Global existence and boundedness of a chemotaxis model with indirect production and general kinetic function

Abstract

This paper is devoted to the chemotaxis model with indirect production and general kinetic function

$$\begin{aligned} \left\{ \begin{aligned}&u_t=\Delta u-\chi \nabla \cdot (u\nabla v)+f(u),\qquad&x\in \Omega ,\,t>0,\\&v_t=\Delta v-v+w,\qquad&x\in \Omega ,\,t>0,\\&\tau w_t+\lambda w=g(u),\qquad&x\in \Omega ,\,t>0, \end{aligned} \right. \end{aligned}$$

in a bounded domain $\Omega \subset \mathbb {R}^n(n\le 3)$ with smooth boundary $\partial \Omega$, where $\chi , \tau , \lambda$ are given positive parameters, f and g are known functions. We find several explicit conditions involving the kinetic function f, g, the parameters $\chi$, $\lambda$, and the initial data $\Vert u_0\Vert _{L^1(\Omega )}$ to ensure the global-in-time existence and uniform boundedness for the corresponding 2D/3D Neumann initial-boundary value problem. Particularly, when $f\equiv 0$, and g is a linear function, the global bounded classical solutions to the corresponding 2D Neumann initial-boundary value problem with arbitrarily large initial data and chemotactic sensitivity are established. Our results partially extend the results of Hu and Tao (Math Models Methods Appl Sci 26:2111–2128, 2016), Tao and Winkler (J Eur Math Soc 19:3641–3678, 2017), etc.

1 Introduction and main results

Motivated by the model for mountain pine beetle dispersal proposed by Strohm–Tyson–Powell (see Ref.[1]), the present work devotes to the chemotaxis model with indirect signal production and general kinetic function

$$\begin{aligned} \left\{ \begin{aligned}&u_t=\Delta u-\chi \nabla \cdot (u\nabla v)+f(u),\qquad&x\in \Omega ,\,t>0,\\&v_t=\Delta v-v+w,\qquad&x\in \Omega ,\,t>0,\\&\tau w_t=-\lambda w+g(u),\qquad&x\in \Omega ,\,t>0,\\&\partial _\nu u=\partial _\nu v=0,\qquad&x\in \partial \Omega ,\,t>0,\\&u(x,0)=u_0(x),\,v(x,0)=v_0(x),\,\tau w(x,0)=\tau w_0(x),\qquad&x\in \Omega , \end{aligned} \right. \end{aligned}$$

(1.1)

in a bounded domain $\Omega \subset \mathbb {R}^n(n\le 3)$ with smooth boundary $\partial \Omega$, where $\chi ,\,\lambda$ and $\tau$ are given positive parameters, the kinetic functions f and g belong to $C^{1}[0, +\infty )$ satisfying $f(0)\ge 0$, $g(s)\ge 0$ for all $s\ge 0$.

The initial data $u_0(x)$, $v_0(x)$, and $w_0(x)$ are given nonnegative functions satisfying:

$$\begin{aligned} u_0(x)\in C(\overline{\Omega }),\,v_0(x)\in C^1(\overline{\Omega }),\,w_0(x)\in C^{1}(\overline{\Omega }),\,u_0\ge 0,\,v_0\ge 0,\,w_0\ge 0\,\,\text {and}\,\,u_0\not \equiv 0. \end{aligned}$$

(1.2)

System (1.1) describes the spatio-temporal evolution of the mountain pine beetle (MPB) in forest habitat, where u represents the density of flying MPB, w denotes the density of nesting MPB which transited from the flying MPB with transition rate g, and v stands for the concentration of beetle pheromone secreted by nesting MPB, which attracts flying MPB to aggregate. The flying MPB undergo birth and death, and some of them may convert into nesting MPB as aforesaid, all these kinetic behaviors are described by the kinetic function f. Different from the well-known Keller–Segel model, the chemotactic cue, in such model, is not directly produced by flying MPB themselves, but indirectly secreted by nesting MPB. Actually, such indirect signaling mechanism also occurs in the coral fertilization, see [2, 3], etc.

Indeed, the model (1.1) with $\tau =0$ can be simplified as

$$\begin{aligned} \left\{ \begin{aligned}&u_t=\Delta u-\chi \nabla \cdot (u\nabla v)+f(u),\qquad&x\in \Omega ,\,t>0,\\&v_t=\Delta v-v+\frac{1}{\lambda }g(u),\qquad&x\in \Omega ,\,t>0,\\&\partial _\nu u=\partial _\nu v=0,\qquad&x\in \partial \Omega ,\,t>0,\\&u(x,0)= u_{0}(x),\,v(x,0)= v_{0}(x),\qquad&x\in \Omega . \end{aligned} \right. \end{aligned}$$

(1.3)

When $f(u)\equiv 0$ and $g(u)=u$, the model (1.3) becomes the classical Keller–Segel model, the solution of which may blow up in finite time or infinite time if $n\ge 2$ (see, e.g., [4,5,6,7,8]). However, such blowup phenomena may be prevented by some appropriate kinetic function f and g, for example, $g(u)=u$ and $f(u)=\mu u(1-u)$ with any $\mu >0$ in one or two-dimensional space (see [9,10,11], etc.), and $\mu$ appropriately large in multi-dimensional space(see [12,13,14], etc.). Particularly, when $n=3$, for any $\mu >0$, only certain weak global solutions were proved to exist in [15], the problem that the classical solution exists globally or not is still open. For model (1.3) with $g(u)=u$ in two-dimensional smooth bounded domain, Xiang [16] proved that the sub-logistic source f can also prevent the blowup of solutions. Very recently, in [17], the author of this paper studied the three-dimensional initial-boundary value problem (1.3) with function f satisfying $f(s)\le a-bs^\alpha (a\ge 0,\,b>0,\,\alpha \ge 1)$ and g satisfying $g(s)\le Ks^\gamma (K>0,\,\gamma >0)$, wherein, several constraints on parameters $\gamma$ and $\alpha$ ensuring the global existence of classical solutions were obtained.

Now we return to model (1.1). We first remark that the initial-boundary value problem (1.1) with kinetic function $f(u)=\mu u(1-u)$ and $g(u)=u$ in three-dimensional smooth bounded domain was studied by Hu-Tao in [18], it was shown that the global bounded classical solution exists for any $\mu >0$, which strikingly differs from the corresponding two-component chemotaxis-growth system, for the latter, as we aforementioned, the finite-time blowup can be prevented only when $\mu$ is appropriately large. Moreover, the model (1.1) with $f(u)\equiv 0$, $g(u)=u$ and the second equation replaced by the equation $0=\Delta v-\frac{\int \limits _\Omega w(x,t)\mathrm{d}x}{|\Omega |} +w$, was studied by Tao-Winkler in [19], therein, it was also shown that the indirect mechanism of signal production may effectively prevent the occurrence of the finite-time blowup of solutions. Very recently, in [20], Fujie-Senba studied a two-chemical substances chemotaxis system which is a counterpart of model (1.1) with $f(u)\equiv 0$, and the equation $w_t=\Delta w-w+u$ instead of the third equation in (1.1). Their results show that in the lower-dimensional case($n\le 3)$, the diffusivity of the chemical w is helpful to prevent the finite-time and infinite-time blowup of solutions.

To our knowledge, the kinetic function f and g involved in the current mathematical analysis on MPB model (see, for example, [18] and [19]), are expressed by quadratic functions and linear functions, respectively. However, in the original model proposed by Strohm–Tyson–Powell (see Ref.[1]), the kinetic function f and g may not be represented by a quadratic, linear function, that is to say, f and $g^2$ are not necessarily infinite quantities of the same order. It is natural to ask whether there are other forms of functions f and g that make the solution of model (1.1) still global and bounded? The purpose of this paper is to solve this problem. Our main results are as follows.

Theorem 1.1

Let $n=2$, $\chi >0$, $\lambda >0$, $\tau >0$, and suppose that the initial data $(u_0,v_0,w_0)$ satisfies (1.2), the kinetic function f and g belong to $C^{1}[0,+\infty )$ satisfying $f(0)\ge 0$, $g(s)\ge 0$ for all $s\ge 0$. If one of the following conditions holds:

• $(i) \exists \alpha >0, \,s.t.\,\sup _{s\ge 0}\{f(s)+\alpha s\}<\infty ,\,\, \liminf _{s\rightarrow \infty }\Big \{-\frac{f(s)\ln s}{g^2(s)}\Big \}=:\mu \in (\frac{2\chi ^2C_{GN}^4m_1}{\lambda ^2},\infty ]$,

• $(ii) \exists \alpha >0, \,s.t.\,\sup _{s\ge 0}\{f(s)+\alpha s\}<\infty \,\text {or}\, f\equiv 0,\, \lim _{s\rightarrow \infty }\frac{g(s)}{s}=0$,

• $(iii) \exists \alpha >0, \,s.t.\,\sup _{s\ge 0}\{f(s)+\alpha s\}<\infty \,\text {or} \,f\equiv 0,\,\limsup _{s\rightarrow \infty }\frac{g(s)}{s}=\gamma \in (0,\infty ) \, \text {satisfying} \,\,(\frac{4\chi \gamma }{\lambda }+\frac{\chi ^2}{\lambda ^2})C_{GN}^4m_1<1$,

where $C_{GN}$ is the Gagliardo–Nirenberg constant given in (3.1) and $m_1$ is a positive constant given in (2.5), then the initial-boundary value problem (1.1) has a unique global-in-time classical solution which is uniformly bounded in the following sense: there exists $C>0$ such that

$$\begin{aligned} \Vert u\Vert _{L^\infty (\Omega )}+\Vert v\Vert _{W^{1,\infty }(\Omega )}+\Vert w\Vert _{L^\infty (\Omega )}\le C \quad \text {for all}\,\,x\in \Omega \,\, \text {and}\,\,t>0. \end{aligned}$$

Remark 1.2

We remark that our results are independent of the explicit value of $\tau$. The condition (i) implies that f(s) and $g^2(s)$ do not have to be infinite quantities of the same order as s goes to infinity. Particularly, when $g(s)=s$, the condition (i) can be rewritten as

$$\begin{aligned} \liminf _{s\rightarrow \infty }\Big \{-\frac{f(s)\ln s}{s^2}\Big \}=:\mu \in (\frac{2\chi ^2C_{GN}^4m_1}{\lambda ^2},\infty ], \end{aligned}$$

which is consistent with that in [16].

Remark 1.3

We can get a by-product from (ii) that even if $f(s)\equiv 0$, the sub-linear g(s), the typical example is $g(s)=s^a\ln ^b(s+1)$ with $a<1,\,b\in \mathbb {R}$ for $s\ge 0$, still can suppress the blowup of model (1.1). Moreover, we see that if $f(s)\equiv 0$ which implies $m_1=\Vert u_0\Vert _{L^1(\Omega )}$ in (2.5), $g(s)=s$, then (iii) can be rewritten as

$$\begin{aligned} (\frac{4\chi }{\lambda }+\frac{\chi ^2}{\lambda ^2})C_{GN}^4\Vert u_0\Vert _{L^1(\Omega )}<1, \end{aligned}$$

(1.4)

this implies that for any large initial mass $\Vert u_0\Vert _{L^1(\Omega )}$ and chemotactic sensitivity $\chi$, so long as the decay rate $\lambda$ is sufficiently large to satisfy (1.4), then the classical solution to (1.1) still exists globally.

In the three-dimensional setting, the global classical solution to model (1.1) can be constructed under some stronger conditions.

