Some results on the nonlinear Klein-Gordon-Maxwell equations

Transcription

1 Some results on the nonlinear Klein-Gordon-Maxwell equations Alessio Pomponio Dipartimento di Matematica, Politecnico di Bari, Italy Granada, Spain, 2011

2 A solitary wave is a solution of a field equation whose energy travels as a localized packet and which preserves this localization in time.

3 A solitary wave is a solution of a field equation whose energy travels as a localized packet and which preserves this localization in time. A soliton is a solitary wave which exhibits some strong form of stability so that it behaves as a relativistic particle. Solitons arise in several mathematical physics contests, such as the field theory, the nonlinear optics and the fluid mechanics.

4 One of the equations which produces soliton solutions is the nonlinear Klein-Gordon equation: 2 ψ t 2 ψ + m2 ψ ψ p 2 ψ = 0 (KG) where ψ = ψ(x, t): R 3 R C is a field function, p > 2, and m is the mass.

5 One of the equations which produces soliton solutions is the nonlinear Klein-Gordon equation: 2 ψ t 2 ψ + m2 ψ ψ p 2 ψ = 0 (KG) where ψ = ψ(x, t): R 3 R C is a field function, p > 2, and m is the mass. Existence and stability results for (KG) have been obtained by several authors: Bellazzini, Benci, Berestycki, Bonanno, Coleman, Fortunato, Glazer, Lions, Martin, Micheletti, Shatah, Sinibaldi, Strauss...

6 The lagrangian density relative to (KG) is: [ L = 1 ] ψ 2 2 t ψ 2 m 2 ψ p ψ p.

7 Let us now assign the electromagnetic field (E, H),

8 Let us now assign the electromagnetic field (E, H), which is described by the guage potentials φ: R 3 R R, A: R 3 R R 3, by means of the following equations ( E = φ + A ), t H = A.

9 The interaction between the electromagnetic field (E, H) and the field ψ is described by the rule of minimal coupling :

10 The interaction between the electromagnetic field (E, H) and the field ψ is described by the rule of minimal coupling : formally in the lagrangian density L the ordinary derivatives t, are substituted by Weyl covariant derivatives,

11 The interaction between the electromagnetic field (E, H) and the field ψ is described by the rule of minimal coupling : formally in the lagrangian density L the ordinary derivatives t, are substituted by Weyl covariant derivatives, namely t t + ieφ, iea, e being the electrical charge.

12 Therefore L becomes [ L 0 = 1 ] ψ 2 t + ieφψ 2 ψ ieaψ 2 m 2 ψ p ψ p.

13 Therefore L becomes [ L 0 = 1 ψ 2 t + ieφψ 2 ψ ieaψ 2 m 2 ψ 2 ] + 1 p ψ p. If we set ψ(x, t) = u(x, t)e is(x,t), then the lagrangian density becomes u, S R L 0 = 1 2 {u2 t u 2 [ S ea 2 (S t + eφ) 2 + m 2 ]u 2 } + 1 p u p.

14 To the lagrangian density L 0 we have to add the lagrangian density of the electromagnetic field ( L 1 = 1 ( E 2 H 2) = 1 A 2 2 φ + t 2 A 2 ).

15 To the lagrangian density L 0 we have to add the lagrangian density of the electromagnetic field ( L 1 = 1 ( E 2 H 2) = 1 A 2 2 φ + t The total action is therefore: S tot (u, S, φ, A) = (L 0 + L 1 ) dx dt. 2 A 2 ).

16 The Eulero-Lagrange equations of the functional S tot = S tot (u, S, φ, A) with respect to u, S, φ and A are: [ ( ) ] u + S ea 2 S 2 t + eφ + m 2 u u p 2 u = 0; (1) [( ) ] S t t + eφ u 2 [( S ea) u 2] = 0; (2) ( ) ( ) A S + φ = e t t + eφ u 2 ; (3) ( A) + ( ) A + φ = e ( S ea) u 2. (4) t t

17 The Eulero-Lagrange equations of the functional S tot = S tot (u, S, φ, A) with respect to u, S, φ and A are: [ ( ) ] u + S ea 2 S 2 t + eφ + m 2 u u p 2 u = 0; (1) [( ) ] S t t + eφ u 2 [( S ea) u 2] = 0; (2) ( ) ( ) A S + φ = e t t + eφ u 2 ; (3) ( A) + ( ) A + φ = e ( S ea) u 2. (4) t t Existence and stability results have been obtained by Long (2006) for a sufficiently small charge e.

