Three dimensional vortices in the nonlinear wave equation

Three dimensional vortices in the nonlinear wave equation Marino Badiale - Vieri Benci - Sergio Rolando Abstract We prove the existence of rotating solitary waves (vortices) for the nonlinear Klein-Gordon equation with nonnegative potential, by finding nonnegative cylindrical solutions to the standing equation Z µ u2 −△u + 2 u + λu = g (u) , u ∈ H 1 (RN ) , (†) 2 dx < ∞ , |y| RN |y| where x = (y, z) ∈ Rk × RN −k , N > k ≥ 2 and µ, λ > 0. The nonnegativity of the potential makes the equation suitable for physical models and guarantees the well-posedness of the corresponding Cauchy problem, but it prevents the use of standard arguments in providing the functional associated to (†) with bounded Palais-Smale sequences.

1

Introduction and main results

In this paper we are concerned with the existence of vortices for the nonlinear wave equation ¤ψ + W ′ (ψ) = 0 , (1) 2 where W ′ is (under the standard identification between ¡ iθ C¢and R ) the gradient 1 of a C potential function W : C → R satisfying W e ψ = W (ψ), that is,

W (ψ) = V (|ψ|)

and W ′ (ψ) = V ′ (|ψ|)

ψ |ψ|

for some V ∈ C 1 (R; R) . (2)

Roughly speaking, a vortex is a solitary wave with nonvanishing angular momentum. A solitary wave is a nonsingular solution which travels as a localized packet in such a way that the energy is conserved in time in the region of space occupied by the wave. A solitary wave bears not only the energy ¸ Z · 1 1 2 2 E (ψ) = |∂t ψ| + |∇ψ| + W (ψ) dx (3) 2 R3 2 but also the other integrals of the motion, such as the angular momentum Z ∂t ψ (x × ∇ψ) dx , (4) M (ψ) = Re R3

1

which represent intrinsic properties of particles. In addition, the solitary waves of (1) exhibit all the most characteristic features of relativistic particles, such as the equivalence between mass and energy. Owing to this particle-like behaviour, solitary waves can thus be regarded as a model for extended particles, in contrast with point particles, and they arise in many problems of mathematical physics, such as classical and quantum field theory, nonlinear optics, fluid mechanics, plasma physics and cosmology (see, for instance, [32], [22], [19]). For an introduction to the theory of solitary waves in nonlinear field equations we refer, e.g., to [3], [6], [28]. Here we are interested in the existence of vortices of equation (1) with nonnegative potentials, that is, W ≥0

and

M (ψ) 6= 0 .

Observe that W ≥ 0 implies E ≥ 0, which is an important request for the consistence of physical models related to the equation since the existence of field configurations with negative energy would yield negative masses. Furthermore, the positivity of the energy also provides good a priori estimates for the solutions of the corresponding Cauchy problem and these estimates allow to prove that, under very general assumptions on W , the problem is well posed (cf. [6]). The most natural way for finding solitary waves for (1) is to look for static waves, i.e., time-independent solutions of the form ψ (t, x) = ψ0 (x) , and then to obtain travelling waves by Lorentz transforming. Unfortunately, this forces to assume that W takes negative values, for it is well known, since the renewed paper [16] of Derrik, that W ≥ 0 implies that any finite-energy static solution of (1) is necessarily trivial. Such a difficulty can be overcome by looking for standing waves, namely, finite-energy solutions having the following form: ψ (t, x) = ψ0 (x) e−iω0 t ,

ω0 > 0 .

(5)

In the literature a lot of work has been done in proving the existence of standing waves in the case in which ψ0 (x) ∈ R (we recall, e.g., [9], [10], [25], [26], [27]). Also in the physical literature there are many papers dealing with this topic, among which we recall the pioneering paper of Rosen [23] and the first rigorous existence paper [12]. In physics, the spherically symmetric standing waves have been called Q-balls by Coleman in [13] and this is the name used in all the subsequent papers. From the results of [9] (see also [6]) it follows that, if W satisfies (2) together with (i) V ≥ 0 and V (0) = V ′ (0) = 0 (ii) ∃V ′′ (0) = Ω2 > 0 2

(iii) V (u0 ) < 21 Ω2 u20 for some u0 > 0, then equation (1) has standing waves (5), with ψ0 (x) ∈ R, for every frequency ω0 ∈ (Ω0 , Ω), where ª © Ω0 := inf ω > 0 : V (u) < 12 ω 2 u2 for some u > 0 . However ψ0 (x) ∈ R implies M (ψ) = 0 and so, in order to get vortices, one has to consider complex valued ψ0 ’s. Thus, we are led to make an ansatz of the following form: ψ (t, x) = u (x) ei(k0 θ(x)−ω0 t) ,

u (x) ≥ 0, θ (x) ∈ R/2πZ, ω0 , k0 > 0,

(6)

by which equation (1) is equivalent to the system ( 2 −△u + k0 |∇θ| u − ω02 u + V ′ (u) = 0 u△θ + 2∇u · ∇θ = 0 .

Then, if we denote x = (y, z) = (y1 , y2 , z), assume u (y, z) = u (|y| , z) and choose the angular coordinate in R3 as phase function, that is,  arctan (y2 /y1 ) if y1 > 0    arctan (y2 /y1 ) + π if y1 < 0 θ (x) := (7) π/2 if y1 = 0 and y2 > 0    −π/2 if y1 = 0 and y2 < 0 ,

we get

△θ = 0 ,

∇θ · ∇u = 0 ,

1

2

|∇θ| =

2

|y|

so that the above system reduces to −△u +

k0

2u

|y|

+ V ′ (u) = ω02 u

,

in R3

and direct computations show that (3) and (4) become " Ã ! # ³ ´ Z 1 1 k0 2 i(kθ(x)−ω0 t) 2 2 E u (x) e = |∇u| + + ω0 u + V (u) dx 2 |y|2 R3 2 Z ´ µ ³ M u (x) ei(kθ(x)−ω0 t) = 0, 0, −ω0 k0

R3

¶ u2 dx .

By studying equation (8) we will prove the following result. Theorem 1 Let W : C → R satisfy (2) and assume that (i) V ≥ 0 and V (0) = 0 (ii) V ′ (u) = Ω2 u + O(up−1 ) as u → 0+ , for some Ω2 > 0 and p > 2 3

(8)

(9)

(10)

(iii) V (u0 ) < 21 Ω2 u20 for some u0 > 0. Then equation (1) has nonzero finite-energy classical solutions of the form (6)(7) for every k0 > 0 and ω0 ∈ (Ω0 , Ω), where ª © Ω0 := inf ω > 0 : V (u) < 12 ω 2 u2 for some u > 0 .

Notice that Ω0 < Ω by assumption (iii), so that the interval (Ω0 , Ω) is nonempty. The finite energy and angular momentum of the solutions we find are given by (9) and (10), and the angular momentum does not vanish since u is nonzero. We observe that the assumptions of Theorem 1 are satisfied by the model potential W (ψ) =

1 2 2 β 1 p q Ω |ψ| − |ψ| + |ψ| , 2 p q

Ω, β > 0, q > p > 2,

which is nonnegative provided that β is small enough. In the physical literature, the existence of solitary waves with nonvanishing angular momentum in classical field theory seems to be an interesting open issue, which has been recently addressed in a number of publications (see for instance [29], [14], [11] and the references therein). In particular, the existence of vortices for equation (1) has been investigated in [18] and [30], for very particular potentials. From the mathematical viewpoint, the existence of vortices has been studied in [8] and [4] (see also [5], [15], [7] for related equations and results), but the requirement W ≥ 0 was not permitted by the results there. Nevertheless, a suitable adaptation of the arguments of [4] (see also [20]) allows to prove a version of Theorem 1 also for the limit case ω0 = Ω (see Remark 5 below). Remark 2 Theorem 1 also gives travelling solitary waves with nonvanishing angular momentum, since, by Lorentz invariance, a solution ψv travelling with any vector velocity v can be obtained from a standing one by boosting. For instance, if ψ (t, x) = u (x) ei(k0 θ(x)−ω0 t) is a standing solution and v = (0, 0, v), |v| < 1 , then ψv (x, t) = u (y, γ (z − vt)) ei(k0 θ(x)−ω0 γ(t−vz)) ,

¡ ¢−1/2 γ = 1 − v2 ,

is a solution representing a bump which travels in the z-direction with speed v. Remark 3 The same arguments leading to Theorem 1 also yield the existence of standing and travelling rotating solitary waves for the nonlinear Schr¨odinger equation i∂t ψ = −△ψ + W ′ (ψ) ,

ψ (t, x) ∈ C, (t, x) ∈ R × R3 .

