Weak solutions of semilinear elliptic equations with Leray-Hardy potential and measure data

We study existence and stability of solutions of (E 1) --$\Delta$u + $\mu$ |x| 2 u + g(u) = $\nu$ in $\Omega$, u = 0 on $\partial$$\Omega$, where $\Omega$ is a bounded, smooth domain of R N , N $\ge$ 2, containing the origin, $\mu$ $\ge$ -- (N --2) 2 4 is a constant, g is a nondecreasing function satisfying some integral growth assumption and $\nu$ is a Radon measure on $\Omega$. We show that the situation differs according $\nu$ is diffuse or concentrated at the origin. When g is a power we introduce a capacity framework to find necessary and sufficient condition for solvability.


Introduction
Let Ω ⊂ R N (N ≥ 2) be a bounded, smooth domain containing the origin.We define the elliptic operator with Leray-Hardy potential L by where µ is a real number satisfying If g : R → R is a continuous nondecreasing function such that g(0) ≥ 0, we are interested in the nonlinear Poisson equation where ν is a Radon measure in Ω.When µ = 0, problem (1.3) reduces to which has been extensively studied by numerous authors in the last 30 years.A fundamental contribution is due to Brezis [4], Benilan and Brezis [2], where ν is bounded and the function g : R → R is nondecreasing, positive on (0, +∞) and satisfies the subcritical assumption in dimension N ≥ 3: They obtained the existence, uniqueness and stability of weak solutions for the problem.When N = 2, Vazquez [21] introduced the exponential orders of growth of g defined by and proved that if ν is any bounded measure in Ω with Lebesgue decomposition where ν r is part of ν with no atom, a j ∈ Ω and α j ∈ R satisfy then (1.4) admits a (unique) weak solution.Later on, Baras and Pierre [1] studied (1.4) when g(u) = |u| p−1 u for p > 1 and they discovered that if p ≥ N N −2 the problem is well posed if and only if ν is absolutely continuous with respect to the Bessel capacity c 2,p ′ with p ′ = p p−1 .It is known that by the improved Hardy inequality [8] and Lax-Milgram theorem, the nonhomogeneous problem with f ∈ L 2 (Ω), has a unique solution in H 1 0 (Ω) if µ > µ 0 , or in a weaker space H(Ω) if µ = µ 0 [17].A natural question is to find sharp condition of f for the existence or nonexistence of (1.8) and the difficulty comes from the fact that the Hardy term |x| −2 u may not be locally integrable in Ω.An attempt done by Dupaigne in [17] is to consider problem (1.8) when µ ∈ [µ 0 , 0) and N ≥ 3 in the sense of distributions (1.9) The corresponding semi-linear problem is studied by [3] with this approach.
We adopt here a different point of view in using a different notion of weak solutions.It is known that the equation L µ u = 0 in R N \ {0} has two distinct radial solutions In the remaining of the paper and when there is no ambiguity, we put τ + = τ + (µ), τ 0 + = τ + (µ 0 ), τ − = τ − (µ) and τ 0 − = τ − (µ 0 ).It is noticeable that identity (1.9) cannot be used to express that Φ µ is a fundamental solution, i.e. f = δ 0 since Φ µ is not locally integrable if µ ≥ 2N .Recently, Chen, Quaas and Zhou found in [11] that the function Φ µ is the fundamental solution in the sense that it solves where and With the power-absorption nonlinearity in Ω * = Ω \ {0}, the precise behaviour near 0 of any positive solution of is given in [18] when p > 1.In this paper it appears a critical exponent with the following properties: if p ≥ p * µ any solution of (1.13) can be extended to be in D ′ (Ω).If 1 < p < p * µ any positive solution of (1.13) either satisfies lim where ℓ = ℓ N,p,µ > 0 or there exists k ≥ 0 such that and in that case u ∈ L p loc (Ω; dγ µ ).In view of [11], it implies that u satisfies Furthermore, it is proved in [18] that when µ > µ 0 and g : R → R + is a continuous nondecreasing function satisfying then for any k > 0 there exists a radial solution of satisfying (1.16), where B * 1 := B 1 (0) \ {0}.When µ = µ 0 and N ≥ 3 it is proved in [18] that if there exists b > 0 such that then there a exists a radial solution of (1.19) satisfying (1.16) with γ = (N +2)b