18 If we set ( ) S ρ = e t + eφ u 2, j = e ( S ea) u 2, (2), (3) and (4) become

19 If we set ( ) S ρ = e t + eφ u 2, j = e ( S ea) u 2, (2), (3) and (4) become ρ div j = 0, t (5) div E = ρ, (6) H E = j. t (7)

20 If we set ( ) S ρ = e t + eφ u 2, j = e ( S ea) u 2, (2), (3) and (4) become ρ div j = 0, t (5) div E = ρ, (6) H E = j. t (7) Interpreting ρ as the charge density and j as the current density, then (5) is the continuity equation, while (6) and (7) are two of Maxwell equations, respectively Gauss and Ampere equations.

21 We are looking for standing waves in the electrostatic case,

22 We are looking for standing waves in the electrostatic case, namely u = u(x), S = ωt, φ = φ(x), A = 0, where ω is a constant different from zero.

23 We are looking for standing waves in the electrostatic case, namely u = u(x), S = ωt, φ = φ(x), A = 0, where ω is a constant different from zero. Then (2) and (4) are identically satisfied, while (1) and (3) become

24 We are looking for standing waves in the electrostatic case, namely u = u(x), S = ωt, φ = φ(x), A = 0, where ω is a constant different from zero. Then (2) and (4) are identically satisfied, while (1) and (3) become { u + [m 2 (eφ ω) 2 ]u u p 2 u = 0 in R 3, φ = e(eφ ω)u 2 in R 3. (KGM)

25 We are looking for standing waves in the electrostatic case, namely u = u(x), S = ωt, φ = φ(x), A = 0, where ω is a constant different from zero. Then (2) and (4) are identically satisfied, while (1) and (3) become { u + [m 2 (eφ ω) 2 ]u u p 2 u = 0 in R 3, φ = e(eφ ω)u 2 in R 3. (KGM) Equations (KGM) are called nonlinear Klein-Gordon-Maxwell equations.

26 This problem has been studied by Benci, Cassani, D Aprile, d Avenia, Fortunato, Georgiev, Mugnai, Pisani, Siciliano, Visciglia...

27 This problem has been studied by Benci, Cassani, D Aprile, d Avenia, Fortunato, Georgiev, Mugnai, Pisani, Siciliano, Visciglia... Remark The sign of ω is not relevant for the existence of solutions. Indeed if (u, φ) is a solution of (KGM) with a certain value of ω, then (u, φ) is a solution corresponding to ω. So, without loss of generality, we shall assume ω > 0.

28 We look for solutions of the problem (KGM) with finite energy,

29 We look for solutions of the problem (KGM) with finite energy, namely u H 1 (R 3 ) and φ D 1,2 (R 3 ), as critical points of the functional E : H 1 (R 3 ) D 1,2 (R 3 ) R: E(u, φ) = 1 2 R 3 u 2 φ 2 + [m 2 (eφ ω) 2 ]u 2 1 p R 3 u p.

30 We look for solutions of the problem (KGM) with finite energy, namely u H 1 (R 3 ) and φ D 1,2 (R 3 ), as critical points of the functional E : H 1 (R 3 ) D 1,2 (R 3 ) R: E(u, φ) = 1 2 R 3 u 2 φ 2 + [m 2 (eφ ω) 2 ]u 2 1 p R 3 u p. The functional E is strongly indefinite, namely it is unbounded from above and from below, even modulo compact perturbations.

31 The reduction method For any u H 1 (R 3 ), there exists a unique φ = φ u D 1,2 (R 3 ) such that φ = e(eφ ω)u 2 in R 3.

