QUASILINEAR ELLIPTIC EQUATIONS WITH SINGULAR QUADRATIC

QUASILINEAR ELLIPTIC EQUATIONS WITH SINGULAR QUADRATIC GROWTH TERMS LUCIO BOCCARDO, TOMMASO LEONORI, LUIGI ORSINA, AND FRANCESCO PETITTA Abstract. In this paper we deal with positive solutions for singular quasilinear problems whose model is ( |∇u|2 in Ω, −∆u + (1−u) γ = g u=0

on ∂Ω,

where Ω is a bounded open set of RN , g ≥ 0 is a function in some Lebesgue space, and γ > 0. We prove both existence and nonexistence of solutions depending on the value of γ and on the size of g.

1. Introduction and statement of results Let Ω be an open bounded set of RN (N ≥ 1). We are interested in the study of singular elliptic quasilinear problems of the type ( −div(M (x)∇u) + h(u)|∇u|2 = g in Ω, (1.1) u=0 on ∂Ω, where M (x) = (mij (x)), i, j = 1, . . . , N is a symmetric matrix whose coefficients mij : Ω → R are measurable functions such that: (1.2)

α|ξ|2 ≤ M (x) ξ · ξ

and

|M (x)| ≤ β ,

N

for almost every x in Ω, for every ξ in R , where 0 < α ≤ β. Moreover we suppose that g ≥ 0 is a function in L1 (Ω). As far as the lower order term is concerned, we assume that h(s) is a continuous and strictly increasing function such that (1.3)

h : [0, σ) → R+ ,

σ > 0,

and that (1.4)

lim h(s) = +∞ .

s→σ −

To fix the ideas we will often refer to the following model problem ( |∇u|2 −∆u + (1−u) in Ω, γ = g (1.5) u=0 on ∂Ω, 1

2

L. BOCCARDO, T. LEONORI, L. ORSINA, AND F. PETITTA

where γ > 0, and g belongs to L1 (Ω), g ≥ 0. The peculiarity of this kind of equations is that the lower order term “forces” the solutions to be bounded. This phenomenon, due to the strongly regularizing effect of the lower order term, has already been observed for semilinear equations such as ( −div(M (x)∇u) + h(u) = g in Ω, u=0 on ∂Ω, with h as above. It has been proved that existence occurs even if g ∈ L1 (Ω) (see [7]) and, more in general, if g is a measure (see [15]). Here we are interested in singular lower order terms depending on the gradient. We recall that existence of an H01 (Ω) solution for (1.1) with nonsingular h is nowadays well known: we refer, just to quote some, to the papers [5], [12], [9], [14] [10] and [13], where several existence and nonexistence results are proved. On the other hand, the study of existence and nonexistence for solution to Dirichlet problems as (1.1) with a function h which becomes singular at s = 0 has been recently carried out. We refer to [2], [8], [1] and [16] for a wide account and further references on this problem. We point out that such a problem is essentially different from the one studied in this paper: indeed in such a case, the main difficulty is to deal with the set where the solution u is “small” in order to prove that the lower order term belongs to L1 (Ω). On the contrary it is clear that in our paper we have to deal with the zone where u is “large”, i.e., where u is near σ. Recently, a similar problem has been studied in [17] and [6]. Our aim is to prove both existence and nonexistence results for problem (1.1). We define a solution as follows. Definition 1.1. A function u ∈ H01 (Ω) is a solution for problem (1.1) if 0 ≤ u < σ a.e. in Ω, h(u)|∇u|2 ∈ L1 (Ω), and Z Z Z 2 M (x)∇u · ∇ϕ + h(u)|∇u| ϕ = gϕ, Ω

for every ϕ ∈

H01 (Ω)

Ω

Ω

∞

∩ L (Ω).

Our first existence result is the following. Theorem 1.2. Assume that g ∈ L1 (Ω), g ≥ 0, and suppose that Z sp lim− h(t) dt = +∞ . s→σ

0

Then (1.1) has at least a solution.

SINGULAR QUASILINEAR EQUATIONS

3

Remark 1.3. Observe that Theorem 1.2 covers the √ case γ ≥ 2 in the model problem (1.5). In general, the fact that if h does not belong to L1 ((0, σ)), then there exists a solution is not surprising. Indeed, if we define Z up h(s) ds , Φ(u) = 0

the condition h(u)|∇u|2 in L1 (Ω) can be rewritten as |∇Φ(u)|2 ∈ L1 (Ω) ,

i.e., Φ(u) ∈ H01 (Ω) .

Thus, by Poincaré inequality, to L2 (Ω), and so is almost √ Φ(u) belongs everywhere finite. Since h is not in L1 ((0, σ)), this fact implies that u < σ almost everywhere in Ω. In other words, the condition “u does not touch σ”√is automatically satisfied starting from the formulation. Note that if h belongs to L1 ((0, σ)), then u may be well equal to σ on a set of positive measure without any contradiction, since in this case the function Φ(u) is finite no matter the values taken by u. As we will see, this fact will lead to nonexistence of solutions for (1.1) if the datum g is “large” (recall that we need for a solution to be almost everywhere smaller than σ). Thanks result of Theorem 1.2, we are left to deal with the √ to the 1 case of h in L ((0, σ)). This corresponds to the case 0 < γ < 2 in the model case (1.5). To better understand what happens in this case, let us give a one-dimensional example. Example 1.4. Consider the following problem: ( 0 (s))2 = λ in (−1, 1), −u00 (s) + (u 1−u(s) (1.6) u(−1) = u(1) = 0 , and suppose that (1.7) has a C 2 solution for every λ > 0, with u(s) < 1 almost everywhere in (−1, 1). If λ = 6 there exists an explicit solution: w(s) = 1 − s2 . Let now u be a solution of (1.7) with λ > 6; multiplying the equation by 1 − u we obtain −(1 − u)u00 + (u0 )2 = λ (1 − u) , which can be rewritten as 00 u2 − u− = λ (1 − u) . 2 Defining v = u − u2 /2 (and, correspondingly, z = w − w2 /2), we have ( ( p p −v 00 (s) = λ 1 − 2v(s) −z 00 (s) = 6 1 − 2z(s) and v(−1) = v(1) = 0 , z(−1) = z(1) = 0 .

4


Subtracting the first equation from the second, multiplying by (z − v)+ and integrating on (−1, 1) yields (since λ > 6) Z 1 Z 1 √ √ + 0 2 [ 1 − 2z − 1 − 2v](z − v)+ ds ≤ 0 . |((z − v) ) | ds ≤ λ −1

−1 + 0

Therefore, ((z − v) ) ≡ 0, which implies (z − v)+ ≡ 0. Hence, v(s) ≥ z(s), and so u(s) ≥ w(s) = 1 − s2 in [−1, 1], which implies u(0) = 1. 00 Since u belongs to C 2 , then u(s) = 1 + u 2(0) s2 + o(s2 ), and we can use the equation to deduce, thanks to the de l’Hôpital rule, (u0 (s))2 00 = lim+ −3u00 (s) = −3u00 (0) . λ = lim+ −u (s) + s→0 s→0 1 − u(s) Therefore, u(s) = 1 − λ6 s2 + o(s2 ). Since λ > 6, this fact contradicts u(s) ≥ 1 − s2 , and so (1.7) can not have solutions such that u < 1 almost everywhere for every λ > 6. On the other hand, if λ < 6 then any solution u of (1.7) is smaller than w (with the same proof as above), and so satisfies u(s) ≤ 1 − s2 . If u(0) = 1 then we have u00 (0) = − λ3 and so u(s) = 1 − λ6 s2 + o(s2 ), a fact that contradicts u(s) ≤ 1 − s2 since λ < 6. Therefore, any solution u of (1.7) with λ < 6 satisfies u(s) ≤ u(0) < 1. In other words, we have that any solution u of (1.7) is strictly smaller than 1 if λ < 6, “touches” 1 at the origin if λ > 6, and cannot be almost everywhere smaller than 1 if λ > 6. Note that if we consider (1.7) in (−R, R), we have by rescaling that w(s) = 1 − (s/R)2 is a solution with 6/R2 as datum, and that we have a contradiction with existence of solutions if λ > 6/R2 . In view of the preceding example, if (1.6) is not satisfied one can expect that (1.1) has always solutions if the size of g is small, while no solutions are expected if the size of g is large enough (depending both on h and on the size of Ω). Let us clarify this fact by introducing the following Definition 1.5. Let f in Lp (Ω), p > N2 , f ≥ 0, λ > 0, and consider the following problem ( −div(M (x)∇u) + h(u)|∇u|2 = λ f in Ω, (1.7) u=0 on ∂Ω. We define Λf as follows: Λf = sup{λ > 0 : ∃u ≤ σ − ε, for some ε > 0, u solution of (1.8)} , where we use the convention that sup ∅ = 0.


