Presentation on Multichannel Nonlinear Scattering for Nonintegrable equations by A. Soffer and M. I. Weinstein Presenter : Chulkwang Kwak Facultad de Matem´ aticas Pontificia Universidad Cat´ olica de Chile 2017 Participating School, KAIST August 21–25, 2017 C. Kwak August 21–25, 2017 1 / 31
Part II C. Kwak August 21–25, 2017 2 / 31
Goal Theorem-Asymptotic stability Let Ω η = ( E ∗ , E ∗ + η ), where η is positive and sufficiently small. Then for all E 0 ∈ Ω η and γ 0 ∈ [0 , 2 π ), there exists a positive number ǫ = ǫ ( E 0 , η ) such that if Φ(0) = e iγ 0 ( ψ E 0 + φ 0 ) where � φ 0 � L 1 ( R 3 x ) + � φ 0 � H 1 ( R 3 x ) < ǫ then � t Φ( t ) = e − i 0 E ( s ) ds + iγ ( t ) ( ψ E ( t )+ φ ( t ) ) with ˙ γ ( t ) ∈ L 1 ( R t ) t →±∞ ( E ( t ) , γ ( t )) = ( E ± , γ ± )) E ( t ) , ˙ ( ⇒ ∃ lim C. Kwak August 21–25, 2017 3 / 31
Goal Theorem A. - Asymptotic stability and φ ( t ) is purely dispersive in the sense that �� x � − σ φ ( t ) � L 2 ( R 3 ) = O ( � t � − 3 2 ) for σ > 2, and � φ ( t ) � L 4 ( R 3 ) = O ( � t � − 3 4 ) as | t | → ∞ . C. Kwak August 21–25, 2017 4 / 31
Goal Asymptotic stability theorem is reduced to Proposition A Assume | E 0 − E ∗ | < η and � φ 0 � L 1 + � φ 0 � H 1 < ǫ, for sufficiently small η > 0 and ǫ = ǫ ( η ) > 0. Then, we have 3 2 �� x � − σ φ ( t ) � L 2 � � φ 0 � L 1 + � φ 0 � H 1 , sup � t � (1) t ∈ R 3 4 � φ ( t ) � L 4 � � φ 0 � L 1 + � φ 0 � H 1 sup � t � (2) t ∈ R and 3 ˙ 2 ( | ˙ sup � t � γ ( t ) | + | E ( t ) | ) � � φ 0 � L 1 + � φ 0 � H 1 (3) t ∈ R C. Kwak August 21–25, 2017 5 / 31
Decay estimates Decay estimate Let K = − ∆ + V acting on L 2 ( R 3 ), and assume Hypotheses on V . Also, V satisfies (NR). Let P c ( K ) denote the projection onto the continuous spectral part of K . If 1 /p + 1 /q = 1, 2 ≤ q ≤ ∞ , then � e itK P c ( K ) ψ � L q ≤ C q | t | − (3 / 2 − 3 /q ) � ψ � L p . If ψ is more regular ( ψ ∈ H 1 ), then � e itK P c ( K ) ψ � L q ≤ C q � t � − (3 / 2 − 3 /q ) ( � ψ � L p + � ψ � H 1 ) . A simple consequence is the following local decay estimate Local decay estimate Under the same assumption as in the above theorem, let σ > 3 / 2 − 3 /q . Then �� x � − σ e itK P c ( K ) ψ � L 2 ≤ C q | t | − (3 / 2 − 3 /q ) � ψ � L p . C. Kwak August 21–25, 2017 6 / 31
Introduction of quantities M j ( T ), j = 1 , 2 , 3 We introduce quantities M j ( T ), j = 1 , 2 , 3 corresponding to (1), (2) and (3). Let 0 < T < ∞ . Define 3 2 �� x � − σ φ ( t ) � L 2 , M 1 ( T ) = sup � t � | t |≤ T 3 4 � φ ( t ) � L 4 M 2 ( T ) = sup � t � | t |≤ T and M 3 ( T ) = sup � φ ( t ) � L 2 . | t |≤ T Once we have the uniform bound of M 1 ( T ) and M 2 ( T ) in T , by taking T → ∞ , we can prove (1) and (2). We note that M 3 ( T ) appears in the estimation of M 1 ( T ), and hence we will additionally control M 3 ( T ) by M 1 ( T ) and M 2 ( T ). C. Kwak August 21–25, 2017 7 / 31
