Scattering for the NLS equation

Transcription

1 Scattering for the NLS equation joint work with Thierry Cazenave (UPMC) Ivan Naumkin Université Nice Sophia Antipolis February 2, 2017

2 Introduction. Consider the nonlinear Schrödinger equation with the initial condition i t u = 4 u + λ juj α u (NLS) u (0, ) = u 0 () (IC) in R N, where α > 0 and λ 2 C, λ 6= 0.

3 For α < 4 N 2 the problem (NLS) is locally well-posed in H1 R N : Proposition Assume that α < 4 N 2 and let u 0 2 H 1 R N. Then, there exist a maximal existence time T max = T max (u 0 ) 2 (0, ] and a unique solution u 2 C [0, T max ), H 1 R N of (NLS). Moreover, if T max <, then ku (t)k H 1!, as t " T max. Moreover, in the case when Im λ 0 and α < 4 N, by using energy estimates one proves that all H 1 -solutions are global in time and remain bounded.

4 In the case Im λ > 0, there appears a blowup dichotomy similar to the Fujita blowup result for the nonlinear heat equation. The blowup part of the dichotomy corresponds to the case when α < 2 N : Proposition Let Im λ > 0 and α < 2 N. It follows that there exists δ > 0 such that if u is a nontrivial global H 1 solution of (NLS), then sup 0st for all t > 0. Moreover, kru (s)k L 2 δt 2 N N α α ku (0)k N N +2 L 2, t 2 N α N α sup kru (s)k L 2!, 0st as t!. In particular, every nontrivial H 1 -solution blows up in nite or in nite time. Remark One expects that the blowup occurs in nite time.

5 The global existence part of the dichotomy corresponds to the case when α > 2 N. As far as we know the best global existence result available is: Proposition Set X = H 1 R N \ L R 2 N, jxj 2 dx. Let u 0 2 X satisfy v 0 2 H 2 R N and jj v 0 2 H 1 R N jx j2, where v 0 (x) = ei 4 u 0 (x). Then, if kv 0 k H 2 is su ciently small, for some (explicit) α N > 2/N, for N 4, (α N = 2/N, N 3) and all α N < α < 4/N, the solution of (NLS) is global. Note that as α N > 2/N, there is a gap between α and 2/N.

6 For (NLS), a natural conjecture is that if α > 2/N, there exists "low energy scattering". Actually, this is already known in dimension N 3. However, in higher dimension the best available result is the following: Proposition Let λ 2 C. Set X = H 1 R N \ L R 2 N, jxj 2 dx. Suppose that u 0 2 X satisfy v 0 2 H 2 R N and jj v 0 2 H 1 R N, where jx j2 v 0 (x) = ei 4 u 0 (x). Then, if kv 0 k H 2 is su ciently small, for some (explicit) α N > 2/N and all α N < α < 4/N, there exists u + 2 X such that e it4 u(t)! u + in X as t!. Again, there is a gap between α and 2/N.

7 The di culty that arises clearly appears by using the pseudo-conformal transformation: where s = u (s, y) = (1 bt) N /2 e i jx j 2 4(1 bt) v (t, x), t 1 bt, y = x 1 bt, with b > 0, and (t, x) 2 [0, 1 b ) RN. A global, asymptotically free solution of (NLS) corresponds to a solution of the non autonomous equation i t v = 4 v + λ (1 bt) 4 N α 2 jvj α v, (NANLS) which has a limit (in some space) as t! 1/b. Therefore, it is equivalent to solve this equation on (0, 1 b ) for "small" initial values.

8 Given an initial value v 0, the Cauchy problem for (NANLS) can be reformulated as the integral equation v (t) = e it4 v 0 Z t iλ (1 s) 4 2 N α e i(t s)4 jvj α v (s) ds. 0 Note that the factor (1 bt) 4 2 N α is integrable if α > 2/N. However, it results di cult to use Strichartz estimates in high dimensional cases due to the singularity of (1 bt) 4 2 N α. One can try to estimate the L norm to overcome this di culty, in which case, one needs to control high-order Sobolev norms. This raises the following regularity issue.