Theorem 1.4

Let $n=3$, $\chi >0$, $\lambda >0$, $\tau >0$, and suppose that the initial data $(u_0,v_0,w_0)$ satisfies (1.2), the function f and the function g belong to $C^{1}[0, +\infty )$ satisfying $f(0)\ge 0$, $g(s)\ge 0$ for all $s\ge 0$. Assume one of the following conditions holds:

• $(i) \liminf _{s\rightarrow \infty }\Big \{-\frac{f(s)}{s^2}\Big \}=:\mu \in (0,\infty ],\,\,\limsup _{s\rightarrow \infty }\frac{g(s)}{s}=\gamma \in [0,\infty )$,

• $(ii)\liminf _{s\rightarrow \infty }\Big \{-\frac{f(s)s}{g^3(s)}\Big \}=:\mu \in (0,\infty ],\,\,\limsup _{s\rightarrow \infty }\frac{g(s)}{s}=\infty$.

Then, the initial-boundary value problem (1.1) has a unique global-in-time classical solution which is uniformly bounded in the following sense: there exists $C>0$ such that

$$\begin{aligned} \Vert u\Vert _{L^\infty (\Omega )}+\Vert v\Vert _{W^{1,\infty }(\Omega )}+\Vert w\Vert _{L^\infty (\Omega )}\le C \quad \text {for all}\,\,x\in \Omega \,\, \text {and}\,\,t>0. \end{aligned}$$

Remark 1.5

We remark that the condition (i) implies that system (1.1) with $f(u)=\mu u(1-u)$ and $g(u)=u$ has a global bounded classical solution for any $\mu >0$, which is consistent with the result on global existence obtained in [18]. It also shows that the indirect production mechanism is helpful for the global existence of solutions to model (1.1).

2 Preliminaries

In this section, we first state the local existence of classical solutions of (1.1) which can be established by adapting methods that are well-established in the context of parabolic–parabolic-ODE taxis models(see Refs. [21,22,23], etc). Then, we recall the Gagliardo–Nirenberg interpolation inequality, and establish some estimates independent of the space dimension n, which shall be used frequently in the sequel sections.

Lemma 2.1

Let $\Omega \subset \mathbb {R}^n(n\ge 1)$ be a bounded domain with smooth boundary, $\chi$ and $\lambda$ be positive constants. Assume that the function f and the function g belong to $C^{1}[0, +\infty )$ satisfying $f(0)\ge 0$, $g(0)\ge 0$. Then, for any nonnegative initial data $(u_0(x),\,v_0(x),\,w_0(x))\in C^0(\overline{\Omega })\times C^1(\overline{\Omega })\times C^1(\overline{\Omega })$, the initial-boundary value problem (1.1) has a unique local-in-time nonnegative classical solution (u, v, w), in the sense that

$$\begin{aligned} \begin{aligned}&u\in C^0(\overline{\Omega }\times [0,T^*))\cap C^{2,1}(\overline{\Omega }\times (0,T^*)),\\&v\in C^0(\overline{\Omega }\times [0,T^*))\cap C^{2,1}(\overline{\Omega }\times (0,T^*)),\\&w\in C^{0,1}(\overline{\Omega }\times [0,T^*)). \end{aligned} \end{aligned}$$

Here, $T^*$ denotes the maximal existence time. Moreover, if $T^*<\infty$, then

$$\begin{aligned} \Vert u(\cdot ,t)\Vert _{L^\infty (\Omega )}\rightarrow \infty ,\qquad \text {as}\,\, t\rightarrow T^{*-}. \end{aligned}$$

(2.1)

Lemma 2.2

(Gagliardo–Nirenberg interpolation inequality [24, 25]) Let $\Omega \subset \mathbb {R}^n(n\ge 1)$ be a bounded smooth domain and let $p\ge 1$ and $q\in (0,p)$. Then, there exists a positive constant $C_{GN}=C(p,q,n,\Omega )$ such that

$$\begin{aligned} \Vert u\Vert _{L^p(\Omega )}\le C_{GN}(\Vert \nabla u\Vert _{L^2(\Omega )}^\delta \Vert u\Vert _{L^q(\Omega )}^{(1-\delta )}+\Vert u\Vert _{L^r(\Omega )}),\qquad \forall u\in H^1(\Omega )\cap L^q(\Omega ), \end{aligned}$$

where $r>0$ is arbitrary and $\delta \in (0,1)$ is given by

$$\begin{aligned} \frac{1}{p}=\delta \left( \frac{1}{2}-\frac{1}{n}\right) +\frac{1-\delta }{q}\Leftrightarrow \delta =\frac{\frac{n}{q}-\frac{n}{p}}{1-\frac{n}{2}+\frac{n}{q}}. \end{aligned}$$

At the end of this section, we state two estimates which are independent of the space dimension. The one is the $L^1$ estimate of u which is elementary but very important.

Lemma 2.3

Assume f satisfies

$$\begin{aligned} \sup _{s\ge 0}\{f(s)+\alpha s\}<\infty \end{aligned}$$

(2.2)

with some positive constant $\alpha$, then the solution component u of (1.1) satisfies

$$\begin{aligned} \Vert u(\cdot ,t)\Vert _{L^1(\Omega )}\le \max \left\{ \Vert u_0(x)\Vert _{L^1(\Omega )},\frac{\sup _{s\ge 0}\{f(s)+\alpha s\}|\Omega |}{\alpha }\right\} \qquad \text {for\,\, all}\,\, t\in [0,T^*). \end{aligned}$$

(2.3)

Proof

Integrating the first equation in (1.1) with respect to $x\in \Omega$, we have

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\int \limits _\Omega u=\int \limits _\Omega f(u) \qquad \text {for all}\, t\in (0,T^*), \end{aligned}$$

which together with (2.2) then yields

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\int \limits _\Omega u+\alpha \int \limits _\Omega u=\int \limits _\Omega (f(u)+\alpha u)\le \sup _{s\ge 0}\{f(s)+\alpha s\}|\Omega | \qquad \text {for all}\, t\in (0,T^*), \end{aligned}$$

from this, one can easily get the desired estimate (2.5). $\square$

Remark 2.4

If $f(u)\equiv 0$, then the solution component u of (1.1) is mass conserved, i.e.,

$$\begin{aligned} \Vert u(\cdot ,t)\Vert _{L^1(\Omega )}=\Vert u_0(x)\Vert _{L^1(\Omega )}\qquad \text {for\,\, all}\,\, t\in [0,T^*). \end{aligned}$$

(2.4)

For the sake of simplicity , we denote

$$\begin{aligned} m_1:= \left\{ \begin{aligned}&\max \left\{ \Vert u_0(x)\Vert _{L^1(\Omega )},\frac{\sup _{s\ge 0}\{f(s)+\alpha s\}|\Omega |}{\alpha }\right\} ,\quad \text {when}\,\,\sup _{s\ge 0}\{f(s)+\alpha s\}<\infty \, \text {with some}\,\alpha >0,\\&\Vert u_0(x)\Vert _{L^1(\Omega )},\quad \text {when}\,\,f(u)\equiv 0, \end{aligned} \right. \end{aligned}$$

(2.5)

and then we can deduce from (2.3) and (2.4) that the solution component u of (1.1) satisfies

$$\begin{aligned} \Vert u(\cdot ,t)\Vert _{L^1(\Omega )}\le m_1\qquad \text {for\,\, all}\,\, t\in [0,T^*). \end{aligned}$$

The other estimate independent of the space dimension is the following coupled estimate of $\int \limits _\Omega u\ln u$ and $\int \limits _\Omega |\nabla v|^2$ as well as $\int \limits _\Omega w^2$, which is crucial for our purpose.

Lemma 2.5

For any $\epsilon >0$, the solution (u, v, w) of (1.1) satisfies

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&\Big \{\int \limits _\Omega u\ln u+\frac{\chi ^2}{4\epsilon }\int \limits _\Omega |\nabla v|^2+\frac{\tau \chi ^2}{2\lambda \epsilon }\int \limits _\Omega w^2\Big \}+\int \limits _\Omega \frac{|\nabla u|^2}{u}+\int \limits _\Omega u\ln u+\frac{\chi ^2}{2\epsilon }\int \limits _\Omega |\nabla v|^2+ \frac{\chi ^2}{4\epsilon }\int \limits _\Omega w^2\\&\le 2\epsilon \int \limits _\Omega u^2+\frac{\chi ^2}{2\lambda ^2\epsilon }\int \limits _\Omega g^2(u)+\int \limits _\Omega f(u)(\ln u+1)+K(\epsilon )\qquad \text {for all}\,\,\,t\in (0,T^*), \end{aligned} \end{aligned}$$

(2.6)

where

$$\begin{aligned} K(\epsilon ):=\sup _{s>0}\{s\ln s-\epsilon s^2\} \end{aligned}$$

is a nonnegative finite number depending on $\epsilon$.

Proof

Testing the first equation in (1.1) by $\ln u+1$, one can easily get

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\int \limits _\Omega u\ln u+\int \limits _\Omega \frac{|\nabla u|^2}{u}&=-\chi \int \limits _\Omega u\Delta v+\int \limits _\Omega f(u)(\ln u+1)\\&\le \epsilon \int \limits _\Omega u^2+\frac{\chi ^2}{4\epsilon }\int \limits _\Omega |\Delta v|^2+\int \limits _\Omega f(u)(\ln u+1) \end{aligned} \end{aligned}$$

(2.7)

for all $t\in (0,T^*)$, where, in the last inequality, we used Young’s inequality with $\epsilon$.

In order to deal with the integral $\int \limits _\Omega |\Delta v|^2$ in (2.7), we multiply the second equation in (1.1) by $-\Delta v$, integrate by parts over $\Omega$ and then use Young’s inequality to get

$$\begin{aligned} \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\int \limits _\Omega |\nabla v|^2+\int \limits _\Omega |\nabla v|^2+\frac{1}{2}\int \limits _\Omega |\Delta v|^2\le \frac{1}{2}\int \limits _\Omega w^2\quad \text {for all}\,\,t\in (0,T^*). \end{aligned}$$

(2.8)

To cope with the integral on the right of (2.8), testing the third equation in (1.1) by 2w we have

$$\begin{aligned} \tau \frac{\mathrm{d}}{\mathrm{d}t}\int \limits _\Omega w^2+2\lambda \int \limits _\Omega w^2=2\int \limits _\Omega wg(u)\quad \text {for all}\,\,t\in (0,T^*). \end{aligned}$$

Applying Young’s inequality to the integral on the right, we have

$$\begin{aligned} \int \limits _\Omega wg(u)\le \frac{\lambda }{2}\int \limits _\Omega w^2+\frac{1}{2\lambda }\int \limits _\Omega g^2(u) \end{aligned}$$

and then, we can obtain

$$\begin{aligned} \tau \frac{\mathrm{d}}{\mathrm{d}t}\int \limits _\Omega w^2+\lambda \int \limits _\Omega w^2\le \frac{1}{\lambda }\int \limits _\Omega g^2(u)\quad \text {for all}\,\,t\in (0,T^*). \end{aligned}$$

(2.9)

A linear combination $(2.7)+\frac{\chi ^2}{2\epsilon }\times (2.8)+\frac{\chi ^2}{2\lambda \epsilon }\times (2.9)$ then yields

