# Existence of Solutions to a Class of Semilinear Elliptic Problem with Nonlinear Singular Terms and Variable Exponent

Authors: Ying Chu; Yanchao Gao; Wenjie Gao
Journal: Journal of Function Spaces
 (2016)

---

## Abstract

The authors of this paper prove the existence and regularity results for the homogeneous Dirichlet boundary value problem to the equation-
d
i
v
(
M
(
x
)
∇
u
n
)
=
f
(
x
)
/
u
α
(
x
) with f
∈
L
m
(
Ω
)
(
m
⩾
1
) and α
(
x
)
>
0. The results show the dependence of the summability of f in some Lebesgue spaces and on the values of α
(
x
).


---

## Body

## 1. Introduction

In this paper, we study the existence of solutions for the following semilinear elliptic problem with nonlinear singular terms and variable exponent:(1)
-
d
i
v
M
x
∇
u
=
f
x
u
α
x
,
x
∈
Ω
,
u
=
0
,
x
∈
∂
Ω
,where Ω is a bounded domain in R
N
(
N
⩾
2
) with smooth boundary ∂
Ω, α
(
x
) is a continuous function on Ω
¯, α
(
x
)
>
0, α
+
=
s
u
p
x
∈
Ω
¯
α
(
x
), α
-
=
i
n
f
x
∈
Ω
¯
α
(
x
), f is a nonnegative function belonging to the Lebesgue space L
m
(
Ω
), for some suitable m
⩾
1, and M is a bounded positive definite matrix; that is, there exist 0
<
γ
⩽
β such that(2)
γ
η
2
⩽
M
x
η
·
η
,
M
x
⩽
β
,for every η in R
N, for almost every x in Ω.Problem (1) has been widely applied in many areas, such as the contexts of chemical heterogeneous catalysts, non-Newtonian fluids, and also the theory of heat conduction in electrically conducting materials; see [1–4] for detailed discussion.Problem (1) has been extensively studied in the past. In [5], Lazer and Mckenna dealt with model (1) with f, a continuous function; they proved that the solution was in H
0
1
(
Ω
) if and only if α
<
3, while it was not in C
1
(
Ω
¯
) if α
>
1.Later, Lair and Shaker in [6] studied the existence of solutions to the elliptic equation(3)
-
Δ
u
=
f
x
u
-
α
,
x
∈
Ω
,
u
=
0
,
x
∈
∂
Ω
.They proved that problem (3) with 0
<
α
<
1 has a unique weak positive solution in H
0
1
(
Ω
) if f
(
x
) is a nonnegative nontrivial function in L
2
(
Ω
).Moreover, the results of Lair and Shaker were generalized by Shi and Yao (see [7]); they studied the following problem:(4)
-
Δ
u
=
f
x
,
u
,
x
∈
Ω
,
u
=
φ
,
x
∈
∂
Ω
,where Ω is a bounded domain in R
N, N
⩾
2, φ
⩾
0 may take the value 0 on ∂
Ω, and f
(
x
,
s
) is possibly singular near s
=
0
. They proved the existence and the uniqueness of positive solutions without assuming monotonicity or strict positivity on f
(
x
,
s
).Recently, Boccardo and Orsina in [8] studied the existence, regularity, and nonexistence of solutions for the following problem:(5)
-
d
i
v
M
x
∇
u
=
f
x
u
α
,
x
∈
Ω
,
u
=
0
,
x
∈
∂
Ω
.They discussed the dependence of the results on the summability of f and the values of α. For the other results of singular elliptic equations, see [9, 10]. In this paper, we generalize the results in [8] to the case when α is a variable exponent by applying the method of regularization, Schauder fixed point theorem, the integrability of solution to the approximate problem with n
=
1, and a necessary compactness argument to overcome some difficulties arising from the singular terms with variable exponent.


