In this paper, we consider the quasilinear parabolic-elliptic-elliptic attraction-repulsion system
{ut=∇⋅(D(u)∇u)−χ∇⋅(u∇v)+ξ∇⋅(u∇w),x∈Ω,t>0,0=Δv−μ1(t)+f1(u),x∈Ω,t>0,0=Δw−μ2(t)+f2(u),x∈Ω,t>0
under homogeneous Neumann boundary conditions in a smooth bounded domain Ω⊂Rn, n≥2. The nonlinear diffusivity D and nonlinear signal productions f1,f2 are supposed to extend the prototypes
D(s)=(1+s)m−1, f1(s)=(1+s)γ1, f2(s)=(1+s)γ2, s≥0,γ1,γ2>0,m∈R.
We proved that if γ1>γ2 and 1+γ1−m>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∇⋅(u∇v)+ξ∇⋅(u∇w),x∈Ω,t>0,0=Δv−μ1(t)+f1(u),x∈Ω,t>0,0=Δw−μ2(t)+f2(u),x∈Ω,t>0
under homogeneous Neumann boundary conditions in a smooth bounded domain Ω⊂Rn, n≥2. The nonlinear diffusivity D and nonlinear signal productions f1,f2 are supposed to extend the prototypes
D(s)=(1+s)m−1, f1(s)=(1+s)γ1, f2(s)=(1+s)γ2, s≥0,γ1,γ2>0,m∈R.
We proved that if γ1>γ2 and 1+γ1−m>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),x∈Ω,t>0,τvt=Δv−αv+βu,x∈Ω,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=Δv−M+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),x∈Ω,t>0,0=Δv−M+g(u),x∈Ω,t>0, | (1.2) |
where M:=1|Ω|∫Ωg(u(x,t))dx. Recently, when D(u)=u−p,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=Δu−χ∇⋅(u∇v)+ξ∇⋅(u∇w),x∈Ω,t>0,τ1vt=Δv+αu−βv,x∈Ω,t>0,τ2wt=Δw+δu−γw,x∈Ω,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 ξγ−χα>0 when n≥2, and the solution would blow up in finite time provided ξγ−χα<0,α=β when n=2. Then, there is a blow-up solution when χα−ξγ>0,δ≥β or χαδ−ξγβ>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 ξγ−χα≥0, and Zhong et al. [27] obtained the global existence of weak solution when ξγ−χα≥0 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=Δu−χ∇⋅(ϕ(u)∇v)+ξ∇⋅(ψ(u)∇w),x∈Ω,t>0,τ1vt=Δv−μ1(t)+f(u),x∈Ω,t>0,τ2wt=Δw−μ2(t)+g(u),x∈Ω,t>0, | (1.4) |
where Ω∈Rn(n≥2) 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∇⋅(u∇v)+ξ∇⋅(u∇w),x∈Ω,t>0,0=Δv−μ1(t)+f1(u),x∈Ω,t>0,0=Δw−μ2(t)+f2(u),x∈Ω,t>0,∂u∂ν=∂v∂ν=∂w∂ν=0,x∈∂Ω,t>0,u(x,0)=u0(x),x∈Ω, | (1.5) |
in a bounded domain Ω⊂Rn,n≥2 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,ξ≥0, 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
D∈C2([0,∞)). | (1.6) |
(I2) The function fi is nonnegative and nondecreasing and satisfies
fi∈⋃θ∈(0,1)Cθloc([0,∞))∩C1((0,∞)) | (1.7) |
with i∈{1,2}.
(I3) The initial datum
u0∈⋃θ∈(0,1)Cθ(¯Ω) is nonnegative and radially decreasing,∂u0∂ν=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 n≥2, 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)m−1, f1(u)≥k1(1+u)γ1, f2(u)≤k2(1+u)γ2 for all u≥0, |
with m∈R, k1,k2,γ1,γ2,d>0 and
γ1>γ2 and 1+γ1−m>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)u0≥M−ε, |
then the corresponding solution of the system (1.5) blows up in finite time.
