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

doi:10.12677/AAM.2022.117517

Advances in Applied Mathematics
Vol. 11 No. 07 ( 2022 ), Article ID: 54129 , 9 pages
10.12677/AAM.2022.117517

带阻尼的三维热带气候模型的适定性研究

陈贤

●How to Cite this Article

浙江师范大学，浙江金华

收稿日期：2022年6月25日；录用日期：2022年7月20日；发布日期：2022年7月27日

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 $1 \leq α \leq \frac{5}{2}$ ，如果 $β$ ， $γ$ ， $δ$ 满足 $β \geq 4$ ， $γ \geq \frac{α + 3}{α}$ ， $δ \geq 1$ 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。

关键词

热带气候模型，适定性，阻尼性，强解

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

Xian Chen

Zhejiang Normal University, Jinhua Zhejiang

Received: Jun. 25^th, 2022; accepted: Jul. 20^th, 2022; published: Jul. 27^th, 2022

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 $1 \leq α \leq \frac{5}{2}$ , if $β$ , $γ$ , $δ$ satisfies $β \geq 4$ , $γ \geq \frac{α + 3}{α}$ , $δ \geq 1$ , then the existence and uniqueness of a damped 3-D tropical climate model is strong.

Keywords:Tropical Climate Model, Well-Posedness, Damping, Strong Solution

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

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：

${\begin{cases} u_{t} + Λ^{2 α} u + (u \cdot \nabla) u + {| u |}^{β - 1} u + \nabla π + d i v (v \otimes v) = 0, \\ v_{t} - Δ v + (u \cdot \nabla) v + {| v |}^{γ - 1} v + \nabla ψ + (v \cdot \nabla) u = 0, \\ ψ_{t} - Δ ψ + (u \cdot \nabla) ψ + {| ψ |}^{δ - 1} ψ + d i v v = 0, \\ d i v u = 0, \\ (u, v, ψ) (x, 0) = (u_{0}, v_{0}, ψ_{0}) \end{cases}$

其中向量场 $u = u (x, t) \in R^{3}$ 和 $v = v (x, t) \in R^{3}$ 分别表示速度的正压模式和第一斜压模式。 $π = π (x, t)$ 和 $ψ = ψ (x, t)$ 分别表示压强和温度。分数拉普拉斯算子 $Λ = {(- Δ)}^{\frac{1}{2}}$ ， $α > 0$ 和 $β, γ, δ \geq 0$ 为实参数。

需要指出的是，当 $ψ = 0$ ， $α = 1$ 和 $d i v v = 0$ 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 $β \geq \frac{7}{2}$ 情况下，强解是唯一的，在 $\frac{7}{2} \leq β \leq 5$ 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 $β \geq 3$ 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 $(u_{0} (x), v_{0} (x), ψ_{0} (x)) \in H^{1} (R^{3})$ 满足 $\nabla \cdot u = 0$ 。对于 $1 \leq α \leq \frac{5}{2}$ ，如果 $β, γ, δ$ 满足

$β \geq 4, γ \geq \frac{α + 3}{α}, δ \geq 1.$

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，

$\begin{array}{l} u \in L^{\infty} (0, T; H^{1} (R^{3})) \cap L^{2} (0, T; H^{α + 1} (R^{3})) \cap L^{β + 1} (0, T; H^{β + 1} (R^{3})) \\ v \in L^{\infty} (0, T; H^{1} (R^{3})) \cap L^{2} (0, T; H^{2} (R^{3})) \cap L^{γ + 1} (0, T; H^{γ + 1} (R^{3})) \\ ψ \in L^{\infty} (0, T; H^{1} (R^{3})) \cap L^{2} (0, T; H^{2} (R^{3})) \cap L^{δ + 1} (0, T; H^{δ + 1} (R^{3})) \end{array}$

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 $(u, v, ψ)$ ，经过分部积分，并使用 $\nabla \cdot u = 0$ ，得到以下能量估计

$\begin{array}{l} {‖ u (t) ‖}_{L^{2}}^{2} + {‖ v (t) ‖}_{L^{2}}^{2} + {‖ ψ (t) ‖}_{L^{2}}^{2} + 2 \int_{0}^{T} ({‖ Λ^{α} u (t) ‖}_{L^{2}}^{2} + {‖ \nabla v (t) ‖}_{L^{2}}^{2} + {‖ \nabla ψ (t) ‖}_{L^{2}}^{2} \\ + {‖ u ‖}_{L^{β + 1}}^{β + 1} + {‖ v ‖}_{L^{γ + 1}}^{γ + 1} + {‖ ψ ‖}_{L^{δ + 1}}^{δ + 1}) d s = {‖ u_{0} ‖}_{L^{2}}^{2} + {‖ v_{0} ‖}_{L^{2}}^{2} + {‖ ψ_{0} ‖}_{L^{2}}^{2} . \end{array}$

