1 Introduction
In this paper, we will study the following nonlinear elliptic equation:
(1.1)
-
div
(
A
(
|
x
|
)
∇
u
)
+
V
(
|
x
|
)
u
=
K
(
|
x
|
)
f
(
u
)
in
ℝ
N
,
where N≥3, f:ℝ→ℝ is a continuous nonlinearity satisfying f(0)=0, and V≥0, A,K>0 are given radial potentials. When A=1, the differential operator is the usual laplacian, and this kind of problem has been much studied in recent years, with different sets of hypotheses on the nonlinearity f and the potentials V,K. Much work has been devoted in particular to problems in which such potentials can be vanishing or divergent at 0 and ∞ because this prevents the use of standard embeddings between Sobolev spaces of radial functions, and new embedding and compactness results must be proved (see for example [1, 2, 3, 4, 9, 10, 11, 12, 13, 14, 19, 21, 22] and the references therein). The case in which the potential A is not trivial has been studied in [20, 15, 18] for the p-laplacian equation, in [16, 17] for bounded domains, and in [23] for exterior domains. The typical result obtained in these works says, roughly speaking, that given a suitable asymptotic behavior at 0 and ∞ for the potentials, there is a suitable range of the exponent q such that if f behaves like the power tq-1, then problem (1.1) has a radial solution.
In this paper, we study problem (1.1) using the ideas introduced in [5, 6, 7]. The main novelty of this approach is that the nonlinearity f is not a pure power as before, but has different power-like behaviors at zero and infinity. The typical example is f(t)=min{tq1-1,tq2-1}. Also, we do not introduce hypotheses on the asymptotic behavior of V,K, but on their ratio. The typical result is the existence of two intervals ℐ1,ℐ2 such that if q1∈ℐ1 and q2∈ℐ2, and f as above, then problem (1.1) has a radial solution. When ℐ1∩ℐ2≠∅, it is possible to choose q1=q2=q∈ℐ1∩ℐ2 so that f(t)=tq-1, and we get results similar to those already known in literature. The main technical device for our results is given by compact embeddings of Sobolev spaces of radial functions in sums of Lebesgue spaces. We refer to [8] for an introduction to sums of Lebesgue spaces and to the main results we shall use in this paper. The paper is organized as follows: After the introduction, in Section 2 we define the main function spaces we shall use, and prove some preliminary embedding results. In Section 3, we introduce some sufficient conditions for compactness of the embeddings, and in Section 4 we prove compactness. In Section 5, we apply the previous results to get existence and multiplicity results for (1.1). Finally, in Section 6 we give some concrete examples that, we hope, could help the reader to understand what is new in our results. Notice that the main hypotheses of our results are introduced at the beginning of Section 2, while the main results for (1.1) are Theorems 5.3 and 5.4.
Notations.
We end this introductory section by collecting some notations used in the paper.
For every R>0, we set BR:={x∈ℝN:|x|<R}.
ω
N
is the (N-1)-dimensional measure of the surface ∂B1={x∈ℝN:|x|=1}.
For any subset A⊆ℝN, we denote Ac:=ℝN∖A. If A is Lebesgue measurable, |A| stands for its measure.
By → and ⇀ we respectively mean strong and weak convergence.
The arrow ↪ denotes continuous embeddings.
C
c
∞
(
Ω
)
is the space of the infinitely differentiable real functions with compact support in the open set Ω⊆ℝN, and Cc,r∞(ℝN) is the radial subspace of Cc∞(ℝN). If B is a ball with center in 0, then Cc,r∞(B) is the radial subspace of Cc∞(B).
If 1≤p≤∞, then Lp(A) and Llocp(A) are the usual real Lebesgue spaces (for any measurable set A⊆ℝN). If ρ:A→(0,+∞) is a measurable function, then Lp(A,ρ(z)dz) is the real Lebesgue space with respect to the measure ρ(z)dz (dz stands for the Lebesgue measure on ℝN).
2 Hypotheses and Pointwise Estimates
Assume N≥3. Let V, K and A be three potentials satisfying the following hypotheses:
A
:
(
0
,
+
∞
)
→
(
0
,
+
∞
)
is a continuous function such that there exist real numbers 2-N<a0,a∞≤2 and c0,c∞>0 satisfying
c
0
≤
lim inf
r
→
0
+
A
(
r
)
r
a
0
≤
lim sup
r
→
0
+
A
(
r
)
r
a
0
<
+
∞
,
c
∞
≤
lim inf
r
→
+
∞
A
(
r
)
r
a
∞
≤
lim sup
r
→
+
∞
A
(
r
)
r
a
∞
<
+
∞
.
V
:
(
0
,
+
∞
)
→
[
0
,
+
∞
)
belongs to Lloc1(0,+∞).
K
:
(
0
,
+
∞
)
→
(
0
,
+
∞
)
belongs to Llocs(0,+∞) for some
s
>
max
{
2
N
N
-
a
0
+
2
,
2
N
N
-
a
∞
+
2
}
.
For any q>1, we define the weighted Lebesgue space LKq=Lq(ℝN,K(|x|)dx) whose norm is
∥
u
∥
L
K
q
=
(
∫
ℝ
N
K
(
|
x
|
)
|
u
|
q
𝑑
x
)
1
q
.
Definition 2.1.
For q1,q2>1, we define the sum space ℒK=LKq1+LKq2 as
ℒ
K
=
L
K
q
1
+
L
K
q
2
=
{
u
=
u
1
+
u
2
:
u
i
∈
L
K
q
i
}
with norm
∥
u
∥
ℒ
K
=
inf
{
max
{
∥
u
1
∥
L
K
q
1
,
∥
u
2
∥
L
K
q
2
}
:
u
=
u
1
+
u
2
,
u
i
∈
L
K
q
i
}
.
We refer to [8] for a treatment of such spaces.
We are now going to prove some pointwise estimates for functions in Cc,r∞(ℝN), which are the starting point of our arguments. In this paper, when dealing with a radial function u, we will often write, with a little abuse of notation, u(x)=u(|x|)=u(r) for |x|=r.
Remark 2.2.
It is easy to check that hypothesis [A] implies that, for each R>0, there exist C0=C0(R)>0 and C∞=C∞(R)>0 such that
A
(
|
x
|
)
≥
C
0
|
x
|
a
0
for all
0
<
|
x
|
≤
R
,
A
(
|
x
|
)
≥
C
∞
|
x
|
a
∞
for all
|
x
|
≥
R
.
Lemma 2.3.
Assume hypothesis [A]. Fix R0>0. Then there exists a constant C=C(N,R0,a∞)>0 such that, for each u∈Cc,r∞(RN), there holds
|
u
(
x
)
|
≤
C
|
x
|
-
N
+
a
∞
-
2
2
(
∫
B
R
0
c
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
)
1
2
for
R
0
≤
|
x
|
<
+
∞
.
Proof.
If u∈Cc,r∞(ℝN) and |x|=r≥R0, we have
-
u
(
r
)
=
∫
r
∞
u
′
(
s
)
𝑑
s
.
Using hypothesis [A], we obtain
|
u
(
r
)
|
≤
∫
r
∞
|
u
′
(
s
)
|
𝑑
s
=
∫
r
∞
|
u
′
(
s
)
|
s
N
+
a
∞
-
1
2
s
-
N
+
a
∞
-
1
2
𝑑
s
≤
(
∫
r
∞
|
u
′
(
s
)
|
2
s
N
-
1
s
a
∞
𝑑
s
)
1
2
(
∫
r
∞
s
-
(
N
+
a
∞
-
1
)
𝑑
s
)
1
2
=
(
ω
N
)
-
1
2
(
∫
B
r
c
|
x
|
a
∞
|
∇
u
|
2
𝑑
x
)
1
2
(
∫
r
∞
s
-
(
N
+
a
∞
-
1
)
𝑑
s
)
1
2
≤
(
C
∞
(
R
0
)
)
-
1
2
(
ω
N
)
-
1
2
(
∫
B
r
c
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
)
1
2
(
∫
r
∞
s
-
(
N
+
a
∞
-
1
)
𝑑
s
)
1
2
.
