Investigation of the Weak Solvability of One Viscoelastic Fractional Voigt Model

Abstract

This article is devoted to the investigation of the weak solvability to the initial boundary value problem, which describes the viscoelastic fluid motion with memory. The memory of the fluid is considered not at a constant position of the fluid particle (as in most papers on this topic), but along the trajectory of the fluid particle (which is more physical). This leads to the appearance of an unknown function z, which is the trajectory of fluid particles and is determined by the velocity v of a fluid particle. However, in this case, the velocity v belongs to $L_2(0,T;V^1)$, which does not allow the use of the classical Cauchy Problem solution. Therefore, we use the theory of regular Lagrangian flows to correctly determine the trajectory of the particle. This paper establishes the existence of weak solutions to the considered problem. For this purpose, the topological approximation approach to the study of mathematical hydrodynamics problems, constructed by Prof. V. G. Zvyagin, is used.

1. Introduction

The motion of an incompressible homogeneous fluid with a constant density (for simplicity we will assume $\rho =1$) in a bounded domain $\mathsf{\Omega }\subset R^n$, $n=2,3$ and on the time interval $[0,T]$, $T\ge 0$, is described by the following system (see [1]):

$$\frac{\mathsf{\partial }v}{\mathsf{\partial }t}+\sum _{i=1}^n{v}_i\frac{\mathsf{\partial }v}{\mathsf{\partial }{x}_i}+\mathrm{grad} p=\mathrm{Div} \sigma +f;$$

(1)

$$\mathrm{div} v=0.$$

(2)

This system of differential equations is written in Cauchy form. Here, $v=(v_1,...,v_n)$ is an unknown vector function of velocity, p is an unknown function of fluid pressure, f is a known vector function of the density of external forces, $\sigma$ is an unknown stress rate tensor. The sign $\mathrm{Div}$ for the tensor $A=\left(a_{ij}\right)$ denotes the vector $\left({\displaystyle \sum _{i=1}^n}\frac{\mathsf{\partial }{a}_{1i}(t,x)}{\mathsf{\partial }{x}_i},...,{\displaystyle \sum _{i=1}^n}\frac{\mathsf{\partial }{a}_{ni}(t,x)}{\mathsf{\partial }{x}_i}\right)$. The equality (2) denotes the condition of fluid incompressibility.

Formally, the flow of all kinds of fluid is described by the system (1) and (2). However, in this system, the number of equations is less than the number of unknowns. For correct formulation, a rheological relation is added to the system of Equations (1) and (2), which usually connects the stress rate tensor and the strain rate tensor $\mathcal{E}\left(v\right)={\left\{{\mathcal{E}}_{ij}\right\}}_{i,j=1}^n,$ ${\mathcal{E}}_{ij}=\frac{1}{2}(\frac{\mathsf{\partial }{v}_i}{\mathsf{\partial }{x}_j}+\frac{\mathsf{\partial }{v}_j}{\mathsf{\partial }{x}_i})$ (see [2]).

An extended class of viscoelastic fluids is described by rheological relations of the form (see [3]):

$$(1+\sum _{i=1}^L{\lambda }_i\frac{d^i}{d{t}^i})\sigma =2(\mu +\sum _{i=1}^M{\mathsf{\varkappa }}_i\frac{d^i}{d{t}^i})\mathcal{E},$$

(3)

here $L,M,\mu ,{\lambda }_i,{\mathsf{\varkappa }}_i$ are some positive constants.

Among the rheological relations (3), some special cases stand out. A rheological relation of the form $\sigma =2\mu \mathcal{E}+2\mathsf{\varkappa }\frac{d\mathcal{E}}{dt}$ is called the Kelvin–Voigt rheological relation. Substituting this relation into the system (1) and (2), we obtain:

$$\frac{\mathsf{\partial }v}{\mathsf{\partial }t}+\sum _{i=1}^n{v}_i\frac{\mathsf{\partial }v}{\mathsf{\partial }{x}_i}-\mu ▵v-\mathsf{\varkappa }\frac{\mathsf{\partial }▵v}{\mathsf{\partial }t}-2\mathsf{\varkappa }\mathrm{Div}\left(\sum _{i=1}^n{v}_i\frac{\mathsf{\partial }\mathcal{E}\left(v\right)}{\mathsf{\partial }{x}_i}\right)+\mathrm{grad}p=f;$$

(4)

$$\mathrm{div} v=0.$$

(5)

This mathematical model has been studied by many authors, in particular, A. P. Okolkov [4], V. G. Zvyagin and M. V. Turbin [5], V. A. Pavlovskaya [6], V.V. Pukhnachev and O.A. Frolovskaya [7], S.N. Antontsov, H.D. De Oliveira and K. Khompysh [8] and many others.

Another important special case of the rheological relation (3) is the relation of the form $\sigma +\lambda \frac{d\sigma }{dt}=2\mu \mathcal{E}+2\mathsf{\varkappa }\frac{d\mathcal{E}}{dt}$. Substituting this relationship into the system of Equations (1) and (2), we obtain:

$$\frac{\mathsf{\partial }v}{\mathsf{\partial }t}+\sum _{i=1}^n{v}_i\frac{\mathsf{\partial }v}{\mathsf{\partial }{x}_i}+\mathrm{grad}p=\mathrm{Div}\sigma +f;$$

(6)

$$\sigma +\lambda (\frac{\mathsf{\partial }\sigma }{\mathsf{\partial }t}+\sum _{i=1}^n{v}_i\frac{\mathsf{\partial }\sigma }{\mathsf{\partial }{x}_i})=2\mu \mathcal{E}+2\mathsf{\varkappa }(\frac{\mathsf{\partial }\mathcal{E}\left(v\right)}{\mathsf{\partial }t}+\sum _{i=1}^n{v}_i\frac{\mathsf{\partial }\mathcal{E}\left(v\right)}{\mathsf{\partial }{x}_i});$$

(7)

$$\mathrm{div}v=0.$$

(8)

This mathematical model is called the Jeffries–Oldroyd model (see [9,10]). The history and description of mathematical models of this type are well presented in the review papers [11,12]. Mathematical results for this model are presented in the review [13]. Weak global time solvability is established independently in [14,15]. The uniqueness of the weak solution has not been proven.

In the paper [16], the equivalence of a weak solution to the initial boundary value problem (6)–(8) with the initial condition

$$v{\mid }_{t=0}=v_0,\sigma {\mid }_{t=0}={\sigma }_0,$$

(9)

and with the boundary condition

$$v{\mid }_{[0,T]\times \mathsf{\partial }\mathsf{\Omega }}=0$$

(10)

and a weak solution for the following integro–differential problem:

$$\frac{\mathsf{\partial }v}{\mathsf{\partial }t}+\sum _{i=1}^n{v}_i\frac{\mathsf{\partial }v}{\mathsf{\partial }{x}_i}-{\mu }_1▵v-{\mu }_2\mathrm{Div}{\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}\mathcal{E}\left(v\right)(s,z(s;t,x))ds+\mathrm{grad}p=f;$$

(11)

$$\mathrm{div}v=0;$$

(12)

$$z(\tau ;t,x)=x+{\int }_t^{\tau }v(s,z(s;t,x\left)\right)ds,0\le t,\tau \le T,x\in \overline{\mathsf{\Omega }};$$

(13)

$$v{\mid }_{[0,T]\times \mathsf{\partial }\mathsf{\Omega }}=0,v{\mid }_{t=0}=v_0.$$

(14)

is established. In other words, in the paper [16], it was proven that if the pair $(v,\sigma )$ is a weak solution to the problem (6)–(10), then a pair $(v,z)$ is a weak solution to the problem (11)–(14), where z is a regular Lagrangian flow, associated to v (the description of the RLF will be given below), and vice versa.

Let us briefly return again to the rheological relation (3). In general, all rheological relations are divided into two groups: differential and integral. The Kelvin–Voigt and Jeffreys–Oldroyd mathematical models belong to the first group. The paper [16] shows the connection between the Jeffreys–Oldroyd model and one integral model. Integrated models better take into account the effects of creep and relaxation, which are inherent in a viscoelastic medium. Thus, in recent years, there has been interest in models with fractional derivatives in the rheological relation (see [17]). The integral model takes into account all previous states of the viscoelastic medium, no matter how far they are from the current moment in time. In particular, the paper [18] considers a mechanical model of parallel connection of Newtonian elements $\left({\mu }_1\dot{e}\right)$ and elements of the Scott Blair model (${\mu }_2{D}_{0t}^{\alpha }e$, see [19]), where e is the strain tensor. For this model, Scott Blair considered Caputo’s fractional derivative $D_0^{\alpha }\phi =I_{0t}^{1-\alpha }\frac{d}{dt}\phi \left(\tau \right)=\frac{1}{\mathsf{\Gamma }(1-\alpha )}{\int }_0^t{(t-\tau )}^{-\alpha }\frac{d}{dt}\phi \left(\tau \right)d\tau$ order $\alpha \in (0,1)$ for the function $\phi \left(t\right),t\in [0,T]$ (see [20]). Here, $\mathsf{\Gamma }(1-\alpha )$ is the Euler gamma function.

Thus, we obtain the following rheological relation $\sigma ={\mu }_1\dot{\mathcal{E}}+{\mu }_2{D}_{0t}^{\alpha }e$ which is called fractional Voigt relation. Passing from the strain tensor e to the strain rate tensor $\mathcal{E}\left(v\right)=\dot{e}$ (see [21]): $\sigma ={\mu }_1\mathcal{E}\left(v\right)+{\mu }_2{I}_{0t}^{1-\alpha }\mathcal{E}\left(v\right)={\mu }_1\mathcal{E}\left(v\right)+{\mu }_2\frac{1}{\mathsf{\Gamma }(1-\alpha )}{\int }_0^t{(t-s)}^{-\alpha }\mathcal{E}\left(v\right)(s,x)ds$, here $I_{0t}^{1-\alpha }$ fractional integral. The following mathematical model is obtained, substituting the Voigt rheological relation into (1) and (2) and taking into account the memory of the medium not at point x, but on the trajectory of the particles of the medium:

$$\frac{\mathsf{\partial }v}{\mathsf{\partial }t}+\sum _{i=1}^n{v}_i\frac{\mathsf{\partial }v}{\mathsf{\partial }{x}_i}-{\mu }_1▵v$$

$$-{\mu }_2\frac{1}{\mathsf{\Gamma }(1-\alpha )}\mathrm{Div}{\int }_0^t{(t-s)}^{-\alpha }\mathcal{E}\left(v\right)(s,z(s;t,x))ds+\mathrm{grad}p=f;$$

(15)

$$\mathrm{div}v=0;$$

(16)

$$z(\tau ;t,x)=x+{\int }_t^{\tau }v(s,z(s;t,x\left)\right)ds;$$

(17)

$$v{\mid }_{[0,T]\times \mathsf{\partial }\mathsf{\Omega }},v{\mid }_{t=0}=v_0..$$

(18)

The weak solvability of this initial boundary value problem is studied in the papers [21,22,23]. In the article [24], the problem of the existence of optimal feedback control for the corresponding model is studied. A large number of mathematical papers devoted to close boundary value problems have been written (this topic is relevant, see, for example, [25,26]).

Note, that the rheological relation must include a combination of theoretical and experimental data. A brief review of the theoretical results above is given. Experimental data for these rheological relations are also obtained. It is known that the viscosity and density of the resulting solution will practically not change, if a small amount of polymer is added to water, which cannot be said from its rheological properties. Polymer additives are fixed reduction in frictional resistance [27], this fact stimulated a series of experimental papers to study aqueous polymer solution motion in pipes and in the boundary layer under laminar and turbulent flow regimes [28,29,30,31,32]. A detailed bibliography devoted to aqueous polymer solution flow in pipes is contained in [33]. In such fluids, the equilibrium state is established with some delay, characteristic of the relaxation time, and not instantly with a change in external conditions. The internal restructuring processes explain this delay. St. Petersburg scientists conducted experiments and proved that these rheological relations are acceptable for weakly concentrated aqueous polymer solutions; for example, solutions of polyethylene oxide and polyacrylamide, solutions of polyacrylamide and guar gum (see [34,35]). Therefore, this article is considering a model also called the aqueous polymer solution motion model.

