# Generalized Solutions for Nonlocal Elliptic Equations and Systems with Nonlinear Singularities

Authors: Youtao Wang; Guangcun Lu
Journal: Abstract and Applied Analysis
 (2015)

---

## Abstract

We use the topological degree method to study the existence of solutions for nonlocal elliptic equations (systems) with a strong singular nonlinearity.


---

## Body

## 1. Introduction and Main Results

Givens
∈
(
0,1
), an integer n
>
2
s, and a bounded open set Ω of R
n with Lipschitz boundary, let K
:
R
n
∖
{
0
}
→
(
0
,
+
∞
) be a function satisfying the following properties:(i)
γ
K
∈
L
1
(
R
n
) with γ
(
x
)
=
m
i
n
{
|
x
|
2
,
1
}.
(ii)
There existsθ
>
0 such that K
(
x
)
≥
θ
|
x
|
-
(
n
+
2
s
) for any x
∈
R
n
∖
{
0
}.
(iii)
K
(
x
)
=
K
(
-
x
) for any x
∈
R
n
∖
{
0
}.The so-called nonlocal elliptic operatorL
K is defined by (1)
L
K
u
x
=
1
2
∫
R
n
u
x
+
y
+
u
x
-
y
-
2
u
x
K
y
d
y
,
x
∈
R
n
.In particular, when K
(
x
)
=
x
-
(
n
+
2
s
), L
K is equal to the fractional Laplace operator -
(
-
Δ
)
s (up to normalization factors).For a Carathéodory functionf
:
Ω
×
R
→
R, the following problem(2)
L
K
u
+
f
x
,
u
=
0
in
Ω
,
u
=
0
in
R
n
∖
Ωand its special case(3)
-
Δ
s
u
=
f
x
,
u
in
Ω
,
u
=
0
in
R
n
∖
Ωhave been widely studied under various contexts; see a recent survey [1] for details.


### 1.1. Previous Work

Motivated by the work of Caffarelli and Silvestre [2], several authors have considered an equivalent problem of (3) by means of an auxiliary variable; see [2–6]. Precisely, let (
x
,
y
) denote the points in C
Ω
≔
Ω
×
(
0
,
∞
)
⊂
R
+
n
+
1 and ∂
L
C
Ω
=
∂
Ω
×
(
0
,
∞
). Take α
=
2
s and X
0
α
(
C
Ω
) as the completion of C
0
∞
(
Ω
×
(
0
,
∞
)
) with respect to the norm (4)
z
X
0
α
C
Ω
=
κ
α
∫
C
Ω
y
1
-
α
∇
z
2
d
x
d
y
1
/
2
,where κ is a normalization constant. For w
∈
X
0
α
(
C
Ω
), let (5)
L
α
w
≔
-
div
⁡
y
1
-
α
∇
w
,
∂
w
∂
ν
α
≔
κ
α
lim
y
→
0
+
⁡
y
1
-
α
∂
w
∂
yand consider the problem 3
′
L
α
w
=
0
in
C
Ω
,
w
=
0
in
∂
L
C
Ω
,
∂
w
∂
ν
α
=
f
x
,
w
in
Ω
×
y
=
0
.An energy solution to this problem is a function w
∈
X
0
α
(
C
Ω
) such that (6)
κ
α
∫
C
Ω
y
1
-
α
∇
w
,
∇
φ
d
x
d
y
=
∫
Ω
f
x
,
w
φ
d
x
∀
φ
∈
X
0
α
C
Ω
.Such an energy solution w yields a function u
=
w
(
·
,
0
) in the sense of traces, which belongs to the space H
0
α
/
2
(
Ω
) and is a weak solution of (3). The converse is also true. The reader may refer to [2–6] for dealing with (3) with this method. In particular, Stinga and Torrea [6] generalized the arguments and results in [2] to the fractional powers L
σ, 0
<
σ
<
1, of a linear second order partial differential operator L that is nonnegative, densely defined, and self-adjoint in L
2
(
Ω
,
d
η
) with a positive measure d
η on Ω.Servadei and Valdinoci developed a variational framework to study the problem (2) in a series of papers [7–11]. They introduced the following Hilbert space (
X
0
(
Ω
,
K
)
,
〈
·
,
·
〉
0
,
Ω
,
K
) in [7, 8]. Let Q
≔
R
2
n
∖
(
C
Ω
×
C
Ω
), where C
Ω
=
R
n
∖
Ω, and let X
(
Ω
,
K
) be the space of all Lebesgue measurable functions u
:
R
n
→
R such that u
|
Ω
∈
L
2
(
Ω
) and that the map (7)
Q
∋
x
,
y
⟼
u
x
-
u
y
K
x
-
y
is  in
L
2
Q
.
X
(
Ω
,
K
) is a Banach space endowed with the so-calledGagliardo norm
(8)
u
Ω
,
K
=
∫
Ω
u
x
2
d
x
+
∫
Q
u
x
-
u
y
2
K
x
-
y
d
x
d
y
1
/
2and is contained in H
s
(
R
n
). Consider the subspace of X
(
Ω
,
K
): (9)
X
0
Ω
,
K
=
u
∈
X
Ω
,
K
∣
u
=
0
a.e.  in
R
n
∖
Ω
.It was proved in [12, Theorem 6] that this space is the closure of C
0
∞
(
Ω
) in X
(
Ω
,
K
). Clearly, the space X
0
(
Ω
,
K
) depends on K. In fact, when K
(
x
)
=
|
x
|
-
(
n
+
2
s
), X
0
(
Ω
,
K
)
=
{
u
∈
H
s
(
R
n
)
∣
u
=
0
a
.
e
.
i
n
R
n
∖
Ω
}  ([11, Lemma 7-b]).  X
0
(
Ω
,
K
) can be endowed with a Hilbert space structure given by the inner product(10)
u
,
v
0
,
Ω
,
K
=
∫
Q
u
x
-
u
y
v
x
-
v
y
K
x
-
y
d
x
d
y([8, Lemma 7]) and contains C
0
2
(
Ω
) ([7, Lemma 11]). Denote by ·
0
,
Ω
,
K the induced norm of the inner product in (10). This norm is equivalent to the restriction of ·
Ω
,
K to X
0
(
Ω
,
K
). Call u
∈
X
0
(
Ω
,
K
) aweak solution of the problem (2) if u satisfies (11)
∫
R
2
n
u
x
-
u
y
ϕ
x
-
ϕ
y
K
x
-
y
d
x
d
y
=
∫
Ω
f
x
,
u
x
ϕ
x
d
xfor all ϕ
∈
X
0
(
Ω
,
K
). Define F
(
x
,
t
)
=
∫
0
t
‍
f
(
x
,
τ
)
d
τ for the above Carathéodory function f. Suppose that there exist a
1
>
0, a
2
>
0 and q
∈
(
2
,
2
s
∗
), 2
s
∗
=
2
n
/
(
n
-
2
s
), such that(12)
f
x
,
t
≤
a
1
+
a
2
t
q
-
1
a.e.
x
∈
Ω
,
t
∈
R
.Then, the functional J
:
X
0
(
Ω
,
K
)
→
R defined by(13)
J
u
=
1
2
∫
Q
u
x
-
u
y
2
K
x
-
y
d
x
d
y
-
∫
Ω
F
x
,
u
x
d
xis of class C
1 and the critical points of J are exactly the weak solutions of (2). Condition (12) is always assumed in the proofs of present several existence results on (2) via variational methods [7–11, 13–15]. Except for [13] u
=
0 is also assumed to be a solution in all other works. For studies of (3), there is a great deal of literature; see [2, 4, 16] and references therein.However, all previous results cannot include the following case:Ω
=
B
n
(
0,1
)
=
{
x
∈
R
n
∣
|
x
|
<
1
} and f
(
x
,
t
)
=
1
/
|
x
|
-
1
+
h
(
t
), where h
∈
C
1
(
R
) such that h
(
t
)
=
cos
⁡
t for |
t
|
≪
1 and that h
(
t
)
=
sin
⁡
t for |
t
|
≫
1.


