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Global boundedness for a chemotaxis-competition system with signal dependent sensitivity and loop

  • Received: 01 February 2021 Revised: 01 April 2021 Published: 26 May 2021
  • Primary: 35A01, 35K55; Secondary: 92C17

  • In this work, the fully parabolic chemotaxis-competition system with loop

    {tu1=d1Δu1(u1χ11(v1)v1)(u1χ12(v2)v2)+μ1u1(1u1a1u2),tu2=d2Δu2(u2χ21(v1)v1)(u2χ22(v2)v2)+μ2u2(1u2a2u1),tv1=d3Δv1λ1v1+h1(u1,u2),tv2=d4Δv2λ2v2+h2(u1,u2)

    is considered under the homogeneous Neumann boundary condition, where xΩ,t>0, ΩRn(n3) is a bounded domain with smooth boundary. For any regular nonnegative initial data, it is proved that if the parameters μ1,μ2 are sufficiently large, then the system possesses a unique and global classical solution for n3. Specifically, when n=2, the global boundedness can be attained without any constraints on μ1,μ2.

    Citation: Chun Huang. Global boundedness for a chemotaxis-competition system with signal dependent sensitivity and loop[J]. Electronic Research Archive, 2021, 29(5): 3261-3279. doi: 10.3934/era.2021037

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  • In this work, the fully parabolic chemotaxis-competition system with loop

    {tu1=d1Δu1(u1χ11(v1)v1)(u1χ12(v2)v2)+μ1u1(1u1a1u2),tu2=d2Δu2(u2χ21(v1)v1)(u2χ22(v2)v2)+μ2u2(1u2a2u1),tv1=d3Δv1λ1v1+h1(u1,u2),tv2=d4Δv2λ2v2+h2(u1,u2)

    is considered under the homogeneous Neumann boundary condition, where xΩ,t>0, ΩRn(n3) is a bounded domain with smooth boundary. For any regular nonnegative initial data, it is proved that if the parameters μ1,μ2 are sufficiently large, then the system possesses a unique and global classical solution for n3. Specifically, when n=2, the global boundedness can be attained without any constraints on μ1,μ2.



    In this paper, we consider the following initial boundary value problem

    {tu1=d1Δu1(u1χ11(v1)v1)(u1χ12(v2)v2)+μ1u1(1u1a1u2),xΩ,t>0,tu2=d2Δu2(u2χ21(v1)v1)(u2χ22(v2)v2)+μ2u2(1u2a2u1),xΩ,t>0,tv1=d3Δv1λ1v1+h1(u1,u2),xΩ,t>0,tv2=d4Δv2λ2v2+h2(u1,u2),xΩ,t>0,u1ν=u2ν=v1ν=v2ν=0,xΩ,t>0,u1(x,0)=u10(x),u2(x,0)=u20(x),v1(x,0)=v10(x),v2(x,0)=v20(x),xΩ (1)

    in a bounded domain ΩRn with smooth boundary Ω, where ν represents differentiation with respect to the outward normal on Ω, d1,d2,d3,d4,μ1,μ2,λ1,λ2,a1,a2 are positive constants. This model comes from [5], it describes the chemotactical communication named EGF/CSF-1 paracrine invasion loop, which might be a target to control or prevent metastasis with therapeutic methods. u1,u2 represent the densities of macrophages and tumor cells, vi(i=1,2) denote the concentration of the chemical. The chemotactic sensitivity functions χij(i,j=1,2) are smooth and positive. Let the initial data u10,u20,v10 and v20 satisfy

    0u10C0(ˉΩ),0u20C0(ˉΩ),0v10W1,(ˉΩ),0v20W1,(ˉΩ). (2)

    For the case χ12=χ21=0, h1=h1(u2), h2=h2(u1), which describes the situation that tumor cells and macrophages mutually attract each other trough chemotactic signals. In such a case, the global solvability, boundedness and asymptotic behavior have been investigated intensively, for instance, Wang et al. detected the boundedness of solutions for n3 in [17], also, they explored the asymptotic behavior of solutions for any n1. Choosing h1(u2)=u2,h2(u1)=u1, χ11,χ22 are two constants, when μ1=μ2=0, the global boundedness and blow-up of solutions have been considered in [6,11,20]. When μ1,μ20, for the fully parabolic case, the global boundedness and large time behavior for n2 and n=3 were detected in [3] and [8] respectively; as for the parabolic-elliptic case, for all n1, the global boundedness and asymptotic behavior were obtained in [21,12]; afterwards, the results in [21,12] were partially improved by Wang et al. in [18].

    When χij(i,j=1,2) are constants, h1=α1u1+β1u2, h2=α2u1+β2u2. Without respect to the kinetic terms, Espejo et al. derived the simultaneous blow-up phenomenon in [4] for the parabolic-elliptic case of (1) in the whole space R2. Considering the Lotka-Volterra-type competition, whether the parabolic-elliptic case or the fully parabolic case of (1), the global dynamics of solutions were detected, it was found that the solution of (1) is globally bounded without any requirement on the size of the parameters for the fully parabolic case in the lower dimensions n2 [14], while the largeness of parameters μ1,μ2 is needed to guarantee the global solvability of (1) for n=3 [15], and the global solution of this system exponentially approaches to a steady state for all n1 [14], specifically, the system was shown to exhibit the large population densities phenomenon in [16], that is, the solution exhibits unbounded peculiarity for the proper choice of initial data. As for the parabolic elliptic case, in [13], the global boundedness result were established for n2 under the condition that χ11μ1,χ12μ1,χ21μ2 and χ22μ2 are suitably small, moreover, the large time behavior of solution was derived.

