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Research article Special Issues

Runoff forecasting model based on variational mode decomposition and artificial neural networks


  • Received: 30 September 2021 Accepted: 29 November 2021 Published: 13 December 2021
  • Accurate runoff forecasting plays a vital role in water resource management. Therefore, various forecasting models have been proposed in the literature. Among them, the decomposition-based models have proved their superiority in runoff series forecasting. However, most of the models simulate each decomposition sub-signals separately without considering the potential correlation information. A neoteric hybrid runoff forecasting model based on variational mode decomposition (VMD), convolution neural networks (CNN), and long short-term memory (LSTM) called VMD-CNN-LSTM, is proposed to improve the runoff forecasting performance further. The two-dimensional matrix containing both the time delay and correlation information among sub-signals decomposing by VMD is firstly applied to the CNN. The feature of the input matrix is then extracted by CNN and delivered to LSTM with more potential information. The experiment performed on monthly runoff data investigated from Huaxian and Xianyang hydrological stations at Wei River, China, demonstrates the VMD-superiority of CNN-LSTM to the baseline models, and robustness and stability of the forecasting of the VMD-CNN-LSTM for different leading times.

    Citation: Xin Jing, Jungang Luo, Shangyao Zhang, Na Wei. Runoff forecasting model based on variational mode decomposition and artificial neural networks[J]. Mathematical Biosciences and Engineering, 2022, 19(2): 1633-1648. doi: 10.3934/mbe.2022076

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  • Accurate runoff forecasting plays a vital role in water resource management. Therefore, various forecasting models have been proposed in the literature. Among them, the decomposition-based models have proved their superiority in runoff series forecasting. However, most of the models simulate each decomposition sub-signals separately without considering the potential correlation information. A neoteric hybrid runoff forecasting model based on variational mode decomposition (VMD), convolution neural networks (CNN), and long short-term memory (LSTM) called VMD-CNN-LSTM, is proposed to improve the runoff forecasting performance further. The two-dimensional matrix containing both the time delay and correlation information among sub-signals decomposing by VMD is firstly applied to the CNN. The feature of the input matrix is then extracted by CNN and delivered to LSTM with more potential information. The experiment performed on monthly runoff data investigated from Huaxian and Xianyang hydrological stations at Wei River, China, demonstrates the VMD-superiority of CNN-LSTM to the baseline models, and robustness and stability of the forecasting of the VMD-CNN-LSTM for different leading times.



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

    {nt+un=Δn(nc)+γnμn2,in Q(0,T)×Ω,ct+uc=Δcc+n+f,in Q,ut+uu=Δuπ+nφ,in Q,u=0,in Q,nν=cν=0,u=0,on (0,T)×Ω,n(x,0)=n0(x),c(x,0)=c0(x),u(x,0)=u0(x),in Ω, (1.1)

    where ΩR2 is a bounded domain with smooth boundary Ω. ν is the outward normal vector to Ω, and γ, μ are positive constants. n, c denote the bacterial density, the oxygen concentration, respectively. u, π 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 U. We observe that in the subdomains where f0 we inject oxygen, and conversely where f0 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 ΩR3 without the term uu.

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

    nt+un=Δn(nc)+n(γμn), (1.2)
    ct+uc=Δcc+n, (1.3)
    ut=Δuπ+nφ, (1.4)
    u=0, (1.5)
    nν=cν=0,u=0. (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 μ>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 nd, cd and ud, and at the final moment T is as close as possible to a desired state nΩ, cΩ and uΩ. The main difficulties for treating the problem are caused by the nonlinearity of uu. 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.

    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 ΩR2. 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 0l<k. Suppose that 1q, r, p>0 and lka1 such that

    1plN=a(1qkN)+(1a)1r. (2.1)

    Then, for any uWk,q(Ω)Lr(Ω), there exist two positive constants C1 and C2 depending only on Ω, q, k, r and N such that the following inequality holds

    DluLpc1DkuaLqu1aLr+c2uLr

    with the following exception: If 1<q< and klNq is a non-negative integer, the (2.1) holds only for a satisfying lka<1.