### 1.2. Main Results of This Paper

We will use the topological degree theory developed by [17] to study generalized solutions of problem (2).Our result can apply to the example just mentioned. Without special statements, we write(14)
r
=
2
n
n
+
2
s
,
2
s
∗
=
2
n
n
-
2
s(the latter plays the role of a critical Sobolev exponent).Theorem 1.
For an integern
≥
2 let K
:
R
n
∖
{
0
}
→
(
0
,
+
∞
) satisfy (i)–(iii) and let f
:
Ω
×
R
→
R be a Carathéodory function verifying the following conditions: there exist positive numbers α, β, q
∈
(
2
,
2
s
∗
), p
∈
[
1,2
) and functions a
∈
L
loc
⁡
r
(
Ω
) and c
∈
L
1
(
Ω
) such that(15)
f
x
,
t
≤
a
x
+
α
t
q
-
1
∀
x
,
t
∈
Ω
×
R
,
(16)
Ω
∋
x
⟼
b
x
≔
f
x
,
0
b
e
l
o
n
g
s
t
o
L
r
Ω
,
(17)
-
f
x
,
t
t
≥
-
β
t
p
-
c
x
∀
x
,
t
∈
Ω
×
R
.Then, problem (2) has at least one generalized solution u in X
0
(
Ω
,
K
); that is, it satisfies (18)
∫
R
2
n
u
x
-
u
y
ϕ
x
-
ϕ
y
K
x
-
y
d
x
d
y
=
∫
Ω
f
x
,
u
x
ϕ
x
d
x
∀
ϕ
∈
C
0
∞
Ω
.In particular, it must have a nontrivial generalized solution if b is not identically zero.Corollary 2.
Under the assumptions of Theorem1, let 1
<
ν
<
2, l
>
2
n
/
2
n
-
(
n
-
2
s
)
ν and let G
∈
L
loc
⁡
l
(
Ω
). If either λ
G
≤
0 or (
λ
G
)
+
≔
max
⁡
{
0
,
λ
G
} belongs to L
κ
(
Ω
) with κ
>
2
/
(
2
-
ν
), then (19)
L
K
u
+
λ
G
x
u
ν
-
2
u
+
f
x
,
u
=
0
i
n
Ω
,
u
=
0
i
n
R
n
∖
Ω
,has at least one nontrivial solution in X
0
(
Ω
,
K
) provided b is not identically zero.Corollary 3.
Under the assumptions of Theorem1, let 2
≤
ν
<
2
s
∗ and G
∈
L
loc
⁡
l
(
Ω
) with l
≥
r
ϑ and ϑ
>
n
+
2
s
/
n
+
2
s
-
(
ν
-
1
)
(
n
-
2
s
). If λ
∈
R is such that λ
G
≤
0, then (20)
L
K
u
+
λ
G
x
u
ν
-
2
u
+
f
x
,
u
=
0
i
n
Ω
,
u
=
0
i
n
R
n
∖
Ω
,has at least one nontrivial solution in X
0
(
Ω
,
K
) provided b is not identically zero.In particular, this corollary includes the example at the end of Section1.2. See Example 1 for more general cases.Our methods can also be used to study the case of nonlocal elliptic operator systems. LetK
1
,
K
2
:
R
n
∖
{
0
}
→
(
0
,
+
∞
) be functions satisfying conditions (i)–(iii). Given two Carathéodory functions f
i
:
Ω
×
R
→
R, i
=
1,2, consider the following problem:(21)
L
K
1
u
+
f
1
x
,
v
=
0
in
Ω
,
L
K
2
v
+
f
2
x
,
u
=
0
in
Ω
,
u
=
v
=
0
in
R
n
∖
Ω
.Call (
u
,
v
)
∈
X
0
(
Ω
,
K
1
)
×
X
0
(
Ω
,
K
2
) ageneralized solution of system (21) if(22)
∫
R
2
n
u
x
-
u
y
ϕ
x
-
ϕ
y
K
1
x
-
y
d
x
d
y
+
∫
R
2
n
v
x
-
v
y
ψ
x
-
ψ
y
·
K
2
x
-
y
d
x
d
y
-
∫
Ω
f
1
x
,
v
ϕ
x
d
x
-
∫
Ω
f
2
x
,
u
ψ
x
d
x
=
0for every (
ϕ
,
ψ
)
∈
C
0
∞
(
Ω
)
×
C
0
∞
(
Ω
). Here is the second main result.Theorem 4.
Under the above assumptions, suppose also that there exist positive numbersα
j, β
j, q
j
∈
(
2
,
2
s
∗
), 1
≤
p
j
,
p
^
j
<
2, and functions a
j
∈
L
l
o
c
r
(
Ω
) and c
j
∈
L
1
(
Ω
) such that(23)
f
j
x
,
t
≤
a
j
x
+
α
i
t
q
j
-
1
∀
x
,
t
∈
Ω
×
R
,
(24)
Ω
∋
x
⟼
b
j
x
≔
f
j
x
,
0
b
e
l
o
n
g
s
t
o
L
r
Ω
,
(25)
-
f
j
x
,
t
h
≥
-
β
j
t
p
j
/
2
h
p
^
j
/
2
-
c
j
x
∀
x
,
t
,
h
∈
Ω
×
R
×
R
.Then, problem (21) has at least one generalized solution in X
0
(
Ω
,
K
1
)
×
X
0
(
Ω
,
K
2
). In particular, it must have a nontrivial generalized solution if one of b
1 and b
2 is not identically zero.Similarly, consider the following problem:(26)
-
Δ
u
x
+
∑
j
=
1
n
f
j
x
,
v
x
∂
u
∂
x
j
x
+
f
0
x
,
v
x
+
a
1
x
=
0
,
-
Δ
v
x
+
∑
j
=
1
n
g
j
x
,
u
x
∂
v
∂
x
j
x
+
g
0
x
,
u
x
+
a
2
x
=
0or(27)
-
Δ
u
x
+
∑
j
=
1
n
f
j
x
,
v
x
∂
v
∂
x
j
x
+
f
0
x
,
v
x
+
a
1
x
=
0
,
-
Δ
v
x
+
∑
j
=
1
n
g
j
x
,
u
x
∂
u
∂
x
j
x
+
g
0
x
,
u
x
+
a
2
x
=
0
.Assume(A)
f
i
,
g
i
:
Ω
×
R
n
→
R are Carathéodory functions, i
=
0
,
…
,
n;
(B)
there exist constants(28)
r
0
∈
2
n
n
+
2
,
∞
,
r
i
∈
n
,
∞
,
n
-
2
2
n
r
i
<
1
s
i
<
∞
,
i
=
1
,
…
,
n
,
and measurable functionsb
i
,
d
i
∈
L
loc
⁡
r
i
(
Ω
), i
=
0,1
,
…
,
n
,
a
1
,
a
2
∈
L
r
0
(
Ω
), such that (29)
f
0
x
,
0
=
g
0
x
,
0
=
0
,
∀
x
∈
Ω
,
f
i
x
,
t
≤
b
i
x
+
k
i
t
s
i
,
∀
x
,
t
∈
Ω
×
R
,
i
=
0
,
…
,
n
,
g
i
x
,
t
≤
d
i
x
+
l
i
t
s
i
,
∀
x
,
t
∈
Ω
×
R
,
i
=
0
,
…
,
n
;
(C)
there exist measurable functionsc
(
x
)
,
c
^
(
x
)
∈
L
1
(
Ω
) and constants 1
<
q
1
,
q
2
,
q
^
1
,
q
^
2
<
2 such that, for any (
x
,
t
1
,
t
2
,
z
)
∈
Ω
×
R
×
R
×
R
n, (30)
-
1
2
z
2
-
k
t
1
q
1
/
2
t
2
q
^
1
/
2
c
x
≤
∑
i
=
1
n
f
i
x
,
t
2
z
i
+
f
0
x
,
t
2
+
a
1
x
t
1
,
-
1
2
z
2
-
k
^
t
1
q
2
/
2
t
2
q
^
2
/
2
c
^
x
≤
∑
i
=
1
n
g
i
x
,
t
2
z
i
+
g
0
x
,
t
2
+
a
2
x
t
1
.
Combing the proof of [17] and that of Theorem 4, we can prove the following.Theorem 5.
Under the conditions (A), (B), and (C), ifa
1 or a
2 is not zero, then the equation systems (26) and (27) have at least a nontrivial generalized solution (
u
,
v
)
∈
W
0
1,2
(
Ω
)
×
W
0
1,2
(
Ω
).Finally, let us point out that the corresponding results of Theorems1 and 4 can be also proved if the operator L
K is replaced by L
σ in [6, (1.10)]. They will be given in other places.The arrangements of this paper are as follows. In Section2, we give some necessary preliminaries. The proof of Theorem 1 will be completed in Section 3. In Section 4, we will prove Corollaries 2 and 3 and give an example. Theorem 4 will be proved in Section 5.


