ELLIPTIC PROBLEMS INVOLVING THE 1 LAPLACIAN AND A SINGULAR LOWER ORDER TERM

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1 ELLIPTIC PROBLEMS INVOLVING THE 1 LAPLACIAN AND A SINGULAR LOWER ORDER TERM V. DE CICCO, D. GIACHETTI AND S. SEGURA DE LEÓN Abstract. This paper is concerned with the Dirichlet ( ) problem or an equation involving the 1 Laplacian operator 1 u := div Du and having a singular Du term o the type (x) u γ. Here L N () is nonnegative, 0 < γ 1 and is a bounded domain with Lipschitz continuous boundary. We prove an existence result or a concept o solution conveniently deined. The solution is obtained as limit o solutions o p Laplacian type problems. Moreover, when (x) > 0 a.e., the solution satisies those eatures that might be expected as well as a uniqueness result. We also give explicit 1 dimensional examples that show that, in general, uniqueness does not hold. We remark that the Anzellotti theory o L divergence measure vector ields must be extended to deal with this equation. 1. Introduction In the present paper we deal with the Dirichlet problem or equations involving the 1 Laplacian and a singular lower order term. This equation, at least in the case > 0 a.e. in, looks like (1) ( Du div Du ) = (x) u γ in, u = 0 on, where R N is a bounded open set with Lipschitz boundary, 0 < γ 1 and is a unction belonging to L N (). Actually we are interested to the case o 0. The natural space to study problems where the 1 Laplacian appears is BV (), the space o unctions u L 1 () whose distributional derivatives are Radon measures with inite total variation. We point out that problems involving a lower order singular term like that appearing in problem (1) have already been studied in the Laplacian or the p Laplacian setting. There is a wide literature dealing with problem (2) p u = (x) u γ in, u = 0 on or p > 1 and 0. The problem (2) or p = 2 was initially proposed in 1960 in the pioneering work [21] by Fulks and Maybee as a model or several physical situations. This problem was then studied by many authors, among which we will quote the works o Stuart [43], Crandall, Rabinowitz and Tartar [17], Lazer and McKenna [30], Boccardo and Orsina [9], Coclite and Coclite [16], Arcoya and Key words and phrases. 1 Laplacian, singular lower order terms Mathematics Subject Classiication: MSC 2010: 35J75, 35D30, 35A01, 35J60, 35A02. 1

2 2 V. DE CICCO, D. GIACHETTI AND S. SEGURA DE LEÓN Moreno Mérida [7], Oliva and Petitta [38], and Giachetti, Martinez Aparicio and Murat in [22], [23], [24] and [25]. In particular, in [9] the authors studied the problem in the ramework o weak solutions in the sense o distributions or belonging to Lebesgue spaces, or to the space o Radon measures, and they prove existence and regularity as well as non existence results. On the other hand, in [22], [23] and [24] the authors deal with airly general singular problems and they ind weak solutions belonging, or γ 1, to the energy space. For a variational approach to the problem (2) in the case p > 1, see or instance Canino and Degiovanni [11]. Due to the methods we are going to use in our paper (approximation o our problem with problems driven by the p Laplacian), we quote now some papers dealing with existence results or problem (2) in the case p > 1. We reer to Giacomoni, Schindler and Takac [26], Perera and Silva [40], Mohammed [37], Loc Hoang and Schmitt [31], De Cave [18], Canino, Sciunzi and Trombetta [10] and Mi [36]. Early papers devoted to the Dirichlet problem or equations involving the 1 Laplacian operator include [29, 27, 2, 8, 19, 15]. The interest in this setting comes out rom an optimal design problem in the theory o torsion and rom the level set ormulation o the Inverse Mean Curvature Flow. On the other hand, it also appears in the variational approach to image restoration. Indeed, total variation minimizing models have become one o the most popular and successul methodology or image restoration since the introduction o the ROF model by Rudin, Osher and Fatemi in [41]. In this paper a variational problem involving the total variation operator is considered. It was designed with the explicit goal o preserving sharp discontinuities (edges) in images while removing noise and other unwanted ine scale detail. The 1 Laplacian operator emerges through the subdierential o the total variation. Since that paper, there has been a burst in the application o the total variation regularization to many dierent image processing problems which include inpainting, blind deconvolution or multichannel image segmentation (or a review on this topic we reer to [13], see also [4]). To deal with the 1 Laplacian, a irst diicult occurs in deining the quotient Du Du, being Du a Radon measure. It was overcome by Andreu, Ballester, Caselles and Mazón in [2] through the Anzellotti theory o pairings o L divergence measure vector ields and the gradient o a BV unction (see [4]). In their deinition appears a vector ield z L (; R N ) such that z 1 and (z, Du) = Du, so that z plays the role o the above rate. Since the Dirichlet boundary condition does not hold in the usual trace orm, they also introduce in [2] a weak ormulation: [z, ν] sign ( u), where [z, ν] stands or the weak trace on o the normal component o z. It is worth observing that the related problem having no singular lower order term (3) ( Du div Du ) = (x) in, u = 0 on, with L N (), has particular eatures and there exists a BV solution only i N is small enough (see [15, 34]).

3 THE 1 LAPLACIAN EQUATION WITH A SINGULAR LOWER ORDER TERM 3 Our main results show that the term (x) u has a regularizing eect. Indeed, in γ Theorem 4.5 below, we prove that there exists a bounded solution to problem (1), in the sense o Deinition 4.1, or every nonnegative datum L N (). Moreover, in Theorem 5.1, we see the improved eatures satisied by the solution when is strictly positive, which lead to uniqueness (Theorem 5.3). In summary, the bigger the data, better eatures has the solution. Being the source term singular on the set {u = 0}, this set is crucial. I p = 1, we are able to prove (as in the p > 1 case) that {x : u(x) = 0} {x : (x) = 0}, except or a set o zero Lebesgue measure and we will prove that this set has locally inite perimeter. The conditions satisied by solutions or non strictly positive data include (a) u γ L1 loc(), (b) (div z)χ {u>0} = u γ in D (), where z is the vector ield deined above and χ {u>0} is the precise representative (in the sense o the BV unction) o the characteristic unction χ {u>0} o the set {u > 0}. We will prove that {u > 0} is a set o locally inite perimeter and so χ {u>0} is a locally BV unction. Note that, by using that the equation (b) can be written as div ( zχ {u>0} ) = ( div z ) χ {u>0} + (z, Dχ {u>0} ), div ( zχ {u>0} ) + Dχ{u>0} = (x) u γ, where the let hand side is a sum o an operator in divergence orm and an additional term Dχ {u>0}, which is a measure concentrated on the reduced boundary {u > 0} (or equivalently on the reduced boundary {u = 0}). Since we do not know that the measure div z has inite total variation, we cannot directly apply Anzellotti s theory. As a consequence, one o our tasks in analyzing problem (1) will be slightly extend it. The proo o our results is obtained passing to the limit in singular problems with principal part the p Laplacian, p > 1 as p tends to 1. To this aim, we need preliminarly to prove that the corresponding singular approximating problems admit a weak solution (see Theorem 3.3). Next we will show that the amily (u p ) p>1 is bounded in the BV norm, hence there exists a unction u BV () such that, up to a subsequence, and u p u in L 1 () u p Du weakly as measures. Similarly, or the vector ield z, we have that ( u p p 2 u p ) p is bounded in L q (; R N ) or any 1 q <. Hence there exists a vector ield z satisying u p p 2 u p z, weakly in L q (; R N ), 1 q <. Such unctions u and z are those satisying (a) and (b) as well as the other conditions o Deinition 4.1.

