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Research article Special Issues

Blow-up and boundedness in quasilinear attraction-repulsion systems with nonlinear signal production


  • In this paper, we consider the quasilinear parabolic-elliptic-elliptic attraction-repulsion system

    {ut=(D(u)u)χ(uv)+ξ(uw),xΩ,t>0,0=Δvμ1(t)+f1(u),xΩ,t>0,0=Δwμ2(t)+f2(u),xΩ,t>0

    under homogeneous Neumann boundary conditions in a smooth bounded domain ΩRn, n2. The nonlinear diffusivity D and nonlinear signal productions f1,f2 are supposed to extend the prototypes

    D(s)=(1+s)m1, f1(s)=(1+s)γ1, f2(s)=(1+s)γ2, s0,γ1,γ2>0,mR.

    We proved that if γ1>γ2 and 1+γ1m>2n, then the solution with initial mass concentrating enough in a small ball centered at origin will blow up in finite time. However, the system admits a global bounded classical solution for suitable smooth initial datum when γ2<1+γ1<2n+m.

    Citation: Ruxi Cao, Zhongping Li. Blow-up and boundedness in quasilinear attraction-repulsion systems with nonlinear signal production[J]. Mathematical Biosciences and Engineering, 2023, 20(3): 5243-5267. doi: 10.3934/mbe.2023243

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  • In this paper, we consider the quasilinear parabolic-elliptic-elliptic attraction-repulsion system

    {ut=(D(u)u)χ(uv)+ξ(uw),xΩ,t>0,0=Δvμ1(t)+f1(u),xΩ,t>0,0=Δwμ2(t)+f2(u),xΩ,t>0

    under homogeneous Neumann boundary conditions in a smooth bounded domain ΩRn, n2. The nonlinear diffusivity D and nonlinear signal productions f1,f2 are supposed to extend the prototypes

    D(s)=(1+s)m1, f1(s)=(1+s)γ1, f2(s)=(1+s)γ2, s0,γ1,γ2>0,mR.

    We proved that if γ1>γ2 and 1+γ1m>2n, then the solution with initial mass concentrating enough in a small ball centered at origin will blow up in finite time. However, the system admits a global bounded classical solution for suitable smooth initial datum when γ2<1+γ1<2n+m.



    Chemotaxis is the property of cells to move in an oriented manner in response to an increasing concentration of chemo-attractant or decreasing concentration of chemo-repellent, where the former is referred to as attractive chemotaxis and the later to repulsive chemotaxis. To begin with, it is important to study the quasilinear Keller-Segel system as follows

    {ut=(D(u)u)χ(ϕ(u)v),xΩ,t>0,τvt=Δvαv+βu,xΩ,t>0, (1.1)

    subject to homogeneous Neumann boundary conditions, where the functions D(u) and ϕ(u) denote the strength of diffusion and chemoattractant, respectively, and the function u=u(x,t) idealizes the density of cell, v=x(x,t) represents the concentration of the chemoattractant. Here the attractive (repulsive) chemotaxis corresponds to χ>0 (χ<0), and |χ|R{0} measures the strength of chemotactic response. The parameters τ{0,1}, and α,β>0 denote the production and degradation rates of the chemical. The above system describes the chemotactic interaction between cells and one chemical signal (either attractive or repulsive), and it has been investigated quite extensively on the existence of global bounded solutions or the occurrence of blow-up in finite time in the past four decades. In particular, the system (1.1) is the prototypical Keller-Segel model [1] when D(u)=1,ϕ(u)=u. In the case τ=1, there are many works to show that the solution is bounded [2,3,4,5], and blow-up in finite time [6,7,8,9,10,11]. If the cell's movement is much slower than the chemical signal diffusing, the second equation of (1.1) is reduced to 0=ΔvM+u, where M:=1|Ω|Ωu(x,t)dx and the simplified system has many significant results [12,13,14,15].

    For further information concerning nonlinear signal production, when the chemical signal function is denoted by g(u), authors derived for more general nonlinear diffusive system as follows

    {ut=(D(u)u)(ϕ(u)v),xΩ,t>0,0=ΔvM+g(u),xΩ,t>0, (1.2)

    where M:=1|Ω|Ωg(u(x,t))dx. Recently, when D(u)=up,ϕ(u)=u and g(u)=ul, it has been shown that all solutions are global and uniformly bounded if p+l<2n, whereas p+l>2n implies that the solution blows up in finite time [16]. What's more, there are many significant works [17,18,19] associated with this system.

    Subsequently, the attraction-repulsion system has been introduced in ([20,21]) as follows

    {ut=Δuχ(uv)+ξ(uw),xΩ,t>0,τ1vt=Δv+αuβv,xΩ,t>0,τ2wt=Δw+δuγw,xΩ,t>0, (1.3)

    subject to homogeneous Neumann boundary conditions, where χ,ξ,α,β,δ,γ>0 are constants, and the functions u(x,t),v(x,t) and w(x,t) denote the cell density, the concentration of the chemoattractant and chemorepellent, respectively. The above attraction-repulsion chemotaxis system has been studied actively in recent years, and there are many significant works to be shown as follows.

    For example, if τ1=τ2=0, Perthame [22] investigated a hyperbolic Keller-Segel system with attraction and repulsion when n=1. Subsequently, Tao and Wang [23] proved that the solution of (1.3) is globally bounded provided ξγχα>0 when n2, and the solution would blow up in finite time provided ξγχα<0,α=β when n=2. Then, there is a blow-up solution when χαξγ>0,δβ or χαδξγβ>0,δ<β for n=2 [24]. Moreover, Viglialoro [25] studied the explicit lower bound of blow-up time when n=2. In another hand, if τ1=1,τ2=0, Jin and Wang [26] showed that the solution is bounded when n=2 with ξγχα0, and Zhong et al. [27] obtained the global existence of weak solution when ξγχα0 for n=3. Furthermore, if τ1=τ2=1, Liu and Wang [28] obtained the global existence of solutions, and Jin et al. [29,30,31] also showed a uniform-in-time bound for solutions. In addition, there are plenty of available results of the attraction-repulsion system with logistic terms [32,33,34,35,36,37,38,39,40], and for further information concerning (1.3) based on the nonlinear signal production, it was used to model the aggregation patterns formed by some bacterial chemotaxis in [41,42,43].

