带阻尼的三维热带气候模型的适定性研究

Well-Posed Study of a Three-Dimensional Tropical Climate Model with Damping

摘要: 本文考虑了带阻尼的三维热带气候模型。利用能量估计的方法，证明了带阻尼的三维热带气候模型对于，如果β，γ，δ满足β≥4，，δ≥1时，那么带阻尼的三维热带气候模型强解是存在的，且是唯一的。

Abstract: This paper considers a three-dimensional tropical climate model with damping. Using the method of energy estimation, a damped three-dimensional tropical climate model is demonstrated. For , if β , γ , δ satisfies β≥4 , , δ≥1 , then the existence and uniqueness of a damped 3-D tropical climate model is strong.

1. 引言

在本文中，我们考虑以下带阻尼的三维热带气候模型：

$\{\begin{array}{l}{u}_t+{\Lambda }^{2\alpha }u+\left(u\cdot \nabla \right)u+{\left|u\right|}^{\beta -1}u+\nabla \pi +div\left(v\otimes v\right)=0,\\ v_t-\Delta v+\left(u\cdot \nabla \right)v+{\left|v\right|}^{\gamma -1}v+\nabla \psi +\left(v\cdot \nabla \right)u=0,\\ {\psi }_t-\Delta \psi +\left(u\cdot \nabla \right)\psi +{\left|\psi \right|}^{\delta -1}\psi +divv=0,\\ divu=0,\\ \left(u,v,\psi \right)\left(x,0\right)=\left(u_0,v_0,{\psi }_0\right)\end{array}$

其中向量场 $\text{u}=\text{u}\left(x,t\right)\in R^3$ 和 $\text{v}=\text{v}\left(x,t\right)\in R^3$ 分别表示速度的正压模式和第一斜压模式。 $\pi =\pi \left(x,t\right)$ 和 $\psi =\psi \left(x,t\right)$ 分别表示压强和温度。分数拉普拉斯算子 $\Lambda ={\left(-\Delta \right)}^{\frac{1}{2}}$， $\alpha >0$ 和 $\beta ,\gamma ,\delta \ge 0$ 为实参数。

需要指出的是，当 $\psi =\text{0}$， $\alpha =1$ 和 $divv=0$ 时，系统(1.1)可简化为带有阻尼的三维磁流体动力学(MHD)型方程。对于阻尼磁流体系统，Ye [1] 首先得到了全局强解的存在性和唯一性。然后，在 [2] [3] 中，证明了含阻尼的三维磁流体方程的全局适定性和衰减。当v = 0时，带阻尼的mhd型方程简化为带阻尼

的Navier-Stokes方程。Cai和Jiu [4] 首先证明了带阻尼的Navier-Stokes方程在任何 $\beta \ge \frac{\text{7}}{\text{2}}$ 情况下，强解是唯一的，在 $\frac{\text{7}}{\text{2}}\le \beta \le \text{5}$ 的情况下带阻尼的Navier-Stokes方程具有全局强解。Zhou [5] 改进了这一结果，

证明了 $\beta \ge \text{3}$ 时全局存在强解。随后，有许多结果致力于三维Navier-Stokes方程的阻尼(例如， [6] [7] [8] [9])。

让我们简单回顾一下关于三维热带气候模式的一些成果。当系统(1.1)没有任何阻尼项时，Wang et al. [10] 在初始数据很小的情况下考虑了正则性准则和全局存在性。在 [11] [12] 中，作者建立了分阶耗散的三维热带气候模式的全局正则性。我们可以看到 [13] - [19] 关于二维热带气候模式的全局正则性问题的一些结果。本文利用阻尼项证明了系统(1.1)具有唯一的全局强解。值得一提的是，由于缺乏v的自由发散条件，热带气候模式的结果与三维MHD方程的结果存在差异。

阻尼项在一定程度上是好项，可以增加正则性，我们为了提高三维热带气候模型的正则性，为此我们增加阻尼项，研究带阻尼的三维热带气候模式的适定性。我们的主要结果可以陈述如下：

定理1.1. 让 $\left({\text{u}}_{\text{0}}\left(x\right),{\text{v}}_{\text{0}}\left(x\right),{\psi }_{\text{0}}\left(x\right)\right)\in H^1\left(R^3\right)$ 满足 $\nabla \cdot u=0$。对于 $\text{1}\le \alpha \le \frac{\text{5}}{\text{2}}$，如果 $\beta ,\gamma ,\delta$ 满足

$\beta \ge 4,\gamma \ge \frac{\alpha +3}{\alpha },\delta \ge 1.$

那么系统(1.1)有一个唯一的全局强解u满足，对于任意T > 0，

$\begin{array}{l}u\in L^{\infty }\left(0,T;H^1\left(R^3\right)\right)\cap L^2\left(0,T;H^{\alpha +1}\left(R^3\right)\right)\cap L^{\beta +1}\left(0,T;H^{\beta +1}\left(R^3\right)\right)\\ v\in L^{\infty }\left(0,T;H^1\left(R^3\right)\right)\cap L^2\left(0,T;H^2\left(R^3\right)\right)\cap L^{\gamma +1}\left(0,T;H^{\gamma +1}\left(R^3\right)\right)\\ \psi \in L^{\infty }\left(0,T;H^1\left(R^3\right)\right)\cap L^2\left(0,T;H^2\left(R^3\right)\right)\cap L^{\delta +1}\left(0,T;H^{\delta +1}\left(R^3\right)\right)\end{array}$

