Multiple solutions for a quasilinear Schrödinger equation

J. Differential Equations 254 (2013) 2015–2032

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Journal of Differential Equations www.elsevier.com/locate/jde

Multiple solutions for a quasilinear Schrödinger equation Xiang-Dong Fang a,1 , Andrzej Szulkin b,∗,2 a b

School of Mathematical Sciences, Dalian University of Technology, 116024 Dalian, PR China Department of Mathematics, Stockholm University, 106 91 Stockholm, Sweden

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 20 August 2012 Revised 12 November 2012 Available online 28 November 2012

In this paper we consider the quasilinear Schrödinger equation

Keywords: Quasilinear Schrödinger equation Multiplicity of solutions Nehari manifold

where g and V are periodic in x1 , . . . , x N and g is odd in u, subcritical and satisfies a monotonicity condition. We employ the approach developed in Szulkin and Weth (2009, 2010) [15,16] and obtain infinitely many geometrically distinct solutions. © 2012 Elsevier Inc. All rights reserved.

  −u + V (x)u −  u 2 u = g (x, u ),

x ∈ RN ,

1. Introduction The quasilinear Schrödinger equation

  −u + V (x)u −  u 2 u = g (x, u ),

x ∈ RN

(1.1)

has been studied recently by several authors, see [1–5,7–11,13,17,19] and the references there. This is a special case of a more general equation which models certain phenomena in physics, see e.g. [7,9,11] for an explanation. In most of the papers mentioned above the existence of positive solutions has been considered, under different assumptions on V and g. A major difficulty here is that the natural functional corresponding to (1.1) is not well defined for all u ∈ H 1 (R N ) if N  2, see Section 2. Despite this problem the existence of a positive solution has

* 1 2

Corresponding author. E-mail addresses: [email protected] (X.-D. Fang), [email protected] (A. Szulkin). Supported by the China Scholarship Council and NSFC 11171047. Supported in part by the Swedish Research Council.

0022-0396/$ – see front matter © 2012 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jde.2012.11.017

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X.-D. Fang, A. Szulkin / J. Differential Equations 254 (2013) 2015–2032

been proved in [9] by using a constrained minimization argument. In [7], by a change of variables, the quasilinear problem (1.1) was transformed to a semilinear one and an Orlicz space framework was used. A slightly different change of variables was used in [1,2], and in [2] a simpler and shorter proof of the results in [7], which does not use Orlicz spaces, was given (see also [13]). In [5,17], Eq. (1.1) with ε 2 in front of u and (u 2 )u, has been studied, with g respectively of subcritical and critical growth. It was shown that there exists a positive solution u ε which concentrates at a local minimum of V as ε → 0. In a recent paper [10] existence of multibump solutions was shown for a quasilinear Schrödinger equation which is more general than (1.1). Instead of transforming the problem to a semilinear one (which does not seem to be possible in this case), the authors have worked with the natural functional which, after a truncation, is defined and continuous on H 1 (R N ). Since this truncated functional is not of class C 1 , nonsmooth critical point theory was used. To the best of our knowledge, except for [10], there are no results about the existence of infinitely many solutions for Eq. (1.1) (however, see [11] where infinitely many solutions were found in the ODE case). Motivated by the arguments in [2,13,17], we also make a change of variables, and then we use the method developed in [15,16] in order to obtain such result. We consider the problem

  −u + V (x)u −  u 2 u = g (x, u ), Setting G (x, u ) :=

u 0





u ∈ H 1 RN .

(1.2)

g (x, s) ds, we suppose that V and g satisfy the following assumptions:

( V ) V is continuous, 1-periodic in xi , 1  i  N, and there exists a constant a0 > 0 such that V (x)  a0 for all x ∈ R N . ( g 1 ) g is continuous, 1-periodic in xi , 1  i  N, and | g (x, u )|  a(1 + |u | p −1 ) for some a > 0 and 4 < p < 2 · 2∗ , where 2∗ := 2N /( N − 2) if N  3, 2∗ := ∞ if N = 1 or 2. ( g 2 ) g (x, u ) = o(u ) uniformly in x as u → 0. ( g 3 ) G (x, u )/u 4 → ∞ uniformly in x as |u | → ∞. ( g 4 ) u → g (x, u )/u 3 is positive for u = 0, nonincreasing on (−∞, 0) and nondecreasing on (0, ∞). Note that ( g 2 ), ( g 4 ) imply g (x, u ) = O (u 3 ) as u → 0. Let ∗ denote the action of Z N on H 1 (R N ) given by

(k ∗ u )(x) := u (x − k),

k ∈ ZN .

(1.3)

It follows from ( V ) and ( g 1 ) that if u 0 is a solution of (1.2), then so is k ∗ u 0 for all k ∈ Z N . Set

  O(u 0 ) := k ∗ u 0 : k ∈ Z N . O(u 0 ) is called the orbit of u 0 with respect to the action of Z N , and it is called a critical orbit for a functional F if u 0 is a critical point of F and F is Z N -invariant, i.e., F (k ∗ u ) = F (u ) for all k ∈ Z N and all u (then of course all points of O (u 0 ) are critical). Two solutions u 1 , u 2 of (1.2) are said to be geometrically distinct if O (u 1 ) = O (u 2 ). The first main result of this paper, which we prove in Section 3, is the following Theorem 1.1. Suppose that ( V ), ( g 1 )–( g 4 ) are satisfied and g is odd in u. Then (1.2) admits infinitely many pairs ±u of geometrically distinct solutions. In Section 4 we let g (x, u ) = q(x)u 3 , that is, we consider the problem

