一类化学活性流体方程组解的存在唯一性

doi:10.12677/aam.2024.134119

期刊菜单

一类化学活性流体方程组解的存在唯一性
Existence and Uniqueness of Solutions for a Class of Chemically Active Fluid Equations

DOI: 10.12677/aam.2024.134119, PDF, HTML, XML, 下载: 35 浏览: 52 科研立项经费支持
作者: 王长佳, 苏日娜^*：长春理工大学数学与统计学院，吉林长春
关键词: 非牛顿流；化学活性流体；存在性；唯一性；Non-Newtonian Fluids； Chemically Active Fluid； Existence； Uniqueness

摘要: 本文拟在三维光滑有界区域Ω中，考虑一类稳态非牛顿化学活性流体运动方程组的第一边值问题。在外力项某一范数适当小的条件下，用迭代方法证明了当指数

时方程组正则解的存在唯一性。

Abstract: In this paper, we consider the first boundary value problem for a class of steady non-Newtonian chemically active fluid equations in a three-dimensional smooth bounded domain Ω. When exponent

, the existence and uniqueness of the regular solutions for the problem was proved by iterative method under the condition that the norm of the external force term is properly small.

文章引用：王长佳, 苏日娜. 一类化学活性流体方程组解的存在唯一性[J]. 应用数学进展, 2024, 13(4): 1292-1307. https://doi.org/10.12677/aam.2024.134119

1. 引言

热溶质对流，又称双扩散流，是指浮力仅由温度梯度和浓度梯度同时产生的密度差异引起的流动，在海洋学、大气科学和研究恒星大气不稳定性等不同领域有许多应用。自然界中有许多热溶质对流的例子，例如，在海洋学中，水流可以由热和盐度梯度共同驱动，由此产生的热溶质对流已被证明会导致盐泉和盐指的形成。自然界中热溶质对流的其他例子包括地壳中的地幔流动和由温度和水浓度差异驱动的大气流动。对于牛顿不可压缩流体，Boussinesq模型假设密度随温度和浓度在Navier-Stokes动量方程的波动项中呈线性变化，加上温度和浓度的输运方程，得到一个紧密耦合的高度非线性偏微分方程组。

[1] 证明了一个化学活性流体运动的强局部解的存在唯一性并得到了解的估计。随后， [2] 基于一个标准Galerkin近似方程组的完整系统证明了前文所得解的经典正则性。 [3] 证明了 $ℝ^{n} (n \geq 2)$ 上有界区域上化学活性流体运动的弱解的存在性。 [4] 利用谱Galerkin方法，在不假设外力随时间衰减的情况下，证明了化学反应流体运动的强解在时间上的整体存在性，并得到了解的一致时间估计。 [5] 证明了Navier-Stokes耦合系统和反应扩散方程(温度和质量分数)的整体解的存在性。

当不考虑温度变化(即 $T = 0$ )时， [6] 研究了非牛顿不可压缩化学反应流体的非定常运动的非线性偏微分方程组，证明了非稳态模型整体弱解的存在性。当化学反应不发生时(即 $ψ = 0$ )，方程组退化成为非牛顿Boussinesq系统。由于Boussinesq方程组中温度和速度耦合所产生的困难，所以与不可压非牛顿流体相比，有关这类方程组各类初边值问题解的研究还较少。 [7] 研究了一类修正的Boussinesq近似的周期初值问题和初值问题，对于 $p > 2 n / (n + 2)$ 证明了其弱解的存在性，唯一性和正则性。随后， [8] 证明了 $p \geq 2$ 时弱解的存在性，以及 $p > (1 + 2 n) / (n + 2)$ 时弱解的唯一性及正则性。 [9] 考虑一类稳态不可压缩非牛顿Boussinesq方程组的第一边值问题。在外力项某一范数适当小的条件下，证明了当指数 $p \in (1, 2)$ 时方程组正则解的存在唯一性。 [10] 在三维空间中讨论一类各向异性非牛顿Boussinesq方程组的初边值问题，证明了弱解的存在性。 [11] 在三维光滑有界区域中研究了奇异情况下的非牛顿Boussinesq方程组的周期初边值问题，在外力项适当小的情况下，证明了正则解的存在性。目前，关于Boussinesq方程组的研究主要集中在牛顿流体，关于非牛顿Boussinesq方程组的研究还比较少。

本文拟在三维光滑有界区域 $Ω$ 中考虑一类非牛顿化学活性流体方程组正则解的存在唯一性，在Oberbeck-Boussinesq下，该类模型可描述如下：

${\begin{cases} - ν d i v [{(1 + | D u |)}^{p - 2} D u] + \frac{1}{ρ} \nabla P + (u \cdot \nabla) u = [β_{T} (T - T_{c}) + β_{ψ} (ψ - ψ_{c})] g + Q_{1}, x \in Ω, \\ - α Δ T + (u \cdot \nabla) T = Q_{2}, x \in Ω, \\ - D Δ ψ + (u \cdot \nabla) ψ = Q_{3}, x \in Ω, \\ d i v u = 0, x \in Ω, \\ {u |}_{\partial Ω} = 0, {T |}_{\partial Ω} = 0, {ψ |}_{\partial Ω} = 0. \end{cases}$ (1)

这里 $1 < p < 2$ ， $D u = (\nabla u + \nabla u^{T}) / 2$ ，未知函数 $u = (u_{1}, u_{2}, u_{3}) \in ℝ^{3}$ 为流体速度， $P \in ℝ$ 为压力， $T \in ℝ$ 为反应物–产物混合物的温度， $ψ \in ℝ$ 为一步不可逆放热反应中反应物的浓度。 $Q_{1} (x)$ ， $Q_{2} (x)$ ， $Q_{3} (x)$ 及 $g (x)$ 是给定源函数(通常， $g = g e$ ，其中g是自由落体加速度， $e = (0, 0, 1)$ )。 $T_{c}$ 和 $ψ_{c}$ 为特征温度和浓度， $ν$ 为运动粘度， $ρ$ 为平均密度，D为扩散系数， $ρ$ 为导热系数。假设 $β_{T}$ ， $T_{c}$ ， $β_{ψ}$ 和 $ψ_{c}$ 为常数。

