1-D cubic NLS with several Diracs as initial data and consequences

Transcription

1 1-D cubic NLS with several Diracs as initial data and consequences Valeria Banica (Univ. Pierre et Marie Curie) joint work with Luis Vega (BCAM) Roma, September 2017

2 1/20 Plan of the talk The 1-D cubic NLS with rough data The 1-D cubic NLS with several Dirac data (existence result and Talbot effect) About the binormal flow and vortex filament dynamics The results transferred to the binormal flow

3 Results for 1-D cubic NLS with rough data The 1-D cubic NLS iu t + u xx ± u 2 u = 0 is well-posed in H s, s 0 (Ginibre-Velo 79, Cazenave-Weissler 90). If s < 0 it is ill-posed (Kenig-Ponce-Vega 01, Christ-Colliander-Tao 03). Well-posedness holds for data with Fourier transform in L p spaces (Vargas-Vega 01, Grünrock 05, Christ 07). Methods of proving existence : fixed points arguments relying on Strichartz type spaces. /20

4 2/20 Results for 1-D cubic NLS with Dirac data For aδ 0 as initial data, the 1-D cubic NLS is ill-posed: when looking for a (unique) solution, by using Galilean invariance, one obtains e ia2 log t which has no limit at t = 0 (Kenig-Ponce-Vega 01). A natural change to do is to consider the perturbed cubic 1DNLS ) iψ t + ψ xx ± ( ψ 2 a2 ψ = 0, t a t e i x2 4t and get as an explicit solution a t e i x2 4t = ae it δ 0 (x). The problem is however ill-posed, as smooth perturbations of the solution a t e i x2 4t at time t = 1 behave near t = 0 as e ia2 log t f (x) (B.-Vega 09).

5 Some notations For a sequence {α k } and s 0 we denote {α k } 2 l 2,s := k Z(1 + k ) 2s α k 2, {α k } 2 l We consider distributions u = k Z α k δ k.,s := sup(1 + k ) 2s α k 2. k Z Their Fourier transform on R writes û(ξ) = k Z α k e ikξ, and in particular û is 2π periodic. Imposing {α k } l 2,s translates into û H s (0, 2π). We denote HpF s := {u S (R), û(x + 2π) = û(x), û H s (0, 2π)}. /20

6 4/20 Result for 1-D cubic NLS with several Dirac data Theorem (B.-Vega 17) Let T > 0, s > 1 2, s 1 2 < s s, 0 < γ < 1 and {α k} l 2,s. We consider the 1-D cubic NLS equation: { i t u + u ± ( u 2 M 2πt )u = 0, u t=0 = k Z α kδ k, with M = k Z α k 2. There exists ɛ 0 = ɛ 0 (T ) > 0 such that if {α k } l 2,s local solution u C([0, T ]; HpF s ). Moreover, this solution is unique of the form ɛ 0 then we have a u(t, x) = k Z(α k + R k (t))e it δ k (x), with {R k } satisfying the decay t γ {R k (t)} l 2,s + t { t R k (t)} l, s < C.

7 5/20 Result for 1-D cubic NLS with several Dirac data Remarks: the theorem is a generalization of a result of Kita 2006 valid for subcubic nonlinearities. the proof goes as follows: plugging the ansatz u(t, x) = k Z A k(t)e it δ k (x) into the equation leads to a discrete system on {A k (t)}, we solve this discrete system by a fixed point argument, in which we treat separately the nonresonant and the resonant part. the resonant part is related to the system i a k (t) = 1 2πt a k(t)( j a j (t) 2 M), which has only the constant in time solutions for M = j a j(0) 2. (without the extra-factor M in the initial equation, that is treating directly the cubic equation, one ends up with the resonant system above without M, so we get a k (t) = e i j a j (0) 2 2π log t a k (0) which has no limit at t = 0).

8 6/20 The proof: the nonlinearity action on the ansatz We denote N (u) = u 2 u. Plugging u(t) = k Z A k(t)e it δ k into the equation we get i t A k (t)e it δ k = ±N ( j Z k Z As Kita we can rewrite the nonlinear term: N ( j Z A j (t)e it δ j )(x) = N ( j Z A j (t)e it δ j ) M 2πt ( A k (t)e it δ k ). k Z A j (t) (x j) 2 ei 4t t and use the 2π periodicity of j Z A j2 i j(t)eij e 4t : = k Z = ei x 2 4t t t k Z e ik x 2t e i k 2 4t 2π 2πt 0 1 2π 2π 0 e ikθ N ( j Z e ikθ N ( j Z ) = ei x 2 4t t t N ( j2 i x A j (t)e ij e 4t )( 2t ), j Z A j (t)e ijθ e i j2 4t ) dθ A k (t)e ijθ e i j2 4t ) dθ (e it δ k )(x).