4 4 V. DE CICCO, D. GIACHETTI AND S. SEGURA DE LEÓN Let us remark that i 0 the solutions are 1 harmonic unctions; i the boundary datum is continuous and the domain satisies some additional geometrical assumptions, the solution is unique (see Sternberg, Williams and Ziemer in [42]) and it is the limit o p harmonic unctions (see Juutinen in [28]). I is not strictly positive and not identically zero there are 1 dimensional examples o non uniqueness (see Section 6). The plan o this paper is the ollowing. Section 2 is dedicated to preliminaries, we introduce our notation, some properties o the space BV () and an extension o the Anzellotti theory, including a Green ormula. The starting point o our main result is studied in Section 3, which is concerned with solutions to problem (2). Section 4 is devoted to the study o existence or general nonnegative data, while Section 5 deals with strictly positive data. Finally, Section 6 provides some explicit 1 dimensional solutions that show that, in general, uniqueness does not hold. Acknowledgements This research has been partially supported by the Spanish Ministerio de Economía y Competitividad and FEDER, under project MTM P. This paper was started during a stay o the third author in Sapienza Università di Roma supported with a grant as visiting proessor by Istituto Nazionale di Alta Matematica Francesco Severi. The authors also thank Francesco Petitta or some useul suggestions. 2. Preliminaries In this Section we will introduce some notation and auxiliary results which will be used throughout this paper. In what ollows, we will consider N 2, and H N 1 (E) will denote the (N 1) dimensional Hausdor measure o a set E and E its Lebesgue measure. In this paper, will always denote an open bounded subset o R N with Lipschitz boundary. Thus, an outward normal unit vector ν(x) is deined or H N 1 almost every x. We will make use o the usual Lebesgue and Sobolev spaces, denoted by L q () and W 1,p 0 (), respectively. On the other hand, Lebesgue spaces with respect to a Radon measure µ are denoted by L q (, µ). We recall that or a Radon measure µ in and a Borel set A the measure µ A is deined by (µ A)(B) = µ(a B) or any Borel set B. I a measure µ is such that µ = µ A or a certain Borel set A, the measure µ is said to be concentrated on A. The truncation unction will be use throughout this paper. Given k > 0, it is deined by T k (s) = min{ s, k} sign (s), or all s R. Moreover we will denote by G k (s) the unction deined by or all s R. G k (s) = s T k (s) 2.1. The energy space. The space o all unctions o inite variation, that is the space o those u L 1 () whose distributional gradient is a Radon measure with

5 THE 1 LAPLACIAN EQUATION WITH A SINGULAR LOWER ORDER TERM 5 inite total variation, is denoted by BV (). This is the natural energy space to study the problems we are interested in. It is endowed with the norm deined by u = u dx + Du, or any u BV (). We recall that the notion o trace can be extended to any u BV () and this act allows us to interpret it as the boundary values o u and to write u. Moreover, the trace deines a linear bounded operator BV () L 1 ( ) which is onto. Using the trace, we have available an equivalent norm, which we will use in the sequel. It is given by u BV () = u dh N 1 + Du. For every u BV (), the Radon measure Du is decomposed into its absolutely continuous and singular parts with respect to the Lebesgue measure: Du = D a u + D s u. We denote by S u the set o all x such that x is not a Lebesgue point o u, that is, x \S u i there exists ũ(x) such that 1 lim ρ 0 B ρ (x) B ρ(x) u(y) ũ(x) dy = 0. We say that x is an approximate jump point o u i there exist two real numbers u + (x) > u (x) and ν u (x) S N 1 such that 1 lim ρ 0 B ρ + u(y) u + (x) dy = 0, (x, ν u (x)) B ρ + (x,ν u(x)) 1 lim ρ 0 Bρ u(y) u (x) dy = 0, (x, ν u (x)) where and B ρ (x,ν u(x)) B + ρ (x, ν u (x)) = {y B ρ (x) : y x, ν u (x) > 0} B ρ (x, ν u (x)) = {y B ρ (x) : y x, ν u (x) < 0}. We denote by J u the set o all approximate jump points o u. By the Federer Vol pert Theorem [1, Theorem 3.78], we know that S u is countably H N 1 rectiiable and H N 1 (S u \J u ) = 0. The precise representative u : \(S u \J u ) R o u is deined as equal to ũ on \S u and equal to u +u + 2 on J u. It is well known (see or instance [1, Corollary 3.80]) that i ρ is a symmetric molliier, then the molliied unctions u ρ ɛ pointwise converges to u in its domain. A compactness result in BV () will be used in what ollows. It states that every sequence that is bounded in BV () has a subsequence which strongly converges in L 1 () to a certain u BV () and the subsequence o gradients weakly converges to Du in the sense o measures. To pass to the limit we will oten apply that some unctionals deined on BV () are lower semicontinuous with respect to the convergence in L 1 (). The most important are the unctionals deined by u Du

6 6 V. DE CICCO, D. GIACHETTI AND S. SEGURA DE LEÓN and u Du + u dh N 1. In the same way, it yields that each ϕ C0() 1 with ϕ 0 deines a unctional u ϕ Du, which is lower semicontinuous in L 1 (). For urther inormation on unctions o bounded variation, we reer to [1, 20, 44] A generalized Green ormula. The theory o L divergence measure vector ields is due to Anzellotti [5] and to Chen and Frid [14]. In spite o their dierent points o view, both approaches introduce, under some hypotheses, the dot product o a bounded vector ield z, whose divergence is a Radon measure, and the gradient Dv o v BV () through a pairing (z, Dv) which deines a Radon measure. However, they dier in handling this concept. They also deine the normal trace o a vector ield through the boundary and establish a generalized Gauss Green ormula. From now on, we denote by DM () the space o all vector ields z L (; R N ) whose divergence in the sense o distribution is a Radon measure with inite total variation, i.e., z DM () i and only i div z is a inite Radon measure belonging to W 1, (). Moreover, DM loc() stands or those vector ields z L (; R N ) which belong to DM (ω) or every open ω. To deine (z, Dv) in Anzellotti s theory, we need some compatibility conditions, such as div z L 1 () and v BV () L (), or div z a Radon measure with inite total variation and v BV () L () C(). Anzellotti s deinition o (z, Dv) can be extended to the case which div z is a Radon measure with inite total variation and v BV () L () (see [35, Appendix A] and [12, Section 5]). We are ollowing these papers or a slightly urther extension to the spaces DM loc() and BV loc (). The starting point is also the ollowing result proved in [14]. Proposition 2.1. For every z DM (), the measure µ = div z is absolutely continuous with respect to H N 1, that is, µ H N 1. Consider now µ = div z with z DM loc() and let v BV loc (). Since the precise representative v is equal H N 1 a.e. to the Borel unction lim ɛ 0 v ρ ɛ, then one deduces rom Proposition 2.1, that v is equal µ a.e. to a Borel unction so that the precise representative o every BV unction is µ measurable. Assume that v L 1 (, µ) and deine a distribution by the ollowing expression (4) (z, Dv), ϕ := v ϕ dµ vz ϕ dx, ϕ C0 (). Every term is well deined since v BV loc () L 1 (, µ) and z L (, R N ). We point out that the deinition o (z, Dv) depends on the precise representative o v; so that i we choose another representative, the distribution will become dierent. We next see that this distribution is actually a Radon measure having locally inite total variation. The proos are similar to those in [35] or [12]. Proposition 2.2. Let v BV loc () L 1 (, µ) and z DM loc(). distribution (z, Dv) deined previously satisies (5) (z, Dv), ϕ ϕ z L (U) Dv U Then the