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&\Big \{\int \limits _\Omega u\ln u+\frac{\chi ^2}{4\epsilon }\int \limits _\Omega |\nabla v|^2+\frac{\tau \chi ^2}{2\lambda \epsilon }\int \limits _\Omega w^2\Big \}+\int \limits _\Omega \frac{|\nabla u|^2}{u}+\frac{\chi ^2}{2\epsilon }\int \limits _\Omega |\nabla v|^2+ \frac{\chi ^2}{4\epsilon }\int \limits _\Omega w^2\\&\le \epsilon \int \limits _\Omega u^2+\frac{\chi ^2}{2\lambda ^2\epsilon }\int \limits _\Omega g^2(u)+\int \limits _\Omega f(u)(\ln u+1)\quad \text {for all}\,\,t\in (0,T^*). \end{aligned} \end{aligned}$$

(2.10)

Since

$$\begin{aligned} \lim _{s\rightarrow 0^+}s\ln s=0\,\,\text {and}\,\,\lim _{s\rightarrow +\infty }\frac{s\ln s}{s^2}=0, \end{aligned}$$

for the same $\epsilon$ in (2.7), we have

$$\begin{aligned} K(\epsilon ):=\sup _{s>0}\{s\ln s-\epsilon s^2\}\in [0,\infty ), \end{aligned}$$

and then

$$\begin{aligned} s\ln s+\epsilon s^2<2\epsilon s^2+K(\epsilon ). \end{aligned}$$

Adding $\int \limits _\Omega u\ln u$ to both sides of (2.10), we then obtain (2.6). $\square$

3 Proof of Theorem 1.1: 2D case

In this section, we assume the space dimension $n=2$. Firstly, we use Lemma 2.2 and Lemma 2.3 to give an estimate which will be used frequently in the sequel of this section:

$$\begin{aligned} \begin{aligned} \int \limits _\Omega u^2=\Vert \sqrt{u}\Vert _{L^4}^4&\le \big \{C_{GN}(\Vert \sqrt{u}\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla \sqrt{u}\Vert _{L^2}^{\frac{1}{2}}+\Vert \sqrt{u}\Vert _{L^2})\big \}^4\\&\le 8C_{GN}^4(\Vert \sqrt{u}\Vert _{L^2}^{2}\Vert \nabla \sqrt{u}\Vert _{L^2}^{2}+\Vert \sqrt{u}\Vert _{L^2}^{4})\\&\le 8C_{GN}^4(\Vert u\Vert _{L^1}\Vert \nabla \sqrt{u}\Vert _{L^2}^{2}+\Vert u\Vert _{L^1}^{2})\\&\le 2C_{GN}^4m_1\int \limits _\Omega \frac{|\nabla u|^2}{u}+8C_{GN}^4m_1^{2}. \end{aligned} \end{aligned}$$

(3.1)

Next, we are ready to use the above estimate and Lemma 2.5 to establish the uniform bounds of $\int \limits _\Omega u\ln u$, $\int \limits _\Omega |\nabla v|^2$, as well as $\int \limits _\Omega w^2$.

Lemma 3.1

Let $n=2$, and assume, besides (2.2), f and g satisfy

$$\begin{aligned} \liminf _{s\rightarrow \infty }\Big \{-\frac{f(s)\ln s}{g^2(s)}\Big \}=:\mu \in (\frac{2\chi ^2C_{GN}^4m_1}{\lambda ^2},\infty ]. \end{aligned}$$

(3.2)

Here, $C_{GN}$ and $m_1$ are the positive constants given in (3.1) and (2.5), respectively. Then, there exists a positive constant C such that

$$\begin{aligned} \int \limits _\Omega u\ln u+\int \limits _\Omega |\nabla v|^2+\int \limits _\Omega w^2\le C\qquad \text {for all}\,\,\,t\in (0,T^*). \end{aligned}$$

Proof

Taking $\epsilon =\frac{1}{4C_{GN}^4m_1}$ in (2.6) and using (3.1) then entails

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&\Big \{\int \limits _\Omega u\ln u+\chi ^2C_{GN}^4m_1\int \limits _\Omega |\nabla v|^2+\frac{2\chi ^2\tau C_{GN}^4m_1}{\lambda }\int \limits _\Omega w^2\Big \}\\&\quad +\int \limits _\Omega u\ln u+2\chi ^2C_{GN}^4m_1\int \limits _\Omega |\nabla v|^2+ \chi ^2C_{GN}^4m_1\int \limits _\Omega w^2\\&\le \frac{2\chi ^2C_{GN}^4m_1}{\lambda ^2}\int \limits _\Omega g^2(u)+\int \limits _\Omega f(u)(\ln u+1)+K(\frac{1}{4C_{GN}^4m_1})+4m_1 \end{aligned} \end{aligned}$$

(3.3)

for all $t\in (0,T^*)$. Since $\mu \in (\frac{2\chi ^2C_{GN}^4m_1}{\lambda ^2}, \infty ]$, we can deduce from the definition of $\mu$ in (3.2) that

$$\begin{aligned} \exists s_1\gg 1,\,\text { s.t.}\,\,\, f(s)\ln s+\frac{2\chi ^2C_{GN}^4m_1}{\lambda ^2}g^2(s)<0\,\,\text {for all}\,\, s\ge s_1. \end{aligned}$$

Therefore, we have

$$\begin{aligned} \begin{aligned}&\frac{2\chi ^2C_{GN}^4m_1}{\lambda ^2}\int \limits _\Omega g^2(u)+\int \limits _\Omega f(u)(\ln u+1)\\&=\int \limits _{\{u< s_1\}} [\frac{2\chi ^2C_{GN}^4m_1}{\lambda ^2}g^2(u)+f(u)\ln u]+\int \limits _{\{u\ge s_1\}} [\frac{2\chi ^2C_{GN}^4m_1}{\lambda ^2}g^2(u)+f(u)\ln u]+\int \limits _\Omega f(u)\\&< \sup _{0<s<s_1}[\frac{2\chi ^2C_{GN}^4m_1}{\lambda ^2}g^2(s)+f(s)\ln s]|\Omega |+\sup _{s>0}f(s)|\Omega |<\infty \quad \text {for all}\,\,\,t\in (0,T^*). \end{aligned} \end{aligned}$$

(3.4)

Here, the last inequality holds due to $f, g\in C^1[0, +\infty )$, $f(0)\ge 0$, $\sup _{s>0}f(s)<\infty$ implied by (2.2).

Let

$$\begin{aligned} y(t):=\int \limits _\Omega u\ln u+\chi ^2C_{GN}^4m_1\int \limits _\Omega |\nabla v|^2+\frac{2\chi ^2\tau C_{GN}^4m_1}{\lambda }\int \limits _\Omega w^2+e^{-1}|\Omega |, \end{aligned}$$

then $y(t)\ge 0$ due to $s\ln s\ge -e^{-1}$; moreover, we can see from (3.3) and (3.4) that y(t) satisfies

$$\begin{aligned} y'(t)+c_1y(t)\le c_2\quad \text {for all}\,\,\,t\in (0,T^*), \end{aligned}$$

where

$$\begin{aligned} c_1:= & {} \min \{1, \frac{\lambda }{2\tau }\},\\ c_2:= & {} \Big ( \sup _{0<s<s_1}[\frac{2\chi ^2C_{GN}^4m_1}{\lambda ^2}g^2(s)+f(s)\ln s]|+\sup _{s>0}f(s)+e^{-1}\Big )|\Omega |+K(\frac{1}{4C_{GN}^4m_1})+4m_1. \end{aligned}$$

This obviously yields

$$\begin{aligned} y(t)\le \Big (y(0)-\frac{c_2}{c_1}\Big )e^{-c_1t}+\frac{c_2}{c_1}\le \max \{y(0), \frac{c_2}{c_1}\}, \end{aligned}$$

and thus, we obtained the desired estimate. $\square$

We emphasize that if $\mu \in (0, \frac{2\chi ^2C_{GN}^4m_1}{\lambda ^2}]$, then the proof of the above lemma is invalid. To achieve the similar a priori estimates, some additional constraints on the signal secretion function g will be required.

Lemma 3.2

Let $n=2$, and assume besides (2.2), g satisfies

$$\begin{aligned} \limsup _{s\rightarrow \infty }\frac{g(s)}{s}=:\gamma \in [0,\infty ). \end{aligned}$$

(3.5)

If $\gamma =0$, or $\gamma \in (0,\infty )$ and $(\frac{4\chi \gamma }{\lambda }+\frac{\chi ^2}{\lambda ^2})C_{GN}^4m_1<1$, then there exists a positive constant C such that

$$\begin{aligned} \int \limits _\Omega u\ln u+\int \limits _\Omega |\nabla v|^2+\int \limits _\Omega w^2\le C\qquad \text {for all}\,\,\,t\in (0,T^*). \end{aligned}$$

Proof

We divide the proof into two steps. Step 1. Let $\gamma =0$ in (3.5). In this case, for any $\epsilon >0$, in view of the definition of $\gamma$ in (3.5), we have

$$\begin{aligned} \exists s_\epsilon \gg 1,\,\text { s.t.}\,\,\, \frac{g^2(s)}{s^2}\le \frac{2\lambda ^2}{\chi ^2}\epsilon ^2,\,\,\forall s\ge s_\epsilon , \end{aligned}$$

i.e.,

$$\begin{aligned} \exists s_\epsilon \gg 1,\,\text { s.t.}\,\,\, \frac{\chi ^2}{2\lambda ^2\epsilon }g^2(s)\le \epsilon s^2,\,\,\forall s\ge s_\epsilon . \end{aligned}$$

(3.6)

Moreover, for the above $s_\epsilon$, by simple computation, we can get the following elementary inequalities

$$\begin{aligned} 1<\ln s+1\le s, \,\,\forall s\ge s_\epsilon . \end{aligned}$$

Noting that

$$\begin{aligned} \sup _{s>0}f(s)\ge 0, \end{aligned}$$

then we further obtain

$$\begin{aligned} (\ln s+1)f(s)\le (\ln s+1)\sup _{s>0} f(s)\le s\sup _{s>0} f(s),\,\,\forall s\ge s_\epsilon . \end{aligned}$$

(3.7)

Combining (3.6) and (3.7) then yields

$$\begin{aligned} \frac{\chi ^2}{2\lambda ^2\epsilon }g^2(s)+(\ln s+1)f(s)\le \epsilon s^2+ s\sup _{s>0} f(s),\,\,\forall s\ge s_\epsilon . \end{aligned}$$

Therefore, the three integrals on the right-hand side of (2.6) can be estimated as follows:

$$\begin{aligned} \begin{aligned}&2\epsilon \int \limits _\Omega u^2+\frac{\chi ^2}{2\lambda ^2\epsilon }\int \limits _\Omega g^2(u)+\int \limits _\Omega f(u)(\ln u+1)\\&=2\epsilon \int \limits _\Omega u^2+\int \limits _{\{u<s_\epsilon \}} (\frac{\chi ^2}{2\lambda ^2\epsilon }g^2(u)+ f(u)(\ln u+1))+\int \limits _{\{u\ge s_\epsilon \}} (\frac{\chi ^2}{2\lambda ^2\epsilon }g^2(u)+ f(u)(\ln u+1))\\&\le 3\epsilon \int \limits _\Omega u^2+\sup _{u<s_\epsilon }(\frac{\chi ^2}{2\lambda ^2\epsilon }g^2(u)+ f(u)(\ln u+1))|\Omega |+\sup _{s>0}f(s)\int \limits _\Omega u\\&\le 3\epsilon \int \limits _\Omega u^2+\sup _{u<s_\epsilon }(\frac{\chi ^2}{2\lambda ^2\epsilon }g^2(u)+ f(u)(\ln u+1))|\Omega |+\sup _{s>0}f(s)m_1\,\quad \,\text {for all}\,\,t\in (0, T^*). \end{aligned} \end{aligned}$$