(11)

Actually we stated the result for the nonlinear Klein-Gordon equation (1) because it is for this equation that, as already mentioned, the assumption W ≥ 0 has special importance on physical grounds. 4

According to the previous discussion, the proof of Theorem 1 relies on finding nonnegative symmetric solutions to equation (8) with suitable integrability properties. In fact we will perform this study in a more general situation, that is, we will study the existence of nontrivial solutions to the following problem:  µ  −△u + 2 u + λu = g (u) in RN    |y|    u (y, z) = u (|y| , z) ≥ 0 in RN (12)  Z  2   u  u ∈ H 1 ¡RN ¢ ,   2 dx < ∞ N R |y| where x = (y, z) ∈ Rk × RN −k with N > k ≥ 2, the nonlinearity g : R → R is continuous and such that g (0) = 0, and µ, λ > 0 are real constants. More precisely, we introduce the spaces o n ¡ ¢ R u2 dx < ∞ , Hs := {u ∈ H : u (y, z) = u (|y| , z)} H := u ∈ H 1 RN : RN |y| 2

and look for weak solutions in the sense of the following definition: we name weak solution to problem (12) any nonnegative u ∈ Hs such that Z Z Z Z uh ∇u · ∇h dx + µ g (u) h dx for all h ∈ H . uh dx = dx + λ 2 RN |y| RN RN RN (13) Regarding the nonlinearity, we will assume that Rt (g0) there exists t0 > 0 such that 0 0 g (s) ds > 12 λt20 (g1) there exists p ∈ (2, 2∗ ) such that g (t) = O(tp−1 ) as t → 0+

(where 2∗ := 2N/ (N − 2)) together with one of the following conditions: (g2) g (β) = 0 for some β > β0 := inf{t > 0 : (g3) g (t) = O(tp−1 ) as t → +∞ .

Rt 0

g (s) ds > 21 λt2 }

We point out that the exponent p is the same in both hypotheses (g1) and (g3). The relationship between (8) and (12) is clear: writing V (|ψ|) =

1 2 2 Ω |ψ| − G (|ψ|) , 2

equation (8) reduces to the equation of (12) with λ = Ω2 − ω02 and g = G′ . The main difficulty in solving problem (12) is that we do not assume the well known superlinear condition due to Ambrosetti and Rabinowitz [1], namely σG (t) ≤ G′ (t) t for some σ > 2 and all t ∈ R, because, together with (g0), it implies σ G (|ψ|) ≥ (const.) |ψ| for |ψ| large 5

and thus forces W to take negative values. Hence a different approach is needed in order to provide the existence of a bounded Palais-Smale sequence for the Euler functional associated to (12). Our existence result is the following. Theorem 4 Let N > k ≥ 2 and µ, λ > 0. Assume that g ∈ C (R; R) satisfies (g0), (g1) and at least one of hypotheses (g2) and (g3). Then problem (12) has at least a nonzero weak solution, which satisfies kukL∞ (RN ) ≤ β if (g2) holds. Let us remark that Theorem 4 actually applies to more general situations than the ones needed to deduce Theorem 1. For instance it also admits pure power nonlinearities, or, more generally, nonlinearities which may satisfy the Ambrosetti-Rabinowitz condition. Remark 5 By asking p > 2∗ in (g1) and replacing p with a different exponent q ∈ (2, 2∗ ) in (g3), the arguments of [4, Section 4] together with the ones of Section 3 below allow to prove Theorem 4 also for λ = 0. We conclude this introductory section by collecting the notations of most frequent use throughout the paper. • Given N, k ∈ N, N > k ≥ 2, we shall always write x = (y, z) ∈ Rk × RN −k . • O (k) is the orthogonal group of Rk . • By u (y, z) = u (|y| , z) we always mean u (y, z) = u (Ry, z) for all R ∈ O (k) and almost every (y, z) ∈ Rk × RN −k . • For any r ∈ R we set r+ := (|r| + r) /2 and r− := (|r| − r) /2, so that r = r+ − r− with r+ , r− ≥ 0. • |A| and χA respectively denote the d-dimensional Lebesgue measure and the characteristic function of any measurable set A ⊆ Rd , d ≥ 1. • By → and ⇀ we respectively mean strong and weak convergence in a Banach space E, whose dual space is denoted by E ′ . The open ball Br (u0 ) := {u ∈ E : ku − u0 kE < r} shall be simply denoted by Br when E = RN and u0 = 0. • ֒→ denotes continuous embeddings. • Cc∞ (A) is the space of the infinitely differentiable (real or complex) functions with compact support in the open set A ⊆ Rd , d ≥ 1. • If 1 ≤ p ≤ ∞ then Lp (A) and Lploc (A) are the usual Lebesgue spaces (for any measurable set A ⊆ Rd , d ≥ 1). • 2∗ := 2N/ (N − 2), N ≥ 3, is the critical exponent for the Sobolev embedding. ∗ • H 1 (RN ) = {u ∈ L2 (RN ) : ∇u ∈ L2 (RN )} and D1,2 (RN ) = {u ∈ L2 (RN ) : ∇u ∈ L2 (RN )} are the usual Sobolev spaces.

2

Preliminaries

In this section we study the functional framework in which problem (12) can be cast into a variational formulation. In particular, Subsection 2.1 is devoted to a brief description of some weighted Sobolev spaces naturally related to problem 6

(12), while in Subsection 2.2 we derive a variational principle (Proposition 7) which allows us to recover weak solutions of problem (12) as critical points of a suitable functional, of which we also give some relevant properties (Lemmas 8 and 9). Throughout the section we assume N > k ≥ 2 and µ, λ > 0.

2.1

Weighted Sobolev spaces −2

Since H = H 1 (RN ) ∩ L2 (RN ; |y| dx) and the convergences in both H 1 (RN ) −2 and L2 (RN ; |y| dx) (which denotes the Lebesgue space with respect to the −2 measure |y| dx) imply almost everywhere convergence in RN , H and Hs are Hilbert spaces with respect to the norm defined by Z Z Z u2 2 2 kuk := |∇u| dx + µ dx + λ u2 dx for all u ∈ H , (14) 2 N N N |y| R R R which is induced by the inner product Z Z Z uv uv dx (u | v) := ∇u · ∇v dx + µ 2 dx + λ RN RN RN |y|

for all u, v ∈ H .

Clearly Hs ֒→ H ֒→ H 1 (RN ) and, by well known embeddings of H 1 (RN ), one has that H ֒→ Lp (RN ) for 2 ≤ p ≤ 2∗ and H ֒→ Lp (A) for A ⊂ RN bounded and 1 ≤ p ≤ 2∗ . In particular, the latter embedding is compact if p < 2∗ and thus it assures that weak convergence in H implies almost everywhere convergence in RN . The fact that the last integral in (14) behaves differently from the others in front of rescalings will yield a role to the larger Hilbert spaces n o ¡ ¢ ¡ ¢ R −2 u2 X := D1,2 RN ∩ L2 (RN ; |y| dx) = u ∈ D1,2 RN : RN |y| 2 dx < ∞ Xs := {u ∈ X : u (y, z) = u (|y| , z)}

endowed with the inner product Z Z (u | v)X := ∇u · ∇v dx + µ RN

which induces the norm given by Z Z 2 2 kukX := |∇u| dx + µ RN

uv RN

RN

2 dx

|y| u2

2 dx

|y|

for all u, v ∈ X ,

for all u ∈ X .

