Ground states of Nonlinear Schr¨odinger Equations with Potentials Vanishing at Infinity Antonio AMBROSETTI SISSA, via Beirut 2-4, 34014 Trieste - Italy
[email protected] Veronica FELLI Univ. Milano Bicocca, Dip. Mat. e Appl., 20126 Milano - Italy
[email protected] Andrea MALCHIODI via Beirut 2-4, 34014 Trieste - Italy
[email protected] Key words: Nonlinear Schr¨ odinger Equations, Weighted Sobolev spaces.
Abstract. We deal with a class on nonlinear Schr¨ odinger equations (NLS) with potentials V (x) ∼ |x|−α , 0 < α < 2, −β and K(x) ∼ |x| , β > 0. Working in weighted Sobolev spaces, the existence of ground states vε belonging to W 1,2 (RN ) is proved under the assumption that p satisfies (3). Furthermore, it is shown that vε are spikes concentrating at a minimum of A = V θ K −2/(p−1) , where θ = (p + 1)/(p − 1) − 1/2.
1
Introduction
This paper deals with existence of ground state solutions of stationary Nonlinear Schr¨odinger Equations of the form 2 p N −ε ∆v + V (x)v = K(x)v , x ∈ R , (NLS) v ∈ W 1,2 (RN ), v(x) > 0, lim|x|→∞ v(x) = 0 . N +2 Hereafter, N ≥ 3 and 1 < p < N −2 . A solution of (N LS) is called a ground state if it is a Mountain-Pass critical point of the corresponding Euler functional, and hence its Morse index is 1. If u is a solution of (N LS), then ψ(x, t) = exp i λ ε−1 t u(x),
represents a standing wave of the Nonlinear Schr¨odinger Equation (1)
iε
∂ψ = −ε2 ∆ψ + (V (x) − λ)ψ − K(x)|ψ|p−1 ψ, ∂t
where ε(= ~) is the Planck constant and i is the imaginary unit. 1
One of the main purposes of this paper is to look for solutions vε of (NLS) which satisfy the following properties (i) vε ∈ W 1,2 (RN ); (ii) vε is a ground state. As for (i), let us point out that standing waves v which have finite L2 norm are the most relevant from the physical point of view since they correspond to bound states. Moreover, if v ∈ W 1,2 (RN ), one can prove that lim|x|→∞ v(x) = 0 (see the proof of Theorem 16), which implies that solutions are well localized in space. On the other hand, concerning (ii), the interest in searching ground states relies on the fact that a standing wave is possibly orbitally stable provided it corresponds to a ground state of (NLS), in the sense specified before, see e.g. [14, 17]. A lot of work has been devoted to the existence of solutions of (NLS), both for ε = 1 and for ε tending to zero. In the latter case, as a specific feature of the nonlinear (focusing) model, solutions concentrate at points with a soliton profile. We limit ourselves to cite few recent papers [5, 6, 8, 9, 11, 12, 15, 16], referring to their bibliography for a broader list of works, although still not exhaustive. However, at our knowledge, it is everywhere assumed (with the only exception of [18, 20]) that lim inf |x|→∞ V (x) > 0. The main new feature of the present paper is that we will be concerned with potentials V such that lim|x|→∞ V = 0. Our main results are Theorems 1 and 3. The former deals with existence of ground states of (NLS), the latter with concentration. Roughly, (NLS) has a ground state which concentrates at a global minimum of the auxiliary potential A := V θ K −2/(p−1) , where θ = (p + 1)/(p − 1) − N/2, provided (i) V (x) ∼ |x|−α with 0 < α < 2, (ii) K(x) ∼ |x|−β with β > 0, (iii) σ < p < (N + 2)/(N − 2), where σ is a number depending upon α and β, and defined in (2). Some comments on the proof and the preceding assumptions are in order. Dealing with a potential V which decays to zero at infinity, the methods used in the preceding papers cannot be employed. First of all, variational theory in W 1,2 (RN ), as in [11, 12], cannot be used. Nor one can apply perturbation methods, as in [6, 16], since the spectrum of the linear operator −∆ + V is [0, +∞), see [10]. To overcome this difficulty, we frame our problem in a class of weighted Sobolev spaces Hε , discussed in [19], consisting of the functions u on RN for which Z ε2 |∇u(x)|2 + V (x)u2 (x) dx < ∞. RN
R In these spaces the nonlinear term RN Kup+1 dx is well-defined if (i − ii − iii) hold. Moreover, under these conditions the Euler functional satisfies the Palais-Smale compactness condition on Hε and this allows us to find in a straightforward way a positive Mountain-Pass solution vε ∈ Hε , see Theorem 13. It is worth pointing out that for such a result it suffices to assume that (i) holds with 0 < α ≤ 2. However we are interested in solutions which belong to W 1,2 and which decay to zero at infinity. To achieve these conditions we prove first some careful integral estimates for solutions in Hε . The proof of the concentration phenomenon also relies on some sharp pointwise decay estimates and on appropriate bounds of the energy of the Mountain-Pass solutions vε , uniformly with respect to ε. These estimates require α to be smaller than 2 and represent one of the main novelty of the arguments in the present paper.
