# Asymptotic Behavior of the Bifurcation Diagrams for Semilinear Problems with Application to Inverse Bifurcation Problems

Authors: Tetsutaro Shibata
Journal: International Journal of Differential Equations
 (2015)

---

## Abstract

We consider the nonlinear eigenvalue problemu
″
(
t
)
+
λ
f
(
u
(
t
)
)
=
0,  u
(
t
)
>
0,  t
∈
I
=
:
(
-
1,1
),  u
(
1
)
=
u
(
-
1
)
=
0, where f
(
u
) is a cubic-like nonlinear term and λ
>
0 is a parameter. It is known by Korman et al. (2005) that, under the suitable conditions on f
(
u
), there exist exactly three bifurcation branches λ
=
λ
j
(
ξ
) (j
=
1,2
,
3), and these curves are parameterized by the maximum norm ξ of the solution u
λ corresponding to λ. In this paper, we establish the precise global structures for λ
j
(
ξ
) (j
=
1,2
,
3), which can be applied to the inverse bifurcation problems. The precise local structures for λ
j
(
ξ
) (j
=
1,2
,
3) are also discussed. Furthermore, we establish the asymptotic shape of the spike layer solution u
2
(
λ
,
t
), which corresponds to λ
=
λ
2
(
ξ
), as λ
→
∞.


