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Testo completo

(1)

fractional Laplacian in bounded domains

Dimitri Mugnai and Dayana Pagliardini

Abstract

First, we study existence results for a linearly perturbed elliptic prob- lem driven by the fractional Laplacian. Then, we show a multiplicity result when the perturbation parameter is close to the eigenvalues. This latter result is obtained by exploiting the topological structure of the sub- levels of the associated functional, which permits to apply a critical point theorem of mixed nature due to Marino and Saccon.

Keywords: fractional Laplacian, eigenvalues, (∇)−condition, ∇−theorem.

2000AMS Subject Classification: 35R11, 35J65, 35J20, 35J61.

1 Introduction

In this paper we consider the problem

( (−∆) 1/2 u = λu + g(x, u) in Ω

u = 0 on ∂Ω, (1)

where Ω is a bounded domain of R N , N ≥ 2, λ ∈ R and g : Ω × R → R is a given function; more precise details will be given below. Of course, problem (1) is a possible fractional counterpart of problem

( −∆u = λu + g(x, u) in Ω

u = 0 on ∂Ω, (2)

which has been object of an extensive study in the last four decades, essentially with the aid of variational methods, when N ≥ 3.

We will not go into details about existence and multiplicity results for (2) according to different assumptions on g, since the bibliography would be huge, but we focus on multiplicity results with very general assumptions on g. The first result for g(x, t) ∼ |t| p−2 t was established by Ambrosetti–Rabinowitz in [1], where the authors assumed that

• g : Ω × R → R is a continuous function;

1

(2)

• there exist a 1 , a 2 > 0 and p ∈ 

2, N −2 2N 

such that for all (x, t) ∈ Ω × R,

|g(x, t)| ≤ a 1 + a 2 |t| p−1 ; (3)

• g(x, t) = o(|t|) for t → 0 uniformly in Ω;

• there exists R ≥ 0 and µ > 2 such that for all |t| > R and all x ∈ Ω 0 < µG(x, t) ≤ g(x, t)t, (4) where G(x, t) = R t

0 g(x, σ)dσ.

Let us note that, as a consequence of (4), we get that the existence of c 1 , c 2 > 0 such that for all (x, t) ∈ Ω × R

G(x, t) ≥ c 1 |t| µ − c 2 . (5)

Under this assumptions, in [1] it is proved that problem (2) has two nontrivial solutions when λ = 0, though the same proof holds if λ < λ 1 , where λ 1 is the first eigenvalue of −∆ in Ω subject to homogeneous Dirichlet boundary conditions.

Almost twenty years later, in [25], assuming g : R → R of class C 1 , Wang proved that problem (2) has three nontrivial solutions under the same related assumptions. Since then, thousands of papers have given other multiplicity results weakening the assumptions on the superlinear and subcritical g (see the recent paper by Mugnai–Papageorgiou [14] for more general operators).

However, not much was done for λ > λ 1 . The first result in proving Wang’s result for λ > λ 1 and close to an eigenvalue can be found in Mugnai [12], whose result is complemented by Rabinowitz–Su–Wang in [18]. However, while in [18]

g : Ω × R → R is assumed to be of class C 1 , in [12] the following assumptions for a superlinear and subcritical g are made:

• g : Ω × R → R is a Carath´eodory function;

• there exist a 1 , a 2 > 0 and p ∈ (2, N −2 2N ) such that (3) holds for all t ∈ R and for a.e. x ∈ Ω;

• g(x, t) = o(|t|) for t → 0 uniformly in Ω;

• condition (4) holds for all t 6= 0 and for a.e. x in Ω with µ = p.

Let us remark that with these weak assumptions, inequality (4) does not imply (5), and for this one has to assume that

• there exists c 1 > 0 such that for all t ∈ R and for a.e. x in Ω, we have G(x, t) ≥ c 1 |t| p ,

see Mugnai [10].

(3)

Going back to problem (1), some remarks are needed. The operator (−∆) 1/2 we consider is the spectral square root of the Laplacian, which should not be confused with the integro–differential operator defined, up to a constant, as

−(−∆) s u(x) = Z

R

n

u(x + y) + u(x − y) − 2u(x)

|y| n+2s dy, x ∈ R n .

Indeed, in this case the homogeneous Dirichlet “boundary conditions” should be interpreted as u ≡ 0 in R N \ Ω (see [4], [19], [20] and [21]). In fact, in [19]

the authors show that these two operators, though often denoted in the same way, are really different, with eigenvalues and eigenfunctions behaving quite differently.

As already said, we will consider the spectral square root of the Laplacian, defined according to the following procedure (see Cabr´ e–Tan [2]). Let H 1/2 (Ω) denote the Sobolev space of order 1/2, defined as

H 1/2 (Ω) :=



u ∈ L 2 (Ω) : Z

Z

|u(x) − u(y)| 2

|x − y| N +1 dxdy < ∞



with norm

kuk 2 H

1/2

(Ω) :=

Z

u 2 dx + Z

Z

|u(x) − u(y)| 2

|x − y| N +1 dxdy.

