Concentration Around a Sphere for a Singularly Perturbed Schr¨ odinger Equation.∗ Marino Badiale
†
Teresa D’Aprile‡
February 26, 2001
Abstract In this paper we study the existence of concentrated solutions for the following nonlinear elliptic equation −h2 ∆v + V (x)v = |v|p−2 v where v : RN → R, N ≥ 3 and 2 < p < N2N −2 . We assume that the potential V is radially symmetric and bounded below away from zero. Existence results are established provided that h is sufficiently small and we are able to find positive solutions with spherical symmetry which exhibit a concentration behaviour near a sphere centred in zero. The proofs of our results are variational and rely on a constrained minimization method; furthermore a penalization-type technique is developed and permits to single out the desired solutions by means of a suitable modification of the variational problem.
1
Introduction
In recent years a great amount of works has been devoted to the study of concentrated solutions for the following nonlinear Schr¨ odinger equation: ih
h2 ∂ψ =− ∆ψ + V (x)ψ − γ|ψ|p−2 ψ, ∂t 2m
(1.1)
where • h > 0, γ > 0, • ψ : RN → C, V : RN → R, • N ≥ 1, p > 2. Here ∆ is the classical Laplace operator and i the imaginary unit. Looking for standing waves of (1.1), i.e. solutions of the form ψ(x, t) = exp(−iEt/h)v(x), where v is a real valued function, the equation for v becomes − h2 ∆v + V (x)v = |v|p−2 v ∗
(1.2)
The first author was supported by M.U.R.S.T., “Variational and Nonlinear Differential Equations”. The second author was supported by M.U.R.S.T., “Equazioni Differenziali e Calcolo delle Variazioni”. † Scuola Normale Superiore, Piazza dei Cavalieri 7, 56126 PISA (Italy), e-mail:
[email protected] ‡ Scuola Normale Superiore, Piazza dei Cavalieri 7, 56126 PISA (Italy), e-mail:
[email protected]
1
where we have assumed γ = 2m = 1 and the parameter E has been absorbed by V. By making the change of variables x → hx, (1.2) can be rewritten as − ∆u + Vh (x)u = |u|p−2 u
(1.3)
where Vh (x) = V (hx) and u(x) = v(hx). The first result in this line, at our knowledge, is due to Floer and Weinstein ([18]). These authors considered the one-dimensional case and constructed for small h > 0, via a Lyapunov-Schmidt reduction, a family of solutions concentrating around any nondegenerate critical point of the potential V , under the condition that V is bounded and p = 4. In [25] and [26] Oh generalized this result to higher dimensions when 2 < p < N2N −2 (N ≥ 3) and V exhibits “mild oscillations” at infinity. Variational methods based on variants of Mountain-Pass Lemma are used in [27] to get existence results for (1.2) where V lies in some class of highly oscillatory V ’s which are not allowed in [25]-[26]. Under the condition lim inf |x|→+∞ V (x) > inf x∈RN V (x) in [29] Wang established that these mountain-pass solutions concentrate at global minimum points of V as h → 0+ ; moreover a point at which a sequence of solutions concentrates must be critical for V. This line of research has been extensively pursued in a set of papers by Del Pino and Felmer ([15]-[17]). We also recall the nonlinear finite dimensional reduction used in [1] and a recent paper by Grossi ([19]). The most complete and general results for this kind of problems seem due to Del Pino and Felmer ([16]) and Li ([20]). As for radial positive bound states, in [29] Wang has worked on equation (1.2) in case that V is radial: he proved the existence of a family {vh } of positive radial solutions with least energy among all nontrivial radial solutions; such a family must concentrate at the origin as h → 0+ . Concerning solutions for equation (1.2) in bounded domains with Dirichlet or Neumann boundary conditions, we recall, among many others, the results obtained in [22], [23], [24] and [30]. In most of the above examples the typical result is that of finding, for small values of h, solutions that look like one or more spikes located near a finite set of critical points of the potential V . A natural question arises: is it possible to find solutions with a more complicated set of concentration, for example, solutions concentrating near a curve or a hypersurface? To answer such a question, a first idea would be to study radial problems and radial solutions; of course, a radial solution concentrating at a point different from the origin must concentrate in all the sphere centred in zero and containing that point. This idea has been first pursued in [14] where the following nonlinear elliptic system has been studied: − h2 ∆v + V (x)v − hp ∆p v + W 0 (v) = 0
(1.4)
where v : RN → RN +1 , N ≥ 3, p > N, the potential V is positive and radial and W is an appropriate singular function satisfying a suitable symmetric property. This kind of equations have been introduced in a set of recent papers (see [4]-[11], [2]-[3] and [13]) where the authors look for soliton-like solutions, i.e. solutions whose energy is finite and which preserve their shape after interactions. Provided that h is sufficiently small, in [14] it was proved that solutions with a certain spherical symmetry exist and exhibit a concentration behaviour near a sphere centred in zero as h → 0+ . However an analogous result does not seem achieved for radial solutions of equation (1.2) even though it is apparently easier if compared with the system (1.4); the reason is essentially due to the fact that in some relevant features equation (1.2) appears much more complicated 2
then the system (1.4). Indeed the hypotheses on the nonlinearity W allow to get solutions as local minimum (instead of generic critical) points of the associated energy functional, and the p-laplacian forces to study the problem in the Sobolev space W 1,p , whose functions are more regular then those of W 1,2 . For these and other reasons, it is not trivial to extend the results in [14] to problem (1.2). Main result of this paper. The main result of this work is theorem 5.2; here we are unable to give the precise statement since for this we would need some definitions and notations which will be provided in the course of the paper. So, for the reader’s sake, we confine ourselves to give an idea. Roughly speaking, theorem 5.2 states, at least for small h > 0, the existence of a positive radially symmetric solution vh to equation (1.2). Furthermore such family {vh } exhibits the following concentration behaviour as h → 0+ : for every sequence hn → 0+ there exists a subsequence, still denoted by hn , such that vhn has a sphere of local maximum points {x ∈ RN | |x| = rhn } with rhn → r > 0 as n → +∞, while vhn vanishes to zero away from the sphere {x ∈ RN | |x| = r}. We also give a roughly idea of the main ingredients which are needed to get this result. As first thing we capture our solutions by usual constrained minimization techniques. We also have to use a truncation-penalization argument, which consists in the introduction of a modified equation for which solutions are studied and soon after are proved to solve our original equation (1.2); this idea is in the spirit of that used in [15], [16] and [17]. The main idea of this paper is the introduction of an auxiliary functional Er (u) by which we define the real function r → Er∗ (1) (see section 4). This function is relevant because the radii r of the spheres of concentration are given as the minimum points of the function r → rN −1 Er∗ (1). We point out that in these elliptic singularly perturbed problems the points of concentration of solutions are always obtained as critical points of some kind of “hidden” functional. In this case it seems that rN −1 Er∗ (1) is the right “hidden” functional. We notice that our results are obtained under the hypothesis that the potential V satisfies assumption a) and, in addition, (5.27) and (5.28) which are not very easy to understand. In the remarks 5.1 and 5.2 we give an idea on the way to get potentials satisfying such hypotheses. The idea, roughly speaking, is that V (which is radial) must be in a suitable interval [r0 , r1 ] much greater than in the line (r1 , +∞). So there is a sort of barrier, in [r0 , r1 ], which prevents the minimizing sequences to concentrate in the origin. The concentration radius r, which corresponds to an absolute minimum of rN −1 Er∗ (1) in (r1 , +∞), results as a sort of balance between the barrier given by V and the natural tendency of the radial solutions to concentrate near the origin. We notice that, when rN −1 Er∗ (1) has only one absolute minimum point in (r1 , +∞), then we have concentration at that radius, while, when rN −1 Er∗ (1) has many absolute minimum points in (r1 , +∞), then we know that we have concentration at one of them, but we are not able to locate exactly the concentration radius. Main hypotheses. Throughout this paper we always make the following assumptions: a) V ∈ C 1 (RN , R), V0 ≡ inf x∈RN V (x) > 0 and V is a radially symmetric function, i.e. V(|x|) = V (x) ∀x ∈ RN ; b) N ≥ 3; c) 2 < p < 2∗ =
2N N −2 .
Organization of the paper. The paper is organized as follows: after the introduction, in section 2 we describe the general functional setting in which we work and we point out some 3
of its properties. Section 3 is devoted to the study of the behaviour of suitable sequences of functions and a sort of “splitting” lemma is established. In section 4 we introduce the auxiliary functionals we said above, and we prove their main properties. Finally in section 5 we explain our truncation-penalization argument and then, via several lemmas, we prove the main results of this paper (theorems 5.1 and 5.2). Notations We fix the following notations we will use from now on. • x y is the standard scalar product between x, y ∈ RN . • |x| is the Euclidean norm of x ∈ RN . • H 1 (RN ) or H 1 (Ω), with Ω ⊂ RN an open set, are the standard Sobolev spaces. • For any U ⊂ RN , U its closure and ∂U its boundary. Furthermore χU denotes the characteristic function of U , while by meas (U ) we intend the Lebesgue measure of U . • ωN is the surface measure of the unit sphere S N −1 of RN . • If x ∈ RN and r > 0, then Br (x) or B(x, r) is the open ball with centre in x and radius r. • Given R2 > R1 > 0, by C(R1 , R2 ) we denote the annulus in RN centred in 0 and with internal radius R1 and external radius R2 , i.e. C(R1 , R2 ) = {x ∈ RN | R1 < |x| < R2 }.
2
Functional Setting
In order to obtain solutions of equation (1.2) or (1.3) we choose a suitable Hilbert space: for every h > 0 let Hh denote the subspace of H 1 (RN ) consisting of functions u such that kukHh ≡
Z
RN
2
2
|∇u| + Vh (x)|u|
dx
1/2
< +∞.
(2.5)
The space Hh can also be defined as the closure of C0∞ (RN ) with respect to the norm (2.5). By hypothesis it is V0 ≡ inf x∈RN V (x) > 0, then Hh is embedded continuously into H 1 (RN ). Since we are interested in finding solutions with spherical symmetry, we denote by Hh,rad the closed subspace of Hh formed by the radial functions: Hh,rad = {u ∈ Hh | u is a radially symmetric function}. The functions in Hh,rad are continuous for x 6= 0 and decay to zero at infinity; furthermore we recall a useful result concerning a compactness property which is an easy consequence of a lemma due to Strauss ([28]). For the proof we refer to [12] (Theorem A.I’, pg. 341).
