Electronic Journal of Differential Equations, Vol. 2021 (2021), No. 87, pp. 1–15.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

SOLUTIONS AND EIGENVALUES OF LAPLACE’S EQUATION

ON BOUNDED OPEN SETS

GUANGCHONG YANG, KUNQUAN LAN

Abstract. We obtain solutions for Laplace’s and Poisson’s equations on

bounded open subsets of Rn (n ≥ 2), via Hammerstein integral operators
involving kernels and Green’s functions, respectively. The new solutions are

different from the previous ones obtained by the well-known Newtonian poten-

tial kernel and the Newtonian potential operator. Our results on eigenvalue
problems of Laplace’s equation are different from the previous results that use

the Newtonian potential operator and require n ≥ 3. As a special case of the

eigenvalue problems, we provide a result under an easily verifiable condition
on the weight function when n ≥ 3. This result cannot be obtained by using

the Newtonian potential operator.

1. Introduction

The Newtonian potential kernel and Newtonian operator have been used to study
the following three problems:

(i) solutions of Laplace’s equation ∆u(x) = 0 in Rn \ {0},
(ii) solutions of Poisson’s equation −∆u(x) = v(x) in Ω, and

(iii) eigenvalue problems of Laplace’s equation −∆u(x) = µg(x)u(x) in Ω.

It is shown in [3, p.21-22], [4, p.17], and [15, Lemma 2.1 (P3)] that a solution
Ψ(·, 0) is a harmonic function in Rn \ {0} for n ∈ N with n ≥ 2; that is, Ψ(·, 0)
is a solution of Laplace’s equation and belongs to C2(Rn \ {0}), where Ψ is the
Newtonian potential kernel, and has singularities. We refer the reader to [12, 13]
for a study on the Newtonian potential, and to [2, 6, 14] for a study on fractional
differential equations, where the related operators involve singularities.

When Ω is a bounded connected open subset in Rn with n ≥ 2 and v ∈ Cµ(Ω), it
is showed in [4, Lemma 4.2 ] that Lv is a solution of the Poisson’s equation, where
L is the Newtonian potential operator. This result is generalized to the case that
Ω is a bounded open subset of Rn in [15, Theorem 2.3].

Eigenvalue problems of the Laplace’s equation can be solved by the result on the
weight Newtonian potential operator [15, Theorem 2.4] together with the Krein-
Rutman theorem. These eigenvalue results can be used to study the existence
of positive classical solutions of nonlinear Poisson’s equations, see [15, Section 3].
However, the Newtonian potential kernel alone cannot be applied to the eigenvalue

2020 Mathematics Subject Classification. 35J05, 31A05, 31B05, 35J08, 47A75.
Key words and phrases. Eigenvalue; Laplace’s equation; Poisson’s equation; Green’s function;

Hammerstein integral operator.
©2021. This work is licensed under a CC BY 4.0 license.

Submitted July 12, 2021. Published October 18, 2021.

1



2 G. YANG, K. LAN EJDE-2021/??

problems with n = 2 because the Newtonian potential kernel changes sign when
n = 2.

In this article, we study the above three problems via different approaches and
deal with the case n ≥ 2 in a unified setting. We only assume that Ω is a bounded
open subset in Rn, and the connectedness on Ω̄ and the smoothness on the boundary
of Ω are not required. Some previous results on existence of classical or weak
solutions of some linear or nonlinear elliptic boundary value problems required the
connectedness and smoothness, for example, see [1, p.633] and [7, 8, 11, 12, 13].

First, we study the Laplace’s equation in B̄ρ and in Ω̄, and obtain solutions of
the Laplace’s equations via Hammerstein integral operators Sδ,ρ with newly defined
kernels Φδ. We give properties of Φδ and Sδ,ρ including the domain of Φδ and
compactness of Sδ,ρ.

Next, we study solutions and nonnegative solutions of the Poisson’s equation in
Ω and give these solutions via Hammerstein integral operators Lδ,ρ with Green’s
functions kδ := Ψ+Φδ for v ∈ Cµ(Ω). These solutions Lδ,ρv are obviously different
from those given by Lv when n ≥ 3 and are new when n = 2.

Finally, we consider the eigenvalue problems of Laplace’s equation in Ω and allow
n = 2. We prove that the spectral radius of the linear integral operator Lg with
the kernel kδg is the eigenvalue of the Laplace’s equation in Ω. As a special case,
when n ≥ 3, we obtain a simple condition on g which ensures that the spectral
radius of Lg is the the eigenvalue of the Laplace’s equation.

In Section 2 of this paper, we provide some results on the Newtonian potential
kernel and the Newtonian potential operator, some of them are new. These results
will be used in Sections 3-5. In Section 3, we introduce the kernel Φδ and study its
properties. The properties of integral operator with the kernel Φδ are given. The
integral operator is then used to give solutions of the Laplace’s equation, and its
compactness will be used to obtain compactness of the integral operators involving
the Green’s functions kδ in Sections 4 and 5. In Section 4, we give solutions of the
Poisson’s equation via integral operators involving the Green’s functions kδ. The
result will be useful for studying the existence of nonzero nonnegative solutions of
the nonlinear Poisson’s equation. In Section 5, we show that the spectral radii of
the integral operators involving the weight Green’s functions are the eigenvalues of
the Laplace’s equations when n ≥ 2.

2. Newtonian potential operator

Let n ∈ N with n ≥ 2 and let Rn be the Euclidean Banach space with norm
|x| =

√∑n
i=1 x

2
i and inner product x · y =

∑n
i=1 xiyi. We always assume that Ω is

a bounded open subset in Rn. Note that Ω is required neither to be a connected
set in Rn nor to have any smoothness on ∂Ω.

We consider the Newtonian potential operator L defined by

(Lv)(x) =

∫
Ω

Ψ(x, y)v(y) dy for x ∈ Rn, (2.1)

where Ψ : Rn × Rn \ {(x, x) : x ∈ Rn} → R is the Newtonian potential kernel
defined by

Ψ(x, y) := −Γ(|x− y|) =

{
− 1

2π ln |x− y| if n = 2,
1

n(n−2)ωn|x−y|n−2 if n ≥ 3,
(2.2)



EJDE-2021/?? SOLUTIONS AND EIGENVALUES OF LAPLACE’S EQUATION 3

where Γ : (0,∞)→ R is defined by

Γ(u) =

{
lnu
2ω2

if n = 2,

− 1
n(n−2)ωnun−2 if n ≥ 3,

(2.3)

where ω2 = π, ωn = 2πn/2

nΓ0(n/2) is the volume of the unit ball in Rn for n ≥ 3, and Γ0

is the Gamma function Γ0(u) =
∫∞

0
su−1e−s ds.