Introducing the cylinder C := Ω × (0, ∞) with lateral boundary ∂ L C = ∂Ω × [0, ∞), set

H 0,L 1 (C) := v ∈ H 1 (C) : v = 0 a.e. on ∂ L C ,

and denote by tr Ω the trace operator on Ω × {0} for functions in H 0,L 1 (C), tr Ω v(x, y) := v(x, 0) for all (x, y) ∈ C.

Then, from standard results, we know that

V 0 (Ω) := u = tr Ω v : v ∈ H 0,L 1 (C) ⊂ H 1/2 (Ω), but a characterization of V 0 is available:

Proposition 1.1 ([2], Proposition 2.1).

V 0 (Ω) =



u ∈ H 1/2 (Ω) : Z

u 2

d(x) dx < ∞



= (

u ∈ L 2 (Ω) : u =

X

k=1

α k ϕ k satisfies

X

k=1

α 2 k λ 1/2 k < ∞ )

.

Here (λ k , ϕ k ) k is the Dirichlet spectral decomposition of −∆ in Ω, (ϕ k ) k being

an orthonormal basis in L 2 (Ω), and d(x) := dist(x, ∂Ω).

(4)

From these preliminaries, by [2, Proposition 2.2], for u = P ∞

k=1 α k ϕ k , we define

(−∆) 1/2 u :=

X

k=1

α k λ 1/2 k ϕ k . (6)

With this definition in hand, the purpose of this paper is to prove a multi- plicity result for problem (1), in a situation similar to that described above for (2), and, in view of the previous considerations, we assume that

(g 1 ) g : Ω × R → R is a Carath´eodory function;

(g 2 ) there exist a 1 , a 2 > 0 and p ∈ 

2, N −1 2N 

such that (3) holds for all t ∈ R and for a.e. x ∈ Ω;

(g 3 ) g(x, t) = o(|t|) for t → 0 uniformly in Ω;

(g 4 ) 0 < pG(x, t) = p Z t

0

g(x, σ)dσ ≤ g(x, t)t for all t 6= 0 and for a.e. x ∈ Ω;

(g 5 ) there exists c 1 > 0 such that G(x, t) ≥ c 1 |t| p for all t ∈ R and for a.e.

x ∈ Ω.

Remark 1.2. Note that (g 3 ) implies that G(x, t) = o(|t| 2 ) as t → 0 uniformly in Ω and from (g 2 ) we get that |G(x, t)| ≤ a 1 |t| + a 2

p |t| p for all t ∈ R and for a.e.

x ∈ Ω. As a consequence, u = 0 solves problem (1), and we look for nontrivial solutions.

In the case g(x, t) = |t| p−2 t, 2 < p < N −1 2N , an existence result for λ = 0 is proved in Cabr´ e–Tan [2]. The proof therein can be immediately extended to the case λ < λ 1 , but we are not aware of existence results for λ ≥ λ 1 , nor for general nonlinearities g. For this, we give our first result:

Theorem 1.3. If (g 1 ) − (g 5 ) hold, then for every λ ∈ R problem (1) has one nontrivial solution.

However, the previous result is standard, and we only give it for a complete description of the existence setting, as a counterpart of the result in Servadei–

Valdinoci [20], when the fractional Laplacian is represented by a nonlocal inte- gral operator, already introduced in [4, 21].

On the other hand, our main interest is providing a multiplicity theorem, which is much more involved. This result relies on the application of a criti- cal point theorem of mixed type proved by Marino–Saccon in [7], recalled in the Appendix, together with an additional linking theorem and a fine estimate of critical levels. More precisely, we prove the following multiplicity result in Wang’s direction, the main result of this paper:

Theorem 1.4. Assume (g 1 ) − (g 5 ). Then, for all i ∈ N, i ≥ 2, there exists

δ i > 0 such that problem (1) has at least three nontrivial solutions for all λ ∈

(λ i − δ i , λ i ).

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We conclude recalling that the ∇–theorem we will employ has been exten- sively used in several contexts, in order to prove multiplicity results of different problems, such as elliptic problems of second and fourth order, variational in- equalities and reversed variational inequalities, see, for instance, [5, 8, 11, 12, 13, 15, 22, 23, 24], and [9], where the analogous multiplicity result of this pa- per is considered when the underlying operator is the nonlocal one studied in [4, 20, 21].

2 Extended problem, preliminary lemmas and proof of Theorem 1.3

We know from [2] that solving problem (1) is equivalent to solving its extension in the cylinder C = Ω × (0, ∞)

 

 

∆v = 0 in C,

v = 0 on ∂ L C,

∂v

∂ν = λv + g(x, v) in Ω × {0},

(7)

where ∂ L C = ∂Ω×(0, ∞) is the cylinder lateral surface and ν is the outer normal at C in Ω × {0}. Let us briefly recall the relation between (7) and (1).

First, we will look for weak solutions to (7). For this, we note that the Sobolev space H 0,L 1 (C) introduced above is an Hilbert space when endowed with the norm

kvk =

Z

C

|Dv| 2 dxdy

 1/2 , induced by the inner product

hv, wi = Z

C

Dv · Dw dxdy.