2. 预备知识与主要结论

本文用 $L^{q} (Ω) (1 < q < \infty)$ 表示通常的Lebesgue空间，其范数表示为 ${‖ \cdot ‖}_{q, Ω}$ ， $W^{m . q} (Ω)$ 表示通常的Sobolev空间，其范数表示为 ${‖ \cdot ‖}_{m, q; Ω}$ ；空间 $W_{0}^{m, q} (Ω)$ 表示 $C_{0}^{\infty} (Ω)$ 在范数 ${‖ \cdot ‖}_{m, q; Ω}$ 意义下的闭包； $W_{0}^{- 1, q} (Ω)$ 表示 $W_{0}^{m, q} (Ω)$ 的对偶空间，其范数表示为 ${‖ \cdot ‖}_{- 1, q; Ω}$ ；令 $M ≐ {u | u \in C_{o}^{\infty} (Ω), d i v u = 0}$ ，用 $V_{q} (Ω)$ 表示M在范数 ${‖ \cdot ‖}_{1, q}$ 意义下的闭包，特别地，当 $q = 2$ 时，简记为V。用 $C^{m, γ} (\bar{Ω}) (m \in ℕ^{+}, γ \in (0, 1))$ 表示通常意义下的Hölder空间，其上范数表示为 ${‖ \cdot ‖}_{C^{m, γ} (\bar{Ω})}$ 。

定义1如果以下积分等式成立：

$\begin{array}{l} \int_{Ω} ν {(1 + | D u |)}^{p - 2} D u D φ d x = \int_{Ω} [β_{T} (T - T_{c}) + β_{ψ} (ψ - ψ_{c})] g φ d x \\ + \int_{Ω} Q_{1} φ d x - \int_{Ω} (u \cdot \nabla) u \cdot φ d x, \forall φ \in V (Ω), \end{array}$ (2)

$\int_{Ω} α \nabla T \cdot \nabla η d x = - \int_{Ω} (u \cdot \nabla) T \cdot η d x + \int_{Ω} Q_{2} η d x, \forall η \in C_{0}^{\infty} (Ω),$ (3)

$\int_{Ω} D \nabla ψ \cdot \nabla ϕ d x = - \int_{Ω} (u \cdot \nabla) ψ \cdot ϕ d x + \int_{Ω} Q_{3} ϕ d x, \forall ϕ \in C_{0}^{\infty} (Ω),$ (4)

则称 $u \in V_{q}$ ， $T \in W_{0}^{1, 2} (Ω)$ ， $ψ \in W_{0}^{1, 2} (Ω)$ 为问题(1)的弱解。

注1应用deRham定理 [12] ，下文把 $(u, T, ψ)$ 和 $(u, P, T, ψ)$ 都称为问题(1)的解。

引理1 [13] 对任意 $q \geq 1$ ，存在常数 $c_{3}$ ，使对所有 $v \in V_{q} (Ω)$ ，成立 ${‖ v ‖}_{q} + {‖ \nabla v ‖}_{q} \leq c_{3} {‖ D v ‖}_{q}$ 。

引理2 [14] 若g满足 $\nabla g \in W^{- 1, q} (Ω)$ ，则 $g \in L^{q} (Ω)$ ， ${‖ g ‖}_{L_{#}^{q}} \leq c_{4} {‖ \nabla g ‖}_{- 1, q}$ ，其中 $L_{#}^{q} = L^{q} / ℝ$ 。

引理3 [15] 对任意给定实数 $ξ, η \geq 0$ 及 $1 < p < 2$ ，成立不等式：

$| \frac{1}{{(1 + ξ)}^{2 - p}} - \frac{1}{{(1 + η)}^{2 - p}} | \leq (2 - p) | ξ - η |$ .

引理4 [16] 对任意张量D，定义 $S (D) \equiv {(1 + | D |)}^{p - 2} D$ ，则有常数 $c_{5}$ ，使对任意一对张量 $D_{1}$ 和 $D_{2}$ ，有：

$(S (D_{1}) - S (D_{2})) \cdot (D_{1} - D_{2}) \geq c_{5} \frac{{| D_{1} - D_{2} |}^{2}}{{(1 + | D_{1} | + | D_{2} |)}^{2 - p}}$ .

本文的主要结论如下：

定理1假设 $p \in (1, 2)$ ， $r > 3$ ， $q > 3$ ， $γ_{0} = 1 - 3 / q$ 。设 $Ω \in ℝ^{3}$ 是光滑有界域， $g, Q_{1} \in L^{q} (Ω)$ ， $Q_{2}, Q_{3} \in L^{2} (Ω)$ 。若 ${‖ g ‖}_{q}$ ， ${‖ Q_{1} ‖}_{q}$ ， ${‖ Q_{2} ‖}_{2}$ ， ${‖ Q_{3} ‖}_{2}$ 足够小，且满足(6)及(21)，那么问题(1)至少有一个解 $(u, P, T, ψ)$ 使得 $u \in C^{1, γ} (\bar{Ω})$ ， $P \in C^{0, γ} (\bar{Ω})$ ， $T \in W^{2, 2} (Ω)$ ， $ψ \in W^{2, 2} (Ω)$ ，并且

${‖ u ‖}_{C^{1, γ} (\bar{Ω})} + {‖ P ‖}_{C^{0, γ} (\bar{Ω})} + {‖ T ‖}_{2, 2} + {‖ ψ ‖}_{2, 2} \leq 2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{4} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q} .$

其中K，K₂，K₃，K₄是正常数。此外，若 $\frac{6}{5} < p < 2$ ，解是唯一的。

3. 定理1的证明

3.1. 近似解的构造与一致估计

首先，令 $u^{- 1} = P^{- 1} = T^{- 1} = ψ^{- 1} = 0$ ，对 $m \geq 0$ ，寻求 $(u^{m}, P^{m}, T^{m}, ψ^{m})$ 满足如下边值问题：

