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RESONANCE INVOLVING THE p-LAPLACIAN

C. O. ALVES, P. C. CARRI ˜AO∗∗ AND O. H. MIYAGAKI∗∗∗

Abstract. In this paper we will investigate the existence of multiple solu- tions for the problem

(P) −∆pu+g(x, u) =λ1h(x)|u|p−2u, in Ω, uH01,p(Ω) where ∆pu= div

|∇u|p−2∇u

is the p-Laplacian operator, ΩIRN is a bounded domain with smooth boundary,h and g are bounded functions, N1 and 1< p <∞. Using the Mountain Pass Theorem and the Ekeland Variational Principle, we will show the existence of at least three solutions for (P).

1. Introduction

In this paper, we will investigate the existence of multiple solutions for the problem

−∆pu+g(x, u) = λ1h(x)|u|p−2u, in Ω, u= 0 on ∂Ω,

(P)

where ∆pu =div(|∇u|p−2∇u) is the p-Laplacian operator, 1 < p < ∞, N 1, Ω is a bounded domain with smooth boundary,

g: Ω×IR→IR is bounded continuous function satisfying g(x,0) = 0,

(G1)

and its primitive denoted by

1991Mathematics Subject Classification. Primary: 35A05, 35A15, 35J20.

Key words and phrases. Radial solutions, Critical Sobolevexponents, Palais-Smale con- dition, Mountain Pass Theorem.

Supported in part by Fapemig/ Brazil.

∗∗ Supported in part by Fapemig and CNPq/Brazil.

∗∗∗ Supported in part by Fapemig and CNPq/ Brazil.

Received: March 18, 1998.

c

1996 Mancorp Publishing, Inc.

191

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G(x, s) = s 0

g(x, t)dt is assumed to be bounded, (G2)

λ1 is the first eigenvalue of the following eigenvalue problem with weight −∆pu = λ1h(x)|u|p−2u, in Ω,

u= 0 on ∂Ω, (PA)

where

(h) 0≤h∈L(Ω) with h >0 on subset of Ω with positive measure.

We recall that λ1 is simple, isolated and it is the unique eigenvalue with positive eigenfunction Φ1 (see [1] or [2]). There are many papers treating problem (P) with h = 1, among others, we would like to mention Lazer

& Landesman [3], Ahmad, Lazer & Paul [4], De Figueiredo & Gossez [5], Amann, Ambrosetti & Mancini [6], Ambrosetti & Mancini [7], Thews [8], Bartolo, Benci & Fortunato [9], Ward [10], Arcoya & Ca˜nada [11], Costa &

Silva [12], Fu [13], Gon¸calves & Miyagaki [14] when p = 2, and Boccardo, Dr´abek & Kuˇcera [15], Anane & Gossez [16], Ambrosetti & Arcoya [17], Arcoya & Orsina [18], Fu & Sanches [19] when p= 2.

We shall show in this paper, the existence of multiple solutions for problem (P), by using similar arguments explored in [14] and [19]. Combining a version of the Mountain Pass Theorem due to Ambrosetti & Rabinowitz (see [20] and [25]) and the Ekeland variational principle (see [21, Theorem 4.1]), we will find nontrivial critical points of Euler- Lagrange functional associated to (P) given by

I(u) = 1 p

|∇u|p−λ1 p

h|u|p+

G(x, u) , u∈H01,p(Ω), (1)

which are weak solutions of (P).

Hereafter, we will denoted by and | |p the usual norms on the spaces H01,p(Ω) and Lp(Ω) respectively, and byW the closed subspace

W =

u∈H01,p(Ω)/

h u|Φ1|p−2Φ1 = 0

.

We can easily prove thatW is a complementary subspace ofΦ1.Therefore we have the following direct sum (see e.g. Br´ezis [22])

H01,p(Ω) =Φ1 ⊕W.

We will be denoted by λ2, the following real number λ2= inf

u∈W

|∇u|p ;

h|u|p= 1

,

and we remind that this value is the second eigenvalue of the p-Laplacian (see [23] or [24]).