2
As for the assumptions, we point out that if V ∼ |x|−α with 0 < α ≤ 2, then (iii) cannot be eliminated if we want to find ground states. For more details concerning this claim, we refer to Proposition 15 in Section 4. Concerning assumption (ii), see also Remark 14-(i). As already pointed out, the only papers dealing with equations on RN with potentials vanishing at infinity are [18] and [20]. The former deals with an eigenvalue problem in the radial case. In the latter, weighted Sobolev spaces have also been used. For more details, see Remark 14-(ii − iii) later on. However, in both the aforementioned papers neither results concerning the fact that the solutions belong to W 1,2 (RN ) are given, nor concentration is proved. The rest of the paper is organized as follows. Section 2 contains our assumptions and main results. Section 3 is devoted to discuss the weighted spaces Hε (including an embedding Theorem from [19]), as well as to prove some uniform integral estimates that are used in the sequel. In Section 4 we deal with the main existence result, Theorem 1. We first prove (see Theorem 13) that in Hε the Mountain-Pass Theorem applies in a direct way for any 0 < α ≤ 2; next, we assume that 0 < α < 2 and prove some exponential decay for the above Mountain-Pass critical points which allows us to show that they give rise to ground states of (NLS), see Theorem 16. Finally in Section 5 we prove that these ground states are spikes concentrating at a minimum of A. This result is achieved by using the preceding decay estimates, jointly with an uniform bound on the energy of the Mountain-Pass critical points found before. Notation. Throughout the paper we will use the following notation: - BR is the sphere {x ∈ RN : |x| < R}; - W k,p (Ω), W k,p (RN ), are the usual Sobolev spaces; - Lp (Ω), Lp (RN ) are the usual Lebesgue spaces. - c, c1 , . . ., C, C1 , . . . denote possibly different constants; - h1 ∼ h2 means that h1 and h2 are of the same order as ε → 0.
2
Assumptions and main results
In order to find solutions of (NLS) we will make the following assumptions on V and K (V ) V : RN → R is smooth and there exist α, a, A > 0 such that a ≤ V (x) ≤ A, 1 + |x|α and, respectively (K) K : RN → R is smooth and ∃ β, k > 0 : 0 < K(x) ≤
k . 1 + |x|β
In order to prove existence of ground states of (NLS) as well as their concentration properties we assume a suitable bound on p involving α and β. Let ( 4β N +2 if 0 < β < α N −2 − α(N −2) , (2) σ = σN,α,β = 1 otherwise. Our main existence result is the following.