  −u + V (x)u −  u 2 u = q(x)u 3 , and we assume that





u ∈ H 1 RN ,

(1.4)

X.-D. Fang, A. Szulkin / J. Differential Equations 254 (2013) 2015–2032

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( Q ) q is continuous, 1-periodic in xi , 1  i  N, and there exists a constant b0 > 0 such that q(x)  b0 for all x ∈ R N . Our second main result is Theorem 1.2. Suppose that ( V ) and ( Q ) are satisfied. Then (1.4) admits infinitely many pairs ±u of geometrically distinct solutions. Remark 1.1. In [10] it was assumed that g and ∂ g /∂ u are continuous, g satisfies ( g 1 ), ( g 2 ) and the Ambrosetti–Rabinowitz type condition ug (x, u )  pG (x, u ) > 0 for some p > 4 and all u = 0. Here our ( g 3 ) is weaker and g need not be differentiable, but instead we need a monotonicity condition ( g 4 ) which does not follow from the hypotheses in [10]. A simple example of a function satisfying our assumptions but not conditions in [10] is g (u ) = u 3 ln(1 + |u |). Another example is the function g in our Theorem 1.2. Also, our proofs are different and much simpler. On the other hand, g in [10] need not be odd and the equation is more general than (1.1) (and our solutions are not of multibump type). We would also like to point out that in one of the main results of a recent paper [19] the existence of a positive solution has been shown under assumptions which are somewhat stronger than in our Theorem 1.1. It will be easily seen from our arguments that we may also obtain such a solution under the assumptions of Theorem 1.1 as well as under the assumptions of Theorem 1.2. More precisely, the functional I introduced in the next section attains a minimum on the Nehari manifold M (cf. [15,16]) and if u is a minimizer, then so is |u |. Hence the conclusion. As we have mentioned, we do not assume the Ambrosetti–Rabinowitz condition. Different ways of weakening or replacing this condition have been studied extensively in recent years, starting with [6]. Notation. C , C 1 , C 2 , . . . will denote different positive constants whose exact value is inessential. | A | is the Lebesgue measure of a measurable set A ⊂ R N . B ρ ( y ) := {x ∈ R N : |x − y | < ρ }. The usual norm in the Lebesgue space L p (Ω) is denoted by u p ,Ω , and by u p if Ω = R N . E denotes the Sobolev space H 1 (R N ) and S is the unit sphere in E. It follows from ( V ) that



1/2



2

|∇ u | + V (x)u

u :=

2



RN

is an equivalent norm in E. It is more convenient for our purposes than the standard one and will be used henceforth. For a functional F we put









F d := u: F (u )  d ,

F c := u: F (u )  c ,





F cd := u: c  F (u )  d .

2. Preliminary results In this section we introduce a variational framework associated with problem (1.2). We observe that (1.2) is formally the Euler–Lagrange equation associated with the energy functional

J (u ) :=

1



2 RN



2



2

1 + 2u |∇ u | +

1



 2

V (x)u −

2 RN

G (x, u ).

(2.1)

RN

It is difficult to apply variational methods in order to study the functional J because, unless N = 1, J is not defined for all u in the space E ≡ H 1 (R N ) which is natural for the problem. To overcome this difficulty, we employ an argument developed in [1,2] (see also [13]). As we have mentioned in the introduction, a somewhat earlier and slightly different variant of this argument may be found in [7]. We make a change of variables v := f −1 (u ), where f is defined by

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X.-D. Fang, A. Szulkin / J. Differential Equations 254 (2013) 2015–2032

f (t ) =

1

on [0, +∞)

(1 + 2 f 2 (t ))1/2

f (t ) = − f (−t ) on (−∞, 0].

and

Below we summarize the properties of f . Proofs may be found in [2–4,17]. Lemma 2.1. The function f satisfies the following properties:

(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12)

f is uniquely defined, C ∞ and invertible; | f (t )|  1 for all t ∈ R; | f (t )|  |t | for all t ∈ R; f (t )/t√→ 1 as t → 0; f (t )/ t → 21/4 as t → +∞; f (t )/2  t f (t )  f (t ) for all t > 0; | f (t )|  21/4 |t |1/2 for all t ∈ R; f 2 (t ) − f (t ) f (t )t  0 for all t ∈ R; 1/ 2 there exists a positive √ constant C such that | f (t )|  C |t | for |t |  1 and | f (t )|  C |t | for |t |  1; | f (t ) f (t )| < 1/ 2 for all t ∈ R; the function f (t ) f (t )t −1 is strictly decreasing for t > 0; the function f p (t ) f (t )t −1 is strictly increasing for p  3 and t > 0.

In [4,17] it is stated that the functions in (11) and (12) are respectively decreasing and increasing but it is easy to see from the proofs there that they are strictly decreasing and strictly increasing. Consider the functional

I ( v ) :=

1

 |∇ v |2 +

2

1



2

RN

 V (x) f 2 ( v ) −

RN





G x, f ( v ) .

(2.2)

RN

Then I is well defined on E and I ∈ C 1 ( E , R) under the hypotheses ( V ), ( g 1 ) and ( g 2 ). Note also that ( V ), ( g 1 ) imply I is invariant with respect to the action of Z N given by (1.3). It is easy to see that





I ( v ), w =



 ∇ v∇ w +

RN



−

V (x) f ( v ) f ( v ) w

RN





g x, f ( v ) f ( v ) w

(2.3)

RN

for all v , w ∈ E and the critical points of I are weak solutions of the problem

  − v + V (x) f ( v ) f ( v ) = g x, f ( v ) f ( v ),

v ∈ E.

