Compactness of a nonlinear eigenvalue problem with a singular nonlinearity

Citation preview

Compactness of a nonlinear eigenvalue problem with a singular nonlinearity Pierpaolo ESPOSITO∗

January 19, 2006

Abstract λf (x) N We study the Dirichlet boundary value problem −∆u = (1−u) 2 on a bounded domain Ω ⊂ R . For 2 ≤ N ≤ 7 we characterize compactness for solutions sequence in terms of spectral informations. As a byproduct, we give an uniqueness result for λ close to 0 and λ∗ in the class of all weak solutions with finite Morse index, λ∗ being the extremal value associated to the nonlinear eigenvalue problem.

Keywords: Compactness, Morse index, Critical Parameter, Electrostatic MEMS. AMS subject classification: 35J60, 35B40, 35J20.

1

Introduction

Let us consider the following problem:  λf (x)    −∆u = (1 − u)2 0 0 : ∃ u solution of (1.1)}, and characterized uλ as the unique semi-stable solution of (1.1) (in the sense that the linearized operator is a positive operator, see also [12]). Moreover, they raised out the special role of dimension N = 7 for problem (1.1) by means of some energy estimates: the minimal branch is always compact for 1 ≤ N ≤ 7 and in general, this is not true anymore for N ≥ 8. For 1 ≤ N ≤ 7 as λ → λ∗ the minimal branch uλ converges to u∗ , the so-called extremal solution, the unique solution of (1.1) with λ = λ∗ . By [4] λ∗ is a turning point and a second branch Uλ of solutions for (1.1) comes out for λ close to λ∗ from u∗ (Uλ ia non degenerate solution with Morse index 1). In [5] we developed a different approach to face non compactness phenomena, based on some fine asymptotic analysis, and proved compactness along the first branch of unstable solutions Uλ of (1.1) in the same low dimensions (the compactness of the minimal branch was also recovered). In particular, this branch cannot never approach λ = 0 for 2 ≤ N ≤ 7. For 2 ≤ N ≤ 7 and f (x) = 1, Joseph and Lundgren in [10] showed that the bifurcation diagram of (1.1) on the ball has exactly the following form: f(x) ≡ 1 and 2 ≤ N ≤ 7 1 ||u||

∞

0

0

λ

*

*

λ

λ

Figure 1: Plots of kuk∞ versus λ in the case f (x) = 1 on the unit ball and 2 ≤ N ≤ 7, where λ∗ =

2(3N −4) . 9

Namely, there are infinitely turning points oscillating around the value λ∗ = 2(3N9 −4) , the solutions number of (1.1) going to +∞ as λ approaches λ∗ . For 2 ≤ N ≤ 7 and f (x) = |x|α , α ≥ 0, numerically the diagram +α−4) (see the thorough discussion in [7]). above still holds on the ball for λ∗ = (2+α)(3N 9 Problem (1.1) presents in general a rich structure of the solutions set. The strong compactness properties of Theorem 1.1 below let us guess that a rigourous analysis could be pushed much more further than the analysis of the first two branches. Theorems 1.2-1.4 below show that some features of the bifurcation diagram on the ball hold for general domains. We could expect that in the solutions set of (1.1) on a general domain there is always a continuous path having the shape of the diagram on the ball. For such an existence result, let us remark the difficulty related to the absence of a good variational principle. An hopeful approach could be based on the analysis directly along the bifurcation diagram: any branch is characterized by a fixed Morse index and, when an eigenvalue of the linearized operator along the branch crosses zero, we have a turning 2

point and the diagram turns into a new branch of higher Morse index (by [4] this the case for example of the first turning point λ∗ ). The following compactness result is the first step in this direction: ¯ be such that: Theorem 1.1. Assume 2 ≤ N ≤ 7. Let f ∈ C(Ω) ! Ã k Y αi |x − pi | g(x) , g(x) ≥ C > 0 in Ω, f (x) =

(1.2)

i=1

for some points pi ∈ Ω and exponents αi ≥ 0. Let {λn }n∈N be a sequence such that λn → λ ∈ [0, λ∗ ] and let un be an associated solution such that: sup m(un , λn ) < +∞.

(1.3)

n∈N

Then, sup k un k∞ < 1. Moreover, if in addition µ1,n := µ1,λn (un ) < 0, then necessarily λ > 0. n∈N

2λf (x) Here and in the sequel, µk,λ (u) denotes the k-th eigenvalue of Lu,λ = −∆ − (1−u) 3 with the convention that eigenvalues are repeated according to their multiplicities, and the Morse index m(u, λ) is the number of negative eigenvalues of Lu,λ .

Theorem 1.1 is quite general and, as we will see below, leads to many interesting properties because it gives a deep relation between compactness and Morse index, a more natural tool in the study of the bifurcation diagram. In particular, it provides a characterization of blow up sequences un (in the sense of blow up of (1 − un )−1 ) and some interesting uniqueness result. We have indeed the following characterization (to be compared with [1, 2] in the context of polinomial subcritical nonlinearities): ¯ be as in (1.2). Let {λn }n∈N be a sequence such that Theorem 1.2. Assume 2 ≤ N ≤ 7. Let f ∈ C(Ω) λn → λ ∈ [0, λ∗ ] and let un be an associated solution. Then, 1. max un → 1 as n → +∞ Ω

2.

R ³ Ω

f (x) (1−un )3

´ N2

→ +∞ as n → +∞

3. m(un , λn ) → +∞ as n → +∞ are equivalent. The following uniqueness result was strongly expected to be true (see the bifurcation diagram): ¯ be as in (1.2). For any fixed k ∈ N there exists δ > 0 Theorem 1.3. Assume 2 ≤ N ≤ 7. Let f ∈ C(Ω) small so that 1. for λ ∈ (0, δ) the minimal solution uλ is the unique solution u of (1.1) with m(u, λ) ≤ k; 2. for λ ∈ (λ∗ − δ, λ∗ ) uλ and Uλ are the unique solutions u of (1.1) with m(u, λ) ≤ k. The equivalence among points (1) and (2) in Theorem 1.2 was already proved in [7]. A weaker form of Theorem 1.3, point (1), was already shown in [7] as uniqueness in the class of solutions of bounded energy: R ³ f (x) ´ N2 ≤ k for some k > 0. Ω (1−u)3

Finally, based on a simple degree argument, we get the existence of a solutions sequence un whose Morse index blows up (equivalently, by Theorem 1.2 the sequence blows up pointwise: max un → 1 as n → +∞): Ω

3

¯ be as in (1.2). There exist a sequence {λn }n∈N and Theorem 1.4. Assume 2 ≤ N ≤ 7. Let f ∈ C(Ω) associated solution un of (1.1) so that m(un , λn ) → +∞

as n → +∞.

The paper is organized as follows. In Section 2 we extend some basic regularity results of [7] to solutions of a non homogeneous Dirichlet version of (1.1) with finite Morse index and finite singular set. Following the approach of [5] for the first branch Uλ of unstable solutions, in Section 3 we describe some blow up techniques for a possible blowing up sequence un (i.e. max un → 1 as n → +∞) to get a strong pointwise estimate Ω

on the R.H.S. of (1.1). This provides the uniform convergence of un to a limit singular solution u0 in Ω with finite Morse index and finite singular set, which does not exist according to the regularity statements of Section 2. In Section 3 we give proofs of Theorems 1.2-1.4. For reader’s convenience, in Appendix we briefly sketch the proof of some results already proved in [5].

Acknowledgements: I would like to thank Professor N. Ghoussoub for his friendly and generous hospitality during my stay at UBC. Nevertheless, I can’t forget to express him the gratitude for introducing me in the study of equation (1.1) and for the several stimulating discussions about it.

2

Regularity properties

In this Section we establish some basic regularity results for the following boundary value problem:  f (x)  in Ω, −∆u = (1 − u)2  u=u ¯ on ∂Ω,

(2.1)

¯ satisfies (1.2) and u ¯ is such that 0 ≤ u where f ∈ C(Ω) ¯ ∈ C 1 (Ω) ¯ ≤ k¯ uk∞ < 1. Solutions u of (2.1) are to f (x) f (x) −1 be considered in the following H 1 (Ω)-weak sense: u − u ¯ ∈ H01 (Ω), (1−u) (Ω) and −∆u = (1−u) 2 ∈ H 2 in −1 −1 1 H (Ω), where H (Ω) is the dual space of H0 (Ω).