3
Theorem 1 Let (V ), (K) hold, with 0 < α < 2 and β > 0, respectively, and suppose p satisfies (3)
σ (N + 2)/(N − 2). For more details we refer to Proposition 15. (ii) If 0 < β < α, then σ > 1 and the range of p in (3) is smaller than the usual one 1 < p < N +2 (N + 2)/(N − 2). If β = 0, we would have σ = N −2 and the interval of admissible p would be empty. (iii) When α = 2 we can still find a solution in Hε but not in W 1,2 (RN ), see Theorem 13. Concerning semiclassical states of (NLS) we show the following concentration behavior. Theorem 3 Let the same assumptions as in Theorem 1 hold. Then, the preceding ground state vε p+1 − N2 . More precisely, vε has a concentrates at a global minimum x∗ of A = V θ K −2/(p−1) , with θ = p−1 ∗ unique maximum xε for which xε → x as ε → 0, and x − xε ∗ vε (x) = U + ωε (x), as ε → 0, ε 2 where ωε (x) → 0 in Cloc (RN ) and L∞ (RN ) as ε → 0 and U ∗ is the unique positive radial solution of
−∆U ∗ + V (x∗ )U ∗ = K(x∗ )(U ∗ )p . The proofs of the preceding two theorems will be carried out in the rest of the paper.
3
Some weighted Sobolev spaces
As anticipated in the Introduction, we will work in a class of weighted Sobolev spaces. Precisely, let us set, for all ε > 0, Z 2 1,2 N Hε = {u ∈ D (R ) : ε |∇u(x)|2 + V (x)u2 (x) dx < ∞}. RN
Hε is a Hilbert space with scalar product and norm, respectively Z 2 kuk2ε = ε |∇u(x)|2 + V (x)u2 (x) dx, N ZR 2 (u|v)ε = ε ∇u(x) · ∇v(x) + V (x)u(x)v(x) dx. RN
Set H = Hε=1 with norm k · kH . Remark 4 Since V is positive and uniformly bounded, it follows that W 1,2 (RN ) ⊂ Hε for all ε > 0.
4
Denote by LqK the weighted space of measurable u : RN → R such that q1 Z K(x)|u(x)|q dx < ∞. |u|q,K = RN
Hε and
LqK
are particular cases of weighted spaces discussed in [19], where the following result is proved.
Theorem 5 Let N ≥ 3 and suppose that (V ), (K) hold with α ∈ (0, 2] and β > 0, respectively. Then for all ε > 0, Hε ⊂ Lp+1 provided K N +2 σ≤p≤ , N −2 and there is Cε > 0 such that |u|q,K ≤ Cε kukε ,
(4)
∀ u ∈ Hε .
Furthermore, the embedding of Hε into LqK is compact if (3) holds. In view of the preceding Theorem we will assume in the sequel that p, α and β always satisfy (3). Remarks 6 (i) If a ≤ V (x) ≤ A, namely when α = 0, we have Hε = W 1,2 (RN ) and Theorem 5 implies provided (β > 0 and) (3) holds. that W 1,2 (RN ) is compactly embedded in Lp+1 K (ii) If V ∼ (1 + |x|α )−1 and K ∼ (1 + |x|β )−1 , with 0 < α ≤ 2 and β > 0, it is proved in [19] that the N +2 N growth restriction σ ≤ p ≤ N −2 is a necessary condition for Hε to be embedded into LK (R ), see also Proposition 15. In the rest of this section we will prove some integral estimates for functions in Hε uniform with respect to ε. We anticipate that, as a byproduct, we will deduce a proof of the embedding result stated in Theorem 5, see Remark 10 below. Proposition 7 Let 0 < α < 2 and let p satisfy (3). Then for all δ > 0 there exists R > 0 such that, for all R ≥ R and all u ∈ Hε with supp{u} ∩ BR = ∅, one has Z N (5) K|u|p+1 ≤ δ ε− 2 (p−1) kukε(p+1 . RN
