一类带不连续值的二维趋化系统解的全局存在性

Global Existence of a Two-Dimension Chemotaxis System with Discontinuous Data

摘要: 本文研究了描述肿瘤血管生成的二维趋化系统解的适定性和大时间行为。证明了当时间趋于无穷时，该系统的解收敛到一个常平衡态。同以往的结果相比，我们研究了不连续初值解的适定性。利用有效粘性通量和加权能量估计，得到了解的一系列先验估计，通过这些先验估计可以得到弱解的全局存在性。以往的研究都是关于连续初值的研究，得到的都是经典解。而实际情况中不连续的情形更加普遍，而本文的结果表明，在不连续初始值的情况下，我们能够得到二维趋化系统的全局弱解。

Abstract: This paper is concerned with the well-posedness and large-time behavior of a two-dimensional PDEODE hybrid chemotaxis system describing the initiation of tumor angiogenesis. We prove the solution converge to the constant equilibrium when the time tends to infinity. In contrast to the ex-isting related results, where continuous initial data is imposed, we are able to prove the asymptotic stability for discontinuous initial data with large oscillations. The key ingredient in our proof is the use of the so-called “effective viscous flux”, which enables us to obtain the desired energy estimates and regularity. The previous results are all about the continuous initial data and the classical solutions are obtained. However, discontinuity is more common in practice, and the results of this paper show that we can obtain the global weak solution of the two-dimensional chemotaxis system under the discontinuous initial data.

1. 前言和主要的结果

1.1. 背景

本文研究以下趋化模型

$\{\begin{array}{l}{u}_t=\Delta u-\nabla \cdot \left(\xi u\nabla \ln c\right),\\ c_t=-\mu uc+ϵ\Delta u\text{.}\end{array}$ (1)

该模型描述了在血管形成期间，血管内皮细胞生长因子与血管内皮细胞之间的关系 [1] [2] 。其中 $u\left(x,t\right)$ 表示血管内皮细胞的密度， $c\left(x,t\right)$ 表示血管内皮细胞生长因子的浓度。参数 $\xi >0$ 代表趋化系数， $\mu$ 表示趋化血管内皮细胞生长因子的消耗率， $ϵ$ 表示生长因子的扩散率。

模型(1)中，第一个方程含有一个对数敏感函数 $\ln c$ ，为了消去对数奇性，可以通过如下Cole-Hopf变换 [3] [4]

$v=-\frac{1}{\mu }\nabla \left(\ln c\right)=-\frac{1}{\mu }\frac{\nabla c}{c},$ (2)

和伸缩变换 $\tilde{t}=ϵ\mu t$ ， $\tilde{x}=\sqrt{ϵ\mu }$ ， $\tilde{v}=\sqrt{\frac{ϵ}{\mu }}v$ ，将(1.1)转换为如下系统(为了方便起见，还是用 $t,v,x$ 分别代替 $\tilde{t},\tilde{v},\tilde{x}$ )

$\{\begin{array}{l}{u}_t-\nabla \cdot \left(uv\right)=\Delta u,\hfill \\ v_t-\nabla \left(-ϵ{\left|v\right|}^2+u\right)=ϵ\Delta v,\hfill \\ \left(u,v\right)\left(x,0\right)=\left(u_0,v_0\right)=\left(u_0,v_0\right)\left(x\right),\hfill \end{array}$ (3)

无穷远状态为

$\left(u,v\right)\left(\pm \infty ,t\right)=\left(\bar{u},0\right),$ (4)

其中 $\bar{u}>0$ 。

对于模型(1)已经有了大量的研究，比如在一维空间中该系统行波解稳定性的研究 [5] [6] [7] [8] ，关于解的全局存在性以及大时间行为的研究 [9] - [14] 。对于高维无界区域的情形，研究者在文献 [1] 中给出了该模型在二维空间非线性行波解的稳定性，在文献 [15] [16] 中得到了经典解的爆破准则，在文献 [4] 中得到了该系统的全局适定性以及解的大时间行为。对于高维有界区域 $\Omega \subset {ℝ}^d\left(d\ge 3\right)$ ，在Neuman边界条件下，研究者在文献 [16] 中得到了具有指数衰减的全局解。在文献 [11] 中，研究者在二维空间研究了模型(1)的边界层问题。关于模(1)的型其它研究请参考文献 [17] [18] 。

以上的研究大都是关于连续初值的研究，在连续初始值的情况下，可以要求初始值足够光滑，这时得到的都是经典解。而在实际应用中不连续的情形更加普遍(比如分片光滑的初值就是不连续的)，因此研究带不连续初值的趋化模型是一个重要而有意义的题。本文将研究二维趋化系统在不连续初值下解的存在性问题，进而考虑解的大时间行为，这有助于研究该模型在不连续初值下解的性态。

首先我们先介绍弱解的定义：

定义1.1我们说 $\left(u,v\right)$ 是系统(3)~(4)的弱解，如果 $\left(u,v\right)$ 满足，对于所有的测试函数 $\varphi \in C_0^{\infty }\left({ℝ}^2\times \left[0,\infty \right)\right)$ 满足以下关系

${\displaystyle {\int }_{{ℝ}^2}{u}_0{\varphi }_0}+{\displaystyle {\int }_0^{\infty }{\displaystyle {\int }_{{ℝ}^2}\left(u{\varphi }_t-\nabla u\cdot \nabla \varphi \right)\text{d}x\text{d}t}}={\displaystyle {\int }_0^{\infty }{\displaystyle {\int }_{{ℝ}^2}uv\cdot \nabla \varphi \text{d}t}}$

和

${\displaystyle {\int }_{{ℝ}^2}{v}_0^j{\varphi }_0\text{d}x}+{\displaystyle {\int }_0^{\infty }{\displaystyle {\int }_{{ℝ}^2}\left(v^j{\varphi }_t-u{\varphi }_{x_j}+ϵ{\varphi }_{x_j}{\left|v^j\right|}^2-ϵ{\varphi }_{x_j{x}_j}{v}^j\right)\text{d}x\text{d}t}}=0,$

其中 $i,j=1,2$ ， ${\varphi }_0\left(x\right)=\varphi \left(x,0\right)$ 。

1.2. 主要结果

定理1.2如果 $4<p_0<\infty$ 如果初始值满足

$u_0-1\in L^2\left({ℝ}^2\right),v_0\in L^2\left({ℝ}^2\right)\cap L^{p_0}\left({ℝ}^2\right),v_0\ge 0,{\nabla }^{\perp }{v}_0=0,$ (5)

${\nabla }^{\perp }=\left({\partial }_2,-{\partial }_1\right)$ 是旋度算子。然后对于任意的 $M>0$ ， ${\Vert v_0\Vert }_{L^{p_0}\left({ℝ}^2\right)}\le M$ ，那么存在 $\eta >0$ ，使得

${\Vert u_0-1\Vert }_{L^2\left({ℝ}^2\right)}^2+{\Vert v_0\Vert }_{L^2\left({ℝ}^2\right)}^2={\theta }_0\le \eta ,$

那么Cauchy问题(3)-(4)有一个全局的弱解 $\left(u,v\right)$ ，使得

$\{\begin{array}{l}u-1\in L^{\infty }\left(\left[0,\infty \right);L^2\left({ℝ}^2\right)\right)\cap C\left(\left(0,\infty \right);C\left({ℝ}^2\right)\right),\nabla v,\nabla u\in L^2\left(\left[0,\infty \right);L^2\left({ℝ}^2\right)\right)\\ v\in L^{\infty }(\left[0,\infty \right);L^2\left({ℝ}^2\right)\cap L^{p_0}(\; ℝ\; 2\; )\end{array}$

当 $2<p_1$ ， $p_2\le \infty$ ，则

${\Vert u\left(x,t\right)-1\Vert }_{L^{p_1}\left({ℝ}^2\right)}\to 0,{\Vert v\left(x,t\right)\Vert }_{L^{p_2}\left({ℝ}^2\right)}\to 0,t\to \infty .$

2. 预备知识

下面介绍一些概念：

$H^k\left({ℝ}^2\right)$ 代表 ${ℝ}^2$ 上指标为k的Sobolev空间，其范数定义为

${\Vert f\Vert }_{H^k\left({ℝ}^2\right)}:={\left({\displaystyle \underset{j=0}{\overset{k}{\sum }}{\displaystyle {\int }_{{ℝ}^2}{\left|{\partial }_x^{j}f\right|}^2\text{d}x}}\right)}^{\frac{1}{2}},$

旋度算子

${\nabla }^{\perp }=\left({\partial }_2,-{\partial }_1\right),$ (7)

${\theta }_0={\Vert u_0-1\Vert }_{L^2}^2+{\Vert v_0\Vert }_{L^2}^2\text{.}$ (8)

其中 ${\theta }_0<1$ ，下面我们将简写 ${\Vert \cdot \Vert }_{L^2}$ 为 $\Vert \cdot \Vert$ 。首先介绍以下两个引理 [5] [16] 。

引理2.1令 $s>\frac{d}{2}$ ， $\left(u_0-1,v_0\right)\in H^s\left({ℝ}^d\right)$ 。那么存在一个最大时间 $T_{\text{*}}=\left({\Vert u_0-1\Vert }_{H^s},{\Vert v_0\Vert }_{H^s}\right)>0$ 使得Cauchy问题(3)~(4)，存在一个唯一解使得