## 1.1. Previous Work

Motivated by the work of Caffarelli and Silvestre [2], several authors have considered an equivalent problem of (3) by means of an auxiliary variable; see [2–6]. Precisely, let (
x
,
y
) denote the points in C
Ω
≔
Ω
×
(
0
,
∞
)
⊂
R
+
n
+
1 and ∂
L
C
Ω
=
∂
Ω
×
(
0
,
∞
). Take α
=
2
s and X
0
α
(
C
Ω
) as the completion of C
0
∞
(
Ω
×
(
0
,
∞
)
) with respect to the norm (4)
z
X
0
α
C
Ω
=
κ
α
∫
C
Ω
y
1
-
α
∇
z
2
d
x
d
y
1
/
2
,where κ is a normalization constant. For w
∈
X
0
α
(
C
Ω
), let (5)
L
α
w
≔
-
div
⁡
y
1
-
α
∇
w
,
∂
w
∂
ν
α
≔
κ
α
lim
y
→
0
+
⁡
y
1
-
α
∂
w
∂
yand consider the problem 3
′
L
α
w
=
0
in
C
Ω
,
w
=
0
in
∂
L
C
Ω
,
∂
w
∂
ν
α
=
f
x
,
w
in
Ω
×
y
=
0
.An energy solution to this problem is a function w
∈
X
0
α
(
C
Ω
) such that (6)
κ
α
∫
C
Ω
y
1
-
α
∇
w
,
∇
φ
d
x
d
y
=
∫
Ω
f
x
,
w
φ
d
x
∀
φ
∈
X
0
α
C
Ω
.Such an energy solution w yields a function u
=
w
(
·
,
0
) in the sense of traces, which belongs to the space H
0
α
/
2
(
Ω
) and is a weak solution of (3). The converse is also true. The reader may refer to [2–6] for dealing with (3) with this method. In particular, Stinga and Torrea [6] generalized the arguments and results in [2] to the fractional powers L
σ, 0
<
σ
<
1, of a linear second order partial differential operator L that is nonnegative, densely defined, and self-adjoint in L
2
(
Ω
,
d
η
) with a positive measure d
η on Ω.Servadei and Valdinoci developed a variational framework to study the problem (2) in a series of papers [7–11]. They introduced the following Hilbert space (
X
0
(
Ω
,
K
)
,
〈
·
,
·
〉
0
,
Ω
,
K
) in [7, 8]. Let Q
≔
R
2
n
∖
(
C
Ω
×
C
Ω
), where C
Ω
=
R
n
∖
Ω, and let X
(
Ω
,
K
) be the space of all Lebesgue measurable functions u
:
R
n
→
R such that u
|
Ω
∈
L
2
(
Ω
) and that the map (7)
Q
∋
x
,
y
⟼
u
x
-
u
y
K
x
-
y
is  in
L
2
Q
.
X
(
Ω
,
K
) is a Banach space endowed with the so-calledGagliardo norm
(8)
u
Ω
,
K
=
∫
Ω
u
x
2
d
x
+
∫
Q
u
x
-
u
y
2
K
x
-
y
d
x
d
y
1
/
2and is contained in H
s
(
R
n
). Consider the subspace of X
(
Ω
,
K
): (9)
X
0
Ω
,
K
=
u
∈
X
Ω
,
K
∣
u
=
0
a.e.  in
R
n
∖
Ω
.It was proved in [12, Theorem 6] that this space is the closure of C
0
∞
(
Ω
) in X
(
Ω
,
K
). Clearly, the space X
0
(
Ω
,
K
) depends on K. In fact, when K
(
x
)
=
|
x
|
-
(
n
+
2
s
), X
0
(
Ω
,
K
)
=
{
u
∈
H
s
(
R
n
)
∣
u
=
0
a
.
e
.
i
n
R
n
∖
Ω
}  ([11, Lemma 7-b]).  X
0
(
Ω
,
K
) can be endowed with a Hilbert space structure given by the inner product(10)
u
,
v
0
,
Ω
,
K
=
∫
Q
u
x
-
u
y
v
x
-
v
y
K
x
-
y
d
x
d
y([8, Lemma 7]) and contains C
0
2
(
Ω
) ([7, Lemma 11]). Denote by ·
0
,
Ω
,
K the induced norm of the inner product in (10). This norm is equivalent to the restriction of ·
Ω
,
K to X
0
(
Ω
,
K
). Call u
∈
X
0
(
Ω
,
K
) aweak solution of the problem (2) if u satisfies (11)
∫
R
2
n
u
x
-
u
y
ϕ
x
-
ϕ
y
K
x
-
y
d
x
d
y
=
∫
Ω
f
x
,
u
x
ϕ
x
d
xfor all ϕ
∈
X
0
(
Ω
,
K
). Define F
(
x
,
t
)
=
∫
0
t
‍
f
(
x
,
τ
)
d
τ for the above Carathéodory function f. Suppose that there exist a
1
>
0, a
2
>
0 and q
∈
(
2
,
2
s
∗
), 2
s
∗
=
2
n
/
(
n
-
2
s
), such that(12)
f
x
,
t
≤
a
1
+
a
2
t
q
-
1
a.e.
x
∈
Ω
,
t
∈
R
.Then, the functional J
:
X
0
(
Ω
,
K
)
→
R defined by(13)
J
u
=
1
2
∫
Q
u
x
-
u
y
2
K
x
-
y
d
x
d
y
-
∫
Ω
F
x
,
u
x
d
xis of class C
1 and the critical points of J are exactly the weak solutions of (2). Condition (12) is always assumed in the proofs of present several existence results on (2) via variational methods [7–11, 13–15]. Except for [13] u
=
0 is also assumed to be a solution in all other works. For studies of (3), there is a great deal of literature; see [2, 4, 16] and references therein.However, all previous results cannot include the following case:Ω
=
B
n
(
0,1
)
=
{
x
∈
R
n
∣
|
x
|
<
1
} and f
(
x
,
t
)
=
1
/
|
x
|
-
1
+
h
(
t
), where h
∈
C
1
(
R
) such that h
(
t
)
=
cos
⁡
t for |
t
|
≪
1 and that h
(
t
)
=
sin
⁡
t for |
t
|
≫
1.


## 1.2. Main Results of This Paper

We will use the topological degree theory developed by [17] to study generalized solutions of problem (2).Our result can apply to the example just mentioned. Without special statements, we write(14)
r
=
2
n
n
+
2
s
,
2
s
∗
=
2
n
n
-
2
s(the latter plays the role of a critical Sobolev exponent).Theorem 1.
For an integern
≥
2 let K
:
R
n
∖
{
0
}
→
(
0
,
+
∞
) satisfy (i)–(iii) and let f
:
Ω
×
R
→
R be a Carathéodory function verifying the following conditions: there exist positive numbers α, β, q
∈
(
2
,
2
s
∗
), p
∈
[
1,2
) and functions a
∈
L
loc
⁡
r
(
Ω
) and c
∈
L
1
(
Ω
) such that(15)
f
x
,
t
≤
a
x
+
α
t
q
-
1
∀
x
,
t
∈
Ω
×
R
,
(16)
Ω
∋
x
⟼
b
x
≔
f
x
,
0
b
e
l
o
n
g
s
t
o
L
r
Ω
,
(17)
-
f
x
,
t
t
≥
-
β
t
p
-
c
x
∀
x
,
t
∈
Ω
×
R
.Then, problem (2) has at least one generalized solution u in X
0
(
Ω
,
K
); that is, it satisfies (18)
∫
R
2
n
u
x
-
u
y
ϕ
x
-
ϕ
y
K
x
-
y
d
x
d
y
=
∫
Ω
f
x
,
u
x
ϕ
x
d
x
∀
ϕ
∈
C
0
∞
Ω
.In particular, it must have a nontrivial generalized solution if b is not identically zero.Corollary 2.
Under the assumptions of Theorem1, let 1
<
ν
<
2, l
>
2
n
/
2
n
-
(
n
-
2
s
)
ν and let G
∈
L
loc
⁡
l
(
Ω
). If either λ
G
≤
0 or (
λ
G
)
+
≔
max
⁡
{
0
,
λ
G
} belongs to L
κ
(
Ω
) with κ
>
2
/
(
2
-
ν
), then (19)
L
K
u
+
λ
G
x
u
ν
-
2
u
+
f
x
,
u
=
0
i
n
Ω
,
u
=
0
i
n
R
n
∖
Ω
,has at least one nontrivial solution in X
0
(
Ω
,
K
) provided b is not identically zero.Corollary 3.
Under the assumptions of Theorem1, let 2
≤
ν
<
2
s
∗ and G
∈
L
loc
⁡
l
(
Ω
) with l
≥
r
ϑ and ϑ
>
n
+
2
s
/
n
+
2
s
-
(
ν
-
1
)
(
n
-
2
s
). If λ
∈
R is such that λ
G
≤
0, then (20)
L
K
u
+
λ
G
x
u
ν
-
2
u
+
f
x
,
u
=
0
i
n
Ω
,
u
=
0
i
n
R
n
∖
Ω
,has at least one nontrivial solution in X
0
(
Ω
,
K
) provided b is not identically zero.In particular, this corollary includes the example at the end of Section1.2. See Example 1 for more general cases.Our methods can also be used to study the case of nonlocal elliptic operator systems. LetK
1
,
K
2
:
R
n
∖
{
0
}
→
(
0
,
+
∞
) be functions satisfying conditions (i)–(iii). Given two Carathéodory functions f
i
:
Ω
×
R
→
R, i
=
1,2, consider the following problem:(21)
L
K
1
u
+
f
1
x
,
v
=
0
in
Ω
,
L
K
2
v
+
f
2
x
,
u
=
0
in
Ω
,
u
=
v
=
0
in
R
n
∖
Ω
.Call (
u
,
v
)
∈
X
0
(
Ω
,
K
1
)
×
X
0
(
Ω
,
K
2
) ageneralized solution of system (21) if(22)
∫
R
2
n
u
x
-
u
y
ϕ
x
-
ϕ
y
K
1
x
-
y
d
x
d
y
+
∫
R
2
n
v
x
-
v
y
ψ
x
-
ψ
y
·
K
2
x
-
y
d
x
d
y
-
∫
Ω
f
1
x
,
v
ϕ
x
d
x
-
∫
Ω
f
2
x
,
u
ψ
x
d
x
=
0for every (
ϕ
,
ψ
)
∈
C
0
∞
(
Ω
)
×
C
0
∞
(
Ω
). Here is the second main result.Theorem 4.
Under the above assumptions, suppose also that there exist positive numbersα
j, β
j, q
j
∈
(
2
,
2
s
∗
), 1
≤
p
j
,
p
^
j
<
2, and functions a
j
∈
L
l
o
c
r
(
Ω
) and c
j
∈
L
1
(
Ω
) such that(23)
f
j
x
,
t
≤
a
j
x
+
α
i
t
q
j
-
1
∀
x
,
t
∈
Ω
×
R
,
(24)
Ω
∋
x
⟼
b
j
x
≔
f
j
x
,
0
b
e
l
o
n
g
s
t
o
L
r
Ω
,
(25)
-
f
j
x
,
t
h
≥
-
β
j
t
p
j
/
2
h
p
^
j
/
2
-
c
j
x
∀
x
,
t
,
h
∈
Ω
×
R
×
R
.Then, problem (21) has at least one generalized solution in X
0
(
Ω
,
K
1
)
×
X
0
(
Ω
,
K
2
). In particular, it must have a nontrivial generalized solution if one of b
1 and b
2 is not identically zero.Similarly, consider the following problem:(26)
-
Δ
u
x
+
∑
j
=
1
n
f
j
x
,
v
x
∂
u
∂
x
j
x
+
f
0
x
,
v
x
+
a
1
x
=
0
,
-
Δ
v
x
+
∑
j
=
1
n
g
j
x
,
u
x
∂
v
∂
x
j
x
+
g
0
x
,
u
x
+
a
2
x
=
0or(27)
-
Δ
u
x
+
∑
j
=
1
n
f
j
x
,
v
x
∂
v
∂
x
j
x
+
f
0
x
,
v
x
+
a
1
x
=
0
,
-
Δ
v
x
+
∑
j
=
1
n
g
j
x
,
u
x
∂
u
∂
x
j
x
+
g
0
x
,
u
x
+
a
2
x
=
0
.Assume(A)
f
i
,
g
i
:
Ω
×
R
n
→
R are Carathéodory functions, i
=
0
,
…
,
n;
(B)
there exist constants(28)
r
0
∈
2
n
n
+
2
,
∞
,
r
i
∈
n
,
∞
,
n
-
2
2
n
r
i
<
1
s
i
<
∞
,
i
=
1
,
…
,
n
,
and measurable functionsb
i
,
d
i
∈
L
loc
⁡
r
i
(
Ω
), i
=
0,1
,
…
,
n
,
a
1
,
a
2
∈
L
r
0
(
Ω
), such that (29)
f
0
x
,
0
=
g
0
x
,
0
=
0
,
∀
x
∈
Ω
,
f
i
x
,
t
≤
b
i
x
+
k
i
t
s
i
,
∀
x
,
t
∈
Ω
×
R
,
i
=
0
,
…
,
n
,
g
i
x
,
t
≤
d
i
x
+
l
i
t
s
i
,
∀
x
,
t
∈
Ω
×
R
,
i
=
0
,
…
,
n
;
(C)
there exist measurable functionsc
(
x
)
,
c
^
(
x
)
∈
L
1
(
Ω
) and constants 1
<
q
1
,
q
2
,
q
^
1
,
q
^
2
<
2 such that, for any (
x
,
t
1
,
t
2
,
z
)
∈
Ω
×
R
×
R
×
R
n, (30)
-
1
2
z
2
-
k
t
1
q
1
/
2
t
2
q
^
1
/
2
c
x
≤
∑
i
=
1
n
f
i
x
,
t
2
z
i
+
f
0
x
,
t
2
+
a
1
x
t
1
,
-
1
2
z
2
-
k
^
t
1
q
2
/
2
t
2
q
^
2
/
2
c
^
x
≤
∑
i
=
1
n
g
i
x
,
t
2
z
i
+
g
0
x
,
t
2
+
a
2
x
t
1
.
Combing the proof of [17] and that of Theorem 4, we can prove the following.Theorem 5.
Under the conditions (A), (B), and (C), ifa
1 or a
2 is not zero, then the equation systems (26) and (27) have at least a nontrivial generalized solution (
u
,
v
)
∈
W
0
1,2
(
Ω
)
×
W
0
1,2
(
Ω
).Finally, let us point out that the corresponding results of Theorems1 and 4 can be also proved if the operator L
K is replaced by L
σ in [6, (1.10)]. They will be given in other places.The arrangements of this paper are as follows. In Section2, we give some necessary preliminaries. The proof of Theorem 1 will be completed in Section 3. In Section 4, we will prove Corollaries 2 and 3 and give an example. Theorem 4 will be proved in Section 5.


