On some conjectures proposed by Haim Brezis

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On some conjectures proposed by Ha¨ım Brezis Pierpaolo Esposito

∗

27th January 2003

Dipartimento di Matematica, Universit` a di Roma ”Tor Vergata”, Via della Ricerca Scientifica, 00133 Roma, Italy (e-mail:[email protected])

Abstract O. Druet solved in [4] two conjectures proposed by Ha¨ım Brezis in [1] about ”low” dimension phenomena for some elliptic problem with critical Sobolev exponent. In [4], the proof of one of the two conjectures is reduced to an asymptotic analysis which is carried over with very general technics involving pointwise estimates. We propose here a different and simpler approach in the blow-up analysis based on integral estimates and on a careful expansion of the energy functional.

1

Introduction

Let Ω be a smooth bounded domain in RN , ¯ We consider the problem function in Ω.  p   −∆u + au = u u>0   u=0

N ≥ 3, and a(x) be a continuous

in Ω in Ω on ∂Ω,

(P)

N +2 where p = N −2 is the critical Sobolev exponent. Equation (P) can be regarded as the minimization problem for the functional R R |∇u|2 + Ω a(x)u2 Ja (u) = Ω R , u ∈ H01 (Ω) \ {0}. (1) 2
p+1 p+1 |u| Ω ∗ Supported

by MIUR, national project Variational Methods and Nonlinear Differential Equa-

tions.

1

In fact, by standard elliptic regularity theory, any minimum ua for Ja provides a smooth solution for problem (P), up to rescaling the Lagrange multiplier and changing ua into |ua |. Remarking that the solvability of (P) implies the coercivity of the operator −∆ + a on H01 (Ω), i.e. the first eigenvalue λ1 is positive, we can assume from now on that −∆ + a is coercive on H01 (Ω). Denoting Sa the infimum of Ja , it is well known that Sa ≤ S, S = S0 being the Sobolev constant, and Sa < S implies Sa achieved( see for instance [2]). For N ≥ 4, in [2] it is also proved that a(x) negative somewhere implies Sa < S and, since S is never attained, the properties (i) a(x) negative somewhere (ii) Sa < S (iii) Sa is achieved are equivalent. So the problem turns out to be completely characterized by the local nature of a(x). On the other hand, in dimension N = 3, the global nature of a(x) becomes significative: in [2] it was discussed the particular case of the unit ball and a(x) ≡ const. Some notations are in order to state the general result: let us define Ga (x, y), x ∈ Ω \ {y}, the Green function in y ∈ Ω of −∆ + a in Ω with Dirichlet boundary condition, as the distributional solution for ( −∆x Ga (x, y) + a(x)Ga (x, y) = δy in Ω Ga (x, y) = 0 on ∂Ω, 1 where δy is the Dirac measure in y, and let us set Ha (x, y) = Ga (x, y) − ω2 |x−y| the regular part of the Green function Ga (x, y), where Ha (x, y) ∈ C(Ω × Ω) is a distributional solution for ( −∆x Ha (x, y) + a(x)Ga (x, y) = 0 in Ω 1 Ha (x, y) = − ω2 |x−y| on ∂Ω.

For the general case, the following result was conjectured in [1] and proved in [4]: ¯ be such Theorem 1.1. Let Ω be a bounded domain in R3 and let a(x) ∈ C(Ω) that −∆ + a is coercive. The properties: (i) ∃ y ∈ Ω sucht that Ha (y, y) > 0 (ii) Sa < S (iii) Sa is achieved are equivalent. 2

