# Existence of Positive Solutions of One-Dimensional Prescribed Mean Curvature Equation

Authors: Ruyun Ma; Lingfang Jiang
Journal: Mathematical Problems in Engineering
 (2014)

---

## Abstract

We consider the existence of positive solutions of one-dimensional prescribed mean curvature equation−
(
u
′
/
1
+
u
′
2
)
′
=
λ
f
(
u
), 0
<
t
<
1, u
(
t
)
>
0, t
∈
(
0
,
1
), u
(
0
)
=
u
(
1
)
=
0 where λ
>
0 is a parameter, and f
:
[
0
,
∞
)
→
[
0
,
∞
) is continuous. Further, when f satisfies m
a
x
{
u
p
,
u
q
}
≤
f
(
u
)
≤
u
p
+
u
q, 0
<
p
≤
q
<
+
∞, we obtain the exact number of positive solutions. The main results are based upon quadrature method.


---

## Body

## 1. Introduction and Preliminaries

In this paper, we are interested in the existence of positive solution of one-dimensional prescribed mean curvature equation(1)
-
(
u
′
1
+
u
′
2
)
′
=
λ
f
(
u
)
,
0
<
t
<
1
,
u
(
t
)
>
0
,
t
∈
(
0,1
)
,
u
(
0
)
=
u
(
1
)
=
0
,
where λ
>
0 is a parameter and f satisfies max
⁡
⁡
{
u
p
,
u
q
}
≤
f
(
u
)
≤
u
p
+
u
q, 0
<
p
≤
q
<
+
∞.The existence of positive solutions of such type of prescribed mean curvature equations both in one and in higher dimension has been discussed in the last decades by several authors (see Habets and Omari [1, 2], Li and Liu [3], Pan [4], Obersnel and Omari [5, 6], and others [7–12]) in connection with various configurations of nonlinearities.Habets and Omari [1] considered the case that the nonlinearity f
(
u
)
=
u
p; they proved that if 0
<
p
<
1, then there exist λ
* and λ
* with 0
<
λ
*
<
λ
* such that (1) has exactly one positive solution for λ
∈
(
0
,
λ
*
]
∪
{
λ
*
}, exactly two solutions for λ
∈
(
λ
*
,
λ
*
), and no positive solution for λ
>
λ
*; if p
=
1, then there exists λ
* with 0
<
λ
*
<
π
2 such that (1) has exactly one positive solution for λ
∈
(
λ
*
,
π
2
) and no positive solution for λ
∈
(
0
,
λ
*
]
∪
[
π
2
,
+
∞
); if p
>
1, then there exist λ
*
>
0 and λ
*
>
0 such that (1) has no positive solution for λ
∈
(
0
,
λ
*
] and exactly one positive solution for λ
>
λ
*.Li and Liu [3] studied the case that the nonlinearity f
(
u
)
=
u
p
+
u
q, 0
<
p
<
q
<
+
∞; they proved that if 1
<
p
<
q
<
+
∞, then (1) has at most one solution for any λ
>
0; there exist 0
<
λ
1
<
λ
2
<
+
∞ such that (1) has no positive solution for 0
<
λ
<
λ
1 and exactly one positive solution for λ
>
λ
2; if 0
<
p
<
q
<
1, then (1) has at most two solutions for any λ
>
0; there exists 0
<
λ
1
<
λ
2
<
+
∞ such that (1) has exactly one positive solution for 0
<
λ
<
λ
1 and no positive solution for λ
>
λ
2.Inspired by [1, 3], we naturally consider (1) with nonlinearity f satisfying
(2)
max
⁡
{
u
p
,
u
q
}
≤
f
(
u
)
≤
u
p
+
u
q
,
0
<
p
≤
q
<
+
∞
.DenoteF
λ
(
s
)
=
λ
∫
0
s
‍
f
(
t
)
d
t, α
(
λ
)
=
F
λ
-
1
(
1
).Consider the following assumptions:(H1)
f
:
[
0
,
∞
)
→
[
0
,
∞
) is continuous, and f
′
(
s
)
≥
0, s
>
0;
(H2)
max
⁡
⁡
{
u
p
,
u
q
}
≤
f
(
u
)
≤
u
p
+
u
q, for 0
<
p
≤
q
<
∞, u
>
0;
(H3)
for fixedλ
>
0, f
(
s
)
-
s
f
′
(
s
)
≤
0 for s
∈
(
0
,
α
(
λ
)
);
(H4)
for fixedλ
>
0, f
(
s
)
-
3
s
f
′
(
s
)
≥
0, s
∈
(
0
,
α
(
λ
)
).We will prove the following.Theorem 1.
Assume that (H1)–(H3) hold, and, if1
<
p
≤
q
<
∞, then the following conclusions hold:(1)
for anyλ
>
0, (1) has at most one solution;
(2)
there exists0
<
λ
1
<
λ
2
<
+
∞ such that (1) has no solution for 0
<
λ
<
λ
1 and has exactly one solution for λ
>
λ
2.Theorem 2.
Assume that (H1), (H2), and (H4) hold, and, if0
<
p
≤
q
<
1, then the following conclusions hold:(1)
for anyλ
>
0, (1) has at most two solutions;
(2)
there exists0
<
λ
1
<
λ
2
<
+
∞ such that (1) has exactly one solution for 0
<
λ
<
λ
1 and has no solution for λ
>
λ
2.Remark 3.
Iff
(
u
)
=
max
⁡
⁡
{
u
p
,
u
q
} and p
=
q, then (2) reduces to f
(
u
)
=
u
p, which has been considered by Habets and Omari [1]; f
(
u
)
=
u
p
+
u
q for 0
<
p
<
q
<
1 and 1
<
p
<
q
<
+
∞, which has been studied by Li and Liu [3], and the case 0
<
p
<
1
<
q
<
+
∞ has been considered by Zhang and Feng [7].Remark 4.
The problem concerning exact number of solutions of semilinear equation(3)
-
u
′′
=
λ
f
(
u
)
,
u
>
0
,
t
∈
(
0,1
)
,
u
(
0
)
=
u
(
1
)
=
0
,
has been discussed by several authors with f
(
u
)
=
u
p
+
u
q and the related problems (see [13–16]).


## 2. Quadrature Technique

Letu
(
t
) be a solution of problem (1); then it is well known that u
(
t
) takes its maximum at c
=
1
/
2, u
(
t
) is symmetric with respect to c, u
′
(
t
)
>
0 for 0
≤
t
<
c, and u
′
(
t
)
<
0 for c
<
t
≤
1. Hence, problem (1) is equivalent to the following problem defined on [
0
,
c
]:
(4)
-
(
u
′
1
+
u
′
2
)
′
=
λ
f
(
u
)
,
u
(
t
)
>
0
,
t
∈
(
0
,
c
)
,
u
(
0
)
=
u
′
(
c
)
=
0
.