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

    Lemma 2.2. Let ΩR2 be a bounded domain with smooth boundary. Then for all non-negative uH1(Ω), there holds

    u3L3(Ω)δu2H1(Ω)(u+1)log(u+1)L1(Ω)+p(δ1)uL1(Ω),

    where δ is any positive number, and p() is an increasing function.

    We first consider the existence of solutions to the linear problem of system (1.1). Assume functions u0H1(Ω), ˆuL4(0,T;L4(Ω)),ˆnL2(0,T;L2(Ω)), and consider

    {utΔu+ˆuu=π+ˆnφ,in Q,u=0,in Q,u=0,on Ω,u(x,0)=u0(x),in Ω. (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 u0H1(Ω), ˆuL4(0,T;L4(Ω)), ˆnL2(0,T;H1(Ω)),φL(Q), and u be the solution of the problem (2.2), then uL(0,T;H1(Ω))L2(0,T;H2(Ω)) and utL2(0,T;L2(Ω)).

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

    12ddtΩu2dx+Ω|u|2dx+Ωu2dx=Ωˆnφudx+Ωu2dxˆnL2uL2+u2L2C(ˆn2L2+u2L2).

    By Gronwall's inequality, we have

    u2L2+T0u2H1dτC(T0ˆn2L2dτ+u02L2).

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

    ut+Au+P(ˆuu)=P(ˆnφ),

    where A:=PΔ is called Stokes operator, which is an unbounded self-adjoint positive operator in L2 with compact inverse, for more properties of Stokes operator, we refer to [10]. Note that u=0, that is Pu=u, PΔu=Δu, Put=ut. So, in following calculations, we ignore the projection operator P. Multiplying this equation by Δu, and integrating it over Ω, we get

    12ddtΩ|u|2dx+Ω|Δu|2dx+Ω|u|2dx=ΩP(ˆuu)ΔudxΩP(ˆnφ)Δudx+Ω|u|2dx.

    For the terms on the right, we have

    ΩP(ˆuu)ΔudxΩP(ˆnφ)Δudx+Ω|u|2dxˆuL4uL4ΔuL2+ˆnL2ΔuL2+u2L2ˆuL4u1/2L2Δu3/2L2+ˆuL4uL2ΔuL2+ˆnL2ΔuL2+u2L212Δu2L2+C(ˆu4L4+ˆu2L4+1)u2L2+ˆn2L2.

    Therefore, we get

    ddtu2L2+u2H1C(ˆu4L4+ˆu2L4+1)u2L2+Cˆn2L2+C.

    By Gronwall's inequality, we derive

    u2L2+T0u2H1dτC.

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

    T0Ω|ut|2dxdtC.

    Summing up, we complete the proof.

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

    {ctΔc+uc+c=ˆn++f,in Q,cν=0,on (0,T)×Ω,c(x,0)=c0(x),in Ω. (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 c0H2(Ω), ˆnL2(0,T;H1(Ω)), fL2(0,T;H1(Ω)), u be the solution of the problem (2.2), and c be the solution of (2.3). Then cL((0,T),H2(Ω))L2((0,T),H3(Ω)) and ctL2(0,T;L2(Ω)).

    Proof. Multiplying the first equation of (2.3) by c, and integrating it over Ω, we infer from Ωc(uc)=12Ωc2udx=0 that

    12ddtΩc2dx+Ω|c|2dx+Ωc2dxˆnL2cL2+fL2cL2.

    Therefore, we have

    c2L2+c2H1C(c02L2+t0(ˆn2L2+f2L2)dτ).

    Multiplying the first equation of (2.3) by Δc, and integrating it over Ω, we get

    12ddtΩ|c|2dx+Ω|Δc|2dx+Ω|c|2dx=ΩucΔcdxΩΔcˆndxΩΔcfdx.

    Using the Young inequality and the Hölder inequality, we obtain

    ΩucΔcdxΩΔcˆndxΩΔcfdxuL4cL4ΔcL2+ˆnL2ΔcL2+fL2ΔcL2CuH1(c12L2Δc12L2+cL2)ΔcL2+ˆnL2ΔcL2+fL2ΔcL2=CuH1c12L2Δc32L2+CcL2ΔcL2+ˆnL2ΔcL2+fL2ΔcL212Δc2L2+Cu4H1c2L2+C(ˆn2L2+f2L2).