32 The reduction method For any u H 1 (R 3 ), there exists a unique φ = φ u D 1,2 (R 3 ) such that φ = e(eφ ω)u 2 in R 3. Moreover φ u 0; φ u ω e, on the set {x R3 u(x) 0}.

33 The map Φ The map is continuously differentiable. Φ : u H 1 (R 3 ) φ u D 1,2 (R 3 )

34 The map Φ The map Φ : u H 1 (R 3 ) φ u D 1,2 (R 3 ) is continuously differentiable. The map Φ is continuous for the weak topology. More precisely:

35 The map Φ The map Φ : u H 1 (R 3 ) φ u D 1,2 (R 3 ) is continuously differentiable. The map Φ is continuous for the weak topology. More precisely: u n u 0 in H 1 (R 3 ) = φ un φ u0 in D 1,2 (R 3 ), up to subsequences.

36 The reduction method (u, φ) H 1 (R 3 ) D 1,2 (R 3 ) is a solution of (KGM);

37 The reduction method (u, φ) H 1 (R 3 ) D 1,2 (R 3 ) is a solution of (KGM); φ = φ u ;

38 The reduction method (u, φ) H 1 (R 3 ) D 1,2 (R 3 ) is a solution of (KGM); φ = φ u ; u H 1 (R 3 ) is a critical point of the functional I : H 1 (R 3 ) R, where I(u) = E(u, φ u ).

39 The reduction method (u, φ) H 1 (R 3 ) D 1,2 (R 3 ) is a solution of (KGM); φ = φ u ; u H 1 (R 3 ) is a critical point of the functional I : H 1 (R 3 ) R, where I(u) = E(u, φ u ). In particular, we have I(u) = 1 2 R 3 u 2 + (m 2 ω 2 )u 2 + eωφ u u 2 1 p R 3 u p.

40 Previous results Benci & Fortunato (2002): existence if 0 < ω < m and 4 < p < 6;

41 Previous results Benci & Fortunato (2002): existence if 0 < ω < m and 4 < p < 6; D Aprile & Mugnai (2004): existence if 0 < ω < m if 4 p < 6, p 2 0 < ω < 2 m if 2 < p < 4;

42 Previous results Benci & Fortunato (2002): existence if 0 < ω < m and 4 < p < 6; D Aprile & Mugnai (2004): existence if 0 < ω < m if 4 p < 6, p 2 0 < ω < 2 m if 2 < p < 4; D Aprile & Mugnai (2004): non-existence if 0 < ω m and p 6 or p 2.

43 The case 4 p < 6 These strange relationships between p, m and ω are related to the boundedness of Palais-Smale sequences.

44 The case 4 p < 6 These strange relationships between p, m and ω are related to the boundedness of Palais-Smale sequences. Let (u n ) n be a Palais-Smale sequence, namely I(u n ) b and I (u n ) 0.

45 The case 4 p < 6 These strange relationships between p, m and ω are related to the boundedness of Palais-Smale sequences. Let (u n ) n be a Palais-Smale sequence, namely I(u n ) b and I (u n ) 0. To prove the boundedness of Palais-Smale sequences, we have to compute pi(u n ) I (u n )[u n ]

46 The case 4 p < 6 These strange relationships between p, m and ω are related to the boundedness of Palais-Smale sequences. Let (u n ) n be a Palais-Smale sequence, namely We have I(u n ) b and I (u n ) 0. pi(u n ) I (u n )[u n ] p 2 ( = u n 2 + (m 2 ω 2 )u 2 n R 3 2 ) + p 4 2 eωφ u n u 2 n + e 2 φ 2 u n u 2 n

47 The case 4 p < 6 These strange relationships between p, m and ω are related to the boundedness of Palais-Smale sequences. Let (u n ) n be a Palais-Smale sequence, namely I(u n ) b and I (u n ) 0. Since φ un 0 and ω < m, we have pi(u n ) I (u n )[u n ] p 2 ( = u n 2 + (m 2 ω 2 ) R 3 2 }{{} >0 ) + p 4 2 eωφ u n u 2 n + e 2 φ 2 u n u 2 n }{{}}{{} 0 0 u 2 n

48 The case 4 p < 6 These strange relationships between p, m and ω are related to the boundedness of Palais-Smale sequences. Let (u n ) n be a Palais-Smale sequence, namely We have I(u n ) b and I (u n ) 0. c 1 + c 2 u n pi(u n ) I (u n )[u n ] c 3 u n 2 and so (u n ) n is bounded in H 1 (R 3 ).