The Newtonian potential operator was studied, for example, in [15], where Ω is
only required to be a bounded open subset in Rn, and in [3, 4], where Ω is a domain
(i.e. a connected open subset) in Rn with suitable smoothness on ∂Ω.

Notation. For ρ > 0 and x ∈ Rn, let Bρ(x) = {y ∈ Rn : |x − y| < ρ}, B̄ρ(x) =
{y ∈ Rn : |x− y| ≤ ρ} and ∂Bρ(x) = {y ∈ Rn : |x− y| = ρ}. We write

Bρ = Bρ(0), B̄ρ = B̄ρ(0), ∂Bρ = ∂Bρ(0).

Let p, q ∈ [1,∞] be the conjugate indices, that is, they satisfy

1/p+ 1/q = 1; (2.4)

and if p =∞, then q = 1; and if p = 1, then q =∞. Hence, if p ∈ (n/2,∞], then

q ∈

{
[1,∞) if n = 2,

[1, n
n−2 ) if n ≥ 3.

(2.5)

The following results on the Newtonian potential kernel can be found in [15,
Lemma 2.1] and will be used to prove Theorem 3.8 in Section 3.

Lemma 2.1. The Newtonian potential kernel Ψ has the following properties.

(1) Ψ(·, y) ∈ C∞(Rn \ {y}) for each y ∈ Rn.

(2) ∆xΨ(x, y) :=
∑n
i=1

∂2Ψ(x,y)
∂x2
i

= 0 for x, y ∈ Rn with x 6= y.

(3) If q satisfies (2.5), then Ψ(x, ·) ∈ Lq(Ω) for each x ∈ Rn.

Let D be a nonempty subset of Rn. We denote by F (D) the set of all the
functions from D → R. Let D1 ⊂ D be a nonempty subset and f ∈ F (D). We still
use f to denote the restriction of f on D1. For µ ∈ (0, 1), we denote by Cµ(D) the
vector space of all locally µ-Hölder continuous functions on D, see [1, p.629]. If D
is bounded and closed, then we denote by C(D̄) the Banach space of all continuous
functions from D̄ to R with the maximum norm ‖ · ‖.

We denote by Lp(Ω) and Lp+(Ω) the Banach space of functions for which the pth
power of the absolute values are Lebesgue integrable with norm ‖ · ‖Lp(Ω), and its
positive cone of all the nonnegative functions in Lp(Ω), respectively.

Proposition 2.2. If p ∈ (n/2,∞], then L maps Lp(Ω) into F (Rn).

Proof. Let n ≥ 2, p ∈ (n/2,∞] and v ∈ Lp(Ω). Let q satisfy (2.5). Let x ∈ Rn and
v ∈ Lp(Ω). By Lemma 2.1(3), for each x ∈ Ω̄, we have

|Lv(x)| ≤
∫

Ω

∣∣Ψ(x, y)
∣∣|v(y)| dy ≤

(∫
Ω

|Ψ(x, y)|q dy
)1/q

‖v‖Lp(Ω) <∞.

Hence, Lv ∈ F (Rn). �

Proposition 2.2 is given in the proof of [15, Theorem 2.1]. By Proposition 2.2, if
p ∈

(
n/2,∞], then L maps Lp(Ω) into F (D) for each nonempty subset D in Rn.

We need the following known results from [15, Theorems 2.1, 2.2 and 2.3].



4 G. YANG, K. LAN EJDE-2021/??

Lemma 2.3. The operator L defined in (2.1) has the following properties.

(1) If p ∈ (n/2,∞], then L maps Lp(Ω) into C(Ω̄).
(2) If p ∈ (n,∞], then L maps Lp(Ω) to C1(Rn).
(3) L maps Cµ(Ω) into C2(Ω).
(4) If v ∈ Cµ(Ω), then −∆(Lv)(x) = v(x) for each x ∈ Ω.

Remark 2.4. We note that the following question has not been solved yet.
Is L : Lp(Ω)→ C(Ω̄) compact for p ∈ (n/2,∞]?

We generalize Lemma 2.3(1) (that is, [15, Theorem 2.1]) from C(Ω̄) to C(Rn).

Theorem 2.5. If p ∈ (n/2,∞], then L maps Lp(Ω) into C(Rn).

Proof. Let D be a nonempty bounded open subset in Rn satisfying Ω ⊂ D. It
suffices to prove Lv ∈ C(D̄). We define a Hammerstein integral operator

(L∗v̄)(x) =

∫
D

Ψ(x, y)v̄(y) dy for x ∈ Rn.

By Lemma 2.3(1), for p ∈ (n/2,∞], L∗ maps Lp(D) into C(D̄). Let v ∈ Lp(Ω).
We define a function v̄ : D̄ → R by

v̄(y) =

{
v(y) if y ∈ Ω,

0 if y ∈ D̄ \ Ω.

Then

(L∗v̄)(x) =

∫
D

Ψ(x, y)v̄(y) dy =

∫
Ω

Ψ(x, y)v̄(y) dy +

∫
D\Ω

Ψ(x, y)v̄(y) dy

=

∫
Ω

Ψ(x, y)v(y) dy = (Lv)(x) for each x ∈ Rn.

This implies

(Lv)(x) = (L∗v̄)(x) for each x ∈ D̄.

This, together with L∗v̄ ∈ C(D̄), implies Lv ∈ C(D̄). �

Lemma 2.6 ([15, Lemma 2.4]). Let p ∈ (n/2,∞] and g ∈ Lp+(Ω). Then

lim
x→τ

∫
Ω

∣∣Ψ(x, y)−Ψ(τ, y)
∣∣g(y) dy = 0 for each τ ∈ Ω̄.

