Determination of the 3D Navier-Stokes equations with damping

Wei Shi; Xinguang Yang; Xingjie Yan; Wei Shi; Xinguang Yang; Xingjie Yan

doi:10.3934/era.2022197

Electronic Research Archive

2022, Volume 30, Issue 10: 3872-3886. doi: 10.3934/era.2022197

Previous Article Next Article

Research article Special Issues

Determination of the 3D Navier-Stokes equations with damping

1.
Department of Mathematics and Information Science, Henan Normal University, Xinxiang 453007, China
2.
Department of Mathematics, China University of Mining and Technology, Xuzhou 221008, China

Academic Editor: Baowei Feng

Received: 14 July 2022 Revised: 15 August 2022 Accepted: 21 August 2022 Published: 25 August 2022

This paper is concerned the determination of trajectories for the three-dimensional Navier-Stokes equations with nonlinear damping subject to periodic boundary condition. By using the energy estimate of Galerkin approximated equation, the finite number of determining modes and asymptotic determined functionals have been shown via the Grashof numbers for the non-autonomous and autonomous damped Navier-Stokes fluid flow respectively.

Keywords:

Citation: Wei Shi, Xinguang Yang, Xingjie Yan. Determination of the 3D Navier-Stokes equations with damping[J]. Electronic Research Archive, 2022, 30(10): 3872-3886. doi: 10.3934/era.2022197

Related Papers:

[1]	Tingting Du, Zhengang Wu . Some identities of the generalized bi-periodic Fibonacci and Lucas polynomials. AIMS Mathematics, 2024, 9(3): 7492-7510. doi: 10.3934/math.2024363
[2]	Tingting Du, Zhengang Wu . Some identities involving the bi-periodic Fibonacci and Lucas polynomials. AIMS Mathematics, 2023, 8(3): 5838-5846. doi: 10.3934/math.2023294
[3]	Hong Kang . The power sum of balancing polynomials and their divisible properties. AIMS Mathematics, 2024, 9(2): 2684-2694. doi: 10.3934/math.2024133
[4]	Utkal Keshari Dutta, Prasanta Kumar Ray . On the finite reciprocal sums of Fibonacci and Lucas polynomials. AIMS Mathematics, 2019, 4(6): 1569-1581. doi: 10.3934/math.2019.6.1569
[5]	Waleed Mohamed Abd-Elhameed, Omar Mazen Alqubori . New expressions for certain polynomials combining Fibonacci and Lucas polynomials. AIMS Mathematics, 2025, 10(2): 2930-2957. doi: 10.3934/math.2025136
[6]	Kritkhajohn Onphaeng, Prapanpong Pongsriiam . Exact divisibility by powers of the integers in the Lucas sequence of the first kind. AIMS Mathematics, 2020, 5(6): 6739-6748. doi: 10.3934/math.2020433
[7]	Ala Amourah, B. A. Frasin, G. Murugusundaramoorthy, Tariq Al-Hawary . Bi-Bazilevič functions of order $ \vartheta +i\delta $ associated with $ (p, q)- $ Lucas polynomials. AIMS Mathematics, 2021, 6(5): 4296-4305. doi: 10.3934/math.2021254
[8]	Can Kızılateş, Halit Öztürk . On parametric types of Apostol Bernoulli-Fibonacci, Apostol Euler-Fibonacci, and Apostol Genocchi-Fibonacci polynomials via Golden calculus. AIMS Mathematics, 2023, 8(4): 8386-8402. doi: 10.3934/math.2023423
[9]	Abdulmtalb Hussen, Mohammed S. A. Madi, Abobaker M. M. Abominjil . Bounding coefficients for certain subclasses of bi-univalent functions related to Lucas-Balancing polynomials. AIMS Mathematics, 2024, 9(7): 18034-18047. doi: 10.3934/math.2024879
[10]	Waleed Mohamed Abd-Elhameed, Amr Kamel Amin, Nasr Anwer Zeyada . Some new identities of a type of generalized numbers involving four parameters. AIMS Mathematics, 2022, 7(7): 12962-12980. doi: 10.3934/math.2022718

Abstract

1. Introduction

Fibonacci polynomials and Lucas polynomials are important in various fields such as number theory, probability theory, numerical analysis, and physics. In addition, many well-known polynomials, such as Pell polynomials, Pell Lucas polynomials, Tribonacci polynomials, etc., are generalizations of Fibonacci polynomials and Lucas polynomials. In this paper, we extend the linear recursive polynomials to nonlinearity, that is, we discuss some basic properties of the bi-periodic Fibonacci and Lucas polynomials.

The bi-periodic Fibonacci $\left \{ f_{n} \left (t\right) \right \}$ and Lucas $\left \{ l_{n} \left (t \right) \right \}$ polynomials are defined recursively by

$\begin{equation*} f_{0}\left (t \right ) = 0, \quad f_{1}\left (t\right ) = 1, \quad {f_{n}\left ( t\right ) = } \begin{cases} ayf_{n-1}\left ( t \right ) +f_{n-2}\left ( t \right ) \quad & n \equiv 0 \pmod{2}, \\ byf_{n-1}\left (t \right ) +f_{n-2}\left (t \right ) \quad & n \equiv 1 \pmod{2}, \end{cases} \quad n\ge 2, \end{equation*}$

and

$\begin{equation*} l_{0}\left ( t \right ) = 2, \quad l_{1}\left (t \right ) = at, \quad {l_{n}\left ( t \right ) = } \begin{cases} byl_{n-1}\left ( t \right ) +l_{n-2}\left ( t \right ) \quad & n \equiv 0 \pmod{2}, \\ ayl_{n-1}\left ( t \right ) +l_{n-2}\left ( t \right ) \quad & n \equiv 1 \pmod{2}, \end{cases} \quad n\ge 2, \end{equation*}$

where $a$ and $b$ are nonzero real numbers. For $t = 1$ , the bi-periodic Fibonacci and Lucas polynomials are, respectively, well-known bi-periodic Fibonacci $\left \{f_{n} \right \}$ and Lucas $\left \{l_{n} \right \}$ sequences. We let