## 2. Preliminaries

Firstly, we review the topological degree theory for mappings of class(
B
+
) developed in [17]. Let H be a Hilbert space with inner product 〈
·
,
·
〉 and let {
E
n
}
n be a strictly increasing sequence of finite dimensional subspace of H such that E
=
∪
n
=
1
∞
E
n is dense in H. Denote by P
n the orthogonal projection from H onto E
n for every integer n
∈
N. Let G be an open bounded set in E and let g be a mapping from G
¯
E, the closure of G in E, into H. Put (31)
G
n
=
G
∩
E
n
∀
n
∈
N
,
g
n
x
=
P
n
g
x
∀
n
∈
N
,
x
∈
G
n
¯
.Since H and E induce equivalent topologies on all finite dimensional spaces E
n the subset G
n
⊂
E
n has the same closure in E
n, E, and H, denoted by G
n
¯.Definition 6.
Under the above assumptions,g is said to be ofclass
(
B
+
) on G
¯
E if and only if the following conditions are satisfied: (a)
g
n
:
G
n
¯
→
E
n is a continuous mapping for each n
∈
N.
(b)
There is not any sequence{
x
n
k
}
k in E such that the sequence {
x
n
k
}
k is weakly convergent in H
,
x
n
k
∈
∂
E
n
k
G
n
k
,
〈
g
(
x
n
k
+
1
)
,
x
n
k
+
1
〉
≤
0 and 〈
g
(
x
n
k
+
1
)
,
v
〉
=
0 for all k
∈
N and v in E
n
k.Lemma 7 (see [17, Lemma 2.3]).
LetH
,
{
E
n
}
n
,
E
,
G
,
g and {
g
n
}
n be as in Definition 6. Assume that g is of class (
B
+
) on G
¯
E. Then, there exists an integer n
0 such that the Leray-Schauder degree deg
⁡
(
g
n
,
G
n
,
0
) is defined and (32)
deg
⁡
g
n
,
G
n
,
0
=
deg
⁡
g
n
0
,
G
n
0
,
0
∀
n
≥
n
0
.It follows that(33)
deg
⁡
g
,
G
,
0
≔
lim
n
→
∞
⁡
deg
⁡
g
n
,
G
n
,
0is defined. It was the topological degree of g on G at  0 in [17]. The corresponding versions with usual properties of the Leray-Schauder degree were given in [17, Theorem 2.1]. In particular, the identity map Id is of class (B
+), and deg
⁡
(
Id
,
G
,
0
)
=
1 if 0
∈
G. Moreover, the following proposition is key for the proof of our main results.Proposition 8 (see [17, Corollary 2.1]).
LetH
,
{
E
n
}
n
,
E and G be as in Definition 6. Let g be a mapping from G
¯
E into H such that g
m is continuous on G
m
¯
E
m for any m
∈
N. Suppose that G contains 0 and(34)
g
x
,
x
>
0
,
∀
x
∈
∂
E
G
.Then there is a weakly Cauchy sequence {
x
n
}
n in G such that (35)
lim
n
→
∞
⁡
g
x
n
,
v
=
0
,
∀
v
∈
E
.Next, we need the following results on the spaceX
0
(
Ω
,
K
).Lemma 9.
(
a
)  X
0
(
Ω
,
K
) and X
(
Ω
,
K
) are continuously embedded in H
s
(
R
n
) and H
s
(
Ω
), respectively  ([8, Lemma 5]).
(
b
) If Ω
⊂
R
n is a bounded open subset with continuous boundary, the embedding X
0
(
Ω
,
K
)
↪
L
p
(
R
n
) is compact for any p
∈
[
1
,
2
s
∗
)  ([8, Lemma 8] and [11, Lemma 9-a]).
(
c
) The embedding X
0
(
Ω
,
K
)
↪
L
p
(
R
n
) is continuous for p
=
2
s
∗ ([11, Lemma 9-b]).
(
d
) The embedding H
s
(
R
n
)
↪
L
p
(
R
n
) is continuous for any p
∈
[
1
,
2
s
∗
] ([18, Theorem 6.5]).
(
e
) If Ω is an open set in R
n of class C
0,1 with bounded boundary, then there exist continuous embeddings W
1
,
p
(
Ω
)
↪
W
s
,
p
(
Ω
) and W
0
1
,
p
(
Ω
)
↪
W
0
s
,
p
(
Ω
) for any p
∈
[
1
,
∞
) and s
∈
(
0,1
) ([18, Proposition 2.2]).Lemma 10.
LetΩ be a bounded open set in R
n with boundary of class C
0,1. Then, the space X
0
(
Ω
,
K
) is separable. Furthermore, there exists a sequence {
v
m
}
m in C
0
∞
(
Ω
) such that {
v
m
}
m is a maximal orthogonal set of X
0
(
Ω
,
K
).Proof.
By Proposition 9(f) of [9], there exists a Hilbert basis {
e
k
}
k
≥
1 of X
0
(
Ω
,
K
), which implies separability of X
0
(
Ω
,
K
). So ∪
m
=
1
∞
{
r
1
e
1
+
⋯
+
r
m
e
m
∣
r
i
∈
Q
,
i
=
1
,
…
,
m
} is a dense countable subset in X
0
(
Ω
,
K
). Let {
f
m
}
m
≥
1 denote this countable set. Since C
0
∞
(
Ω
) is dense in X
0
(
Ω
,
K
) by [12, Theorem 6], for each m
∈
N we can take f
m
,
k
∈
C
0
∞
(
Ω
) such that f
m
-
f
m
,
k
0
,
Ω
,
K
<
1
/
k
∀
k
∈
N. Then, {
f
m
,
k
∣
(
m
,
k
)
∈
N
×
N
} is also dense in X
0
(
Ω
,
K
). Let {
h
l
∣
l
∈
N
} be a maximal subset of {
f
m
,
k
∣
(
m
,
k
)
∈
N
×
N
} such that any finite elements in {
h
l
∣
l
∈
N
} are linearly independent. Then, S
p
a
n
(
{
h
l
∣
l
∈
N
}
)
=
S
p
a
n
(
{
f
m
,
k
∣
(
m
,
k
)
∈
N
×
N
}
) is dense in X
0
(
Ω
,
K
). Making the Hilbert-Schmidt orthogonalization procedure for {
h
m
∣
m
∈
N
}, we obtain an orthogonal set {
e
m
∣
m
∈
N
}, which is also a maximal in X
0
(
Ω
,
K
).