本文的结构如下：第一部分我们给出关于三维热带气候模型的相关进展和研究现状，并且给出主要结果；第二部分，我们给出主要结果的证明步骤。

2. 定理1.1的证明

在这一节中，我们证明定理1.1。首先，我们对系统(1.1)给出了一个先验 $L^2$ 估计。将(1.1)分别乘以 $\left(\text{u},\text{v},\psi \right)$，经过分部积分，并使用 $\nabla \cdot u=0$，得到以下能量估计

$\begin{array}{l}{\Vert u\left(t\right)\Vert }_{L^2}^2+{\Vert v\left(t\right)\Vert }_{L^2}^2+{\Vert \psi \left(t\right)\Vert }_{L^2}^2+2{\displaystyle {\int }_0^T({\Vert {\Lambda }^{\alpha }u\left(t\right)\Vert }_{L^2}^2}+{\Vert \nabla v\left(t\right)\Vert }_{L^2}^2+{\Vert \nabla \psi \left(t\right)\Vert }_{L^2}^2\\ +{\Vert u\Vert }_{L^{\beta +1}}^{\beta +1}+{\Vert v\Vert }_{L^{\gamma +1}}^{\gamma +1}+{\Vert \psi \Vert }_{L^{\delta +1}}^{\delta +1})\text{d}s={\Vert u_0\Vert }_{L^2}^2+{\Vert v_0\Vert }_{L^2}^2+{\Vert {\psi }_0\Vert }_{L^2}^2.\end{array}$

我们在哪里使用了下面的事实

$\begin{array}{l}{\displaystyle \underset{R^3}{\int }\nabla \psi \cdot v\text{d}x}+{\displaystyle \underset{R^3}{\int }\left(\nabla \cdot v\right)\psi \text{d}x}=0,\\ {\displaystyle \underset{R^3}{\int }div\left(v\otimes v\right)\cdot u\text{d}x}+{\displaystyle \underset{R^3}{\int }\left(v\cdot \nabla \right)u\cdot v\text{d}x=0}\end{array}$

将(1.1)分别乘以 $-\Delta u$ 、 $-\Delta v$ 和 $-\Delta \psi$，在 $R^3$ 中积分后相加，得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert \nabla u\Vert }_{L^2}^2+{\Vert \nabla v\Vert }_{L^2}^2+{\Vert \nabla \psi \Vert }_{L^2}^2\right)+{\Vert {\Lambda }^{\alpha +1}u\Vert }_{L^2}^2+{\Vert \Delta v\Vert }_{L^2}^2+{\Vert \Delta \psi \Vert }_{L^2}^2\\ +{\Vert {\left|u\right|}^{\frac{\beta -1}{2}}\nabla u\Vert }_{L^2}^2+{\Vert {\left|v\right|}^{\frac{\gamma -1}{2}}\nabla v\Vert }_{L^2}^2+{\Vert {\left|\psi \right|}^{\frac{\delta -1}{2}}\nabla \psi \Vert }_{L^2}^2+\frac{4\left(\beta -1\right)}{{\left(\beta +1\right)}^2}{\Vert \nabla {\left|u\right|}^{\frac{\beta +1}{2}}\Vert }_{L^2}^2\\ +\frac{4\left(\gamma -1\right)}{{\left(\gamma +1\right)}^2}{\Vert \nabla {\left|v\right|}^{\frac{\gamma +1}{2}}\Vert }_{L^2}^2+\frac{4\left(\delta -1\right)}{{\left(\delta +1\right)}^2}{\Vert \nabla {\left|\psi \right|}^{\frac{\delta +1}{2}}\Vert }_{L^2}^2\\ ={\displaystyle \underset{R^3}{\int }\left(u\cdot \nabla \right)u}\cdot \Delta u\text{d}x+{\displaystyle \underset{R^3}{\int }div\left(v\otimes v\right)}\cdot \Delta u\text{d}x+{\displaystyle \underset{R^3}{\int }\left(u\cdot \nabla \right)v}\cdot \Delta v\text{d}x\\ +{\displaystyle \underset{R^3}{\int }\left(v\cdot \nabla \right)u}\cdot \Delta v\text{d}x+{\displaystyle \underset{R^3}{\int }\nabla \psi }\cdot \Delta v\text{d}x+{\displaystyle \underset{R^3}{\int }\left(u\cdot \nabla \right)\psi }\cdot \Delta \psi \text{d}x+{\displaystyle \underset{R^3}{\int }\left(\nabla \cdot v\right)\cdot \Delta \psi \text{d}x}\\ =I_1+I_2+I_3+I_4+I_5+I_6+I_7,\end{array}$

首先，应用Young不等式、Hölder不等式和Gagliardo-Nirenberg不等式， $I_1$ 可以估计如下。

$\begin{array}{c}{I}_1={\displaystyle \underset{R^3}{\int }\left(u\cdot \nabla \right)u\cdot \Delta u\text{d}x}\\ ={\displaystyle \underset{R^3}{\int }{\Lambda }^{1-\alpha }\left(u\cdot \nabla u\right)\cdot {\Lambda }^{\alpha +1}u\text{d}x}\\ ={\displaystyle \underset{R^3}{\int }{\Lambda }^{2-\alpha }\left(u\otimes u\right)\cdot {\Lambda }^{\alpha +1}u\text{d}x}\\ \le {\Vert {\Lambda }^{\alpha +1}u\Vert }_{L^2}{\Vert {\Lambda }^{2-\alpha }\left(u\otimes u\right)\Vert }_{L^2}\end{array}$