5

Remark 1.6. The definition of Λf may seem strange, since we require the solutions to (1.8) to be strictly smaller than σ, a fact that is not true for the solution u = 1 − s2 obtained in Example 1.4 for λ = 6. This restriction is purely technical, is needed for the proof of Theorem 1.8 below, and, as in the case of the example above, may yield that Λf is an actual supremum, rather than a maximum. Note indeed that in Example 1.4 we have {λ > 0 : ∃u ≤ σ − ε, for some ε > 0, u solution of (1.7)} = (0, 6) . See also Section 6 for further remarks on Λf . We have the following existence result for λ small (without growth assumptions on h). Theorem 1.7. Let f ∈ Lp (Ω), p > N2 , f ≥ 0. Then there exists Λ such that for every λ < Λ problem (1.8) has a solution (smaller than σ − ε for some ε = ε(λ) > 0). Therefore, Λf > 0. In the particular case in which the principal part is the laplacian, we can be much more precise. Theorem 1.8. Let f be as in Theorem 1.7 and let M (x) ≡ I, the identity matrix. Then (1.8) has a unique solution uλ for every λ < Λf . Moreover, uλ ≤ uµ if λ < µ. In order to prove the preceding result, we will compare solutions of (1.8), with M (x) ≡ I, that correspond to different values of λ > 0. In the general case, comparison principles for H01 (Ω) solutions of (1.8) are difficult to obtain, and to our knowledge the only results in this direction are contained in [3] and [4] (see also Remark 3.1 below). √ If, in addition, h belongs to L1 ((0, σ))) we will show that Λf < +∞. In order to give a sharper statement of this result, let us define λ1 (f ) and ϕ1 (f ) as the first eigenvalue and the first eigenfunction of −div(M (x)∇u) in Ω with weight f 0 in Lp (Ω), p > N2 , i.e., ( −div(M (x)∇ϕ1 (f )) = λ1 (f )f (x)ϕ1 (f ) in Ω, ϕ1 (f ) = 0 on ∂Ω. We recall that ϕ1 (f ) belongs to L∞ (Ω), that ϕ1 (f ) > 0 in Ω, and that Z Z (1.8) M (x)∇u · ∇u ≥ λ1 (f ) f (x) u2 , ∀u ∈ H01 (Ω) . Ω

Ω

We also define (α is given by (1.2)): Z s Z s 1 (1.9) H(s) = h(t) dt , and ψ(u) = e−H(t) dt . α 0 0

6


Our first nonexistence result is the following. Theorem 1.9. Let f be in Lp (Ω), p > N2 , f 0. Suppose that h belongs to L1 ((0, σ)), i.e., that Z σ 1 h(s) ds < +∞ , H(σ) = α 0 and let λ > λ1 (f )eH(σ) ψ(σ). Then there exists no weak solution for (1.8). Therefore, Λf < +∞. √ In view of Theorem 1.9, we are left with the case h in L1 ((0, σ)), but h not in L1 ((0, σ)) (i.e. 1 ≤ γ < 2 in (1.5)). In this case we have to make additional assumptions on h, M and f . Theorem 1.10. Let f be in Lp (Ω), p > N2 . Let M (x) ≡ I, and suppose that h and f are such that ( there exists γ ∈ [1, 2) : lim− (σ − s)γ h(s) = C > 0 , s→σ (1.10) there exists ρ > 0 : f (x) ≥ ρ in Ω . Then (1.8) has no solutions for λ large enough, so that Λf < +∞. As a consequence of theorems 1.2, 1.7, 1.8, 1.9 and 1.10, we have a complete picture for the model example (1.5): i) if γ ≥ 2, (1.5) has a solution in H01 (Ω) ∩ L∞ (Ω) for every g in L1 (Ω), g ≥ 0; ii) if g = λ f , with f in Lp (Ω), p > N2 , (1.5) has a solution for every λ < Λf ; iii) if g = λ f , with f in Lp (Ω), p > N2 , and 0 < γ < 2, then Λf < +∞, and (1.5) has no solutions for λ > Λf . The plan of the paper is as follows. In Section 2 we will prove Theorem 1.2 by approximating (1.1) with a sequence of nonsingular problems, and in Section 3 we will prove Theorem 1.7 and Theorem 1.8. In Section 4 we will deal with nonexistence results, proving theorems 1.9 and 1.10. In order to prove this latter result, we will transform (1.8) into a semilinear problem which has solutions for every λ > 0, and prove (using one-dimensional analysis and super- and sub-solution techniques) that if λ is large enough these solutions have “flat” zones of nonzero measure which correspond to zones where u ≡ σ. The final Section 5 will be devoted to the study of the asymptotic behaviour of sequences of solutions of approximating problems if the limit problem (1.1) has no solution, while in Section 6 we will study the case λ = Λf .

7


Notation. We will use the following notation throughout the paper: if k > 0 we define Tk (s) = max(−k, min(s, k)) ,

Gk (s) = s − Tk (s) ,

and by εn we indicate any quantity that tends to 0 as n tends to infinity. 2. Proof of Theorem 1.2 Our approach to prove the existence of a solution of (1.1) is by approximation. We will consider the sequence {un } of solutions of ( −div(M (x)∇un ) + hn (un )|∇un |2 = gn in Ω, (2.1) un = 0 on ∂Ω, where gn = Tn (g), and hn is defined as  0 if s < 0,    nh( 1 )s if 0 ≤ s < 1 , n n (2.2) hn (s) = 1  ≤ s < σ and h(s) ≤ n, h(s) if  n   n if 0 ≤ s < σ and h(s) > n or if s ≥ σ. hn (s)

n

σ

1 n

s

Remark 2.1. The definition of hn (s) in [0, n1 ) is needed to have that hn (s) s ≥ 0 for every s in R, and that hn (s) is continuous at 0. If h(0) = 0, there is no need to define hn (s) as above on [0, n1 ): it is enough to take hn (s) = h(s) on this set. Since hn (s) s ≥ 0, and hn is bounded, by a result of [5] there exists a solution un of (2.1), i.e., a function un in H01 (Ω) such that hn (un )|∇un |2 belongs to L1 (Ω), and such that Z Z Z 2 M (x)∇un · ∇ϕ + hn (un )|∇un | ϕ = gn ϕ , (2.3) Ω

Ω

Ω

8


for every ϕ ∈ H01 (Ω) ∩ L∞ (Ω). Moreover, since gn belongs to L∞ (Ω), and since hn (s) s ≥ 0, we have that un belongs to L∞ (Ω). In order to prove Theorem 1.2, we begin by proving some properties of the sequence {un }. Proposition 2.2. Consider the sequence {un } of solutions of (2.1) with g ∈ L1 (Ω), g ≥ 0. Then: i) {un } is nonnegative and bounded in H01 (Ω); consequently, it weakly converges (up to subsequences) to some function u in H01 (Ω); ii) Z hn (un )|∇un |2 ≤ kgkL1 (Ω) ; (2.4) Ω

iii) 0 ≤ u ≤ σ, almost everywhere in Ω; iv) for every k in (0, σ), Tk (un ) strongly converges to Tk (u) in H01 (Ω); v) un strongly converges to u in H01 (Ω); vi) If meas({u = σ}) = 0, then h(un )|∇un |2 converges to h(u)|∇u|2 almost everywhere in Ω. Proof. i) First of all, observe that since gn ≥ 0, the fact that hn (s) s ≥ 0 implies, by standard arguments, that un ≥ 0 a.e. in Ω. Choosing ϕ = Tσ (un ) as test function in (2.3) (as in [9]) we obtain Z Z Z 2 M (x)∇un · ∇Tσ (un ) + hn (un )|∇un | Tσ (un ) = Tσ (un )gn . Ω

Ω

Ω

Using (1.2), the fact that gn ≤ g, and since both Tσ (un ) = σ and hn (un ) = n on the set {un ≥ σ}, we have Z Z 2 α |∇un | + nσ |∇un |2 ≤ σkgkL1 (Ω) , {un ≤σ}

{un ≥σ}

which implies that {un } is bounded in H01 (Ω). Therefore, there exists u in H01 (Ω) such that (up to subsequences) un converges to u weakly in H01 (Ω). ii) Let ε > 0, let 0 ≤ k < σ, and choose ϕ = 1ε Tε (Gk (un )) as test function in (2.3). We obtain Z 1 M (x)∇un · ∇Tε (Gk (un )) ε Ω Z Z 1 1 2 + hn (un )|∇un | Tε (Gk (un )) = Tε (Gk (un ))gn . ε Ω Ω ε


9

Dropping the first (nonnegative) term we have, since Tε (Gk (un )) = ε where un ≥ k + ε, and since 0 ≤ gn ≤ g, Z Z Z 2 hn (un )|∇un | ≤ gn ≤ g. {un ≥k+ε}

{un ≥k}

{un ≥k}

+

Taking the limit as ε tends to 0 , we have Z Z 2 hn (un )|∇un | ≤ (2.5) {un ≥k}

g,

{un ≥k}

which then gives (2.4) taking k = 0. Since −div(M (x)∇un ) = gn − hn (un )|∇un |2 , and the right hand side is bounded in L1 (Ω) as a consequence of (2.4) and of the assumptions on g, we obtain from a result of [11] that (up to subsequences) ∇un converges to ∇u almost everywhere in Ω. iii) From (2.5), and the fact that hn is increasing, we deduce Z Z 1 2 (2.6) |∇un | ≤ g. hn (k) Ω {un ≥k} Choosing k = σ we have Z (2.7)

1 |∇un | ≤ n {un ≥σ} 2

Z g. Ω

Letting n tend to infinity, and using Fatou lemma together with the almost everywhere convergence of ∇un , we have Z |∇u|2 = 0, {u>σ}

which implies 0 ≤ u ≤ σ almost everywhere in Ω. iv) Let 0 < k < σ, and 2

ϕη (s) = seηs ,

η > 0.