---

## Body

## 1. Introduction

We consider the following nonlinear eigenvalue problem:(1)
u
′′
(
t
)
+
λ
f
u
t
=
0
,
t
∈
I
=
:
(
-
1,1
)
,
(2)
u
(
t
)
>
0
,
t
∈
I
,
(3)
u
(
1
)
=
u
(
-
1
)
=
0
,
where f
(
u
) is a cubic-like nonlinear term and λ
>
0 is a parameter. We assume the following conditions (A.1)–(A.3), which have been introduced in [1].(A.1)
f
(
u
) is a C
2-function on [
0
,
∞
) and has three positive roots at 0
<
a
<
b
<
c, and
(4)
f
(
u
)
>
0
on
⁡
[
0
,
a
)
∪
(
b
,
c
)
,
f
(
u
)
<
0
on
⁡
(
a
,
b
)
∪
(
c
,
∞
)
,
(5)
∫
a
c
‍
f
(
s
)
d
s
>
0
.
(A.2)
There exists a constanta
<
β
<
c such that
(6)
f
′′
(
u
)
>
0
for
0
≤
u
<
β
,
f
′′
(
u
)
<
0
for
u
>
β
.
(A.3)
Letτ
∈
(
b
,
c
) satisfy f
′
(
τ
)
=
0. Then
(7)
∫
a
τ
‍
f
(
s
)
d
s
≤
0
. The typical example of f
(
u
) which satisfies (A.1)–(A.3) is f
(
u
)
=
(
u
-
a
)
(
u
-
b
)
(
c
-
u
) with c
>
2
b
-
a and the area of the negative hump of f
(
u
) is nearly equal to that of the positive hump. For example, if we choose b
<
(
a
+
c
)
/
2 appropriately, then f
(
u
)
=
(
u
-
a
)
(
u
-
b
)
(
c
-
u
) satisfies (A.1)–(A.3).Nonlinear elliptic eigenvalue problems have been studied by many authors. We refer to [2–8] and the references therein. Among other things, (1)–(3) have been investigated by many authors. We refer to [1, 9–12] and the references therein. In particular, the following basic properties of the structure of bifurcation diagram for (1)–(3) have been proved in [1, Theorem 3.1].Theorem 1 (see [1, Theorem 3.1]).
Assume (A.1)–(A.3). Then there exists a criticalλ
0 such that (1)–(3) have exactly one positive solution for 0
<
λ
<
λ
0, exactly two positive solutions for λ
=
λ
0, and exactly three positive solutions for λ
>
λ
0. Furthermore, all solutions lie on two smooth solution curves, and λ is parameterized by ξ
:
=
∥
u
λ
∥
∞ as λ
=
λ
(
ξ
). One of the curves, referred to as the lower curve λ
1
(
ξ
), satisfies λ
1
(
0
)
=
0 and increases in 0
≤
ξ
<
a, and
(8)
λ
1
(
ξ
)
⟶
∞
a
s
ξ
⟶
a
,
(9)
u
1
(
λ
,
t
)
⟶
a
f
o
r
t
∈
I
a
s
λ
⟶
∞
,
where u
1
(
λ
,
t
) is a solution of (1)–(3) corresponding to λ
=
λ
1
(
ξ
). (Note that λ
→
∞ is equivalent to ξ
→
a.) The upper curve, which consists of two branches λ
j
(
ξ
) (j
=
2,3), is a parabola-like curve with exactly one turn to the right at ξ
=
ξ
0
∈
(
b
,
c
). Furthermore,
(10)
λ
2
(
ξ
)
⟶
∞
a
s
ξ
⟶
θ
0
,
(11)
u
2
λ
,
0
⟶
θ
0
,
u
2
λ
,
t
⟶
a
t
∈
I
∖
0
a
s
λ
⟶
∞
,
(12)
λ
3
(
ξ
)
⟶
∞
a
s
ξ
⟶
c
,
(13)
u
3
(
λ
,
t
)
⟶
c
(
t
∈
I
)
a
s
λ
⟶
∞
,
where u
j
(
λ
,
t
) is a solution of (1)–(3) corresponding to λ
=
λ
j
(
ξ
) (j
=
2,3) and θ
0
∈
(
b
,
c
) is a constant which satisfies
(14)
∫
a
θ
0
‍
f
(
s
)
d
s
=
0
.(See Figure1 for the bifurcation diagram.)Figure 1As we see from (9), (11), and (13), one of the most interesting facts in Theorem 1 is the difference between the asymptotic shapes of u
j
(
λ
,
t
) (j
=
1,3) and u
2
(
λ
,
t
) for λ sufficiently large (see Figures 2 and 3). This difference comes from the fact that u
j
(
λ
,
t
) (j
=
1,3) is stable and u
2
(
λ
,
t
) is unstable. This drives us to the question whether these facts give effect on the asymptotic behavior of the bifurcation branches or not.Figure 2Figure 3The purpose of this paper is to establish the precise asymptotic formulas forλ
j
(
ξ
) (j
=
1,2
,
3) as ξ
→
a
,
θ
0
,
c, respectively, to clarify how the difference of the asymptotic shapes of solutions corresponding to these three curves gives effect to the asymptotic formulas for λ
j
(
ξ
) (j
=
1,3) and λ
2
(
ξ
). Furthermore, we establish the asymptotic widths t
a
,
λ and t
d
,
λ of the spike layer of solution u
2
(
λ
,
t
) as ξ
→
θ
0.Finally, we establish the precise asymptotic formulas forλ
1
(
ξ
) as ξ
→
0 and λ
j
(
ξ
) as ξ
→
ξ
0 (j
=
2,3). By using the asymptotic formulas above, we propose new inverse bifurcation problems.Now we state our main result. LetF
(
u
)
:
=
∫
0
u
‍
f
(
s
)
d
s.Theorem 2.
Assume that0
<
ξ
<
a. Let δ
ξ
>
0 satisfy log
⁡
δ
ξ
=
δ
ξ
log
⁡
(
a
-
ξ
). Then as ξ
→
a,
(15)
λ
1
(
ξ
)
=
1
f
′
a
-
log
⁡
a
-
ξ
+
O
log
⁡
δ
ξ
2
.Theorem 3.
Assume thatξ
>
θ
0. Let δ
ξ
>
0 satisfy log
⁡
δ
ξ
=
δ
ξ
log
⁡
(
ξ
-
θ
0
).
(i) Asξ
→
θ
0,
(16)
λ
2
(
ξ
)
=
1
f
′
a
-
log
⁡
ξ
-
θ
0
+
O
log
⁡
δ
ξ
2
.
(ii) Lett
a
,
λ
∈
(
0,1
) satisfy u
2
(
λ
,
t
a
,
λ
)
=
a. Then as λ
→
∞ (namely, ξ
→
θ
0),
(17)
t
a
,
λ
=
1
2
+
O
ξ
-
θ
0
1
/
2
log
⁡
ξ
-
θ
0
.
(iii) Letd
∈
(
a
,
θ
0
) be an arbitrary fixed constant. Furthermore, let t
d
,
λ
∈
(
0,1
) satisfy u
2
(
λ
,
t
d
,
λ
)
=
d. Then as λ
→
∞ (namely, ξ
→
θ
0),
(18)
t
d
,
λ
=
1
2
λ
C
d
+
O
ξ
-
θ
0
1
/
2
,
where
(19)
C
d
:
=
∫
d
θ
0
‍
1
F
θ
0
-
F
θ
d
θ
.Theorem 4.
Assume thatξ
<
c. Let δ
ξ
>
0 satisfy log
⁡
δ
ξ
=
δ
ξ
log
⁡
(
c
-
ξ
). Then as ξ
→
c,
(20)
λ
3
(
ξ
)
=
1
f
′
c
-
log
⁡
c
-
ξ
+
O
log
⁡
δ
ξ
2
.Now, we establish the local asymptotic behavior ofλ
1
(
ξ
) and λ
j
(
ξ
) (j
=
2,3) near ξ
=
0 and ξ
=
ξ
0, respectively, where ξ
0
∈
(
θ
0
,
1
) satisfy
(21)
2
∫
0
1
‍
1
F
ξ
0
-
F
ξ
0
s
d
s
=
ξ
0
∫
0
1
‍
F
ξ
0
-
F
ξ
0
s
-
3
/
2
f
ξ
0
-
s
f
ξ
0
s
d
s
.Theorem 5.
(i) Asξ
→
0,
(22)
λ
1
ξ
=
ξ
2
f
ξ
2
+
∑
k
=
1
∞
‍
L
k
ξ
ξ
k
2
,
where {
L
k
(
ξ
)
} (k
=
1,2
,
…) are bounded functions of ξ determined explicitly.
(ii) Assume thatf
(
u
) is analytic near u
=
0. Then as ξ
→
0,
(23)
λ
1
ξ
=
2
f
0
ξ
+
∑
k
=
1
∞
‍
M
k
ξ
ξ
k
,
where {
M
k
(
ξ
)
} (k
=
1,2
,
…) are bounded functions of ξ determined explicitly.Theorem 6.
Letj
=
2,3. Then as ξ
→
ξ
0,
(24)
λ
j
ξ
=
λ
0
+
∑
k
=
1
∞
‍
J
k
ξ
ξ
-
ξ
0
k
,
where {
J
k
(
ξ
)
} (k
=
1,2
,
…) are bounded functions determined explicitly.Finally, we apply our results above to the inverse bifurcation problems.Theorem 7.
Letk
=
1,2. Assume that f
k
(
u
)
=
(
u
-
a
k
)
(
u
-
b
k
)
(
c
k
-
u
) satisfy (A.1)–(A.3), where 0
<
a
k
<
b
k
<
c
k are unknown constants. Let λ
j
(
k
,
ξ
) (j
=
1,3) be the bifurcation branches corresponding to f
k
(
u
). Further, let θ
k
,
0 be the number defined in (14) for f
k
(
u
). Finally, for η
:
=
a
1
-
ξ
1
=
a
2
-
ξ
2
>
0 and ζ
:
=
c
1
-
ξ
1
=
c
2
-
ξ
2
>
0, let
(25)
μ
1
k
,
η
:
=
λ
1
k
,
ξ
k
,
μ
3
k
,
ζ
:
=
λ
3
k
,
ξ
k
.
(i) Suppose that(26)
lim
⁡
η
→
0
μ
1
1
,
η
μ
1
2
,
η
=
1
,
(27)
λ
1
1
,
ξ
-
λ
1
2
,
ξ
=
o
ξ
2
a
s
ξ
⟶
0
.
Then, a
1
=
a
2, b
1
=
b
2, and c
1
=
c
2.
(ii) Suppose that(28)
lim
⁡
η
→
0
μ
1
1
,
η
μ
1
2
,
η
=
lim
⁡
ζ
→
0
μ
3
1
,
ζ
μ
3
2
,
ζ
=
1
,
(29)
λ
1
1
,
ξ
-
λ
1
2
,
ξ
=
o
ξ
a
s
ξ
⟶
0
.
Then, a
1
=
a
2, b
1
=
b
2, and c
1
=
c
2.
(iii) Suppose that (28) and θ
1,0
=
θ
2,0 hold. Then, a
1
=
a
2, b
1
=
b
2, and c
1
=
c
2.Theorem7(i) implies that if we assume that the unknown nonlinear term f
(
u
) is a cubic, then the precise information about the local asymptotic behavior along with the rough global asymptotics of one branch determines the unknown f
(
u
). Theorem 7(ii) implies that the unknown f
(
u
) is determined by only the asymptotic behavior of two branches.Our methods to prove Theorems2–4 are based on the precise calculation of the time map. We prove Theorems 5 and 6 by the method developed in [13]. By using Theorems 2, 4, and 5, we prove Theorem 7.


