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Optimal control for the coupled chemotaxis-fluid models in two space dimensions

  • Received: 01 August 2021 Revised: 01 September 2021 Published: 26 October 2021
  • Primary: 92C17, 49J20; Secondary: 49K20, 35K51

  • This paper deals with a distributed optimal control problem to the coupled chemotaxis-fluid models. We first explore the global-in-time existence and uniqueness of a strong solution. Then, we define the cost functional and establish the existence of Lagrange multipliers. Finally, we derive some extra regularity for the Lagrange multiplier.

    Citation: Yunfei Yuan, Changchun Liu. Optimal control for the coupled chemotaxis-fluid models in two space dimensions[J]. Electronic Research Archive, 2021, 29(6): 4269-4296. doi: 10.3934/era.2021085

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  • This paper deals with a distributed optimal control problem to the coupled chemotaxis-fluid models. We first explore the global-in-time existence and uniqueness of a strong solution. Then, we define the cost functional and establish the existence of Lagrange multipliers. Finally, we derive some extra regularity for the Lagrange multiplier.



    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 I5, we have

    I5=ΩnΔcdxΩfΔcdxC(n2L2+f2L2)+14Δc2L2.

    Along with I4 and I5, we conclude

    ddtΔc2L2+Δc2L2+Δc2L2C(1+Δc2L2+Δu2L2+n2L2+f2L2). (3.10)

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

    ddt(n2L2+Δc2L2)+Δc2L2+(1Cδ12)n2H1+(1δ)Δc2L2C(1+Δc2L2+Δu2L2+n2L2+f2L2).

    By choosing δ small enough and using (3.3) and (3.5), we have

    n2L2+Δc2L2+t0nH1dτ+t0ΔcH1dτC.

    The proof is complete.

    Lemma 3.4. Assume fL2(0,T;H1(Ω)), let (n,c,u) be a local solution to (3.1). Then, it holds that

    n2L2+t0n2H2dτC. (3.11)

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

    12ddtΩ|n|2dx+Ω|Δn|2dx+Ω|n|2dx=ΩunΔndx+Ω(nc)Δndx+(1+γ)Ω|n|2dx+μΩn2Δndx=I6+I7+I8.

    For the term I6, with the estimate (3.3), we have

    I6=ΩunΔndx=12Ωu(n)2dxuL2n2L4uL2(n12L2Δn12L2+nL2)2δΔn2L2+Cn2L2.

    For the term I7, taking (3.8) into considering, we conduct that

    I7=Ω(nc)Δndx=Ω(nc+nΔc)ΔndxΔnL2(nL3cL6+nCΔcL2)CΔnL2(nH13cH1+nH43ΔcL2)CnH2nH43cH2Cn53H2n13L2cH2δn2H2+C(δ)n2L2c6H2δn2H2+C.

    For the term I8, thanks to the nonnegativity of n, we see that

    I8=(1+γ)Ω|n|2dx+μΩn2Δndx=(1+γ)Ω|n|2dx2μΩ|n|2ndx(1+γ)n2L2.

    Combine the estimates about I6, I7 and I8, it follows that

    ddtn2L2+(14δ)n2H2Cn2L2+C.

    By taking δ small enough, we get

    n2L2+t0n2H2dτC.

    Therefore, this proof is complete.

    Lemma 3.5. The operator F:Xu×XnXu×Xn, is continuous.

    Proof. Let {(ˆnm,ˆum)}mN be a sequence of Xu×Xn, Then, with Lemmas 2.3, 2.4 and 2.5 in hand, we conduct that {(nm,um)=F(ˆnm,ˆum)}mN is bounded in Yu×Yn. Taking the compactness of Yu×Yn in Xu×Xn into consider, we see that F is a compact operator, which means there exists a subsequence of {F(ˆnm,ˆum)}mN, for convenience, still denoted as {F(ˆnm,ˆum)}mN, and exists an element (ˆn,ˆu) in Yu×Yn such that

    F(ˆnm,ˆum)(ˆn,ˆu) weakly in Yu×Yn and strongly in Xu×Xn.

    Let m and take the limit, it is clear that (n,u)=F(ˆnm,ˆum) and (ˆnm,ˆum)=(ˆn,ˆu), this means that F(ˆnm,ˆum)=(ˆnm,ˆum). Since uniqueness of limit, the map F is continuous.

