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Article

# Spatial Decay Estimates and Continuous Dependence for the Oldroyd Fluid

by
Yuanfei Li
School of Data Science, Guangzhou Huashang College, Guangzhou 511300, China
Mathematics 2024, 12(8), 1240; https://doi.org/10.3390/math12081240
Submission received: 1 February 2024 / Revised: 14 April 2024 / Accepted: 15 April 2024 / Published: 19 April 2024
(This article belongs to the Special Issue Applications of Differential Equations in Sciences)

## Abstract

:
This article investigates the Oldroyd fluid, which is widely used in industrial and engineering environments. When the Oldroyd fluid passes through a three-dimensional semi-infinite cylinder, the asymptotic properties of the solutions are established. On this basis, we also studied the continuous dependence of the viscosity coefficient.
MSC:
35B40; 35Q30; 76D05

## 1. Introduction

In this paper, we let R denote that
$R = ( x 1 , x 2 , x 3 ) | ( x 1 , x 2 ) ∈ D , x 3 > 0 ,$
where $D ⊂$$( x 1 , x 2 )$-plane, and D is bounded (see Figure 1). We also require that D has a smooth boundary $∂ D$.
Let $u i , p ,$ and $q i ( i = 1 , 2 , 3 )$ denote the velocity, the pressure, and viscoelastic variables of the fluid, respectively. These variables satisfy the following Oldroyd fluid equations
(see [1]):
$u i , t − μ Δ u i − λ Δ q i + p , i = 0 , x ∈ R , t > 0 ,$
$u i , i = 0 , x ∈ R , t > 0 ,$
$q i , t + γ q i − u i = 0 , x ∈ R , t > 0 ,$
$u i , u i , j , q i , q i , j → 0 , as x 3 → ∞ .$
In (1)–(4) and the following, we use commas for derivation, repeated English subscripts for summation from 1 to 3, and repeated Greek subscripts for summation from 1 to 2, e.g., $u i , i = ∑ i = 1 3 ∂ u i ∂ x i$ and $u α , α = ∑ i = 1 2 ∂ u α ∂ x α$.
On the boundary, Equations (1)–(4) satisfy
$u i ( x , 0 ) = q i ( x , 0 ) = 0 , x ∈ R ,$
$u i ( x 1 , x 2 , 0 , t ) = g i ( x 1 , x 2 , t ) , ( x 1 , x 2 ) ∈ D , t > 0 ,$
$q i ( x 1 , x 2 , 0 , t ) = h i ( x 1 , x 2 , t ) , ( x 1 , x 2 ) ∈ D , t > 0 ,$
$u i = 0 , ∂ q i ∂ n = 0 , x ∈ ∂ D × { x 3 ≥ 0 } , t > 0 ,$
where $g i$ and $h i$ are given functions, and $μ , γ ,$ and $λ$ are positive constants.
Because viscoelastic fluid is widely used in real life, this type of fluid has attracted more and more attention. The Oldroyd fluid Equations (1)–(3) are not only clearly recorded in [2], but also discussed in [1]. Meanwhile, Oskolkov and Shadiev [2] and Christov and Jordan [3] have proved the existence of the solution to Equations (1)–(3) under different conditions. In the context of industrial and engineering applications, Oldroyd fluids always pass through a cylinder. Therefore, it is necessary to study the properties and continuous dependence of solutions to Oldroyd fluids on a cylinder.
The first purpose of the present paper is to study the spatial decay results of the solutions to Equations (1)–(4) when $x 3 → ∞$. In fact, Keiller [4] studied the spatial decay of steady perturbations of plane Poiseuille flow for the Oldroyd-B equations as $t → ∞$. For such a type of study, one can also see [5]. After the earlier work of Boley [6], the spatial decay results of fluid equations with spatial variables in a cylinder, which can be thought of as decay results of the Saint-Venant type, have been paid full attention. For more of such Saint-Venant type results, one can see [7,8,9,10,11,12,13,14,15,16]. Compared with references [4,5], the innovation of this paper is the use of the methods of references [9,11] to further extend the attenuation results to a semi-infinite cylinder. We demonstrate that the solution decays exponentially on the semi-infinite cylinder, indicating that the velocity and viscoelastic variables of the fluid decay exponentially with distance from the finite end to the infinite. This is the first result that we establish in the present paper.
The second purpose is to study the continuous dependence of the solutions to Equations (1)–(4) on the viscosity coefficient. When a small disturbance occurs in the viscosity coefficient, we study what kind of disturbance will occur in the solutions of the equation. In past decades, studies on structural stability have received a lot of attention, and a large number of papers have emerged, including those of Liu and Zheng [17], Avalos and Lesiecka [18], Meyvacı [19], Quintanilla [1,20,21], Liu et al. [22,23,24,25,26], Hameed and Harfash [27], Scott et al. [28,29], Li et al. [30,31,32,33], Ciarletta and Straughan [34], and Franchi and Straughan [35]. Straughan [1] proved that Kelvin–Voigt fluid depended continuously on the coefficients of the fluid. Research results have been focused on bounded regions in both two-dimensional and three-dimensional spaces.
In fact, the study of structural stability in the cylindrical region is equally important. Due to its widespread application in practice, it has gradually begun to receive attention (see [36,37,38]). This paper will continue the research in this field. We want to extend the results of [1] to the case of the semi-infinite cylinder. Taking the viscosity coefficient as an example, we demonstrate how to derive the continuous dependence results of the solutions on the other coefficients in Equations (1)–(4). In this process, we adopted methods of energy estimation and prior estimates.