By test functions computations( see [13]), one gets i) ⇒ ii). The fact that ii) ⇒ iii) is well-known and is already present for instance in [2]. Brezis, in [1], proposed the two following conjectures: (a) the implication iii) ⇒ ii) does hold; (b) the implication ii) ⇒ i) does hold. Druet proved these two conjectures and thus Theorem 1.1 in [4]. The proof of the conjecture (a) makes use of the minimality condition D 2 Ja (ua ) ≥ 0. Thanks to (a), the proof that ii) ⇒ i) is reduced in [4] to some asymptotic analysis in the following way: let a ∈ C(Ω) such that Sa < S. By continuity and monotonicity of Sa+δ with respect to δ > 0, we get Sa+δ < S for 0 < δ < δ0 and Sa+δ = S for δ ≥ δ0 , δ0 some positive real number. Since Sa+δ < S for 0 < δ < δ0 , there exists ua+δ a smooth positive function achieving Sa+δ for 0 < δ < δ0 and satisfying R 3 the renormalization Ω u6a+δ = (Sa+δ ) 2 . By uniform coercivity of −∆ + (a + δ) R on H01 (Ω), it is easily seen that supδ∈(0,δ0 ) Ω |∇ua+δ |2 < +∞ and then, up to a subsequence, we can assume ua+δ ⇀ u as δ → δ0 in H01 (Ω). Since, by (a), Sa+δ0 is not achieved, u ≡ 0 and then we have ua+δ ⇀ 0 as δ → δ0 weakly but not strongly in H01 (Ω). Hence ua+δ must blow-up as δ → δ0 . In view of the compactness of the embedding of H01 (Ω) into L2 (Ω), we have Z Z 3 3 lim |∇ua+δ |2 = lim u6a+δ = (Sa+δ0 ) 2 = S 2 δ→δ0

Ω

δ→δ0

Ω

3

and then we obtain |∇ua+δ |2 ⇀ S 2 δy0 as δ → δ0 weakly in the sense of mea¯ see [14]). In [4], Druet carried over an asymptotic analysis sures for some y0 ∈ Ω( based on pointwise estimates of ua+δ as δ → δ0 and proved that y0 ∈ Ω and Ha+δ0 (y0 , y0 ) = 0. This proves (b) thanks to the monotonicity Ha (y, y) > Ha+δ0 (y, y). The aim of this paper is to give an alternative and more direct proof that y 0 ∈ Ω and Ha+δ0 (y0 , y0 ) ≥ 0 by exploiting integral estimates and a careful expansion of Sa+δ = Ja+δ (ua+δ ). Moreover, the same computations lead to the implication i) ⇒ ii): in particular we get Ha+δ0 (y0 , y0 ) = 0. Hence, assuming ii) ⇔ iii), we prove the equivalence i) ⇔ ii). Related problems can be found in [3], [6], [8], [9], [10], [11], [12] and [5], [7] concerning the Euclidean and Riemannian case respectively.

3

2

Expansion of the energy functional

Replacing a(x) + δ0 with a(x), we are led to study the asymptotic behaviour for {uδ }, where uδ achieves Sa−δ and is a smooth positive solution of ( −∆uδ + (a − δ)uδ = u5δ in Ω (2) uδ = 0 on ∂Ω 3 ¯ such that |∇uδ |2 ⇀ S 2 δy0 as δ → 0+ weakly in the sense of measures, y0 ∈ Ω. 1 Moreover, we assume −∆ + a coercive on H0 (Ω).

Let us define for (ǫ, y) ∈ (0, +∞) × Ω Uǫ,y (x) = ǫ

− 21

U



x−y ǫ



1

, U (x) =

34

1

(1 + |x|2 ) 2

and P : H 1 (Ω) → H01 (Ω) the orthogonal projection defined for ϕ ∈ H 1 (Ω) as Z Z ∇P ϕ∇ψ = ∇ϕ∇ψ ∀ ψ ∈ H01 (Ω), Ω

Ω

R where H01 (Ω) is endowed with the inner product < u, v >= Ω ∇u∇v and the induced norm || · ||. 1 1 Let us set ψǫ,y = Uǫ,y −P Uǫ,y , fǫ,y = ψǫ,y +3 4 ω2 ǫ 2 H(·, y) and H(x, y) := H0 (x, y), where H0 (x, y) is the regular part of the Green function of −∆ in Ω with Dirichlet boundary condition. We have the following properties( see the appendix in [10]): 1

0 ≤ P Uǫ,y ≤ Uǫ,y , ||ψǫ,y ||∞ = O( 5

||fǫ,y ||∞ = O

ǫ2 d3

!

 ǫ 1 ǫ2 ) , ||ψǫ,y ||L6 (Ω) = O ( ) 2 d d

, sup dH(y, y) < 0 , sup |∇H(x, y)| = O( y∈Ω

x∈Ω

1 ) d2

where d = dist(y, ∂Ω). We follow now [10]: for δ small, we can decompose uδ in the form uδ = αδ P Uǫδ ,yδ + wδ for αδ , ǫδ > 0, yδ ∈ Ω, wδ ∈ Tδ such that αδ → 1 , ǫδ → 0 , yδ → y0 ,

ǫδ → 0 , wδ → 0 in H01 (Ω) as δ → 0, dδ 4

(3)

(4)

where Tδ = Span {P Uǫδ ,yδ ,

∂P Uǫδ ,yδ ∂ǫ

,

∂P Uǫδ ,yδ ∂yi

: i = 1, .., 3}⊥ .

