带Logistic源的奇异趋化模型解的整体存在性

Global Existence of Solutions to a Singular Chemotaxis System with Logistic Source

摘要: 本文考虑带有齐次Numann边界的抛物–抛物趋化系统：，，其中，为有界光滑区域。证明得到若及且，则系统存在整体古典解。

Abstract: This paper deals with the chemotaxis system with signal-dependent sensitivity and logistic source under homogenous Neumann boundary condition: , , where  and  is a bounded smooth domain. If with and , then the system admits a globally bounded classical solution.

1. 引言

趋化性是指细胞占据一个空间，由其中不均匀分布的物质产生的化学信号刺激细胞的运动。对细胞的不同研究和实验一直在探索细胞如何引导其自然运动，并根据化学梯度刺激的强度随机改变其运动路线。趋化系统在1970年由Keller和Segel提出，它描述了细胞受化学信号刺激所产生的聚集现象。这一系统在过去几年得到了广泛的研究，相关的趋化模型及生物学背景概述参阅 [1]。我们考虑如下的带有奇异灵敏度的趋化模型：

$\{\begin{array}{l}{u}_t=\Delta u-\chi \nabla \cdot \left(\frac{u}{v}\nabla v\right)+f\left(u\right),x\in \Omega ,t>0,\\ v_t=\Delta v-v+u,x\in \Omega ,t>0,\end{array}$ (1.1)

其中 $\frac{\chi }{v}$ 是由韦伯–费希纳定律确定的，刻画细胞向化学信号运动的趋化强度。下面介绍关于系统(1.1)的相关结论。若 $f\left(u\right)\equiv 0$，当 $n=2$ 时，如果 $\chi <{\chi }_0$ ( ${\chi }_0>1.01$ )或者当 $n\ge 3$ 时，如果 $\chi <\sqrt{\frac{2}{n}}$，则解整体有界 [2] [3]。若 $f\left(u\right)=ru-\mu u^k$， $r,\mu >0$，当 $\chi$ 相对于r恰当小，系统(1.1)存在一个全局有界古典解 [4]。

对于如下带有奇异灵敏度的模型：

$\{\begin{array}{l}{u}_t=\Delta u-\chi \nabla \cdot \left(\frac{u}{v^{\alpha }}\nabla v\right)+f\left(u\right),x\in \Omega ,t>0,\\ v_t=\Delta v-uv,x\in \Omega ,t>0,\end{array}$ (1.2)

其中 $\chi >0$。该系统可以追溯到Keller和Segel研究的流动的大肠杆菌游动带的形成。 $-uv$ 表示细胞u在趋化过程中氧气v被消耗，这自然表明v在时间上保持有界。关于解的整体存在性，人们对它的研究不如对其信号产生的研究那么广泛。 $f\left(u\right)=ru-\mu u^k$ 的存在有助于推导适当的先验估计，并确保经典解或最终正则化的弱解的全局存在。若 $\alpha \in \left(0,1\right]$ 且 $f\left(u\right)\equiv 0$，当 $\chi >0$，若 ${\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}<{\chi }^{\frac{1}{\alpha -1}}$ 和 ${\displaystyle {\int }_{\Omega }{u}_0}<\frac{1-\chi {\Vert v_0\Vert }^{1-\alpha }}{2C_{GN}^4}$，则系统(1.2)存在整体解 [5]。若 $\alpha =1$ 且 $f\left(u\right)\equiv ru-\mu u^2$，对于 $n\ge 2$，当 $0<\chi <\sqrt{\frac{2}{n}}$ 且 $\mu >\frac{n-2}{n}$ 时，系统 (1.2)存在全局经典解；对于 $n=1$，当 $\chi <2$ 且 $\mu >0$ 时，系统(1.2)存在整体有界解 [6]，另外，在二维情况下， [7] 证明了 $\chi >1$ 及 $\mu >0$ 时古典解的整体存在。

本文将考虑如下的带有奇异灵敏度和logistic源的趋化系统：

$\{\begin{array}{l}{u}_t=\Delta u-\chi \nabla \cdot \left(\frac{u}{v^{\alpha }}\nabla v\right)+ku-\mu u^2,x\in \Omega ,t>0,\\ v_t=\Delta v-uv,x\in \Omega ,t>0,\\ {\partial }_{\nu }u={\partial }_{\nu }v=0,x\in \partial \Omega ,t>0,\\ u\left(\cdot ,0\right)=u_0,v\left(\cdot ,0\right)=v_0,x\in \Omega ,\end{array}$ (1.3)

