Semilinear equations with exponential nonlinearity and measure data

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Semilinear equations with exponential nonlinearity and measure data

Équations semi linéaires avec non linéarité exponentielle et données mesures

Daniele Bartolucci

, Fabiana Leoni

, Luigi Orsina

, Augusto C. Ponce

aDipartimento di Matematica, Università di Roma “Tre”, Largo S. Leonardo Murialdo 1, 00146 Roma, Italy bDipartimento di Matematica, Università di Roma “La Sapienza”, Piazza A. Moro 2, 00185 Roma, Italy cLaboratoire Jacques-Louis Lions, université Pierre et Marie Curie, boîte courier 187, 75252 Paris cedex 05, France dRutgers University, Department of Mathematics, Hill Center, Busch Campus, 110 Frelinghuysen Rd., Piscataway, NJ 08854, USA

Received 27 July 2004; accepted 15 December 2004 Available online 22 April 2005

Abstract

We study the existence and non-existence of solutions of the problem −u+e^u−1=µ inΩ,

u=0 on∂Ω, (0.1)

whereΩ is a bounded domain inR^N,N3, andµis a Radon measure. We prove that ifµ4πH^N−2, then (0.1) has a unique solution. We also show that the constant 4πin this condition cannot be improved.

2005 Elsevier SAS. All rights reserved.

Résumé

Nous étudions l’existence et la non existence des solutions de l’équation −u+e^u−1=µ dansΩ,

u=0 sur∂Ω, (0.2)

* Corresponding author.

E-mail addresses: bartolu@mat.uniroma3.it, bartoluc@mat.uniroma1.it (D. Bartolucci), leoni@mat.uniroma1.it (F. Leoni), orsina@mat.uniroma1.it (L. Orsina), ponce@ann.jussieu.fr, augponce@math.rutgers.edu (A.C. Ponce).

0294-1449/$ – see front matter 2005 Elsevier SAS. All rights reserved.

doi:10.1016/j.anihpc.2004.12.003

oùΩest un domaine borné dansR^N,N3, etµest une mesure de Radon. Nous démontrons que siµvérifieµ4πH^N−2, alors le problème (0.2) admet une unique solution. Nous montrons que la constante 4π dans cette condition ne peut pas être améliorée.

2005 Elsevier SAS. All rights reserved.

MSC: 35J60; 35B05

Keywords: Nonlinear elliptic equations; Exponential nonlinearity; Measure data

1. Introduction

LetΩ⊂R^N,N2, be a bounded domain with smooth boundary. We consider the problem −u+e^u−1=µ inΩ,

u=0 on∂Ω, (1.1)

whereµ∈M(Ω), the space of bounded Radon measures inΩ. We say that a functionuis a solution of (1.1) if u∈L¹(Ω), e^u∈L¹(Ω)and the following holds:

−

Ω

uζ+

Ω

(e^u−1)ζ=

Ω

ζdµ ∀ζ∈C₀²(Ω). (1.2)

HereC₀²(Ω) denotes the set of functionsζ ∈C²(Ω) such thatζ=0 on∂Ω. A measureµis a good measure for problem (1.1) if (1.1) has a solution. We shall denote byGthe set of good measures. Problem (1.1) has been recently studied by Brezis, Marcus and Ponce in [1], where the general case of a continuous nondecreasing nonlinearity g(u), withg(0)=0, is dealt with. Applying Theorem 1 of [1] tog(u)=e^u−1, it follows that for everyµ∈M(Ω) there exists a largest good measureµfor (1.1), which we shall denote byµ^∗.

In the caseN=2, the set of good measures for problem (1.1) has been characterized by Vázquez in [9]. More precisely, a measureµis a good measure if and only ifµ({x})4πfor everyx inΩ. Note that anyµ∈M(Ω) can be decomposed as

µ=µ₀+^∞

i=1

α_iδ_xi,

withµ0({x})=0 for everyx inΩ, andδx_i is the Dirac mass concentrated atxi. Using Vázquez’s result, it is not difficult to check that (see [1, Example 5])

µ^∗=µ₀+ ∞ i=1

min{4π, α_i}δ_xi.