7 THE 1 LAPLACIAN EQUATION WITH A SINGULAR LOWER ORDER TERM 7 or all open set U and or all ϕ C 0 (U). Proo. I U is an open set and ϕ C0 (U), then it was proved in [35] that (6) (z, DT k (v)), ϕ ϕ z L (U) DT k (v) ϕ z L (U) Dv holds or every k > 0. On the other hand, (z, DT k (v)), ϕ = T k (v) ϕ dµ U T k (v)z ϕ dx. We may let k in each term on the right hand side, due to v L 1 (, µ) and v L 1 (). Thereore, and so (6) implies (5). lim (z, DT k(v)), ϕ = (z, Dv), ϕ, k Corollary 2.3. The distribution (z, Dv) is a Radon measure. It and its total variation (z, Dv) are absolutely continuous with respect to the measure Dv and (z, Dv) (z, Dv) z L (U) Dv B B holds or all Borel sets B and or all open sets U such that B U. In particular, i v BV (), then the measure (z, Dv) has inite total variation. Moreover, going back to (4), we can conclude that the ollowing proposition holds. Proposition 2.4. Let z DM loc() and let v BV loc () L (). Then zv DM loc(). Moreover the ollowing ormula holds in the sense o measures (7) div (zv) = (div z)v + (z, Dv). On the other hand, or every z DM (), a weak trace on o the normal component o z is deined in [5] and denoted by [z, ν]. Moreover, it satisies (8) [z, ν] L ( ) z L (). We explicitly point out that i z DM () and v BV () L (), then (9) v[z, ν] = [vz, ν] holds (see [12, Lemma 5.6] or [3, Proposition 2]). Having the dot product (z, Dv) and the normal component [z, ν], we may already prove our irst Green ormula. Proposition 2.5. Let z DM (), v BV () and assume that v L 1 (, µ). With the above deinitions, the ollowing Green ormula holds (10) v dµ + (z, Dv) = [z, ν]v dh N 1. Proo. Applying the Green ormula proved in [35, Theorem A.1] or in [12, Theorem 5.3], we obtain (11) T k (v) dµ + (z, DT k (v)) = [z, ν]t k (v) dh N 1, B U

8 8 V. DE CICCO, D. GIACHETTI AND S. SEGURA DE LEÓN or every k > 0. Note that the same argument appearing in the proo o Proposition 2.2 leads to (z, DT k (v)) = (z, Dv). lim k We may take limits in the other terms since v L 1 (, µ) and v L 1 ( ). Hence, letting k go to in (11), we get (10). Observe that we may apply (10) to a vector ield z DM () and the constant v 1. Since (z, Dv) = 0, we obtain (12) div z = [z, ν] dh N 1. This act and (7) are enough to prove the Green ormula we will apply in what ollows. Proposition 2.6. Let z DM loc() and set µ = div z. Let v BV () L () be such that that v L 1 (, µ). Then vz DM () and the ollowing Green ormula holds (13) v dµ + (z, Dv) = [vz, ν] dh N 1. Proo. Taking into account that v L 1 (, µ) implies that div (vz) = v div z + (z, Dv) is a Radon measure with inite variation, and so vz DM () (observe that we assume v L ()). It ollows now rom (12) that (13) holds. We will deal with z DM loc() such that the product vz DM () or some v BV () L (). To iner the Dirichlet boundary condition, we will use the ollowing result. Proposition 2.7. Let z DM loc() and v BV () L (). I vz DM (), then (14) [vz, ν] v z L () H N 1 a.e. on. Proo. We irst claim that, (15) φ[vz, ν] L ( ) φv L ( ) z L () holds or every nonnegative φ L ( ). To this end, let ϕ BV () L () satisy ϕ 0 and ϕ = φ. Applying (8) and (9), we may manipulate as ollows φ[vz, ν] L ( ) = [ϕvz, ν] L ( ) ϕvz L () ϕv L () z L (). By [5, Lemma 5.5], we ind a sequence (ϕ n ) n in W 1,1 () C() L () such that (1) ϕ n = φ. (2) ϕ n L () = φ L ( ). (3) ϕ n (x) = 0, i dist(x, ) > 1 n, or all n N. It ollows rom φ[vz, ν] L ( ) ϕ n v L () z L () or all n N that (15) holds true.

9 THE 1 LAPLACIAN EQUATION WITH A SINGULAR LOWER ORDER TERM 9 To prove (14), assume to get a contradiction that there exist ɛ > 0 and a measurable set E such that H N 1 (E) > 0 and Letting φ = χ E, we deduce which contradicts (15). [vz, ν] v z L () + ɛ, on E. φ[vz, ν] L ( ) φv L ( ) z L () + ɛ, 3. Weak solution or p Laplacian type problems For every p > 1 let us consider the ollowing problem p (u) = (x) (16) u γ in, u = 0 on. Remark 3.1. Let us note that the deinition o the unction (x) u on the right hand γ side o (16) needs to be precised. In all the paper, we will intend that the unction F (x, s) = (x) u, deined in [0, + [ with values in [0, + ], is F (x, 0) := 0 on γ the set {x : (x) = 0}. In this Section we will prove that or every p > 1 the problem (16) admits a weak solution in the ollowing sense. Deinition 3.2. A unction u W 1,p 0 () is a weak solution o problem (16) i or every open set ω there exists c ω > 0 such that u c ω a.e. in ω and or every v W 1,p 0 (). u p 2 u v dx = u γ v dx Theorem 3.3. For any ixed p > 1 and 0 < γ 1, and or any L N (), with 0, there exists a bounded unique weak solution o problem (16) in the sense o Deinition 3.2. Proo. By Theorems 4.1 and 4.4 o [18] there exists u W 1,p 0 () L () distributional solution to problem (16), i.e. a unction u such that u c ω a.e. in ω or every open set ω and (17) u p 2 u ϕ dx = u γ ϕ dx or every ϕ C0 (). Given v W 1,p 0 (), we will prove that (18) u p 2 u v dx = u γ v dx. Observe that we may assume v 0 without loss o generality. Consider a sequence ϕ n C 0 () such that ϕ n 0 and (19) ϕ n v strongly in W 1,p 0 (). We take or every η > 0 the unction ρ η (v ϕ n ),