    Theorem 3.1. Let u0H1(Ω), n0H1(Ω), c0H2(Ω) with n00 in Ω, and fL2(0,T;H1(Ω)), then (1.1) exists unique strong solution (n,c,u). Moreover, there exists a positive C constant such that

    nL(0,T;H1(Ω))+nL2(0,T;H2(Ω))+ntL2(0,T;L2(Ω))+cL(0,T;H2(Ω))+cL2(0,T;H3(Ω))+ctL2(0,T;L2(Ω))+uL(0,T;H1(Ω))+uL2(0,T;H2(Ω))+utL2(0,T;L2(Ω))C. (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=n1n2, c=c1c2 and u=u1u2, where (ni,ci,ui) is the strong solution of the system, where i=1,2. Thus, we obtain the following system

    ntΔn+u1n+un2=(n1c)(nc2)+γnμn(n1+n2),in (0,T)×ΩQ, (3.13)
    ctΔc+u1c+uc2+c=n,in (0,T)×ΩQ, (3.14)
    utΔu+u1u+uu2=nφ,in (0,T)×ΩQ, (3.15)
    u=0,in (0,T)×ΩQ, (3.16)
    nν=cν=0,u=0,on (0,T)×Ω, (3.17)
    n0(x)=c0(x)=u0(x)=0,in Ω. (3.18)

    Taking the L2-inner product with n for the (3.13) implies

    12ddtΩn2dx+Ω|n|2dx+Ωn2dxΩun2ndx+Ωn1cndx+Ωnc2ndx+(1+γ)Ωn2dx=I9+I10+I11+I12.

    For the term I9, due to the estimates (3.3) and (3.8), we have

    I9=Ωun2ndxn2L2uL4nL4Cn2L2uH1(n12L2n12L2+nL2)δ3n2L2+Cn2L2.

    For the term I10, with the estimate (3.8) and (3.11), we get

    I10=Ωn1cndxnL2n1L4cL4CnL2n1H1cH1δ3n2L2+C.

    For the term I11,

    I11=Ωnc2ndxnL2c2L4nL4nL2c2H1nH1δ3n2L2+C.

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

    ddtn2L2+nH1δn2L2+Cn2L2+C. (3.19)

    Taking the L2-inner product with c for the (3.14) implies

    12ddtΩc2dx+Ω|c|2dx+Ωc2dx=Ωu1ccdxΩuc2cdx+Ωncdxc2L4u1L2+uL2c2L4cL4+nL2cL2C(c12L2c12L2+cL2)2u1L2+(c12L2c12L2+cL2)uL2c2H1+nL2cL2δc2L2+Cc2L2.

    Then, we get

    ddtc2L2+cH1δc2L2+Cc2L2. (3.20)

    Taking the L2-inner product with c for the (3.15) implies

    12Ωu2dx+Ω|u|2dx=Ωnφudx.

    Straightforward calculations yield

    ddtu2L2+uH1C(u2L2+n2L2). (3.21)

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

    ddt(n2L2+c2L2+u2L2)+(nH1+cH1+uH1)δ(n2L2+c2L2+u2L2)+(n2L2+c2L2+u2L2)+C.

    By choosing δ small enough, we get

    ddt(n2L2+c2L2+u2L2)C(n2L2+c2L2+u2L2)+C.

    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 UL2(0,T;H1(Ωc)) is a nonempty, closed and convex set, where control domain ΩcΩ, and ΩdΩ 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 nd, cd and ud, and at the final moment T is as close as possible to a desired state nΩ, cΩ and uΩ. We consider the optimal control problem as follows

    Minimize the cost functional

    J(n,c,u,f)=β12nnd2L2(Qd)+β22ccd2L2(Qd)+β32uud2L2(Qd)+β42n(T)nΩ2L2(Ωd)+β52c(T)cΩ2L2(Ωd)+β62u(T)uΩ2L2(Ωd)+β72f(x,t)2L2(Qc), (4.1)

    subject to the system (1.1). Moreover, the nonnegative constants βi,i=1,2,,7 are given but not all zero, the functions nd, cd, ud represents the desired states satisfying

    ndL2(Qd),cdL2(Qd),udL2(Qd),nΩL2(Ωc),cΩL2(Ωc),uΩL2(Ωc),fU.

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

    Sad={s=(n,c,u,f)H:s is a strong solution of (1.1)}.

    The function space H is given by

    H=Yn×Yc×Yu×U,

    where Yc=L(0,T;H2(Ω))L2(0,T;H3(Ω)).

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

    Theorem 4.1. Suppose fU is satisfied, and n00, then the optimal control problem (4.1) admits a solution (ˉn,ˉc,ˉu,ˉf)Sad.

