Existence Result for a Class of Time-Fractional Nonstationary Incompressible Navier–Stokes–Voigt Equations

Abstract

We are devoted in this work to dealing with a class of time-fractional nonstationary incompressible Navier–Stokes–Voigt equation involving the Caputo fractional derivative. By exploiting the properties of the operators in the equation, we use the Rothe method to show the existence of weak solutions to the equation by verifying all the conditions of the surjectivity theorem for nonlinear weakly continuous operators.

1. Introduction

The Navier–Stokes (NS) equations are fundamental in fluid dynamics, describing the conservation of momentum in viscous, incompressible fluids. However, for fluids with viscoelastic properties, such as polymer solutions or blood, the NS equations alone may not be sufficient to accurately capture the fluid’s behavior. The Navier–Stokes–Voigt (NSV) equations are a modified form of the NS equations that incorporate the effects of fluid viscoelasticity through the introduction of the Voigt model (see, e.g., [1,2,3,4,5]). The NSV equations extend the NS equations by incorporating the Voigt model, which is a simple representation of viscoelasticity. This model combines the elastic and viscous components of the fluid’s stress tensor, allowing for a more accurate description of fluid flows with viscoelastic characteristics. The NSV equations have several applications in fluid dynamics, particularly in the study of complex fluids and their interactions with solid boundaries. They can be used to simulate flows in biological systems, such as blood flow through blood vessels, or to analyze the behavior of polymer solutions in industrial processes.

The fractional Navier–Stokes–Voigt (FNSV) equations are the generalization of the classical NSV equations that incorporate the concept of fractional calculus, and they are also the natural generalization of the fractional Navier–Stokes (FNS) equations. For more details about the theory of FNS equations, we refer to [6,7,8,9,10,11,12].

There are several aspects of the significance of the FNSV equations. First, by incorporating fractional derivatives, the FNSV equations introduce a degree of memory and nonlocality into the fluid dynamics, enabling it to describe these phenomena more precisely. This enhanced description allows researchers to gain deeper insights into the underlying physics of fluid motion. Second, the study of the FNSV equations not only advance fluid mechanics but also contribute to the development of fractional calculus and nonlinear partial differential equations. By delving into the theoretical underpinnings of these equations, researchers can uncover new mathematical properties and relationships that enrich our understanding of fluid dynamics and its applications. Third, the significance of the FNSV equations transcends individual disciplines, fostering interdisciplinary collaboration among mathematicians, physicists, engineers, and other experts. This collaboration enables the integration of diverse perspectives and expertise, leading to more comprehensive and innovative solutions to complex fluid dynamics problems. Fourth, the equations have broad application potentials, including modeling blood flow, modeling drug delivery, investigating the rheological properties of viscoelastic materials, simulating groundwater flow, and analyzing the impact of complex fluids on aircraft design and performance.

Now, we introduce our problem. Let $\Omega$ be a bounded and connected domain in ${\mathbb{R}}^3$ with Lipschitz continuous boundary $\Gamma =\partial \Omega$, and $Q=\Omega \times (0,T)$. The following time-fractional Navier–Stokes–Voigt equation discussed in this work has the following form:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{}_0{}^{C}D_t^{q}u-{\nu }_0\Delta u-{\alpha }^2\Delta \left({}_0{}^{C}D_t^{q}u\right)+(u·\nabla )u+\nabla p=f,\mathrm{in}Q,\hfill \\ \nabla ·u=0,\mathrm{in}Q,\hfill \\ u(x,t)=0,\mathrm{on}\partial \Omega \times (0,T),\hfill \\ u(x,0)=u_0\left(x\right),\mathrm{in}\Omega .\hfill \end{array}\right.\end{array}$$

(1)

Here, ${}_0{}^{C}D_t^{q}u$ is the Caputo time-fractional derivative of velocity field u, $(u·\nabla )vs.=\left({\sum }_{j=1}^3{u}_j\right.{\left.{\displaystyle \frac{\partial v_i}{\partial x_j}}\right)}_{i=1}^3$ is the nonlinear convective term, $p=p(x,t)$ is the unknown pressure occupying $\Omega$, ${\nu }_0>0$ stands for the constant viscosity of the fluid, $\alpha$ is a length-scale parameter characterizing the elasticity of the fluid, and $f=f(x,t)$ is a given external volume force of density occupying $\Omega$.

In this work, we first give its variational formulation, which can be transferred to an operator equation, in Section 2 and then present an existence result on solutions to the equation in Section 3. The main novelties of this paper are shown in the following: First, instead of the FNS equations and integral NSV equations widely investigated in the literature (see [1,2,13]), we deal with FNSV equations, which are more general than the previous equations. Second, we use Rothe’s method to prove the existence result based on the surjective theorem of nonlinear weakly continuous operators (see [14,15,16]), which are more general and easier to verify than the ones in previous contributions. Third, the existence result for our problem improves many existence results in previous works, for instance [13], and the main problem can be generalized to a more complex form such as a mixed FNSV equation. Note that the weak solution of the integral NSV equation is unique. However, in this work, we do not investigate the uniqueness of the weak solution of the FNSV equation, which can be our future work.

2. Variational Formulation

Let X be a Banach space with the norm ${\parallel ·\parallel }_X$, the dual space $X^{\ast }$, and the duality pairing ${\langle ·,·\rangle }_X$. For $T>0$, $p>1$, let $L^p(0,T;X)$ and $C\left(\right[0,T];X)$ be the spaces of all p-th integral and continuous functions from $[0,T]$ into X with usual norms, respectively.

Definition 1

([17,18,19,20]). Let $g:[0,\infty )\to X$ be a function. The fractional integral and the Caputo fractional derivative of order $q\in (0,1)$ for g are defined by, respectively,

$${}_{0}I_t^{q}g\left(t\right)={\displaystyle \frac{1}{\Gamma \left(q\right)}}{\int }_0^t{(t-s)}^{q-1}g\left(s\right)ds,t>0,$$

$${}_0{}^{C}D_t^{q}g\left(t\right)={\displaystyle \frac{1}{\Gamma (1-q)}}{\int }_0^t{(t-s)}^{-q}{g}^{\prime }\left(s\right)ds={}_{0}I_t^{1-q}{g}^{\prime }\left(t\right),t>0,$$

where Γ is the gamma function.

Next, we show the variational formulation of problem (1).