It has been shown in [2] that if v ∈ E is a critical point of the functional I , then u = f ( v ) ∈ E and u is a solution of (1.2). Let

   M := v ∈ E \ {0}: I ( v ), v = 0 .

(2.4)

Recall that M is called the Nehari manifold. We do not know whether M is of class C 1 under our assumptions and therefore we cannot use minimax theory directly on M. To overcome this difficulty, we employ an argument developed in [15,16].

X.-D. Fang, A. Szulkin / J. Differential Equations 254 (2013) 2015–2032

2019

3. Proof of Theorem 1.1 We assume that ( V ) and ( g 1 )–( g 4 ) are satisfied from now on. First, ( g 1 ) and ( g 2 ) imply that for each ε > 0 there is C ε > 0 such that

g (x, u )  ε |u | + C ε |u | p −1 for all u ∈ R. Lemma 3.1. G (x, u )  0 and

1 g (x, u )u 4

(3.1)

 G (x, u ).

This follows immediately from ( g 2 ) and ( g 4 ). For t > 0, let

h(t ) := I (tu ) =



t2

|∇ u |2 +

2 RN

1



 V (x) f 2 (tu ) −

2 RN





G x, f (tu ) .

RN

Lemma 3.2. For each u = 0 there is a unique t u > 0 such that h (t ) > 0 for 0 < t < t u and h (t ) < 0 for t > t u . Moreover, tu ∈ M if and only if t = t u .

ε sufficiently small we obtain

Proof. By (3.1) and Lemma 2.1-(7), for

h(t ) 



t2

|∇ u |2 +

2

1



t

V (x) f 2 (tu ) −

2

RN

RN



2



RN

f 2 (tu ) −

2

Cε



p

RN



|∇ u |2 − C 1 t p /2

2



ε

f (tu ) p

RN

|u | p /2 .

RN

Since p > 4 and u is not constant, h(t ) > 0 whenever t > 0 is small enough. Using Lemma 2.1-(3), we have

h(t ) 

t2

 |∇ u |2 +

2

t2



t

RN



2

|∇ u | +

2



RN

t

2



V (x)u 2 −

2

RN 2



RN



2

V (x)u − t

2



G x, f (tu )



2 u =0

RN

G (x, f (tu )) f 4 (tu )

·

f 4 (tu )

(tu )2

· u2 .

It follows from ( g 3 ), Lemma 2.1-(5) and Fatou’s lemma that the last integral on the right-hand side above tends to infinity with t. Hence h(t ) → −∞ as t → ∞ and h has a positive maximum. The condition h (t ) = 0 is equivalent to



 2

|∇ u | = RN

u =0

g (x, f (tu )) f (tu ) tu

−

V (x) f (tu ) f (tu ) tu

Let

Z (s) :=

g (x, f (s)) f (s) s

−

V (x) f (s) f (s) s

.

 u2 .

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By ( g 4 ) and Lemma 2.1-(12) the function

g (x, f (s)) f (s)

=

s

g (x, f (s)) f 3 (s)

·

f (s) f 3 (s) s

is strictly increasing for s > 0. Hence also s → Z (s) is strictly increasing according to Lemma 2.1-(11). So there exists a unique t u > 0 such that h (t u ) = 0 and the first conclusion follows. The second conclusion is an immediate consequence of the fact that h (t ) = t −1  I (tu ), tu . 2 Lemma 3.3. (1) There exists ρ > 0 such that c := infM I  inf S ρ I > 0, where S ρ := {u ∈ E: u = ρ }.

(2) u 2  2c for all u ∈ M.

Proof. (1) We claim that there are C 1 , ρ > 0 such that





 |∇ u |2 + V (x) f 2 (u )  C 1 u 2 whenever u  ρ .

(3.2)

RN

If the claim is false, then for all n there exists un = 0 such that un → 0 in E and V (x) f (un ))  2

1 u n 2 . n

un u n .

Let v n :=





2

|∇ v n | +

R N (|∇ un |

2

+

Then

V (x) v n2



 2 f (u n ) 1 V (x) − 1 v n2  . 2

 +

RN

RN

n

un

Going if necessary to a subsequence, un → 0 a.e. Since un → 0 in L 2 (R N ), for every |{x ∈ R N : |un (x)| > ε }| → 0 as n → ∞. Hence by the Hölder inequality,





ε > 0 the measure

  (r −2)/r v n2  x ∈ R N : un (x) > ε v n r2 → 0,

(3.3)

|un |>ε

where r = 2∗ if N  3 and r > 2 if N = 1 or 2. Now it follows from Lemma 2.1-(4) that the second integral above goes to 0. So v n = 1 and v n → 0 in E, a contradiction. Using (3.1), Lemma 2.1-(3),(7) and the Sobolev inequality, it is easy to see that







G x, f (u )  RN



ε

f 2 (u ) +

2

Cε



p

RN

f (u ) p  C 2 ε u 2 + C 3 C ε u p /2 ,

RN

so choosing ε small enough and using the claim we have, for sufficiently small ρ , I (u )  C 4 u 2 − C 5 u p /2 and inf S ρ I > 0. The inequality infM I  inf S ρ I is a consequence of Lemma 3.2 since for every u ∈ M there is s > 0 such that su ∈ S ρ (and I (t u u )  I (su )). (2) For u ∈ M, by Lemma 2.1-(3) we have

c

1

 |∇ u |2 +

2 RN

1



 V (x) f 2 (u ) −

2 RN

RN





G x, f (u )

X.-D. Fang, A. Szulkin / J. Differential Equations 254 (2013) 2015–2032





1 2

RN



1

V (x) f 2 (u )

2

RN





1

|∇ u |2 +

2021

2



1

|∇ u |2 +

V (x)u 2 =

2

RN

1 2

u 2 .