49 The case 2 < p < 4 We have pi(u n ) I (u n )[u n ] p 2 ( = u n 2 + (m 2 ω 2 )u 2 n R 3 2 ) + p 4 2 eωφ u n u 2 n + e 2 φ 2 u n u 2 n

50 The case 2 < p < 4 We have pi(u n ) I (u n )[u n ] p 2 ( = u n 2 + (m 2 ω 2 ) R 3 2 }{{} >0 ) + p 4 2 eωφ u n u 2 n + e 2 φ 2 u n u 2 n }{{}}{{} 0 0 u 2 n

51 The case 2 < p < 4 Since eφ un ω, where u n 0, we have pi(u n ) I (u n )[u n ] p 2 ( ) = u n 2 + (m 2 ω 2 )u 2 n + p 4 R eωφ u n u 2 n + e 2 φ 2 u n u 2 n p 2 R 3 2 u n 2 + (p 2)m2 2ω 2 u 2 n 2

52 The case 2 < p < 4 p 2 If we require ω < m, we have 2 pi(u n ) I (u n )[u n ] p 2 ( ) = u n 2 + (m 2 ω 2 )u 2 n + p 4 R eωφ u n u 2 n + e 2 φ 2 u n u 2 n p 2 R 3 2 u n 2 + (p 2)m2 2ω 2 u 2 n }{{ 2 } >0

53 The case 2 < p < 4 We have c 1 + c 2 u n pi(u n ) I (u n )[u n ] c 3 u n 2 and so (u n ) n is bounded in H 1 (R 3 ).

54 The case 2 < p < 4 If we set g 0 (p) = p 2 2, then, by D Aprile & Mugnai (2004), there exists a nontrivial solution (u, φ) H 1 (R 3 ) D 1,2 (R 3 ) of (KGM) if 0 < ω < g 0 (p)m.

55 The case 2 < p < 4 If we set g 0 (p) = p 2 2, then, by D Aprile & Mugnai (2004), there exists a nontrivial solution (u, φ) H 1 (R 3 ) D 1,2 (R 3 ) of (KGM) if 0 < ω < g 0 (p)m. Let g(p) = { (p 2)(4 p) if 2 < p < 3, 1 if 3 p < 4.

56 For any p (2, 4), we have that g 0 (p) < g(p) p Figure 1: g 0 pfi dash, g pfi solid

57 The case 2 < p < 4 Theorem (A. Azzollini, L. Pisani, A.P.) Let p (2, 4). Assume that 0 < ω < g(p) m, then (KGM) admits a nontrivial weak solution (u, φ) H 1 (R 3 ) D 1,2 (R 3 ).

58 Sketch of the proof The proof relies on the monotonicity trick introduced by Struwe and by Jeanjean. We look for the critical points of the functional I λ (u) = 1 u 2 + (m 2 ω 2 )u 2 + eωφ u u 2 λ u p, 2 R 3 p R 3 for λ [δ, 1]. If λ = 1, then I 1 = I.

59 Sketch of the proof The proof relies on the monotonicity trick introduced by Struwe and by Jeanjean. We look for the critical points of the functional I λ (u) = 1 u 2 + (m 2 ω 2 )u 2 + eωφ u u 2 λ u p, 2 R 3 p R 3 for λ [δ, 1]. If λ = 1, then I 1 = I. We work in H 1 r (R 3 ) := {u H 1 (R 3 ) u is radially symmetric }.