Note, that using the ideas of the fractional Voigt model, it is possible to consider a close initial boundary value problem, which will connect the weak solutions of these two, on the one hand, completely different mathematical models of Kelvin–Voigt and Jeffreys–Oldroyd. Namely, this paper proposes considering the following initial boundary value problem:

$$\frac{\mathsf{\partial }v}{\mathsf{\partial }t}+\sum _{i=1}^n{v}_i\frac{\mathsf{\partial }v}{\mathsf{\partial }{x}_i}-{\mu }_0▵v$$

$$-{\mu }_1\frac{1}{\mathsf{\Gamma }(1-\alpha )}\mathrm{Div}{\int }_0^t{e^{\frac{-(t-s)}{\lambda }}(t-s)}^{-\alpha }\mathcal{E}\left(v\right)(s,z(s;t,x))ds+\nabla p=f;$$

(19)

$$\mathrm{div}v=0;$$

(20)

$$z(\tau ;t,x)=x+{\int }_t^{\tau }v(s,z(s;t,x\left)\right)ds;$$

(21)

$$v{\mid }_{\mathsf{\Gamma }}=0,(t,x)\in \mathsf{\Gamma }=[0,T]\times \mathsf{\partial }\mathsf{\Omega },v\left(0\right)=v_0,x\in \mathsf{\Omega }.$$

(22)

Here, ${\mu }_0>0,{\mu }_1\ge 0$, $0<\alpha <1$, $\lambda >0$ are constants. We have the Kelvin–Voigt mathematical model for $\lambda =1$, and we have the mathematical model (11)–(14) for $\lambda =0$, the weak solutions of which are equivalent to the weak solutions of the Jeffreys–Oldroyd model. Thus, the problem (19)–(22) connects the solutions of two completely different mathematical models. Note that a special case of this initial boundary value problem under consideration, based on the solvability proved in this paper, was announced in [36].

The purpose of this paper is to establish the existence of weak solutions to the initial boundary value problem (19)–(22). For this purpose, we use the topological approximation approach to the study of mathematical hydrodynamics problems. This approach was proposed by Prof. V.G. Zvyagin (see [37,38]) and further developed in his papers and the papers of his students ([5,15,21,22,23,24] and others). The main idea of this approach for the problem under study is as follows. (1) A family of auxiliary problems that approximate the original one (based on the addition of the Laplace operator to the second degree) is considered. (2) The solvability of the introduced auxiliary problems is studied in better function spaces than the original ones. To do this, an operator interpretation is given, the properties of operators are studied, a priori estimates are established for solutions to a family of auxiliary problems (the estimates depend on the approximation parameter, and estimates obtained in the function spaces of the auxiliary family), the theory of topological degree is applied to the resulting operator equality. (3) Next, to solve the auxiliary problem, estimates are established in the original function spaces (the estimates do not depend on the approximation parameter). Based on the obtained estimates, a passage to the limit is made. The passage to the limit proves the existence of a weak solution to the initial boundary value problem under study.

Note that other approaches to studying initial boundary value problems solvability (Galerkin method, method of semigroup theory, iterative methods), as a rule, are based on such useful properties of operators as positive definiteness, self-adjointness and others determined by the linear part of the equation, which is not always obtained and depends on the initial conditions. In particular, A.P. Oskolkov in [4] recognized the impossibility of using the modified Galerkin method in the study of the special case of this initial boundary value. Thus, the use of the topological approximation approach in this problem is the decisive point for obtaining solvability.

The work consists of several sections. In the second section, the functional spaces, the necessary information from the theory of regular Lagrangian flows is given, the definition of a weak solution to the problem under study is introduced, and the main result of the work is formulated. The third section is devoted to a family of auxiliary problems. This section introduces the concept of solving an auxiliary problem, gives an operator interpretation of the problem, studies the properties of operators, proves a priori estimates for solutions to a family of auxiliary problems, and proves the existence theorem for solutions to the auxiliary problem based on the theory of topological degree of condensing operators. The fourth section is devoted to the passage to the limit, which completes the proof of the main result of this article. At the end of the paper, the obtained results and planned further research are noted, a list of references is provided.

2. Preliminaries

Let us denote the set of measurable vector functions $v:\mathsf{\Omega }\to R^n$, summable with p-th degree by $L_p\left(\mathsf{\Omega }\right)$, $1\le p<\infty$. Let $W_p^m\left(\mathsf{\Omega }\right)$, $m\ge 1$, $p\ge 1$ is Sobolev space. We will also use the space $C_0^{\infty }\left(\mathsf{\Omega }\right)$ of infinitely differentiable vector functions from $\mathsf{\Omega }$ to $R^n$ with compact support in $\mathsf{\Omega }$. Denote by $\mathcal{V}$ the set $\{v:v\in C_0^{\infty }\left(\mathsf{\Omega }\right),\mathrm{div}v=0\}$. Let us denote the closure of $\mathcal{V}$ in the norm of $L_2\left(\mathsf{\Omega }\right)$ by $V^0$, $V^1$ is closure of $\mathcal{V}$ in the norm of $W_2^1\left(\mathsf{\Omega }\right)$, and $V^2$ is space $W_2^2\left(\mathsf{\Omega }\right)\cap V^1$.

Let us introduce a scale of spaces $V^{\beta }$, $\beta \in R$. Consider the Lere projector $P:L_2\left(\mathsf{\Omega }\right)\to V^0$ and the operator $A=-P\mathsf{\Delta }$, defined on $D\left(A\right)=V^2$, which can be continued to $V^0$ to a closed operator, which is a self-adjoint positive operator with a compact inverse. Let $0<{\lambda }_1\le {\lambda }_2\le \cdots \le {\lambda }_k\le \cdots$ —the eigenvalues of the operator A. By virtue of Hilbert’s theorem on the spectral decomposition of compact operators, the eigenfunctions $\left\{e_j\right\}$ of the operator A form an orthonormal basis in $V^0$. Denote the set of finite linear combinations composed of $e_j$ by

$$E_{\infty }=\{v=\sum _{j=1}^N{v}_j{e}_j:v_j\in R,N\in N\},$$

and define the space $V^{\beta }$, $\beta \in R$, as the completion of $E_{\infty }$ by the norm

$${\parallel v\parallel }_{V^{\beta }}=(\sum _{k=1}^{\infty }{\lambda }_k^{\beta }|v_k{{|}^2)}^{\frac{1}{2}},\mathrm{where} v=\sum _{k=1}^{\infty }{v}_k{e}_k.$$

(23)

On the space $V^{\beta }$, $\beta >-1/2$, norm (23) is equivalent to the usual norm ${\parallel ·\parallel }_{W_2^{\beta }\left(\mathsf{\Omega }\right)}$ of the space $W_2^{\beta }\left(\mathsf{\Omega }\right)$. Moreover, according to the norms in the spaces $V^1$, $V^2$ and $V^3$ can be defined as follows:

$${\parallel v\parallel }_{V^1}={({\int }_{\mathsf{\Omega }}\nabla v\left(x\right):\nabla v\left(x\right)\mathrm{d}x)}^{\frac{1}{2}},{\parallel v\parallel }_{V^2}={\left({\int }_{\mathsf{\Omega }}\mathsf{\Delta }v\left(x\right)\mathsf{\Delta }v\left(x\right)\mathrm{d}x\right)}^{\frac{1}{2}},$$

$${\parallel v\parallel }_{V^3}={({\int }_{\mathsf{\Omega }}\nabla \mathsf{\Delta }v\left(x\right):\nabla \mathsf{\Delta }v\left(x\right)\mathrm{d}x)}^{\frac{1}{2}}.$$

Here, the sign “:” for matrices A and B means $A:B={\displaystyle \sum _{i,j=1}^n}{a}_{ij}{b}_{ij}$.

Next, $V^{-\beta }$ be the space conjugate to $V^{\beta }$.

We consider solutions to this problem in space:

$$W_1=\{v\in L_2(0,T;V^1)\cap L_{\infty }(0,T;V^0),v^{\prime }\in L_{4/3}(0,T;V^{-1})\}$$

with the norm ${\parallel v\parallel }_{W_1}={\parallel v\parallel }_{L_2(0,T;V^1)}+{\parallel v\parallel }_{L_{\infty }(0,T;V^0)}+{\parallel v^{\prime }\parallel }_{L_{4/3}(0,T;V^{-1})}$.

In this paper, we will use the theory of a regular Lagrangian flow (see [39,40]).

Definition 1.

The function $z(\tau ;t,x)$, $(\tau ;t,x)\in [0,T]\times [0,T]\times \overline{\mathsf{\Omega }}$ satisfying the following conditions is called the regular Lagrangian flow (RLF) associated by v:

1.

for a. e. x and any $t\in [0,T]$ the function $\gamma \left(t\right)=z(\tau ;t,x)$ is absolutely continuous and satisfies the equation

$$z(\tau ;t,x)=x+{\int }_t^{\tau }v(s,z(s;t,x\left)\right)ds,t,\tau \in [0,T];$$

2.

for any $t,\tau \in [0,T]$ and every Borel set $B\subset \overline{\mathsf{\Omega }}$$m\left(z\right(\tau ;t,B\left)\right)=m\left(B\right)$, where m is the Lebesgue measure in $R^n$;

3.

for $t_i\in [0,T],i=\overline{1,3}$, and a. e. $x\in \overline{\mathsf{\Omega }}$ the following relation is valid:

$$z(t_3;t_1,x)=z(t_3;t_2,z(t_2;t_1,x)).$$

We consider some results about RLF.

Theorem 1.

Let $v\in L_1(0,T;W_p^1\left(\mathsf{\Omega }\right)),1\le p\le +\infty ,\mathrm{div}v(t,x)=0$ and ${v|}_{[0,T]\times \mathsf{\partial }\mathsf{\Omega }}=0$. Then, there exists a unique RLF $z\in C(D;L)$, associated to v

$$z(\tau ;t,\overline{\mathsf{\Omega }})\subset \overline{\mathsf{\Omega }},\frac{\mathsf{\partial }}{\mathsf{\partial }\tau }z(\tau ;t,x)=v(\tau ,z(\tau ;t,x)),\tau \in [0,T],x\in \mathsf{\Omega },$$

here $C(D,L)$ is Banach space of continuous functions on $D=[0,T]\times [0,T]$ with values in L (metric space of vector functions measurable on Ω).

Theorem 2.

Let $v,v^m\in L_1(0,T;W_1^p\left(\mathsf{\Omega }\right)),m=1,2,...$, for some $p>1$. Let $\mathrm{div}v(t,x)=0,$  ${\mathrm{div}}^{m}v(t,x)=0,v{{|}_{[0,T]\times \mathsf{\partial }\mathsf{\Omega }}=v^m|}_{[0,T]\times \mathsf{\partial }\mathsf{\Omega }}=0$. Let inequalities

$$\parallel v_x{\parallel }_{L_1(0,T;L_p\left(\mathsf{\Omega }\right))}+{\parallel v\parallel }_{L_1(0,T;L_1\left(\mathsf{\Omega }\right))}\le C_1,\parallel v_x^m{\parallel }_{L_1(0,T;L_p\left(\mathsf{\Omega }\right))}+{\parallel v^m\parallel }_{L_1(0,T;L_1\left(\mathsf{\Omega }\right))}\le C_2$$

be fulfilled. Let the sequence $v^m$ converges to v in $L_1\left(Q_T\right)$ as $m\to +\infty$. Let $z(\tau ;t,x)$ and $z^m(\tau ;t,x)$ be RLF associated to v and $v^m$, respectively. Then, the sequence of $z^m$ converges to z in Lebesgue measure in $[0,T]\times \mathsf{\Omega }$ for $t\in [0,T]$.

We give the definition of a weak solution to the problem (19)–(22).

Definition 2.

Let us take $f\in L_2(0,T;V^{-1})$. A function $v\in W_1$, satisfying for any $\phi \in V^1$ and a. e. $t\in (0,T)$ identity

$$\langle v^{\prime },\phi \rangle -{\int }_{\mathsf{\Omega }}\sum _{i=1}^n{v}_{i}v\frac{\mathsf{\partial }\phi }{\mathsf{\partial }{x}_i}dx+{\mu }_0{\int }_{\mathsf{\Omega }}\nabla v:\nabla \phi dx$$

$$+{\mu }_1\frac{1}{\mathsf{\Gamma }(1-\alpha )}{\int }_{\mathsf{\Omega }}{\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }\mathcal{E}\left(v\right)(s,z(s;t,x))ds\mathcal{E}\left(\phi \right)dx=\langle f,\phi \rangle ,$$

(24)

and initial condition $v\left(0\right)=v_0$ is a weak solution to problem (19)–(22). Here, z is the RLF associated with v.

The main result of this article is the following theorem:

Theorem 3.

Let us take $f\in L_2(0,T;V^{-1})$, $v\left(0\right)\in V^0$. Then, the problem (19)–(22) has at least one weak solution $v\in W_1$.

As described above, the proof will be based on the topological approximation approach. To do this, we formulate a family of auxiliary problems with a small parameter.