我们在哪里使用了下面的事实

$\begin{array}{l} \int_{R^{3}} \nabla ψ \cdot v d x + \int_{R^{3}} (\nabla \cdot v) ψ d x = 0, \\ \int_{R^{3}} d i v (v \otimes v) \cdot u d x + \int_{R^{3}} (v \cdot \nabla) u \cdot v d x = 0 \end{array}$

将(1.1)分别乘以 $- Δ u$ 、 $- Δ v$ 和 $- Δ ψ$ ，在 $R^{3}$ 中积分后相加，得到

$\begin{array}{l} \frac{1}{2} \frac{d}{d t} ({‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla v ‖}_{L^{2}}^{2} + {‖ \nabla ψ ‖}_{L^{2}}^{2}) + {‖ Λ^{α + 1} u ‖}_{L^{2}}^{2} + {‖ Δ v ‖}_{L^{2}}^{2} + {‖ Δ ψ ‖}_{L^{2}}^{2} \\ + {‖ {| u |}^{\frac{β - 1}{2}} \nabla u ‖}_{L^{2}}^{2} + {‖ {| v |}^{\frac{γ - 1}{2}} \nabla v ‖}_{L^{2}}^{2} + {‖ {| ψ |}^{\frac{δ - 1}{2}} \nabla ψ ‖}_{L^{2}}^{2} + \frac{4 (β - 1)}{{(β + 1)}^{2}} {‖ \nabla {| u |}^{\frac{β + 1}{2}} ‖}_{L^{2}}^{2} \\ + \frac{4 (γ - 1)}{{(γ + 1)}^{2}} {‖ \nabla {| v |}^{\frac{γ + 1}{2}} ‖}_{L^{2}}^{2} + \frac{4 (δ - 1)}{{(δ + 1)}^{2}} {‖ \nabla {| ψ |}^{\frac{δ + 1}{2}} ‖}_{L^{2}}^{2} \\ = \int_{R^{3}} (u \cdot \nabla) u \cdot Δ u d x + \int_{R^{3}} d i v (v \otimes v) \cdot Δ u d x + \int_{R^{3}} (u \cdot \nabla) v \cdot Δ v d x \\ + \int_{R^{3}} (v \cdot \nabla) u \cdot Δ v d x + \int_{R^{3}} \nabla ψ \cdot Δ v d x + \int_{R^{3}} (u \cdot \nabla) ψ \cdot Δ ψ d x + \int_{R^{3}} (\nabla \cdot v) \cdot Δ ψ d x \\ = I_{1} + I_{2} + I_{3} + I_{4} + I_{5} + I_{6} + I_{7}, \end{array}$

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。

$\begin{matrix} I_{1} = \int_{R^{3}} (u \cdot \nabla) u \cdot Δ u d x \\ = \int_{R^{3}} Λ^{1 - α} (u \cdot \nabla u) \cdot Λ^{α + 1} u d x \\ = \int_{R^{3}} Λ^{2 - α} (u \otimes u) \cdot Λ^{α + 1} u d x \\ \leq {‖ Λ^{α + 1} u ‖}_{L^{2}} {‖ Λ^{2 - α} (u \otimes u) ‖}_{L^{2}} \end{matrix}$

$\begin{matrix} \leq {‖ Λ^{α + 1} u ‖}_{L^{2}} {‖ Λ^{2 - α} u ‖}_{L^{\frac{2 (β + 1)}{β - 1}}} {‖ u ‖}_{L^{β + 1}} \\ \leq {‖ Λ^{α + 1} u ‖}_{L^{2}}^{1 + θ} {‖ \nabla u ‖}_{L^{2}}^{1 - θ} {‖ u ‖}_{L^{β + 1}} \\ \leq \frac{1}{2} {‖ Λ^{α + 1} u ‖}_{L^{2}}^{2} + C {‖ u ‖}_{L^{β + 1}}^{\frac{2}{1 - θ}} {‖ \nabla u ‖}_{L^{2}}^{2} \end{matrix}$