10 10 V. DE CICCO, D. GIACHETTI AND S. SEGURA DE LEÓN where ρ η is a standard convolution kernel and v ϕ n := in{v, ϕ n }. By taking it as test unction in (17), we get (20) u p 2 u (ρ η (v ϕ n )) dx = u γ (ρ η (v ϕ n )) dx. We want to pass to the limit as η 0 and we get ρ η (v ϕ n ) v ϕ n strongly in W 1,p 0 (). This implies or the let hand side o (17) that (21) u p 2 u (ρ η (v ϕ n )) dx u p 2 u (v ϕ n ) dx. As ar as the right hand side o (20), let us observe that or any η > 0 supp (ρ η (v ϕ n )) K n, where K n is a compact set contained in, and that we have, choosing ϕ 1 on K n in (17), u γ L1 (K n ). Moreover, or any η > 0 and, as η 0, and so ρ η (v ϕ n ) L v ϕ n L ρ η (v ϕ n ) v ϕ n a.e., ρ η (v ϕ n ) v ϕ n w L. We conclude that, as η 0, (22) u γ (ρ η (v ϕ n )) dx By (20), (21) and (22) we obtain (23) u p 2 u (v ϕ n ) dx = u γ (v ϕ n) dx. u γ (v ϕ n) dx. Now, we are going to pass to the limit in (23), as n +. Since we have (24) v ϕ n v u p 2 u (v ϕ n ) dx We are going now to prove that u γ (v ϕ n) dx as n +. Indeed and u γ (v ϕ n) u γ v in W 1,p (), u p 2 u v dx. u γ v dx, a.e. 0 u γ (v ϕ n) u γ v.

11 THE 1 LAPLACIAN EQUATION WITH A SINGULAR LOWER ORDER TERM 11 Then by Lebesgue dominated convergence theorem, it is suicient to prove that (25) u γ v L1 (). We will prove that there exists a constant C > 0, independent o n, satisying or every n N (26) u γ ϕ n dx C. Indeed, by (17) we have u γ ϕ n dx = u p 2 u ϕ n dx u p 1 ϕ n dx u p dx + ϕ n p dx C, where the last inequality is due to (19) and to the act that u W 1,p 0 (). This prove (26). By (19), (26) and Fatou s Lemma, (25) ollows. Thereore ormula (18) holds. Let us now prove the uniqueness o the solution. Assuming that there exist two solutions u 1, u 2, the uniqueness ollows taking u 1 u 2 as test unction in the two equations satisied by u 1 and u 2 and observing that the term u is nonincreasing γ in the u variable. 4. Main result This section is devoted to solve problem ( ) Du div = (x) (27) Du u γ in, u = 0 on, or nonnegative data L N () and 0 < γ 1. We begin by introducing the notion o solution to this problem. Deinition 4.1. Let L N () with 0 a.e.. We say that u BV () L (), u 0 a.e., is a weak solution o problem (27) i there exists z DM loc() with z 1 such that (a) u γ L1 loc(), (b) χ {u>0} BV loc (), (c) (div z)χ {u>0} = u γ, in D (), (d) (z, Du) = Du, (z, Dχ {u>0} ) = Dχ {u>0} as measures in, (e) u + [uz, ν] = 0 H N 1 a.e. on. Remark 4.2. Since χ {u>0} BV loc (), it ollows rom Proposition 2.4 that zχ {u>0} DM loc() and the ollowing equality holds (28) div ( zχ {u>0} ) = ( div z ) χ {u>0} + (z, Dχ {u>0} ).

12 12 V. DE CICCO, D. GIACHETTI AND S. SEGURA DE LEÓN Then, since (z, Dχ {u>0} ) = Dχ {u>0}, the equation (c) is equivalent to div ( zχ {u>0} ) + Dχ{u>0} = u γ. We recall that the term Dχ {u>0} is a measure concentrated on the reduced boundary {u > 0} (or equivalently the reduced boundary {u = 0}). Moreover Dχ {u>0} coincides locally with the perimeter o the H N 1 -rectiiable set {u = 0}, i.e. or every open set ω we have that Dχ {u>0} (ω) coincides with the perimeter o {u = 0} ω. Remark 4.3. Let us point out that, by u γ ϕ dx < +, ϕ C 0 (), any solution to (27) satisies which means that {x : u(x) = 0, (x) > 0} = 0, (29) {x : u(x) = 0} {x : (x) = 0}, except or a set o zero Lebesgue measure. Note that this implies that i > 0 a.e. in, then u > 0 a.e. in. Moreover, i 0, then or every solution u we have u 0 and i 0, then u 0 is a solution. A urther remark is in order. It ollows rom (29) and by the act that (x, s) (x) s is a Carathéodory unction on [0, + [ with values in [0, + ] (see Remark γ 3.1), that (30) u γ = u γ χ {u>0} a.e. Remark 4.4. We explicitly observe that there is a variational ormulation o solution to (27), see (31) below. This ormulation looks like that o renormalized solution. Theorem 4.5. For every nonnegative L N () with 0 there is a weak solution to problem (27). Furthermore, or every nondecreasing and Lipschitz continuous unction h : [0, + [ [0, + [ such that h(0) = 0 and or every ϕ C0 (), (31) Moreover ϕ Dh(u) + h(u)z ϕ dx = (32) h(u) + [h(u)z, ν] = 0 holds H N 1 a.e. on. Proo. The proo will be divided in several steps. Step 1. Approximating problems. For every p > 1 let us consider the ollowing problem (33) p (u p ) = (x) u γ p h(u)ϕ dx. uγ in, u p = 0 on.

13 THE 1 LAPLACIAN EQUATION WITH A SINGULAR LOWER ORDER TERM 13 By Theorem 3.3 there exists u p W 1,p 0 () L () weak solution to problem (33), i.e. it satisies (34) u p p 2 u p v dx = v dx u γ p or every v W 1,p 0 (). Moreover or every open set ω there exists c ω > 0 such that u p c ω a.e. in ω. Step 2. BV estimate. Taking the test unction u p in problem (34), and by using the Hölder inequality we get (35) u p p dx = u 1 γ p dx N γ N 1 N up 1 γ N. N 1 Thus, applying the Sobolev and the Young inequalities we have (36) u p N S u p dx S u p p S(p 1) dx + N 1 p p S N γ N 1 N up 1 γ N + S. N 1 Since 1 γ < 1, it ollows that (u p ) is bounded in L N N 1 (). Thereore, by (35) and (36), we have (37) u p p dx M, or certain constant which does not depend on p. Thanks to Young s inequality, it implies that u p dx 1 u p p dx + p 1 M +. p p Having in mind that u p = 0, we obtain that (u p ) is bounded in BV (). Hence, there exists a unction u such that, (38) u BV (), and, up to a subsequence, (39) u p u in L q () or every 1 q < N N 1, (40) u p u in L N N 1 () and u p u pointwise a.e. in. Moreover (41) u 0 a.e., and u p Du weakly as measures. Step 3. L estimate. Taking the test unction G k (u p ) = (u p k) +, with k ]0, + [, in problem (34), and by using the Hölder inequality we get G k (u p ) p dx = G k (u p ) dx u γ p 1 k γ G k (u p )dx N k γ G k (u p ) N. N 1