## 3. Proof of Theorem1

Take an increasing sequence of open subsets ofΩ, {
Ω
k
}
k, such that each of them has C
0,1-boundary and that (36)
Ω
k
¯
⊂
Ω
k
+
1
∀
k
∈
N
,
Ω
=
⋃
k
=
1
∞
Ω
k
.By Lemma 10, we may choose a sequence {
v
1
,
m
}
m in C
0
∞
(
Ω
1
) such that {
v
1
,
m
}
m is a maximal orthogonal set of X
0
(
Ω
1
,
K
). Then, we can find a sequence {
v
2
,
m
}
m in C
0
∞
(
Ω
2
)
∖
C
0
∞
(
Ω
1
,
K
) such that {
v
1
,
m
,
v
2
,
m
∣
m
∈
N
} is a maximal orthogonal set of X
0
(
Ω
2
,
K
). By the mathematical induction, it is easy to find the set {
v
k
,
m
∣
k
,
m
∈
N
} in C
0
∞
(
Ω
) such that {
v
j
,
m
∣
m
∈
N
,
j
=
1
,
…
,
k
} is a maximal orthogonal set of X
0
(
Ω
k
,
K
) for every k
∈
N. Let us rewrite the countable set {
v
k
,
m
∣
k
,
m
∈
N
} as a sequence {
e
k
}
k. Let E
m be the vector subspace of X
0
(
Ω
,
K
) spanned by {
e
1
,
…
,
e
m
}, and E
=
∪
m
E
m. For conveniences we set H
=
X
0
(
Ω
,
K
) and denote by P
m the orthogonal projection from H onto E
m.Lemma 11.
(
a
)  E is dense in X
0
(
Ω
,
K
).
(
b
) For each u
∈
C
0
∞
(
Ω
), there are k
∈
N and a sequence {
u
m
}
m in E such that the supports of all u
m are contained in Ω
k and that u
m
→
u in X
0
(
Ω
,
K
) as m
→
∞.
(
c
) For every m
∈
N and for every given u
∈
X
0
(
Ω
,
K
), there exists a unique T
m
(
u
) in X
0
(
Ω
,
K
) such that (37)
T
m
u
,
v
0
,
Ω
,
K
=
-
∫
Ω
m
f
x
,
u
x
v
x
d
x
∀
v
∈
X
0
Ω
,
K
.Moreover, if v
∈
C
0
∞
(
Ω
k
), then (38)
T
m
u
,
v
0
,
Ω
,
K
=
T
k
u
,
v
0
,
Ω
,
K
∀
m
≥
k
.
(
d
) Suppose that a sequence {
u
k
}
k
⊂
X
0
(
Ω
,
K
) weakly converges to u in X
0
(
Ω
,
K
). Then, {
T
m
(
u
k
)
}
k weakly converges to T
m
(
u
) in X
0
(
Ω
,
K
) for m
=
1,2
,
….
(
e
) For every given u
∈
E (the support of u must be contained in some Ω
m
0 by the construction of E), there exists a unique T
(
u
)
∈
X
0
(
Ω
,
K
) such that (39)
T
u
,
v
0
,
Ω
,
K
=
-
∫
Ω
f
x
,
u
x
v
x
d
x
∀
v
∈
X
0
Ω
,
K
,
T
u
,
v
0
,
Ω
,
K
=
T
m
u
,
v
0
,
Ω
,
K
∀
v
∈
X
0
Ω
,
K
,
∀
m
≥
m
0
.
f  P
m
∘
(
T
E
m
)
:
E
m
→
E
m is continuous for every m
∈
N.
(
g
) There exists a constant C
>
0 such that (40)
u
+
T
u
,
u
0
,
Ω
,
K
≥
u
0
,
Ω
,
K
2
1
-
C
u
0
,
Ω
,
K
p
-
2
-
c
L
1
Ω
u
0
,
Ω
,
K
-
2
∀
u
∈
E
∖
0
.Proof.
(a) Since{
e
k
}
k is a maximal orthogonal set of X
0
(
Ω
,
K
), E is dense in X
0
(
Ω
,
K
).
(b) For a givenu
∈
C
0
∞
(
Ω
), by the choices of {
Ω
m
}
m, there exists k
∈
N such that the support of u is contained in Ω
k. Let {
v
j
,
m
∣
m
∈
N
,
j
=
1
,
…
,
k
} be the maximal orthogonal set of X
0
(
Ω
k
,
K
) as constructed above. Then, S
p
a
n
(
{
v
j
,
m
∣
m
∈
N
,
j
=
1
,
…
,
k
}
) is dense in X
0
(
Ω
k
,
K
). Hence, we can find a sequence {
u
m
}
m
≥
1 in E
∩
C
0
∞
(
Ω
k
) such that u
m
-
u
0
,
Ω
,
K
→
0 as m
→
∞.
(c) By (16), we can write f
(
x
,
t
)
=
b
(
x
)
+
f
0
(
x
,
t
)
∀
(
x
,
t
)
∈
Ω
×
R. Then, f
0
(
x
,
0
)
=
0
∀
x
∈
Ω, and (15) implies (41)
f
0
x
,
t
≤
a
x
+
b
x
+
α
t
q
-
1
∀
x
,
t
∈
Ω
×
R
.Note that 2
s
∗
<
2
n
/
(
n
-
2
) and 1
/
r
+
1
/
2
s
∗
=
1. Moreover, since q
∈
(
2
,
2
s
∗
) we get (42)
2
n
n
+
2
s
=
r
<
r
q
-
1
<
r
2
s
∗
-
1
=
2
n
n
+
2
s
2
n
n
-
2
s
-
1
=
2
n
n
-
2
s
=
2
s
∗
.For given m
∈
N, by [19, Theorem 3.2.4] (see also [20, page 30]) we have a continuous mapping F
m from L
r
(
q
-
1
)
(
Ω
) into L
r
(
Ω
m
), where (43)
F
m
u
x
=
-
f
0
x
,
u
x
∀
x
∈
Ω
m
.For u
,
v
∈
X
0
(
Ω
,
K
), we have u
∈
L
r
(
q
-
1
)
(
Ω
) and v
∈
L
2
s
∗
(
Ω
) by Lemma 9(b) and (c). Thus, F
m
(
u
)
∈
L
r
(
Ω
m
) and (44)
∫
Ω
m
f
0
x
,
u
x
v
x
d
x
=
∫
Ω
m
F
m
u
v
x
d
x
≤
F
m
u
L
r
Ω
m
v
L
2
s
∗
.Using Lemma 9(c) again, there exists a constant C
>
0 such that v
L
2
s
∗
≤
C
v
0
,
Ω
,
K for all v
∈
X
0
(
Ω
,
K
). It follows that (45)
-
∫
Ω
m
f
0
x
,
u
x
v
x
d
x
-
∫
Ω
b
x
v
x
d
x
≤
C
F
m
u
L
r
Ω
m
+
b
L
r
Ω
v
0
,
Ω
,
K
.Hence, Riesz representative theorem yields a unique T
m
(
u
)
∈
X
0
(
Ω
,
K
) such that(46)
T
m
u
,
v
0
,
Ω
,
K
=
∫
Ω
m
F
m
u
v
d
x
-
∫
Ω
b
x
v
x
∀
v
∈
X
0
Ω
,
K
.
Ifv
∈
C
0
∞
(
Ω
k
), for each integer m
≥
k we deduce (47)
T
m
u
,
v
0
,
Ω
,
K
=
∫
Ω
m
F
m
u
v
d
x
-
∫
Ω
b
x
v
x
=
∫
Ω
k
F
k
u
v
d
x
-
∫
Ω
b
x
v
x
=
T
k
u
,
v
0
,
Ω
,
K
.
(d) Let{
u
k
}
k be a sequence weakly converging to u in X
0
(
Ω
,
K
). Since (48)
1
<
r
q
-
1
<
2
s
∗
<
2
n
n
-
2
,from Lemma 9(b), we deduce that {
u
k
}
k converges to u in L
r
(
q
-
1
)
(
Ω
). Then, the continuity of the map F
m
:
L
r
(
q
-
1
)
(
Ω
)
→
L
r
(
Ω
m
) implies that {
F
m
(
u
k
)
}
k converges to F
m
(
u
) in L
r
(
Ω
m
). Moreover, for v
∈
X
0
(