Lemma 2.7. Let p ∈ (n/2,∞] and g ∈ Lp+(Ω). Then the operator Lg defined by

Lgv(x) =

∫
Ω

Ψ(x, y)g(y)v(y) dy (2.6)

is a compact linear operator from C(Ω̄) to C(Ω̄).

Proof. (i) By Lemma 2.1(3), for each x ∈ Ω̄, we have

|Lgv(x)| ≤
∫

Ω

∣∣Ψ(x, y)
∣∣g(y)|v(y)| dy ≤

(∫
Ω

|Ψ(x, y)|q dy
)1/q

‖g‖Lp(Ω)‖v‖C(Ω̄) <∞.

By a proof similar to that of [5, Lemma 2.1] and applying Lemma 2.6, Lg maps
C(Ω̄) to C(Ω̄) and is compact. �

Note that the linear operator Lg in Lemma 2.7 is different from [15, Theorem
2.4], where the kernel is |Ψ(x, y)|.



EJDE-2021/?? SOLUTIONS AND EIGENVALUES OF LAPLACE’S EQUATION 5

3. Solutions of Laplace’s equation

In this section, we study solutions of the Laplace’s equation

∆u(x) = 0 for each x ∈ B̄ρ. (3.1)

A function u : B̄ρ → R is said to be a (classical) solution of (3.1) if u ∈ C(B̄ρ)∩
C2(Bρ) and u satisfies (3.1).

If Ω is a bounded open subset in Rn and Ω̄ ⊂ Bρ, then a solution of (3.1) is a
solution of the Laplace’s equation

∆u(x) = 0 for x ∈ Ω̄. (3.2)

Recall that a function u : Ω→ R is said to be a harmonic function in Ω if u ∈ C2(Ω)
and u satisfies ∆u(x) = 0 for x ∈ Ω, see [3, p.20] or [4, p.13]. Hence, every solution
of (3.2) is a harmonic function in Ω.

It is well known that Ψ(·, 0) is a harmonic function in Rn \ {0}, see [3, p.21-22],
[4, p.17], and [15, Lemma 2.1 (P3)]. At the end of this section, we shall provide
other harmonic functions via a Hammerstein integral operator.

Notation. For each δ > 0, let

yδ = δ2|y|−2y for y ∈ Rn \ {0}, (3.3)

Dδ = D1 ∪ (D2)δ, (3.4)

where D1 = {(x, 0) ∈ Rn × Rn : x ∈ Rn} and

(D2)δ =
{

(x, y) ∈ Rn × (Rn \ {0}) : y ∈ Rn \ {0} and x 6= yδ
}
,

(D0)δ = {(x, x) : x ∈ B̄δ}, (3.5)

r = max{|x| : x ∈ Ω̄}, δ > r, ρ ∈ [δ, δ2/r). (3.6)

For x, y ∈ Rn, we denote by (Dδ)1(x) and (Dδ)2(y) the cross sections of Dδ at x
and y, respectively. Then

(Dδ)1(x) = {y ∈ Rn : (x, y) ∈ Dδ}, (Dδ)2(y) = {x ∈ Rn : (x, y) ∈ Dδ}. (3.7)

Lemma 3.1. Both (Dδ)1(x) and (Dδ)2(y) are open subsets in Rn for x, y ∈ Rn.

Proof. It is easy to verify that for each x ∈ Rn,

(Dδ)1(x) = {0} ∪ {y ∈ Rn \ {0} : yδ 6= x}, (3.8)

(Dδ)2(y) =

{
Rn if y = 0 ∈ Rn,
Rn \ {yδ} if y ∈ Rn \ {0}.

(3.9)

It follows from (3.8) that (Dδ)1(0) = Rn and

Rn \ (Dδ)1(x) = {y ∈ Rn \ {0} : yδ = x} for x ∈ Rn \ {0}.
It is easy to verify that the set on the right-hand side of the above equation is a
closed set in Rn. Hence, (Dδ)1(x) is open in Rn for each x ∈ Rn. By (3.9), it is
obvious that (Dδ)2(y) is open in Rn for each y ∈ Rn. �

Let x ∈ Rn. For each i ∈ In := {1, · · · , n}, we write

x = (x1, · · · , xi−1, xi, xi+1, · · · , xn) = (xi, x̂i),

where x̂i = (x1, · · · , xi−1, xi+1, · · · , xn). For (x, y) ∈ Rn ×Rn and i ∈ In, we write

(x, y) = (xi, x̂i, y) = (x, yi, ŷi).



6 G. YANG, K. LAN EJDE-2021/??

We denote by Dδ(x̂i, y) and Dδ(x, ŷi) the cross sections of Dδ at (x̂i, y) and (x, ŷi),
respectively. By Lemma 3.1, we see that the cross sections Dδ(x̂i, y) and Dδ(x, ŷi)
are open in R. These open cross sections will be implicitly used in some partial
derivatives such as in the proof of Theorem 3.8.

The following result provides some useful subsets of Dδ.

Proposition 3.2. (1) If δ ≥ r, then (B̄δ2/r × Ω) ∪ (Bδ2/r × Ω̄) ⊂ Dδ.

(2) If δ > 0, then (B̄δ × B̄δ) \ (D0)δ ⊂ Dδ.

Proof. (1) Let δ ≥ r and (x, y) ∈ B̄δ2/r × Ω. If y = 0, then

(x, y) = (x, 0) ∈ D1 ⊂ Dδ.

If y 6= 0, then |x| ≤ δ2/r and y ∈ Ω. Since Ω is open, we have 0 < |y| < r. Hence,∣∣∣δ2|y|−2y
∣∣∣ = δ2/|y| > δ2/r ≥ |x|.

This implies x 6= δ2|y|−2y = yδ. By (3.4), (x, y) ∈ (D2)δ ⊂ Dδ.
Let (x, y) ∈ Bδ2/r × Ω̄. If y = 0, then (x, y) = (x, 0) ∈ D1 ⊂ Dδ. If y 6= 0, then

|x| < δ2/r and 0 < |y| ≤ r. Hence, we have∣∣δ2|y|−2y
∣∣ = δ2/|y| ≥ δ2/r > |x|.

