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Article

Strong Solution for a Nonlinear Non-Newtonian Shear Thickening Fluid

College of Science, Liaoning University of Technology, Jinzhou 121001, China
*
Author to whom correspondence should be addressed.
Mathematics 2025, 13(5), 878; https://doi.org/10.3390/math13050878
Submission received: 11 February 2025 / Revised: 2 March 2025 / Accepted: 4 March 2025 / Published: 6 March 2025
(This article belongs to the Special Issue Mathematical Modeling for Fluid Mechanics)

Abstract

:
This paper consider a nonlinear shear thickening fluid in one dimensional bounded interval. The model illustrates that the movement of the compressible fluid is driven by non-Newtonian gravity, and represents a more realistic phenomenon. The well-posedness of strong solution was proved by considering the influence of damping term. The essential difficulty lies in the equation’s significant nonlinearity and the initial state may allow for vacuum.

1. Introduction

We study a 1D compressible non-Newtonian fluid with damping term represented by the following form
ρ t + ( ρ u ) x = 0 ( ρ u ) t + ( ρ u 2 ) x + ρ Φ x u x 2 + μ 0 p 2 2 u x x + π x = ϑ ρ u ε Φ x 2 + 1 Φ x 2 + ε 2 q 2 Φ x x = 4 π g ρ 1 Ω Ω ρ d x π = A ρ γ , A > 0 , γ > 1
with the initial and boundary condition
ρ , u | t = 0 = ρ 0 , u 0 , x Ω u | Ω = Φ | Ω = 0 , t 0 , T
where p > 2 , 1 < q < 2 and Ω = 1 , 1 , Ω T = Ω × 0 , T . Variables u, ρ , Φ , π respectively represent the velocity (m/s), density (kg/m3), gravitational potential (m2/s2) and pressure, μ 0 is given constants and μ 0 > 0 , γ is adiabatic gas index, constant ϑ > 0 is the frictional force, without loss of generality, throughout the paper we take ϑ = 1 .
Fluid dynamics has become a popular topic of research at home and abroad. The Navier-Stokes (NS) equations have important applications in fluid dynamics, capturing the behavior of viscous fluids, whether compressible or incompressible. These equations facilitate the portrayal of fluid dynamics across both spatial and temporal dimensions.
There are many results regarding Newtonian fluid solutions. The well-posedness of weak solutions for incompressible viscous fluids under inconsistent or unclear boundary conditions were discussed in [1]. Reference [2] analyzed the suitability of global solution for incompressible NS equations when viscosity coefficient changes in relation to fluid density. In [3], the local existence of solution for compressible NS equation was obtained. Based on this, ref. [4] examined the well-posedness of classical solution of compressible NS equation that the initial density ρ 0 x is non-negative and may be zero. Reference [5] extended the global existence result of solutions in R 3 from [6] for a case of γ 3 / 2 .
In recent decades, there has also been a lot of attention paid in the investigation of non-Newtonian flow. In [7], non-Newtonian fluid was divided into shear thinning fluid and shear thickening fluid by different ranges of the value of p. In [8], Yuan et al. discussed the existence of local solutions of 1-D non-Newtonian fluid in which viscous term is singular and completely nonlinear. Reference [9] proved the existence of solution of a non-Newtonian fluid-particle interaction model. For other relevant results, please make readers reference to [10,11,12,13,14,15,16] and the references quoted therein. This study advances the field of civil engineering and environmental sciences by introducing gravity and damping terms that reveal the complex behavior of shear flows under different conditions, thus providing a more realistic model than previous studies focusing on ideal fluids.
This paper focus on the shear thickening model with gravitational potential and damping in the case of p > 2 , 1 < q < 2 . The damping term is usually expressed as internal friction or external resistance of the fluid, and model (1) is more reflective of the real physical phenomena. In this paper, the physical modeling of the fluid faces many complexities due to its nonlinear behavior which is affected by gravity, damping, and multiscale. In addition, the boundary conditions need to be rationalized and the coupling needs to be taken into account.

Main Result

Theorem 1.
Let p > 2 , 1 < q < 2 , and the initial condition ( ρ 0 , u 0 ) satisfy ρ 0 H 1 ( Ω ) and u 0 H 0 1 ( Ω ) H 2 ( Ω ) . In addition, the compatibility conditions for almost everywhere x Ω are met as follows
u 0 x 2 + μ 0 p 2 2 u 0 x x + π x ( ρ 0 ) = ρ 0 1 2 g , for some g L 2 ( Ω )
Then there exists a unique strong solution ρ , u , Φ to problem (1) and (2) for a T 0 , + that follows
ρ C ( 0 , T ; H 1 ( Ω ) ) , u C ( 0 , T ; H 0 1 ( Ω ) ) L ( 0 , T ; H 2 ( Ω ) ) Φ L ( 0 , T ; H 2 ( Ω ) ) , ρ t C ( 0 , T ; L 2 ( Ω ) ) u t L 2 ( 0 , T ; H 0 1 ( Ω ) ) , ρ u t L ( 0 , T ; L 2 ( Ω ) ) Φ t L ( 0 , T ; H 1 ( Ω ) ) , u x 2 + μ 0 p 2 2 u x x L 2 0 , T ; L 2 Ω

