# Remarks on the Systems of Semilinear Fractional Rayleigh-Stokes Equation

Authors: Le Dinh Long
Journal: Advances in Mathematical Physics
 (2021)

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## 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.


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## Body

## 1. Introduction

LetΩ⊂ℝN (N≥1) be a smooth domain with the boundary ∂Ω, and T>0 is a given time. In this paper, we study the initial value problem for systems of Rayleigh- Stokes problem as follows
(1)∂tu+−Δβu−d∂tαΔu=Gu,v,x,t∈Ω×0,T,∂tv+−Δβv−d∂tαΔv=Hu,v,x,t∈Ω×0,T,ux,t=0,x∈∂Ω,ux,0=φx,x∈Ω,vx,0=θx,x∈Ω.where φ,θ is Cauchy input data. Some functions G,H called the source data which are defined later. Here, ∂t=∂/∂t, and ∂tα is the Riemann-Liouville fractional derivative of order 0<α<1 given by [1, 2]:
(2)∂tαwx,t=1Γ1−α∂∂t∫0tt−s−αwx,sds,where Γ. 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(3)∂tXt+1+α∂tβAXt=ft,Xt+γtB.Qht,t∈0,T,Xt∂D=0,X0=x0,t∈0,T.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
(4)∂∂t−Δ−ù∂α∂tαΔu=Ft,u+σt,uẆt,onJ×X,where Wt,.t∈J¯ 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.LetA=−Δ. The domain DAs for s≥0 is a Banach space equipped with the norm
(5)hDAs≔∑n=1∞h,φn2λn2s1/2,h∈DAs.The definition of the negative fractional powerA−s can be found in [33]. Its domain DA−s is a Hilbert space endowed with the dual inner product .,.−s,s taken between DA−s and DAs. This generates the norm
(6)hDA−s=∑n=1∞h,φn−s,s2λn−2s1/2.A coupleu,v of functions ux,t,vx,t:Q¯T⟶ℝ,Q¯T=Ω¯×0,T are called a function of two variables x,t(7)u,v:Q¯T⟶ℝ2,u,vx,t=ux,t,vx,t.Here, the norm ofu,v∈X×X (for any space X) is defined
(8)u,vX2=uX+vX.Theorem 1.
Letφ,θ∈HmΩ×HmΩ. Let G,H satisfies G0,0=H0,0=0 and the globally Lipschitz function
(9)Gv1,v2−Gv¯1,v¯2≤Lgv1−v¯1L2Ω+v2−v¯2L2Ω,Hv1,v2−Hv¯1,v¯2≤Lhv1−v¯1L2Ω+v2−v¯2L2Ω,for any v1,v¯1,v2,v¯2∈L2Ω. Then, problem (1) has a unique solution u,v belongs to L∞0,T;L2Ω×L∞0,T;L2Ω and regularity estimates hold
(10)u,vL∞0,T;L2Ω×L∞0,T;L2Ω≤D1φHmΩ+θHmΩ.Proof.
Assume that the mild solutionu is described by a Fourier series
(11)ux,t=∑jux,t,ejxejx,vx,t=∑jvx,t,ejxejx.
Thanks to the results of [24], we deduce that the solution of Problem (1) with the initial condition ux,0=φx is given by
(12)ujt=∫0∞e−νtKj,α,β,νdνφj+∫0t∫0∞e−νt−sKj,α,β,νdνGux,s,vx,s,ejxds,where K is represented as follows
(13)Kj,α,β,ν=dπλjβsinαπνα−ν+λjβdναcosαπ+λjβ2+λjβdναsinαπ2·
Hence, the mild solution of Problem (1) is given by
(14)ux,t=∑j∫0∞e−νtKj,α,β,νdνφjejx+∑j∫0t∫0∞e−νt−sKj,α,β,νdνGux,s,vx,s,ejxdsejx,(15)vx,t=∑j∫0∞e−νtKj,α,β,νdνθjejx+∑j∫0t∫0∞e−νt−sKj,α,β,νdνHux,s,vx,s,ejxdsejx.
We get the following estimate(16)∫0∞e−νtKj,α,β,νdν≤Cα,β,d1+λjβt1−α·Now, we continue to show that (1) has a unique mild solution.
For anya≥0, denote by La∞0,T;HmΩ2 the function space L∞0,T;HmΩ2 associated with the norm
(17)χ1,χ2a,m≔sup0≤t≤Texp−atχ1·,tHmΩ+sup0≤t≤Texp−atχ2·,tHmΩ,for any
(18)χ1,χ2∈L∞0,T;L2Ω×L∞0,T;L2Ω.
