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Global existence of solutions to the chemotaxis system with logistic source under nonlinear Neumann boundary conditions

Minh Le
Abstract

We consider classical solutions to the chemotaxis system with logistic source f(u):=auμu2f(u):=au-\mu u^{2} under nonlinear Neumann boundary conditions uν=|u|p\frac{\partial u}{\partial\nu}=|u|^{p} with p>1p>1 in a smooth convex bounded domain Ωn\Omega\subset\mathbb{R}^{n} where n2n\geq 2. This paper aims to show that if p<32p<\frac{3}{2}, and μ>0\mu>0, n=2n=2, or μ\mu is sufficiently large when n3n\geq 3, then the parabolic-elliptic chemotaxis system admits a unique positive global-in-time classical solution that is bounded in Ω×(0,)\Omega\times(0,\infty). The similar result is also true if p<32p<\frac{3}{2}, n=2n=2, and μ>0\mu>0 or p<75p<\frac{7}{5}, n=3n=3, and μ\mu is sufficiently large for the parabolic-parabolic chemotaxis system.

1 Introduction

We are concerned in this paper with solutions to the chemotaxis model as follows:

{ut=Δuχ(uv)+auμu2xΩ,t(0,Tmax),τvt=Δv+αuβvxΩ,t(0,Tmax),\begin{cases}u_{t}=\Delta u-\chi\nabla\cdot(u\nabla v)+au-\mu u^{2}\qquad&x\in{\Omega},\,t\in(0,T_{\rm max}),\\ \tau v_{t}=\Delta v+\alpha u-\beta v\qquad&x\in{\Omega},\,t\in(0,T_{\rm max}),\end{cases} (KS)

in a smooth, convex, bounded domain Ωn\Omega\subset\mathbb{R}^{n} where α,β,a,μ>0\alpha,\beta,a,\mu>0, τ0\tau\geq 0, and χ\chi\in\mathbb{R}. The system (KS) is complemented with the nonnegative initial conditions in C2+γ(Ω)C^{2+\gamma}(\Omega), where γ(0,1)\gamma\in(0,1), not identically zero:

u(x,0)=u0(x),v(x,0)=v0(x),xΩ,u(x,0)=u_{0}(x),\qquad v(x,0)=v_{0}(x),\qquad x\in\Omega, (1.1)

and the nonlinear Neumann boundary conditions

uν=|u|p,vν=0,xΩ,t(0,Tmax).\frac{\partial u}{\partial\nu}=|u|^{p},\qquad\frac{\partial v}{\partial\nu}=0,\qquad x\in\partial\Omega,\,t\in(0,T_{\rm max}). (1.2)

where p>1p>1 and ν\nu is the outward normal vector. The PDE system in (KS) is used in mathematical biology to model the mechanism of chemotaxis, that is, the movement of an organism in response to a chemical stimulus. Here u(x,t),v(x,t)u(x,t),v(x,t) represent the cell density and the concentration of chemical substances at position xx at the time tt, respectively ([16]). Historically, the system (KS) without the logistic source under homogeneous Neumann boundary condition well-known as the most simplified Keller-Segel system has been intensively studied in many various directions such as global boundedness solutions, blow-up solutions, steady states, and blow-up sets. A significant milestone in the derivation of the model, along with many interesting results, can be found in [12]. Additionally, [3] provides a comprehensive summary of the developed techniques and results accumulated over decades to address chemotaxis systems. For a more detailed exploration of the derivation of the system (KS), consider referring to [11], which offers a broad survey of variations of chemotaxis models and their corresponding biological backgrounds. Furthermore, in two spatial dimension, the simplest form has been the focus of numerous research papers due to its intriguing mathematical property, the critical mass phenomena. This property states that if the initial mass is less than a certain number, solutions exist globally and remain bounded in time, while if the mass is larger than that value, solutions blow up in finite time. There are many research papers focusing on finding the precise value of critical mass. In two spatial dimensional domain, global existence and boundedness of solutions are studied in [7], and [5] for the sub-critical mass, and finite time blow-up solutions for the super-critical mass is investigated in [25, 26, 31, 32]. However, in higher spatial dimensional domain, this critical mass property is no longer true. In fact, it is proved in [37] that finite time blow-up solutions can be constructed in a smooth bounded domain regardless of how small the mass is.
We now shift our attention to the homogeneous Neumann boundary condition. In addition to the biological chemotaxis phenomena, the logistic term, auμu2au-\mu u^{2}, introduced in the evolution equation for uu plays a role in describing the growth of the population. Specifically, the term auau, with aa\in\mathbb{R}, is the growth rate of population and the term μu2-\mu u^{2} models additional overcrowding effects. It was investigated in [36] that the quadratic degradation term μu2-\mu u^{2} can prevent blow-up solutions. In fact, it was proven that if μ>n2nχα\mu>\frac{n-2}{n}\chi\alpha, and τ=0\tau=0, then the solutions exist globally and remain bounded at all time in a convex bounded domain with smooth boundary Ωn\Omega\subset\mathbb{R}^{n}, where n2n\geq 2. This result was later improved in [13, 15, 42] that μ=n2nχα\mu=\frac{n-2}{n}\chi\alpha can prevent blow-up when τ=0\tau=0. In a two-dimensional space with τ=1\tau=1, the system (KS) possesses a unique classical solution which is nonnegative and bounded in Ω×(0,)\Omega\times(0,\infty) ( see e.g. [27, 28]). These results were later improved in [43, 44] by replacing the logistic sources by sub-logistic ones such as auμu2lnp(u+e)au-\mu\frac{u^{2}}{\ln^{p}(u+e)} for p(0,1)p\in(0,1). In a higher-dimensional space with τ=1\tau=1, the similar results can be found in [39] for the parabolic-parabolic model with an additional largeness assumption of μ\mu. In addition to the global classical solutions, the existence global weak solutions results for any arbitrary μ>0\mu>0 were also obtained in [36] for the parabolic-elliptic models and in [18] for the parabolic-parabolic models in a three-dimensional system. Furthermore, interested readers are referred to [14, 17, 19, 20, 23, 35, 38, 45, 46] to study more about qualitative and quantitative works of chemotaxis systems with logistic sources.
The problem becomes more interesting and challenging if the homogeneous Neumann boundary condition is replaced by the nonlinear Neumann boundary condition. The method in this paper to obtain global boundedness results is first to establish a L1L^{1} estimate, then for Lp0L^{p_{0}} for some p0>1p_{0}>1, and finally apply a Moser-type iteration to obtain for LL^{\infty}. Although this approach has been widely applied in treating global boundedness problems for reaction-diffusion equations ([1, 2, 8]), or for chemotaxis systems ([39]), the main difficulties rely heavily on tedious integral estimations. Unlike the homogeneous Neumann boundary conditions, it is not even straightforward to see whether the total mass of the cell density function is globally bounded or not due to the nonlinear boundary term. In fact, most of the technical challenges in this paper are to deal with the nonlinear boundary term. Fortunately, the Sobolev’s trace inequality enables us to solve a part of a problem:
Main Question: ”What is the largest value pp so that logistic damping still avoids blow-up?”
This types of question for nonlinear parabolic equations has been intensively studied in 1990s. To be more precise, if we consider χ=0\chi=0, our problem is similar to the following PDE:

{Ut=ΔUμUQxΩ,t(0,Tmax),Uν=UPxΩ,t(0,Tmax),U(x,0)=U0(x)xΩ¯.\begin{cases}U_{t}=\Delta U-\mu U^{Q}\qquad&x\in{\Omega},\,t\in(0,T_{\rm max}),\\ \frac{\partial U}{\partial\nu}=U^{P}\qquad&x\in{\Omega},\,t\in(0,T_{\rm max}),\\ U(x,0)=U_{0}(x)&x\in{\bar{\Omega}}.\end{cases} (NBC)

where Ω\Omega is a smooth bounded domain in n\mathbb{R}^{n}, Q,P>1Q,P>1, μ>0\mu>0 and U0W1,(Ω)U_{0}\in W^{1,\infty}(\Omega) is a nonnegative function. The study concerning the global existence was first investigated in [6], and then improved in [29] for n2n\geq 2. Particularly, it was shown that P=Q+12P=\frac{Q+1}{2} is critical for the blow up in the following sense:

  1. 1.

    if P<Q+12P<\frac{Q+1}{2} then all solutions of (NBC) exist globally and are globally bounded,

  2. 2.

    If P>Q+12P>\frac{Q+1}{2} (or P=Q+12P=\frac{Q+1}{2} and μ\mu is sufficiently small ) then there exist initial functions U0U_{0} such that the corresponding solutions of (NBC) blow-up in LL^{\infty}-norm.

In comparison to our problem, we have Q=2Q=2 and P=32P=\frac{3}{2} is the critical power. Indeed, we also obtain the similar critical power p=32p=\frac{3}{2} as in Theorem 1.1. Notice that the local existence of positive solution was not mentioned in [6, 29, 30], and it is not clear for us to define UPU^{P} without knowing UU is nonnegative, so the presence of absolute sign in (1.2) is necessary to obtain local positive solutions from nonnegative, not identically zero initial data.
Heuristically, the analysis diagram can be presented as follows. In case τ=0\tau=0, by substituting Δv=αuβv-\Delta v=\alpha u-\beta v into the first equation of (KS), we obtain

ut=Δu+auχuv+(χμ)u2χuv.u_{t}=\Delta u+au-\chi\nabla u\cdot\nabla v+(\chi-\mu)u^{2}-\chi uv.

If μ\mu is sufficiently large, then solutions might be bounded globally since the nonlinear term (χμ)u2(\chi-\mu)u^{2} might dominate other terms including the nonlinear boundary term. In case τ=1\tau=1, we cannot substitute Δv=βvαu\Delta v=\beta v-\alpha u directly into the first equation of (KS); however, we still have some certain controls of vv by uu from the second equation of (KS) thanks to Sobolev inequality. We expect that this intuition should be true in lower spacial dimension and ”weaker” nonlinear boundary terms since the critical Sobolev exponent decreases if the spacial dimension increases. Indeed, our analysis does not work for n4n\geq 4 since we do not have enough rooms to control other positive nonlinear terms by using the term (χμ)u2(\chi-\mu)u^{2}. One can also find similar ideas on sub-logistic source preventing 2D blow-up in [43].
We summarize the main results to answer a part of the main question. Let us begin with the following theorem for the parabolic-elliptic case.

Theorem 1.1.

Let Ω\Omega be a bounded, convex domain with smooth boundary in n\mathbb{R}^{n} where n2n\geq 2, and τ=0\tau=0. If μ>n2nχα\mu>\frac{n-2}{n}\chi\alpha, and 1<p<321<p<\frac{3}{2} or μ=n2nχα\mu=\frac{n-2}{n}\chi\alpha with n3n\geq 3 and 1<p<1+1n1<p<1+\frac{1}{n} then the system (KS) with initial conditions (1.1) and boundary condition (1.2) possesses a unique positive classical solution which remains bounded in Ω×(0,)\Omega\times(0,\infty).

Remark 1.1.

It is an open question whether there exists a classical finite time blow-up solution if p32p\geq\frac{3}{2}.

Remark 1.2.

The proof of borderline boundedness in Theorem 1.1 when μ=n2nχα\mu=\frac{n-2}{n}\chi\alpha is adopted and modified from the arguments in [13, 15, 42]. However, applying Lemma 3.6 to overcome challenges in boundary integral estimations was not possible. Instead, we had to derive an alternative and improved estimation to handle the boundary term, which necessitated the condition of p<1+1np<1+\frac{1}{n}.

The next theorem is for the parabolic-parabolic system in a two-dimensional space:

Theorem 1.2.

Let Ω\Omega be a bounded, convex domain with smooth boundary, and τ=1\tau=1, n=2n=2, 1<p<321<p<\frac{3}{2}, then the system (KS) with initial conditions (1.1) and boundary condition (1.2) possesses a unique positive classical solution which remains bounded in Ω×(0,)\Omega\times(0,\infty).

In three-dimensional space, we prove the following theorem for the parabolic-parabolic case.

Theorem 1.3.

Let Ω\Omega be a bounded, convex domain with smooth boundary, and τ=1\tau=1, n=3n=3, 1<p<751<p<\frac{7}{5}, then there exists μ0>0\mu_{0}>0 such that for every μ>μ0\mu>\mu_{0}, the system (KS) with initial conditions (1.1) and boundary condition (1.2) possesses a unique positive classical solution which remains bounded in Ω×(0,)\Omega\times(0,\infty).

Remark 1.3.

Here we expect p=75p=\frac{7}{5} may not be the threshold of global boundedness and blow-up solutions, but rather the limitation of our analysis tools.

Remark 1.4.

We leave the open question whether for every n4n\geq 4, there exists pn>1p_{n}>1 such that if 1<p<pn1<p<p_{n} solutions remain bounded in Ω×(0,)\Omega\times(0,\infty).

Remark 1.5.

In case p=1p=1, one may adopt and modify the proof of Theorem 1.1, 1.2, and 1.3 to obtain similar results.

The paper is organized as follows. The local well-possedness of solutions toward the system (KS) including the short-time existence, positivity and uniqueness are established in Section 2. In Section 3, we recall some basic inequalities and provide some essential estimates on the boundary of the solutions, which will be needed in the sequel sections. Section 4 is devoted to establishing the LlogLL\log L, and LrL^{r} bounds for solutions. In Section 5, a Moser-type iteration scheme is applied to obtain an LL^{\infty} bound for solutions from an LrL^{r} bound with a sufficiently large rr. Finally, the main theorems are proved in Section 6.

2 Local well-posedness

In this section, we prove the short-time existence, uniqueness and positivity of solutions to the system (KS) under certain conditions of initial data. Although the proof just follows a basic fixed point argument, however we cannot find any suitable reference for our system. For the sake of completeness, here we will provide a proof, which is a modification of the proof of Theorem 1.1 in [10]. First, we introduce some notations used throughout this section. We follow some definitions of Holder continuous spaces given in [22]. For kk\in\mathbb{N}, and γ(0,1]\gamma\in(0,1], we define the following norms and seminorms:

[f]k+γ,ΩT\displaystyle[f]_{k+\gamma,\Omega_{T}} :=|β|+2j=ksup(x,t)(y,s)ΩT|DxβDtj(f(x,t)f(y,s))||xy|γ+|ts|γ2,\displaystyle:=\sum_{|\beta|+2j=k}\sup_{(x,t)\neq(y,s)\in\Omega_{T}}\frac{|D^{\beta}_{x}D^{j}_{t}\left(f(x,t)-f(y,s)\right)|}{|x-y|^{\gamma}+|t-s|^{\frac{\gamma}{2}}},
fk+γ,ΩT\displaystyle\left\langle f\right\rangle_{k+\gamma,\Omega_{T}} :=|β|+2j=k1sup(x,t)(x,s)ΩT|DxβDtj(f(x,t)f(x,s))||ts|1+γ2,\displaystyle:=\sum_{|\beta|+2j=k-1}\sup_{(x,t)\neq(x,s)\in\Omega_{T}}\frac{|D^{\beta}_{x}D^{j}_{t}\left(f(x,t)-f(x,s)\right)|}{|t-s|^{\frac{1+\gamma}{2}}},
|f|k+γ,ΩT\displaystyle|f|_{k+\gamma,\Omega_{T}} :=[f]k+γ,ΩT+fk+γ,ΩT,\displaystyle:=[f]_{k+\gamma,\Omega_{T}}+\left\langle f\right\rangle_{k+\gamma,\Omega_{T}},
fCk+γ(ΩT)\displaystyle\left\|f\right\|_{C_{k+\gamma}(\Omega_{T})} :=sup(x,t)ΩT|f(x,t)|+|f|k+γ,ΩT,\displaystyle:=\sup_{(x,t)\in\Omega_{T}}|f(x,t)|+|f|_{k+\gamma,\Omega_{T}},

where ΩT:=Ω×(0,T)\Omega_{T}:=\Omega\times(0,T). We write fCk+γ(ΩT)f\in C^{k+\gamma}(\Omega_{T}) if fCk+γ(ΩT)<\left\|f\right\|_{C_{k+\gamma}(\Omega_{T})}<\infty. Now let us recall a useful result for the linear model. We consider the linear second order elliptic equation of non-divergence form:

Lu:=utaijDiju+biDiu+cu=f in ΩT.Lu:=u_{t}-a^{ij}D_{ij}u+b^{i}D_{i}u+cu=f\text{ in }\Omega_{T}. (2.1)

Assume that there exists Λλ>0\Lambda\geq\lambda>0 such that

λ|ξ|2aij(x,t)ξiξjΛ|ξ|2,(x,t)ΩT,ξn,\lambda|\xi|^{2}\leq a^{ij}(x,t)\xi_{i}\xi_{j}\leq\Lambda|\xi|^{2},\qquad(x,t)\in\Omega_{T},\xi\in\mathbb{R}^{n}, (2.2)

where aij,bi,cCγ(ΩT¯)(0<γ<1)a^{ij},b^{i},c\in C^{\gamma}(\bar{\Omega_{T}})(0<\gamma<1) and

1λ{i,jaijCγ(Ω¯T)+ibiCγ(Ω¯T)+cCγ(Ω¯T)}Λγ.\frac{1}{\lambda}\left\{\sum_{i,j}\left\|a^{ij}\right\|_{C^{\gamma}(\bar{\Omega}_{T})}+\sum_{i}\left\|b^{i}\right\|_{C^{\gamma}(\bar{\Omega}_{T})}+\left\|c\right\|_{C^{\gamma}(\bar{\Omega}_{T})}\right\}\leq\Lambda_{\gamma}. (2.3)
Theorem 2.1 ([22], p. 79, Theorem 4.31).