2. A Priori Estimates

In this part, we intend to demonstrate local existence of strong solution. Let ( ρ , u ) be a smooth solution of (1) and (2). It can be obtained that
Ω ρ ( t ) x ̣ = Ω ρ 0 x ̣ : = m 0 ( > 0 )
where ρ 0 δ with 0 < δ 1 . We get the estimate of u 0 x x L 2 . Assume that the boundary value issue has a smooth solution u 0 H 0 1 ( Ω ) H 2 ( Ω )
u 0 x 2 + μ 0 p 2 2 u 0 x x + π x ( ρ 0 ) = ρ 0 1 2 g u 0 | Ω = 0
From (3), we have
u 0 x 2 p 1 + μ 0 u 0 x 2 + μ 0 p 4 2 u 0 x x u 0 x 2 + μ 0 p 2 2 u 0 x x μ 0 p 2 2 u 0 x x
then
u 0 x x L 2 μ 0 2 p 2 π x ( ρ 0 ) ( ρ 0 ) 1 2 g L 2 μ 0 2 p 2 ( A γ | ρ 0 x | L 2 | ρ 0 | L γ 1 + | ρ 0 | L 1 2 | g | L 2 ) C
where a constant C depends only on M 0 , M 0 is denoted as
M 0 = 1 + ρ 0 H 1 + u 0 H 0 1 H 2 + g L 2 + μ 0 + μ 0 1
Define function
Ψ ( t ) = sup 0 s t ( 1 + ρ ( s ) H 1 + u ( s ) W 0 1 , p + ρ u t ( s ) L 2 )
Next, after estimating each term of Ψ ( t ) we will show that it is locally bounded.