From simplicity and isolation of λ1 (see [1] or [2]), we have 0< λ1 < λ2 and by definition of λ2 it follows that

h|w|p 1 λ2

|∇w|p , ∀w∈W.

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In this work, we will impose the following condition g(x, t)t→0, as |t| → ∞, ∀x∈Ω, (G3)

which appeared in [7] for p = 2 and [17] for the general case p > 1. This condition together with the assumptions on the limits

T(x) = lim inf

|t|→∞ G(x, t) and S(x) = lim sup

|t|→∞ G(x, t), ∀x∈Ω, imply that problem (P) is in the class of the strongly resonance problem in the sense of Bartolo-Benci & Fortunato [9].

The following condition means a nonresonance with higher eigenvalues

G(x, t)≥ λ1−λ2 p

h(x)|t|p, ∀x∈Ω, ∀t∈IR.

(G4)

In addition to (G3) which is a behaviour ofgat infinity, we assume a condi- tion of the behaviour of Gat origin

there exist 0< δ and 0< m < λ1 such that G(x, t)≥ mph(x)|t|p, for |t|< δ, ∀x∈Ω.

(G5)

Our main result is the following:

Theorem 1. Assume conditions (h), (G1)-(G5). Then, problem (P) has at least three solutions u1, u2 and u3, with

I(u1), I(u2)<0and I(u3)>0, provided that the following conditions hold

there exist t, t+ ∈IR with t <0< t+ such that

G(x, t±Φ1)T(x)dx <0, (G6)

and

S(x)dx≤0.

(G7)

Remark 1. Theorem 1 improves in some sense the main result proved in [14], since the proof given in [14] works only in Hilbert space framework.

2. Preliminary Results

In this section, we will state and prove some results required in the proof of Theorem 1. We recall that I : H01,p(Ω) →IR is said to satisfy Palais- Smale condition at the level c∈IR ((PS)c in short), if any sequence {un} ⊂ H01,p(Ω) such that

I(un)→c and I(un)0, possesses a convergent subsequence inH01,p(Ω).

Lemma 1. Assume (h), (G1) and (G2). Then I satisfies the (PS)c condi- tion∀c <T(x)dx.

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Proof. We are going to adapt some arguments used in [16, p.1148]. First of all, we shall show that {un} is bounded. Assume that{un} is unbounded, therefore, up to subsequence, we have

un → ∞.

Letting

vn= un un, (*n)

we can assume that there exists v∈H01,p(Ω) such that vn vinH01,p(Ω)

and

vn→v in Ls(Ω), for 1≤s < p= Np N−p.

Now, we will show that v= 0 and that there exists γ ∈IR such that v(x) =γΦ1(x), ∀x∈Ω.

From (1) and choosing tn=un,we obtain I(un)un

tpn =

|∇vn|p−λ1

h|vn|p+ 1 tpn

g(x, un)un. (2)

Using (G1) together with the fact that

n→∞lim

I(un)un

tpn = 0, we get

h|v|p = 1 λ1 (3)

and thereforev= 0.

Using the weak convergencevn v,we know that v ≤1.

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By (3) and (4), it follows thatv is an eigenfunction forλ1.Then there exists γ ∈IR such that

v(x) =γΦ1(x), ∀x∈Ω.

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In particular,

un

un →γΦ1,∀x∈Ω, which implies

|un(x)| → ∞, ∀x∈Ω, and by (G2) and Fatou’s lemma, we have

(6) lim infn→∞

G(x, un(x))dx

lim infn→∞ G(x, un(x))dx

T(x)dx.

Now, using the inequality

c+on(1) =I(un)

G(x, un(x))dx, (7)

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we have by (6) that

c≥

T(x)dx,

which contradicts the hypothesis on the level c, then{un} is bounded.

Let u∈H01,p(Ω) be a function such thatun u,using a similar arguments explored in [18], we can conclude that

un→u in H01,p(Ω), and Lemma 1 follows.

