On the shape of the solutions of some semilinear elliptic problems

On the shape of the solutions of some semilinear elliptic problems ∗ Massimo Grossi†

Riccardo Molle‡

May 9, 2001

Abstract This paper deals with semilinear elliptic problems of Dirichlet type, in star shaped domains. An abstract result is stated, which gives sufficient conditions for a positive solution of the problem to have strictly star shaped superlevels. Moreover, it turns out that the maximum point is the only critical point for these solutions. Then we apply the result to the “single-peak” solutions of some widely studied problems. First a nonlinear and subcritical elliptic equation is considered, when the nonlinearity approaches the critical one. Then Schr¨odinger type problems are studied. Finally, the case when the potential is constant is also analyzed, on a bounded domain. Key words Starshaped domains. Asymptotic behavior. Maximum principles. A. M. S. subject classification 2000 35B38, 35B50, 35J10, 35J60 ∗

First author is supported by M.U.R.S.T., project “Variational methods and nonlinear differential equations”. Second author is supported by M.U.R.S.T., project “Metodi variazionali e topologici nello studio di fenomeni non lineari”. † Dipartimento di Matematica, Universit`a di Roma La Sapienza” P.le Aldo Moro, 2 00185 Roma. ‡ Dipartimento di Matematica, Universit`a Di Roma ”Tor Vergata”, Via della Ricerca Scientifica - 00133 Roma.

1

1

Introduction

In this paper we study the shape of the level sets of solutions of    −∆u = f (x, u)

u>0   u=0

in Ω in Ω on ∂Ω,

(1.1)

where Ω is a smooth (possibly unbounded) domain of IRN , N ≥ 2 and f ∈ C 1 (IR+ , IR). This topic is widely studied in literature. Let us start by recalling the celebrated Gidas-Ni-Nirenberg theorem ( see [13] or also [3], for a more general statement). Theorem 1.1 (see [13] or [3]) Let us consider a smooth solution of the following problem   in Ω  −∆u = f (u) u>0 in Ω (1.2)   u=0 on ∂Ω, where Ω ⊂ IRN is a smooth and bounded domain which is convex in the directions x1 , . . . , xN and symmetric with respect to the planes xi = x0i . Here f : IR+ → IR is a locally lipschitz continuous function. Then (x − x0 ) · ∇u < 0 ∀x ∈ Ω \ {x0 },

(1.3)

where x0 = (x01 , . . . , x0N ). In other words, (1.3) states that the superlevel sets of the solutions of (1.2) are strictly starshaped with respect to x0 . In particular x0 is the only critical point for u. In this paper we deal with convex (or only starshaped) domains Ω which are not necessarily symmetric and we again try to obtain inequality (1.3). In this case there exist some results similar to Theorem 1.1 if n = 2 (see for example [1, 5, 6, 8, 19] and the references therein). On the other hand very little is known if the dimension of the space is greater than two. Here we consider smooth solutions u of (1.1) in starshaped domains Ω ⊂ IRN , N ≥ 2; under some restrictions on the nonlinearity f and the solution u we again obtain inequality (1.3). In particular we study “low energy solutions”, i.e. solutions with an energy level close to the minimal one. A more precise definition will be provided for any fixed nonlinearity. 2

The paper is organized as follows. In Section 2 we state a general result which holds under several assumptions involving f and u. We point out that this theorem holds whether Ω is bounded or not and for f depending on x. In the other sections we give some applications of the theorem. In Section 3 we consider a problem close to the critical one in dimension n ≥ 3. In Section 4 we study standing wave solutions arising from the nonlinear Schr¨odinger equation. Finally Section 5 deals with a singularly perturbed subcritical problem in a bounded domain.

2

An abstract result

Theorem 2.1 Let Ω be a smooth domain in IRN , with N ≥ 1, and f ∈ ¯ satisfies: C 1 (Ω × IR+ ). Suppose that u ∈ C 3 (Ω) ∩ C 1 (Ω)    −∆u = f (x, u)  

u>0 u=0

in Ω in Ω on ∂Ω.