(3.8)

Setting

$$\begin{aligned} L(\epsilon ):=\sup _{u<s_\epsilon }(\frac{\chi ^2}{2\lambda ^2\epsilon }g^2(u)+ f(u)(\ln u+1))|\Omega |+\sup _{s>0}f(s)m_1. \end{aligned}$$

It is not hard to see that $L(\epsilon )$ is a nonnegative constant depending on $\epsilon$, and then, we can combine (2.6) and (3.8) to obtain

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&\Big \{\int \limits _\Omega u\ln u+\frac{\chi ^2}{4\epsilon }\int \limits _\Omega |\nabla v|^2+\frac{\tau \chi ^2}{2\lambda \epsilon }\int \limits _\Omega w^2\Big \}+\int \limits _\Omega \frac{|\nabla u|^2}{u}+\int \limits _\Omega u\ln u+\frac{\chi ^2}{2\epsilon }\int \limits _\Omega |\nabla v|^2+ \frac{\chi ^2}{4\epsilon }\int \limits _\Omega w^2\\&\le 3\epsilon \int \limits _\Omega u^2+L(\epsilon )+K(\epsilon )\qquad \text {for all}\,\,\,t\in (0,T^*). \end{aligned} \end{aligned}$$

Invoking the inequality (3.1) once again, we can take $\epsilon =\frac{1}{6C_{GN}^4m_1}$ to further deduce

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&\Big \{\int \limits _\Omega u\ln u+\frac{3\chi ^2C_{GN}^4m_1}{2}\int \limits _\Omega |\nabla v|^2+\frac{3\tau \chi ^2C_{GN}^4m_1}{\lambda }\int \limits _\Omega w^2\Big \}\\&+\int \limits _\Omega u\ln u+3\chi ^2C_{GN}^4m_1\int \limits _\Omega |\nabla v|^2+ \frac{3\chi ^2C_{GN}^4m_1}{2}\int \limits _\Omega w^2\\&\le L(\frac{1}{6C_{GN}^4m_1})+K(\frac{1}{6C_{GN}^4m_1})+4m_1 \end{aligned} \end{aligned}$$

(3.9)

for all $t\in (0,T^*)$. Similar to the proof of Lemma 3.1, we can use the ODE comparison argument to get the desired estimate.

Step 2. Let $\gamma \in (0,\infty )$ in (3.5). In this case, for any $\epsilon >0$, we have

$$\begin{aligned} \exists s_\epsilon \gg 1,\,\text { s.t.}\,\,\, g^2(s)\le (\gamma ^2+\epsilon ) s^2,\,\,\forall s\ge s_\epsilon . \end{aligned}$$

Similar to the previous proof, we have

$$\begin{aligned} \begin{aligned}&2\epsilon \int \limits _\Omega u^2+\frac{\chi ^2}{2\lambda ^2\epsilon }\int \limits _\Omega g^2(u)+\int \limits _\Omega f(u)(\ln u+1)\\&=2\epsilon \int \limits _\Omega u^2+\int \limits _{\{u<s_\epsilon \}} (\frac{\chi ^2}{2\lambda ^2\epsilon }g^2(u)+ f(u)(\ln u+1))+\int \limits _{\{u\ge s_\epsilon \}} (\frac{\chi ^2}{2\lambda ^2\epsilon }g^2(u)+ f(u)(\ln u+1))\\&\le (2\epsilon +\frac{\chi ^2(\gamma ^2+\epsilon )}{2\lambda ^2\epsilon })\int \limits _\Omega u^2+\sup _{u<s_\epsilon }(\frac{\chi ^2}{2\lambda ^2\epsilon }g^2(u)+ f(u)(\ln u+1))|\Omega |+\sup _{s>0}f(s)\int \limits _\Omega u\\&\le (2\epsilon +\frac{\chi ^2(\gamma ^2+\epsilon )}{2\lambda ^2\epsilon })\int \limits _\Omega u^2+L(\epsilon ) \end{aligned} \end{aligned}$$

(3.10)

for all $t\in (0,T^*)$. Setting

$$\begin{aligned} F(\epsilon ):=2\epsilon +\frac{\chi ^2(\gamma ^2+\epsilon )}{2\lambda ^2\epsilon },\,\epsilon \in (0,\infty ), \end{aligned}$$

by the elementary inequality, we have

$$\begin{aligned} \begin{aligned} F(\epsilon )&=2\epsilon +\frac{\chi ^2\gamma ^2}{2\lambda ^2\epsilon }+\frac{\chi ^2}{2\lambda ^2}\\&\ge \frac{2\chi \gamma }{\lambda }+\frac{\chi ^2}{2\lambda ^2}\quad \text {for all}\,\,\epsilon \in (0,\infty ), \end{aligned} \end{aligned}$$

where the equality holds iff $\epsilon =\frac{\chi \gamma }{2\lambda }$. With this $\epsilon$, we can combine (2.6), (3.10) as well as (3.1) once again to deduce

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&\Big \{\int \limits _\Omega u\ln u+\frac{\chi \lambda }{2 \gamma }\int \limits _\Omega |\nabla v|^2+\frac{\tau \chi }{\gamma }\int \limits _\Omega w^2\Big \}+\int \limits _\Omega \frac{|\nabla u|^2}{u}+\int \limits _\Omega u\ln u+\frac{\chi \lambda }{\gamma }\int \limits _\Omega |\nabla v|^2+ \frac{\chi \lambda }{2\gamma }\int \limits _\Omega w^2\\&\le (\frac{2\chi \gamma }{\lambda }+\frac{\chi ^2}{2\lambda ^2})(2C_{GN}^4m_1\int \limits _\Omega \frac{|\nabla u|^2}{u}+8C_{GN}^4m_1^{2})+L(\frac{\chi \gamma }{2\lambda })+K(\frac{\chi \gamma }{2\lambda })\quad \text {for all}\,\,t\in (0, T*). \end{aligned} \end{aligned}$$

Since $(\frac{4\chi \gamma }{\lambda }+\frac{\chi ^2}{\lambda ^2})C_{GN}^4m_1<1$, we then further have

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&\left\{ \int \limits _\Omega u\ln u+\frac{\chi \lambda }{2 \gamma }\int \limits _\Omega |\nabla v|^2+\frac{\tau \chi }{\gamma }\int \limits _\Omega w^2+e^{-1}\right\} +\int \limits _\Omega u\ln u+\frac{\chi \lambda }{\gamma }\int \limits _\Omega |\nabla v|^2+ \frac{\chi \lambda }{2\gamma }\int \limits _\Omega w^2\\&\le 8C_{GN}^4m_1^{2}(\frac{2\chi \gamma }{\lambda }+\frac{\chi ^2}{2\lambda ^2})+L(\frac{\chi \gamma }{2\lambda })+K(\frac{\chi \gamma }{2\lambda })\quad \text {for all}\,\,t\in (0, T*). \end{aligned} \end{aligned}$$

This along with the ODE comparison argument once again then yields the desired estimate. We thus complete the proof of this lemma. $\square$

Remark 3.3

Along the proof of Lemma 3.2, one can easily conclude that: If $f\equiv 0$ which implies $m_1=\Vert u_0\Vert _{L^1(\Omega )}$, then both the conditions

$$\begin{aligned} \lim _{s\rightarrow \infty }\frac{g(s)}{s}=0 \end{aligned}$$

and

$$\begin{aligned} \limsup _{s\rightarrow \infty }\frac{g(s)}{s}=\gamma \in (0,\infty )\,\, \text {satisfying}\,\,\,(\frac{4\chi \gamma }{\lambda }+\frac{\chi ^2}{\lambda ^2})C_{GN}^4\Vert u_0\Vert _{L^1(\Omega )}<1 \end{aligned}$$

are enough to guarantee the uniform boundedness of $\int \limits _\Omega u\ln u+\int \limits _\Omega |\nabla v|^2+\int \limits _\Omega w^2$.

Corollary 3.4

Let $n=2$, $\alpha >0$, $C_{GN}>0$ given in (3.1). If one of the following conditions holds

• $(i) \sup _{s\ge 0}\{f(s)+\alpha s\}<\infty ,\,\, \liminf _{s\rightarrow \infty }\Big \{-\frac{f(s)\ln s}{g^2(s)}\Big \}=:\mu \in (\frac{2\chi ^2C_{GN}^4m_1}{\lambda ^2},\infty ]$,

• $(ii)\sup _{s\ge 0}\{f(s)+\alpha s\}<\infty \,\text {or} f\equiv 0,\, \lim _{s\rightarrow \infty }\frac{g(s)}{s}=0$,

• $(iii)\sup _{s\ge 0}\{f(s)+\alpha s\}<\infty \,\text {or} f\equiv 0,\,\limsup _{s\rightarrow \infty }\frac{g(s)}{s}=\gamma \in (0,\infty ),\,\,(\frac{4\chi \gamma }{\lambda }+\frac{\chi ^2}{\lambda ^2})C_{GN}^4m_1<1$,

then for any $2<q<\infty$ there exists $C=C(q)>0$ such that the solution component v of (1.1) fulfills

$$\begin{aligned} \Vert v(\cdot ,t)\Vert _{W^{1,q}(\Omega )}\le C\qquad \text {for all}\,\,\,t\in (0,T^*). \end{aligned}$$

(3.11)

Proof

From Lemma 3.1 and Lemma 3.2, the uniform bound of $\Vert w\Vert _{L^2(\Omega )}$ is obtained, the parabolic regularity (Ref. Lemma 4.1 of [26] and Lemma 1 of [27]) then ensures (3.11). $\square$

At the end of this section, we state the proof of Theorem 1.1.

Proof of Theorem 1.1.

By the standard testing procedure, we see that for any $p>1$ there holds

$$\begin{aligned} \frac{d}{dt}\int \limits _\Omega u^p+p(p-1)\int \limits _\Omega u^{p-2}|\nabla u|^2=\chi p(p-1)\int \limits _\Omega u^{p-1}\nabla u\cdot \nabla v+p\int \limits _\Omega u^{p-1} f(u) \end{aligned}$$

(3.12)

for all $t\in (0, T^*)$. Applying the Young inequality to the first term on the right of (3.12) then yields

$$\begin{aligned} \begin{aligned} \chi p(p-1)\int \limits _\Omega u^{p-1}\nabla u\cdot \nabla v&\le \frac{p(p-1)}{2}\int \limits _\Omega u^{p-2}|\nabla u|^2+\frac{\chi ^2p(p-1)}{2}\int \limits _\Omega u^p|\nabla v|^2\\&\le \frac{p(p-1)}{2}\int \limits _\Omega u^{p-2}|\nabla u|^2+\epsilon \int \limits _\Omega u^{p+1}+C(\epsilon ,\chi ,p)\int \limits _\Omega |\nabla v|^{2(p+1)}\\&\le \frac{p(p-1)}{2}\int \limits _\Omega u^{p-2}|\nabla u|^2+\epsilon \int \limits _\Omega u^{p+1}+C(\epsilon ,\chi ,p,|\Omega |) \end{aligned} \end{aligned}$$