${\begin{cases} - ν d i v [{(1 + | D u^{m - 1} |)}^{p - 2} D u^{m}] + \frac{1}{ρ} \nabla P^{m} = [β_{T} (T^{m} - T_{c}) + β_{ψ} (ψ^{m} - ψ_{c})] g + Q_{1} - (u^{m - 1} \cdot \nabla) u^{m - 1}, x \in Ω, \\ - α Δ T^{m} = - (u^{m - 1} \cdot \nabla) T^{m - 1} + Q_{2}, x \in Ω, \\ - D Δ ψ^{m} = - (u^{m - 1} \cdot \nabla) ψ^{m - 1} + Q_{3}, x \in Ω, \\ d i v u^{m} = 0, x \in Ω, \\ {u^{m} |}_{\partial Ω} = 0, {T^{m} |}_{\partial Ω} = 0, {ψ^{m} |}_{\partial Ω} = 0. \end{cases} $ (5)

命题1假设 $p \in (1, 2)$ ， $q > 3$ ， $γ_{0} = 1 - 3 / q$ ， $Ω$ 是光滑有界域， $g, Q_{1} \in L^{q} (Ω)$ ， $Q_{2}, Q_{3} \in L^{2} (Ω)$ 。则对 $\forall m \in ℕ$ ，问题(5)均存在弱解 $(u^{m}, P^{m}, T^{m}, ψ^{m}) \in C^{1, γ} (\bar{Ω}) \times C^{0, γ} (\bar{Ω}) \times W^{2, 2} (Ω) \times W^{2, 2} (Ω), \forall γ < γ_{0}$ 。

进一步，若 ${‖ g ‖}_{q}$ ， ${‖ Q_{1} ‖}_{q}$ ， ${‖ Q_{2} ‖}_{2}$ ， ${‖ Q_{3} ‖}_{2}$ 满足

${‖ g ‖}_{q} < 1$ ， $K {‖ Q_{1} ‖}_{q} + K_{2} {‖ Q_{2} ‖}_{2} + K_{3} {‖ Q_{3} ‖}_{2} + K_{4} {‖ g ‖}_{q} < \frac{1}{4 K_{1}}$ (6)

则对 $\forall m \in ℕ$ 有：

${‖ u^{m} ‖}_{C^{1, γ_{0}} (\bar{Ω})} + {‖ P^{m} ‖}_{C^{0, γ_{0}} (\bar{Ω})} + {‖ T^{m} ‖}_{2, 2} + {‖ ψ^{m} ‖}_{2, 2} \leq 2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q}$ . (7)

证明令 $I_{m} = {‖ u^{m} ‖}_{C^{1, γ_{0}}} + {‖ P^{m} ‖}_{C^{0, γ_{0}}} + {‖ T^{m} ‖}_{2, 2} + {‖ ψ^{m} ‖}_{2, 2}$ .

首先，显然 $T^{0}$ 满足

${\begin{cases} - α Δ T^{0} = Q_{2}, x \in Ω, \\ {T^{0} |}_{\partial Ω} = 0. \end{cases}$

其中 $Q_{2} \in L^{2} (Ω)$ 。因此由椭圆型方程理论可知，该问题存在解 $T^{0} \in W^{2, 2} (Ω)$ ，且有

${‖ T^{0} ‖}_{2, 2} \leq \frac{c_{2}}{α} {‖ Q_{2} ‖}_{2}$ (8)

其次，函数 $ψ^{0}$ 满足

${\begin{cases} - D Δ ψ^{0} = Q_{3}, x \in Ω, \\ {ψ^{0} |}_{\partial Ω} = 0. \end{cases}$

其中 $Q_{3} \in L^{2} (Ω)$ 。同理可知存在解 $ψ^{0} \in W^{2, 2} (Ω)$ ，并且

${‖ ψ^{0} ‖}_{2, 2} \leq \frac{c_{2}}{D} {‖ Q_{3} ‖}_{2}$ (9)

然后，函数 $u^{0}$ 满足

${\begin{cases} - ν Δ u^{0} + \frac{1}{ρ} \nabla P^{0} = [β_{T} (T^{0} - T_{c}) + β_{ψ} (ψ^{0} - ψ_{c})] g + Q_{1}, x \in Ω, \\ d i v u^{0} = 0, x \in Ω, \\ {u^{0} |}_{\partial Ω} = 0. \end{cases}$

其中 $g \in L^{q} (Ω)$ ， $Q_{1} \in L^{q} (Ω)$ 。由于

${‖ [β_{T} (T^{0} - T_{c}) + β_{ψ} (ψ^{0} - ψ_{c})] g + Q_{1} ‖}_{q} \leq C β_{T} {‖ T^{0} ‖}_{2, 2} {‖ g ‖}_{q} + C β_{ψ} {‖ ψ^{0} ‖}_{2, 2} {‖ g ‖}_{q} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} + {‖ Q_{1} ‖}_{q},$

从而 $[β_{T} (T^{0} - θ_{c}) + β_{ψ} (ψ^{0} - ψ_{c})] g + Q_{1} \in L^{q} (Ω)$ ，由文献 [15] 中的定理3.1知，该问题存在解 $(u^{0}, P^{0}) \in C^{1, γ_{0}} (\bar{Ω}) \times C^{0, γ_{0}} (\bar{Ω})$ ，且满足如下估计

$\begin{matrix} {‖ u^{0} ‖}_{C^{1, γ_{0}}} + {‖ P^{0} ‖}_{C^{0, γ_{0}}} \leq \tilde{c} ({‖ u_{0} ‖}_{1, 2} + {‖ [β_{T} (T^{0} - T_{c}) + β_{ψ} (ψ^{0} - ψ_{c})] g + Q_{1} ‖}_{q}) \\ \leq \tilde{c} [{‖ u_{0} ‖}_{1, 2} + C β_{T} {‖ T^{0} ‖}_{2, 2} {‖ g ‖}_{q} + C β_{ψ} {‖ ψ^{0} ‖}_{2, 2} {‖ g ‖}_{q} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} + {‖ Q_{1} ‖}_{q}], \end{matrix}$ (10)

其中 $\tilde{c} = c_{1} > 1, r > 2$ 。

对上述方程组中第一个等式两端同乘 $u_{0}$ ，并在 $Ω$ 上积分可得

$ν \int_{Ω} {| D u^{0} |}^{2} d x = \int_{Ω} [β_{T} (T^{0} - T_{c}) + β_{ψ} (ψ^{0} - ψ_{c})] g \cdot u^{0} d x + \int_{Ω} Q_{1} \cdot u^{0} d x$ .