Let the assumptions (2.2), (2.3) be in force, and ΩC2+γ(0<γ<1).\partial\Omega\in C^{2+\gamma}(0<\gamma<1). Let fCγ(ΩT¯)f\in C^{\gamma}(\bar{\Omega_{T}}), gC1+γ(Ω¯T)g\in C^{1+\gamma}(\bar{\Omega}_{T}) and u0C2+γ(Ω¯)u_{0}\in C^{2+\gamma}(\bar{\Omega}) satisfying the first order compatibility condition:

u0ν=g(x,0) on Ω.\frac{\partial u_{0}}{\partial\nu}=g(x,0)\text{ on }\partial\Omega. (2.4)

Then there exists a unique solution uC2+γ(Ω¯T)u\in C^{2+\gamma}(\bar{\Omega}_{T}) to the problem (2.1) with the Neumann boundary condition uν=g\frac{\partial u}{\partial\nu}=g on Ω×(0,T)\partial\Omega\times(0,T). Moreover, there exists a constant CC independent of gg and u0u_{0} such that

uC2+γ(Ω¯T)C(1λfCγ(Ω¯T)+gC1+γ(Ω¯T)+u0C2+γ(Ω¯)).\left\|u\right\|_{C^{2+\gamma}(\bar{\Omega}_{T})}\leq C\left(\frac{1}{\lambda}\left\|f\right\|_{C^{\gamma}(\bar{\Omega}_{T})}+\left\|g\right\|_{{C^{1+\gamma}(\bar{\Omega}_{T})}}+\left\|u_{0}\right\|_{C^{2+\gamma}(\bar{\Omega})}\right). (2.5)

where CC is dependent only on n,γ,Λ/λ,Λγn,\gamma,\Lambda/\lambda,\Lambda_{\gamma} and Ω\Omega.

This estimate, together with Leray-Schauder fixed point argument is the main tools to prove the following theorem.

Theorem 2.2.

If nonnegative functions u0,v0u_{0},v_{0} are in C2+γ(Ω¯)C^{2+\gamma}(\bar{\Omega}) such that

u0ν=|u0|1+γ on Ω,\frac{\partial u_{0}}{\partial\nu}=|u_{0}|^{1+\gamma}\qquad\text{ on }\partial\Omega, (2.6)

where γ(0,1)\gamma\in(0,1). Then there exists T>0T>0 such that problem (KS) admits a unique nonnegative solution u,vu,v in C2+γ(Ω¯T)C^{2+\gamma}(\bar{\Omega}_{T}). Moreover, if u0,v0u_{0},v_{0} are not identically zero in Ω\Omega then u,vu,v are strictly positive in ΩT¯\bar{\Omega_{T}}.

Remark 2.1.

The convexity assumption of domain Ω\Omega is not necessary in this theorem.

Remark 2.2.

By substituting γ=p1\gamma=p-1 into Theorem 2.2, we obtain local existence and uniqueness of positive solutions in Theorem 1.1, 1.2, and 1.3.

Proof.

From now to the end of this proof, we will use CC as a universal notation for constants different from time to time. Firstly, the short-time existence of classical solution will be proved by a fixed point argument. Let uC1+γ(Ω¯T)u\in C^{1+\gamma}(\bar{\Omega}_{T}) be such that u(x,0)=u0(x)u(x,0)=u_{0}(x) in Ω\Omega. Then, the functions u0u_{0} and g(x,t)=|u(x,t)|1+γg(x,t)=|u(x,t)|^{1+\gamma} satisfy condition (2.4), and gC1+γ(Ω¯T)g\in C^{1+\gamma}(\bar{\Omega}_{T}). We assume T<1T<1, and consider the set of functions given by

BT(R):={uC1+γ(Ω¯T) such that uC1+γ(Ω¯T)R}.B_{T}(R):=\left\{u\in C^{1+\gamma}(\bar{\Omega}_{T})\text{ such that }\left\|u\right\|_{C^{1+\gamma}(\bar{\Omega}_{T})}\leq R\right\}.

Now we define the map

A:BT(R)C1+γ(Ω¯T)A:B_{T}(R)\longrightarrow C^{1+\gamma}(\bar{\Omega}_{T})

where Au:=UAu:=U is a solution of

{Ut=ΔUχ(uV)+auμu2xΩ,t(0,Tmax),τVt=ΔV+αuβVxΩ,t(0,Tmax),\begin{cases}U_{t}=\Delta U-\chi\nabla\cdot(u\nabla V)+au-\mu u^{2}\qquad&x\in{\Omega},\,t\in(0,T_{\rm max}),\\ \tau V_{t}=\Delta V+\alpha u-\beta V\qquad&x\in{\Omega},\,t\in(0,T_{\rm max}),\end{cases} (2.7)

under Neumann boundary condition:

Uν=|u|1+γ,Vν=0,xΩ,t(0,Tmax),\frac{\partial U}{\partial\nu}=|u|^{1+\gamma},\qquad\frac{\partial V}{\partial\nu}=0,\qquad x\in\partial\Omega,\,t\in(0,T_{\rm max}), (2.8)

and initial data (U(x,0),V(x,0))=(u0(x),v0(x))\left(U(x,0),V(x,0)\right)=\left(u_{0}(x),v_{0}(x)\right) in Ω\Omega. We first prove that AA sends bounded sets into relative compact sets of C1+γ(Ω¯T)C^{1+\gamma}(\bar{\Omega}_{T}). Indeed, the inequality (2.5) implies there exists R>0R^{\prime}>0 independent of TT such that AuC2+γ(Ω¯T)R\left\|Au\right\|_{C^{2+\gamma}(\bar{\Omega}_{T})}\leq R^{\prime} for all uu in BT(R)B_{T}(R). As bounded sets in C2+γ(Ω¯T)C^{2+\gamma}(\bar{\Omega}_{T}) are relatively compact in C1+γ(Ω¯T)C^{1+\gamma}(\bar{\Omega}_{T}). We claim that AA is continuous. In fact, let unuu_{n}\to u in C1+γ(Ω¯T)C^{1+\gamma}(\bar{\Omega}_{T}), we need to prove Un:=AunU:=AuU_{n}:=Au_{n}\to U:=Au in C1+γ(Ω¯T)C^{1+\gamma}(\bar{\Omega}_{T}). Now we can see that UnUU_{n}-U satisfies

{(UnU)t=Δ(UnU)+fn,xΩ,t(0,Tmax),τ(VnV)t=Δ(VnV)+α(unu)β(VnV)xΩ,t(0,Tmax),\begin{cases}(U_{n}-U)_{t}=\Delta(U_{n}-U)+f_{n},\,\qquad&x\in{\Omega},\,t\in(0,T_{\rm max}),\\ \tau(V_{n}-V)_{t}=\Delta(V_{n}-V)+\alpha(u_{n}-u)-\beta(V_{n}-V)\qquad&x\in{\Omega},\,t\in(0,T_{\rm max}),\end{cases} (2.9)

where fn:=χ(unVnuV)+un(aμun)u(aμu)f_{n}:=-\chi\nabla\cdot(u_{n}\nabla V_{n}-u\nabla V)+u_{n}(a-\mu u_{n})-u(a-\mu u). One can verify that fnf_{n} satisfies the assumptions of Theorem 2.1. Plus, the boundary condition

(UnU)ν=|un|1+γ|u|1+γ,(VnV)ν=0,xΩ,t(0,Tmax).\frac{\partial(U_{n}-U)}{\partial\nu}=|u_{n}|^{1+\gamma}-|u|^{1+\gamma},\qquad\frac{\partial(V_{n}-V)}{\partial\nu}=0,\qquad x\in\partial\Omega,\,t\in(0,T_{\rm max}). (2.10)

We claim that VnVV_{n}\to V in C2+γ(ΩT¯)C^{2+\gamma}(\bar{\Omega_{T}}) for τ0\tau\geq 0. Indeed, when τ>0\tau>0, we make use of (2.5) and when τ=0\tau=0, we apply Schauder type estimate for elliptic equation to obtain that VnVV_{n}\to V in C2+γ(ΩT¯)C^{2+\gamma}(\bar{\Omega_{T}}). This leads to fn0f_{n}\to 0 in Cγ(Ω¯T)C^{\gamma}(\bar{\Omega}_{T}), combine with inequality (2.5) entail that UnUU_{n}\to U in C2+γ(Ω¯T)C^{2+\gamma}(\bar{\Omega}_{T}). In order to apply the Leray-Schauder fixed point theorem we just have to prove that if TT is sufficiently small, and R2(1+d(Ω)1γ)u0C2+γ(Ω¯)R\geq 2(1+d(\Omega)^{1-\gamma})\left\|u_{0}\right\|_{C^{2+\gamma}(\bar{\Omega})}, then A(BT(R))BT(R)A(B_{T}(R))\subset B_{T}(R). Indeed,

|Au(x,t)||Au(x,0)|+tDtAuC0(Ω¯)u0C0(Ω¯)+TR|Au(x,t)|\leq|Au(x,0)|+t\left\|D_{t}Au\right\|_{C^{0}(\bar{\Omega})}\leq\left\|u_{0}\right\|_{C^{0}(\bar{\Omega})}+TR^{\prime}
|Au(x,t)Au(x,s)||ts|1+γ2DtAuC0(Ω¯T)|ts|1γ2RT1γ2,\frac{|Au(x,t)-Au(x,s)|}{|t-s|^{\frac{1+\gamma}{2}}}\leq\left\|D_{t}Au\right\|_{C^{0}(\bar{\Omega}_{T})}|t-s|^{\frac{1-\gamma}{2}}\leq R^{\prime}T^{\frac{1-\gamma}{2}},

and,

|DxAu(x,t)DxAu(y,s)||xy|γ+|ts|γ2\displaystyle\frac{|D_{x}Au(x,t)-D_{x}Au(y,s)|}{|x-y|^{\gamma}+|t-s|^{\frac{\gamma}{2}}} |ts|1γ2Dx2AuC0(Ω¯T)+|xy|1γ(sγ2+D2u0C0(Ω¯))\displaystyle\leq|t-s|^{1-\frac{\gamma}{2}}\left\|D^{2}_{x}Au\right\|_{C^{0}(\bar{\Omega}_{T})}+|x-y|^{1-\gamma}(s^{\frac{\gamma}{2}}+\left\|D^{2}u_{0}\right\|_{C^{0}(\bar{\Omega})})
T1γ2R+d(Ω)1γTγ2R+d(Ω)1γD2u0C0(Ω¯).\displaystyle\leq T^{1-\frac{\gamma}{2}}R^{\prime}+d(\Omega)^{1-\gamma}T^{\frac{\gamma}{2}}R^{\prime}+d(\Omega)^{1-\gamma}\left\|D^{2}u_{0}\right\|_{C^{0}(\bar{\Omega})}.

These above estimates imply that

AuC1+γ(Ω¯T)R2+RT+RT1γ2+T1γ2R+d(Ω)1γTγ2R.\left\|Au\right\|_{C^{1+\gamma}(\bar{\Omega}_{T})}\leq\frac{R}{2}+R^{\prime}T+R^{\prime}T^{\frac{1-\gamma}{2}}+T^{1-\frac{\gamma}{2}}R^{\prime}+d(\Omega)^{1-\gamma}T^{\frac{\gamma}{2}}R^{\prime}.

Since RR^{\prime} is independent of TT for all T<1T<1, we can choose TT sufficiently small as to have

RT+RT1γ2+T1γ2R+d(Ω)1γTγ2RR2.R^{\prime}T+R^{\prime}T^{\frac{1-\gamma}{2}}+T^{1-\frac{\gamma}{2}}R^{\prime}+d(\Omega)^{1-\gamma}T^{\frac{\gamma}{2}}R^{\prime}\leq\frac{R}{2}.

This further implies that

AuC1+γ(Ω¯T)R for all uBT(R).\left\|Au\right\|_{C^{1+\gamma}(\bar{\Omega}_{T})}\leq R\text{ for all }u\in B_{T}(R).

Thus AA has a fixed point in BT(R)B_{T}(R). Now if uu is a fixed point of AA, uC2+γ(Ω¯T)u\in C^{2+\gamma}(\bar{\Omega}_{T}) and it is a solution of (KS).
Secondly, the nonnegativity of solutions will be proved by the truncation method: Letting

ϕ:=min{u,0}\phi:=\min\left\{u,0\right\}

and ψ(t):=12Ωϕ2𝑑x\psi(t):=\frac{1}{2}\int_{\Omega}\phi^{2}\,dx, we see that ψ\psi is continuously differentiable with the derivative

ψ(t)\displaystyle\psi^{\prime}(t) =Ω|ϕ|2+aΩϕ2+Ωϕ|u|1+γ𝑑S+χΩϕϕvμΩϕ3\displaystyle=-\int_{\Omega}|\nabla\phi|^{2}+a\int_{\Omega}\phi^{2}+\int_{\partial\Omega}\phi|u|^{1+\gamma}\,dS+\chi\int_{\Omega}\phi\nabla\phi\cdot\nabla v-\mu\int_{\Omega}\phi^{3}
Ω|ϕ|2+aΩϕ2+χΩϕϕvμΩϕ3.\displaystyle\leq-\int_{\Omega}|\nabla\phi|^{2}+a\int_{\Omega}\phi^{2}+\chi\int_{\Omega}\phi\nabla\phi\cdot\nabla v-\mu\int_{\Omega}\phi^{3}. (2.11)

We make use of Young’s inequality combined with the global boundedness of |v||\nabla v| in Ω¯T\bar{\Omega}_{T} to obtain

χΩϕϕvϵΩ|ϕ|2+CΩϕ2,\displaystyle\chi\int_{\Omega}\phi\nabla\phi\cdot\nabla v\leq\epsilon\int_{\Omega}|\nabla\phi|^{2}+C\int_{\Omega}\phi^{2}, (2.12)

for some C>0C>0. We also have μΩϕ3CΩϕ2-\mu\int_{\Omega}\phi^{3}\leq C\int_{\Omega}\phi^{2}, where C=μsupΩT|u(x,t)|C=\mu\sup_{\Omega_{T}}|u(x,t)|. This together with (2), (2.12) implies that ψ(t)Cψ(t)\psi^{\prime}(t)\leq C\psi(t) for all 0<t<T0<t<T. By Gronwall’s inequality and the initial condition ψ(0)=0\psi(0)=0, we imply that ψ0\psi\equiv 0 or u0u\geq 0.
Thirdly, we will prove that if u00u_{0}\not\equiv 0 then uu is strictly positive in ΩT¯\bar{\Omega_{T}} by a contradiction proof. Suppose that there exists (x0,t0)Ω¯T(x_{0},t_{0})\in\bar{\Omega}_{T} such that minΩ¯Tu(x,t)=u(x0,t0)=0\min_{\bar{\Omega}_{T}}u(x,t)=u(x_{0},t_{0})=0. By the strong parabolic maximum principle, we obtain (x0,t0)Ω×(0,T)(x_{0},t_{0})\in\partial\Omega\times(0,T). However, it is a contradiction due to Hopf’s lemma:

0>uν(x0,t0)=|u(x0,t0)|1+γ=0.0>\frac{\partial u}{\partial\nu}(x_{0},t_{0})=|u(x_{0},t_{0})|^{1+\gamma}=0.

