A SELECTION CRITERION OF SOLUTIONS TO A SYSTEM OF EIKONAL EQUATIONS GISELLA CROCE AND

Keywords: everywhere solutions, eikonal equation, functions of bounded variation, direct methods

Key words and phrases. almost

Introduction

In this article we deal with the following system of eikonal equations:

(1.1)

     ∂u ∂x i = 1 , i = 1, . . . , N, a.e. in Ω u = 0 , on ∂Ω ,
where Ω is an open bounded connected Lipschitz domain of R N . System (1.1) arises in several nonlinear models in mechanics and material science and the problem of the existence of solutions has been widely studied. With no attempt to be exhaustive here we recall some results that have motivated our study.

Examining the pyramidal construction, introduced in [START_REF] Cellina | On minima of a functional of the gradient, sufficient conditions[END_REF] and [START_REF] Friesecke | A necessary and sufficient condition for nonattainment and formation of microstructure almost everywhere in scalar variational problems[END_REF], one can easily see that there exist infinitely many W 1,∞ 0 (Ω) solutions of (1.1). Indeed, if Q ⊂ R N is a hyperrectangle oriented in such a way that the exterior normal to its faces is contained in the set

E = {x = (x 1 , . . . , x N ) ∈ R N : |x 1 | = • • • = |x N |},
then the distance function in the l 1 -norm from the boundary of Q, d 1 (•, ∂Q), solves problem (1.1) in Q. In a general domain Ω, Vitali's theorem allows us to cover Ω, up to a Lebesgue measure zero set, by a countable union of domains Q i as before. Then the function defined by d 1 (•, ∂Q i ) in Q i , i ∈ N, and 0 elsewhere, is a W 1,∞ (Ω) solution to system (1.1). Since there are infinitely many Vitali's coverings of Ω, problem (1.1) admits infinitely many W 1,∞ (Ω) solutions. As a consequence, it is an interesting question to select and characterize a particular class of solutions or better one single solution.

Let S(Ω) denote the set of the W 1,∞ (Ω) solutions to problem (1.1). One possible strategy to select a specific function in S(Ω) can be developed using the theory of viscosity solution, introduced by Crandall and Lions (see [START_REF] Crandall | Viscosity solutions of Hamilton-Jacobi equations[END_REF][START_REF] Dacorogna | Implicit partial differential equations[END_REF]). In this approach one takes advantage of the various useful properties enjoyed by viscosity solutions, such as maximality, uniqueness and explicit formulas. To do that, note that system (1.1) is equivalent to (1.2) F (Du) = 0 , a.e. in Ω u = 0 , on ∂Ω , where F : R N → R is any continuous function which is zero if and only if |x i | = 1 for every i = 1, . . . , N . One therefore investigates the existence of a viscosity solution to problem (1.2) with any suitably chosen F . It is well known that for N = 1, u(x) = d 1 (x, ∂Ω) is the unique viscosity solution to problem (1.2). In higher dimensions, some geometrical conditions on the domain come into play. Indeed in [START_REF] Cardaliaguet | Geometric restrictions for the existence of viscosity solutions[END_REF] and [START_REF] Pisante | Sufficient conditions for the existence of viscosity solutions for nonconvex Hamiltonians[END_REF] it is proved that there exists a viscosity solution to (1.2) if and only if Ω is a hyperrectangle such that the normals to each face lie in E. In this case, d 1 (•, ∂Ω) is a viscosity solution. Although the previous result is in some sense negative, it was nonetheless put to use in [START_REF] Dacorogna | Viscosity solutions, almost everywhere solutions and explicit formulas[END_REF] to select a special solution to problem (1.1) (and to more general problems). There the authors construct a Vitali covering of Ω made up of domains Ω i admitting a viscosity solution and define the relative viscosity solution over each of these sets. The covering is built in a recursive way, with the idea of taking, at every step, the largest possible hyperrectangle.

Another strategy to characterize a class of functions in S(Ω) is to use a variational method, that is, to choose a meaningful functional over S(Ω) and to optimize it. The minimizers or maximizers, supposing they exist, would be selected solutions to problem (1.1). There is an evident difficulty to apply this method, since the set S(Ω) is not convex. For example the natural functionals

v → Ω |v| p , p ≥ 1,
have in general neither a minimizer nor a maximizer over S(Ω). Indeed any minimizing sequence converges to 0, which does not belong to S(Ω), and the limit function of maximizing sequences is plus or minus the distance from ∂Ω in the l 1 -norm, which usually do not verify (1.1).

In [START_REF] Dacorogna | Numerical methods for the solution of a system of eikonal equations with Dirichlet boundary conditions[END_REF] we find the first attempt of selection through a variational method. The authors study numerically a variational problem over the set of non-negative solutions to problem (1.1): they obtain a maximizing sequence for the problem (1.3) sup

Ω u, u ≥ 0, u ∈ S(Ω)
through the numerical minimization of

J (u) := - Ω u + 1 2 Ω |∇u| 2 + ε 2 Ω |∆u| 2 + 1 2ε N i=1 Ω ∂u ∂x i -1 2 .
Unfortunately, as we said above, there is in general no optimal solution to the variational problem (1.3). Nevertheless, according to the seminal idea presented in [START_REF] Dacorogna | Numerical methods for the solution of a system of eikonal equations with Dirichlet boundary conditions[END_REF] of selecting "regular solutions", it seems to be natural to minimize in some way the discontinuity set of the gradient of the solutions. With this aim in mind let us consider the functions v ∈ S(Ω) such that the distributional gradient of ∂v ∂xi , i = 1, . . . , N, has no Cantor part locally, that is, ∂v ∂xi is a SBV loc (Ω) function (see section 2 for further details). If J ∂v ∂x i denotes the jump set of ∂v ∂xi , i = 1, . . . , N , one could try to minimize the functional

J t (v) := H t N i=1 J ∂v ∂x i over E(Ω) = v ∈ S(Ω) : ∂v ∂x i ∈ SBV loc (Ω), i = 1, . . . , N ,
for some t ≥ N -1. As shown in [START_REF] Champion | A particular class of solutions of a system of eikonal equations[END_REF], for t = N -1, this variational problem is not well-posed, since J N -1 (v) can be infinite for every v ∈ E(Ω), in general. Indeed the jump set of Dv could have a fractal behaviour near the boundary of Ω. From the other hand, H t J ∂v ∂x i = 0 for every i = 1, . . . , N , for every function in E(Ω) and for every t > N -1 (see Section 3). The above negative result on the variational problem

inf H N -1 N i=1 J ∂v ∂x i , v ∈ E(Ω)
leads the authors of [START_REF] Champion | A particular class of solutions of a system of eikonal equations[END_REF] to consider a weighted H N -1 measure of the jumps. Using a given increasing sequence Ω n ⊂ Ω of polyhedra whose boundary is composed of a finite number of hyperplanes with normals lying in E, they define a C 0 function h : Ω → R + and minimize

v → N i=1 J ∂v ∂x i h(x)dH N -1 (x)
over E(Ω). It is obvious that this selection criterion depends on the definition of the function h and in turn, h depends on the particular sequence of sets Ω n . At the same time it is clear that the candidate functional should depend on some geometric properties of the domain Ω. Our aim in this paper is to find an "optimal" weighted measure of the jump set of Dv. We propose a family of weights which depends only on the distance from the boundary of Ω and then minimize the corresponding variational problem.

We assume that Ω ⊂ R 2 is a compatible domain, according to Definition 2.1. Observe that any convex, open, bounded, Lipschitz subset of R 2 such that ∂Ω is composed of a finite number of C 1 curves is compatible. A polygon is a compatible domain, as well. This hypothesis is motivated in section 3 through some enlightening examples. Our main result is the following Theorem 1.1. Let Ω be a compatible domain of R 2 . Let H : R + → R + be a continuous, increasing function such that

1 0 H(t) t dt < +∞ . Let F (v) = 2 i=1 Ω H(d 1 (x, ∂Ω))d D ∂v ∂x i .
Then the variational problem

(1.4) inf{F (v), v ∈ E(Ω)} has a solution.
We point out that the most difficult part of the proof of Theorem 1.1 consists in showing that the variational problem (1.4) is well-posed, that is, defining a function v ∈ E(Ω) such that F (v) is finite. This will be done in Proposition 4.4. In section 3 we sketch the proof of Proposition 4.4 in the special case Ω = (0, 1) × (0, 1) where the computations are much easier than in the general case; we motivate there the hypothesis on H. After proving that F is finite for at least one function in E(Ω), we show the existence of a minimizer using the direct methods of the calculus of variations, as in [START_REF] Champion | A particular class of solutions of a system of eikonal equations[END_REF]. We finally remark that the generalization of our result to higher dimensions is not immediate, because in that case Proposition 4.4 is much more complicated. This is why we state our main result in R 2 . However, we plan to address the problem in higher dimensions in a future work.

We will call T f a triangular domain. The class of all triangular domains will be denoted by T . We will write T instead of T f if the function f will be clear by the context.

In the sequel a special role is played by the family of lines in the plane with slope ±1. We will denote by ℓ ± the line with slope ±1, passing through (0, 0). Let L be the set of lines parallel to ℓ + or ℓ -. We define

S := {[x, y] : [x, y] ⊂ ℓ , ℓ ∈ L } ,
where [x, y] stands for the segment joining the two points x = (x 1 , x 2 ), y = (y 1 , y 2 ). We will denote by P the class of bounded Lipschitz domains whose boundary can be written as an at most countable union of segments lying in S; P f ⊂ P will be the class of domains whose boundary can be written as a finite union of segments in S; finally, P ∞ = P \ P f . For a given set A ⊂ R 2 , with • A we denote the relative interior of A with respect to the topology induced by the euclidean metric on R 2 . Let

R = 1 √ 2 -1 √ 2 1 √ 2 1 √ 2
and V = -1 0 0 1 be the π 4 counterclockwise rotation and the reflexion with respect to the vertical axis respectively. We define

T = {R 2k+1 (T ), V(R 2k+1 (T )), k = 0, 1, 2, 3, T ∈ T } . Let us consider the set E := {ν ∈ S 1 : |ν 1 | = |ν 2 |}. For a given rectifiable curve of class C 1 , γ ⊂ R 2 ,
we denote by N : γ → S 1 the Gauss map. Definition 2.1. We say that γ is an admissible boundary curve if the set N -1 (E) ⊂ γ has a finite number of connected components. We will denote by Γ the class of admissible boundary curves. A bounded connected Lipschitz set Ω ⊂ R 2 will be called compatible domain if its boundary ∂Ω is the union of a finite number of admissible boundary curves.

We recall that an open bounded set Ω ⊂ R 2 is Lipschitz if for every p ∈ ∂Ω there exist a radius r > 0 and a map h p : B r (p) → B 1 (0) such that h p is a bijection,

h p , h -1 p are Lipschitz functions, h p (∂Ω ∩ B r (p)) = Q 0 and h p (Ω ∩ B r (p)) = Q + , where B r (p) = {x ∈ R 2 : ||x -p|| ≤ r}, Q 0 = {x = (x 1 , x 2 ) ∈ B 1 (0) : x 2 = 0} and Q + = {x = (x 1 , x 2 ) ∈ B 1 (0) : x 2 > 0}. We remark that if Ω is Lipschitz, then H 1 (∂Ω) is finite.
We observe that any domain in P f is compatible, since any segment σ ∈ S is an admissible boundary curve, being

N -1 (E) = • σ connected. Any convex, open, bounded, Lipschitz subset of R 2 such that ∂Ω is composed of a finite number of C 1 curves is compatible. A polygon is also a compatible domain. Remark 2.2.
If Ω is a compatible domain, then it can be covered by a polygonal set in P f and a finite number of domains in T with mutually disjoint interior, i.e.,

Ω = P ∪ N j=1 W j (T j ),
where P ∈ P f , N ∈ N, T j ∈ T and W j is a linear transformation of the form R 2kj +1 or V(R 2kj +1 ), with k j ∈ {0, 1, 2, 3}.

As already explained in the introduction, in this article we minimize a weighted measure of the discontinuity set of the gradient of the solutions to problem (1.1).

To do that, we will use the definition and some results on the spaces BV (Ω) and SBV (Ω) of functions of bounded variation that we recall here. For simplicity we will restrict ourselves to the case where Ω is a subset of R 2 . We refer to [START_REF] Alberti | A note on the theory of SBV functions[END_REF][START_REF] Ambrosio | Functions of bounded variation and free discontinuity problems[END_REF][START_REF] Evans | Measure theory and fine properties of functions[END_REF][START_REF] Federer | Geometric measure theory[END_REF] for further details.

Definition 2.3 (BV, SBV function).

A BV (Ω) function is an L 1 (Ω) function such that its distributional gradient is a finite Radon measure. A SBV (Ω) function is a BV (Ω) function such that its gradient can be decomposed in the following way:

Dw = D a w + D j w = ∇w L 2 + (w + -w -)ν w H 1 ⌊J w = ∇wL 2 + [w]ν w H 1 ⌊J w
where D a w is absolutely continuous with respect to the Lebesgue measure L 2 in R 2 with density ∇w, D j w = [w]ν w H 1 ⌊J w is the jump part of Dw, H 1 is the onedimensional Hausdorff measure, w + and w -denote the upper and lower approximate limits of w, J w the jump set of w and ν w its generalized normal.

