具有间接信号吸收和Logistic源的生物趋化模型解的有界性

Boundedness of Solution for the Chemotaxis Model with Indirect Signal Absorption and Logistic Source

摘要: 本文考虑具有二维间接信号吸收的拟线性趋化模型：其中Ω∈Rⁿ(n=2)是一个有界区域且具有光滑边界，μ,l>0，非线性扩散系数D(u)和趋化敏感系数S(u)分别满足D(u)≥（u+1）^m-1，S(u)≤（u+1）^q-1且D(⋅),S(⋅)∈C^1+l([0,∞))。本文利用能量方法和半群理论证明在和0 < q ≤ 2的条件下，该生物趋化模型的解全局有界，其中C,λ0为正常数。

Abstract: In this paper, we consider the following two-dimensional quasilinear chemotaxis model with in-direct signal absorption: on a bounded domain Ω∈Rn(n=2), with smooth boundary , μ and l are positive constants, the nonlinear diffusivity D(u) and chemosensitivity S(u) are supposed to satisfy D(u)≥（u+1）^m-1, S(u)≤（u+1）^q-1 and D(⋅),S(⋅)∈C^1+l([0,∞)). Finally, we use the energy method and the semigroup theory to prove that the solution of the biologicalchemotaxis model is globally bounded under the conditions and 0 < q ≤ 2, where C,λ₀ are the positive constants.

1. 引言

趋化现象是自然界中常见的现象，它描述的是细胞或细菌沿化学信号浓度梯度方向的定向运动 [1] [2] [3]。在过去几年里，一些学者致力于研究趋化消耗模型

$\{\begin{array}{l}{u}_t=\nabla \cdot \left(D\left(u\right)\nabla u\right)-\nabla \cdot \left(S\left(u\right)\nabla v\right)+f\left(u\right),x\in \Omega \times \left(0,T\right),\hfill \\ v_t=\Delta v-uv,x\in \Omega \times \left(0,T\right),\hfill \\ \frac{\partial u}{\partial \upsilon }=\frac{\partial v}{\partial \upsilon }=0,x\in \partial \Omega \times \left(0,T\right),\hfill \\ u\left(x,0\right)=u_0\left(x\right),v\left(x,0\right)=v_0\left(x\right),x\in \Omega ,\hfill \end{array}$

其中， $\Omega \in R^n$ 中表示具有光滑边界 $\partial \Omega$ 的有界区域， $u=u\left(x,t\right)$ 代表细胞密度， $v=\left(v,t\right)$ 代表化学信号物质的浓度， $D\left(u\right)$ 是扩散系数， $S\left(u\right)$ 是趋化敏感系数， $f\left(u\right)$ 是Logistic源，表示细胞的增殖和死亡， $-uv$ 是化学诱导剂的消耗。在 $D\left(u\right)=1$， $S\left(u\right)=u$， $f\left(u\right)=0$ 的情况下，当 $n\le 2$ 时， ${\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}$ 无小性限制，Tao在文献 [4] 证明了上述模型存在全局经典解。当 $n=3$ 时，初值足够光滑，Tao和Winkler在文献 [5] 中

讨论了上述趋化模型全局弱解的存在性。当 $n\ge 3$， ${\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}\le \frac{1}{6\left(n+1\right)\chi }$， $t\to \infty$ 时，模型的全局经典解

$\left(u,v\right)$ 收敛于 $\left({\bar{u}}_0,0\right)$，其中 ${\bar{u}}_0=\frac{1}{\bar{\Omega }}{\displaystyle {\int }_{\Omega }u\left(x,0\right)\text{d}x}$。在 $D\left(u\right)=1$， $S\left(u\right)=u$， $f\left(u\right)=\kappa u-\mu u^2$ 的情况下，Lankeit

和Wang在文献 [6] 中证明了当在 $\mu$ 足够大时该模型存在全局经典解，在 $\mu >0$ 时该模型存在经典弱解。

Zheng在文献 [7] 中证明了 $D\left(u\right)\ge C_D{\left(u+1\right)}^{m-1}$， $S\left(u\right)=u$， $C_D>0$， $m>\{\begin{array}{l}1-\frac{\mu }{\chi \left[1+{\lambda }_0{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}{2}^3\right]},N\le 2\hfill \\ 1,N\ge 3\hfill \end{array}$

