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Multiplicity of Solutions to Elliptic Problems Involving the 1-Laplacian with a Critical Gradient Term

  • Boumediene Abdellaoui , Andrea Dall’Aglio EMAIL logo and Sergio Segura de León
Published/Copyright: April 21, 2017

Abstract

In the present paper we study the Dirichlet problem for an equation involving the 1-Laplacian and a total variation term as reaction.We prove a strong multiplicity result. Namely, we show that for any positive Radon measure concentrated in a set away from the boundary and singular with respect to a certain capacity, there exists an unbounded solution, and measures supported on disjoint sets generate different solutions.These results can be viewed as the analogue for the 1-Laplacian operator of some known multiplicity results which were first obtained by Ireneo Peral, to whom this article is dedicated, and his collaborators.

1 Introduction and Statement of the Main Result

The starting point of this paper lies in the paper [6] by Andreu, Dall’Aglio and Segura de León.In that paper, existence and uniqueness results for the following problem were obtained:

(1.1) { u - div ( D u | D u | ) = | D u | + f ( x ) in Ω , u = 0 on Ω ,

where Ω is a bounded open subset of N. The main result of that article provides a bounded solution for every datum f(x) belonging to Lm(Ω), with m>N (see [6, Theorem 1]). Knowing that for small enough data the solution is the trivial one (see [22, Theorem 4.2]), it was further shown that this is the unique bounded solution (see [6, Proposition 4]).

On the other hand, in [14], the second and third author studied the problem

(1.2) { - div ( D u | D u | ) = | D u | + f ( x ) in Ω , u = 0 on Ω ,

showing a criterion on the datum to determine when the unique solution is the trivial one.The question that is dealt in the present paper is whether there exists some other solution (which must be unbounded). We will focus on problem (1.2) for the sake of simplicity.

In the p-Laplacian setting (p>1) problem (1.2) has been studied in [3, 1, 2].The main result of the present paper can be seen as a translation of those to our setting.

Recall that Abdellaoui, Dall’Aglio and Peral proved in [3] the following multiplicity result.

  1. Let fLm(Ω) be nonnegative and small enough, where m>N2.For any positive Radon measure μ which is concentrated on a set which has zero capacity, there exists a solution uμ to the problem

    { - Δ u μ = | u μ | 2 + f ( x ) in Ω , u μ = 0 on Ω ,

    which is unbounded near the set where the measure is concentrated.

This kind of result was later developed and extended by Abdel Hamid and Bidaut-Veron to the case of the p-Laplacian, with p>1 (see [1, 2]).It must be noted that the results in those papers rely on a Cole–Hopf change of unknown, which does not work in the case p=1.The absence of any kind of Cole–Hopf change of unknown is indeed one of the biggest difficulties to deal with multiplicity in our framework.Another difficulty comes from the definition of solution to problems involving the 1-Laplacian operator (as introduced in [5] by Andreu, Ballester, Caselles and Mazón, see also [7]).Indeed, the suitable energy space is BV(Ω), the space of all functions of bounded variation, and this notion relies on a bounded vector field 𝐳 which plays the role of Du|Du| in the sense that it satisfies 𝐳1 and (𝐳,Du)=|Du| (where (𝐳,Du) stands for a type of dot product of 𝐳 and Du). Moreover, the equation holds, and so -div𝐳=|Du|+f is just a Radon measure.In our computations, we need to manipulate products of the form (𝐳,Du).According to the Anzellotti theory, (𝐳,Du) is well defined as a Radon measure whenever div𝐳 is a Radon measure and uBV(Ω)C(Ω)L(Ω).This last condition can be relaxed to avoid the continuity of u, namely, div𝐳 is a Radon measure and uBV(Ω)L(Ω) (see [12], see also [23, 11]).Nevertheless, this is not enough to deal with unbounded solutions.For this reason, we have to extend the Anzellotti theory (see Section 3 below).

Despite all these difficulties, we can prove the multiplicity of solutions for the 1-Laplacian.Being a very singular operator, the result we obtain is not as sharp as the one stated for the Laplacian.Indeed, measures are assumed to be singular with respect to a stronger capacity and must be zero near the boundary of Ω.Nevertheless, we can prove the following result.

Theorem 1.1

Let fLm(Ω), with m>N, be nonnegative and small enough to satisfy

(1.3) f m < ( N - m ( N - 1 ) N ) ( m - 1 ) / m | Ω | [ ( N - 1 ) / N ] - 1 / m S N , 1 ,

where SN,1 denotes the best constant in the Sobolev embedding BV(Ω)LN/(N-1)(Ω).Let μ be a positive Radon measure which is concentrated on a set A of zero W1,q-capacity. Assume also that the distance from A to Ω is positive.Consider a renormalized solution vp of

(1.4) { - Δ p ( v p ) = f ( x ) ( 1 + v p p - 1 ) p - 1 + μ in Ω , v p = 0 on Ω ,

and set up=(p-1)log(1+vpp-1).Then the “sequence” (up) converges, as p goes to 1, to an unbounded solution uμ of the problem

(1.5) { - div ( D u μ | D u μ | ) = | D u μ | + f ( x ) in Ω , u μ = 0 on Ω ,

in the sense of Definition 4.1 below.This solution is, however, bounded on every subset of Ω having positive distance from the set A, satisfies eδuμBV(Ω) for all 0<δ<1, and

(1.6) - div ( e u μ 𝐳 ) = e u μ f + μ in 𝒟 ( Ω ) ,

where zL(Ω;RN) is the vector field appearing in Definition 4.1 below.

This result implies a strong multiplicity of solutions.Indeed, we can take two singular measures μ1, μ2 concentrated on two disjoint closed sets A1, A2; for instance, two Dirac deltas δx1, δx2.Each of them generates an unbounded solution uμ1, uμ2 of (1.5), and the two are different from each other, since uμ1 is unbounded near A1, and bounded near A2, and vice versa.

The plan of this paper is the following.The next section is devoted to preliminaries, i.e., we introduce our notation, the notions of capacity and renormalized solution as well as auxiliary results on BV-functions.Section 3 is devoted to extend the Anzellotti theory of L-divergence-measure vector fields to our needs. The multiplicity Theorem 1.1 is proved in Section 4, while the last section deals with radial explicit solutions.

2 Preliminaries

2.1 General Notation

From now on, we fix an integer N2.The symbol N-1(E) stands for the (N-1)-dimensional Hausdorff measure of a set EN, and |E| for its Lebesgue measure.Moreover, Ω will always denote an open subset of N with Lipschitz boundary.Thus, an outward normal unit vector ν(x) is defined for N-1-almost every xΩ.

The space of all C-functions having compact support in Ω is denoted by C0(Ω).The symbol Lq(Ω), with 1q, denotes the usual Lebesgue space with respect to Lebesgue measure and q=qq-1 is the conjugate of q.We will denote by W01,q(Ω) the usual Sobolev space of measurable functions having weak gradient in Lq(Ω;N) and zero trace on Ω.The dual space of W01,q(Ω) will be denoted by W-1,q(Ω); we recall that its elements can be written as divF for some FLq(Ω;N).Finally, if 1p<N, we will denote by p*=NpN-p its Sobolev conjugate exponent and by SN,p the best constant in the Sobolev embedding, i.e.,

( Ω | u | p * ) p / p * S N , p Ω | u | p .

The truncation function will be use throughout this paper.Given k>0, it is defined by

T k ( s ) = min { | s | , k } sign ( s )

for all s.Moreover, we set Gk(s)=s-Tk(s) for all s.

2.2 Capacity

Let 1p<N.For every compact set KΩ, we define its p-capacity with respect to Ω as

cap 1 , p ( K , Ω ) = inf { Ω | u | p : u C 0 ( Ω ) , u χ K }

(we will use the convention that inf=+).For any open set UΩ, its p-capacity is then defined by

cap 1 , p ( U , Ω ) = sup { cap 1 , p ( K , Ω ) : K is a compact subset of U } .