    In summary, for the two-species and two-stimuli chemotaxis system, most of the results are focusing on the case that the chemotactic sensitivity functions are constants and the signal production is linear. Therefore, the objective in the present study is to investigate the global boundedness of solutions for (1) when χij,(i,j=1,2),h1,h2 are general functions. Our work is motivated by the method in [17], but in contrast, the existence of the chemical signalling loop in our model makes the computations and analysis fairly subtle.

    We shall suppose throughout this paper that the functions χij(s),hi(s,τ)(i,j=1,2) satisfy the following conditions:

    (H1) χij(s)C1+θ([0,)),i,j=1,2, for some θ>0.

    (H2) 0<χ11(s),χ12(s)K1 for some K1>0; 0<χ21(s),χ22(s)K2 for some K2>0.

    (H3) hi(s,τ)C1+θ([0,)×[0,)),i=1,2, for some θ>0.

    (H4) hi(0,0)=0 and 0<hi(s,τ)s,hi(s,τ)τ<Chi, with Chi>0,i=1,2.

    From the above (H3) and (H4), a straight calculation yields

    hi(s,τ)hi(0,0)=Chi(s+τ)fori=1,2. (3)

    Now we state our main results as follows.

    Theorem 1.1. Let ΩRn(n3) be a smoothly bounded domain, and let d1,d2,d3, d4,μ1,μ2,λ1,λ2, a1,a2 be positive constants. Assume that χij,hi(i,j=1,2) satisfy (H1)-(H4), μ1,μ2 satisfy

    μ1max{2K21d1η1+(2n+8)C2h1d3+2K21d3+2K21d4+2,2K21d1η1+(2n+8)C2h2d4+2K21d3+2K21d4+2}, (4)
    μ2max{2K22d2η2+(2n+8)C2h1d3+2K22d3+2K22d4+2,2K22d2η2+(2n+8)C2h2d4+2K22d3+2K22d4+2} (5)

    with

    η1=2(d1+d3)2d1d3+2(d1+d4)2d1d4+2C2h1+2C2h2+1d1,η2=2(d2+d3)2d2d3+2(d2+d4)2d2d4+2C2h1+2C2h2+1d2. (6)

    Then for all u10,u20,v10 and v20 satisfying (2), the classical solution (u1,u2,v1,v2) of (1) is unique and globally bounded in the sense that

    u1(,t)L(Ω)+v1(,t)L(Ω)+u2(,t)L(Ω)+v2(,t)L(Ω)C

    for all t0, with some constant C>0 that is independent of t.

    Corollary 1. Let ΩR2 be a smoothly bounded domain, and let d1,d2,d3,d4,μ1,μ2, λ1,λ2, a1,a2 be positive constants. Assume that χij,hi(i,j=1,2) satisfy (H1)-(H4), and χij(i,j=1,2) fulfill

    |χij(s)|Lforalls0 (7)

    with some L>0. Then for all u10,u20,v10 and v20 satisfying (2), the classical solution (u1,u2,v1,v2) of (1) is globaly bounded.

    Remark 1. It is obvious that there exist functions χij,hi(i,j=1,2) which satisfy (H1)-(H4), such as, we can choose the standard chemotactic sensitivity functions χij(s)=c0(1+cs)2 with c0,c>0, and choose hi=c1u1+c2u2 with c1,c2>0.

    In this paper, we deal with the quasilinear chemotaxis-competition system with loop. First, we give the local existence and some properties to prepare for the later work. Next, under the condition that μ1,μ2 are sufficiently large, we establish the global boundedness result when n3. At last, for the case n=2, we obtain the boundedness result without any requirement on the size of μ1,μ2.

    As a preliminary, we first give the local existence and some important estimates of solutions for (1).

    Lemma 2.1. Let ΩRn(n1) be a smoothly bounded domain, and let χij,hi(i,j=1,2) satisfy (H1)-(H4). Assume that the initial data u10,u20,v10,v20 satisfy (2). Then there exists a maximal Tmax(0,] such that the system (1) has a unique nonnegative classical solution (u1,u2,v1,v2)

    u1,u2C0(¯Ω×[0,Tmax))C2,1(¯Ω×(0,Tmax)),v1,v2C0(¯Ω×[0,Tmax))C2,1(¯Ω×(0,Tmax)),

    which satisfies

    eitherTmax=,or||u1(,t)||L(Ω)+||u2(,t)||L(Ω)astTmax.

    Besides, the solution fulfills

    Ωu1(x,t)dxm1:=max{Ωu10(x)dx,|Ω|}forallt(0,Tmax) (8)

    and

    Ωu2(x,t)dxm2:=max{Ωu20(x)dx,|Ω|}forallt(0,Tmax) (9)

    as well as

    t+τtΩu21(x,t)dxdsκ1:=m1+m1μ1forallt[0,Tmaxτ) (10)

    and

    t+τtΩu22(x,t)dxdsκ2:=m2+m2μ2forallt[0,Tmaxτ), (11)

    where τ=min{1,Tmax2}.

    Proof. The local existence of classical solution to (1) can be shown by using well-established methods for chemotaxis problems in [19]. And the relation (8)-(11) can be directly derived by a similar method in [15].

    Next, we recall the following lemma (see Lemma 3.4 in [9] or Lemma 2.3 in [1]), which is significant for our latter proof.