14 14 V. DE CICCO, D. GIACHETTI AND S. SEGURA DE LEÓN Thus, the Sobolev and the Young inequalities imply G k (u p ) N S G k (u p ) dx S G k (u p ) p dx + N 1 p Now, by choosing k satisying S N k γ S(p 1) p S N k γ G k (u p ) N + S(p 1). N 1 < 1, we get G k (u p ) N C(p 1), N 1 with C independent o p. By applying Fatou s Lemma we deduce G k (u) N N 1 lim in G k(u p ) N p 1 N 1 We conclude that 0 u k a.e. in and so u L (). lim in C(p 1) = 0. p 1 Step 4. L 1 loc estimate or the singular term. We are going to prove in this step that there exists a constant C > 0, independent o p, satisying (42) ϕ dx C, ϕ C0 (), 1 < p p 0. u γ p Fixed ϕ C0 (), with ϕ 0, by (37) we have u γ ϕ dx = u p p 2 u p ϕ dx u p p 1 ϕ dx p ( ) u p p dx + ϕ p dx M + ϕ p0 + 1 dx < +. This implies that (42) holds or every ϕ C0 (), even i it changes sign. By this estimate and Fatou s Lemma we get (43) u γ ϕ dx < +, ϕ C 0 (). Step 5. Vector ield z. Now, we want to ind a vector ield z DM loc() with z 1, to play the Du role o Du. In this Step, we will ollow the argument o [33]. Fix 1 q < and consider 1 < p < q. It ollows rom (37), that u p p 2 q u p dx = u p q(p 1) dx Hence, ( u p p dx ) q p 1 q p M q p 1 q p. u p p 2 u p q M 1 p 1 q 1 p (1 + M) 1 q (1 + ) 1 q and so the sequence ( u p p 2 u p ) p is bounded in L q (; R N ). Then there exists z q L q (; R N ) such that, up to subsequences, u p p 2 u p z q, weakly in L q (; R N ).

15 THE 1 LAPLACIAN EQUATION WITH A SINGULAR LOWER ORDER TERM 15 A diagonal argument shows that there exists a vector ield z (independent o q) satisying (44) u p p 2 u p z, weakly in L q (; R N ), 1 q <. Moreover, by applying the lower semicontinuity o the q norm, the previous estimate u p p 2 u p q M 1 p 1 q 1 p implies z q 1 q, q <, so that, letting q go to, we obtain z L (; R N ) and z 1. Using ϕ C0 (), with ϕ 0, as a unction test in (34), we arrive at (45) u p p 2 u p ϕ dx = ϕ dx, u γ p and when we take p 1, using (41), Fatou s Lemma and (44) it becomes z ϕ dx u γ ϕ dx. Thereore, (46) div z u γ in D (), so div z is a nonnegative Radon measure. By (42) and (45) we have that 0 ϕ div z = z ϕ dx < + holds or every ϕ C0 () satisying ϕ 0. This implies that the total variation o div z is locally inite. Thereore z DM loc(). Let us note that the total variation o div z is only locally bounded, since by (42) the term u is bounded only in γ L1 p loc () (see also (45)). Step 6. The ollowing equation (47) u div z = u 1 γ holds in D () and the unction u belongs to L 1 (, div z). Firstly, let us prove that by (46) we have (48) u div z u 1 γ, in D (), namely (49) u ϕ div z u 1 γ ϕ ϕ C 0 (). Indeed we can apply (46) using the test unction (u ρ ɛ )ϕ, where ρ ɛ is a standard molliier and ϕ C0 (). We get (50) (u ρ ɛ )ϕ div z u γ (u ρ ɛ)ϕ dx. We recall that, since u L () then we have also that u u H N 1 -a.e. This implies that u u div z-a.e. since div z << H N 1 (this gives also that u L 1 loc (, div z) ). By Proposition 3.64 (b) and Proposition 3.69 (b) o [1] we have that u ρ ɛ u H N 1 a.e.

16 16 V. DE CICCO, D. GIACHETTI AND S. SEGURA DE LEÓN and thereore div z-a.e. We can pass to the limit in both sides o the inequality (50) by dominated convergence theorem (with respect to the measure div z on the let hand side and the Lebesgue measure on the right hand side), since u ρ ɛ u in and since u ϕ L 1 () by (43). Thereore (48) holds. γ Now, in order to prove the opposite inequality, let ϕ C0 (), with ϕ 0, and take u p ϕ as a unction test in (34). Then ϕ u p p dx + u p u p p 2 u p ϕ dx = u 1 γ p ϕ dx. Applying Young s inequality, it yields ϕ u p dx + u p u p p 2 u p ϕ dx 1 ϕ u p p dx + u p u p p 2 u p ϕ dx + p 1 ϕ dx p p u 1 γ p ϕ dx + p 1 ϕ dx. p To pass to the limit on the let hand side, we apply the lower semicontinuity, jointly with (39) and (44). Assume irst that 0 < γ < 1. On the right hand side, we point out that, as p 1, by (40) (51) u 1 γ p u 1 γ in L N so that N 1 1 N 1 γ () L N 1 (), (52) lim u p 1 1 γ p ϕ dx = u 1 γ ϕ dx. It leads to (53) ϕ Du + uz ϕ dx u 1 γ ϕ dx. Note that i γ = 1 the irst integral on the right hand side does note depend on p. Taking into account (7) and (5), we deduce by (53) that u ϕ div z = ϕ(z, Du) + uz ϕ dx ϕ Du + uz ϕ dx u 1 γ ϕ dx. Thereore, (54) u div z u 1 γ, in D () and by (48) and (54) we conclude that (47) holds. This implies that u div z L 1 (), so that u L 1 (, div z). Step 7. The equality (z, Du) = Du, as measures on, holds. It is a straightorward consequence o (53) and (47). Indeed, consider ϕ C0 () with ϕ 0. Then we have ϕ Du + u z ϕ dx u 1 γ ϕ dx = u ϕ div z. Thereore by Proposition 2.4 ϕ Du u z ϕ dx u ϕ div z = ϕ(z, Du).

