A nonlinear fractional Rayleigh–Stokes equation under nonlocal integral conditions

*Correspondence: vnguyen@snu.ac.kr 4Department of Civil and Environmental Engineering, Seoul National University, Seoul, South Korea Full list of author information is available at the end of the article Abstract In this paper, we study the fractional nonlinear Rayleigh–Stokes equation under nonlocal integral conditions, and the existence and uniqueness of the mild solution to our problem are considered. The ill-posedness of the mild solution to the problem recovering the initial value is also investigated. To tackle the ill-posedness, a regularized solution is constructed by the Fourier truncation method, and the convergence rate to the exact solution of this method is demonstrated.


Introduction
Most of fluids in the real world, such as in food products (mayonnaise, mustard, chocolate, ketchup, butter, cheese, yogurt, etc.), in natural substances (honey, magma, lava, gums, etc.), in biology (blood, semen, synovia, mucus, etc.), in industry (paint, glue, lubricant, ink, molten polymer, etc.), in cosmetics (soap solution, toothpaste, cream, silicone, nail polish, etc.) are treated as non-Newtonian fluids. Therefore, the study on non-Newtonian fluids is a substantial subject in science and industrial applications. As the Rayleigh-Stokes problem for an edge, the first problem of Stokes for a non-Newtonian fluid flow past an impulsively started flat plate has received much attention because of its practical importance [1,2]. For a second grade fluid, the equation of motion is of higher order than the Navier-Stokes equation, because it exhibits all properties of viscoelastic fluids.
Recently, fractional calculus has encountered much success in the description of constitutive relations of viscoelastic fluids. The starting point of the fractional derivative model of a viscoelastic fluid is usually a classical differential equation which is modified by replacing the time derivative of an integer order with a fractional calculus operator. This generalization allows one to define precisely noninteger order integrals or derivatives.
The forward problems for equation (1.1) have been examined in a plenty of studies. For instance, Zierep and Fetecau [19] discussed the energetic balance in the Rayleigh-Stokes problem for a Maxwell fluid for several initial and/or boundary conditions. Fetecau and Zierep [20] found the exact solutions both for the Stokes' problem and for the Rayleigh-Stokes problem within the context of the fluids of second grade. Bazhlekova et al. [21] presented an introduction about an analysis of the Rayleigh-Stokes problem for a generalized second-grade fluid. The exact solution of the Rayleigh-Stokes problem for a generalized second grade fluid in a porous half-space with a heated flat plate was considered by Xue et al. [22]. Exact solutions of the Rayleigh-Stokes equation in the case of homogeneous initial and boundary conditions was considered by Zhao and Yang [23]. The backward problems have currently been studied by many mathematicians. Ngoc et al. [24], for instance, pondered the inverse problem for the nonlinear fractional Rayleigh-Stokes equations. Equation (1.1) associated with Gaussian random noise was examined by Triet et al. [25], and so forth.
To the best of our knowledge, there are still very few studies on the Rayleigh-Stokes equation accompanied with nonlocal integral conditions. In comparison with the initial condition u(x, 0) = f (x) or the final condition u(x, T) = g(x), the nonlocal conditions are of more significant complexity. Furthermore, it is emphasized that we cannot apply Parseval's equality to obtain stable estimates in the L p space. To tackle this limitation, we need to develop additional techniques and Sobolev embedding in our study.
This manuscript is structured as follows. An introduction of preliminary results is described in Sect. 2; the regularity of the mild solution to the problem in a linear case is illustrated in Sect. 3. In Sect. 4, the problem recovering the initial value and the convergence of the regularized solution are detailed. The regularity of the mild solution to the problem in the nonlinear case is mentioned in Sect. 5. Eventually, the conclusion is presented in Sect. 6.