Clearly H = X ∩ L2 (RN ) ֒→ X ֒→ D1,2 (RN ) and Hs = Xs ∩ L2 (RN ) ֒→ Xs ֒→ ∗ X. Moreover, by well known embeddings of D1,2 (RN ), one has X ֒→ L2 (RN ) and X ֒→ Lp (A) with compact embedding if A ⊂ RN is bounded and 1 ≤ p < 2∗ (which also assures that weak convergence in X implies almost everywhere convergence in RN ). The following density result slightly clarifies the relation between the above spaces and will be used in Lemma 27. 7

Lemma 6 H and Hs are dense in X and Xs respectively. Proof. Fix any u ∈ X and consider a radial truncating sequence {ϕn } ⊂ Cc∞ (RN ) such that 0 ≤ ϕn ≤ 1 on RN , ϕn = 1 in Bn and ϕn = 0 in RN \ B2n , |∇ϕn | ≤ C/n for some constant C > 0. Clearly ϕn u ∈ H and ϕn u ∈ Hs if u ∈ Xs . Then, by dominated convergence, one readily obtains Z

RN

and Z

RN

2

(ϕn u − u) 2

|y|

2

|∇ (ϕn u − u)| dx

dx =

Z

2

(ϕn − 1) u2 2

|y|

RN

≤ 2 Z =

Z

dx = o (1)n→∞

2

RN

RN

|∇ϕn | u2 dx + 2

Z

2

RN

2

(ϕn − 1) |∇u| dx

2

|∇ϕn | u2 dx + o (1)n→∞ . ∗

Finally, by H¨older inequality and the fact that u ∈ L2 (RN ), we infer that Z Z Z C2 2 2 2 2 u2 dx |∇ϕn | u dx = |∇ϕn | u dx ≤ 2 n B2n \Bn RN B2n \Bn ÃZ !2/2∗ C2 2/N 2∗ ≤ |B2n \ Bn | |u| dx n2 B2n \Bn ÃZ !2/2∗ ¡ ¢ ∗ 2/N 2 ≤ C 2 2N − 1 |u| dx = o (1)n→∞ . RN \Bn

Hence kϕn u − ukX = o (1) as n → ∞. ¥

2.2

Variational approach

Let g ∈ C (R; R) satisfy the hypotheses of Theorem 4. Set χ := χ(0,β) if (g2) holds, χ := χ(0,+∞) otherwise. Then define f (t) := χ (t) g (t)

and F (t) :=

Z

t

f (s) ds

0

for all t ∈ R .

(15)

So, in any case, from (g0) one deduces that (F0) there exists t0 > 0 such that F (t0 ) > 21 λt20 . Moreover, since f : R → R is continuous, it is easy to check that our assumptions imply that (f ) ∃M > 0, p ∈ (2, 2∗ ) such that |f (t)| ≤ M p |t| which yields in turn 8

p−1

for all t ∈ R ,

p

(F) |F (t)| ≤ M |t| for all t ∈ R

(where p ∈ (2, 2∗ ) and M > 0).

Thanks to (f ), (F) and the continuous embedding H ֒→ Lp (RN ), the functional I : H → R defined by Z 1 2 F (u) dx for all u ∈ H (16) I (u) := kuk − 2 RN is of class C 1 on H and has Fr´echet derivative I ′ (u) ∈ H ′ at any u ∈ H given by Z ′ I (u) h = (u | h) − f (u) h dx for all h ∈ H . (17) RN

We now show that the set of weak solutions to problem (12) equals the set of critical points of the functional J := I|Hs : Hs → R defined as the restriction of I to Hs , which is obviously such that J ∈ C 1 (Hs ; R) and J ′ (u) h = I ′ (u) h for all u, h ∈ Hs . Proposition 7 Every critical point of J is a weak solution to problem (12) and, if (g2) holds, it satisfies u ≤ β almost everywhere in RN . Proof. Let u ∈ Hs be such that J ′ (u) h = 0 for all h ∈ Hs . Then, by virtue of the principle of symmetric criticality [21], u is a critical point of I, i.e., I ′ (u) h = 0 for all h ∈ H. Now, using h = u− ∈ Hs as test function in (17) and observing that f (u) u− vanishes almost everywhere in RN , one readily obtains ku− k = 0, that is, u ≥ 0. If f = χ(0,+∞) g, this implies f (u) = g (u) and thus (13) holds. Otherwise, if f = χ(0,β) g, we compute (17) for h = (u − β)+ ∈ Hs and, since f (u) (u − β)+ = 0, we get 0

=

Z

RN

≥

Z

RN

∇u · ∇ (u − β)+ dx + µ Z ∇u · ∇ (u − β)+ dx =

Z

u (u − β)+ RN

RN

2

|y|

dx + λ

Z

RN

u (u − β)+ dx

¯ ¯ ¯∇ (u − β) ¯2 dx . +

This implies (u − β)+ = 0, i.e., u ≤ β, which yields f (u) = g (u) and thus proves (13) again. ¥ We conclude this subsection with some lemmas which will be useful in the following. The first one shows that (F0) forces J to take negative values. Lemma 8 Let ξ > 0 be such that F (t0 ) > (1/2 + ξ) λt20 , where pt0 > 0 is given by condition (F0). Set Aµ,λ,ξ := {(y, z) ∈ Rk × RN −k : |y| > µ/2λξ}. Then there exists u ¯ ∈ Cc∞ (Aµ,λ,ξ ) ∩ Hs such that J (¯ u) < 0. p Proof. Let R > Rµ,λ,ξ := 1 + µ/2λξ and consider two functions ϕR , ψR ∈ Cc∞ ((0, +∞)) such that 9

• ϕR (t) ≡ t0 on (R, 4R), ϕR (t) ≡ 0 on [0, R − 1) ∪ (4R + 1, +∞), 0 ≤ ϕR ≤ t0 on [0, +∞) • ψR (t) ≡ 1 on [0, 5R), ψR (t) ≡ 0 on (5R + 1, +∞), 0 ≤ ψR ≤ 1 on [0, +∞) ′ • supR>Rµ,λ,ξ kϕ′R kL∞ ((0,+∞)) < ∞, supR>Rµ,λ,ξ kψR kL∞ ((0,+∞)) < ∞ .

Set uR (x) := ϕR (|y|) ψR (|x|) for all x = (y, z) ∈ Rk × RN −k and C0 := supR>Rµ,λ,ξ k∇uR kL∞ (RN ) < ∞. Clearly uR ∈ Cc∞ (Aµ,λ,ξ ) ∩ Hs and 0 ≤ uR ≤ t0 . We now show that J (uR ) < 0 for R large enough. Letting © ª A1 := x ∈ RN : |x| < 5R + 1, R − 1 < |y| < 4R + 1 © ª A2 := x ∈ RN : |x| < 5R, R < |y| < 4R ⊂ A1 © ª A4 := x ∈ RN : 5R ≤ |x| ≤ 5R + 1 © ª A5 := x ∈ RN : |x| < 5R, R − 1 ≤ |y| ≤ R or 4R ≤ |y| ≤ 4R + 1 A3 := A1 \ A2 ⊂ A4 ∪ A5 , one has uR (x) ≡ 0 on RN \A1 , uR (x) ≡ t0 on A2 and |y| > R−1 > Rµ,λ,ξ −1 = p µ/2λξ for all x = (y, z) ∈ A1 . Hence ! Z Z Ã 1 λ 2 µ 2 2 J (uR ) = |∇uR | dx + 2 uR + 2 uR − F (uR ) dx 2 A3 2 |y| A1 ¶ ¶ Z Z µµ 1 1 2 2 ≤ |∇uR | dx + + ξ λuR − F (uR ) dx 2 A4 ∪A5 2 A1 ¶ ¶ µµ Z 1 1 2 2 C (|A4 | + |A5 |) + + ξ λt0 − F (t0 ) dx + ≤ 2 0 2 A2 ¶ ¶ Z µµ 1 + ξ λu2R − F (uR ) dx . (18) + 2 A3 In order to estimate each term of (18) as R → +∞, denote by C any positive N N constant independent from R. First, |A4 | = C((5R + 1) −(5R) ) = CRN −1 + N −1 o(R ) as R → +∞. Second, as |x| < 5R implies © ª © |z| < 5R, one has A ª5 ⊂ y ∈ Rk : R − 1 ≤ |y| ≤ R or 4 R ≤ |y| ≤ 4R + 1 × z ∈ RN −k : |z| ≤ 5R , so k k k that |A5 | ≤ C(Rk −(R − 1) +(4R + 1) −(4R) )RN −k = CRN −1 +o(RN −1 ) as k R → +∞. Third, for any yR ∈ R such that |yR | = 2R one has BR (yR , 0) ⊂ A2 , so that (1/2 + ξ) λt20 − F (t0 ) < 0 imply ¶ ¶ µµ ¶ ¶ Z µµ 1 1 + ξ λt20 − F (t0 ) dx ≤ + ξ λt20 − F (t0 ) |BR (yR , 0)| 2 2 A2 = −CRN .