Ω
,
K
), we have v
∈
L
2
s
∗
(
Ω
). Recall that r
-
1
+
(
2
s
∗
)
-
1
=
1 and L
2
s
∗
(
Ω
)
=
(
L
r
(
Ω
)
)
∗. We deduce that (49)
∫
Ω
m
v
x
F
m
u
k
x
d
x
⟶
∫
Ω
m
v
x
F
m
u
x
d
x
as
k
⟶
∞
,and hence l
i
m
k
→
∞
〈
T
m
(
u
k
)
,
v
〉
0
,
Ω
,
K
=
〈
T
m
(
u
)
,
v
〉
0
,
Ω
,
K by (46).
(e) Sincef
0
(
x
,
0
)
=
0
∀
x
∈
Ω, and S
u
p
p
(
u
)
⊂
Ω
m
0, by (46) we deduce that (50)
-
∫
Ω
f
x
,
u
x
v
x
d
x
=
-
∫
Ω
f
0
x
,
u
x
v
x
d
x
-
∫
Ω
b
x
v
x
d
x
=
-
∫
Ω
m
f
0
x
,
u
x
v
x
d
x
-
∫
Ω
b
x
v
x
d
x
=
T
m
u
,
v
0
,
Ω
,
K
∀
v
∈
X
0
,
∀
m
≥
m
0
.This shows that X
0
(
Ω
,
K
)
∋
v
↦
-
∫
Ω
‍
f
(
x
,
u
(
x
)
)
v
(
x
)
d
x is a continuous linear functional. Using the Riesz representative theorem again we obtain a unique T
(
u
)
∈
X
0
(
Ω
,
K
) such that (51)
T
u
,
v
0
,
Ω
,
K
=
-
∫
Ω
f
x
,
u
x
v
x
d
x
∀
v
∈
X
0
.Clearly, 〈
T
(
u
)
,
v
〉
0
,
Ω
,
K
=
〈
T
m
(
u
)
,
v
〉
0
,
Ω
,
K
∀
v
∈
X
0 for all m
≥
m
0.
(f) By the construction ofE
m, we have an integer m
0
∈
N such that each u
∈
E
m has a support contained in Ω
m
0. Let {
u
k
}
k
⊂
E
m converge to u
∈
E
m. By (d), {
T
n
(
u
k
)
}
k weakly converges to T
n
(
u
) in X
0
(
Ω
,
K
) for every n
∈
N. Then, (e) implies that 〈
T
(
u
k
)
,
v
〉
0
,
Ω
,
K
→
〈
T
(
u
)
,
v
〉
0
,
Ω
,
K
∀
v
∈
X
0
(
Ω
,
K
) as k
→
∞. In particular, since v
∈
E
m satisfies P
m
v
=
v, we have (52)
P
m
∘
T
u
k
,
v
0
,
Ω
,
K
=
T
u
k
,
P
m
v
0
,
Ω
,
K
⟶
T
u
,
P
m
v
0
,
Ω
,
K
=
P
m
∘
T
u
,
v
X
0
,
Ω
,
K
.This shows that {
P
m
∘
T
(
u
k
)
}
k weakly converges to P
m
T
(
u
) in E
m. However, the strong converge and the weak ones on finitely dimensional space E
m are equivalent. Hence, P
m
∘
T
(
u
k
)
→
P
m
∘
T
(
u
) as m
→
∞.
(g) Since1
≤
p
<
2
<
2
s
∗, by Lemma 9(b) and (c) there is a constant C
>
0 such that u
L
p
(
Ω
)
≤
C
u
0
,
Ω
,
K
∀
u
∈
X
0
(
Ω
,
K
). It follows from this and (17) that (53)
u
+
T
u
,
u
0
,
Ω
,
K
=
u
0
,
Ω
,
K
2
-
∫
Ω
f
x
,
u
x
u
x
d
x
≥
u
0
,
Ω
,
K
2
-
∫
Ω
β
u
p
+
c
d
x
=
u
0
,
Ω
,
K
2
-
β
∫
Ω
u
p
d
x
-
c
L
1
=
u
0
,
Ω
,
K
2
-
β
u
L
p
p
-
c
L
1
≥
u
0
,
Ω
,
K
2
-
β
C
p
u
0
,
Ω
,
K
p
-
c
L
1
.This leads to (g).Proof of Theorem1.
Letβ, C, and c be as above. Since 1
≤
p
<
2, we have R
>
0 such that(54)
1
-
β
C
p
R
p
-
2
-
c
L
1
R
-
2
>
1
4
.Let G
=
{
u
∈
E
:
u
0
,
Ω
,
K
<
R
}. Define g
:
G
¯
E
→
X
0
(
Ω
,
K
) by (55)
g
u
=
u
+
T
u
∀
u
∈
G
¯
E
.Let us prove that g is of class (
B
+
) on G
¯
E. Note that G
n
=
G
n
∩
E
n has the same closure in E
n, E, and H, denoted by G
n
¯. Let g
n
:
G
n
¯
→
E
n be defined by g
n
(
u
)
=
P
n
(
g
(
u
)
)
=
P
n
u
+
P
n
∘
T
(
u
)
=
u
+
P
n
∘
T
(
u
) for each n
∈
N. Suppose that a sequence {
u
k
}
k
⊂
G
n
¯ converges to u
∈
G
n
¯. Lemma 11(f) implies P
n
∘
T
(
u
k
)
→
P
n
∘
T
(
u
) in E
n. Hence, g
n is continuous. By the proof of Lemma 11(g) and (54) we deduce that (56)
g
u
,
u
0
,
Ω
,
K
=
u
+
T
u
,
u
0
,
Ω
,
K
≥
R
2
4
∀
u
∈
∂
E
G
=
u
∈
E
:
u
0
,
Ω
,
K
=
R
.This implies that Definition 6(b) is satisfied. Hence, g is of class (
B
+
) on G
¯
E. Moreover, it also shows that g satisfies the conditions of Proposition 8. Thus, we have a weakly Cauchy sequence {
u
n
}
n
⊂
G such that(57)
lim
n
→
∞
⁡
g
u
n
,
v
0
,
Ω
,
K
=
0
∀
v
∈
E
.Let u be the weak limit of {
u
n
}
n in X
0
(
Ω
,
K
). For a given v
∈
E, the support of it is contained in some Ω
k, and thus (58)
g
u
n
,
v
0
,
Ω
,
K
=
u
n
+
T
u
n
,
v
0
,
Ω
,
K
=
u
n
+
T
m
u
n
,
v
0
,
Ω
,
K
=
u
n
+
T
k
u
n
,
v
0
,
Ω
,
K
∀
m
≥
k
,
∀
n
∈
Nby Lemma 11(c). For each fixed n
∈
N, there exists n
~
∈
N such that u
n
∈
C
0
∞
(
Ω
n
~
). Hence, Lemma 11(e) yields (59)
g
u
n
,
v
0
,
Ω
,
K
=
u
n
+
T
u
n
,
v
0
,
Ω
,
K
=
u
n
+
T
m
u
n
,
v
0
,
Ω
,
K
=
u
n
+
T
k
u
n
,
v
0
,
Ω
,
K
∀
m
≥
max
⁡
k
,
n
~
.Taking n
→
∞ in both sides of 〈
g
(
u
n
)
,
v
〉
0
,
Ω
,
K
=
〈
u
n
+
T
k
(
u
n
)
,
v
〉
0
,
Ω
,
K and using (57) and Lemma 11(d), we deduce(60)
u
+
T
m
u
,
v
0
,
Ω
,
K
=
0
,
∀
m
≥
k
.
For any givenv
∈
C
0
∞
(
Ω
), by Lemma 11(b), we have an integer k and a sequence {
v
l
}
l
⊂
E such that S
u
p
p
(
v
l
)
⊂
Ω
k for any l
∈
N and that v
l
→
v in X
0
(
Ω
,
K
). So (60) leads to (61)
u
+
T
m
u
,
v
0
,
Ω
,
K
=
lim
l
→
∞
⁡
u
+
T
m
u
,
v
l
0
,
Ω
,
K
=
0
∀
m
≥
k
;that is, 〈
u
,
v
〉
0
,
Ω
,
K
-
∫
Ω
m
‍
f
(
x
,
u
(
x
)
)
v
(
x
)
d
x
=
0
∀
m
≥
k. Letting m
→
∞, we get (62)
u
,
v
0
,
Ω
,
K
-
∫
Ω
f
x
,
u
x
v
x
d
x
=
0
,which shows that u is a generalized solution. Note that u might be zero! But u
=
0 is not a solution if b
=
f
(
·
,
0
) takes nonzero values on a nonzero measure set. The proof is completed.