Finally, given a Borelian subset BΩ, the definition is extended by setting

cap 1 , p ( B , Ω ) = inf { cap 1 , p ( U , Ω ) : U is an open subset of Ω , B U } .

We point out that the p-capacity is not a Radon measure, although it is an outer measure.Using the definition of capacity, it is easy to see that 1<p<q and cap1,q(A,Ω)=0 imply that cap1,p(A,Ω)=0 as well as N-1(A)=0.

2.3 Radon Measures

We recall that a Radon measure is a distribution of order 0 and that every positive distribution T, which is a distribution satisfying T,φ0 for all nonnegative φC0(Ω), is a nonnegative Radon measure.Given a Radon measure μ, we denote by |μ| its total variation. The Lebesgue spaces with respect to μ are denoted by Lq(Ω,μ), where 1q.

For a Radon measure μ in Ω and a Borel set AΩ,the measure μA is defined by (μA)(B)=μ(AB) for any Borel set BΩ.If a measure μ is such that μ=μA for a certain Borel set A, then the measure μ is said to be concentrated on A.

Let μ be a Radon measure in Ω.We say that μ is singular with respect to the p-capacity if it is concentrated on a subset EΩ such that cap1,p(E,Ω)=0,and we say that it is absolutely continuous with respect to the p-capacity if cap1,p(E,Ω)=0 implies μ(E)=0.Although the p-capacity is not a measure, every Radon measure μ can be decomposed as μ=μa+μs, where μa is absolutely continuous and μs is singular, with respect to the p-capacity.Moreover, thanks to [10, Theorem 2.1], every Radon measure μ which is absolutely continuous with respect to the p-capacity can be written as μ=f-divF, where fL1(Ω) and FLp(Ω;N).

2.4 Definition of Renormalized Solution

Two definitions of solution must be considered, those to problem (1.4) and to problem (1.5).We point out that the definition of a solution to problem (1.5) relies on the theory of L-divergence-measure vector fields, which will be studied in the next section, so we postpone the definition of a solution to problem (1.5) to Section 4.We now introduce the concept of renormalized solution to problem (1.4);we refer to [15] for a detailed study of this concept.

Given the measure μ, we decompose it as μ=μ0+μs+-μs-, where μ0 is absolutely continuous with respect to the p-capacity, while μs+ and μs- are two nonnegative measures which are concentrated on two disjoint subsets of zero p-capacity.

Let f(x) be a function in L1(Ω) and h(s) a real continuous function.A measurable function v:Ω is a renormalized solution to the problem

{ - Δ p ( v ) = f ( x ) h ( v ) + μ in Ω , v = 0 on Ω ,

if the following conditions hold:

  1. The function v is finite almost everywhere and Tk(v)W01,p(Ω) for all k>0.(As a consequence, a generalized gradient v can be defined, see [9, Lemma 2.1].)

  2. The gradient satisfies |v|p-1Lq(Ω) for every q<NN-1.

  3. f h ( v ) L 1 ( Ω ) .

  4. For every function SW1,() such that S has compact support in (consequently the limits S(+)=lims+S(s) and S(-)=lims-S(s) exist), we have

    Ω S ( v ) φ | v | p + Ω S ( v ) | v | p - 2 v φ
    = Ω f ( x ) h ( v ) S ( v ) φ + Ω S ( v ) φ 𝑑 μ 0 + S ( + ) Ω φ 𝑑 μ s + - S ( - ) Ω φ 𝑑 μ s -

    for all φW1,r(Ω)L(Ω), with r>N, such that S(v)φW01,p(Ω).

2.5 BV -Functions

The space BV(Ω) of functions of bounded variation isdefined as the space of functions uL1(Ω) whosedistributional gradient Du is a vector valued Radon measure onΩ with finite total variation.This space is a Banach space with the norm defined by

u BV = Ω | u | d x + | D u | ( Ω ) .

We recall that the notion of trace can be extended to any uBV(Ω) and this fact allows us to interpret it as the boundary values of u and to write u|Ω.Moreover, it holds that the trace is a linear bounded operator BV(Ω)L1(Ω) which is onto.Using the trace, an equivalent norm in BV(Ω) can be defined by

u = Ω | u | d N - 1 + | D u | ( Ω ) .

For every uBV(Ω), the Radon measure Du can be decomposed into its absolutely continuous and singular parts with respect to the Lebesgue measure: Du=Dau+Dsu.So, for each measurable set E, we have Dau(E)=Eu(x)𝑑x, where u is the Radon–Nikodým derivative of the measure Dau with respect to the Lebesgue measure.

We denote by Su the set of all xΩ such that x isnot an approximate continuity point of u and by Ju the set of approximate jump points ofu.By the Federer–Vol’pert theorem, [4, Theorem 3.78], we know that Su is countably N-1-rectifiable and N-1(SuJu)=0.Moreover, DuJu=(u+-u-)νuN-1Ju,where u+ and u- denote the “one-sided” limits of u at a jump point.Using Su and Ju, we may split Dsu in two parts: the jumppart Dju and the Cantor part Dcu, defined by

D j u = D s u J u and D c u = D s u ( Ω S u ) ,

respectively.Thereby,

D j u = ( u + - u - ) ν u N - 1 J u .

If x is an approximate continuity point of u,we define the approximate limit of u by u~(x).The precise representativeu*:Ω(SuJu) of u is defined as equal to u~ on ΩSu and equal to u++u-2 on Ju.It is well known (see, for instance, [4, Corollary 3.80]) that if ρ is a symmetric mollifier, then the mollified functions uρϵ converges pointwise to u* in its domain.

To pass to the limit, we will often apply that some functionals defined on BV(Ω) are lower semicontinuous with respect to the convergence in L1(Ω).We recall that the functional defined by

u | D u | ( Ω ) + Ω | u | d N - 1

is lower semicontinuous with respect to the convergence in L1(Ω).Similarly, if we fix φC01(Ω), withφ0, the functional defined by

u Ω φ d | D u |

is lower semicontinuous in L1(Ω).

For further information concerning functions of bounded variation we refer to [4] or [25].

3 Extending Anzellotti’s Theory

In this section we will study some properties involvingdivergence-measure vector fields and functions of boundedvariation. Our aim is to extend the Anzellotti theory introduced in [8].

Following [12], we define 𝒟(Ω) as the space of all vector fields 𝐳L(Ω;N) whose divergence in the sense of distributions is a Radon measure with finite total variation.

The theory of L-divergence-measure vector fields is due to Anzellotti [8] and, independently, to Chen and Frid [12].In spite of their different points of view, both approaches introduce the normal trace of a vector field through the boundary and establish the same generalizedGauss–Green formula.Both also define the pairing (𝐳,Du), where 𝐳𝒟(Ω) and u is a certain BV-function, as a Radon measure.However, they differ in handling this concept.In the present paper we will need the “dot product” to be defined for every uBV(Ω) and every 𝐳𝒟(Ω) satisfying a certain condition (see Corollary 3.4 below).We begin by recalling a result proved in [12].

Proposition 3.1

For every zDM(Ω), the measure μ=divz is absolutely continuous with respect to HN-1.As a consequence, |μ| is also absolutely continuous with respect to HN-1.

Consider now μ=div𝐳 with 𝐳𝒟(Ω) and let uBV(Ω).Then the precise representative u* of u is equal N-1-a.e. to a Borel function, that is, to limϵ0ρϵu, where (ρϵ) is a symmetric mollifier.Then one deduces from Proposition 3.1 that u* is equal μ-a.e. to a Borel function.So, given uBV(Ω), u* is always μ-measurable.Moreover, uBV(Ω)L(Ω) implies uL(Ω,μ)L1(Ω,μ).

