完全可压缩MHD方程组一组重要的先验估计 A Set of Important Priori Estimates for Full Compressible MHD Equations

doi:10.12677/PM.2022.127135

Pure Mathematics
Vol. 12 No. 07 ( 2022 ), Article ID: 54222 , 11 pages
10.12677/PM.2022.127135

完全可压缩MHD方程组一组重要的先验估计

王传宝，陈菲，王帅

●How to Cite this Article

青岛大学数学与统计学院，山东青岛

收稿日期：2022年6月18日；录用日期：2022年7月20日；发布日期：2022年7月27日

摘要

本篇论文在研究Navier-Stokes方程组的基础上，进一步对耦合项和磁场项分析得到能量估计，从而得到完全可压缩磁流体力学(Magnetohydrodynamics, MHD)方程组的解的先验估计。特别地，此结果是完全可压缩MHD方程组在更低的正则空间下建立强解的整体存在唯一性的重要步骤。

关键词

完全可压缩MHD方程组，先验估计，柯西问题

A Set of Important Priori Estimates for Full Compressible MHD Equations

Chuanbao Wang, Fei Chen, Shuai Wang

School of Mathematics and Statistics, Qingdao University, Qingdao Shandong

Received: Jun. 18^th, 2022; accepted: Jul. 20^th, 2022; published: Jul. 27^th, 2022

ABSTRACT

This paper is based on Navier-Stokes equations. The coupling terms and magnetic field terms are further analyzed to obtain energy estimates so that we can get a priori estimates of full compressible Magnetohydrodynamics (MHD) equations. What’s more, our results play an important role in establishing the global existence and uniqueness of strong solutions in the space of lower regularity.

Keywords:Full Compressible MHD Equations, Priori Estimates, Cauchy Problem

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

本篇论文研究完全可压缩磁流体力学(MHD)方程组的柯西问题，主要如下表示：

${\begin{cases} ρ_{t} + div (ρ u) = 0, \\ ρ u_{t} + ρ u \cdot \nabla u + \nabla P = μ Δ u + (μ + λ) \nabla div u + (\nabla \times B) \times B, \\ C_{ν} ρ (θ_{t} + u \cdot \nabla θ) + P div u = κ Δ θ + 2 μ {| D (u) |}^{2} + λ {(div u)}^{2} + ν {| \nabla \times B |}^{2}, \\ B_{t} - (\nabla \times u) \times B = - \nabla \times (ν \nabla \times B), div B = 0, \\ {(ρ, u, θ, B) |}_{t = 0} = (ρ_{0}, u_{0}, θ_{0}, B_{0}), x \in R^{3}, \\ (ρ, u, θ, B) \to (1, 0, 1, 0), | x | \to \infty, t > 0, \end{cases}$ (1)

其中， $(x, t) \in R^{3} \times (0, + \infty)$ 。 $ρ, u = (u_{1}, u_{2}, u_{3}), θ, B = (B_{1}, B_{2}, B_{3})$ 分别代表流体的密度、速度、温度以及磁场。 $μ$ 和 $λ$ 是粘性系数并满足 $μ \geq 0$ 与 $2 μ + 3 λ \geq 0$ 。 $κ$ 和 $ν$ 分别表示热传导和磁场扩散系数。此外， $γ > 1$ 为绝热指数，理想气体常数 $R > 0$ ，内能e和比热容常数 $C_{ν} > 0$ ，压力P可以表示为

$P = R ρ θ = (γ - 1) ρ e, e = \frac{R θ}{γ - 1} = C_{ν} θ,$

粘性压力张量 $D (u)$ 表达式为

$D (u) = \frac{(\nabla u) + {(\nabla u)}^{'}}{2} .$

在连续性假设下，流体的运动满足质量守恒、动量守恒和能量守恒。从方程组的结构看，磁流体力学(MHD)方程组是Navier-Stokes方程组耦合磁场形成的新的方程组，也使得在研究这两种方程组问题时使用类似的方法成为可能。但是由于速度场和磁场的强耦合，磁场方程和温度方程的出现，导致MHD方程组的处理与分析要更加复杂，这也使得MHD方程组成为近些年来偏微分方程研究的重点与热点问题。

关于降低此类方程组解的正则性的相关研究，也引起了许多学者的广泛关注。在研究等熵情况下解的存在性的基础上(参见 [1])，Huang和Li [2] 建立了在真空以及大震荡初值情形下Navier-Stokes方程组的经典解的整体存在唯一性，并且得到了弱解的全局存在性。后来，对比Valli [3] 研究完全可压缩Navier-Stokes方程组结果的基础上，并且借助于 [4] 中的先验估计，当初始速度和初始温度在比H²正则性更低的空间中，Xu和Zhang [4] 建立了整体解的存在唯一性。近期，基于 [5] [6] 的结果，Zhang [7] 进一步研究了等熵可压缩MHD方程组的相关问题，降低了方程组中关于初始速度的正则性。

本文的主要内容是建立关于完全可压缩MHD方程组(1)的先验估计。基于本文的研究成果，后续我们将进一步研究其整体适定性。即关于方程组(1)中的初始速度，初始磁场和初始温度，在这三项的低正则空间中，建立全局解的存在唯一性。其主要难点，进行方程组(1)的随体导数，有效粘性通量和涡度的能量估计，以及低阶的先验估计的导出。

记初始能量为：

$E_{0} = \int^{} (\frac{1}{2} (ρ_{0} {| u_{0} |}^{2} + {| B_{0} |}^{2}) + R (ρ_{0} \log ρ_{0} - ρ_{0} + 1) + C_{ν} ρ_{0} (θ_{0} - \log θ_{0} - 1)) d x,$

方程组(1)的随体导数，有效粘性通量和涡度的定义分别如下：

${\begin{cases} \dot{f} ≜ f_{t} + u \cdot \nabla f, \\ G ≜ (2 μ + λ) div u - R (ρ θ - 1) + \nabla^{- 1} [(\nabla \times B) \times B], \\ ω ≜ \nabla \times u + μ^{- 1} Δ^{- 1} [\nabla \times ((\nabla \times B) \times B)], \end{cases}$ (2)