这里我们使用了以下的Gagliardo-Nirenberg不等式：

${‖ Λ^{2 - α} u ‖}_{L^{\frac{2 (β + 1)}{β - 1}}} \leq C {‖ Λ^{α + 1} u ‖}_{L^{2}}^{θ} {‖ \nabla u ‖}_{L^{2}}^{1 - θ}$

在这里

$\frac{β - 1}{2 (β + 1)} = \frac{1 - α}{3} + (\frac{1}{2} - \frac{α}{3}) θ + \frac{1 - θ}{2}$

通过直接计算，得到

$θ = \frac{(5 - 2 α) (β + 1) - 3 (β - 1)}{2 α (β + 1)}$

如果下列限制成立

$\frac{2}{1 - θ} \leq β + 1$

然后我们得到 $β \geq \frac{4}{2 α - 1}$

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：

$\begin{array}{l} I_{2} + I_{3} + I_{4} \\ = \int_{R^{3}} \nabla \cdot (v \otimes v) \cdot Δ u d x + \int_{R^{3}} (u \cdot \nabla) v \cdot Δ v d x + \int_{R^{3}} (v \cdot \nabla) u \cdot Δ v d x \\ = \int_{R^{3}} (v_{j} \partial_{i} v_{i} + v_{i} \partial_{i} v_{j}) \partial_{k} \partial_{k} u_{j} d x + \int_{R^{3}} u_{i} \partial_{i} v_{j} \partial_{k} \partial_{k} v_{j} d x + \int_{R^{3}} v_{i} \partial_{i} u_{j} \partial_{k} \partial_{k} v_{j} d x \\ = \int_{R^{3}} (v_{j} \partial_{i} v_{i} + v_{i} \partial_{i} v_{j}) \partial_{k} \partial_{k} u_{j} d x + \int_{R^{3}} (\partial_{k} \partial_{k} u_{i} \partial_{i} v_{j} + \partial_{k} u_{i} \partial_{k} \partial_{i} v_{j}) d x + \int_{R^{3}} v_{i} \partial_{i} u_{j} \partial_{k} \partial_{k} v_{j} d x \\ \leq \int_{R^{3}} | v | | \nabla v | | \nabla^{2} u | d x + \int_{R^{3}} | v | | \nabla u | | \nabla^{2} v | d x \\ = J_{1} + J_{2} \end{array}$

接下来，我们将估计 $J_{1}$ 和 $J_{2}$ 。

$\begin{matrix} J_{1} = \int_{R^{3}} | v | | \nabla v | | \nabla^{2} u | d x \\ = \int_{R^{3}} {| v |}^{\frac{γ - 1}{2}} | \nabla v | {| v |}^{\frac{3 - γ}{2}} | Δ u | d x \\ \leq {‖ {| v |}^{\frac{γ - 1}{2}} \nabla v ‖}_{L^{2}} {‖ v ‖}_{L^{γ + 1}}^{\frac{3 - γ}{2}} {‖ Δ u ‖}_{L^{\frac{γ + 1}{γ - 1}}} \\ \leq {‖ {| v |}^{\frac{γ - 1}{2}} \nabla v ‖}_{L^{2}} {‖ v ‖}_{L^{γ + 1}}^{\frac{3 - γ}{2}} {‖ \nabla u ‖}_{L^{2}}^{1 - λ_{1}} {‖ Λ^{1 + α} u ‖}_{L^{2}}^{λ_{1}} \\ \leq \frac{1}{2} {‖ {| v |}^{\frac{γ - 1}{2}} \nabla v ‖}_{L^{2}}^{2} + \frac{1}{2} {‖ Λ^{1 + α} u ‖}_{L^{2}}^{2} + C {‖ v ‖}_{L^{γ + 1}}^{\frac{3 - γ}{1 - λ_{1}}} {‖ \nabla u ‖}_{L^{2}}^{2} \end{matrix}$

这里我们使用了Gagliardo-Nirenberg不等式：

${‖ Δ u ‖}_{L^{\frac{γ + 1}{γ - 1}}} \leq C {‖ Λ^{α + 1} u ‖}_{L^{2}}^{λ_{1}} {‖ \nabla u ‖}_{L^{2}}^{1 - λ_{1}}$

在这里

$\frac{γ - 1}{γ + 1} = \frac{1}{3} + (\frac{1}{2} - \frac{α}{3}) λ_{1} + \frac{1 - λ_{1}}{2}, λ_{1} \in [\frac{1}{α}, 1]$

