Dirichlet problems involving the Hardy-Leray operators with multiple polars

Huyuan Chen; Xiaowei Chen

doi:10.1515/anona-2022-0320

Artikel Open Access

Dirichlet problems involving the Hardy-Leray operators with multiple polars

Huyuan Chen und Xiaowei Chen

Veröffentlicht/Copyright: 4. August 2023

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Abstract

Our aim of this article is to study qualitative properties of Dirichlet problems involving the Hardy-Leray operator ℒ V ≔ − Δ + V , where V ( x ) = ∑ i = 1 m μ i ∣ x − A i ∣ 2 , with μ i ≥ − ( N − 2 ) 2 4 being the Hardy-Leray potential containing the polars’ set A m = { A i : i = 1 , … , m } in R N ( N ≥ 2 ). Since the inverse-square potentials are critical with respect to the Laplacian operator, the coefficients { μ i } i = 1 m and the locations of polars { A i } play an important role in the properties of solutions to the related Poisson problems subject to zero Dirichlet boundary conditions. Let Ω be a bounded domain containing A m . First, we obtain increasing Dirichlet eigenvalues:

ℒ V u = λ u in Ω , u = 0 on ∂ Ω ,

and the positivity of the principle eigenvalue depends on the strength μ i and polars’ setting. When the spectral does not contain the origin, we then consider the weak solutions of the Poisson problem

( E ) ℒ V u = ν in Ω , u = 0 on ∂ Ω ,

when ν belongs to L p ( Ω ) , with p > 2 N N + 2 in the variational framework, and we obtain a global weighted L ∞ estimate when p > N 2 . When the principle eigenvalue is positive and ν is a Radon measure, we build a weighted distributional framework to show the existence of weak solutions of problem ( E ) . Moreover, via this weighted distributional framework, we can obtain a sharp assumption of ν ∈ C γ ( Ω ¯ \ A m ) for the existence of isolated singular solutions for problem ( E ) .

Keywords: Hardy-Leray operator; Dirichlet eigenvalues; Poisson problem; multiple polars

MSC 2010: 35J75; 35P15; 35B44

1 Introduction and main results

Let Ω be a bounded, connected, and C 2 domain in R N with the integer N ≥ 2 ,

A m = { A i : i = 1 , … , m } ⊂ Ω

and

(1.1) ℒ V = − Δ + V , V ( x ) = ∑ i = 1 m μ i ∣ x − A i ∣ 2 ,

where the integer m ≥ 1 and

μ → ≔ ( μ 1 , … , μ m ) ∈ [ μ 0 , + ∞ ) m with μ 0 = − ( N − 2 ) 2 4 .

When A 1 = { 0 } , the Hardy operator ℒ V = reduces to

ℒ μ ≔ − Δ + μ ∣ x ∣ − 2 ,

and from the improved Hardy inequality [6], it is known that the operator ℒ μ is then positive definite for μ ∈ [ μ 0 , + ∞ ) . When μ ≥ μ 0 and μ ≠ 0 , the Hardy potential has homogeneity − 2 , which is critical from both mathematical and physical viewpoints. The Hardy-Leray problems have great applications to various fields as molecular physics [34], quantum cosmological models such as the Wheeler-de-Witt equation (see, e.g., [4]), and combustion models [28]. The Hardy inequalities are essential tools in the study of Hardy problems (see references [6,38] for the various Hardy inequalities with a single polar, [21,37] for basic properties, its applications in nonlinear field equations [2], and in controllability [36]. Isolated singularities are classified for semilinear elliptic equations with Hardy-Leray potentials in [17–19]). More general equations with various singular potentials could be seen [5,8,29–31].

From the Lax-Milgram theorem, the non-homogeneous problem

(1.2) ℒ μ u = f in Ω , u = 0 on ∂ Ω ,

with f ∈ L 2 ( Ω ) , has a unique solution in H 0 1 ( Ω ) if μ > μ 0 , or in a weaker Hilbert space if μ = μ 0 (see [5,20]). When f belongs to L 1 or a space of Radon measures, the distributional framework has to be involved to solve Problem (1.2), naturally,

(1.3) ∫ Ω u ℒ μ ξ d x = ∫ Ω f ξ d x , ∀ ξ ∈ C c ∞ ( Ω ) .

However, this distributional framework cannot be used to characterize the singularities at the polar. It is well known that the equation ℒ μ u = 0 in R N \ { 0 } has two distinct radial solutions:

(1.4) Φ μ ( x ) = ∣ x ∣ τ − ( μ ) if μ > μ 0 − ∣ x ∣ − N − 2 2 ln ∣ x ∣ if μ = μ 0 and Γ μ ( x ) = ∣ x ∣ τ + ( μ ) ,

with

(1.5) τ − ( μ ) = − N − 2 2 − μ − μ 0 and τ + ( μ ) = − N − 2 2 + μ − μ 0 .

It is noted that Identity (1.3) cannot be used to express that Φ μ is a fundamental solution, i.e. ℒ μ Φ μ = δ 0 . Moreover, Φ μ is not locally integrable at the origin if μ ≥ 2 N . To overcome this difficulty, Chen et al. in [13] adopted a weighted distributional framework, which shows that the function Φ μ is a fundamental solution of Hardy operator ℒ μ , i.e.

∫ R N Φ μ ℒ μ ∗ ξ d γ μ ( x ) = c μ ξ ( 0 ) , ∀ ξ ∈ C c 1 , 1 ( R N ) ,

where

d γ μ ( x ) = Γ μ ( x ) d x , ℒ μ ∗ ξ = − Δ ξ − 2 τ + ( μ ) ∣ x ∣ 2 ⟨ x , ∇ ξ ⟩

and c μ is the normalized constant defined by:

(1.6) c μ = 2 μ − μ 0 ∣ S N − 1 ∣ if μ > μ 0 , ∣ S N − 1 ∣ if μ = μ 0 ,

where S N − 1 is the unit sphere in R N and ∣ S N − 1 ∣ is the volume of the unit sphere. We remark that the new metric d γ μ ( x ) is the key point to fit the singularity of Φ μ , and then the isolated singularity of non-homogeneous Problem (1.2) could be fully classified under this distributional setting. Precisely, for a given “source” f , regular in Ω \ { 0 } , the solutions of the Dirichlet problem

ℒ μ u = f in Ω \ { 0 } , u = 0 on ∂ Ω

with isolated singularities at the origin are classified fully in [13] via the new formulation of distributional identity associated with some specific weight (also see [17] for semilinear Hardy problem). Extensive treatments of the associated semilinear problems are developed in [1,15,16] via an introduction of a notion of very weak solutions and a decomposition of Radon measures related to the metric d γ μ .

When A m contains multiple points, the Hardy-Leray potential V is used to describe the interaction between the electric dipoles in molecular physics, where A m represents the location of m nuclei (see Lévy-Leblond [34]). In the molecular system with the interactions between the electrons and the fixed nuclei, the potential V is related to Coulomb forces and it is described by the Hartree-Fock model (see [9]). Meanwhile, the multiple polars make the situation much intricate in the view point of partial differential equation. Our aim in this article is to show the role of the interaction among polars and their configuration in the related Dirichlet eigenvalue problem. It is shown in [11] that there exists ϱ m ∈ ( 0 , π 2 ) such that

(1.7) ∫ R N ( ∣ ∇ u ∣ 2 + ν m u 2 ) d x ≥ ( N − 2 ) 2 4 ∑ i = 1 m ∫ R N u 2 ( x ) ∣ x − A i ∣ 2 d x for any u ∈ C c ∞ ( R N ) ,

where

d 0 = 1 4 min { dist ( A m , ∂ Ω ) , ∣ A i − A j ∣ : A i , A j ∈ A m , i ≠ j }

and

(1.8) ν m = ϱ m + ( N − 2 ) 2 4 ( m + 1 ) d 0 2 > 0 .

For more related Hardy inequalities with multiple polars, we refer to [3,10]. This type of inequality provides basic tools for the study of the nonlinear Schrödinger equations involving Hardy operators with multiple polars ([22–27,33] and references therein).

We start our analysis in this article with the eigenvalues of Hardy-Leray operators with multiple polars subject to the zero Dirichlet boundary conditions. In fact, the spectral of elliptic operators with singular potentials has been studied extensively ([7,12,35] and references therein). It is known that the coefficients μ → and the configuration of A m determine the related Hardy inequalities, which could also be characterized by the positivity of the principle eigenvalue.

Let H V , 0 1 ( Ω ) be the closure of C c ∞ ( Ω ) under the norm of

‖ u ‖ m = ∫ Ω ( ∣ ∇ u ∣ 2 + ( V + 4 ν m ) u 2 ) d x ,

which is a Hilbert space with the inner product

⟨ u , v ⟩ m = ∫ Ω ( ∇ u ⋅ ∇ v + ( V + 4 ν m ) u v ) d x .

We first study the eigenvalues for the Hardy-Leray operator with multiple polars in bounded domain

(1.9) ℒ V u = λ u in Ω , u ∈ H V , 0 1 ( Ω )

in the variational sense that

⟨ u , v ⟩ m = ( λ + 4 ν m ) ∫ Ω u v d x for any v ∈ H V , 0 1 ( Ω ) .

We have the following.

Proposition 1.1

Let N ≥ 2 , then Problem (1.9) admits a sequence of real eigenvalues

− 4 ν m < λ 1 ( μ → ) < λ 2 ( μ → ) ≤ ⋯ ≤ λ j ( μ → ) ≤ λ j + 1 ( μ → ) ≤ ⋯ .

Denote by ϕ j the eigenfunction associating with λ j ( μ → ) , j ∈ N . These eigenvalues consist of the discrete spectral of ℒ V , denoting σ d ( ℒ V ) .

Moreover, we have the following properties:

λ j ( μ → ) = min { ‖ u ‖ m 2 : u ∈ H 0 , j , ‖ u ‖ L 2 ( Ω ) = 1 } − 4 ν m , where
H 0 , 1 = H V , 0 1 ( Ω ) , H 0 , j ≔ { u ∈ H V , 0 1 ( Ω ) : ⟨ u , ϕ m ⟩ m = 0 f o r m = 1 , … , j − 1 } f o r j > 1 ;
{ ϕ j : j ∈ N } is an orthonormal basis of L 2 ( Ω ) ;
λ 1 ( μ → ) is simple and lim j → ∞ λ j ( μ → ) = + ∞ .

Proposition 1.1 provides qualitative properties for the eigenvalues of (1.9). Moreover, we provide the examples for the case λ 1 ( μ → ) < 0 , when ∑ i = 1 m μ i < − μ 0 and the polars are located in a small ball of Ω in Section 2. These properties indicate the importance of the interaction among polars and their configuration in the eigenvalue problem.

Our second aim is to obtain weak solutions for Poisson problem with various kind sources and related estimates. As mentioned earlier, the function Γ μ , obtained by the ordinary differential equation (ODE) approach, plays an essential role in the derivation of distributional identity and the classification of measure data in [15,16]. Thanks to the multiple polars, it fails to obtain the counterpart of Γ μ by the ODE method, since this type of function cannot be expressed by any source in the distributional senses. To this end, let ϒ : B R ( 0 ) → ( 0 , + ∞ ) be a function in C 2 ( B R ( 0 ) \ A m ) such that

ϒ ( x ) = Γ μ i ( x − A i ) for any x ∈ B d 0 ( A i ) \ { A i } and i = 1 , … m ,

where R > 0 such that Ω ⊂ B R 2 ( 0 ) . It is worth noting that the choice of ϒ is to make it non-vanish in Ω ¯ \ A m and to keep the property of Γ μ i in the neighborhood of the polar A i for all i = 1 , … , m .

Proposition 1.2

(i) Let 0 ∉ σ d ( ℒ V ) and f ∈ L p ( Ω ) , with either

p > 2 N N + 2 with N ≥ 2

p ≥ 2 N N + 2 if N ≥ 3 and μ i > μ 0 f o r a l l i = 1 , … , m ,

then the homogeneous problem

(1.10) ℒ V u = f in Ω , u ∈ H V , 0 1 ( Ω )

has a unique weak solution u f in the sense that

(1.11) ∫ Ω ( ∇ u f ⋅ ∇ v + V u f v ) d x = ∫ Ω f v d x for all v ∈ H V , 0 1 ( Ω ) .

Moreover, assume more that λ 1 ( μ ) > 0 , then u f ≥ 0 a.e. in Ω if f ≥ 0 a.e. in Ω .