The first regularity result we give is already contained in [7] for u¯ = 0. We extend it to cover non homogeneous boundary values and we slightly improve the original statement (for N = 2). The following holds: Proposition 2.1. Let N ≥ 2. Let u be a H 1 (Ω)-weak solution of (2.1) so that

Then, 0 ≤ u ≤ kuk∞ < 1.

N f ∈ L 2 (Ω). 3 (1 − u)

(2.2)

Proof: Thanks to (2.2), we want to show that kuk∞ ≤ 1 ,

(1 − u)−1 ∈ Lp (Ω)

(2.3)

for any p > 1. Fix p > 1 and let us introduce Tk u = min{u, 1 − k}, the truncated function of u at level 1 − k, 0 < k < 1. Let us first discuss the case N = 2. For k small, take (1 − Tk u)−1 − (1 − u ¯)−1 ∈ H01 (Ω) as a test function for (2.1): Z Z Z ´ |∇Tk u|2 ∇u∇¯ u f (x) ³ −1 −1 (2.4) (1 − T u) − (1 − u ¯ ) = + k 2 (1 − u ¯)2 (1 − u)2 Ω (1 − Tk u) ZΩ ZΩ ∇u∇¯ u f (x) + < +∞ ≤ 2 (1 − u ¯ ) |1 − u|3 Ω Ω 4

³ ´ 1−¯ u by means of (2.2). Since log 1−T ∈ H01 (Ω), by Moser-Trudinger inequality and (2.4) we get that for any u k p > 1: ¶ ¶p µ 2 Z Z µ Z 1−u ¯ 2 p 1−u ¯ −p )| ≤ C, ≤ C exp |∇ log( (1 − Tk u) ≤ C 1 − Tk u 16π Ω 1 − Tk u Ω Ω where C denotes various positive constants depending only on p. Taking the limit as k → 0, we get the R validity of Ω (1 − min{u, 1})−p ≤ C for any p > 1. Then, {u > 1} is a set of zero measure and the previous estimate implies the validity of (2.3). The case N ≥ 3 is more involved. By (1.2) and (2.2) we easily derive that |{|1 − u| ≤ ε}| → 0 as ε → 0, where | · | stands for the Lebesgue measure. Hence, by (2.2) let us fix some ε > 0 small so that Z

{|1−u|≤ε}

µ

f (x) |1 − u|3

¶ N2

≤

µ

p+1 SN 2p2

¶ N2

,

(2.5)

2N

where SN is the Sobolev constant for the embedding of H01 (Ω) into L N −2 (Ω). For k small, take (1 − Tk u)−p−1 − (1 − u ¯)−p−1 ∈ H01 (Ω) as a test function for (2.1) and by (2.2) we get: Z Z Z ´ ∇u∇¯ u f (x) ³ |∇Tk u|2 −p−1 −p−1 (1 − T u) − (1 − u ¯ ) = (p + 1) + (2.6) (p + 1) k 2 (1 − Tk u)p+2 ¯)p+2 Ω (1 − u Ω (1 − u) Z Ω ´2 f (x) ³ −p −p 2 − (1 − u 2 (1 − T u) ≤2 + C, ¯ ) k 3 Ω |1 − u| because of the following relation for δ = 1: ´2 ³ p p ¯)− 2 + Cδ (1 − u ¯)−p , (1 − Tk u)−p ≤ (1 + δ) (1 − Tk u)− 2 − (1 − u

δ > 0.

Since (1 − Tk u)−1 ≤ |1 − u|−1 , by Sobolev embedding and (2.6) then we get that: Z Z ´¯ ´¯ ¯ ³ ¯ ³ p 2 ¯∇ (1 − Tk u)− p2 ¯2 + C ¯∇ (1 − Tk u)− p2 − (1 − u ¯)− 2 ¯ ≤ 2 Ω Ω Z ´2 p p2 f (x) ³ −p ≤ (1 − Tk u) 2 − (1 − u ¯)− 2 + C 3 p + 1 Ω |1 − u| Z ´2 p p f (x) ³ p2 ≤ (1 − Tk u)− 2 − (1 − u ¯)− 2 + C 3 p + 1 {|1−u|≤ε} |1 − u| ! N2 Z ÃZ ´¯ ¯ ³ p f (x) N p2 −1 2 ¯∇ (1 − Tk u)− p2 − (1 − u ¯)− 2 ¯ + C, ( SN )2 ≤ 3 p+1 |1 − u| {|1−u|≤ε} Ω

where C denotes various positive constants depending only on ε and p. Hence, by (2.5) we get that Z ³Z ³ ´ N2N ´¯ ´ NN−2 ¯ ³ p −2 2 −1 −p −p ¯∇ (1 − Tk u)− p2 − (1 − u 2 2 (1 − Tk u) − (1 − u ¯) ¯)− 2 ¯ ≤ C, ≤ SN Ω

and in turn

Ω

Z

Ω

pN

(1 − Tk u)− N −2 ≤ C,

where C > 0 does not depend on k. Taking the limit as k → 0, as before we get the validity of (2.3). 5

(2.7)

Now we can conclude. By (2.3) we get that the R.H.S. in (2.1) is in Lp (Ω) for any p > 1. By elliptic regularity ¯ and then, property (2.3) implies kuk∞ < 1. Indeed, if u(x0 ) = 1 for some x0 ∈ Ω, then theory, u ∈ C 1 (Ω) (1 − u(x))2 ≤ C|x − x0 |2 contradicting (2.3). By maximum principle (in a weak form), 0 ≤ u ≤ kuk∞ < 1.¥ By means of some energy estimates (in the spirit of [7]) Proposition 2.1 applies now to semi-stable solutions of (2.1): Proposition 2.2. Let 2 ≤ N ≤ 7. Let u be a H 1 (Ω)-weak solution of (2.1) so that kuk∞ ≤ 1 and ¶ Z µ 2f (x) 2 2 φ ≥ 0 , ∀ φ ∈ H01 (Ω). |∇φ| − (1 − u)3 Ω

(2.8)

Then, 0 ≤ u ≤ kuk∞ < 1. Proof: In order to show that (2.8) gives energy estimates sufficiently good for 2 ≤ N ≤ 7, we adapt the argument in [7] to cover non homogeneous boundary values. Let us remark that (2.1) on u − u ¯ gives ¶ µZ Z Z Z Z f (x) f (x) 1 f (x) f (x) ∇u∇(u − u ¯) + ≤ ≤C (1 − u ¯) = C +C 2 2 2 Ω (1 − u)2 Ω 1−u Ω (1 − u) Ω Ω (1 − u) R f (x) and then, Ω (1−u) ¯ gives that 2 < +∞. Now, (2.8) on u − u µZ ¶ Z Z Z Z 2f (x) f (x) 4f (x) f (x) 2 2 |∇(u − u ¯)| + ≤ 2C (1 − u ¯) ≤ C + (u − u ¯) 3 2 (1 − u)3 Ω 1−u Ω (1 − u) Ω Ω (1 − u) ¶ µZΩ Z Z 2f (x) ∇u∇(u − u ¯) ≤ C. +4 = C |∇(u − u ¯)|2 + Ω Ω Ω 1−u

√ Fix 0 < p < 4 + 2 6 and introduce, as in the previous proof, Tk u = min{u, 1 − k}, 0 < k < 1. For k small, taking (1 − Tk u)−p−1 − (1 − u ¯)−p−1 ∈ H01 (Ω) as a test function in (2.1) we get: ¶ Z Z µ ´ |∇Tk u|2 f (x) ³ ∇u∇¯ u (p + 1) = − (1 − Tk u)−p−1 − (1 − u ¯)−p−1 . (2.9) p+2 p+2 2 (1 − Tk u) (1 − u ¯) Ω Ω (1 − u)

Moreover, by (2.8) we have: Z ´2 f (x) ³ −p −p 2 − (1 − u 2 ¯ ) (1 − T u) 2 k 3 Ω (1 − u)

