1 Introduction
In this article, we consider the following elliptic problem involving both critical Choquard and discontinuous and singular nonlinearities. Precisely, we deal with
(
P
λ
)
−
Δ
u
=
λ
∫
Ω
|
u
(
y
)
|
2
μ
*
|
x
−
y
|
μ
d
y
|
u
|
2
μ
*
−
2
u
+
χ
{
u
<
a
}
u
−
γ
,
u
>
0
in
Ω
,
u
≡
0
on
∂
Ω
,
where Ω is a bounded domain of
R
n
with smooth boundary ∂Ω, γ > 0, n ≥ 3, a > 0, λ > 0,
2
μ
*
=
(
2
n
−
μ
)
/
(
n
−
2
)
, 0 < μ < n and χ
A
denotes the characteristic function of a set A.
An important obstacle in investigating this class of problems is that the corresponding energy functional is nondifferentiable due to discontinuous nonlinearity. Therefore, we utilize the idea of generalized gradients as explained in the important work of Clarke ([1]), which was later applied to the setting of partial differential equations by Chang [2].
The occurrence of discontinuous nonlinearities arises in the modeling of several physical issues, including the obstacle problem, the seepage surface problem, and the Elenbass equation, for further information, see [3], [4]. These significant applications have driven a long series of investigations on problems involving such nonlinearities. We mention the pioneering work of Badiale and Tarentello [5], where existence and multiplicity results are established in the situation of critical growth and discontinuous nonlinearities in
R
n
with n ≥ 3. We also quote further papers that consider different varieties of diffusion operators and nonlinearities, see [6], [7], [8], [9].
The problems involving Choquard-type nonlinearity are widely studied since these problems found their applications in various physical phenomena. First, Pekar [10] used such kind of nonlinearities to describe the quantum mechanics of a polaron at rest whereas Choquard [11] described the model of an electron trapped in its own hole using such nonlinearity. One of the initial studies of problems involving Choquard nonlinearities using variational methods was conducted by Lieb [11] wherein he established the existence and uniqueness of a positive radial ground state of the following problem
−
Δ
u
+
V
0
u
=
I
2
∗
|
u
|
2
u
in
R
3
,
where
I
μ
(
x
)
=
A
μ
|
x
|
n
−
μ
with
A
μ
=
Γ
n
−
μ
2
2
μ
Γ
μ
2
π
n
2
. Without any attempt to provide the complete list, we refer to [12], [13], [14], [15], [16], and the references therein for the study of Choquard problems using variational methods.
The problems involving singular nonlinearities have a very long history. These type of problems has numerous applications in the physical world such as in the study of non-Newtonian flows in porous media and heterogeneous catalysts. One of the seminal breakthroughs in the study of such problems was the work of Crandall, Rabinowitz and Tartar [17]. By applying the method of sub-supersolutions to the nonsingular approximated problem and then passing to the limit, the authors proved the existence of a solution to a class of elliptic PDEs involving a singular nonlinearity. Following this pioneering work, a significant amount of study has been conducted on elliptic singular equations about the existence and qualitative properties of solutions. In this regard, we refer to the survey articles [18], [19] and references therein.
The investigation of singular problems in combination with critical growth nonlinearities was pioneered by Haitao [20] wherein the author considered the following problem:
(1.1)
−
Δ
u
=
λ
u
−
γ
+
u
p
,
u
>
0
in
Ω
,
u
=
0
on
∂
Ω
,
where
Ω
⊂
R
n
(
n
≥
3
)
is a smooth bounded domain, γ ∈ (0, 1) and
1
<
p
≤
n
+
2
n
−
2
. Using monotone iterations and the mountain pass lemma, the author proved the existence and multiplicity results for the maximal range of parameter λ (i.e. global multiplicity). Later in [7], [21], the authors studied such problems for the higher singular cases, i.e. with γ ∈ (1, 3). Finally, Hirano, Saccon and Shioji in [22] handled problem (1.1) for any γ > 0 and showed the existence of
L
loc
1
solutions u satisfying
(
u
−
ϵ
)
+
∈
H
0
1
(
Ω
)
for all ϵ > 0 using variational methods and nonsmooth analysis arguments. We also mention the work [23] where the authors studied a doubly nonlocal critical singular problem in the spirit of [22] and obtained the existence, multiplicity and regularity results.
The stirring motivation to consider the problem
(
P
λ
)
are the works [7], [8], where the authors discussed the problem involving critical nonlinearity with singular and discontinuous nonlinearities for n = 2 and n ≥ 3 respectively. More precisely, in [8] the authors considered the following problem
−
Δ
u
=
λ
χ
{
u
<
a
}
u
−
γ
+
u
2
*
−
1
,
u
>
0
in
Ω
,
u
=
0
on
∂
Ω
,
for 0 < γ < 3, λ > 0, 2* = 2n/(n − 2) is the critical Sobolev constant and obtain the existence and multiplicity results for a suitable range of λ.
Following the above discussion, we considered the problem
(
P
λ
)
in the present work. The novelty features of this work are doubled with the presence of a nonlocal and critical Hartree-type nonlinearity and a more singular nonlinearity (by considering higher values of γ > 1 with respect to former contributions). This brings additional technical difficulties and forces us to follow a new approximation approach. Based on the notion of weak solutions given in the next section, we prove the following existence and multiplicity result:
Theorem 1.1.
For any a > 0, there exists Λ
a
> 0 such that
(
P
λ
)
has no solution for any λ > Λ
a
.
(
P
λ
)
admits at least one (minimal) solution v
λ
for any λ ∈ (0, Λ
a
) and γ > 0. Moreover for any
ω
∈
(
max
{
γ
+
1
4
,
1
}
,
∞
)
,
v
λ
ω
∈
H
0
1
(
Ω
)
. In addition,
v
λ
∈
H
0
1
(
Ω
)
if and only if γ < 3.
Further if we take 0 < γ < 3 and μ < min{4, n}, then
(
P
λ
)
admits at least two solutions for any λ ∈ (0, Λ
a
).
We also discuss the boundary behaviour and Hölder regularity results of weak solutions. We have the following result in this direction.
Theorem 1.2.
Let u be a weak solution of
(
P
λ
)
and ϕ
γ
be given by Definition 2.3. Then u ∈ L
∞(Ω) and C
1
ϕ
γ
≤ u ≤ C
2
ϕ
γ
for some positive constants C
1 and C
2. Moreover, the following assertions hold:
When γ > 1,
u
(
x
)
∈
C
2
γ
+
1
(
Ω
̄
)
.
When γ = 1,
u
(
x
)
∈
C
β
(
Ω
̄
)
, for all 0 < β < 1.
When γ < 1,
u
(
x
)
∈
C
1,1
−
γ
(
Ω
̄
)
.
The problems involving discontinuous nonlinearities but without singular term are tackled using variational techniques and the generalized gradient theory for locally Lipschitz functional. However, the presence of a singular term makes the associated energy functional neither differentiable nor locally Lipschitz in
H
0
1
(
Ω
)
which prohibits the use of both techniques directly. To overcome these difficulties we first considered the regularized problem
(
P
λ
,
ϵ
)
(see Section 4). The use of this regularization makes the associated energy functional differentiable and thus allow the use of suitable variational methods.
We begin our analysis by studying the purely singular discontinuous problem
(
S
λ
)
(see Section 3). To this attempt, we consider the regularized problem
(
S
λ
,
ϵ
)
. The analysis of
(
S
λ
,
ϵ
)
is divided into two cases depending up on the parameter γ i.e., when (a) 0 < γ < 3 and (b) γ ≥ 3. For Case (a), we applied Perron’s method and show the existence of the unique solution to
(
S
λ
,
ϵ
)
in
H
0
1
(
Ω
)
∩
L
∞
(
Ω
)
. Concerning Case (b), we make use of monotone methods to obtain the existence of a unique solution of
(
S
λ
,
ϵ
)
. The existence of the minimal weak solution of
(
S
λ
)
is then obtained as the limit of solutions of the regularized problem. After studying the purely singular problem, we then show the existence of a weak solution of
(
P
λ
)
for a suitable range of λ taking advantage of the construction of suitable sub and supersolutions. Under the restriction μ < min{4, n}, this solution is then shown to be the local minimum of the energy functional in
H
0
1
(
Ω
)
topology. Then the existence of a second solution is obtained by investigating the translated problem associated to
(
P
λ
)
. The associated energy functional is locally Lipschitz which leads to the use of generalized gradients technique. We further employ Ekeland variational principle and the concentration-compactness principle to get the existence of a second solution. We point out here that the nonsmooth analysis arguments as performed in [22] cannot be used here because of the discontinuous term.
Turning to the structure of the paper, in Section 2 we collect the preliminaries required in the subsequent sections. In Section 3 we study the purely singular discontinuous problem. In Section 4, we obtain the existence of first solution. In Section 5, the existence of the second solution is discussed that achieves the proof of Theorem 1.1. Finally, in Section 6 we discuss the regularity of the solutions and prove Theorem 1.2.
Notations:
Throughout the paper, we will use the following notations:
δ(x) ≔ dist(x, ∂Ω) and d
Ω = diam(Ω);
We denote positive constants by M, M
1, M
2, …;
We denote the standard norm on
L
p
(
R
n
)
by |⋅|
p
.
3 The purely singular discontinuous problem
In order to prove the existence results for
(
P
λ
)
, we translate the problem by the minimal solution to the purely singular problem:
(
S
λ
)
-
Δ
u
=
λ
χ
{
u
<
a
}
u
−
γ
,
u
>
0
in
Ω
,
u
=
0
on
∂
Ω
.
We first study the existence of weak solutions to
(
S
λ
)
. We remark here that the minimal solution of
(
S
λ
)
plays the barrier function by below to
(
P
λ
)
providing the control of the singular nonlinearity term while applying variational methods to get solutions to
(
P
λ
)
. We have the following existence and regularity result for
(
S
λ
)
.
Proposition 3.1.
There exists a weak minimal solution u
λ
of
(
S
λ
)
for any γ > 0. Furthermore, we have u
λ
∼ ϕ
γ
near ∂Ω. Moreover regularity results as stated in Theorem 1.2 hold.
The main idea is to solve an approximating regular problem that admits a unique solution which is a strict subsolution to
(
S
λ
)
and then pass through the limit. The approximating regular problem is obtained by replacing χ
{u<a} with the continuous function χ
ϵ
(u − a) where
(3.1)
χ
ϵ
(
t
)
=
χ
(
−
∞
,
−
ϵ
)
(
t
)
−
t
ϵ
−
1
χ
[
−
ϵ
,
0
)
(
t
)
,
t
∈
R
.
So we consider the following problem
(
S
λ
,
ϵ
)
−
Δ
v
=
λ
χ
ϵ
(
v
−
a
)
v
−
γ
in
Ω
,
v
=
0
on
∂
Ω
,
v
>
0
in
Ω
.
We have the following existence results concerning
(
S
λ
,
ϵ
)
.
Proposition 3.2.
We have
Let 0 < γ < 3. Then there exists an ϵ
0 = ϵ
0(a) such that for 0 < ϵ < ϵ
0 there exists a unique solution
u
λ
,
ϵ
∈
H
0
1
(
Ω
)
∩
L
∞
(
Ω
)
of the problem
(
S
λ
,
ϵ
)
. Also the map ϵ → u
λ,ϵ
is nonincreasing. Moreover, we have u
λ,ϵ
∼ ϕ
γ
for any ϵ ∈ (0, ϵ
0).
Let γ ≥ 3. Then there exists unique solution
u
λ
,
ϵ
∈
H
loc
1
(
Ω
)
of
(
S
λ
,
ϵ
)
such that for any compact K ⋐ Ω, there exists M(K) > 0 which satisfies u
λ,ϵ
≥ M(K) > 0 a.e. in K. Moreover
(3.2)
u
λ
,
ϵ
i
s
u
n
i
f
o
r
m
l
y
b
o
u
n
d
e
d
i
n
L
∞
(
Ω
)
,
and
(3.3)
u
λ
,
ϵ
ℓ
i
s
u
n
i
f
o
r
m
l
y
b
o
u
n
d
e
d
i
n
H
0
1
(
Ω
)
with
ℓ
>
γ
+
1
4
.
Also the map ϵ → u
λ,ϵ
is nonincreasing and u
λ,ϵ
∼ ϕ
γ
independent of ϵ.
We will consider the above two cases separately. First we consider the case when 0 < γ < 3. In this case, the proof is based on the classical sub-supersolution method in a variational setting called Perron’s method. In this direction, inspired by Haitao [20], we establish the following result.
Lemma 3.3.
Let
0
<
ϵ
<
a
2
. Suppose that
u
̲
,
u
̄
∈
H
0
1
(
Ω
)
∩
L
∞
(
Ω
)
are weak subsolution and supersolution of
(
S
λ
,
ϵ
)
, respectively, such that
0
<
u
̲
≤
u
̄
in Ω and
u
̲
≥
M
(
K
)
>
0
for every K ⋐ Ω, for some constant M(K). Then there exists a solution
u
∈
H
0
1
(
Ω
)
∩
L
∞
(
Ω
)
of
(
S
λ
,
ϵ
)
satisfying
u
̲
≤
u
≤
u
̄
in Ω.
Proof.
For sake of clarity, we give only the proof for γ ≠ 1. We introduce the energy functional associated to
(
S
λ
,
ϵ
)
:
S
λ
ϵ
,
a
(
u
)
=
1
2
‖
u
‖
2
−
λ
∫
Ω
H
ϵ
(
u
)
d
x
,
where
(3.4)
H
ϵ
(
t
)
=
0
if
u
≤
0
,
(
1
−
γ
)
−
1
t
1
−
γ
if
0
<
t
<
a
2
,
(
1
−
γ
)
−
1
(
a
/
2
)
1
−
γ
+
∫
a
/
2
t
χ
ϵ
(
s
−
a
)
s
−
γ
d
t
if
t
≥
a
/
2
.
Observe that the map χ
ϵ
(t) lies in [0,1], is continuous and nonincreasing. Also we see that the map χ
ϵ
(t)t
−γ
is nonincreasing. Concerning H
ϵ
(t), we observe that H
ϵ
(t) ≤ (1 − γ)−1
t
1−γ
for t > 0. Now let us consider the conical shell set defined as:
M
=
v
∈
H
0
1
(
Ω
)
:
u
̲
≤
v
≤
u
̄
in
Ω
.
Clearly,
M
≠
∅
is closed and convex. Also
S
λ
ϵ
,
a
is weakly sequentially lower semicontinuous over
M
. Indeed, it is enough to show that
S
λ
a
,
ϵ
is sequentially lower semicontinuous. Let
{
v
k
}
k
∈
N
⊂
M
be such that v
k
→ v in
H
0
1
(
Ω
)
. Now observe that
|
H
ϵ
(
v
k
)
|
≤
(
1
−
γ
)
−
1
(
a
/
2
)
1
−
γ
−
+
|
(
1
−
γ
)
−
1
|
u
̲
1
−
γ
if
γ
>
1
,
(
1
−
γ
)
−
1
(
a
/
2
)
1
−
γ
−
+
|
(
1
−
γ
)
−
1
|
u
̄
1
−
γ
if
0
<
γ
<
1
.
Using the fact that
0
<
u
̲
≤
u
̄
∈
H
0
1
(
Ω
)
∩
L
∞
(
Ω
)
and Ω is bounded, we conclude that sequence |H
ϵ
(v
k
)| is bounded by a L
1(Ω) function (thanks to γ < 3). Thus by using the dominated convergence theorem and continuity of the norm, we obtain the required claim. Therefore there exists
u
∈
M
such that
S
λ
ϵ
,
a
(
u
)
=
inf
v
∈
M
S
λ
ϵ
,
a
(
v
)
.
Next, we show that u is the required weak solution of
(
S
λ
,
ϵ
)
. For this, let
φ
∈
C
c
∞
(
Ω
)
and κ > 0. We define
η
κ
=
u
̄
if
u
+
κ
φ
≥
u
̄
,
u
+
κ
φ
if
u
̲
≤
u
+
κ
φ
≤
u
̄
,
u
̲
if
u
+
κ
φ
≤
u
̲
.