Define the following two inner products of ${\left(L^2(\Omega )\right)}^3$ and ${\left(H_0^1(\Omega )\right)}^3$, respectively,

$${(u,v)}_1:={\int }_{\Omega }\sum _{j=1}^3{u}_j{v}_{j}ds,u=(u_1,u_2,u_3),v=(v_1,v_2,v_3)\in {\left(L^2(\Omega )\right)}^3$$

$${(u,v)}_2:={\int }_{\Omega }\sum _{j=1}^3\nabla u_j·\nabla v_{j}ds,u=(u_1,u_2,u_3),v=(v_1,v_2,v_3)\in {\left(H_0^1(\Omega )\right)}^3,$$

and the corresponding norms ${\parallel v\parallel }_1:=\sqrt{{(v,v)}_1},{\parallel v\parallel }_2:=\sqrt{{(v,v)}_2}$. Consider the following:

$$V_0=\{v\in {\left(C_0^{\infty }(\Omega )\right)}^3:\nabla ·v=0\}.$$

Denote by H the closure of $V_0$ in ${\left(L^2(\Omega )\right)}^3$ and by V the closure of $V_0$ in ${\left(H_0^1(\Omega )\right)}^3$. Clearly, H and V are Hilbert spaces with scalar products ${(·,·)}_1$ and ${(·,·)}_2$. It is easy to see that $V\subset H\subset V^{\ast }$ and the embeddings are dense, continuous, and compact. For $q\in (0,1)$, let $p>\frac{1}{q}$ and

$$\mathcal{V}=L^p(0,T;V),{\mathcal{V}}^{\ast }=L^q(0,T;V^{\ast })and\mathcal{W}=\{v\in \mathcal{V}:{}_0{}^{C}D_t^{q}v\in \mathcal{V}\}.$$

The dual pairing between ${\mathcal{V}}^{\ast }$ and $\mathcal{V}$ is given by

$${\langle v^{\ast },v\rangle }_{\mathcal{V}}:={\int }_0^T{\langle v^{\ast }\left(t\right),v\left(t\right)\rangle }_{V}dt,v\in \mathcal{V},v^{\ast }\in {\mathcal{V}}^{\ast }.$$

It is essential to define the trilinear form $b:V\times V\times V\to \mathbb{R}$ by

$$b(u,v,w):=\sum _{i,j}^3{\int }_{\Omega }{u}_i{\displaystyle \frac{\partial v_j}{\partial x_i}}{w}_{j}dx.$$

Lemma 1

([21,22]). The following properties hold:

(i) 

$$b(y,z,w)=-b(y,w,z),\forall y,z,w\in V;$$

(ii) 

$$b(y,z,z)=0,\forall y,z\in V;$$

(iii) 

$$\begin{array}{c}\hfill \left|b(y,z,w)\right|\le \left\{\begin{array}{c}{C\parallel y\parallel }_1^{\frac{1}{4}}{\parallel y\parallel }_2^{\frac{3}{4}}{\parallel z\parallel }_2{\parallel w\parallel }_1^{\frac{1}{4}}{\parallel w\parallel }_2^{\frac{3}{4}},\hfill \\ {C\parallel y\parallel }_2{\parallel z\parallel }_2{\parallel w\parallel }_2,\hfill \end{array}\forall y,z,w\in V\right.\end{array}$$

and in particular,

$$\begin{array}{c}\hfill \left|b(y,z,y)\right|\le {C\parallel y\parallel }_1^{{\displaystyle \frac{1}{2}}}{\parallel y\parallel }_2^{{\displaystyle \frac{3}{2}}}{\parallel z\parallel }_2,\forall y,z\in V.\end{array}$$

We now give the definition of weak solutions to problem (1).

Definition 2

([1,2]). A function $u\in \mathcal{W}$ is called a weak solution to problem (1) if

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{({}_0{}^{C}D_t^{q}u\left(t\right),v)}_1+{\nu }_0{(u\left(t\right),v)}_2+{\alpha }^2{({}_0{}^{C}D_t^{q}u\left(t\right),v)}_2+b(u\left(t\right),u\left(t\right),v)\hfill \\ ={\langle f\left(t\right),v\rangle }_V,\forall v\in V,a.e.t\in (0,T),\hfill \\ u\left(0\right)=u_0.\hfill \end{array}\right.\end{array}$$

We next give an equivalent formulation for Definition 2, which is in fact an operator equation. To this aim, we introduce two linear bounded operators, $D,E:V\to V^{\ast }$, and a nonlinear operator, $F:V\to V^{\ast }$, defined by, respectively,

$$\begin{array}{c}\hfill {\langle Du,v\rangle }_V:={(u,v)}_1,\forall u,v\in V,\end{array}$$

$$\begin{array}{c}\hfill {\langle Eu,v\rangle }_V:={(u,v)}_2,\forall u,v\in V,\end{array}$$

$$\begin{array}{c}\hfill {\langle Fu,v\rangle }_V:=b(u,u,v),\forall u,v\in V.\end{array}$$

Now, we have an equivalent formulation with Definition 2.

Definition 3.

A function $u\in \mathcal{W}$ is called a weak solution to problem (1) on $(0,T)$ if

$$\begin{array}{c}\hfill \left\{\begin{array}{c}D\left({}_0{}^{C}D_t^{q}u\left(t\right)\right)+{\nu }_{0}Eu\left(t\right)+{\alpha }^{2}E\left({}_0{}^{C}D_t^{q}u\left(t\right)\right)+Fu\left(t\right)=f\left(t\right),in{V}^{\ast },\hfill \\ u\left(0\right)=u_0,inV.\hfill \end{array}\right.\end{array}$$

Lemma 2

([2,14,22]). The operators $D,E$ are linear and bounded, and there exist constants $c_i>0(i=1,2,\cdots ,5)$ such that

$$\begin{array}{c}\hfill {\parallel Dv\parallel }_{V^{\ast }}\le c_1{\parallel v\parallel }_V{,\parallel Ev\parallel }_{V^{\ast }}\le c_2{\parallel v\parallel }_V{,\parallel Fv\parallel }_{V^{\ast }}\le c_3{\parallel v\parallel }_V^2,\forall v\in V,\end{array}$$

$$\begin{array}{c}\hfill {\langle Dv,v\rangle }_V\ge c_4{\parallel v\parallel }_V^2,{\langle Ev,v\rangle }_V\ge c_5{\parallel v\parallel }_V^2,{\langle Fv,v\rangle }_V=0,\forall v\in V.\end{array}$$

3. Existence of Solutions

We first present the main result of this work.