Let x0 be a maximum point for u and assume that Ω is starshaped with respect to x0 . If there exist an open set ω ⊂ Ω and a constant α > 0 such that: i) ii) iii) iv) v)

(x − x0 ) · ∇u(x) < 0 ∀x ∈ ω \ {x0 }, (x − x0 ) · ∇x f (x, s) + (2 + α)f (x, s) − αsfs0 (x, s) ≤ 0 ∀s ≥ 0, ∀x ∈ ω, (x − x0 ) · ∇u(x) + αu(x) < 0 ∀x ∈ ∂ω, [fs0 (x, u)]+ ∈ LN/2 (Ω \ ω), [(x − x0 ) · ∇u + αu]+ ∈ H 1 (Ω \ ω), λ1 (−∆ − [fs0 (x, u)]+ ) > 0 in H01 (Ω \ ω),

then (x − x0 ) · ∇u(x) < 0

∀x ∈ Ω \ {x0 }.

In particular, x0 is the only critical point for u in Ω and the superlevel sets are strictly starshaped with respect to x0 . Remark 2.2 We do not require a priori that x0 is an absolute maximum point for u, but this is a consequence of Theorem 2.1. Proof Arguing by contradiction, let us suppose that there exists y¯ ∈ Ω\{x0 } such that (¯ y − x0 ) · ∇u(¯ y ) ≥ 0; by assumption (i) y¯ is not in ω. 3

Set w(x) = (x − x0 ) · ∇u + αu,

(2.1)

where α is the same constant as in the assumptions of the theorem. It turns out that w(¯ y ) > 0. Now, let us call D the connected component of the set {x ∈ Ω | w(x) > 0} containing y¯. By assumption (iii), ω ∩ D = ∅. Furthermore, by Hopf Lemma ¯ ∩ ∂Ω, and since Ω is starshaped with respect to x0 , we have that if z ∈ D then w(z) = 0. So, by assumption (iv), w ∈ H01 (D). Then, using assumption (ii), we get that w verifies: (

−∆w ≤ fs0 (x, u)w ≤ [fs0 (x, u)]+ w w ∈ H01 (D)

in D

and this contradicts (v). q.e.d. Remark 2.3 If Ω is bounded, the assumption (iv) is not necessary, since it follows at once from the regularity of u and of f Let us consider a smooth domain D in IRN and a function k in LN/2 (D). Recall that the operator L = −∆ − k satisfies the maximum principle in D ¯ such that if, given a function u ∈ C 2 (D) ∩ C(D) i) Lu ≥ 0 in D ii) u ≥ 0 on ∂D, it must be u ≥ 0 in D. Remark 2.4 Berestycki, Nirenberg and Varadhan in [4] have proved that, if Ω is bounded, then L satisfies the maximum principle if, and only if, λ1 (L) > 0 in H01 (D). So, when Ω is bounded, assumption (v) is equivalent to the assumption that −∆ − [fs0 (x, u)]+ satisfies the maximum principle in Ω \ ω. A condition, easier to verify, in which assumption (v) occurs is stated in the following lemma. Lemma 2.5 Given a smooth domain D in IRN , if k ∈ LN/2 (D) verifies kkkN/2 < S, then λ1 (−∆ − k) > 0 in H01 (D). 4

Proof Suppose, by contradiction, that there exists a function w ∈ H01 (D), R R 2 not identically zero, such that D |∇w| ≤ D kw2 . Then it yields Z D

|∇w|2 ≤ kkkN/2 kwk22∗ ≤ kkkN/2 ·

Z 1 k∇wk22 < |∇w|2 , S D

which is impossible. q.e.d.

3

The critical case

In this section we apply Theorem 2.1 to front a semilinear and subcritical elliptic equation, with Dirichlet boundary condition, in a convex domain. In particular we consider the case when the nonlinearity approach the critical one. Let Ω be a smooth, bounded domain in IRN , with N ≥ 3. For ε > 0 consider the problem: (Pε )

 N +2 −ε   −∆u = N (N − 2)u N −2

u>0   u=0

in Ω in Ω on ∂Ω.

Theorem 3.1 Let uε be a solution of (Pε ) such that: |∇uε |2 lim R Ω ∗ 2/2∗ = S, 2 ε→0 ( Ω uε ) R

(3.1)

where S is the best constant in the Sobolev embedding and 2∗ = 2N/(N − 2). If Ω is convex then there exists ε¯ such that, denoted by xε a point where uε (xε ) = kuε k∞ , for every 0 < ε < ε¯ it occurs: (x − xε ) · ∇uε (x) < 0

∀x ∈ Ω \ {xε }.