由引理1及Hölder不等式，有

$ν \int_{Ω} {| D u^{0} |}^{2} d x = ν {‖ D u^{0} ‖}_{2}^{2} \geq \frac{ν}{c_{3}^{2}} {‖ u^{0} ‖}_{1, 2}^{2}$ ，

$| \int_{Ω} [β_{T} (T^{0} - T_{c}) + β_{ψ} (ψ^{0} - ψ_{c})] g \cdot u^{0} d x | \leq [C β_{T} {‖ T^{0} ‖}_{2, 2} {‖ g ‖}_{q} + C β_{ψ} {‖ ψ^{0} ‖}_{2, 2} {‖ g ‖}_{q} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q}] {‖ u^{0} ‖}_{1, 2}$ ，

$| \int_{Ω} Q_{1} \cdot u^{0} d x | \leq {‖ Q_{1} ‖}_{2} {‖ u^{0} ‖}_{2} \leq {‖ Q_{1} ‖}_{q} {‖ u^{0} ‖}_{1, 2}$ .

联立不等式，可得

${‖ u^{0} ‖}_{1, 2} \leq \frac{c_{3}^{2}}{ν} [C β_{T} {‖ T^{0} ‖}_{2, 2} {‖ g ‖}_{q} + C β_{ψ} {‖ ψ^{0} ‖}_{2, 2} {‖ g ‖}_{q} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} + {‖ Q_{1} ‖}_{q}]$ .

代入(10)中，可以得到

${‖ u^{0} ‖}_{C^{1, γ_{0}}} + {‖ P^{0} ‖}_{C^{0, γ_{0}}} \leq c_{1} (\frac{c_{3}^{2}}{ν} + 1) [\frac{C c_{2} β_{T}}{α} {‖ Q_{2} ‖}_{q} {‖ g ‖}_{q} + \frac{C c_{2} β_{ψ}}{D} {‖ Q_{3} ‖}_{q} {‖ g ‖}_{q} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} + {‖ Q_{1} ‖}_{q}] .$ (11)

最后，结合式(8)，(9)及(11)可得

$\begin{matrix} I_{0} = {‖ u^{0} ‖}_{C^{1, γ_{0}}} + {‖ P^{0} ‖}_{C^{0, γ_{0}}} + {‖ T^{0} ‖}_{2, 2} + {‖ ψ^{0} ‖}_{2, 2} \\ \leq c_{1} (\frac{c_{3}^{2}}{ν} + 1) (\frac{C c_{2} β_{T}}{α} {‖ Q_{2} ‖}_{2} + \frac{C c_{2} β_{ψ}}{D} {‖ Q_{3} ‖}_{2} + β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} + c_{1} (\frac{c_{3}^{2}}{ν} + 1) {‖ Q_{1} ‖}_{q} + \frac{c_{2}}{α} {‖ Q_{2} ‖}_{2} + \frac{c_{2}}{D} {‖ Q_{3} ‖}_{2} . \end{matrix}$ (12)

设 $m \geq 1$ ，假设 $(u^{m}, P^{m}, T^{m}, ψ^{m}) \in C^{1, γ} (\bar{Ω}) \times C^{0, γ} (\bar{Ω}) \times W^{2, 2} (Ω) \times W^{2, 2} (Ω)$ 是问题(5)的解。

对 $m + 1$ ，由于 $T^{m} \in W^{2, 2} (Ω)$ ， $u^{m} \in C^{1, γ} (\bar{Ω})$ ， $Q_{2} \in L^{2} (Ω)$ ，可得到估计 ${‖ (u^{m} \cdot \nabla) T^{m} ‖}_{2} \leq C {‖ u^{m} ‖}_{C^{1, γ_{0}}} {‖ T^{m} ‖}_{2, 2}$ ，从而有 $- \frac{1}{α} (u^{m} \cdot \nabla) T^{m} + \frac{1}{α} Q_{2} \in L^{2} (Ω)$ 。从而由椭圆型方程理论可知，问题

${\begin{cases} - α Δ T^{m + 1} = - (u^{m} \cdot \nabla) T^{m} + Q_{2}, x \in Ω, \\ {T^{m + 1} |}_{\partial Ω} = 0. \end{cases}$

存在解 $T^{m + 1} \in W^{2, 2} (Ω)$ ，且有

${‖ T^{m + 1} ‖}_{2, 2} \leq c_{2} {‖ - \frac{1}{α} (u^{m} \cdot \nabla) T^{m} + \frac{Q_{2}}{α} ‖}_{2} \leq \frac{c_{2}}{α} (C {‖ u^{m} ‖}_{C^{1, γ_{0}}} {‖ T^{m} ‖}_{2, 2} + {‖ Q_{2} ‖}_{2}) .$ (13)

然后，对于边值问题

${\begin{cases} - D Δ ψ^{m + 1} = - (u^{m} \cdot \nabla) ψ^{m} + Q_{3}, x \in Ω, \\ {ψ^{m + 1} |}_{\partial Ω} = 0, \end{cases}$

同样可得弱解 $ψ^{m + 1} \in W^{2, 2} (Ω)$ ，使得

${‖ ψ^{m + 1} ‖}_{2, 2} \leq \frac{c_{2}}{D} (C {‖ u^{m} ‖}_{C^{1, γ_{0}}} {‖ ψ^{m} ‖}_{2, 2} + {‖ Q_{3} ‖}_{2}) .$ (14)

接着，由于 $g \in L^{q} (Ω)$ ， $Q_{1} \in L^{q} (Ω)$ ，而

$\begin{array}{l} {‖ [β_{T} (T^{m + 1} - T_{c}) + β_{ψ} (ψ^{m + 1} - ψ_{c})] g + Q_{1} + (u^{m} \cdot \nabla) u^{m} ‖}_{q} \\ \leq C β_{T} {‖ T^{m + 1} ‖}_{2, 2} {‖ g ‖}_{q} + C β_{ψ} {‖ ψ^{m + 1} ‖}_{2, 2} {‖ g ‖}_{q} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} + {‖ Q_{1} ‖}_{q} + C {‖ u^{m} ‖}_{C^{1, γ_{0}}}^{2}, \end{array}$