$\begin{equation*} {\varsigma \left ( n \right ) = } \begin{cases} 0 \quad & n \equiv 0 \pmod{2}, \\ 1 \quad & n \equiv 1 \pmod{2}, \end{cases} \quad n\ge 2. \end{equation*}$

In ^[1], the scholars give the Binet formulas of the bi-periodic Fibonacci and Lucas polynomials as follows:

$\begin{equation} f_{n}\left (t \right ) = \frac{a^{\varsigma \left (n+1 \right) }}{ \left( ab \right )^{\left \lfloor \frac{n}{2} \right \rfloor } }\left (\frac{\sigma^{n} \left ( t\right ) -\tau ^{n}\left ( t \right ) }{\sigma\left (t \right ) -\tau \left ( t \right ) }\right), \end{equation}$

(1.1)

and

$\begin{equation} l_{n}\left (t \right ) = \frac{a^{\varsigma \left (n \right) }}{ \left( ab \right )^{\left \lfloor \frac{n+1}{2} \right \rfloor } }\left (\sigma ^{n}\left (t\right ) +\tau ^{n}\left ( t\right ) \right), \end{equation}$

(1.2)

where $n\ge 0$ , $\sigma\left (t \right)$ , and $\tau \left (t \right)$ are zeros of $\lambda ^{2}-abt\lambda -ab$ . This is $\sigma \left (t \right) = \frac{abt +\sqrt{a^{2} b^{2} t^{2}+4ab} }{2}$ and $\tau \left (t \right) = \frac{abt -\sqrt{a^{2} b^{2} t^{2}+4ab} }{2}$ . We note the following algebraic properties of $\sigma\left (t \right)$ and $\tau \left (t \right)$ :

$\begin{equation*} \sigma \left ( t \right ) +\tau \left (t \right ) = abt, \quad \sigma \left (t \right ) -\tau \left ( t \right ) = \sqrt{a^{2} b^{2} t^{2} +4ab}, \quad \sigma\left ( t \right ) \tau \left ( t\right ) = -ab. \end{equation*}$

Many scholars studied the properties of bi-periodic Fibonacci and Lucas polynomials; see ^[2,3,4,5,6]. In addition, many scholars studied the power sums problem of second-order linear recurrences and its divisible properties; see ^[7,8,9,10].

Taking $a = b = 1$ and $t = 1$ , we obtain the Fibonacci $\left \{ F_{n} \right \}$ or Lucas $\left \{ L_{n} \right \}$ sequence. Melham ^[11] proposed the following conjectures:

Conjecture 1. Let $m\ge 1$ be an integer, then the sum

$\begin{equation*} L_{1} L_{3} L_{5}\cdots L_{2m+1}\sum\limits_{k = 1}^{n} F_{2k}^{2m+1} \end{equation*}$

can be represented as $\left (F_{2n+1}-1 \right)^{2}R_{2m-1}\left (F_{2n+1} \right)$ , including $R_{2m-1}\left (t \right)$ as a polynomial with integer coefficients of degree $2m-1$ .

Conjecture 2. Let $m\ge 1$ be an integer, then the sum

$\begin{equation*} L_{1} L_{3} L_{5}\cdots L_{2m+1}\sum\limits_{k = 1}^{n} L_{2k}^{2m+1} \end{equation*}$

can be represented as $\left (L_{2n+1}-1 \right)Q_{2m}\left (L_{2n+1} \right)$ , where $Q_{2m}\left (t \right)$ is a polynomial with integer coefficients of degree $2m$ .

In ^[12], the authors completely solved the Conjecture $2$ and discussed the Conjecture $1$ . Using the definition and properties of bi-periodic Fibonacci and Lucas polynomials, the power sums problem and their divisible properties are studied in this paper. The results are as follows:

Theorem 1. We get the identities

$\begin{align} & \sum\limits_{k = 1}^{n}f_{2k}^{2m+1} \left (t\right ) = \frac{a^{2m+1} }{b\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits _{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m+1}{m-j}\left ( \frac{f_{\left ( 2n+1 \right )\left ( 2j+1\right ) }\left ( t \right ) - f_{2j+1}\left ( t \right ) }{l_{2j+1} \left ( t \right ) } \right ), \end{align}$

(1.3)

$\begin{align} & \sum\limits_{k = 1}^{n}f_{2k+1}^{2m+1} \left ( t \right ) = \frac{\left ( ab \right ) ^{m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits_{j = 0}^{m} \binom{2m+1}{m-j} \left ( \frac{f_{\left ( 2n+2\right )\left (2j+1 \right ) }\left (t\right ) -f_{2\left ( 2j+1 \right ) }\left (t\right ) }{l_{2j+1}\left ( t \right ) } \right ), \end{align}$

(1.4)

$\begin{align} & \sum\limits_{k = 1}^{n}l_{2k}^{2m+1} \left (t \right ) = \sum\limits_{j = 0}^{m} \binom{2m+1}{m-j} \left ( \frac{l_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left (t \right ) -l_{2j+1}\left ( t\right ) }{l_{2j+1}\left ( t\right ) } \right ), \end{align}$

(1.5)

$\begin{align} &\sum\limits_{k = 1}^{n}l_{2k+1}^{2m+1} \left (t\right ) = \frac{ a ^{m+1} }{b^{m+1} } \sum\limits_{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m+1}{m-j} \left ( \frac{l_{\left ( 2n+2\right )\left ( 2j+1\right ) }\left ( t \right ) -l_{2\left ( 2j+1 \right ) }\left ( t \right ) }{l_{2j+1}\left ( t \right ) } \right ), \end{align}$

(1.6)

where $n$ and $m$ are positive integers.

