A constrained optimal control framework for vehicle platoons with delayed communication

A M Ishtiaque Mahbub; Behdad Chalaki; Andreas A. Malikopoulos; A M Ishtiaque Mahbub; Behdad Chalaki; Andreas A. Malikopoulos

doi:10.3934/nhm.2023043

Networks and Heterogeneous Media

2023, Volume 18, Issue 3: 982-1005. doi: 10.3934/nhm.2023043

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Research article Special Issues

A constrained optimal control framework for vehicle platoons with delayed communication

University of Delaware, 130 Academy Street, Newark, DE-19716, USA

Received: 22 November 2021 Revised: 02 May 2022 Accepted: 11 January 2023 Published: 21 March 2023
58F15, 58F17, 53C35

Vehicle platooning using connected and automated vehicles (CAVs) has attracted considerable attention. In this paper, we address the problem of optimal coordination of CAV platoons at a highway on-ramp merging scenario. We present a single-level constrained optimal control framework that optimizes the fuel economy and travel time of the platoons while satisfying the state, control, and safety constraints. We also explore the effect of delayed communication among the CAV platoons and propose a robust coordination framework to enforce lateral and rear-end collision avoidance constraints in the presence of bounded delays. We provide a closed-form analytical solution to the optimal control problem with safety guarantees that can be implemented in real time. Finally, we validate the effectiveness of the proposed control framework using a high-fidelity commercial simulation environment.

Keywords:

Citation: A M Ishtiaque Mahbub, Behdad Chalaki, Andreas A. Malikopoulos. A constrained optimal control framework for vehicle platoons with delayed communication[J]. Networks and Heterogeneous Media, 2023, 18(3): 982-1005. doi: 10.3934/nhm.2023043

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Abstract

1. Introduction

In this paper, we study the coupled chemotaxis-fluid models with the initial-bounary conditions

$\begin{align} \left\{\begin{aligned} &n_{t}+u \cdot \nabla n = \Delta n-\nabla \cdot(n\nabla c)+\gamma n-\mu n^{2}, &&\text {in } Q \equiv(0, T) \times \Omega , \\ &c_{t}+u \cdot \nabla c = \Delta c-c+n+f, &&\text{in } Q, \\ &u_{t}+u \cdot \nabla u = \Delta u-\nabla \pi+n \nabla \varphi, &&\text{in } Q, \\ &\nabla \cdot u = 0, &&\text{in } Q, \\ &\frac{\partial n}{\partial \nu} = \frac{\partial c}{\partial \nu} = 0,\quad u = 0, &&\text{on } (0, T) \times \partial\Omega, \\ &n(x,0) = n_0(x), c(x,0) = c_0(x), u(x,0) = u_0(x), &&\text{in }\Omega, \end{aligned} \right. \end{align}$

(1.1)

where $\Omega \subset \mathbb{R}^2$ is a bounded domain with smooth boundary $\partial \Omega$ . $\nu$ is the outward normal vector to $\partial \Omega$ , and $\gamma$ , $\mu$ are positive constants. $n$ , $c$ denote the bacterial density, the oxygen concentration, respectively. $u$ , $\pi$ are the fluid velocity and the associated pressure. Here, the function $f$ denotes a control that acts on chemical concentration, which lies in a closed convex set $\mathcal{U}$ . We observe that in the subdomains where $f \geq 0$ we inject oxygen, and conversely where $f \leq 0$ we extract oxygen.

In order to understand the development of system (1.1), let us mention some previous contributions in this direction. Jin [11] dealed with the time periodic problem of (1.1) in spatial dimension $n = 2,3$ . Jin [12] also obtained the existence of large time periodic solution in $\Omega \subset R^3$ without the term $u\cdot\nabla u$ .

Espejo and Suzuki [6] discussed the chemotaxis-fluid model

$\begin{align} &n_{t}+u \cdot \nabla n = \Delta n-\nabla \cdot(n \nabla c)+n(\gamma-\mu n), \end{align}$

(1.2)

$\begin{align} &c_{t}+u \cdot \nabla c = \Delta c-c+n, \end{align}$

(1.3)

$\begin{align} &u_{t} = \Delta u-\nabla \pi+n \nabla \varphi, \end{align}$

(1.4)

$\begin{align} &\nabla \cdot u = 0 , \end{align}$

(1.5)

$\begin{align} &\frac{\partial n}{\partial \nu} = \frac{\partial c}{\partial \nu} = 0,\quad u = 0. \end{align}$

(1.6)

They proved the global existence of weak solution. Tao and Winkler [17] proved the existence of global classical solution and the uniform boundedness. Tao and Winkler [18] also obtained the global classical solution and uniform boundedness under the condition of $\mu > 23$ .

The optimal control problems governed by the coupled partial differential equations is important. Colli et al. [4] studied the distributed control problem for a phase-field system of conserved type with a possibly singular potential. Liu and Zhang [14] considered the optimal control of a new mechanochemical model with state constraint. Chen et al. [3] studied the distributed optimal control problem for the coupled Allen-Cahn/Cahn-Hilliard equations. Recently, Guillén-González et al. [9] studied a bilinear optimal control problem for the chemo-repulsion model with the linear production term. The existence, uniqueness and regularity of strong solutions of this model are deduced. They also derived the first-order optimality conditions by using a Lagrange multipliers theorem. Frigeri et al. [8] studied an optimal control problem for two-dimensional nonlocal Cahn-Hilliard-Navier-Stokes systems with degenerate mobility and singular potential. Some other results can be found in [2,5,13,15,19].

In this paper, we discuss the optimal control problem for (1.1). We adjust the external source $f$ , so that the bacterial density $n$ , oxygen concentration $c$ and fluid velocity $u$ are as close as possible to a desired state $n_d$ , $c_d$ and $u_d$ , and at the final moment $T$ is as close as possible to a desired state $n_{\Omega}$ , $c_{\Omega}$ and $u_{\Omega}$ . The main difficulties for treating the problem are caused by the nonlinearity of $u\cdot \nabla u$ . Our method is based on fixed point method and Simon's compactness results. We overcome the above difficulties and derive first-order optimality conditions by using a Lagrange multipliers theorem.

2. Basic estimates of linearized problem

In this section, we will construct the existence and some priori estimates of the linearized problem for the chemotaxis-Navier-Stokes system in a bounded domain $\Omega \subset \mathbb{R}^2$ . The proofs in this section will be established for a detailed framework.

In the following lemmas we will state the Gagliardo-Nirenberg interpolation inequality [7].

Lemma 2.1. Let $l$ and $k$ be two integers satisfying $0 \leqslant l<k$ . Suppose that $1 \leqslant q$ , $r \leqslant \infty$ , $p>0$ and $\frac{l}{k} \leqslant a \leqslant 1$ such that

$\begin{align} \frac{1}{p}-\frac{l}{N} = a\left(\frac{1}{q}-\frac{k}{N}\right)+(1-a) \frac{1}{r}. \end{align}$

(2.1)

Then, for any $u \in W^{k, q}(\Omega) \cap L^{r}(\Omega)$ , there exist two positive constants $C_1$ and $C_2$ depending only on $\Omega$ , $q$ , $k$ , $r$ and $N$ such that the following inequality holds

$\begin{align*} \left\|D^{l} u\right\|_{L^{p}} \leqslant c_{1}\left\|D^{k} u\right\|_{L^{q}}^{a}\|u\|_{L^{r}}^{1-a}+c_{2}\|u\|_{L^{r}} \end{align*}$

with the following exception: If $1<q<\infty$ and $k-l-\frac{N}{q}$ is a non-negative integer, the (2.1) holds only for $a$ satisfying $\frac{l}{k} \leq a<1$ .

The following log-interpolation inequality has been proved by [1].

Lemma 2.2. Let $\Omega \subset \mathbb{R}^{2}$ be a bounded domain with smooth boundary. Then for all non-negative $u\in H^1(\Omega)$ , there holds

$\begin{align*} \|u\|_{L^{3}(\Omega)}^{3} \leq \delta\|u\|_{H^{1}(\Omega)}^{2}\|(u+1) \log (u+1)\|_{L^{1}(\Omega)}+p\left(\delta^{-1}\right)\|u\|_{L^{1}(\Omega)}, \end{align*}$

where $\delta$ is any positive number, and $p(\cdot)$ is an increasing function.

We first consider the existence of solutions to the linear problem of system (1.1). Assume functions $u_0\in H^1(\Omega)$ , $\hat{u} \in L^{4}\left(0,T; L^{4}(\Omega)\right), \hat{n} \in L^{2}\left(0,T; L^{2}(\Omega)\right)$ , and consider

$\begin{align} \left\{\begin{aligned} &u_{t}-\Delta u+\hat{u} \cdot \nabla u = -\nabla \pi+\hat{n} \nabla \varphi, &&\text{in } Q, \\ &\nabla \cdot u = 0, &&\text{in } Q, \\ &u = 0, &&\text{on } \partial\Omega, \\ &u(x,0) = u_0(x), &&\text{in }\Omega. \end{aligned} \right. \end{align}$

(2.2)

By using fixed point method, the existence of solutions can be easily obtained. Therefore, we ignore the process of proof and just give the regularity estimate.

Lemma 2.3. Let $u_0\in H^1(\Omega)$ , $\hat{u} \in L^{4}\left(0,T; L^{4}(\Omega)\right)$ , $\hat{n} \in L^{2}\left(0,T; H^{1}(\Omega)\right), \nabla\varphi\in L^\infty(Q)$ , and $u$ be the solution of the problem (2.2), then $u \in L^{\infty}\left(0, T; H^{1}(\Omega)\right) \cap L^{2}\left(0, T; H^{2}(\Omega)\right)$ and $u_t\in L^2(0,T; L^2(\Omega))$ .

Proof. Multiplying the first equation of (2.2) by $u$ , and integrating it over $\Omega$ , we get

$\begin{align*} &\frac{1}{2} \frac{d}{d t} \int_{\Omega} u^{2} d x+\int_{\Omega}|\nabla u|^{2} d x+\int_{\Omega} u^{2} d x \\ = &\int_{\Omega} \hat{n} \nabla \varphi \cdot u d x+\int_{\Omega} u^{2} d x \\ \leq& \|\hat{n}\|_{L^{2}}\|u\|_{L^{2}}+\|u\|^2_{L^{2}} \\ \leq&C(\|\hat{n}\|^2_{L^{2}}+\|u\|^2_{L^{2}}). \end{align*}$

By Gronwall's inequality, we have

$\begin{align*} \|u\|^2_{L^2} +\int_0^T\| u\|_{H^1}^{2} d\tau \leq C \left(\int_{0}^T\|\hat{n}\|_{L^{2}}^{2} d \tau+\|u_0\|_{L^2}^{2}\right). \end{align*}$

Operating the Helmholtz projection operator $P$ to the first equation of (2.2), we know

$\begin{align*} u_{t}+A u+P(\hat{u} \cdot \nabla u) = P(\hat{n} \nabla \varphi), \end{align*}$

where $A: = -P \Delta$ is called Stokes operator, which is an unbounded self-adjoint positive operator in $L^2$ with compact inverse, for more properties of Stokes operator, we refer to [10]. Note that $\nabla\cdot u = 0$ , that is $P u = u$ , $P \Delta u = \Delta u$ , $P u_{t} = u_{t}$ . So, in following calculations, we ignore the projection operator $P$ . Multiplying this equation by $\Delta u$ , and integrating it over $\Omega$ , we get

$\begin{align*} &\frac{1}{2} \frac{d}{d t} \int_{\Omega}|\nabla u|^{2} d x+\int_{\Omega}|\Delta u|^{2} d x + \int_{\Omega}|\nabla u|^{2} d x \\ = &\int_{\Omega} P(\hat{u} \nabla u) \Delta u d x-\int_{\Omega} P(\hat{n} \nabla \varphi) \Delta u d x+ \int_{\Omega}|\nabla u|^{2} d x. \end{align*}$

For the terms on the right, we have

$\begin{align*} &\int_{\Omega} P(\hat{u} \nabla u) \Delta u d x- \int_{\Omega} P(\hat{n} \nabla \varphi) \Delta u d x+\int_{\Omega}|\nabla u|^{2} d x \\ \leq&\|\hat{u}\|_{L^{4}}\|\nabla u\|_{L^{4}}\|\Delta u\|_{L^{2}}+\|\hat{n}\|_{L^{2}}\|\Delta u\|_{L^{2}}+\|\nabla u\|^2_{L^2} \\ \leq&\|\hat{u}\|_{L^{4}}\|\nabla u\|_{L^{2}}^{1 / 2}\|\Delta u\|_{L^{2}}^{3 / 2}+\|\hat{u}\|_{L^{4}}\|\nabla u\|_{L^{2}}\|\Delta u\|_{L^{2}}+\|\hat{n}\|_{L^{2}}\|\Delta u\|_{L^{2}}+\|\nabla u\|^2_{L^2} \\ \leq &\frac{1}{2}\|\Delta u\|_{L^{2}}^{2}+C\left(\|\hat{u}\|_{L^{4}}^{4}+\|\hat{u}\|_{L^{4}}^{2}+1\right)\|\nabla u\|_{L^{2}}^{2}+\|\hat{n}\|_{L^{2}}^{2}. \end{align*}$

Therefore, we get

$\begin{align*} \frac{d}{d t} \|\nabla u\|_{L^2}^{2} +\| \nabla u\|_{H^1}^{2} \leq C\left(\|\hat{u}\|_{L^{4}}^{4}+\|\hat{u}\|_{L^{4}}^{2}+1\right)\|\nabla u\|_{L^{2}}^{2}+C\|\hat{n}\|_{L^{2}}^{2}+C. \end{align*}$

By Gronwall's inequality, we derive

$\begin{align*} \|\nabla u\|_{L^2}^{2} +\int_0^T\| \nabla u\|_{H^1}^{2}d\tau \leq C. \end{align*}$

Multiplying the first equation of (2.2) by $u_t$ , and combining with above inequality, we have

$\int_0^T\int_\Omega|u_t|^2dxdt\leq C.$

Summing up, we complete the proof.

For the above solution $u$ , we consider the following linear problem

$\begin{align} \left\{\begin{aligned} &c_{t}-\Delta c+u \cdot \nabla c+c = \hat{n}_++f, &&\text{in } Q, \\ &\frac{\partial c}{\partial \nu} = 0,&&\text{on } (0, T) \times \partial\Omega, \\&c(x,0) = c_0(x), &&\text{in }\Omega. \end{aligned} \right. \end{align}$

(2.3)

Along with fixed point method, the existence of solutions can be easily obtained. Thus we omit the proof and only give the regularity estimate.

Lemma 2.4. Let $c_0\in H^2(\Omega)$ , $\hat{n} \in L^{2}\left(0,T; H^{1}(\Omega)\right)$ , $f \in L^{2}\left(0,T; H^{1}(\Omega)\right)$ , $u$ be the solution of the problem (2.2), and $c$ be the solution of (2.3). Then $c \in L^{\infty}\left((0, T), H^{2}(\Omega)\right) \cap L^{2}\left((0, T), H^{3}(\Omega)\right)$ and $c_t\in L^2(0,T; L^2(\Omega))$ .