Proof. proof
The proof is carried out in several steps. First, let us introduce some quantities we need in the
(i) the sequence of radii Rn,ε defined by
Rn,ε = εRn ,
2−α n Rn = a1/2
(ii) the sequence of continuous functions ψn,ε : R+ 7→ [0, 1] if 1 r−R1,ε ψ1,ε (r) = − R2,ε −R1,ε + 1 if 0 if
2 2−α
;
satisfying 0 ≤ r ≤ Rn−1,ε , R1,ε ≤ r ≤ R2,ε , r ≥ R2,ε ,
and for n ≥ 2
ψn,ε (r) =
0
r−Rn,ε Rn,ε −Rn−1,ε
+1
r−Rn,ε − Rn+1,ε −R +1 n,ε 0
5
if
0 ≤ r ≤ Rn−1,ε ,
if
Rn−1,ε ≤ r ≤ Rn,ε ,
if
Rn,ε ≤ r ≤ Rn+1,ε ,
if
r ≥ Rn+1,ε ;
(iii) a sequence of sets An,ε An,ε = {x ∈ RN : Rn−1,ε ≤ |x| ≤ Rn+1,ε }. Note that A1,ε is a ball, An,ε is an annulus for n ≥ 2, and that the ψn,ε ’s have been chosen in such a way that X u(x) = ψn,ε (|x|)u(x). n
These cut-off functions are useful to estimate integrals over RN by means of a discrete sum of integrals on the annuli An,ε . Lemma 8 There exists c > 0 such that Z Z ∞ X 1 |ψn,ε u|p+1 , Kup+1 ≤ c β 1 + R A RN n,ε n,ε n=1
∀ u ∈ Hε , ∀ε ∈ (0, 1].
Proof. On An,ε one has u = ψn−1,ε u + ψn,ε u + ψn+1,ε u (with abuse of notation we are taking A0,n = ∅), which implies ! Z Z Z Z p+1 p+1 p+1 p+1 Ku ≤ sup K |ψn,ε u| + |ψn,ε u| + |ψn,ε u| . An,ε
An,ε
An−1,ε
An,ε
An+1,ε
Since the width of An,ε is small with respect to Rn,ε then there exists c2 = c2 (K) such that sup K ≤ c2
(6)
An,ε
The last two formulas imply Z Kup+1 ≤ c2
≤ c2
1+
β Rn+1,ε
1
Z
β 1 + Rn−1,ε Z 1
An,ε
+
1
β 1 + Rn+1,ε
1 1+
β Rn,ε
≤ c2
|ψn,ε u|p+1 +
An−1,ε
1 β 1 + Rn−1,ε
.
Z
1 β 1 + Rn,ε
|ψn,ε u|p+1
An,ε
! |ψn,ε u|p+1
.
An+1,ε
Summing over all the integers n, the Lemma follows. R Next, we estimate each term An,ε |ψn,ε u|p+1 . Let γ satisfy γ(2∗ − 2) = p − 1, or equivalently 2∗ γ = (p − 1)N/2. Then one has Z Z |ψn,ε u|p+1 =
An,ε
|ψn,ε u|2
∗
γ+2−2γ
.
An,ε
Using the H¨older’s inequality we find Z
|ψn,ε u|p+1 ≤ c1
"Z
An,ε
|ψn,ε u|2
∗
#γ "Z
An,ε
#1−γ |ψn,ε u|2
.
An,ε
∗
From the embedding of D1,2 into L2 we infer "Z Z p+1
|ψn,ε u|
(7)
≤ c2
An,ε
2
#2∗ γ/2 "Z
#1−γ 2
|∇(ψn,ε u)|
|ψn,ε u|
An,ε
.
An,ε
From (6) and (7) we get Z (8)
Ku An,ε
p+1
≤ c3
1 β 1 + Rn,ε
"Z
2
#2∗ γ/2 "Z
|∇(ψn,ε u)| An,ε
|ψn,ε u| An,ε
We now show 6
#1−γ 2
.
Lemma 9 There holds
Z
Z
2
|∇u|2 + ε−2 V u2
|∇(ψn,ε u)| ≤ c4 An,ε
An,ε
Proof. First we evaluate |∇(ψn,ε u)|2 = |u∇ψn,ε + ψn,ε ∇u|2 ≤ 2u2 |∇ψn,ε |2 + 2|∇u|2 .