由此可推出

$Δ G = div (ρ \dot{u}), μ Δ u = \nabla \times (ρ \dot{u}) .$ (3)

关于以上三个物理量的定义，可参考 [2] [8] [9]。

本文的结构如下：首先介绍证明需要用到的两个引理；重点在第二部分，建立时间独立的先验估计；最后，会给出本文的主要结论。

2. 预备知识

2.1. Gagliardo-Nirenberg不等式

引理2.1 (参见 [10]) 假设 $1 \leq k, s, t \leq \infty, 0 \leq r < p$ 以及 $\frac{r}{p} \leq α \leq 1$ 且满足

$\frac{1}{k} - \frac{r}{q} = α (\frac{1}{s} - \frac{p}{q}) + (1 - α) \frac{1}{t},$ (4)

其中 $p - r - \frac{q}{s} \in 0 \cup Z^{+}$ 。则存在依赖于 $p, q, k, r, s, t$ 的常数C，使得对于任意的 $φ \in W^{p, s} (R^{q}) \cap L^{t} (R^{q})$ ，

${‖ \nabla^{r} φ ‖}_{L^{k}} \leq C {‖ \nabla^{p} φ ‖}_{L^{s}}^{α} {‖ φ ‖}_{L^{t}}^{(1 - α)} .$ (5)

2.2. 随体导数，有效粘性通量和涡度的L^p估计

接着介绍从参考文献 [5] [7] 推广而得的 $L^{p}$ 估计，将用于后续的证明。

引理2.2令 $(ρ, u, θ, B)$ 是方程组(1)的一组光滑解，存在仅依赖于 $μ, λ, ν$ 和R的正常数C，使得对于任意的 $p \in [2, 6]$ ，得到以下估计：

$({‖ \nabla G ‖}_{L^{p}} + {‖ \nabla ω ‖}_{L^{p}}) \leq C {‖ ρ \dot{u} ‖}_{L^{p}},$ (6)

$({‖ G ‖}_{L^{p}} + {‖ ω ‖}_{L^{p}}) \leq C {‖ ρ \dot{u} ‖}_{L^{p}}^{(3 p - 6) / (2 p)} {({‖ ρ θ - 1 ‖}_{L^{2}} + {‖ \nabla u ‖}_{L^{2}} + {‖ {| B |}^{2} ‖}_{L^{2}})}^{(6 - p) / (2 p)},$ (7)

${‖ \nabla u ‖}_{L^{p}} \leq C ({‖ G ‖}_{L^{p}} + {‖ ω ‖}_{L^{p}} + {‖ ρ θ - 1 ‖}_{L^{p}} + {‖ {| B |}^{2} ‖}_{L^{p}}),$ (8)

${‖ \nabla u ‖}_{L^{p}} \leq C {‖ \nabla u ‖}_{L^{2}}^{(6 - p) / (2 p)} {({‖ ρ \dot{u} ‖}_{L^{2}} + {‖ ρ θ - 1 ‖}_{L^{6}} + {‖ B \cdot \nabla B ‖}_{L^{2}})}^{(3 p - 6) / (2 p)} .$ (9)

3. 先验估计

通过引理2.2的结论以及参考文献 [4] 中第三部分的相关证明，最终我们可以得到一组低阶的时间独

立的先验估计，具体验证过程如下：令 $ψ (t) ≜ \min {1, t}$ ，接着定义 $Π_{i} (T) (i = 1, 2, 3, 4)$ 分别如下：

$Π_{1} (T) = \sup_{t \in [0, T]} ({‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla B ‖}_{L^{2}}^{2}) + \int_{0}^{T} \int (ρ {| \dot{u} |}^{2} + {| \dot{B} |}^{2}) d x d t,$

$Π_{2} (T) = \sup_{t \in [0, T]} {‖ \sqrt{ρ} (θ - 1) ‖}_{L^{2}}^{2} + \int_{0}^{T} ({‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla B ‖}_{L^{2}}^{2} + {‖ \nabla θ ‖}_{L^{2}}^{2}) d t,$

$\begin{matrix} Π_{3} (T) = \sup_{t \in [0, T]} (ψ ({‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla B ‖}_{L^{2}}^{2}) + ψ^{2} ({‖ \sqrt{ρ} \dot{u} ‖}_{L^{2}}^{2} + {‖ \dot{B} ‖}_{L^{2}}^{2}) + ψ^{2} {‖ \nabla θ ‖}_{L^{2}}^{2}) \\ + \int_{0}^{T} \int (ψ (ρ {| \dot{u} |}^{2} + {| \dot{B} |}^{2}) + ψ^{2} ({| \nabla \dot{u} |}^{2} + {| \nabla \dot{B} |}^{2}) + ψ^{2} ρ {(\dot{θ})}^{2}) d x d t, \end{matrix}$

$Π_{4} (T) = \sup_{t \in [0, T]} ψ^{4} \int ρ {| \dot{θ} |}^{2} d x + \int_{0}^{T} ψ^{4} {| \nabla \dot{θ} |}^{2} d x d t .$

命题3.1对于给定的正常数 $\bar{ρ} > 2$ ， $\bar{θ} > 0$ 以及 $Ω > 0$ (不必小)，假设初值 $(ρ_{0}, u_{0}, θ_{0}, B_{0})$ 满足：

${\begin{cases} 0 < \inf ρ_{0} \leq \sup ρ_{0} \leq \bar{ρ}, \\ 0 < \inf θ_{0} \leq \sup θ_{0} \leq \bar{θ}, \\ {‖ \nabla u_{0} ‖}_{L^{2}}, {‖ \nabla B_{0} ‖}_{L^{2}} \leq Ω . \end{cases}$ (10)

若存在仅依赖于 $μ, λ, κ, ν, R, γ, \bar{ρ}, \bar{θ}$ 和 $Ω$ 的正常数Q和 $ζ$ ，使得初始能量满足 $E_{0} \leq ζ$ 且方程组(1)的一组光滑解 $(ρ, u, θ, B)$ 满足