## 2. Preliminaries

Firstly, we give the definition of weak solutions to problem (1).Definition 1.
A functionu
∈
H
0
1
(
Ω
) is called a weak solution of problem (1), if the following identity holds:(6)
∫
Ω
M
x
∇
u
∇
φ
d
x
=
∫
Ω
f
u
α
x
φ
d
x
,
∀
φ
∈
C
0
1
Ω
.In order to prove our results, we will consider the following approximation problem:(7)
-
d
i
v
M
x
∇
u
n
=
f
n
u
n
+
1
/
n
α
x
,
x
∈
Ω
,
u
n
=
0
,
x
∈
∂
Ω
,where n
∈
N, f
n
(
x
)
=
m
i
n
{
f
(
x
)
,
n
}.Lemma 2.
Problem (7) has a nonnegative solution u
n in H
0
1
(
Ω
)
⋂
‍
L
∞
(
Ω
).Proof.
Letn
∈
N be fixed and ω a function in L
2
(
Ω
). It is not difficult to prove that the following problem has a unique solution v
∈
H
0
1
(
Ω
)
⋂
‍
L
∞
(
Ω
) (see [11, 12]):(8)
-
d
i
v
M
x
∇
v
=
f
n
ω
+
1
/
n
α
x
,
x
∈
Ω
,
v
=
0
,
x
∈
∂
Ω
.So, for any ω
∈
L
2
(
Ω
), we define the mapping Γ
:
L
2
(
Ω
)
→
H
0
1
(
Ω
)
⋂
‍
L
∞
(
Ω
) as Γ
(
ω
)
=
v. Taking v as a test function, we have, using (2),(9)
γ
∫
Ω
∇
v
2
d
x
⩽
∫
Ω
M
x
∇
v
·
∇
v
d
x
=
∫
Ω
f
n
v
ω
+
1
/
n
α
x
d
x
⩽
n
1
+
α
+
∫
Ω
v
d
x
.By Poincaré inequality (on the left hand side) and Hölder’s inequality (on the right hand side), we get that(10)
∫
Ω
v
2
d
x
⩽
C
n
1
+
α
+
∫
Ω
v
2
d
x
1
/
2
, for some constant C independent of ω. This implies that (11)
v
L
2
Ω
⩽
C
n
1
+
α
+
.Therefore, the ball of L
2
(
Ω
) of radius C
n
1
+
α
+ is invariant under the mapping Γ
. Since the embedding H
0
1
(
Ω
)
↪
L
2
(
Ω
) is compact, we obtain that Γ is a compact operator and v
L
2
(
Ω
)
⩽
C. It is also easy to prove that Γ is continuous on L
2
(
Ω
), so by Schauder’s fixed point theorem, we get that there exists a function u
n
∈
H
0
1
(
Ω
), for every fixed n
∈
N, such that u
n
=
S
(
u
n
); that is, problem (7) has a solution. Since f
n
/
(
u
n
+
1
/
n
)
α
(
x
)
⩾
0, the maximum principle implies that u
n
⩾
0. Since the right hand side of (7) belongs to L
∞
(
Ω
), the result of Theorem  4.2 in [13] implies that u
n
∈
L
∞
(
Ω
).Lemma 3.
The sequenceu
n is increasing with respect to n, u
n
>
0 in Ω, and for every Ω
′
⊂
⊂
Ω there exists C
Ω
′
>
0 (independent of n) such that(12)
u
n
x
⩾
C
Ω
′
>
0
f
o
r
e
v
e
r
y
x
∈
Ω
′
,
f
o
r
e
v
e
r
y
n
∈
N
.Proof.
Due to0
⩽
f
n
⩽
f
n
+
1 and α
(
x
)
>
0, we have that (13)
-
d
i
v
M
x
∇
u
n
=
f
n
u
n
+
1
/
n
α
x
⩽
f
n
+
1
u
n
+
1
/
n
+
1
α
x
,
-
d
i
v
M
x
∇
u
n
+
1
=
f
n
+
1
u
n
+
1
+
1
/
n
+
1
α
x
, so that(14)
-
d
i
v
M
x
∇
u
n
-
u
n
+
1
⩽
f
n
+
1
1
u
n
+
1
/
n
+
1
α
x
-
1
u
n
+
1
+
1
/
n
+
1
α
x
=
f
n
+
1
u
n
+
1
+
1
/
n
+
1
α
x
-
u
n
+
1
/
n
+
1
α
x
u
n
+
1
/
n
+
1
α
x
u
n
+
1
+
1
/
n
+
1
α
x
. Choosing (
u
n
-
u
n
+
1
)
+
=
m
a
x
{
u
n
-
u
n
+
1
,
0
} as a test function, observing that (15)
u
n
+
1
+
1
n
+
1
α
x
-
u
n
+
1
n
+
1
α
x
u
n
-
u
n
+
1
+
⩽
0
, and applying (2), we get that (16)
0
⩽
γ
∫
Ω
∇
u
n
-
u
n
+
1
+
2
d
x
⩽
0
.This implies that (
u
n
-
u
n
+
1
)
+
=
0 a.e. in Ω; that is, u
n
⩽
u
n
+
1 for every n
∈
N. Since the sequence {
u
n
} is increasing with respect to n, we only need to prove that (12) holds for u
1. Applying Lemma 2, we know that u
1
∈
L
∞
(
Ω
); that is, there exists a constant C (depending only on Ω and N) such that (17)
u
1
L
∞
Ω
⩽
C
f
1
L
∞
Ω
⩽
C
,and then (18)
-
d
i
v
M
x
∇
u
1
=
f
1
u
1
+
1
α
x
⩾
f
1
C
+
1
α
x
.Due to f
1
/
(
C
+
1
)
α
(
x
)
⩾
0, f
1
/
(
C
+
1
)
α
(
x
)
≢
0, the strong maximum principle implies that u
1
>
0 in Ω and (12) holds for u
1. The monotonicity of u
n implies that (12) holds for u
n.Remark 4.
Ifu
n and v
n are two solutions of (7), following the lines of the proof of the first part in Lemma 3 we may show that u
n
⩽
v
n. By symmetry, this implies that the solution of (7) is unique.Lemma 5.
The solutionu
1 to problem (7) with n
=
1 satisfies(19)
∫
Ω
u
1
-
r
d
x
<
∞
,
∀
r
<
1
.Proof.
Bym
i
n
{
f
(
x
)
,
1
}
/
(
u
1
+
1
)
α
(
x
)
⩽
1 and Lemma  2.2 in [14], we know that there exists 0
<
β
<
1 such that u
1
∈
C
1
,
β
(
Ω
¯
) and u
1
C
1
,
β
⩽
C, which implies that the gradient of u
1 exists everywhere; then Hopf Lemma in [15] shows that ∂
u
1
(
x
)
/
∂
ν
>
0 in Ω
¯, where ν is the outward unit normal vector of ∂
Ω at x. Moreover, following the lines of the proof of Lemma in [5], we get that (20)
∫
Ω
u
1
r
d
x
<
∞
,
i
f
f
r
>
-
1
.We know clearly that the estimates onu
n depend on f and α
(
x
), and we will discuss it in different cases.