9 The proof: the discrete system The family e it (x k) 2 δ k (x) = ei 4t t i t A k (t) = ± e i k 2 4t 2πt 2π 0 is an orthogonal basis of L 2 (0, 2πt), so e ikθ N ( j Z A k (t)e ijθ e i j2 4t ) dθ M 2πt A k(t). Now we develop the cubic power and get i t A k (t) = ± 1 e i k 2 j 2 1 +j 2 2 j3 3 4t A j1 (t)a j2 (t)a j3 (t) M 2πt 2πt A k(t). k j 1+j 2 j 3=0 We split the summation indices into the following two sets: NR k = {(j 1, j 2, j 3 ) Z 3, k j 1 + j 2 j 3 = 0, k 2 j j 2 2 j 2 3 0}, Res k = {(j 1, j 2, j 3 ) Z 3, k j 1 + j 2 j 3 = 0, k 2 j j 2 2 j 2 3 = 0}. As we are in one dimension, the second set is simply Res k = {(k, j, j), (j, j, k), j Z}. /20

10 The proof: the fixed point framework Finally the system writes i t A k (t) = A k(t) 2πt ( j A j (t) 2 M)+ (j 1,j 2,j 3) NR k e i k 2 j 2 1 +j2 2 j3 3 4t 2πt A j1 (t)a j2 (t)a j3 (t). We want to obtain the existence of A k (t) = α k + R k (t), with {R k } X γ := {{f k } C([0, T ]; l 2,s ) C 1 (]0, T ]; l, s ), {f k } X γ < }, where {f k } X γ := sup 0 t<t t γ {f k (t)} l 2,s + t { t f k (t)} l, s. We shall prove that the operator Φ : {R k } {Φ k ({R j })} defined as Φ k ({R j })(t) := i t 0 α k + R k (τ) ( 2πτ j (α j + R j (τ)) 2 α j 2 ) dτ /20 +i t 0 (j 1,j 2,j 3) NR k e i k 2 j 2 1 +j2 2 j2 3 4τ 2πτ (α j1 +R j1 (τ))(α j2 + R j2 (τ))(α j3 +R j3 (τ)) dτ is a contraction in X γ on a small ball of radius δ, to be chosen later.

11 The proof: the resonant part /20 The resonant part Φ R k j, and then in time: Φ R k ({R j })(t) C C t 0 t 0 we perform Cauchy-Schwarz in the summation in α k + R k (τ) ( τ j α j R j (τ) + j α k + R k (τ) (ɛ 0 δτ γ + δ 2 τ 2γ ) dτ τ t R j (τ)) 2 ) dτ C α k (ɛ 0 δt γ + δ 2 t 2γ R k (τ) 2 ) + C( 0 τ 1+γ (ɛ 2 0δ 2 τ 2γ + δ 4 τ 4γ ) dτ) 1 γ 2 t 2. Then Φ R ({R k })(t) 2 l := (1 + k ) 2s Φ R 2,s k ({R j }) 2 k t k Cɛ 2 0(ɛ 2 0δ 2 t 2γ +δ 4 t 4γ )+C (1 + k )2s R k (τ) 2 (ɛ 2 0δ 2 τ 2γ +δ 4 τ 4γ ) dτt γ 0 τ 1+γ Cδ 2 t 2γ (ɛ ɛ 2 0δ 2 t 2γ + δ 4 t 4γ ) Cδ 2 t 2γ (ɛ δ 2 t 2γ ) 2 δ2 t 2γ 10. The t Φ R ({R k }(t)) 2 l, s is treated similarly, by using l 2,s l 2, s l, s.

