# Non-Single-Valley Solutions forp-Order Feigenbaum’s Type Functional Equation f

Authors: Min Zhang
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

This work deals with Feigenbaum’s functional equationf
(
φ
(
x
)
)
=
φ
p
(
f
(
x
)
), φ
(
0
)
=
1, 0
≤
φ
(
x
)
≤
1, x
∈
[
0
,
1
], where p
≥
2 is an integer, φ
p is the p-fold iteration of φ, and f
(
x
) is a strictly increasing continuous function on [
0
,
1
] that satisfies f
(
0
)
=
0, f
(
x
)
<
x, (
x
∈
(
0
,
1
]
). Using a constructive method, we discuss the existence of non-single-valley continuous solutions of the above equation.


---

## Body

## 1. Introduction

In 1978, Feigenbaum [1, 2] and independently Couliet and Tresser [3] introduced the notion of renormalization for real dynamical systems. In 1992, Sullivan [4] proved the uniqueness of the fixed point for the period-doubling renormalization operator. This fixed point of renormalization satisfies a functional equation known as the Cvitanović-Feigenbaum equation:
(1)
g
(
x
)
=
-
1
λ
g
(
g
(
-
λ
x
)
)
,
0
<
λ
<
1
,
g
(
0
)
=
1
,
-
1
≤
g
(
x
)
≤
1
,
x
∈
[
-
1,1
]
.As mentioned above, this equation and its solution play an important role in the theory initiated by Feigenbaum [1, 2]. However, it is difficult to find an exact solution of the above equation in general. This problem can be studied in classes of smooth functions or of continuous functions. In classes of smooth functions, the existence of smooth solutions for (1) has been established in [5–8] and references therein. As far as we know, continuous solutions of (1) in classes of continuous functions have been relatively little researched. In this direction, we refer to [9, 10]. In particular, Yang and Zhang [9] demonstrated the existence of a single-valley continuous solution for the following equation:
(2)
φ
(
x
)
=
1
λ
φ
(
φ
(
λ
x
)
)
,
0
<
λ
<
1
,
φ
(
0
)
=
1
,
0
≤
φ
(
x
)
≤
1
,
x
∈
[
0,1
]
,
which is called the second type of Feigenbaum’s functional equations. In the last years, a number of authors considered the more general equation
(3)
φ
(
x
)
=
1
λ
φ
p
(
λ
x
)
,
0
<
λ
<
1
,
φ
(
0
)
=
1
,
0
≤
φ
(
x
)
≤
1
,
x
∈
[
0,1
]
,
where p
≥
2 is an integer and φ
p is the p-fold iteration of φ. It is easy to see that (2) is a special case of (3). For p large enough, Eckmann et al. [11] showed that there exists a solution of (3) similar to the function φ
(
x
)
=
|
1
-
2
x
2
|. For any p
≥
2, Zhang et al. [12] and Liao et al. [13] proved that (3) has single-valley-extended continuous solutions.In the present paper, we will consider Feigenbaum’s functional equations(4)
f
(
φ
(
x
)
)
=
φ
p
(
f
(
x
)
)
,
φ
(
0
)
=
1
,
0
≤
φ
(
x
)
≤
1
,
x
∈
[
0,1
]
,
where f
(
x
) is a strictly increasing continuous function on [
0,1
] that satisfies f
(
0
)
=
0, f
(
x
)
<
x, (
x
∈
(
0,1
]
). We will prove the existence of single-valley-extended non-single-valley continuous solutions of (4) by the constructive method. Obviously, let f
(
x
)
=
λ
x; then (4) is (3).