This paper is devoted to the study of problem (1.1) in the caseN3. First of all, let us recall that ifµis a good measure, then (1.1) has a unique solutionu (see [1, Corollary B.1]). This solution can be either obtained as the limit of the sequence(u_n)of solutions of

−u_n+min{e^un−1, n} =µ inΩ,

u_n=0 on∂Ω,

or as the limit of a sequence(v_n)of solutions of −v_n+e^vn−1=µ_n inΩ,

v_n=0 on∂Ω,

withµ_n=ρ_n∗µ, where(ρ_n)is a sequence of mollifiers. Ifµis not a good measure, then both sequences(u_n)and (v_n)converge to the solutionu^∗of problem (1.1) with datumµ^∗(see [1]). It has also been proved in [1] that the setGof good measures is convex and closed with respect to the strong topology inM(Ω). Moreover, it is easy to see that ifνµandµ∈G, thenν∈G.

Before stating our results, let us briefly recall the definitions of Hausdorff measure and Hausdorff dimension of a set. Lets0, and letA⊂R^Nbe a Borel set. Givenδ >0, let

H^s_δ(A)=inf

i

ω_sr_i^s: K⊂

i

B_riwithr_i< δ, ∀i

,

where the infimum is taken over all coverings ofAwith open ballsB_riof radiusr_i< δ, andω_s=π^s/2/ (s/2+1).

We define the (spherical)s-dimensional Hausdorff measure inR^Nas H^s(A)=lim

δ↓0H^s_δ(A),

and the Hausdorff dimension ofAas dim_H(A)=inf

s0:H^s(A)=0 .

Given a measureµinM(Ω), we say that it is concentrated on a Borel setE⊂Ω ifµ(A)=µ(E∩A) for every Borel setA⊂Ω. Given a measureµinM(Ω), and a Borel setE⊂Ω, the measureµ E is defined by µ E(A)=µ(E∩A)for every Borel setA⊂Ω.

One of our main results is the following

Theorem 1. Letµ∈M(Ω). Ifµ4πH^N−2, that is, ifµ(A)4πH^N−2(A)for every Borel setA⊂Ωsuch that H^N−2(A) <∞, then there exists a unique solutionuof (1.1).

As a corollary of Theorem 1, we have

Corollary 1. Letµ∈M(Ω). Ifµ4πH^N−2, thenµ^∗=µ.

The proof of Theorem 1 relies on a decomposition lemma for Radon measures (see Section 3 below) and on the following sharp estimate concerning the exponential summability for solutions of the Laplace equation. We denote by M^N/2(Ω)the Morrey space with exponent ^N₂ equipped with the norm · N/2(see Definition 1 below).

Theorem 2. Letf be a function in M^N/2(Ω), and letube the solution of −u=f inΩ,

u=0 on∂Ω. (1.3)

Then, for every 0< α <2N ωN, it holds

Ω

e^{((2N ωN−α)/f N/2)|u|}(N ω_N)²

α diam(Ω)^N. (1.4)

This theorem is the counterpart in the caseN3 of a result proved, forN=2 andf ∈L¹(Ω), by Brezis and Merle in [2]. Note that, forN=2, the space M^N/2(Ω)coincides withL¹(Ω).

As a consequence of Theorem 1, we have that the set of good measuresGcontains all measuresµwhich satisfy µ4πH^N−2. IfN=2, then the result of Vázquez states that the converse is also true. In our case, that isN3, this is false. After this work was completed, A.C. Ponce found explicit examples of good measures which are not

4πH^N−2(see [7, Theorems 2 and 3]). The existence of such measures was conjectured by L. Véron in a personal communication.

We now present some necessary conditions a measureµ∈Ghas to satisfy. We start with the following

Theorem 3. Letµ∈M(Ω). Ifµ(A) >0 for some Borel setA⊂Ω such that dim_H(A) < N−2, then (1.1) has no solution.

Observe that in the case of dimensionN=2, no measureµsatisfies the assumptions of Theorem 3.

As a consequence of Theorem 3 we have

Corollary 2. Letµ∈M(Ω). Ifµ⁺is concentrated on a Borel setA⊂Ωwith dim_H(A) < N−2, thenµ^∗= −µ⁻. The next theorem, which is one of the main results of this paper, states that there exists no solution of (1.1) ifµ is strictly larger than 4πH^N−2on an(N−2)-rectifiable set.

Theorem 4. Letµ∈M(Ω). Assume there existε >0 and an(N−2)-rectifiable setE⊂Ω, withH^N−2(E) >0, such thatµ E(4π+ε)H^N−2 E. Then, (1.1) has no solution.

Corollary 3. Assumeµ=α(x)H^N−2 E, whereE⊂Ω is(N−2)-rectifiable andαisH^N−2 E-integrable.

Then,µ^∗=min{4π, α(x)}H^N−2 E.