3. Approximative Problem

Consider the family equations ($0\le \xi \le 1$) with a small parameter $\theta >0$:

$$\theta \frac{\mathsf{\partial }{\mathsf{\Delta }}^{2}v}{\mathsf{\partial }t}+\frac{\mathsf{\partial }v}{\mathsf{\partial }t}+\xi \sum _{i=1}^n{v}_i\frac{\mathsf{\partial }v}{\mathsf{\partial }{x}_i}-{\mu }_0▵v$$

$$-\frac{\xi {\mu }_1}{\mathsf{\Gamma }(1-\alpha )}\mathrm{Div}{\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }\mathcal{E}(s,z(s;t,x))ds+\nabla p=\xi f,$$

(25)

$$\mathrm{div}v(t,x)=0,t\in [0,T],x\in \mathsf{\Omega };$$

(26)

$$z(\tau ;t,x)=x+{\int }_t^{\tau }v(s,z(s;t,x\left)\right)ds,t,\tau \in [0,T],x\in \mathsf{\Omega };$$

(27)

$$v(0,x)=v_0\left(x\right),x\in \mathsf{\Omega };v(t,x){\mid }_{\mathsf{\Gamma }}=0,(t,x)\in \mathsf{\Gamma }=\mathsf{\partial }\mathsf{\Omega };▵v{\mid }_{\mathsf{\Gamma }}=0.$$

(28)

Consider another function space for this family:

$$W_2=\{v\in C([0,T];V^3),v^{\prime }\in L_2(0,T;V^3)\}$$

with the norm ${\parallel v\parallel }_{W_2}={\parallel v\parallel }_{C(0,T;V^3)}+{\parallel v^{\prime }\parallel }_{L_2(0,T;V^3)}$.

Let us formulate the definition of a weak solution for the auxiliary problem.

Definition 3.

Let us take $f\in L_2(0,T;V^{-1})$. A function $v\in W_2$, satisfying for any $\phi \in V^1$ and a. e. $t\in (0,T)$ identity

$$\langle v^{\prime },\phi \rangle -\xi {\int }_{\mathsf{\Omega }}\sum _{i,j=1}^n{v}_i{v}_j\frac{\mathsf{\partial }{\phi }_j}{\mathsf{\partial }{x}_i}dx+{\mu }_0{\int }_{\mathsf{\Omega }}\nabla v:\nabla \phi dx-\xi \theta {\int }_{\mathsf{\Omega }}\nabla \mathsf{\Delta }{v}^{\prime }:\nabla \phi dx$$

$$+\frac{{\mu }_1\xi }{\mathsf{\Gamma }(1-\alpha )}{\int }_{\mathsf{\Omega }}{\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }\mathcal{E}(s,z(s;t,x))\mathcal{E}\left(\phi \right)dsdx=\xi \langle f,\phi \rangle$$

(29)

and initial condition $v(0,·)=v_0$ is a weak solution to problem (25)–(28). Here, z is the RLF associated to v.

To study the weak solvability of problem (25)–(28), let us move to the operator interpretation in (29). For this introduce operators:

$$J:V^3\to V^{-1},\langle Jv,\phi \rangle ={\int }_{\mathsf{\Omega }}v\phi dx,v\in V^3,\phi \in V^1;$$

$$A:V^1\to V^{-1},\langle Av,\phi \rangle ={\int }_{\mathsf{\Omega }}\nabla v:\nabla \phi dx,v\in V^1,\phi \in V^1;$$

$$A_2:V^3\to V^{-1},\langle A_{2}v,\phi \rangle =-{\int }_{\mathsf{\Omega }}\nabla \mathsf{\Delta }v:\nabla \phi dx,v\in V^3,\phi \in V^1;$$

$$B:V^1\times [0,T]\times [0,T]\times \overline{\mathsf{\Omega }}\to V^{-1},$$

$$(B(v,z)\left(t\right),\phi )=({\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\beta }\mathcal{E}\left(v\right)(s,z(s;t,x))ds,\mathcal{E}\left(\phi \right)),$$

$$v\in V^1,z\in [0,T]\times [0,T]\times \overline{\mathsf{\Omega }},\phi \in V^1,t\in (0,T);$$

$$K:L_4\left(\mathsf{\Omega }\right)\to V^{-1},\langle K\left(v\right),\phi \rangle ={\int }_{\mathsf{\Omega }}\sum _{i,j=1}^n{v}_i{v}_j\frac{\mathsf{\partial }{\phi }_j}{\mathsf{\partial }{x}_i}dx,v\in L_4\left(\mathsf{\Omega }\right),\phi \in V^1.$$

We obtain the operator equation:

$$J{v}^{\prime }-\theta A_2{v}^{\prime }+{\mu }_{0}Av+\frac{{\mu }_1\xi }{\mathsf{\Gamma }(1-\alpha )}B(v,z)-\xi K\left(v\right)=\xi f.$$

(30)

We introduce the operators to study the weak solvability of the operator Equation (30) satisfying the condition (28) for a fixed $0\le \xi \le 1$:

$$L:W_2\to L_2(0,T;V^{-1})\times V^3,L\left(v\right)=((J+\epsilon A_2)v^{\prime }+{\mu }_{0}Av,v{|}_{t=0});$$

$$C:W_2\to L_2(0,T;V^{-1})\times V^3,C\left(v\right)=(K\left(v\right),0);$$

$$G:W_2\to L_2(0,T;V^{-1})\times V^3,G\left(v\right)=({\displaystyle \frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}B(v,z),0).$$

We obtain the following equivalent problem

$$L\left(v\right)=\xi (C\left(v\right)-G\left(v\right)+(f,v_0)).$$

The following properties are valid for the operators introduced above:

Lemma 1.

1.

For any $v\in L_2(0,T;V^1)$ the function $Av$ belongs to $L_2(0,T;V^{-1})$, the operator $A:L_2([0,T];V^1)\to L_2(0,T;V^{-1})$ is continuous and the following estimates hold:

$${\parallel Av\parallel }_{V^{-1}}\le {\parallel v\parallel }_{V^1}{;\parallel Av\parallel }_{L_2(0,T;V^{-1})}\le {\parallel v\parallel }_{L_2(0,T;V^1)}.$$

2.

For any function $v\in L_p(0,T;V^3)$, $1\le p<\infty$, the function $(J+\theta A_2)v$ belongs to $L_p(0,T;V^{-1})$ and operator $(J+\theta A_2):L_p(0,T;V^3)\to L_p(0,T;V^{-1})$ is continuous and reversible, and the following estimate hold:

$${\theta \parallel v\parallel }_{L_p(0,T;V^3)}\le \parallel (J+\theta A_2){v\parallel }_{L_p(0,T;V^{-1})}\le C_3(1+\theta ){\parallel v\parallel }_{L_p(0,T;V^3)}.$$

Moreover, the inverse operator ${(J+\theta A_2)}^{-1}:L_p(0,T;V^{-1})\to L_p(0,T;V^3)$ is continuous and for any $w\in L_p(0,T;V^{-1})$ estimate hold

$$\parallel {(J+\theta A_2)}^{-1}{w\parallel }_{L_p(0,T;V^3)}\le \frac{1}{\theta }{\parallel w\parallel }_{L_p(0,T;V^{-1})}.$$

(31)

3.

The operator $L:W_2\to L_2(0,T;V^{-1})\times V^3$ is invertible and its inverse operator $L^{-1}:L_2(0,T;V^{-1})\times V^3\to W_2$ is a continuous operator.

Proof of Lemma 1.

These properties of operators are well known (see, for example, [38], Section 7.2). □

Lemma 2.

For any $v\in L_2(0,T;V^1)$, $z^m\in [0,T]\times [0,T]\times \overline{\mathsf{\Omega }}$, and $B(v,z)\in L_2(0,T;V^{-1})$ and mapping $B:L_2(0,T;V^1)\times [0,T]\times [0,T]\times \overline{\mathsf{\Omega }}\to L_2(0,T;V^{-1})$ is continuous and bounded. Moreover, for any fixed $z\in [0,T]\times [0,T]\times \overline{\mathsf{\Omega }}$, for any u, $v\in L_2(0,T;V^1)$ estimate hold

$${\parallel B(v,z)-B(u,z)\parallel }_{k,L_2(0,T;V^{-1})}\le C_4{T}^{1/2-\alpha }\sqrt{\frac{\lambda }{2+2k\lambda }}{\parallel v-u\parallel }_{k,L_2(0,T;V^1)}.$$

(32)

Proof of Lemma 2.

The continuity and boundedness of the operator B are proved in a similar way [41] Lemma 2.2.

Let us obtain an estimate (32).

Let $\overline{v}\left(t\right)=e^{-kt}v\left(t\right)$, $\overline{u}\left(t\right)=e^{-kt}u\left(t\right)$. For any $\phi \in L_2(0,T,V^1)$ we have:

$$\langle e^{-kt}B(v,z)\left(t\right)-e^{-kt}B(u,z)\left(t\right),\phi \left(t\right)\rangle$$

$$={\int }_0^T{\int }_{\mathsf{\Omega }}{\int }_0^t{e}^{-(t-s)(1/\lambda +k)}{(t-s)}^{-\alpha }{\mathcal{E}}_{ij}(\overline{v}-\overline{u})(s,z(s;t,x))ds{\mathcal{E}}_{ij}\left(\phi \right)\left(t\right)dxdt.$$

We obtained the following term using the Holder inequality and estimate

$${\int }_0^t{(t-s)}^{-\alpha }{\phi \left(s\right)ds\parallel }_{L_p(0,T)}\le C_4{T}^{1-\alpha }{\parallel \phi \left(s\right)\parallel }_{L_p(0,T)},\phi \left(s\right)\in L_p(0,T),1\le p<\infty .$$

$$\begin{array}{c}\langle e^{-kt}B(v,z)\left(t\right)-e^{-kt}B(u,z)\left(t\right),\phi \left(t\right)\rangle \\ \le {\int }_0^T{\int }_0^t{e}^{-(t-s)(1/\lambda +k)}{(t-s)}^{-\alpha }{\left({\int }_{\mathsf{\Omega }}{\mathcal{E}}^2(\overline{v}-\overline{u})(s,z(s;t,x))dx\right)}^{1/2}\times \\ \times {\left({\int }_{\mathsf{\Omega }}{\mathcal{E}}^2\left(\phi \right)(t,x)dx\right)}^{1/2}dsdt\\ ={\int }_0^T{\int }_0^t{e}^{-(t-s)(1/\lambda +k)}{(t-s)}^{-\alpha }{\left({\int }_{\mathsf{\Omega }}{\mathcal{E}}^2(\overline{v}-\overline{u})(s,z(s;t,x))dx\right)}^{1/2}{\parallel \phi \parallel }_{V^1}dsdt\\ \le {\int }_0^T{\int }_0^t{e}^{-(t-s)(1/\lambda +k)}{(t-s)}^{-\alpha }\parallel \overline{v}-\overline{u}{\parallel }_{V^1}{\parallel \phi \parallel }_{V^1}dsdt\\ \le {\int }_0^T\left({\int }_0^t{e}^{-2(t-s)(1/\lambda +k)}ds{)}^{1/2}\right({\int }_0^t{[(t-s)}^{-\alpha }\parallel \overline{v}-\overline{u}{\parallel }_{V^1}{]}^{2}ds{)}^{1/2}{\parallel \phi \parallel }_{V^1}dt\\ \le \left({\int }_0^T{\int }_0^t{e}^{-2(t-s)(1/\lambda +k)}{ds\parallel \phi \parallel }_{V^1}^{2}dt{)}^{1/2}\right({\int }_0^T{\int }_0^t{(t-s)}^{-2\alpha }{\parallel \overline{v}-\overline{u}\parallel }_{V^1}^{2}dsdt{)}^{1/2}\\ \le \sqrt{C_4}{T}^{1/2-\alpha }({\int }_0^T\parallel \overline{v}-\overline{u}{\parallel }_{V^1}^{2}dt{)}^{1/2}({\int }_0^T{\int }_0^t{e}^{-2(t-s)(1/\lambda +k)}{ds\parallel \phi \parallel }_{V^1}^2{dt)}^{1/2}\\ =\sqrt{C_4}{T}^{1/2-\alpha }\parallel \overline{v}-\overline{u}{\parallel }_{L_2(0,T;V^1)}({\int }_0^T{\int }_0^t{e}^{-2(t-s)(1/\lambda +k)}{ds\parallel \phi \parallel }_{V^1}^2{dt)}^{1/2}.\end{array}$$

Consider the last integral:

$$\begin{array}{c}({\int }_0^T{\int }_0^t{e}^{-2(t-s)(1/\lambda +k)}{ds\parallel \phi (t,·)\parallel }_{V^1}^{2}dt{)}^{1/2}=(\frac{\lambda }{2(1+k\lambda )}{\int }_0^{T}1-e^{-2t(1/\lambda +k)}{\parallel \phi (t,·)\parallel }_{V^1}^{2}dt{)}^{1/2}\\ ⩽(\frac{\lambda }{2(1+k\lambda )}{\int }_0^T{\parallel \phi (t,·)\parallel }_{V^1}^{2}dt{)}^{1/2}=(\frac{\lambda }{2(1+k\lambda )}{)}^{1/2}{\parallel \phi \parallel }_{L_2(0,T;V^1)}.\end{array}$$

Thus, we obtained the following estimate:

$$\begin{array}{c}\langle e^{-kt}B(v,z)\left(t\right)-e^{-kt}B(u,z)\left(t\right),\phi \left(t\right)\rangle \le C_4{T}^{1/2-\alpha }\sqrt{\frac{\lambda }{2+2k\lambda }}\parallel \overline{v}-\overline{u}{\parallel }_{L_2(0,T;V^1)}{\parallel \phi \parallel }_{L_2(0,T;V^1)}.\end{array}$$

Finally, we obtain the estimate (32). □

Next, we need definitions measure of non-compactness and L –condensing operators. Let us give their formulations.