$\left(u-1,v\right)\in L^{\infty }\left(\left(0,T_{\text{*}}\right],H^s\left({ℝ}^d\right)\right)\text{.}$

引理2.2令 $s>\frac{d}{2}+1$ ，且 $\left(u_0-1,v_0\right)\in H^s\left({ℝ}^d\right)$ ， $\left(u,v\right)$ 是引理2.1中得到的解，如果

${\displaystyle {\int }_0^T{\Vert v\Vert }_{L^q}^{\frac{2q}{q-d}}\text{d}t}<\infty \left(d<q<\infty \right),$ (9)

那么解可以延拓到 $T_{\text{*}}$ 之外。

作变量替换 $\tilde{u}=u-1$ ，系统(3)可转化为

$\{\begin{array}{l}{\tilde{u}}_t-\Delta \tilde{u}=\nabla \cdot \left(\tilde{u}v\right)+\nabla \cdot v,\\ v_t-\nabla \tilde{u}=ϵ\left(\nabla {\left|v\right|}^2+\Delta v\right),\\ \left(\tilde{u},v\right)\left(x,0\right)=\left(u_0-1,v_0\right)\left(x\right)\text{.}\end{array}$ (10)

定义有效粘性通量 $F$ 为：

$F=\nabla \tilde{u}+\left(\tilde{u}+1\right)v\text{.}$ (11)

那么(10)中的第一个方程可以写成

$\nabla \cdot F={\tilde{u}}_t\text{.}$ (12)

由此可以得到关于F的估计：

引理2.3 [19] 如果 $\left(\tilde{u},\text{v}\right)$ 是(10)的光滑解，那么存在常数C使得

${\Vert \nabla F\Vert }_{L^p}\le C\left({\Vert {\tilde{u}}_t\Vert }_{L^p}\right)+{\Vert {\nabla }^{\perp }\tilde{u}\cdot v\Vert }_{L^p}\left(p>1\right)\text{.}$

引理2.4 [20] 令 $1\le q$ ， $r\le \infty$ 且 $0<a\le 1$ 使得

$\frac{1}{p}=a\left(\frac{1}{q}-\frac{1}{2}\right)+\left(1-a\right)\frac{1}{r},$

那么对于 $u\in W^{1,p}\left({ℝ}^2\right)\cap L^r\left({ℝ}^2\right)$ ，然后存在一个常数只依赖于 $q,r,n$ 的C，使得

${\Vert u\Vert }_{L^p}\le C{\Vert \nabla u\Vert }_{L^q\left({ℝ}^2\right)}^a{\Vert u\Vert }_{L^r\left({ℝ}^2\right)}^{1-a}\text{.}$ (13)

3. 渐进解的先验估计

首先对初值进行磨光

${\tilde{u}}_0^{\delta }=j^{\delta }\ast {\bar{u}}_0,v_0^{\delta }=j^{\delta }\ast v_0\text{.}$

考虑以下渐进系统

$\{\begin{array}{l}{\tilde{u}}_t^{\delta }-\Delta {\tilde{u}}^{\delta }=\nabla \cdot \left({\tilde{u}}^{\delta }{v}^{\delta }\right)+\nabla \cdot v^{\delta }\\ v_t^{\delta }-\nabla {\tilde{u}}^{\delta }=ϵ\left(-\nabla {\left({\left|v\right|}^{\delta }\right)}^2+\Delta v^{\delta }\right),\end{array}$ (14)

其中初始值为

$\left({\tilde{u}}_0^{\delta },v_0^{\delta }\right)\in H^3(\; ℝ\; 2\; )$

且

${\Vert {\tilde{u}}_0^{\delta }\Vert }^2+{\Vert v_0^{\delta }\Vert }^2\le {\Vert {\tilde{u}}_0\Vert }^2+{\Vert v_0\Vert }^2={\theta }_0\text{.}$

为方便起见，下面用 $\left(\tilde{u},v\right)$ 代替 $\left({\tilde{u}}^{\delta },v^{\delta }\right)$ 。令 $T>0$ ， $\left(\tilde{u},v\right)$ 是(14)在 ${ℝ}^2\times \left[0,T\right]$ 上的光滑解，令 $\sigma \left(t\right)=\min \left\{1,t\right\}$ ，定义

$\{\begin{array}{l}{A}_1\left(T\right)=\underset{t\in \left[0,T\right]}{\sup }\left({\Vert \tilde{u}\Vert }^2+{\Vert v\Vert }^2\right)+{\displaystyle {\int }_0^T{\Vert \nabla \tilde{u}\Vert }^2\text{d}t}+ϵ{\displaystyle {\int }_0^T{\Vert \nabla v\Vert }^2\text{d}t},\\ A_2\left(T\right)=\underset{t\in \left[0,T\right]}{\sup }\left(\sigma {\Vert \nabla \tilde{u}\Vert }^2+{\sigma }^2{\Vert {\tilde{u}}_t\Vert }^2+{\sigma }^2{\Vert v_t\Vert }^2+\sigma ϵ{\Vert \nabla v\Vert }^2\right)+{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla {\tilde{u}}_t\Vert }^2\text{d}t\\ +ϵ{\displaystyle {\int }_0^T{\sigma }^2{\Vert \nabla v\Vert }^2\text{d}t}+{\displaystyle {\int }_0^T\sigma {\Vert v_t\Vert }^2\text{d}t}+{\displaystyle {\int }_0^T\sigma {\Vert {\tilde{u}}_t\Vert }^2\text{d}t},\\ A_3\left(T\right)=\underset{t\in \left[0,T\right]}{\sup }{\Vert v\Vert }_{L^4}^4+{\displaystyle {\int }_0^T{\Vert v\Vert }_{L^4}^4\text{d}t}\text{.}\end{array}$ (15)

下面我们将通过先验假设的方法来得到解的先验估计，我们首先假设

$A_1\left(T\right)\le 2{\theta }_0,A_2\left(T\right)\le 2{\theta }_0^{\frac{1}{2}},A_3\left(T\right)\le 2{\theta }_0^{{\eta }_0},\underset{t\in \left[0,T\right]}{\sup }{\Vert v\Vert }_{L^{p_0}}\le 6M,$ (16)

其中

${\eta }_0=\frac{p_0-4}{2\left(p_0-2\right)}\in \left(0,\frac{1}{2}\right)\text{.}$ (17)

下面我们从基本能量估计开始。

引理3.1如果定理1.2中的条件满足， $\left(\tilde{u},v\right)$ 是满足初值条件并且满足(16)的光滑解，则有下面不等式成立

${\Vert \tilde{u}\Vert }^2+{\Vert v\Vert }^2+{\displaystyle {\int }_0^T{\Vert \nabla \tilde{u}\Vert }^2\text{d}t}+ϵ{\displaystyle {\int }_0^T{\Vert \nabla v\Vert }^2\text{d}t}\le \frac{3}{2}{\theta }_0,$ (18)

证明 用 $\tilde{u}$ 乘以(10)式中的第一个方程，用 $\sigma v$ 乘以(10)中的第二个方程，把两式所得结果相加，然后在 ${ℝ}^2$ 上作积分，则有然后用类似文献 [19] 引理3.1的方法，容易得到所需的结果。

接下来我们将给出解的一阶能量估计，得到解的一阶能量估计时需要利用加权能量的技巧。

引理3.2如果定理1.2中的条件满足， $\left(\tilde{u},v\right)$ 是满足初值条件并且满足(16)的光滑解，则有

$\begin{array}{l}\sigma {\Vert \nabla \tilde{u}\Vert }^2+\sigma {\Vert {\tilde{u}}_t\Vert }^2+\sigma ϵ{\Vert v_t\Vert }^2+\sigma ϵ{\Vert \nabla v\Vert }^2+{\Vert v_t\Vert }^2\text{d}t\\ +{\displaystyle {\int }_0^T\sigma {\Vert {\tilde{u}}_t\Vert }^2\text{d}t}+ϵ{\displaystyle {\int }_0^T{\sigma }^2{\Vert \nabla v_t\Vert }^2\text{d}t}+\frac{1}{4}{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla {\tilde{u}}_t\Vert }^2\text{d}t\le {\theta }_0^{\frac{1}{2}},\end{array}$ (19)

其中 $\sigma =\sigma \left(t\right)=\min \left\{1,t\right\}$ 。

证明 第一步：用 $\sigma {\tilde{u}}_t$ ， $\sigma v_t$ ，分别乘以(14)中第一个方程，第二个方程，然后在 ${ℝ}^2\times \left(0,T\right]$ 作积分，可得到

$\frac{1}{2}\sigma {\Vert \nabla \tilde{u}\Vert }^2+{\displaystyle {\int }_0^T\sigma {\Vert {\tilde{u}}_t\Vert }^2\text{d}t}=\frac{1}{2}{\displaystyle {\int }_0^{\sigma \left(T\right)}{\Vert \nabla \tilde{u}\Vert }^2\text{d}t}-{\displaystyle {\int }_0^T\sigma }{\displaystyle {\int }_{{ℝ}^2}\tilde{u}v\nabla {\tilde{u}}_t\text{d}x\text{d}t}-{\displaystyle {\int }_0^T\sigma }{\displaystyle {\int }_{{ℝ}^2}v\nabla {\tilde{u}}_t\text{d}x\text{d}t}$ (20)