Theorem 1.2. Let n≥2, Ω⊂Rn 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)m−1, f1(u)≤k1(1+u)γ1, f2(u)=k2(1+u)γ2 for all u≥0, |
with m∈R, k1,k2,γ1,γ2,d>0 and
γ2<1+γ1<2n+m. | (1.10) |
Then for each u0∈⋃θ∈(0,1)Cθ(¯Ω), u0≥0 with ∂u0∂ν=0 on ∂Ω, and the system (1.5) admits a unique global classical solution (u,v,w) with
u,v,w∈C2,1(¯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 Ω⊂Rn with n≥2 be a bounded domain with smooth boundary. Assume that D fulfills (1.6), f1,f2 satisfy (1.7) and u0∈⋃θ∈(0,1)Cθ(¯Ω) with ∂u0∂ν=0 on ∂Ω as well as u0≥0, then there exist Tmax∈(0,∞] and a classical solution (u,v,w) to (1.5) uniquely determined by
{u∈C0(¯Ω×[0,Tmax))∩C2,1(¯Ω×(0,Tmax)),v∈∩q>nL∞((0,Tmax);W1,q(Ω))∩C2,0(¯Ω×(0,Tmax)),w∈∩q>nL∞((0,Tmax);W1,q(Ω))∩C2,0(¯Ω×(0,Tmax)). |
In addition, the function u≥0 in Ω×(0,Tmax) and if Tmax<∞ then
limt↗Tmaxsup‖u(⋅,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=χv−ξw, then the system (1.5) is rewritten as
{ut=∇⋅(D(u)∇u)−∇⋅(u∇h),x∈Ω,t>0,0=Δh−μ(t)+f(u),x∈Ω,t>0,∂u∂ν=∂h∂ν=0,x∈∂Ω,t>0,u(x,0)=u0(x),x∈Ω, | (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
{rn−1ut=(rn−1D(u)ur)r−(rn−1uhr)r,r∈(0,R),t>0,0=(rn−1hr)r−rn−1μ(t)+rn−1f(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)=n∫s1n0ρn−1u(ρ,t)dρ,s=rn∈[0,Rn], t∈(0,Tmax), |
then
Us(t)=u(s1n,t), Uss(t)=1ns1n−1ur(s1n,t), | (2.5) |
and
Ut(s,t)=n2s2−2nD(Us)Uss−sμ(t)Us+Us⋅∫s0f(Us(σ,t))dσ. | (2.6) |
Proof. Firstly, integrating the second equation of (2.4) over (0,r), we have
rn−1hr(r,t)=rnnμ(t)−∫r0ρn−1f(u(ρ,t))dρ, |
so
s1−1nhr(s1n,t)=snμ(t)−1n∫s0f(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)=1ns1n−1ur(s1n,t),∀s∈(0,Rn), t∈(0,Tmax), |
as well as
Ut(s,t)=n∫s1n0ρn−1ut(ρ,t)dρ=n2s2−2nD(Us)Uss−ns1−1nUshr=n2s2−2nD(Us)Uss−sμ(t)Us+Us⋅∫s0f(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)=nωn∫Ωu(⋅,t)=nMωn,t∈[0,Tmax),|f(s)|,f1(s),f2(s)≤C0,0≤s≤A,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)r−r1−n∫r0ρn−1f(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
rn−1hr=μ(t)⋅∫r0ρn−1dρ−∫r0ρn−1f(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 u≥C∗=max{0,(k2ξk1χ)1γ1−γ2−1}, and split
∫r0ρn−1f(u(ρ,t))dρ=∫r0χ{u(⋅,t)≥C∗}(ρ)⋅ρn−1f(u(ρ,t))dρ+∫r0χ{u(⋅,t)<C∗}(ρ)⋅ρn−1f(u(ρ,t))dρ. |
Combining these we have
hr=1nμ(t)r−r1−n∫r0χ{u(⋅,t)≥C∗}(ρ)⋅ρn−1f(u(ρ,t))dρ−r1−n∫r0χ{u(⋅,t)<C∗}(ρ)⋅ρn−1f(u(ρ,t))dρ≤1nμ(t)r−r1−n∫r0χ{u(⋅,t)<C∗}(ρ)⋅ρn−1f(u(ρ,t))dρ≤1nμ(t)r+C0r1−n∫r0χ{u(⋅,t)<C∗}(ρ)⋅ρn−1dρ≤1nμ(t)r+C0r1−n∫r0ρn−1dρ≤1n(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 u0∈C2 ([0,∞)) and f∈C2([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+n−1rhr=μ(t)−f(u), | (2.11) |
and from (2.4) we obtain
urt=((D(u)ur)r+n−1rD(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)=n−1r,a2(r,t)=−hr,b(r,t)=−n−1r2D(u)−μ(t)−hrr+f(u)+uf′(u), |
for all r∈(0,R) and t∈(0,T). Moreover, we have hr≤rn(μ(t)+C0) by (2.8) and from (2.11) such that
−hrr=n−1rhr−μ(t)+f(u)≤n−1nμ(t)+n−1nC0−μ(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 u0r≤0, 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+b−c3)≤(D(u)y)rr+a1(D(u)y)r+a2yr+by+εec3t(c1+c2−c3),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):y≤0 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 D≥0 in [0,∞), not only y(⋅,t0) but also z:(0,R)→R,r⟼D(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
0≤yt(r0,t0)≤zrr(r0)+a1(r0,t0)zr(r0)+a2(r0,t0)yr(r0,t0)+b(r0,t0)y(r0,t0)+εec3t0(c1+c2−c3)≤−c32εec3t0<0, |
since we have
c1+c2≤c32. |
So that t0=T, implying y≤0 in [0,R]×[0,T] and hence ur≤εec3t in [0,R]×[0,T]. Letting first ε↘0 and then T↗Tmax, this proves that ur≤0 in [0,R]×[0,Tmax), and we have Uss≤0 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):=∫s00s−γ(s0−s)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)≥nM−s0(1−γ)(2−γ)ωn⋅s2−γ0}. | (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⋅(nM−4s02γ(3−γ)). | (3.3) |
Proof. If (3.3) was false for some t∈Sϕ such that U(s02,t)<1ωn⋅(nM−4s02γ(3−γ)), then necessarily δ:=4s02γ(3−γ)<nM. By the monotonicity of U(⋅,t) we would obtain that U(s,t)<nM−δωn for all s∈(0,s02). Since U(s,t)<nMωn for all s∈(0,Rn), we have
ϕ(t)<nM−δωn⋅∫s020s−γ(s0−s)ds+nMωn⋅∫s0s02s−γ(s0−s)ds=nMωn⋅∫s00s−γ(s0−s)ds−δωn⋅∫s020s−γ(s0−s)ds |
=nMωn⋅s2−γ0(1−γ)(2−γ)−δωn⋅2γ(3−γ)s2−γ04(1−γ)(2−γ). |
In view of the definition of Sϕ, we find that nM−s0<nM−2γ(3−γ)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 s0≤Rn6. Then the function μ(t) has property that
μ(t)≤C1+12s∫s0f(Us(σ,t))dσ for all s∈(0,s0) and any t∈Sϕ, | (3.4) |
where C1=χ2C0+C0+C23+C3=13(χ2C0+C0+χk1(γ1−γ2)2γ2(2ξk2γ2χk1γ1)γ1γ1−γ2)+χf1(2δωns0).