Let v
=
u
′
/
1
+
u
′
2. If u
(
t
) is a solution of (4) with ρ
=
u
(
c
), then (
u
,
v
) is a solution of the following problem defined on [
0
,
c
]:
(5)
u
′
=
v
1
-
v
2
,
v
′
=
-
λ
f
(
u
)
,
u
(
0
)
=
0
,
u
(
c
)
=
ρ
,
v
(
c
)
=
0
.
From H
(
t
)
=
1
-
v
2
(
t
)
-
1
-
F
λ
(
u
) satisfying
(6)
d
H
(
t
)
d
t
=
-
v
(
t
)
1
-
v
2
(
t
)
(
v
′
(
t
)
+
λ
f
(
u
)
)
=
0
and H
(
c
)
=
-
F
λ
(
ρ
), we see that
(7)
F
λ
(
ρ
)
-
F
λ
(
u
(
t
)
)
=
1
-
1
1
+
u
′
2
(
t
)
.
Therefore,
(8)
u
′
=
(
F
λ
(
ρ
)
-
F
λ
(
u
)
)
[
2
-
(
F
λ
(
ρ
)
-
F
λ
(
u
)
)
]
1
-
(
F
λ
(
ρ
)
-
F
λ
(
u
)
)
.
Then
(9)
1
-
(
F
λ
(
ρ
)
-
F
λ
(
u
)
)
(
F
λ
(
ρ
)
-
F
λ
(
u
)
)
[
2
-
(
F
λ
(
ρ
)
-
F
λ
(
u
)
)
]
d
u
=
d
t
.
Integrating (9) from 0 to c leads to
(10)
∫
0
ρ
‍
1
-
(
F
λ
(
ρ
)
-
F
λ
(
u
)
)
(
F
λ
(
ρ
)
-
F
λ
(
u
)
)
[
2
-
(
F
λ
(
ρ
)
-
F
λ
(
u
)
)
]
d
u
=
1
2
.
By substituting t
=
0 in (7), we get that F
λ
(
ρ
)
<
1 and 0
<
ρ
<
α
(
λ
)
=
F
λ
-
1
(
1
).LetT be defined on Σ
=
{
(
λ
,
ρ
)
:
λ
∈
(
0
,
∞
)
,
ρ
∈
(
0
,
α
(
λ
)
]
} by
(11)
T
(
λ
,
ρ
)
=
∫
0
ρ
‍
1
-
(
F
λ
(
ρ
)
-
F
λ
(
u
)
)
(
F
λ
(
ρ
)
-
F
λ
(
u
)
)
[
2
-
(
F
λ
(
ρ
)
-
F
λ
(
u
)
)
]
d
u
.
For simplicity, we denote
(12)
ξ
=
ξ
λ
(
ρ
,
t
)
=
F
λ
(
ρ
)
-
F
λ
(
ρ
t
)
,
ξ
′
=
∂
∂
ρ
ξ
λ
(
ρ
,
t
)
,
ξ
′′
=
∂
2
∂
ρ
2
ξ
λ
(
ρ
,
t
)
,
Δ
F
=
F
λ
(
ρ
)
-
F
λ
(
u
)
,
Δ
f
¯
=
λ
ρ
f
(
ρ
)
-
λ
u
f
(
u
)
,
Δ
f
¯
′
=
λ
ρ
2
∂
∂
ρ
f
(
ρ
)
-
λ
u
2
∂
∂
u
f
(
u
)
.
It follows that
(13)
T
(
λ
,
ρ
)
=
ρ
∫
0
1
‍
1
-
ξ
ξ
(
2
-
ξ
)
d
t
.Lemma 5.
T
(
λ
,
ρ
) has continuous derivatives up to the second order on Σ with respect to ρ and
(14)
T
ρ
′
(
λ
,
ρ
)
=
∫
0
1
‍
ξ
(
1
-
ξ
)
(
2
-
ξ
)
-
ρ
ξ
′
(
ξ
(
2
-
ξ
)
)
3
/
2
d
t
=
1
ρ
∫
0
ρ
‍
Δ
F
(
1
-
Δ
F
)
(
2
-
Δ
F
)
-
Δ
f
¯
[
Δ
F
(
2
-
Δ
F
)
]
3
/
2
d
u
,
T
ρ
ρ
′′
(
λ
,
ρ
)
=
3
ρ
∫
0
1
‍
(
1
-
ξ
)
(
ρ
ξ
′
)
2
(
ξ
(
2
-
ξ
)
)
5
/
2
d
t
-
∫
0
1
‍
2
ξ
′
+
ρ
ξ
′′
(
ξ
(
2
-
ξ
)
)
3
/
2
d
t
=
1
ρ
2
∫
0
ρ
‍
(
(
3
(
1
-
Δ
F
)
Δ
f
¯
2
-
(
Δ
f
¯
′
+
2
Δ
f
¯
)
Δ
F
(
2
-
Δ
F
)
)
(
Δ
f
¯
2
-
(
Δ
f
¯
′
+
2
Δ
f
¯
)
)
×
[
Δ
F
(
2
-
Δ
F
)
]
-
5
/
2
)
d
u
.Proof.
From Habets and Omari [1], it is easy to obtain the results.Lemma 6.
Let (H1) and (H2) hold;α
(
λ
) is strictly decreasing on (
0
,
∞
) with respect to λ, and
(15)
lim
⁡
λ
→
0
α
(
λ
)
=
+
∞
,
lim
⁡
λ
→
+
∞
α
(
λ
)
=
0
.Proof.
Since(16)
λ
1
p
+
1
t
p
+
1
≤
F
λ
(
t
)
≤
λ
(
1
p
+
1
+
1
q
+
1
)
t
p
+
1
,
t
<
1
,
λ
t
q
+
1
q
+
1
≤
F
λ
(
t
)
≤
λ
(
1
p
+
1
+
1
q
+
1
)
t
q
+
1
,
t
>
1
,
we have
(17)
(
λ
p
+
1
+
λ
q
+
1
)
-
1
/
(
p
+
1
)
t
1
/
(
p
+
1
)
≤
F
λ
-
1
(
t
)
≤
(
λ
p
+
1
)
-
1
/
(
p
+
1
)
t
1
/
(
p
+
1
)
,
t
<
λ
p
+
1
+
λ
q
+
1
,
(
λ
p
+
1
+
λ
q
+
1
)
-
1
/
(
q
+
1
)
t
1
/
(
q
+
1
)
≤
F
λ
-
1
(
t
)
≤
(
λ
q
+
1
)
-
1
/
(
p
+
1
)
t
1
/
(
p
+
1
)
,
t
>
λ
q
+
1
.
Then
(18)
(
λ
p
+
1
+
λ
q
+
1
)
-
1
/
(
p
+
1
)
≤
F
λ
-
1
(
1
)
≤
(
λ
p
+
1
)
-
1
/
(
p
+
1
)
,
ii
1
<
λ
p
+
1
+
λ
p
+
1
,
(
λ
p
+
1
+
λ
q
+
1
)
-
1
/
(
q
+
1
)
≤
F
λ
-
1
(
1
)
≤
(
λ
q
+
1
)
-
1
/
(
p
+
1
)
,
1
>
λ
q
+
1
.