Finally, taking into account that, by continuity, supR>Rµ,λ,ξ kuR kL∞ (RN ) ≤ t0 ° ° implies supR>Rµ,λ,ξ °(1/2 + ξ) λu2R − F (uR )°L∞ (RN ) < ∞, we have ¶ ¶ Z µµ 1 + ξ λu2R − F (uR ) dx ≤ C |A3 | ≤ C (|A4 | + |A5 |) 2 A3 10

¡ ¢ ≤ CRN −1 + o RN −1 R→+∞ .

¡ ¢ Therefore, plugging into (18), we get J (uR ) ≤ C RN −1 − RN + o(RN −1 ) as R → +∞, and the proof is thus complete. ¥ The next lemma assures that weak limits of criticizing sequences are actually critical points for J. Lemma 9 For any h ∈ Hs the mapping J ′ (·) h : Hs → R is sequentially weakly continuous. Proof. Of course we need only consider the nonlinear term of the mapping. So we fix h ∈ H 1 (RRN ) and show the sequential weak continuity on H 1 (RN ) of assuming un ⇀ u in H 1 (RN ), the mapping u 7→ RN Rf (u) h dx. Accordingly, R we need to show that RN f (un ) h dx → RN f (u) h dx. With a view to arguing by density, let ϕ ∈ Cc∞ (RN ) and let R > 0 be such that supp ϕ ⊂ BR . On one hand condition (f ) assures the continuity of the Nemytski˘ı operator f : Lp−1 (BR¡) → ¢L1 (BR ). On the other hand, from the compactness of the em1 N ֒→ Lp−1 (BR ) we deduce that un → u in Lp−1 (BR ). Hence Rbedding H R |f (un ) − f (u)| |ϕ| dx = o (1)n→∞ . Then, by H¨older inequality, the continRN uous embedding H 1 (RN ) ֒→ Lp (RN ) and the boundedness of {un } in H 1 (RN ), there exists a constant C > 0 (independent from ϕ and n) such that Z

RN

≤

|f (un ) − f (u)| |h| dx Z

Z

|f (un ) − f (u)| |h − ϕ| dx + |f (un ) − f (u)| |ϕ| dx RN Z ≤ |f (un )| |h − ϕ| dx + |f (u)| |h − ϕ| dx + o (1)n→∞ RN RN Z Z p−1 p−1 ≤ Mp |un | |h − ϕ| dx + M p |u| |h − ϕ| dx + o (1)n→∞ RN RN ³ ´ (p−1)/p (p−1)/p ≤ M p kun kLp (RN ) + kukLp (RN ) kh − ϕkLp (RN ) + o (1)n→∞ ZR

N

≤ C kh − ϕkH 1 (RN ) + o (1)n→∞

¡ ¢ and the density of Cc∞ RN in H 1 (RN ) allows us to conclude. ¥

3

Existence of bounded Palais-Smale sequences

Assume N > k ≥ 2 and µ, λ > 0, and let g ∈ C (R; R) satisfy the hypotheses of Theorem 4. The aim of this section is to prove that the functional J defined in Subsection 2.2 as the restriction of I to Hs admits a bounded Palais-Smale sequence at its mountain-pass level c > 0 (see (21) below), that is, a bounded sequence {wn } ⊂ Hs such that J (wn ) → c and J ′ (wn ) → 0 in Hs′ . 11

Throughout the section we let u ¯ ∈ Cc∞ (Aµ,λ,ξ ) ∩ Hs be the map given in Lemma 8. Moreover we denote ¶ Z µ λ 2 1 2 for all u ∈ Hs F (u) − u dx F (u) := kukX − J (u) = 2 2 RN and set ut := u(t−1 ·) for every u ∈ Hs and t > 0. Notice that ut ∈ Hs and ! Z Ã −1 2 ¯ ¯2 ° t °2 u(t x) 2 −1 −2 ¯∇u(t x)¯ + µ °u ° = t dx = tN −2 kukX (19) X −1 y|2 N |t R and

¡ ¢ F ut =

2

Z

RN

µ

¡

F u(t

−1

¶ ¢ λ −1 2 x) − u(t x) dx = tN F (u) . 2 2

(20)

2

Similarly kut − v t k = tN −2 ku − vkX + λtN ku − vkL2 (R3 ) , so that the mapping u 7→ ut is continuous from Hs into itself. ¯ Lemma 10 Let {tn } ⊂ (0, +∞) be a sequence such that tn → 1. Then u ¯tn → u in Hs . ¯ > 0 be such that supp u Proof. p Let R ¯ ⊂ Aµ,λ,ξ ∩ BR¯ and set A := {x ∈ B2R¯ : 1 |y| > 2 µ/2λξ} ⊃ Aµ,λ,ξ ∩ BR¯ , so that, for n large enough, both u ¯ and u ¯tn −1 −1 tn −1 tn ∞ u(tn ·) → ¯ and ∇¯ u = tn ∇¯ ¯(tn ·) → u ¯ =u belong to Cc (A). Moreover, u ∇¯ u almost everywhere in RN . Then, as n → ∞, we conclude k¯ utn − u ¯kL2 (RN ) = k¯ utn − u ¯kL2 (A) = o (1), k∇ (¯ utn − u ¯)kL2 (RN ) = k∇ (¯ utn − u ¯)kL2 (A) = o (1) and 2

(¯ utn − u ¯)

Z

RN

2

|y|

dx =

Z

2

(¯ utn − u ¯)

A

2

|y|

8λξ dx ≤ µ

by dominated convergence. ¥

Z

A

¡

¯ u ¯tn − u

¢2

dx = o (1)

u) (recall Henceforth we fix a ε∗ > 0 such that J(u) < 0 for all u ∈ Bε∗ (¯ from Lemma 8 that J (¯ u) < 0) and, by Lemma 10, a threshold t∗ ∈ (0, 1) such u) for all t ∈ (t∗ , 1). that u ¯t ∈ Bε∗ (¯ Let us now introduce the mountain-pass level c := inf

max

γ∈Γ u∈γ([0,1])

J (u) , Γ := {γ ∈ C ([0, 1] ; Hs ) : γ (0) = 0, γ (1) = u ¯} . (21)

By condition (F) one readily has c > 0. Indeed, the continuity of the embedding Hs ֒→ Lp (RN ) assures the existence of a constant M0 > 0 such that Z 1 1 2 2 p F (u) dx ≥ kuk − M0 kuk for all u ∈ Hs J (u) = kuk − 2 2 RN and thus of some l > 0 and ρ ∈ (0, k¯ uk) such that J (u) > l for all u ∈ Hs with kuk = ρ. 12

The existence of a Palais-Smale sequence at level c then follows from standard deformation arguments, but, as we do not assume the already mentioned Ambrosetti-Rabinowitz condition, such a sequence is not necessarily bounded. The existence of a bounded Palais-Smale sequence is not a trivial problem and the rest of the section is devoted to this issue. The arguments we use derive from the ones of [2]. Lemma 11 For all t ∈ (t∗ , 1) one has c = inf γ∈Γ maxu∈γ([0,1]) J (ut ). Proof. Letting t ∈ (t∗ , 1) and ct := inf γ∈Γ maxu∈γ([0,1]) J (ut ), we show that c ≤ ct and ct ≤ c. For any δ > 0, fix γ1 ∈ Γ such that maxu∈γ1 ([0,1]) J (ut ) ≤ ct + δ and set ( t γ1 (2s) if 0 ≤ s ≤ 1/2 γ¯1 (s) := t 2 (1 − s) u ¯ + 2 (s − 1/2) u ¯ if 1/2 ≤ s ≤ 1 . Then γ¯1 ∈ C ([0, 1] ; Hs ), γ¯1 (0) = 0 and γ¯1 (1) = u ¯, i.e., γ¯1 ∈ Γ. Moreover u) and thus J (¯ γ1 (s)) < 0 for all s ∈ [1/2, 1]. u) implies γ¯1 (s) ∈ Bε∗ (¯ u ¯t ∈ Bε∗ (¯ Hence we get c ≤ =

max

J (u) = max J (¯ γ1 (s)) = max J (¯ γ1 (s)) s∈[0,1] s∈[0,1/2] ¡ ¢ t t max J(γ1 (2s) ) = max J(γ1 (s) ) = max J ut

u∈¯ γ1 ([0,1]) s∈[0,1/2]

s∈[0,1]

u∈γ1 ([0,1])