    Proof. Along with Theorem 3.1, we conduct that Sad, then there exists the minimizing sequence {(nm,cm,um,fm)}mNSad such that

    limm+J(nm,cm,um,fm)=inf(n,c,u,f)SadJ(n,c,u,f). (4.2)

    According to the definition of Sad, for each mN there exists (nm,cm,um,fm) satisfying

    {nmt+umnm=Δnm(nmcm)+γnmμn2m,in Q,cmt+umcm=Δcmcm+nm+fm,in Q,umt+umum=Δumπ+nmφ,in Q,um=0,in Q,nmν|Ω=cmν|Ω=0,um|Ω=0,nm(0)=n0,cm(0)=c0,um(0)=u0,in Ω. (4.3)

    Observing that U is a closed convex subset of L2(0,T;H1(Ωc)). According to the definition of Sad, we deduce that there exists (ˉn,ˉc,ˉu,ˉf) bounded in H such that, for subsequence of (nm,cm,um,fm)mN, for convenience, still denoted by (nm,cm,um,fm), as m+

    nmˉn, weakly in L2(0,T;H2(Ω)) and weakly*  in L(0,T;H1(Ω)),cmˉc, weakly in L2(0,T;H3(Ω)) and weakly*  in L(0,T;H2(Ω)),umˉu, weakly in L2(0,T;H2(Ω)) and weakly*  in L(0,T;H1(Ω)),fmˉf, weakly in L2(0,T;H1(Ωc)), and ˜fU.

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

    nmˉn, strongly in C([0,T];L2(Ω))L2(0,T;H1(Ω)),cmˉc, strongly in C([0,T];H1(Ω))L2(0,T;H2(Ω)),umˉu, strongly in C([0,T];L2(Ω))L2(0,T;H1(Ω)).

    Since (nmcm)=nmcm+nmΔcm is bounded in L2(0,T;L2(Ω)), then

    (nmcm)χ, weakly in L2(0,T;L2(Ω)).

    Recalling that

    nmcmˉnˉc, weakly in L(0,T;L2(Ω)).

    Therefore, we get that χ=(ˉnˉc). Owing to (ˉn,ˉc,ˉu,ˉf)H, we see that (ˉn,ˉc,ˉu,ˉf) is solution of the system (1.1), along with (4.2) implies that

    limm+J(nm,cm,um,fm)=inf(u,c,u,f)SadJ(u,c,u,f)J(ˉn,ˉc,ˉu,ˉf).

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

    J(ˉn,ˉc,ˉu,ˉf)lim infm+J(nm,cm,um,fm).

    Therefore, this implies that (ˉn,ˉc,ˉu,ˉf) 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

    minJ(s) subject to sS={sH:G(s)N}, (5.1)

    where J:XR is a functional, G:XY is an operator, X and Y are Banach spaces, and nonempty closed convex set H is subset of X and nonempty closed convex cone N with vertex at the origin in Y.

    A+ denotes its polar cone

    A+={ρX:ρ,aX0,aA}.

    We consider the following Banach spaces

    X=Vn×Vc×Vu×L2(0,T;H1(Ωc)),Y=L2(Q)×L2(0,T;H1(Ω))×L2(Q)×H1(Ω)×H2(Ω)×H1(Ω),

    where

    Vn={nYn:nν on (0,T)×Ω},Vc={nYc:cν on (0,T)×Ω},Vu={nYu:u=0 on (0,T)×Ω and u=0 in (0,T)×Ω}

    and the operator G=(G1,G2,G3,G4,G5,G6):XY, where

    G1:XL2(Q),G2:XL2(0,T;H1(Ω)),G3:XL2(Q),G4:XH1(Ω),G5:XH2(Ω),G6:XH1(Ω),

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

    {G1=nt+unΔn+(nc)γn+μn2,G2=ct+ucΔc+cnf,G3=ut+uuΔu+πnφ,G4=n(0)n0,G5=c(0)c0,G6=u(0)u0. (5.2)

    The function spaces are given as follows

    H=Vn×Vc×Vu×U.

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

    minJ(s) subject to sSad={sH:G(s)=0}. (5.3)

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

    Lemma 5.1. The functional J:XR is Fréchet differentiable and the Fréchet derivative of J in ˉs=(ˉn,ˉc,ˉu,ˉf)X in the direction r=(˜n,˜c,˜u,˜f) is given by

    J(ˉs)[r]=β1T0Ωd(ˉnnd)˜ndxdt+β2T0Ωd(ˉccd)˜cdxdt+β3T0Ωd(ˉuud)˜u(T)dxdt+β4Ωd(ˉn(T)nΩ)˜n(T)dx+β5Ωd(ˉc(T)cΩ)˜cdx+β6Ωd(ˉu(T)uΩ)˜u(T)dx+β7T0Ωdˉf˜fdxdt. (5.4)

    Lemma 5.2. The operator G:XY is continuous-Fréchet differentiable and the Fréchet derivative of J in ˉs=(ˉn,ˉc,ˉu,ˉf)X in the direction r=(˜n,˜c,˜u,˜f), is the linear operator