$0 < ρ \leq 2 \bar{ρ}, Π_{1} (ψ (T)) \leq 3 Q, Π_{i} (T) \leq 2 E_{0}^{1 / 2 i} (i = 2, 3, 4),$ (11)

则可获得如下估计：

$0 < ρ \leq 3 \bar{ρ} / 2, Π_{1} (ψ (T)) \leq 2 Q, Π_{2} (T) \leq E_{0}^{1 / 2 i} (i = 2, 3, 4) .$ (12)

证明：详见下列引理3.2与引理3.3。

另外在进行接下来的证明之前，我们规定 $E_{0} \leq 1$ 并且C是大于0的正常数。

引理3.1 (参见 [2] [5])在命题1的条件下，存在仅依赖于 $μ, λ, κ, ν, R, γ$ 和 $\bar{ρ}$ 的正常数 $C (\bar{ρ})$ ，使得 $(ρ, u, θ, B)$ 是方程组(1)在 $R^{3} \times (0, T]$ 的一组光滑解，且满足 $0 < ρ \leq 2 \bar{ρ}$ ，则对于任意的 $t \in (0, T]$ 可得到如下估计：

$\sup_{t \in [0, T]} ({‖ \sqrt{ρ} u ‖}_{L^{2}}^{2} + {‖ B ‖}_{L^{2}}^{2} + {‖ ρ - 1 ‖}_{L^{2}}^{2}) \leq C (\bar{ρ}) E_{0},$ (13)

和

${‖ (θ - 1) (\cdot, t) ‖}_{L^{2}} \leq C (\bar{ρ}) E_{0}^{1 / 2} + C (\bar{ρ}) E_{0}^{1 / 3} {‖ \nabla θ (\cdot, t) ‖}_{L^{2}} .$ (14)

3.1. $Π_{1} (T)$ 的估计

引理3.2在命题1的条件下，存在仅依赖于 $μ, λ, κ, ν, R, γ, \bar{ρ}$ 和 $Ω$ 的正常数Q与 $ζ_{1}$ ，并且满足 $Q \geq Ω + 1$ 与 $ζ_{1} \leq 1$ ，使得 $(ρ, u, θ, B)$ 是方程组(1)在 $R^{3} \times (0, T]$ 上的一组光滑解满足：

$0 < ρ \leq 2 \bar{ρ}, Π_{2} (T) \leq 2 E_{0}^{1 / 4},$ (15)

则对于任意的 $Π_{1} (ψ (t)) \leq 3 Q$ 与 $E_{0} \leq ζ_{1}$ 可获得以下估计：

$Π_{1} (ψ (T)) \leq 2 Q .$ . (16)

证明：首先由方程组(2)₂易得

$div u = \frac{1}{2 μ + λ} [R (ρ θ - 1) + G - \nabla^{- 1} [(\nabla \times B) \times B]],$ (17)

接着将(1)₂乘以 $2 u_{t}$ ，(1)₄乘以 $2 B_{t}$ ，分别在 $R^{3}$ 上积分后相加，分部积分之后可得

$\begin{array}{l} \frac{d}{d t} \int^{} (μ {| \nabla u |}^{2} + (μ + λ) {(div u)}^{2} + ν {| \nabla B |}^{2}) d x + \int^{} (ρ {| u_{t} |}^{2} + {| B_{t} |}^{2}) d x \\ \leq \int^{} (- 2 \nabla P \cdot u_{t} + 2 ρ {| u \cdot \nabla u |}^{2} + 2 ((\nabla \times B) \times B) \cdot u_{t}) d x \\ + 2 \int^{} (B \cdot \nabla u - u \cdot \nabla B - B div u) \cdot B_{t} d x \end{array}$

$\begin{array}{l} = 2 R \frac{d}{d t} \int^{} (ρ θ - 1) div u d x - \frac{R^{2}}{2 μ + λ} \frac{d}{d t} \int^{} {(ρ θ - 1)}^{2} d x \\ + 2 \frac{d}{d t} \int^{} (B \cdot \nabla B - \frac{1}{2} \nabla ({| B |}^{2})) \cdot u d x - \frac{2}{2 μ + λ} \int^{} P_{t} G d x \\ + \frac{2}{2 μ + λ} \int^{} P_{t} \cdot \nabla^{- 1} [(\nabla \times B) \times B] d x + 2 \int^{} ρ {| u \cdot \nabla u |}^{2} d x \\ - 2 \int^{} {(B \cdot \nabla B - \frac{1}{2} \nabla ({| B |}^{2}))}_{t} \cdot u d x + 2 \int^{} (B \cdot \nabla u - u \cdot \nabla B - B div u) \cdot B_{t} d x \\ ≜ \sum_{i = 1}^{8} A_{i} \end{array}$ (18)

然后，利用(15)，(5)和(11)，对于 $p \in [2, 6]$ 可得

$\begin{matrix} {‖ ρ θ - 1 ‖}_{L^{p}} \leq {‖ ρ (θ - 1) + (ρ - 1) ‖}_{L^{p}} \\ \leq {‖ ρ (θ - 1) ‖}_{L^{2}}^{(6 - p) / 2 p} {‖ ρ (θ - 1) ‖}_{L^{6}}^{(3 p - 2) / 2 p} + {‖ ρ - 1 ‖}_{L^{p}} \\ \leq C (\bar{ρ}) E_{0}^{(6 - p) / 16 p} {‖ \nabla θ ‖}_{L^{2}}^{(3 p - 2) / 2 p} + C (\bar{ρ}) E_{0}^{1 / p}, \end{matrix}$ (19)

再结合(9)和(5)，可推出

${‖ \nabla u ‖}_{L^{6}} \leq C ({‖ \sqrt{ρ} \dot{u} ‖}_{L^{2}} + {‖ \nabla θ ‖}_{L^{2}} + E_{0}^{1 / 6} + {‖ B \cdot \nabla B ‖}_{L^{2}}),$ (20)

由椭圆系统的 $L^{2}$ 标准估计与(1)₄，可得出

${‖ \nabla B ‖}_{L^{6}} \leq C ({‖ \dot{B} ‖}_{L^{2}} + {‖ B \cdot \nabla u ‖}_{L^{2}}),$ (21)