From now on we will omit the dependence on δ. We need to estimate the rate of convergence of α and w. First we prove Lemma 2.1. There exists C > 0 such that Z Z Z Z 4 2 2 2 |∇v|2 , ∀ v ∈ Tδ . Uǫ,y v ≥ C a(x)v − 5 |∇v| +

(5)

Ω

Ω

Ω

Ω

for δ small. Proof The proof is based on a well known inequality( see [10]): Z Z Z 4 2 4 2 |∇v| − 5 Uǫ,y v ≥ |∇v|2 , ∀ v ∈ Tδ . 7 Ω Ω Ω

(6)

We proceed in the following way. Setting   Z Z 4 2 2 , U v a(x)v − 5 1 + Cδ = inf R ǫ,y v∈Tδ :

Ω

|∇v|2 =1

Ω

Ω

we have that Cδ is attained if Cδ < 1. In fact, let Cδ < 1 and let vn be a minimizing sequence: up to a subsequence, we can assume that vn ⇀ vδ as n → +∞ weakly R R R 4 vδ2 = Cδ . in H01 (Ω) and vδ ∈ Tδ , Ω |∇vδ |2 ≤ 1 and 1 + Ω a(x)vδ2 − 5 Ω Uǫ,y Since Cδ < 1, we get vδ 6= 0 and the inequality Z Z Z Z 4 2 2 2 Uǫ,y vδ ≤ Cδ a(x)vδ − 5 |∇vδ |2 |∇vδ | + Ω

Ω

Ω

Ω

holds. Hence we obtain Ω |∇vδ |2 = 1 and Cδ is achieved by vδ . Now we show that lim inf δ→0 Cδ > 0. Otherwise, we find a subsequence Cδ → L ≤ 0 such that Cδ is achieved by some smooth vδ satisfying  4   −(1 − Cδ )∆vδ + a(x)vδ − 5Uǫ,y vδ = 0 in Ω vδ = 0 on ∂Ω   R 2 |∇vδ | = 1. Ω R

Up to a subsequence, vδ ⇀ v as δ → 0 weakly in H01 (Ω). Hence v solves −(1 −
′ L)∆v + av = 0 in H01 (Ω) . So there holds Z Z Z Z 0 = (1 − L) |∇v|2 + a(x)v 2 ≥ |∇v|2 + a(x)v 2 Ω

Ω

Ω

5

Ω

and, by coercivity, vδ ⇀ v ≡ 0 weakly in H01 (Ω). So we get Z Z Z 3 3 2 4 2 2 |∇vδ |2 = Cδ = |∇vδ | − 5 Uǫ,y vδ + o(||vδ || ) ≥ 7 7 Ω Ω Ω in view of the compactness of the embedding of H01 (Ω) into L2 (Ω) and this contradicts L ≤ 0. Finally, we set C = 21 lim inf δ→0 Cδ > 0. From this Lemma, we derive now the exact behaviour of w: Lemma 2.2. We have the estimate ||w|| = O

ǫ

d

1

+ ǫ2



(7)

and there holds the formula Z Z Z Z ǫ 4 . a(x)P Uǫ,y w + o Uǫ,y w2 = − a(x)w2 − 5 |∇w|2 + d Ω Ω Ω Ω Proof The function w satisfies the equation ( 5 5 −∆w = (αP Uǫ,y + w) − αUǫ,y − (a(x) − δ) (αP Uǫ,y + w) w=0

(8)

in Ω on ∂Ω.

(9)

By Sobolev inequality, multiplying (9) for w and integrating by parts we get Z Z Z Z 4 Uǫ,y w2 = −α (a(x) − δ)P Uǫ,y w a(x)w2 − 5 |∇w|2 + Ω

Ω

Ω

Ω

  Z
4 2 2 Uǫ,y |ψǫ,y ||w| +o ||w|| + O ||w||||ψǫ,y ||L6 (Ω) +

(10)

Ω

5 w =< P Uǫ,y , w >= 0. because Ω Uǫ,y By (3) and Sobolev inequality we get  ! 56 Z Z 1  ǫ  12 24 2 ǫ 4 5 Uǫ,y ψǫ,y w = O (||w||)  Uǫ,y + d d Ω Bd (y)  ǫ =O ||w|| d

R

and

Z

Ω

P Uǫ,y |w| = O(||w||)

Z

Ω

6 5 Uǫ,y

 65

Z

Ω\Bd (y)

 1  = O ǫ 2 ||w|| .