    We turn our eyes into a multi-dimensional attraction-repulsion system

    {ut=Δuχ(ϕ(u)v)+ξ(ψ(u)w),xΩ,t>0,τ1vt=Δvμ1(t)+f(u),xΩ,t>0,τ2wt=Δwμ2(t)+g(u),xΩ,t>0, (1.4)

    where ΩRn(n2) is a bounded domain with smooth boundary, μ1(t)=1|Ω|Ωf(u)dx,μ2(t)=1|Ω|Ωg(u)dx and τ1,τ2{0,1}. Later on, the system (1.4) has attracted great attention of many mathematicians. In particular, when ϕ(u)=ψ(u)=u,f(u)=uk and g(u)=ul, Liu and Li [44] proved that all solutions are bounded if k<2n, while blow-up occurs for k>l and k>2n in the case τ1=τ2=0.

    Inspired by the above literature, we are devoted to deal with the quasilinear attraction-repulsion chemotaxis system

    {ut=(D(u)u)χ(uv)+ξ(uw),xΩ,t>0,0=Δvμ1(t)+f1(u),xΩ,t>0,0=Δwμ2(t)+f2(u),xΩ,t>0,uν=vν=wν=0,xΩ,t>0,u(x,0)=u0(x),xΩ, (1.5)

    in a bounded domain ΩRn,n2 with smooth boundary, where ν denotes outward normal derivatives on Ω. The function u(x,t) denotes the cell density, v(x,t) represents the concentration of an attractive signal (chemo-attractant), and w(x,t) is the concentration of a repulsive signal (chemo-repellent). The parameters satisfy χ,ξ0, which denote the strength of the attraction and repulsion, respectively. Here μ1(t)=1|Ω|Ωf1(u(x,t))dx, μ2(t)=1|Ω|Ωf2(u(x,t))dx, and f1,f2 are nonnegative Hölder continuous functions.

    In the end, we propose the following assumptions on D,f1,f2 and u0 for the system (1.5).

    (I1) The nonlinear diffusivity D is positive function satisfying

    DC2([0,)). (1.6)

    (I2) The function fi is nonnegative and nondecreasing and satisfies

    fiθ(0,1)Cθloc([0,))C1((0,)) (1.7)

    with i{1,2}.

    (I3) The initial datum

    u0θ(0,1)Cθ(¯Ω) is nonnegative and radially decreasing,u0ν=0 on Ω. (1.8)

    The goal of the article is twofold. On the one hand, we need to find out the mutual effect of the nonlinear diffusivity D(u) and the nonlinear signal production fi(u)(i=1,2). On the other hand, we need to make a substantial step towards the dynamic of blowing up in finite time. Hence, we draw our main results concerning (1.5) read as follows.

    Theorem 1.1. Let n2, R>0 and Ω=BR(0)Rn be a ball, and suppose that the function D satisfies (1.6) and f1,f2 are assumed to fulfill (1.7) as well as

    D(u)d(1+u)m1, f1(u)k1(1+u)γ1, f2(u)k2(1+u)γ2 for all u0,

    with mR, k1,k2,γ1,γ2,d>0 and

    γ1>γ2 and 1+γ1m>2n. (1.9)

    For any M>0 there exist ε=ε(γ1,M,R)(0,M) and r=r(γ1,M,R)(0,R) such that if u0 satisfies (1.8) with

    Ωu0=M and Br(0)u0Mε,

    then the corresponding solution of the system (1.5) blows up in finite time.

    Theorem 1.2. Let n2, ΩRn be a smooth bounded domain, and suppose that the function D satisfies (1.6) and f1,f2 are assumed to fulfill (1.7) as well as

    D(u)d(1+u)m1, f1(u)k1(1+u)γ1, f2(u)=k2(1+u)γ2 for all u0,

    with mR, k1,k2,γ1,γ2,d>0 and

    γ2<1+γ1<2n+m. (1.10)

    Then for each u0θ(0,1)Cθ(¯Ω), u00 with u0ν=0 on Ω, and the system (1.5) admits a unique global classical solution (u,v,w) with

    u,v,wC2,1(¯Ω×(0,))C0(¯Ω×[0,)).

    Furthermore, u,v and w are all non-negative and bounded.

    The structure of this paper reads as follows: In section 2, we will show the local-in-time existence of a classical solution to the system (1.5) and some lemmas that we will use later. In section 3, we will prove Theorem 1.1 by establishing a superlinear differential inequality. In section 4, we will solve the boundedness of u in L and prove Theorem 1.2.

    Firstly, we state one result concerning local-in-time existence of a classical solution to the system (1.5). Then, we denote some new variables to transfer the original equations in (1.5) to a new system according to the ideas in [19,20,21,22,23,24,25]. In addition, in order to prove the main result, we will state some lemmas which will be needed later.

    Lemma 2.1. Let ΩRn with n2 be a bounded domain with smooth boundary. Assume that D fulfills (1.6), f1,f2 satisfy (1.7) and u0θ(0,1)Cθ(¯Ω) with u0ν=0 on Ω as well as u00, then there exist Tmax(0,] and a classical solution (u,v,w) to (1.5) uniquely determined by

    {uC0(¯Ω×[0,Tmax))C2,1(¯Ω×(0,Tmax)),vq>nL((0,Tmax);W1,q(Ω))C2,0(¯Ω×(0,Tmax)),wq>nL((0,Tmax);W1,q(Ω))C2,0(¯Ω×(0,Tmax)).

    In addition, the function u0 in Ω×(0,Tmax) and if Tmax< then

    limtTmaxsupu(,t)L(Ω)=. (2.1)

    Moreover,

    Ωv(,t)=0,Ωw(,t)=0 for all t(0,Tmax). (2.2)

    Finally, the solution (u,v,w) is radially symmetric with respect to |x| if u0 satisfies (1.8).

    Proof. The proof of this lemma needs to be divided into four steps. Firstly, the method to solve the local time existence of the classical solution to the problem (1.5) is based on a standard fixed point theorem. Next, we will use the standard extension theorem to obtain (2.1). Then, we are going to use integration by parts to deduce (2.2). Finally, we would use the comparison principle to conclude that the solution is radially symmetric. For the details, we refer to [45,46,47,48].