≤ ct + δ

which yields c ≤ ct since δ is arbitrary. Note that this also implies ct > 0. Conversely, for any δ > 0 we fix γ2 ∈ Γ such that maxu∈γ2 ([0,1]) J (u) ≤ c + δ and set ( 1/t γ2 (2s) if 0 ≤ s ≤ 1/2 γ¯2 (s) := 1/t 2 (1 − s) u ¯ + 2 (s − 1/2) u ¯ if 1/2 ≤ s ≤ 1 . Then γ¯2 ∈ C ([0, 1] ; Hs ), γ¯2 (0) = 0 and γ¯2 (1) = u ¯, i.e., γ¯2 ∈ Γ. Moreover, ¡ ¢t t if s ∈ [1/2, 1], one has γ¯2 (s) = 2 (1 − s) u ¯1/t + 2 (s − 1/2) u ¯ = 2 (1 − s) u ¯+ t t t t u) and thus J(¯ γ2 (s) ) < u) implies γ¯2 (s) ∈ Bε∗ (¯ 2 (s − 1/2) u ¯ , so that u ¯ ∈ Bε∗ (¯ 0. Hence we get ¡ ¢ t t ct ≤ max J ut = max J(¯ γ2 (s) ) = max J(¯ γ2 (s) ) u∈¯ γ2 ([0,1])

=

=

s∈[0,1]

max J((γ2 (2s)

s∈[0,1/2]

max

u∈γ2 ([0,1])

s∈[0,1/2]

1/t t

) ) = max J (γ2 (2s)) = max J(γ2 (s)) s∈[0,1/2]

s∈[0,1]

J (u) ≤ c + δ

and the conclusion ensues from the arbitrariness of δ. ¥ Hereafter, by Lemma 11, we assume that to any t ∈ (t∗ , 1) there corresponds a path γt ∈ Γ such that ¢ ¡ ¡ ¢ (22) max J ut ≤ c + 1 − tN , u∈γt ([0,1])

13

by which we define the set © ¡ ¢ª Λt := u ∈ γt ([0, 1]) : J (u) ≥ c − 1 − tN .

Note that maxu∈γt ([0,1]) J (u) ≥ c implies Λt 6= ∅ (indeed, as 1 − tN > 0 and J is continuous, Λt even contains a continuous piece of γt ([0, 1])). 2

−2 Lemma 12 For every t ∈ (t∗ , 1) and u ∈ Λt one has kukX ≤ (c + 2) N/tN . ∗

Proof. Fix any t ∈¡ (t∗ , 1) ¢and u ∈ Λt . By (22) and the definition of Λt one has J (ut ) − J (u) ≤ 2 1 − tN . On the other hand, from (19) and (20) it follows that ´ ° ¡ ¢ ¡ ¢ 1 ³° °ut °2 − kuk2 − F ut + F (u) J ut − J (u) = X X 2µ ¶ ° ° ° ¡ ¢ ¡ ¢ 1 ° °ut °2 − 1 °ut °2 − F ut + 1 F ut = N −2 N X X 2 t t µ ¶ ¶ µ ° ° ¡ ¢ 1 1 1 2 1 − N −2 °ut °X − 1 − N F ut = 2 t t N −2 ° ° N ¡ ¢ 1−t °ut °2 + 1 − t F ut = − N −2 N X 2t t µ ¶ ° ¡ t¢ 1 − tN 1 t2 − tN ° t °2 ° = − u X +F u tN 2 1 − tN µ ¶ ° ¡ ¢ 1 ° 1 − tN °ut °2 + F ut ≥ − , X tN 2∗

where ¡for the ¢final we have taken into account that the mapping ¡ inequality ¢ t 7→ − t2 − tN / 1 − tN is decreasing for t > 0 and tends to −2/2∗ as t → 1. 2 So we obtain − (1/2∗ ) kut kX + F (ut ) ≤ 2tN , whence ° ¡ ¢ 1 ° °2 ¡ ¢ 1 ° °2 1 ° °2 1 ° °ut °2 − 2tN J ut = °ut °X − F ut ≥ °ut °X − ∗ °ut °X − 2tN = X 2 2 2 N

and therefore 2

kukX =

1 tN −2

° t °2 °u ° ≤ X

¢ ¢ N (c + 2) N ¡ ¡ t¢ N ¡ J u + 2tN ≤ N −2 c + 1 + tN ≤ , −2 tN −2 t tN ∗

where (22) has been used again in order to estimate J (ut ) ≤ c + 1 − tN . ¥

The aim of the section will be accomplished in Proposition 14 and Corollary 15, where we take advantage of the following well known deformation lemma (see [31, Lemma 2.3], here written for the space Hs , our functional J and its mountain-pass level c). Lemma 13 Let S ⊂ Hs and ε, δ > 0 be such that kJ ′ (u)kHs′ ≥ 8ε/δ for all u ∈ S2δ satisfying |J (u) − c| ≤ 2ε, where S2δ := {v ∈ Hs : inf h∈S kv − hk ≤ 2δ}. Then there exists η ∈ C ([0, 1] × Hs ; Hs ) such that 14

• η (τ, u) = u provided that τ = 0 or |J (u) − c| > 2ε or u ∈ / S2δ • J (η (1, u)) ≤ c − ε provided that J (u) ≤ c + ε and u ∈ S • η (τ, ·) is an homeomorphism of Hs for every τ ∈ [0, 1] • J (η (·, u)) is nonincreasing for every u ∈ Hs . Proposition 14 There exists a Palais-Smale sequence {wn } ⊂ Hs for J at level c such that (c + 2) N 2 (23) sup kwn kX ≤ 1 + 2 N −2 . n t∗ −2 for sake of brevity. First we observe that Proof. Set c∗ := 2N (c + 2) /tN ∗ t limt→1− supu∈Λt |J (u ) − J (u)| = 0. Indeed, for all ¡t ∈ (t∗¢, 1) and u ∈ Λt , the definition of Λt and (20) yield J (ut ) − J (u) ≤ 2 1 − tN = o (1)t→1 and −tN F (u) = −F (ut ) ≤ J (ut ) ≤ c + 1 − tN by inequality (22), whence, by (19), (20) and Lemma 12, we get

¡ ¢ J (u) − J ut

¡ ¢ 1 − tN −2 2 kukX − 1 − tN F (u) 2 ¡ ¢ c∗ ¡ ¢ c + 1 − tN ≤ 1 − tN −2 + 1 − tN = o (1)t→1 . 4 tN

=

Now, for every m ≥ 1, define ¾ ½ 1 1 2 Um := u ∈ Hs : kukX ≤ c∗ + , |J (u) − c| ≤ m m

tm and choose tm ∈ (t∗ , 1) such that 1 − tN m ≤ 1/32m and J (u) ≤ J (u ) + 1/32m for all u ∈ Λtm . Then for every u ∈ Λtm the inequality of Lemma 12 holds and one has (recall the definition of Λtm )

¢ ¡ J (u) ≥ c − 1 − tN m ≥c−

1 32m

and (by (22), with t = tm )

¡ ¢ J (u) ≤ J utm +

¢ ¡ 1 1 1 ≤ c + 1 − tN ≤c+ , m + 32m 32m 16m

(24)

whence Λtm ⊆ Um and Um is not empty. For sake of contradiction, assume that ½ ¾ 8 1 1 ∃m ¯ > max ∀u ∈ Um kJ ′ (u)kH ′ ≥ √ , (25) ¯ s c∗ 8c m ¯ 2

and apply = {h ∈ Hs : khkX ≤ c∗ /2}, ε = 1/16m ¯ and √ Lemma 13 with S √ ¯ (so that 8ε/δ = 1/ m). ¯ Note that δ = 1/2 m ¾ ½ √ √ ¯ S2δ = S1/ m ¯ = v ∈ Hs : min kv − hk ≤ 1/ m h∈S

15

because S is convex and closed in Hs , and observe that if u ∈ S1/√m ¯ satisfies |J (u) − c| ≤ 1/8m ¯ then u ∈ Um (and thus the last inequality of (25) holds), ¯ √ because there exists h ∈ S such that ku − hkX ≤ ku − hk ≤ 1/ m ¯ and thus r r 1 1 1 c∗ + √ ≤ c∗ + , kukX ≤ khkX + √ ≤ 2 m ¯ m ¯ m ¯ where the assumption m ¯ > 8/c∗ has been used to derive the last inequality. So there exists an homeomorphism Φ : Hs → Hs (namely Φ := η (1, ·) of Lemma 13) such that (i) Φ (u) = u if |J (u) − c| ≥ c (recall that c > 1/8m ¯ = 2ε) 2