17 THE 1 LAPLACIAN EQUATION WITH A SINGULAR LOWER ORDER TERM 17 The arbitrariness o ϕ implies that Du (z, Du), as measures on. On the other hand, since z 1, the opposite inequality holds, i.e. Du (z, Du), as measures on. This concludes the proo o Step 7. Step 8. The boundary condition u + [uz, ν] = 0 holds H N 1 a.e. on. By taking u p as test unction in (34) we have u p p dx = u 1 γ p dx. On the other hand, by the act that u p = 0 on and by the Young inequality we get u p dx + u p dh N 1 1 u p p dx + p 1 p p u 1 γ p dx + p 1 p. We may use the lower semicontinuity o the unctional on the let hand side to pass to the limit as p 1 and obtain by (51) and (47) Du + u dh N 1 u 1 γ dx = u div z. Applying the Green ormula (see Proposition 2.6) on the right hand side we have Du + u dh N 1 (z, Du) [uz, ν] dh N 1. Then, by Step 7, we arrive at (u + [uz, ν]) dh N 1 0. Since [uz, ν] u z u, we conclude that u + [uz, ν] = 0 H N 1 a.e. on, which concludes the proo o the Step 8. Step 9. Variational ormulation. Consider a Lipschitz continuous and nondecreasing unction h : [0, + [ [0, + [ such that h(0) = 0. Then by (38) we have that h(u) BV (). Following the argument o Step 6, we have (55) h(u) div z u γ h(u), in D (). In order to prove the opposite inequality, let ϕ C 0 (), with ϕ 0, and taking h(u p )ϕ as test unction in (34) we obtain ϕ u p p h (u p ) dx + h(u p ) u p p 2 u p ϕ dx = u γ h(u p)ϕ dx.

18 18 V. DE CICCO, D. GIACHETTI AND S. SEGURA DE LEÓN Applying Young s inequality, it yields (56) ϕ h(u p ) dx + h(u p ) u p p 2 u p ϕ dx = ϕ u p h (u p ) dx + h(u p ) u p p 2 u p ϕ dx 1 ϕ u p p h (u p ) dx + h(u p ) u p p 2 u p ϕ dx + p 1 h (u p )ϕ dx p p h(u p )ϕ dx + p 1 p L ϕ dx, u γ p where L denotes the Lipschitz constant o h. Moreover, it ollows rom (57) h(u p ) Lu p, the convergence (39) and Vitali s Theorem that, as p 1, (58) h(u p ) h(u), in L q () or every 1 q < N N 1. By (57) we also get E u γ p h(u p )ϕ dx L E u 1 γ p ϕ dx or every Borel measurable set E. Thereore, by (52) and Vitali s Theorem we obtain (59) lim p 1 u γ h(u p )ϕ dx = h(u)ϕ dx. p uγ The lower semicontinuity, jointly with (56), (58) and (59), leads to (60) ϕ Dh(u) + h(u)z ϕ dx h(u)ϕ dx. uγ By (7) we deduce that h(u) div z ϕ = which implies that ϕ(z, Dh(u)) + h(u)z ϕ dx ϕ Dh(u) + h(u)z ϕ dx (61) h(u) div z u γ h(u), in D (). By (55) and (61), we conclude that (62) h(u) div z = u γ h(u), in D (). Next, we deduce that (63) (z, Dh(u)) = Dh(u). h(u)ϕ dx uγ We point out that we cannot apply [5, Proposition 2.8] since u is not continuous (it is proved or pairings such as z DM () and u BV () C() L ()).

19 THE 1 LAPLACIAN EQUATION WITH A SINGULAR LOWER ORDER TERM 19 Thus, we must ocus on our equation and repeat the same arguments as in Step 7. In act, let us consider ϕ C0 () with ϕ 0. Then by (60) and (62) we have ϕ Dh(u) + h(u) z ϕ dx h(u)ϕ dx = ϕh(u) div z. uγ Thereore by (4) ϕ Dh(u) The arbitrariness o ϕ implies that h(u) z ϕ dx h(u) ϕ div z = ϕ(z, Dh(u)). Dh(u) (z, Dh(u)), as measures on. On the other hand, since z 1, the opposite inequality holds, i.e. Dh(u) (z, Dh(u)), as measures on. This proves that (z, Dh(u)) = Dh(u). By (63), (62) and (7) we have (64) Dh(u) div ( h(u)z ) = u γ h(u), in D () and then (31) holds. Next, take h(u p ) as test unction in (34) to obtain u p p h (u p ) dx = h(u p ) dx. u γ p The boundary condition and Young s inequality yield (65) h(u p ) dx + h(u p ) dh N 1 = u p h (u p ) dx 1 u p p h (u p ) dx + p 1 h (u p ) dx p p h(u p ) dx + p 1 p Having in mind (57), we deduce h(u p ) dx = lim p 1 u γ p u γ p h(u) dx. uγ h (u p ) dx. On the let hand side o (65), one can use the lower semicontinuity o the unctional to arrive at Dh(u) + h(u) dh N 1 h(u) dx. uγ Applying (64) and Green s ormula (13), this inequality becomes Dh(u) + h(u) dh N 1 Dh(u) [h(u)z, ν] dh N 1, wherewith h(u) + [h(u)z, ν] dh N 1 0. Hence, h(u) + [h(u)z, ν] = 0 holds H N 1 a.e. on, i.e. (32) is proved. Step 10. The unction χ {u>0} belongs to BV loc ().

20 20 V. DE CICCO, D. GIACHETTI AND S. SEGURA DE LEÓN Let (h n ) be a sequence o Lipschitz continuous and nondecreasing unctions h n : [0, + [ [0, + [ such that h n (0) = 0 and h n (u) χ {u>0} in L 1 (). (For instance, we may consider the truncation h n (s) = nt 1/n (s + ).) By (64), or every n and ϕ C0(), 1 with ϕ 0, we have ϕ Dh n (u) h n (u)z ϕ dx + u γ h n(u)ϕ dx. On account o (43), this act implies that or every ω there exists a constant C such that Dh n (u) (ω) C. Hence or every ω Dχ {u>0} (ω) lim in n + Dh n(u) (ω) C, and χ {u>0} belongs to BV loc (). Then by Proposition 2.4 we have that zχ {u>0} DM loc() and by (7) the ollowing equality holds div ( zχ {u>0} ) = ( div z ) χ {u>0} + (z, Dχ {u>0} ) in D (). Step 11. u satisies the inequality div ( ) zχ {u>0} + Dχ{u>0} u, in D (). γ By (64) or every n and ϕ C0(), 1 with ϕ 0, we have ϕ Dh n (u) + h n (u)z ϕ dx = By the lower semicontinuity we have ϕ Dχ {u>0} + χ {u>0} z ϕ dx u γ h n(u)ϕ dx. u γ ϕ dx. Step 12. u satisies the equation ( div z ) χ {u>0} = u in D (). γ Since χ {u>0} BV loc (), we have that (30) and (46) imply Since z 1, the inequality ( div z ) χ {u>0} u γ in D (). (66) (z, Dχ {u>0} ) Dχ {u>0} holds. Moreover, by (4), (66) and Step 11 we have that ( div z ) χ {u>0} = div ( zχ {u>0} ) + (z, Dχ{u>0} ) div ( zχ {u>0} ) + Dχ{u>0} u γ. This concludes the Step. Step 13. (z, Dχ {u>0} ) = Dχ {u>0} By (66) we need to prove the opposite inequality (z, Dχ {u>0} ) Dχ {u>0}. We note that by Steps 11 and 12 or every ϕ C0 () with ϕ 0 we have ϕ Dχ {u>0} + Thereore by (4) we get ϕ Dχ {u>0} χ {u>0} z ϕ dx u γ ϕ dx = ϕχ {u>0} div z. χ {u>0} z ϕ dx χ {u>0} ϕ div z = ϕ(z, Dχ {u>0} ).