Observe that
η
κ
=
u
+
κ
φ
−
φ
κ
+
φ
κ
∈
M
, where
φ
κ
=
(
u
+
κ
φ
−
u
̄
)
+
and
φ
κ
=
(
u
+
κ
φ
−
u
̲
)
−
. Since u is a minimizer of
S
λ
ϵ
,
a
over
M
, we have
(3.5)
0
≤
lim
t
→
0
S
λ
ϵ
,
a
(
u
+
t
(
η
κ
−
u
)
)
−
S
λ
ϵ
,
a
(
u
)
t
=
∫
Ω
∇
u
∇
(
η
κ
−
u
)
d
x
−
λ
∫
Ω
(
η
κ
−
u
)
χ
ϵ
(
u
−
a
)
u
−
γ
d
x
.
Using the definition of η
κ
, φ
κ
and φ
κ
, from (3.5) we have
(3.6)
∫
Ω
∇
u
∇
φ
d
x
−
λ
∫
Ω
χ
ϵ
(
u
−
a
)
u
−
γ
φ
d
x
≥
1
κ
G
κ
−
G
κ
,
where
G
κ
=
∫
Ω
∇
u
∇
φ
κ
d
x
−
λ
∫
Ω
χ
ϵ
(
u
−
a
)
u
−
γ
φ
κ
d
x
,
and
G
κ
=
∫
Ω
∇
u
∇
φ
κ
d
x
−
λ
∫
Ω
χ
ϵ
(
u
−
a
)
u
−
γ
φ
κ
d
x
.
Next we will give estimates of
G
κ
and
G
κ
. First we give:
Estimate of
G
κ
: Setting Ω
κ
= {φ
κ
> 0}, we have
1
κ
∫
Ω
∇
u
∇
φ
κ
d
x
=
1
κ
∫
Ω
∇
(
u
−
u
̄
)
∇
φ
κ
d
x
+
1
κ
∫
Ω
∇
u
̄
∇
φ
κ
d
x
=
1
κ
∫
Ω
κ
∇
(
u
−
u
̄
)
∇
(
u
+
κ
φ
−
u
̄
)
d
x
+
1
κ
∫
Ω
∇
u
̄
∇
φ
κ
d
x
(3.7)
≥
∫
Ω
κ
∇
(
u
−
u
̄
)
∇
φ
d
x
+
1
κ
∫
Ω
∇
u
̄
∇
φ
κ
d
x
=
o
(
1
)
+
1
κ
∫
Ω
∇
u
̄
∇
φ
κ
d
x
.
Using (3.7) and employing the facts that
u
̄
is a supersolution of
(
S
λ
,
ϵ
)
, χ
ϵ
(t)t
γ
is nonincreasing and
(
χ
ϵ
(
u
̄
−
a
)
u
̄
−
γ
−
χ
ϵ
(
u
−
a
)
u
−
γ
)
φ
∈
L
1
(
Ω
)
, together with dominated convergence theorem we obtain the following
(3.8)
1
κ
G
κ
≥
o
(
1
)
+
1
κ
∫
Ω
∇
u
̄
∇
φ
κ
d
x
−
λ
∫
Ω
χ
ϵ
(
u
−
a
)
u
−
γ
φ
κ
d
x
=
o
(
1
)
+
1
κ
∫
Ω
∇
u
̄
∇
φ
κ
d
x
−
λ
∫
Ω
χ
ϵ
(
u
̄
−
a
)
u
̄
−
γ
φ
κ
d
x
+
λ
κ
∫
Ω
χ
ϵ
(
u
̄
−
a
)
u
̄
−
γ
φ
κ
d
x
−
∫
Ω
χ
ϵ
(
u
−
a
)
u
−
γ
φ
κ
d
x
≥
o
(
1
)
+
λ
κ
∫
Ω
κ
(
χ
ϵ
(
u
̄
−
a
)
u
̄
−
γ
−
χ
ϵ
(
u
−
a
)
u
−
γ
)
(
u
−
u
̄
)
d
x
+
λ
∫
Ω
κ
χ
ϵ
(
u
̄
−
a
)
u
̄
−
γ
−
χ
ϵ
(
u
−
a
)
u
−
γ
φ
d
x
≥
o
(
1
)
.
In the similar fashion, we see that
(3.9)
1
κ
G
κ
≤
o
(
1
)
.
Using (3.8) and (3.9), we conclude from (3.6) that
0
≤
∫
Ω
∇
u
∇
φ
d
x
−
λ
∫
Ω
χ
ϵ
(
u
−
a
)
u
−
γ
φ
d
x
,
and the claim follows using the arbitrariness of φ.□
Proof of Proposition 3.2 (1):
In view of Lemma 3.3, we construct an ordered pair of sub- and supersolution
u
̲
and
u
̄
, respectively, of
(
S
λ
,
ϵ
)
for ϵ small enough. Choose 0 < ϵ
0 < a. Then for 0 < ϵ < ϵ
0, we see that χ
ϵ
(t − a)t
−γ
→ ∞ uniformly as t → 0. Thus we can find θ > 0 sufficiently small so that
λ
1
θ
|
e
1
|
∞
≤
λ
χ
ϵ
(
θ
|
e
1
|
∞
−
a
)
(
θ
|
e
1
|
∞
)
−
γ
.
Now using the fact that χ
ϵ
(t − a)t
−γ
is nonincreasing, we obtain by taking
u
̲
=
θ
e
1
that
−
Δ
u
̲
=
λ
1
(
θ
e
1
)
≤
λ
χ
ϵ
(
θ
|
e
1
|
∞
−
a
)
(
θ
|
e
1
|
∞
)
−
γ
≤
λ
χ
ϵ
(
u
̲
−
a
)
u
̲
−
γ
.
For supersolution, we take
u
̄
=
u
̂
, where
u
̂
is the solution of the purely singular problem (replacing the discontinuity term by 1). Finally, we choose θ small enough so that
0
<
u
̲
≤
u
̄
a.e in Ω. Applying Lemma 3.3 we obtain the existence of u
λ,ϵ
. It is easy to see that u
λ,ϵ
∼ ϕ
γ
. Uniqueness of u
λ,ϵ
for γ < 3 follows from the nonincreasing nature of t → χ
ϵ
(t − a)t
−γ
on (0, + ∞). Finally, we prove that the map ϵ → u
λ,ϵ
is nonincreasing. For this, let ϵ
1 < ϵ
2 and
u
λ
,
ϵ
1
,
u
λ
,
ϵ
2
be the corresponding solutions. Observe that for 0 < ϵ
1 < ϵ
2,
u
λ
,
ϵ
1
is a supersolution for
(
S
λ
,
ϵ
2
)
. Hence from uniqueness and Lemma 3.3, we get
u
λ
,
ϵ
2
≤
u
λ
,
ϵ
1
.□
Next we consider the case γ ≥ 3. We first show the validity of a weak comparison principle which will be used to get the uniqueness of the solution. Here we adopt the ideas developed in [24].
For any
k
∈
N
, we define the real valued function g
k
(τ) by
g
k
(
τ
)
=
max
−
λ
χ
ϵ
(
τ
−
a
)
τ
−
γ
,
−
k
if
τ
>
0
,
−
k
if
τ
≤
0
.
Now consider the real valued function G
k
(τ) defined by
G
k
′
(
τ
)
=
g
k
(
τ
)
,
G
k
(
1
)
=
0
.
Finally, we define the functional
Φ
k
:
H
0
1
(
Ω
)
→
[
−
∞
,
+
∞
]
by
Φ
k
(
v
)
=
1
2
‖
v
‖
2
+
∫
Ω
G
k
(
v
)
d
x
,
v
∈
H
0
1
(
Ω
)
.
Let u be a fixed supersolution of
(
S
λ
,
ϵ
)
and consider w as the minimum of the functional Φ
k
on the convex set
M
≔
v
∈
H
0
1
(
Ω
)
:
0
≤
v
≤
u
a.e. in
Ω
.
By following proof of Proposition 5.1 in [25], Page no. 16], it follows that
(3.10)
∫
Ω
∇
w
∇
(
v
−
w
)
d
x
≥
−
∫
Ω
G
k
′
(
w
)
(
v
−
w
)
d
x
for
v
∈
w
+
(
H
0
1
(
Ω
)
∩
L
c
∞
(
Ω
)
)
and
0
≤
v
≤
u
,
where
L
c
∞
(
Ω
)
denotes the space of L
∞ functions with compact support in Ω.
Lemma 3.4.
We have
(3.11)
∫
Ω
∇
w
∇
v
d
x
≥
−
∫
Ω
G
k
′
(
w
)
v
d
x
for
v
∈
C
c
∞
(
Ω
)
with
v
≥
0
in
Ω
.
Proof.
To prove this, let us consider
ψ
∈
C
c
∞
(
R
)
with 0 ≤ ψ ≤ 1 defined in
R
such that ψ(t) = 1 for t ∈ [−1, 1] and ψ(t) = 0 for t ∈ (−∞, 2] ∪ [2, ∞). Then for any
φ
∈
C
c
∞
(
Ω
)
with φ ≥ 0 in Ω, we set
φ
k
≔
ψ
w
k
φ
,
φ
k
,
t
=
min
{
w
+
t
φ
k
,
u
}
,
with
k
∈
N
and t > 0. We have
φ
k
,
t
∈
w
+
(
H
0
1
(
Ω
)
∩
L
c
∞
(
Ω
)
)
and w ≤ φ
k,t
≤ u, so by (3.10) we have
(3.12)
∫
Ω
∇
w
∇
(
φ
k
,
t
−
w
)
d
x
≥
−
∫
Ω
G
k
′
(
w
)
(
φ
k
,
t
−
w
)
d
x
.
Now using (3.12), we have
‖
φ
k
,
t
−
w
‖
2
+
∫
Ω
G
k
′
(
φ
k
,
t
)
−
G
k
(
w
)
(
φ
k
,
t
−
w
)
d
x
(3.13)
=
∫
Ω
∇
φ
k
,
t
∇
(
φ
k
,
t
−
w
)
d
x
−
∫
Ω
∇
w
∇
(
φ
k
,
t
−
w
)
d
x
+
∫
Ω
G
k
′
(
φ
k
,
t
)
(
φ
k
,
t
−
w
)
d
x
−
∫
Ω
G
k
′
(
w
)
(
φ
k
,
t
−
w
)
d
x
≤
∫
Ω
∇
φ
k
,
t
∇
(
φ
k
,
t
−
w
)
d
x
+
∫
Ω
G
k
′
(
φ
k
,
t
)
(
φ
k
,
t
−
w
)
d
x
=
∫
Ω
∇
φ
k
,
t
∇
(
φ
k
,
t
−
w
−
t
φ
k
)
d
x
+
∫
Ω
G
k
′
(
φ
k
,
t
)
(
φ
k
,
t
−
w
−
t
φ
k
)
d
x
+
t
∫
Ω
∇
φ
k
,
t
∇
φ
k
d
x
+
t
∫
Ω
G
k
′
(
φ
k
,
t
)
φ
k
d
x
.
Now using the definition of φ
k,t
, we easily see that
(3.14)
∫
Ω
∇
φ
k
,
t
∇
(
φ
k
,
t
−
w
−
φ
k
)
d
x
=
∫
Ω
∇
u
∇
(
φ
k
,
t
−
w
−
φ
k
)
d
x
.
Using (3.14) in (3.13), we get
(3.15)
‖
φ
k
,
t
−
w
‖
2
+
∫
Ω
G
k
′
(
φ
k
,
t
)
−
G
k
(
w
)
(
φ
k
,
t
−
w
)
d
x
≤
∫
Ω
∇
u
∇
(
φ
k
,
t
−
w
−
t
φ
k
)
d
x
+
∫
Ω
G
k
′
(
u
)
(
φ
k
,
t
−
w
−
t
φ
k
)
d
x
+
t
∫
Ω
∇
φ
k
,
t
∇
φ
k
d
x
+
∫
Ω
G
k
′
(
φ
k
,
t
)
φ
k
d
x
.
Now by the definition of G
k
, it follows that u is also a supersolution to the equation
−
Δ
z
=
−
G
k
′
(
z
)
, so by observing φ
k,t
− w − tφ
k
≤ 0, we deduce from (3.15) that
‖
φ
k
,
t
−
w
‖
2
+
∫
Ω
G
k
′
(
φ
k
,
t
)
−
G
k
(
w
)
(
φ
k
,
t
−
w
)
d
x
≤
t
∫
Ω
∇
φ
k
,
t
∇
φ
k
d
x
+
t
∫
Ω
G
k
′
(
φ
k
,
t
)
φ
k
d
x
.
Exploiting again the fact that φ
k,t
− w ≤ tφ
k
, by simple computations we deduce that
∫
Ω
∇
φ
k
,
t
∇
φ
k
d
x
+
∫
Ω
G
k
′
(
φ
k
,
t
)
φ
k
d
x
≥
−
∫
Ω
|
G
k
′
(
φ
k
,
t
)
−
G
k
(
w
)
‖
φ
k
|
d
x
.
Now passing to the limit t → 0 and employing Lebesgue Dominated convergence theorem, we see that
∫
Ω
∇
w
∇
φ
k
d
x
+
∫
Ω
G
k
′
(
w
)
φ
k
d
x
≥
0
.
Finally the proof holds by tending k to infinity.□
Next we prove a weak comparison principle from which the uniqueness of the solution follows. Precisely, we have:
Theorem 3.5.
Let γ > 0, v be a subsolution to
(
S
λ
,
ϵ
)
such that
(
v
−
ν
)
+
∈
H
0
1
(
Ω
)
for every ν > 0 and let u be a supersolution to
(
S
λ
,
ϵ
)
. Then v ≤ u a.e. in Ω.
Proof.
Let w be as in Lemma 3.4. Since
w
∈
H
0
1
(
Ω
)
and is nonnegative, for any ν > 0 supp(v − w − ν)+ is contained in supp(v − ν)+. From this, we conclude that
(
v
−
w
−
ν
)
+
∈
H
0
1
(
Ω
)
for any
ν
>
0
.
Now using standard density arguments, we obtain from (3.11) that
(3.16)
∫
Ω
∇
w
∇
K
τ
(
(
v
−
w
−
ν
)
+
)
d
x
≥
−
∫
Ω
G
k
′
(
w
)
K
τ
(
(
v
−
w
−
ν
)
+
)
d
x
,
where for a fixed τ > 0, K
τ
(t) ≔ min{t, τ} for t ≥ 0 and K
τ
(−t) ≔ − K
τ
(t) for t < 0. Let now
ψ
k
∈
C
c
∞
(
Ω
)
such that
ψ
k
→
(
v
−
w
−
ν
)
+
∈
H
0
1
(
Ω
)
and set
ψ
̃
τ
,
k
≔
K
τ
(
min
(
v
−
w
−
ν
)
+
,
ψ
k
+
)
.
It follows that
ψ
̃
τ
,
k
∈
H
0
1
(
Ω
)
∩
L
c
∞
(
Ω
)
and by a density argument
∫
Ω
∇
v
∇
ψ
̃
τ
,
k
d
x
≤
∫
Ω
λ
χ
ϵ
(
v
−
a
)
v
γ
ψ
̃
τ
,
k
d
x
.
Passing to the limit as k → ∞, we obtain
(3.17)
∫
Ω
∇
v
∇
K
τ
(
(
v
−
w
−
ν
)
+
)
d
x
≤
∫
Ω
λ
χ
ϵ
(
v
−
a
)
v
γ
K
τ
(
(
v
−
w
−
ν
)
+
)
d
x
.
Choosing ν > 0 such that ν
−γ
< k, from (3.16) and (3.17) we deduce that
‖
K
τ
(
(
v
−
w
−
ν
)
+
)
‖
2
≤
∫
Ω
λ
χ
ϵ
(
v
−
a
)
v
γ
+
G
k
′
(
w
)
K
τ
(
(
v
−
w
−
ν
)
+
)
d
x
=
∫
Ω
(
−
G
k
′
(
v
)
+
G
k
′
(
w
)
)
K
τ
(
(
v
−
w
−
ν
)
+
)
d
x
≤
0
.