从(2.12)我们可以直接计算出 $λ_{1} = \frac{11 - γ}{2 α (γ + 1)}$ ， $γ \geq \frac{11 - 2 α}{2 α + 1}$ 和 $α \geq 1$ 。

如果下列限制成立

$\frac{3 - γ}{1 - λ_{1}} \leq γ + 1$

然后，我们可以得到 $γ \geq \frac{4 α + 11}{4 α + 1}$ 。

同理， $J_{2}$ 可以估算如下：

$\begin{matrix} J_{2} = \int_{R^{3}} | v | | \nabla u | | \nabla^{2} v | d x \\ \leq {‖ v ‖}_{L^{γ + 1}} {‖ \nabla u ‖}_{L^{\frac{2 γ + 1}{γ - 1}}} {‖ Δ v ‖}_{L^{2}} \\ \leq C {‖ v ‖}_{L^{γ + 1}}^{2} {‖ \nabla u ‖}_{L^{\frac{2 γ + 1}{γ - 1}}}^{2} + \frac{1}{4} {‖ Δ u ‖}_{L^{2}}^{2} \\ \leq C {‖ v ‖}_{L^{γ + 1}}^{2} {‖ \nabla u ‖}_{L^{2}}^{2 (1 - λ_{2})} {‖ Λ^{1 + α} u ‖}_{L^{2}}^{2 λ_{2}} + \frac{1}{4} {‖ Δ v ‖}_{L^{2}}^{2} \\ \leq C {‖ v ‖}_{L^{γ + 1}}^{\frac{2}{1 - λ_{2}}} {‖ \nabla u ‖}_{L^{2}}^{2} + \frac{1}{4} {‖ Λ^{1 + α} u ‖}_{L^{2}}^{2} + \frac{1}{4} {‖ Δ v ‖}_{L^{2}}^{2} \end{matrix}$

这里我们使用了Gagliardo-Nirenberg不等式：

${‖ \nabla u ‖}_{L^{\frac{2 (γ + 1)}{γ - 1}}} \leq C {‖ Λ^{α + 1} u ‖}_{L^{2}}^{λ_{2}} {‖ \nabla u ‖}_{L^{2}}^{1 - λ_{2}} .$

通过直接计算，我们有 $λ_{2} = \frac{3}{α (γ + 1)}$ 。如果下列限制成立

$\frac{2}{1 - λ_{2}} \leq γ + 1$

然后，我们可以得到 $γ \geq \frac{α + 3}{α}$ 。对于 $I_{5}$ 和 $I_{7}$ ，我们可以得到

$\begin{matrix} I_{5} + I_{7} = \int_{R^{3}} \nabla ψ \cdot Δ v d x + \int_{R^{3}} (\nabla \cdot v) Δ ψ d x \\ \leq \int_{R^{3}} | \nabla ψ | | Δ v | d x \\ \leq \frac{1}{4} {‖ Δ v ‖}_{L^{2}}^{2} + C {‖ \nabla ψ ‖}_{L^{2}}^{2} . \end{matrix}$

应用soblove不等式， $I_{6}$ 可以估计如下

$\begin{matrix} I_{6} = \int_{R^{3}} (u \cdot \nabla) ψ Δ ψ d x \\ \leq {‖ u ‖}_{L^{β + 1}} {‖ \nabla ψ ‖}_{L^{\frac{2 β + 1}{β - 1}}} {‖ Δ ψ ‖}_{L^{2}} \\ \leq \frac{1}{4} {‖ Δ ψ ‖}_{L^{2}}^{2} + C {‖ u ‖}_{L^{β + 1}}^{\frac{2 (β + 1)}{β - 2}} {‖ \nabla ψ ‖}_{L^{2}}^{2} . \end{matrix}$