时，对于任意足够光滑的初值都存在一个经典有界解。

不同于一般的直接信号吸收的趋化模型，近几年关于间接信号吸收的生物趋化模型

$\{\begin{array}{l}{u}_t=\nabla \cdot \left(D\left(u\right)\nabla u\right)-\nabla \cdot \left(S\left(u\right)\nabla v\right)+\mu \left(u-u^2\right),x\in \Omega \times \left(0,T\right),\hfill \\ v_t=\Delta v-vw,x\in \Omega \times \left(0,T\right),\hfill \\ w_t=-\delta w+u,x\in \Omega \times \left(0,T\right),\hfill \\ \frac{\partial u}{\partial \upsilon }=\frac{\partial v}{\partial \upsilon }=0,x\in \partial \Omega \times \left(0,T\right),\hfill \\ u\left(x,0\right)=u_0\left(x\right),v\left(x,0\right)=v_0\left(x\right),w\left(x,0\right)=w_0\left(x\right),x\in \Omega ,\hfill \end{array}$ (1.1)

其中 $\Omega \in R^n$ 具有光滑边界的有界区域， $n\in {\rm N}$， $\delta >0$ 是给定的参数。2019年，Fuest在文献 [8] 中讨论当

$D\left(u\right)=1$， $S\left(u\right)=u$， $\mu =0$， $n\le 2$ 或 $n\ge 3$， ${\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}\le \frac{1}{3n}$ 时该模型存在唯一的全局经典解。2020年，

Liu，Li和Huang在文献 [9] 中证明了具有Logistics源的间接信号吸收模型，当 $D\left(u\right)=1$， $S\left(u\right)=u$， $n\ge 4$，

$\mu$ 足够大时，模型的解有全局存在性和有界性，当 $\mu =0$， $n\ge 3$ 和 $0\le {\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}\le \frac{\pi }{\sqrt{n}}$ 时上述模型存在唯

一的全局经典解。Zheng等人在文献 [10] 中证明了当 $n=2$，非线性扩散系数 $D\left(u\right)$ 和趋化敏感系数 $S\left(u\right)$ 满足 $D\left(u\right)\ge u^m$， $S\left(u\right)\le u^q$，当 $m>\max \left\{1,2q-3\right\}$ 时，上述模型的解全局有界。受文献 [7]， [9]， [10] 和 [11] 的启发，本文主要利用能量方法证明趋化模型(1.1)在 $\delta =1$， $n=2$， $\mu >0$ 的条件下，非线性扩散系数 $D\left(u\right)$ 和趋化敏感系数 $S\left(u\right)$ 满足

$D\left(u\right)\ge {\left(u+1\right)}^{m-1}$， $S\left(u\right)\le {\left(u+1\right)}^{q-1}$ 且 $D(\cdot ),S(\cdot )\in C^{1+l}\left(\left[0,\infty \right)\right)$， $l>0$ (1.2)

$m>q-1-\frac{\mu }{4\cdot \left(1+\bar{C}{\lambda }_0{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}^3{2}^3\right)}$ 且 $1<q\le 2$ (1.3)

成立，其中 $\bar{C},{\lambda }_0$ 为正常数，则该拟线性趋化模型的解全局有界。主要结论如下：

定理1：设 $n=2$，初值满足 $u_0\in C^0\left(\bar{\Omega }\right)$， $v_0\in W^{1,\infty }\left(\Omega \right)$， $w_0\in C^1\left(\bar{\Omega }\right)$ 且 $u_0\ge 0$， $v_0\ge 0$， $w_0\ge 0$，则存在一个非负函数 $\left(u,v,w\right)$ ：

$u\in C^0\left(\bar{\Omega }\right)\times \left[0,T_{\max }\right)\cap C^{2,1}\left(\bar{\Omega }\times \left(0,T_{\max }\right)\right)$，

$v\in C^0\left(\bar{\Omega }\right)\times \left[0,T_{\max }\right)\cap C^{2,1}\left(\bar{\Omega }\times \left(0,T_{\max }\right)\right)$，

$w\in C^{0,1}\left(\bar{\Omega }\right)\times \left[0,T_{\max }\right)$，

是模型(1.1)的经典解。当 $D\left(u\right)$ 和 $S\left(u\right)$ 满足(1.2)， $m>q-1-\frac{\mu }{4\cdot \left(1+\bar{C}{\lambda }_0{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}^3{2}^3\right)}$ 且 $1<q\le 2$ 时，则对

任意的 $t\in \left(0,\infty \right)$，存在一个常数 $C>0$，使得

${\Vert u\left(\cdot ,t\right)\Vert }_{L^{\infty }\left(\Omega \right)}\le C$， ${\Vert \nabla v\left(\cdot ,t\right)\Vert }_{L^{\infty }\left(\Omega \right)}\le C$ 和 ${\Vert w\left(\cdot ,t\right)\Vert }_{L^{\infty }\left(\Omega \right)}\le C$，

