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Research article

Global existence of classical solutions for the 2D chemotaxis-fluid system with logistic source

  • Received: 12 December 2021 Revised: 17 January 2022 Accepted: 25 January 2022 Published: 09 February 2022
  • MSC : 35K55, 35Q92, 35Q35, 92C17

  • In this paper, we consider the incompressible chemotaxis-Navier-Stokes equations with logistic source in spatial dimension two. We first show a blow-up criterion and then establish the global existence of classical solutions to the system for the Cauchy problem under some rough conditions on the initial data.

    Citation: Yina Lin, Qian Zhang, Meng Zhou. Global existence of classical solutions for the 2D chemotaxis-fluid system with logistic source[J]. AIMS Mathematics, 2022, 7(4): 7212-7233. doi: 10.3934/math.2022403

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  • In this paper, we consider the incompressible chemotaxis-Navier-Stokes equations with logistic source in spatial dimension two. We first show a blow-up criterion and then establish the global existence of classical solutions to the system for the Cauchy problem under some rough conditions on the initial data.



    The coupled chemotaxis-Navier-Stokes model with logistic source terms is as following: QT=(0,T]×R2

    {nt+un=DnΔn(nχ(c)c)+κnμn2,(x,t)QT,ct+uc=DcΔcg(c)n,(x,t)QT,ut+(u)uP=DuΔu+nΦ,(x,t)QT,u=0,(x,t)QT. (1.1)

    Here the unknows are the concentration of bacteria n=n(x,t):QTR+; the oxygen concentration c=c(x,t):QTR+; the given vector field u=u(x,t):QTR2; and the pressure of the fluid P=P(x,t):QTR. We denote the corresponding diffusion coefficients for the cells, the oxygen and the fluid by Dn,Dc and Du.

    Apart from that, χ(c) represents the chemotactic sensitivity, and g(c) is the consumption rate of oxygen. In [19], it is shown that the functions g(c) and χ(c) are constants at large c and rapidly approach zero below some critical c. Moreover, the experimentalists in [19] used multiples of the Heaviside step function to model χ() and g(). The scalar valued function Φ is also given.

    The model (1.1) was proposed in [12], which in order to describe the movement of bacteria in response to the presence of a chemical signal substance (oxygen or another nutrient) in their fluid environment. κ is the effective growth rate of the population and μ controlling death by overcrowding. For more the logistic term related, see [10,11,24,35].

    In recent years, the study of the dynamics of solutions to (1.1) has attracted many researchers. Finite time blow-up phenomena is one of the most important dynamical problems in (1.1). One can refer to [3,30]. It is worth pointing that the logistic term κnμn2 of (1.1) can avoid blow-up phenomena in [9].

    To make the system (1.1) be well-posed, we add some initial conditions as follows:

    (n,c,u)|t=0=(n0(x),c0(x),u0(x)),xR2. (1.2)

    Besides, we also have the corresponding chemotaxis-only subsystem of (1.1) on letting u0, that is, the system

    {nt=DnΔn(nχ(c)c)+κnμn2,ct=DcΔcc+n, (1.3)

    is a variant of the classical Keller-Segel model. Up to now, one notices that the homogeneity of model (1.3) may be undermined by the cross-diffusive term (nχ(c)c) and even enforce blow-up of solutions [1]. Although we are facing great challenges, there are still numerous analytic results which mainly fixed attention on the local and global solvability of corresponding initial(-boundary)-value problems in either bounded or unbounded domains Ω, under diverse technical conditions on χ() and g().

    As for initial boundary value problem (1.3) in higher space dimensions, Winkler [22] proved the global existence of a small solution to (1.3) and that it asymptotically behaves like the solution of a decoupled system of linear parabolic equations. On the other hand, a result concerning that blow-up behavior occurring for some radially symmetric positive initial data in higher dimension was recently obtained in [23]. And one can see a relevant reference in [25].

    If system (1.3) is coupled with Navier-Stokes equations, we have the following model

    {nt+un=DnΔn(nχ(c)c)+κnμn2,ct+uc=DcΔcc+n,ut+uu=DuΔuPnϕ+f(x,t). (1.4)

    As far as we know, results on smooth global solvability could be established for the two-dimensional version of (1.4) whenever μ is positive ([18]), while a similar statement could be derived when μ23 in three dimension at least for a Stokes simplification of (1.4) in which the nonlinear convective term uu is neglected [17]. In a recent paper [31], Winkler reveals that whenever ω>0, requiring that

    κmin{μ,μ32+ω}<η

    with some η=η(ω)>0, and that f satisfies a suitable assumption on ultimate smallness, is sufficient to ensure that each of these generalized solutions becomes eventually smooth and classical.