2

RN

Lemma 3.4. I is coercive on M, i.e., I (u ) → ∞ as u → ∞, u ∈ M. Proof. We employ a similar argument as in [15], Proposition 2.7. Arguing by contradiction, let (un ) ⊂ M be a sequence such that un → ∞ and I (un )  d for some d. Let v n := un / un . Then v n  v in E and v n (x) → v (x) a.e. in R N up to a subsequence. Choose yn ∈ R N so that



 v n2 = max y ∈R N

B 1 ( yn )

v n2 .

(3.4)

B 1 ( y)

Since I and M are invariant with respect to the action of Z N given by (1.3), we may assume that ( yn ) is bounded in R N . If

 v n2 → 0 as n → ∞,

(3.5)

B 1 ( yn )

then it follows that v n → 0 in L r (R N ) for 2 < r < 2∗ by P.L. Lions’ lemma (cf. [18], Lemma 1.21). Using (3.1) and Lemma 2.1-(3),(7) we see that RN G (x, f (sv n )) → 0 for all s ∈ R. We have by Lemma 2.1-(9)

d  I (un )  I (sv n )

=

s2



|∇ v n |2 +

2

1

s2



RN



|∇ v n |2 +

2

2

=

s



|∇ v n |2 +

2 RN

−



s

2

RN

|sv n |1



min 1, C

2





V (x) v n2





V (x)(sv n )2 −

2



G x, f (sv n )

RN



2



C2 2

C s



V (x) v n2 − |sv n |1

2 2



G x, f (sv n )

RN



C 2 s2

RN 2



V (x) f 2 (sv n ) −

2

RN







G x, f (sv n )

RN

 − C1

(sv n )

RN

p /2

 −





G x, f (sv n ) →

s2 2





min 1, C 2 .

RN

Taking sufficiently large s we obtain a contradiction. Hence (3.5) cannot hold and since v n → v in L 2loc (R N ), v = 0. Since |un (x)| → ∞ if v (x) = 0, it follows from Lemma 2.1-(5), ( g 3 ) and Fatou’s lemma that

 RN

G (x, f (un ))

u n

2

 = RN

G (x, f (un )) f 4 (u

n)

·

f 4 (u n ) un2

· v n2 → ∞.

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X.-D. Fang, A. Szulkin / J. Differential Equations 254 (2013) 2015–2032

Thus

0

I (u n )

u n

2

1



2

 −

G (x, f (un ))

un 2

RN

→ −∞,

2

a contradiction. This completes the proof.

Lemma 3.5. If V is a compact subset of E \ {0}, then there exists R > 0 such that I  0 on (R+ V ) \ B R (0). Proof. We may assume without loss of generality that V ⊂ S. Arguing by contradiction, suppose there exist un ∈ V and w n = tn un such that I ( w n )  0 and tn → ∞ as n → ∞. Passing to a subsequence, we may assume that un → u ∈ S. Since | w n (x)| → ∞ if u (x) = 0, it follows from ( g 3 ), Lemma 2.1-(5) and Fatou’s lemma that



G (x, f ( w n )) tn2

RN

 =

G (x, f ( w n ))un2 w n2

RN



=

G (x, f ( w n )) f 4(w

RN

n)

·

f 4(wn) w n2

· un2 → ∞.

By Lemma 2.1-(3),

0

a contradiction.

I (wn) tn2



1 2

 − RN

G (x, f ( w n )) tn2

→ −∞,

2

Recall that S is the unit sphere in E and define the mapping m : S → M by setting

m( w ) := t w w , where t w is as in Lemma 3.2. Note that m( w ) = t w . Lemmas 3.6 and 3.7 below are taken from [16] (see Proposition 8 and Corollary 10 there). That the hypotheses in [16] are satisfied is a consequence of Lemmas 3.2, 3.3 and 3.5 above. Indeed, if h(t ) = I (t w ) and w ∈ S, then h (t ) > 0 for 0 < t < t w and h (t ) < 0 for t > t w by Lemma 3.2, t w  δ > 0 by Lemma 3.3 and t w  R for w ∈ V ⊂ S by Lemma 3.5. Lemma 3.6. The mapping m is a homeomorphism between S and M, and the inverse of m is given by m−1 (u ) = uu . We shall consider the functional Ψ : S → R given by

  Ψ ( w ) := I m( w ) . Lemma 3.7. (1) Ψ ∈ C 1 ( S , R) and



      Ψ ( w ), z = m( w ) I m( w ) , z for all z ∈ T w ( S ).

X.-D. Fang, A. Szulkin / J. Differential Equations 254 (2013) 2015–2032

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(2) If ( w n ) is a Palais–Smale sequence for Ψ , then (m( w n )) is a Palais–Smale sequence for I . If (un ) ⊂ M is a bounded Palais–Smale sequence for I , then (m−1 (un )) is a Palais–Smale sequence for Ψ . (3) w is a critical point of Ψ if and only if m( w ) is a nontrivial critical point of I . Moreover, the corresponding values of Ψ and I coincide and inf S Ψ = infM I . (4) If I is even, then so is Ψ . From now on we assume that the nonlinearity g = g (x, u ) is odd in u. As a consequence of Lemma 3.7 we see as in Remark 2.12 of [15] that since m, m−1 are equivariant and I , Ψ are invariant with respect to the action of Z N given by (1.3), there is a one-to-one correspondence between the critical orbits of I |M and Ψ . Hence the proof of Theorem 1.1 will be completed once we show that Ψ has infinitely many critical orbits. Lemma 3.8. The mapping m−1 defined in Lemma 3.6 is Lipschitz continuous. Proof. We use an argument from [15]. By Lemma 3.3-(2) we have, for all u , v ∈ M,

        −1   m (u ) − m−1 ( v ) =  u − v  =  u − v + ( v − u ) v   u v   u u v   

2

u

u − v 

2 c

u − v .