Preliminary results
We recall the spectral problem as follows: admitting the eigenvalues 0 < λ 1 ≤ λ 2 ≤ · · · ≤ λ n ≤ . . . with λ n → ∞ as n → ∞ (see [26]). The corresponding eigenfunctions φ n ∈ H 1 0 ( ). For all s ≥ 0, the operator A s (here A = -) also possesses the following representation: Consider on D(A s ) the norm (noting that λ 1 > 0) By duality, we can set D(A -s ) = (D(A s )) * by identifying (L 2 ( )) * = L 2 ( ) and using the socalled Gelfand triple (see [27]). Then D(A -s ) is a Hilbert space with the following norm: wherein ·, · denotes the duality bracket between D(A -s ) and D(A). For any p ≥ 0, we define the space where ·, · is the inner product in L 2 ( ), then H( ) is a Hilbert space with the norm Suppose that problem (1.1) has a solution u which admits the form u(x, t) = ∞ n=1 u(x, t), φ n (x) φ n (x). From [21,24], we deduce that the solution of problem (1.1) with the initial condition u(x, 0) = u 0 (x) is given by where S n,α (t) is given by
Proof The detailed interpretation can be found in [24]. We briefly present some main steps of the proof. We have e -zt λ n sin(απ)z α (-z + λ n μz α cos(απ)) 2 + (λ n μz α sin(απ)) 2 dz. (2.13) In terms of the left-hand side of inequality (2.12), we have (2.14) It is worth noting that With regard to the right-hand side of inequality (2.12), we have For the rest of paper, we give the definition of a mild solution to problem (1.1)-(1.2) in the subsequent notion.

The regularity of the mild solution to problem (1.1)-(1.2) in the linear case
In this section, we consider the regularity of the mild solution of problem (1.1)-(1.2) under the condition that the source term is linear.
Proof We will estimate the terms A 1 (x, t), A 2 (x, t), and A 3 (x, t) in two cases: ξ 1 , ξ 2 > 0 and ξ 1 = 0, ξ 2 > 0. Part i: In the case of ξ 1 , ξ 2 > 0. Thanks to Parseval's equality, we have This implies Similarly, the term A 2 (·, t) D (A 2 ) can be assessed as follows: Put another way, the integral Conjoining (3.5), (3.6), and (3.7), we arrive at In other words, we have Using Parseval's equality again, we can estimate the term A 3 (·, t) as follows: The latter estimation allows us to deduce that Combining (3.4), (3.9), and (3.11) helps us rest assured that Part ii: In the case of ξ 1 = 0, ξ 2 > 0. Paserval's equality gives us It should be noted that We get (3.14) Therefore, we immediately obtain that In terms of A 2 (·, t), by computations similar to above, we have This implies

The problem recovering the initial value
In this section, we are inclined to deliberate on the following problem: (4.1) Our primary aim in this part is to recover the initial data u(x, 0) = V (x). To achieve this goal, we firstly go to prove the point that the problem is ill-posed in L 2 (0, T). For convenience of readers, we presume that F = 0. From the formula of the mild solution (2.10), we have that Proof We ponder on the linear operator K : L 2 (D) → L 2 (D) as follows: where It is worth mentioning that m(x, τ ) = m(τ , x). Therefore, the operator K is self-adjoint and the compactness of K is presented as follows.
We define the finite rank operator K L In that This allows us to conclude that K is a compact operator. In conjunction with (4.3), we have Combining the latter conclusion and using Kirsch [28], we deduce that the problem of recovering the initial value V from (4.7) is ill-posed. To ensure mathematical clarity, we provide an example as follows. If we choose the input data g j (x) = φ j (x) 8) and the initial data regarding g j is In the next step, we assess the initial data V j in respect of L 2 norm Conjoining (4.8) with (4.10), we arrive at the conclusion that the solution of problem (4.1) is instable.
With the aim of putting forward the next theory, we give an assumption as follows. Presume that g ∈ L p (D) and F ∈ L ∞ (0, T; L p (D)) are noisy data, and presume gg L p (D) + F -F L ∞ (0,T;L p (D)) ≤ . (4.11) Using the Fourier truncation method, we provide a construction of regularized solution to problem (4.1) as follows: Theorem 4.2 Let (g , F ) satisfy (4.11), and let b, δ be such that Suppose that there exists β such that V ∈ D A β . (4.14) Then we have that the following estimation holds: Remark 4.1 On account of λ K ∼ K 2/d , we need to choose K such that In fact, we choose K = (r-1)d/2(δ-b) for 0 < r < 1, and then the error is of order Proof To begin with, we define the function W (x) in the following way:

S n,α (ts)F n u(s) ds dt φ n (x). (4.16)
As the next step, we ponder the term V -V D(A δ ) and present an estimation of it with regard to the D(A δ ) norm for δ > 0. Thanks to the triangle inequality, we obtain We denote the two terms B 1 and B 2 as follows: It is worth noting that δ > b, the term B 1 can be assessed in the following way: (4.21) Similarly, the term B 2 can be estimated as follows: The latter estimation and the previous one allow us to have that (4.23) Regarding the term W -V D(A δ ) , we have Conjoining (4.23) and (4.24), we come to the conclusion that The Sobolev embedding D(A δ ) → L 2d d-4δ ( ) allows us to rest assured that