As
∫
r
∞
s
-
(
N
+
a
∞
-
1
)
𝑑
s
=
r
-
(
N
+
a
∞
-
2
)
N
+
a
∞
-
2
,
we obtain
|
u
(
r
)
|
≤
C
r
-
N
+
a
∞
-
2
2
(
∫
B
R
0
c
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
)
1
2
,
where
C
=
(
C
∞
(
R
0
)
)
-
1
2
(
ω
N
)
-
1
2
(
1
N
+
a
∞
-
2
)
1
2
=
C
(
N
,
R
0
,
a
∞
)
,
and this is our assertion. ∎
Lemma 2.4.
Assume hypothesis [A]. Fix R0>0. Then there exists a constant C=C(N,R0,a0)>0 such that, for each u∈Cc,r∞(BR0), there holds
|
u
(
x
)
|
≤
C
|
x
|
-
N
+
a
0
-
2
2
(
∫
B
R
0
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
)
1
2
for
0
<
|
x
|
<
R
0
.
Proof.
Let u∈Cc,r∞(BR0) and take |x|=r<R0. Since u(R0)=0, we have
-
u
(
r
)
=
u
(
R
0
)
-
u
(
r
)
=
∫
r
R
0
u
′
(
s
)
𝑑
s
.
The same arguments as in the proof of Lemma 2.3 yield
|
u
(
r
)
|
≤
∫
r
R
0
|
u
′
(
s
)
|
𝑑
s
≤
(
∫
r
R
0
|
u
′
(
s
)
|
2
s
N
-
1
s
a
0
𝑑
s
)
1
2
(
∫
r
R
0
s
-
(
N
+
a
0
-
1
)
𝑑
s
)
1
2
≤
(
ω
N
)
-
1
2
(
∫
B
R
0
∖
B
r
|
x
|
a
0
|
∇
u
|
2
𝑑
x
)
1
2
(
∫
r
R
0
s
-
(
N
+
a
0
-
1
)
𝑑
s
)
1
2
≤
(
ω
N
)
-
1
2
(
C
0
(
R
0
)
)
-
1
2
(
∫
B
R
0
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
)
1
2
(
1
N
+
a
0
-
2
)
1
2
r
-
N
+
a
0
-
2
2
,
that is,
|
u
(
r
)
|
≤
C
r
-
N
+
a
0
-
2
2
(
∫
B
R
0
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
)
1
2
,
where
C
=
(
ω
N
)
-
1
2
(
C
0
(
R
0
)
)
-
1
2
(
1
N
+
a
0
-
2
)
1
2
=
C
(
N
,
R
0
,
a
0
)
,
which is the assertion. ∎
Corollary 2.5.
Assume hypothesis [A]. Fix R0>0. Then there exists a constant C=C(N,R0,a0,a∞)>0 such that, for each u∈Cc,r∞(RN), there holds
|
u
(
x
)
|
≤
C
|
x
|
-
N
+
a
0
-
2
2
(
∫
B
R
0
+
1
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
+
∫
B
R
0
c
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
)
1
2
for
0
<
|
x
|
<
R
0
.
Proof.
Let u∈Cc,r∞(ℝN). Take a radial function ρ∈Cc,r∞(ℝN) such that ρ(x)∈[0,1], ρ≡1 in BR0, and ρ(x)≡0 if |x|≥R0+12. Hence ρu∈Cc,r∞(BR0+1), so we can apply the previous lemma (in the ball BR0+1) and get
|
ρ
(
x
)
u
(
x
)
|
≤
C
|
x
|
-
N
+
a
0
-
2
2
(
∫
B
R
0
+
1
A
(
|
x
|
)
|
∇
ρ
u
|
2
𝑑
x
)
1
2
.
If |x|<R0, then ρ(x)=1. Hence
|
u
(
x
)
|
≤
C
|
x
|
-
N
+
a
0
-
2
2
(
∫
B
R
0
+
1
A
(
|
x
|
)
|
∇
ρ
u
|
2
𝑑
x
)
1
2
.
We also have
|
∇
ρ
u
|
2
≤
(
ρ
|
∇
u
|
+
|
u
|
|
∇
ρ
|
)
2
≤
2
(
ρ
2
|
∇
u
|
2
+
u
2
|
∇
ρ
|
2
)
.
Hence, for x∈BR0,
|
u
(
x
)
|
≤
C
|
x
|
-
N
+
a
0
-
2
2
2
(
∫
B
R
0
+
1
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
+
∫
B
R
0
+
1
A
(
|
x
|
)
u
2
|
∇
ρ
|
2
𝑑
x
)
1
2
.
We also have
∫
B
R
0
+
1
A
(
|
x
|
)
u
2
|
∇
ρ
|
2
𝑑
x
≤
C
∫
B
R
0
+
1
∖
B
R
0
A
(
|
x
|
)
u
2
𝑑
x
,
where the constant C=max|∇ρ|2 depends on ρ, and hence on R0. We can now apply Lemma 2.3 in the domain BR0c, and we get
u
2
(
y
)
≤
C
|
y
|
-
N
+
a
∞
-
2
2
∫
B
R
0
c
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
for |y|>R0. Recalling that A is bounded in BR0+1∖BR0, we get, for y∈BR0+1∖BR0,
A
(
|
y
|
)
u
2
(
y
)
≤
C
|
y
|
-
N
+
a
∞
-
2
2
∫
B
R
0
c
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
.
Hence, integrating with respect to y∈BR0+1∖BR0, we get
∫
B
R
0
+
1
∖
B
R
0
A
(
|
y
|
)
u
2
(
y
)
𝑑
y
≤
C
∫
B
R
0
c
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
,
where C=C(N,a∞,R0). Pasting all together, we get, for |x|<R0,
|
u
(
x
)
|
≤
C
|
x
|
-
N
+
a
0
-
2
2
2
(
∫
B
R
0
+
1
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
+
∫
B
R
0
+
1
A
(
|
x
|
)
u
2
|
∇
ρ
|
2
𝑑
x
)
1
2
≤
C
|
x
|
-
N
+
a
0
-
2
2
(
∫
B
R
0
+
1
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
+
∫
B
R
0
c
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
)
1
2
,
which is the assertion. ∎
We now introduce another function space.
Definition 2.6.
Set
S
A
=
{
u
∈
C
c
,
r
∞
(
ℝ
N
)
:
∫
ℝ
N
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
<
+
∞
}
,
which is a subspace of Cc,r∞(ℝN). We define on SA the norm
∥
u
∥
A
=
(
∫
ℝ
N
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
)
1
2
.
Definition 2.7.
Let 2-N<a0,a∞≤2. We define the following real numbers:
p
0
:=
2
N
N
+
a
0
-
2
,
p
∞
:=
2
N
N
+
a
∞
-
2
.
Notice that p0,p∞≥2.
The main property of SA is given by the following lemma.
Lemma 2.8.
Consider A satisfying hypothesis [A]. For each R0>0, we have the continuous embeddings
S
A
↪
L
p
0
(
B
R
0
)
𝑎𝑛𝑑
S
A
↪
L
p
∞
(
B
R
0
c
)
.
Proof.
Let us fix R0>0 and C>0 such that A(|x|)≥C|x|a∞ for |x|≥R0. Take u∈SA. We want to estimate ∫BR0c|u|p∞𝑑x. With an integration by parts, and using Lemma 2.3, we obtain
∫
B
R
0
c
|
u
|
p
∞
𝑑
x
=
ω
N
∫
R
0
∞
r
N
-
1
|
u
(
r
)
|
p
∞
𝑑
r
≤
2
ω
N
N
+
a
∞
-
2
∫
R
0
∞
r
N
|
u
(
r
)
|
p
∞
-
1
|
u
′
(
r
)
|
𝑑
r
≤
2
ω
N
N
+
a
∞
-
2
(
∫
R
0
∞
r
a
∞
|
u
′
(
r
)
|
2
r
N
-
1
𝑑
r
)
1
2
(
∫
R
0
∞
r
(
N
-
N
-
1
+
a
∞
2
)
2
|
u
(
r
)
|
2
(
p
∞
-
1
)
𝑑
r
)
1
2
≤
C
2
ω
N
1
2
N
+
a
∞
-
2
(
∫
B
R
0
c
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
)
1
2
(
∫
R
0
∞
r
N
-
1
|
u
(
r
)
|
p
∞
r
2
-
a
∞
|
u
(
r
)
|
p
∞
-
2
𝑑
r
)
1
2
≤
C
p
∞
-
2
2
2
ω
N
1
2
N
+
a
∞
-
2
(
∫
B
R
0
c
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
)
p
∞
4
(
∫
R
0
∞
r
N
-
1
|
u
(
r
)
|
p
∞
𝑑
r
)
1
2
≤
C
(
∫
B
R
0
c
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
)
p
∞
4
(
∫
B
R
0
c
|
u
|
p
∞
𝑑
x
)
1
2
,
where C=C(N,R0,a∞) may change from line to line. From this we obtain
(
∫
B
R
0
c
|
u
|
p
∞
𝑑
x
)
1
p
∞
≤
C
(
∫
B
R
0
c
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
)
1
2
≤
C
∥
u
∥
A
.