## 3. The Case0
<
α
-
⩽
α
(
x
)
⩽
α
+
<
1

In this case, we obtain a priori estimates onu
n in H
0
1
(
Ω
) only if f is more regular than L
1
(
Ω
), and we have the following results.Lemma 6.
Letu
n be the solution of (7) with 0
<
α
-
⩽
α
(
x
)
⩽
α
+
<
1, and suppose that f
∈
L
m
(
Ω
) with m
=
2
N
/
N
+
2
+
(
N
-
2
)
α
-
=
(
2
∗
/
1
-
α
-
)
′. Then the sequence {
u
n
} is bounded in H
0
1
(
Ω
).Proof.
Choosingu
n as a test function in (7), by Hölder’s inequality, (2), and the fact that f
n
⩽
f, we get(21)
γ
∫
Ω
∇
u
n
2
d
x
⩽
∫
Ω
f
n
u
n
u
n
+
1
/
n
α
x
d
x
⩽
∫
Ω
f
u
n
1
-
α
x
d
x
⩽
∫
Ω
f
u
n
1
-
α
-
d
x
+
∫
Ω
f
u
n
1
-
α
+
d
x
⩽
f
L
m
Ω
∫
Ω
u
n
1
-
α
-
m
′
d
x
1
/
m
′
+
f
L
m
Ω
∫
Ω
u
n
1
-
α
+
m
′
d
x
1
/
m
′
⩽
f
L
m
Ω
∫
Ω
u
n
1
-
α
-
m
′
d
x
1
/
m
′
+
f
L
m
Ω
∫
Ω
1
1
-
α
-
/
α
+
-
α
-
d
x
α
+
-
α
-
/
1
-
α
-
∫
Ω
u
n
1
-
α
-
m
′
d
x
1
-
α
+
/
1
-
α
-
1
/
m
′
=
f
L
m
Ω
∫
Ω
u
n
1
-
α
-
m
′
d
x
1
/
m
′
+
Ω
α
+
-
α
-
/
1
-
α
-
m
′
f
L
m
Ω
∫
Ω
u
n
1
-
α
-
m
′
d
x
1
-
α
+
/
1
-
α
-
m
′
.By the assumption of m, we have (
1
-
α
-
)
m
′
=
2
∗, and using Sobolev Embedding Theorem (on the left hand side), we have that(22)
γ
S
∫
Ω
u
n
2
∗
d
x
2
/
2
∗
⩽
γ
∫
Ω
∇
u
n
2
d
x
⩽
f
L
m
Ω
∫
Ω
u
n
2
∗
d
x
1
/
m
′
+
Ω
α
+
-
α
-
/
1
-
α
-
m
′
f
L
m
Ω
∫
Ω
u
n
2
∗
d
x
1
-
α
+
/
1
-
α
-
m
′
;that is, (23)
γ
S
∫
Ω
u
n
2
∗
d
x
2
/
2
∗
⩽
f
L
m
Ω
∫
Ω
u
n
2
∗
d
x
1
-
α
-
/
2
∗
+
Ω
α
+
-
α
-
/
1
-
α
-
m
′
f
L
m
Ω
∫
Ω
u
n
2
∗
d
x
1
-
α
+
/
2
∗
.Since 1
-
α
+
<
1
-
α
-
<
2, (22) yields the boundedness of u
n in L
2
∗
(
Ω
). By this estimate and (22), the conclusion follows.Once we have the boundedness ofu
n, we can prove an existence result for (1).Theorem 7.
Suppose thatf is a nonnegative function in L
m
(
Ω
)
(
f
≢
0
), with m
=
2
N
/
N
+
2
+
(
N
-
2
)
α
-
=
2
∗
/
1
-
α
-
′, and let 0
<
α
-
⩽
α
(
x
)
⩽
α
+
<
1. Then problem (1) has a solution u
∈
H
0
1
(
Ω
) satisfying (6).Proof.
Sinceu
n is bounded in H
0
1
(
Ω
) by Lemma 6 and u
n converges to u pointwise in Ω (by Lemma 3), then we know that there exists u
∈
H
0
1
(
Ω
) such that (24)
u
n
⇀
u
w
e
a
k
l
y
i
n
H
0
1
Ω
a
n
d
s
t
r
o
n
g
l
y
i
n
L
2
Ω
,
u
n
⟶
u
a
.
e
.
i
n
Ω
,
∇
u
n
⇀
∇
u
w
e
a
k
l
y
i
n
L
2
Ω
. So we have that(25)
l
i
m
n
→
∞
∫
Ω
M
x
∇
u
n
·
∇
φ
d
x
=
∫
Ω
M
x
∇
u
·
∇
φ
d
x
,
∀
φ
∈
C
0
1
Ω
.Since u
n satisfies (12), we get that (26)
0
⩽
f
n
φ
u
n
+
1
/
n
α
x
⩽
φ
L
∞
Ω
f
m
i
n
C
Ω
′
α
-
,
C
Ω
′
α
+
,where Ω
′
=
{
x
:
φ
≠
0
}. Then by Lebesgue Dominated Convergence Theorem, we have that(27)
l
i
m
n
→
∞
∫
Ω
f
n
φ
u
n
+
1
/
n
α
x
d
x
=
∫
Ω
f
φ
u
α
x
d
x
.Since u
n is a solution of (7), this implies that (28)
∫
Ω
M
x
∇
u
n
∇
φ
d
x
=
∫
Ω
f
n
φ
u
n
+
1
/
n
α
x
d
x
,
∀
φ
∈
C
0
1
Ω
.Letting n
→
∞, combining (25) with (27), we get that (29)
∫
Ω
M
x
∇
u
·
∇
φ
d
x
=
∫
Ω
f
φ
u
α
x
d
x
,which proves that (1) has a solution u in H
0
1
(
Ω
).The summability ofu depends on the summability of f, which is proved in the next Lemma.Lemma 8.
Suppose thatf
∈
L
m
(
Ω
), m
⩾
2
N
/
N
+
2
+
(
N
-
2
)
α
-, and let 0
<
α
-
⩽
α
(
x
)
⩽
α
+
<
1. Then the solution u of (1) given by Theorem 7 is such that(i)
ifm
>
N
/
2, then u
∈
L
∞
(
Ω
);
(ii)
if2
N
/
N
+
2
+
(
N
-
2
)
α
-
⩽
m
<
N
/
2, then u
∈
L
s
(
Ω
), s
=
N
m
(
1
+
α
-
)
/
N
-
2
m.Proof.
To prove (i), letk
>
1 and define G
k
(
s
)
=
(
s
-
k
)
+. Taking G
k
(
u
n
) as a test function in (7), using (2), we get (30)
γ
∫
Ω
∇
G
k
u
n
2
d
x
⩽
∫
Ω
M
x
∇
G
k
u
n
·
∇
G
k
u
n
d
x
=
∫
Ω
f
n
G
k
u
n
u
n
+
1
/
n
α
x
d
x
.Since G
k
(
u
n
)
≠
0, it follows that(31)
γ
∫
Ω
∇
G
k
u
n
2
d
x
⩽
∫
Ω
f
G
k
u
n
d
x
.Starting from inequality (31), Theorem  4.2 in [13] shows that there exists a constant C (independent of n), such that (32)
u
n
L
∞
Ω
⩽
C
f
L
m
Ω
,which implies that u belongs to L
∞
(
Ω
).
To prove (ii), noting that ifm
=
2
N
/
N
+
2
+
(
N
-
2
)
α
-, s
=
2
N
/
N
-
2
=
2
∗, since u
∈
H
0
1
(
Ω
), the result when m
=
2
N
/
N
+
2
+
(
N
-
2
)
α
- is true by Sobolev Embedding Theorem. If 2
N
/
N
+
2
+
(
N
-
2
)
α
-
<
m
<
N
/
2, letting δ
>
1 and choosing u
n
2
δ
-
1 as a test function in (7), using Hölder’s inequality, we get that(33)
γ
2
δ
-
1
∫
Ω
∇
u
n
2
u
n
2
δ
-
2
d
x
⩽
∫
Ω
f
u
n
2
δ
-
1
u
n
+
1
/
n
α
x
d
x
⩽
∫
x
∈
Ω
,
u
n
⩾
1
f
u
n
2
δ
-
1
u
n
α
-
d
x
+
∫
x
∈
Ω
,
u
n
<
1
f
u
n
2
δ
-
1
u
n
α
+
d
x
⩽
∫
Ω
f
u
n
2
δ
-