    Lemma 2.2. Let T>0, 0fL1loc([0,T)), y(t) be a nonnegative absolutely continuous function on [0,T). Assume that there exist a>0,b>0 such that

    t+τtf(s)dsbforallt[0,Tτ)

    and

    y(t)+ay(t)f(t)foralmostallt(0,T),

    then

    y(t)max{y(0)+b,ba+2b}forallt(0,T),

    where τ=min{1,T2}.

    Based on Lemma 2.1 and Lemma 2.2, we can now derive some basic properties of v1,v2.

    Lemma 2.3. Let ΩRn(n1) be a smooth and bounded domain, λi>0, ai>0, μi>0, dj>0 (i=1,2,j=1,2,3,4). Assume that h1,h2 satisfy (H3)-(H4). Then there exist M1,M2,N1,N2>0 such that the solution of (1) satisfies

    Ω|v1(x,t)|2dxM1,Ω|v2(x,t)|2dxM2forallt[0,Tmax), (12)
    t+τtΩ|Δv1(x,s)|2dxdsN1,t+τtΩ|Δv2(x,s)|2dxdsN2, (13)

    for all t[0,Tmaxτ), where τ=min{1,Tmax2}.

    Proof. Multiplying the third equation in (1) by Δv1 and integrating the result equation over Ω, we have

    12ddtΩ|v1|2dx+d3Ω|Δv1|2dx+λ1Ω|v1|2dx=Ωh1(u1,u2)Δv1dxd32Ω|Δv1|2dx+C2h1d3Ωu21dx+C2h1d3Ωu22dxforallt(0,Tmax),

    therefore,

    ddtΩ|v1|2dx+d3Ω|Δv1|2dx+2λ1Ω|v1|2dx2C2h1d3Ωu21dx+2C2h1d3Ωu22dx (14)

    for all t(0,Tmax). By denoting

    y1(t):=Ω|v1(x,t)|2dx,f1(t)
    =2C2h1d3Ωu21dx+2C2h1d3Ωu22dxforallt(0,Tmax),

    we obtain

    y1(t)+2λ1y1(t)+d3Ω|Δv1(x,t)|2dxf1(t)forallt(0,Tmax). (15)

    Lemma 2.1 and the definition of f1 entail that

    t+τtf1(s)dsc1 (16)

    with some c1>0, then it follows from Lemma 2.2 that

    y1(t)=Ω|v1(x,t)|2dxc2 (17)

    with some c2>0. Whence by an integration of (15) over (t,t+τ) we find

    y1(t+τ)+2λ1t+τty1(s)ds+d3t+τtΩ|Δv1(x,s)|2dxdsy1(t)+t+τtf1(s)dsc1+c2forallt[0,Tmaxτ),

    along with the nonnegativity of y1, we conclude (13). The other inequalities in (12) and (13) can be obtained by the similar method.

    To improve the condition that warrants the global boundedness of solution for (1) when n=2, we recall the following generalization of Gagliardo-Nirenberg inequality which is given in Lemma A.5 of [10].

    Lemma 2.4. Let ΩR2 be a smooth and bounded domain. Then for all φW1,2(Ω), one can find C>0 such that for any ϵ>0 there exists Cϵ>0 with the property that

    φ3L3(Ω)ϵφ2L2(Ω)φln|φ|L1(Ω)+Cφ3L1(Ω)+Cϵ, (18)

    The following lemma plays an important role in the proof of Corollary 1, and the proof is similar to Lemma 2.5 in [14].

    Lemma 2.5. Let ΩRn(n1) be a smooth and bounded domain, (u1,u2,v1,v2) be the classical solution of (1). Assume that χij,hi(i=1,2,j=1,2,3,4) satisfy (H1)-(H4). Let p1 be such that n2<pn and

    supt(0,Tmax)(u1(,t)Lp(Ω)+u2(,t)Lp(Ω))<. (19)

    Then Tmax=, and

    supt>0(u1(,t)L(Ω)+u2(,t)L(Ω)+v1(,t)L(Ω)+v2(,t)L(Ω))<. (20)

    In this section, we first prove the global boundedness of solutions for n3 under the condition that μ1,μ2 are sufficiently large; next, we remove the requirement on the largeness of parameters μ1,μ2 when n=2.

    To prepare our analysis, we establish several differential inequalities in the following two lemmas..

    Lemma 3.1. Let ΩRn(n1) be a smooth and bounded domain, λi>0, μi>0, dj>0 (i=1,2, j = 1, 2, 3, 4). Assume that χij,hi(i=1,2,j=1,2,3,4) satisfy (H1)-(H4). Then for any classical solution (u1,u2,v1,v2) of (1) we have

    ddtΩu21dx+d1Ω|u1|2dx2K21d1Ωu21|v1|2dx+2K21d1Ωu21|v2|2dx+2μ1Ωu21(1u1)dx, (21)
    ddtΩu22dx+d2Ω|u2|2dx2K22d2Ωu22|v1|2dx+2K22d2Ωu22|v2|2dx+2μ2Ωu22(1u2)dx (22)

    for all t(0,Tmax). And

    ddtΩ|v1|4dx+4λ1Ω|v1|4dx+d3Ω||v1|2|2dx(2n+8)C2h1d3Ω(u21+u22)|v1|2dx+2d3Ω|v1|2|v1|2νdSforallt(0,Tmax) (23)

    as well as

    ddtΩ|v2|4dx+4λ2Ω|v2|4dx+d4Ω||v2|2|2dx(2n+8)C2h2d4Ω(u21+u22)|v2|2dx+2d4Ω|v2|2|v2|2νdSforallt(0,Tmax). (24)