2
. In fact this condition is independent of b > 0, by contrast the case N = 2 and µ = 0 where the introduction of the exponential order of growth of g is a necessity.Moreover, it is easy to see that u satisfies In view of these results and identity (1.10), we introduce a definition of weak solutions adapted to the operator L µ in a measure framework.Since Γ µ is singular at 0 if µ < 0, there is need of defining specific set of measures and we denote by M(Ω * ; Γ µ ), the set of Radon measures ν in Ω * such that If ν ∈ M + (Ω * ), we define its natural extension, with the same notation since there is no ambiguity, as a measure in Ω by Since the idea is to use the weight Γ µ in the expression of the weak solution, the expression Γ µ ν has to be defined properly if τ + < 0. We denote by M(Ω; Γ µ ) the set of measures ν on Ω which coincide with the above natural extension of ν⌊ is defined acoordingly.Notice that the Dirac mass at 0 does not belong to M(Ω; Γ µ ) although it is a limit of {ν n } ⊂ M(Ω; Γ µ ).We detote by M(Ω; Γ µ ) the set of measures which can be written under the form where ν⌊ Ω * ∈ M(Ω; Γ µ ) and k ∈ R. Before stating our main theorem we make precise the notion of weak solution used in this article.We denote Ω * := Ω \ {0}, ρ(x) = dist(x, ∂Ω) and where L * µ is given by (1.10) and c µ is defined in (1.12).
A measure for which problem (1.3) admits a solution is a g-good measure.In the regular case we prove the following and g : R → R be a Hölder continuous nondecreasing function such that g(r)r ≥ 0 for any r ∈ R. Then for any ν ∈ L 1 (Ω, dγ µ ), problem (1.3) has a unique weak solution u ν such that for some c 1 > 0, Furthermore, if u ν ′ is the solution of (1.3) with right-hand side ν ′ ∈ L 1 (Ω, dγ µ ), there holds and for all ξ ∈ X µ (Ω), ξ ≥ 0.
Furthermore, if ν⌊ Ω * = 0, condition (1.5) can be replaced by the following weaker one Normally, the estimates on the Green kernel plays an essential role for approximating the solution of elliptic problems with absorption and Radon measure.However, we have banned the estimates on the Green kernel for Hardy operators due to luck them for µ ≥ µ 0 , and our main idea is to separate the measure ν * in M(Ω; Γ µ ) and the Dirac mass at the origin, and then to glue the solutions with above measures respectively.This requires a very week new assumption: the weak ∆ 2 -condition.
In the previous result, it is noticeable that if k = 0 (resp.ν⌊ Ω * = 0) only condition (1.5) (resp.condition (1.32)) is needed.In the two cases the weak ∆ 2 -condition is unnecessary.In the power case where g(u) = |u| p−1 u := g p (u), the following result follows from Theorem B and C.
µ , p * 0 in the case k = 0 and ν⌊ Ω * = 0. We remark that p * µ is the sharp exponent for existence of (1.32) when ν⌊ Ω * = 0, while the critical exponent becomes p * 0 when k = 0 and ν has atom in Ω \ {0}.The supercritical case of equation (1.33) corresponds to the fact that not all measures are g p -good and the case where k = 0 is already treated.
ǫ is absolutely continuous with respect to the c 2,p ′ -Bessel capacity.Finally we characterize the compacts removable sets in Ω.
can be extended a weak solution of the same equation in whole Ω if and only if µ .The rest of this paper is organized as follows.In Section 2, we build the framework for weak solutions of (1.3) involving L 1 data.Section 3 is devoted to solve existence and uniqueness of weak solution of (1.3), where the absorption is subcritical and ν is a related Radon measure.Finally, we deal with the super critical case in Section 4 by characterized by Bessel Capacity.