## 2. Basic Definitions and Lemmas

In this section, we will give some characterizations of single-valley-extended non-single-valley continuous solutions of (4); they will be proved in the Appendix.Definition 1.
One callsφ a single-valley-extended continuous solution of (4) if (i)  φ is a continuous solution of (4); (ii) there exists α
∈
(
f
(
1
)
,
1
) such that φ is strictly decreasing on [
f
(
1
)
,
α
] and strictly increasing on [
α
,
1
].Definition 2.
One callsφ a single-valley-extended non-single-valley continuous solution of (4) if (i)  φ is a single-valley-extended continuous solution of (4); (ii)  φ has at least an extreme point on (
0
,
f
(
1
)
).In the following, we always letλ
=
f
(
1
)
=
φ
p
-
1
(
1
) and define the sets
(5)
Δ
k
=
[
f
k
+
1
(
1
)
,
f
k
(
1
)
]
,
Δ
k
1
=
[
f
k
(
α
)
,
f
k
(
1
)
]
,
Δ
k
2
=
[
f
k
+
1
(
1
)
,
f
k
(
α
)
]
,
Δ
k
α
=
[
f
k
+
1
(
α
)
,
f
k
(
α
)
]
,
f
k
+
1
(
α
)
,
f
k
(
α
)
∀
k
≥
0
.
Obviously,
(6)
Δ
k
=
Δ
k
1
∪
Δ
k
2
,
Δ
k
α
=
Δ
k
+
1
1
∪
Δ
k
2
,
and, from the fact that {
f
k
(
1
)
} and {
f
k
(
α
)
} are, respectively, strictly decreasing and lim
⁡
k
→
+
∞
f
k
(
1
)
=
0, lim
⁡
k
→
+
∞
f
k
(
α
)
=
0, then
(7)
[
0,1
]
=
⋃
k
=
0
+
∞
‍
Δ
k
=
⋃
k
=
0
+
∞
‍
(
Δ
k
1
∪
Δ
k
2
)
,
[
0
,
α
]
=
⋃
k
=
0
+
∞
‍
Δ
k
α
=
⋃
k
=
0
+
∞
‍
(
Δ
k
+
1
1
∪
Δ
k
2
)
.Lemma 3.
Suppose thatφ
(
x
) is a single-valley-extended non-single-valley continuous solution of (4) and α is the extreme point of φ in (
λ
,
1
). Then the following conclusions hold:(i)
φ
(
x
) has a unique minimum point α with φ
(
α
)
=
0;
(ii)
0 is a recurrent but not periodic point of φ;
(iii)
λ is an extreme point of φ and φ
(
λ
)
>
λ;
(iv)
φ
(
x
) has a unique fixed point β
=
φ
(
β
) on [
0,1
], and
(8)
φ
p
-
1
(
1
)
=
f
(
1
)
=
λ
<
β
<
α
;
(v)
forx
∈
[
0
,
λ
] and 0
≤
i
≤
p
-
1, then φ
i
(
x
)
=
α if and only if x
=
f
(
α
) and i
=
p
-
1.Lemma 4.
Suppose thatφ
(
x
) is a single-valley-extended non-single-valley continuous solution of (4). Let J
=
[
0
,
λ
], J
0
=
φ
(
J
), and J
i
=
φ
i
(
J
0
); then the following conclusions hold:(i)
J
0
,
J
1
,
…
,
J
p
-
2
⊂
(
λ
,
1
] are pairwise disjoint;
(ii)
for alli
=
0,1
,
…
,
p
-
2, then φ
i
:
J
0
↦
J
i is a homeomorphism.Lemma 5.
Suppose thatφ
(
x
) is a single-valley-extended non-single-valley continuous solution of (4). Then for all n
≥
1, φ has infinite many extreme points f
n
(
1
) and f
n
(
α
); f
n
(
1
) is the maximum point of φ on [
f
n
(
α
)
,
α
]; f
n
(
α
) is the minimum point of φ on [
0
,
f
n
(
1
)
].Lemma 6.
Suppose thatφ
(
x
) is a single-valley-extended non-single-valley continuous solution of (4). Then the equation φ
p
-
1
(
x
)
=
f
(
x
) has only one solution x
=
1 on (
φ
(
f
(
α
)
)
,
1
].Lemma 7.
Letφ
1, φ
2 be two single-valley-extended non-single-valley continuous solutions of (4). If
(9)
φ
1
(
x
)
=
φ
2
(
x
)
,
x
∈
[
λ
,
1
]
,
then φ
1
(
x
)
=
φ
2
(
x
) on [
0,1
].