In Theorem 4 (and also in Corollary 3), the assumption thatEis(N−2)-rectifiable is important. In fact, one can find(N−2)-unrectifiable setsF ⊂Ω, with 0<H^N−2(F ) <∞, such thatν=αH^N−2 F is a good measure for everyα >0 (see [7]).

As a consequence of the previous results, we can derive some information onµ^∗. To this extent, letµ∈M(Ω).

Since e^u−1 is bounded foru <0,µ⁻will play no role in the existence-nonexistence theory for (1.1). Therefore, we only have to deal withµ⁺, which we recall can be uniquely decomposed as

µ⁺=µ₁+µ₂+µ₃, (1.5)

where

µ₁(A)=0 for every Borel setA⊂Ωsuch thatH^N−2(A) <∞, (1.6) µ₂=α(x)H^N−2 E for some Borel setE⊂Ω, and someH^N−2-measurableα, (1.7) µ₃(Ω\F )=0 for some Borel setF ⊂ΩwithH^N−2(F )=0. (1.8) By a result of Federer (see [4] and also [6, Theorem 15.6]), the setEcan be uniquely decomposed as a disjoint unionE=E₁∪E₂, whereE₁is(N−2)-rectifiable andE₂is purely(N−2)-unrectifiable. In particular,

µ₂=α(x)H^N−2 E₁+α(x)H^N−2 E₂. (1.9)

Combining Corollaries 1–3, we establish the following

Theorem 5. Givenµ∈M(Ω), decomposeµ⁺as in (1.5)–(1.9). Then,

µ^∗=(µ₁)^∗+(µ₂)^∗+(µ₃)^∗−µ⁻. (1.10)

In addition,

(µ₁)^∗=µ₁, (1.11) (µ2)^∗= α(x)H^N−2 E1

_∗

+ α(x)H^N−2 E2

_∗

, (1.12)

α(x)H^N−2 E_1∗

=min

4π, α(x)

H^N−2 E₁, (1.13)

α(x)H^N−2 E2

_∗

min

4π, α(x)

H^N−2 E2, (1.14)

(µ₃)^∗(A)=0 for every Borel setA⊂Ωwith dim_H(A) < N−2. (1.15) In view of the examples presented in [7], one can find measuresµ0 for which equality in (1.14) fails and such that(µ₃)^∗(F ) >0 for some Borel setF⊂Ω, withH^N−2(F )=0.

The plan of the paper is as follows. In the next section we will prove Theorem 2. In Section 3 we will present a decomposition result for Radon measures. Theorem 1 will then be proved in Section 4. Theorems 3 and 4 will be established in Section 5. The last section will be devoted to the proof of Theorem 5 and Corollaries 1–3.

2. Proof of Theorem 2

We first recall the definition of the Morrey space M^p(Ω); see [5].

Definition 1. Letp1 be a real number. We say that a functionf ∈L¹(Ω)belongs to the Morrey space M^p(Ω) if

f p=sup

B_r

1 r^{N (1−1/p)}

Ω∩Br

f (y)dy <+∞,

where the supremum is taken over all open ballsBr⊂R^N.

The following theorem is well-known (for the proof, see for example [5, Section 7.9]).

Theorem 6. Letf ∈M^p(Ω)for somep^N₂, and letube the solution of −u=f inΩ,

u=0 on∂Ω.

Ifp >^N₂, thenu belongs toL^∞(Ω). Ifp=^N₂, then e^β|u| is uniformly bounded inL¹(Ω)norm, for everyβ <

β₀=2N ω_N/(e f

N/2).

Theorem 2 in the Introduction improves the upper boundβ₀given in [5]. It turns out that the constant ^{2N ω}_f ^N

N/2

is sharp. Indeed we have the following

Example 1. LetE= {x=(x₁, x₂, . . . , x_N)∈R^N: x₁=x₂=0},and letµ=4πH^N−2 E.Defineµ_n=ρ_n∗µ, where(ρ_n)is a sequence of mollifiers, and letu_nbe the solution of

−u_n=µ_n inB₂(0), u_n=0 on∂B₂(0).

By standard elliptic estimates,u_n→uinW₀^1,q(B₂(0)), for everyq < _N^N₋₁and a.e., whereuis the solution of −u=4πH^N−2 E inB₂(0),

u=0 on∂B₂(0).