≤ ≤

Z

´¯ ¯ ³ p 2 ¯∇ (1 − Tk u)− p2 − (1 − u ¯)− 2 ¯ (2.10) Ω ¶ Z µ p2 |∇Tk u|2 ∇u∇¯ u (1 + δ) − + C, 4 (1 − Tk u)p+2 (1 − u ¯)p+2 Ω

for some C > 0 depending on p and δ > 0. Inserting (2.9) into (2.10), we get that Z Z ´2 ´ p2 (1 + δ) f (x) ³ f (x) ³ −p −p −p−1 −p−1 2 − (1 − u 2 ¯ ) ≤ 2 (1 − T u) (1 − T u) − (1 − u ¯ ) +C k k 3 4(p + 1) Ω (1 − u)2 Ω (1 − u) Z ´2 f (x) ³ p2 (1 + δ)2 −p −p 2 − (1 − u 2 (1 − T u) ¯ ) +C ≤ k 4(p + 1) Ω (1 − u)3 √ 2 (1+δ)2 > 0 for δ small and in view of (2.7), where C > 0 does not depend on k. Since p < 4 + 2 6, 2 − p4(p+1) then, Z f (x) (1 − Tk u)−p ≤ C (1 − u)3 Ω 6

for some C > 0 not depending on k. Taking the limit as k → 0, we get that Z f (x) ≤C 3+p Ω (1 − u) √ √ for any 0 < p < 4 + 2 6. Since 3N 2 < 3 + (4 + 2 6) for 2 ≤ N ≤ 7, we get the validity of (2.2) and hence, applying Proposition 2.1 the proof is complete. ¥ We conclude the Section providing a regularity result for solutions of (2.1) with finite Morse index and finite singular set. We have that: ¯ be a H 1 (Ω)-weak solution of (2.1) so that kuk∞ ≤ 1 and the Theorem 2.3. Let 2 ≤ N ≤ 7. Let u ∈ C(Ω) singular set S = {x ∈ Ω : u(x) = 1} is a finite set. Assume that u has finite Morse index: there exists a finite dimensional space T ⊂ H01 (Ω) so that ¶ Z Z µ 2f (x) 2 ⊥ 1 ∇φ∇ψ = 0 ∀ ψ ∈ T }. (2.11) φ ≥ 0 , for any φ ∈ T = {φ ∈ H (Ω) : |∇φ|2 − 0 (1 − u)3 Ω Ω Then, S = ∅. Proof: Assume by contradiction that S 6= ∅ and fix x0 ∈ S. We show now that for δ small ¶ Z µ 2f (x) 2 2 φ ≥ 0 , for any φ ∈ H01 (Bδ ), |∇φ| − (1 − u)3 Bδ

(2.12)

where Bδ := Bδ (x0 ). By contradiction, if (2.12) were false, then there exists δ0 > 0 small and φ0 ∈ C0∞ (Bδ0 ) such that ¶ µ Z 2f (x) 2 2 |∇φ0 | − φ < 0. (1 − u)3 0 B δ0

(2.13)

We can assume that φ0 = 0 in Bδ for some 0 < δ < δ0 small. Indeed, let us replace φ0 with a truncated function φδ with 0 < δ < δ02 small enough, and so that (2.13) is still true while φδ = 0 in Bδ . Letting 0 < δ < δ02 , set φδ = χδ φ0 , where χδ is a cut-off function defined as:   0 |x − x0 | ≤ δ ,   ¶  µ √ log |x − x0 | χδ (x) = 2 1 − δ ≤ |x − x0 | ≤ δ ,  log δ  √   1 |x − x0 | ≥ δ .

By Fatou’s Lemma, we have:

Z

Bδ0

2f (x) 2 φ ≤ lim inf δ→0 (1 − u)3 0

For the gradient term, we have the expansion: Z Z Z |∇φδ |2 = φ20 |∇χδ |2 + B δ0

B δ0

B δ0

7

Z

Bδ0

2f (x) 2 φ . (1 − u)3 δ

χ2δ |∇φ0 |2 + 2

Z

B δ0

χδ φ0 ∇χδ ∇φ0 .

(2.14)

The following estimates hold: Z Z 0≤ φ20 |∇χδ |2 ≤ 4kφ0 k2∞

δ≤|x−x0 |≤ δ

Bδ0

and ¯ ¯2

√

Z

Bδ0

1 C ≤ 2 2 log 1δ |x − x0 | log δ

¯ 4kφ0 k∞ k∇φ0 k∞ χδ φ0 ∇χδ ∇φ0 ¯ ≤ log 1δ

and provide by Lebesgue’s Theorem: Z

B δ0

Combining (2.14)-(2.15), we get that: Z

|∇φδ |2 →

Bδ0

µ

Z

B δ0

|∇φδ |2 −

|∇φ0 |2

Z

B1 (0)

as δ → 0.

2f (x) 2 φ (1 − u)3 δ

¶

1 , |x| (2.15)

0 sufficiently small. In this way, we find 0 < δ1 < δ0 small and φ0 ∈ C0∞ (Bδ0 \ Bδ1 ) such that (2.13) holds. Since by contradiction we are assuming that (2.12) is false for any δ > 0, we can iterate now the argument to find a strictly decreasing sequence δn and φn ∈ C0∞ (Bδn \ Bδn+1 ) such that: ¶ µ Z 2f (x) 2 2 < 0. φ |∇φn | − (1 − u)3 n Bδ n Since {φn }n∈N are mutually hortogonal having disjoint supports, we have found an infinite dimensional set M = Span {φn : n ∈ N} ⊂ H01 (Ω) so that ¶ Z µ 2f (x) 2 2 |∇φ| − φ < 0 ∀ φ ∈ M, (1 − u)3 Ω contradicting (2.11). Hence, (2.12) is proved. 1 (Ω \ S). We fix δ > 0 small so that u ∈ C 1 (∂Bδ ) and By elliptic regularity theory, we get that u ∈ Cloc ¯δ ) satisfying 0 ≤ u ¯ ∈ C 1 (B ¯ ≤ k¯ uk∞,Bδ < 1. Since max u < 1. Then, we extend it on Bδ as a function u ∂Bδ

(2.12) holds on Bδ , we can apply Proposition 2.2 to get that kuk∞,Bδ < 1 contradicting u(x0 ) = 1. Hence, S = ∅ and kuk∞ < 1. ¥

3

Compactness issues

In this Section we turn to the compactness result stated in Theorem 1.1. We follow the approach developed in [5] to prove compactness of the second branch of solutions. We extend the argument to deal with higher branches along which blow up could occur at many finitely points (not only the maximum point as for the second branch). ¯ is in the form (1.2), and let (un )n be a solutions sequence of (1.1) Let 2 ≤ N ≤ 7. Assume that f ∈ C(Ω) ∗ associated to λn → λ ∈ [0, λ ]. Since we want to show that sup kun k∞ < 1, by contradiction and up to n∈N

a subsequence, we will assume all along the Section that un (xn ) = max un → 1− as n → +∞, xn being a Ω

maximum point of un .

8

3.1

A blow-up approach

Let yn ∈ Ω be a sequence of points so that un (yn ) → 1− as n → +∞. Set µn = 1 − un (yn ). As we will see later, for our purposes it is not restrictive to assume that µ3n λ−1 n → 0 and yn → p as n → +∞. Depending on the location of p and the rate of |yn − p|, the length scale to see around yn some non trivial limit profile is the following:  3 −1  if p ∈ Z µ 2 λn 2   n3 − 1 α αi +2 − 2i 2 2 µn λn |yn − pi | if p = pi , µ−3 → +∞ as n → +∞ (3.1) rn = n λn |yn − pi | 1 3  −  −3 αi +2  µn2+αi λn 2+αi if lim inf µ λn |yn − pi | < +∞, n

n→+∞

where Z = {p1 , . . . , pk } is the zero set of the potential f (x). Only to give an idea, let us establish the following rough correspondence: the first situation in the definition of rn corresponds to a blow up at some point outside Z, the second one to a “slow” blow up at some pi ∈ Z, while the third one is a “fast” blow at some pi ∈ Z. Let us now introduce the following rescaled function around yn : Un (y) =

1 − un (rn y + yn ) , µn

y ∈ Ωn =

Ω − yn . rn

Since Un (0) = 1 by construction, in order to get a limit profile equation we should add a condition avoiding vanishing on compact sets of Ωn . Let us remark that, for xn the maximum point of un and εn = 1 − un (xn ), the associated rescaled function Un satisfies: Un ≥ Un (0) = 1 in Ωn . Proposition 3.1. Assume that µ3n λ−1 n (dist (yn , ∂Ω))

−2

→0

as n → +∞

(3.2)

and Un ≥ C > 0

in Ωn ∩ BRn (0),

(3.3)

1 (RN ), where U is a solution for some Rn → +∞ as n → +∞. Then, up to a subsequence, Un → U in Cloc of the equation: ( |y + y0 |γ in RN , ∆U = s (3.4) U2 U (y) ≥ C > 0 in RN ,

where s > 0, γ ∈ {0, α1 , . . . , αk } and y0 ∈ RN (depending on the type of blow up). Moreover, there exists a function φn ∈ C0∞ (Ω) such that: ¶ Z µ 2λn f (x) 2 2 |∇φn | − 0.