It follows that lim
⁡
λ
→
0
α
(
λ
)
=
∞, lim
⁡
λ
→
+
∞
α
(
λ
)
=
0.
From the fact thatF
λ
(
α
(
λ
)
)
=
1, we have λ
∫
0
α
(
λ
)
‍
f
(
s
)
d
s
=
1, and ∫
0
α
(
λ
)
‍
f
(
s
)
d
s
+
λ
f
(
α
(
λ
)
)
α
′
=
0; it follows from the assumption (H1) that
(19)
α
′
=
-
∫
0
α
‍
f
(
s
)
d
s
λ
f
(
α
)
<
0
.
Then α
(
λ
) is strictly decreasing on (
0
,
∞
).Lemma 7.
Let (H1)–(H3) hold, if1
<
p
≤
q
<
+
∞ and λ
>
0. Then T
(
λ
,
ρ
) has the following properties:(i)
lim
⁡
ρ
→
0
⁡
T
(
λ
,
ρ
)
=
+
∞;
(ii)
T
(
λ
,
α
(
λ
)
)
>
0;
(iii)
T
ρ
′
(
λ
,
ρ
)
=
(
∂
/
∂
ρ
)
T
(
λ
,
ρ
)
<
0, for ρ
∈
(
0
,
α
(
λ
)
].Proof .
(i) From (H2),(20)
F
λ
(
ρ
)
-
F
λ
(
ρ
t
)
=
λ
∫
ρ
t
ρ
‍
f
(
s
)
d
s
≤
ρ
p
+
1
λ
p
+
1
(
1
-
t
p
+
1
)
+
ρ
q
+
1
λ
q
+
1
(
1
-
t
q
+
1
)
;
therefore,
(21)
T
(
λ
,
ρ
)
=
ρ
∫
0
1
‍
1
-
ξ
ξ
(
2
-
ξ
)
d
t
≥
ρ
∫
0
1
‍
(
(
1
-
ξ
)
(
[
ρ
p
+
1
λ
p
+
1
(
1
-
t
p
+
1
)
+
ρ
q
+
1
ρ
p
+
1
λ
p
+
1
×
λ
q
+
1
(
1
-
t
q
+
1
)
]
(
ρ
p
+
1
λ
p
+
1
]
)
-
1
/
2
(
[
λ
p
+
1
×
(
2
-
ξ
)
[
ρ
p
+
1
λ
p
+
1
(
1
-
t
p
+
1
)
)
-
1
/
2
)
d
t
=
ρ
(
1
-
p
)
/
2
∫
0
1
‍
(
(
1
-
ξ
)
(
[
λ
p
+
1
(
1
-
t
p
+
1
)
+
ρ
q
-
p
λ
p
+
1
λ
q
+
1
×
λ
q
+
1
(
1
-
t
q
+
1
)
]
(
(
[
λ
p
+
1
[
λ
p
+
1
(
1
-
t
p
+
1
)
+
ρ
q
-
p
×
(
2
-
ξ
)
)
-
1
/
2
)
d
t
.
Hence, from p
>
1, then lim
⁡
ρ
→
0
T
(
λ
,
ρ
)
=
+
∞.
(ii) SinceF
λ
(
α
(
λ
)
t
)
>
0, for t
∈
(
0,1
), thus
(22)
T
(
λ
,
α
(
λ
)
)
=
α
(
λ
)
∫
0
1
‍
F
λ
(
α
(
λ
)
t
)
ξ
(
2
-
ξ
)
d
t
>
0
.
(iii) From Lemma5, we know that T
ρ
′
(
λ
,
ρ
)
=
∫
0
1
‍
(
(
ξ
(
1
-
ξ
)
(
2
-
ξ
)
-
ρ
ξ
′
)
/
(
ξ
(
2
-
ξ
)
)
3
/
2
)
d
t, so
(23)
ξ
(
1
-
ξ
)
(
2
-
ξ
)
-
ρ
ξ
′
≤
2
ξ
-
ρ
ξ
′
=
2
F
λ
(
ρ
)
-
ρ
λ
f
(
ρ
)
-
[
2
F
λ
(
ρ
t
)
-
λ
ρ
t
f
(
ρ
t
)
]
=
H
(
λ
,
ρ
)
-
H
(
λ
,
ρ
t
)
,
hiii
t
∈
(
0,1
)
,
ρ
∈
(
0
,
α
(
λ
)
]
,
where H
(
λ
,
s
)
=
2
F
λ
(
s
)
-
λ
s
f
(
s
)
,
s
∈
(
0
,
α
(
λ
)
].
From the assumption (H3), we have that(
∂
/
∂
s
)
H
(
λ
,
s
)
=
λ
f
(
s
)
-
λ
s
f
′
(
s
)
<
0, s
∈
(
0
,
α
(
λ
)
]; thus ξ
(
1
-
ξ
)
(
2
-
ξ
)
-
ρ
ξ
′
≤
2
ξ
-
ρ
ξ
′
<
0.Lemma 8.
Assume that (H1) and (H2) hold, and letη
(
λ
)
=
T
(
λ
,
α
(
λ
)
). Then η
(
λ
) has the following properties:(i)
for0
<
p
≤
q
<
+
∞, η
(
λ
) is continuous in (
0
,
+
∞
);
(ii)
for0
<
p
≤
q
<
+
∞, lim
⁡
λ
→
0
η
(
λ
)
=
+
∞, lim
⁡
λ
→
∞
η
(
λ
)
=
0.Proof.
(i) Applyingη
(
λ
)
=
T
(
λ
,
α
(
λ
)
)
=
α
(
λ
)
∫
0
1
‍
(
F
λ
(
α
t
)
/
1
-
F
λ
2
(
α
t
)
)
d
t
=
∫
0
α
(
λ
)
‍
(
F
λ
(
s
)
/
1
-
F
λ
2
(
s
)
)
d
s, we may prove the following (26), (29) and (30) hold. For simplicity, we denote α
(
λ
) by α.
Ifα
≤
1, then λ
s
p
+
1
/
(
p
+
1
)
≤
F
λ
(
s
)
≤
λ
s
p
+
1
/
(
p
+
1
)
+
λ
s
q
+
1
/
(
q
+
1
), s
∈
(
0
,
α
), and from F
λ
(
α
(
λ
)
)
=
1, we get
(24)
1
-
F
λ
2
(
s
)
=
F
λ
2
(
α
(
λ
)
)
-
F
λ
2
(
s
)
=
(
F
λ
(
α
(
λ
)
)
-
F
λ
(
s
)
)
×
(
F
λ
(
α
(
λ
)
)
+
F
λ
(
s
)
)
≥
λ
2
(
α
p
+
1
p
+
1
-
s
p
+
1
p
+
1
)
×
(
α
p
+
1
p
+
1
+
s
p
+
1
p
+
1
)
=
λ
2
(
α
2
p
+
2
(
p
+
1
)
2
-
s
2
p
+
2
(
p
+
1
)
2
)
,
s
∈
(
0
,
α
)
.
