A Variational Analysis of a Gauged Nonlinear Schrödinger Equation

A Variational Analysis of a Gauged Nonlinear Schrödinger Equation Alessio Pomponio, joint work with David Ruiz Dipartimento di Meccanica, Matematica e Management, Politecnico di Bari Variational and Topological Methods in Nonlinear Phenomena Alghero, June 24-28, 2013.

We are concerned with a planar gauged Nonlinear Schrödinger 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, namely (A 0, A 1, A 2 ) = (A 0, A), and D µ = µ + ia µ is the covariant derivative (µ = 0, 1, 2).

The field equations, in non-relativistic notation, are B = A, E = A 0 t A, where (E, B) is the electromagnetic field. The natural and obvious equation of gauge field dynamic is the Maxwell equation: µ F µν = j ν, where F µν = µ A ν ν A µ, and j µ is the conserved matter current, j 0 = φ 2, j i = 2Im ( φd i φ).

A modified gauge field equation has been introduced adding the so-called Chern-Simons term into the previous equation [Deser, Hangen, Jackiw, Schonfeld, Templeton, in the 80s]:

A modified gauge field equation has been introduced adding the so-called Chern-Simons term into the previous equation [Deser, Hangen, Jackiw, Schonfeld, Templeton, in the 80s]: µ F µν + 1 2 κɛναβ F αβ = j ν, where κ is a parameter that measures the strength of the Chern-Simons term, ɛ ναβ is the Levi-Civita tensor, and super-indices are related to the Minkowski metric with signature (1, 1, 1).

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( φd 2 φ), 0 A 2 2 A 0 = Im( φd 1 φ), 1 A 2 2 A 1 = 1 2 φ 2, (1) where the unknowns are (φ, A 0, A 1, A 2 ).

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( φd 2 φ), 0 A 2 2 A 0 = Im( φd 1 φ), 1 A 2 2 A 1 = 1 2 φ 2, (1) where the unknowns are (φ, A 0, A 1, A 2 ). As usual in Chern-Simons theory, problem (1) is invariant under gauge transformation, for any arbitrary C -function χ. φ φe iχ, A µ A µ µ χ,

This model was first proposed and studied by Jackiw & Pi, in several papers in the beginning of 90s, and sometimes has received the name of Chern-Simons-Schrödinger equation.

This model was first proposed and studied by Jackiw & Pi, in several papers in the beginning of 90s, and sometimes has received the name of Chern-Simons-Schrödinger equation. The initial value problem, as well as global existence and blow-up, has been addressed in [Bergé, de Bouard & Saut, 1995; Huh, 2009-2013] for the case p = 3.

The existence of stationary states for (1) and general p > 1 has been studied recently in [Byeon, Huh & Seok, JFA 2012].

The existence of stationary states for (1) and general p > 1 has been studied recently in [Byeon, Huh & Seok, JFA 2012]. They look for the classical solutions, that is, solutions (φ, A 0, A 1, A 2 ) of (1) in C 2 (R 2 ) C 1 (R 2 ) C 1 (R 2 ) C 1 (R 2 ). Moreover they seek the standing wave solutions of the form φ(t, x) = u( x )e iωt, A 0 (x) = k( x ), A 1 (t, x) = x 2 x 2 h( x ), A 2(t, x) = x 1 x 2 h( x ), where ω > 0 is a given frequency and u, k, h are real valued functions on [0, ) and h(0) = 0.

With this ansatz, (1) becomes u + ωu + A 0 u + A 2 1 u + A2 2 u u p 1 u = 0, 1 A 0 = A 2 u 2, 2 A 0 = A 1 u 2, 1 A 2 2 A 1 = 1 2 u2.

With this ansatz, (1) becomes u + ωu + A 0 u + A 2 1 u + A2 2 u u p 1 u = 0, 1 A 0 = A 2 u 2, 2 A 0 = A 1 u 2, 1 A 2 2 A 1 = 1 2 u2. First, they solve A 0, A 1, A 2 in terms of u. The ansatz implies 1 s h (s) = 1 2 u2 (s).