3.1 Preservation of the Norm

We point out that every div𝐳, with 𝐳𝒟(Ω), defines a functional on W01,1(Ω) by

div 𝐳 , u W - 1 , ( Ω ) , W 0 1 , 1 ( Ω ) = - Ω 𝐳 u .

To express this functional in terms of an integral with respect to the measure μ=div𝐳, we need a Meyers–Serrin type theorem (see [4, Theorem 3.9] for its extension to BV-functions and [23, Lemma A.1]).

Since

- Ω 𝐳 φ = Ω φ 𝑑 μ

holds for every φC0(Ω), it is easy to obtain this equality for every φW01,1(Ω)C(Ω).Now, given uW01,1(Ω)L(Ω), by [23, Lemma A.1], we may find a sequence (un)n in W01,1(Ω)C(Ω) satisfying unu* in L1(Ω,μ) and unu in L1(Ω;N).Thus,

- Ω 𝐳 u n = Ω u n 𝑑 μ

for every n.Letting n go to infinity, we have that

- Ω 𝐳 u = Ω u * 𝑑 μ ,

and so

div 𝐳 , u W - 1 , ( Ω ) , W 0 1 , 1 ( Ω ) = Ω u * 𝑑 μ

for every uW01,1(Ω)L(Ω).Then the norm of this functional is given by

μ W - 1 , ( Ω ) = sup { | Ω u * d μ | : Ω | u | 1 } .

We have seen that μ=div𝐳 can be extended from W01,1(Ω) to BV(Ω)L(Ω).Next, we will prove that this extension preserves the norm.

Theorem 3.2

If zDM(Ω), then μ=divz can beextended to BV(Ω)L(Ω) in such a way that

μ W - 1 , ( Ω ) = sup { | Ω u * d μ | : u BV ( Ω ) L ( Ω ) , | D u | ( Ω ) + Ω | u | d N - 1 1 } .

Proof.

Since we already know that BV(Ω)L(Ω) is a subset of L1(Ω,μ), all we have to prove is

(3.1) | Ω u * d μ | μ W - 1 , ( Ω ) ( | D u | ( Ω ) + Ω | u | d N - 1 )

for all uBV(Ω)L(Ω).This inequality will be proved in two steps.Step 1.Assume first that uW1,1(Ω)L(Ω). Applying [8, Lemma 5.5], we find a sequence (wn)n in W1,1(Ω)C(Ω) such that

w n | Ω = u | Ω , Ω | w n | Ω | u | d N - 1 + 1 n , Ω | w n | 1 n ,

w n ( x ) = 0 if dist(x,Ω)>1n and wn(x)0 for all xΩ.Then we have

| Ω ( u * - w n * ) 𝑑 μ | = | μ , ( u - w n ) W - 1 , ( Ω ) , W 0 1 , 1 ( Ω ) |
μ W - 1 , ( Ω ) Ω | u - w n |
μ W - 1 , ( Ω ) ( Ω | u | + Ω | u | d N - 1 + 1 n ) .

Hence,

| Ω u * 𝑑 μ | | Ω ( u * - w n * ) 𝑑 μ | + | Ω w n * 𝑑 μ |
(3.2) μ W - 1 , ( Ω ) ( Ω | u | d x + Ω | u | d N - 1 + 1 n ) + | Ω w n * d μ | .

Since the sequence (wn)n tends pointwise to 0 and it is uniformly boundedin L(Ω), by Lebesgue’s theorem, limnΩwn*𝑑μ=0.Then , taking the limit in (3.2), we obtain (3.1).Step 2.In the general case, we apply [23, Lemma A.1] andfind a sequence (un) in W1,1(Ω)C(Ω)L(Ω)which approximates u as in the Meyers–Serrin type theorem.Then, from

| Ω u n * d μ | μ W - 1 , ( Ω ) ( Ω | u n | + Ω | u | d N - 1 ) for all n ,

it follows that (3.1) holds.∎

Corollary 3.3

Let zDM(Ω) satisfy divz=ν+f for a certain Radon measure ν and a certain fLN(Ω).If either ν0 or ν0, then μ=divz can be extended to BV(Ω) and

μ W - 1 , ( Ω ) = sup { | Ω u * d μ | : u BV ( Ω ) , | D u | ( Ω ) + Ω | u | d N - 1 1 } .

Moreover, BV(Ω)L1(Ω,μ).

Proof.

Consider uBV(Ω), write u+=max{u,0} and, for every k>0, apply the previous result to Tk(u+).Then

| Ω ( T k ( u + ) ) * 𝑑 μ | μ W - 1 , ( Ω ) ( | D T k ( u + ) | ( Ω ) + Ω T k ( u + ) 𝑑 N - 1 )
(3.3) μ W - 1 , ( Ω ) ( | D u + | ( Ω ) + Ω u + 𝑑 N - 1 ) .

On the other hand, observe that u* is a ν-measurable function, so we obtain

Ω ( T k ( u + ) ) * 𝑑 μ = Ω T k ( u + ) * 𝑑 ν + Ω T k ( u + ) f

for every k>0.We may apply Levi’s theorem and Lebesgue’s theorem to deduce that

lim k + Ω ( T k ( u + ) ) * 𝑑 ν = Ω ( u + ) * 𝑑 ν and lim k + Ω T k ( u + ) f = Ω u + f .

Thus,

lim k + Ω ( T k ( u + ) ) * 𝑑 μ = Ω ( u + ) * 𝑑 μ .

Now, taking the limit when k goes to in (3.3) yields

(3.4) | Ω ( u + ) * 𝑑 μ | μ W - 1 , ( Ω ) ( | D u + | ( Ω ) + Ω u + 𝑑 N - 1 ) .

Assume, for instance, that ν0.Since Ω(u+)*𝑑μ-=Ωu+f-, we already have that (u+)* is μ--integrable.Hence, as a consequence of (3.4), we deduce that (u+)* is also μ+-integrable and so μ-integrable.

Since we may prove a similar inequality to u-=max{-u,0}, adding both inequalities we deduce thatu* is μ-integrable and that

| Ω u * d μ | μ W - 1 , ( Ω ) ( | D u | ( Ω ) + Ω | u | d N - 1 )

holds true.∎

3.2 A Green Formula

Let 𝐳𝒟(Ω) and uBV(Ω).Assume that div𝐳=ν+f, with ν a Radon measure satisfying either ν0 or ν0, and fLN(Ω).In the spirit of [8], we define the followingdistribution on Ω.For every φC0(Ω), we write

( 𝐳 , D u ) , φ = - Ω u * φ 𝑑 μ - Ω u 𝐳 φ ,

where μ=div𝐳.Note that the previous subsection implies that every term in the above definition has sense.The following result can be proved by applying [23, Proposition A.1] to the truncations Tk(u) and then let k go to infinity.

Proposition 3.4

Let z and u be as above.The distribution (z,Du) defined previously satisfies

| ( 𝐳 , D u ) , φ | φ 𝐳 L ( U ) U d | D u |

for all open set UΩ and for all φC0(U).As a consequence, the distribution (z,Du) is actually a Radon measure. Both (z,Du) and its total variation |(z,Du)| are absolutelycontinuous with respect to the measure |Du|.

On the other hand, for every 𝐳𝒟(Ω), a weak trace on Ω of the normal component of 𝐳 is defined in [8] and denoted by [𝐳,ν].

Proposition 3.5

Let z and u be as above.With the above definitions, the following Green formula holds:

(3.5) Ω u * 𝑑 μ + Ω d ( 𝐳 , D u ) = Ω [ 𝐳 , ν ] u 𝑑 N - 1 ,

where μ=divz.

Proof.