12 The proof: the non-resonant part On the non-resonant part operator Φ NR we shall perform an integration by parts to get advantage of the non-resonant phase (without the phase gain we have an issue for the discrete summations ; this is the only reason of adding the derivative in time in the definition of the space X γ ) For instance the boundary term Φ NR,B k ({R j })(t) for the IBP is t (j 1,j 2,j 3) NR k e i k 2 j 2 1 +j2 2 j2 3 4t π(k 2 j j 2 2 j 2 3 )(α j 1 + R j1 (t))(α j2 + R j2 (t))(α j3 + R j3 (t)). On the non-resonant set 1 k 2 j j 2 2 j 2 3 = 2 j 1 j 2 k j 1, so Φ NR,B k ({R j })(t) Ct j 1,j 2 Z (α j1 + R j1 (t))(α j2 + R j2 (t))(α j3 + R j3 (t)). (1 + j 1 j 2 )(1 + k j 1 ) 0/20

13 The proof: example of term in the non-resonant part By Cauchy-Schwarz in the summation in j 1, j 2 we get Φ NR,B ({R k })(t) 2 Ct 2 (1+ j 1 ) 2s (1+ j 2 ) 2s (α j1 +R j1 (t))(α j2 +R j2 (t)) 2 j 1,j 2 Z j 1,j 2 Z α k j1+j 2 + R k j1+j 2 (t) 2 (1 + j 1 ) 2s (1 + j 2 ) 2s (1 + j 1 j 2 ) 2 (1 + k j 1 ) 2. Therefore we control Φ NR,B ({R k })(t) 2 l 2,s Ct2 (ɛ δ 2 t 2γ ) 3, provided that the following sum is finite k,j 1,j 2 Z (1 + k ) 2s (1 + j 1 ) 2s (1 + j 2 ) 2s (1 + k j 1 + j 2 ) 2s (1 + j 1 j 2 ) 2 (1 + k j 1 ) 2. 1/20

14 12/20 Example of summation estimation Lemma For 0 < s 1 2 < s the following sum is finite: k,j 1,j 2 Z (1 + k ) 2s (1 + j 1 ) 2 s (1 + j 2 ) 2s (1 + j 1 j 2 ) 2 (1 + k j 1 ) 2 (1 + k j 1 + j 2 ) 2s. We split the summation in k into nine regions, in terms of the comparison of k with j 1 and with j 1 j 2. We denote, for j 1, j 2 fixed, the two series of three exhaustive regions on Z: B 1 = { k 1 2 j1 j2 }, B2 = { 1 2 j1 j2 < k 3 2 j1 j2 }, B3 = { 3 j1 j2 < k }, 2 C 1 = { k 1 2 j1 }, C2 = { 1 2 j1 < k 3 2 j1 }, C3 = { 3 j1 < k }. 2 We split the sum as follows: k,j 1,j 2 Z = j 1,j 2 Z ( k B 1 B 3 + k B 2 C 1 + k B 2 C 2 + k B 2 C 3 ).

15 Example of summation estimation On B 1 B 3 we have j 1,j 2 Z,k B 1 B 3 C (1+ k ) 2s (1+ k j 1+j 2 ) 2s < C so, as 2s > 1, 2s (1 + k ) (1 + j 1 ) 2 s (1 + j 2 ) 2s (1 + j 1 j 2 ) 2 (1 + k j 1 ) 2 (1 + k j 1 + j 2 ) 2s j 1,j 2 Z 1 1 (1 + j 2 ) 2s (1 + j 1 j 2 ) 2 (1 + k j <. k B 1 B 1 ) 2 3 On C 1 we have k 1 2 j 1 so, as 2s > 1 and 2 s + 2 2s > 1, j 1,j 2 Z,k B 2 C 1 C j 1,j 2 Z 2s (1 + k ) (1 + j 1 ) 2 s (1 + j 2 ) 2s (1 + j 1 j 2 ) 2 (1 + k j 1 ) 2 (1 + k j 1 + j 2 ) 2s (1 + j 1 ) 2s (1 + j 1 ) 2 s+2 (1 + j 2 ) 2s (1 + j 1 j 2 ) 2 C j 1 Z (1 + k j 1 + j 2 ) 2s k B 2 C (1 + j 1 ) 2 s+2 2s (1 + j 1 j <. 2 ) 2 j 2 Z 3/20