Theorem 1.

If $f\in C([0,T];V^{\ast })$ and $u_0\in V$, then problem (1) has a weak solution.

The proof of Theorem 1 will be divided into several steps by applying the Rothe method (see [23,24]). We shall construct a sequence of lemmas. First, we introduce an auxiliary problem associated with problem (1). Then we establish the existence of a solution to the Rothe problem corresponding to the auxiliary problem by exploiting the surjectivity result for nonlinear weakly continuous operators. Next, the Rothe sequences of piecewise approximations are estimated. Finally, we obtain the existence of solutions to problem (1), which are the limit of the Rothe sequences.

Let $w\left(t\right)={}_0{}^{C}D_t^{q}u\left(t\right)$ for a.e. $t\in (0,T)$. Then

$$u\left(t\right)=u_0+{}_{0}I_t^{q}w\left(t\right)\mathrm{for} \mathrm{all}t\in [0,T].$$

(2)

It is not hard to see that $u\in \mathcal{W}$ defined by (2) is a weak solution to problem (1) if and only if $w\in \mathcal{V}$ solves the following problem:

Problem 1.

Find $w\in \mathcal{V}$ such that

$$(D+{\alpha }^{2}E)w\left(t\right)+({\nu }_{0}E+F)(u_0+{}_{0}I_t^{q}w\left(t\right))=f\left(t\right)for a.e.t\in (0,T).$$

Therefore, in the sequel, it is sufficient to establish the result on the existence of solutions to Problem 1.

Now, we introduce the Rothe method for Problem 1 [23,24,25,26]. Let $N\in \mathbb{N}$ be fixed, $\tau ={\displaystyle \frac{T}{N}}$ denote the time step, and $t_k=k\tau$ for $k=1,2,\dots ,N$. For any $g\in \mathcal{V}$, define its approximation and the fractional integral ${}_{0}I_t^{q}g\left(t\right)$ by, respectively,

$$g_{\tau }^k={\displaystyle \frac{1}{\tau }}{\int }_{t_{k-1}}^{t_k}g\left(s\right)ds\mathrm{for}k=1,\dots ,N,$$

$$\begin{array}{ccc}\hfill {}_0\tilde{I}_{t_k}^{q}g\left(t_k\right)& =& {\displaystyle \frac{1}{\Gamma \left(q\right)}}\sum _{j=1}^k{\int }_{t_{j-1}}^{t_j}{(t_k-s)}^{q-1}g\left(t_j\right)ds\hfill \\ & =& {\displaystyle \frac{{\tau }^q}{\Gamma (q+1)}}\sum _{j=1}^{k}g\left(t_j\right)\left({(k-j+1)}^q-{(k-j)}^q\right)\mathrm{for}k=1,\dots ,N.\hfill \end{array}$$

The following problem is called the Rothe problem for Problem 1.

Problem 2.

Find a sequence ${\left\{w_{\tau }^k\right\}}_{k=1}^N\subset V$ such that

$$(D+{\alpha }^{2}E)w_{\tau }^k+({\nu }_{0}E+F)v_{\tau }^k=f_{\tau }^k$$

(3)

for $k=1,\dots ,N$, where

$$v_{\tau }^k=u_0+{\displaystyle \frac{{\tau }^q}{\Gamma (q+1)}}\sum _{j=1}^k{w}_{\tau }^j\left({(k-j+1)}^q-{(k-j)}^q\right).$$

Lemma 3.

There exists ${\tau }_0>0$ such that for all $\tau \in (0,{\tau }_0)$, Problem 2 has a solution ${\left\{w_{\tau }^k\right\}}_{k=1}^N\subset V$.

Proof. 

Clearly, Problem 2 is equivalent to the following problem:

find $w_{\tau }^k\in V$ such that

$$\Phi \left(w_{\tau }^k\right)=f_{\tau }^k\mathrm{for}k=1,\dots ,N,$$

(4)

where $\Phi :V\to V^{\ast }$ is given, without loss of generality, by

$$\Phi \left(v\right)=(D+{\alpha }^{2}E)v+({\nu }_{0}E+F)\left(u_0+{\displaystyle \frac{{\tau }^q}{\Gamma (q+1)}}\sum _{j=1}^{k-1}{w}_{\tau }^j\left({(k-j+1)}^q-{(k-j)}^q\right)+{\displaystyle \frac{{\tau }^q}{\Gamma (q+1)}}v\right).$$

We will prove that Equation (4) is surjective for small values of $\tau$ by proving that $\Phi$ is weakly continuous and coercive (see [27], Theorem 1.2).

Since $D,E,F$ are linear and bounded, they are weakly continuous. From ([27], Proposition 2.6), we deduce that $\Phi$ is weakly continuous. It remains to be proved that $\Phi$ is coercive. Denote

$$z_{\tau }=u_0+{\displaystyle \frac{{\tau }^q}{\Gamma (q+1)}}\sum _{j=1}^{k-1}{w}_{\tau }^j\left({(k-j+1)}^q-{(k-j)}^q\right),$$

$$K_{\tau }=\parallel u_0{\parallel }_V+{\displaystyle \frac{{\tau }^q}{\Gamma (q+1)}}\sum _{j=1}^{k-1}{\parallel w_{\tau }^j\parallel }_V\left({(k-j+1)}^q-{(k-j)}^q\right),$$

(5)

$$C_{\tau }={\displaystyle \frac{{\tau }^q}{\Gamma (q+1)}},v_{\tau }=z_{\tau }+C_{\tau }v,v\in V.$$