    G(ˉs)[r]=(G1(ˉs)[r],G2(ˉs)[r],G3(ˉs)[r],G4(ˉs)[r],G5(ˉs)[r],G6(ˉs)[r])

    defined by

    {G1(ˉs)[r]=˜ntΔ˜n+ˉu˜n+˜uˉn+(ˉn˜c)+(˜nˉc)γ˜n+2μ˜nˉn,inQ,G2(ˉs)[r]=˜ctΔ˜c+ˉu˜c+˜uˉc+˜c˜n˜f,inQ,G3(ˉs)[r]=˜utΔ˜u+ˉu˜u+˜uˉu˜nφ,inQ,˜u=0,inQ,˜nν=˜cν=0,˜u=0,on(0,T)×Ω,˜n(0)=˜n0,˜c(0)=˜c0,˜u(0)=˜u0,inΩ.

    Lemma 5.3. Let ˉs=(ˉn,ˉc,ˉu,ˉf)Sad, then ˉs is a regular point.

    Proof. For any fixed (ˉn,ˉc,ˉu,ˉf)Sad, we set (gn,gc,gu,˜n0,˜c0,˜u0)Y. Since 0C(ˉf), it suffices to show the existence of (˜n,˜c,˜u)Yn×Yc×Yu such that

    {˜ntΔ˜n+ˉu˜n+˜uˉn+(ˉn˜c)+(˜nˉc)γ˜n+2μ˜nˉn=gn,in Q,˜ctΔ˜c+ˉu˜c+˜uˉc+˜c˜n=gc,in Q,˜utΔ˜u+ˉu˜u+˜uˉu˜nφ=gu,in Q,˜u=0,in Q,˜nν=˜cν=0,˜u=0,on (0,T)×Ω,˜n(0)=˜n0,˜c(0)=˜c0,˜u(0)=˜u0,in Ω. (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:(˙n,˙u)Xn×Xu(˜n,˜u)Yn×Yu with (˜n,˜c,˜u) solving the decoupled problem:

    {˜ntΔ˜n+ˉu˜n+˜uˉn+(ˉn˜c)+(˜nˉc)γ˜n+2μ˙nˉn=gn,in Q,˜ctΔ˜c+ˉu˜c+˜uˉc+˜c˙n=gc,in Q,˜utΔ˜u+ˉu˜u+˙uˉu˙nφ=gu,in Q. (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:Xn×XuXn×Xu is well-defined and compact.

    Similar to the proof of Theorem 3.1, (˜n,˜u) solves the coupled problem (ˉn,ˉc,ˉu,ˉf)Sad, and we set (gn,gc,gu,˜n0,˜c0,˜u0)Y. Since 0C(ˉf), it suffices to show the existence of (˜n,˜c,˜u)Yn×Yc×Yu such that

    {˜ntΔ˜n+˜n=ˉu˜n˜uˉn(ˉn˜c)(˜nˉc)+α(γ+1)˜n2μ˜nˉn+αgn,in Q,˜ctΔ˜c+˜c=ˉu˜c˜uˉc+α˜n+αgc,in Q,˜utΔ˜u=ˉu˜u˜uˉu+α˜nφ+αgu,in Q, (5.7)

    complemented by the corresponding Neumann boundary and initial conditions.

    Taking the L2-inner product with ˜u for the third equation of (5.7) implies

    12Ω˜u2dx+Ω|˜u|2dx=αΩ˜nφ˜udx+αΩ˜ugudx.

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

    ddt˜u2L2+˜u2H1C(˜n2L2+gu2L2)+C˜u2L2. (5.8)

    Taking the L2-inner product with ˜c for the second equation of (5.7) implies

    12Ω˜c2dx+Ω|˜c|2dx+Ω˜c2dx=Ω˜uˉc˜cdx+αΩ˜n˜cdx+αΩgc˜cdx.

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

    ddt˜c2L2+˜c2H1C(˜n2L2+gc2L2)+C˜c2L2. (5.9)

    Taking the L2-inner product with Δ˜c for the second equation of (5.7) implies

    12Ω|˜c|2dx+Ω|Δ˜c|2dx+Ω|˜c|2dx=Ω˜uˉcΔ˜cdx+Ωˉu˜cΔ˜cdxαΩ˜nΔ˜cdxαΩgcΔ˜cdx=J1+J2+J3.

    For the term J1

    J1=Ω˜uˉcΔ˜cdxΔ˜cL2ˉcL4˜uL416Δ˜c2L2+Cˉc2H1˜u2H1.

    For the term J2, we see that

    J2=Ωˉu˜cΔ˜cdx=12Ωˉu|˜c|2dxˉuL2˜c2L4ˉuL2(˜c12L2Δ˜c12L2+˜cL2)16Δ˜c2L2+C˜c2L2.