Let us give the following operator(19)Qχ1,χ2t=Q1χ1,χ2t,Q2χ1,χ2t,where Q1 and Q2 are defined by the following
(20)Q1χ1,χ2t=∑j∫0∞e−νtKj,α,β,νdνφjejx+∑j∫0t∫0∞e−νt−sKj,α,β,νdνGχ1x,s,χ2x,s,ejxdsejx,Q2χ1,χ2t=∑j∫0∞e−νtKj,α,β,νdνθjejx+∑j∫0t∫0∞e−νt−sKj,α,β,νdνHχ1x,s,χ2x,s,ejxdsejx.
From two equality as above, ifχ1,χ2=0,0, we find that two following equality
(21)Q1χ1,χ2t=∑j∫0∞e−νtKj,α,β,νdνφjejx,Q2χ1,χ2t=∑j∫0∞e−νtKj,α,β,νdνθjejx.
Hence, we get that for anya>0(22)exp−atQ1χ1,χ2tHmΩ+exp−atQ2χ1,χ2tHmΩ=e−at∑jλj2m∫0∞e−νtKj,α,β,νdν2φj2+e−at∑jλj2m∫0∞e−νtKj,α,β,νdν2θj2.
Since the fact that(23)∫0∞e−νtKj,α,β,νdν≤Cα,β,d,we know that
(24)∑jλj2m∫0∞e−νtKj,α,β,νdν2φj2≤Cα,β,d2∑jλj2mφj2,(25)∑jλj2m∫0∞e−νtKj,α,β,νdν2θj2≤Cα,β,d2∑jλj2mθj2.
Combining (22), (24), and (25), we find that
(26)exp−atQ1χ1,χ2tHmΩ+exp−atQ2χ1,χ2tHmΩ≤Cα,β,dφHmΩ+θHmΩ,which allows us to deduce that
(27)Qχ1,χ2∈La∞0,T;L2Ω2,ifχ1,χ2=0,0.
Let two functionsχ1,χ2 and χ¯1,χ¯2 in the space La∞0,T;L2Ω2. Then, using Parseval’s equality, we get
(28)Q1χ1,χ2t−Q1χ¯1,χ¯2tHmΩ≤∫0t∑jλj2m∫0∞e−νt−sKj,α,β,νdνGχ1x,s,χ2x,s−Gχ¯1x,s,χ¯2x,s,ejx2ds.
Noting that(29)∫0∞e−νt−sKj,α,β,νdν≤Cα,β,dλj−βt−sα−1.
Hence, and noting thatλjm−β≤λ1m−β, we find that
(30)e−atQ1χ1,χ2t−Q1χ¯1,χ¯2tHmΩ≤Cα,β,dλ1m−βe−at∫0tt−sα−1Gχ1x,s,χ2x,s−Gχ¯1x,s,χ¯2x,sL2Ωds≤Cα,β,dλ1m−βLg∫0tt−sα−1e−at−se−asχ1s−χ¯1sL2Ω+χ2s−χ¯2sL2Ωds.
Let us emphasize that for0≤s≤T then
(31)e−asχ1s−χ¯1sL2Ω+χ2s−χ¯2sL2Ω≤C1,msup0≤t≤Texp−atχ1·,tL2Ω+C1,msup0≤t≤Texp−atχ2·,tL2Ω=C1,mχ1,χ2a,m.
Combining (30) and (31), we find that
(32)e−atQ1χ1,χ2t−Q1χ¯1,χ¯2tHmΩ≤C1α,β,d,m∫0tt−sα−1e−at−sdsχ1,χ2a,m,where we denote
(33)C1α,β,d,m=Cα,β,dC1,mλ1m−βLg.
Next, we need to deal with the integral quantityInt1=∫0tt−sα−1e−at−sds. By change variable s=tγ, we find that
(34)Int1=∫01t−tγα−1exp−at1−γtdγ=∫01tα1−γα−1exp−at1−γdγ=at1−γα2exp−at1−γtaα/2∫011−γα/2−1dγ.
Thank the inequalityy≤ey for y≥0, we know that
(35)at1−γα/2≤expat1−γ.
Combining with the fact that∫011−γα/2−1dγ=2/α, we deduce that the following inequality
(36)∫0tt−sα−1e−at−sds≤2αTaα/2.
This inequality together with (32) leads to
(37)e−atQ1χ1,χ2t−Q1χ¯1,χ¯2tHmΩ≤2C1α,β,d,mαTaα/2χ1,χ2a,m.
By a similar way, we also get that(38)e−atQ2χ1,χ2t−Q2χ¯1,χ¯2tHmΩ≤2C1α,β,d,mαTaα/2χ1,χ2a,m.
Therefore, we can deduce that(39)Qχ1,χ2−Qχ¯1,χ¯2a,m≤4C1α,β,d,mαTaα/2χ1,χ2a,m.