Following the “Dirichlet to Neumann” approach in a bounded domain Ω of R N (cfr. [2]), for a given u ∈ V 0 (Ω) we consider the harmonic extension v of u, i.e. the solution of the problem

 

 

∆v = 0 in C, v = 0 on ∂ L C, v = u in Ω × {0},

which is well defined by [2, Lemma 2.8]. Now, by [2, Proposition 2.2], we can give a definition of (−∆) 1/2 : V 0 (Ω) → V 0 (Ω) , equivalent to (6), as the operator that maps the Dirichlet datum u in the Neumann value of its harmonic extension

(−∆) 1/2 u = ∂v

∂ν (·, 0),

(6)

and hence, if v solves (7), then u = tr Ω v solves (1), see [2, Proposition 2.2]. For this reason, from now on, we will look for weak solutions to (7).

Of course, problem (7) is variational and its solutions are critical points of the C 1 functional f λ : H 0,L 1 (C) → R defined by

f λ (u) = 1 2

Z

C

|Du| 2 dxdy − λ 2 Z

u 2 dx − Z

G(x, u)dx.

Before attacking functional f λ , we recall some other tools we will use in the following. Recalling that (λ k , ϕ k ) k denote the eigenvalues and the associated eigenfunctions of −∆ with homogeneous Dirichlet condition on ∂Ω with (ϕ k ) k

an orthonormal basis of L 2 (Ω), from [2], we have

H 0,L 1 (C) = n

v(x, y) =

X

k=1

b k ϕ k (x)e −λ

1/2k

y (x, y) ∈ C :

X

k=1

λ 1/2 k b 2 k < ∞ o .

Setting e k (x, y) = ϕ k (x)e −λ

1/2k

y , we observe that the e k are orthogonal in H 0,L 1 (C). Now, for every integer i ≥ 1, put

H i = span(e 1 , · · · , e i ) and H i = span(e i+1 , · · · ).

Then, a simple calculation gives the proof of the following inequalities:

Proposition 2.1. If u ∈ H i , then Z

C

|Du| 2 dxdy ≤ λ i

Z

u 2 dx. (8)

Proposition 2.2 (Constrained Poincar´ e inequality). If u ∈ H i , then Z

C

|Du| 2 dxdy ≥ λ i+1 Z

u 2 dx. (9)

We will also use the continuous inclusions (see [2, Lemma 2.4]) H 0,L 1 (C) ,→ L r (Ω) for all r ∈ h

1, N −1 2N i

, (10)

and the compact ones (see [2, Lemma 2.5])

H 0,L 1 (C) ,→ L r (Ω) for all r ∈ h

1, N −1 2N 

. (11)

Now, we have all the ingredients to look for critical points of functional f λ , i.e. to solve problem (7). As usual, the first step in applying variational methods is the following result:

Proposition 2.3. If c ∈ R, then f λ satisfies the (P S) c condition, namely:

every sequence (u n ) n such that f λ (u n ) → c and f λ 0 (u n ) → 0 as n → ∞ has a

converging subsequence.

(7)

Proof. Let (u n ) n ⊂ H 0,L 1 (C) be a Palais–Smale sequence at level c ∈ R. Then, taken k ∈ (2, p), there exis M, N > 0 such that

kf λ (u n ) − f λ 0 (u n )u n ≤ M + N ku n k for all n ∈ N. (12) On the other hand, by (g 4 ) and (g 5 ), we get

kf λ (u n ) − f λ 0 (u n )u n

=  k 2 − 1



ku n k 2 − λ  k 2 − 1

 Z

u 2 n dx + Z

[g(x, u n )u n − kG(x, u n )]dx

≥  k 2 − 1



ku n k 2 − λ  k 2 − 1

 Z

u 2 n dx + (p − k) Z

G(x, u n ) dx

≥  k 2 − 1



ku n k 2 − λ  k 2 − 1

 Z

u 2 n dx + (p − k)c 1

Z

|u n | p dx.

(13) By the Young inequality, for every ε > 0 there exists D ε > 0 such that

u 2 n ≤ ε|u n | p + D ε . Hence, (13) implies that

kf λ (u n )−f λ 0 (u n )u n ≥  k 2 − 1



ku n k 2 +



(p − k)c 1 − ε|λ|  k 2 − 1

 Z

|u n | p dx−D ε . Choosing ε sufficiently small, we finally get

kf λ (u n ) − f λ 0 (u n )u n ≥  k 2 − 1



ku n k 2 − D ε ,

and by (12), we get that (u n ) n is bounded. Then, we can assume that u n * u in H 0,L 1 (C) and u n → u in L r (Ω) for all r ∈ h

1, N −1 2N  .

Since f λ 0 (u n )(u n − u) → 0 as n → ∞, we immediately get that that u n

converges strongly to u, i.e. (P S) c holds.

At this point we can prove Theorem 1.3:

Proof of Theorem 1.3. First of all we observe that, from the Remark 1.2, f λ (0) = 0 and by (g 2 ) and (g 3 ), we get that, given  > 0, there exists C  > 0 such that

G(x, s) ≤ |u| 2 + C  |s| p (14) for a.e. x ∈ Ω and all s ∈ R.

Now, we have to distinguish two cases: λ < λ 1 and λ ∈ [λ i , λ i+1 ), for some i ∈ N.

First case: λ < λ 1 . We want to apply the Mountain Pass Theorem (see [1]).