6 Uǫ,y

! 23  

We insert these estimates in (10): using the coercivity (5), first we get (7) and in turn we obtain (8). 6

We are now in position to prove Theorem 1.1 expanding the energy functional. Proof ( of Theorem 1.1) With the aid of (7) and (8), we expand now R R |∇uδ |2 + Ω (a(x) − δ)u2δ Ω Sa−δ = R 6
13 u Ω δ R R R R 4 2 α2 Ω |∇P Uǫ,y |2 + Ω a(x)P Uǫ,y w2 + Ω a(x)P Uǫ,y w + o( dǫ ) + 5 Ω Uǫ,y =  R  13 R R ǫ 4 2 5 6 −6 α6 Ω Uǫ,y Ω Uǫ,y ψǫ,y + 15 Ω Uǫ,y w + o( d ) because, as in Lemma 2.2, Z

 ǫ ||w|| = o d d

ǫ

4 Uǫ,y wψǫ,y = O

Ω

and similarly ǫ

 ǫ ||ψǫ,y ||L6 (Ω) = o . d d Ω R R R 3 3 5 6 ψǫ,y + o( dǫ ), we obtain = S 2 + o( dǫ ) and Ω |∇P Uǫ,y |2 = S 2 − Ω Uǫ,y Since Ω Uǫ,y Z  Z Z ǫ − 12 5 2 Sa−δ = S + S Uǫ,y ψǫ,y + a(x)P Uǫ,y + a(x)P Uǫ,y w + o . d Ω Ω Ω Z

4 2 Uǫ,y ψǫ,y =O

By (4) and a Taylor expansion for H(x, y) we get 5

ψǫ,y (x)

ǫ2 = −3 ω2 ǫ H(y, y) + O ǫ |x − y| sup |∇H(x, y)| + 3 d x∈Ω ! 1 5 1 1 ǫ 2 |x − y| ǫ 2 + = −3 4 ω2 ǫ 2 H(y, y) + O d2 d3 1 2

1 4

and hence Z 5 Uǫ,y ψǫ,y

=

Z

Ω

Ω

1 2

5 Uǫ,y

"

1

1 2

1 4

−3 ω2 ǫ H(y, y) + O

3

= −3 2 ω2 ǫH(y, y)

Z

R3

If d → 0, then

because

dx R3 (1+|x|2 ) 52

R

Z

Ω

dx 5

(1 + |x|2 ) 2

ǫ + o( ). d

1 1 ǫ Sa−δ = S − S − 2 3 2 ω22 ǫH( y, y) + o( ) d = ω32 and

2 a(x)P Uǫ,y

+

Z

ǫ a(x)P Uǫ,y w = O(ǫ) = o( ). d Ω 7

5

ǫ 2 |x − y| ǫ 2 + 3 d2 d

!

!#

Since supy∈Ω dH(y, y) < 0( see (4)), we conclude Sa−δ > S. A contradiction. So we can exclude the boundary concentration: y0 ∈ Ω. The expansion for Sa−δ becomes " 2  Z 1 P Uǫ,y − 21 a(x) Sa−δ = S + S ǫ −3 2 ω22 H(y0 , y0 ) + 1 ǫ2 Ω     Z P Uǫ,y w + o(ǫ). a(x) + 1 1 ǫ2 ǫ2 Ω

By (4) we get P Uǫ,y ǫ

1 2

1

=

34 (ǫ2

+ |x −

1

1 y|2 ) 2

+ 3 4 ω2 H(x, y) −

fǫ,y ǫ

1

→ 3 4 ω2 G(x, y0 ) as δ → 0

1 2

in Ls (Ω) for any s < 32 and uniformly in any compact set of Ω \ {0}, where G(x, y) := G0 (x, y). Moreover there holds Z Z 1 1 32 32 → a(x) as δ → 0. a(x) 2 ǫ + |x − y|2 |x − y0 |2 Ω Ω

By (7), the functions w ˜ = w1 are uniformly bounded in H01 (Ω) and solve the ǫ2 equation   5   5   PU PU U −∆w ˜ + (a(x) − δ) α 1ǫ,y + w ˜ = ǫ2 α 1ǫ,y + w ˜ − ǫ2 α ǫ,y in Ω 1 ǫ2 ǫ2 ǫ2  w ˜=0 on ∂Ω.