Since the limitation(40)lima⟶+∞4C1α,β,d,mαTaα/2=0,we know that there exists a positive a∗ such that 4C1α,β,d,m/αT/a∗α/2<1. Thus, we can deduce that Q is contractive on La∗∞0,T;HmΩ2. Applying Banach fixed point theorem, we get that Q has a fixed point u,v, so, the function u,v is also the unique solution of (1). Since (14), we find that
(41)u.,tHmΩ≤∑j∫0∞e−νtKj,α,β,νdνφjejxHmΩ+∑j∫0t∫0∞e−νt−sKj,α,β,νdνGux,s,vx,s,ejxdsejxHmΩ≤∑jλj2m∫0∞e−νtKj,α,β,νdν2φj2+∫0t∑jλj2m∫0∞e−νt−sKj,α,β,νdνGux,s,vx,s,ejx2ds.
By looking at (24) and (29), we follow from (41) that
(42)u.,tHmΩ≤Cα,β,dφHmΩ+Cα,β,dλ1m−β∫0tt−sα−1Gux,s,vx,sL2Ωds,where we have used that G0,0=0 and
(43)Gux,s,vx,sL2Ω≤G0,0+Lgu.,sL2Ω+v.,sL2Ω.
Hence, we derive that(44)u.,tHmΩ≤Cα,β,dφHmΩ+Cα,β,dλ1m−βLg∫0tt−sα−1u.,sL2Ω+v.,sL2Ωds.
By a similar way, we also obtain that the following estimate(45)v.,tHmΩ≤Cα,β,dθHmΩ+Cα,β,dλ1m−βLg∫0tt−sα−1u.,sL2Ω+v.,sL2Ωds.
Combining (44) and (45), we get that
(46)u.,tHmΩ+v.,tHmΩ≤Cα,β,dφHmΩ+θHmΩ+2Cα,β,dλ1m−βLg∫0tt−sα−1u.,sL2Ω+v.,sL2Ωds.
By applying Gronwall’s inequality as in [34], we obtain that
(47)u.,tHmΩ+v.,tHmΩ≤Cα,β,dφHmΩ+θHmΩEα,12Cα,β,dλ1m−βLg.
Hence, we can deduce thatu,v belongs to L∞0,T;L2Ω×L∞0,T;L2Ω, and furthermore, we also derive that
(48)u,vL∞0,T;L2Ω×L∞0,T;L2Ω≤D1φHmΩ+θHmΩ,where we set
(49)D1=Cα,β,dEα,12Cα,β,dλ1m−βLg.☐Theorem 2.
Letφ,θ∈Hm+2Ω×Hm+2Ω. Then, Problem (1) has a unique solution u,v∈Cα0,T;HmΩ×Cα0,T;HmΩ.Proof.
Fork>0, we set the following space
(50)Ck0,T;HmΩ×Ck0,T;HmΩ=v1,v2∈C0,T;HmΩ×C0,T;HmΩsup0≤t<s≤Tv1.,t−v1.sHmΩt−sk+sup0≤t<s≤Tv2.,t−v2.sHmΩt−sk<∞.
It is easy to see that(51)ux,t′−ux,t=∑j∫0∞e−νt′Kj,α,β,νdν−∫0∞e−νtKj,α,β,νdνφjejx+∑j∫tt~∫0∞e−νsKj,α,β,νdνGux,t′−s,vx,t′−s,ejxdsejx+∑j∫0t∫0∞e−νsKj,α,β,νdνGux,t′−s,vx,t′−s−Gux,t−s,vx,t−s,ejxdsejx=J1x,t′−t+J2x,t′−t+J3x,t′−t.
First, we look at the second termJ2x,t′−t. Using the inequality, we find that
(52)J2.,t′−tHmΩ≤∫tt~∑j∫0∞e−νsKj,α,β,νdνGux,t′−s,vx,t′−s,ejxHmΩds=∫tt′∑jλj2m∫0∞e−νsKj,α,β,νdν2Gux,t′−s,vx,t′−s,ejx2·
Using (16), we obtain that
(53)∑jλj2m∫0∞e−νsKj,α,β,νdν2Gux,t′−s,vx,t′−s,ejx2≤Cα,β,dλ1m−βsα−1Gux,t′−s,vx,t′−sL2Ω≤Cα,β,dλ1m−βLgsα−1u.,t′−sL2Ω+v.,t′−sL2Ω≤Cα,β,dλ1m−βLgsα−1u,vL∞0,T;L2Ω×L∞0,T;L2Ω≤D1¯sα−1.
Combining (52) and (53) and using the inequality c+dα≤cα+dα for any 0<α<1, we get that
(54)J2.,t′−tHmΩ≤D1¯∫tt′sα−1ds=D1¯t′α−tαα≤D1¯t′−tαα·
Next, we consider the second termJ2x,t′−t. It is easy to observe that
(55)J1x,t′−t=∑j∫0∞e−νt′Kj,α,β,νdν−∫0∞e−νtKj,α,β,νdνφjejx=∫tt′∑jK¯s,j,α,β,νφjejxds.