Thus, u>0u>0 and by similar arguments we also have v>0v>0.
Finally, the uniqueness of classical solutions will be proved by a contradiction proof. Assuming (u1,v1(u_{1},v_{1} and (u2,v2)(u_{2},v_{2}) are two positive classical solutions of the system (KS). Let U:=u1u2U:=u_{1}-u_{2}, V:=v1v2V:=v_{1}-v_{2}, then (U,V)(U,V) is a solution of the following system:

{Ut=ΔU+F,xΩ,t(0,Tmax),τVt=ΔV+γUβVxΩ,t(0,Tmax),\begin{cases}U_{t}=\Delta U+F,\,\qquad&x\in{\Omega},\,t\in(0,T_{\rm max}),\\ \tau V_{t}=\Delta V+\gamma U-\beta V\qquad&x\in{\Omega},\,t\in(0,T_{\rm max}),\end{cases} (2.13)

where F:=χ(u1v1u2v2)+f(u1)f(u2)F:=-\chi\nabla(u_{1}\nabla v_{1}-u_{2}\nabla v_{2})+f(u_{1})-f(u_{2}), and the boundary condition

Uν=|u1|1+γ|u2|1+γ,Vν=0,xΩ,t(0,Tmax).\frac{\partial U}{\partial\nu}=|u_{1}|^{1+\gamma}-|u_{2}|^{1+\gamma},\qquad\frac{\partial V}{\partial\nu}=0,\qquad x\in\partial\Omega,\,t\in(0,T_{\rm max}). (2.14)

By mean value theorem, there exists z(x,t)z(x,t) between u1(x,t)u_{1}(x,t) and u2(x,t)u_{2}(x,t) such that

u1(x,t)u2(x,t)=(u1(x,t)u2(x,t))f(z(x,t)).u_{1}(x,t)-u_{2}(x,t)=(u_{1}(x,t)-u_{2}(x,t))f^{\prime}(z(x,t)).

Multiplying the first equation of (2.13) by UU implies

12ddtΩU2𝑑x\displaystyle\frac{1}{2}\frac{d}{dt}\int_{\Omega}U^{2}\,dx =Ω|U|2+ΩU(|u1|1+γ|u2|1+γ)𝑑S\displaystyle=-\int_{\Omega}|\nabla U|^{2}+\int_{\partial\Omega}U(|u_{1}|^{1+\gamma}-|u_{2}|^{1+\gamma})\,dS
+χΩ(u1v1u2v2)U+ΩU2f(z).\displaystyle+\chi\int_{\Omega}(u_{1}\nabla v_{1}-u_{2}\nabla v_{2})\cdot\nabla U+\int_{\Omega}U^{2}f^{\prime}(z). (2.15)

We make use of the global boundedness property of u1,u2u_{1},u_{2} in Ω¯T\bar{\Omega}_{T}, thereafter apply Sobolev’s trace theorem, and finally Young’s inequality to have

ΩU(|u1|1+γ|u2|1+γ)𝑑SCΩU2𝑑SϵΩ|U|2+C(ϵ)ΩU2.\displaystyle\int_{\partial\Omega}U(|u_{1}|^{1+\gamma}-|u_{2}|^{1+\gamma})\,dS\leq C\int_{\partial\Omega}U^{2}\,dS\leq\epsilon\int_{\Omega}|\nabla U|^{2}+C(\epsilon)\int_{\Omega}U^{2}. (2.16)

Since,

u1v1u2v2=Uv1+u2V,\displaystyle u_{1}\nabla v_{1}-u_{2}\nabla v_{2}=U\nabla v_{1}+u_{2}\nabla V,

we have

χΩU(u1v1u2v2)\displaystyle\chi\int_{\Omega}\nabla U\cdot(u_{1}\nabla v_{1}-u_{2}\nabla v_{2}) CΩU|U|+|U||V|\displaystyle\leq C\int_{\Omega}U|\nabla U|+|\nabla U||\nabla V|
ϵΩ|U|2+CΩU2+CΩ|V|2.\displaystyle\leq\epsilon\int_{\Omega}|\nabla U|^{2}+C\int_{\Omega}U^{2}+C\int_{\Omega}|\nabla V|^{2}. (2.17)

We also have ΩU2f(z)CΩU2\int_{\Omega}U^{2}f^{\prime}(z)\leq C\int_{\Omega}U^{2} where C=supmin{u1,u2}zmax{u1,u2}|f(z)|C=\sup_{\min{\{u_{1},u_{2}\}}\leq z\leq\max{\{u_{1},u_{2}\}}}|f^{\prime}(z)|. Multiplying the second equation of (2.13) by VV, and applying Young’s inequality, we obtain

ddtΩV2+Ω|V|2CΩU2.\displaystyle\frac{d}{dt}\int_{\Omega}V^{2}+\int_{\Omega}|\nabla V|^{2}\leq C\int_{\Omega}U^{2}. (2.18)

From (2) to (2.18), we obtain

ddt{ΩU2+ΩV2}C{ΩU2+ΩV2}.\displaystyle\frac{d}{dt}\left\{\int_{\Omega}U^{2}+\int_{\Omega}V^{2}\right\}\leq C\left\{\int_{\Omega}U^{2}+\int_{\Omega}V^{2}\right\}. (2.19)

The initial conditions and Gronwall’s inequality imply that UV0U\equiv V\equiv 0, and thus there is a unique solution to the system (KS). ∎

3 Preliminaries

We collect some useful tools that will frequently be used in the sequel. Let us begin with an extended version of the Gagliardo-Nirenberg interpolation inequality, which was established in [21].

Lemma 3.1.

Let Ω\Omega be a bounded and smooth domain of n\mathbb{R}^{n} with n1n\geq 1. Let r1r\geq 1, 0<qp<0<q\leq p<\infty, s1s\geq 1. Then there exists a constant CGN>0C_{GN}>0 such that

fLp(Ω)pCGN(fLr(Ω)pafLq(Ω)p(1a)+fLs(Ω)p)\left\|f\right\|^{p}_{L^{p}(\Omega)}\leq C_{GN}\left(\left\|\nabla f\right\|_{L^{r}(\Omega)}^{pa}\left\|f\right\|^{p(1-a)}_{L^{q}(\Omega)}+\left\|f\right\|^{p}_{L^{s}(\Omega)}\right)

for all fLq(Ω)f\in L^{q}(\Omega) with f(Lr(Ω))n\nabla f\in(L^{r}(\Omega))^{n}, and a=1q1p1q+1n1r[0,1]a=\frac{\frac{1}{q}-\frac{1}{p}}{\frac{1}{q}+\frac{1}{n}-\frac{1}{r}}\in[0,1].

Consequently, the next lemma is derived as follows:

Lemma 3.2.

If Ω\Omega be a bounded and smooth domain of n\mathbb{R}^{n} with n1n\geq 1, and fW1,2(Ω)f\in W^{1,2}(\Omega) then there exists a positive constant CC depending only on Ω\Omega such that the following inequality

Ωf2CηΩ|f|2+Cηn2(Ω|f|)2\displaystyle\int_{\Omega}f^{2}\leq C\eta\int_{\Omega}|\nabla f|^{2}+\frac{C}{\eta^{\frac{n}{2}}}\left(\int_{\Omega}|f|\right)^{2} (3.1)

holds for all η(0,1)\eta\in(0,1).

Proof.

The Lemma follows from Lemma 3.1 by choosing p=r=2p=r=2 and q=s=1q=s=1 and Young’s inequality. ∎

The following lemma gives estimates on solutions of the parabolic equations. For more details, see Lemma 2.12.1 in [9].

Lemma 3.3.

Let p1p\geq 1 and q1q\geq 1 satisfy

{q<npnp,when p<n,q<,when p=n,q=,when p>n.\begin{cases}q&<\frac{np}{n-p},\qquad\text{when }p<n,\\ q&<\infty,\qquad\text{when }p=n,\\ q&=\infty,\qquad\text{when }p>n.\\ \end{cases}

Assuming g0W1,q(Ω)g_{0}\in W^{1,q}(\Omega) and gg is a classical solution to the following system

{gt=Δgζ1g+ζ2fin Ω×(0,T),gν=0on Ω×(0,T),g(,0)=g0in Ω\begin{cases}g_{t}=\Delta g-\zeta_{1}g+\zeta_{2}f&\text{in }\Omega\times(0,T),\\ \frac{\partial g}{\partial\nu}=0&\text{on }\partial\Omega\times(0,T),\\ g(\cdot,0)=g_{0}&\text{in }\Omega\end{cases} (3.2)

for some T(0,]T\in(0,\infty]. If fL((0,T);Lp(Ω))f\in L^{\infty}\left((0,T);L^{p}(\Omega)\right), then gL((0,T);W1,q(Ω))g\in L^{\infty}\left((0,T);W^{1,q}(\Omega)\right).

The following lemma provides estimates on the boundary (see Lemma 5.3 in [24]):

Lemma 3.4.

Assume that Ω\Omega is a convex bounded domain, and that fC2(Ω¯)f\in C^{2}(\bar{\Omega}) satisfies fν=0\frac{\partial f}{\partial\nu}=0 on Ω\partial\Omega. Then

|f|2ν0 on Ω.\frac{\partial|\nabla f|^{2}}{\partial\nu}\leq 0\qquad\text{ on }\partial\Omega.

The next lemma giving an useful estimate will later be applied in Section 4. Interested readers are referred to [34, 44] for more details about the proof.

Lemma 3.5.

Let Ω2\Omega\subset\mathbb{R}^{2} be a bounded domain with smooth boundary, and let p>1p>1 and 1r<p1\leq r<p. Then there exists C>0C>0 such that for each η>0\eta>0, one can pick C(η)>0C(\eta)>0 such that

uLp(Ω)pηuL2(Ω)pruln|u|Lr(Ω)r+CuLr(Ω)p+C(η)\left\|u\right\|^{p}_{L^{p}(\Omega)}\leq\eta\left\|\nabla u\right\|^{p-r}_{L^{2}(\Omega)}\left\|u\ln{|u|}\right\|^{r}_{L^{r}(\Omega)}+C\left\|u\right\|^{p}_{L^{r}(\Omega)}+C(\eta) (3.3)

holds for all uW1,2(Ω)u\in W^{1,2}(\Omega).

The following lemma providing estimates on the boundary will be useful in Section 4.

Lemma 3.6.

If r12r\geq\frac{1}{2}, p(1,32)p\in(1,\frac{3}{2}), and gC1(Ω¯),g\in C^{1}(\bar{\Omega}), then for every ϵ>0\epsilon>0, there exists a constant C=C(ϵ,Ω,p,r)C=C(\epsilon,\Omega,p,r) such that

Ω|g|p+2r1ϵΩ|g|2r+1+ϵΩ|gr|2+C.\int_{\partial\Omega}|g|^{p+2r-1}\leq\epsilon\int_{\Omega}|g|^{2r+1}+\epsilon\int_{\Omega}|\nabla g^{r}|^{2}+C. (3.4)
Proof.

Let ϕ:=|g|r\phi:=|g|^{r}, we have ϕ2+p1rW1,1(Ω)\phi^{2+\frac{p-1}{r}}\in W^{1,1}(\Omega). Trace theorem, W1,1(Ω)L1(Ω)W^{1,1}(\Omega)\to L^{1}(\partial\Omega), yields

Ωϕ2+p1r\displaystyle\int_{\partial\Omega}\phi^{2+\frac{p-1}{r}} c1Ωϕ2+p1r+(2+p1r)c1Ωϕ1+p1r|ϕ|\displaystyle\leq c_{1}\int_{\Omega}\phi^{2+\frac{p-1}{r}}+(2+\frac{p-1}{r})c_{1}\int_{\Omega}\phi^{1+\frac{p-1}{r}}|\nabla\phi|
c1Ωϕ2+p1r+3c1Ωϕ1+p1r|ϕ|\displaystyle\leq c_{1}\int_{\Omega}\phi^{2+\frac{p-1}{r}}+3c_{1}\int_{\Omega}\phi^{1+\frac{p-1}{r}}|\nabla\phi| (3.5)

where c1=c1(n,Ω)>0c_{1}=c_{1}(n,\Omega)>0. By Young’s inequality, the following holds for all ϵ>0\epsilon>0

3c1Ωϕ1+p1r|ϕ|ϵΩ|ϕ|2+c12ϵΩϕ2+2(p1)r.\displaystyle 3c_{1}\int_{\Omega}\phi^{1+\frac{p-1}{r}}|\nabla\phi|\leq\epsilon\int_{\Omega}|\nabla\phi|^{2}+\frac{c_{1}^{2}}{\epsilon}\int_{\Omega}\phi^{2+\frac{2(p-1)}{r}}. (3.6)

We apply Young’s inequality again to obtain a further estimate

c1Ωϕ2+p1r+c12ϵΩϕ2+2(p1)rϵΩϕ2+1r+c2.\displaystyle c_{1}\int_{\Omega}\phi^{2+\frac{p-1}{r}}+\frac{c_{1}^{2}}{\epsilon}\int_{\Omega}\phi^{2+\frac{2(p-1)}{r}}\leq\epsilon\int_{\Omega}\phi^{2+\frac{1}{r}}+c_{2}. (3.7)

where c2c_{2} depending on ϵ,p,r,n,Ω\epsilon,p,r,n,\Omega. We complete the proof of (3.4) by collecting (3),(3.6) and (3.7) together. ∎

The following lemma is similar to the previous lemma, however, it is necessary to trace the dependency of η,n\eta,n, and pp due to the involved technique of iteration used in Section 5.

Lemma 3.7.

If p(1,32)p\in(1,\frac{3}{2}), and gC1(Ω¯),g\in C^{1}(\bar{\Omega}), then for every η(0,12)\eta\in(0,\frac{1}{2}), there exists a constant c=c(n,Ω,p)>0c=c(n,\Omega,p)>0 such that

Ω|g|p+2r1ηΩ|g|2r+1+ηΩ|gr|2+cηn+22p3(Ω|g|r)2.\int_{\partial\Omega}|g|^{p+2r-1}\leq\eta\int_{\Omega}|g|^{2r+1}+\eta\int_{\Omega}|\nabla g^{r}|^{2}+c\eta^{\frac{n+2}{2p-3}}\left(\int_{\Omega}|g|^{r}\right)^{2}. (3.8)
Proof.