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

$\begin{array}{l} \frac{d}{d t} ({‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla v ‖}_{L^{2}}^{2} + {‖ \nabla ψ ‖}_{L^{2}}^{2}) + {‖ Λ^{α + 1} u ‖}_{L^{2}}^{2} + {‖ Δ v ‖}_{L^{2}}^{2} + {‖ Δ ψ ‖}_{L^{2}}^{2} \\ + {‖ {| u |}^{\frac{β - 1}{2}} \nabla u ‖}_{L^{2}}^{2} + {‖ {| v |}^{\frac{γ - 1}{2}} \nabla v ‖}_{L^{2}}^{2} + {‖ {| ψ |}^{\frac{δ - 1}{2}} \nabla ψ ‖}_{L^{2}}^{2} + \frac{4 (β - 1)}{{(β + 1)}^{2}} {‖ \nabla {| u |}^{\frac{β + 1}{2}} ‖}_{L^{2}}^{2} \\ + \frac{4 (γ - 1)}{{(γ + 1)}^{2}} {‖ \nabla {| v |}^{\frac{γ + 1}{2}} ‖}_{L^{2}}^{2} + \frac{4 (δ - 1)}{{(δ + 1)}^{2}} {‖ \nabla {| ψ |}^{\frac{δ + 1}{2}} ‖}_{L^{2}}^{2} \\ \leq C (1 + {‖ u ‖}_{L^{β + 1}}^{β + 1} + {‖ v ‖}_{L^{γ + 1}}^{γ + 1}) ({‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla v ‖}_{L^{2}}^{2} + {‖ \nabla ψ ‖}_{L^{2}}^{2}) . \end{array}$

在这里，我们使用了 $β \geq 4$ ， $γ \geq \frac{α + 3}{α}$ 去确保 $\frac{2 (β + 1)}{β - 2} \leq β + 1$ 、(2.8)、(2.13)及(2.16)。根据Gronwall不等式和(2.1)，我们

$\begin{array}{l} {‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla v ‖}_{L^{2}}^{2} + {‖ \nabla ψ ‖}_{L^{2}}^{2} + \int_{0}^{t} ({‖ Λ^{α + 1} u ‖}_{L^{2}}^{2} + {‖ Δ v ‖}_{L^{2}}^{2} + {‖ Δ ψ ‖}_{L^{2}}^{2} + {‖ {| u |}^{\frac{β - 1}{2}} \nabla u ‖}_{L^{2}}^{2} + {‖ {| v |}^{\frac{γ - 1}{2}} \nabla v ‖}_{L^{2}}^{2} + {‖ {| ψ |}^{\frac{δ - 1}{2}} \nabla ψ ‖}_{L^{2}}^{2}) \\ \leq C (t, {‖ u_{0} ‖}_{H^{1}}, {‖ v_{0} ‖}_{H^{1}}, {‖ ψ_{0} ‖}_{H^{1}}) \end{array}$

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 $(u_{1}, v_{1}, ψ_{1})$ 和 $(u_{2}, v_{2}, ψ_{2})$ 是方程组(1.1)的两个解，它们的解 $(u_{0}, v_{0}, ψ_{0})$ 相同。我们定义 $(\bar{u}, \bar{v}, \bar{ψ}) = (u_{1} - u_{2}, v_{1} - v_{2}, ψ_{1} - ψ_{2})$ 。然后我们可以得到

${\begin{cases} \partial_{t} \bar{u} - Λ^{2 α} \bar{u} + {| u_{1} |}^{α - 1} u_{1} - {| u_{2} |}^{α - 1} u_{2} + \nabla \bar{π} = - [(u_{2} \cdot \nabla \bar{u} + \bar{u} \cdot \nabla u_{1}) + \nabla \cdot (v_{2} \otimes \bar{v}) + \nabla \cdot (\bar{v} \otimes v_{1})], \\ \partial_{t} \bar{v} = - Δ \bar{v} + {| v_{1} |}^{β - 1} v_{1} - {| v_{2} |}^{β - 1} v_{2} = - (u_{2} \cdot \nabla \bar{v} + \bar{u} \cdot \nabla v_{1}) - \nabla \bar{ψ} - (v_{2} \cdot \nabla \bar{u} + \bar{v} \cdot \nabla u_{1}), \\ \partial_{t} \tilde{ψ} - Δ \bar{ψ} + {| ψ_{1} |}^{γ - 1} ψ_{1} - {| ψ_{2} |}^{γ - 1} ψ_{2} = - (u_{2} \cdot \nabla \bar{ψ} + \bar{u} \cdot \nabla ψ_{1}) - \nabla \cdot \bar{v}, \\ \nabla \cdot \bar{u} = 0. \end{cases}$