21 THE 1 LAPLACIAN EQUATION WITH A SINGULAR LOWER ORDER TERM 21 The arbitrariness o ϕ implies that as measures on. This concludes the proo. Dχ {u>0} (z, Dχ {u>0} ), Remark 4.6. We recall that the boundary condition used, as a rule, in the Dirichlet problem or equations involving the 1 Laplacian is [z, ν] sign ( u) holds on, which requires that z DM (). This is not the case in our situation since we just have z DM loc() and thereore the trace [z, ν] is not deined and the usual boundary condition has no sense. On the contrary, the above boundary condition h(u) + [h(u)z, ν] = 0, satisied or every Lipschitz continuous and nondecreasing unction h : [0, + [ [0, + [ such that h(0) = 0, is completely meanigul since h(u)z DM () by Proposition 2.6. Remark 4.7. We point out that i k > 0, then χ {u>k} BV (). This act is a consequence o the inequality u γ h(g ku) k γ h(g ku) L 1 (), valid or every Lipschitz continuous and nondecreasing h : [0, + [ [0, + [ such that h(0) = 0, since we may apply the variational ormulation to h(g k (s)) and ollow the argument o Step 9. Moreover, we can deduce the ollowing acts as well: div (χ {u>k} z) + Dχ {u>k} = u γ χ {u>k} in D () ; (z, Dχ {u>k} ) = Dχ {u>k} as measures in ; χ {u>k} + [χ {u>k} z, ν] = 0 holds H N 1 a.e. on. Remark 4.8. The BV -estimate proven in Step 2 depends only on the act that γ > 0 (see (36)). We point out that in the non singular setting (when γ = 0) this act does not hold, unless is small enough. This act agrees with [34], where it is seen that only i W 1, 1 there is a BV estimate. Remark 4.9. The L estimate proven in Step 3 is not a consequence o Stampacchia s procedure. We remark that it depends on the singularity (since it goes to 0 at ininity) and the act that we are dealing with the 1-Laplacian operator. 5. The case o strictly positive In this section we will assume that (x) > 0 a.e. in and we will prove those speciic eatures holding in this case, in particular, we will see a uniqueness result. We will see in Section 6, uniqueness does not hold or which vanishes on a set o positive Lebesgue measure. Remark that 0 must be studied in a dierent way, the solutions being 1-harmonic unctions. I 0, then the approximate solution u p vanishes and so the solution u ounded in Theorem 4.5 becomes the trivial solution. Moreover, applying the results in [42] (which require some additional geometrical assumptions) we obtain that this trivial solution is the only one. Theorem 5.1. Let L N () be such that (x) > 0 or almost all x. Then the solution u we have ound in Theorem 4.5 to problem (27) satisies (a) u(x) > 0 or almost all x,

22 22 V. DE CICCO, D. GIACHETTI AND S. SEGURA DE LEÓN (b) u γ L1 (), and there exists z DM () with div z L 1 () and z 1 such that (c) div z = u γ in D (), (d) (z, Du) = Du as measures in, (e) [z, ν] sign ( u) H N 1 a.e. on. Proo. Since u is a solution, we already know that (d) holds. Condition (a) is a consequence o Remark 4.3, and it yields condition (c) by Deinition 4.1. Observe that assuming condition (b), we deduce in a straightorward way that z DM () (with div z L 1 ()) and so [z, ν] has sense and (e) holds. Hence, only condition (b) remains to be proved. We will see it in two steps. Step 1. For every nonnegative v W 1,1 0 (), we have z v dx = u v dx. γ In this Step we use an argument close to those used in Section 3, where existence and uniqueness o approximating problems (16) are proved. We repeat these arguments or the sake o completeness (note that here p = 1). Consider a sequence ϕ n C0 () such that ϕ n 0 and (67) ϕ n v strongly in W 1,1 0 (). We take or every η > 0 the unction ρ η (v ϕ n ), where ρ η is a standard convolution kernel and v ϕ n := in{v, ϕ n }. By taking it as test unction in (c), we get (68) z (ρ η (v ϕ n )) dx = u γ (ρ η (v ϕ n )) dx. We want to pass to the limit as η 0 using ρ η (v ϕ n ) v ϕ n This implies or the let hand side o (68) that (69) z (ρ η (v ϕ n )) dx strongly in W 1,1 (). z (v ϕ n ) dx As ar as the right hand side o (68) is concerned, let us observe that, or η > 0 small enough, supp (ρ η (v ϕ n )) K n, where K n is a compact set contained in. Moreover, it ollows rom u L 1 γ loc () (see (43)), that u γ L1 (K n ). Furthermore, or any η > 0 and, as η 0, and so ρ η (v ϕ n ) L v ϕ n L ρ η (v ϕ n ) v ϕ n a.e., ρ η (v ϕ n ) v ϕ n w L.

23 THE 1 LAPLACIAN EQUATION WITH A SINGULAR LOWER ORDER TERM 23 We conclude that, as η 0, (70) u γ (ρ η (v ϕ n )) dx By (68), (69) and (70) we obtain (71) z (v ϕ n ) dx = u γ (v ϕ n) dx. u γ (v ϕ n) dx. Now, we are going to pass to the limit in (71), as n +. Since we have v ϕ n v z (v ϕ n ) dx We are going now to prove that u γ (v ϕ n) dx as n +. Indeed and u γ (v ϕ n) u γ v in W 1,1 (), 0 u γ (v ϕ n) u γ v. z v dx. u γ v dx, Then by Lebesgue dominated convergence theorem, it is suicient to prove that (72) u γ v L1 (). Indeed, by condition (c) we have u γ ϕ n dx = z ϕ n dx z a.e. ϕ n dx C, where the last inequality is due to (67). It ollows rom (67) and Fatou s Lemma that (72) holds. Thereore, Step 1 is proved. Step 2. The inequality u v dx γ v + v dhn 1 dx holds or every nonnegative v W 1,1 () L (). Applying [5, Lemma 5.5], we ind a sequence (w n ) n in W 1,1 () C() such that (1) w n = v. (2) w n dx v dh N n. (3) w n (x) 0, or all x. Since v w n W 1,1 0 (), Step 1 becomes u γ v w n dx = z v w n dx z v dx + z w n dx v + v dh N n.