    Combining this and above inequalities, we conclude

    ddtc2L2+c2H1Cu4H1c2L2+C(ˆn2L2+f2L2).

    We therefore verify that

    c2L2+t0c2H1C(t0ˆn2L2dτ+t0f2L2dτ).

    Applying to the first equation of (2.3), multiplying it by Δc, and integrating over Ω give

    12ddtΩ|Δc|2dx+Ω|Δc|2dx+Ω|Δc|2dx=Ω(uc)ΔcdxΩˆn+ΔcdxΩfΔcdx.

    For the terms on the right, we obtain

    Ω(uc)ΔcdxΩˆn+ΔcdxΩfΔcdxΔcL2(uL4ΔcL4+uL4cL4)+ˆnL2ΔcL2+fL2ΔcL2ΔcL2(uL4Δc12L2Δc12L2+uL4ΔcL2+u12L2Δu12L2c12L2Δc12L2+uL2c12L2Δc12L2+u12L2Δu12L2cL2+uL2cL2)+ˆnL2ΔcL2+fL2ΔcL212Δc2L2+C(1+Δc2L2+Δu2L2+ˆn2L2+f2L2).

    Straightforward calculations yield

    Δc2L2+t0Δc2H1dτC(1+t0ˆn2H1dτ+t0f2H1dτ).

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

    T0Ω|ct|2dxdtC,

    and thereby precisely arrive at the conclusion.

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

    {ntΔn+un+n=(nc)+(1+γ)ˆn+μˆn+n,in Q,nν|Ω=0,n(x,0)=n0(x),in Ω. (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 0n0H1(Ω), ˆnL2(0,T;H1(Ω))L4(0,T;L4(Ω)), and u, c, n are the solutions of the problem (2.2), (2.3) and (2.4), respectively. Then n0, nL(0,T;H1(Ω))L2(0,T;H2(Ω)) and ntL2(0,T;L2(Ω)).

    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

    ddtA(t)ndxA(t)nνds+A(t)ndx=(1+γ)A(t)ˆn+dxμA(t)ˆn+ndx.

    Since nν0 on {n<0}, from this we deduce that the right hand side is nonnegative. Integrating this equality on [0,t] gives

    A(t)ndxdτ+t0A(t)ndxdτ=0.

    Then, we get n0.

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

    12ddtΩn2dx+Ω(n2+|n|2)dx+μΩˆn+n2dx=Ωncndx+(1+γ)Ωnˆn+dxnL4cL4nL2+(1+γ)ˆnL2nL2C(n12L2n12L2+nL2)cH2nL2+(1+γ)ˆnL2nL2C(n2L2c4H2+n2L2c2H2+ˆnL2)+12n2H1.

    So, we derive that

    n2L2+T0n2H1dtC(1+T0ˆn2L2dt).

    Multiplying the first equation of (2.4) by Δn, and integrating it over Ω, we get

    12ddtΩ|n|2dx+Ω|Δn|2dx+Ω|n|2dx=ΩunΔndx+Ω((nc)Δn(1+γ)ˆn+Δn+μˆn+nΔn)dxuL4nL4ΔnL2+nL4ΔcL4ΔnL2+nL4cL4ΔnL2+(1+γ)ˆnL2ΔnL2+μnL4ˆnL4ΔnL2CuH1(n12L2Δn12L2+nL2)ΔnL2+nL4(Δc12L2Δc12L2+ΔcL2)ΔnL2+μnL4ˆnL4ΔnL2+(n12L2Δn12L2+nL2)cH1ΔnL2+(1+γ)ˆnL2ΔnL212Δn2L2+C(n2L2+n4L4+Δc4L2+Δc2L2+ˆn2L2+ˆn4L4)12Δn2L2+C(1+n2L2+n4L2+n2L2n2L2+Δc2L2+ˆn2L2+ˆn4L4).

    Straightforward calculations yield

    n2L2+T0Ω(|Δn|2+|n|2+ˆn+|n|2)dxdtC.