Definition 4.

A non-negative real function ψ defined on a subset of a Banach space F is called a measure of non-compactness if for any subset $\mathcal{M}$ of this space the following properties hold:

1.

$\psi \left(\overline{co}\mathcal{M}\right)=\psi \left(\mathcal{M}\right)$;

2.

for any two sets ${\mathcal{M}}_1$ and ${\mathcal{M}}_2$ from ${\mathcal{M}}_1\subset {\mathcal{M}}_2$ it follows that $\psi \left({\mathcal{M}}_1\right)\le \psi \left({\mathcal{M}}_2\right)$.

In our article, we use Kuratowski non-compactness measures.

• $\psi \left(\mathcal{M}\right)=0$, if $\mathcal{M}$ is a relatively compact subset;
• $\psi (\mathcal{M}\cup K)=\psi \left(\mathcal{M}\right)$, if K is a relatively compact set.

Definition 5.

Let X be a bounded subset of a Banach space and $L:X\to F$ is mapping from X to a Banach space F. A mapping $g:X\to F$ is called L – condensing if $\psi \left(g\right(\mathcal{M}\left)\right)<\psi \left(L\right(\mathcal{M}\left)\right)$ for any set $\mathcal{M}\subseteq X$ such that $\psi \left(g\right(\mathcal{M}\left)\right)\ne 0$.

Let ${\gamma }_k$ be the measure of non-compactness of Kuratowski in the space $L_2(0,T;V^{-1})$ with the norm ${\parallel v\parallel }_{k,L_2(0,T;V^{-1})}=({\int }_0^T{\parallel v\parallel }_{V^{-1}}^2{e}^{-kt}{dt)}^{\frac{1}{2}}$. Then, the following Lemma holds.

Lemma 3.

The mapping $B:W_2\to L_2(0,T;V^{-1})$ is L –condensing with respect to the measure of non-compactness of the Kuratowski ${\gamma }_k$.

Proof of Lemma 3.

The proof is similar in [21,23] with the estimate (32) from Lemma 2. □

The following Lemma is satisfied, due to the properties above:

Lemma 4.

1.

The mapping $K:L_4\left(\mathsf{\Omega }\right)\to V^{-1}$ is continuous and has the following estimate:

$${\parallel K\left(v\right)\parallel }_{V^{-1}}\le C_5{\parallel v\parallel }_{L_4\left(\mathsf{\Omega }\right)}^2.$$

2.

For any $v\in L_4(0,T;L_4\left(\mathsf{\Omega }\right))$ the map $K:L_4(0,T;L_4\left(\mathsf{\Omega }\right))\to L_2(0,T;V^{-1})$ is continuous and the function $K\left(v\right)\in L_2(0,T;V^{-1})$.

3.

For any function $v\in W_2$, the function $K\left(v\right)\in L_2(0,T;V^{-1})$ and the mapping $K:W_2\to L_2(0,T;V^{-1})$ is compact.

Proof of Lemma 4.

This Lemma is proved in this same way as in [5]. □

Let us prove the existence of a priori estimates for solutions to the auxiliary family of equations.

Lemma 5.

Let $f\in L_2(0,T;V^{-1})$, $v_0\in V^3$. The following estimates are held for any solution to the (30).

$${\parallel v\parallel }_{L_2(0,T;V^1)}\le C_6(\parallel v_0{\parallel }_{V^0}+\sqrt{\theta }\parallel v_0{\parallel }_{V^2}+{\parallel f\parallel }_{L_2(0,T;V^{-1})});$$

(33)

$${\parallel v\parallel }_{C([0,T];V^0)}\le C_7(\parallel v_0{\parallel }_{V^0}+\sqrt{\theta }\parallel v_0{\parallel }_{V^2}+{\parallel f\parallel }_{L_2(0,T;V^{-1})});$$

(34)

$${\theta \parallel v\parallel }_{C([0,T];V^2)}^2\le C_8(\parallel v_0{\parallel }_{V^0}^2+\theta \parallel v_0{\parallel }_{V^2}^2+{\parallel f\parallel }_{L_2(0,T;V^{-1})}^2),$$

(35)

where the constants $C_6$, $C_7$, $C_8$ do not depend on ε and ξ.

Proof of Lemma 5.

Let $v\in W_2$ be the solution of the operator Equation (30). Then, for any $\phi \in V^1$ and a. e. $t\in (0,T)$ equality (29) holds. Since it is valid for all $\phi \in V^1$, we take $\phi =\overline{v}$, where $\overline{v}\left(t\right)=e^{-kt}v$. Then,

$${\int }_{\mathsf{\Omega }}{v}^{\prime }\overline{v}dx-\xi {\int }_{\mathsf{\Omega }}\sum _{i,j=1}^n{v}_i{v}_j\frac{\mathsf{\partial }{\overline{v}}_j}{\mathsf{\partial }{x}_i}dx+{\mu }_0{\int }_{\mathsf{\Omega }}\nabla \left(v\right):\nabla \left(\overline{v}\right)dx$$

$$+\frac{{\mu }_1\xi }{\mathsf{\Gamma }(1-\alpha )}{\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }(\mathcal{E}\left(v\right)(s,z(s;t,x))ds,\mathcal{E}\left(\overline{v}\right))$$

$$-\theta {\int }_{\mathsf{\Omega }}\nabla \mathsf{\Delta }{v}^{\prime }\left(t\right):\nabla \overline{v}\left(t\right)dx=\xi \langle f,\overline{v}\rangle .$$

Let us make the change of variable $v=e^{kt}\overline{v}$ and consider all the terms on the left side:

$${\int }_{\mathsf{\Omega }}{v}^{\prime }\overline{v}dx={\int }_{\mathsf{\Omega }}{\left(e^{kt}\overline{v}\right)}^{\prime }\overline{v}dx=e^{kt}{\int }_{\mathsf{\Omega }}{\overline{v}}^{\prime }\overline{v}dx+k{e}^{kt}{\int }_{\mathsf{\Omega }}\overline{v}\overline{v}dx$$

$$=\frac{e^{kt}}{2}{\int }_{\mathsf{\Omega }}\frac{\mathsf{\partial }\left(\overline{v}\overline{v}\right)}{\mathsf{\partial }t}dxk{e}^{kt}\parallel \overline{v}{\parallel }_{V^0}^2=\frac{e^{kt}}{2}\frac{d}{dt}\parallel \overline{v}{\parallel }_{V^0}^2+k{e}^{kt}{\parallel \overline{v}\parallel }_{V^0}^2.$$

$${\int }_{\mathsf{\Omega }}\sum _{i,j=1}^n{v}_i{v}_j\frac{\mathsf{\partial }\overline{v_j}}{\mathsf{\partial }{x}_i}dx=\frac{e^{kt}}{2}{\int }_{\mathsf{\Omega }}\sum _{i,j=1}^n\overline{v_i}\frac{\mathsf{\partial }\overline{v_j}\overline{v_j}}{\mathsf{\partial }{x}_i}dx=-\frac{e^{kt}}{2}{\int }_{\mathsf{\Omega }}\sum _{i=1}^n\frac{\mathsf{\partial }\overline{v_i}}{\mathsf{\partial }{x}_i}\sum _{j=1}^n\overline{v_j}\overline{v_j}$$

$$=-\frac{e^{kt}}{2}{\int }_{\mathsf{\Omega }}\mathrm{div} v\sum _{j=1}^n\overline{v_j}\overline{v_j}=0$$

Consider the following term:

$$-\theta {\int }_{\mathsf{\Omega }}\nabla \mathsf{\Delta }{v}^{\prime }:\nabla \overline{v}dx=-\theta {\int }_{\mathsf{\Omega }}\nabla \mathsf{\Delta }{\left(e^{kt}\overline{v}\right)}^{\prime }:\nabla \overline{v}dx=-\theta k{e}^{kt}{\int }_{\mathsf{\Omega }}\nabla \mathsf{\Delta }\overline{v}:\nabla \overline{v}dx$$

$$-\theta e^{kt}{\int }_{\mathsf{\Omega }}\nabla \mathsf{\Delta }{\overline{v}}^{\prime }:\nabla \overline{v}dx=\theta k{e}^{kt}{\int }_{\mathsf{\Omega }}\mathsf{\Delta }\overline{v}\mathsf{\Delta }\overline{v}dx+\frac{\theta e^{kt}}{2}{\int }_{\mathsf{\Omega }}\frac{\mathsf{\partial }}{\mathsf{\partial }t}\left(\mathsf{\Delta }\overline{v}\mathsf{\Delta }\overline{v}\right)dx$$

$$=\theta k{e}^{kt}\parallel \overline{v}{\parallel }_{V^2}^2+\frac{\theta e^{kt}}{2}\frac{d}{dt}{\parallel \overline{v}\parallel }_{V^2}^2.$$

Finally, we consider the last term:

$$e^{kt}{\mu }_0\underset{\mathsf{\Omega }}{\int }\nabla \left(\overline{v}\right):\nabla \left(\overline{v}\right)dx=e^{kt}{\mu }_0{\parallel \overline{v}\parallel }_{V^1}^2.$$

As a result, we obtain:

$$\frac{e^{kt}}{2}\frac{d}{dt}\parallel \overline{v}{\parallel }_{V^0}^2+k{e}^{kt}\parallel \overline{v}{\parallel }_{V^0}^2+{\mu }_0{e}^{kt}\parallel \overline{v}{\parallel }_{V^1}^2+\theta k{e}^{kt}\parallel \overline{v}{\parallel }_{V^2}^2+\frac{\theta e^{kt}}{2}\frac{d}{dt}{\parallel \overline{v}\parallel }_{V^2}^2$$

$$=-\frac{{\mu }_1\xi }{\mathsf{\Gamma }(1-\alpha )}{\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }(\mathcal{E}\left(\overline{v}\right)(s,z(s;t,x))ds,\mathcal{E}\left(\overline{v}\right))=e^{kt}\xi \langle f,\overline{v}\rangle .$$

Using Cauchy inequality below, we will estimate the right side of the equality.

$$bc\le \frac{\delta b^2}{2}+\frac{c^2}{2\delta }$$

for $\delta =1/{\mu }_0$, we obtain:

$$\xi e^{kt}\langle f,\overline{v}\rangle \le e^{kt}{\parallel f\parallel }_{V^{-1}}\parallel \overline{v}{\parallel }_{V^1}\le \frac{e^{kt}}{2{\mu }_0}{\parallel f\parallel }_{V^{-1}}^2+\frac{{\mu }_0{e}^{kt}}{2}{\parallel \overline{v}\parallel }_{V^1}^2.$$

Multiplying both sides of the equality by $e^{-kt}$, for almost all $t\in (0,T)$ we have

$$\frac{1}{2}\frac{d}{dt}\parallel \overline{v}{\parallel }_{V^0}^2+k\parallel \overline{v}{\parallel }_{V^0}^2+\frac{{\mu }_0}{2}\parallel \overline{v}{\parallel }_{V^1}^2+\frac{\theta }{2}\frac{d}{dt}\parallel \overline{v}{\parallel }_{V^2}^2+\theta k{\parallel \overline{v}\parallel }_{V^2}^2$$

$$\le -\frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}|\left(e^{-kt}{\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }(\mathcal{E}\left(e^{-kt}\overline{v}\right)(s,z(s;t,x))ds,\mathcal{E}\left(\overline{v}\right))\right)|+\frac{1}{2{\mu }_0}{\parallel f\parallel }_{V^{-1}}^2.$$

Integrating the last inequality over t from 0 to $\tau$, where $\tau \in [0,T]$, we obtain

$$\frac{1}{2}\parallel \overline{v}{\parallel }_{V^0}^2+\frac{\theta }{2}\parallel \overline{v}{\parallel }_{V^2}^2+k{\int }_0^{\tau }\parallel \overline{v}{\parallel }_{V^0}^{2}dt+\frac{{\mu }_0}{2}{\int }_0^{\tau }\parallel \overline{v}{\parallel }_{V^1}^{2}dt+\theta k{\int }_0^{\tau }{\parallel \overline{v}\parallel }_{V^2}^{2}dt$$

$$\le \frac{1}{2}\parallel v_0{\parallel }_{V^0}^2+\frac{1}{2{\mu }_0}{\int }_0^{\tau }{\parallel f\parallel }_{V^{-1}}^{2}dt+\frac{\theta }{2}{\parallel v_0\parallel }_{V^2}^2$$

$$+\frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}{\int }_0^{\tau }|(e^{-kt}{\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }(\mathcal{E}(e^{-kt}\overline{v})(s,z(s;t,x))ds,\mathcal{E}\left(\overline{v}\right)))|dt.$$