$\begin{array}{c}\le {\Vert {\Lambda }^{\alpha +1}u\Vert }_{L^2}{\Vert {\Lambda }^{2-\alpha }u\Vert }_{L^{\frac{2(\beta +1)}{\beta -1}}}{\Vert u\Vert }_{L^{\beta +1}}\\ \le {\Vert {\Lambda }^{\alpha +1}u\Vert }_{L^2}^{1+\theta }{\Vert \nabla u\Vert }_{L^2}^{1-\theta }{\Vert u\Vert }_{L^{\beta +1}}\\ \le \frac{1}{2}{\Vert {\Lambda }^{\alpha +1}u\Vert }_{L^2}^2+C{\Vert u\Vert }_{L^{\beta +1}}^{\frac{2}{1-\theta }}{\Vert \nabla u\Vert }_{L^2}^2\end{array}$

这里我们使用了以下的Gagliardo-Nirenberg不等式：

${\Vert {\Lambda }^{2-\alpha }u\Vert }_{L^{\frac{2\left(\beta +1\right)}{\beta -1}}}\le C{\Vert {\Lambda }^{\alpha +1}u\Vert }_{L^2}^{\theta }{\Vert \nabla u\Vert }_{L^2}^{1-\theta }$

在这里

$\frac{\beta -1}{2\left(\beta +1\right)}=\frac{1-\alpha }{3}+\left(\frac{1}{2}-\frac{\alpha }{3}\right)\theta +\frac{1-\theta }{2}$

通过直接计算，得到

$\theta =\frac{\left(5-2\alpha \right)\left(\beta +1\right)-3\left(\beta -1\right)}{2\alpha \left(\beta +1\right)}$

如果下列限制成立

$\frac{2}{1-\theta }\le \beta +1$

然后我们得到 $\beta \ge \frac{\text{4}}{\text{2}\alpha -\text{1}}$

通过Hölder不等式、Gagliardo-Nirenberg不等式和Young不等式， $I_2$ 、 $I_3$ 和 $I_4$ 的项可以估计如下：

$\begin{array}{l}{I}_2+I_3+I_4\\ ={\displaystyle \underset{R^3}{\int }\nabla \cdot \left(v\otimes v\right)\cdot \Delta u\text{d}x+}{\displaystyle \underset{R^3}{\int }\left(u\cdot \nabla \right)v\cdot \Delta v\text{d}x}+{\displaystyle \underset{R^3}{\int }\left(v\cdot \nabla \right)u\cdot \Delta v\text{d}x}\\ ={\displaystyle \underset{R^3}{\int }\left(v_j{\partial }_i{v}_i+v_i{\partial }_i{v}_j\right){\partial }_k{\partial }_k{u}_j\text{d}x}+{\displaystyle \underset{R^3}{\int }{u}_i{\partial }_i{v}_j{\partial }_k{\partial }_k{v}_j\text{d}x}+{\displaystyle \underset{R^3}{\int }{v}_i{\partial }_i{u}_j{\partial }_k{\partial }_k{v}_j\text{d}x}\\ ={\displaystyle \underset{R^3}{\int }\left(v_j{\partial }_i{v}_i+v_i{\partial }_i{v}_j\right){\partial }_k{\partial }_k{u}_j\text{d}x}+{\displaystyle \underset{R^3}{\int }\left({\partial }_k{\partial }_k{u}_i{\partial }_i{v}_j+{\partial }_k{u}_i{\partial }_k{\partial }_i{v}_j\right)\text{d}x}+{\displaystyle \underset{R^3}{\int }{v}_i{\partial }_i{u}_j{\partial }_k{\partial }_k{v}_j\text{d}x}\\ \le {\displaystyle \underset{R^3}{\int }\left|v\right|\left|\nabla v\right|\left|{\nabla }^{2}u\right|\text{d}x}+{\displaystyle \underset{R^3}{\int }\left|v\right|\left|\nabla u\right|\left|{\nabla }^{2}v\right|\text{d}x}\\ =J_1+J_2\end{array}$

接下来，我们将估计 $J_1$ 和 $J_2$。

$\begin{array}{c}{J}_1={\displaystyle \underset{R^3}{\int }\left|v\right|\left|\nabla v\right|\left|{\nabla }^{2}u\right|\text{d}x}\\ ={\displaystyle \underset{R^3}{\int }{\left|v\right|}^{\frac{\gamma -1}{2}}\left|\nabla v\right|{\left|v\right|}^{\frac{3-\gamma }{2}}\left|\Delta u\right|\text{d}x}\\ \le {\Vert {\left|v\right|}^{\frac{\gamma -1}{2}}\nabla v\Vert }_{L^2}{\Vert v\Vert }_{L^{\gamma +1}}^{\frac{3-\gamma }{2}}{\Vert \Delta u\Vert }_{L^{\frac{\gamma +1}{\gamma -1}}}\\ \le {\Vert {\left|v\right|}^{\frac{\gamma -1}{2}}\nabla v\Vert }_{L^2}{\Vert v\Vert }_{L^{\gamma +1}}^{\frac{3-\gamma }{2}}{\Vert \nabla u\Vert }_{L^2}^{1-{\lambda }_1}{\Vert {\Lambda }^{1+\alpha }u\Vert }_{L^2}^{{\lambda }_1}\\ \le \frac{1}{2}{\Vert {\left|v\right|}^{\frac{\gamma -1}{2}}\nabla v\Vert }_{L^2}^2+\frac{1}{2}{\Vert {\Lambda }^{1+\alpha }u\Vert }_{L^2}^2+C{\Vert v\Vert }_{L^{\gamma +1}}^{\frac{3-\gamma }{1-{\lambda }_1}}{\Vert \nabla u\Vert }_{L^2}^2\end{array}$