16 Example of summation estimation On B 2 C 2 we have k 3 2 j 1 and j 1 < 3 j 1 j 2 so j 1,j 2 Z,k B 2 C 2 C j 1,j 2 Z 2s (1 + k ) (1 + j 1 ) 2 s (1 + j 2 ) 2s (1 + j 1 j 2 ) 2 (1 + k j 1 ) 2 (1 + k j 1 + j 2 ) 2s (1 + j 1 ) 2s (1 + j 1 ) 2 s (1 + j 2 ) 2s (1 + j 1 j 2 ) 2 C j 1,j 2 Z (1 + k j 1 ) 2 k B 2 C <. (1 + j 1 ) 2 s+2 2s (1 + j 2 ) 2s On B 2 C 3 we have 3 2 j 1 < k 3 2 j 1 j ( j 1 + j 2 ): j 1,j 2 Z,k B 2 C 3 C j 1,j 2 Z C j 1,j 2 Z 2s (1 + k ) (1 + j 1 ) 2 s (1 + j 2 ) 2s (1 + j 1 j 2 ) 2 (1 + k j 1 ) 2 (1 + k j 1 + j 2 ) 2s (1 + j 1 ) 2s + (1 + j 2 ) 2s (1 + j 1 ) 2 s+2 (1 + j 2 ) 2s (1 + j 1 j 2 ) 2 (1 + k j 1 + j 2 ) 2s k B 2 C C (1 + j 1 ) 2 s+2 2s (1 + j 2 ) 2s (1 + j j 1,j 2 Z 1 ) 2 s+2 (1 + j 1 j <. 2 ) 2 4/20

17 15/20 A Talbot effect The linear and nonlinear Schrödinger evolution on the torus of functions with bounded variation was proved to present Talbot effect features (Berry, Klein; Oskolkov; Kapitanski, Rodnianski; Taylor ; Erdogan, Tzirakis 96-13). Here we place ourselves in a more singular setting on R. A consequence of the Theorem is that the solution u(t) of the modified cubic NLS with initial data u 0 = k Z α kδ k such that {α k } l 2,s ɛ 0 behaves for small times like the linear evolution e it u 0. We compute first the linear evolution e it u 0 which display a Talbot effect.

18 A Talbot effect for linear evolutions of several Diracs Proposition Let t = 1 p 2π q with q odd. Let u 0 be such that û 0 is 2π periodic and û 0 located modulo 2π only in a neighborhood of zero of radius less than π p. For a given x R we denote l x Z and 0 m x < q the unique numbers such that x l x m x q [0, 1 q ), and we define Then for some θ mx R ξ x := πq p (x l x m x q ) [0, π p ). e it u 0 (x) = 1 q û 0 (ξ x ) e it ξ2 x +ix ξx +iθmx. 6/20 The data u 0 = k Z δ k enters the above setting of the 2π periodicity in Fourier and localization in Fourier, as û 0 = u 0 = k Z δ k. Therefore e it u 0 (x) = 0 for x / Z + Z q, and is a Dirac mass otherwise, which is a Talbot effect. This kind of data does not enter our nonlinear framework.

19 17/20 A Talbot effect for nonlinear evolutions of several Diracs If moreover û 0 is located modulo 2π only in a neighborhood of zero of radius less than η π p with 0 < η < 1, then the previous linear evolution vanishes for x at distance larger than η q from Z + Z q. Proposition Let u 0 such that û 0 is a 2π periodic, located modulo 2π only in a neighborhood of zero of radius less than η π p with 0 < η < 1 and having Fourier coefficients {α k } satisfying {α k } l 2,s ɛ 0. Let u(t, x) be the local in time solution obtained in the Theorem. Then for t = 1 p 2π q with q odd and for all x at distance larger than η q from Z + Z q the function u(t, x) almost vanish for small times, in the sense: u(t, x) = k Z R k (t)e it δ k (x), with {R k } satisfying the decay t γ {R k (t)} l 2,s + t { t R k (t)} l, s < C.

20 A model for one vortex filament dynamics In a 3D homogeneous incompressible inviscid fluid a vortex filament is a a vortex tube with infinitesimal cross section: the vorticity is a singular measure supported along a curve in R 3 that moves with the flow. The binormal flow is the oldest, simpler and richer model for one vortex filament dynamics (Da Rios 1906, Arms-Hama 1965 using a truncated Biot-Savart s law). It imposes the evolution in time of a R 3 -curve χ(t) by χ t = χ s χ ss = c b. 8/20 Frenet s system Hasimoto 72 (direct computations) (...Madelung 1 ) The filament function u(t, s) = c(t, s)e i s 0 τ(t,s)ds satisfies the 1D NLS iu t + u ss ( u 2 A(t) ) u = 0, with A(t) in terms of curvature and torsion (c, τ)(t, 0).

21 18/20 A model for one vortex filament dynamics In a 3D homogeneous incompressible inviscid fluid a vortex filament is a a vortex tube with infinitesimal cross section: the vorticity is a singular measure supported along a curve in R 3 that moves with the flow. The binormal flow is the oldest, simpler and richer model for one vortex filament dynamics (Da Rios 1906, Arms-Hama 1965 using a truncated Biot-Savart s law). It imposes the evolution in time of a R 3 -curve χ(t) by χ t = χ s χ ss = c b. Frenet s system Hasimoto 72 (direct computations) (...Madelung 1 ) The filament function u(t, s) = c(t, s)e i s 0 τ(t,s)ds satisfies the 1D NLS iu t + u ss ( u 2 A(t) ) u = 0, with A(t) in terms of curvature and torsion (c, τ)(t, 0).