    For the convenience of analysis and in order to prove Theorem 1.1, we set h=χvξw, then the system (1.5) is rewritten as

    {ut=(D(u)u)(uh),xΩ,t>0,0=Δhμ(t)+f(u),xΩ,t>0,uν=hν=0,xΩ,t>0,u(x,0)=u0(x),xΩ, (2.3)

    where μ(t)=χμ1(t)ξμ2(t)=1|Ω|Ωf(u(x,t))dx and f(u)=χf1(u)ξf2(u).

    For the same reason, we will convert the system (2.3) into a scalar equation. Let us assume Ω=BR(0) with some R>0 is a ball and the initial data u0=u0(r) with r=|x|[0,R] satisfies (1.8). In the radial framework, the system (2.3) can be transformed into the following form

    {rn1ut=(rn1D(u)ur)r(rn1uhr)r,r(0,R),t>0,0=(rn1hr)rrn1μ(t)+rn1f(u),r(0,R),t>0,ur=hr=0, r=R,t>0,u(r,0)=u0(r),r(0,R). (2.4)

    Lemma 2.2. Let us introduce the function

    U(s,t)=ns1n0ρn1u(ρ,t)dρ,s=rn[0,Rn], t(0,Tmax),

    then

    Us(t)=u(s1n,t), Uss(t)=1ns1n1ur(s1n,t), (2.5)

    and

    Ut(s,t)=n2s22nD(Us)Usssμ(t)Us+Uss0f(Us(σ,t))dσ. (2.6)

    Proof. Firstly, integrating the second equation of (2.4) over (0,r), we have

    rn1hr(r,t)=rnnμ(t)r0ρn1f(u(ρ,t))dρ,

    so

    s11nhr(s1n,t)=snμ(t)1ns0f(u(σ1n,t))dσ,s(0,Rn), t(0,Tmax).

    Then, a direct calculation yields

    Us(s,t)=u(s1n,t),s(0,Rn), t(0,Tmax),

    and

    Uss(s,t)=1ns1n1ur(s1n,t),s(0,Rn), t(0,Tmax),

    as well as

    Ut(s,t)=ns1n0ρn1ut(ρ,t)dρ=n2s22nD(Us)Ussns11nUshr=n2s22nD(Us)Usssμ(t)Us+Uss0f(Us(σ,t))dσ

    for all s(0,Rn) and t(0,Tmax).

    Furthermore, by a direct calculation and (1.7), we know that the functions U and f satisfy the following results

    {Us(s,t)=u(s1n,t)>0,s(0,Rn),t(0,Tmax),U(0,t)=0,U(Rn,t)=nωnΩu(,t)=nMωn,t[0,Tmax),|f(s)|,f1(s),f2(s)C0,0sA,C0=C0(A)>0, (2.7)

    where ωn=n|B1(0)| and A is a positive constant.

    Lemma 2.3. Suppose that (1.7), (1.8) and (2.7) hold, then we have

    hr(r,t)=1nμ(t)rr1nr0ρn1f(u(ρ,t))dρfor all r(0,R),t(0,Tmax).

    In particular,

    hr(r,t)1n(μ(t)+C0)r. (2.8)

    Proof. By integration the second equation in (2.4), we obtain that

    rn1hr=μ(t)r0ρn1dρr0ρn1f(u(ρ,t))dρ for all r(0,R),t(0,Tmax).

    According to (1.9), we can easily get that f(u)0 if uC=max{0,(k2ξk1χ)1γ1γ21}, and split

    r0ρn1f(u(ρ,t))dρ=r0χ{u(,t)C}(ρ)ρn1f(u(ρ,t))dρ+r0χ{u(,t)<C}(ρ)ρn1f(u(ρ,t))dρ.

    Combining these we have

    hr=1nμ(t)rr1nr0χ{u(,t)C}(ρ)ρn1f(u(ρ,t))dρr1nr0χ{u(,t)<C}(ρ)ρn1f(u(ρ,t))dρ1nμ(t)rr1nr0χ{u(,t)<C}(ρ)ρn1f(u(ρ,t))dρ1nμ(t)r+C0r1nr0χ{u(,t)<C}(ρ)ρn1dρ1nμ(t)r+C0r1nr0ρn1dρ1n(μ(t)+C0)r,

    so we complete this proof.

    To show the existence of a finite-time blow-up solution of (2.4), we need to prove that Uss is nonpositive by the following lemma. The proof follows the strategy in [48].

    Lemma 2.4. Suppose that D,f and u0 satisfy (I1),(I2) and (I3) respectively. Then

    ur(r,t)0 for all r(0,R),t(0,Tmax). (2.9)

    Moreover,

    Uss(s,t)0 for all r(0,R),t(0,Tmax). (2.10)

    Proof. Without loss of generality we may assume that u0C2 ([0,)) and fC2([0,)). Applying the regularity theory in ([49,50]), we all know that u and ur belong to C0([0,R]×[0,T))C2,1((0,R)×(0,T)) and we fixed T(0,Tmax). From (2.4), we have for r(0,R) and t(0,T)

    hrr+n1rhr=μ(t)f(u), (2.11)

    and from (2.4) we obtain

    urt=((D(u)ur)r+n1rD(u)ur+uf(u)uμ(t)urhr)r=(D(u)ur)rr+a1(D(u)ur)r+a2urr+bur,

    for all r(0,R) and t(0,T), where

    a1(r,t)=n1r,a2(r,t)=hr,b(r,t)=n1r2D(u)μ(t)hrr+f(u)+uf(u),

    for all r(0,R) and t(0,T). Moreover, we have hrrn(μ(t)+C0) by (2.8) and from (2.11) such that

    hrr=n1rhrμ(t)+f(u)n1nμ(t)+n1nC0μ(t)+f(u)f(u)+C0for all r(0,R) and t(0,T),

    then setting c1:=sup(r,t)(0,R)×(0,T)(2f(u)+uf(u)+C0), we obtain

    b(r,t)c1for all r(0,R) and t(0,T),

    and we introduce

    c2:=sup(r,t)(0,R)×(0,T)((D(u))rr+a1(D(u))r)<,

    and set c3=2(c1+c2+1). Since ur(r,t)=0 for r{0,R},t(0,T) (because u is radially symmetric) and u0r0, the function y:[0,R]×[0,T]R, (r,t)ur(r,t)εec3t belongs to C0([0,R]×[0,T]) and fulfills