Using estimate (32) for $u=0$, we obtain:

$$\frac{1}{2}\parallel \overline{v}{\parallel }_{V^0}^2+\frac{\theta }{2}\parallel \overline{v}{\parallel }_{V^2}^2+k{\int }_0^{\tau }\parallel \overline{v}{\parallel }_{V^0}^{2}dt+\frac{{\mu }_0}{2}{\int }_0^{\tau }\parallel \overline{v}{\parallel }_{V^1}^{2}dt+\theta k{\int }_0^{\tau }{\parallel \overline{v}\parallel }_{V^2}^{2}dt$$

$$\le \frac{1}{2}\parallel v_0{\parallel }_{V^0}^2+{\displaystyle \frac{{\mu }_1{C}_4{T}^{1/2-\alpha }\sqrt{\frac{\lambda }{2+2k\lambda }}}{\mathsf{\Gamma }(1-\alpha )}}\parallel \overline{v}{\parallel }_{L_2(0,T;V^1)}^2+\frac{1}{2{\mu }_0}{\parallel f\parallel }_{L_2(0,T;V^{-1})}^2+\frac{\theta }{2}{\parallel v_0\parallel }_{V^2}^2.$$

Take k large enough that ${\displaystyle \frac{{\mu }_1{C}_4{T}^{1/2-\alpha }\sqrt{\frac{\lambda }{2+2k\lambda }}}{\mathsf{\Gamma }(1-\alpha )}}\le {\mu }_0/4$. Let us estimate each term on the left side:

$$\frac{{\mu }_0}{2}{\int }_0^{\tau }\parallel \overline{v}{\parallel }_{V^1}^{2}dt\le \frac{1}{2}\parallel v_0{\parallel }_{V^0}^2+\frac{\theta }{2}\parallel v_0{\parallel }_{V^2}^2+\frac{1}{2{\mu }_0}{\parallel f\parallel }_{L_2(0,T;V^{-1})}^2+\frac{{\mu }_0}{4}{\parallel \overline{v}\parallel }_{L_2(0,T;V^1)}^2,$$

$$\frac{\theta }{2}\parallel \overline{v}{\parallel }_{V^2}^2\le \frac{1}{2}\parallel v_0{\parallel }_{V^0}^2+\frac{\theta }{2}\parallel v_0{\parallel }_{V^2}^2+\frac{1}{2{\mu }_0}{\parallel f\parallel }_{L_2(0,T;V^{-1})}^2+\frac{{\mu }_0}{4}{\parallel \overline{v}\parallel }_{L_2(0,T;V^1)}^2,$$

$$\frac{1}{2}\parallel \overline{v}{\parallel }_{V^0}^2\le \frac{1}{2}\parallel v_0{\parallel }_{V^0}^2+\frac{\theta }{2}\parallel v_0{\parallel }_{V^2}^2+\frac{1}{2{\mu }_0}{\parallel f\parallel }_{L_2(0,T;V^{-1})}^2+\frac{{\mu }_0}{4}{\parallel \overline{v}\parallel }_{L_2(0,T;V^1)}^2.$$

We take maximum on the left side over $\tau \in [0,T]$, since the right side of the inequality under consideration does not depend on $\tau$

$$\frac{{\mu }_0}{2}\parallel \overline{v}{\parallel }_{L_2(0,T;V^1)}^2\le \frac{1}{2{\mu }_0}{\parallel f\parallel }_{L_2(0,T;V^{-1})}^2+\frac{{\mu }_0}{4}\parallel \overline{v}{\parallel }_{L_2(0,T;V^1)}^2+\frac{1}{2}\parallel v_0{\parallel }_{V^0}^2+\frac{\theta }{2}{\parallel v_0\parallel }_{V^2}^2,$$

$$\frac{\theta }{2}\parallel \overline{v}{\parallel }_{C([0,T];V^2)}^2\le \frac{1}{2{\mu }_0}{\parallel f\parallel }_{L_2(0,T;V^{-1})}^2+\frac{{\mu }_0}{4}\parallel \overline{v}{\parallel }_{L_2(0,T;V^1)}^2+\frac{1}{2}\parallel v_0{\parallel }_{V^0}^2+\frac{\theta }{2}{\parallel v_0\parallel }_{V^2}^2,$$

$$\frac{1}{2}\parallel \overline{v}{\parallel }_{C([0,T];V^0)}^2\le \frac{1}{2{\mu }_0}{\parallel f\parallel }_{L_2(0,T;V^{-1})}^2+\frac{{\mu }_0}{4}\parallel \overline{v}{\parallel }_{L_2(0,T;V^1)}^2+\frac{1}{2}\parallel v_0{\parallel }_{V^0}^2+\frac{\theta }{2}{\parallel v_0\parallel }_{V^2}^2.$$

From here, the required estimates (33)–(35) follow immediately. □

Lemma 6.

Let $f\in L_2(0,T;V^{-1})$, $v_0\in V^3$. Then, for any $v\in W_2$ operator Equation (30) estimates hold:

$$\theta \parallel v^{\prime }{\parallel }_{L_2(0,T;V^3)}\le C_9(1+\frac{1}{\theta })(\parallel v_0{\parallel }_{V^0}^2+{\parallel f\parallel }_{L_2(0,T;V^{-1})}^2)+C_9\sqrt{\theta }\parallel v_0{\parallel }_{V^2}+C_9{\parallel v_0\parallel }_{V^2}^2;$$

(36)

$${\parallel v\parallel }_{C([0,T];V^3)}\le \parallel v_0{\parallel }_{V^3}+\frac{C_9{T}^{\frac{1}{2}}}{\theta }(1+\frac{1}{\theta })(\parallel v_0{\parallel }_{V^0}^2+{\parallel f\parallel }_{L_2(0,T;V^{-1})}^2)$$

$$+\frac{C_9{T}^{\frac{1}{2}}}{\sqrt{\theta }}\parallel v_0{\parallel }_{V^2}+\frac{C_9{T}^{\frac{1}{2}}}{\theta }{\parallel v_0\parallel }_{V^2}^2;$$

(37)

$$\parallel v^{\prime }{\parallel }_{L_{4/3}(0,T;V^{-1})}\le C_{10}(\parallel v_0{\parallel }_{V^0}^2+\theta \parallel v_0{\parallel }_{V^2}^2+{\parallel f\parallel }_{L_2(0,T;V^{-1})}^2+1);$$

(38)

$$\theta \parallel v^{\prime }{\parallel }_{L_{4/3}(0,T;V^3)}\le C_{11}(\parallel v_0{\parallel }_{V^0}^2+\theta \parallel v_0{\parallel }_{V^2}^2+{\parallel f\parallel }_{L_2(0,T;V^{-1})}^2+1);$$

(39)

where the constants $C_9$, $C_{10}$, $C_{11}$ do not depend on θ, v, ξ.

Proof of Lemma 6.

Let $v\in W_2$ is solution to (30). Then, it satisfies the following operator equation

$$J{v}^{\prime }+\theta A_2{v}^{\prime }+{\mu }_{0}Av+\xi {\displaystyle \frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}B(v,z)-\xi K\left(v\right)=\xi f.$$

Therefore,

$$\parallel (J+\theta A_2)v^{\prime }{\parallel }_{L_2(0,T;V^{-1})}={\parallel \xi f-{\mu }_{0}Av-{\displaystyle \frac{\xi {\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}B(v,z)-\xi K\left(v\right)\parallel }_{L_2(0,T;V^{-1})}.$$

Let us estimate the right side, using the estimate (32) for $u=0$.

$$\parallel \xi f-{\mu }_{0}Av-{\displaystyle \frac{\xi {\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}{B(v,z)+\xi K\left(v\right)\parallel }_{L_2(0,T;V^{-1})}$$

$$\le {\parallel f\parallel }_{L_2(0,T;V^{-1})}+{\displaystyle \frac{{\mu }_1{C}_4{T}^{1-\alpha }}{\mathsf{\Gamma }(1-\alpha )}}{\parallel v\parallel }_{L_2(0,T;V^1)}+{\mu }_0{\parallel v\parallel }_{L_2(0,T;V^1)}+{\parallel K\left(v\right)\parallel }_{L_2(0,T;V^{-1})}.$$

(40)

Let us separately estimate the value ${\parallel K\left(v\right)\parallel }_{L_2(0,T;V^{-1})}$. Using an estimate of the operator K, as well as the continuity of the embedding $V^2\subset L_4\left(\mathsf{\Omega }\right)$, we have:

$${\parallel K\left(v\right)\parallel }_{L_2(0,T;V^{-1})}=({\int }_0^T{\parallel K\left(v\right)\parallel }_{V^{-1}}^2{dt)}^{\frac{1}{2}}\le C_5({\int }_0^T{\parallel v\left(t\right)\parallel }_{L_4\left(\mathsf{\Omega }\right)}^4{dt)}^{\frac{1}{2}}$$

$$\le C_{12}({\int }_0^T{\parallel v\left(t\right)\parallel }_{V^2}^4{dt)}^{\frac{1}{2}}\le C_{12}{T}^{\frac{1}{2}}\underset{t\in [0,T]}{max}{\parallel v\left(t\right)\parallel }_{V^2}^2=C_{12}{T}^{\frac{1}{2}}{\parallel v\parallel }_{C([0,T];V^2)}^2.$$

Let us rewrite (40) in the form:

$$\parallel \xi f-{\mu }_{0}Av-{\displaystyle \frac{\xi {\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}{B(v,z)+\xi K\left(v\right)\parallel }_{L_2(0,T;V^{-1})}$$

$$\le C_{13}{(\parallel f\parallel }_{L_2(0,T;V^{-1})}+{\parallel v\parallel }_{L_2(0,T;V^1)}+C_{12}{T}^{\frac{1}{2}}{\parallel v\parallel }_{C([0,T];V^2)}^2).$$

From a priori estimates (33) and (35) it follows that

$$\parallel (J+\theta A_2)v^{\prime }{\parallel }_{L_2(0,T;V^{-1})}\le C_9(1+\frac{1}{\theta })(\parallel v_0{\parallel }_{V^0}^2+{\parallel f\parallel }_{L_2(0,T;V^{-1})}^2)$$

$$+C_9\sqrt{\theta }\parallel v_0{\parallel }_{V^2}+C_9{\parallel v_0\parallel }_{V^2}^2.$$

In order to obtain a lower estimate, we use estimate (31). We obtain:

$$\theta \parallel v^{\prime }{\parallel }_{L_2(0,T;V^3)}\le {\parallel (J+\theta A_2)v^{\prime }\parallel }_{L_2(0,T;V^{-1})}$$

$$\le C_9(1+\frac{1}{\theta })(\parallel v_0{\parallel }_{V^0}^2+{\parallel f\parallel }_{L_2(0,T;V^{-1})}^2)+C_9\sqrt{\theta }\parallel v_0{\parallel }_{V^2}+C_9{\parallel v_0\parallel }_{V^2}^2.$$

Therefore, inequality (36) is proved.

Let us pass to estimate (37). Let us represent the function $v\in W_2$ as follows:

$$v=v_0-{\int }_0^t{v}^{\prime }\left(s\right)ds.$$

Then,

$${\parallel v\parallel }_{V^1}\le \parallel v_0-{\int }_0^t{v}^{\prime }{\left(s\right)\parallel }_{V^1}ds\le \parallel v_0{\parallel }_{V^3}+\sqrt{T}{\parallel v^{\prime }\parallel }_{L_2(0,T;V^1)}.$$

We pass to the maximum at $\tau \in [0,T]$ on the left side since the right side of the resulting inequality does not depend on t. Then, using (36), we obtain

$$\underset{t\in [0,T]}{max}{\parallel v\left(t\right)\parallel }_{V^3}\le \parallel v_0{\parallel }_{V^3}+\frac{C_9{T}^{\frac{1}{2}}}{\theta }(1+\frac{1}{\theta })(\parallel v_0{\parallel }_{V^0}^2+{\parallel f\parallel }_{L_2(0,T;V^{-1})}^2)$$

$$+\frac{C_9{T}^{\frac{1}{2}}}{\sqrt{\theta }}\parallel v_0{\parallel }_{V^2}+\frac{C_9{T}^{\frac{1}{2}}}{\theta }{\parallel v_0\parallel }_{V^2}^2.$$

Thus, estimate (37) is established.