## 2. Proof of Theorem2

In this section, letξ
<
a and let ξ
→
a. In what follows, C denotes various positive constants independent of λ
≫
1. We know that if (
λ
,
u
)
∈
R
+
×
C
2
(
I
¯
) satisfies (1)–(3), then
(30)
u
t
=
u
-
t
,
0
≤
t
≤
1
,
(31)
u
0
=
max
⁡
-
1
≤
t
≤
1
u
t
,
(32)
u
′
t
<
0
,
0
<
t
≤
1
.
We parameterize the solution pair by using the L
∞ norm of the solution ξ
=
u
λ
∞ such as (
λ
,
u
)
=
(
λ
1
(
ξ
)
,
u
ξ
) (0
≤
ξ
<
a). By (1), for t
∈
I
¯,
(33)
u
ξ
′′
t
+
λ
f
u
ξ
t
u
ξ
′
t
=
0
.
This implies that, for t
∈
I
¯,
(34)
d
d
t
1
2
u
ξ
′
t
2
+
λ
F
u
ξ
t
=
0
.
By this, (31), and putting t
=
0, for -
1
≤
t
≤
1, we obtain
(35)
u
ξ
′
t
2
+
2
λ
F
u
ξ
t
=
constant
=
2
λ
F
ξ
.
This implies that
(36)
u
ξ
′
t
2
=
2
λ
F
ξ
-
F
u
ξ
t
.
By this and (32), for 0
≤
t
≤
1, we obtain
(37)
-
u
ξ
′
t
=
2
λ
F
ξ
-
F
u
ξ
t
.
By this, for 0
≤
t
≤
1,
(38)
λ
=
1
2
∫
0
1
‍
-
u
ξ
′
t
F
ξ
-
F
u
ξ
t
d
t
=
1
2
∫
0
ξ
‍
1
F
ξ
-
F
(
θ
)
d
θ
.
By mean value theorem, for 0
<
θ
<
ξ, we have
(39)
F
ξ
-
F
θ
=
f
ξ
ξ
-
θ
-
1
2
f
′
ξ
θ
ξ
-
θ
2
,
where θ
<
ξ
θ
<
ξ. We know that f
′
(
ξ
θ
)
<
0, since, by (A.2), f
′
(
0
)
<
f
′
(
y
)
<
f
′
(
a
)
<
0 for 0
<
y
<
a. Let δ
:
=
δ
ξ
>
0 defined in Theorem 2. We easily see that δ
ξ
→
0 as ξ
→
a. Furthermore, by Taylor expansion, as ξ
→
a,
(40)
f
ξ
=
f
′
a
ξ
-
a
+
O
ξ
-
a
2
,
a
-
ξ
=
δ
ξ
1
/
δ
ξ
≪
δ
ξ
.
By (38) and (39), we obtain
(41)
λ
=
1
2
∫
0
ξ
‍
1
f
ξ
ξ
-
θ
-
f
′
ξ
θ
ξ
-
θ
2
/
2
d
θ
=
1
2
∫
ξ
-
δ
ξ
‍
1
f
ξ
ξ
-
θ
-
f
′
ξ
θ
ξ
-
θ
2
/
2
d
θ
+
∫
0
ξ
-
δ
‍
1
f
ξ
ξ
-
θ
-
f
′
ξ
θ
ξ
-
θ
2
/
2
d
θ
:
=
1
2
Z
1
+
Z
2
.
We write ξ
(
t
)
:
=
ξ
ξ
-
t (0
<
t
<
δ). Then,
(42)
Z
1
=
∫
0
δ
‍
1
f
ξ
t
+
f
′
ξ
t
2
/
2
d
t
+
∫
0
δ
‍
1
f
ξ
t
+
f
′
ξ
t
t
2
/
2
d
t
-
∫
0
δ
‍
1
f
ξ
t
+
f
′
ξ
t
2
/
2
d
t
:
=
Z
11
+
Z
12
.
Then,
(43)
Z
11
=
2
f
′
ξ
log
⁡
f
′
ξ
t
+
f
ξ
2
f
′
ξ
+
f
′
ξ
2
t
2
+
2
f
′
ξ
f
ξ
t
2
f
′
ξ
+
f
′
ξ
2
t
2
+
2
f
′
ξ
f
ξ
t
t
=
0
t
=
δ
=
2
f
′
ξ
-
log
⁡
f
ξ
+
D
,
where
(44)
D
≔
log
⁡
f
′
ξ
δ
+
f
ξ
+
f
′
ξ
2
δ
2
+
2
f
′
ξ
f
ξ
δ
=
O
log
⁡
δ
by (40). We put
(45)
X
:
=
f
ξ
t
+
1
2
f
′
ξ
t
2
,
Y
:
=
f
ξ
t
+
1
2
f
′
ξ
t
t
2
.
We note that ξ
-
δ
<
ξ
(
t
)
<
ξ. Then by (41) and (43),
(46)
Z
12
=
1
2
∫
0
δ
‍
f
′
ξ
-
f
′
ξ
t
t
2
X
Y
X
+
Y
d
t
≤
C
∫
0
δ
‍
f
′′
ξ
ξ
-
ξ
t
X
d
t
≤
C
δ
∫
0
δ
‍
1
f
ξ
t
+
f
′
ξ
t
2
/
2
d
t
≤
C
δ
-
log
⁡
f
ξ
+
D
.
We next calculate Z
2. We have
(47)
Z
2
=
∫
δ
ξ
‍
1
f
ξ
t
+
f
′
ξ
t
2
/
2
d
t
≤
C
∫
δ
ξ
‍
1
t
d
t
≤
-
C
log
⁡
δ
.
By (40)–(44), (46), (47), and Taylor expansion, we obtain
(48)
λ
=
1
f
′
ξ
-
log
⁡
f
ξ
+
D
1
+
O
δ
+
O
-
log
⁡
δ
=
1
f
′
a
1
+
O
a
-
ξ
1
+
O
δ
×
-
log
⁡
a
-
ξ
+
O
log
⁡
δ
+
O
log
⁡
δ
=
1
f
′
a
-
log
⁡
⁡
a
-
ξ
+
O
δ
log
⁡
⁡
a
-
ξ
+
O
log
⁡
δ
=
1
f
′
a
-
log
⁡
a
-
ξ
+
O
(
log
⁡
δ
)
.
This implies (15). Thus, the proof is complete.