其中 $\bar{C},{\lambda }_0$ 为正常数。

注：文献 [10] 证明了当 $n=2$ 时， $m>\max \left\{1,2q-3\right\}$ 或 $m=0,0<q<\frac{3}{2}$，模型存在全局经典解。与文献 [10] 的结果相对比，本文证明了当 $m>q-1-\frac{\mu }{4\cdot \left(1+\bar{C}{\lambda }_0{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}^3{2}^3\right)}$， $1<q\le 2$ 时，获得了模型解的全局有界性，推广了文献 [10] 的结果。

2. 解的全局有界性

为了证明定理1的结果，先给出一个必要的引理。

引理1：设 $n=2$， $m>q-1-\frac{\mu }{4\cdot \left(1+\bar{C}{\lambda }_0{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}^3{2}^3\right)}$， $1<q\le 2$， ${\lambda }_0,\mu >0$。令 $\left(u,v,w\right)$ 是模型(1.1)

的解，则对任意 $p>1$， $t\in \left(0,T_{\max }\right)$，存在一个正常数C使得

${\displaystyle {\int }_{\Omega }{\left(u\left(\cdot ,t\right)+1\right)}^p}\le C$。

证明：第一步：当 $1<p<2$ 时，对趋化模型(1.1)的第一个方程乘 ${\left(u+1\right)}^{p-1}$ 并在 $\Omega$ 上积分，

$\begin{array}{l}\frac{1}{p}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^p\text{d}x}+\left(p-1\right){\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{m+p-3}}{\left|\nabla u\right|}^2\text{d}x\\ \le -\frac{p+1}{p}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^p\text{d}x}-{\displaystyle {\int }_{\Omega }\nabla \left(S\left(u\right)\nabla v\right)}{\left(u+1\right)}^{p-1}\text{d}x\\ +\frac{p+1}{p}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^p\text{d}x}+\mu {\displaystyle {\int }_{\Omega }\left(u-u^2\right){\left(u+1\right)}^{p-1}\text{d}x}\end{array}$ (2.1)

对(2.1)右端第二项运用Young’s不等式可得

$\begin{array}{c}-{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p-1}}\nabla \left(S\left(u\right)\nabla v\right)\text{d}x=\left(p-1\right){\displaystyle {\int }_{\Omega }S\left(u\right)}{\left(u+1\right)}^{p-2}\nabla u\cdot \nabla v\text{d}x\\ =\left(p-1\right){\displaystyle {\int }_{\Omega }\nabla \Psi \left(u\right)}\cdot \nabla v\text{d}x\\ =-\left(p-1\right){\displaystyle {\int }_{\Omega }\Psi \left(u\right)\Delta v}\\ \le \left(p-1\right){\displaystyle {\int }_{\Omega }\Psi \left(u\right)}\left|\Delta v\right|\\ \le \frac{\left(p-1\right)}{p+q-2}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+q-2}\left|\Delta v\right|\text{d}x}\\ \le \left(p-1\right){\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+q-1}\text{d}x}+\left(p-1\right){\displaystyle {\int }_{\Omega }{\left|\Delta v\right|}^{p+q-1}\text{d}x}+C_1\end{array}$ (2.2)

其中 $\Psi \left(u\right)={\displaystyle {\int }_0^{u}S\left(\sigma \right){\left(1+\sigma \right)}^{p-2}\cdot \nabla \sigma }$。

再对(2.1)右端第三项和第四项运用 ${\left(u+1\right)}^{\beta }\le 2^{\beta }\left(u^{\beta }+1\right)$ 可得

$\begin{array}{l}\frac{p+1}{p}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^p\text{d}x}+\mu {\displaystyle {\int }_{\Omega }\left(u-u^2\right){\left(u+1\right)}^{p-1}\text{d}x}\\ \le \frac{p+1}{p}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^p\text{d}x}+\mu {\displaystyle {\int }_{\Omega }{\left(u+1\right)}^p\text{d}x}+\mu {\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p-1}\text{d}x}-\frac{\mu }{4}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+1}\text{d}x}\\ \le \left({\epsilon }_1-\frac{\mu }{4}\right){\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+1}\text{d}x}+C_2\end{array}$，