    {yt=(D(u)(y+εec3t))rr+a1(D(u)(y+εec3t))r+a2yr+b(y+εec3t)c3εec3t=(D(u)y)rr+a1(D(u)y)r+a2yr+by+εec3t((D(u))rr+a1(D(u))r+bc3)(D(u)y)rr+a1(D(u)y)r+a2yr+by+εec3t(c1+c2c3),in (0,R)×(0,T),y<0,on {0,R}×(0,T),y(,0)<0,in (0,R). (2.12)

    By the estimate for y(,0) in (2.12) and continuity of y, the time t0:=sup{t(0,T):y0 in [0,R]×(0,T)}(0,T] is defined. Suppose that t0<T, then there exists r0[0,R] such that y(r0,t0)=0 and y(r,t)0 for all r[0,R] and t[0,t0]; hence, yt(r0,t0)0. As D0 in [0,), not only y(,t0) but also z:(0,R)R,rD(u(r,t0))y(r,t0) attains its maximum 0 at r0. Since the second equality in (2.12) asserts r0(0,R), we conclude zrr(r0)0,zr(r0)=0 and yr(r0,t0)=0. Hence, we could obtain the contradiction

    0yt(r0,t0)zrr(r0)+a1(r0,t0)zr(r0)+a2(r0,t0)yr(r0,t0)+b(r0,t0)y(r0,t0)+εec3t0(c1+c2c3)c32εec3t0<0,

    since we have

    c1+c2c32.

    So that t0=T, implying y0 in [0,R]×[0,T] and hence urεec3t in [0,R]×[0,T]. Letting first ε0 and then TTmax, this proves that ur0 in [0,R]×[0,Tmax), and we have Uss0 because of (2.5).

    In this section our aim is to establish a function and to select appropriate parameters such that the function satisfies ODI, which means finiteness of Tmax by counter evidence. Firstly, we introduce a moment-like functional as follows

    ϕ(t):=s00sγ(s0s)U(s,t)ds,t[0,Tmax), (3.1)

    with γ(,1) and s0(0,Rn). As a preparation of the subsequent analysis of ϕ, we denote

    Sϕ:={t(0,Tmax)|ϕ(t)nMs0(1γ)(2γ)ωns2γ0}. (3.2)

    The following lemma provides a lower bound for U.

    Lemma 3.1. Let γ(,1) and s0(0,Rn), then

    U(s02,t)1ωn(nM4s02γ(3γ)). (3.3)

    Proof. If (3.3) was false for some tSϕ such that U(s02,t)<1ωn(nM4s02γ(3γ)), then necessarily δ:=4s02γ(3γ)<nM. By the monotonicity of U(,t) we would obtain that U(s,t)<nMδωn for all s(0,s02). Since U(s,t)<nMωn for all s(0,Rn), we have

    ϕ(t)<nMδωns020sγ(s0s)ds+nMωns0s02sγ(s0s)ds=nMωns00sγ(s0s)dsδωns020sγ(s0s)ds
    =nMωns2γ0(1γ)(2γ)δωn2γ(3γ)s2γ04(1γ)(2γ).

    In view of the definition of Sϕ, we find that nMs0<nM2γ(3γ)δ4, which contradicts our definition of δ.

    An upper bound for μ is established by the following lemma.

    Lemma 3.2. Let γ(,1) and s0>0 such that s0Rn6. Then the function μ(t) has property that

    μ(t)C1+12ss0f(Us(σ,t))dσ for all s(0,s0) and any tSϕ, (3.4)

    where C1=χ2C0+C0+C23+C3=13(χ2C0+C0+χk1(γ1γ2)2γ2(2ξk2γ2χk1γ1)γ1γ1γ2)+χf1(2δωns0).

    Proof. First for any fixed tSϕ, we may invoke Lemma 3.1 to see that

    U(s02,t)nMδωn,

    and thus, as UnMωn,

    U(s0,t)U(s02,t)s02nMωnnMδωns02=2δωns0.

    However, by concavity of U(,t), as asserted by Lemma 2.4,

    U(s0,t)U(s02,t)s02Us(s0,t)Us(s,t) for all s(s0,Rn).

    Then let s0(0,Rn), we know that

    μ(t)=1Rns00f(Us(σ,t))dσ+1RnRns0f(Us(σ,t))dσ=1Rns0f(Us(σ,t))dσ+1Rns0sf(Us(σ,t))dσ+1RnRns0f(Us(σ,t))dσ,t(0,Tmax). (3.5)

    Since γ1>γ2 and Young's inequality such that ξf2(u)ξk2(1+u)γ2χk12(1+u)γ1+C2χ2f1(u)+C2 with C2=χk1(γ1γ2)2γ2(2ξk2γ2χk1γ1)γ1γ1γ2 for u0, then for all s(0,Rn) and t(0,Tmax) we show that

    χ2f1(Us(s,t))C2f(Us(s,t))χf1(Us(s,t)). (3.6)

    Accordingly, by the monotonicity of Us(,t) along with (1.7) and (3.6), we have

    s0sf(Us(σ,t))dσs0sχf1(Us(σ,t))dσs0sχf1(Us(s,t))dσs0χf1(Us(s,t)),s(0,s0), t(0,Tmax).

    Since the condition of (2.7) implies that

    s0f(Us(σ,t))dσ=s0χ{Us(,t)1}(σ)f(Us(σ,t)dσ+s0χ{Us(,t)<1}(σ)f(Us(σ,t)dσs0χ{Us(,t)1}(σ)(χ2f1(Us(σ,t))C2)dσC0ss0χ{Us(,t)1}(σ)χ2f1(Us(σ,t))dσ(C0+C2)s=s0χ2f1(Us(σ,t))dσs0χ{Us(,t)<1}(σ)χ2f1(Us(σ,t))dσ(C0+C2)ss0χ2f1(Us(s,t))dσχ2C0s(C0+C2)ssχ2f1(Us(s,t))(χ2C0+C0+C2)s.

    Therefore, we obtain

    s0sf(Us(σ,t))dσ2s0ss0f(Us(σ,t))dσ+2(χ2C0+C0+C2)s0.