    For the term J3, we get

    J3=αΩ˜nΔ˜cdxαΩgcΔ˜cdx16Δ˜c2L2+C(˜n2L2+gc2L2).

    Therefore, combining J1, J2 and J3, we have

    ddt˜c2L2+˜c2H1C˜c2L2+C(˜n2L2+gc2L2). (5.10)

    Taking the L2-inner product with ˜n for the first equation of (5.7) implies

    ddtΩ˜n2dx+Ω|˜n|2dx+Ω˜n2dx=Ω˜uˉn˜ndx+Ω˜nˉn˜cdx+Ω˜n˜nˉcdx+α(γ+1)Ω˜n2dx+2μΩˉn˜n2dx+αΩ˜ngndx=J4+J5+J6+J7.

    For the term J4, by Gagliardo-Nirenberg interpolation inequality, we have

    J4=Ω˜uˉn˜ndx˜uL4ˉnL2˜nL4C(˜u12L2˜u12L2+˜uL2)ˉnL2˜nH1δ˜n2H1+C˜uL2˜uL2+C˜u2L2δ˜n2H1+δ˜u2L2+C˜u2L2.

    For the term J5,

    J5=Ω˜nˉn˜cdx˜nL2ˉnL4˜cL4˜nL2ˉnH1(˜c12L2Δ˜c12L2+˜cL2)δ˜n2L2+˜cL2Δ˜cL2+C˜c2L2δ˜n2L2+δΔ˜cL2+C˜c2L2.

    For the term J6,

    J6=Ω˜n˜nˉcdx˜n2L4ΔˉcL2(˜n12L2˜n12L2+˜nL2)ΔˉcL2δ˜n2L2+C˜n2L2+C.

    For the term J7,

    J7=α(γ+1)Ω˜n2dx+2μΩˉn˜n2dx+αΩ˜ngndx(γ+1)˜n2L2+gnL2˜nL2+ˉnL2˜n2L4(γ+1)˜n2L2+gnL2˜nL2+ˉnL2(˜n12L2˜n12L2+˜nL2)δ˜nL2+C˜n2L2+Cgn2L2.

    Therefore, by choosing δ small enough, from J4, J5, J6 and J7, it follows that

    ddt˜n2L2+˜n2H1C(˜n2L2+˜c2L2+˜u2L2)+δΔ˜cL2+δ˜u2L2+Cgn2L2. (5.11)

    By choosing δ small enough and combining (5.8)-(5.11), we get

    ddt(˜n2L2+˜c2H1+˜u2L2)+˜n2H1+˜c2H2+˜u2H1C(gn2L2+gc2L2+gu2L2)+C(˜n2L2+˜c2H1+˜u2L2).

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

    ˜n2L2+˜c2H1+˜u2L2+t0˜n2H1dτ+t0˜c2H2dτ+t0˜u2H1dτC. (5.12)

    Taking the L2-inner product with Δ˜u for the third equation of (5.7) implies

    12ddtΩ|˜u|2dx+Ω|Δ˜u|2dx=Ωˉu˜uΔ˜udx+Ω˜uˉuΔ˜udxαΩ˜nφΔ˜udxαΩguΔ˜udx=J8+J9+J10.

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

    J8=Ωˉu˜uΔ˜udxˉuL4˜uL4Δ˜uL2ˉuH1(˜u12L2Δ˜u12L2+˜uL2)Δ˜uL2δΔ˜u2L2+C˜u2L2

    and

    J9=Ω˜uˉuΔ˜udxΔ˜uL2ˉuL4˜uL4CΔ˜uL2ˉuH1(˜u12L2˜u12L2+˜uL2)δΔ˜u2L2+C˜u2L2.

    For the term J10, we deduce

    J10=αΩ˜nφΔ˜udxαΩguΔ˜udxδΔ˜u2L2+C(˜n2L2+gu2L2).

    By choosing δ small enough, with the estimates J8, J9 and J10, we have

    ddt˜u2L2+Δ˜u2L2C˜u2L2+Cgu2L2. (5.13)

    Applying to the first equation of (5.7), multiplying it by Δ˜c, and integrating over Ω give

    12ddtΩ|Δc|2dx+Ω|Δc|2dx+Ω|Δc|2dx=Ω(ˉu˜c)Δ˜cdxΩ(˜uˉc)Δ˜cdx+αΩ˜nΔ˜cdx+αΩgcΔ˜cdx=J11+J12+J13.

    For the first term J11, we have

    J11=Ω(ˉu˜c)Δ˜cdx=Ωˉu˜cΔ˜cdxΩˉuΔ˜cΔ˜cdxΔ˜cL2ˉuL4˜cL4+Δ˜cL2ˉuL4Δ˜cL4Δ˜cL2(ˉu12L2Δˉu12L2+ˉuL2)(ˉc12L2Δˉc12L2+ˉcL2)+Δ˜cL2ˉuH1(Δ˜c12L2Δ˜c12L2+Δ˜cL2)δΔ˜c2L2+CΔˉu2L2+CΔ˜c2L2.