3. Existence of two solutions (Ekeland’s principle) We will denote by Q± the following sets

Q+={tΦ1+w, t≥0 and w∈W} and Q ={tΦ1+w, t≤0 andw∈ W}. It is easy to see that

∂Q+=∂Q=W.

Lemma 2. If conditions (h),(G2) and (G6) hold, then functional I is bound- ed from below on H01,p(Ω). Moreover, the infimum is negative on Q+ and Q.

Proof. From condition (G2), its easy to see thatI is bounded from below on H01,p(Ω).

Using condition (G6), we have I(t±Φ1) =

G(x, t±Φ1)

T(x)dx <0, therefore

u∈Qinf±I(u)<0.

Remark 2. Using condition (G4) and the definition of the number λ2, we remark that

I(w) 1 p

|∇w|p−λ2

p

h(x)|w|p 0, ∀w∈W.

Therefore Lemma 2 implies that if the infimum of I on Q± is achieved by, for example, u±0 ∈Q±, we can assume that

u±0 ∈Q±\W.

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This fact is very important when we are working with Ekeland’s variational principle.

Theorem 2. If conditions (h),(G1),(G2),(G4) and (G6) hold, then there exist u1 ∈Q+and u2 ∈Q solutions of (P), such that

I(u1), I(u2) <0.

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Proof. From the proof of Lemma 2 we can conclude that

u∈Qinf±I(u)

G(x, t±Φ1)

T(x)dx <0.

If

u∈Qinf±I(u) =

G(x, t±Φ1) =I(t±Φ1)

T(x)dx <0, occurs we can take u1 =t+Φ1 and u2 =tΦ1.Otherwise if

u∈Qinf±I(u)<

G(x, t±Φ1)

T(x)dx,

holds using the Ekeland’s variational principle and the same argument ex- plored in [14], we can show that there exist sequences {un} ⊂ Q+ and {vn} ⊂Q satisfying

I(un) inf

u∈Q+I(u) and I(un)0, and

I(vn) inf

u∈QI(u) and I(vn)0.

By Lemma 1, there exist u1 and u2 such that

un→u1 and vn→u2 inH01,p(Ω).

Therefore, u1 and u2 are solutions of (P) verifying I(u1) = inf

u∈Q+I(u)<0 and I(u2) = inf

u∈QI(u)<0,

which implies from Remark 2 that u1 ∈Q+ and u2 ∈Q. This completes the proof of Theorem 2.

4. Existence of a third solution (Mountain Pass) Using condition (G5) and arguing as in [14], we can easily show that

G(x, t)≥ m

p h(x)|t|p−C|t|σ, ∀x∈Ω, t∈IR (9)

where p < σ < p and C is a constant independent ofx.

By (9), we have that

I(u) m 1

|∇u|p−C

|u|σ, and then

I(u)≥ m

1 up+o(u), as u →0.

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Using (G6), we obtain

I(t±Φ1)<0,

which together with (10) imply that there exist r, ρ > 0 and e=t+Φ1 such that

I(u)≥r >0, for u ≤ρ and I(e)<0.

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Therefore, using a version of the Mountain Pass Theorem without a sort of Palais-Smale condition [25, Theorem 6], there exists a sequence {un} ⊂ H01,p(Ω) satisfying

I(un)→c≥r >0 and I(un)H1,p

0 (Ω)(1 +un)→0.

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Remark 3. We recall the sequence obtained in (11) was introduced by Ce- rami in [26].

Theorem 3. If conditions (h),(G1) -(G3) and (G5)-(G7) hold, then prob- lem (P) has a solution u3,with

I(u3)>0.

Proof. Let {un} ⊂ H01,p(Ω) be the sequence obtained in (11); then arguing as in Lemma 1, if {un} is unbounded, we can assume that

|un(x)| → ∞, ∀x∈Ω.