(ii) Assume that μ → ≥ 0 (i.e. μ i ≥ 0 for all i = 1 , … , m ) and f ϒ − 1 ∈ L p ( Ω ) with

p > N 2 .

Let u f be a unique weak solution of the homogeneous Problem (1.10), then there exists c 1 > 0 independent of f such that

(1.12) ∥ u f ϒ − 1 ∥ L ∞ ( Ω ) ≤ c 1 ‖ f ϒ − 1 ‖ L p ( Ω ) .

We remark that if λ i ( μ → ) = 0 , Poisson Problem (1.10) has infinitely many weak solutions in H V , 0 1 ( Ω ) with the restriction ∫ Ω f ϕ i d x = 0 , and it has no weak solution if ∫ Ω f ϕ i d x ≠ 0 . The part ( i i ) in Proposition 1.2 is proved by using the well-known De Giorgi method, and this L ∞ estimate is a very important property.

It is known that one important approach to characterize singular solutions of elliptic equations by involving the Dirac mass source in some suitable distributional sense, we refer the readers to [14,39,40] and a survey in [41] for general second-order elliptic equation with Radon measures. Our next purpose is to consider the isolated singular solutions of

(1.13) ℒ V u = f in Ω \ A m , u = 0 on ∂ Ω ,

where the source f is regular in Ω \ A m and able to be singular at A m . Our approach is to consider problem

(1.14) ℒ V u = f + ∑ i = 1 m k i δ A i in Ω , u = 0 on ∂ Ω ,

in a weighted distributional sense for which Dirac mass is well posed at the polars. To this end, we let

υ 0 ( x ) = ℒ V ϒ ( x ) for x ∈ Ω \ A m ,

and then, υ 0 is a continuous function in Ω ¯ \ Σ and

∣ υ 0 ∣ ≤ c 2 ϒ in Ω \ A m ,

where

c 2 = max sup x ∈ Ω \ ( ∪ i = 1 m B d 0 ( A i ) ) υ 0 ( x ) ϒ ( x ) , sup x ∈ B d 0 ∗ ( A i ) ∑ j ≠ i m ∣ μ j ∣ ∣ x − A j ∣ 2 with i = 1 , … , m

and B d 0 ∗ ( A i ) = B d 0 ( A i ) \ { A i } .

Definition 1.3

Here, the weak solution u of (1.14) in the d V x -distributional sense, that is, u ∈ L 1 ( Ω , ζ m − 1 d V x ) satisfies the distributional identity

(1.15) ∫ Ω u ℒ V ∗ ( ξ ) d V x = ∫ Ω f ξ d V x + ∑ i = 1 m k i ξ ( A i ) , ∀ ξ ∈ X V ( Ω ) ,

where

(1.16) ζ m ( x ) = ∏ i = 1 m ∣ x − A i ∣ ,

(1.17) d V x = ϒ ( x ) d x , ℒ V ∗ ξ = 1 ϒ ℒ μ ( ϒ ξ ) ,

and the test functions’ space

(1.18) X V ( Ω ) = { ξ ∈ C 0 ( Ω ¯ ) ∩ C 1 ( Ω ¯ ∗ ) : ζ m ℒ V ∗ ξ ∈ L ∞ ( Ω ) } ,

with

Ω ¯ ∗ ≔ Ω ¯ \ A m .

Note that C 0 1 , 1 ( Ω ¯ ) is a dense subset of X V ( Ω ) and for i = 1 , … , m

ℒ V ∗ ξ = − Δ ξ − 2 τ + ( μ i ) ∣ x − A i ∣ 2 ⟨ x − A i , ∇ ξ ⟩ + ∑ j ≠ i m μ j ∣ x − A j ∣ 2 ξ in B d 0 ∗ ( A i ) ,

and if g is C 1 ( Ω ¯ ) , the bounded solution u g of

ℒ V ∗ u = g in Ω \ A m , u = 0 on ∂ Ω ,

belongs to X V ( Ω ) .

Theorem 1.4

Let λ 1 ( μ → ) > 0 and f ∈ C loc γ ( Ω ¯ \ A m ) , with γ ∈ ( 0 , 1 ) .

If
(1.19) f ∈ L 1 ( Ω , d V x ) i.e. ∫ Ω ∣ f ∣ d V x < + ∞ ,
then for any k = ( k 1 , … , k m ) ∈ R m , Problem (1.14) admits a unique weak solution u k , which is a classical solution of (1.13).
If
(1.20) lim ∣ x − A i ∣ → 0 f ( x ) ∣ x − A i ∣ τ i = 0 for all i = 1 , 2 , … , m
with some τ i < 2 − τ − ( μ i ) . Then, we have the asymptotic behavior
(1.21) lim ∣ x − A i ∣ → 0 + u k ( x ) Φ μ i − 1 ( x − A i ) = c μ i − 1 k i .
If f ≥ 0 and for some i 0
(1.22) lim r → 0 + ∫ B d 0 ( A i 0 ) \ B r ( A i 0 ) f d V x = + ∞ ,
then Problem (1.14) has no nonnegative solutions.
Homogeneous problem
(1.23) ℒ V u = 0 i n Ω \ A m , u = 0 o n ∂ Ω , lim ∣ x − A i ∣ → 0 + u ( x ) Φ μ i ( x − A i ) = 0 for a l l i = 1 , … , m
has only zero classical solution.

We remark that the nonexistence under Assumption (1.22) shows the sharpness of Assumption (1.19) for the existence, and the uniqueness follows by Kato’s inequality under the setting λ 1 ( μ → ) > 0 . In the classical sense, Problem (1.13) has a unique solution under the restriction that

lim ∣ x − A i ∣ → 0 + u ( x ) Φ μ i − 1 ( x − A i ) = 0 for all i = 1 , … , m

by the Liouville results Theorem 1.4 (iii).

In view of distribution Definition 1.3, we extend a definition of weak solutions adapted to the operator ℒ V in a framework for general Radon measures. Since ϒ is singular at A i if μ i < 0 , we have to give a specific set of measures and denote by M ( Ω ∗ ; ϒ ) , the set of Radon measures ν in Ω ∗ ≔ Ω \ A m such that

(1.24) ∫ Ω ∗ ξ ϒ d ∣ ν ∣ ≔ sup ∫ Ω ∗ ζ ϒ d ∣ ν ∣ : ζ ∈ C c ( Ω ∗ ) , 0 ≤ ζ ≤ ξ < + ∞ .

Denote by ν ∈ M + ( Ω ∗ ; ϒ ) the related positive cone, i.e. the set of nonnegative measures in M ( Ω ∗ ; ϒ ) . In fact, for ν ∈ M ( Ω ∗ ; ϒ ) , we have the Jordan decomposition

ν = ν + − ν − , ν ± ∈ M + ( Ω ∗ ; ϒ ) .

We denote by M ¯ ( Ω ; ϒ ) the set of measures that can be written under the form:

(1.25) ν = ν ⌊ Ω ∗ + ∑ i = 1 m k i δ A i ,

where ν ⌊ Ω ∗ ∈ M ( Ω ∗ ; ϒ ) and k i ∈ R , δ A i is the Dirac mass at A i . Before stating our main theorem, we make precise the notion of weak solutions.

Definition 1.5

For given ν ∈ M ¯ ( Ω ; ϒ ) such that ν = ν ⌊ Ω ∗ + ∑ i = 1 m k i δ A i , with k i ∈ R , we say that u is a weak solution of

(1.26) ℒ V u = ν in Ω , u = 0 on ∂ Ω ,

if u ∈ L 1 ( Ω , ζ m − 1 d V x ) and

(1.27) ∫ Ω u ℒ V ∗ ξ d V ( x ) = ∫ Ω ξ ϒ d ν + ∑ i = 1 m k i ξ ( A i ) for all ξ ∈ X V ( Ω ) ,

where ζ m is given in (1.16).

Theorem 1.6

Let λ 1 ( μ → ) > 0 and ν ∈ M ¯ ( Ω ; ϒ ) , then (1.26) admits a unique weak solution u ∈ L 1 ( Ω , ζ m − 1 d V x ) .

Our results and the proofs are basic in the area of non-uniformly elliptic equations, so it is conducive to postgraduate studying. The rest of this article is organized as follows. In Section 2, we show the qualitative properties of the eigenvalues of ℒ V and comparison principle, maximum principle. Section 3 is devoted to asymptotic behavior at the polars. Section 4 is addressed to Kato’s type inequalities and to prove Theorem 1.4. We classify distributional solutions of Poisson problems involving L 1 sources and Radon measures in Section 5.

2 Eigenvalues

2.1 Basic results

Theorem 2.1

Let N ≥ 2 , μ → ∈ [ μ 0 , + ∞ ) m , and Ω be a bounded, connected and C 2 domain in R N containing A m .

Then, the embedding H V , 0 1 ( Ω ) ↪ W 0 1 , p ( Ω ) is continuous for any p ∈ [ 1 , 2 ) .
Assume more that N ≥ 3 , μ i > μ 0 for all i = 1 , … , m , then
H V , 0 1 ( Ω ) = H 0 1 ( Ω ) .
If μ i = μ 0 for some i = 1 , … , m , then
H V , 0 1 ( Ω ) ⊋ H 0 1 ( Ω ) .
Assume more that N = 2 , μ i ≥ 0 ( μ 0 = 0 ) for all i = 1 , … , m and ∑ i = 1 m μ i > 0 , then
H V , 0 1 ( Ω ) ⊊ H 0 1 ( Ω ) .

Proof

Let η 0 : R N → [ 0 , 1 ] be a radially symmetric C ∞ function such that

η 0 ( x ) = 1 in B 1 2 ( 0 ) and η 0 ( x ) = 0 in R N \ B 1 ( 0 ) .

Let u ∈ H V , 0 1 ( Ω ) . For i = 1 , … , m , we denote

u i ( x ) = u ( x ) η 0 ( d 0 − 1 ( x − A i ) ) , ∀ x ∈ Ω

and

u m + 1 = u − ∑ i = 1 m u i in Ω .

Note that in B d 0 ( A i )

V + 4 ν m = μ i ∣ x − A i ∣ − 2 + ∑ j = 1 , j ≠ i m μ j ∣ x − A j ∣ 2 + 4 ϱ m + ( N − 2 ) 2 4 ( m + 1 ) d 0 2 ≥ μ i ∣ x − A i ∣ − 2 − ( N − 2 ) 2 4 ( m − 1 ) d 0 − 2 + 4 ϱ m + ( N − 2 ) 2 4 ( m + 1 ) d 0 2 ≥ μ i ∣ x − A i ∣ − 2 ,

where we use μ i ≥ μ 0 . Then ∈ = ⊊ we have that

∫ Ω ( ∣ ∇ u i ∣ 2 + ( V + 4 ν m ) u i 2 ) d x ≥ ∫ Ω ∣ ∇ u i ∣ 2 + μ i u i 2 ∣ x − A i ∣ 2 d x .

If μ i = μ 0 and N ≥ 3 , we have that u i ∈ H A i , μ 0 ( Ω ) with i = 1 , … , m , where H A i , μ 0 ( Ω ) is a Hilbert space of the closure of C c ∞ ( Ω ) under the norm of

‖ v ‖ A i , μ 0 = ∫ Ω ∣ ∇ v ∣ 2 + μ 0 v 2 ∣ x − A i ∣ 2 d x

and the inner product

⟨ u , v ⟩ A i , μ 0 = ∫ Ω ∇ u ⋅ ∇ v + μ 0 u v ∣ x − A i ∣ 2 d x .

It is known from [38, Theorem 2.2] that H A i , μ 0 ( Ω ) ↪ W 0 1 , p ( Ω ) is continuous for any p ∈ [ 1 , 2 ) .

If μ i > μ 0 for N ≥ 3 or μ i ≥ 0 for N = 2 , we have that

∫ Ω ( ∣ ∇ u i ∣ 2 + ( V + 4 ν m ) u i 2 ) d x ≥ ∫ Ω ∣ ∇ u i ∣ 2 + μ i u i 2 ∣ x − A i ∣ 2 d x ≥ 1 − μ i μ 0 ∫ Ω ∣ ∇ u i ∣ 2 d x ,

where 1 − μ i μ 0 > 0 . Thus, we see that u i ∈ H 0 1 ( Ω ) .