    For system (1.1), when κ=μ=0, [6] demonstrates that there exists a global weak solutions for the 2D chemotaxis-Stokes equations, i.e., the nonlinear convective term uu is removed in the fluid equation of (1.1), by making use of quasi-energy functionals associated with (1.1), under the following conditions on χ() and g():

    χ(c)>0,χ(c)0,g(0)=0,g(c)>0,d2dc2(g(c)χ(c))<0, (1.5)

    which were relaxed in [8,13],

    χ(c)0,χ(c)0,g(c)0,g(c)0,
    d2dc2(g(c)χ(c))0,χ(c)g(c)+χ(c)g(c)χ(c)>0, (1.6)

    for the fully chemotaxis-Navier-Stokes system (1.1) without any smallness conditions on the initial data.

    Furthermore, for the smooth initial data, Chae et al. establish the global existence of smooth solutions [3] by assuming that χ(c) and g(c) satisfy

    χ(c)0,χ(c)0,g(c)0,g(c)0,g(0)=0, (1.7)

    and that there exists a constant η such that

    supc0|χ(c)ηg(c)|<ϵfor a sufficiently smallϵ>0, (1.8)

    which is taken away in [4]. For the full 3D chemotaxis-Navier-Stokes system, the global classical solution near constant steady states and the global weak solutions under the special situation that χ() precisely coincides with a fixed multiple of g() are constructed in [6] and [3], respectively.

    In the case of κ=μ=0, Winkler establishes the existence of global classical solutions in 2D bounded convex domains Ω in [26]. Then he [27] asserts that a solution of the two-dimensional chemotaxis-Navier-Stokes system stabilizes to the spatially uniform equilibrium (ˉn,0,0) with ¯n0=1|Ω|Ωn0(x)dx in the sense of L(Ω). If κ0, μ>0, in a bounded smooth domain ΩR3, J. Lankeit [12] shows that after some waiting time weak solutions constructed become smooth and finally converge to the semi-trivial steady state (κμ,0,0). More related paper on the bounded domain case, for the 2D case, one may refer to [16,20,21] and for the 3D case, one also refer to [2,16,21,28,29]. Additionally, under some strong structural assumptions on χ and g, the global existence of weak solutions is proved as well as their eventual smoothness and stabilization to the 3D version of the chemotaxis-Navier-Stokes system is established in [28] and [29].

    Recently, Duan et al. [7] established the global existence of weak solutions and classical solutions for both the Cauchy problem and the initial-boundary value problem supplemented with some initial data. In addition, Wu et al. [32] improved the results by taking more careful calculations than those of [7].

    The main purpose of this paper is to demonstrate the global existence of classical solutions to (1.1) for the Cauchy problem. We first introduce the corresponding blowup criterion.

    Now, it is position to state our main theorems in this article.

    Theorem 1.1. Let m3. Assume that χ(), g()Cm(R) with g(0)=0, and that lΦL(R2)< for 1|l|m.

    (A1) Then there exists T>0, the maximal time of existence, such that, if the initial data (n0,c0,u0)Hm1(R2)×Hm(R2)×Hm(R2;R2), then there exists a unique classical solution (n,c,u) to system (1.1)-(1.2) satisfying for any T<T

    (n,c,u)L(0,T;Hm1(R2)×Hm(R2)×Hm(R2;R2))

    and

    (n,c,u)L2(0,T;Hm1(R2)×Hm(R2)×Hm(R2;R2)).

    (A2) Moreover, if the maximal time of existence T<, then

    T0c(τ)2L(R2)dτ= (1.10)

    The proof of A1 in Theorem 1.1 is standard, one can refer to [3,33]. We will focus on proving part A2.

    Theorem 1.2. Let m3. Under the assumptions of Theorem 1.1, then system (1.1)-(1.2) admits a unique global-in-time classical solution (n,c,u) satisfying for any T>0

    (n,c,u)L(0,T;Hm1(R2)×Hm(R2)×Hm(R2;R2))

    and

    (n,c,u)L2(0,T;Hm1(R2)×Hm(R2)×Hm(R2;R2)).

    Before we are going to prove the main results, we give an important proposition, which can be found in [14].

    Let

    X0={n0L1L2, n0>0,c0L2, c0L1L, c0>0,u0L2}.

    Proposition 1.1. Let the triple (n0,c0,u0)X0, ΦL(R2) and χ(), g()Cm(R) with m3 and g(0)=0. Then, system (1.1) has a unique global solution (n,c,u) such that

    nL(R+;L1(R2))Lloc(R+;L2(R2))L2loc(R+;H1(R2));cL(R+;L1(R2)L(R2))Lloc(R+;H1(R2))L2loc(R+;H2(R2));uL(R+;L1(R2))L2loc(R+;H1(R2)).