This proves the second embedding. For the first one, we start by assuming that C=C0(R0+1)>0 is such that A(|x|)≥C|x|a0 for 0<|x|=r≤R0+1. We want to estimate the integral ∫BR0|u|p0𝑑x. Let us take the same radial cut-off function ρ(x) as in Corollary 2.5. We have ρu∈Cc,r∞(BR0+1) and we can employ Lemma 2.4. With the same computations used in the previous case, we have
∫
B
R
0
+
1
|
ρ
u
|
p
0
𝑑
x
≤
C
(
∫
B
R
0
+
1
A
(
|
x
|
)
|
∇
(
ρ
u
)
|
2
𝑑
x
)
p
0
4
(
∫
B
R
0
+
1
|
ρ
u
|
p
0
𝑑
x
)
1
2
,
where C=C(N,R0,a0). From this we derive
∫
B
R
0
+
1
|
ρ
u
|
p
0
𝑑
x
≤
C
(
∫
B
R
0
+
1
A
(
|
x
|
)
|
∇
(
ρ
u
)
|
2
𝑑
x
)
p
0
2
.
Using the continuity of A in the compact set BR0+1∖BR0¯ and thanks to Lemma 2.3, we obtain
∫
B
R
0
+
1
A
(
|
x
|
)
|
∇
(
ρ
u
)
|
2
𝑑
x
≤
C
∫
B
R
0
+
1
A
(
|
x
|
)
(
|
∇
u
|
2
ρ
2
+
|
∇
ρ
|
2
u
2
)
𝑑
x
≤
C
(
∫
B
R
0
+
1
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
+
∫
B
R
0
+
1
∖
B
R
0
|
∇
ρ
|
2
u
2
𝑑
x
)
≤
C
(
∥
u
∥
A
2
+
∥
u
∥
A
2
∫
B
R
0
+
1
∖
B
R
0
|
x
|
-
(
N
+
a
∞
-
2
)
𝑑
x
)
≤
C
∥
u
∥
A
2
,
where C=C(N,R0,a∞). Thus
∫
B
R
0
+
1
|
ρ
u
|
p
0
𝑑
x
≤
C
∥
u
∥
A
p
0
holds, where C=C(N,R0,a0,a∞). As a consequence, we get
∫
B
R
0
|
u
|
p
0
𝑑
x
=
∫
B
R
0
|
ρ
u
|
p
0
𝑑
x
≤
∫
B
R
0
+
1
|
ρ
u
|
p
0
𝑑
x
≤
C
∥
u
∥
A
p
0
,
and hence
(2.1)
(
∫
B
R
0
|
u
|
p
0
𝑑
x
)
1
p
0
≤
C
∥
u
∥
A
.
This concludes the proof. ∎
We now want to introduce the completion of SA with respect to ∥⋅∥A.
Definition 2.9.
D
A
is the space of all measurable functions u:ℝN→ℝ such that u∈Lp0(BR)∩Lp∞(BRc) for all R>0 and for which there is a sequence {un}n⊂SA such that the following conditions hold:
u
n
→
u
in Lp0(BR) and in Lp∞(BRc) for all R>0.
{
u
n
}
n
is a Cauchy sequence with respect to ∥⋅∥A.
Of course, DA is a linear space. From the previous results we deduce the following two lemmas, which say that DA is the completion of SA with respect to ∥⋅∥A. The arguments are essentially standard, so we will skip the details.
Lemma 2.10.
Assume hypothesis [A]. Let u∈DA. Then u has weak derivatives Diu in the open set Ω=RN∖{0} (i=1,…,N) and Diu∈Lloc2(Ω) holds.
If {un}n is any sequence in SA as in the definition of u∈DA, then
∫
ℝ
N
A
(
|
x
|
)
|
∇
u
-
∇
u
n
|
2
𝑑
x
→
0
.
In particular, ∫RNA(|x|)|∇u|2𝑑x<+∞,
∥
u
∥
A
=
(
∫
ℝ
N
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
)
1
2
is a norm on DA, and ∥u∥A=limn∥un∥A.
Proof.
The proof is a simple exercise on weak derivatives, and we leave it to the reader. ∎
Lemma 2.11.
Assume [A]. Consider the space DA endowed with the norm ∥⋅∥A. Then the following conditions hold:
For any
R
0
>
0
, there is a constant
C
=
C
(
N
,
R
0
,
a
0
,
a
∞
)
>
0
such that, for all
u
∈
D
A
,
|
u
(
x
)
|
≤
C
|
x
|
-
N
+
a
0
-
2
2
∥
u
∥
A
for all
x
∈
B
R
0
,
|
u
(
x
)
|
≤
C
|
x
|
-
N
+
a
∞
-
2
2
∥
u
∥
A
for all
x
∈
B
R
0
c
.
We have the continuous embeddings
D
A
↪
L
p
0
(
B
R
0
)
𝑎𝑛𝑑
D
A
↪
L
p
∞
(
B
R
0
c
)
.
D
A
, endowed with the norm
∥
⋅
∥
A
, is a Hilbert space.
Proof.
As first thing we notice that if u∈DA, then we can assume that we have a sequence {un}n⊂SA satisfying the properties stated in the definition of DA, and also such that un(x)→u(x) for a.e. x∈ℝN. It is then easy to get the estimates in (i): if u∈DA and {un}n⊂SA is as before, we have that the estimates hold for un, and then we can pass to the limit to get that they hold also for u, with the same constants. The same argument gives (ii).
To prove completeness, let {un}n⊂DA be a Cauchy sequence. By definition and Lemma 2.10, for each n we get vn∈SA such that
∥
u
n
-
v
n
∥
A
≤
1
n
.
This implies that also {vn}n is a Cauchy sequence in SA. By Lemma 2.8, we get that {vn}n converges in Lp0(BR) and in Lp∞(BRc) for all R>0. Let us fix R=1. By standard results, we get a subsequence {vnk}k such that vnk(x)→w1(x) a.e. in B1 and vnk(x)→w2(x) a.e. in B1c. Let us define the measurable function u:ℝN→ℝ by u(x)=w1(x) in B1, and u(x)=w2(x) in B1c. Hence vnk(x)→u(x) a.e. in ℝN. For any fixed R>0, we know that {vn}n converges in Lp0(BR), say to v∈Lp0(BR). Thus also {vnk}k converges to v, so a subsequence of {vnk}k converges to v a.e. in BR. But {vnk}k converges a.e. to u, so u=v in BR. Hence u∈Lp0(BR) and {vn}n converges to u in Lp0(BR). With the same argument we obtain that u∈Lp∞(BR) and {vn}n converges to u in Lp∞(BRc).
We have obtained a measurable function u and a sequence {vn}n⊂SA such that, for all R>0, we have u∈Lp0(BR), u∈Lp∞(BR), and vn→u in Lp0(BR) and in Lp∞(BR). Furthermore, {vn}n is a Cauchy sequence in SA. By definition, we have u∈DA and, by Lemma 2.10, we have ∥u-vn∥A→0. To conclude, we just have to show that ∥u-un∥A→0. But this is obvious because
∥
u
-
u
n
∥
A
≤
∥
u
-
v
n
∥
A
+
∥
v
n
-
u
n
∥
A
≤
1
n
+
∥
u
-
v
n
∥
A
.
We have proved that DA is a Banach space. It is also a Hilbert space because the norm is induced by the inner product (u,v)A=∫ℝNA(|x|)∇u⋅∇vdx. ∎
We now define another function space.
Definition 2.12.
Set
X
=
{
u
∈
D
A
:
∫
ℝ
N
V
(
|
x
|
)
|
u
|
2
𝑑
x
<
+
∞
}
with norm
∥
u
∥
2
=
∥
u
∥
X
2
=
∫
ℝ
N
A
(
|
x
|
)
|
∇
u
|
2
+
∫
ℝ
N
V
(
|
x
|
)
|
u
|
2
𝑑
x
.