Moreover, we see that u m + 1 has support in Ω \ ( ∪ i = 1 m B d 0 ∕ 2 ( A i ) ) and for x ∈ Ω \ ( ∪ i = 1 m B d 0 ∕ 2 ( A i ) )

V + 4 ν m ≥ − ( N − 2 ) 2 4 ∑ j = 1 m ∣ x − A j ∣ − 2 + 4 ϱ m + ( N − 2 ) 2 4 ( m + 1 ) d 0 2 ≥ − ( N − 2 ) 2 4 m d 0 2 − 2 + 4 ϱ m + ( N − 2 ) 2 4 ( m + 1 ) d 0 2 > 0 .

This means u m + 1 ∈ H 0 1 ( Ω ) , and obviously, u m + 1 ∈ W 0 1 , p ( Ω ) for any p ∈ ( 1 , 2 ] .

As a consequence, we have that u = ∑ i = 1 m + 1 u i ∈ W 0 1 , p ( Ω ) for any p ∈ [ 1 , 2 ) and u ∈ H 0 1 ( Ω ) if N ≥ 3 and μ i > μ 0 for all i = 1 , … , m or if N = 2 and μ i ≥ 0 for all i = 1 , … , m .

Inversely, let u ∈ H 0 1 ( Ω ) , then in B d 0 ( A i ) ,

V + 4 ν m = μ i ∣ x − A i ∣ − 2 + ∑ j = 1 , j ≠ i m μ j ∣ x − A j ∣ − 2 + 4 ϱ m + ( N − 2 ) 2 4 ( m + 1 ) d 0 2 ≤ μ i ∣ x − A i ∣ − 2 + d 0 − 2 ∑ j = 1 , j ≠ i m ∣ μ j ∣ + 4 ϱ m + ( N − 2 ) 2 4 ( m + 1 ) d 0 2 ,

denoting a m = d 0 − 2 ∑ j = 1 , j ≠ i m ∣ μ j ∣ + 4 ϱ m + ( N − 2 ) 2 4 ( m + 1 ) d 0 2 , and we have that

∫ Ω ( ∣ ∇ u i ∣ 2 + ( V + 4 ν m ) u i 2 ) d x ≤ ∫ Ω ∣ ∇ u i ∣ 2 + μ i u i 2 ∣ x − A i ∣ 2 d x + a m ∫ Ω u 2 d x ≤ 1 − μ i μ 0 ∫ Ω ∣ ∇ u i ∣ 2 d x + a m ∫ Ω u 2 d x ,

where 1 − μ i μ 0 > 0 for μ i > μ 0 and N ≥ 3 . Therefore, we have that u ∈ H V , 0 1 ( Ω ) if μ i > μ 0 . In Ω \ ( ∪ i = 1 m B d 0 ∕ 2 ( A i ) ) , we see that

V + 4 ν m ≤ ∑ j = 1 m ∣ μ j ∣ ∣ x − A j ∣ 2 + 4 ϱ m + ( N − 2 ) 2 4 ( m + 1 ) d 0 2 ≤ ∑ j = 1 m ∣ μ j ∣ d 0 2 − 2 + 4 ϱ m + ( N − 2 ) 2 4 ( m + 1 ) d 0 2 .

This means u m + 1 ∈ H V , 0 1 ( Ω ) .

As a consequence, we have that H V , 0 1 ( Ω ) ≅ H 0 1 ( Ω ) if N ≥ 3 and μ i > μ 0 for all i = 1 , … , m .

If μ i = μ 0 for some i = 1 , … , m , then ∣ x − A i ∣ τ + ( μ i ) η 0 ( d 0 − 1 ( x − A i ) ) is a function in H V , 0 1 ( Ω ) but not in H 0 1 ( Ω ) .

When N = 2 , we observe that H V , 0 1 ( Ω ) ⊂ H 0 1 ( Ω ) . Now, we prove H V , 0 1 ( Ω ) ≠ H 0 1 ( Ω ) if there is μ i 0 > 0 . For τ ∈ ( − 1 2 , 0 ) , letting v τ = η 0 ( d 0 − 1 ( x − A i 0 ) ) ln τ ∣ x − A i 0 ∣ ,

v τ ∈ H 0 1 ( Ω ) .

Since v τ 2 ∣ ⋅ − A i 0 ∣ 2 is not integrable near A i 0 , then v τ ∉ H V , 0 1 ( Ω ) , which completes the proof.□

Lemma 2.2

For u ∈ H V , 0 1 ( Ω ) , we also have ∣ u ∣ ∈ H V , 0 1 ( Ω ) , and

(2.1) ⟨ ∣ u ∣ , ∣ u ∣ ⟩ m = ⟨ u , u ⟩ m .

Proof

When H V , 0 1 ( Ω ) = H 0 1 ( Ω ) , this lemma is well known. For the convenience, we provide the details of the proof.

Given u ∈ H V , 0 1 ( Ω ) , we decompose ∣ u ∣ = u + + u − , where u + = max { u , 0 } and u − = − min { u , 0 } . Now, we show ∣ u ∣ ∈ H V , 0 1 ( Ω ) . Given ε > 0 , we denote

F ε ( t ) = t 2 + ε 2 − ε if t ≥ 0 , 0 if t < 0 ,

and then ∣ F ε ′ ∣ ≤ 1 thanks to

F ε ′ ( t ) = t t 2 + ε 2 if t ≥ 0 , 0 if t < 0 .

Observe that F ε ( u ) ∈ H V , 0 1 ( Ω ) ,

u + = lim ε → 0 + F ε ( u ) , a . e .

and

∫ Ω F ε ( u ) D ϕ d x = − ∫ { u > 0 } u D u u 2 + ε 2 ϕ d x .

Using the dominated convergence theorem, passing to the limit, we obtain that u + ∈ H V , 0 1 ( Ω ) . Therefore, we have that u − ∈ H V , 0 1 ( Ω ) so is ∣ u ∣ .

Moreover, we see that

⟨ ∣ u ∣ , ∣ u ∣ ⟩ m = ⟨ u + , u + ⟩ m + ⟨ u − , u − ⟩ m = ⟨ u , u ⟩ m ,

since { u > 0 } ∩ { u < 0 } = ∅ . We complete the proof.□

Corollary 2.3

Let N ≥ 2 , μ → ∈ [ μ 0 , + ∞ ) m and Ω be a bounded domain in R N containing A m . Then, the embedding H V , 0 1 ( Ω ) ↪ L q ( Ω ) is compact for any q ∈ 1 , 2 N N − 2 if N ≥ 3 and for any q ∈ ( 1 , + ∞ ) if N = 2 .

Proof

If N ≥ 3 , for any q ∈ 1 , 2 N N − 2 , there exists p ∈ ( 1 , 2 ) such that p N N − p > q , since

lim p → 2 p N N − p = 2 N N − 2 .

From Theorem 2.1, we have that H V , 0 1 ( Ω ) ↪ W 0 1 , p ( Ω ) is continuous. From the standard Sobolev embedding theorems, our result follows the compact embedding W 0 1 , p ( Ω ) ↪ L q ( Ω ) for any q ∈ ( 1 , p N N − p ) . If N = 2 , for any q ∈ ( 1 , + ∞ ) , there exists p ∈ ( 1 , 2 ) such that p N N − p > q , since

lim p → 2 p N N − p = + ∞ .

Then, our argument follows.□

Lemma 2.4

Let f ∈ L 2 ( Ω ) , then the homogeneous problem

(2.2) ( ℒ V + 4 ν m ) u = f i n Ω , u = 0 o n ∂ Ω

has a unique weak solution u ˜ f ∈ H V , 0 1 ( Ω ) in the sense that

(2.3) ⟨ u ˜ f , v ⟩ m = ∫ Ω f v d x for all v ∈ H V , 0 1 ( Ω ) .

Moreover, (i) u ˜ f ≥ 0 if f ≥ 0 ;

(ii) if f ∈ C loc γ ( Ω \ A m ) with γ ∈ ( 0 , 1 ) is nonnegative and

u ˜ f ( z 0 ) = 0 f o r s o m e p o i n t z 0 ∈ Ω \ A m ,

then u ˜ f ≡ 0 and f ≡ 0 in Ω \ A m .

Proof

Let

L ( f ) = ∫ Ω f v d x ,

then

∫ Ω u f d x ≤ ‖ f ‖ L 2 ( Ω ) ‖ u ‖ L 2 ( Ω ) ≤ ‖ f ‖ L 2 ( Ω ) ‖ u ‖ m ,

so L : H V , 0 1 ( Ω ) → R is a bound linear functional. If 0 ∉ σ d ( ℒ V ) , it follows by the Fredholm alternative theorem that Problem (2.2) has a unique weak solution u ˜ f ∈ H V , 0 1 ( Ω ) in the sense that

⟨ u ˜ f , v ⟩ m = ∫ Ω f v d x for any v ∈ H V , 0 1 ( Ω ) .

Moreover, ( i ) u ˜ f ≥ 0 a.e. in Ω if f is nonnegative if we take ( u ˜ f ) − as a test function. If f is nonnegative and nontrivial, then u ≥ 0 and u ≢ 0 .

(ii) If f is Hölder continuous in Ω \ A m , then u ˜ f is a nonnegative classical solution of

( ℒ V + 4 ν m ) u = f in Ω \ A m , u = 0 on ∂ Ω ,

and if { x ∈ Ω \ A m : u ˜ f ( x ) = 0 } is nonempty, it follows by the Hopf lemma that

{ x ∈ Ω \ A m : u ˜ f ( x ) = 0 } = Ω \ A m .

We complete the proof.□

Now, we are in a position to prove Proposition 1.1.

Proof of Proposition 1.1

Part 1. The functional

ℐ m ( u ) = ∫ Ω ( ∣ ∇ u ∣ 2 + ( V + 4 ν m ) u 2 ) d x for u ∈ H V , 0 1 ( Ω )

is weakly lower semicontinuous. Moreover, let

ℳ 1 ≔ { u ∈ H V , 0 1 ( Ω ) , ‖ u ‖ L 2 ( Ω ) = 1 } .

Then, we have that

ℐ m ( u ) < + ∞ for u ∈ ℳ 1 .

Put λ ˜ 1 ( μ → ) ≔ inf u ∈ ℳ 1 ℐ m ( u ) . By Corollary 2.3, the embedding H V , 0 1 ( Ω ) ↪ L 2 ( Ω ) is compact; thus, it follows that λ ˜ 1 ( μ → ) is attained by a function ϕ 1 ∈ ℳ 1 . Consequently, there exists a Lagrange multiplier λ ∈ R such that

⟨ ϕ 1 , w ⟩ m = ℐ ′ ( ϕ 1 ) w = λ ∫ Ω ϕ 1 w d x for all w ∈ H V , 0 1 ( Ω ) .

Choosing φ = ϕ 1 yields to λ = λ ˜ 1 ( μ → ) > 0 ; hence, ϕ 1 is an eigenfunction of (1.9) corresponding to the eigenvalue

λ 1 ( μ → ) ≔ λ ˜ 1 ( μ → ) − 4 ν m > − 4 ν m .

Next, we proceed inductively and assume that ϕ 2 , … , ϕ k ∈ H V , 0 1 ( Ω ) and λ ˜ 2 ( μ → ) ≤ ⋯ ≤ λ ˜ k ( μ → ) are already given for some k ∈ N with the properties that for j = 2 , … , k , where the function ϕ j is a minimizer of ℐ within the set

ℳ j ≔ { u ∈ H V , 0 1 ( Ω ) : ‖ u ‖ L 2 ( Ω ) = 1 , ⟨ ϕ n , u ⟩ m = 0 for n = 1 , … j − 1 } .

Take

λ ˜ j ( μ → ) = inf ϕ ∈ ℳ j ℐ ( ϕ ) = ℐ ( ϕ j )

and

(2.4) ⟨ ϕ j , φ ⟩ m = λ ˜ j ( μ → ) ∫ Ω ϕ j φ d x for all φ ∈ H V , 0 1 ( Ω ) .

Again by the compact embedding, the value λ k + 1 ( μ → ) is attained by a function ϕ k + 1 ∈ ℳ k + 1 . Consequently, there exists a Lagrange multiplier λ ∈ R such that

(2.5) ⟨ ϕ k + 1 , φ ⟩ m = λ ∫ Ω ϕ k + 1 φ d x for all φ ∈ ℳ k + 1 ( Ω ) .

Choosing φ = ϕ k + 1 , we have that λ = λ ˜ k + 1 ( μ → ) . Moreover, for j = 1 , … , k , we have, by (2.4) and the definition of ℳ k + 1 ( Ω ) , that

⟨ ϕ k + 1 , ϕ j ⟩ m = ⟨ ϕ j , ϕ k + 1 ⟩ m = λ ˜ j ( μ → ) ∫ Ω ϕ j ϕ k + 1 d x = 0 = λ ˜ k + 1 ( μ → ) ∫ Ω ϕ j ϕ k + 1 d x .