Applying the Green formula proved in [23], we obtain

(3.6) Ω ( T k ( u ) ) * 𝑑 μ + Ω d ( 𝐳 , D T k ( u ) ) = Ω [ 𝐳 , ν ] T k ( u ) 𝑑 N - 1

for every k>0.In the proof of the previous proposition, we have seen that

lim k Ω d ( 𝐳 , D T k ( u ) ) = Ω d ( 𝐳 , D u ) .

We may take limits in the other terms since u*L1(Ω,μ) and uL1(Ω).Hence, letting k go to in (3.6), we get (3.5).∎

4 Multiplicity of Solutions

In this section we will assume that f is a nonnegative function belonging to Lm(Ω), with m>N.We also assume that (1.3) holds.The constant on the right-hand side of (1.3) is obtained in the proof of Theorem 1.1 (see step 1 below).It could also been deduced from an argument by Grenon in [19, 18], checking the dependence on p>1 of every involved constant and letting p go to 1.(It should be mentioned that Grenon assumes p>2NN+1, but this hypothesis can be removed.)We point out that this procedure leads to the same constant.

It is worth showing the translation of condition (1.3) to the N-norm. Indeed, Hölder’s inequality implies

(4.1) f N f m | Ω | [ 1 / N ] - [ 1 / m ] < ( m - N N ( m - 1 ) ) ( m - 1 ) / m S N , 1 - 1 S N , 1 - 1 .

So, [14, Theorem 4.1] implies that the only bounded solution to problem (1.5) is the trivial one.We now turn to define a solution to problem (1.5) following the concept introduced in [5].

Definition 4.1

Given fLm(Ω), with m>N, we say that u is a solution of problem (1.5) if uBV(Ω) is such that the jump part satisfies Dju=0 and there exists a vector field 𝐳𝒟(Ω), with 𝐳1, satisfying

- div 𝐳 = | D u | + f in 𝒟 ( Ω ) ,
(4.2) ( 𝐳 , D u ) = | D u | as measures in Ω ,
(4.3) [ 𝐳 , ν ] sign ( - u ) N - 1 -a.e. on Ω .

By Proposition 3.4, identity (4.2) has sense.Heuristically, identity (4.2), jointly with 𝐳1, leads to 𝐳=Du|Du|, while (4.3) is a weak formulation of the Dirichlet boundary condition.

The proof of our main theorem below uses the following elementary technical result:

Lemma 4.2

If α>0 and 1<p<2, then for all s0,

( e α s / p - 1 ) p e α s - 1 2 ( e α s / p - 1 ) p + 1 .

Proof of Theorem 1.1.

Fix a positive Radon measure μ satisfying the following:

  1. μ is concentrated in a set A,

  2. dist ( A ¯ , Ω ) > 0 ,

  3. there exists q>1 such that capq(A,Ω)=0.

Since our aim is to let p go to 1, we may take q very close to 1.For instance, we may take q2.

The proof of Theorem 1.1 will be developed in several stages.Step 1: Problems with measure datum.For any 1<p<q, consider the problem

(4.4) { - Δ p ( v p ) = f ( x ) ( 1 + v p p - 1 ) p - 1 + μ in Ω , v p = 0 on Ω .

This problem has been studied in [19, Theorem 1.1] and [2, Theorem 6.2].Then (4.1) implies

lim sup p 1 f N / p S N , p lim p 1 f N | Ω | ( p - 1 ) / N S N , p = f N S N , 1 < 1 ,

and so fN/pSN,p<1 for p close enough to 1.Hölder’s and Sobolev’s inequalities imply

Ω f | w | p f N / p S N , p Ω | w | p

for all wW01,p(Ω).As a consequence, we have

1 < 1 f N / p S N , p inf { Ω | w | p Ω f | w | p : w W 0 1 , p ( Ω ) , Ω f | w | p 0 } .

Thus, for p close enough to 1, we may apply [2, Theorem 6.2] and find a renormalized solution to (4.4).Since the data f and μ are nonnegative, it follows that vp0.

Taking 1-1/(1+Tk(vp)p-1)p-1 as test function in the renormalized formulation of (4.4), we obtain

{ v p < k } | v p | p ( 1 + v p p - 1 ) p Ω f ( 1 + v p p - 1 ) p - 1 + μ ( Ω )
Ω f + Ω f ( v p p - 1 ) p - 1 + μ ( Ω )
f 1 + f m ( v p p - 1 ) p - 1 m + μ ( Ω ) .

Passing to the limit as k goes to +, we have

(4.5) Ω | v p | p ( 1 + v p p - 1 ) p f 1 + f m ( v p p - 1 ) p - 1 m + μ ( Ω ) .

Thus, to go on, we need an estimate of vpp-1m, not depending on p.This is a consequence of the regularity of renormalized solutions.Indeed, observe that we may choose the truncate Tk(vp) as test function in the renormalized formulation of (4.4), and deduce that

{ v p < k } | v p | p k Ω f ( 1 + v p p - 1 ) p - 1 + k μ ( Ω ) k M p ,

where

M p = f 1 + f m ( v p p - 1 ) p - 1 m + μ ( Ω ) .

An application of [9, Lemma 4.1] leads to

| { v p p - 1 > k } | ( S N , p M p k ) N / ( N - p ) for all k > 0 .

In the setting of Marcinkiewicz spaces (see, for instance, [16, Appendix: Singular Integrals]), this inequality states that [vpp-1]N/(N-p)SN,pMp, and so

[ v p m ( p - 1 ) ] N / [ m ( N - p ) ] = [ v p p - 1 ] N / ( N - p ) m ( S N , p M p ) m .

On the other hand, having in mind [16, inequality (6.5), Appendix], we deduce

Ω | v p p - 1 | m N N - m ( N - p ) | Ω | 1 - m ( N - p ) / N [ v p m ( p - 1 ) ] N / m ( N - p ) N N - m ( N - p ) | Ω | 1 - m ( N - p ) / N ( S N , p M p ) m .

Hence, recalling the definition of Mp,

(4.6) v p p - 1 m ( N N - m ( N - p ) ) 1 / m | Ω | [ 1 / m ] - [ ( N - p ) / N ] S N , p ( f 1 + f m v p p - 1 m ( p - 1 ) p - 1 + μ ( Ω ) ) .

Note that a bound for the norm vpp-1m can be obtained if

f m < ( N - m ( N - p ) N ) ( m - 1 ) / m ( p - 1 ) p - 1 | Ω | [ ( N - p ) / N ] - [ 1 / m ] S N , p .

The limit, as p goes to 1, of the right-hand side is straightforward (since limp1SN,p=SN,1) and is given by

( N - m ( N - 1 ) N ) ( m - 1 ) / m | Ω | [ ( N - 1 ) / N ] - [ 1 / m ] S N , 1 .

Hence, our hypothesis (1.3) allows us to rearrange inequality (4.6) and obtain a bound for the norm vpp-1m for p close enough to 1.So, there is no loss of generality in assuming that q is small enough to perform the above manipulations for all 1<p<q.Thus, we have found a bound (independent of p) of the right-hand side of (4.5), that is, there exists M>0 such that

f 1 + f m ( v p p - 1 ) p - 1 m + μ ( Ω ) M for all 1 < p < q ,

and so

(4.7) Ω | v p | p ( 1 + v p p - 1 ) p M for all 1 < p < q .

Step 2: Problems having gradient terms. In this step, we are considering the problems

(4.8) { - Δ p ( u p ) = | u p | p + f ( x ) in Ω , u p = 0 on Ω .

According to the results in [1, 2], the function up=(p-1)log(1+vpp-1) belongs to W01,p(Ω) and it is a solution to (4.8). In terms of these new functions, estimate (4.7) becomes

(4.9) Ω | u p | p M for all 1 < p < q .

Applying Young’s inequality, it follows that

Ω | u p | 1 p Ω | u p | p + p - 1 p | Ω | M + | Ω | for all 1 < p < q .