We shall handle the solutions v of problem (1.1) belonging to

E(Ω) = {v ∈ S(Ω) : v xi ∈ SBV loc (Ω), i = 1, 2} where v xi stands for ∂v ∂xi , i = 1, 2. Therefore (2.1) D (v xi ) = 2 ν vx i H 1 ⌊J vx i on ω ∀ i = 1, 2
for any open subset ω ⊂ ω ⊂ Ω. We recall moreover the following result:

Lemma 2.4. If P : R 2 → R is the projection (x 1 , x 2 ) → x 2 , and E is a measurable set of R 2 then R H 0 (E ∩ P -1 {y})dy ≤ H 1 (E).
Proof. It is sufficient to apply Theorem 2.10.25 of [START_REF] Federer | Geometric measure theory[END_REF] with

f = P , X = R 2 , Y = R, k = 0 and m = 1.
We end this section with some results on compactness and lower semi-continuity in the space of functions of bounded variation. Definition 2.5 (BV norm). The BV norm of a BV (Ω) function w is defined by

w BV (Ω) = w L 1 (Ω) + |Dw|(Ω) . Definition 2.6 (weak* convergence in BV ). Let (u n ) n , u ∈ BV (Ω). We say that (u n ) n weakly* converges to u in BV (Ω) if u n → u in L 1 (Ω) and the measures Du n weakly* converge to the measure Du in M(Ω, R 2 ), that is, lim n→∞ Ω ϕdDu n = Ω ϕdDu ∀ ϕ ∈ C 0 (Ω) .
Theorem 2.7 (compactness for SBV functions). Let Ω be a bounded open subset of R 2 with H 1 (∂Ω) < +∞. Let (u n ) n be a sequence of functions in SBV (Ω) and assume that: i) the functions u n are uniformly bounded in BV (Ω); ii) the gradients ∇u n are equi-integrable;

iii) there exists a function

f : [0, ∞) → [0, ∞] such that f (t)/t → ∞ as t → 0 + and Ju n f ([u n ])dH 1 ≤ C < ∞ ∀n ∈ N.
Then there exists a subsequence (u n k ) k and a function u ∈ SBV (Ω) such that u n k → u weakly* in BV (Ω), with the Lebesgue and jump parts of the derivatives converging separately, i.e., D a u n k → D a u and

D j u n k → D j u weakly* in M(Ω, R 2 ).
Theorem 2.8 (semicontinuity in BV ). Let Ω be a bounded open subset of R 2 . Let (u n ) n be a sequence of functions in BV (Ω) such that u n → u weakly* in BV (Ω).

Then Ω f (x)d|D j u|(x) ≤ lim inf n→∞ Ω f (x)d|D j u n |(x)
for any non-negative continuous function f : Ω → [0, +∞[ .

Statement of the main result and remarks

As already explained in the introduction, our strategy to select some particular or special solutions to (1.1) is based on a variational method, in other words we search for minimizers of a suitable and meaningful functional defined on E(Ω). We will motivate the choice of the functional and the conditions imposed on the domain Ω with the aid of some enlightening examples. Our main result is the following Theorem 3.1. Let Ω be a compatible domain and H : R + → R + be an increasing, continuous function such that

(3.1) 1 0 H(t) t dt < ∞ . Let F (v) = 2 i=1 Ω H(d 1 (x, ∂Ω))d(|D v xi |)(x) .
Then the variational problem

(3.2) inf{F (v) , v ∈ E(Ω)} has a solution.
One could ask why we do not minimize the simpler functionals

H t J vx 1 ∪ J vx 2 , t ≥ 1 over E(Ω).
To give the answer we will distinguish the cases t = 1 and t > 1. The first case was partially studied in [START_REF] Champion | A particular class of solutions of a system of eikonal equations[END_REF], where it was proved the following theorem for the functional

G Ω (v) := 2 i=1 Ω d(|Dv xi |)(x) .
Theorem 3.2. Let P ∈ P f . Then the variational problem

inf{G P (v), v ∈ E(P )} has a solution.
If Ω does not belong to P f , the minimization of G Ω may be not well-posed. Indeed we recall that in [START_REF] Champion | A particular class of solutions of a system of eikonal equations[END_REF] it was proved that if Ω = (0, 1) × (0, 1) then G Ω (v) is infinite for every v ∈ E(Ω). In Examples 3.3 and 3.4 we consider the case where Ω is a domain in P ∞ .

Example 3.3. We define a set Ω ∈ P ∞ such that for every v ∈ E(Ω) G Ω (v) = ∞.
We start fixing some notations. For a function f : [a, b] → R + we will denote by S f the set

S f = {(x 1 , x 2 ) ∈ R 2 : a ≤ x 1 ≤ b, 0 ≤ x 2 ≤ f (x 1 )}.
For a given N ∈ N, with the symbol g N (a,b) we will denote the step function

g N (a,b) = N -1 j=0 (b -a) N -j N χ [a+j b-a N ,a+(j+1) b-a N ) .
Let us consider the sequence of real numbers t 0 = 0, t n = n i=1

1 2 i = 2 n -1
2 n , and define for n ∈ N the functions

h n = 1 2 n χ [tn-1,tn) , g n = g Nn (tn-1,tn) , h = +∞ n=1 h n , f = +∞ n=1 (h n + g n ) ,
where N n will be chosen later. Let Ω be R(

• S f ).
We can easily see that, if we set

C n = R(S gn+ 1 2 n ), then • C i ∩ • C j = ∅ and R(S f ) = R(S h ) ∪ ∞ n=1 C n .
Thus the claim will be proved if we show that for every n ∈ N

H 1 J vx 1 ∩ • C n + H 1 J vx 2 ∩ • C n ≥ 1 .
Let σ n be the third middle part of the segment

R (t n-1 , f (t n-1 )) , R (t n , f (t n )) ; then H 1 (σ n ) = √ 2 3 2 n . For any 0 < t < H 1 (σn) 4
consider the segment σ t n = τ -t (σ n ), where τ w , w ∈ R, denotes the translation in the direction (0, w). Clearly we have

σ t n ⊂ •
Cn. Now, let r t (s) be a parametrization for σ t n . If w t (s) denotes the restriction of a function v ∈ E(Ω) to the segment σ t n , we have that

|w t (s)| ≤ d 1 (r t (s), ∂Ω) = d 1 (r t (s), ∂C n ) ≤ t + h n , h n = √ 2 2 n+1 N n .
This implies that w ′ t has at least H 1 (σn) 2(t+hn) jumps. Therefore, by Lemma 2.4

H 1 J vx 1 ∩ • Cn ≥ H 1 (σn ) 4 0 H 1 (σ n ) 2(t + h n ) dt ≥ H 1 (σn ) 4 0 H 1 (σ n ) 2(t + h n ) -1 dt = H 1 (σ n ) 2 ln H 1 (σ n ) 4h n + 1 - H 1 (σ n ) 4 = √ 2 3 2 n+1 ln √ 2 12 N n + 1 - √ 2 3 • 2 n+2 .
A suitable choice of N n implies the claim.

Example 3.4. We are going to define a domain Ω ∈ P ∞ and a function v ∈ E(Ω) such that G Ω (v) is finite. Let t n be defined as in the previous example. For any

n ∈ N choose h n < 1 2 n and consider the domain Ω = R( • S f ) with f defined by f = ∞ n=1 h n χ (tn-1,tn) . Let R n = [t n-1 , t n ] × [0, h n ] and define the function v(x) = d 1 (x, ∂R n ) if x ∈ R n . Clearly v is a solution to problem (1.1); moreover R(S f ) = ∪ ∞ n=1 R(R n ) and H 1 (J vx i ∩ R(R n )) ≤ 4 h n + 1 2 n ≤ 8 2 n .
It follows the following estimate:

G Ω (v) ≤ H 1 (J vx 1 ) + H 1 (J vx 2 ) ≤ ∞ n=1 H 1 (J vx 1 ∩ R(R n )) + H 1 (J vx 2 ∩ R(R n ) ≤ C ∞ n=1 1 2 n < ∞ ,
where C denotes a positive constant independent of n.

We now pass to the case t > 1. The reason why we do not use the functionals

H t J vx 1 ∪ J vx 2 for some t > 1, to select a solution to problem (1.1) is clear: for every v ∈ E(Ω), H t J vx 1 ∪ J vx 2 = 0 , ∀ t > 1.
Indeed, for a given v ∈ E(Ω)

H 1 J vx i ∩ ω < ∞ ,
for every ω ⊂ ω ⊂ Ω and for i = 1, 2. Recall now that Hausdorff measures have the property that if E is any measurable set such that H r (E) < +∞, then H r+ε (E) = 0 for every ε > 0 (see [START_REF] Evans | Measure theory and fine properties of functions[END_REF]). This implies that H t J vx i ∩ ω = 0 for every t > 1. By definition of measure H t J vx i = 0 for every t > 1. We remark that this argument can be applied also if we work in dimension greater than 2.

The previous observations suggest that, if one wants to isolate the most "regular" functions in E(Ω), he is in some sense obliged to use a weighted H 1 measure. In this article we define a weight depending only on the distance to the boundary.

In the following example we are going to motivate hypothesis (3.1) on H and to illustrate the ideas behind the proof of Theorem 3.1, in the case where Ω is a square. Due to the simple geometry of the domain, we can perform several steps of the proof avoiding almost all the technical difficulties that we need to deal with in the general case and that are addressed in the next section.

Example 3.5. Let Ω = (-1, 1) × (-1, 1). We are going to define a "reasonable" function v ∈ E(Ω) and impose that F (v) is finite. Since the weight H in the definition of F depends only on the distance from the boundary, our aim is to investigate how fast the discontinuity of ∂v ∂xi , i = 1, 2 develops near the boundary. With this information we will be able to define an appropriate weight H.

We define v as follows. Let R be the triangle with vertices in (-1, 1), (1, 1) and (0, 0). Then Ω is union of the counter-clockwise ± π 2 and π-rotations of R. Therefore it is sufficient to define v in R. We fill R with squares belonging to P f ; consequently they can be identified once we know the position of the upper vertex and the length of the diagonal. Let Ω 0 be the square in P f with upper vertex in (0, 1) and length of the diagonal equal to 1. For n ∈ N and i = 1, . . . , 2 n-1 let Q i n be the square in P f with upper vertex in 2i-1 2 n , 1 and length of the diagonal equal to 1 2 n . We set

Q -i n = V(Q i n )
. With this notation we can define the following covering of R made up by squares with mutually disjoint interior:

R = ∞ n=0 Ω n , Ω n := 2 n-1 i=1 Q i n ∪ Q -i n .
Observe that the north corner of each square belongs to γ = [(-1, 1), [START_REF] Alberti | A note on the theory of SBV functions[END_REF][START_REF] Alberti | A note on the theory of SBV functions[END_REF]]. Now we define the solution v as

v 0 (0) = d 1 (x, ∂Ω 0 )χ Ω0 , v i n (x) = d 1 (x, ∂Q i n )χ Q i n , v -i n (x) = d 1 (x, ∂Q -i n )χ Q -i n , v n (x) = 2 n-1 i=1 [v i n (x) + v -i n (x)] ; v(x) = ∞ n=0 v n (x).
It is clear from the definition that v ∈ E(Ω) and that the distributional gradient of Dv is supported on the sides and the diagonals of the squares we used to fill R. In order to estimate them, let us fix some notations. Let

R m = (x 1 , x 2 ) ∈ R : 1 m + 1 < x 2 ≤ 1 m .
We denote by:

• S the union of the sides of the squares Q ±i n , • D v the union of the vertical diagonals of the squares Q ±i n , • D h the union of the horizontal diagonals of the squares Q ±i n , for n ∈ N and i = 0, . . . , 2 n-1 . Therefore F (v) is finite if the three following quantities

D h H(d 1 (x, γ))dH 1 , Dv H(d 1 (x, γ))dH 1 , S H(d 1 (x, γ))dH 1
are finite. The first quantity is very simple to estimate, if one remarks that the horizontal diagonals appear only at heigth 1 2 n , n ∈ N, and their total length is 1 for every n ∈ N. Therefore

D h H(d 1 (x, γ))dH 1 ≤ ∞ n=1 H 1 2 n .
Using the Cauchy's condensation criterion, the last series is finite if and only if

(3.3) ∞ n=1 H 1 n 1 n < ∞ .
To estimate the sides and the vertical diagonals we will use a different strategy. We observe that the number N m of squares which intersect R m is bounded by m + 1. Indeed if x 0 is the north corner of a square intersecting R m , the distance from x 0 to the next north corner of a square intersecting R m is at least 2 m+1 . Therefore

N m = H 1 (γ) 2 m+1 = m + 1 .
Now, let Q be a square intersecting R m . The the H 1 measure of the intersection of the vertical diagonal of Q with R m is less or equal to 1 m -1 m+1 = 1 m(m+1) and the H 1 measure of the intersection of one of the sides of Q with R m is bounded by

√ 2 1 m -1 m+1 = √ 2 m(m+1) . Therefore Dv H(d 1 (x, γ))dH 1 ≤ ∞ m=1 H 1 m N m 1 m(m + 1) ≤ ∞ m=1 H 1 m 1 m and S H(d 1 (x, γ))dH 1 ≤ 2 ∞ m=1 H 1 m N m √ 2 m(m + 1) ≤ 2 √ 2 ∞ m=1 H 1 m 1 m .
Hence we find the same condition on H as in (3.3):

∞ m=1 H 1 m 1 m < ∞ .

Proof of the main result

In this section we are going to prove Theorem 3.1. The proof can be divided into two parts: in the first we show that the variational problem (3.2) is well-posed (see Proposition 4.4); in the second one we prove that there exists a minimizer of the functional F in E(Ω).

We are going to concentrate on the first step, that is, we are going to construct a function v ∈ E(Ω) such that F (v) is finite. For this purpose, recalling that Ω is a compatible domain, according to Remark 2.2, we have W j (T j )

where P ∈ P f , T j ∈ T and W j (T j ) = R 2kj +1 (T j ) or V(R 2kj +1 (T j )) for some

k j ∈ {0, 1, 2, 3}.
With the aim of setting v in each domain W j (T j ) we will define a special countable covering of the interior of a triangular domain made up by squares. We start by introducing three operators defined in T . For a given

T = {(x 1 , x 2 ) : a ≤ x 1 ≤ b, h(b) ≤ x 2 ≤ h(x 1 )} ∈ T , let x 0 1 be such that h(x 0 1 ) = x 0 1 + h(b) -a and define q(T ) := (a, x 0 1 ) × (h(b), h(b) + x 0 1 -a); u(T ) := {(x 1 , x 2 ) ∈ T : a < x 1 < x 0 1 , h(b) + x 0 1 < x 2 < h(x 1 )}; r(T ) := {(x 1 , x 2 ) ∈ T : a + x 0 1 < x 1 < b, h(b) < x 2 < h(x 1 )}.
We explicitly observe that u and r have values in T while q maps any triangular domain to a square in R(P f ). be the set of all the m-permutations of the two letters u and r. For σ ∈ S m , using the notation

σ(T ) = α 1 • α 2 • • • • • α m (T ), we set Q m,σ T = q(σ(T )) ; σ ∈ S m .
We finally define the following family of squares contained in T :

Q(T ) := {Q m,σ T : m ∈ N ∪ {0} , σ ∈ S m }.
Remark 4.2. We remark that Q(T ) is a covering of • T composed of squares with mutually disjoint interiors belonging to R(P f ). It may be useful to think of Q(T ) as being constructed in steps, starting from m = 1 and adding at step m the squares Q m,σ T with σ ∈ S m . Since the cardinality of S m , ♯(S m ), is equal to 2 m , we add 2 m squares at the m-th step. Therefore the first steps of the construction are:

Step 0. We start with Q 0 T = q(T ) ; Step 1.

we add Q 1,u T = q(u(T )) and

Q 1,r T = q(r(T )) ; Step 2. we add Q 2,(u,u) T = q(u(u(T ))), Q 2,(u,r) T = q(u(r(T ))), Q 2,(r,u) T = q(r(u(T ))), Q 2,(r,r) T = q(r(r(T )))
Step 3. we add Q 3,(u,u,u) T = q(u(u(u(T )))), Q 3,(u,u,r) T = q(u(u(r(T )))),

• • • • • • In the sequel, when ♯(S m ) will play a role, we will use for

Q m,σ T the notation Q m,k T , k ∈ {1, . . . , 2 m }.
We are now in a position to define the candidate function v : Ω → R + such that F (v) is finite, which is the natural generalization of the construction made in Example 3.5.