This and (3.4), imply (x, y) ∈ (D2)δ ⊂ Dδ.
(2) Let (x, y) ∈ (B̄δ × B̄δ) \ (D0)δ. Then |x| ≤ δ, |y| ≤ δ and x 6= y. If y = 0,

then (x, y) = (x, 0) ∈ D1 ⊂ Dδ. If y 6= 0, then x 6= yδ. In fact, if not, then x = yδ.
Since x 6= y, we have yδ 6= y. Hence, |y| 6= δ. This, together with |y| ≤ δ, implies
|y| < δ. By x = yδ and |y| < δ, we have

|x| = |yδ| = δ2/|y| > δ2/δ = δ ≥ |x|,
a contradiction. Hence, (x, y) ∈ (D2)δ ⊂ Dδ. �

Remark 3.3. In the proof of Proposition 3.2, we see that we need the hypothesis
that Ω is open to prove B̄δ2/r × Ω ⊂ Dδ, but the inclusion Bδ2/r × Ω̄ ⊂ Dδ holds
for any bounded subset Ω ⊂ Rn.

Corollary 3.4. (1) If (3.6) holds, then

Ω̄× Ω̄ ⊂ B̄δ × Ω̄ ⊂ B̄ρ × Ω̄ ⊂ Bδ2/r × Ω̄ ⊂ Dδ. (3.10)

(2) If δ ≥ r, then

(Ω̄× Ω̄) \ (D0)δ ⊂ (B̄δ × Ω̄) \ (D0)δ ⊂ (B̄δ × B̄δ) \ (D0)δ ⊂ Dδ.

Proof. Since δ ≥ r, we have Ω̄ ⊂ B̄δ. The results (1) and (2) follow from Proposition
3.2 (1) and (2), respectively. �

For each δ > 0, we define a function ηδ : Rn × Rn → R by

ηδ(x, y) =
(
δ−1|x||y|

)2 − 2x · y + δ2. (3.11)

Lemma 3.5. For δ > 0, the function ηδ in (3.11) has the following properties.

(i) ηδ ∈ C∞(Rn × Rn).

(ii) ηδ(x, y) =

{
δ2 if (x, 0) ∈ Rn × Rn,
δ−2|y|2|x− yδ|2 if (x, y) ∈ Rn × (Rn \ {0}).

(iii) ηδ(x, y) > 0 for each (x, y) ∈ Dδ and ηδ(x, y) = 0 for each (x, y) ∈ Rn \Dδ.
(iv) ηδ(x, y) = δ−2

(
δ2 − |x|2

)(
δ2 − |y|2

)
+ |x− y|2 for x, y ∈ Rn.



EJDE-2021/?? SOLUTIONS AND EIGENVALUES OF LAPLACE’S EQUATION 7

Proof. (i) By (3.11), we see that ηδ : Rn × Rn → R is continuous, and

ηδ(x, y) = δ−2
[ n∑
i=1

y2
i

][ n∑
i=1

x2
i

]
− 2
[ n∑
i=1

xiyi

]
+ δ2.

Differentiating both sides of the above equation implies that for each i ∈ In,

∂ηδ(x, y)

∂xi
= 2
[
δ−2|y|2xi − yi

]
,

∂2ηδ(x, y)

∂x2
i

= 2δ−2|y|2

and
∂ηδ(x, y)

∂yi
= 2
[
δ−2|x|2yi − xi

]
,

∂2ηδ(x, y)

∂y2
i

= 2δ−2|x|2.

Moreover, all other partial derivatives are 0. It is easy to see that all of the partial
derivatives are continuous on Rn × Rn. Hence, the result (i) holds.

(ii) Let x ∈ Rn and y ∈ Rn. If y = 0, then by (3.11), we have ηδ(x, y) = δ2. If
y 6= 0, then by (3.11), we have

ηδ(x, y) =
(
δ−1|x||y|

)2 − 2x · y + δ2 =
|y|2

δ2

(
|x|2 − 2x · yδ + |yδ|2

)
=
|y|2

δ2
|x− yδ|2

and the result (ii) holds.
(iii) The result follows from the result (ii).
(iv) Since

|x− y|2 = |x|2 − 2x · y + |y|2 for x, y ∈ Rn,
for x, y ∈ Rn we have

ηδ(x, y) =
(
δ−1|x||y|

)2
+ δ2 − 2x · y =

(
δ−1|x||y|

)2
+ δ2 + |x− y|2 − |x|2 − |y|2

= δ−2
(
δ2 − |x|2

)(
δ2 − |y|2

)
+ |x− y|2

and the result holds. �

From Lemma 3.5, we obtain the following result.

Corollary 3.6. Let δ > 0 and y ∈ Rn. Then (y, y) ∈ Dδ if and only if |y| 6= δ.

Proof. Let y ∈ Rn. It is easy to verify that if y ∈ Rn \ {0}, then

y = yδ if and only if |y| = δ.

Assume that (y, y) ∈ Dδ. If y = 0, then |y| = 0 6= δ. If y 6= 0, then by Lemma 3.5
(ii), y 6= yδ. It follows that |y| 6= δ. Conversely, if |y| 6= δ, then y = 0 or y 6= 0 and
y 6= yδ. It follows from Lemma 3.5 (ii) that (y, y) ∈ Dδ. �

With ρ > 0 and the function ηδ defined in (3.11), we define a kernel function
Φδ : Dδ → R by

Φδ(x, y) = Γ(
√
ηδ(x, y)). (3.12)

By Lemma 3.5 (iii), ηδ(x, y) > 0 only when (x, y) ∈ Dδ. This, together with (2.3),

implies that Γ(
√
ηδ(x, y)) exists only when (x, y) ∈ Dδ. Hence, Dδ is the natural

domain of Φδ.

Remark 3.7. By Corollary 3.4, if δ > r, then Ω̄× Ω̄ ⊂ Dδ. By Proposition 3.2 (1)
with δ = r and Ω = Br and Lemma 3.5 (iv) with δ = r, x = y and |x| = δ, we see
that

Br ×Br ⊂ Br × B̄r ⊂ Dr, B̄r × B̄r 6⊂ Dr.



8 G. YANG, K. LAN EJDE-2021/??

Hence, the natural domain of Φδ contains Ω̄ × Ω̄ if δ > r, but the natural domain
of Φr does not contain B̄r × B̄r.

Theorem 3.8. For δ > 0, the function Φδ has the following properties.