Proof. First for any fixed t∈Sϕ, we may invoke Lemma 3.1 to see that
U(s02,t)≥nM−δωn, |
and thus, as U≤nMωn,
U(s0,t)−U(s02,t)s02≤nMωn−nM−δωns02=2δωns0. |
However, by concavity of U(⋅,t), as asserted by Lemma 2.4,
U(s0,t)−U(s02,t)s02≥Us(s0,t)≥Us(s,t) for all s∈(s0,Rn). |
Then let s0∈(0,Rn), we know that
μ(t)=1Rn∫s00f(Us(σ,t))dσ+1Rn∫Rns0f(Us(σ,t))dσ=1Rn∫s0f(Us(σ,t))dσ+1Rn∫s0sf(Us(σ,t))dσ+1Rn∫Rns0f(Us(σ,t))dσ,∀t∈(0,Tmax). | (3.5) |
Since γ1>γ2 and Young's inequality such that ξf2(u)≤ξk2(1+u)γ2≤χk12(1+u)γ1+C2≤χ2f1(u)+C2 with C2=χk1(γ1−γ2)2γ2(2ξk2γ2χk1γ1)γ1γ1−γ2 for u≥0, then for all s∈(0,Rn) and t∈(0,Tmax) we show that
χ2f1(Us(s,t))−C2≤f(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))dσ≤∫s0sχf1(Us(σ,t))dσ≤∫s0sχf1(Us(s,t))dσ≤s0χf1(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)dσ≥∫s0χ{Us(⋅,t)≥1}(σ)⋅(χ2f1(Us(σ,t))−C2)dσ−C0s≥∫s0χ{Us(⋅,t)≥1}(σ)⋅χ2f1(Us(σ,t))dσ−(C0+C2)s=∫s0χ2f1(Us(σ,t))dσ−∫s0χ{Us(⋅,t)<1}(σ)⋅χ2f1(Us(σ,t))dσ−(C0+C2)s≥∫s0χ2f1(Us(s,t))dσ−χ2C0s−(C0+C2)s≥sχ2f1(Us(s,t))−(χ2C0+C0+C2)s. |
Therefore, we obtain
∫s0sf(Us(σ,t))dσ≤2s0s∫s0f(Us(σ,t))dσ+2(χ2C0+C0+C2)s0. |
Since (3.5) we have for all s∈(0,s0)
1Rn∫s0f(Us(σ,t))dσ+1Rn∫s0sf(Us(σ,t))dσ≤1Rn∫s0f(Us(σ,t))dσ+2s0Rns∫s0f(Us(σ,t))dσ+2(χ2C0+C0+C2)s0Rn, | (3.7) |
where s0≤Rn6 such that 1Rn≤16s0≤16s, 2s0Rns≤13s and s0Rn≤16 for all s∈(0,s0). Finally, we estimate the last summand of (3.5)
1Rn∫Rns0f(Us(σ,t))dσ≤1Rn∫Rns0χf1(Us(σ,t))dσ≤χf1(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<2−2n, |
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)≥n2∫s00s2−2n−γ(s0−s)UssD(Us)ds+12∫s00s−γ(s0−s)Us⋅{∫s0f(Us(σ,t))dσ}ds−C1∫s00s1−γ(s0−s)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)=n2s2−2nD(Us)Uss−sμ(t)Us+Us⋅∫s0f(Us(σ,t))dσ≥n2s2−2nUssD(Us)+12Us⋅∫s0f(Us(σ,t))dσ−C1sUs. |
Notice ϕ(t) conforms ϕ(t)=∫s00s−γ(s0−s)U(s,t)ds. So (3.9) is a direct consequence.
