Multiple positive solutions for equations involving critical Sobolev exponent in R N ∗
C. O. Alves
Abstract This article concerns with the problem
−div(|∇u|m−2∇u) =λhuq+um∗−1, in RN.
Using the Ekeland Variational Principle and the Mountain Pass Theorem, we show the existence of λ∗ > 0 such that there are at least two non- negative solutions for eachλin (0, λ∗).
1 Introduction
In this work, we study the existence of solutions for the problem (P)
−∆mu=λhuq+um∗−1, RN u≥0, u6= 0, u∈D1,m(RN)
where ∆mu= div (|∇u|m−2∇u),λ >0,N > m≥2,m∗=N m/(N−m), 0<
q < m−1,his a nonnegative function withLΘ(RN) with Θ = N m−(q+1)(NN m −m), and
D1,m(RN) ={u∈Lm∗(RN)| ∂u
∂xi ∈Lm(RN)} endowed with the normkuk= R
|∇u|m1/m
.
The case q= 0, m= 2 was studied by Tarantello [20], and a more general case withm≥2 by Cao, LI & Zhou [5]. In these two references, [5] and [20], it is proved that (P) has multiple solutions. In the casem= 2,h∈Lp(RN) with p1 ≤p ≤p2 and 1 < q < 2∗−1, Pan [18] proved the existence of a positive solution for (P). In the more general case, m≥2, h∈LΘ(RN), Gon¸calves &
Alves [10] proved the existence of a positive solution for (P).
∗1991 Mathematics Subject Classifications: 35J20, 35J25.
Key words and phrases: Mountain Pass Theorem, Ekeland Variational Principle.
c1997 Southwest Texas State University and University of North Texas.
Submitted April 22,1997. Published August 19, 1997.
1
By a solution to (P), we mean a functionu∈D1,m(RN),u≥0 and u6= 0 satisfying
Z
|∇u|m−2∇u∇Φ =λ Z
huqΦ + Z
um∗−1Φ, ∀Φ∈D1,m(RN).
Hereafter, R
, D1,m, Lp and |.|p stand for R
RN, D1,m(RN), Lp(RN) and |.|Lp respectively.
In the search of solutions we apply minimizing arguments to the energy functional
I(u) = 1 m
Z
|∇u|m− λ q+ 1
Z
h u+q+1
− 1 m∗
Z
u+m∗
(1) associated to (P), whereu+(x) = max{u(x),0}. Note that the conditionh∈LΘ implies thatI∈C1 D1,m,R
.
To show the existence of at least two critical points of the energy functional, we shall use the Ekeland Variational Principle [8], and the Mountain Pass Theo- rem of Ambrosetti & Rabinowitz [2] without the Palais-Smale condition. Using the Ekeland Variational Principle, we obtain a solutionu1 withI(u1)<0, and by the Mountain Pass Theorem we prove the existence of a second solutionu2
with I(u2) >0. Techniques for finding the solutions u1 and u2 are borrowed from Cao, Li & Zhou [5]. Then we combine these techniques with arguments developed by Chabrowski [6], Noussair, Swanson & Jianfu [17], Jianfu & Xip- ing [12], Azorero & Alonzo [9], Gon¸calves & Alves [10] and Alves, Gon¸calves &
Miyagaki [1] to obtain the following result
Theorem 1 There exists a constant λ∗ > 0, such that (P) has at least two solutions,u1 andu2, satisfying
I(u1)<0< I(u2) ∀λ∈(0, λ∗).
2 Preliminary Results
In this section we establish some results needed for the proof of Theorem 1.
Definition. A sequence{un} ⊂D1,mis called a (P S)csequence, ifI(un)→c andI0(un)→0.
Lemma 1 If {un}is a (P S)c sequence, then {un}is bounded, and {u+n} is a (P S)c sequence.
Proof. Using the hypothesis that {un} is a (P S)c sequence, there exist no
andM >0 such that I(un)− 1
m∗I0(un)un≤M+kunk ∀n≥no. (2) Now, using (1) and the H¨older’s inequality, we have
I(un)− 1
m∗I0(un)un≥ 1
N kunkm+c1kunkq+1 (3) where c1 is a constant depending of N, m, q,khkΘ and Θ. It follows from (2) and (3) that{un} is bounded. Now, we shall show that{u+n}is a also (P S)c
sequence. Since{un}is bounded, the sequenceu−n =un−u+n is also bounded.