Now, we will prove (38). Let $v\in W_2$ is solution of the (30). Then,

$$\parallel v^{\prime }{\parallel }_{L_{4/3}(0,T;V^{-1})}\le {\parallel \xi f-{\mu }_{0}Av-{\displaystyle \frac{\xi {\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}B(v,z)-\theta A^2{v}^{\prime }+K\left(v\right)\parallel }_{L_{4/3}(0,T;V^{-1})}$$

$$\le {\parallel f\parallel }_{L_{4/3}(0,T;V^{-1})}+{\mu }_0{\parallel Av\parallel }_{L_{4/3}(0,T;V^{-1})}+{\displaystyle \frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}{\parallel B(v,z)\parallel }_{L_{4/3}(0,T;V^{-1})}$$

$$+\theta \parallel A^2{v}^{\prime }{\parallel }_{L_{4/3}(0,T;V^{-1})}+{\parallel K\left(v\right)\parallel }_{L_{4/3}(0,T;V^{-1})}.$$

(41)

Now, we will consider the terms on the right side of the last inequality. Let us start by estimating the operator ${\parallel K\left(v\right)\parallel }_{L_{4/3}(0,T;V^{-1})}$. To do this, we will use the following inequality for $n=3$

$${\parallel u\parallel }_{L_4\left(\mathsf{\Omega }\right)}\le 2^{\frac{1}{2}}{\parallel u\parallel }_{L_2\left(\mathsf{\Omega }\right)}^{\frac{1}{4}}{\parallel \nabla u\parallel }_{L_2\left(\mathsf{\Omega }\right)}^{\frac{3}{4}},u\in V^1,$$

and estimate of operator K, we obtain (for the case $n=2$ the proof is similar):

$${\parallel K\left(v\right)\parallel }_{L_{4/3}(0,T;V^{-1})}=({\int }_0^T{\parallel K\left(v\right)\parallel }_{V^{-1}}^{\frac{4}{3}}{dt)}^{\frac{3}{4}}\le C_5({\int }_0^T{\parallel v\parallel }_{L_4\left(\mathsf{\Omega }\right)}^{\frac{8}{3}}{dt)}^{\frac{3}{4}}$$

$$\le 2C_5({\int }_0^T{\parallel v\parallel }_{L_2\left(\mathsf{\Omega }\right)}^{\frac{2}{3}}{\parallel \nabla v\parallel }_{L_2\left(\mathsf{\Omega }\right)}^2{dt)}^{\frac{3}{4}}\le C_{14}({\int }_0^T{\parallel v\parallel }_{V^0}^{\frac{2}{3}}{\parallel v\parallel }_{V^1}^2{dt)}^{\frac{3}{4}}$$

$$\le C_{14}{\parallel v\parallel }_{C([0,T];V^0)}^{\frac{1}{2}}({\int }_0^T{\parallel v\parallel }_{V^1}^2{dt)}^{\frac{3}{4}}=C_{14}{\parallel v\parallel }_{C([0,T];V^0)}^{\frac{1}{2}}{\parallel v\parallel }_{L_2(0,T;V^1)}^{\frac{3}{2}}.$$

We obtain the following estimate, using Holder’s inequality:

$${\parallel Av\parallel }_{L_{4/3}(0,T;V^{-1})}=({\int }_0^T{\parallel Av\parallel }_{V^{-1}}^{\frac{4}{3}}{dt)}^{\frac{3}{4}}\le ({\int }_0^T{\parallel v\parallel }_{V^1}^{\frac{4}{3}}{dt)}^{\frac{3}{4}}$$

$$\le T^{\frac{1}{4}}({\int }_0^T{\parallel v\parallel }_{V^1}^2{dt)}^{\frac{1}{2}}=T^{\frac{1}{4}}{\parallel v\parallel }_{L_2(0,T;V^1)}.$$

Let us estimate the following operator, using the estimate (32) for $u=0$ and Holder’s inequality:

$${\parallel B(v,z)\parallel }_{L_{4/3}(0,T;V^{-1})}=({\int }_0^T{\parallel B(v,z)\parallel }_{V^{-1}}^{\frac{4}{3}}{dt)}^{\frac{3}{4}}\le T^{\frac{1}{4}}({\int }_0^T{\parallel B(v,z)\parallel }_{V^{-1}}^2{dt)}^{\frac{1}{2}}$$

$$=T^{\frac{1}{4}}{\parallel B(v,z)\parallel }_{L_2(0,T;V^{-1})}\le T^{\frac{1}{4}}{T}^{1-\alpha }{C}_{10}{\parallel v\parallel }_{L_2(-0,T;V^1)}.$$

Finally, consider the last term.

$$\theta \parallel A^2{v}^{\prime }{\parallel }_{L_{4/3}(0,T;V^{-1})}=\theta ({\int }_0^T\parallel A^2{v}^{\prime }{\parallel }_{V^{-1}}^{\frac{4}{3}}{dt)}^{\frac{3}{4}}\le \theta ({\int }_0^T\parallel v^{\prime }{\parallel }_{V^3}^{\frac{4}{3}}{dt)}^{\frac{3}{4}}\le \theta {\parallel v^{\prime }\parallel }_{L_{4/3}(0,T;V^3)}.$$

Let us estimate the right part, using the left part of the estimate (31) for $p=4/3$.

$$\theta \parallel v^{\prime }{\parallel }_{L_{4/3}(0,T;V^3)}\le {\parallel f\parallel }_{L_{4/3}(0,T;V^{-1})}-{\mu }_0{\parallel Av\parallel }_{L_{4/3}(0,T;V^{-1})}$$

$$+{\parallel K\left(v\right)\parallel }_{L_{4/3}(0,T;V^{-1})}+{\mu }_1{\parallel B(v,z)\parallel }_{L_{4/3}(0,T;V^{-1})}.$$

Thus,

$$\theta \parallel A^2{v}^{\prime }{\parallel }_{L_{4/3}(0,T;V^{-1})}\le \theta \parallel v^{\prime }{\parallel }_{L_{4/3}(0,T;V^3)}\le {\parallel f\parallel }_{L_{4/3}(0,T;V^{-1})}$$

$$+{\mu }_0{\parallel Av\parallel }_{L_{4/3}(0,T;V^{-1})}+{\displaystyle \frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}{\parallel B(v,z)\parallel }_{L_{4/3}(0,T;V^{-1})}+{\parallel K\left(v\right)\parallel }_{L_{4/3}(0,T;V^{-1})}.$$

So, from (41), the estimates of our operators above, and a priori estimates (33) and (34), we obtain

$$\parallel v^{\prime }{\parallel }_{L_{4/3}(0,T;V^{-1})}\le {2(\parallel f\parallel }_{L_{4/3}(0,T;V^{-1})}+{\mu }_0{\parallel Av\parallel }_{L_{4/3}(0,T;V^{-1})}$$

$$+{\parallel K\left(v\right)\parallel }_{L_{4/3}(0,T;V^{-1})}+{\displaystyle \frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}{\parallel B(v,z)\parallel }_{L_{4/3}(0,T;V^{-1})})$$

$$\le C_{15}{(\parallel f\parallel }_{L_2(0,T;V^{-1})}+{\parallel v\parallel }_{L_2(0,T;V^1)}+{\parallel v\parallel }_{C([0,T];V^0)}^{\frac{1}{2}}{\parallel v\parallel }_{L_2(0,T;V^1)}^{\frac{3}{2}})$$

$$\le C_{16}{\left(\right(\parallel f\parallel }_{L_2(0,T;V^{-1})}+\parallel v_0{\parallel }_{V^0}+\sqrt{\theta }\parallel v_0{\parallel }_{V^2})$$

$$+{(\parallel f\parallel }_{L_2(0,T;V^{-1})}+\parallel v_0{\parallel }_{V^0}+\sqrt{\theta }\parallel v_0{\parallel }_{V^2}{)}^{\frac{1}{2}}{(\parallel f\parallel }_{L_2(0,T;V^{-1})}+\parallel v_0{\parallel }_{V^0}+\sqrt{\theta }\parallel v_0{\parallel }_{V^2}{)}^{\frac{3}{2}})$$

$$\le C_{16}{(\parallel f\parallel }_{L_2(0,T;V^{-1})}+1+\parallel v_0{\parallel }_{V^0}+\sqrt{\theta }\parallel v_0{{\parallel }_{V^2})}^2$$

$$\le 4C_{16}{(\parallel f\parallel }_{L_2(0,T;V^{-1})}^2+1+\parallel v_0{\parallel }_{V^0}+\sqrt{\theta }\parallel v_0{\parallel }_{V^2}).$$

Then, we obtain inequality (38), where $C_{10}=4C_{16}$.

Finally, again applying the estimates for our operators, a priori estimates (33) and (34) for the right side of (41), we obtain

$$\theta \parallel v^{\prime }{\parallel }_{L_{4/3}(0,T;V^3)}\le {2(\parallel f\parallel }_{L_{4/3}(0,T;V^{-1})}+{\mu }_0{\parallel Av\parallel }_{L_{4/3}(0,T;V^{-1})}$$

$$+{\displaystyle \frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}{\parallel B(v,z)\parallel }_{L_{4/3}(0,T;V^{-1})}{)+\parallel K\left(v\right)\parallel }_{L_{4/3}(0,T;V^{-1})}$$

$$\le C_{17}{(\parallel f\parallel }_{L_2(0,T;V^{-1})}+{\parallel v\parallel }_{L_2(0,T;V^1)}+{\parallel v\parallel }_{C([0,T];V^0)}^{\frac{1}{2}}{\parallel v\parallel }_{L_2(0,T;V^1)}^{\frac{3}{2}})$$

$$\le C_{18}{(\parallel f\parallel }_{L_2(0,T;V^{-1})}+\parallel v_0{\parallel }_{V^0}+\sqrt{\theta }\parallel v_0{\parallel }_{V^2})$$

$$+{\left(\right(\parallel f\parallel }_{L_2(0,T;V^{-1})}+\parallel v_0{\parallel }_{V^0}+\sqrt{\theta }\parallel v_0{\parallel }_{V^2}{)}^{\frac{1}{2}}{(\parallel f\parallel }_{L_2(0,T;V^{-1})}+\parallel v_0{\parallel }_{V^0}+\sqrt{\theta }\parallel v_0{\parallel }_{V^2}{)}^{\frac{3}{2}})$$

$$\le C_{18}{(\parallel f\parallel }_{L_2(0,T;V^{-1})}+1+\parallel v_0{\parallel }_{V^0}+\sqrt{\theta }\parallel v_0{\parallel }_{V^2}{\left)\right)}^2$$

$$\le 4C_{18}{(\parallel f\parallel }_{L_2(0,T;V^{-1})}^2+1+\parallel v_0{\parallel }_{V^0}+\sqrt{\theta }\parallel v_0{\parallel }_{V^2}\left)\right).$$

Thus, inequality (39) is established, where $C_{11}=4C_{18}$. □

Thus, from Lemmas 5 and 6 we obtain the following result.

Lemma 7.

Let $f\in L_2(0,T;V^{-1})$, $v_0\in V^3$. Then, for any $v\in W_2$ operator Equation (30) there is an estimate:

$${\parallel v\parallel }_{W_2}\le C_{19},$$

where $C_{19}$ depends on θ.

Theorem 4.

Let $f\in L_2(0,T;V^{-1})$, $v_0\in V^3$. Then, there is at least one solution $v\in W_2$ of the approximative problem (25)–(28) for $\xi =1$.

Proof of Theorem 4.

Let us prove this theorem, using the theory of topological degree of condensing vector fields.

Here, we recall some elementary facts from the topological degree theory (see [42,43]).

Let E be a normed space and B be the class of open bounded subset of E. Denote by $\mathsf{\Sigma }$ the set of triples $(I-k,D,p)$ where I is identity operator in E, $p\in E,D\in B,k:\overline{D}\to E$ is L—condensing operator, $p\notin (I-k)(\mathsf{\partial }D)$. Here $\mathsf{\partial }D$ denotes the boundary of D.

Theorem 5.

There exists a unique map $d:\mathsf{\Sigma }\to \mathbb{Z}$ satisfying the following four conditions (axioms).

1.

(Normalization). For any $D\in B$ such that $0\in D$, one has

$$d(I,D,0)=1.$$

2.

(Additivity). For any $(I-k,D,p)\in \mathsf{\Sigma }$ and all open sets $D_1,D_2\subset D$ such that $p\notin f(\overline{D}\setminus (D_1\bigcup D_2))$ one has

$$d((I-k){\mid }_{\overline{D_1}},D_1,p)+d((I-k){\mid }_{\overline{D_2}},D_2,p)=d(I-k,D,p).$$

3.

(Homotopic invariance). Let $D\in B, D\ne Ø$, and let $h:[0,1]\times \overline{D}\to E$ be a compact operator. Assume that $p\ne x-h(t,x)$ for $t\in [0,T],x\in \mathsf{\partial }D$. Such h is called a homotopy. Then, $d(I-h_t,D,p)$ does not depend on $t\in [0,T]$ where $h_t:\overline{D}\to E,h_t\left(x\right)=h(t,x)$.

4.

(Homogeneity). For any $(I-k,D,p)\in \mathrm{\Sigma }$ such that $D\ne Ø$, one has

$$d(f,D,p)=d(f-p,D,0).$$

The most important property of the degree, which we will use in this proof, is

Theorem 6.

Let $(I-k,D,p)\in \mathrm{\Sigma }$ and $d(I-k,D,p)\ne 0$. Then, the equation

$$x-k\left(x\right)=p$$

has a solution $x_0\in D$.

From Lemma 6, it follows that all solutions of the equation

$$L\left(v\right)-\xi C\left(v\right)+\xi G\left(v\right)=\xi (f,v_0).$$

(42)

lie in the ball $B_R\subset W_2$ with center at zero and radius $R=C_{19}+1$. From Lemma 1, the operator $L:W_2\to L_2(0,T;V^{-1})\times V^3$ is invertible. Then, from (42)

$$v=\xi L^{-1}(C\left(v\right)-G\left(v\right)+(f,v_0)).$$

From Lemmas 1, 3 and 4, the operator $L^{-1}:L_2(0,T;V^{-1})\times V^3\to W_2$ is continuous, the operator $(C\left(v\right)-G\left(v\right)+(f,v_0)):W_2\to L_2(0,T;V^{-1})\times V^3$ is L — condensing with respect to the Kuratowski measure of non-compactness ${\gamma }_k$. Therefore, the operator $L^{-1}(C\left(v\right)-G\left(v\right)+(f,v_0)):W_2\to W_2$ is also L — condensing.

