Standing waves for a Gauged Nonlinear Schr odinger equation David - PowerPoint PPT Presentation

Standing waves for a Gauged Nonlinear Schr¨ odinger equation David Ruiz Departamento de An´ alisis Matem´ atico, Universidad de Granada Granada, 2-6 February, 2015.

Outline The problem 1 The limit functional 2 Main results 3

The problem Consider a planar gauged Nonlinear Schr¨ odinger Equation: iD 0 φ + ( D 1 D 1 + D 2 D 2 ) φ + | φ | p − 1 φ = 0. Here t ∈ R , x = ( x 1 , x 2 ) ∈ R 2 , φ : R × R 2 → C is the scalar field, A µ : R × R 2 → R are the components of the gauge potential and D µ = ∂ µ + iA µ is the covariant derivative ( µ = 0, 1, 2 ).

The problem Consider a planar gauged Nonlinear Schr¨ odinger Equation: iD 0 φ + ( D 1 D 1 + D 2 D 2 ) φ + | φ | p − 1 φ = 0. Here t ∈ R , x = ( x 1 , x 2 ) ∈ R 2 , φ : R × R 2 → C is the scalar field, A µ : R × R 2 → R are the components of the gauge potential and D µ = ∂ µ + iA µ is the covariant derivative ( µ = 0, 1, 2 ). In Chern-Simons theory, a modified gauge field equation has been introduced [Hagen, Jackiw, Schonfeld, Templeton, in the ’80s]; see also [Tarantello, PNLDE 2007.] ∂ µ F µν + 1 2 κǫ ναβ F αβ = j ν , F µν = ∂ µ A ν − ∂ ν A µ . Here κ ∈ R is the Chern-Simons constant and ǫ ναβ is the Levi-Civita tensor. Moreover, j µ is the conserved matter current, j 0 = | φ | 2 , j i = 2Im ( ¯ φ D i φ ) .

At low energies, the Maxwell term becomes negligible and can be dropped, giving rise to: 1 2 κǫ ναβ F αβ = j ν . See [Jackiw & Pi, ’90s]. Taking for simplicity κ = 2 , we arrive to the system  iD 0 φ + ( D 1 D 1 + D 2 D 2 ) φ + | φ | p − 1 φ = 0,   ∂ 0 A 1 − ∂ 1 A 0 = Im ( ¯  φ D 2 φ ) , (1) ∂ 0 A 2 − ∂ 2 A 0 = − Im ( ¯ φ D 1 φ ) ,    ∂ 1 A 2 − ∂ 2 A 1 = 1 2 | φ | 2 ,

As usual in Chern-Simons theory, problem (1) is invariant under gauge transformation, φ → φ e i χ , A µ → A µ − ∂ µ χ , for any arbitrary C ∞ -function χ . The initial value problem for p = 3 , as well as global existence and blow-up, has been addressed in [Berg´ e, de Bouard & Saut, 1995; Huh, 2009-2013; Liu-Smith-Tataru 2013; Oh-Pusateri, preprint; Liu-Smith, preprint; Chen-Smith, preprint]. The existence of standing waves for (1) and general p > 1 has been studied in [Byeon, Huh & Seok, 2012 and preprint]. They look for vortex solutions , i.e., solutions in the form:

φ ( t , x ) = u ( r ) e i ( N θ + ω t ) , A 0 ( x ) = A 0 ( | x | ) , A 1 ( t , x ) = − x 2 A 2 ( t , x ) = x 1 | x | 2 h ( | x | ) , | x | 2 h ( | x | ) . Here ( r , θ ) are polar coordinates, h is a positive function and N ∈ N is the order of the vortex at 0 .

φ ( t , x ) = u ( r ) e i ( N θ + ω t ) , A 0 ( x ) = A 0 ( | x | ) , A 1 ( t , x ) = − x 2 A 2 ( t , x ) = x 1 | x | 2 h ( | x | ) , | x | 2 h ( | x | ) . Here ( r , θ ) are polar coordinates, h is a positive function and N ∈ N is the order of the vortex at 0 . With this ansatz they obtain the nonlocal equation: � � ω + ( h u ( | x | ) − N ) 2 u = | u | p − 1 u , − ∆ u + + A 0 ( | x | ) ( P ) | x | 2 with � + ∞ � r h ( s ) − N s 2 u 2 ( s ) ds , u 2 ( s ) ds . h u ( r ) = A 0 ( r ) = s 0 r Moreover, any solution satisfies that u ( | x | ) ∼ | x | N around the origin.