    Proof. Multiplying the first equation in (1) by u1 and integrating by parts over Ω, in light of (H1)-(H2) and the Young inequality, we can see that

    12ddtΩu21dx+d1Ω|u1|2dx=Ωu1χ11(v1)v1u1dx+Ωu1χ12(v2)v2u1dx+μ1Ωu21(1u1a1u2)dxK1Ωu1v1u1dx+K1Ωu1v2u1dx+μ1Ωu21(1u1a1u2)dxd12Ω|u1|2dx+K21d1Ωu21|v1|2dx+K21d1Ωu21|v2|2dx+μ1Ωu21(1u1)dx

    for all t(0,Tmax), which directly yields (21). Similarly, we can derive (22).

    To derive (23), in light of the third equation in (1) and the identity 2v1Δv1=Δ|v1|22|D2v1|2, it follows that

    14ddtΩ|v1|4dx=Ω(v1)3v1tdx=d3Ω|v1|2(12Δ|v1|2|D2v1|2)dxλ1Ω|v1|4Ωh1(u1,u2)(|v1|2v1)dxd32Ω|v1|2|v1|2νdSd32Ω||v1|2|2dxd3Ω|v1|2|D2v1|2dxλ1Ω|v1|4dxΩh1(u1,u2)|v1|2Δv1dxΩh1(u1,u2)v1|v1|2dx (25)

    for all t(0,Tmax), here, in view of (H4) and the relation |Δv1|2n|D2v1|2, one obtains

    Ωh1(u1,u2)|v1|2Δv1dxnCh1Ω(u1+u2)|v1|2|D2v1|dxd3Ω|v1|2|D2v1|2dx+nC2h12d3(u21+u22)|v1|2dxforallt(0,Tmax), (26)

    and

    Ωh1(u1,u2)v1|v1|2dxCh1Ω(u1+u2)|v1||v1|2dxd34Ω||v1|2|2dx+C2h1d3(2u21+2u22)|v1|2dxforallt(0,Tmax). (27)

    Consequently, plugging (26) and (27) into (25), we arrive at (23). In addition, (24) can be established in a same manner.

    Lemma 3.2. Let ΩRn(n1) be a smooth and bounded domain, λi>0, μi>0, dj>0 (i=1,2, j = 1, 2, 3, 4). Assume that χij,hi(i=1,2,j=1,2,3,4) satisfy (H1)-(H4). Then for any classical solution (u1,u2,v1,v2) of (1) we have

    ddtΩu1|v1|2dx+Ωu1|v1|2dx3d38Ω||v1|2|2dx+(2(d1+d3)2d3+C2h1)Ω|u1|2dx+C2h1Ω|u2|2dx+(μ12λ1+1)Ωu1|v1|2dx+(2K21d3μ1+2)Ωu21|v1|2dx+2K21d3Ωu21|v2|2dx+d3Ωu1|v1|2νdSforallt(0,Tmax), (28)
    ddtΩu1|v2|2dx+Ωu1|v2|2dx3d48Ω||v2|2|2dx+(2(d1+d4)2d4+C2h2)Ω|u1|2dx+C2h2Ω|u2|2dx+(μ12λ2+1)Ωu1|v2|2dx+(2K21d4μ1+2)Ωu21|v2|2dx+2K21d4Ωu21|v1|2dx+d4Ωu1|v2|2νdSforallt(0,Tmax), (29)
    ddtΩu2|v1|2dx+Ωu2|v1|2dx3d38Ω||v1|2|2dx+(2(d2+d3)2d3+C2h1)Ω|u2|2dx+C2h1Ω|u1|2dx+(μ22λ1+1)Ωu2|v1|2dx+(2K22d3μ2+2)Ωu22|v1|2dx+2K22d3Ωu22|v2|2dx+d3Ωu2|v1|2νdSforallt(0,Tmax), (30)
    ddtΩu2|v2|2dx+Ωu2|v2|2dx3d48Ω||v2|2|2dx+(2(d2+d4)2d4+C2h2)Ω|u2|2dx+C2h2Ω|u1|2dx+(μ22λ2+1)Ωu2|v2|2dx+(2K22d4μ2+2)Ωu22|v2|2dx+2K22d4Ωu22|v1|2dx+d4Ωu2|v2|2νdSforallt(0,Tmax). (31)

    Proof. Notice that the estimates for (28)-(31) are similar, thereupon, we only consider the priori estimates for (28). Utilizing the first equation and the third equation in (1), one finds

    ddtΩu1|v1|2dx=Ω|v1|2u1tdx+2Ωu1v1v1t=d1Ωu1|v1|2dx+Ωu1χ11(v1)v1|v1|2dx+Ωu1χ12(v2)v2|v1|2dx+μ1Ωu1|v1|2(1u1a1u2)dx+d3Ωu1(Δ|v1|22|D2v1|2)dx2λ1Ωu1|v1|2dx+2Ωu1(h1(u1,u2)u1u1+h1(u1,u2)u2u2)v1dx(d1+d3)Ωu1|v1|2dx+Ωu1χ11(v1)v1|v1|2dx+Ωu1χ12(v2)v2|v1|2dx+(μ12λ1)Ωu1|v1|2dxμ1Ωu21|v1|2dx+d3Ωu1|v1|2νdS+2Ωh1(u1,u2)u1u1u1v1dx+2Ωh1(u1,u2)u2u1u2v1dx (32)

    for all t(0,Tmax), where we have used the identity 2v1Δv1=Δ|v1|22|D2v1|2. To estimate the right side term of (32), we apply the Young inequality and the conditions (H2), (H4) to obtain