这里我们使用了Gagliardo-Nirenberg不等式：

${\Vert \Delta u\Vert }_{L^{\frac{\gamma +1}{\gamma -1}}}\le C{\Vert {\Lambda }^{\alpha +1}u\Vert }_{L^2}^{{\lambda }_1}{\Vert \nabla u\Vert }_{L^2}^{1-{\lambda }_1}$

在这里

$\frac{\gamma -1}{\gamma +1}=\frac{1}{3}+\left(\frac{1}{2}-\frac{\alpha }{3}\right){\lambda }_1+\frac{1-{\lambda }_1}{2},{\lambda }_1\in \left[\frac{1}{\alpha },1\right]$

从(2.12)我们可以直接计算出 ${\lambda }_1=\frac{11-\gamma }{2\alpha \left(\gamma +1\right)}$， $\gamma \ge \frac{11-2\alpha }{2\alpha +1}$ 和 $\alpha \ge \text{1}$。

如果下列限制成立

$\frac{3-\gamma }{1-{\lambda }_1}\le \gamma +1$

然后，我们可以得到 $\gamma \ge \frac{\text{4}\alpha +\text{11}}{\text{4}\alpha +\text{1}}$。

同理， $J_2$ 可以估算如下：

$\begin{array}{c}{J}_2={\displaystyle \underset{R^3}{\int }\left|v\right|\left|\nabla u\right|\left|{\nabla }^{2}v\right|\text{d}x}\\ \le {\Vert v\Vert }_{L^{\gamma +1}}{\Vert \nabla u\Vert }_{L^{\frac{2\gamma +1}{\gamma -1}}}{\Vert \Delta v\Vert }_{L^2}\\ \le C{\Vert v\Vert }_{L^{\gamma +1}}^2{\Vert \nabla u\Vert }_{L^{\frac{2\gamma +1}{\gamma -1}}}^2+\frac{1}{4}{\Vert \Delta u\Vert }_{L^2}^2\\ \le C{\Vert v\Vert }_{L^{\gamma +1}}^2{\Vert \nabla u\Vert }_{L^2}^{2\left(1-{\lambda }_2\right)}{\Vert {\Lambda }^{1+\alpha }u\Vert }_{L^2}^{2{\lambda }_2}+\frac{1}{4}{\Vert \Delta v\Vert }_{L^2}^2\\ \le C{\Vert v\Vert }_{L^{\gamma +1}}^{\frac{2}{1-{\lambda }_2}}{\Vert \nabla u\Vert }_{L^2}^2+\frac{1}{4}{\Vert {\Lambda }^{1+\alpha }u\Vert }_{L^2}^2+\frac{1}{4}{\Vert \Delta v\Vert }_{L^2}^2\end{array}$

这里我们使用了Gagliardo-Nirenberg不等式：

${\Vert \nabla u\Vert }_{L^{\frac{2\left(\gamma +1\right)}{\gamma -1}}}\le C{\Vert {\Lambda }^{\alpha +1}u\Vert }_{L^2}^{{\lambda }_2}{\Vert \nabla u\Vert }_{L^2}^{1-{\lambda }_2}.$

通过直接计算，我们有 ${\lambda }_{\text{2}}=\frac{\text{3}}{\alpha \left(\gamma +\text{1}\right)}$。如果下列限制成立

$\frac{2}{1-{\lambda }_2}\le \gamma +1$

然后，我们可以得到 $\gamma \ge \frac{\alpha +\text{3}}{\alpha }$。对于 $I_5$ 和 $I_7$，我们可以得到

$\begin{array}{c}{I}_5+I_7={\displaystyle \underset{R^3}{\int }\nabla \psi \cdot \Delta v\text{d}x}+{\displaystyle \underset{R^3}{\int }\left(\nabla \cdot v\right)\Delta \psi \text{d}x}\\ \le {\displaystyle \underset{R^3}{\int }\left|\nabla \psi \right|\left|\Delta v\right|\text{d}x}\\ \le \frac{1}{4}{\Vert \Delta v\Vert }_{L^2}^2+C{\Vert \nabla \psi \Vert }_{L^2}^2.\end{array}$

应用soblove不等式， $I_6$ 可以估计如下

$\begin{array}{c}{I}_6={\displaystyle \underset{R^3}{\int }\left(u\cdot \nabla \right)\psi \Delta \psi \text{d}x}\\ \le {\Vert u\Vert }_{L^{\beta +1}}{\Vert \nabla \psi \Vert }_{L^{\frac{2\beta +1}{\beta -1}}}{\Vert \Delta \psi \Vert }_{L^2}\\ \le \frac{1}{4}{\Vert \Delta \psi \Vert }_{L^2}^2+C{\Vert u\Vert }_{L^{\beta +1}}^{\frac{2\left(\beta +1\right)}{\beta -2}}{\Vert \nabla \psi \Vert }_{L^2}^2.\end{array}$