    The rest of this article is organized as follows. Let's briefly establish a blow-up criterion in Section 2, discuss the Cauchy problem in Section 3.

    Now, we show the proof of the blow-up criterion for the fluid chemotaxis equations.

    Proof. Above all, we take the L2 estimate of n into account. Multiplying n on both sides of (1.1)1 and integrating in spaces, by the fact that χ is continuous and c is uniformly bounded, we deduce

    12ddtn2L2+Dnn2L2+μn3L3Cχ(c)ncL2nL2+κn2L2Cc2Ln2L2+14n2L2+κn2L2.

    Next, testing Δc to both sides of (1.1)2 and making use of the integration by parts, it follows that

    12ddtc2L2+DcΔc2L2(uc)(u)cL2cL2+Cg(c)nL2ΔcL2CuLc2L2+Cn2L2+14Δc2L2.

    Similary, taking the L2 inner product Δu to both sides of (1.1)3 and using the integration by parts, we also deduce

    12ddtu2L2+DuΔu2L2((u)u)(u)uL2uL2+CnL2ΔuL2CuLu2L2+Cn2L2+CΔu2L2.

    From all the estimates, it means that

    ddt(n2L2+c2L2+u2L2)+(n2L2+Δc2L2+Δu2L2)C(1+uL+c2L)(n2L2+c2L2+u2L2).

    An application of Gronwall's inequality yields that

    sup(n2L2+c2L2+u2L2)+T0(n2L2+Δc2L2+Δu2L2)dtC(n02L2+c02L2+u02L2)exp(T01+uL+c2L).

    Observe that nL(0,T;L2) and nL2(0,T;L2) are uniformly bounded if T0uL+c2Ldt is bounded. Meanwhile, we notice that nLqxLt and nq2L2xL2t for all 2<q<, and, owing to the following deduction

    1qddtnqLq+Dnnq22L2+μR2nq+1dxCR2|nχ(c)cnq1|dx+κR2nqdxCc2LnqLq+12nq22L2+κnqLq.

    Collecting the above inequality, it ensures that n(t)LqC, with C is independent of q. Then, letting q, nLxLt is obtained.

    Additionally, we consider the estimate in the space (n,c,u)H1×H2×H2. It follows that

    12ddtn2L2+DnΔn2L2+2μn12n2L2C(un)(u)nL2nL2+Cχ(c)ncL22nL2+Cχ(c)nΔcL22nL2+Cχ(c)nccL22nL2+κn2L2CuLn2L2+CnL2cL2nL2+CnLΔcL22nL2+CnLcLcL22nL2+κn2L2CuLn2L2+Cn2L2c2L+182n2L2+Cn2LΔc2L2+182n2L2+Cn2Lc2Lc2L2+182n2L2+κn2L2.

    In conjunction with Young's inequality and Gronwall's inequality, we conclude

    supn2L2+T02n2L2dt(n02L2+Cn2L(0,T;L)(T0Δc2L2dt+c2L(0,T;L2)T0c2Ldt))×exp(T01+uL+c2Ldt).

    Afterwards, nH1xLtH2xL2t. For the H2 estimate of c, we get

    12ddtΔc2L2+DcΔc2L2CuLΔc2L2+CΔuL2cLΔcL2+Cg(c)ncL2ΔcL2+Cg(c)ncL2ΔcL2CuLΔc2L2+CΔuL2cLΔcL2+CcL2nLΔcL2+CnL2ΔcL2CuLΔc2L2+CΔu2L2c2L+18Δc2L2+Cc2L2n2L+18Δc2L2+Cn2L2+18Δc2L2,

    and

    12ddtΔu2L2+DuΔu2L2Cuu2L2+CΔnL2ΔuL2Cuu2L2+CΔn2L2+12Δu2L2.

    Then we get cH2xLtH3xL2t and uH2xLtH3xL2t by Gronwall's inequality. In what follows, we show the estimate in the space (n,c,u)H2×H3×H3. In a similar way, it indicates

    12ddtn2H2+Dnn2H2CuLnH2nH2+CuLnH1nH2Cχ(c)ncH2nH2+κn2H2+CnLn2H2Cu2Ln2H2+Cu2Ln2H1+Cχ(c)nc2H2+κn2H2+CnLn2H2+12n2H2.

    We control

    χ(c)ncH2CnH2χ(c)cH2,

    and

    2(χ(c)c)L2C3cL2+C2cL2cL+Cc3L6.

    We also obtained cH2xLtH3xL2t. Consequently, in view of Young's inequality and Gronwall's inequality, it follows that

    supn2H2+T0n2H2dtn02H2exp(C+T01+uL+c2Ldt).