60 Sketch of the proof

61 Sketch of the proof For a.e. λ [δ, 1], there exists a bounded Palais-Smale sequence for I λ ;

62 Sketch of the proof For a.e. λ [δ, 1], there exists a bounded Palais-Smale sequence for I λ ; for a.e. λ [δ, 1], there exists u λ, critical point of I λ and u λ p C 1, for some C 1 > 0;

63 Sketch of the proof For a.e. λ [δ, 1], there exists a bounded Palais-Smale sequence for I λ ; for a.e. λ [δ, 1], there exists u λ, critical point of I λ and u λ p C 1, for some C 1 > 0; for a.e. λ [δ, 1], u λ satisfies the Pohozaev identity;

64 Sketch of the proof For a.e. λ [δ, 1], there exists a bounded Palais-Smale sequence for I λ ; for a.e. λ [δ, 1], there exists u λ, critical point of I λ and u λ p C 1, for some C 1 > 0; for a.e. λ [δ, 1], u λ satisfies the Pohozaev identity; there exists C > 0 such that u λ C, for a.e. λ [δ, 1];

65 Sketch of the proof For a.e. λ [δ, 1], there exists a bounded Palais-Smale sequence for I λ ; for a.e. λ [δ, 1], there exists u λ, critical point of I λ and u λ p C 1, for some C 1 > 0; for a.e. λ [δ, 1], u λ satisfies the Pohozaev identity; there exists C > 0 such that u λ C, for a.e. λ [δ, 1]; u λ ū in H 1 r (R 3 ) and u λ ū in L p (R 3 ), as λ 1, with ū a critical point of I;

66 Sketch of the proof For a.e. λ [δ, 1], there exists a bounded Palais-Smale sequence for I λ ; for a.e. λ [δ, 1], there exists u λ, critical point of I λ and u λ p C 1, for some C 1 > 0; for a.e. λ [δ, 1], u λ satisfies the Pohozaev identity; there exists C > 0 such that u λ C, for a.e. λ [δ, 1]; u λ ū in H 1 r (R 3 ) and u λ ū in L p (R 3 ), as λ 1, with ū a critical point of I; ū 0, since ū p C 1 ;

67 Sketch of the proof For a.e. λ [δ, 1], there exists a bounded Palais-Smale sequence for I λ ; for a.e. λ [δ, 1], there exists u λ, critical point of I λ and u λ p C 1, for some C 1 > 0; for a.e. λ [δ, 1], u λ satisfies the Pohozaev identity; there exists C > 0 such that u λ C, for a.e. λ [δ, 1]; u λ ū in H 1 r (R 3 ) and u λ ū in L p (R 3 ), as λ 1, with ū a critical point of I; ū 0, since ū p C 1 ; (ū, φū) is a nontrivial solution of (KGM).

68 Ground state solutions If (u, φ u ) H 1 (R 3 ) D 1,2 (R 3 ) is a solution of (KGM), then E(u, φ u ) = I(u) 0.

69 Ground state solutions If (u, φ u ) H 1 (R 3 ) D 1,2 (R 3 ) is a solution of (KGM), then E(u, φ u ) = I(u) 0. We are interested in the existence of ground state solutions for the problem (KGM),

70 Ground state solutions If (u, φ u ) H 1 (R 3 ) D 1,2 (R 3 ) is a solution of (KGM), then E(u, φ u ) = I(u) 0. We are interested in the existence of ground state solutions for the problem (KGM), namely, we are looking for (u 0, φ 0 ) H 1 (R 3 ) D 1,2 (R 3 ) such that

71 Ground state solutions If (u, φ u ) H 1 (R 3 ) D 1,2 (R 3 ) is a solution of (KGM), then E(u, φ u ) = I(u) 0. We are interested in the existence of ground state solutions for the problem (KGM), namely, we are looking for (u 0, φ 0 ) H 1 (R 3 ) D 1,2 (R 3 ) such that (u 0, φ 0 ) is a nontrivial solution of (KGM);

72 Ground state solutions If (u, φ u ) H 1 (R 3 ) D 1,2 (R 3 ) is a solution of (KGM), then E(u, φ u ) = I(u) 0. We are interested in the existence of ground state solutions for the problem (KGM), namely, we are looking for (u 0, φ 0 ) H 1 (R 3 ) D 1,2 (R 3 ) such that (u 0, φ 0 ) is a nontrivial solution of (KGM); E(u 0, φ 0 ) E(u, φ), for any (u, φ) solutions of (KGM).