## 3. Constructive Method of Solutions

In this section, we will prove constructively the existence of single-valley-extended non-single-valley continuous solutions of (4).Theorem 8.
Fix a strictly increasing continuous functionf
(
x
) on [
0,1
] with f
(
0
)
=
0, f
(
x
)
<
x
(
x
∈
(
0,1
]
). Denote f
(
1
)
=
λ. If φ
0
(
x
) is a continuous function on [
λ
,
1
] and satisfies the following conditions,(i)
there exists anα
∈
(
λ
,
1
) such that φ
0
(
α
)
=
0 and φ
0 is strictly decreasing on [
λ
,
α
] and strictly increasing on [
α
,
1
];
(ii)
φ
0
p
-
1
(
1
)
=
f
(
1
)
=
λ, φ
0
p
(
λ
)
=
f
(
φ
0
(
1
)
);
(iii)
there exists anα
0
∈
(
λ
,
1
] with φ
0
p
-
1
(
α
0
)
=
0, φ
0
(
λ
)
>
α
0; denote J
0
=
[
α
0
,
1
], J
i
=
φ
0
i
(
J
0
); then
(a)
J
0
,
J
1
,
…
,
J
p
-
2
⊂
(
λ
,
1
] are pairwise disjoint,
(b)
for alli
=
0,1
,
…
,
p
-
2, then φ
0
i
:
J
0
↦
J
i is a homeomorphism,
(c)
α is an endpoint of J
p
-
2 and α
=
φ
0
p
-
2
(
α
0
);
(iv)
the equationφ
0
p
-
1
(
x
)
=
f
(
x
) has only one solution x
=
1 on (
α
0
,
1
],then there exists a uniquely single-valley-extended non-single-valley continuous function φ
(
x
) satisfying the equation
(10)
f
(
φ
(
x
)
)
=
φ
p
(
f
(
x
)
)
,
x
∈
[
0,1
]
,
φ
(
x
)
=
φ
0
(
x
)
,
x
∈
[
λ
,
1
]
.
And φ has infinitely many extreme points. Conversely, if φ
0 is the restriction on [
λ
,
1
] of a single-valley-extended non-single-valley continuous solution to (4), then conditions (
i
)
–
(
i
v
) above must hold.Proof.
Suppose thatφ
0 satisfies conditions (i)–(iv). Define ψ
+
=
φ
0
p
-
1
|
[
α
0
,
1
]. By condition (iii), we have that ψ
+ is a homeomorphism. And by ψ
+
(
α
0
)
=
φ
0
p
-
1
(
α
0
)
=
0
≤
ψ
+
(
1
) we know that ψ
+ is strictly increasing.
Firstly, we defineφ on Δ
k by induction. Obviously, φ
=
φ
0 is well defined on Δ
0. Suppose that φ
(
x
) is well defined as φ
k
(
x
) on Δ
k and strictly increasing and decreasing, respectively, on Δ
k
1 and Δ
k
2 for all k
≤
m, where m
≥
0 is a certain integer. Let
(11)
φ
m
+
1
(
x
)
=
ψ
+
-
1
(
f
(
φ
m
(
f
-
1
(
x
)
)
)
)
,
(
x
∈
Δ
m
+
1
)
;
then φ
(
x
) is well defined as φ
m
+
1
(
x
) on Δ
m
+
1 and strictly increasing and decreasing, respectively, on Δ
m
+
1
1 and Δ
m
+
1
2. Thereby φ
(
x
) is well defined as a continuous function φ
k
(
x
) on Δ
k and strictly increasing and decreasing, respectively, on Δ
k
1 and Δ
k
2 for all k
≥
0 by induction. And
(12)
ψ
+
(
φ
k
+
1
(
f
(
x
)
)
)
=
f
(
φ
k
(
x
)
)
,
x
∈
Δ
k
.
Secondly, we prove thatφ
k and φ
k
+
1 have the same value on the common endpoint f
k
+
1
(
1
) of Δ
k and Δ
k
+
1 for all k
≥
0. From condition (ii) we have
(13)
ψ
+
(
φ
0
(
λ
)
)
=
φ
0
p
-
1
(
φ
0
(
λ
)
)
=
φ
0
p
(
λ
)
=
f
(
φ
0
(
1
)
)
.
And letting m
=
0, x
=
λ in (11) we get
(14)
φ
1
(
f
(
1
)
)
=
ψ
+
-
1
(
f
(
φ
0
(
1
)
)
)
=
φ
0
(
λ
)
=
φ
0
(
f
(
1
)
)
.
That is, φ
0 and φ
1 have the same value on the common endpoint f
(
1
)
=
λ of Δ
0 and Δ
1. Suppose that
(15)
φ
m
(
f
m
(
1
)
)
=
φ
m
-
1
(
f
m
(
1
)
)
,
where m
≥
1 is a certain integer. Let x
=
f
m
+
1
(
1
) in (11); then we have
(16)
φ
m
+
1
(
f
m
+
1
(
1
)
)
=
ψ
+
-
1
(
f
(
φ
m
(
f
m
(
1
)
)
)
)
=
ψ
+
-
1
(
f
(
φ
m
-
1
(
f
m
(
1
)
)
)
)
=
φ
m
(
f
m
+
1
(
1
)
)
.
That is, φ
k and φ
k
+