    In what follows, for the estimate of c, we further have

    12ddtc2H3+Dcc2H3CuLcH3cH3+CuH3cLcH3+Cg(c)n2H2+14c2H3Cu2Lc2H3+Cu2H3c2L+C(c2H2n2H2+c2H1c2Ln2H2)+12c2H3.

    For the estimate of u, we thereby obtain

    12ddtu2H3+Duu2H3CuLuH3uH3+CnH2uH3Cu2Lu2H3+Cn2H2+12u2H3.

    Applying Gronwall's inequality, it implies (c,u)(H3xLtH4xL2t)×(H3xLtH4xL2t). Let us demonstrate Hm1×Hm×Hm estimates. We've proved that the case m=2,3 and 4, and therefore we consider the case m5. Operating α(|α|m1) and multiplying αn on both sides of (1.1)1 and integrating in spaces, we thereby obtain

    12ddtn2Hm1+Dnn2Hm1CuLnHm1nHm1+CuHm1nLnHm1+Cχ(c)ncHm1nHm1+κn2Hm1+CnLn2Hm1Cu2Ln2Hm1+Cu2Hm1n2L+Cχ(c)nc2Hm1+κn2Hm1+CnLn2Hm1+12n2Hm1.

    The estimate for the case m=4 was already obtained, thus cL(0,T;L) is bounded. Subsequently, we derive

    χ(c)Hm2C(1+cL)χ(c)Hm3.

    the classical product lemma on each step of iteration shows

    χ(c)Hm2C(1+cL)m1.

    Therefore we also obtain

    χ(c)ncHm1C(1+cHm+cmL)nHm1

    applying the product lemma. Add up the inequality above and we get

    ddtn2Hm1+n2Hm1Cu2Ln2Hm1+Cu2Hm1n2L+C(1+cHm+cmL)2n2Hm1+κn2Hm1+CnLn2Hm1.

    Similarly, for the Hm estimate of c, we get

    12ddtc2Hm+Dcc2HmCuLcHmcHm+CuHmcLcHm+C(1+c2m2L)n2Hm1+14c2HmCu2Lc2Hm+Cu2Hmc2L+C(1+c2m2L)n2Hm1+12c2Hm.

    For the estimate of u, we also deduce that

    12ddtu2Hm+Duu2HmCuLuHmuHm+CnHm1uHmCu2Lu2Hm+Cn2Hm1+12u2Hm.

    Finally, by summing up all the above estimates and utilizing Gronwall's inequality, the fact (n,c,u)(Hm1xLtHmxL2t)×(HmxLtHm+1xL2t)×(HmxLtHm+1xL2t) is obtained.

    As a matter of fact that cL is solely responsible for nL2xLt and nL2xL2t by the above process. That is

    ddtn2L2+n2L2+μR2n3dxCR2|nχ(c)cn|dx+κn2L2Cc2Ln2L2+12n2L2+κn2L2.

    This yields uL2xLt and uL2xL2t by

    ddtu2L2+u2L2CnL2uL2Cn2L2+Cu2L2.

    Furthermore, we obtain nLqxLt and nq2L2xL2t for all 2<q<

    ddtnqLq+nq22L2+μR2nq+1dxCqR2|nχ(c)cnq1|dx+κRdnqdxCqc2LnqLq+12nq22L2+κnqLq.

    Besides, we note that cL2xLt and 2cL2xL2t. Actually,

    ddtc2L2+2c2L2CcLuL22cL2+CnL22cL2Cc2Lu2L2+Cn2L2+122c2L2.

    Finally, we denote vorticity as ω:=×u; where ω=1u22u1 in two dimensions. Then, let us set n=(2n,1n). We investigate the vorticity equation

    ωtΔω+uω=nΦ,

    We notice that ωL2xLt and ωL2xL2t, on account of

    ddtω2L2+ω2L2CnL2ωL2Cn2L2+Cω2L2.

    In addition, we note that ωL2xLt and 2ωL2xL2t. In fact, testing Δω, this implies that

    ddtω2L2+2ω2L2uL4ωL4ΔωL2+nL2ΔωL2Cu12L2u12L2ω12L22ω32L2+nL2ΔωL2Cu2L2u2L2ω2L2+Cn2L2+122ω2L2.

    Hence, via embedding, we get

    T0uLdtT0uH2dtCT0ωH2dt<.

    This implies the desired result.

    In this section, we will show that the local classical solutions can be extended at any time T>0.

    Proof of Theorem 1.2. The process is similar to the proof of Theorem 1.2 in [7]. Using contradictory methods, supposing that the maximal time T is finite, we will prove

    T0c(τ)2L(R2)dτ<,

    which contradicts the extensibility criterion (1.10). Actually, according to the fact

    c(τ)2L(R2)˜CGNcL2(R2)3cL2(R2)˜CGN2(c2L2(R2)+3c2L2(R2)),

    where ˜CGN is a positive constant resulted from the Gagliardo-Nirenberg inequality.

