ON ELLIPTIC EQUATIONS IN R N WITH CRITICAL EXPONENTS ∗
C.O. Alves, J.V. Goncalves, & O.H. Miyagaki
Abstract
In this note we use variational arguments –namely Ekeland’s Principle and the Mountain Pass Theorem– to study the equation
−∆u+a(x)u=λuq+u2∗−1 inRN.
The main concern is overcoming compactness difficulties due both to the unboundedness of the domainRN, and the presence of the critical expo- nent 2∗= 2N/(N−2).
1 Introduction
In this note we use variational methods to explore existence of weak solutions for the problem
(∗)
−∆u+a(x)u=λuq+u2∗−1 in RN Ra(x)u2<∞, R
|∇u|2<∞ u≥0, u6≡0
whereais a nonnegative L∞loc function,λ≥0, 0< q≤1 and 2∗ is the critical exponent, 2∗= 2N/(N−2), forN ≥3.
This problem has been explored by many authors including Br´ezis & Niren- berg [6], Ambrosetti-Br´ezis & Cerami [1], Guedda & Veron [9] (see also their references) for the case of elliptic equations in bounded domains. As far as unbounded domains are concerned we recall the work by Benci & Cerami [12], Noussair-Swanson & Jianfu [3], Jianfu & Xiping [14], Egnell [7], Azorero &
Alonso [4], Miyagaki [10].
∗1991 Mathematics Subject Classifications: 35J20, 35K20.
Key words and phrases: Elliptic equations, unbounded domains, critical exponents, variational methods.
c1996 Southwest Texas State University and University of North Texas.
Submitted: August 7, 1996. Published October 22, 1996.
Partially supported by CNPq/Brasil.
1
In this work we shall assume the following condition ona.
(a1) a(x)>0, x∈BRco and Z
Bc
Ro
1 a <∞.
Our main results are the following:
Theorem 1 Let 0 < q < 1and assume (a1). Then there exists λ∗ > 0such that(∗)has a solution for 0< λ < λ∗,N ≥3.
Theorem 2 Let N ≥ 4 and q = 1. Assume (a1) and a(x) = 0, x ∈ B2r0
for somer0∈(0, R0/2). Then there is λ∗ >0such that (∗)has a solution for 0< λ < λ∗.
These theorems complement the results in [10] in the sense that here the function a is allowed to vanish on a ball BRo. Actually in Theorem 2, we requirea to vanish onB2r0. In addition we consider the case 0< q ≤1 while in [10],a >0 is continuous and q∈(1,2∗).
2 Preliminaries
Let
E=
u∈ D1,2 | Z
au2<∞
with inner product and norm given by hu, vi=
Z
(∇u∇v+auv), kuk2= Z
|∇u|2+au2 .
Recall thatD1,2 is the closure ofCo∞with respect to the gradient normkuk21= R|∇u|2. Moreover
D1,2=n
u∈L2∗ |∂iu∈L2o and the norm
kuk0 ≡ |u|L2∗+|∇u|L2
is equivalent to theD1,2 norm. In additionD1,2→L2∗.
The following lemma is a variant of a result by Willem & Omana [13] and by Costa [2].
Lemma 1 Assume (a1). Then E → Ls for 1 ≤ s ≤ 2∗ and E ,→ Ls for 1≤s <2∗.
We shall look for the critical points of the functional I(u) =1
2 Z
|∇u|2+a|u|2
− 1 q+ 1
Z
λuq+1+ − 1 2∗
Z u2+∗ in the Hilbert spaceE.
Using standard techniques we can show that I ∈ C1(E,R), and that its derivative is given by
hI0(u), vi= Z
(∇u∇v+auv)−λ Z
uq+v−Z
u2+∗−1v . Therefore, the critical points ofI are the weak solutions of (∗).
The following auxiliary result concerns the geometry ofI.
Lemma 2 Ifasatisfies(a1)and if0< q≤1then there existsλ∗>0such that if0< λ < λ∗ then
(i) I(u)≥r, kuk=ρ, for some r, ρ >0 If in additionφ≥0, φ6≡0, andφ∈E then
(ii) I(tφ)→ −∞ as t→ ∞
(iii) I(tφ)<0, for small t >0, and0< q <1.
3 Proofs
For the sake of completeness, we present a proof of Lemma 3, which is based on the proof in [2].
Proof of Lemma 3. At first letR > Ro. Then we have Z
BRc
|u|= Z
BcR
a1/2|u| a1/2 ≤ Z
BcR
1 a
!1/2 Z
BRc
a|u|2
!1/2
≤Ckuk which shows that
|u|L1≤Ckuk, u∈E.