Lemma 2.1 For every h > 0 the injection Hh,rad ⊂ Ls (RN ) is compact for s ∈ 2, N2N −2 . For sake of simplicity in what follows we will adopt the following convention: for every u ∈ Hh,rad we denote by ϕu the real function defined by ϕu : (0, +∞) → R,
ϕu (s) = u(x), |x| = s. 4
(2.6)
In is obvious that ϕu ∈ H 1 (a, +∞) for every a > 0. By using the homogeneity of equations (1.2) and (1.3), the solutions of our problem will be captured by means of a constrained minimization method: more precisely for every h > 0 consider the functional Eh (u) =
Z
RN
|∇u|2 + Vh (x)|u|2 dx ≡ ωN
Z
+∞
0
sN −1 (|ϕ0u (s)|2 + Vh (s)|ϕu (s)|2 ) ds ≡ kuk2Hh ,
where, with obvious notation, Vh (s) = V(hs). There results Eh ∈ C 1 (Hh,rad , R), and Eh is weakly-lower semicontinuous and coercive. Now for every α ∈ (0, 1] introduce the set
Mh (α) = u ∈ Hh,rad hN −1
Z
|u|p dx = α .
RN
It is standard to check that, if u0 ∈ Mh (α) minimizes the functional Eh on the set Mh (α), i.e. N −1
1/(p−2)
Eh (u0 ) = inf u∈Mh (α) Eh (u), then, letting u = Eh (u0α)h u0 , we obtain a solution of (1.3). This is essentially the method we will employ in the next sections.
3
Preliminaries
This section is devoted to establish some preliminary results concerning the study of the behaviour of sequences {uhn }, with uhn ∈ Hhn ,rad , whose energy grows like N1−1 . Toward hn our aims the following lemma constitutes a crucial step; it uses in an essential way the radial symmetry of the functions in Hhn ,rad . We notice that throughout this paper factors like hN −1 , tN −1 or rN −1 appear anywhere. This is because we work with radial functions u and with their “one-dimensional sections” ϕu . When passing from u to ϕu and viceversa a factor with power N − 1 appears of course in all the relevant integral quantities. Lemma 3.1 Let hn → 0+ an arbitrary sequence and uhn ∈ Hhn ,rad such that Z
lim sup hnN −1 n→+∞ RN
|∇uhn |2 + |uhn |2 dx ≤ L < +∞.
For every n ∈ N and x ∈ RN set ϕn (|x|) ≡ |uhn (x)|. Then for all t > 0 it holds lim sup n→+∞
Z
+∞
t/hn
|ϕ0n |2 + |ϕn |2 dr ≤
L . ωN tN −1
(3.7)
Furthermore for large n there results: ∀s, s0 ∈
t , +∞ : hn
|ϕn (s) − ϕn (s0 )| ≤
2L ωN tN −1
1/2
As a consequence the sequence {ϕn } is equi-uniformly continuous in ing sense: for every η > 0 there exists δ > 0 such that ∀n ∈ N,
∀s, s0 ∈
t , +∞ : hn
|s − s0 |1/2 .
h
t hn , +∞
in the follow-
|s − s0 | ≤ δ ⇒ |ϕn (s) − ϕn (s0 )| ≤ η.
Proof. An immediate computation leads to Z
−1 lim sup hN n n→+∞ RN
2
2
|∇uhn | + |uhn |
dx = lim sup ωN n→+∞
5
Z
0
+∞
(hn s)N −1 |ϕ0n |2 + |ϕn |2 ds ≥
≥ ωN lim sup n→+∞
Z
+∞ t hn
(hn s)N −1 |ϕ0n |2 + |ϕn |2 ds ≥ ωN tN −1 lim sup n→+∞
It is obvious that each ϕn belongs to the class H 1
t hn , +∞
Z
t hn
|ϕn (s) − ϕn (s )| ≤
Z
+∞
t/hn
|ϕ0n (r)|2 dr
!1/2
0 1/2
|s − s |
≤
|ϕ0n |2 + |ϕn |2 ds.
. Then from the standard theory
of the one-dimensional Sobolev spaces we deduce that for all s, s0 ∈ n 0
+∞
2L ωN tN −1
t hn , +∞
1/2
and for large
|s − s0 |1/2 .
(3.8) 2
A second preliminary result we will need is given in the following lemma inspired by the “splitting lemma” of [11] in the spirit of the concentration compactness principle of [21]. Lemma 3.2 Let hn → 0+ an arbitrary sequence and uhn ∈ Hhn ,rad such that Z
−1 lim sup hN n n→+∞ RN
|∇uhn |2 + |uhn |2 dx < +∞,
sup |uhn (x)| > ε
|x|≥ ht
∀n ∈ N
(3.9) (3.10)
n
and |uhn (x)| ≤ ε
t ∀x ∈ B 0, hn
∀n ∈ N,
(3.11)
for some t, ε > 0. Then, setting ϕn (|x|) ≡ |uhn (x)|, there exist ` ∈ N, τ > 0, R1 , . . . , R` > 0 and ` sequences of positive numbers {rh1 n }, . . . , {rh` n }, with rhi n > htn , such that, up to subsequence {hn rhi n } is bounded ∀i = 1, . . . , `; (3.12) ϕn (rhi n ) > ε
∀ n ∈ N,
rhi n is the maximum point for ϕn
∀i = 1, . . . , `;
in (0, +∞) \
[
(rhj n − Rj , rhj n + Rj );
(3.13) (3.14)
j
∀ s 6∈
` [
(rhi n − Ri , rhi n + Ri ) :
ϕn (s) ≤ ε;
(3.15)
i=1
|rhi n − rhj n | → +∞ as n → +∞ ∀i, j ∈ {1, . . . , `} with i 6= j; ε ϕn (s) > ∀s ∈ (rhi n − τ, rhi n + τ ) ∀i ∈ {1, . . . , `}. 2
(3.16)
Proof. For sake of clarity we divide the proof into 3 steps. Step 1. Let {rhn } be a sequence of positive numbers such that rhn ≥
t hn
∀n ∈ N.
Then for every η > 0 and γ > 0 there exists R > γ such that, eventually passing to a subsequence, ϕn (r) ≤ η f or |r − rhn | = R.
6
We argue by contradiction and assume the existence η > 0 and γ > 0 such that, for every R > γ, there exists nR ∈ N verifying ∀ n ≥ nR
∃ shn > 0
with |shn − rhn | = R,
s.t. ϕn (shn ) > η.
Let us fix R1 > γ arbitrarily and take n1 ∈ N such that ∃ s1hn > 0
∀ n ≥ n1 Since rhn ≥ can assume
t hn ,
it is rhn −
t 2hn
with |s1hn − rhn | = R1
s.t. |ϕn (s1hn )| > η.
→ +∞ as n → +∞; then provided n1 is sufficiently large, we s1hn ≥
t +1 2hn
∀n ≥ n1 .
Now choose R2 > R1 + 1: there exists n2 ∈ N, n2 > n1 , such that ∀ n ≥ n2
∃ s2hn > 0
with |s2hn − rhn | = R2
and moreover s2hn ≥
t +1 2hn
s.t. |ϕn (s2hn )| > η
∀n ≥ n2 .
We easily infer that, for every k ≥ 2, we can choose Rk > Rk−1 + 1 and we obtain the existence of nk ∈ N, nk > nk−1 , such that ∀ n ≥ nk
∃ skhn > 0
with |skhn − rhn | = Rk
and skhn ≥
t +1 2hn
s.t. |ϕn (skhn )| > η
∀n ≥ nk .
We claim that ∃δ > 0
such that ∀ k ∈ N, ∀ n ≥ nk :
Bδ (skhn )
⊂
η r > 0 ϕn (r) > .
Indeed, according to lemma 3.1, the sequence {ϕn } is equi-uniformly continuous in hence there exists δ > 0 such that ∀ n ∈ N,
∀ s, s0 ∈
t , +∞ : 2hn
(3.17)
2
|s − s0 | ≤ δ ⇒ |ϕn (s) − ϕn (s0 )| ≤
η . 2
h
t 2hn , +∞
,
(3.18)
Without loss of generality we may assume δ ≤ 12 . Now take k ∈ N, n ≥ nk and s > 0 with |skhn − s| ≤ δ; by construction we have s, skhn ∈ [ 2ht n , +∞); then by (3.18) we have ϕn (s) ≥ ϕn (skhn ) − |ϕn (skhn ) − ϕn (s)| > η −
η η = 2 2
and this proves (3.17). Furthermore for all k, k 0 ∈ N with k > k 0 0 0 k shn − skhn ≥ |skhn − rhn | − |skhn − rhn | = Rk − Rk0 ≥ Rk − Rk−1 > 1.
This implies that for all k ∈ N and for all n ≥ nk the intervals {[sihn − δ, sihn + δ]}i=1,...,k are disjoint and k [ t η [sihn − δ, sihn + δ] ⊂ , +∞ ∩ r > 0 ϕn (r) > . 2hn 2 i=1 7
nR
o
+∞ Since k is arbitrary, this would mean that the sequence t/(2h ϕ2n dr is not bounded which n) is in contradiction with (3.7) of lemma 3.1: step 1 is proved.
Step 2. Let γ > 0 and {rhn } ⊂ (0, +∞) such that, rhn >
t , hn
ϕn (rhn ) ≥ γ
∀n ∈ N.
Then the sequence {hn rhn } is bounded. Lemma 3.1 assures that there exists δ > 0 such that ∀n ∈ N,
t ∀s, s ∈ , +∞ : 2hn 0
Assume, without loss of generality,
t hn
|s − s0 | ≤ δ ⇒ |ϕn (s) − ϕn (s0 )| ≤ t 2hn
−δ >
(rhn − δ, rhn + δ) ⊂
γ . 2
(3.19)
for all n ∈ N, so that there results:
t , +∞ 2hn
∀n ∈ N.
(3.20)
Combining (3.19) and (3.20) we have ∀n ∈ N,
∀s ∈ (rhn − δ, rhn + δ) :
ϕn (s) ≥
γ . 2
Now assume by contradiction that, up to a subsequence, hn rhn → +∞
as
n → +∞.
Then for every n ∈ N we easily compute −1 hN n
Z
RN
−1 |uhn |2 dx ≥ hN n
Z
C(rhn −δ,rhn +δ)
|uhn |2 dx ≥
γ 2 N −1 ωN hn (rhn + δ)N − (rhn − δ)N . 4 N
An elementary calculus shows that
−1 lim hN (rhn + δ)N − (rhn − δ)N = lim 2N (hn rhn )N −1 δ = +∞, n
n→+∞
n→+∞
which implies −1 lim hN n
n→+∞
Z
RN
|uhn |2 dx = +∞,
in contradiction with (3.9). This contradiction permits us to conclude. Step 3. End of the proof. For sake of clarity the proof of this step is divided into two parts. In the first part, with an iterative procedure, we’ll prove the existence of ` ∈ N, R1 , . . . , R` > 0 and ` sequences of positive numbers {rh1 n }, . . . , {rh` n } such that, up to subsequence, (3.12), (3.13), (3.14) and (3.15) are satisfied. In the second part we shall show how it is possible to select `0 of the ` sequences {rhi n } (with `0 ≤ `) for which, provided we increase the associated Ri , all the requirements of the lemma are easily deduced. For every n ∈ N let rh1 n ≥ 0 be the maximum point of ϕn . According to (3.10) and (3.11) it is t rh1 n > and ϕn (rh1 n ) > ε. hn 8
Last step implies that the sequence {hn rh1 n } is bounded. Using step 1 we infer the existence of R1 > 1 such that, eventually passing to a subsequence, ϕn (r) ≤ ε
for |r − rh1 n | = R1 .