和

$\frac{1}{2}\sigma ϵ{\Vert \nabla v\Vert }^2+{\displaystyle {\int }_0^T\sigma {\Vert v_t\Vert }^2\text{d}t}=-ϵ{\displaystyle {\int }_0^T\sigma }{\displaystyle {\int }_{{ℝ}^2}\left(v_t\nabla {\left|v\right|}^2\right)\text{d}x\text{d}t}+ϵ{\displaystyle {\int }_0^{\sigma \left(T\right)}{\Vert \nabla v\Vert }^2\text{d}t}+{\displaystyle {\int }_0^T\sigma }{\displaystyle {\int }_{{ℝ}^2}{v}_t\nabla \tilde{u}\text{d}x\text{d}t}\text{.}$ (21)

其中上式用了 $\frac{1}{2}{\displaystyle {\int }_0^T{\sigma }_t}{\Vert \nabla \tilde{u}\Vert }^2\text{d}t=\frac{1}{2}{\displaystyle {\int }_0^{\sigma \left(T\right)}{\Vert \nabla \tilde{u}\Vert }^2\text{d}t}$ ， $\frac{1}{2}{\displaystyle {\int }_0^T{\sigma }_t}{\Vert \nabla v\Vert }^2\text{d}t=\frac{1}{2}{\displaystyle {\int }_0^{\sigma \left(T\right)}{\Vert \nabla v\Vert }^2\text{d}t}$ 。

首先处理(20)式右边的第二项，根据G-N不等式有

$\begin{array}{l}-{\displaystyle {\int }_0^T\sigma }{\displaystyle {\int }_{{ℝ}^2}\tilde{u}v\nabla {\tilde{u}}_t\text{d}x\text{d}t}\\ \le \lambda {\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla {\tilde{u}}_t\Vert }^2\text{d}t+C{\displaystyle {\int }_0^T{\Vert \tilde{u}v\Vert }^2\text{d}t}\\ \le \lambda {\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla {\tilde{u}}_t\Vert }^2\text{d}t+C{\displaystyle {\int }_0^T{\Vert \tilde{u}\Vert }_{L^4}^2{\Vert v\Vert }_{L^4}^2\text{d}t}\\ \le \lambda {\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla {\tilde{u}}_t\Vert }^2\text{d}t+C{\displaystyle {\int }_0^T{\Vert \nabla \tilde{u}\Vert }^2\text{d}t}+C{\displaystyle {\int }_0^T{\Vert \tilde{u}\Vert }^2{\Vert v\Vert }_{L^4}^4\text{d}t}\\ \le \lambda {\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla {\tilde{u}}_t\Vert }^2\text{d}t+C{\theta }_0.\end{array}$ (22)

对于(21)中右边最后一项，通过计算可得

$\begin{array}{c}-{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^2}v\nabla {\tilde{u}}_t\text{d}x\text{d}t}}=-{\displaystyle {\int }_0^T{\left(\sigma {\displaystyle {\int }_{{ℝ}^2}v\nabla \tilde{u}\text{d}x}\right)}_t\text{d}t}+{\displaystyle {\int }_0^{\sigma \left(T\right)}{\displaystyle {\int }_{{ℝ}^2}v\nabla \tilde{u}\text{d}x\text{d}t}}+{\displaystyle {\int }_0^T\sigma }{\displaystyle {\int }_{{ℝ}^2}{v}_t\nabla \tilde{u}\text{d}x\text{d}t}\\ =:I_1+I_2+I_3,\end{array}$ (23)

由Cauchy不等式，(15)，(16)，及 $0\le \sigma \le 1$ ，则有

$I_1=-\sigma {\displaystyle {\int }_{{ℝ}^2}v\nabla \tilde{u}\text{d}x}\le \frac{\sigma }{4}{\Vert \nabla \tilde{u}\Vert }^2+\sigma {\Vert v\Vert }^2\le \frac{\sigma }{4}{\Vert \nabla \tilde{u}\Vert }^2+2{\theta }_0,$ (24)

$\begin{array}{c}{I}_2={\displaystyle {\int }_0^{\sigma \left(T\right)}{\displaystyle {\int }_{{ℝ}^2}v\nabla \tilde{u}\text{d}x\text{d}t}}\le \frac{1}{2}{\displaystyle {\int }_0^{\sigma \left(T\right)}{\Vert \nabla \tilde{u}\Vert }^2\text{d}t}+\frac{1}{2}{\displaystyle {\int }_0^{\sigma \left(T\right)}{\Vert v\Vert }^2\text{d}t}\\ \le \frac{1}{2}{\displaystyle {\int }_0^T{\Vert \nabla \tilde{u}\Vert }^2\text{d}t}+\frac{1}{2}\underset{t\in \left[0,T\right]}{\sup }{\Vert v\Vert }^2\le 2{\theta }_0.\end{array}$ (25)

$\begin{array}{c}{I}_3={\displaystyle {\int }_0^T\sigma }{\displaystyle {\int }_{{ℝ}^2}{v}_t\nabla \tilde{u}\text{d}x\text{d}t}\\ ={\displaystyle {\int }_0^T\sigma }{\displaystyle {\int }_{{ℝ}^2}{\left|\nabla \tilde{u}\right|}^2\text{d}x\text{d}t}+ϵ{\displaystyle {\int }_0^T\sigma }{\displaystyle {\int }_{{ℝ}^2}\nabla {\left|v\right|}^2\nabla \tilde{u}\text{d}x\text{d}t}+ϵ{\displaystyle {\int }_0^T\sigma }{\displaystyle {\int }_{{ℝ}^2}\Delta v\nabla \tilde{u}\text{d}x\text{d}t},\end{array}$

对于 $I_3$ 中的第二项，由Cauchy不等式以及(16)有

$ϵ{\displaystyle {\int }_0^T\sigma }{\displaystyle {\int }_{{ℝ}^2}\nabla {\left|v\right|}^2\nabla \tilde{u}\text{d}x\text{d}t}\le C{\displaystyle {\int }_0^T\sigma }{\displaystyle {\int }_{{ℝ}^2}ϵ\left|v\right|\left|\nabla v\right|\left|\nabla \tilde{u}\right|\text{d}x\text{d}t}\le C{\theta }_0^{\frac{1}{2}+{\eta }_0}.$

$I_3$ 中的第三项，由(14)中的第一个等式有

$I_3={\displaystyle {\int }_0^T\sigma }{\displaystyle {\int }_{{ℝ}^2}{v}_t\nabla \tilde{u}\text{d}x\text{d}t}\le C{\theta }^{\frac{1}{2}+{\eta }_0}\text{.}$ (26)

由(20)，(22)，(23)，(24)，(25)，(26)得

$\frac{1}{4}\sigma {\Vert \nabla \tilde{u}\Vert }^2+{\displaystyle {\int }_0^T\sigma }{\Vert {\tilde{u}}_t\Vert }^2\text{d}t\le \lambda {\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla {\tilde{u}}_t\Vert }^2\text{d}t+C{\theta }_0^{\frac{1}{2}+{\eta }_0}\text{.}$ (27)

并且

$\begin{array}{c}{\displaystyle {\int }_0^T\sigma }{\displaystyle {\int }_{{ℝ}^2}\nabla u{v}_t\text{d}x\text{d}t}\le \frac{1}{8}{\displaystyle {\int }_0^T\sigma }{\displaystyle {\int }_{{ℝ}^2}{\Vert v_t\Vert }^2\text{d}x\text{d}t}+C{\displaystyle {\int }_0^T\sigma }{\displaystyle {\int }_{{ℝ}^2}{\left|\nabla \tilde{u}\right|}^2\text{d}x\text{d}t}\\ \le \frac{1}{8}{\displaystyle {\int }_0^T\sigma }{\displaystyle {\int }_{{ℝ}^2}{\Vert v_t\Vert }^2\text{d}x\text{d}t}+C{\theta }_0^{\frac{1}{2}+{\eta }_0},\end{array}$ (28)

$\begin{array}{c}ϵ{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^2}\sigma \nabla {\left|v\right|}^2{v}_t\text{d}x\text{d}t}}\le ϵ{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^2}\sqrt{\sigma }\left|\nabla v\right|\left|\sqrt{\sigma }\right|\left|v\right|\left|v_t\right|\text{d}x\text{d}t}}\\ \le ϵ\sqrt{\sigma }{\Vert v\Vert }_{L^{\infty }}\left(\frac{1}{4}{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^2}\sigma {\left|v_t\right|}^2\text{d}x\text{d}t}}+{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^2}{\left|\nabla v\right|}^2\text{d}x\text{d}t}}\right)\\ \le \frac{ϵ}{4}{\displaystyle {\int }_0^T\sigma {\Vert v_t\Vert }^2\text{d}t}+C{\theta }_0^{\frac{1}{2}+{\eta }_0},\end{array}$ (29)