The space X is a Hilbert space with respect to the norm ∥⋅∥X. We will write (u,v)X for the scalar product in X, that is,
(
u
,
v
)
X
=
∫
ℝ
N
A
(
|
x
|
)
∇
u
⋅
∇
v
d
x
+
∫
ℝ
N
V
(
|
x
|
)
u
v
𝑑
x
.
We look for weak solutions of equation (1.1) in the space X. This means that a solution (1.1) is a function u∈X such that, for all h∈X,
∫
ℝ
N
A
(
|
x
|
)
∇
u
∇
h
d
x
+
∫
ℝ
N
V
(
|
x
|
)
u
h
𝑑
x
-
∫
ℝ
N
K
(
|
x
|
)
f
(
u
)
h
𝑑
x
=
0
holds. We will obtain such weak solutions by standard variational methods, that is, we will introduce (in Section 5) a functional on X whose critical points are weak solutions. To get such critical points we need, as usual, some compactness properties for the functional, which we will derive from the compactness of suitable embeddings. So in the following sections we will prove that the space X is compactly embedded in LKq1+LKq2 for suitable q1,q2. The following lemma is a step to obtain such compact embeddings.
Lemma 2.13.
Assume hypotheses [A], [V] and [K]. Let 1<q<∞ and 0<r<R. Then there exists a constant C=C(N,a0,a∞,r,R,q,s)>0 such that for any u,h∈X we have
∫
B
R
∖
B
r
K
(
|
x
|
)
|
u
|
q
-
1
|
h
|
𝑑
x
C
∥
K
(
|
⋅
|
)
∥
L
s
(
B
R
∖
B
r
)
≤
{
(
∫
B
R
∖
B
r
|
u
|
2
𝑑
x
)
q
-
1
2
∥
h
∥
i
f
q
≤
q
~
,
(
∫
B
R
∖
B
r
|
u
|
2
𝑑
x
)
q
~
-
1
2
∥
u
∥
q
-
q
~
∥
h
∥
i
f
q
>
q
~
,
where q~=2(1+1N-1s)-a0N.
Proof.
We denote by σ the conjugate exponent of p0, i.e. σ=2NN-a0+2. Thanks to Hölder’s inequality, initially applied with p0>1 and then with sσ>1, we obtain
∫
B
R
∖
B
r
K
(
|
x
|
)
|
u
|
q
-
1
|
h
|
𝑑
x
≤
(
∫
B
R
∖
B
r
K
(
|
x
|
)
σ
|
u
|
(
q
-
1
)
σ
𝑑
x
)
1
σ
(
∫
B
R
∖
B
r
|
h
|
p
0
𝑑
x
)
1
p
0
≤
(
(
∫
B
R
∖
B
r
K
(
|
x
|
)
s
𝑑
x
)
σ
s
(
∫
B
R
∖
B
r
|
u
|
(
q
-
1
)
σ
(
s
σ
)
′
𝑑
x
)
1
(
s
σ
)
′
)
1
σ
C
∥
h
∥
≤
C
∥
K
(
|
⋅
|
)
∥
L
s
(
B
R
∖
B
r
)
∥
h
∥
(
∫
B
R
∖
B
r
|
u
|
2
q
-
1
q
~
-
1
𝑑
x
)
q
~
-
1
2
,
where we use the following computations:
1
(
s
σ
)
′
=
1
-
σ
s
=
s
(
N
-
a
0
+
2
)
-
2
N
s
(
N
-
a
0
+
2
)
,
and hence
(
s
σ
)
′
=
s
(
N
-
a
0
+
2
)
s
(
N
-
a
0
+
2
)
-
2
N
,
so that
σ
(
s
σ
)
′
=
2
N
N
-
a
0
+
2
s
(
N
-
a
0
+
2
)
s
(
N
-
a
0
+
2
)
-
2
N
=
2
s
N
s
N
+
2
s
-
2
N
-
a
0
s
=
2
q
~
-
1
.
If q=q~, the proof is over.
If q<q~, we apply Hölder’s inequality again, with conjugate exponents q~-1q-1>1 and q~-1q~-q, obtaining
∫
B
R
∖
B
r
K
(
|
x
|
)
|
u
|
q
-
1
|
h
|
𝑑
x
≤
C
∥
K
(
|
⋅
|
)
∥
L
s
(
B
R
∖
B
r
)
∥
h
∥
(
|
B
R
∖
B
r
|
q
~
-
q
q
~
-
1
(
∫
B
R
∖
B
r
|
u
|
2
𝑑
x
)
q
-
1
q
~
-
1
)
q
~
-
1
2
=
C
∥
K
(
|
⋅
|
)
∥
L
s
(
B
R
∖
B
r
)
∥
h
∥
(
∫
B
R
∖
B
r
|
u
|
2
𝑑
x
)
q
-
1
2
.
If q>q~, we have q-1q~-1>1. Thanks to Lemma 2.3, we obtain
∫
B
R
∖
B
r
K
(
|
x
|
)
|
u
|
q
-
1
|
h
|
𝑑
x
≤
C
∥
K
(
|
⋅
|
)
∥
L
s
(
B
R
∖
B
r
)
∥
h
∥
(
∫
B
R
∖
B
r
|
u
|
2
q
-
1
q
~
-
1
-
2
|
u
|
2
𝑑
x
)
≤
C
∥
K
(
|
⋅
|
)
∥
L
s
(
B
R
∖
B
r
)
∥
h
∥
(
(
C
∥
u
∥
r
N
+
a
∞
-
2
2
)
2
q
-
q
~
q
~
-
1
∫
B
R
∖
B
r
|
u
|
2
𝑑
x
)
q
~
-
1
2
≤
C
∥
K
(
|
⋅
|
)
∥
L
s
(
B
R
∖
B
r
)
∥
h
∥
∥
u
∥
q
-
q
~
(
∫
B
R
∖
B
r
|
u
|
2
𝑑
x
)
q
~
-
1
2
.
In all the previous computations, C may vary, depending only on N,a0,a∞,r,R,q,s. ∎
Lemma 2.14.
Consider A satisfying hypothesis [A]. Let E be a smooth bounded open set such that E¯⊂RN∖{0}. Then the embedding
D
A
↪
L
2
(
E
)
is continuous and compact.
Proof.
The continuity of the embedding is obvious thanks to Lemma 2.11 and the fact that p0≥2. We prove now the compactness of the embedding. Let {un}n be a bounded sequence in DA. By continuity of the embedding, we obtain
∥
u
n
∥
L
2
(
E
)
≤
C
.
Moreover, as the function A(x) is continuous and strictly positive in the compact set E¯, there holds
∫
E
|
∇
u
n
|
2
𝑑
x
≤
C
∫
E
A
(
|
x
|
)
|
∇
u
n
|
2
𝑑
x
≤
C
∥
u
n
∥
A
2
≤
C
.
Thus, {un}n is bounded also in the space H1(E). Thanks to Rellich’s theorem, {un}n has a convergent subsequence in L2(E), and this gives our assertion. ∎
4 Compactness of Embeddings
We start this section by proving the following two lemmas, which give the most important technical steps for our compactness results. For future purposes, these two lemmas are stated in a form which is a little more general than needed in the present paper.
Lemma 4.1.
Let N≥3 and R0>0. Assume [A], [V], [K]. If we assume
Λ
:=
ess
sup
x
∈
B
R
0
K
(
|
x
|
)
|
x
|
α
V
(
|
x
|
)
β
<
+
∞
for some
0
≤
β
≤
1
and
α
∈
ℝ
,
then for all u,h∈X, for all q>max{1,2β} and for all 0<R<R0 we have
∫
B
R
K
(
|
x
|
)
|
u
|
q
-
1
|
h
|
𝑑
x
≤
{
Λ
∥
u
∥
q
-
1
C
(
∫
B
R
|
x
|
α
-
ν
(
q
-
1
)
N
-
a
0
+
2
(
1
-
2
β
+
a
0
β
)
2
N
𝑑
x
)
N
-
a
0
+
2
(
1
-
2
β
+
a
0
β
)
2
N
∥
h
∥
,
0
≤
β
≤
1
2
,
Λ
∥
u
∥
q
-
1
C
(
∫
B
R
|
x
|
α
-
ν
(
q
-
2
β
)
1
-
β
𝑑
x
)
1
-
β
∥
h
∥
,
1
2
<
β
<
1
,
Λ
∥
u
∥
q
-
2
C
(
∫
B
R
|
x
|
2
α
-
2
ν
(
q
-
2
)
V
(
|
x
|
)
|
u
|
2
𝑑
x
)
1
2
∥
h
∥
,
β
=
1
,
where ν:=N+a0-22 and C=C(N,R0,a0,a∞).