In particular, the maximum point xε is the only critical point and the superlevels are strictly starshaped. Remark 3.2 It is well known that (Pε ) has at least one solution and that the solutions are smooth. In particular, by using the mountain pass Lemma (see [2]), it is standard to see that there exists a “minimal energy” solution, which verifies (3.1). 5

Remark 3.3 Han in [17] and Rey in [27] have studied the behavior of a sequence of solutions of (Pε ) that verifies (3.1). In particular they have shown that solutions of (Pε ) concentrate at a critical point of the Robin function. Proof of Theorem 3.1 We will show that, for ε sufficiently small, there exist N +2 ωε and αε such that Theorem 2.1 applies for f (x, s) = N (N − 2)s N −2 −ε . N +2 Let us denote pε = N − ε. Fixed αε = pε2−1 , it turns out at once that −2 (ii) occurs. To find ωε , we proceed by steps. Step 1

∀R > 0 ∃¯ ε1 s. t. if 0 < ε < ε¯1 then (x − xε ) · ∇uε (x) < 0

∀x ∈ B xε ,

!

R 1/αε

kuε k∞

\ {xε }.

Arguing by contradiction, let us suppose that there exist R0 , a sequence εn → 0 and a sequence {zn }n in Ω such that: (zn − xεn ) · ∇uεn (zn ) ≥ 0, |zn − xεn |
0 vε (0) = 1 and 0 < vε ≤ 1

in Ωε in Ωε in Ωε .

Moreover, recall that xε → x0 ∈ Ω and that kuε k∞ → ∞, as ε → 0 (see [17]). Now, using classical regularity results for elliptic equations and since, by [7] 1 U (y) = (1 + |y|2 )(N −2)/2

6

is the only solution for  N −2    −∆u = N (N − 2)u N +2

in IRN in IRN ∀y ∈ IRN ,

u>0    1 = u(0) ≥ u(y) we get

2 . uεn −→ U, as n → ∞ in Cloc

(3.5)

εn Set yn = kuεn k1/α (zn − xεn ), from (3.3) it follows that yn ∈ B(0, R0 ), ∞ hence, up to a subsequence, there exists y0 ∈ B(0, 2R0 ) such that yn → y0 as n → ∞. If y0 6= 0 then it follows from (3.5) that

yn · ∇vεn (yn ) → y0 · ∇U (y0 ) < 0, as n → ∞, and this is a contradiction, since, by (3.2), we have: yn · ∇vεn (yn ) =

1 kuεn k∞

(zn − xεn ) · ∇uεn (zn ) ≥ 0.

(3.6)

So it must be y0 = 0, but also this cannot happen. In fact, consider the function ϕn (t) = vεn (tyn ). It yields that ϕ has a maximum at 0, since xεn is a maximum for uεn , and another critical point in [0,1], by (3.6). So there exists a value t¯ ∈ [0, 1] such that ϕ00 (t¯) = 0. Now let n → ∞, from (3.5) and from the assumption that y0 = 0, it follows that 0 is a degenerate critical point for U and this is not true. Step 2

∀R > 1 ∃¯ ε2 s. t. if 0 < ε < ε¯2 then (x − xε ) · ∇uε (x) + αε uε (x) < 0

∀x ∈ ∂B xε ,

R 1/αε

kuε k∞

!

.

In the same notation of the previous step, we have that, for every R > 0: N − 2 1 − |y|2 as ε → ∞, uniformly on ∂B(0, R). 2 (1 + |y|2 )N/2 (3.7) If R > 1, from (3.7) it follows that, for ε sufficiently small,

y·∇vε (y)+αε vε (y) →

y · ∇vε (y) + αε vε (y) < 0, 7

∀y ∈ ∂B(0, R)

and this is equivalent to our claim. ¯ s. t. ∀R > R ¯ ∃¯ Step 3 ∃R ε3 s. t. if 0 < ε < ε¯3 then "Z

 1/αε

Ω\B(xε ,R/kuε k∞

N/2 2)pε upεε −1

N (N −

)

#2/N

< S.

Define  

¯ = inf R R 

Z

U

2∗

!