Theorem 2. We get the identities

$\begin{align} & \sum\limits_{k = 1}^{n}f_{2k}^{2m} \left ( t \right ) = \frac{a^{2m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits_{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m}{m-j} \frac{f_{2j\left ( 2n+1 \right ) }\left (t \right ) }{f_{2j}\left (t\right ) } -\frac{a^{2m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} }\binom{2m}{m}\left ( -1 \right )^{m}\left ( n+ \frac{1}{2 } \right ) , \end{align}$

(1.7)

$\begin{align} & \sum\limits_{k = 1}^{n}f_{2k+1}^{2m} \left ( t \right ) = \frac{\left ( ab \right ) ^{m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits_{j = 0}^{m} \binom{2m}{m-j} \left ( \frac{f_{2j\left ( 2n+2\right ) }\left ( t \right ) -f_{4j }\left ( t \right ) }{f_{2j}\left ( t \right ) } \right )-\frac{\left ( ab \right ) ^{m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} }\binom{2m}{m}n , \end{align}$

(1.8)

$\begin{align} & \sum\limits_{k = 1}^{n}l_{2k}^{2m} \left ( t \right ) = \sum\limits_{j = 0}^{m} \binom{2m}{m-j} \frac{f_{2j\left ( 2n+1 \right ) }\left ( t \right ) }{l_{2j+1}\left (t \right ) }-2^{2m-1}-\binom{2m}{m}\left ( n+\frac{1}{2} \right ) , \end{align}$

(1.9)

$\begin{align} &\sum\limits_{k = 1}^{n}l_{2k+1}^{2m} \left ( t \right ) = \frac{ a ^{m} }{b^{m} } \sum\limits_{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m}{m-j} \left ( \frac{f_{2j\left ( 2n+2\right ) }\left ( t \right ) -f_{4j }\left (t \right ) }{f_{2j}\left ( t \right ) } \right ) -\frac{a^{m} }{b^{m} } \binom{2m}{m} \left ( -1 \right )^{m}n , \end{align}$

(1.10)

where $n$ and $m$ are positive integers.

As for application of Theorem 1, we get the following:

Corollary 1. We get the congruences:

$\begin{equation} bl_{1} \left ( t \right ) l_{3} \left ( t \right ) \cdots l_{2m+1}\left ( t \right )\sum\limits _{k = 1}^{n}f_{2k}^{2m+1}\left ( t \right ) \equiv 0\pmod{f_{2n+1}\left ( t \right ) -1}, \end{equation}$

(1.11)

and

$\begin{equation} al_{1} \left ( t\right ) l_{3} \left (t \right ) \cdots l_{2m+1}\left ( t \right )\sum\limits _{k = 1}^{n}l_{2k}^{2m+1}\left ( t \right ) \equiv 0\pmod{l_{2n+1}\left (t \right ) -at}, \end{equation}$

(1.12)

where $n$ and $m$ are positive integers.

Taking $t = 1$ in Corollary 1, we have the following conclusions for bi-periodic Fibonacci $\left \{ f_{n} \right \}$ and Lucas $\left \{ l_{n} \right \}$ sequences.

Corollary 2. We get the congruences:

$\begin{equation} bl_{1} l_{3} \cdots l_{2m+1} \sum\limits _{k = 1}^{n}f_{2k}^{2m+1} \equiv 0\pmod{f_{2n+1} -1}, \end{equation}$

(1.13)

and

$\begin{equation} al_{1} l_{3} \cdots l_{2m+1} \sum\limits _{k = 1}^{n}l_{2k}^{2m+1} \equiv 0\pmod{l_{2n+1} -a }, \end{equation}$

(1.14)

where $n$ and $m$ are nonzero real numbers.

Taking $a = b = 1$ and $t = 1$ in Corollary 1, we have the following conclusions for bi-periodic Fibonacci $\left \{ F_{n} \right \}$ and Lucas $\left \{ L_{n} \right \}$ sequences.

Corollary 3. We get the congruences:

$\begin{equation} L_{1} L_{3} \cdots L_{2m+1} \sum\limits _{k = 1}^{n}F_{2k}^{2m+1} \equiv 0\pmod{F_{2n+1} -1}, \end{equation}$

(1.15)

and

$\begin{equation} L_{1} L_{3} \cdots L_{2m+1} \sum\limits _{k = 1}^{n}L_{2k}^{2m+1} \equiv 0\pmod{L_{2n+1} -1 }, \end{equation}$

(1.16)

where $n$ and $m$ are nonzero real numbers.

2. Proofs of theorems

To begin, we will give several lemmas that are necessary in proving theorems.

Lemma 1. We get the congruence

$\begin{equation*} f_{\left ( 2n+1 \right )\left ( 2j+1\right ) } \left ( t \right ) -f_{2j+1 } \left ( t \right ) \equiv 0 \pmod{f_{2n+1 } \left ( t \right )-1}, \end{equation*}$

where $n$ and $m$ are nonzero real numbers.