By the arbitrariness of τ we deduce that
v
≤
w
+
ν
≤
u
+
ν
a.e. in
Ω
and the conclusion follows letting ν → 0.□
Regarding the existence of solutions, we use the classical approach of regularizing the singular nonlinearities u
−γ
by
u
+
1
k
−
γ
and derive uniform a priori estimates for the weak solution of the regularized problem. More precisely, we study the following approximated problem
(
S
λ
,
ϵ
,
k
)
−
Δ
u
=
λ
χ
ϵ
(
u
−
a
)
u
+
1
k
γ
,
u
>
0
in
Ω
,
u
=
0
on
∂
Ω
.
Lemma 3.6.
For any
k
∈
N
\
{
0
}
and γ > 0, there exists a unique nonnegative weak solution
u
λ
,
k
,
ϵ
∈
H
0
1
(
Ω
)
of the problem
(
S
λ
,
ϵ
,
k
)
in the sense that
(3.18)
∫
Ω
∇
u
λ
,
k
,
ϵ
∇
v
d
x
=
λ
∫
Ω
χ
ϵ
(
u
λ
,
k
,
ϵ
−
a
)
u
λ
,
k
,
ϵ
+
1
k
γ
v
d
x
for all
v
∈
H
0
1
(
Ω
)
.
Moreover,
The solution
u
λ
,
k
,
ϵ
∈
C
1
,
α
(
Ω
̄
)
for every α ∈ (0, 1) and u
λ,k,ϵ
> 0 in Ω.
The sequence
{
u
λ
,
k
,
ϵ
}
k
∈
N
is monotonically increasing in the sense that u
λ,k+1,ϵ
≥ u
λ,k,ϵ
for all
k
∈
N
.
For every compact set K ⋐ Ω and
k
∈
N
, there exists a constant M(K) > 0 independent of k such that u
λ,k,ϵ
≥ M(K) > 0.
u
λ,k,ϵ
is uniformly bounded in L
∞(Ω) both in k and ϵ.
u
λ
,
k
,
ϵ
ω
is bounded in
H
0
1
(
Ω
)
with
ω
>
γ
+
1
4
, independently of both k and ϵ.
Proof.
Proof of parts (i), (ii), and (iii) are standard and hence omitted. We first give the proof of part (v). Since
u
λ
,
k
,
ϵ
∈
L
∞
(
Ω
)
∩
H
0
1
(
Ω
)
and positive, for any ν > 0 and ω > 0,
(
u
λ
,
k
,
ϵ
+
ν
)
ω
−
ν
ω
belongs to
H
0
1
(
Ω
)
and so by taking it as a test function in (3.18) with
ν
∈
(
0
,
1
k
)
and ω ∈ [γ, ∞), we obtain
(3.19)
∫
Ω
∇
u
λ
,
k
,
ϵ
∇
(
u
λ
,
k
,
ϵ
+
ν
)
ω
d
x
≤
λ
∫
Ω
χ
ϵ
(
u
λ
,
k
,
ϵ
(
x
)
−
a
)
u
λ
,
k
,
ϵ
+
1
k
γ
(
u
λ
,
k
,
ϵ
+
ν
)
ω
d
x
≤
λ
∫
Ω
(
u
λ
,
k
,
ϵ
+
ν
)
ω
−
γ
d
x
.
Passing ν → 0 in (3.19) via Fatou’s Lemma, we obtain
(3.20)
4
ω
(
ω
+
1
)
2
∫
Ω
|
∇
u
λ
,
k
,
ϵ
ω
+
1
2
|
2
d
x
≤
λ
∫
Ω
(
u
λ
,
k
,
ϵ
)
ω
−
γ
d
x
.
In order to estimate the R.H.S term of (3.20), we choose ω that satisfies (according to boundary behaviour of u
k,ϵ
):
ω
>
γ
−
1
2
.
Now using the boundary behaviour of u
λ,k,ϵ
(3.21)
∫
Ω
u
λ
,
k
,
ϵ
ω
−
γ
d
x
≤
λ
∫
Ω
e
1
2
(
ω
−
γ
)
γ
+
1
d
x
<
∞
,
since
2
(
ω
−
γ
)
γ
+
1
>
−
1
. Combining (3.20) and (3.21), we obtain the required conclusion.
We now turn our attention to the L
∞ estimates (assertion iv). For this fix
p
>
n
2
. Now take
ϕ
m
(
u
λ
,
k
,
ϵ
)
≔
(
u
λ
,
k
,
ϵ
−
m
)
+
with m ≥ 1 as a test function in (3.18) and using Sobolev embeddings and Hölder inequality, we get
(3.22)
∫
T
m
|
ϕ
m
(
u
λ
,
k
,
ϵ
)
|
2
n
n
−
2
d
x
n
−
2
2
n
≤
M
∫
T
m
|
∇
ϕ
m
(
u
λ
,
k
,
ϵ
)
|
2
d
x
=
M
∫
Ω
∇
u
λ
,
k
,
ϵ
∇
ϕ
m
(
u
λ
,
k
,
ϵ
)
d
x
≤
M
∫
Ω
χ
ϵ
(
u
λ
,
k
,
ϵ
−
a
)
u
λ
,
k
,
ϵ
+
1
k
γ
ϕ
m
(
u
λ
,
k
,
ϵ
)
d
x
≤
M
∫
T
m
χ
ϵ
(
u
λ
,
k
,
ϵ
−
a
)
ϕ
m
(
u
λ
,
k
,
ϵ
)
d
x
≤
M
|
χ
ϵ
(
u
λ
,
k
,
ϵ
−
a
)
|
p
∫
T
m
|
ϕ
m
(
u
λ
,
k
,
ϵ
)
|
2
n
n
−
2
d
x
n
−
2
2
n
|
T
m
|
1
−
n
−
2
2
n
−
1
p
≤
M
|
Ω
|
1
p
∫
T
m
|
ϕ
m
(
u
λ
,
k
,
ϵ
)
|
2
n
n
−
2
d
x
n
−
2
2
n
|
T
m
|
1
−
n
−
2
2
n
−
1
p
where T
m
≔ {x ∈ Ω : u
λ,k,ϵ
≥ m}. Let j > m ≥ 1, then T
j
⊂ T
m
and ϕ
m
(u
λ,k,ϵ
) ≥ j − m for x ∈ T
j
. Using above facts, from (3.22), we obtain
|
j
−
m
‖
T
j
|
n
−
2
2
n
≤
∫
T
j
|
ϕ
m
(
u
λ
,
k
,
ϵ
)
|
2
n
n
−
2
d
x
n
−
2
2
n
≤
∫
T
m
|
ϕ
m
(
u
λ
,
k
,
ϵ
)
|
2
n
n
−
2
d
x
n
−
2
2
n
≤
M
‖
Ω
|
1
p
|
T
m
|
1
−
n
−
2
2
n
−
1
p
which further implies
|
T
j
|
≤
M
|
Ω
|
2
n
(
n
−
2
)
p
|
T
m
|
2
n
n
−
2
1
−
n
−
2
2
n
−
1
p
|
j
−
m
|
2
n
n
−
2
.
Since
p
>
n
2
, we have that
2
n
n
−
2
1
−
n
−
2
2
n
−
1
p
>
1
.
Thus by [25], Lemma B.1], there exists m
0 such that |T
m
| = 0 for all m ≥ m
0. This completes the proof.□
Proof of Proposition 3.2 (2):
Let γ ≥ 3 and u
λ,k,ϵ
be the weak solution of the problem
(
S
λ
,
ϵ
,
k
)
in the sense that it satisfies (3.18). Now from Lemma 3.6, we know that
u
λ
,
k
,
ϵ
ω
+
1
2
is uniformly bounded in
H
0
1
(
Ω
)
with
ω
>
γ
−
1
2
. Since γ ≥ 3, we have ω > 1. This together with the fact that for every compact subset K ⋐ Ω there exists M = M(K) independent of k such that 0 < M ≤ u
λ,k,ϵ
(x) for x ∈ K, we get that u
λ,k,ϵ
is uniformly bounded in
H
loc
1
(
Ω
)
. Precisely,
∫
K
|
∇
u
λ
,
k
,
ϵ
|
2
d
x
≤
M
−
(
ω
−
1
)
∫
K
u
λ
,
k
,
ϵ
(
ω
−
1
)
|
∇
u
λ
,
k
,
ϵ
|
2
d
x
≤
4
M
−
(
ω
−
1
)
ω
+
1
∫
K
|
∇
u
λ
,
k
,
ϵ
ω
+
1
2
|
2
d
x
≤
M
1
,
where M
1 is independent of k. Then there exists a
u
λ
,
ϵ
∈
H
loc
1
(
Ω
)
such that
(3.23)
u
λ
,
k
,
ϵ
⇀
u
λ
,
ϵ
,
u
λ
,
k
,
ϵ
→
u
λ
,
ϵ
in
L
loc
r
(
Ω
)
for
1
≤
r
<
2
*
and a.e in
Ω
.
Now by using the weak convergence property we are able to pass to the limit in the left hand side of (3.18), i.e. for any
v
∈
H
loc
1
(
Ω
)
with K = supp(v) ⋐Ω,
(3.24)
∫
Ω
∇
u
λ
,
k
,
ϵ
∇
v
d
x
→
∫
Ω
∇
u
λ
,
ϵ
∇
v
d
x
as
k
→
∞
.
Finally using the facts that 0 ≤ χ
ϵ
(u
λ,k,ϵ
− a) ≤ 1, M
1(K) ≤ u
λ,1,ϵ
≤ u
λ,k,ϵ
a.e. in K and from Lebesgue dominated convergence theorem, we get
(3.25)
∫
Ω
χ
ϵ
(
u
λ
,
k
,
ϵ
−
a
)
u
λ
,
k
,
ϵ
+
1
k
γ
v
d
x
→
∫
Ω
χ
ϵ
(
u
λ
,
ϵ
−
a
)
u
λ
,
ϵ
γ
v
d
x
.
Passing to the limit in (3.18) and using (3.24) and (3.25), we see that u
ϵ
is a weak solution of
(
S
λ
,
ϵ
)
. For uniqueness of the solution, using Theorem 3.5, it is sufficient to show that
(
u
λ
,
ϵ
−
ν
)
+
∈
H
0
1
(
Ω
)
for every ν > 0. From Lemma 3.6, we have
u
λ
,
k
,
ϵ
ω
∈
H
0
1
(
Ω
)
, where
ω
>
γ
+
1
4
≥
1
. Let
φ
m
∈
C
c
1
(
Ω
)
such that φ
m
converges to
u
λ
,
k
,
ϵ
ω
in
H
0
1
(
Ω
)
and set
ψ
m
≔
φ
m
1
ω
−
ν
+
.
Clearly ψ
m
is uniformly bounded in
H
0
1
(
Ω
)
and converges a.e. to
(
u
λ
,
k
,
ϵ
−
ν
)
+
. Therefore we obtain that
(
u
λ
,
k
,
ϵ
−
ν
)
+
∈
H
0
1
(
Ω
)
and hence
(
u
λ
,
ϵ
−
ν
)
+
∈
H
0
1
(
Ω
)
. This proves uniqueness. Next we prove that the map ϵ → u
λ,ϵ
is nonincreasing. For this, let ϵ
1 < ϵ
2 and
u
λ
,
ϵ
1
,
u
λ
,
ϵ
2
be the corresponding solutions. Arguing by contradiction, suppose that there exists F ⊂ Ω with positive measure such that
w
=
u
λ
,
ϵ
1
−
u
λ
,
ϵ
2
<
0
a.e on F. Using the facts that
u
λ
,
ϵ
1
,
u
λ
,
ϵ
2
are solutions of
(
S
λ
,
ϵ
1
)
and
(
S
λ
,
ϵ
2
)
, respectively, λ > 0 and for t
1 < t
2 and ϵ
1 < ϵ
2,
χ
ϵ
1
(
t
1
−
a
)
≥
χ
ϵ
2
(
t
2
−
a
)
, by taking v = w
−, we get
∫
Ω
|
∇
w
−
|
2
d
x
=
−
λ
∫
Ω
χ
ϵ
1
(
u
λ
,
ϵ
1
−
a
)
u
λ
,
ϵ
1
−
γ
−
χ
ϵ
2
(
u
λ
,
ϵ
2
−
a
)
u
λ
,
ϵ
2
−
γ
v
d
x
≤
0
,
and hence w ≥ 0 a.e. in Ω, which yields a contradiction. Lastly we check the validity of (3.2) and (3.3). From Lemma 3.6, we have
u
λ
,
k
,
ϵ
ω
is uniformly bounded in
H
0
1
(
Ω
)
for
ω
>
γ
+
1
4
. This implies the existence of
ψ
∈
H
0
1
(
Ω
)
such that
u
λ
,
k
,
ϵ
ω
⇀
ψ
in
H
0
1
(
Ω
)
and
u
λ
,
k
,
ϵ
ω
→
ψ
un L
r
(Ω) for every 1 ≤ r < 2* a.e. in Ω. This together with (3.23) imply that
ψ
=
u
λ
,
ϵ
ω
. Thus we have
‖
u
λ
,
ϵ
ω
‖
≤
lim inf
k
→
∞
‖
u
λ
,
k
,
ϵ
ω
‖
2
≤
M
.
From Lemma 3.6, we know that u
λ,k,ϵ
is uniformly bounded in L
∞(Ω) both in k and ϵ and hence u
λ,ϵ
is uniformly bounded in L
∞(Ω). This completes the proof.□
Lastly, we give the
Proof of Proposition 3.1:
We divide the proof into two cases:
Case A: 0 < γ < 3. In this case, since u
λ,ϵ
is a solution of
(
S
λ
,
ϵ
)
, using Proposition 3.2, Remark 2.4 and Hardy’s inequality we find,
‖
u
λ
,
ϵ
‖
2
=
λ
∫
Ω
χ
ϵ
(
u
λ
,
ϵ
−
a
)
u
λ
,
ϵ
1
−
γ
d
x
≤
C
∫
Ω
u
λ
,
ϵ
ϕ
γ
−
γ
d
x
≤
M
∫
Ω
|
∇
u
λ
,
ϵ
|
2
d
x
1
2
=
M
‖
u
λ
,
ϵ
‖
.
Thus,
{
u
λ
,
ϵ
}
ϵ
is a bounded sequence in
H
0
1
(
Ω
)
. Let u
λ,ϵ
⇀ u
λ
in
H
0
1
(
Ω
)
and a.e in Ω. From the lower bound u
λ,ϵ
≥ Mϕ
γ
, we obtain that
χ
ϵ
(
u
λ
,
ϵ
−
a
)
u
λ
,
ϵ
−
γ
is a bounded sequence in
L
loc
∞
(
Ω
)
. Then using elliptic regularity theory,
{
u
λ
,
ϵ
}
ϵ
is a bounded sequence in
C
loc
α
(
Ω
)
for some α > 0 and hence, u
λ,ϵ
→ u
λ
uniformly on compact subsets of Ω. Let
ψ
∈
H
0
1
(
Ω
)
be arbitrary. Using Remark 2.4, the estimate
χ
ϵ
(
u
λ
,
ϵ
−
a
)
u
λ
,
ϵ
−
γ
ψ
≤
M
ϕ
γ
−
γ
ψ
and the weak convergence of u
λ,ϵ
⇀ u
λ
, we obtain that u
λ
solves
(
S
λ
)
.
Case B: γ ≥ 3. The boundedness of the sequence
{
u
λ
,
ϵ
}
ϵ
follows using the same arguments as in the Proof of Proposition 3.2 (2). Let u
λ,ϵ
⇀ u
λ
in
H
0
1
(
Ω
)
and a.e. in Ω. Using the fact that for any compact set K ⋐ Ω, 0 < M(K) ≤ u
λ,ϵ
for every ϵ > 0 and following the similar arguments of Case A, we conclude that u
λ
is a weak solution of
(
S
λ
)
.