收集以上的估计(2.5)~(2.18)并将它们应用到(2.4)中，有下列结果成立

$\begin{array}{l}\frac{\text{d}}{\text{d}t}\left({\Vert \nabla u\Vert }_{L^2}^2+{\Vert \nabla v\Vert }_{L^2}^2+{\Vert \nabla \psi \Vert }_{L^2}^2\right)+{\Vert {\Lambda }^{\alpha +1}u\Vert }_{L^2}^2+{\Vert \Delta v\Vert }_{L^2}^2+{\Vert \Delta \psi \Vert }_{L^2}^2\\ +{\Vert {\left|u\right|}^{\frac{\beta -1}{2}}\nabla u\Vert }_{L^2}^2+{\Vert {\left|v\right|}^{\frac{\gamma -1}{2}}\nabla v\Vert }_{L^2}^2+{\Vert {\left|\psi \right|}^{\frac{\delta -1}{2}}\nabla \psi \Vert }_{L^2}^2+\frac{4\left(\beta -1\right)}{{\left(\beta +1\right)}^2}{\Vert \nabla {\left|u\right|}^{\frac{\beta +1}{2}}\Vert }_{L^2}^2\\ +\frac{4\left(\gamma -1\right)}{{\left(\gamma +1\right)}^2}{\Vert \nabla {\left|v\right|}^{\frac{\gamma +1}{2}}\Vert }_{L^2}^2+\frac{4\left(\delta -1\right)}{{\left(\delta +1\right)}^2}{\Vert \nabla {\left|\psi \right|}^{\frac{\delta +1}{2}}\Vert }_{L^2}^2\\ \le C\left(1+{\Vert u\Vert }_{L^{\beta +1}}^{\beta +1}+{\Vert v\Vert }_{L^{\gamma +1}}^{\gamma +1}\right)\left({\Vert \nabla u\Vert }_{L^2}^2+{\Vert \nabla v\Vert }_{L^2}^2+{\Vert \nabla \psi \Vert }_{L^2}^2\right).\end{array}$

在这里，我们使用了 $\beta \ge \text{4}$， $\gamma \ge \frac{\alpha +\text{3}}{\alpha }$ 去确保 $\frac{2\left(\beta +1\right)}{\beta -2}\le \beta +1$ 、(2.8)、(2.13)及(2.16)。根据Gronwall不等式和(2.1)，我们

$\begin{array}{l}{\Vert \nabla u\Vert }_{L^2}^2+{\Vert \nabla v\Vert }_{L^2}^2+{\Vert \nabla \psi \Vert }_{L^2}^2+{\displaystyle \underset{0}{\overset{t}{\int }}({\Vert {\Lambda }^{\alpha +1}u\Vert }_{L^2}^2+{\Vert \Delta v\Vert }_{L^2}^2+{\Vert \Delta \psi \Vert }_{L^2}^2+{\Vert {\left|u\right|}^{\frac{\beta -1}{2}}\nabla u\Vert }_{L^2}^2}+{\Vert {\left|v\right|}^{\frac{\gamma -1}{2}}\nabla v\Vert }_{L^2}^2+{\Vert {\left|\psi \right|}^{\frac{\delta -1}{2}}\nabla \psi \Vert }_{L^2}^2)\\ \le C\left(t,{\Vert u_0\Vert }_{H^1},{\Vert v_0\Vert }_{H^1},{\Vert {\psi }_0\Vert }_{H^1}\right)\end{array}$

这样我们就完成了强解存在的证明。

接下来，我们将证明定理1.1中构造的强解的唯一性。假设 $\left({\text{u}}_{\text{1}},{\text{v}}_{\text{1}},{\psi }_{\text{1}}\right)$ 和 $\left({\text{u}}_{\text{2}},{\text{v}}_{\text{2}},{\psi }_{\text{2}}\right)$ 是方程组(1.1)的两个解，它们的解 $\left({\text{u}}_{\text{0}},{\text{v}}_{\text{0}},{\psi }_{\text{0}}\right)$ 相同。我们定义 $\left(\text{<mover accent="true"> <mi> u </mi> <mo> ¯ </mo> </mover>},\text{<mover accent="true"> <mi> v </mi> <mo> ¯ </mo> </mover>},\bar{\psi }\right)=\left(u_1-u_2,v_1-v_2,{\psi }_1-{\psi }_2\right)$。然后我们可以得到