其中 $\chi ,\alpha ,k,\mu >0$。 $\Omega \in {ℝ}^2\left(n\ge 2\right)$ 是光滑有界区域。初始值满足

$u_0\subset C^0\left(\bar{\Omega }\right),v_0\subset W^{1,\infty }\left(\Omega \right),u_0\ge 0,v_0\ge 0.$ (1.4)

我们将证明系统(1.3)古典解的整体存在性。

定理1.1. 设 $n\ge 2$ 及 $\alpha \in \left(0,1\right]$。若 ${\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}<{\chi }^{\frac{1}{1-\alpha }}\sqrt{\frac{n}{2}}$ 且 $\mu >\frac{n-2}{2n}$，则系统存在整体古典解。

2. 古典解的整体存在性

在这一章节中，我们主要证明古典解的整体存在性。首先陈述方程(1.3)古典解的局部存在性，证明的具体细节见文献 [7]。

引理2.1. 令 $\alpha >0$，初值 $\left(u_0,v_0\right)$ 满足(1.4)。则方程(1.3)存在 $\left(u,v\right)\in {\left(C^0\left(\bar{\Omega }\times \left(0,T_{\max }\right)\right)\cap C^{2,1}\left(\bar{\Omega }\times \left(0,T_{\max }\right)\right)\right)}^2$ 在古典解意义下满足系统(1.4)。此外， $T_{\max }=\infty$ 或者 $\lim {\sup }_{t\to T_{\max }}{\Vert u\Vert }_{L^{\infty }\left(\Omega \right)}=\infty$。

引理2.2. 令 $\alpha >0$，则 ${\Vert v\left(\cdot ,t\right)\Vert }_{L^{\infty }\left(\Omega \right)}\le {\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}$， $t\in \left[0,T_{\max }\right)$。

证明. 根据 $u,v$ 的正性，由比较原理得证。

我们通过变换 $w=-\ln \frac{v}{{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}}$ 的方法来估计v的逐点下界。

引理2.3. 令 $\alpha \in \left(0,1\right]$，如果对于 $p>\frac{n}{2}$，存在 $L_1\left(t\right)>0$ 使得 ${\Vert u\Vert }_{L^p\left(\Omega \right)}\le L_1\left(t\right)$， $t\in \left[0,T_{\max }\right)$，那么存在 $d\left(t\right)>0$，使得

$v\ge {\Vert v\Vert }_{L^{\infty }\left(\Omega \right)}{\text{e}}^{-d\left(t\right)},\Omega \times \left(0,T_{\max }\right).$ (2.1)

证明. 令 $w=-\ln \frac{v}{{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}}$，则

$\{\begin{array}{l}{w}_t=\Delta w-{\left|\nabla w\right|}^2+u,\Omega \times \left(0,T_{\max }\right),\\ {\partial }_{\nu }w=0,\partial \Omega \times \left(0,T_{\max }\right),\\ w\left(\cdot ,0\right)=w_0:=-\ln \frac{v}{{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}},\Omega .\end{array}$ (2.2)

通过常数变易法，有

$\begin{array}{c}w={\text{e}}^{t\Delta }{w}_0+{\displaystyle {\int }_0^t{\text{e}}^{\left(t-s\right)\Delta }{\left(u-\left|\nabla w\right|\right)}^2}\\ \le {\text{e}}^{t\Delta }{w}_0+{\displaystyle {\int }_0^t{\text{e}}^{\left(t-s\right)\Delta }u\text{d}s},t\in \left(0,T_{\max }\right).\end{array}$

利用半群估计( [8], Lemma 1.3(i))，知存在 $c_1>0$ 使得

$\begin{array}{c}{\Vert w\Vert }_{L^{\infty }\left(\Omega \right)}\le {\Vert {\text{e}}^{t\Delta }{w}_0\Vert }_{L^{\infty }\left(\Omega \right)}+{\displaystyle {\int }_0^t{\Vert {\text{e}}^{\left(t-s\right)\Delta }u\Vert }_{L^{\infty }\left(\Omega \right)}\text{d}s}\\ \le {\Vert w_0\Vert }_{L^{\infty }\left(\Omega \right)}+c_1{\displaystyle {\int }_0^t\left(1+{\left(t-s\right)}^{-\frac{n}{2p}}\right){\text{e}}^{-{\lambda }_1\left(t-s\right)}{\Vert u\Vert }_{L^p\left(\Omega \right)}\text{d}s},t\in \left(0,T_{\max }\right).\end{array}$