## 2. The Main Theorems

We first list some lemmas.
Lemma 1
(See [7]). If $χ | ∂ D = 0 ,$ then
$λ 1 ∫ D χ 2 d A ≤ ∫ D χ , α χ , α d A ,$
where $λ 1$ is the smallest positive eigenvalue of
$Δ φ + λ φ = 0 , in D , φ = 0 , on ∂ D .$
Lemma 2
(See [39]). Assume that $∂ R$ is the Lipschitz boundary of R. If $∫ R v d x = 0$, then $∃ φ = ( φ 1 , φ 2 , φ | 3 )$ and $φ | ∂ R = 0$ such that
$φ i , i = v , i n R ,$
and
$∫ R φ i , j φ i , j d x ≤ k 1 ∫ R [ φ j , j ] 2 d x , k 1 > 0 .$
This lemma in the case of two dimensions has been established by [40].
We consider the identity
$∫ 0 t ∫ R z e − ω η ( ξ − z ) u i , η − μ Δ u i − λ Δ q i + p , i u i , η d x d η = 0 ,$
where $R z = ( x 1 , x 2 , x 3 ) | ( x 1 , x 2 ) ∈ D , x 3 ≥ z > 0$. By using the divergence theorem and (5), (8), we have
$1 2 e − ω t μ ∫ R z ( ξ − z ) u i , j u i , j d x + ∫ 0 t ∫ R z e − ω η ( ξ − z ) u i , η u i , η + 1 2 ω μ u i , j u i , j d x d η = − μ ∫ 0 t ∫ R z e − ω η u i , η u i , 3 d x d η − λ ∫ 0 t ∫ R z e − ω η u i , η q i , 3 d x d η + ∫ 0 t ∫ R z e − ω η u 3 , η p d x d η − λ ∫ 0 t ∫ R z e − ω η ( ξ − z ) q i , j u i , j η d x d η = − μ ∫ 0 t ∫ R z e − ω η u i , η u i , 3 d x d η − λ ∫ 0 t ∫ R z e − ω η u i , η q i , 3 d x d η + ∫ 0 t ∫ R z e − ω η u 3 , η p d x d η − λ e − ω t ∫ R z ( ξ − z ) q i , j u i , j d x − λ ω ∫ 0 t ∫ R z e − ω η ( ξ − z ) q i , j u i , j d x d η + λ ∫ 0 t ∫ R z e − ω η ( ξ − z ) q i , j η u i , j d x d η ,$
where ω is a positive constant.
By (2), we have the following identity:
$∫ 0 t ∫ R z e − ω η ( ξ − z ) q i , j η + γ q i , j − u i , j q i , j η d x d η = 0 .$
From (11), it follows that
$1 2 e − ω t γ ∫ R z ( ξ − z ) q i , j q i , j d x + ∫ 0 t ∫ R z e − ω η ( ξ − z ) 1 2 ω γ q i , j q i , j + q i , j η q i , j η d x d η = ∫ 0 t ∫ R z e − ω η ( ξ − z ) u i , j q i , j η d x d η .$
If we let
$E ( z , t ) = 1 2 e − ω t ∫ R z ( ξ − z ) μ u i , j u i , j + δ 1 γ q i , j q i , j d x + ∫ 0 t ∫ R z e − ω η ( ξ − z ) u i , η u i , η + 1 2 ω μ u i , j u i , j + 1 2 ω γ δ 1 q i , j q i , j + δ 1 q i , j η q i , j η d x d η ,$
then from (10) and (12), we have
$E ( z , t ) = − μ ∫ 0 t ∫ R z e − ω η u i , η u i , 3 d x d η − λ ∫ 0 t ∫ R z e − ω η u i , η q i , 3 d x d η + ∫ 0 t ∫ R z e − ω η u 3 , η p d x d η − λ e − ω t ∫ R z ( ξ − z ) q i , j u i , j d x − λ ω ∫ 0 t ∫ R z e − ω η ( ξ − z ) q i , j u i , j d x d η + ( λ + δ 1 ) ∫ 0 t ∫ R z e − ω η ( ξ − z ) q i , j η u i , j d x d η ≐ ∑ i = 1 6 A i ( z , t ) ,$
where $δ 1 > 0$.
Based on the energy function $E ( z , t )$, we can obtain the spatial decay theorem.