Supposing λ > 0 (the other case being easier), by (14), the Poincar´ e inequality

(8)

(8) for i = 0 and by the continuous embedding of H 0,L 1 (C) in L 2 (Ω) and in L p (Ω), we have that

f λ (u) ≥ 1

2 kuk 2 − λ

1 kuk 2 − 

2 kuk 2 L

2

(Ω) − C  kuk p L

p

(Ω)

≥ 1 2

 1 − λ

λ 1 − c

λ 1

 kuk 2 − Ckuk p ,

for some absolute constant c, C > 0. Now, choosing  enough small, we can suppose A = 1

2

 1 − λ

λ 1 − c

λ 1



> 0 and taking ρ sufficiently small we have that inf

kuk=ρ f λ (u) ≥ α > 0 for some α > 0. Moreover, if u 6= 0 and t > 0, by (g 5 ),

f λ (tu) = t 2

2 kuk 2 − λ

2 t 2 kuk 2 2 − Z

G(x, tu) dx

≤ t 2

2 kuk 2 − λ

2 t 2 kuk 2 2 − c 1 t p kuk p p → −∞,

since p > 2. Recalling that (P S) c -condition holds for all c ∈ R, the Mountain Pass Theorem implies that (1) has a nontrivial solution.

Second case: λ ∈ [λ i , λ i+1 ) for some i ∈ N. In this case we want to apply the Linking Theorem (see the Appendix). If u ∈ H i , from (8) and (g 4 ), we have

f λ (u) ≤  λ i

2 − λ 2

 Z

u 2 dx − Z

G(x, u) dx ≤ 0.

Moreover, by (14) and (9), if u ∈ H i , f λ (u) ≥ 1

2 kuk 2 − λ 2λ i+1

kuk 2 − Z

G(x, u) dx

≥ 1 2

 1 − λ

λ i+1

 kuk 2 −  2

Z

u 2 dx − C 

Z

u p dx > 0.

Thus, by (10), we obtain f λ (u) ≥ 1

2

 1 − λ

λ i+1

− c

2

 kuk 2 − Ckuk p

for some c, C > 0 and for all u ∈ H i . Choosing  and ρ > 0 small enough, we get

inf

u∈H⊥i , kuk=ρ

f λ (u) ≥ α > 0.

Finally, if u ∈ H i and t > 0, by (g 5 ) we have f λ (u + te i+1 ) = 1

2 ku + te i+1 k 2 − λ 2 Z

(u + te i+1 ) 2 dx − Z

G(x, u + te i+1 ) dx

≤ 1

2 ku + te i+1 k 2 − λ 2 Z

(u + te i+1 ) 2 dx − c 1

Z

|u + te i+1 | p dx → −∞

(9)

as t → ∞, since all norms are equivalent in H i ⊕ (e i+1 ).

Recalling that (P S) c -condition holds for all c ∈ R, the Linking Theorem implies that problem (1) has a nontrivial solution.

3 Proof of Theorem 1.4

In this section we prove Theorem 1.4, the main contribution of this paper. For this, from now on, we assume that there exist i ≤ j in N such that λ i−1 < λ i =

· · · = λ j < λ j+1 .

Let us start introducing some notations: if i < j in N, we introduce the following sets:

S j + (ρ) = {u ∈ H j : kuk = ρ}, and

T i,j (R) = n

u ∈ H j : kuk = R o

∪ n

u ∈ H i : kuk ≤ R o . We can now state our first Lemma.

Lemma 3.1. Assume i, j ≥ 2 are such that λ i−1 < λ i = · · · = λ j < λ j+1 and λ ∈ (λ i−1 , λ j ). Then there exist R and ρ, R > ρ > 0, such that

sup f λ (T i−1,j (R)) < inf f λ (S i−1 + (ρ)).

Proof. From (14), (9) and by the compact embedding in L 2 (Ω) and in L p (Ω) given in (11), we have the existence of ρ > 0 such that

inf f λ (S + i−1 (ρ)) > 0.

Moreover, it is clear that f λ (H i−1 ) ≤ 0. We conclude the proof by showing that

lim

kuk→∞

u∈H

j

f λ (u) = −∞.

Such a result easily follows from (g 5 ) and from inequality (8). Indeed, if u ∈ H j , then

f λ (u) ≤ 1

2 kuk 2 − λ

j kuk 2 − c 1 kuk p p , and since all norms in H j are equivalent, the lemma follows.

Now take

a ∈ 

sup f λ (T i−1,j (R), inf f λ (S i−1 + (ρ)) 

and b > sup f λ (B j (R)), where B j (R) is the ball in H j with radius R. Then

Lemma 3.2. Suppose there exist integers i, j ≥ 2 such that λ i−1 < λ i = · · · =

λ j < λ j+1 .Then, for every δ > 0 there exists  0 > 0 such that ∀ λ ∈ [λ i−1 +

δ, λ j+1 − δ], the only critical point u of f λ in H i−1 ⊕ H j such that f λ (u) ∈

[− 0 ,  0 ], is the trivial one.