Definition 4.3 (of the candidate function).

Let w be a minimizer of the functional G P . For a fixed j ∈ {1, . . . , N }, let

v j m (x) = σ∈Sm d 1 x, ∂ W j (Q m,σ Tj ) χ Q m,σ T j (x) ; v j (x) = m∈N v j m (x).
We define

v(x) = w(x)χ P (x) + N j=1 v j (x).
In the following we will prove that problem (3.2) is well-posed. We remark that in Ω \ P , Dv x1 and Dv x2 are supported on the sides of each square W j (Q m,σ Tj ) and on its vertical and horizontal diagonals respectively. For this reason, we denote by D + T and D - T the union of all the diagonals of the squares in Q(T ) parallel to ℓ + and ℓ -respectively, and by S T the union of all the sides of the squares in Q(T ). Proof. First we note that according to (4.1) we can estimate F (v) as follows:

F (v) = 2 i=1 P H(d 1 (x, ∂Ω))d|D w xi |(x) + 2 i=1 Ω\P H(d 1 (x, ∂Ω))d|D v xi (x)|(x) ≤ 2 i=1 P C d|D w xi |(x) + 2 i=1 N j=1 Wj (Tj ) H(d 1 (x, ∂Ω))d|D v xi (x)|(x)
where we have used that for some positive constant C, H(d 1 (x, ∂Ω)) ≤ C for all x ∈ Ω.

The first term of the last estimate is finite, due to the choice of w. Therefore it is sufficient to estimate

2 i=1 N j=1 Wj (Tj ) H(d 1 (x, ∂Ω))d|D v xi (x)|(x) .
The construction of v, the properties of the rigid motion W j and the duality between the metrics d 1 and d ∞ imply that we can restrict ourselves to prove that there exists a positive constant M such that (4.2)

ST j ∪D - T j ∪D + T j H(d ∞ (x, W -1 j (∂Ω)))dH 1 ≤ M ∀ j ∈ {1, . . . , N }.
The last estimate is a consequence of Lemmata 4.8, 4.6 and 4.7.

In the next lemmata we aim to prove (4.2). We will estimate separately the integral on the sets S Tj , D - Tj and D + Tj . From now on we will work on a single triangular domain T j ; since the proofs are independent of j, we will simply write T instead of T j . Up to a dilatation and a rigid motion, we can assume in the sequel that

T = {(x 1 , x 2 ) : 0 ≤ x 1 ≤ 1, 0 ≤ x 2 ≤ h(x 1 )} .
with h(1) = 0, h(0) ≤ 1 and h ′ (t) < 0 for every t ∈ (0, 1). We will denote by γ the curve W -1 j (∂Ω) and we observe explicitly that the graph of the function h is a proper subset of γ. We define

L n = x ∈ T : 1 n + 1 < d ∞ (x, γ) ≤ 1 n , for n ∈ N. Let Q ∈ Q(T )
having a not empty intersection with L n and assume that Q touches γ in (x 1 , h(x 1 )). Let e n (Q), n n (Q), d n (Q) denote respectively the length of the intersection of the east side, north side, diagonal parallel to ℓ + of Q with L n . Then it is easily to verify that

(4.3) e n (Q) ≤ h x 1 + 1 n + 1 -h x 1 + 1 n + 1 n(n + 1) , (4.4) n n (Q) ≤ h -1 h (x 1 ) + 1 n + 1 -h -1 h (x 1 ) + 1 n + 1 n(n + 1) and (4.5) d n (Q) ≤ 1 n(n + 1)
.

Lemma 4.5. Let N n be the number of squares in Q(T ) which intersect L n . Then there exists a positive constant c depending only on γ such that

(4.6) N n ≤ c(n + 1).
Proof. Let (x 0 1 , x 0 1 ) be the north-east corner of q(T ). Define

x j 1 = h -1 x 0 1 - j 1 + n , j = 1, . . . , m ≤ (n + 1) .
Fix i ∈ {0, . . . , m} and note that if on the portion of γ lying between (x i 1 , h(x i 1 )) and (x i+1 1 , h(x i+1 1 )) there are two points, say p and q, which are north-east vertex of squares,

Q p and Q q in Q(T ), then L n ∩ Q p or L n ∩ Q q is empty.
Consequently the number of squares in r(T ) which intersect L n can be at most n + 1. With the same argument one can prove that the same estimate holds for the number of squares in u(T ) which intersect L n . Lemma 4.6. There exists a positive constant M such that

D + T H(d ∞ (x, γ))dH 1 ≤ M.
Proof. The proof easily follows from (4.5), (4.6) and the hypotheses on H. Indeed one has

D + T H(d ∞ (x, γ))dH 1 ≤ C ∞ n=1 H 1 n 1 n < +∞ ,
for some positive constant C independent of n.

Lemma 4.7. There exists a positive constant M such that (4.7)

D - T H(d ∞ (x, γ))dH 1 ≤ M.
Proof. If L i denotes the total length of the diagonals parallel to ℓ -of the squares added at the i-th step of the construction of v, then (4.8)

L i ≤ 2H 1 (γ).
Indeed L i is nothing but the length of the non flat parts of the graph of a piecewise affine function defined on the interval (0, 1) and whose derivative is 0 or -1 everywhere but on a finite number of points.

The proof is divided into several steps. We will denote by r n the length of the side of

Q n,σ T with σ = (r, . . . , r n ) and R n = n i=1 r i ; u n will denote the length of the side of Q n,σ T with σ = (u, . . . , u n ).
Step 1. Let h ∈ C 1 ((0, 1)). Assume that there exists 0 < ε < 1 such that -1 + ε < h ′ (t) < 0 for every t ∈ (0, 1). Let d - n be the intersection of L n with the diagonal parallel to ℓ -of a square

Q ∈ Q(T ). Let x Q 1 = min (x1,x2)∈d - n x 1 . Assume that (x Q 1 , x Q 2 ) ∈ d - n . Then h x Q 1 + 1 n - 1 n ≤ x Q 2 ≤ h x Q 1 + 1 n + 1 - 1 n + 1 , since (x Q 1 , x Q 2 ) ∈ L n . The intersection between d - n , belonging to the line x 2 = -x 1 + x Q 1 + x Q 2 ,
and the lower boundary of L n , that is,

x 2 = h x 1 + 1 n -1 n , gives x 1 -x Q 1 = x Q 2 + 1 n -h x 1 + 1 n ≤ h x Q 1 + 1 n + 1 - 1 n + 1 + 1 n -h x 1 + 1 n .
Lagrange's theorem implies that

x 1 -x Q 1 ≤ |h ′ (ξ)| x 1 -x Q 1 + 1 n - 1 n + 1 + 1 n - 1 n + 1 for some ξ ∈ x Q 1 + 1 n+1 , x 1 + 1 n .
Using the hypothesis on h we get

x 1 -x Q 1 ≤ (1 -ε) x 1 -x Q 1 + 1 n 2 + 1 n 2 ,
that is,

x 1 -x Q 1 ≤ 2 -ε ε 1 n 2 . This implies that H 1 (d - n ) ≤ √ 2 2 -ε ε 1 n 2 for a given square Q such that Q ∩ L n = ∅.
Using estimate (4.6) on the number of squares intersecting L n , the previous inequality gives

D - T H(d ∞ (x, γ))dH 1 ≤ c 1 (ε, γ) ∞ n=1 H 1 n 1 n 2 (n + 1) ≤ c 2 (ε, γ) ∞ n=1 H 1 n 1 n .
where c 1 (ε, γ) and c 2 (ε, γ) denote two positive constants depending only on γ and ε. Hypothesis (3.1) on H implies that the last sum is finite.

The case where -1 + ε < h ′ (t) < 0 for every t ∈ (0, 1) can be handled in a similar way.

Step 2. Assume that there exists 0 < ε < 1 such that |h ′ (t) + 1| ≤ ε for every t ∈ [0, 1]. We are going to prove by induction that the length of the side l i,k , k = 1, . . . , 2 i of any square Q i,k T added at the i-th step of the covering of Definition 4.1 satisfies:

(4.9) 1 (2 + ε) i ≤ l i,k ≤ 1 (2 -ε) i ∀ k = 1, . . . , 2 i .
To to that, observe that the length of the side l of q( T ), for a given domain T

= {(x 1 , x 2 ) : a < x 1 < b, c < x 2 < h(x 1 )}, can be estimated by (4.10) max{b -a, h(a) -c} 2 + ε ≤ l ≤ min{b -a, h(a) -c} 2 -ε .
Indeed, it is sufficient to compute the intersections between the lines

x 2 = x 1 + c-a and x 2 = (-1 ± ε)(x 1 -b) + c or x 2 = (-1 ± ε)(x 1 -a) + h(a)
. This implies that the length l 0,1 of the side of q(T ) satisfies

(4.11) 1 2 + ε ≤ l 0,1 ≤ 1 2 -ε .
Now, suppose that estimate (4.9) holds for i-1. At step i we add 2 i squares and any of these is confined in a domain belonging to T with one of his sides that coincides with the side of one of the squares added in the previous step. Thus, to prove that estimate (4.9) holds for i, it suffices to use (4.10) with

1 (2+ε) i-1 ≤ l ≤ 1 (2-ε) i-1 . We need to estimate the l ∞ -distance from γ of the diagonal d -parallel to ℓ - of a square Q ∈ Q(T ). Assume that the length of the side of Q is l. Let x Q = (x Q 1 , x Q 2 )
∈ γ be the north-east corner of Q. Then the l ∞ -distance from γ of d -is smaller than the l ∞ -distance of d -from the piecewise affine function

x 2 (x 1 ) = (-1 + ε)(x 1 -x Q 1 ) + x Q 2 , if x 1 > x Q 1 (-1 + ε)(x 1 -x Q 1 ) + x Q 2 , if x 1 ≤ x Q 1 .
This gives (4.12)

d ∞ (d -, γ) ≤ 1 + ε 2 -ε l.
Using the previous estimates and (4.8) we obtain

D - T H(d ∞ (x, γ)) dH 1 ≤ ∞ n=1 L n H 1 + ε (2 -ε) n+1 ≤ 2H 1 (γ) ∞ n=1 H 1 + ε (2 -ε) n+1 .
In order to prove that the last sum is finite, we start by observing that, by the monotonicity of H,

H 1 + ε (2 -ε) n+1 ≤ H 2 [(2 -ε) n ]
.

We now let

u n = [(2 -ε) n ] and a n = H 2 n
1 n and we apply the Schlömilch's generalization of the condensation criterion for series to deduce the desired convergence. We recall that, if a n is a positive non increasing sequence of real numbers and u n a strictly increasing sequence of natural numbers such that for some positive constant C (4.13)

u n+1 -u n u n -u n-1 ≤ C ∀ n ∈ N then ∞ n=0 a n is finite if and only if ∞ n=0 (u n+1 -u n )a
c(ε)[(2 -ε) n ] ≤ u n+1 -u n ≤ (2 -ε) n+1 + 1 -(2 -ε) n ≤ [(2 -ε) n ] + 1 (1 -ε) + 1 ≤ [(2 -ε) n ] + 2 -ε
for some positive constant c(ε) independent of n. This implies that hypothesis (4.13) is satisfied. Therefore there exists a positive constant C such that

∞ n=0 H 2 [(2 -ε) n ] = ∞ n=0 H 2 [(2 -ε) n ] 1 [(2 -ε) n ] [(2 -ε) n ] ≤ C + 1 c(ε) ∞ n=n(ε) (u n+1 -u n )a un < ∞ .
Step 3. Assume that h ∈ C 1 ([0, 1]) and fix 0 < ε < 1. The uniform continuity of h ′ implies that there exists δ > 0 such that if |t -s| < δ then |h ′ (t)h ′ (s)| < ε/4. Let n ′ ∈ N be such that the length of the side of Q n ′ ,σ T is less than δ for every σ ∈ S n ′ . Let (x σ 1 , x σ 2 ) be the north-east corner of

Q n ′ ,σ T . For any σ ∈ S n ′ , if |h ′ (x σ 1 ) + 1| ≤ ε/2, then |h ′ (x 1 ) + 1| ≤ |h ′ (x 1 ) -h ′ (x σ 1 )| + |h ′ (x σ 1 ) + 1| < ε and if |h ′ (x σ 1 ) + 1| > ε/2, then |h ′ (x 1 ) + 1| ≥ |h ′ (x σ 1 ) + 1| -|h ′ (x 1 ) -h ′ (x σ 1 )| > ε 4 . Let N (n ′ ) = 1 + n ′ i=0 2 i = 2 n ′ +1
. Using (4.8) we have

D - T H(d ∞ (x, γ)) dH 1 ≤ n ′ H 1 (γ) + N (n ′ ) i=1 D - T i H(d ∞ (x, γ))
where, up to dilatations, T i , with i = 1 . . . N (n ′ ), are triangular domains satisfying the hypotheses of Step 1 or 2. This implies that D - T H(d ∞ (x, γ)) dH 1 is finite.