9 The smoothness condition. In order to prove the existence of solutions in H s R N to (NLS) the nonlinear term must be su ciently smooth. For example, in the simplest perturbation argument it is required to control s derivatives of the nonlinearity juj α u. This leads to the condition [s] < α (C1) if α is not an even integer. There is an issue here! Even is u is very smooth, juj α u need not to be smooth enough. For example, consider u (x) = x 1 e jx j2 is analytic, but if 0 < α < 1 2, then juj α u /2 H 2 R N.

10 Of course condition (C1) might be a purely technical condition which one should be able to remove with a better argument. In fact, one can sometimes improve condition (C1) by using the fact that there is one time derivative by two space derivatives. However, there is still the condition [s] < 2α. (C2) Is a condition like (C2) necessary?

11 The answer is a rmative: Proposition Let 0 < α < 1, λ 2 Cf0g and suppose s > 3 + N 2 + α. There exists (arbitrary small) u 0 2 Cc R N such that there is no T > 0 for which there exists a solution u 2 C [0, T ], H s R N of (NLS). Remark In particular, s 3 + N 2 + α is a necessary condition for (NLS) to be locally well posed in H s R N, even in an arbitrarily small ball. Remark The optimal condition for H s regularity is unknown.

12 What can we do? We want to show the existence of regular solutions. But, as we saw, there is a regularity issue for solving (NLS) in H s if, for example s is "large" compared to α. So, what can we do? We note that the di culty appears when the solution u = 0, because in this point, the map u! juj α u is not smooth enough. Then, one way to avoid the problem is to construct solutions that do not vanish. This is not a completely trivial for (NLS) because there is no comparison principles, as for instance, for the heat equation. Nevertheless, something can be done!

13 Local existence of smooth solutions for NLS. We will apply a contraction mapping argument to show existence of smooth solutions that do not vanish. For that purpose, we need to construct an appropriate space. As we want the solutions to be in L 2, they need to decay su ciently fast when jxj!. For simplicity we consider the initial data u 0 (x) = hxi n := 1 + jxj 2 n 2 with n > 0 su ciently large. Let u be solution of iu t + 4u = 0, u (0) = u 0. (LSE) Let us nd a space where, for some T > 0, we can prove the estimate inf x 2R N hxin ju (t, x)j η (T ), uniformly for t 2 [0, T ].

14 Integrating (LSE) we have Then u (t) = u 0 + i Z t 0 4u. inf hxin ju (t, x)j 1 t khxi n 4 uk L x 2R N ((0,t)R N ). It is not possible to use Sobolev s embedding H s,! L for s > N 2 to estimate khxi n 4 uk L ((0,t)R N ). This would require hxi n 4 u 0 2 L 2, which fails if N 4.

15 However, we note hxi n 4 k u 0 C hxi 2 2k 2 L 2 if k is large enough. Therefore, we use Taylor s formula u (t) = m (it) j 4 j u 0 + i m+1 Z t (t s) m 4 m+1 u (s) ds j=0 j! m! 0 with m large, and we estimate the weighted L norms by Sobolev theorem together with energy estimates.

16 We de ne the space to work with. Fix k > N 2, n > N 2 + 1, 2m > k + n + 1, set J = 2m k + n and de ne X by X = fu 2 H J R N ; hxi n D β u 2 L R N for 0 jβj 2m hxi n D β u 2 L 2 R N for 2m + 1 jβj 2m k, hxi J jβj D β u 2 L 2 R N for 2m k < jβj Jg. We equip X with the obvious norm. Sobolev s embedding and energy estimates yield e it4 u C (1 + jtj) m+n+1 kuk X. X

17 Note that u 0 = hxi n 2 X and u (t) = e it4 u 0 satis es Then, khxi n 4 u (t)k L C (1 + jtj) m+n+1. inf hxin ju (t, x)j 1 t khxi n 4 uk L x 2R N ((0,t)R N ) 1 2, if jtj is small. Therefore, we can control the linear part from below.

18 We now need to control juj α u X. We have to calculate the derivatives of juj α u. Since juj α = (uu) α 2 we see that D β (juj α u) contains the term A = juj α D β u and terms of the form B = juj α 2p D ρ u p j=1 D γ 1,j ud γ 2,j u where γ + ρ = β, 1 p jγj, jγ 1,j + γ 2,j j 1, p j=0 (γ 1,j + γ 2,j ) = γ.