Theorem 1.
Let $( u i , q i , p )$ be the solutions of Equations (1)–(8). If $g i ∈ C ( ∂ D × [ 0 , ∞ ) )$ and $∫ D g 3 d A = 0$, then the following inequality holds:
$1 2 e − ω t ∫ R z ( ξ − z ) μ u i , j u i , j + δ 1 γ q i , j q i , j d x + ∫ 0 t ∫ R z e − ω η ( ξ − z ) u i , η u i , η + 1 2 ω μ u i , j u i , j + 1 2 ω γ δ 1 q i , j q i , j + δ 1 q i , j η q i , j η d x d η ≤ 1 m 1 Q ( 0 , t ) e − m 1 z$
where $m 1 > 0$, and $Q ( 0 , t ) > 0$ is a computable function that depends only on t.
Remark 1.
Obviously, $e − m 1 z → 0$ as $z → ∞$. Therefore Theorem 1 shows that the solutions of (1)–(8) decay exponentially to zero as $z → ∞$.
We suppose that $u i , q i ,$ and p are the solutions of (1)–(8), and $u i ∗ , q i ∗ ,$ and $p ∗$ are the solutions of (1)–(8) with $γ = γ ∗$. Let
$v i = u i − u i ∗ , Σ i = q i − q i ∗ , π = p − p ∗ , γ ˜ = γ − γ ∗ ,$
Then, $v i , Σ i ,$ and π satisfy
$v i , t − μ Δ v i − λ Δ Σ i + π , i = 0 , x ∈ R , t > 0 ,$
$v i , i = 0 , x ∈ R , t > 0 ,$
$Σ i , t + γ ˜ q i + γ ∗ Σ i − v i = 0 , x ∈ R , t > 0 ,$
$v i , v i , j , Σ i , Σ i , j , → 0 , a s x 3 → ∞$
and the following conditions
$v i ( x , 0 ) = Σ i ( x , 0 ) = 0 , x ∈ R ,$
$v i ( x 1 , x 2 , 0 , t ) = 0 , ( x 1 , x 2 ) ∈ D , t > 0 ,$
$Σ i ( x 1 , x 2 , 0 , t ) = 0 , ( x 1 , x 2 ) ∈ D , t > 0 ,$
$v i = 0 , ∂ Σ i ∂ n = 0 , x ∈ ∂ D × { x 3 ≥ 0 } , t > 0 .$
The continuous dependence theorem can be written as follows.
Theorem 2.
Let $u i$ and $q i$ be the solutions of (1)–(8) and $u i ∗$ and $q i ∗$ be the solutions of (1)–(8) with $γ = γ ∗$. If $∫ D g 3 d A = 0$, then
$( u i , q i ) → ( u i ∗ , q i ∗ ) , as γ → γ ∗ .$
Specifically,
$1 4 ω μ ∫ 0 t ∫ R z ( ξ − z ) e − ω η v i , j v i , j d x d η + ∫ 0 t ∫ R z ( ξ − z ) e − ω η v i , η v i , η d x d η + 1 4 δ 2 ω γ ∗ ∫ 0 t ∫ R z ( ξ − z ) e − ω η Σ i , j Σ i , j d x d η + 1 2 δ 2 ∫ 0 t ∫ R z ( ξ − z ) e − ω η Σ i , j η Σ i , j η d x d η ≤ 2 δ 2 ω γ ∗ δ 1 Q ( 0 , t ) γ ˜ 2 1 m 2 + 2 m 1 e − m 2 z + 4 m 2 δ 2 ω γ ∗ m 1 ( m 1 − m 2 ) δ 1 Q ( 0 , t ) γ ˜ 2 e − m 1 z − e − m 2 z ,$
or
$1 4 ω μ ∫ 0 t ∫ R z ( ξ − z ) e − ω η v i , j v i , j d x d η + ∫ 0 t ∫ R z ( ξ − z ) e − ω η v i , η v i , η d x d η + 1 4 δ 2 ω γ ∗ ∫ 0 t ∫ R z ( ξ − z ) e − ω η Σ i , j Σ i , j d x d η + 1 2 δ 2 ∫ 0 t ∫ R z ( ξ − z ) e − ω η Σ i , j η Σ i , j η d x d η ≤ 2 δ 2 ω γ ∗ δ 1 Q ( 0 , t ) γ ˜ 2 1 m 2 + 2 m 1 e − m 2 z + 4 m 2 δ 2 ω γ ∗ m 1 δ 1 Q ( 0 , t ) γ ˜ 2 z e − m 2 z ,$
where $m 2$ and $δ 2$ are positive constants.