Step 4. Assume that h ∈ C 1 ((0, 1)) and lim

t→0 h ′ (t) = -∞ or lim t→1 h ′ (t) = -∞.
Surely there exists n ′′ ∈ N such that the intervals [0,

u n ′′ ], [R n ′′ , 1] do not contain any x 1 such that h ′ (x 1 ) = -1. Let N (n ′′ ) = n ′′ i=0 2 i -2 = 2 n ′′ +1 -3.
Using (4.8) we get the following estimate:

D - T H(d ∞ (x, γ)) dH 1 ≤ n ′′ H 1 (γ) + N (n ′′ ) i=1 D - T i H(d ∞ (x, γ))
where, up to dilatations, T i , with i = 1 . . . N (n ′′ ), are triangular domains satisfying the hypotheses of Step 3. This implies that

D - T H(d ∞ (x, γ)) dH 1 is finite. Lemma 4.8. There exists a positive constant M such that ST H(d ∞ (x, γ))dH 1 ≤ M.
Proof. First we observe that, since T is a triangular domain, up to inverting the coordinate axes, we can assume that we are in one of the following cases:

Case 1. There exist two constants c 1 , c 2 > 0 such that -c 1 ≤ h ′ (t) ≤ -c 2 < 0 for every t ∈ [0, 1].

Case 2. There exists a constant c 1 < 0 such that c 1 ≤ h ′ (t) < 0 for every t ∈ (0, 1) and h ′ (1) = 0. Case 3. There exists a constant c 1 < 0 such that c 1 ≤ h ′ (t) < 0 for every t ∈ (0, 1) and lim t→1 h ′ (t) = -∞.

Step 1. Assume that we are in the hypotheses of Case 1. We observe that h and h -1 are Lipschitz functions, say with Lipschitz constant c; then (4.3) and (4.4) together with estimate (4.6) imply that

ST H(d ∞ (x, γ))dH 1 ≤ ∞ n=1 H 1 n c n(n + 1) c(n + 1)
where c depends only on γ. The last sum is finite due to hypothesis (3.1) on H.

Step 2. Assume that we are in the hypotheses of Case 2. It is sufficient to estimate

ST ∩r(T ) H(d ∞ (x, γ))dH 1 , since Step 1 implies that ST ∩u(T ) H(d ∞ (x, γ))dH 1
is bounded. To this purpose, we define the following sequence of points. Let x 0 1 ∈ (0, 1) be such that h(x 0 1 ) = x 0 1 . We set

x j 1 = h -1 x 0 1 - j n , j = 1, . . . , [x 0 1 n] =: M n .
There exists at most one point (x j 1 , h(x j 1 )) with

x j 1 ≤ xj 1 < x j+1 1 , j = 1, . . . , M n -1, which is the north-east vertex of a square Q j ∈ Q(T ) such that L n ∩ Q j is not empty. For such a square, since |h ′ | is bounded, one has (4.14) e n (Q j ) ≤ 1 + sup x1∈[0,1] |h ′ (x 1 )| 1 n 2 ≤ C 1 1 n 2
for some positive constant C 1 . As well

n n (Q j ) ≤ 1 n(n + 1)   1 + sup x2∈[h(x j 1 )+ 1 n+1 ,h(x j 1 )+ 1 n ] |(h -1 ) ′ (x 2 )|   ≤ 1 n 2 1 + sup x2∈[h(x j+1 1
),h(x j-1

1

)]

1 |h ′ (h -1 (x 2 ))| ≤ 1 n 2 1 + sup x2∈[h(x j+1 1
),h(x j 1 )]

1 |h ′ (h -1 (x 2 ))| + sup x2∈[h(x j 1 ),h(x j-1 1 )] 1 |h ′ (h -1 (x 2 ))| .
We remark that Mn j=1

1 n sup x2∈[h(x j 1 ),h(x j-1 1 )] 1 |h ′ (h -1 (x 2 ))| is a particular Riemann sum for x 0 1 -1 n 1 n 1 |h ′ (h -1 (x 2 ))| dx 2
which is finite. Therefore there exists n 0 ∈ N such that (4.15)

Mn j=1 1 n sup x2∈[h(x j 1 ),h(x j-1 1 )] 1 |h ′ (h -1 (x 2 ))| ≤ x 0 1 0 1 |h ′ (h -1 (x 2 ))| dx 2 + 1 ≤ C 2
for every n ≥ n 0 , for some positive constant C 2 independent of n. Estimates (4.14) and (4.15) give

S∩r(T ) H(d ∞ (x, γ))dH 1 ≤ ≤ ∞ n=n0 H 1 n Mn j=1 2 n 2 sup x2∈[h(x j 1 ),h(x j-1 1 )] 1 |h ′ (h -1 (x 2 ))| + C 1 + 1 n 2 + + n0 n=1 H 1 n H 1 (S ∩ r(T ) ∩ L n ) ≤ C 2 ∞ n=1 H 1 n 1 n + (C 1 + 1) ∞ n=1 H 1 n 1 n + C 3 ,
where we have used that for n ≤ n 0 , H

1 n ≤ H 1 n 0 and H 1 (S ∩ r(T ) ∩ L n ) is finite, since v xi , i = 1, 2 is SBV loc (T )
. The last sum is finite due to hypothesis (3.1) on H.

Step 3. Assume that we are in the hypotheses of Case 3. As in the previous step it is sufficient to estimate

ST ∩r(T ) H(d ∞ (x, γ))dH 1 .
Let x 0 1 ∈ (0, 1) be such that h(x 0 1 ) = x 0 1 . Let us consider the following sequence of points:

x j 1 = x 0 1 + j n , j = 1, . . . , M n = [n(1 -x 0 1 )] .
There exists at most one point (x j 1 , h(x j 1 )) with x j 1 ≤ xj 1 < x j+1 1 , j = 1, . . . , M n -1, which is the north-east vertex of a square Q j ∈ Q(T ) such that L n ∩ Q j is not empty.

For such a square, since |(h -1 ) ′ | is bounded, one has (4.16)

n n (Q j ) ≤ 1 + sup x1∈[0,h(0)] (h -1 ) ′ (x 1 ) 1 n 2 ≤ C 1 1 n 2
for some positive constant C 1 . As well for every n ≥ n 0 , for some positive constant C 2 independent of n. Therefore, arguing as in the previous step, we have

e n (Q j ) ≤ 1 n(n + 1)   1 + sup x1∈[x j 1 + 1 n+1 ,x j 1 + 1 n ] |h ′ (x 1 )|   ≤ 1 n 2 1 + sup x1∈[x j 1 ,x j+2 1 ] |h ′ (x 1 )| ≤ 1 n 2 1 + sup x2∈[x j 1 ,x j+1
S∩r(T ) H(d ∞ (x, γ))dH 1 ≤ ∞ n=n0 H 1 n Mn j=1 2 n 2 sup x1∈[x j 1 ,x j+1 1 ] |h ′ (x 1 )| + C 1 + 1 n 2 + n0 n=1 H 1 n H 1 (S ∩ r(T ) ∩ L n ) ≤ C 2 ∞ n=1 H 1 n 1 n + (C 1 + 1) ∞ n=1 H 1 n 1 n + C 3 < ∞ ,
where C 3 denotes a positive constant independent of n.

We have therefore shown that the functional F is well-defined. We are now in position to show Theorem 3.1. The proof follows quite easily from the direct methods of the calculus of variations. Note that this technique was already used in [START_REF] Champion | A particular class of solutions of a system of eikonal equations[END_REF]. We start recalling a lemma proved in [START_REF] Champion | A particular class of solutions of a system of eikonal equations[END_REF]. Proof. (of Theorem 3.1) Let (v n ) n ⊂ E(Ω) be a minimizing sequence. Lemma 4.9 assures that (v n ) n is uniformly bounded in L ∞ (Ω). Moreover v n is uniformly Lipschitz in Ω since Ω is Lipschitz and |∇v n | ≤ √ 2 a.e. in Ω for every n. Therefore, up to a subsequence, v n → v ∞ in C 0 (Ω) and v n → v ∞ weakly* in W 1,∞ (Ω) for some v ∞ ∈ W 1,∞ 0 (Ω). We are now going to show that v ∞ belongs to E(Ω). Since (v n ) n is a minimizing sequence, there exists C > 0 such that F (v n ) ≤ C for every n ∈ N. For a fixed m ∈ N we can say that Let us fix i ∈ {1, 2}. Note that ∂v n ∂xi ∈ SBV (B m ) and takes only the two values ±1 for every n. This implies that

C ≥ F (v n ) ≥ α(m)
∂v n ∂x i BV (Bm) = L 2 (B m ) + 2H 1 (J v n x i ∩ B m ) ≤ L 2 (B m ) + 2C.
We can apply Theorem 2.7 to the sequence ∂v n ∂x i n :

• hypothesis i) has been verified in the previous estimate;

• hypothesis ii) is verified as ∇ ∂v n ∂xi = 0 a.e. in Ω; • hypothesis iii) can be verified choosing f ≡ 1: in this way Therefore lim inf n→∞ F (v n ) ≥ F (v ∞ ), i.e., v ∞ minimizes F .

J v n x i ∩Bm f ∂v n ∂x i dH 1 (x) = H 1 (J v n x i ∩ B m ) ≤ C.

Definition 4 . 1 (

 41 of the covering of T ). Let for m ∈ N S m := {σ = (α 1 , . . . , α m ) : α i ∈ {u, r}} ,

Proposition 4 . 4 .

 44 The function v ∈ E(Ω) defined in 4.3 satisfies F (v) < +∞.

  un is finite. In our case we observe that the convergence of the series ∞ n=0 a n is assured by hypothesis (3.1) on H. Moreover for every n ≥ n(ε)

1 ]

 1 t)|dt + 1 ≤ C 2

Lemma 4 . 9 .

 49 Let Ω be an open bounded connected subset of R N with Lipschitz boundary. Then -d 1 (•, ∂Ω) ≤ v ≤ d 1 (•, ∂Ω) on Ω for every function v ∈ S(Ω).

  n xi |)(x), ∀ n ∈ N where B m = x ∈ Ω : d 1 (x, ∂Ω) > 1 m .

H(d 1

 1 (x, ∂Ω) d(|Dv ∞ xi |)(x) = Ω H(d 1 (x, ∂Ω) d(|Dv ∞ xi |)(x) .

  Consequently∂v n ∂x i → g i weak* in BV (B m ) for someg i ∈ SBV (B m ) . Since ∂v n ∂x i → ∂v ∞ ∂x i weak* in L ∞ (Ω), we infer that ∂v ∞ ∂x i ∈ SBV (B m ). Moreover ∂v n ∂x i → ∂v ∞ ∂x i in L 1 (B m) and so ∂v ∞ ∂x i = 1 a.e. in B m . This being true for every i and for every B m , we deduce that v ∞ belongs to E(Ω).To show that v ∞ is a minimizer of F , we remark that lim inf

	Bm
	Bm

n→∞ Bm

H(d 1 (x, ∂Ω))d(|Dv n xi |)(x) ≥ Bm H(d 1 (x, ∂Ω))d(|Dv ∞ xi |)(x)

for every B m , due to Theorem 2.8. This implies that lim inf n→∞ Ω H(d 1 (x, ∂Ω) d(|Dv n xi |)(x) ≥ sup

Acknowledgements We would like to thank T. Champion for some useful discussions. Part of this work was completed during a visit of the first author to Seconda Università di Napoli and of the second author to the Hokkaido University whose hospitality is gratefully acknowledged.


Gisella Croce
email: gisella.croce@univ-lehavre.fr

AN ELLIPTIC PROBLEM WITH DEGENERATE COERCIVITY AND A SINGULAR QUADRATIC GRADIENT LOWER ORDER TERM

Keywords: 1991 Mathematics Subject Classification. 35J15, 35J25, 35J66, 35J70, 35J75 Nonlinear elliptic problems, Dirichlet condition, degenerate coercivity, distributional solution, singular lower order term, quadratic growth

In this paper we study a Dirichlet problem for an elliptic equation with degenerate coercivity and a singular lower order term with natural growth with respect to the gradient. The model problem is

where Ω is an open bounded set of R N , N ≥ 3 and p, θ > 0. The source f is a positive function belonging to some Lebesgue space. We will show that, even if the lower order term is singular, it has some regularizing effects on the solutions, when p > θ -1 and θ < 2.

Introduction

In this paper we study the following problem:

(1)

   -div b(x) (1 + |u|) p ∇u + B |∇u| 2 |u| θ = f in Ω, u = 0 on ∂Ω,
where Ω is an open bounded set of R N , N ≥ 3, B, p > 0 and θ > 0. We assume that b : Ω → R is a measurable function such that for some positive constants α and β

(2) α ≤ b(x) ≤ β for a.e. x ∈ Ω .

Moreover f is a positive function belonging to some Lebesgue space L m (Ω), with m ≥ 1. We point out three characteristics of this problem: the operator A(v) = -div b(x) (1 + |v|) p ∇v is defined on H 1 0 (Ω) but is not coercive on this space when v is large, as proved in [20]. The lower order term has a quadratic growth with respect to the gradient and is singular in the variable u. As we will see, existence and summability of solutions to problem (1) depend on these features.

It is known that the degenerate coercivity has in some sense a bad effect on the summability of the solutions to problem (3)

-div (a(x, u)∇u) = f in Ω, u = 0 on ∂Ω, as proved in [9]. There f ∈ L m (Ω) was not assumed to be positive, a : Ω × R → R was a Carathéodory function such that α (1 + |s|) p ≤ a(x, s) ≤ β, for p ∈ (0, 1) and α, β > 0. Apart from the case where m > N 2 , the summability of the solutions is lower than the summability of the solutions to elliptic coercive problems. Indeed, in [9] it is shown that if 2N N + 2 -p(N -2) < m < N 2 there exists a

H 1 0 (Ω)∩L r (Ω) distributional solution, with r = N m(1 -p) N -2m ; if N N + 1 -p(N -1) < m < 2N N + 2 -p(N -2)
, there exists a W 1,s 0 (Ω) distributional solution, with s = N m(1 -p) N -m(1 + p)

. For p > 1 the authors prove a non-existence result for constant sources f . Note that a bad effect on the regularity of the solutions appears even when the right hand side of (3) is an element of H -1 (Ω), such as -div(F ), with F ∈ L 2 (Ω). As a matter of fact, in this case the solutions are in general not in H 1 0 (Ω) (see [16]). The presence of lower order terms can have a regularizing effect on the solutions. In [7] and [14] three kinds of lower order terms are considered for elliptic problems with degenerate coercivity, with no restriction on p. In the first paper the author analyses a lower order term defined by a Carathéodory function g : Ω × R × R N → R N with the following properties. There exists d ∈ L 1 (Ω), two positive constants µ 1 , µ 2 > 0 and a continuous increasing real function h such that g(x, s, ξ)s ≥ 0, µ 1 |ξ| 2 ≤ |g(x, s, ξ)| when |s| ≥ µ 2 and |g(x, s, ξ)| ≤ d(x)h(|s|)|ξ| 2 . It is proved that for a L 1 (Ω) source there exists a H 1 0 (Ω) distributional solution to -div (a(x, u)∇u) + g(x, u, ∇u) = f in Ω, u = 0 on ∂Ω .