当 $p>\frac{n}{2}$ 时，式子 ${\displaystyle {\int }_0^{\infty }\left(1+{\left(t-s\right)}^{-\frac{n}{2p}}\right){\text{e}}^{-{\lambda }_1\left(t-s\right)}{\Vert u\Vert }_{L^p\left(\Omega \right)}\text{d}s}$ 是可积的，又因为 ${\Vert u\Vert }_{L^p\left(\Omega \right)}<L_1\left(t\right)$，所以可得

${\Vert w\Vert }_{L^{\infty }\left(\Omega \right)}\le {\Vert w_0\Vert }_{L^{\infty }\left(\Omega \right)}+c_1{L}_1\left(t\right){\displaystyle {\int }_0^t\left(1+{\left(t-s\right)}^{-\frac{n}{2p}}\right){\text{e}}^{-{\lambda }_1\left(t-s\right)}\text{d}s}=:d\left(t\right).$

由w的定义可知 $v\ge {\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}{\text{e}}^{-d\left(t\right)}$ 于 $\Omega \times \left(0,T_{\max }\right)$。

引理2.4. 令 $\alpha \in \left(0,1\right]$。当 $p>\frac{n}{2}\left(n\ge 2\right)$ 时，如果 ${\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}<{\chi }^{\frac{1}{1-\alpha }}\sqrt{\frac{n}{2}}$ 并且 $\mu >\frac{n-2}{2n}$，则存在 $L_2\left(t\right)>0$ 使得

${\Vert u\Vert }_{L^p\left(\Omega \right)}\le L_2\left(t\right),t\in \left[0,T_{\max }\right).$ (2.3)

证明. 对于 $p>\frac{n}{2}$， $r>0$，由方程(1.3)得到

$\begin{array}{c}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }{u}^p{v}^{-r}\text{d}x}=p{\displaystyle {\int }_{\Omega }{u}^{p-1}{v}^{-r}{u}_t\text{d}x}-r{\displaystyle {\int }_{\Omega }{u}^p{v}^{-r-1}{v}_t\text{d}x}\\ =p{\displaystyle {\int }_{\Omega }{u}^{p-1}{v}^{-r}\Delta u\text{d}x}-p\chi {\displaystyle {\int }_{\Omega }{u}^{p-1}{v}^{-r}\nabla \cdot \left(\frac{u}{v_{\alpha }}\nabla v\right)\text{d}x}\\ -r{\displaystyle {\int }_{\Omega }{u}^p{v}^{-r-1}\Delta v}+pk{\displaystyle {\int }_{\Omega }{u}^p{v}^{-r}\text{d}x}+\left(r-\mu p\right){\displaystyle {\int }_{\Omega }{u}^{p+1}{v}^{-r}\text{d}x}\end{array}$

$\begin{array}{l}=-p\left(p-1\right){\displaystyle {\int }_{\Omega }{u}^{p-2}{v}^{-r}{\left|\nabla u\right|}^2\text{d}x}-pr\chi {\displaystyle {\int }_{\Omega }{u}^p{v}^{-r-\alpha -1}{\left|\nabla v\right|}^2\text{d}x}\\ -r\left(r+1\right){\displaystyle {\int }_{\Omega }{u}^p{v}^{-r-2}{\left|\nabla v\right|}^2\text{d}x}+2pr{\displaystyle {\int }_{\Omega }{u}^{p-1}{v}^{-r-1}\nabla u\cdot \nabla v\text{d}x}\\ +p\left(p-1\right)\chi {\displaystyle {\int }_{\Omega }{u}^p{v}^{-r-\alpha }\nabla u\cdot \nabla v\text{d}x}+pk{\displaystyle {\int }_{\Omega }{u}^p{v}^{-r}\text{d}x}\\ +\left(r-\mu p\right){\displaystyle {\int }_{\Omega }{u}^{p+1}{v}^{-r}\text{d}x},t\in \left(0,T_{\max }\right).\end{array}$ (2.4)