    (d1+d3)Ωu1|v1|2dx+Ωu1χ11(v1)v1|v1|2dx+Ωu1χ12(v2)v2|v1|2dxd38Ω||v1|2|2dx+2(d1+d3)2d3Ω|u1|2dx+d38Ω||v1|2|2dx+2K21d3Ωu21|v1|2dx+d38Ω||v1|2|2dx+2K21d3Ωu21|v2|2dxforallt(0,Tmax) (33)

    and

    2Ωh1(u1,u2)u1u1u1v1dx+2Ωh1(u1,u2)u2u1u2v1dxC2h1Ω|u1|2dx+Ωu21|v1|2dx+C2h1Ω|u2|2dx+Ωu21|v1|2dx (34)

    for all t(0,Tmax). Inserting (33) and (34) into (32), one can immediately get (28). Analogously, we can obtain (29)-(31).

    With Lemma 3.1 and Lemma 3.2 at hand, now, relying on a series of estimates, under an additional largeness assumption on μ1,μ2, we can attain the boundedness of Ωu21dx, Ωu22dx, Ω|v1|4dx and Ω|v2|4dx.

    Lemma 3.3. Let ΩRn(n1) be a smooth and bounded domain, λi>0, μi>0, dj>0 (i=1,2, j = 1, 2, 3, 4). Assume that χij,hi(i=1,2,j=1,2,3,4) satisfy (H1)-(H4), and (u1,u2,v1,v2) is a classical solution of (1). Then if μ1,μ2 satisfy (4) and (5) in Theorem 1.1, one can find C1>0 such that

    Ωu21dx+Ωu22dx+Ω|v1|4dx+Ω|v2|4dxC1forallt(0,Tmax). (35)

    Proof. It follows from Lemma 3.1 and Lemma 3.2 that

    ddt(η1Ωu21+η2Ωu22+Ω|v1|4+Ω|v2|4+Ωu1|v1|2+Ωu1|v2|2+Ωu2|v1|2+Ωu2|v2|2)+Ωu21+Ωu22+Ω|u1|2+Ω|u2|2+4λ1Ω|v1|4+4λ2Ω|v2|4+d34Ω||v1|2|2+d44Ω||v2|2|2+Ωu1|v1|2+Ωu1|v2|2+Ωu2|v1|2+Ωu2|v2|2(2η1μ1+1)Ωu212η1μ1Ωu31+(2η2μ2+1)Ωu222η2μ2Ωu32+(μ12λ1+1)Ωu1|v1|2+(μ12λ2+1)Ωu1|v2|2+(μ22λ1+1)Ωu2|v1|2+(μ22λ2+1)Ωu2|v2|2+2d3Ω|v1|2|v1|2ν+2d4Ω|v2|2|v2|2ν+d3Ωu1|v1|2ν+d4Ωu1|v2|2ν+d3Ωu2|v1|2ν+d4Ωu2|v2|2ν+(2K21d1η1+(2n+8)C2h1d3+2K21d3+2K21d4+2μ1)Ωu21|v1|2+(2K21d1η1+(2n+8)C2h2d4+2K21d3+2K21d4+2μ1)Ωu21|v2|2+(2K22d2η2+(2n+8)C2h1d3+2K22d3+2K22d4+2μ2)Ωu22|v1|2+(2K22d2η2+(2n+8)C2h2d4+2K22d3+2K22d4+2μ2)Ωu22|v2|2. (36)

    Making use of the Young inequality, for any ϵ1,ϵ2,ϵ3,ϵ4(0,1), it is clear that

    (μ12λ1+1)Ωu1|v1|2dxϵ1Ω|v1|4dx+(μ12λ1+1)24ϵ1Ωu21dx,(μ12λ2+1)Ωu1|v2|2dxϵ2Ω|v2|4dx+(μ12λ2+1)24ϵ2Ωu21dx,(μ22λ1+1)Ωu2|v1|2dxϵ3Ω|v1|4dx+(μ22λ1+1)24ϵ3Ωu22dx,(μ22λ2+1)Ωu2|v2|2dxϵ4Ω|v2|4dx+(μ22λ2+1)24ϵ4Ωu22dx. (37)

    According to Lemma 4.2 in [7], there exists C>0 such that

    |vi|2νC|vi|2,i=1,2,forallt(0,Tmax),xΩ. (38)

    And thanks to the boundary trace embedding:

    W1,2(Ω)↪↪W12,2(Ω)L2(Ω), (39)

    which warrants that for any ϵ5>0, one can find C(ϵ5)>0 such that

    Ωϕ2dSϵ5Ω|ϕ|2dx+C(ϵ5)(Ω|ϕ|dx)2 (40)

    holds for all ϕW1,2(Ω). Thereupon, invoking Young's inequality, (38) and (40), we can see that