In both cases, any solution to
(
S
λ
)
is a supersolution to
(
S
λ
,
ϵ
)
. According to Theorem 3.5, u
λ
is the minimal solution to
(
S
λ
)
. Since u
λ,ϵ
∼ ϕ
γ
independent of ϵ, we have u
λ
∼ ϕ
γ
. Finally, we prove the Hölder regularity results using the boundary behaviour following the proof of [26], Theorem 1.1]. For this, let G(x, y) be the associated Green’s function of −Δ in Ω. We consider the following three cases:
Case 1. When 0 < γ < 1. By Green’s formula (see [27]), we have
(3.26)
u
λ
(
x
)
=
∫
Ω
(
−
G
(
x
,
y
)
)
λ
χ
{
u
λ
(
y
)
<
a
}
u
λ
−
γ
(
y
)
d
y
,
x
∈
Ω
,
and
(3.27)
∇
u
λ
(
x
)
=
∫
Ω
(
−
G
x
(
x
,
y
)
)
λ
χ
{
u
λ
(
y
)
<
a
}
u
λ
−
γ
(
y
)
d
y
,
x
∈
Ω
.
Now let x
1, x
2 ∈ Ω such that |x
1 − x
2| < ρ. Then by [26], Lemma 3.2], we can find a path σ(t): [0, 1] → Ω satisfying
(3.28)
|
σ
′
(
t
)
|
≤
C
1
|
x
1
−
x
2
|
.
Now
(3.29)
|
∇
u
(
x
1
)
−
∇
u
(
x
2
)
|
≤
λ
∫
Ω
|
G
x
(
x
1
,
y
)
−
G
x
(
x
2
,
y
)
|
u
λ
−
γ
(
y
)
d
y
=
λ
∫
Ω
\
B
R
(
x
1
)
|
G
x
(
x
1
,
y
)
−
G
x
(
x
2
,
y
)
|
u
λ
−
γ
(
y
)
d
y
+
λ
∫
B
R
(
x
1
)
|
G
x
(
x
1
,
y
)
−
G
x
(
x
2
,
y
)
|
u
λ
−
γ
(
y
)
d
y
=
A
1
+
A
2
(
say
)
,
where R = (C
1 + 4)|x
1 − x
2|, C
1 > 0 as in (3.28). Now, using [26], Proposition 3.1], we have
(3.30)
A
1
≤
λ
∫
Ω
\
B
R
(
x
1
)
∫
0
1
|
G
x
x
(
σ
(
t
)
,
y
)
|
⋅
|
σ
′
(
t
)
|
d
t
u
−
γ
(
y
)
d
y
≤
λ
C
1
|
x
1
−
x
2
|
∫
Ω
\
B
R
(
x
1
)
∫
0
1
min
{
|
σ
(
t
)
−
y
|
,
δ
(
y
)
}
|
σ
(
t
)
−
y
|
n
+
1
d
t
u
−
γ
(
y
)
d
y
≤
M
|
x
1
−
x
2
|
∫
Ω
\
B
R
(
x
1
)
min
{
|
x
1
−
y
|
,
δ
(
y
)
}
|
x
1
−
y
|
n
+
1
u
−
γ
(
y
)
d
y
,
here we have used
|
σ
(
t
)
−
y
|
≥
|
x
1
−
y
|
−
|
x
1
−
σ
(
t
)
|
≥
(
C
1
+
4
)
|
x
1
−
x
2
|
−
c
|
x
1
−
x
2
|
≥
4
|
x
1
−
x
2
|
and
|
σ
(
t
)
−
y
|
≤
|
x
1
−
y
|
+
|
x
1
−
σ
(
t
)
|
≤
2
(
C
1
+
4
)
|
x
1
−
x
2
|
≤
2
C
1
+
4
C
1
+
4
|
x
1
−
y
|
.
Now using u
λ
∼ ϕ
γ
, we obtain from (3.30)
(3.31)
A
1
≤
M
|
x
1
−
x
2
|
∫
Ω
\
B
R
(
x
1
)
min
{
|
x
1
−
y
|
,
δ
(
y
)
}
|
x
1
−
y
|
n
+
1
δ
−
γ
(
y
)
d
y
≤
M
|
x
1
−
x
2
|
∫
Ω
\
B
R
(
x
1
)
1
|
x
1
−
y
|
n
+
γ
d
y
=
M
|
x
1
−
x
2
|
∫
R
∞
r
−
(
n
+
γ
)
⋅
r
n
−
1
d
r
≤
M
|
x
1
−
x
2
|
R
−
γ
=
M
|
x
1
−
x
2
|
1
−
γ
.
Again using [26], Proposition 3.1] and the fact that u
λ
∼ ϕ
γ
, we have
(3.32)
A
2
≤
λ
∫
B
R
(
x
1
)
|
G
x
(
x
1
,
y
)
−
G
x
(
x
2
,
y
)
|
u
−
γ
(
y
)
d
y
≤
λ
∫
B
R
(
x
1
)
|
G
x
(
x
1
,
y
)
|
u
−
γ
(
y
)
d
y
+
∫
B
R
+
|
x
1
−
x
2
|
(
x
2
)
|
G
x
(
x
2
,
y
)
|
u
−
γ
(
y
)
d
y
≤
M
∫
B
R
(
x
1
)
min
{
|
x
1
−
y
|
,
δ
(
y
)
}
|
x
1
−
y
|
n
δ
−
γ
(
y
)
d
y
+
M
∫
B
R
+
|
x
1
−
x
2
|
(
x
2
)
min
{
|
x
2
−
y
|
,
δ
(
y
)
}
|
x
2
−
y
|
n
δ
−
γ
(
y
)
d
y
≤
M
∫
B
R
(
x
1
)
|
x
1
−
y
|
−
n
−
γ
+
1
d
y
+
M
∫
B
R
+
|
x
1
−
x
2
|
(
x
2
)
|
x
2
−
y
|
−
n
−
γ
+
1
d
y
≤
M
∫
0
R
+
|
x
1
−
x
2
|
r
−
γ
d
r
≤
M
|
x
1
−
x
2
|
1
−
γ
.
In view of (3.29), (3.30) and (3.31), we conclude that
u
λ
∈
C
1,1
−
γ
(
Ω
̄
)
as required.
Case 2. When γ = 1. In this case, since u
λ
∼ ϕ
γ
, we have for some D > 0
(3.33)
u
λ
−
1
(
x
)
≤
M
δ
−
1
(
x
)
(
D
−
log
δ
(
x
)
)
−
1
2
,
x
∈
Ω
.
For any x
1, x
2 ∈ Ω, let d = max{δ(x
1), δ(x
2)}, say, d = δ(x
1) ≥ d(x
2). If
|
x
1
−
x
2
|
≤
1
2
d
, by the interior estimate of the elliptic equation (see [27]), u
λ
∼ ϕ
γ
and (3.33), we have
|
∇
u
λ
(
y
)
|
≤
M
(
D
−
log
d
)
1
2
+
D
−
log
1
2
d
−
1
2
,
for all
y
∈
B
d
/
2
(
x
1
)
,
and so
(3.34)
|
u
λ
(
x
1
)
−
u
λ
(
x
2
)
|
≤
M
|
x
1
−
x
2
|
[
D
−
log
|
x
1
−
x
2
|
]
1
2
.
If
|
x
1
−
x
2
|
≥
1
2
d
, by u
λ
∼ ϕ
γ
, we also have
(3.35)
|
u
(
x
1
)
−
u
(
x
2
)
|
≤
M
|
x
1
−
x
2
|
[
D
−
log
|
x
1
−
x
2
|
]
1
2
.
In view of (3.34) and (3.35), we conclude that
u
∈
C
β
(
Ω
̄
)
for any β ∈ (0, 1).
Similarly, we can prove the case γ > 1 with straightforward modifications. This finishes the proof.□
4 Existence of a first solution
In this section, we establish the existence of the first solution of the problem
(
P
λ
)
. Here again, we follow the regularizing techniques as in the last section. Define
Λ
a
=
sup
{
λ
>
0
:
(
P
λ
)
has
a
t
least one solution
}
.
We have
Proof.
Let
(
P
λ
)
admit a solution v
λ
. Multiplying
(
P
λ
)
by e
1, we get
(4.1)
λ
1
∫
Ω
v
λ
e
1
d
x
=
λ
∫
Ω
∫
Ω
v
λ
2
μ
*
(
y
)
v
λ
2
μ
*
−
1
(
x
)
e
1
(
x
)
|
x
−
y
|
μ
d
x
d
y
+
∫
Ω
χ
{
v
λ
<
a
}
v
λ
−
γ
e
1
(
x
)
d
x
≥
λ
∫
Ω
1
d
Ω
μ
∫
Ω
v
λ
2
μ
*
(
y
)
d
y
v
λ
2
μ
*
−
1
(
x
)
+
χ
{
v
λ
<
a
}
v
λ
−
γ
(
x
)
e
1
(
x
)
d
x
.
Now for any m > 0 by noting the superlinear nature of the map
t
→
m
t
2
μ
*
−
1
+
χ
{
t
<
a
}
t
−
γ
at infinity, we guarantee the existence of a constant M = M(a) > 0 such that
m
t
2
μ
*
−
1
+
χ
{
t
<
a
}
t
−
γ
>
M
t
. Employing this observation in (4.1), we conclude that
λ
1
∫
Ω
v
λ
e
1
d
x
≥
λ
M
∫
Ω
v
λ
e
1
d
x
.
This implies Λ
a
< ∞. Next we show that 0 < Λ
a
. For this, we consider the following singular problem without the jump discontinuity
(4.2)
−
Δ
u
=
λ
u
−
γ
+
∫
Ω
u
2
μ
*
(
y
)
|
x
−
y
|
μ
d
y
u
2
μ
*
−
1
in
Ω
,
u
>
0
in
Ω
,
u
=
0
on
∂
Ω
.
It is well known (see Theorems 1.1, 2.2 and 2.5 in [17]) that there exists a unique
w
λ
∈
C
0
(
Ω
̄
)
∩
C
2
(
Ω
)
solving the following singular problem for all λ > 0:
−
Δ
w
=
λ
w
−
γ
,
w
>
0
in
Ω
,
w
=
0
on
∂
Ω
.
Also by Dini’s Theorem, we have that w
λ
→ 0 uniformly in Ω as λ → 0+. Clearly, w
λ
is a subsolution to (4.2). Next let
z
∈
H
0
1
(
Ω
)
solve
−
Δ
z
=
1
,
z
>
0
in
Ω
,
z
=
0
in
∂
Ω
.
Define z
λ
= w
λ
+ z. Next we claim that there exists
λ
̂
>
0
small such that for
λ
<
λ
̂
, z
λ
is a supersolution to (4.2). The choice of
λ
̂
is such that
λ
̂
(
|
w
λ
+
z
|
∞
)
2
2
μ
*
−
1
M
̂
≤
1
, where
M
̂
is such that
∫
Ω
d
y
|
x
−
y
|
μ
<
M
̂
. Note that the choice of such
M
̂
is possible since Ω is bounded. Then for
λ
<
λ
̂
, we have
−
Δ
z
λ
=
λ
w
λ
−
γ
+
1
≥
λ
z
λ
−
γ
+
(
|
w
λ
+
z
|
∞
)
2
2
μ
*
−
1
M
̂
≥
λ
z
λ
−
γ
+
∫
Ω
z
λ
2
μ
*
(
y
)
z
λ
2
μ
*
−
1
(
x
)
|
x
−
y
|
μ
d
y
.
This completes the proof of the claim. Let
K
λ
=
u
∈
H
0
1
(
Ω
)
:
w
λ
≤
u
≤
z
λ
in
Ω
. Clearly,
K
λ
is a closed convex (hence weakly closed) set in
H
0
1
(
Ω
)
. Now, we define the following iterative scheme for all
λ
<
λ
̂
:
u
0
=
w
λ
,
−
Δ
u
k
−
λ
u
k
−
γ
=
λ
∫
Ω
u
k
−
1
2
μ
*
(
y
)
u
k
−
1
2
μ
*
−
1
(
x
)
|
x
−
y
|
μ
d
y
,
u
k
>
0
in
Ω
,
u
k
=
0
on
∂
Ω
,
k
=
1,2,3
,
…
The above scheme is well defined as we can solve for u
k
is closed convex set
K
λ
. Using the monotonicity of the operator −Δu − λu
−γ
, we have that the sequence
{
u
k
}
k
∈
N
is nondecreasing and w
λ
≤ u
k
≤ z
λ
for all k. In particular,
{
u
k
}
k
∈
N
is uniformly bounded in
C
α
(
Ω
̄
)
. By the Ascoli-Arzela theorem,
u
k
→
u
̄
λ
∈
C
0
(
Ω
)
as k → ∞ and
w
λ
≤
u
̄
λ
≤
z
λ
. Now following the arguments as in the proof of Proposition 3.2, we conclude that
u
̄
λ
is a solution of (4.2). Noting that
|
u
̄
λ
|
∞
→
0
for λ → 0,
u
̄
λ
solves
(
P
λ
)
for small λ and hence Λ
a
> 0.□
Now we consider the following perturbed regular problem associated to
(
P
λ
)
,
(
P
λ
,
ϵ
)
−
Δ
u
=
λ
∫
Ω
u
2
μ
*
(
y
)
|
x
−
y
|
μ
d
y
u
2
*
−
1
+
χ
ϵ
(
u
−
a
)
u
−
γ
,
u
≡
0
on
∂
Ω
,
u
>
0
in
Ω
,
where χ
ϵ
is defined as in (3.1). The formal energy functional J
λ,ϵ
associated to
(
P
λ
,
ϵ
)
is defined as
J
λ
,
ϵ
(
u
)
=
1
2
‖
u
‖
2
−
λ
∫
Ω
H
ϵ
(
u
)
d
x
−
λ
2
2
μ
*
∫
Ω
∫
Ω
|
u
(
y
)
|
2
μ
*
|
u
(
x
)
|
2
*
|
x
−
y
|
μ
d
x
d
y
,
where H
ϵ
is defined as in (3.4).
We now have the following existence result:
Lemma 4.2.
(
P
λ
)
admits a solution v
λ
for all λ ∈ (0, Λ
a
). Moreover, v
λ
∼ ϕ
λ
and satisfies the Sobolev regularity result as stated in Theorem 1.1.
Proof.
We first show that given any 0 < ϵ < ϵ
0(a), where ϵ
0 is obtained in Proposition 3.2 and λ ∈ (0, Λ
a
), the approximating problem
(
P
λ
,
ϵ
)
admits a solution v
λ,ϵ
. Let u
λ,ϵ
be the solution of
(
S
λ
,
ϵ
)
as obtained in Proposition 3.2. Given any λ ∈ (0, Λ
a
), there exists
λ
̄
>
λ
such that
(
P
λ
̄
)
admits a solution
v
̄
and by the definition of χ
ϵ
,
v
̄
is a supersolution of
(
P
λ
,
ϵ
)
. Then by Theorem 3.5, we see that
u
λ
,
ϵ
≤
v
̄
. Now the existence of a solution v
λ,ϵ
of
(
P
λ
,
ϵ
)
is obtained as a local minimizer of J
λ,ϵ
over the convex set
M
ϵ
=
u
∈
H
0
1
(
Ω
)
:
u
λ
,
ϵ
≤
u
≤
v
̄
. As v
λ,ϵ
solves
(
P
λ
,
ϵ
)
, it is easy to check that
{
v
λ
,
ϵ
}
ϵ
is bounded in
H
0
1
(
Ω
)
(thanks to
v
̄
∈
L
∞
(
Ω
)
) and hence weakly converges to some
v
λ
∈
H
0
1
(
Ω
)
. Then following the convergence arguments of Proposition 3.1, we conclude that v
λ
is a solution of
(
P
λ
)
.
Now since |v
λ,ϵ
|∞ ≤ M, where M is independent of ϵ (because that same is true for u
λ,ϵ
by Proposition 3.2), we see that u
λ,ϵ
≤ v
λ,ϵ
≤ w, where w is the solution of
−
Δ
w
=
λ
w
−
γ
+
K
,
w
>
0
in
Ω
,
w
=
0
on
∂
Ω
,
for some appropriate K. Also since u
λ,ϵ
∼ ϕ
γ
and w ∼ ϕ
γ
, we conclude that v
λ,ϵ
∼ ϕ
γ
for every ϵ and hence v
λ
∼ ϕ
γ
. Finally, the Sobolev regularity results for v
λ,ϵ
follows on the similar lines of the proof of Lemma 3.6 item v) using the fact that v
λ,ϵ
∈ L
∞(Ω) independent of ϵ and for v
λ
using the arguments as in the Proof of Proposition 3.2 (2).□
Next following the proof of Proposition 3.1 and Lemma 3.4 of [7], we have the following lemma:
Lemma 4.3.