其中， $C_2=\frac{1}{p+1}{\left({\epsilon }_1\frac{p+1}{p}\right)}^{-p}{\left(\frac{p+1}{p}+2\mu \right)}^{p+1}\left|\Omega \right|+\frac{2}{p+1}{\left(\frac{p+1}{p-1}\right)}^{-\frac{\left(p-1\right)}{2}}\mu \left|\Omega \right|$。

对(2.2)右端第一项运用Young’s不等式可得

$\left(p-1\right){\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+q-1}}dx\le \left(p-1\right){{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+1}+\left(p-1\right)\left|\Omega \right|}}^{\frac{p+1}{2-q}}$

最后整理可得

$\begin{array}{l}\frac{1}{p}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^p\text{d}x}+\left(p-1\right){\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{m+p-3}}{\left|\nabla u\right|}^2\text{d}x\\ \le -\frac{p+1}{p}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^p\text{d}x}+\left(p-1\right){\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+1}\text{d}x}\\ +\left(p-1\right){\displaystyle {\int }_{\Omega }{\left|\Delta v\right|}^{p+q-1}}+\left({\epsilon }_1-\frac{\mu }{4}\right){\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+1}\text{d}x}+C_3\end{array}$ (2.3)

对任意 $t\in \left(s_0,T_{\max }\right)$，对(2.3)运用常数变易法可得

$\begin{array}{l}\frac{1}{p}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^p\text{d}x}\\ \le \frac{1}{p}{\text{e}}^{-\left(p+1\right)\left(t-s_0\right)}{\Vert u\left(s_0\right)\Vert }_{L^p\left(\Omega \right)}^p+\left({\epsilon }_1+\left(p-1\right)-\frac{\mu }{4}\right){\displaystyle {\int }_{s_0}^t{\text{e}}^{-\left(p+1\right)\left(t-s\right)}}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+1}\text{d}x\text{d}s}\\ +\left(p-1\right){\displaystyle {\int }_{s_0}^t{\text{e}}^{-\left(p+1\right)\left(t-s\right)}}{\displaystyle {\int }_{\Omega }{\left|\Delta v\right|}^{p+q-1}\text{d}x\text{d}s}+C_3{\displaystyle {\int }_{s_0}^t{\text{e}}^{-\left(p+1\right)\left(t-s\right)}\text{d}s}\\ \le \left({\epsilon }_1+\left(p-1\right)-\frac{\mu }{4}\right){\displaystyle {\int }_{s_0}^t{\text{e}}^{-\left(p+1\right)\left(t-s\right)}}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+1}\text{d}x\text{d}s}\\ +\left(p-1\right){\displaystyle {\int }_{s_0}^t{\text{e}}^{-\left(p+1\right)\left(t-s\right)}}{\displaystyle {\int }_{\Omega }{\left|\Delta v\right|}^{p+q-1}\text{d}x\text{d}s}+C_4\end{array}$ (2.4)

其中 $C_4=\frac{1}{p}{\text{e}}^{-\left(p+1\right)\left(t-s_0\right)}{\Vert u\left(s_0\right)\Vert }_{L^p\left(\Omega \right)}^p+C_3{\displaystyle {\int }_{s_0}^t{\text{e}}^{-\left(p+1\right)\left(t-s\right)}\text{d}s}$。

对任意的 $s_0\in \left(0,T_{\max }\right)$， $s_0\le 1$，令 $t\in \left(s_0,T_{\max }\right)$，并对模型(1.1)第二个方程变形可得

$v_t-\Delta v+v=-vw+v$。

对(2.4)右端第二项运用文献 [6] 的引理3.3和文献 [7] 的引理2.3可得

$\begin{array}{l}\left(p-1\right){\displaystyle {\int }_{s_0}^t{\text{e}}^{-\left(p+1\right)\left(t-s\right)}}{\displaystyle {\int }_{\Omega }{\left|\Delta v\right|}^{p+q-1}\text{d}x\text{d}s}\\ \le \left(p-1\right){\text{e}}^{-\left(p+1\right)t}{\lambda }_0\left[{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}^{p+q-1}{2}^{p+q-1}{\displaystyle {\int }_{s_0}^t{\text{e}}^{\left(p+1\right)s}{\displaystyle {\int }_{\Omega }\left(w^{p+q-1}+1\right)\text{d}x\text{d}s}+{\text{e}}^{\left(p+1\right)s_0}{\Vert v\left(s_0,t\right)\Vert }_{L^{2,p+q-1}\left(\Omega \right)}^{p+q-1}}\right]\end{array}$ (2.5)