    2d3Ω|v1|2|v1|2ν+2d4Ω|v2|2|v2|2ν+d3Ω(u1+u2)|v1|2ν+d4Ω(u1+u2)|v2|2ν2d3CΩ|v1|4+2d4CΩ|v2|4+d3CΩ(u1+u2)|v1|2+d4CΩ(u1+u2)|v2|2C(2d3+2d23)Ω|v1|4+C(2d4+2d24)Ω|v2|4+2CΩu21+2CΩu22ˆϵΩ||v1|2|2+C1(ˆϵ)(Ω|v1|2)2+ˆϵΩ||v2|2|2+C2(ˆϵ)(Ω|v2|2)2+ˆϵΩ|u1|2+C3(ˆϵ)(Ωu1)2+ˆϵΩ|u2|2+C4(ˆϵ)(Ωu2)2, (41)

    where 0<ˆϵ<min{d34,d44,1}. Inserting (37) and (41) into (36), utilizing (4) and (5), we can see that

    ddt(η1Ωu21+η2Ωu22+Ω|v1|4+Ω|v2|4+Ωu1|v1|2+Ωu1|v2|2+Ωu2|v1|2+Ωu2|v2|2)+Ωu21dx+Ωu22+(1ˆϵ)Ω|u1|2+(1ˆϵ)Ω|u2|2+(4λ1ϵ1ϵ3)Ω|v1|4+(4λ2ϵ2ϵ4)Ω|v2|4+(d34ˆϵ)Ω||v1|2|2+(d44ˆϵ)Ω||v2|2|2+Ω(u1+u2)(|v1|2+|v2|2)(2η1μ1+1+(μ12λ1+1)24ϵ1+(μ12λ2+1)24ϵ2)Ωu212η1μ1Ωu31+(2η2μ2+1+(μ22λ1+1)24ϵ3+(μ22λ2+1)24ϵ4)Ωu222η2μ2Ωu32 (42)
    +C1(ˆϵ)(Ω|v1|2)2+C2(ˆϵ)(Ω|v2|2)2+C3(ˆϵ)(Ωu1)2+C4(ˆϵ)(Ωu2)2

    for all t(0,Tmax) and ϵi(0,1),(i=1,2,3,4). Let

    Y(t):=η1Ωu21dx+η2Ωu22dx+Ω|v1|4dx+Ω|v2|4dx+Ωu1|v1|2dx+Ωu1|v2|2dx+Ωu2|v1|2dx+Ωu2|v2|2dx, (43)

    selecting ϵi,(i=1,2,3,4) satisfying

    ϵ1+ϵ3<2λ1,ϵ2+ϵ4<2λ2,

    applying Young's inequality, Lemma 2.1, Lemma 2.3 and (42), we can find δ>0 and C1>0 such that

    ddtY(t)+δY(t)C1forallt(0,Tmax), (44)

    which with the ODE comparison principle means that (35) holds for all t(0,Tmax).

    On the basis of the L2 bound for u1,u2, we can now establish the Lp estimates for u1,u2.

    Lemma 3.4. Let ΩRn(n3) be a smooth and bounded domain, λi>0, μi>0, dj>0 (i=1,2, j = 1, 2, 3, 4). Assume that χij,hi(i=1,2,j=1,2,3,4) satisfy (H1)-(H4), and (u1,u2,v1,v2) is a classical solution of (1). If there exists C2>0 such that

    Ωu21dx+Ωu22dx+Ω|v1|4dx+Ω|v2|4dxC2forallt(0,Tmax). (45)

    Then for all p>1, one can find a positive constant C3 such that

    u1Lp(Ω)+u2Lp(Ω)C3forallt(0,Tmax). (46)

    Proof. Testing the first equation of (1) by up11 to obtain

    1pddtΩup1=d1(p1)Ωup21|u1|2+(p1)Ωup11χ11(v1)v1u1+(p1)Ωup11χ12(v2)v2u1+μ1Ωup1(1u1a1u2)d1(p1)Ωup21|u1|2+d1(p1)4Ωup21|u1|2+p1d1K21Ωup1|v1|2+d1(p1)4Ωup21|u1|2+p1d1K21Ωup1|v2|2+μ1Ωup1(1u1)d1(p1)2Ωup21|u1|2+p1d1K21Ωup1(|v1|2+|v2|2)+μ1Ωup1(1u1) (47)

    for all t(0,Tmax). Since (45) entails that

    Ω|vi|4dxC2forallt(0,Tmax),i=1,2, (48)

    with C2>0, and thus we apply the H¨older inequality to obtain

    p1d1K21Ωup1(|v1|2+|v2|2)dxp1d1K21(Ωu2p1dx)12(Ω|v1|4dx)12+p1d1K21(Ωu2p1dx)12(Ω|v2|4dx)12C3(Ωu2p1dx)12forallt(0,Tmax) (49)

    with C3>0, to estimate the term (Ωu2p1dx)12, fixing

    ˆλ:=pn2n41n2+pn2,

    due to the fact that n3, this guarantees that ˆλ(0,1), then we can employ the Gagliardo-Nirenberg inequality, Lemma 2.1 and Young's inequality to get

    C3(Ωu2p1dx)12=C3up212L4(Ω)C4(up212ˆλL2(Ω)up212(1ˆλ)L2p(Ω)+up212L2p(Ω))C5(up212ˆλL2(Ω)+1)d1(p1)2Ωup21|u1|2dx+C6 (50)

    with C4,C5,C6>0. A combination of (47), (49) and (50) yields

    1pddtΩup1dx+Ωup1dxμ1Ωup1(1u1)dx+Ωup1dx+C6C7 (51)

    with C7>0. Therefore, we apply a comparison argument to establish the boundedness of u1Lp(Ω). Applying the same arguments as above, we can easily obtain the boundedness of u2Lp(Ω). This completes the proof of this lemma.