Proof.
Assume 0<R<R0. We study the various cases separately. We will apply Lemma 2.4 several times. Notice that the constants involved depend on R0 but not on R.
(i) If β=0, we apply Hölder’s inequality with conjugate exponents p0=2NN+a0-2 and σ=2NN-a0+2. We apply also (2.1) and Lemma 2.4, and get
1
Λ
∫
B
R
K
(
|
x
|
)
|
u
|
q
-
1
|
h
|
𝑑
x
≤
∫
B
R
|
x
|
α
|
u
|
q
-
1
|
h
|
𝑑
x
≤
(
∫
B
R
(
|
x
|
α
|
u
|
q
-
1
)
2
N
N
-
a
0
+
2
𝑑
x
)
N
-
a
0
+
2
2
N
(
∫
B
R
|
h
|
p
0
𝑑
x
)
1
p
0
≤
∥
u
∥
q
-
1
C
(
∫
B
R
|
x
|
α
-
ν
(
q
-
1
)
N
-
a
0
+
2
2
N
𝑑
x
)
N
-
a
0
+
2
2
N
∥
h
∥
.
(ii) If 0<β<12, then it is possible to apply Hölder’s inequality with conjugate exponents 1β and 11-β:
1
Λ
∫
B
R
K
(
|
x
|
)
|
u
|
q
-
1
|
h
|
𝑑
x
≤
∫
B
R
|
x
|
α
V
(
|
x
|
)
β
|
u
|
q
-
1
|
h
|
𝑑
x
=
∫
B
R
|
x
|
α
|
u
|
q
-
1
|
h
|
1
-
2
β
V
(
|
x
|
)
β
|
h
|
2
β
𝑑
x
≤
(
∫
B
R
(
|
x
|
α
|
u
|
q
-
1
|
h
|
1
-
2
β
)
1
1
-
β
𝑑
x
)
1
-
β
(
∫
B
R
V
(
|
x
|
)
|
h
|
2
𝑑
x
)
β
≤
(
∫
B
R
(
|
x
|
α
|
u
|
q
-
1
|
h
|
1
-
2
β
)
1
1
-
β
𝑑
x
)
1
-
β
∥
h
∥
2
β
.
We apply again Hölder’s inequality with exponent 1-β1-2βp0>1. Its conjugate exponent is given by the formula
1
(
1
-
β
1
-
2
β
p
0
)
′
=
1
-
1
-
2
β
p
0
(
1
-
β
)
=
p
0
(
1
-
β
)
-
(
1
-
2
β
)
p
0
(
1
-
β
)
=
2
N
N
+
a
0
-
2
(
1
-
β
)
-
(
1
-
2
β
)
2
N
N
+
a
0
-
2
(
1
-
β
)
=
2
N
(
1
-
β
)
-
(
N
+
a
0
-
2
)
(
1
-
2
β
)
2
N
(
1
-
β
)
=
N
-
a
0
+
2
(
1
-
2
β
+
a
0
β
)
2
N
(
1
-
β
)
.
We obtain that
1
Λ
∫
B
R
K
(
|
x
|
)
|
u
|
q
-
1
|
h
|
𝑑
x
≤
[
(
∫
B
R
(
|
x
|
α
1
-
β
|
u
|
q
-
1
1
-
β
)
(
1
-
β
1
-
2
β
p
0
)
′
𝑑
x
)
1
/
(
1
-
β
1
-
2
β
p
0
)
′
(
∫
B
R
|
h
|
p
0
𝑑
x
)
1
-
2
β
(
1
-
β
)
p
0
]
1
-
β
∥
h
∥
2
β
≤
∥
u
∥
q
-
1
[
(
∫
B
R
(
|
x
|
α
1
-
β
-
ν
q
-
1
1
-
β
)
(
1
-
β
1
-
2
β
p
0
)
′
𝑑
x
)
1
/
(
1
-
β
1
-
2
β
p
0
)
′
C
∥
h
∥
1
-
2
β
1
-
β
]
1
-
β
∥
h
∥
2
β
=
∥
u
∥
q
-
1
C
(
∫
B
R
|
x
|
α
-
ν
(
q
-
1
)
N
-
a
0
+
2
(
1
-
2
β
+
a
0
β
)
2
N
𝑑
x
)
N
-
a
0
+
2
(
1
-
2
β
+
a
0
β
)
2
N
∥
h
∥
,
where we used (2.1) and Lemma 2.3.
(iii) If β=12, there follows
1
Λ
∫
B
R
K
(
|
x
|
)
|
u
|
q
-
1
|
h
|
𝑑
x
≤
∫
B
R
|
x
|
α
|
u
|
q
-
1
V
(
|
x
|
)
1
2
|
h
|
𝑑
x
≤
(
∫
B
R
|
x
|
2
α
|
u
|
2
(
q
-
1
)
𝑑
x
)
1
2
(
∫
B
R
V
(
|
x
|
)
|
h
|
2
𝑑
x
)
1
2
≤
∥
u
∥
q
-
1
C
(
∫
B
R
|
x
|
2
α
-
2
ν
(
q
-
1
)
𝑑
x
)
1
2
∥
h
∥
.
(iv) If 12<β<1, then we can apply Hölder’s inequality first with conjugate exponents equal to 2, then with 12β-1 and 12(1-β). We get
1
Λ
∫
B
R
K
(
|
x
|
)
|
u
|
q
-
1
|
h
|
𝑑
x
≤
∫
B
R
|
x
|
α
V
(
|
x
|
)
β
|
u
|
q
-
1
|
h
|
𝑑
x
≤
∫
B
R
|
x
|
α
V
(
|
x
|
)
2
β
-
1
2
|
u
|
q
-
1
V
(
|
x
|
)
1
2
|
h
|
𝑑
x
≤
(
∫
B
R
|
x
|
2
α
V
(
|
x
|
)
2
β
-
1
|
u
|
2
(
q
-
1
)
𝑑
x
)
1
2
(
∫
B
R
V
(
|
x
|
)
|
h
|
2
𝑑
x
)
1
2
≤
(
∫
B
R
|
x
|
2
α
|
u
|
2
(
q
-
2
β
)
V
(
|
x
|
)
2
β
-
1
|
u
|
2
(
2
β
-
1
)
𝑑
x
)
1
2
∥
h
∥
≤
(
(
∫
B
R
|
x
|
α
1
-
β
|
u
|
q
-
2
β
1
-
β
𝑑
x
)
2
(
1
-
β
)
(
∫
B
R
V
(
|
x
|
)
|
u
|
2
𝑑
x
)
2
β
-
1
)
1
2
∥
h
∥
≤
C
∥
u
∥
q
-
2
β
(
(
∫
B
R
|
x
|
α
1
-
β
-
ν
q
-
2
β
1
-
β
𝑑
x
)
2
(
1
-
β
)
(
∫
B
R
V
(
|
x
|
)
|
u
|
2
𝑑
x
)
2
β
-
1
)
1
2
∥
h
∥
=
C
∥
u
∥
q
-
2
β
(
∫
B
R
|
x
|
α
-
ν
(
q
-
2
β
)
1
-
β
𝑑
x
)
1
-
β
(
∫
B
R
V
(
|
x
|
)
|
u
|
2
𝑑
x
)
2
β
-
1
2
∥
h
∥
≤
C
(
∫
B
R
|
x
|
α
-
ν
(
q
-
2
β
)
1
-
β
𝑑
x
)
1
-
β
∥
u
∥
q
-
1
∥
h
∥
.