22 18/20 A model for one vortex filament dynamics In a 3D homogeneous incompressible inviscid fluid a vortex filament is a a vortex tube with infinitesimal cross section: the vorticity is a singular measure supported along a curve in R 3 that moves with the flow. The binormal flow is the oldest, simpler and richer model for one vortex filament dynamics (Da Rios 1906, Arms-Hama 1965 using a truncated Biot-Savart s law). It imposes the evolution in time of a R 3 -curve χ(t) by χ t = χ s χ ss = c b. Frenet s system Hasimoto 72 (direct computations) (...Madelung 1 ) The filament function u(t, s) = c(t, s)e i s 0 τ(t,s)ds satisfies the 1D NLS iu t + u ss ( u 2 A(t) ) u = 0, with A(t) in terms of curvature and torsion (c, τ)(t, 0).

23 Binormal flow results 9/20 Conversely, for A(t) and u s.t. iu t + u ss construct a solution of the binormal flow. Examples: Lines: u(t, s) = 0, A(t) = 0. Circles: u(t, s) = 1, A(t) = 1. Helices: u(t, s) = e itn2 e ins, A(t) = 1. ( u 2 A(t) ) u = 0, one can Travelling waves: u(t, s) = e itn2 e ins cosh(s 2Nt), A(t) = 1, (Hasimoto 72, Hopfinger-Browand 81). Self-similar solutions u(t, s) = a x 2 ei 4t t for A(t) = a 2 t and perturbations (physicists 70-80, Guttierez-Rivas-Vega 03, Guttierez-Vega 04, Banica-Vega 08-15). χ(0) admits a corner at x = 0 u 0 presents a Dirac mass at x = 0. Local well-posedness for (c, τ) in Sobolev spaces (Hasimoto 72, Nishiyama-Tani 94-97, Koiso 97-08), for currents for a weak formulation, with analysis at the level of the frame (Jerrard-Smets 11), for curves with a corner and curvature in weighted space (B.-Vega 15).

24 19/20 Binormal flow results Conversely, for A(t) and u s.t. iu t + u ss construct a solution of the binormal flow. Examples: Lines: u(t, s) = 0, A(t) = 0. Circles: u(t, s) = 1, A(t) = 1. Helices: u(t, s) = e itn2 e ins, A(t) = 1. ( u 2 A(t) ) u = 0, one can Travelling waves: u(t, s) = e itn2 e ins cosh(s 2Nt), A(t) = 1, (Hasimoto 72, Hopfinger-Browand 81). Self-similar solutions u(t, s) = a x 2 ei 4t t for A(t) = a 2 t and perturbations (physicists 70-80, Guttierez-Rivas-Vega 03, Guttierez-Vega 04, Banica-Vega 08-15). χ(0) admits a corner at x = 0 u 0 presents a Dirac mass at x = 0. Local well-posedness for (c, τ) in Sobolev spaces (Hasimoto 72, Nishiyama-Tani 94-97, Koiso 97-08), for currents for a weak formulation, with analysis at the level of the frame (Jerrard-Smets 11), for curves with a corner and curvature in weighted space (B.-Vega 15).

25 20/20 Corners interaction through the binormal flow A non-closed curve with one corner and curvature in weighted space smoothens instantaneously is an oscillating way (B.-Vega 15). This is in link with the Kelvin waves observed in vortex reconnections. A planar regular polygon with M sides is expected to evolve through the binormal flow to skew polygons with Mq sides at times of type p q (numerical simulations Grinstein-De Vore 96, Jerrard-Smets 15 and integration of the Frenet frame at rational times De la Hoz-Vega 15). At infinitesimal times evidence is given for the evolution to be the superposition of the evolutions of each initial corner (De la Hoz-Vega 17). Here the framework is of a broken line, for instance with two corners. The Theorem says that the curve gets through the binormal flow instantaneously smooth. Moreover, for infinitesimal times the evolution is as a superposition of the evolutions of each initial corner. The smoothening is in an oscillating way: the Proposition insures that at times of type p q the curvature of χ(t) displays concentrations near the locations x such that xq Z, and almost straight segments are between.