Lemma 3.4. Let s0∈(0,Rn6], and γ∈(−∞,1) satisfying γ<2−2n. Then J1(t) in (3.9) satisfies
J1(t)≥−I, | (3.10) |
where
I:={−n2dm(2−2n−γ)∫s00s1−2n−γ(s0−s),m<0,n2d(2−2n−γ)∫s00s1−2n−γ(s0−s)ln(Us+1),m=0,n2dm(2−2n−γ)∫s00s1−2n−γ(s0−s)(Us+1)m,m>0, | (3.11) |
for all t∈Sϕ.
Proof. Since D∈C2([0,∞)), suppose that
G(τ)=∫τ0D(δ)dδ, |
then
0<G(τ)≤d∫τ0(1+δ)m−1dδ≤{−dm,m<0,dln(τ+1),m=0,dm(τ+1)m,m>0. |
Here integrating by parts we obtain
J1(t)=n2∫s00s2−2n−γ(s0−s)dG(Us)=n2s2−2n−γ(s0−s)G(Us)|s00+n2∫s00s2−2n−γG(Us)ds−n2(2−2n−γ)∫s00s1−2n−γ(s0−s)G(Us)ds. |
Hence a direct calculation yields
J1(t)≥{n2dm(2−2n−γ)∫s00s1−2n−γ(s0−s),m<0,−n2d(2−2n−γ)∫s00s1−2n−γ(s0−s)ln(Us+1),m=0,−n2dm(2−2n−γ)∫s00s1−2n−γ(s0−s)(Us+1)m,m>0, |
for all t∈Sϕ. We conclude (3.10).
Lemma 3.5. Assume that γ∈(−∞,1) satisfying γ<2−2n and s0∈(0,Rn6]. Then we have
J2(t)+J3(t)≥k1χ4∫s00s1−γ(s0−s)U1+γ1sds−C4∫s00s1−γ(s0−s)Usds | (3.12) |
for all t∈Sϕ, where C4=C1+(χ2C0+C0+C2)2.
Proof. Since Lemma 3.2 we have
∫s0f(Us(σ,t))dσ≥s2χf1(Us(s,t))−(χ2C0+C0+C2)sfor all s∈(0,s0) and t∈(0,Tmax). |
Therefore,
J2(t)=12∫s00s−γ(s0−s)Us⋅{∫s0f(Us(σ,t))dσ}ds≥χ4∫s00s1−γ(s0−s)Usf1(Us(s,t))ds−(χ2C0+C0+C2)2∫s00s1−γ(s0−s)Usds≥k1χ4∫s00s1−γ(s0−s)U1+γ1sds−(χ2C0+C0+C2)2∫s00s1−γ(s0−s)Usds, |
where f1(Us(s,t))≥k1(1+Us)γ1≥k1(Us)γ1. Combining these inequalities we can deduce (3.12).
Lemma 3.6. Let γ1>max{0,m−1}. For any γ∈(−∞,1) satisfying
γ∈min{2−2n⋅1+γ1γ1, 2−2n⋅1+γ11+γ1−m}, | (3.13) |
and s0∈(0,Rn6], the function ϕ:[0,Tmax)→R defined in (3.1) satisfies
ϕ′(t)≥{Cψ(t)−Cs3−γ−2n⋅1+γ1γ10,m≤1,Cψ(t)−Cs3−γ−2n⋅1+γ11+γ1−m0,m>1, | (3.14) |
with C>0 for all t∈Sϕ, where ψ(t):=∫s00s1−γ(s0−s)U1+γ1sds.