Preliminaries

We provide several valuable lemmas to consider for subsequent use in proving the main Theorem.
Lemma 1.
u x x L 2 C Ψ max q q 1 , 3 t
where C is only dependent on M 0 .
Proof. 
This can be seen from (1) that
u x 2 + μ 0 p 2 2 u x x = ρ u + ρ u t + ρ u u x + ρ Φ x + π x
Since μ 0 p 2 2 u x x u x 2 + μ 0 p 2 2 u x x , then
u x x μ 0 2 p 2 ρ u + ρ u t + ρ u u x + ρ Φ x + π x
By applying L 2 norm to the aforementioned inequality that obtain
u x x L 2 C ρ u + ρ u u x + ρ u t + ρ Φ x + π x L 2 C ρ L u L p + ρ L u x L p u L + ρ L 1 2 ρ u t L 2 + ρ L Φ x x L 2 + π x L 2
We deal with | Φ x x | L 2 , multiply (1) by Φ , integrat it with respect to x to give
Ω ε Φ x 2 + 1 Φ x 2 + ε 2 q 2 Φ x x Φ d x = 4 π g Ω ρ Φ d x m 0 Ω Φ d x
integrating by parts, using embedding theorem and Young’s inequality as follows
ε 2 q 2 Ω Φ x q d x Ω ε Φ x 2 + 1 Φ x 2 + ε 2 q 2 Φ x 2 d x = 4 π g Ω ρ Φ d x m 0 Ω Φ d x 8 π g m 0 Φ L 8 π g m 0 Φ x L q 1 q Φ x L q q + 1 p ( 8 π g m 0 ) p
and so
Ω Φ x q d x C m 0 , 1 < q < 2
It can also be obtained from (1) that
Φ x x = ε Φ x 2 + 1 Φ x 2 + ε q 2 Φ x 2 + ε 2 4 π g ρ m 0 Φ x 2 + ε ε Φ x 2 + 1 2 q 1 ε 2 Φ x 2 1 q 1 Φ x 2 + ε 2 q 2 4 π g ρ m 0 1 q 1 Φ x 2 q + 1 4 π g ρ m 0
Take L 2 norm on both side, use Young’s inequality that obtain
| Φ x x | L 2 C Ψ 1 q 1 ( t )
Together with the above estimates and (5), then Lemma 1 is established. □
Lemma 2.
sup 0 t T ρ H 1 2 C exp C 0 t Ψ max q q 1 + 1 , 3 d s
where a positive constant C is only dependent on M 0 .
Proof. 
To begin with, multiplying (1) by ρ , integrating it with regard to x, using integration by parts, Sobolev embedding theorem that
1 2 d d t ρ L 2 2 = Ω ρ u ρ x d x = 1 2 Ω u ρ 2 x d x = 1 2 Ω u x ρ 2 d x 1 2 ρ L 2 2 u x L C ρ L 2 2 u x x L 2
Thus can be obtained
d d t ρ L 2 2 C ρ L 2 2 u x x L 2
Taking the derivative of x in (1), multiplying by ρ x and integrating with regard to x gives
d d t Ω ρ x 2 d x = Ω 3 2 u x ρ x 2 + u x x ρ x ρ d x 3 2 Ω u x ρ x 2 + u x x ρ x ρ d x 3 2 u x L ρ x L 2 2 + u x x L 2 ρ x L 2 ρ L C u x x L 2 ρ x H 1 2 + u x x L 2 ρ x H 1 2 C u x x L 2 ρ x H 1 2
By combining (8) and (9), Gronwall’s inequality, then Lemma 2 holds. □
Lemma 3.
0 t ρ u t ( s ) L 2 2 d s + u x L p p C 1 + 0 t Ψ max 2 p 2 q + 1 , 2 γ + 3 ( s ) d s
where C is only dependent on M 0 .
Proof. 
Multiplying (1) by u t , integrating it on Ω T that
Ω T ρ u t 2 d x d s + Ω T u x 2 + μ 0 p 2 2 u x u x t d x d s = Ω π u x ( t ) d x Ω π u x ( 0 ) d x + Ω T u t ρ u ρ u u x ρ Φ x π t u x d x d s
First calculating the following components
Ω u x 2 + μ 0 p 2 2 u x u x t d x = 1 2 Ω u x 2 + μ 0 p 2 2 u x t 2 d x = 1 2 d d t Ω 0 u x 2 s + μ 0 p 2 2 d s d x
and
0 u x 2 s + μ 0 p 2 2 d s = 2 p u x 2 + μ 0 p 2 μ 0 p 2 2 p u x p μ 0 p 2
By (1), we can derive π t = γ π u x π x u
As the aforementioned is substituted into (11), and combine Young’s inequality and Sobolev inequality gives
0 t ρ u t L 2 2 d s + 1 p u x L p p C + Ω π u x ( t ) d x + Ω T u t ρ u ρ u u x ρ Φ x π t u x d x d s C + Ω π u x ( t ) d x + Ω T ρ u u t + ρ u u x u t + ρ Φ x u t + γ π u x 2 + π x u u x d x d s C + k = 1 6 l k
Additionally, it can be obtained from (1) that
π ( t ) H 1 + ρ ( t ) L C ρ ( t ) L γ 1 ρ ( t ) H 1 + ρ ( t ) H 1 C Ψ γ ( t )
Then we estimate each l k , ( k = 1 , , 6 ) as follows.
l 1 = Ω π u x t d x C α π ( t ) L 2 p p 1 + 1 2 u x ( t ) L p p C α π ( t ) L 2 p p 1 Ω π ( t ) 2 d x Ω π ( 0 ) 2 d x + 0 t s Ω ( π ( s ) ) 2 d x d s Ω π ( 0 ) 2 d x + 2 Ω T ρ x u ρ u x π ( ρ ) π d x d s C + C 0 t ρ ( s ) L 2 u x ( s ) L p ρ ( s ) L γ 1 π ( s ) L d s C 1 + 0 t Ψ 2 γ + 1 s d s l 2 = Ω T ρ u u t d x d s C α 0 t ρ ( s ) L u s L p 2 d s 0 t Ψ 3 t d s l 3 = Ω T ρ u u x u t d x d s C α 0 t ρ ( s ) L u x ( s ) L p 2 u s L 2 d s C α 0 t ρ ( s ) L u x ( s ) L p 4 d s 0 t Ψ 5 t d s l 4 = Ω T ρ Φ x u t d x d s C α 0 t ρ ( s ) L 2 Φ x x ( s ) L 2 2 d s 0 t Ψ q + 1 q 1 t d s l 5 = Ω T γ π u x 2 d x d s C α 0 t π ( s ) L u x ( s ) L p 2 d s 0 t Ψ γ + 2 t d s l 6 = Ω T π x u u x d x d s C α 0 t π x ( s ) L 2 u s L u x ( s ) L p d s C α 0 t π x ( s ) L 2 u x s L p 2 u x ( s ) L p d s 0 t Ψ γ + 3 t d s