Then we have

$$\begin{array}{ccc}& & {\langle ({\nu }_{0}E+F)v_{\tau },v\rangle }_V={\langle ({\nu }_{0}E+F)v_{\tau },{\displaystyle \frac{1}{C_{\tau }}}(v_{\tau }-z_{\tau })\rangle }_V\hfill \\ & & ={\displaystyle \frac{1}{C_{\tau }}}{\langle ({\nu }_{0}E+F)v_{\tau },v_{\tau }\rangle }_V-{\displaystyle \frac{1}{C_{\tau }}}{\langle ({\nu }_{0}E+F)v_{\tau },z_{\tau }\rangle }_V\hfill \\ & & \ge {\displaystyle \frac{c_5{\nu }_0}{C_{\tau }}}{\parallel v_{\tau }\parallel }_V^2-{\displaystyle \frac{1}{C_{\tau }}}{\langle ({\nu }_{0}E+F)v_{\tau },z_{\tau }\rangle }_V\hfill \\ & & \ge {\displaystyle \frac{c_5{\nu }_0}{C_{\tau }}}\parallel v_{\tau }{\parallel }_V^2-{\displaystyle \frac{1}{C_{\tau }}}\parallel z_{\tau }{\parallel }_V{\parallel ({\nu }_{0}E+F)v_{\tau }\parallel }_{V^{\ast }}\hfill \\ & & \ge {\displaystyle \frac{c_5{\nu }_0}{C_{\tau }}}\parallel v_{\tau }{\parallel }_V^2-{\displaystyle \frac{1}{C_{\tau }}}\parallel z_{\tau }{\parallel }_V(c_2{\nu }_0\parallel v_{\tau }{\parallel }_V+c_3\parallel v_{\tau }{\parallel }_V^2)\hfill \\ & & ={\displaystyle \frac{c_5{\nu }_0}{C_{\tau }}}\parallel v_{\tau }{\parallel }_V^2-{\displaystyle \frac{1}{C_{\tau }}}\parallel z_{\tau }{\parallel }_V(c_2{\nu }_0\parallel z_{\tau }+C_{\tau }{v\parallel }_V+c_3\parallel z_{\tau }+C_{\tau }v{\parallel }_V^2)\hfill \\ & & \ge {\displaystyle \frac{c_5{\nu }_0}{C_{\tau }}}\parallel v_{\tau }{\parallel }_V^2-{\displaystyle \frac{1}{C_{\tau }}}\parallel z_{\tau }{\parallel }_V(c_2{\nu }_0\parallel z_{\tau }{\parallel }_V+c_2{\nu }_0{C}_{\tau }{\parallel v\parallel }_V+c_3{\parallel z_{\tau }\parallel }_V^2\hfill \\ & & +2c_3{C}_{\tau }\parallel z_{\tau }{\parallel }_V{\parallel v\parallel }_V+c_3{C}_{\tau }^2{\parallel v\parallel }_V^2)\hfill \\ & & \ge {\displaystyle \frac{c_5{\nu }_0}{C_{\tau }}}\parallel v_{\tau }{\parallel }_V^2-c_3{C}_{\tau }\parallel z_{\tau }{\parallel }_V{\parallel v\parallel }_V^2-(c_2{\nu }_0+2c_3\parallel z_{\tau }{\parallel }_V{)\parallel v\parallel }_V\hfill \\ & & -{\displaystyle \frac{1}{C_{\tau }}}\parallel z_{\tau }{\parallel }_V(c_2{\nu }_0\parallel z_{\tau }{\parallel }_V+c_3\parallel z_{\tau }{\parallel }_V^2).\hfill \end{array}$$

This gives

$$\begin{array}{ccc}& & {\langle \Phi v,v\rangle }_V\hfill \\ & \ge & (c_4+c_5{\alpha }^2){\parallel v\parallel }_V^2+{\displaystyle \frac{c_5{\nu }_0}{C_{\tau }}}\parallel v_{\tau }{\parallel }_V^2-c_3{C}_{\tau }\parallel z_{\tau }{\parallel }_V{\parallel v\parallel }_V^2-(c_2{\nu }_0+2c_3\parallel z_{\tau }{\parallel }_V{)\parallel v\parallel }_V\hfill \\ & & -{\displaystyle \frac{1}{C_{\tau }}}\parallel z_{\tau }{\parallel }_V(c_2{\nu }_0\parallel z_{\tau }{\parallel }_V+c_3\parallel z_{\tau }{\parallel }_V^2)\hfill \\ & \ge & (c_4+c_5{\alpha }^2){\parallel v\parallel }_V^2-{\displaystyle \frac{c_3{\tau }^q}{\Gamma (q+1)}}{K}_{\tau }{\parallel v\parallel }_V^2-(c_2{\nu }_0+2c_3{K}_{\tau }){\parallel v\parallel }_V\hfill \\ & & -{\displaystyle \frac{\Gamma (q+1)}{{\tau }^q}}{K}_{\tau }(c_2{\nu }_0{K}_{\tau }+c_3{K}_{\tau }^2)\hfill \\ & \ge & (c_4+c_5{\alpha }^2-{\displaystyle \frac{c_3{\tau }^q{K}_{\tau }}{\Gamma (q+1)}}{)\parallel v\parallel }_V^2-(c_2{\nu }_0+2c_3{K}_{\tau }){\parallel v\parallel }_V\hfill \\ & & -{\displaystyle \frac{\Gamma (q+1)}{{\tau }^q}}{K}_{\tau }(c_2{\nu }_0{K}_{\tau }+c_3{K}_{\tau }^2).\hfill \end{array}$$

It is easily seen that we can find sufficiently small ${\tau }_0>0$ such that

$$c_4+c_5{\alpha }^2-{\displaystyle \frac{c_3{\tau }^q{K}_{\tau }}{\Gamma (q+1)}}>0,\forall \tau \in (0,{\tau }_0).$$

Thus, $\Phi$ is coercive for all $\tau \in (0,{\tau }_0)$. The proof is complete.  □

Lemma 4.

There exists ${\tau }_0>0$ such that for all $\tau \in (0,{\tau }_0)$,

$$\begin{array}{ccc}& & \underset{k=1,\dots ,N}{max}{\parallel w_{\tau }^k\parallel }_V\le C,\hfill \end{array}$$

(6)

$$\begin{array}{ccc}& & \underset{k=1,\dots ,N}{max}{\parallel v_{\tau }^k\parallel }_V\le C\hfill \end{array}$$

(7)

for a constant $C>0$ independent of τ.

Proof. 