Proof. We prove it by complete induction for $j\ge 0$ . This clearly holds when $j = 0$ . If $j = 1$ , we note that $abf_{3\left (2n+1 \right) } \left (t \right) = \left (a^{2} b^{2} t^{2} +4ab \right) f_{2n+1}^{3}\left (t \right)-3abf_{2n+1} \left (t\right)$ and we obtain

$\begin{equation*} \begin{split} & f_{3\left ( 2n+1 \right ) }\left ( t \right )-f_{3 }\left ( t \right ) = \left ( abt^{2}+4 \right ) f_{2n+1}^{3}\left ( t\right )-3f_{2n+1} \left (t \right )-\left ( abt^{2}+4 \right )f_{1}^{3} \left (t \right )+3f_{1} \left (t\right ) \\ & = \left ( abt^{2}+4 \right )\left ( f_{2n+1}\left ( t \right )-f_{1}\left ( t \right )\right )\left ( f_{2n+1}^{2}\left (t \right )+ f_{2n+1}\left ( t \right )f_{1}\left ( t \right )+ f_{1}^{2}\left ( t \right ) \right )-3\left ( f_{2n+1} \left ( t\right ) -f_{1}\left ( t \right ) \right ) \\ & = \left ( abt^{2}+4 \right )\left ( f_{2n+1}\left ( t\right )-1\right )\left ( f_{2n+1}^{2}\left (t \right )+ f_{2n+1}\left ( t \right )f_{1}\left ( t \right )+ f_{1}^{2}\left ( t \right ) \right )-3\left ( f_{2n+1} \left ( t \right ) -1 \right ) \\ & \equiv 0 \pmod{f_{2n+1}\left (t \right ) -1 } . \end{split} \end{equation*}$

This is obviously true when $j = 1$ . Assuming that Lemma 1 holds if $j = 1, 2, \dots, k$ , that is,

$\begin{equation*} f_{\left ( 2n+1 \right )\left ( 2j+1\right ) } \left ( t \right ) -f_{2j+1 } \left ( t \right ) \equiv 0 \pmod{f_{2n+1 } \left ( t \right )-1}. \end{equation*}$

If $j = k+1 \ge 2$ , we have

$\begin{equation*} l_{2\left ( 2n+1 \right ) } \left ( t \right ) f_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left ( t \right ) = f_{\left ( 2n+1 \right )\left ( 2j+3 \right ) }\left ( t \right ) +abf_{\left ( 2n+1 \right )\left ( 2j-1 \right ) }\left ( t \right ), \end{equation*}$

and

$\begin{equation*} \begin{split} abl_{2\left ( 2n+1 \right ) } \left ( t \right ) = \left ( a^{2} b^{2} t^{2} +4ab \right )f_{2n+1}^{2}\left ( t \right )-2ab \equiv \left ( a^{2} b^{2} t^{2} +4ab \right )f_{1}^{2}\left ( t \right )-2ab \pmod{f_{2n+1}\left ( t \right ) -1 }. \end{split} \end{equation*}$

We have

$\begin{equation*} \begin{split} &f_{\left ( 2n+1 \right )\left ( 2k+3 \right ) } \left (t\right ) -f_{ 2k+3 } \left ( t\right )\\ & = l_{2\left ( 2n+1 \right ) }\left ( t \right ) f_{\left ( 2n+1 \right )\left ( 2k+1 \right ) }\left ( t \right )-abf_{\left (2n+1\right )\left ( 2k-1 \right ) } \left ( t \right ) -l_{2} \left ( t \right )f_{2k+1}\left ( t \right )+abf_{2k-1} \left ( t \right ) \\ & \equiv \left ( \left ( abt^{2}+4 \right )f_{ 1}^{2} \left (t \right ) -2 \right )f_{\left ( 2n+1 \right )\left ( 2k+1 \right ) }\left (t \right )-abf_{\left ( 2n+1 \right )\left ( 2k-1 \right ) } \left (t \right ) \\ &\quad -\left ( \left ( abt^{2}+4 \right )f_{ 1}^{2} \left ( t \right ) -2 \right ) f_{2k+1} \left ( t \right )+abf_{2k-1}\left ( t \right ) \\ & \equiv \left ( \left ( abt^{2}+4 \right )f_{ 1}^{2} \left ( t \right ) -2 \right )\left ( f_{\left ( 2n+1 \right ) \left ( 2k+1 \right ) }\left ( t \right )-f_{2k+1}\left ( t\right ) \right ) -ab\left ( f_{\left ( 2n+1 \right )\left ( 2k-1 \right ) } \left (t \right ) -f_{2k-1}\left ( t \right ) \right ) \\ & \equiv 0 \pmod{f_{2n+1}\left (t \right ) -1 } . \end{split} \end{equation*}$

This completely proves Lemma 1. □

Lemma 2. We get the congruence

$\begin{equation*} al_{\left ( 2n+1 \right )\left ( 2j+1\right ) } \left ( t \right ) -al_{2j+1 } \left ( t \right ) \equiv 0 \pmod{l_{2n+1 } \left ( t \right )-at}, \end{equation*}$

where $n$ and $m$ are nonzero real numbers.

Proof. We prove it by complete induction for $j\ge 0$ . This clearly holds when $j = 0$ . If $j = 1$ , we note that $al_{3\left (2n+1 \right) } \left (t \right) = b l_{2n+1}^{3}\left (t \right)+3al_{2n+1} \left (t \right)$ and we obtain

$\begin{equation*} \begin{split} & a l_{3\left ( 2n+1 \right ) }\left (t \right )-al_{3 }\left ( t \right ) = b l_{2n+1}^{3}\left ( t \right )+3al_{2n+1} \left ( t \right )-bl_{1}^{3} \left (t \right )-3al_{1} \left ( t \right ) \\ & = \left ( l_{2n+1}\left ( t \right )-l_{1}\left ( t \right )\right )\left ( bl_{2n+1}^{2}\left ( t \right )+ bl_{2n+1}\left ( t \right )l_{1}\left ( t \right )+ bl_{1}^{2}\left (t \right ) \right )-3a\left ( l_{2n+1} \left ( t \right ) -l_{1}\left ( t\right ) \right ) \\ & = \left ( l_{2n+1}\left ( t \right )-at\right )\left ( bl_{2n+1}^{2}\left ( t \right )+ bayl_{2n+1}\left ( t \right )+ ba^{2}t^{2} \right )-3a\left ( l_{2n+1} \left ( t \right ) -at \right )\\ & \equiv 0 \pmod{l_{2n+1}\left ( t \right ) -at } . \end{split} \end{equation*}$

