Stability and Instability of Standing Waves for the Nonlinear Fractional Schrödinger Equation. Shihui Zhu (joint with J. Zhang)

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1 and of Standing Waves the Fractional Schrödinger Equation Shihui Zhu (joint with J. Zhang) Department of Mathematics, Sichuan Normal University & IMS, National University of Singapore

2 P1 iu t ( + k 2 ) s u + ( 1 x γ u 2 )u = 0, x R N, (1) u = u(t, x) : [0, T ) R N C, 0 < T + ; ( + k 2 ) s u = F 1 [( ξ 2 + k 2 ) s F[u](ξ)], s > 0; ( 1 x γ u 2 )(x) = u(y) 2 x y γ dy; 0 < γ < 4s. Boson star, fractional quantum mechanics etc.

3 Known P2 In particular, when N = 3, s = 1 2 Boson star equation. and γ = 1, Eq.(1) is the Fröhlich, Lenzmann, Lewin( ): well-posedness, orbital stability, singularity of blow-up solutions. Bao and Dong(2011): efficient and accurate numerical methods to compute the ground states and dynamics.

4 Known P3 For general 0 < s < 1, Laskin(2000,2002): expanding the Feynman path integral from the Brownian-like to the Lévylike quantum mechanical paths. Cho, Hwang, Hajaiej and Ozawa(2012): local well-posedness, existence of blow-up solutions. Guo, Huo, Huang( ): global well-posedness.

5 Known P4 Existence and stability of standing waves the classic nonlinear Schrödinger equation(i.e.s = 1) Strauss(1977): existence of standing waves. Cazenave and Lions(1982): orbital stability of standing waves. Weinstein(1983): strongly instability of standing waves. Existence and stability of standing waves the fourth-order nonlinear Schrödinger equation(i.e.s = 2) Zhu, Zhang and Yang(2010): existence of standing waves and ground state. Fibich, Baruch and Mandelbaum(2011): numerical studies standing waves. Levandosky(1998): orbital stability of standing waves.

6 Problems and Arguments P5 Problems 1. Existence of standing waves (1). 2. of these standing waves. Arguments Weinstein(1983): blow-up argument Gérard(1998) Hmidi & Keraani(2005): profile decomposition Cazenave and Lions(1982), Zhang(2000): variational argument

7 Local Well-posedness P6 Denote H s := {v S (R N ) (1 + ξ 2 ) s v(ξ) 2 dξ < + } and Impose E(v) := 1 2 v( ) s vdx 1 4 v( 1 x γ v 2 )vdx. u(0, x) = u 0 H s. (2) Proposition 1 (Cho, Hwang, Hajaiej and Ozawa, 2012) Let N 2, 0 < s < 1 and 0 < γ 2s. If u 0 H s, then! u(t, x) of (1)- (2) on I = [0, T ) such that u(t, x) C(I; H s ) C 1 (I; H s ). Moreover, all t I, u(t, x) satisfies the following conservation laws. (i) Conservation of mass u(t) 2 = u 0 2. (ii) Conservation of energy E(u(t)) = E(u 0)

8 Profile Decomposition P7 Proposition 2 Let N 2 and 0 < s < 1. If {v n} + n=1 is a bounded sequence in H s, then there exists a subsequence of {v n} + n=1 (still denoted {v n} + n=1 ), a family {xj n} + j=1 of sequences in RN and a family {V j } + j=1 of H s functions satisfying the following. (i) For every k j, x k n x j n + as n +. (ii) For every l 1 and every x R N, v n(x) can be decomposed as with lim l + v n(x) = Moreover, we have, as n +, l V j (x x j n) + vn(x) l j=1 lim sup vn l p = 0 every p (2, n + v n 2 2 = l V j vn l o(1), j=1 2N (N 2s) + ). l v n 2 Ḣ = V j 2 s Ḣ + vn l 2 s Ḣ + o(1). s j=1

9 Orbital of Standing Waves P8 Let k = 0, N 2 and M > 0. following variational problem For any 0 < γ < 2s, we define the d M := v(x) 2 v(y) 2 dxdy is the energy func- where E(v) := 1 2 ξ 2s v 2 dξ 1 4 tional. Define the set inf {v H s v 2 2 =M} E(v) (3) x y γ S M := {v H s v is the minimizer of the variational problem (3)}. From the Euler-Lagrange Theorem, we see that any v S M, there exists ω R such that (4) ( ) s v + ωv ( 1 x γ v 2 )v = 0, v H s.