L 1 data
Throughout this section we assume N ≥ 2 and µ ≥ µ 0 and in what follows, we denote by c i with i ∈ N a generic positive constant.We first recall some classical comparison results for Hardy operator L µ .The next lemma is proved in [11,Lemma 2.1], and in [16 As an immediate consequence we have (2.1) in the sense of Definition 1.1, it satisfies also If u is a weak solution of (2.2) there holds ) The following form of Kato's inequality, proved in [11, Proposition 2.1], plays an essential role in the obtention a priori estimates and uniqueness of weak solution of (1.3). and for some τ > τ − .Let u f be the solution of (2.9) Then there holds: and the embedding then G is a convex nonnegative function.If ρν ∈ L 2 (Ω) we define the functional J ν in the space H 1 µ,0 (Ω) by (2.15) The functional J is strictly convex, lower semicontinuous and coercive in H 1 µ,0 (Ω), hence it admits a unique minimum u which satisfies and (Ω, ρdγ µ ) and denote by {u n } the sequence of the corresponding minimizing problem in H 1 µ,0 (Ω).By Proposition 2.1 we have that, for any ξ ∈ X µ (Ω), (2.18) We denote by η 0 the solution of Its existence is proved in [11, Lemma 2.2], as well as the estimate 0 ≤ η 0 ≤ c 6 ρ for some c 6 > 0. Since g is monotone, we obtain from (2.18) Hence {u n } is a Cauchy sequence in L 1 (Ω, dγ µ ).Let η 1 solve the equation (we can always assume that Ω ⊂ B 1 ).As in the proof of [11, Lemma 2.2], for any x 0 ∈ Ω there exists r 0 > 0 such that B r 0 (x 0 ) ⊂ Ω and for t > 0 small enough w t,x 0 (x) = t(r 2 0 − |x − x 0 | 2 ) is a subsolution of (2.19), hence of (2.21).Therefore η 1 exists.Using again the density of C ∞ 0 (Ω) in H 1 µ,0 (Ω) and integrating on Ω \ B ǫ and letting ǫ → 0, we obtain as a variant of (2.20) Hence {u n } is a Cauchy sequence in L 1 (Ω, |x| −1 dγ µ ) with limit u and {g(u n )} is a Cauchy sequence in L 1 (Ω, ρdγ µ ) with limit g(u).Then (2.17) holds.As for (1.28) it is a consequence of (2.18) and (1.29) is proved similarly.

The subcritical case
In this section as well as in the next one we always assume that N ≥ 3 and µ ≥ µ 0 , or N = 2 and µ > 0, since the case N = 2, µ = 0, which necessitates specific tools, has already been completely treated in [21].We recall that the set M(Ω * ; Γ µ ) of Radon measures is defined in introduction as the set of measures in Ω * satisfying (1.22), and any positive measure ν ∈ M(Ω * ; Γ µ ) is naturaly extended by formula (1.23) as a positive measure in Ω.The space M(Ω; where ν⌊ Ω * ∈ M(Ω * ; Γ µ ).

The linear equation
This solution is denoted by G µ [ν], and this defines the Green operator of L µ in Ω with homogeneous Dirichlet conditions.
Proof.By linearity and using the result of [11] on fundamental solution, we can assume that k = 0 and ν ≥ 0. Let {ν n } ⊂ L 1 (Ω, ρdγ µ ) be a sequence such that ν n ≥ 0 and and by Proposition 2.1, we may let u n be the unique, nonnegative weak solution of with n ∈ N.There holds Then u n ≥ 0 and using the function η 1 defined in the proof of Theorem A for test function, we have c which implies that {u n } is bounded in L 1 (Ω, 1 |x| dγ µ (x)).For any ǫ > 0 sufficiently small, set the test function ξ in {ζ ∈ X µ (Ω) : ζ = 0 in B ǫ }, then we have that Note that in Ω \ B ǫ , the operator L * µ is uniformly elliptic and the measure dγ µ is equivalent to the Hausdorff measure dx, then [24, Corollary 2.8] could be applied to obtain that for some c > 0 independent of n but dependent of O ′ , That is, {u n } is uniformly bounded in W 1,q loc (Ω \ {0}).As a consequence, by the arbitrary of ǫ, there exist a subsequence, still denoting {u n } n and u such that u n → u a.e. in Ω.
Meanwhile, we deduce from Fatou's lemma, We next claim that u n → u in L 1 (Ω, |x| −1 dγ µ ).Let ω ⊂ Ω be a Borel set and ψ ω be the solution of Then ψ ω ≤ η 1 , thus it is uniformly bounded.Assuming that Ω ⊂ B 1 , clearly ψ ω is bounded from above by the solution Ψ ω of and by standard rearrangement, sup Therefore {u n } is uniformly integrable for the measure |x| −1 dγ µ .Letting n → ∞ in (3.4) implies the claim.