19 Let α > 0 and assume, in addition, n > N 2α. Suppose that u 2 X satis es η inf hxin ju (x)j 1 (EB) x 2R N for some η > 0. Using (EB) we estimate A and B to get juj α u X C (1 + η kuk X ) 2J kuk α+1 X. (NE) Similar estimate holds for the di erence juj α u jvj α v X. We now combine e it4 u C (1 + jtj) m+n+1 kuk X X with (NE) and apply a simple contraction mapping argument on C ([0, T ], X ) to obtain:

20 Theorem Let α > 0, λ 2 C, and let X the space de ned above. If u 0 2 X satis es inf x 2R N hxin ju 0 (x)j > 0 there exist T > 0 and a unique solution u 2 C ([ (NLS). T, T ], X ) of Remark We note that the above local existence result holds in any space dimension and for any α > 0.

21 Existence of scattering for NLS. Note that integrating (NLS) we get u (t) = u 0 + i Then we estimate Z t 0 4u Z t iλ juj α u. 0 inf x 2R N hxi n ju (t, x)j 1 t khxi n 4 uk L ((0,t)R N ) t jλj hxi n juj α u L ((0,t)R N ). We used the last estimate to obtain lower bounds for the local problem. However, it is not convenient for the studying the global problem.

22 To study the scattering problem we apply the pseudo-conformal transform to (NLS) and consider (NANLS). Integrating we have v (t) = v 0 + i Z t 0 4v Z t iλ (1 bt) 4 2 N α jvj α v, 0 where t 2 [0, 1 4 N α b ) and hence since (1 bt) 2 2 L 1 0, 1 b for α > 2 N we deduce inf x 2R N hxi n jv (t, x)j inf x 2R N hxi n jv 0 j 1 b khxi n 4 uk L ((0,t)R N ) + hxi n jvj α v L. ((0,t)R N ) Estimating khxi n 4 uk L ((0,t)R N ) and hxi n jvj α v L ((0,t)R N ) similarly to the case of the local result and taking b > 0 big enough we prove the estimate from below η inf x 2R N hxin jv (x)j 1.

23 Making use of the fact (1 bt) 4 2 N α! 0 as b!, by L1 (0, b 1 ) a contraction argument we show that if b > 0 is large enough, there exists a solution v 2 C [0, 1 b ], X of the equation (NANLS) with the initial condition v (0) = v 0. Since u (t, x) = (1 + bt) N 2 e i bjx j 2 4(1+bt) t v 1 + bt, x 1 + bt solves (NLS) on [0, ), this proves the following result.

24 Theorem Let α > 2 N, λ 2 C, let X be as above and set = H 1 R N \ L R 2 N, jxj 2 dx b 2 R, v 0 2 X and bjx j2. Suppose u 0 = ei 4 v 0, where inf x 2R N hxin jv 0 (x)j > 0. If b > 0 is su ciently large, there exists a unique, global solution u 2 C ([0, ), ) \ L (0, ) R N of (NLS). Moreover u scatters, i.e. there exists u + 2 such that e it4 u (t)! u + in as t!. In addition, sup t0 (1 + t) N 2 ku (t)k L <.

25 Scattering-critical case. The critical case corresponds to α = 2 N. The criticality clearly appears by transforming (NLS) by the pseudo-conformal transformation and analyzing the corresponding (NANLS), that in this case takes the form: i t v = 4 v + λ (1 bt) 1 jvj 2 N v. Of course (1 bt) 1 /2 L 1 0, 1 b and therefore, the method used to treat the case α > 2 N cannot be applied to the case α = 2 N. For the original (NLS) (1 bt) 1 /2 L 1 means that the nonlinear part behaves as the linear ow as t!.

26 Note that in the case when Im λ > 0, the number α = 2 N is also critical in the sense of the Fujita s type blowup dichotomy for (NLS). It is remarkable that these two criticality numbers coincide! Recall that when Im λ > 0 and α < 2 N every nontrivial H1 -solution blows up in nite or in nite time. We expect that this is also true for α = 2 N. In the case when Im λ 0 and α < 4 N all H1 -solutions are global. In particular, one can consider the existence of scattering problem for α = 2 N.