Now we distinguish two cases: either (a1) for n sufficiently large ∀r 6∈ (rh1 n − R1 , rh1 n + R1 ) :
ϕn (r) ≤ ε
or (b1) up to a subsequence, sup
ϕn (r) > ε.
1 −R ,r 1 +R ) r6∈(rh 1 h 1 n
n
In the case (a1) the first part of the step is proved with ` = 1. Consider case (b1). Let rh2 n be the maximum point of ϕn in (0, +∞) \ (rh1 n − R1 , rh1 n + R1 ); we have rh2 n >
t hn
and ϕn (rh2 n ) > ε.
Again from step 2 we deduce the boundedness of the sequence {hn rh2 n }. By applying step 1 we can choose R2 > 1 verifying, up to a subsequence, ϕn (r) ≤ ε
for |r − rh2 n | = R2 .
Again we have another alternative: either (a2) for n sufficiently large ∀r 6∈ (rh1 n − R1 , rh1 n + R1 ) ∪ (rh2 n − R2 , rh2 n + R2 ) :
ϕn (r) ≤ ε
or (b2) up to a subsequence, sup
ϕn (r) > ε.
1 −R ,r 1 +R )∪(r 2 −R ,r 2 +R ) r6∈(rh 1 h 1 2 h 2 h n
n
n
n
If case (a2) holds true, the first part of the step is proved with ` = 2. In the case (b2) we consider the maximum point of ϕn in (0, +∞)\ (rh1 n − R1 , rh1 n + R1 ) ∪ (rh2 n − R2 , rh2 n + R2 ) and we repeat the same argument used in the case (b1). This alternative process terminates in a finite number of steps, in the sense that there exists ` ∈ N such that for every `0 < ` (b`0 ) holds true while (a`) occurs. Indeed by lemma 3.1 we get the existence of δ ∈ (0, 12 ) such that ∀n ∈ N,
∀s, s0 ∈
t , +∞ : 2hn
ε |s − s0 | ≤ δ ⇒ |ϕn (s) − ϕn (s0 )| < . 2
Provided that n is sufficiently large, it is
t 2hn , +∞
t hn
−δ >
t 2hn
so that (rhi n − δ, rhi n + δ) ⊂
. Then by (3.21) we can infer that for every n ∈ N and for all i = 1, . . . , ` ∀s ∈ (rhi n − δ, rhi n + δ) :
ε ϕn (s) ≥ ϕn (rhi n ) − |ϕn (rhi n ) − ϕn (s)| > . 2 9
(3.21)
By construction Ri > 1, hence we have |rhi n − rhj n | > 1 for i 6= j; then, since δ < 21 , we deduce that the intervals {(rhi n − δ, rhi n + δ)}i=1,...,` are disjoint and there results ` [
(rhi n
−
δ, rhi n
+ δ) ⊂
i=1
ε t , +∞ ∩ s ≥ 0 ϕn (s) > . 2hn 2
Hence (3.7) of lemma 3.1 put a bound from above to the number `. It remains to prove how the choice of the rhi n can be refined in such a way as to verify (3.16). Indeed it is sufficient to consider the following partition of the set {1, . . . , `}: 0
` [
{1, . . . , `} =
Ik
k=1
with `0 ≤ ` where the subsets Ik verify ∀k ∈ {1, . . . , `0 }
∀i, j ∈ Ik ,
i 6= j :
{rhi n − rhj n } is bounded
and ∀h, k ∈ {1, . . . , `0 }, h 6= k, ∀i ∈ Ih , ∀j ∈ Ik : |rhi n − rhj n | → +∞
as
n → +∞.
Obviously such partition {Ik }k∈{1,...,`0 } is unique. Now for every k ∈ {1, . . . , `0 } choose ik ∈ Ik b k > 0 sufficiently large such that ϕn (rhikn ) = supi∈Ik ϕn (rhi n ); hence it makes sense to choose R such that [ b k ) ∀n ∈ N. b k , r ik + R (rhi n − Ri , rhi n + Ri ) ⊂ (rhikn − R hn i∈Ik
b1, . . . , R b 0 and the `0 sequences r i1 , . . . , r i`0 satisfy all the Then the `0 positive numbers R ` hn hn requirements of the lemma. 2
Last lemma represents a crucial step for the proof of the main result of this paper in section 5.
4
Construction of the Auxiliary Functional
The object is now to define a suitable auxiliary functional which will prove very useful for the achievements of our results. First consider the set Υ ≡ {u ∈ H 1 | u radial, u ≡ 0 in a neighborhood of 0, u compactly supported}. We need u = 0 near the origin to avoid the possible non regularity of u in zero. Obviously it is Υ ⊂ Hh,rad for all h > 0. By using the convention introduced in (2.6), for every r > 0 and α ∈ (0, 1] put Z
Υr (α) = u ∈ Υ ωN rN −1
+∞
0
|ϕu (s)|p ds = α .
Then for r > 0 and u ∈ Υ the auxiliary functional Er is defined as Er (u) = ωN
Z
0
+∞
|ϕ0u (s)|2 + V(r)|ϕu (s)|2 ds.
We notice that Er is well defined in Υ. 10
Finally for every r > 0 and α ∈ (0, 1] set Er∗ (α) =
inf
u∈Υr (α)
Er (u).
In order to study the properties of the functional Er we will need the elementary results given by the following two lemmas. Lemma 4.1 Let A ⊂ R a measurable set, {fn } ⊂ C(A, R) and gn : A → R two sequences of functions verifying gn is measurable,
a.e. in RN ,
gn (x) ≥ 0
and fn → f
unif ormly in A as n → +∞
for some f ∈ C(A, R) with inf A |f (x)| = ν > 0. Then there results lim sup n→+∞
Z
A
fn gn dx = lim sup n→+∞
Z
f gn dx,
A
lim inf
n→+∞
Z
fn gn dx = lim inf
A
n→+∞
Z
f gn dx.
A
Proof. It is sufficient to prove the first equality; the second will follow by replacing fnR with −fn and f with −f . First we notice that f has constant sign. Then if the sequence { A gn } R R is unbounded it is obvious that both lim supn→+∞ A fn gn dx and lim supn→+∞ A f gn dx are equal to +∞ if f > 0 and to −∞ if f < 0. Thus it remains to consider the case when {gn } R is bounded in L1 (A, R); let M ≡ supn∈N A gn dx. Fixed ε > 0 arbitrarily, we can choose n sufficiently large so that |fn (x) − f (x)| ≤ ε for every x ∈ A, and this implies lim sup n→+∞
Z
A
|fn − f |gn dx ≤ εM.
The arbitrariness of ε yields Z
lim
n→+∞
A
(fn − f )gn dx = 0.
The conclusion follows from elementary computations.
2
Lemma 4.2 For every r > 0 and α ∈ (0, 1] it is: Er∗ (α) = α2/p Er∗ (1). Proof. Fix u ∈ Υr (α) and take t = α−1/p . Then it is tu ∈ Υr (1) and so Er∗ (1) ≤ Er (tu) = t2 Er (u) = α−2/p Er (u). By take the infimum on the right side for u ∈ Υr (α) we obtain the inequality Er∗ (1) ≤ α−2/p Er∗ (α). In order to obtain the opposite inequality we take u ∈ Υr (1) arbitrarily and put t = α1/p so that tu ∈ Υr (α) and we repeat the above arguments. 2 The usefulness of having introduced the auxiliary functional Er will be clear in the next lemma which establishes a relation between the behaviour of the two functionals Eh and Er with respect to suitable sequences of functions in the set Υ.
11
Lemma 4.3 Let hn → 0 an arbitrary sequence and consider un ∈ Υ ∩ Mhn (αn ) with {αn } ⊂ (0, 1] such that un ≡ 0
in
RN \ C(rn − R, rn + R)
(4.22)
αn → α
(4.23)
and hn rn → r,
n → +∞
as
for some rn , R, r > 0 and α ∈ (0, 1]. Then there results −1 lim inf hN Ehn (un ) ≥ rN −1 Er∗ (α). n n→+∞
−1 E (u ) < +∞. According Proof. In order to avoid triviality, assume lim inf n→+∞ hN hn n n to (4.23) it is rn → +∞ as n → +∞ and then it makes sense to assume rn − R > 0 for x in C(R, 3R) and wn ≡ 0 every n ∈ N. For every n ∈ N define wn (x) ≡ un x + (rn − 2R) |x| everywhere else. Observe that each wn belongs to Υ, so that Er (wn ) is well defined. Taking into account of the notations introduced in (2.6) we infer
ϕwn (s) = ϕun (s + rn − 2R) ∀s ≥ 0,
∀n ∈ N.
Then it follows lim inf
n→+∞
Z
3R
n→+∞
R
−1 ≤ lim inf hN n n→+∞
≤
1 rN −1 V
ωN 0 Then compute
−1 lim inf hN n n→+∞
−1 lim inf hN Ehn (un ) n→+∞ n
=
=
|ϕwn (s)|2 ds = lim inf
Z
RN
1 hn (rn − R)
N −1 Z
Z
rn +R
rn −R
rn +R
rn −R
Vhn (x)|un |2 dx ≤
|ϕun (s)|2 ds ≤
sN −1 |ϕun (s)|2 ds ≤
1 ωN
rN −1 V
−1 lim inf hN Ehn (un ) < +∞. n
0 n→+∞
Z rn +R N −1 ωN lim inf hn sN −1 |ϕ0un (s)|2 n→+∞ rn −R
Z 3R N −1 ωN lim inf hn (s n→+∞ R
+ V(hn s)|ϕun (s)|2 ds =
+ rn − 2R)N −1 |ϕ0wn (s)|2 + Vhn (s + rn − 2R)|ϕwn (s)|2 ds ≥
Z 3R N −1 ≥ ωN lim inf hn (s + rn − 2R)N −1 |ϕ0wn (s)|2 n→+∞ R Z 3R +ωN lim inf hnN −1 (s + rn − 2R)N −1 (Vhn (s + rn − n→+∞ R
+ V(r)|ϕwn (s)|2 ds+ 2R) − V(r)) |ϕwn (s)|2 ds.
−1 (s+r −2R)N −1 → r N −1 uniformly for s ∈ (R, 3R); hence It’s immediate to prove that hN hn n lemma 4.1 applies in the first integral and gives −1 lim inf hN Ehn (un ) ≥ rN −1 lim inf Er (wn )+ n n→+∞
−1 +ωN lim inf hN n n→+∞
Z
3R
R
n→+∞
(s + rn − 2R)N −1 (Vhn (s + rhn − 2R) − V(r)) |ϕwn (s)|2 ds.