To establish property (3.5), it will be crucial the knowledge of the linear instability for solutions of (3.4) in low dimensions: Theorem 3.2. ([5]) Assume either 1 ≤ N ≤ 7 or N ≥ 8, γ > (

√ 3N −14−4 6 √ . 4+2 6

|y|γ in RN , U2 U (y) ≥ C > 0 in RN . ∆U =

9

Let U be a solution of

(3.6)

Then, µ1 (U ) = inf

½Z

RN

¡

Moreover, if N ≥ 8 and 0 ≤ γ ≤ µ1 (U ) ≥ 0.

2|y|γ 2 ¢ |∇φ| − φ ; φ ∈ C0∞ (RN ) and U3 2

√ 6 3N −14−4 √ , 4+2 6

Z

2

φ =1

RN

¾

< 0.

(3.7)

then there exists at least a solution U of (3.6) such that

For the sake of completeness, we will sketch the proof of Theorem 3.2 in the Appendix. Proof (of Proposition 3.1): First of all, let us remark that (3.2) implies µ3n λ−1 n → 0 as n → +∞. Then, rn → 0 and by (3.2) Ωn → RN as n → +∞. Introduce the following notation   k Y fi (x) =  |x − pj |αj  g(x). (3.8) j=1, j6=i

The function Un satisfies ∆Un =

fn (y) 2 Un

in Ωn , where fn (y) is given by:

 if p ∈ /Z   f (rrnny + yn ) yn −pi αi αi +2 | f (r y + y ) if p = pi , µ−3 → +∞ as n → +∞ y + | i n n n λn |yn − pi | |yn −pi | |yn −pi | fn = y −p −3 α  i +2  |y + nrn i |αi fi (rn y + yn ) if lim sup µn λn |yn − pi | < +∞,

(3.9)

n→+∞

and p =

αi +2 lim yn . Only in the latter situation lim sup µ−3 < +∞, up to a subsequence n λn |yn − pi |

n→+∞

n→+∞

assume that

y n − pi → y0 rn

as n → +∞.

(3.10)

Let R > 0. For n large, decompose Un = Un1 + Un2 , where Un2 satisfies: ½ ∆Un2 = ∆Un in BR (0) , Un2 = 0 on ∂BR (0) . Since (3.3) implies 0 ≤ ∆Un ≤ CR on BR (0), by elliptic regularity theory we get that Un2 is uniformly bounded in C 1,β (BR (0)), β ∈ (0, 1). Up to a subsequence, we get that Un2 → U 2 in C 1 (BR (0)). Since Un1 = Un ≥ C on ∂BR (0), by harmonicity Un1 ≥ C in BR (0). Since Un (0) = 1, by Harnack inequality we get: µ ¶ ¡ ¢ sup Un1 ≤ CR inf Un1 ≤ CR Un1 (0) = CR 1 − Un2 (0) ≤ CR 1 + sup |Un2 (0)| < ∞ . BR/2 (0)

BR/2 (0)

n∈N

Hence, Un1 is uniformly bounded in C 1,β (BR/4 (0)), β ∈ (0, 1). Up to a further subsequence, we get that Un1 → U 1 in C 1 (BR/4 (0)) and then, Un → U 1 + U 2 in C 1 (BR/4 (0)), for any R > 0. By a diagonal process 1 and up to a subsequence, we find that Un → U in Cloc (RN ). According to the three situations described in the definition (3.9) of fn , the function U ≥ C > 0 is a solution of (3.4) with s = f (p), γ = 0 and s = fi (pi ), γ = 0 and s = fi (pi ), γ = α, y0 as in (3.10) respectively. Set f (y) := lim fn (y) = s|y + y0 |γ . n→+∞

Since 2 ≤ N ≤ 7 and s > 0, by Theorem 3.2 we get that µ1 (U ) < 0 and then, we find φ ∈ C0∞ (RN ) so that: ¶ Z µ 2f (y) 2 < 0. φ |∇φ|2 − U3 10

− N 2−2

Define now φn (x) = rn Z µ Ω

|∇φn |2 −

n φ( x−y rn ). We have that:

2λn f (x) 2 φ (1 − un )3 n

¶

=

Z µ

|∇φ|2 −

2fn (y) 2 φ Un3

¶

→

Z µ

|∇φ|2 −

2f (y) 2 φ U3

¶

0. Recall that in this situation rn = µn get easily that on Ωn ∩ BRn (0): 0 ≤ ∆Un ≤ C(

(3.11) 1 − 2+α

λn

i

. By (3.11), we

yn − pi ¯¯ α3i |yn − pi | 2αi ¯¯ ) 3 y+ . rn rn

Given R > 0, then 0 ≤ ∆Un ≤ CR on BR (0) for n large. Arguing as in the proof of Proposition 3.1, up to a 1 subsequence, we get that Un → U in Cloc (RN ), where U ∈ C 1 (RN ) ∩ C 2 (RN \ {−y0 }) is a solution of the equation  fi (pi )  ∆U = |y + y0 |αi in RN \ {−y0 } , 2 αU i  U (y) ≥ C|y + y | 3 in RN . 0

By Hopf Lemma, we have that U (−y0 ) > 0. Indeed, let B some ball so that −y0 ∈ ∂B and assume by contradiction that U (−y0 ) = 0. Since ¯ , U (y) > U (−y0 ) in B, −∆U + c(y)U = 0 in B , U ∈ C 2 (B) ∩ C(B) αi

0| and c(y) = fi (pi ) |y+y ≥ 0 is a bounded function, by Hopf Lemma we get that ∂ν U (−y0 ) < 0, where ν is U3 the unit outward normal of B. Hence, U becomes negative in a neighborhood of −y0 in contradiction with the positivity of U . Hence, U (−y0 ) > 0. The argument now goes as in the proof of Proposition 3.1.

3.2

A pointwise estimate

Let us assume now the validity of (1.3), namely m(un , λn ) ≤ k for any n ∈ N and some k ∈ N. This information, combined with Proposition 3.1, will permit us to control the blow up behavior of un . Indeed, the following pointwise estimate on un is available: ¯ be as in (1.2). Let un be a solution of (1.1) associated Theorem 3.3. Assume 2 ≤ N ≤ 7. Let f ∈ C(Ω) to λn ∈ [0, λ∗ ]. Assume that λn → λ and un (xn ) = max un → 1 as n → +∞. Then, up to a subsequence, Ω

there exist constants C > 0, N0 ∈ N and m-sequences x1n , . . . , xm n , m ≤ k, such that 1

α

2

1 − un (x) ≥ Cλn3 d(x) 3 dn (x) 3 ,

∀ x ∈ Ω , ∀ n ≥ N0 ,

(3.12)

where d(x)βα := min{|x − pi |αi : i = 1, . . . , k}β , β ∈ R, is a weighted distance function from Z and dn (x) = min{|x − xin | : i = 1, . . . , m} is the distance function from {x1n , . . . , xm n }. 11

More precisely, letting rni be associated to xin by means of (3.1), for any i, j = 1, . . . , m, i 6= j, there holds: εin := 1 − un (xin ) → 0 , Uni (y) =

rni + rnj 1 − un (rni y + xin ) 1 → U i (y) in Cloc (RN ) , →0 i εn |xin − xjn |

(3.13)

as n → +∞, where U i satisfies an equation of type (3.4). In addition, there exist m-sequences of test ∞ functions φ1n , . . . , φm n ∈ C0 (Ω) so that ¶ Z µ 2λn f (x) i 2 i 2 (φ ) < 0 , Supp φin ⊂ BM rni (xin ) , ∀ i = 1, . . . , m, (3.14) |∇φn | − (1 − un )3 n Ω for some M > 0 large. Proof: Let εn = 1 − un (xn ), where xn is a maximum point of un . Since 0≤

λn f (x) λn ≤ 2 k f k∞ , (1 − un )2 εn

then we get that: ε2n λ−1 n → 0 as n → +∞.

