# Infinitely Many Eigenfunctions for Polynomial Problems: Exact Results

Authors: Yi-Chou Chen
Journal: Journal of Applied Mathematics
 (2015)

---

## Abstract

LetF
x
,
y
=
a
s
x
y
s
+
a
s
-
1
x
y
s
-
1
+
⋯
+
a
0
x be a real-valued polynomial function in which the degree s of y in F
x
,
y is greater than or equal to 1. For any polynomial y
x, we assume that T
:
R
x
→
R
x is a nonlinear operator with T
y
x
=
F
x
,
y
x. In this paper, we will find an eigenfunction y
x
∈
R
x to satisfy the following equation: F
x
,
y
x
=
a
y
x for some eigenvalue a
∈
R and we call the problem F
x
,
y
x
=
a
y
x a fixed point like problem. If the number of all eigenfunctions in F
x
,
y
x
=
a
y
x is infinitely many, we prove that (i) any coefficients of F
x
,
y
,
a
s
x
,
a
s
-
1
x
,
…
,
a
0
x, are all constants in R and (ii) y
x is an eigenfunction in F
x
,
y
x
=
a
y
x if and only if y
x
∈
R.


---

## Body

## 1. Introduction and Preliminaries

Lenstra [1] investigated that
(1)
F
x
,
y
x
=
0
in which F
(
x
,
y
) is a polynomial function over the algebraic rational number field Q
(
α
) (where α is an algebraic number). He found a polynomial y
=
y
(
x
)
∈
Q
(
α
)
[
x
] satisfying the polynomial equation
(2)
F
x
,
y
x
=
x
.
Further, Tung [2] extended (2) to solve polynomial solutions (near solutions) y
(
x
)
∈
K
[
x
] (K is a field) for the following equation:
(3)
F
x
,
y
x
=
a
x
m
,
where a
∈
K is a constant depending on the polynomial solution y
(
x
) and m
∈
N a given nonnegative integer.Moreover, Lai and Chen [3–5] extended (3) to solve y
(
x
)
∈
R
[
x
] satisfying the polynomial equation as the form
(4)
F
x
,
y
x
=
a
p
m
x
,
x
∈
R
,
where a
∈
R, m
∈
N, p
(
·
) is an irreducible polynomial in x
∈
R, and the polynomial function F
(
x
,
y
)
:
R
×
R
→
R is written by
(5)
F
(
x
,
y
)
=
∑
i
=
0
s
‍
a
i
x
y
i
with
s
≥
1
,
where s denotes the degree deg
⁡
y
F of y in F
(
x
,
y
).Recently, Chen and Lai [6, 7] research a quasicoincidence problem in which an arbitrary nonzero polynomial function f
(
x
)
∈
R
[
x
] is given as follows:
(6)
F
x
,
y
x
=
a
f
x
,
where a is a constant.Definition 1 (Chen and Lai, [6]).
A polynomial functiony
=
y
(
x
) satisfying (6) is called a quasicoincidence solution corresponding to some real number a. This number a is called a quasicoincidence value corresponding to the polynomial solutions y
=
y
(
x
).In this paper, we define a fixed point like problem in which they
(
x
)
∈
R
[
x
] is replaced by the arbitrary polynomial f
(
x
)
∈
R
[
x
] throughout this paper. Then we restate (6) as the following equation:
(7)
F
x
,
y
x
=
a
y
(
x
)
.
It is a new developed fixed point like problem. We call the polynomial (7) as afixed point like problem. The number of all eigenfunctions in (7) may be infinitely many, or finitely many, or not solvable.Since there may exist many eigenfunctions corresponding to the eigenvaluea, for convenience, we use the following notations to represent different situations:(1)
E
function, the set of all eigenfunctions “y
(
x
)” satisfying (7);
(2)
E
value, the set of all eigenvalues “a” satisfying (7);
(3)
E
function
(
a
), the set of all eigenfunctions “y
(
x
)” corresponding to a fixed eigenvalue “a”. For each a
∈
R, the cardinal number of E
function
(
a
), denoted by E
function
(
a
), satisfies the following condition:
(8)
E
function
(
a
)
≤
deg
⁡
y
F
(
x
,
y
)
.In Section2, we derive some properties of eigenfunctions of F
(
x
,
y
). If (7) has infinitely many eigenfunctions, the concerned properties are described in Section 3.Throughout the paper, we denote the polynomial function by(9)
F
x
,
y
=
a
s
x
y
s
+
a
s
-
1
x
y
s
-
1
+
⋯
+
a
1
x
y
+
a
0
x
=
∑
i
=
0
s
‍
a
i
(
x
)
y
i
with s
≥
1. Moreover, we may assume that a
0
(
x
) is nonzero. Since a
0
(
x
)
=
0, problem (7) may become
(10)
a
s
x
y
s
x
+
a
s
-
1
x
y
s
-
1
x
+
⋯
+
a
1
x
y
x
=
a
y
x
.
Moreover, if problem (7) has infinitely many eigenfunctions, dividing y
(
x
) by both sides of the above equation, then there may exist infinitely many nonzero eigenfunctions y
(
x
) satisfying
(11)
a
s
x
y
s
-
1
x
+
a
s
-
1
x
y
s
-
2
x
+
⋯
+
a
1
x
=
a
for some a
∈
R. Therefore, this problem becomes a special case of (3).