接着，利用Hölder不等式和Young不等式可以发现，

${‖ B \cdot \nabla u ‖}_{L^{2}} \leq C {‖ \nabla B ‖}_{L^{2}} {‖ \nabla u ‖}_{L^{2}}^{1 / 2} {‖ \nabla u ‖}_{L^{6}}^{1 / 2} \leq C (ξ) ({‖ \nabla B ‖}_{L^{2}}^{3} + {‖ \nabla u ‖}_{L^{2}}^{3}) + ξ {‖ \nabla u ‖}_{L^{6}},$

类似地，

${‖ B \cdot \nabla B ‖}_{L^{2}} \leq C (ξ) {‖ \nabla B ‖}_{L^{2}}^{3} + ξ {‖ \nabla B ‖}_{L^{6}},$

其中 $0 < ξ < 1$ ，而且通过能量估计得到的“ ${‖ \nabla u ‖}_{L^{6}}$ ”与“ ${‖ \nabla B ‖}_{L^{6}}$ ”能够被左端吸收。综上所得，即有

$({‖ \nabla u ‖}_{L^{6}} + {‖ \nabla B ‖}_{L^{6}}) \leq C ({‖ \sqrt{ρ} \dot{u} ‖}_{L^{2}} + {‖ \dot{B} ‖}_{L^{2}} + {‖ \nabla θ ‖}_{L^{2}} + E_{0}^{1 / 6} + {‖ \nabla u ‖}_{L^{2}}^{3} + {‖ \nabla B ‖}_{L^{2}}^{3}) .$ (22)

紧接着，从方程组(1)可以推出

$\frac{P_{t}}{γ - 1} = κ Δ θ + (2 μ {| D (u) |}^{2} + λ {(div u)}^{2}) + ν {| \nabla \times B |}^{2} - P div u - \frac{div (P u)}{γ - 1},$ (23)

做好这些准备工作之后，我们开始分析(18)的右端各项。类似于 [2] 中(3.27)的推导，由(23)，(15)，(5)，(6)，(7)，(19)，(22)和(13)可以得到

$\begin{matrix} | A_{4} | \leq C \int^{} P (| \nabla u | | G | + | u | | \nabla G |) d x + C \int^{} (| \nabla θ | | \nabla G | + {| \nabla u |}^{2} | G | + {| \nabla B |}^{2} | G |) d x \\ \leq C \int^{} ρ (| \nabla u | | G | + | u | | \nabla G |) d x + C \int^{} ρ (θ - 1) (| \nabla u | | G | + | u | | \nabla G |) d x \\ + C {‖ \nabla G ‖}_{L^{2}} {‖ \nabla θ ‖}_{L^{2}} + C {‖ \nabla G ‖}_{L^{2}} ({‖ \nabla B ‖}_{L^{2}}^{3 / 2} {‖ \nabla B ‖}_{L^{6}}^{1 / 2} + {‖ \nabla u ‖}_{L^{2}}^{3 / 2} {‖ \nabla u ‖}_{L^{6}}^{1 / 2}) \\ \leq C (\bar{ρ}) ({‖ ρ θ - 1 ‖}_{L^{2}} + {‖ \nabla u ‖}_{L^{2}} + {‖ B ‖}_{L^{2}} + {‖ \nabla B ‖}_{L^{2}}^{3}) {‖ \nabla u ‖}_{L^{2}} + C {‖ \nabla G ‖}_{L^{2}} ({‖ ρ u ‖}_{L^{2}} + {‖ \nabla θ ‖}_{L^{2}}) \\ + C (\bar{ρ}) {‖ ρ (θ - 1) ‖}_{L^{2}}^{1 / 2} {‖ \nabla θ ‖}_{L^{2}}^{1 / 2} {‖ \nabla G ‖}_{L^{2}} {‖ \nabla u ‖}_{L^{2}} + C {‖ \nabla G ‖}_{L^{2}} ({‖ \nabla B ‖}_{L^{2}}^{3 / 2} {‖ \nabla B ‖}_{L^{6}}^{1 / 2} + {‖ \nabla u ‖}_{L^{2}}^{3 / 2} {‖ \nabla u ‖}_{L^{6}}^{1 / 2}) \end{matrix}$

$\begin{array}{l} \leq C (\bar{ρ}, ξ) E_{0}^{1 / 4} + ξ {‖ \nabla G ‖}_{L^{2}}^{2} + C (\bar{ρ}, ξ) ({‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla B ‖}_{L^{2}}^{2}) + C (\bar{ρ}, ξ) {‖ \nabla G ‖}_{L^{2}} {‖ \nabla u ‖}_{L^{2}}^{2} \\ + C (\bar{ρ}, ξ) {‖ \nabla θ ‖}_{L^{2}}^{2} + ξ ({‖ \sqrt{ρ} \dot{u} ‖}_{L^{2}}^{2} + {‖ \dot{B} ‖}_{L^{2}}^{2}) + C (\bar{ρ}, ξ) ({‖ \nabla u ‖}_{L^{2}}^{6} + {‖ \nabla B ‖}_{L^{2}}^{6}) \\ \leq C (\bar{ρ}) ξ ({‖ \sqrt{ρ} \dot{u} ‖}_{L^{2}}^{2} + {‖ \dot{B} ‖}_{L^{2}}^{2}) + C (\bar{ρ}, ξ) ({‖ \nabla u ‖}_{L^{2}}^{6} + {‖ \nabla B ‖}_{L^{2}}^{6}) \\ + C (\bar{ρ}, ξ) ({‖ \nabla θ ‖}_{L^{2}}^{2} + {‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla B ‖}_{L^{2}}^{2} + 1), \end{array}$ (24)