    Since (3.5) we have for all s(0,s0)

    1Rns0f(Us(σ,t))dσ+1Rns0sf(Us(σ,t))dσ1Rns0f(Us(σ,t))dσ+2s0Rnss0f(Us(σ,t))dσ+2(χ2C0+C0+C2)s0Rn, (3.7)

    where s0Rn6 such that 1Rn16s016s, 2s0Rns13s and s0Rn16 for all s(0,s0). Finally, we estimate the last summand of (3.5)

    1RnRns0f(Us(σ,t))dσ1RnRns0χf1(Us(σ,t))dσχf1(2δωns0)=C3. (3.8)

    Together with (3.5), (3.7) and (3.8) imply (3.4).

    Lemma 3.3. Assume that γ(,1) satisfying

    γ<22n,

    and s0(0,Rn6]. Then the function ϕ:[0,Tmax)R defined by (3.1) belongs to C0([0,Tmax))C1((0,Tmax)) and satisfies

    ϕ(t)n2s00s22nγ(s0s)UssD(Us)ds+12s00sγ(s0s)Us{s0f(Us(σ,t))dσ}dsC1s00s1γ(s0s)Usds=:J1(t)+J2(t)+J3(t), (3.9)

    for all t[0,Tmax), where C1 is defined in Lemma 3.2.

    Proof. Combining (2.6) and (3.4) we have

    Ut(s,t)=n2s22nD(Us)Usssμ(t)Us+Uss0f(Us(σ,t))dσn2s22nUssD(Us)+12Uss0f(Us(σ,t))dσC1sUs.

    Notice ϕ(t) conforms ϕ(t)=s00sγ(s0s)U(s,t)ds. So (3.9) is a direct consequence.

    Lemma 3.4. Let s0(0,Rn6], and γ(,1) satisfying γ<22n. Then J1(t) in (3.9) satisfies

    J1(t)I, (3.10)

    where

    I:={n2dm(22nγ)s00s12nγ(s0s),m<0,n2d(22nγ)s00s12nγ(s0s)ln(Us+1),m=0,n2dm(22nγ)s00s12nγ(s0s)(Us+1)m,m>0, (3.11)

    for all tSϕ.

    Proof. Since DC2([0,)), suppose that

    G(τ)=τ0D(δ)dδ,

    then

    0<G(τ)dτ0(1+δ)m1dδ{dm,m<0,dln(τ+1),m=0,dm(τ+1)m,m>0.

    Here integrating by parts we obtain

    J1(t)=n2s00s22nγ(s0s)dG(Us)=n2s22nγ(s0s)G(Us)|s00+n2s00s22nγG(Us)dsn2(22nγ)s00s12nγ(s0s)G(Us)ds.

    Hence a direct calculation yields

    J1(t){n2dm(22nγ)s00s12nγ(s0s),m<0,n2d(22nγ)s00s12nγ(s0s)ln(Us+1),m=0,n2dm(22nγ)s00s12nγ(s0s)(Us+1)m,m>0,

    for all tSϕ. We conclude (3.10).

    Lemma 3.5. Assume that γ(,1) satisfying γ<22n and s0(0,Rn6]. Then we have

    J2(t)+J3(t)k1χ4s00s1γ(s0s)U1+γ1sdsC4s00s1γ(s0s)Usds (3.12)

    for all tSϕ, where C4=C1+(χ2C0+C0+C2)2.

    Proof. Since Lemma 3.2 we have

    s0f(Us(σ,t))dσs2χf1(Us(s,t))(χ2C0+C0+C2)sfor all s(0,s0) and t(0,Tmax).

    Therefore,

    J2(t)=12s00sγ(s0s)Us{s0f(Us(σ,t))dσ}dsχ4s00s1γ(s0s)Usf1(Us(s,t))ds(χ2C0+C0+C2)2s00s1γ(s0s)Usdsk1χ4s00s1γ(s0s)U1+γ1sds(χ2C0+C0+C2)2s00s1γ(s0s)Usds,

    where f1(Us(s,t))k1(1+Us)γ1k1(Us)γ1. Combining these inequalities we can deduce (3.12).

    Lemma 3.6. Let γ1>max{0,m1}. For any γ(,1) satisfying

    γmin{22n1+γ1γ1, 22n1+γ11+γ1m}, (3.13)

    and s0(0,Rn6], the function ϕ:[0,Tmax)R defined in (3.1) satisfies

    ϕ(t){Cψ(t)Cs3γ2n1+γ1γ10,m1,Cψ(t)Cs3γ2n1+γ11+γ1m0,m>1,  (3.14)

    with C>0 for all tSϕ, where ψ(t):=s00s1γ(s0s)U1+γ1sds.

    Proof. From (3.10) and (3.12) we have

    ϕ(t)k1χ4ψ(t)IC4s00s1γ(s0s)Usds, (3.15)

    for all tSϕ and I is given by (3.11). In the case m<0,

    n2dm(22nγ)s00s12nγ(s0s)dsn2dm(22nγ)s0s00s12nγds=n2dms32nγ0.

    If m=0, we use the fact that ln(1+x)x<1 for any x>0 and Hölder's inequality to estimate

    n2d(22nγ)s00s12nγ(s0s)ln(Us+1)ds=n2d(22nγ)s00[s1γ(s0s)U1+γ1s]11+γ1s12nγ1γ1+γ1(s0s)111+γ1ln(1+Us)Usdsn2d(22nγ){s00s1γ(s0s)U1+γ1sds}11+γ1{s00(s12nγ1γ1+γ1(s0s)γ11+γ1)1+γ1γ1ds}γ11+γ1n2d(22nγ)sγ11+γ10{s00s(12nγ)γ12nγ1ds}γ11+γ1ψ11+γ1(t)=C5ψ11+γ1(t)s(32nγ)γ12n1+γ10,

    for all tSϕ with C5:=n2d(22nγ)(12γ2n1+γ1γ1)γ11+γ1>0 by (3.13). In the case m>0, by using the elementary inequality (a+b)α2α(aα+bα) for all a,b>0 and every α>0, we obtain

    n2dm(22nγ)s00s12nγ(s0s)(Us+1)mds2mn2dm(22nγ)s00s12nγ(s0s)Umsds+2mn2dm(22nγ)s00s12nγ(s0s)ds, (3.16)

    for all tSϕ, and we first estimate the second term on the right of (3.16)

    2mn2dm(22nγ)s00s12nγ(s0s)ds2mn2dms32nγ0.