(ii) J (Φ (u)) ≤ c − 1/16m ¯ if kukX ≤ c∗ /2 and J (u) ≤ c + 1/16m ¯ (iii) J (Φ (u)) ≤ J (u) for every u ∈ Hs , by which we define the path γ := Φ ◦ γtm¯ ∈ C ([0, 1] ; Hs ). By (i) one has u) = u ¯, since γ (0) = Φ(γtm¯ (0)) = Φ (0) = 0 and γ (1) = Φ(γtm¯ (1)) = Φ (¯ |J (¯ u) − c| = |J (¯ u)| + c > c (recall that J (¯ u) < 0). Hence γ ∈ Γ. We finally deduce the contradiction which assures that the hypothesis (25) is false and thus concludes the proof. Let u∗ ∈ γtm¯ ([0, 1]) be such that J (Φ (u∗ )) = maxu∈γtm¯ ([0,1]) J (Φ (u)) = maxv∈γ([0,1]) J (v). On one hand, if u∗ ∈ γtm¯ ([0, 1])\ ¢ ¡ ). Λtm¯ then J (Φ (u∗ )) ≤ J (u∗ ) < c − 1 − tN m ¯ (by (iii) and the definition of Λtm ¯ ¯ and Lemma 12 On the other hand, if u∗ ∈ Λtm¯ then (24) holds (with m = m) 2 gives ku∗ kX ≤ c∗ /2 (recall that tm ¯ by ¯ ∈ (t∗ , 1)), whence J (Φ (u∗ )) ≤ c − 1/16m (ii). Therefore, in any case one obtains maxv∈γ([0,1]) J (v) < c, which contradicts the definition (21) of c. ¥ Corollary 15 The sequence {wn } of Proposition 14 is bounded in Hs . p

Proof. Since J (wn ) → c and (F) implies F (wn ) ≤ M |wn | almost everywhere, there exists a constant C1 > 0 such that Z Z λ p 2 C1 ≥ J (wn ) ≥ w dx − M |wn | dx for all n . 2 RN n N R Setting q := (2∗ − 2) / (2∗ − p) and q ′ := q/ (q − 1) = (2∗ − 2) / (p − 2), from H¨older and Sobolev inequalities one infers that there exists a second constant C2 > 0 such that µZ ¶1/q µZ ¶1/q′ Z Z p 2/q 2∗ /q ′ 2∗ 2 |wn | dx = |wn | |wn | dx ≤ wn dx |wn | dx RN

RN

≤

2/q C2 kwn kL2 (RN )

RN

2∗ /q ′ kwn kX

RN

. 2

Therefore, by (23), there exists C3 > 0 such that C1 ≥ (λ/2) kwn kL2 (RN ) − 2/q

C3 kwn kL2 (RN ) for all n, where 2/q < 2 since q > 1. Hence no diverging subsequence is allowed for {kwn kL2 (RN ) } and the proof is thus complete. ¥ 16

4

Proof of Theorem 4

This section is devoted to the proof of Theorem 4, which relies on the application of a version of the concentration-compactness principle due to Solimini [24]. Accordingly, in order to state his result, we preliminarly introduce a group of rescaling operators, of which we also remark some basic properties. As usual, we assume N > k ≥ 2 and let µ, λ > 0. Definition 16 Let t > 0 and x ∈ RN . For any u ∈ Lp (RN ) with 1 < p < ∞ we define ¡ ¢ Tt,x u := t−(N −2)/2 u t−1 · + x .

Clearly Tt,x u ∈ Lp (RN ) for all u ∈ Lp (RN ) and in particular Tt,x u ∈ D1,2 (RN ) if u ∈ D1,2 (RN ). Moreover, by direct computations, it is easy to see that the ∗ linear operator u ∈7→ Tt,x u is an isometry of both L2 (RN ) and D1,2 (RN ). Notice that −1 Tt,x = T1/t,−tx

and

Tt1 ,x1 Tt2 ,x2 = Tt1 t2 ,x1 /t2 +x2 .

(26)

Remark 17 For any z˜ = (0, z) ∈ Rk × RN −k and t > 0, direct computations easily show that the linear operators u 7→ Tt,˜z u and u 7→ T1,˜z u are isometries of X and H respectively. Moreover Tt,˜z u ∈ Xs if u ∈ Xs and T1,˜z u ∈ Hs if u ∈ Hs . The next proposition is proved in [4]. Proposition 18 Let 1 < p < ∞ and assume that {tn } ⊂ (0, +∞) and {xn } ⊂ RN are such that tn → t 6= 0 and xn → x. Then Ttn ,xn un ⇀ Tt,x u in Lp (RN ) if un ⇀ u in Lp (RN ). Corollary 19 Let {tn } ⊂ (0, +∞) and {˜ zn } ⊂ {0} × RN −k ⊂ RN be such that tn → t 6= 0 and z˜n → z˜. Then Ttn ,˜zn un ⇀ Tt,˜z u in Xs (up to a subsequence) if un ⇀ u in Xs . Proof. From the boundedness of {un }, by Remark 17 we deduce that also {Ttn ,˜zn un } is bounded ¢converges ¢ in Xs . Hence (up to a subsequence) it weakly ∗ ¡ ∗ ¡ in Xs and L2 RN . On the other hand Ttn ,˜zn un ⇀ Tt,˜z u in L2 RN , because ∗ un ⇀ u in L2 (RN ) and Proposition 18 applies. ¥ We are here in position to recall the above mentioned result of Solimini [24], which is the following. Theorem 20 If {vn } ⊂ D1,2 (RN ) is bounded, then, up to a subsequence, either ∗ exist {tn } ⊂ (0, +∞) and {xn } ⊂ RN such that vn → 0 in L2 (RN ) ¡or there ¢ 2∗ N and v 6= 0. Ttn ,xn vn ⇀ v in L R

Let us now turn to the proof of Theorem 4, which will be divided in several lemmas. Accordingly, we hereafter assume that all the hypotheses of the theorem are satisfied. 17

The starting point is the Palais-Smale sequence {wn } ⊂ Hs provided by Proposition 14, which, we recall, is bounded in Hs by Corollary 15 and satisfies J (wn ) → c > 0 and J ′ (wn ) → 0 in Hs′ . As {wn } is bounded in D1,2 (RN ), it must satisfy one of the alternatives allowed by Theorem 20. The following lemma shows that the first one cannot occur. ∗

Lemma 21 The sequence {wn } does not converge to 0 in L2 (RN ). ∗

Proof. For sake of contradiction, assume that wn → 0 in L2 (RN ). By H¨older inequalities and the boundedness of {wn } in L2 (RN ), one infers that wn → 0 in Lq (RN ) for all q ∈ (2, 2∗ ]. Hence (f ) and (F) imply Z Z p |wn | dx → 0 |f (wn ) wn | dx ≤ M p RN

RN

and

Z

RN

|F (wn )| dx ≤ M

Z

p

RN

|wn | dx → 0 .

On the other hand, J ′ (wn ) wn → 0 since J ′ (wn ) → 0 in Hs′ and {wn } is bounded in Hs . Therefore we conclude Z 2 ′ kwn k = J (wn ) wn + f (wn ) wn dx → 0 RN

and thus J (wn ) =

1 2 kwn k − 2

Z

RN

F (wn ) dx → 0 ,

which contradicts J (wn ) → c > 0. ¥ ∗

Corollary 22 There exist {tn } ⊂ (0, +∞), {xn } ⊂ RN and w ∈ L2 (RN ), ∗ w 6= 0, such that (up to a subsequence) Ttn ,xn wn ⇀ w in L2 (RN ). Proof. Apply Theorem 20 and use Lemma 21. ¥ Now we can easily exploit the z-translation invariance of J to improve the result of Corollary 22. To this end, we set xn =: (yn , zn ), y˜n := (yn , 0) and z˜n := (0, zn ), so that xn = y˜n + z˜n . Lemma 23 Let un := T1,˜zn wn . Then {un } is bounded in Hs , J ′ (un ) → 0 in ∗ Hs′ and Ttn ,˜yn un ⇀ w in L2 (RN ). Proof. The boundedness of {un } follows from the one of {wn }, since the operators T1,˜zn are isometries of Hs . Moreover, by (17) and easy computations, one gets J ′ (un ) h = J ′ (wn ) T1,−˜zn h for all h ∈ Hs , so that kJ ′ (un )kHs′ = kJ ′ (wn )kH ′ because also T1,−˜zn are isometries of Hs . Finally, recalling (26), s ∗ Ttn ,˜yn un = Ttn ,˜yn T1,˜zn wn = Ttn ,xn wn ⇀ w in L2 (RN ). ¥ 18