This proves that the summability of the gradient of the solutions is much larger than that one of the solutions of problem (3). It is even larger than the summability of the gradient of the solutions to elliptic coercive problems with L 1 (Ω) sources, which is L s (Ω) for every s < N N -1 (see [10] for example). We remark moreover that the lower order term gives the existence of a solution for p ≥ 1; for these values of p, (3) has no solution.

In a previous article [14] we consider two kinds of lower order terms h(u). For h(u) = |u| q-1 u, with q > p + 1, we stablish the existence of a distributional solution u ∈ W 1,t 0 (Ω) ∩ L q (Ω) , t < 2q p + 1 + q , for any L 1 (Ω) source f . If f ∈ L m (Ω), m > 1 and q ≥ p + 1 m -1 then there exists a distributional solution u in H 1 0 (Ω) ∩ L qm (Ω) .

If p + 1 2m -1 < q < p + 1 m -1
, there exists a distributional solution u in W 1, 2qm p+1+q 0

(Ω) such that |u| qm ∈ L 1 (Ω). These results show that if q is sufficiently large, there exists a distributional solution for any source; this is not the case for problem (3).

The second lower order term analysed in [14] is h(u), where h : [0, s 0 ) → R is a continuous, increasing function such that h(0) = 0 and lim s→s - 0 h(s) = +∞ for some s 0 > 0. The regularizing effects of this lower order term are even better than the previous one. Indeed for a positive L 1 (Ω) source, there exists a bounded H 1 0 (Ω) solution.

In the literature we find several papers about elliptic coercive problems with lower order terms having a quadratic growth with respect to the gradient (see [6,10,11,12,8] for example and the references therein), that is, for problem

-div(M (x)∇u) + g(u)|∇u| 2 = f in Ω, u = 0 on ∂Ω.
In these works it is assumed that M : Ω → R N 2 is a bounded elliptic Carathéodory map, so that there exists α > 0 such that α|ξ| 2 ≤ M (x)ξ • ξ for every ξ ∈ R N . Various assumptions are made on g. With no attempt of being exhaustive, we will describe some recent results where a singular g has been considered, namely

g(u) = 1 |u| θ .
The case where 0 < θ ≤ 1, introduced in [2, 3, 4], has been studied in [2,3,4,8,[START_REF] Boccardo | Existence results for quasilinear elliptic and parabolic problems with quadratic gradient terms and sources[END_REF][START_REF] Giachetti | An elliptic problem with a lower order term having singular behaviour[END_REF]. From this body of literature one can deduce that for a

positive source f ∈ L m (Ω), if 2N 2N -θ(N -2) ≤ m < N 2 there exists a strictly positive solution u ∈ H 1 0 (Ω) ∩ L (2-θ)m * * (Ω); if 1 < m < 2N 2N -θ(N -2)
then the solution u belongs to W 1,q 0 (Ω), with q = N m(2 -θ) N -mθ . The authors of [5] consider the general case θ < 2, assuming that f is a strictly positive function on every compactly contained subset of Ω. They prove that if f ∈ L 2N N +2 (Ω) there exists a positive H 1 0 (Ω) solution. Finally, in [START_REF] Giachetti | An elliptic problem with a lower order term having singular behaviour[END_REF] the lower order term is taken to be λu + µ |∇u| 2 |u| θ χ {u>0} , where χ {u>0} denotes the characteristic function of the set {u > 0}, λ > 0 and µ ∈ R.

In this paper we consider the same lower order term as above in an elliptic problem defined by an operator with degenerate coercivity. We will see that if 0 < θ < 2, then |∇u| 2 |u| θ has a regularizing effect, even if it is singular in u. We are going to state our results. We will distinguish the cases 0 < θ < 1 and 1 ≤ θ < 2.

Theorem 1.1. Let 0 < θ < 1. Assume that f is a positive function belonging to L m (Ω), with m ≥ 2N 2N -θ(N -2)
. Then there exists a function u ∈ H 1 0 (Ω), strictly positive on Ω, such that

|∇u| 2 u θ ∈ L 1 (Ω) and (4)
Ω b(x) (1 + u) p ∇u • ∇ϕ + B Ω |∇u| 2 u θ ϕ = Ω f ϕ , for every ϕ ∈ H 1 0 (Ω) ∩ L ∞ (Ω).
In the case where m <

2N 2N -θ(N -2) = 2 * θ ′
, we are able to prove the existence of an infinite energy solution, belonging to W 1,σ 0 (Ω), with σ = mN (2 -θ) N -θm (smaller than 2).

Theorem 1.2. Let 0 < θ < 1. Assume that f is a positive function belonging to

L m (Ω), with N 2N -θ(N -1) < m < 2N 2N -θ(N -2)
. Then there exists a function u ∈ W 1,σ 0 (Ω), strictly positive on Ω, such that

|∇u| 2 u θ ∈ L 1 (Ω) and
(5)

Ω b(x) (1 + u) p ∇u • ∇ϕ + B Ω |∇u| 2 u θ ϕ = Ω f ϕ ,
for every ϕ ∈ C 1 0 (Ω). In the case where 1 ≤ θ < 2, we are able to prove the same results as in the case 0 < θ < 1, under a stronger hypothesis on f .

Theorem 1.3. Let 1 ≤ θ < 2 and p > θ -1. Assume that f ∈ L m (Ω), with m ≥ 2N 2N -θ(N -2)
, and satisfies ess inf {f (x) : x ∈ ω} > 0 for every ω ⊂⊂ Ω. Then there exists a function u ∈ H 1 0 (Ω), strictly positive on Ω, such that

|∇u| 2 u θ ∈ L 1 (Ω) and Ω b(x) (1 + u) p ∇u • ∇ϕ + B Ω |∇u| 2 u θ ϕ = Ω f ϕ for every ϕ ∈ H 1 0 (Ω) ∩ L ∞ (Ω). Theorem 1.4. Let 1 ≤ θ < 2 and p > θ -1. Assume that f ∈ L m (Ω), with N 2N -θ(N -1) < m < 2N 2N -θ(N -2)
, and satisfies ess inf {f (x) : x ∈ ω} > 0 for every ω ⊂⊂ Ω. Then there exists a function u ∈ W 1,σ 0 (Ω), strictly positive on Ω, such that

|∇u| 2 u θ ∈ L 1 (Ω) and Ω b(x) (1 + u) p ∇u • ∇ϕ + B Ω |∇u| 2 u θ ϕ = Ω f ϕ for every ϕ ∈ C 1 0 (Ω).
We remark that if θ < N N -1 we are able to prove the existence of solutions when the source f belongs to L 1 (Ω). We would like to point out the regularizing effects of the lower order term, in the case where p > θ-1 and 0 < θ < 2. Our results furnish H 1 0 (Ω) solutions for less summable sources than for problem (3), since

2N 2N -θN + 2θ < 2N N (1 -p) + 2(p + 1)
.

Even in the case where the source f is less summable, we get a better regularity of solutions than for problem (3

): indeed σ = mN (2 -θ) N -θm ≥ N m(1 -p) N -m(1 + p) , as m ≤ N 2
and p ≤ θ -1. In the case where 0 < p ≤ θ -1, we are able to prove the existence of a solution to problem (1) with the same regularity as the solutions of problem (3).

Theorem 1.5. Let 1 ≤ θ < 2 and 0 < p ≤ θ -1. Assume that f ∈ L m (Ω) and satisfies ess inf {f (x) : x ∈ ω} > 0 for every ω ⊂⊂ Ω.

(1) If m > N 2 , then there exists a strictly positive

H 1 0 (Ω) ∩ L ∞ (Ω) solution to problem (1). (2) If 2N N + 2 -p(N -2) ≤ m < N 2
, then there exists a strictly positive H 1 0 (Ω)∩ L r (Ω) solution to problem (1), where r = N m(1 -p) N -2m .

(

) If N N + 1 -p(N -1) < m < 2N N + 2 -p(N -2) 3
, then there exists a strictly positive W 1,s 0 (Ω) solution to problem (1), where s

= N m(1 -p) N -m(1 + p) . Moreover |∇u| 2 u θ ∈ L 1 (Ω).
In the case where θ ≥ 2, the situations changes. Indeed we will prove a nonexistence result of finite energy solutions. Let λ 1 (f ) denote the first positive eigenvalue of -∆u = λf u in Ω, u = 0 on ∂Ω, where f in L q (Ω), with q > N 2 . Using a result of [5], it is quite easy to prove the following Theorem 1.6. Let f ≥ 0, f ≡ 0, be a L q (Ω) function, with q > N 2 . If either θ > 2, or θ = 2 and λ 1 (f ) > β Bα , then there is no H 1 0 (Ω) solution to problem (1).

A priori estimates

To prove the existence of solutions to problem (1) we use the following approximating problems:

     -div b(x) (1 + |T n (u n )|) p ∇u n + B u n |∇u n | 2 (|u n | + 1 n ) θ+1 = T n (f ) in Ω, u n = 0 on ∂Ω,
where, for n ∈ N and s ∈ R T n (s) = max{-n, min{n, s}} .

These problems are well-posed due to the following result proved in [6,11,12].

Theorem 2.1. Let f be a bounded function. Let M : Ω × R → R N 2 be a Carathédory function such that there exist two positive constants α 0 an β 0 such that M (x, s)ξ • ξ ≥ α 0 |ξ| 2 , |M (x, s)| ≤ β 0 for a.e. x ∈ Ω, for every (s, ξ) ∈ R × R N . Let g(s) be a Carathéodory function such that g(s)s ≥ 0, |g(s)| ≤ γ(s), where γ is a continuous, non-negative and increasing function. Then there exists a H 1 0 (Ω) bounded solution to

-div(M (x, u)∇u) + g(u)|∇u| 2 = f in Ω, u = 0 on ∂Ω .
By Theorem 2.1 the solutions u n of the above approximating problems are bounded H 1 0 (Ω) non-negative functions, since f is assumed to be positive and the lower order term has the same sign as u n . This implies that u n satisfies (6)

     -div b(x) (1 + T n (u n )) p ∇u n + B u n |∇u n | 2 (u n + 1 n ) θ+1 = T n (f ) in Ω, u n = 0 on ∂Ω.
We are now going to prove some a priori estimates. The next lemma gives a control of the lower order term.

Lemma 2.2. Let u n be the solutions to problems (6). Then it results

(7) B Ω u n |∇u n | 2 (u n + 1 n ) θ+1 ≤ Ω f .
Proof. Let us consider T h (u n ) h , h > 0, as a test function in (6). We have, dropping the non-negative operator term,

B Ω |∇u n | 2 u n (u n + 1 n ) θ+1 T h (u n ) h ≤ Ω f T h (u n ) h .
It is now sufficient to pass to the limit as h → 0, using Fatou's lemma and the fact that

T h (u n ) h → 1 as h → 0.
We prove now two a priori estimates on u n , which are true for every p > 0 and θ ∈ (0, 2). In the sequel C will denote a positive constant independent of n; µ(E) will be the Lebesgue measure of a set

E ⊂ R N . Lemma 2.3. Let 0 < θ < 2. Let f be a positive function belonging to L m (Ω), with m ≥ 2N 2N -θ(N -2)
. Then the solutions u n to problems (6) are uniformly bounded in H 1 0 (Ω). Thus there exists a function u ∈ H 1 0 (Ω) such that, up to a subsequence, u n → u weakly in H 1 0 (Ω) and a.e. in Ω. Proof. The assertion follows by proving that the solutions u n to problems (6) are uniformly bounded in H 1 0 (Ω). If we take (u n + 1) θ -1 as a test function in problem (6) we obtain

B Ω |∇u n | 2 (u n + 1 n ) θ+1 u n (u n + 1) θ ≤ B Ω |∇u n | 2 (u n + 1 n ) θ+1 u n + Ω f u θ n + C ,
dropping the positive operator term. We can estimate the right hand side using (7) in order to get

B Ω |∇u n | 2 (u n + 1 n ) θ+1 u n (u n + 1) θ ≤ Ω f u θ n + C .
By working in {u n ≥ 1}, the previous inequality gives

B 2 {un≥1} |∇u n | 2 ≤ Ω f u θ n + C ≤ {un≥1} f u θ n + C ≤ C {un≥1} f (u n -1) θ + C .
We use the Sobolev inequality in the left hand side and the Hölder inequality

with exponent 2 * θ in the last term, recalling that f belongs to L m (Ω) with m ≥ 2N 2N -θ(N -2) = 2 * θ ′ . Thus (8) S B 2    {un≥1} (u n -1) 2 *    2 2 * ≤ B 2 {un≥1} |∇u n | 2 ≤ C    {un≥1} (u n -1) 2 *    θ 2 * + C .
Since we are assuming θ < 2, we deduce that

{un≥1} (u n -1) 2 * ≤ C . It follows from (8) that (9) {un≥1} |∇u n | 2 ≤ C .
Let us search for the same kind of estimate in {u n < 1}. Taking T 1 (u n ) as a test function in problem (6), we get

(10) α 2 p {un<1} |∇T 1 (u n )| 2 ≤ α {un<1} |∇T 1 (u n )| 2 (1 + u n ) p ≤ Ω f T 1 (u n ) ≤ Ω f
using hypothesis (2) and dropping the non-negative lower order term. As a consequence of estimates ( 9) and (10), u n is uniformly bounded in H 1 0 (Ω). By compactness, there exists a function u ∈ H 1 0 (Ω) such that, up to a subsequence,

u n → u weakly in H 1 0 (Ω) and a.e. in Ω. Lemma 2.4. Let 0 < θ < 2. Let f be a positive function belonging to L m (Ω), with N 2N -θ(N -1) < m < 2N 2N -θ(N -2)
. Then the solutions u n to problems ( 6)

are uniformly bounded in W 1,σ 0 (Ω), σ = mN (2 -θ) N -θm
. Thus there exists a function u ∈ W 1,σ 0 (Ω) such that, up to a subsequence, u n → u weakly in W 1,σ 0 (Ω) and a.e. in Ω.

