmin , Tmax ).
(5.61)
By (5.19), we have Sω (u1 (t)) = lim Sω (e−itn ϕ 1 ), 1
n→∞
this together with (5.3), (5.56) and (5.61) shows Sω (wn1 ) → 0, as n → ∞.
(5.62)
Hence, Lemma 2.6 together with (5.58) and (5.62) shows wn1 H 1 → 0, as n → ∞.
(5.63)
un (0) − e−itn ϕ 1 H 1 → 0, as n → ∞.
(5.64)
We see from (5.7) and (5.63) that 1
1 = −∞, T 1 = ∞. Assume for a contradiction that T 1 < ∞. Let Now, we shall show Tmin max max 1 1 ) such that t T 1 and put {tn } be a sequence in (Tmin , Tmax n max 1 1 u˜ n (t) = u1 (t + tn ) with I˜n = (Tmin − tn , Tmax − tn ).
We see that {u˜ n } satisfies u˜ n (t) ∈ Aω,+ for t ∈ I˜n , and Sω (u˜ n ) = m∗ω , u˜ n
(d+2)(p−1) 2(d+2) 2 ∩Lt,x d (I˜n ×Rd )
Lt,x
= ∞.
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
Then, we can apply the above argument as deriving (5.64) to this sequence and find that there 1 (Rd ), a sequence {τ } with τ = lim τ ∈ [−∞, ∞] such that exists a non-trivial ψ ∈ Hrad n ∞ n n→∞
lim u1 (tn ) − e−iτn ψ H 1 = lim u˜ n (0) − e−iτn ψ H 1 = 0.
n→∞
n→∞
This together with the Strichartz estimate yields ∇ (eit u1 (tn ) − ei(t−τn ) ψ)
2(d+2)
Lt,x d
(R×Rd )
→ 0, as n → ∞.
(5.65)
Case 1. τ∞ = ±∞. By the dispersive estimate for the free solution, for any compact interval I , we have → 0, as n → ∞, ∇ ei(t−τn ) ψ 2(d+2) Lt,x d
(I ×Rd )
this together with (5.65) yields ∇ eit u1 (tn )
2(d+2)
Lt,x d
(I ×Rd )
→ 0, as n → ∞.
(5.66)
Case 2. τ∞ ∈ R. For any interval I with τ∞ ∈ I and |I | 1, we have by (5.65), lim ∇ eit u1 (tn ) 2(d+2) = ∇ ei(t−τ∞ ) ψ 2(d+2) 1. n→∞
(I ×Rd )
Lt,x d
Lt,x d
(I ×Rd )
(5.67)
1 , which is a Then, Proposition 3.1 together with (5.66), (5.67) implies that u1 exists beyond Tmax 1 1 contradiction. Thus, Tmax = ∞. Similarly, we have Tmin = −∞. Therefore, u1 is a global solution and it is just the desired critical element uc satisfying (5.51) and (5.52). 2
We now show the trajectory of the critical element is precompact in the energy space 1 (Rd ). Hrad Proposition 5.6 (Compactness of the critical element). Let uc be the critical element in Proposi1 (Rd ). tion 5.5, then {uc (t) : t ∈ R} is precompact in Hrad Proof. For {tn } ⊂ R, if tn → t ∗ ∈ R, as n → ∞, then we see by the continuity of uc (t) in t that uc (tn ) → uc (t ∗ ) in H 1 (Rd ), as n → ∞. If tn → ∞. Applying the above argument as deriving (5.64) to uc (t + tn ), there exist tn ∈ R 1 (Rd ) that and φ ∈ Hrad
uc (tn ) − e−itn φ → 0 in H 1 (Rd ), as n → ∞. (i) If tn → −∞, then we have
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
∇ eit uc (tn )
= ∇ eit φ
2(d+2) Lt,x d ([0,∞)×Rd )
2(d+2)
Lt,x d
([−tn ,∞)×Rd )
2929
+ on (1) → 0, as n → ∞.
Hence, we can solve (1.1) for t > tn globally by iteration with small Strichartz norm when n large enough, which contradicts uc
2(d+2)
Lt,x d
(d+2)(p−1) 2 ([0,∞)×Rd )
∩Lt,x
= ∞.