同理，

$\begin{matrix} | A_{5} | \leq C (\bar{ρ}) ξ ({‖ \sqrt{ρ} \dot{u} ‖}_{L^{2}}^{2} + {‖ \dot{B} ‖}_{L^{2}}^{2}) + C (\bar{ρ}, ξ) ({‖ \nabla u ‖}_{L^{2}}^{6} + {‖ \nabla B ‖}_{L^{2}}^{6}) \\ + C (\bar{ρ}, ξ) ({‖ \nabla θ ‖}_{L^{2}}^{2} + {‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla B ‖}_{L^{2}}^{2} + 1) . \end{matrix}$

接下来再结合 [2] 中(3.28)的推导过程，利用(5)和(22)可以得到

$\begin{matrix} A_{6} \leq C (\bar{ρ}) {‖ u ‖}_{L^{6}}^{2} {‖ \nabla u ‖}_{L^{2}} {‖ \nabla u ‖}_{L^{6}} \\ \leq C (\bar{ρ}) ξ ({‖ \sqrt{ρ} \dot{u} ‖}_{L^{2}}^{2} + {‖ \dot{B} ‖}_{L^{2}}^{2}) + C (\bar{ρ}, ξ) ({‖ \nabla u ‖}_{L^{2}}^{6} + {‖ \nabla B ‖}_{L^{2}}^{6}) \\ + C (\bar{ρ}, ξ) ({‖ \nabla θ ‖}_{L^{2}}^{2} + {‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla B ‖}_{L^{2}}^{2} + 1), \end{matrix}$

现在分析最后两项。由分部积分，(2)₁，(5)，(11)，(15)与(22)，可以得到

$\begin{matrix} A_{7} + A_{8} \leq C \int^{} (| B | | \nabla u | + | u | | \nabla B |) | B_{t} | d x \\ \leq C \int^{} | B | | \nabla u | (| \dot{B} | + | u | | \nabla B |) + | u | | \nabla B | (| \dot{B} | + | u | | \nabla B |) d x \\ \leq C {‖ B ‖}_{L^{6}} {‖ \nabla u ‖}_{L^{2}}^{1 / 2} {‖ \nabla u ‖}_{L^{6}}^{1 / 2} {‖ \dot{B} ‖}_{L^{2}} + C {‖ B ‖}_{L^{6}} {‖ u ‖}_{L^{6}} {‖ \nabla u ‖}_{L^{2}} {‖ \nabla B ‖}_{L^{6}} \\ + C {‖ u ‖}_{L^{6}} {‖ \nabla B ‖}_{L^{2}}^{1 / 2} {‖ \nabla B ‖}_{L^{6}}^{1 / 2} {‖ \dot{B} ‖}_{L^{2}} + C {‖ u ‖}_{L^{6}}^{2} {‖ \nabla B ‖}_{L^{2}} {‖ \nabla B ‖}_{L^{6}} \\ \leq C (\bar{ρ}) ξ ({‖ \sqrt{ρ} \dot{u} ‖}_{L^{2}}^{2} + {‖ \dot{B} ‖}_{L^{2}}^{2}) + C (\bar{ρ}, ξ) ({‖ \nabla u ‖}_{L^{2}}^{6} + {‖ \nabla B ‖}_{L^{2}}^{6}) \\ + C (\bar{ρ}, ξ) ({‖ \nabla θ ‖}_{L^{2}}^{2} + {‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla B ‖}_{L^{2}}^{2} + 1) . \end{matrix}$

综上所得，即有

$\begin{matrix} \sum_{i = 4}^{8} A_{i} \leq C (\bar{ρ}) ξ ({‖ \sqrt{ρ} \dot{u} ‖}_{L^{2}}^{2} + {‖ \dot{B} ‖}_{L^{2}}^{2}) + C (\bar{ρ}, ξ) ({‖ \nabla u ‖}_{L^{2}}^{6} + {‖ \nabla B ‖}_{L^{2}}^{6}) \\ + C (\bar{ρ}, ξ) ({‖ \nabla θ ‖}_{L^{2}}^{2} + {‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla B ‖}_{L^{2}}^{2} + 1) . \end{matrix}$ (25)

然后将(25)的结果代入(18)中，令 $ξ$ 足够地小，在 $(0, t)$ 上进行积分后，再关于 $t \in (0, ψ (T))$ 取上确界，由(15)就可以推出

$\begin{array}{l} \sup_{t \in [0, ψ (T)]} ({‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla B ‖}_{L^{2}}^{2}) + \int_{0}^{ψ (T)} \int (ρ {| \dot{u} |}^{2} + {| \dot{B} |}^{2}) d x d t \\ \leq C + C Ω + C (\bar{ρ}) E_{0}^{1 / 4} + C (\bar{ρ}) E_{0}^{1 / 4} \sup_{t \in [0, ψ (T)]} ({‖ \nabla u ‖}_{L^{2}}^{4} + {‖ \nabla B ‖}_{L^{2}}^{4}) \\ \leq Q + C (\bar{ρ}) E_{0}^{1 / 4} \sup_{t \in [0, ψ (T)]} ({‖ \nabla u ‖}_{L^{2}}^{4} + {‖ \nabla B ‖}_{L^{2}}^{4}) ， \end{array}$ (26)

其中

$Q = C Ω + C (\bar{ρ}) + 1$ ,

并且Q仅依赖于 $μ, λ, κ, ν, R, γ, \bar{ρ}$ 和 $Ω$ 。最后令 $ζ_{1} = \min {1, {(9 C (\bar{ρ}) Q)}^{- 4}}$ ，即可证明出引理2。

3.2. $Π_{2} (T)$ 的估计

引理3.3：在命题1的条件下，存在仅依赖于 $μ, λ, κ, ν, R, γ, \bar{ρ}, \bar{θ}$ 和 $Ω$ 的正常数 $ζ_{2}$ ，使得 $(ρ, u, θ, B)$ 是方程组(1)在 $R^{3} \times (0, T]$ 的一组光滑解，并且Q如引理2所表示，则对于 $E_{0} \leq ζ_{2}$ 可获得以下估计：

$Π_{2} (T) \leq E_{0}^{1 / 4}$ .