通过杨氏不等式 得

$\begin{array}{l}2pr{\displaystyle {\int }_{\Omega }{u}^{p-1}{v}^{-r-1}\nabla u\cdot \nabla v\text{d}x}+p\left(p-1\right)\chi {\displaystyle {\int }_{\Omega }{u}^p{v}^{-r-\alpha }\nabla u\cdot \nabla v\text{d}x}\\ ={\displaystyle {\int }_{\Omega }\left[2pr+p\left(p-1\right)\chi v^{1-\alpha }\right]u^{p-1}{v}^{-r-1}\nabla u\cdot \nabla v\text{d}x}\\ \le p\left(p-1\right){\displaystyle {\int }_{\Omega }{u}^{p-2}{v}^{-r}{\left|\nabla u\right|}^2\text{d}x}+{\displaystyle {\int }_{\Omega }\left[2pr+\left(p-1\right)\chi v^{1-\alpha }\right]u^{p-1}{v}^{-r-1}\nabla u\cdot \nabla v\text{d}x}\\ \le p\left(p-1\right){\displaystyle {\int }_{\Omega }{u}^{p-2}{v}^{-r}{\left|\nabla u\right|}^2\text{d}x}+{\displaystyle {\int }_{\Omega }\frac{{\left[2pr+p\left(p-1\right)\chi v^{1-\alpha }\right]}^2}{4p\left(p-1\right)}}{\displaystyle {\int }_{\Omega }{u}^p{v}^{-r-2}{\left|\nabla v\right|}^2\text{d}x}.\end{array}$ (2.5)

由估计(2.4)和(2.5)可得

$\begin{array}{l}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }{u}^p{v}^{-r}\text{d}x}\le {\displaystyle {\int }_{\Omega }\left\{\frac{{\left[2pr+p\left(p-1\right)\chi v^{1-\alpha }\right]}^2}{4p\left(p-1\right)}-r\left(r+1\right)-v^{1-\alpha }pr\chi \right\}}{\displaystyle {\int }_{\Omega }{u}^p{v}^{-r-2}{\left|\nabla v\right|}^2\text{d}x}\\ +pk{\displaystyle {\int }_{\Omega }{u}^p{v}^{-r}\text{d}x}+\left(r-\mu p\right){\displaystyle {\int }_{\Omega }{u}^{p+1}{v}^{-r}\text{d}x},t\in \left(0,T_{\max }\right).\end{array}$ (2.6)

定义

$h\left(r\right):=\frac{{\left[2pr+p\left(p-1\right)\chi v^{1-\alpha }\right]}^2}{4p\left(p-1\right)}-r\left(r+1\right)-v^{1-\alpha }pr\chi ,r>0.$

当 $r\in \left(r_{-},r_{+}\right)\left(r_{\pm }=\frac{p-1}{2}\left(1\pm \sqrt{1-p{\chi }^2{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}^{2-2\alpha }}\right)\right)$，有

$\begin{array}{c}h\left(r\right)=\frac{1}{p-1}\left[r^2-\left(p-1\right)r+\frac{p{\left(p-1\right)}^2{\chi }^2{v}^{2-2\alpha }}{4}\right]\\ \le \frac{1}{p-1}\left[r^2-\left(p-1\right)r+\frac{p{\left(p-1\right)}^2{\chi }^2{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}^{2-2\alpha }}{4}\right]\\ <0.\end{array}$

如果 ${\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}<{\chi }^{\frac{1}{1-\alpha }}\sqrt{\frac{n}{2}}$ 并且 $\mu >\frac{n-2}{2n}$，则存在 $p\in \left(\frac{n}{2},\frac{1}{{\chi }^2{\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}^{2-2\alpha }}\right)$ 使得 $r\in \left(r_{-},\min \left\{r_{+},\mu p\right\}\right)$。因此，由式(2.6)得

$\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }{u}^p{v}^{-r}\text{d}x}\le pk{\displaystyle {\int }_{\Omega }{u}^p{v}^{-r}\text{d}x},t\in \left(0,T_{\max }\right).$

进而 有

${\displaystyle {\int }_{\Omega }{u}^p{v}^{-r}\text{d}x}\le {\text{e}}^{pkt}{\displaystyle {\int }_{\Omega }{u}_0^p{v}_0^{-r}\text{d}x}:=C_2\left(t\right),t\in \left(0,T_{\max }\right).$

综上所述，可得

$\begin{array}{c}{\Vert u\Vert }_{L^p\left(\Omega \right)}={\left({\displaystyle {\int }_{\Omega }{u}^p{v}^{-r}{v}^r\text{d}x}\right)}^{\frac{1}{p}}\le {\left({\displaystyle {\int }_{\Omega }{u}^p{v}^{-r}{\Vert v^r\Vert }_{L^{\infty }\left(\Omega \right)}\text{d}x}\right)}^{\frac{1}{p}}\\ \le {\Vert v\Vert }_{L^{\infty }\left(\Omega \right)}^{\frac{r}{p}}\left({\left({\displaystyle {\int }_{\Omega }{u}^p{v}^{-r}\text{d}x}\right)}^{\frac{1}{p}}\right)\le {\Vert v\Vert }_{L^{\infty }\left(\Omega \right)}^{\frac{r}{p}}{C}_2{\left(t\right)}^{\frac{1}{p}}=:L_2\left(t\right).\end{array}$

证明完成。

证明定理1.1.