The function ϕη has the following property: (2.8)

αϕ0η (s) −

βh(k) α |ϕη (s)| ≥ , α 2

∀η >

2 2

β 2 h2 (k) , ∀s ∈ R . 4α4

h (k) Hence, we fix η > β 4α , and choose ϕ = ϕη (Tk (un ) − Tk (u)) as test 4 function in (2.3). We obtain (for the sake of brevity, we omit the arguments from ϕη and ϕ0η ) Z M (x)∇un · ∇(Tk (un ) − Tk (u))ϕ0η Ω Z Z (2.9) 2 + hn (un )|∇un | ϕη = ϕη gn . Ω

Ω

10


Since un converges to u a.e., and since |Tk (un ) − Tk (u)| ≤ 2k, we have Z ϕη (Tk (un ) − Tk (u))gn = εn . Ω

Since Tk (un ) → Tk (u) a.e. and weakly in H01 (Ω) and since ϕ0η is bounded, it follows that Z − M (x)∇Tk (u) · ∇(Tk (un ) − Tk (u))ϕ0η = εn . Ω

Thus adding such a quantity on both sides of (2.9), dropping the nonnegative term Z hn (un )|∇un |2 ϕη , {un >k}

and using (1.2), we have (recall that un = Tk (un ) + Gk (un )) Z M (x)∇(Tk (un ) − Tk (u)) · ∇(Tk (un ) − Tk (u))ϕ0η Ω Z hn (un )|∇un |2 ϕη + {un ≤k} Z ≤ εn + β |∇Gk (un )||∇Tk (u)||ϕ0η (k − Tk (u))| . Ω

We note that, since hn ≤ h, since hn is increasing, and by (1.2), Z Z h(k) 2 hn (un )|∇un | ϕη ≤ M (x)∇Tk (un ) · ∇Tk (un )|ϕη | . α {un ≤k}

Ω

Since we have (by (1.2)) Z M (x)∇Tk (un ) · ∇Tk (un )|ϕη | Ω Z = M (x)∇(Tk (un ) − Tk (u)) · ∇(Tk (un ) − Tk (u))|ϕη | + εn ΩZ ≤ β |∇(Tk (un ) − Tk (u))|2 |ϕη | + εn , Ω

and

Z

|∇Gk (un )||∇Tk (u)||ϕ0η (k − Tk (u))| = εn ,

Ω

we conclude that, using (1.2), Z βh(k) 2 0 |∇(Tk (un ) − Tk (u))| αϕη − |ϕη | ≤ εn . α Ω Using (2.8), we therefore deduce that Z α |∇(Tk (un ) − Tk (u))|2 ≤ εn , 2 Ω


11

and thus Tk (un ) strongly converges to Tk (u) in H01 (Ω). v) Since ∇un converges to ∇u almost everywhere in Ω, it is enough to prove that {|∇un |2 } is equiintegrable. Let E ⊂ Ω be measurable, and let ε > 0. Using (2.6), we have, for 0 < k < σ, Z Z Z Z 1 2 2 2 |∇Gk (un )| ≤ |∇Gk (un )| = |∇un | ≤ g. hn (k) Ω E Ω {un ≥k} By the assumptions on h, there exist kε > 0 and nε > 0 such that for every n ≥ nε we have Z ε |∇Gkε (un )|2 ≤ . 2 E Once kε is fixed, by iv) and by Vitali convergence theorem we have Z ε |∇Tkε (un )|2 ≤ , ∀ n ∈ N , 2 E if the measure of E is small enough. This concludes the proof of v), since un = Tkε (un ) + Gkε (un ). vi) Even if we already know that both un and ∇un converge almost everywhere in Ω, the fact that h(0) can be strictly positive means that we have to be careful when dealing with the set {u = 0}. Anyway, as we are going to show, the presence of the quadratic gradient term will allow us to conclude using a result by G. Stampacchia (see [21]). Define Ω0 = {x ∈ Ω : u(x) < σ}; by assumption, and by iii), we have meas(Ω) = meas(Ω0 ). Let Eu = {x ∈ Ω0 : un (x) 6→ u(x)} , E∇u = {x ∈ Ω0 : ∇un (x) 6→ ∇u(x)} , so that meas(Eu ∪ E∇u ) = 0. Thus, if we define Ω00 = Ω0 \(Eu ∪ E∇u ), we have that meas(Ω00 ) = meas(Ω). Define now F+ = {x ∈ Ω00 : u(x) > 0} ,

F0 = {x ∈ Ω00 : u(x) = 0, ∇u(x) = 0} .

Since, by a result by G. Stampacchia, we have that ∇u = 0 almost everywhere on the set {u = 0}, we have meas(F+ ∪ F0 ) = meas(Ω00 ) = meas(Ω) . Now, if x ∈ F+ , then hn (un (x))|∇un (x)|2 tends to h(u(x))|∇u(x)|2 , while if x ∈ F0 , then, since hn (un (x)) is bounded and ∇un (x) converges to ∇u(x) = 0, we have lim hn (un (x))|∇un (x)|2 = 0 = h(u(x))|∇u(x)|2 .

n→+∞

12


Therefore, hn (un (x))|∇un (x)|2 converges to h(u(x))|∇u(x)|2 in F+ ∪F0 , i.e., almost everywhere in Ω. Observe that the results of Proposition 2.2 are not enough to prove the existence of a solution for (1.1). Indeed, to pass to the limit in (2.3) we need the strong compactness of the lower order term in L1 (Ω). In order to prove this fact, we need an information about the measure of the set in which u is close to σ. Lemma 2.3. Let {un } be a sequence of solutions of (2.1), with g ∈ L1 (Ω), g ≥ 0, and suppose that ( ∀δ > 0 ∃τ > 0 , ∃n0 > 0 : (2.10) meas({σ − τ ≤ un ≤ σ + τ }) ≤ δ , ∀n ≥ n0 . Then hn (un )|∇un |2 is strongly compact in L1 (Ω). Proof. We already know, by Proposition 2.2, that, up to subsequences, un almost everywhere converges in Ω to some function u, with u ≤ σ almost everywhere. We begin to prove that, under assumption (2.10), meas({u = σ}) = 0. Indeed, we have lim inf χ{σ−τ ≤un ≤σ+τ } (x) ≥ χ{σ−τ σ}

and the last integral tends to zero as n tends to infinity since u ≤ σ almost everywhere in Ω (by iii) of Proposition 2.2). Let now ε > 0;


13

since g belongs to L1 (Ω), there exists δε > 0 such that Z ε meas(E) < δε ⇒ g< . 2 E Let kε < σ be such that (2.10) holds true with δε ; therefore, Z ε g < , ∀n ≥ n0 . 2 {kε ≤un ≤σ} Once kε < σ is fixed, we have that Tkε (un ) is strongly compact in H01 (Ω) by iv) of Proposition 2.2; therefore, we can choose meas(E) small enough so that Z ε h(kε ) |∇Tkε (un )|2 ≤ , 2 E uniformly with respect to n. By applying Vitali theorem the conclusion then follows. We can now prove √ Theorem 1.2. In view 1of Lemma 2.3, we are going to prove that if h does not belong to L ((0, σ)), then (2.10) holds true. Proof of Theorem 1.2. Thanks to Proposition 2.2 we have Z hn (un )|∇un |2 ≤ kgkL1 (Ω) . (2.11) Ω

Rsp Defining Φn (s) = 0 hn (t)dt, we can write the above inequality as Z |∇Φn (un )|2 ≤ kgkL1 (Ω) . Ω

From Poincaré inequality we then deduce, for every τ > 0, and for some C > 0, Z Z 2 C |Φn (un )| ≤ C |Φn (un )|2 ≤ kgkL1 (Ω) . {σ−τ ≤un ≤σ+τ }

Ω

Thus, since Φn is increasing, we have meas({σ − τ ≤ un ≤ σ + τ }) ≤

kgkL1 (Ω) . C|Φn (σ − τ )|2

If n is large enough, then hn (s) = h(s) on [0, σ − τ ], so that meas({σ − τ ≤ un ≤ σ + τ }) ≤

kgkL1 (Ω) , C|Φ(σ − τ )|2

Rt p where Φ(t) = 0 h(s) ds. Since Φ is unbounded on [0, σ) by (1.6), (2.10) follows from the above inequality.

14


Remark 2.4. For the sake of simplicity, we chose to present the existence result of Theorem 1.2 for problem (1.1), even if several generalizations could be possible. For instance, in the proof we have never used the linearity of the principal part with respect to the gradient, so that it is easy to see that there exists a solution for ( A(u) + H(x, u, ∇u) = g in Ω u=0 on ∂Ω, where A(u) = −div(a(x, ∇u)) is a pseudomonotone operator (see [19] for more details), and H is such that H(x, s, 0) ≡ 0, and h1 (s)|ξ|2 ≤ H(x, s, ξ) ≤ h2 (s)|ξ|2 , with h1 and h2 continuous, increasing functions such that (1.3), (1.4), and (1.6) hold true. 3. Proof of theorems 1.7 and 1.8 Let {un } be the sequence of solutions of ( −div(M (x)∇un ) + hn (un )|∇un |2 = λf (3.1) un = 0

in Ω, on ∂Ω,

where hn (s) has been defined in (2.2). Such solutions exist by a result in [5] and are, thanks to the assumptions on f , in H01 (Ω)∩L∞ (Ω) for every fixed n. In order to prove Theorem 1.7, we will use the summability assumption on f to prove that the sequence {un } of solutions of (3.1) is uniformly bounded in L∞ (Ω) by a constant (depending on λ) which can be made strictly smaller than σ by choosing λ small. So that, roughly speaking, we deal with an equation which is no longer singular. This will allow us to apply Lemma 2.3 to pass to the limit in (3.1). Proof of Theorem 1.7. Let k > 0, and choose Gk (un ) as test function in (3.1). Using the fact that un ≥ 0, and that hn (s) is nonnegative, we have Z Z M (x)∇un · ∇Gk (un ) ≤ λ f Gk (un ). Ω

Ω

Then we use (1.2) to deduce, thanks to a classical result by G. Stampacchia (see [20]), that there exists a constant C0 > 0 (depending only on Ω, α, p and N ) such that (3.2)

kun kL∞ (Ω) ≤ λ C0 kf kLp (Ω) .