## 2. Some Lemmas and a Former Theorem

Throughout this paper, we consider (7) for the polynomial function (9).Lemma 2.
Lety
(
x
)
∈
E
function. Then
(12)
y
x
=
d
p
x
f
o
r
s
o
m
e
d
∈
R
,
and this p
(
x
) is divisible a
0
(
x
) and is denoted by p
(
x
)
∣
a
0
(
x
).Proof.
Sincey
(
x
)
∈
E
function, we have F
(
x
,
y
(
x
)
)
=
a
y
(
x
) for some a
∈
R. This means
(13)
a
s
(
x
)
y
s
(
x
)
+
a
s
-
1
(
x
)
y
s
-
1
(
x
)
+
⋯
+
a
1
(
x
)
y
(
x
)
+
a
0
(
x
)
=
a
y
(
x
)
for some a
∈
R. It leads to
(14)
y
(
x
)
a
s
x
y
s
-
1
x
+
a
s
-
1
x
y
s
-
2
x
+
⋯
+
a
1
x
-
a
=
-
a
0
x
;
then y
(
x
) is a factor of a
0
(
x
).In Lemma2, any eigenfunction is a factor p
(
x
) of a
0
(
x
). Thus we define a class of this factor as follows.Notation 1.
Letp
(
x
)
∈
R
[
x
], and we denote Φ
(
p
(
x
)
)
=
{
α
p
(
x
)
:
α
∈
R
}.According to Notation1, it is obvious that for any p
(
x
) in R
[
x
], we have the cardinal number
(15)
Φ
p
x
=
∞
.For convenience, we explain the relations ofE
function and Φ
(
p
(
x
)
) in the following lemma.Lemma 3.
Consider(16)
E
function
=
⋃
p
x
∣
a
0
x
⁡
Φ
p
x
∩
E
function
.Proof.
For anyy
(
x
)
∈
E
function, by Lemma 2, we have y
(
x
)
∣
a
0
(
x
). That is,
(17)
y
x
∈
Φ
p
x
for some factor p
(
x
) of a
0
(
x
). It follows that
(18)
E
function
⊆
⋃
p
x
∣
a
0
x
⁡
Φ
p
x
and we obtain
(19)
E
function
=
⋃
p
(
x
)
∣
a
0
(
x
)
⁡
Φ
p
x
∩
E
function
.We will use the definitions of “the pigeonhole principle,” which concert with Grimaldi [8] and the above relation can be explained as the following lemma.Lemma 4.
Suppose that the cardinal numberE
function
=
∞; there exists a factor p
(
x
) of a
0
(
x
) such that the cardinal number
(20)
Φ
p
x
∩
E
function
=
∞
.Proof.
By Lemma3, we obtain
(21)
∞
=
E
function
=
⋃
p
(
x
)
∣
a
0
(
x
)
⁡
Φ
p
x
∩
E
function
≤
∑
p
(
x
)
∣
a
0
(
x
)
‍
Φ
p
x
∩
E
function
.
Since the number of all factor p
(
x
) of a
0
(
x
) is at most 2
deg
⁡
a
0
(
x
), by pigeonhole’s principle, it yields
(22)
Φ
p
x
∩
E
function
=
∞
for some factor p
(
x
) of a
0
(
x
).In order to solve the problem (7), [6, Lemma 3 and Theorem 11] are needed as follows.Lemma 5 (see [6, Lemma 3]).
Assume that the number of all quasicoincidence solutions (defined in Definition1) is infinitely many; then, for any two quasicoincidence solutions y
1
(
x
) and y
2
(
x
), we have
(23)
y
1
x
-
y
2
x
=
λ
g
x
for some constant λ
∈
R and some factor g
(
x
) of f
(
x
).Theorem 6 (see [6, Theorem 11]).
Assume that the number of all quasicoincidence solutions (defined in Definition1) is infinitely many; then
(24)
F
x
,
y
=
∑
i
=
0
s
‍
c
i
f
x
g
i
x
y
-
y
1
x
i
for some y
1
(
x
)
∈
R
[
x
], g
(
x
) is a factor of f
(
x
), and c
i
∈
R for i
=
0,1
,
…
,
s.