S N (N + 2)


R

!N/2   

¯ and let R > R. It is straightforward to see that Z 1/α Ω\B(xε ,R/kuε k∞ ε )

ε −1)N/2 (x)dx u(p ε

=

Z Ωε \{|y|>R}

vε(pε −1)N/2 (y)dy

and we claim that Z Ωε \{|y|>R}

vε(pε −1)N/2 (y)dy

−→

Z

∗

U 2 (y)dy

as ε → 0.

{|y|>R}

In fact, at first by (3.5) we have vε (y) → U (y), as ε → 0, for every y in IR ; furthermore assumption (3.1) implies the estimate vε (y) ≤ C/|y|(N −2) for a suitable constant C, as it is shown in [17]. N

To complete the proof, choose R > 1 in such a way that Step 3 is fulfilled, ε set ε¯ = min{¯ ε1 , ε¯2 , ε¯3 } and, for 0 < ε < ε¯, fix ωε = B(xε , R/kuε k1/α ∞ ). Now, Step 1 guarantees that (i) of Theorem 2.1 holds and Step 2 implies (iii). Assumption (iv) holds because Ω is bounded and, finally, assumption (v) follows from Step 3, taking into account Lemma 2.5. q.e.d.

4

Applications to nonlinear Schr¨ odinger equations

This section deals with a model class of nonlinear and subcritical Schr¨odinger equations in IRN (for a more general domain, see Remark 4.5). In particular we study the stationary solutions (see [12, 24, 25], for example). 8

It is known that these equations, for every “stable” critical point x0 of the potential, have a positive solution uh , that “concentrates” in x0 , as h goes to 0 (see Definitions 4.3 and 4.1). Furthermore uh has only one maximum point xh (see [11, 20, 28]). We will show that, actually, xh is the only critical point for uh and that the superlevels of uh are strictly starshaped with respect to xh . Define the problem  2 p   −h ∆u + V (x)u = u

(Ph )  u > 0  u ∈ H 1 (IRN ),

in IRN in IRN

where h > 0, N ≥ 1, p > 1 and p < (N + 2)/(N − 2) if N ≥ 3. Moreover, on the potential V assume: (V0 ) V ∈ C 2 (IRN ), inf IRN V = γ > 0; (V1 ) |∇V (x)| ≤ M, for some M > 0. Now, before giving the definition of single-peak solution for problem (Ph ) and of stable critical point for the potential V , let us introduce some notation. Denote by wµ the unique solution of: in IRN (P¯µ )  v > 0 in IRN  lim|y|→∞ v(y) = 0, v(0) = kvk∞  p   −∆v + µv = v

and define the “energy” of the solution wµ as 1Z µZ 1 Z 2 |∇wµ | + wµ − wµp+1 . Eµ = N N N 2 IR 2 IR p + 1 IR Let us recall that 0 is a nondegenerate maximum point for wµ . Furthermore wµ is a radial function and, if we denote by ωµ (|x|) = wµ (x), the function ω(r) is strictly decreasing and verifies: lim ωµ (r) · r(N −1)/2 er = Kµ > 0,

(4.1)

√ lim ωµ0 (r) · r(N −1)/2 er = −Kµ µ

(4.2)

r→∞

r→∞

for a constant Kµ (see [13, 18]). 9

¯ a family of solutions of (Ph ), for Definition 4.1 Let uh ∈ C 2 (Ω) ∩ C 1 (Ω) N small h, and let xh ∈ IR be such that kuh k∞ = uh (xh ). Then uh is said a family of single-peak solutions near a point x0 ∈ IRN if i) xh −→ h Rx0 as h → 0, −N 1 ii) h (h−2 |∇uh |2 + V (x)u2h ) − 2 Ω

1 R p+1 Ω

i

up+1 → EV (x0 ) as h → 0. h

Lemma 4.2 If uh is a family of single-peak solutions of (Ph ), then, for h sufficiently small, xh is the only local maximum point and uh (x) −→ 0, as h → 0, for any x ∈ Ω \ x0 ;

(4.3)

!