This is obviously true when $j = 1$ . Assuming that Lemma 2 holds if $j = 1, 2, \dots, k$ , that is,

$\begin{equation*} al_{\left ( 2n+1 \right )\left ( 2j+1\right ) } \left (t \right ) -al_{2j+1 } \left ( t \right ) \equiv 0 \pmod{l_{2n+1 } \left ( t \right )-at}. \end{equation*}$

If $j = k+1 \ge 2$ , we have

$\begin{equation*} l_{2\left ( 2n+1 \right ) } \left (t \right ) l_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left ( t\right ) = l_{\left ( 2n+1 \right )\left ( 2j+3 \right ) }\left ( t\right ) +l_{\left ( 2n+1 \right )\left ( 2j-1 \right ) }\left ( t \right ), \end{equation*}$

and

$\begin{equation*} \begin{split} & al_{2\left ( 2n+1 \right ) } \left ( t \right ) = bl_{2n+1}^{2}\left ( t \right )+2a \equiv bl_{1}^{2}\left ( t \right )+2a \pmod{l_{2n+1}\left ( t \right ) -at }. \end{split} \end{equation*}$

We have

$\begin{equation*} \begin{split} &al_{\left ( 2n+1 \right )\left ( 2k+3 \right ) } \left ( t\right ) -al_{\left ( 2k+3 \right ) } \left ( t \right )\\ & = a\left ( l_{2\left ( 2n+1 \right ) }\left ( t \right ) l_{\left ( 2n+1 \right )\left ( 2k+1 \right ) }\left (t \right )-l_{\left (2n+1\right )\left ( 2k-1 \right ) } \left ( t \right ) \right )-a\left ( l_{2} \left (t \right )l_{2k+1}\left ( t \right )-l_{2k-1} \left (t \right ) \right ) \\ & \equiv \left ( bl_{ 1}^{2} \left ( t \right ) +2a \right )l_{\left ( 2n+1 \right )\left ( 2k+1 \right ) }\left ( t \right )-al_{\left ( 2n+1 \right )\left ( 2k-1 \right ) } \left ( t \right ) -\left ( bl_{ 1}^{2} \left ( t \right ) +2a \right ) l_{2k+1} \left ( t \right )+al_{2k-1}\left ( t \right ) \\ & \equiv \left ( abt^{2} +2 \right )\left ( al_{\left ( 2n+1 \right ) \left ( 2k+1 \right ) }\left ( t\right )-al_{2k+1}\left ( t \right ) \right ) - \left ( al_{\left ( 2n+1 \right )\left ( 2k-1 \right ) } \left ( t\right ) -al_{2k-1}\left (t \right ) \right ) \\ & \equiv 0 \pmod{l_{2n+1}\left ( t \right ) -at } . \end{split} \end{equation*}$

This completely proves Lemma 2. □

Proof of Theorem 1. We only prove (1.3), and the proofs for other identities are similar.

$\begin{equation*} \begin{split} \sum\limits _{k = 1}^{n}f_{2k}^{2m+1}\left ( t \right )& = \sum\limits_{k = 1}^{n}\left (\frac{a^{\varsigma \left ( 2k+1 \right ) } }{\left (ab \right )^{ \left \lfloor \frac{2k}{2} \right \rfloor} }\cdot \left ( \frac{\sigma ^{2k}\left ( t \right ) - \tau ^{2k} \left (t \right ) }{\sigma\left ( t \right ) - \tau \left (t \right ) } \right ) \right ) ^{2m+1} \\ & = \frac{a^{2m+1} }{\left ( \sigma\left (t \right ) -\tau \left ( t \right ) \right )^{2m+1} }\sum\limits _{k = 1}^{n}\frac{\left ( \sigma ^{2k}\left (t \right ) - \tau ^{2k} \left ( t \right ) \right )^{2m+1} }{\left ( ab \right )^{\left ( 2m+1 \right )k } } \\ & = \frac{a^{2m+1} }{\left ( \sigma\left (t\right ) -\tau \left ( t \right ) \right )^{2m+1} }\sum\limits _{k = 1}^{n}\sum\limits _{j = 0}^{2m+1}\left ( -1 \right )^{j} \binom{2m+1}{j} \frac{ \sigma^{2k\left ( 2m+1-j \right ) } \left ( t \right ) \tau ^{2kj} \left ( t \right ) }{\left ( ab \right )^{\left ( 2m+1 \right )k } } \\ & = \frac{a^{2m+1} }{\left ( \sigma\left ( t \right ) -\tau \left (t \right ) \right )^{2m+1} }\sum\limits _{j = 0}^{2m+1}\left ( -1 \right )^{j} \binom{2m+1}{j} \left ( \frac{1-\frac{\sigma ^{2n\left ( 2m+1-2j \right ) }\left (t \right ) }{\left ( ab \right )^{\left ( 2m+1-2j \right )n } } }{\frac{\left ( ab \right )^{2m+1-2j} }{\sigma ^{2\left ( 2m+1-2j \right ) }\left (t \right ) } -1} \right )\\ & = \frac{a^{2m+1} }{\left (\sigma\left (t \right ) -\tau \left (t\right ) \right )^{2m+1} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m+1}{j} \left ( \frac{1-\frac{\sigma ^{2n\left ( 2m+1-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2m+1-2j \right )n } } }{\frac{\left ( ab \right )^{2m+1-2j} }{\sigma ^{2\left ( 2m+1-2j \right ) }\left (t \right ) } -1} - \frac{1-\frac{\sigma ^{2n\left ( 2j-1-2m \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2j-1-2m\right )n } } }{\frac{\left ( ab \right )^{2j-1-2m} }{\sigma ^{2\left ( 2j-1-2m \right ) }\left ( t \right ) } -1}\right )\\ & = \frac{a^{2m+1} }{\left ( \sigma\left ( t \right ) -\tau \left ( t \right ) \right )^{2m+1} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m+1}{j} \left ( \frac{\frac{\sigma^{2 \left ( 2m+1-2j \right ) }\left ( t\right ) }{\left ( ab \right )^{ 2m+1-2j } }- \frac{\sigma ^{\left ( 2n+2 \right ) \left ( 2m+1-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( n+1 \right ) \left ( 2m+1-2j \right ) } } +1- \frac{\sigma ^{2n \left ( 2j-1-2m \right ) }\left (t \right ) }{\left ( ab \right )^{ \left ( 2j-1-2m \right ) n }} }{1- \frac{\sigma^{2 \left ( 2m+1-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{ \left ( 2m+1-2j \right ) }}} \right )\\ & = \frac{a^{2m+1} }{\left ( \sigma\left ( t \right ) -\tau \left (t \right ) \right )^{2m+1} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m+1}{j}\\ & \quad \times \left ( \frac{\sigma^{2m+1-2j}\left ( t \right ) - \tau ^{2m+1-2j}\left ( t \right ) -\frac{\sigma^{\left ( 2n+1 \right ) \left ( 2m+1-2j \right ) }\left ( t \right )}{\left ( ab \right )^{\left ( 2m+1-2j \right )n } } +\frac{\tau ^{\left ( 2n+1 \right ) \left ( 2m+1-2j \right ) }\left (t \right )}{\left ( ab \right )^{ \left ( 2m+1-2j \right )n } }}{-\sigma ^{2m+1-2j}\left ( t \right ) -\tau ^{2m+1-2j}\left ( t \right ) } \right ) \\ & = \frac{a^{2m+1} }{b\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits _{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m+1}{m-j}\left ( \frac{f_{\left ( 2n+1 \right )\left ( 2j+1\right ) }\left (t \right ) - f_{2j+1}\left (t \right ) }{l_{2j+1} \left ( t \right ) } \right ) . \end{split} \end{equation*}$