Bounds of M j ( T ) Lemma A - M 2 ( T ) bound Under the assumptions in Theorem A. and definitions of M 1 and M 2 , we have M 2 ( T ) ≤ C 2 ( � φ 0 � L 1 + � φ 0 � H 1 ) + C 2 ( ψ E , ∂ E ψ E )( M 1 ( T ) + M 2 ( T ) 2 + M 2 ( T ) 3 ) + C ′ 2 M 2 ( T ) 3 , where C 2 ( ψ E , ∂ E ψ E ) → 0 as E → E ∗ . Lemma B - M 1 ( T ) bound Under the assumptions in Theorem A. and definitions of M 1 , M 2 and M 3 , we have M 1 ( T ) ≤ C 1 ( � φ 0 � L 1 + � φ 0 � H 1 ) 1 ( M 2 ( T ) 2 + M 2 ( T ) 3 + M 3 ( T ) M 2 ( T ) 2 ) , + C ′ whenever 0 < | E − E ∗ | ≪ 1. C. Kwak August 21–25, 2017 8 / 31
Bounds of M j ( T ) Lemma C - M 3 ( T ) bound Under the assumptions in Theorem A. and definitions of M 1 , M 2 and M 3 , we have M 3 ( T ) 2 ≤ C 3 ( ψ E , ∂ E ψ E )( M 1 ( T ) 2 + M 2 ( T ) 2 + M 2 ( T ) 4 ) + C ′ 3 M 2 ( T ) 4 , where C 3 ( ψ E , ∂ E ψ E ) → 0 as E → E ∗ . C. Kwak August 21–25, 2017 9 / 31
Control of ˙ E and ˙ γ We first control | ˙ E | and | ˙ γ | . From E ( t ) = � ∂ E ψ E , ψ E 0 � − 1 Im � F 2 , ψ E 0 � ˙ and γ ( t ) = −� ψ E , ψ E 0 � − 1 Re � F 2 , ψ E 0 � , ˙ where F 2 = F 2 , lin + F 2 , nl F 2 , lin = (2 ψ 2 E − ψ 2 E 0 ) φ + ψ 2 E φ F 2 , nl = 2 ψ E | φ | 2 + ψ E φ 2 + | φ | 2 φ, by H¨ older inequality, we have | ˙ E | ≤ |� ∂ E ψ E , ψ E 0 �| − 1 |� F 2 , ψ E 0 �| ≤ |� ∂ E ψ E , ψ E 0 �| − 1 � �� x � σ (3 ψ 2 E + ψ 2 E 0 ) ψ E 0 � L 2 �� x � − σ φ � L 2 � + � 3 ψ E ψ E 0 � L 2 � φ � 2 L 4 + � ψ E 0 � L 4 � φ � 3 L 4 C. Kwak August 21–25, 2017 10 / 31
Control of ˙ E and ˙ γ and γ | ≤ |� ψ E , ψ E 0 �| − 1 |� F 2 , ψ E 0 �| | ˙ ≤ |� ψ E , ψ E 0 �| − 1 � �� x � σ (3 ψ 2 E + ψ 2 E 0 ) ψ E 0 � L 2 �� x � − σ φ � L 2 � + � 3 ψ E ψ E 0 � L 2 � φ � 2 L 4 + � ψ E 0 � L 4 � φ � 3 . L 4 Using the definitions of M 1 ( T ) and M 2 ( T ), 2 ( M 1 ( T ) + M 2 ( T ) 2 + M 2 ( T ) 3 ) | ˙ E ( t ) | ≤ C E ( ψ E , ψ E 0 ) � t � − 3 and 2 ( M 1 ( T ) + M 2 ( T ) 2 + M 2 ( T ) 3 ) γ ( t ) | ≤ C γ ( ψ E , ψ E 0 ) � t � − 3 | ˙ C. Kwak August 21–25, 2017 11 / 31
Proof of Proposition A We assume that Lemmas A, B and C hold true. We remove M 3 ( T ) in M 1 ( T ) ≤ C 1 ( � φ 0 � L 1 + � φ 0 � H 1 ) 1 ( M 2 ( T ) 2 + M 2 ( T ) 3 + M 3 ( T ) M 2 ( T ) 2 ) , + C ′ by using M 3 ( T ) 2 ≤ C 3 ( ψ E , ∂ E ψ E )( M 1 ( T ) 2 + M 2 ( T ) 2 + M 2 ( T ) 4 ) + C ′ 3 M 2 ( T ) 4 . Then we have 1 ( M 2 ( T ) 2 + M 2 ( T ) 4 ) M 1 ( T ) ≤ C 1 ( � φ 0 � L 1 + � φ 0 � H 1 ) + C ′ (4) for 0 < | E − E ∗ | ≪ 1. Substitution of (4) into M 2 ( T ) ≤ C 2 ( � φ 0 � L 3 + � φ 0 � H 1 ) 4 + C 2 ( ψ E , ∂ E ψ E )( M 1 ( T ) + M 2 ( T ) 2 + M 2 ( T ) 3 ) + C ′ 2 M 2 ( T ) 3 , yields M 2 ( T ) ≤ C 1 ( � φ 0 � L 1 + � φ 0 � H 1 ) + C 2 M 2 ( T ) 2 + C 3 M 2 ( T ) 3 + C 4 M 2 ( T ) 4 . C. Kwak August 21–25, 2017 12 / 31