## 3. The Proof of Theorem 1

In this section, we first state some lemmas that are related to (1)–(8). Using these lemmas, we can obtain Theorem 1.
Lemma 3.
Suppose that $u i , q i ,$ and p are the solutions of (1)–(8) with $∫ D g 3 d A = 0$. Then,
$E ( z , t ) ≤ 1 m 1 − ∂ ∂ z E ( z , t ) ,$
where $δ 1 = 16 λ 2 γ μ , ω ≥ 8 ( λ + δ 1 ) 2 μ δ 1$, and
$1 m 1 = 2 k 2 λ 1 + k 2 μ 2 ω + k 2 λ 2 γ δ 1 ω + k 2 μ 2 λ 1 + k 2 λ 2 λ 1 γ + μ 2 ω + λ 2 γ δ 1 ω$
Proof.
From (13), we also have
$− ∂ ∂ z E ( z , t ) = 1 2 e − ω t ∫ R z μ u i , j u i , j + δ 1 γ q i , j q i , j d x + ∫ 0 t ∫ R z e − ω η u i , η u i , η + 1 2 ω μ u i , j u i , j + 1 2 ω γ δ 1 q i , j q i , j + δ 1 q i , j η q i , j η d x d η .$
Next, we bound each $A i ( i = 1 , 2 , … , 6 )$ in terms of $− ∂ ∂ z E ( z , t )$. By Hölder’s and Young’s inequalities, we have
$A 1 ( z , t ) ≤ μ ∫ 0 t ∫ R z e − ω η u i , 3 u i , 3 d x d η ∫ 0 t ∫ R z e − ω η u i , η u i , η d x d η 1 2$
$≤ μ 2 ω 1 2 ω μ ∫ 0 t ∫ R z e − ω η u i , 3 u i , 3 d x d η + ∫ 0 t ∫ R z e − ω η u i , η u i , η d x d η ,$
$A 2 ( z , t ) ≤ λ 2 γ δ 1 ω 1 2 ω γ δ 1 ∫ 0 t ∫ R z e − ω η q i , 3 q i , 3 d x d η + ∫ 0 t ∫ R z e − ω η u i , η u i , η d x d η ,$
$A 4 ( z , t ) ≤ λ γ μ δ 1 1 2 μ e − ω t ∫ R z ( ξ − z ) u i , j u i , j d x + 1 2 γ δ 1 e − ω t ∫ R z ( ξ − z ) q i , j q i , j d x , A 5 ( z , t ) ≤ λ γ μ δ 1 [ 1 2 ω μ ∫ 0 t ∫ R z e − ω η ( ξ − z ) u i , j u i , j d x d η$
$+ 1 2 ω γ δ 1 ∫ 0 t ∫ R z e − ω η ( ξ − z ) q i , j q i , j d x d η ] , A 6 ( z , t ) ≤ λ + δ 1 2 ω μ δ 1 [ δ 1 ∫ 0 t ∫ R z e − ω η ( ξ − z ) q i , j η q i , j η d x d η$
$+ 1 2 ω μ ∫ 0 t ∫ R z e − ω η ( ξ − z ) u i , j u i , j d x d η ] .$
Next, we will bound $A 3 ( z , t )$. To do this, we note that
$∫ D z u 3 , t d A = ∫ D u 3 , t d A + ∫ 0 z ∫ D ξ u 3 , 3 t d A d ξ = ∫ D u 3 , t d A − ∫ 0 z ∫ D ξ u α , α t d A d ξ = ∫ D g 3 , t d A .$
In view of $∫ D g 3 d A = 0 ,$ we have $∫ D z u 3 , t d A = 0$. Through using Lemma 2, $∃ φ = ( φ 1 , φ 2 , φ 3 )$ such that
$φ i , i = u 3 , t , in R , φ i = 0 , on ∂ R .$
Therefore, we have
$A 3 ( z , t ) = ∫ 0 t ∫ z ∞ ∫ D ξ e − ω η φ i , i p d x d η = − ∫ 0 t ∫ z ∞ ∫ D ξ e − ω η φ i p , i d x d η = ∫ 0 t ∫ z ∞ ∫ D ξ e − ω η φ i u i , η − μ Δ u i − λ Δ q i d x d η = ∫ 0 t ∫ z ∞ ∫ D ξ e − ω η φ i u i , η d x d η + μ ∫ 0 t ∫ z ∞ ∫ D ξ e − ω η φ i , j u i , j d x d η + λ ∫ 0 t ∫ z ∞ ∫ D ξ e − ω η φ i , j q i , j d x d η ≐ ∑ i = 1 3 A 3 i ( z , t ) .$
Using the Schwarz inequality and Lemmas 1 and 2, we have
$A 31 ( z , t ) ≤ ∫ 0 t ∫ R z e − ω η u i , η u i , η d x d η ∫ 0 t ∫ R z e − ω η φ i φ i d x d η 1 2 ≤ 1 λ 1 ∫ 0 t ∫ R z e − ω η u i , η u i , η d x d η ∫ 0 t ∫ R z e − ω η φ i , β φ i , β d x d η 1 2 ≤ k 1 λ 1 ∫ 0 t ∫ R z e − ω η u i , η u i , η d x d η ∫ 0 t ∫ R z e − ω η φ i , i 2 d x d η 1 2 ≤ k 1 λ 1 ∫ 0 t ∫ R z e − ω η u i , η u i , η d x d η ∫ 0 t ∫ R z e − ω η u 3 , η 2 d x d η 1 2 ≤ k 1 λ 1 ∫ 0 t ∫ R z e − ω η u i , η u i , η d x d η ,$
$A 32 ( z , t ) ≤ μ ∫ 0 t ∫ R z e − ω η u i , j u i , j d x d η ∫ 0 t ∫ R z e − ω η φ i , j φ i , j d x d η 1 2 ≤ k 1 μ 2 ω 1 2 ω μ ∫ 0 t ∫ R z e − ω η u i , j u i , j d x d η + ∫ 0 t ∫ R z e − ω η u 3 , η 2 d x d η ,$
$A 33 ( z , t ) ≤ k 1 λ 2 γ δ 1 ω 1 2 ω δ 1 γ ∫ 0 t ∫ R z e − ω η q i , j q i , j d x d η + ∫ 0 t ∫ R z e − ω η u 3 , η 2 d x d η .$
Inserting (31)–(33) into (30) and in view of (23), we obtain
$A 3 ( z , t ) ≤ k 1 λ 1 + k 1 μ 2 ω + k 1 λ 2 γ δ 1 ω − ∂ ∂ z E ( z , t ) .$
Since
$δ 1 = 16 λ 2 γ μ , ω ≥ 8 ( λ + δ 1 ) 2 μ δ 1 ,$
by inserting (24)–(28) and (34) into (14), we can obtain Lemma 3.
We also need the bound of $E ( 0 , t )$. Therefore, we bound $E ( 0 , t )$ using known data. □