Thus, the vector field $v-\xi L^{-1}(C\left(v\right)-G\left(v\right)+(f,v_0))$ is non-degenerate on the boundary of the ball $B_R$. By Theorem 6 a topological degree is defined for this vector field $\mathrm{deg}(I-\xi L^{-1}(C-G+f),B_R,0)$. Using properties of homotopy invariance and degree normalization, we obtain

$$\mathrm{deg}(I-\xi L^{-1}(C-G+f),B_R,0)=\mathrm{deg}(I,B_R,0)=1.$$

By Theorem 6, the existence of at least one solution of the $v\in W_2$ Equation (30) is obtained. It proves that weak solutions exist to the initial boundary value problem (25)–(28). □

4. Passage to the Limit

We carry out the passage to the limit in the approximative problem (25)–(28) at $\xi =1$. Since the space $V^3$ is dense in $V^0$, for any $v_0^{*}\in V^0$ there is a sequence $v_0^m\in V^3$ converging to $v_0^{*}$ in $V^0$. We put $v_0^m\equiv 0$, ${\theta }_m=1/m$, when $v_0^{*}\equiv 0$, then $\parallel v_0^{*}{\parallel }_{V^0}\ne 0$, then from some number $\parallel v_0^m{\parallel }_{V^2}\ne 0$. And we consider ${\theta }_m=1/(m\parallel v_0^m{\parallel }_{V^2}^2)$. By virtue of our choice, the resulting sequence $\left\{{\theta }_m\right\}$ converges to zero at $m\to \infty$. And ${\theta }_m{\parallel v_0^m\parallel }_{V^2}^2\le 1$.

By Theorem 4, for each ${\theta }_m$ and $v_0^m$ there is a solution $v_m\in W_2\subset W_1$ approximative problem (25)–(28), when $\xi =1$. Thus, every solution $v_m$ for any $\phi \in V^1$ and a. e. $t\in (0,T)$ satisfying the identity

$$\langle {\left(v^m\right)}^{\prime },\phi \rangle -{\int }_{\mathsf{\Omega }}\sum _{i,j=1}^n{v}_i^m{v}_j^m\frac{\mathsf{\partial }{\phi }_j}{\mathsf{\partial }{x}_i}dx+{\mu }_0{\int }_{\mathsf{\Omega }}\nabla v^m:\nabla \phi dx-{\theta }_m{\int }_{\mathsf{\Omega }}\nabla \mathsf{\Delta }{\left(v^m\right)}^{\prime }:\nabla \phi dx$$

$$+{\displaystyle \frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}({\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }\mathcal{E}\left(v^m\right)(s,z^m(s;t,x))ds,\mathcal{E}\left(\phi \right))=\langle f,\phi \rangle .$$

(43)

and the initial condition

$$v(0,x)=v_0^m.$$

Thus, from estimates (33), (34), (38) and (39) we obtain that

$$\parallel v^m{\parallel }_{L_2(0,T;V^1)}^2\le C_{20}{(\parallel f\parallel }_{L_2(0,T;V^{-1})}^2+1),$$

(44)

$$\parallel v^m{\parallel }_{C([0,T];V^0)}^2\le C_{21}{(\parallel f\parallel }_{L_2(0,T;V^{-1})}^2+1),$$

(45)

$$\parallel {\left(v^m\right)}^{\prime }{\parallel }_{L_{4/3}(0,T;V^{-1})}\le C_{22}{(\parallel f\parallel }_{L_2(0,T;V^{-1})}^2+1),$$

(46)

$$\theta \parallel {\left(v^m\right)}^{\prime }{\parallel }_{L_{4/3}(0,T;V^3)}\le C_{23}{(\parallel f\parallel }_{L_2(0,T;V^{-1})}^2+1),$$

(47)

where the constants $C_{20}-C_{23}$ do not depend on $\theta$. Due to embedding $C([0,T];V^0)\subset L_{\infty }(0,T;V^0)$ and the estimates (44)–(47) (passing to a subsequence if necessary), we obtain that

$$v^m\to v^{*}\mathrm{weak}\mathrm{in}{L}_2(0,T;V^1)\mathrm{at}m\to \infty ,$$

$$v^m\to v^{*}*-\mathrm{weakly}\mathrm{in}{L}_{\infty }(0,T;V^0)\mathrm{at}m\to \infty ,$$

$${\left(v^m\right)}^{\prime }\to {\left(v^{*}\right)}^{\prime }\mathrm{weak}\mathrm{in}{L}_{4/3}(0,T;V^{-1})\mathrm{at}m\to \infty ,$$

and that the limit function $v^{*}$ belongs to the space $W_1.$

Consider (21) for the limit function $v^{*}$. Note that the limit function $v^{*}\in W_1$ satisfies the conditions of Theorem 1. Therefore, $z^{*}(\tau ;t,x)$ associated with $v^{*}$. Denote by $z^m(\tau ;t,x)$ the RLF associated to $v^m$. From a priori estimate of Lemma 6 and Theorem 2 we have:

Lemma 8.

The sequence $z^m(\tau ;t,x)$ converges in $(\tau ,x)$ Lebesgue measure on $[0,T]\times \mathsf{\Omega }$ to $z(\tau ;t,x)$ for $t\in [0,T]$.

The proof of the solvability of the initial boundary value problem (19)–(22) is divided into two parts. The first part will be devoted to the passage to the limit in the approximative problem (25)–(28) with a smooth function $\phi$ from $V^1$, then for an arbitrary function $\phi \in V^1$.

Let the function $\phi$ from $V^1$ be smooth. Let us pass to the limit in each term (43). ${\left(v^m\right)}^{\prime }\to {\left(v^{*}\right)}^{\prime }$ in $L_{4/3}(0,T;V^{-1})$.

By the definition of weak convergence of $v^m\to v^{*}$ in $L_2(0,T;V^1)$ and for any $\phi \in V^1$, we obtain

$${\mu }_0{\int }_{\mathsf{\Omega }}\nabla v^m:\nabla \phi dx\to {\mu }_0{\int }_{\mathsf{\Omega }}\nabla v^{*}:\nabla \phi dx$$

as $m\to \infty$,

$$\langle {\left(v^m\right)}^{\prime },\phi \rangle \to \langle {\left(v^{*}\right)}^{\prime },\phi \rangle .$$

Next, using estimate (47), without loss of generality and, if necessary, passing to a subsequence, we have that there exists a function $u\in L_{4/3}(0,T;V^3)$ such that

$${\theta }_m{\left(v^m\right)}^{\prime }\to u\mathrm{weak}\mathrm{in}{L}_{4/3}(0,T;V^3)\mathrm{at}m\to \infty .$$

Then,

$${\theta }_m\langle \nabla \mathsf{\Delta }{\left(v^m\right)}^{\prime },\nabla \phi \rangle \to \langle \nabla \mathsf{\Delta }u,\nabla \phi \rangle ,\mathrm{at}m\to \infty .$$

However, the sequence ${\theta }_m{\left(v^m\right)}^{\prime }$ converges to zero in the sense of distributions on $[0,T]$ in $V^{-3}$. Indeed, for any compactly supported smooth scalar function $\psi$ and $\phi \in V^3$, we obtain

$$\underset{m\to \infty }{lim}|{\theta }_m{\int }_0^T{\int }_{\mathsf{\Omega }}\nabla \mathsf{\Delta }{\left(v^m\right)}^{\prime }:\nabla \phi dx\psi \left(t\right)dt|=\underset{m\to \infty }{lim}{\theta }_m|{\int }_0^T{\int }_{\mathsf{\Omega }}\mathsf{\Delta }{\left(v^m\right)}^{\prime }\mathsf{\Delta }\phi dx\psi \left(t\right)dt|$$

$$=\underset{m\to \infty }{lim}{\theta }_m|{\int }_0^T{\int }_{\mathsf{\Omega }}\nabla {\left(v^m\right)}^{\prime }:\nabla \mathsf{\Delta }\phi dx\psi \left(t\right)dt|$$

$$=\underset{m\to \infty }{lim}{\theta }_m\underset{m\to \infty }{lim}|{\int }_0^T{\int }_{\mathsf{\Omega }}\nabla {\left(v^m\right)}^{\prime }:\nabla \mathsf{\Delta }\phi dx\psi \left(t\right)dt|$$

$$=\underset{m\to \infty }{lim}{\theta }_m\underset{m\to \infty }{lim}|{\int }_{\mathsf{\Omega }}({\int }_0^T\nabla {\left(v^m\right)}^{\prime }\psi \left(t\right)dt):\nabla \mathsf{\Delta }\phi dx|$$

$$=\underset{m\to \infty }{lim}{\theta }_m\underset{m\to \infty }{lim}|{\int }_{\mathsf{\Omega }}({\int }_0^T\nabla v^m\frac{\mathsf{\partial }\psi \left(t\right)}{\mathsf{\partial }t}dt):\nabla \mathsf{\Delta }\phi dx|$$

$$=\underset{m\to \infty }{lim}{\theta }_m\underset{m\to \infty }{lim}|{\int }_0^T{\int }_{\mathsf{\Omega }}\nabla v^m:\nabla \mathsf{\Delta }\phi dx\frac{\mathsf{\partial }\psi \left(t\right)}{\mathsf{\partial }t}dt|.$$

Since $v^m$ converges weakly to $v^{*}$ in $L_2(0,T;V^1)$ and hence converges to $v^{*}$ in the sense of distributions, then

$$\underset{m\to \infty }{lim}{\theta }_m\underset{m\to \infty }{lim}|{\int }_0^T{\int }_{\mathsf{\Omega }}\nabla v^m:\nabla \mathsf{\Delta }\phi dx\frac{\mathsf{\partial }\psi \left(t\right)}{\mathsf{\partial }t}dt|$$

$$=|{\int }_0^T{\int }_{\mathsf{\Omega }}\nabla v^{*}:\nabla \mathsf{\Delta }\phi dx\frac{\mathsf{\partial }\psi \left(t\right)}{\mathsf{\partial }t}dt|\underset{m\to \infty }{lim}{\theta }_m=0.$$

Thus, due to the uniqueness of the weak limit,

$${\theta }_m\langle \nabla \mathsf{\Delta }{\left(v^m\right)}^{\prime },\nabla \phi \rangle \to 0\mathrm{at}m\to \infty .$$

Now, let us show that

$${\displaystyle \frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}({\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }\mathcal{E}(v^m)(s,z^m(s;t,x))ds,\mathcal{E}(\phi ))$$

$$\to {\displaystyle \frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}({\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }\mathcal{E}(v^{*})(s,z^{*}(s;t,x))ds,\mathcal{E}(\phi )).$$

(48)

Consider the difference

$${\displaystyle \frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}({\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }\mathcal{E}(v^m)(s,z^m(s;t,x))ds,\mathcal{E}(\phi ))$$

$$-{\displaystyle \frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}({\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }\mathcal{E}(v^{*})(s,z^{*}(s;t,x))ds,\mathcal{E}(\phi ))$$

$$={\displaystyle \frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}({\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }$$

$$\times {\int }_{\mathsf{\Omega }}[\mathcal{E}(v^m)(s,z^m(s;t,x))-\mathcal{E}(v^{*})(s,z^m(s;t,x))]:\mathcal{E}(\phi )dxds)$$

$$+{\displaystyle \frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}({\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }$$

$$\times {\int }_{\mathsf{\Omega }}[\mathcal{E}(v^{*})(s,z^m(s;t,x))-\mathcal{E}(v^{*})(s,z^{*}(s;t,x))]:\mathcal{E}(\phi )dxds)=Z_1^m+Z_2^m.$$

(1) Let us first show that $Z_1^m\to 0$ as $m\to \infty$.