Hence, (2.5) holds with λ = λ ˜ k + 1 ( μ → ) for all φ ∈ H V , 0 1 ( Ω ) . Inductively, we have now constructed a normalized sequence ( ϕ k ) k in H V , 0 1 ( Ω ) and a nondecreasing sequence { λ ˜ k ( μ → ) } k in R such that Property ( i ) holds and such that ϕ k is an eigenfunction of (1.9) corresponding to λ = λ ˜ k ( μ → ) − 4 ν m for each k ∈ N .

Part 2: we show property ( i i i ) , i.e. lim k → + ∞ λ k ( μ → ) = + ∞ . Supposing by contradiction that c ≔ lim k → ∞ λ k ( μ → ) < + ∞ , we deduce that ⟨ ϕ k , ϕ k ⟩ m ≤ c for every k ∈ N . Hence, the sequence ( ϕ k ) is bounded in H V , 0 1 ( Ω ) , and therefore, by Rellich compactness theorem, ( ϕ k ) contains a convergent subsequence ( ϕ k j ) j in L 2 ( Ω ) . However, this is impossible since the functions { ϕ k j } j ∈ N are L 2 ( Ω ) -orthogonal.

Part 3: we prove that { ϕ k : k ∈ N } is an orthonormal basis of L 2 ( Ω ) . By contradiction, we assume that there exists v ∈ H V , 0 1 ( Ω ) , with ‖ v ‖ L 2 ( Ω ) = 1 and ∫ Ω v ϕ k d x = 0 for any k ∈ N . Since lim k → ∞ λ k ( μ → ) = + ∞ , there exists an integer k 0 > 0 such that

ℐ m ( v ) < λ ˜ k 0 ( μ → ) = inf u ∈ ℳ k 0 ℐ m ( u ) ,

which, by definition of ℳ k 0 , implies that ∫ Ω v ϕ k d x ≠ 0 for some k ∈ { 1 , … , k 0 − 1 } , which leads to a contradiction. We conclude that H V , 0 1 ( Ω ) is contained in the L 2 -closure of the span of { ϕ k : k ∈ N } . Since H V , 0 1 ( Ω ) is dense in L 2 ( Ω ) , we conclude that the span of { ϕ k : k ∈ N } is dense in L 2 ( Ω ) , and hence, { ϕ k : k ∈ N } is an orthonormal basis of L 2 ( Ω ) .

It is standard to improve the regularities of ϕ k away from A m , that is, we have that ϕ i is a classical solution of

( ℒ V + 4 ν m ) u = λ ˜ k ( μ → ) u in Ω \ A m , u = 0 on ∂ Ω .

Now, we prove that ϕ 1 is positive in Ω \ A m . By Lemma 2.2, ∣ ϕ 1 ∣ is a first eigenfunction of ℐ m and ∣ ϕ 1 ∣ verifies that

( ℒ V + 4 ν m ) u ≥ 0 in Ω \ A m , u = 0 on ∂ Ω .

By Lemma 2.4 part ( i i ) , it holds that

∣ ϕ 1 ∣ > 0 in Ω \ A m .

which means ϕ 1 cannot change signs and we can choose it as positive.

Finally, we prove the simplicity of λ ˜ 1 ( μ → ) . Let ϕ ˜ 1 be another first eigenfunction of ℐ m ,

Φ ( x ) = ϕ ˜ 1 ( x 0 ) ϕ ( x ) − ϕ 1 ( x 0 ) ϕ ˜ 1 ( x ) for x ∈ Ω \ A m ,

where x 0 ∈ Ω \ A m is fixed. Then, Φ is a first eigenfunction of ℐ m , so is ∣ Φ ∣ , however, ∣ Φ ∣ ( x 0 ) ∣ = 0 . So we can have

Φ ≡ 0 in Ω \ A m .

That is,

ϕ ˜ 1 = ϕ 1 ( x 0 ) ˜ ϕ 1 ( x 0 ) ϕ 1 in Ω \ A m .

Let

λ k ( μ → ) = λ ˜ k ( μ → ) − 4 ν m ,

which is the eigenvalue of ℒ V in Ω subject to the zero boundary condition.□

2.2 More properties

Corollary 2.5

Assume that two bounded domains Ω 1 and Ω 2 satisfy that
A m ⊂ Ω 1 ⊂ Ω 2 .
Let λ 1 ( μ → , Ω i ) be the first eigenvalue of (1.9), then λ 1 ( μ → , Ω 1 ) ≥ λ 1 ( μ → , Ω 2 ) .
Let μ → ( 1 ) ≥ μ → ( 2 ) (i.e., μ i ( 1 ) ≥ μ i ( 2 ) for all i = 1 , … , m ) and λ 1 ( μ → ( j ) ) be the first eigenvalue of (1.9)with μ → = μ → ( j ) for j = 1 , 2 , then
λ 1 ( μ → ( 1 ) ) ≥ λ 1 ( μ → ( 2 ) ) .

Proof

Let ϕ 1 be the eigenfunction of (1.9) with respect to λ 1 ( μ → , Ω 1 ) such that ‖ ϕ 1 ‖ L 2 ( Ω 1 ) = 1 , then by zero extension, we have that ϕ 1 ∈ H V , 0 1 ( Ω 2 ) , then
λ 1 ( μ → , Ω 2 ) = inf u ∈ H V , 0 1 ( Ω 2 ) , ‖ u ‖ L 2 = 1 ∫ Ω 2 ( ∣ ∇ u ∣ 2 + V u 2 ) d x ≤ ∫ Ω 2 ( ∣ ∇ ϕ 1 ∣ 2 + V ϕ 1 2 ) d x = λ 1 ( μ → , Ω 1 ) .
Let V j be the Hardy potential corresponding to μ → ( j ) with j = 1 , 2 , then
V 1 ≥ V 2 ,
from the definition of the first eigenvalue, we see that
λ 1 ( μ → ( 1 ) ) = inf u ∈ H V , 0 1 ( Ω ) , ‖ u ‖ L 2 = 1 ∫ Ω ( ∣ ∇ u ∣ 2 + V 1 u 2 ) d x ≥ ∫ Ω ( ∣ ∇ ϕ 1 ∣ 2 + V 2 ϕ 1 2 ) d x = λ 1 ( μ → ( 2 ) ) ,

which completes the proof.□

Let

u ϑ ( r ) = r 2 − N 2 cos ( ϑ ln r )

and

ϑ m = − ( N − 2 ) 2 4 − ∑ i = 1 m μ i if ∑ i = 1 m μ i < − ( N − 2 ) 2 4 .

Direct computation shows that, letting r = ∣ x ∣ ,

(2.6) − Δ u ϑ ( ∣ x ∣ ) + ∑ i = 1 m μ i ∣ x ∣ 2 u ϑ ( ∣ x ∣ ) = ϑ 2 + ( N − 2 ) 2 4 + ∑ i = 1 m μ i u ϑ ( ∣ x ∣ ) ∣ x ∣ 2 = ( ϑ 2 − ϑ m 2 ) u ϑ ( ∣ x ∣ ) ∣ x ∣ 2 .

For n ∈ N , we denote

r n = e π 2 − n π 1 ϑ m and R n = e 3 π 2 − n π 1 ϑ m .

Proposition 2.6

Assume that N ≥ 3 ,

∑ i = 1 m μ i < μ 0 a n d B r n 0 ( 0 ) ⊂ Ω ,

for some n 0 ∈ Z , then there exists r ¯ ∈ ( 0 , r n 0 − 1 ) such that A m ⊂ B r ¯ ( 0 ) and

λ 1 ( μ → ) < 0 .

Proof

Let

u n ( x ) = u ϑ ( ∣ x ∣ ) for e π 2 − n π 1 ϑ ≤ ∣ x ∣ ≤ e 3 π 2 − n π 1 ϑ

and

u n ( x ) = 0 for ∣ x ∣ < e π 2 − n π 1 ϑ or ∣ x ∣ > e 3 π 2 − n π 1 ϑ .

Then, there exists n 0 ∈ Z such that

r n 0 − 1 < e π 2 1 ϑ < e 3 π 2 1 ϑ ≤ R n 0

and u ϑ ∈ H V , 0 1 ( Ω ) .

Direct computation shows that

λ 1 ( μ → ) ≤ ∫ Ω ( − Δ u ϑ + V ( x ) u ϑ ) u ϑ d x ‖ u ϑ ‖ L 2 ( Ω ) 2 = ∫ Ω ( − Δ u ϑ + ∑ i = 1 m μ i ∣ x ∣ 2 u ϑ ) u ϑ d x ‖ u ϑ ‖ L 2 ( Ω ) 2 + 1 ‖ u ϑ ‖ L 2 ( Ω ) 2 ∫ B e 3 π 2 1 ϑ ( 0 ) \ B e π 2 1 ϑ ( 0 ) V ( x ) − ∑ i = 1 m μ i ∣ x ∣ 2 u ϑ 2 ( x ) d x < ( ϑ 2 − ϑ m 2 ) ∫ B e 3 π 2 1 ϑ ( 0 ) \ B e π 2 1 ϑ ( 0 ) u ϑ 2 ( ∣ x ∣ ) ∣ x ∣ 2 d x 1 ‖ u ϑ ‖ L 2 ( Ω ) 2 + sup e π 2 1 ϑ ≤ ∣ x ∣ ≤ e 3 π 2 1 ϑ V ( x ) − ∑ i = 1 m μ i ∣ x ∣ 2 ≤ ( ϑ 2 − ϑ m 2 ) e − 3 π ϑ + sup e π 2 1 ϑ ≤ ∣ x ∣ ≤ e 3 π 2 1 ϑ ∑ i = 1 m μ i ∣ x ∣ 2 − ∣ x − A i ∣ 2 ∣ x − A i ∣ 2 ∣ x ∣ 2 ≤ 0 ,

if r ¯ > 0 small enough, since for e π 2 1 ϑ ≤ ∣ x ∣ ≤ e 3 π 2 1 ϑ , there holds

∑ i = 1 m μ i ∣ x ∣ 2 − ∣ x − A i ∣ 2 ∣ x − A i ∣ 2 ∣ x ∣ 2 ≤ ∑ i = 1 m ∣ μ i ∣ 2 ∣ r ¯ ∣ R n 0 r n 0 − 1 4 → 0 + as r ¯ → 0 + .

We complete the proof.□

We remark that Proposition 2.6 shows the negative principle eigenvalue provided ∑ i = 1 m μ i < − ( N − 2 ) 2 4 and the polars concentrate in a small ball.

Next, we consider the case of two point polars: A 2 = { A 1 , A 2 } and μ 0 ≤ μ 1 < μ 2 ≤ 0 . Let

β i = μ i μ 0 with i = 1 , 2

and

r 2 = β 1 + β 2 − β 1 β 2 β 1 + β 2 ∣ A 1 − A 2 ∣ .

Direct computation shows that r 2 > 1 2 ∣ A 1 − A 2 ∣ since 0 ≤ β 2 < β 1 ≤ 1 .

Proposition 2.7

Assume that N ≥ 3 , A 2 = { A 1 , A 2 } , A 0 = β 1 β 1 + β 2 A 1 + β 2 β 1 + β 2 A 2 ,

μ 0 ≤ μ 1 < μ 2 ≤ 0 a n d Ω ⊂ B r 2 ( A 0 ) ,

then

λ 1 ( μ → ) ≥ 0 .

Proof

From [11, Proposition 3.1], we have that

∫ Ω ∣ ∇ w ∣ 2 d x + ∫ Ω μ 1 ∣ x − A 1 ∣ 2 + μ 2 ∣ x − A 2 ∣ 2 w 2 d x ≥ 0 for any w ∈ C 0 2 ( Ω ) .

By the density of C 0 2 ( Ω ) ⊂ H V , 0 1 ( Ω ) , we have that

λ 1 ( μ → ) ≥ 0 .

We complete the proof.□

3 Poisson problem

3.1 Comparison principle and Liouville result

In this subsection, we consider the basic property of the operator ℒ V associated with the principle eigenvalue.

Proposition 3.1

(i) Let λ 1 ( μ → ) > 0 , O be a bounded open set such that

O ¯ ⊂ Ω ¯ \ A m ,

then ℒ V verifies the comparison principle, that is, if

u , v ∈ C 1 , 1 ( O ) ∩ C ( O ¯ )

satisfying

ℒ V u ≥ ℒ V v in O a n d u ≥ v on ∂ O ,

then

u ≥ v in O .