Lemma 4.
Assume that $u i , q i ,$ and p are the solutions of (1)–(8) with $∫ D g 3 d A = 0$. Then,
$1 2 e − ω t ∫ R μ u i , j u i , j + δ 1 γ q i , j q i , j d x + ∫ 0 t ∫ R e − ω η u i , η u i , η + 1 2 ω μ u i , j u i , j + 1 2 ω γ δ 1 q i , j q i , j + δ 1 q i , j η q i , j η d x d η ≤ Q ( 0 , t ) ,$
and
$E ( 0 , t ) ≤ 1 m 1 Q ( 0 , t ) ,$
where $Q ( 0 , t )$ is a positive computable function.
Proof.
We let $z = 0$ in Lemma 3, and then we can have
$E ( 0 , t ) ≤ 1 m 1 − ∂ ∂ z E ( 0 , t ) .$
Therefore, to bound $E ( 0 , t )$, it is necessary to seek a bound for $− ∂ ∂ z E ( 0 , t )$. In (23), by choosing $z = 0$, we have
$− ∂ ∂ z E ( 0 , t ) = 1 2 e − ω t ∫ R μ u i , j u i , j + δ 1 γ q i , j q i , j d x + ∫ 0 t ∫ R e − ω η u i , η u i , η + 1 2 ω μ u i , j u i , j + 1 2 ω γ δ 1 q i , j q i , j + δ 1 q i , j η q i , j η d x d η .$
From (14), we have
$− ∂ ∂ z E ( 0 , t ) = − μ ∫ 0 t ∫ D e − ω η u i , η u i , 3 d A d η − λ ∫ 0 t ∫ D e − ω η u i , η q i , 3 d A d η + ∫ 0 t ∫ D e − ω η u 3 , η p d A d η − λ e − ω t ∫ R q i , j u i , j d x − λ ω ∫ 0 t ∫ R e − ω η q i , j u i , j d x d η + ( λ + δ 1 ) ∫ 0 t ∫ R e − ω η q i , j η u i , j d x d η ≐ ∑ i = 1 6 Θ i ( 0 , t ) ,$
Now, we let $G$ denote
$G ( x , t ) = g ( x 1 , x 2 , 0 , t ) e − σ 1 x 3 ,$
where $σ 1$ is a positive constant, and $g α , α − σ 1 g 3 = 0$.
Using Equation (1) and the divergence theorem, we have
$Θ 1 ( 0 , t ) + Θ 2 ( 0 , t ) + Θ 3 ( 0 , t ) = − μ ∫ 0 t ∫ D e − ω η G i , η u i , 3 d A d η − λ ∫ 0 t ∫ D e − ω η G i , η q i , 3 d A d η + ∫ 0 t ∫ D e − ω η G 3 , η p d A d η = λ ∫ 0 t ∫ R e − ω η ( q i , j G i , η ) , j d x d η + μ ∫ 0 t ∫ R e − ω η ( u i , j G i , η ) , j d x d η − ∫ 0 t ∫ R e − ω η ( p G i , η ) , i d x d η = μ ∫ 0 t ∫ R e − ω η u i , j G i , j η d x d η + λ ∫ 0 t ∫ R e − ω η q i , j G i , j η d x d η − ∫ 0 t ∫ R e − ω η u i , η G i , η d x d η ≐ ϱ 1 ( 0 , t ) + ϱ 2 ( 0 , t ) + ϱ 3 ( 0 , t ) .$
Combining Hölder’s inequality and Young’s inequality in (39), we obtain
$ϱ 1 ( 0 , t ) ≤ 1 8 μ ω ∫ 0 t ∫ R e − ω η u i , j u i , j d x d η + 2 ω μ ∫ 0 t ∫ R e − ω η G i , j η G i , j η d x d η ,$
$ϱ 2 ( 0 , t ) ≤ 1 8 ω γ δ 1 ∫ 0 t ∫ R e − ω η q i , j q i , j d x d η + 2 γ ω δ 1 λ 2 ∫ 0 t ∫ R e − ω η G i , j η G i , j η d x d η ,$
$ϱ 3 ( 0 , t ) ≤ 1 4 ∫ 0 t ∫ R e − ω η u i , η u i , η d x d η + ∫ 0 t ∫ R e − ω η G i , η G i , η d x d η .$
Inserting (40)–(42) into (39), we have
$Θ 1 ( 0 , t ) + Θ 2 ( 0 , t ) + Θ 3 ( 0 , t ) ≤ 1 4 − ∂ ∂ z E ( 0 , t ) + 1 4 Q ( 0 , t ) ,$
where
$1 4 Q ( 0 , t ) = 2 ω μ + 2 γ ω δ 1 λ 2 ∫ 0 t ∫ R e − ω η G i , j η G i , j η d x d η + ∫ 0 t ∫ R e − ω η G i , η G i , η d x d η .$
Combining Hölder’s inequality and Young’s inequality again, we obtain
$Θ 4 ( 0 , t ) ≤ λ γ μ δ 1 1 2 μ e − ω t ∫ R u i , j u i , j d x + 1 2 γ δ 1 e − ω t ∫ R q i , j q i , j d x ,$
$Θ 5 ( 0 , t ) ≤ λ γ μ δ 1 1 2 ω μ ∫ 0 t ∫ R e − ω η u i , j u i , j d x d η + 1 2 ω γ δ 1 ∫ 0 t ∫ R e − ω η q i , j q i , j d x d η ,$
$Θ 6 ( 0 , t ) ≤ λ + δ 1 2 ω μ δ 1 δ 1 ∫ 0 t ∫ R e − ω η q i , j η q i , j η d x d η + 1 2 ω μ ∫ 0 t ∫ R e − ω η u i , j u i , j d x d η .$
Since
$δ 1 = 16 λ 2 γ μ , ω ≥ 8 ( λ + δ 1 ) 2 μ δ 1 ,$
we have
$Θ 4 ( 0 , t ) + Θ 5 ( 0 , t ) + Θ 6 ( 0 , t ) ≤ 1 2 − ∂ ∂ z E ( 0 , t ) .$
Inserting (43) and (47) into (37), we have
$− ∂ ∂ z E ( 0 , t ) ≤ 3 4 − ∂ ∂ z E ( 0 , t ) + 1 4 Q ( 0 , t ) .$
Combining (35), (36), and (48), we can obtain Lemma 4.
From Lemma 3, it follows that
$∂ ∂ z E ( z , t ) e m 1 z ≤ 0 .$
Integrating (49) from 0 to z, we have
$E ( z , t ) ≤ E ( 0 , t ) e − m 1 z .$
Combining (50) and Lemma 4, we can obtain Theorem 1. □