Proof. From (3.10) and (3.12) we have
ϕ′(t)≥k1χ4ψ(t)−I−C4∫s00s1−γ(s0−s)Usds, | (3.15) |
for all t∈Sϕ and I is given by (3.11). In the case m<0,
−n2dm(2−2n−γ)∫s00s1−2n−γ(s0−s)ds≤−n2dm(2−2n−γ)s0∫s00s1−2n−γds=−n2dms3−2n−γ0. |
If m=0, we use the fact that ln(1+x)x<1 for any x>0 and Hölder's inequality to estimate
n2d(2−2n−γ)∫s00s1−2n−γ(s0−s)ln(Us+1)ds=n2d(2−2n−γ)∫s00[s1−γ(s0−s)U1+γ1s]11+γ1⋅s1−2n−γ−1−γ1+γ1(s0−s)1−11+γ1ln(1+Us)Usds≤n2d(2−2n−γ){∫s00s1−γ(s0−s)U1+γ1sds}11+γ1⋅{∫s00(s1−2n−γ−1−γ1+γ1(s0−s)γ11+γ1)1+γ1γ1ds}γ11+γ1≤n2d(2−2n−γ)sγ11+γ10{∫s00s(1−2n−γ)γ1−2nγ1ds}γ11+γ1ψ11+γ1(t)=C5ψ11+γ1(t)s(3−2n−γ)γ1−2n1+γ10, |
for all t∈Sϕ with C5:=n2d(2−2n−γ)⋅(12−γ−2n⋅1+γ1γ1)γ11+γ1>0 by (3.13). In the case m>0, by using the elementary inequality (a+b)α≤2α(aα+bα) for all a,b>0 and every α>0, we obtain
n2dm(2−2n−γ)∫s00s1−2n−γ(s0−s)(Us+1)mds≤2mn2dm(2−2n−γ)∫s00s1−2n−γ(s0−s)Umsds+2mn2dm(2−2n−γ)∫s00s1−2n−γ(s0−s)ds, | (3.16) |
for all t∈Sϕ, and we first estimate the second term on the right of (3.16)
2mn2dm(2−2n−γ)∫s00s1−2n−γ(s0−s)ds≤2mn2dms3−2n−γ0. |
Since γ1>m−1 and by Hölder's inequality we deduce that
2mn2dm(2−2n−γ)∫s00s1−2n−γ(s0−s)Umsds=2mn2dm(2−2n−γ)∫s00s(1−γ)⋅m1+γ1(s0−s)m1+γ1Ums⋅s1−2n−γ−(1−γ)⋅m1+γ1(s0−s)1−m1+γ1ds≤2mn2dm(2−2n−γ){∫s00[s(1−γ)⋅m1+γ1(s0−s)m1+γ1Ums]1+γ1mds}m1+γ1×{∫s00[s1−2n−γ−(1−γ)⋅m1+γ1(s0−s)1−m1+γ1]1+γ11+γ1−mds}1+γ1−m1+γ1≤2mn2dm(2−2n−γ)ψm1+γ1(t)s1+γ1−m1+γ10⋅{∫s00s(1+γ1−m)(1−γ)−2n(1+γ1)1+γ1−mds}1+γ1−m1+γ1≤C6ψm1+γ1(t)s(1+γ1−m)(3−γ)−2n(1+γ1)1+γ10, |
for all t∈Sϕ with C6=2mn2dm(2−2n−γ)(12−γ−2n⋅1+γ11+γ1−m)1+γ1−m1+γ1>0 where γ<2−2n⋅1+γ11+γ1−m from (3.13).
Next, we can estimate the third expression on the right of (3.15) as follows
C4∫s00s1−γ(s0−s)Usds=C4∫s00[s1−γ(s0−s)U1+γ1s]11+γ1⋅s1−γ−1−γ1+γ1(s0−s)1−11+γ1ds≤C4{∫s00s1−γ(s0−s)U1+γ1sds}11+γ1⋅{∫s00[s1−γ−1−γ1+γ1(s0−s)1−11+γ1]1+γ1γ1ds}γ11+γ1≤C4ψ11+γ1(t)sγ11+γ10{∫s00s1−γds}γ11+γ1=C7ψ11+γ1(t)s(3−γ)γ11+γ10, |
where C7=C4(12−γ)γ11+γ1 for all t∈Sϕ. By (3.15) and collecting the estimates above we have
ϕ′(t)≥{k1χ4ψ(t)+n2dms3−2n−γ0−C7ψ11+γ1(t)s(3−γ)γ11+γ10,m<0, t∈Sϕ,k1χ4ψ(t)−C5ψ11+γ1(t)s(3−2n−γ)γ1−2n1+γ10−C7ψ11+γ1(t)s(3−γ)γ11+γ10,m=0, t∈Sϕ,k1χ4ψ(t)−2mn2dms3−2n−γ0−C6ψm1+γ1(t)s(1+γ1−m)(3−γ)−2n(1+γ1)1+γ10−C7ψ11+γ1(t)s(3−γ)γ11+γ10,m>0, t∈Sϕ. |
If m=0, by Young's inequality we can find positive constants C8,C9 such that
C5ψ11+γ1(t)s(3−2n−γ)γ1−2n1+γ10≤k1χ16ψ(t)+C8s3−γ−2n⋅1+γ1γ10,∀t∈Sϕ, |
while as m>0 we have
C6ψm1+γ1(t)s(1+γ1−m)(3−γ)−2n(1+γ1)1+γ10≤k1χ16ψ(t)+C9s3−γ−2n⋅1+γ11+γ1−m0,∀t∈Sϕ. |
On the other hand, we use Young's inequality again
C7ψ11+γ1(t)s(3−γ)γ11+γ10≤k1χ16ψ(t)+C10s3−γ0,∀t∈Sϕ. |
In the case m<0, because of s0∈(0,Rn6], we have
s3−2n−γ0=s3−γ−2n⋅1+γ1γ10⋅s2nγ10≤(Rn6)2nγ1s3−γ−2n⋅1+γ1γ10, |
when m>0 we have
s3−2n−γ0=s3−γ−2n⋅1+γ11+γ1−m0⋅s2mn(1+γ1−m)0≤(Rn6)2mn(1+γ1−m)s3−γ−2n⋅1+γ11+γ1−m0. |
All in all, we have
ϕ′(t)≥{k1χ8ψ(t)−C11s3−γ−2n⋅1+γ1γ10−C10s3−γ0,m≤0,k1χ8ψ(t)−C12s3−γ−2n⋅1+γ11+γ1−m0−C10s3−γ0,m>0, | (3.17) |