则有

$\sigma {\Vert \nabla \tilde{u}\Vert }^2+\sigma ϵ{\Vert \nabla v\Vert }^2+{\displaystyle {\int }_0^T\sigma {\Vert v_t\Vert }^2\text{d}t}+{\displaystyle {\int }_0^T\sigma {\Vert {\tilde{u}}_t\Vert }^2\text{d}t}\le \lambda {\displaystyle {\int }_0^T{\sigma }^2{\Vert \nabla {\tilde{u}}_t\Vert }^2\text{d}t}+C{\theta }_0^{\frac{1}{2}+{\eta }_0}\text{.}$ (30)

步骤2. 估计 ${\displaystyle {\int }_0^T{\sigma }^2{\Vert \nabla {\tilde{u}}_t\Vert }^2\text{d}t}$ ，对时间求导得

$\{\begin{array}{l}{\tilde{u}}_{tt}-\Delta {\tilde{u}}_t=\nabla \cdot {\left(\tilde{u}v\right)}_t+\nabla \cdot v_t\\ v_{tt}-\nabla {\tilde{u}}_t=2ϵ{\left(\nabla {\left|v\right|}^2\right)}_t+ϵ\Delta v_t\text{.}\end{array}$ (31)

所得方程第一个方程乘以 ${\sigma }^2{\tilde{u}}_t$ ，第二个方程乘以 ${\sigma }^2{v}_t$ ，把所得结果相加并在作积分可得

$\begin{array}{l}\frac{{\sigma }^2}{2}{\Vert {\tilde{u}}_t\Vert }^2+\frac{{\sigma }^2}{2}\Vert v_t\Vert +{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla {\tilde{u}}_t\Vert }^2\text{d}t+ϵ{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla v_t\Vert }^2\text{d}t\\ \le {\displaystyle {\int }_0^{\sigma \left(T\right)}\sigma }{\Vert {\tilde{u}}_t\Vert }^2\text{d}t+{\displaystyle {\int }_0^{\sigma \left(T\right)}\sigma }{\Vert v_t\Vert }^2\text{d}t+ϵ{\displaystyle {\int }_0^{\sigma \left(T\right)}{\sigma }^2}{\displaystyle {\int }_{{ℝ}^2}{v}_t}\left(\nabla {\left|v\right|}^2\right)\text{d}x\text{d}t+{\displaystyle {\int }_0^T{\sigma }^2}{\displaystyle {\int }_{{ℝ}^2}\left|{\left(\tilde{u}v\right)}_t\right|\left|\nabla {\tilde{u}}_t\right|\text{d}x\text{d}t}\\ \le {\displaystyle {\int }_0^T\sigma }{\Vert {\tilde{u}}_t\Vert }^2\text{d}t+{\displaystyle {\int }_0^{\sigma \left(T\right)}\sigma }{\Vert v_t\Vert }^2\text{d}t+{\displaystyle {\int }_0^T{\sigma }^2}{\displaystyle {\int }_{{ℝ}^2}\left|\tilde{u}{v}_t\nabla {\tilde{u}}_t\right|\text{d}x\text{d}t}+{\displaystyle {\int }_0^T{\sigma }^2}{\displaystyle {\int }_{{ℝ}^2}\left|{\tilde{u}}_{t}v\nabla {\tilde{u}}_t\right|\text{d}x\text{d}t}+ϵ\left|{\displaystyle {\int }_0^{\sigma \left(T\right)}{\sigma }^2}{\displaystyle {\int }_{{ℝ}^2}{v}_t}{\left(\nabla {\left|v\right|}^2\right)}_t\text{d}x\text{d}t\right|.\end{array}$ (32)

因此有

$\begin{array}{l}\frac{{\sigma }^2}{2}{\Vert {\tilde{u}}_t\Vert }^2+\frac{{\sigma }^2}{2}{\Vert v_t\Vert }^2+{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla {\tilde{u}}_t\Vert }^2\text{d}t+ϵ{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla v_t\Vert }^2\text{d}t\\ \le \frac{1}{4}{\theta }_0^{\frac{1}{2}}+{\displaystyle {\int }_0^{\sigma \left(T\right)}\sigma }{\Vert v_t\Vert }^2\text{d}t+{\displaystyle {\int }_0^T\sigma }{\Vert {\tilde{u}}_t\Vert }^2\text{d}t+{\displaystyle {\int }_0^T{\sigma }^2}{\displaystyle {\int }_{{ℝ}^2}\left|\tilde{u}{v}_t\nabla {\tilde{u}}_t\right|\text{d}x\text{d}t}+{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^2}{\sigma }^2{\tilde{u}}_{t}v\nabla {\tilde{u}}_t\text{d}x\text{d}t}}.\end{array}$ (34)

下面估计(34)右边第四项，由

$\begin{array}{c}{\displaystyle {\int }_0^T{\sigma }^2}{\displaystyle {\int }_{{ℝ}^2}\left|\tilde{u}{v}_t\nabla {\tilde{u}}_t\right|\text{d}x\text{d}t}\le \frac{1}{4}{\displaystyle {\int }_0^T{\sigma }^2}{\displaystyle {\int }_{{ℝ}^2}{\left|\nabla {\tilde{u}}_t\right|}^2\text{d}x\text{d}t}+{\displaystyle {\int }_0^T{\sigma }^2}{\displaystyle {\int }_{{ℝ}^2}{\left|\tilde{u}{v}_t\right|}^2\text{d}x\text{d}t}\\ \le \frac{1}{4}{\displaystyle {\int }_0^T{\sigma }^2}{\displaystyle {\int }_{{ℝ}^2}{\left|\nabla {\tilde{u}}_t\right|}^2\text{d}x\text{d}t}+{\left(\sqrt{\sigma }{\Vert \tilde{u}\Vert }_{L^{\infty }}\right)}^2{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^2}\sigma }}{\left|v_t\right|}^2\text{d}t\text{d}x\\ \le \frac{1}{4}{\displaystyle {\int }_0^T{\sigma }^2}{\displaystyle {\int }_{{ℝ}^2}{\left|\nabla {\tilde{u}}_t\right|}^2\text{d}x\text{d}t}+C{\theta }_0^{\frac{1}{2}+{\eta }_0}\text{.}\end{array}$ (35)

下面估计(34)中右边最后一项，由(13)和(16)可得

$\begin{array}{c}{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^2}{\tilde{u}}_{t}v\nabla u_t\text{d}x\text{d}t}}\le \frac{1}{4}{\displaystyle {\int }_0^T{\sigma }^2{\Vert \nabla {\tilde{u}}_t\Vert }^2\text{d}t}+{\displaystyle {\int }_0^T{\sigma }^2{\Vert {\tilde{u}}_{t}v\Vert }^4\text{d}t}\\ \le \frac{1}{4}{\displaystyle {\int }_0^T{\sigma }^2{\Vert \nabla {\tilde{u}}_t\Vert }^2\text{d}t}+{\displaystyle {\int }_0^T{\sigma }^2}{\Vert {\tilde{u}}_t\Vert }_{L^4}^2{\Vert v\Vert }_{L^4}^2\text{d}t\\ \le \frac{1}{4}{\displaystyle {\int }_0^T{\sigma }^2{\Vert \nabla {\tilde{u}}_t\Vert }^2\text{d}t}+C{\displaystyle {\int }_0^T{\sigma }^2}\Vert {\tilde{u}}_t\Vert \Vert \nabla {\tilde{u}}_t\Vert {\Vert v\Vert }_{L^4}^2\text{d}t\\ \le \frac{1}{2}{\displaystyle {\int }_0^T{\sigma }^2{\Vert \nabla {\tilde{u}}_t\Vert }^2\text{d}t}+C{\displaystyle {\int }_0^T{\sigma }^2}{\Vert {\tilde{u}}_t\Vert }^2{\Vert v\Vert }_{L^4}^4\text{d}t\\ \le \frac{1}{2}{\displaystyle {\int }_0^T{\sigma }^2{\Vert \nabla {\tilde{u}}_t\Vert }^2\text{d}t}+C{\theta }_0^{{\eta }_0}{\displaystyle {\int }_0^T{\sigma }^2}{\Vert {\tilde{u}}_t\Vert }^2\text{d}t.\end{array}$ (36)

根据 $0\le \sigma \le 1$ ， ${\theta }_0\le 1$ ，可得

$\begin{array}{l}\frac{1}{2}{\sigma }^2{\Vert {\tilde{u}}_t\Vert }^2+\frac{1}{2}{\sigma }^2{\Vert v_t\Vert }^2+\frac{1}{4}{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla {\tilde{u}}_t\Vert }^2\text{d}t+ϵ{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla v_t\Vert }^2\text{d}t\\ \le C\lambda {\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla {\tilde{u}}_t\Vert }^2\text{d}t+{\displaystyle {\int }_0^T\sigma }{\Vert v_t\Vert }^2\text{d}t+C{\theta }_0^{\frac{1}{2}+{\eta }_0}+\frac{1}{4}{\theta }_0^{\frac{1}{2}},\end{array}$

选取 $\left(C+4\right)\lambda \le \frac{1}{8}$ ，可得

$\begin{array}{l}\sigma {\Vert \nabla \tilde{u}\Vert }^2+\sigma ϵ{\Vert \nabla v\Vert }^2+{\sigma }^2{\Vert {\tilde{u}}_t\Vert }^2+{\sigma }^2{\Vert v_t\Vert }^2+100{\displaystyle {\int }_0^T\sigma }{\Vert v_t\Vert }^2\text{d}t\\ +{\displaystyle {\int }_0^T\sigma }{\Vert {\tilde{u}}_t\Vert }^2\text{d}t+\frac{1}{4}{\displaystyle {\int }_0^T{\sigma }^2}\Vert \nabla {\tilde{u}}_t\Vert \text{d}t+ϵ{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla v_t\Vert }^2\text{d}t\le C{\theta }_0^{\frac{1}{2}+{\eta }_0}+\frac{1}{2}{\theta }_0^{\frac{1}{2}}\text{.}\end{array}$ (37)