(Recall that M depends on Ω, m, μ and f, but not on p.)This BV-estimate implies (see [4]) that there exists uBV(Ω) satisfying (up to subsequences)

(4.10) u p u pointwise a.e. in Ω and strongly in L r ( Ω ) ,  1 r < N N - 1 .

On the other hand, estimate (4.9) is the starting point in [5] to get a suitable vector field 𝐳.So, following [5] (see also [21, Theorem 3.5]), we get 𝐳L(Ω;N) satisfying 𝐳1 and

(4.11) | u p | p - 2 u p 𝐳 weakly in L s ( Ω ; N ) ,  1 s < .

Step 3: Passing to the limit in (4.8).Let φC0(Ω) be nonnegative.Taking φ as test function in (4.8), we have

Ω | u p | p - 2 u p φ = Ω φ | u p | p + Ω f φ .

Applying Young’s inequality, we have

Ω φ | u p | + Ω f φ 1 p Ω φ | u p | p + p - 1 p Ω φ + Ω f φ Ω | u p | p - 2 u p φ + p - 1 p Ω φ ,

and appealing to the lower-semicontinuity on the left-hand side, we deduce

Ω φ d | D u | + Ω f φ Ω 𝐳 φ .

Therefore,

(4.12) | D u | + f - div 𝐳 in 𝒟 ( Ω ) ,

and so div𝐳 is a Radon measure.It has finite total variation since it is the distributional limit of Δp(up) and this sequence is bounded in L1(Ω), due to (4.9) and equation (4.8).

On the other hand, -div𝐳 is a Radon measure which is above an LN-function.Thus, we may apply the results of Section 3, and so we have at our disposal a Green’s formula involving the Radon measure (𝐳,Du) (Proposition 3.5).Step 4: Passing to the limit in (4.4).We need another estimate, this one to get the convergence of eup.Fix 0<δ<1 and k>0, and take eδTk(up)-1 as test function in the weak formulation of (4.8). Dropping nonnegative terms, we obtain

δ Ω e δ T k ( u p ) | T k ( u p ) | p = Ω ( e δ T k ( u p ) - 1 ) | u p | p + Ω f ( e δ T k ( u p ) - 1 ) Ω ( e δ T k ( u p ) - 1 ) | u p | p - Ω f .

Then, rearranging and taking into account (4.9), we get

Ω e δ T k ( u p ) | u p | p - δ Ω e δ T k ( u p ) | T k ( u p ) | p Ω | u p | p + Ω f M + Ω f ,

from which follows

( 1 - δ ) Ω e δ T k ( u p ) | u p | p M + Ω f

for all 0<δ<1 and all k>0.Thanks to Levi’s monotone convergence theorem, we may let k go to + and get

( 1 - δ ) Ω | ( e δ u p - 1 ) | p M + Ω f

for all 0<δ<1.Hence, as a consequence of Young’s inequality, we have a W01,1-estimate of eδup-1, and on account of (4.10), we have eδuBV(Ω) for all 0<δ<1 as well as

e δ u p e δ u pointwise a.e. in Ω and strongly in L r ( Ω ) ,  1 r < N N - 1 .

A straightforward consequence of the last convergence is

(4.13) e u p e u strongly in L r ( Ω ) ,  1 r < N N - 1 .

(Nevertheless, we do not claim that euBV(Ω), see Remark 4.3 below.)

Now, recalling that every renormalized solution is a distributional solution as well, we take φC0(Ω) as test function in (4.4).Then

Ω | v p | p - 2 v p φ = Ω f ( 1 + v p p - 1 ) p - 1 φ + Ω φ d μ ,

which in terms of up becomes

(4.14) Ω e u p | u p | p - 2 u p φ = Ω f e u p φ + Ω φ 𝑑 μ .

Our next aim is to let p go to 1.To this end, we are analyzing each term in (4.14).On the left-hand side, we apply (4.13) and (4.11) to pass to the limit. On the other hand, (4.13) implies that eupeu in Lm(Ω), and so we may also pass to the limit on the right-hand side. We conclude that

(4.15) - div ( e u 𝐳 ) = f e u + μ in 𝒟 ( Ω ) .

We note that our assumptions on μ imply that μcap1,q and so μL1(Ω)+W-1,q(Ω), by [10, Theorem 2.1].We deduce from (4.15) that euLq(Ω); in particular, we have uL(Ω).Step 5: Dju=0.This fact is proved in [6] for a bounded solution to problem (1.5) through [6, Lemma 2].The only modification of that proof we need in our setting is to choose λ>3 and take λ-T1(u) as test function in Un.Step 6: The equation -divz=|Du|+f holds as measures.First we claim that

(4.16) - e u div 𝐳 e u | D u | + e u f + μ

holds as measures.(Here we do not mean that eu is integrable with respect to the positive Radon measures -div𝐳 and |Du|.)To prove our claim for any k>0, our starting point is (𝐳,DeTk(u))|DeTk(u)| together with the equality

- e T k ( u ) div 𝐳 = ( 𝐳 , D e T k ( u ) ) - div ( e T k ( u ) 𝐳 ) .

Then, by the chain rule for BV-functions with no jump part (see [4]),

- e T k ( u ) div 𝐳 | D e T k ( u ) | - div ( e T k ( u ) 𝐳 ) = e T k ( u ) | D T k ( u ) | - div ( e T k ( u ) 𝐳 ) .

We now choose a nonnegative φC0(Ω), obtaining

- Ω φ e T k ( u ) d ( div 𝐳 ) Ω φ e T k ( u ) d | D T k u | + Ω e T k ( u ) 𝐳 φ .

Applying Levi’s monotone convergence theorem to the measures -div𝐳 and |Du|, we get

- Ω φ e u d ( div 𝐳 ) Ω φ e u d | D u | + Ω e u 𝐳 φ = Ω φ e u d | D u | + Ω φ e u f + Ω φ 𝑑 μ ,

due to (4.15).Therefore, (4.16) is proved.

Next, we will study these measures concentrated on the sets {u<k}A, with k>0. Having in mind that μ(ΩA)=0, it follows that

- e u div 𝐳 ( { u < k } A ) e u | D u | ( { u < k } A ) + e u f χ ( { u < k } A ) .

Observing that every term is finite, we deduce

- div 𝐳 ( { u < k } A ) | D u | ( { u < k } A ) + f χ ( { u < k } A ) .

Letting k go to infinity yields

- div 𝐳 ( { u < } A ) | D u | ( { u < } A ) + f χ ( { u < } A ) .

We point out that {u=+}Su satisfies N-1({u=+})=0, and so it is a null set with respect to all the involved measures.Since cap1,q(A,Ω)=0, and so N-1(A)=0, a similar consequence is seen for A.Thus,

- div 𝐳 | D u | + f ,

and this inequality and (4.12) lead to the desired equality.Step 7: (z,Du)=|Du| as measures.Given φC0(Ω), with φ0, take e-upφ as test function in (4.8) to get

2 Ω e - u p φ | u p | p + Ω f e - u p φ = Ω e - u p | u p | p - 2 u p φ .

It is straightforward that then

2 Ω φ | ( e - u p ) | p + Ω f e - u p φ Ω e - u p | u p | p - 2 u p φ ,

and, by Young’s inequality, we have

2 Ω φ | ( e - u p ) | + Ω f e - u p φ Ω e - u p | u p | p - 2 u p φ + 2 p - 1 p Ω φ .

The lower semicontinuity of the total variation leads to

2 Ω φ d | D ( e - u ) | + Ω f e - u φ Ω e - u 𝐳 φ = - Ω φ d ( div ( e - u 𝐳 ) ) = Ω φ e - u d | D u | + Ω φ e - u f - Ω φ d ( 𝐳 , D ( e - u ) ) ,

since

div ( e - u 𝐳 ) = e - u div 𝐳 + ( 𝐳 , D ( e - u ) ) in 𝒟 ( Ω ) .