(i) Φδ : Dδ → R is continuous.
(ii) Φδ(·, y) ∈ C∞

(
(Dδ)2(y)

)
for each y ∈ Rn.

(iii) ∆xΦδ(x, y) = 0 for (x, y) ∈ Dδ.

Proof. (i) By Lemma 3.5 (i), ηδ : Rn×Rn → R is continuous. By (2.3), Γ : (0,∞)→
R is continuous. These, together with Lemma 3.5 (iii), imply that Φδ : Dδ → R is
continuous.

(ii) By (2.3), (3.11) and (3.12), we have for (x, y) ∈ Dδ,

Φδ(x, y) =


Γ(δ) if y = 0,

(2π)−1 ln(δ−1|y|)−Ψ(x, yδ) if y 6= 0, n = 2,

−(δ|y|−1)n−2Ψ(x, yδ) if y 6= 0, n ≥ 3.

(3.13)

This, together with Lemma 2.1 (1), implies that the result (ii) holds.
(iii) By (3.13), we have for each i ∈ In,

∂2Φδ(x, y)

∂x2
i

=


0 if y = 0,

−∂
2Ψ(x,yδ)
∂x2
i

if y 6= 0, n = 2,

−(δ|y|−1)n−2 ∂
2Ψ(x,yδ)
∂x2
i

if y 6= 0, n ≥ 3.

It follows that for (x, y) ∈ Dδ,

∆xΦδ(x, y) =


0 if y = 0,

−
∑n
i=1

∂2Ψ(x,yδ)
∂x2
i

if y 6= 0, n = 2,

−(δ|y|−1)n−2
∑n
i=1

∂2Ψ(x,yδ)
∂x2
i

if y 6= 0, n ≥ 3.

(3.14)

Note that (x, y) ∈ Dδ with y 6= 0 implies x 6= yδ. By Lemma 2.1 (2),

∆xΨ(x, yδ) =

n∑
i=1

∂2Ψ(x, yδ)

∂x2
i

= 0 for (x, y) ∈ Dδ with y 6= 0.

This and (3.14) imply ∆xΦδ(x, y) = 0 for (x, y) ∈ Dδ. �

Corollary 3.9. If (3.6) holds, then the following assertions hold.

(i) Φδ ∈ C∞(B̄ρ × Ω̄).
(ii) ∆xΦδ(x, y) = 0 for (x, y) ∈ B̄ρ × Ω̄.

Proof. (i) By Corollary 3.4, we have B̄ρ × Ω̄ ⊂ Dδ. By Lemma 3.5 (i) and (iii),
ηδ ∈ C∞(B̄ρ × Ω̄). Since Γ, y ∈ C∞(0,∞), where y(x) =

√
x for x ∈ (0,∞). It

follows that ΦδΓ(y(ηδ)) ∈ C∞(B̄ρ × Ω̄).
(ii) Since B̄ρ × Ω̄ ⊂ Dδ, by Theorem 3.8 (ii), the result (ii) holds. �

Let δ and ρ satisfy (3.6). With the kernel Φδ given in (3.12), we study the
Hammerstein integral operator

(Sδ,ρv)(x) =

∫
Ω

Φδ(x, y)v(y) dy for x ∈ B̄ρ, (3.15)

where v : Ω→ R is a function.

Theorem 3.10. If (3.6) holds, then the following assertions hold.



EJDE-2021/?? SOLUTIONS AND EIGENVALUES OF LAPLACE’S EQUATION 9

(1) Sδ,ρ maps L1(Ω) to C∞(B̄ρ).
(2) Sδ,ρ : L1(Ω)→ C(B̄ρ) is compact.

Proof. (1) By Corollary 3.9 (i), Φδ ∈ C∞(B̄ρ × Ω̄). This, (3.15), and the Leibniz
integral rule (see [9, Lemma 2.2, p.226]), imply Sδ,ρv ∈ C∞(B̄ρ) for v ∈ L1(Ω).

(2) By Corollary 3.9 (i), Φδ : B̄ρ × Ω̄ → R is continuous. The result (2) follows
from [5, Lemma 2.1]. �

Theorem 3.11. If (3.6) holds, then for each v ∈ L1(Ω), the function uδ,ρ : B̄ρ → R
defined by

uδ,ρ(x) = (Sδ,ρv)(x) =

∫
Ω

Φδ(x, y)v(y) dy for each x ∈ B̄ρ

is a solution of (3.1).

Proof. By Theorem 3.10, we have

uδ,ρ = Sδ,ρv ∈ C(B̄ρ) ∩ C2(Bρ).

By Lemma 3.9 (i), Φδ ∈ C∞(B̄ρ × Ω̄). By [9, Lemma 2.2, p.226] and Lemma 3.9
(i), we have for each x ∈ B̄ρ,

∆uδ,ρ(x) = ∆

∫
Ω

Φδ(x, y)v(y)dy =

∫
Ω

∆xΦδ(x, y)v(y)dy. (3.16)

By Corollary 3.9 (ii), we have

∆xΦδ(x, y) = 0 for (x, y) ∈ B̄ρ × Ω̄.

This and (3.16), imply that uδ,ρ satisfies (3.1). �

As a special case of Theorem 3.11, we give solutions of the Laplace’s equation
(3.2).

Corollary 3.12. For v ∈ L1(Ω) and δ > r, the function uδ : Ω̄→ R defined by

uδ(x) =

∫
Ω

Φδ(x, y)v(y) dy for each x ∈ Ω̄

is a solution of (3.2).

This corollary provides harmonic functions via functions in L1(Ω), where Ω is
not necessarily a domain. As mentioned in the Introduction, Ψ(·, 0) is a harmonic
function in Rn \ {0} given in [3, p.21-22], [4, p.17], and [15, Lemma 2.1 (P3)].

4. Solutions to Poisson’s equation

In this section, we study solutions of the Poisson’s equation

−∆u(x) = v(x) for each x ∈ Ω, (4.1)

where Ω is a bounded open subset in Rn, n ≥ 2 and v ∈ Cµ(Ω).