Consequently, if λ < Λ = kf kLpσ(Ω) C0 , (3.2) implies (2.10), and so both Proposition 2.2 and Lemma 2.3 can be applied, yielding the existence of a solution of (1.8) for λ small. Therefore, Λf > 0.


15

Proof of Theorem 1.8. We recall that now M (x) ≡ I. We are going to prove that (1.8) has a solution for every λ < Λf . Let 0 < λ < Λf , and let µ in (λ, Λf ) be such that (1.8) has a solution v such that 0 ≤ v ≤ σ − ε for some ε > 0. Define ( h(s) if 0 ≤ s ≤ σ − ε, h(s) = h(σ − ε) if s > σ − ε. Since h is bounded and f ≥ 0, by the results of [5] there exists a nonnegative function u, solution in H01 (Ω) ∩ L∞ (Ω) of ( −∆u + h(u)|∇u|2 = λ f in Ω, u=0 on ∂Ω. Clearly, v is a solution of the above problem with µ instead of λ. We now follow the lines of [3]: let k > 0 be fixed, and choose e−H(u) (ψ(u)− ψ(v))+ as test function in the equation for u, and e−H(v) (ψ(u) −ψ(v))+ in the equation for v, where (as in (1.10) with α = 1) Z s Z s e−H(t) dt. H(s) = h(t) dt , and ψ(s) = 0

0

We obtain Z

∇u · ∇(ψ(u) − ψ(v))+ e−H(u) Ω Z − h(u)|∇u|2 (ψ(u) − ψ(v))+ e−H(u) ZΩ + h(u)|∇u|2 (ψ(u) − ψ(v))+ e−H(u) ZΩ = λ f e−H(u) (ψ(u) − ψ(v))+ ,

(3.3)

Ω

which can be rewritten as Z Z + ∇ψ(u) · ∇(ψ(u) − ψ(v)) = λ f e−H(u) (ψ(u) − ψ(v))+ . Ω

Ω

Analogously, we obtain Z Z + ∇ψ(v) · ∇(ψ(u) − ψ(v)) = µ f e−H(v) (ψ(u) − ψ(v))+ . Ω

Ω

Subtracting the two identities, and recalling that λ < µ, we obtain Z Z + 2 |∇(ψ(u) − ψ(v)) | ≤ λ f (ψ(u) − ψ(v))+ (e−H(u) − e−H(v) ). Ω

Ω

16


Since H and ψ are increasing, the right hand side is negative, and so Z

|∇(ψ(u) − ψ(v))+ |2 = 0 ,

Ω

which implies u ≤ v. Since u ≤ v ≤ σ − ε, we have h(u) = h(u) and so u is a solution of (1.8). To prove uniqueness of solutions for fixed λ, let u and v be solutions of (1.8) for the same λ. If H and ψ are as in (1.10) (with α = 1), using e−H(u) (ψ(u) − ψ(v))+ and e−H(v) (ψ(u) − ψ(v))+ as test functions, and reasoning as above, one proves that u ≤ v; exchanging the roles of u and v yields the reverse inequality, so that u = v. To prove that if λ < µ, then uλ ≤ uµ , use e−H(uλ ) (ψ(uλ )−ψ(uµ ))+ and e−H(uµ ) (ψ(uλ )−ψ(uµ ))+ as test functions, and reason as above. Remark 3.1. We want to stress that the fact that M (x) ≡ I has been used in (3.3) in order to cancel two equal terms: this is the key ingredient in the proof of the comparison principle for solutions of (1.8). In the general case of a symmetric matrix M , comparison results are few and partial: we refer the reader to the papers [3] and [4].

4. Nonexistence of solutions In this section we are going to prove Theorem 1.9 and Theorem 1.10. We recall that we are dealing with solutions of

(4.1)

( −div(M (x)∇u) + h(u)|∇u|2 = λ f u=0

in Ω, on ∂Ω.

We begin with the proof of Theorem 1.9, i.e., with the case of h in L1 ((0, σ)). Proof of Theorem 1.9. Suppose by contradiction that there exists a solution u ∈ H01 (Ω) such that 0 ≤ u < σ a.e., and choose e−H(u) ϕ1 (f ) as test function in the weak formulation of (4.1), where, as in (1.10), 1 H(s) = α

Z

s

h(t) dt . 0


17

We obtain Z

(4.2)

M (x)∇u · ∇ϕ1 (f )e−H(u) Ω Z e−H(u) − M (x)∇u · ∇u h(u)ϕ1 (f ) α ZΩ + |∇u|2 e−H(u) h(u)ϕ1 (f ) ZΩ = λ f e−H(u) ϕ1 (f ) . Ω

Thus, if we define (again as in (1.10)) Z s ψ(s) = e−H(t) dt , 0

using the definition of λ1 (f ) and ϕ1 (f ), as well as the symmetry of M , we deduce that the first term in (4.2) can be written as Z Z −H(u) M (x)∇u · ∇ϕ1 (f )e = M (x)∇ϕ1 (f ) · ∇ψ(u) Z Ω Ω = λ1 (f ) f ϕ1 (f )ψ(u) . Ω

Therefore, using (1.2) and dropping nonnegative terms, we have, from (4.2), Z f ϕ1 (f )[λ1 (f )ψ(u) − λe−H(u) ] ≥ 0 Ω

Consider now the function Θ : [0, σ) → R+ defined by Θ(s) = λ1 (f )ψ(s) − λe−H(s) . Since both ψ(0) = 0 and e−H(0) = 1, we have Θ(0) = −λ < 0. Moreover, h(s) 0 Θ (s) = λ1 (f ) + λ e−H(s) ≥ 0 , α so that Θ(s) ≤ Θ(σ) = λ1 (f )ψ(σ) − λe−H(σ) Therefore, if λ > λ1 (f )eH(σ) ψ(σ) we have Z 0≤ f ϕ1 (f )[λ1 (f )ψ(u) − λe−H(u) ] < 0 , Ω

which is a contradiction.

18


We turn now (4.1) under assumption (1.11), which √ to the study of 1 implies that h belongs to L ((0, σ)), while h itself does not. Since M ≡ I, we are going to deal with solutions of ( −∆u + h(u)|∇u|2 = λ f in Ω, (4.3) u=0 on ∂Ω. As in (1.10), we define Z s h(t) dt , H(s) =

Z and ψ(s) =

0

s

e−H(t) dt .

0

By the assumptions on h, H is unbounded on (0, σ), while ψ is bounded. We define L = ψ(σ), so that 0 ≤ ψ(s) ≤ L, and ψ will be increasing (hence invertible) from [0, σ] to [0, L]. Let now u ∈ H01 (Ω) be a solution of (4.3), with 0 ≤ u < σ almost everywhere in Ω. Defining v = ψ(u), we have ∇v = e−H(u) ∇u ,

∆v = e−H(u) [∆u − h(u)|∇u|2 ] .

If we set g(s) = e−H(ψ

(4.4)

−1 (s))

,

we have that v ∈ H01 (Ω) is a solution of ( −∆v = λ f g(v) in Ω, (4.5) v=0 on ∂Ω. Lemma 4.1. The function g : [0, L] → R defined by (4.4) is such that: i) g(0) = 1 and g(L) = 0; ii) g is decreasing; iii) lim− g 0 (s) = −∞; s→L

iv) L

Z 0

dt qR L t

< +∞ .

g(s) ds

A Example 4.2. In the particular case h(s) = 1−s (with A > 0), it is easy to see that A g(s) = [1 − (1 + A)s] 1+A .

Proof. Since ψ −1 (0) = 0, we have g(0) = 1. On the other hand, since both ψ −1 (L) = σ and H is unbounded, we have g(L) = 0, so that i) is proved. Furthermore, using that (ψ −1 (s))0 = eH(ψ

−1 (s))

,


19

we have (4.6)

g 0 (s) = −h(ψ −1 (s)) e−H(ψ

−1 (s))

(ψ −1 (s))0 = −h(ψ −1 (s)) ,

which implies both ii) and iii). As for iv), it is easy to see, using the definition of g, and by changing variables, that Z σ Z L e−H(t) dt dt qR qR < +∞ ⇐⇒ < +∞ . σ −2H(s) L 0 0 e ds g(s) ds t t We are going to prove that there exists D > 0 such that Rσ h(t) t e−2H(s) ds (4.7) lim = D. t→σ − e−2H(t) Once we prove this, we will have that Z σp Z σ e−H(t) dt qR < +∞ ⇐⇒ h(t) dt < +∞ , σ −2H(s) 0 0 e ds t and the latter result is true by (1.11) since 1 ≤ γ < 2. By (1.11), we have Rσ R σ −2H(s) h(t) t e−2H(s) ds e ds lim− = C lim− t . −2H(t) γ −2H(t) t→σ t→σ (σ − t) e e Applying the de l’Hôpital rule, we have R σ −2H(s) e ds −e−2H(t) lim− t = lim , t→σ (σ − t)γ e−2H(t) t→σ − −[γ(σ − t)γ−1 + 2(σ − t)γ h(t)]e−2H(t) 1 and the latter limit is equal (again by (1.11)) to 2C (if γ > 1) or (if γ = 1). Therefore, (4.7) holds, and iv) is proved.