Here, we set the following function(56)K¯t,j,α,β,ν=ddt∫0∞e−νtKj,α,β,νdν.
In order to give the further process, we use one result in Theorem1 [24]. If f∈H2Ω, then, we get the following estimate
(57)∑jK¯s,j,α,β,νfj2≤DΩ,d,T,N2s−2αfH2Ω2red·.
Set the functionf0x=∑jλjmφjejx. Since the fact that φ∈H2+mΩ, we know that f0 belongs to the space H2Ω. Hence, using Parseval’s equality, we get that
(58)J1.,t′−tHmΩ≤∫tt′∑jλj2mK¯s,j,α,β,ν2φj2ds≤∫tt′∑jK¯s,j,α,β,νfj02ds≤DΩ,d,T,N∫tt′sα−1dsφH2+mΩ=DΩ,d,T,Nt′α−tααφH2+mΩ≤DΩ,d,T,Nt′−tααφH2+mΩ.
Next, we treat the third termJ3x,t′−t. Using the inequality, we find that
(59)J3.,t′−tHmΩ≤Cα,β,dλ1m−β∫0tsα−1Gux,t′−s,vx,t′−s−Gux,t−s,vx,t−sL2Ωds≤Cα,β,dλ1m−β∫0tt−sα−1Gux,s+t′−t,vx,s+t′−t−Gux,s,vx,sL2Ωds≤Cα,β,dλ1m−βLg∫0tt−sα−1u.,s+t′−t−u.,sL2Ω−v.,s+t′−t−v.,sL2Ωds.
Combining (51), (52), (58), and (59), we obtain that
(60)u.,t′−u.,tHmΩ≤J1.,t′−tHmΩ+J2.,t′−tHmΩ+J3.,t′−tHmΩ≤DΩ,d,T,Nt′−tααφH2+mΩ+D1¯t′−tαα+Cα,β,dλ1m−βLg∫0tt−sα−1u.,s+t′−t−u.,sL2Ω−v.,s+t′−t−v.,sL2Ωds.
By a similar way as above, we also obtain that(61)v.,t′−v.,tHmΩ≤DΩ,d,T,Nt′−tααθH2+mΩ+D1¯t′−tαα+Cα,β,dλ1m−βLh∫0tt−sα−1u.,s+t′−t−u.,sL2Ω−v.,s+t′−t−v.,sL2Ωds.
Combining (60) and (61), we derive that
(62)u.,t′−u.,tHmΩ+v.,t′−v.,tHmΩ≤DΩ,d,T,Nt′−tααφH2+mΩ+θH2+mΩ+2D1¯t′−tαα+C¯∫0tt−sα−1u.,s+t′−t−u.,sHmΩ−v.,s+t′−t−v.,sHmΩds.
Here, we setC¯=Cm2Cα,β,dλ1m−βLg+Lh. Let h>0 fixed and let the following function
(63)Yρt=e−ρtu.,t+h−u.,tHmΩ+e−ρtv.,t+h−v.,tHmΩ.
From some above observations, we can deduce that(64)eρtYρt≤DΩ,d,T,NhααφH2+mΩ+θH2+mΩ+2D1¯hαα+C¯eρt∫0te−ρt−st−sα−1e−ρsu.,s+h−u.,sHmΩ−v.,s+h−v.,sHmΩds≤DΩ,d,T,NhααφH2+mΩ+θH2+mΩ+2D1¯hαα+C¯eρt∫0te−ρt−st−sα−1dsmaxs∈0,TYρs.
From (36), we get that
(65)∫0tt−sα−1e−ρt−sds≤2αTρα/2.
This together with (64) that
(66)Yρt≤DΩ,d,T,NhααφH2+mΩ+θH2+mΩ+2D1¯hαα+C¯Tρα/2maxs∈0,TYρsred·.
This implies that(67)maxt∈0,TYρt≤DΩ,d,T,NhααφH2+mΩ+θH2+mΩ+2D1¯hαα+C¯Tρα/2maxs∈0,TYρs.
SinceC¯T/ρα/2⟶0 when ρ⟶+∞, we can choose ρ∗>0 such that
(68)C¯Tρα/2<1/2.
Hence, we follow from (67) and (64) that
(69)maxt∈0,TYρ∗t≤2DΩ,d,T,NhααφH2+mΩ+θH2+mΩ+4D1¯hαα,which allows us to conclude that
(70)u.,t+h−u.,tHmΩ+v.,t+h−v.,tHmΩ≤2eρ∗tDΩ,d,T,NhααφH2+mΩ+θH2+mΩ+eρ∗t4D1¯hααred·.
This inequality says that(71)u,v∈Cα0,T;HmΩ×Cα0,T;HmΩ.☐



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*Source: 6880435-2021-06-28.xml*