## 3. Proof of Theorem3

In this section, letξ
→
θ
0 with θ
0
<
ξ. We write δ
:
=
δ
ξ defined in the statement in Theorem 3. By (38), we have
(49)
λ
=
1
2
∫
0
ξ
‍
1
F
ξ
-
F
θ
d
θ
=
1
2
∫
0
a
‍
1
F
ξ
-
F
θ
d
θ
+
∫
a
θ
0
‍
1
F
ξ
-
F
θ
d
θ
+
∫
θ
0
ξ
‍
1
F
ξ
-
F
θ
d
θ
:
=
1
2
I
+
I
I
+
I
I
I
.Lemma 8.
Asξ
→
θ
0,
(50)
I
=
2
f
′
a
-
1
2
log
⁡
ξ
-
θ
0
+
O
log
⁡
δ
.Proof.
We have(51)
I
=
∫
a
-
δ
a
‍
1
F
ξ
-
F
θ
d
θ
+
∫
0
a
-
δ
‍
1
F
ξ
-
F
θ
d
θ
:
=
W
1
+
W
2
.
We first calculate W
1. For a
-
δ
<
θ
<
a, by (14) and Taylor expansion, we have
(52)
F
ξ
-
F
θ
=
∫
θ
ξ
‍
f
s
d
s
=
∫
a
θ
0
‍
f
s
d
s
+
∫
θ
0
ξ
‍
f
s
d
s
+
∫
θ
a
‍
f
s
d
s
=
∫
θ
0
ξ
‍
f
s
d
s
+
∫
θ
a
‍
f
s
d
s
=
f
θ
0
ξ
-
θ
0
1
+
O
ξ
-
θ
0
+
1
2
f
′
a
1
+
O
δ
a
-
θ
2
.
By (51) and (52), we obtain
(53)
W
1
=
∫
a
-
δ
a
f
θ
0
ξ
-
θ
0
1
+
O
ξ
-
θ
0
f
′
a
1
+
O
δ
a
-
θ
2
2
‍
+
f
′
a
1
+
O
δ
a
-
θ
2
2
-
1
/
2
d
θ
=
∫
0
δ
‍
f
θ
0
ξ
-
θ
0
1
+
O
ξ
-
θ
0
f
′
a
1
+
O
δ
y
2
2
+
f
′
a
1
+
O
δ
y
2
2
-
1
/
2
d
y
:
=
∫
0
δ
‍
1
D
1
+
D
2
y
2
d
y
,
where
(54)
D
1
:
=
f
θ
0
ξ
-
θ
0
1
+
O
ξ
-
θ
0
,
D
2
:
=
1
2
f
′
a
1
+
O
δ
.
Then we obtain
(55)
W
1
=
1
D
2
log
⁡
⁡
2
D
2
y
+
2
D
2
D
2
y
2
+
D
1
0
δ
=
1
D
2
log
⁡
⁡
2
D
2
δ
+
2
D
2
D
2
δ
2
+
D
1
-
log
⁡
2
D
1
D
2
=
1
D
2
-
1
2
log
⁡
⁡
ξ
-
θ
0
+
O
log
⁡
δ
=
2
f
′
a
1
+
O
δ
-
1
2
log
⁡
⁡
ξ
-
θ
0
+
O
log
⁡
δ
=
2
f
′
a
-
1
2
log
⁡
ξ
-
θ
0
+
O
log
⁡
δ
.
We next calculate W
2. For 0
<
θ
<
a
-
δ, by (A.2) and (14), we have
(56)
F
ξ
-
F
θ
=
∫
θ
ξ
‍
f
s
d
s
=
∫
θ
a
‍
f
s
d
s
+
∫
a
ξ
‍
f
s
d
s
≥
∫
θ
a
‍
f
(
s
)
d
s
≥
C
f
′
a
a
-
θ
.
By this, we obtain
(57)
W
2
≤
∫
0
a
-
δ
‍
1
f
′
a
a
-
θ
d
s
≤
C
.
This along with (55) implies (50). Thus the proof is complete.Lemma 9.
Asξ
→
θ
0,
(58)
I
I
=
2
f
′
a
-
1
2
log
⁡
ξ
-
θ
0
+
O
log
⁡
δ
.Proof.
We have(59)
I
I
=
∫
a
θ
0
‍
1
F
ξ
-
F
θ
d
θ
=
∫
a
a
+
δ
‍
1
F
ξ
-
F
θ
d
θ
+
∫
a
+
δ
θ
0
‍
1
F
ξ
-
F
(
θ
)
d
θ
:
=
V
1
+
V
2
.
We first calculate V
1. By (14) and Taylor expansion, for a
<
θ
<
a
+
δ, we obtain
(60)
F
ξ
-
F
θ
=
∫
θ
ξ
‍
f
s
d
s
=
∫
θ
0
ξ
‍
f
s
d
s
+
∫
a
θ
0
‍
f
s
d
s
-
∫
a
θ
‍
f
s
d
s
=
f
θ
0
ξ
-
θ
0
1
+
O
ξ
-
θ
0
+
1
2
f
′
a
1
+
O
δ
a
-
θ
2
.
By this and (55), we obtain
(61)
V
1
=
∫
a
a
+
δ
‍
1
F
ξ
-
F
θ
d
θ
=
W
1
=
1
f
′
a
-
1
2
log
⁡
ξ
-
θ
0
+
O
log
⁡
δ
.
We next calculate V
2. Let an arbitrary 0
<
ϵ
≪
1 be fixed. Then,
(62)
V
2
=
∫
a
+
δ
θ
0
‍
1
F
ξ
-
F
θ
d
θ
=
∫
θ
0
-
ϵ
θ
0
‍
1
F
ξ
-
F
θ
d
θ
+
∫
a
+
δ
a
+
ϵ
‍
1
F
ξ
-
F
θ
d
θ
+
∫
a
+
ϵ
θ
0
-
ϵ
‍
1
F
ξ
-
F
(
θ
)
d
θ
:
=
V
21
+
V
22
+
V
23
.
We calculate V
21. For θ
0
-
ϵ
<
θ
<
θ
0, we obtain
(63)
F
ξ
-
F
θ
=
∫
θ
ξ
‍
f
s
d
s
=
∫
θ
0
ξ
‍
f
s
d
s
+
∫
θ
θ
0
‍
f
s
d
s
≥
∫
θ
θ
0
‍
f
s
d
s
≥
C
f
θ
0
θ
0
-
θ
.
By this,
(64)
V
21
≤
∫
θ
0
-
ϵ
θ
0