Using the Green representation formula, and setting ρ(x)=dist(x, E), one can prove that u(x) behaves as

−2 lnρ(x),for anyx in a suitable neighborhood ofE∩B₁(0).Moreover, it is easy to verify that µ_n

N/2→2N ω_N asn→ ∞. Then, by Fatou’s lemma

lim inf

n→+∞

B2(0)

e^{(2N ωN/ µn N/2)un}

B2(0)

e^u= +∞.

We now turn to the proof of Theorem 2. We start with the following well-known Lemma 1. Letf:[0, d] →R⁺be aC¹-function, and

g(r)= sup

t∈[0,r]f (t ).

Then,gis absolutely continuous on[0, d], and its derivative satisfies the following inequality:

0g(r) f(r)₊

a.e., (2.1)

wheres⁺=max{s,0}is the positive part ofs∈R.

Proof. First of all, observe that sincef is continuous, then so isg. We now prove that, for everyx < yin[0, d], there existx˜y˜in[x, y]such that

0g(y)−g(x)

f (y)˜ −f (x)˜ ₊

. (2.2)

Indeed, ifg(y)=g(x), then it is enough to choosex˜=xandy˜=y. Ifg(y) > g(x), then let us define

˜

x=max

zx: g(z)=g(x)

and y˜=min

zy: g(z)=g(y) .

Clearly, sinceg is nondecreasing, we havex˜y˜. In order to prove (2.2), simply observe thatf (x)˜ =g(x)and f (y)˜ =g(y). Indeed, if for examplef (x)˜ =g(x), then it must bef (x) < g(x), and this implies that˜ g(z)=g(x) for somez > x, thus contradicting the definition ofx˜.

Sincef is absolutely continuous, (2.2) implies thatgis absolutely continuous, as required, so thatg(r)exists for almost everyr. We now establish (2.1). Starting from (2.2), and applying the mean value problem tof, we have that there existsξ˜∈ [ ˜x,y˜]such that

0g(y)−g(x)

f (y)˜ −f (x)˜ ₊

= f(˜ξ )₊

(y˜− ˜x) f(˜ξ )₊

(y−x).

Dividing byy−x, and lettingy→x, the result follows. 2 Proof of Theorem 2. We split the proof into two steps:

Step 1. Givenf ∈C_c^∞(Ω),f0, let

v(x)= 1

N (N−2)ω_N

Ω

1

|x−y|^N−2− 1 d^N−2

f (y)dy ∀x∈Ω, (2.3)

wheredis the diameter ofΩ. Then, for every 0< α <2N ω_N, it holds

Ω

e^{((2N ωN−α)/f N/2)v(x)}dx(N ω_N)²

α d^N. (2.4)

Let us set ν(x, r)=

B_r(x)

f (y)dy ∀x∈Ω.

In particular,

ν(x, r)ω_Nr^N f _L∞ and ν(x, r)=

∂Br(x)

f (y)dσ (y)N ω_Nr^N−1 f _L∞, (2.5)

wheredenotes the derivative with respect toranddσis the(N−1)-dimensional measure on∂B_r(x). Then,

v(x)= 1

N (N−2)ω_N d

0

1

r^N−2− 1 d^N−2

∂B_r(x)

f (y)dσ (y)

dr

= 1

N (N−2)ω_N d

0

1

r^N−2− 1 d^N−2

ν(x, r)dr.

Integrating by parts, we have

v(x)= 1

N (N−2)ω_N 1

r^N−2− 1 d^N−2

ν(x, r)

^d

0

+ 1 N ω_N

d

0

ν(x, r) r^N−1 dr.

By (2.5), lim

r→0

ν(x, r) r^N−2 =0, and so

v(x)= 1 N ω_N

d

0

ν(x, r) r^N−1 dr.

Define now

ψ (x, r)= sup

t∈[0,r]

ν(x, t ) t^N−2 .

It follows from Lemma 1 thatψ (x,·)is absolutely continuous. Then, integrating by parts,

v(x) 1 N ωN

d

0

ψ (x, r)

r dr= − 1 N ωN

d

0

ln

d r

ψ (x, r)dr

= − 1

N ω_Nψ (x, r)ln d

r ^d

0

+ 1 N ω_N

d

0

ln d

r

ψ(x, r)dr.

By (2.5), lim

r→0ψ (x, r)ln d

r

=0,

and then, observing thatψ (x, d)ν(x, d)/d^N−2= f

L¹/d^N−2>0, v(x) 1

N ωN

d

0

ln d

r

ψ(x, r)dr= d

0

ψ (x, d) N ωN

ln d

r

ψ(x, r) ψ (x, d)dr.