For any 0 < λ < Λ
a
and 0 < μ ≤ min{n, 4},
J
λ
(
v
λ
)
=
min
v
∈
M
0
J
λ
(
v
)
, where
M
0
=
u
∈
H
0
1
(
Ω
)
:
u
λ
≤
u
≤
v
̄
and u
λ
is as in Proposition 3.1.
Now we claim that v
λ
is a local minimum of J
λ
in
H
0
1
(
Ω
)
. We have
Theorem 4.4.
Let a > 0 and 0 < μ ≤ min{n, 4}. Then for λ ∈ (0, Λ
a
), v
λ
is a local minimum of J
λ
in
H
0
1
(
Ω
)
.
Proof.
We assume that v
λ
is not a local minimum of J
λ
in
H
0
1
(
Ω
)
and derive a contradiction. Let
{
v
k
}
⊂
H
0
1
(
Ω
)
be such that v
k
→ v
λ
in
H
0
1
(
Ω
)
and J
λ
(v
k
) < J
λ
(v
λ
). For
v
̲
=
u
λ
and solution
v
̄
of
(
P
λ
̄
)
where
0
<
λ
<
λ
̄
<
Λ
a
, define
z
k
=
max
{
v
̲
,
min
{
v
k
,
v
̄
}
}
,
w
̄
k
=
(
v
k
−
v
̄
)
+
,
w
̲
k
=
(
v
k
−
v
̲
)
−
,
A
̄
k
=
supp
(
w
̄
k
)
and
A
̲
k
=
supp
(
w
̲
k
)
.
Claim A:
|
A
̄
k
|
,
|
A
̲
k
|
and
‖
w
̄
k
‖
→
0
as k → ∞.
The proof of the claim can be proved on the similar lines of the proof of Theorem 2.2 of [8] and hence omitted. Now note that
z
k
∈
M
0
=
u
∈
H
0
1
(
Ω
)
:
v
̲
≤
u
≤
v
̄
and
v
k
=
z
k
−
w
̲
k
+
w
̄
k
. Now
(4.3)
J
λ
(
v
k
)
=
J
λ
(
z
k
)
+
1
2
∫
A
̄
k
|
∇
v
k
|
2
−
|
∇
v
̄
|
2
d
x
+
1
2
∫
A
̲
k
|
∇
v
k
|
2
−
|
∇
v
̲
|
2
d
x
−
λ
∫
A
̄
k
(
H
(
v
k
)
−
H
(
v
̄
)
)
d
x
−
λ
∫
A
̲
k
(
H
(
v
k
)
−
H
(
v
̲
)
)
d
x
−
λ
2
2
μ
*
∫
Ω
∫
Ω
v
k
2
μ
*
(
x
)
v
k
2
μ
*
(
y
)
−
z
k
2
μ
*
(
x
)
z
k
2
μ
*
(
y
)
|
x
−
y
|
μ
d
x
d
y
=
J
λ
(
z
k
)
+
1
2
∫
A
̄
k
|
∇
w
̄
k
|
2
d
x
+
∫
A
̄
k
∇
v
̄
∇
w
̄
k
d
x
+
1
2
∫
A
̲
k
|
∇
w
̲
k
|
2
d
x
−
∫
A
̲
k
∇
v
̲
∇
w
̲
k
d
x
−
λ
∫
A
̄
k
(
H
(
v
̄
+
w
̄
k
)
−
H
(
v
̄
)
)
d
x
−
λ
2
2
μ
*
∫
Ω
∫
Ω
(
v
k
2
μ
*
(
x
)
−
z
k
2
μ
*
(
x
)
)
v
k
2
μ
*
(
y
)
|
x
−
y
|
μ
d
x
d
y
−
λ
∫
A
̲
k
(
H
(
v
̲
−
w
̲
k
)
−
H
(
v
̲
)
)
d
x
−
λ
2
2
μ
*
∫
Ω
∫
Ω
(
v
k
2
μ
*
(
x
)
−
z
k
2
μ
*
(
x
)
)
z
k
2
μ
*
(
y
)
|
x
−
y
|
μ
d
x
d
y
.
Employing the facts that
v
̲
and
v
̄
are respectively the sub- and supersolutions of
(
P
λ
)
we obtain from (4.3)
J
λ
(
v
k
)
≥
J
λ
(
z
k
)
+
I
k
+
J
k
,
where
I
k
=
1
2
∫
A
̄
k
|
∇
w
̄
k
|
2
d
x
+
λ
∫
A
̄
k
χ
{
v
̄
<
a
}
v
̄
−
γ
w
̄
k
−
H
(
v
̄
+
w
̄
k
)
−
H
(
v
̄
)
d
x
+
λ
2
∫
Ω
∫
A
̄
k
v
̄
2
μ
*
(
y
)
v
̄
2
μ
*
−
1
(
x
)
w
̄
k
(
x
)
−
1
2
μ
*
(
v
̄
+
w
̄
k
)
2
μ
*
(
x
)
−
v
̄
2
μ
*
(
x
)
v
k
2
μ
*
(
y
)
+
z
k
2
μ
*
(
y
)
|
x
−
y
|
μ
d
x
d
y
and
J
k
=
1
2
∫
A
̲
k
|
∇
w
̲
k
|
2
d
x
−
λ
∫
A
̲
k
χ
{
v
̲
<
a
}
v
̲
−
γ
w
̲
k
+
H
(
v
̲
−
w
̲
k
)
−
H
(
v
̲
)
d
x
−
λ
2
∫
Ω
∫
A
̲
k
v
̲
2
μ
*
(
y
)
v
̲
2
μ
*
−
1
(
x
)
w
̲
k
(
x
)
−
1
2
μ
*
(
v
̲
−
w
̲
k
)
2
μ
*
(
x
)
−
v
̲
2
μ
*
(
x
)
v
k
2
μ
*
(
y
)
+
z
k
2
μ
*
(
y
)
|
x
−
y
|
μ
d
x
d
y
.
Now we claim that I
k
, J
k
≥ 0 for large k which is a contradiction to our assumption that J
λ
(v
k
) < J
λ
(v
λ
) for all k. We only show that I
k
≥ 0. The case of J
k
≥ 0 runs in a similar fashion.
Dividing
A
̄
k
into three subdomains, viz,
A
̄
k
∩
{
x
∈
Ω
:
a
<
v
̄
(
x
)
}
,
A
̄
k
∩
{
x
∈
Ω
:
v
̄
(
x
)
≤
a
≤
(
v
̄
+
w
̄
k
)
(
x
)
}
and
A
̄
k
∩
{
x
∈
Ω
:
(
v
̄
+
w
̄
k
)
(
x
)
<
a
}
, one can check that the second integral on the right hand side of I
k
is nonnegative. Now using the fact that
z
k
≤
v
̄
and the mean value theorem, we obtain, for some θ = θ(x) ∈ (0, 1) that
I
k
,
1
=
∫
Ω
∫
A
̄
k
v
̄
2
μ
*
(
y
)
v
̄
2
μ
*
−
1
(
x
)
w
̄
k
(
x
)
−
1
2
μ
*
(
v
̄
+
w
̄
k
)
2
μ
*
(
x
)
−
v
̄
2
μ
*
(
x
)
z
k
2
μ
*
(
y
)
|
x
−
y
|
μ
d
x
d
y
≥
∫
Ω
v
̄
2
μ
*
(
y
)
|
x
−
y
|
μ
d
y
∫
A
̄
k
v
̄
2
μ
*
−
1
(
x
)
w
̄
k
(
x
)
−
1
2
μ
*
(
v
̄
+
w
̄
k
)
2
μ
*
(
x
)
−
v
̄
2
μ
*
(
x
)
d
x
=
−
∫
Ω
v
̄
2
μ
*
(
y
)
|
x
−
y
|
μ
d
y
∫
A
̄
k
v
̄
+
θ
w
̄
k
2
μ
*
−
1
(
x
)
−
v
̄
2
μ
*
−
1
(
x
)
w
̄
k
(
x
)
d
x
≥
−
∫
Ω
v
̄
2
μ
*
(
y
)
|
x
−
y
|
μ
d
y
∫
A
̄
K
(
v
̄
+
w
̄
k
)
2
μ
*
−
2
w
̄
k
2
d
x
≥
−
M
∫
Ω
∫
A
̄
k
v
̄
2
μ
*
(
y
)
v
̄
2
μ
*
−
2
(
x
)
+
w
̄
k
2
μ
*
−
2
(
x
)
w
̄
k
2
(
x
)
|
x
−
y
|
μ
d
x
d
y
.
Now by the application of Hardy–Littlewood–Sobolev and Hölder’s inequalities, we obtain
(4.4)
I
k
,
1
≥
M
1
|
v
̄
|
2
*
2
μ
*
∫
A
̄
k
v
̄
2
*
d
x
2
μ
*
−
2
2
μ
*
‖
w
̄
k
‖
2
+
‖
w
̄
k
‖
2
μ
*
.
Also
I
k
,
2
=
∫
Ω
∫
A
̄
k
v
̄
2
μ
*
(
y
)
v
̄
2
μ
*
−
1
(
x
)
w
̄
k
(
x
)
|
x
−
y
|
μ
d
x
d
y
−
1
2
μ
*
∫
Ω
\
A
̄
k
∫
A
̄
k
(
v
̄
+
w
̄
k
)
2
μ
*
(
x
)
−
v
̄
2
μ
*
(
x
)
v
̄
2
μ
*
(
y
)
|
x
−
y
|
μ
d
x
d
y
−
1
2
μ
*
∫
A
̄
k
∫
A
̄
k
(
v
̄
+
w
̄
k
)
2
μ
*
(
x
)
−
v
̄
2
μ
*
(
x
)
v
k
2
μ
*
(
y
)
|
x
−
y
|
μ
d
x
d
y
≥
∫
Ω
∫
A
̄
k
v
̄
2
μ
*
(
y
)
v
̄
2
μ
*
−
1
(
x
)
w
̄
k
(
x
)
−
1
2
μ
*
(
v
̄
+
w
̄
k
)
2
μ
*
(
x
)
−
v
̄
2
μ
*
(
x
)
|
x
−
y
|
μ
d
x
d
y
−
1
2
μ
*
∫
A
̄
k
∫
A
̄
k
(
v
̄
+
w
̄
k
)
2
μ
*
(
x
)
−
v
̄
2
μ
*
(
x
)
v
k
2
μ
*
(
y
)
|
x
−
y
|
μ
d
x
d
y
.
Now the first integral in the last inequality can be estimated like I
k,1. In the following we will estimate the second integral. Again using the mean value theorem, Hardy–Littlewood–Sobolev and Hölder’s inequalities, we have
(4.5)
1
2
μ
*
∫
A
̄
k
∫
A
̄
k
(
v
̄
+
w
̄
k
)
2
μ
*
(
x
)
−
v
̄
2
μ
*
(
x
)
v
k
2
μ
*
(
y
)
|
x
−
y
|
μ
d
x
d
y
≤
∫
A
̄
k
∫
A
̄
k
(
v
̄
+
w
̄
k
)
2
μ
*
−
1
(
x
)
w
̄
k
(
x
)
v
k
2
μ
*
(
y
)
|
x
−
y
|
μ
d
x
d
y
≤
M
‖
v
̄
‖
L
2
*
(
A
̄
k
)
+
‖
w
̄
k
‖
L
2
*
(
A
̄
k
)
2
μ
*
−
1
2
*
2
‖
w
̄
k
‖
2
2
*
=
o
k
(
1
)
.
Now using and Claim A, we deduce that I
k
≥ 0. This completes the proof of the theorem.□
5 Existence of a second solution
This section is devoted to obtain a second solution for
(
P
λ
)
for λ ∈ (0, Λ
a
). Here we restrict ourselves to the case 0 < γ < 3. We obtain the second solution by translating the problem to the solution v
λ
obtained in the previous section. Precisely, we consider the following problem
(
P
̂
λ
)
−
Δ
w
=
λ
χ
{
w
+
v
λ
<
a
}
(
w
+
v
λ
)
−
γ
−
χ
{
v
λ
<
a
}
v
λ
−
γ
+
λ
∫
Ω
(
w
+
v
λ
)
2
μ
*
(
y
)
(
w
+
v
λ
)
2
μ
*
−
1
(
x
)
−
v
λ
2
μ
*
(
y
)
v
λ
2
μ
*
−
1
(
x
)
|
x
−
y
|
μ
dy
in
Ω
,
w
>
0
in
Ω
,
w
=
0
on
∂
Ω
.
Clearly, if
w
λ
∈
H
0
1
(
Ω
)
weakly solves
(
P
̂
λ
)
, then w
λ
+ v
λ
weakly solves
(
P
λ
)
. Let us define, for x ∈ Ω,
f
(
x
,
s
)
=
χ
{
s
+
v
λ
<
a
}
(
s
+
v
λ
)
−
γ
−
χ
{
v
λ
<
a
}
v
λ
−
γ
χ
R
+
(
s
)
.
Let
F
(
x
,
t
)
=
∫
0
t
f
(
x
,
s
)
d
s
. Now the energy functional
G
λ
:
H
0
1
(
Ω
)
→
R
associated with
(
P
̂
λ
)
is given as:
G
λ
(
w
)
=
1
2
‖
w
‖
2
−
λ
∫
Ω
F
(
x
,
w
)
d
x
−
λ
2
2
μ
*
∫
Ω
∫
Ω
(
w
+
v
λ
)
2
μ
*
(
y
)
(
w
+
v
λ
)
2
μ
*
(
x
)
|
x
−
y
|
μ
d
x
d
y
−
λ
∫
Ω
∫
Ω
v
λ
2
μ
*
(
y
)
v
λ
2
μ
*
−
1
(
x
)
w
(
x
)
|
x
−
y
|
μ
d
x
d
y
.
Proposition 5.1.
The map
G
λ
is locally Lipschitz.
Proof.
The proof is similar to the proof of [8], Proposition 3.1] and hence omitted.□
Remark 5.2.
Note that
J
λ
w
+
+
v
λ
=
J
λ
(
v
λ
)
+
G
λ
(
w
)
−
1
2
‖
w
−
‖
2
for any
w
∈
H
0
1
(
Ω
)
. Therefore, since v
λ
is a local minimum of J
λ
, it follows that 0 is a local minimum of
G
λ
in
H
0
1
(
Ω
)
-topology.
Definition 5.3.
Let
Φ
:
H
0
1
(
Ω
)
→
R
be a locally Lipschitz map. The generalized derivative of Φ at u in the direction of v (denoted by Φ0(u, v)) is defined as:
Φ
0
(
u
,
v
)
=
lim sup
h
→
0
,
t
↓
0
Φ
(
u
+
h
+
t
v
)
−
Φ
(
u
+
h
)
t
,
u
,
v
∈
H
0
1
(
Ω
)
.
We say that u is ‘generalized’ critical point of Φ if Φ0(u, v) ≥ 0 for all
v
∈
H
0
1
(
Ω
)
.
Remark 5.4.
From [7], Definition 4.1], for w ≥ 0 and
ψ
∈
H
0
1
(
Ω
)
, we have the following inequality:
(5.1)
G
λ
0
(
w
,
ψ
)
=
∫
Ω
∇
(
v
λ
+
w
)
∇
ψ
d
x
−
λ
∫
Ω
∫
Ω
(
v
λ
+
w
)
2
μ
*
(
y
)
(
v
λ
+
w
)
2
μ
*
−
1
(
x
)
ψ
(
x
)
|
x
−
y
|
μ
d
x
d
y
−
λ
∫
Ω
z
ψ
(
v
λ
+
w
)
−
γ
ψ
d
x
,
for some measurable function
z
ψ
∈
[
χ
{
v
λ
+
w
<
a
}
,
χ
{
v
λ
+
w
≤
a
}
]
.