将(2.21)式分别与 $(\bar{u}, \bar{v}, \bar{ψ})$ 做 $L^{2}$ 内积，将结果相加，我们得到

$\begin{array}{l} \frac{1}{2} \frac{d}{d t} {‖ (\bar{u}, \bar{v}, \bar{ψ}) ‖}_{L^{2}}^{2} + {‖ Λ^{α} \bar{u} ‖}_{L^{2}}^{2} + {‖ \nabla (\bar{v}, \bar{ψ}) ‖}_{L^{2}}^{2} + \int_{R^{3}} ({| u_{1} |}^{α - 1} u_{1} - {| u_{2} |}^{α - 1} u_{2}) \cdot \bar{u} d x \\ + \int_{R^{3}} ({| v_{1} |}^{β - 1} v_{1} - {| v_{2} |}^{β - 1} v_{2}) \cdot \bar{v} d x + \int_{R^{3}} ({| ψ_{1} |}^{γ - 1} ψ_{1} - {| ψ_{2} |}^{γ - 1} ψ_{2}) \cdot \bar{ψ} d x \\ \leq \int_{R^{3}} {| \bar{u} |}^{2} | \nabla u_{1} | d x + \int_{R^{3}} | Δ \bar{u} | (v_{1}, v_{2}) | \tilde{v} | d x + \int_{R^{3}} | \nabla v_{1} | | \bar{u} | | \bar{v} | d x + \int_{R^{3}} {| \bar{v} |}^{2} | \nabla v_{1} | d x + \int_{R^{3}} | \nabla ψ_{1} | | \bar{u} | | \bar{ψ} | d x \\ = K_{1} + K_{2} + \cdot \cdot \cdot + K_{5} . \end{array}$

应用Hölder不等式，当 $β \geq 1$ 时，我们可以得到

$\begin{array}{l} \int_{R^{3}} ({| u_{1} |}^{β - 1} u_{1} - {| u_{2} |}^{β - 1} u_{2}) \cdot \bar{u} d x \\ = \int_{R^{3}} ({| u_{1} |}^{β - 1} u_{1} - {| u_{2} |}^{β - 1} u_{2}) \cdot (u_{1} - u_{2}) d x \\ = \int_{R^{3}} {| u_{1} |}^{β + 1} d x - \int_{R^{3}} {| u_{1} |}^{β - 1} u_{1} u_{2} d x - \int_{R^{3}} {| u_{2} |}^{β - 1} u_{2} u_{1} d x + \int_{R^{3}} {| u_{2} |}^{β + 1} d x \\ \geq {‖ u_{1} ‖}_{L^{β + 1}}^{β + 1} - {‖ u_{1} ‖}_{L^{β + 1}}^{β} {‖ u_{2} ‖}_{L^{β + 1}} - {‖ u_{2} ‖}_{L^{β + 1}}^{β} {‖ u_{1} ‖}_{L^{β + 1}} + {‖ u_{2} ‖}_{L^{β + 1}}^{β + 1} \\ = ({‖ u_{1} ‖}_{L^{β + 1}}^{β} - {‖ u_{2} ‖}_{L^{β + 1}}^{β}) ({‖ u_{1} ‖}_{L^{β + 1}} - {‖ u_{2} ‖}_{L^{β + 1}}) \geq 0. \end{array}$

同样，对于 $γ, δ \geq 1$ ，我们有

$\begin{array}{l} \int_{R^{3}} ({| v_{1} |}^{β - 1} v_{1} - {| v_{2} |}^{β - 1} v_{2}) \bar{v} d x \geq 0, \\ \int_{R^{3}} ({| ψ_{1} |}^{β - 1} ψ_{1} - {| ψ_{2} |}^{β - 1} ψ_{2}) \bar{ψ} d x \geq 0. \end{array}$

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

$\begin{matrix} K_{1} = \int_{R^{3}} {| \bar{u} |}^{2} | \nabla u_{1} | d x \\ \leq C {‖ \bar{u} ‖}_{L^{4}}^{2} {‖ \nabla u_{1} ‖}_{L^{2}} \\ \leq C {‖ \bar{u} ‖}_{L^{2}}^{2 (1 - \frac{3}{4 α})} {‖ Λ^{α} \bar{u} ‖}_{L^{2}}^{\frac{3}{2 α}} {‖ \nabla u_{1} ‖}_{L^{2}} \\ \leq \frac{1}{4} {‖ Λ^{α} \bar{u} ‖}_{L^{2}}^{2} + C {‖ \bar{u} ‖}_{L^{2}}^{2} {‖ \nabla u_{1} ‖}_{L^{2}}^{\frac{4 α}{4 α - 3}} \\ \leq \frac{1}{4} {‖ Λ^{α} \bar{u} ‖}_{L^{2}}^{2} + C {‖ \bar{u} ‖}_{L^{2}}^{2} {‖ \nabla u_{1} ‖}_{L^{2}}^{4} \end{matrix}$