Definition 4.1. Let Ω̄ ⊂ D. A function u : D → R is said to be a (classical)
solution of (4.1) if u ∈ C2(Ω) ∩ C(Ω̄) and u satisfies (4.1). A solution u of (4.1) is
said to be nonnegative if u ∈ P , where P is the cone in C(Ω̄) defined by

P = {u ∈ C(Ω̄) : u(x) ≥ 0 for x ∈ Ω̄}. (4.2)



10 G. YANG, K. LAN EJDE-2021/??

A classical result [4, Lemma 4.2 ] shows that if Ω is a bounded connected open
in Rn and v ∈ Cµ(Ω), then Lv is a solution of (4.1). This result was generalized
to the case that Ω is a bounded open subset in Rn in [15, Theorem 2.3 (2)]. In
the following, we provide other solutions involving an integral operator with the
Green’s function in bounded open subsets in Rn. We show that some of solutions
u of (4.1) satisfy the Dirichlet boundary condition

u(x) = 0 for x ∈ ∂Bδ, (4.3)

where δ is the same as in (3.6).
For each δ > 0, we define a function kδ : Dδ \ {(x, x) : x ∈ Rn} → R by

kδ(x, y) = Ψ(x, y) + Φδ(x, y), (4.4)

where Ψ and Φδ are the same as in (2.2) and (3.12). If Ω is a domain (a connected
open subset in Rn), then following [4, p.19], kδ : Ω̄ × Ω̄ \ {(x, x) : x ∈ Rn} → R
is called the (Dirichlet) Green’s function for Ω. When Ω = Bδ, the expression of
the Green’s function kδ is given and studied in [4, (2.23), p.19], where G(x, y) =
−kδ(x, y). In the following, we study the Green’s function kδ in (4.4) for a general
bounded open subset Ω in Rn.

Lemma 4.2. The Green’s function kδ in (4.4) has the following properties.

(i) kδ : Dδ \ {(x, x) : x ∈ Rn} → R is continuous.
(ii) If δ > r, then the following assertions hold.

(ii.1) kδ(x, y) ≥ 0 for (x, y) ∈ B̄δ × B̄δ \ (D0)δ.
(ii.2) kδ(x, y) > 0 for (x, y) ∈ Bδ ×Bδ \ (D0)δ.
(ii.3) kδ(x, y) = 0 for (x, y) ∈ (∂Bδ × B̄δ) ∪ (B̄δ × ∂Bδ) \ (D0)δ.

Proof. (i) By (2.2), Ψ is continuous at (x, y) ∈ Rn × Rn with x 6= y. By Theorem
3.8 (i), Φδ : Dδ → R is continuous. The result follows.

(ii.1) Let (x, y) ∈ B̄δ × B̄δ with x 6= y. Then |x| ≤ δ and |y| < δ. By Lemma 3.5
(ii), we have

ηδ(x, y) ≥ |x− y|2,
√
ηδ(x, y) ≥ |x− y|.

Because Γ is increasing on (0,∞), we have

kδ(x, y) = Γ(
√
ηδ(x, y))− Γ(|x− y|) ≥ 0.

(ii.2) Let (x, y) ∈ Bδ ×Bδ with x 6= y. By Lemma 3.5 (iv), we have

ηδ(x, y) > |x− y|2,
√
ηδ(x, y) > |x− y|.

Because Γ is increasing on (0,∞), we have

kδ(x, y) = Γ(
√
ηδ(x, y))− Γ(|x− y|) > 0.

(ii.3) Let (x, y) ∈ (∂Bδ × B̄δ) ∪ (B̄δ × ∂Bδ) with x 6= y. By Lemma 3.5 (iv), we

have ηδ(x, y) = |x− y|2 and
√
ηδ(x, y) = |x− y|. Hence,

kδ(x, y) = Γ(
√
ηδ(x, y))− Γ(|x− y|) = 0

and the result holds. �

With (3.6), we define an integral operator Lδ,ρ by

(Lδ,ρv)(x) = (Lv)(x) + (Sδ,ρv)(x) =

∫
Ω

kδ(x, y)v(y) dy for x ∈ B̄ρ. (4.5)



EJDE-2021/?? SOLUTIONS AND EIGENVALUES OF LAPLACE’S EQUATION 11

Theorem 4.3. Assume that (3.6) holds. Then the operator Lδ,ρ in (4.5) has the
following properties.

(1) If p ∈ (n/2,∞], then Lδ,ρ maps Lp(Ω) into C(B̄ρ). Moreover, for each
v ∈ Lp(Ω), (Lδ,δv)(x) = 0 for x ∈ ∂Bδ.

(2) If p ∈ (n,∞], then Lδ,ρ maps Lp(Ω) to C1(B̄ρ).
(3) Lδ,ρ maps Cµ(Ω) into C2(Ω).

Proof. (1) Since p ∈ (n/2,∞], then by Theorem 2.5, L maps Lp(Ω) into C(Rn).
Since δ > r and ρ ∈ [δ, δ2/r), by Theorem 3.10 (1), Sδ,ρ maps L1(Ω) to C∞(B̄ρ).
It follows from (4.5) that Lδ,ρv ∈ C∞(B̄ρ) for v ∈ Lp(Ω). By Lemma 4.2 (3),

kδ(x, y) = 0 for (x, y) ∈ (∂Bδ × B̄δ) \ (D0)δ.

Hence, by (4.5) with ρ = δ, for each v ∈ Lp(Ω) we have

(Lδ,δv)(x) =

∫
Ω

kδ(x, y)v(y) dy = 0 for each x ∈ B̄δ.

(2) Since p ∈ (n,∞], by Lemma 2.3 (2), L maps Lp(Ω) to C1(Rn). By (4.5),
Lδ,ρv ∈ C1(B̄ρ) for v ∈ Lp(Ω).

(3) By Lemma 2.3 (3), L maps Cµ(Ω) into C2(Ω). By Theorem 3.10 (1), Sδ,ρ
maps L1(Ω) to C∞(B̄ρ). Since Ω ⊂ B̄ρ, by (4.5), Lδ,ρv ∈ C2(Ω) for v ∈ Cµ(Ω). �

Theorem 4.4. Assume that (3.6) holds. Then the following assertions hold.

(i) If v ∈ Cµ(Ω), then Lδ,ρv is a solution of (4.1). Moreover, Lδ,δv is a
solution of (4.1) subject to (4.3).