Remark 5.5.
Suppose for some nontrivial nonnegative
w
λ
∈
H
0
1
(
Ω
)
we have
G
λ
(
w
λ
,
ψ
)
≥
0
for all
ψ
∈
H
0
1
(
Ω
)
, i.e., w
λ
is a generalized critical point of
G
λ
. Then we claim that
(5.2)
λ
∫
Ω
(
v
λ
+
w
λ
)
2
μ
*
(
y
)
(
v
λ
+
w
λ
)
2
μ
*
−
1
(
x
)
|
x
−
y
|
μ
d
y
≤
−
Δ
(
v
λ
+
w
λ
)
≤
λ
∫
Ω
(
v
λ
+
w
λ
)
2
μ
*
(
y
)
(
v
λ
+
w
λ
)
2
μ
*
−
1
(
x
)
|
x
−
y
|
μ
d
y
+
(
v
λ
+
w
λ
)
−
γ
.
Indeed, since w
λ
≥ 0 and
G
λ
0
(
w
λ
,
ψ
)
≥
0
, using (5.1), we have for any
ψ
∈
H
0
1
(
Ω
)
(5.3)
0
≤
G
λ
0
(
w
λ
,
ψ
)
=
∫
Ω
∇
(
v
λ
+
w
λ
)
∇
ψ
d
x
−
λ
∫
Ω
∫
Ω
(
v
λ
+
w
λ
)
2
μ
*
(
y
)
(
v
λ
+
w
λ
)
2
μ
*
−
1
(
x
)
|
x
−
y
|
μ
ψ
d
x
d
y
−
λ
∫
Ω
z
ψ
(
v
λ
+
w
λ
)
−
γ
ψ
d
x
.
Taking ψ ≥ 0 in (5.3), we have
λ
∫
Ω
∫
Ω
(
v
λ
+
w
)
2
μ
*
(
y
)
(
v
λ
+
w
)
2
μ
*
−
1
(
x
)
|
x
−
y
|
μ
ψ
d
x
d
y
+
λ
∫
Ω
z
ψ
(
v
λ
+
w
)
−
γ
ψ
d
x
≤
∫
Ω
∇
(
v
λ
+
w
)
∇
ψ
d
x
.
Since z
ψ
≥ 0 and given that ψ ≥ 0, we have
(5.4)
λ
∫
Ω
∫
Ω
(
v
λ
+
w
)
2
μ
*
(
y
)
(
v
λ
+
w
)
2
μ
*
−
1
(
x
)
|
x
−
y
|
μ
ψ
d
x
d
y
≤
∫
Ω
∇
(
v
λ
+
w
)
∇
ψ
d
x
.
Next let us consider
φ
∈
H
0
1
(
Ω
)
which is nonpositive, so that ψ = −φ ≥ 0. Again using (5.3), we have
λ
∫
Ω
∫
Ω
(
v
λ
+
w
)
2
μ
*
(
y
)
(
v
λ
+
w
)
2
μ
*
−
1
(
x
)
|
x
−
y
|
μ
φ
d
x
d
y
+
λ
∫
Ω
z
φ
(
v
λ
+
w
)
−
γ
φ
d
x
≤
∫
Ω
∇
(
v
λ
+
w
)
∇
φ
d
x
.
Multiplying by −1 on both sides and using the fact that z
φ
∈ [0, 1], we get
λ
∫
Ω
∫
Ω
(
v
λ
+
w
)
2
μ
*
(
y
)
(
v
λ
+
w
)
2
μ
*
−
1
(
x
)
ψ
(
x
)
|
x
−
y
|
μ
d
x
d
y
+
λ
∫
Ω
(
v
λ
+
w
)
−
γ
ψ
d
x
≥
∫
Ω
∇
(
v
λ
+
w
)
∇
ψ
d
x
.
Since ψ = −φ is any arbitrary nonnegative function in
H
0
1
(
Ω
)
, the previous expression implies that
(5.5)
−
Δ
(
v
λ
+
w
λ
)
≤
λ
∫
Ω
(
v
λ
+
w
λ
)
2
μ
*
(
y
)
(
v
λ
+
w
λ
)
2
μ
*
−
1
(
x
)
|
x
−
y
|
μ
d
y
+
(
v
λ
+
w
λ
)
−
γ
weakly
.
Combining (5.4) and (5.5), we have the validity of (5.2). Hence the claim.
Note that −Δ(v
λ
+ w
λ
) is a positive distribution and hence it is given by a positive, regular Radon measure say ν. Then using (5.2), we can show that ν is absolutely continuous with respect to the Lebesgue measure. Now by Radon Nikodyn theorem there exists a locally integrable function g such that −Δ(v
λ
+ w
λ
) = g and hence
g
∈
L
loc
p
(
Ω
)
for some p > 1. Now using Lemma B.2 of [28] and elliptic regularity, we can conclude that
v
λ
+
w
λ
∈
W
loc
2
,
q
(
Ω
)
for all q < ∞ and for almost every x ∈ Ω, using (5.2) we have −Δ(v
λ
+ w
λ
) > 0. In particular,
(5.6)
−
Δ
(
v
λ
+
w
λ
)
>
0
for a.e on
{
x
∈
Ω
:
(
v
λ
+
w
λ
)
(
x
)
=
a
}
.
On the other hand, we have −Δ(v
λ
+ w
λ
) = 0 a.e on the set {x ∈ Ω: (v
λ
+ w
λ
)(x) = a}. This contradicts (5.6) unless the Lebesgue measure of the set {x ∈ Ω: (v
λ
+ w
λ
)(x) = a} is zero. Therefore
z
ψ
=
χ
{
v
λ
+
w
λ
<
a
}
a.e. in Ω for any
ψ
∈
H
0
1
(
Ω
)
and hence v
λ
+ w
λ
is a second solution for
(
P
λ
)
.
Our next target is to show the existence of a generalized critical point for
G
λ
which gives us the second solution of
(
P
λ
)
. We will employ the Mountain Pass theorem and Ekeland variational principle to this end. We define
X
+
=
u
∈
H
0
1
(
Ω
)
:
u
≥
0
a.e in
Ω
. Since 0 is a local minimum of
G
λ
, there exists a κ
0 > 0 such that
G
λ
(
0
)
≤
G
λ
(
u
)
for ‖u‖ ≤ κ
0. Then the following two cases arise:
ZA (Zero Altitude):
inf
G
λ
(
w
)
:
‖
w
‖
=
κ
,
w
∈
X
+
=
G
λ
(
0
)
=
0
for all κ ∈ (0, κ
0).
MP (Mountain Pass): There exists κ
1 ∈ (0, κ
0) such that
inf
{
G
λ
(
w
)
=
κ
1
,
w
∈
X
+
}
>
G
λ
(
0
)
.
Lemma 5.6.
Let ZA hold for some λ ∈ (0, Λ
a
). Then there exists a nontrivial ‘generalized’ critical point w
λ
∈ X
+ for
G
λ
.
Proof.
Fix κ ∈ (0, κ
0). Then there exists a sequence
{
v
k
}
k
∈
N
⊂
X
+
with ‖v
k
‖ = κ and
G
λ
(
v
k
)
≤
1
k
. Fix
0
<
q
<
1
2
min
{
κ
0
−
κ
,
κ
}
and define
R
=
{
w
∈
X
+
:
κ
−
q
≤
‖
w
‖
≤
κ
+
q
}
. Note that
R
is closed and
G
λ
is Lipschitz continuous on
R
(in view of Proposition 5.1). Thus by Ekeland’s variational Principle, there exists
{
ℓ
k
}
k
∈
N
⊂
R
such that the following holds:
G
λ
(
ℓ
k
)
≤
G
λ
(
v
k
)
≤
1
k
,
G
λ
(
ℓ
k
)
≤
G
λ
(
l
)
+
1
k
‖
l
−
ℓ
k
‖
for all
l
∈
R
.
We note that
(5.7)
κ
−
1
k
=
‖
v
k
‖
−
1
k
≤
‖
ℓ
k
‖
≤
‖
v
k
‖
+
1
k
=
κ
+
1
k
.
Therefore, for ℵ ∈ X
+ we can choose ϵ > 0 sufficiently small such that
ℓ
k
+
ϵ
(
ℵ
−
ℓ
k
)
∈
R
for all large k. Then by item 3 above, we get
G
λ
(
ℓ
k
+
ϵ
(
ℵ
−
ℓ
k
)
)
−
G
λ
(
ℓ
k
)
ϵ
≥
−
1
k
‖
ℵ
−
ℓ
k
‖
.
Letting ϵ → 0+, we conclude
G
λ
0
(
ℓ
k
,
ℵ
−
ℓ
k
)
≥
−
1
k
‖
ℵ
−
ℓ
k
‖
for all
ℵ
∈
X
+
.
From Remark 5.4, for any ℵ ∈ X
+, there exists
z
k
ℵ
−
ℓ
k
∈
[
χ
{
v
λ
+
ℓ
k
<
a
}
,
χ
{
v
λ
+
ℓ
k
≤
a
}
]
such that
(5.8)
∫
Ω
∇
(
v
λ
+
ℓ
k
)
∇
(
ℵ
−
ℓ
k
)
d
x
−
λ
∫
Ω
∫
Ω
(
v
λ
+
ℓ
k
)
2
μ
*
(
y
)
(
v
λ
+
ℓ
k
)
2
μ
*
−
1
(
x
)
(
ℵ
−
ℓ
k
)
(
x
)
|
x
−
y
|
μ
d
x
d
y
−
λ
∫
Ω
z
k
ℵ
−
ℓ
k
(
v
λ
+
ℓ
k
)
−
γ
(
ℵ
−
ℓ
k
)
d
x
≥
−
1
k
‖
ℵ
−
ℓ
k
‖
.
Since
{
ℓ
k
}
k
∈
N
is bounded in
H
0
1
(
Ω
)
, we may assume ℓ
k
⇀ w
λ
∈ X
+ weakly in
H
0
1
(
Ω
)
as well as a.e. in Ω. In the following we show that w
λ
is a solution of
(
P
̂
λ
)
. For
φ
∈
C
c
∞
(
Ω
)
, set
φ
k
,
ϵ
=
(
ℓ
k
+
ϵ
φ
)
−
and
ℓ
=
ℓ
k
+
ϵ
φ
+
φ
k
,
ϵ
=
(
l
k
+
ϵ
φ
)
+
.
Thus ℓ ∈ X
+ and with this choice of ℓ, (5.8) becomes
(5.9)
∫
Ω
∇
(
v
λ
+
ℓ
k
)
∇
(
ϵ
φ
+
φ
k
,
ϵ
)
d
x
−
λ
∫
Ω
∫
Ω
(
v
λ
+
ℓ
k
)
2
μ
*
(
y
)
(
v
λ
+
ℓ
k
)
2
μ
*
−
1
(
x
)
(
ϵ
φ
+
φ
k
,
ϵ
)
(
x
)
|
x
−
y
|
μ
d
x
d
y
−
λ
∫
Ω
z
k
ϵ
φ
+
φ
ϵ
,
k
(
v
λ
+
ℓ
k
)
−
γ
(
ϵ
φ
+
φ
k
,
ϵ
)
d
x
≥
−
1
k
‖
ϵ
φ
+
φ
k
,
ϵ
‖
,
where now in view of the fact {x ∈ Ω: (ϵφ + φ
k,ϵ
)(x) ≤ 0} = {x ∈ Ω: φ(x) ≤ 0}, we have
(5.10)
z
k
ϵ
φ
+
φ
k
,
ϵ
=
χ
{
v
λ
+
ℓ
k
<
a
}
+
χ
{
v
λ
+
ℓ
k
=
a
}
∩
{
φ
≤
0
}
.
For a fixed ϵ, we now let k → ∞ and show that we can pass to the required limits in each of the terms in (5.9). We have
φ
k
,
ϵ
⇀
(
w
λ
+
ϵ
φ
)
−
in
H
0
1
(
Ω
)
. It can be shown as in [18], Lemma 7.5.2] that
(5.11)
∫
Ω
∇
(
v
λ
+
ℓ
k
)
∇
(
ϵ
φ
+
φ
k
,
ϵ
)
d
x
≤
∫
Ω
∇
(
v
λ
+
w
λ
)
∇
(
ϵ
φ
+
(
w
λ
+
ϵ
φ
)
−
)
d
x
+
o
k
(
1
)
.
Clearly
z
k
ϵ
φ
+
φ
k
,
ϵ
is bounded in Ω and hence
z
k
ϵ
φ
+
φ
k
,
ϵ
→
z
̃
weak* in L
∞(Ω). Since
(
ϵ
φ
+
φ
k
,
ϵ
)
(
v
λ
+
ℓ
k
)
−
γ
→
(
ϵ
φ
+
(
w
λ
+
ϵ
φ
)
−
)
(
v
λ
+
w
λ
)
−
γ
in
L
1
(
Ω
)
,
we conclude that
(5.12)
∫
Ω
z
k
ϵ
φ
+
φ
k
,
ϵ
(
ϵ
φ
+
φ
k
,
ϵ
)
(
v
λ
+
ℓ
k
)
−
γ
d
x
→
∫
Ω
z
̃
(
ϵ
φ
+
(
w
λ
+
ϵ
φ
)
−
)
(
v
λ
+
w
λ
)
−
γ
d
x
.
Now it is standard that
(5.13)
∫
Ω
∫
Ω
(
v
λ
+
ℓ
k
)
2
μ
*
(
y
)
(
v
λ
+
ℓ
k
)
2
μ
*
−
1
(
x
)
φ
(
x
)
|
x
−
y
|
μ
d
x
d
y
→
∫
Ω
∫
Ω
(
v
λ
+
w
λ
)
2
μ
*
(
y
)
(
v
λ
+
w
λ
)
2
μ
*
−
1
(
x
)
φ
(
x
)
|
x
−
y
|
μ
d
x
d
y
.
Again since 0 ≤ φ
k,ϵ
≤ ϵ|φ|, we see that
(5.14)
∫
Ω
∫
Ω
(
v
λ
+
ℓ
k
)
2
μ
*
(
y
)
(
v
λ
+
ℓ
k
)
2
μ
*
−
1
(
x
)
φ
k
,
ϵ
|
x
−
y
|
μ
d
x
d
y
→
∫
Ω
∫
Ω
(
v
λ
+
w
λ
)
2
μ
*
(
y
)
(
v
λ
+
w
λ
)
2
μ
*
−
1
(
x
)
(
w
λ
+
ϵ
φ
)
−
|
x
−
y
|
μ
d
x
d
y
.
Combining , we conclude that
(5.15)
0
≤
∫
Ω
∇
(
v
λ
+
w
λ
)
∇
(
ϵ
φ
+
(
w
λ
+
ϵ
φ
)
−
)
d
x
−
λ
∫
Ω
z
̃
(
v
λ
+
w
λ
)
−
γ
(
ϵ
φ
+
(
w
λ
+
ϵ
φ
)
−
)
d
x
−
λ
∫
Ω
∫
Ω
(
v
λ
+
w
λ
)
2
μ
*
(
y
)
(
v
λ
+
w
λ
)
2
μ
*
−
1
(
x
)
(
ϵ
φ
+
(
w
λ
+
ϵ
φ
)
−
)
(
x
)
|
x
−
y
|
μ
d
x
d
y
.
Note that
0
≤
z
̃
≤
1
a.e. in Ω and
z
̃
depends only upon φ. Rewriting the above equation using the fact that v
λ
solves
(
P
λ
)
in the weak sense, we obtain
(5.16)
∫
Ω
∇
(
v
λ
+
w
λ
)
∇
φ
d
x
−
λ
∫
Ω
∫
Ω
(
v
λ
+
w
λ
)
2
μ
*
(
y
)
(
v
λ
+
w
λ
)
2
μ
*
−
1
(
x
)
φ
(
x
)
|
x
−
y
|
μ
d
x
d
y
−
λ
∫
Ω
z
̃
(
v
λ
+
w
λ
)
−
γ
φ
d
x
≥
−
1
ϵ
∫
Ω
∇
w
λ
∇
(
w
λ
+
ϵ
φ
)
−
d
x
+
λ
ϵ
∫
Ω
(
z
̃
(
v
λ
+
w
λ
)
−
γ
−
χ
{
v
λ
<
a
}
v
λ
−
γ
)
(
w
λ
+
ϵ
φ
)
−
d
x
+
λ
ϵ
∫
Ω
∫
Ω
(
(
v
λ
+
w
λ
)
2
μ
*
(
y
)
(
v
λ
+
w
λ
)
2
μ
*
−
1
(
x
)
−
v
λ
2
μ
*
(
y
)
v
λ
2
μ
*
−
1
(
x
)
)
(
w
λ
+
ϵ
φ
)
−
(
x
)
|
x
−
y
|
μ
d
x
d
y
.