‍
1
C
f
θ
0
θ
0
-
θ
d
θ
≤
C
ϵ
.
We next calculate V
22. For a
+
δ
<
θ
<
a
+
ϵ,
(65)
F
ξ
-
F
θ
=
∫
a
ξ
‍
f
s
d
s
-
∫
a
θ
‍
f
s
d
s
≥
-
∫
a
θ
‍
f
s
d
s
≥
C
f
′
a
θ
-
a
.
By this, we obtain
(66)
V
22
≤
C
∫
a
+
δ
a
+
ϵ
‍
1
θ
-
a
d
θ
≤
C
ϵ
.
Finally, we calculate V
23. For a
+
ϵ
<
θ
<
θ
0
-
ϵ, by (14), we obtain
(67)
F
ξ
-
F
θ
=
∫
θ
0
ξ
‍
f
s
d
s
+
∫
θ
θ
0
‍
f
s
d
s
≥
∫
θ
θ
0
‍
f
s
d
s
≥
C
.
By this, we obtain
(68)
V
23
≤
∫
a
+
ϵ
θ
0
-
ϵ
‍
1
F
ξ
-
F
θ
d
θ
≤
C
.
By (62), (64), (66), and (68), we obtain
(69)
V
2
≤
C
.
By this, (59), and (61), we obtain (58). Thus, the proof is complete.Lemma 10.
Asξ
→
θ
0,
(70)
I
I
I
=
O
ξ
-
θ
0
1
/
2
.Proof.
By (A.3) and (14), we know that τ
<
θ
0
<
ξ. Then for θ
0
<
θ
<
ξ, we have
(71)
F
(
ξ
)
-
F
(
θ
)
=
∫
θ
ξ
‍
f
s
d
s
≥
C
f
θ
0
ξ
-
θ
.
By this, for λ
≫
1, we obtain
(72)
I
I
I
≤
C
∫
θ
0
ξ
‍
1
f
θ
0
ξ
-
θ
d
θ
=
2
f
θ
0
ξ
-
θ
0
.
Thus, the proof is complete.Proof of Theorem3(i).
Theorem3(i) follows directly from Lemmas 8–10. Thus, the proof is complete.Proof of Theorem3(ii).
By (37) and (49), we have
(73)
t
a
,
λ
=
∫
0
t
a
,
λ
‍
-
u
2
′
λ
,
t
2
λ
F
ξ
-
F
u
2
λ
,
t
d
t
=
1
2
λ
∫
a
ξ
‍
1
F
ξ
-
F
θ
d
θ
=
1
2
λ
∫
a
θ
0
‍
1
F
ξ
-
F
θ
d
θ
+
∫
θ
0
ξ
‍
1
F
ξ
-
F
θ
d
θ
:
=
1
2
λ
I
I
+
I
I
I
.
By this, Lemmas 8–10, and direct calculation, we obtain Theorem 3(ii). Thus, the proof is complete.Proof of Theorem3(iii).
By (37) and (49), we have
(74)
t
d
,
λ
=
1
2
λ
∫
d
ξ
‍
1
F
ξ
-
F
θ
d
θ
=
1
2
λ
∫
d
θ
0
‍
1
F
ξ
-
F
θ
d
θ
+
∫
θ
0
ξ
‍
1
F
(
ξ
)
-
F
θ
d
θ
=
1
2
λ
∫
d
θ
0
‍
1
F
θ
0
-
F
θ
d
θ
+
∫
θ
0
ξ
‍
1
F
(
ξ
)
-
F
θ
d
θ
+
∫
d
θ
0
‍
1
F
ξ
-
F
θ
-
1
F
θ
0
-
F
θ
d
θ
:
=
1
2
λ
C
d
+
I
I
I
+
J
.
We have only to calculate J. We fix a constant ϵ
>
0 sufficiently small. Then we have
(75)
J
:
=
J
1
+
J
2
=
∫
θ
0
-
ϵ
θ
0
F
(
θ
0
)
-
F
ξ
‍
×
F
ξ
-
F
θ
F
θ
0
-
F
θ
×
F
ξ
-
F
θ
+
F
θ
0
-
F
θ
-
1
d
θ
+
∫
d
θ
0
-
ϵ
‍
F
(
θ
0
)
-
F
(
ξ
)
×
F
ξ
-
F
θ
F
θ
0
-
F
θ
F
ξ
-
F
θ
+
F
θ
0
-
F
θ
×
F
ξ
-
F
θ
+
F
θ
0
-
F
θ
-
1
d
θ
.
By (A.2), for θ
0
<
ξ, we have
(76)
0
<
F
(
ξ
)
-
F
θ
0
≤
f
θ
0
ξ
-
θ
0
.
Further, since d
<
θ
<
θ
0
-
ϵ
<
θ
0
<
ξ, we have
(77)
F
ξ
-
F
θ
≥
C
,
F
(
θ
0
)
-
F
θ
≥
C
.
By this and (76), we obtain
(78)
J
2
≤
C
ξ
-
θ
0
.
We next calculate J
1. For θ
0
-
ϵ
<
θ
<
θ
0, we obtain
(79)
F
ξ
-
F
θ
≥
f
θ
0
ξ
-
θ
,
F
θ
0
-
F
θ
≥
f
θ
0
θ
0
-
θ
.
By (76) and (79), we obtain
(80)
J
1
≤
∫
θ
0
-
ϵ
θ
0
‍
F
ξ
-
F
θ
0
F
ξ
-
F
θ
F
θ
0
-
F
θ
d
θ
≤
C
ξ
-
θ
0
∫
θ
0
-
ϵ
θ
0
‍
1
ξ
-
θ
θ
0
-
θ
d
θ
=
C
ξ
-
θ
0
∫
0
ϵ
‍
1
ξ
-
θ
0
+
t
t
d
t
≤
C
ξ
-
θ
0
∫
0
ϵ
‍
1
x
2
+
ξ
-
θ
0
d
x
=
C
ξ
-
θ
0
tan
-
1
ϵ
ξ
-
θ
0
≤
C
ξ
-
θ
0
.
By Lemma 10, (74), (75), (78), and (80), we obtain Theorem 3(iii). Thus, the proof is complete.