    Similarly, for the term J12,

    J12=Ω(˜uˉc)Δ˜cdx=Ω˜uˉcΔ˜cdxΩ˜uΔˉcΔ˜cdxΔ˜cL2˜uL4ˉcL4+˜uL4ΔˉcL4Δ˜cL2CΔ˜cL2(˜u12L2Δ˜u12L2+˜uL2)ˉcH1+(˜u12L2˜u12L2+˜uL2)(Δˉc12L2Δˉc12L2+ΔˉcL2)Δ˜cL2δΔ˜c2L2+δΔ˜u2L2+CΔˉc2L2+C˜u2L2.

    For the rest term J13, we see

    J13=αΩ˜nΔ˜cdx+αΩgcΔ˜cdxδΔ˜c2L2+C(˜n2L2+gc2L2).

    By choosing δ small enough, we get

    ddtΔ˜c2L2+Δ˜c2H1C(˜n2L2+Δ˜c2L2+˜u2L2)+CΔˉu2L2+δΔ˜u2L2+CΔˉc2L2+Cgc2L2. (5.14)

    From (5.13) and (5.14), along with δ small enough, it follows that

    ddt(˜u2L2+Δ˜c2L2)+Δ˜u2L2+Δ˜c2H1C(˜u2L2+Δ˜c2L2)+(˜n2L2+Δˉu2L2+Δˉc2L2+gc2L2)+Cgu2L2.

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

    ˜u2L2+Δ˜c2L2+t0Δ˜u2L2dτ+t0Δ˜c2H1dτC.

    Taking the L2-inner product with Δ˜n for the first equation of (5.7) implies

    12ddtΩ|˜n|2dx+Ω|Δ˜n|2dx+Ω|˜n|2dx=Ωˉu˜nΔ˜ndxΩ˜uˉnΔ˜ndxΩ(˜nˉc)Δ˜ndxΩ(ˉn˜c)Δ˜ndxα(1+γ)Ω˜nΔ˜ndx+2μΩ˜nˉnΔ˜ndxαΩgnΔ˜ndx=J14+J15+J16+J17+J18.

    With the Gagliardo-Nirenberg interpolation inequality in hand, we can estimate J14 as follows

    J14=Ωˉu˜nΔ˜ndxˉuL4˜nL4Δ˜nL2CˉuH1(˜n12L2Δ˜n12L2+˜nL2)Δ˜nL2δΔ˜n2L2+C˜n2L2.

    Similar to above estimates, we see

    J15=Ω˜uˉnΔ˜ndx˜uL4ˉnL4Δ˜nL2C˜uH1ˉnH1Δ˜nL2δΔ˜nL2+Cˉn2H1.

    Similarly, we derive

    J16=Ω(˜nˉc)Δ˜ndx=Ω˜nˉcΔ˜ndxΩ˜nΔˉcΔ˜ndx˜nL4ˉcL4Δ˜nL2+˜nL4ΔˉcL4Δ˜nL2(˜n12L2Δ˜n12L2+˜nL2)ˉcH1Δ˜nL2+(˜n12L2˜n12L2+˜nL2)(Δˉc12L2Δˉc12L2+ΔˉcL2)Δ˜nL2δΔ˜n2L2+C˜n2L2+CΔˉc2L2+C

    and

    J17=Ω(ˉn˜c)Δ˜ndx=Ωˉn˜cΔ˜ndxΩˉnΔ˜cΔ˜ndxˉnL4˜cL4Δ˜nL2+ˉnL4Δ˜cL4Δ˜nL2(ˉn12L2Δˉn12L2+ˉnL2)˜cH1Δ˜nL2+ˉnH1(Δ˜c12L2Δ˜c12L2+Δ˜cL2)Δ˜nL2δΔ˜n2L2+CΔ˜c2L2+C.

    For the rest terms, we know

    J18=α(1+γ)Ω˜nΔ˜ndx+2μΩ˜nˉnΔ˜ndxαΩgnΔ˜ndx(1+γ)˜nL2Δ˜nL2+2μ˜nL4ˉnL4Δ˜nL2+gnL2Δ˜nL2(1+γ)˜nL2Δ˜nL2+C(˜n12L2˜n12L2+˜nL2)ˉnH1Δ˜nL2+gnL2Δ˜nL2δΔ˜n2L2+C˜n2L2+Cgn2L2.