(v) If β=1, then the hypothesis q>max{1,2β} implies q>2. Thus, we have
1
Λ
∫
B
R
K
(
|
x
|
)
|
u
|
q
-
1
|
h
|
𝑑
x
≤
∫
B
R
|
x
|
α
V
(
|
x
|
)
|
u
|
q
-
1
|
h
|
𝑑
x
≤
∫
B
R
|
x
|
α
V
(
|
x
|
)
1
2
|
u
|
q
-
1
V
(
|
x
|
)
1
2
|
h
|
𝑑
x
≤
(
∫
B
R
|
x
|
2
α
V
(
|
x
|
)
|
u
|
2
(
q
-
1
)
𝑑
x
)
1
2
(
∫
B
R
V
(
|
x
|
)
|
h
|
2
𝑑
x
)
1
2
≤
(
∫
B
R
|
x
|
2
α
|
u
|
2
(
q
-
2
)
V
(
|
x
|
)
|
u
|
2
𝑑
x
)
1
2
∥
h
∥
≤
C
∥
u
∥
q
-
2
(
∫
B
R
|
x
|
2
α
-
2
ν
(
q
-
2
)
V
(
|
x
|
)
|
u
|
2
𝑑
x
)
1
2
∥
h
∥
.
The proof is now concluded. ∎
Lemma 4.2.
Let N≥3 and R0>0. Assume [A], [V], [K]. Assume also that
Λ
:=
ess
sup
x
∈
B
R
0
c
K
(
|
x
|
)
|
x
|
α
V
(
|
x
|
)
β
<
+
∞
for some
0
≤
β
≤
1
and
α
∈
ℝ
.
Then for all u,h∈X, for all q>max{1,2β} and for all R>R0 we have
∫
B
R
c
K
(
|
x
|
)
|
u
|
q
-
1
|
h
|
𝑑
x
≤
{
Λ
∥
u
∥
q
-
1
C
(
∫
B
R
c
|
x
|
α
-
ν
(
q
-
1
)
N
-
a
∞
+
2
(
1
-
2
β
+
a
∞
β
)
2
N
𝑑
x
)
N
-
a
∞
+
2
(
1
-
2
β
+
a
∞
β
)
2
N
∥
h
∥
,
0
≤
β
≤
1
2
,
Λ
∥
u
∥
q
-
1
C
(
∫
B
R
c
|
x
|
α
-
ν
(
q
-
2
β
)
1
-
β
𝑑
x
)
1
-
β
∥
h
∥
,
1
2
<
β
<
1
,
Λ
∥
u
∥
q
-
2
C
(
∫
B
R
c
|
x
|
2
α
-
2
ν
(
q
-
2
)
V
(
|
x
|
)
|
u
|
2
𝑑
x
)
1
2
∥
h
∥
,
β
=
1
,
where ν:=N+a∞-22 and C=C(N,R0,a0,a∞).
The proof of Lemma 4.2 is the same as that of Lemma 4.1, and we will skip it.
Definition 4.3.
For α∈ℝ, β∈[0,1] and a∈(2-N,2], we define the functions α*(a,β) and q*(a,α,β) as follows:
α
*
(
a
,
β
)
:=
max
{
2
β
-
1
-
N
2
-
a
β
+
a
2
,
-
(
1
-
β
)
N
}
=
{
2
β
-
1
-
N
2
-
a
β
+
a
2
if
0
≤
β
≤
1
2
,
-
(
1
-
β
)
N
if
1
2
≤
β
≤
1
,
q
*
(
a
,
α
,
β
)
:=
2
α
-
2
β
+
N
+
a
β
N
+
a
-
2
.
Theorem 4.4.
Let N≥3. Assume [A], [V], [K]. Assume also that there exists R0>0 such that
ess
sup
x
∈
B
R
0
K
(
|
x
|
)
|
x
|
α
0
V
(
|
x
|
)
β
0
<
+
∞
for some
0
≤
β
0
≤
1
and
α
0
>
α
*
(
a
0
,
β
0
)
.
Then limR→0+S0(q1,R)=0 for any q1∈R such that
max
{
1
,
2
β
0
}
<
q
1
<
q
*
(
a
0
,
α
0
,
β
0
)
.
Proof.
Let u,h∈X satisfy ∥u∥=∥h∥=1. Take R such that 0<R<R0. We apply Lemma 4.1 with α=α0, β=β0 and h=u. Recall that ν=N+a0-22.
(i) If 0≤β0≤12, we obtain
∫
B
R
K
(
|
x
|
)
|
u
|
q
1
𝑑
x
=
∫
B
R
K
(
|
x
|
)
|
u
|
q
1
-
1
|
u
|
𝑑
x
≤
C
(
R
2
α
0
-
4
β
0
+
2
N
+
2
a
0
β
0
-
(
N
+
a
0
-
2
)
q
1
N
-
a
0
+
2
(
1
-
2
β
0
+
a
0
β
0
)
N
)
N
-
a
0
+
2
(
1
-
2
β
0
+
a
0
β
0
)
2
N
since
α
0
-
ν
(
q
1
-
1
)
N
-
a
0
+
2
(
1
-
2
β
0
+
a
0
β
0
)
2
N
+
N
=
2
α
0
-
4
β
0
+
2
N
+
2
a
0
β
0
-
(
N
+
a
0
-
2
)
q
1
N
-
a
0
+
2
(
1
-
2
β
0
+
a
0
β
0
)
N
,
and it is easy to check that
2
α
0
-
4
β
0
+
2
N
+
a
0
β
0
-
(
N
+
a
0
-
2
)
q
1
=
(
N
+
a
0
-
2
)
(
q
*
(
a
0
,
α
0
,
β
0
)
-
q
1
)
>
0
and
N
-
a
0
+
2
(
1
-
2
β
0
+
a
0
β
0
)
≥
N
>
0
.
(ii) If 12<β0<1, we get
∫
B
R
K
(
|
x
|
)
|
u
|
q
1
𝑑
x
=
∫
B
R
K
(
|
x
|
)
|
u
|
q
1
-
1
|
u
|
𝑑
x
≤
C
(
R
2
α
0
-
(
N
+
a
0
-
2
)
(
q
1
-
2
β
0
)
2
(
1
-
β
0
)
+
N
)
1
-
β
0
because
α
0
-
ν
(
q
1
-
2
β
0
)
1
-
β
0
+
N
=
α
0
-
N
+
a
0
-
2
2
(
q
1
-
2
β
0
)
1
-
β
0
+
N
=
2
α
0
-
(
N
+
a
0
-
2
)
(
q
1
-
2
β
0
)
2
(
1
-
β
0
)
+
N
=
N
+
a
0
-
2
2
(
1
-
β
0
)
(
q
*
(
a
0
,
α
0
,
β
0
)
-
q
1
)
>
0
.
(iii) Finally, if β0=1, it holds that
∫
B
R
K
(
|
x
|
)
|
u
|
q
1
𝑑
x
=
∫
B
R
K
(
|
x
|
)
|
u
|
q
1
-
1
|
u
|
𝑑
x
≤
C
R
2
α
0
-
(
N
+
a
0
-
2
)
(
q
1
-
2
)
2
because
2
α
0
-
2
ν
(
q
1
-
2
)
=
2
α
0
-
2
N
+
a
0
-
2
2
(
q
1
-
2
)
=
2
α
0
-
(
N
+
a
0
-
2
)
(
q
1
-
2
)
=
(
N
+
a
0
-
2
)
(
q
*
(
a
0
,
α
0
,
1
)
-
q
1
)
>
0
.
Hence, in any of the previous cases there exists a constant δ=δ(N,a0,α0,β0,q1)>0 such that
𝒮
0
(
q
1
,
R
)
≤
C
R
δ
→
0
as
R
→
0
,
from which our assertion follows. ∎
Theorem 4.5.
Let N≥3. Assume [A], [V], [K]. Assume also that there exists R0>0 such that
ess
sup
|
x
|
>
R
0
K
(
|
x
|
)
|
x
|
α
∞
V
(
|
x
|
)
β
∞
<
+
∞
for some
0
≤
β
∞
≤
1
and
α
∞
∈
ℝ
.
Then limR→+∞S∞(q2,R)=0 for each q2∈R such that
q
2
>
max
{
1
,
2
β
∞
,
q
*
(
a
∞
,
α
∞
,
β
∞
)
}
.
Proof.
Let u∈X satisfy ∥u∥=1. Consider R≥R0. We can apply Lemma 4.2 with α=α∞ and β=β∞. The arguments are the same as in Theorem 4.4, so we will skip the details.