□

Proof of Theorem 2. We only prove (1.7), and the proofs for other identities are similar.

$\begin{equation*} \begin{split} \sum\limits _{k = 1}^{n}f_{2k}^{2m }\left ( t \right )& = \sum\limits_{k = 1}^{n}\left (\frac{a^{\varsigma \left ( 2k +1\right ) } }{\left (ab \right )^{ \left \lfloor \frac{2k}{2} \right \rfloor} }\cdot \left ( \frac{\sigma ^{2k}\left ( t\right ) - \tau ^{2k} \left ( t \right ) }{\sigma\left ( t\right ) - \tau \left (t \right ) } \right ) \right ) ^{2m} \\ & = \frac{a^{2m} }{\left ( \sigma\left (t \right ) -\tau \left (t \right ) \right )^{2m} }\sum\limits _{k = 1}^{n}\frac{\left ( \sigma^{2k}\left ( t\right ) - \tau ^{2k} \left ( t \right ) \right )^{2m} }{\left ( ab \right )^{ 2mk } } \\ & = \frac{a^{2m} }{\left ( \sigma\left ( t \right ) -\tau \left (t \right ) \right )^{2m} }\sum\limits _{k = 1}^{n}\sum\limits _{j = 0}^{2m}\left ( -1 \right )^{j} \binom{2m}{j} \frac{ \sigma^{2k\left ( 2m-j \right ) } \left (t \right ) \tau ^{2kj} \left ( t \right ) }{\left ( ab \right )^{2mk } }\\ & = \frac{a^{2m} }{\left (\sigma\left (t \right ) -\tau \left ( t \right ) \right )^{2m} }\sum\limits _{j = 0}^{2m}\left ( -1 \right )^{j} \binom{2m }{j} \left ( \frac{1-\frac{\sigma ^{2n\left ( 2m-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2m-2j \right )n } } }{\frac{\left ( ab \right )^{2m-2j} }{\sigma ^{2\left ( 2m-2j \right ) }\left ( t \right ) } -1} \right ) \end{split} \end{equation*}$