By Lemma 9 and the preceding one, the sequence {un } weakly converges in Hs to some critical point u ∈ Hs of J. The proof of Theorem 4 is thus accomplished if we show that u 6= 0, which is the aim of the next lemmas. The removal of translations from the rescalings Ttn ,˜yn is the first step in that direction and it is the topic of Lemma 25, where we will take advantage of the following elementary proposition (see [4] for a proof). Proposition 24 Let {ξn } ⊂ Rk be such that limn→∞ |ξn | = +∞ and fix R > 0. Then for any m ∈ N, m ≥ 2, there exists nm ∈ N such that for any n > nm one can find g1 , ..., gm ∈ O (k) satisfying the condition i 6= j ⇒ BR (gi ξn ) ∩ BR (gj ξn ) = ∅ . Hereafter we denote Tt := Tt,0 for any t > 0. Lemma 25 There exists v ∈ Xs , v 6= 0, such that (up to a subsequence) Ttn un ⇀ v in Xs . Proof. Set Tn := Ttn ,˜yn and vn := Ttn un for brevity, and recall from Lemma ∗ 23 that Tn un ⇀ w 6= 0 in L2 (RN ). From Remark 17 we get vn ∈ Xs and kvn kX = kun kX , so that (up to a subsequence) we can assume vn ⇀ v in Xs . If v 6= 0 the proof is completed. So, for sake of contradiction, assume vn ⇀ 0 in ∗ Xs (and thus in L2 (RN )). First, we deduce that lim |tn y˜n | = +∞ .

(27)

n→∞

Otherwise, up ¡to a¢subsequence, tn y˜n → y˜0 ∈ Rk × {0} and T1,−tn y˜n Tn un ⇀ ∗ T1,−˜y0 w in L2 RN by Proposition 18. But, since T1,−tn y˜n Tn = Ttn , this means ∗ vn ⇀ T1,−˜y0 w 6= 0 in L2 (RN ), which is a contradiction. Now we observe that w 6= 0 implies that there exist δ > 0 and A ⊆ RN with |A| = 6 0 such that either w > δ or w < −δ almost everywhere in A. Then, fixing R > 0 such that |BR ∩ A| > 0, by weak convergence we obtain ¯ ¯Z ¯ ¯Z ¯ ¯ ¯ ¯ ¯ ≥ δ |BR ∩ A| > 0 . ¯ ¯ ¯ (28) wχ dx → T u χ dx BR ∩A n n BR ∩A ¯ ¯ ¯ ¯ RN

RN

vn = T1,tn y˜n vn and hence On the other hand, Tn un = Tn Tt−1 n ¯ ¯Z Z Z ¯ ¯ ¯ ¯ ≤ |T u | dx = |vn | dx T u χ dx n n ¯ N n n BR ∩A ¯ BR BR (tn y˜n ) R ÃZ !1/2∗ 2∗

≤ C

BR (tn y˜n )

|vn |

dx

(29)

for some constant C > 0 which only depends on R. From (28) and (29) it follows that Z 2∗ lim inf |vn | dx > 0 n→∞

BR (tn y˜n )

19

and hence, up to a subsequence, we can assume Z 2∗ inf |vn | dx > ε0 for some ε0 > 0 . n

(30)

BR (tn y˜n )

This will yield a contradiction. Indeed, using (27), from Proposition 24 it readily follows that for every m ∈ N, m ≥ 2, there exists nm ∈ N such that for any n > nm one can find g1 , ..., gm ∈ O (k) satisfying the condition i 6= j ⇒ BR (tn (gi yn , 0)) ∩ BR (tn (gj yn , 0)) = ∅ . As a consequence, using (30) and the fact that vn ∈ Xs , we get Z m Z m Z X X 2∗ 2∗ 2∗ |vn | dx > mε0 |vn | dx = |vn | dx ≥ RN

i=1

BR (tn (gi yn ,0))

i=1

BR (tn y˜n )

for every natural numbers m ≥ 2 and n > nm . This finally implies Z 2∗ |vn | dx → +∞ RN

which is a contradiction, since kvn kL2∗ (RN ) = kTtn un kL2∗ (RN ) = kun kL2∗ (RN ) ∗ and {un } is bounded in L2 (RN ). ¥ In order to apply Corollary 19 with a view to concluding that {un } has a nonzero weak limit in Xs (and thus in Hs ), we need to check that the dilation parameters {tn } are actually well-behaved. This is the content of the next lemmas. Lemma 26 There exists c1 > 0 such that tn ≥ c1 for all n, i.e. inf n tn > 0.

Proof. Recall from Lemma 25 that Ttn unR⇀ v 6= 0 in Xs , and fix φ ∈ Cc∞ (RN ) and R > 0 such that supp φ ⊂ BR and RN vφ dx 6= 0. Then Ttn un ⇀ v in ∗ L2 (BR ) (because Xs ֒→ L2 (RN ) ֒→ L2 (BR )) and Z Z Z Z (Ttn un ) φ dx = (Ttn un ) φ dx → vφ dx = vφ dx 6= 0 . RN

BR

RN

BR

On the other hand, {un } is bounded in L2 (RN ) and we get ¯Z ¯ µZ ¶1/2 ¯ ¯ 2 ¯ ¯ u ) φ dx (T ≤ kφk u | dx |T tn n L2 (RN ) ¯ N tn n ¯ R RN µZ ¯ ¶1/2 ¯ ¯ −(N −2)/2 ¡ −1 ¢¯2 = kφkL2 (RN ) un tn x ¯ dx ¯tn RN

= kφkL2 (RN )

µ

t2n

µ

Z

RN

u2n dx

¶1/2

¶

≤ tn kφkL2 (RN ) sup kun kL2 (RN ) . n

As a conclusion, no vanishing subsequence is allowed for {tn } and the thesis follows. ¥ 20

Lemma 27 There exists c2 > 0 such that tn ≤ c2 for all n, i.e. supn tn < +∞. Proof. Recall from Lemma 25 that Ttn un ⇀ v in Xs and kvkX > 0. Then, by Lemma 6, there exists v˜ ∈ Hs such that (v | v˜)X > 0, so that ¢ ¡ ¢ ¡ un | T1/tn v˜ X = un | Tt−1 v˜ X = (Ttn un | v˜)X → (v | v˜)X > 0 (31) n

by (26) and Remark 17. For sake of contradiction, up we now ° ° to a subsequence v kX and v˜ = T1/tn v˜ ∈ Xs with °T1/tn v˜°X = k˜ assume tn → +∞. Then Tt−1 n Z

RN

¯ ¯ ¯T1/t v˜¯2 dx = n

Z

RN

Z ¯ ¯2 ¯ (N −2)/2 ¯ −2 v˜ (tn x)¯ dx = tn ¯tn

2

RN

|˜ v | dx → 0 ,

ª © ∗ which implies that T1/tn v˜ is bounded in Hs and L2 (RN ), and thus it converges to zero in Lq (RN ) for all q ∈ [2, 2∗ ) by interpolation. Hence (recall Lemma 23) J ′ (un ) T1/tn v˜ → 0 and Z ¯ ¯ ° ° ¯un T1/t v˜¯ dx ≤ kun k 2 N °T1/t v˜° 2 N n n L (R ) L (R ) RN ¶ µ ° ° ≤ sup kun kL2 (RN ) °T1/tn v˜°L2 (RN ) → 0 . n

Moreover, by (f ) and the boundedness of {un } in Lp (RN ), one has Z Z ¯ ¯ ¯ p−1 ¯¯ ¯f (un ) T1/t v˜¯ dx ≤ M p |un | T1/tn v˜¯ dx n RN