(10)

Proof. Assume by contradiction that there exist δ > 0, λ n ∈ [λ i−1 + δ, λ j+1 − δ]

and (u n ) n in H i−1 ⊕ H j \{0} such that

f λ

n

(u n ) −→ 0 if n → ∞ and such that for all z ∈ H i−1 ⊕ H j

Z

C

Du n · Dz dxdy − λ n

Z

u n z dx − Z

g(x, u n )z dx = 0, (15) where

f λ

n

(u n ) = 1 2

Z

C

|Du n | 2 dxdy − λ n

2 Z

u 2 n dx − Z

G(x, u n ) dx.

Up to a subsequence, we can assume that λ n → λ in [λ i−1 + δ, λ j+1 − δ]. Choose z = u n in (15). Then, by (g 4 ) we obtain

0 = Z

C

|Du n | 2 dxdy − λ n Z

u 2 n dx − Z

g(x, u n )u n dx

= 2f λ

n

(u n ) + Z

[2G(x, u n ) − g(x, u n )u n ] dx

≤ 2f λ

n

(u n ) + (2 − p) Z

G(x, u n ) dx.

In particular, we deduce that

n→∞ lim Z

G(x, u n ) dx = 0. (16)

Now, let v n in H i−1 and w n in H j be such that u n = v n + w n for every n ∈ N, and choose z = v n − w n in (15). Then

Z

C

|Dv n | 2 dxdy − λ n

Z

v 2 n dx −

Z

C

|Dw n | 2 dxdy − λ n

Z

w n 2 dx



= Z

g(x, u n )(w n − v n ) dx ∀ n ∈ N.

By (8) and (9), we get the existence of c > 0 independent of n, (c = δ/λ j+1 ), such that

cku n k 2 ≤ Z

g(x, u n )(v n − w n ) dx ∀ n ∈ N. (17) Here we used the fact that v n and w n are orthogonal, so that ku n k 2 = kv n k 2 + kw n k 2 for every n ∈ N.

Moreover, by the H¨ older inequality, we get

Z

g(x, u n )(v n − w n ) dx

Z

|g(x, u n )| p/(p−1) dx

 1−1/p

·

Z

|v n − w n | p dx

 1/p .

(18)

(11)

By the Sobolev embedding and by the compact embedding in (11), there exists an universal constant γ p > 0 such that

kv n − w n k L

p

(Ω) ≤ γ p kv n − w n k = γ p ku n k. (19) In this way, since u n 6= 0, (17), (18) and (19) imply that there exists c 0 > 0 such that

ku n k ≤ c 0

Z

|g(x, u n )| p/(p−1) dx

 (p−1)/p

∀ n ∈ N. (20)

Up to a subsequence, there are two possibilities: either ku n k → ∞, or ku n k is bounded.

First case: ku n k → ∞. Without loss of generality, we can suppose that there exists u ∈ H i−1 ⊕ H j such that u n

ku n k * u in H 0,L 1 (C) and u n

ku n k → u in L r (Ω) for every r ∈ h

1, N −1 2N 

, see (11). First of all, (16) implies that

0 ← 2 f λ

n

(u n )

ku n k 2 −→ 1 − λ Z

u 2 dx as n → ∞, so that u 6≡ 0. Moreover (g 2 ) and (g 5 ) imply

Z

|g(x, u n )| p/(p−1) dx ≤ a 0 1 + a 0 2 Z

G(x, u n ) dx.

But the last quantity is bounded by (16), while (20) leads to a contradiction.

Second case: ku n k is bounded. As before, we can suppose that there exists u ∈ H i−1 ⊕ H j such that u n * u in H i−1 ⊕ H j and u n → u in L r (Ω) for every r ∈ h

1, N −1 2N 

. Moreover (16) and (g 5 ) imply u = 0.

If u n → 0, then by (20) and (g 3 ), we would have

1 ≤ lim

n→∞ c 0

Z

|g(x, u n )| p/(p−1) dx

 (p−1)/p

ku n k = 0,

which is absurd.

So, there should exist σ > 0 such that ku n k ≥ σ for all n ∈ N; but also in this case, since u n → 0 in L p (Ω), from (20) we would obtain

σ ≤ lim

n→∞ c 0

Z

|g(x, u n )| p/(p−1) dx

 (p−1)/p

= 0, that is absurd, as well.

The lemma is completely proved.

For the following result we denote by P : H 0,L 1 (C) −→ span(e i , · · · , e j ) and

Q : H 0,L 1 (C) −→ H i−1 ⊕ H i the orthogonal projections.

(12)

Lemma 3.3. Suppose there exist integers i, j ≥ 2 such that λ i−1 < λ i = · · · = λ j < λ j+1 . Let λ ∈ R and (u n ) n in H 0,L 1 (C) be such that (f λ (u n )) n is bounded, P u n → 0 and Q∇f λ → 0 as n → ∞. Then (u n ) n is bounded.

Proof. Assume by contradiction that (u n ) n in unbounded. Then we can suppose that there exists u in H 0,L 1 (C) such that u n

ku n k * u in H 0,L 1 (C) and u n ku n k → u in L r (Ω) for every r ∈ h

1, N −1 2N 

as n → ∞.

Note that u n = P u n +Qu n , P u n → 0 and Q∇f λ (u n ) → 0, where ∇f λ (u n ) = V n is such that

f λ 0 (u n )v = Z

C

V n · Dv dxdy ∀ v ∈ H 0,L 1 (C).