Dirac masses
We assume that g : R → R is a continuous nondecreasing function such that rg(r) ≥ 0 for all r ∈ R. The next lemma dealing with problem (ii) N ≥ 3, µ = µ 0 and g satisfies (1.32).
Proof.Without loss of generality we assume B R ⊂ Ω ⊂ B 1 for some R ∈ (0, 1).(i) The case µ > µ 0 .It follows from [18, Theorem 3.1] that for any k ∈ R there exists a radial vanishing respectively on ∂B 1 and ∂B R and satisfying R and the extension of ṽk,R by 0 in Ω * is a subsolution of (3.13) in Ω * and it is still smaller than v k,1 in Ω * .By the well known method on super and subsolutions (see e.g. [26, Theorem 1.4.6]),there exists a function u in Ω * satisfying ṽk,R ≤ u ≤ v k,1 in Ω * and By standard methods in the study of isolated singularities (see e.g.[18], [23], and [14] and [15] for various extensions) For any ǫ > 0 and ξ ∈ X µ (Ω), Using (1.12), we obtain (ii)The case µ = µ 0 .In [18, Theorem 3.2] it is proved that if for some b > 0 there holds then there exists a solution of (1.19) satisfying (1.16) with γ = (N +2)b

2
. Actually we claim that the finiteness of this integral is independent of the value of b.To see that, set s = t N−2 N+2 , then We infer that for ǫ > 0 there exists s ǫ > 2 and τ ǫ = s ǫ ln s ǫ such that which implies the claim.Next we prove as in case (i) the existence of v k,1 (resp.
vanishing respectively on ∂B 1 and ∂B R and satisfying We end the proof as above.
Remark.It is important to notice that conditions (1.18) and (1.32) (or equivalently (1.20)) are also necessary for the existence of radial solutions in a ball, hence their are also necessary for the existence of non radial solutions of the Dirichlet problem (3.12).
For any ξ ∈ C 1,1 c (Ω * ) and ǫ small enough so that supp (ξ) ⊂ Ω ǫ , there holds There exists u νσ = lim ǫ→0 u νσ,ǫ and it satisfies the identity Using the maximum principle and Lemma 3.1, there holds in a neighborhood of 0, and u νσ is also bounded by cΦ µ in this neighborhood.This implies that Φ Now for the first inegral term in (3.26), we have where Using the fact that ξ(x) → ξ(0) and ∇ξ(x) → ∇ξ(0) we easily infer that I n , II n and III n to 0 when n → ∞, the most complicated case being the case when µ = µ 0 , which is the justification of introducing the explicit cut-off function ℓ n .Therefore (3.24) is still valid if it is assumed that ξ ∈ C 1,1 c (Ω).This means that u νσ is a weak solution of Furthermore u νσ is unique and u νσ is a decreasing function of σ with limit u when σ → 0. Taking η 1 as test function, we have Using monotone convergence theorem we infer that u νσ → u in L 1 (Ω, |x| −1 dγ µ ) and g(u νσ ) → g(u ν ) in L 1 (Ω, dγ µ ).Hence u = u ν is the weak solution of (3.22).