We first prove estimate (6). For all $1\le k\le N$, we have

$$\begin{array}{c}\hfill (D+{\alpha }^{2}E)w_{\tau }^k+({\nu }_{0}E+F)v_{\tau }^k=f_{\tau }^k.\end{array}$$

(8)

Multiplying the above equation by $w_{\tau }^k$, we can obtain

$$\begin{array}{ccc}& & {\langle f_{\tau }^k,w_{\tau }^k\rangle }_V\hfill \\ & =& {\langle (D+{\alpha }^{2}E)w_{\tau }^k,w_{\tau }^k\rangle }_V+{\langle ({\nu }_{0}E+F)v_{\tau }^k,w_{\tau }^k\rangle }_V\hfill \\ & \ge & (c_4+c_5{\alpha }^2-{\displaystyle \frac{c_3{\tau }^q{K}_{\tau }}{\Gamma (q+1)}})\parallel w_{\tau }^k{\parallel }_V^2+{\displaystyle \frac{{\alpha }^2\Gamma (q+1)}{{\tau }^q}}\parallel v_{\tau }^k{\parallel }_V^2-(c_2{\nu }_0+2c_3{K}_{\tau }){\parallel w_{\tau }^k\parallel }_V\hfill \\ & & -{\displaystyle \frac{\Gamma (q+1)}{{\tau }^q}}{K}_{\tau }(c_2{\nu }_0{K}_{\tau }+c_3{K}_{\tau }^2).\hfill \end{array}$$

Hence, we obtain

$$\begin{array}{ccc}& & (c_4+c_5{\alpha }^2-{\displaystyle \frac{c_3{\tau }^q{K}_{\tau }}{\Gamma (q+1)}}){\parallel w_{\tau }^k\parallel }_V^2\hfill \\ & \le & ({\alpha }_f+c_2{\nu }_0+2c_3{K}_{\tau }){\parallel w_{\tau }^k\parallel }_V+{\displaystyle \frac{\Gamma (q+1)}{{\tau }^q}}\left(K_{\tau }(c_2{\nu }_0{K}_{\tau }+c_3{K}_{\tau }^2)-{\alpha }^2{\parallel v_{\tau }^k\parallel }_V^2\right).\hfill \end{array}$$

We can choose sufficiently small ${\tau }_0>0$ such that

$$c_4+c_5{\alpha }^2-{\displaystyle \frac{c_3{\tau }^q{K}_{\tau }}{\Gamma (q+1)}}>{\displaystyle \frac{c_4+c_5{\alpha }^2}{2}},\forall \tau \in (0,{\tau }_0).$$

With the above $\tau$, the following two cases are discussed:

(i)

$c_2{\nu }_0\parallel u_0{\parallel }_V^2+c_3\parallel u_0{\parallel }_V^3\ge {\alpha }^2{\parallel v_{\tau }^k\parallel }_V^2$;

(ii)

$c_2{\nu }_0\parallel u_0{\parallel }_V^2+c_3\parallel u_0{\parallel }_V^3<{\alpha }^2{\parallel v_{\tau }^k\parallel }_V^2$.

Let (i) hold. We have

$$\begin{array}{ccc}& & (c_4+c_5{\alpha }^2){\parallel w_{\tau }^k\parallel }_V^2\hfill \\ & \le & \langle (D+{\alpha }^{2}E)w_{\tau }^k,w_{\tau }^k\rangle \hfill \\ & =& \langle f_{\tau }^k,w_{\tau }^k\rangle -\langle ({\nu }_{0}E+F)v_{\tau }^k,w_{\tau }^k\rangle \hfill \\ & \le & \parallel f_{\tau }^k{\parallel }_{V^{\ast }}\parallel w_{\tau }^k{\parallel }_V+\parallel ({\nu }_{0}E+F)v_{\tau }^k{\parallel }_{V^{\ast }}{\parallel w_{\tau }^k\parallel }_V\hfill \\ & \le & {\alpha }_f\parallel w_{\tau }^k{\parallel }_V+(c_2{\nu }_0\parallel v_{\tau }^k{\parallel }_V+c_3\parallel v_{\tau }^k{\parallel }_V^2)\parallel w_{\tau }^k{\parallel }_V,\hfill \end{array}$$

and hence

$$\begin{array}{c}\hfill (c_4+c_5{\alpha }^2)\parallel w_{\tau }^k{\parallel }_V\le {\alpha }_f+c_2{\nu }_0\parallel v_{\tau }^k{\parallel }_V+c_3{\parallel v_{\tau }^k\parallel }_V^2,\end{array}$$

which implies that $\left\{w_{\tau }^k\right\}$ is bounded since $\left\{v_{\tau }^k\right\}$ is bounded.

Let (ii) hold. We have

$$\begin{array}{ccc}& & \parallel w_{\tau }^k{\parallel }_V^2\hfill \\ & \le & {\displaystyle \frac{2}{c_4+c_5{\alpha }^2}}({\alpha }_f+c_2{\nu }_0+2c_3{K}_{\tau }){\parallel w_{\tau }^k\parallel }_V\hfill \\ & & +{\displaystyle \frac{2\Gamma (q+1)}{(c_4+c_5{\alpha }^2){\tau }^q}}\left(K_{\tau }(c_2{\nu }_0{K}_{\tau }+c_3{K}_{\tau }^2)-{\alpha }^2{\parallel v_{\tau }^k\parallel }_V^2\right)\hfill \\ & =& {\displaystyle \frac{2}{c_4+c_5{\alpha }^2}}({\alpha }_f+c_2{\nu }_0+2c_3{K}_{\tau }){\parallel w_{\tau }^k\parallel }_V\hfill \\ & & +{\displaystyle \frac{2\Gamma (q+1)}{(c_4+c_5{\alpha }^2){\tau }^q}}(\left(\parallel u_0{\parallel }_V+{\displaystyle \frac{{\tau }^q}{\Gamma (q+1)}}\sum _{j=1}^{k-1}{\parallel w_{\tau }^j\parallel }_V\left({(k-j+1)}^q-{(k-j)}^q\right)\right)\hfill \\ & & \times (c_2{\nu }_0\left(\parallel u_0{\parallel }_V+{\displaystyle \frac{{\tau }^q}{\Gamma (q+1)}}\sum _{j=1}^{k-1}{\parallel w_{\tau }^j\parallel }_V\left({(k-j+1)}^q-{(k-j)}^q\right)\right)\hfill \\ & & +c_3{\left(\parallel u_0{\parallel }_V+{\displaystyle \frac{{\tau }^q}{\Gamma (q+1)}}\sum _{j=1}^{k-1}{\parallel w_{\tau }^j\parallel }_V\left({(k-j+1)}^q-{(k-j)}^q\right)\right)}^2)-{\alpha }^2{\parallel v_{\tau }^k\parallel }_V^2)\hfill \\ & =& {\displaystyle \frac{2}{c_4+c_5{\alpha }^2}}({\alpha }_f+c_2{\nu }_0+2c_3{K}_{\tau }){\parallel w_{\tau }^k\parallel }_V\hfill \\ & & +{\displaystyle \frac{2\Gamma (q+1)}{(c_4+c_5{\alpha }^2){\tau }^q}}(c_2{\nu }_0\parallel u_0{\parallel }_V^2+c_3\parallel u_0{\parallel }_V^3-{\alpha }^2{\parallel v_{\tau }^k\parallel }_V^2\hfill \\ & & +{\displaystyle \frac{{\tau }^q}{\Gamma (q+1)}}(m_1+{\displaystyle \frac{{\tau }^q}{\Gamma (q+1)}}(c_2{\nu }_0+3c_3\parallel u_0{\parallel }_V)\sum _{j=1}^{k-1}{\parallel w_{\tau }^j\left({(k-j+1)}^q-{(k-j)}^q\right)\parallel }_V\hfill \\ & & +c_3{\displaystyle \frac{{\tau }^{2q}}{{\Gamma }^2(q+1)}}{\left(\sum _{j=1}^{k-1}{\parallel w_{\tau }^j\parallel }_V\left({(k-j+1)}^q-{(k-j)}^q\right)\right)}^2)\hfill \\ & & \times \sum _{j=1}^{k-1}{\parallel w_{\tau }^j\parallel }_V\left({(k-j+1)}^q-{(k-j)}^q\right))\hfill \\ & \le & {\displaystyle \frac{2}{c_4+c_5{\alpha }^2}}({\alpha }_f+c_2{\nu }_0+2c_3{K}_{\tau })\parallel w_{\tau }^k{\parallel }_V+{\displaystyle \frac{2}{c_4+c_5{\alpha }^2}}(m_1+{\displaystyle \frac{{\tau }^q}{\Gamma (q+1)}}(c_2{\nu }_0+3c_3\parallel u_0{\parallel }_V)\hfill \\ & & \times \sum _{j=1}^{k-1}{\parallel w_{\tau }^j\parallel }_V\left({(k-j+1)}^q-{(k-j)}^q\right)+\hfill \\ & & c_3{\displaystyle \frac{{\tau }^{2q}}{{\Gamma }^2(q+1)}}{\left(\sum _{j=1}^{k-1}{\parallel w_{\tau }^j\parallel }_V\left({(k-j+1)}^q-{(k-j)}^q\right)\right)}^2)\hfill \\ & & \times \sum _{j=1}^{k-1}{\parallel w_{\tau }^j\parallel }_V\left({(k-j+1)}^q-{(k-j)}^q\right),\hfill \end{array}$$