where 0 < α 1 , substituting the above estimates into (12), we then succeed in proving Lemma 3. □
Lemma 4.
ρ u t t L 2 2 + 0 t u x t L 2 2 d s C 1 + 0 t Ψ max 4 q 2 q 1 , 8 s d s
where C is only dependent on M 0 .
Proof. 
Taking derivative of t in (1), multiplying by u t , integrating on Ω that obtains
1 2 d d t Ω ρ u t 2 d x + Ω u x 2 + μ 0 p 2 2 u x t u x t d x = Ω π t u x t d x + Ω ρ u x u u t u u x Φ x ρ u t u x ρ Φ x t + ρ u t u t d x
It follows that
u x 2 + μ 0 p 2 2 u x t u x t = u x 2 p 1 + μ 0 u x 2 + μ 0 p 4 2 u x t 2 μ 0 p 2 2 u x t 2
Consequently, (14) may be reformulated as
1 2 d d t Ω ρ u t 2 d x + Ω u x t 2 d x C Ω π x u u x t + γ π u x u x t + ρ u u x u t + ρ x u t u 2 + ρ x u x u t u 2 + ρ u u t u x 2 + 2 ρ u u t u x t + ρ x u u t Φ x + ρ u x u t Φ x + ρ u t 2 u x + ρ u t 2 + ρ u t Φ x t d x = j = 1 12 I j
Taking advantage of the embedding theorem and Young’s inequality yields
I 1 = Ω π x u c d x π x L 2 u L u x t L 2 C Ψ 2 ( γ + 1 ) t + 1 6 u x t L 2 2 I 2 = Ω γ π u x u x t d x C π L u x L p u x t L 2 C Ψ 2 ( γ + 1 ) t + 1 6 u x t L 2 2 I 3 = Ω ρ u t u u x d x ρ L 1 2 ρ u t L 2 u x L p u x x L 2 C Ψ max q q 1 + 3 , 6 t I 4 = Ω ρ x u t u 2 d x ρ x L 2 u x L p 2 u x t L 2 C Ψ 6 t + 1 6 u x t L 2 2 I 5 = Ω ρ x u x u t u 2 d x ρ x L 2 u x L p u x L p 2 u x t L 2 C Ψ 8 t + 1 6 u x t L 2 2 I 6 = Ω ρ u u t u x 2 d x ρ L 1 2 u x L p 2 u x L p ρ u t L 2 C Ψ 5 t
I 7 = 2 Ω ρ u t u u x t d x 2 ρ L 1 2 ρ u t L 2 u L u x t L 2 C Ψ 5 t + 1 6 u x t L 2 2 I 8 = Ω ρ x u Φ x u t d x ρ x L 2 u x L p Φ x x L 2 u x t L 2 C Ψ 4 q 2 q 1 t + 1 6 u x t L 2 2 I 9 = Ω ρ u t Φ x u x d x ρ L 1 2 ρ u t L 2 Φ x x L 2 u x L p C Ψ 2 q 1 q 1 t I 10 = Ω ρ u t 2 u x d x ρ u t L 2 2 u x x L 2 C Ψ max 3 q 2 q 1 , 5 t I 11 = Ω ρ u t 2 d x ρ u t L 2 2 C Ψ 2 t I 12 = Ω ρ u t Φ x t d x ρ L 1 2 ρ u t L 2 Φ x t L 2
In order to evaluate I 12 , we must estimate Φ x t L 2 , taking the derivative of t in (1), multiplying by Φ t , integrating with regard to x, we get
q 1 ε 2 q 2 Ω Φ x t 2 d x q 1 Ω ε Φ x 2 + 1 Φ x 2 + ε 2 q 2 Φ x t 2 d x Ω ε Φ x 2 + 1 Φ x 2 + ε 2 q 2 Φ x t Φ x t d x = Ω ε Φ x 2 + 1 Φ x 2 + ε 2 q 2 Φ x x t Φ t d x = 4 π g Ω ρ t Φ t d x C ρ t L 2 Φ t L 2 ε Φ x t L 2 2 + C ρ t L 2 2
By (1), we get
ρ t k L 2 ρ k L u x k 1 L 2 + ρ x k L 2 u k 1 L C Ψ 2 ( t )
We thus obtain
Φ x t L 2 C Ψ 2 t
Combining these results gives
I 12 = Ω ρ u t Φ x t d x ρ L 1 2 ρ u t L 2 Φ x t L 2 Ψ 4 t
Next, it is obtained by substituting I j j = 1 , 2 , , 12 into (16), integrating with regard to t τ , t , we make it to the conclusion that
ρ u t t L 2 2 + τ t u x t L 2 2 d s ρ u t τ L 2 2 + C τ t Ψ max 4 q 2 q 1 , 8 s d s
To estimate ρ u t t L 2 2 , we have to determine lim τ 0 ρ u t L 2 2 , multiply (1) by u t , integrate it over Ω , we have
Ω ρ u t 2 d x C Ω ρ u 2 u x 2 + ρ 1 π x u x 2 + μ 0 p 2 2 u x x 2 + ρ Φ x 2 + ρ u 2 d x
Because of the smoothness of ρ , u , Φ that
lim τ 0 Ω ρ u 2 u x 2 + ρ 1 π x u x 2 + μ 0 p 2 2 u x x 2 + ρ Φ x 2 + ρ u 2 x , τ d x = Ω ρ 0 u 0 2 u 0 x 2 + ρ 0 1 π x ρ 0 u 0 x 2 + μ 0 p 2 2 u 0 x x 2 + ρ 0 Φ x 2 + ρ 0 u 0 2 d x ρ 0 L u 0 L 2 u 0 x L 2 2 + g L 2 2 + ρ 0 L Φ x L 2 2 + ρ 0 L u 0 L 2 2 C
We take the limit of τ and substitute it into (19), then Lemma 4 holds. □
With the help of above Lemmas and the definition of Ψ , we deduce that
| u x | L p ( Ω ) p + | ρ | H 1 ( Ω ) + | ρ u t | L 2 ( Ω ) + 0 t | ρ u t ( s ) | L 2 ( Ω ) 2 d s + 0 t | u x t | L 2 ( Ω ) 2 d s C exp C ˜ 0 t Ψ max 4 q 2 q 1 , 8 s d s
where C and C ˜ are only dependence on M 0 and are positive constants. By the above inequality, there is a time T 1 > 0 with the following result
e s s sup 0 t T 1 ρ H 1 + ρ u t L 2 + ρ t L 2 + u W 0 1 , p H 2 + Φ H 2 + 0 T 1 u x t L 2 2 d s C