In [Byeon, Huh & Seok, 2012 and preprint] it is shown that ( P ) is indeed the Euler-Lagrange equation of the energy functional: � � I ω ( u ) = 1 1 � |∇ u | 2 + ω u 2 � R 2 | u | p + 1 dx dx − p + 1 2 R 2 � � | x | � 2 � u 2 ( x ) + 1 u 2 ( s ) s ds − 2 N dx | x | 2 8 R 2 0 That functional is defined in the space: � � u 2 ( x ) � u ∈ H 1 r ( R 2 ) : H = | x | 2 dx < + ∞ . R 2 It can be proved that I ω is well-defined and C 1 .

A useful inequality In [Byeon, Huh & Seok 2012 and preprint], it is proved that, for any u ∈ H , � R 2 | u ( x ) | 4 dx � 1 2 � � � 1 � � | x | � � � 2 2 u 2 R 2 |∇ u ( x ) | 2 dx u 2 ( s ) s ds − 2 N � 2 dx . | x | 2 R 2 0 Furthermore, the equality is attained by the family of functions: √ � � 8 λ ( N + 1 ) | λ x | N u λ = ∈ H : λ ∈ ( 0, + ∞ ) . 1 + | λ x | 2 ( N + 1 )

Byeon-Huh-Seok results If p > 3 , I ω is unbounded from below and exhibits a mountain-pass geometry. The case p = 3 is special: static solutions can be found via the minimizers of the previous inequality. Alternatively, one can pass, via a self-dual equation, to a singular Liouville equation in R 2 . If 1 < p < 3 solutions are found as minimizers on a L 2 -sphere if N = 0 . Hence, ω comes out as a Lagrange multiplier, and it is not controlled. In general, the global behavior of the energy functional I ω is not studied for 1 < p < 3 . This is the main purpose of this talk.

On the boundedness from below of I ω Theorem Let N ∈ N , p ∈ ( 1, 3 ) . There exists ω 0 ( p ) > 0 such that: If 0 < ω < ω 0 , then I ω is unbounded from below. If ω > ω 0 , then I ω is bounded from below and coercive. If ω = ω 0 , then I ω 0 is bounded from below, not coercive and ´ ınf I ω 0 < 0 . The threshold value ω 0 has an explicit expression, and it is independent of N . A. Pomponio and D. R., 2015. Y. Jiang, A. Pomponio and D.R., preprint.

The limit functional Let u a fixed function, and define u ρ ( r ) = u ( r − ρ ) . Let us now estimate I ω ( u ρ ) as ρ → + ∞ . � + ∞ ( 2 π ) − 1 I ω ( u ρ ) = 1 ρ | 2 + ω u 2 ( | u ′ ρ ) r dr 2 0 � + ∞ � � r � 2 u 2 ρ ( r ) + 1 0 su 2 ρ ( s ) ds − 2 N dr 8 r 0 � + ∞ 1 | u ρ | p + 1 r dr . − p + 1 0

The limit functional Let u a fixed function, and define u ρ ( r ) = u ( r − ρ ) . Let us now estimate I ω ( u ρ ) as ρ → + ∞ . � + ∞ ( 2 π ) − 1 I ω ( u ρ ) ∼ 1 − ∞ ( | u ′ | 2 + ω u 2 )( r + ρ ) dr 2 � + ∞ � � r � 2 u 2 ( r ) + 1 − ∞ ( s + ρ ) u 2 ( s ) ds − 2 N dr 8 r + ρ − ∞ � + ∞ 1 − ∞ | u | p + 1 ( r + ρ ) dr . − p + 1

The limit functional Let u a fixed function, and define u ρ ( r ) = u ( r − ρ ) . Let us now estimate I ω ( u ρ ) as ρ → + ∞ . � + ∞ ( 2 π ) − 1 I ω ( u ρ ) ∼ 1 − ∞ ( | u ′ | 2 + ω u 2 ) ρ dr 2 � + ∞ � � r � 2 u 2 ( r ) + 1 − ∞ ρ u 2 ( s ) ds − 2 N dr 8 ρ − ∞ � + ∞ 1 − ∞ | u | p + 1 ρ dr . − p + 1