(3.15)

If (3.15) were false, the right hand side of (1.1) were uniformly bounded (up to a subsequence) and by elliptic ¯ (up to a further subsequence). In particular, since εn → 0 as n → +∞, regularity theory, un → u in C 1 (Ω) we get that λn → 0 as n → +∞. Hence, u would be an harmonic function so that u = 0 on ∂Ω, max u = 1. Ω

A contradiction.

As needed in Proposition 3.1, (3.15) now implies: −2 ε3n λ−1 → 0 as n → +∞. n (dist (xn , ∂Ω))

(3.16)

In order to prove it, we will use the following Lemma and we refer to Appendix for the proof: Lemma 3.4. Let An be a bounded domain in RN so that An → T as n → +∞, where T is an hyperspace so that 0 ∈ T and dist (0, ∂T ) = 1. Let hn be a function on An and Wn be a solution of:  hn (y)   in An ,  ∆Wn = Wn2 (3.17) Wn (y) ≥ C > 0 in An ,    Wn (0) = 1,

for some C > 0. Assume that sup k hn k∞ < +∞ and ∂An ∩ B2 (0) is smooth. Then, either n∈N

min

Wn ≤ C

(3.18)

∂ν Wn ≤ 0,

(3.19)

∂An ∩B2 (0)

or min

∂An ∩B2 (0)

where ν is the unit outward normal of An .

12

−2 Indeed, by contradiction and up to a subsequence, assume that ε3n λ−1 n dn → δ > 0 as n → +∞, where dn := dist (xn , ∂Ω). In view of (3.15) we get dn → 0 as n → +∞. Introduce a different rescaled function Wn : Ω − xn 1 − un (dn y + xn ) , y ∈ An = . Wn (y) = εn dn Since dn → 0, we get that An → T1 as n → +∞. The function Wn solves problem (3.17) with hn (y) = λn d2n ε3 f (dn y + xn ) and C = Wn (0) = 1. We have that for n large: n

k hn k∞ ≤

2 λn d2n k f k∞ ≤ k f k∞ . ε3n δ

Since Wn = ε1n → +∞ on ∂An , by Lemma 3.4 we get that (3.19) must hold. A contradiction to Hopf Lemma applied to un . Hence, the validity of (3.16). Let rn be associated to xn according to (3.1). Up to a subsequence, Proposition 3.1 gives: 1 − un (rn y + xn ) 1 → U (y) in Cloc (RN ) as n → +∞, εn where U satisfies an equation of type (3.4), and provides the existence of φn ∈ C0∞ (Ω) such that (3.5) holds with Supp φn ⊂ BM rn (xn ), M > 0.

Let now x1n = xn , ε1n = εn , rn1 = rn , U 1 = U and φ1n = φn . If (3.12) is true for some subsequence of un with x1n , we take m = 1 and the proof is done. Otherwise, we proceed by an inductive method. Indeed, assume that, up to a subsequence, we have already found l-sequences x1n , . . . , xln , associated rn1 , . . . , rnl (defined by (3.1)) and test functions φ1n , . . . , φln ∈ C0∞ (Ω) so that (3.13)-(3.14) holds at l-th step. If (3.12) holds for some subsequence of un with x1n , . . . , xln , we take m = l and the proof is done. Otherwise, up to a subsequence, l+1 we will show the existence of xl+1 and φl+1 so that (3.13)-(3.14) are still true at (l + 1)-th step. Since n , rn n 1 (3.13)-(3.14) at l-th step imply that φn , . . . , φln have mutually disjoint compact supports, by (3.5) we get that m(un , λn ) ≥ l. Then, by (1.3) the inductive process must stop after a finite number of steps, say m steps, with m ≤ k, and (3.12) holds with x1n , . . . , xm n In order to complete the proof, we need to show how the induction process works. Assume that (3.13)-(3.14) hold at l-th step and (3.12) is not true for any subsequence of un with x1n , . . . , xln . Let xl+1 ∈ Ω be such that n ³ ´ ´ ³ 1 1 ¡ ¢ 2 2 −3 − −α −α 3 d (xl+1 )− 3 3 d (x)− 3 1 − u (x) 1 − un (xl+1 →0 (3.20) λn 3 d(xl+1 n n n n n ) = λn min d(x) n ) x∈Ω

as n → +∞, where dn (x) is the distance function from {x1n , . . . , xln }. Let εl+1 := 1 − un (xl+1 n n ).

Formula (3.20) gives a lot of informations about the blow up around xl+1 n . First of all, it can be rewritten in the more convenient form: 3

−1

2 2 (εl+1 n ) λn

l+1 i |xl+1 n − xn | |xn − pj |

The inductive assumption gives of

rnj

i j rn +rn |xin −xjn |

we get for |y| ≤ R and n ≥ nR :

αj 2

→ 0 as n → +∞ , ∀ i = 1, . . . , l, j = 1, . . . , k.

(3.21)

→ 0 as n → +∞ for any i, j = 1, . . . , l, i 6= j. Then, by definition

³ ´ α 2 −1 λn 3 d(rnj y + xjn )− 3 dn (rnj y + xjn )− 3 1 − un (rnj y + xjn )  −2 j j j −α if xjn → p ∈ /Z   d(rnjy + xn ) j3 |y| 3 Uαn (y) 2 xn −pi − i rn − j j if xn → pi ∈ Z, (εjn )−3 λn |xjn − pi |αi +2 → +∞ y+ j | 3 |y| 3 Un (y) | j =   |xn −pij| −1 |xjn −pi | − αi − 2 j |y + (rn ) (xn − pi )| 3 |y| 3 Un (y) if (εjn )−3 λn |xjn − pi |αi +2 ≤ C 13

j 1−un (rn y+xjn ) εjn

1 → U j (y) in Cloc (RN ) as n → +∞ for any j = 1, . . . , l. Associating (eventually) to the limit point y0 as in (3.10), we get that: ³ ´ α 2 −1 λn 3 d(rnj y + xjn )− 3 dn (rnj y + xjn )− 3 1 − un (rnj y + xjn )  2 α if xjn → p ∈ /Z  d(p)− 3 |y|− 3 U j (y) 2 − j → |y| 3 U (y) if xjn → pi ∈ Z, (εjn )−3 λn |xjn − pi |αi +2 → +∞ αi  2 |y + y0 |− 3 |y|− 3 U j (y) if (εjn )−3 λn |xjn − pi |αi +2 ≤ C

for any j = 1, . . . , l. By inductive assumption, we have that Unj (y) = xjn

uniformly for |y| ≤ R as n → +∞. Since U j is bounded away from zero, then (3.20) gives also that xl+1 n cannot asymptotically lie in balls centered at xin of radius ≈ rni , i = 1, . . . , l, namely: rni →0 − xin |

|xl+1 n

as n → +∞ , ∀ i = 1, . . . , l.