我们使用了Gagliardo-Nirenberg不等式：

${‖ \bar{u} ‖}_{L^{4}} \leq C {‖ \bar{u} ‖}_{L^{2}}^{1 - \frac{3}{4 α}} {‖ Λ^{α} \bar{u} ‖}_{L^{2}}^{\frac{3}{4 α}}$

同样的， $K_{4}$ 也可以这样估计

$\begin{matrix} K_{4} = \int_{R^{3}} {| \bar{v} |}^{2} | \nabla u_{1} | d x \\ \leq C {‖ \bar{v} ‖}_{L^{4}}^{2} {‖ \nabla u_{1} ‖}_{L^{2}} \\ \leq C {‖ \bar{v} ‖}_{L^{2}}^{\frac{1}{2}} {‖ \nabla \bar{v} ‖}_{L^{2}}^{\frac{3}{2}} {‖ \nabla u_{1} ‖}_{L^{2}}^{4} \\ \leq \frac{1}{4} {‖ \nabla \bar{v} ‖}_{L^{2}}^{2} + C {‖ \bar{v} ‖}_{L^{2}}^{2} {‖ \nabla u_{1} ‖}_{L^{2}}^{4} \end{matrix}$

接下来，我们将估计 $K_{2}$

$\begin{matrix} K_{2} = \int_{R^{3}} | \nabla \bar{u} | (v_{1}, v_{2}) | \bar{v} | d x \\ \leq C {‖ \nabla \bar{u} ‖}_{L^{\frac{6}{5 - 2 α}}} {‖ (v_{1}, v_{2}) ‖}_{L^{4}} {‖ \bar{v} ‖}_{L^{\frac{12}{4 α - 1}}} \\ \leq C {‖ Λ^{α} \bar{u} ‖}_{L^{2}} {‖ (v_{1}, v_{2}) ‖}_{L^{4}} {‖ \tilde{v} ‖}_{L^{\frac{12}{4 α - 1}}} \\ \leq \frac{1}{4} {‖ Λ^{α} \bar{u} ‖}_{L^{2}}^{2} + C {‖ (v_{1}, v_{2}) ‖}_{L^{4}}^{2} {‖ \bar{v} ‖}_{L^{\frac{12}{4 α - 1}}}^{2} \\ \leq \frac{1}{4} {‖ Λ^{α} \bar{u} ‖}_{L^{2}}^{2} + C {‖ (v_{1}, v_{2}) ‖}_{L^{4}}^{\frac{1}{2}} {‖ \nabla (v_{1}, v_{2}) ‖}_{L^{2}}^{\frac{3}{2}} {‖ \bar{v} ‖}_{L^{2}}^{2 (1 - θ_{1})} {‖ \nabla \bar{v} ‖}_{L^{2}}^{2 θ_{1}} \\ \leq \frac{1}{4} {‖ Λ^{α} \bar{u} ‖}_{L^{2}}^{2} + \frac{1}{4} {‖ \nabla \bar{v} ‖}_{L^{2}}^{2} + C {‖ (v_{1}, v_{2}) ‖}_{L^{2}}^{\frac{1}{2 (1 - θ_{1})}} {‖ \nabla (v_{1}, v_{2}) ‖}_{L^{2}}^{\frac{3}{2 (1 - θ_{1})}} {‖ \bar{v} ‖}_{L^{2}}^{2} . \end{matrix}$

这里我们使用了Gagliardo-Nirenberg不等式：

${‖ \bar{v} ‖}_{L^{\frac{12}{4 α - 1}}} \leq C {‖ \bar{v} ‖}_{L^{2}}^{1 - θ_{1}} {‖ \nabla \bar{v} ‖}_{L^{2}}^{θ_{1}}$

通过直接计算，我们可以得到 $θ_{1} = \frac{7 - 4 α}{4}$ 和 $\frac{1}{2 (1 - θ_{1})} \leq 2$ ， $\frac{3}{2 (1 - θ_{1})} \leq 6$ 。所以我们有

$K_{2} \leq \frac{1}{4} {‖ Λ^{α} \bar{u} ‖}_{L^{2}}^{2} + \frac{1}{4} {‖ \nabla \bar{v} ‖}_{L^{2}}^{2} + C {‖ (v_{1}, v_{2}) ‖}_{L^{2}}^{2} {‖ \nabla (v_{1}, v_{2}) ‖}_{L^{2}}^{6} {‖ \bar{v} ‖}_{L^{2}}^{2}$