从而 $[β_{T} (T^{m + 1} - T_{c}) + β_{ψ} (ψ^{m + 1} - ψ_{c})] g + Q_{1} - (u^{m} \cdot \nabla) u^{m} \in L^{q} (Ω)$ 。由文献 [15] 可知，问题

${\begin{cases} - ν d i v [{(1 + | D u^{m} |)}^{p - 2} D u^{m + 1}] + \frac{1}{ρ} \nabla P^{m + 1} = [β_{T} (T^{m + 1} - T_{c}) + β_{ψ} (ψ^{m + 1} - ψ_{c})] g + Q_{1} - (u^{m} \cdot \nabla) u^{m}, x \in Ω, \\ d i v u^{m + 1} = 0, x \in Ω, \\ {u^{m + 1} |}_{\partial Ω} = 0. \end{cases} $

有弱解 $(u^{m + 1}, P^{m + 1}) \in C^{1, γ} (\bar{Ω}) \times C^{0, γ} (\bar{Ω})$ ，且有

$\begin{array}{l} {‖ u^{m + 1} ‖}_{C^{1, γ_{0}}} + {‖ P^{m + 1} ‖}_{C^{0, γ_{0}}} \\ \leq \tilde{c} ({‖ u^{m + 1} ‖}_{1, 2} + {‖ [β_{T} (T^{m + 1} - T_{c}) + β_{ψ} (ψ^{m + 1} - ψ_{c})] g + Q_{1} - (u^{m} \cdot \nabla) u^{m} ‖}_{q}) \\ \leq \tilde{c} ({‖ u^{m + 1} ‖}_{1, 2} + C β_{T} {‖ T^{m + 1} ‖}_{2, 2} {‖ g ‖}_{q} + C β_{ψ} {‖ ψ^{m + 1} ‖}_{2, 2} {‖ g ‖}_{q} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} + {‖ Q_{1} ‖}_{q} + C {‖ u^{m} ‖}_{C^{1, γ_{0}}}^{2}) . \end{array}$ (15)

其中 $\tilde{c} = c_{1} {(1 + {‖ u^{m} ‖}_{C^{1, γ_{0}}})}^{r}, c_{1} > 1, r > 2$ 。

对上述方程组的第一个等式两边同乘 $u^{m + 1}$ ，并在 $Ω$ 上积分可得

$\begin{array}{l} \int_{Ω} ν {(1 + | D u^{m} |)}^{p - 2} {| D u^{m + 1} |}^{2} d x \\ = \int_{Ω} [β_{T} (T^{m + 1} - T_{c}) + β_{ψ} (ψ^{m + 1} - ψ_{c})] g u^{m + 1} d x + \int_{Ω} Q_{1} u^{m + 1} d x - \int_{Ω} (u^{m} \cdot \nabla) u^{m} \cdot u^{m + 1} d x . \end{array}$

由引理1，可得

$\int_{Ω} ν {(1 + | D u^{m} |)}^{p - 2} {| D u^{m + 1} |}^{2} d x \geq ν {(1 + {‖ D u^{m} ‖}_{\infty})}^{p - 2} {‖ D u^{m + 1} ‖}_{2}^{2} \geq \frac{1}{c_{3}^{2}} ν {(1 + {‖ u^{m} ‖}_{C^{1, γ_{0}}})}^{p - 2} {‖ u^{m + 1} ‖}_{1, 2}^{2},$

由Hölder不等式，可得

$\begin{array}{l} | \int_{Ω} [β_{T} (T^{m + 1} - T_{c}) + β_{ψ} (ψ^{m + 1} - ψ_{c})] g \cdot u^{m + 1} d x + \int_{Ω} Q_{1} \cdot u^{m + 1} d x - \int_{Ω} (u^{m} \cdot \nabla) u^{m} \cdot u^{m + 1} d x | \\ \leq {‖ [β_{T} (T^{m + 1} - T_{c}) + β_{ψ} (ψ^{m + 1} - ψ_{c})] g ‖}_{q} {‖ u^{m + 1} ‖}_{2} + {‖ Q_{1} ‖}_{q} {‖ u^{m + 1} ‖}_{2} + {‖ (u^{m} \cdot \nabla) u^{m} ‖}_{q} {‖ u^{m + 1} ‖}_{2} \\ \leq C β_{T} {‖ T^{m + 1} ‖}_{2, 2} {‖ g ‖}_{q} {‖ u^{m + 1} ‖}_{1, 2} + C β_{ψ} {‖ ψ^{m + 1} ‖}_{2, 2} {‖ g ‖}_{q} {‖ u^{m + 1} ‖}_{1, 2} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} {‖ u^{m + 1} ‖}_{1, 2} \\ + {‖ Q_{1} ‖}_{q} {‖ u^{m + 1} ‖}_{1, 2} + C {‖ u^{m} ‖}_{C^{1, γ_{0}}}^{2} {‖ u^{m + 1} ‖}_{1, 2} . \end{array}$

联立不等式，有

$\begin{array}{l} {‖ u^{m + 1} ‖}_{1, 2} \\ \leq \frac{c_{3}^{2}}{ν} {(1 + {‖ u^{m} ‖}_{C^{1, γ_{0}}})}^{2 - p} [C β_{T} {‖ T^{m + 1} ‖}_{2, 2} {‖ g ‖}_{q} + C β_{ψ} {‖ ψ^{m + 1} ‖}_{2, 2} {‖ g ‖}_{q} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} + {‖ Q_{1} ‖}_{q} + C {‖ u^{m} ‖}_{C^{1, γ_{0}}}^{2}] . \end{array}$