So
(25)
α
∫
0
1
‍
F
λ
(
α
t
)
1
-
F
λ
2
(
α
t
)
d
t
=
∫
0
α
‍
F
λ
(
s
)
1
-
F
λ
2
(
s
)
d
s
≤
∫
0
α
‍
(
λ
s
p
+
1
/
(
p
+
1
)
)
+
(
λ
s
q
+
1
/
(
q
+
1
)
)
λ
2
(
(
α
2
p
+
2
/
(
p
+
1
)
2
)
-
(
s
2
p
+
2
/
(
p
+
1
)
2
)
)
d
s
=
α
∫
0
1
‍
(
λ
(
α
t
)
p
+
1
/
(
p
+
1
)
)
+
(
λ
(
α
t
)
q
+
1
/
(
q
+
1
)
)
λ
2
(
(
α
2
p
+
2
/
(
p
+
1
)
2
)
-
(
(
α
t
)
2
p
+
2
/
(
p
+
1
)
2
)
)
d
t
≤
α
∫
0
1
‍
t
p
+
1
+
α
q
-
p
t
q
+
1
1
-
t
2
p
+
2
d
t
.
Since α
≤
1, we have from (25) that
(26)
α
∫
0
1
‍
t
p
+
1
+
α
q
-
p
t
q
+
1
1
-
t
2
p
+
2
d
t
≤
2
∫
0
1
‍
t
p
+
1
1
-
t
2
p
+
2
d
t
.
If α
>
1, then λ
s
p
+
1
/
(
p
+
1
)
≤
F
λ
(
s
)
≤
(
λ
s
p
+
1
/
(
p
+
1
)
)
+
(
λ
s
q
+
1
/
(
q
+
1
)
), s
∈
(
0,1
) and (
λ
s
q
+
1
/
(
q
+
1
)
)
≤
F
λ
(
s
)
≤
(
λ
s
p
+
1
/
(
p
+
1
)
)
+
(
λ
s
q
+
1
/
(
q
+
1
)
), s
∈
(
1
,
α
), and, from the fact that F
λ
(
α
)
=
1, we get that if s
∈
(
0,1
), then
(27)
1
-
F
λ
2
(
s
)
=
F
λ
2
(
α
)
-
F
λ
2
(
s
)
=
(
F
λ
(
α
)
-
F
λ
(
s
)
)
(
F
λ
(
α
)
+
F
λ
(
s
)
)
≥
λ
2
(
1
p
+
1
-
s
p
+
1
p
+
1
+
α
q
+
1
q
+
1
-
1
q
+
1
)
×
(
1
p
+
1
+
α
q
+
1
q
+
1
+
s
p
+
1
p
+
1
-
1
q
+
1
)
≥
λ
2
(
α
q
+
1
q
+
1
-
s
p
+
1
q
+
1
)
(
α
q
+
1
q
+
1
+
s
p
+
1
q
+
1
)
=
λ
2
(
α
2
q
+
2
(
q
+
1
)
2
-
s
2
p
+
2
(
q
+
1
)
2
)
.
And, if s
∈
(
1
,
α
), then
(28)
1
-
F
λ
2
(
s
)
=
F
λ
2
(
α
)
-
F
λ
2
(
s
)
=
(
F
λ
(
α
)
-
F
λ
(
s
)
)
(
F
λ
(
α
)
+
F
λ
(
s
)
)
≥
λ
2
(
α
q
+
1
q
+
1
-
s
q
+
1
q
+
1
)
×
(
1
p
+
1
+
α
q
+
1
q
+
1
-
1
q
+
1
+
1
p
+
1
+
s
q
+
1
q
+
1
-
1
q
+
1
)
≥
λ
2
(
α
q
+
1
q
+
1
-
s
q
+
1
q
+
1
)
(
α
q
+
1
q
+
1
+
s
q
+
1
q
+
1
)
=
λ
2
1
(
q
+
1
)
2
(
α
2
q
+
2
-
s
2
q
+
2
)
;
therefore,
(29)
α
∫
0
1
‍
F
λ
(
α
t
)
1
-
F
λ
2
(
α
t
)
d
t
=
∫
0
α
‍
F
λ
(
s
)
1
-
F
λ
2
(
s
)
d
s
≤
∫
0
1
‍
(
λ
s
p
+
1
/
(
p
+
1
)
)
+
(
λ
s
q
+
1
/
(
q
+
1
)
)
λ
2
(
(
α
2
q
+
2
/
(
q
+
1
)
2
)
-
(
s
2
p
+
2
/
(
q
+
1
)
2
)
)
d
s
+
∫
1
α
‍
(
λ
(
s
p
+
1
/
(
p
+
1
)
)
)
+
(
λ
(
s
q
+
1
/
(
q
+
1
)
)
)
λ
2
(
(
α
2
q
+
2
/
(
q
+
1
)
2
)
-
(
s
2
q
+
2
/
(
q
+
1
)
2
)
)
d
s
≤
∫
0
α
‍
(
λ
s
p
+
1
/
(
p
+
1
)
)
+
(
λ
s
q
+
1
/
(
q
+
1
)
)
λ
2
(
(
α
2
q
+
2
/
(
q
+
1
)
2
)
-
(
s
2
p
+
2
/
(
q
+
1
)
2
)
)
d
s
=
α
∫
0
1
‍
(
(
α
t
)
p
+
1
/
(
p
+
1
)
)
+
(
(
α
t
)
q
+
1
/
(
q
+
1
)
)
(
α
2
q
+
2
/
(
q
+
1
)
2
)
-
(
(
α
t
)
2
p
+
2
/
(
q
+
1
)
2
)
d
t
=
α
∫
0
1
‍
(
α
p
-
q
t
p
+
1
/
(
p
+
1
)
)
+
(
t
q
+
1
/
(
p
+
1
)
)
(
1
/
(
q
+
1
)
2
)
-
(
α
2
p
-
2
q
t
2
p
+
2
/
(
q
+
1
)
2
)
d
t
≤
α
q
+
1
p
+
1
∫
0
1
‍
2
t
p
+
1
1
-
t
2
p
+
2
d
t
.