Proof. Multiplying the first equation of (2.3) by $c$ , and integrating it over $\Omega$ , we infer from $\int_{\Omega} c(u \cdot \nabla c) = -\frac{1}{2} \int_{\Omega} c^{2} \nabla \cdot u d x = 0$ that

$\begin{align*} \frac{1}{2} \frac{d}{d t} \int_{\Omega} c^{2} d x+\int_{\Omega}|\nabla c|^{2} d x+\int_{\Omega} c^{2} d x \leq \|\hat{n}\|_{L^{2}}\|c\|_{L^{2}}+\|f\|_{L^{2}}\|c\|_{L^{2}}. \end{align*}$

Therefore, we have

$\begin{align*} \|c\|^2_{L^2} +\| c\|^2_{H^1} \leq C(\|c_0\|^2_{L^2}+ \int_0^t(\|\hat{n}\|^2_{L^{2}}+\|f\|^2_{L^{2}})d\tau). \end{align*}$

Multiplying the first equation of (2.3) by $-\Delta c$ , and integrating it over $\Omega$ , we get

$\begin{align*} &\frac{1}{2}\frac{d}{d t}\int_{\Omega} |\nabla c|^2 d x+\int_{\Omega} |\Delta c|^2 d x +\int_{\Omega} |\nabla c|^2 d x \\ = &\int_{\Omega}u\nabla c\Delta c d x- \int_{\Omega}\Delta c\hat{n} d x- \int_{\Omega}\Delta c f d x. \end{align*}$

Using the Young inequality and the Hölder inequality, we obtain

$\begin{align*} &\int_{\Omega}u\nabla c\Delta c d x- \int_{\Omega}\Delta c\hat{n} d x- \int_{\Omega}\Delta c f d x \\ \leq & \|u\|_{L^{4}}\|\nabla c\|_{L^{4}}\|\Delta c\|_{L^{2}}+\|\hat{n}\|_{L^{2}}\|\Delta c\|_{L^{2}}+\|f\|_{L^{2}}\|\Delta c\|_{L^{2}} \\ \leq&C\|u\|_{H^{1}}(\|\nabla c\|_{L^{2}}^{\frac{1}{2}}\|\Delta c\|^{\frac{1}{2}}_{L^{2}}+\|\nabla c\|_{L^{2}})\|\Delta c\|_{L^{2}}+\|\hat{n}\|_{L^{2}}\|\Delta c\|_{L^{2}}+\|f\|_{L^{2}}\|\Delta c\|_{L^{2}} \\ = &C\|u\|_{H^{1}}\|\nabla c\|_{L^{2}}^{\frac{1}{2}}\|\Delta c\|^{\frac{3}{2}}_{L^{2}}+C\|\nabla c\|_{L^{2}}\|\Delta c\|_{L^{2}}+\|\hat{n}\|_{L^{2}}\|\Delta c\|_{L^{2}}+\|f\|_{L^{2}}\|\Delta c\|_{L^{2}} \\ \leq&\frac{1}{2}\|\Delta c\|^{2}_{L^{2}}+C\|u\|^4_{H^{1}}\|\nabla c\|_{L^{2}}^{2}+C(\|\hat{n}\|^2_{L^2}+\|f\|^2_{L^2}). \end{align*}$

Combining this and above inequalities, we conclude

$\begin{align*} \frac{d}{d t}\|\nabla c\|^2_{L^2} +\|\nabla c\|^2_{H^1} \leq C\|u\|^4_{H^{1}}\|\nabla c\|_{L^{2}}^{2}+C(\|\hat{n}\|^2_{L^2}+\|f\|^2_{L^2}). \end{align*}$

We therefore verify that

$\begin{align*} \|\nabla c\|^2_{L^2}+\int_{0}^{t}\|\nabla c\|^2_{H^1}\leq C( \int_{0}^{t}\|\hat{n}\|^2_{L^2}d\tau+\int_{0}^{t}\|f\|^2_{L^2}d\tau). \end{align*}$

Applying $\nabla$ to the first equation of (2.3), multiplying it by $\nabla \Delta c$ , and integrating over $\Omega$ give

$\begin{align*} &\frac{1}{2} \frac{d}{d t} \int_{\Omega}|\Delta c|^{2} d x+\int_{\Omega}|\nabla \Delta c|^{2} d x+\int_{\Omega}|\Delta c|^{2} d x \nonumber \\ = &\int_{\Omega} \nabla(u \nabla c) \nabla \Delta c d x- \int_{\Omega} \nabla \hat{n}_{+} \nabla \Delta c d x- \int_{\Omega} \nabla f \nabla \Delta c d x. \end{align*}$

For the terms on the right, we obtain

$\begin{align*} &\int_{\Omega} \nabla(u \nabla c) \nabla \Delta c d x- \int_{\Omega} \nabla \hat{n}_{+} \nabla \Delta c d x- \int_{\Omega} \nabla f \nabla \Delta c d x\nonumber \\ \leq&\|\nabla \Delta c\|_{L^{2}}\left(\|u\|_{L^{4}}\|\Delta c\|_{L^{4}}+\|\nabla u\|_{L^{4}}\|\nabla c\|_{L^{4}}\right)+\|\nabla \hat{n}\|_{L^{2}}\|\nabla \Delta c\|_{L^{2}} \nonumber \\ &+\|\nabla f\|_{L^{2}}\|\nabla \Delta c\|_{L^{2}} \nonumber \\ \leq&\|\nabla \Delta c\|_{L^{2}}(\|u\|_{L^{4}}\|\Delta c\|_{L^{2}}^{\frac{1}{2}}\|\nabla \Delta c\|_{L^{2}}^{\frac{1}{2}}+\|u\|_{L^{4}}\|\Delta c\|_{L^{2}} \nonumber \\ &+\|\nabla u\|_{L^{2}}^{\frac{1}{2}}\|\Delta u\|_{L^{2}}^{\frac{1}{2}}\|\nabla c\|_{L^{2}}^{\frac{1}{2}}\|\Delta c\|_{L^{2}}^{\frac{1}{2}}+\|\nabla u\|_{L^{2}}\|\nabla c\|_{L^{2}}^{\frac{1}{2}}\|\Delta c\|_{L^{2}}^{\frac{1}{2}} \nonumber \\ &+\|\nabla u\|_{L^{2}}^{\frac{1}{2}}\|\Delta u\|_{L^{2}}^{\frac{1}{2}}\|\nabla c\|_{L^2}+\|\nabla u\|_{L^2}\|\nabla c\|_{L^2}) +\|\nabla \hat{n}\|_{L^{2}}\|\nabla \Delta c\|_{L^{2}} \nonumber \\ &+\|\nabla f\|_{L^{2}}\|\nabla \Delta c\|_{L^{2}} \nonumber \\ \leq& \frac{1}{2}\|\nabla \Delta c\|_{L^{2}}^{2}+C(1+\|\Delta c\|_{L^{2}}^{2}+\|\Delta u\|_{L^{2}}^{2}+\|\nabla \hat{n}\|_{L^{2}}^{2}+\|\nabla f\|_{L^{2}}^{2}). \end{align*}$

Straightforward calculations yield

$\begin{align*} \|\Delta c\|_{L^{2}}^{2}+\int_{0}^{t}\|\Delta c\|_{H^{1}}^{2} d \tau \leq C\left(1+\int_{0}^{t}\|\hat{n}\|_{H^{1}}^{2}d\tau+\int_{0}^{t}\|f\|_{H^{1}}^{2} d \tau\right). \end{align*}$

Multiplying the first equation of (2.3) by $c_t$ , and combining with above inequality, we have

$\int_0^T\int_\Omega|c_t|^2dxdt\leq C,$

and thereby precisely arrive at the conclusion.

With above solutions $u$ and $c$ in hand, we deal with the following linear problem.

$\begin{align} \left\{\begin{aligned} &n_{t}-\Delta n+u \cdot \nabla n+n = - \nabla \cdot(n \nabla c)+(1+\gamma) \hat{n}_{+} -\mu \hat{n}_{+} n, &&\text{in } Q, \\ &\left.\frac{\partial n}{\partial \nu}\right|_{\partial \Omega} = 0, \\&n(x,0) = n_0(x), &&\text{in }\Omega. \end{aligned} \right. \end{align}$

(2.4)

By a similar argument as the above two problems, the existence of solutions can be easily obtained. Therefore, we only give the regularity estimate.

Lemma 2.5. Suppose $0\leq n_0\in H^1(\Omega)$ , $\hat{n} \in L^{2}\left(0,T; H^{1}(\Omega)\right) \cap L^{4}\left(0,T; L^{4}(\Omega)\right)$ , and $u$ , $c$ , $n$ are the solutions of the problem (2.2), (2.3) and (2.4), respectively. Then $n \geq 0$ , $n\in L^{\infty}(0, T; H^1(\Omega)) \cap L^2(0, T; H^2(\Omega))$ and $n_t\in L^2(0,T; L^2(\Omega))$ .

Proof. Firstly, we verify the nonnegativity of $n$ . We examine the set $A(t) = \{x : n(x, t)<0\}$ . Along with (2.4), we get

$\begin{align*} \frac{d}{d t} \int_{A(t)} n d x-\int_{\partial A(t)} \frac{\partial n}{\partial \nu} d s+\int_{A(t)} n d x = (1+\gamma) \int_{A(t)} \hat{n}_{+} d x-\mu \int_{A(t)} \hat{n}_{+} n d x. \end{align*}$

Since $\frac{\partial n}{\partial \nu} \geq 0$ on $\partial\{n<0\}$ , from this we deduce that the right hand side is nonnegative. Integrating this equality on $[0,t]$ gives

$\begin{align*} \int_{A(t)} n d x d \tau+ \int_{0}^{t} \int_{A(t)} n d x d \tau = 0 . \end{align*}$

Then, we get $n \geq 0$ .

Next, multiplying the first equation of (2.4) by $n$ , and integrating it over $\Omega$ , we get

$\begin{align*} &\frac{1}{2} \frac{d}{d t} \int_{\Omega} n^{2} d x+\int_{\Omega}\left(n^{2}+|\nabla n|^{2}\right) d x+\mu \int_{\Omega} \hat{n}_{+} n^{2} d x \\ = & \int_{\Omega} n\nabla c \nabla n d x+\left(1+\gamma\right) \int_{\Omega} n \hat{n}_{+} d x \\ \leq& \|n\|_{L^{4}}\|\nabla c\|_{L^{4}}\|\nabla n\|_{L^2}+(1+\gamma) \|\hat{n}\|_{L^{2}}\|n\|_{L^{2}} \\ \leq& C(\|n\|_{L^2}^{\frac{1}{2}}\|\nabla n\|_{L^2}^{\frac{1}{2}}+\|n\|_{L^2})\| c\|_{H^{2}}\|\nabla n\|_{L^2}+(1+\gamma) \|\hat{n}\|_{L^{2}}\|n\|_{L^{2}} \\ \leq& C(\|n\|_{L^2}^{2}\| c\|_{H^{2}}^{4}+\|n\|_{L^2}^{2}\| c\|_{H^{2}}^{2}+\|\hat{n}\|_{L^{2}})+\frac{1}{2}\|n\|_{H^{1}}^{2}. \end{align*}$

So, we derive that

$\begin{align*} \|n\|_{L^{2}}^{2}+\int_{0}^{T}\|n\|_{H^{1}}^{2} d t \leq C \left(1+\int_{0}^{T}\|\hat{n}\|_{L^{2}}^{2} d t\right). \end{align*}$

Multiplying the first equation of (2.4) by $-\Delta n$ , and integrating it over $\Omega$ , we get

$\begin{align*} &\frac{1}{2} \frac{d}{d t} \int_{\Omega}|\nabla n|^{2} d x+\int_{\Omega}|\Delta n|^{2} d x+\int_{\Omega}|\nabla n|^{2} d x \nonumber \\ = &\int_{\Omega} u \nabla n \Delta n d x+\int_{\Omega}\left(\nabla \cdot(n \nabla c) \Delta n-(1+\gamma) \hat{n}_{+} \Delta n+\mu \hat{n}_{+} n \Delta n\right) d x \nonumber \\ \leq&\|u\|_{L^{4}}\|\nabla n\|_{L^{4}}\|\Delta n\|_{L^{2}}+\|n\|_{L^{4}}\|\Delta c\|_{L^{4}}\|\Delta n\|_{L^{2}}+\|\nabla n\|_{L^{4}}\|\nabla c\|_{L^{4}}\|\Delta n\|_{L^{2}} \nonumber \\ &+(1+\gamma)\|\hat{n}\|_{L^{2}}\|\Delta n\|_{L^{2}}+\mu\|n\|_{L^{4}}\|\hat{n}\|_{L^{4}}\|\Delta n\|_{L^{2}} \nonumber \\ \leq &C\|u\|_{H^1}(\|\nabla n\|_{L^{2}}^{\frac{1}{2}}\|\Delta n\|_{L^{2}}^{\frac{1}{2}}+\|\nabla n\|_{L^{2}})\|\Delta n\|_{L^{2}} \\ &+\|n\|_{L^{4}}\left(\|\Delta c\|_{L^{2}}^{\frac{1}{2}}\|\nabla \Delta c\|_{L^{2}}^{\frac{1}{2}}+\|\Delta c\|_{L^{2}}\right)\|\Delta n\|_{L^{2}} +\mu\|n\|_{L^{4}}\|\hat{n}\|_{L^{4}}\|\Delta n\|_{L^{2}} \nonumber \\ &+(\|\nabla n\|_{L^{2}}^{\frac{1}{2}}\|\Delta n\|_{L^{2}}^{\frac{1}{2}}+\|\nabla n\|_{L^{2}})\|\nabla c\|_{H^{1}}\|\Delta n\|_{L^{2}}+(1+\gamma)\|\hat{n}\|_{L^{2}}\|\Delta n\|_{L^{2}} \nonumber \\ \leq& \frac{1}{2}\|\Delta n\|_{L^{2}}^{2}+C\left(\|\nabla n\|_{L^{2}}^{2}+\|n\|_{L^{4}}^{4}+\|\Delta c\|_{L^{2}}^{4}+\|\nabla \Delta c\|_{L^{2}}^{2}+\|\hat{n}\|_{L^{2}}^{2}+\|\hat{n}\|_{L^{4}}^{4}\right) \nonumber \\ \leq& \frac{1}{2}\|\Delta n\|_{L^{2}}^{2}+C(1+\|\nabla n\|_{L^{2}}^{2}+\|n\|_{L^{2}}^{4}+\|n\|_{L^{2}}^{2}\|\nabla n\|_{L^{2}}^{2}+\|\nabla \Delta c\|_{L^{2}}^{2} \nonumber \\ &+\|\hat{n}\|_{L^{2}}^{2}+\|\hat{n}\|_{L^{4}}^{4}). \end{align*}$

Straightforward calculations yield

$\begin{align*} \|\nabla n\|_{L^{2}}^{2}+\int_{0}^{T} \int_{\Omega}\left(|\Delta n|^{2}+|\nabla n|^{2}+\hat{n}_{+}|\nabla n|^{2}\right) d x d t \leq C. \end{align*}$

Multiplying the first equation of (2.4) by $n_t$ , and combining with above inequality, we have

$\int_0^T\int_\Omega|n_t|^2dxdt\leq C.$

The proof is complete.

Introduce the spaces

$\begin{align*} &X_{u} = L^{4}\left(0,T; L^{4}(\Omega)\right), \quad X_{n} = L^{4}\left(0,T; L^{4}(\Omega)\right) \cap L^{2}\left(0,T; H^{1}(\Omega)\right), \\ &Y_{u} = L^{\infty}\left(0, T; H^{1}(\Omega)\right) \cap L^{2}\left(0, T; H^{2}(\Omega)\right), \\ &Y_{n} = L^{\infty}(0, T; H^1(\Omega)) \cap L^2(0, T; H^2(\Omega)). \end{align*}$

Define a map

$\begin{align*} &\mathcal{F}: X_{u} \times X_{n}\rightarrow X_{u} \times X_{n}, \\ &\mathcal{F}(\hat{u}, \hat{n}) = (u, n), \end{align*}$

where the $(u, n)$ is the solution of the decoupled linear problem

$\begin{align*} \left\{\begin{aligned} &n_{t}-\Delta n+u \cdot \nabla n+n = - \nabla \cdot(n \nabla c)+(1+\gamma) \hat{n}_{+} -\mu \hat{n}_{+} n, &&\text {in } (0, T) \times \Omega \equiv Q, \\ &c_{t}-\Delta c+u \cdot \nabla c+c = \hat{n}_++f, &&\text{in } (0, T) \times \Omega \equiv Q, \\ &u_{t}-\Delta u+\hat{u} \cdot \nabla u = -\nabla \pi+\hat{n} \nabla \varphi, &&\text{in } (0, T) \times \Omega \equiv Q, \\ &\nabla \cdot u = 0, &&\text{in } (0, T) \times \Omega \equiv Q, \\ &\frac{\partial n}{\partial \nu} = \frac{\partial c}{\partial \nu} = 0,\quad u = 0, &&\text{on } (0, T) \times \partial\Omega, \\ &n(x,0) = n_0(x), c(x,0) = c_0(x), u(x,0) = u_0(x), &&\text{in }\Omega. \end{aligned} \right. \end{align*}$

Next, we use fixed point method to prove the local existence of solutions of the problem (1.1).

Lemma 2.6. The map $\mathcal{F}: X_{u} \times X_{n}\rightarrow X_{u} \times X_{n}$ is well defined and compact.

Proof. Let $(\hat{n},\hat{u})\in X_{u} \times X_{n}$ , by Lemmas 2.3, 2.4, 2.5 we deduce that $(n,u) = \mathcal{F}(\hat{n},\hat{u})$ is bounded in $Y_u\times Y_n$ . Note that the embeddings $H^{2}(\Omega) \hookrightarrow H^{1}(\Omega)$ is compact and interpolating between $L^{\infty}(0, T; H^1(\Omega))$ and $L^2(0, T; H^2(\Omega))$ . It is easy to get that $u$ is bounded in $L^{4}\left(0,T; L^{4}(\Omega)\right)$ and $n$ is bounded in $L^{4}\left(0,T; L^{4}(\Omega)\right) \cap L^{2}\left(0,T; H^{1}(\Omega)\right)$ . Therefore, the operator $\mathcal{F}:X_{u} \times X_{n}\rightarrow X_{u} \times X_{n}$ is a compact operator.

3. Existence and uniqueness of strong solution of system

From Lemma 2.6, $(n,u)\in Y_n\times Y_u$ satisfies pointwisely a.e. in $Q$ the following problem

$\begin{array}{l} \left\{\begin{array}{l} n_{t}-\Delta n+u \cdot \nabla n+n = - \nabla \cdot(n \nabla c) \\ +\alpha(1+\gamma) n -\mu n^2, &&\text {in } Q, \\ c_{t}-\Delta c+u \cdot \nabla c+c = n+\alpha f, &&\text{in } Q, \\ u_{t}-\Delta u+u \cdot \nabla u = -\nabla \pi+\alpha n \nabla \varphi, &&\text{in } Q, \\ \nabla \cdot u = 0, &&\text{in } Q, \\ \frac{\partial n}{\partial \nu} = \frac{\partial c}{\partial \nu} = 0,\quad u = 0, &&\text{on } (0, T) \times \partial\Omega, \\ n(x,0) = n_0(x), c(x,0) = c_0(x), u(x,0) = u_0(x), &&\text{in }\Omega. \end{array} \right. \end{array}$

(3.1)

In order to prove the existence of solution, we first give some a priori estimates.