73 Ground state solutions Theorem (A. Azzollini, A.P.) The problem (KGM) has a ground state solution if 4 p < 6 and 0 < ω < m; p 2 2 < p < 4 and 0 < ω < m 2.

74 Ground state solutions Theorem (A. Azzollini, A.P.) The problem (KGM) has a ground state solution if 4 p < 6 and 0 < ω < m; p 2 2 < p < 4 and 0 < ω < m 2. Remark We do not know if theorem holds also requiring only 0 < ω < g(p)m, if 2 < p < 4.

75 Sketch of the proof We seek critical points of the functional I constrained on the Nehari manifold: N = { u H 1 (R 3 ) \ {0} I (u), u = 0 },

76 Sketch of the proof We seek critical points of the functional I constrained on the Nehari manifold: N = { u H 1 (R 3 ) \ {0} I (u), u = 0 }, all the nontrivial critical points of I belong to N ;

77 Sketch of the proof We seek critical points of the functional I constrained on the Nehari manifold: N = { u H 1 (R 3 ) \ {0} I (u), u = 0 }, all the nontrivial critical points of I belong to N ; N is a natural constraint, namely each critical point of I N is a critical point of I.

78 Sketch of the proof There exists C > 0 such that, for all u N, we get I(u) C.

79 Sketch of the proof There exists C > 0 such that, for all u N, we get I(u) C. If σ := inf I(u) > 0; u N

80 Sketch of the proof There exists C > 0 such that, for all u N, we get I(u) C. If σ := inf I(u) > 0; u N our aim is to find ū N such that I(ū) = σ, which would imply that (ū, φū) is a ground state solution of (KGM).

81 Sketch of the proof There exists (u n ) n N, a Palais-Smale sequence at livel σ, namely such that I(u n ) σ, I (u n ) 0.

82 Sketch of the proof There exists (u n ) n N, a Palais-Smale sequence at livel σ, namely such that I(u n ) σ, I (u n ) 0. (u n ) n is bounded in H 1 (R 3 );

83 Sketch of the proof There exists (u n ) n N, a Palais-Smale sequence at livel σ, namely such that I(u n ) σ, I (u n ) 0. (u n ) n is bounded in H 1 (R 3 ); (u n ) n is not vanishing,

84 Sketch of the proof There exists (u n ) n N, a Palais-Smale sequence at livel σ, namely such that I(u n ) σ, I (u n ) 0. (u n ) n is bounded in H 1 (R 3 ); (u n ) n is not vanishing, namely there exist C > 0, r > 0 and a sequence (ξ n ) n R 3 such that u 2 n C, for any n; B r(ξ n)

85 Sketch of the proof There exists (u n ) n N, a Palais-Smale sequence at livel σ, namely such that I(u n ) σ, I (u n ) 0. (u n ) n is bounded in H 1 (R 3 ); (u n ) n is not vanishing, namely there exist C > 0, r > 0 and a sequence (ξ n ) n R 3 such that u 2 n C, for any n; B r(ξ n) if v n := u n ( + ξ n ) N, then (v n ) n is bounded Palais-Smale sequence in H 1 (R 3 ) and: B r v 2 n C, for any n.

86 Sketch of the proof There exists v 0 H 1 (R 3 ) such that v n v 0 in H 1 (R 3 ) and, by the weak topology continuity of Φ, φ vn φ v0 in D 1,2 (R 3 );

87 Sketch of the proof There exists v 0 H 1 (R 3 ) such that v n v 0 in H 1 (R 3 ) and, by the weak topology continuity of Φ, φ vn φ v0 in D 1,2 (R 3 ); by the non vanishing, v 0 0 and φ v0 0;

88 Sketch of the proof There exists v 0 H 1 (R 3 ) such that v n v 0 in H 1 (R 3 ) and, by the weak topology continuity of Φ, φ vn φ v0 in D 1,2 (R 3 ); by the non vanishing, v 0 0 and φ v0 0; I (v 0 ) = 0 and, so, v 0 N ;