1
-
α
-
d
x
+
∫
Ω
f
u
n
2
δ
-
1
-
α
+
d
x
⩽
f
L
m
Ω
∫
Ω
u
n
2
δ
-
1
-
α
-
m
′
d
x
1
/
m
′
+
f
L
m
Ω
∫
Ω
u
n
2
δ
-
1
-
α
+
m
′
d
x
1
/
m
′
=
f
L
m
Ω
∫
Ω
u
n
2
δ
-
1
-
α
-
m
′
d
x
1
/
m
′
+
Ω
α
+
-
α
-
/
2
δ
-
1
-
α
-
m
′
∫
Ω
u
n
2
δ
-
1
-
α
-
m
′
d
x
2
δ
-
1
-
α
+
/
2
δ
-
1
-
α
-
m
′
.By Sobolev inequality (on the left hand side), we have that(34)
∫
Ω
∇
u
n
2
u
n
2
δ
-
2
d
x
=
1
δ
2
∫
Ω
∇
u
n
δ
2
d
x
⩾
S
δ
2
∫
Ω
u
n
2
∗
δ
d
x
2
/
2
∗
,where S is the constant of the Sobolev embedding; combining with (33) and (34), we have that(35)
S
γ
2
δ
-
1
δ
2
∫
Ω
u
n
2
∗
δ
d
x
2
/
2
∗
⩽
f
L
m
Ω
∫
Ω
u
n
2
δ
-
1
-
α
-
m
′
d
x
1
/
m
′
+
Ω
α
+
-
α
-
/
2
δ
-
1
-
α
-
m
′
∫
Ω
u
n
2
δ
-
1
-
α
-
m
′
d
x
2
δ
-
1
-
α
+
/
2
δ
-
1
-
α
-
m
′
.We choose δ in such a way that 2
∗
δ
=
(
2
δ
-
1
-
α
-
)
m
′; that is, (36)
δ
=
1
+
α
-
m
N
-
2
2
N
-
2
m
,which yields that δ
>
1 if and only if 2
N
/
N
+
2
+
(
N
-
2
)
α
-
<
m
<
N
/
2, and that 2
∗
δ
=
N
m
(
1
+
α
-
)
/
N
-
2
m
=
s. Therefore, (35) becomes (37)
∫
Ω
u
n
s
d
x
2
/
2
∗
⩽
δ
2
S
γ
2
δ
-
1
f
L
m
Ω
∫
Ω
u
n
s
d
x
1
/
m
′
+
Ω
α
+
-
α
-
/
2
δ
-
1
-
α
-
m
′
∫
Ω
u
n
s
d
x
2
δ
-
1
-
α
+
/
2
δ
-
1
-
α
-
m
′
,which implies that(38)
∫
Ω
u
n
s
d
x
1
+
α
+
/
2
∗
δ
⩽
δ
2
S
γ
2
δ
-
1
f
L
m
Ω
∫
Ω
u
n
s
d
x
α
+
-
α
-
/
2
∗
δ
+
Ω
α
+
-
α
-
/
2
∗
δ
.Using Young’s inequality on the right hand side in (38), we have that (39)
∫
Ω
u
n
s
d
x
1
+
α
+
/
2
∗
δ
⩽
δ
2
S
γ
2
δ
-
1
f
L
m
Ω
ε
∫
Ω
u
n
s
d
x
1
+
α
+
/
2
∗
δ
+
ε
-
α
+
-
α
-
/
1
+
α
-
+
Ω
α
+
-
α
-
/
2
∗
δ
,where ε
=
S
γ
(
2
δ
-
1
)
/
2
δ
2
f
L
m
(
Ω
)
. Thus, we get that(40)
∫
Ω
u
n
s
d
x
1
+
α
+
/
2
∗
δ
⩽
2
δ
2
f
L
m
Ω
S
γ
2
δ
-
1
2
δ
2
f
L
m
Ω
S
γ
2
δ
-
1
α
+
-
α
-
/
1
+
α
-
+
Ω
α
+
-
α
-
/
2
∗
δ
.Therefore, we know that u
n is bounded in L
s
(
Ω
), and so does u
∈
L
s
(
Ω
).Theorem 9.
Suppose thatf
∈
L
m
(
Ω
), (
1
+
α
+
)
N
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
) ⩽ m
<
2
N
/
N
+
2
+
(
N
-
2
)
α
-, and 0
<
α
-
⩽
α
(
x
)
⩽
α
+
<
1. Then problem (1) has a solution u in W
0
1
,
q
(
Ω
), q
=
N
m
(
1
+
α
-
)
/
N
-
m
(
1
-
α
-
).Proof.
The lines of our proof are that if we can prove thatu
n is bounded in W
0
1
,
q
(
Ω
) (with q as in the statement), the existence of a solution u in W
0
1
,
q
(
Ω
) of (1) will be proved by passing to the limit in (7) as in the proof of Theorem 7. To prove that u
n is bounded in W
0
1
,
q
(
Ω
), we begin by proving that it is bounded in L
s
(
Ω
), with s
=
N
m
(
1
+
α
-
)
/
N
-
2
m. To attain this goal, we choose u
n
2
δ
-
1 as a test function in (7) as in the statement of Lemma 8, where 1
+
α
+
/
2
⩽
δ
<
1; however ∇
u
n
2
δ
-
1 will be singular at u
n
=
0, and therefore, we choose (
u
n
+
ε
)
2
δ
-
1
-
ε
2
δ
-
1 as a test function in (7), where ε
<
1
/
n for n fixed; by (2) and f
n
⩽
f, we have that(41)
γ
2
δ
-
1
∫
Ω
∇
u
n
2
u
n
+
ε
2
δ
-
2
d
x
⩽
∫
Ω
f
u
n
+
ε
2
δ
-
1
-
α
-
d
x
+
∫
Ω
f
u
n
+
ε
2
δ
-
1
-
α
+
d
x
.By Sobolev Embedding Theorem (H
0
1
(
Ω
)
↪
L
2
∗
(
Ω
)) on the left hand side, it follows that(42)
∫
Ω
∇
u
n
2
u
n
+
ε
2
δ
-
2
d
x
=
∫
Ω
∇
u
n
+
ε
δ
-
ε
δ
2
δ
2
d
x
⩾
S
δ
2
∫
Ω
u
n
+
ε
δ
-
ε
δ
2
∗
d
x
2
/
2
∗
,where S is the best constant of the Sobolev Embedding Theorem. Combining (41) with (42), we have that(43)
S
γ
2
δ
-
1
δ
2
∫
Ω
u
n
+
ε
δ
-
ε
δ
2
∗
d
x
2
/
2
∗
⩽
∫
Ω
f
u
n
+
ε
2
δ
-
1
-
α
-
d
x
+
∫
Ω
f
u
n
+
ε
2
δ
-
1
-
α
+
d
x
.Using Hölder’s inequality on the right hand side, we get (44)
S
γ
2
δ
-
1
δ
2
∫
Ω
u
n
+
ε
δ
-
ε
δ
2
∗
d
x
2
/
2
∗
⩽
f
L
m
Ω
∫
Ω
u
n
+
ε
2
δ
-
1
-
α
-
m
′
d
x
1
/
m
′
+
Ω
α
+
-
α
-
/
2
δ
-
1
-
α
-
m
′
f
L
m
Ω
∫
Ω
u
n
+
ε
2
δ
-
1
-
α
-
m
′
d
x
2
δ
-
1
-
α
+
/
2
δ
-
1
-
α
-
m
′
.Letting ε
→
0, we get (35); that is, (45)
∫
Ω
u
n
2
∗
δ
d
x
2
/
2
∗
⩽
δ
2
S
γ
2
δ
-
1
f
L
m
Ω
∫
Ω
u
n
2
δ
-
1
-
α
-
m
′
d
x
1
/
m
′
+
Ω
α
+
-
α
-
/
2
δ
-
1
-
α
-
m
′
∫
Ω
u
n
2
δ
-
1
-
α
-
m
′
d
x
2
δ
-
1
-
α
+
/
2
δ
-
1
-
α
-
m
′
, where δ is chosen in such a way that 2
∗
δ
=
(
2
δ
-
1
-
α
-
)
m
′; that is, (46)
δ
=
1
+
α
-
N
-
2
m
2
N
-
2
m
.
Ifm
=
(
1
+
α
+
)
N
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
), choosing δ
=
1
+
α
+
/
2 in (43), and letting ε
→
0, we have that (47)
∫
Ω
u
n
2
∗
δ
d
x
2
/
2
∗
⩽
δ
2
S
γ
2
δ
-
1
∫
Ω
f
u
n
2
δ
-
1
-
α
-
d
x
+
∫
Ω
f
d
x
.Using Hölder’s inequality and Young’s inequality, we get (48)
∫
Ω
u
n
2
∗
δ
d
x