We use the same notations as in the proof of Lemma 3.6 and assume ϵ(0,1)\epsilon\in(0,1). By Young’s inequality,

abϵas+s1s(sϵ)11sbss1<ϵas+(sϵ)11sbss1,ab\leq\epsilon a^{s}+\frac{s-1}{s}(s\epsilon)^{\frac{1}{1-s}}b^{\frac{s}{s-1}}<\epsilon a^{s}+(s\epsilon)^{\frac{1}{1-s}}b^{\frac{s}{s-1}},

for any a,b0a,b\geq 0, ϵ(0,12)\epsilon\in(0,\frac{1}{2}) and s>1s>1, we obtain

c1Ωϕ2+p1rc1δ1Ωϕ2+1r+c1(p1δ1)p12pΩϕ2,\displaystyle c_{1}\int_{\Omega}\phi^{2+\frac{p-1}{r}}\leq c_{1}\delta_{1}\int_{\Omega}\phi^{2+\frac{1}{r}}+c_{1}\left(\frac{p-1}{\delta_{1}}\right)^{\frac{p-1}{2-p}}\int_{\Omega}\phi^{2}, (3.9)

and

c12ϵΩϕ2+2p2rc12ϵδ2Ωϕ2+1r+c12ϵ(2p2δ2)2p232pΩϕ2.\displaystyle\frac{c_{1}^{2}}{\epsilon}\int_{\Omega}\phi^{2+\frac{2p-2}{r}}\leq\frac{c_{1}^{2}}{\epsilon}\delta_{2}\int_{\Omega}\phi^{2+\frac{1}{r}}+\frac{c_{1}^{2}}{\epsilon}\left(\frac{2p-2}{\delta_{2}}\right)^{\frac{2p-2}{3-2p}}\int_{\Omega}\phi^{2}. (3.10)

Choosing δ1=ϵ2c1\delta_{1}=\frac{\epsilon}{2c_{1}} and δ2=ϵ22c12\delta_{2}=\frac{\epsilon^{2}}{2c_{1}^{2}}, and then combining with (3), (3.6), (3.9), and (3.10), we obtain

Ωϕ2+p1rϵΩ|ϕ|2+ϵΩϕ2+1r+(c3ϵ2p12p3+c4ϵp1p2),Ωϕ2,\displaystyle\int_{\partial\Omega}\phi^{2+\frac{p-1}{r}}\leq\epsilon\int_{\Omega}|\nabla\phi|^{2}+\epsilon\int_{\Omega}\phi^{2+\frac{1}{r}}+(c_{3}\epsilon^{\frac{2p-1}{2p-3}}+c_{4}\epsilon^{\frac{p-1}{p-2}}),\int_{\Omega}\phi^{2}, (3.11)

where c3,c4>0c_{3},c_{4}>0 independent of r,ϵr,\epsilon. Notice that

c3ϵ2p12p3+c4ϵp1p2=ϵ2p12p3(c3+c4ϵ1(2p)(32p))c5ϵ2p12p3,\displaystyle c_{3}\epsilon^{\frac{2p-1}{2p-3}}+c_{4}\epsilon^{\frac{p-1}{p-2}}=\epsilon^{\frac{2p-1}{2p-3}}(c_{3}+c_{4}\epsilon^{\frac{1}{(2-p)(3-2p)}})\leq c_{5}\epsilon^{\frac{2p-1}{2p-3}},

where c5:=c3+c4c_{5}:=c_{3}+c_{4}. This, together with (3.11) implies that

Ωϕ2+p1rϵΩ|ϕ|2+ϵΩϕ2+1r+c5ϵ2p12p3Ωϕ2.\displaystyle\int_{\partial\Omega}\phi^{2+\frac{p-1}{r}}\leq\epsilon\int_{\Omega}|\nabla\phi|^{2}+\epsilon\int_{\Omega}\phi^{2+\frac{1}{r}}+c_{5}\epsilon^{\frac{2p-1}{2p-3}}\int_{\Omega}\phi^{2}. (3.12)

We apply Lemma 3.2 with η=min{1,c51ϵ232p}\eta=\min\left\{1,c_{5}^{-1}\epsilon^{\frac{2}{3-2p}}\right\} to have

c5ϵ2p12p3Ωϕ2ϵΩ|ϕ|2+c6ϵn+22p3(Ωϕ)2,\displaystyle c_{5}\epsilon^{\frac{2p-1}{2p-3}}\int_{\Omega}\phi^{2}\leq\epsilon\int_{\Omega}|\nabla\phi|^{2}+c_{6}\epsilon^{\frac{n+2}{2p-3}}\left(\int_{\Omega}\phi\right)^{2}, (3.13)

where c6>0c_{6}>0 independent of r,ϵr,\epsilon. Collect (3.12), (3.13) and substitute η=2ϵ\eta=2\epsilon proves (3.8). ∎

Lemma 3.8.

If p(1,75)p\in(1,\frac{7}{5}), n=3n=3, and (u,v)(u,v) are in C1(Ω¯×(0,Tmax))C^{1}(\bar{\Omega}\times(0,T_{\rm max})) and

Ω|v(,t)|2A\displaystyle\int_{\Omega}|\nabla v(\cdot,t)|^{2}\leq A (3.14)

holds for all t(0,Tmax)t\in(0,T_{\rm max}), then for every ϵ>0\epsilon>0, there exists a constant C=C(ϵ,Ω,p,A)C=C(\epsilon,\Omega,p,A) such that

Ωup|v|2ϵΩu3+|u|2+u2|v|2+||v|2|2+C.\int_{\partial\Omega}u^{p}|\nabla v|^{2}\leq\epsilon\int_{\Omega}u^{3}+|\nabla u|^{2}+u^{2}|\nabla v|^{2}+\left|\nabla|\nabla v|^{2}\right|^{2}+C. (3.15)

holds for all t(0,Tmax)t\in(0,T_{\rm max}).

Proof.

By trace theorem W1,1(Ω)L1(Ω)W^{1,1}(\Omega)\to L^{1}(\partial\Omega),

Ω|u|p|v|2c1Ω|u|p|v|2+c1Ω|u|p||v|2|+c1pΩ|u|p1|u||v|2\displaystyle\int_{\partial\Omega}|u|^{p}|\nabla v|^{2}\leq c_{1}\int_{\Omega}|u|^{p}|\nabla v|^{2}+c_{1}\int_{\Omega}|u|^{p}\left|\nabla|\nabla v|^{2}\right|+c_{1}p\int_{\Omega}|u|^{p-1}|\nabla u||\nabla v|^{2} (3.16)

where c1=c1(Ω)>0c_{1}=c_{1}(\Omega)>0. Apply Young’s inequality yields

c1Ω|u|p|v|2\displaystyle c_{1}\int_{\Omega}|u|^{p}|\nabla v|^{2} ϵ2Ωu2|v|2+2p2(pϵc1)2ppΩ|v|2\displaystyle\leq\frac{\epsilon}{2}\int_{\Omega}u^{2}|\nabla v|^{2}+\frac{2-p}{2}\left(\frac{p}{\epsilon c_{1}}\right)^{\frac{2-p}{p}}\int_{\Omega}|\nabla v|^{2}
ϵ2Ωu2|v|2+c2\displaystyle\leq\frac{\epsilon}{2}\int_{\Omega}u^{2}|\nabla v|^{2}+c_{2} (3.17)

where c2:=2p2A(pϵc1)2ppc_{2}:=\frac{2-p}{2}A\left(\frac{p}{\epsilon c_{1}}\right)^{\frac{2-p}{p}}. Note that 2p<32p<3, we apply Young’s inequality to obtain

c1Ω|u|p||v|2|\displaystyle c_{1}\int_{\Omega}|u|^{p}\left|\nabla|\nabla v|^{2}\right| ϵ2Ω||v|2|2+12ϵΩu2p\displaystyle\leq\frac{\epsilon}{2}\int_{\Omega}\left|\nabla|\nabla v|^{2}\right|^{2}+\frac{1}{2\epsilon}\int_{\Omega}u^{2p}
ϵ2Ω||v|2|2+ϵΩu3+c3\displaystyle\leq\frac{\epsilon}{2}\int_{\Omega}\left|\nabla|\nabla v|^{2}\right|^{2}+\epsilon\int_{\Omega}u^{3}+c_{3} (3.18)

where c3=c3(ϵ,Ω,p)c_{3}=c_{3}(\epsilon,\Omega,p). By Young’s inequality,

c1pΩ|u|p1|u||v|2\displaystyle c_{1}p\int_{\Omega}|u|^{p-1}|\nabla u||\nabla v|^{2} ϵ2Ωu2|v|2+c4Ω|u|23p|v|2\displaystyle\leq\frac{\epsilon}{2}\int_{\Omega}u^{2}|\nabla v|^{2}+c_{4}\int_{\Omega}|\nabla u|^{\frac{2}{3-p}}|\nabla v|^{2}
ϵ2Ωu2|v|2+ϵΩ|u|2+c5Ω|v|62p2p\displaystyle\leq\frac{\epsilon}{2}\int_{\Omega}u^{2}|\nabla v|^{2}+\epsilon\int_{\Omega}|\nabla u|^{2}+c_{5}\int_{\Omega}|\nabla v|^{\frac{6-2p}{2-p}} (3.19)

where c4,c5c_{4},c_{5} are positive and dependent on ϵ,Ω,p\epsilon,\Omega,p. Here, we use the condition 1<p<751<p<\frac{7}{5} to obtain 3p2p<83\frac{3-p}{2-p}<\frac{8}{3}. By Young’s inequality,

c5Ω|v|62p2pηΩ|v|163+c6\displaystyle c_{5}\int_{\Omega}|\nabla v|^{\frac{6-2p}{2-p}}\leq\eta\int_{\Omega}|\nabla v|^{\frac{16}{3}}+c_{6} (3.20)

where c6=c6(η,ϵ,p,Ω)c_{6}=c_{6}(\eta,\epsilon,p,\Omega). In light of Gagliardo-Nirenberg inequality,

|v|2L83(Ω)\displaystyle\left\||\nabla v|^{2}\right\|_{L^{\frac{8}{3}}(\Omega)} cGN|v|2L2(Ω)34|v|2L1(Ω)14+cGN|v|2L1(Ω)\displaystyle\leq c_{GN}\left\|\nabla|\nabla v|^{2}\right\|^{\frac{3}{4}}_{L^{2}(\Omega)}\left\||\nabla v|^{2}\right\|^{\frac{1}{4}}_{L^{1}(\Omega)}+c_{GN}\left\||\nabla v|^{2}\right\|_{L^{1}(\Omega)}
cGNA14|v|2L2(Ω)34+cGNA.\displaystyle\leq c_{GN}A^{\frac{1}{4}}\left\|\nabla|\nabla v|^{2}\right\|^{\frac{3}{4}}_{L^{2}(\Omega)}+c_{GN}A. (3.21)

Hence

ηΩ|v|163\displaystyle\eta\int_{\Omega}|\nabla v|^{\frac{16}{3}} η(cGNA14|v|2L2(Ω)34+cGNA)83\displaystyle\leq\eta\left(c_{GN}A^{\frac{1}{4}}\left\|\nabla|\nabla v|^{2}\right\|^{\frac{3}{4}}_{L^{2}(\Omega)}+c_{GN}A\right)^{\frac{8}{3}}
25/3(cGNA14)83ηΩ||v|2|2+25/3(cGNA)83η.\displaystyle\leq 2^{5/3}(c_{GN}A^{\frac{1}{4}})^{\frac{8}{3}}\eta\int_{\Omega}\left|\nabla|\nabla v|^{2}\right|^{2}+2^{5/3}(c_{GN}A)^{\frac{8}{3}}\eta. (3.22)

Choosing η\eta such that 25/3(cGNA14)83η=ϵ{2^{5/3}(c_{GN}A^{\frac{1}{4}})^{\frac{8}{3}}}\eta=\epsilon, and plugging into (3), (3.20), and (3) respectively, we obtain

c1pΩ|u|p1|u||v|2ϵΩ||v|2|2+ϵ2Ωu2|v|2+ϵΩ|u|2+c7\displaystyle c_{1}p\int_{\Omega}|u|^{p-1}|\nabla u||\nabla v|^{2}\leq\epsilon\int_{\Omega}\left|\nabla|\nabla v|^{2}\right|^{2}+\frac{\epsilon}{2}\int_{\Omega}u^{2}|\nabla v|^{2}+\epsilon\int_{\Omega}|\nabla u|^{2}+c_{7} (3.23)

where c7=c6+25/3(cGNA)83ηc_{7}=c_{6}+2^{5/3}(c_{GN}A)^{\frac{8}{3}}\eta. We finally complete the proof of (3.15) by substituting (3), (3) and (3.23) into (3.16). ∎

4 A priori estimates

Let us first give a priori estimate for the parabolic-elliptic system,

Lemma 4.1.

If μ>0\mu>0 and p(1,32)p\in(1,\frac{3}{2}), for all r(1,χα(χαμ)+)r\in(1,\frac{\chi\alpha}{(\chi\alpha-\mu)_{+}}) then there exists c=c(r,u0Lr(Ω))>0c=c(r,\left\|u_{0}\right\|_{L^{r}(\Omega)})>0 such that

u(,t)Lr(Ω)c,t(0,Tmax).\left\|u(\cdot,t)\right\|_{L^{r}(\Omega)}\leq c,\qquad\forall t\in(0,T_{\rm max}). (4.1)
Proof.

Multiplying the first equation in the system (KS) by u2r1u^{2r-1} yields

12rddtΩu2r\displaystyle\frac{1}{2r}\frac{d}{dt}\int_{\Omega}u^{2r} =Ωu2r1ut\displaystyle=\int_{\Omega}u^{2r-1}u_{t}
=Ωu2r1[Δuχ(uv)+f(u)]\displaystyle=\int_{\Omega}u^{2r-1}\left[\Delta u-\chi\nabla(u\nabla v)+f(u)\right]
=2r1r2Ω|ur|2χ2r12rΩu2rΔv+Ωf(u)u2r1+Ωu2r+p1𝑑S\displaystyle=-\frac{2r-1}{r^{2}}\int_{\Omega}|\nabla u^{r}|^{2}-\chi\frac{2r-1}{2r}\int_{\Omega}u^{2r}\Delta v+\int_{\Omega}f(u)u^{2r-1}+\int_{\partial\Omega}u^{2r+p-1}\,dS
=2r1r2Ω|ur|2+2r12rΩu2r(χαuχβv)\displaystyle=-\frac{2r-1}{r^{2}}\int_{\Omega}|\nabla u^{r}|^{2}+\frac{2r-1}{2r}\int_{\Omega}u^{2r}(\chi\alpha u-\chi\beta v)
+Ωu2r+p1𝑑S+aΩu2rμΩu2r+1.\displaystyle+\int_{\partial\Omega}u^{2r+p-1}\,dS+a\int_{\Omega}u^{2r}-\mu\int_{\Omega}u^{2r+1}. (4.2)

Since v0v\geq 0, we have

ddtΩu2r\displaystyle\frac{d}{dt}\int_{\Omega}u^{2r} 2(2r1)rΩ|ur|2[2rμχα(2r1)]Ωu2r+1\displaystyle\leq-\frac{2(2r-1)}{r}\int_{\Omega}|\nabla u^{r}|^{2}-\left[2r\mu-\chi\alpha(2r-1)\right]\int_{\Omega}u^{2r+1}
+2rΩu2r+p1𝑑S+2raΩu2r.\displaystyle+2r\int_{\partial\Omega}u^{2r+p-1}\,dS+2ra\int_{\Omega}u^{2r}. (4.3)

By Lemma 3.6, we obtain

2rΩu2r+p1𝑑S2rϵΩ|ur|2+2rϵΩu2r+1+c2.\displaystyle 2r\int_{\partial\Omega}u^{2r+p-1}\,dS\leq 2r\epsilon\int_{\Omega}|\nabla u^{r}|^{2}+2r\epsilon\int_{\Omega}u^{2r+1}+c_{2}. (4.4)

We make use of Young’s inequality to obtain

(2ra+1)Ωu2rϵΩu2r+1+c3.\displaystyle(2ra+1)\int_{\Omega}u^{2r}\leq\epsilon\int_{\Omega}u^{2r+1}+c_{3}. (4.5)

Collecting (4), (4.4) and (4.5), we have

ddtΩu2r+Ωu2r\displaystyle\frac{d}{dt}\int_{\Omega}u^{2r}+\int_{\Omega}u^{2r} [2rϵ2(2r1)r]Ω|ur|2\displaystyle\leq\left[2r\epsilon-\frac{2(2r-1)}{r}\right]\int_{\Omega}|\nabla u^{r}|^{2}
[2rμχα(2r1)2ϵ]Ωu2r+1+c4.\displaystyle-\left[2r\mu-\chi\alpha(2r-1)-2\epsilon\right]\int_{\Omega}u^{2r+1}+c_{4}. (4.6)

If χα(χαμ)+>2r>1\frac{\chi\alpha}{(\chi\alpha-\mu)_{+}}>2r>1, then selecting ϵ=min{2r1r2,2rμχα(2r1)2}\epsilon=\min\left\{\frac{2r-1}{r^{2}},\frac{2r\mu-\chi\alpha(2r-1)}{2}\right\} and plugging into (4), we deduce

ddtΩu2r+Ωu2rc2\displaystyle\frac{d}{dt}\int_{\Omega}u^{2r}+\int_{\Omega}u^{2r}\leq c_{2} (4.7)

This yields (4.1), hence the proof is complete. ∎

In the parabolic-parabolic case τ=1\tau=1, the following lemma gives us a priori bounds for solution of (KS) with initial data (1.1) and the boundary condition (1.2).

Lemma 4.2.

If 1<p<321<p<\frac{3}{2}, and (u,v)(u,v) is a classical solution to (KS) with initial data (1.1) and the boundary condition (1.2) without the convexity assumption of Ω\Omega, and n2n\geq 2 then there exists a positive constant CC such that

Ω(u(,t)+1)ln(u(,t)+1)+Ω|v(,t)|2C\int_{\Omega}(u(\cdot,t)+1)\ln{(u(\cdot,t)+1)}+\int_{\Omega}|\nabla v(\cdot,t)|^{2}\leq C (4.8)

for all t(0,Tmax)t\in(0,T_{\rm max}).

Proof.