89 Sketch of the proof There exists v 0 H 1 (R 3 ) such that v n v 0 in H 1 (R 3 ) and, by the weak topology continuity of Φ, φ vn φ v0 in D 1,2 (R 3 ); by the non vanishing, v 0 0 and φ v0 0; I (v 0 ) = 0 and, so, v 0 N ; by the weak lower semicontinuity if H 1 -norm, we get I(v 0 ) = σ;

90 Sketch of the proof There exists v 0 H 1 (R 3 ) such that v n v 0 in H 1 (R 3 ) and, by the weak topology continuity of Φ, φ vn φ v0 in D 1,2 (R 3 ); by the non vanishing, v 0 0 and φ v0 0; I (v 0 ) = 0 and, so, v 0 N ; by the weak lower semicontinuity if H 1 -norm, we get I(v 0 ) = σ; (v 0, φ v0 ) is a ground state solution for the problem (KGM).

91 The zero-mass case We are interested in the limit case ω = m,

92 The zero-mass case We are interested in the limit case ω = m, namely { u + (2eωφ e 2 φ 2 )u f (u) = 0 in R 3 φ = e(eφ ω)u 2 in R 3. (KGM 0 )

93 The zero-mass case We are interested in the limit case ω = m, namely { u + (2eωφ e 2 φ 2 )u f (u) = 0 in R 3 φ = e(eφ ω)u 2 in R 3. (KGM 0 ) If we assume f (0) = 0, the first equation in (KGM 0 ) has the form of a nonlinear Schrödinger equation with a potential vanishing at infinity.

94 The zero-mass case We are interested in the limit case ω = m, namely { u + (2eωφ e 2 φ 2 )u f (u) = 0 in R 3 φ = e(eφ ω)u 2 in R 3. (KGM 0 ) If we assume f (0) = 0, the first equation in (KGM 0 ) has the form of a nonlinear Schrödinger equation with a potential vanishing at infinity. Indeed, if φ D 1,2 (R 3 ) (and is radial), we have lim x (2eωφ e2 φ 2 ) = 0.

95 The zero-mass case We are interested in the limit case ω = m, namely { u + (2eωφ e 2 φ 2 )u f (u) = 0 in R 3 φ = e(eφ ω)u 2 in R 3. (KGM 0 ) If we assume f (0) = 0, the first equation in (KGM 0 ) has the form of a nonlinear Schrödinger equation with a potential vanishing at infinity. Indeed, if φ D 1,2 (R 3 ) (and is radial), we have lim x (2eωφ e2 φ 2 ) = 0. So we are in the so-called zero-mass case for nonlinear field equations (see Berestycki & Lions (1983)).

96 The zero-mass case In order to get solutions, we need some stronger hypotheses on f, which force it to be inhomogeneous, with a supercritical growth near the origin and subcritical one at infinity.

97 The zero-mass case In order to get solutions, we need some stronger hypotheses on f, which force it to be inhomogeneous, with a supercritical growth near the origin and subcritical one at infinity. Let f : R R satisfies: (f1) f C 1 (R, R); (f2) t R \ {0} : αf (t) f (t)t; (f3) t R : f (t) C 1 min( t p, t q ); (f4) t R : f (t) C 2 min( t p 1, t q 1 ); with 4 < α p < 6 < q and C 1, C 2, positive constants.

98 The zero-mass case In order to get solutions, we need some stronger hypotheses on f, which force it to be inhomogeneous, with a supercritical growth near the origin and subcritical one at infinity. Let f : R R satisfies: (f1) f C 1 (R, R); (f2) t R \ {0} : αf (t) f (t)t; (f3) t R : f (t) C 1 min( t p, t q ); (f4) t R : f (t) C 2 min( t p 1, t q 1 ); with 4 < α p < 6 < q and C 1, C 2, positive constants. Namely { c1 t f (t) p, for t 1, c 2 t q, for t 1.

99 The zero-mass case Theorem (A. Azzollini, L. Pisani, A.P.) Assume that f satisfies the above hypotheses, then there exists a couple (u 0, φ 0 ) D 1,2 (R 3 ) D 1,2 (R 3 ) which is a nontrivial weak solution of (KGM 0 ).