Let Ω
ϵ
= {x ∈ Ω: w
λ
+ ϵφ ≤ 0}. Note that |Ω
ϵ
| → 0 as ϵ → 0 and hence
−
∫
Ω
∇
w
λ
∇
(
w
λ
+
ϵ
φ
)
−
d
x
=
∫
Ω
ϵ
|
∇
w
λ
|
2
d
x
+
ϵ
∫
Ω
ϵ
∇
w
λ
∇
φ
d
x
≥
o
(
ϵ
)
.
Note that the last term in the RHS of (5.16) is nonnegative. Using the fact that
v
λ
−
γ
φ
∈
L
1
(
Ω
)
, we see that
∫
Ω
(
z
̃
(
v
λ
+
w
λ
)
−
γ
−
χ
{
v
λ
<
a
}
v
λ
−
γ
)
(
w
λ
+
ϵ
φ
)
−
d
x
≤
∫
Ω
ϵ
2
v
λ
−
γ
(
w
λ
+
ϵ
φ
)
−
d
x
≤
2
ϵ
∫
Ω
ϵ
v
λ
−
γ
|
φ
|
d
x
=
o
(
ϵ
)
.
Letting ϵ → 0 in (5.16), it can be seen that given any
φ
∈
H
0
1
(
Ω
)
, there exists
z
̃
=
z
̃
φ
with
0
≤
z
̃
≤
1
such that
0
≤
∫
Ω
∇
(
v
λ
+
w
λ
)
∇
φ
d
x
−
λ
∫
Ω
∫
Ω
(
v
λ
+
w
λ
)
2
μ
*
(
y
)
(
v
λ
+
w
λ
)
2
μ
*
−
1
(
x
)
φ
(
x
)
|
x
−
y
|
μ
d
x
d
y
−
λ
∫
Ω
z
̃
(
v
λ
+
w
λ
)
−
γ
φ
d
x
.
Moreover since
z
̃
is obtained as the weak* limit of
z
k
ϵ
φ
+
φ
k
,
ϵ
, we get that
z
̃
=
1
a.e. in {v
λ
+ w
λ
< a} and
z
̃
=
0
a.e. in {v
λ
+ w
λ
> a}. Therefore from Remark 5.5, w
λ
is a generalized critical point of
G
λ
. It remains to show that w
λ
≢ 0. Note that if
G
λ
(
w
λ
)
≠
0
we are done. So assume
G
λ
(
w
λ
)
=
0
. From (5.7), we see that
‖
ℓ
k
‖
≥
κ
2
for all large k. Thus it is sufficient to show that ℓ
k
→ w
λ
strongly in
H
0
1
(
Ω
)
. Taking ℵ = w
λ
in (5.8), we get
(5.17)
∫
Ω
∇
(
v
λ
+
w
λ
)
∇
(
w
λ
−
ℓ
k
)
d
x
−
λ
∫
Ω
∫
Ω
(
v
λ
+
ℓ
k
)
2
μ
*
(
y
)
(
v
λ
+
ℓ
k
)
2
μ
*
−
1
(
x
)
(
w
λ
−
ℓ
k
)
(
x
)
|
x
−
y
|
μ
d
x
d
y
−
λ
∫
Ω
z
k
w
λ
−
ℓ
k
(
v
λ
+
ℓ
k
)
−
γ
(
w
λ
−
ℓ
k
)
d
x
+
1
k
‖
w
λ
−
ℓ
k
‖
≥
‖
w
λ
−
ℓ
k
‖
2
.
Now for any measurable set E ⊂ Ω, as v
λ
≥ M
1
ϕ
γ
and ℓ
k
∈ X
+, thanks to Hardy’s inequality, we have
(5.18)
∫
E
z
k
w
λ
−
ℓ
k
|
ℓ
k
−
w
λ
|
(
v
λ
+
ℓ
k
)
−
γ
d
x
≤
M
∫
E
|
ℓ
k
−
w
λ
|
v
λ
γ
d
x
≤
M
∫
E
|
ℓ
k
−
w
λ
|
δ
(
x
)
2
γ
1
+
γ
d
x
≤
M
∫
E
|
ℓ
k
−
w
λ
|
δ
(
x
)
δ
(
x
)
1
−
γ
1
+
γ
d
x
≤
M
‖
ℓ
k
−
w
λ
‖
‖
δ
(
x
)
1
−
γ
1
+
γ
‖
L
2
(
E
)
.
Since ℓ
k
→ w
λ
pointwise a.e. in Ω, by Vitali’s convergence theorem,
(5.19)
∫
E
z
k
w
λ
−
ℓ
k
|
ℓ
k
−
w
λ
|
(
v
λ
+
ℓ
k
)
−
γ
d
x
→
0
as
k
→
∞
.
Also using Brezis–Lieb Lemma, we have
(5.20)
∫
Ω
∫
Ω
(
v
λ
+
ℓ
k
)
2
μ
*
(
y
)
(
v
λ
+
ℓ
k
)
2
μ
*
−
1
(
x
)
(
w
λ
−
ℓ
k
)
(
x
)
|
x
−
y
|
μ
d
x
d
y
=
∫
Ω
∫
Ω
(
v
λ
+
ℓ
k
)
2
μ
*
(
y
)
(
v
λ
+
ℓ
k
)
2
μ
*
−
1
(
x
)
(
w
λ
+
v
λ
)
(
x
)
|
x
−
y
|
μ
d
x
d
y
−
∫
Ω
∫
Ω
(
v
λ
+
ℓ
k
)
2
μ
*
(
y
)
(
v
λ
+
ℓ
k
)
2
μ
*
(
x
)
|
x
−
y
|
μ
d
x
d
y
=
−
∫
Ω
∫
Ω
(
w
λ
−
ℓ
k
)
2
μ
*
(
y
)
(
w
λ
−
ℓ
k
)
2
μ
*
(
x
)
|
x
−
y
|
μ
d
x
d
y
+
o
k
(
1
)
.
Now using (5.19) and (5.20) in (5.17), we get
(5.21)
‖
w
λ
−
ℓ
k
‖
2
−
λ
∫
Ω
∫
Ω
(
w
λ
−
ℓ
k
)
2
μ
*
(
y
)
(
w
λ
−
ℓ
k
)
2
μ
*
(
x
)
|
x
−
y
|
μ
d
x
d
y
≤
o
k
(
1
)
.
Again taking ℵ = 2ℓ
k
in (5.8) and using the fact that v
λ
solves
(
P
λ
)
, we obtain
(5.22)
−
1
k
‖
ℓ
k
‖
≤
∫
Ω
∇
v
λ
∇
ℓ
k
d
x
+
∫
Ω
|
∇
ℓ
k
|
2
d
x
−
λ
∫
Ω
∫
Ω
(
v
λ
+
ℓ
k
)
2
μ
*
(
y
)
(
v
λ
+
ℓ
k
)
2
μ
*
−
1
(
x
)
ℓ
k
(
x
)
|
x
−
y
|
μ
d
x
d
y
−
λ
∫
Ω
z
k
ℓ
k
l
k
(
v
λ
+
ℓ
k
)
−
γ
d
x
=
‖
ℓ
k
‖
2
−
λ
∫
Ω
∫
Ω
(
(
v
λ
+
ℓ
k
)
2
μ
*
(
y
)
(
v
λ
+
ℓ
k
)
2
μ
*
−
1
(
x
)
−
v
λ
2
μ
*
(
y
)
v
λ
2
μ
*
−
1
(
x
)
)
ℓ
k
|
x
−
y
|
μ
d
x
d
y
+
λ
∫
Ω
(
χ
{
v
λ
<
a
}
v
λ
−
γ
−
z
k
ℓ
k
(
v
λ
+
ℓ
k
)
−
γ
)
ℓ
k
d
x
=
‖
w
λ
‖
2
+
‖
ℓ
k
−
w
λ
‖
2
−
λ
∫
Ω
∫
Ω
(
(
v
λ
+
ℓ
k
)
2
μ
*
(
y
)
(
v
λ
+
ℓ
k
)
2
μ
*
−
1
(
x
)
−
v
λ
2
μ
*
(
y
)
v
λ
2
μ
*
−
1
(
x
)
)
ℓ
k
(
x
)
|
x
−
y
|
μ
d
x
d
y
+
λ
∫
Ω
(
χ
{
v
λ
<
a
}
v
λ
−
γ
−
z
k
ℓ
k
(
v
λ
+
ℓ
k
)
−
γ
)
ℓ
k
d
x
+
o
k
(
1
)
.
Now as w
λ
solves
(
P
̂
λ
)
, we have
‖
w
λ
‖
2
=
λ
∫
Ω
∫
Ω
(
(
v
λ
+
w
λ
)
2
μ
*
(
y
)
(
v
λ
+
w
λ
)
2
μ
*
−
1
(
x
)
−
v
λ
2
μ
*
(
y
)
v
λ
2
μ
*
−
1
(
x
)
)
w
λ
(
x
)
|
x
−
y
|
μ
d
x
d
y
+
λ
∫
Ω
(
χ
{
v
λ
+
w
λ
<
a
}
(
v
λ
+
w
λ
)
−
γ
−
χ
{
v
λ
<
a
}
v
λ
−
γ
)
w
λ
d
x
.
Using this identity in (5.22), we get
(5.23)
−
1
k
‖
ℓ
k
‖
≤
‖
ℓ
k
−
w
λ
‖
2
−
λ
∫
Ω
∫
Ω
(
(
v
λ
+
ℓ
k
)
2
μ
*
(
y
)
(
v
λ
+
ℓ
k
)
2
μ
*
−
1
(
x
)
−
v
λ
2
μ
*
(
y
)
v
λ
2
μ
*
−
1
(
x
)
)
ℓ
k
(
x
)
|
x
−
y
|
μ
d
x
d
y
+
λ
∫
Ω
(
χ
{
v
λ
+
w
λ
<
a
}
(
v
λ
+
w
λ
)
−
γ
−
χ
{
v
λ
<
a
}
v
λ
−
γ
)
w
λ
d
x
+
λ
∫
Ω
(
χ
{
v
λ
<
a
}
v
λ
−
γ
−
z
k
ℓ
k
(
v
λ
+
ℓ
k
)
−
γ
)
ℓ
k
d
x
+
λ
∫
Ω
∫
Ω
(
(
v
λ
+
w
λ
)
2
μ
*
(
y
)
(
v
λ
+
w
λ
)
2
μ
*
−
1
(
x
)
−
v
λ
2
μ
*
(
y
)
v
λ
2
μ
*
−
1
(
x
)
)
w
λ
(
x
)
|
x
−
y
|
μ
d
x
d
y
+
o
k
(
1
)
.
Now as ℓ
k
→ w
λ
pointwise a.e. in Ω and |{x ∈ Ω: (v
λ
= w
λ
)(x) = a}| = 0, using estimates similar to (5.18) we have
∫
Ω
(
χ
{
v
λ
+
w
λ
<
a
}
(
v
λ
+
w
λ
)
−
γ
−
χ
{
v
λ
<
a
}
v
λ
−
γ
)
w
λ
d
x
+
∫
Ω
(
χ
{
v
λ
<
a
}
v
λ
−
γ
−
z
k
ℓ
k
(
v
λ
+
ℓ
k
)
−
γ
)
ℓ
k
d
x
=
o
k
(
1
)
.
Using above estimate and Brezis–Lieb lemma, we obtain form (5.23)
(5.24)
o
k
(
1
)
≤
‖
ℓ
k
−
w
λ
‖
2
−
λ
∫
Ω
∫
Ω
(
w
λ
−
ℓ
k
)
2
μ
*
(
y
)
(
w
λ
−
ℓ
k
)
2
μ
*
(
x
)
|
x
−
y
|
μ
d
x
d
y
.
Also as
G
λ
(
ℓ
k
)
≤
1
k
, we have
G
λ
(
ℓ
k
)
=
1
2
‖
ℓ
k
‖
2
−
λ
∫
Ω
F
(
x
,
ℓ
k
)
d
x
−
λ
2
2
μ
*
∫
Ω
∫
Ω
(
ℓ
k
+
v
λ
)
2
μ
*
(
y
)
(
ℓ
k
+
v
λ
)
2
μ
*
(
x
)
|
x
−
y
|
μ
d
x
d
y
−
λ
∫
Ω
∫
Ω
v
λ
2
μ
*
(
y
)
v
λ
2
μ
*
−
1
(
x
)
ℓ
k
(
x
)
|
x
−
y
|
μ
d
x
d
y
≤
1
k
.
From the fact that ℓ
k
⇀ w
λ
weakly in
H
0
1
(
Ω
)
, this implies
(5.25)
1
2
‖
ℓ
k
−
w
λ
‖
2
−
λ
2
2
μ
*
∫
Ω
∫
Ω
(
w
λ
−
ℓ
k
)
2
μ
*
(
y
)
(
w
λ
−
ℓ
k
)
2
μ
*
(
x
)
|
x
−
y
|
μ
d
x
d
y
+
G
λ
(
w
λ
)
+
λ
∫
Ω
F
(
x
,
w
λ
)
d
x
−
λ
∫
Ω
F
(
x
,
ℓ
k
)
d
x
≤
o
k
(
1
)
.
Now using the Hardy’s inequality and Vitali’s convergence theorem as in (5.18) one can check that
∫
Ω
F
(
x
,
ℓ
k
)
d
x
→
∫
Ω
F
(
x
,
w
λ
)
d
x
as k → ∞. Also as
G
λ
(
w
λ
)
=
0
, (5.25) implies
(5.26)
1
2
‖
ℓ
k
−
w
λ
‖
2
−
λ
2
2
μ
*
∫
Ω
∫
Ω
(
w
λ
−
ℓ
k
)
2
μ
*
(
y
)
(
w
λ
−
ℓ
k
)
2
μ
*
(
x
)
|
x
−
y
|
μ
d
x
d
y
≤
o
k
(
1
)
.
Now from (5.21), (5.24) and (5.26) we get
1
2
−
1
2
2
μ
*
‖
ℓ
k
−
w
λ
‖
≤
o
k
(
1
)
and hence ℓ
k
→ w
λ
in
H
0
1
(
Ω
)
.□
Next we consider the case (MP). Here to deal with the critical nonlinearity, we use the following Talenti functions to study the critical level:
V
ϵ
(
x
)
=
S
(
n
−
μ
)
(
2
−
n
)
4
(
n
−
μ
+
2
)
(
C
(
n
,
μ
)
)
2
−
n
2
(
n
−
μ
+
2
)
ϵ
ϵ
2
+
|
x
|
2
n
−
2
2
,
0
<
ϵ
<
1
.
Fix any y ∈ Ω
a
= {x ∈ Ω: v
λ
(x) < a}. Let r > 0 such that B
4r
(y) ⊂ Ω. Now define
ψ
∈
C
c
∞
(
Ω
)
such that 0 ≤ η ≤ 1 in
R
n
, η ≡ 1 in B
r
(y) and η ≡ 0 in
R
n
\
B
2
r
(
y
)
. For each ϵ > 0 and
x
∈
R
n
, we define w
ϵ
(x) = ψ(x)V
ϵ
(x). In the following, we set the notation
‖
u
‖
H
L
2
2
μ
*
=
∫
Ω
∫
Ω
|
u
(
y
)
|
2
μ
*
|
u
(
x
)
|
2
μ
*
|
x
−
y
|
μ
d
x
d
y
.
Proposition 5.7.