Or
(30)
α
(
λ
)
∫
0
1
‍
F
λ
(
α
t
)
1
-
F
λ
2
(
α
t
)
d
t
=
∫
0
α
‍
F
λ
(
s
)
1
-
F
λ
2
(
s
)
d
s
≤
∫
0
1
‍
(
λ
s
p
+
1
/
(
p
+
1
)
)
+
(
λ
s
q
+
1
/
(
q
+
1
)
)
λ
2
(
(
α
2
q
+
2
/
(
q
+
1
)
2
)
-
(
s
2
p
+
2
/
(
q
+
1
)
2
)
)
d
s
+
∫
1
α
‍
(
λ
(
s
p
+
1
/
(
p
+
1
)
)
)
+
(
λ
(
s
q
+
1
/
(
q
+
1
)
)
)
λ
2
(
(
α
2
q
+
2
/
(
q
+
1
)
2
)
-
(
s
2
q
+
2
/
(
q
+
1
)
2
)
)
d
s
≤
∫
0
α
‍
(
λ
(
s
p
+
1
/
(
p
+
1
)
)
)
+
(
λ
(
s
q
+
1
/
(
q
+
1
)
)
)
λ
2
(
(
α
2
q
+
2
/
(
q
+
1
)
2
)
-
(
s
2
q
+
2
/
(
q
+
1
)
2
)
)
d
s
=
α
∫
0
1
‍
(
λ
(
(
α
t
)
p
+
1
/
(
p
+
1
)
)
)
+
(
λ
(
(
α
t
)
q
+
1
/
(
q
+
1
)
)
)
λ
2
(
(
α
2
q
+
2
/
(
q
+
1
)
2
)
-
(
(
α
t
)
2
q
+
2
/
(
q
+
1
)
2
)
)
d
t
≤
α
q
+
1
p
+
1
∫
0
1
‍
2
t
p
+
1
1
-
t
2
p
+
2
d
t
.
It follows from (26), (29), and (30); we get that η
(
λ
) is continuous in (
0
,
∞
) and (i) holds. (ii) From (25) and the fact that 1
=
F
λ
(
α
(
λ
)
), the following holds.
Ifα
≤
1, then
(31)
λ
α
p
+
1
p
+
1
+
λ
α
q
+
1
q
+
1
≥
1
=
F
λ
(
α
)
≥
λ
α
p
+
1
p
+
1
;
thus, if s
∈
(
0
,
α
), then
(32)
1
-
F
λ
2
(
s
)
=
F
λ
2
(
α
)
-
F
λ
2
(
s
)
=
(
F
λ
(
α
)
-
F
λ
(
s
)
)
(
F
λ
(
α
)
+
F
λ
(
s
)
)
≤
λ
2
(
α
p
+
1
p
+
1
-
s
p
+
1
p
+
1
+
α
q
+
1
q
+
1
-
s
q
+
1
q
+
1
)
×
(
α
p
+
1
p
+
1
+
s
p
+
1
p
+
1
+
α
q
+
1
q
+
1
+
s
q
+
1
q
+
1
)
≤
4
λ
2
(
α
2
p
+
2
(
p
+
1
)
2
-
s
2
p
+
2
(
p
+
1
)
2
)
;
from (31), we obtain
(33)
α
(
λ
)
∫
0
1
‍
F
λ
(
α
t
)
1
-
F
λ
2
(
α
t
)
d
t
=
∫
0
α
‍
F
λ
(
s
)
1
-
F
λ
2
(
s
)
d
s
≥
∫
0
α
‍
s
p
+
1
/
(
p
+
1
)
4
(
(
α
2
p
+
2
/
(
p
+
1
)
2
)
-
(
s
2
p
+
2
/
(
p
+
1
)
2
)
)
d
s
=
α
∫
0
1
‍
(
α
t
)
p
+
1
/
(
p
+
1
)
4
(
(
α
2
p
+
2
/
(
p
+
1
)
2
)
-
(
(
α
t
)
2
p
+
2
/
(
p
+
1
)
2
)
)
d
t
=
1
2
α
∫
0
1
‍
t
p
+
1
1
-
t
2
p
+
2
d
t
.
If α
≥
1, then (
λ
s
q
+
1
/
(
q
+
1
)
)
≤
(
λ
s
p
+
1
/
(
p
+
1
)
)
≤
F
λ
(
s
)
≤
(
λ
s
p
+
1
/
(
p
+
1
)
)
+
(
λ
s
q
+
1
/
(
q
+
1
)
) for s
∈
(
0,1
) and λ
(
s
q
+
1
/
(
q
+
1
)
)
≤
F
λ
(
s
)
≤
(
λ
s
p
+
1
/
(
p
+
1
)
)
+
(
λ
s
q
+
1
/
(
q
+
1
)
) for s
∈
(
1
,
α
(
λ
)
), and, from the fact that F
λ
(
α
(
λ
)
)
=
1, we get the following.
Ifs
∈
(
0,1
), then
(34)
1
-
F
λ
2
(
s
)
=
F
λ
2
(
α
(
λ
)
)
-
F
λ
2
(
s
)
=
(
F
λ
(
α
)
-
F
λ
(
s
)
)
(
F
λ
(
α
(
λ
)
)
+
F
λ
(
s
)
)
≤
λ
2
(
α
p
+
1
p
+
1
-
s
p
+
1
p
+
1
+
α
q
+
1
q
+
1
-
s
q
+
1
q
+
1
)
×
(
α
p
+
1
p
+
1
+
α
q
+
1
q
+
1
+
s
p
+
1
p
+
1
+
s
q
+
1
q
+
1
)
≤
6
λ
2
(
α
q
+
1
p
+
1
-
s
q
+
1
p
+
1
)
(
α
q
+
1
p
+
1
+
s
q
+
1
p
+
1
)
=
6
λ
2
(
α
2
q
+
2
(
p
+
1
)
2
-
s
2
q
+
2
(
p
+
1
)
2
)
,
and, if s
∈
(
1
,
α
), then
(35)
1
-
F
λ
2
(
s
)
=
F
λ
2
(
α
(
λ
)
)
-
F
λ
2
(
s
)
=
(
F
λ
(
α
)
-
F
λ
(
s
)
)
(
F
λ
(
α
)
+
F
λ
(
s
)
)
≤
λ
2
(
α
p
+
1
p
+
1
-
s
p
+
1
p
+
1
+
α
q
+
1
q
+
1
-
s
q
+
1
q
+
1
)
×
(
α
p
+
1
p
+
1
+
α
q
+
1
q
+
1
+
s
p
+
1
p
+
1
+
s
q
+
1
q
+
1
)
≤
4
λ
2
(
α
q
+
1
q
+
1
-
s
q
+
1
q
+
1
)
(
α
q
+
1
p
+
1
+
s
q
+
1
p
+
1
)
≤
4
λ
2
1
(
p
+
1
)
2
(
α
2
q
+
2
-
s
2
q
+
2
)
.