3. Convergence of Approximate Solution

In this part, we follow the idea to construct approximate solution by a standard iteration argument in [8], and prove the approximative solution strongly convergence to the initial value problem. We first define a function u 0 = 0 , assume u k 1 where k 1 . The following problem has a unique smooth solution ρ k , u k , Φ k .
ρ t k + ρ x k u k 1 + ρ k u x k 1 = 0 ρ k u t k + ρ k u k 1 u x k + ρ k Φ x k u x k 2 + μ 0 p 2 2 u x k x + π x k = ρ k u k ε Φ x k 2 + 1 Φ x k 2 + ε 2 q 2 Φ x k x = 4 π g ρ k m 0
with the initial and boundary condition
ρ k , u k , Φ k | t = 0 = ρ 0 , u 0 , Φ 0 u k | Ω = Φ k | Ω = 0
where ρ k , u k , Φ k represents the density, velocity and gravitational potential at the k-th time step, respectively. μ 0 represents the coefficient of viscosity. π x k represents the partial derivative of pressure with respect to spatial position for the k-th time step.
During this iteration process, decomposition of the non-linear coupled system into a series of decoupled problems, every one of which allows for smooth solutions. Meanwhile, the following estimate holds
e s s sup 0 t T 1 ρ k H 1 + ρ k u t k L 2 + u k W 0 1 , p H 2 + Φ k H 2 + ρ t k L 2 + 0 T 1 u x t k L 2 2 d s C
where a normal constant C depends only on M 0 and not on k.
Through iteration, we can first get a smooth solution ρ k that satisfies the initial value problem that follows
ρ t k + u k 1 ρ x k + u x k 1 ρ k = 0 ρ k | t = 0 = ρ 0
The previously mentioned problem has a unique solution ρ k for a known smooth function u k 1 . From the method of characteristics that
d d t ρ k = ρ k u x k 1 d d t x = u k 1 x , t
x | t = 0 = x 0 , ρ k | t = 0 = ρ 0 δ
From (23) and (24), it is possible to derive
x ( t ) = x 0 + 0 t u k 1 x s , s d s : = X x 0 , t
And from (23), one can find that d ln ρ k = u x k 1 d t . That means
ρ k x , t = ρ 0 X x 0 , t exp 0 t u x k 1 X x 0 , s , s d s
with t 0 , T 1 , applying the Sobolev inequality, yields
ρ k x , t C exp 0 T 1 u x k 1 · , s L d s > 0
We demonstrate that ρ k , u k , Φ k strongly converges to ρ , u , Φ . Initially, define as follows variables
ρ ¯ k + 1 = ρ k + 1 ρ k , u ¯ k + 1 = u k + 1 u k , Φ ¯ k + 1 = Φ k + 1 Φ k
The following equations are satisfied by ρ ¯ k + 1 , u ¯ k + 1 , Φ ¯ k + 1
ρ ¯ t k + 1 + ( ρ k u ¯ k ) x + ( ρ ¯ k + 1 u k ) x = 0 ρ k + 1 u ¯ t k + 1 + ρ k + 1 u k u ¯ x k + 1 + L p u x k + 1 x + L p u x k x + π x k + 1 π x k
= ρ ¯ k + 1 u k u t k u k u x k Φ x k u ¯ k + 1 ρ k + 1 Φ ¯ x k + 1 + u ¯ k u x k ρ k u ¯ k + 1
ε Φ x k + 1 2 + 1 Φ x k + 1 2 + ε 2 q 2 Φ x k + 1 x ε Φ x k 2 + 1 Φ x k 2 + ε 2 q 2 Φ x k x = 4 π g ρ ¯ k + 1
where L p u x k + 1 = u x k + 1 2 + μ 0 p 2 2 u x k + 1 .
Multiplying (25) by ρ ¯ k + 1 , integrating with regard to x that get
d d t ρ ¯ k + 1 L 2 2 C ρ ¯ k + 1 L 2 2 u x k L + ρ ¯ k + 1 L 2 ρ k H 1 u ¯ x k L 2 C ρ ¯ k + 1 L 2 2 u x x k L 2 + C ς ρ k H 1 2 ρ ¯ k + 1 L 2 2 + ς u ¯ x k L 2 2 B ς k ( t ) ρ ¯ k + 1 L 2 2 + ς u ¯ x k L 2 2
where B ς k ( t ) = C u x x k L 2 + C ς ρ k H 1 2 , for 1 k K , 0 t T 1 and a constant C ς is only dependent on M 0 , and then combining (22) as follows: 0 t B ς k ( s ) d s C 1 + t