## 4. Proofs of Corollaries and Examples

Proof of Corollary2.
Letf
~
(
x
,
t
)
=
λ
G
(
x
)
|
t
|
ν
-
2
t
+
f
(
x
,
t
). It suffices to check that f
~ satisfies (15)–(17). Clearly, we can assume λ
≠
0. Let ρ
=
l
/
r. Since 1
<
ν
<
2, we have (
n
-
2
s
)
ν
>
n
-
2
s and thus (63)
l
>
2
n
2
n
-
n
-
2
s
ν
>
2
n
n
+
2
s
=
r
.Then, ρ
>
1 and(64)
1
<
ν
-
1
ρ
′
=
ν
-
1
ρ
ρ
-
1
=
l
l
-
r
<
n
+
2
s
n
-
2
s
=
2
s
∗
-
1
.By Young’s inequality, we obtain(65)
λ
G
x
t
ν
-
2
t
=
λ
G
x
t
ν
-
1
≤
λ
ρ
G
x
ρ
+
λ
ρ
′
t
ν
-
1
ρ
′
.Note that G
∈
L
loc
⁡
l
(
Ω
) and r
ρ
=
l imply |
G
|
ρ
∈
L
loc
⁡
r
(
Ω
). Let q
~
-
1
=
m
a
x
{
q
-
1
,
ν
-
1
ρ
′
}, which sits in (
1
,
2
s
∗
-
1
). From these and (15), it follows that (66)
f
~
x
,
t
≤
a
x
+
λ
ρ
G
x
ρ
+
λ
ρ
′
t
ν
-
1
ρ
′
+
α
t
q
-
1
≤
a
x
+
λ
ρ
G
x
ρ
+
C
+
α
~
t
q
~
-
1
a.e.
x
∈
Ω
,
∀
t
∈
Rfor some constants C
>
0 and α
~
>
0, where Young’s inequality is used again. So f
~ satisfies (15). Moreover, b
~
(
x
)
≔
f
~
(
x
,
0
)
=
f
(
x
,
0
)
=
b
(
x
); that is, f
~ satisfies (16).
Finally, let us check that (17) holds for f
~. If λ
G
≤
0, then (67)
-
f
~
x
,
t
t
=
-
λ
G
x
t
ν
-
f
x
,
t
t
≥
-
β
t
p
-
c
x
.For another case, observe that (68)
-
f
~
x
,
t
t
=
-
λ
G
x
t
ν
-
f
x
,
t
t
≥
-
β
t
p
-
c
x
-
λ
G
+
x
t
ν
.Since κ
>
2
/
(
2
-
ν
)
>
1, we may choose a real number σ in (
2
/
(
2
-
ν
)
,
κ
). Let σ
′
=
σ
/
(
σ
-
1
). By Young’s inequality, we have (69)
λ
G
+
x
t
ν
≤
1
σ
λ
G
+
x
σ
+
1
σ
′
t
ν
σ
′
.Note that |
(
λ
G
)
+
|
σ
∈
L
1
(
Ω
) since σ
<
κ. Moreover, 1
<
ν
σ
′ and (70)
2
2
-
ν
<
σ
⟺
2
-
ν
2
>
1
σ
⟺
2
ν
>
σ
σ
-
1
=
σ
′
⟺
ν
σ
′
<
2
.Let p
~
=
m
a
x
{
p
,
ν
σ
′
}, which belongs to [
1,2
). Using Young’s inequality, we can derive (71)
-
β
t
p
-
1
σ
′
t
ν
σ
′
≥
-
β
~
t
p
~
-
C
∀
t
∈
Rfor some constants β
~
>
0 and C
>
0. Hence, for a.e. x
∈
Ω and all t
∈
R, we have (72)
-
f
~
x
,
t
t
=
-
λ
G
x
t
ν
-
f
x
,
t
t
≥
-
β
~
t
p
~
-
c
~
x
,where c
~
(
x
)
≔
c
(
x
)
+
1
/
σ
|
(
λ
G
)
+
(
x
)
|
σ
+
C belongs to L
1
(
Ω
) as above. This shows that (17) is true for f
~. The desired conclusion follows from Theorem 1 immediately.Proof of Corollary3.
Letf
~
(
x
,
t
)
=
λ
G
(
x
)
|
t
|
ν
-
2
t
+
f
(
x
,
t
). Since λ
G
≤
0, we see that f
~ satisfies (17) from the above proof. It remains to prove that f
~ satisfies (15)-(16). Let ϑ
'
=
ϑ
/
(
ϑ
-
1
). Note that(73)
ϑ
>
n
+
2
s
n
+
2
s
-
ν
-
1
n
-
2
s
⟺
n
+
2
s
-
ν
-
1
n
-
2
s
n
+
2
s
=
1
-
ν
-
1
n
-
2
s
n
+
2
s
>
1
ϑ
⟺
1
-
1
ϑ
=
ϑ
-
1
ϑ
>
ν
-
1
n
-
2
s
n
+
2
s
⟺
ϑ
ν
-
1
ϑ
-
1
=
ν
-
1
ϑ
'
<
2
s
∗
-
1
=
n
+
2
s
n
-
2
s
.By Young’s inequality, we have(74)
G
x
t
ν
-
1
≤
G
x
ϑ
ϑ
+
t
ν
-
1
ϑ
'
ϑ
'
.Now, l
≥
r
ϑ implies l
/
ϑ
≥
r and |
G
|
ϑ
∈
L
loc
⁡
l
/
ϑ
(
Ω
) because G
∈
L
loc
⁡
l
(
Ω
). We obtain |
G
|
ϑ
∈
L
loc
⁡
r
(
Ω
). As in the proof of Corollary 2, using Young’s inequality, we may derive from this and (73)-(74) that f
~ satisfies (15) and (16).Example 1.
Letn
≥
2 and Ω
⊂
R
n be as above. Consider(75)
L
K
u
+
G
x
u
+
h
u
=
0
in
Ω
,
u
=
0
in
R
n
∖
Ω
,where G
≤
0 belongs to L
loc
⁡
l
(
Ω
) with l
∈
(
n
+
2
s
/
4
s
,
n
/
2
s
), h
∈
C
(
R
) is absolutely continuous, h
(
0
)
≠
0, s
u
p
t
∈
R
h
(
t
)
t
<
∞, and |
h
′
(
t
)
|
≤
e
1
+
e
2
|
t
|
ρ, a.e., t
∈
R, 0
≤
ρ
<
4
s
/
n
-
2
s. Then, (75) has a nontrivial generalized solution.
In fact, takingν
=
2 in Corollary 3, we should require ϑ
>
n
+
2
s
/
4
s. Since l
∈
(
n
+
2
s
/
4
s
,
n
/
2
s
), there is sufficiently small ϵ
>
0 such that (76)
1
+
ϵ
<
n
+
2
s
4
s
+
ϵ
<
l
<
n
2
s
<
n
2
s
+
r
ϵ
=
r
n
+
2
s
4
s
+
r
ϵ
.This means that we can take ϑ
=
n
+
2
s
/
4
s
+
ϵ. Moreover, 1
≤
ρ
+
1
<
1
+
4
s
/
n
-
2
s
=
2
s
∗, and (77)
h
t
-
h
0
=
∫
0
t
h
′
τ
d
τ
≤
e
1
t
+
e
2
ρ
+
1
t
ρ
+
1
≤
e
1
ρ
ρ
+
1
+
e
1
+
e
2
ρ
+
1
t
ρ
+
1
,
-
f
x
,
t
t
=
-
G
x
t
2
-
h
t
t
≥
-
sup
t
∈
R
⁡
h
t
t
>
-
∞
.Hence, (15)–(17) are satisfied for f
(
x
,
t
)
=
G
(
x
)
t
+
h
(
t
).