With this ansatz, (1) becomes u + ωu + A 0 u + A 2 1 u + A2 2 u u p 1 u = 0, 1 A 0 = A 2 u 2, 2 A 0 = A 1 u 2, 1 A 2 2 A 1 = 1 2 u2. First, they solve A 0, A 1, A 2 in terms of u. The ansatz implies 1 s h (s) = 1 2 u2 (s). From the condition h(0) = 0, we get and so h(r) = r 0 s 2 u2 (s) ds, A 1 (x) = x x 2 s x 2 0 2 u2 (s) ds, A 2 (x) = x x 1 s x 2 0 2 u2 (s) ds.

For what concerns A 0, we have k (s) = h(s) u 2 (s). s By integrating both sides from r to, we obtain A 0 (x) = k( x ) = ξ + x h(s) u 2 (s) ds, s where ξ is an arbitrary constant which is just the value of k at infinity.

Therefore we need only to solve, in R 2, the equation: ( u + ω + ξ + h2 ( x ) + x 2 + x ) h(s) u 2 (s) ds u = u p 1 u. s

Therefore we need only to solve, in R 2, the equation: ( u + ω + ξ + h2 ( x ) + x 2 + x ) h(s) u 2 (s) ds u = u p 1 u. s Let us show that the constant ω + ξ is a gauge invariant of the stationary solutions of the problem.

Since problem (1) is invariant under gauge transformation, φ φe iχ, A µ A µ µ χ, for any arbitrary C function χ, if ( (φ, A 0, A 1, A 2 ) = u( x )e iωt, k( x ), x ) 2 x 2 h( x ), x 1 x 2 h( x ), is a solution of (1), then, taking χ = ct, ( ( φ, Ã 0, A 1, A 2 ) = u( x )e i(ω+c)t, k( x ) c, x ) 2 x 2 h( x ), x 1 x 2 h( x ), is a solution of (1), too.

The equation We will take ξ = 0 in what follows, that is, lim A 0 (x) = 0, x + which was assumed in [Jackiw & Pi, Bergé, de Bouard & Saut].

The equation We will take ξ = 0 in what follows, that is, lim A 0 (x) = 0, x + which was assumed in [Jackiw & Pi, Bergé, de Bouard & Saut]. Our aim is to solve, in R 2, the nonlocal equation: ( u + ω + h2 ( x ) + x 2 + x ) h(s) u 2 (s) ds u = u p 1 u, s (P) where h(r) = r 0 s 2 u2 (s) ds.

In [Byeon, Huh, Seok, JFA 2012] it is shown that (P) is indeed the Euler-Lagrange equation of the energy functional: I ω : H 1 r (R 2 ) R, defined as I ω (u) = 1 ( u 2 + ωu 2) dx 2 R 2 + 1 u 2 ( (x) x ) 2 8 R 2 x 2 su 2 (s) ds dx 1 u p+1 dx. 0 p + 1 R 2

Since ( 1 x ) 2 x 2 su 2 (s) ds = C ( ) 2 0 x 2 u 2 (x) dx C u 4 L B(0, x ) 4 (R 2 ),

Since ( 1 x ) 2 x 2 su 2 (s) ds = C ( ) 2 0 x 2 u 2 (x) dx C u 4 L B(0, x ) 4 (R 2 ), then, for any u H 1 r (R 2 ), u 2 (x) R 2 x 2 ( x 2 su 2 (s) ds) dx < +. 0

Since ( 1 x ) 2 x 2 su 2 (s) ds = C ( ) 2 0 x 2 u 2 (x) dx C u 4 L B(0, x ) 4 (R 2 ), then, for any u H 1 r (R 2 ), u 2 (x) R 2 x 2 ( x 2 su 2 (s) ds) dx < +. 0 Therefore the functional is well defined in H 1 r (R 2 ).