The regularity of the mild solution to problem (1.1)-(1.2) in the nonlinear case
In this section, we concentrate on examining the subsequent nonlinear problem (5.1) (i) Let 1/2 < α < 1, and presume 0 < s < 1. If ξ 1 , ξ 2 > 0, g belongs to D(A s ), and F is a global Lipschitz source function satisfying for T is small enough, then problem (5.1) has a unique solution (ii) Let 1/2 < α < 1, and presume 0 < s < 1. If ξ 1 = 0, ξ 2 > 0, g ∈ D(A s+α ), and F is a global Lipschitz source function satisfying for T is small enough, then problem (5.1) has a unique solution u ∈ L ∞ 0, T; D A s .

Moreover, we have
Proof Part i. Using (2.11) in Sect. 2, we obtain For convenience of calculations in the next steps, we posit With the purpose of proving the point that nonlinear equation (5.4) has a unique solution in L ∞ (0, T; D(A s )), we posit the following mapping: Using the triangle inequality again, we have We will show that the mapping I is a contraction mapping in L ∞ (0, T; D(A s )). Now, we give estimations of the terms A 2 (w 1 ) -A 2 (w 2 ) L ∞ (0,T;D(A s )) and A 3 (w 1 ) -A 3 (w 2 ) L ∞ (0,T;D(A s )) as follows.
Step 1: In terms of the term A 2 (w 1 ) -A 2 (w 2 ) L ∞ (0,T;D(A s )) . Firstly, we have It is noteworthy that Thanks to the inequality S n,α (t) ≤ C 1 (μ,α) 1+λ n (t-s) 1-α (see Lemma (2.1)), we get T 0 t 0 S n,α (t) F n s, w 1 (s) -F n s, w 2 (s) ds On account of s < 1, we have that Using the global Lipschitz property of F, we arrive at The Sobolev embedding D(A s ) → L 2 (D) allows us to deduce that This implies that, for any t, 0 ≤ t ≤ T, we have In other words, We can see that the right-hand side of (5.14) is independent of t, as a result we reach the conclusion that Step 2: In terms of the term A 3 (w 1 )(·, t) -A 3 (w 2 )(·, t) L ∞ (0,T;D(A s )) .
Thanks to Parseval's equality and Holder's inequality, we can assess the term A 3 (w 1 )(·, t) -A 3 (w 2 )(·, t) L ∞ (0,T;D(A s )) in the following way: It is worth noting that α > 1/2, we get We can estimate the term T 0 ∞ n=1 λ 2s-2 n (F n (s, w 1 (s)) -F n (s, w 2 (s))) 2 ds as follows: where we used the global Lipschitz property of F function and note that s < 1.
Conjoining the latter estimate and the previous one, we claim that Put another way, Suppose that T is small enough such that we can claim that I is a contraction mapping in L ∞ (0, T; D(A s ). Furthermore, it should be noted that if w 1 = 0, then This implies The latter estimation helps us assert that It is worth noting that M 9 √ T + M 10 T α-1 < 1, we get We break the back of the proof of Part i. Part ii: From (2.11) in Sect. 2, we get where B j (j = 1, 2, 3) are defined as follows: and F n x, u(s) = F x, s, u(s) , φ n (x) .
Firstly, we posit the following function: Similar to Part i, we go to prove the point that J is a contraction mapping in L ∞ (0, t; D(A s )).
In view of the triangle inequality, we have The term B 2 (w 1 ) -B 2 (w 2 ) L ∞ (0,T;D(A s )) can be estimated in the following way: S n,α (ts) F n s, w 1 (s) -F n s, w 2 (s) ds dt 2 . (5.30) It is important to note that S n,α (t) ≥ C 2 (μ,α,t) λ n and v(t) ≥ Nt -γ , we get  The latter estimation allows us to deduce that if w 1 (x, t) ∈ L ∞ (0, T; D(A s )), then J w 1 (x, t) ∈ L ∞ 0, T; D A s .
Thanks to the Banach fixed point theorem, we come to the conclusion that J w = w has a unique solution w ∈ L ∞ (0, T; D(A s )). Furthermore, in a similar way to Part i, we take w = 0, and we immediately get We get the proof out of the way.

Conclusion
In this paper, we examined the fractional nonlinear Rayleigh-Stokes equation under nonlocal integral conditions. The existence and uniqueness are considered using the Fourier truncation method. The convergence rate between the obtained solution and the regularized solution is demonstrated.