    Now, we just need to verify that

    T0R2|c(x,τ)|2dxdτ<, (3.1)

    and

    T0R2|3c(x,τ)|2dxdτ<. (3.2)

    According to Proposition 1, we obtain (3.1).

    As for (3.2), applying Δ to the (1.1)2, multiplying Δc with the resulted equation, and integrating over R2, we have

    12ddtR2|Δc|2dx+DcR2|Δc|2dx=R2((uc)+g(c)nc+g(c)n)Δcdx:=I1+I2+I3.

    For I1, we get

    I11DcR2|u|2|c|2dx+1DcR2|u|2|D2c|2dx+Dc4R2|Δc|2dx1Dcu2L3(R2)c2L6(R2)+1Dcu2L(R2)D2c2L2(R2)+Dc4R2|Δc|2dx.

    Similarly, let M=c0L, we conclude

    I21Dcsup0sMg2(s)R2|n|2|c|2dx+Dc8R2|Δc|2dx1Dcsup0sMg2(s)n2L3(R2)c2L6(R2)+Dc8R2|Δc|2dx,

    and

    I31Dcsup0sMg2(s)R2|n|2dx+Dc8R2|Δc|2dx1Dcsup0sMg2(s)R2|n|2dx+Dc8R2|Δc|2dx.

    Collecting I1I3, we obtain

    ddtR2|Δc|2dx+DcR2|Δc|2dxC2(u2H2(R2)+n2H1(R2))c2H1(R2)+C2R2|n|2dx,

    with C2 is a positive constant depending on the Sobolev's embedding and the initial data. The Gronwall's inequality implies that

    c(t)2H1(R2)+DcT02c(τ)2H1(R2)dτc02H2(R2)eC2T0(n(τ)2H1(R2)+u(τ)2H2(R2))dτ+C2T0n(τ)2H1(R2)dτ,

    for all t(0,T). Therefore we have verified (3.2) provided that

    T0(n2H1(R2)+u2H1(R2))(τ)dτ<.

    According to Proposition 1, we verify the above inequality. For convenience of readers, we give a sketch of the proof. To begin with, taking the L2-inner product with n for equation (1.1)1, we get

    12ddtR2n2dx+DnR2|n|2dx+μR2n3dx=R2nχ(c)cndx+κR2n2dxDn4R2|n|2dx+sup0sMχ2(s)DnR2n2|c|2dx+κR2n2dx.

    Next, we control R2n2|c|2dx of the above inequality

    R2n2|c|2dxn2L4(R2)c2L4(R2)CGNnL2(R2)n2L2(R2)cL(R2)ΔcL2(R2)Dn4R2|n|2dx+C3R2|Δc|2dxR2n2dx,

    with C3=sup0sMχ4(s)C2GNM2D3n, by using Sobolev's embedding, Young's inequality and the boundedness of c. Therefore, we have

    ddtR2n2dx+DnR2|n|2dx(2C3R2|Δc|2dx+κ)R2n2dx.

    In view of Gronwall's inequality, we deduce

    R2n2(x,t)dx+DnT0R2|n(x,τ)|2dxC4,

    for all t(0,T) and some positive constant C4 depending on the initial data and the maximal time T, where we also used the inequality

    T0R2|Δc(x,τ)|2dxdτ<,

    by recalling Proposition 1.1. A direct integration on [0,T] implies that

    T0n(τ)2H1(R2)dτ<. (3.3)

    Similarly, we also have

    T0u(τ)2H1(R2)dτ<. (3.4)

    Let's first investigate the integrability of the second derivative of u. For convenience, let ω:=u be the vorticity of u and then the vorticity equation as follow:

    ωt+(u)ω=DuΔω+(nΦ).

    Next, a direct energy method follows that

    12ddtR2|ω|2dx+DuR2|ω|2dx=R2ω(nΦ)dxDu2R2|ω|2dx+12DuR2n2|Φ|2dx.

    This leads to the vorticity estimate

    ddtR2|ω|2dx+DuR2|ω|2dxC5R2n2dx,

    where C5 is a positive constant depending on Φ. In conjunction with Gronwall's inequality and (3.3), we infer

    R2|ω|2(x,t)dx+DuT0R2|ω(x,τ)|2dxdτC6,

    for all t(0,T) with C6:=R2|ω|2(x,0)dx+C4C5T. Application of the above inequality and the Biot-Savart law, we derive

    T02u(τ)2L2(R2)dτ<. (3.5)

    Sum up (3.3)-(3.5), it implies the desired result. This completes the Proof of Theorem 1.2.