Lemma 3.1. Let $(n,c,u)$ be a local solution to (3.1). Then, it holds that

$\begin{align} &\|n\|_{L^1}+\int_0^t(\|n\|_{L^1}+\|n\|_{L^2})d\tau \leq C, \end{align}$

(3.2)

$\begin{align} &\|\nabla u\|_{L^{2}}^{2}+\int_{0}^{t}\|\nabla u\|_{H^{1}}^{2} d \tau \leq C, \end{align}$

(3.3)

$\begin{align} &\|\nabla c\|^2_{L^2}+\int_{0}^{t}\|\nabla c\|^2_{H^1}d\tau \leq C. \end{align}$

(3.4)

Proof. With Lemma 2.5 in hand, we get $n\geq 0$ . Integrating the first equation of (3.1) over $\Omega$ , we see that

$\begin{align*} \frac{d}{d t} \int_{\Omega} n d x+\int_{\Omega} n d x+\mu \int_{\Omega} n^{2} d x = \alpha(1+\gamma) \int_{\Omega} n d x \leq \frac{\mu}{2} \int_{\Omega} n^{2} d x+C. \end{align*}$

Solving this differential inequality, we obtain that

$\begin{align*} \|n\|_{L^1}+\int_0^t(\|n\|_{L^1}+\|n\|_{L^2})d\tau \leq C. \end{align*}$

Multiplying the third equation of (3.1) by $u$ , and integrating it over $\Omega$ , we get

$\begin{align*} \frac{1}{2} \frac{d}{d t} \int_{\Omega} u^{2} d x+\int_{\Omega}|\nabla u|^{2} d x+\int_{\Omega} u^{2} d x = &\alpha\int_{\Omega} n \nabla \varphi \cdot u d x+\int_{\Omega} u^{2} d x \\ \leq& \|n\|_{L^{2}}\|u\|_{L^{2}}+\|u\|^2_{L^{2}} \leq C(\|n\|^2_{L^{2}}+\|u\|^2_{L^{2}}). \end{align*}$

Therefore, we see that

$\begin{align*} \|u\|^2_{L^{2}}+\int_{0}^t\|u\|_{H^1} d\tau \leq C. \end{align*}$

By the Gagliardo-Nirenberg interpolation inequality, we deduce that

$\begin{align*} \int_{0}^{t}\|u\|_{L^{4}}^{4} d \tau \leq & C \int_{0}^{t}(\|u\|_{L^{2}}^{2}\|\nabla u\|_{L^{2}}^{2} d+\|u\|_{L^{2}}^{2}) \tau \\ \leq& \|u\|_{L^{2}}^{2} \int_{0}^{t}\|\nabla u\|_{L^{2}}^{2} d \tau+ \int_{0}^{t}\|u\|_{L^{2}}^{2}d \tau \\ \leq& C. \end{align*}$

Multiplying the third equation of (3.1) by $\Delta u$ , and integrating it over $\Omega$ , we get

$\begin{align*} \frac{d}{d t} \|\nabla u\|_{L^2}^{2} +\| \nabla u\|_{H^1}^{2} \leq C\left(\|u\|_{L^{4}}^{4}+\|u\|_{L^{4}}^{2}+1\right)\|\nabla u\|_{L^{2}}^{2}+C\|n\|_{L^{2}}^{2}+C. \end{align*}$

Thus, we know

$\begin{align*} \|\nabla u\|_{L^{2}}^{2}+\int_{0}^{t}\|\nabla u\|_{H^{1}}^{2} d \tau \leq C. \end{align*}$

Multiplying the second equation of (3.1) by $c$ , and integrating it over $\Omega$ , we have

$\begin{align*} \frac{1}{2} \frac{d}{d t} \int_{\Omega} c^{2} d x+\int_{\Omega}|\nabla c|^{2} d x+\int_{\Omega} c^{2} d x \leq \|n\|_{L^{2}}\|c\|_{L^{2}}+\alpha\|f\|_{L^{2}}\|c\|_{L^{2}}. \end{align*}$

Then, we have

$\begin{align*} \|c\|_{L^2}+\int_{0}^{t}\|c\|_{H^1}d\tau\leq C. \end{align*}$

Multiplying the second equation of (3.1) by $-\Delta c$ , and integrating it over $\Omega$ , we get

$\begin{align*} \frac{d}{d t}\|\nabla c\|^2_{L^2} +\|\nabla c\|^2_{H^1} \leq C\|u\|^4_{H^{1}}\|\nabla c\|_{L^{2}}^{2}+C(\|n\|^2_{L^2}+\|f\|^2_{L^2}). \end{align*}$

Further, we have

$\begin{align*} \|\nabla c\|^2_{L^2}+\int_{0}^{t}\|\nabla c\|^2_{H^1}d\tau \leq C. \end{align*}$

The proof is complete.

Lemma 3.2. Let $(n,c,u)$ be a local solution to (3.1). Then, it holds that

$\begin{align} \|(n+1)\ln(n+1)\|_{L^1}+\|\nabla c\|^2_{L^2}+\|\nabla c\|^2_{H^1}\leq C. \end{align}$

(3.5)

Proof. We rewrite the first equation of (3.1) as

$\begin{align*} &\frac{d}{d t}(n+1)+u \cdot \nabla(n+1)-\Delta(n+1) \\ = &-\nabla \cdot\left((n+1) \cdot \nabla c\right)+\Delta c +\alpha(1+\gamma) n -\mu n^2. \end{align*}$

Multiplying the above equation by $\ln(n + 1)$ and integrating the equation, we have

$\begin{align*} &\frac{d}{d t} \int_{\Omega}(n+1) \ln (n+1) d x+4 \int_{\Omega}|\nabla \sqrt{n+1}|^{2} d x \\ \leq &\int_{\Omega} \nabla(n+1) \cdot \nabla c d x+\int_{\Omega} \Delta c \ln (n+1) d x+\alpha(1+\gamma)\int_{\Omega}n\ln (n+1)d x \\ = &I_1+I_2+I_3. \end{align*}$

For $I_1$ , integrating by parts and using Young's inequality with small $\delta$ , we get

$\begin{align*} I_1 = -\int_{\Omega} n \Delta c d x \leq \|n\|_{L^2}\|\Delta c\|_{L^2} \leq \delta\|\Delta c\|^2_{L^2}+C\|n\|^2_{L^2}. \end{align*}$

For the term $I_2$ , we have

$\begin{align*} I_2 = &\int_{\Omega} \Delta c \ln (n+1) d x\leq \delta\|\Delta c\|^2_{L^2}+C\|\ln(n+1)\|^2_{L^2} \\ \leq& \delta\|\Delta c\|^2_{L^2}+C\int_{\Omega}(n+1)\ln(n+1) d x. \end{align*}$

For the rest term $I_3$ , straightforward calculations yield

$\begin{align*} I_3 = \alpha(1+\gamma)\int_{\Omega}n\ln (n+1)d x \leq (1+\gamma)\int_{\Omega}(n+1)\ln (n+1)d x. \end{align*}$

Combining $I_1$ , $I_2$ with $I_3$ , we conduct that

$\begin{align} &\frac{d}{d t} \int_{\Omega}(n+1) \ln (n+1) d x+4 \int_{\Omega}|\nabla \sqrt{n+1}|^{2} d x \\ \leq& \delta\|\Delta c\|^2_{L^2}+C\int_{\Omega}(n+1)\ln (n+1)d x+C\|n\|^2_{L^2}. \end{align}$

(3.6)

Multiplying the second equation of (3.1) by $\Delta c$ , and integrating it over $\Omega$ , we get

$\begin{align*} &\frac{1}{2}\frac{d}{d t}\int_{\Omega} |\nabla c|^2 d x+\int_{\Omega} |\Delta c|^2 d x +\int_{\Omega} |\nabla c|^2 d x \\ = &\int_{\Omega}u\nabla c\Delta c d x- \int_{\Omega}\Delta c n d x- \alpha\int_{\Omega}\Delta c f d x. \end{align*}$

Straightforward calculations yield

$\begin{align} \frac{d}{d t}\|\nabla c\|^2_{L^2} +\|\nabla c\|^2_{H^1} \leq C\|\nabla c\|_{L^{2}}^{2}+C(\|n\|^2_{L^2}+\|f\|^2_{L^2}). \end{align}$

(3.7)

Combing (3.6) and (3.7), it follows that

$\begin{align*} &\frac{d}{d t} \int_{\Omega}(n+1) \ln (n+1) d x+\frac{d}{d t}\|\nabla c\|^2_{L^2}+(1-\delta)\|\nabla c\|^2_{H^1} +4 \int_{\Omega}|\nabla \sqrt{n+1}|^{2} d x \\ \leq&C\int_{\Omega}(n+1)\ln (n+1)d x+C(\|f\|^2_{L^2}+\|n\|^2_{L^2}). \end{align*}$

Taking $\delta$ small enough, and solving this differential inequality, we obtain that

$\begin{align*} \|(n+1)\ln(n+1)\|_{L^1}+\|\nabla c\|^2_{L^2}+\|\nabla c\|^2_{H^1}\leq C. \end{align*}$

The proof is complete.

Lemma 3.3. Assume $f\in L^2(0,T;H^1(\Omega))$ , let $(n,c,u)$ be a local solution to (3.1). Then, it holds that

$\begin{align} \|n\|^2_{L^2}+\|\Delta c\|^2_{L^2}+\int_{0}^t\|n\|_{H^1}d\tau +\int_{0}^t\|\Delta c\|_{H^1}d\tau\leq C. \end{align}$

(3.8)

Proof. Taking the $L^2$ -inner product with $n$ for the first equation of (3.1) implies

$\begin{align*} &\frac{1}{2} \frac{d}{d t} \int_{\Omega} n^{2} d x+\int_{\Omega}\left(n^{2}+|\nabla n|^{2}\right) d x+\mu \int_{\Omega} n^{3} d x \\ = & \int_{\Omega} n\nabla c \nabla n d x+\alpha\left(1+\gamma\right) \int_{\Omega} n^2 d x \\ = & -\frac{1}{2}\int_{\Omega} n^2\Delta c d x +\alpha\left(1+\gamma\right) \int_{\Omega} n^2 d x. \end{align*}$

Here, we note that

$\begin{align*} \left|\int_{\Omega} n^{2} \Delta c d x\right| \leq& \|n\|_{L^{3}}^{2}\|\Delta c\|_{L^{3}} \nonumber \\ \leq&C\|n\|_{L^{3}}^{2}(\|\nabla\Delta c\|_{L^{2}}^{\frac{2}{3}}\|\nabla c\|_{L^{2}}^{\frac{1}{3}}+\|\nabla c\|_{L^{2}}) \nonumber \\ \leq&C\|n\|_{L^{3}}^{2}(\|\nabla \Delta c\|_{L^{2}}^{\frac{2}{3}}+1). \end{align*}$

From Lemma 2.2 and (3.2), it follows that

$\begin{align*} &-\frac{\chi}{2}\int_{\Omega} n^{2} \Delta c d x \nonumber \\ \leq& C\left(\delta\|n\|_{H^{1}}^{2}\|(n+1) \log (n+1)\|_{L^{1}}+p\left(\delta^{-1}\right)\|n\|_{L^{1}}\right)^{\frac{2}{3}}(\|\nabla \Delta c\|_{L^{2}}^{\frac{2}{3}}+1) \nonumber \\ \leq& C\left(\delta\|n\|_{H^{1}}^{2}+p\left(\delta^{-1}\right)\right)^{\frac{2}{3}}(\|\nabla \Delta c\|_{L^{2}}^{\frac{2}{3}}+1) \nonumber \\ \leq&C(\delta \|n\|_{H^{1}}^{\frac{4}{3}} \|\nabla\Delta c \|_{L^2}^{\frac{2}{3}}+\delta \|n\|_{H^{1}}^{\frac{4}{3}}+p^{\frac{2}{3}}\left(\delta^{-1}\right)\|\nabla\Delta c\|_{L^2}^{\frac{2}{3}}+p^{\frac{2}{3}}\left(\delta^{-1}\right)) \nonumber \\ \leq&\delta\|\nabla\Delta c\|^2_{L^2}+C\delta^\frac{1}{2}\|n\|^2_{H^{1}}+C_\delta^{-1 / 2} p\left(\delta^{-1}\right). \end{align*}$

As an immediate consequence

$\begin{align} \frac{d}{d t}\|n\|^2_{L^2}+\| n\|^2_{H^1} \leq \delta\|\nabla\Delta c\|^2_{L^2}+C\delta^\frac{1}{2}\|n\|^2_{H^{1}}+C\|n\|^2_{L^2}. \end{align}$

(3.9)

Applying $\nabla$ to the first equation of (3.1), multiplying it by $\nabla \Delta c$ , and integrating over $\Omega$ give

$\begin{align*} &\frac{1}{2} \frac{d}{d t} \int_{\Omega}|\Delta c|^{2} d x+\int_{\Omega}|\nabla \Delta c|^{2} d x+\int_{\Omega}|\Delta c|^{2} d x \nonumber \\ = &\int_{\Omega} \nabla(u \nabla c) \nabla \Delta c d x- \int_{\Omega} \nabla n \nabla \Delta c d x- \int_{\Omega} \nabla f \nabla \Delta c d x = I_4+I_5. \end{align*}$

For $I_4$ , by using the Gagliardo-Nirenberg interpolation inequality, we get

$\begin{align*} I_4 = &\int_{\Omega} \nabla(u \nabla c) \nabla \Delta c d x \nonumber \\ \leq&\|\nabla \Delta c\|_{L^{2}}\left(\|u\|_{L^{4}}\|\Delta c\|_{L^{4}}+\|\nabla u\|_{L^{4}}\|\nabla c\|_{L^{4}}\right) \nonumber \\ \leq&\|\nabla \Delta c\|_{L^{2}}(\|u\|_{L^{4}}\|\Delta c\|_{L^{2}}^{\frac{1}{2}}\|\nabla \Delta c\|_{L^{2}}^{\frac{1}{2}}+\|u\|_{L^{4}}\|\Delta c\|_{L^{2}} \nonumber \\ &+\|\nabla u\|_{L^{2}}^{\frac{1}{2}}\|\Delta u\|_{L^{2}}^{\frac{1}{2}}\|\nabla c\|_{L^{2}}^{\frac{1}{2}}\|\Delta c\|_{L^{2}}^{\frac{1}{2}}+\|\nabla u\|_{L^{2}}\|\nabla c\|_{L^{2}}^{\frac{1}{2}}\|\Delta c\|_{L^{2}}^{\frac{1}{2}} \nonumber \\ &+\|\nabla u\|_{L^{2}}^{\frac{1}{2}}\|\Delta u\|_{L^{2}}^{\frac{1}{2}}\|\nabla c\|_{L^2}+\|\nabla u\|_{L^2}\|\nabla c\|_{L^2}) \nonumber \\ \leq& \frac{1}{4}\|\nabla \Delta c\|_{L^{2}}^{2}+C(1+\|\Delta c\|_{L^{2}}^{2}+\|\Delta u\|_{L^{2}}^{2}). \end{align*}$

For the term $I_5$ , we have

$\begin{align*} I_5 = &- \int_{\Omega} \nabla n \nabla \Delta c d x- \int_{\Omega} \nabla f \nabla \Delta c d x \\ \leq&C(\|\nabla n\|^2_{L^2}+\|\nabla f\|^2_{L^2})+\frac{1}{4}\|\nabla \Delta c\|_{L^{2}}^{2}. \end{align*}$

Along with $I_4$ and $I_5$ , we conclude

$\begin{align} &\frac{d}{d t}\|\Delta c\|^2_{L^2}+\|\nabla\Delta c\|^2_{L^2}+ \|\Delta c\|^2_{L^2} \\ \leq & C(1+\|\Delta c\|_{L^{2}}^{2}+\|\Delta u\|_{L^{2}}^{2}+\|\nabla n\|^2_{L^2}+\|\nabla f\|^2_{L^2}) . \end{align}$

(3.10)

Combining (3.9) and (3.10), it follows that

$\begin{align*} &\frac{d}{d t}(\|n\|^2_{L^2}+\|\Delta c\|^2_{L^2})+ \|\Delta c\|^2_{L^2}+(1-C\delta^\frac{1}{2})\| n\|^2_{H^1}+(1-\delta)\|\nabla\Delta c\|^2_{L^2} \\ \leq&C(1+\|\Delta c\|_{L^{2}}^{2}+\|\Delta u\|_{L^{2}}^{2}+\|\nabla n\|^2_{L^2}+\|\nabla f\|^2_{L^2}) . \end{align*}$

By choosing $\delta$ small enough and using (3.3) and (3.5), we have

$\begin{align*} \|n\|^2_{L^2}+\|\Delta c\|^2_{L^2}+\int_{0}^t\|n\|_{H^1}d\tau +\int_{0}^t\|\Delta c\|_{H^1}d\tau\leq C. \end{align*}$

The proof is complete.