1 have the same value on the common endpoint f
k
+
1
(
1
) of Δ
k and Δ
k
+
1 for all k
≥
0 by induction. Therefore, we can let
(17)
φ
(
x
)
=
{
1
,
(
x
=
0
)
,
φ
k
(
x
)
,
(
x
∈
Δ
k
)
.
Since φ
k is continuous on Δ
k and increasing and decreasing, respectively, on Δ
k
1 and Δ
k
2 for k
≥
0 and (14), (15), and (16), we have that φ is a non-single-valley continuous function and has infinitely many extreme points on (
0,1
].
Thirdly, we prove thatφ is continuous at x
=
0 as follows. It is trivial that {
f
k
(
α
)
} is strictly decreasing and lim
⁡
k
→
∞
f
k
(
α
)
=
0. We prove {
φ
k
(
f
k
(
α
)
)
}
|
k
=
1
∞ is strictly increasing on [
α
0
,
1
] by induction as follows. Since f
(
φ
1
(
f
(
α
)
)
)
>
0 and from (11), we get
(18)
φ
2
(
f
2
(
α
)
)
=
ψ
+
-
1
(
f
(
φ
1
(
f
(
α
)
)
)
)
>
ψ
+
-
1
(
0
)
=
ψ
+
-
1
(
f
(
φ
0
(
α
)
)
)
=
φ
1
(
f
(
α
)
)
.
Suppose that φ
m
(
f
m
(
α
)
)
>
φ
m
-
1
(
f
m
-
1
(
α
)
), where m
≥
2 is a certain integer. Therefore, by (11) and the fact that ψ
+
-
1
∘
f is strictly increasing, we have that
(19)
φ
m
+
1
(
f
m
+
1
(
α
)
)
=
ψ
+
-
1
(
f
(
φ
m
(
f
m
(
α
)
)
)
)
>
ψ
+
-
1
(
f
(
φ
m
-
1
(
f
m
-
1
(
α
)
)
)
)
=
φ
m
(
f
m
(
α
)
)
.
Thereby, {
φ
k
(
f
k
(
α
)
)
}
|
k
=
1
∞ is strictly increasing in [
α
0
,
1
] by induction. Let lim
⁡
k
→
∞
φ
k
(
f
k
(
α
)
)
=
γ; then γ
∈
[
α
0
,
1
]. From (12), we have that
(20)
φ
0
p
-
1
(
φ
k
+
1
(
f
k
+
1
(
α
)
)
)
=
ψ
+
(
φ
k
+
1
(
f
k
+
1
(
α
)
)
)
=
f
(
φ
k
(
f
k
(
α
)
)
)
.
Let k
→
∞; we get φ
0
p
-
1
(
γ
)
=
f
(
γ
). By condition (iv), we know γ
=
1
=
φ
(
0
). This proves that φ is continuous at x
=
0. Thereby, φ is a continuous function on [
0,1
]. We have that φ
(
x
) satisfies (10) by (12) and φ
(
x
) is unique from Lemma 7.
Obviously, ifφ
0 is the restriction on [
λ
,
1
] of a single-valley-extended non-single-valley continuous solution to (4), then conditions (i)–(iv) must hold by the lemmas in Section 2.Example 9.
Letφ
0
(
x
)
:
[
1
/
4,1
]
↦
[
0,1
] be defined by
(21)
φ
0
(
x
)
=
{
-
13
8
x
+
39
32
,
(
1
4
≤
x
≤
3
4
)
,
x
-
3
4
,
(
3
4
≤
x
≤
1
)
.
Obviously, φ
0 satisfies the conditions of Theorem 8 with p
=
2, f
(
x
)
=
x
/
4, and λ
=
f
(
1
)
=
1
/
4, α
=
3
/
4. Hence it is the restriction to [
1
/
4,1
] of a single-valley-extended continuous solution φ to (4). Since φ
0 has the minimum point α
=
3
/
4 and φ
0
(
1
/
4
)
=
13
/
16
>
3
/
4, φ is a single-valley-extended non-single-valley continuous solution. Its graph is depicted in Figure 1.Figure 1
The graph of non-single-valley solution.Example 10.
Letφ
0
(
x
)
:
[
1
/
5,1
]
↦
[
0,1
] be defined by
(22)
φ
0
(
x
)
=
{
-
109
25
x
+
218
125
,
(
1
5
≤
x
≤
2
5
)
,
x
-
2
5
,
(
2
5
≤
x
≤
1
)
.
Obviously, φ
0 satisfies the conditions of Theorem 8 with p
=
3, f
(
x
)
=
x
2
/
5, and λ
=
f
(
1
)
=
1
/
5
,
α
=
2
/
5. Hence it is the restriction to [
1
/
5,1
] of a single-valley-extended continuous solution φ to (4). Since φ
0 has the minimum point α
=
2
/
5 and φ
0
(
1
/
5
)
=
109
/
125
>
2
/
5, φ
0
2
(
1
/
5
)
=
59
/
125
>
2
/
5, φ is a single-valley-extended non-single-valley continuous solution. Its graph is similar to Figure 1.



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*Source: 731863-2014-07-10.xml*