$\{\begin{array}{l}{\partial }_t\bar{u}-{\Lambda }^{2\alpha }\bar{u}+{\left|u_1\right|}^{\alpha -1}{u}_1-{\left|u_2\right|}^{\alpha -1}{u}_2+\nabla \bar{\pi }=-\left[\left(u_2\cdot \nabla \bar{u}+\bar{u}\cdot \nabla u_1\right)+\nabla \cdot \left(v_2\otimes \bar{v}\right)+\nabla \cdot \left(\bar{v}\otimes v_1\right)\right],\\ {\partial }_t\bar{v}=-\Delta \bar{v}+{\left|v_1\right|}^{\beta -1}{v}_1-{\left|v_2\right|}^{\beta -1}{v}_2=-\left(u_2\cdot \nabla \bar{v}+\bar{u}\cdot \nabla v_1\right)-\nabla \bar{\psi }-\left(v_2\cdot \nabla \bar{u}+\bar{v}\cdot \nabla u_1\right),\\ {\partial }_t\tilde{\psi }-\Delta \bar{\psi }+{\left|{\psi }_1\right|}^{\gamma -1}{\psi }_1-{\left|{\psi }_2\right|}^{\gamma -1}{\psi }_2=-\left(u_2\cdot \nabla \bar{\psi }+\bar{u}\cdot \nabla {\psi }_1\right)-\nabla \cdot \bar{v},\\ \nabla \cdot \bar{u}=0.\end{array}$

将(2.21)式分别与 $\left(\text{<mover accent="true"> <mi> u </mi> <mo> ¯ </mo> </mover>},\text{<mover accent="true"> <mi> v </mi> <mo> ¯ </mo> </mover>},\bar{\psi }\right)$ 做 $L^2$ 内积，将结果相加，我们得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \left(\bar{u},\bar{v},\bar{\psi }\right)\Vert }_{L^2}^2+{\Vert {\Lambda }^{\alpha }\bar{u}\Vert }_{L^2}^2+{\Vert \nabla \left(\bar{v},\bar{\psi }\right)\Vert }_{L^2}^2+{\displaystyle \underset{R^3}{\int }\left({\left|u_1\right|}^{\alpha -1}{u}_1-{\left|u_2\right|}^{\alpha -1}{u}_2\right)\cdot \bar{u}\text{d}x}\\ +{\displaystyle \underset{R^3}{\int }\left({\left|v_1\right|}^{\beta -1}{v}_1-{\left|v_2\right|}^{\beta -1}{v}_2\right)\cdot \bar{v}\text{d}x}+{\displaystyle \underset{R^3}{\int }\left({\left|{\psi }_1\right|}^{\gamma -1}{\psi }_1-{\left|{\psi }_2\right|}^{\gamma -1}{\psi }_2\right)\cdot \bar{\psi }\text{d}x}\\ \le {\displaystyle \underset{R^3}{\int }{\left|\bar{u}\right|}^2\left|\nabla u_1\right|\text{d}x}+{\displaystyle \underset{R^3}{\int }\left|\Delta \bar{u}\right|}\left(v_1,v_2\right)\left|\tilde{v}\right|\text{d}x+{\displaystyle \underset{R^3}{\int }\left|\nabla v_1\right|}\left|\bar{u}\right|\left|\bar{v}\right|\text{d}x+{\displaystyle \underset{R^3}{\int }{\left|\bar{v}\right|}^2\left|\nabla v_1\right|\text{d}x}+{\displaystyle \underset{R^3}{\int }\left|\nabla {\psi }_1\right|}\left|\bar{u}\right|\left|\bar{\psi }\right|\text{d}x\\ =K_1+K_2+\cdot \cdot \cdot +K_5.\end{array}$

应用Hölder不等式，当 $\beta \ge \text{1}$ 时，我们可以得到

$\begin{array}{l}{\displaystyle \underset{R^3}{\int }\left({\left|u_1\right|}^{\beta -1}{u}_1-{\left|u_2\right|}^{\beta -1}{u}_2\right)\cdot \bar{u}\text{d}x}\\ ={\displaystyle \underset{R^3}{\int }\left({\left|u_1\right|}^{\beta -1}{u}_1-{\left|u_2\right|}^{\beta -1}{u}_2\right)\cdot \left(u_1-u_2\right)\text{d}x}\\ ={\displaystyle \underset{R^3}{\int }{\left|u_1\right|}^{\beta +1}\text{d}x}-{\displaystyle \underset{R^3}{\int }{\left|u_1\right|}^{\beta -1}{u}_1{u}_2\text{d}x}-{\displaystyle \underset{R^3}{\int }{\left|u_2\right|}^{\beta -1}{u}_2{u}_1\text{d}x}+{\displaystyle \underset{R^3}{\int }{\left|u_2\right|}^{\beta +1}\text{d}x}\\ \ge {\Vert u_1\Vert }_{L^{\beta +1}}^{\beta +1}-{\Vert u_1\Vert }_{L^{\beta +1}}^{\beta }{\Vert u_2\Vert }_{L^{\beta +1}}-{\Vert u_2\Vert }_{L^{\beta +1}}^{\beta }{\Vert u_1\Vert }_{L^{\beta +1}}+{\Vert u_2\Vert }_{L^{\beta +1}}^{\beta +1}\\ =\left({\Vert u_1\Vert }_{L^{\beta +1}}^{\beta }-{\Vert u_2\Vert }_{L^{\beta +1}}^{\beta }\right)\left({\Vert u_1\Vert }_{L^{\beta +1}}-{\Vert u_2\Vert }_{L^{\beta +1}}\right)\ge 0.\end{array}$