选取 ${\theta }_0$ ，使得 $C{\theta }_0^{{\eta }_0}\le \frac{1}{2}$ ，则有

$\begin{array}{l}\sigma {\Vert \nabla \tilde{u}\Vert }^2+\sigma ϵ{\Vert \nabla v\Vert }^2+{\sigma }^2{\Vert {\tilde{u}}_t\Vert }^2+{\sigma }^2{\Vert v_t\Vert }^2+100{\displaystyle {\int }_0^T\sigma }{\Vert v_t\Vert }^2\text{d}t\\ +{\displaystyle {\int }_0^T\sigma }{\Vert {\tilde{u}}_t\Vert }^2\text{d}t+\frac{1}{4}{\displaystyle {\int }_0^T{\sigma }^2}\Vert \nabla {\tilde{u}}_t\Vert \text{d}t+ϵ{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla v_t\Vert }^2\text{d}t\le {\theta }_0^{\frac{1}{2}}\text{.}\end{array}$

因此完成引理3.2的证明。

接下来，我们将给出 $\tilde{u}\left(x,t\right)$ 的估计，从而得到 $\tilde{u}\left(x,t\right)$ 的大时间行为。

引理3.3 [19] 如果定理1.2中的条件满足， $\left(\tilde{u},v\right)$ 是满足初值条件并且满足(16)的光滑解，则有

${\sigma }^4{\Vert \tilde{u}\Vert }_{L^{\infty }}^4\le C{\theta }_0^{\frac{1}{2}+{\eta }_0},$

且 $t\in \left(\sigma \left(T\right),T\right]$ ，则

$-\frac{1}{4}\le \tilde{u}\left(x,t\right)\le \frac{1}{4}\text{.}$ (38)

证明：引理3.3的证明可参考文献 [19] 。

以下引理是关于v的估计。

引理3.4如果定理1.2中的条件满足， $\left(\tilde{u},v\right)$ 是满足初值条件并且满足(16)的光滑解，那么

$\underset{t\in \left[0,T\right]}{\sup }{\displaystyle {\int }_{{ℝ}^2}{v}^4\text{d}x}+{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^2}{v}^4\text{d}x\text{d}t}}\le {\theta }_0^{{\eta }_0},$

其中 ${\eta }_0=\frac{p_0-4}{2\left(p_0-2\right)}$ 。

证明：由 $v_t=\nabla \tilde{u}+ϵ\Delta v+ϵ\nabla {\left|v\right|}^2$ ，有

$\Delta v-ϵ\nabla {\left|v\right|}^2+\left(\tilde{u}+1\right)v=F,$ (39)

两边同时乘以 ${\left|v\right|}^{2}v$ ，然后在 ${ℝ}^2$ 作积分，则

$\frac{1}{4}{\left({\displaystyle {\int }_{{ℝ}^2}{v}^4\text{d}x}\right)}_t+{\displaystyle {\int }_{{ℝ}^2}\left(\tilde{u}+1\right)v^4\text{d}x}=ϵ{\displaystyle {\int }_{{ℝ}^2}{\left|v\right|}^{2}v\nabla {\left|v\right|}^2\text{d}x}+ϵ{\displaystyle {\int }_{{ℝ}^2}\Delta v\cdot {\left|v\right|}^{2}v\text{d}x}+{\displaystyle {\int }_{{ℝ}^2}F}{\left|v\right|}^{2}vdx\text{.}$

在 $\left[\sigma \left(T\right),T\right]$ 作积分，可得

$\begin{array}{l}\frac{1}{4}{\displaystyle {\int }_{{ℝ}^2}{v}^4\text{d}x}+\frac{3}{4}{\displaystyle {\int }_{\sigma \left(T\right)}^T{\displaystyle {\int }_{{ℝ}^2}{v}^4\text{d}x\text{d}t}}\\ \le \underset{t\in \left[0,\sigma \left(T\right)\right]}{\sup }\left(\frac{1}{4}{\displaystyle {\int }_{{ℝ}^2}{v}^4\text{d}x}\right)+{\displaystyle {\int }_{\sigma \left(T\right)}^T\left|F\right|{\left|v\right|}^3\text{d}x\text{d}t}+Cϵ{\displaystyle {\int }_{\sigma \left(T\right)}^T{\Vert v\Vert }_{L^{\infty }}^2}{\displaystyle {\int }_{{ℝ}^2}{\left|v\right|}^2\left|\nabla v\right|\text{d}t\text{d}x}-3ϵ{\displaystyle {\int }_{\sigma \left(T\right)}^T{\displaystyle {\int }_{{ℝ}^2}{\left|v\right|}^2{\left|\nabla v\right|}^2\text{d}x\text{d}t}}\\ \le \underset{t\in \left[0,\sigma \left(T\right)\right]}{\sup }\left({\displaystyle {\int }_{{ℝ}^2}{v}^4\text{d}x}\right)+Cϵ{\displaystyle {\int }_{\sigma \left(T\right)}^T{\displaystyle {\int }_{{ℝ}^2}\left({\left|v\right|}^4+{\left|\nabla v\right|}^2\right)\text{d}t\text{d}x}}+\frac{1}{4}{\displaystyle {\int }_{\sigma \left(T\right)}^T{\displaystyle {\int }_{{ℝ}^2}{v}^4\text{d}x\text{d}t}}+C{\displaystyle {\int }_{\sigma \left(T\right)}^T{\left|F\right|}^4\text{d}x\text{d}t}\\ \le \underset{t\in \left[0,\sigma \left(T\right)\right]}{\sup }\left(\frac{1}{4}{\displaystyle {\int }_{{ℝ}^2}{v}^4\text{d}x}\right)+C{\theta }_0^{\frac{1}{2}+{\eta }_0}+\frac{1}{4}{\displaystyle {\int }_{\sigma \left(T\right)}^T{\displaystyle {\int }_{{ℝ}^2}{v}^4\text{d}x\text{d}t}}+C{\displaystyle {\int }_{\sigma \left(T\right)}^T{\displaystyle {\int }_{{ℝ}^2}{\left|F\right|}^4\text{d}x\text{d}t}}\text{.}\end{array}$ (40)

由G-N不等式(13)有

${\displaystyle {\int }_{\sigma \left(T\right)}^T{\Vert F\Vert }_{L^4}^4\text{d}t}\le C{\displaystyle {\int }_{\sigma \left(T\right)}^T{\Vert F\Vert }^2{\Vert \nabla F\Vert }^2\text{d}t}\le C\left(\underset{\sigma \left(T\right)\le t\le T}{\sup }{\Vert F\Vert }^2\right){\displaystyle {\int }_{\sigma \left(T\right)}^T{\Vert \nabla F\Vert }^2\text{d}t}.$ (41)

由于

$\begin{array}{c}\Vert F\Vert \le \Vert \nabla \tilde{u}+\tilde{u}v+v\Vert \\ \le \Vert \nabla \tilde{u}\Vert +\Vert \tilde{u}v\Vert +\Vert v\Vert \\ \le \Vert \nabla \tilde{u}\Vert +{\Vert \tilde{u}\Vert }_{L^4}{\Vert v\Vert }_{L^4}+\Vert v\Vert \\ \le \Vert \nabla \tilde{u}\Vert +C{\Vert \tilde{u}\Vert }^{\frac{1}{2}}{\Vert \nabla \tilde{u}\Vert }^{\frac{1}{2}}{\Vert v\Vert }_{L^4}+\Vert v\Vert \end{array}$

$\begin{array}{l}\le \Vert \nabla \tilde{u}\Vert +C\Vert \tilde{u}\Vert {\Vert v\Vert }_{L^4}^2+\Vert v\Vert \\ \le \Vert \nabla \tilde{u}\Vert +C{\theta }_0^{\frac{1}{4}+\frac{{\eta }_0}{2}}+C{\theta }_0^{\frac{1}{2}}\\ \le \Vert \nabla \tilde{u}\Vert +C{\theta }_0^{\frac{1}{4}+\frac{{\eta }_0}{2}},\end{array}$ (42)

当 $\sigma \left(T\right)\le t\le T$ ， $\sigma =1$ ，则

$\underset{\sigma \left(T\right)\le t\le T}{\sup }{\Vert F\Vert }^2\le 4\sigma {\Vert \nabla \tilde{u}\Vert }^2+C{\left({\theta }_0^{\frac{1}{4}+{\eta }_0}\right)}^2\le 4{\theta }_0^{\frac{1}{2}}+C{\theta }_0^{\frac{1}{2}+{\eta }_0}\le C{\theta }_0^{\frac{1}{2}}.$ (43)