A useful inequality In [Byeon, Huh & Seok], it is proved that, for any u H 1 r (R 2 ), u(x) 4 dx R 2 ( ) 1 ( 2 u(x) 2 2 u 2 ( x 2 12 dx R 2 R 2 x 2 su (s)ds) dx) 2. 0

A useful inequality In [Byeon, Huh & Seok], it is proved that, for any u H 1 r (R 2 ), u(x) 4 dx R 2 ( ) 1 ( 2 u(x) 2 2 u 2 ( x 2 12 dx R 2 R 2 x 2 su (s)ds) dx) 2. 0 Furthermore, the equality is attained by a continuum of functions { } 8l u l = 1 + lx 2 H1 r (R 2 ) : l (0, + ).

Byeon-Huh-Seok results: case p > 3 In this case the local nonlinearity dominates the nonlocal term.

Byeon-Huh-Seok results: case p > 3 In this case the local nonlinearity dominates the nonlocal term. I ω is unbounded from below.

Byeon-Huh-Seok results: case p > 3 In this case the local nonlinearity dominates the nonlocal term. I ω is unbounded from below. I ω exhibits a mountain-pass geometry.

Byeon-Huh-Seok results: case p > 3 In this case the local nonlinearity dominates the nonlocal term. I ω is unbounded from below. I ω exhibits a mountain-pass geometry. The existence of a solution is not so direct, since for p (3, 5) the (PS) property is not known to hold: we do not know if (PS)-sequences are bounded. This problem is bypassed by using a constrained minimization taking into account the Nehari and Pohozaev identities, in the spirit of [Ruiz, JFA 2006] for the Schrödinger-Poisson equation.

Byeon-Huh-Seok results: case p = 3 This is a special case:

Byeon-Huh-Seok results: case p = 3 This is a special case: static solutions can be found by passing to a self-dual equation, which leads to a Liouville equation that can be solved explicitly.

Byeon-Huh-Seok results: case p = 3 This is a special case: static solutions can be found by passing to a self-dual equation, which leads to a Liouville equation that can be solved explicitly. Any standing wave solutions (φ, A 0, A 1, A 2 ) of the previous type with φ > 0 have the following form: (φ, A 0, A 1, A 2 ) = ( ( 8le iωt 1 + lx 2, 2l 1 + lx 2 where l > 0 is an arbitrary real constant. ) 2 ω, 2l 2 x 2 1 + lx 2, 2l 2 x 1 1 + lx 2 ),

Byeon-Huh-Seok results: case 1 < p < 3 In this case, the nonlocal term prevails over the local nonlinearity.

Byeon-Huh-Seok results: case 1 < p < 3 In this case, the nonlocal term prevails over the local nonlinearity. Solutions are found as minimizers on a L 2 -sphere.

Byeon-Huh-Seok results: case 1 < p < 3 In this case, the nonlocal term prevails over the local nonlinearity. Solutions are found as minimizers on a L 2 -sphere. The value ω comes out as a Lagrange multiplier, and it is not controlled. By the gauge invariance, this is not a problem if we are looking for solutions (φ, A 0, A 1, A 2 ) of the entire system (1). For what concerns the single equation (P), a solution u is found only for a particular value of ω.

Analogy with Schrödinger-Poisson equation?

Analogy with Schrödinger-Poisson equation? The nonlinear Schrödinger-Poisson equation is { u + u + λφu = u p 1 u in R 3, φ = u 2 in R 3, (2) here λ > 0. If ū is a critical point of the functional and if I λ (u) = 1 ( u 2 + u 2 )dx + λ 2 R 3 4 1 u p+1 dx, p + 1 R 3 then (ū, φū) is a solution of (2). φū = 1 x ū 2, R 3 ( ) 1 x u 2 u 2 dx

Analogy with Schrödinger-Poisson equation? First difference: while the critical exponent for problem (P) is p = 3, for (2) it is p = 2.