(ii) If tn → ∞, then we have ∇ eit uc (tn )
2(d+2)
Lt,x d
((−∞,0]×Rd )
= ∇ eit φ
2(d+2)
Lt,x d
((−∞,−tn ]×Rd )
+ on (1) → 0, as n → ∞.
Hence, uc can solve (1.1) for t < tn when n large enough with diminishing Strichartz norm, which contradicts uc
2(d+2)
(d+2)(p−1) 2 ((−∞,0])×Rd )
∩Lt,x
Lt,x d
= ∞.
1 (Rd ). Thus tn is bounded, which implies that tn is precompact, so is uc (tn ) in Hrad Similar argument makes sense when tn → −∞, we will omit the proof. 2
We define for R > 0, E˜R (u(t)) =
|∇u(t)|2 + |u(t)|
2(d+2) d
+ |u(t)|p+1 dx,
|x|≥R
then by the compactness of the critical element, we have Corollary 5.7. Let uc be the critical element in Proposition 5.5, then for any > 0, there exists R0 () > 0 such that E˜R0 (uc (t)) ≤ E(uc ), ∀ t ∈ R. 6. Extinction of the critical element In this section, we prove the non-existence of the critical element by deriving a contradiction from Proposition 5.6 and the virial identity in the radial case. For a bounded real-valued function φ ∈ C ∞ (Rd ), we can define the virial quantity: VR (t) =
φR (x)|u(t, x)|2 dx, where φR (x) = R 2 φ( |x| R ), ∀ R > 0.
Rd
Then, for u ∈ C(I ; H 1 (Rd )), we have
(6.1)
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
VR (t) = 2R VR (t) = 4
x φ ( |x| R ) |x| · ∇u(t, x) u(t, x) dx,
·
Rd
(6.2)
∂j ∂k φR (x)∂j u(t, x)∂k u(t, x) dx −
−
2 φR (x)|u(t, x)|2 dx
2(p−1) p+1
φR (x)|u(t, x)|p+1 dx +
4 d+2
φR (x)|u(t, x)|
2(d+2) d
dx.
(6.3)
Theorem 6.1. There does not exist the radial critical element uc of (1.1) in Proposition 5.5. Proof. Let the weight function φ in (6.1) be a smooth, radial function satisfying 0 ≤ φ ≤ 1, and φ(x) =
|x| ≤ 1, |x| ≥ 2.
|x|2 , 0,
On the one hand, by (6.2), we have ∀ R > 0, |VR (t)| R, t ∈ R.
(6.4)
On the other hand, by (6.3), we have VR (t) = 4
φR (r)|∇uc (t, x)|2 dx −
Rd
−
2 φR (x)|uc (t, x)|2 dx
2(p−1) p+1
p+1
φR (x)|uc (t, x)|
dx +
4 d+2
φR (x)|uc (t, x)|
2(d+2) d
dx
≥ 8K(uc (t)) −
|uc (t)|2 dx
C R2 R≤|x|≤2R
|∇uc (t)|2 + |uc (t)|p+1 + |uc (t)|
−C
2(d+2) d
dx.
(6.5)
R≤|x|≤2R
By Lemma 2.9 and Lemma 2.6, we have 8K(uc (t)) ≥ min
d(p−1)−4 d(p−1)
∇uc (t) 2L2
+
d d+2 uc (t)
2(d+2) d 2(d+2) L d
, δ mω − Sω (uc (t))
E(uc (t)). Thus, choosing > 0 small enough and R = R() sufficiently large, then by Corollary 5.7, we get VR (t) E(uc (t)) = E(u0 ), this together with (6.4) implies for any T > 0,
X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
2931
T T · E(u0 ) VR (t) dt = |VR (T ) − VR (0)| R. 0
Taking T large enough, we obtain a contradiction unless uc ≡ 0, which is impossible due to = ∞. 2 uc 2(d+2) (d+2)(p−1) Lt,x d
∩Lt,x
2
(R×Rd )
7. Blow-up We will show the blow-up result in Theorem 1.3. Let the weight function φ in (6.1) be a smooth, radial function ([29]) satisfying φ(r) = r 2 for r ≤ 1, φ (r) ≤ 2 and φ(r) is constant for r ≥ 3. By (6.3), we have VR (t) ≤
4 +
2|∇u|2 −
d(p−1) p+1 p+1 |u|
+
2(d+2) 2d d d+2 |u|
|u(t, x)|2 dx + C
C R2 R≤|x|≤3R
= 8K(u) +
dx
|u|p+1 + |u|
2(d+2) d
dx (7.1)
R≤|x|≤3R
|u|2 dx + C
C R2 R≤|x|≤3R
|u(t)|p+1 + |u(t)|
2(d+2) d
dx.