对(2.5)右端运用Young’s不等式可得

${\displaystyle {\int }_{\Omega }{w}^{p+q-1}\text{d}x}<{\displaystyle {\int }_{\Omega }{w}^{p+1}\text{d}x}+C_5$ (2.6)

对模型(1.1)第三个方程两边同乘以 $w^p$ 并在 $\Omega$ 上积分，

$\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }{w}^{p+1}}+\frac{\left(p+1\right)}{2}{\displaystyle {\int }_{\Omega }{w}^{p+1}}\text{d}x\le \left(p+1\right)C_6{\displaystyle {\int }_{\Omega }{u}^{p+1}}\text{d}x$ (2.7)

再对(2.7)运用常数变易法可得

${\displaystyle {\int }_{\Omega }{w}^{p+1}}\le {\text{e}}^{-\frac{\left(p+1\right)}{2}\left(t-s_0\right)}{\Vert w\left(s_0\right)\Vert }_{L^{p+1}}^{p+1}+\left(p+1\right)C_6{\displaystyle {\int }_{s_0}^t{\text{e}}^{-\frac{p+1}{2}\left(t-s\right)}}{\displaystyle {\int }_{\Omega }{u}^{p+1}}\text{d}x\text{d}s$ (2.8)

将(2.8)代入到(2.6)中可得

${\displaystyle {\int }_{\Omega }{w}^{p+q-1}}\text{d}x\le C_7{\displaystyle {\int }_{\Omega }{u}^{p+1}}\text{d}x\text{d}s+C_8<C_7{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+1}}\text{d}x\text{d}s+C_8$ (2.9)

其中 $\bar{C}=C_7=\left(p+1\right)C_6{\displaystyle {\int }_{s_0}^t{\text{e}}^{-\frac{p+1}{2}\left(t-s\right)}}\text{d}s$， $C_8={\text{e}}^{-\frac{\left(p+1\right)}{2}\left(t-s_0\right)}{\Vert w\left(s_0\right)\Vert }_{L^{p+1}}^{p+1}+C_5$。

整理(2.4)，(2.5)和(2.9)可得

$\begin{array}{c}\frac{1}{p}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^p}\le \left({\epsilon }_1+\left(p-1\right)+\bar{C}\left(p-1\right){\lambda }_0{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}^{p+q-1}{2}^{p+q-1}-\frac{\mu }{4}\right){\displaystyle {\int }_{s_0}^t{\text{e}}^{-\left(p+1\right)\left(t-s\right)}}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+1}}\text{d}x\text{d}s\\ +C_9\left(p-1\right){\lambda }_0{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}^{p+q-1}{2}^{p+q-1}{\text{e}}^{-\left(p+1\right)t}{\displaystyle {\int }_{s_0}^t{\displaystyle {\int }_{\Omega }{\text{e}}^{\left(p+1\right)s}}\text{d}x\text{d}s}\\ +\left(p-1\right){\lambda }_0{\text{e}}^{-\left(p+1\right)\left(t-s_0\right)}{\Vert v\left(s_0,t\right)\Vert }_{L^{2,p+q-1}\left(\Omega \right)}^{p+q-1}\end{array}$

令 $p=p_0:=1+\frac{\mu }{4\cdot \left(1+\bar{C}{\lambda }_0{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}^3{2}^3\right)}>1$，当 $p<2,1<q\le 2$ 时，有

$\mu =4\cdot \left(\left(p_0-1\right)+\bar{C}\left(p_0-1\right){\lambda }_0{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}^3{2}^3\right)>4\cdot \left(\left(p_0-1\right)+\bar{C}\left(p_0-1\right){\lambda }_0{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}^{p_0+q-1}{2}^{p_0+q-1}\right)$

当 $0<{\epsilon }_1<\frac{\mu }{4}-\left(p_0-1\right)+\bar{C}\left(p_0-1\right){\lambda }_0{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}^{p_0+q-1}{2}^{p_0+q-1}$ 时，存在一个正常数 $C_{10}$，使得

${\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p_0}}\text{d}x\le C_{10}$

再令 $p<\frac{2p_0}{{\left(2-p_0\right)}^{+}},\alpha >\frac{1}{2}$，则

$p<\frac{1}{\frac{1}{p_0}-\frac{1}{2}+\frac{2}{2}\left(\alpha -\frac{1}{2}\right)}\le \frac{2p_0}{{\left(2-p_0\right)}^{+}}$ (2.10)

由常数变易法和Hölder’s不等式可得

$v\left(t\right)={\text{e}}^{-\tau \left(A+1\right)}v\left(s_0\right)+{\displaystyle {\int }_{s_0}^t{\text{e}}^{-\left(t-s\right)\left(A+1\right)}}\left(-v\left(s\right)w\left(s\right)+v\left(s\right)\right)\text{d}s$ (2.11)