Therefore, for any 0< α <2N ω_N, e^{((2N ωN−α)/f N/2)v(x)}exp

d 0

2N ωN−α f N/2

ψ (x, d) N ωN

ln d

r

ψ(x, r) ψ (x, d)dr

.

Since ^ψ_ψ(x,d)^(x,r)dris a probability measure on(0, d), Jensen’s inequality implies

e^{((2N ωN−α)/f N/2)v(x)} d

0

d r

((2N ω_N−α)/f _N/2)(ψ(x,d)/N ω_N)

ψ(x, r) ψ (x, d)dr.

Clearly,

ψ (x, d)sup

y∈Ω

ψ (y, d)= f

N/2 and ψ (x, d) f

L¹

d^N−2. Thus,

e^{((2N ωN−α)/f N/2)v(x)}d^{N−α/N ωN} f L¹

d

0

ψ(x, r)

r^{2−α/N ωN} dr. (2.6)

Now, by (2.1) we have ψ(x, r)

ν(x, r) r^N−2

₊

ν(x, r) r^N−2 , so that

Ω

ψ(x, r)dx 1 r^N−2

Ω ∂B_r(x)

f (y)dσ (y)

dx= 1 r^N−2

Ω ∂B_r(0)

f (y+x)dσ (y)

dx

= 1 r^N−2

∂Br(0) Ω

f (y+x)dx

dσ (y)N ω_Nr f

L¹.

Hence, from (2.6),

Ω

e^{((2N ωN−α)/f N/2)v(x)}dxN ω_Nd^{N−α/N ωN} d

0

dr

r^{1−α/N ωN} =(N ωN)² α d^N

which is (2.4). This concludes the proof of Step 1.

Step 2. Proof of Theorem 2 completed.

Letf ∈M^N/2(Ω). Clearly, it suffices to prove the theorem forf 0. By extendingf to be identically zero outsideΩ, we have

B_r

f (y)dy f

N/2r^N−2 for every ballBr⊂R^N. (2.7)

Let(ρ_n)⊂C_c^∞(B₁),ρ_n0, be a sequence of mollifiers. Take(ζ_n)⊂C_c^∞(Ω)to be such that 0ζ_n1 inΩ, and ζ_n(x)=1 ifd(x, ∂Ω)_n¹. Setf_n=ζ_n(ρ_n∗f ). We claim that

f_n

N/2 f

N/2 ∀n1. (2.8)

In fact, given any ballB_r(z)⊂R^N, we have

Br(z)

f_n(x)dx

Br(z)

(ρ_n∗f )(x)dx=

Br(z) R^N

ρ_n(x−y)f (y)dy

dx=

R^N Br(z−t )

f (y)dy

ρ_n(t )dt.

Since (2.7) holds, we get

B_r(z)

f_n(x)dx f

N/2r^N−2

R^N

ρ_n(t )dt= f

N/2r^N−2,

which is precisely (2.8).

Letu_nbe the unique solution of −u_n=f_n inΩ,

u_n=0 on∂Ω.

We shall denote byv_n the function given by (2.3), withf replaced byf_n. Note that, by the standard maximum principle, 0u_nv_ninΩ,∀n1. Given 0< α <2N ω_N, it follows from (2.8) and the previous step that

Ω

e^{((2N ωN−α)/f N/2)un(x)}dx

Ω

e^{((2N ωN−α)/fn N/2)vn(x)}dx(N ω_N)²

α d^N ∀n1. (2.9)

Sincefn→f inL¹(Ω), standard elliptic estimates imply thatun→uinL¹(Ω)and a.e. Thus, asn→ ∞in (2.9), it follows from Fatou’s lemma that e^{((2N ωN−α)/f N/2)u}∈L¹(Ω)and

Ω

e^{((2N ωN−α)/f N/2)u(x)}dx(N ωN)² α d^N.

This concludes the proof of the theorem. 2

3. A useful decomposition result

Our goal in this section is to establish the following:

Lemma 2. Letµ∈M(R^N),µ0. Givenδ >0, there exists an open setA⊂R^N such that (a) µ(Br\A)2N ωNr^N−2for every ballBr⊂R^Nwith 0< r < δ;

(b) for every compact setK⊂A, µ N2δ(K)

4πH^N_δ⁻²(K),

whereN_2δ(K)denotes the neighborhood ofKof radius 2δ.