通过常数变易法，有

$\begin{array}{c}u={\text{e}}^{t\Delta }{u}_0+{\displaystyle {\int }_0^t{\text{e}}^{\left(t-s\right)\Delta }\nabla \cdot \left(\frac{u}{v^{\alpha }}\nabla v\right)\text{d}s}+{\displaystyle {\int }_0^t{\text{e}}^{\left(t-s\right)\Delta }\left(ku-\mu u^2\right)\text{d}s}\\ \le {\text{e}}^{t\Delta }{u}_0+{\displaystyle {\int }_0^t{\text{e}}^{\left(t-s\right)\Delta }\nabla \cdot \left(\frac{u}{v^{\alpha }}\nabla v\right)\text{d}s}+c_1{T}_{\max },t\in \left(0,T_{\max }\right)\end{array}$ (2.7)

其中 $c_1:={\sup }_{\xi >0}\left(k\xi -\mu {\xi }^2\right)$。

若 $p_1>\frac{n}{2}$，定义

$p_{\mathcal{l}+1}=\{\begin{array}{l}\frac{n{p}_{\mathcal{l}}}{2\left(n-p_{\mathcal{l}}\right)},p_{\mathcal{l}}\le n,\\ \infty ,p_{\mathcal{l}}>n,\end{array}\mathcal{l}=1,2,\cdots .$

当 $p_{\mathcal{l}}\le n$。为了去估计 ${\Vert u\Vert }_{L^{p_{\mathcal{l}+1}}\left(\Omega \right)}$，需要去控制 ${\displaystyle {\int }_0^t{\Vert {\text{e}}^{\left(t-s\right)\Delta }\nabla \cdot \left(\frac{u}{v^{\alpha }}\nabla v\right)\Vert }_{L^{p_{\mathcal{l}+1}}\left(\Omega \right)}\text{d}s}$。由半群估计( [8], Lemma 1.3(iv))和引理2.3，存在 $c_2,C_3\left(t\right)>0$ 使得

${\displaystyle {\int }_0^t{\Vert {\text{e}}^{\left(t-s\right)\Delta }\nabla \cdot \left(\frac{u}{v^{\alpha }}\nabla v\right)\Vert }_{L^{p_{\mathcal{l}+1}}\left(\Omega \right)}\text{d}s}\le c_2{C}_3\left(t\right){\displaystyle {\int }_0^t\left(1+{\left(t-s\right)}^{-\frac{1}{2}-\frac{n}{2}\left(\frac{1}{\theta }-\frac{1}{p_{\mathcal{l}+1}}\right)}{\text{e}}^{-{\lambda }_1\left(t-s\right)}\right)}{\Vert u\nabla v\Vert }_{L^{\theta }\left(\Omega \right)}\text{d}s.$ (2.8)

如果 $\eta <\frac{n{p}_{\mathcal{l}}}{n-p_{\mathcal{l}}}$，由( [9], Lemma 2.4(ii))，存在 $c_3>0$ 满足

${\Vert \nabla v\Vert }_{L^{\eta }\left(\Omega \right)}\le c_3\left(1+{\sup }_{t\in \left(0,T_{\max }\right)}{\Vert u\Vert }_{L^{p_{\mathcal{l}}}\left(\Omega \right)}\right).$ (2.9)

取 $\theta <\frac{n{p}_{\mathcal{l}}}{2n-p_{\mathcal{l}}}$ 接近 $\frac{n{p}_{\mathcal{l}}}{2n-p_{\mathcal{l}}}$，则 $-\frac{1}{2}-\frac{n}{2}\left(\frac{1}{\theta }-\frac{1}{p_{\mathcal{l}+1}}\right)<1$ 和 $\frac{p_{\mathcal{l}}\theta }{p_{\mathcal{l}}-\theta }<\frac{n{p}_{\mathcal{l}}}{n-p_{\mathcal{l}}}$。从而，根据(2.3)及(2.7)~(2.9)并利用Hölder不等式，可得