(3.13)

with $\epsilon$ to be determined later, where in the last inequality, we used the uniform boundedness for $\Vert \nabla v\Vert _{L^{2(p+1)}(\Omega )}$ implied in (3.11). If f satisfies (2.2), then the last term on the right of (3.12) can be estimated as

$$\begin{aligned} \begin{aligned} p\int \limits _\Omega u^{p-1} f(u)&=p\int \limits _\Omega u^{p-1} (f(u)+\alpha u-\alpha u)\\&\le p\sup _{s>0}(f(s)+\alpha s)\int \limits _\Omega u^{p-1}-\alpha p\int \limits _\Omega u^p\\&\le -\frac{\alpha p}{2}\int \limits _\Omega u^p+C(p,\sup _{s>0}(f(s)+\alpha s),|\Omega |). \end{aligned} \end{aligned}$$

This in conjunction with (3.12) and (3.13) then yields

$$\begin{aligned} \begin{aligned}&\frac{d}{dt}\int \limits _\Omega u^p+\frac{p(p-1)}{2}\int \limits _\Omega u^{p-2}|\nabla u|^2+\frac{\alpha p}{2}\int \limits _\Omega u^p\\&\le \epsilon \int \limits _\Omega u^{p+1}+C(\epsilon ,\chi ,p,\sup _{s>0}(f(s)+\alpha s),|\Omega |)\qquad \text {for all}\,\,t\in (0,T^*). \end{aligned} \end{aligned}$$

(3.14)

We further apply the Gagliardo–Nirenberg interpolation inequality (see Lemma 2.2) to $\int \limits _\Omega u^{p+1}$ to get

$$\begin{aligned} \begin{aligned} \int \limits _\Omega u^{p+1}=\Vert u^{\frac{p}{2}}\Vert _{L^{\frac{2(p+1)}{p}}(\Omega )}^{\frac{2(p+1)}{p}}&\le (2C_{GN})^{\frac{2(p+1)}{p}}\Big (\Vert u^{\frac{p}{2}}\Vert _{L^{\frac{2}{p}}(\Omega )}^{\frac{2}{p}}\Vert \nabla u^{\frac{p}{2}}\Vert _{L^2(\Omega )}^2+\Vert u^{\frac{p}{2}}\Vert _{L^{\frac{2}{p}}(\Omega )}^{\frac{2(p+1)}{p}}\Big )\\&=(2C_{GN})^{\frac{2(p+1)}{p}}\Big (\Vert u\Vert _{L^1(\Omega )}\Vert \nabla u^{\frac{p}{2}}\Vert _{L^2(\Omega )}^2+\Vert u\Vert _{L^1(\Omega )}^{p+1}\Big )\\&\le (2C_{GN})^{\frac{2(p+1)}{p}}\Big (m_1\Vert \nabla u^{\frac{p}{2}}\Vert _{L^2(\Omega )}^2+m_1^{p+1}\Big )\\&=(2C_{GN})^{\frac{2(p+1)}{p}}\Big (\frac{p^2}{4}m_1\int \limits _\Omega u^{p-2}|\nabla u|^2+m_1^{p+1}\Big ). \end{aligned} \end{aligned}$$

(3.15)

Therefore, we can take $\epsilon =(2C_{GN})^{\frac{-2(p+1)}{p}}\frac{2(p-1)}{pm_1}$ in (3.14) to deduce

$$\begin{aligned} \frac{d}{dt}\int \limits _\Omega u^p+\frac{\alpha p}{2}\int \limits _\Omega u^p \le C(\chi ,p,\Vert u_0\Vert _{L^1(\Omega )}, \sup _{s>0}(f(s)+\alpha s),|\Omega |)\qquad \text {for all}\,\,t\in (0,T^*). \end{aligned}$$

(3.16)

An ODE comparison argument then entails

$$\begin{aligned} \int \limits _\Omega u^p\le C(\chi ,p,\Vert u_0\Vert _{L^1(\Omega )}, \sup _{s>0}(f(s)+\alpha s),|\Omega |)\qquad \text {for all}\,\,t\in (0,T^*). \end{aligned}$$

If $f\equiv 0$, we then combine (3.12) and (3.13) to get

$$\begin{aligned} \frac{d}{dt}\int \limits _\Omega u^p+\frac{p(p-1)}{2}\int \limits _\Omega u^{p-2}|\nabla u|^2\le \epsilon \int \limits _\Omega u^{p+1}+C(\epsilon ,\chi ,p,|\Omega |)\qquad \text {for all}\,\,t\in (0,T^*). \end{aligned}$$

(3.17)

Adding $\int \limits _\Omega u^p$ to both sides of (3.17) and using Young’s inequality then entails

$$\begin{aligned} \begin{aligned} \frac{d}{dt}\int \limits _\Omega u^p+\int \limits _\Omega u^p+\frac{p(p-1)}{2}\int \limits _\Omega u^{p-2}|\nabla u|^2&\le \int \limits _\Omega u^p+\epsilon \int \limits _\Omega u^{p+1}+C(\epsilon ,\chi ,p,|\Omega |)\\&\le 2\epsilon \int \limits _\Omega u^{p+1}+C(\epsilon ,\chi ,p,|\Omega |)\qquad \text {for all}\,\,t\in (0,T^*). \end{aligned} \end{aligned}$$

(3.18)

Using (3.15) again, we can take $\epsilon =(2C_{GN})^{\frac{-2(p+1)}{p}}\frac{(p-1)}{pm_1}$ in (3.18) to deduce

$$\begin{aligned} \frac{d}{dt}\int \limits _\Omega u^p+\int \limits _\Omega u^p \le C(\chi ,p,\Vert u_0\Vert _{L^1(\Omega )},|\Omega |)\qquad \text {for all}\,\,t\in (0,T^*). \end{aligned}$$

(3.19)

As before, one can use an ODE comparison argument to get

$$\begin{aligned} \int \limits _\Omega u^p\le C(\chi ,p,\Vert u_0\Vert _{L^1(\Omega )},|\Omega |)\qquad \text {for all}\,\,t\in (0,T^*). \end{aligned}$$

All in all, whether $f\equiv 0$ or f satisfies (2.2), for any $p>1$, we can find a positive constant C such that

$$\begin{aligned} \Vert u\Vert _{L^p(\Omega )}\le C\qquad \text {for all}\,\,t\in (0,T^*). \end{aligned}$$

(3.20)

This and the uniform bound for $\nabla v$ in $L^q(\Omega )$ for any $q\in (0, \infty )$ allow us to invoke [28, Appendix A] to derive

$$\begin{aligned} \Vert u\Vert _{L^\infty (\Omega )}\le c_3 \qquad \text {for all}\,\,t\in (0,T^*), \end{aligned}$$

which in conjunction with the extensibility criterion 2.1 and the parabolic regularity theorem then proves Theorem 1.1. $\square$

4 Proof of Theorem 1.4: 3D case

In this section, we assume $n=3$. Similar to the two-dimensional case, we shall first invoke (2.6) and some constraints on f and g to establish the uniform boundedness for $\int \limits _\Omega u\ln u+\int \limits _\Omega |\nabla v|^2+\int \limits _\Omega w^2$. We remark that, in the three-dimensional setting, the Gagliardo–Nirenberg inequality (3.1) becomes:

$$\begin{aligned} \begin{aligned} \int \limits _\Omega u^2=\Vert \sqrt{u}\Vert _{L^4}^4&\le \big \{C_{GN}(\Vert \sqrt{u}\Vert _{L^2}^{\frac{1}{4}}\Vert \nabla \sqrt{u}\Vert _{L^2}^{\frac{3}{4}}+\Vert \sqrt{u}\Vert _{L^2})\big \}^4\\&\le 8C_{GN}^4(\Vert \sqrt{u}\Vert _{L^2}\Vert \nabla \sqrt{u}\Vert _{L^2}^{3}+\Vert \sqrt{u}\Vert _{L^2}^{4})\\&\le 8C_{GN}^4(\Vert u\Vert _{L^1}^{\frac{1}{2}}\Vert \nabla \sqrt{u}\Vert _{L^2}^{3}+\Vert u\Vert _{L^1}^{2})\\&\le C_{GN}^4m_1^{\frac{1}{2}}\Big (\int \limits _\Omega \frac{|\nabla u|^2}{u}\Big )^{\frac{3}{2}}+8C_{GN}^4m_1^{2}, \end{aligned} \end{aligned}$$

therefore, unlike the case of $n=2$, we can’t use the integral $\int \limits _\Omega \frac{|\nabla u|^2}{u}$ on the right-hand side of (2.6) to control the integral $\int \limits _\Omega u^2$ on the left-hand side.

Lemma 4.1

Assume f and g belong to $C^{1}[0, +\infty )$ satisfying $f(0)\ge 0$ and

$$\begin{aligned} \left\{ \begin{aligned}&\liminf _{s\rightarrow \infty }\Big \{-\frac{f(s)\ln s}{s^2}\Big \}=:\mu \in (\frac{2\chi \gamma }{\lambda },\infty ],\\&\limsup _{s\rightarrow \infty }\frac{g(s)}{s}=:\gamma \in (0,\infty ). \end{aligned} \right. \end{aligned}$$

(4.1)

Then, there exits a positive constant C such that

$$\begin{aligned} \int \limits _\Omega u\ln u+\int \limits _\Omega |\nabla v|^2+\int \limits _\Omega w^2\le C\qquad \text {for all}\,\,\,t\in (0,T^*). \end{aligned}$$

Proof

For any $\epsilon >0$, let us define

$$\begin{aligned} F(\epsilon ):=2\epsilon +\frac{\chi ^2\gamma ^2}{2\lambda ^2\epsilon }. \end{aligned}$$

It is not hard to see

$$\begin{aligned} \min _{\epsilon >0} F(\epsilon )=F(\frac{\chi \gamma }{2\lambda })=\frac{2\chi \gamma }{\lambda }. \end{aligned}$$

Let $\epsilon = \frac{\chi \gamma }{2\lambda }$ in (2.6), then we have

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&\Big \{\int \limits _\Omega u\ln u+\frac{\chi \lambda }{2\gamma }\int \limits _\Omega |\nabla v|^2+\frac{\tau \chi }{\gamma }\int \limits _\Omega w^2\Big \}\\&\quad +\int \limits _\Omega \frac{|\nabla u|^2}{u}+\int \limits _\Omega u\ln u+\frac{\chi \lambda }{\gamma }\int \limits _\Omega |\nabla v|^2+ \frac{\chi \lambda }{2\gamma }\int \limits _\Omega w^2\\&\le \int \limits _\Omega [\frac{\chi \gamma }{\lambda } u^2+\frac{\chi }{\lambda \gamma } g^2(u)+ f(u)(\ln u+1)]+K(\frac{\chi \gamma }{2\lambda }),\,\,t\in (0,T^*). \end{aligned} \end{aligned}$$

(4.2)

We first infer that there exists a positive constant $C_1$ such that

$$\begin{aligned} \frac{\chi }{\lambda \gamma }g^2(s)+\frac{\chi \gamma }{\lambda }s^2+f(s)(\ln s+1)\le C_1\quad \text {for all}\,\,s>0. \end{aligned}$$

(4.3)

Actually, from the first assumption in (4.1) we see

$$\begin{aligned} \limsup _{s\rightarrow \infty } \frac{f(s)(\ln s+1)}{s^2}=-\mu . \end{aligned}$$

This in conjunction with the second assumption in (4.1) then entails

$$\begin{aligned} \limsup _{s\rightarrow \infty }\frac{\frac{\chi }{\lambda \gamma }g^2(s)+\frac{\chi \gamma }{\lambda }s^2+f(s)(\ln s+1)}{s^2}\le \frac{\chi }{\lambda \gamma }\gamma ^2+\frac{\chi \gamma }{\lambda }-\mu =\frac{2\chi \gamma }{\lambda }-\mu <0, \end{aligned}$$

and thus, there exists $s_0\gg 1$ such that

$$\begin{aligned} \phi (s):=\frac{\chi }{\lambda \gamma }g^2(s)+\frac{\chi \gamma }{\lambda }s^2+f(s)(\ln s+1)<0, s\in (s_0,\infty ). \end{aligned}$$