Let denote y(t):=Ω(u(,t)+1)ln(u(,t)+1)+Ω|v(,t)|2y(t):=\int_{\Omega}(u(\cdot,t)+1)\ln{(u(\cdot,t)+1)}+\int_{\Omega}|\nabla v(\cdot,t)|^{2}, we have

y(t)\displaystyle y^{\prime}(t) =Ω[Δuχ(uv)+f(u)][ln(u+1)+1]\displaystyle=\int_{\Omega}\left[\Delta u-\chi\nabla\cdot(u\nabla v)+f(u)\right]\left[\ln{(u+1)+1}\right]
+2Ωv(Δv+αuβv)\displaystyle+2\int_{\Omega}\nabla v\cdot\nabla\left(\Delta v+\alpha u-\beta v\right)
:=I1+I2.\displaystyle:=I_{1}+I_{2}. (4.9)

By integration by parts, I1I_{1} can be rewritten as

I1\displaystyle I_{1} =Ω|u|2u+1+χΩuu+1uv+aΩu[ln(u+1)+1]\displaystyle=-\int_{\Omega}\frac{|\nabla u|^{2}}{u+1}+\chi\int_{\Omega}\frac{u}{u+1}\nabla u\cdot\nabla v+a\int_{\Omega}u\left[\ln{(u+1)+1}\right]
μΩu2[ln(u+1)+1]+Ωup[ln(u+1)+1]𝑑S\displaystyle-\mu\int_{\Omega}u^{2}\left[\ln{(u+1)+1}\right]+\int_{\partial\Omega}u^{p}\left[\ln{(u+1)+1}\right]\,dS (4.10)

By integration by parts, Cauchy-Schwarz inequality and elementary inequality ln(u+1)u\ln(u+1)\leq u, we have

χΩuu+1uv\displaystyle\chi\int_{\Omega}\frac{u}{u+1}\nabla u\cdot\nabla v =χΩ(uln(u+1))v\displaystyle=\chi\int_{\Omega}\nabla\left(u-\ln(u+1)\right)\cdot\nabla v
=χΩ(uln(u+1))Δv12Ω(Δv)2+χ2Ωu2.\displaystyle=-\chi\int_{\Omega}\left(u-\ln(u+1)\right)\Delta v\leq\frac{1}{2}\int_{\Omega}(\Delta v)^{2}+\chi^{2}\int_{\Omega}u^{2}. (4.11)

One can verify that there exists c1(μ,a)>0c_{1}(\mu,a)>0 satisfying

u[ln(u+1)+1]μ4au2[ln(u+1)+1]+c1,u\left[\ln{(u+1)+1}\right]\leq\frac{\mu}{4a}u^{2}\left[\ln{(u+1)+1}\right]+c_{1},

hence,

aΩu[ln(u+1)+1]μ4Ωu2[ln(u+1)+1]+c1.a\int_{\Omega}u\left[\ln{(u+1)+1}\right]\leq\frac{\mu}{4}\int_{\Omega}u^{2}\left[\ln{(u+1)+1}\right]+c_{1}. (4.12)

In light of Sobolev’s trace theorem, W1,1(Ω)L1(Ω)W^{1,1}(\Omega)\hookrightarrow L^{1}(\partial\Omega), there exists c2(Ω)>0c_{2}(\Omega)>0 such that

Ωup[ln(u+1)+1]𝑑S\displaystyle\int_{\partial\Omega}u^{p}\left[\ln{(u+1)+1}\right]\,dS c2Ωup[ln(u+1)+1]+pc2Ωup1|u|[ln(u+1)+1]\displaystyle\leq c_{2}\int_{\Omega}u^{p}\left[\ln{(u+1)+1}\right]+pc_{2}\int_{\Omega}u^{p-1}|\nabla u|\left[\ln{(u+1)+1}\right]
+c2Ωupu+1|u|[ln(u+1)+1].\displaystyle+c_{2}\int_{\Omega}\frac{u^{p}}{u+1}|\nabla u|\left[\ln{(u+1)+1}\right]. (4.13)

By Young’s inequality, we have

pc2Ωup1|u|[ln(u+1)+1]14Ω|u|2u+1+pc2Ωu2p2(u+1)[ln(u+1)+1]2,\displaystyle pc_{2}\int_{\Omega}u^{p-1}|\nabla u|\left[\ln{(u+1)+1}\right]\leq\frac{1}{4}\int_{\Omega}\frac{|\nabla u|^{2}}{u+1}+pc_{2}\int_{\Omega}u^{2p-2}(u+1)\left[\ln{(u+1)+1}\right]^{2}, (4.14)

and

c2Ωupu+1|u|[ln(u+1)+1]14Ω|u|2u+1+c22Ωu2pu+1[ln(u+1)+1]2.\displaystyle c_{2}\int_{\Omega}\frac{u^{p}}{u+1}|\nabla u|\left[\ln{(u+1)+1}\right]\leq\frac{1}{4}\int_{\Omega}\frac{|\nabla u|^{2}}{u+1}+c^{2}_{2}\int_{\Omega}\frac{u^{2p}}{u+1}\left[\ln{(u+1)+1}\right]^{2}. (4.15)

By the similar argument as in (4.12), there exists c3(p,Ω,μ)>0c_{3}(p,\Omega,\mu)>0 such that

c2Ωup[ln(u+1)+1]\displaystyle c_{2}\int_{\Omega}u^{p}\left[\ln{(u+1)+1}\right] +pc2Ωu2p2(u+1)[ln(u+1)+1]2\displaystyle+pc_{2}\int_{\Omega}u^{2p-2}(u+1)\left[\ln{(u+1)+1}\right]^{2}
+c22Ωu2pu+1[ln(u+1)+1]2μ4Ωu2[ln(u+1)+1]+c3.\displaystyle+c^{2}_{2}\int_{\Omega}\frac{u^{2p}}{u+1}\left[\ln{(u+1)+1}\right]^{2}\leq\frac{\mu}{4}\int_{\Omega}u^{2}\left[\ln{(u+1)+1}\right]+c_{3}. (4.16)

From (4) to (4), we obtain

Ωup[ln(u+1)+1]𝑑S12Ω|u|2u+1+μ4Ωu2[ln(u+1)+1]+c3.\int_{\partial\Omega}u^{p}\left[\ln{(u+1)+1}\right]\,dS\leq\frac{1}{2}\int_{\Omega}\frac{|\nabla u|^{2}}{u+1}+\frac{\mu}{4}\int_{\Omega}u^{2}\left[\ln{(u+1)+1}\right]+c_{3}. (4.17)

Now, we handle I2I_{2} as follows:

I2\displaystyle I_{2} =2Ω(Δv)22βΩ|v|2+2αΩuv.\displaystyle=-2\int_{\Omega}(\Delta v)^{2}-2\beta\int_{\Omega}|\nabla v|^{2}+2\alpha\int_{\Omega}\nabla u\cdot\nabla v. (4.18)

By integration by part and Young’s inequality, we have

2αΩuv12Ω(Δv)2+2α2Ωu2.\displaystyle 2\alpha\int_{\Omega}\nabla u\cdot\nabla v\leq\frac{1}{2}\int_{\Omega}(\Delta v)^{2}+2\alpha^{2}\int_{\Omega}u^{2}. (4.19)

One can verify that there exists c4(α,β,χ,Ω)>0c_{4}(\alpha,\beta,\chi,\Omega)>0 such that

(χ2+2α2)Ωu2+2βΩ(u+1)ln(u+1)μ4Ωu2[ln(u+1)+1]+c4.(\chi^{2}+2\alpha^{2})\int_{\Omega}u^{2}+2\beta\int_{\Omega}(u+1)\ln{(u+1)}\leq\frac{\mu}{4}\int_{\Omega}u^{2}\left[\ln{(u+1)+1}\right]+c_{4}. (4.20)

Collecting (4), (4.12), (4.17) and from (4.18) to (4.20), we obtain

y(t)+2βy(t)12Ω|u|2u+1μ4Ωu2[ln(u+1)+1]+c5c5,t(0,Tmax),y^{\prime}(t)+2\beta y(t)\leq-\frac{1}{2}\int_{\Omega}\frac{|\nabla u|^{2}}{u+1}-\frac{\mu}{4}\int_{\Omega}u^{2}\left[\ln{(u+1)+1}\right]+c_{5}\leq c_{5},\qquad\forall t\in(0,T_{\rm max}), (4.21)

where c5:=c1+c3+c4c_{5}:=c_{1}+c_{3}+c_{4}. This, together with the Gronwall’s inequality, yields

y(t)e2βty(0)+c52β(1e2βt)Cy(t)\leq e^{-2\beta t}y(0)+\frac{c_{5}}{2\beta}(1-e^{-2\beta t})\leq C

where C:=max{y(0),c52β}C:=\max\left\{y(0),\frac{c_{5}}{2\beta}\right\}, and the proof of (4.8) is complete. ∎

The following lemma gives an L2L^{2}-bound in two-dimensional space for the parabolic-parabolic system.

Lemma 4.3.

If τ=1\tau=1, n=2n=2, 1<p<321<p<\frac{3}{2}, and (u,v)(u,v) is a classical solution to (KS) with initial data (1.1) and the boundary condition then there exists a positive constant CC such that

Ωu2(,t)+Ω|v(,t)|4C\int_{\Omega}u^{2}(\cdot,t)+\int_{\Omega}|\nabla v(\cdot,t)|^{4}\leq C (4.22)

for all t(0,Tmax)t\in(0,T_{\rm max}).

Proof.

Let denote

ϕ(t):=12Ωu2+14Ω|v|4,\phi(t):=\frac{1}{2}\int_{\Omega}u^{2}+\frac{1}{4}\int_{\Omega}|\nabla v|^{4},

we have

ϕ(t)\displaystyle\phi^{\prime}(t) =Ωu[Δuχ(uv)+f(u)]\displaystyle=\int_{\Omega}u\left[\Delta u-\chi\nabla\cdot(u\nabla v)+f(u)\right]
+Ω|v|2v(Δv+αuβv)\displaystyle+\int_{\Omega}|\nabla v|^{2}\nabla v\cdot\nabla\left(\Delta v+\alpha u-\beta v\right)
:=J1+J2.\displaystyle:=J_{1}+J_{2}. (4.23)

By integration by parts, we obtain

J1\displaystyle J_{1} =Ω|u|2+χΩuuv+aΩu2μΩu3+Ωup+1𝑑S.\displaystyle=-\int_{\Omega}|\nabla u|^{2}+\chi\int_{\Omega}u\nabla u\cdot\nabla v+a\int_{\Omega}u^{2}-\mu\int_{\Omega}u^{3}+\int_{\partial\Omega}u^{p+1}\,dS. (4.24)

By Young’s inequality, we have

χΩuuv+aΩu212Ω|u|2+χ2Ωu2|v|2.\displaystyle\chi\int_{\Omega}u\nabla u\cdot\nabla v+a\int_{\Omega}u^{2}\leq\frac{1}{2}\int_{\Omega}|\nabla u|^{2}+\chi^{2}\int_{\Omega}u^{2}|\nabla v|^{2}. (4.25)

In light of Sobolev’s trace theorem, W1,1(Ω)L1(Ω)W^{1,1}(\Omega)\hookrightarrow L^{1}(\partial\Omega), there exists c1:=c1(Ω)>0c_{1}:=c_{1}(\Omega)>0 such that

Ωup+1𝑑Sc1Ωup+1+c1(p+1)Ωup|u|.\displaystyle\int_{\partial\Omega}u^{p+1}\,dS\leq c_{1}\int_{\Omega}u^{p+1}+c_{1}(p+1)\int_{\Omega}u^{p}|\nabla u|. (4.26)

Since 1<p<321<p<\frac{3}{2}, we apply Young’s inequality to obtain

Ωup+1𝑑S\displaystyle\int_{\partial\Omega}u^{p+1}\,dS μ4Ωu3+14Ω|u|2+c1(p+1)2Ωu2p+c2\displaystyle\leq\frac{\mu}{4}\int_{\Omega}u^{3}+\frac{1}{4}\int_{\Omega}|\nabla u|^{2}+{c_{1}(p+1)^{2}}\int_{\Omega}u^{2p}+c_{2}
14Ω|u|2+μ2Ωu3+c3,\displaystyle\leq\frac{1}{4}\int_{\Omega}|\nabla u|^{2}+\frac{\mu}{2}\int_{\Omega}u^{3}+c_{3}, (4.27)

where c2,c3>0c_{2},c_{3}>0 depending only on p,μ,Ωp,\mu,\Omega. To deal with J2J_{2}, we make use of the following pointwise identity

vΔv=12Δ(|v|2)|D2v|2\displaystyle\nabla v\cdot\nabla\Delta v=\frac{1}{2}\Delta(|\nabla v|^{2})-|D^{2}v|^{2}

to obtain

J2\displaystyle J_{2} =12Ω||v|2|2Ω|v|2|D2v|2\displaystyle=-\frac{1}{2}\int_{\Omega}|\nabla|\nabla v|^{2}|^{2}-\int_{\Omega}|\nabla v|^{2}|D^{2}v|^{2}
+αΩ|v|2vu\displaystyle+\alpha\int_{\Omega}|\nabla v|^{2}\nabla v\cdot\nabla u
βΩ|v|4+12Ω|v|2ν|v|2.\displaystyle-\beta\int_{\Omega}|\nabla v|^{4}+\frac{1}{2}\int_{\partial\Omega}\frac{\partial|\nabla v|^{2}}{\partial\nu}|\nabla v|^{2}. (4.28)

Applying Lemma 3.4 and the pointwise inequality (Δv)22|D2v|2(\Delta v)^{2}\leq 2|D^{2}v|^{2} to (4), we deduce

J2\displaystyle J_{2} 12Ω||v|2|2βΩ|v|4\displaystyle\leq-\frac{1}{2}\int_{\Omega}|\nabla|\nabla v|^{2}|^{2}-\beta\int_{\Omega}|\nabla v|^{4}
12Ω|v|2|Δv|2+αΩ|v|2vu.\displaystyle-\frac{1}{2}\int_{\Omega}|\nabla v|^{2}|\Delta v|^{2}+\alpha\int_{\Omega}|\nabla v|^{2}\nabla v\cdot\nabla u. (4.29)

By integral by parts and Young’s inequality, we obtain

αΩ|v|2vu\displaystyle\alpha\int_{\Omega}|\nabla v|^{2}\nabla v\cdot\nabla u =αΩu|v|2vαΩu|v|2Δv\displaystyle=-\alpha\int_{\Omega}u\nabla|\nabla v|^{2}\cdot\nabla v-\alpha\int_{\Omega}u|\nabla v|^{2}\Delta v
14Ω||v|2|2+14Ω|v|2|Δv|2+2α2Ωu2|v|2\displaystyle\leq\frac{1}{4}\int_{\Omega}|\nabla|\nabla v|^{2}|^{2}+\frac{1}{4}\int_{\Omega}|\nabla v|^{2}|\Delta v|^{2}+2\alpha^{2}\int_{\Omega}u^{2}|\nabla v|^{2} (4.30)

Collecting from (4) to (4) yields

ϕ+4βϕ14Ω|u|212Ω||v|2|2μ2Ωu3+c4Ωu2|v|2+c5Ωu2+c3\displaystyle\phi^{\prime}+4\beta\phi\leq-\frac{1}{4}\int_{\Omega}|\nabla u|^{2}-\frac{1}{2}\int_{\Omega}|\nabla|\nabla v|^{2}|^{2}-\frac{\mu}{2}\int_{\Omega}u^{3}+c_{4}\int_{\Omega}u^{2}|\nabla v|^{2}+c_{5}\int_{\Omega}u^{2}+c_{3} (4.31)

where c4,c5c_{4},c_{5} are positive constants depending on α,β,χ,a\alpha,\beta,\chi,a. By Young’s inequality

c4Ωu2|v|2c4ϵΩ|v|6+c4ϵΩu3\displaystyle c_{4}\int_{\Omega}u^{2}|\nabla v|^{2}\leq c_{4}\epsilon\int_{\Omega}|\nabla v|^{6}+\frac{c_{4}}{\sqrt{\epsilon}}\int_{\Omega}u^{3} (4.32)

By Gagliardo-Nirenberg inequality for n=2n=2 and (4.8), there exists cGN>0c_{GN}>0 such that