## 5. Proof of Theorem4

Consider the product Hilbert spaceH
=
X
0
(
Ω
,
K
1
)
×
X
0
(
Ω
,
K
2
) equipped with inner product(78)
u
,
v
H
=
u
1
,
u
2
,
v
1
,
v
2
H
=
u
1
,
v
1
X
0
,
Ω
,
K
1
+
u
2
,
v
2
X
0
,
Ω
,
K
2for u
=
(
u
1
,
u
2
)
,
v
=
(
v
1
,
v
2
)
∈
H. The induced norm is (79)
u
,
v
H
=
u
0
,
Ω
,
K
1
2
+
v
0
,
Ω
,
K
2
2
1
/
2
.LetΩ
=
∪
k
=
1
∞
Ω
k and E
=
∪
m
E
m be as in Section 3. For every integer m
∈
N let E
m
=
∪
k
+
l
≤
m
(
E
k
×
E
l
) and E
=
∪
m
E
m. Denote by P
m the orthogonal projection from E onto E
m. Corresponding to Lemma 11, we have the following.Lemma 13.
(
a
)  E is dense in H.
(
b
) For each u
∈
C
0
∞
(
Ω
)
×
C
0
∞
(
Ω
), there are k
∈
N and a sequence {
u
m
}
m in E such that the supports of all u
m are contained in Ω
k
×
Ω
k and that u
m
→
u in H as m
→
∞.
(
c
) For every m
∈
N, and for every given u
∈
H, there exists a unique T
m
(
u
) in H such that (80)
T
m
u
,
v
H
=
-
∫
Ω
m
f
1
x
,
u
2
x
v
1
x
d
x
-
∫
Ω
m
f
2
x
,
u
1
x
v
2
x
d
xfor any v
=
(
v
1
,
v
2
)
∈
H. Moreover, if v
=
(
v
1
,
v
2
)
∈
C
0
∞
(
Ω
k
)
×
C
0
∞
(
Ω
k
), then 〈
T
m
(
u
)
,
v
〉
H
=
〈
T
k
(
u
)
,
v
〉
H
∀
m
≥
k.
(
d
) Suppose that a sequence {
u
k
}
k
⊂
H weakly converges to u in H. Then, {
T
m
(
u
k
)
}
k weakly converges to T
m
(
u
) in H for m
=
1,2
,
….
(
e
) For every given u
∈
E (the support of u must be contained in some Ω
m
0
×
Ω
m
0 by the construction of E), there exists a unique T
(
u
)
∈
H such that (81)
T
u
,
v
H
=
-
∫
Ω
f
1
x
,
u
2
x
v
1
x
d
x
-
∫
Ω
f
2
x
,
u
1
x
v
2
x
d
x
,
T
u
,
v
H
=
T
m
u
,
v
∀
v
∈
H
Ω
,
∀
m
≥
m
0
.
(
f
)  P
m
∘
(
T
E
m
)
:
E
m
→
E
m is continuous for every m
∈
N.
(
g
) There exists a constant C
>
0 such that (82)
u
+
T
u
,
u
H
≥
u
H
2
1
-
c
1
L
1
Ω
+
c
2
L
1
Ω
u
H
-
2
-
C
u
H
p
1
-
2
+
u
H
p
^
1
-
2
+
u
H
p
2
-
2
+
u
H
p
^
2
-
2for all u
∈
E
∖
{
0
}
.Proof.
(a) and (b) follow from Lemma11(a) and (b) immediately.
(c) By (24), we can write f
j
(
x
,
t
)
=
b
j
(
x
)
+
f
0
,
j
(
x
,
t
)
∀
(
x
,
t
)
∈
Ω
×
R as in the proof of Lemma 11(c); then, f
0
,
j
(
x
,
0
)
=
0
∀
x
∈
Ω, and (83)
f
0
,
j
x
,
t
≤
a
j
x
+
b
j
x
+
α
j
t
q
j
-
1
∀
x
,
t
∈
Ω
×
R
,
j
=
1,2
.Moreover, for every given m
∈
N and j
=
1,2, we have continuous mappings (84)
L
r
q
j
-
1
Ω
∋
u
⟼
F
m
,
j
u
∈
L
r
Ω
m
,where F
m
,
j
(
u
)
(
x
)
=
f
0
,
j
(
x
,
u
(
x
)
) for x
∈
Ω
m, j
=
1,2. Recall that 2
s
∗
=
2
s
∗. For u
1
,
v
1
∈
X
0
(
Ω
,
K
1
)
,
u
2
,
v
2
∈
X
0
(
Ω
,
K
2
), by Lemma 9(b) and (c), we have (85)
v
j
∈
L
2
s
∗
Ω
m
,
u
j
∈
L
r
q
i
-
1
Ω
,
i
,
j
=
1,2
.Thus, F
m
,
1
(
u
2
)
∈
L
r
(
Ω
m
), F
m
,
2
(
u
1
)
∈
L
r
(
Ω
m
), and (86)
-
∫
Ω
m
f
0,1
x
,
u
2
x
v
1
x
d
x
=
∫
Ω
m
F
m
,
1
u
2
v
1
x
d
x
≤
F
m
,
1
u
2
L
r
Ω
m
v
1
L
2
s
∗
,
-
∫
Ω
m
f
0,2
x
,
u
1
x
v
2
x
d
x
=
∫
Ω
m
F
m
,
2
u
1
v
2
x
d
x
≤
F
m
,
2
u
1
L
r
Ω
m
v
2
L
2
s
∗
.Using Lemma 9(c) again, there exist constants C
>
0 such that v
j
L
2
s
∗
≤
C
v
j
0
,
Ω
,
K
j for all v
=
(
v
1
,
v
2
)
∈
H, j
=
1,2. It follows that (87)
-
∫
Ω
m
f
0,1
x
,
u
2
x
v
1
x
d
x
-
∫
Ω
b
1
x
v
1
x
d
x
≤
C
F
m
,
1
u
2
L
r
Ω
m
+
b
1
L
r
Ω
v
1
0
,
Ω
,
K
1
,
-
∫
Ω
m
f
0,2
x
,
u
1
x
v
2
x
d
x
-
∫
Ω
b
2
x
v
2
x
d
x
≤
C
F
m
,
2
u
1
L
r
Ω
m
+
b
2
L
r
Ω
v
2
0
,
Ω
,
K
2
.By the Riesz representative theorem, for each j
∈
{
1,2
}, we have a unique T
m
j
(
u
)
∈
X
0
,
Ω
,
K
j such that(88)
T
m
1
u
,
v
1
0
,
Ω
,
K
1
=
-
∫
Ω
m
f
1
x
,
u
2
v
1
x
d
x
,
(89)
T
m
2
u
,
v
2
0
,
Ω
,
K
2
=
-
∫
Ω
m
f
2
x
,
u
1
v
2
x
d
x
.Setting T
m
(
u
)
≔
(
T
m
1
(
u
)
,
T
m
2
(
u
)
), we obtain(90)
T
m
u
,
v
H
=
T
m
1
u
,
v
1
0
,
Ω
,
K
1
+
T
m
2
u
,
v
2
0
,
Ω
,
K
2
=
-
∫
Ω
m
f
1
x
,
u
2
v
1
x
d
x
-
∫
Ω
m
f
2
x
,
u
1
v
2
x
d
x
.Another claim can be proved as that of Lemma 11(c).
(d) Letu
k
=
(
u
k
1
,
u
k
2
) for each k
∈
N. Then, {
u
k
1
}
k
⊂
X
0
,
Ω
,
K
1 weakly converges to u
1 in X
0
(
Ω
,
K
1
) and {
u
k
2
}
k
⊂
X
0
,
Ω
,
K
2 weakly converges to u
2 in X
0
(
Ω
,
K
2
). For each m
∈
N, by Lemma 11(d), {
T
m
1
(
u
k
)
}
k weakly converges to T
m
1
(
u
) in X
0
(
Ω
,
K
1
) and {
T
m
2
(
u
k
)
}
k weakly converges to T
m
2
(
u
) in X
0
(
Ω
,
K
2
). Thus, {
T
m
(
u
k
)
}
k weakly converges to T
m
(
u
) in H for m
=
1,2
,
….
(e) Sincef
0
,
j
(
x
,
0
)
=
0
∀
x
∈
Ω, j
=
1,2, and S
upp
⁡
(
u
)
⊂
Ω
m
0
×
Ω
m
0, for any integer m
≥
m
0 and v
=
(
v
1
,
v
2
)
∈
H, we derive from (89) that (91)
-
∫
Ω
f
1
x
,
u
2
x
v
1
x
d
x
=
-
∫
Ω
f
0,1
x
,
u
2
x
v
1
x
d
x
-
∫
Ω
b
1
x
v
1
x
d
x
=
-
∫
Ω
m
f
0,1
x
,
u
2
x
v
1
x
d
x
-
∫
Ω
b
1
x
v
1
x
d
x
=
T
m
1
u
,
v
1
0
,
Ω
,
K
1
,
-
∫
Ω
f
2
x
,
u
1
x
v
2
x
d
x
=
-
∫
Ω
f
0,2
x
,
u
1
x
v
2
x
d
x
-
∫
Ω
b
2
x
v
2
x
d
x
=
-
∫
Ω
m
f
0,2
x
,
u
1
x
v
2
x
d
x
-
∫
Ω
b
2
x
v
2
x
d
x
=
T
m
2
u
,
v
2
0
,
Ω
,
K
2
.These show that (92)
X
0
Ω
,
K
1
∋
v
1
⟼
-
∫
Ω
f
1
x
,
u
2
x
v
1
x
d
x
,
X
0
Ω
,
K
2
∋
v
2
⟼
-
∫
Ω
f
2
x
,
u
1
x
v
2
x
d
xare two continuous linear functionals. Using the Riesz representative theorem again we obtain a unique T
1
(
u
)
∈
X
0
(
Ω
,
K
1
)
,
T
2
(
u
)
∈
X
0
(
Ω
,
K
2
) such that (93)
T
1
u
,
v
1
0
,
Ω
,
K
1
=
-
∫
Ω
f
1
x
,
u
2
x
v
1
x
d
x
,
T
2
u
,
v
2
0
,
Ω
,
K
2
=
-
∫
Ω
f
2
x
,
u
1
x
v
2
x
d
xfor all v
=
(
v
1
,
v
2
)
∈
H. Set T
(
u
)
≔
(
T
1
(
u
)
,
T
2
(
u
)
); then, (94)
T
u
,
v
H
=
-
∫
Ω
f
1
x
,
u
2
x
v
1
x
d
x
-
∫
Ω
f
2
x
,
u
1
x
v
2
x
d
x
,
∀
v
∈
H
.Clearly, 〈
T
(
u
)
,
v
〉
H
=
〈
T
m
(
u
)
,
v
〉
H
∀
v
∈
H for all m
≥
m
0.
(f) follows the above (e) and Lemma11(f) directly.
(g) Since1
≤
p
j
,
p
^
j
<
2
<
2
s
∗, by Lemma 9(b) and (c) there is a constant C
0
>
0 such that (95)
u
i
L
p
^
j
Ω
+
u
i
L
p
j
Ω
≤
C
0
u
i
0
,
Ω
,
K
i
∀
u
=
u
1
,
u
2
∈
H
,
i
,
j
=
1,2
.It follows from this and (25) that (96)
u
+
T
u
,
u
H
=
u
H
2
-
∫
Ω
f
1
x
,
u
2
x
u
1
x
d
x
-
∫
Ω
f
2
x
,
u
1
x
u
2
x
d
x
≥
u
H
2
-
∫
Ω
β
1
u
2
p
1
/
2
u
1
p
^
1
/
2
+
c
1
d
x
-
∫
Ω
β
2
u
1
p
2
/
2
u
2
p
^
2
/
2
+
c
2
d
x
≥
u
H
2
-
c
1
L
1
-
c
2
L
1
-
β
1
2
∫
Ω
u
2
p
1
+
u
1
p
^
1
d
x
-
β
2
2
·
∫
Ω
u
1
p
2
+
u
2
p
^
2
d
x
≥
u
H
2
-
β
1
+
β
2
·
u
1
L
p
^
1
p
^
1
+
u
1
L
p
2
p
2
+
u
2
L
p
1
p
1
+
u
2
L
p
^
2
p
^
2
-
c
1
L
1
-
c
2
L
1
≥
u
H
2
-
C
u
1
0
,
Ω
,
K
1
p
^
1
+
u
1
0
,
Ω
,
K
1
p
2
+
u
2
0
,
Ω
,
K
2
p
1
+
u
2
0
,
Ω
,
K
2
p
^
2
-
c
1
L
1
-
c
2
L
1
≥
u
H
2
-
C
u
H
p
1
+
u
H
p
^
1
+
u
H
p
2
+
u
H
p
^
2
-
c
1
L
1
-
c
2
L
1
.Here, C
=
(
β
1
+
β
2
)
m
a
x
{
C
0
p
1
,
C
0
p
^
1
,
C
0
p
2
,
C
0
p
^
2
}. This leads to (g).Proof of Theorem4.
We replace the spaceX
0
(
Ω
,
K
) in the proof of Theorem 1 by H. Since p
j
<
2
,
p
^
j
<
2 for j
=
1,2, as in (54), we have R
>
0 such that(97)
1
-
C
R
p
1
-
2
+
R
p
^
1
-
2
+
R
p
2
-
2
+
R
p
^
2
-
2
-
c
1
L
1
+
c
1
L
1
R
-
2
>
1
4
.Then repeating the proof of Theorem 1, we get a u
=
(
u
1
,
u
2
)
∈
H such that 〈
u
+
T
(
u
)
,
v
〉
H
=
0 for any v
=
(
ϕ
,
ψ
)
∈
C
0
∞
(
Ω
)
×
C
0
∞
(
Ω
); namely, (22) holds.



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*Source: 823143-2015-06-23.xml*