Note that $f(0)\ge 0$, f and g belong to $C^{1}[0 +\infty )$ imply that $\sup _{s\in (0,s_0]}\phi (s)$ is finite. Then, we have

$$\begin{aligned} \frac{\chi }{\lambda \gamma }g^2(s)+\frac{\chi \gamma }{\lambda }s^2+f(s)(\ln s+1)\le \max \{\sup _{s\in (0,s_0]}\phi (s), 0\}\quad \text {for all}\,\,s>0, \end{aligned}$$

and thus we obtain (4.3). Substituting (4.3) into (4.2) then entails

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&\Big \{\int \limits _\Omega u\ln u+\frac{\chi \lambda }{2\gamma }\int \limits _\Omega |\nabla v|^2+\frac{\tau \chi }{\gamma }\int \limits _\Omega w^2\Big \}\\&\quad +\int \limits _\Omega \frac{|\nabla u|^2}{u}+\int \limits _\Omega u\ln u+\frac{\chi \lambda }{\gamma }\int \limits _\Omega |\nabla v|^2+ \frac{\chi \lambda }{2\gamma }\int \limits _\Omega w^2\\&\le C_1|\Omega |+K(\frac{\chi \gamma }{2\lambda }),\,\,t\in (0,T^*). \end{aligned} \end{aligned}$$

This along with the ODE comparison argument once again, then yields the uniform bound for $\int \limits _\Omega u\ln u+\frac{\chi \lambda }{2\gamma }\int \limits _\Omega |\nabla v|^2+\frac{\tau \chi }{\gamma }\int \limits _\Omega w^2$, and thus we complete the proof of this lemma. $\square$

Next lemma shows that if g is a sub-linear function, then the range of $\mu$ in (4.1) can be expanded.

Lemma 4.2

Let $n=3$, f and g satisfy

$$\begin{aligned} \left\{ \begin{aligned}&\liminf _{s\rightarrow \infty }\Big \{-\frac{f(s)\ln s}{s^2}\Big \}=:\mu \in (0,\infty ],\\&\lim _{s\rightarrow \infty }\frac{g(s)}{s}=0. \end{aligned} \right. \end{aligned}$$

(4.4)

Then, there exits a positive constant C such that

$$\begin{aligned} \int \limits _\Omega u\ln u+\int \limits _\Omega |\nabla v|^2+\int \limits _\Omega w^2\le C\qquad \text {for all}\,\,\,t\in (0,T^*). \end{aligned}$$

Proof

We first consider $\mu \in (0,\infty )$ in (4.4). In this case, for any $\epsilon >0$, we deduce from (4.4) that

$$\begin{aligned} \exists s_\epsilon \gg 1,\,\text { s.t.}\,\,\, g^2(s)\le \frac{2\lambda ^2}{\chi ^2}\epsilon ^2 s^2\,\,\text {and}\,\,-f(s)\ln s>(\mu -\epsilon )s^2,\,\,\text {for all}\, s\ge s_\epsilon . \end{aligned}$$

Thus,

$$\begin{aligned} 2\epsilon s^2+\frac{\chi ^2}{2\lambda ^2\epsilon } g^2(s)+f(s)\ln s\le (4\epsilon -\mu )s^2,\,\text {for all}\, s\ge s_\epsilon , \end{aligned}$$

and then,

$$\begin{aligned} \begin{aligned}&2\epsilon \int \limits _\Omega u^2+\frac{\chi ^2}{2\lambda ^2\epsilon }\int \limits _\Omega g^2(u)+\int \limits _\Omega f(u)(\ln u+1)\\&=\int \limits _{\{u\ge s_\epsilon \}} [2\epsilon u^2+\frac{\chi ^2}{2\lambda ^2\epsilon } g^2(u)+f(u)\ln u]+\int \limits _{\{u<s_\epsilon \}} [2\epsilon u^2+\frac{\chi ^2}{2\lambda ^2\epsilon } g^2(u)+f(u)\ln u]+\int \limits _\Omega f(u)\\&\le (4\epsilon -\mu )\int \limits _{\{u\ge s_\epsilon \}} u^2+\sup _{u<s_\epsilon }[\frac{\chi ^2}{2\lambda ^2\epsilon }g^2(u)+ f(u)\ln u]|\Omega |+\sup _{s>0}f(s)|\Omega |\,\quad \,\text {for all}\,\,t\in (0, T*). \end{aligned} \end{aligned}$$

This in conjunction with (2.6) and letting $\epsilon =\frac{\mu }{4}$ then yields

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&\Big \{\int \limits _\Omega u\ln u+\frac{\chi ^2}{\mu }\int \limits _\Omega |\nabla v|^2+\frac{2\tau \chi ^2}{\lambda \mu }\int \limits _\Omega w^2\Big \}+\int \limits _\Omega \frac{|\nabla u|^2}{u}+\int \limits _\Omega u\ln u+\frac{2\chi ^2}{\mu }\int \limits _\Omega |\nabla v|^2+ \frac{\chi ^2}{\mu }\int \limits _\Omega w^2\\&\le \sup _{u<s_{\frac{\mu }{4}}}[\frac{2\chi ^2}{\lambda ^2\mu }g^2(u)+ f(u)\ln u]|\Omega |+\sup _{s>0}f(s)|\Omega | +K(\frac{\mu }{4})\,\quad \,\text {for all}\,\,t\in (0, T*). \end{aligned} \end{aligned}$$

for all $t\in (0,T^*)$. Again the conditions that, f, g belong to $C^{1}[0 +\infty )$, $f(0)\ge 0$, $\sup _{s>0}f(s)<\infty$, imply

$$\begin{aligned} N_{\mu }:=\sup _{u<s_{\frac{\mu }{4}}}[\frac{2\chi ^2}{\lambda ^2\mu }g^2(u)+ f(u)\ln u]|\Omega |+\sup _{s>0}f(s)|\Omega | +K(\frac{\mu }{4})<\infty . \end{aligned}$$

Writing

$$\begin{aligned} y(t):=\int \limits _\Omega u\ln u+\frac{\chi ^2}{\mu }\int \limits _\Omega |\nabla v|^2+\frac{2\tau \chi ^2}{\lambda \mu }\int \limits _\Omega w^2, \,t\in [0,T^*), \end{aligned}$$

then we have

$$\begin{aligned} y'(t)+\min \{1,\frac{\lambda }{2\tau }\}y(t)\le N_{\mu },\,\,t\in (0,T^*), \end{aligned}$$

which implies the desired estimate.

If $\mu =\infty$ in (4.4), along the proof of above, just need to use a finite positive constant $\widetilde{\mu }$ instead of $\mu$, then the uniform bound for $\int \limits _\Omega u\ln u+\int \limits _\Omega |\nabla v|^2+\int \limits _\Omega w^2$ also can be constructed. $\square$

The third lemma in this section shows that even if g is a super-linear function, the previous a prior estimates still may be valid.

Lemma 4.3

Let $n=3$, f and g satisfy

$$\begin{aligned} \left\{ \begin{aligned}&\liminf _{s\rightarrow \infty }\Big \{-\frac{f(s)\ln s}{g^2(s)}\Big \}=:\mu \in (0,\infty ],\\&\lim _{s\rightarrow \infty }\frac{g(s)}{s}=\infty . \end{aligned} \right. \end{aligned}$$

(4.5)

Then, there exits a positive constant C such that

$$\begin{aligned} \int \limits _\Omega u\ln u+\int \limits _\Omega |\nabla v|^2+\int \limits _\Omega w^2\le C\qquad \text {for all}\,\,\,t\in (0,T^*). \end{aligned}$$

Proof

Let

$$\begin{aligned} F(\epsilon ,s):=2\epsilon s^2+\frac{\chi ^2}{2\lambda ^2\epsilon }g^2(s)+f(s)\ln s. \end{aligned}$$

We first claim that:

$$\begin{aligned} \exists \epsilon >0 \,\,s.t.\, \sup _{s\in (0, \infty )}F(\epsilon ,s)<\infty . \end{aligned}$$

(4.6)

Based on this, the uniform bound on $\int \limits _\Omega u\ln u+\int \limits _\Omega |\nabla v|^2+\int \limits _\Omega w^2$ can be established along the same procedure as the proof of Lemma 4.2.

Actually, if $\mu \in (0,\infty )$ in (4.5), using the properties of the upper and lower limits, we can deduce from (4.5) that

$$\begin{aligned} \begin{aligned} \limsup _{s\rightarrow \infty }[\frac{2\epsilon s^2}{g^2(s)}+\frac{\chi ^2}{2\lambda ^2\epsilon }+\frac{f(s)\ln s}{g^2(s)}]&\le 2\epsilon \lim _{s\rightarrow \infty }\frac{ s^2}{g^2(s)}+\frac{\chi ^2}{2\lambda ^2\epsilon }+\limsup _{s\rightarrow \infty }\frac{f(s)\ln s}{g^2(s)}\\&=\frac{\chi ^2}{2\lambda ^2\epsilon }-\mu . \end{aligned} \end{aligned}$$

Let $\epsilon =\frac{\chi ^2}{\lambda ^2\mu }$, then we have

$$\begin{aligned} \limsup _{s\rightarrow \infty }[\frac{2\chi ^2 s^2}{\lambda ^2\mu g^2(s)}+\frac{\mu }{2}+\frac{f(s)\ln s}{g^2(s)}]\le -\frac{\mu }{2}<0. \end{aligned}$$

Therefore,

$$\begin{aligned} \limsup _{s\rightarrow \infty }F(\frac{\chi ^2}{\lambda ^2\mu },s)=\limsup _{s\rightarrow \infty }g^2(s)\Big (\frac{2\chi ^2 s^2}{\lambda ^2\mu g^2(s)}+\frac{\mu }{2}+\frac{f(s)\ln s}{g^2(s)}\Big )<0 \end{aligned}$$

(4.7)

which ensures that $F(\frac{\chi ^2}{\lambda ^2\mu },s)$ has a finite upper bound.

If $\mu =\infty$ in (4.5), then for any $\epsilon$ there still holds

$$\begin{aligned} \limsup _{s\rightarrow \infty }[\frac{2\epsilon s^2}{g^2(s)}+\frac{\chi ^2}{2\lambda ^2\epsilon }+\frac{f(s)\ln s}{g^2(s)}]<0, \end{aligned}$$

and thus

$$\begin{aligned} \limsup _{s\rightarrow \infty }F(\epsilon ,s)<0, \end{aligned}$$

$F(\epsilon ,s)$ still has a finite upper bound. This shows that the claim (4.6) is true. $\square$

With the uniform boundedness on $\Vert w\Vert _{L^2(\Omega )}$, we can use the parabolic regularity (Ref. Lemma 4.1 of [26] and Lemma 1 of [27]) once again, to obtain the uniform bound for $\Vert v(\cdot ,t)\Vert _{W^{1,q}(\Omega )}$ with $q\in (2, 6)$, which will be used in the sequel to establish the uniform bound for $\Vert w\Vert _{L^3(\Omega )}$.