2

Below we shall use the notation





K := w ∈ S: Ψ ( w ) = 0 ,





K d := w ∈ K : Ψ ( w ) = d .

Choose a subset F of K such that F = −F and each orbit O ( w ) ⊂ K has a unique representative in F . According to the remark preceding Lemma 3.8 we must show that the set F is infinite. Arguing indirectly, assume

F is a finite set.

(3.6)

In Lemma 3.13 below we shall show that Palais–Smale sequences have a certain discreteness property. First we need some preparations. The following result has been proved in Lemma 2.13 of [15]: Lemma 3.9. κ := inf{ v − w : v , w ∈ K , v = w } > 0. Lemma 3.10. For every C > 0 there exists δ > 0 such that

d ( f (t ) f (t )) dt

 δ > 0 as |t |  C .

Proof. Since

 d f (t ) d f (t ) 1 f (t ) f (t ) = = = 2 1 / 2 2 3 / 2 dt dt (1 + 2 f (t )) (1 + 2 f (t )) (1 + 2 f 2 (t ))2 and f is bounded for |t |  C , the conclusion follows.

2

Lemma 3.11. If {un1 }, {un2 } are bounded in E, then there exists C > 0, depending only on the bound on un1 and un2 , such that

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X.-D. Fang, A. Szulkin / J. Differential Equations 254 (2013) 2015–2032





 1  ∇ u − u 2 2 + n n

RN

 C

          V (x) f un1 f un1 − f un2 f un2 un1 − un2

RN

 1      ∇ u − u 2 2 + V (x) u 1 − u 2 2 ≡ C u 1 − u 2 2 . n n n n n n

RN

Proof. We may assume un1 = un2 (otherwise the conclusion is trivial). Set

w n :=

un1 − un2

un1

gn (x) :=

and

− un2

f (un1 ) f (un1 ) − f (un2 ) f (un2 ) un1 − un2

.

We argue by contradiction and assume that un1 , un2 may be found so that



 2

V (x) gn (x) w n2 → 0.

|∇ w n | + RN

(3.7)

RN

Since f (t ) f (t ) is strictly increasing by Lemma 3.10, gn (x) is positive if w n (x) = 0. Hence



 |∇ w n |2 → 0 and

RN

V (x) w n2 → 1.

(3.8)

RN

For a given C 1 > 0, let A n := {x ∈ R N : |un1 (x)|  C 1 or |un2 (x)|  C 1 }, B n := R N \ A n . Then for each ε > 0, C 1 may be chosen so that | An |  ε . It follows from Lemma 3.10, (3.7) and the Mean Value Theorem that



 V (x) w n2

δ Bn

Choosing

V (x) gn (x) w n2 → 0.



(3.9)

Bn

ε small enough and arguing as in (3.3) (with the same r), we obtain 

V (x) w n2  C 2 ε (r −2)/r 

1 2

An

because C 2 is independent of ε . This and (3.9) contradict (3.8). Lemma 3.12. If {un1 }, {un2 } are bounded in E, then for each bound on un1 and un2 , such that

2

ε > 0 there exists C ε > 0, depending only on the

       hn (x) u 1 − u 2  ε u 1 − u 2  + C ε u 1 − u 2  , n n n n n n p /2 RN

where





hn (x) := g x, f un1

 







f un1 − g x, f un2

 



f un2 = hn1 (x) − hn2 (x).

X.-D. Fang, A. Szulkin / J. Differential Equations 254 (2013) 2015–2032

2025

Proof. By (3.1), Lemma 2.1-(6),(3),(7), the Hölder and the Sobolev inequalities, we have

    (ε | f (un1 )| + C ε | f (un1 )| p −1 ) f (un1 )un1 1 h1 (x) u 1 − u 2  · un − un2 n n n 1 u RN

n

RN

 



ε

RN





f 2 (un1 )

|un1 |

|un1 − un2 | + C ε

| f (un1 )| p 1 un − un2 1 |u n |



( p −2)/2 1   1 1 u − u 2 ε u u − u 2 + 2 p/4 C ε u 1 n

n

n

n

n

n

RN

   ( p −2)/2  1  u − u 2   C 1 ε un1 − un2  + 2 p /4 C ε un1  p /2 n n p /2      C 1 ε u 1 − u 2  + C 2 C ε u 1 − u 2  . n

n

n

n p /2

Here it should be understood that the integrands in the first two lines are zero if un1 (x) = 0. Similarly we have

       h2 (x) u 1 − u 2  C 1 ε u 1 − u 2  + C 2 C ε u 1 − u 2  . n n n n n n n p /2 RN

Since C 1 , C 2 depend only on the bound on un1 and un2 but are independent of ε and the particular choice of un1 , un2 , we may replace C 1 ε by ε /2 and C 2 C ε by C ε /2. The conclusion follows. 2 Lemma 3.13. Let d  c. If ( v n1 ), ( v n2 ) ⊂ Ψ d are two Palais–Smale sequences for Ψ , then either v n1 − v n2 → 0 as n → ∞ or lim supn→∞ v n1 − v n2  ρ (d) > 0, where ρ (d) depends on d but not on the particular choice of Palais–Smale sequences. Proof. Our argument is an adaptation of the proof of Lemma 2.14 in [15]. Let un1 := m( v n1 ) and un2 := m( v n2 ). Then (un1 ), (un2 ) are Palais–Smale sequences for I and since (un1 ), (un2 ) ⊂ I d , these sequences are bounded. We consider two cases. Case 1: un1 − un2 p /2 → 0 as n → ∞. It follows from Lemmas 3.11 and 3.12 that for each n large enough,