Proof of Theorem B
The idea is to glue altogether two solutions one with the Dirac mass and the other with the measure in Ω * , this is the reason why the weak ∆ 2 condition is introduced.
Then there exists a unique weak solution to ( Proof.Following the notations of Lemma 3.3, we set ν σ = χ Bσ ν⌊ Ω * and denote by U νσ the solution of It is a positive function and there holds Since the mapping σ → U νσ is decreasing, then there exists U = lim σ→0 U νσ and U satisfies (3.39).
As a consequence U νσ → U in L 1 (Ω, |x| −1 dγ µ ) as σ → 0. We take η 1 for test function in the weak formulation of (3.39), then By the monotone convergence theorem we obtain the identity and the fact that g(U νσ ) → g(U ) in L 1 (Ω, ρdγ µ ).Going to the limit as σ → 0 in the weak formulation of (3.38), we infer that U = u ν is the solution of (1.3).
Remark.In the course of the proof we have used the following result which is independent of any assumption on g but for the monotonicity: If {ν n } ⊂ M + (Ω; Γ µ ) is an increasing sequence of g-good measures converging to a measure ν ∈ M + (Ω; Γ µ ), then ν is a g-good measure, {u νn } converges to u ν in L 1 (Ω, |x| −1 dγ µ ) and {g(u νn )} converges to g(u ν ) in L 1 (Ω, ρdγ µ ).
Proof.The proof is similar to the one of [7,Prop. 4.1].Observe that u ν,k ↓ u * and the sequence {u ν,k } is uniformly integrable in L 1 (Ω, |x| −1 dγ µ ).By Fatou's lemma u satisfies Hence u * is a subsolution of (1.3) and by construction it is the largest of all nonnegative subsolutions.The mapping is a positive distribution, hence a measure ν * , called the reduced measure of ν.
Proof of Theorem E. Assume that ν ≥ 0. By Lemma 4.2 and Remark at the end of Section 3.5 the following assertions are equivalent: and since u νσ (x) ≤ c|x| τ + if |x| ≤ σ 2 (4.7) holds in D ′ (Ω).This implies that u ∈ L p (Ω) and |x| −2 u νσ ∈ L α (B σ 2 ) for any α < N (2−τ + ) + .Using [1] the measure ν σ is absolutely continuous with respect to the c 2,p ′ -Bessel capacity.If E ⊂ Ω is a Borel set such that c 2,p ′ (E) = 0, then Conversely, if ν is nonnegative and absolutely continuous with respect to the c 2,p ′ -Bessel capacity, then so is ν σ = χ B c σ ν.For 0 ≤ ǫ ≤ σ 2 we consider the problem Since µ |x| 2 is bounded in Ω ǫ and ν σ is absolutely continuous with respect to the c 2,p ′ capacity there exists a solution u νσ,ǫ thanks to [1], unique by monotonicity.Now the mapping ǫ → u νσ,ǫ is decreasing.We use the method developed in Lemma 3.1, when ǫ → 0, we know that u νσ,ǫ increase to some u σ which is dominated by G[ν σ ] and satisfies Because u σ ≤ G[ν σ ] and ν σ = 0 in B σ , there holds u(x) ≤ c ′ 11 Γ µ (x) in B σ 2 , and then u σ is a solution in Ω and u = u νσ .Letting σ → 0, we conclude as in Lemma 3.1 that u νσ converges to u ν which is the weak solution of If ν is a signed measure absolutely continuous with respect to the c 2,p ′ -capacity, so are ν + and ν − .Hence there exists solutions u ν + and u ν − .For 0 < ǫ < σ 2 we construct u νσ,ǫ with the property that −u −ν − σ ,ǫ ≤ u νσ,ǫ ≤ u ν + σ ,ǫ , we let ǫ → 0 and derive the existence of u νσ which is eventually the weak solution of and satisfies Hence u = u ν and ν is a good solution.
Proof of Theorem F. Part 1.Without loss of generality we can assume that Ω is a bounded smooth domain.Let K ⊂ Ω be compact.If 0 ∈ K and p < p * µ there exists a solution u kδ 0 , hence K is not removable.If 0 / ∈ K and c 2,p ′ (K) > 0, there exists a capacitary measure ν K ∈ W −2,p (Ω) ∩ M + (Ω) with support in K.This measure is g p -good by Theorem E, hence K is not removable.However, the condition p > p 0 is ensured when µ < 0 since p ≥ p * µ > p 0 .We consider a sequence {η n } ⊂ S(R N ) such that 0 ≤ η n ≤ 1, η n = 0 on a neighborhood of K and such that η n W 2,p ′ → 0 when n → ∞.Such a sequence exists by [19]  However the condition p > p 0 is not necessary in order the left-hand side of (4.16) be bounded, since we have Finally, if u is a signed solution, then |u| is a subsolution.For ǫ > 0 we set K ǫ = {x ∈ R N : dist (x, K) ≤ ǫ}.If ǫ is small enough K ǫ ⊂ Ω.Let v := v ǫ be the solution of such that {v ǫn } converges to v locally uniformly in Ω \ K and in the C 2 loc (Ω \ K)-topology.This implies that v is a positive solution of (1.34) in Ω \ K. Hence it is a solution in Ω.This implies that u ∈ L p (Ω) and |u(x)| ≤ v(x) ≤ c 14 Γ µ (x) in Ω * .We conclude as in the nonnegative case that u is a weak solution in Ω.

Lemma 3 . 2
12) is an extension of[18, Theorem 3.1, Theorem 3.2].Actually it was quoted in this article as Remark 3.1 and Remark 3.2 and we give here their proof.Notice also that when N ≥ 3 and µ = µ 0 we give a more complete result that[18, Theorem 3.2].Let k ∈ R and g : R → R be a continuous nondecreasing function such that rg(r) ≥ 0 for all r ∈ R. Then problem (3.12) admits a unique solution u := u kδ 0 if one of the following conditions is satisfied: (i) N ≥ 2, µ > µ 0 and g satisfies (1.18);