1 2C+1

Since solutions v of (4.5) which are almost everywhere smaller than L correspond to solutions u of (4.3) which are almost everywhere smaller than σ via the transformation u = ψ −1 (v), we are now going to consider problem (4.5) on its own. Theorem 4.3. For every λ > 0 there exists a unique solution vλ of (4.5), with vλ in H01 (Ω), and 0 ≤ vλ ≤ L. Proof. Define

  if s < 0,  1 g(s) = g(s) if 0 ≤ s ≤ L,   0 if s > L, and consider the problem ( −∆w = λ f g(w) in Ω, w=0 on ∂Ω.

20


It is easy to see, by using a fixed point argument, that for every λ > 0 there exists a solution w in H01 (Ω), which is nonnegative since λ f g(w) ≥ 0 in Ω. Taking (w − L)+ as test function in the weak formulation, and using the definition of g as well as the fact that f ≥ 0, we have w ≤ L. Hence, by definition of g, g(w) = g(w), and so w is a solution of (4.5). Uniqueness of solutions then follows from the next lemma. Lemma 4.4. Let v and v in H 1 (Ω) be such that −∆v ≤ λ f g(v) ,

−∆v ≥ λ f g(v) ,

in H −1 (Ω) and (v − v) ≤ 0 on ∂Ω. Then v ≤ v. Proof. Subtracting the above inequalities and choosing (v − v)+ as test function yields Z Z + 2 |∇(v − v) | ≤ λ 0≤ f (g(v) − g(v))(v − v)+ ≤ 0 , Ω

Ω

since, by ii) of Lemma 4.1, g is decreasing. Therefore, (v − v)+ = 0, i.e., v ≤ v. 4.1. Construction of unidimensional “flat” solutions. To prove that (4.3) has no solutions for λ large enough, we are going to prove that (4.5) has solutions v such that meas({v = L}) > 0 if λ is large enough. In order to deal with the general N -dimensional case, we are going to study the one-dimensional equation first, with f ≡ 1. Therefore, we are going to fix R > 0 and consider the problem ( −v 00 (s) = λ g(v(s)) in (−R, R), (4.8) v(±R) = 0 . We know from Theorem 4.3 that a solution v of (4.8) exists for every λ and for every R, with 0 ≤ v ≤ L. In order to study the properties of the solutions v as λ changes, we are going to study the solutions of a “shooting” problem. Define g(s) ≡ 1 for s < 0, let 0 < ` < L, and consider the solution v` of the Cauchy problem ( −v`00 (s) = λ g(v` (s)) for s ≥ 0, (P` ) v` (0) = ` , v`0 (0) = 0 . Since g is Lipschitz continuous around `, then there exists a unique solution v` of (P` ) (at least locally near 0). Note that (PL ) is singular, since g 0 (L) = −∞ by iii) of Lemma 4.1. Thus, (PL ) has a maximal solution, which is v L (s) ≡ L, and a minimal solution vL (s), such that any other solution v satisfies vL (s) ≤ v(s) ≤ v L (s). We are going to


21

prove that (PL ) has infinitely many solutions not identically equal to L. Theorem 4.5. Let v` be the solution of (P` ). Then: i) v`0 (s) < 0 for every s > 0, and v` is uniquely defined for every s ≥ 0; ii) for every ` < L there exists R` > 0 such that v` (R` ) = 0. Furthermore, Z ` 1 dt qR (4.9) R` = √ ; ` 2λ 0 g(s) ds t iii) as ` increases to L, v` increases; therefore, R` increases, and converges to the finite value Z L 1 dt qR RL = √ ; L 2λ 0 g(s) ds t

iv) as ` increases to L, v` uniformly converges on [0, s], for every s > 0, to vL , the minimal solution of (PL ), with vL (s) < L for every s > 0, and vL (RL ) = 0. Furthermore, vL is the only solution of (PL ) such that vL (s) < L for every s > 0; v) if w is a solution (not identically equal to L) of (PL ), then ( L if 0 ≤ s ≤ R, (4.10) w(s) = vL (s − R) if s > R, for some R > 0. Proof. i) The proof is easy: since v`00 (0) = −λg(`) < 0, we have v`0 (s) < 0 for s > 0 and sufficiently small. If there exists s such that v`0 (s) = 0, then v` is decreasing in [0, s], so that (since g is decreasing as well), g(v` (s)) ≥ g(`) for every s in [0, s]. But then we get a contradiction integrating the equation: Z s Z s 0 00 0 0 = v` (s) − v` (0) = v` (t)dt = −λ g(v` (t))dt ≤ −g(`)s < 0 . 0

0

v`0 (s)

Since < 0, v` stays away from L (where g is no longer Lipschitz continuous). Therefore, v` is unique and exists for every s ≥ 0. ii) If we suppose that v` (s) 6= 0 for every s ≥ 0, then v` (s) > 0 for every s ≥ 0. Since v` is decreasing by i), there exists (4.11)

lim v` (s) = T ,

s→+∞

22


with 0 ≤ T < `. Therefore, from the equation we get that lim v`00 (s) = −λ g(T ) < 0 ,

s→+∞

which implies lim

s→+∞

v`0 (s)

s

Z

v`00 (t) dt = −∞ .

= lim

s→+∞

0

This, however, yields s

Z

v`0 (t) dt + ` = −∞ ,

lim v` (s) = lim

s→+∞

s→+∞

0

a contradiction with (4.11). Thus, there exists R` > 0 such that v` (R` ) = 0. We now multiply the equation by v`0 and integrate on [0, s]. We obtain, using the initial conditions on v` and v`0 , Z s 1 0 2 − [v` (s)] = λ g(v` (t)) v`0 (t) dt = λ[G(v` (s)) − G(`)] , 2 0 where we have defined Z G(s) =

s

g(t) dt . 0

Therefore, taking into account that v`0 (s) < 0, p (4.12) v`0 (s) = − 2λ[G(`) − G(v` (s))] . Since v`0 (s) < 0 for every s > 0, we p have G(v` (s)) 6= G(`) for every s > 0. Therefore, we can divide by G(`) − G(v` (s)) and integrate between 0 and R` to obtain Z R` √ v`0 (s) ds p = − 2λ R` . G(`) − G(v` (s)) 0 Setting t = v` (s) in the first integral, recalling that v` (0) = `, that v` (R` ) = 0, and the definition of G, we have Z ` √ dt qR = 2λ R` , ` 0 g(s) ds t which then gives (4.9). Observe that the integral is singular at t = `, but that, since g is decreasing, s Z ` Z ` dt 1 dt ` qR √ ≤p =2 . ` g(`) `−t g(`) 0 0 g(s) ds t


23

iii) To prove that v` increases as ` increases, let ` < `0 and let v` and v`0 be the solutions of (P` ) and (P`0 ) respectively. Since v` (0) < v`0 (0), then v` ≤ v`0 in a neighbourhood of the origin. If there exists s such that v` (s) = v`0 (s) then, subtracting the equations, multiplying by v` − v`0 , and integrating on [0, s] yields Z s Z s 00 00 0 − (v` − v`0 )(v` − v` ) = λ (g(v` ) − g(v`0 ))(v` − v`0 ) , 0

0

and the right hand side is negative since g is decreasing. Integrating by parts the left hand side (and observing that v`0 (0) − v`0 0 (0) = 0, and that v` (s) − v`0 (s) = 0), we obtain Z s 0≤ (v`0 (t) − v`0 0 (t))2 dt ≤ 0 , 0

which implies

v`0 (s)

≡ s

Z v` (s) = ` +

v`0 0 (s)

on [0, s], and so Z s 0 v` (t) dt = ` + v`0 0 (t) dt = ` − `0 + v`0 (s) ,

0

0

a contradiction with the assumptions v` (s) = v`0 (s) and ` < `0 . Thus, v` increases with `, which implies that R` increases as well; therefore, there exists RL = lim− R` . `→L

Using (4.9), we have Z ` Z L Z L χ[0,`) (t)dt 1 1 1 dt qR qR =√ =√ θ` (t)dt, R` = √ ` ` 2λ 0 2λ 0 2λ 0 g(s)ds g(s)ds t t where we have defined χ[0,`) (t) θ` (t) = qR . ` g(s) ds t We clearly have, for every 0 ≤ t < L, χ[0,L) (t) lim− θ` (t) = qR = θL (t) , L `→L g(s) ds t so that the proof of iii) will be complete if we prove that Z L Z L (4.13) lim− θ` (t) dt = θL (t) dt . `→L

0

0

24

L. BOCCARDO, T. LEONORI, L. ORSINA, AND F. PETITTA 1 ), 2n

Define Ln = L(1 −

and

χ[0,L ) (t) . θn (t) = θLn (t) = qR n Ln g(s) ds t If m < n, in [0, Lm ) we have, since g(s) ≥ 0, 1 1 ≥ qR = θn (t) , θm (t) = qR Lm Ln g(s) ds g(s) ds t t so that m < n ⇒ θm (t) ≥ θn (t) in [Lm−1 , Lm ) .