Analogy with Schrödinger-Poisson equation? First difference: while the critical exponent for problem (P) is p = 3, for (2) it is p = 2. The case 1 < p < 3 for problem (P) is similar to the case 1 < p < 2 for (2)? In particular, in [Ruiz, JFA 2006] it is proved that if 1 < p < 2, for λ small enough, the functional I λ is bounded from below and it possesses at least two critical points: one is the minimum, at negative level, the other is a mountain-pass critical point, at positive level.

On the boundedness from below of Iω

On the boundedness from below of I ω Theorem (A.P. & D. Ruiz) Let p (1, 3). There exists ω 0 such that: if ω (0, ω 0 ), then I ω is unbounded from below;

On the boundedness from below of I ω Theorem (A.P. & D. Ruiz) Let p (1, 3). There exists ω 0 such that: if ω (0, ω 0 ), then I ω is unbounded from below; if ω = ω 0, then I ω0 is bounded from below, not coercive and inf I ω0 < 0; if ω > ω 0, then I ω is bounded from below and coercive. ω 0 has an explicit expression: ω 0 = 3 p 3 + p 3 p 1 2(3 p) 2 2 3 p ( m 2 (3 + p) p 1 ) p 1 2(3 p), with m = + ( ( )) 2 2 p 1 p + 1 cosh2 2 r 1 p dr.

Rough sketch of the proof I ω is coercive when the problem is posed on a bounded domain.

Rough sketch of the proof I ω is coercive when the problem is posed on a bounded domain. There exists a minimizer u n on the ball B(0, n) with Dirichlet boundary conditions.

Rough sketch of the proof I ω is coercive when the problem is posed on a bounded domain. There exists a minimizer u n on the ball B(0, n) with Dirichlet boundary conditions. To prove boundedness of u n, the problem is the possible loss of mass at infinity as n +. We need to study the behavior of those masses. If unbounded, the sequence u n behaves as a soliton, if u n is interpreted as a function of a single real variable.

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 +.

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 +. (2π) 1 I ω (u ρ ) = 1 2 + 1 8 + 0 + 0 1 p + 1 ( u ρ 2 + ωu 2 ρ)r dr u 2 ( ρ(r) r 2 su 2 r ρ(s) ds) dr + 0 0 u ρ p+1 r dr.

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 +. (2π) 1 I ω (u ρ ) = 1 2 + 1 8 + ρ + 1 p + 1 ( u 2 + ωu 2 )(r + ρ) dr ρ + u 2 ( (r) r r + ρ ρ 2 (s + ρ)u 2 (s) ds) dr ρ u p+1 (r + ρ) dr.

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 +. (2π) 1 I ω (u ρ ) 1 2 + 1 8 + + 1 p + 1 ( u 2 + ωu 2 )(r + ρ) dr + u 2 ( (r) r r + ρ 2 (s + ρ)u 2 (s) ds) dr u p+1 (r + ρ) dr.

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 +. (2π) 1 I ω (u ρ ) 1 2 + 1 8 + + 1 p + 1 ( u 2 + ωu 2 )(r + ρ) dr + u 2 ( (r) r r + ρ u p+1 (r + ρ) dr. (s + ρ)u 2 (s) ds) 2 dr

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 +. (2π) 1 I ω (u ρ ) 1 2 + 1 8 + + 1 p + 1 ( u 2 + ωu 2 )ρ dr + u 2 ( (r) r ρ u p+1 ρ dr. ρu 2 (s) ds) 2 dr

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 + (2π) 1 I ω (u ρ ) ρ 2 + 1 8 + 1 p + 1 ( u 2 + ωu 2 ) dr + ( r u 2 (r) 2 u 2 (s) ds) dr ] u p+1 dr.

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 + (2π) 1 I ω (u ρ ) ρ 2 + 1 24 ( u 2 + ωu 2 ) dr ( + 1 p + 1 + ) 3 u 2 (r)dr ] u p+1 dr.