Then
(36)
α
∫
0
1
‍
F
λ
(
α
t
)
1
-
F
λ
2
(
α
t
)
d
t
=
∫
0
α
‍
F
λ
(
s
)
1
-
F
λ
2
(
s
)
d
s
≥
∫
0
1
‍
λ
s
p
+
1
/
(
p
+
1
)
6
λ
2
(
(
α
2
q
+
2
/
(
p
+
1
)
2
)
-
(
s
2
q
+
2
/
(
p
+
1
)
2
)
)
d
s
+
∫
1
α
‍
λ
(
s
q
+
1
/
(
q
+
1
)
)
4
λ
2
(
(
α
2
q
+
2
/
(
p
+
1
)
2
)
-
(
s
2
q
+
2
/
(
p
+
1
)
2
)
)
d
s
≥
∫
0
α
‍
λ
s
q
+
1
/
(
q
+
1
)
6
λ
2
(
(
α
2
q
+
2
/
(
p
+
1
)
2
)
-
(
s
2
q
+
2
/
(
p
+
1
)
2
)
)
d
s
=
α
∫
0
1
‍
λ
(
α
t
)
q
+
1
/
(
q
+
1
)
6
λ
2
(
(
α
2
q
+
2
/
(
p
+
1
)
2
)
-
(
(
α
t
)
2
q
+
2
/
(
p
+
1
)
2
)
)
d
t
=
α
∫
0
1
‍
(
α
t
)
q
+
1
/
(
q
+
1
)
6
(
(
α
2
q
+
2
/
(
p
+
1
)
2
)
-
(
(
α
t
)
2
q
+
2
/
(
p
+
1
)
2
)
)
d
t
=
α
p
+
1
6
(
q
+
1
)
∫
0
1
‍
t
q
+
1
1
-
t
2
q
+
2
d
t
.
It follows from (26), (29), (33), (36), and Lemma 6 that lim
⁡
λ
→
0
⁡
η
(
λ
)
=
+
∞ and lim
⁡
λ
→
∞
⁡
η
(
λ
)
=
0.Lemma 9.
Let (H1), (H2), and (H4) hold; assume that0
<
p
≤
q
<
1. Then T
(
λ
,
ρ
) has the following properties, for fixed λ
>
0
:
(i)
lim
⁡
ρ
→
0
⁡
T
(
λ
,
ρ
)
=
0;
(ii)
lim
⁡
ρ
→
0
⁡
(
∂
/
∂
ρ
)
T
(
λ
,
ρ
)
=
+
∞;
(iii)
T
(
λ
,
α
(
λ
)
)
>
0;
(iv)
(
∂
2
/
∂
ρ
2
)
T
(
λ
,
ρ
)
<
0, for ρ
∈
(
0
,
α
(
λ
)
).Proof.
(i) Sinceρ
→
0, we choose 0
<
ρ
<
1; thus from (H2), we get
(37)
F
λ
(
ρ
)
-
F
λ
(
ρ
t
)
=
λ
∫
ρ
t
ρ
‍
f
(
s
)
d
s
≥
ρ
p
+
1
λ
p
+
1
(
1
-
t
p
+
1
)
;
therefore,
(38)
T
(
λ
,
ρ
)
=
ρ
∫
0
1
‍
1
-
ξ
ξ
(
2
-
ξ
)
d
t
≤
ρ
∫
0
1
‍
1
-
ξ
ρ
p
+
1
(
λ
/
(
p
+
1
)
)
(
1
-
t
p
+
1
)
(
2
-
ξ
)
d
t
=
ρ
(
1
-
p
)
/
2
∫
0
1
‍
1
-
ξ
(
λ
/
(
p
+
1
)
)
(
1
-
t
p
+
1
)
(
2
-
ξ
)
d
t
.
Hence, from p
<
1, we have lim
⁡
ρ
→
0
⁡
T
(
λ
,
ρ
)
=
0.
(ii) From Lemma5, we have that
(39)
∂
∂
ρ
T
(
λ
,
ρ
)
=
T
ρ
′
(
λ
,
ρ
)
=
∫
0
1
‍
ξ
(
1
-
ξ
)
(
2
-
ξ
)
-
ρ
ξ
′
(
ξ
(
2
-
ξ
)
)
3
/
2
d
t
;
according to the assumption (H2), we have
(40)
ξ
=
F
λ
(
ρ
)
-
F
λ
(
ρ
t
)
=
λ
∫
ρ
t
ρ
‍
f
(
s
)
d
s
≤
λ
ρ
p
+
1
(
1
-
t
p
+
1
)
p
+
1
+
λ
ρ
q
+
1
(
1
-
t
q
+
1
)
q
+
1
,
and, from the assumption (H4), that is, f
(
ρ
)
≥
3
ρ
f
′
(
ρ
)
≥
2
ρ
f
′
(
ρ
), ρ
∈
(
0
,
α
), we have ρ
ξ
′
≤
(
3
/
2
)
ξ; thus
(41)
ξ
(
1
-
ξ
)
(
2
-
ξ
)
-
ρ
ξ
′
≥
ξ
(
1
-
ξ
)
(
2
-
ξ
)
-
3
2
ξ
=
ξ
(
(
1
-
ξ
)
(
2
-
ξ
)
-
3
2
)
.
Thus
(42)
∂
∂
ρ
T
(
λ
,
ρ
)
=
∫
0
1
‍
ξ
(
1
-
ξ
)
(
2
-
ξ
)
-
ρ
ξ
′
(
ξ
(
2
-
ξ
)
)
3
/
2
d
t
≥
∫
0
1
‍
ξ
(
(
1
-
ξ
)
(
2
-
ξ
)
-
(
3
/
2
)
)
(
ξ
(
2
-
ξ
)
)
3
/
2
d
t
≥
∫
0
1
‍
(
(
(
1
-
ξ
)
(
2
-
ξ
)
-
(
3
2
)
)
(
+
λ
ρ
q
+
1
(
1
-
t
q
+
1
)
q
+
1
)
-
1
/
2
×
(
λ
ρ
p
+
1
(
1
-
t
p
+
1
)
p
+
1
+
λ
ρ
q
+
1
(
1
-
t
q
+
1
)
q
+
1
)
-
1
/
2
(
+
λ
ρ
q
+
1
(
1
-
t
q
+
1
)
q
+
1
)
-
1
/
2
×
(
2
-
ξ
)
-
3
/
2
)
d
t
.
Therefore, lim
⁡
ρ
→
0
⁡
(
∂
/
∂
ρ
)
T
(
λ
,
ρ
)
=
+
∞.
(iii) It is proved in the same way as in Lemma7(ii).
(iv) From Lemma5, we get that
(43)
T
ρ
ρ
′′
(
λ
,
ρ
)
=
3
ρ
∫
0
1
‍
(
1
-
ξ
)
(
ρ
ξ
′
)
2
(
ξ
(
2
-
ξ
)
)
5
/
2
d
t
-
∫
0
1
‍
2
ξ
′
+
ρ
ξ
′′
(
ξ
(
2
-
ξ
)
)
3
/
2
d
t
.
From (H1) where ξ
′′
≥
0, using the assumption (H4) and the fact that ξ
≤
1, we arrive at
(44)
3
ρ
(
1
-
ξ
)
(
ξ
′
)
2
-
(
2
ξ
′
+
ρ
ξ
′′
)
(
ξ
(
2
-
ξ
)
)
=
3
ρ
(
ξ
′
)
2
-
3
ρ
ξ
(
ξ
′
)
2
-
(
2
ξ
′
+
ρ
ξ
′′
)
(
2
ξ
-
ξ
2
)
=
3
ρ
(
ξ
′
)
2
-
3
ρ
ξ
(
ξ
′
)
2
-
4
ξ
ξ
′
+
2
ξ
2
ξ
′
-
ρ
ξ
′′
(
2
ξ
-
ξ
2
)
<
ξ
′
(
3
ρ
ξ
′
-
4
ξ
)
(
1
-
ξ
)
-
ρ
ξ
′′
(
2
ξ
-
ξ
2
)
<
0
.
