#### Abstract

In this paper, we study the Cauchy problem for a system of Rayleigh-Stokes equations. In this system of equations, we use derivatives in the classical Riemann-Liouville sense. This system has many applications in some non-Newtonian fluids. We obtained results for the existence, uniqueness, and frequency of the solution. We discuss the stability of the solutions and find the solution spaces. Our main technique is to use the Banach mapping theorem combined with some techniques in Fourier analysis.

#### 1. Introduction

Let () be a smooth domain with the boundary , and is a given time. In this paper, we study the initial value problem for systems of Rayleigh- Stokes problem as follows where is Cauchy input data. Some functions called the source data which are defined later. Here, , and is the Riemann-Liouville fractional derivative of order given by [1, 2]: where is the Gamma function. As far as we know, there are currently several definitions for fraction derivatives and fraction integrals, such as Riemann-Liouville, Caputo, Hadamard, Riesz, and Griinwald-Letnikov. Some works are attracting the attention of the community, like Debbouche and his group [3–5], Karapinar et al. [6–13], Benchohra et al. [14–16], Inc et al. [17–20].

And some very interesting results about unchanged level derivatives are surveyed by.

In fluid dynamics, the Rayleigh problem is the first Stokes problem, which determines the flow generated by the sudden motion of an infinitely long plate from the resting state, named after Lord Rayleigh and Sir George Stokes. This problem is considered to be one of the simplest problems with the correct solution for the Navier-Stokes equation. In recent times, with the development of fractions, a number of authors such as Shen et al. [21] investigated Rayleigh-Stokes, which is a more general form than the classical model. The fractional Rayleigh-Stokes equation (1) has applications in non-Newtonian behavior of fluids [21], and other applications of this equation can be given in [21, 22]. We list some papers on fractional Rayleigh stokes in the following. (i)The initial and boundary values for the Rayleigh-Stokes problem in the case of homogeneity have been explored in a number of interesting papers; see for example [23–27] and its references(ii)The authors in [21, 28, 29] used the Fourier transform and the fractional Laplace transform to obtain the exact solution(iii)Numerical solutions for Problem (1) has been studied by many authors in [4, 23, 24, 30, 31](iv)In [32], the authors concerned with the following problem for a following stochastic Rayleigh-Stokes equation

The existence and uniqueness of mild solution in each case are established separately by applying a standard method that is Banach fixed point theorem. In [22], Caraballo et al. investigated the following time-fractional Rayleigh-Stokes stochastic equation where represents a standard Wiener process.

To the best of the author’s knowledge, the problem of the system of equations for a fractional Rayleigh-Stokes with a nonlinear source, i.e., Problem (1), has yet to be studied. The goal of this paper is to develop a theory of the existence and regularity estimate for the mild solution to the Problem (1). Our main technique is to use the Banach mapping theorem combined with some techniques in Fourier analysis.

#### 2. The Existence and Regularity of the Solution

In this section, we consider the existence and mild solution of Problem (1). Before going into the main theorem of this section, we briefly discuss spectral, eigenvalues, and related functional spaces on the Laplacian operator.

Let . The domain for is a Banach space equipped with the norm

The definition of the negative fractional power can be found in [33]. Its domain is a Hilbert space endowed with the dual inner product taken between and . This generates the norm

A couple of functions are called a function of two variables

Here, the norm of (for any space ) is defined

Theorem 1. *Let . Let satisfies and the globally Lipschitz function
for any . Then, problem (1) has a unique solution belongs to and regularity estimates hold
*

*Proof. *Assume that the mild solution is described by a Fourier series
Thanks to the results of [24], we deduce that the solution of Problem (1) with the initial condition is given by
where is represented as follows
Hence, the mild solution of Problem (1) is given by
We get the following estimate
Now, we continue to show that (1) has a unique mild solution.

For any , denote by the function space associated with the norm
for any
Let us give the following operator
where and are defined by the following
From two equality as above, if , we find that two following equality
Hence, we get that for any Since the fact that
we know that
Combining (22), (24), and (25), we find that
which allows us to deduce that
Let two functions and in the space . Then, using Parseval’s equality, we get
Noting that
Hence, and noting that , we find that
Let us emphasize that for then
Combining (30) and (31), we find that
where we denote
Next, we need to deal with the integral quantity . By change variable , we find that
Thank the inequality for , we know that
Combining with the fact that , we deduce that the following inequality
This inequality together with (32) leads to
By a similar way, we also get that
Therefore, we can deduce that
Since the limitation
we know that there exists a positive such that Thus, we can deduce that is contractive on . Applying Banach fixed point theorem, we get that has a fixed point , so, the function is also the unique solution of (1). Since (14), we find that
By looking at (24) and (29), we follow from (41) that
where we have used that and
Hence, we derive that
By a similar way, we also obtain that the following estimate
Combining (44) and (45), we get that
By applying Gronwall’s inequality as in [34], we obtain that
Hence, we can deduce that belongs to , and furthermore, we also derive that
where we set
☐

Theorem 2. *Let . Then, Problem (1) has a unique solution .*

*Proof. *For , we set the following space
It is easy to see that
First, we look at the second term . Using the inequality, we find that
Using (16), we obtain that
Combining (52) and (53) and using the inequality for any , we get that
Next, we consider the second term . It is easy to observe that
Here, we set the following function
In order to give the further process, we use one result in Theorem 1 [24]. If , then, we get the following estimate
Set the function . Since the fact that , we know that belongs to the space . Hence, using Parseval’s equality, we get that
Next, we treat the third term . Using the inequality, we find that
Combining (51), (52), (58), and (59), we obtain that
By a similar way as above, we also obtain that
Combining (60) and (61), we derive that
Here, we set Let fixed and let the following function
From some above observations, we can deduce that
From (36), we get that
This together with (64) that
This implies that
Since when , we can choose such that
Hence, we follow from (67) and (64) that
which allows us to conclude that
This inequality says that
☐

#### Data Availability

No data were used to support this study.

#### Conflicts of Interest

The authors declare that they have no competing interests.

#### Authors’ Contributions

Both authors contributed equally and significantly in writing this paper. Four authors read and approved the final manuscript.