Proof. Given a sequence of open sets(A_k)_k0, for eachk1 we let R_k=sup

r∈ [0, δ): µ(B_r\A_k−1)2N ω_Nr^N−2for some ballB_r⊂R^N

. (3.1)

We now construct the sequence(A_k)inductively as follows. LetA₀=φ. We have two possibilities. IfR₁=0, then we takeA_k=φfor everyk1. Otherwise,R₁>0 and there existsr₁∈(^R₂¹, R₁]andx₁∈R^N such that

µ B_r1(x₁)

2N ω_Nr₁^N−2.

Let A₁=B_r1(x₁). IfR₂=0, then we letA_k=φ for every k2. AssumeR₂>0. In this case, we may find r₂∈(^R₂², R₂]andx₂∈R^Nsuch that

µ B_r2(x₂)\A₁

2N ω_Nr₂^N−2.

Proceeding by induction, we obtain a sequence of ballsB_r1(x₁), B_r2(x₂), . . .and open sets

A_k=B_r1(x₁)∪ · · · ∪B_rk(x_k) (3.2) such that

R_k

2 < rkRk (3.3)

and

µ B_rk(x_k)\A_k−1

2N ω_Nr_k^N−2 ∀k1. (3.4)

Note thatR_k→0 ask→ ∞. In fact, by (3.3) and (3.4) we have N ω_N

2^N−3 ∞ k=1

R_k^N−22N ω_N ∞ k=1

r_k^N−2∞

k=1

µ B_rk(x_k)\A_k−1

=µ

k

B_rk(x_k)

µ _M.

In particular,

kR_k^N−2<∞, which implies the desired result.

Let A=

∞ j=1

Aj= ∞ k=1

Br_k(xk).

We claim thatAsatisfies (a) and (b).

Proof of (a). GivenB_r⊂R^Nsuch that 0< r < δ, letk1 be sufficiently large so thatR_k< r. By the definition ofR_k, we haveµ(B_r\A_k)2N ω_Nr^N−2. SinceA_k⊂A, we haveB_r\A⊂B_r\A_k and the result follows.

Proof of (b). Given a compact setK⊂A, let J=

j1:Br_j(xj)∩K=φ . In particular,

K⊂

j∈J

B_rj(x_j).

Moreover, sincerj< δ, we haveBr_j(xj)⊂N2δ(K)for everyj∈J. Thus, µ N_2δ(K)

µ

j∈J

B_rj(x_j)

µ

j∈J

B_rj(x_j)\A_j−1

=

j∈J

µ B_rj(x_j)\A_j−1

2N ω_N

j∈J

r_j^N−22N ω_N

ω_N−2H^N_δ⁻²(K).

Since 2N ω_N/ω_N−2=4π, we get µ N_2δ(K)

4πH^N_δ⁻²(K).

This concludes the proof of Lemma 2. 2

4. Proof of Theorem 1

We first observe that, as a consequence of Theorem 2, we have the following Proposition 1. Letµ∈M(Ω)be such that

µ⁺(Ω∩B_r)2N ω_Nr^N−2 for every ballB_r⊂R^N. Then,µis a good measure for (1.1).

Proof. Sinceµµ⁺, it is enough to show thatµ⁺is a good measure. Thus, without loss of generality, we may assume thatµ0. Moreover, extendingµto be identically zero outsideΩ, we may also assume thatµ∈M(R^N) and

µ(B_r)2N ω_Nr^N−2 for every ballB_r ⊂R^N. We shall split the proof of Proposition 1 into two steps:

Step 1. Assume there existsε >0 such that

µ(Br)2N ωN(1−ε)r^N−2 for every ballBr⊂R^N. Then,µis a good measure.

Let (ρ_n)⊂C_c^∞(B₁), ρ_n0, be a sequence of mollifiers. Set µ_n=ρ_n∗µ. Proceeding as in the proof of Theorem 2, Step 2, we have

µ_{n N/2}2N ω_N(1−ε) ∀n1.

Letvnbe the unique solution of −vn=µn inΩ,

vn=0 on∂Ω.

Applying Theorem 2 toα=2N ω_N− µ_{n N/2}2N ω_Nε >0,we conclude that

Ω

e^vnC ∀n1, (4.1)

for some constantC >0 independent ofn. By standard elliptic estimatesv_n→va.e., wherevis a solution for −v=µ inΩ,

v=0 on∂Ω.

Hence, by Fatou’s lemma and (4.1), it follows that e^v∈L¹(Ω). Since

−v+e^v−1=µ+e^v−1 inΩ,

µ+e^v−1 is a good measure. In particular,µµ+e^v−1 andv0,imply thatµis a good measure as well.