    Since γ1>m1 and by Hölder's inequality we deduce that

    2mn2dm(22nγ)s00s12nγ(s0s)Umsds=2mn2dm(22nγ)s00s(1γ)m1+γ1(s0s)m1+γ1Umss12nγ(1γ)m1+γ1(s0s)1m1+γ1ds2mn2dm(22nγ){s00[s(1γ)m1+γ1(s0s)m1+γ1Ums]1+γ1mds}m1+γ1×{s00[s12nγ(1γ)m1+γ1(s0s)1m1+γ1]1+γ11+γ1mds}1+γ1m1+γ12mn2dm(22nγ)ψm1+γ1(t)s1+γ1m1+γ10{s00s(1+γ1m)(1γ)2n(1+γ1)1+γ1mds}1+γ1m1+γ1C6ψm1+γ1(t)s(1+γ1m)(3γ)2n(1+γ1)1+γ10,

    for all tSϕ with C6=2mn2dm(22nγ)(12γ2n1+γ11+γ1m)1+γ1m1+γ1>0 where γ<22n1+γ11+γ1m from (3.13).

    Next, we can estimate the third expression on the right of (3.15) as follows

    C4s00s1γ(s0s)Usds=C4s00[s1γ(s0s)U1+γ1s]11+γ1s1γ1γ1+γ1(s0s)111+γ1dsC4{s00s1γ(s0s)U1+γ1sds}11+γ1{s00[s1γ1γ1+γ1(s0s)111+γ1]1+γ1γ1ds}γ11+γ1C4ψ11+γ1(t)sγ11+γ10{s00s1γds}γ11+γ1=C7ψ11+γ1(t)s(3γ)γ11+γ10,

    where C7=C4(12γ)γ11+γ1 for all tSϕ. By (3.15) and collecting the estimates above we have

    ϕ(t){k1χ4ψ(t)+n2dms32nγ0C7ψ11+γ1(t)s(3γ)γ11+γ10,m<0, tSϕ,k1χ4ψ(t)C5ψ11+γ1(t)s(32nγ)γ12n1+γ10C7ψ11+γ1(t)s(3γ)γ11+γ10,m=0, tSϕ,k1χ4ψ(t)2mn2dms32nγ0C6ψm1+γ1(t)s(1+γ1m)(3γ)2n(1+γ1)1+γ10C7ψ11+γ1(t)s(3γ)γ11+γ10,m>0, tSϕ.

    If m=0, by Young's inequality we can find positive constants C8,C9 such that

    C5ψ11+γ1(t)s(32nγ)γ12n1+γ10k1χ16ψ(t)+C8s3γ2n1+γ1γ10,tSϕ,

    while as m>0 we have

    C6ψm1+γ1(t)s(1+γ1m)(3γ)2n(1+γ1)1+γ10k1χ16ψ(t)+C9s3γ2n1+γ11+γ1m0,tSϕ.

    On the other hand, we use Young's inequality again

    C7ψ11+γ1(t)s(3γ)γ11+γ10k1χ16ψ(t)+C10s3γ0,tSϕ.

    In the case m<0, because of s0(0,Rn6], we have

    s32nγ0=s3γ2n1+γ1γ10s2nγ10(Rn6)2nγ1s3γ2n1+γ1γ10,

    when m>0 we have

    s32nγ0=s3γ2n1+γ11+γ1m0s2mn(1+γ1m)0(Rn6)2mn(1+γ1m)s3γ2n1+γ11+γ1m0.

    All in all, we have

    ϕ(t){k1χ8ψ(t)C11s3γ2n1+γ1γ10C10s3γ0,m0,k1χ8ψ(t)C12s3γ2n1+γ11+γ1m0C10s3γ0,m>0, (3.17)

    for all tSϕ with C11=C8n2dm(Rn6)2nγ1 and C12=C9+2mn2dm(Rn6)2mn(1+γ1m). When 0<m1, we have 1+γ11+γ1m1+γ1γ1 such that s3γ2n1+γ11+γ1m0=s3γ2n1+γ1γ10s2n(1+γ1γ11+γ11+γ1m)0(Rn6)2(1m)(1+γ1)nγ1(1+γ1m)s3γ2n1+γ1γ10. In the case m1

    s3γ0=s2n1+γ1γ10s3γ2n1+γ1γ10(Rn6)2n1+γ1γ1s3γ2n1+γ1γ10,

    and if m>1 we have

    s3γ0=s2n1+γ11+γ1m0s3γ2n1+γ11+γ1m0(Rn6)2n1+γ1(1+γ1m)s3γ2n1+γ11+γ1m0.

    Thus (3.17) turns into (3.14).

    Next, we need to build a connection between ϕ(t) and ψ(t). Let us define

    Sψ:={t(0,Tmax)|ψ(t)s3γ0}. (3.18)

    Lemma 3.7. Let γ(,1) satisfying γ>1γ1 and (3.13). Then for any choice of s0(0,Rn6], the following inequality

    ϕ(t){Csγ1(3γ)0ϕ1+γ1(t)Cs3γ2n1+γ1γ10,m1,Csγ1(3γ)0ϕ1+γ1(t)Cs3γ2n1+γ11+γ1m0,m>1,  (3.19)

    holds for all tSϕSψ with C>0.