(3.22)

Finally, the choice of xl+1 as a minimum point in (3.20) gives that: n 1 − un (βn y + xl+1 n ) ≥ l+1 εn

µ

α d(βn y + xl+1 n ) α d(xl+1 n )

¶ 31 µ

dn (βn y + xl+1 n ) l+1 dn (xn )

¶ 32

,

(3.23)

for any sequence βn . Indeed, (3.20) implies εl+1 n

³ ¢´ α 2¡ −α l+1 32 3 3 d (x)− 3 1 − u (x) ≤ d(xl+1 n n n ) dn (xn ) min d(x) x∈Ω

≤

α l+1 32 3 d(xl+1 n ) dn (xn ) d(βn y

³ ´ 2 −α 3 d (β y + xl+1 )− 3 + xl+1 1 − un (βn y + xl+1 n n n ) n n )

and then, (3.23) must hold. Here and in the sequel of the proof, the crucial point to establish the validity of (3.3) (or (3.11)) for suitable rescaled functions around xl+1 is exactly given by the validity of (3.23). By n 3 −1 (3.21), we get that in particular (εl+1 ) λ → 0 as n → +∞. We need now to discuss all the possible types n n of blow up at xl+1 . n 1st Case Assume that xl+1 →q∈ / Z. Then, |xl+1 − pj | ≥ C > 0 for any j = 1, . . . , k which reduces (3.21) n n to: 3 − 21 2 (εl+1 n ) λn → 0 as n → +∞ , ∀ i = 1, . . . , l. (3.24) i |xl+1 n − xn | l+1 In order to apply Proposition 3.1 to xl+1 n , we need to show that (3.2) holds for xn : ¡ ¢−2 3 −1 (εl+1 dist (xl+1 → 0 as n → +∞. n ) λn n , ∂Ω)

3 −1 −2 We proceed exactly as in the proof of (3.16). By contradiction, up to a subsequence, assume that (εl+1 n ) λn d n → l+1 δ > 0 as n → +∞, where dn = dist (xn , ∂Ω) (do not confuse dn with dn (x)), and then by (3.24): 3

−1

3

−1

2 2 2 2 dn dn 2 (εl+1 (εl+1 n ) λn n ) λn √ = ≤ →0 1 3 − 2 |xl+1 − xi | i i |xl+1 δ |xl+1 n − xn | n n − xn | n (εl+1 n ) 2 λn

Let Mn =

³

dn (xl+1 n ) dn

´ 12

as n → +∞ , ∀ i = 1, . . . , l.

→ +∞ as n → +∞. We introduce the rescaling Wn of un in the form:

Wn (y) =

1 − un (dn y + xl+1 n ) l+1 εn

for y ∈ An = 14

Ω − xl+1 n ∩ BMn (0). dn

Since dn → 0 and Mn → +∞, we get that An → T1 as n → +∞. Since dn y + xl+1 n , y ∈ An , is uniformly far away from Z, by (3.23) we get for Wn (here, βn is exactly dn ): ¡ C0 dn Mn ¢ 32 ≥ Wn (y) ≥ C0 1 − l+1 2 dn (xn ) for n large and for any y ∈ An . Hence, the function Wn solves problem (3.17) with hn (y) = C0 2 .

xl+1 n ) and C =

1 εl+1 n

+

Since

k hn k∞ ≤ and Wn =

λn d2n f (dn y 3 (εl+1 n )

2 λn d2n k f k∞ ≤ k f k∞ 3 δ (εl+1 ) n

→ +∞ on ∂An ∩ B2 (0), Lemma 3.4 provides that (3.19) must hold, contradicting Hopf

Lemma for un . Hence, (3.2) holds for xl+1 n . −1

3

l+1 2 2 Associated to xl+1 = (εl+1 be defined according to (3.1). By (3.24) we get that n , let rn n ) λn

rnl+1 l+1 |xn − xin |

→0

as n → +∞ , ∀ i = 1, . . . , l,

(3.25)

´ 12 ³ d (xl+1 n ) → +∞ as n → +∞. Since rnl+1 y + xl+1 / Z, and then, Rn = nrl+1 n , |y| ≤ Rn , is uniformly close to q ∈ n by (3.23) we get that: Unl+1 (y) :=

¡ rl+1 Rn ¢ 2 1 − un (rnl+1 y + xl+1 C0 n ) ≥ C0 1 − n l+1 3 ≥ l+1 2 εn dn (xn )

1 for n large and for |y| ≤ Rn . Up to a subsequence, Proposition 3.1 provides Unl+1 → U l+1 in Cloc (RN ) as n → +∞, U l+1 being a solution of an equation of type (3.4), and some φl+1 ∈ C0∞ (Ω) such that (3.5) holds n l+1 l+1 with Supp φn ⊂ BM rnl+1 (xn ), M > 0. By (3.25), combined with (3.22), we get that (3.13)-(3.14) are still true at the (l + 1)-th step, as needed.

2nd Case Assume that xl+1 → pj with the following rate: n αj +2 −3 λn |xl+1 → +∞ as n → +∞. (εl+1 n ) n − pj | 3

−1

αj

l+1 2 2 Let rnl+1 = (εl+1 − pj |− 2 according to (3.1). By (3.21) we get that (3.25) still holds and then, n ) λn |xn ´ 12 ³ l+1 l+1 |x −p | d (xn ) } → +∞ as n → +∞. Since rnl+1 y + xl+1 Rn = min{ nrl+1 j , nrl+1 n , |y| ≤ Rn , is uniformly close to n n pj ∈ Z, estimate (3.23) implies:

Unl+1 (y) :=

¡ 1 − un (rnl+1 y + xl+1 rl+1 Rn ¢ 2 1 rnl+1 Rn ¢ α3j ¡ n ) 1 − n l+1 3 ≥ ≥ 1 − l+1 l+1 2 εn |xn − pj | dn (xn )

1 (RN ) as for n large and |y| ≤ Rn . Up to a subsequence, Proposition 3.1 provides Unl+1 → U l+1 in Cloc n → +∞, where U l+1 solves an equation of type (3.4), and the existence of φl+1 ∈ C0∞ (Ω) such that (3.5) n l+1 l+1 holds, Supp φn ⊂ BM rnl+1 (xn ) for some M > 0. Finally, (3.22) with (3.25) gives that (3.13)-(3.14) are still true at the (l + 1)-th step, also in this second case.

3rd Case Assume that xl+1 → pj and n −3 αj +2 (εl+1 λn |xl+1 ≤ C. n ) n − pj |

15

By (3.21) we get that for any i = 1, . . . , l: 3 − 21 ³ ´1 2 |xl+1 (εl+1 n − pj | n ) λn l+1 −3 l+1 αj +2 2 (ε ) λ |x − p | → 0 as n → +∞. = αj n j n n l+1 i i |xl+1 n − xn | |xl+1 n − xn | |xn − pj | 2 3

1 − 2+α

2+αj λn Let rnl+1 = (εl+1 n )

rnl+1 = i |xl+1 n − xn |

Ã

j

according to (3.1). By (3.21) and (3.26) we get that for any i = 1, . . . , l: −1

3

2 2 (εl+1 n ) λn

|xl+1 n

−

xin |

|xl+1 n

− pj |

providing the validity of (3.25). Let Rn = ( uniformly close to pj ∈ Z, by (3.23) we get: Unl+1 (y) :=

(3.26)

αj 2

2 ! 2+α j µ

1 dn (xl+1 n ) 2 ) l+1 rn

1 − un (rnl+1 y + xl+1 n ) l+1 εn

≥ ≥

|xl+1 n − pj | l+1 |xn − xin |

αj ¶ 2+α

j

→0

as n → +∞,

(3.27)

→ +∞ as n → +∞. Since rnl+1 y + xl+1 n , |y| ≤ Rn , is

¡ |rnl+1 y + xl+1 ¢ αj ¡ n − pj | 3 |xl+1 n

µ

1 2

1−

rnl+1 Rn ¢ 32 dn (xl+1 n )

− pj | ¶− α3j α |xl+1 − p | xl+1 j n n − pj 3j | |y + l+1 l+1 rn rn

for n large and |y| ≤ Rn . We use now Proposition 3.1 in combination with Remark 3.1 to get that, up to 1 a subsequence, Unl+1 → U l+1 in Cloc (RN ) as n → +∞ and U l+1 is a solution of an equation of type (3.4). ⊂ BM rnl+1 (xl+1 Moreover, we find φl+1 ∈ C0∞ (Ω) such that (3.5) holds and Supp φl+1 n n n ), M > 0. Since (3.22) together with (3.25) gives the validity of (3.13)-(3.14) at the (l + 1)-th step, the induction scheme also works in this last case and the proof of Theorem 3.3 is complete. ¥

3.3

Compactness of unstable branches

We are now in position to give the proof of Theorem 1.1. The essential ingredient will be the pointwise estimate of Theorem 3.3. The contradiction will come out from the regularity result of Theorem 2.3. Proof (of Theorem 1.1): By contradiction, up to a subsequence, let us assume that max un → 1 as Ω

n → +∞. Up to a further subsequence, Theorem 3.3 gives the existence of m-sequences x1n , . . . , xm n so that ¯ as n → +∞ and the following pointwise estimate holds: xin → xi ∈ Ω 1