where $m_1={\overline{b}}_0+2{\overline{b}}_1\parallel u_0{\parallel }_V+3{\overline{b}}_2{\parallel u_0\parallel }_V^2$. It follows that

$$\begin{array}{ccc}& & \parallel w_{\tau }^k{\parallel }_V\hfill \\ & \le & {\displaystyle \frac{2}{c_4+c_5{\alpha }^2}}({\alpha }_f+c_2{\nu }_0+2c_3{K}_{\tau })+\left({\displaystyle \frac{2}{c_4+c_5{\alpha }^2}}\right(m_1+{\displaystyle \frac{{\tau }^q}{\Gamma (q+1)}}(c_2{\nu }_0+3c_3\parallel u_0{\parallel }_V)\hfill \\ & & \times \sum _{j=1}^{k-1}{\parallel w_{\tau }^j\parallel }_V\left({(k-j+1)}^q-{(k-j)}^q\right)+c_3{\displaystyle \frac{{\tau }^{2q}}{{\Gamma }^2(q+1)}}\hfill \\ & & \times {\left(\sum _{j=1}^{k-1}{\parallel w_{\tau }^j\parallel }_V\left({(k-j+1)}^q-{(k-j)}^q\right)\right)}^2)\hfill \\ & & \times \sum _{j=1}^{k-1}{\parallel w_{\tau }^j\parallel }_V\left({(k-j+1)}^q-{(k-j)}^q\right){)}^{{\displaystyle \frac{1}{2}}}\hfill \\ & \le & {\displaystyle \frac{2}{c_4+c_5{\alpha }^2}}({\alpha }_f+c_2{\nu }_0+2c_3{K}_{\tau })+{\displaystyle \frac{1}{c_4+c_5{\alpha }^2}}\sum _{j=1}^{k-1}{\parallel w_{\tau }^j\parallel }_V\left({(k-j+1)}^q-{(k-j)}^q\right)\hfill \\ & & +{\displaystyle \frac{1}{2}}(m_1+{\displaystyle \frac{{\tau }^q}{\Gamma (q+1)}}(c_2{\nu }_0+3c_3\parallel u_0{\parallel }_V)\sum _{j=1}^{k-1}{\parallel w_{\tau }^j\parallel }_V\left({(k-j+1)}^q-{(k-j)}^q\right)\hfill \\ & & +c_3{\displaystyle \frac{{\tau }^{2q}}{{\Gamma }^2(q+1)}}{\left(\sum _{j=1}^{k-1}{\parallel w_{\tau }^j\parallel }_V\left({(k-j+1)}^q-{(k-j)}^q\right)\right)}^2)\hfill \\ & \le & m_2+m_3\sum _{j=1}^{k-1}{\parallel w_{\tau }^j\parallel }_V\left({(k-j+1)}^q-{(k-j)}^q\right)\hfill \\ & & +m_4{\left(\sum _{j=1}^{k-1}{\parallel w_{\tau }^j\parallel }_V\left({(k-j+1)}^q-{(k-j)}^q\right)\right)}^2,\hfill \end{array}$$

where

$$m_2={\displaystyle \frac{2}{c_4+c_5{\alpha }^2}}({\alpha }_f+c_2{\nu }_0+2c_3\parallel u_0{\parallel }_V)+{\displaystyle \frac{1}{2}}{m}_1,$$

$$m_3={\displaystyle \frac{4c_3{T}^q}{(c_4+c_5{\alpha }^2)\Gamma (q+1)}}+{\displaystyle \frac{T^q}{2\Gamma (q+1)}}(c_2{\nu }_0+3c_3\parallel u_0{\parallel }_V),m_4={\displaystyle \frac{c_3{T}^{2q}}{{\Gamma }^2(q+1)}}.$$