同样，对于 $\gamma ,\delta \ge 1$，我们有

$\begin{array}{l}{\displaystyle \underset{R^3}{\int }\left({\left|v_1\right|}^{\beta -1}{v}_1-{\left|v_2\right|}^{\beta -1}{v}_2\right)\bar{v}\text{d}x}\ge 0,\\ {\displaystyle \underset{R^3}{\int }\left({\left|{\psi }_1\right|}^{\beta -1}{\psi }_1-{\left|{\psi }_2\right|}^{\beta -1}{\psi }_2\right)\bar{\psi }\text{d}x}\ge 0.\end{array}$

应用Hölder, Gagliardo-Nirenberg和Young不等式， $K_{\text{1}}$ 和 $K_{\text{4}}$ 可以估计如下

$\begin{array}{c}{K}_1={\displaystyle \underset{R^3}{\int }{\left|\bar{u}\right|}^2\left|\nabla u_1\right|\text{d}x}\\ \le C{\Vert \bar{u}\Vert }_{L^4}^2{\Vert \nabla u_1\Vert }_{L^2}\\ \le C{\Vert \bar{u}\Vert }_{L^2}^{2\left(1-\frac{3}{4\alpha }\right)}{\Vert {\Lambda }^{\alpha }\bar{u}\Vert }_{L^2}^{\frac{3}{2\alpha }}{\Vert \nabla u_1\Vert }_{L^2}\\ \le \frac{1}{4}{\Vert {\Lambda }^{\alpha }\bar{u}\Vert }_{L^2}^2+C{\Vert \bar{u}\Vert }_{L^2}^2{\Vert \nabla u_1\Vert }_{L^2}^{\frac{4\alpha }{4\alpha -3}}\\ \le \frac{1}{4}{\Vert {\Lambda }^{\alpha }\bar{u}\Vert }_{L^2}^2+C{\Vert \bar{u}\Vert }_{L^2}^2{\Vert \nabla u_1\Vert }_{L^2}^4\end{array}$

我们使用了Gagliardo-Nirenberg不等式：

${\Vert \bar{u}\Vert }_{L^4}\le C{\Vert \bar{u}\Vert }_{L^2}^{1-\frac{3}{4\alpha }}{\Vert {\Lambda }^{\alpha }\bar{u}\Vert }_{L^2}^{\frac{3}{4\alpha }}$

同样的， $K_{\text{4}}$ 也可以这样估计

$\begin{array}{c}{K}_4={\displaystyle \underset{R^3}{\int }{\left|\bar{v}\right|}^2\left|\nabla u_1\right|\text{d}x}\\ \le C{\Vert \bar{v}\Vert }_{L^4}^2{\Vert \nabla u_1\Vert }_{L^2}\\ \le C{\Vert \bar{v}\Vert }_{L^2}^{\frac{1}{2}}{\Vert \nabla \bar{v}\Vert }_{L^2}^{\frac{3}{2}}{\Vert \nabla u_1\Vert }_{L^2}^4\\ \le \frac{1}{4}{\Vert \nabla \bar{v}\Vert }_{L^2}^2+C{\Vert \bar{v}\Vert }_{L^2}^2{\Vert \nabla u_1\Vert }_{L^2}^4\end{array}$

接下来，我们将估计 $K_{\text{2}}$

$\begin{array}{c}{K}_2={\displaystyle \underset{R^3}{\int }\left|\nabla \bar{u}\right|}\left(v_1,v_2\right)\left|\bar{v}\right|\text{d}x\\ \le C{\Vert \nabla \bar{u}\Vert }_{L^{\frac{6}{5-2\alpha }}}{\Vert \left(v_1,v_2\right)\Vert }_{L^4}{\Vert \bar{v}\Vert }_{L^{\frac{12}{4\alpha -1}}}\\ \le C{\Vert {\Lambda }^{\alpha }\bar{u}\Vert }_{L^2}{\Vert \left(v_1,v_2\right)\Vert }_{L^4}{\Vert \tilde{v}\Vert }_{L^{\frac{12}{4\alpha -1}}}\\ \le \frac{1}{4}{\Vert {\Lambda }^{\alpha }\bar{u}\Vert }_{L^2}^2+C{\Vert \left(v_1,v_2\right)\Vert }_{L^4}^2{\Vert \bar{v}\Vert }_{L^{\frac{12}{4\alpha -1}}}^2\\ \le \frac{1}{4}{\Vert {\Lambda }^{\alpha }\bar{u}\Vert }_{L^2}^2+C{\Vert \left(v_1,v_2\right)\Vert }_{L^4}^{\frac{1}{2}}{\Vert \nabla \left(v_1,v_2\right)\Vert }_{L^2}^{\frac{3}{2}}{\Vert \bar{v}\Vert }_{L^2}^{2\left(1-{\theta }_1\right)}{\Vert \nabla \bar{v}\Vert }_{L^2}^{2{\theta }_1}\\ \le \frac{1}{4}{\Vert {\Lambda }^{\alpha }\bar{u}\Vert }_{L^2}^2+\frac{1}{4}{\Vert \nabla \bar{v}\Vert }_{L^2}^2+C{\Vert \left(v_1,v_2\right)\Vert }_{L^2}^{\frac{1}{2\left(1-{\theta }_1\right)}}{\Vert \nabla \left(v_1,v_2\right)\Vert }_{L^2}^{\frac{3}{2\left(1-{\theta }_1\right)}}{\Vert \bar{v}\Vert }_{L^2}^2.\end{array}$