Let n ≥ 3, 0 < μ < n then the following holds:
‖
w
ϵ
‖
2
≤
S
H
,
L
2
n
−
μ
n
−
μ
+
2
+
O
(
ϵ
n
−
2
)
.
‖
w
ϵ
‖
H
L
2
2
μ
*
≤
S
H
,
L
2
n
−
μ
n
−
μ
+
2
+
O
(
ϵ
n
)
.
‖
w
ϵ
‖
H
L
2
2
μ
*
≥
S
H
,
L
2
n
−
μ
n
−
μ
+
2
−
O
(
ϵ
n
)
.
Proof.
For proof of part (i), we refer to [29], Lemma 1.46]. For (ii) and (iii), see [30], Proposition 2.8].□
Lemma 5.8.
The following holds:
If μ < min{4, n} then for all κ < 1,
‖
v
λ
+
t
w
ϵ
‖
H
L
2
2
μ
*
≥
‖
v
λ
‖
H
L
2
2
μ
*
+
‖
t
w
ϵ
‖
H
L
2
2
μ
*
+
M
̃
t
2
2
μ
*
−
1
∫
Ω
∫
Ω
w
ϵ
2
μ
*
(
y
)
w
ϵ
2
μ
*
−
1
(
x
)
v
λ
(
x
)
|
x
−
y
|
μ
d
x
d
y
+
2
2
μ
*
t
∫
Ω
∫
Ω
v
λ
2
μ
*
(
y
)
v
λ
2
μ
*
−
1
(
x
)
w
ϵ
(
x
)
|
x
−
y
|
μ
d
x
d
y
−
O
ϵ
2
n
−
μ
4
κ
.
There exists a constant T
0 > 0 such that
∫
Ω
∫
Ω
w
ϵ
2
μ
*
(
y
)
w
ϵ
2
μ
*
−
1
(
x
)
v
λ
(
x
)
|
x
−
y
|
μ
d
x
d
y
≥
C
̃
T
0
ϵ
n
−
2
2
.
Proof.
For a proof, see the proof of [31], Lemma 4.2].□
Next we have the following lemma:
Lemma 5.9.
There exists ϵ
0 and R
0 ≥ 1 such that
G
λ
(
R
w
ϵ
)
<
G
λ
(
0
)
=
0
for all ϵ ∈ (0, ϵ
0) and R ≥ R
0.
G
λ
(
t
R
0
w
ϵ
)
<
1
2
n
−
μ
+
2
2
n
−
μ
S
H
,
L
2
n
−
μ
n
−
μ
+
2
λ
n
−
2
n
−
μ
+
2
for all t ∈ (0, 1] and ϵ ∈ (0, ϵ
0).
Proof.
Noting that for w ∈ X
+,
J
λ
(
v
λ
+
w
)
=
J
λ
(
v
λ
)
+
G
λ
(
w
)
, this is equivalent to show that
J
λ
(v
λ
+ Rw
ϵ
) < J
λ
(v
λ
) for all ϵ ∈ (0, ϵ
0) and R ≥ R
0.
J
λ
(
v
λ
+
t
R
0
w
ϵ
)
<
J
λ
(
v
λ
)
+
1
2
n
−
μ
+
2
2
n
−
μ
S
H
,
L
2
n
−
μ
n
−
μ
+
2
λ
n
−
2
n
−
μ
+
2
for all t ∈ (0, 1] and ϵ ∈ (0, ϵ
0).
Now using the fact that v
λ
solves
(
P
λ
)
, first we estimate J
λ
(v
λ
+ tRw
ϵ
) as follows
J
λ
(
v
λ
+
t
R
w
ϵ
)
=
1
2
∫
Ω
|
∇
(
v
λ
+
t
R
w
ϵ
)
|
2
d
x
−
λ
∫
Ω
F
(
v
λ
+
t
R
w
ϵ
)
d
x
−
λ
2
2
μ
*
‖
v
λ
+
t
R
w
ϵ
‖
H
L
2
2
μ
*
=
1
2
∫
Ω
|
∇
v
λ
|
2
d
x
+
R
2
t
2
2
∫
Ω
|
∇
w
ϵ
|
2
d
x
+
t
R
∫
Ω
∇
v
λ
∇
w
ϵ
d
x
−
λ
2
2
μ
*
‖
v
λ
+
t
R
w
ϵ
‖
H
L
2
2
μ
*
−
λ
∫
Ω
F
(
v
λ
+
t
R
w
ϵ
)
d
x
=
1
2
∫
Ω
|
∇
v
λ
|
2
d
x
+
R
2
t
2
2
∫
Ω
|
∇
w
ϵ
|
2
d
x
+
λ
t
R
∫
Ω
χ
{
v
λ
<
a
}
v
λ
−
γ
w
ϵ
d
x
−
λ
2
2
μ
*
‖
v
λ
+
t
R
w
ϵ
‖
H
L
2
2
μ
*
+
λ
t
R
∫
Ω
∫
Ω
v
λ
2
μ
*
(
y
)
v
λ
2
μ
*
−
1
(
x
)
w
ϵ
(
x
)
|
x
−
y
|
μ
d
x
d
y
−
λ
∫
Ω
F
(
v
λ
+
t
R
w
ϵ
)
d
x
≤
J
λ
(
v
λ
)
+
1
2
‖
t
R
w
ϵ
‖
2
−
λ
2
2
μ
*
‖
t
R
w
ϵ
‖
H
L
2
2
μ
*
−
λ
M
̃
t
2
2
μ
*
−
1
2
2
μ
*
∫
Ω
∫
Ω
w
ϵ
2
μ
*
(
y
)
w
ϵ
2
μ
*
−
1
(
x
)
v
λ
(
x
)
|
x
−
y
|
μ
d
x
d
y
+
λ
∫
Ω
χ
{
v
λ
<
a
}
v
λ
−
γ
w
ϵ
+
F
(
v
λ
)
−
F
(
v
λ
+
t
R
w
ϵ
)
d
x
+
O
ϵ
2
n
−
μ
4
κ
.
Using Proposition 5.7 and Lemma 5.8, we obtain
(5.27)
J
λ
(
v
λ
+
t
R
w
ϵ
)
≤
J
λ
(
v
λ
)
+
t
2
R
2
2
S
H
,
L
2
n
−
μ
n
−
μ
+
2
+
O
ϵ
n
−
2
2
−
λ
t
2
2
μ
*
R
2
2
μ
*
2
2
μ
*
S
H
,
L
2
n
−
μ
n
−
μ
+
2
−
O
(
ϵ
n
)
+
O
ϵ
2
n
−
μ
4
κ
+
λ
∫
Ω
χ
{
v
λ
<
a
}
v
λ
−
γ
w
ϵ
+
F
(
v
λ
)
−
F
(
v
λ
+
t
R
w
ϵ
)
d
x
−
λ
M
̃
t
2
2
μ
*
−
1
2
2
μ
*
T
0
ϵ
n
−
2
2
.
Finally by estimating the last integral in (5.27) using the similar lines of [8], Lemma 3.2] (Page no. 1671–1672), we conclude by taking
κ
=
2
2
μ
*
that
J
λ
(
v
λ
+
t
R
w
ϵ
)
≤
J
λ
(
v
λ
)
+
t
2
R
2
2
S
H
,
L
2
n
−
μ
n
−
μ
+
2
+
O
ϵ
n
−
2
2
−
λ
t
2
2
μ
*
R
2
2
μ
*
2
2
μ
*
S
H
,
L
2
n
−
μ
n
−
μ
+
2
−
O
(
ϵ
n
)
+
O
ϵ
2
n
−
μ
4
κ
−
λ
M
̃
t
2
2
μ
*
−
1
2
2
μ
*
T
0
ϵ
n
−
2
2
+
o
ϵ
n
−
2
2
.
Now the Lemma follows using the arguments as in [23], Lemma 6.4].□
Lemma 5.10.
Let (MP) hold. Then there exists a solution w
λ
∈ X
+ of
(
P
̂
λ
)
and hence a second solution of
(
P
λ
)
.
Proof.
Define a complete metric space (M, d) as
M
=
ζ
∈
C
(
[
0,1
]
,
X
+
)
:
ζ
(
0
)
=
0
,
‖
ζ
(
1
)
‖
>
κ
1
,
G
λ
(
ζ
(
1
)
)
<
0
,
d
(
ζ
,
η
)
=
max
t
∈
[
0,1
]
‖
ζ
(
t
)
−
η
(
t
)
‖
.
From (i) of Lemma 5.9, if R is chosen large, it is clear that M is non-empty. Let
c
0
=
inf
ζ
∈
M
max
t
∈
[
0,1
]
G
λ
(
ζ
(
t
)
)
.
Then (ii) of Lemma 5.9 and (MP) implies that
(5.28)
0
<
c
0
<
1
2
n
−
μ
+
2
2
n
−
μ
S
H
,
L
2
n
−
μ
n
−
μ
+
2
λ
n
−
2
n
−
μ
+
2
.
Define
Φ
(
ζ
)
=
max
t
∈
[
0,1
]
G
λ
(
ζ
(
t
)
)
,
ζ
∈
M
.
By applying Ekeland’s variational principle to the above functional we get a sequence
{
ζ
k
}
k
∈
N
⊂
M
such that
max
t
∈
[
0,1
]
G
λ
(
ζ
k
(
t
)
)
<
c
0
+
1
k
.
max
t
∈
[
0,1
]
G
λ
(
ζ
k
(
t
)
)
≤
max
t
∈
[
0,1
]
G
λ
(
ζ
(
t
)
)
+
1
k
max
t
∈
[
0,1
]
‖
ζ
(
t
)
−
ζ
k
(
t
)
‖
for all ζ ∈ M.
Setting
Γ
k
=
{
t
∈
[
0,1
]
:
G
λ
(
ζ
k
(
t
)
)
=
max
s
∈
[
0,1
]
G
λ
(
ζ
k
(
s
)
)
}
we obtain by arguing as in [5], Page no. 659] t
k
∈ Γ
k
such that for w
k
= ζ
k
(t
k
) and ℵ ∈ X
+, we have
(5.29)
G
λ
0
w
k
,
ℵ
−
w
k
max
{
1
,
‖
ℵ
−
w
k
‖
}
≥
−
1
k
and
(5.30)
G
λ
(
w
k
)
→
c
0
as
k
→
∞
.
From (5.30) and using the fact that F(w
k
) ≤ 0, we have
(5.31)
c
0
+
o
k
(
1
)
≥
1
2
‖
w
k
‖
2
−
λ
2
2
μ
*
‖
w
k
+
v
λ
‖
H
L
2
2
μ
*
−
λ
∫
Ω
∫
Ω
v
λ
2
μ
*
(
y
)
v
λ
2
μ
*
−
1
(
x
)
w
k
(
x
)
|
x
−
y
|
μ
d
x
d
y
.
Again substituting ℵ = 2w
k
+ v
λ
in (5.29), by Remark 5.4 we obtain
z
k
2
w
k
+
v
λ
=
z
̃
k
∈
[
χ
{
w
k
+
v
λ
<
a
}
,
χ
{
w
k
+
v
λ
≤
a
}
]
such that
(5.32)
‖
w
k
+
v
λ
‖
2
−
λ
‖
w
k
+
v
λ
‖
H
L
2
2
μ
*
−
λ
∫
Ω
z
̃
k
(
w
k
+
v
λ
)
1
−
γ
d
x
≥
−
1
n
max
{
1
,
‖
w
k
+
v
λ
‖
}
.
From (5.31) and (5.32) we derive
1
2
‖
w
k
‖
2
−
1
2
μ
*
‖
w
k
‖
2
≤
M
1
+
M
2
‖
w
k
‖
,
where M
1, M
2 are positive constants. This implies that
{
w
k
}
k
∈
N
is a bounded sequence and hence w
k
⇀ w
λ
weakly in
H
0
1
(
Ω
)
. As in the zero altitude case we see that w
λ
solves
(
P
̂
λ
)
. Now we claim that w
k
→ w
λ
in
H
0
1
(
Ω
)
and that w
λ
≢ 0. Without loss of generality we assume
G
λ
(
w
λ
)
≠
0
.
As ‖w
k
‖ ≤ M, from (5.30), for ℵ ∈ X
+ we have
G
λ
0
(
w
k
,
ℵ
−
w
k
)
≥
−
M
1
k
(
1
+
‖
ℵ
‖
)
=
o
k
(
1
)
. Then as in zero altitude case, we get
(5.33)
‖
w
k
−
w
λ
‖
−
λ
‖
w
k
−
w
λ
‖
2
2
μ
*
=
o
k
(
1
)
.
Also using Brezis–Lieb, we have
(5.34)
G
λ
(
w
k
)
=
1
2
‖
w
k
‖
2
−
λ
∫
Ω
F
(
w
k
)
d
x
−
λ
2
2
μ
*
‖
w
k
+
v
λ
‖
H
L
2
2
μ
*
−
λ
∫
Ω
∫
Ω
v
λ
2
μ
*
(
y
)
v
λ
2
μ
*
−
1
(
x
)
w
k
(
x
)
|
x
−
y
|
μ
d
x
d
y
=
1
2
‖
w
k
−
w
λ
‖
2
+
1
2
‖
w
λ
‖
2
+
∫
Ω
∇
(
w
k
−
w
λ
)
∇
w
λ
d
x
−
λ
∫
Ω
F
(
w
k
)
d
x
−
λ
2
2
μ
*
‖
w
k
+
v
λ
‖
H
L
2
2
μ
*
−
λ
∫
Ω
∫
Ω
v
λ
2
μ
*
(
y
)
v
λ
2
μ
*
−
1
(
x
)
w
k
(
x
)
|
x
−
y
|
μ
d
x
d
y
=
1
2
‖
w
k
−
w
λ
‖
2
−
λ
2
2
μ
*
‖
w
k
−
v
λ
‖
H
L
2
2
μ
*
+
1
2
‖
w
λ
‖
2
−
λ
∫
Ω
F
(
w
k
)
d
x
−
λ
2
2
μ
*
‖
w
λ
+
v
λ
‖
H
L
2
2
μ
*
−
λ
∫
Ω
∫
Ω
v
λ
2
μ
*
(
y
)
v
λ
2
μ
*
−
1
(
x
)
w
k
(
x
)
|
x
−
y
|
μ
d
x
d
y
+
o
k
(
1
)
=
1
2
‖
w
k
−
w
λ
‖
2
−
λ
2
2
μ
*
‖
w
k
−
v
λ
‖
H
L
2
2
μ
*
+
G
λ
(
w
λ
)
+
λ
∫
Ω
(
F
(
w
λ
)
−
F
(
w
k
)
)
d
x
+
o
k
(
1
)
=
1
2
‖
w
k
−
w
λ
‖
2
−
λ
2
2
μ
*
‖
w
k
−
v
λ
‖
H
L
2
2
μ
*
+
G
λ
(
w
λ
)
+
o
k
(
1
)
.
Now as
G
λ
(
w
λ
)
=
0
, using (5.28), (5.30), (5.33) and (5.34), we get
(5.35)
‖
w
k
−
w
λ
‖
2
=
2
(
2
n
−
μ
)
n
−
μ
+
2
c
0
+
o
k
(
1
)
<
S
H
,
L
2
n
−
μ
n
−
μ
+
2
λ
n
−
2
n
−
μ
+
2
+
o
k
(
1
)
.
Also by Hardy–Littlewood–Sobolev inequality, we have
(5.36)
‖
w
k
−
w
λ
‖
2
1
−
λ
S
H
,
L
−
2
μ
*
‖
w
k
−
w
λ
‖
2
2
μ
*
−
2
≤
‖
w
k
−
w
λ
‖
2
−
λ
‖
w
k
−
w
λ
‖
H
L
2
2
μ
*
=
o
k
(
1
)
.
Thus combining (5.35) and (5.36), we obtain w
k
→ w
λ
in
H
0
1
(
Ω
)
. This completes the proof.□
We are now ready to give the
Proof of Theorem 1.1:
The existence of the first solution v
λ
for all λ ∈ (0, Λ
a
) and γ > 0 follows from Lemma 4.2. The existence of second solution w
λ
for the same range of λ and 0 < γ < 3 follows from Lemmata 5.6 and 5.10 keeping in view Remark 5.5.□