Therefore, (
∂
2
/
∂
ρ
2
)
T
(
λ
,
ρ
)
<
0, for ρ
∈
(
0
,
α
).Lemma 10.
Assume that0
<
p
≤
q
<
1. Denote w
(
λ
)
=
sup
⁡
⁡
{
T
(
λ
,
ρ
)
∣
0
<
ρ
<
α
(
λ
)
}. Then(i)
w
(
λ
) is strictly decreasing in (
0
,
+
∞
);
(ii)
w
(
λ
) is continuous in (
0
,
+
∞
);
(iii)
lim
⁡
λ
→
0
⁡
w
(
λ
)
=
+
∞, lim
⁡
λ
→
+
∞
⁡
w
(
λ
)
=
0.Proof.
(i) From Lemma9, we get
(45)
w
(
λ
)
=
max
⁡
{
T
(
λ
,
ρ
)
∣
0
<
ρ
<
α
(
λ
)
}
.
Let 0
<
λ
1
<
λ
2
<
+
∞. From the definition of ξ
=
ξ
λ
(
ρ
,
t
) and Lemma 6, we obtain
(46)
α
(
λ
2
)
<
α
(
λ
1
)
.
And, for 0
<
ρ
≤
α
(
λ
2
) and 0
<
t
<
1,
(47)
ξ
λ
1
(
ρ
,
t
)
=
λ
1
∫
ρ
t
ρ
‍
f
(
s
)
d
s
<
ξ
λ
2
(
ρ
,
t
)
=
λ
2
∫
ρ
t
ρ
‍
f
(
s
)
d
s
.
It follows that, for 0
<
ρ
≤
α
(
λ
2
),
(48)
ρ
∫
0
1
‍
1
-
ξ
λ
2
(
ρ
,
t
)
ξ
λ
2
(
ρ
,
t
)
(
2
-
ξ
λ
2
(
ρ
,
t
)
)
d
t
<
ρ
∫
0
1
‍
1
-
ξ
λ
1
(
ρ
,
t
)
ξ
λ
1
(
ρ
,
t
)
(
2
-
ξ
λ
1
(
ρ
,
t
)
)
d
t
;
thus,
(49)
T
(
λ
2
,
ρ
)
<
T
(
λ
1
,
ρ
)
.
Hence, w
(
λ
2
)
<
w
(
λ
1
), which implies that w
(
λ
) is strictly decreasing in (
0
,
+
∞
).
(ii) Letλ
0
∈
(
0
,
+
∞
). Since F
λ
0
(
α
(
λ
0
)
)
=
1, there exists δ
>
0 such that, for 0
<
λ
≤
λ
0
+
δ and 0
<
ρ
≤
α
(
λ
0
)
+
δ,
(50)
F
λ
(
ρ
)
=
λ
∫
0
ρ
‍
f
(
s
)
d
s
<
(
λ
0
+
δ
)
∫
0
α
(
λ
0
)
+
δ
‍
f
(
s
)
d
s
<
3
2
.
It follows that ξ
=
ξ
λ
(
ρ
,
t
)
<
3
/
2, for 0
<
ρ
≤
α
(
λ
0
)
+
δ, t
∈
[
0,1
]. Further,
(51)
ξ
λ
(
ρ
,
t
)
=
λ
∫
ρ
t
ρ
‍
f
(
s
)
d
s
≥
λ
ρ
p
+
1
(
1
-
t
p
+
1
)
p
+
1
,
ρ
≤
1
,
ξ
λ
(
ρ
,
t
)
=
λ
∫
ρ
t
ρ
‍
f
(
s
)
d
s
≥
λ
ρ
q
+
1
(
1
-
t
q
+
1
)
q
+
1
,
hhhhhhhhhhhhhh
ρ
>
1
,
ρ
t
≥
1
,
ξ
λ
(
ρ
,
t
)
=
λ
∫
ρ
t
ρ
‍
f
(
s
)
d
s
≥
λ
ρ
q
+
1
(
1
-
t
p
+
1
)
q
+
1
,
hhhhhhhhhhhhhh
ρ
>
1
,
ρ
t
≤
1
.
Define
(52)
ψ
1
(
λ
,
ρ
)
=
∫
0
1
‍
1
-
ξ
λ
(
ρ
,
t
)
∫
ρ
-
p
t
ρ
-
p
‍
f
(
ρ
p
+
1
s
)
d
s
(
2
-
ξ
λ
(
ρ
,
t
)
)
d
t
,
hhhhhhhhhhhhhhhhhhhhhhhhhhh
ρ
≤
1
,
ψ
2
(
λ
,
ρ
)
=
∫
0
1
‍
1
-
ξ
λ
(
ρ
,
t
)
∫
ρ
-
q
t
ρ
-
q
‍
f
(
ρ
q
+
1
s
)
d
s
(
2
-
ξ
λ
(
ρ
,
t
)
)
d
t
,
hhhhhhhhhhhhhhhhhhhhhhhhhhh
ρ
>
1
.
Thus, if ρ
≤
1, then
(53)
ψ
1
(
λ
,
ρ
)
=
∫
0
1
‍
1
-
ξ
λ
(
ρ
,
t
)
∫
ρ
-
p
t
ρ
-
p
‍
f
(
ρ
p
+
1
s
)
d
s
(
2
-
ξ
λ
(
ρ
,
t
)
)
d
t
≤
∫
0
1
‍
1
-
ξ
λ
(
ρ
,
t
)
∫
ρ
-
p
t
ρ
-
p
‍
ρ
p
(
p
+
1
)
s
p
d
s
(
2
-
ξ
λ
(
ρ
,
t
)
)
d
t
≤
2
∫
0
1
‍
1
-
ξ
λ
(
ρ
,
t
)
∫
ρ
-
p
t
ρ
-
p
‍
ρ
p
(
p
+
1
)
s
p
d
s
d
t
≤
2
p
+
1
∫
0
1
‍
1
-
ξ
λ
(
ρ
,
t
)
1
-
t
p
+
1
d
t
≤
2
p
+
1
∫
0
1
‍
1
1
-
t
p
+
1
d
t
,
and, if ρ
>
1, then
(54)
ψ
2
(
λ
,
ρ
)
=
∫
0
1
‍
1
-
ξ
λ
(
ρ
,
t
)
∫
ρ
-
q
t
ρ
-
q
‍
f
(
ρ
q
+
1
s
)
d
s
(
2
-
ξ
λ
(
ρ
,
t
)
)
d
t
≤
∫
0
1
‍
1
-
ξ
λ
(
ρ
,
t
)
∫
ρ
-
q
t
ρ
-
q
‍
ρ
q
(
q
+
1
)
s
q
d
s
(
2
-
ξ
λ
(
ρ
,
t
)
)
d
t
≤
2
∫
0
1
‍
1
-
ξ
λ
(
ρ
,
t
)
∫
ρ
-
q
t
ρ
-
q
‍
ρ
q
(
q
+
1
)
s
q
d
s
d
t
≤
2
q
+
1
∫
0
1
‍
1
-
ξ
λ
(
ρ
,
t
)
1
-
t
p
+
1
d
t
≤
2
q
+
1
∫
0
1
‍
1
1
-
t
p
+
1
d
t
.