It is natural to define the limit functional J ω : H 1 (R) R, J ω (u) = 1 2 + 1 p + 1 ( u 2 + ωu 2) dr + 1 ( + 24 + u p+1 dr. ) 3 u 2 dr

It is natural to define the limit functional J ω : H 1 (R) R, We have J ω (u) = 1 2 + 1 p + 1 ( u 2 + ωu 2) dr + 1 ( + 24 + u p+1 dr. ) 3 u 2 dr I ω (u ρ ) 2πρ J ω (u), as ρ +.

It is natural to define the limit functional J ω : H 1 (R) R, We have J ω (u) = 1 2 + 1 p + 1 ( u 2 + ωu 2) dr + 1 ( + 24 + u p+1 dr. ) 3 u 2 dr I ω (u ρ ) 2πρ J ω (u), as ρ +. Of course inf J ω < 0 inf I ω =.

It is natural to define the limit functional J ω : H 1 (R) R, We have J ω (u) = 1 2 + 1 p + 1 ( u 2 + ωu 2) dr + 1 ( + 24 + u p+1 dr. ) 3 u 2 dr I ω (u ρ ) 2πρ J ω (u), as ρ +. Of course We will show inf J ω < 0 inf I ω =. inf J ω < 0 inf I ω =.

The limit functional Proposition Let p (1, 3) and ω > 0. Then: a) J ω is coercive and attains its infimum;

The limit functional Proposition Let p (1, 3) and ω > 0. Then: a) J ω is coercive and attains its infimum; b) 0 is a local minimum of J ω ;

The limit functional Proposition Let p (1, 3) and ω > 0. Then: a) J ω is coercive and attains its infimum; b) 0 is a local minimum of J ω ; c) there exists ω 0 > 0 such that min J ω < 0 if and only if ω (0, ω 0 ).

The limit problem The critical points of J ω are solutions of the nonlocal equation u + ωu + 1 ( + 2 u 2 (s) ds) u = u p 1 u, in R. (P ) 4

The limit problem The critical points of J ω are solutions of the nonlocal equation u + ωu + 1 ( + 2 u 2 (s) ds) u = u p 1 u, 4 in R. (P ) We can find the explicit solutions of problem (P ).

The limit problem The critical points of J ω are solutions of the nonlocal equation u + ωu + 1 ( + 2 u 2 (s) ds) u = u p 1 u, in R. (P ) 4 We can find the explicit solutions of problem (P ). Observe that (P ) can be written as [ u + ω + 1 ( + ) ] 2 u 2 (s) ds u = u p 1 u, in R. 4

One dimensional nonlinear Schrödinger equation For any k > 0, let w k H 1 (R) be the unique positive radial solution of: w k + kw k = w p k, in R. (3)

One dimensional nonlinear Schrödinger equation For any k > 0, let w k H 1 (R) be the unique positive radial solution of: w k + kw k = w p k, in R. (3) Any solution of (3) is of the form u(r) = ±w k (r ξ), for some ξ R. Moreover, w k (r) = k 1 p 1 w 1 ( kr), where w 1 (r) = ( ( )) 1 2 p 1 p + 1 cosh2 2 r 1 p.

One dimensional nonlinear Schrödinger equation For any k > 0, let w k H 1 (R) be the unique positive radial solution of: w k + kw k = w p k, in R. (3) Any solution of (3) is of the form u(r) = ±w k (r ξ), for some ξ R. Moreover, where w 1 (r) = In what follows we define w k (r) = k 1 p 1 w 1 ( kr), ( ( )) 1 2 p 1 p + 1 cosh2 2 r 1 p. m = + w 2 1 dr.

Proposition u is a nontrivial solution of (P ) if and only if u(r) = w k (r ξ) for some ξ R and k a root of k = ω + 1 4 m2 k 5 p p 1, k > 0. (4)

Proposition u is a nontrivial solution of (P ) if and only if u(r) = w k (r ξ) for some ξ R and k a root of k = ω + 1 4 m2 k 5 p p 1, k > 0. (4) Moreover, there exists ω 1 > ω 0 > 0 such that if ω > ω 1, (4) has no solution and there is no nontrivial solution of (P ); if ω = ω 1, (4) has only one solution k 0 and w k0 (r) is the only nontrivial solution of (P ); if ω (0, ω 1 ), (4) has two solutions k 1 (ω) < k 2 (ω) and w k1 (r), w k2 (r) are the only two nontrivial solutions of (P ).