(ii) Let λ 1 ( μ → ) > 0 , f ∈ L p ( Ω ) with p > 2 N N + 2 and u f be a weak solution of

(3.1) ℒ V u = f i n Ω , u = 0 o n ∂ Ω ,

in the sense that

(3.2) ⟨ u , v ⟩ m = 4 ν m ∫ Ω u v d x + ∫ Ω f v d x for any v ∈ H V , 0 1 ( Ω ) .

If f ≥ 0 a.e. in Ω , we have that u f ≥ 0 a.e. in Ω .

In particular, if f = 0 a.e. in Ω , so is u f .

Proof

( i ) Since O ¯ ⊂ Ω ¯ \ A m , there is no polar in O ¯ . Let w = u − v and w − = min { w , 0 } , then w ≥ 0 on ∂ O by the assumption that u ≥ v on ∂ O and then w − = 0 on ∂ O . We will prove that w − ≡ 0 on O . If O − ≔ { x ∈ O : w ( x ) < 0 } is not empty, then it is a bounded C 1 , 1 domain in O . From Hardy inequality [11, Inequality (3.5)], there holds,

0 = ∫ O − − Δ w − + ∑ i = 1 m μ i ∣ x − A i ∣ 2 w − w − d x ≥ ∫ O − ∣ ∇ w − ∣ 2 + ∑ i = 1 m μ i ∣ x − A i ∣ 2 w − 2 d x ≥ λ 1 ( μ → , O − ) ∫ O − w − 2 d x ,

where λ 1 ( μ → , O − ) ≥ λ 1 ( μ → , Ω ) > 0 . Then, w − = 0 a.e. on O − , which is impossible with the definition of O − .

( i i ) Note that u f ∈ H V , 0 1 ( Ω ) , so are ( u f ) ± , where ( u f ) ± = ± max { ± u f , 0 } . Use ( u f ) − as a test function, we have that

0 ≥ ∫ Ω f ( u f ) − d x = ∫ Ω ∣ ∇ ( u f ) − ∣ 2 + ∑ i = 1 m μ i ∣ x − A i ∣ 2 ( u f ) − 2 d x ≥ λ 1 ( μ → ) ∫ Ω ( u f ) − 2 d x ,

which implies ( u f ) − = 0 a.e. on Ω , where by Corollary 2.3

∫ Ω f ( u f ) − d x ≤ ∫ Ω ∣ u f ∣ p ′ d x 1 p ′ ∫ Ω ∣ f ∣ p d x 1 p ≤ ‖ u ‖ m ‖ f ‖ L p ( Ω )

with p ′ = p p − 1 < 2 N N − 2 if p > 2 N N + 2 and N ≥ 3 ; p ′ ∈ ( 1 , + ∞ ) if N = 2 .

In the case that f = 0 a.e. in Ω , use ( u f ) ± as test functions, respectively, we have that u f = 0 a.e. in Ω .□

Our next result is the Liouville theorem of related Poisson problem without source in the classical sense, which plays an important role in the obtention of classification of isolated singular solutions of (1.13).

Proposition 3.2

Let λ 1 ( μ → ) > 0 , then Problem (1.23) has only zero classical solution.

In order to show the nonexistence, we need the following lemma.

Lemma 3.3

Let λ 1 ( μ → ) > 0 and f ∈ C 1 ( Ω ¯ ) be nonnegative, then problem

(3.3) ℒ V u = f i n Ω \ A m , u = 0 o n ∂ Ω , lim x → A i u ( x ) Φ μ i ( x − A i ) = 0 f o r a l l i = 1 , … , m

has a unique nonnegative classical solution u f ∈ H V , 0 1 ( Ω ) . Moreover, we have that for all i = 1 , … , m

(3.4) limsup ∣ x − A i ∣ → 0 + u f ( x ) Γ μ i ( x − A i ) < + ∞ ,

and it verifies the distributional identity:

(3.5) ∫ Ω u f ℒ V ∗ ξ d V x = ∫ Ω f ξ d V x for a n y ξ ∈ C 0 1 , 1 ( Ω ) .

Proof

The existence of weak solution to (3.3) follows by the Ekeland variational principle for the functional

ℐ 0 ( u ) = 1 2 ‖ u ‖ m 2 − 4 ν m ∫ Ω u 2 d x − ∫ Ω f u d x ≥ 1 2 λ 1 ( μ → ) ‖ u ‖ L 2 2 − ‖ f ‖ L 2 ( Ω ) ‖ u ‖ L 2 ( Ω )

by using the assumption that λ 1 ( μ → ) > 0 . The critical point u f of ℐ 0 verifies that

(3.6) ⟨ u f , v ⟩ m = 4 ν m ∫ Ω u f v d x + ∫ Ω f v d x for any v ∈ H V , 0 1 ( Ω ) .

By Proposition 3.1 part ( i i ) u f ≥ 0 a.e. in Ω . Away from the polars, it is standard to obtain that the solution u f is C 2 locally in Ω \ A m , see, for example, [32].

We prove the singularity of (3.4). Let

W i ( x ) = Γ μ i ( x − A i ) − 1 2 ∣ x − A i ∣ τ , x ∈ B d 1 ( A i ) \ { A i } ,

where τ + ( μ i ) − 1 < τ < τ + ( μ i ) , then there exists d 1 ∈ ( 0 , d 0 ] such that for x ∈ B d 1 ( A i ) \ { A i }

ℒ V W i ( x ) ≥ − c τ ∣ x − A i ∣ τ + ( μ j ) − 2 + ∑ j = 1 , j ≠ i m μ j W i ( x ) ∣ x − A j ∣ 2 ≥ 1 ,

that is,

ℒ V W i ( x ) ≥ 1 in B d 1 ( A i ) \ { A i } .

Let

Ω n = Ω \ ∪ j = 1 m B r n ( A j ) ¯ ,

where

r n = 2 − n d 1 .

We observe that Poisson problem

(3.7) ℒ V u = f in Ω n , u = 0 on ∂ Ω n

has a unique positive classical solution u n , which verifies that

∫ Ω n ( ∣ ∇ u n ∣ 2 + V u n 2 ) d x = ∫ Ω n f v d x for any v ∈ H 0 1 ( Ω n ) .

Moreover, there exists c 3 > 0 independent of n such that

‖ u n ‖ m ≤ c 3 .

By the comparison principle, we have that

u n ≤ u f in Ω n ,

and the mapping n → u n is increasing; thus, the limit exists, denoting

u ∞ = lim n → + ∞ u n ,

and u ∞ is a classical solution of (3.3), satisfying that

‖ u ∞ ‖ m ≤ c 3 .

Passing to the limit, we have that

⟨ u ∞ , v ⟩ m = 4 ν m ∫ Ω u ∞ v d x + ∫ Ω f v d x for any v ∈ H V , 0 1 ( Ω ) .

This means u ∞ = u f by the uniqueness in weak solution of (3.3).

Note that there exists t i ≥ 1 such that

t i W i ≥ u f on ∂ B d 1 ( A i ) .

Therefore, we obtain that for all i = 1 , … , m and all n ∈ N

t i W i ≥ u f ≥ u n in B d 1 ( A i ) \ { A i } ,

which implies that for all i = 1 , … , m

u f = u ∞ ≤ t i W i in B d 1 ( A i ) \ { A i }

and (3.4) follows, then u f ∈ L 1 ( Ω , d V x ) .

From Lemma 4.9 in [18] with h ≡ 1 , there exists c 4 > 0 such that for all i = 1 , … , m

(3.8) ∣ ∇ u 0 ( x ) ∣ ≤ c 4 ∣ x − A i ∣ τ + ( μ i ) − 1 , ∀ x ∈ B d 1 ( A i ) \ { A i } .

Multiply by ϒ ξ and integrate (3.3) over Ω n ; then, we obtain that

(3.9) ∫ Ω n u 0 ℒ V ∗ ξ d V x = ∫ Ω n f ξ d V x + ∑ i = 1 m ∫ ∂ B r n ( A i ) ( ∇ u 0 ⋅ n → ϒ ξ − ∇ ( ϒ ξ ) ⋅ n → u 0 ) d ω ( x ) ,

where n → is the unit normal vector of ∂ B r n ( A i ) pointing inward and d ω is the volume of ∂ B r n ( A i ) . For any ξ ∈ C 0 2 ( Ω ) , we note that

(3.10) ℒ V ∗ ξ ∈ C ( Ω ¯ \ A m ) and ∣ ℒ V ∗ ξ ∣ ≤ c 5 ‖ ξ ‖ C 1.1 ϒ in Ω \ A m ,

which, together with (3.4), we derive that

∫ Ω n u 0 ℒ V ∗ ξ d V x → ∫ Ω u 0 ℒ V ∗ ξ d V x , ∫ Ω n f ξ d V x → ∫ Ω f ξ d V x as n → + ∞ .

If μ i > μ 0 for all i = 1 , … , m and on ∂ B r n ( A i )

∇ u 0 ⋅ x ∣ x ∣ ϒ ξ ≤ c 6 ‖ ξ ‖ L ∞ ( Ω ) r n 2 τ + ( μ ) − 1 and ∇ ( ϒ ξ ) ⋅ x ∣ x ∣ u μ ≤ c 6 ‖ ξ ‖ C 1 ( Ω ) r n 2 τ + ( μ ) − 1 ,

due to 2 τ + ( μ ) − 1 > 1 − N , then we obtain that

lim r → 0 + ∫ ∂ B r n ( A i ) ( ∇ u 0 ⋅ n → ϒ ξ − ∇ ( ϒ ξ ) ⋅ n → u 0 ) d ω ( x ) = 0 .

Thus, (3.9) implies the distributional identity (3.5).

If μ i = μ 0 for some i = 1 , … , m . Here, we deal with the case μ 0 < 0 (i.e., N ≥ 3 ).

Let μ i ( 1 ) ≥ μ i ( 2 ) ∈ ( μ 0 , + ∞ ) , denoting the solution u f , j of (3.3) replaced μ → by μ → ( j ) with j = 1 , 2 . Note that

f = − Δ u f , 1 + V u f , 1 ≥ − Δ u f , 2 + V u f , 2 .

By the comparison principle, we have that

u f , 1 ≥ u f , 2 if μ i ( 1 ) ≥ μ i ( 2 ) .

Now, we take a decreasing sequence { μ → ( j ) } j ∈ N ⊂ ( μ 0 , + ∞ ) m such that μ → ( j ) → μ → as j → + ∞ , the corresponding solutions u f , j of (3.3) replaced μ → by μ → ( j ) . Thus, we have that

0 ≤ u f , j ≤ u f , j + 1 ≤ ⋯ ≤ u f in Ω \ A m ,

and by the uniqueness and regularity results, we see that

u f , j → u f in C 2 ( Ω \ A m ) as j → + ∞ .

Thanks to u f ∈ L 1 ( Ω , d V x ) , the dominate monotonicity convergence theorem implies that

u f , j → u f as j → + ∞ a.e. in Ω and in L 1 ( Ω , d V x ) .

Passing to the limit of identity (3.9) as j → + ∞ , (3.5) holds for any μ → ∈ [ μ 0 , + ∞ ) m .□

Now, we are in a position to prove Proposition 3.2.

Proof of Proposition 3.2

Note that for μ ≥ μ 0 , we have that τ − ( μ ) ≤ 2 − N 2 ≤ τ + ( μ ) . Let

(3.11) v 0 ( x ) = ∣ x ∣ 2 − N 2 ln + R 0 ∣ x ∣ 1 2 in R N \ { 0 } ,

where R 0 > 0 such that Ω ⊂ B R 0 ( 0 ) . Direct computation shows that

− Δ v 0 ( x ) + μ ∣ x ∣ 2 v 0 ( x ) = ( N − 2 ) 2 4 + μ ∣ x ∣ − 2 v 0 ( x ) + 1 4 ∣ x ∣ − 2 ln + R 0 ∣ x ∣ − 2 v 0 ( x ) for x ∈ B R 0 ( 0 ) .