Now using the interpolation inequality
|u|s≤ |u|α1|u|1r−α, α+1−α
r = 1
s, 1≤s≤r≤2∗, 0≤α≤1 and the embeddingE→L2∗, we infer thatE→Ls, 1≤s≤2∗.
On the other hand, for sufficiently largeR >0 we have Z
BRc
1 a< . So ifun*0 inE, then for largen
Z
BRc |un| ≤C Z
BRc
1 a ≤ . Using compact Sobolev embeddings we also have
un→0 inL1(BR)
so thatun → 0 in L1. Using again the interpolation inequality stated above, one concludes the proof of Lemma 3.
Proof of Lemma 4.
Verification of(i). From the continuous embedding in Lemma 3, we have I(u) ≥ 12kuk2−λcq+1q+1kukq+1−S−22∗∗/2kuk2∗
≥ kukq+1
1
2kuk2−(q+1)−λcq+1q+1 −S−22∗/2∗ kuk2∗−(q+1) whereS is the best constant for the embeddingD1,2→L2∗, that is
S= inf
( R|∇u|2
R |u|2∗2/2∗ | u∈ D1,2, u6≡0 )
.
Letting
Q(t)≡ 1
2t2−(q+1)−S−2∗/2
2∗ t2∗−(q+1), t≥0, there isρ >0 such that
maxt≥0 Q(t) =Q(ρ)>0.
Takingkuk=ρand λ∗= cq+1
q+1Q(ρ) we get (i).
Verification of(ii). Takingφ6≡ 0,φ≥0, φ∈E we have I(tφ) = t2
2kφk2− tq+1 q+ 1λ
Z
φq+1−t2∗ 2∗
Z φ2∗ which gives (ii).
Verification of(iii). It is clear from the expression ofI(tφ) above taking into account that 0< q <1.
Proof of Theorem 1. By the proof of lemma 4,Iis bounded from below on Bρ. By the Ekeland Principle [8], there existsu∈Bρ such that
I(u)≤inf
B¯ρ I+ and
I(u)< I(u) +ku−uk, u6≡ u. Now since 0< q <1 it follows that
I(tφ)<0, for smallt >0, φ6≡ 0, andφ∈ Co∞. Again by Lemma 4
∂BinfρI≥r >0 and inf
Bρ
I <0.
Choose >0 such that
0< < inf
∂BρI−inf
Bρ
I.
Hence
I(u)< inf
∂Bρ
I so that
u∈Bρ. Hence letting
F(u)≡I(u) +ku−uk we notice thatuis a point of minimum of F onBρand so
I(u+δv)−I(u)
δ +kvk ≥ 0
which by passing to the limit asδ→0 gives that hI0(u), vi+kvk ≥0
and hencekI0(u)k ≤. Therefore, there is a sequence un∈Bρ such that I(un)→c∗≡inf
Bρ
I <0 and I0(un)→0.
Since of courseun is bounded,
un* u∗ in E
and
un →u∗ a.e. in RN. Now passing to the limit in
o(1) = Z
(∇un∇φ+aunφ)−λ Z
uqn+φ− Z
u2n+∗−1φ, φ∈E we infer thatI0(u∗) = 0 showing thatu∗ is a solution of problem (∗).
In order to show thatu∗6≡ 0, we follow the arguments in [6]. Assume that u∗≡0 and that
kunk2→`≥0.
UsingI0(un)→0 we have
kunk2− Z
u2n+∗ =o(1) so thatR
u2n+∗ →`and from the expression c∗+o(1) =λ
1 2− 1
q+ 1 Z
uq+1n+ + 1 N
Z u2n+∗ we infer that
c∗= ` N which is impossible.
Proof of Theorem 2. By Lemma 4 and the Mountain Pass Theorem, there exists a sequenceun in E such that
I(un)→c and I0(un)→0 where
c= inf
γ∈Γ max
0≤t≤1I(γ(t)), c≥r and
Γ ={γ∈C([0,1], X) | γ(0) = 0, γ(1) =e} wheree∈E satisfiesI(e)≤0.
Claim. There ise≡eλ such that 0< c < N1SN2,0< λ < λ∗. From the expression
hI0(un), uni −2∗I(un) =
1−2∗ 2
kunk2+λ 2∗
2 −1 Z
u2n+
one shows, by takingλ∗>0 smaller than the one found in lemma 4, thatun is bounded. So that, passing to subsequences,
un * uin E andun →ua.e. inRN for someu∈ E.
Remark. I(un+)→c andI0(un+)→0.
Indeed,un− is also bounded so that o(1) =hI0(un), un−i=
Z
(|∇un−|2+au2n−). Moreover, ifφ∈E then
hI0(un+), φi = hun+, φi −λ Z
un+φ− Z
u2n+∗−1φ
= hI0(un), φi − hun−, φi so that,I0(un+)→0.On the other hand,
I(un)−1 2 Z
(|∇un−|2+au2n−)
= 1
2 Z
(|∇un+|2+au2n+)−λ 2 Z
u2n+− 1 2∗
Z u2n+∗
= I(un+),
which givesI(un+)→c. So we may assume thatun≥0 and thusu≥0.