The map ψ Let us evaluate J ω on the curve k w k. We have [ p 5 ψ(k) = J ω (w k ) = m 2(3 + p) k 3+p 2(p 1) + ω ] 2 k 5 p 2(p 1) + m2 24 k 3(5 p) 2(p 1).

The map ψ Let us evaluate J ω on the curve k w k. We have [ p 5 ψ(k) = J ω (w k ) = m 2(3 + p) k 3+p 2(p 1) + ω ] 2 k 5 p 2(p 1) + m2 24 k 3(5 p) 2(p 1). Then: [ d 7 3p 5 p ψ(k) = m k 2(p 1) k + ω + 1 ] dk 4(p 1) 4 m2 k 5 p p 1.

The map ψ Let us evaluate J ω on the curve k w k. We have [ p 5 ψ(k) = J ω (w k ) = m 2(3 + p) k 3+p 2(p 1) + ω ] 2 k 5 p 2(p 1) + m2 24 k 3(5 p) 2(p 1). Then: [ d 7 3p 5 p ψ(k) = m k 2(p 1) k + ω + 1 ] dk 4(p 1) 4 m2 k 5 p p 1. The roots of (4) are exactly the critical points of ψ.

The map ψ If ω > ω 1, ψ is positive and increasing without critical points.

The map ψ If ω > ω 1, ψ is positive and increasing without critical points. If ω = ω 1, ψ is still positive and increasing, but it has an inflection point at k = k 0.

The map ψ If ω > ω 1, ψ is positive and increasing without critical points. If ω = ω 1, ψ is still positive and increasing, but it has an inflection point at k = k 0. If ω (ω 0, ω 1 ), ψ has a local maximum and minimum attained at k 1 and k 2, respectively, and w k2 is a local minimizer of J ω with J ω (w k2 ) > 0. If ω = ω 0, ψ(k 2 ) = 0. Observe then, in this case, the minimum of J ω0 is 0, and is attained at 0 and w k2.

The threshold value ω 0 In order to get the value of ω 0, observe that J ω0 (w k2 ) = 0.

The threshold value ω 0 In order to get the value of ω 0, observe that J ω0 (w k2 ) = 0. Therefore, ω 0 > 0 solves: and we infer that d ψ(k) = 0, dk ψ(k) = 0. ω 0 = 3 p 3 + p 3 p 1 2(3 p) 2 2 3 p ( m 2 (3 + p) p 1 ) p 1 2(3 p).

The values ω 0 (p) < ω 1 (p), for p (1, 3). We cannot obtain a more explicit expression of m depending on p.

The values ω 0 (p) < ω 1 (p), for p (1, 3). We cannot obtain a more explicit expression of m depending on p. For some specific values of p, m can be explicitly computed, and hence ω 0 and ω 1. For instance, if p = 2, m = 6, ω 1 = 2 9 3 and ω 0 = 2 5 15.

On the boundedness from below of I ω Theorem (A.P. & D. Ruiz) Let p (1, 3). We have: if ω (0, ω 0 ), then I ω is unbounded from below; if ω = ω 0, then I ω0 is bounded from below, not coercive and inf I ω0 < 0; if ω > ω 0, then I ω is bounded from below and coercive.

Sketch of the proof: case ω > ω0

Sketch of the proof: case ω ω 0 I ω is coercive when the problem is posed on a bounded domain.

Sketch of the proof: case ω ω 0 I ω is coercive when the problem is posed on a bounded domain. There exists u n a minimizer for I ω H 1 0,r (B(0,n)). Moreover, I ω (u n ) inf I ω, as n +.