涉及 $K_{3}$ 和 $K_{5}$ 的项可以估计如下

$\begin{matrix} K_{3} + K_{5} = \int_{R^{3}} | \nabla (v_{1}, ψ_{1}) | | \bar{u} | | (\bar{v}, \bar{ψ}) | d x \\ \leq C {‖ \nabla (v_{1}, ψ_{1}) ‖}_{L^{2}} {‖ \bar{u} ‖}_{L^{4}} {‖ (\bar{v}, \bar{ψ}) ‖}_{L^{4}} \\ \leq C {‖ \nabla (v_{1}, ψ_{1}) ‖}_{L^{2}} {‖ \bar{u} ‖}_{L^{2}}^{1 - θ_{2}} {‖ Λ^{α} \bar{u} ‖}_{L^{2}}^{θ_{2}} {‖ (\bar{v}, \bar{ψ}) ‖}_{L^{2}}^{\frac{1}{4}} {‖ \nabla (\bar{v}, \bar{ψ}) ‖}_{L^{2}}^{\frac{3}{4}} \\ \leq \frac{1}{4} {‖ Λ^{α} \bar{u} ‖}_{L^{2}}^{2} + C {‖ \nabla (\bar{v}, \bar{ψ}) ‖}_{L^{2}}^{\frac{2}{2 - θ_{2}}} {‖ \bar{u} ‖}_{L^{2}}^{\frac{2 (1 - θ_{2})}{2 - θ_{2}}} {‖ (\bar{v}, \bar{ψ}) ‖}_{L^{2}}^{\frac{1}{2}} {‖ \nabla (v_{1}, ψ_{1}) ‖}_{L^{2}}^{\frac{3}{2}} \\ \leq \frac{1}{4} {‖ Λ^{α} \bar{u} ‖}_{L^{2}}^{2} + \frac{1}{4} {‖ \nabla (\bar{v}, \bar{ψ}) ‖}_{L^{2}}^{2} + C {‖ \bar{u} ‖}_{L^{2}}^{\frac{8 (1 - θ_{2})}{5 - 4 θ_{2}}} {‖ (\bar{v}, \bar{ψ}) ‖}_{L^{2}}^{\frac{2}{5 - 4 θ_{2}}} {‖ \nabla (v_{1}, ψ_{1}) ‖}_{L^{2}}^{\frac{8}{5 - 4 θ_{2}}} \end{matrix}$

通过直接计算，我们可以得到 $θ_{2} = \frac{3}{4 α}$ 和 $\frac{8 (1 - θ_{2})}{5 - 4 θ_{2}} \leq 2$ ， $\frac{8}{5 - 4 θ_{2}} \leq 4$ ， $\frac{2}{5 - 4 θ_{2}} \leq 2$ 。所以我们有

$K_{3} + K_{5} \leq \frac{1}{4} {‖ Λ^{α} \bar{u} ‖}_{L^{2}}^{2} + \frac{1}{4} {‖ \nabla (\bar{v}, \bar{ψ}) ‖}_{L^{2}}^{2} + C {‖ (\bar{u}, \bar{v}, \bar{ψ}) ‖}_{L^{2}}^{2} {‖ \nabla (v_{1}, ψ_{1}) ‖}_{L^{2}}^{4} .$

将(2.24)~(2.31)估计应用到(2.22)中，可以得到

$\frac{d}{d t} {‖ (\bar{u}, \bar{v}, \bar{ψ}) ‖}_{L^{2}}^{2} \leq C ({‖ (v_{1}, v_{2}) ‖}_{L^{2}}^{2} {‖ \nabla (v_{1}, v_{2}) ‖}_{L^{2}}^{6} + {‖ \nabla (u_{1}, v_{1}, ψ_{1}) ‖}_{L^{2}}^{4}) {‖ (\bar{u}, \bar{v}, \bar{ψ}) ‖}_{L^{2}}^{2}$

应用Grönwall不等式，通过 $H^{1}$ -估计，可以得到

${‖ (\bar{u}, \bar{v}, \bar{ψ}) ‖}_{L^{2}}^{2} = 0$

这样完成了定理1.1唯一性部分的证明。

文章引用

陈贤. 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[J]. 应用数学进展, 2022, 11(07): 4933-4941. https://doi.org/10.12677/AAM.2022.117517

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