## 4. Proof of Theorem4

Letξ
<
c and let ξ
→
c in this section. The proof of Theorem 4 is similar to that of Theorem 2. Let δ
:
=
δ
ξ defined in the statement of Theorem 4. By (38), we have
(81)
λ
=
1
2
∫
0
ξ
‍
1
F
ξ
-
F
θ
d
θ
=
1
2
∫
ξ
-
δ
ξ
‍
1
F
ξ
-
F
θ
d
θ

+

∫
0
ξ
-
δ
‍
1
F
ξ
-
F
θ
d
θ
:
=
1
2
(
K
1
+
K
2
)
.
By the same argument as that to obtain Z
1 in (41), if ξ is close to c, we obtain
(82)
K
1
=
2
f
′
ξ
(
-
log
⁡
f
(
ξ
)
+
O
(
log
⁡
δ
)
)
(
1
+
O
(
δ
)
)
.
Now, we calculate K
2. We have
(83)
K
2
=
1
2
∫
0
a
‍
1
F
ξ
-
F
(
θ
)
d
θ
+
∫
a
θ
0
‍
1
F
(
ξ
)
-
F
(
θ
)
d
θ
+
∫
θ
0
ξ
-
δ
‍
1
F
ξ
-
F
θ
d
θ
:
=
1
2
K
21
+
K
22
+
K
23
.
We calculate K
21. Since ξ
→
c, by (4) and (5), for 0
<
θ
<
a, we have F
(
ξ
)
-
F
(
θ
)
≥
C. By this, we obtain
(84)
K
21
≤
∫
0
a
‍
1
C
d
s
≤
C
.
We calculate K
22. Since ξ
→
c, by (4), for a
<
θ
<
θ
0, we have
(85)
F
ξ
-
F
θ
=
∫
θ
ξ
‍
f
s
d
s
=
∫
θ
θ
0
‍
f
s
d
s
+
∫
θ
0
ξ
‍
f
s
d
s
≥
∫
θ
0
ξ
‍
f
(
s
)
d
s
≥
C
.
This implies that K
22
≤
C. Finally, we calculate K
23. Let θ
0
<
θ
<
ξ
-
δ. By (6), we have |
f
′
(
θ
)
|
<
|
f
′
(
y
)
|
<
|
f
′
(
c
)
| for θ
<
y
<
c. By this and Taylor expansion, we obtain
(86)
F
ξ
-
F
θ
=
f
(
ξ
)
(
ξ
-
θ
)
+
1
2
f
′
ξ
θ
ξ
-
θ
2
≥
C
ξ
-
θ
2
,
where θ
<
ξ
θ
<
ξ. By this, we obtain
(87)
K
23
≤
∫
θ
0
ξ
-
δ
‍
1
C
ξ
-
θ
2
d
θ
≤
∫
θ
0
ξ
-
δ
‍
1
ξ
-
θ
d
θ
≤
C
-
log
⁡
δ
.
By (81), (82), (83), (87), and the same calculation as that in (48), we obtain Theorem 4. Thus, the proof is complete.