Now as in the proof of Theorem 1 one shows thatusatisfies the equation in (∗). Again arguing as in [6] we assume thatu≡0. Then
kunk2→` for some`≥0 and using the facts that
I(un)→c, I0(un)→0
we infer thatc≥ N1SN/2, contradicting 0< c < N1SN/2given by the Claim.
Proof of the Claim. (Arguments adapted from [6].) Consider the cut-off functionφ∈ Co∞such that
φ≡1 on Br0, φ≡0 on RN\B2r0. Now consider the function
w(x) = [N(N−2)](N−2)/4
(+|x|2)(N−2)/2 , x∈RN, >0 which satisfies
−∆w=w2∗−1 inRN. It is well known (see e.g. Talenti [5], Aubin [11] ) that
kwk21 =|w|22∗∗ = SN/2.
Let
ψ=φw
and letv∈ Co∞ given by
v= ψ
Rψ2∗1/2∗. Now it can be shown (see e.g. [6], [10]) thatX≡R
|∇v|2 satisfies X≤S+O((N−2)/2).
Moreover there is somet>0 such that
maxt≥0 I(tv) =I(tv)
and d
dtI(tv)|t=t= 0.
which gives
0< t< X1/(2∗−2)≡t0.
Notice thata= 0 onB2r0 andv= 0 onRN\B2r0. Moreovert≥d0≡d0(r0) for some d0 > 0. Otherwise sinceX is bounded, if t → 0, thenI(tv)→ 0 contradicting
I(tv) = max
t≥0 I(tv)≥r >0 given by lemma 4 (i). On the other hand
I(tv) = t2 2
Z
|∇v|2−t2∗ 2∗ −λt2
2 Z
v2
≤ t2
2t20∗−2−t2∗ 2∗
−λt2 2
Z
B2r0
v2. Now recalling that as a function oft,
t2
2t20∗−2−t2∗ 2∗
increases on the interval (0, t0) we get I(tv) ≤ t20∗
1 2− 1
2∗
−λt2 2
Z
B2r0
v2
≤ 1
Nt20∗−λd20 2
Z
B2r0
v2
≤ 1 N
h
S+O
(N−2)/2i2∗/(2∗−2)
−λd20 2
Z
B2r0
v2
= 1
N h
S+O
(N−2)/2 iN/2
−λd20 2
Z
B2r0
v2.
Using the inequality
(b+c)α≤bα+α(b+c)α−1c b, c≥0, α≥1 withb=S,c=O (N−2)/2
andα=N/2 we get I(tv)≤ 1
NSN/2+O((N−2)/2)−c0λ Z
B2r0
v2.
Therefore,
I(tv)≤ 1
NSN/2+(N−2)/2 (
M−c0λ(2−N)/2 Z
B2r0
v2 )
, wherec0=d20/2 and M is a positive constant.
We shall show that (N−2)/2
(
M−c0λ(N−2)/2 Z
B2r0
v2 )
<0, for small >0. So that
I(tv)< 1 NSN/2 and hence
0< c < 1 NSN/2. Noticing that
d1≤ Z
B2r0
ψ2∗≤d2, for some d1, d2>0, (see [6]), it follows by a change of variables that
I(tv)≤ 1
NSN/2+(N−2)/2 (
M−c0λ(4−N)/2
Z r0−1/2 0
sN−1ds (1 +s2)N−2
) .
We are going to consider separately the casesN = 4 andN ≥5.
Case N = 4. We have I(tv) ≤ 1
4S2+ (
M−c0λ
Z r0−1/2 0
s3ds (1 +s2)2
)
≤ 1
4S2+n
M−c0λln(r0−1/2)o . Now since
c0λln(r0−1/2)→ ∞as→0 we infer that
I(tv)<1
4S2, for small >0.
Case N≥5. Noticing that c0λ(4−N)/2
Z r0−1/2 0
sN−1ds
(1 +s2)N−2 → ∞as→0 we infer that
I(tv)< 1
NSN/2for small >0, which concludes the proof of this claim.
References
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C.O. Alves
Dep. Mat. Univ. Fed. Paraiba
58109-970 - Campina Grande(PB), Brasil E-mail [email protected]
J.V. Goncalves
Dep. Mat. Univ. Bras´ılia 70.910-900 Brasilia(DF), Brasil E-mail [email protected] O.H. Miyagaki
Dep. Mat. Univ. Fed. Vi¸cosa 36570-000 Vi¸cosa(MG), Brasil E-mail [email protected]