Proof. The assertion follows by proving that the solutions u n to problems (6) are

uniformly bounded in W 1,σ 0 (Ω). Take (u n +1) θ+2γ -1, with γ = 2 * -θm ′ 2m ′ -2 * , as a test function in problems (6). Note that γ < 0: indeed 2 * -θm ′ < 0 and 2m ′ -2 * > 0, since m < N 2 . Moreover, θ + 2γ = 2 * (2 -θ) 2m ′ -2 * > 0, as θ < 2.
Dropping the non-negative operator term and using estimate (7), we get

B Ω |∇u n | 2 (u n + 1 n ) θ+1 u n (u n + 1) 2γ+θ ≤ Ω f (u n + 1) 2γ+θ + C .
By working in {u n ≥ 1} the previous inequality gives (11

) B 2(γ + 1) 2 {un≥1} ∇ (u n + 1) γ+1 -2 γ+1 2 ≤ B 2 {un≥1} |∇u n | 2 (u n + 1) 2γ ≤ {un≥1} f (u n + 1) 2γ+θ + {un≤1} f (u n + 1) 2γ+θ + C ≤ {un≥1} f (u n + 1) 2γ+θ + C .
The Hölder inequality on the right hand side and the Sobolev inequality on the left one imply

(12) S    {un≥1} [(u n + 1) γ+1 -2 γ+1 ] 2 *    2 2 * ≤ C {un≥1} |∇u n | 2 (u n + 1) 2γ ≤ C + C    {un≥1} (u n + 1) (2γ+θ)m ′    1 m ′ .
We remark that the choice of γ is equivalent to require (γ + 1)2 * = (2γ + θ)m ′ ; moreover 2 2 * ≥ 1 m ′ , due to the hypotheses on m and θ. Hence [START_REF] Boccardo | Existence results for quasilinear elliptic and parabolic problems with quadratic gradient terms and sources[END_REF] {un≥1}

(u n + 1) (γ+1)2 * = {un≥1} (u n + 1) (2γ+θ)m ′ ≤ C ∀ n ∈ N .
Now, with σ = mN (2 -θ) N -θm as in the statement, and recalling that γ < 0, let us write

{un≥1} |∇u n | σ = {un≥1} |∇u n | σ (u n + 1) -γσ (u n + 1) -γσ .
The Hölder inequality with exponent 2 σ and estimates (12) and ( 13) give

(14) {un≥1} |∇u n | σ ≤    {un≥1} |∇u n | 2 (u n + 1) -2γ    σ 2    {un≥1} (u n + 1) -γσ 2 2-σ    2-σ 2 ≤ C since -γ 2σ 2 -σ = (γ+1)2 * .
It remains to analyse the behaviour of ∇u n on {u n ≤ 1}.

Taking T 1 (u n ) as a test function in (6) and dropping the non-negative the lower order term, we get

α 2 p {un≤1} |∇T 1 (u n )| 2 ≤ α {un≤1} |∇T 1 (u n )| 2 (1 + u n ) p ≤ Ω f T 1 (u n ) ≤ Ω f
by hypothesis (2). This last estimate and (14) imply that u n is uniformly bounded in W 1,σ 0 (Ω). Since σ > 1, there exists a function u ∈ W 1,σ 0 (Ω) such that, up to a subsequence, u n → u weakly in W 1,σ 0 (Ω) and a.e. in Ω.

In the following lemma, we will assume some hypotheses on p. This will give, in some cases, some better estimates than Lemmata 2.3 and 2.4.

Lemma 2.5. Let 0 < p < 1. Let f ∈ L m (Ω), r = N m(1 -p) N -2m and s = N m(1 -p) N -m(1 + p) . (1) If m > N 2
, the solutions of (6) are uniformly bounded in

H 1 0 (Ω) ∩ L ∞ (Ω). Thus there exists a function u ∈ H 1 0 (Ω) ∩ L ∞ (Ω) such that, up to a subse- quence, u n → u weakly in H 1 0 (Ω) and a.e. in Ω. (2) If 2N N + 2 -p(N -2) ≤ m < N 2
, the solutions of (6) are uniformly bounded in H 1 0 (Ω) ∩ L r (Ω). Thus there exists a function u ∈ H 1 0 (Ω) ∩ L r (Ω) such that, up to a subsequence, u n → u weakly in H 1 0 (Ω) and a.e. in Ω.

(3) If N N + 1 -p(N -1) < m < 2N N + 2 -p(N -2)
, the solutions of (6) are uniformly bounded in W 1,s 0 (Ω). Thus there exists a function u ∈ W 1,s 0 (Ω) such that, up to a subsequence, u n → u weakly in W 1,s 0 (Ω) and a.e. in Ω.

Proof. In problems (6) consider as a test function the same test functions as in [9]. With this choice, the lower order term is non-negative and we can take into account only the term given by the operator. Therefore one can follow the same proofs as in [9] to get the above estimates.

Remark 1. Let p > θ -1. Lemmata 2.3 and 2.4 give a further uniform estimate on u n than Lemma 2.5. Indeed, if one chooses u n as a test function in (6), then, by hypothesis (2)

Ω |∇u n | 2 α (1 + |u n |) p + Bu 2 n (u n + 1 n ) θ+1 ≤ Ω f u n .
If p > θ -1, the lower order term has a leading role in the left hand side of the previous inequality.

We are going to prove the a.e. convergence of the gradients of u n . We will follow the same technique as in [8]. Remark that a similar technique was used for elliptic degenerate problems in [1].

Lemma 2.6. Let u n be the solutions to problems (6) and u be the function found in Lemmata 2.3, or 2.4 or 2.5, according to the summability of f . Up to a subsequence, ∇u n converges to ∇u a.e. in Ω.

Proof. Let h, k > 0. In the sequel C will denote a constant independent of n, h, k. Let us consider T h (u n -T k (u)) as a test function in problems (6). Then

Ω b(x) (1 + T n (u n )) p ∇u n • ∇T h (u n -T k (u)) ≤ h Ω f + B Ω |∇u n | 2 u n (u n + 1 n ) θ+1 h .
By estimate (7) on the right hand side and by hypothesis (2) on the left one, we get

Ω ∇u n • ∇T h (u n -T k (u)) (1 + T n (u n )) p ≤ Ch .
Then we can write

{|un-T k (u))|≤h} |∇(u n -T k (u))| 2 (1 + u n ) p ≤ Ω ∇(u n -T k (u)) • ∇T h (u n -T k (u)) (1 + T n (u n )) p ≤ Ch - Ω ∇T k (u) • ∇T h (u n -T k (u)) (1 + T n (u n )) p .
At the limit as n → ∞ one has lim sup

n→∞ {|un-T k (u)|≤h} |∇T h (u n -T k (u))| 2 (1 + u n ) p ≤ Ch . Since u n ≤ h + k in {|u n -T k (u)| ≤ h}, we get (15) lim sup n→∞ {|un-T k (u)|≤h} |∇T h (u n -T k (u))| 2 ≤ Ch(1 + h + k) p .
We recall that u n is uniformly bounded in W 1,η 0 (Ω), where η equals 2 or σ or s, according to the statements of Lemmata 2.3, 2.4 and 2.5. Let q ∈ (1, η). We can write

Ω |∇(u n -u)| q = {|un-u|≤h,|u|≤k} |∇(u n -u)| q + {|un-u|≤h,|u|>k} |∇(u n -u)| q + {|un-u|>h} |∇(u n -u)| q .
Using the Hölder inequality with exponent 2 q on the first term of the right hand side and exponent η q on the other ones, we have

Ω |∇(u n -u)| q ≤ ≤ C    {|un-u|≤h,|u|≤k} |∇(u n -u)| 2    q 2 +C µ({|u| > k}) 1-q η + µ({|u n -u| > h}) 1-q η ,
where we have used that u n is uniformly bounded in W 1,η 0 (Ω) to estimate the last two terms. By [START_REF] Giachetti | An elliptic problem with a lower order term having singular behaviour[END_REF] the limit as n → ∞ gives lim sup

n→∞ Ω |∇(u n -u)| q ≤ [Ch(1 + k + h) p ] q 2 + Cµ({|u| > k}) 1-q η .
The limit as h → 0 implies lim sup

n→∞ Ω |∇(u n -u)| q ≤ Cµ({|u| > k}) 1-q η .
At the limit as k → +∞, µ({|u| > k}) converges to 0. Therefore ∇u n → ∇u in L q (Ω). Up to a subsequence, ∇u n → ∇u a.e. in Ω.

3. Existence results in the case 0 < θ < 1

To prove the existence of solutions to problem (1), the key point is to prove that the function u found by compactness in the lemmata of Section 2 is strictly positive. In the case 0 < θ < 1, we use a technique similar to that in [8].

Proposition 1. Let 0 < θ < 1. Let u n and u be as in Lemma 2.6. Then u > 0.

Proof. We define, for s ≥ 0,

H n (s) = s 0 t(1 + T n (t)) p α(t + 1 n ) θ+1 dt , H(s) = s 0 (1 + t) p αt θ dt .
Observe that H is well-defined, since θ < 1. We choose e -BHn(un) φ, where φ is a positive C ∞ 0 (Ω) function, as a test function in (6). This gives

Ω b(x) (1 + T n (u n )) p ∇u n • ∇φ e -BHn(un) - Ω T n (f )e -BHn(un) φ = = B Ω b(x) (1 + T n (u n )) p e -BHn(un) |∇u n | 2 φH ′ n (u n ) -B Ω e -BHn(un) |∇u n | 2 u n ( 1 n + u n ) θ+1 φ ≥ B Ω α (1 + T n (u n )) p e -BHn(un) |∇u n | 2 φH ′ n (u n ) -B Ω e -BHn(un) |∇u n | 2 u n ( 1 n + u n ) θ+1
φ by hypothesis (2). The last quantity is positive, due to the choice of H n and φ. As a consequence

Ω b(x) (1 + T n (u n )) p ∇u n • ∇φ e -BHn(un) ≥ Ω T n (f )e -BHn(un) φ ≥ Ω T 1 (f )e -BHn(un) φ .

Now, we set

P n (s) = s 0 e -BHn(t) (1 + T n (t)) p dt , P (s) = s 0 e -BH(t) (1 + t) p dt .
With these definitions, we remark that we have just proved that the inequality (un) holds distributionally. Observe that for every n ∈ N,

-div(b(x)∇(P n (u n ))) ≥ T 1 (f )e -BHn
P n (u n ) ∈ H 1 0 (Ω), since P ′ n is bounded and u n ∈ H 1 0 (Ω). Let z n be the H 1 0 (Ω) solution to -div(b(x)∇z n ) = T 1 (f )e -BHn(un) ; let z be the H 1 0 (Ω) solution to -div(b(x)∇z) = T 1 (f )e -BH(u) . Then -div(b(x)∇(P n (u n ))) ≥ -div(b(x)∇z n ) .
The comparison principle in H 1 0 (Ω) implies that P n (u n (x)) ≥ z n (x) for a.e. x ∈ Ω. Up to a subsequence, z n → z weakly in H 1 0 (Ω) and a.e. in Ω. At the limit a.e. in Ω, as n → +∞, we have P (u) ≥ z. By the strong maximum principle z > 0 and so P (u) > 0. Since P is strictly increasing, u > 0 in Ω.

Corollary 1. Let 0 < θ < 1. Let u n and u be as in Lemma 2.6. Then |∇u| 2 u θ ∈ L 1 (Ω).

Proof. We pass to the limit in (7). The a.e. convergence of u n to u (see Lemmata 2.3, 2.4 and 2.5), the a.e. convergence of ∇u n to ∇u (see Lemma 2.6) and Proposition 1 imply

B Ω |∇u| 2 u θ ≤ Ω f by Fatou's lemma.
We are going to prove Theorem 1.1.

Proof. We are going to prove that the function u found in Lemma 2.3, and studied in Lemma 2.6, Proposition 1 and Corollary 1, is a weak solution to problem (1). We use the same technique as in [8]. We will prove that (4) holds true for every positive and bounded ϕ ∈ H 1 0 (Ω). The general case follows from the fact that every such function ϕ can be written as ϕ + -ϕ -with ϕ ± bounded, positive and belonging to H 1 0 (Ω). We pass to the limit as n → ∞ in

Ω b(x) (1 + T n (u n )) p ∇u n • ∇ϕ + B Ω |∇u n | 2 u n (u n + 1 n ) 1+θ ϕ = Ω T n (f )ϕ ,
where ϕ is a positive bounded H 1 0 (Ω) function. Regarding the first term we observe that b(x) (1 + T n (u n )) p ∇ϕ strongly converges to b(x) (1 + u) p ∇ϕ in L 2 (Ω) and ∇u n weakly converges to ∇u in L 2 (Ω). For the second one we use the a.e. convergence of ∇u n , proved in Lemma 2.6. Fatou's lemma implies ( 16)

Ω b(x) (1 + u) p ∇u • ∇ϕ + B Ω |∇u| 2 u θ ϕ ≤ Ω f ϕ .
The proof of the opposite inequality is more delicate. To this aim, we define, for n ∈ N and s ≥ 0,

H 1 n (t) = t 0 B(1 + s) p α(s + 1 n ) θ ds , H 0 (t) = t 0 B(1 + s) p αs θ ds . H 0 is well-posed, since θ < 1. Let us consider v = e -H 1 n (un) e H 1 j (Tj (u)) ϕ ,
where j ∈ N and ϕ is a positive bounded H 1 0 (Ω) function, as a test function in (6). Then

Ω b(x) (1 + T n (u n )) p ∇u n • ∇ϕ e -H 1 n (un) e H 1 j (Tj (u)) + B α Ω b(x) (1 + T n (u n )) p ϕ e -H 1 n (un) e H 1 j (Tj (u)) ∇u n • ∇T j (u) (T j (u) + 1 j ) θ (T j (u) + 1) p = Ω T n (f )e -H 1 n (un) e H 1 j (Tj (u)) ϕ+ B α Ω b(x)|∇u n | 2 (1 + T n (u n )) p (1 + u n ) p ( 1 n + u n ) θ ϕ e -H 1 n (un) e H 1 j (Tj (u)) -B Ω |∇u n | 2 u n ( 1 n + u n ) θ+1 e -H 1 n (un) e H 1 j (Tj (u)) ϕ .
Note that by hypothesis (2) and inequality

u n + 1 1 + T n (u n ) p ≥ 1 > u n u n + 1 n ,
the sum of the last two terms is non-negative. At the limit as n → ∞ we have

Ω b(x) (1 + u) p ∇u • ∇ϕ e -H0(u) e H 1 j (Tj (u)) + B α Ω b(x) (1 + u) p ϕ e -H0(u) e H 1 j (Tj (u)) ∇u • ∇T j (u) (T j (u) + 1 j ) θ (T j (u) + 1) p ≥ Ω f e -H0(u) e H 1 j (Tj (u)) ϕ + B α Ω b(x)|∇u| 2 u θ ϕ e -H0(u) e H 1 j (Tj (u)) -B Ω |∇u| 2 u θ e -H0(u) e H 1 j (Tj (u)) ϕ ,
using the weak convergence of u n to u in H 1 0 (Ω) in the left hand side and Fatou's lemma in the right one. Now we pass to the limit as j → ∞, using that e -H0(u) e H 1 j (Tj (u)) ≤ 1 and Corollary 1. We obtain

(17) Ω b(x) (1 + u) p ∇u • ∇ϕ ≥ Ω f ϕ -B Ω ϕ |∇u| 2 u θ .
Inequalities ( 16) and (17) imply that

Ω b(x) (1 + u) p ∇u • ∇ϕ + B Ω ϕ |∇u| 2 u θ = Ω f ϕ
for every positive and bounded ϕ ∈ H 1 0 (Ω). We are going to prove Theorem 1.2.