It follows that the singular integral in (53) and (54) converges uniformly in [
0
,
λ
0
+
δ
]
×
[
0
,
α
(
λ
0
)
+
δ
], and hence ψ
1
(
λ
,
ρ
) and ψ
2
(
λ
,
ρ
) are continuous in [
0
,
λ
0
+
δ
]
×
[
0
,
α
(
λ
0
)
+
δ
]. By (i), we have
(55)
lim
⁡
λ
→
λ
0
-
w
(
λ
)
≥
w
(
λ
0
)
≥
lim
⁡
λ
→
λ
0
+
w
(
λ
)
.
We will see that the equalities hold in (55). If this does not hold, then we may assume, for instance,
(56)
lim
⁡
λ
→
λ
0
-
w
(
λ
)
>
w
(
λ
0
)
.
Choose an increasing sequence {
λ
n
}
⊂
(
0
,
λ
0
) satisfying lim
⁡
n
→
∞
⁡
λ
n
=
λ
0 and then choose ρ
n
∈
(
0
,
α
(
λ
n
)
) such that
(57)
α
(
λ
n
)
<
α
(
λ
0
)
+
δ
,
w
(
λ
n
)
=
T
(
λ
n
,
ρ
n
)
.
Since lim
⁡
n
→
∞
⁡
α
(
λ
n
)
=
α
(
λ
0
), we may assume that there exists ρ
0
∈
[
0
,
α
(
λ
0
)
] such that lim
⁡
n
→
∞
ρ
n
=
ρ
0; by the continuity of ψ
1
(
λ
,
ρ
) and ψ
2
(
λ
,
ρ
), we have
(58)
lim
⁡
n
→
∞
w
(
λ
n
)
=
lim
⁡
n
→
∞
T
(
λ
n
,
ρ
n
)
=
lim
⁡
n
→
∞
ρ
n
1
-
p
λ
n
ψ
1
(
λ
n
,
ρ
n
)
=
ρ
0
1
-
p
λ
0
ψ
1
(
λ
0
,
ρ
0
)
=
T
(
λ
0
,
ρ
0
)
≤
w
(
λ
0
)
,
ρ
0
≤
1
,
or
(59)
lim
⁡
n
→
∞
w
(
λ
n
)
=
lim
⁡
n
→
∞
T
(
λ
n
,
ρ
n
)
=
lim
⁡
n
→
∞
ρ
n
1
-
q
λ
n
ψ
2
(
λ
n
,
ρ
n
)
=
ρ
0
1
-
q
λ
0
ψ
2
(
λ
0
,
ρ
0
)
=
T
(
λ
0
,
ρ
0
)
≤
w
(
λ
0
)
,
hhhhhhhhhhhhhhhhhhhhhhhhhhh
ρ
0
>
1
,
which contradicts (56). Therefore, lim
⁡
λ
→
λ
0
-
⁡
w
(
λ
)
=
w
(
λ
0
). Similarly, we have lim
⁡
λ
→
λ
0
+
⁡
w
(
λ
)
=
w
(
λ
0
). So w
(
λ
) is continuous at λ
0.
(iii) Note thatlim
⁡
λ
→
+
∞
⁡
α
(
λ
)
=
0, for any λ
>
0, 0
<
ρ
≤
α
(
λ
), and 0
<
t
<
1, and, from (55), we have
(60)
lim
⁡
λ
→
+
∞
w
(
λ
)
=
lim
⁡
λ
→
+
∞
max
⁡
0
<
ρ
≤
α
(
λ
)
T
(
λ
,
ρ
)
≤
lim
⁡
λ
→
+
∞
2
ρ
×
∫
0
1
(
1
ρ
p
+
1
(
1
-
t
p
+
1
)
p
+
1
}
)
-
1
/
2
×
(
λ
max
⁡
⁡
{
ρ
q
+
1
(
1
-
t
q
+
1
)
q
+
1
,
ρ
p
+
1
(
1
-
t
p
+
1
)
p
+
1
}
)
-
1
/
2
)
d
t
≤
lim
⁡
λ
→
+
∞
2
ρ
∫
0
1
‍
1
λ
(
ρ
2
(
1
-
t
q
+
1
)
/
(
q
+
1
)
)
d
t
≤
lim
⁡
λ
→
+
∞
2
(
q
+
1
)
∫
0
1
‍
1
λ
(
1
-
t
p
+
1
)
d
t
=
0
.
Hence, lim
⁡
λ
→
+
∞
⁡
w
(
λ
)
=
0.
Fixλ
1
>
0 and 0
<
ρ
1
<
α
(
λ
1
). For λ
<
λ
1, the definition of w
(
λ
) and the fact that α
(
λ
1
)
<
α
(
λ
) imply w
(
λ
)
≥
T
(
λ
,
ρ
1
).
Sincelim
⁡
λ
→
0
ξ
λ
(
ρ
1
,
t
)
=
0,
(61)
lim
⁡
λ
→
0
T
(
λ
,
ρ
1
)
=
lim
⁡
λ
→
0
ρ
1
∫
0
1
‍
1
-
ξ
λ
(
ρ
1
,
t
)
ξ
λ
(
ρ
1
,
t
)
(
2
-
ξ
λ
(
ρ
1
,
t
)
)
d
t
≥
lim
⁡
λ
→
0
ρ
1
∫
0
1
(
1
-
ξ
λ
(
ρ
1
,
t
)
)
×
(
λ
{
ρ
1
q
+
1
(
1
-
t
q
+
1
)
q
+
1
+
ρ
1
p
+
1
(
1
-
t
p
+
1
)
p
+
1
}
{
+
ρ
1
p
+
1
(
1
-
t
p
+
1
)
p
+
1
}
×
(
2
-
ξ
λ
(
ρ
1
,
t
)
)
)
-
1
/
2
d
t
=
+
∞
.
Hence, lim
⁡
λ
→
0
⁡
w
(
λ
)
=
+
∞.


## 3. The Proofs of the Main Results

Proof of Theorem1.
Similar to the proof of Theorem 1.1 [3], it follows from Lemmas 7 and 8 that the results are easy to prove.Proof of Theorem2.
Similar to the proof of Theorem 1.2 [3], it follows from Lemmas 8, 9, and 10 that the results are easy to prove.



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*Source: 610926-2014-04-13.xml*