On the other hand Vhn (s + rhn − 2R) → V(r) uniformly for s ∈ (R, 3R), hence for every fixed ε > 0 and for n large enough it is lim inf hnN −1 n→+∞
Z
3R
R
(s + rn − 2R)N −1 |Vhn (s + rhn − 2R) − V(r)| |ϕwn |2 dx ≤ 12
≤
Z 3R N −1 ε lim inf hn (s n→+∞ R
N −1
+ rn − 2R)
2
|ϕwn | ds = εr
N −1
lim inf
n→+∞
Z
3R
R
|ϕwn |2 ds.
In the last inequality we have used again lemma 4.1. In the first part of the proof we have R obtained lim inf n→+∞ R3R |ϕwn |2 ds < +∞; then from the arbitrariness of ε we easily deduce Z 3R N −1 lim inf hn (s n→+∞ R
+ rn − 2R)N −1 (Vhn (s + rn − 2R) − V(r)) |ϕwn |2 dx = 0
and hence we obtain −1 lim inf hN Ehn (un ) ≥ rN −1 lim inf Er (wn ). n n→+∞
Now observe that by hypothesis it is α = lim αn = n→+∞
lim hN −1 n→+∞ n
= ωN lim hnN −1 n→+∞
Z
3R
R
(4.24)
n→+∞
Z
R
RN
|un |p dx =
p
RN
|un | dx =
αn −1 , hN n
by which we deduce
−1 ωN lim hN n→+∞ n
Z
rn +R
rn −R
sN −1 |ϕun (s)|p ds =
(s + rn − 2R)N −1 |ϕwn (s)|p ds = ωN rN −1 lim
Z
3R
n→+∞ R
|ϕwn (s)|p ds.
−1 (s + r − 2R)N −1 → r N −1 uniformly for Last equality follows from lemma 4.1 since hN n n s ∈ (R, 3R). For every n ∈ N choose tn ∈ (0, +∞) such that tn wn ∈ Υr (α); more precisely, R 3R N −1 p since by the above computations it is ωN r R |ϕwn (s)| ds → α, there results
tn = α
1/p
ωN r
N −1
Z
3R
R
p
|ϕwn (s)| ds
!−1/p
→1
as
n → +∞.
This implies −2 ∗ Er (wn ) = t−2 n Er (tn wn ) ≥ tn Er (α) ∀n ∈ N
by which, taking the limit as n → +∞ we get lim inf Er (wn ) ≥ Er∗ (α). n→+∞
By inserting last inequality in (4.24) we conclude −1 lim inf hN Ehn (un ) ≥ rN −1 Er∗ (α). n n→+∞
The desired conclusion follows.
2
Then we conclude this section with the following lemma. We will use it later (Remarks 5.1 and 5.2) to show how to get potentials V satisfying the hypotheses of our main results. Lemma 4.4 The following statements hold: i) The function r > 0 7−→ Er∗ (1) is continuous. ii) limr→+∞ rN −1 Er∗ (1) = +∞. iii) For every a, M > 0 there exists σ = σ(a, M ) > 0 independent on V such that for every r ∈ (0, a): Er∗ (1) < M ⇒ Er∗ (1) ≥ V(r)σ. 13
Proof. i) Fix b > a > 0 arbitrarily and take r ∈ [a, b] and u ∈ Υr (1). For every r ∈ [a, b] put tr =
(N −1)/p r r
. As easy computation shows that tr u ∈ Υr (1). Hence we have Er∗ (1)
≤ Er (tr u) =
t2r Er (u)
2(N −1)/p
r r
=
Er (u).
This implies: Er∗ (1)
sup r∈[a,b]
=
2(N −1)/p
r a
ωN
Z
≤
+∞
0
2(N −1)/p
r a
sup Er (u) = r∈[a,b]
|ϕ0u (s)|2
!
2
sup V(r) |ϕu (s)|
+
!
ds.
r∈[a,b]
So we deduce that the function r ∈ [a, b] 7−→ Er∗ (1) is bounded; then choose M > 0 such that Er∗ (1) < M for all r ∈ [a, b]. Hence the definition of Er implies that each Er∗ (1) can be also written in the form Er∗ (1)
≡ inf
+∞
Z Er (u) u ∈ Υr (1),
0
M |ϕu | ds ≤ ω N V0 2
Thus taken s, s0 ∈ (a, b) arbitrarily, choose u ∈ Υs0 (1) with 0 (N −1)/p s s
R
RN
.
|ϕu |2 ds ≤
M ωN V0 .
We have
u ∈ Υs (1) and it is Es∗ (1)
≤
≤ Es
2(N −1)/p
s0 s
s0 s
(N −1)/p !
u
Es0 (u) +
s0 s
=
s0 s
2(N −1)/p
2(N −1)/p
Es (u) =
|V(s) − V(s0 )|
M . V0
By taking the infimum on the right side we get Es∗ (1) ≤
s0 s
2(N −1)/p
Es∗0 (1) +
s0 s
2(N −1)/p
|V(s) − V(s0 )|
M . V0
Taking the limit for s0 → s and using the continuity of V: Es∗ (1) ≤ lim inf Es∗0 (1). 0 s →s
By changing the role of s and s0 we obtain the symmetric inequality Es∗0 (1)
≤
s s0
2(N −1)/p
Es∗ (1)
+
s s0
2(N −1)/p
|V(s) − V(s0 )|
M V0
by which Es∗ (1) ≥ lim sup Es∗0 (1) s0 →s
and the proof of i) ends. ii) Taking into account of the continuous immersion H 1 (0, +∞) ⊂ Lp (0, +∞) we infer the existence of a constant A > 0 such that Z 0
+∞
|ϕ|p ds
1/p
≤A
Z 0
+∞
|ϕ0 |2 + |ϕ|2 ds 14
1/2
∀ϕ ∈ H 1 (0, +∞).
Fix r > 0 arbitrarily and take u ∈ Υr (1). We have 1 2/p
+∞
Z
= r2(N −1)/p
ωN
0
|ϕu |p ds
2/p
≤ A2 r2(N −1)/p 1 V0
≤ A2 r2(N −1)/p max 1,
Z
0
+∞
|ϕ0u |2 + |ϕu |2 ds ≤
Er (u).
By taking the infimum for u ∈ Υr (1) we deduce rN −1 Er∗ (1)
1 ≥ max 1, V0
−1
1
r(p−2)(N −1)/p → +∞
2/p ωN A2
as
r → +∞.
iii) Fix a, M > 0 arbitrarily and take r ≤ a such that Er∗ (1) < M . By the definition of Er there follows n o Er∗ (1) ≡ inf Er (u) u ∈ Υr (1) ∩ Γ∗ (4.25)
where the set Γ∗ is given by Γ∗ ≡ {u ∈ Υ | dimensional Sobolev spaces yields |ϕu (s) − ϕu (s0 )| ≤
s
Z
|ϕ0u |2 ds
s0
1/2
R +∞ 0
|s − s0 |1/2 <
|ϕ0u |2 ds ≤
s
M ωN }.
M |s − s0 |1/2 ωN
The theory of the one-
∀u ∈ Γ∗ ,
∀s > s0 > 0,
by which it follows that the set of functions {ϕu | u ∈ Γ∗ } is equi-uniformly continuous in (0, +∞). This fact provides the existence of σ > 0 such that 1 ∀s, s0 ∈ (0, +∞), |s − s0 | < σ ⇒ |ϕu (s) − ϕu (s0 )| ≤ . 2
∀u ∈ Γ∗ ,
(4.26)
Now take u ∈ Γ∗ and distinguish two cases: either a) sups≥0 |ϕu (s)| > 1, or b) for every s ≥ 0 it is |ϕu (s)| ≤ 1. If a) holds true, there is su ∈ (0, +∞) verifying ϕu (su ) > 1. From (4.26) there follows |ϕu (s)| > 12 ∀s ∈ (su − σ, su + σ), and hence σ Er (u) ≥ ωN V(r) . 2 Consider case b). Then there results:
Er (u) ≥ ωN V(r)
Z
+∞
2
|ϕu | ds ≥ ωN V(r)
0
We have proved that if u ∈
Γ∗
Z
+∞
0
|ϕu |p ds = V(r)
1 1 ≥ V(r) N −1 . rN −1 a
it is
Er (u) ≥ V(r) min
σ 1 , 2 aN −1
By using (4.25), we obtain the thesis.
∀r ∈ (0, a). 2
Now we have in our hands all the instruments to prove in the next section the main existence and concentration result of this work. 15
5
Concentration Around a Sphere
In this section we solve the problem of finding a family of solutions {vh } to equation (1.2) exhibiting a concentration behaviour around a sphere {x ∈ RN | |x| = r}. Toward this end we use two devices. To get existence of solutions we introduce a truncation-penalization argument. So we modify the equation and the energy functional and we find, by a standard constrained minimization method, a solution for the modified equation. Soon after we show that, at least for small h, these solutions actually solve the original problem. This is for existence. To study concentration we use the auxiliary functional we have introduced in the previous section. Suppose assumptions a)-c) hold and assume r1 > r0 > 0 such that r1N −1 Er∗1 (1) > inf rN −1 Er∗ (1)
(5.27)
r≥r1
and inf
r0 ≤r≤r1
V(r) ≥
inf
r≥r1
2p/(p−2)
rN −1 Er∗ (1)
p V0
4/(p−2)
1 max 1, V0
26
.
2(N +1) 2 ωN
(5.28)
r0
As we said in the introduction, these hypotheses are quite involved. However we show in remarks 5.1 and 5.2 how to get some examples of potentials V satisfying them. In order to achieve our results we introduce a suitable penalization of equation (1.3) so that the concentration inside the ball B(0, r1 ) is avoided and an adequate balance in the concentration is achieved. 1/(p−2)
Let γ =
V0 p inf r≥r1 (rN −1 Er∗ (1))
and set
fγ (u) =
p−2 u |u|
if |u| ≤ γ
γ p−2 u
Now define
if |u| > γ.
gγ (x, ξ) = χB(0,r0 ) fγ (ξ) + 1 − χB(0,r0 ) |ξ|p−2 ξ. An immediate computation leads to |ξ| ≤ γ ⇒ gγ (x, ξ)ξ ≤ ξ 2 γ p−2
∀x ∈ RN .
(5.29)
Let us denote Gγ (x, ξ) = 0ξ gγ (x, τ ) dτ ≡ χB(0,r0 ) (x) 0ξ fγ (τ ) dτ + 1 − χB(0,r0 ) (x) It is easy to check that gγ is a Caratheodory function; then the functional R
R
u ∈ Hh,rad →
Z
RN
1 p p |ξ| .