The limit functional Let u a fixed function, and define u ρ ( r ) = u ( r − ρ ) . Let us now estimate I ω ( u ρ ) as ρ → + ∞ . � + ∞ � 1 − ∞ ( | u ′ | 2 + ω u 2 ) dr ( 2 π ) − 1 I ω ( u ρ ) ∼ ρ 2 � + ∞ � � r � 2 + 1 − ∞ u 2 ( r ) − ∞ u 2 ( s ) ds dr 8 � + ∞ � 1 − ∞ | u | p + 1 dr − . p + 1

The limit functional Let u a fixed function, and define u ρ ( r ) = u ( r − ρ ) . Let us now estimate I ω ( u ρ ) as ρ → + ∞ . � + ∞ � 1 − ∞ ( | u ′ | 2 + ω u 2 ) dr ( 2 π ) − 1 I ω ( u ρ ) ∼ ρ 2 � � + ∞ � 3 + 1 − ∞ u 2 ( r ) dr 24 � + ∞ � 1 − ∞ | u | p + 1 dr − . p + 1

It is natural to define the limit functional J ω : H 1 ( R ) → R , � + ∞ � � + ∞ � 3 J ω ( u ) = 1 dr + 1 � | u ′ | 2 + ω u 2 � − ∞ u 2 dr 2 24 − ∞ � + ∞ 1 − ∞ | u | p + 1 dr . − p + 1 We have I ω ( u ρ ) ∼ 2 πρ J ω ( u ) , as ρ → + ∞ . Then, ´ ınf J ω < 0 ⇒ ´ ınf I ω = − ∞ .

It is natural to define the limit functional J ω : H 1 ( R ) → R , � + ∞ � � + ∞ � 3 J ω ( u ) = 1 dr + 1 � | u ′ | 2 + ω u 2 � − ∞ u 2 dr 2 24 − ∞ � + ∞ 1 − ∞ | u | p + 1 dr . − p + 1 We have I ω ( u ρ ) ∼ 2 πρ J ω ( u ) , as ρ → + ∞ . Then, ´ ınf J ω < 0 ⇒ ´ ınf I ω = − ∞ . We will actually show that ´ ınf J ω < 0 ⇔ ´ ınf I ω = − ∞ .

The limit functional Proposition Let p ∈ ( 1, 3 ) and ω > 0 . Then J ω is coercive and attains its infimum. The proof of the coercivity is based on the Gagliardo-Nirenberg inequality: � u � L 4 ( R ) � C � u ′ � 1/4 L 2 ( R ) � u � 3/4 L 2 ( R ) . Hence � � + ∞ � 3 � � + ∞ � � + ∞ − ∞ u 4 dr � C − ∞ | u ′ | 2 dr + − ∞ u 2 dr . 2

The limit problem The Euler-Lagrange equation of the functional J ω is: � � 2 � � � + ∞ ω + 1 − u ′′ + u = | u | p − 1 u , − ∞ u 2 ( s ) ds in R . (2) 4 � �� � k

The limit problem The Euler-Lagrange equation of the functional J ω is: � � 2 � � � + ∞ ω + 1 − u ′′ + u = | u | p − 1 u , − ∞ u 2 ( s ) ds in R . (2) 4 � �� � k Then u = ± w k up to translations, where √ 1 p − 1 w 1 ( w k ( r ) = k kr ) , and 1 � � p − 1 �� 2 1 − p p + 1 cosh 2 w 1 ( r ) = r . 2

Therefore, � � + ∞ � 2 k = ω + 1 = ω + 1 5 − p − ∞ w k ( r ) 2 dr 4 m 2 k p − 1 , 4 where � + ∞ − ∞ w 1 ( r ) 2 dr . m =

Proposition u is a nontrivial solution of (2) if and only if u ( r ) = ± w k ( r − ξ ) for some ξ ∈ R and k is a root of the equation k = ω + 1 5 − p p − 1 , k > 0. 4 m 2 k (3)