## 3. Main Theorems

In this section, we considerF
(
x
,
y
)
=
a
y
(
x
) for the polynomial function F
(
x
,
y
) defined in (9).We investigate the fixed point like problem of simple polynomial functionsF
(
x
,
y
) with s
=
1 at first. Theorems 7 and 8 describe the necessary and sufficient results of these simple functions.Theorem 7.
LetF
(
x
,
y
) be a polynomial function with deg
⁡
y
F
=
1 as the form F
(
x
,
y
)
=
a
1
(
x
)
y
+
a
0
(
x
) for some a
1
(
x
), a
0
(
x
)
∈
R
[
x
]. If the cardinal number E
function
=
∞, then (i)
a
1
(
x
)
∈
R;
(ii)
any eigenfunctiony
(
x
) of (7) is of the form
(25)
y
(
x
)
=
λ
a
0
(
x
)
for someλ
∈
R.Proof.
SinceE
function
=
∞, by Lemma 4, there exists a factor p
(
x
) of a
0
(
x
) such that
(26)
Φ
p
x
∩
E
function
=
∞
.
There exist two different eigenfunctions y
1
(
x
)
,
y
2
(
x
)
∈
Φ
(
p
(
x
)
)
∩
E
function with
(27)
y
1
x
=
α
1
p
x
,
y
2
x
=
α
2
p
x
,
for different constants α
1
,
α
2
∈
R. Since y
1
(
x
)
,
y
2
(
x
)
∈
E
function, we have
(28)
F
x
,
y
1
x
=
a
y
1
x
,
F
x
,
y
2
x
=
b
y
2
x
,
where a, b
∈
E
value. It follows that
(29)
F
x
,
y
1
x
=
a
1
x
y
1
x
+
a
0
x
=
a
y
1
x
,
F
x
,
y
2
x
=
a
1
x
y
2
x
+
a
0
x
=
b
y
2
x
.
By (27) and (29), we have
(30)
a
1
x
α
1
p
x
+
a
0
x
=
a
α
1
p
x
,
a
1
x
α
2
p
x
+
a
0
x
=
b
α
2
p
x
.
By (30), we get
(31)
a
1
x
α
1
p
x
-
α
2
p
x
=
a
α
1
p
x
-
b
α
2
p
x
.
Since α
1
≠
α
2 and p
(
x
) is nonzero, it follows that
(32)
a
1
(
x
)
=
a
α
1
-
b
α
2
α
1
-
α
2
∈
R
.
For any y
(
x
)
∈
E
function, we have
(33)
F
x
,
y
x
=
a
1
x
y
x
+
a
0
x
=
a
y
x
.
By (32), we let a
1
(
x
)
=
a
1
~
∈
R, (33) becomes
(34)
a
1
~
y
x
+
a
0
x
=
a
y
x
,
and it follows that
(35)
a
0
x
=
a
-
a
1
~
y
x
.
Owing to a
0
(
x
)
≠
0, then we obtain
(36)
y
x
=
λ
~
a
0
x
,
where λ
~
=
1
/
a
-
a
1
~.The following theorem is the sufficient conditions of Theorem7.Theorem 8.
LetF
(
x
,
y
) be a polynomial function with deg
⁡
y
F
=
1 as the form F
(
x
,
y
)
=
a
1
(
x
)
y
+
a
0
(
x
) for some a
1
(
x
), a
0
(
x
)
∈
R
[
x
]. If (i)
a
1
(
x
)
∈
R,
(ii)
any eigenfunctiony
(
x
) of (7) is of the form
(37)
y
x
=
λ
a
0
x
,
for someλ
∈
R, then E
function
=
∞.Proof.
By (i), we leta
1
(
x
)
=
a
1
∈
R, then F
(
x
,
y
(
x
)
)
=
a
y
(
x
) for some a
∈
R. This implies
(38)
a
1
y
x
+
a
0
x
=
a
y
x
and then y
(
x
)
=
1
/
a
-
a
1
a
0
(
x
) is an eigenfunction of (7) for any constant a
≠
a
1. It follows that
(39)
∞
=
1
a
-
a
1
a
0
(
x
)
:
a
∈
R
-
a
1
≤
E
function
;
then E
function
=
∞.In Theorems7 and 8, problem (7) with deg
⁡
y
F
=
1 is introduced. In the following theorems, we deal with (7) with deg
⁡
y
F
≥
2 when the number of all eigenfunctions is infinitely many.Theorem 9.