10 Orbital of Standing Waves P9 Theorem 1 Let k = 0, N 2 and M > 0. Assume 0 < s < 1 and 0 < γ < 2s. Then arbitrary ε > 0, there exists δ > 0 such that any u 0 H s, if the initial data u 0 satisfies inf u 0 v H s < δ, v S M then the corresponding solution u(t, x) of the Cauchy problem (1)-(2) is such that inf u(t, x) v(x) H s < ε v S M all t > 0, where S M is defined in (4).

11 Sketch of the Proof of Theorem 1 P10 Key: The existence of minimizer of Variational Problem (3) i.e. Let N 2, 0 < γ < 2s and M > 0. Then, d M := inf E(v) = min E(v). {v H s v 2 2 =M} {v H s v 2 2 =M} Firstly, the variational problem (3) is well defined. Using the Hölder inequality with 1 = γ + 2s γ, we deduce that 2s 2s v(x) 2 x y dx γ C x s v(x + y) γ s 2 v 2s γ s 2 C v γ ṡ H s Thus, ( 1 v(x) x γ v(x) 2 ) v(x) 2 2 dx 2s γ s v 2. x y γ dx v(y) 2 2 C v γ ṡ H 4s γ s s v 2. From Young inequality, C(ε, γ, s, M) > 0 such that all 0 < ε < 1 2 E(v) 1 2 v 2 Ḣ C v 4s γ s s 2 v γ ṡ ( 1 H s 2 ε) v 2 Ḣ C(ε, γ, s, v 2). s (5)

12 Sketch of the Proof of Theorem 1 P11 Secondly, there exists C 0 > 0, d M C 0 < 0. Indeed, take v n = ρ N 2 n R(ρ nx), where ρ n > 0 and R is a function such that v n 2 2 = R 2 2 = M. We deduce that E(v n) = 1 2 ξ 2s v n 2 dξ 1 vn(x) 2 v n(y) 2 dxdy 4 x y γ ξ 2s R 2 dξ ργ n R(x) 2 R(y) 2 4 dxdy. = ρ2s n 2 x y γ Since 0 < γ < 2s, we can choose ρ n > 0 sufficiently small such that there exists C 0 > 0 such that E(v n) C 0 < 0.

13 Sketch of the Proof of Theorem 1 P12 Thirdly, we prove the existence of minimizers of (3). Take the minimizing sequence {v n} + n=1 Hs of (3) satisfying that as n +, E(v n) d M and v n 2 2 M. (6) Then, n large enough, E(v n) satisfies E(v n) < d M + 1. From (5), any 0 < ε < 1 and n 1 large enough, we deduce that 2 ( 1 2 ε) vn 2 Ḣ s d M C(ε, γ, s, M). Hence, {v n} + n=1 is bounded in Hs. Apply Proposition 2, we see that v n 2 2 = v n(x) = l V j (x x j n) + vn, l (7) j=1 l Vn j 2 2+ vn l 2 2+o(1), v n 2 Ḣ = s j=1 vn(x) 2 v n(y) 2 x y γ dxdy = l j=1 l Vn j 2 Ḣ + vn l 2 s Ḣ +o(1), s j=1 V j n (x) 2 V j n (y) 2 x y γ dxdy+ v l n (x) 2 x