Proof of Proposition A This can be rewritten as M 2 ( T ) f ( M 2 ( T )) ≤ L, where f ( x ) = 1 − C 2 x − C 3 x 2 − C 4 x 3 and L = C 1 ( � φ 0 � L 1 + � φ 0 � H 1 ). For positive constants C 2 , C 3 , C 4 , we can know that there exists x 0 > 0 such that x 0 f ( x 0 ) = sup xf ( x ) x> 0 and xf ( x ) is increasing on (0 , x 0 ). Let | E 0 − E ∗ | = 2 η , where η > 0 will be chosen sufficiently small such that 1 C E ( ψ E , ψ E 0 ) , C γ ( ψ E , ψ E 0 ) ≤ η 2 and ηf ( η ) ≤ x 0 f ( x 0 ) . 2 We choose C 1 ǫ ≤ ηf ( η ). C. Kwak August 21–25, 2017 13 / 31
Proof of Proposition A If � φ 0 � L 1 ( R 3 x ) + � φ 0 � H 1 ( R 3 x ) ≤ ǫ, we know L ≤ ηf ( η ) ≤ x 0 f ( x 0 ) 2 and M 2 (0) = � φ 0 � L 4 < ǫ ≤ η. By the continuity of M 2 ( T ), we have M 2 ( T ) ≤ η and therefore M 1 ( T ) ≤ Cη for some C > 0. Hence, we have from 2 ( M 1 ( T ) + M 2 ( T ) 2 + M 2 ( T ) 3 ) | ˙ E ( t ) | ≤ C E ( ψ E , ψ E 0 ) � t � − 3 and 2 ( M 1 ( T ) + M 2 ( T ) 2 + M 2 ( T ) 3 ) γ ( t ) | ≤ C γ ( ψ E , ψ E 0 ) � t � − 3 | ˙ that | ˙ 3 2 � t � − 3 E | ≤ C E η (5) 2 and 3 2 � t � − 3 2 . | ˙ γ | ≤ C γ η (6) C. Kwak August 21–25, 2017 14 / 31
Proof of Proposition A Integration of (5) and (6) yields � T | ˙ γ ( t ) | dt ≤ � 3 2 , E ( t ) | + | ˙ Cη (7) − T where � C is independent of T and η . By choosing η sufficiently small, (7) ensures that � t | ˙ 3 E ( s ) | ds ≤ � 2 < η, | E ( t ) − E 0 | ≤ Cη | t | ≤ T, 0 and thus sup { t : | E ( t ) − E 0 | < η } = ∞ . Taking T → ∞ , we have M 1 ( ∞ ) ≤ η and M 2 ( ∞ ) ≤ Cη. C. Kwak August 21–25, 2017 15 / 31
Proof of Lemma A We first consider � φ � L 4 . Recall � t φ ( t ) = U ( t, 0) P c ( H ( E 0 )) φ 0 − i U ( t, s ) P c ( H ( E 0 )) F ( s ) ds, 0 where F = F 1 + F 2 , γψ E − i ˙ F 1 = ˙ E∂ E ψ E , F 2 = F 2 , lin + F 2 , nl , F 2 , lin = (2 ψ 2 E − ψ 2 E 0 ) φ + ψ 2 E φ, F 2 , nl = 2 ψ E | φ | 2 + ψ E φ 2 + | φ | 2 φ. We use � e itK P c ( K ) ψ � L q ≤ C q � t � − (3 / 2 − 3 /q ) ( � ψ � L p + � ψ � H 1 ) for the linear part and � e itK P c ( K ) ψ � L q ≤ C q | t | − (3 / 2 − 3 /q ) � ψ � L p for the nonlinear part to obtain that C. Kwak August 21–25, 2017 16 / 31
Proof of Lemma A � t � φ ( t ) � L 4 ≤ � U ( t, 0) φ 0 � L 4 + � U ( t, s ) P c ( H ( E 0 )) F � L 4 ds 0 ≤ C � t � − 3 4 ( � φ 0 � L 3 + � φ 0 � H 1 ) 4 � t � | t − s | − 3 3 + | ˙ + C ′ | ˙ γ ( t ) |� ψ E � L E ( t ) |� ∂ E ψ E � L 4 4 4 3 (8) 0 � + � (3 ψ 2 E + ψ 2 3 + � 3 ψ E | φ | 2 � L 3 + �| φ | 2 φ � L E 0 ) φ � L ds 4 4 4 3 ≤ C 1 � t � − 3 4 ( � φ 0 � L 1 + � φ 0 � H 1 ) � t | t − s | − 3 4 [ I + II + III + IV + V ] ds. + C 2 0 C. Kwak August 21–25, 2017 17 / 31
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