So we obtain

V n = u n − K(λu n + g(x, u n )),

where K : L 2 (Ω) −→ H 0,L 1 (C) is the operator defined by K(u) = v, with v weak solution of

 

 

∆v = 0 in C, v = 0 on ∂ L C,

∂v

∂ν = u in Ω × {0}.

(21)

In particular,

hQ∇f λ (u n ), u n i = h∇f λ (u n ), u n i − hP ∇f λ (u n ), u n i

= ku n k 2 − λ Z

u 2 n dx − Z

g(x, u n )u n dx

− Z

C

D(P (u n − K(λu n + g(x, u n )))) · Du n dxdy.

But for every z ∈ H 0,L 1 (C), P z is a smooth function and DP u n = P Du n , because u ∈ span(e i · · · e j ) and P z⊥Qz. In this way the last integral in the previous equation is equal to

Z

C

|DP u n | 2 dxdy − λ Z

|P u n | 2 dx − Z

g(x, u n )P u n dx.

As a consequence,

hQ∇f λ (u n ), u n i = 2f λ (u n ) + 2 Z

G(x, u n ) dx

− Z

g(x, u n )u n dx

− kP u n k 2 + λ Z

|P u n | 2 dx + Z

g(x, u n )P u n dx.

(22)

(13)

Now, observe that (g 2 ) implies that

n→∞ lim Z

|g(x, u n )P u n | dx ku n k p−1 = 0,

since |g(x, u n )P u n | ≤ kP u n k (a 1 + a 2 |u n | p−1 ) and kP u n k → 0 as n → ∞ by assumption. So, starting from (22), using (g 4 ) and dividing by ku n k p−1 (p − 1 >

1), we get

n→∞ lim Z

G(x, u n ) dx ku n k p−1 = 0.

Moreover, (g 5 ) implies that

n→∞ lim Z

|u n | p dx ku n k p−1 = 0, and so u ≡ 0. Now, dividing 2f λ (u n ) by ku n k 2 , we have

n→∞ lim Z

G(x, u n ) dx ku n k 2 = 1

2 (23)

and so, by (g 5 ), there exists a constant c > 0 such that Z

|u n | p dx ≤ cku n k 2 . (24) Now, let us show that

n→∞ lim Z

|g(x, u n )P u n | dx

ku n k 2 = 0. (25)

Indeed, (g 2 ) and H¨ older’s inequality imply that

n→∞ lim Z

|g(x, u n )P u n | dx ku n k 2

≤ kP u n k ku n k 2

 a 1 + a 2

Z

|u n | p−1 dx



≤ kP u n k

"

a 1

ku n k 2 + a 0 2 ku n k 2/p

 R

Ω |u n | p dx ku n k 2

 (p−1)/p # ,

and equality (25) follows from (24) and the fact that p > 2.

(14)

In this way (22), (g 4 ) and (25) imply

n→∞ lim Z

G(x, u n ) dx ku n k 2 = 0, which contradicts (23).

Now, by Lemma 3.2 and Lemma 3.3, we can prove the following result:

Proposition 3.4. Suppose there exist integers i, j ≥ 2 such that λ i−1 < λ i =

· · · = λ j < λ j+1 . Then, for all δ > 0 there exists  0 > 0 such that ∀ λ ∈ [λ i−1 + δ, λ j+1 −δ] and ∀  0 ,  00 ∈ (0,  0 ) with  0 <  00 , condition (∇)(f λ , H i−1 ⊕H j ,  0 ,  00 ) holds (see the Appendix).

Before proving Proposition 3.4, we give a property of the operator K defi- nited in (21):

Lemma 3.5. K : L 2 (Ω) −→ H 0,L 1 (C) is compact.

Proof. Let (u n ) n ⊂ L 2 (Ω) be a bounded sequence and set v n := K(u n ) for all n ∈ N. Since, by (21), we have

kv n k 2 ≤ kuk L

2

(Ω) kvk L

2

(Ω) ,

we immediately get that (v n ) n is bounded. Thus, we can suppose that there exists v ∈ H 0,L 1 (C) such that v n * v in H 0,L 1 (C) and v n → v in L 2 (Ω) as n → ∞.

In this way, since Z

C

|Dv n | 2 dxdy − Z

C

Dv n · Dv dx − Z

u n (v n − v) dx = 0

for all n ∈ N, exploiting the previous convergences and recalling that (u n ) n is bounded in L 2 (Ω), we immediately have that

kv n k 2 → kvk 2 as n → ∞, that is, v n → v in H 0,L 1 (C) as n → ∞.

Proof of Proposition 3.4. Assume by contradiction that there exists δ > 0 such that for all  0 > 0 there exist λ ∈ [λ i−1 + δ, λ j+1 − δ] and  0 ,  00 ∈ (0,  0 ), such that condition (∇)(f λ , H i−1 ⊕ H j ,  0 ,  00 ) doesn’t hold.

