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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/(N2).

1 Introduction

In this note we use variational methods to explore existence of weak solutions for the problem

(∗)



−∆u+a(x)u=λuq+u21 in RN Ra(x)u2<∞, R

|∇u|2<∞ u≥0, u6≡0

whereais a nonnegative Lloc 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

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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 ofCowith 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.

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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|1rα, α+1−α

r = 1

s, 1≤s≤r≤2, 0≤α≤1 and the embeddingE→L2, we infer thatE→Ls, 1≤s≤2.

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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) ≥ 12kuk2λcq+1q+1kukq+1S−22/2kuk2

≥ kukq+1

1

2kuk2(q+1)λcq+1q+1S22∗/2 kuk2(q+1) whereS is the best constant for the embeddingD1,2→L2, that is

S= inf

( R|∇u|2

R |u|22/2 | u∈ D1,2, u6≡0 )

.

Letting

Q(t)≡ 1

2t2(q+1)−S2/2

2 t2(q+1), t≥0, there isρ >0 such that

maxt0 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).

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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

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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 thatu6≡ 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

γΓ max

0t1I(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 −2I(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.

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Remark. I(un+)→c andI0(un+)→0.

Indeed,un is also bounded so that o(1) =hI0(un), uni=

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∈ Cosuch that

φ≡1 on Br0, φ≡0 on RN\B2r0. Now consider the function

w(x) = [N(N−2)](N2)/4

(+|x|2)(N2)/2 , x∈RN, >0 which satisfies

−∆w=w21 inRN. It is well known (see e.g. Talenti [5], Aubin [11] ) that

kwk21 =|w|22 = SN/2.

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Let

ψ=φw

and letv∈ Co given by

v= ψ

21/2. Now it can be shown (see e.g. [6], [10]) thatX≡R

|∇v|2 satisfies X≤S+O((N2)/2).

Moreover there is somet>0 such that

maxt0 I(tv) =I(tv)

and d

dtI(tv)|t=t= 0.

which gives

0< t< X1/(22)≡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

t0 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

2t202−t2 2

−λt2 2

Z

B2r0

v2. Now recalling that as a function oft,

t2

2t202−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

(N2)/2i2/(22)

−λd20 2

Z

B2r0

v2

= 1

N h

S+O

(N2)/2 iN/2

−λd20 2

Z

B2r0

v2.

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Using the inequality

(b+c)α≤bα+α(b+c)α1c b, c≥0, α≥1 withb=S,c=O (N2)/2

andα=N/2 we get I(tv)≤ 1

NSN/2+O((N2)/2)−c0λ Z

B2r0

v2.

Therefore,

I(tv)≤ 1

NSN/2+(N2)/2 (

M−c0λ(2N)/2 Z

B2r0

v2 )

, wherec0=d20/2 and M is a positive constant.

We shall show that (N2)/2

(

M−c0λ(N2)/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+(N2)/2 (

M−c0λ(4N)/2

Z r0−1/2 0

sN1ds (1 +s2)N2

) .

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(r01/2)o . Now since

c0λln(r01/2)→ ∞as→0 we infer that

I(tv)<1

4S2, for small >0.

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Case N≥5. Noticing that c0λ(4N)/2

Z r0−1/2 0

sN1ds

(1 +s2)N2 → ∞as→0 we infer that

I(tv)< 1

NSN/2for small >0, which concludes the proof of this claim.

References

[1] A. Ambrosetti, H. Br´ezis & G. Cerami, Combined effects of concave and convex nonlinearities in some elliptic problems, Preprint.

[2] D. Costa,On a nonlinear elliptic problem in RN, Preprint.

[3] E. S. Noussair, C. A. Swanson & Y. Jianfu,Positive finite energy solutions of critical semilinear elliptic problems, Canadian J. Math. 44(1992) 1014- 1029.

[4] J. G. Azorero & I. P. Alonzo,Multiplicity of solutions for elliptic problems with critical exponents with a nonsymmetric term, Trans. Amer. Math.

Society 323(1991) 877-895.

[5] G. Talenti, Best constant in Sobolev inequality, Ann. Math. 110 (1976) 353-372.

[6] H. Br´ezis & L.Nirenberg,Some Variational problems with lack of compact- ness.Proc. Symp. Pure Math. (AMS) 45(1986) 165-201.

[7] H. Egnell, Existence and nonexistence results for m-Laplace equations in- volving critical Sobolev exponents, Arch. Rat. Mech. Anal. 104(1988) 57-77.

[8] I. Ekeland, On the variational principle. J. Math. Anal. App. 47(1974) 324-353.

[9] M. Guedda & L. Veron, Quasilinear elliptic equations involving critical Sobolev exponents, Non. Anal. 12(1989) 879-902.

[10] O. Miyagaki,On a class of semilinear elliptic problems in IRn with critical growth, CMS Technical Report, Univ. Wisconsin (1994).

[11] T. Aubin,Problemes isoperimetriques et espaces de Sobolev, J. Diff. Geom- etry 11(1976) 573-598.

[12] V. Benci & G. Cerami,Existence of positive solutions of the equation−∆u+

a(x)u=u(n+2)/(n2)in Rn, J. Funct. Anal. 88(1990) 90-117.

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[13] W. Omana & M. Willem, Homoclinic orbits for a class of Hamiltonian systems, Diff. Int. Equations 5(1992) 1115-1120.

[14] Y. Jianfu & Z. Xiping, On the existence of nontrivial solution of a quasi- linear elliptic boundary value problem for unbounded domains, Acta Math.

Sci. 7(1987) 341-359.

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]

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