By simplifying and applying the chain rule, we deduce

Ω φ d | D ( e - u ) | - Ω φ d ( 𝐳 , D ( e - u ) ) Ω φ d | D ( e - u ) | = Ω φ e - u d | D u | ,

from where -(𝐳,D(e-u))=|D(e-u)| follows.As a consequence of the definition of the pairing of a vector field and a gradient, it follows that

( 𝐳 , D ( 1 - e - u ) ) = - ( 𝐳 , D ( e - u ) ) = | D ( 1 - e - u ) | .

Finally, thanks to [20, Proposition 2.2], we are done.Step 8: L-estimate near the boundary.Recall that, on account of our hypothesis dist(A¯,Ω)>0, we may apply [13, Theorem 4.3] and deduce that each solution up is bounded in any closed subset of ΩA¯.Let BR denote a ball of radius R>0 such that B¯RA¯= and |BR|<1.We explicitly point out that BR(NΩ) can be a non-null set, since we want to prove regularity up to the boundary.In what follows, we will write

A k , R p = { x B R Ω : u p ( x ) > k } and A k , R = { x B R Ω : u ( x ) > k } .

Consider φC0(BR), with 0φ1, 1<α<NN-1 and k>0.Our aim is to prove that

(4.17) B R Ω | ( e α G k ( u p ) / p - 1 ) | p φ p C A k , R p | φ | p ( e α G k ( u p ) - 1 ) + C f m | A k . R p | 1 / m ,

where C is a constant which does not depend on p.

To this end, take (e(α-1)Gk(up)-e-Gk(up))φp as a test function in problem (4.4), written in terms of up.Then we have

Ω e u p | G k ( u p ) | p [ ( α - 1 ) e ( α - 1 ) G k ( u p ) + e - G k ( u p ) ] φ p
p Ω e u p | G k ( u p ) | p - 1 ( e ( α - 1 ) G k ( u p ) - e - G k ( u p ) ) φ p - 1 | φ | + Ω f e u p ( e ( α - 1 ) G k ( u p ) - e - G k ( u p ) ) φ p .

Having in mind that we are actually integrating on the set {up>k}, we have eup=ek+Gk(up).Thus, dividing the last inequality by ek, we obtain

Ω | G k ( u p ) | p [ ( α - 1 ) e α G k ( u p ) + 1 ] φ p
(4.18) p Ω | G k ( u p ) | p - 1 ( e α G k ( u p ) - 1 ) φ p - 1 | φ | + Ω f ( e α G k ( u p ) - 1 ) φ p .

We are now analyzing the first term on the right-hand side of (4.18).Observe that Young’s inequality implies

p Ω | G k ( u p ) | p - 1 ( e α G k ( u p ) - 1 ) φ p - 1 | φ | ( p - 1 ) Ω | G k ( u p ) | p ( e α G k ( u p ) - 1 ) φ p + Ω | φ | p ( e α G k ( u p ) - 1 ) ,

and so one term can be absorbed by the left-hand side of (4.18) becoming

( α - p ) Ω | G k ( u p ) | p ( e α G k ( u p ) - 1 ) φ p Ω | φ | p ( e α G k ( u p ) - 1 ) + Ω f ( e α G k ( u p ) - 1 ) φ p .

Since there is no loss of generality in assuming 1<p<α+12 and 1<p<2, we may let α-p>α-12 and αp<α2. Furthermore, easy manipulations lead to

( α - 1 2 α 2 ) Ω | ( e α G k ( u p ) / p - 1 ) | p φ p ( α - p 2 α p ) Ω | ( e α G k ( u p ) / p - 1 ) | p φ p
(4.19) Ω | φ | p ( e α G k ( u p ) - 1 ) + Ω f ( e α G k ( u p ) - 1 ) φ p .

We point out that in every term of (4.19), we are integrating on the set Ak,Rp.Applying Lemma 4.2 to the last term of (4.19), we get

(4.20) Ω f ( e α G k ( u p ) - 1 ) φ p 2 A k , R p f ( e α G k ( u p ) / p - 1 ) p φ p + 2 A k , R p f φ p .

To estimate the right-hand side of (4.20), we will apply the Hölder and Sobolev inequalities.(We explicitly point out that it follows from upW01,p(BR)L(BR) that (eαGk(up)/p-1)φ belongs to W01,p(BR)L(BR), even though αp>1.)Performing those manipulations, we obtain

A k , R p f ( e α G k ( u p ) / p - 1 ) p φ p
f m [ A k , R p ( e α G k ( u p ) / p - 1 ) p * φ p * ] p / p * | A k , R p | [ 1 / m ] - [ p / p * ]
f m S N , p [ A k , R p | ( ( e α G k ( u p ) / p - 1 ) φ ) | p ] | B R | [ 1 / m ] - [ p / p * ]
f m S N , p 2 p - 1 [ A k , R p φ p | ( e α G k ( u p ) / p - 1 ) | p + A k , R p ( e α G k ( u p ) / p - 1 ) p | φ | p ] | B R | 1 / m - p / p *
(4.21) 2 ( S N , 1 + 1 ) f m [ A k , R p φ p | ( e α G k ( u p ) / p - 1 ) | p + A k , R p ( e α G k ( u p ) / p - 1 ) p | φ | p ] | B R | 1 / m - p / p * .

Here we have estimated the constant taking p close enough to 1.Now we set δ=12(1N-1m)>0 and note that

1 m - p p * = p N - 1 m > δ .

Recalling that |BR|<1, we deduce that |BR|1/m-p/p*|BR|δ.Going back to (4.21), we have

A k , R p f ( e α G k ( u p ) / p - 1 ) p φ p 2 ( S N , 1 + 1 ) f m [ A k , R p φ p | ( e α G k ( u p ) / p - 1 ) | p + A k , R p ( e α G k ( u p ) / p - 1 ) p | φ | p ] | B R | δ
2 ( S N , 1 + 1 ) f m [ A k , R p φ p | ( e α G k ( u p ) / p - 1 ) | p + A k , R p ( e α G k ( u p ) - 1 ) | φ | p ] | B R | δ ,

where the last inequality is due to Lemma 4.2.This inequality implies that (4.19) is transformed in

( α - 1 2 α 2 ) A k , R p | ( e α G k ( u p ) / p - 1 ) | p φ p A k , R p | φ | p ( e α G k ( u p ) - 1 ) + 4 ( S N , 1 + 1 ) f m | B R | δ [ A k , R p φ p | ( e α G k ( u p ) / p - 1 ) | p
+ A k , R p ( e α G k ( u p ) - 1 ) | φ | p ] + 2 A k , R p f φ p .

Now R>0 is chosen small enough to have 4fm|BR|δ<α-14α2.Hence, we find a constant C>0 independent of p satisfying

α - 1 4 α 2 A k , R p | ( e α G k ( u p ) / p - 1 ) | p φ p C A k , R p | φ | p ( e α G k ( u p ) - 1 ) + 2 A k , R p f φ p .

To finish the proof of (4.17), it is enough to apply Hölder’s inequality.

The next step is to let p go to 1 in (4.17).Applying Young’s inequality, it follows that

B R Ω | ( e α G k ( u p ) / p - 1 ) | φ 1 p B R Ω | ( e α G k ( u p ) / p - 1 ) | p φ p + p - 1 p | B R | p / ( p - 1 )
C p B R Ω | φ | p ( e α G k ( u p ) - 1 ) + C p f m | A k . R p | 1 / m + p - 1 p .