那么由(13)有

$\begin{array}{c}{\displaystyle {\int }_{\sigma \left(T\right)}^T{\Vert \nabla F\Vert }^2\text{d}t}\le C{\displaystyle {\int }_{\sigma \left(T\right)}^T\sigma }{\Vert {\tilde{u}}_t\Vert }^2\text{d}t+C{\displaystyle {\int }_{\sigma \left(T\right)}^T\sigma }{\Vert \nabla \tilde{u}\Vert }_{L^4}^2{\Vert v\Vert }_{L^4}^2\text{d}t\\ \le C{\theta }_0^{\frac{1}{2}}+C{\displaystyle {\int }_{\sigma \left(T\right)}^T\sigma }{\Vert \nabla \tilde{u}\Vert }_{L^4}^2{\Vert v\Vert }_{L^4}^2\text{d}t.\end{array}$ (44)

由 ${\displaystyle {\int }_0^T{\sigma }^2{\Vert \nabla \tilde{u}\Vert }_{L^4}^4\text{d}t}\le C{\theta }_0^{{\eta }_0}$ 和(16)可得

$C{\displaystyle {\int }_{\sigma \left(T\right)}^T\sigma {\Vert \nabla \tilde{u}\Vert }_{L^4}^2{\Vert v\Vert }_{L^4}^2\text{d}t}\le C{\displaystyle {\int }_{\sigma \left(T\right)}^T{\sigma }^2{\Vert \nabla \tilde{u}\Vert }_{L^4}^4\text{d}t}+C{\displaystyle {\int }_{\sigma \left(T\right)}^T{\Vert v\Vert }_{L^4}^4\text{d}t}\le C{\theta }_0^{{\eta }_0},$ (45)

因此

${\displaystyle {\int }_{\sigma \left(T\right)}^T{\Vert \nabla F\Vert }^2\text{d}t}\le C{\theta }_0^{{\eta }_0},$ (46)

由(41)，(44)，(46)得

${\displaystyle {\int }_{\sigma \left(T\right)}^T{\Vert F\Vert }_{L^4}^4\text{d}t}\le C{\theta }_0^{\frac{1}{2}+{\eta }_0}\text{.}$

那么

$\underset{t\in \left[\sigma \left(T\right),T\right]}{\sup }{\displaystyle {\int }_{{ℝ}^2}{v}^4\text{d}x}+{\displaystyle {\int }_{\sigma \left(T\right)}^T{v}^4\text{d}x\text{d}t}\le C\left(M\right){\theta }_0^{4{\eta }_0}+C{\theta }_0^{\frac{1}{2}+{\eta }_0}\text{.}$ (47)

当 $0\le t\le \sigma \left(T\right)$ ，由插值不等式，(3.3)可得

${\Vert v\Vert }_{L^p}\le {\Vert v\Vert }^{\frac{2\left(p_0-p\right)}{p\left(p_0-2\right)}}{\Vert v\Vert }_{L^{p_0}}^{\frac{p_0\left(p-2\right)}{p\left(p_0-2\right)}}\le C{\theta }_0^{\frac{p_0-p}{p\left(p_0-2\right)}}{M}^{\frac{p_0\left(p-2\right)}{p\left(p_0-2\right)}}$

则

${\Vert v\Vert }_{L^4}\le C{\theta }_0^{\frac{p_0-4}{4\left(p_0-2\right)}}{M}^{\frac{p_0\left(p-2\right)}{4\left(p_0-2\right)}}\le C\left(M\right){\theta }_0^{{\eta }_0},$ (48)

因此

$\underset{t\in \left[0,\sigma \left(T\right)\right]}{\sup }{\displaystyle {\int }_{{ℝ}^2}{v}^4\text{d}x}\le 2\underset{t\in \left[0,\sigma \left(T\right)\right]}{\sup }{\displaystyle {\int }_{{ℝ}^2}{v}^4\text{d}x}\le C\left(M\right){\theta }_0^{4{\eta }_0}\text{.}$ (49)

由(47)，(48)，(49)取 $C\left(M\right){\theta }_0^{3{\eta }_0}\le \frac{1}{2}$ ， $C{\theta }_0^{\frac{1}{2}}\le \frac{1}{2}$ ，则

$\underset{t\in \left[0,T\right]}{\sup }{\displaystyle {\int }_{{ℝ}^2}{v}^4\text{d}x}+{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^2}{v}^4\text{d}t\text{d}x}}\le C\left(M\right){\theta }_0^{4{\eta }_0}+\frac{1}{2}{\theta }_0^{\frac{1}{2}+{\eta }_0}\le {\theta }_0^{{\eta }_0}\text{.}$

引理3.5如果定理1.2中的条件满足， $\left(\tilde{u},v\right)$ 是满足初值条件并且满足(16)的光滑解，那么

$\underset{t\in \left[0,T\right]}{\sup }{\Vert v\Vert }_{L^{p_0}}\le 3M$ . (50)

证明：用 ${\sigma }^{\sigma }{\left|v\right|}^{p_0-2}\left|v\right|$ 乘以(3.27)，然后在 上作积分，则有

$\begin{array}{l}\frac{1}{p_0}{\left({\displaystyle {\int }_{{ℝ}^2}{\sigma }^{p_0}{\left|v\right|}^{p_0}\text{d}x}\right)}_t+{\displaystyle {\int }_{{ℝ}^2}\left(\tilde{u}+1\right){\left(\sigma \left|v\right|\right)}^{p_0}\text{d}x}\\ =ϵ{\displaystyle {\int }_{{ℝ}^2}{\sigma }^{p_0}{\left|v\right|}^{p_0-2}v\nabla {\left|v\right|}^2\text{d}x}+ϵ{\displaystyle {\int }_{{ℝ}^2}{\sigma }^{p_0}{\left|v\right|}^{p_0-2}v\Delta v\text{d}x}\\ +{\displaystyle {\int }_{{ℝ}^2}{\sigma }^{p_0}\left|\text{F}\right|{\left|v\right|}^{p_0-2}\text{d}x}+{\displaystyle {\int }_{{ℝ}^2}{\sigma }^{p_0-1}}{\left|v\right|}^{p_0}\text{d}x.\end{array}$ (51)

将上式在 $\left[0,T\right]$ 上积分有

$\begin{array}{c}{\displaystyle {\int }_{{ℝ}^2}{\sigma }^{p_0}}{\left|v\right|}^{p_0}\text{d}x\le Cϵ{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^2}{\sigma }^{p_0}{\left|v\right|}^{p_0}\left|\nabla v\right|\text{d}x\text{d}t}}+Cϵ{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^2}{\sigma }^{p_0}}}{\left|\nabla v\right|}^2{\left|v\right|}^{p_0-2}\text{d}x\text{d}t\\ +{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^2}{\sigma }^{p_0}}}\left|F\right|{\left|v\right|}^{p_0-1}\text{d}x\text{d}t+{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^2}{\sigma }^{p_0-1}}}{\left|v\right|}^{p_0}\text{d}x\text{d}t\\ \le ϵ{\displaystyle {\int }_0^T{\sigma }^{\frac{p_0-2}{2}}}{\Vert v\Vert }_{L^{\infty }}^{p_0-2}{\displaystyle {\int }_{{ℝ}^2}{\left|v\right|}^2\left|\nabla v\right|\text{d}x\text{d}t}+Cϵ{\displaystyle {\int }_0^T\left({\sigma }^{\frac{p_0-2}{2}}{\Vert v\Vert }_{L^{\infty }}^{p_0-2}\right)}{\displaystyle {\int }_{{ℝ}^2}{\left|\nabla v\right|}^2\text{d}x\text{d}t}\\ +{\displaystyle {\int }_0^T\left({\Vert v\Vert }_{L^{\infty }}^{p_0-4}\right){\sigma }^{\frac{p_0-4}{2}}}{\displaystyle {\int }_{{ℝ}^2}\left|v\right|\text{d}x\text{d}t}+{\displaystyle {\int }_0^T\left|\sigma F\right|{\left|v\right|}^{p_0-1}\text{d}x\text{d}t}\\ \le ϵ\left({\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^2}{v}^4}}+{\left|\nabla v\right|}^2\text{d}x\text{d}t\right)+C{\theta }_0^{\frac{1}{2}+{\eta }_0}+C{\theta }_0^{{\eta }_0}+{\displaystyle {\int }_0^T\left|\sigma F\right|{\left|\sigma v\right|}^{p_0-1}\text{d}x\text{d}t}.\end{array}$ (52)

由不等式

$ab\le ϵa^q+{\left(ϵq\right)}^{-\frac{r}{q}}{r}^{-1}{b}^r,a,b\ge 0,ϵ>0,q,r>0,\frac{1}{q}+\frac{1}{r}=1,$

有

$\begin{array}{l}{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^2}\left|\sigma F\right|{\left|\sigma v_0\right|}^{p_0-1}\text{d}x\text{d}t}}\\ \le \frac{1}{2}{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^2}{\left|\sigma v\right|}^{p_0}\text{d}x\text{d}t}}+\frac{1}{p_0}{\left(\frac{2p_0-2}{p_0}\right)}^{p_0-1}{\displaystyle {\int }_0^T{\left|\sigma F\right|}^{p_0}\text{d}x\text{d}t}\\ \le \frac{1}{2}{\displaystyle {\int }_0^T{\sigma }^{\frac{p_0-4}{2}}}{\Vert v\Vert }_{L^{\infty }}^{p_0-4}{\displaystyle {\int }_{{ℝ}^2}{\left|v\right|}^4\text{d}t\text{d}x}+\frac{1}{p_0}{\left(\frac{2p_0-2}{p_0}\right)}^{p_0-1}{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^2}{\left|\sigma F\right|}^{p_0}\text{d}x\text{d}t}},\end{array}$