代入(15)中，有

$\begin{array}{l} {‖ u^{m + 1} ‖}_{C^{1, γ_{0}}} + {‖ P^{m + 1} ‖}_{C^{0, γ_{0}}} \\ \leq c_{1} {(1 + {‖ u^{m} ‖}_{C^{1, γ_{0}}})}^{r} {\frac{c_{3}^{2}}{ν} {(1 + {‖ u^{m} ‖}_{C^{1, γ_{0}}})}^{2 - p} [C β_{T} {‖ T^{m + 1} ‖}_{2, 2} {‖ g ‖}_{q} + C β_{ψ} {‖ ψ^{m + 1} ‖}_{2, 2} {‖ g ‖}_{q} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} \\ \overset{}{} + {‖ Q_{1} ‖}_{q} + C {‖ u^{m} ‖}_{C^{1, γ_{0}}}^{2}] + C β_{T} {‖ T^{m + 1} ‖}_{2, 2} {‖ g ‖}_{q} + C β_{ψ} {‖ ψ^{m + 1} ‖}_{2, 2} {‖ g ‖}_{q} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} + {‖ Q_{1} ‖}_{q} + C {‖ u^{m} ‖}_{C^{1, γ_{0}}}^{2}} \\ \leq c_{1} (\frac{c_{3}^{2}}{ν} + 1) {(1 + {‖ u^{m} ‖}_{C^{1, γ_{0}}})}^{r + 2 - p} [\frac{C c_{2} β_{T}}{α} {‖ g ‖}_{q} (C {‖ u^{m} ‖}_{C^{1, γ_{0}}} {‖ T^{m} ‖}_{2, 2} + {‖ Q_{2} ‖}_{2}) \\ + \frac{C c_{2} β_{ψ}}{D} {‖ g ‖}_{q} (C {‖ u^{m} ‖}_{C^{1, γ_{0}}} {‖ ψ^{m} ‖}_{2, 2} + {‖ Q_{3} ‖}_{2}) + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} + {‖ Q_{1} ‖}_{q} + C {‖ u^{m} ‖}_{C^{1, γ_{0}}}^{2}] . \end{array}$ (16)

联合(13)，(14)和(16)，有

$\begin{matrix} I_{m + 1} = {‖ u^{m + 1} ‖}_{C^{1, γ_{0}}} + {‖ P^{m + 1} ‖}_{C^{0, γ_{0}}} + {‖ T^{m + 1} ‖}_{2, 2} + {‖ ψ^{m + 1} ‖}_{2, 2} \\ \leq c_{1} (\frac{c_{3}^{2}}{ν} + 1) {(1 + I_{m})}^{r + 2 - p} [\frac{C c_{2} β_{T}}{α} {‖ g ‖}_{q} {‖ Q_{2} ‖}_{2} + \frac{C c_{2} β_{ψ}}{D} {‖ g ‖}_{q} {‖ Q_{3} ‖}_{2} + {‖ Q_{1} ‖}_{q} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} \\ + (\frac{C c_{2} β_{T}}{α} {‖ g ‖}_{q} + \frac{C c_{2} β_{ψ}}{D} {‖ g ‖}_{q} + C) I_{m}^{2}] + (\frac{C c_{2}}{α} + \frac{C c_{2}}{D}) I_{m}^{2} + \frac{c_{2}}{α} {‖ Q_{2} ‖}_{2} + \frac{c_{2}}{D} {‖ Q_{3} ‖}_{2} . \end{matrix}$ (17)

下面用归纳法证明估计式(7)。对任意 $t \geq 0$ ，构造函数

$\begin{array}{l} F (t) = c_{1} (\frac{c_{3}^{2}}{ν} + 1) {(1 + t)}^{r + 2 - p} [\frac{C c_{2} β_{T}}{α} {‖ g ‖}_{q} {‖ Q_{2} ‖}_{2} + \frac{C c_{2} β_{ψ}}{D} {‖ g ‖}_{q} {‖ Q_{3} ‖}_{2} + {‖ Q_{1} ‖}_{q} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} \\ + (\frac{C c_{2} β_{T}}{α} {‖ g ‖}_{q} + \frac{C c_{2} β_{ψ}}{D} {‖ g ‖}_{q} + C) t^{2}] + (\frac{C c_{2}}{α} + \frac{C c_{2}}{D}) t^{2} + \frac{c_{2}}{α} {‖ Q_{2} ‖}_{2} + \frac{c_{2}}{D} {‖ Q_{3} ‖}_{2} - t . \end{array}$

考察 $F (t)$ 在 $[0, 1]$ 上根的情况。对 $0 \leq t \leq 1$ ，有

$\begin{array}{l} F (t) \leq K [\frac{C c_{2} β_{T}}{α} {‖ g ‖}_{q} {‖ Q_{2} ‖}_{2} + \frac{C c_{2} β_{ψ}}{D} {‖ g ‖}_{q} {‖ Q_{3} ‖}_{2} + {‖ Q_{1} ‖}_{q} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} + (\frac{C c_{2} β_{T}}{α} {‖ g ‖}_{q} \\ + \frac{C c_{2} β_{ψ}}{D} {‖ g ‖}_{q} + C) t^{2}] + (\frac{C c_{2}}{α} + \frac{C c_{2}}{D}) t^{2} + \frac{c_{2}}{α} {‖ Q_{2} ‖}_{2} + \frac{c_{3}}{D} {‖ Q_{3} ‖}_{2} - t \\ \leq [K (\frac{C c_{2} β_{T}}{α} + \frac{C c_{2} β_{ψ}}{D} + C) + \frac{C c_{2}}{α} + \frac{C c_{2}}{D}] t^{2} - t + K {‖ Q_{1} ‖}_{q} + \frac{K C c_{2} β_{T} + c_{2}}{α} {‖ Q_{2} ‖}_{2} \\ + \frac{K C c_{2} β_{ψ} + c_{2}}{D} {‖ Q_{3} ‖}_{2} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} . \end{array}$

其中 $K = c_{1} (\frac{c_{3}^{2}}{ν} + 1) \cdot 2^{r + 2 - p} > 1$ 。简记 $K_{1} = K (\frac{C c_{2} β_{T}}{α} + \frac{C c_{2} β_{ψ}}{D} + C) + \frac{C c_{2}}{α} + \frac{C c_{2}}{D}$ ， $K_{2} = \frac{K C c_{2} β_{T} + c_{2}}{α}$ ， $K_{3} = \frac{K C c_{2} β_{ψ} + c_{2}}{D}$ ， $K_{4} = β_{T} T_{c} + β_{ψ} ψ_{c}$ ，则