$\begin{array}{c}{\Vert w\left(\cdot ,t\right)\Vert }_{L^{p_0}\left(\Omega \right)}\le {\Vert w_0\Vert }_{L^{p_0}\left(\Omega \right)}+{\displaystyle {\int }_0^t{\text{e}}^{-\left(t-s\right)}{\Vert u\left(\cdot ,s\right)\Vert }_{L^{p_0}\left(\Omega \right)}}\text{d}s\\ \le {\Vert w_0\Vert }_{L^{p_0}\left(\Omega \right)}+{\Vert u\left(\cdot ,s\right)\Vert }_{L^{p_0}\left(\Omega \right)}{\left(\frac{p_0}{p_0-1}\right)}^{-\frac{p_0-1}{p_0}}\le C_{11}\end{array}$ (2.12)

其中 $\tau \in \left[0,s_0\right]$，再运用文献 [6] 的引理3.3和(2.11)可得

$\begin{array}{l}{\Vert {\left(A+1\right)}^{\alpha }v\left(t\right)\Vert }_{L^p\left(\Omega \right)}\\ \le C_{12}{s}_0^{-\alpha -\frac{2}{2}\left(1-\frac{1}{p}\right)}{\Vert v\left(s_0,t\right)\Vert }_{L^1\left(\Omega \right)}+C_{12}{\displaystyle {\int }_{s_0}^t{\left(t-s\right)}^{-\alpha -\frac{2}{2}\left(\frac{1}{p_0}-\frac{1}{p}\right)}{\text{e}}^{-\mu \left(t-s\right)}}{\Vert -v\left(s\right)w\left(s\right)+v\left(s\right)\Vert }_{L^{p_0}\left(\Omega \right)}\text{d}s\\ \le C_{13}{\displaystyle {\int }_0^{+\infty }{\sigma }^{-\alpha -\frac{2}{2}\left(\frac{1}{p_0}-\frac{1}{p}\right)}}{\text{e}}^{-\mu \sigma }\text{d}\sigma +C_{14}{s}_0^{-\alpha -\frac{2}{2}\left(1-\frac{1}{p}\right)}\end{array}$ (2.13)

由(2.10)，(2.13)和文献 [10] 的引理3.1可知对所有的 $t\in \left(0,T_{\max }\right),p\in \left[1,\frac{2p_0}{{\left(2-p_0\right)}^{+}}\right)$ 有

${\displaystyle {\int }_{\Omega }{\left|\nabla v\right|}^p}\le C_{15}$。

第二步：对任意的 $p>1$，在模型(1.1)的第一个方程两边同乘 ${\left(u+1\right)}^{p-1}$ 并在 $\Omega$ 上积分并运用Young’s不等式可得

$\begin{array}{l}\frac{1}{p}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^p\text{d}x}+\left(p-1\right){\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{m+p-3}}{\left|\nabla u\right|}^2\text{d}x\\ \le \frac{p-1}{2}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{m+p-3}{\left|\nabla u\right|}^2\text{d}x}+\frac{p-1}{2}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+2q-m-3}{\left|\nabla v\right|}^2\text{d}x}-\frac{\mu }{2}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+1}\text{d}x+C}\end{array}$ (2.14)

令 $1<l_0<\frac{2p_0}{2{\left(2-p_0\right)}^{+}}$，对(2.14)右端第二项运用Hölder’s不等式可得

$\begin{array}{l}\frac{p-1}{2}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+2q-m-3}{\left|\nabla v\right|}^2\text{d}x}\\ \le \frac{p-1}{2}{\left({\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{\frac{l_0}{l_0-1}\left(p+2q-m-3\right)}}\right)}^{\frac{l_0-1}{l_0}}{\left({\displaystyle {\int }_{\Omega }{\left|\nabla v\right|}^{2l_0}}\right)}^{\frac{1}{l_0}}\text{d}x\\ \le C{\Vert {\left(u+1\right)}^{\frac{m+p-1}{2}}\Vert }_{L^{2\frac{l_0}{l_0-1}\frac{p+2q-m-3}{m+p-1}}}^{2\frac{p+2q-m-3}{m+p-1}}\end{array}$ (2.15)