Gγ (hx, u) dx
belongs to the class C 1 (Hh,rad , R) and its differential is not zero in every point except for zero. Hence if for every h > 0 we define the set Nh =
Z u ∈ Hh,rad p
RN
Gγ (hx, u) dx =
1 hN −1
,
by the implicit function theorem the set Nh is a C 1 -submanifold of Hh,rad . For every h > 0 define the infimum value ch as follows: ch = inf Eh (u). u∈Nh
The following lemma provides an estimate of the values ch . 16
Lemma 5.1 lim suph→0+ hN −1 ch ≤ inf r≥r1 rN −1 Er∗ (1). Proof. Consider r ≥ r1 arbitrarily and w ∈ Υr (1). For every h > 0, define vh : RN → R by setting if |x| ≤ hr , 0 vh (x) = w x − r x if |x| > hr . h |x| By using the convention (2.6) we can write
r h
ϕvh (s) = ϕw s −
∀s ≥
r . h
It is obvious that vh ∈ Hh,rad for every h > 0; then compute h
N −1
Z
p
RN
|vh | dx = ωN h
N −1
+∞
Z
N −1
s
r/h
r N −1
by which, since hN −1 s + supported, we have lim h
h→0+
|ϕvh (s)| ds = ωN h
Z
|vh | dx = ωN r
N −1
Z
p
RN
N −1
Z
+∞
0
r s+ h
N −1
|ϕw (s)|p ds
→ rN −1 uniformly on compact sets and ϕw is compactly
h
N −1
p
N −1
Z
+∞
0
|ϕw (s)|p ds = 1.
Now for every h > 0 put
th = h
p
RN
|vh | dx
−1/p
→1
h → 0+ .
as
Notice that vh (x) = 0 if |x| ≤ rh1 so that it is pGγ (hx, th vh ) = |th vh |p in RN , and hence there results: Z Z 1 p Gγ (hx, th vh ) dx = |th vh |p dx = N −1 , h RN RN i.e. th vh ∈ Nh . Hence we obtain lim sup hN −1 ch ≤ lim sup hN −1 Eh (th vh ) = h→0+
h→0+
= lim sup t2h hN −1 h→0+
r/h
h→0+
= ωN lim sup t2h hN −1 h→0+
≤
Z
+∞
s+
0
Z
ωN lim sup t2h hN −1 0 h→0+ Z
+ωN lim sup t2h hN −1 0 h→0+
RN
+∞
Z
= ωN lim sup t2h hN −1
Z
+∞
|∇vh |2 + Vh (x)|vh |2 dx =
sN −1 |ϕ0vh (s)|2 + Vh (s)|ϕvh (s)|2 ds =
r h
+∞
N −1
r s+ h
r s+ h
|ϕ0w (s)|2 + Vh s +
N −1
N −1
N −1
lim sup h h→0+
ch ≤ r
N −1
ωN
Z
0
+∞
ds ≤
|ϕ0w (s)|2 + V(r)|ϕw (s)|2 ds+
Vh
Notice that t2h hN −1 s + hr → rN −1 and Vh s + while ϕw is compactly supported, hence we deduce N −1
r |ϕw (s)|2 h
r h
r s+ h
− V(r) |ϕw (s)|2 ds.
→ V(r) uniformly on compact sets,
|ϕ0w (s)|2 + V(r)|ϕw (s)|2 ds ≡ rN −1 Er (w). 17
By taking the infimum on the right side of last inequality first for w ∈ Υr (1) and then for r ≥ r1 we achieve the desired conclusion. 2 Next object is to minimize the energy functional Eh in the manifold Nh . Lemma 5.2 Assume that hypotheses a)-c) hold. Then for every h > 0 the infimum ch is attained in the set Nh by a function uh ≥ 0. Proof. Fix h > 0 arbitrarily and consider uhk a minimizing sequence in Nh for the functional Eh ; it has obviously bounded energy. Then up to a subsequence we have uhk * uh
weakly in Hh,rad
k → +∞,
as
for some uh ∈ Hh,rad and, from lemma 2.1, uhk → uh
in Lp (RN , R)
as
k → +∞.
(5.30)
By (5.30) we deduce p
Z
RN \B(0,
→
r0 ) h
Z
Gγ (hx, uhk ) dx p
RN \B(0,
r0 ) h
|uh | dx = p
=
Z
RN \B(0,
Z
RN \B(0,
r0 ) h
r0 ) h
|uhk |p dx →
Gγ (hx, uh ) dx.
(5.31)
On the other hand an easy computation shows that 1 1 |Gγ (hx, ξ)| ≤ γ p + γ p−2 |ξ|2 p 2
∀x ∈ B 0,
r0 , ∀ξ ∈ R. h
Then the Nemitski operator u ∈ H 1 (B(0, rh0 )) 7→ Gγ (hx, u) ∈ L1 (B(0, rh0 )) is compact, by which, up to a subsequence, Z
B(0,
r0 ) h
Gγ (hx, uhk ) dx →
Z
B(0,
r0 ) h
Gγ (hx, uh ) dx.
(5.32)
Combining (5.31) and (5.32) we conclude p
Z
RN
Gγ (hx, uh ) dx = p lim
Z
k→+∞ RN
Gγ (hx, uhk ) dx =
1 hN −1
.
In other words we have proved that the manifold Nh is weakly closed, which implies uh ∈ Nh . By the weakly lower semicontinuity we obtain ch ≤ Eh (uh ) ≤ lim inf Eh (uhk ) = ch , k→+∞
i.e. uh is the desired minimizing function. We notice that since Eh (u) = Eh (|u|) we may assume that uh ≥ 0 for every h > 0. 2 By applying the Lagrange multiplier rule there exists λh ∈ R such that uh solves the − ∆uh + Vh (x)uh = λh gγ (hx, uh ).
(5.33)
Using the homogeneity of the functions which are involved in the definition of gγ we get the following two inequalities: 2Gγ (hx, ξ) ≤ gγ (hx, ξ)ξ ≤ pGγ (hx, ξ) ∀ξ ∈ R, ∀x ∈ RN . 18
(5.34)
By multiplying both members of (5.33) for uh and integrating by parts one obtains ch ≡
Z
RN
2
2
|∇uh | + Vh (x)|uh |
dx = λh
Z
RN
gγ (hx, uh )uh dx ≥ λh 2
Z
RN
Gγ (hx, uh ) dx =
2λh . phN −1
Lemma 5.1 assures that the family {λh } is bounded at least for small h, more precisely lim sup λh ≤ h→0+
p inf rN −1 Er∗ (1). 2 r≥r1
(5.35)
In the same way, using the first inequality of (5.34), we obtain, ch ≤ λh p
Z
RN
Gγ (hx, u) dx =
λh , N h −1
by which lim inf λh ≥ lim inf hN −1 ch . h→0+
(5.36)
h→0+
Now we want to study the behaviour of the family {uh } as h → 0+ . To this aim we need the result provided by the next two lemmas. Lemma 5.3
Assume that hypotheses a)-c) hold. Let uh ≥ 0 as in lemma 5.2 and fix
a ∈ (0, γ], with γ ≡
V0 p inf r≥r1 (r N −1 Er∗ (1))
1/(p−2)
. Then for small h there results:
r0 uh (x) ≤ a f or |x| = ⇒ uh (x) ≤ a hn
r0 ∀x ∈ B 0, . h
Proof. Since each uh solves (5.33), we can choose as a test function ψh (x) =
(uh − a)+ ≡ max{uh − a, 0}
0
if x ∈ B 0, rh0 ,
if x ∈ RN \ B 0, rh0 .
By hypothesis we easily infer that ψh is continuous on the sphere {x ∈ RN | |x| = rh0 }, hence it is continuous everywhere. It is standard to check that ψh ∈ Hh,rad . By construction there r0 results ∇uh≡ ∇ψh a.e. in B 0, h . Hence, taking into account that gγ (hx, ξ) = fγ (ξ) for x ∈ B 0, rh0 , after integration by parts, one gets Z
B (0,
r0 h
)
|∇ψh |2 + Vh (x)uh ψh − λh fγ (uh )ψh dx = 0.
The object is to show how last equality implies ψh ≡ 0 at least for small h. To this aim first observe that it holds uh ψh = |ψh |2 + aψh , hence we can write Z
B (0,
r0 h
)
|∇ψh |2 + Vh (x)|ψh |2 + aVh (x)ψh − λh fγ (uh )ψh dx = 0.
A direct computation shows that fγ (uh )ψh =
p−1 uh ψh ≤ γ p−2 (|ψh |2 + aψh )
γ p−2 uh ψh = γ p−2 (|ψh |2 + aψhn ) 19
if |uh | ≤ γ, if |uh | > γ.
(5.37)
Hence in every case there results fγ (uh )ψh ≤ γ p−2 (|ψh |2 + aψh ); so taking into account of (5.37) we can write Z
r 0, h0
B(
)
|∇ψh |2 + Vh (x) − λh γ p−2 (ψh2 + aψh ) dx ≤ 0.
(5.38)
The choice of γ gives Vh − λh γ
p−2
≥ V0
1 1 − λh p inf r≥r1 (rN −1 Er∗ (1))
by which, using (5.35), at least for h sufficiently small, Vh − λh γ p−2 ≥
V0 > 0. 3
Then the integral in (5.38) is nonnegative, hence it must be zero. We conclude in particular ψh ≡ 0, by which r0 uh ≤ a ∀x ∈ B 0, h and the thesis follows. 2 Lemma 5.4 Assume that hypotheses a)-c) hold and, in addition, (5.28). Let uh ≥ 0 as in lemma 5.2 and consider a generic sequence hn → 0+ . Then, for large n, there results: r1 ∀x ∈ B 0, hn
∃rn >
uhn (x) ≤ γ ≡
:
V0 p inf r≥r1 (rN −1 Er∗ (1))
1/(p−2)
;
r1 such that uhn (x) > γ f or |x| = rn . hn
Proof. For every n ∈ N let ϕn : [0, +∞) → R such that |x| = s.
ϕn (s) = uhn (x),
We begin by proving the first part. According to lemma 5.3 it is sufficient to prove that for large n it is r0 r1 ∀r ∈ , : ϕn (r) ≤ γ. (5.39) hn hn We argue by contradiction and suppose that, up to a subsequence ∀n ∈ N
∃rn ∈
r0 r1 , hn hn
1 dx ≤ max 1, V0
:
ϕn (rn ) > γ.
From lemma 5.1 we obtain Z
lim sup hnN −1 n→+∞ RN
= max 1,
2
2
|∇uhn | + |uhn |
1 V0
−1 lim sup hN chn ≤ max 1, n n→+∞
1 V0
−1 lim sup hN Ehn (uhn ) = n n→+∞
inf
r≥r1
rN −1 Er∗ (1) .