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

    T0Ω|nt|2dxdtC.

    The proof is complete.

    Introduce the spaces

    Xu=L4(0,T;L4(Ω)),Xn=L4(0,T;L4(Ω))L2(0,T;H1(Ω)),Yu=L(0,T;H1(Ω))L2(0,T;H2(Ω)),Yn=L(0,T;H1(Ω))L2(0,T;H2(Ω)).

    Define a map

    F:Xu×XnXu×Xn,F(ˆu,ˆn)=(u,n),

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

    {ntΔn+un+n=(nc)+(1+γ)ˆn+μˆn+n,in (0,T)×ΩQ,ctΔc+uc+c=ˆn++f,in (0,T)×ΩQ,utΔu+ˆuu=π+ˆnφ,in (0,T)×ΩQ,u=0,in (0,T)×ΩQ,nν=cν=0,u=0,on (0,T)×Ω,n(x,0)=n0(x),c(x,0)=c0(x),u(x,0)=u0(x),in Ω.

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

    Lemma 2.6. The map F:Xu×XnXu×Xn is well defined and compact.

    Proof. Let (ˆn,ˆu)Xu×Xn, by Lemmas 2.3, 2.4, 2.5 we deduce that (n,u)=F(ˆn,ˆu) is bounded in Yu×Yn. Note that the embeddings H2(Ω)H1(Ω) is compact and interpolating between L(0,T;H1(Ω)) and L2(0,T;H2(Ω)). It is easy to get that u is bounded in L4(0,T;L4(Ω)) and n is bounded in L4(0,T;L4(Ω))L2(0,T;H1(Ω)). Therefore, the operator F:Xu×XnXu×Xn is a compact operator.

    From Lemma 2.6, (n,u)Yn×Yu satisfies pointwisely a.e. in Q the following problem

    {ntΔn+un+n=(nc)+α(1+γ)nμn2,in Q,ctΔc+uc+c=n+αf,in Q,utΔu+uu=π+αnφ,in Q,u=0,in Q,nν=cν=0,u=0,on (0,T)×Ω,n(x,0)=n0(x),c(x,0)=c0(x),u(x,0)=u0(x),in Ω. (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

    nL1+t0(nL1+nL2)dτC, (3.2)
    u2L2+t0u2H1dτC, (3.3)
    c2L2+t0c2H1dτC. (3.4)

    Proof. With Lemma 2.5 in hand, we get n0. Integrating the first equation of (3.1) over Ω, we see that

    ddtΩndx+Ωndx+μΩn2dx=α(1+γ)Ωndxμ2Ωn2dx+C.

    Solving this differential inequality, we obtain that

    nL1+t0(nL1+nL2)dτC.

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

    12ddtΩu2dx+Ω|u|2dx+Ωu2dx=αΩnφudx+Ωu2dxnL2uL2+u2L2C(n2L2+u2L2).

    Therefore, we see that

    u2L2+t0uH1dτC.

    By the Gagliardo-Nirenberg interpolation inequality, we deduce that

    t0u4L4dτCt0(u2L2u2L2d+u2L2)τu2L2t0u2L2dτ+t0u2L2dτC.

    Multiplying the third equation of (3.1) by Δu, and integrating it over Ω, we get

    ddtu2L2+u2H1C(u4L4+u2L4+1)u2L2+Cn2L2+C.

    Thus, we know

    u2L2+t0u2H1dτC.

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

    12ddtΩc2dx+Ω|c|2dx+Ωc2dxnL2cL2+αfL2cL2.

    Then, we have

    cL2+t0cH1dτC.

    Multiplying the second equation of (3.1) by Δc, and integrating it over Ω, we get

    ddtc2L2+c2H1Cu4H1c2L2+C(n2L2+f2L2).

    Further, we have

    c2L2+t0c2H1dτC.

    The proof is complete.

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

    (n+1)ln(n+1)L1+c2L2+c2H1C. (3.5)

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

    ddt(n+1)+u(n+1)Δ(n+1)=((n+1)c)+Δc+α(1+γ)nμn2.