Suppose that the cardinal numberE
function
=
∞ and deg
⁡
y
⁡
F
(
x
,
y
)
≥
2. Then the polynomial F
(
x
,
y
) can be represented as
(40)
F
x
,
y
=
∑
i
=
0
s
‍
β
i
y
i
for some constants β
i
∈
R.Proof.
SinceE
function
=
∞, by Lemma 4, there exists a factor p
(
x
) of a
0
(
x
) satisfying
(41)
Φ
p
x
∩
E
function
=
∞
.
Let y
1
(
x
) be an eigenfunction in Φ
(
p
(
x
)
)
∩
E
function such that
(42)
F
x
,
y
1
x
=
a
1
y
1
(
x
)
for some eigenvalue a
1
∈
R. By Remainder Theorem, we get
(43)
F
x
,
y
=
y
-
y
1
x
F
1
x
,
y
+
a
1
y
1
x
,
where F
1
(
x
,
y
) is the quotient and a
1
y
1
(
x
) is the remainder.
From the above identity and considering any eigenfunctiony
(
x
) in Φ
(
p
(
x
)
)
∩
E
function
/
{
y
1
(
x
)
} with F
(
x
,
y
(
x
)
)
=
a
y
(
x
), we substitute (43) by taking y
=
y
(
x
) above and it becomes
(44)
a
y
x
=
F
x
,
y
x
=
y
x
-
y
1
x
F
1
x
,
y
x
+
a
1
y
1
x
.
Since y
1
(
x
), y
(
x
)
∈
Φ
(
p
(
x
)
), we have
(45)
y
1
x
=
λ
1
p
x
,
(46)
y
(
x
)
=
λ
p
(
x
)
for some different constants λ
1 and λ. Substituting (45) and (46) into (44), it becomes
(47)
a
λ
p
x
=
λ
p
x
-
λ
1
p
x
F
1
x
,
y
x
+
a
1
λ
1
p
x
and it leads to
(48)
F
1
x
,
y
x
=
a
λ
-
a
1
λ
1
λ
-
λ
1
∈
R
for any eigenfunction y
(
x
)
∈
Φ
(
p
(
x
)
)
∩
E
function
/
{
y
1
(
x
)
}.
By (48), there exist infinitely many quasicoincidence solutions in Φ
(
p
(
x
)
)
∩
E
function
/
{
y
1
(
x
)
} to satisfy
(49)
F
1
x
,
y
=
a
f
x
with f
(
x
)
=
1. This problem is a quasicoincidence problem; then by Theorem 6, we have
(50)
F
1
x
,
y
=
∑
i
=
0
s
-
1
‍
c
i
f
x
g
i
x
y
-
y
1
x
i
.
Moreover, since f
(
x
)
=
1 and f
(
x
)
/
g
i
(
x
)
∈
R
[
x
] for any i
=
0,1
,
2
,
…
,
s
-
1, it implies that g
(
x
)
∈
R and by Lemma 5, any y
2
x
,
y
3
(
x
)
∈
Φ
(
p
(
x
)
)
∩
E
function
/
{
y
1
(
x
)
}, we have
(51)
y
2
(
x
)
-
y
3
(
x
)
=
d
g
(
x
)
=
d
′
∈
R
.
By definitions of Φ
(
p
(
x
)
), y
2
(
x
), and y
3
(
x
) can also be represented as
(52)
y
2
x
=
λ
2
p
x
,
y
3
(
x
)
=
λ
3
p
(
x
)
for some λ
2
,
λ
3
∈
R. By (51), it follows that
(53)
y
2
x
-
y
3
x
=
λ
2
-
λ
3
p
x
∈
R
.
Moreover, by (53), this implies that p
(
x
)
∈
R and by (45), y
1
(
x
)
∈
R, say, y
1
(
x
)
=
b
1 and (50) implies that
(54)
F
1
x
,
y
=
∑
i
=
0
s
-
1
‍
c
i
y
-
b
1
i
=
∑
i
=
0
s
-
1
‍
d
i
y
i
for some d
i
∈
R, i
=
0,1
,
2
,
…
,
s
-
1.
By (54), (43) implies that
(55)
F
x
,
y
=
y
-
y
1
x
F
1
x
,
y
+
a
1
y
1
x
=
y
-
b
1
∑
i
=
0
s
-
1
d
i
y
i
+
a
1
b
1
=
∑
i
=
0
s
β
i
y
i
.Conversely, ifF
(
x
,
y
) can be expressed as in Theorem 9, then the cardinal number E
function
=
∞; this problem becomes the sufficient conditions of Theorem 9.Theorem 10.