14 Sketch of the Proof of Theorem 1 P13 Thus, we have E(v n) = l E(V j (x x j n)) + E(vn) l + o(1). (8) j=1 Take the scaling transmation V j ρ j = ρ jv j (x x j n) with ρ j = M V j (x x j n) 2 1. We have Vρ j j 2 2 = M. We deduce that as n + and l + d M E(v n) = l ( E(V j ρj ) + ρ2 j 1 ρ 2 4 j=1 j l d M ρ + inf 2 j 1 ρ 2 j j 1 4 j=1 d M + C 0 ( M V j ( V j (x x j n ) 2 V j (y x j n ) 2 l j=1 ) 1 + o(1), x y γ ) dxdy V j (x x j n ) 2 V j (y x j n ) 2 dxdy x y γ + E(vn) l + o(1 ) + vl n 2 2 M dm where C 0 > 0. Theree, there exists only one term V j0 0 in the decomposition (7) such that V j = M. V j 0 is the minimizer of (3).

15 Strong of Standing Waves P14 Theorem 2 Let k = 0 and N 2. Assume 0 < γ 2 = s < 1. Then, the ground state solitary waves e iωt Q(x) of the fractional nonlinear Schrödinger equation (1) are unstable in the following sense: For arbitrary ε > 0, there exist the radial initial data {u 0,n} + n=1 Hs 0 with s 0 = max{2s, 2s+1 2 } satisfying x u 0,n L 2, x u 0,n L 2, u 0,n Q H s < ε (9) and the corresponding solution {u n(t, x)} + n=1 blows up in the finite time, where Q is the ground state solution of ( ) s 1 Q + Q ( x 2s Q 2 )Q = 0. (10)

16 Sketch of the Proof of Theorem 2 P15 Proposition 3 Let N 2 and 0 < s < 1. Then, v 2 dx 2 N Proof. ( For all v H s, we have ) 1 vx( ) 1 s 2 ( xvdx v( ) s vdx) 1 2, v H s. v 2 dx = 2 N = 2 N 2 N 2 N F[ (xv)]f 1 [v]dξ F[v] ξ ξ F[v]dξ ξ s F[v] ξ 1 s ξ F[v] dξ ( v( ) s vdx ) 1 2 ( vx( ) 1 s xvdx ) 1 2.

17 Sketch of the Proof of Theorem 2 P16 Proposition 4 (see [Cho, Hwang, Kwon and Lee, 2012]) Let k = 0, N 2, 0 < s < 1 and γ = 2s. Assume that u 0 H s 0 with s 0 = max{2s, γ+1 2 } is radial symmetric, and x u 0 L 2 and x u 0 L 2. If u(t, x) is the solution of the Cauchy problem (1)-(2), then all t I (the maximal time interval), ux( ) 1 s udx is nonnegative and ux( ) 1 s xudx 2sE(u 0)t 2 + Ct + C.

18 Sketch of the Proof of Theorem 2 P17 Let {c n} + n=1 C\{0} be such that cn > 1 and lim cn = 1, and n + ρn = 1. We take the initial data {ρ n} + n=1 R+ be such that lim n + u 0,n = c nρ N 2 n Q(ρ nx), where Q is the ground state solution of (10). We see that all n 1, u 0,n 2 > Q 2 and and lim u 0,n 2 = lim cn Q 2 = Q 2 n + n + lim u 0,n Ḣs = lim cn n + n + ρs n Q Ḣs = Q Ḣs. Hence, from the Brézis-Lieb Lemma, ε > 0, u 0,n Q H s sufficiently large. On the other hand, we see that < ε as n is E(u 0,n ) = ( cn 2 c n 4 )ρ 2s n Q 2 2 Ḣ s C 0 < 0.

19 Sketch of the Proof of Theorem 2 P18 Then, u nx( ) 1 s xu ndx 2sC 0 t 2 + Ct + C, which implies that there exists 0 < T < + such that lim t T u nx( ) 1 s xu ndx = 0. Finally, using the conservation of mass and applying the inequality in Proposition 3 to u n, we see that all time t u 0,n 2 2 = un(t) ( N unx( ) 1 s xu ) ndx 1 2 ( u n( ) s u ndx) ( N unx( ) 1 s xu ) ndx 1 2 u n(t) Ḣs. Theree, there exists 0 < T < + such that lim un(t) t T Ḣ s = +.

20 Thanks!

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