Ω|v|6\displaystyle\int_{\Omega}|\nabla v|^{6} cGN(Ω||v|2|2)(Ω|v|2)+cGN(Ω|v|2)3\displaystyle\leq c_{GN}\left(\int_{\Omega}|\nabla|\nabla v|^{2}|^{2}\right)\left(\int_{\Omega}|\nabla v|^{2}\right)+c_{GN}\left(\int_{\Omega}|\nabla v|^{2}\right)^{3}
c6(Ω||v|2|2)+c7,\displaystyle\leq c_{6}\left(\int_{\Omega}|\nabla|\nabla v|^{2}|^{2}\right)+c_{7}, (4.33)

where c6,c7c_{6},c_{7} are positive constants depending on cGNc_{GN} and supt(0,Tmax)Ω|v|2\sup_{t\in(0,T_{\rm max})}\int_{\Omega}|\nabla v|^{2}. We make use of (3.3) for n=2n=2 and (4.8) to obtain

Ωu3\displaystyle\int_{\Omega}u^{3} ϵ(Ω|u|2)(Ωu|lnu|)+C(Ωu)3+c(ϵ)\displaystyle\leq\epsilon\left(\int_{\Omega}|\nabla u|^{2}\right)\left(\int_{\Omega}u|\ln u|\right)+C\left(\int_{\Omega}u\right)^{3}+c(\epsilon)
c8ϵΩ|u|2+c9,\displaystyle\leq c_{8}\epsilon\int_{\Omega}|\nabla u|^{2}+c_{9}, (4.34)

where c8:=supt(0,Tmax)Ωu|lnu|c_{8}:=\sup_{t\in(0,T_{\rm max})}\int_{\Omega}u|\ln u| and c9>0c_{9}>0 depending on ϵ\epsilon and supt(0,Tmax)Ωu\sup_{t\in(0,T_{\rm max})}\int_{\Omega}u. In light of Young’s inequality

c5Ωu2μ4Ωu3+c10.\displaystyle c_{5}\int_{\Omega}u^{2}\leq\frac{\mu}{4}\int_{\Omega}u^{3}+c_{10}. (4.35)

where c10>0c_{10}>0 depending on c5,μ,|Ω|c_{5},\mu,|\Omega|. Combining from (4.31) to (4.35), we have

ϕ+4βϕ\displaystyle\phi^{\prime}+4\beta\phi (c4c8ϵ14)Ω|u|2+(c4c6ϵ14)Ω||v|2|2+c11,\displaystyle\leq\left(c_{4}c_{8}\sqrt{\epsilon}-\frac{1}{4}\right)\int_{\Omega}|\nabla u|^{2}+\left(c_{4}c_{6}\epsilon-\frac{1}{4}\right)\int_{\Omega}|\nabla|\nabla v|^{2}|^{2}+c_{11}, (4.36)

where c11>0c_{11}>0 depending on ϵ\epsilon. Choosing ϵ\epsilon sufficiently small and substituting into (4.36), we obtain

ϕ+4βϕc11.\phi^{\prime}+4\beta\phi\leq c_{11}. (4.37)

This, together with Gronwall’s inequality yields ϕ(t)C:=max{ϕ(0),c114β}\phi(t)\leq C:=\max\left\{\phi(0),\frac{c_{11}}{4\beta}\right\} for all t(0,Tmax)t\in(0,T_{\rm max}), and the proof of Lemma 4.3 is complete. ∎

The next lemma is the key step in the proof of the parabolic-parabolic system in three-dimensional space.

Lemma 4.4.

Let (u,v)(u,v) be a classical solution to (KS) in a convex bounded domain Ω\Omega with smooth boundary. If τ=1\tau=1, n=3n=3, 1<p<751<p<\frac{7}{5} and μ\mu is sufficiently large, then there exists a positive constant CC such that

Ωu2(,t)+Ω|v(,t)|4C\int_{\Omega}u^{2}(\cdot,t)+\int_{\Omega}|\nabla v(\cdot,t)|^{4}\leq C (4.38)

for all t(0,Tmax)t\in(0,T_{\rm max}).

Remark 4.1.

When we look at the proof of Theorem 1.2 carefully, n=2n=2 is utilized to estimate Ωu2|v|2\int_{\Omega}u^{2}|\nabla v|^{2}. For n=3n=3, in order to eliminate this term we borrow the idea as in [33, 41] by introducing an extra term Ωu|v|2\int_{\Omega}u|\nabla v|^{2}, in the function ϕ\phi for Theorem 1.2. Note that the term ut|v|2u_{t}|\nabla v|^{2} will introduce μu2|v|2-\mu u^{2}|\nabla v|^{2}, and a sufficiently large parameter μ\mu will help the estimates.

Proof.

Let call ψ(t):=Ωu2+Ω|v|4+13Ωu|v|2\psi(t):=\int_{\Omega}u^{2}+\int_{\Omega}|\nabla v|^{4}+\frac{1}{3}\int_{\Omega}u|\nabla v|^{2}, we have

ψ(t)\displaystyle\psi^{\prime}(t) =2Ωu[Δuχ(uv)+f(u)]\displaystyle=2\int_{\Omega}u\left[\Delta u-\chi\nabla\cdot(u\nabla v)+f(u)\right]
+4Ω|v|2v(Δv+αuβv)\displaystyle+4\int_{\Omega}|\nabla v|^{2}\nabla v\cdot\nabla\left(\Delta v+\alpha u-\beta v\right)
+23Ωuv(Δv+αuβv)\displaystyle+\frac{2}{3}\int_{\Omega}u\nabla v\cdot\nabla\left(\Delta v+\alpha u-\beta v\right)
+23Ω|v|2[Δuχ(uv)+f(u)]\displaystyle+\frac{2}{3}\int_{\Omega}|\nabla v|^{2}\left[\Delta u-\chi\nabla\cdot(u\nabla v)+f(u)\right]
:=K1+K2+K3+K4.\displaystyle:=K_{1}+K_{2}+K_{3}+K_{4}. (4.39)

By integration by parts, K1K_{1} can be written as:

K1\displaystyle K_{1} =2Ω|u|2+2Ωup+1𝑑S+2χΩuuv\displaystyle=-2\int_{\Omega}|\nabla u|^{2}+2\int_{\partial\Omega}u^{p+1}\,dS+2\chi\int_{\Omega}u\nabla u\cdot\nabla v
+2aΩu22μΩu3.\displaystyle+2a\int_{\Omega}u^{2}-2\mu\int_{\Omega}u^{3}. (4.40)

By Lemma 3.6, we obtain

2Ωup+1𝑑S2ϵ1Ωu3+2ϵ1Ω|u|2+2c1.\displaystyle 2\int_{\partial\Omega}u^{p+1}\,dS\leq 2\epsilon_{1}\int_{\Omega}u^{3}+2\epsilon_{1}\int_{\Omega}|\nabla u|^{2}+2c_{1}. (4.41)

By Young’s inequality, we have

2χΩuuvχϵ1Ω|u|2+χϵ1Ωu2|v|2.\displaystyle 2\chi\int_{\Omega}u\nabla u\cdot\nabla v\leq\chi\epsilon_{1}\int_{\Omega}|\nabla u|^{2}+\frac{\chi}{\epsilon_{1}}\int_{\Omega}u^{2}|\nabla v|^{2}. (4.42)

We also have

2aΩu22aϵ1Ωu3+c2,\displaystyle 2a\int_{\Omega}u^{2}\leq 2a\epsilon_{1}\int_{\Omega}u^{3}+c_{2}, (4.43)

where c2>0c_{2}>0 depending on ϵ1\epsilon_{1}. From (4) to (4.43), we obtain

K1\displaystyle K_{1} [(2+χ)ϵ12]Ω|u|2+2[(a+1)ϵ1μ]Ωu3+χϵ1Ωu2|v|2+c3,\displaystyle\leq\left[(2+\chi)\epsilon_{1}-2\right]\int_{\Omega}|\nabla u|^{2}+2\left[(a+1)\epsilon_{1}-\mu\right]\int_{\Omega}u^{3}+\frac{\chi}{\epsilon_{1}}\int_{\Omega}u^{2}|\nabla v|^{2}+c_{3}, (4.44)

where c3=2+c2c_{3}=2+c_{2}. We choose ϵ1=12min{22+χ,μa+1}\epsilon_{1}=\frac{1}{2}\min\left\{\frac{2}{2+\chi},\frac{\mu}{a+1}\right\} and substitute into (4.44) to obtain

K1\displaystyle K_{1} Ω|u|2μΩu3+χϵ1Ωu2|v|2+c3.\displaystyle\leq-\int_{\Omega}|\nabla u|^{2}-\mu\int_{\Omega}u^{3}+\frac{\chi}{\epsilon_{1}}\int_{\Omega}u^{2}|\nabla v|^{2}+c_{3}. (4.45)

To deal with K2K_{2}, we use similar estimates to (4) and (4) in estimating J2J_{2} in Lemma 4.3 to obtain

K2+4βΩ|v|4\displaystyle K_{2}+4\beta\int_{\Omega}|\nabla v|^{4} 2(αϵ21)Ω||v|2|2+(2αϵ2+3α2)Ωu2|v|2.\displaystyle\leq 2(\alpha\epsilon_{2}-1)\int_{\Omega}|\nabla|\nabla v|^{2}|^{2}+\left(\frac{2\alpha}{\epsilon_{2}}+3\alpha^{2}\right)\int_{\Omega}u^{2}|\nabla v|^{2}. (4.46)

By substituting ϵ2=12α\epsilon_{2}=\frac{1}{2\alpha} into (4.46), we deduce

K2+4βΩ|v|4\displaystyle K_{2}+4\beta\int_{\Omega}|\nabla v|^{4} Ω||v|2|2+7α2Ωu2|v|2\displaystyle\leq-\int_{\Omega}|\nabla|\nabla v|^{2}|^{2}+7\alpha^{2}\int_{\Omega}u^{2}|\nabla v|^{2} (4.47)

To deal with K3K_{3}, we make use of the following identity

vΔv=12Δ(|v|2)|D2v|2.\displaystyle\nabla v\cdot\nabla\Delta v=\frac{1}{2}\Delta(|\nabla v|^{2})-|D^{2}v|^{2}.

Rewriting K3K_{3} as

K3\displaystyle K_{3} =13Ωu|v|223Ωu|D2v|2\displaystyle=-\frac{1}{3}\int_{\Omega}\nabla u\cdot\nabla|\nabla v|^{2}-\frac{2}{3}\int_{\Omega}u|D^{2}v|^{2}
+2α3Ωuvu2β3Ωu|v|2+13Ωu|v|2ν.\displaystyle+\frac{2\alpha}{3}\int_{\Omega}u\nabla v\cdot\nabla u-\frac{2\beta}{3}\int_{\Omega}u|\nabla v|^{2}+\frac{1}{3}\int_{\partial\Omega}u\frac{\partial|\nabla v|^{2}}{\partial\nu}. (4.48)

We drop the last term due to Lemma 3.4, neglect the second term and apply Cauchy-Schwartz inequality

abϵa2+14ϵb2,ab\leq\epsilon a^{2}+\frac{1}{4\epsilon}b^{2},

to the third and the forth terms with a sufficiently small ϵ\epsilon to obtain

K3+2β3Ωu|v|2\displaystyle K_{3}+\frac{2\beta}{3}\int_{\Omega}u|\nabla v|^{2} 13Ω||v|2|2+α28Ωu3+13Ω|u|2.\displaystyle\leq\frac{1}{3}\int_{\Omega}|\nabla|\nabla v|^{2}|^{2}+\frac{\alpha^{2}}{8}\int_{\Omega}u^{3}+\frac{1}{3}\int_{\Omega}|\nabla u|^{2}. (4.49)

By integration by parts, K4K_{4} can be rewritten as:

K4\displaystyle K_{4} =13Ω|v|2u+13Ωup|v|2\displaystyle=-\frac{1}{3}\int_{\Omega}\nabla|\nabla v|^{2}\cdot\nabla u+\frac{1}{3}\int_{\partial\Omega}u^{p}|\nabla v|^{2}
+χ3Ω|v|2uv+a3Ωu|v|2μ3Ωu2|v|2.\displaystyle+\frac{\chi}{3}\int_{\Omega}\nabla|\nabla v|^{2}\cdot u\nabla v+\frac{a}{3}\int_{\Omega}u|\nabla v|^{2}-\frac{\mu}{3}\int_{\Omega}u^{2}|\nabla v|^{2}. (4.50)

By Cauchy-Schwarz inequality, we obtain

13Ω|v|2u16Ω||v|2|2+13Ω|u|2.\displaystyle-\frac{1}{3}\int_{\Omega}\nabla|\nabla v|^{2}\cdot\nabla u\leq\frac{1}{6}\int_{\Omega}\left|\nabla|\nabla v|^{2}\right|^{2}+\frac{1}{3}\int_{\Omega}|\nabla u|^{2}. (4.51)

In light of Lemma 3.8 and Lemma 4.2, the following inequality

13Ωup|v|213Ω|u|2+u2|v|2+||v|2|2+u3+c1\displaystyle\frac{1}{3}\int_{\partial\Omega}u^{p}|\nabla v|^{2}\leq\frac{1}{3}\int_{\Omega}|\nabla u|^{2}+u^{2}|\nabla v|^{2}+\left|\nabla|\nabla v|^{2}\right|^{2}+u^{3}+c_{1} (4.52)

holds for all t(0,Tmax)t\in(0,T_{\rm max}), with some positive constant c1c_{1}. By Young’s inequality, we obtain

χ3Ω|v|2uv16Ω||v|2|2+χ23Ωu2|v|2,\displaystyle\frac{\chi}{3}\int_{\Omega}\nabla|\nabla v|^{2}\cdot u\nabla v\leq\frac{1}{6}\int_{\Omega}\left|\nabla|\nabla v|^{2}\right|^{2}+\frac{\chi^{2}}{3}\int_{\Omega}u^{2}|\nabla v|^{2}, (4.53)

and

a3Ωu|v|213Ωu2|v|2+c2,\displaystyle\frac{a}{3}\int_{\Omega}u|\nabla v|^{2}\leq\frac{1}{3}\int_{\Omega}u^{2}|\nabla v|^{2}+c_{2}, (4.54)

with some positive constant c2c_{2}. Combining from (4) to (4.54), we obtain

K4\displaystyle K_{4} 23Ω|u|2+||v|2|2+χ2μ+23Ωu2|v|2+13Ωu3+c3,\displaystyle\leq\frac{2}{3}\int_{\Omega}|\nabla u|^{2}+\left|\nabla|\nabla v|^{2}\right|^{2}+\frac{\chi^{2}-\mu+2}{3}\int_{\Omega}u^{2}|\nabla v|^{2}+\frac{1}{3}\int_{\Omega}u^{3}+c_{3}, (4.55)

with some positive constant c3c_{3}. Collecting (4.45), (4.47), (4.49), and (4.55), we have

ψ+4βΩ|v|4+2β3Ωu|v|2\displaystyle\psi^{\prime}+4\beta\int_{\Omega}|\nabla v|^{4}+\frac{2\beta}{3}\int_{\Omega}u|\nabla v|^{2} (7α2+χϵ1+χ2μ+23)Ωu2|v|2\displaystyle\leq\left(7\alpha^{2}+\frac{\chi}{\epsilon_{1}}+\frac{\chi^{2}-\mu+2}{3}\right)\int_{\Omega}u^{2}|\nabla v|^{2}
+(13μ)Ωu3+α28Ωu2+c5.\displaystyle+\left(\frac{1}{3}-\mu\right)\int_{\Omega}u^{3}+\frac{\alpha^{2}}{8}\int_{\Omega}u^{2}+c_{5}. (4.56)

This leads to

ψ+2βψ\displaystyle\psi^{\prime}+2\beta\psi (7α2+χϵ1+χ2μ+23)Ωu2|v|2\displaystyle\leq\left(7\alpha^{2}+\frac{\chi}{\epsilon_{1}}+\frac{\chi^{2}-\mu+2}{3}\right)\int_{\Omega}u^{2}|\nabla v|^{2}
+(13μ)Ωu3+16β+α28Ωu2+c5.\displaystyle+\left(\frac{1}{3}-\mu\right)\int_{\Omega}u^{3}+\frac{16\beta+\alpha^{2}}{8}\int_{\Omega}u^{2}+c_{5}. (4.57)

By Young’s inequality, for every ϵ>0\epsilon>0, there exists a positive constant c6=c6(ϵ)c_{6}=c_{6}(\epsilon) such that:

16β+α28Ωu2ϵΩu3+c6.\displaystyle\frac{16\beta+\alpha^{2}}{8}\int_{\Omega}u^{2}\leq\epsilon\int_{\Omega}u^{3}+c_{6}. (4.58)

Therefore, we need to choose μ0\mu_{0} sufficiently large such that

{7α2+χϵ1+χ2μ0+23013+ϵμ00μ02(a+1)2+χ.\displaystyle\begin{cases}7\alpha^{2}+\frac{\chi}{\epsilon_{1}}+\frac{\chi^{2}-\mu_{0}+2}{3}&\leq 0\\ \frac{1}{3}+\epsilon-\mu_{0}&\leq 0\\ \mu_{0}&\geq\frac{2(a+1)}{2+\chi}.\end{cases}

Therefore, if μ>μ0\mu>\mu_{0}, where

μ0:=max{13,2(a+1)2+χ,3(χ2+χ+7α2+χ2+22)}\displaystyle\mu_{0}:=\max\left\{\frac{1}{3},\frac{2(a+1)}{2+\chi},3\left(\frac{\chi}{2+\chi}+7\alpha^{2}+\frac{\chi^{2}+2}{2}\right)\right\} (4.59)

and then (4) yields

ψ+2βψc6.\displaystyle\psi^{\prime}+2\beta\psi\leq c_{6}.