Proof. We are going to prove that the function u found in Lemma 2.4 and studied in Lemma 2.6, Proposition 1 and Corollary 1, is a weak solution to problem (1). We use the same technique as in [11,21].

We first prove (5) for every positive C 1 0 (Ω) function ϕ. With the same argument as in the previous theorem (i.e., using Fatou's lemma) one can prove that (18)

Ω b(x) (1 + u) p ∇u • ∇ϕ + B Ω |∇u| 2 u θ ϕ ≤ Ω f ϕ .
To prove the opposite inequality, we slightly modify the previous proof, since we no longer have uniform estimates of u n in H 1 0 (Ω). Observe that, however, T k (u n ) is uniformly bounded in H 1 0 (Ω). Indeed, it is sufficient to consider T k (u n ) as a test function in (6): we obtain

{un≤k} |∇T k (u n )| 2 ≤ Ck(1 + k) p ∀ n ∈ N by hypothesis (2). We will use, for k ∈ N and s ∈ R R k (s) =    1, s ≤ k k + 1 -s, k ≤ s ≤ k + 1 0, s > k + 1 ,
to define a test function. We set, for t ≥ 0,

H 1 n (t) = t 0 B(1 + s) p α(s + 1 n ) θ ds , H 0 (t) = t 0 B(1 + s) p αs θ ds .
This is possible, since θ < 1. We consider

v = e -H 1 n (un) e H 1 j (Tj (u)) R k (u n )ϕ ,
where ϕ is a positive C 1 0 (Ω) function and j ∈ N , as a test function in (6). Then

Ω b(x) (1 + T n (u n )) p ∇u n • ∇ϕ e -H 1 n (un) e H 1 j (Tj (u)) R k (u n ) + B α Ω b(x) (1 + T n (u n )) p ϕ e -H 1 n (un) e H 1 j (Tj (u)) ∇u n • ∇T j (u) (T j (u) + 1 j ) θ (T j (u) + 1) p R k (u n ) = Ω T n (f )e -H 1 n (un) e H 1 j (Tj (u)) ϕR k (u n )+ {k≤un≤k+1} b(x)|∇u n | 2 (1 + T n (u n )) p ϕ e -H 1 n (un) e H 1 j (Tj (u)) + B α Ω b(x)|∇u n | 2 (1 + T n (u n )) p (1 + u n ) p ( 1 n + u n ) θ ϕ e -H 1 n (un) e H 1 j (Tj (u)) R k (u n ) -B Ω |∇u n | 2 u n ( 1 n + u n ) θ+1 e -H 1 n (un) e H 1 j (Tj (u)) R k (u n )ϕ .
The sum of the last two terms is positive, since b(x) ≥ α by hypothesis (2) and by inequality

u n + 1 1 + T n (u n ) p ≥ 1 ≥ u n u n + 1 n . Dropping the non-negative term {k≤un≤k+1} b(x)|∇u n | 2 (1 + T n (u n )) p ϕ e -H 1 n (un) e H 1 j
(Tj (uj )) , at the limit as n → ∞ we have, by Fatou's lemma, the weak convergence of u n in W 1,σ 0 (Ω) and the weak convergence of

T k (u n ) in H 1 0 (Ω), Ω b(x) (1 + u) p ∇u • ∇ϕ e -H0(u) e H 1 j (Tj (u)) R k (u)+ + B α Ω b(x) (1 + u) p ϕ e -H0(u) e H 1 j (Tj (u)) ∇u • ∇T j (u) (T j (u) + 1 j ) θ (T j (u) + 1) p R k (u) ≥ Ω f e -H0(u) e H 1 j (Tj (u)) ϕ R k (u) + B α Ω b(x)|∇u| 2 u θ ϕ e -H0(u) e H 1 j (Tj (u)) R k (u) -B Ω |∇u| 2 u θ e -H0(u) e H 1 j (Tj (u)) R k (u)ϕ .
As in the previous proof, it is now sufficient to pass to the limit as j → ∞ first, using that e -H0(u) e H 1 j (Tj (u)) ≤ 1 and Corollary 1, and then to the limit as k → ∞, using that R k (u) tends to 1. We thus obtain ( 19)

Ω b(x) (1 + u) p ∇u • ∇ϕ ≥ Ω f ϕ -B Ω |∇u| 2 u θ ϕ .
Inequalities ( 18) and ( 19) imply that ( 20)

Ω b(x) (1 + u) p ∇u • ∇ϕ + B Ω |∇u| 2 u θ ϕ = Ω f ϕ
for every positive ϕ ∈ C 1 0 (Ω). Now, let ϕ any C 1 0 (Ω) function. We define ϕ ε ± = ρ ε * ϕ ± as the convolution of a mollifier ρ ε , for ε > 0, with ϕ ± . Then ϕ ε ± is a positive C 1 0 (Ω) function, for ε sufficiently small. By (20) we have

Ω b(x) (1 + u) p ∇u • ∇(ϕ ε --ϕ ε + ) + B Ω |∇u| 2 u θ (ϕ ε --ϕ ε -) = Ω f (ϕ ε --ϕ ε -) .
Since ϕ ε --ϕ ε -→ ϕ uniformly in Ω and in W 1,q 0 (Ω) for every q ≥ 1, as ε → 0, the result follows.

4. Existence results in the case 1 ≤ θ < 2 As in the above case, we need to prove that the function u found in Section 2 is not 0 in Ω. To this aim, we are going to prove that for every ω ⊂⊂ Ω there exists a positive constant c ω such that the solutions u n to problems (6) satisfy u n ≥ c ω in ω for every n ∈ N. We will follow a similar technique to that one in [5]. The following theorem, proved in [18] (and in [5]), will be useful to us.

Theorem 4.1. Let B : Ω × R → R be a Carathéodory function such that for every ω ⊂⊂ Ω there exists m ω > 0 such that B(x, s) ≥ m ω l(s) for a.e. x ∈ Ω and for every s ≥ 0. Assume that l : R + → R + is a continuous increasing function such that l(s)/s is increasing for s sufficiently large and for some t 0 > 0 Then for every ω ⊂⊂ Ω there exists a constant C ω > 0 such that every sub-

solution v ∈ H 1 loc (Ω) of -div(b(x)∇v) + B(x, v) = 0 such that v + ∈ L ∞ loc (Ω) and B(x, v + ) ∈ L 1 loc (Ω) satisfies v ≤ C ω in ω.
Remark 2. We recall that a sub-solution of -div(b(x)∇v)

+ l(v)g(x) = 0 is a W 1,1 loc (Ω) function such that Ω b(x)∇v • ∇φ + Ω l(v)g(x)φ ≤ 0
for every C ∞ c (Ω) positive function φ. Remark 3. In the literature condition ( 21) is called the Keller-Osserman condition, due to the papers [17, 19] on semilinear equations.

Proposition 2. Let 1 ≤ θ < 2. Let u n be the solutions of (6). Then for every ω ⊂⊂ Ω there exists a strictly positive constant c ω such that u n ≥ c ω in ω for every n ∈ N.

Proof. Step 1. Let u n be a H 1 0 (Ω) ∩ L ∞ (Ω) solution to (6). We perform a change of variable in order to get a sub-solution of an elliptic semi-linear problem, as in Theorem 4.1.

We set a n (s

) = 1 (1 + T n (s)) p . Then u n satisfies, distributionally, -div (b(x)a n (u n )∇u n ) + B u θ n |∇u n | 2 ≥ T 1 (f ) , that is, (22) -div(b(x)∇u n )a n (u n ) -a ′ n (u n )b(x)|∇u n | 2 + B u θ n |∇u n | 2 ≥ T 1 (f ) . Let k n (t) = t 1 B αr θ a n (r)
dr and ψ n (s) =

1 s e -kn(t) a β α n (t)dt . We remark that (23) ψ ′ n (s) = -a β α n (s)e -kn(s) , ψ ′′ n (s) ψ ′ n (s) = β α a ′ n (s) a n (s) - B αs θ a n (s)
.

We define

v n = ψ n (u n ). Then div(b(x)∇v n ) = div(b(x)ψ ′ n (u n )∇u n ) = ψ ′ n (u n )div(b(x)∇u n ) + b(x)ψ ′′ n (u n )|∇u n | 2 and therefore -a n (u n )div(b(x)∇u n ) = -a n (u n ) div(b(x)∇v n ) ψ ′ n (u n ) + a n (u n )b(x) ψ ′′ n (u n ) ψ ′ n (u n ) |∇u n | 2 .
By inequality (22) we have

T 1 (f ) ≤ -a n (u n ) div(b(x)∇v n ) ψ ′ n (u n ) +a n (u n )b(x) ψ ′′ n (u n ) ψ ′ n (u n ) |∇u n | 2 -a ′ n (u n )b(x)|∇u n | 2 + B u θ n |∇u n | 2 .
Using that a ′ n (s) ≤ 0,

ψ ′′ n (s) ψ ′ n (s)
≤ 0 and hypothesis (2) we obtain

T 1 (f ) ≤ -a n (u n ) div(b(x)∇v n ) ψ ′ n (u n ) +a n (u n )α ψ ′′ n (u n ) ψ ′ n (u n ) |∇u n | 2 -a ′ n (u n )β|∇u n | 2 + B u θ n |∇u n | 2 .
Due to (23)

T 1 (f ) ≤ -a n (u n ) div(b(x)∇v n ) ψ ′ n (u n ) . Observing that ψ ′ n (s) = -a β α n (s)e -kn(s) ≤ 0, v n satisfies 0 ≥ -div(b(x)∇v n ) + T 1 (f )e -kn(ψ -1 n (vn)) a β α -1 n (ψ -1 n (v n )) .
Step 2. We now study, for s ≥ 0

s → e -kn(ψ -1 n (s)) a β α -1 n (ψ -1 n (s)) . We remark that ψ -1 n (s) ≤ 1, since s ≥ 0 = ψ n (1) and ψ n is decreasing. Therefore (24) a β α -1 n (ψ -1 n (s)) ≥ a β α -1 n (1) = a 0 ,
as a n is decreasing.

Recalling that

ψ n (s) = 1 s e -kn(t) a β α n (t)dt , k n (t) = t 1 B αr θ a n (r) dr and a n (s) = a 1 (s) , s ≤ 1 a n (s) ≤ a 1 (s) , s > 1 it is not difficult to prove that (25) ψ n (s) ≥ ψ 1 (s) ,
distinguishing the cases s ≤ 1 and s > 1. Now, inequality (25) and the fact that

ψ n is decreasing imply that ψ -1 n (s) ≤ ψ -1 1 (s) for every s ≥ 0. Recalling that ψ -1 n (s) ≤ 1 and a n (s) = a 1 (s) ≥ 0 for s ≥ 1, we deduce easily that (26) e -kn(ψ -1 n (s)) ≥ e -k1(ψ -1 1 (s)) . Due to (24) and (26), v n satisfies 0 ≥ -div(b(x)∇v n ) + B(x, v n ) with B(x, s) = T 1 (f )a 0 l(s) , s ≥ 0 0 , s ≤ 0 , where l(s) = e -k1(ψ -1 1 (s)) -1, s ≥ 0.
Step 3. We are going to prove that l satisfies the hypotheses of Theorem 4.1. We observe that l is continuous and increasing, since ψ -1 1 is decreasing and k 1 is increasing. We claim that l(s)/s is increasing for s sufficiently large. This is equivalent to prove that Y (t) = l(ψ 1 (t)) ψ 1 (t) is decreasing for small positive t. Now,

Y ′ (t) < 0 if and only if (27) l ′ (ψ 1 (t))ψ 1 (t) - t 1 l ′ (ψ 1 (s))ψ ′ 1 (s)ds > 0 .
We remark that l ′ (ψ

1 (s)) = B αs θ a β α +1 1 (s)
. Let w 0 ∈ (0, 1) be such that

h(t) = l ′ (ψ 1 (t)) = B αt θ a β α +1 1 (t)
is decreasing in (0, w 0 ]. Therefore We have proved that

l ′ (ψ 1 (t))ψ 1 (t) - t 1 l ′ (ψ 1 (s))ψ ′ 1 (s)ds = 1 t e -k1(s) a β α 1 (s) [h(t) -h(s)] ds
l ′ (ψ 1 (t))ψ 1 (t) - t 1 l ′ (ψ 1 (s))ψ ′ 1 (s)ds ≥ M 1 h(t) -M 2 .
If t is sufficiently small, the last quantity is positive, since h is decreasing for small positive t. Therefore (27) holds.