Lemma 3.4. Assume $f\in L^2(0,T;H^1(\Omega))$ , let $(n,c,u)$ be a local solution to (3.1). Then, it holds that

$\begin{align} \|\nabla n\|^2_{L^2} +\int_{0}^t\|n\|^2_{H^2} d\tau \leq C. \end{align}$

(3.11)

Proof. Taking the $L^2$ -inner product with $-\Delta n$ for the first equation of (3.1) implies

$\begin{align*} &\frac{1}{2} \frac{d}{d t} \int_{\Omega}|\nabla n|^{2} d x+\int_{\Omega}|\Delta n|^{2} d x+\int_{\Omega}|\nabla n|^{2} d x \\ = &\int_{\Omega}u\nabla n \Delta n d x+\int_{\Omega} \nabla \cdot(n\nabla c)\Delta n d x+(1+\gamma)\int_{\Omega}|\nabla n|^2 d x+\mu\int_{\Omega}n^2\Delta n d x \\ = &I_6+I_7+I_8. \end{align*}$

For the term $I_6$ , with the estimate (3.3), we have

$\begin{align*} I_6 = &\int_{\Omega}u\nabla n \Delta n d x = -\frac{1}{2}\int_{\Omega}\nabla u(\nabla n)^2 d x\leq \|\nabla u\|_{L^2}\|\nabla n\|^2_{L^4} \\ \leq& \|\nabla u\|_{L^2}(\|\nabla n\|^{\frac{1}{2}}_{L^2}\|\Delta n\|^{\frac{1}{2}}_{L^2}+\|\nabla n\|_{L^2})^2 \\ \leq& \delta\|\Delta n\|^{2}_{L^2}+C\|\nabla n\|_{L^2}^2. \end{align*}$

For the term $I_7$ , taking (3.8) into considering, we conduct that

$\begin{align*} I_7 = &\int_{\Omega} \nabla \cdot(n\nabla c)\Delta n d x \\ = &\int_{\Omega}(\nabla n\nabla c+n\Delta c)\Delta n d x \\ \leq&\|\Delta n\|_{L^{2}}\left(\|\nabla n\|_{L^{3}}\|\nabla c\|_{L^{6}}+\|n\|_{\mathcal{C}}\|\Delta c\|_{L^{2}}\right) \nonumber \\ \leq& C\|\Delta n\|_{L^{2}}\left(\|\nabla n\|_{H^{\frac{1}{3}}}\|\nabla c\|_{H^{1}}+\|n\|_{H^{\frac{4}{3}}}\|\Delta c\|_{L^{2}}\right) \nonumber \\ \leq& C \|n\|_{H^{2}}\|n\|_{H^{\frac{4}{3}}}\|c\|_{H^{2}} \leq C\|n\|_{H^{2}}^{\frac{5}{3}}\|n\|_{L^{2}}^{\frac{1}{3}}\|c\|_{H^{2}} \nonumber \\ \leq& \delta\| n\|_{H^{2}}^{2}+C(\delta) \|n\|_{L^{2}}^{2}\|c\|_{H^{2}}^{6}\leq \delta\| n\|_{H^{2}}^{2}+C. \end{align*}$

For the term $I_8$ , thanks to the nonnegativity of $n$ , we see that

$\begin{align*} I_8 = &(1+\gamma)\int_{\Omega}|\nabla n|^2 d x+\mu\int_{\Omega}n^2\Delta n d x \\ = &(1+\gamma)\int_{\Omega}|\nabla n|^2 d x-2\mu\int_{\Omega}|\nabla n|^2 n d x \\ \leq&(1+\gamma)\|\nabla n\|^2_{L^2}. \end{align*}$

Combine the estimates about $I_6$ , $I_7$ and $I_8$ , it follows that

$\begin{align*} \frac{d}{d t}\|\nabla n\|^2_{L^2}+(1-4\delta)\|n\|^2_{H^2}\leq C\|\nabla n\|^2_{L^2}+C. \end{align*}$

By taking $\delta$ small enough, we get

$\begin{align*} \|\nabla n\|^2_{L^2} +\int_{0}^t\|n\|^2_{H^2} d\tau \leq C. \end{align*}$

Therefore, this proof is complete.

Lemma 3.5. The operator $\mathcal{F}:X_{u} \times X_{n}\rightarrow X_{u} \times X_{n}$ , is continuous.

Proof. Let $\{(\hat{n}_m,\hat{u}_{m})\}_{m\in \mathbb{N}}$ be a sequence of $X_{u} \times X_{n}$ , Then, with Lemmas 2.3, 2.4 and 2.5 in hand, we conduct that $\{(n_m,u_m) = \mathcal{F}(\hat{n}_m,\hat{u}_{m})\}_{m\in \mathbb{N}}$ is bounded in $Y_u\times Y_n$ . Taking the compactness of $Y_u\times Y_n$ in $X_{u} \times X_{n}$ into consider, we see that $\mathcal{F}$ is a compact operator, which means there exists a subsequence of $\{\mathcal{F}(\hat{n}_m,\hat{u}_{m})\}_{m\in \mathbb{N}}$ , for convenience, still denoted as $\{\mathcal{F}(\hat{n}_m,\hat{u}_{m})\}_{m\in \mathbb{N}}$ , and exists an element $(\hat{n},\hat{u})$ in $Y_u\times Y_n$ such that

$\begin{align*} \mathcal{F}(\hat{n}_m,\hat{u}_{m})\rightarrow (\hat{n},\hat{u}) \text{ weakly in } Y_u\times Y_n \text{ and strongly in } X_{u} \times X_{n}. \end{align*}$

Let $m\rightarrow \infty$ and take the limit, it is clear that $(n,u) = \mathcal{F}(\hat{n}_m,\hat{u}_{m})$ and $(\hat{n}_m,\hat{u}_{m}) = (\hat{n},\hat{u})$ , this means that $\mathcal{F}(\hat{n}_m,\hat{u}_{m}) = (\hat{n}_m,\hat{u}_{m})$ . Since uniqueness of limit, the map $\mathcal{F}$ is continuous.

Theorem 3.1. Let $u_0\in H^1(\Omega)$ , $n_0\in H^1(\Omega)$ , $c_0\in H^2(\Omega)$ with $n_0\geq 0$ in $\Omega$ , and $f\in L^2(0,T;H^1(\Omega))$ , then (1.1) exists unique strong solution $(n,c,u)$ . Moreover, there exists a positive $C$ constant such that

$\begin{align} \|n\|&_{L^{\infty}(0,T;H^1(\Omega))}+\|n\|_{L^2(0,T;H^2(\Omega))}+\|n_t\|_{L^2(0,T;L^2(\Omega))}+\|c\|_{L^{\infty}(0,T;H^2(\Omega))} \\ &+\|c\|_{L^{2}(0,T;H^3(\Omega))}+\|c_t\|_{L^2(0,T;L^2(\Omega))} +\|u\|_{L^{\infty}(0,T;H^1(\Omega))} \\ &+\|u\|_{L^2(0,T;H^2(\Omega))}+\|u_t\|_{L^2(0,T;L^2(\Omega))}\leq C. \end{align}$

(3.12)

Proof. From Lemmas 3.1, 3.3 and 3.4, it is easy to verify the existence of solution and (3.11). Therefore, we will prove the uniqueness of the solution in the following part. For convenience, we set $n = n_1-n_2$ , $c = c_1-c_2$ and $u = u_1-u_2$ , where $(n_i, c_i, u_i)$ is the strong solution of the system, where $i = 1, 2$ . Thus, we obtain the following system

$\begin{align} n_{t}-\Delta n+u_1 \cdot \nabla n+u\nabla n_2 = - \nabla \cdot(n_1 \nabla c) \\ -\nabla(n\nabla c_2)+\gamma n -\mu n(n_1+n_2), &&\text {in } (0, T) \times \Omega \equiv Q, \end{align}$

(3.13)

$\begin{align} &c_{t}-\Delta c+u_1 \cdot \nabla c+u\nabla c_2+c = n, &&\text{in } (0, T) \times \Omega \equiv Q, \end{align}$

(3.14)

$\begin{align} &u_{t}-\Delta u+u_1 \cdot \nabla u+u\cdot \nabla u_2 = n \nabla \varphi, &&\text{in } (0, T) \times \Omega \equiv Q, \end{align}$

(3.15)

$\begin{align} &\nabla \cdot u = 0, &&\text{in } (0, T) \times \Omega \equiv Q, \end{align}$

(3.16)

$\begin{align} &\frac{\partial n}{\partial \nu} = \frac{\partial c}{\partial \nu} = 0,\quad u = 0, &&\text{on } (0, T) \times \partial\Omega, \end{align}$

(3.17)

$\begin{align} &n_0(x) = c_0(x) = u_0(x) = 0, &&\text{in }\Omega. \end{align}$

(3.18)

Taking the $L^2$ -inner product with $n$ for the (3.13) implies

$\begin{align*} &\frac{1}{2}\frac{d}{d t}\int_{\Omega}n^2 d x+\int_{\Omega}|\nabla n|^2 d x+\int_{\Omega} n^2 d x \\ \leq&-\int_{\Omega}u \nabla n_2 n d x+\int_{\Omega}n_1\nabla c \nabla n d x +\int_{\Omega} n\nabla c_2 \nabla n d x+(1+\gamma) \int_{\Omega} n^2 d x \\ = &I_9+I_{10}+I_{11}+I_{12}. \end{align*}$

For the term $I_9$ , due to the estimates (3.3) and (3.8), we have

$\begin{align*} I_9 = &-\int_{\Omega}u \nabla n_2 n d x \leq \|\nabla n_2\|_{L^2}\|u\|_{L^4}\|n\|_{L^4} \\ \leq &C\|\nabla n_2\|_{L^2}\|u\|_{H^1}(\|n\|^{\frac{1}{2}}_{L^2}\|\nabla n\|^\frac{1}{2}_{L^2}+\|n\|_{L^2}) \\ \leq &\frac{\delta}{3} \|\nabla n\|^2_{L^2}+C\|n\|^2_{L^2}. \end{align*}$

For the term $I_{10}$ , with the estimate (3.8) and (3.11), we get

$\begin{align*} I_{10} = &\int_{\Omega}n_1\nabla c \nabla n d x \leq \|\nabla n\|_{L^2}\|n_1\|_{L^4}\|\nabla c\|_{L^4} \\ \leq&C\|\nabla n\|_{L^2}\|n_1\|_{H^1}\|\nabla c\|_{H^1} \\ \leq&\frac{\delta}{3}\|\nabla n\|^2_{L^2}+C. \end{align*}$

For the term $I_{11}$ ,

$\begin{align*} I_{11} = &\int_{\Omega} n\nabla c_2 \nabla n d x \leq\|\nabla n\|_{L^2}\|\nabla c_2\|_{L^4}\|n\|_{L^4} \\ \leq& \|\nabla n\|_{L^2}\|\nabla c_2\|_{H^1}\|n\|_{H^1} \\ \leq&\frac{\delta}{3}\|\nabla n\|^2_{L^2}+C. \end{align*}$

With the use of estimates $I_i(i = 9,10, 11,12)$ , we have

$\begin{align} \frac{d}{d t}\|n\|^2_{L^2}+\|n\|_{H^1} \leq \delta\|\nabla n\|^2_{L^2}+ C\|n\|^2_{L^2}+C. \end{align}$

(3.19)

Taking the $L^2$ -inner product with $c$ for the (3.14) implies

$\begin{align*} &\frac{1}{2}\frac{d}{ d t}\int_{\Omega}c^2 d x+\int_{\Omega}|\nabla c|^2 d x+ \int_{\Omega}c^2 d x \nonumber \\ = &-\int_{\Omega}u_1\nabla c c d x-\int_{\Omega}u\nabla c_2 c d x+ \int_{\Omega}n c d x \nonumber \\ \leq &\|c\|^2_{L^4}\|\nabla u_1\|_{L^2}+\|u\|_{L^2}\|\nabla c_2\|_{L^4}\|c\|_{L^4}+\|n\|_{L^2}\|c\|_{L^2} \nonumber \\ \leq& C(\|c\|^{\frac{1}{2}}_{L^2}\|\nabla c\|^{\frac{1}{2}}_{L^2}+\|c\|_{L^2})^2\|\nabla u_1\|_{L^2}+(\|c\|^{\frac{1}{2}}_{L^2}\|\nabla c\|^{\frac{1}{2}}_{L^2}+\|c\|_{L^2})\|u\|_{L^2}\|\nabla c_2\|_{H^1} \nonumber \\ &+\|n\|_{L^2}\|c\|_{L^2} \nonumber \\ \leq&\delta\|\nabla c\|^2_{L^2}+C\|c\|^2_{L^2}. \end{align*}$

Then, we get

$\begin{align} \frac{d}{d t}\|c\|^2_{L^2}+\|c\|_{H^1}\leq \delta \|\nabla c\|^2_{L^2}+C\|c\|^2_{L^2}. \end{align}$

(3.20)

Taking the $L^2$ -inner product with $c$ for the (3.15) implies

$\begin{align*} \frac{1}{2} \int_{\Omega}u^2 d x+ \int_{\Omega}|\nabla u|^2 d x = \int_{\Omega} n \nabla \varphi u d x . \end{align*}$

Straightforward calculations yield

$\begin{align} \frac{d}{d t}\|u\|^2_{L^2}+\|u\|_{H^1}\leq C (\|u\|^2_{L^2}+\|n\|^2_{L^2}). \end{align}$

(3.21)

Then, a combination of (3.19), (3.20) and (3.21) yields

$\begin{align*} &\frac{d}{d t}(\|n\|^2_{L^2}+\|c\|^2_{L^2}+\|u\|^2_{L^2})+(\|n\|_{H^1}+\|c\|_{H^1}+\|u\|_{H^1}) \\ \leq&\delta(\|\nabla n\|^2_{L^2}+\|\nabla c\|^2_{L^2}+\|\nabla u\|^2_{L^2})+(\|n\|^2_{L^2}+\|c\|^2_{L^2}+\|u\|^2_{L^2})+C. \end{align*}$

By choosing $\delta$ small enough, we get

$\begin{align*} \frac{d}{d t}(\|n\|^2_{L^2}+\|c\|^2_{L^2}+\|u\|^2_{L^2}) \leq C(\|n\|^2_{L^2}+\|c\|^2_{L^2}+\|u\|^2_{L^2})+C. \end{align*}$

Applying Gronwall's lemma to the resulting differential inequality, we finally obtain the uniqueness of the solution.

4. Existence of an optimal control

In this section, we will prove the existence of the optimal solution of control problem. The method we use for treating this problem was inspired by some ideas of Guillén-González et al [9]. Assume $\mathcal{U}\subset L^2(0,T;H^1(\Omega_c))$ is a nonempty, closed and convex set, where control domain $\Omega_c\subset \Omega$ , and $\Omega_d\subset \Omega$ is the observability domain. We adjust the external source $f$ , so that the bacterial density $n$ , oxygen concentration $c$ and fluid velocity $u$ are as close as possible to a desired state $n_d$ , $c_d$ and $u_d$ , and at the final moment $T$ is as close as possible to a desired state $n_{\Omega}$ , $c_{\Omega}$ and $u_{\Omega}$ . We consider the optimal control problem as follows

Minimize the cost functional

$\begin{align} J(n,c,u,f) = &\frac{\beta_1}{2}\|n-n_d\|_{L^2(Q_d)}^2+\frac{\beta_2}{2}\|c-c_d\|_{L^2(Q_d)}^2 +\frac{\beta_3}{2}\|u-u_d\|_{L^2(Q_d)}^2 \\ &+\frac{\beta_4}{2}\|n(T)-n_{\Omega}\|_{L^2(\Omega_d)}^2+\frac{\beta_5}{2}\|c(T)-c_{\Omega}\|_{L^2(\Omega_d)}^2 \\ &+\frac{\beta_6}{2}\|u(T)-u_{\Omega}\|_{L^2(\Omega_d)}^2+\frac{\beta_7}{2}\|f(x,t)\|_{L^2(Q_c)}^2, \end{align}$

(4.1)

subject to the system (1.1). Moreover, the nonnegative constants $\beta_i,i = 1,2,\cdots,7$ are given but not all zero, the functions $n_d$ , $c_d$ , $u_d$ represents the desired states satisfying

$\begin{align*} &n_d\in L^2(Q_d), c_d\in L^2(Q_d), u_d\in L^2(Q_d), \\ &n_{\Omega}\in L^2(\Omega_c), c_{\Omega}\in L^2(\Omega_c), u_{\Omega}\in L^2(\Omega_c), \;f\in \mathcal{U}. \end{align*}$

The set of admissible solutions of optimal control problem (4.1) is defined by

$\begin{align*} \mathcal{S}_{a d} = \{s = (n,c,u,f) \in \mathcal{H}: s \text { is a strong solution of }(1.1)\}. \end{align*}$

The function space $\mathcal{H}$ is given by

$\begin{align*} \mathcal{H} = Y_n\times Y_c \times Y_u\times\mathcal{U}, \end{align*}$

where $Y_c = L^{\infty}\left(0, T; H^{2}(\Omega)\right) \cap L^{2}\left(0, T; H^{3}(\Omega)\right)$ .

Now, we prove the existence of a global optimal control for problem (1.1).

Theorem 4.1. Suppose $f\in \mathcal{U}$ is satisfied, and $n_0\geq 0$ , then the optimal control problem (4.1) admits a solution $(\bar{n},\bar{c},\bar{u},\bar{f})\in \mathcal{S}_{a d}$ .