Next, we multiply (26) by u ¯ k + 1 , integrate with regard to x, make use of Young’s inequality and together with (25), we get
1 2 d d t Ω ρ k + 1 u ¯ k + 1 2 d x + Ω u ¯ k + 1 L p u x k + 1 x + L p u x k x d x = Ω ρ ¯ k + 1 u k u t k u k u x k Φ x k u ¯ k + 1 ρ k + 1 Φ ¯ x k + 1 + u ¯ k u x k u ¯ k + 1 u ¯ k + 1 π x k + 1 π x k d x C Ω ρ ¯ k + 1 u k u t k u k u x k Φ x k u ¯ k + 1 + ρ k + 1 Φ ¯ x k + 1 + u ¯ k u x k u ¯ k + 1 + u ¯ k + 1 π x k + 1 π x k d x
Let ν ( s ) = s 2 + μ 0 p 2 2 s that
ν ( s ) = s 2 + μ 0 p 2 2 s = s 2 p 1 + μ 0 s 2 + μ 0 p 4 2 μ 0 p 2 2
The following is an estimation for the second term of (29).
Ω L p u x k + 1 x + L p u x k x u ¯ k + 1 d x = Ω 0 1 σ θ u x k + 1 + 1 θ u x k d θ u ¯ x k + 1 2 d x μ 0 p 2 2 Ω u ¯ x k + 1 2 d x
Similarly, let (27) be multiplied by Φ ¯ k + 1 and integrate it with regard to x to get
Ω ε Φ x k + 1 2 + 1 Φ x k + 1 2 + ε 2 q 2 Φ x k + 1 x ε Φ x k 2 + 1 Φ x k 2 + ε 2 q 2 Φ x k x Φ ¯ k + 1 d x = 4 π g Ω ρ ¯ k + 1 Φ ¯ k + 1 d x
Let ϖ s = ε s 2 + 1 s 2 + ε 2 q 2 s , ϖ s = ε s 2 + 1 s 2 + ε 2 q 2 s q 1 s 2 + ε 2 q 2 , then (32) can be written as
Ω ϖ s x k + 1 x ϖ s x k x Φ ¯ k + 1 d x = Ω 0 1 ϖ θ Φ x k + 1 + 1 θ Φ x k d θ Φ ¯ x k + 1 2 d x C Ω Φ ¯ x k + 1 2 d x
Combining the above formulas and H o ¨ lder’s inequality as follows
Φ ¯ x k + 1 L 2 2 C ρ ¯ k + 1 L 2 2
Substituting (31) and (33) into (29) to get
d d t Ω ρ k + 1 u ¯ k + 1 2 d x + μ 0 p 2 2 Ω u ¯ x k + 1 2 d x C ρ ¯ k + 1 L 2 u k L 2 u ¯ x k + 1 L 2 + ρ ¯ k + 1 L 2 u x t k L 2 u ¯ x k + 1 L 2 + π k + 1 π k L 2 u ¯ x k + 1 L 2 + ρ k + 1 H 1 1 2 u ¯ x k L 2 u x x k L 2 ρ k + 1 u ¯ k + 1 L 2 + ρ ¯ k + 1 L 2 u x k L p u x x k L 2 u ¯ x k + 1 L 2 + ρ k + 1 H 1 Φ ¯ x k + 1 L 2 u ¯ x k + 1 L 2 + ρ ¯ k + 1 L 2 Φ x x k L 2 u ¯ x k + 1 L 2 D ς k ( t ) ρ ¯ k + 1 L 2 2 + C ρ k + 1 u ¯ k + 1 L 2 2 + ς u ¯ x k + 1 L 2 2
where D ς k ( t ) = C + C u x t k L 2 2 is defined such that for each k 1 , 0 t T 1 , exists 0 t D ς k ( t ) d s C + C t .
Combining the inequalities (28), (33) and (34) gives us
d d t ρ ¯ k + 1 L 2 2 + ρ k + 1 u ¯ k + 1 L 2 2 + u ¯ x k + 1 L 2 2 + Φ ¯ x k + 1 L 2 2 E ς k ( t ) ρ ¯ k + 1 L 2 2 + C ρ k + 1 u ¯ k + 1 L 2 2 + ς u ¯ x k L 2 2
where E ς k ( t ) is only dependent on B ς k and D ς k . By using the uniform estimate (22) for 0 t T 1 and 1 k K to get
0 t E ς k ( t ) d s C + C ς t
By utilizing Gronwall’s inequality, integrating over 0 , t 0 , T 1 , (35) which can be expressed as
ρ ¯ k + 1 L 2 2 + ρ k + 1 u ¯ k + 1 L 2 2 + 0 t u ¯ x k + 1 L 2 2 + Φ ¯ x k + 1 L 2 2 d s C exp C ς t 0 t u ¯ x k L 2 2 + ρ k u ¯ k L 2 2 d s
That means we can choose constants ς and T small enough that T < T 1 , C exp C ς T < 1 2 , utilizing Gronwall’s inequality
k = 1 K sup 0 t T ρ ¯ k + 1 L 2 2 + ρ k + 1 u ¯ k + 1 L 2 2 + 0 T u ¯ x k + 1 L 2 2 + Φ ¯ x k + 1 L 2 2 d t < C
Then, from (38), the following strongly converges as k +
ρ k ρ L 0 , T ; L 2 Ω
u k u L 0 , T ; L 2 Ω L 2 0 , T ; H 0 1 Ω
Φ k Φ L 0 , T ; H 2 Ω
The uniform estimates are satisfied by ρ , u , Φ owing to the lower semi-continuity of norms
e s s sup 0 t T 1 ρ H 1 + ρ u t L 2 + u W 0 1 , p H 2 + ρ t L 2 + Φ H 2 + 0 T u x t L 2 2 d s C