We are going to study the last condition on l, that is, the existence of a positive t 0 such that (28)

+∞ t0 dt t 0 l(s)ds < ∞ .
Using the change of variable τ = ψ -1 1 (s) we get

t 0 l(s)ds = t 0 [e -k1(ψ -1 1 (s)) -1]ds = 1 ψ -1 1 (t) [e -k1(τ ) -1]a β α 1 (τ )e -k1(τ ) dτ .
It is easy to see that e -k1(τ ) -1 ≥ 1 2 e -k1(τ ) for τ ≤ τ 0 sufficiently small. Moreover a 1 (τ ) ≥ 1 2 , for τ ≤ 1. Therefore it suffices to find t 0 sufficiently large (t 0 > ψ 1 (τ 0 )) such that

+∞ t0 dt 1 ψ -1 1 (t) e -2k1(τ ) dτ < ∞ .
The last integral can be estimated, using the change w = ψ -1 1 (t) and the fact that a 1 (s) ≤ 1, in the following way:

0 ψ -1 1 (t0) ψ ′ 1 (w)dw 1 w e -2k1(τ ) dτ = ψ -1 1 (t0) 0 e -k1(w) a β α 1 (w)dw 1 w e -2k1(τ ) dτ ≤ ψ -1 1 (t0) 0 dw
w0 w e 2[k1(w)-k1(τ )] dτ where w 0 is chosen in such a way that k ′ 1 is decreasing in (0, w 0 ]. We observe that

1 0 k ′ 1 (t)dt < ∞, as θ < 2.
Hence it suffices to prove that there exists a strictly positive constant c such that

k ′ 1 (w) w0 w e 2[k1(w)-k1(τ )] dτ ≥ c . Now, since k ′ 1 (τ ) is decreasing in (0, w 0 ], k ′ 1 (w) w0 w e 2[k1(w)-k1(τ )] dτ ≥ w0 w k ′ 1 (τ )e 2[k1(w)-k1(τ )] dτ = 1 2 - 1 2 e 2[k1(w)-k1(w0)] .
Observe that e 2k1(w) → 0 as w → 0, since k 1 (w) =

w 1 B αt θ a 1 (t)
dt → -∞ as w → 0, by hypothesis θ ≥ 1. Therefore (28) is proved.

Step 4. Theorem 4.1 applies and gives, for every ω ⊂⊂ Ω, the existence of a constant C ω > 0 such that v n ≤ C ω . Recalling that ψ n (s) ≥ ψ 1 (s) by (25), we have

C ω ≥ v n = ψ n (u n ) ≥ ψ 1 (u n ). Since ψ 1 is decreasing, u n ≥ ψ -1 1 (C ω ) = c ω > 0 in every ω ⊂⊂ Ω.
Corollary 2. Let 1 ≤ θ < 2. Let u n and u be as in Lemma 2.6. Then |∇u| 2 u θ ∈ L 1 (Ω).

Proof. As in the proof of Corollary 1, we pass to the limit in (7) using the a.e. convergence of u n to u (see Lemmata 2.3, 2.4 and 2.5), the a.e. convergence of ∇u n to ∇u (see Lemma 2.6) and Proposition 2.

Corollary 3. For every ω ⊂⊂ Ω there exists a positive constant cω such that

u n (u n + 1 n ) 1+θ ≤ cω ∀ x ∈ ω . Proof. It is sufficient to observe that in every subset ω ⊂⊂ Ω u n (u n + 1 n ) 1+θ ≤ 1 u θ n ≤ 1 c θ ω = cω , since u n ≥ c ω > 0 in ω by Proposition 2.
As in [5] we prove the strong convergence of T k (u n ) in H 1 loc (Ω). This will be used to compute the limit of the lower order term in problems (6).

Lemma 4.2. Let u n be the solutions to problems (6) and u be the function found in Lemmata 2.3, 2.4, 2.5. Then, up to a subsequence,

T k (u n ) → T k (u) in H 1 loc (Ω). Proof. We are going to prove that lim n→∞ Ω |∇(T k (u n ) -T k (u))| 2 φ = 0 for all positive φ ∈ C ∞ c (Ω). Let ϕ λ (s) = se λs 2 , λ > 0.
As in [11], we will consider as a test function ϕ λ (T k (u n ) -T k (u))φ, where λ will be chosen later. In the sequel ε(n) will denote any quantity converging to 0, as n → ∞. From (6) we get (29)

Ω b(x) (1 + T n (u n )) p ∇u n • ∇(T k (u n ) -T k (u))ϕ ′ λ (T k (u n ) -T k (u))φ +B Ω u n |∇u n | 2 (u n + 1 n ) θ+1 ϕ λ (T k (u n ) -T k (u))φ = - Ω b(x) (1 + T n (u n )) p ∇u n • ∇φ ϕ λ (T k (u n ) -T k (u)) + Ω T n (f )ϕ λ (T k (u n ) -T k (u))φ . It is not difficult to prove that Ω T n (f )ϕ λ (T k (u n )-T k (u))φ → 0 , Ω b(x) (1 + T n (u n )) p ∇u n •∇φ ϕ λ (T k (u n )-T k (u)) → 0 ,
as n → ∞. Indeed for the first limit one can use the Lebesgue Theorem. For the second one it is sufficient to observe that ∇u n converges weakly in some Sobolev space given by the statements of Lemmata 2.3, 2.4 and 2.5 and b(x)

(1 + T n (u n )) p ∇φ ϕ λ (T k (u n )- T k (u)
) is uniformly bounded with respect to n.

We are going to treat the left hand side of (29). We choose ω φ ⊂⊂ Ω, with suppφ ⊂ ω φ . Then

B Ω u n |∇u n | 2 (u n + 1 n ) θ+1 ϕ λ (T k (u n )-T k (u))φ ≥ -Bc ω φ Ω |∇T k (u n )| 2 |ϕ λ (T k (u n )-T k (u))|φ
by Corollary 3. We deduce from (29) that ( 30)

Ω b(x) (1 + T n (u n )) p ∇u n • ∇(T k (u n ) -T k (u))ϕ ′ λ (T k (u n ) -T k (u))φ -Bc ω φ Ω |∇T k (u n )| 2 |ϕ λ (T k (u n ) -T k (u))|φ ≤ ε(n) .
We remark that

{un≥k} b(x) (1 + T n (u n )) p ∇u n • ∇(T k (u n ) -T k (u))ϕ ′ λ (T k (u n ) -T k (u))φ = ε(n) .
Hence inequality (30) is equivalent to

(31) {un≤k} b(x) (1 + T n (u n )) p ∇u n • ∇(T k (u n ) -T k (u))ϕ ′ λ (T k (u n ) -T k (u))φ -Bc ω φ Ω |∇T k (u n )| 2 |ϕ λ (T k (u n ) -T k (u))|φ ≤ ε(n) . Remark that {un≤k} b(x) (1 + T n (u n )) p ∇T k (u)•∇(T k (u n )-T k (u))ϕ ′ λ (T k (u n )-T k (u))φ → 0 , n → ∞ .
Adding the above quantity in both sides of (31) we get

{un≤k} b(x) (1 + T n (u n )) p ∇(u n -T k (u)) • ∇(T k (u n ) -T k (u))ϕ ′ λ (T k (u n ) -T k (u))φ -Bc ω φ Ω |∇T k (u n )| 2 |ϕ λ (T k (u n ) -T k (u))|φ ≤ ε(n) .
By hypothesis (2) on b, we obtain (32

) {un≤k} α (1 + k) p |∇(T k (u n ) -T k (u))| 2 ϕ ′ λ (T k (u n ) -T k (u))φ -Bc ω φ Ω |∇T k (u n )| 2 |ϕ λ (T k (u n ) -T k (u))|φ ≤ ε(n) .
It is easy to prove that

Ω |∇T k (u n )| 2 |ϕ λ (T k (u n )-T k (u))|φ ≤ {un≤k} 2|∇(T k (u n )-T k (u))| 2 |ϕ λ (T k (u n )-T k (u))|φ+ε(n) .
We deduce from (32) that the quantity (33

) {un≤k} α (1 + k) p ϕ ′ λ (T k (u n ) -T k (u)) -2Bc ω φ |ϕ λ (T k (u n ) -T k (u))| |∇(T k (u n )-T k (u))| 2 φ
tends to 0. Now, ϕ λ has the following property: for every a, b > 0, aϕ

′ λ (s) - b|ϕ λ (s)| ≥ a 2 if λ > b 2 4a 2 . Therefore there exists λ > 0 such that α (1 + k) p ϕ ′ λ (s) -2Bc ω φ |ϕ λ (s)| ≥ α 2(1 + k) p ∀ s ∈ R .
Applying this inequality to the quantity (33), the statement of the theorem is proved.

We are now going to prove Theorems 1.3 and 1.4 in a unique proof. As we will see the only difference is the choice of the test functions ϕ. Theorem 1.5 can be proved with the same technique.

Proof. By Lemmata 2.3 and 2.4 the solutions u n to (6) are uniformly bounded in H 1 0 (Ω) and W 1,σ 0 (Ω) respectively; moreover ∇u n converges to ∇u a.e. in Ω up to a subsequence, by Lemma 2.6. The solutions u n satisfy

Ω b(x) (1 + T n (u n )) p ∇u n • ∇ϕ + B Ω |∇u n | 2 u n (u n + 1 n ) 1+θ ϕ = Ω T n (f )ϕ .
For the proof of Theorem 1.3 we consider for ϕ a bounded H 1 0 (Ω) function. For the proof of Theorem 1.4, ϕ is a C 1 0 (Ω) function. To compute the limit of the first term in the case where u n weakly converges to u in H 1 0 (Ω) (Theorem 1.3) it is sufficient to use that b(x) (1 + T n (u n )) p ∇ϕ strongly converges to b(x) (1 + u) p ∇ϕ in (L 2 (Ω)) N for every ϕ ∈ H 1 0 (Ω) ∩ L ∞ (Ω). In the case where u n weakly converges to u in W 1,σ 0 (Ω), with σ < 2 (Theorem 1.4), one uses that b(x) (1 + T n (u n )) p ∇ϕ strongly converges to b(x) (1 + u) p ∇ϕ in (L r (Ω)) N for every r ≥ 1 and for every ϕ ∈ C 1 0 (Ω).

To compute the limit of Ω |∇u n | 2 u n (u n + 1 n ) 1+θ ϕ we will use the same technique as in [5]. We are going to prove that

|∇u n | 2 u n (u n + 1 n ) 1+θ is equi-integrable. Let E ⊂⊂ ω ⊂⊂ Ω. Then E |∇u n | 2 u n (u n + 1 n ) 1+θ ≤ E∩{un≤k} |∇u n | 2 u n (u n + 1 n ) 1+θ + E∩{un≥k} |∇u n | 2 u n (u n + 1 n ) 1+θ ≤ cω E∩{un≤k} |∇T k (u n )| 2 + {un≥k} |∇u n | 2 u n (u n + 1 n ) 1+θ ,
where we have used Corollary 3 to estimate the first term. Now, if we choose T 1 (u n -T k-1 (u n )) in problems (6) we have, dropping the non-negative operator term, (34)

B {un≥k} |∇u n | 2 u n (u n + 1 n ) θ+1 ≤ {un≥k-1}
f .

Observe that there exists a constant C > 0 such that µ({u n ≥ k -1}) ≤ C k -1 , as u n are uniformly bounded in L 1 (Ω). This implies that the right hand side of (34) converges to 0 as k → ∞, uniformly with respect to n. We deduce that there exists k 0 > 1 such that (35)

{un≥k} |∇u n | 2 u n (u n + 1 n ) θ+1 ≤ ε 2 ∀ k ≥ k 0 , ∀ n ∈ N .
Moreover, since T k (u n ) → T k (u) in H 1 loc (Ω) by Lemma 4.2, there exist n ε , δ ε such that for every E ⊂⊂ Ω with µ(E) < δ ε we have

E∩{un≤k} |∇T k (u n )| 2 = E |∇T k (u n )| 2 ≤ ε 2c ω ∀ n ≥ n ε .
This and (35) imply that |∇u n | 2 u n (u n + 1 n ) 1+θ is equi-integrable. Now, recall that

|∇u n | 2 u n (u n + 1 n ) 1+θ converges a.e. to
|∇u| 2 u θ , belonging to L 1 (Ω) by Corollary 2. By Vitali's theorem we have the result.

A non-existence result in the case θ ≥ 2

We are going to prove Theorem 1.6 about the non-existence of finite energy solutions to problem (1) when θ ≥ 2. We will use the following result of [5]: Theorem 5.1. Let M (x, s) be a N × N matrix whose entries are Carathéodory functions m ij : Ω × R → R, for every i, j = 1, . . . , N . Assume that there exist two positive constants α 1 , β 1 such that M (x, s)ξ • ξ ≥ α 1 |ξ| 2 and |M (x, s)| ≤ β 1 for a.e.

x ∈ Ω, and for all (s, ξ) ∈ R × R N . Let g : Ω × (0, +∞) → R + be a Carathéodory function such that for some constants s 0 , Λ > 0 and θ ≥ 2 it holds g(x, s) ≥ Λ s θ ∀ s ∈ (0, s 0 ] . Let f ≥ 0, f ≡ 0, be a L q (Ω) function, with q > N 2 . If one of the following conditions holds:

(1) θ > 2 (2) θ = 2 and λ 1 (f ) > β1 Λα1 , then there is no H 1 0 (Ω) solution to problem -div (M (x, u)∇u) + g(x, u)|∇u| 2 = f in Ω, u = 0 on ∂Ω. It is easy to prove that g(s)s θ → 1, as s → 0 + . Hence for every fixed 0 < ε < B there exists s ε > 0 such that Bg(s) ≥ B-ε s θ

for every s ∈ (0, s ε ]. Theorem 5.1 therefore applies to problem (36). We deduce that there is no H 1 0 (Ω) solution to problem (36) if either θ > 2, or θ = 2 and λ 1 (f ) > β (B-ε)α , for every 0 < ε < B. As a consequence there is no H 1 0 (Ω) solution to problem (1) if either θ > 2, or θ = 2 and λ 1 (f ) > β Bα .

1

  (s) [h(t) -h(s)] dsdue to the choice of w 0 . Let

  -div (b(x)∇v) + Bg(v)|∇v| 2 = f in Ω,

Acknowledgments

The author thanks L. Orsina and M.M. Porzio for very helpful discussions on the subject of this paper and the anonymous referee for his suggestion to include Theorem 1.6 in the paper. Part of this work was done during a visit to La Sapienza Università di Roma whose hospitality is gratefully acknowledged.