Proof. Along with Theorem 3.1, we conduct that $\mathcal{S}_{a d} \neq \emptyset$ , then there exists the minimizing sequence $\{(n_m,c_m,u_m,f_m)\}_{m\in \mathbb{N}}\in \mathcal{S}_{ad}$ such that

$\begin{align} \lim\limits_{m\rightarrow +\infty}J(n_m,c_m,u_m,f_m) = \inf\limits_{(n,c,u,f)\in \mathcal{S}_{ad}}J(n,c,u,f). \end{align}$

(4.2)

According to the definition of $\mathcal{S}_{ad}$ , for each $m\in \mathbb{N}$ there exists $(n_m,c_m,u_m,f_m)$ satisfying

$\begin{align} \left\{\begin{aligned} &n_{mt}+u_{m} \cdot \nabla n_{m} = \Delta n_{m}-\nabla \cdot(n_{m} \cdot \nabla c_{m})+\gamma n_{m} -\mu n_{m}^{2}, &&\text {in } Q, \\ &c_{mt}+u_{m} \cdot \nabla c_{m} = \Delta c_{m}-c_{m}+n_{m}+f_{m}, &&\text{in } Q, \\ &u_{mt}+u_{m} \cdot \nabla u_{m} = \Delta u_{m}-\nabla \pi+n_{m} \nabla \varphi, &&\text{in } Q, \\ &\nabla \cdot u_{m} = 0, &&\text{in } Q, \\ &\frac{\partial n_{m}}{\partial \nu}\Big|_{\partial\Omega} = \frac{\partial c_{m}}{\partial \nu}\Big|_{\partial\Omega} = 0,\quad u_{m}|_{\partial\Omega} = 0, \\ &n_{m}(0) = n_0, c_{m}(0) = c_0, u_{m}(0) = u_0,&&\text{in }\Omega. \end{aligned} \right. \end{align}$

(4.3)

Observing that $\mathcal{U}$ is a closed convex subset of $L^2(0,T;H^1(\Omega_c))$ . According to the definition of $\mathcal{S}_{ad}$ , we deduce that there exists $(\bar{n},\bar{c},\bar{u},\bar{f})$ bounded in $\mathcal{H}$ such that, for subsequence of $(n_m,c_m,u_m, f_m)_{m\in \mathbb{N}}$ , for convenience, still denoted by $(n_m,c_m,u_m,f_m)$ , as $m\rightarrow +\infty$

$\begin{align*} &n_{m} \rightarrow \bar{n}, \text { weakly in } L^{2}\left(0, T ; H^{2}(\Omega)\right) \text { and weakly* } \text { in } L^{\infty}\left(0, T ; H^{1}(\Omega)\right), \\ &c_{m} \rightarrow \bar{c}, \text { weakly in } L^{2}\left(0, T ; H^{3}(\Omega)\right) \text { and weakly* } \text { in } L^{\infty}\left(0, T ; H^{2}(\Omega)\right), \\ &u_{m} \rightarrow \bar{u}, \text { weakly in } L^{2}\left(0, T ; H^{2}(\Omega)\right) \text { and weakly* } \text { in } L^{\infty}\left(0, T ; H^{1}(\Omega)\right), \\ &f_{m} \rightarrow \bar{f}, \text { weakly in } L^2(0,T;H^1(\Omega_c)), \text { and } \tilde{f} \in \mathcal{U}. \end{align*}$

According to the Aubin-Lions lemma [16] and the compact embedding theorems, we obtain

$\begin{align*} &n_{m} \rightarrow \bar{n}, \quad \text { strongly in } C\left([0, T] ; L^{2}(\Omega)\right) \cap L^{2}\left(0, T ; H^{1}(\Omega)\right), \\ &c_{m} \rightarrow \bar{c}, \quad \text{ strongly in } C\left([0, T] ; H^{1}(\Omega)\right) \cap L^{2}\left(0, T ; H^{2}(\Omega)\right), \\ &u_{m} \rightarrow \bar{u}, \quad \text { strongly in } C\left([0, T] ; L^{2}(\Omega)\right) \cap L^{2}\left(0, T ; H^{1}(\Omega)\right). \end{align*}$

Since $\nabla \cdot\left(n_{m} \nabla c_{m}\right) = \nabla n_{m} \cdot \nabla c_{m}+n_{m} \Delta c_{m}$ is bounded in $L^2(0,T;L^2(\Omega))$ , then

$\begin{align*} \nabla \cdot\left(n_{m} \nabla c_{m}\right)\rightarrow \chi, \text{ weakly in } L^2(0,T;L^2(\Omega)). \end{align*}$

Recalling that

$\begin{align*} n_{m} \nabla c_{m}\rightarrow \bar{n}\nabla\bar{c}, \text{ weakly in } L^{\infty}(0,T; L^2(\Omega)). \end{align*}$

Therefore, we get that $\chi = \nabla(\bar{n}\nabla\bar{c})$ . Owing to $(\bar{n},\bar{c},\bar{u},\bar{f})\in \mathcal{H}$ , we see that $(\bar{n},\bar{c},\bar{u},\bar{f})$ is solution of the system (1.1), along with (4.2) implies that

$\begin{align*} \lim\limits_{m\rightarrow +\infty}J(n_m,c_m,u_m,f_m) = \inf\limits_{(u,c,u,f)\in \mathcal{S}_{ad}}J(u,c,u,f)\leq J(\bar{n},\bar{c},\bar{u},\bar{f}). \end{align*}$

On the other hand, we deduce from the weak lower semicontinuity of the cost functional

$\begin{align*} J(\bar{n},\bar{c},\bar{u},\bar{f})\leq \liminf\limits_{m\rightarrow +\infty} J(n_m,c_m,u_m,f_m). \end{align*}$

Therefore, this implies that $(\bar{n},\bar{c},\bar{u},\bar{f})$ is an optimal pair for problem (1.1).

5. The first-order necessary optimality condition

In order to derive the first-order necessary optimality conditions for a local optimal solution of problem (4.1). To this end, we will use a result on existence of Lagrange multipliers in Banach spaces ([20]). First, we discuss the following problem

$\begin{align} \min J(s) \text{ subject to } s\in \mathcal{S} = \{s\in \mathcal{H}:G(s)\in \mathcal{N}\}, \end{align}$

(5.1)

where $J:X\rightarrow \mathbb{R}$ is a functional, $G:X\rightarrow Y$ is an operator, $X$ and $Y$ are Banach spaces, and nonempty closed convex set $\mathcal{H}$ is subset of $X$ and nonempty closed convex cone $\mathcal{N}$ with vertex at the origin in $Y$ .

$A^+$ denotes its polar cone

$\begin{align*} A^+ = \{\rho\in X^{\prime}:\langle \rho,a\rangle_{X^{\prime}}\geq 0, \forall a \in A\}. \end{align*}$

We consider the following Banach spaces

$\begin{align*} &X = V_n\times V_c\times V_u\times L^2(0,T;H^1(\Omega_c)), \\ &Y = L^2(Q)\times L^2(0,T;H^1(\Omega))\times L^2(Q)\times H^1(\Omega)\times H^2(\Omega) \times H^1(\Omega), \end{align*}$

where

$\begin{align*} &V_n = \{n \in Y_n:\frac{\partial n}{\partial \nu}\text{ on } (0,T)\times \partial \Omega \}, \\ &V_c = \{n \in Y_c:\frac{\partial c}{\partial \nu}\text{ on } (0,T)\times \partial \Omega \}, \\ &V_u = \{n \in Y_u: u = 0 \text{ on } (0,T)\times \partial \Omega \text{ and } \nabla \cdot u = 0 \text{ in } (0,T)\times \Omega\} \end{align*}$

and the operator $G = (G_1,G_2,G_3,G_4,G_5,G_6):X\rightarrow Y$ , where

$\begin{align*} &G_1: X \rightarrow L^2(Q),&& G_2: X \rightarrow L^2(0,T;H^1(\Omega)),&&&G_3: X \rightarrow L^2(Q), \\ &G_4: X \rightarrow H^1(\Omega),&&G_5: X \rightarrow H^2(\Omega),&&&G_6: X \rightarrow H^1(\Omega), \end{align*}$

which are defined at each point $s = (n,c,u,f)\in X$ by

$\begin{align} \left\{\begin{aligned} &G_1 = n_{t}+u \cdot \nabla n-\Delta n+\nabla \cdot(n \cdot \nabla c)-\gamma n+\mu n^{2}, \\ &G_2 = c_{t}+u \cdot \nabla c-\Delta c+c-n-f, \\ &G_3 = u_{t}+u \cdot \nabla u-\Delta u+\nabla \pi-n \nabla \varphi, \\ &G_4 = n(0)-n_0, \\ &G_5 = c(0)-c_0, \\ &G_6 = u(0)-u_0. \end{aligned} \right. \end{align}$

(5.2)

The function spaces are given as follows

$\begin{align*} \mathcal{H} = V_n\times V_c \times V_u \times \mathcal{U}. \end{align*}$

We see that $\mathcal{H}$ is a closed convex subset of $X$ and $\mathcal{N} = \{0\}$ , and rewrite the optimal control problem

$\begin{align} \min J(s) \text{ subject to } s\in \mathcal{S}_{ad} = \{s\in \mathcal{H}:G(s) = 0\}. \end{align}$

(5.3)

Taking the differentiability of $J$ and $G$ into consider, it follows that

Lemma 5.1. The functional $J:X\rightarrow R$ is Fréchet differentiable and the Fréchet derivative of $J$ in $\bar{s} = (\bar{n},\bar{c},\bar{u},\bar{f})\in X$ in the direction $r = (\tilde{n},\tilde{c},\tilde{u}, \tilde{f})$ is given by

$\begin{align} J^{\prime}(\bar{s})[r] = &\beta_1\int_{0}^T\int_{\Omega_d}(\bar{n}-n_d)\tilde{n}d x d t+\beta_2\int_{0}^T\int_{\Omega_d}(\bar{c}-c_d)\tilde{c}d x d t \\ &+\beta_3\int_{0}^T\int_{\Omega_d}(\bar{u}-u_d)\tilde{u}(T)d x d t +\beta_4\int_{\Omega_d} (\bar{n}(T)-n_{\Omega})\tilde{n}(T)d x \\ &+\beta_5\int_{\Omega_d} (\bar{c}(T)-c_{\Omega})\tilde{c}d x+\beta_6\int_{\Omega_d} (\bar{u}(T)-u_{\Omega})\tilde{u}(T)d x \\ &+\beta_7\int_{0}^T\int_{\Omega_d}\bar{f}\tilde{f} d x d t. \end{align}$

(5.4)

Lemma 5.2. The operator $G:X\rightarrow Y$ is continuous-Fréchet differentiable and the Fréchet derivative of $J$ in $\bar{s} = (\bar{n},\bar{c},\bar{u},\bar{f})\in X$ in the direction $r = (\tilde{n},\tilde{c},\tilde{u}, \tilde{f})$ , is the linear operator

$\begin{align*} G^{\prime}(\bar{s})[r] = (G_1^{\prime}(\bar{s})[r],G_2^{\prime}(\bar{s})[r],G_3^{\prime}(\bar{s})[r],G_4^{\prime}(\bar{s})[r],G_5^{\prime}(\bar{s})[r],G_6^{\prime}(\bar{s})[r]) \end{align*}$

defined by

$\begin{align*} \left\{\begin{aligned} &G_1^{\prime}(\bar{s})[r] = \tilde{n}_{t}-\Delta \tilde{n}+\bar{u} \cdot \nabla \tilde{n}+\tilde{u}\nabla \bar{n}+\nabla \cdot(\bar{n} \nabla \tilde{c}) \\ &+\nabla(\tilde{n}\nabla \bar{c})-\gamma \tilde{n} +2\mu \tilde{n}\bar{n}, &&\mathit{\text{in}}\;Q , \\ &G_2^{\prime}(\bar{s})[r] = \tilde{c}_{t}-\Delta \tilde{c}+\bar{u} \cdot \nabla \tilde{c}+\tilde{u}\cdot\nabla \bar{c}+\tilde{c}- \tilde{n}-\tilde{f},&&\mathit{\text{in}}\;Q , \\ &G_3^{\prime}(\bar{s})[r] = \tilde{u}_{t}-\Delta \tilde{u}+\bar{u} \cdot \nabla \tilde{u}+\tilde{u}\cdot \nabla \bar{u}-\tilde{n} \nabla \varphi,&&\mathit{\text{in}}\;Q , \\ &\nabla\cdot \tilde{u} = 0,&&\mathit{\text{in}}\;Q , \\ &\frac{\partial \tilde{n}}{\partial \nu} = \frac{\partial \tilde{c}}{\partial \nu} = 0, \tilde{u} = 0, &&\mathit{\text{on}}\;(0,T)\times\partial\Omega, \\ &\tilde{n}(0) = \tilde{n}_0, \tilde{c}(0) = \tilde{c}_0,\tilde{u}(0) = \tilde{u}_0, &&\mathit{\text{in}}\;\Omega. \end{aligned} \right. \end{align*}$

Lemma 5.3. Let $\bar{s} = (\bar{n},\bar{c},\bar{u},\bar{f})\in \mathcal{S}_{ad}$ , then $\bar{s}$ is a regular point.

Proof. For any fixed $(\bar{n},\bar{c},\bar{u},\bar{f})\in \mathcal{S}_{ad}$ , we set $(g_n,g_c,g_u,\tilde{n}_0, \tilde{c}_0, \tilde{u}_0)\in Y$ . Since $0\in \mathcal{C}(\bar{f})$ , it suffices to show the existence of $(\tilde{n}, \tilde{c}, \tilde{u})\in Y_n\times Y_c\times Y_u$ such that

$\begin{align} \left\{\begin{aligned} &\tilde{n}_{t}-\Delta \tilde{n}+\bar{u} \cdot \nabla \tilde{n}+\tilde{u}\nabla \bar{n}+\nabla \cdot(\bar{n} \nabla \tilde{c}) \\ &+\nabla(\tilde{n}\nabla \bar{c})-\gamma \tilde{n} +2\mu \tilde{n}\bar{n} = g_n, &&\text{in }Q , \\ &\tilde{c}_{t}-\Delta \tilde{c}+\bar{u} \cdot \nabla \tilde{c}+\tilde{u}\cdot\nabla \bar{c}+\tilde{c}- \tilde{n} = g_c,&&\text{in }Q , \\ &\tilde{u}_{t}-\Delta \tilde{u}+\bar{u} \cdot \nabla \tilde{u}+\tilde{u}\cdot \nabla \bar{u}-\tilde{n} \nabla \varphi = g_u,&&\text{in }Q , \\ &\nabla\cdot \tilde{u} = 0,&&\text{in }Q , \\ &\frac{\partial \tilde{n}}{\partial \nu} = \frac{\partial \tilde{c}}{\partial \nu} = 0, \tilde{u} = 0, &&\text{on }(0,T)\times\partial\Omega, \\ &\tilde{n}(0) = \tilde{n}_0, \tilde{c}(0) = \tilde{c}_0,\tilde{u}(0) = \tilde{u}_0, &&\text{in }\Omega. \end{aligned} \right. \end{align}$

(5.5)

Next, we use Leray-Schauder's fixed point method to prove the existence of solutions of the problem (5.5), the operator $T:(\dot{n},\dot{u})\in X_n\times X_u\rightarrow (\tilde{n},\tilde{u})\in Y_n\times Y_u$ with $(\tilde{n}, \tilde{c},\tilde{u})$ solving the decoupled problem:

(5.6)

The system (5.6) is complemented by the corresponding Neumann boundary and initial conditions. Similar to the proof of Lemmas 2.3, 2.4, 2.5 and 2.6, we conduct that operator $T:X_n\times X_u\rightarrow X_n\times X_u$ is well-defined and compact.

Similar to the proof of Theorem 3.1, $(\tilde{n},\tilde{u})$ solves the coupled problem $(\bar{n},\bar{c},\bar{u},\bar{f})$ $\in \mathcal{S}_{ad}$ , and we set $(g_n,g_c,g_u,\tilde{n}_0, \tilde{c}_0, \tilde{u}_0)\in Y$ . Since $0\in \mathcal{C}(\bar{f})$ , it suffices to show the existence of $(\tilde{n}, \tilde{c}, \tilde{u})\in Y_n\times Y_c\times Y_u$ such that

$\begin{align} \left\{\begin{aligned} &\tilde{n}_{t}-\Delta \tilde{n}+\tilde{n} = -\bar{u} \cdot \nabla \tilde{n}-\tilde{u}\cdot\nabla \bar{n}-\nabla \cdot(\bar{n} \nabla \tilde{c}) \\ &-\nabla(\tilde{n}\nabla \bar{c})+\alpha(\gamma+1) \tilde{n} -2\mu \tilde{n}\bar{n}+\alpha g_n, &&\text{in }Q , \\ &\tilde{c}_{t}-\Delta \tilde{c}+\tilde{c} = -\bar{u} \cdot \nabla \tilde{c}-\tilde{u}\cdot\nabla \bar{c}+ \alpha \tilde{n}+\alpha g_c,&&\text{in }Q , \\ &\tilde{u}_{t}-\Delta \tilde{u} = -\bar{u} \cdot \nabla \tilde{u}-\tilde{u}\cdot \nabla \bar{u}+\alpha\tilde{n} \nabla \varphi+\alpha g_u,&&\text{in }Q, \end{aligned} \right. \end{align}$

(5.7)

complemented by the corresponding Neumann boundary and initial conditions.