4. Proof of the Main Theorem

4.1. Existence

Assume that smooth solution u 0 δ H 0 1 Ω H 2 Ω satisfies the boundary conditions for 0 < δ 1 , ρ 0 δ = ρ 0 + δ that follows
u 0 x δ 2 + μ 0 p 2 2 u 0 x δ x = π x ( ρ 0 δ ) ( ρ 0 δ ) 1 2 g δ u 0 δ ( 0 ) = u 0 δ ( 1 ) = 0
where there exists g δ C 0 Ω that allows
g δ L 2 g L 2 , lim δ 0 g δ g L 2 = 0
The subsequence ρ 0 δ j of ρ 0 δ and the subsequence u 0 δ j of u 0 δ are both present. The following strong convergence can be determined as δ j 0
u 0 x δ j 2 + μ 0 p 2 2 u 0 x δ j x u 0 x 2 + μ 0 p 2 2 u 0 x x i n L 2 Ω
π x ρ 0 δ j ρ 0 δ j 1 2 g δ j π x ρ 0 ρ 0 1 2 g i n L 2 Ω
Therefore, ρ 0 , u 0 satisfies u 0 x 2 + μ 0 p 2 2 u 0 x x = π x ( ρ 0 ) ( ρ 0 ) 1 2 g . A unique solution to the following problem exists for a T 0 , +
ρ t + ( ρ u ) x = 0 ( ρ u ) t + ( ρ u 2 ) x + ρ Φ x u x 2 + μ 0 p 2 2 u x x + π x = ρ u ε Φ x 2 + 1 Φ x 2 + ε 2 q 2 Φ x x = 4 π g ρ m 0 ρ , u , Φ t = 0 = ρ 0 δ , u 0 δ , Φ 0 δ u Ω = Φ Ω = 0
and the uniform estimate is satisfied by ρ δ , u δ , Φ δ
e s s sup 0 t T 1 ρ δ H 1 + ρ δ u t δ L 2 + u δ W 0 1 , p H 2 + ρ t δ L 2 + Φ δ H 2 + 0 T u x t δ L 2 2 d s C
That which follows strong convergence exists as δ 0
ρ δ ρ L 0 , T ; L 2 Ω
u δ u L 0 , T ; L 2 Ω L 2 0 , T ; H 0 1 Ω
Φ δ Φ L 0 , T ; H 2 Ω
The uniform estimations are satisfied by ρ , u , Φ according to the lower semi-continuity of norms
e s s sup 0 t T 1 ρ H 1 + ρ u t L 2 + u W 0 1 , p H 2 + ρ t L 2 + Φ H 2 + 0 T u x t L 2 2 d s C