Corollary 4.4

Let $n=3$. If one of the following conditions holds

• $(i) \liminf _{s\rightarrow \infty }\Big \{-\frac{f(s)\ln s}{s^2}\Big \}\in (\frac{2\chi \gamma }{\lambda },\infty ],\,\limsup _{s\rightarrow \infty }\frac{g(s)}{s}=\gamma \in (0,\infty )$,

• $(ii)\liminf _{s\rightarrow \infty }\Big \{-\frac{f(s)\ln s}{s^2}\Big \}\in (0,\infty ],\,\lim _{s\rightarrow \infty }\frac{g(s)}{s}=0$,

• $(iii) \liminf _{s\rightarrow \infty }\Big \{-\frac{f(s)\ln s}{g^2(s)}\Big \}\in (0,\infty ],\,\lim _{s\rightarrow \infty }\frac{g(s)}{s}=\infty$,

then for any $2<q<6$ there exists $C=C(q)>0$ such that the solution component v of (1.1) fulfills

$$\begin{aligned} \Vert v(\cdot ,t)\Vert _{W^{1,q}(\Omega )}\le C\qquad \text {for all}\,\,\,t\in (0,T^*). \end{aligned}$$

(4.8)

Remark 4.5

We emphasize that for the typical type $g(s)=s$, the condition (i) can be rewritten as

$$\begin{aligned} \liminf _{s\rightarrow \infty }\Big \{-\frac{f(s)\ln s}{s^2}\Big \}\in (\frac{2\chi }{\lambda },\infty ], \end{aligned}$$

which cover some typical examples of f, such as $f(s)=as-\frac{bs^2}{\ln ^\beta (s+1)}$ with $a\in \mathbb {R}$, $b>0$, $\beta <1$; $f(s)=as-\frac{bs^2}{\ln (\ln (s+e))}$ with $b>0$; $f(s)=as-\frac{bs^2}{\ln (s+1)}$ with $a\in \mathbb {R}$, $b>\frac{2\chi }{\lambda }$; $f(s)=as-bs^\beta$ with $a\in \mathbb {R}$, $b>0$, and $\beta \ge 2$.

On the basis of the uniform boundedness of $\Vert \nabla v\Vert _{L^4(\Omega )}$ implied in (4.8), we can further establish the following coupled estimate on $\int \limits _\Omega u^2$, $\int \limits _\Omega |\nabla v|^2$ and $\int \limits _\Omega w^3$.

Lemma 4.6

Let the assumptions in Corollary 4.4 be satisfied. Then, for any $\epsilon >0$, there exists a positive constant $c(\epsilon )$ depending on $\epsilon$ such that

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&\Big \{\int \limits _\Omega u^2+\int \limits _\Omega |\nabla v|^4+\frac{\tau \epsilon }{\lambda }\int \limits _\Omega w^3\Big \}+\min \{1,\frac{\lambda }{\tau }\}\Big \{\int \limits _\Omega u^2+\int \limits _\Omega |\nabla v|^4+\frac{\tau \epsilon }{\lambda }\int \limits _\Omega w^3\Big \}\\&\le 2\epsilon \int \limits _\Omega u^3+2\int \limits _\Omega f(u)u+\frac{4\epsilon }{\lambda ^3}\int \limits _\Omega g^3(u)+c(\epsilon )\quad \text {for all}\,\,t\in (0,T^*). \end{aligned} \end{aligned}$$

(4.9)

Proof

Testing the first equation in (1.1) by 2u entails

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\int \limits _\Omega u^2&=-2\int \limits _\Omega |\nabla u|^2+2\chi \int \limits _\Omega u\nabla u\cdot \nabla v+2\int \limits _\Omega f(u)u\\&\le -2\int \limits _\Omega |\nabla u|^2+2\int \limits _\Omega |\nabla u|^2+\frac{\chi ^2}{2}\int \limits _\Omega u^2|\nabla v|^2+2\int \limits _\Omega f(u)u\\&\le \epsilon \int \limits _\Omega u^3+\frac{\chi ^6}{54\epsilon ^2}\int \limits _\Omega |\nabla v|^6+2\int \limits _\Omega f(u)u \quad \text {for all}\,\,t\in (0,T^*) \end{aligned} \end{aligned}$$

(4.10)

with $\epsilon >0$. To deal with the integral $\int \limits _\Omega |\nabla v|^6$, we first use the second equation in (1.1) and the pointwise identity $2\nabla v\cdot \nabla \Delta v=\Delta |\nabla v|^2-2|D^2v|^2$ to attain

$$\begin{aligned} \begin{aligned} (|\nabla v|^2)_t&=2\nabla v\cdot \nabla (\Delta v-v+w)\\&=\Delta |\nabla v|^2-2|D^2v|^2-2|\nabla v|^2+2\nabla w\cdot \nabla v \end{aligned} \end{aligned}$$

Testing the above equation by $2|\nabla v|^2$ and using the Young inequality as well as the fact that $|\Delta v|^2\le 3|D^2v|^2$, then yields

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\int \limits _\Omega |\nabla v|^4+4\int \limits _\Omega |\nabla v|^4+\int \limits _\Omega |\nabla |\nabla v|^2|^2\le \epsilon \int \limits _\Omega w^3+\frac{1372}{27\epsilon ^2}\int \limits _\Omega |\nabla v|^6+2\int \limits _{\partial \Omega }|\nabla v|^2\frac{\partial |\nabla v|^2}{\partial \nu } \end{aligned}$$

(4.11)

for all $t\in (0, T^*)$, where $\epsilon >0$ is the same as in (4.10). Moreover, by using the one-sided pointwise inequality $\frac{\partial |\nabla v|^2}{\partial \nu }\le C_\Omega |\nabla v|^2$ with $C_\Omega$ representing a upper bound for the curvatures of $\partial \Omega$(cf., Ref.[29]), and recalling the embedding $W^{1.2}(\Omega )\hookrightarrow \hookrightarrow W^{\frac{1}{2},2}(\Omega )\hookrightarrow L^2(\partial \Omega )$(cf., Ref.[30, Proposition 4.22(ii) and Theorem 4.24(i)]), as well as the uniform boundedness of $\int \limits _\Omega |\nabla v|^4$ implied in (4.8), the boundary integral in (4.11) can be estimated

$$\begin{aligned} \begin{aligned} 2\int \limits _{\partial \Omega }|\nabla v|^2\frac{\partial |\nabla v|^2}{\partial \nu }&\le 2C_\Omega \int \limits _{\partial \Omega }|\nabla v|^4\\&\le 2C_\Omega (\frac{1}{4C_\Omega }\int \limits _{\Omega }|\nabla |\nabla v|^2|^2+c_1\int \limits _\Omega |\nabla v|^4)\\&\le \frac{1}{2}\int \limits _{\Omega }|\nabla |\nabla v|^2|^2+c_2\quad \text {for all}\,\,t\in (0, T^*). \end{aligned} \end{aligned}$$

(4.12)

Substituting (4.12) into (4.11) then yields

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\int \limits _\Omega |\nabla v|^4+4\int \limits _\Omega |\nabla v|^4+\frac{1}{2}\int \limits _\Omega |\nabla |\nabla v|^2|^2\le \epsilon \int \limits _\Omega w^3+\frac{1372}{27\epsilon ^2}\int \limits _\Omega |\nabla v|^6+c_2\,\,\text {for all}\,\,t\in (0,T^*). \end{aligned}$$

(4.13)

To deal with $\int \limits _\Omega w^3$, we can test the third equation in (1.1) by $3w^2$ and then use Young’s inequality to obtain

$$\begin{aligned} \tau \frac{\mathrm{d}}{\mathrm{d}t}\int \limits _\Omega w^3+2\lambda \int \limits _\Omega w^3\le \frac{4}{\lambda ^2}\int \limits _\Omega g^3(u)\,\,\text {for all}\,\,t\in (0,T^*). \end{aligned}$$

(4.14)

A combination $(4.10)+(4.13)+(4.14)\times \frac{\epsilon }{\lambda }$ then yields

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&\Big \{\int \limits _\Omega u^2+\int \limits _\Omega |\nabla v|^4+\frac{\tau \epsilon }{\lambda }\int \limits _\Omega w^3\Big \}+4\int \limits _\Omega |\nabla v|^4+\frac{1}{2}\int \limits _\Omega |\nabla |\nabla v|^2|^2+\epsilon \int \limits _\Omega w^3\\&\le \epsilon \int \limits _\Omega u^3+(\frac{\chi ^6}{54\epsilon ^2}+\frac{1372}{27\epsilon ^2})\int \limits _\Omega |\nabla v|^6+2\int \limits _\Omega f(u)u+\frac{4\epsilon }{\lambda ^3}\int \limits _\Omega g^3(u)+c_2 \end{aligned} \end{aligned}$$

(4.15)

for all $t\in (0,T^*)$. Invoking the Gagliardo–Nirenberg inequality and the Young inequality as well as the uniform bound on $\Vert \nabla v\Vert _{L^4(\Omega )}$ implied in (4.8) once again, we have

$$\begin{aligned} \begin{aligned} (\frac{\chi ^6}{54\epsilon ^2}+\frac{1372}{27\epsilon ^2})\int \limits _\Omega |\nabla v|^6&=(\frac{\chi ^6}{54\epsilon ^2}+\frac{1372}{27\epsilon ^2})\Vert |\nabla v|^2\Vert _{L^3(\Omega )}^3\\&\le 4C_{GN}^3(\frac{\chi ^6}{54\epsilon ^2}+\frac{1372}{27\epsilon ^2})(\Vert \nabla |\nabla v|^2\Vert _{L^2(\Omega )}^{\frac{2}{3}}\Vert |\nabla v|^2\Vert _{L^2(\Omega )}^{\frac{3}{2}}+\Vert |\nabla v|^2\Vert _{L^2(\Omega )}^{3})\\&\le \frac{1}{2}\int \limits _\Omega |\nabla |\nabla v|^2|^2+c(\epsilon ) \end{aligned} \end{aligned}$$

for all $t\in (0,T^*)$. Upon substitution into (4.15) and adding the term $\int \limits _\Omega u^2$ on both sides of the resulted inequality, then yields

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&\Big \{\int \limits _\Omega u^2+\int \limits _\Omega |\nabla v|^4+\frac{\tau \epsilon }{\lambda }\int \limits _\Omega w^3\Big \}+4\int \limits _\Omega |\nabla v|^4+\epsilon \int \limits _\Omega w^3+\int \limits _\Omega u^2\\&\le \epsilon \int \limits _\Omega u^3+2\int \limits _\Omega f(u)u+\frac{4\epsilon }{\lambda ^3}\int \limits _\Omega g^3(u)+\int \limits _\Omega u^2+c(\epsilon )\quad \text {for all}\,\,t\in (0,T^*), \end{aligned} \end{aligned}$$

which in conjunction with the Young inequality: $u^2\le \epsilon u^3+\epsilon ^{-2}$, then implies (4.9). $\square$

To control the right-hand side of (4.9), we need to explore some additional conditions on f and g. It is worth noting that such conditions should be compatible with the previous conditions stated in Corollary 4.4.