    Proof. We first split

    U(s,t)=s0Us(σ,t)dσ=s0χ{Us(,t)<1}(σ)Us(σ,t)dσ+s0χ{Us(,t)1}(σ)Us(σ,t)dσs+s0χ{Us(,t)1}(σ){σ1γ(s0σ)U1+γ1s}11+γ1σ1γ1+γ1(s0σ)11+γ1dσs+(s0s)11+γ1ψ11+γ1(t){s0σ1γ1+γ11+γ1γ1dσ}γ11+γ1=s+(γ1γ+γ11)γ11+γ1(s0s)11+γ1sγ+γ111+γ1ψ11+γ1(t), (3.20)

    for all s(0,s0) and t(0,Tmax) where γ1γ+γ11>0. According to the definition of Sψ, we can find

    s(s0s)11+γ1sγ+γ111+γ1ψ11+γ1(t)=s2γ1+γ1(s0s)11+γ1ψ11+γ1(t)s2γ1+γ10s11+γ10(s3γ0)11+γ1=1, (3.21)

    for all s(0,s0) and tSψ. Combining (3.20) and (3.21) we have

    U(s,t)C1sγ+γ111+γ1(s0s)11+γ1ψ11+γ1(t),

    where C1=1+(γ1γ+γ11)γ11+γ1 for all s(0,s0) and tSψ. Invoking Hölder's inequality, we get

    ϕ(t)=s00sγ(s0s)U(s,t)dsC1s00sγ+γ+γ111+γ1(s0s)111+γ1dsψ11+γ1(t)C1sγ11+γ10s00sγ+γ+γ111+γ1dsψ11+γ1(t)=C2sγ1(3γ)1+γ10ψ11+γ1(t), (3.22)

    where C2=C11+γ1γ1(2γ) for all s(0,s0) and tSψ. Employing these conclusion we deduce (3.19).

    These preparations above will enable us to establish a superlinear ODI for ϕ as mentioned earlier, and we prove our main result on blow-up based on a contradictory argument.

    Proof of Theorem 1.1. Step 1. Assume on the contrary that Tmax=+, and we define the function

    S:={T(0,+)|ϕ(t)>nMs0ωn(1γ)(2γ)s2γ0 for all t[0,T]}. (3.23)

    Let us choose s0>0 such that

    s0min{Rn6,nM2,nMγ12(1γ)ωn[(C3+1)(1+γ1)1]}, (3.24)

    where M and ωn were defined in (2.7) and C3=(γ1γ+γ11)γ11+γ1 has been mentioned in (3.20). Then we pick 0<ε(γ1,M,R)=ε<s0n and s(γ1,M,R)(0,s0) wtih r(γ1,M,R)=(s)1n(0,R) such that

    U(s,0)U(s,0)=nωnBr(0)u0dxnωn(Mε),s(s,Rn).

    Therefore it is possible to estimate

    ϕ(0)=s00sγ(s0s)U(s,0)dss00sγ(s0s)U(s,0)ds>nωn(Ms0n)s00sγ(s0s)ds=nMs0ωn(1γ)(2γ)s2γ0. (3.25)

    Then S is non-empty and denote T=supS(0,]. Next, we need to prove (0,T)SϕSψ. Note that

    ϕ(t)>nMs0ωn(1γ)(2γ)s2γ0,t(0,T), (3.26)

    we obtain (0,T)Sϕ. From (3.20) we have

    ϕ(t)s00sγ(s0s)[s+C3sγ+γ111+γ1(s0s)11+γ1ψ11+γ1(t)]dss0s00s1γds+C3s00sγ+γ+γ111+γ1(s0s)γ11+γ1ψ11+γ1(t)ds=s3γ02γ+C3(1+γ1)γ1(2γ)sγ1(3γ)1+γ10ψ11+γ1(t).

    It follows from (3.24) and (3.26) that

    ϕ(t)nM2(1γ)(2γ)ωns2γ0for all t(0,T). (3.27)

    Then

    C3(1+γ1)γ1(2γ)sγ1(3γ)1+γ10ψ11+γ1(t)nM2(1γ)(2γ)ωns2γ0s3γ02γ.

    Note that (3.24) implies

    nMγ12C3(1γ)ωn(1+γ1)s0γ1C3(1+γ1)1,

    then we have

    ψ(t)[(nMs2γ02(1γ)(2γ)ωns3γ02γ)γ1(2γ)C3(1+γ1)sγ1(3γ)1+γ10]1+γ1[nMγ12C3(1γ)ωn(1+γ1)s0γ1C3(1+γ1)]1+γ1s3γ0s3γ0.

    Therefore, (0,T)SϕSψ.

    Step 2. Applying Lemma 3.7 we can find γ(,1) and C1,C2>0 such that for all s0(0,Rn6]

    ϕ(t){C1sγ1(3γ)0ϕ1+γ1(t)C2s3γ2n1+γ1γ10,m1,C1sγ1(3γ)0ϕ1+γ1(t)C2s3γ2n1+γ11+γ1m0,m>1, 

    for all tSϕSψ and with (3.22) we have

    ψ(t)C3sγ1(3γ)0ϕ1+γ1(t),tSψ.

    To specify our choice of s0, for given M>0 we choose s0(0,Rn6] small enough such that

    s0nM2, (3.28)

    and also

    sγ10<TC1γ14(nM2ωn(1γ)(2γ))γ1, (3.29)

    as well as

    s1+γ10C3(nM2ωn(1γ)(2γ))1+γ1. (3.30)

    From (3.23), (3.28) and (3.30) we have

    ψ(t)C3sγ1(3γ)0ϕ1+γ1(t)>C3(nMs0ωn(1γ)(2γ)1s0)1+γ1s3γ0s3γ0,tSψ,

    which shows that SSϕSψ. Since 1+γ1m>2n, we have (1+γ1)(12n(1+γ1m))>0 if m>1 so that we can choose s0 sufficiently small satisfying (3.28)(3.30) such that

    s(1+γ1)(12n(1+γ1m))0C12C2(nM2ωn(1γ)(2γ))1+γ1,

    while in the case m1, the condition γ1>m1+2n2n which infers that (1+γ1)(12nγ1)>0 and we select s0 small enough fulfilling (3.28)(3.30) such that

    s(1+γ1)(12nγ1)0C12C2(nM2ωn(1γ)(2γ))1+γ1.

    It is possible to obtain

    C12sγ1(3γ)0ϕ1+γ1(0)C2s3γ2n1+γ11+γ1m0C12C2(nM2ωn(1γ)(2γ))1+γ1s(1+γ1)+2n1+γ1(1+γ1m)01,m>1,

    and we have

    C12sγ1(3γ)0ϕ1+γ1(0)C2s3γ2n1+γ1γ10C12C2(nM2ωn(1γ)(2γ))1+γ1s(1+γ1)+2n1+γ1γ101,m1.

    All in all, for any mR, we apply an ODI comparison argument to obtain that

    ϕ(t)C12sγ1(3γ)0ϕ1+γ1(t),t(0,T).

    By a direct calculation we obtain

    1γ1(1ϕγ1(t)1ϕγ1(0))C12sγ1(3γ)0t,t(0,T).