Sketch of the proof: case ω ω 0 I ω is coercive when the problem is posed on a bounded domain. There exists u n a minimizer for I ω H 1 0,r (B(0,n)). Moreover, I ω (u n ) inf I ω, as n +. If u n is bounded, then I ω (u n ) is also bounded and therefore inf I ω is finite. In what follows we assume that u n is an unbounded sequence.

Sketch of the proof: case ω ω 0 Let ξ n R be the largest center of mass of u n : we have that ξ n u n 2.

Sketch of the proof: case ω ω 0 Let ξ n R be the largest center of mass of u n : we have that ξ n u n 2. Define ψ n : [0, + ] [0, 1] be a smooth function such that ψ n (r) = { 0, if r ξn 3 u n, 1, if r ξ n + 2 u n.

Sketch of the proof: case ω ω 0 Let ξ n R be the largest center of mass of u n : we have that ξ n u n 2. Define ψ n : [0, + ] [0, 1] be a smooth function such that ψ n (r) = { 0, if r ξn 3 u n, 1, if r ξ n + 2 u n. Let s estimate I ω (u n ) with I ω (ψ n u n ) and I ω ((1 ψ n )u n ): I ω (u n ) I ω (u n ψ n ) + I ω (u n (1 ψ n )) + c u n (1 ψ n ) 2 L 2 (R 2 ) + O( u n ).

Sketch of the proof: conclusion for ω > ω 0 Since I ω (u n ψ n ) = 2πξ n J ω (u n ψ n ) + O( u n ).

Sketch of the proof: conclusion for ω > ω 0 Since we have I ω (u n ψ n ) = 2πξ n J ω (u n ψ n ) + O( u n ). I ω (u n ) 2πξ n J ω (u n ψ n ) + I ω (u n (1 ψ n )) + c u n (1 ψ n ) 2 L 2 (R 2 ) + O( u n ).

Sketch of the proof: conclusion for ω > ω 0 Since we have I ω (u n ψ n ) = 2πξ n J ω (u n ψ n ) + O( u n ). I ω (u n ) 2πξ n J ω (u n ψ n ) + I ω (u n (1 ψ n )) + c u n (1 ψ n ) 2 L 2 (R 2 ) + O( u n ). Since u n ψ n H1 (R) c, we can prove that J ω (u n ψ n ) c > 0.

Sketch of the proof: conclusion for ω > ω 0 Since I ω (u n ψ n ) = 2πξ n J ω (u n ψ n ) + O( u n ). we have I ω (u n ) 2πξ n J ω (u n ψ n ) + I ω (u n (1 ψ n )) + c u n (1 ψ n ) 2 L 2 (R 2 ) + O( u n ). Since u n ψ n H1 (R) c, we can prove that J ω (u n ψ n ) c > 0. Since ξ n u n 2, we infer that I ω (u n ) > I ω (u n (1 ψ n )). Contradiction with the definition of u n, which proves that inf I ω >.

Sketch of the proof: conclusion for ω > ω 0 Let us now show that I ω is coercive.

Sketch of the proof: conclusion for ω > ω 0 Let us now show that I ω is coercive. Take u n H 1 r (R 2 ) an unbounded sequence, and assume that I ω (u n ) is bounded from above.

Sketch of the proof: conclusion for ω > ω 0 Let us now show that I ω is coercive. Take u n H 1 r (R 2 ) an unbounded sequence, and assume that I ω (u n ) is bounded from above. u n L2 (R 2 ) +.

Sketch of the proof: conclusion for ω > ω 0 Let us now show that I ω is coercive. Take u n H 1 r (R 2 ) an unbounded sequence, and assume that I ω (u n ) is bounded from above. u n L2 (R 2 ) +. Then for any ω 0 < ω < ω, I ω (u n ) = I ω (u n ) + ω ω u n 2 2 L 2 (R 2 ), a contradiction, since we know that I ω is bounded from below.