Hence by lemma 3.1 we get, at least for large n, ∀s, s0 ∈
r0 , +∞ : hn
1/2 N −1 E ∗ (1) inf r r≥r1 r 1 |ϕn (s)−ϕn (s0 )| ≤ 2 max 1, |s−s0 |1/2 . N −1
20
V0
ω N r0
Now choose δ > 0 verifying
1 2 max 1, V0
1/2 N −1 E ∗ (1) inf r≥r1 r r γ δ 1/2 = . N −1
2
ω N r0
Then fix s ∈ (rn − δ, rn + δ) and compute ϕn (s) ≥ ϕn (rn ) − |ϕn (s) − ϕn (rn )| ≥ γ − Let us put In = (rn − δ, rn + δ) ∩
rn ∈
r0 r1 hn , h n
−1 ≥ ω N hN n
≥ ωN
inf
r0 ≤s≤r1
r0 r1 hn , hn
; since
r1 hn
r0 hn
−
→ +∞ as n → +∞ and since
it is obvious that for large n it is meas(In ) ≥ δ. This implies
−1 hN chn ≥ hnN −1 n
γ γ = . 2 2
inf
r0 ≤s≤r1
Z
RN
−1 Vhn (x)u2hn dx ≥ ωN hN n
Z
V(s)
V(s) r0N −1 δ
γ2 4
In
sN −1 ϕ2n ds ≥ ωN
= ωN r0N −1
inf
r0 ≤s≤r1
In
sN −1 Vhn (s)ϕ2n ds ≥
inf
r0 ≤s≤r1
V(s)
Z
V(s) r0N −1
16
In
ϕ2n ds ≥
−1 N −1 E ∗ (1) inf r r≥r r 1 1 2 max 1, . N −1
γ4
Z
V0
ω N r0
By the definition of γ and by using assumption (5.28) we obtain for n sufficiently large hnN −1 chn ≥ 2 inf
r≥r1
rN −1 Er∗ (1) ,
but taking the limit as n → +∞ we achieve a contradiction with lemma 5.1. Hence (5.39) follows. We go on with the proof of the second part of the lemma. Assume by absurd that, eventually passing to a subsequence, it is uhn (x) ≤ γ for every r > hr1n , which implies, taking into account of the first part of the proof, uhn (x) ≤ γ
∀x ∈ RN .
By using (5.29) it follows: gγ (hn x, uhn )uh ≤ u2hn γ p−2 . Multiplying both members of (5.33) for uhn and integrating by parts we easily obtain Z
RN
|∇uhn |2 + Vhn (x)|uhn |2 dx = λhn
Z
RN
gγ (hn x, uhn )uhn dx ≤ λhn γ p−2
Z
RN
u2hn dx
by which Z
RN
|∇uhn |2 + (Vhn (x) − λhn γ p−2 )|uhn |2 dx ≤ 0.
By repeating the same arguments we have used at the end of lemma 5.3 at least for large n we get Vhn (x) − λhn γ p−2 ≥ V30 > 0; so the integral in the above inequality is strictly positive for n sufficiently large and this is a contradiction. 2 Now we are able to give the main result of this section.
21
Theorem 5.1 Assume that hypotheses a)-c) hold and, in addition, (5.27) and (5.28). For every h > 0 we set x (5.40) vh (x) = uh h where uh ∈ Nh is the minimizing function for ch , i.e. Eh (uh ) = ch . Then vh ≥ 0 and the family {vh } exhibits the following concentration behaviour as h → 0+ : for each sequence hn → 0+ there exists a subsequence, still denoted by hn , such that vhn has a sphere of local
maximum points {x ∈ RN | |x| = rhn } with vhn (x) > γ ≡
V0 p inf r≥r1 (r N −1 Er∗ (1))
1/(p−2)
for
|x| = rhn and rhn → r > r 1 ,
−1 rN −1 Er∗ (1) = inf rN −1 Er∗ (1) = lim hN chn ; n n→+∞
r≥r1
also, for every δ > 0, it holds: vhn → 0
n → +∞
as
o n |x| − r ≥ δ .
unif ormly in the set
Proof. We focus our attention on a generic sequence hn → 0+ . For every n ∈ N let ϕn : [0, +∞) → R such that ϕn (s) = uhn (x), |x| = s. The object is to apply lemma 3.2 to the sequence {uhn } with γ in place of ε and r1 in place of t. Indeed we have −1 lim sup hN n n→+∞
Z
RN
|∇uhn |2 + |uhn |2 dx ≤ max 1,
1 V0
inf rN −1 Er∗ (1) < +∞.
r≥r1
Furthermore lemma 5.3 and 5.4 assure that hypotheses (3.10) and (3.11) are satisfied too. Hence we get the existence of ` ∈ N, τ > 0, R1 , . . . , R` > 0 and ` sequences of positive numbers {rh1 n }, . . . , {rh` n }, with rhi n > hr1n , such that, up to subsequence, {hn rhi n } is bounded
∀i = 1, . . . , `;
(5.41)
ϕn (rhi n ) > γ; rhi n is the maximum point for ϕn
in (0, +∞) \
(5.42) [
(rhj n − Rj , rhj n + Rj );
(5.43)
j
∀ r 6∈
` [
(rhi n − Ri , rhi n + Ri ) :
ϕn (r) ≤ γ;
(5.44)
i=1
|rhi n − rhj n | → +∞ as n → +∞ ∀i, j ∈ {1, . . . , `} with i 6= j; (5.45) γ ϕn (s) > ∀s ∈ (rhi n − τ, rhi n + τ ) ∀i ∈ {1, . . . , `}. (5.46) 2 Before going on we fix some notations. Consider R > 2 maxi=1,...,` Ri and n sufficiently large such that, according to (5.45), it is C(rhi n − R, rhi n + R) ∩ C(rhj n − R, rhj n + R) = ∅ for i 6= j; then it makes sense to consider ηnR ∈ C ∞ (RN , R) a radial function so that ηnR
N
≡ 1 on R \
` [
C(rhi n
−
R, rhi n
+ R),
ηnR
≡ 0 in
i=1
` [
i=1
0 ≤ ηnR ≤ 1, 22
|∇ηnR | ≤
c R
C
rhi n
R R − , rhi n + , 2 2
where c is a constant independent of R and n. Crucial steps in the proof of the theorem are the following 4 claims. Claim 1. For every η > 0 there exists Rη > 2 maxi=1,...,` Ri such that for every R > Rη and large n, Z −1 2 2 R hN |∇u | + u hn n hn ηn dx < η RN
and s 6∈
` [
(rhi n − R, rhi n + R) ⇒ ϕn (s) ≤ η.
i=1
Fix r > 0 arbitrarily and consider I an interval with length r; by the theory of the one-dimensional Sobolev spaces the immersion H 1,2 (I) ⊂ L∞ (I) is continuous and then ∀ϕ ∈ H 1,2 (I)
kϕkL∞ (I) ≤ AkϕkH 1,2 (I)
where A is a constant depending only on r but not on the location of I on real line. Then take R > 2 maxi=1,...,` Ri and, by using ηnR uhn as a test function in (5.33), we get Z
RN
|∇uhn |2 ηnR
+
Vhn (x)u2hn ηnR
Notice that the definition of by which, using (5.29),
ηnR
−
λhn gγ (hn x, uhn )uhn ηnR
and (5.44) imply that
ηnR
dx = −
Z
RN
h∇uhn , ∇ηnR iuhn dx.
≡ 0 in the set {x ∈
RN
(5.47) | uhn (x) > γ}
gγ (hn x, uhn )uhn ηnR ≤ u2hn γ p−2 ηnR . Combining last inequality together with (5.47) we deduce Z
RN
2
|∇uhn | + (Vhn (x) − λhn γ
p−2
)u2hn
ηnR dx
≤−
Z
RN
h∇uhn , ∇ηnR iuhn dx.
From the choice of γ we immediately infer Vhn (x) − λhn γ p−2 ≥ inequality in (5.48) and using H¨older inequality we have Z
RN
c V0 |∇uhn | + u2hn ηnR dx ≤ 3 R
2
Z
RN
c |∇uhn |uhn dx ≤ R
V0 3
Z
RN
(5.48)
for large n. Inserting last
2
1/2 Z
|∇uhn | dx
RN
u2hn
dx
−1 there results: By multiplying both members for hN n −1 hN n
Z
RN
V0 c −1 |∇uhn | + u2hn ηnR dx ≤ hN n 3 R
2
Z
RN
2
|∇uhn | dx
1/2
−1 hN n
Z
RN
u2hn
dx
1/2
.
(5.49) Notice that by lemma 5.1 and by the definition of Eh it follows −1 lim sup hN n n→+∞
Z
RN
−1 |∇uhn |2 ≤ lim sup hN Ehn (uhn ) = lim sup hnN −1 chn < +∞. n n→+∞
n→+∞
In the same way −1 lim sup hN n n→+∞
Z
RN
23
u2hn < +∞.
1/2
.
−1 2 N −1 2 The boundedness of the terms hN n RN |∇uhn | dx and hn RN uhn dx in (5.49) give the first part of the claim. We go on with the proof of the second part. Given η ∈ (0, γ), for R > 0 sufficiently big and for large n it is
R
−1 hN n
Z
RN
R
|∇uhn |2 + u2hn ηnR dx < r0N −1 ωN
η2 . A2
Setting τnR (|x|) = ηnR (x), the above inequality can be rewritten in terms of the functions ϕn in the following way: +∞
Z
r0 /hn
−1 hN n ≤ N r0 −1
Z
+∞
|ϕ0n |2 + ϕ2n τnR ds ≤
N −1
s
0
|ϕ0n |2
+
ϕ2n
−1 hN n N −1 r0
τnR ds
Z
+∞
r0 /hn
sN −1 |ϕ0n |2 + ϕ2n τnR ds ≤
−1 hN n = ωN r0N −1
Z
RN
|∇uhn |2 + u2hn ηnR dx <
η2 . A2 (5.50)
Suppose n sufficiently large such that the following statements holds: |rhi n − rhj n | > r
rhi n −
r0 >r hn
∀i, j = 1, . . . , `, i 6= j,
(5.51)
where r > 0 has been introduced at the beginning of the step. The first inequality of (5.51) follows directly from (5.45); the second holds true provided that n is sufficiently large since S rhi n − hr0n ≥ hr1n − hr0n → +∞. Now consider s ∈ ( hr0n , +∞)\ `i=1 (rhi n −R, rhi n +R); by using (5.51) S we can find an interval I with length r such that s ∈ I ⊂ ( hr0n , +∞) \ `i=1 (rhi n − R, rhi n + R), and then Z ϕn (s) ≤ kϕn kL∞ (I) ≤ Akϕn kH 1,2 (I) = A
I
|ϕ0n |2 + ϕ2n ds
1/2
;
notice that by construction it is τnR ≡ 1 in I, so from (5.50) we get ϕn (s) ≤ A
Z I
|ϕ0n |2
+
ϕ2n
τnR ds
1/2
≤A
Z
+∞
r0 /hn
|ϕ0n |2
+
ϕ2n
τnR ds
!1/2
≤ η.