因此

${\displaystyle {\int }_{{ℝ}^2}{\left|v\right|}^{p_0}\text{d}x}\le C{\theta }_0^{{\eta }_0}+{\left(\frac{2p_0-2}{p_0}\right)}^{p_0-1}{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^2}{\left|\sigma F\right|}^{p_0}\text{d}x\text{d}t}}.$

由G-N不等式(13)可得

$\begin{array}{c}{\displaystyle {\int }_0^T{\Vert \sigma F\Vert }_{L^{p_0}}^{p_0}\text{d}t}\le C{\displaystyle {\int }_0^T{\Vert \sigma F\Vert }^2{\Vert \sigma \nabla F\Vert }^{p_0-2}\text{d}t}\\ \le C\underset{t\in \left[0,T\right]}{\sup }{\Vert \sigma F\Vert }^2{\Vert \sigma \nabla F\Vert }^{p_0-4}{\displaystyle {\int }_0^T{\Vert \sigma \nabla F\Vert }^2\text{d}}t.\end{array}$ (53)

由

$\underset{t\in \left[0,\sigma \left(T\right)\right]}{\sup }{\sigma }^{\frac{1}{2}}\Vert F\Vert \le C{\sigma }^{\frac{1}{2}}\Vert \nabla \tilde{u}\Vert +C{\sigma }^{\frac{1}{2}}{\theta }_0^{\frac{1}{4}}\le C{\theta }_0^{\frac{1}{4}}$

和

${\displaystyle {\int }_0^T\sigma {\Vert \nabla F\Vert }^2\text{d}t}\le C{\theta }_0^{{\eta }_0},$

可得

$\underset{t\in \left[0,T\right]}{\sup }{\Vert \sigma F\Vert }^2\le \underset{t\in \left[0,T\right]}{\sup }\sigma {\Vert F\Vert }^2\le C{\theta }_0^{\frac{1}{2}}\text{.}$

由(10)，(13)可得

$\begin{array}{c}{\Vert \nabla \tilde{u}\Vert }_{L^4}\le {\Vert F-\tilde{u}v-v\Vert }_{L^4}\\ \le {\Vert F\Vert }_{L^4}+{\Vert \tilde{u}v\Vert }_{L^4}+{\Vert v\Vert }_{L^4}\\ \le C\left({\Vert F\Vert }^{\frac{1}{2}}{\Vert \nabla F\Vert }^{\frac{1}{2}}+{\Vert \tilde{u}\Vert }_{L^{\infty }}{\Vert v\Vert }_{L^4}+{\Vert v\Vert }_{L^4}\right)\\ \le \left({\Vert F\Vert }^{\frac{1}{2}}{\Vert \nabla F\Vert }^{\frac{1}{2}}+{\Vert \tilde{u}\Vert }_{L^4}^{\frac{1}{2}}{\Vert \nabla \tilde{u}\Vert }_{L^4}^{\frac{1}{2}}{\Vert v\Vert }_{L^4}+{\Vert v\Vert }_{L^4}\right)\text{.}\end{array}$

利用上式和(13)可得

下面对 $I_4$ 做估计，由(13)和Hölder不等式可得

$\Vert \nabla F\Vert \le C\Vert {\tilde{u}}_t\Vert +C\Vert {\nabla }^{\perp }\tilde{u}\cdot v\Vert \le C\Vert {\tilde{u}}_t\Vert +C{\Vert \nabla \tilde{u}\Vert }_{L^4}{\Vert v\Vert }_{L^4}\text{.}$

由(15)，(16)，(43)及以上不等式，可得

$\begin{array}{c}{I}_4\le C{\displaystyle {\int }_0^T\sigma }\left({\Vert \nabla \tilde{u}\Vert }^2+{\theta }_0^{\frac{1}{2}+{\eta }_0}\right)\sigma \left({\Vert {\tilde{u}}_t\Vert }^2+{\Vert \nabla \tilde{u}\Vert }_{L^4}{\Vert v\Vert }_{L^4}^2\right)\text{d}t\\ \le \sigma \left(A_2\left(T\right)+{\theta }_0^{\frac{1}{2}+{\eta }_0}\right){\displaystyle {\int }_0^T\sigma }\left({\Vert u_t\Vert }^2+{\Vert \nabla \tilde{u}\Vert }_{L^4}^2{\Vert v\Vert }_{L^4}^2\right)\text{d}t\\ \le C{\theta }_0^{\frac{1}{2}}{\displaystyle {\int }_0^T\sigma }{\Vert {\tilde{u}}_t\Vert }^2\text{d}t+C{\theta }_0^{\frac{1}{2}}{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla \tilde{u}\Vert }_{L^4}^4\text{d}t+C{\theta }_0^{\frac{1}{2}}{A}_3\left(T\right)\\ \le C{\theta }_0^{\frac{1}{2}}{A}_2\left(T\right)+C{\theta }_0^{\frac{1}{2}}{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla \tilde{u}\Vert }_{L^4}^4\text{d}t+C{\theta }_0^{\frac{1}{2}}{A}_3\left(T\right)\\ \le C{\theta }_0+C{\theta }_0^{\frac{1}{2}}{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla \tilde{u}\Vert }_{L^4}^4\text{d}t+C{\theta }_0^{\frac{1}{2}+{\eta }_0}\\ \le C{\theta }_0^{\frac{1}{2}}+C{\theta }_0^{\frac{1}{2}}{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla \tilde{u}\Vert }_{L^4}^4\text{d}t.\end{array}$

以下估计 $I_5$ ，由(13)，(16)，(17)可得

$\begin{array}{c}{I}_5=C{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \tilde{u}\Vert }_{L^4}^2{\Vert \nabla \tilde{u}\Vert }_{L^4}^2\text{d}t\\ \le \frac{1}{4}{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \tilde{u}\Vert }_{L^4}^4\text{d}t+C{A}_3^2\left(T\right){\displaystyle {\int }_0^T{\displaystyle {\int }_0^T{\sigma }^2}}{\Vert \tilde{u}\Vert }_{L^4}^4\text{d}t\\ \le \frac{1}{4}{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla \tilde{u}\Vert }_{L^4}^4\text{d}t+C{A}_3{\left(T\right)}^2{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \tilde{u}\Vert }^2{\Vert \nabla \tilde{u}\Vert }^2\text{d}t\\ \le \frac{1}{4}{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla \tilde{u}\Vert }_{L^4}^4\text{d}t+C{\theta }_0^{\frac{1}{2}+3{\eta }_0}\\ \le \frac{1}{4}{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla \tilde{u}\Vert }_{L^4}^4\text{d}t+C{\theta }_0^{{\eta }_0}\text{.}\end{array}$

取 ${\theta }_0$ 使得 $C{\theta }_0^{\frac{1}{2}}\le \frac{1}{4}$ ，那么

${\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla \tilde{u}\Vert }_{L^4}^4\text{d}t\le C{\theta }_0^{{\eta }_0}+C{\theta }_0^{\frac{1}{2}}\le C{\theta }_0^{{\eta }_0}\text{.}$

$\begin{array}{c}{\displaystyle {\int }_0^T\sigma }{\Vert \nabla F\Vert }^2\text{d}t\le C{\displaystyle {\int }_0^T\sigma }{\Vert {\tilde{u}}_t\Vert }^2\text{d}t+C{\displaystyle {\int }_0^T\sigma }{\Vert \nabla \tilde{u}\Vert }_{L^4}^2{\Vert v\Vert }_{L^4}^2\text{d}t\\ \le {\displaystyle {\int }_0^T\sigma }{\Vert {\tilde{u}}_t\Vert }^2\text{d}t+C{\displaystyle {\int }_0^T\left({\sigma }^2{\Vert \nabla \tilde{u}\Vert }_{L^4}^4+{\Vert v\Vert }_{L^4}^4\right)\text{d}t}\\ \le C{\theta }_0^{\frac{1}{2}}+C{\theta }_0^{{\eta }_0}\le C{\theta }_0^{{\eta }_0}\text{.}\end{array}$

由于

$\underset{t\in \left[0,T\right]}{\sup }\Vert \sigma \nabla F\Vert \le C{\theta }_0^{\frac{{\eta }_0}{2}}+C\sigma \Vert {\tilde{u}}_t\Vert +C{\sigma }^{\frac{1}{2}}\Vert \nabla \tilde{u}\Vert \le C\sigma \left({\theta }_0^{\frac{{\eta }_0}{2}}+{\theta }_0^{\frac{1}{4}}\right)\le C{\theta }_0^{\frac{{\eta }_0}{2}},$

那么

$\underset{t\in \left[0,T\right]}{\sup }{\Vert \sigma \nabla F\Vert }^{p_0-4}\le C{\theta }_0^{\frac{{\eta }_0\left(p_0-4\right)}{2}}\text{.}$

把以上三个不等式代入(53)可得

${\displaystyle {\int }_0^T{\Vert \sigma F\Vert }_{L^{p_0}}^{p_0}\text{d}t}\le C{\theta }_0^{\frac{{\eta }_0\left(p_0-2\right)+1}{2}}=C{\theta }_0^{\frac{p_0-2}{4}},$ (54)