2

C un1 − un2  



 1  ∇ u − u 2 2 + n n

RN





V (x) gn (x) un1 − un2

RN

   

  = I un1 , un1 − un2 − I un2 , un1 − un2 +      2ε  u 1 − u 2  + C ε  u 1 − u 2  n

n

n



ε > 0 and

2



hn (x) un1 − un2



RN

n p /2 ,

where gn , hn are as in Lemmas 3.11 and 3.12, and C does not depend on the choice of un1 − un2 → 0 and Lemma 3.8 implies v n1 − v n2 = m−1 (un1 ) − m−1 (un2 ) → 0.

ε . Therefore

Case 2: un1 − un2 p /2 does not go to 0 as n → ∞. It follows from Lemma 1.21 in [18] that there exist ε > 0 and yn ∈ R N such that

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X.-D. Fang, A. Szulkin / J. Differential Equations 254 (2013) 2015–2032





un1 − un2

2





= max y ∈R N

B 1 ( yn )

un1 − un2

2

 ε for all n

(3.10)

B 1 ( y)

after passing to a subsequence. Since m, m−1 and I , Ψ are equivariant with respect to the action of Z N given by (1.3), we may assume that the sequence ( yn ) is bounded in R N . Passing to a subsequence once more, there exist u 1 , u 2 and α 1 , α 2 such that

un1  u 1 ,

un2  u 2 ,

un1 → α 1 ,

un2 → α 2

and I (u 1 ) = I (u 2 ) = 0. According to (3.10), u 1 = u 2 and by Lemma 3.3-(2),

√

2c  α i  ν (d) < ∞,





where ν (d) := sup u : u ∈ I d ∩ M and i = 1, 2

(that ν (d) < ∞ is a consequence of Lemma 3.4). Suppose u 1 , u 2 = 0. Then u 1 , u 2 ∈ M and v 1 := m−1 (u 1 ) ∈ K , v 2 := m−1 (u 2 ) ∈ K , v 1 = v 2 . Hence





 1      un un2   u 1 u2    = β1 v 1 − β2 v 2 ,  − −   1 2  1 2 α α u n u n

lim inf v n1 − v n2  = lim inf  n→∞

n→∞

where

β1 :=

u 1

α1

√ 

2c

ν (d)

and

β2 :=

u 2

α2

√ 

2c

ν (d)

.

Since v 1 = v 2 = 1, it is easy to see from the inequalities above that

√  1     κ 2c  1 2 1 2 2    lim inf v n − v n  β1 v − β2 v  min{β1 , β2 } v − v  n→∞ ν (d)

(3.11)

(κ is the constant in Lemma 3.9). Hence (3.11) implies lim infn→∞ v n1 − v n2  ρ (d) > 0, where depends only on d (via ν (d)). If u 2 = 0, then u 1 = 0 and

ρ (d)

√  1   1   un un2  u 1 2c 2    lim inf v n − v n = lim inf 1 − .  1  2  n→∞ n→∞ u ν ( d) α u n n The case u 1 = 0 is similar.

2

It is well known that Ψ admits a pseudo-gradient vector field H : S \ K → T S (see e.g. [12], Remark A.17-(iv) or [14], p. 86). Moreover, since Ψ is even, we may assume H is odd. Let η : G → S \ K be the flow defined by

⎧ ⎨ d ⎩

dt





η(t , w ) = − H η(t , w ) ,

η(0, w ) = w ,

where

  G := (t , w ): w ∈ S \ K , T − ( w ) < t < T + ( w )

(3.12)

X.-D. Fang, A. Szulkin / J. Differential Equations 254 (2013) 2015–2032

2027

and ( T − ( w ), T + ( w )) is the maximal existence time for the trajectory t → η(t , w ). Note that η is odd in w because H is and t → Ψ (η(t , w )) is strictly decreasing by the properties of a pseudogradient. Let P ⊂ S, δ > 0 and define





U δ ( P ) := w ∈ S: dist( w , P ) < δ .

η which will be needed in the proof of Theorem 1.1.

Below we summarize the properties of Ψ and

Lemma 3.14. Let d  c. Then for every δ > 0 there exists ε = ε (δ) > 0 such that (a) Ψdd−+εε ∩ K = K d and (b) limt → T + ( w ) Ψ (η(t , w )) < d − ε for w ∈ Ψ d+ε \ U δ ( K d ). Part (a) is an immediate consequence of (3.6) and (b) has been proved in [15], see Lemmas 2.15 and 2.16 there. The argument is exactly the same except that S + should be replaced by S. We point out that an important role in the proof of Lemma 3.14 is played by the discreteness property of the Palais–Smale sequences expressed in Lemma 3.13. Proof of Theorem 1.1. Our argument is borrowed from [15] where it appears in the proof of Theorem 1.2. Let

Σ : = { A ⊂ S: A = A , A = − A }. Recall [12,14] that for A ⊂ Σ , the Krasnoselskii genus γ ( A ) is by definition the smallest integer k for which there exists an odd mapping A → Rk \ {0}. If there is no such mapping for any k, then γ ( A ) := +∞. Moreover, γ (∅) := 0. Define



ck := inf d ∈ R:







γ Ψ d  k , k  1.