## 4. The Proof of Theorem 2

From Theorem 1, we have the following result
$∫ 0 t ∫ R z e − ω η ( ξ − z ) q i , j q i , j d x d η ≤ 2 ω γ ∗ m 1 δ 1 Q ( 0 , t ) e − m 1 z .$
From Lemma 4, it can be followed that
$∫ 0 t ∫ R e − ω η q i , j q i , j d x d η ≤ 1 ω γ ∗ δ 1 Q ( 0 , t ) .$
We let $F ( z , t )$ denote
$F ( z , t ) = − μ ∫ 0 t ∫ R z e − ω η v i , η v i , 3 d x d η − λ ∫ 0 t ∫ R z e − ω η v i , η Σ i , 3 d x d η + ∫ 0 t ∫ R z e − ω η π v 3 , η d x d η .$
By using the divergence theorem and (15)–(18), (22), we obtain
$F ( z , t ) = − λ ∫ 0 t ∫ R z e − ω η ( ξ − z ) , j v i , η Σ i , j d x d η − μ ∫ 0 t ∫ R z e − ω η ( ξ − z ) , j v i , η v i , j d x d η + ∫ 0 t ∫ R z e − ω η ( ξ − z ) , i π v i d x d η = ∫ 0 t ∫ R z e − ω η ( ξ − z ) v i , η v i , η d x d η + μ ∫ 0 t ∫ R z e − ω η ( ξ − z ) v i , j η v i , j d x d η + λ ∫ 0 t ∫ R z e − ω η ( ξ − z ) v i , j η Σ i , j d x d η = 1 2 ω μ ∫ 0 t ∫ R z e − ω η ( ξ − z ) v i , j v i , j d x d η + 1 2 e − ω t μ ∫ R z ( ξ − z ) v i , j v i , j d x + λ e − ω t ∫ R z ( ξ − z ) Σ i , j v i , j d x + ∫ 0 t ∫ R z e − ω η ( ξ − z ) v i , η v i , η d x d η − λ ∫ 0 t ∫ R z e − ω η ( ξ − z ) Σ i , j η v i , j d x d η + ω λ ∫ 0 t ∫ R z e − ω η ( ξ − z ) Σ i , j v i , j d x d η .$
We begin with
$∫ 0 t ∫ R z ( ξ − z ) e − ω η Σ i , j η + γ ˜ q i , j + γ ∗ Σ i , j − v i , j Σ i , j η d x d η = 0 .$
Using (17), we have from (55) that
$∫ 0 t ∫ R z ( ξ − z ) e − ω η Σ i , j η Σ i , j η d x d η + 1 2 e − ω t γ ∗ ∫ R z ( ξ − z ) Σ i , j Σ i , j d x − ∫ 0 t ∫ R z ( ξ − z ) e − ω η Σ i , j η v i , j d x d η + 1 2 ω γ ∗ ∫ 0 t ∫ R z ( ξ − z ) e − ω η Σ i , j Σ i , j d x d η = − γ ˜ ∫ 0 t ∫ R z ( ξ − z ) e − ω η q i , j Σ i , j η d x d η .$
Now, we let
$H ( z , t ) = 1 2 e − ω t μ ∫ R z ( ξ − z ) v i , j v i , j d x + 1 2 ω μ ∫ 0 t ∫ R z e − ω η ( ξ − z ) v i , j v i , j d x d η + ∫ 0 t ∫ R z e − ω η ( ξ − z ) v i , η v i , η d x d η + ∫ 0 t ∫ R z ( ξ − z ) e − ω η Σ i , j η Σ i , j η d x d η + 1 2 e − ω t γ ∗ ∫ R z ( ξ − z ) Σ i , j Σ i , j d x + 1 2 ω γ ∗ ∫ 0 t ∫ R z ( ξ − z ) e − ω η Σ i , j Σ i , j d x d η + λ e − ω t ∫ R z ( ξ − z ) Σ i , j v i , j d x + ω λ ∫ 0 t ∫ R z e − ω η ( ξ − z ) Σ i , j v i , j d x d η − λ ∫ 0 t ∫ R z e − ω η ( ξ − z ) Σ i , j η v i , j d x d η − ∫ 0 t ∫ R z ( ξ − z ) e − ω η Σ i , j η v i , j d x d η .$
Combining (53) and (56), we can have
$H ( z , t ) = − λ ∫ 0 t ∫ R z e − ω η v i , η Σ i , 3 d x d η − μ ∫ 0 t ∫ R z e − ω η v i , η v i , 3 d x d η + ∫ 0 t ∫ R z e − ω η π v 3 , η d x d η − γ ˜ δ 2 ∫ 0 t ∫ R z ( ξ − z ) e − ω η q i , j Σ i , j η d x d η ≐ ∑ i = 1 4 I i ( z , t ) .$
It follows from (57) that
$− ∂ ∂ z H ( z , t ) = 1 2 e − ω t μ ∫ R z v i , j v i , j d x + 1 2 ω μ ∫ 0 t ∫ R z e − ω η v i , j v i , j d x d η + ∫ 0 t ∫ R z e − ω η v i , η v i , η d x d η + 1 2 δ 2 ω γ ∗ ∫ 0 t ∫ R z e − ω η Σ i , j Σ i , j d x d η + δ 2 ∫ 0 t ∫ R z e − ω η Σ i , j η Σ i , j η d x d η + 1 2 δ 2 e − ω t γ ∗ ∫ R z Σ i , j Σ i , j d x + λ e − ω t ∫ R z Σ i , j v i , j d x + ω λ ∫ 0 t ∫ R z e − ω η Σ i , j v i , j d x d η − ( λ + δ 2 ) ∫ 0 t ∫ R z e − ω η Σ i , j η v i , j d x d η ,$