Sketch of the proof: conclusion for ω = ω0

Sketch of the proof: conclusion for ω = ω 0 We reach a contradiction unless J ω0 (u n ψ n ) 0.

Sketch of the proof: conclusion for ω = ω 0 We reach a contradiction unless J ω0 (u n ψ n ) 0. This implies that ψ n u n ( ξ n ) w k2, where w k2 nontrivial minimum of J ω0. is the

Sketch of the proof: conclusion for ω = ω 0 We reach a contradiction unless J ω0 (u n ψ n ) 0. This implies that ψ n u n ( ξ n ) w k2, where w k2 is the nontrivial minimum of J ω0. With this extra information, we have a better estimate: I ω0 (u n ) 2πξ n J ω0 (u n ψ n ) + I ω0 ( un (1 ψ n ) ) + c u n (1 ψ n ) 2 L 2 (R 2 ) + O(1).

On the solutions of (P)

On the solutions of (P) Theorem (On the boundedness of I ω ) Let p (1, 3). We have: if ω (0, ω 0 ), then I ω is unbounded from below; if ω = ω 0, then I ω0 is bounded from below, not coercive and inf I ω0 < 0; if ω > ω 0, then I ω is bounded from below and coercive. Theorem (A.P. & D. Ruiz) For almost every ω (0, ω 0 ], (P) admits a positive solution. Moreover, there exist ω > ω > ω 0 such that: if ω > ω, then (P) has no solutions different from zero; if ω (ω 0, ω), then (P) admits at least two positive solutions: one of them is a global minimizer for I ω and the other is a mountain-pass solution.

Sketch of the proof: case ω (0, ω 0 ]

Sketch of the proof: case ω (0, ω 0 ] Performing the rescaling u u ω = ω u( ω ), we get [ 1 ( I ω (u ω ) = ω u 2 + u 2) dx + 1 u 2 ( (x) x 2 2 R 2 8 R 2 x 2 su (s)ds) 2 dx 0 ] ω p 3 2 u p+1 dx. p + 1 R 2

Sketch of the proof: case ω (0, ω 0 ] Performing the rescaling u u ω = ω u( ω ), we get [ 1 ( I ω (u ω ) = ω u 2 + u 2) dx + 1 u 2 ( (x) x 2 2 R 2 8 R 2 x 2 su (s)ds) 2 dx 0 ] ω p 3 2 u p+1 dx. p + 1 R 2 The geometrical assumptions of the Mountain Pass Theorem are satisfied and we can apply the monotonicity trick [Struwe, Jeanjean], finding a solution for almost every ω (0, ω 0 ].

Sketch of the proof: case ω > ω 0 Recall that, for any u H 1 r (R 2 ), u(x) 4 dx R 2 ( ) 1 ( 2 u(x) 2 2 u 2 ( x 2 12 dx R 2 R 2 x 2 su (s)ds) dx) 2. 0

Sketch of the proof: case ω > ω 0 Recall that, for any u H 1 r (R 2 ), u(x) 4 dx R 2 ( ) 1 ( 2 u(x) 2 2 u 2 ( x 2 12 dx R 2 R 2 x 2 su (s)ds) dx) 2. 0 Let u be a solution of (P). We multiply (P) by u and integrate. By this inequality, we get 0 1 u 2 dx + (ωu 2 + 34 ) 4 u4 u p+1 dx. R 2 R 2

Sketch of the proof: case ω > ω0

Sketch of the proof: case ω > ω 0 Since inf I ω0 < 0, there exists ω > ω 0 such that inf I ω < 0 for ω (ω 0, ω).

Sketch of the proof: case ω > ω 0 Since inf I ω0 < 0, there exists ω > ω 0 such that inf I ω < 0 for ω (ω 0, ω). Being I ω coercive and weakly lower semicontinuous, the infimum is attained (at negative level). If ω (ω 0, ω), the functional satisfies the geometrical assumptions of the Mountain Pass Theorem. Since I ω is coercive, (PS) sequences are bounded.