4.2. Uniqueness

To demonstrate the uniqueness of the solution, we first assume ρ , u , Φ , ρ ¯ , u ¯ , Φ ¯ are two solutions of equation satisfying the same initial conditions. Equations (1) and (2) are then combined to yield
ρ u t + ρ u 2 x + ρ Φ x u x 2 + μ 0 p 2 2 u x x + π x = ρ u
ρ ¯ u ¯ t + ρ ¯ u ¯ 2 x + ρ ¯ Φ ¯ x u ¯ x 2 + μ 0 p 2 2 u ¯ x x + π ¯ x = ρ ¯ u ¯
Subtracting the two equations, multiplying both sides of by u u ¯ , integrating with regard to Ω T
1 2 d d t Ω ρ u u ¯ 2 d x Ω u x 2 + μ 0 p 2 2 u x x u ¯ x 2 + μ 0 p 2 2 u ¯ x x u u ¯ d x Ω ρ ρ ¯ u Φ ¯ x u ¯ u ¯ x u ¯ t u u ¯ + π π ¯ u u ¯ x + ρ u u ¯ Φ x Φ ¯ x + ρ u ¯ x u u ¯ 2 + ρ u u ¯ 2 d x
This paper assume that h ( s ) = s 2 + μ 0 p 2 2 s , we can get that
h ( s ) = p 1 s 2 + μ 0 s 2 + μ 0 p 4 2 μ 0 p 2 2
So there is
Ω u u ¯ u x 2 + μ 0 p 2 2 u x x u ¯ x 2 + μ 0 p 2 2 u ¯ x x d x = Ω 0 1 h θ u x + 1 θ u ¯ x d θ u x u ¯ x 2 d x μ 0 p 2 2 Ω u x u ¯ x 2 d x
Then, similarly, we have
Ω ε Φ x 2 + 1 Φ x 2 + ε 2 q 2 Φ x x ε Φ ¯ x 2 + 1 Φ ¯ x 2 + ε 2 q 2 Φ ¯ x x Φ Φ ¯ d x = 4 π g Ω ρ ρ ¯ Φ Φ ¯ d x
Thus, we can derive
Ω ε Φ x 2 + 1 Φ x 2 + ε 2 q 2 ε Φ ¯ x 2 + 1 Φ ¯ x 2 + ε 2 q 2 Φ x Φ ¯ x 2 d x = 4 π g Ω ρ ρ ¯ Φ Φ ¯ d x C ρ ρ ¯ L 2 2 + ε Φ x Φ ¯ x L 2 2
We conclude that
Φ x Φ ¯ x L 2 2 C ρ ρ ¯ L 2 2
Then substitute the above equation into (42) and use Young’s inequality
1 2 d d t Ω ρ u u ¯ 2 d x + Ω u x u ¯ x 2 d x C ρ ρ ¯ L 2 u u ¯ x L 2 u x L + ρ ρ ¯ L 2 u u ¯ x L 2 u ¯ x t L 2 + ρ ρ ¯ L 2 u ¯ x u ¯ x x L 2 u u ¯ x L 2 + π π ¯ L 2 u u ¯ x L 2 + ρ ρ ¯ L 2 u u ¯ x L 2 Φ ¯ x x L 2 + ρ L u u ¯ x L 2 Φ x Φ ¯ x x L 2 + ρ u u ¯ L 2 2 u ¯ x x L 2 + ρ u u ¯ L 2 2 C ρ ρ ¯ L 2 2 u x L 2 2 + u ¯ x t L 2 2 + 1 + C π π ¯ L 2 2 + C ρ u u ¯ L 2 2
According to Equation (41), we know that when t < T 1 ,
0 t C u x L 2 2 + u ¯ x t L 2 2 + 1 s d s C 1 + t
In addition, we can also obtain from equation (1)
ρ ρ ¯ t + ρ ¯ x u u ¯ + ρ ¯ u x u ¯ x + ρ ρ ¯ u x + ρ ρ ¯ x u = 0
Multiply (47) by ρ ρ ¯ , integrate it over Ω gives
d d t Ω ρ ρ ¯ 2 d x C ρ ¯ L u x u ¯ x L 2 ρ ρ ¯ L 2 + ρ ¯ x L 2 u u ¯ L ρ ρ ¯ L 2 + ρ ρ ¯ L 2 2 u x L C ρ ρ ¯ L 2 2 + C ε u u ¯ x L 2
Similarly we have π t = π x u γ π u x , π ¯ t = π ¯ x u ¯ γ π ¯ u ¯ x , then
π π ¯ t + π π ¯ x u + π ¯ x u u ¯ + γ π π ¯ u x + γ π ¯ u x u ¯ x = 0
Next, multiplying by π π ¯ , integrating on Ω gets
d d t Ω π π ¯ 2 d x C u x L π π ¯ L 2 2 + u u ¯ L π ¯ x L 2 π π ¯ L 2 + C u u ¯ x L 2 π ¯ L π π ¯ L 2 C π π ¯ L 2 2 + C ε u u ¯ x L 2
From Equations (46), (48) and (50), integrating over t, we obtain
Ω u x u ¯ x 2 d x + ρ ρ ¯ L 2 2 + ρ u u ¯ L 2 2 + π π ¯ L 2 2 0 t C s ρ ρ ¯ L 2 2 + ρ u u ¯ L 2 2 + π π ¯ L 2 2 d s
Applying Gronwall’s inequality, we get
ρ ρ ¯ L 2 2 + ρ u u ¯ L 2 2 + π π ¯ L 2 2 = 0
Therefore
ρ = ρ ¯ , u = u ¯ , Φ = Φ ¯ .

5. Conclusions

This paper investigated a generalized 1D shear thickening non-Newtonian fluids under the condition that p > 2 , 1 < q < 2 . Considering the case that the model has non-Newtonian potential and damping term, the well-posedness result for strong solution was obtained. This study has important implications in flow behavior and engineering applications, especially in optimizing industrial fluid transport processes and in the fields of geology and environmental sciences. Its limitations are mainly due to the complexity of modeling non-Newtonian fluids, and the coupling of potential and damping terms that may complicate the existential study of the model.
Future studies can introduce multidimensional models to explore the fluid behavior in different dimensions. Meanwhile, further investigation of the effects of different damping terms on the flow and modeling them accordingly will help to deepen our understanding.

Author Contributions

Writing—review and editing, Y.S., L.J. and Y.C. All authors have read and agreed to the published version of the manuscript.

Funding

This work is supported by the National Natural Science Foundation of China (No. 62203200) and the Science and Technology Joint Foundation of Liaoning Province (No. 2024-MSLH-199).

Data Availability Statement

No data were created or analyzed in this study.

Conflicts of Interest

The authors declare no conflicts of interest.

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Song, Y.; Jin, L.; Chen, Y. Strong Solution for a Nonlinear Non-Newtonian Shear Thickening Fluid. Mathematics 2025, 13, 878. https://doi.org/10.3390/math13050878

AMA Style

Song Y, Jin L, Chen Y. Strong Solution for a Nonlinear Non-Newtonian Shear Thickening Fluid. Mathematics. 2025; 13(5):878. https://doi.org/10.3390/math13050878

Chicago/Turabian Style

Song, Yukun, Lin Jin, and Yang Chen. 2025. "Strong Solution for a Nonlinear Non-Newtonian Shear Thickening Fluid" Mathematics 13, no. 5: 878. https://doi.org/10.3390/math13050878

APA Style

Song, Y., Jin, L., & Chen, Y. (2025). Strong Solution for a Nonlinear Non-Newtonian Shear Thickening Fluid. Mathematics, 13(5), 878. https://doi.org/10.3390/math13050878

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