Take  0 > 0 as given by Lemma 3.2, and take a sequence (u n ) n in H 0,L 1 (C) such that f λ (u n ) ∈ [ 0 ,  00 ] for every n ∈ N, d(u n , H i−1 ⊕ H j ) → 0 and Q∇f λ (u n ) → 0 as n → ∞, i.e. (u n ) is a sequence which makes (∇)(f λ , H i−1 ⊕ H j ,  0 ,  00 ) false.

Then, by Lemma 3.3, we get that (u n ) n is bounded and we can assume that u n * u in H 0,L 1 (C) and u n → u in L r (Ω) for all r ∈ h

1, N −1 2N 

as n → ∞.

(15)

Now,

Q∇f λ (u n ) = u n − P u n − K(λu n + g(x, u n )),

and we know that g(x, u n ) → g(x, u) in L p/(p−1) (Ω) as n → ∞ by (g 2 ). More- over, Q∇f λ (u n ) → 0 and P u n as n → ∞. Thus, by Lemma 3.5, we find that u n → u in H i−1 ⊕ H j and u is a critical point of f λ on H i−1 ⊕ H j . Therefore, by Lemma 3.2, we know that u = 0, while 0 <  0 ≤ f λ (u n ) for all n ∈ N, which contradicts the fact that f λ is continuous. Hence, the claim follows.

In order to apply the ∇-Theorem (see the Appendix), at this point we have only to show that sup f λ (B j (R)) is small enough. In order to do that, we need the following Lemma.

Lemma 3.6.

lim

λ→λ

j

sup f λ (H j ) = 0.

Proof. Assume by contradiction that there exist  > 0 two and sequences µ n → λ j , (u n ) n in H j , such that

sup f µ

n

(H j ) = f µ

n

(u n ) ≥  ∀ n ∈ N.

Note that (u n ) n is well defined, since f µ

n

attains a maximum its H j thanks to condition (g 5 ).

If (u n ) n is bounded, we can assume that u n → u in H j . In this way, by continuity,

 ≤ 1 2 Z

C

|Du| 2 dxdy − λ j

2 Z

u 2 dx − Z

G(x, u) dx ≤ 0, by (8) and (g 5 ), that is an absurd. So we can assume that ku n k → ∞.

In this case, by (g 5 ) and the Sobolev inequality, there exists γ p > 0 such that

0 <  ≤ f µ

n

(u n ) ≤ 1

2 ku n k 2 − µ n

2λ j

ku n k 2 − c 1 γ p ku n k p , (26) and since all norms are equivalent in H j , the right hand side of the last inequality would tend to −∞, which is absurd again.

In order to prove Theorem 1.4, first we need some preliminary results.

Theorem 3.7. Suppose there exist integers i, j ≥ 2 such that λ i−1 < λ i =

· · · = λ j < λ j+1 . Then, there exist δ 1 > 0,  0 ,  00 > 0 such that for every λ ∈ (λ j − δ 1 , λ j ) problem (1) has at least two non trivial solutions u 1 , u 2 with f λ (u i ) ∈ [ 0 ,  00 ], i = 1, 2.

Proof. Take δ 0 > 0 and find  0 as in Proposition 3.4. Fix  0 <  00 <  0 . Then,

by Lemma 3.6, there exists δ 1 ≤ δ 0 such that, if λ ∈ (λ j − δ 1 , λ j ), we have

sup f λ (H j ) <  00 and by Proposition 3.4 condition (∇) − (f λ , H i−1 ⊕ H j ,  0 ,  00 )

holds. Moreover, since λ < λ j , the topological situation of Lemma 3.1 is satis-

fied.

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By the ∇-Theorem (see Theorem 4.3 in the Appendix), there exist two critical points u 1 , u 2 of f λ such that f λ (u i ) ∈ [ 0 ,  00 ], i = 1, 2. In particular u 1

and u 2 are nontrivial solutions of (1), since f λ (0) = 0.

In order to prove the existence of a third nontrivial solution, let us prove the following Lemma.

Lemma 3.8. Suppose there exist integers i, j ≥ 2 such that λ i−1 < λ i = · · · = λ j < λ j+1 . Then, there exists δ i > 0, ρ 1 > 0 and R 1 > ρ 1 such that for all λ ∈ (λ i − δ i , λ i ),

inf f λ (S j + (ρ 1 )) > sup f λ (T i,j+1 (R 1 )).

In particular there exists a critical point u 3 of f λ such that f λ (u 3 ) ≥ inf f λ (S j +1 )).

Proof. Take λ ∈ [λ i−1 , λ i ). By (9) and (g 5 ) we get that for all τ > 0 there exist ρ 1 > 0 such that, if v ∈ H j and kvk = ρ 1 , then

f λ (v) ≥ 1 2

 1 − λ

λ j+1

− τ



ρ 2 1 . (27)

Choosing τ small enough, we get that C = 1 − λ j λ j+1

− τ > 0, so that (27) reads

f λ (v) ≥ Cρ 2 1 .

By (g 5 ) and (8), as in (26), we get that f λ (u) → −∞ if u ∈ H j+1 and kuk → ∞, since all the norms in H j+1 are equivalent. By Lemma 3.6, there exists δ i > 0 such that ∀ λ in (λ i − δ i , λ i ), we have

sup f λ (H j ) < Cρ 2 1 .

Finally, we choose δ i < δ 1 , where δ 1 is the one found in Theorem 3.7.