    Therefore, Taking δ small enough and together with J14J18, we see that

    ddt˜n2L2+˜n2H1C(˜n2L2+ˉn2H1+Δˉc2L2+Δ˜c2L2+gn2L2)+C.

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

    ˜n2L2+t0˜n2H1dτC.

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

    Theorem 5.1. Assume that ˉs=(ˉn,ˉc,ˉu,ˉf)Sad be an optimal solution for the control problem (5.3). Then, there exist Lagrange multipliers (λ,η,ρ,ξ,φ,ω)L2(Q)×(L2(0,T;H1(Ω)))×L2(Q)×(H1(Ω))×(H2(Ω))×(H1(Ω)) such that for all (˜n,˜c,˜u,˜f)Vn×Vc×Vu×C(ˉf) has

    β1T0Ωd(ˉnnd)˜ndxdt+β2T0Ωd(ˉccd)˜cdxdt+β3T0Ωd(ˉuud)˜udxdt+β4Ωd(ˉn(T)nΩ)˜n(T)dx+β5Ωd(ˉc(T)cΩ)˜c(T)dxT0Ω(˜ntΔ˜n+ˉu˜n+˜uˉn+(ˉn˜c)+(˜nˉc)γ˜n+2μ˜nˉn)λdxdtT0Ω(˜ctΔ˜c+ˉu˜c+˜uˉc+˜c˜n)ηdxdt+β7T0Ωd˜fˉfdxdtT0Ω(˜utΔ˜u+ˉu˜u+˜uˉu˜nφ)ρdxdt+Ω˜n(0)ξdx+Ω˜c(0)φdx+Ω˜u(0)ωdx+β6Ωd(ˉu(T)uΩ)˜u(T)dx+T0Ω˜fηdxdt0, (5.15)

    where C(ˉf)={θ(fˉf):θ0,fU}.

    Proof. With the Lemma 5.3 in hand, we get that ˉsSad is a regular point. Then, togather with Theorem 3.1 in [20], it follows that there exist Lagrange multipliers (λ,η,ρ,ξ,φ,ω)L2(Q)×(L2(0,T;H1(Ω)))×L2(Q)×(H1(Ω))×(H2(Ω))×(H1(Ω)) such that

    J(ˉs)[r]G1(ˉs)[r],λG2(ˉs)[r],ηG3(ˉs)[r],ρG4(ˉs)[r],ξG5(ˉs)[r],φG6(ˉs)[r],ω0,

    for all r=(˜n,˜c,˜u,˜f)Vn×Vc×Vu×C(ˉf). Hence, the proof follows from Lemmas 5.1 and 5.2.

    Corollary 5.1. Assume that ˉs=(ˉn,ˉc,ˉu,ˉf)Sad be an optimal solution for the control problem (5.3). Then, there exist Lagrange multipliers (λ,η,ρ)L2(Q)×(L2(0,T;H1(Ω)))×L2(Q), satisfying

    T0Ω(˜ntΔ˜n+ˉu˜n+(˜nˉc)γ˜n+2μ˜nˉn)λdxdtT0Ω˜nηdxdtT0Ω˜nφρdxdt=β1T0Ωd(ˉnnd)˜ndxdt, (5.16)
    T0Ω(˜ctΔ˜c+ˉu˜c+˜c)ηdxdt+T0Ω(ˉn˜c)λdxdt=β2T0Ωd(ˉccd)˜cdxdt, (5.17)
    T0Ω(˜utΔ˜u+ˉu˜u+˜uˉu)ρdxdt+T0Ω˜uˉnλdxdt+T0Ω˜uˉcηdxdt=β3T0Ωd(ˉuud)˜udxdt, (5.18)

    which corresponds to the linear system

    {λtΔλ+ˉuλλˉcγλ+2μλˉnηφρ=β1(ˉnnd),ηtΔη+ˉuη+η+(ˉnλ)=β2(ˉccd),ρtΔρ+(ˉu)ρ+(ρT)ˉu+λˉn+ηˉc=β3(ˉuud), (5.19)

    subject to the following boundary and final conditions

    {ρ=0,inQ,λν=ην,ρ=0,on(0,T)×Ω,λ(T)=β4(ˉn(T)nΩ),η(T)=β5(ˉc(T)cΩ),ρ(T)=β5(ˉc(T)cΩ),inΩ,

    and the following identities hold

    T0Ωd(β7ˉf+η)(fˉf)dxdt0,fU. (5.20)

    Proof. By taking (˜c,˜u,˜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 (˜n,˜c,˜u)=(0,0,0), as an immediate consequence we obtain

    β7T0˜fˉfdxdt+T0˜fηdxdt0,˜fC(ˉf).

    By choosing ˜f=fˉfC(ˉf) for all ˉfU, 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

    λ2H1+η2L2+ρ2L2+t0λ2H2dτ+t0η2H1dτ+t0ρ2H1dτC.