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

    ddtΩ(n+1)ln(n+1)dx+4Ω|n+1|2dxΩ(n+1)cdx+ΩΔcln(n+1)dx+α(1+γ)Ωnln(n+1)dx=I1+I2+I3.

    For I1, integrating by parts and using Young's inequality with small δ, we get

    I1=ΩnΔcdxnL2ΔcL2δΔc2L2+Cn2L2.

    For the term I2, we have

    I2=ΩΔcln(n+1)dxδΔc2L2+Cln(n+1)2L2δΔc2L2+CΩ(n+1)ln(n+1)dx.

    For the rest term I3, straightforward calculations yield

    I3=α(1+γ)Ωnln(n+1)dx(1+γ)Ω(n+1)ln(n+1)dx.

    Combining I1, I2 with I3, we conduct that

    ddtΩ(n+1)ln(n+1)dx+4Ω|n+1|2dxδΔc2L2+CΩ(n+1)ln(n+1)dx+Cn2L2. (3.6)

    Multiplying the second equation of (3.1) by Δc, and integrating it over Ω, we get

    12ddtΩ|c|2dx+Ω|Δc|2dx+Ω|c|2dx=ΩucΔcdxΩΔcndxαΩΔcfdx.

    Straightforward calculations yield

    ddtc2L2+c2H1Cc2L2+C(n2L2+f2L2). (3.7)

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

    ddtΩ(n+1)ln(n+1)dx+ddtc2L2+(1δ)c2H1+4Ω|n+1|2dxCΩ(n+1)ln(n+1)dx+C(f2L2+n2L2).

    Taking δ small enough, and solving this differential inequality, we obtain that

    (n+1)ln(n+1)L1+c2L2+c2H1C.

    The proof is complete.

    Lemma 3.3. Assume fL2(0,T;H1(Ω)), let (n,c,u) be a local solution to (3.1). Then, it holds that

    n2L2+Δc2L2+t0nH1dτ+t0ΔcH1dτC. (3.8)

    Proof. Taking the L2-inner product with n for the first equation of (3.1) implies

    12ddtΩn2dx+Ω(n2+|n|2)dx+μΩn3dx=Ωncndx+α(1+γ)Ωn2dx=12Ωn2Δcdx+α(1+γ)Ωn2dx.

    Here, we note that

    |Ωn2Δcdx|n2L3ΔcL3Cn2L3(Δc23L2c13L2+cL2)Cn2L3(Δc23L2+1).

    From Lemma 2.2 and (3.2), it follows that

    χ2Ωn2ΔcdxC(δn2H1(n+1)log(n+1)L1+p(δ1)nL1)23(Δc23L2+1)C(δn2H1+p(δ1))23(Δc23L2+1)C(δn43H1Δc23L2+δn43H1+p23(δ1)Δc23L2+p23(δ1))δΔc2L2+Cδ12n2H1+C1/2δp(δ1).

    As an immediate consequence

    ddtn2L2+n2H1δΔc2L2+Cδ12n2H1+Cn2L2. (3.9)

    Applying to the first equation of (3.1), multiplying it by Δc, and integrating over Ω give

    12ddtΩ|Δc|2dx+Ω|Δc|2dx+Ω|Δc|2dx=Ω(uc)ΔcdxΩnΔcdxΩfΔcdx=I4+I5.

    For I4, by using the Gagliardo-Nirenberg interpolation inequality, we get

    I4=Ω(uc)ΔcdxΔcL2(uL4ΔcL4+uL4cL4)ΔcL2(uL4Δc12L2Δc12L2+uL4ΔcL2+u12L2Δu12L2c12L2Δc12L2+uL2c12L2Δc12L2+u12L2Δu12L2cL2+uL2cL2)14Δc2L2+C(1+Δc2L2+Δu2L2).

    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.

    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).

    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:

    \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 \dot{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}- \dot{n} = g_c,&&\text{in }Q , \\ &\tilde{u}_{t}-\Delta \tilde{u}+\bar{u} \cdot \nabla \tilde{u}+\dot{u}\cdot \nabla \bar{u}-\dot{n} \nabla \varphi = g_u,&&\text{in }Q. \end{aligned} \right. \end{align} (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.

    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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