2
/
2
∗
⩽
δ
2
S
γ
2
δ
-
1
f
L
m
Ω
∫
Ω
u
n
2
δ
-
1
-
α
-
m
′
d
x
1
/
m
′
+
∫
Ω
f
d
x
⩽
δ
2
S
γ
2
δ
-
1
f
L
m
Ω
ε
∫
Ω
u
n
2
∗
δ
d
x
2
/
2
∗
+
ε
-
2
∗
/
2
m
′
-
2
∗
+
∫
Ω
f
d
x
,where ε
=
S
γ
(
2
δ
-
1
)
/
2
δ
2
f
L
m
(
Ω
). Thus we have that (49)
∫
Ω
u
n
2
∗
δ
d
x
2
/
2
∗
⩽
2
δ
2
S
γ
2
δ
-
1
f
L
m
Ω
2
δ
2
f
L
m
Ω
S
γ
2
δ
-
1
2
∗
/
2
m
′
-
2
∗
+
∫
Ω
f
d
x
.Therefore we obtain that u
n is bounded in L
N
(
1
+
α
+
)
/
N
-
2
(
Ω
), where N
(
1
+
α
+
)
/
N
-
2 is the value of s for m
=
(
1
+
α
+
)
N
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
).
If(
1
+
α
+
)
N
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
)
<
m
<
2
N
/
N
+
2
+
(
N
-
2
)
α
-, it is clear that the inequality on m holds true if and only if 1
+
α
+
/
2
<
δ
<
1, starting from (35) and arguing as in the proof of Lemma 8, we also get that u
n is bounded in L
s
(
Ω
) with s
=
N
m
(
1
+
α
-
)
/
N
-
2
m.
The right hand side of (41) is bounded with respect to n (and ε, which we take smaller than 1) by using the estimate on u
n in L
s
(
Ω
) and the choice of δ.
Sinceδ
<
1, (50)
∫
Ω
∇
u
n
2
u
n
+
ε
2
-
2
δ
d
x
=
∫
Ω
∇
u
n
2
u
n
+
ε
2
δ
-
2
d
x
⩽
C
.
Ifq
=
N
m
(
1
+
α
-
)
/
N
-
m
(
1
-
α
-
)
<
2, by Hölder’s inequality, we have that(51)
∫
Ω
∇
u
n
q
d
x
=
∫
Ω
∇
u
n
q
u
n
+
ε
1
-
δ
q
u
n
+
ε
1
-
δ
q
d
x
⩽
∫
Ω
∇
u
n
q
u
n
+
ε
1
-
δ
q
2
/
q
d
x
q
/
2
∫
Ω
u
n
+
ε
2
1
-
δ
q
/
2
-
q
d
x
1
-
q
/
2
=
∫
Ω
∇
u
n
2
u
n
+
ε
2
1
-
δ
d
x
q
/
2
∫
Ω
u
n
+
ε
2
1
-
δ
q
/
2
-
q
d
x
1
-
q
/
2
⩽
C
∫
Ω
u
n
+
ε
2
1
-
δ
q
/
2
-
q
d
x
1
-
q
/
2
.
The choice ofδ and the value of q are such that 2
(
1
-
δ
)
q
/
2
-
q
=
s, so that the right hand side of (51) is bounded with respect to n and ε. Hence, u
n is bounded in W
0
1
,
q
(
Ω
).Theorem 10.
Suppose thatf
∈
L
m
(
Ω
), 1
/
2
δ
-
α
+
<
m < (
(
1
+
α
+
)
N/(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
)
) with 1
+
α
-
/
2 < δ
<
1
+
α
+
/
2, and 0
<
α
-
⩽
α
(
x
)
⩽
α
+
<
1. Then problem (1) has a solution u in W
0
1
,
q
(
Ω
), q
=
N
m
(
1
+
α
-
)
/
N
-
m
(
1
-
α
-
).Proof.
The lines of our proof are similar to that in the proof of Theorem9. We also begin by proving that u
n is bounded in L
s
(
Ω
), with s
=
N
m
(
1
+
α
-
)
/
N
-
2
m. To this aim, we also choose (
u
n
+
ε
)
2
δ
-
1
-
ε
2
δ
-
1 as a test function in (7), where 1
+
α
-
/
2
<
δ
<
1
+
α
+
/
2, ε
<
1
/
n for n fixed. Since f
n
⩽
f and (2), we have that (52)
γ
2
δ
-
1
∫
Ω
∇
u
n
2
u
n
+
ε
2
δ
-
2
d
x
⩽
∫
Ω
f
u
n
+
ε
2
δ
-
1
-
α
-
d
x
+
∫
Ω
f
u
n
+
ε
2
δ
-
1
-
α
+
d
x
.
Using Sobolev Embedding Theorem (H
0
1
(
Ω
)
↪
L
2
∗
(
Ω
)) on the left hand side, it follows that (53)
S
γ
2
δ
-
1
δ
2
∫
Ω
u
n
+
ε
δ
-
ε
δ
2
∗
d
x
2
/
2
∗
⩽
∫
Ω
f
u
n
+
ε
2
δ
-
1
-
α
-
d
x
+
∫
Ω
f
u
n
+
ε
2
δ
-
1
-
α
+
d
x
,where S is the constant of the Sobolev Embedding Theorem.
Using Hölder’s inequality and Lemma5 on the right hand side, we get (54)
S
γ
2
δ
-
1
δ
2
∫
Ω
u
n
+
ε
δ
-
ε
δ
2
∗
d
x
2
/
2
∗
⩽
∫
Ω
f
u
n
+
ε
2
δ
-
1
-
α
-
d
x
+
∫
Ω
f
u
n
+
ε
-
2
δ
+
α
+
+
1
d
x
⩽
∫
Ω
f
u
n
+
ε
2
δ
-
1
-
α
-
d
x
+
∫
Ω
f
u
1
-
2
δ
+
α
+
+
1
d
x
⩽
f
L
m
Ω
∫
Ω
u
n
+
ε
2
δ
-
1
-
α
-
m
′
d
x
1
/
m
′
+
f
L
m
Ω
u
1
2
δ
-
1
-
α
+
L
m
′
Ω
⩽
f
L
m
Ω
∫
Ω
u
n
+
ε
2
δ
-
1
-
α
-
m
′
d
x
1
/
m
′
+
C
f
L
m
Ω
.
Lettingε
→
0, we obtain that(55)
∫
Ω
u
n
2
∗
δ
d
x
2
/
2
∗
⩽
δ
2
S
γ
2
δ
-
1
f
L
m
Ω
∫
Ω
u
n
2
δ
-
1
-
α
-
m
′
d
x
1
/
m
′
+
C
,where δ is chosen in such a way that 2
∗
δ
=
(
2
δ
-
1
-
α
-
)
m
′; that is (56)
δ
=
1
+
α
-
N
-
2
m
2
N
-
2
m
.
If1
<
m
<
(
1
+
α
+
)
N
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
), it is clear that the inequality on m holds true if and only if 1
+
α
-
/
2
<
δ
<
1
+
α
+
/
2, and arguing as to the case m
=
(
1
+
α
+
)
N
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
) in the proof of Theorem 9, we also obtain that u
n is bounded in L
s
(
Ω
), with s
=
N
m
(
1
+
α
-
)
/
N
-
2
m.
Sinceδ
<
1, (57)
∫
Ω
∇
u
n
2
u
n
+
ε
2
-
2
δ
d
x
=
∫
Ω
∇
u
n
2
u
n
+
ε
2
δ
-
2
d
x
⩽
C
.
Ifq
=
N
m
(
1
+
α
-
)
/
N
-
m
(
1
-
α
-
)
<
2, similarly to the proof of Theorem 9, we have by Hölder’s inequality that (58)
∫
Ω
∇
u
n
q
d
x
⩽
C
∫
Ω
u
n
+
ε
2
1
-
δ
q
/
2
-
q
d
x
1
-
q
/
2
.
Since the choice ofδ and the value of q, the right hand side of the above inequality is bounded with respect to n and ε. Hence, u
n is bounded in W
0
1
,
q
(
Ω
).