    Q. Zhang was partially supported by the Natural Science Foundation of Hebei Province [grant number A2020201014 and A2019201106]; the Second Batch of Young Talents of Hebei Province; Nonlinear Analysis Innovation Team of Hebei University.

    All authors declare that there is no interests in this paper.

    In this appendix, we will give a sketch proof of local existence.

    We first introduce the dynamic partition of the unity to define Besov spaces. One may check [15] for more details. Let φC0(Rd) be supported in C={ξRd,34|ξ|83} such that

    qZφ(2qξ)=1,forξ0.

    Defining χ(ξ)=1qNφ(2qξ). For fS, we set Littlewood-Paley operators as follows

    Δ1f=χ(D)f;qN,Δqf=φ(2qD)fandqZ,˙Δqf=φ(2qD)f.

    The following low-frequency cut-off will be also used:

    Sqf=1qq1Δqfand˙Sqf=qq1˙Δqf.

    Now, let us recall the definition of the Besov space. For sR, 1p,r, we define the homogenous Besov space ˙Bsp,r as the set of tempered distributions of fS/P such that

    f˙Bsp,r:=(qZ2qsr˙Δqfrp)1r<,

    where P is the polynomial space. The inhomogeneous space Bsp,r is the set of tempered distribution f such that

    fBsp,r:=(q12qsrΔqfrp)1r<.

    It is worthwhile to remark that Bs2,2 and Bs, coincide with the usual Sobolev spaces Hs and the usual Hölder space Cs for sRZ, respectively.

    In our study, we require the space-time Besov spaces as following manner: for T>0 and n1, we denote by LρTBsp,r the set of all tempered distribution f satisfying

    fLρTBsp,r(qZ2qsr˙ΔqfrLp(Rd))1rLρT<.

    Lemma A.1. [15] Let 1pq. Assume that fLp, then there exists a constant C independent of f, j such that

    suppˆf{|ξ|C2j}αfqC2j|α|+dj(1p1q)fp,
    suppˆf{1C2j|ξ|C2j}fpC2j|α|sup|β|=|α|βfp.

    Lemma A.2. [15] There exists a constant C>0 such that for S>0, we have

    uvHsCuvHs+CuHsv.

    Lemma A.3. [5] Let u be a solution of the transport equation

    {ut+vu=0u(x,0)=u0

    and define Rq:=vΔquΔq(vu), 1pp1, 1r and sR such that

    s>dmin(1p1,1p)(ors>1dmin(1p1,1p)ifdivv=0).

    There exists a sequence cqr(Z) such that cqr=1 and a constant C depending only on d,r,s,p, and p1, which satisfy

    qZ,2qsRqpCcqZ(t)uBsp,r,

    with

    Z(t):={vBdp1p1,L,ifs<1+dp1,vBs1p1,r,ifeithers>1+dp1ors=1+dp1forr=1.

    Theorem A.4. Let s2. Assume that χ(), g()Cs(R) with g(0)=0, and that lΦL(R2)< for 1|l|s.

    (A1) Then there exists T>0, the maximal time of existence, such that, if the initial data (n0,c0,u0)Hs1(R2)×Hs(R2)×Hs(R2;R2), then there exists a unique classical solution (n,c,u) to system (1.1)-(1.2) satisfying for any T<T

    (n,c,u)L(0,T;Hs(R2)×Hs+1(R2)×Hs+1(R2;R2))

    and

    (n,c,u)L2(0,T;Hs(R2)×Hs+1(R2)×Hs+1(R2;R2)).

    Proof. We construct the following regularized system:

    {nkt+uknk=Δnk(nkχ(ck)ck)+nk(nk)2,kN,ckt+ukck=Δckg(ck)nk,ukt+(uk)ukPk=Δuk+nkΦ,uk=0,(nk,ck,uk)|t=0=(Skn0,Skc0,Sku0). (A.1)

    Step 1. Uniform estimates.

    Taking the operation Δq with q1 on the first equation of (A.1), we have

    Δqnkt+Δq(uknk)=ΔΔqnkΔq(nkχ(ck)ck)+ΔqnkΔq(nk)2. (A.2)

    Multiplying (A.2) by Δqnk and integrating by parts gives

    12ddtΔqnk22+Δqnk22=R2Δq(uknk)ΔqnkdxR2Δq(nkχ(ck)ck)Δqnkdx+R2ΔqnkΔqnkdxR2Δq(nk)2ΔqnkdxΔq(uknk)2Δqnk2+Δq(nkck)2Δqnk2+Δqnk22+Δq(nk)22Δqnk2.