27 Suppose that Im λ 0 and α = 2 N. In addition to the problem we had in the case α > 2 N, now we must deal with a non-integrable term λ (1 bt) 1 jvj N 2 v. First we aim to prove global existence of solutions for (NANLS) in X. Given ` 2 N, we set kuk 1,` = sup khi n D β uk L 0jβj` and kuk 2,` = kuk 3,` = sup khi n D β uk L 2 2m+1jβj` sup khi J `D β uk L 2. 2m+3+kjβj`

28 Let 0 < σ < (4J + 2α + 1) J and set 0 = σ 0 < σ < σ j < σ k σ J < 1, 1 j < k J. Given 0 < T < 1 b and v 2 C ([0, T ], X ) let Φ 2,T = Φ 1,T = Φ 3,T = sup 0t<T sup 0t<T sup 0t<T Φ 4,T = sup (1 bt) σ j kvk 1,j 0j2m sup (1 bt) σ j kvk 2,j 2m+1j2m+2+k sup (1 bt) σ j kvk 3,j 2m+3+kjJ sup 0t<T (1 bt) σ 1 inf x 2R N hxi n jv(t, x)j and set Ψ T = maxfφ 1,T, Φ 2,T, Φ 3,T, Φ 4,T g.

29 We have Theorem Suppose Im λ 0. Given any K > 0, there exists b 0 > 1 such that if v 0 2 X satis es 1 kv 0 k X + inf jv 0 (x)j K x 2R Nhxin then for every b b 0 the local solution v of (NANLS) on the maximal interval [0, T max ) with 0 < T max 1 b satis es T max = 1 b and Ψ T 4K, 0 < T < 1 b.

30 In order to control Φ 4,T = sup 0t<T (1 bt) σ 1 inf x 2R N hx i n jv (t,x )j we use jvj t = i (v v v v) + Im λ(1 bt) 1 jvj α+1. 2jvj Setting w(t, x) = hxi n jv(t, x)j we deduce w t = hxi n i (v v v v) + Im λ(1 bt) 1 hxi nα w α+1 2jvj so that 1 α t 1 w = hxi n i α + Im λ(1 bt) 1 hxi nα. (v v v v ) 1 2jv j w α+1

31 Integrating, we estimate C R t w (t,x ) α w (0,x ) α 0 ds + αj Im λj R t ds (1 bs) σ 2 w (s,x ) α bs. Since 1 w (0,x ) C and 1 w (t,x ) C (1 bt) σ 1 we get 1 w(t, x) α C + C Z t Then we estimate Φ 4,T. 0 ds αj Im λj + (1 bs) σ jlog (1 bt)j. 2+(α+1)σ 1 b

32 Once we have the global existence, we study the long-time asymptotics. We have: Theorem Suppose Im λ 0. Let v 2 C ([0, 1 b ), X ) be the solution of (NANLS). i) If Im λ = 0, there exists h 2 L (R N ) with hi n h 2 L (R N ), such that khi n (v(t, ) e i ϕ(t,) h())k L C (1 bt) minfα,1 σ J g is true, where ϕ(t, x) = λ b jh(x)jα log(1 bt) and 0 < σ J < 1.

33 ii) If Im λ < 0, then for some b 1 b 0 and all b b 1, there exist f 0, hi n Λ 2 L, with kf 0 k L 1 4, such that the asymptotics hi n v (t) Ξ (t, ) e i Re λ α Im λ log Ξ α (t,) Λ () L C (1 bt)1 σ J where Ξ (t, x) = 1 + jv 0 j α αj Im λj 1/α j log(1 bt)j + f 0 b is valid.

34 Since u (t, x) = (1 + bt) N 2 e i bjx j 2 4(1+bt) t v 1 + bt, x 1 + bt solves (NLS) on [0, ), we have proved: Theorem bjx j2 Suppose that u 0 = ei 4 v 0, where b 2 R and v 0 2 X satis es inf x 2R N hxi n jv 0 (x)j > 0. If b > 0 is su ciently large, there exists a unique, global solution u 2 C ([0, ), Σ) \ L ((0, ) R N ) of (NLS). Moreover, the following properties hold. i) If Im λ = 0, there is h 2 L (R N ) with hi n h 2 L (R N ), such that the following holds u(t, ) (1 + bt) N 2 e i Φ(t,) h C hti N 1 + bt L 2 δ where 0 < δ < minfα, 1g and Φ (t, x) := bjxj2 4(1 + bt) λ b x h 1 + bt α ln (1 + bt).