for all t∈Sϕ with C11=C8−n2dm(Rn6)2nγ1 and C12=C9+2mn2dm(Rn6)2mn(1+γ1−m). When 0<m≤1, we have 1+γ11+γ1−m≤1+γ1γ1 such that s3−γ−2n⋅1+γ11+γ1−m0=s3−γ−2n⋅1+γ1γ10s2n(1+γ1γ1−1+γ11+γ1−m)0≤(Rn6)2(1−m)(1+γ1)nγ1(1+γ1−m)s3−γ−2n⋅1+γ1γ10. In the case m≤1
s3−γ0=s2n⋅1+γ1γ10⋅s3−γ−2n⋅1+γ1γ10≤(Rn6)2n⋅1+γ1γ1⋅s3−γ−2n⋅1+γ1γ10, |
and if m>1 we have
s3−γ0=s2n⋅1+γ11+γ1−m0⋅s3−γ−2n⋅1+γ11+γ1−m0≤(Rn6)2n⋅1+γ1(1+γ1−m)⋅s3−γ−2n⋅1+γ11+γ1−m0. |
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)≥s3−γ0}. | (3.18) |
Lemma 3.7. Let γ∈(−∞,1) satisfying γ>1−γ1 and (3.13). Then for any choice of s0∈(0,Rn6], the following inequality
ϕ′(t)≥{Cs−γ1(3−γ)0ϕ1+γ1(t)−Cs3−γ−2n⋅1+γ1γ10,m≤1,Cs−γ1(3−γ)0ϕ1+γ1(t)−Cs3−γ−2n⋅1+γ11+γ1−m0,m>1, | (3.19) |
holds for all t∈Sϕ∩Sψ 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)dσ≤s+∫s0χ{Us(⋅,t)≥1}(σ)⋅{σ1−γ(s0−σ)U1+γ1s}11+γ1⋅σ−1−γ1+γ1(s0−σ)−11+γ1dσ≤s+(s0−s)−11+γ1ψ11+γ1(t)⋅{∫s0σ−1−γ1+γ1⋅1+γ1γ1dσ}γ11+γ1=s+(γ1γ+γ1−1)γ11+γ1(s0−s)−11+γ1sγ+γ1−11+γ1ψ11+γ1(t), | (3.20) |
for all s∈(0,s0) and t∈(0,Tmax) where γ1γ+γ1−1>0. According to the definition of Sψ, we can find
s(s0−s)−11+γ1sγ+γ1−11+γ1ψ11+γ1(t)=s2−γ1+γ1(s0−s)11+γ1ψ−11+γ1(t)≤s2−γ1+γ10⋅s11+γ10⋅(s3−γ0)−11+γ1=1, | (3.21) |
for all s∈(0,s0) and t∈Sψ. Combining (3.20) and (3.21) we have
U(s,t)≤C1sγ+γ1−11+γ1(s0−s)−11+γ1ψ11+γ1(t), |
where C1=1+(γ1γ+γ1−1)γ11+γ1 for all s∈(0,s0) and t∈Sψ. Invoking Hölder's inequality, we get
ϕ(t)=∫s00s−γ(s0−s)U(s,t)ds≤C1∫s00s−γ+γ+γ1−11+γ1(s0−s)1−11+γ1ds⋅ψ11+γ1(t)≤C1sγ11+γ10∫s00s−γ+γ+γ1−11+γ1ds⋅ψ11+γ1(t)=C2sγ1(3−γ)1+γ10⋅ψ11+γ1(t), | (3.22) |
where C2=C11+γ1γ1(2−γ) for all s∈(0,s0) and t∈Sψ. 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)>nM−s0ωn(1−γ)(2−γ)⋅s2−γ0 for all t∈[0,T]}. | (3.23) |
Let us choose s0>0 such that
s0≤min{Rn6,nM2,nMγ12(1−γ)ωn[(C3+1)(1+γ1)−1]}, | (3.24) |
where M and ωn were defined in (2.7) and C3=(γ1γ+γ1−1)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)=nωn∫Br⋆(0)u0dx≥nωn(M−ε),∀s∈(s⋆,Rn). |
Therefore it is possible to estimate
ϕ(0)=∫s00s−γ(s0−s)U(s,0)ds≥∫s00s−γ(s0−s)U(s⋆,0)ds>nωn(M−s0n)∫s00s−γ(s0−s)ds=nM−s0ωn(1−γ)(2−γ)⋅s2−γ0. | (3.25) |
Then S is non-empty and denote T=supS∈(0,∞]. Next, we need to prove (0,T)⊂Sϕ∩Sψ≠∅. Note that
ϕ(t)>nM−s0ωn(1−γ)(2−γ)⋅s2−γ0,∀t∈(0,T), | (3.26) |
we obtain (0,T)⊂Sϕ. From (3.20) we have
ϕ(t)≤∫s00s−γ(s0−s)[s+C3sγ+γ1−11+γ1(s0−s)−11+γ1ψ11+γ1(t)]ds≤s0∫s00s1−γds+C3∫s00s−γ+γ+γ1−11+γ1(s0−s)γ11+γ1ψ11+γ1(t)ds=s3−γ02−γ+C3(1+γ1)γ1(2−γ)sγ1(3−γ)1+γ10⋅ψ11+γ1(t). |
It follows from (3.24) and (3.26) that
ϕ(t)≥nM2(1−γ)(2−γ)ωn⋅s2−γ0for all t∈(0,T). | (3.27) |
Then
C3(1+γ1)γ1(2−γ)sγ1(3−γ)1+γ10⋅ψ11+γ1(t)≥nM2(1−γ)(2−γ)ωn⋅s2−γ0−s3−γ02−γ. |
Note that (3.24) implies
nMγ12C3(1−γ)ωn(1+γ1)s0−γ1C3(1+γ1)≥1, |
then we have
ψ(t)≥[(nMs2−γ02(1−γ)(2−γ)ωn−s3−γ02−γ)⋅γ1(2−γ)C3(1+γ1)s−γ1(3−γ)1+γ10]1+γ1≥[nMγ12C3(1−γ)ωn(1+γ1)s0−γ1C3(1+γ1)]1+γ1⋅s3−γ0≥s3−γ0. |
Therefore, (0,T)⊂Sϕ∩Sψ≠∅.