    Multiplying 22qs on both sides of the above inequality, then taking the 1 norm, using Hölder's inequality and Young's inequality together with Lemma 4.2, we get

    12ddtnk2Hs+nk2Hs+1uknkHsnkHs+nkckHsnkHs+1+nk2Hs+(nk)2HsnkHsCukHsnkHs+1nkHs+CnkHsckHs+1nkHs+1+nk2Hs+Cnk2HsnkHsCuk2Hsnk2Hs+18nk2Hs+1+Cnk2Hsck2Hs+1+18nk2Hs+1+nk2Hs+C(nk4Hs+nk2Hs).

    from which we have

    ddtnk2Hs+nk2Hs+1C(uk2Hsnk2Hs+nk2Hsck2Hs+1+nk2Hs+nk4Hs). (A.3)

    In a similarly way to (A.3), we obtain

    12ddtck2Hs+1+ck2Hs+2Cuk2Hs+1ck2Hs+1+18ck2Hs+2+Cck2Hs+1+18nk2Hs+1.

    it follows that

    ddtck2Hs+1+ck2Hs+2C(uk2Hs+1ck2Hs+1+ck2Hs+1)+18nk2Hs+1. (A.4)

    Applying Δq with q1 to the third equation of (A.1) yields

    Δqukt+(uk)ΔqukΔqPk=(uk)ΔqukΔq((uk)uk)+Δq(nkΦ).

    Multiplying the above equality with Δquk yields

    12ddtΔquk22+Δquk22=Rd((uk)ΔqukΔq((uk)uk))Δqukdx+RdΔq(nkΦ)Δqukdx (uk)ΔqukΔq((uk)uk)2Δquk2+Δq(nkΦ)2Δquk2.

    Multiplying 22q(s+1) on both side of the above inequality and taking the l1 norm, we obtain

    ddtuk2Hs+1+uk2Hs+2ukuk2Hs+1+18nk2Hs+1+uk2Hs+1C(uk4Hs+1+uk2Hs+1)+18nk2Hs+1. (A.5)

    Summing up (A.3)-(A.5), we obtain

    ddt(nk2Hs+ck2Hs+1+uk2Hs+1)+nk2Hs+1+ck2Hs+2+uk2Hs+2C(nk2Hs+ck2Hs+1+uk2Hs+1)(1+nk2Hs+ck2Hs+1+uk2Hs+1)(1+nk2Hs+ck2Hs+1+uk2Hs+1)2.

    We conclude from the Gronwall inequality that

    1+nk2Hs+ck2Hs+1+uk2Hs+11+nk02Hs+ck02Hs+1+uk02Hs+11C(1+nk02Hs+ck02Hs+1+uk02Hs+1)t.

    We can choose

    T=12C(1+nk02Hs+ck02Hs+1+uk02Hs+1)>0.

    such that

    supt[0,T](nk(t)2Hs+ck(t)2Hs+1+uk(t)2Hs+1)+t0(nk2Hs+1+ck2Hs+2+uk2Hs+2)(τ)dτ2(1+nk02Hs+ck02Hs+1+uk02Hs+1). (A.6)

    Step 2. Compactness.

    From (A.6), we obtain

    nkL([0,T],Hs)L2([0,T],Hs+1),ckL([0,T],Hs+1)L2([0,T],Hs+2),ukL([0,T],Hs+1)L2([0,T],Hs+2).

    In order to show the convergence, we also need uniform boundedness for tnk, tck and tuk. From the first equation of (A.1), we know

    tnkLtH1ΔnkLtH1+uknkLtH1+(nkck)LtH1+nkLtH1+(nk)2LtH1nkLtHs+ukLtHs+1nkLtHs+nkLtHsckLtHs+1+nkLtHs+nk2LtHsC.

    Similarly, we have

    tckLtH1ΔckLtH1+ukckLtH1+cknkLtH1ckLtHs+1+ukLtHs+1ckLtHs+1+ckLtHs+1nkLtHsC.

    and

    tukLtH1ΔukLtH1+(uk)ukLtH1+nkΦLtH1ukLtHs+1+uk2LtHs+1+nkLtHsC.

    Since L2 is locally compactly embedded in H1, we apply the Aubin-Lions Lemma to conclude that, up to an extraction of subsequence, the approximate solution sequence (nk,ck,uk) strongly converges in L([0,T];H1) to some function (n,c,u) such that

    nkL([0,T];Hs)L2([0,T],Hs+1),ckL([0,T];Hs+1)L2([0,T],Hs+2),ukL([0,T];Hs+1)L2([0,T],Hs+2).

    Using the above estimates, it is easy to pass the limit in the approximate system (A.1) and (n,c,u) solve (1.1) in the sense of distribution. By a classical deduction [34], we get nC([0,T];Hs), cC([0,T];Hs+1) and uC([0,T];Hs+1).