35 ii) If Im λ < 0, then for some b 1 b 0 and all b b 1, there exist f 0, Λ 2 L, with kf 0 k L 1 4 and hin Λ 2 L, such that the asymptotics t u (t) Ξ 1 + bt, 1 + bt Λ 1 + bt e i Φ(t,) L C hti N 2 where Φ (t, x) := b jxj2 4 (1 + bt) + Re λ t α Im λ log Ξ α 1 + bt, x 1 + bt δ and Ξ (t, x) = 1 + jv 0 j α αj Im λj 1/α j log(1 bt)j + f 0 b is valid.

36 Proof of asymptotic expansion. Suppose that Im λ = 0. We start from jvj t = i (v v v v) + Im λ(1 bt) 1 jvj α+1. 2jvj Integrating the last equation on (s, t) with 0 s t < 1 b and using that Ψ T C, we obtain khi n (jv(t)j C Z t s jv(s)j)k L Z t s khi n vk L (1 bτ) σ J dτ C (t s) 1 σ J. Then there exists w 2 L such that hi n w 2 L and khi n (jv(t)j w)k L C (1 bt) 1 σ J for 0 t < 1 b. We write equation (NANLS) in the form iv t = v + λ(1 bt) 1 w α v + λ(1 bt) 1 (jvj α w α )v.

37 Setting v(t, x) = g(t, x) exp i λ b w α log(1 bt) (A1) we obtain ig t = e i λ b w α log(1 bt) [ v + λ(1 bt) 1 (jvj α w α )v]. Integrating on (s, t) with 0 s < t < 1 b yields We conclude that khi n (g (t) g (s))k L R t s khin vk L + R t s (1 bτ) 1 khi n (jvj α w α )vk L. khi n (g(t) g(s))k L C (t s) minfα,1 σ J g. Therefore, there exists h 2 L with hi n h 2 L such that khi n (g(t) h)k L C (1 bt) minfα,1 σ J g. Then asymptotic expansion for Im λ = 0 follows.

38 Next consider the case Im λ < 0. We again use jvj t = i (v v v v) + Im λ(1 bt) 1 jvj α+1. 2jvj Dividing the last equation over jvj α+1 and integrating we have where L := i 1 jvj α = 1 jv 0 j α + (v v v v ). Then 2jv j jvj α = αj Im λj j log(1 b jv 0 j α bt)j Z t 1 + jv 0 j α αj=λj b j log(1 bt)j + f (t) 0 L 1 jvj α+1 with f (t) := jv 0 j α R t 0 L 1. Since Ψ jv j α+1 T C, f (t) is convergent in L (R N ), as t! 1/b. Let f 0 := jv 0 j α R 1/b L 1. Taking b 0 jv j α+1 large enough we assure kf 0 k L 1 4.

39 We deduce jv (t)j α vf α L C (1 bt) 1 σ J, for t 0 t < 1/b, where v f =! jv 0 j α 1/α 1 + jv 0 j α αj=λj. b j log(1 bt)j + f 0

40 Next, we introduce the decomposition with and v (t, x) = w (t, x) ψ (t, x) e i θ(t,x ), ψ (t, x) = v f (1 + f 0 ) 1/α jv 0 j Z t θ (t, x) = <λ (1 bτ) 1 vf α (τ) dτ = <λ 0 α=λ log into (NANLS) to obtain the following equation for w (t, x) jv 0 j α vf α (1 + f 0 ) iw t (t, x) = ψ 1 (t, x) e i θ(t,x ) λ(1 bt) 1 (jvj α v α f ) v v.

41 We deduce that there exists w f 2 L such that kw (t) w f k L C (1 bt) 1 σ J. Then, hin v (t) v f (1 + f 0 ) 1/α jv 0 j e i <λ α=λ log jv 0 j α v α f (1+f 0 ) w f! L C (1 bt) 1 σ J. Denoting Λ = (1 + f 0 ) 1/α e i <λ α=λ log(1+f0) w f and observing that Λ 2 L we attain the asymptotic expansion for Im λ < 0.