We have proved that for n sufficiently large s∈
` [ r0 , +∞ \ (rhi n − R, rhi n + R) ⇒ ϕn (s) ≤ η. hn i=1
On the other hand we have assumed η ≤ γ, by which, taking into account of lemma 5.3, we obtain the second part of the claim. Claim 2. For every δ > 0 there exists R(δ) > 2 maxi=1,...,` Ri such that, for every R ≥ R(δ) and for large n it holds −1 hN n
Z
RN
|(1 −
ηnR )uhn |p dx
≥
−1 hN n
Z
RN
|uhn |p − δ
and −1 −1 hN Ehn ((1 − ηnR )uhn ) ≤ hN Ehn (uhn ) + δ. n n
24
Fix δ ∈ (0, 1) arbitrarily. According to last step for R > 8 maxi=1,...,` Ri sufficiently big and for large n Z −1 hN n
and s 6∈
RN
` [
rhi n
i=1
Now observe hnN −1
Z
RN
|∇uhn |2 + u2hn ηnR/2 dx < δ
(5.52)
R R − , rhi n + 4 4
(5.53)
|ηnR/2 uhn |p dx
⇒ ϕn (s) ≤ δ.
Z
−1 hN n
=
RN
uh ηnR/2 n δ
p
δ p dx.
R i R 4 , rhn + 4 ), while, by construction, R/2 is ηn uhn ≤ δ everywhere, so we get
Notice that by (5.53) it is un ≤ δ in RN \ ∪`i=1 C(rhi n − R/2
ηn
≡ 0 in ∪`i=1 C(rhi n − R4 , rhi n + R4 ); in other words, it −1 hN n
Z
RN
−1 |ηnR/2 uhn |p dx ≤ hN n
Z
RN
ηnR/2 u2hn δ p−2 dx,
by which, taking into account of (5.52), −1 hN n
Z
RN
|ηnR/2 uhn |p dx ≤ δ p−1 ≤ δ. R/2
On the other hand by construction for every x ∈ RN there results 1 − ηnR (x) = 1 or ηn (x) = R/2 1; this implies |ηn uhn |p + |(1 − ηnR )uhn |p ≥ |uhn |p and then we conclude −1 hN n
Z
RN
|(1 −
ηnR )uhn |p dx
≥
Z
−1 hN n
−1 ≥ hN n
p
RN
Z
RN
|uhn | dx −
−1 hN n
Z
RN
|ηnR/2 uhn |p dx ≥
|uhn |p dx − δ
and the first part of the claim follows. As regards the second part, for sake of simplicity put CnR
=
` [
C(rhi n
−
R, rhi n
` [
!
+ R) \
i=1
C
rhi n
i=1
R R − , rhi n + 2 2
!
.
In other words CnR is the set where ηnR is different from 0 and 1. Observe −1 hN n
Z
R Cn
2 N −1 ∇ (1 − ηnR )uhn dx ≤ hn
c2 |∇uhn | dx+ 2 R R Cn
Z
2
c |uhn | dx+2 R R Cn
Z
2
!
Z
R Cn
|∇uhn ||uhn | dx .
H¨older inequality yields −1 hN n
Z
R Cn
c2 |∇uhn | dx + 2 R R Cn
Z
2 −1 ∇ (1 − ηhRn )uhn dx ≤ hN n
c +hnN −1 2 R
Z
R Cn
|∇uhn |2 dx
2
!1/2
Z
R Cn
|uhn | dx +
!1/2
Z
R Cn
|uhn |2 dx
−1 2 N −1 by which, since by lemma 5.1 supn∈N hN R |∇uhn | dx < +∞ and supn∈N hn n Cn +∞, it follows that for some constant C1 it holds:
R
−1 hN n
Z
R Cn
|∇((1 −
ηnR )uhn )|2 dx
≤
−1 hN n
Z
25
R Cn
!
2
2
|∇uhn | dx + C1
1 1 + 2 . R R
R
R Cn
|uhn |2 dx <
(5.54)
By (5.54), the thesis of claim 1 easily follows. Indeed take R(δ) > 0 sufficiently large such 1 1 < δ. From (5.54) it immediately follows that for every R ≥ R(δ) and that C1 R(δ) + R(δ) 2 for large n Z Z −1 hN n
R Cn
−1 |∇((1 − ηnR )uhn )|2 dx ≤ hN n
|∇uhn |2 dx + δ.
R Cn
On the other hand by construction we have (1 − ηnR )uhn is equal to 0 or to uhn in RN \ CnR . Thus one computes −1 hN n
Z
RN
−1 +hN n
Z
−1 |∇((1 − ηnR )uhn )|2 = hN n
Z
RN \CnR
|∇((1 −
ηnR )uhn )|2 dx
R Cn
≤
|∇((1 − ηnR )uhn )|2 dx+ Z
hnN −1
RN
|∇uhn |2 dx + δ.
(5.55)
Taking into account that 0 ≤ 1 − ηhRn ≤ 1 we also have Z
RN
Vhn (x)|(1 − ηhRn )uhn |2 dx ≤
Z
RN
Vhn (x)|uhn |2 dx.
(5.56)
Combining (5.55) and (5.56) we obtain −1 −1 hN Ehn ((1 − ηnR )uhn ) ≤ hN Ehn (uhn ) + δ n n
and this completes the proof of claim 2. By the inequality rhi n > hr1n for every i = 1, . . . , ` and by (5.41) we get the existence of r1 , . . . , r` ≥ 0 verifying, up to subsequence, hn rhi n → ri
as
n → +∞,
ri ≥ r1 ∀i = 1, . . . , `.
(5.57)
−1 ∗ −1 c . Er1 = inf r>r1 rN −1 Er∗ = limn→+∞ hN Claim 3. There results ` = 1 and rN hn n 1
According to last claim, let R1 > 2 maxi=1,...,` Ri such that for large n −1 hN n
Z
RN
|(1 −
ηnR1 )uhn |p dx
≥
−1 hN n
Z
RN
1 |uhn |p − , 2
(5.58)
1 2
(5.59)
C(rhi n − R1 , rhi n + R1 ) ⇒ uhn (x) ≤ γ.
(5.60)
−1 hnN −1 Ehn ((1 − ηnR1 )uhn ) ≤ hN Ehn (uhn ) + n
and, by claim 1, x 6∈
` [
i=1
Because of the inequality rhi n >
r1 hn ,
we have rhi n −
r0 hn
→ +∞. Hence, taking into account
of (5.45), it makes sense to assume C rhi n − R1 , rhi n + R1 ∩ C rhj n − R1 , rhj n + R1 = ∅ for
i 6= j and C rhi n − R1 , rhi n + R1
⊂ RN \ B 0, hr0n . In particular, since by construction
1 − ηnR1 ≡ 0 in RN \ ∪`i=1 C rhi n − R1 , rhi n + R1 , we can write (1 − ηnR1 )uhn = we have put for every i = 1, . . . , `, ein ≡ (1 − ηnR1 )uhn χC (ri u
hn
26
i +R . −R1 ,rh 1) n
P`
ein i=1 u
where
ein ∈ Hhn ,rad . The object is to apply lemma 4.3 to each sequence u ein . There results: u To this aim notice that (5.60) and the definition of Gγ leads to pGγ (hn x, uhn ) = |uhn |p and pGγ (hn x, (1 − ηnR1 )uhn ) = |(1 − ηnR1 )uhn |p ; then, since uhn ∈ Nhn and using (5.58), we get for large n
1 −1 = hN n 2 ≤
−1 hN n
Z
RN
|(1 −
Z
pGγ (hn x, uhn ) dx −
RN
ηnR1 )uhn |p dx
≤
Z
−1 hN n
1 −1 = hN n 2 p
RN
|uhn | dx =
Z
RN
−1 hN n
|uhn |p −
Z
RN
1 ≤ 2
pGγ (hn x, uhn ) dx = 1,
by which −1 hN n
Z
|(1 − ηnR1 )uhn |p dx ∈
RN
1 ,1 . 2
(5.61)
i ) for some αi ∈ (0, 1]. Up to a subsequence ein ∈ Mhn (αn Now fix i = 1, . . . , `; we have u n i we may assume αn → αi ∈ [0, 1]. What we are going to prove now is that αi > 0: indeed by (5.46) we obtain
αni = −1 hN n
Z
RN
ein |p dx |u
= ωN
Z
i +R rh 1 n
i −R rh 1
sN −1 |ϕn (s)|p ds ≥ ωN (rhi n − R1 )N −1
n
γp min{2R1 , 2τ }. 2p p
−1 (r i − R )N −1 γ min{2R , 2τ } ≥ As a consequence lim inf n→+∞ αni ≥ ωN lim inf n→+∞ hN 1 1 n hn 2p p −1 γ i and gives e ωN r N min{2R , 2τ } > 0. Hence lemma 4.3 applies to the sequence u 1 n i 2p −1 ∗ −1 ein ) ≥ r N lim inf hN Ehn (u Eri (αi ) ∀i ∈ {1, . . . , `}. n i n→+∞
Furthermore, by (5.61), ` X
αi = lim
n→+∞
i=1
` X
αni = lim
n→+∞
i=1
−1 = lim hN n n→+∞
Z
RN
` X
−1 hN n
Z
i=1
eihn |p dx = |u
RN
|(1 − ηnR1 )uhn |p dx ∈
1 ,1 . 2
(5.62)
Combining lemma 5.1 together with (5.59) and the above inequalities one computes −1 −1 inf rN −1 Er∗ (1) ≥ lim sup hnN −1 chn ≥ lim inf hN chn = lim inf hN Ehn (uhn ) ≥ n n
r≥r1
n→+∞
n→+∞
n→+∞
−1 ≥ lim inf hN Ehn ((1 − ηnR1 )uhn ) − n n→+∞
≥
` X i=1
−1 eihn ) lim inf hN Ehn (u n→+∞ n
1 ≥ 2
` 1 X 1 − ≥ rN −1 Er∗i (αi ) − . 2 i=1 i 2
Hence lemma 4.2 allows us to infer −1 inf rN −1 Er∗i (1) ≥ lim sup hN chn ≥ lim inf hnN −1 chn ≥ n
r≥r1
n→+∞
n→+∞
` X
1 2/p −1 ∗ αi rN Eri (1) − . i 2 i=1
(5.63)
e1hn }, . . . , {u e`hn } and of the ` positive numbers α1 , . . . , α` The construction of the ` sequences {u depends on the particular R1 > 2 maxi=1,...,` Ri chosen at the beginning of the claim. To
27
e1,1 e1,` emphasize this fact we denote them as {u hn }, . . . , {u hn } and α1,1 , . . . , α1,` . We proceed by choosing, according to last claim, R2 > 2R1 such that for large n −1 hN n
Z
RN
−1 |(1 − ηnR2 )uhn |p dx ≥ hN n
Z
RN
1 |uhn |p − , 4
(5.64)
1 −1 −1 Ehn (uhn ) + , hN Ehn ((1 − ηnR2 )uhn ) ≤ hN n n 4
(5.65)
and, by claim 1, x 6∈
` [
C(rhi n − R2 , rhi n + R2 ) ⇒ uhn (x) ≤ γ.