Taking the $L^2$ -inner product with $\tilde{u}$ for the third equation of (5.7) implies

$\begin{align*} \frac{1}{2}\int_{\Omega} \tilde{u}^2 d x+\int_{\Omega} |\nabla \tilde{u}|^2 d x = \alpha\int_{\Omega} \tilde{n}\nabla \varphi \tilde{u} d x +\alpha \int_{\Omega} \tilde{u} g_u d x. \end{align*}$

By the Poincaré inequality and Young's inequality, we have

$\begin{align} \frac{d}{d t}\|\tilde{u}\|^2_{L^2}+\|\tilde{u}\|^2_{H^1}\leq C(\|\tilde{n}\|^2_{L^2}+\|g_u\|^2_{L^2})+C\|\tilde{u}\|^2_{L^2}. \end{align}$

(5.8)

Taking the $L^2$ -inner product with $\tilde{c}$ for the second equation of (5.7) implies

$\begin{align*} &\frac{1}{2}\int_{\Omega} \tilde{c}^2 d x+\int_{\Omega} |\nabla \tilde{c}|^2 d x+\int_{\Omega} \tilde{c}^2 d x \\ = &\int_{\Omega} \tilde{u}\nabla \bar{c} \tilde{c} d x+ \alpha\int_{\Omega} \tilde{n} \tilde{c} d x +\alpha\int_{\Omega} g_c \tilde{c} d x. \end{align*}$

With the Poincaré inequality and Young's inequality in hand, we see that

$\begin{align} \frac{d}{d t}\|\tilde{c}\|^2_{L^2}+\|\tilde{c}\|^2_{H^1}\leq C(\|\tilde{n}\|^2_{L^2}+\|g_c\|^2_{L^2})+C\|\tilde{c}\|^2_{L^2}. \end{align}$

(5.9)

Taking the $L^2$ -inner product with $-\Delta \tilde{c}$ for the second equation of (5.7) implies

$\begin{align*} &\frac{1}{2}\int_{\Omega} |\nabla \tilde{c}|^2 d x+\int_{\Omega} |\Delta \tilde{c}|^2 d x+\int_{\Omega} |\nabla \tilde{c}|^2 d x \\ = &\int_{\Omega} \tilde{u}\nabla \bar{c} \Delta \tilde{c} d x+\int_{\Omega} \bar{u}\nabla \tilde{c} \Delta \tilde{c} d x- \alpha\int_{\Omega} \tilde{n} \Delta \tilde{c} d x -\alpha\int_{\Omega} g_c \Delta \tilde{c} d x \\ = &J_1+J_2+J_3. \end{align*}$

For the term $J_1$

$\begin{align*} J_1 = &\int_{\Omega} \tilde{u}\nabla \bar{c} \Delta \tilde{c} d x \leq \|\Delta \tilde{c}\|_{L^2}\|\nabla \bar{c}\|_{L^4}\|\tilde{u}\|_{L^4} \\ \leq &\frac{1}{6}\|\Delta \tilde{c}\|^2_{L^2}+C\|\nabla \bar{c}\|^2_{H^1}\|\tilde{u}\|^2_{H^1}. \end{align*}$

For the term $J_2$ , we see that

$\begin{align*} J_2 = & \int_{\Omega} \bar{u}\nabla \tilde{c} \Delta \tilde{c} d x = - \frac{1}{2} \int_{\Omega} \nabla \bar{u}|\nabla \tilde{c}|^2 d x \\ \leq&\|\nabla\bar{u}\|_{L^2}\|\nabla \tilde{c}\|^2_{L^4} \\ \leq& \|\nabla\bar{u}\|_{L^2}(\|\nabla \tilde{c}\|^{\frac{1}{2}}_{L^2}\|\Delta \tilde{c}\|^{\frac{1}{2}}_{L^2}+\|\nabla \tilde{c}\|_{L^2}) \\ \leq&\frac{1}{6}\|\Delta \tilde{c}\|^2_{L^2}+C\|\nabla \tilde{c}\|^2_{L^2}. \end{align*}$

For the term $J_3$ , we get

$\begin{align*} J_3 = &- \alpha\int_{\Omega} \tilde{n} \Delta \tilde{c} d x -\alpha\int_{\Omega} g_c \Delta \tilde{c} d x \\ \leq&\frac{1}{6}\|\Delta \tilde{c}\|^2_{L^2}+C(\|\tilde{n}\|^2_{L^2}+\|g_c\|^2_{L^2}). \end{align*}$

Therefore, combining $J_1$ , $J_2$ and $J_3$ , we have

$\begin{align} \frac{d}{d t}\|\nabla \tilde{c}\|^2_{L^2}+\|\nabla \tilde{c}\|^2_{H^1} \leq C\|\nabla \tilde{c}\|^2_{L^2}+ C(\|\tilde{n}\|^2_{L^2}+\|g_c\|^2_{L^2}). \end{align}$

(5.10)

Taking the $L^2$ -inner product with $\tilde{n}$ for the first equation of (5.7) implies

$\begin{align*} &\frac{d}{d t}\int_{\Omega} \tilde{n}^2 d x+\int_{\Omega} |\nabla\tilde{n}|^2 d x+ \int_{\Omega} \tilde{n}^2 d x \\ = &-\int_{\Omega}\tilde{u}\nabla\bar{n}\tilde{n}d x+\int_{\Omega}\nabla \tilde{n}\bar{n}\nabla\tilde{c} d x+\int_{\Omega}\nabla \tilde{n}\tilde{n}\nabla\bar{c} d x+\alpha (\gamma +1)\int_{\Omega} \tilde{n}^2 d x \\ &+2\mu \int_{\Omega}\bar{n}\tilde{n}^2 d x+\alpha \int_{\Omega}\tilde{n} g_n d x \\ = &J_4+J_5+J_6+J_7. \end{align*}$

For the term $J_4$ , by Gagliardo-Nirenberg interpolation inequality, we have

$\begin{align*} J_4 = &-\int_{\Omega}\tilde{u}\nabla\bar{n}\tilde{n}d x\leq \|\tilde{u}\|_{L^4}\|\nabla\bar{n}\|_{L^2}\|\tilde{n}\|_{L^4} \\ \leq& C (\|\nabla\tilde{u}\|^{{\frac{1}{2}}}_{L^2}\|\tilde{u}\|^{\frac{1}{2}}_{L^2}+\|\tilde{u}\|_{L^2}) \|\nabla\bar{n}\|_{L^2}\|\tilde{n}\|_{H^1} \\ \leq &\delta \|\tilde{n}\|^2_{H^1}+C\|\nabla\tilde{u}\|_{L^2}\|\tilde{u}\|_{L^2}+C\|\tilde{u}\|^2_{L^2} \\ \leq &\delta \|\tilde{n}\|^2_{H^1}+\delta \|\nabla\tilde{u}\|^2_{L^2}+C\|\tilde{u}\|^2_{L^2}. \end{align*}$

For the term $J_5$ ,

$\begin{align*} J_5 = &\int_{\Omega}\nabla \tilde{n}\bar{n}\nabla\tilde{c} d x \leq \|\nabla\tilde{n}\|_{L^2} \|\bar{n}\|_{L^4}\|\nabla\tilde{c}\|_{L^4} \\ \leq &\|\nabla\tilde{n}\|_{L^2}\|\bar{n}\|_{H^1}(\|\nabla\tilde{c}\|^{\frac{1}{2}}_{L^2}\|\Delta\tilde{c}\|^{\frac{1}{2}}_{L^2}+\|\nabla\tilde{c}\|_{L^2}) \\ \leq&\delta \|\nabla\tilde{n}\|^2_{L^2}+\|\nabla\tilde{c}\|_{L^2}\|\Delta\tilde{c}\|_{L^2}+C\|\nabla\tilde{c}\|^2_{L^2} \\ \leq& \delta \|\nabla\tilde{n}\|^2_{L^2}+\delta \|\Delta\tilde{c}\|_{L^2} +C\|\nabla\tilde{c}\|^2_{L^2}. \end{align*}$

For the term $J_6$ ,

$\begin{align*} J_6 = &\int_{\Omega}\nabla \tilde{n}\tilde{n}\nabla\bar{c} d x\leq \|\tilde{n}\|^2_{L^4}\|\Delta \bar{c}\|_{L^2} \\ \leq&(\|\tilde{n}\|^{\frac{1}{2}}_{L^2}\|\nabla \tilde{n}\|^{\frac{1}{2}}_{L^2}+\| \tilde{n}\|_{L^2})\|\Delta \bar{c}\|_{L^2} \\ \leq&\delta \|\nabla \tilde{n}\|^2_{L^2}+C\|\tilde{n}\|^2_{L^2}+C. \end{align*}$

For the term $J_7$ ,

$\begin{align*} J_7 = &\alpha (\gamma +1)\int_{\Omega} \tilde{n}^2 d x+2\mu \int_{\Omega}\bar{n}\tilde{n}^2 d x+\alpha \int_{\Omega}\tilde{n} g_n d x \\ \leq &(\gamma +1)\|\tilde{n}\|^2_{L^2}+\|g_n\|_{L^2}\|\tilde{n}\|_{L^2}+\|\bar{n}\|_{L^2}\|\tilde{n}\|^2_{L^4} \\ \leq& (\gamma +1)\|\tilde{n}\|^2_{L^2}+\|g_n\|_{L^2}\|\tilde{n}\|_{L^2}+\|\bar{n}\|_{L^2}(\|\tilde{n}\|^{\frac{1}{2}}_{L^2}\|\nabla\tilde{n}\|^{\frac{1}{2}}_{L^2}+\|\tilde{n}\|_{L^2} ) \\ \leq& \delta\|\nabla\tilde{n}\|_{L^2}+ C\|\tilde{n}\|^2_{L^2}+C\|g_n\|^2_{L^2}. \end{align*}$

Therefore, by choosing $\delta$ small enough, from $J_4$ , $J_5$ , $J_6$ and $J_7$ , it follows that

$\begin{align} &\frac{d}{d t}\|\tilde{n}\|^2_{L^2}+\|\tilde{n}\|^2_{H^1} \\ \leq& C(\|\tilde{n}\|^2_{L^2}+\|\nabla\tilde{c}\|^2_{L^2}+\|\tilde{u}\|^2_{L^2}) +\delta \|\Delta\tilde{c}\|_{L^2} +\delta \|\nabla\tilde{u}\|^2_{L^2}+C\|g_n\|^2_{L^2}. \end{align}$

(5.11)

By choosing $\delta$ small enough and combining (5.8)-(5.11), we get

$\begin{align*} &\frac{d}{d t}(\|\tilde{n}\|^2_{L^2}+\|\tilde{c}\|^2_{H^1}+\|\tilde{u}\|^2_{L^2})+\|\tilde{n}\|^2_{H^1}+\|\tilde{c}\|^2_{H^2}+\|\tilde{u}\|^2_{H^1} \\ \leq& C(\|g_n\|^2_{L^2}+\|g_c\|^2_{L^2}+\|g_u\|^2_{L^2})+C(\|\tilde{n}\|^2_{L^2}+\|\tilde{c}\|^2_{H^1}+\|\tilde{u}\|^2_{L^2}). \end{align*}$

Applying Gronwall's lemma to the resulting differential inequality, we obatin

$\begin{align} \|\tilde{n}\|^2_{L^2}+\|\tilde{c}\|^2_{H^1}+\|\tilde{u}\|^2_{L^2} +\int_0^t\|\tilde{n}\|^2_{H^1}d\tau+\int_0^t\|\tilde{c}\|^2_{H^2}d\tau+\int_0^t\|\tilde{u}\|^2_{H^1}d\tau\leq C. \end{align}$

(5.12)

Taking the $L^2$ -inner product with $-\Delta\tilde{u}$ for the third equation of (5.7) implies

$\begin{align*} &\frac{1}{2} \frac{d}{d t}\int_{\Omega} |\nabla\tilde{u}|^2 d x+ \int_{\Omega} |\Delta\tilde{u}|^2 d x \\ = &\int_{\Omega}\bar{u}\cdot\nabla \tilde{u} \Delta\tilde{u} d x+\int_{\Omega}\tilde{u}\cdot\nabla\bar{u}\Delta \tilde{u} d x-\alpha \int_{\Omega} \tilde{n} \nabla \varphi \Delta \tilde{u} d x-\alpha \int_{\Omega}g_u \Delta \tilde{u} d x \\ = &J_8+J_9+J_{10}. \end{align*}$

With the use of the Gagliardo-Nirenberg interpolation inequality, we derive

$\begin{align*} J_8 = &\int_{\Omega}\bar{u}\cdot\nabla \tilde{u} \Delta\tilde{u} d x \leq \|\bar{u}\|_{L^4} \|\nabla\tilde{u}\|_{L^4}\|\Delta\tilde{u}\|_{L^2} \\ \leq&\|\bar{u}\|_{H^1}(\|\nabla\tilde{u}\|^{\frac{1}{2}}_{L^2}\|\Delta\tilde{u}\|^{\frac{1}{2}}_{L^2}+\|\nabla\tilde{u}\|_{L^2})\|\Delta\tilde{u}\|_{L^2} \\ \leq& \delta\|\Delta \tilde{u}\|^2_{L^2}+C\|\nabla\tilde{u}\|^2_{L^2} \end{align*}$

and

$\begin{align*} J_9 = &\int_{\Omega}\tilde{u}\cdot\nabla\bar{u}\Delta \tilde{u} d x\leq \|\Delta \tilde{u}\|_{L^2}\|\nabla\bar{u}\|_{L^4}\|\tilde{u}\|_{L^4} \\ \leq& C\|\Delta \tilde{u}\|_{L^2}\|\nabla\bar{u}\|_{H^1}(\|\nabla\tilde{u}\|^{\frac{1}{2}}_{L^2}\|\tilde{u}\|^{\frac{1}{2}}_{L^2}+\|\tilde{u}\|_{L^2}) \\ \leq& \delta\|\Delta \tilde{u}\|^2_{L^2}+C\|\nabla\tilde{u}\|^2_{L^2}. \end{align*}$

For the term $J_{10}$ , we deduce

$\begin{align*} J_{10} = &\alpha \int_{\Omega} \tilde{n} \nabla \varphi \Delta \tilde{u} d x-\alpha \int_{\Omega}g_u \Delta \tilde{u} d x \\ \leq&\delta\|\Delta \tilde{u}\|^2_{L^2}+C(\|\tilde{n}\|^2_{L^2}+\|g_u\|^2_{L^2}) . \end{align*}$

By choosing $\delta$ small enough, with the estimates $J_8$ , $J_9$ and $J_{10}$ , we have

$\begin{align} \frac{d}{d t}\|\nabla\tilde{u}\|^2_{L^2}+\|\Delta \tilde{u}\|^2_{L^2}\leq C\|\nabla\tilde{u}\|^2_{L^2}+C\|g_u\|^2_{L^2}. \end{align}$

(5.13)

Applying $\nabla$ to the first equation of (5.7), multiplying it by $\nabla \Delta \tilde{c}$ , and integrating over $\Omega$ give

$\begin{align*} &\frac{1}{2}\frac{d}{ d t}\int_{\Omega} |\Delta c|^2 d x + \int_{\Omega}|\nabla\Delta c|^2 d x+\int_{\Omega} |\Delta c|^2 d x \\ = &-\int_{\Omega}\nabla (\bar{u}\nabla\tilde{c})\nabla\Delta\tilde{c}d x-\int_{\Omega}\nabla(\tilde{u}\nabla\bar{c})\nabla\Delta\tilde{c} d x+\alpha\int_{\Omega}\nabla \tilde{n}\nabla\Delta\tilde{c} d x \\ &+\alpha \int_{\Omega}\nabla g_c\nabla\Delta\tilde{c} d x \\ = &J_{11}+J_{12}+J_{13}. \end{align*}$

For the first term $J_{11}$ , we have

$\begin{align*} J_{11} = &-\int_{\Omega}\nabla (\bar{u}\nabla\tilde{c})\nabla\Delta\tilde{c}d x = - \int_{\Omega}\nabla \bar{u}\nabla\tilde{c} \nabla\Delta\tilde{c}d x-\int_{\Omega}\bar{u}\Delta\tilde{c} \nabla\Delta\tilde{c}d x \\ \leq& \|\nabla\Delta\tilde{c}\|_{L^2}\|\nabla \bar{u}\|_{L^4}\|\nabla\tilde{c}\|_{L^4}+\|\nabla\Delta\tilde{c}\|_{L^2}\|\bar{u}\|_{L^4}\|\Delta\tilde{c}\|_{L^4} \\ \leq &\|\nabla\Delta\tilde{c}\|_{L^2} (\|\nabla \bar{u}\|^{\frac{1}{2}}_{L^2}\|\Delta \bar{u}\|^{\frac{1}{2}}_{L^2}+\|\nabla \bar{u}\|_{L^2})(\|\nabla \bar{c}\|^{\frac{1}{2}}_{L^2}\|\Delta \bar{c}\|^{\frac{1}{2}}_{L^2}+\|\nabla \bar{c}\|_{L^2}) \\ &+\|\nabla\Delta\tilde{c}\|_{L^2}\|\bar{u}\|_{H^1}(\|\nabla\Delta\tilde{c}\|^\frac{1}{2}_{L^2}\|\Delta\tilde{c}\|^\frac{1}{2}_{L^2}+\|\Delta\tilde{c}\|_{L^2}) \\ \leq&\delta\|\nabla\Delta\tilde{c}\|^2_{L^2}+C\|\Delta \bar{u}\|^2_{L^2}+C\|\Delta \tilde{c}\|^2_{L^2}. \end{align*}$