由于 $l_0>1,p>\max \left\{2q-m-3,p_0-\frac{p_0}{l_0}-2q+m+3\right\}$，所以

$\frac{p_0}{m+p-1}\le \frac{l_0}{l_0-1}\frac{p+2q-m-3}{m+p-1}<\infty$，

对(2.15)右端运用文献 [7] 的引理2.1可得

$\begin{array}{l}C{\Vert {\left(u+1\right)}^{\frac{m+p-1}{2}}\Vert }_{L^{2\frac{l_0}{l_0-1}\frac{p+2q-m-3}{m+p-1}}}^{2\frac{p+2q-m-3}{m+p-1}}\\ \le C{\left({\Vert \nabla {\left(u+1\right)}^{\frac{m+p-1}{2}}\Vert }_{L^2\left(\Omega \right)}^{{\mu }_1}{\Vert {\left(u+1\right)}^{\frac{m+p-1}{2}}\Vert }_{L^{\frac{2p_0}{m+p-1}}\left(\Omega \right)}^{1-{\mu }_1}+{\Vert {\left(u+1\right)}^{\frac{m+p-1}{2}}\Vert }_{L^{\frac{2p_0}{m+p-1}}\left(\Omega \right)}\right)}^{2\frac{p+2q-m-3}{m+p-1}}\\ \le C\left({\Vert \nabla {\left(u+1\right)}^{\frac{m+p-1}{2}}\Vert }_{L^2\left(\Omega \right)}^{2{\mu }_1^{\frac{p+2q-m-3}{m+p-1}}}+1\right)\\ \le C\left({\Vert \nabla {\left(u+1\right)}^{\frac{m+p-1}{2}}\Vert }_{L^2\left(\Omega \right)}^{2\frac{l_0\left(p+2q-m-3\right)-p_0\left(l_0-1\right)}{l_0\left(m+p-1\right)}}+1\right)\end{array}$

其中， ${\mu }_1=\frac{2\frac{m+p-1}{2p_0}-\frac{2\left(l_0-1\right)\left(m+p-1\right)}{2l_0\left(p+2q-m-3\right)}}{1-\frac{2}{2}+\frac{2\left(m+p-1\right)}{2p_0}}=\left(m+p-1\right)\frac{\frac{2}{2p_0}-\frac{2\left(l_0-1\right)}{2l_0\left(p+2q-m-3\right)}}{1-\frac{2}{2}+\frac{2\left(m+p-1\right)}{2p_0}}\in \left(0,1\right)$。

由 $p=p_0:=1+\frac{\mu }{4\cdot \left(1+\bar{C}{\lambda }_0{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}^3{2}^3\right)}>1$， $l_0<\frac{2p_0}{2{\left(2-p_0\right)}^{+}}$ 和 $m>q-1-\frac{\mu }{4\cdot \left(1+\bar{C}{\lambda }_0{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}^3{2}^3\right)}$ 可知

$\frac{l_0\left(p+2q-m-3\right)-p_0\left(l_0-1\right)}{l_0\left(m+p-1\right)}<1$。

最后，利用Young’s不等式整理可得

$\frac{1}{p}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^p\text{d}x}+\frac{\left(p-1\right)}{4}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{m+p-3}}{\left|\nabla u\right|}^2\text{d}x+\frac{\mu }{2}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+1}\text{d}x}\le C$ (2.16)

对(2.16)在 $\left(0,t\right)$ 上积分可得对任意的 $p>1$，有

${\displaystyle {\int }_{\Omega }{\left(u+1\right)}^p\text{d}x}\le C$。

定理1的证明：首先，假设 $\left(u,v,w\right)$ 是模型(1.1)的解，对任意的 $t\in \left(0,T_{\max }\right)$，存在常数 $C>0$，由一阶常微分方程理论和Hölder’s不等式可得

${\Vert w\left(\cdot ,t\right)\Vert }_{L^p\left(\Omega \right)}\le {\Vert w_0\Vert }_{L^p\left(\Omega \right)}+{\displaystyle {\int }_0^t{\text{e}}^{-\left(t-s\right)}{\Vert u\left(\cdot ,s\right)\Vert }_{L^p\left(\Omega \right)}}\text{d}s\le {\Vert w_0\Vert }_{L^p\left(\Omega \right)}+{\Vert u\left(\cdot ,s\right)\Vert }_{L^p\left(\Omega \right)}{\left(\frac{p}{p-1}\right)}^{-\frac{p-1}{p}}\le C_{16}$ (2.17)