Applying Gronwall’s inequality, we see that

ψ(t)max{ψ(0),c62β}\displaystyle\psi(t)\leq\max\left\{\psi(0),\frac{c_{6}}{2\beta}\right\} (4.60)

and thereby conclude the proof. ∎

Remark 4.2.

μ0\mu_{0} defined as in (4) is not sharp. We leave the open question to obtain an optimal formula μ0\mu_{0}.

5 Global boundedness

In this section, we show that if uu is uniformly bounded in time under Lr0(Ω)\left\|\cdot\right\|_{L^{r_{0}}(\Omega)}, then it is also uniformly bounded in time under L(Ω)\left\|\cdot\right\|_{L^{\infty}(\Omega)}. A method due to [1, 2] based on Moser-type iterations is implemented to prove the following theorem.

Theorem 5.1.

Let r0>n2r_{0}>\frac{n}{2} and (u,v)(u,v) be a classical solution of (KS) on Ω×(0,Tmax)\Omega\times(0,T_{\rm max}) with maximal existence time Tmax(0,]T_{\rm max}\in(0,\infty]. If

supt(0,Tmax)u(,t)Lr0(Ω)<,\sup_{t\in(0,T_{\rm max})}\left\|u(\cdot,t)\right\|_{L^{r_{0}}(\Omega)}<\infty,

then

supt(0,Tmax)(u(,t)L(Ω)+v(,t)W1,(Ω))<.\sup_{t\in(0,T_{\rm max})}\left(\left\|u(\cdot,t)\right\|_{L^{\infty}(\Omega)}+\left\|v(\cdot,t)\right\|_{W^{1,\infty}(\Omega)}\right)<\infty.
Remark 5.1.

One can also follow the argument as in [30] to prove the above theorem.

Remark 5.2.

The Ln2+L^{\frac{n}{2}+}-criterion for homogeneous Neumann boundary conditions has been studied in [4] for general chemotaxis systems and in [40] for the fully parabolic chemotaxis system, both with and without a logistic source. However, Theorem 5.1 not only addresses nonlinear Neumann boundary conditions, but also employs a different analysis approach compared to [4, 40]. Instead of utilizing the semigroup estimate, the analysis in Theorem 5.1 relies on the LpL^{p}-regularity theory for parabolic equations.

The following lemma helps us to establish the iteration process from LrL^{r} to LL^{\infty} when r>nr>n.

Lemma 5.2.

Let (u,v)(u,v) be a classical solution of (KS) on (0,Tmax)(0,T_{\rm max}) and

Ur:=max{u0L(Ω),supt(0,Tmax)u(,t)LrΩ)}.U_{r}:=\max\left\{\|u_{0}\|_{L^{\infty}(\Omega)},\sup_{t\in(0,T_{\rm max})}\|u(\cdot,t)\|_{L^{r}{\Omega)}}\right\}.

If supt(0,Tmax)u(,t)LrΩ)<\sup_{t\in(0,T_{\rm max})}\|u(\cdot,t)\|_{L^{r}{\Omega)}}<\infty for some r>nr>n, then there exists constants A,B>0A,B>0 independent of rr such that

U2r(ArB)12rUr.\displaystyle U_{2r}\leq(Ar^{B})^{\frac{1}{2r}}U_{r}. (5.1)
Proof.

The main idea is first to establish an inequality as follow:

ddtΩu2r+Ωu2rArB(Ωur)2\displaystyle\frac{d}{dt}\int_{\Omega}u^{2r}+\int_{\Omega}u^{2r}\leq Ar^{B}\left(\int_{\Omega}u^{r}\right)^{2} (5.2)

and thereafter apply Moser iteration technique. Therefore, the dependency of all constants in terms of rr will be traced carefully. Multiplying the first equation in the system (KS) by u2r1u^{2r-1} we obtain

12rddtΩu2r\displaystyle\frac{1}{2r}\frac{d}{dt}\int_{\Omega}u^{2r} =Ωu2r1ut\displaystyle=\int_{\Omega}u^{2r-1}u_{t}
=Ωu2r1[Δuχ(uv)+auμu2]\displaystyle=\int_{\Omega}u^{2r-1}\left[\Delta u-\chi\nabla(u\nabla v)+au-\mu u^{2}\right]
=2r1r2Ω|ur|2𝑑x+J,\displaystyle=-\frac{2r-1}{r^{2}}\int_{\Omega}|\nabla u^{r}|^{2}\,dx+J, (5.3)

where

J\displaystyle J :=Ωu2r1[χ(uv)+auμu2]\displaystyle:=\int_{\Omega}u^{2r-1}\left[-\chi\nabla(u\nabla v)+au-\mu u^{2}\right]
=χ2r12rΩu2rv+aΩu2r\displaystyle=\chi\frac{2r-1}{2r}\int_{\Omega}\nabla u^{2r}\cdot\nabla v+a\int_{\Omega}u^{2r}
μΩu2r+1+Ωup+2r1𝑑S\displaystyle-\mu\int_{\Omega}u^{2r+1}+\int_{\partial\Omega}u^{p+2r-1}\,dS
=χ2r1rΩururv+aΩu2r\displaystyle=\chi\frac{2r-1}{r}\int_{\Omega}u^{r}\nabla u^{r}\cdot\nabla v+a\int_{\Omega}u^{2r}
μΩu2r+1+Ωup+2r1𝑑S.\displaystyle-\mu\int_{\Omega}u^{2r+1}+\int_{\partial\Omega}u^{p+2r-1}\,dS. (5.4)

By Lemma 3.7, for any ϵ(0,1)\epsilon\in(0,1), we obtain

Ωup+2r1𝑑SϵΩ|ur|2+ϵΩu2r+1+c1(1ϵ)a1(Ωur)2.\displaystyle\int_{\partial\Omega}u^{p+2r-1}\,dS\leq\epsilon\int_{\Omega}|\nabla u^{r}|^{2}+\epsilon\int_{\Omega}u^{2r+1}+c_{1}\left(\frac{1}{\epsilon}\right)^{a_{1}}\left(\int_{\Omega}u^{r}\right)^{2}. (5.5)

where a1=n+232pa_{1}=\frac{n+2}{3-2p}. We apply Young’s inequality to the first two terms of the right hand side of (5) and combine with (5.5) to have

J\displaystyle J 3ϵΩ|ur|2+(2r1)24r2ϵχ2Ωu2r|v|2\displaystyle\leq 3\epsilon\int_{\Omega}|\nabla u^{r}|^{2}+\frac{(2r-1)^{2}}{4r^{2}\epsilon}\chi^{2}\int_{\Omega}u^{2r}|\nabla v|^{2}
+aΩu2r+(ϵμ)Ωu2r+1+c1(1ϵ)a1(Ωur)2,\displaystyle+a\int_{\Omega}u^{2r}+(\epsilon-\mu)\int_{\Omega}u^{2r+1}+c_{1}\left(\frac{1}{\epsilon}\right)^{a_{1}}\left(\int_{\Omega}u^{r}\right)^{2}, (5.6)

for all ϵ(0,1)\epsilon\in(0,1). Lemma 3.3 asserts that vv is in L((0,T);W1,(Ω))L^{\infty}\left((0,T);W^{1,\infty}(\Omega)\right). Thus,

sup0<t<TvL2=c2<,\sup_{0<t<T}\left\|\nabla v\right\|^{2}_{L^{\infty}}=c_{2}<\infty,

It follows from (5) and (5) that

ddtΩu2r+Ωu2r\displaystyle\frac{d}{dt}\int_{\Omega}u^{2r}+\int_{\Omega}u^{2r} 2r(2r1r2+3ϵ)Ω|ur|2+2r(ϵμ)Ωu2r+1\displaystyle\leq 2r\left(-\frac{2r-1}{r^{2}}+3\epsilon\right)\int_{\Omega}|\nabla u^{r}|^{2}+2r(\epsilon-\mu)\int_{\Omega}u^{2r+1}
+[(2r1)22rϵχ2c2+2ra+1]Ωu2r+2rc1(1ϵ)a1(Ωur)2.\displaystyle+\left[\frac{(2r-1)^{2}}{2r\epsilon}\chi^{2}c_{2}+2ra+1\right]\int_{\Omega}u^{2r}+2rc_{1}\left(\frac{1}{\epsilon}\right)^{a_{1}}\left(\int_{\Omega}u^{r}\right)^{2}. (5.7)

Substituting ϵ=min{r13r2,μ}\epsilon=\min\left\{\frac{r-1}{3r^{2}},\mu\right\} into (5) and noticing that 1ϵ<6r+μ1\frac{1}{\epsilon}<6r+\mu^{-1}, we obtain

ddtΩu2r+Ωu2r2Ω|ur|2+c3r2Ωu2r+2rc1(6r+μ1)a1(Ωur)2,\displaystyle\frac{d}{dt}\int_{\Omega}u^{2r}+\int_{\Omega}u^{2r}\leq-2\int_{\Omega}|\nabla u^{r}|^{2}+c_{3}r^{2}\int_{\Omega}u^{2r}+2rc_{1}\left(6r+\mu^{-1}\right)^{a_{1}}\left(\int_{\Omega}u^{r}\right)^{2}, (5.8)

where c3c_{3} are independent of rr. We apply Lemma 3.2, and substitute into (5.8) to obtain the following inequality for all η(0,1)\eta\in(0,1)

ddtΩu2r+Ωu2r(c3r2η2)Ω|ur|2+(c4r2ηn2+2rc1(6r+μ1)a1)(Ωur)2,\displaystyle\frac{d}{dt}\int_{\Omega}u^{2r}+\int_{\Omega}u^{2r}\leq(c_{3}r^{2}\eta-2)\int_{\Omega}|\nabla u^{r}|^{2}+\left(\frac{c_{4}r^{2}}{\eta^{\frac{n}{2}}}+2rc_{1}\left(6r+\mu^{-1}\right)^{a_{1}}\right)\left(\int_{\Omega}u^{r}\right)^{2}, (5.9)

where c4>0c_{4}>0 independent of r,ηr,\eta. By substituting η=min{1c3r2,1}\eta=\min\left\{\frac{1}{c_{3}r^{2}},1\right\} into this, we obtain the following

ddtΩu2r+Ωu2r[c5rn+2+2rc1(6r+μ1)a1](Ωur)2,\displaystyle\frac{d}{dt}\int_{\Omega}u^{2r}+\int_{\Omega}u^{2r}\leq\left[c_{5}r^{n+2}+2rc_{1}(6r+\mu^{-1})^{a_{1}}\right]\left(\int_{\Omega}u^{r}\right)^{2}, (5.10)

where c5c_{5} independent of rr. Since r>nr>n, there exists positive constants A¯,B\bar{A},B independent of rr such that

c5rn+2+2rc1(6r+μ1)a1A¯rB.c_{5}r^{n+2}+2rc_{1}(6r+\mu^{-1})^{a_{1}}\leq\bar{A}r^{B}.

Indeed, since a1>n+2a_{1}>n+2, we have

c5rn+2+2rc1(6r+μ1)a1[c5na1n1+2c1(6+1μn)a1]ra1+1.c_{5}r^{n+2}+2rc_{1}(6r+\mu^{-1})^{a_{1}}\leq\left[\frac{c_{5}}{n^{a_{1}-n-1}}+2c_{1}\left(6+\frac{1}{\mu n}\right)^{a_{1}}\right]r^{a_{1}+1}.

By substituting this into (5.10), we obtain (5.2). Applying Gronwall’s inequality, we have

Ωu2r(,t)max{A¯rBUr2r,Ωu02r}\int_{\Omega}u^{2r}(\cdot,t)\leq\max\left\{\bar{A}r^{B}U_{r}^{2r},\int_{\Omega}u_{0}^{2r}\right\}

This entails

u(,t)L2r(Ω)max{(A¯rB)12rUr,|Ω|12ru0L(Ω)},\displaystyle\left\|u(\cdot,t)\right\|_{L^{2r}(\Omega)}\leq\max\left\{(\bar{A}r^{B})^{\frac{1}{2r}}U_{r},|\Omega|^{\frac{1}{2r}}\left\|u_{0}\right\|_{L^{\infty}(\Omega)}\right\},

and further implies that

U2r(ArB)12rUr\displaystyle U_{2r}\leq({A}r^{B})^{\frac{1}{2r}}U_{r}

where A=max{A¯,|Ω|nB}{A}=\max\left\{\bar{A},|\Omega|n^{-B}\right\}. The proof of (5.1) is complete. ∎

The next lemma will fill in the gap of the iteration process from LrL^{r} to LnL^{n} when n2<rn\frac{n}{2}<r\leq n.

Lemma 5.3.

Let (u,v)(u,v) be the classical solution to (KS) on Ω×(0,Tmax)\Omega\times(0,T_{\rm max}) with maximal existence time Tmax(0,]T_{\rm max}\in(0,\infty]. If uL((0,Tmax);Lr(Ω))u\in L^{\infty}\left((0,T_{\rm max});L^{r}(\Omega)\right) for some n2<rn\frac{n}{2}<r\leq n, then uL((0,Tmax);L2r(Ω))u\in L^{\infty}\left((0,T_{\rm max});L^{2r}(\Omega)\right).

Proof.