Now, we denote

w 0 ( x ) = ∑ j = 1 m ( Φ μ j ( x − A j ) + v 0 ( x − A j ) ) ,

then there exists d 2 ∈ ( 0 , d 1 ] such that for any j = 1 , … , m

(3.12) ℒ V w 0 ( x ) = 1 4 ∣ x − A j ∣ − 2 ln + R 0 ∣ x − A j ∣ − 2 v 0 ( x − A j ) + ∑ i = 1 , i ≠ j m μ i Φ μ i ( x − A i ) + v 0 ( x − A i ) ∣ x − A i ∣ 2 > 0 for x ∈ B d 2 ( A j ) \ { A j } ,

since ∑ i = 1 , i ≠ j m Φ μ i ( x − A i ) + v 0 ( x − A i ) ∣ x − A i ∣ 2 is bounded in B d 2 ( A j ) and

lim x → A j 1 4 ∣ x − A j ∣ − 2 ln + R 0 ∣ x − A j ∣ − 2 v 0 ( x − A j ) = + ∞ .

There exists t 0 > 0 such that

∣ ℒ V w 0 ∣ ≤ t 0 in Ω ¯ \ ( ∪ j = 1 m B d 2 ( A j ) ¯ ) .

Now, we denote

(3.13) W = w 0 + t 0 u 1 ,

where u 1 is the solution of (3.3) with f = 1 . Then,

(3.14) ℒ V W ≥ 0 in Ω \ A m ,

and for all i = 1 , … , m

lim ∣ x − A i ∣ → 0 + W ( x ) Φ μ i ( x − A i ) = 1 .

Let u be a solution of (1.23), then for any ε > 0 , there exists r ε > 0 converging to zero as ε → 0 such that

u ( x ) ≤ ε W ( x ) for x ∈ ∪ j = 1 m B r ε ( A j ) ¯ \ { A j } .

We see that

u ( x ) = 0 < ε W ( x ) for x ∈ ∂ Ω ,

then by Proposition 3.1, we have that

u ≤ ε W in Ω \ A m .

By the arbitrary of ε , we have that u ≤ 0 in Ω \ A m . The same way can be used to obtain that u ≥ 0 in Ω \ A m . This ends the proof.□

4 Estimates

4.1 Asymptotic

Proposition 4.1

Let 0 ∉ σ d , f be a nonnegative, nontrivial function in C α ( Ω ¯ \ A m ) with α ∈ ( 0 , 1 ) such that

(4.1) ∣ f ( x ) ∣ ≤ c 7 ζ m − 1 ( x ) , ∀ x ∈ Ω \ A m

and ξ 0 be a unique nonnegative solution of

(4.2) ℒ V u = ϒ f in Ω \ A m , u = 0 on ∂ Ω , lim x → A i u ( x ) Φ μ i ( x − A i ) = 0 for all i = 1 , … , m .

Then, for any i = 1 , … , m , denoting r = ∣ x − A i ∣ , there exists l i ≥ 0 such that

(4.3) lim ∣ x − A i ∣ → 0 + ∣ x − A i ∣ − τ + ( μ i ) ξ 0 ( x ) = l i .

Proof

Here, we adopt the Veron’s idea in [39] to obtain the asymptotic behavior at the poles. For simplicity, we assume that A i = 0 . From (3.4), we have that

limsup ∣ x ∣ → 0 ∣ x ∣ − τ + ( μ i ) ξ 0 ( x ) < + ∞ .

Step 1: We use here the spherical coordinates ( r , θ ) in the variable x , and we recall that Δ ′ is the Laplace-Beltrami operator on S N − 1 , then for ( r , θ ) ∈ ( 0 , d 0 ) × S N − 1 ,

r τ + ( μ i ) f ( r , θ ) = ℒ V v 0 = − ∂ r r v 0 − N − 1 r ∂ r v 0 − 1 r 2 Δ ′ v 0 + μ i r 2 v 0 + V ˜ ( r , θ ) v 0 ,

where V ˜ ( r , θ ) = ∑ j ≠ i , j = 1 m μ j ∣ x − A j ∣ , which is bounded in ( 0 , d 0 ) × S N − 1 .

Let

v ˜ 0 ( t , θ ) = r − τ + ( μ i ) v 0 ( r , θ ) with t = ln r ,

then v ˜ 0 is bounded and a straightforward computation shows that

(4.4) ∂ t t v ˜ 0 + ( 2 τ + ( μ i ) + N − 2 ) ∂ t v ˜ 0 + Δ ′ v ˜ 0 = e t ( e t V ˜ ( e t , θ ) v ˜ 0 − e t f ( e t , θ ) ) ≕ e t ψ ( t , θ ) ,

where 2 τ + ( μ i ) + N − 2 > 0 if μ i > μ 0 and 2 τ + ( μ i ) + N − 2 = 0 if μ i = μ 0 . Note that ψ is uniformly bounded in θ ∈ ( − ∞ , ln d 0 ) × S N − 1 , since v ˜ 0 is bounded and f ∈ C α ( Ω ¯ \ A m ) such that ∣ f ( x ) ∣ ≤ c 7 ζ m − 1 ( x ) for x ∈ Ω \ A m .

Step 2. The convergence process. Since v ˜ 0 is bounded in ( − ∞ , r 0 ] × S N − 1 , we obtain that v ˜ 0 is bounded in W 2 , q ( [ T − 1 , T + 1 ] × S N − 1 ) independently of T ≤ ln d 0 − 2 , for any q < ∞ . Hence, v ˜ 0 is bounded in any C 1 , γ ( [ T − 1 , T + 1 ] × S N − 1 ) for any γ ∈ [ 0 , 1 ) . Differentiating the equation and using the standard elliptic equations regularity, we obtain that v is bounded in W 3 , q ( [ T − 1 , T + 1 ] × S N − 1 ) and in C 2 , γ ( [ T − 1 , T + 1 ] × S N − 1 ) . We consider the negative trajectory of v in C 0 1 ( S N − 1 ) defined by:

T − ( v ) = ⋃ t ≤ r 0 − 1 { v ˜ 0 ( t , . ) } .

By the previous estimates and the Arzela-Ascoli theorem, it is a relatively compact subset of C 1 ( S N − 1 ) ; hence, its limit set at − ∞ (or alpha-limit set), denoting A ( T − ( v ) ) , is a non-empty connected compact subset of C 1 ( S N − 1 ) . Integrating (4.4) over S N − 1 yields that

(4.5) ∂ t t w 0 + ( 2 τ + ( μ i ) + N − 2 ) ∂ t w 0 = e t ∫ S N − 1 ψ ( t , θ ) d ω ( θ ) ,

where w 0 ( t ) = ∫ S N − 1 v ˜ 0 ( t , θ ) d ω ( θ ) . Multiplying ∂ t w 0 in (4.5) and integrating in ( − ∞ , s ) , we have that

(4.6) ( 2 τ + ( μ i ) + N − 2 ) ∫ − ∞ s ( ∂ t w 0 ) 2 d t + w t 2 ( s ) 2 ≤ ∫ − ∞ s e t d t ∣ S N − 1 ∣ ‖ ψ ‖ L ∞ ,

which, together with (4.5), implies that w t and w t t is bounded and converge to zero as t → − ∞ .

Since f ∈ C 1 ( Ω ¯ ) and ∣ ∇ v 0 ∣ is bounded by (3.8), then ∂ t ψ is uniformly bounded in ( − ∞ , ln d 0 ) × S N − 1 ; differentiating (4.5) with respect to t , multiplying by ∂ t t w 0 , and integrating on ( − ∞ , ln d 0 ) yields that

(4.7) ( 2 τ + ( μ i ) + N − 2 ) ∫ − ∞ s ( ∂ t t w 0 ) 2 d t + 1 2 ( ∂ t t w 0 ( s ) ) 2 = ∫ − ∞ s e t ∫ S N − 1 ψ ( t , θ ) d ω ( θ ) d t .

As a consequence, ∂ t w 0 and ∂ t t w 0 are uniformly continuous on ( − ∞ , ln d 0 ) , belong to L 2 ( − ∞ , ln d 0 ) , when μ i = μ 0 , (4.4) reduces to

(4.8) ∂ t t v ˜ 0 + Δ ′ v ˜ 0 = e t ψ ( t , θ ) .

From (4.6) and (4.7), there exists c 8 > 0 such that

∣ ∂ t w 0 ∣ + ∣ ∂ t t w 0 ∣ ≤ c 8 e t .

Therefore, we have that

lim t → − ∞ ( ‖ v t ( t , ⋅ ) ‖ L 2 ( S N − 1 ) + ‖ v t t ( t , ⋅ ) ‖ L 2 ( S N − 1 ) ) = 0 ,

and then, the set A ( T − ( v ) ) is a compact connected subset of the set of nonnegative solutions of

(4.9) Δ ′ v ˜ 0 = 0 in S N − 1 ,

whose nonnegative solution could only be nonnegative constants. Thus, in Form (4.4), there exists a sequence t n → − ∞ and l i > 0 such that

(4.10) lim t n → − ∞ v ˜ 0 ( t n , θ ) = l i ≥ 0 uniformly for θ ∈ S N − 1 .

Moreover, from (4.6) and (4.7)

∣ w t t ∣ , ∣ w t ∣ ≤ c 9 t

for some c 9 > 0 , this implies that l i ≥ 0 in (4.10) is unique. Therefore, (4.3) holds, which completes the proof.□

Corollary 4.2

Let λ 1 ( μ → ) > 0 and f ∈ C 1 ( Ω ¯ \ A m ) be a nonnegative, nontrivial function verifying (4.1), then problem

(4.11) ℒ V ∗ u = f in Ω \ A m , u = 0 on ∂ Ω

has a unique bounded nonnegative classical solution v f . Moreover, there exists α i > 0 such that

(4.12) lim ∣ x − A i ∣ → 0 + v f ( x ) = α i .

Proof

In fact, let u f be the unique solution of (3.3), we see that u f = ϒ v f and direct computation shows v f is a bounded solution of (4.11). Vice versa holds. From Proposition 4.1, we can supplement the values of v f at A m such that v f is continuous in Ω ¯ .□

Similarly, we have the following corollary:

Corollary 4.3

Let f be a nonnegative, nontrivial function in C α ( Ω ¯ ) , with α ∈ ( 0 , 1 ) , and ξ 0 be unique nonnegative solution of

(4.13) ( ℒ V + 4 ν m ) u = ϒ f i n Ω , u = 0 on ∂ Ω , lim x → A i u ( x ) Φ μ i ( x − A i ) = 0 f o r a l l i = 1 , … , m .

Then, for any i = 1 , … , m , denoting r i = ∣ x − A i ∣ , there exists l i ≥ 0 such that

(4.14) lim r i → 0 + r i − τ + ( μ i ) ξ 0 ( x ) = l i .

4.2 Poisson problem with L p source

We start this section by De Giorgi estimate, and the following lemma is the essential lemma for De Giorgi method (the book [32]). For the reader’s convenience, we provide the details of the proof.

Lemma 4.4

Let φ : [ k 0 , + ∞ ) → [ 0 , + ∞ ) be a nondecreasing function such that

(4.15) φ ( h ) ≤ M h − t α φ β ( t ) ∀ h > t ≥ k 0 ,

where k 0 ≥ 0 , α > 0 , and β > 1 . Then,

φ ( k 0 + t 0 ) = 0 ,

where

t 0 = M 2 β β − 1 φ ( k 0 ) β − 1 α .

Proof

Let

t n = k 0 + t 0 − t 0 2 n , n = 0 , 1 , 2 , … .

From Assumption (4.15), we have that

φ ( t n + 1 ) ≤ M 2 n + 1 t 0 α φ β ( t n ) , n = 0 , 1 , 2 , … .

Inductively, we assume that

(4.16) φ ( t n ) ≤ φ ( k 0 ) r n for n = 0 , 1 , 2 , … ,

where r > 0 will be determined later. Therefore, we have that

φ ( t n + 1 ) ≤ M 2 n + 1 t 0 α φ β ( t n ) ≤ φ ( k 0 ) r n + 1 M α 2 ( n + 1 ) α t 0 α r n ( β − 1 ) − 1 ( φ ( k 0 ) ) β − 1 .

Now, we choose r = 2 α β − 1 , and we have that

M α 2 ( n + 1 ) α t 0 α r n ( β − 1 ) − 1 ( φ ( k 0 ) ) β − 1 ≤ 1 .

Then, (4.16) holds for n + 1 , so does for any n ∈ N and pass to n → + ∞ , we have that φ ( k 0 + t 0 ) = 0 for such t 0 .□

Lemma 4.5

Assume that μ → ≥ 0 and f ϒ − 1 ∈ L p ( Ω ) , with

p > N 2 .