(ii) If v ∈ P ∩Cµ(Ω), then Lδ,ρv is a nonnegative solution of (4.1). Moreover,
Lδ,δv is a nonnegative solution of (4.1) subject to (4.3).

Proof. (i) By Theorem 4.3 (1) and (3), Lδ,ρv ∈ C2(Ω)∩C(Ω̄) for v ∈ Cµ(Ω). Since
δ > r and ρ ∈ [δ, δ2/r), by Theorem 3.11, we have for v ∈ Cµ(Ω),

∆(Sδ,ρv)(x) = 0 for each x ∈ B̄ρ.

By Lemma 2.3 (4),

−∆(Lv)(x) = v(x) for each x ∈ Ω.

Hence, we have for each x ∈ Ω,

−∆(Lδ,ρv)(x) = −∆(Lv)(x)−∆(Sδ,ρv)(x) = v(x)

and Lδ,ρv is a solution of (4.1). By the last result of Theorem 4.3 (1), the solution
Lδ,δv of (4.1) satisfies (4.3).

(ii) By Lemma 4.2 (1), we have

kδ(x, y) ≥ 0 for (x, y) ∈ B̄δ × B̄δ \ (D0)δ.

Since v ∈ P , it follows that Lδ,ρv(x) ≥ 0 for x ∈ Ω̄. The last result follows from
the last result of Theorem 4.3 (1). �

Theorem 4.4 gives solutions (Lv)+(Sδ,ρv) of (4.1) which are different from those
Lv obtained in [4, Lemma 4.2 ] and [15, Theorem 2.3 (2)].



12 G. YANG, K. LAN EJDE-2021/??

5. Eigenvalues of Laplace’s equations

We study the eigenvalue problem of the Laplace’s equation

−∆u(x) = µg(x)u(x) for x ∈ Ω, (5.1)

where Ω is a bounded open subset in Rn, n ≥ 2 and g : Ω̄→ R is a function.
The eigenvalue problem is to determine that under what conditions on g, there

exist µ > 0 and u ∈ P \ {0} such that (5.1) holds, where P is the same as in (4.2).
If n ≥ 3, p ∈ (n/2,∞] and g ∈ Lp+(Ω), the eigenvalue problem can be solved by

[15, Theorem 2.4], with the well known Krein-Rutman theorem, where the operator
Lg in (2.6) is used. However, the method cannot be applied for n = 2 because the
Newtonian potential kernel Ψ in (2.2) changes sign.

In the following, we study the eigenvalue problem (5.1) using the linear Ham-
merstein integral operator

(Lgv)(x) =

∫
Ω

kδ(x, y)g(y)v(y) dy for x ∈ Ω̄. (5.2)

Proposition 5.1. Let n ≥ 2, p ∈ (n/2,∞], g ∈ Lp+(Ω) and δ > r. Then the linear

integral operator Lg defined by (5.2) is a compact operator from C(Ω̄) to C(Ω̄)
satisfying Lg(P ) ⊂ P .

Proof. (i) Since n ≥ 2, p ∈ (n/2,∞] and g ∈ Lp+(Ω), by Lemma 2.7, Lg is a compact

operator from C(Ω̄) to C(Ω̄). We define an operator (Sδ,ρ)g by

(Sδ,ρ)gv(x) =

∫
Ω

Φδ(x, y)g(y)v(y) dy for x ∈ Ω̄. (5.3)

Since δ > r, by Theorem 3.10 (2) with ρ = δ, Sδ,ρ : L1(Ω)→ C(B̄δ) is compact. It
follows from Ω̄ ⊂ B̄ρ that Sδ,ρ : L1(Ω) → C(Ω̄) is compact. It is obvious that the
map T defined by

(Tv)(x) = g(x)v(x)

is continuous from C(Ω̄) to L1(Ω). Hence, (Sδ,ρ)g = Sδ,ρT is compact from C(Ω̄) to
C(Ω̄). Noting that Lg = Lg + (Sδ,ρ)g, we see that Lg : C(Ω̄)→ C(Ω̄) is compact.
By Corollary 3.4 (2) and Lemma 4.2 (1), we have

(Ω̄× Ω̄) \ (D0)δ ⊂ (B̄δ × B̄δ) \ (D0)δ, kδ(x, y) ≥ 0 for (x, y) ∈ B̄δ × B̄δ \ (D0)δ.

This implies

kδ(x, y) ≥ 0 for (x, y) ∈ (Ω̄× Ω̄) \ (D0)δ. (5.4)

Since g ∈ Lp+(Ω), it follows from (5.2) and (5.4) that

Lgv(x) ≥ 0 for v ∈ P, x ∈ Ω̄,

and Lgv ∈ P for v ∈ P . �

The following Krein-Rutman theorem can be found in [10].

Lemma 5.2. Assume that P is a total cone in a real Banach space X and L :
X → X is a compact linear operator such that L(P ) ⊂ P and r(L) > 0. Then there
exists an eigenvector u ∈ P \ {0} such that r(L)u = Lu.

It is well known that the cone P defined in (4.2) is a total cone in C(Ω̄).



EJDE-2021/?? SOLUTIONS AND EIGENVALUES OF LAPLACE’S EQUATION 13

Theorem 5.3. Let n ≥ 2, p ∈ (n/2,∞], g ∈ Lp+(Ω) and δ > r. Assume that there

exists a measurable set Ω0 ⊂ Ω̄ with meas(Ω0) > 0 such that

γ := inf
{∫

Ω0

kδ(x, y)g(y) dy : x ∈ Ω0

}
> 0. (5.5)

Then the following assertions hold.

(i) r(Lg) > 0, where r(Lg) = limm→∞ m

√
‖Lm

g ‖ is the spectral radius of Lg,

(ii) There exists an eigenvector u ∈ P \ {0} such that

−∆u(x) =
1

r(Lg)
g(x)u(x) for x ∈ Ω, (5.6)

where P is the same as in (4.2).

Proof. (i) The proof is similar to that of [15, Theorem 2.4]. Let u(x) ≡ 1 for x ∈ Ω̄.
Then

(Lgu)(x) =

∫
Ω

kδ(x, y)g(y)u(y) dy ≥
∫

Ω0

kδ(x, y)g(y) dy ≥ γ for x ∈ Ω0.