100 The zero-mass case In this case a completely different approach is required.

101 The zero-mass case In this case a completely different approach is required. We will not find weak solutions of (KGM 0 ) as critical points of some functional E defined on D 1,2 (R 3 ) D 1,2 (R 3 ).

102 The zero-mass case In this case a completely different approach is required. We will not find weak solutions of (KGM 0 ) as critical points of some functional E defined on D 1,2 (R 3 ) D 1,2 (R 3 ). For a fixed u D 1,2 (R 3 ), we do not know if the second equation φ = e(eφ ω)u 2 in R 3 has a (unique) solution φ u D 1,2 (R 3 ).

103 The zero-mass case We could consider a functional E : H 1 (R 3 ) D 1,2 (R 3 ) R whose critical points are finite energy weak solutions;

104 The zero-mass case We could consider a functional E : H 1 (R 3 ) D 1,2 (R 3 ) R whose critical points are finite energy weak solutions; for every u H 1 (R 3 ) we can find φ u D 1,2 (R 3 ), the unique solution of the second equation;

105 The zero-mass case We could consider a functional E : H 1 (R 3 ) D 1,2 (R 3 ) R whose critical points are finite energy weak solutions; for every u H 1 (R 3 ) we can find φ u D 1,2 (R 3 ), the unique solution of the second equation; we could consider the reduced functional I(u) = E(u, φ u ) = 1 u 2 + eωφ u u 2 f (u); 2 R 3 R 3

106 The zero-mass case We could consider a functional E : H 1 (R 3 ) D 1,2 (R 3 ) R whose critical points are finite energy weak solutions; for every u H 1 (R 3 ) we can find φ u D 1,2 (R 3 ), the unique solution of the second equation; we could consider the reduced functional I(u) = E(u, φ u ) = 1 u 2 + eωφ u u 2 f (u); 2 R 3 R 3 the mountain pass geometry in H 1 (R 3 ) is not immediately available for I.

107 The zero-mass case Following a similar approach used by Berestycki & Lions (1983) and Bellazzini, Bonanno & Siciliano (2009), the solution (u 0, φ 0 ) D 1,2 (R 3 ) D 1,2 (R 3 ) will be found as limit of solutions of approximating problems { u + (ε + 2eωφ e 2 φ 2 )u f (u) = 0 in R 3, φ = e(eφ ω)u 2 in R 3, (KGM ε ) with ε > 0.

108 Sketch of the proof For every ε > 0, we can find a mountain pass level solution (u ε, φ uε ) H 1 (R 3 ) D 1,2 (R 3 ) of (KGM ε );

109 Sketch of the proof For every ε > 0, we can find a mountain pass level solution (u ε, φ uε ) H 1 (R 3 ) D 1,2 (R 3 ) of (KGM ε ); there exists C > 0 such that (u ε, φ uε ) D 1,2 D1,2 C;

110 Sketch of the proof For every ε > 0, we can find a mountain pass level solution (u ε, φ uε ) H 1 (R 3 ) D 1,2 (R 3 ) of (KGM ε ); there exists C > 0 such that (u ε, φ uε ) D 1,2 D 1,2 C; there exists (u 0, φ 0 ) D 1,2 (R 3 ) D 1,2 (R 3 ) such that, u ε u 0 and φ uε φ 0 in D 1,2 (R 3 ), as ε 0;

111 Sketch of the proof For every ε > 0, we can find a mountain pass level solution (u ε, φ uε ) H 1 (R 3 ) D 1,2 (R 3 ) of (KGM ε ); there exists C > 0 such that (u ε, φ uε ) D 1,2 D1,2 C; there exists (u 0, φ 0 ) D 1,2 (R 3 ) D 1,2 (R 3 ) such that, u ε u 0 and φ uε φ 0 in D 1,2 (R 3 ), as ε 0; u 0 0, and φ 0 0 and so (u 0, φ 0 ) is a weak nontrivial solution of (KGM 0 ).