(4.14) Let now

θ(t) =

+∞ X

θk+1 (t) χ[Lk ,Lk+1 ) (t) ,

k=0

and observe that, by (4.14), θ(t) ≥

n−1 X

θk+1 (t)χ[Lk ,Lk+1 ) (t) ≥

n−1 X

θn (t)χ[Lk ,Lk+1 ) (t) = θn (t) .

k=0

k=0

Since θn converges to θL , we can apply Lebesgue theorem to prove (4.13) if we prove that Z L (4.15) θ(t) dt < +∞ . 0

We have, by monotone convergence theorem, and recalling the definition of θk (s), Z L +∞ Z Lk+1 +∞ Z Lk+1 X X dt qR θk+1 (t) dt = θ(t) dt = . Lk+1 L L 0 k k k=0 k=0 g(s) ds t Since g is decreasing, we have Z Lk+1 g(s) ds ≥ g(Lk+1 )(Lk+1 − t) , t

so that Z L Z Lk+1 +∞ +∞ √ X X 1 dt L − Lk p √ p k+1 θ(t) dt ≤ =2 . L − t g(L ) g(L ) k+1 0 L k+1 k+1 k k=0 k=0 Recalling the definition of Lk , we then have Z L +∞ √ X 1 p (4.16) θ(t) dt ≤ 2 L . k+1 g(L )2 0 k+1 k=0

25


On the other hand, we have Z L +∞ Z Lk+1 X dt dt qR qR = . L L 0 k=0 Lk g(s) ds g(s) ds t t Once again, since g is decreasing, we have, for t in (Lk , Lk+1 ), Z L g(s) ds ≤ g(t) (L − t) ≤ g(Lk )(L − t) , t

and so L

Z 0

dt qR L t

Therefore, Z

L

0

g(s) ds

≥

+∞ X

1

k=0

p g(Lk )

Z

Lk+1

Lk

√

dt . L−t

√ +∞ √ X L − Lk − L − Lk+1 p ≥2 , g(Lk ) k=0 g(s) ds

dt qR L t

and so, recalling the definition of Lk , Z L +∞ X √ √ dt 1 qR p ≥ 2L( 2 − 1) . k L g(L )2 0 k k=0 g(s) ds t Recalling (4.16), we then have √ Z L Z L dt 2 qR θ(t) dt ≤ √ , L 2−1 0 0 g(s) ds t

and the latter integral is finite by iv) of Lemma 4.1. Thus (4.15) is proved, and the proof of iii) is complete. iv) Let s > 0 be fixed, and consider v` (s) on [0, s]. Since v` increases as ` increases to L, then v` (s) converges to some function vL (s). Since g(s) ≤ 1, then Z s 0 g(v` (t)) dt ≤ λ s ≤ λ s , ∀s ∈ [0, s] , |v` (s)| = λ 0

|v`0 (s)|

so that is uniformly bounded with respect to ` on [0, s]. Since v` (0) = ` is bounded as well, Ascoli-Arzelà theorem implies that v` converges uniformly to vL as ` tends to L. This convergence, and the fact that R` converges to RL , imply that vL (RL ) = 0. Since v` increases to vL , the fact that g is decreasing implies that g(v` (s)) decreases to

26


g(vL (s)). Therefore, and since g(s) ≤ 1, Lebesgue theorem implies, for every s in [0, s], Z s Z s lim− g(v` (t)) dt = g(vL (t)) dt , `→L

0

0

so that for every s in [0, s] there exists Z s Z 0 lim− v` (s) = − lim− λ g(v` (t)) dt = −λ `→L

`→L

0

s

g(vL (t)) dt = w(s) .

0

Since from the equation we have that (v`0 )0 is uniformly bounded (with respect to `), and furthermore v`0 (0) = 0 is bounded as well, a further application of Ascoli-Arzelà theorem implies that v`0 uniformly converges to w on [0, s]. This fact (together with the convergence of v` to vL ) implies that w = vL0 . Therefore, Z s 0 vL (s) = −λ g(vL (t)) dt , 0

which implies (since g(vL (t)) is continuous) that vL is a solution of ( for s ≥ 0, −vL00 (s) = λ g(vL (s)) (PL ) 0 vL (0) = L , vL (0) = 0 . Since vL (RL ) = 0, we have that vL is not identically equal to L. To prove that vL is the minimal solution of (PL ), let w be a solution of (PL ). We then have that w ≥ v` for every ` < L. The proof of this fact can be achieved with the same ideas used to prove that v` ≤ v`0 if ` < `0 in iii). Since vL is the limit of v` , we then have vL ≤ w, as desired. To prove that vL (s) < L for every s > 0, suppose that there exists s > 0 such that vL (s) = L. Then vL (s) ≡ L in [0, s]. Indeed, since vL00 (s) ≤ 0, vL is concave, and so its graph is above the (horizontal) line connecting (0, L) and (s, L), and below the tangent line at (0, L), which (since vL0 (0) = 0) is the same horizontal line. Thus, vL is constantly equal to L on [0, s]. If we define S = sup{s > 0 : vL (s) = L} , we have that 0 < s ≤ S < RL , that vL (S) = L, and that vL0 (S) = 0. Therefore, vL is a solution of ( −vL00 (s) = λg(vL (s)) for s ≥ S, 0 vL (S) = L , vL (S) = 0 , which is decreasing (hence strictly decreasing) for s > S (the proof of this fact is analogous to the one in i)). If we define z(s) = vL (s + S),


27

then z(s) ≤ vL (s), and z is a solution of (PL ). Since vL is the minimal solution of the same problem, then vL (s) ≤ z(s), so that vL ≡ z. This implies that S = 0, a contradiction since S ≥ s > 0. To prove that vL is the only solution of PL such that vL (s) < L for every s > 0, let w be another such solution. Since w has no “flat” zones (being w0 (s) < 0 for every s > 0), then w(RL ) = 0 (just start from the equation and perform the same calculations used to obtain (4.9) for R` ). We therefore have that vL (0) = w(0), and that vL (RL ) = w(RL ). Subtracting the equations satisfied by vL and w, multiplying by vL −w, integrating, using that g is decreasing and that vL (s) ≤ w(s) for every s > 0 since vL is the minimal solution, we have vL0 ≡ w0 , and so vL ≡ w. v) Let w be a solution of (PL ) which is not identically equal to L. Since vL (s) is the unique solution of (PL ) which is different from L for every s > 0, there exists s such that w(s) = L. Reasoning as in iv) (i.e., using the concavity of w), we have that w(s) ≡ L on [0, s]. Define R = sup{s > 0 : w(s) = L} > 0 , and observe that R < +∞ since w is not identically equal to L. Setting z(s) = w(s + R) for s ≥ 0, and reasoning as in iv), one has that z is a solution of (PL ), and that z(s) < L for every s > 0. By uniqueness, z(s) = vL (s), and so w is of the form (4.10). Theorem 4.5 allows us to prove the existence of a flat solution in dimension N = 1. In fact, let now R be fixed, and consider the solution v to the Dirichlet problem (4.8). Since v is symmetric with respect to the origin, then v 0 (0) = 0. Setting ` = v(0), we then have that v = v` (if ` < L), or that v is one of the solutions of (PL ) which are between vL and v L ≡ L (if ` = L). Therefore, it is easy to see that v is almost everywhere smaller than L (i.e., v has no “flat” zones where it is constantly equal to L) if and only if R ≤ RL . In other words, if λ is fixed, and R > RL , then any solution v of (4.8) has a nonzero measure “flat” zone {v = L}. Observe that since Z L 1 dt RL = √ , RL 2λ 0 g(s) ds t 1

then RL = Cλ− 2 , so that, for every R > 0, we have R > RL for λ large enough. Therefore, problem (4.8) has “flat” solutions; hence there are no solutions almost everywhere smaller than σ of the corresponding quasilinear singular problem (4.3) in dimension 1; i.e., Λf ≡1 is finite.

28


4.2. Construction of radial “flat” solutions. Now we turn our attention to the N -dimensional semilinear problem (4.5). We are going to prove that, under the same assumptions on h for which the onedimensional semilinear problem has “flat” solutions for λ large, problem (4.5) has “flat” solutions as well. In order to do that, we are going to prove the following result. Theorem 4.6. Let Ω = BR (0) and f ≡ 1; then, for λ large enough, there exists a subsolution v of (4.5) such that meas({v = L}) > 0. Proof. Let vL (s) be the minimal solution of the one-dimensional Cauchy problem ( −vL00 (s) = g(vL (s)) if s ≥ 0, 0 vL (0) = L , vL (0) = 0 , and let RL be such that vL (RL ) = 0. Let R > 0, and consider the function ( L if 0 ≤ s ≤ R, w(s) = vL (s − R) if s > R. By (4.10), w is a C 2 “flat” solution of the same problem solved by vL , with w(RL + R) = 0. If we consider w as a radial function, we then have w0 (r) w0 (r) 00 −∆w = −w (r) − (N − 1) = g(w(r)) − (N − 1) . r r We are going to prove that there exists R > 0 such that −∆w ≤ 2 g(w) in BRL +R (0). In order to do that, it is enough to prove that w0 (r) ≤ g(w(r)) , r for every R ≤ r ≤ RL + R, since if r < R we have −∆w = 0 = 2g(w). Thus, we have to prove that −(N − 1)

vL0 (r − R) ≤ g(vL (r − R)) , r or, equivalently, that −(N − 1)

−(N − 1)

vL0 (s) ≤ g(vL (s)) , s+R

∀0 ≤ r − R ≤ RL ,

∀0 ≤ s ≤ RL .