    Hence, according to (3.25) and (3.29) we conclude

    t<2C1γ1sγ10(2ωn(1γ)(2γ)nM)γ1T2,

    for all t(0,T). As a consequence, we infer that Tmax must be finite.

    In this section, we are preparing to prove Theorem 1.2 by providing the Lp estimate of u and the Moser-type iteration.

    Lemma 4.1. Let (u,v,w) be a classical solution of the system (1.5) under the condition of Theorem 1.2. Suppose that

    γ2<1+γ1<2n+m. (4.1)

    Then for any p>max{1,2m,γ2}, there exists C=C(p)>0 such that

    Ω(1+u)p(x,t)dxCon (0,Tmax). (4.2)

    Proof. Notice f1(u)k1(1+u)γ1, f2(u)=k2(1+u)γ2 for all u0. Multiplying the first equation of (1.5) by p(1+u)p1 and integrating by parts with the boundary conditions for u,v and w, we have

    ddtΩ(1+u)pdx+p(p1)Ω(1+u)p2D(u)|u|2dx=χp(p1)Ωu(1+u)p2uvdxξp(p1)Ωu(1+u)p2uwdx=χ(p1)Ω(1+u)pΔvdx+χpΩ(1+u)p1Δvdx+ξ(p1)Ω(1+u)pΔwdxξpΩ(1+u)p1Δwdxχ(p1)Ω(1+u)pf1(u)dx+χpΩ(1+u)p1μ1(t)dx+ξ(p1)Ω(1+u)pμ2(t)dxξ(p1)Ω(1+u)pf2(u)dx+ξpΩ(1+u)p1f2(u)dxk1χ(p1)Ω(1+u)p+γ1dx+χpΩ(1+u)p1μ1(t)dx+ξ(p1)Ω(1+u)pμ2(t)dxk2ξ(p1)Ω(1+u)p+γ2dx+k2ξpΩ(1+u)p+γ21dx,t(0,Tmax). (4.3)

    Firstly,

    p(p1)Ω(1+u)p2D(u)|u|2dxdp(p1)Ω(1+u)p+m3|u|2dx=4dp(p1)(p+m1)2Ω|(1+u)p+m12|2dx.

    By Young's inequality and Hölder's inequality, we obtain

    χpΩ(1+u)p1μ1(t)dxC1Ω(1+u)p+γ1dx+C2μp+γ11+γ11(t)=C1Ω(1+u)p+γ1dx+C2(1|Ω|Ωf1(u)dx)p+γ11+γ1C1Ω(1+u)p+γ1dx+C3(Ω(1+u)1+γ1dx)p+γ11+γ1C1Ω(1+u)p+γ1dx+C3{(Ω(1+u)p+γ1dx)1+γ1p+γ1|Ω|p1p+γ1}p+γ11+γ1=C1Ω(1+u)p+γ1dx+C3|Ω|p11+γ1Ω(1+u)p+γ1dx.

    for all t(0,Tmax). Then by Hölder's inequality we obtain

    ξ(p1)Ω(1+u)pμ2(t)dx=k2ξ(p1)|Ω|Ω(1+u)γ2dxΩ(1+u)pdxk2ξ(p1)|Ω|{Ω(1+u)p+γ2dx}γ2p+γ2|Ω|pp+γ2×{Ω(1+u)p+γ2dx}pp+γ2|Ω|γ2p+γ2=k2ξ(p1)Ω(1+u)p+γ2dx,t(0,Tmax).

    Furthermore, by using Young's inequality and (4.1) we have

    k2ξpΩ(1+u)p+γ21dxC4Ω(1+u)p+γ1dx+C5,

    for all t(0,Tmax). Therefore, combining these we conclude

    ddtΩ(1+u)pdx+4dp(p1)(p+m1)2Ω|(1+u)p+m12|2dxC6Ω(1+u)p+γ1dx+C5,t(0,Tmax),

    where C6=C1+C3|Ω|p11+γ1+C4+k1χ(p1). By means of Gagliardo-Nirenberg inequality we can find C7 such that

    C6Ω(1+u)p+γ1dx=C6(1+u)p+m122(p+γ1)p+m1L2(p+γ1)p+m1(Ω)C7(1+u)p+m122(p+γ1)p+m1aL2(Ω)(1+u)p+m122(p+γ1)p+m1(1a)L2p+m1(Ω)+C7(1+u)p+m122(p+γ1)p+m1L2p+m1(Ω)

    for all t(0,Tmax), where

    a=p+m12p+m12(p+γ1)p+m12(121n)(0,1).

    Since 1m+γ1<2n, we have 2(p+γ1)p+m1a<2, and we use Young's inequality to see that for all t(0,Tmax)

    C6Ω(1+u)p+γ1dx2dp(p1)(p+m1)2(1+u)p+m122L2(Ω)+C8.

    In quite a similar manner, we obtain C9=C9(p)>0 fulfilling

    Ω(1+u)pdx2dp(p1)(p+m1)2(1+u)p+m122L2(Ω)+C9for all t(0,Tmax).

    Finally, combining these to (4.3) we obtain

    ddtΩ(1+u)pdx+Ω(1+u)pdxC5+C8+C9for all t(0,Tmax).

    Thus,

    Ω(1+u)pdxmax{Ω(1+u0)pdx,C5+C8+C9}for all t(0,Tmax).

    We have done the proof.

    Under the condition of Lemma 4.1 we can use the above information to prove Theorem 1.2.

    Proof of Theorem 1.2. From Lemma 4.1, we let p>max{γ1n,γ2n,1}. By the elliptic Lp-estimate to the two elliptic equations in (1.5), we get that for all t(0,Tmax) there exists some C10(p)>0 such that

    v(,t)w2,pγ1(Ω)C10(p),w(,t)w2,pγ2(Ω)C10(p), (4.4)

    and hence, by the Sobolev embedding theorem, we get

    v(,t)C1(¯Ω)C10(p),w(,t)C1(¯Ω)C10(p). (4.5)

    Now the Moser iteration technique ([3,51]) ensures that u(,t)L(Ω)C for any t(0,Tmax).

    This concludes by Lemma 2.1 that Tmax=.

    The paper is supported by the Research and Innovation Team of China West Normal University (CXTD2020–5).

    The authors declare that there is no conflict of interest.



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