Assume that(56)
F
x
,
y
=
∑
i
=
0
s
‍
β
i
y
i
for some β
i
∈
R for i
=
0,1
,
…
,
s; then
(57)
|
E
function
|
=
∞
.Proof.
For anyy
(
x
)
=
c
∈
R,
(58)
F
x
,
y
x
=
∑
i
=
0
s
c
i
y
i
x
=
∑
i
=
0
s
‍
c
i
c
i
(
this
is
a
constant
)
=
a
c
,
for some a
=
∑
i
=
0
s
c
i
c
i
/
c
∈
R. Then R
⊆
E
function and then |
E
function
|
=
∞.In fact, ifE
function
=
∞, then E
function
=
R and we prove it as follows.Theorem 11.
If|
E
function
|
=
∞, we have
(59)
E
function
=
R
.Proof.
SinceE
function
=
∞, by the proof of Theorem 10, we have
(60)
R
⊆
E
function
.
Conversely, considering anyy
(
x
)
∈
E
function, we have
(61)
F
x
,
y
x
=
∑
i
=
0
s
‍
β
i
y
i
x
=
a
y
x
for some eigenvalue a
∈
R. By Lemma 2, we have y
(
x
)
∣
β
0; this implies y
(
x
)
∈
R and then E
function
⊆
R. This proof is completed.From Theorems9 and 11, we can easily obtain the following two corollaries.Corollary 12.
LetF
(
x
,
y
)
=
∑
i
=
0
s
‍
a
i
(
x
)
y
i
∈
R
[
x
,
y
], s
≥
2, with a
j
(
x
)
∉
R for some j, then |
E
function
|
<
∞.Proof.
This result can be immediately obtained from Theorem9.Corollary 13.
If there exists an eigenfunctiony
(
x
)
∈
E
function with y
(
x
)
∉
R, then |
E
function
|
<
∞.Proof.
This result can be immediately obtained from Theorem11.From Theorems7, 8, and 10 and Corollary 12, we provide some examples of fixed point like problem (7) for some a
∈
R, which have infinitely many eigenfunctions and do not have infinitely many eigenfunctions as follows.Example 14.
In the following examples, we by the form ofF
(
x
,
y
), we can decide whether the number of all eigenfunctions in (7) is infinitely many or not.(1)
If(
x
,
y
)
=
x
y
+
1, there do not exist infinitely many eigenfunctions (Theorem 7).
(2)
IfF
(
x
,
y
)
=
x
y
+
x, there do not exist infinitely many eigenfunctions (Theorem 7).
(3)
IfF
(
x
,
y
)
=
y
+
x, there exist infinitely many eigenfunctions and
(62)
E
function
=
{
λ
x
:
λ
∈
R
}
(
Theorem
8
)
.
(4)
IfF
(
x
,
y
)
=
-
y
2
+
7
y
+
1, there exist infinitely many eigenfunctions and
(63)
E
function
=
R
(
Theorem
10
)
.
(5)
IfF
(
x
,
y
)
=
x
y
2
+
x
y
+
1, there do not exist infinitely many eigenfunctions (Corollary 12).
(6)
IfF
(
x
,
y
)
=
∑
i
=
0
s
‍
c
i
y
i
+
x, s
≥
2, for any constants c
i
∈
R, there do not exist infinitely many eigenfunctions (Corollary 12).We would like to provide one open problem as follows.Further Development. For a real-valued polynomial function F
:
R
×
R
→
R, if |
E
function
|
<
∞, can we find a co-NP complete algorithm to solve all eigenfunctions y
=
y
(
x
) satisfying (7)?



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*Source: 516159-2015-02-23.xml*