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Using (11), we have

on(1) =I(un)(un) =unp−λ1|un|pp+

g(x, un)undx, and then

0≤ unp−λ1|un|pp ≤ −

|g(x, un)un|+on(1).

Combining (12), (G3) with the inequality above, we conclude that unp−λ1|un|pp0.

Now, using the equality c+on(1) =I(un) = 1

p

unp−λ1|un|pp+

G(x, un(x))dx, together with Fatou Lemma and (G7) we obtain

c≤lim sup

n→∞

G(x, un(x))dx

S(x)dx≤0,

which is a contradiction , because c >0 by (11). Then {un}is bounded.

Letu3 ∈H01,p(Ω) be such that

un u3. (13)

By a similar argument explored in [18], we have that un→u3 in H01,p(Ω), (14)

and consequently

I(u3) =c≥r >0 and I(u3) = 0, which shows that u3 is a solution of problem (P).

5. Proof of Theorem 1

Theorem 1 is an immediate consequence of Theorems 2 and 3.

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6. Example

Making Ω = (0,1), p = 2,and h = 1,we shall give an elementary example of a nonlinearity g verifying the set of assumptions.

We recall that λn=n2, n= 1,2, ...are eigenvalues of (PA) and Φ1= sinπx is the first eigenvalue of (PA).

Letg: Ω×IR→IR defined by

g(x, s) =R(x)g1(s), where g1:IR→IR is given by

g1(s) =

s, for 0≤s≤1,

2−s, for 1< s≤5, s−8 for 5< s≤8 +230, 8 +

30−s, for 8 +230 < s≤8 + 0 for s≥8 + 30,

−g(−s) for s≤0,30,

and R: Ω→IR is defined by R(x) =

4x+ 1, for 0≤x≤ 12,

−4x+ 5, for 12 ≤x≤1.

Then

G1(s) = s 0

g1(t)dt and G(x, s) = s 0

g(x, t)dt=R(x)G1(s) and

S(x) =T(x) =−R(x) 2 .

By the definition ofg,it is easy to see that it verifies the conditions (Gi) for i= 6.

Thus, we shall prove thatG satisfies (G6), fort+= 8.

Indeed, observe that

G(x,1(x)) =G(1−x,1(1−x)), x∈Ω, that is, the function above is symmetric with respect to x= 12. Then,

1 0

G(x,1(x))dx = 2

12

0

R(x)G1(8Φ1(x))dx

= 2

12

0

4x G1(8Φ1(x))dx+ 2

12

0

G1(8Φ1(x))dx

I1+I2.

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Now, we shall estimate each integralsIj,(j = 1,2). Since G1(2 +

2) = 0, choosingα0∈IR such that 8Φ10) = 2 +

2,which satisfies 0< α0< 16,we obtain

I1 2

α0

0

4x G1(8Φ1(x))dx0 < 4 3. On the other hand,

I2 = 2(

16

0

+

13

16

+

12

13

)G1(8Φ1(x))dx

2(

16

0

G1(2)dx+

13

16

G1(4)dx+

12

13

G1(6)dx)

≤ −2.

Therefore, 1 0

G(x,1(x))dx=I1+I2 <−2 3 <

1 0

T(x)dx.

Analogously for t=−8.This proves thatGsatisfies (G6).

Acknowledgments. The authors are grateful to the referee for his helpful remarks. Also, the first and the third author wish to thank the Departa- mento de Matem´atica, Universidade Federal de Minas Gerais, for the warm hospitality.

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C. O. Alves

Departamento de matem´atica e Estat´ıstica Universidade Federal da Para´ıba

58109-970 Campina Grande (PB)-BRAZIL E-mail address: [email protected]

P. C. Carri˜ao

Departamento de Matem´atica

Universidade Federal de Minas Gerais 31270-010 Belo Horizonte (MG)-BRAZIL E-mail address: [email protected]

O. H. Miyagaki

Departamento de Matem´atica Universidade Federal de Vic¸osa 36571-000 Vic¸osa (MG)-BRAZIL E-mail address: [email protected]

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