Denote the integral over $\mathsf{\Omega }$ in $Z_1^m$ by I:

$$I={\int }_{\mathsf{\Omega }}[\mathcal{E}(v^m)(s,z^m(s;t,x))-\mathcal{E}(v^{*})(s,z^m(s;t,x))]:\mathcal{E}(\phi )dx.$$

Let us make a change of variables $x=z^m(t;s,y)$ in I (note that the reverse change is $y=z^m(s;t,x)$):

$$I={\int }_{\mathsf{\Omega }}[\mathcal{E}(v^m)(s,y)-\mathcal{E}(v^{*})(s,y)]:\mathcal{E}(\phi )(z^m(t;s,y))dy.$$

Let us write $Z_1^m$ in the following form and continue further decomposition:

$$Z_1^m={\displaystyle \frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}({\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }$$

$$\times {\int }_{\mathsf{\Omega }}[\mathcal{E}(v^m)(s,y)-\mathcal{E}(v^{*})(s,y)]:\mathcal{E}(\phi )(z^m(t;s,y))dyds)$$

$$={\displaystyle \frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}({\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }$$

$$\times {\int }_{\mathsf{\Omega }}[\mathcal{E}(v^m)(s,y)-\mathcal{E}(v^{*})(s,y)]:[\mathcal{E}(\phi )(z^m(t;s,y))\mathcal{E}(\phi )(z^{*}(t;s,y))]dyds)$$

$$+{\displaystyle \frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}({\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }$$

$$\times {\int }_{\mathsf{\Omega }}[\mathcal{E}(v^m)(s,y)-\mathcal{E}(v^{*})(s,y)]:\mathcal{E}(\phi )(z^{*}(t;s,y))dyds)=Z_{11}^m+Z_{12}^m.$$

(a) We obtain, that $Z_{12}^m\to 0$ for $m\to \infty$ due to the weak convergence of $v^m$ to $v^{*}$ in space $L_2(0,T;V^1)$.

(b) Using the Holder and Cauchy–Bunyakovsky inequalities, we obtain

$$|Z_{11}^m{|}^2\le C_{24}({\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }{\parallel v^m(s,·)-v^{*}(s,·)\parallel }_{V^1}$$

$$\times \parallel {\phi }_x(z^m(t;s,·))-{\phi }_x(z^{*}(t;s,·)){{\parallel }_{V^0}ds)}^2$$

$$\le C_{25}\parallel v^m(s,·)-v^{*}{(s,·)\parallel }_{L_2(0,T;V^1)}{\int }_0^T{\parallel {\phi }_x(z^m(t;s,·))-{\phi }_x(z^{*}(t;s,·))\parallel }_{V^0}ds.$$

(49)

Denote the second multiplier in the last inequality by ${\mathsf{\Phi }}_m(s):$

$${\mathsf{\Phi }}_m(s)={\int }_0^T{\parallel {\phi }_x(z^m(t;s,·))-{\phi }_x(z^{*}(t;s,·))\parallel }_{V^0}ds.$$

Let us show the convergence of ${\mathsf{\Phi }}_m(s)\to 0$ for $m\to \infty$ for all $s\in [0,T].$ Note that

$${\mathsf{\Phi }}_m(s)={\int }_0^T{\int }_{\mathsf{\Omega }}{|{\phi }_x(z^m(t;s,y))-{\phi }_x(z^{*}(t;s,y))|}^{2}dyds.$$

Let $\epsilon >0$ is a fairly small number. The continuity of ${\phi }_x$ in $\overline{\mathsf{\Omega }}$ means that there exists $\delta (\epsilon )$ such that if $|x^{″}-x^{\prime }|\le \delta (\epsilon )$, then

$$|{\phi }_x(x^{″})-{\phi }_x(x^{\prime })|\le \epsilon .$$

(50)

Since the sequence $z^m(t;s,y)$ converges to $z^{*}(t;s,y)$ in Lebesgue measure in $(t,y)$, hence for $\delta (\epsilon )$ there exists a number $N=N(\delta (\epsilon ))$ such that $m\ge N$ satisfies the following inequality

$$m(\{(t,y):|z^m(t;s,y)-z^{*}(t;s,y)|\ge \delta (\epsilon )\})\le \epsilon .$$

(51)

Define

$$Q(>\delta (\epsilon ))=\{(t,y)\in Q_T:|z^m(t;s,y)-z^{*}(t;s,y)|>\delta (\epsilon )\};$$

$$Q(\le \delta (\epsilon ))=\{(t,y)\in Q_T:|z^m(t;s,y)-z^{*}(t;s,y)|\le \delta (\epsilon )\}.$$

Then,

$${\mathsf{\Phi }}_m(s)\le C_{26}({\int }_{Q(>\delta (\epsilon ))}{|{\phi }_x(z^m(t;s,y))-{\phi }_x(z^{*}(t;s,y))|}^{2}dyds$$

$$+{\int }_{Q(\le \delta (\epsilon ))}{|{\phi }_x(z^m(t;s,y))-{\phi }_x(z^{*}(t;s,y))|}^{2}dyds)=C_{26}({\mathsf{\Phi }}_m^1(s)+{\mathsf{\Phi }}_m^2(s)).$$

(52)

For ${\mathsf{\Phi }}_m^2(s)$, due to (50) we have $|z^m(t;s,y)-z^{*}(t;s,y)|\le \delta (\epsilon )$. Therefore

$${\mathsf{\Phi }}_m^2(s)\le {\int }_{Q(\le \delta (\epsilon ))}{\epsilon }^{2}dyds=C_{27}{\epsilon }^2.$$

(53)

For ${\mathsf{\Phi }}_m^1(s)$, due to (51), we have $m(Q(>\delta (\epsilon )))\le \epsilon$. Therefore

$${\mathsf{\Phi }}_m^1(s)\le C_{28}\parallel {\phi }_x{\parallel }_{C(\mathsf{\Omega })}{\int }_{Q(>\delta (\epsilon ))}dyds=C_{28}\epsilon {\parallel {\phi }_x\parallel }_{C(\mathsf{\Omega })}.$$

(54)

Thus, from (52)–(54) it follows that for small $\epsilon >0$ and $m\ge N(\delta (\epsilon ))$ the inequality

$${\mathsf{\Phi }}_m(s)\le C_{29}\epsilon .$$

Hence, the convergence of ${\mathsf{\Phi }}_m(s)\to 0$ is obtained for $m\to \infty$ for all $s\in [0,T]$. Consider the right side of inequality (49). Due to the limitation of the first term (because $v^m\in L_2(0,T;V^1)$) and convergence to 0 of the second multiplier at $m\to \infty$, we obtain that $Z_{11}^m\to 0$ at $m\to \infty$.

Thus, we have proved that $Z_1^m\to 0$ as $m\to \infty$.

(2) Now, let us show that $Z_2^m\to 0$ for $m\to \infty$. Consider smooth and finite on $[0,T]\times \mathsf{\Omega }$ function $\tilde{v}(t,x)$ such that $\parallel v^{*}-\tilde{v}{\parallel }_{L_2(0,T;V^1)}\le \epsilon$ for a sufficiently small $\epsilon >0$. Now, let us estimate $Z_2^m$ in three integrals

$$|Z_2^m|\le C_{30}({\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }{\int }_{\mathsf{\Omega }}{\parallel v^{*}(s,z^m(s;t,x))-\tilde{v}(s,z^m(s;t,x))\parallel }_{V^1}ds$$

$$+{\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }{\int }_{\mathsf{\Omega }}{\parallel \tilde{v}(s,z^m(s;t,x))-\tilde{v}(s,z^{*}(s;t,x))\parallel }_{V^1}ds$$

$$+{\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }{\int }_{\mathsf{\Omega }}\parallel \tilde{v}(s,z^{*}(s;t,x))-v^{*}(s,z^{*}(s;t,x)){\parallel }_{V^1}ds)$$

$$=C_{30}(Z_{21}^m+Z_{22}^m+Z_{23}^m).$$

In the norms under the integrals $Z_{21}^m$ and $Z_{23}^m$, it is necessary to make a change of variables:

$$\parallel v^{*}(s,z_m(s;t,x))-\tilde{v}(s,z^m(s;t,x)){\parallel }_{V^1}={\parallel v^{*}(s,y)-\tilde{v}(s,y)\parallel }_{V^1};$$

$$\parallel \tilde{v}(s,z^{*}(s;t,x))-v^{*}(s,z^{*}(s;t,x)){\parallel }_{V^1}={\parallel \tilde{v}(s,y)-v^{*}(s,y)\parallel }_{V^1}.$$

Then, we obtain

$$Z_{21}^m+Z_{23}^m=C_{31}({\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }\parallel v^{*}(s,·)-\tilde{v}(s,·){\parallel }_{V^1}ds)\le C_{27}\epsilon .$$

Let us also estimate $Z_{22}^m$

$$Z_{22}^m\le C_{31}({\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }({\int }_{\mathsf{\Omega }}|{\tilde{v}}_x(s,z^m(s;t,·))-{\tilde{v}}_x(s,z^{*}(s;t,·)){|}^{2}dx{)}^{1/2}ds).$$

By Lemma 8, $z^m(s;t,x)$ converges to $z(s;t,x)$ and the function ${\tilde{v}}_x(t,x)$ is bounded and smooth, so by Lebesgue’s theorem, we obtain the convergence of $Z_2^m\to 0$ for $m\to \infty$. Thus, convergence (48) was proved.

After implementing all the calculations, it was found that the function $v^{*}$ for a smooth function $\phi$ from $V^1$ satisfies the equality:

$$\langle {(v^{*})}^{\prime },\phi \rangle -\underset{\mathsf{\Omega }}{\int }\sum _{i,j=1}^n{v}_i^{*}{v}_j^{*}\frac{\mathsf{\partial }{\phi }_j}{\mathsf{\partial }{x}_i}dx+{\mu }_0{\int }_{\mathsf{\Omega }}\nabla v^{*}:\nabla \phi dx$$

$$+{\displaystyle \frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}\left({\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }\mathcal{E}(v^{*})(s,z^{*}(s;t,x))ds,\mathcal{E}(\phi )\right)=\langle f,\phi \rangle .$$

(55)

The following estimate was obtained, since for the sequence $\left\{v^m\right\}$ there are a priori estimates (44)–(47) and the properties of weak convergence for $v^{*}$ are satisfied:

$$\parallel v^{*}{\parallel }_{L_{\infty }(0,T;V^0)}+\parallel v^{*}{\parallel }_{L_2(0,T;V^1)}+\parallel v^{*}{\parallel }_{L_{4/3}(0,T;V^{-1})}\le C_{28}{(\parallel f\parallel }_{L_2(0,T;V^{-1})}+1).$$

Whence it follows that $v^{*}\in W_1$. Thus, we have proved the passage to the limit for a smooth function $\phi$ from $V^1$.

We prove this passage to the limit for an arbitrary trial function $\phi$ from $V^1$. Let us represent (55) for a smooth $\phi$ in the following form:

$$[G_1,\phi ]-[G_2,\phi ]=0,$$

(56)

where

$$[G_1,\phi ]=\langle v^{\prime },\phi \rangle -{\int }_{\mathsf{\Omega }}\sum _{i,j=1}^n{v}_i{v}_j\frac{\mathsf{\partial }{\phi }_j}{\mathsf{\partial }{x}_i}dx+{\mu }_0{\int }_{\mathsf{\Omega }}\nabla v:\nabla \phi dx$$

$$+{\displaystyle \frac{{\mu }_1}{\mathsf{\Gamma }(1-\alpha )}}({\int }_0^t{e}^{\frac{-(t-s)}{\lambda }}{(t-s)}^{-\alpha }\mathcal{E}(v)(s,z(s;t,x))ds,\mathcal{E}(\phi ));$$

$$[G_2,\phi ]=\langle f,\phi \rangle .$$

Lemma 9.

Let the function φ is smooth. Then,

$$|[G_1,\phi ]|\le C_{32}{\parallel \phi \parallel }_{V^1},|[G_2,\phi ]|\le C_{33}{\parallel \phi \parallel }_{V^1}.$$

(57)

Proof of this Lemma repeats absolutely all the calculations that we carried out obtaining a priori estimates and I part of this Theorem.

Since the set of smooth functions is dense in $V^1$, for $\phi \in V^1$ there exists a sequence of smooth functions ${\phi }^l\in V^1$ such that $|{\phi }^l{-\phi |}_{V^1}\to 0$ for $l\to \infty$. Due to (56), we obtain

$$\begin{array}{c}[G_1,\phi ]-[G_2,\phi ]=[G_1,\phi -{\phi }^l]-[G_2,\phi -{\phi }^l]+[G_1,{\phi }^l]-[G_2,{\phi }^l]\\ =[G_1,\phi -{\phi }^l]-[G_2,\phi -{\phi }^l].\end{array}$$

From the last equality and estimates (57) we obtain

$$|[G_1,\phi ]-[G_2,\phi ]|\le C_{34}|\phi -{\phi }^l|.$$

Taking into account the last inequality and moving to the limit at $l\to \infty$ inequality (55) for $\phi ={\phi }^l$ we obtain equality (55) for an arbitrary $\phi \in V^1$, which completes the proof of the existence of weak solutions to the initial boundary value problem (19)–(22).

5. Conclusions

This paper is devoted to a mathematical model describing the motion of the viscoelastic fluid. The features of the model under consideration are taking into account the memory of the fluid (a term was introduced containing the Caputo fractional derivative), the Voigt type relation described in the first part of the article, and consideration of the fluid along the trajectory of particle motion. All these features describe the fluid flow process more accurately and physically.

The main result of this paper is the solution’s existence for the mathematical model under consideration. The results obtained provided an opportunity for further study of this mathematical model. The authors propose the following future research directions for the model under consideration—(1) application of the method proposed by one of the reviewers, described in [44] for the numerical analysis of the obtained solutions; (2) studying the solvability of this problem with turbulence; (3) the investigation of attractors for this problem and so forth.