$F (t) \leq K_{1} t^{2} - t + K {‖ Q_{1} ‖}_{q} + K_{2} {‖ Q_{2} ‖}_{2} + K_{3} {‖ Q_{3} ‖}_{2} + K_{4} {‖ g ‖}_{q} ≜ G (t) .$

注意到当且仅当 $Δ = 1 - 4 K_{1} (K {‖ Q_{1} ‖}_{q} + K_{2} {‖ Q_{2} ‖}_{2} + K_{3} {‖ Q_{3} ‖}_{2} + K_{4} {‖ g ‖}_{q}) > 0$ 时，函数 $G (t)$ 有两个正根 $s_{1} < s_{2}$ ，此时 $0 < s_{1} = \frac{1 - \sqrt{Δ}}{2 K_{1}} < 1$ 。因为 $F (0) > 0$ ， $F (t) \leq G (t)$ ，在 $t \in [0, 1]$ 时存在 $t_{1} (0 < t_{1} < s_{1})$ ，使 $F (t_{1}) = 0$ ，即

$\begin{array}{l} c_{1} (\frac{c_{3}^{2}}{ν} + 1) {(1 + t_{1})}^{r + 2 - p} [\frac{C c_{2} β_{T}}{α} {‖ g ‖}_{q} {‖ Q_{2} ‖}_{2} + \frac{C c_{2} β_{ψ}}{D} {‖ g ‖}_{q} {‖ Q_{3} ‖}_{2} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} + {‖ Q_{1} ‖}_{q} \\ + (\frac{C c_{2} β_{T}}{α} {‖ g ‖}_{q} + \frac{C c_{2} β_{ψ}}{D} {‖ g ‖}_{q} + C) t_{1}^{2}] + (\frac{C c_{2}}{α} + \frac{C c_{2}}{D}) t_{1}^{2} + \frac{c_{2}}{α} {‖ Q_{2} ‖}_{2} + \frac{c_{2}}{D} {‖ Q_{3} ‖}_{2} - t_{1} = 0. \end{array}$

因为 $t_{1} > 0$ ，则

$c_{1} (\frac{c_{3}^{2}}{ν} + 1) [\frac{C c_{2} β_{T}}{α} {‖ g ‖}_{q} {‖ Q_{2} ‖}_{2} + \frac{C c_{2} β_{ψ}}{D} {‖ g ‖}_{q} {‖ Q_{3} ‖}_{2} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} + {‖ Q_{1} ‖}_{q}] + \frac{c_{2}}{α} {‖ Q_{2} ‖}_{2} + \frac{c_{2}}{D} {‖ Q_{3} ‖}_{2} - t_{1} \leq 0.$

结合(12)，有

$t_{1} \geq c_{1} (\frac{c_{3}^{2}}{ν} + 1) [\frac{C c_{2} β_{T}}{α} {‖ g ‖}_{q} {‖ Q_{2} ‖}_{2} + \frac{C c_{2} β_{ψ}}{D} {‖ g ‖}_{q} {‖ Q_{3} ‖}_{2} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} + {‖ Q_{1} ‖}_{q}] + \frac{c_{2}}{α} {‖ Q_{2} ‖}_{2} + \frac{c_{2}}{D} {‖ Q_{3} ‖}_{2} \geq I_{0} .$

如果假设 $I_{m} \leq t_{1}$ ，通过不等式(17)，及 $F (t_{1}) = 0$ ，可得

$\begin{array}{l} I_{m + 1} \leq c_{1} (\frac{c_{3}^{2}}{ν} + 1) {(1 + I_{m})}^{r + 2 - p} [\frac{C c_{2} β_{T}}{α} {‖ g ‖}_{q} {‖ Q_{2} ‖}_{2} + \frac{C c_{2} β_{ψ}}{D} {‖ g ‖}_{q} {‖ Q_{3} ‖}_{2} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} \\ + {‖ Q_{1} ‖}_{q} + (\frac{C c_{2} β_{T}}{α} {‖ g ‖}_{q} + \frac{C c_{2} β_{ψ}}{D} {‖ g ‖}_{q} + C) I_{m}^{2}] + (\frac{C c_{2}}{α} + \frac{C c_{2}}{D}) I_{m}^{2} + \frac{c_{2}}{α} {‖ Q_{2} ‖}_{2} + \frac{c_{2}}{D} {‖ Q_{3} ‖}_{2} \\ \leq c_{1} (\frac{c_{3}^{2}}{ν} + 1) {(1 + t_{1})}^{r + 2 - p} [\frac{C c_{2} β_{T}}{α} {‖ g ‖}_{q} {‖ Q_{2} ‖}_{2} + \frac{C c_{2} β_{ψ}}{D} {‖ g ‖}_{q} {‖ Q_{3} ‖}_{2} + (β_{T} T_{c} + β_{ψ} ψ_{c}) {‖ g ‖}_{q} \\ + {‖ Q_{1} ‖}_{q} + (\frac{C c_{2} β_{T}}{α} {‖ g ‖}_{q} + \frac{C c_{2} β_{ψ}}{D} {‖ g ‖}_{q} + C) t_{1}^{2}] + (\frac{C c_{2}}{α} + \frac{C c_{2}}{D}) t_{1}^{2} + \frac{c_{2}}{α} {‖ Q_{2} ‖}_{2} + \frac{c_{2}}{D} {‖ Q_{3} ‖}_{2} \\ \leq F (t_{1}) + t_{1} = t_{1} . \end{array}$

综上， $I_{m} \leq t_{1} < s_{1} < \frac{1}{2 K_{1}} < 2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q} \leq 1, (m \in ℕ)$ ，命题得证。