Since Lgu ∈ P , for x ∈ Ω0 we have

L 2
g u(x) =

∫
Ω

kδ(x, y)g(y)[Lgu(y)] dy ≥
∫

Ω0

kδ(x, y)g(y)[Lgu(y)] dy ≥ γ2.

Repeating the process implies Lm
g u(x) ≥ γm and r(Lg) ≥ γ.

(ii) It is well known that P is a total cone in C(Ω̄). The result follows from
Lemma 5.2, Proposition 5.1 and the result (i). �

Theorem 5.3 is different from [15, Theorem 2.4], where kδ is replaced by |Ψ|.
The condition (5.5) depends on Green’s function kδ. In the following, we provide a
sufficient condition for (5.5) with n ≥ 3 to hold, which is independent of kδ and is
easily verified. To do that, we first prove the following result.

Lemma 5.4. Let n ≥ 2 and 0 < σ < δ <∞. Then the following assertions hold.

(i) Φδ(x, y) ≥ Γ
(
δ−1
√
δ2 − σ2

)
for (x, y) ∈ B̄σ × B̄σ.

(ii) If n ≥ 3, then

kδ(x, y) ≥ Γ
(
δ−1
√
δ2 − σ2

)
for (x, y) ∈ B̄σ × B̄σ \ (D0)δ.

(iii) If n = 2, then

kδ(x, y) ≥ 1

4π
ln
[
1 +

δ−2
(
δ2 − σ2

)2
|x− y|2

]
for (x, y) ∈ B̄σ × B̄σ \ (D0)δ.

Proof. By Lemma 3.5 (iv), for x, y ∈ B̄σ we have

ηδ(x, y) = δ−2
(
δ2 − |x|2

)(
δ2 − |y|2

)
+ |x− y|2

≥ δ−2
(
δ2 − σ2

)(
δ2 − σ2

)
+ |x− y|2

= δ−2
(
δ2 − σ2

)2
+ |x− y|2.

(5.7)

(i) By (5.7), we have

ηδ(x, y) ≥ δ−2
(
δ2 − σ2

)
for (x, y) ∈ B̄σ × B̄σ.

Since Γ is increasing on (0,∞), we have

Φδ(x, y) = Γ
(√

ηδ(x, y)
)
≥ Γ

(
δ−1
√
δ2 − σ2

)
for (x, y) ∈ B̄σ × B̄σ.



14 G. YANG, K. LAN EJDE-2021/??

(ii) Since n ≥ 3, by (2.2) we have

Ψ(x, y) =
1

n(n− 2)ωn

1

|x− y|n−2
≥ 0 for (x, y) ∈ B̄δ × B̄δ \ (D0)δ.

This, (4.4) and the result (i), imply

kδ(x, y) ≥ Φδ(x, y) ≥ Γ
(
δ−1
√
δ2 − σ2

)
for x, y ∈ B̄σ × B̄σ \ (D0)δ.

(iii) Since n = 2, by (2.2) and (4.4), for (x, y) ∈ Dδ \ (D0)δ, we have

kδ(x, y) = Ψ(x, y) + Φδ(x, y) = − 1

2π
ln |x− y|+ 1

2π
ln Γ

(√
ηδ(x, y)

)
=

1

4π

[
ln ηδ(x, y)− ln |x− y|2

]
=

1

4π
ln
ηδ(x, y)

|x− y|2
.

This and (5.7), imply that

kδ(x, y) =
1

4π
ln
ηδ(x, y)

|x− y|2
≥ 1

4π
ln
δ−2
(
δ2 − σ2

)2
+ |x− y|2

|x− y|2

≥ 1

4π
ln
[
1 +

δ−2
(
δ2 − σ2

)2
|x− y|2

]
for (x, y) ∈ B̄σ × B̄σ \ (D0)δ.

and result (iii) holds. �

Corollary 5.5. Let n ≥ 3, p ∈ (n/2,∞] and g ∈ Lp+(Ω) with
∫

Ω
g(y) dy > 0. Then

r(Lg) > 0 and there exists an eigenvector u ∈ P \ {0} such that (5.6) holds.

Proof. Let r < δ and σ ∈ [r, δ). By Lemma 5.4 (ii), we have

kδ(x, y) ≥ Γ
(
δ−1
√
δ2 − σ2

)
for (x, y) ∈ B̄σ × B̄σ \ (D0)δ.

Since σ ≥ r, we have Ω̄ ⊂ B̄r ⊂ B̄σ. Hence,∫
Ω

kδ(x, y)g(y) dy ≥ Γ
(
δ−1
√
δ2 − σ2

) ∫
Ω

g(y) dy > 0 for x ∈ Ω̄.

This implies

γ = inf
{∫

Ω

kδ(x, y)g(y) dy : x ∈ Ω̄
}
≥ Γ

(
δ−1
√
δ2 − σ2

) ∫
Ω

g(y) dy > 0.

The results follow from Theorem 5.3 (ii). �

By Lemma 5.4 (iii), we see that when n = 2, kδ has no positive lower bound on
the set B̄σ× B̄σ \ (D0)δ since |x− y|2 may tend to zero on B̄σ× B̄σ \ (D0)δ. Hence,
it is not clear whether Corollary 5.5 holds when n = 2.

Acknowledgments. The author would like to thank the reviewers and handling
editor very much for providing valuable comments.

G. Yang was supported by the Applied Basic Research Project of Sichuan Province
(No. 2018JY0169). K. Lan was supported by the Natural Sciences and Engineering
Research Council of Canada under grant No. 135752-2018.



EJDE-2021/?? SOLUTIONS AND EIGENVALUES OF LAPLACE’S EQUATION 15

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Nonlinear Anal. 196 (2020), 111811, 20 pp.

Guangchong Yang

College of Applied Mathematics, Chengdu University of Information Technology,
Chengdu, Sichuan 610225, China

Email address: gcyang@cuit.edu.cn

Kunquan Lan

Department of Mathematics, Ryerson University, Toronto, Ontario, Canada M5B 2K3

Email address: klan@ryerson.ca


	1. Introduction
	2. Newtonian potential operator
	3. Solutions of Laplace's equation
	4. Solutions to Poisson's equation
	5. Eigenvalues of Laplace's equations
	Acknowledgments

	References