## 4. The Case1
<
α
-
⩽
α
(
x
)
⩽
α
+

The case1
<
α
-
⩽
α
(
x
)
⩽
α
+ has many analogies with the case 0
<
α
-
<
α
+
<
1. In this case, we can also prove that u
n is bounded in H
0
1
(
Ω
) only if f is more regular than L
1
(
Ω
) and α
+ and α
- are close to 1.Lemma 11.
Suppose thatf
∈
L
m
(
Ω
)
(
m
>
1
), and let u
n be the solution of (7) with 1
<
α
-
<
α
+
<
2
-
1
/
m. Then u
n is bounded in H
0
1
(
Ω
).Proof.
Takingu
n as a test function in (7), by (2), we obtain that (59)
γ
∫
Ω
∇
u
n
2
d
x
⩽
∫
Ω
f
u
n
α
x
-
1
d
x
.Using Lemmas 2 and 3, we know that u
n
⩾
u
1 and there exists a constant M
>
0 s.t. u
1
⩽
M. Hence (
M
/
u
1
)
α
(
x
)
-
1
⩽
(
M
/
u
1
)
α
+
-
1, and we have (60)
γ
∫
Ω
∇
u
n
2
d
x
⩽
∫
Ω
f
u
n
α
x
-
1
d
x
⩽
∫
Ω
f
u
1
α
x
-
1
d
x
⩽
1
+
M
α
+
-
α
-
∫
Ω
f
u
1
α
+
-
1
d
x
.Using Hölder’s inequality on the right hand side, and Lemma 5, we obtain (61)
γ
∫
Ω
∇
u
n
2
d
x
⩽
1
+
M
α
+
-
α
-
∫
Ω
f
u
1
α
+
-
1
d
x
⩽
1
+
M
α
+
-
α
-
f
L
m
Ω
∫
Ω
1
u
1
α
+
-
1
m
′
d
x
1
/
m
′
⩽
C
1
+
M
α
+
-
α
-
f
L
m
Ω
.Therefore, u
n is bounded in H
0
1
(
Ω
).Once we have the boundedness ofu
n, we can prove the following existence theorem along the lines of Theorem 7.Theorem 12.
Suppose thatf
∈
L
m
(
Ω
)
(
m
>
1
) and 1
<
α
-
<
α
+
<
2
-
1
/
m. Then problem (1) has a solution u in H
0
1
(
Ω
).The summability ofu can be proved along the lines of Lemma 8 with little changes.Lemma 13.
Suppose thatf
∈
L
m
(
Ω
)
(
m
>
1
) and 1
<
α
-
<
α
+
<
2
-
1
/
m. Then the solution u of (1) given by Theorem 12 is such that(i)
ifm
>
N
/
2, then u
∈
L
∞
(
Ω
);
(ii)
ifN
(
1
+
α
+
)
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
)
⩽
m
<
N
/
2, then u
∈
L
s
(
Ω
), s
=
N
m
(
1
+
α
-
)
/
N
-
2
m.Proof.
The proof of (i) is similar to the proof of Lemma8(i); we omit the details here.
To prove (ii) we chooseu
n
2
δ
-
1 as a test function with δ
⩾
1
+
α
+
/
2 in (7); similarly to the proof of Lemma 8, we obtain that(62)
S
γ
2
δ
-
1
δ
2
∫
Ω
u
n
2
∗
δ
d
x
2
/
2
∗
⩽
∫
Ω
f
u
n
2
δ
-
1
-
α
-
d
x
+
∫
Ω
f
u
n
2
δ
-
1
-
α
+
d
x
.
Ifm
=
N
(
1
+
α
+
)
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
), choosing δ
=
1
+
α
+
/
2 in (62), by Hölder’s inequality, we get (63)
S
γ
2
δ
-
1
δ
2
∫
Ω
u
n
2
∗
δ
d
x
2
/
2
∗
⩽
f
L
m
Ω
∫
Ω
u
n
2
δ
-
1
-
α
-
m
′
d
x
1
/
m
′
+
Ω
1
-
1
/
m
f
L
m
Ω
.
We chooseδ in such a way that 2
∗
δ
=
(
2
δ
-
1
-
α
-
)
m
′; that is, δ
=
(
1
+
α
-
)
m
(
N
-
2
)
/
2
(
N
-
2
m
). Since m
=
N
(
1
+
α
+
)
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
) being 2
/
2
∗
>
1
/
m
′, it follows that u
n is bounded in L
N
(
1
+
α
+
)
/
N
-
2
(
Ω
).
IfN
(
1
+
α
+
)
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
)
<
m
<
N
/
2, starting from inequality (62) and Hölder’s inequality, we have that (64)
S
γ
2
δ
-
1
δ
2
∫
Ω
u
n
2
∗
δ
d
x
2
/
2
∗
⩽
f
L
m
Ω
∫
Ω
u
n
2
δ
-
1
-
α
-
m
′
d
x
1
/
m
′
+
Ω
α
+
-
α
-
/
2
δ
-
1
-
α
-
m
′
∫
Ω
u
n
2
δ
-
1
-
α
-
m
′
d
x
2
δ
-
1
-
α
+
/
2
δ
-
1
-
α
-
m
′
.
We also chooseδ in such a way that 2
∗
δ
=
(
2
δ
-
1
-
α
-
)
m
′, which yields that δ
>
1
+
α
+
/
2, if and only if m
>
N
(
1
+
α
+
)
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
), and that 2
∗
δ
=
s. So, since 2
/
2
∗
>
1
/
m
′ being m
<
N
/
2, we have the boundedness of u
n in L
s
(
Ω
), and so does u
∈
L
s
(
Ω
).Moreover, we can only prove that a positive power ofu
n is bounded in H
0
1
(
Ω
) only if f is more regular than L
1
(
Ω
) and α
+ is close to α
- and we only have the boundedness of u
n in H
l
o
c
1
(
Ω
).Lemma 14.
Suppose thatf
∈
L
m
(
Ω
)
(
m
>
1
), and let u
n be the solution of (7) with 1
<
α
-
⩽
α
(
x
)
⩽
α
+ and α
+
-
α
-
<
1
-
1
/
m. Then u
n
1
+
α
-
/
2 is bounded in H
0
1
(
Ω
), and u
n is bounded in H
l
o
c
1
(
Ω
) and in L
s
(
Ω
), with s
=
N
(
1
+
α
-
)
/
N
-
2.Proof.
Takingu
n
α
- as a test function in (7), since u
n
α
-
/
(
u
n
+
1
/
n
)
α
-
⩽
1 and f
n
⩽
f, by Lemma 5 and (2), we get that (65)
α
-
γ
∫
Ω
∇
u
n
2
u
n
α
-
-
1
d
x
⩽
∫
Ω
f
u
n
α
-
u
n
+
1
/
n
α
-
d
x
+
∫
Ω
f
u
n
α
-
u
n
α
+
d
x
⩽
Ω
1
-
1
/
m
f
L
m
Ω
+
f
L
m
Ω
∫
Ω
1
u
n
α
+
-