Let u f be a unique weak solution of the homogeneous Problem (1.10), then there exists c 10 > 0 independent of f such that

sup x ∈ Ω \ A m u f ( x ) ϒ − 1 ( x ) ≤ c 10 ‖ f ϒ − 1 ‖ L p ( Ω ) .

Proof

Given k > 0 , let S k be the function defined for t ∈ R by

S k ( t ) = max { 0 , t − k } .

Let u f ( x ) = v f ( x ) ϒ and from (1.11) with v = ϒ S k ( v f ) ∈ H V , 0 1 ( Ω ) ∩ H V , 0 1 ( { v f > k } ) , we have that

(4.17) ∫ Ω ∣ ∇ u f ∣ 2 d x + ∫ Ω ∑ i = 1 m μ i ( S k ( v f ) + k ) S k ( v f ) ϒ 2 ∣ x − A i ∣ 2 d x = ∫ Ω f S k ( v f ) ϒ d x .

For μ → ∈ [ 0 , + ∞ ) m , it follows from Corollary 2.3 that

∫ Ω ∣ ∇ u f ∣ 2 d x + ∫ Ω ∑ i = 1 m μ i ( S k ( v f ) + k ) S k ( v f ) ϒ 2 ( x ) ∣ x − A i ∣ 2 d x ≥ ∫ Ω ∣ ∇ ( S k ( v f ) ϒ ) ∣ 2 + ∑ i = 1 m μ i S k 2 ( v f ) ϒ 2 ∣ x − A i ∣ 2 d x + ∫ Ω ∑ i = 1 m μ i k S k ( v f ) ϒ 2 ∣ x − A i ∣ 2 d x ≥ c 11 ∫ Ω ( S k ( v f ) ϒ ) q d x 2 q ,

where c 11 > 0 , and we choose q = 2 p p − 1 . Thanks to p > N 2 , we have that

2 < q < 2 ∗ .

Therefore, there exists k 1 > k 0 such that for k ≥ k 1 ,

(4.18) ∫ Ω ∣ ∇ u f ∣ 2 d x + ∫ Ω ∑ i = 1 m μ i ( S k ( v f ) + k ) S k ( v f ) ϒ 2 ∣ x − A i ∣ 2 d x ≥ c 11 2 ∫ { v f > k } ( S k ( v f ) ϒ ) q d x 2 q .

The right-hand side of (4.17) has the following estimate:

∫ Ω f S k ( v f ) ϒ d x ≤ ∫ { v f > k } f S k ( v f ) ϒ d x ≤ ∫ { v f > k } f q ′ d x 1 q ′ ∫ { v f > k } ( S k ( v f ) ϒ ) q d x 1 q .

It then follows from (4.17) that

(4.19) ∫ { v f > k } ( S k ( v f ) ϒ ) q d x 1 q ≤ c 12 ∫ { v f > k } f q ′ d x 1 q ′ ,

where q ′ = q q − 1 < 2 < q .

Moreover, we see that { v f > h } ⊂ { v f > k } if h > k ,

∫ { v f > k } ( S k ( v f ) ϒ ) q d x ≥ ∫ { v f > h } ( S k ( v f ) ϒ ) q d x ≥ ( h − k ) q ∫ { v f > h } ϒ q d x

and

∫ { v f > k } f q ′ d x = ∫ { v f > k } ( f ϒ − 1 ) q ′ ϒ q ′ d x ≤ ∫ { v f > k } ( f ϒ − 1 ) q q − 2 d x 1 − q ′ q ∫ { v f > k } ϒ q d x q ′ q ,

where 1 − q ′ q = 1 q − 1 .

Thus, letting φ ( h ) = ∫ { v f > h } ϒ q d x and p = q q − 2 , we obtain that

(4.20) φ ( h ) ≤ c 13 ∫ { v f > k } ( f ϒ − 1 ) q q − 2 d x 1 − q ′ q ( h − k ) − q φ ( k ) q ′ q ,

and now we apply Lemma 4.4, and there exists t 0 > 0 such that

0 = φ ( k + t 0 ) = ∫ { v f > k + t 0 } ϒ q d x ,

which implies that ∣ { v f > k + t 0 } ∣ = 0 . We complete the proof.□

Proof of Proposition 1.2

Part (i): Let

L ( f ) = ∫ Ω f v d x ,

then

∫ Ω u f d x ≤ ‖ f ‖ L p ( Ω ) ‖ u ‖ L p ′ ( Ω ) ≤ ‖ f ‖ L p ( Ω ) ‖ u ‖ m ,

so L : H V , 0 1 ( Ω ) is a bound linear functional. If 0 ∉ σ d ( ℒ V ) , it follows by the Fredholm alternative theorem that Problem (1.10) has a unique weak solution u f ∈ H V , 0 1 ( Ω ) .

If λ 1 ( μ → ) > 0 , the existence of weak solution of (1.10) follows also by the Ekeland variational principle with the functional

ℐ 0 ( u ) = 1 2 ‖ u ‖ m 2 − 4 ν m ∫ Ω u 2 d x − ∫ Ω u f d x ≥ 1 2 λ 1 ( μ ) ‖ u ‖ L 2 2 − ‖ f ‖ L p ( Ω ) ‖ u ‖ L p ′ ( Ω ) .

The critical point u 0 of ℐ 0 verifies that

⟨ u 0 , v ⟩ m = 4 ν m ∫ Ω u 0 v d x + ∫ Ω f v d x for any v ∈ H V , 0 1 ( Ω ) ,

which is equivalent to (3.6). Moreover, we have u 0 ≥ 0 a.e. in Ω if f is nonnegative.

Part (ii): It follows by Lemma 4.5 directly.□

5 Weighted distributional solutions

5.1 Kato’s inequality

The following proposition is the Kato’s type estimate. For the uniformly elliptic second-order operators, the Kato’s inequalities could see the survey [41].

Proposition 5.1

Let λ 1 ( μ → ) > 0 and f ∈ L 1 ( Ω , ρ d V x ) , then there exists a unique d V x -distributional solution u ∈ L 1 ( Ω , ζ m − 1 d V x ) of the problem

(5.1) ℒ V u = f i n Ω , u = 0 o n ∂ Ω ,

that is,

(5.2) ∫ Ω u ℒ V ∗ ( ξ ) d V x = ∫ Ω f ξ d V x , ∀ ξ ∈ X V ( Ω ) .

Moreover, we have that for any ξ ∈ C 0 1.1 ( Ω ) , ξ ≥ 0 ,

(5.3) ∫ Ω ∣ u ∣ ℒ V ∗ ( ξ ) d V x ≤ ∫ Ω sign ( u ) f ξ d V x

and

(5.4) ∫ Ω u + ℒ V ∗ ( ξ ) d V x ≤ ∫ Ω sign + ( u ) f ξ d V x .

Proof of Proposition 5.1

Uniqueness. Let w be a d V x -distributional solution of

(5.5) ℒ V w = 0 in Ω , w = 0 on ∂ Ω .

By Corollary 4.2, for any Borel subset O of Ω ,

(5.6) ℒ V ∗ u = ζ n in Ω , u = 0 on ∂ Ω

has a unique bounded solution η ω , n ≥ 0 , where ζ n : Ω ¯ ↦ [ 0 , 1 ] is a C 1 ( Ω ¯ ) function such that ζ n → χ O in L ∞ ( Ω ) as n → ∞ . From Corollary 4.2, there exists α i > 0 such that

lim ∣ x − A i ∣ → 0 + η ω , n ( x ) = α i ,

so after adding the value at A m for function η ω , n , it belongs to the test function space X V ( Ω ) .

By Lemma 5.3, we have that ∫ Ω w ϒ ζ n d x = 0 . Passing to the limit as n → ∞ , we have that

∫ O w ϒ d x = 0 ,

which implies that w = 0 .

Existence and estimate (5.3). For σ > 0 , we define an even convex function ϕ σ as:

(5.7) ϕ σ ( t ) = ∣ t ∣ − σ 2 if ∣ t ∣ ≥ σ , t 2 2 σ if ∣ t ∣ < σ ∕ 2 .

Then, for any t ∈ R , ∣ ϕ σ ′ ( t ) ∣ ≤ 1 , ϕ σ ( t ) → ∣ t ∣ and ϕ σ ′ ( t ) → sign ( t ) when σ → 0 + . Let { f n } be a sequence of functions in C 1 ( Ω ¯ ) such that

lim n → ∞ ∫ Ω ∣ f n − f ∣ ρ d γ μ = 0 .

Let u n be the corresponding solution to (5.1) with right-hand side f n , then for any σ > 0 and ξ ∈ C 0 1.1 ( Ω ) , ξ ≥ 0 , we have that

(5.8) ∫ Ω ϕ σ ( u n ) ( − Δ ) ( ϒ ξ ) d x = ∫ Ω ϒ ξ ( − Δ ) ϕ σ ( u n ) d x = ∫ Ω ξ ϕ σ ′ ( u n ) ( − Δ ) u n d V x − ∫ Ω ξ ϕ σ ′ ′ ( u n ) ∣ ∇ u n ∣ 2 d V x ≤ ∫ Ω ξ ϕ σ ′ ( u n ) f n d V x − ∫ Ω V u n ϕ σ ′ ( u n ) ξ d V x .

Put σ → 0 and we obtain that

∫ Ω ∣ u n ∣ ( − Δ ) ( ϒ ξ ) d x ≤ ∫ Ω ξ sign ( u n ) f n d V x − ∫ Ω V ∣ u n ∣ ξ d x ,

that is,

(5.9) ∫ Ω ∣ u n ∣ ℒ V ∗ ( ξ ) d V x ≤ ∫ Ω ξ ∣ f n ∣ d V x .

Taking ξ = ξ 0 , the solution of (4.2), we have that

(5.10) ∫ Ω ∣ u n ∣ d V x ≤ c 14 ∫ Ω ∣ f n ∣ ρ d V x .

Similarly,

(5.11) ∫ Ω ∣ u n − u m ∣ d V x ≤ c 14 ∫ Ω ∣ f n − f m ∣ ρ d V x .

Therefore, { u n } is a Cauchy sequence in L 1 ( Ω , ρ d V x ) and its limit u is a d V x -distributional solution of (5.1). Passing to the limit as n → ∞ in (5.9), we obtain (5.3). Inequality (5.4) is proved by replacing ϕ σ by ϕ ˜ σ , which is zero on ( − ∞ , 0 ) and ϕ σ on [ 0 , ∞ ) .□

We remark that the assumption λ 1 ( μ → ) > 0 is essential for Kato’s inequality.

5.2 Isolated singular solutions

Lemma 5.2

Let ν = ∑ i = 1 m k i δ A i , then (5.1) admits a unique weak solution u ν ∈ L 1 ( Ω , ζ m − 1 d V x ) , which is a classical solution of

(5.12) ℒ V u = 0 i n Ω \ A m , u = 0 o n ∂ Ω , lim x → A i u ( x ) Φ μ i ( x − A i ) = c μ i k i f o r a l l i = 1 , … , m .

Proof

The uniqueness follows from Kato’s inequality in Proposition 5.1.

Existence. Let

w m ( x ) = ∑ i = 1 m k i Φ μ i ( x − A i ) η 0 ( x ) , x ∈ R N \ A m ,

where η 0 : [ 0 , + ∞ ) → [ 0 , 1 ] be a C 2 function such that

(5.13) η 0 = 1 if dist ( x , ∂ Ω ) > d 0 2 and η 0 = 0 if dist ( x , ∂ Ω ) < d 0 4 .

Direct computation shows that

g 0 ( x ) = ℒ V w m ( x ) − ∑ i = 1 m k i δ A i for x ∈ Ω ,

and then,

∣ g 0 ( x ) ∣ ≤ c 15 ζ m − 2 ( x ) , ∀ x ∈ Ω \ A m .

From Lemma 5.3, the problem

ℒ V u = − g 0 in Ω \ A m , u = 0 on ∂ Ω , lim x → A i u ( x ) Φ μ i ( x − A i ) = 0 for all i = 1 , … , m

has a unique solution w f and

lim ∣ x − A i ∣ → 0 ∣ w f ( x ) ∣ Φ μ i ( x − A i ) = 0 .

As a consequence, w m + w f is the unique weak solution of (5.1) with ν = ∑ i = 1 m k i δ A i .□

5.3 Singular classical solutions

This subsection is denoted to consider the classical solution of

(5.14) ℒ V u = f in Ω \ A m , u = 0 on ∂ Ω , lim x → A i u ( x ) Φ μ i ( x − A i ) = 0 for all i = 1 , … , m ,

where f is a regular source in Ω \ A m .