(i) If 0≤β∞≤12, with considerations similar to those used for β0, we find
∫
B
R
K
(
|
x
|
)
|
u
|
q
2
𝑑
x
=
∫
B
R
K
(
|
x
|
)
|
u
|
q
2
-
1
|
u
|
𝑑
x
≤
C
(
R
2
α
∞
-
4
β
∞
+
2
N
+
2
a
∞
β
∞
-
(
N
+
a
∞
-
2
)
q
2
N
-
a
∞
+
2
(
1
-
2
β
∞
+
a
∞
β
∞
)
N
)
N
-
a
∞
+
2
(
1
-
2
β
∞
+
a
∞
β
∞
)
2
N
since
2
α
∞
-
4
β
∞
+
2
N
+
2
a
∞
β
∞
-
(
N
+
a
∞
-
2
)
q
2
<
0
,
N
-
a
∞
+
2
(
1
-
2
β
∞
+
a
∞
β
∞
)
≥
N
>
0
.
(ii) On the other hand, if 12<β∞<1, we have
∫
B
R
K
(
|
x
|
)
|
u
|
q
2
𝑑
x
=
∫
B
R
K
(
|
x
|
)
|
u
|
q
2
-
1
|
u
|
𝑑
x
≤
C
(
R
2
α
∞
-
(
N
+
a
∞
-
2
)
(
q
2
-
2
β
∞
)
2
(
1
-
β
∞
)
+
N
)
1
-
β
∞
as
2
α
∞
-
(
N
+
a
∞
-
2
)
(
q
2
-
2
β
∞
)
2
(
1
-
β
∞
)
+
N
<
0
.
(iii) Finally, if β∞=1, we obtain
∫
B
R
K
(
|
x
|
)
|
u
|
q
2
𝑑
x
=
∫
B
R
K
(
|
x
|
)
|
u
|
q
2
-
1
|
u
|
𝑑
x
≤
C
R
2
α
∞
-
(
N
+
a
∞
-
2
)
(
q
2
-
2
)
2
because
2
α
∞
-
(
N
+
a
∞
-
2
)
(
q
2
-
2
)
<
0
.
In each of the previous cases, there exists δ=δ(N,a,α∞,β∞,q2)>0 such that
𝒮
∞
(
q
2
,
R
)
≤
C
R
-
δ
→
0
as
R
→
∞
,
from which our assertion follows. ∎
From the previous theorems we easily derive our main compactness result.
Theorem 4.6.
Assume N≥3. Assume [A], [V], [K]. Moreover, assume the hypotheses of the two previous theorems, that is: there are R1,R2>0, α0,α∞∈R, β0,β∞∈[0,1] such that
ess
sup
|
x
|
<
R
1
K
(
|
x
|
)
|
x
|
α
0
V
(
|
x
|
)
β
0
<
+
∞
,
𝑎𝑛𝑑
ess
sup
|
x
|
≥
R
2
K
(
|
x
|
)
|
x
|
α
∞
V
(
|
x
|
)
β
∞
<
+
∞
.
Thus, for q1 and q2 such that
q
1
∈
ℐ
1
=
(
max
{
1
,
2
β
0
}
,
q
*
(
a
0
,
α
0
,
β
0
)
)
,
q
2
∈
ℐ
2
=
(
max
{
1
,
2
β
∞
,
q
*
(
a
∞
,
α
∞
,
β
∞
)
}
,
+
∞
)
,
the embedding
X
↪
L
K
q
1
(
ℝ
N
)
+
L
K
q
2
(
ℝ
N
)
is continuous and compact.
5 Applications: Existence Results
We now use these results on compact embeddings to obtain existence and multiplicity results for nonlinear elliptic equations. We will deal with equation (1.1), and we will assume hypotheses [A],[V],[K]. As to the nonlinearity f, we will assume the following hypotheses:
f
:
ℝ
→
ℝ
is a continuous function, and there are constants q1,q2>2 and M>0 such that
|
f
(
t
)
|
≤
M
min
{
|
t
|
q
1
-
1
,
|
t
|
q
2
-
1
}
for all
t
∈
ℝ
Define F(t)=∫0tf(s)𝑑s; then there is θ>2 such that 0≤θF(t)≤f(t)t for all t. Furthermore, there is t0>0 such that F(t0)>0.
The simplest example of a function satisfying (f1), (f2) is given by
f
(
t
)
=
min
{
t
q
1
-
1
,
t
q
2
-
1
}
if t≥0, and f(t)=-f(-t) if t≤0 (or also f(t)=0 if t≤0), with q1,q2>2. Notice that if q1≠q2, there is no pure power function, i.e. f(t)=tq, satisfying (f1). However, we do not assume q1≠q2, so pure power functions are included in our results when the hypotheses will allow to choose q1=q2.
We define the functional I:X→ℝ by
I
(
u
)
=
1
2
∫
ℝ
N
A
(
|
x
|
)
|
∇
u
|
2
𝑑
x
+
1
2
∫
ℝ
N
V
(
|
x
|
)
u
2
𝑑
x
-
∫
ℝ
N
K
(
|
x
|
)
F
(
u
)
𝑑
x
.
Theorem 5.1.
Assume the hypotheses of Theorem 4.6. Assume (f1), (f2) with qi∈Ii, where the intervals Ii are given in Theorem 4.6. Then I is a C1 functional on X whose differential is given by
I
′
(
u
)
h
=
∫
ℝ
N
A
(
|
x
|
)
∇
u
∇
h
d
x
+
∫
ℝ
N
V
(
|
x
|
)
u
h
𝑑
x
-
∫
ℝ
N
K
(
|
x
|
)
f
(
u
)
h
𝑑
x
for all u,h∈X.
Proof.
We know that the embedding of X in LKq1+LKq2 is continuous. By the previous results and [8, Proposition 3.8], we also know that the functional
Φ
(
u
)
=
∫
ℝ
N
K
(
|
x
|
)
F
(
u
)
𝑑
x
is of class C1 on LKq1+LKq2, with differential given by
Φ
′
(
u
)
h
=
∫
ℝ
N
K
(
|
x
|
)
f
(
u
)
h
𝑑
x
.
Obviously, the quadratic part of I is C1, with differential given by
h
→
∫
ℝ
N
A
(
|
x
|
)
∇
u
∇
h
d
x
+
∫
ℝ
N
V
(
|
x
|
)
f
(
u
)
h
𝑑
x
.
The assertion follows easily. ∎
Theorem 5.2.
Assume the hypotheses of Theorem 4.6. Assume (f1), (f2) with qi∈Ii, where the intervals Ii are given in Theorem 4.6. Then I:X→R satisfies the Palais–Smale condition.
Proof.
Assume that {un}n is a sequence in X such that I(un) is bounded and I′(un)→0 in the dual space X′. We have to prove that {un}n has a converging subsequence. For this, notice that from the hypotheses we derive, for a suitable positive constant C,
C
+
C
∥
u
n
∥
≥
I
(
u
n
)
-
1
θ
I
′
(
u
n
)
u
n
=
(
1
2
-
1
θ
)
∥
u
n
∥
2
+
∫
ℝ
N
(
1
θ
f
(
u
n
)
u
n
-
F
(
u
n
)
)
𝑑
x
≥
(
1
2
-
1
θ
)
∥
u
n
∥
2
,
and this implies that {un}n is bounded. So we can assume, up to a subsequence, un⇀u in X and un→u in LKq1(ℝN)+LKq2(ℝN). Now we have
∥
u
n
-
u
∥
2
=
(
u
n
,
u
n
-
u
)
X
-
(
u
,
u
n
-
u
)
X
=
I
′
(
u
n
)
(
u
n
-
u
)
+
Φ
′
(
u
n
)
(
u
n
-
u
)
-
(
u
,
u
n
-
u
)
X
.
Of course (u,un-u)X→0 because un⇀u in X. We also have that I′(un)→0 in X′ while un-u is bounded in X, so I′(un)(un-u)→0. Lastly, we know that Φ is C1 in the space LKq1(ℝN)+LKq2(ℝN), and un→u in that space, so Φ′(un) is bounded (as a sequence in the dual space) and un-u→0, so Φ′(un)(un-u)→0. Hence we get ∥un-u∥2→0, which is the assertion. ∎
Theorem 5.3.
Assume the hypotheses of Theorem 4.6. Assume (f1), (f2) with qi∈Ii, where the intervals Ii are given in Theorem 4.6. Then I:X→R has a nonnegative and nontrivial critical point.
Proof.