24 24 V. DE CICCO, D. GIACHETTI AND S. SEGURA DE LEÓN Then Fatou s Lemma implies v dx lim in uγ n u γ v w n dx v + v dh N 1. In particular, chosing v(x) = 1 or all x, we obtain u γ L 1 (), as desired. As a consequence o Theorem 5.1 (b) and (c), every v BV () L () can be taken as test unction in our problem. Corollary 5.2. Let L N () be such that (x) > 0 or almost all x. Then (73) (z, Dv) v[z, ν] dh N 1 = u γ v dx holds or every v BV () L (). Proo. Fix v BV () L (). It ollows rom Theorem 5.1 (b) and (c) that div z L 1 (), so that we may apply Anzellotti s theory to the case div z L 1 () and v BV () L (). This gives or every v BV () L () v div z dx = u γ v dx, which ollows by (c) o Theorem 5.1. Then equation (73) is a consequence o Green s ormula (13) and (9). Theorem 5.3. For every L N (), with (x) > 0 a.e., there exists at most a solution o (27) in the sense o Deinition 4.1. Proo. Assume to get a contradiction that u 1 and u 2 are positive solutions to problem (27). Then, or i = 1, 2, u i (x) > 0 or almost all x and there exist vector ields z i satisying all the requirements o Deinition 4.1. Applying Corollary 5.2 to the dierence o the solutions and taking conditions (d) and (e) in Theorem 5.1 into account, we obtain Du 1 (z 1, Du 2 ) + (u 1 + u 2 [z 1, ν]) dh N 1 = u γ (u 1 u 2 ) dx 1 Du 2 (z 2, Du 1 ) + (u 2 + u 1 [z 2, ν]) dh N 1 = (u 1 u 2 ) dx. u γ 2 Adding both equations and rearranging its terms, we get [ Du1 (z 2, Du 1 ) ] [ + Du2 (z 1, Du 2 ) ] + (u 1 + u 1 [z 2, ν]) dh N 1 + (u 2 + u 2 [z 1, ν]) dh N 1 ( = u γ ) 1 u γ (u1 u 2 ) dx = (u γ 1 2 u 1 u uγ 2 )(u 1 u 2 ). 2 [ Since Du1 (z 2, Du 1 ) ] [ 0, Du2 (z 1, Du 2 ) ] 0, (u 1 +u 1 [z 2, ν]) dh N 1 0 and (u 2 + u 2 [z 1, ν]) dh N 1 0, we iner that 0 u 1 u 2 (u γ 1 uγ 2 )(u 1 u 2 ),

25 THE 1 LAPLACIAN EQUATION WITH A SINGULAR LOWER ORDER TERM 25 whose integrand is nonpositive, so that this term vanishes. It ollows that the integrand vanishes as well. Thereore, (u γ 1 uγ 2 )(u 1 u 2 ) = 0 and so u 1 = u 2 in. 6. Explicit 1 dimensional solutions In this Section we will show some explicit solutions taken =] 1, 1[ and γ = 1. We restrict to the one dimensional case or the sake o simplicity. Indeed, in any dimension and considering any γ ]0, 1], one may think that the constant unction u(x) = 1/γ W 1, () is a solution to our problem reasoning as ollows. Since then /u γ = / W 1, (), and applying [34, Theorem 4.3 and Remark 4.7] we get a vector ield z DM satisying z 1 and div z = u in D (). Obviously, γ (z, Du) = 0 = Du holds as measures in. Unortunately, the condition [z, ν] sign ( u) H N 1 a.e. on is not guaranteed. Nevertheless, when =] 1, 1[, it is easy to deine a unction z which satisy the weak orm o boundary condition, just choosing z such that z( 1) = 1 i u( 1) 0 and z(1) = 1 i u(1) 0. In what ollows we deal with the problem ( u ) = (74) u, in ] 1, 1[ ; u u( 1) = 0 = u(1). It is worth recalling the conditions o being a solution in this 1 dimensional setting. As just mentioned, the boundary condition becomes z( 1) = 1 i u( 1) 0 and z(1) = 1 i u(1) 0. On the other hand, the condition z χ {u>0} = /u implies that z is a unction and so z does not jump (at least where u > 0). Finally, it ollows rom (z, u ) = u that z(x) = 1 i u increases at x and z(x) = 1 i u decreases at x. Example 6.1. Set (x) = χ ] 1/2,1/2[ (x). From the previous argument, we already know that a solution to (74) is given by u 1 (x) = 1 2 with 1, i 1 < x < 1 2 ; z 1 (x) = 2x, i 1 2 x 1 2 ; 1, i 1 2 < x < 1. The same unction z 1 allows us to check that any u satisying u nondecreasing in ] 1, 1 2 ], u(x) = 1 2 in ] 1 2, 1 2 [ and u nonincreasing in [ 1 2, 1[, is a solution to (74). For instance, u 2 (x) = 1 2 χ ] 1 2, 1 2 [ (x) deines a solution. We conclude that uniqueness does not hold. Observe that, or u 2 = 1 2 χ ] 1 2, 1 2 [, other choices o the auxiliary unction are possible, namely: 2(x + 1), i 1 < x < 1 2 ; z 2 (x) = 2x, i 1 2 x 1 2 ; 2(x 1), i 1 2 < x < 1.

26 26 V. DE CICCO, D. GIACHETTI AND S. SEGURA DE LEÓN Remark 6.2. We point out that the unction z is not unique. Moreover, what is really important in our equation is the identity z χ {u>0} = u. We remark that in the previous example z 1 = z 1χ {u>0}, while z 2 z 2χ {u>0}. Furthermore, one may wonder i the identity (zχ {u>0} ) = u holds as well. The answer is negative since one can easily check that, or solution u = 1 2 χ ] 1 2, 1 2 [, it yields (z 1 χ {u>0} ) = 2χ ] 1 2, 1 2 [ + δ 1/2 + δ 1/2 = z 1χ {u>0} + δ 1/2 + δ 1/2 z 1χ {u>0}. Note that the terms δ 1/2 and δ 1/2 are the measure Dχ {u>0} concentrated in {u > 0}. Example 6.3. Consider the unction :] 1, 1[ R deined by x 1 2, i 1 2 x < 1 ; (x) = 0, i 1 2 < x < 1 2 ; x 1 2, i 1 < x 1 2. It is straightorward that a solution o (74) is given by the constant unction u 1 (x) = 1 8 with 4x 2 + 4x 1, i 1 2 x < 1 ; z 1 (x) = 0, i 1 2 < x < 1 2 ; 4x 2 + 4x + 1, i 1 < x 1 2. Another solution is given by u 2 (x) = 1 16 χ {>0}(x) with 8x 2 + 8x 1, i 1 2 x < 1 ; z 2 (x) = 2x, i 1 2 < x < 1 2 ; 8x 2 + 8x + 1, i 1 < x 1 2. In this case, z 2χ {u2>0} = /u 2 holds, and z 2χ {u2>0} z 2. Observe that this example shows that there is not uniqueness even where { > 0}. There are still other solutions to (74), or instance, u 3 (x) = 1 16 χ ] 1, 1 4 [ ] 1 4,1[ (x) with 8x 2 + 8x 1, i 1 2 x < 1 ; z 3 (x) = 1, i 1 4 x < 1 2 ; 4x, i 1 4 < x < 1 4 ; 1, i 1 2 < x 1 4 ; 8x 2 + 8x + 1, i 1 < x 1 2.

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