    By virtue of Heat equation theory, we can prove the time differentiation. For example, we consider the following equations:

    utΔu=f(x,t),thenut=Δuf(x,t).

    Suppose uLtHs and fLtHs2, we obtain utLtHs2. For the arbitrariness of s, we have utLtHs, s>0. In a similar way, we get

    t(ut)Δut=ft

    and uttLtHs2. Thus we show the time differentiation.

    Step 3. Uniqueness.

    Let us consider the two solutions (n1,c1,u1) and (n2,c2,u2) associated with the same initial data and satisfy (1.1). We use the notation δn=n1n2, δc=c1c2 and δu=u1u2. Then we have

    {tδn+δun1+u2δn=Δδn(δnχ(c1)c1)(n2(χ(c1)χ(c2))c1)n2χ(c2)(δc)+δnn1δnn2δn,tδc+δuc1+u2δc=Δδcn1(g(c1)g(c2))g(c2)δn,tδu+(δu)u1+(u2)δu(P1P2)=Δδu+δnΦ. (A.7)

    Taking the L2-inner product of the first equation with δn, we have

    12ddtδn(t)22+δn(t)22=R2(δun1)δndxR2(δnχ(c1)c1)δndxR2(n2(χ(c1)χ(c2))c1)δndxR2n2χ(c2)δcδndx+R2δnδndxR2n1δnδndxR2n2δnδndxC(δu22+δn22n12Hs)+Cδn22c12Hs+1+Cδc22n22Hsc12Hs+1+18δn22+Cn22Hsδc22+Cδn22+n1Hsδn22+n2Hsδn22.

    from which we get

    ddtδn(t)22+δn(t)22C(δu22+δn22n12Hs+δn22c12Hs+1+n22Hsδc22+δn22+n1Hsδn22+n2Hsδn22+δc22n22Hsc12Hs+1). (A.8)

    Next, taking the L2-inner product of the second equation of Eqs.(A.7) with δc, we know

    12ddtδc(t)22+δc(t)22=R2(δuc1)δcdxR2(g(c1)g(c2))n1δcdxR2δng(c2)δcdxC(δu22+δc22c12Hs+1+δc22n1Hs+δn22+δc22).

    Hence we get

    ddtδc(t)22+δc(t)22C(δu22+δc22c12Hs+1+δc22n1Hs+δn22+δc22). (A.9)

    Taking i on both sides of the second equation of Eqs (A.7) yields

    tiδc+u2icΔiδc=i(δuc1)iu2δci(n1(g(c1)g(c2)))i(g(c2)δn).

    Multiplying the above equation with iδc and integrating with respect to space variable, we obtain

    12ddtδc(t)22+Δδc(t)22=iRdi(δuc1)iδcdxiRdiu2δciδcdxiRdi(n1(g(c1)g(c2)))iδcdxiRdi(g(c2)δn)iδcdxRd(δuc1)ΔδcdxRd(δc)u2δcdx+CRdn1δcΔδcdx+Rdg(c2)δnΔδcdxCδu22c12Hs+1+18Δδc22+Cδc22u2Hs+1+Cδc22n12Hs+18Δδc22+Cδn22+18Δδc22.

    Then we get

    ddtδc(t)22+Δδc(t)22C(δu22c12Hs+1+δc22u2Hs+1+δc22n12Hs+δn22). (A.10)

    Performing the L2-inner product of the third equation of system (A.7) with δu, we get

    12ddtδu(t)22+δu(t)22=Rd((δu)u1)δudx+RdδnΦδudxCδu22u1Hs+1+C(δn22+δu22).

    Such that

    ddtδu(t)22C(δu22u1Hs+1+δn22+δu22). (A.11)

    From ((A.8)-(A.11), we have

    ddt(δn(t)22+δc(t)22+δc(t)22+δu(t)22)+δn22+δc22+Δδc22+δu(t)22C(δu22+δn22n12Hs+δn22c12Hs+1+n22Hsδc22+δn22+n1Hsδn22+n2Hsδn22+δc22n22Hsc12Hs+1+δu22+δc22c12Hs+1+δc22n1Hs+δn22+δc22+δu22c12Hs+1+δc22u2Hs+1+δc22n12Hs+δn22+δu22u1Hs+1).

    Consequently, we have

    ddt(δn(t)22+δc(t)22+δc(t)22+δu(t)22)CF(t)(δn22+δc22+δc22+δu22)

    where

    F(t)=1+n1Hs+n2Hs+u1Hs+1+u2Hs+1+n12Hs+n22Hs+c12Hs+1+c22Hs+1+n22Hsc12Hs+1.

    By the above estimates, we know that F(t) is integrable. Therefore, we finally obtain the uniqueness by using the Gronwall inequality.



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