Firstly, to have a nonnegative solution, we assume as usual f(t)=0 for t≤0. To prove the theorem, we apply the standard Mountain Pass lemma. We have proven that I satisfies the Palais–Smale condition, so it is enough to prove that it has the usual Mountain Pass geometry, that is, we have to prove the following two conditions:
There are ρ,α>0 such that I(u)≥α for all ∥u∥=ρ.
There is v∈X such that ∥v∥≥ρ and I(u)≤0.
As for (i), let us take 0<R1<R2 such that
𝒮
0
(
q
1
,
R
1
)
<
+
∞
and
𝒮
∞
(
q
2
,
R
2
)
<
+
∞
,
which is possible because qi∈ℐi. Then, using the definition of 𝒮0 and 𝒮∞, Lemma 2.13 and the embedding of X in L2(BR2∖BR1), we get
∫
B
R
1
K
(
|
x
|
)
|
u
|
q
1
≤
c
∥
u
∥
q
1
,
∫
B
R
2
c
K
(
|
x
|
)
|
u
|
q
2
≤
c
∥
u
∥
q
2
,
∫
B
R
2
∖
B
R
1
K
(
|
x
|
)
|
u
|
q
1
≤
c
∥
u
∥
q
i
.
Hence,
|
∫
ℝ
N
K
(
|
x
|
)
F
(
u
)
𝑑
x
|
≤
∫
B
R
1
K
(
|
x
|
)
F
(
u
)
𝑑
x
+
∫
B
R
2
∖
B
R
1
K
(
|
x
|
)
F
(
u
)
𝑑
x
+
∫
ℝ
N
∖
B
R
2
K
(
|
x
|
)
F
(
u
)
𝑑
x
≤
c
∫
B
R
1
K
(
|
x
|
)
|
u
|
q
1
𝑑
x
+
c
∫
B
R
2
∖
B
R
1
K
(
|
x
|
)
|
u
|
q
1
𝑑
x
+
c
∫
ℝ
N
∖
B
R
2
K
(
|
x
|
)
|
u
|
q
2
𝑑
x
≤
c
∥
u
∥
q
1
+
c
∥
u
∥
q
2
,
so that
I
(
u
)
≥
1
2
∥
u
∥
2
-
c
∥
u
∥
q
1
-
c
∥
u
∥
q
2
.
Here c is a constant, independent from u, which may change from line to line. From the last inequality (i) follows easily.
To get (ii), we start with remarking that, from (f2), there is c>0 such that F(t)≥ctθ for all t≥t0. The potential K is not zero a.e., and from this fact it is easy to deduce that there are δ∈(0,1) and a measurable subset Aδ⊂(δ,1δ) such that |Aδ|>δ and K(r)>δ in Aδ. Now take a function φ∈C∞(ℝ) such that 0≤φ(r)≤1 for all r, φ(r)=1 for r∈(δ,1δ), φ(r)=0 for r≤12δ, and r≥1+1δ. Define now ψ(x)=φ(|x|) for x∈ℝN. As φ∈Cc∞(ℝ∖{0}), we have ψ∈Cc∞(ℝN∖{0}), and furthermore ψ is radial, so ψ∈X. Define Ωδ={x∈ℝN:|x|∈Aδ}. Hence, if we take λ>t0, we get
K
(
|
x
|
)
F
(
λ
ψ
(
x
)
)
≥
δ
c
λ
θ
in
Ω
δ
,
and K(|x|)F(λψ(x))≥0 for all x, so that
∫
ℝ
N
K
(
|
x
|
)
F
(
λ
ψ
(
x
)
)
≥
∫
Ω
δ
K
(
|
x
|
)
F
(
λ
ψ
(
x
)
)
≥
c
δ
λ
θ
|
Ω
δ
|
=
C
δ
λ
θ
,
where Cδ>0 depends only on δ and N. We then get
I
(
λ
ψ
)
≤
λ
2
∥
ψ
∥
2
-
C
δ
λ
θ
,
so I(λψ)→-∞ as λ→+∞, and this gives the result. ∎
As I satisfies the Palais–Smale condition, arguing as in the proof of [8, Theorem 1.2], we also get a result of existence of infinity solutions.
Theorem 5.4.
Assume the hypotheses of Theorem 4.6. Assume (f1), (f2) with qi∈Ii, where the intervals Ii are given in Theorem 4.6. Assume furthermore the following two assumptions:
There exists
m
>
0
such that
F
(
t
)
≥
m
min
{
t
q
1
,
t
q
2
}
for all
t
>
0
.
Then I:X→R has a sequence {un}n of critical points such that I(un)→+∞.
6 Examples
In this section, we give some examples that could help to understand what is new (and what is not) in our results. We will make a comparison, in some concrete cases, between our results and those of Su and Wang [20]. They study a p-laplacian equation, so we compare their results with ours only in the case p=2. Our problem is also linked to those studied in [15, 18], but in the following examples we assume A(r)=min{rα,rβ} with α≠β, and this rules out the results of [15, 18], in which A(r)=rα for some α∈ℝ. Su and Wang [20] define three functions q*,q*,q** which depend on the asymptotic behavior of the potentials, and find existence of solutions for, say, f(t)=tq1-1+tq2-1 when qi∈(q*,q*) or when qi>max{q*,q**} (i=1,2).
Example 1.
Let us choose the functions A,V,K as follows:
A
(
r
)
=
min
{
r
2
,
r
3
/
2
}
,
V
(
r
)
=
min
{
1
,
1
r
1
/
2
}
,
K
(
r
)
=
max
{
r
1
/
2
,
r
3
/
2
}
.
It is simple to verify that in this case the results of [20] do not apply because if we compute the functions q* and q*, then q*=4N+62N-1 and q*=2N+1N (and q** is not defined), so that q*<q* while q*>q* is a needed hypothesis. To apply our results, we can choose β0=β∞=0, α0=12, α∞=32, a0=2 and a∞=32. We then get q*(a0,α0,β0)=2N+1N>2 and q*(a∞,α∞,β∞)=4N+62N-1. Hence, if we choose
2
<
q
1
<
2
N
+
1
N
<
4
N
+
6
2
N
-
1
<
q
2
and f(t)=min{tq1-1,tq2-1}, we can apply our existence results. Notice that in this case ℐ1∩ℐ2=∅.
Example 2.
Assume N≥6 and choose the functions A,V,K as follows:
A
(
r
)
=
max
{
1
r
2
,
1
r
3
}
,
V
(
r
)
=
1
r
4
,
K
(
r
)
=
min
{
1
,
1
r
2
}
.
In this case, following [20], the computations give q*=2NN-5 and q*=2(N-2)N-4, and one has q*<q* for N≥3, so in this case one gets existence for qi∈(2(N-2)N-4,2NN-5). To apply our results, we can choose β0=β∞=α0=0, α∞=-2, a0=-3 and a∞=-2. We then get q*(a0,α0,β0)=2NN-5 and q*(a∞,α∞,β∞)=2(N-2)N-4, so
ℐ
1
∩
ℐ
2
=
(
2
(
N
-
2
)
N
-
4
,
2
N
N
-
5
)
,
which is the same interval. Hence, for pure power functions we obtain exactly the same result as in [20], while we can not treat functions like f(t)=tq1-1+tq2-1. On the other hand, we are free to choose 2<q1<2(N-2)N-4<q2 and f(t)=min{tq1-1,tq2-1}, and such a function does not satisfy the hypotheses of [20] because it satisfies (f2) with θ=q1<2(N-2)N-4, which is not allowed in [20].
Example 3.
Assume again N≥6 and choose the functions A,V,K as follows:
A
(
r
)
=
max
{
1
r
2
,
1
r
3
}
,
V
(
r
)
=
e
2
r
,
K
(
r
)
=
e
r
.
In this case, the results of [20] do not apply because of the exponential growth of the potential K. We can choose a0=-3, a∞=-2, β0=α0=α∞=0 and β∞=12, and we get, as before,
q
*
(
a
0
,
α
0
,
β
0
)
=
2
N
N
-
5
and
q
*
(
a
∞
,
α
∞
,
β
∞
)
=
2
(
N
-
2
)
N
-
4
,
so again we get existence of solutions for functions like f(t)=min{tq1-1,tq2-1} and for the same range of exponents qi as above. In particular, we can choose f(t)=tq-1 for
q
∈
(
2
(
N
-
2
)
N
-
4
,
2
N
N
-
5
)
.
and
Federica Zaccagni