这里我们使用了Gagliardo-Nirenberg不等式：

${\Vert \bar{v}\Vert }_{L^{\frac{12}{4\alpha -1}}}\le C{\Vert \bar{v}\Vert }_{L^2}^{1-{\theta }_1}{\Vert \nabla \bar{v}\Vert }_{L^2}^{{\theta }_1}$

通过直接计算，我们可以得到 ${\theta }_1=\frac{7-4\alpha }{4}$ 和 $\frac{1}{2\left(1-{\theta }_1\right)}\le 2$， $\frac{3}{2\left(1-{\theta }_1\right)}\le 6$。所以我们有

$K_2\le \frac{1}{4}{\Vert {\Lambda }^{\alpha }\bar{u}\Vert }_{L^2}^2+\frac{1}{4}{\Vert \nabla \bar{v}\Vert }_{L^2}^2+C{\Vert \left(v_1,v_2\right)\Vert }_{L^2}^2{\Vert \nabla \left(v_1,v_2\right)\Vert }_{L^2}^6{\Vert \bar{v}\Vert }_{L^2}^2$

涉及 $K_{\text{3}}$ 和 $K_{\text{5}}$ 的项可以估计如下

$\begin{array}{c}{K}_3+K_5={\displaystyle \underset{R^3}{\int }\left|\nabla \left(v_1,{\psi }_1\right)\right|}\left|\bar{u}\right|\left|\left(\bar{v},\bar{\psi }\right)\right|\text{d}x\\ \le C{\Vert \nabla \left(v_1,{\psi }_1\right)\Vert }_{L^2}{\Vert \bar{u}\Vert }_{L^4}{\Vert \left(\bar{v},\bar{\psi }\right)\Vert }_{L^4}\\ \le C{\Vert \nabla \left(v_1,{\psi }_1\right)\Vert }_{L^2}{\Vert \bar{u}\Vert }_{L^2}^{1-{\theta }_2}{\Vert {\Lambda }^{\alpha }\bar{u}\Vert }_{L^2}^{{\theta }_2}{\Vert \left(\bar{v},\bar{\psi }\right)\Vert }_{L^2}^{\frac{1}{4}}{\Vert \nabla \left(\bar{v},\bar{\psi }\right)\Vert }_{L^2}^{\frac{3}{4}}\\ \le \frac{1}{4}{\Vert {\Lambda }^{\alpha }\bar{u}\Vert }_{L^2}^2+C{\Vert \nabla \left(\bar{v},\bar{\psi }\right)\Vert }_{L^2}^{\frac{2}{2-{\theta }_2}}{\Vert \bar{u}\Vert }_{L^2}^{\frac{2\left(1-{\theta }_2\right)}{2-{\theta }_2}}{\Vert \left(\bar{v},\bar{\psi }\right)\Vert }_{L^2}^{\frac{1}{2}}{\Vert \nabla \left(v_1,{\psi }_1\right)\Vert }_{L^2}^{\frac{3}{2}}\\ \le \frac{1}{4}{\Vert {\Lambda }^{\alpha }\bar{u}\Vert }_{L^2}^2+\frac{1}{4}{\Vert \nabla \left(\bar{v},\bar{\psi }\right)\Vert }_{L^2}^2+C{\Vert \bar{u}\Vert }_{L^2}^{\frac{8\left(1-{\theta }_2\right)}{5-4{\theta }_2}}{\Vert \left(\bar{v},\bar{\psi }\right)\Vert }_{L^2}^{\frac{2}{5-4{\theta }_2}}{\Vert \nabla \left(v_1,{\psi }_1\right)\Vert }_{L^2}^{\frac{8}{5-4{\theta }_2}}\end{array}$

通过直接计算，我们可以得到 ${\theta }_{\text{2}}=\frac{\text{3}}{\text{4}\alpha }$ 和 $\frac{8\left(1-{\theta }_2\right)}{5-4{\theta }_2}\le 2$， $\frac{8}{5-4{\theta }_2}\le 4$， $\frac{2}{5-4{\theta }_2}\le 2$。所以我们有

$K_3+K_5\le \frac{1}{4}{\Vert {\Lambda }^{\alpha }\bar{u}\Vert }_{L^2}^2+\frac{1}{4}{\Vert \nabla \left(\bar{v},\bar{\psi }\right)\Vert }_{L^2}^2+C{\Vert \left(\bar{u},\bar{v},\bar{\psi }\right)\Vert }_{L^2}^2{\Vert \nabla \left(v_1,{\psi }_1\right)\Vert }_{L^2}^4.$

将(2.24)~(2.31)估计应用到(2.22)中，可以得到

$\frac{\text{d}}{\text{d}t}{\Vert \left(\bar{u},\bar{v},\bar{\psi }\right)\Vert }_{L^2}^2\le C\left({\Vert \left(v_1,v_2\right)\Vert }_{L^2}^2{\Vert \nabla \left(v_1,v_2\right)\Vert }_{L^2}^6+{\Vert \nabla \left(u_1,v_1,{\psi }_1\right)\Vert }_{L^2}^4\right){\Vert \left(\bar{u},\bar{v},\bar{\psi }\right)\Vert }_{L^2}^2$

应用Grönwall不等式，通过 $H^1$ -估计，可以得到

${\Vert \left(\bar{u},\bar{v},\bar{\psi }\right)\Vert }_{L^2}^2=0$

这样完成了定理1.1唯一性部分的证明。