RN

≤ Mp

µZ

p

RN

µ

|un | dx p−1

¶(p−1)/p µZ

≤ M p sup kun kLp (RN ) n

Therefore we obtain ¡

¢ un | T1/tn v˜ X = J ′ (un ) T1/tn v˜ − λ

Z

RN

RN

¶

¯ ¯ ¯T1/t v˜¯p dx n

¶1/p

° ° °T1/t v˜° p N → 0 . n L (R )

un T1/tn v˜ dx +

Z

RN

f (un ) T1/tn v˜ dx → 0

which contradicts (31). ¥

We are now able to easily conclude the proof of Theorem 4. Proof of Theorem 4. By the last Lemmas 26 and 27, up to a subsequence we can assume tn → t 6= 0. Thus, from Ttn un ⇀ v 6= 0 in Xs (Lemma 25) we deduce un ⇀ Tt−1 v 6= 0 in Xs (up to a subsequence) by Corollary 19. Therefore, recalling from Lemma 23 that {un } is bounded in Hs and J ′ (un ) → 0 in Hs′ , by the embedding Hs ֒→ Xs and Lemma 9 one infers that un ⇀ Tt−1 v in Hs and Tt−1 v ∈ Hs is a nonzero critical point for J. The conclusion then follows from Proposition 7. ¥

21

5

Proof of Theorem 1

In this section we give the proof of Theorem 1, which follows from Theorem 4 together with an extendibility argument aimed at removing of the singularity of ∇θ on the plane y = 0, where θ is the angular coordinate given by (7). Let W : C → R satisfy (2) and assume that all the hypotheses of Theorem 1 hold. Let k0 > 0 and ω0 ∈ (Ω0 , Ω), and set G (s) :=

1 2 2 Ω s − V (s) 2

for all s ∈ R .

In order to apply Theorem 4 with N = 3, k = 2, µ = k0 , λ = Ω2 − ω02 and g = G′ , one readily checks that (g0) and (g1) are satisfied. Moreover, V ≥ 0 implies g (s) lim sup q−1 < +∞ for every q > 2 . (32) s→+∞ s So, if g (s) ≥ 0 for s > 0 large, then (32) with q = p assures that (g3) holds. Otherwise, if there exists a sequence {sn } such that sn → +∞ and g (sn ) < 0, it is not difficult to deduce (g2) from (g0) and (g1). Therefore Theorem 4 provides equation (8) with a nonzero nonnegative solution u ∈ Hs in the following weak sense: Z Z Z Z uh ′ 2 dx + V (u) h dx = ω uh dx for all h ∈ H . ∇u · ∇h dx + k0 0 2 R3 R3 R3 R3 |y| (33) Note that, either if (g2) holds or if (g3) holds, one has that V ′ (u) = Ω2 u − g (u) ∈ L1loc (R3 ).

(34)

Moreover, according to definitions 16 and 15, we have # Z Z " ¢ 1 k0 2 ¡ 2 2 2 2 I (u) = G (u) dx , |∇u| + 2 u + Ω − ω0 u dx − 2 R3 |y| R3 so that (9) becomes Z ´ ³ E u (x) ei(kθ(x)−ω0 t) = I (u) + ω02

R3

u2 dx < ∞ .

Now we set ψ0 (x) := u (x) eik0 ϑ(x)

¡ ¢ for all x = (y, z) ∈ O := R2 \ {0} × R . 2

2

−2

Notice that ∇ψ0 = eik0 ϑ (∇u + ik0 u∇ϑ) implies |∇ψ0 | = |∇u| + k0 |y| u ∈ L1 (R3 ), so that ψ0 ∈ H 1 (R3 ). In order to conclude the proof of Theorem 1, we need the following two lemmas.

22

Lemma 28 The mapping ψ0 satisfies −△ψ0 + W ′ (ψ0 ) = ω02 ψ0

(35)

in the distributional sense on O. Proof. Since θ ∈ C ∞ (O; R/2πZ), the claim of the lemma is equivalent to Z Z Z ¡ ¢ ∇ψ0 · ∇ e−ik0 ϑ ϕ dx + W ′ (ψ0 ) e−ik0 ϑ ϕ dx = ω02 ψ0 e−ik0 ϑ ϕ dx (36) O

O

O

for all ϕ ∈ Cc∞ (O; C), where W ′ (ψ0 ) = V ′ (u) eik0 ϑ by (2). Writing ϕ = ϕ1 +iϕ2 with ϕ1 , ϕ2 ∈ Cc∞ (O; R), we readily get Z Z Z Z W ′ (ψ0 ) e−ik0 ϑ ϕ dx = V ′ (u) ϕ dx = V ′ (u) ϕ1 dx + i V ′ (u) ϕ2 dx O

R3

O

and

Z

O

−ik0 ϑ

ψ0 e

ϕ dx =

Z

uϕ dx =

Z

R3

O

uϕ1 dx + i

R3

Z

uϕ2 dx . R3

On the other hand, denoting ξ · η = ξ1 η1 + ξ2 η2 + ξ3 η3 for any ξ, η ∈ C3 , one has Z ¡ ¢ ∇ψ0 · ∇ e−ik0 ϑ ϕ dx O

= = = = =

Z

(∇u + ik0 u∇ϑ) · (∇ϕ − ik0 ϕ∇ϑ) dx ZO ³ Z ´ 2 ∇u · ∇ϕ + k0 uϕ |∇ϑ| dx + ik0 u∇ϑ · ∇ϕ dx O O ! Z Z Ã uϕ div (u∇ϑ) ϕ dx ∇u · ∇ϕ + k0 2 dx − ik0 |y| O O ! Z Ã uϕ ∇u · ∇ϕ + k0 2 dx |y| O ! ! Z Ã Z Ã uϕ2 uϕ1 ∇u · ∇ϕ2 + k0 2 dx ∇u · ∇ϕ1 + k0 2 dx + i |y| |y| R3 R3

where we have taken into account that ∇u · ∇ϑ = 0 and div (u∇ϑ) = ∇u · ∇ϑ + u△ϑ = 0. Hence, observing that Cc∞ (O; R) ⊂ H, one concludes that (36) holds thanks to (33). ¥ Lemma 29 The mapping ψ0 satisfies (35) in the distributional sense on R3 . Proof. Let ϕ ∈ Cc∞ (R3 ; C) and take {ηn } ⊂ C ∞ (R3 ; R) such that ηn → 1 almost everywhere in R3 and 23

• 0 ≤ ηn ≤ 1, ηn (y, z) = 0 for |y| ≤ 1/n and ηn (y, z) = 1 for |y| ≥ 2/n • |∇ηn | ≤ (const.) n on R3 . Clearly ηn ϕ ∈ Cc∞ (O; C) and |∇ηn (y, z)| = 0 for |y| ≤ 1/n or |y| ≥ 2/n. Then Lemma 28 gives Z Z Z Z ϕ ∇ψ0 · ∇ηn dx + ηn ∇ψ0 · ∇ϕ dx + W ′ (ψ0 ) ϕηn dx = ψ0 ϕηn dx . R3

R3

R3

R3

(37) Setting An := {(y, z) ∈ supp ϕ : 1/n < |y| < 2/n}, we have |An | ≤ (const.) /n2 and ¯Z ¯ Z ¯ ¯ ¯ ¯ ≤ (const.) kϕk ∞ 3 n ϕ ∇ψ · ∇η dx |∇ψ0 | dx 0 n L (R ) ¯ 3 ¯ R An µZ ¶1/2 2 ≤ (const.) |∇ψ0 | dx → 0. An

Passing to the limit in (37) and using the Lebesgue’s dominated convergence theorem for the other terms (recall that |W ′ (ψ0 )| = |V ′ (u)| ∈ L1loc (R3 ) by (2) and (34)), the claim follows. ¥ Proof of Theorem 1. Set ψ (t, x) := ψ0 (x) e−iω0 t for all x ∈ O and t ∈ R. Since Lemma 29, together with standard elliptic regularity arguments (see for example [17]), yields that ψ0 defines a classical solution to (35) on R3 , a straightforward substitution proves that ψ is actually a classical solution of (1) on R3 × R. ¥

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Marino Badiale, Sergio Rolando Dipartimento di Matematica Universit`a degli Studi di Torino Via Carlo Alberto 10 10123 Torino, Italia [email protected], [email protected] Vieri Benci Dipartimento di Matematica Applicata “U. Dini” Universit`a degli Studi di Pisa Via Bonanno 25/b 56126 Pisa, Italia [email protected] Research of M. Badiale supported by MIUR, project “Metodi variazionali e equazioni differenziali non lineari” Research of V. Benci supported by MIUR, project “Metodi variazionali e topologici nello studio di fenomeni non lineari” 26