where $δ 2$ is a positive constant.
In view of Hölder’s and Young’s inequalities, we obtain
$λ e − ω t ∫ R z Σ i , j v i , j d x ≥ − 1 2 e − ω t μ ∫ R z v i , j v i , j d x − λ 2 2 μ e − ω t ∫ R z Σ i , j Σ i , j d x , ω λ ∫ 0 t ∫ R z e − ω η Σ i , j v i , j d x d η ≥ − 1 8 ω μ ∫ 0 t ∫ R z e − ω η v i , j v i , j d x d η − 2 λ 2 ω μ ∫ 0 t ∫ R z e − ω η Σ i , j Σ i , j d x d η , ( λ + δ 2 ) ∫ 0 t ∫ R z e − ω η Σ i , j η v i , j d x d η ≥ − 1 8 ω μ ∫ 0 t ∫ R z e − ω η v i , j v i , j d x d η − 2 ( λ + δ 2 ) 2 ω μ ∫ 0 t ∫ R z e − ω η Σ i , j η Σ i , j η d x d η .$
Inserting the above inequalities into (56) and choosing $δ 2 ≥ max { λ γ ∗ μ , 4 λ 2 γ ∗ μ } , ω = 4 ( λ + δ 2 ) 2 μ δ 2$, we obtain
$− ∂ ∂ z H ( z , t ) ≥ 1 4 ω μ ∫ 0 t ∫ R z e − ω η v i , j v i , j d x d η + ∫ 0 t ∫ R z e − ω η v i , η v i , η d x d η + 1 4 δ 2 ω γ ∗ ∫ 0 t ∫ R z e − ω η Σ i , j Σ i , j d x d η + 1 2 δ 2 ∫ 0 t ∫ R z e − ω η Σ i , j η Σ i , j η d x d η .$
Next, we use similar methods to those in (24), (25), and (34) to obtain
$I 1 ( z , t ) ≤ μ ω 1 4 ω μ ∫ 0 t ∫ R z e − ω η v i , 3 v i , 3 d x d η + ∫ 0 t ∫ R z e − ω η v i , η v i , η d x d η ,$
$I 2 ( z , t ) ≤ λ γ ∗ δ 2 ω 1 4 ω γ ∗ δ 2 ∫ 0 t ∫ R z e − ω η Σ i , 3 Σ i , 3 d x d η + ∫ 0 t ∫ R z e − ω η v i , η v i , η d x d η , I 3 ( z , t ) ≤ k 2 λ 1 + k 2 μ ω + k 2 λ γ ∗ δ 1 ω − ∂ ∂ z F ( z , t ) ,$
$I 4 ( z , t ) ≤ − 1 4 δ 2 ∫ 0 t ∫ R z ( ξ − z ) e − ω η Σ i , j η Σ i , j η d x d η + δ 2 γ ˜ 2 ∫ 0 t ∫ R z ( ξ − z ) e − ω η q i , j q i , j d x d η .$
Combining (60)–(63), we have
$H ( z , t ) ≤ 1 2 m 2 − ∂ ∂ z H ( z , t ) + 1 2 H ( z , t ) + δ 2 γ ˜ 2 ∫ 0 t ∫ R z ( ξ − z ) e − ω η q i , j q i , j d x d η ,$
where
$1 2 m 2 = μ ω + λ γ ∗ δ 2 ω + k 2 λ 1 + k 2 μ ω + k 2 λ γ ∗ δ 1 ω .$
Combining (51) and (64), we have
$H ( z , t ) ≤ 1 m 2 − ∂ ∂ z H ( z , t ) + 4 δ 2 ω γ ∗ m 1 δ 1 Q ( 0 , t ) γ ˜ 2 e − m 1 z ,$
or
$∂ ∂ z H ( z , t ) e m 2 z ≤ 4 m 2 δ 2 ω γ ∗ m 1 δ 1 Q ( 0 , t ) γ ˜ 2 e − ( m 1 − m 2 ) z ,$
Integrating (61) from 0 to z, we have
$H ( z , t ) ≤ H ( 0 , t ) e − m 2 z + 4 m 2 δ 2 ω γ ∗ m 1 ( m 1 − m 2 ) δ 1 Q ( 0 , t ) γ ˜ 2 e − m 1 z − e − m 2 z , i f m 1 ≠ m 2 ,$
$H ( z , t ) ≤ H ( 0 , t ) e − m 2 z + 4 m 2 δ 2 ω γ ∗ m 1 δ 1 Q ( 0 , t ) γ ˜ 2 z e − m 2 z , i f m 1 = m 2 .$
Now, we choose $z = 0$ in (65) to obtain
$H ( 0 , t ) ≤ 1 m 2 − ∂ ∂ z H ( 0 , t ) + 4 δ 2 ω γ ∗ m 1 δ 1 Q ( 0 , t ) γ ˜ 2 ,$
From (68), we have
$− ∂ ∂ z H ( 0 , t ) = − μ ∫ 0 t ∫ D e − ω η v i , η v i , 3 d A d η − λ ∫ 0 t ∫ D e − ω η v i , η Σ i , 3 d A d η + ∫ 0 t ∫ D e − ω η π v 3 , η d A d η − γ ˜ δ 2 ∫ 0 t ∫ R e − ω η q i , j Σ i , j η d x d η .$
From (60), we obtain
$− ∂ ∂ z H ( 0 , t ) ≥ 1 4 ω μ ∫ 0 t ∫ R e − ω η v i , j v i , j d x d η + ∫ 0 t ∫ R e − ω η v i , η v i , η d x d η + 1 4 δ 2 ω γ ∗ ∫ 0 t ∫ R e − ω η Σ i , j Σ i , j d x d η + 1 2 δ 2 ∫ 0 t ∫ R e − ω η Σ i , j η Σ i , j η d x d η .$