## 5. Proofs of Theorems5 and 6

We first characterizeξ
=
ξ
0. Let λ
(
ξ
) be the curve consisting of λ
2
(
ξ
) and λ
3
(
ξ
). We know that ξ
0 is determined by λ
′
(
ξ
0
)
=
0. By this and (38), we obtain
(88)
λ
ξ
ξ
=
1
2
∫
0
1
‍
1
F
ξ
-
F
ξ
s
d
s
-
ξ
2
2
∫
0
1
‍
f
(
ξ
)
-
s
f
(
ξ
s
)
F
ξ
-
F
ξ
s
3
/
2
d
s
.
By this, we see that 0
<
ξ
0
<
1 satisfies the following equation:
(89)
2
∫
0
1
‍
1
F
(
ξ
0
)
-
F
(
ξ
0
s
)
d
s
=
ξ
0
∫
0
1
‍
f
(
ξ
)
-
s
f
(
ξ
s
)
F
ξ
-
F
ξ
s
3
/
2
d
s
.Proof of Theorem6.
We study the asymptotic behavior ofλ
j
(
ξ
) (j
=
2,3) as ξ
→
ξ
0. We put η
=
ξ
-
ξ
0 and consider the case where |
η
|
≪
1. For 0
<
s
<
1, we put
(90)
G
(
ξ
,
s
)
:
=
F
(
ξ
)
-
F
(
ξ
s
)
-
(
F
(
ξ
0
)
-
F
(
ξ
0
s
)
)
.
Then,
(91)
F
ξ
-
F
ξ
s
=
F
ξ
0
-
F
ξ
0
s
+
G
ξ
,
s
=
F
ξ
0
-
F
ξ
0
s
1
+
G
ξ
,
s
F
ξ
0
-
F
ξ
0
s
:
=
F
ξ
0
-
F
ξ
0
s
1
+
H
ξ
,
s
η
.
We show that, for 0
<
s
<
1 and η sufficiently small,
(92)
H
ξ
,
s
≤
C
.
Let an arbitrary 0
<
ϵ
≪
1 be fixed. First, we consider the case 1
-
ϵ
<
s
<
1. By (A.2), we have
(93)
F
ξ
0
-
F
ξ
0
s
=
∫
ξ
0
s
ξ
0
‍
f
(
s
)
d
s
≥
f
(
ξ
0
)
ξ
0
(
1
-
s
)
.
We put K
(
ξ
,
s
)
=
F
(
ξ
)
-
F
(
ξ
s
). By mean value theorem,
(94)
G
ξ
,
s
=
K
ξ
,
s
-
K
ξ
0
,
s
≤
f
ξ
~
1
-
s
f
ξ
~
1
s
η
≤
f
ξ
~
1
-
f
ξ
~
1
s
η
+
f
ξ
~
1
s
η
1
-
s
≤
ξ
~
1
f
′
ξ
~
2
η
1
-
s
+
f
ξ
~
1
s
η
1
-
s
,
where ξ
0
~
ξ
~
1
~
ξ
~
2
~
ξ. By this and (93), we obtain (92). Next, let 0
<
s
≤
1
-
ϵ. Since ξ
0
>
θ
0 (cf. Figure 1), we have
(95)
F
ξ
0
-
F
ξ
0
s
≥
C
ϵ
>
0
.
Furthermore,
(96)
G
ξ
,
s
≤
F
ξ
-
F
ξ
0
+
F
ξ
s
-
F
ξ
0
s
≤
C
η
.
By this and (95), we obtain (92). For |
x
|
≪
1, we know that
(97)
1
+
x
-
1
/
2
=
1
+
∑
k
=
1
∞
‍
2
k
-
1
!
!
k
!
2
k
-
1
k
x
k
,
where (
2
k
-
1
)
!
!
=
(
2
k
-
1
)
(
2
k
-
3
)
⋯
3
·
1 for k
≥
1. By this, (38), and (91), for j
=
2,3, we obtain
(98)
λ
j
ξ
=
ξ
2
∫
0
1
‍
1
F
(
ξ
)
-
F
(
ξ
s
)
d
s
=
ξ
2
∫
0
1
‍
1
F
ξ
0
-
F
ξ
0
s
1
+
H
ξ
,
s
η
-
1
/
2
d
s
=
ξ
2
∫
0
1
‍
1
F
ξ
0
-
F
ξ
0
s
×
1
+
∑
k
=
1
∞
‍
2
k
-
1
!
!
k
!
2
k
-
1
k
H
ξ
,
s
k
η
k
d
s
=
ξ
0
2
∫
0
1
‍
1
F
ξ
0
-
F
(
ξ
0
s
)
+
η
2
∫
0
1
‍
1
F
(
ξ
0
)
-
F
(
ξ
0
s
)
+
ξ
0
+
η
2
∑
k
=
1
∞
‍
∫
0
1
‍
1
F
ξ
0
-
F
ξ
0
s
2
k
-
1
‼
k
!
2
k
×
-
1
k
H
ξ
,
s
k
η
k
d
s
=
λ
j
(
ξ
0
)
+
∑
k
=
1
∞
‍
H
k
(
ξ
)
η
k
.
By this, we obtain Theorem 6. Thus, the proof is complete.Proof of Theorem5.
We writeλ
=
λ
1
(
ξ
). For 0
<
s
<
1 and 0
<
ξ
≪
1, we put
(99)
D
(
ξ
,
s
)
:
=
F
ξ
-
F
(
ξ
s
)
-
ξ
f
ξ
1
-
s
ξ
2
f
ξ
1
-
s
2
.
By Taylor expansion,
(100)
F
(
ξ
)
-
F
(
ξ
s
)
-
ξ
f
(
ξ
)
(
1
-
s
)
~
-
ξ
2
f
′
(
0
)
2
1
-
s
2
.
By (99) and (100), for 0
<
s
<
1 and 0
<
ξ
≪
1, we obtain
(101)
F
(
ξ
)
-
F
(
ξ
s
)
=
ξ
f
(
ξ
)
1
-
s
1
+
D
(
ξ
,
s
)
ξ
(
1
-
s
)
,
D
ξ
,
s
≤
C
.
By this, (97), and (98), we obtain
(102)
λ
1
ξ
=
ξ
2
∫
0
1
‍
1
F
ξ
-
F
ξ
s
d
s
=
ξ
2
f
ξ
∫
0
1
‍
1
1
-
s
1
+
D
ξ
,
s
ξ
1
-
s
-
1
/
2
d
s
=
ξ
2
f
ξ
×
∫
0
1
‍
1
1
-
s
×
1
+
∑
k
=
1
∞
‍
2
k
-
1
‼
k
!
2
k
-
1
k
D
ξ
,
s
k
1
-
s
k
ξ
k
d
s
=
ξ
2
f
ξ
2
+
∑
k
=
1
∞
‍
L
k
ξ
ξ
k
,
where
(103)
L
k
(
ξ
)
=
∫
0
1
‍
2
k
-
1
!
!
k
!
2
k
-
1
k
D
ξ
,
s
k
1
-
s
k
-
1
/
2
d
s
.
Thus, we obtain Theorem 5(i). Theorem 5(ii) is obtained by Theorem 5(i) and Taylor expansion of f
(
u
) near u
=
0. Thus, the proof is complete.