R≤|x|≤3R
Since u is radial, we have the following radial Sobolev inequalities p+1
u Lp+1 (|x|≥R) ≤ 2(d+2) d 2(d+2) L d (|x|≥R)
u
≤
p+3
C (d−1)(p−1) 2 R
C 2(d−1) R d
p−1
u L22(|x|≥R) ∇u L22(|x|≥R) , 2(d+1)
2
d u L2 (|x|≥R) ∇u Ld 2 (|x|≥R) ,
this together with (7.1), the mass conservation and Young’s inequality shows ∀ > 0, there exists R large enough that VR (t) ≤ 8K(u) + ∇u(t) 2L2 + . By K(u) < 0, energy, mass conservation and Lemma 2.7, we see K(u(t)) < − (mω − Sω (u(t))) , ∀ t ∈ Imax , thus ∇u(t) 2L2 < So we have by (7.2),
p+1 d(p−1) 2(p+1) u(t) Lp+1
− (mω − Sω (u)).
(7.2)
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X. Cheng et al. / J. Differential Equations 261 (2016) 2881–2934
VR (t) ≤ 8K(u) + ∇u(t) 2L2 + = 16Sω (u) − 8ω u 2L2 + < 16Sω (u) − 8ω u 2L2 +
p+1 16−4d(p−1) u(t) Lp+1 p+1
16−4d(p−1) p+1
+ ∇u(t) 2L2 + p+1 + d(p−1) 2(p+1) u(t) Lp+1 − (mω − Sω (u)) + .
Here we take > 0 small enough such that 16−4d(p−1) + p+1 We also note by K(u) < 0 and Proposition 2.4, mω ≤ Hω (u(t)) = ω2 u(t) 2L2 +
d(p−1) 2(p+1)
< 0.
p+1 d(p−1)−4 4(p+1) u(t) Lp+1 ,
so 4(p+1) d(p−1)−4
p+1 mω − ω2 u(t) 2L2 ≤ u(t) Lp+1 .
Thus, we have VR (t) ≤ 16Sω (u) − 8ω u 2L2 − (mω − Sω (u)) + d(p−1) 4(p+1) 2 ω + 16−4d(p−1) + − u(t) m . ω 2 p+1 2(p+1) d(p−1)−4 2 L By Sω (u) < mω and the energy, mass conservation, we see there exists δ1 > 0 small enough such that Sω (u) ≤ (1 − δ1 )mω . Then, we get VR (t) ≤ 16(1 − δ1 )mω − 8ω u 2L2 − (mω − Sω (u)) + 4(p+1) 2 ω + − 4d(p−1)−16 + d(p−1) p+1 2(p+1) d(p−1)−4 mω − 2 u L2 ω 4(p+1) d(p−1) d(p−1) 4(p+1) 2 = − 16δ1 + δ1 − d(p−1)−4 2(p+1) mω − 2(p+1) · d(p−1)−4 · 2 u L2 + 2d(p−1) mω + . ≤ − 16δ1 + δ1 − d(p−1)−4 We can take > 0 small enough that VR (t) ≤ −4δ1 mω , which implies that u must blow up in finite time. Remark 2. The blowup is shown for p ≤ 5, which leads to the restriction of the blowup result to d ≥ 2. This is a technical restriction. See also [15,29,30] for some related discussion. Acknowledgments The authors are grateful to Ze Li and Zihua Guo for pointing out a mistake in the original version of the article, which has now been corrected. X. Cheng is grateful to K. Nakanishi for discussing on the combined nonlinear Schrödinger equation in Central China Normal University in 2015. X. Cheng has been partially supported by the NSFC grant of China (No. 11526072) and also “the Fundamental Research Funds for the Central Universities” (No. 2014B14214). C. Miao has been partially supported by the NSFC grant of China (No. 11231006, No. 11371059). C. Miao is also supported by Beijing Center of Mathematics and Information Interdisciplinary
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