β uh (x) ≤ α exp − |x − xh | , h

(4.4)

for suitable positive constants α, β. For the proof of inequality (4.4) see, for example, [10], while (4.3) is an immediate consequence of (4.4). Definition 4.3 Let x0 be a critical point of V , we say that x0 is C 1 -stable if, for all ε > 0, there exists δ > 0 such that, if G ∈ C 1 (IRN ) verifies max (|G(x) − V (x)| + |∇G(x) − ∇V (x)|) < δ,

|x−x0 | 0  v ∈ H 1 (IRN ),

in IRN in IRN

where v(y) = u(xh + hy). Let us write, in the following, vh (y) = uh (xh + hy) the related solutions. 10

By the corresponding property of uh , it is easily seen that vh has only a maximum in 0. Furthermore, taking into account the limit problem (P¯V (x0 ) ), from Lemma 4.2 it follows that lim vh (y) = wV (x0 ) (y)

h→0

2 in Cloc (IRN ), L∞ (IRN ) and H 1 (IRN ).

(4.5)

Now we can state the main result of this section, that say something else about the shape of the solutions uh . Theorem 4.4 Assume that V satisfies (V0 ) and (V1 ) and that x0 is a C 1 stable critical point of V . If inf {(x − x0 ) · ∇V (x) + 2V (x)} = C > 0,

IRN

(4.6)

¯ > 0 such that, called xh the maximum point of the solution then there exists h uh , ¯ (x − xh ) · ∇uh < 0, ∀x ∈ IRN \ {xh }, if 0 < h < h. (4.7) Proof

It is straightforward that to show (4.7) is equivalent to show y · ∇vh (y) < 0, ∀y ∈ IRN \ {0},

where vh is the solution of (P˜h ) corresponding to uh . We will verify that, for small h, the assumptions of Theorem 2.1 hold, for Ω = IRN and f (x, s) = sp − V (xh + hy)s. Fixed α = 2/(p − 1), we want to define ωh = B(0, R), for a suitable R. ¯ 1 s. t. if 0 < h < h ¯ 1 then Step 1 ∀R > 0 ∃h y · ∇vh (y) < 0

∀y ∈ B(0, R) \ {0}.

Arguing as in Step 1 of the proof of Theorem 3.1, this is a direct conse2 quence of the Cloc -convergence stated in (4.5), taking into account that 0 is a nondegenerate critical point for wV (x0 ) . ¯ 2 s. t. if 0 < h < h ¯ 2 then Step 2 ∃h hy · ∇V (xh + hy) + 2V (xh + hy) >

C >0 2

This follows at once by assumptions (4.6) and (V1 ). 11

∀y ∈ IRN .

¯ 3 s. t. if 0 < h < h ¯ 3 then ¯ 1 s. t. ∀R > R ¯ 1 ∃h ∃R

Step 3

y · ∇vh (y) + αvh (y) < 0

∀y ∈ ∂B(0, R).

From (4.1) and (4.2) it follows that q ωV0 (x0 ) (r) = − V (x0 ) < 0. r→∞ ω V (x0 ) (r)

(4.8)

lim

Note that y · ∇wV (x0 ) (y) + αwV (x0 ) (y) = |y| ωV0 (x0 ) (|y|) + αωV (x0 ) (|y|) ωV0 (x0 ) (r) α = |y| ωV (x0 ) (|y|) + ; ωV (x0 ) (r) |y| ¯ 1 such that so, by (4.8), we get that there exists R "

#

y · ∇wV (x0 ) (y) + αwV (x0 ) (y) < 0,

¯1. if y > R

(4.9)

2 Now it is easily seen that Step 3 follows from the Cloc -convergence stated in (4.5). ¯ 4 s. t. if 0 < h < h ¯ 4 then ¯ 2 s. t. ∀R > R ¯ 2 ∃h Step 4 ∃R

"Z



IRN \B(0,R)

N/2 pvhp−1 (y)

#2/N

dx

< S.