又因为 $p>n$ 时， $\theta \in \left[1,\infty \right]$ 可得

${\Vert \nabla v\left(\cdot ,t\right)\Vert }_{L^{\theta }\left(\Omega \right)}\le C_{17}{\Vert \nabla v_0\Vert }_{L^{\infty }\left(\Omega \right)}+C_{18}{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}{\Vert w\left(\cdot ,t\right)\Vert }_{L^p\left(\Omega \right)}{\displaystyle {\int }_0^{\infty }{\left(1+s^{\varphi }\right)}^{-\lambda s}\text{d}s}\le C$ (2.18)

其中 $\varphi >-1$，即可证明 ${\Vert \nabla v\left(\cdot ,t\right)\Vert }_{L^{\infty }\left(\Omega \right)}\le C$。

其次，当 $m>q-1-\frac{\mu }{4\cdot \left(1+\bar{C}{\lambda }_0{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}^3{2}^3\right)}$， $1<q\le 2$。当 $D\left(u\right)$ 和 $S\left(u\right)$ 满足(1.2)，对任意的 $p>1$，

模型(1.1)两边同乘 ${\left(u+1\right)}^{p-1}$ 并在 $\Omega$ 上积分可得

$\begin{array}{l}\frac{1}{p}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^p\text{d}x}+\left(p-1\right){\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{m+p-3}}{\left|\nabla u\right|}^2\text{d}x\\ \le \frac{p-1}{4}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{m+p-3}}{\left|\nabla u\right|}^2\text{d}x+\left(p-1\right)C_{19}^2{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+2q-m-3}\text{d}x}+\mu {\displaystyle {\int }_{\Omega }\left(u-u^2\right){\left(u+1\right)}^{p-1}\text{d}x}\\ \le \frac{p-1}{4}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{m+p-3}}{\left|\nabla u\right|}^2\text{d}x+C_{20}p{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+2q-m-3}\text{d}x-}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^p\text{d}x}-\frac{\mu }{4}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+1}\text{d}x}\end{array}$ (2.19)

其中 $C_{20}=C_{19}^2+2\mu +1$， $q-1-\frac{\mu }{4\cdot \left(1+\bar{C}{\lambda }_0{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}^3{2}^3\right)}<m<2q-3$，对上式右端第二项运用文献 [7] 的引理2.1和Young’s不等式可得

$\begin{array}{l}{C}_{20}p{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^{p+2q-m-3}}\text{d}x\\ \le C_{21}\left({\Vert \nabla {\left(u+1\right)}^{\frac{m+p-1}{2}}\Vert }_{L^2\left(\Omega \right)}^{\frac{2\left(p+2q-m-3\right)}{m+p-1}{\varsigma }_1}{\Vert {\left(u+1\right)}^{\frac{m+p-1}{2}}\Vert }_{L^1\left(\Omega \right)}^{\frac{2\left(p+2q-m-3\right)}{m+p-1}1-{\varsigma }_1}+{\Vert {\left(u+1\right)}^{\frac{m+p-1}{2}}\Vert }_{L^1\left(\Omega \right)}^{\frac{2\left(p+2q-m-3\right)}{m+p-1}}\right)\\ \le C_{22}{\Vert \nabla {\left(u+1\right)}^{\frac{m+p-1}{2}}\Vert }_{L^2\left(\Omega \right)}^2+C_{23}{\Vert {\left(u+1\right)}^{\frac{m+p-1}{2}}\Vert }_{L^1\left(\Omega \right)}^{\frac{2\left(p+2q-m-3\right)}{m+p-1}}\end{array}$ (2.20)

整理(2.19)和(2.20)可得

$\frac{1}{p}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^p\text{d}x}+{\displaystyle {\int }_{\Omega }{\left(u+1\right)}^p\text{d}x}+C_{20}{{\displaystyle {\int }_{\Omega }\left|\nabla {\left(u+1\right)}^{\frac{m+p-1}{2}}\right|}}^2\text{d}x\le C_{24}$， (2.21)

再运用Gronwall不等式可得

${\Vert u\left(\cdot ,t\right)\Vert }_{L^p\left(\Omega \right)}\le C_{24}$。

再对上式运用标准的Alikakos-Moser迭代即可得到

${\Vert u\left(\cdot ,t\right)\Vert }_{L^{\infty }\left(\Omega \right)}\le C$。

最后，对模型(1.1)的第三个方程求一阶线性常微分方程的解，显然可得

${\Vert w\left(\cdot ,t\right)\Vert }_{L^{\infty }\left(\Omega \right)}\le C$。

从而定理1得证。

基金项目

新疆维吾尔自治区自然科学基金项目(NO. 2018D01C004)。

参考文献

NOTES

^*通讯作者。