α
-
m
′
d
x
⩽
Ω
1
-
1
/
m
f
L
m
Ω
+
C
f
L
m
Ω
. Since (66)
∫
Ω
∇
u
n
2
u
n
α
-
-
1
d
x
=
4
1
+
α
-
2
∫
Ω
∇
u
n
1
+
α
-
/
2
2
d
x
, we have that (67)
4
α
-
γ
1
+
α
-
2
∫
Ω
∇
u
n
1
+
α
-
/
2
2
d
x
⩽
C
+
Ω
1
-
1
/
m
f
L
m
Ω
. Thus, we have that u
n
1
+
α
-
/
2 is bounded in H
0
1
(
Ω
)
.
Applying Sobolev Embedding Theorem tou
n
1
+
α
-
/
2
(
H
0
1
(
Ω
)
↪
L
2
∗
(
Ω
)
), we have that (68)
S
∫
Ω
u
n
1
+
α
-
/
2
2
∗
d
x
2
/
2
∗
⩽
∫
Ω
∇
u
n
1
+
α
-
/
2
2
d
x
,where S is the best constant of the Sobolev embedding. Since the boundedness of u
n
1
+
α
-
/
2 in H
0
1
(
Ω
), we thus have the boundedness of u
n in L
s
(
Ω
).
To prove the boundedness ofu
n in H
l
o
c
1
(
Ω
), we choose u
n
φ
2 as a test function in (7), where φ
∈
C
0
1
(
Ω
), Ω
′
=
{
x
∈
Ω
,
φ
≠
0
}. By (2) and (12), we have that (69)
γ
∫
Ω
∇
u
n
2
φ
2
d
x
+
2
∫
Ω
M
x
∇
u
n
·
∇
φ
u
n
φ
d
x
⩽
∫
Ω
M
x
∇
u
n
∇
u
n
φ
2
d
x
=
∫
Ω
f
n
u
n
φ
2
u
n
+
1
/
n
α
x
d
x
⩽
∫
Ω
f
n
φ
2
u
n
α
x
-
1
d
x
⩽
∫
Ω
f
n
φ
2
C
Ω
′
α
x
-
1
d
x
⩽
1
m
i
n
C
Ω
′
α
+
-
1
,
C
Ω
′
α
-
-
1
∫
Ω
f
n
φ
2
d
x
.By Young’s inequality, we get that (70)
2
β
∫
Ω
∇
u
n
·
∇
φ
u
n
φ
d
x
⩽
γ
2
∫
Ω
∇
u
n
2
φ
2
d
x
+
2
β
2
γ
∫
Ω
∇
φ
2
u
n
2
d
x
.Since u
n is bounded in L
s
(
Ω
) (where s
⩾
2), by Hölder inequality, we obtain that (71)
γ
2
∫
Ω
∇
u
n
2
φ
2
d
x
⩽
1
m
i
n
C
Ω
′
α
+
-
1
,
C
Ω
′
α
-
-
1
∫
Ω
f
n
φ
2
d
x
+
2
β
2
γ
∫
Ω
∇
φ
2
u
n
2
d
x
⩽
φ
L
∞
Ω
2
m
i
n
C
Ω
′
α
+
-
1
,
C
Ω
′
α
-
-
1
∫
Ω
f
d
x
+
2
β
2
γ
∇
φ
L
∞
Ω
2
∫
Ω
u
n
2
d
x
⩽
Ω
1
-
1
/
m
φ
L
∞
Ω
2
m
i
n
C
Ω
′
α
+
-
1
,
C
Ω
′
α
-
-
1
f
L
m
Ω
+
2
β
2
γ
∇
φ
L
∞
Ω
2
Ω
1
-
2
/
s
∫
Ω
u
n
s
d
x
2
/
s
⩽
C
f
,
φ
,and hence u
n is bounded in H
l
o
c
1
(
Ω
).Once we have the boundedness ofu
n, we can prove the following existence theorem along the lines of Theorem 7.Theorem 15.
Suppose thatf is a nonnegative function in L
m
(
Ω
)
(
m
>
1
), (
f
≢
0
), 1
<
α
-
⩽
α
(
x
)
⩽
α
+ and α
+
-
α
-
<
1
-
1
/
m. Then problem (1) has a solution u in H
l
o
c
1
(
Ω
). Furthermore, u
1
+
α
-
/
2belongs to H
0
1
(
Ω
).The summability ofu can be proved along the lines of Lemma 8 with little changes.Lemma 16.
Suppose thatf
∈
L
m
(
Ω
), 1
<
α
-
⩽
α
(
x
)
⩽
α
+, and α
+
-
α
-
<
1
-
1
/
m. Then the solution u of (1) given by Theorem 15 is such that(i)
ifm
>
N
/
2, then u
∈
L
∞
(
Ω
);
(ii)
ifN
(
1
+
α
+
)
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
)
⩽
m
<
N
/
2, then u
∈
L
s
(
Ω
), s
=
N
m
(
1
+
α
-
)
/
N
-
2
m.Proof.
The proof of (i) is similar to the proof of Lemma8(i); we omit the details here.
To prove (ii), we chooseu
n
2
δ
-
1
(
δ
⩾
1
+
α
+
/
2
) as a test function in (7); applying (2) and Sobolev Embedding Theorem, we have that(72)
S
γ
2
δ
-
1
δ
2
∫
Ω
u
n
2
∗
δ
d
x
2
/
2
∗
⩽
∫
Ω
f
u
n
2
δ
-
1
-
α
-
d
x
+
∫
Ω
f
u
n
2
δ
-
1
-
α
+
d
x
.
Ifm
=
N
(
1
+
α
+
)
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
), we choose δ
=
1
+
α
+
/
2; using Hölder inequality, we have that (73)
S
γ
2
δ
-
1
δ
2
∫
Ω
u
n
2
∗
δ
d
x
2
/
2
∗
⩽
f
L
m
Ω
∫
Ω
u
n
2
δ
-
1
-
α
-
m
′
d
x
1
/
m
′
+
Ω
1
-
1
/
m
f
L
m
Ω
.We also choose δ in such a way that 2
∗
δ
=
(
2
δ
-
1
-
α
-
)
m
′, since m
=
N
(
1
+
α
+
)
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
) being 2
/
2
∗
>
1
/
m
′, and we have the boundedness of u
n in L
N
(
1
+
α
+
)
/
N
-
2
(
Ω
) which is the value of s for m
=
N
(
1
+
α
+
)
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
).
IfN
(
1
+
α
+
)
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
)
<
m
<
N
/
2, starting from (72) using Hölder inequality, we get that (74)
S
γ
2
δ
-
1
δ
2
∫
Ω
u
n
2
∗
δ
d
x
2
/
2
∗
⩽
f
L
m
Ω
∫
Ω
u
n
2
δ
-
1
-
α
-
m
′
d
x
1
/
m
′
+
f
L
m
Ω
Ω
α
+
-
α
-
/
2
δ
-
1
-
α
-
m
′
∫
Ω
u
n
2
δ
-
1
-
α
-
m
′
d
x
2
δ
-
1
-
α
+
/
2
δ
-
1
-
α
-
m
′
.
We also chooseδ in such a way that 2
∗
δ
=
(
2
δ
-
1
-
α
-
)
m
′, which yields that δ
>
1
+
α
+
/
2, if and only if m
>
N
(
1
+
α
+
)
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
), and that 2
∗
δ
=
s. Since 2
/
2
∗
>
1
/
m
′ being m
<
N
/
2, we have the boundedness of u
n in L
s
(
Ω
), and so does u
∈
L
s
(
Ω
).