Thanks to α<NN-1, we may use that eαGk(up)-1 converges to eαGk(u)-1 in L1(BRΩ).This fact and the fact that upu pointwise in BRΩ allow us to pass to the limit on the right-hand side.On the left-hand side, we deduce that eαGk(up)/p-1 converges to eαGk(u)-1 in L1(BRΩ) (it is enough to realize that |eαGk(up)/p-1||eαGk(up)-1| and use a variant of the dominated convergence theorem and apply the lower semicontinuity of the total variation).Therefore, we conclude that

(4.22) B R Ω φ d | D ( e α G k ( u ) - 1 ) | C B R Ω | φ | ( e α G k ( u ) - 1 ) + C f m | A k . R | 1 / m .

To obtain a Caccioppoli type inequality, consider 0<ρ<R and a function φC0(BR) such that 0φ1 and φ1 in Bρ, the ball concentric with BR and having radius ρ.We may assume that |φ|2R-ρ.Then (4.22) becomes

(4.23) B ρ Ω d | D ( e α G k ( u ) - 1 ) | C R - ρ B R Ω ( e α G k ( u ) - 1 ) + C f m | A k . R | 1 / m .

This Caccioppoli inequality will allow us to apply Stampacchia’s theorem.To begin with, consider B(R+ρ)/2, the ball concentric with BR but having radius R+ρ2, and take the function ηC0(B(R+ρ)/2) satisfying 0η1, η1 in Bρ and |η|2R-ρ.We do not know that eαGk(u)-1BV(BRΩ) yet, and so we cannot apply Sobolev’s inequality.Instead, we will consider a suitable truncation.Indeed, we have

B ρ Ω ( e α G k ( T h ( u ) ) - 1 ) B ( R + ρ ) / 2 Ω ( e α G k ( T h ( u ) ) - 1 ) η
| A k , R | 1 / N [ B ( R + ρ ) / 2 Ω ( e α G k ( T h ( u ) ) - 1 ) N / ( N - 1 ) η N / ( N - 1 ) ] ( N - 1 ) / N
| A k , R | 1 / N S N , 1 B ( R + ρ ) / 2 Ω d | D ( ( e α G k ( u ) - 1 ) η ) | ,

and the monotone convergence theorem gives us the desired inequality:

B ρ Ω ( e α G k ( u ) - 1 ) | A k , R | 1 / N S N , 1 B ( R + ρ ) / 2 Ω d | D ( ( e α G k ( u ) - 1 ) η ) | .

Hence, we deduce from inequality (4.23) that

B ρ Ω ( e α G k ( u ) - 1 ) | A k , R | 1 / N S N , 1 [ B ( R + ρ ) / 2 Ω η d | D ( e α G k ( u ) - 1 ) | + B ( R + ρ ) / 2 Ω ( e α G k ( u ) - 1 ) | η | ]
| A k , R | 1 / N S N , 1 [ B ( R + ρ ) / 2 Ω d | D ( e α G k ( u ) - 1 ) | + 2 R - ρ B ( R + ρ ) / 2 Ω ( e α G k ( u ) - 1 ) ]
(4.24) C R - ρ | A k , R | 1 / N B R Ω ( e α G k ( u ) - 1 ) + C | A k , R | 1 / N + 1 / m .

To apply Stampacchia’s procedure, take 0<h<k and observe that the following hold:

B R Ω ( e α G k ( u ) - 1 ) B R Ω ( e α G h ( u ) - 1 ) ,
| A k , R | 1 k - h B R Ω G h ( u ) 1 α ( k - h ) B R Ω ( e α G h ( u ) - 1 ) .

Therefore, inequality (4.24) yields

B ρ Ω ( e α G k ( u ) - 1 ) C α 1 / N ( k - h ) 1 / N ( R - ρ ) [ B R Ω ( e α G h ( u ) - 1 ) ] 1 + 1 / N
+ C α 1 / N + 1 / m ( k - h ) 1 / N + 1 / m [ B R Ω ( e α G h ( u ) - 1 ) ] 1 / m + 1 / N .

We point out that k-h, R-ρ and BRΩ(eαGh(u)-1) can be taken as small as we want in Stampacchia’s procedure.Thus, we may unify all the exponents obtaining

B ρ Ω ( e α G k ( u ) - 1 ) C ( k - h ) 1 + 1 / N ( R - ρ ) [ B R Ω ( e α G h ( u ) - 1 ) ] 1 / m + 1 / N .

Applying Stampacchia’s theorem (see [24, Lemma 5.1], and observe that the last exponent is larger than 1) to

φ ( h , R ) = B R Ω ( e α G h ( u ) - 1 ) ,

we get a k0 such that φ(k,R)=0 for all k>k0, and so uL(BRΩ).

Therefore, we have seen that uL(BRΩ) for every ball satisfying B¯RA¯= and |BR|<1.Finally, by a compactness argument, u is bounded in a strip around Ω.Step 9: Boundary condition.As a consequence of step 8, we know that there exists a strip around the boundary Ω where u is bounded.Let ϕC(Ω) be a nonnegative function.Then there exists a nonnegative φC(Ω¯) such that φ|Ω=ϕ and φ vanishes outside that strip.For instance, we may consider φ=φ1φ2, where φ1 is the solution to the Dirichlet problem for the Laplace equation with datum ϕ, and φ2 vanishes outside that strip and satisfies φ2|Ω1.In other words, we search a smooth function satisfying φ|Ω=ϕ and such that u is bounded in suppφ.

Now, we may follow the same argument used in [6, Theorem 1, step 10] to get

[ 𝐳 , ν ] = sign ( - u ) N - 1 -a.e. on Ω .

Therefore, we have proved that u is actually a solution to problem (1.5).∎

Remark 4.3

We explicitly point out that euBV(Ω) for every nontrivial measure μ.Indeed, if euBV(Ω), then the following manipulations would hold:

f e u + μ = - div ( e u 𝐳 ) = - e u div ( 𝐳 ) - ( 𝐳 , D ( e u ) ) = - e u div ( 𝐳 ) - | D ( e u ) ) |
= - e u div ( 𝐳 ) - ( e u ) | D u | = e u [ - div ( 𝐳 ) - | D u | ] = e u f ,

wherewith μ=0.Compare this argument with [3, Remark 2.11].

Remark 4.4

Assume for a moment that the function e-u is μ-measurable.Having in mind step 7 and [20, Proposition 2.2], we have (𝐳,DeTk(u))=|DeTk(u)| for any k>0.We are able to see, redoing the same calculations, that (4.16) becomes an equality.This fact and the fact that -div𝐳=|Du|+f imply that the measure e-uμ vanishes.Even though this argument does not work, it suggest that μ is concentrated on the set {u=+}.Compare this note with [3, Remark 2.16].

Remark 4.5

It is worth noting that we can recover the singular measure from the solution to (1.5) we have found.The argument is very similar to that of [3] for p=2.To check it, take eu/(1+ϵu)-1, with ϵ>0, as test function in (1.5).Then, since u is a solution to problem (1.5), it follows that

| D u | ( Ω ) + Ω | e u / ( 1 + ϵ u ) - 1 | 𝑑 N - 1 = Ω e u / ( 1 + ϵ u ) ( 1 - 1 ( 1 + ϵ u ) 2 ) d | D u | + Ω f ( x ) ( e u / ( 1 + ϵ u ) - 1 ) .

Since uBV(Ω) and uL(Ω), we deduce that the left-hand side is bounded.Thus,

Ω e u / ( 1 + ϵ u ) ( 1 - 1 ( 1 + ϵ u ) 2 ) d | D u | M for all ϵ > 0 ,

and so there exists a measure μ such that, up to subsequences,

| D u | e u / ( 1 + ϵ u ) ( 1 - 1 ( 1 + ϵ u ) 2 ) μ weakly in the sense of measures.

Now, it is easy to check that this measure satisfies equation (1.6).In fact, eu/(1+ϵu)BV(Ω) and

- div ( e u / ( 1 + ϵ u ) 𝐳 ) = - e u / ( 1 + ϵ u ) div 𝐳 - ( 𝐳 , D e u / ( 1 + ϵ u ) ) = | D u | e u / ( 1 + ϵ u ) ( 1 - 1 ( 1 + ϵ u ) 2 ) + f e u / ( 1 + ϵ u ) .