Define (

Z

¯ 2 = inf R R

{|y|>R}

(p−1)N/2 wV (x0 ) (y)dy

) N/2

< (S/p)

¯ 2 < ∞ by (4.1)). (R From (4.4) and (4.5), it follows that: kvhp−1 kLN/2 (IRN \B(0,R)) −→ kwVp−1 (x0 ) kLN/2 (IRN \B(0,R)) ,

as h → 0,

and so we get Step 4. ¯ 1, h ¯ 2, h ¯ 3, h ¯ 4 } and set ωh = Now choose R > max{R¯1 , R¯2 }, fix h < min{h B(0, R). We claim that ωh has the desired properties. In fact, Steps 1-3 guarantee that assumptions (i)-(iii) of Theorem 2.1, respectively, hold. 12

To verify assumption (iv), we remark first that [pvhp−1 − V (xh + hy)]+ ∈ LN/2 (IRN ), since vh vanishes at infinity and V verifies (V0 ). Now we want to prove that [y · ∇vh + αvh ]+ ∈ H 1 (IRN ). Let us consider the function ψ = y · ∇vh . iIf we show that ψ ∈ H 1 (IRN ), then we have done. From Green’s representation formula (see, for example [22], §20) applied to problem (P˜h ), it follows that ∇vh , hence ψ, has an exponential decay at infinity. Furthermore, by computation, we have −∆ψ = −(V (xh +hy)−pvhp−1 )ψ+2vhp −(hy·∇V (xh +hy)+2V (xh +hy))vh in IRN . (4.10) N 1 From (4.10) it turns out that ψ ∈ H (IR ), taking into account the behavior at infinity of ψ and hypothesis (V1 ). To verify also assumption (v), observe first that 



λ1 −∆ − [p(vhp−1 − V (hy)]+ ≥ λ1 (−∆ − pvhp−1 ) in H01 (IRN \ B(0, R)), by monotonicity. Then, by Step 4 and Lemma 2.5, we have that −∆ − pvhp−1 is strictly positive in H01 (IRN \ B(0, R)). q.e.d. Remark 4.5 For sake of simplicity, in this section we have studied problem (Ph ) on IRN , referring, for example, to the family of single-peak solutions provided by Li in [20] near a C 1 stable critical point of the potential. We could also consider problem (Ph ) on a generic smooth domain Ω, bounded or unbounded. In this case, del Pino and Felmer in [11] have proved that, under assumption (V0 ), there is a single-peak family u¯h of solutions of (Ph ), corresponding to every topological nontrivial critical level c¯ of the potential V . The solution u¯h has only a maximum xh and concentrate around a critical point x0 of V , such that V (x0 ) = c¯. Under the same assumptions of Theorem 4.4, stated in Ω, and if Ω is strictly starshaped with respect to x0 , then we can conclude by an analogous proof that (x−xh )·∇¯ uh < 0 for every x ∈ Ω\{xh }, for small h.

13

5

The subcritical case

This section is devoted to study the problem  2 p   −ε ∆u + u = u

(P˜ε )  u > 0  u=0

in Ω in Ω on ∂Ω,

where Ω ⊂ IRN , with N ≥ 1, is a smooth bounded domain, ε > 0, p > 1 and p < (N + 2)/(N − 2) if N ≥ 3. Contrary to the previous section, now we do not have a potential, whose stable critical points generate a family of single peak solutions, as ε → 0. Nevertheless, Ni and Wei in [23] have showed that the solutions uε which minimizes the functional Jε (u) =

1Z 2 1 Z p+1 (ε |∇u|2 + u2 ) − u 2 Ω p+1 Ω

(the ground state solution), are actually single peak solutions and, called xε the maximum point of uε , it is verified xε → x0 , where x0 satisfies distΩ (x0 , ∂Ω) ≥ distΩ (x, ∂Ω) for any x ∈ Ω where distΩ (·, ∂Ω) = miny∈∂Ω ky − ·kIRN . In [16] it was proved that any “stable” critical point x0 of the distance function generates a family of single peak solutions which concentrates at x0 . Given such a family of single peak solutions an a domain strictly starshaped with respect to x0 , we will show that the superlevels of these solutions are strictly starshaped and that the maximum point is the only critical point. In Remark 5.2 we will speak about another way to state this result. Theorem 5.1 Let uε a family of single peak solutions of (P˜ε ). Called xε the maximum point of uε and x0 the limit point of xε , if Ω is strictly starshaped with respect to x0 , then there exists ε¯ > 0 such that (x − xε ) · ∇uε < 0 ∀x ∈ Ω \ {xε }, for every 0 < ε < ε¯.