$\begin{equation*} \begin{split} & = \frac{a^{2m} }{\left (\sigma\left (t \right ) -\tau \left ( t \right ) \right )^{2m} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m}{j} \left ( \frac{1-\frac{\sigma ^{2n\left ( 2m-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2m-2j \right )n } } }{\frac{\left ( ab \right )^{2m-2j} }{\sigma ^{2\left ( 2m-2j \right ) }\left ( t \right ) } -1} +\frac{1-\frac{\sigma ^{2n\left ( 2j-2m \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2j-2m\right )n } } }{\frac{\left ( ab \right )^{2j-2m} }{\sigma ^{2\left ( 2j-2m \right ) }\left (t\right ) } -1}\right )\\ & \quad +\frac{a^{2m} }{\left ( \sigma \left ( t \right ) -\tau \left (t \right ) \right )^{2m} } \left ( -1 \right ) ^{m+1}\binom{2m}{m}n\\ & = \frac{a^{2m} }{\left (\sigma\left ( t \right ) -\tau \left (t \right ) \right )^{2m} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m}{j} \left ( \frac{\frac{\sigma ^{2 \left ( 2m -2j \right ) }\left ( t \right ) }{\left ( ab \right )^{ 2m -2j } }- \frac{\sigma ^{\left ( 2n+2 \right ) \left ( 2m -2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( n+1 \right ) \left ( 2m -2j \right ) } } -1+\frac{\sigma^{2n \left ( 2j -2m \right ) }\left ( t \right ) }{\left ( ab \right )^{ \left ( 2j -2m \right ) n }} }{1- \frac{\sigma^{2 \left ( 2m -2j \right ) }\left ( t \right ) }{\left ( ab \right )^{ 2m-2j }}} \right )\\ & \quad +\frac{a^{2m} }{\left (\sigma \left ( t \right ) -\tau \left ( t \right ) \right )^{2m} } \left ( -1 \right ) ^{m+1}\binom{2m}{m}n \\ & = \frac{a^{2m } }{\left ( \sigma\left ( t \right ) -\tau \left ( t \right ) \right )^{2m } }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m }{j} \left ( \frac{\sigma^{2m -2j}\left ( t \right ) - \tau ^{2m -2j}\left (t\right ) -\frac{\sigma^{\left ( 2n+1 \right ) \left ( 2m -2j \right ) }\left (t \right )}{\left ( ab \right )^{n\left ( 2m -2j \right ) } } +\frac{\tau ^{\left ( 2n+1 \right ) \left ( 2m-2j \right ) }\left (t \right )}{\left ( ab \right )^{n\left ( 2m-2j \right ) } }}{ \tau ^{2m -2j}\left ( t \right )-\sigma ^{2m -2j}\left ( t \right ) } \right ) \\ & \quad +\frac{a^{2m} }{\left ( \sigma \left (t \right ) -\tau \left ( t \right ) \right )^{2m} } \left ( -1 \right ) ^{m+1}\binom{2m}{m}n\\ & = \frac{a^{2m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits _{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m}{m-j}\left ( \frac{f_{2j\left ( 2n+1 \right ) }\left ( t \right ) - f_{ 2j}\left (t \right ) }{f_{ 2j} \left (t \right ) } \right ) +\frac{a^{2m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \left ( -1 \right ) ^{m+1}\binom{2m}{m}n . \end{split} \end{equation*}$

□

Proof of Corollary 1. First, from the definition of $f_{n}\left (t \right)$ and binomial expansion, we easily prove $\left (f_{2n+1}\left (t \right)-1, a^{2} b^{2} t^{2} +4ab \right) = 1$ . Therefore, $\left (f_{2n+1}\left (t \right)-1, \left (a^{2} b^{2} t^{2} +4ab \right)^{m} \right) = 1$ . Now, we prove (1.11) by Lemma 1 and (1.3):

$\begin{equation*} \begin{split} &b l_{1} \left ( t\right ) l_{3} \left ( t \right ) \cdots l_{2m+1}\left ( t \right )\sum\limits _{k = 1}^{n}f_{2k}^{2m+1}\left ( t\right ) \\ & = l_{1} \left ( t\right ) l_{3} \left (t \right ) \cdots l_{2m+1}\left ( t \right ) \left ( \frac{a^{2m+1} }{\left ( \sigma\left (t \right ) -\tau \left (t \right ) \right )^{2m} } \sum\limits_{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m+1}{m-j} \left ( \frac{f_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left ( t \right ) -f_{2j+1}\left ( t \right ) }{l_{2j+1}\left (t\right ) } \right ) \right )\\ & \equiv 0\pmod{f_{2n+1}\left ( t \right ) -1}. \end{split} \end{equation*}$

Now, we use Lemma 2 and (1.5) to prove (1.12):

$\begin{equation*} \begin{split} &al_{1} \left ( t\right ) l_{3} \left ( t \right ) \cdots l_{2m+1}\left ( t\right )\sum\limits _{k = 1}^{n}l_{2k}^{2m+1}\left ( t \right ) \\ & = l_{1} \left ( t \right ) l_{3} \left (t \right ) \cdots l_{2m+1}\left ( t \right ) \left ( \sum\limits_{j = 0}^{m} \binom{2m+1}{m-j} \left ( \frac{al_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left ( t\right ) -al_{2j+1}\left (t \right ) }{l_{2j+1}\left ( t \right ) } \right ) \right )\\ & \equiv 0\pmod{l_{2n+1}\left ( t \right ) -at}. \end{split} \end{equation*}$

□

3. Conclusions

In this paper, we discuss the power sums of bi-periodic Fibonacci and Lucas polynomials by Binet formulas. As corollaries of the theorems, we extend the divisible properties of the sum of power of linear Fibonacci and Lucas sequences to nonlinear Fibonacci and Lucas polynomials. An open problem is whether we extend the Melham conjecture to nonlinear Fibonacci and Lucas polynomials.

Use of AI tools declaration

The authors declare that they did not use Artificial Intelligence (AI) tools in the creation of this paper.

Acknowledgments

The authors would like to thank the editor and referees for their helpful suggestions and comments, which greatly improved the presentation of this work. All authors contributed equally to the work, and they have read and approved this final manuscript. This work is supported by Natural Science Foundation of China (12126357).