In this way, the classical Linking Theorem (see the Appendix) shows the existence of a critical point u 3 of f λ such that f λ (u 3 ) ≥ Cρ 2 1 .

Note that, although the topological structure found in Lemma 3.8 is equal to the one of Lemma 3.1, it is not possible to apply ∇-Theorem again, since it is not clear if (∇) − 

f λ , H j ⊕ H j+1 , Cρ 2 1 , sup f λ (B j+1 (R 1 ))  holds.

Now, we are ready to prove Theorem 1.4, which becomes an obvious corollary of all the previous results.

Proof of Theorem 1.4. Take δ i as given in Lemma 3.8. Then the critical point u 3 found there is different from the critical points u 1 , u 2 found in Theorem 3.7, since

f λ (u i ) ≤ sup f λ (H j ) < Cρ 2 1 ≤ f λ (u 3 ),

i = 1, 2.

(17)

4 Appendix

Though it is a well-known result in critical point theory, we recall the Linking Theorem (see [16, Theorem 5.3] or [17, Theorem 1.1]), in order to state an a priori estimate on critical values which we will use to prove Theorem 1.4.

Theorem 4.1 (Linking Theorem). Let X be a Banach space such that X = X 1 ⊕ X 2 with dim X 1 < ∞, and let f : X → R be of class C 1 . If

i) f (0) = 0;

ii) there exist ρ, α > 0 such that inf f (S ρ ∩ X 2 ) ≥ α and ∃ R > ρ, e ∈ X 2 with kek = 1 such that sup f (Σ R ) ≤ 0, where S ρ denotes the sphere in X of radius ρ,

Σ R = ∂ X

1

⊕span(e) ∆ R and

∆ R = {u + te : u ∈ X 1 , t > 0, ku + tek ≤ R}.

iii) (P S) β holds, where

β = inf

h∈H sup

u∈∆

R

f (h(u)) and

H = {h ∈ C(∆ R , X) : h

R

= Id}.

Then, β is a critical value for f , and α ≤ β ≤ sup

u∈∆

R

f (u).

Definition 4.2 (see [7]). Let X be a Hilbert space, f : X −→ R a C 1 function, M a closed subspace of X, a, b ∈ R ∪ {−∞, +∞}.

We say that the condition (∇) − (f, M, a, b) holds if there exists γ > 0 such that

inf n

kP M ∇f (u)k : a ≤ f (u) ≤ b, d(u, M ) ≤ γ o

> 0,

where P M : X −→ M is the orthogonal projection of X on M and d(u, M ) =

z∈M inf d(u, z) denotes the distance of u from the subspace M .

This means that, if the condition above holds, then f |

M

cannot have critical points u with a ≤ f (u) ≤ b with some uniformity.

Theorem 4.3 (∇-Theorem, see [7]). Let X be a Hilbert space and let X i , i = 1, 2, 3 be three subspaces of X such that X = X 1 ⊕ X 2 ⊕ X 3 with dim X i < ∞ for i = 1, 2. Denote with P i : X −→ X i the orthogonal projection of X on X i . Let f : X −→ R be a function of class C 1 .

Let ρ, ρ 0 , ρ 00 , ρ 1 be such that ρ 1 > 0, 0 ≤ ρ 0 < ρ < ρ 00 and define

∆ = n

u ∈ X 1 ⊕ X 2 : ρ 0 ≤ kP 2 uk ≤ ρ 00 , kP 1 uk ≤ ρ 1

o

,

(18)

T = ∂ X

1

⊕X

2

∆, S 23 (ρ) = n

u ∈ X 2 ⊕ X 3 : kuk = ρ o and

B 23 (ρ) = n

u ∈ X 2 ⊕ X 3 : kuk ≤ ρ o . Suppose that

a 0 := sup f (T ) < inf f (S 23 (ρ)) =: a 00 .

Let a and b be such that a 0 < a < a 00 e b > sup f (∆). Suppose that condition (∇) − (f, X 1 ⊕ X 3 , a, b) holds and (P S) c holds for all c ∈ [a, b]. Then, f has at least two critical points in f −1 ([a, b]). Moreover, if

inf f (B 23 (ρ)) > −∞

and (P S) c holds for all c ∈ [a 1 , b] with

a 1 < inf f (B 23 (ρ)), then f has another critical level in [a 1 , a 0 ].

Note that, in our context, the critical level in [a 1 , a 0 ] could be 0, so that no other nontrivial solution is obtained.

Acknowledgement

The authors would like the thank the anonymous referee for her/his careful reading of the manuscript and her/his comments, which improved the final presentation.

References

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Dimitri Mugnai 1 : Dipartimento di Matematica e Informatica, University of Perugia, Via Vanvitelli 1, 06123 Perugia - Italy

e-mail: dimitri.mugnai@unipg.it

Dayana Pagliardini: Scuola Normale Superiore, Piazza dei Cavalieri 7, 56126 Pisa- Italy

e-mail: dayana.pagliardini@sns.it

1

D. M. is member of the Gruppo Nazionale per l’Analisi Matematica, la Probabilit` a e le

loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM), and is

supported by the GNAMPA project Systems with irregular operators

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