## 6. Proof of Theorem7

Proof of Theorem7(i).
Letf
(
u
)
=
(
u
-
a
)
(
u
-
b
)
(
c
-
u
) with a
=
a
j
,
b
=
b
j
,
c
=
c
j (j
=
1,2). By (26) and Theorem 2, we obtain f
′
(
a
1
)
=
f
′
(
a
2
). This implies that
(104)
(
c
1
-
a
1
)
(
b
1
-
a
1
)
=
(
c
2
-
a
2
)
(
b
2
-
a
2
)
.
By direct calculation,
(105)
F
ξ
-
F
ξ
s
=
-
1
4
ξ
4
1
-
s
1
+
s
+
s
2
+
s
3
+
1
3
a
+
b
+
c
ξ
3
1
-
s
1
+
s
+
s
2
-
1
2
(
a
b
+
b
c
+
c
a
)
ξ
2
(
1
-
s
)
(
1
+
s
)
+
a
b
c
ξ
(
1
-
s
)
.
Let A
:
=
(
a
b
+
b
c
+
c
a
)
/
(
2
a
b
c
). By using (38), (105), and the calculation in Section 5, we obtain
(106)
λ
1
(
ξ
)
=
ξ
2
a
b
c
∫
0
1
‍
1
1
-
s
1
-
A
1
+
s
ξ
+
O
ξ
2
-
1
/
2
d
s
=
ξ
2
a
b
c
∫
0
1
‍
1
1
-
s
1
+
1
2
A
ξ
1
+
s
+
O
ξ
2
d
s
=
ξ
2
a
b
c
2
+
5
3
A
ξ
+
O
ξ
2
.
This implies that
(107)
λ
1
(
ξ
)
=
2
a
b
c
ξ
1
+
5
3
A
ξ
+
O
ξ
2
.
By this, (27), (28), and Theorem 2, we obtain
(108)
a
1
b
1
c
1
=
a
2
b
2
c
2
,
(109)
a
1
b
1
+
b
1
c
1
+
c
1
a
1
=
a
2
b
2
+
b
2
c
2
+
c
2
a
2
.
By (104), (108), (109), and direct calculation, we easily obtain a
1
=
a
2, b
1
=
b
2, and c
1
=
c
2. Thus, the proof is complete.Proof of Theorem7(ii).
By Theorem2 and (28), we obtain f
′
(
a
1
)
=
f
′
(
a
2
) and f
′
(
c
1
)
=
f
′
(
c
2
). By this, we obtain (104) and
(110)
(
c
1
-
a
1
)
(
c
1
-
b
1
)
=
(
c
2
-
a
2
)
(
c
2
-
b
2
)
>
0
.
By this and (104), we obtain (
c
1
-
a
1
)
2
=
(
c
2
-
a
2
)
2. By this and (104), we have
(111)
p
:
=
b
1
-
a
1
=
b
2
-
a
2
,
q
:
=
c
1
-
a
1
=
c
2
-
a
2
.
By (29) and (107), we obtain (108). By (108) and (111), we obtain
(112)
a
1
(
a
1
+
p
)
(
a
1
+
q
)
=
a
2
(
a
2
+
p
)
(
a
2
+
q
)
.
By this, we obtain a
1
=
a
2. By this and (111), we obtain b
1
=
b
2 and c
1
=
c
2. Thus, the proof is complete.Proof of Theorem7(iii).
Letθ
0
:
=
θ
1,0
=
θ
2,0. We put
(113)
G
(
t
)
:
=
∫
0
t
‍
y
(
y
-
p
)
(
q
-
y
)
d
y
.
For j
=
1,2, by (14) and (111), we obtain
(114)
0
=
∫
a
j
θ
0
‍
x
-
a
j
x
-
b
j
c
j
-
x
d
x
=
∫
a
j
θ
0
‍
x
-
a
j
x
-
a
j
+
p
a
j
+
q
-
x
d
x
=
∫
0
θ
0
-
a
j
‍
y
y
-
p
q
-
y
d
y
.
We study the shape of G
(
t
) for t
≥
0. Since c
+
a
>
2
b by (5), we have 2
p
<
q. By this and direct calculation, we see that G
(
0
)
=
0 and it decreases for 0
<
t
<
p and attains its minimum at t
=
p, and G
(
p
)
<
0. Then G
(
t
) increases for p
<
t
<
q and attains its maximum at t
=
q, and G
(
q
)
=
q
3
(
q
-
2
p
)
/
12
>
0. Then, G
(
t
) decreases for t
>
q and G
(
t
)
→
-
∞ as t
→
∞. Therefore, there exists p
<
t
1
<
q
<
t
2 such that G
(
t
1
)
=
G
(
t
2
)
=
0. We assume that a
1
<
a
2 and we derive a contradiction. In this case, by (114), we obtain θ
0
-
a
2
=
t
1 and θ
0
-
a
1
=
t
2. This implies that θ
0
=
a
1
+
t
2
>
a
1
+
q
=
c
1. This is a contradiction, since b
1
<
θ
0
<
c
1. If we assume that a
2
<
a
1, we obtain the same contradiction as above. Consequently, we obtain a
1
=
a
2. By this and (111), we obtain b
1
=
b
2 and c
1
=
c
2. Thus, the proof is complete.



---

*Source: 138629-2015-01-06.xml*