Thus the ck ’s are those numbers at which the sets Ψ d change genus and it is easy to see that ck  ck+1 . Let k  1 and set d := ck . By Lemma 3.9, K d is either empty or a discrete set, hence γ ( K d ) = 0 or 1. By the continuity property of the genus there exists δ > 0 such that γ ( U ) = γ ( K d ), where U := U δ ( K d ) and δ < κ2 . For such δ , choose ε > 0 so that the conclusions of Lemma 3.14 hold. Then

for each w ∈ Ψ d+ε \ U there exists t ∈ [0, T + ( w )) such that Ψ (η(t , w )) < d − ε . Let e = e ( w ) be the infimum of the time for which Ψ (η(t , w ))  d − ε . Since d − ε is not a critical value of Ψ , it is easy to see by the implicit function theorem that e is a continuous mapping and since Ψ is even, e (− w ) = e ( w ). Define a mapping h : Ψ d+ε \ U → Ψ d−ε by setting h( w ) := η(e ( w ), w ). Then h is odd and continuous, so it follows from the properties of the genus and the definition of ck that









γ Ψ d +ε  γ ( U ) + γ Ψ d −ε  γ ( U ) + k − 1 = γ ( K d ) + k − 1. If γ ( K d ) = 0, then γ (Ψ d+ε )  k − 1, contrary to the definition of ck . So γ ( K d ) = 1 and K d = ∅. If ck+1 = ck = d, then γ ( K d ) > 1 (see e.g. [12], Proposition 8.5 or [14], Lemma II.5.6). Since this is impossible, we must have ck+1 > ck and K ck = ∅ for all k  1, a contradiction to (3.6). The proof is complete. 2

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4. Proof of Theorem 1.2 In this section we look for critical points of the functional I : E → R given by

I (u ) :=



1

|∇ u |2 +

2

1

 V (x) f 2 (u ) −

2

RN



1

q(x) f 4 (u ).

4

RN

RN

Note that it follows from Lemma 2.1-(3),(7) that

f (t )  2|t |s for all t ∈ R, 1  s  1.

(4.1)

2

Let







 2

E := u ∈ E:

|∇ u | < RN

q(x)u

2

.

RN

Since q  b0 > 0, it is easy to see that E = ∅. Recall that h(t ) := I (tu ), t > 0. Lemma 4.1. (1) For each u ∈ E there is a unique t u > 0 such that h (t ) > 0 for 0 < t < t u and h (t ) < 0 for t > t u . Moreover, tu ∈ M if and only if t = t u . (2) If u ∈ / E , then tu ∈ / M for any t > 0. Proof. (1) For each u ∈ E , we have, by the Lebesgue dominated convergence theorem and Lemma 2.1(5),(4)

lim

t →∞

I (tu ) t2

= lim

1

t →∞

=

1



  |∇ u |2 +

2 RN



2

u =0



|∇ u |2 − RN

V (x)

f 2 (tu )

(tu )2

−

q(x) f 4 (tu ) 2



(tu )2

 u2



q(x)u 2 < 0

RN

and

lim

t →0

I (tu ) t2

1

= lim

t →0

=

1



  |∇ u |2 +

2 RN



 2

|∇ u | +

2 RN

V (x)

u =0

V (x)u

f 2 (tu )

(tu )2

−

q(x) f 4 (tu ) 2

(tu )2

 |∇ u |2 =

> 0.

(4.2)

RN

RN

u =0

q(x) f 3 (tu ) f (tu ) tu

 u2

 2

Hence h has a positive maximum. The condition h (t ) = 0 is equivalent to





−

V (x) f (tu ) f (tu ) tu

 u2 .

X.-D. Fang, A. Szulkin / J. Differential Equations 254 (2013) 2015–2032

2029

Let

Z (s) :=

q(x) f 3 (s) f (s) s

−

V (x) f (s) f (s) s

.

As in the proof of Lemma 3.2 we see that s → Z (s) is strictly increasing, hence the first conclusion. The second conclusion follows because as previously we have h (t ) = t −1  I (tu ), tu . (2) If tu ∈ M for some t > 0, then  I (tu ), u  = 0 and therefore



 |∇ u |2 =

q(x)

f 3 (tu ) f (tu ) tu

u =0

RN



q(x)

<

f 3 (tu ) f (tu ) tu

u =0

− V (x)

f (tu ) f (tu ) tu

 u2

 u2 

q(x)u 2 ,

RN

where the first inequality follows from the fact that f (t ) f (t )/t is positive for t = 0 and the second one from Lemma 2.1-(7),(10). Hence u ∈ E . 2 Lemma 4.2. (1) There exists ρ > 0 such that c := infM I  inf S ρ I > 0.

(2) u 2  2c for all u ∈ M.

Proof. (1) Using (4.1) with r = 4s ∈ (2, min{2∗ , 4}) and the Sobolev inequality we obtain

1



 q(x) f 4 (u )  4

4 RN

q(x)|u |r  C 2 u r .

RN

Hence it follows from (3.2) that inf S ρ I > 0 if ρ is small enough. That infM I  inf S ρ I is seen as in Lemma 3.3. (2) The argument is the same as in Lemma 3.3-(2). 2 We do not know whether I is coercive on M. However, we prove Lemma 4.3. All Palais–Smale sequences (un ) ⊂ M are bounded. Proof. Suppose un → ∞ and set v n := un / un . As in the proof of Lemma 3.4 we may assume that (3.4) holds with ( yn ) bounded and we see that (3.5) implies v n → 0 in L r (R N ) for all 2 < r < 2∗ . By (4.1),





G x, f (sv n ) :=

1 4

q(x) f 4 (sv n )  4sr q(x)| v n |r ,

where we can choose r ∈ (2, min{2∗ , 4}). So



1 4

RN

 q(x) f 4 (sv n )  C sr

| v n |r → 0. RN

Now the same argument as in the proof of Lemma 3.4 shows that, up to a subsequence, v n → v in L 2loc (R N ) and a.e., and v = 0.