Conflict of interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

References

[1]	A. M. Alghamdi, S. Gala, M. A. Ragusa, Regularity criterion for weak solutions to the Navier-Stokes involving one velocity and one vorticity components, Siberian Electron. Math. Rep., 19 (2022), 309–315. https://doi.org/10.33048/semi.2022.19.025 doi: 10.33048/semi.2022.19.025
[2]	A. Choucha, S. Boulaaras, D. Ouchenane, Exponential decay and global existence of solutions of a singular nonlocal viscoelastic system with distributed delay and damping terms, Filomat, 35 (2021), 795–826. https://doi.org/10.2298/FIL2103795C doi: 10.2298/FIL2103795C
[3]	C. S. Dou, Z. S. Zhao, Analytical solution to 1D compressible Navier-Stokes equations, J. Funct. Spaces, 2021 (2021), 6339203. https://doi.org/10.1155/2021/6339203 doi: 10.1155/2021/6339203
[4]	C. Foias, O. Manley, R. Rosa, R. Temam, Navier-Stokes Equations and Turbulence, Cambridge University Press, Cambridge, 2001.
[5]	J. L. Lions, Quelques Méthodes de Résolution des Problémes aux Limites Non Linéaires, Dunod, Gauthier-Villars, Paris, 1969.
[6]	J. C. Robinson, J. L. Rodrigo, W. Sadowski, The Three-Dimensional Navier-Stokes Equations, Cambridge University Press, Cambridge, 2016.
[7]	R. Temam, Navier-Stokes Equations and Nonlinear Functional Analysis, Philadelphia, PA, 1995.
[8]	J. M. Ball, Continuity properties and global attractors of generalized semiflows and the Navier-Stokes equations, J. Nonlinear Sci., 7 (1997), 475–502. https://doi.org/10.1007/s003329900037 doi: 10.1007/s003329900037
[9]	A. Cheskidov, C. Foias, On global attractors of the 3D Navier-Stokes equations, J. Differ. Equations, 231 (2006), 714–754. https://doi.org/10.1016/j.jde.2006.08.021 doi: 10.1016/j.jde.2006.08.021
[10]	X. Cai, Q. Jiu, Weak and strong solutions for the incompressible Navier-Stokes equations with damping, J. Math. Anal. Appl., 343 (2008), 799–809. https://doi.org/10.1016/j.jmaa.2008.01.041 doi: 10.1016/j.jmaa.2008.01.041
[11]	F. Li, B. You, Pullback exponential attractors for the three dimensional non-autonomous Navier-Stokes equations with nonlinear damping, Discrete Contin. Dyn. Syst. Ser. B, 25 (2020), 55–80. https://doi.org/10.3934/dcdsb.2019172 doi: 10.3934/dcdsb.2019172
[12]	F. Li, B. You, Y. Xu, Dynamics of weak solutions for the three dimensional Navier-Stokes equations with nonlinear damping, Discrete Contin. Dyn. Syst. Ser. B, 23 (2018), 4267–4284. https://doi.org/10.3934/dcdsb.2018137 doi: 10.3934/dcdsb.2018137
[13]	D. Pardo, J. Valero, Á. Giménez, Global attractors for weak solutions of the three-dimensional Navier-Stokes equations with damping, Discrete Contin. Dyn. Syst. Ser. B, 24 (2019), 3569–3590. https://doi.org/10.3934/dcdsb.2018279 doi: 10.3934/dcdsb.2018279
[14]	X. L. Song, Y. R. Hou, Attractors for the three-dimensional incompressible Navier-Stokes equations with damping, Discrete Contin. Dyn. Syst., 31 (2011), 239–252. https://doi.org/10.3934/dcds.2011.31.239 doi: 10.3934/dcds.2011.31.239
[15]	C. Foias, O. P. Manley, R. Temam, Y. M. Trève, Asymptotic analysis of the Navier-Stokes equations, Physica D, 9 (1983), 157–188. https://doi.org/10.1016/0167-2789(83)90297-X doi: 10.1016/0167-2789(83)90297-X
[16]	D. A. Jones, E. S. Titi, Upper bounds on the number of determining modes, nodes, and volume elements for the Navier-Stokes equations, Indiana Univ. Math. J., 42 (1993), 875–887. https://doi.org/10.1512/iumj.1993.42.42039 doi: 10.1512/iumj.1993.42.42039
[17]	P. Constantin, P. Foias, R. Temam, On the dimension of the attractors in two-dimensional turbulence, Physica D, 30 (1988), 284–296. https://doi.org/10.1016/0167-2789(88)90022-X doi: 10.1016/0167-2789(88)90022-X
[18]	R. Selmi, A. Châabani, Well-posedness, stability and determining modes to 3D Burgers equation in Gevrey class, Z. Angew. Math. Phys., 71 (2020), 162. https://doi.org/10.1007/s00033-020-01389-3 doi: 10.1007/s00033-020-01389-3
[19]	J. C. Kuang, Applied Inequalities (Changyong Budengshi), 2 $^{\text{nd}}$ edition, Hunan Education Publishing House, 1993.
[20]	B. L. Guo, P. C. Zhu, Partial regularity of suitable weak solutions to the system of the incompressible Non-Newtonian fluids, J. Differ. Equations, 178 (2002), 281–297. https://doi.org/10.1006/jdeq.2000.3958 doi: 10.1006/jdeq.2000.3958
[21]	V. Kalantarov, A. Kostianko, S. Zelik, Determining functionals and finite-dimensional reduction for dissipative PDEs revisited, preprint, arXiv: 2111.04125.

Reader Comments

Your name:*

Email:*
© 2022 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0)

通讯作者: 陈斌, bchen63@163.com

1.
沈阳化工大学材料科学与工程学院沈阳 110142

Electronic Research Archive

1 1.7

Metrics

Article views(1979) PDF downloads(108) Cited by(4)

Preview PDF

Download XML

Export Citation

Article outline

Show full outline

Electronic Research Archive

Determination of the 3D Navier-Stokes equations with damping

Related Papers:

Abstract

1. Introduction

2. Proofs of theorems

3. Conclusions

Use of AI tools declaration

Acknowledgments

Conflict of interest

References

Reader Comments

通讯作者: 陈斌, bchen63@163.com

Metrics

Other Articles By Authors

Catalog

Electronic Research Archive

Determination of the 3D Navier-Stokes equations with damping

Related Papers:

Abstract

1. Introduction

2. Proofs of theorems

3. Conclusions

Use of AI tools declaration

Acknowledgments

Conflict of interest

References

Reader Comments

通讯作者: 陈斌, bchen63@163.com

Metrics

Other Articles By Authors

Related pages

Tools

Export File

Citation

Format

Content

Catalog