证明：首先，分别将(1)₂乘以 $u$ ，(1)₄乘以 $B$ ，然后将这两式在 $R^{3}$ 上积分后相加，再由(13)与(14)可得

$\begin{array}{l} \frac{d}{d t} \int^{} (\frac{1}{2} (ρ {| u |}^{2} + {| B |}^{2}) + R (ρ \log ρ + 1 - ρ)) d x \\ + \int^{} (μ {| \nabla u |}^{2} + (μ + λ) {(div u)}^{2} + ν {| \nabla B |}^{2}) d x \\ = R \int^{} ρ (θ - 1) div u d x \\ \leq C (\bar{ρ}) ({‖ θ - 1 ‖}_{L^{2}} + {‖ ρ - 1 ‖}_{L^{2}}) {‖ \nabla u ‖}_{L^{2}} \\ \leq C (\bar{ρ}) (E_{0}^{1 / 2} + E_{0}^{1 / 3} {‖ \nabla θ ‖}_{L^{2}}) {‖ \nabla u ‖}_{L^{2}} \\ \leq C (\bar{ρ}) E_{0}^{2 / 3} + C (\bar{ρ}) E_{0}^{1 / 3} ({‖ \nabla θ ‖}_{L^{2}}^{2} + {‖ \nabla u ‖}_{L^{2}}^{2}) . \end{array}$ (27)

同样地，将(1)₃乘以 $θ - 1$ 后在 $R^{3}$ 上积分，可以得到

$\begin{array}{l} \frac{R}{2 (γ - 1)} \frac{d}{d t} \int^{} ρ {(θ - 1)}^{2} d x + κ {‖ \nabla θ ‖}_{L^{2}}^{2} \\ \leq C (\bar{ρ}) \int^{} θ | θ - 1 | | div u | d x + C \int^{} | θ - 1 | {| \nabla u |}^{2} d x + C \int^{} | θ - 1 | {| \nabla B |}^{2} d x \\ ≜ \sum_{i = 1}^{3} B_{i} . \end{array}$ (28)

其中从(11)可以推出

$\sup_{t \in [0, T]} ({‖ \nabla u ‖}_{L^{2}} + {‖ \nabla B ‖}_{L^{2}}) \leq Π_{1} (ψ (T)) + Π_{3} (T) \leq C (\bar{ρ}, Ω),$ (29)

进而可以结合 [2] 中的(3.33)式，得到 $B_{1}$ 的能量估计，即

$B_{1} \leq C (\bar{ρ}, Ω) E_{0}^{1 / 2} + C (\bar{ρ}, Ω) E_{0}^{1 / 6} ({‖ \nabla θ ‖}_{L^{2}}^{2} + {‖ \nabla u ‖}_{L^{2}}^{2}) .$

下一步便是分析 $B_{2}$ 与 $B_{3}$ 。从(14)，(5)，(11)，(22)，(29)得到

$\begin{matrix} B_{2} + B_{3} \leq C {‖ θ - 1 ‖}_{L^{2}}^{1 / 2} {‖ θ - 1 ‖}_{L^{6}}^{1 / 2} ({‖ \nabla u ‖}_{L^{2}} {‖ \nabla u ‖}_{L^{6}} + {‖ \nabla B ‖}_{L^{2}} {‖ \nabla B ‖}_{L^{6}}) \\ \leq C (\bar{ρ}, Ω) (E_{0}^{1 / 4} {‖ \nabla θ ‖}_{L^{2}}^{1 / 2} + E_{0}^{1 / 6} {‖ \nabla θ ‖}_{L^{2}}) \\ \times ({‖ \sqrt{ρ} \dot{u} ‖}_{L^{2}} + {‖ \dot{B} ‖}_{L^{2}} + {‖ \nabla θ ‖}_{L^{2}} + E_{0}^{1 / 6} + {‖ \nabla u ‖}_{L^{2}}^{3} + {‖ \nabla B ‖}_{L^{2}}^{3}) \\ \leq C (\bar{ρ}, Ω, ξ) E_{0}^{1 / 3} ({‖ \sqrt{ρ} \dot{u} ‖}_{L^{2}}^{2} + {‖ \dot{B} ‖}_{L^{2}}^{2} + 1) \\ + C (\bar{ρ}, Ω) (E_{0}^{1 / 6} + ξ) ({‖ \nabla θ ‖}_{L^{2}}^{2} + {‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla B ‖}_{L^{2}}^{2}) . \end{matrix}$ (30)

然后将 $B_{1} - B_{3}$ 的估计代入(28)，便能得出

$\begin{array}{l} \frac{R}{2 (γ - 1)} \frac{d}{d t} \int^{} ρ {(θ - 1)}^{2} d x + κ {‖ \nabla θ ‖}_{L^{2}}^{2} \\ \leq C (\bar{ρ}, Ω, ξ) E_{0}^{1 / 3} ({‖ \sqrt{ρ} \dot{u} ‖}_{L^{2}}^{2} + {‖ \dot{B} ‖}_{L^{2}}^{2} + 1) + C (\bar{ρ}, Ω) (E_{0}^{1 / 6} + ξ) ({‖ \nabla θ ‖}_{L^{2}}^{2} + {‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla B ‖}_{L^{2}}^{2}) . \end{array}$ (31)

再进一步结合(27)与(31)，可得到

$\begin{array}{l} \int^{} (\frac{1}{2} (ρ {| u |}^{2} + {| B |}^{2}) + R (ρ \log ρ + 1 - ρ) + \frac{R}{2 (γ - 1)} ρ {(θ - 1)}^{2}) d x \\ + \int^{} (μ {| \nabla u |}^{2} + (μ + λ) {(div u)}^{2} + ν {| \nabla B |}^{2} + κ {| \nabla θ |}^{2}) d x \\ \leq C (\bar{ρ}, Ω, ξ) E_{0}^{1 / 3} ({‖ \sqrt{ρ} \dot{u} ‖}_{L^{2}}^{2} + {‖ \dot{B} ‖}_{L^{2}}^{2} + 1) \\ + C (\bar{ρ}, Ω) (E_{0}^{1 / 6} + ξ) ({‖ \nabla θ ‖}_{L^{2}}^{2} + {‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla B ‖}_{L^{2}}^{2}) . \end{array}$ (32)