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Let

X.-D. Fang, A. Szulkin / J. Differential Equations 254 (2013) 2015–2032

ϕ ∈ C 0∞ (RN ). Then  I (un ), ϕ  → 0 and hence    f (u n ) f (u n ) f 3 (u n ) f (u n ) ∇ v n ∇ ϕ + V (x) vnϕ − q(x) v n ϕ → 0. un

RN

un

RN

Since |un (x)| → ∞ if v (x) = 0 and f (t ) f (t ) → 2−1/2 as t → ∞, we get using Lemma 2.1-(5) that

f (u n ) f (u n ) un

→ 0 and

f 3 (u n ) f (u n ) un

→1

for such x. By the Lebesgue dominated convergence theorem we therefore have



 ∇ v∇ϕ =

RN

q(x) v ϕ .

RN

So v = 0 and − v = q(x) v. This is impossible because − − q has absolutely continuous spectrum. The proof is complete. 2 Lemma 4.4. If V is a compact subset of E , then there exists R > 0 such that I  0 on (R+ V ) \ B R (0). Proof. We argue by contradiction as at the beginning of the proof of Lemma 3.5 and we obtain un ∈ V ⊂ S, w n = tn un , where un → u, tn → ∞ and I ( w n )  0. We have

0

I (tn un )

=

tn2

1



  |∇ un |2 +

2

V (x)

un =0

RN

f 2 (tn un )

(tn un )2

−

q(x) f 4 (tn un ) 2

(tn un )2



 un2 .

It follows from Lemma 2.1-(5) and the Lebesgue dominated convergence theorem that

 

 |∇ un |2 +

0

un =0

RN



→

f 2 (tn un )

(tn un )2

−

q(x) f 4 (tn un ) 2

(tn un )2



|∇ u |2 −

RN

a contradiction.

V (x)

q(x)u 2 < 0,

RN

2

Let U := E ∩ S and define the mapping m : U → M by setting

m( w ) := t w w , where t w is as in Lemma 4.1. Lemma 4.5. U is an open subset of S. Proof. Obvious because E is open in E.

2

Lemma 4.6. Assume un ∈ U , un → u 0 ∈ ∂ U . If tn un ∈ M, then tn → ∞.

un2

X.-D. Fang, A. Szulkin / J. Differential Equations 254 (2013) 2015–2032

2031

Proof. If (tn ) is bounded, we may assume tn → t 0 and since M is bounded away from 0 by Lemma 4.2-(2), t 0 = 0. So





0 = I (tn un ), un → I (t 0 u 0 ), u 0 and t 0 u 0 ∈ M. Hence u 0 ∈ U .

2

The following two lemmas correspond to Lemma 3.6 and Lemma 3.7 in the preceding section. Lemma 4.7. The mapping m is a homeomorphism between U and M, and the inverse of m is given by m−1 (u ) = uu . We consider the functional Ψ : U → R given by

  Ψ ( w ) := I m( w ) . Lemma 4.8. (1) Ψ ∈ C 1 (U , R) and

 

    Ψ ( w ), z = m( w ) I m( w ) , z for all z ∈ T w (U ). (2) If ( w n ) is a Palais–Smale sequence for Ψ , then (m( w n )) is a Palais–Smale sequence for I . If (un ) ⊂ M is a bounded Palais–Smale sequence for I , then (m−1 (un )) is a Palais–Smale sequence for Ψ . (3) w is a critical point of Ψ if and only if m( w ) is a nontrivial critical point of I . Moreover, the corresponding values of Ψ and I coincide and infU Ψ = infM I . (4) Ψ is even (because I is). A remark is in order here. Lemmas 4.7 and 4.8 are taken from [16] again. Although in [16] they are formulated for U = S, it is easy to see that they also hold if U is a proper open subset of S provided the other hypotheses are satisfied. That this is the case follows from Lemmas 4.1 and 4.4. Let p ∈ (4, 2 · 2∗ ). Then for each ε > 0 there exists C ε such that |q(x)u 3 |  ε |u | + C ε |u | p −1 , hence Lemma 3.12 holds for g (x, u ) = q(x)u 3 and so does Lemma 3.13 (we use Lemma 4.3 to assure boundedness of Palais–Smale sequences). Proof of Theorem 1.2. The argument is the same as in Theorem 1.1. However, there are two details which need to be clarified. Let η be the flow given by (3.12). If T + ( w ) < ∞, then limt → T + ( w ) η(t , w ) exists (cf. [15], Lemma 2.15, Case 1) but unlike the situation in [15], this limit may be a point w 0 ∈ ∂ U . This possibility is ruled out by Lemma 4.6. Now we see as before that the conclusions of Lemma 3.14 hold. Finally, we need to show that U contains sets of arbitrarily large genus. Since the spectrum of − − q in L 2 (R N ) is absolutely continuous and q  b0 > 0, E ∪ {0} contains an infinite-dimensional subspace E 0 . Hence E 0 ∩ S ⊂ U and γ ( E 0 ∩ S ) = ∞. 2 Acknowledgments The second author would like to thank Louis Jeanjean for helpful comments on an earlier version of this paper and for pointing out the reference [5]. The authors would like to thank the referee for pointing out the reference [1].

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