(9)
From the definition of Rn,ε we get α |Rn+1,ε − Rn,ε |2 = ε2 |Rn+1 − Rn |2 ≥ cε2 Rn+1,ε . −α As above, Rn+1,ε ≤ c5 inf An,ε V , and we deduce
|∇ψn,ε |2 ≤ c6 |Rn+1,ε − Rn,ε |−2 ≤ c7 ε−2 V,
x ∈ An,ε .
Substituting in (9) and integrating over An,ε , the lemma follows. Proof of Proposition 7 completed. Lemma 9 together with (8) yields Z (10)
Ku
p+1
≤c
An,ε
#2∗ γ/2 "Z |∇u|2 + ε−2 V u2
"Z
1 β 1 + Rn,ε
An,ε
#1−γ 2
|ψn,ε u|
.
An,ε
Let M, s > 0 and let θ, θ0 be any pair of conjugate exponents (M, s, θ, θ0 will be fixed appropriately later). For brevity, let us set Z S = Sn,ε = |∇u|2 + ε−2 V u2 An,ε
Z T = Tn,ε
|ψn,ε u|2
= An,ε
so that (10) becomes Z
Kup+1 ≤ c
An,ε
1 1+
β Rn,ε
S2
∗
γ/2
· T 1−γ .
Since S2
∗
γ/2
·
T (1−γ) β 1 + Rn,ε
= M εs S 2
∗
γ/2
· M −1 ε−s
0
T (1−γ) β 1 + Rn,ε
≤
0 0 T (1−γ)θ 1 θ sθ 2∗ γθ/2 1 M ε S + 0 M −θ ε−sθ θ 0 , θ θ β 1 + Rn,ε
we get Z (11)
Kup+1 ≤ c1 M θ εsθ S 2
∗
γθ/2
An,ε
0
T (1−γ)θ
0
+ c2 M −θ ε−sθ
1+
β Rn,ε
0
θ 0 .
Now we choose s, θ satisfying 2∗ γθ = p + 1,
θ(s − 2∗ γ) = −sθ0 .
Then S
2∗ γθ/2
"Z =
#2∗ γθ/2 "Z 2 −2 2 −(p+1) |∇u| + ε V u =ε
An,ε
An,ε
7
#(p+1)/2 2 2 2 ε |∇u| + V u ,
and hence θ sθ
(12)
M ε S
2∗ γθ/2
θ −(p−1)N/2
#(p+1)/2 2 2 2 . ε |∇u| + V u
"Z
=M ε
An,ε
On the other hand we also have T (1−γ)θ
0
β (1 + Rn,ε )θ0
≤
#(p+1)/2
"Z
1
u
β (1 + Rn,ε )θ0
An,ε
α (p+1)/2 (1 + Rn,ε )
=
"Z
β (1 + Rn,ε )θ0
An,ε
α (p+1)/2 (1 + Rn,ε )
≤
"Z
β Rn,ε )θ0
(1 +
2
#(p+1)/2 u2 α 1 + Rn,ε #(p+1)/2 V u2
.
An,ε
Inserting the above inequality and (12) into (11), and taking into account that −sθ0 = θ(s − 2∗ γ) = −(p − 1)N/2, we infer Z
Kup+1 ≤ c3 ε−(p−1)N/2 ×
An,ε
M θ
#(p+1)/2 #(p+1)/2 "Z α (p+1)/2 2 0 (1 + Rn,ε ) . V u2 · |ε ∇u|2 + V u2 + M −θ β )θ0 (1 + Rn,ε An,ε An,ε
"Z
Now, let us remark that α (p+1)/2 ) (1 + Rn,ε
(1 +
β )θ0 Rn,ε
0
−βθ +α(p+1)/2 ∼ Rn,ε → 0,
(Rn,ε → ∞),
since p > σ implies that −βθ0 + α(p + 1)/2 < 0. Then, given δ > 0, we can choose M, R > 0 such that Mθ