Step 2. Applying Lemma 3.7 we can find γ∈(−∞,1) and C1,C2>0 such that for all s0∈(0,Rn6]
ϕ′(t)≥{C1s−γ1(3−γ)0ϕ1+γ1(t)−C2s3−γ−2n⋅1+γ1γ10,m≤1,C1s−γ1(3−γ)0ϕ1+γ1(t)−C2s3−γ−2n⋅1+γ11+γ1−m0,m>1, |
for all t∈Sϕ∩Sψ and with (3.22) we have
ψ(t)≥C3s−γ1(3−γ)0ϕ1+γ1(t),∀t∈Sψ. |
To specify our choice of s0, for given M>0 we choose s0∈(0,Rn6] small enough such that
s0≤nM2, | (3.28) |
and also
sγ10<TC1γ14(nM2ωn(1−γ)(2−γ))γ1, | (3.29) |
as well as
s1+γ10≤C3(nM2ωn(1−γ)(2−γ))1+γ1. | (3.30) |
From (3.23), (3.28) and (3.30) we have
ψ(t)≥C3s−γ1(3−γ)0ϕ1+γ1(t)>C3(nM−s0ωn(1−γ)(2−γ)⋅1s0)1+γ1⋅s3−γ0≥s3−γ0,∀t∈Sψ, |
which shows that S⊂Sϕ∩Sψ. Since 1+γ1−m>2n, we have (1+γ1)(1−2n(1+γ1−m))>0 if m>1 so that we can choose s0 sufficiently small satisfying (3.28)–(3.30) such that
s(1+γ1)(1−2n(1+γ1−m))0≤C12C2(nM2ωn(1−γ)(2−γ))1+γ1, |
while in the case m≤1, the condition γ1>m−1+2n≥2n which infers that (1+γ1)(1−2nγ1)>0 and we select s0 small enough fulfilling (3.28)–(3.30) such that
s(1+γ1)(1−2nγ1)0≤C12C2(nM2ωn(1−γ)(2−γ))1+γ1. |
It is possible to obtain
C12s−γ1(3−γ)0ϕ1+γ1(0)C2s3−γ−2n⋅1+γ11+γ1−m0≥C12C2(nM2ωn(1−γ)(2−γ))1+γ1⋅s−(1+γ1)+2n⋅1+γ1(1+γ1−m)0≥1,∀m>1, |
and we have
C12s−γ1(3−γ)0ϕ1+γ1(0)C2s3−γ−2n⋅1+γ1γ10≥C12C2(nM2ωn(1−γ)(2−γ))1+γ1⋅s−(1+γ1)+2n⋅1+γ1γ10≥1,∀m≤1. |
All in all, for any m∈R, we apply an ODI comparison argument to obtain that
ϕ′(t)≥C12s−γ1(3−γ)0ϕ1+γ1(t),∀t∈(0,T). |
By a direct calculation we obtain
−1γ1(1ϕγ1(t)−1ϕγ1(0))≥C12s−γ1(3−γ)0t,∀t∈(0,T). |
Hence, according to (3.25) and (3.29) we conclude
t<2C1γ1sγ10(2ωn(1−γ)(2−γ)nM)γ1≤T2, |
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{1,2−m,γ2}, there exists C=C(p)>0 such that
∫Ω(1+u)p(x,t)dx≤Con (0,Tmax). | (4.2) |
Proof. Notice f1(u)≤k1(1+u)γ1, f2(u)=k2(1+u)γ2 for all u≥0. Multiplying the first equation of (1.5) by p(1+u)p−1 and integrating by parts with the boundary conditions for u,v and w, we have