α

2

1 − un (x) ≥ Cλn3 d(x) 3 dn (x) 3

for any x ∈ Ω and n ≥ N0 , for some C > 0 and N0 ∈ N large, where d(x) and dn (x) = min{|x − xin | : i = 1, . . . , m}. Therefore, we get that in Ω:

(3.28) α 3

1

= min{|x − pi |αi : i = 1, . . . , k} 3

1

0≤

λn3 λn f (x) f (x) ≤ C −2 2α 4 . 2 (1 − un ) d(x) 3 dn (x) 3

(3.29)

Since by (1.2) ¯ f (x) ¯ α ¯ ¯ ≤ |x − pi | 3i kfi k∞ ≤ C 2α 3 d(x)

for |x − pi | ≤ δ and fi as in (3.8), we get that is uniformly bounded in Ls (Ω), for any 1 < s

f (x)

2α

d(x) 3 < 3N 4 .

is a bounded function on Ω. Hence, by (3.29)

λn f (x) (1−un )2

By elliptic regularity theory and Sobolev embeddings, 16

¯ where u0 ∈ C(Ω) ¯ is a up to a subsequence, we get that un → u0 weakly in H01 (Ω) and strongly in C(Ω), H 1 (Ω)-weak solution of the equation:  λf (x)  in Ω , −∆u0 = (3.30) (1 − u0 )2  u0 = 0 on ∂Ω. The uniform convergence of un also gives: max u0 = 1. Since any weak harmonic function in H01 (Ω) is Ω

identically zero, we have that λ > 0 in (3.30). Since λ > 0, by (3.28) we get that u0 < 1 in Ω \ S, where S = {x1 , . . . , xm , p1 , . . . , pk } is a finite set. Since ¶ ¶ Z µ Z µ 2λf (x) 2 2λn f (x) 2 2 |∇φ|2 − → |∇φ| − φ φ (1 − un )3 (1 − u0 )3 Ω Ω for any φ ∈ C0∞ (Ω), by (1.3) we get that u0 has a finite Morse index according to definition (2.11). Hence, Theorem 2.3 implies ku0 k∞ < 1, contradicting max u0 = 1. Ω

If we also assume that µ1,n < 0, then λ > 0. Indeed, if λn → 0, then by compactness and elliptic regularity ¯ where u0 is an harmonic function so that u0 = 0 on ∂Ω. Then, theory, we would get un → u0 in C 1 (Ω), ¯ Hence, µ1,n → µ1,0 (0) > 0 as n → +∞, contradicting µ1,n < 0. u0 = 0 and un → 0 in C 1 (Ω). ¥

4

Some consequences

In the last Section, we derive some consequences of Theorem 1.1. First, let us prove the characterization of blow up stated in Theorem 1.2. Proof (of Theorem 1.2): R ³

(1) ⇒ (2) Assume that max un → 1 as n → +∞. If

Ω

Ω

3N 4

f (x) (1−un )3

´ N2

≤ C < ∞ along a subsequence, the

R.H.S. of (1.1) would be uniformly bounded in L . By elliptic regularity theory and Sobolev embeddings, ¯ where u0 is a H 1 (Ω)-weak solution of (1.1) with λ = un → u0 weakly in H01 (Ω) and strongly in C(Ω), N ³ ´ R 2 f (x) < ∞. By Proposition 2.1 we get ku0 k∞ < 1 and, by uniform convergence, lim λn and Ω (1−u 3 0) n→+∞ R ³ f (x) ´ N2 → +∞ as n → +∞. kun k∞ → ku0 k∞ < 1 as n → +∞. A contradiction. Hence, necesssarily Ω (1−u 3 n) (2) ⇒ (1) The viceversa is trivial as it follows by the following inequality: Z µ Ω

f (x) (1 − un )3

¶ N2

≤

µ

kf k∞ (1 − kun k∞ )3

¶ N2

|Ω|,

where | · | stands for the Lebesgue measure. (1) ⇒ (3) Assume that max un → 1 as n → +∞. By Theorem 1.1 m(un , λn ) → +∞ as n → +∞. Ω

(3) ⇒ (1) Since as before get that

f (x) (1−un )3

≤

kf k∞ (1−kun k∞ )3 ,

µk,λn (un ) ≥ µk (Ln ) ,

by the variational characterization of the eigenvalues we Ln := −∆ − 17

2λn kf k∞ , (1 − kun k∞ )3

where µk (Ln ) stands for the k-th eigenvalue of the operator Ln . Therefore, point (3) implies that the number 2λn kf k∞ of negative eigenvalues of Ln blows up as n → +∞. Hence, (1−ku 3 → +∞ as n → +∞. It implies the n k∞ ) validity of point (1). ¥ Now, we establish the uniqueness result contained in Theorem 1.3. Proof (of Theorem 1.3): (1) Let λn → 0 as n → +∞ and associated solutions un of (1.1) so that m(un , λn ) ≤ k, k ∈ N. Theorem 1.1 implies that µ1,n ≥ 0 along a subsequence. By the characterization of the minimal solution uλ as the only semi-stable solution, we get that un = uλn for n large. Hence, necessarily there exists δ = δk > 0 so that uλ is the unique solution u of (1.1) with m(u, λ) ≤ k for any λ ∈ (0, δ).

(2) Let λn → λ∗ as n → +∞ and associated solutions un with m(un , λn ) ≤ k, for some k ∈ N. By Theorem ¯ for any 1.1 we get that sup kun k∞ < 1. By elliptic regularity theory, un is uniformly bounded in C 1,β (Ω) n∈N

¯ as n → +∞, where u0 is a C 1 (Ω)-solution ¯ β ∈ (0, 1). Up to a subsequence, un → u0 in C 1 (Ω) of (1.1) with λ = λ∗ . In [7] it is proved that equation (1.1) admits for λ = λ∗ an unique solution, the extremal solution ¯ as n → +∞. By [4], in a C 1 -small neighborhood of u∗ problem (1.1) has only u∗ . Then, un → u∗ in C 1 (Ω) the two solutions uλ , Uλ for λ close to λ∗ . Hence, either un = uλn or un = Uλn and the uniqueness result follows. ¥ Finally, we conclude this Section by showing the existence of a solutions sequence whose Morse index blows up. Proof (of Theorem 1.4): Let us define the solution set V as V = {(λ, u) ∈ [0, +∞) × E : u is a solution of (1.1)}, ¯ is endowed with the usual norm. By contradiction and in view of the equivalence of where E = C01 (Ω) Theorem 1.2, let us assume that sup max u ≤ 1 − 2δ, (4.1) (λ,u)∈V

Ω

1 2 ).

Hence, V is a compact set in [0, +∞) × E. By Theorem 1.3 we can fix λ1 , λ2 ∈ (0, λ∗ ), for some δ ∈ (0, λ1 < λ2 , so that (1.1) possesses: - for λ1 only the non degenerate solution uλ1 with m(uλ1 , λ1 ) = 0; - for λ2 has only the two non degenerate solutions uλ2 , Uλ2 with m(uλ2 , λ2 ) = 0, m(Uλ2 , λ2 ) = 1. Let us define the projection of V onto E: U = {u ∈ E : ∃ λ so that (λ, u) ∈ U }, and let us consider a δ-neighborhood of U in E: Uδ = {u ∈ E : distE (u, U) ≤ δ}. Let us remark that by (4.1) we get: sup max u ≤ 1 − δ.

u∈Uδ

Ω

−2

Let us regularize the nonlinearity (1 − u)

in the following way: ½ (1 − u)−2 if u ≤ 1 − δ gδ (u) = δ −2 if u ≥ 1 − δ, 18

in such a way that, for any fixed λ, problem (1.1) in Uδ is equivalent to find a zero of the map Tλ = Id − Kλ : E → E, where Kλ (u) = −∆−1 (λf (x)gδ (u)) is a compact operator and ∆−1 is the laplacian resolvent with homogeneous Dirichlet boundary condition. We can define the Leray-Schauder degree dλ of Tλ on Uδ with respect to zero, since by definition of U (the set of all solutions) ∂Uδ does not contain any solution of (1.1) for any value of λ. Since dλ is well defined for any λ ∈ [0, λ∗ ], by omotopy dλ1 = dλ2 . To get a contradiction, let us now compute dλ1 , dλ2 . Since the only zero of Tλ1 in Uδ is uλ1 with Morse index zero, we have that dλ1 = 1. While, Tλ2 has in Uδ exactly two zeroes uλ2 , Uλ2 with Morse index zero, ¥ one respectively, and hence, dλ2 = 1 − 1 = 0. This contradicts dλ1 = dλ2 . The proof is complete.