Moreover, for $w_{\tau }^1$, we have

$$\begin{array}{ccc}& & {\alpha }_f{\parallel w_{\tau }^1\parallel }_V\hfill \\ & \ge & {\langle f_{\tau }^1,w_{\tau }^1\rangle }_V\hfill \\ & =& {\langle (D+{\alpha }^{2}E)w_{\tau }^1,w_{\tau }^1\rangle }_V+{\langle ({\nu }_{0}E+F)v_{\tau }^1,w_{\tau }^1\rangle }_V\hfill \\ & \ge & c_4+c_5{\alpha }^2{\parallel w_{\tau }^1\parallel }_V^2+{\displaystyle \frac{\Gamma (q+1)}{{\tau }^q}}{〈A{v}_{\tau }^1,v_{\tau }^1〉}_V-{\displaystyle \frac{\Gamma (q+1)}{{\tau }^q}}{〈({\nu }_{0}E+F)v_{\tau }^1,u_0〉}_V\hfill \\ & \ge & (c_4+c_5{\alpha }^2)\parallel w_{\tau }^1{\parallel }_V^2+{\displaystyle \frac{{\alpha }^2\Gamma (q+1)}{{\tau }^q}}{\parallel v_{\tau }^1\parallel }_V^2\hfill \\ & & -{\displaystyle \frac{\Gamma (q+1)}{{\tau }^q}}{\parallel u_0\parallel }_V\left(c_2{\nu }_0\parallel u_0+{\displaystyle \frac{{\tau }^q}{\Gamma (q+1)}}{w}_{\tau }^1{\parallel }_V+c_3{\parallel u_0+{\displaystyle \frac{{\tau }^q}{\Gamma (q+1)}}{w}_{\tau }^1\parallel }_V^2\right)\hfill \\ & \ge & (c_4+c_5{\alpha }^2-{\displaystyle \frac{c_3{\tau }^q{\parallel u_0\parallel }_V}{\Gamma (q+1)}})\parallel w_{\tau }^1{\parallel }_V^2-(c_2{\nu }_0\parallel u_0{\parallel }_V+2c_3\parallel u_0{\parallel }_V^2)\parallel w_{\tau }^1{\parallel }_V\hfill \\ & & -{\displaystyle \frac{\Gamma (q+1)}{{\tau }^q}}\parallel u_0{\parallel }_V(c_2{\nu }_0\parallel u_0{\parallel }_V+c_3\parallel u_0{\parallel }_V^2)+{\displaystyle \frac{{\alpha }^2\Gamma (q+1)}{{\tau }^q}}{\parallel v_{\tau }^1\parallel }_V^2,\hfill \end{array}$$

and hence

$$\begin{array}{ccc}& & (c_4+c_5{\alpha }^2-{\displaystyle \frac{c_3{\tau }^q{\parallel u_0\parallel }_V}{\Gamma (q+1)}}){\parallel w_{\tau }^1\parallel }_V^2\hfill \\ & \le & ({\alpha }_f+c_2{\nu }_0\parallel u_0{\parallel }_V+2c_3\parallel u_0{\parallel }_V^2)\parallel w_{\tau }^1{\parallel }_V\hfill \\ & & +{\displaystyle \frac{\Gamma (q+1)}{{\tau }^q}}(c_2{\nu }_0\parallel u_0{\parallel }_V^2+c_3\parallel u_0{\parallel }_V^3-{\alpha }^2\parallel v_{\tau }^1{\parallel }_V^2)\hfill \\ & \le & ({\alpha }_f+c_2{\nu }_0\parallel u_0{\parallel }_V+2c_3\parallel u_0{\parallel }_V^2)\parallel w_{\tau }^1{\parallel }_V.\hfill \end{array}$$

Obviously, $c_4+c_5{\alpha }^2-{\displaystyle \frac{c_3{\tau }^q{\parallel u_0\parallel }_V}{\Gamma (q+1)}}>{\displaystyle \frac{c_4+c_5{\alpha }^2}{2}}$ for all $\tau \in (0,{\tau }_0)$. Then

$$\begin{array}{c}\hfill \parallel w_{\tau }^1{\parallel }_V\le {\displaystyle \frac{2}{c_4+c_5{\alpha }^2}}({\alpha }_f+c_2{\nu }_0\parallel u_0{\parallel }_V+2c_3\parallel u_0{\parallel }_V^2),\end{array}$$

which gives the boundedness of $w_{\tau }^1$. Therefore, we can obtain (6) by induction.

Furthermore, estimate (7) follows from (6). The proof of Lemma 4 is complete.  □

In the final step of the Rothe method, define ${\overline{w}}_{\tau }$, ${\overline{v}}_{\tau }\in \mathcal{V}$, ${\overline{f}}_{\tau }\in {\mathcal{V}}^{\ast }$ by, respectively,

$${\overline{w}}_{\tau }\left(t\right)=w_{\tau }^k,{\overline{v}}_{\tau }\left(t\right)=v_{\tau }^k,{\overline{f}}_{\tau }\left(t\right)=f_{\tau }^k,$$

for a.e. $t\in (t_{k-1},t_k]$ with $k=1,\dots ,N$.

Lemma 5.

There exists $w\in \mathcal{V}$ solving Problem 1, satisfying, for a subsequence,

$${\overline{w}}_{\tau }\to wweakly in\mathcal{V}.$$

Proof. 

Consider the Nemytski operators $\mathcal{D},\mathcal{E},:\mathcal{V}\to {\mathcal{V}}^{\ast },\mathcal{F}:\mathcal{W}\to {\mathcal{V}}^{\ast }$ corresponding to $D,E,F$, respectively, i.e.,

$$\left(\mathcal{D}v\right)\left(t\right)=Dv\left(t\right),\left(\mathcal{E}v\right)\left(t\right)=Ev\left(t\right),\left(\mathcal{F}\overline{v}\right)\left(t\right)=F\overline{v}\left(t\right)\mathrm{for}vs.\in \mathcal{V},\overline{v}\in \mathcal{W}a.e.t\in (0,T).$$

Then Equation (8) is transferred to be

$$({\nu }_0\mathcal{D}+\mathcal{E}){\overline{w}}_{\tau }+({\nu }_0\mathcal{E}+\mathcal{F}){\overline{v}}_{\tau }={\overline{f}}_{\tau }.$$

(9)

It follows from (6) that $\left\{{\overline{w}}_{\tau }\right\}$ is bounded in $\mathcal{V}$. Then we obtain, for a subsequence,

$${\overline{w}}_{\tau }\to w\mathrm{weakly} \mathrm{in}\mathcal{V}.$$

(10)

We are left with the task of determining that w solves Problem 1. Since the operator ${}_{0}I_t^q:\mathcal{V}\to C([0,T];V)$ is linear and continuous, it is also weakly continuous, and hence

$${}_{0}I_t^q{\overline{w}}_{\tau }\to {}_{0}I_t^{q}w\mathrm{weakly} \mathrm{in}C([0,T];V),\mathrm{as}\tau \to 0.$$