Step 2. Proof of the proposition completed.

Letα_n↑1. For everyn1, the measureα_nµsatisfies the assumptions of Step 1. Thus,α_nµ∈G,∀n1. Since α_nµ→µstrongly inM(Ω)andGis closed inM(Ω), we haveµ∈G. 2

We recall the following result:

Lemma 3. Ifµ₁, . . . , µ_k∈M(Ω)are good measures for (1.1), then so is sup_iµ_i.

Proof. Ifk=2, this is precisely [1, Corollary 4]. The general case easily follows by induction onk. 2 We then have a slightly improved version of Proposition 1:

Proposition 2. Letµ∈M(Ω). Assume there existsδ >0 such that µ⁺(Ω∩B_r)2N ω_Nr^N−2 for every ballB_r⊂R^Nwithr∈(0, δ).

Then,µis a good measure for (1.1).

Proof. LetB_δ(x₁), . . . , B_δ(x_k)be a finite covering ofΩ. For eachi=1, . . . , k, letµ_i=µ B_δ(x_i)∈M(Ω). It is easy to see thatµ_i satisfies the assumptions of Proposition 1, so that eachµ_i is a good measure for (1.1). Thus, by the previous lemma, sup_iµ_i∈G. Sinceµsup_iµ_i, we conclude thatµis also a good measure for (1.1). 2

We can now present the

Proof of Theorem 1. As above, sinceµµ⁺, it suffices to show thatµ⁺is a good measure. In particular, we may assume thatµ0. Moreover, it suffices to establish the theorem for a measureµsuch thatµ(4π−ε)H^N−2 for someε >0. The general case follows as in Step 2 of Proposition 1.

We first extendµto be identically zero outsideΩ. By Lemma 2, there exists an open setAˆ₁⊂R^Nsuch that(a) and(b)hold withδ=1 andA= ˆA₁. By induction, given an open setAˆ_k−1⊂R^N, we apply Lemma 2 toµ Aˆ_k−1 andδ_k=¹_k to obtain an open setAˆ_k⊂ ˆA_k−1such that

(a_k) µ Aˆ_k−1(B_r\ ˆA_k)2N ω_Nr^N−2for every ballB_r⊂R^Nwith 0< r <¹_k; (bk) for every compact setK⊂ ˆAk,

µ N_{2/ k}(K)

µ Aˆ_k−1 N_{2/ k}(K)

4πH^N_{1/ k}⁻²(K).

By Proposition 2, each measureµ Ω\ ˆA₁,µ Aˆ₁\ ˆA₂, . . . , µ Aˆ_k−1\ ˆA_k is good. We now invoke Lemma 3 to conclude that

µ Ω\ ˆA_k=sup{µ Ω\ ˆA₁, µ Aˆ₁\ ˆA₂, . . . , µ Aˆ_k−1\ ˆA_k} is a good measure for everyk1. LetAˆ=

kAˆ_k. Sinceµ Ω\ ˆA_k→µ Ω\ ˆAstrongly inM(Ω)and the setG of good measures is closed with respect to the strong topology, we conclude thatµ Ω\ ˆAis also a good measure for (1.1).

We now claim thatµ(A)ˆ =0. In fact, letK⊂ ˆAbe a compact set. In particular,K⊂ ˆA_k. By(b_k), we have µ N2/ k(K)

4πH^N_{1/ k}⁻²(K) ∀k1.

Ask→ ∞, we conclude that

µ(K)4πH^N−2(K). (4.2)

In particular,H^N−2(K) <∞. Recall that, by assumption,

µ(K)4π(1−ε)H^N−2(K). (4.3)

Combining (4.2) and (4.3), we getµ(K)=0. SinceK⊂ ˆAis arbitrary, we conclude thatµ(A)ˆ =0. Therefore, µ=µ Ω\ ˆAand soµis a good measure. This concludes the proof of Theorem 1. 2

5. Proofs of Theorems 3 and 4

In this section we derive some necessary conditions for a measure to be good for problem (1.1). Let us start with a regularity property for solutions of elliptic equations with measure data.