Lemma 4.7

Assume f and g belong to $C^{1}[0,+\infty )$, $f(0)\ge 0$, $g(s)\ge 0$ for all $s\ge 0$ and

$$\begin{aligned} \left\{ \begin{aligned}&\liminf _{s\rightarrow \infty }\Big \{-\frac{f(s)}{s^2}\Big \}=:\mu \in (0,\infty ],\\&\limsup _{s\rightarrow \infty }\frac{g(s)}{s}=\gamma \in [0,\infty ) \end{aligned} \right. \end{aligned}$$

(4.16)

or

$$\begin{aligned} \left\{ \begin{aligned}&\liminf _{s\rightarrow \infty }\Big \{-\frac{f(s)s}{g^3(s)}\Big \}=:\mu \in (0,\infty ],\\&\lim _{s\rightarrow \infty }\frac{g(s)}{s}=\infty . \end{aligned} \right. \end{aligned}$$

(4.17)

Then, there exits a positive constant $\epsilon$ and constant C depending on $\epsilon$ such that

$$\begin{aligned} 2\epsilon s^3+2f(s)s+\frac{4\epsilon }{\lambda ^3} g^3(s)<C\quad \text {for all}\,\,s\ge 0 \end{aligned}$$

(4.18)

Proof

On the basis of (4.16), we have

$$\begin{aligned} \limsup _{s\rightarrow \infty }[2\epsilon +2\frac{f(s)}{s^2}+\frac{4\epsilon }{\lambda ^3} \frac{g^3(s)}{s^3}]\le 2\epsilon +\frac{4\epsilon \gamma ^3}{\lambda ^3}-2\mu <0 \end{aligned}$$

for $\epsilon <\frac{\mu \lambda ^3}{\lambda ^3+2\gamma ^3}$, if $\mu \in (0,\infty )$; and for any $\epsilon$, if $\mu =\infty$. This allows us to deduce that

$$\begin{aligned} \begin{aligned}&\forall \epsilon \in (0, \frac{\mu \lambda ^3}{\lambda ^3+2\gamma ^3}), \exists s_\epsilon \gg 1, \text {such that, for all}\,\,s>s_\epsilon ,\,\text {there holds}\\&2\epsilon s^3+2f(s)s+\frac{4\epsilon }{\lambda ^3} g^3(s)=s^3(2\epsilon +2\frac{f(s)}{s^2}+\frac{4\epsilon }{\lambda ^3} \frac{g^3(s)}{s^3})<0. \end{aligned} \end{aligned}$$

(4.19)

For the above $\epsilon$ and $s_\epsilon$, recalling that f and g belong to $C^{1}[0,+\infty )$, $f(0)\ge 0$, $g(s)\ge 0$ for all $s\ge 0$, we further deduce that

$$\begin{aligned} \sup _{s>0}[2\epsilon s^3+2f(s)s+\frac{4\epsilon }{\lambda ^3} g^3(s)]=\sup _{s\in [0,s_\epsilon ]}[2\epsilon s^3+2f(s)s+\frac{4\epsilon }{\lambda ^3} g^3(s)]=:C(\epsilon )<\infty . \end{aligned}$$

Thus, (4.18) is true.

Similarly, if (4.17) holds, then we can deduce that

$$\begin{aligned} \limsup _{s\rightarrow \infty }[2\epsilon \frac{s^3}{g^3(s)} +2\frac{f(s)s}{g^3(s)}+\frac{4\epsilon }{\lambda ^3}]=\frac{4\epsilon }{\lambda ^3}-2\mu <0 \end{aligned}$$

for $\epsilon <\frac{\mu \lambda ^3}{2}$, if $\mu \in (0,\infty )$; and for any $\epsilon$, if $\mu =\infty$. Similar to the previous case, we can deduce that for any $\epsilon <\frac{\mu \lambda ^3}{2}$, there exists $s_\epsilon \gg 1$, such that

$$\begin{aligned} \sup _{s>0}[2\epsilon s^3+2f(s)s+\frac{4\epsilon }{\lambda ^3} g^3(s)]=\sup _{s\in [0,s_\epsilon ]}[2\epsilon s^3+2f(s)s+\frac{4\epsilon }{\lambda ^3} g^3(s)]=:C(\epsilon )<\infty . \end{aligned}$$

We thus complete the proof of this lemma. $\square$

Remark 4.8

We remark that from $\lim \inf _{s\rightarrow \infty }\{-\frac{f(s)}{s^2}\}\in (0, \infty ]$ we can deduce $\liminf \{-\frac{f(s)\ln s}{s^2}\}=\infty$, this shows that (4.16) is compatible with (4.1) and (4.4); meanwhile, from (4.17) we can also deduce $\lim \inf _{s\rightarrow \infty }\{-\frac{f(s)\ln s}{g^2(s)}\}=\infty$, which shows that (4.17) is compatible with (4.5).

Corollary 4.9

Assume f, g satisfy the same conditions as in Lemma 4.7. Then, there exists a constant $C>0$ such that

$$\begin{aligned} \int \limits _\Omega w^3\le C\qquad \text {for all}\,\,\,t\in (0,T^*), \end{aligned}$$

(4.20)

Proof

Invoking Lemma 4.7, there exists a positive constant $\epsilon$ and $C(\epsilon )$ such that

$$\begin{aligned} \epsilon s^3+2f(s)s+\frac{4\epsilon }{\lambda ^3} g^3(s)+s^2<C(\epsilon ), \text {for all}\,\,s\ge 0. \end{aligned}$$

Fix such $\epsilon$, then from (4.15) we can deduce

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&\Big \{\int \limits _\Omega u^2+\int \limits _\Omega |\nabla v|^4+\frac{\tau \epsilon }{\lambda }\int \limits _\Omega w^3\Big \}+\min \{1,\frac{\lambda }{\tau }\}\Big \{\int \limits _\Omega u^2+\int \limits _\Omega |\nabla v|^4+\frac{\tau \epsilon }{\lambda }\int \limits _\Omega w^3\Big \}\\&\le \int \limits _\Omega [2\epsilon u^3+2f(u)u+\frac{4\epsilon }{\lambda ^3} g^3(u)]+c(\epsilon )\\&\le \sup _{s>0}[2\epsilon s^3+2f(s)s+\frac{4\epsilon }{\lambda ^3} g^3(s)]|\Omega |+c(\epsilon )\\&\le C(\epsilon )\quad \text {for all}\,\,t\in (0, T^*). \end{aligned} \end{aligned}$$

(4.21)

Upon an ODE comparison argument then yields (4.20). $\square$

On the basis of the uniform boundedness on $\Vert w\Vert _{L^3(\Omega )}$, we can once again apply the parabolic regularity to obtain the uniform bound on v in $W^{1,q}(\Omega )$ for any $1<q<\infty$.

Corollary 4.10

For any $1<q<\infty$ there exists $C=C(q)>0$ such that the solution component v of (1.1) fulfills

$$\begin{aligned} \Vert v(\cdot ,t)\Vert _{W^{1,q}(\Omega )}\le C\qquad \text {for all}\,\,\,t\in (0,T^*). \end{aligned}$$

(4.22)

With the uniform boundedness of $\Vert v(\cdot ,t)\Vert _{W^{1,q}(\Omega )}$ for any $q>1$ at hand, we next complete the proof of Theorem 1.4.

Proof of Theorem 1.4

Similar to the proof of Theorem 1.1, by standard testing procedure and using the Young inequality and the uniform boundedness of $\Vert \nabla v\Vert _{L^{2(p+1)}(\Omega )}$ for any $p>2$ implied in (4.22), we can deduce that

$$\begin{aligned} \frac{d}{dt}\int \limits _\Omega u^p\le \epsilon \int \limits _\Omega u^{p+1}+p\int \limits _\Omega u^{p-1} f(u)+c(\epsilon )\quad \text {for all}\,\,t\in (0, T^*) \end{aligned}$$

(4.23)

with $\epsilon >0$ to be determined later. Now, adding the term $\int \limits _\Omega u^p$ on both side of (4.23) yields

$$\begin{aligned} \frac{d}{dt}\int \limits _\Omega u^p+\int \limits _\Omega u^p\le \epsilon \int \limits _\Omega u^{p+1}+\int \limits _\Omega u^p+p\int \limits _\Omega u^{p-1} f(u)+c(\epsilon )\quad \text {for all}\,\,t\in (0, T^*). \end{aligned}$$

(4.24)

Setting

$$\begin{aligned} F(\epsilon , s):=\epsilon s^{p+1}+s^p+ps^{p-1}f(s), \end{aligned}$$

we first claim that

$$\begin{aligned} \exists \epsilon >0,\,\text {such that}\,\, \sup _{s\ge 0}F(\epsilon , s)<\infty . \end{aligned}$$

(4.25)

Indeed, if f and g satisfy (4.16), then we have

$$\begin{aligned} \lim \sup _{s\rightarrow \infty }(\epsilon +\frac{1}{s}+\frac{pf(s)}{s^2})\le \epsilon +p\lim \sup _{s\rightarrow \infty }\frac{f(s)}{s^2}=\epsilon -p\mu . \end{aligned}$$

If $\mu \in (0,\infty )$, then for any $\epsilon <p\mu$, we have $\lim \sup _{s\rightarrow \infty }(\epsilon +\frac{1}{s}+\frac{pf(s)}{s^2})<0$; if $\mu =\infty$, then for any $\epsilon >0$, we have $\lim \sup _{s\rightarrow \infty }(\epsilon +\frac{1}{s^2}+\frac{pf(s)}{s^2})=-\infty$. In a word, there exists $\epsilon >0$, such that

$$\begin{aligned} \lim \sup _{s\rightarrow \infty }(\epsilon +\frac{1}{s}+\frac{pf(s)}{s^2})<0, \end{aligned}$$

(4.26)

which implies

$$\begin{aligned} \lim \sup _{s\rightarrow \infty }F(\epsilon , s)=\lim \sup _{s\rightarrow \infty }s^{p+1}(\epsilon +\frac{1}{s}+\frac{pf(s)}{s^2})<0 \end{aligned}$$

(4.27)

and thus the assertion (4.25) is true. Similarly, if f and g satisfy (4.17), then we have

$$\begin{aligned} \lim \sup _{s\rightarrow \infty }(\frac{\epsilon s^3}{g^3(s)}+\frac{s^2}{g^3(s)}+\frac{pf(s)s}{g^3(s)})\le p\lim \sup _{s\rightarrow \infty }\frac{f(s)s}{g^3(s)}=-p\mu <0, \end{aligned}$$

from this we also have

$$\begin{aligned} \lim \sup _{s\rightarrow \infty }F(\epsilon , s)=\lim \sup _{s\rightarrow \infty }g^3(s)s^{p-2}(\frac{\epsilon s^3}{g^3(s)}+\frac{s^2}{g^3(s)}+\frac{pf(s)s}{g^3(s)})<0, \end{aligned}$$

and thus the assertion (4.25) is also valid.

(4.25) in conjunction with (4.24) then yields

$$\begin{aligned} \begin{aligned} \frac{d}{dt}\int \limits _\Omega u^p+\int \limits _\Omega u^p&\le \epsilon \int \limits _\Omega u^{p+1}+\int \limits _\Omega u^p+p\int \limits _\Omega u^{p-1} f(u)+c(\epsilon )\\&\le \sup _{s\ge 0}F(\epsilon , s)|\Omega |+c(\epsilon )\\&\le C(\epsilon )\quad \text {for all}\,\,t\in (0, T^*). \end{aligned} \end{aligned}$$

(4.28)

This, together with the Gronwall inequality, yields

$$\begin{aligned} \int \limits _\Omega u^p\le C(p)\quad \text {for all}\,\,t\in (0, T^*). \end{aligned}$$

(4.29)

Then, similar to the proof of Theorem 1.1, it follows from (4.29), (4.22), as well as [28, Appendix A], that there exists a constant $C>0$ such that

$$\begin{aligned} \Vert u(\cdot ,t)\Vert _{L^\infty (\Omega )}\le C\quad \text {for all}\,\,t\in (0, T^*). \end{aligned}$$

This in conjunction with Lemma 2.1 proves Theorem 1.4.