Next, we prove the distributional identity for the classical solution of the nonhomogeneous problem with “removable” singularity, i.e., lim ∣ x − A i ∣ → 0 + u ( x ) Φ μ − 1 ( x − A i ) = 0 .

Lemma 5.3

Assume that λ 1 ( μ → ) > 0 , f ∈ C γ ( Ω ¯ \ A m ) for some γ ∈ ( 0 , 1 ) verifies that for some c 16 > 0 , τ i > τ − ( μ i ) with i = 1 , … , m and t < 1

(5.15) i f μ i > μ 0 , ∣ f ( x ) ∣ ≤ c 16 ∣ x − A i ∣ τ i − 2 , ∀ x ∈ B d 1 ( A i ) \ { A i } , i f μ i = μ 0 , ∣ f ( x ) ∣ ≤ c 16 ∣ x − A i ∣ − N + 2 2 ( − ln ∣ x − A i ∣ ) t , ∀ x ∈ B d 1 ( A i ) \ { A i } .

Then, (5.14) has a unique solution u f such that

(5.16) i f μ i > μ 0 , ∣ u f ( x ) ∣ ≤ c 17 ∣ x − A i ∣ min { τ + ( μ i ) , τ i } , ∀ x ∈ B d 1 ( A i ) \ { A i } , i f μ i = μ 0 , ∣ u f ( x ) ∣ ≤ c 17 ∣ x − A i ∣ − N − 2 2 ( − ln ∣ x − A i ∣ ) t

for some c 17 > 0 and all i = 1 , … , m .

Proof

Here, we only have to prove the case f ≥ 0 by the linearity of operator ℒ V . Let { f n } be a sequence of positive functions in C γ ( Ω ¯ ) such that

f n ≤ f n + 1 ≤ ⋯ → f as n → + ∞ in Ω \ A m .

Let u f n be the solution of (5.14) with source f n by Lemma 3.3.

Let

w i ( x ) = ∣ x − A i ∣ min { τ + ( μ i ) , τ i } − ε i ∣ x − A i ∣ τ 0 , ∀ x ∈ R N \ { A i } if μ i > μ 0 , ∣ x − A i ∣ 2 − N 2 ( − ln ∣ x − A i ∣ ) t − ε i ∣ x − A i ∣ τ 0 , ∀ x ∈ R N \ { A i } if μ i = μ 0 < 0 ,

where τ + ( μ i ) − 1 < τ 0 < τ + ( μ i ) and ε i > 0 be small such that w i > 0 in Ω \ A m . Let

w ¯ = ∑ i = 1 m w i ,

and from our choice of τ and τ 0 , there exists d 2 ∈ ( 0 , d 0 ] such that for x ∈ ∪ i = 1 m B d 2 ( A i ) \ { A i }

ℒ V w ¯ ( x ) ≥ c 18 ∣ x − A i ∣ τ − 2 if μ i > μ 0 , c 18 ∣ x − A i ∣ − N + 2 2 ( − ln ∣ x − A i ∣ ) t if μ i = μ 0 < 0 ,

and there exists c 19 > 0 such that

∣ ℒ V w ¯ ( x ) ∣ ≤ c 19 in Ω \ ( ∪ i = 1 m B d 2 ( A i ) ) .

Note that

W ¯ = c 16 c 18 w ¯ + c 16 c 18 c 19 u 1 ,

which is a positive supersolution of (5.14), and then, we have that

0 ≤ u f n ≤ W ¯ .

Thanks to W ¯ ∈ L 1 ( Ω , d V x ) ∩ C 2 ( Ω \ A m ) , the dominate monotonicity convergence theorem and standard interior regularity results imply that the limit of { u f n } as n → + ∞ is a classical solution of Problem (5.14) satisfying (5.16). The uniqueness follows from Lemma 3.2.□

Proof of Theorem 1.4

Part (i): Since f is a function in C loc γ ( Ω ¯ \ A m ) satisfying (1.19), it follows from Proposition 5.1, Problem (1.13) admits a unique weak solution u f , and the approximation of weak solution, we have that u f ∈ C 2 ( Ω \ A m ) , and then, it is a classical solution of Problem (1.13), so is u k ≔ u f + u ν by Lemma 5.2.

Part (ii): for τ < 2 − τ − ( μ i )

lim ∣ x − A i ∣ → 0 f ( x ) ∣ x − A i ∣ τ = 0 for all i = 1 , 2 , … , m ,

then it follows by Lemma 5.3 that

lim ∣ x − A i ∣ → 0 + u f ( x ) Φ μ i − 1 ( x − A i ) = 0 .

Therefore, (1.21) holds true.

Part (iii): We prove Problem (1.13) has no nonnegative solution provided that f is a nonnegative function in C loc γ ( Ω ¯ \ A m ) satisfying (1.22); there exists some i 0 such that

(5.17) lim ∣ x − A i 0 ∣ → 0 + ∫ B d 0 ( A i 0 ) \ B r ( A i 0 ) f Γ μ i 0 d x = + ∞ .

By contradiction, we assume that Problem (1.13) has a nonnegative distributional solution of u f ∈ L 1 ( Ω , ζ m − 1 d V x ) ∩ C 0 2 ( Ω \ A m ) .

Let η i 0 : R N → [ 0 , 1 ] be a C 2 function such that

η i 0 ( x ) = 1 for all x ∈ B d 0 ∕ 4 ( A i 0 ) and η i 0 ( x ) = 0 for x ∈ R N \ B d 0 ∕ 2 ( A i 0 ) .

Set w f = u f η i 0 , then we observe that

− Δ w f + μ i 0 ∣ x − A i 0 ∣ 2 w f = f η i 0 + f ˜ in Ω \ { A i 0 } ,

where

f ˜ = − 2 ∇ η i 0 ⋅ ∇ u f − u f Δ η i 0 − ∑ i = 1 , i ≠ i 0 m μ i 0 ∣ x − A i ∣ 2 u f η i 0 ∈ L 1 ( Ω , ζ m − 1 d V x ) ∩ C 0 2 ( Ω \ { A i 0 } ) ,

since ∑ i = 1 , i ≠ i 0 m μ i 0 ∣ x − A i ∣ 2 u f η i 0 is only singular at A i 0 and 2 ∇ η i 0 ⋅ ∇ u f , u f Δ η i 0 belongs to C 2 ( Ω ¯ ) , with the support in B d 0 ∕ 2 ( A i 0 ) ¯ \ B d 0 ∕ 4 ( A i 0 ) .

Let v f be a distributional solution of

(5.18) − Δ u + μ i 0 ∣ x − A i 0 ∣ 2 u = ∣ f ˜ ∣ in Ω , u = 0 on ∂ Ω

in the sense of (5.2). From Proposition 5.1 and the approximation of weak solution, we have v f ∈ L 1 ( Ω , ζ m − 1 d V x ) ∩ C 0 2 ( Ω \ { A i 0 } ) .

Let

u ˜ f = w f + v f in Ω \ { A i 0 } ,

which is a nonnegative classical solution of

(5.19) − Δ u + μ i 0 ∣ x − A i 0 ∣ 2 u = f + ∣ f ˜ ∣ + f ˜ in Ω \ { A i 0 } , u = 0 on ∂ Ω ,

where f + ∣ f ˜ ∣ + f ˜ ≥ f verifies (5.17). Then, a contradiction is obtained by [13, Theorem 1.3 part (iii)], which states that (5.19) has no positive solution under Assumption (1.22).

Part (iv): It follows by Proposition 3.2 directly.□

5.4 Poisson problem with measure data

Proof of Theorem 1.6

From Lemma 5.2 and the linearity, we only have to show that for ν ∈ M ( Ω ; ϒ ) , (5.1) admits a unique weak solution u ∈ L 1 ( Ω , ζ m − 1 d V x ) . We can assume that ν ≥ 0 . Let { ν n } ⊂ C 1 ( Ω ¯ ) be a sequence such that ν n ≥ 0 and

∫ Ω ξ ϒ ν n d x → ∫ Ω ξ d ( ϒ ν ) for all ξ ∈ X V ( Ω ) ,

then by Proposition 5.1, for any n ∈ N Poisson problem

(5.20) ℒ V u n = ν n in Ω , u n = 0 on ∂ Ω

has a unique, nonnegative weak solution u n such that

(5.21) ∫ Ω u n ℒ V ∗ ξ d V x = ∫ Ω ξ ν n ϒ d x for all ξ ∈ X V ( Ω ) .

Then, using the test function η 1 , the solution of (4.11) with f = ζ m − 1 in Corollary 4.2, there exists c 20 > 0 such that

(5.22) ∫ Ω u n ζ m − 1 d V x = c 20 ∫ Ω η 1 ϒ ν n d x ≤ c 20 ‖ η 1 ‖ L ∞ ( Ω ) ‖ ν ‖ M ( Ω , ϒ ) ,

which implies that { u n } is uniformly bounded in L 1 ( Ω , ζ m − 1 d V x ) .

For any ε > 0 sufficiently small, set the test function ξ in { ζ ∈ X V ( Ω ) : ζ = 0 in ∪ i = 1 m B ε ( A i ) } , then we have that

(5.23) ∫ Ω \ ( ∪ i = 1 m B ε ( A i ) ) u n ℒ V ∗ ξ d V x = ∫ Ω \ ( ∪ i = 1 m B ε ( A i ) ) ξ ν n ϒ d x for all ξ ∈ X V ( Ω ) .

Therefore, for any open sets O , O ′ verifying O ¯ ⊂ O ′ ⊂ O ¯ ′ ⊂ Ω \ ( ∪ i = 1 m B ε ( A i ) ) , there exists c 21 > 0 independent of n such that

‖ u n ‖ L 1 ( O ′ ) ≤ c 21 ‖ ν ‖ M ( Ω , ϒ ) .

Note that in Ω \ ( ∪ i = 1 m B ε ( A i ) ) , the operator ℒ μ ∗ is uniformly elliptic and the measure d V x is equivalent to the Hausdorff measure d x , then [38, Corollary 2.8] could be applied to obtain that for some c 22 , c 23 > 0 independent of n but dependent of O ′ ,

‖ u n ‖ W 1 , q ( O ) ≤ c 22 ‖ u n ‖ L 1 ( O ′ ) + ‖ ν n ˜ ‖ L 1 ( Ω , d V x ) ≤ c 23 ‖ ν ‖ M ( Ω , ϒ ) .

That is, { u n } is uniformly bounded in W loc 1 , q ( Ω \ A m ) .

As a consequence, by the arbitrary of ε , there exists a subsequence, still denoting { u n } n and u such that

u n → u a.e. in Ω as n → + ∞ .

Meanwhile, we deduce from Fatou’s lemma,

(5.24) ∫ Ω u ζ m − 1 d V x ≤ c 24 ∫ Ω η 1 ϒ d ν .

We next claim that u n → u in L 1 ( Ω , ζ m − 1 d V x ) as n → + ∞ . Let O ⊂ Ω be a Borel set and ψ o be the solution of

(5.25) ℒ V ∗ ψ o = ζ m − 1 χ O in Ω , ψ o = 0 on ∂ Ω .

Then, ψ o ≤ η 1 ; thus, it is uniformly bounded.

From Lemma 4.5 with N 2 < p < N , we have that

sup Ω ∣ ψ o ( x ) ∣ ≤ c 25 ∫ O ζ m − p d x → 0 as ∣ O ∣ → 0 .

Taking ξ = ψ o in (5.21) implies that

∫ O u n ζ m − 1 d V ( x ) = ∫ O ν n ϒ ψ o d x ≤ sup Ω ψ o ∫ O ν n ϒ d x → 0 as ∣ O ∣ → 0 .

Therefore, { u n } is uniformly integrable for the measure ζ m − 1 d V x . Passing to the limit of (5.21) as n → ∞ implies the claim.□

Funding information: This work was supported by the Natural Science Foundation of China, No. 12071189, by Jiangxi Province Science Fund for Distinguished Young Scholars, No. 20212ACB211005, by the research project of degree and postgraduate education and teaching reform of Jiangxi Normal University, No: YJG202202, by the Science and Technology Research Project of Jiangxi Provincial Department of Education, No. GJJ200307 and GJJ200325.
Conflict of interest: The authors state no conflict of interest.

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Received: 2022-07-25

Revised: 2023-01-03

Accepted: 2023-03-05

Published Online: 2023-08-04

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https://doi.org/10.1515/anona-2022-0320

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Hardy-Leray operator; Dirichlet eigenvalues; Poisson problem; multiple polars

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