同时此处 $E_{0} \leq ζ_{2, 1} ≜ \min {1, 4 {(C (\bar{ρ}, Ω))}^{- 1} \min {μ, ν, κ}^{6}}$ 以及 $ξ \leq \min {μ, ν, κ} {(4 C (\bar{ρ}, Ω))}^{- 1}$ 。在 $(0, ψ (T))$ 上积分之后由(11)可以得到

$\begin{array}{l} \sup_{t \in [0, ψ (T)]} \int^{} (\frac{1}{2} (ρ {| u |}^{2} + {| B |}^{2}) + {(ρ - 1)}^{2} + \frac{R}{2 (γ - 1)} ρ {(θ - 1)}^{2}) d x \\ + \int_{0}^{ψ (T)} ({‖ \nabla θ ‖}_{L^{2}}^{2} + {‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla B ‖}_{L^{2}}^{2}) d t \leq C (\bar{ρ}, Ω) E_{0}^{1 / 3}, \end{array}$ (33)

其中需注意

$ρ \log ρ + 1 - ρ = {(ρ - 1)}^{2} \int_{0}^{1} \frac{1 - α}{α ρ - 1 + 1} d α \geq \frac{1}{2 (2 \bar{ρ} + 1)} {(ρ - 1)}^{2} .$ .

再之后，由(1)₃与椭圆系统的 $L^{p}$ 标准估计，能够得到

$\begin{matrix} {‖ \nabla^{2} θ ‖}_{L^{2}}^{2} \leq C ({‖ ρ \dot{θ} ‖}_{L^{2}}^{2} + {‖ θ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla u ‖}_{L^{4}}^{4} + {‖ \nabla B ‖}_{L^{4}}^{4}) \\ \leq C (\bar{ρ}) ({‖ \nabla θ ‖}_{L^{2}}^{2} + {‖ \nabla u ‖}_{L^{2}}^{2}) {({‖ \sqrt{ρ} \dot{u} ‖}_{L^{2}} + {‖ \dot{B} ‖}_{L^{2}} + {‖ \nabla θ ‖}_{L^{2}} + 1 + {‖ \nabla u ‖}_{L^{2}}^{3} + {‖ \nabla B ‖}_{L^{2}}^{3})}^{2} \\ + C ({‖ ρ \dot{θ} ‖}_{L^{2}}^{2} + {‖ \nabla u ‖}_{L^{4}}^{4} + {‖ \nabla B ‖}_{L^{4}}^{4}), \end{matrix}$ (34)

此处使用了

$\begin{matrix} {‖ θ \nabla u ‖}_{L^{2}}^{2} \leq C {‖ θ - 1 ‖}_{L^{6}}^{2} ({‖ \nabla u ‖}_{L^{2}} {‖ \nabla u ‖}_{L^{6}}) + C ({‖ \nabla u ‖}_{L^{2}}^{2}) \\ \leq C (\bar{ρ}) ({‖ \nabla θ ‖}_{L^{2}}^{2} + {‖ \nabla u ‖}_{L^{2}}^{2}) {({‖ \sqrt{ρ} \dot{u} ‖}_{L^{2}} + {‖ \dot{B} ‖}_{L^{2}} + {‖ \nabla θ ‖}_{L^{2}} + 1 + {‖ \nabla u ‖}_{L^{2}}^{3} + {‖ \nabla B ‖}_{L^{2}}^{3})}^{2} . \end{matrix}$

此外，从(11)和(22)推得

$\sup_{t \in [0, T]} ψ ({‖ \nabla u ‖}_{L^{6}} + {‖ \nabla B ‖}_{L^{6}}) \leq C (\bar{ρ}) E_{0}^{1 / 12},$

将上式结合(11)和(34)得出

$\begin{array}{l} \sup_{t \in [0, T]} ψ^{4} {‖ \nabla^{2} θ ‖}_{L^{2}}^{2} \\ \leq C (\bar{ρ}) \sup_{t \in [0, T]} ψ^{2} ({‖ \nabla θ ‖}_{L^{2}}^{2} + {‖ \nabla u ‖}_{L^{2}}^{2}) \\ \times \sup_{t \in [0, T]} ψ^{2} {({‖ \sqrt{ρ} \dot{u} ‖}_{L^{2}} + {‖ \dot{B} ‖}_{L^{2}} + {‖ \nabla θ ‖}_{L^{2}} + 1 + {‖ \nabla u ‖}_{L^{2}}^{3} + {‖ \nabla B ‖}_{L^{2}}^{3})}^{2} \\ + \sup_{t \in [0, T]} (ψ^{4} {‖ ρ \dot{θ} ‖}_{L^{2}}^{2} + (ψ {‖ \nabla u ‖}_{L^{2}}) {(ψ {‖ \nabla u ‖}_{L^{6}})}^{3} + (ψ {‖ \nabla B ‖}_{L^{2}}) {(ψ {‖ \nabla B ‖}_{L^{6}})}^{3}) \\ \leq C (\bar{ρ}) E_{0}^{1 / 8}, \end{array}$ (35)

再从(5)和(11)，可验证

$\sup_{t \in [0, T]} ψ^{2} {‖ θ - 1 ‖}_{L^{\infty}} \leq \sup_{t \in [0, T]} ψ^{2} ({‖ \nabla θ ‖}_{L^{2}} + {‖ \nabla^{2} θ ‖}_{L^{2}}) \leq C (\bar{ρ}) E_{0}^{1 / 16} \leq \frac{1}{2},$ (36)

此处 $E_{0} \leq ζ_{2, 2} ≜ \min {1, {(2 C (\bar{ρ}))}^{- 16}}$ 。再令 $E_{0} \leq \min {ζ_{2, 1}, ζ_{2, 2}}$ ，使用(34)能在 $(x, t) \in R^{3} \times [ψ (T), T]$ 上验证

$\frac{1}{2} \leq θ (x, t) \leq \frac{3}{2} .$

接着结合 [2] 中(3.45)的相关结果，分别将(1)₂乘以 $u$ ，(1)₃乘以 $1 - θ^{- 1}$ ，(1)₄乘以 $B$ ，然后将这三式在 $R^{3}$ 上积分后相加，并且利用

