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 [35], Karapinar et al. [613], Benchohra et al. [1416], Inc et al. [1720].

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 [2327] 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.