(5.66)
i=1
Then set, for every i = 1, . . . , `, R2 e2,i u n ≡ (1 − ηn )uhn χC (r i
hn
i +R −R2 ,rh 2) n
∈ Hhn ,rad .
Hence we repeat the same arguments we have used above and, for large n, by (5.66) we deduce pGγ (hn x, uhn ) = |uhn |p and pGγ (hn x, (1 − ηnR2 )uhn ) = |(1 − ηnR2 )uhn |p , then, since uhn ∈ Nhn and using (5.64), we compute 3 −1 = hN n 4
Z
RN
1 −1 pGγ (hn x, uhn ) dx − = hN n 4 −1 ≤ hN n
Z
RN
1 −1 |uhn | − ≤ hN n 4
Z
p
RN
−1 |uhn |p dx = hN n
Z
RN
Z
RN
|(1 − ηnR2 )uhn |p dx ≤
pGγ (hn x, uhn ) dx = 1,
by which hnN −1
Z
RN
|(1 −
ηnR2 )uhn |p dx
3 ∈ ,1 . 4
(5.67)
We notice that, being R2 > 2R1 , by construction there results 1 − ηnR2 ≥ 1 − ηnR1 in RN ; as 2,i e1,i e2,i a consequence for every i = 1, . . . , ` it follows u hn ≥ u hn and so, setting αn > 0 such that 2,i 2,i 2,i en ∈ Mhn (αn ), it is αn → α2,i ≥ α1,i . The analogous of (5.62) and (5.63) for the sequences u e2,i u hn gives ` X
α2,i ∈
i=1
3 ,1 , 4
−1 −1 inf rN −1 Er∗ (1) ≥ lim sup hN chn ≥ lim inf hN chn ≥ n n
r≥r1
n→+∞
n→+∞
` X
1 2/p −1 ∗ α2,i rN Eri (1) − . i 4 i=1
By a recursive argument for every k we find ` numbers αk,1 , . . . , αk,` , with αk,i ≥ αk−1,i > 0 for every i = 1, . . . , ` and verifying ` X i=1
αk,i
1 ∈ 1 − k,1 , 2
−1 −1 inf rN −1 Er∗ (1) ≥ lim sup hN chn ≥ lim inf hN chn ≥ n n
r≥r1
n→+∞
n→+∞
(5.68) ` X
2/p
−1 ∗ αk,i rN Eri (1) − i
i=1
1 . 2k
(5.69)
Now we have in our hand all the instrument to achieve the desired conclusion. Indeed by construction for every i = 1, . . . , ` the sequence {αk,i } ⊂ (0, 1] is nondecreasing, then put βi = sup αk,i = lim αk,i ∈ (0, 1]. k≥1
k→+∞
28
By taking the limit for k → +∞ in (5.68) and (5.69) there results ` X
−1 −1 inf rN −1 Er∗ (1) ≥ lim sup hN chn ≥ lim inf hN chn ≥ n n
βi = 1,
r≥r1
i=1
n→+∞
n→+∞
` X
2/p N −1 ∗ ri Eri (1).
βi
i=1
(5.70) We recall that ri ≥ r1 for all i = 1, . . . , `; then the second of (5.70) gives inf
r≥r1
rN −1 Er∗ (1)
` X
≥
2/p βi
i=1
!
inf rN −1 Er∗ (1).
r≥r1
If by absurd it was ` > 1, then it would result βi ∈ (0, 1) for every i = 1, . . . , `, and P P 2/p consequently `i=1 βi > `i=1 βi = 1, in contradiction with the above inequality. Hence ` must be equal to 1. Again by (5.70) we obtain −1 ∗ −1 Er1 (1), inf rN −1 Er∗ (1) ≥ lim sup hN chn ≥ lim inf hnN −1 chn ≥ rN n 1
r≥r1
n→+∞
n→+∞
N −1 ∗ −1 c which implies inf r≥r1 rN −1 Er∗ (1) = limn→+∞ hN Er1 (1). hn = r 1 n
Claim 4. End of the proof. In order to conclude we have just only to combine the results obtained in the previous claims. Indeed we already know that for every h > 0 the infimum ch is attained by a function uh ∈ Nh and uh ≥ 0 in RN . Furthermore, considered a generic sequence hn → 0+ , (5.42), (5.43), (5.57) and last claim give that, up to a subsequence, uhn has a sphere of local maximum points
n
o
x ∈ RN |x| = rh1 n ;
uhn (x) > γ ∀x ∈ RN with |x| = rh1 n ; hn rh1 n → r1 ≥ r1 ,
−1 ∗ rN Er1 (1) = inf rN −1 Er∗ (1); 1 r≥r1
moreover from claim 1 for every η > 0 there exists R > 0 such that, for large n, uhn (x) < η ∀x 6∈ C(rh1 n − R, rh1 n + R).
Setting vhn (x) = uhn hxn and rhn = hn rh1 n , we have vhn ≥ 0 in RN and, using a rescaling argument, the results above can be rewritten as follows: vhn has a sphere of local maximum points {x ∈ RN | |x| = rhn }, vhn (x) > γ ∀x ∈ RN with |x| = rhn ; rhn → r ≥ r1 , rN −1 Er∗ (1) = inf rN −1 Er∗ (1), r≥r1
lim sup
sup
n→+∞ x6∈C(rhn −δ,rhn +δ)
vhn (x) = 0
∀δ > 0
which exactly correspond to the thesis.
2
Finally we can give the main existence and concentration result for equation (1.2).
29
Theorem 5.2 Assume that hypotheses a)-c) hold and, in addition, (5.27) and (5.28). Then there exists h0 > 0 such that for every h ∈ (0, h0 ) there is a positive solution vbh to equation (1.2). Furthermore the family {vbh } exhibits the following concentration behaviour as h → 0+ : for each sequence hn → 0+ there exists a subsequence, still denoted by hn , such that vbhn has a sphere of local maximum points {x ∈ RN | |x| = rhn } with vbhn (x) > γb ≡ for |x| = rhn and
V0 2p
1/(p−2)
−1 rN −1 Er∗ (1) = inf rN −1 Er∗ (1) = lim hN chn ; n
rhn → r > r 1 ,
n→+∞
r≥r1
also, for every δ > 0, it holds: vbhn → 0
n → +∞
as
unif ormly in the set
n o |x| − r ≥ δ .
Proof. The proof is just only an application of theorem 5.1. We recall that each function uh solves equation (5.33). Hence, by rescaling, we immediately obtain that the function vh related to uh by (5.40), satisfies: −h2 ∆vh + V (x)vh = λh gγ (x, vh ). On the other hand by last theorem we deduce that, for every sequence hn → 0+ up to a subsequence {vhn } concentrate around a sphere of radius major then r1 , so it holds vhn → 0+ uniformly in the ball B(0, r1 ). The arbitrariness of the sequence hn allows us to conclude vh → 0+ uniformly in B(0, r1 ). Hence, provided that h is sufficiently small, we get gγ (x, vh ) ≡ |vh |p−2 vh , so for small h vh provides a solution to equation: − h2 ∆vh + V (x)vh = λh |vh |p−2 vh .
(5.71) 1/(p−2)
Using the homogeneity of equation (5.71) we deduce that vbh ≡ λh vh solves the original N −1 equation (1.2). We notice that by theorem 5.1 it follows limh→0+ h ch = inf r≥r1 rN −1 Er∗ (1), hence, by (5.35) and (5.36), lim sup λh ∈ h→0+
inf
r≥r1
rN −1 Er∗ (1),
p inf rN −1 Er∗ (1) . 2 r≥r1
(5.72)
Now consider a generic sequence hn → 0+ . By (5.72), provided n is sufficiently large, we can assume 1 N −1 ∗ N −1 ∗ λ hn ∈ inf r Er (1), p inf r Er (1) . r≥r1 2 r≥r1 Hence by the definition of vbhn and by theorem 5.1 we easily deduce, up to a subsequence, vbhn has a sphere of local maximum points {x ∈ RN | |x| = rhn },
∀x ∈ RN with |x| = rhn :
1/(p−2)
vbhn (x) > γλhn
≥γ
1/(p−2)
1 inf rN −1 Er∗ (1) 2 r≥r1
rhn → r ≥ r1 , rN −1 Er∗ (1) = inf rN −1 Er∗ (1), r≥r1
∀δ > 0 :
lim sup
sup
n→+∞ x6∈C(rhn −δ,rhn +δ)
30
vbhn (x) ≤
≡
V0 2p
1/(p−2)
;
≤ p inf
r≥r1
1/(p−2)
rN −1 Er∗ (1)
lim sup
sup
n→+∞ x6∈C(rhn −δ,rhn +δ)
vhn (x) = 0.
Finally standard regularity argument shows that vbh is a classical solution of equation (1.2). Now, since vbh ≥ 0, we use the strong maximum principle to conclude that vbh > 0 in RN . 2
Remark 5.1 By i) of lemma 4.4 we immediately obtain that the function r ≥ r1 7→ rN −1 Er∗ (1) is continuous and, by ii), rN −1 Er∗ (1) → +∞ as r → +∞. This fact, combined with (5.27), implies that the infimum inf r≥r1 rN −1 Er∗ (1) is attained by some r > r1 . If such minimum point r is unique, then all the bound states {vbh } we have found in last theorem concentrate at the sphere {x ∈ RN | |x| = r} as h → 0+ in the sense specified in theorem 5.1. In general rhN −1 Er∗h (1) → inf r≥r0 rN −1 Er∗ (1) as h → 0+ , where the circle {x ∈ RN | |x| = rh } is a set of local maxima of vbh for small h. Remark 5.2 We conclude by observing that if we require the function V(s) to be sufficiently big for r0 ≤ s ≤ r1 , then (5.27) and (5.28) holds true. Indeed, fixed r > r1 and V(r) arbitrarily, then we can choose V large enough in [r0 , r1 ] so that inf
r0 ≤r≤r1
≥
2p/(p−2) p 4/(p−2)
V(r) ≥ rN −1 Er∗ (1)
V0
2p/(p−2)
inf rN −1 Er∗ (1)
r≥r1
p V0
4/(p−2)
max 1,
max 1,
1 V0
1 V0
26 2(N +1) 2 ωN
r0
26 2(N +1) 2 ωN
≥
.
r0
Then (5.28) is satisfied. As regards (5.27), apply iii) of lemma 4.4 with a = r1 and M =
r r1
N −1
Er∗ (1) : what we deduce is the existence of σ > 0 independent on V such that: Er∗1 (1)
<
r r1
N −1
Er∗ (1) ⇒ Er∗1 (1) ≥ V(r1 )σ.
This implies that provided V(r1 ) is large enough so as to verify V(r1 )σ > then the inequality Er∗1 (1) >
r r1
N −1
Er∗ (1) is easily satisfied.
31
r r1
N −1
Er∗ (1),
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