5

Appendix

First of all, we give a sketch of proof of Theorem 3.2 and we refer to [5] for the details. Proof (of Theorem 3.2): By contradiction, we assume µ1 (U ) ≥ 0 and then, Z Z |y|γ 2 2 φ , ∀ φ ∈ D1,2 (RN ) . |∇φ| ≥ 2 U3 In particular, by (5.1) we get that Z

(1 +

|y|γ

|y|2 )

N −2 2 +δ

U3

≤C

Z

1 N

(1 + |y|2 ) 2 +δ

< +∞ ,

(5.1)

(5.2)

for any δ > 0. Step 1. We want√to show that (5.1) allows us to perform the following Moser-type iteration scheme: for any 0 < q < 4 + 2 6 and β there holds µ ¶ Z Z 1 1 1 + ≤ C (5.3) γ q (1 + |y|2 )β U q (1 + |y|2 )β−1− 2 U q+3 (provided the second integral is finite). Indeed, let R > 0 and consider a smooth radial cut-off function η so that: 0 ≤ η ≤ 1, η = 1 in BR (0), η = 0 η2 in RN \ B2R (0). Multiplying (3.6) by , q > 0, integrating by parts and using (5.2) we get: 2 (1 + |y| )β−1 U q+1 Z Z Z ´¯2 η 8(q + 1) 2 |y|γ η 2 1 ¯¯ ³ |y|γ η 2 ¯ ≥ − ¯ ¯∇ β−1 (1 + |y|2 )β−1 U q+3 q2 (1 + |y|2 )β−1 U q+3 q Uq (1 + |y|2 ) 2 Z ´ ³ 2(q + 2) 1 η η + . β−1 ∆ β−1 2 q q U (1 + |y|2 ) 2 (1 + |y|2 ) 2 Since 8q + 8 − q 2 > 0 for any 0 < q < q+ , assuming that R|∇η| + R2 |∆η| ≤ C we get that: Z Z 1 |y|γ η 2 ≤ Cq , (1 + |y|2 )β−1 U q+3 (1 + |y|2 )β U q where Cq does not depend on R > 0. Taking the limit as R → +∞, we get the validity of (5.3). 19

Step 2. Let now 1 ≤ N ≤ 7 or N ≥ 8, γ > Z

√ for some 0 < q < q+ = 4 + 2 6.

√ 6 3N −14−4 √ . 4+2 6

We want to show that

1 < +∞ (1 + |y|2 )U q

(5.4)

Given δ > 0, set βi = β0 − i(1 + γ2 ) and qi = q0 + 3i, i ∈ N. By (5.2) and iterating Step 1 two times in view √ of q0 < q1 < q+ = 4 + 2 6 < q2 , we get that: Z 1 < +∞ (5.5) (1 + |y|2 )β2 U q2 √ where β2Z= N −6−3γ + δ, q2 = 9. Fix now q: 0 < q < q+ = 4 + 2 6 < 9. By (5.5) and H¨older inequality we 2 1 2q 18 < +∞ provided − 9−q β2 + 9−q > N or equivalently get that (1 + |y|2 )U q q>

9N − 18 . 6 − 2δ + 3γ

To have (5.6) for some δ > 0 small and q< q+ at the same time, we need to require √

6 3N −14−4 √ . 4+2 6

1 ≤ N ≤ 7 or N ≥ 8, γ > √ 4 + 2 6 such that (5.4) holds.

(5.6) 3N −6 2+γ

< q+ or equivalently

Our assumptions then provide the existence of some 0 < q < q+ =

√ Step 3. To obtain a contradiction, fix 0 < q < 4 + 2 6 such that (5.4) holds. Letting η as before, using equation (3.6) we compute: µ ¶ Z Z Z Z Z ¯ ¡ η ¢¯2 8q + 8 − q 2 q+2 η |y|γ η 2 |∇η|2 ∆η 2 2|y|γ ¯∇ q ¯ − = − + − . q U3 4(q + 1) U q+3 Uq 4(q + 1) Uq U2 U2 Since 8q + 8 − q 2 > 0, by (5.4) we get that: Z Z Z Z ¯ ¡ η ¢¯2 ¢ ¡ |y|γ 1 2|y|γ ³ η ´2 8q + 8 − q 2 ¯∇ ¯ − 0 on ∂An ∩ B2 (0). Let G(y) be the Green function at 0 of the operator −∆ in B2 (0) with homogeneous Dirichlet R boundary condition.R Since ∂ν G < 0 on ∂T and ∂An → ∂T , we get that ∂ν G < 0 on ∂An ∩ B2 (0) and ∂An ∩B2 (0) ∂ν Gdσ → ∂T ∩B2 (0) ∂ν Gdσ < 0. Since G ≥ 0 in B2 (0), ∂ν G ≤ 0 on ∂B2 (0) and the assumptions on Wn , by the representation formula we get: µ ¶Z Z hn (y) ∂ν Gdσ. G − min W 1 = Wn (0) ≥ − n 2 ∂An ∩B2 (0) ∂An ∩B2 (0) An ∩B2 (0) Wn µ ¶ ¯R ¯ n (y) −1 ¯ But ¯ An ∩B2 (0) hW G ≤ C, and then, 1 ≥ −C +C min W for some C > 0 large enough. Hence, 2 n ∂An ∩B2 (0)

n

min

∂An ∩B2 (0)

Wn is uniformly bounded providing the validity of (3.18). The proof is complete.

20

¥

References [1] A. Bahri and P.-L. Lions, Morse index of some min-max critical points. I. Application to multiplicity result. Comm. Pure Appl. Math. 41 (1988), 1027–1037. [2] A. Bahri and P.-L. Lions, Solutions of superlinear elliptic equations and their Morse indices. Comm. Pure Appl. Math. 45 (1992), 1205–1215. [3] X. Cabr´e, Extremal solutions and instantaneous complete blow-up for elliptic and parabolic problems. Preprint (2005). [4] M.G. Crandall and P.H. Rabinowitz, Some continuation and variational methods for positive solutions of nonlinear elliptic eigenvalue problems. Arch. Ration. Mech. Anal. 58 (1975), 207–218. [5] P. Esposito, N. Ghoussoub and Y. Guo, Compactness along the branch of semi-stable and unstable solutions for an elliptic problem with a singular nonlinearity. Submitted (2005). [6] G. Flores, G.A. Mercado and J.A. Pelesko, Dynamics and Touchdown in Electrostatic MEMS. Proceedings of ICMENS 2003 (2003), 182–187. [7] N. Ghoussoub and Y. Guo, On the partial differential equations of electrostatic MEMS devices: stationary case. Submitted (2005). [8] N. Ghoussoub and Y. Guo, On the partial differential equations of electrostatic MEMS devices II: dynamic case. In preparation. [9] Y. Guo, Z. Pan and M.J. Ward, Touchdown and pull-in voltage behavior of a MEMS device with varying dielectric properties. SIAM J. Appl. Math. 66 (2005), 309–338. [10] D.D. Joseph and T.S. Lundgren, Quasilinear Dirichlet problems driven by positive sources. Arch. Ration. Mech. Anal. 49 (1973), 241–268. [11] J.P. Keener and H.B. Keller, Positive solutions of convex nonlinear eigenvalue problems. J. Differential Equations 16 (1974), 103–125. [12] F. Mignot and J.P. Puel, Sur une classe de probl`emes non lin´eaires avec non lin´earit´e positive, croissante, convexe. Comm. Partial Differential Equations 5 (1980), no. 8, 791–836. [13] J.A. Pelesko, Mathematical Modeling of Electrostatic MEMS with Tailored Dielectric Properties. SIAM J. Appl. Math. 62 (2002), no. 3, 888–908. [14] J.A. Pelesko and D.H. Bernstein, Modeling MEMS and NEMS. Chapman Hall and CRC Press (2002).

21