Up to a subsequence, we can assume that w ˜ ⇀ f as δ → 0 weakly in H01 (Ω). Hence ∞ f satisfies for any Φ ∈ C0 (Ω \ {y0 }) Z Z Z 1 4 a(x)G(x, y0 )Φ. (11) a(x)f Φ = −3 ω2 ∇f ∇Φ + Ω

Ω

Ω

H01 (Ω),

Since f ∈ it can be easily seen that (11) holds for any Φ ∈ H01 (Ω) and hence 1 f (x) = 3 4 ω2 (Ha (x, y0 ) − H(x, y0 )). In view of the compactness of the embedding of H01 (Ω) into Ls (Ω) for any 1 ≤ s < 6, we get Z Z Z 1 2 a(x)P Uǫ,y + a(x)P Uǫ,y w = 3 2 ω22 ǫ a(x)G(x, y0 )2 Ω Ω ZΩ 1 +3 4 ω2 ǫ a(x)G(x, y0 )f + o(ǫ). Ω

¯ and G(x, y0 ) ∈ Ls for any 1 ≤ s < 3, we get that G(x, y0 ) Since f ∈ W ∩ C(Ω) is an admissible test function in (11) and then Z Z Z 1 ∆f G(x, y0 ) = −f (y0 ) a(x)G(x, y0 )f = a(x)G(x, y0 )2 + 3 4 ω2 2,s

Ω

Ω

Ω

1

= 3 4 ω2 (H(y0 , y0 ) − Ha (y0 , y0 )) .

8

Finally we get 1

1

Sa−δ = S − 3 2 ω22 S − 2 ǫHa (y0 , y0 ) + o(ǫ) < S and then Ha (y0 , y0 ) ≥ 0. Hence (ii) ⇒ (i). To prove the converse, let us remark that the expansion for Sa−δ = Ja−δ (uδ ), uδ = αδ P Uǫδ ,yδ + wδ , is based on (7) and on Z Z Z Z 1 2 2 4 2 4 a(x)G(x, y0 )f + o (ǫδ ) (8’) |∇wδ | + a(x)wδ − 5 Uǫδ ,yδ wδ = −3 ω2 ǫδ Ω

Ω

Ω

Ω

and not on the equations satisfied by uδ and wδ . Let us assume Ha (y0 , y0 ) > 0 and let us consider test functions in the form uǫ =
R R 1 4 4 P Uǫ,y0 + ǫ 2 f , ǫ > 0. Since Ω Uǫ,y → 0 as ǫ → 0, we get f 2 = O ||f ||2∞ Ω Uǫ,y 0 0 1 that (7) and (8’) hold for w = ǫ 2 f and then the expansion 1

1

Ja (uǫ ) = S − 3 2 ω22 S − 2 ǫHa (y0 , y0 ) + o(ǫ) follows. Hence Sa < S and (i) ⇒ (ii) is proved.

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[7] E. Hebey, and M. Vaugon, The best constant problem in the Sobolev embedding theorem for complete Riemannian manifold, Duke Math. J. 79 (1995), 235-279. [8] E. Hebey, Asymptotic behavior of positive solutions of quasilinear elliptic equations with critical Sobolev growth, Differential Integral Equations 13 (2000), 1073-1080. [9] O. Rey, Proof of the conjectures of H. Br´ezis and L.A. Peletier, Manuscripta Math. 65 (1989), 19-37. [10] O. Rey, The role of the Green’s function in a non-linear elliptic equation involving the critical Sobolev exponent, J. Funct. Analysis 89 (1990), 1-52. [11] F. Robert, Asymptotic behaviour of a nonlinear elliptic equation with critical Sobolev exponent: the radial case, Adv. Differential Equations 6 (2001), no. 7, 821-846. [12] F. Robert, Asymptotic behaviour of a nonlinear elliptic equation with critical Sobolev exponent: the radial case. II, NoDEA Nonlinear Differential Equations Appl. 9 (2002), no. 3, 361-384. [13] R. Schoen, Conformal deformation of a Riemannian metric to constant scalar curvature, J. Differential Geom. 20 (1984), 479-495. [14] M. Struwe, A global compactness result for elliptic boundary value problems involving limiting nonlinearities, Math. Z. 187 (1984), 511-517.

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