$\begin{array}{c}{\Vert u\Vert }_{L^{p_{\mathcal{l}+1}}\left(\Omega \right)}\le c_2{C}_3\left(t\right){\displaystyle {\int }_0^t\left(1+{\left(t-s\right)}^{-\frac{1}{2}-\frac{n}{2}\left(\frac{1}{\theta }-\frac{1}{p_{\mathcal{l}+1}}\right)}{\text{e}}^{-{\lambda }_1\left(t-s\right)}\right){\Vert u\nabla v\Vert }_{L^{\theta }\left(\Omega \right)}\text{d}s}+{\Vert u_0\Vert }_{L^{\infty }\left(\Omega \right)}+c_1{T}_{\max }\\ \le C_4\left(t\right){\left(1+{\sup }_{t\in \left(0,T_{\max }\right)}{\Vert u\Vert }_{L^{p_{\mathcal{l}}}\left(\Omega \right)}\right)}^2{\displaystyle {\int }_0^t\left(1+{\left(t-s\right)}^{-\frac{1}{2}-\frac{n}{2}\left(\frac{1}{\theta }-\frac{1}{p_{\mathcal{l}+1}}\right)}{\text{e}}^{-{\lambda }_1\left(t-s\right)}\right)\text{d}s}+c_4\left(1+T_{\max }\right)\\ \le c_4\left(1+T_{\max }\right)+C_5\left(t\right){\left(1+{\sup }_{t\in \left(0,T_{\max }\right)}{\Vert u\Vert }_{L^{p_1}\left(\Omega \right)}\right)}^{2\mathcal{l}}\\ \le C_6\left(t\right),t\in \left(0,T_{\max }\right)\end{array}$ (2.10)

其中 $c_4,C_4\left(t\right),C_5\left(t\right),C_6\left(t\right)>0$。易知对 $\mathcal{l}\ge 1$ 有 $p_{\mathcal{l}+1}\ge p_{\mathcal{l}l}\ge \frac{n}{2}$，并且当 $p_{\mathcal{l}}\to n$ 时，有 $p_{\mathcal{l}+1}\to \infty$。固存在 $\bar{p}>n$ 满足(2.10)成立。

当 $p_{\mathcal{l}}>n$，由( [10], Lemma 3.3)，知

${\displaystyle {\int }_0^t{\Vert {\text{e}}^{\left(t-s\right)\Delta }\nabla \cdot \left(\frac{u}{v^{\alpha }}\nabla v\right)\Vert }_{L^{\infty }\left(\Omega \right)}\text{d}s}\le c_5{C}_7\left(t\right){\displaystyle {\int }_0^t{\left(t-s\right)}^{-\frac{1}{2}-\frac{n}{2p_{\mathcal{l}}}}{\text{e}}^{-{\lambda }_1\left(t-s\right)}{\Vert u\nabla v\Vert }_{L^{p_{\mathcal{l}}}\left(\Omega \right)}\text{d}s}.$

当 $\eta =\infty$ 时，(2.9)式依然成立，又因为积分 ${\displaystyle {\int }_0^{\infty }{\left(t-s\right)}^{-\frac{1}{2}-\frac{n}{2p_{\mathcal{l}}}}\text{d}s}$ 可积，则存在 $C_8\left(t\right),C_9\left(t\right)>0$ 使得

${\Vert u\Vert }_{L^{\infty }\left(\Omega \right)}\le C_8\left(t\right)\left({\Vert u_0\Vert }_{L^{\infty }\left(\Omega \right)}+{\displaystyle {\int }_0^t\left(1+{\left(t-s\right)}^{-\frac{1}{2}-\frac{n}{2p_{\mathcal{l}}}}\right){\text{e}}^{-{\lambda }_1\left(t-s\right)}\text{d}s}\right)+c_1{T}_{\max }=:C_9\left(t\right).$

证明完毕。

3. 结论

本文通过一系列的不等式放缩行为进而完成了系统(1.3)的解的全局存在性证明，首先对 ${\Vert u\Vert }_{L^p\left(\Omega \right)}$ 进行了估计，最终通过对 ${\Vert u\Vert }_{L^{\infty }\left(\Omega \right)}$ 的估计完成了证明，即定理1.1，若 ${\Vert v_0\Vert }_{L^{\infty }\left(\Omega \right)}<{\chi }^{\frac{1}{1-\alpha }}\sqrt{\frac{n}{2}}$ 且 $\mu >\frac{n-2}{2n}$，则系统(1.3)存在整体古典解。