因此取 ${\theta }_0$ 足够下，当 $C{\theta }_0^{{\eta }_0}+C{\theta }_0^{\frac{p_0-2}{4}}\le {\left(3M\right)}^{p_0}$ 时，

即

$\underset{t\in \left[0,T\right]}{\sup }{\Vert v\Vert }_{L^{p_0}}\le 3M.$

因此完成了引理3.5的证明。

根据引理3.1~3.5，可得

$A_1\left(T\right)\le \frac{3}{2}{\theta }_0,A_2\left(T\right)\le {\theta }_0^{\frac{1}{2}},A_3\left(T\right)\le {\theta }_0^{{\eta }_0},\underset{t\in \left[0,T\right]}{\sup }{\Vert v\Vert }_{L^{p_0}}\le 3M$ . (55)

4. 主定理证明

下面利用爆破准则得到全局解

命题4.1假设 $\left({\tilde{u}}_0^{\delta },v_0^{\delta }\right)\in H^3\left({ℝ}^2\right)$ ，那么系统(16)有一个唯一解，使得 $\left({\tilde{u}}^{\delta },v^{\delta }\right)\in L^{\infty }\left(\left[0,\infty \right),H^3\right)$ 。

证明：通过引理2.1，则存在 $T_{\text{*}}>0$ ，使得满足初始条件的系统(3.1)在 ${ℝ}^2\times \left[0,T^{\text{*}}\right]$ 上的Cauchy问题有唯一解。首先，从(3.3)可知

$\begin{array}{c}{A}_1\left(0\right)\le {\theta }_0,A_2\left(0\right)<2{\theta }_0^{\frac{1}{2}},{\Vert v_0^{\delta }\Vert }_{L^{p_0}}\le M\text{.}\end{array}$

令 $C\left(M\right){\theta }_0^{3{\eta }_0}\le 1$ ，

$A_3\left(0\right)={\Vert v_0^{\delta }\Vert }_{L^4}^4\le C\left(M\right){\theta }_0^{4{\eta }_0}\le {\theta }_0^{{\eta }_0}.$ (56)

因此，存在一个 $T_1>0$ ，使得 $T_1\le T_{\text{*}}$ ，(3.3)成立。

反证，如果 $$ ，由爆破准则，则

${\displaystyle {\int }_0^{T^{\text{*}}}{\Vert v^{\delta }\Vert }_{L^4}^4\text{d}t}<\infty$ .

根据引理3.1-3.5，当 $T=T^{\text{*}}$ 时，上式成立，所以存在 $T^{\text{**}}>T^{\text{*}}$ ，使得

$\left(u^{\delta }-1,v^{\delta }\right)\in L^{\infty }\left(\left[0,T^{\text{**}}\right],H^3\right),$

和当 $T=T^{\text{**}}$ 时(16)成立，与最大存在时刻矛盾。所以根据连续性技巧，可证命题4.1。

从(55)得到下面关于 $\delta$ 的一致估计

$\{\begin{array}{l}\underset{t\in \left[0,T\right]}{\sup }\left({\Vert {\tilde{u}}^{\delta }\Vert }^2+{\Vert v^{\delta }\Vert }^2\right)+ϵ{\Vert \nabla v^{\delta }\Vert }^2+{\displaystyle {\int }_0^T{\Vert \nabla {\tilde{u}}^{\delta }\Vert }^2\text{d}t}+ϵ{\displaystyle {\int }_0^T{\Vert \nabla v^{\delta }\Vert }^2\text{d}t}\le \frac{3{\theta }_0}{2},\\ \underset{t\in \left[0,T\right]}{\sup }\left(\sigma {\Vert \nabla {\tilde{u}}^{\delta }\Vert }^2+{\sigma }^2{\Vert {\tilde{u}}_t^{\delta }\Vert }^2+{\sigma }^2{\Vert v_t^{\delta }\Vert }^2+\sigma ϵ{\Vert \nabla v^{\delta }\Vert }^2\right)+{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla {\tilde{u}}_t^{\delta }\Vert }^2\text{d}t\\ +ϵ{\displaystyle {\int }_0^T{\sigma }^2}{\Vert \nabla v^{\delta }\Vert }^2\text{d}t+{\displaystyle {\int }_0^T\sigma }{\Vert v_t^{\delta }\Vert }^2\text{d}t+{\displaystyle {\int }_0^T\sigma }{\Vert {\tilde{u}}_t^{\delta }\Vert }^2\text{d}t\le {\theta }_0^{\frac{1}{2}},\\ \underset{t\in \left[0,T\right]}{\sup }{\Vert {\tilde{u}}^{\delta }\Vert }_{L^4}^4+{\displaystyle {\int }_0^T{\Vert v^{\delta }\Vert }_{L^4}^4\text{d}t}\le {\theta }_0^{{\eta }_0},\\ \underset{t\in \left[0,T\right]}{\sup }{\Vert v^{\delta }\Vert }_{L^{p_0}}\le 3M\text{.}\end{array}$ (57)

由(13)，(57)，可得

$\sigma {\Vert u^{\delta }-1\Vert }_{L^4}^2\le C\Vert u^{\delta }-1\Vert \sigma \Vert \nabla u^{\delta }\Vert \le C,{\sigma }^2{\Vert \nabla u^{\delta }\Vert }_{L^4}^2\le C\text{.}$ (58)

所以由(4.2)，(4.3)，则有

$\{\begin{array}{l}{u}^{\sigma }-1\in L^{\infty }\left(\left[0,\infty \right),L^2\left({ℝ}^2\right)\right),\left(\nabla v^{\delta },\nabla u^{\delta }\right)\in L^2\left(\left[0,\infty \right),L^2\left({ℝ}^2\right)\right),\\ u^{\delta }-1\in L^{\infty }\left(\left(0,\infty \right),W^{1,4}\left({ℝ}^2\right)\right),u_t^{\delta },v_t^{\delta }\in L^2\left(\left(0,\infty \right),H^1\left({ℝ}^2\right)\right),\\ v\in L^{\infty }\left(\left[0,\infty \right);L\left({ℝ}^2\right)\cap L^{p_0}\left({ℝ}^2\right)\right)\cap L^4\left(\left[0,\infty \right);L^4\left({ℝ}^2\right)\right)\text{.}\end{array}$ (59)

由 $W^{1,4}\left({ℝ}^2\right)$ ↪↪ $C\left({ℝ}^2\right)$ ， $L^2\left({ℝ}^2\right)$ ↪↪ $H^{-1}\left({ℝ}^2\right)$ ，因此Aubin-Lions-Simon引理，可以抽取一个子列 $\left(u^{\delta },v^{\delta }\right)$ ，当 $\delta \to 0$ ，则

$\{\begin{array}{l}{v}^{\delta }\left(\cdot ,t\right)\to v\text{stronglyin}C\left(\left[0,\infty \right),H^{-1}\left({ℝ}^2\right)\right),\\ u^{\delta }\left(\cdot ,t\right)\to u\left(\cdot ,t\right)\text{Stronglyin}C\left(\left[0,\infty \right),C\left({ℝ}^2\right)\right),\\ \nabla u^{\delta }\left(\cdot ,t\right)\to \nabla u\left(\cdot ,t\right)\text{weaklyin}{L}^2\left(\left[0,\infty \right),L\left({ℝ}^2\right)\right),\\ \nabla v^{\delta }\left(\cdot ,t\right)\to \nabla v\left(\cdot ,t\right)\text{weaklyin}{L}^2\left(\left[0,\infty \right),L\left({ℝ}^2\right)\right)\text{.}\end{array}$ (60)

那么

${\displaystyle {\int }_1^{\infty }{\Vert \nabla \tilde{u}\Vert }^2\text{d}t}\le C,$

且

$\begin{array}{c}{\displaystyle {\int }_1^{\infty }{\left|\left({\Vert \nabla \tilde{u}\Vert }^2\right)\right|}_t\text{d}t}\le 2{\displaystyle {\int }_1^{\infty }\Vert \nabla \tilde{u}\Vert \Vert \nabla {\tilde{u}}_t\Vert \text{d}t}\\ \le {\left({\displaystyle {\int }_1^{\infty }{\Vert \nabla \tilde{u}\Vert }^2\text{d}t}\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_1^{\infty }{\sigma }^2}{\Vert \nabla {\tilde{u}}_t\Vert }^2\text{d}t\right)}^{\frac{1}{2}}\\ \le C,\end{array}$

那么

$\Vert \nabla \tilde{u}\Vert \to 0,t\to \infty ,$ (61)

据此可得

${\Vert u-1\Vert }_{L^{\infty }}\to 0,t\to \infty .$ (62)

根据插值不等式，当 $2<p_1\le \infty$ 时，可得

${\Vert u-1\Vert }_{L^{p_1}}\to 0,t\to \infty \text{.}$

类似地

${\displaystyle {\int }_1^{\infty }\left|{\left({\Vert \nabla v\Vert }^2\right)}_t\right|\text{d}t}\le C,$

那么

${\Vert v\Vert }_{L^{p_1}}\to 0,t\to \infty .$

至此完成定理1.2的证明。