ddt∫Ω(1+u)pdx+p(p−1)∫Ω(1+u)p−2D(u)|∇u|2dx=χp(p−1)∫Ωu(1+u)p−2∇u⋅∇vdx−ξp(p−1)∫Ωu(1+u)p−2∇u⋅∇wdx=−χ(p−1)∫Ω(1+u)pΔvdx+χp∫Ω(1+u)p−1Δvdx+ξ(p−1)∫Ω(1+u)pΔwdx−ξp∫Ω(1+u)p−1Δwdx≤χ(p−1)∫Ω(1+u)pf1(u)dx+χp∫Ω(1+u)p−1μ1(t)dx+ξ(p−1)∫Ω(1+u)pμ2(t)dx−ξ(p−1)∫Ω(1+u)pf2(u)dx+ξp∫Ω(1+u)p−1f2(u)dx≤k1χ(p−1)∫Ω(1+u)p+γ1dx+χp∫Ω(1+u)p−1μ1(t)dx+ξ(p−1)∫Ω(1+u)pμ2(t)dx−k2ξ(p−1)∫Ω(1+u)p+γ2dx+k2ξp∫Ω(1+u)p+γ2−1dx,∀t∈(0,Tmax). | (4.3) |
Firstly,
p(p−1)∫Ω(1+u)p−2D(u)|∇u|2dx≥dp(p−1)∫Ω(1+u)p+m−3|∇u|2dx=4dp(p−1)(p+m−1)2∫Ω|∇(1+u)p+m−12|2dx. |
By Young's inequality and Hölder's inequality, we obtain
χp∫Ω(1+u)p−1μ1(t)dx≤C1∫Ω(1+u)p+γ1dx+C2μp+γ11+γ11(t)=C1∫Ω(1+u)p+γ1dx+C2(1|Ω|∫Ωf1(u)dx)p+γ11+γ1≤C1∫Ω(1+u)p+γ1dx+C3(∫Ω(1+u)1+γ1dx)p+γ11+γ1≤C1∫Ω(1+u)p+γ1dx+C3{(∫Ω(1+u)p+γ1dx)1+γ1p+γ1⋅|Ω|p−1p+γ1}p+γ11+γ1=C1∫Ω(1+u)p+γ1dx+C3|Ω|p−11+γ1∫Ω(1+u)p+γ1dx. |
for all t∈(0,Tmax). Then by Hölder's inequality we obtain
ξ(p−1)∫Ω(1+u)pμ2(t)dx=k2ξ(p−1)|Ω|∫Ω(1+u)γ2dx∫Ω(1+u)pdx≤k2ξ(p−1)|Ω|{∫Ω(1+u)p+γ2dx}γ2p+γ2|Ω|pp+γ2×{∫Ω(1+u)p+γ2dx}pp+γ2|Ω|γ2p+γ2=k2ξ(p−1)∫Ω(1+u)p+γ2dx,∀t∈(0,Tmax). |
Furthermore, by using Young's inequality and (4.1) we have
k2ξp∫Ω(1+u)p+γ2−1dx≤C4∫Ω(1+u)p+γ1dx+C5, |
for all t∈(0,Tmax). Therefore, combining these we conclude
ddt∫Ω(1+u)pdx+4dp(p−1)(p+m−1)2∫Ω|∇(1+u)p+m−12|2dx≤C6∫Ω(1+u)p+γ1dx+C5,∀t∈(0,Tmax), |
where C6=C1+C3|Ω|p−11+γ1+C4+k1χ(p−1). By means of Gagliardo-Nirenberg inequality we can find C7 such that
C6∫Ω(1+u)p+γ1dx=C6‖(1+u)p+m−12‖2(p+γ1)p+m−1L2(p+γ1)p+m−1(Ω)≤C7‖∇(1+u)p+m−12‖2(p+γ1)p+m−1⋅aL2(Ω)⋅‖(1+u)p+m−12‖2(p+γ1)p+m−1⋅(1−a)L2p+m−1(Ω)+C7‖(1+u)p+m−12‖2(p+γ1)p+m−1L2p+m−1(Ω) |
for all t∈(0,Tmax), where
a=p+m−12−p+m−12(p+γ1)p+m−12−(12−1n)∈(0,1). |
Since 1−m+γ1<2n, we have 2(p+γ1)p+m−1⋅a<2, and we use Young's inequality to see that for all t∈(0,Tmax)
C6∫Ω(1+u)p+γ1dx≤2dp(p−1)(p+m−1)2‖∇(1+u)p+m−12‖2L2(Ω)+C8. |
In quite a similar manner, we obtain C9=C9(p)>0 fulfilling
∫Ω(1+u)pdx≤2dp(p−1)(p+m−1)2‖∇(1+u)p+m−12‖2L2(Ω)+C9for all t∈(0,Tmax). |
Finally, combining these to (4.3) we obtain
ddt∫Ω(1+u)pdx+∫Ω(1+u)pdx≤C5+C8+C9for all t∈(0,Tmax). |
Thus,
∫Ω(1+u)pdx≤max{∫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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