From the estimate feu/(1+ϵu)feuL1(Ω) and the monotone convergence theorem, we have feu/(1+ϵu)feu in L1(Ω), wherewith we may let ϵ0 proving that

- div ( e u 𝐳 ) = f e u + μ in 𝒟 ( Ω ) .

Remark 4.6

Roughly speaking, Theorem 1.1 states that for each measure μ concentrated in a set N-1-null, we find a solution to problem (1.5).A few words on the map

μ singular measure u solution to (1.5)

are in order.Although we cannot assert that this is an one-to-one map, we can state that not every measure μ leads to the same unbounded solution u.It is enough to choose two singular measures μ1 and μ2 satisfying suppμ1suppμ2=.Applying our theorem to each μi, we obtain an unbounded solution ui to problem (1.5), i=1,2.However, we know that u1 is bounded in suppμ2 and u2 is bounded in suppμ1, and so these solutions are different.

5 Multiplicity of Radial Solutions

In this section we deal with the case of radial solutions in a ball, and so in the following examples we always assume Ω=BR(0) (i.e., the ball centered at the origin having radius R>0), and we search solutions depending on |x|.In what follows, we set ωN=|B1(0)| and denote by δ0 the Dirac measure concentrated in the origin.

Let us begin by dealing with the homogeneous case.

Example 5.1

Assume that f0.Then problem (1.5) has a trivial solution, given by

u ( x ) 0 , with 𝐳 ( x ) 0 .

In the paper [14] it is shown that u(x)0 is the only bounded solution of (1.5).On the other hand, we will now show that (1.5) has infinitely many unbounded radial solutions.

A first kind of solution is the following:

(5.1) u ( x ) = - ( N - 1 ) log ( | x | α R ) , 𝐳 ( x ) = - x | x |

for any choice of α1.Note that this solution is zero on BR only when α=1.

Then another kind of solution is given by

(5.2) u ( x ) = { - ( N - 1 ) log ( | x | ρ ) if 0 < | x | < ρ , 0 if ρ | x | < R

for every ρ such that 0<ρ<R.In this case the vector field 𝐳 is given by

𝐳 ( x ) = { - x | x | if 0 < | x | < ρ , - ρ N - 1 x | x | N if ρ | x | < R .

It is easy to check that both (5.1) and (5.2) are solutions according to Definition 4.1.

All these solutions can be achieved using the procedure of Theorem 1.1. Indeed, consider the singular measure μ=Cδ0, with C>0, and the approximating problems

{ - Δ p ( v p ) = C δ 0 in B R ( 0 ) , v p = 0 on B R ( 0 ) .

It is easy to check that

v p ( x ) = ( C N ω N ) 1 / ( p - 1 ) p - 1 N - p ( 1 | x | ( N - p ) / ( p - 1 ) - 1 R ( N - p ) / ( p - 1 ) ) .

Now set

(5.3) u p ( x ) = ( p - 1 ) log ( 1 + v p p - 1 ) .

We may distinguish two cases according to the size of the constant C.

  1. C N ω N R N - 1 .In this case, it is straightforward that the limit, for p1, is

    u ( x ) = - ( N - 1 ) log ( | x | ) + log ( C N ω N ) ,

    which can be written as (5.1)for α=1R(CNωN)1/(N-1)1.

  2. 0 < C < N ω N R N - 1 .Here limp1up(x) is given by (5.2) with ρ=(CNωN)1/(N-1).

Therefore, both types of solutions (5.1) and (5.2) correspond to different multiples of the Dirac delta centered at the origin.

Let us also note that the solutions up(x) correspond to the unbounded solutions of the following problem:

- Δ p u p = | u p | p in Ω = { x N : | x | < R } ,

exhibited by Ferone and Murat for p>1 in [17, Remark 2.11], i.e.,

u p ( x ) = ( p - 1 ) log ( | x | - ( N - p ) / ( p - 1 ) - m R - ( N - p ) / ( p - 1 ) - m ) ,

where m is any constant satisfying mR(N-p)/(p-1)<1.These solutions are the same as(5.3), where C and m are related by

C = N ω N R N - p ( N - p ) p - 1 ( 1 - m R ( N - p ) / ( p - 1 ) ) p - 1 .

The previous example can be adapted to obtain a multiplicity result in a general open set.

Example 5.2

Assume that Ω is a bounded domain with Lipschitz boundary and f0. Then problem (1.5) has infinitely many nonnegative unbounded solutions. More precisely, for any x0Ω, we can find a solution u(x) which is unbounded near x0.

Fix x0Ω and choose ρ>0 such that Bρ(x0)Ω.We set

u ( x ) = { - ( N - 1 ) log ( | x - x 0 | ρ ) if 0 < | x - x 0 | < ρ , 0 if | x - x 0 | > ρ ,

with the associated vector field

𝐳 ( x ) = { - x - x 0 | x - x 0 | if 0 < | x - x 0 | < ρ , - ρ N - 1 ( x - x 0 ) | x - x 0 | N if | x - x 0 | > ρ .

Example 5.3

In this example, we will show unbounded radial solutions of problem (1.5) with constant datum fλ]0,N-1R] in Ω=BR(0).

In this case, a solution and its correspondent vector field are given by

(5.4) u ( x ) = - ( N - 1 ) log ( | x | α R ) + λ ( | x | - R ) , 𝐳 ( x ) = - x | x |

for any choice of α1.Another type of solution is given, for any choice of ρ]0,R[, by

(5.5) u ( x ) = { - ( N - 1 ) log ( | x | ρ ) + λ ( | x | - ρ ) if 0 < | x | < ρ , 0 if ρ < | x | < R ,

with the associated vector field

𝐳 ( x ) = { - x | x | if 0 < | x | < ρ , - λ N x - ( 1 - λ ρ N ) ρ N - 1 x | x | N if ρ < | x | < R .

The details are left to the reader.Note that |𝐳(x)|1 due to the inequality λN-1RAs in the first examples, these solutions are related to a singular measure of the form Cδ0.In particular, solution (5.4) is obtained for large values of C, while (5.5) corresponds to small values of C.

Example 5.4

In this final example, we will exhibit an unbounded solution to problem (1.1) with f0. A solution of this problem is defined by

u ( x ) = g ( | x | ) , where g ( r ) = ( N - 1 ) e - r r α R e s s 𝑑 s

for every choice of α1, with 𝐳(x)=-x|x|.Another possibility is given by

u ( x ) = { g ( | x | ) if 0 < | x | < ρ , 0 if ρ < | x | < R ,    𝐳 ( x ) = { - x | x | if 0 < | x | < ρ , - ρ N - 1 x | x | N if ρ < | x | < R ,

where

g ( r ) = ( N - 1 ) e - r r ρ e s s 𝑑 s .

It is important to observe that all unbounded solutions exhibited in this section satisfy

e u BV ( B R ( 0 ) ) , e δ u W 1 , 1 ( B R ( 0 ) ) for every δ < 1 .


Dedicated to Ireneo, an inspiring mathematician and, above all, a good friend



Communicated by Antonio Ambrosetti and David Arcoya


Award Identifier / Grant number: MTM2015-70227-P

Funding statement: The third author has been partially supported by the Spanish Ministerio de Economìa y Competitividad and FEDER, under project MTM2015-70227-P.

Acknowledgements

The third author thanks Cristina Trombetti for fruitful discussionsconcerning the subject of this paper.

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Received: 2017-01-31
Revised: 2017-02-21
Accepted: 2017-03-03
Published Online: 2017-04-21
Published in Print: 2017-05-01

© 2017 by De Gruyter

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