Recalling (4.12), written for ` = L and λ = 1, and the definition of G, we have s Z L vL0 (s) = − 2 g(t) dt , vL (s)


so that we have to prove that q R L 2 vL (s) g(t) dt (N − 1) ≤ g(vL (s)) , s+R We are going to prove that there exists R > 0 inequality q R L 2 vL (s) g(t) dt (N − 1) ≤ g(vL (s)) , R holds; that is, R > 0 is such that Z L R2 g 2 (vL (s)) , g(t) dt ≤ 2 2(N − 1) vL (s)

29

∀0 ≤ s ≤ RL . such that the stronger

∀0 ≤ s ≤ RL ,

∀0 ≤ s ≤ RL .

Since vL (s) takes values in [0, L] as s ranges between 0 and RL , finding R such that the above inequality holds amounts to finding C > 0 such that Z L (4.17) g(s) ds ≤ C g 2 (L − t) , ∀0 ≤ t ≤ L . L−t

Define Z

2

L

η(t) = C g (L − t) −

g(s) ds , L−t

so that (4.17) is equivalent to proving that η(t) ≥ 0 for every t in [0, L]. We have η(0) = 0, and η 0 (t) = −2Cg(L − t)g 0 (L − t) − g(L − t) = g(L − t)[−2Cg 0 (L − t) − 1] , so that η 0 (t) > 0 for every t (which then implies η(t) ≥ 0 as desired) if there exists C > 0 such that 1 (4.18) g 0 (L − t) ≤ − , ∀0 ≤ t ≤ L . 2C Indeed, by (4.6) we have g 0 (L − t) = −h(ψ −1 (L − t)) . Now, if h(0) > 0 then such a C exists using (1.4), and since h is strictly increasing. On the contrary, if h(0) = 0, (4.18) fails near 0. In this case, however, we have Z L Z L 2 η(L) = Cg (0) − g(s) ds = C − g(s) ds , 0

0

so that there exists C1 > 0 such that η(L) > 0. Since η is continuous, there exists δ > 0 such that η(s) > 0 for every s in [L − δ, L]. Since h is strictly increasing, there exists C2 > 0 such that (4.18) holds true in

30


[0, L − δ]. Taking C = max(C1 , C2 ) we then have η(t) ≥ 0 for every t, as desired. Thus, if we define R = RL + R, we have found a function w such that w(R) = 0, and such that −∆w ≤ 2 g(w) in BR (0). Let now R > 0 be fixed, and consider v(x) = w(R x/R). It is easy to see that 2

2

2

R 2R 2R −∆v = − 2 ∆w ≤ 2 g(w) = 2 g(v) , R R R

v(|x| = R) = 0 ,

2

so that v is a subsolution of (4.5) for λ =

2R R2

.

Once we know that, for λ large enough, there exists a radial “flat” subsolution in BR , we can give the proof of the nonexistence theorem. Proof of Theorem 1.10. Let x0 in Ω and R > 0 be such that BR (x0 ) ⊂ Ω, and let λ be large enough so that there exists a subsolution v of ( −∆v = λ ρ g(v) in BR (x0 ), v=0 on ∂BR (x0 ), where ρ is given by (1.11), such that meas({v = L}) > 0. Such a λ exists by Theorem 4.6. Define w as v in BR (x0 ) and zero in Ω \ BR (x0 ). Then w is a subsolution of ( −∆v = λ ρ g(v) in Ω, v=0 on ∂Ω, hence, by (1.11), a subsolution of (4.5). If v is the solution of (4.5), we have by Lemma 4.4 that w ≤ v ≤ L. Since meas({w = L}) > 0, we have that meas({v = L}) > 0, and so Λf < +∞. 5. Limit equation for approximating problems In this section we want to describe the behavior of the approximating sequences of solutions to problem (4.1) for those values of λ such that a solution does not exist. To fix the ideas, we will work with the Laplace operator, under assumption (1.11), with h(0) = 0, f ≡ 1, and Ω = BR (0) a ball with radius R > 0. As in Section 2, let {un } be the sequence of solutions in H01 (BR ) ∩ L∞ (BR ) of ( −∆un + hn (un )|∇un |2 = λ in BR , (5.1) un = 0 on ∂BR , where ( h(s) if s < σ and h(s) ≤ n, hn (s) = n if s < σ and h(n) > n, or if s ≥ σ.

31


Since h is strictly increasing, if σn is such that h(σn ) = n, we have ( h(s) if s ≤ σn , hn (s) = n if s > σn , with σn increasing to σ as n tends to infinity. Reasoning as in Section 4, if we define Z s Z s −1 e−Hn (t) dt , gn (s) = e−Hn (ψn (s)) , hn (t) dt , ψn (s) = Hn (s) = 0

0

then the function vn = ψn (un ) is a solution in H01 (BR ) ∩ L∞ (BR ) of ( −∆vn = λ gn (vn ) in BR , vn = 0 on ∂BR . An explicit calculation yields   g(s) if s ≤ ψ(σn ),  gn (s) = g(ψ(σn )) − n(s − ψ(σn )) if ψ(σn ) < s ≤ ψ(σn ) +   n )) 0 if s > ψ(σn ) + g(ψ(σ , n

g(ψ(σn )) , n

where g(s) has been defined in Section 4. Since g is concave, and σn is an increasing sequence of real numbers, gn (s) ≥ gn+1 (s) ≥ g(s), so that (thanks to Lemma 4.4), vn ≥ vn+1 ≥ v, where v is the solution of (4.5) given by Theorem 4.3. It is easy to see (using the boundedness of gn ) that {vn } is bounded in 1 H0 (BR ) so that it converges (the whole sequence, since it is decreasing) n )) to some function w. Since w ≤ vn ≤ ψ(σn )+ g(ψ(σ , and σn tends to σ, n we have 0 ≤ w ≤ ψ(σ) = L. Furthermore, if w(x) < L, then gn (vn (x)) converges to g(w(x)), while if w(x) = L, then vn (x) ≥ L for every n in N, and so 0 ≤ gn (vn (x)) ≤ g(σn ); therefore, gn (vn (x)) converges to g(w(x)) = 0. In other words, w is a solution of (4.5), hence the solution of the same problem. We have therefore proved that if un is the sequence of solutions of (5.1), then vn = ψn (un ) converges to the solution of (4.5). If we suppose to be under the assumptions of Theorem 1.10, the solutions of (4.5) have a “flat” nonzero measure zone ω = {v = L} if λ is large enough; furthermore, since Ω is a ball, and v is radially symmetric, then its interior ω = Br (0) is a ball as well (of radius r for some 0 < r < R). What happens in this case to un ? Using the fact that vn decreases to v, it is easy to see that un converges to some function u which has ω as the “flat” zone {u = σ}. What can we say about the equation? Reasoning as in Proposition 2.2, we have that {un } is bounded in H01 (BR ), that the lower order

32


quasilinear term hn (un )|∇un |2 is bounded in L1 (BR ), that ∇un converges almost everywhere to ∇u, and that Tk (un ) strongly converges to Tk (u) in H01 (BR ) for every 0 < k < σ. Since hn (un )|∇un |2 is bounded in L1 (BR ), it converges weakly∗ in the sense of measures to some nonnegative bounded Radon measure ν. Standard elliptic results then imply that u is the solution of ( −∆u + ν = λ in BR , u=0 on ∂BR . In addition, since Tk (un ) strongly converges to Tk (u) for every 0 < k < σ, we have that hn (un )|∇un |2 χ{un 0 such that meas({σ − τ ≤ uλ ≤ σ + τ }) ≤ δ ,

∀λ < Λf .

In other words, the assumptions of Lemma 2.3 are satisfied, and so h(uλ )|∇uλ |2 is compact in L1 (Ω). Therefore, we can pass to the limit in the equation satisfied by uλ , to have that u is a solution for λ = Λf . What we do not know is whether in the general case the measure of the set {u = σ} is zero or not; i.e., if there exists a solution for λ = Λf . What we can prove is however that the set {u = σ} (where u is as above the limit of uλ as λ increases to Λf ) cannot be empty. Indeed,


35

suppose that {u = σ} = ∅. Since both f and Ω are smooth, standard elliptic results imply that {uλ } is equi-Hölder continuous: one can see that this is true by performing the (lipschitz continuous) change of variable as in Section 4 and use De Giorgi’s theorem to prove that the solutions of the semilinear equation are equi-Hölder continuous. Thus, the convergence of uλ to u is uniform, and so u is continuous. This implies that there exists ε > 0 such that 0 ≤ u ≤ σ − ε in Ω. We claim that if this is the case, than there exist solutions strictly smaller than σ for some λ > Λf (and this contradicts the definition of Λf ). Indeed, ˜ as follows: define h ( h(s) if 0 ≤ s ≤ σ − 2ε , ˜ h(s) = h(σ − 2ε ) if s > σ − 2ε . Then clearly u solves 2 ˜ −∆u + h(u)|∇u| = Λf f ,

and, by the results of [5], there exists a solution v of 2 ˜ −∆v + h(v)|∇v| = (Λf + δ) f ,

with v ≥ u, for every δ > 0. It is easy to see, reasoning as in the proof of Theorem 1.8 (or performing the change of variable as in Section 4), that ku − vkL∞ (Ω) ≤ C δ kf kL∞ (Ω) , so that, if δ is small enough, 0 ≤ v ≤ σ −

3ε , 4

and so v is a solution of

−∆v + h(v)|∇v|2 = (Λf + δ) f , thus contradicting the definition of Λf . Therefore, we have that, as λ tends to Λf , uλ increases to some function u which “touches” σ. If it touches it on a set of positive measure, then u is not a solution, while it is a solution if the measure of {u = σ} is zero. In the first case, it is clear that no solution exists for λ > Λf , but what happens if we have a solution for λ = Λf ? Do solutions exist for λ > Λf ? We do not know what happens in the general case, but we can give an answer in the radial case: there can be only one solution u such that the set {u = σ} is not empty and has zero measure. To simplify the calculations, rewrite the equation as ( 2 = λ f in Ω, −∆u + |∇u| g(u) (6.2) u=0 on ∂Ω,