In view of (20) and (21), we can conclude from (70) that
$− ∂ ∂ z H ( 0 , t ) = − γ ˜ ∫ 0 t ∫ R e − ω η q i , j Σ i , j η d x d η ≤ 1 4 δ 2 ∫ 0 t ∫ R e − ω η Σ i , j η Σ i , j η d x d η + γ ˜ 2 δ 2 ∫ 0 t ∫ R e − ω η q i , j q i , j d x d η .$
Using (52) and (71), we have from (72) that
$− ∂ ∂ z H ( 0 , t ) ≤ 1 2 − ∂ ∂ z H ( 0 , t ) + 1 ω γ ∗ δ 1 δ 2 Q ( 0 , t ) γ ˜ 2 ,$
or
$− ∂ ∂ z H ( 0 , t ) ≤ 2 δ 2 ω γ ∗ δ 1 Q ( 0 , t ) γ ˜ 2 .$
Inserting (73) into (69), we have
$H ( 0 , t ) ≤ 2 δ 2 ω γ ∗ δ 1 Q ( 0 , t ) γ ˜ 2 1 m 2 + 2 m 1 .$
Inserting (74) into (67) and (68), we have
$H ( z , t ) ≤ 2 δ 2 ω γ ∗ δ 1 Q ( 0 , t ) γ ˜ 2 1 m 2 + 2 m 1 e − m 2 z$
$+ 4 m 2 δ 2 ω γ ∗ m 1 ( m 1 − m 2 ) δ 1 Q ( 0 , t ) γ ˜ 2 e − m 1 z − e − m 2 z , i f m 1 ≠ m 2 ,$
$H ( z , t ) ≤ 2 δ 2 ω γ ∗ δ 1 Q ( 0 , t ) γ ˜ 2 1 m 2 + 2 m 1 e − m 2 z + 4 m 2 δ 2 ω γ ∗ m 1 δ 1 Q ( 0 , t ) γ ˜ 2 z e − m 2 z , i f m 1 = m 2 .$
Inequalities (75) and (76) show that $H ( z , t )$ decays exponentially as $z → ∞$. Therefore, integrating (60) from z to , we obtain
$F ( z , t ) ≥ 1 4 ω μ ∫ 0 t ∫ R z ( ξ − z ) e − ω η v i , j v i , j d x d η + ∫ 0 t ∫ R z ( ξ − z ) e − ω η v i , η v i , η d x d η + 1 4 δ 2 ω γ ∗ ∫ 0 t ∫ R z ( ξ − z ) e − ω η Σ i , j Σ i , j d x d η + 1 2 δ 2 ∫ 0 t ∫ R z ( ξ − z ) e − ω η Σ i , j η Σ i , j η d x d η .$
Combining (75), (76), and (77), we can obtain Theorem 2.

## 5. Conclusions

In this article, we used energy estimations and prior estimations to obtain the properties and structural stability of the solution on a cylinder. This can also provide reference for further research on other fluid equation systems.

## Funding

This research was funded by the Research team project of Guangzhou Huashang College (2021HSKT01).

Not applicable.

Not applicable.

## Data Availability Statement

No new data was created in the manuscript.

## Acknowledgments

The author would like to deeply thank all the reviewers for their insightful and constructive comments.

## Conflicts of Interest

The author declares no conflicts of interest.

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Figure 1. Cylindrical pipe R.
Figure 1. Cylindrical pipe R.
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Li, Y. Spatial Decay Estimates and Continuous Dependence for the Oldroyd Fluid. Mathematics 2024, 12, 1240. https://doi.org/10.3390/math12081240

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Li Y. Spatial Decay Estimates and Continuous Dependence for the Oldroyd Fluid. Mathematics. 2024; 12(8):1240. https://doi.org/10.3390/math12081240

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Li, Yuanfei. 2024. "Spatial Decay Estimates and Continuous Dependence for the Oldroyd Fluid" Mathematics 12, no. 8: 1240. https://doi.org/10.3390/math12081240

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