    Proof of Theorem 1.1. In light of Lemma 3.4, it follows from the well known Moser-type iterations and Lemma 2.1 that Theorem 1.1 holds.

    In the above section, the global boundedness of solution is derived under the condition that μ1,μ2 are sufficiently large. In this section, motivated by the method in [2,3,17], we remove the restriction on μ1,μ2 when n=2. The key point of the proof is to establish the boundedness of Ω|v1|4dx and Ω|v1|4dx, to this end, we first establish the boundedness of t+τtΩ|vi|4dx,i=1,2 in the following lemma.

    Lemma 3.5. Let ΩRn(n2) be a smooth and bounded domain, λi>0, μi>0, dj>0 (i=1,2, j = 1, 2, 3, 4). Assume that hi(i=1,2) satisfy (H3)-(H4), and (u1,u2,v1,v2) is a classical solution of (1). Then one can find a constant C4>0 such that

    t+τtΩ|v1|4dxds+t+τtΩ|v2|4dxdsC4forallt(0,Tmax), (52)

    where τ=min{1,Tmax2}.

    Proof. Since n2, we can apply the Gagliardo-Nirenberg inequality to obtain

    Ω|v1|4dx=v14L4(Ω)C(GN)4(Δv1¯λL2(Ω)v11¯λL2(Ω)+v1L2(Ω))4C8(Δv12L2(Ω)+1) (53)

    with C8>0, ¯λ=n4(0,12], where we have used the boundedness result of v1L2(Ω) in Lemma 2.3. Integrating (53) over (t,t+τ), and by virtue of Lemma 2.3, we can compute

    t+τtΩ|v1|4dxdsC9forallt(0,Tmax), (54)

    where C9>0. Similarly, we can derive

    t+τtΩ|v1|4dxdsC10forallt(0,Tmax) (55)

    with C10>0, this readily yields (52).

    In the second step, we derive the boundedness of Ω|uilnui|dx,i=1,2.

    Lemma 3.6. Let ΩRn(n2) be a smooth and bounded domain, λi>0, μi>0, dj>0 (i=1,2, j = 1, 2, 3, 4). Assume that χij(i=1,2,j=1,2) satisfy

    0<χij(s)Lforalls0 (56)

    with L>0, and (u1,u2,v1,v2) is a classical solution of (1). Then there exists C4>0 such that

    Ω|u1lnu1|dxC4,Ω|u2lnu2|dxC4forallt(0,Tmax). (57)

    Proof. Multiplying the first equation in (1) by (1+lnu1) and integrating it over Ω, one obtains

    ddtΩu1lnu1dx=d1Ω|u1|2u1dx+Ωχ11(v1)v1u1dx+Ωχ12(v2)v2u1dx+μ1Ωu1(1u1a1u2)(1+lnu1)=d1Ω|u1|2u1dx+Ω(v11χ11(s)ds)u1dx+Ω(v21χ12(s)ds)u1dx (58)
    +μ1Ωu1(1u1a1u2)(1+lnu1)=d1Ω|u1|2u1dxΩΔ(v11χ11(s)ds)u1dxΩΔ(v21χ12(s)ds)u1dx+μ1Ωu1(1u1a1u2)(1+lnu1)forallt(0,Tmax).

    To estimate ΩΔ(v11χ11(s)ds)u1dx, utilizing the Young inequality and (56), it follows

    ΩΔ(v11χ11(s)ds)u1dx12Ωu21dx+12Ω|Δ(v11χ11(s)ds)|2dx=12u12L2(Ω)+12χ11(v1)|v1|2+χ11(v1)Δv12L2(Ω)12u12L2(Ω)+L2v14L4(Ω)+K21Δv12L2(Ω). (59)

    Like wise, we have

    ΩΔ(v21χ12(s)ds)u1dx12u12L2(Ω)+L2v24L4(Ω)+K21Δv22L2(Ω). (60)

    As for μ1Ωu1(1u1a1u2)(1+lnu1), in view of the boundedness of Ωuidx(i=1,2) in Lemma 2.1, and the inequalities ρ(1ρ)14, ρ(1ρ)lnρ0 as well as ρlnρ1e for all ρ>0, it holds that

    μ1Ωu1(1u1a1u2)(1+lnu1)μ1Ωu1(1u1)dx+μ1Ωu1(1u1)lnu1dxa1μ1Ωu1u2lnu1dxμ1|Ω|4+a1μ1m2eforallt(0,Tmax). (61)

    Plugging (59)-(61) into (58), we can see that

    ddtΩu1lnu1dx+d1Ω|u1|2u1dxu12L2(Ω)+(L+K1)2(v14L4(Ω)+Δv12L2(Ω)+v24L4(Ω)+Δv22L2(Ω))+μ1|Ω|4+a1μ1m2e (62)

    for all t(0,Tmax). According to n2, we can utilize the Gagliardo-Nirenberg inequality to derive

    Ωu1lnu1dxΩu21dx=u1214L4(Ω)C11u1214¯λL2(Ω)u1214(1¯λ)L2(Ω)+C11u1214L2(Ω)C12(Ω|u1|2u1+1) (63)

    where ¯λ=n4(0,12], C12>0. Together with (62) and (63), we arrive at

    ddtΩu1lnu1dx+d1C12Ωu1lnu1dxu12L2(Ω)+(L+K1)2(v14L4(Ω)+Δv12L2(Ω)+v24L4(Ω)+Δv22L2(Ω))+μ1|Ω|4+a1μ1m2e+d1 (64)

    for all t(0,Tmax). Thus, we infer from (10), (13), (52) and Lemma 2.2 that Ω|u1lnu1|dx is bounded. Equally, we can establish the boundedness for Ω|u2lnu2|dx. Whereby the proof is completed.