Similarly, for the term $J_{12}$ ,

$\begin{align*} J_{12} = &-\int_{\Omega}\nabla(\tilde{u}\nabla\bar{c})\nabla\Delta\tilde{c} d x = -\int_{\Omega}\nabla\tilde{u}\nabla\bar{c} \nabla\Delta\tilde{c} d x-\int_{\Omega}\tilde{u}\Delta\bar{c}\nabla\Delta\tilde{c} d x \\ \leq&\|\nabla\Delta\tilde{c}\|_{L^2}\|\nabla \tilde{u}\|_{L^4}\|\nabla \bar{c}\|_{L^4}+\|\tilde{u}\|_{L^4}\|\Delta\bar{c}\|_{L^4}\|\nabla\Delta\tilde{c}\|_{L^2} \\ \leq &C\|\nabla\Delta\tilde{c}\|_{L^2}(\|\nabla \tilde{u}\|^{\frac{1}{2}}_{L^2}\|\Delta \tilde{u}\|^{\frac{1}{2}}_{L^2}+\|\nabla \tilde{u}\|_{L^2})\|\nabla \bar{c}\|_{H^1} \\ &+(\|\tilde{u}\|^{\frac{1}{2}}_{L^2}\|\nabla\tilde{u}\|^{\frac{1}{2}}_{L^2}+\|\tilde{u}\|_{L^2})(\|\Delta\bar{c}\|^{\frac{1}{2}}_{L^2}\|\nabla\Delta\bar{c}\|^{\frac{1}{2}}_{L^2}+\|\Delta\bar{c}\|_{L^2})\|\nabla\Delta\tilde{c}\|_{L^2} \\ \leq&\delta\|\nabla\Delta\tilde{c}\|^2_{L^2}+\delta \|\Delta \tilde{u}\|^2_{L^2}+C \|\nabla\Delta\bar{c}\|^2_{L^2}+C\|\nabla\tilde{u}\|^2_{L^2}. \end{align*}$

For the rest term $J_{13}$ , we see

$\begin{align*} J_{13} = &\alpha\int_{\Omega}\nabla \tilde{n}\nabla\Delta\tilde{c} d x +\alpha \int_{\Omega}\nabla g_c\nabla\Delta\tilde{c} d x \\ \leq &\delta\|\nabla\Delta\tilde{c}\|^2_{L^2}+C(\|\nabla \tilde{n}\|^2_{L^2}+\|\nabla g_c\|^2_{L^2}). \end{align*}$

By choosing $\delta$ small enough, we get

$\begin{align} &\frac{d}{d t}\|\Delta\tilde{c}\|^2_{L^2}+\|\Delta\tilde{c}\|^2_{H^1} \\ \leq& C(\|\nabla \tilde{n}\|^2_{L^2}+\|\Delta \tilde{c}\|^2_{L^2}+\|\nabla\tilde{u}\|^2_{L^2})+ C \|\Delta \bar{u}\|^2_{L^2}+\delta \|\Delta \tilde{u}\|^2_{L^2} \\ &+C \|\nabla\Delta\bar{c}\|^2_{L^2}+C\|\nabla g_c\|^2_{L^2}. \end{align}$

(5.14)

From (5.13) and (5.14), along with $\delta$ small enough, it follows that

$\begin{align*} &\frac{d}{d t}(\|\nabla\tilde{u}\|^2_{L^2}+\|\Delta\tilde{c}\|^2_{L^2})+\|\Delta \tilde{u}\|^2_{L^2}+\|\Delta\tilde{c}\|^2_{H^1} \\ \leq& C(\|\nabla\tilde{u}\|^2_{L^2}+\|\Delta\tilde{c}\|^2_{L^2})+(\|\nabla \tilde{n}\|^2_{L^2}+\|\Delta \bar{u}\|^2_{L^2}+ \|\nabla\Delta\bar{c}\|^2_{L^2}+\|\nabla g_c\|^2_{L^2}) +C\|g_u\|^2_{L^2}. \end{align*}$

Applying Gronwall's lemma to the resulting differential inequality, we know

$\begin{align*} \|\nabla\tilde{u}\|^2_{L^2}+\|\Delta\tilde{c}\|^2_{L^2}+\int_{0}^t \|\Delta \tilde{u}\|^2_{L^2} d\tau+\int_{0}^t \|\Delta\tilde{c}\|^2_{H^1} d\tau\leq C. \end{align*}$

Taking the $L^2$ -inner product with $-\Delta\tilde{n}$ for the first equation of (5.7) implies

$\begin{align*} &\frac{1}{2}\frac{d}{d t}\int_{\Omega} |\nabla \tilde{n}|^2 d x +\int_{\Omega} |\Delta \tilde{n}|^2 d x+\int_{\Omega} |\nabla \tilde{n}|^2 d x \\ = &-\int_{\Omega}\bar{u}\cdot \nabla\tilde{n} \Delta \tilde{n}d x-\int_{\Omega}\tilde{u}\cdot \nabla\bar{n}\Delta \tilde{n}d x-\int_{\Omega} \nabla(\tilde{n}\nabla \bar{c})\Delta\tilde{n} d x-\int_{\Omega} \nabla(\bar{n}\nabla \tilde{c})\Delta\tilde{n} d x \\ &-\alpha(1+\gamma)\int_{\Omega}\tilde{n}\Delta\tilde{n} d x+2\mu\int_{\Omega}\tilde{n}\bar{n} \Delta\tilde{n}d x-\alpha \int_{\Omega}g_n\Delta\tilde{n} d x \\ = &J_{14}+J_{15}+J_{16}+J_{17}+J_{18}. \end{align*}$

With the Gagliardo-Nirenberg interpolation inequality in hand, we can estimate $J_{14}$ as follows

$\begin{align*} J_{14} = &-\int_{\Omega}\bar{u}\cdot \nabla\tilde{n} \Delta \tilde{n}d x\leq \|\bar{u}\|_{L^4}\|\nabla\tilde{n}\|_{L^4}\|\Delta\tilde{n}\|_{L^2} \\ \leq &C\|\bar{u}\|_{H^1}(\|\nabla\tilde{n}\|^{\frac{1}{2}}_{L^2}\|\Delta\tilde{n}\|^{\frac{1}{2}}_{L^2}+\|\nabla\tilde{n}\|_{L^2})\|\Delta\tilde{n}\|_{L^2} \\ \leq &\delta\|\Delta\tilde{n}\|^2_{L^2}+C\|\nabla\tilde{n}\|^{2}_{L^2}. \end{align*}$

Similar to above estimates, we see

$\begin{align*} J_{15} = &-\int_{\Omega}\tilde{u}\cdot \nabla\bar{n}\Delta \tilde{n}d x\leq \|\tilde{u}\|_{L^4}\|\nabla \bar{n}\|_{L^4}\|\Delta\tilde{n}\|_{L^2} \\ \leq&C\|\tilde{u}\|_{H^1}\|\nabla \bar{n}\|_{H^1}\|\Delta\tilde{n}\|_{L^2} \\ \leq& \delta \|\Delta\tilde{n}\|_{L^2}+C\|\nabla \bar{n}\|^2_{H^1}. \end{align*}$

Similarly, we derive

$\begin{align*} J_{16} = &-\int_{\Omega} \nabla(\tilde{n}\nabla \bar{c})\Delta\tilde{n} d x = -\int_{\Omega}\nabla\tilde{n}\nabla \bar{c}\Delta\tilde{n} d x-\int_{\Omega}\tilde{n}\Delta \bar{c}\Delta\tilde{n} d x \\ \leq &\|\nabla\tilde{n}\|_{L^4}\|\nabla\bar{c}\|_{L^4}\|\Delta\tilde{n}\|_{L^2}+\|\tilde{n}\|_{L^4}\|\Delta\bar{c}\|_{L^4}\|\Delta\tilde{n}\|_{L^2} \\ \leq&(\|\nabla\tilde{n}\|^{\frac{1}{2}}_{L^2}\|\Delta\tilde{n}\|^{\frac{1}{2}}_{L^2} +\|\nabla\tilde{n}\|_{L^2})\|\nabla\bar{c}\|_{H^1}\|\Delta\tilde{n}\|_{L^2} \\ &+(\|\tilde{n}\|^{\frac{1}{2}}_{L^2}\|\nabla\tilde{n}\|^{\frac{1}{2}}_{L^2}+\|\tilde{n}\|_{L^2})(\|\Delta\bar{c}\|^{\frac{1}{2}}_{L^2}\|\nabla\Delta\bar{c}\|^{\frac{1}{2}}_{L^2}+\|\Delta\bar{c}\|_{L^2})\|\Delta\tilde{n}\|_{L^2} \\ \leq&\delta\|\Delta\tilde{n}\|^2_{L^2}+C\|\nabla\tilde{n}\|^{2}_{L^2}+C\|\nabla\Delta\bar{c}\|^{2}_{L^2}+C \end{align*}$

and

$\begin{align*} J_{17} = &-\int_{\Omega} \nabla(\bar{n}\nabla \tilde{c})\Delta\tilde{n} d x = -\int_{\Omega} \nabla\bar{n}\nabla \tilde{c}\Delta\tilde{n} d x-\int_{\Omega} \nabla\bar{n}\Delta \tilde{c}\Delta\tilde{n} d x \\ \leq&\|\nabla\bar{n}\|_{L^4}\|\nabla\tilde{c}\|_{L^4}\|\Delta\tilde{n}\|_{L^2}+\|\bar{n}\|_{L^4}\|\Delta\tilde{c}\|_{L^4}\|\Delta\tilde{n}\|_{L^2} \\ \leq &(\|\nabla\bar{n}\|^{\frac{1}{2}}_{L^2}\|\Delta\bar{n}\|^{\frac{1}{2}}_{L^2}+\|\nabla\bar{n}\|_{L^2})\|\nabla \tilde{c}\|_{H^1}\|\Delta\tilde{n}\|_{L^2} \\ &+\|\bar{n}\|_{H^1}(\|\Delta\tilde{c}\|^{\frac{1}{2}}_{L^2}\|\nabla\Delta\tilde{c}\|^{\frac{1}{2}}_{L^2}+\|\Delta\tilde{c}\|_{L^2})\|\Delta\tilde{n}\|_{L^2} \\ \leq &\delta\|\Delta\tilde{n}\|^2_{L^2}+C\|\nabla\Delta\tilde{c}\|^{2}_{L^2}+C. \end{align*}$

For the rest terms, we know

$\begin{align*} J_{18} = &-\alpha(1+\gamma)\int_{\Omega}\tilde{n}\Delta\tilde{n} d x+2\mu\int_{\Omega}\tilde{n}\bar{n} \Delta\tilde{n}d x-\alpha \int_{\Omega}g_n\Delta\tilde{n} d x \\ \leq&(1+\gamma)\|\tilde{n}\|_{L^2}\|\Delta\tilde{n}\|_{L^2}+2\mu\|\tilde{n}\|_{L^4}\|\bar{n}\|_{L^4}\|\Delta\tilde{n}\|_{L^2}+\|g_n\|_{L^2}\|\Delta\tilde{n}\|_{L^2} \\ \leq& (1+\gamma)\|\tilde{n}\|_{L^2}\|\Delta\tilde{n}\|_{L^2}+C(\|\tilde{n}\|^{\frac{1}{2}}_{L^2}\|\nabla\tilde{n}\|^{\frac{1}{2}}_{L^2}+\|\tilde{n}\|_{L^2})\|\bar{n}\|_{H^1}\|\Delta\tilde{n}\|_{L^2} \\ &+\|g_n\|_{L^2}\|\Delta\tilde{n}\|_{L^2} \\ \leq &\delta\|\Delta\tilde{n}\|^2_{L^2}+C\|\nabla\tilde{n}\|^{2}_{L^2}+C\|g_n\|^2_{L^2}. \end{align*}$

Therefore, Taking $\delta$ small enough and together with $J_{14}-J_{18}$ , we see that

$\begin{align*} &\frac{d}{d t}\|\nabla\tilde{n}\|^{2}_{L^2}+ \|\nabla\tilde{n}\|^{2}_{H^1} \\ \leq& C(\|\nabla\tilde{n}\|^{2}_{L^2} +\|\nabla \bar{n}\|^2_{H^1}+ \|\nabla\Delta\bar{c}\|^{2}_{L^2}+\|\nabla\Delta\tilde{c}\|^{2}_{L^2}+\|g_n\|^2_{L^2})+C. \end{align*}$

Applying Gronwall's lemma to the resulting differential inequality, we know

$\begin{align*} \|\nabla\tilde{n}\|^{2}_{L^2}+\int_0^t \|\nabla\tilde{n}\|^{2}_{H^1}d\tau \leq C. \end{align*}$

Therefore, from Leray-Schauder theorem, we derive the existence of solution for (5.5). Along with the regularity of $(\tilde{n}, \tilde{c}, \tilde{u})$ , the uniqueness of solution can easily get, so we omit the process.

Theorem 5.1. Assume that $\bar{s} = (\bar{n},\bar{c},\bar{u},\bar{f})\in \mathcal{S}_{a d}$ be an optimal solution for the control problem (5.3). Then, there exist Lagrange multipliers $(\lambda, \eta,\rho,\xi,\varphi,\omega)\in L^2(Q)\times (L^2(0,T;H^1(\Omega)))^{\prime}\times L^2(Q)\times (H^1(\Omega))^{\prime}\times(H^2(\Omega))^{\prime}\times(H^1(\Omega))^{\prime}$ such that for all $(\tilde{n}, \tilde{c}, \tilde{u},\tilde{f})\in V_n\times V_c\times V_u\times\mathcal{C}(\bar{f})$ has

$\begin{align} &\beta_1\int_{0}^T\int_{\Omega_d} (\bar{n}-n_d)\tilde{n}d x d t + \beta_2\int_{0}^T\int_{\Omega_d} (\bar{c}-c_d)\tilde{c}d x d t + \beta_3\int_{0}^T\int_{\Omega_d} (\bar{u}-u_d)\tilde{u}d x d t \\ &+\beta_4\int_{\Omega_d} (\bar{n}(T)-n_{\Omega})\tilde{n}(T)d x +\beta_5\int_{\Omega_d} (\bar{c}(T)-c_{\Omega})\tilde{c}(T)d x \\ &-\int_{0}^T\int_{\Omega}(\tilde{n}_{t}-\Delta \tilde{n}+\bar{u} \cdot \nabla \tilde{n}+\tilde{u}\cdot\nabla \bar{n}+\nabla \cdot(\bar{n} \nabla \tilde{c}) +\nabla(\tilde{n}\nabla \bar{c})-\gamma \tilde{n} +2\mu \tilde{n}\bar{n})\lambda d x d t \\ &-\int_{0}^T\int_{\Omega}\left(\tilde{c}_{t}-\Delta \tilde{c}+\bar{u} \cdot \nabla \tilde{c}+\tilde{u}\cdot\nabla \bar{c}+\tilde{c}- \tilde{n}\right)\eta d x d t+\beta_7\int_{0}^T\int_{\Omega_d} \tilde{f}\bar{f}d x d t \\ &-\int_{0}^T\int_{\Omega}\left(\tilde{u}_{t}-\Delta \tilde{u}+\bar{u} \cdot \nabla \tilde{u}+\tilde{u}\cdot \nabla \bar{u}-\tilde{n} \nabla \varphi \right) \rho d x d t +\int_{\Omega}\tilde{n}(0)\xi d x+\int_{\Omega}\tilde{c}(0)\varphi d x \\ &+\int_{\Omega}\tilde{u}(0)\omega d x+\beta_6\int_{\Omega_d} (\bar{u}(T)-u_{\Omega})\tilde{u}(T)d x+\int_{0}^T\int_{\Omega}\tilde{f}\eta d x d t \geq 0, \end{align}$

(5.15)

where $\mathcal{C}(\bar{f}) = \{\theta(f-\bar{f}):\theta\geq 0, f\in \mathcal{U}\}$ .

Proof. With the Lemma 5.3 in hand, we get that $\bar{s}\in\mathcal{S}_{ad}$ is a regular point. Then, togather with Theorem 3.1 in [20], it follows that there exist Lagrange multipliers $(\lambda, \eta,\rho,\xi,\varphi,\omega)\in L^2(Q)\times (L^2(0,T;H^1(\Omega)))^{\prime}\times L^2(Q)\times (H^1(\Omega))^{\prime}\times(H^2(\Omega))^{\prime}\times(H^1(\Omega))^{\prime}$ such that

$\begin{align*} &J^{\prime}(\bar{s})[r]-\langle G_1^{\prime}(\bar{s})[r],\lambda \rangle-\langle G_2^{\prime}(\bar{s})[r],\eta \rangle-\langle G_3^{\prime}(\bar{s})[r],\rho \rangle-\langle G_4^{\prime}(\bar{s})[r],\xi \rangle \\ &-\langle G_5^{\prime}(\bar{s})[r],\varphi \rangle -\langle G_6^{\prime}(\bar{s})[r],\omega \rangle \geq 0, \end{align*}$

for all $r = (\tilde{n}, \tilde{c}, \tilde{u},\tilde{f})\in V_n\times V_c\times V_u\times\mathcal{C}(\bar{f})$ . Hence, the proof follows from Lemmas 5.1 and 5.2.