36


where Ω = BR (0) is a ball, and f is a smooth radial function. On the function g we will suppose that it is a positive, C 1 ((0, σ)) function such that Z σ dt 0 p lim− g(s) = 0 , ∃ lim− g (s) ∈ [−∞, 0] , < +∞ , s→σ s→σ g(t) 0 for some σ > 0. The function g(s) = (1 − s)γ obviously satisfies the above assumption if γ < 2 and σ = 1. We are going to prove that if Λ is such that there exists a C 2 radial solution uΛ such that uΛ (0) = σ, and uΛ (r) < σ for every r > 0, then there exist solutions strictly smaller than σ if λ < Λ (i.e., Λf = Λ), and there exists no solution for λ > Λ. Since uΛ is a radial solution, we have −u00Λ (r) − (N − 1)

u0Λ (r) (u0Λ (r))2 + = Λ f (r) , r g(uΛ (r))

∀r > 0 ,

and so, by the de l’Hôpital rule, which can be applied since u0Λ (0) = 0, lim+

r→0

(u0Λ (r))2 = Λ f (0) + N u00Λ (0) . g(uΛ (r))

We know, again by the de l’Hôpital rule, that if it exists (6.3)

lim+

r→0

2u0Λ (r) u00Λ (r) 2u00Λ (r) = lim , g 0 (uΛ (r)) u0Λ (r) r→0+ g 0 (uΛ (r))

then this limit is equal to lim+

r→0

(u0Λ (r))2 . g(uΛ (r))

Since uΛ is a C 2 function, and the origin is a maximum for uΛ , we have lim u00Λ (r) = u00Λ (0) ≤ 0 .

r→0+

Then we may have three possibilities. i) if g 0 (σ) = −∞, then limit (6.3) is equal to zero, and u00Λ (0) = − Λ fN(0) ; 2u00 (0) ii) if g 0 (σ) < 0, then limit (6.3) is equal to g0Λ(σ) , and u00Λ (0) = − NΛ−f (0) 2 ; 0

g 0 (σ)

iii) if g (σ) = 0 then the limit (6.3) is not finite if u00Λ (0) < 0, which will yield that Λ f (0) + N u00Λ (0) = −∞, a contradiction; therefore, if g 0 (σ) = 0, we have u00Λ (0) = 0. If we are in cases i) or ii), we have uΛ (r) = σ − CΛ r2 + o(r2 ) ,


where C is either

f (0) N

or

f (0) 2 . N − g0 (σ)

37

Let now λ < Λ, and let uλ be a

solution of our problem: by Theorem 1.8 we have that uλ ≤ uΛ . If uλ (0) < σ, we have nothing to prove. If uλ (0) = σ, then we can repeat the above proof to have uλ (r) = σ − C λ r2 + o(r2 ) , with C as above. Since λ < Λ, this contradicts the fact that uλ ≤ uΛ , and so uλ (0) has to be smaller than σ. Analogously, we arrive to a contradiction if we suppose that there exists a solution uλ with no “flat” zones for λ > Λ. Thus, in both cases i) and ii) we have that Λf = Λ is a supremum, and there is no solution for λ > Λf . Now we turn our attention to case iii), which is more delicate. In this case, since u00Λ (0) = 0, we have (6.4)

lim+

r→0

(u0Λ (r))2 = Λ f (0) > 0 . g(uΛ (r))

Define now, for s ≤ σ, the bounded function Z s dt p G(s) = , g(t) 0 and observe that (6.4) can be rewritten as 2 d lim+ G(uΛ (r)) = Λ f (0) . r→0 dr Thus, if we define the function vΛ (r) = G(uΛ (r)), then vΛ is apdecreasing function (since uΛ is decreasing) such that vΛ0 (0) = − Λ f (0). Thus, p vΛ (r) = G(σ) − Λ f (0) r + o(r) . Let now λ < Λ. If the solution uλ is such that uλ (0) = σ, we can repeat the above proof to have that vλ (r) = G(uλ (r)) satisfies p vλ (r) = G(σ) − λ f (0) r + o(r) , so that vλ (r) ≥ vΛ (r) in a neighbourhood of 0. This yields that uλ (r) ≥ uΛ (r) in a neighbourhood of 0, and so uλ (r) ≡ uΛ (r) near the origin, a contradiction since λ 6= Λ. Thus, it has to be uλ (0) < σ, as desired. If λ > Λ, the same argument shows that there is no solution uλ with no “flat” zones {uλ = σ}.

38


References [1] D.Arcoya, L. Boccardo, T. Leonori, A. Porretta: , J. Differential Equations, to appear. [2] D. Arcoya, J. Carmona, T. Leonori, P.J. Mart´ınez-Aparicio, L. Orsina, F. Petitta: Existence and nonexistence of solutions for singular quadratic quasilinear equations, J. Differential Equations, 246 (2009) 4006–4042. [3] D. Arcoya, S. Segura de Leon: Uniqueness for some elliptic equations with lower order terms, ESAIM: Control, Optimization and the Calculus of Variations, 16 (2010), 327–336. [4] G. Barles, F. Murat: Uniqueness and maximum principle for quasilinear elliptic equations with quadratic growth conditions, Arch. Rational Mech. Anal., 133 (1995), 77–101. [5] A. Bensoussan, L. Boccardo, F. Murat: On a nonlinear partial differential equation having natural growth terms and unbounded solution, Ann. Inst. H. Poincaré Anal. Nonlin., 5 (1988), 347–364. [6] H. Hamid, M.F. Bidaut-Véron, Correlation between two quasilinear elliptic problems with a source term involving the function or its gradient, C. R. Math. Acad. Sci. Paris, 346 (2008), 1251–1256. [7] L. Boccardo: On the regularizing effect of strongly increasing lower order terms, J. Evol. Equ., 3 (2003), 225–236. [8] L. Boccardo: Dirichlet problems with singular and gradient quadratic lower order terms, ESAIM: Control, Optimization and the Calculus of Variations, 14 (2008), 411–426. [9] L. Boccardo, T. Gallouët: Strongly nonlinear elliptic equations having natural growth terms and L1 data, Nonlinear Anal. TMA, 19 (1992), 573–579. [10] L. Boccardo, T. Gallouët, L. Orsina: Existence and nonexistence of solutions for some nonlinear elliptic equations, J. Anal. Math., 73 (1997), 203–223. [11] L. Boccardo, F. Murat: Almost everywhere convergence of the gradients of solutions to elliptic and parabolic equations, Nonlinear Anal., 19 (1992), 581– 597. [12] L. Boccardo, F. Murat, J.-P. Puel: L∞ -estimate for nonlinear elliptic partial differential equations and application to an existence result, SIAM J. Math. Anal., 23 (1992), 326–333. [13] H. Brezis, M. Marcus, A.C. Ponce: Nonlinear elliptic equations with measures revisited. In: Mathematical Aspects of Nonlinear Dispersive Equations (J. Bourgain, C. Kenig, and S. Klainerman, eds.), Annals of Mathematics Studies, 163, Princeton University Press, Princeton, NJ, 2007, 55–110. [14] H. Brezis, L. Nirenberg: Removable singularities for nonlinear elliptic equations, Topol. Methods Nonlinear Anal., 9 (1997), 201–219. [15] L. Dupaigne, A. Ponce, A. Porretta: Elliptic equations with vertical asymptotes in the nonlinear term, J. Anal. Math., 98 (2006), 349–396. [16] D. Giachetti, F. Murat: An elliptic problem with a lower order term having singular behaviour, Boll. Un. Mat. Ital. B, to appear. [17] D. Giachetti, S. Segura de Leon: Quasilinear stationary problems with a quadratic gradient term having singularities, preprint. [18] T. Leonori: Bounded solutions for some Dirichlet problems with L1 (Ω) data, Boll. U.M.I., 4 (2007) 785–796.


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[19] J. Leray, J. L. Lions: Quelques résultats de Viˇsik sur les problémes elliptiques semi-linéaires par les méthodes de Minty et Browder, Bull. Soc. Math. France, 93 (1965), 97–107. [20] G. Stampacchia: Le probléme de Dirichlet pour les équations elliptiques du second ordre ` a coefficients discontinus, Ann. Inst. Fourier (Grenoble), 15 (1965), 189–258. ´ [21] G. Stampacchia: Equations elliptiques du second ordre á coefficients discontinus. Sém. sur les équat. diff. Collége de France, 1963-64. ` di Roma Lucio Boccardo, Dipartimento di Matematica, Universita “La Sapienza”, P.le Aldo Moro 2, 00185 Roma, Italy E-mail address: [email protected] ´ lisis Matema ´ tico, UniversiTommaso Leonori, Departamento de Ana dad de Granada, Campus Fuentenueva s/n, 18071 Granada, Spain E-mail address: [email protected] ` di Roma “La Luigi Orsina, Dipartimento di Matematica, Universita Sapienza”, P.le Aldo Moro 2, 00185 Roma, Italy E-mail address: [email protected] ´ lisis Matema ´ tico, UniverFrancesco Petitta, Departamento de Ana sitat de Valencia, C/ Dr. Moliner 50, 46100 Burjassot, Valencia, Spain E-mail address: [email protected]