Then
I0(un)u−n →0 and we conclude that
u−n→0. (4) From (4) we achieve that
kunk=u+n+on(1). (5) Therefore, by (4) and (5)
I(un) =I(u+n) +on(1) and
I0(un) =I0(u+n) +on(1),
which consequently implies that{u+n}is a (P S)c sequence. 2
From Lemma 1, it follows that any (P S)c sequence can be considered as a sequence of nonnegative functions.
Lemma 2 If {un}is a (P S)c sequence with un * u inD1,m, then I0(u) = 0, and there exists a constantM >0depending ofN, m, q,|h|Θ andΘ, such that
I(u)≥ −M λΘ
Proof. If{un}is a (P S)c sequence withun* u, using arguments similar to those found in [10], [12] and [17], we can obtain a subsequence, still denoted by un, satisfying
un(x) → u(x) a.e. in RN (6)
∇un(x) → ∇u(x) a.e. in RN (7)
u(x) ≥ 0 a.e. in RN. (8)
From (6), (7) and using the hypothesis that{un}is bounded inD1,m, we get
I0(u) = 0, (9)
which in impliesI0(u)u= 0, and kukm=λ
Z
huq+1+ Z
um∗.
Consequently
I(u) =λ 1
m− 1
q+ 1 Z
huq+1+ 1 N
Z um∗.
Using H¨older and Young Inequalities, we obtain I(u)≥ −1
N |u|mm∗∗−M λΘ+ 1
N |u|mm∗∗=−M λΘ
whereM=M(N, m, q,Θ,khkΘ). 2
For the remaining of this article, we will denote by S the best Sobolev constant for the imbeddingD1,m,→Lm∗.
Lemma 3 Let{un} ⊂D1,m be a(P S)c sequence with c < 1
NSN/m−M λΘ,
whereM >0is the constant given in Lemma 2. Then, there exists a subsequence {unj}that converges strongly in D1,m.
Proof By Lemmas 1 and 2, there is a subsequence, still denoted by{un}and a functionu∈D1,m such thatun* u. Letwn =un−u. Then by a lemma in Brezis & Lieb [3], we have
kwnkm = kunkm− kukm+on(1) (10) kwnkmm∗∗ = |un|mm∗∗− |u|mm∗∗+on(1). (11) Using the Lebesgue theorem (see Kavian [13]), it follows that
Z
huq+1n −→
Z
huq+1. (12)
From (10), (11) and (12), we obtain
kwnkm=|wn|mm∗∗+on(1) (13)
and 1
mkwnkm− 1
m∗ |wn|mm∗∗=c−I(u) +on(1). (14)
Using the hypothesis that{wn}is bounded inD1,m, there existsl≥0 such that kwnkm→l≥0. (15) From (13) and (15), we have
|wn|mm∗∗→l, (16) and using the best Sobolev constantS and recalling that
kwnkm≥S Z
|wn|m∗ m/m∗
, (17)
we deduce from (15), (16) and (17) that
l≥Slm/m∗. (18)
Now, we claim thatl= 0. Indeed, ifl >0,from (18)
l≥SN/m. (19)
By (14), (15) and (16), we have 1
Nl=c−I(u). (20)
From (19), (20) and Lemma 2 we get c≥ 1
NSN/m−M λΘ,
but this result contradicts the hypothesis. Thus,l= 0 and we conclude that un→u in D1,m.
3 Existence of a first solution (Local Minimiza- tion)
Theorem 2 There exists a constantλ∗1>0such that for 0< λ < λ∗1 Problem (P) has a weak solution u1 with I(u1)<0.
Proof. Using arguments similar to those developed in [5], we have I(u)≥
1 m−
kukm+o(kukm)−C()λm/(m−(q+1)),
whereC() is a constant depending on >0. The last inequality implies that for small, there exist constantsγ, ρandλ∗1>0 such that
I(u)≥γ >0, kuk=ρ , and 0< λ < λ∗1.
Using the Ekeland Variational Principle, for the complete metric spaceBρ(0) withd(u, v) =ku−vk, we can prove that there exists a (P S)γosequence{un} ⊂ Bρ(0) with
γo= inf{I(u)|u∈Bρ(0)}.
Choosing a nonnegative function Φ ∈ D1,m\{0}, we have that I(tΦ) < 0 for smallt >0 and consequentlyγo<0.