Corollary 5.1. Assume that $\bar{s} = (\bar{n},\bar{c},\bar{u},\bar{f})\in \mathcal{S}_{a d}$ be an optimal solution for the control problem (5.3). Then, there exist Lagrange multipliers $(\lambda, \eta,\rho)\in L^2(Q)\times (L^2(0,T;H^1(\Omega)))^{\prime}\times L^2(Q)$ , satisfying

$\begin{align} &\int_{0}^T\int_{\Omega}(\tilde{n}_{t}-\Delta \tilde{n}+\bar{u} \cdot \nabla \tilde{n} +\nabla(\tilde{n}\nabla \bar{c})-\gamma \tilde{n} +2\mu \tilde{n}\bar{n})\lambda d x d t -\int_{0}^T\int_{\Omega}\tilde{n}\eta d x d t \\ &-\int_{0}^T\int_{\Omega}\tilde{n} \nabla \varphi \rho d x d t = \beta_1\int_{0}^T\int_{\Omega_d}(\bar{n}-n_d)\tilde{n}d x d t, \end{align}$

(5.16)

$\begin{align} &\int_{0}^T\int_{\Omega}\left(\tilde{c}_{t}-\Delta \tilde{c}+\bar{u} \cdot \nabla \tilde{c}+\tilde{c}\right)\eta d x d t+\int_{0}^T\int_{\Omega} \nabla \cdot(\bar{n} \nabla \tilde{c})\lambda d x d t \\ = &\beta_2\int_{0}^T\int_{\Omega_d} (\bar{c}-c_d)\tilde{c}d x d t, \end{align}$

(5.17)

$\begin{align} &\int_{0}^T\int_{\Omega}\left(\tilde{u}_{t}-\Delta \tilde{u}+\bar{u} \cdot \nabla \tilde{u}+\tilde{u}\cdot \nabla \bar{u} \right) \rho d x d t + \int_{0}^T\int_{\Omega}\tilde{u}\nabla \bar{n} \lambda d x d t \\ &+\int_{0}^T\int_{\Omega} \tilde{u}\cdot\nabla \bar{c}\eta d x d t = \beta_3\int_{0}^T\int_{\Omega_d} (\bar{u}-u_d)\tilde{u}d x d t, \end{align}$

(5.18)

which corresponds to the linear system

$\begin{align} \left\{\begin{aligned} &-\lambda_t-\Delta \lambda +\bar{u}\cdot\nabla \lambda -\nabla \lambda\nabla \bar{c}-\gamma \lambda +2\mu\lambda\bar{n}-\eta - \nabla\varphi\rho \\ & = \beta_1(\bar{n}-n_d), \\ &-\eta_t-\Delta \eta+\bar{u}\cdot\nabla\eta+\eta+\nabla(\bar{n}\nabla\lambda) = \beta_2 (\bar{c}-c_d), \\ &-\rho_t-\Delta \rho+(\bar{u}\cdot\nabla)\rho+(\rho\cdot\nabla^{T})\bar{u}+\lambda \nabla \bar{n}+\eta \nabla \bar{c} = \beta_3(\bar{u}-u_d), \end{aligned} \right. \end{align}$

(5.19)

subject to the following boundary and final conditions

$\begin{align*} \left\{\begin{aligned} &\nabla\cdot \rho = 0, &&\mathit{\text{in}}\; Q, \\ &\frac{\partial \lambda }{\partial \nu} = \frac{\partial \eta }{\partial \nu}, \rho = 0,&& \mathit{\text{on}}\; (0,T)\times \partial\Omega, \\ &\lambda(T) = \beta_4(\bar{n}(T)-n_{\Omega}),\eta(T) = \beta_5 (\bar{c}(T)-c_{\Omega}), \\ &\rho(T) = \beta_5(\bar{c}(T)-c_{\Omega}),&& \mathit{\text{in}}\; \Omega, \end{aligned} \right. \end{align*}$

and the following identities hold

$\begin{align} \int_{0}^T\int_{\Omega_d}(\beta_7\bar{f}+\eta)(f-\bar{f}) d x d t \geq 0, \;\forall f \in\mathcal{U}. \end{align}$

(5.20)

Proof. By taking $(\tilde{c},\tilde{u},\tilde{f}) = (0,0,0)$ in (5.15), then it follows that the equation (5.16) holds. In light of an analogous argument, and in light of the (5.15), it guarantees that (5.17) and (5.18) hold. On the other hand, let $(\tilde{n},\tilde{c},\tilde{u}) = (0,0,0)$ , as an immediate consequence we obtain

$\begin{align*} \beta_7\int_{0}^{T}\tilde{f}\bar{f} d x d t+\int_{0}^{T}\tilde{f}\eta d x d t\geq 0, \quad \forall \tilde{f} \in \mathcal{C}(\bar{f}). \end{align*}$

By choosing $\tilde{f} = f-\bar{f}\in \mathcal{C}(\bar{f})$ for all $\bar{f}\in \mathcal{U}$ , thus we achieve (5.20).

Theorem 5.2. Under the assumptions of Theorem 5.1, system (5.19) has a unique weak solution such that

$\begin{align*} \|\lambda\|^2_{H^1}+\|\eta\|^2_{L^2}+\|\rho\|^2_{L^2}+\int_0^t\|\lambda\|^2_{H^2}d \tau +\int_0^t\|\eta\|^2_{H^1}d \tau+\int_0^t\|\rho\|^2_{H^1} d \tau \leq C. \end{align*}$

Proof. For convenience, we set $\tilde{\lambda} = \lambda(T-t)$ , $\tilde{\eta} = \eta(T-t)$ , $\tilde{\rho} = \rho(T-t)$ , in order to simplify notations, we still write $\lambda$ , $\eta$ , $\rho$ instead of $\tilde{\lambda}$ , $\tilde{\eta}$ , $\tilde{\rho}$ , then the adjoint system (5.19) can be written as follow

$\begin{align} \left\{\begin{aligned} &\lambda_t-\Delta \lambda +\bar{u}\cdot\nabla \lambda -\nabla \lambda\nabla \bar{c}-\gamma \lambda +2\mu\lambda\bar{n}-\eta - \nabla\varphi\rho \\ & = \beta_1(\bar{n}-n_d),&&\text{ in } Q, \\ &\eta_t-\Delta \eta+\bar{u}\cdot\nabla\eta+\eta+\nabla(\bar{n}\nabla\lambda) = \beta_2 (\bar{c}-c_d),&&\text{ in } Q, \\ &\rho_t-\Delta \rho+(\bar{u}\cdot\nabla)\rho+(\rho\cdot\nabla^{T})\bar{u}+\lambda \nabla \bar{n}+\eta \nabla \bar{c} = \beta_3(\bar{u}-u_d),&&\text{ in } Q, \end{aligned} \right. \end{align}$

(5.21)

subject to the following boundary and final conditions

$\begin{align*} \left\{\begin{aligned} &\nabla\cdot \rho = 0, &&\text{ in } Q, \\ &\frac{\partial \lambda }{\partial \nu} = \frac{\partial \eta }{\partial \nu}, \rho = 0,&& \text{ on } (0,T)\times\partial \Omega, \\ &\lambda(0) = \beta_4(\bar{n}(T)-n_{\Omega}),\eta(0) = \beta_5 (\bar{c}(T)-c_{\Omega}), \\ &\rho(0) = \beta_5(\bar{c}(T)-c_{\Omega}),&& \text{ in } \Omega. \end{aligned} \right. \end{align*}$

Following an analogous reasoning as in the proof of Lemma 5.3, we omit the process and just give a number of a priori estimates as follows.

Taking the $L^2$ -inner product with $\lambda$ for the first equation of (5.21) implies

$\begin{align*} &\frac{1}{2}\frac{d}{d t}\int_{\Omega}\lambda^2 d x+ \int_{\Omega}|\nabla\lambda|^2 d x +2\mu\int_{\Omega}\lambda^2\bar{n} d x \\ = &\int_{\Omega} \nabla \lambda \nabla \bar{c}d x+\gamma\int_{\Omega}\lambda^2 d x +\int_{\Omega}\lambda \eta d x+\int_{\Omega}\lambda \nabla \varphi \rho d x+ \beta_1\int_{\Omega}(\bar{n}-n_d) \lambda d x \\ \leq &\|\nabla\lambda\|_{L^2}\|\nabla\bar{c}\|_{L^2}+\gamma\|\lambda\|^2_{L^2}+\|\lambda\|_{L^2}(\|\eta\|_{L^2}+\|\rho\|_{L^2})+\beta_1\|\bar{n}-n_d\|_{L^2}\|\lambda\|_{L^2} \\ \leq& \frac{1}{2}\|\nabla\lambda\|^2_{L^2}+C(\|\lambda\|^2_{L^2}+\|\eta\|^2_{L^2}+\|\rho\|^2_{L^2})+C\|\bar{n}-n_d\|^2_{L^2}. \end{align*}$

Then, we have

$\begin{align} \frac{d}{d t}\|\lambda\|^2_{L^2}+\|\lambda\|^2_{H^1}\leq C(\|\lambda\|^2_{L^2}+\|\eta\|^2_{L^2}+\|\rho\|^2_{L^2})+C\|\bar{n}-n_d\|^2_{L^2}. \end{align}$

(5.22)

Taking the $L^2$ -inner product with $-\Delta\eta$ for the first equation of (5.21) implies

$\begin{align*} &\frac{1}{2}\frac{d}{d t}\int_{\Omega}|\nabla\lambda|^2 d x + \int_{\Omega}|\Delta\lambda|^2 d x \\ = &\int_{\Omega}\bar{u}\cdot \nabla \lambda \Delta\lambda d x -\int_{\Omega}\nabla \lambda \nabla \bar{c}\Delta \lambda d x-\gamma \int_{\Omega}\lambda \Delta \lambda d x +2\mu \int_{\Omega}\lambda \bar{n}\Delta \lambda d x \\ &- \int_{\Omega}\eta \Delta \lambda d x-\int_{\Omega}\nabla \varphi \rho\Delta \lambda d x +\beta_1\int_{\Omega}(\bar{n}-n_d)\Delta\lambda d x \\ \leq&\|\bar{u}\|_{L^4}\|\nabla \lambda\|_{L^4}\|\Delta \lambda \|_{L^2} +\|\nabla \lambda\|_{L^4}\|\nabla \bar{c}\|_{L^4}\|\Delta \lambda \|_{L^2} +\gamma\|\nabla \lambda\|^2_{L^2} \\ &+\| \lambda\|_{L^4}\|\bar{n}\|_{L^4}\|\Delta \lambda \|_{L^2}+\|\eta\|_{L^2}\|\Delta \lambda \|_{L^2}+ \| \rho\|_{L^2} \|\Delta \lambda \|_{L^2} \\ &+\beta_1\|\Delta \lambda \|_{L^2}\|\bar{n}-n_d\|^2_{L^2} \\ \leq& \|\bar{u}\|_{H^1}(\|\nabla \lambda\|^{\frac{1}{2}}_{L^2}\|\Delta \lambda\|^{\frac{1}{2}}_{L^2}+\|\nabla \lambda\|_{L^2})\|\Delta \lambda\|_{L^2}+\gamma\|\nabla \lambda\|^2_{L^2} \\ &+(\|\nabla \lambda\|^{\frac{1}{2}}_{L^2}\|\Delta \lambda\|^{\frac{1}{2}}_{L^2}+\|\nabla \lambda\|_{L^2})\|\nabla\bar{c}\|_{H^1}\|\Delta \lambda\|_{L^2}+\|\eta\|_{L^2}\|\Delta \lambda \|_{L^2} \\ &+\| \rho\|_{L^2} \|\Delta \lambda \|_{L^2}+\beta_1\|\Delta \lambda \|_{L^2}\|\bar{n}-n_d\|^2_{L^2} \\ \leq&\frac{1}{2}\|\Delta \lambda\|^{2}_{L^2}+C(\|\nabla \lambda\|^{2}_{L^2}+\|\eta\|^2_{L^2}+\| \rho\|^2_{L^2} ). \end{align*}$

Thus, we get

$\begin{align} \frac{d}{d t}\|\nabla\lambda\|^2_{L^2}+\|\nabla\lambda\|^2_{H^1}\leq C(\|\nabla\lambda\|^2_{L^2}+\|\eta\|^2_{L^2}+\|\rho\|^2_{L^2})+C\|\bar{n}-n_d\|^2_{L^2}. \end{align}$

(5.23)

Taking the $L^2$ -inner product with $\eta$ for the second equation of (5.21) implies

$\begin{align*} &\frac{1}{2}\frac{d}{d t}\int_{\Omega}\eta^2 d x+ \int_{\Omega}|\nabla\eta|^2 d x+ \int_{\Omega}\eta^2 d x \\ = & \int_{\Omega}\bar{n}\nabla\lambda \nabla \eta d x+\beta_2 \int_{\Omega}\eta (\bar{c}-c_d) d x \\ \leq& \|\bar{n}\|_{L^{4}}\|\nabla \lambda\|_{L^4}\|\nabla \eta\|_{L^2}+\beta_2\|\eta \|_{L^2}\|\bar{c}-c_d\|_{L^2} \\ \leq&\|\bar{n}\|_{H^1}(\|\nabla \lambda\|^{\frac{1}{2}}_{L^2}\|\Delta \lambda\|^{\frac{1}{2}}_{L^2}+\|\nabla \lambda\|_{L^2})\|\nabla \eta\|_{L^2}+\beta_2\|\eta \|_{L^2}\|\bar{c}-c_d\|_{L^2} \\ \leq &\frac{1}{2}\|\nabla \eta\|^2_{L^2}+\delta\|\Delta \lambda\|^2_{L^2}+C\|\nabla \lambda\|_{L^2}+C\|\eta \|^2_{L^2}+C\|\bar{c}-c_d\|^2_{L^2}. \end{align*}$

As an immediate consequence, we obtain

$\begin{align} \frac{d}{d t}\|\eta \|^2_{L^2}+\|\eta \|^2_{H^1} \leq \delta\|\Delta \lambda\|^2_{L^2}+C\|\nabla \lambda\|_{L^2}+C\|\eta \|^2_{L^2}+C\|\bar{c}-c_d\|^2_{L^2}. \end{align}$

(5.24)

Taking the $L^2$ -inner product with $\rho$ for the third equation of (5.21) implies

$\begin{align*} &\frac{1}{2}\frac{d}{d t}\int_{\Omega}\rho^2 d x+ \int_{\Omega}|\nabla\rho|^2 d x \\ = &-\int_{\Omega} (\rho\cdot\nabla^{T})\bar{u} \rho d x-\lambda\int_{\Omega}\nabla \bar{n} \rho d x-\int_{\Omega}\eta \nabla \bar{c} \rho d x+\beta_3\int_{\Omega}(\bar{u}-u_d) \rho d x \\ \leq &\|\rho\|_{L^2}\|\nabla \bar{u}\|_{L^4}\|\rho\|_{L^4}+\lambda \|\nabla \bar{n}\|_{L^2}\|\rho\|_{L^2}+\|\eta \|_{L^2}\|\nabla \bar{c}\|_{L^4}\|\rho\|_{L^4} \\ &+\beta_3\|\rho\|_{L^2}\|\bar{u}-u_d\|_{L^2} \\ \leq& \|\rho\|_{L^2}\|\nabla \bar{u}\|_{H^1}(\|\rho\|^{\frac{1}{2}}_{L^2}\|\nabla\rho\|^{\frac{1}{2}}_{L^2}+\|\rho\|_{L^2})+\lambda \|\nabla \bar{n}\|_{L^2}\|\rho\|_{L^2} \\ &+\|\eta \|_{L^2}\|\nabla \bar{c}\|_{H^1}(\|\rho\|^{\frac{1}{2}}_{L^2}\|\nabla\rho\|^{\frac{1}{2}}_{L^2}+\|\rho\|_{L^2})+\beta_3\|\rho\|_{L^2}\|\bar{u}-u_d\|_{L^2} \\ \leq&\frac{1}{2}\|\nabla\rho\|^{2}_{L^2}+C\|\rho\|^2_{L^2}(\|\nabla \bar{u}\|^2_{H^1}+1)+C\|\eta \|^2_{L^2}+C\|\bar{u}-u_d\|^2_{L^2}. \end{align*}$

Therefore, we see that

$\begin{align} \frac{d}{d t}\|\rho\|^{2}_{L^2}+\|\rho\|^{2}_{H^1}\leq C\|\rho\|^2_{L^2}(\|\nabla \bar{u}\|^2_{H^1}+1)+C\|\eta \|^2_{L^2}+C\|\bar{u}-u_d\|^2_{L^2}. \end{align}$

(5.25)

Combining (5.22)-(5.25) and taking $\delta$ small enough, we have

$\begin{align*} &\frac{d}{ d t}(\|\lambda\|^2_{H^1}+\|\eta\|^2_{L^2}+\|\rho\|^2_{L^2})+ \|\lambda\|^2_{H^2}+ \|\eta\|^2_{H^1} +\|\rho\|^2_{H^1} \\ \leq& C(\|\nabla \bar{u}\|^2_{H^1}+1)( \|\lambda\|^2_{H^1}+\|\eta\|^2_{L^2}+\|\rho\|^2_{L^2})+C\|\bar{n}-n_d\|^2_{L^2} \\ &+C\|\bar{c}-c_d\|^2_{L^2}+C\|\bar{u}-u_d\|^2_{L^2}. \end{align*}$

Applying Gronwall's lemma to the resulting differential inequality, we know

$\begin{align*} \|\lambda\|^2_{H^1}+\|\eta\|^2_{L^2}+\|\rho\|^2_{L^2}+\int_0^t\|\lambda\|^2_{H^2}d \tau +\int_0^t\|\eta\|^2_{H^1}d \tau+\int_0^t\|\rho\|^2_{H^1} d \tau \leq C. \end{align*}$

The proof is complete.

Acknowledgment

The authors would like to express their deep thanks to the referee's valuable suggestions for the revision and improvement of the manuscript.

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