    Proof. For convenience, we set ˜λ=λ(Tt), ˜η=η(Tt), ˜ρ=ρ(Tt), in order to simplify notations, we still write λ, η, ρ instead of ˜λ, ˜η, ˜ρ, then the adjoint system (5.19) can be written as follow

    {λtΔλ+ˉuλλˉcγλ+2μλˉnηφρ=β1(ˉnnd), in Q,ηtΔη+ˉuη+η+(ˉnλ)=β2(ˉccd), in Q,ρtΔρ+(ˉu)ρ+(ρT)ˉu+λˉn+ηˉc=β3(ˉuud), in Q, (5.21)

    subject to the following boundary and final conditions

    {ρ=0, in Q,λν=ην,ρ=0, on (0,T)×Ω,λ(0)=β4(ˉn(T)nΩ),η(0)=β5(ˉc(T)cΩ),ρ(0)=β5(ˉc(T)cΩ), in Ω.

    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 L2-inner product with λ for the first equation of (5.21) implies

    12ddtΩλ2dx+Ω|λ|2dx+2μΩλ2ˉndx=Ωλˉcdx+γΩλ2dx+Ωληdx+Ωλφρdx+β1Ω(ˉnnd)λdxλL2ˉcL2+γλ2L2+λL2(ηL2+ρL2)+β1ˉnndL2λL212λ2L2+C(λ2L2+η2L2+ρ2L2)+Cˉnnd2L2.

    Then, we have

    ddtλ2L2+λ2H1C(λ2L2+η2L2+ρ2L2)+Cˉnnd2L2. (5.22)

    Taking the L2-inner product with Δη for the first equation of (5.21) implies

    12ddtΩ|λ|2dx+Ω|Δλ|2dx=ΩˉuλΔλdxΩλˉcΔλdxγΩλΔλdx+2μΩλˉnΔλdxΩηΔλdxΩφρΔλdx+β1Ω(ˉnnd)ΔλdxˉuL4λL4ΔλL2+λL4ˉcL4ΔλL2+γλ2L2+λL4ˉnL4ΔλL2+ηL2ΔλL2+ρL2ΔλL2+β1ΔλL2ˉnnd2L2ˉuH1(λ12L2Δλ12L2+λL2)ΔλL2+γλ2L2+(λ12L2Δλ12L2+λL2)ˉcH1ΔλL2+ηL2ΔλL2+ρL2ΔλL2+β1ΔλL2ˉnnd2L212Δλ2L2+C(λ2L2+η2L2+ρ2L2).

    Thus, we get

    ddtλ2L2+λ2H1C(λ2L2+η2L2+ρ2L2)+Cˉnnd2L2. (5.23)

    Taking the L2-inner product with η for the second equation of (5.21) implies

    12ddtΩη2dx+Ω|η|2dx+Ωη2dx=Ωˉnληdx+β2Ωη(ˉccd)dxˉnL4λL4ηL2+β2ηL2ˉccdL2ˉnH1(λ12L2Δλ12L2+λL2)ηL2+β2ηL2ˉccdL212η2L2+δΔλ2L2+CλL2+Cη2L2+Cˉccd2L2.

    As an immediate consequence, we obtain

    ddtη2L2+η2H1δΔλ2L2+CλL2+Cη2L2+Cˉccd2L2. (5.24)

    Taking the L2-inner product with ρ for the third equation of (5.21) implies

    12ddtΩρ2dx+Ω|ρ|2dx=Ω(ρT)ˉuρdxλΩˉnρdxΩηˉcρdx+β3Ω(ˉuud)ρdxρL2ˉuL4ρL4+λˉnL2ρL2+ηL2ˉcL4ρL4+β3ρL2ˉuudL2ρL2ˉuH1(ρ12L2ρ12L2+ρL2)+λˉnL2ρL2+ηL2ˉcH1(ρ12L2ρ12L2+ρL2)+β3ρL2ˉuudL212ρ2L2+Cρ2L2(ˉu2H1+1)+Cη2L2+Cˉuud2L2.

    Therefore, we see that

    ddtρ2L2+ρ2H1Cρ2L2(ˉu2H1+1)+Cη2L2+Cˉuud2L2. (5.25)

    Combining (5.22)-(5.25) and taking δ small enough, we have

    ddt(λ2H1+η2L2+ρ2L2)+λ2H2+η2H1+ρ2H1C(ˉu2H1+1)(λ2H1+η2L2+ρ2L2)+Cˉnnd2L2+Cˉccd2L2+Cˉuud2L2.

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

    λ2H1+η2L2+ρ2L2+t0λ2H2dτ+t0η2H1dτ+t0ρ2H1dτC.

    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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