$2 μ {| D (u) |}^{2} + λ {(div u)}^{2} \geq 0,$

以及

$ν \int^{} {| \nabla B |}^{2} - {| \nabla \times B |}^{2} d x \geq 0,$

便可以推出

$\begin{array}{l} \frac{d}{d t} \int^{} (\frac{1}{2} (ρ {| u |}^{2} + {| B |}^{2}) + R (ρ \log ρ + 1 - ρ) + \frac{R}{γ - 1} ρ (θ - \log θ - 1)) d x \\ + ({‖ \nabla θ ‖}_{L^{2}}^{2} + {‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla B ‖}_{L^{2}}^{2}) \leq C (\bar{ρ}) E_{0}, \end{array}$

其中 $R (ρ \log ρ + 1 - ρ)$ 如同(33)中的处理，此外，

$(θ - \log θ - 1) = {(θ - 1)}^{2} \int_{0}^{1} \frac{α}{α (θ - 1) + 1} d x \geq \frac{1}{8} (θ - 1) 1_{θ (\cdot, t) > 2} + \frac{1}{12} {(θ - 1)}^{2} 1_{θ (\cdot, t) < 3} .$

最终在 $[ψ (T), T]$ 上积分便可以得到

$\begin{array}{l} \sup_{t \in [ψ (T), T]} \int^{} (ρ {| u |}^{2} + {| B |}^{2} + {(ρ - 1)}^{2} + \frac{R}{2 (γ - 1)} ρ {(θ - 1)}^{2}) d x \\ + \int_{ψ (T)}^{T} ({‖ \nabla θ ‖}_{L^{2}}^{2} + {‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla B ‖}_{L^{2}}^{2}) d t \leq C (\bar{ρ}) E_{0} . \end{array}$ (37)

最后，令 $E_{0} \leq ζ_{2} ≜ \min {ζ_{2, 1}, ζ_{2, 2}, {(C (\bar{ρ}, Ω))}^{- 12}, {(C (\bar{ρ}))}^{- 4 / 3}}$ 并结合(33)与(37)两式结果，引理3也可证

毕。

另外，还可以进行进一步通过能量估计缩小 $\bar{ρ}$ 的范围以及得出关于 $Π_{3} (T)$ 与 $Π_{4} (T)$ 的先验估计。不过在关于降低正则性的相关证明中，这几项估计并没有使用，所以这里我们便不再详细介绍了，有兴趣的读者可以阅读参考文献 [2] [9] [11] [12] [13] 了解类似证明。

4. 结论

定理4.1 对于给定的正常数 $\bar{ρ} > 2$ ， $\bar{θ} > 1$ 以及 $Ω > 0$ (不必小)，假设方程组(1)的初值 $(ρ_{0}, u_{0}, θ_{0}, B_{0})$ 满足：

${\begin{cases} 0 \leq \inf ρ_{0} \leq \sup ρ_{0} \leq \bar{ρ}, \\ 0 \leq \inf θ_{0} \leq \sup θ_{0} \leq \bar{θ}, \\ {‖ \nabla u_{0} ‖}_{L^{2}}, {‖ \nabla B_{0} ‖}_{L^{2}} \leq Ω . \end{cases}$

则存在仅依赖于 $μ, λ, κ, ν, R, γ, \bar{ρ}, \bar{θ}$ 和 $Ω$ 的正常数Q和 $ζ$ ，使得若初始能量满足：

$E_{0} \leq ζ$ ，

则可获得以下估计：

$0 < ρ \leq 2 \bar{ρ}, \forall (x, t) \in R^{3} \times [0, T],$

$\sup_{t \in [0, T]} ({‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla B ‖}_{L^{2}}^{2}) + \int_{0}^{T} \int (ρ {| \dot{u} |}^{2} + {| \dot{B} |}^{2}) d x d t \leq Q,$

$\begin{array}{l} \sup_{t \in [0, T]} ({‖ \sqrt{ρ} u ‖}_{L^{2}}^{2} + {‖ B ‖}_{L^{2}}^{2} + {‖ ρ - 1 ‖}_{L^{2}}^{2} + {‖ \sqrt{ρ} (θ - 1) ‖}_{L^{2}}^{2}) \\ + \int_{0}^{T} ({‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla B ‖}_{L^{2}}^{2} + {‖ \nabla θ ‖}_{L^{2}}^{2}) d t \leq Q E_{0}^{\frac{1}{4}}, \end{array}$

$\begin{array}{l} \sup_{t \in [0, T]} (ψ ({‖ \nabla u ‖}_{L^{2}}^{2} + {‖ \nabla B ‖}_{L^{2}}^{2}) + ψ^{2} ({‖ \sqrt{ρ} \dot{u} ‖}_{L^{2}}^{2} + {‖ \dot{B} ‖}_{L^{2}}^{2}) + ψ^{2} {‖ \nabla θ ‖}_{L^{2}}^{2}) \\ + \int_{0}^{T} \int (ψ (ρ {| \dot{u} |}^{2} + {| \dot{B} |}^{2}) + ψ^{2} ({| \nabla \dot{u} |}^{2} + {| \nabla \dot{B} |}^{2}) + σ^{2} ρ {(\dot{θ})}^{2}) d x d t \leq Q E_{0}^{\frac{1}{6}}, \end{array}$

$\sup_{t \in [0, T]} ψ^{4} \int ρ {| \dot{θ} |}^{2} d x + \int_{0}^{T} ψ^{4} {| \nabla \dot{θ} |}^{2} d x d t \leq Q E_{0}^{\frac{1}{8}} .$

基金项目

国家自然科学基金资助项目(编号12101345)，山东省自然科学基金资助项目(编号ZR2021QA017)。

文章引用

王传宝,陈菲,王帅. 完全可压缩MHD方程组一组重要的先验估计
A Set of Important Priori Estimates for Full Compressible MHD Equations[J]. 理论数学, 2022, 12(07): 1231-1241. https://doi.org/10.12677/PM.2022.127135

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