14

Proof Also to prove this theorem we are showing that the assumptions of Theorem 2.1 hold for f (x, u) = (up − u)/ε2 , when ε is sufficiently small. Since xε → x0 and Ω is strictly starshaped with respect to x0 , it is easily seen that Ω is strictly starshaped with respect to xε , for small ε. Fixed α = 2/(p − 1), it is clear that (ii) of Theorem 2.1 holds. To set ω, we proceed by steps, in a way similar to the one followed to prove Theorem 4.4. Step 1 ∀R > 0 ∃¯ ε1 s. t. if 0 < ε < ε¯1 then (x − xε ) · ∇uε (x) < 0

∀x ∈ B(xε , εR) \ {xε }.

Define Ωε = (Ω − xε )/ε and vε (y) = uε (xε + εy). Then vε solves  p   −∆v + v = v  

in Ωε in Ωε on ∂Ωε .

v>0 v=0

By classical regularity results, 2 in Cloc ,

vε (y) −→ w1 (y), as ε → 0,

(5.1)

where w1 is the solution of problem (P¯1 ), described in Section 4. Now, arguing as in Step 1 of Theorem 3.1, the claim is proved. ¯ 1 s. t. ∀R > R ¯ 1 ∃¯ Step 2 ∃R ε2 s. t. if 0 < ε < ε¯2 then (x − xε ) · ∇uε (x) + αuε (x) < 0

∀x ∈ ∂B(xε , εR).

The proof of this step is the same proof of Step 3 of Theorem 4.4, taking into account that, for y = (x − xε )/ε, it yields (x − xε ) · ∇uε (x) + αuε (x) = y · ∇vε (y) + αvε (y). Step 3

¯ 2 s. t. ∀R > R ¯ 2 ∃¯ ∃R ε3 s. t. if 0 < ε < ε¯3 then "Z



Ω\B(xε ,εR)

Define ¯ 2 = inf{R | R

Z |y|>R

#2/N

N/2

p p−1 u (x) ε2 ε

(p−1)N/2

w1

15

dx

< S.

(y)dy < (S/p)N/2 }

(5.2)

(5.3)

¯ 2 < ∞ because w1 has an exponential decaying at infinity). (R An easy calculation shows that 

Z Ω\B(xε ,εR)

N/2

1 p−1 u (x) ε2 ε

dx =

Z Ωε \B(0,R)

vε(p−1)N/2 (y)dy.

(5.4)

Ni and Wei in [23] have proved that there exist a, b > 0, not depending on ε, such that vε (y) ≤ a exp(−b|y|), (5.5) hence, using (5.1), we get vεp−1 → w1p−1 in LN/2 (IRN ), as ε → 0, for every ¯ 2 , inequality (5.2) follows from (5.3) and (5.4). p > 1. Then, if R > R To complete the proof, choose R > max{R¯1 , R¯2 } and fix ε < min{¯ ε1 , ε¯2 , ε¯3 }. Setting ωε = B(xε , εR), we claim that ωε has the desired properties. In fact, Steps 1 and 2 guarantee that assumptions (i) and (iii) of Theorem 2.1 hold. Now, to verify assumption (v), observe that 

λ1

1 p 1 − 1)+ ≥ λ1 (−∆ − 2 up−1 −∆ − 2 (pup−1 ε ε ) in H0 (Ω), ε ε 

by monotonicity. Then, we have that −∆ − εp2 up−1 is strictly positive in H01 (Ω), by Step 3 and Lemma 2.5, and this completes the proof. q.e.d. Remark 5.2 Another way to prove Theorem 5.1 is to follows the same steps as in the previous proof, but in such a way that in Ω \ B(xε , εR) it holds uε ≤ k < 1 (this is possible because vε → V in L∞ (IRN ), by (5.1) and (5.5)). Then we can apply to uε on Ω\B(xε , εR) a result stated in [14] (see Theorems 3.1 and 3.3 therein or also [21]) to obtain our statement. Remark 5.3 An easy situation in which Theorem 5.1 applies is when Ω is strictly convex. To see an example in which Ω is not convex and Theorem 5.1 holds, we can consider when Ω is a “dumbbell” and is strictly starshaped with respect to its middle point x0 . In [16] it is proved that there exists a family of single peak solutions that concentrate at x0 , so Theorem 5.1 applies to these solutions.

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e-mail: [email protected], [email protected]

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