Then

$${}_{0}I_t^q{\overline{w}}_{\tau }\left(t\right)\to {}_{0}I_t^{q}w\left(t\right)\mathrm{weakly} \mathrm{in}V,\mathrm{as}\tau \to 0,\forall t\in [0,T].$$

(11)

Moreover, there exists a constant $C>0$ such that $\parallel {\overline{w}}_{\tau }\left(s\right)\parallel \le C$ for a.e. $s\in (0,T)$. For $t\in (t_{k-1},t_k]$, we obtain that

$$\begin{array}{ccc}& & \parallel {\overline{v}}_{\tau }\left(t\right)-u_0-{}_{0}I_t^q{\overline{w}}_{\tau }{\left(t\right)\parallel }_V\hfill \\ & =& {\displaystyle \frac{1}{\Gamma \left(q\right)}}{\parallel {\int }_0^{t_k}{(t_k-s)}^{q-1}{\overline{w}}_{\tau }\left(s\right)ds-{\int }_0^t{(t-s)}^{q-1}{\overline{w}}_{\tau }\left(s\right)ds\parallel }_V\hfill \\ & \le & {\displaystyle \frac{1}{\Gamma \left(q\right)}}{\parallel {\int }_0^t\left({(t_k-s)}^{q-1}-{(t-s)}^{q-1}\right){\overline{w}}_{\tau }\left(s\right)ds\parallel }_V\hfill \\ & & +{\displaystyle \frac{1}{\Gamma \left(q\right)}}{\parallel {\int }_t^{t_k}{(t_k-s)}^{q-1}{\overline{w}}_{\tau }\left(s\right)ds\parallel }_V\hfill \\ & \le & {\displaystyle \frac{C}{\Gamma \left(q\right)}}{\int }_0^t|{(t_k-s)}^{q-1}-{(t-s)}^{q-1}|ds+{\displaystyle \frac{C}{\Gamma \left(q\right)}}{\int }_t^{t_k}{(t_k-s)}^{q-1}ds\hfill \end{array}$$

(12)

$$\begin{array}{ccc}& \le & {\displaystyle \frac{C}{\Gamma (q+1)}}|t_k^q-{(t_k-t)}^q-t^q|+{\displaystyle \frac{C}{\Gamma \left(q\right)}}{(t_k-t)}^q.\hfill \end{array}$$

(13)

We can obtain that ${\overline{v}}_{\tau }\left(t\right)-u_0-{}_{0}I_t^q{\overline{w}}_{\tau }\left(t\right)\to 0$, as $\tau \to 0$ for all $t\in [0,T]$. Then

$${\overline{v}}_{\tau }\left(t\right)\to u_0+{}_{0}I_t^{q}w\left(t\right)\mathrm{weakly} \mathrm{in}V,\mathrm{as}\tau \to 0$$

(14)

for all $t\in [0,T]$.

Furthermore, it is clear that $\mathcal{D},\mathcal{E}$ are weakly continuous. Hence

$$({\nu }_0\mathcal{D}+\mathcal{E})w_{\tau }\to ({\nu }_0\mathcal{D}+\mathcal{E})w\mathrm{weakly} \mathrm{in}{\mathcal{V}}^{\ast },\mathrm{as}\tau \to 0.$$

(15)

It follows from [2,27] that $\mathcal{F}$ is weakly continuous. Then for any $v\in \mathcal{V}$, we infer that

$$\begin{array}{ccc}& & \underset{\tau \to 0}{lim}{\langle ({\nu }_0\mathcal{E}+\mathcal{F}){\overline{v}}_{\tau },v\rangle }_{\mathcal{V}}\hfill \\ & & =\underset{\tau \to 0}{lim}{\int }_0^T{\langle ({\nu }_{0}E+F){\overline{v}}_{\tau }\left(t\right),v\left(t\right)\rangle }_{V}dt\hfill \\ & & ={\int }_0^T\underset{\tau \to 0}{lim}{\langle ({\nu }_{0}E+F){\overline{v}}_{\tau }\left(t\right),v\left(t\right)\rangle }_{V}dt\hfill \\ & & ={\int }_0^T{\langle ({\nu }_{0}E+F)(u_0+{}_{0}I_t^{q}w\left(t\right)),v\left(t\right)\rangle }_{V}dt\hfill \\ & & ={\langle ({\nu }_0\mathcal{E}+\mathcal{F})(u_0+{}_{0}I_t^{q}w),v\rangle }_{\mathcal{V}}.\hfill \end{array}$$

Therefore,

$$({\nu }_0\mathcal{E}+\mathcal{F}){\overline{v}}_{\tau }\to ({\nu }_0\mathcal{E}+\mathcal{F})(u_0+{}_{0}I_t^{q}w)\mathrm{weakly} \mathrm{in}{\mathcal{V}}^{\ast },\mathrm{as}\tau \to 0.$$

(16)

Together with ${\overline{f}}_{\tau }\to f$ in ${\mathcal{V}}^{\ast }$, as $\tau \to 0$ (see [28], Lemma 3.3), from (10), (16), and (15), taking the limit in (9), we finally obtain

$$({\nu }_0\mathcal{D}+\mathcal{E})w+({\nu }_0\mathcal{E}+\mathcal{F})(u_0+{}_{0}I_t^{q}w)=f,$$

which proves that $w\in \mathcal{V}$ is a solution to Problem 1. The proof of Lemma 5 is complete.  □

In the end, combining the above lemmas, we obtain Theorem 1.

By apply the method above, we can also solve the following generalized time-fractional Navier–Stokes–Voigt equation:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\rho }_1{}_0{}^{C}D_t^{q}u-{\rho }_2\Delta u-{\rho }_3\Delta \left({}_0{}^{C}D_t^{q}u\right)+(u·\nabla )u+\nabla p=f,\mathrm{in}Q,\hfill \\ \nabla ·u=0,\mathrm{in}Q,\hfill \\ u(x,t)=0,\mathrm{on}\partial \Omega \times (0,T),\hfill \\ u(x,0)=u_0\left(x\right),\mathrm{in}\Omega .\hfill \end{array}\right.\end{array}$$

(17)

Here, ${\rho }_i\in L(\Omega ;{\mathbb{R}}_{+})(i=1,2,3)$.