Takingλ∗1 >0,such that 0< 1
NSN/m−M λΘ ∀λ∈(0, λ∗1)
from Lemma 3, we obtain a subsequence{unj} ⊂ {un}and u1 ∈D1,m, such that
unj →u in D1,m. Therefore,
I0(u1) = 0 and I(u1) =γo<0,
which completes this proof. 2
4 Existence of a second solution (Mountain Pass)
In this section, we shall use arguments similar to those explored by Cao, Li &
Zhou [5], Chabrowski [6], Noussair, Swanson & Jianfu [17], Jianfu & Xiping [12]
and Gon¸calves & Alves [10] to obtain the following
Theorem 3 There exists a constantλ∗2 >0such that for 0< λ < λ∗2 Problem (P) has a weak solutionu2 with I(u2)>0.
Proof. By arguments found in [5] and [10], we can prove that there exists δ1 > 0 such that for all λ ∈ (0, δ1), the functional I has the Mountain Pass Geometry, that is:
(i) There exist positive constantsr, ρwithI(u)≥r >0 for kuk=ρ (ii) There existse∈D1,m withI(e)<0 andkek> ρ .
Then by [16], there exists a (P S)γ1 sequence{vn}with γ1= inf
g∈Γ max
t∈[0,1]I(g(t)) where
Γ ={g∈C([0,1], D1,m)|g(0) = 0 and g(1) =e}.
Using the next claim, which is a variant of a result found in [5], we can complete the proof of this theorem.
Claim. There existsλ∗2>0 such that for the constantM given by Lemma 2, 0< γ1< 1
NSN/m−M λΘ ∀λ∈(0, λ∗2).
Assuming this claim, by Lemma 3 there exists a subsequence {vnj} ⊂ {vn} and a functionu2∈D1,m such thatvnj →u2. Therefore,
I0(u2) = 0 and I(u2) =γ1>0.
Which concludes the present proof. 2
Verification of the above claim. Forx∈RN, let
Ψ(x) =
N
N−m m−1
m−1(N−m)/m2
h
1 +|x|m/(m−1)i
N−m m
.
Then it is well known that (see [7] or [19])
kΨkm=|Ψ|mm∗∗=SN/m. (21) Letδ2>0 such that
1
NSN/m−M λΘ>0 ∀λ∈(0, δ2). Then from (1) and (21), we have
I(tΨ)≤tm mSN/m, and there existsto∈(0,1) with
sup
0≤t≤to
I(tΨ)< 1
NSN/m−M λΘ ∀λ∈(0, δ2). Moreover, from (1) and (21), we have
I(tΨ) = tm
m −tm∗ m∗
SN/m−λtq+1 q+ 1
Z
hΨq+1, and remarking that
tm m −tm∗
m∗
≤ 1
N ∀t≥0, we obtain
I(tΨ)≤ 1
NSN/m−λtq+1 q+ 1
Z
hΨq+1;
therefore,
sup
t≥to
I(tΨ)≤ 1
NSN/m−λtq+10 q+ 1
Z
hΨq+1.
Now, takingλ >0 such that
−λtq+10 q+ 1
Z
hΨq+1<−M λΘ
that is,
0< λ < tq+10 R hΨq+1 M(q+ 1)
!1/(Θ−1)
=δ3
we deduce that sup
t≥to
I(tΨ)< 1
NSN/m−M λΘ ∀λ∈(0, δ3). Choosingλ∗2 = min{δ1, δ2, δ3}, we have
sup
t≥0
I(tΨ)< 1
NSN/m−M λΘ ∀λ∈(0, λ∗2). and consequently
0< γ1< 1
NSN/m−M λΘ ∀λ∈(0, λ∗2) which proves the claim.
Proof of Theorem 1. Theorem 1 is an immediate consequence of Theorems 2 and 3.
Remark. Using Lemma 3 and the same arguments explored by Azorero &
Alonzo, in the case 0 < q < p [9], we can easily show that for small λ the following problem has infinitely many solutions with negative energy levels.
(P)∗ −∆mu =λh|u|q−1u+|u|m∗−2u, in RN u∈D1,m
This result is obtained using the concept and properties of genus, and working with a truncation of the energy functional associated with (P)∗.
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C. O. Alves
Universidade Federal da Para´ıba - PB - Brazil E-mail address: [email protected]