## 5. The Case0
<
α
-
<
1
<
α
+

Ifα
-
<
1
<
α
+, the boundedness of u
n in H
0
1
(
Ω
) can also be obtained only if f is more regular than L
1
(
Ω
), and the proof has many analogies with the case 0
<
α
-
<
α
+
<
1. We have the following results.Lemma 17.
Suppose thatf
∈
L
m
(
Ω
), with m
=
2
N
/
N
+
2
+
(
N
-
2
)
α
-, and let u
n be the solution of (7) with 0
<
α
-
<
1
<
α
+
<
2
-
1
/
m. Then the sequence {
u
n
} is bounded in H
0
1
(
Ω
).Proof.
We chooseu
n as a test function in (7), by Hölder’s inequality, (2), and Lemma 5, since f
n
⩽
f, we have that (75)
γ
∫
Ω
∇
u
n
2
d
x
⩽
f
L
m
Ω
∫
Ω
u
n
1
-
α
-
m
′
d
x
1
/
m
′
+
f
L
m
Ω
∫
Ω
1
u
1
α
+
-
1
m
′
d
x
1
/
m
′
⩽
f
L
m
Ω
∫
Ω
u
n
1
-
α
-
m
′
d
x
1
/
m
′
+
C
f
L
m
Ω
.
Applying Sobolev Embedding Theorem on the left hand side, we get(76)
S
γ
∫
Ω
u
n
2
∗
d
x
2
/
2
∗
⩽
f
L
m
Ω
∫
Ω
u
n
1
-
α
-
m
′
d
x
1
/
m
′
+
C
f
L
m
Ω
. Let 2
∗
=
(
1
-
α
-
)
m
′, and it follows that (77)
S
γ
∫
Ω
u
n
2
∗
d
x
2
/
2
∗
⩽
f
L
m
Ω
∫
Ω
u
n
2
∗
d
x
1
/
m
′
+
C
f
L
m
Ω
.
By Young’s inequality, we get that(78)
S
γ
∫
Ω
u
n
2
∗
d
x
2
/
2
∗
⩽
f
L
m
Ω
ε
∫
Ω
u
n
2
∗
d
x
2
/
2
∗
+
ε
-
2
∗
/
2
m
′
-
2
∗
+
C
f
L
m
Ω
.
Thus, we have that(79)
∫
Ω
u
n
2
∗
d
x
2
/
2
∗
⩽
f
L
m
Ω
S
γ
-
f
L
m
Ω
ε
ε
-
2
∗
/
2
m
′
-
2
∗
+
C
. We choose ε
=
S
γ
/
2
f
L
m
(
Ω
) to get (80)
∫
Ω
u
n
2
∗
d
x
2
/
2
∗
⩽
2
f
L
m
Ω
S
γ
2
f
L
m
Ω
S
γ
2
∗
/
2
m
′
-
2
∗
+
C
.
So the boundedness ofu
n in L
2
∗
(
Ω
) is obtained; using the estimate and (75) again, we have the estimate of u
n in H
0
1
(
Ω
).Once the boundedness ofu
n in H
0
1
(
Ω
) is obtained, we can prove the following existence theorem.Theorem 18.
Suppose thatf
∈
L
m
(
Ω
) with m
=
2
N
/
N
+
2
+
(
N
-
2
)
α
-, and 0
<
α
-
<
1
<
α
+
<
2
-
1
/
m. Then problem (1) has a solution u in H
0
1
(
Ω
).Lemma 19.
Suppose thatf
∈
L
m
(
Ω
) with m
⩾
2
N
/
N
+
2
+
(
N
-
2
)
α
-, and α
-
<
1
<
α
+
<
2
-
1
/
m. Then the solution u of (1) given by Theorem 18 is such that(i)
ifm
>
N
/
2, then u
∈
L
∞
(
Ω
);
(ii)
if2
N
/
N
+
2
+
(
N
-
2
)
α
-
⩽
m
<
N
/
2, then u
∈
L
s
(
Ω
), s
=
N
m
(
1
+
α
-
)
/
N
-
2
m.Proof.
The proof of (i) is similar to that for Lemma8(i), and we omit the details here.
To prove (ii), ifN
(
1
+
α
+
)
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
)
⩽
m
<
N
/
2, the proof is identical to that for Lemma 13, and we also omit it here.
Ifm
=
2
N
/
N
+
2
+
(
N
-
2
)
α
-, we can prove the results by Sobolev Embedding Theorem.
If2
N
/
N
+
2
+
(
N
-
2
)
α
-
<
m
<
N
(
1
+
α
+
)
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
), we choose 1
<
δ
<
1
+
α
+
/
2, and we use once again u
n
2
δ
-
1 as a test function in (7). Using δ
>
1
>
1
+
α
-
/
2, as well as Hölder’s inequality, Sobolev Embedding Theorem, Lemma 5, and (2), we get (81)
S
γ
2
δ
-
1
δ
2
∫
Ω
u
n
2
∗
δ
d
x
2
/
2
∗
⩽
f
L
m
Ω
∫
Ω
u
n
2
δ
-
1
-
α
-
m
′
d
x
1
/
m
′
+
C
f
L
m
Ω
.
The choice ofδ in such a way that 2
∗
δ
=
(
2
δ
-
1
-
α
-
)
m
′ yields that 1
<
δ
<
1
+
α
+
/
2, if and only if 2
N
/
N
+
2
+
(
N
-
2
)
α
-
<
m
<
N
(
1
+
α
+
)
/
(
1
+
α
-
)
(
N
-
2
)
+
2
(
1
+
α
+
), and that 2
∗
δ
=
s. The choice of m
<
N
/
2 implies that 2
/
2
∗
>
1
/
m
′. Thus we have the boundedness of u
n in L
s
(
Ω
), and so does the limit u in L
s
(
Ω
).



---

*Source: 9794739-2016-06-13.xml*