Lemma 4. Letν∈M(Ω)and letube the solution of the Dirichlet problem −u=ν inΩ,

u=0 on∂Ω. (5.1)

If e^u∈L¹(Ω), thenu⁺belongs toW₀^1,p(Ω)for everyp <2, and u⁺

W₀^1,p C p,measΩ, ν _M, e^u _L1

∀p <2. (5.2)

Proof. Letνn=ρn∗ν, where(ρn)is a sequence of mollifiers, and letunbe the solution of −u_n=ν_n inΩ,

un=0 on∂Ω. (5.3)

Then it is well-known that the sequence(u_n)converges touinW₀^1,q(Ω), for everyq <_N^N₋₁ (see [8]).

UsingT_k(u⁺_n)=min{k,max{u_n,0}}as a test function in (5.3), we have

Ω

∇Tk(u⁺_n)²dx=

Ω

Tk(u⁺_n) νndxk νn L¹k ν _M. Lettingn→ ∞, by weak lower semicontinuity we obtain

Ω

∇T_k(u⁺)²dxk ν _M. (5.4)

On the other hand, assumption e^u∈L¹(Ω)implies, for everyk >0, e^kmeas{u > k}

{u>k}

e^udx e^u _L1,

and so

meas{u > k}e^−k e^u _L1. (5.5)

For everyη >1 we have |∇u⁺|> η

=

|∇u|> η u > k

∪

|∇u|> η 0uk

,

so that, by (5.4) and (5.5), meas

|∇u⁺|> η

meas{u > k} +meas

|∇u|> η 0uk

e^−k e^u _L1+ 1 η²

Ω

∇Tk(u⁺)²dxC

e^−k+ k η²

,

whereC=max{ e^u _L1, ν _M}. Minimizing onk, we find meas

|∇u⁺|> η

C1+2 lnη η² .

Therefore,|∇u⁺| belongs to the Marcinkiewicz space of exponentp, for every p <2.Since Ω is bounded, it follows that|∇u⁺| ∈L^p(Ω), for everyp <2, and that (5.2) holds. 2

Theorem 3 can now be obtained as a consequence of the above results.

Proof of Theorem 3. By inner regularity, it is enough to prove that ifµ∈M(Ω)is a good measure for prob- lem (1.1), thenµ(K)0 for every compact setK⊂Ωwith dim_H(K) < N−2.

By Lemma 3, ifµis a good measure, then so isµ⁺=sup{µ,0}. Letv0 be the solution of problem (1.1) with datumµ⁺. In particular,vsatisfies

Ω

∇v∇ζ+

Ω

(e^v−1)ζ=

Ω

ζdµ⁺ ∀ζ ∈C_c^∞(Ω). (5.6)

Take now a compact setK⊂Ω with dim_H(K) < N−2, and letq be such that 2< q < N−dim_H(K). Then the q-capacity ofKis zero (see e.g. [3]), and there exists a sequence of smooth functionsζn∈C_c^∞(Ω)such that

0ζn1 inΩ, ζn=1 inK, ζn→0 inW₀^1,q(Ω)and a.e. (5.7)

Usingζ_nas test function in (5.6) yields 0µ⁺(K)

Ω

ζ_ndµ⁺=

Ω

∇v∇ζ_n+

Ω

(e^v−1) ζ_n.

Since, by Lemma 4,v∈W₀^1,q(Ω), the right-hand side tends to 0 asn→ ∞. Hence,µ⁺(K)=0, which implies µ(K)0, as desired. 2

Before presenting the proof of Theorem 4, we need some preliminary lemmas. The first one is well-known (see e.g. [3]).

Lemma 5. Iff ∈L¹(R^N), then, for every 0s < N,

rlim→0

1 r^s

B_r(x)

f (y)dy=0 H^s-a.e. inR^N.

In the following, we will denote the angular mean of a functionw∈L¹(R^N)on the sphere centered atx∈R^N with radiusr >0 by

w(x, r)= –

∂Br(x)

wdσ= 1 N ωNr^N−1

∂Br(x)

wdσ. (5.8)

The next result provides an estimate of the asymptotic behavior, asr→0, of the angular mean of a function in terms of its Laplacian.

Lemma 6. Letw∈L¹(R^N)be such thatw∈M(R^N). Setµ= −w. Then, 1

N ω_N lim inf

r→0

µ(B_r(x))

r^N−2 lim inf

r→0

w(x, r)

ln(1/r) lim sup

r→0

w(x, r) ln(1/r) 1

N ω_Nlim sup

r→0

µ(B_r(x)) r^N−2 .

Proof. We claim that, for every 0< r < s <1, we have

w(x, r)− w(x, s)= 1 N ω_N

s

r

µ(B_ρ(x))

ρ^N−1 dρ. (5.9)