    In what follows, we proceed to show that Ωu21dx+Ωu22dx+Ω|v1|4dx+Ω|v2|4dx is bounded for all μ1,μ2>0.

    Lemma 3.7. Let ΩR2 be a smooth and bounded domain, λi>0, μi>0, dj>0 (i=1,2, j=1,2,3,4). Assume that the conditions in Lemma 3.6 are satisfied, and (u1,u2,v1,v2) is a classical solution of (1). Then there exists C5>0 such that

    Ωu21dx+Ωu22dx+Ω|v1|4dx+Ω|v2|4dxC5forallt(0,Tmax). (65)

    Proof. From (21)-(24), we deduce that

    ddt(Ωu21dx+Ωu22dx+Ω|v1|4dx+Ω|v2|4dx)+d1Ω|u1|2dx+d2Ω|u2|2dx+4λ1Ω|v1|4dx+d3Ω||v1|2|2dx+4λ2Ω|v2|4dx+d4Ω||v2|2|2dx+Ω(u21+u22)dx((2n+8)C2h1d3+2K21d1)Ωu21|v1|2dx+((2n+8)C2h1d3+2K22d2)Ωu22|v1|2dx+((2n+8)C2h2d4+2K21d1)Ωu21|v2|2dx+((2n+8)C2h2d4+2K22d2)Ωu22|v2|2dx+2d3Ω|v1|2|v1|2νdS+2d4Ω|v2|2|v2|2νdS+Ω(u21+u22)dx+2μ1Ωu21(1u1)dx+2μ2Ωu22(1u2)dxforallt(0,Tmax). (66)

    To handle the first term on the right side of (66), we notice that n=2, accordingly, in view of (18), the following inequality

    u13L3(Ω)ϵu12L2(Ω)u1ln|u1|L1(Ω)+Cu13L1(Ω)+Cϵ (67)

    holds for all ϵ>0, then, the boundedness of u1ln|u1|L1(Ω) and u1L1(Ω) in Lemma 3.6 and Lemma 2.1 guarantees that

    u13L3(Ω)C13ϵu12L2(Ω)+C13forallt(0,Tmax)andϵ>0 (68)

    with C13>0. Therefore, utilizing the H¨older inequality, the Gagliardo-Nirenberg inequality and the Young inequality, it follows from (12), (67) as well as (68) that

    ((2n+8)C2h1d3+2K21d1)Ωu21|v1|2dx((2n+8)C2h1d3+2K21d1)u12L3(Ω)|v1|2L3(Ω)((2n+8)C2h1d3+2K21d1)u12L3(Ω)(CGN|v1|223L2(Ω)|v1|213L1(Ω)+CGN|v1|2L1(Ω))((2n+8)C2h1d3+2K21d1)u12L3(Ω)(C14|v1|223L2(Ω)+C14)d34|v1|22L2(Ω)+C15u13L3(Ω)+C15d34|v1|22L2(Ω)+d12u12L2(Ω)+C16forallt(0,Tmax) (69)

    with C14,C15,C16>0, where we have fixed ϵ=d12C13C15 when using (68).

    In a similar way, we can find C17,C18,C19>0 such that

    ((2n+8)C2h1d3+2K22d2)Ωu22|v1|2dxd34|v1|22L2(Ω)+d22u22L2(Ω)+C17 (70)

    and

    ((2n+8)C2h2d4+2K21d1)Ωu21|v2|2dxd44|v2|22L2(Ω)+d12u12L2(Ω)+C18 (71)

    as well as

    ((2n+8)C2h2d4+2K22d2)Ωu22|v2|2dxd44|v2|22L2(Ω)+d22u22L2(Ω)+C19 (72)

    for all t(0,Tmax).

    For the term 2d3Ω|v1|2|v1|2νdS+2d4Ω|v2|2|v2|2νdS, (12) and (38)-(40) enable us to see that

    2d3Ω|v1|2|v1|2νdS+2d4Ω|v2|2|v2|2νdSd32|v1|22L2(Ω)+d42|v2|22L2(Ω)+C20forallt(0,Tmax) (73)

    with C20>0.

    What's more, making use of the identity s2(1s)427 for all s>0, the last two terms in (66) can be bounded as

    2μ1Ωu21(1u1)dx+2μ2Ωu22(1u2)dx8|Ω|27(μ1+μ2)forallt(0,Tmax). (74)

    Thus, plugging (69)-(74) into (66), we can conclude that

    ddt˜Y(t)+˜δ˜Y(t)Ω(u21+u22)dx+C21forallt(0,Tmax), (75)

    where ˜Y(t):=Ωu21dx+Ωu22dx+Ω|v1|4dx+Ω|v2|4dx, ˜δ:=min{4λ1,4λ2,1}, C21>0, which combined with (10), (11) and Lemma 2.2 show (65).

    Proof of Corollary 1. A combination of Lemma 2.5 and Lemma 3.7 directly yields Corollary 1.

    We would like to thank the anonymous reviewers for their valuable suggestions and fruitful comments which lead to significant improvement of this work.



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