By Lemma 3.3 we see that vv is in L((0,T);W1,q(Ω))L^{\infty}\left((0,T);W^{1,q}(\Omega)\right) for q<rnnrq<\frac{rn}{n-r} if r<nr<n and any q<q<\infty if r=nr=n. Set

λ:={2nn2 if n31r1 if n=2,\displaystyle\lambda:=\left\{\begin{matrix}\frac{2n}{n-2}&\text{ if }n\geq 3\\ \frac{1}{r-1}&\text{ if }n=2,\end{matrix}\right. (5.11)

and apply Holder’s inequality yields

Ωu2r|v|2(Ωu2r+λ)2r2r+λ(Ω|v|2(2r+λ)λ)λ2r+λ.\int_{\Omega}u^{2r}|\nabla v|^{2}\leq\left(\int_{\Omega}u^{2r+\lambda}\right)^{\frac{2r}{2r+\lambda}}\left(\int_{\Omega}|\nabla v|^{\frac{2(2r+\lambda)}{\lambda}}\right)^{\frac{\lambda}{2r+\lambda}}. (5.12)

Since n2<r<n\frac{n}{2}<r<n, we find that

2(2r+λ)λ<rnnr,\frac{2(2r+\lambda)}{\lambda}<\frac{rn}{n-r},

therefore vv belongs to L((0,T);W1,2(2r+λ)λ(Ω))L^{\infty}\left((0,T);W^{1,\frac{2(2r+\lambda)}{\lambda}}(\Omega)\right). Clearly, vv is in L((0,T);W1,2(2r+λ)λ(Ω))L^{\infty}\left((0,T);W^{1,\frac{2(2r+\lambda)}{\lambda}}(\Omega)\right) when r=nr=n. Thus,

sup0<t<TvL2(2r+λ)λ2=c5<,\sup_{0<t<T}\left\|\nabla v\right\|^{2}_{L^{\frac{2(2r+\lambda)}{\lambda}}}=c_{5}<\infty,

notice that

Ωu2r|Ω|λ2r+λ(Ωu2r+λ)2r2r+λ\int_{\Omega}u^{2r}\leq|\Omega|^{\frac{\lambda}{2r+\lambda}}\left(\int_{\Omega}u^{2r+\lambda}\right)^{\frac{2r}{2r+\lambda}} (5.13)

By the similar argument as in (5) and combine with (5.13), it follows that

ddtΩu2r+Ωu2r\displaystyle\frac{d}{dt}\int_{\Omega}u^{2r}+\int_{\Omega}u^{2r} 2r(2r1r2+3ϵ)Ω|ur|2+2r(ϵμ)Ωu2r+1\displaystyle\leq 2r\left(-\frac{2r-1}{r^{2}}+3\epsilon\right)\int_{\Omega}|\nabla u^{r}|^{2}+2r(\epsilon-\mu)\int_{\Omega}u^{2r+1}
+c6(Ωu2r+λ)2r2r+λ+2rc1(1ϵ)a1(Ωur)2,\displaystyle+c_{6}\left(\int_{\Omega}u^{2r+\lambda}\right)^{\frac{2r}{2r+\lambda}}+2rc_{1}\left(\frac{1}{\epsilon}\right)^{a_{1}}\left(\int_{\Omega}u^{r}\right)^{2}, (5.14)

where c6=[(2r1)22rϵχ2c5+2ra|Ω|λ2r+λ+1]c_{6}=\left[\frac{(2r-1)^{2}}{2r\epsilon}\chi^{2}c_{5}+2ra|\Omega|^{\frac{\lambda}{2r+\lambda}}+1\right]. Substitute ϵ=min{r13r2,μ}\epsilon=\min\left\{\frac{r-1}{3r^{2}},\mu\right\} into this yields

ddtΩu2r+Ωu2r2Ω|ur|2+c6(Ωu2r+λ)2r2r+λ+c7,\frac{d}{dt}\int_{\Omega}u^{2r}+\int_{\Omega}u^{2r}\leq-2\int_{\Omega}|\nabla u^{r}|^{2}+c_{6}\left(\int_{\Omega}u^{2r+\lambda}\right)^{\frac{2r}{2r+\lambda}}+c_{7}, (5.15)

where c7:=2rc1Ur2rc_{7}:=2rc_{1}U_{r}^{2r}. Apply GN inequality, and then Young’s inequality yields

c6(Ωu2r+λ)2r2r+λ\displaystyle c_{6}\left(\int_{\Omega}u^{2r+\lambda}\right)^{\frac{2r}{2r+\lambda}} c6CGN(Ω|ur|2)s(Ωur)2(1s)+c6CGN(Ωur)2\displaystyle\leq c_{6}C_{GN}\left(\int_{\Omega}|\nabla u^{r}|^{2}\right)^{s}\left(\int_{\Omega}u^{r}\right)^{2(1-s)}+c_{6}C_{GN}\left(\int_{\Omega}u^{r}\right)^{2}
c6CGNUr2r(1s)(Ω|ur|2)s+c6CGNUr2r\displaystyle\leq c_{6}C_{GN}U_{r}^{2r(1-s)}\left(\int_{\Omega}|\nabla u^{r}|^{2}\right)^{s}+c_{6}C_{GN}U_{r}^{2r}
Ω|ur|2+c8,\displaystyle\leq\int_{\Omega}|\nabla u^{r}|^{2}+c_{8}, (5.16)

where s=2n(r+λ)(n+2)(2r+λ)(0,1)s=\frac{2n(r+\lambda)}{(n+2)(2r+\lambda)}\in(0,1), and c8=c8(r,n,Ω,Ur)c_{8}=c_{8}(r,n,\Omega,U_{r}). From (5.15) and (5), we obtain

ddtΩu2r+Ωu2rc9,\displaystyle\frac{d}{dt}\int_{\Omega}u^{2r}+\int_{\Omega}u^{2r}\leq c_{9}, (5.17)

where c9=c7+c8c_{9}=c_{7}+c_{8}. Apply Gronwall’s inequality to this yields

supt(0,Tmax)u(,t)L2r(Ω)max{u0L(Ω),c9}.\sup_{t\in(0,T_{\rm max})}\left\|u(\cdot,t)\right\|_{L^{2r}(\Omega)}\leq\max\left\{\left\|u_{0}\right\|_{L^{\infty}(\Omega)},c_{9}\right\}.

and thereby conclude the proof. ∎

We are now ready to prove Theorem 5.1.

Proof of Theorem 5.1.

If r0(n2,n]r_{0}\in\left(\frac{n}{2},n\right], then

supt(0,Tmax)u(,t)L2r0(Ω)<.\sup_{t\in(0,T_{\rm max})}\left\|u(\cdot,t)\right\|_{L^{2r_{0}}(\Omega)}<\infty.

due to Lemma 5.3. Since 2r0>n2r_{0}>n meets the criteria of Lemma 5.2, the following inequality holds

U2k+1r0(A(2kr0)B)12k+1r0U2kr0.\displaystyle U_{2^{k+1}r_{0}}\leq\left(A(2^{k}r_{0})^{B}\right)^{\frac{1}{2^{k+1}r_{0}}}U_{2^{k}r_{0}}. (5.18)

for all integers k1k\geq 1. If r0>nr_{0}>n, then (5.18) is an immediate consequence of Lemma 5.2. We take log\log of the above inequality to obtain

lnU2k+1r0ak+lnU2kr0,\ln U_{2^{k+1}r_{0}}\leq a_{k}+\ln U_{2^{k}r_{0}},

where

ak\displaystyle a_{k} =lnA2k+1r0+Bkln22k+1r0+Blnr02k+1r0.\displaystyle=\frac{\ln A}{2^{k+1}r_{0}}+\frac{Bk\ln 2}{2^{k+1}r_{0}}+\frac{B\ln r_{0}}{2^{k+1}r_{0}}.

One can verify that

k=1ak=ln(A(4r0)B)2r0.\displaystyle\sum_{k=1}^{\infty}a_{k}=\frac{\ln\left(A(4r_{0})^{B}\right)}{2r_{0}}.

Thus, we obtain

U2k+1rA12r0(4r0)B2r0U2r0U_{2^{k+1}r}\leq A^{\frac{1}{2r_{0}}}(4r_{0})^{\frac{B}{2r_{0}}}U_{2r_{0}} (5.19)

for all k1k\geq 1. Send kk\to\infty yields

UA12r0(4r0)B2r0U2r0.U_{\infty}\leq A^{\frac{1}{2r_{0}}}(4r_{0})^{\frac{B}{2r_{0}}}U_{2r_{0}}. (5.20)

This asserts that uL((0,Tmax);L(Ω))u\in L^{\infty}\left((0,T_{\rm max});L^{\infty}(\Omega)\right), and thereafter Lemma 3.3 yields that vL((0,Tmax);W1,(Ω))v\in L^{\infty}\left((0,T_{\rm max});W^{1,\infty}(\Omega)\right). ∎

6 Proof of main theorems

We are now ready to prove our main results. Throughout this section, unless specified otherwise, the notation CC represents constants that may vary from time to time. Let us begin with the proof of the parabolic-elliptic system.

Proof of Theorem 1.1.

In case μ>n2nχα\mu>\frac{n-2}{n}\chi\alpha, we obtain χα(χαμ)+>n2\frac{\chi\alpha}{(\chi\alpha-\mu)_{+}}>\frac{n}{2}. We first apply Lemma 4.1 to have uL((0,Tmax);Lq(Ω))u\in L^{\infty}\left((0,T_{\rm max});L^{q}(\Omega)\right) for any q(n2,χα(χαμ)+)q\in\left(\frac{n}{2},\frac{\chi\alpha}{(\chi\alpha-\mu)_{+}}\right) and thereby conclude that uL((0,Tmax);L(Ω))u\in L^{\infty}\left((0,T_{\rm max});L^{\infty}(\Omega)\right) thank to Theorem 5.1.
When μ=n2nχα\mu=\frac{n-2}{n}\chi\alpha, let w=un4w=u^{\frac{n}{4}}, and apply trace embedding Theorem W1,1(Ω)L1(Ω)W^{1,1}(\Omega)\to L^{1}(\partial\Omega), we have

Ωw2+4(p1)n\displaystyle\int_{\partial\Omega}w^{2+\frac{4(p-1)}{n}} CΩw2+4(p1)n+C(2+4(p1)n)Ωw1+4(p1)n|w|\displaystyle\leq C\int_{\Omega}w^{2+\frac{4(p-1)}{n}}+C\left(2+\frac{4(p-1)}{n}\right)\int_{\Omega}w^{1+\frac{4(p-1)}{n}}|\nabla w|
CΩw2+4(p1)n+3CΩw1+4(p1)n|w|\displaystyle\leq C\int_{\Omega}w^{2+\frac{4(p-1)}{n}}+3C\int_{\Omega}w^{1+\frac{4(p-1)}{n}}|\nabla w|
CΩw2+4(p1)n+ϵ2Ω|w|2+C(ϵ)Ωw2+8(p1)n,\displaystyle\leq C\int_{\Omega}w^{2+\frac{4(p-1)}{n}}+\frac{\epsilon}{2}\int_{\Omega}|\nabla w|^{2}+C(\epsilon)\int_{\Omega}w^{2+\frac{8(p-1)}{n}}, (6.1)

where the last inequality comes from Young’s inequality for any arbitrary ϵ>0\epsilon>0. By Lemma 3.1, we have

Ωw2+8(p1)nC(Ω|w|2)p¯a2(Ωw4n)np¯(1a)4+C(Ωw4n)np¯4,\displaystyle\int_{\Omega}w^{2+\frac{8(p-1)}{n}}\leq C\left(\int_{\Omega}|\nabla w|^{2}\right)^{\frac{\bar{p}a}{2}}\left(\int_{\Omega}w^{\frac{4}{n}}\right)^{\frac{n\bar{p}(1-a)}{4}}+C\left(\int_{\Omega}w^{\frac{4}{n}}\right)^{\frac{n\bar{p}}{4}}, (6.2)

where

p¯\displaystyle\bar{p} =2+8(p1)n,\displaystyle=2+\frac{8(p-1)}{n},
a\displaystyle a =n2(n+4(p1)+2)(n+4(p1))(n22n+4),\displaystyle=\frac{n^{2}(n+4(p-1)+2)}{(n+4(p-1))(n^{2}-2n+4)},
p¯a2\displaystyle\frac{\bar{p}a}{2} =n2+4(p1)n2nn22n+4<1, since p<1+1n.\displaystyle=\frac{n^{2}+4(p-1)n-2n}{n^{2}-2n+4}<1,\qquad\text{ since }p<1+\frac{1}{n}.

We make use of uniformly boundedness of Ωu\int_{\Omega}u and then apply Young’s inequality into (6.2) to obtain:

Ωw2+8(p1)nϵ2Ω|w|2+C(ϵ),\displaystyle\int_{\Omega}w^{2+\frac{8(p-1)}{n}}\leq\frac{\epsilon}{2}\int_{\Omega}|\nabla w|^{2}+C(\epsilon), (6.3)

for any ϵ>0\epsilon>0. We apply Young’s inequality in (6) and then use (6.2) to have

Ωw2+4(p1)nϵΩ|w|2+C(ϵ).\displaystyle\int_{\partial\Omega}w^{2+\frac{4(p-1)}{n}}\leq\epsilon\int_{\Omega}|\nabla w|^{2}+C(\epsilon). (6.4)

This implies that

Ωun2+p1ϵΩ|un4|2+C(ϵ).\displaystyle\int_{\partial\Omega}u^{\frac{n}{2}+p-1}\leq\epsilon\int_{\Omega}|\nabla u^{\frac{n}{4}}|^{2}+C(\epsilon). (6.5)

Substitute r=n4r=\frac{n}{4} and μ=n2nχα\mu=\frac{n-2}{n}\chi\alpha into (4), we have

ddtΩun24(n2)nΩ|un4|2+n2Ωun2+p1𝑑S+na2Ωun2.\displaystyle\frac{d}{dt}\int_{\Omega}u^{\frac{n}{2}}\leq-\frac{4(n-2)}{n}\int_{\Omega}|\nabla u^{\frac{n}{4}}|^{2}+\frac{n}{2}\int_{\partial\Omega}u^{\frac{n}{2}+p-1}\,dS+\frac{na}{2}\int_{\Omega}u^{\frac{n}{2}}. (6.6)

We apply Lemma 3.1 and uniform boundedness of Ωu\int_{\Omega}u to obtain that

Ωun2ϵΩ|un4|2+C(ϵ).\displaystyle\int_{\Omega}u^{\frac{n}{2}}\leq\epsilon\int_{\Omega}|\nabla u^{\frac{n}{4}}|^{2}+C(\epsilon). (6.7)

This, together with (6.5) and (6.6) with ϵ\epsilon sufficiently small implies that

ddtΩun2+Ωun2C.\frac{d}{dt}\int_{\Omega}u^{\frac{n}{2}}+\int_{\Omega}u^{\frac{n}{2}}\leq C.

Thus, by Gronwall’s inequality we obtain that

supt(0,Tmax)Ωun2<.\displaystyle\sup_{t\in(0,T_{\rm max})}\int_{\Omega}u^{\frac{n}{2}}<\infty. (6.8)

For any ϵ(0,1)\epsilon\in(0,1), we choose n4<r<n4+nϵ4χα\frac{n}{4}<r<\frac{n}{4}+\frac{n\epsilon}{4\chi\alpha} and substitute into (4) to obtain that

ddtΩu2r+Ωu2r[2rϵ2(2r1)r]Ω|ur|2+3ϵΩu2r+1+C.\displaystyle\frac{d}{dt}\int_{\Omega}u^{2r}+\int_{\Omega}u^{2r}\leq\left[2r\epsilon-\frac{2(2r-1)}{r}\right]\int_{\Omega}|\nabla u^{r}|^{2}+3\epsilon\int_{\Omega}u^{2r+1}+C. (6.9)

We apply Lemma 3.1 and (6.8) to obtain

Ωu2r+1\displaystyle\int_{\Omega}u^{2r+1} C(Ω|ur|2)(Ωun2)2n+(Ωu)2r+1\displaystyle\leq C\left(\int_{\Omega}|\nabla u^{r}|^{2}\right)\left(\int_{\Omega}u^{\frac{n}{2}}\right)^{\frac{2}{n}}+\left(\int_{\Omega}u\right)^{2r+1}
CΩ|ur|2+C,\displaystyle\leq C\int_{\Omega}|\nabla u^{r}|^{2}+C, (6.10)

where CC is independent of rr. Therefore, we have

ddtΩu2r+Ωu2r[2rϵ+3Cϵ2(2r1)r]Ω|ur|2+C.\displaystyle\frac{d}{dt}\int_{\Omega}u^{2r}+\int_{\Omega}u^{2r}\leq\left[2r\epsilon+3C\epsilon-\frac{2(2r-1)}{r}\right]\int_{\Omega}|\nabla u^{r}|^{2}+C. (6.11)

We now have to choose ϵ\epsilon and r>n4r>\frac{n}{4} such that

4χαn(rn4)<ϵ<2(2r1)r(2r+C),\displaystyle\frac{4\chi\alpha}{n}\left(r-\frac{n}{4}\right)<\epsilon<\frac{2(2r-1)}{r(2r+C)}, (6.12)

which is possible for any rr satisfying

n4<r<n4(4(n2)n2+n2χα+3C+1).\frac{n}{4}<r<\frac{n}{4}\left(\frac{4(n-2)}{\frac{n}{2}+\frac{n}{2\chi\alpha}+3C}+1\right).

This, together with (6.11), (6.12) implies that there exists some r0>n2r_{0}>\frac{n}{2} such that uL((0,Tmax);Lr0(Ω))u\in L^{\infty}\left((0,T_{\rm max});L^{r_{0}}(\Omega)\right). We finally complete the proof by applying Theorem 5.1

Next we prove the main theorems for the parabolic-parabolic system in two- and three-dimensional space.

Proof of Theorem 1.2 and Theorem 1.3.

Theorem 1.2 and Theorem 1.3 are immediate consequences of Lemma 4.3, Lemma 4.4 and Theorem 5.1. ∎

Acknowledgement

The author would like to express deepest appreciation to Professor Zhengfang Zhou for his unwavering encouragement, support, and numerous fruitful discussions. The author is profoundly grateful to Professor Michael Winkler for his valuable comments and suggestions. Lastly, the author would like to express gratitude to the referee for carefully reading the manuscript and providing constructive feedback as well as helpful references.

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