3.2. 近似解的收敛性

对 $\forall j \in ℕ$ ，设 $Q^{j + 1} (x) = u^{j + 1} (x) - u^{j} (x)$ ， $R^{j + 1} (x) = P^{j + 1} (x) - P^{j} (x)$ ， $J^{j + 1} (x) = T^{j + 1} (x) - T^{j} (x)$ ， $W^{j + 1} (x) = ψ^{j + 1} (x) - ψ^{j} (x)$ 。则对 $\forall φ \in C_{0}^{\infty} (Ω)$ ， $\forall η \in C_{0}^{\infty} (Ω)$ ，及 $\forall ϕ \in C_{0}^{\infty} (Ω)$ ，有

$\begin{array}{l} \int_{Ω} ν {(1 + | D u^{j} |)}^{p - 2} D Q^{j + 1} D φ d x \\ = \int_{Ω} ν [{(1 + | D u^{j - 1} |)}^{p - 2} - {(1 + | D u^{j} |)}^{p - 2}] D u^{j} D φ d x + \int_{Ω} (β_{T} J^{j + 1} + β_{ψ} W^{j + 1}) g \cdot φ d x \\ - \int_{Ω} (Q^{j} \cdot \nabla) u^{j} φ d x - \int_{Ω} (u^{j - 1} \cdot \nabla) Q^{j} φ d x + \int_{Ω} \frac{1}{ρ} R^{j + 1} d i v φ d x . \end{array}$ (18)

$\int_{Ω} α \nabla J^{j + 1} \cdot \nabla η d x = - \int_{Ω} (u^{j} \cdot \nabla) J^{j} η d x - \int_{Ω} (Q^{j} \cdot \nabla) T^{j - 1} η d x$ (19)

$\int_{Ω} D \nabla W^{j + 1} \cdot \nabla ϕ d x = - \int_{Ω} (u^{j} \cdot \nabla) W^{j} ϕ d x - \int_{Ω} (Q^{j} \cdot \nabla) ψ^{j - 1} ϕ d x$ (20)

命题2 假设命题1的条件均满足， ${u^{m}}$ ， ${P^{m}}$ ， ${T^{m}}$ ， ${ψ^{m}}$ 为相应的序列。若

$\begin{array}{l} (2 - p + 2 c_{3}^{2} + 2 c_{3}) (2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q}) \\ \times {(1 + 2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q})}^{2 - p} < 1. \end{array}$ (21)

其中K在命题1中定义。则级数 $\sum_{m} Q^{m}$ ， $\sum_{m} J^{m}$ 及 $\sum_{m} W^{m}$ 分别收敛到 $W^{1, 2} (Ω)$ 中的函数Q，J和W，级数 $\sum_{m} R^{m}$ 收敛到 $L^{2} (Ω)$ 中的函数R。

证明先证明如下估计：

$\begin{array}{l} {‖ D Q^{1} ‖}_{2} + {‖ J^{1} ‖}_{1, 2} + {‖ W^{1} ‖}_{1, 2} \leq \frac{1}{ν} {(1 + 2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q})}^{2 - p} (2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} \\ {+ 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q})}^{2} (2 ν - p ν + C c_{3} + \frac{C c_{3} β_{T}}{α} {‖ g ‖}_{q} + \frac{C c_{3} β_{ψ}}{D} {‖ g ‖}_{q} + \frac{C}{α} + \frac{C}{D}); \end{array}$

b) 若对 $j \geq 1$ ，有

$\begin{array}{l} {‖ D Q^{j} ‖}_{2} + {‖ J^{j} ‖}_{1, 2} + {‖ W^{j} ‖}_{1, 2} \\ \leq \frac{(4 ν - 2 p ν + 2 c_{3}^{2} + 2 c_{3} + \frac{2 C c_{3} β_{T}}{α} {‖ g ‖}_{q} + \frac{2 C c_{3} β_{ψ}}{D} {‖ g ‖}_{q} + \frac{C β_{T}}{α} {‖ g ‖}_{q} + \frac{C c_{3} β_{T}}{α} {‖ g ‖}_{q}}{2 - p + 2 c_{3}^{2} + 2 c_{3}} \\ \frac{+ \frac{C β_{ψ}}{D} {‖ g ‖}_{q} + \frac{C c_{3} β_{ψ}}{D} {‖ g ‖}_{q}) (2 ν - p ν + C c_{3} + \frac{C c_{3} β_{T}}{α} {‖ g ‖}_{q} + \frac{C c_{3} β_{ψ}}{D} {‖ g ‖}_{q} + \frac{C}{α} + \frac{C}{D})}{2 - p + 2 c_{3}^{2} + 2 c_{3}} \\ \times (1 + 2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q}) \cdot {(2 - p + 2 c_{3}^{2} + 2 c_{3}) (2 K {‖ Q_{1} ‖}_{q} \overset{}{} \\ {+ 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q}) {(1 + 2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q})}^{2 - p}}}^{j}, \end{array}$

那么，

$\begin{array}{l} {‖ D Q^{j + 1} ‖}_{2} + {‖ J^{j + 1} ‖}_{1, 2} + {‖ W^{j + 1} ‖}_{1, 2} \\ \leq \frac{(4 ν - 2 p ν + 2 c_{3}^{2} + 2 c_{3} + \frac{2 C c_{3} β_{T}}{α} {‖ g ‖}_{q} + \frac{2 C c_{3} β_{ψ}}{D} {‖ g ‖}_{q} + \frac{C β_{T}}{α} {‖ g ‖}_{q} + \frac{C c_{3} β_{T}}{α} {‖ g ‖}_{q}}{2 - p + 2 c_{3}^{2} + 2 c_{3}} \\ \frac{+ \frac{C β_{ψ}}{D} {‖ g ‖}_{q} + \frac{C c_{3} β_{ψ}}{D} {‖ g ‖}_{q}) (2 ν - p ν + C c_{3} + \frac{C c_{3} β_{T}}{α} {‖ g ‖}_{q} + \frac{C c_{3} β_{ψ}}{D} {‖ g ‖}_{q} + \frac{C}{α} + \frac{C}{D})}{2 - p + 2 c_{3}^{2} + 2 c_{3}} \\ \times (1 + 2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q}) \cdot {(2 - p + 2 c_{3}^{2} + 2 c_{3}) (2 K {‖ Q_{1} ‖}_{q} \overset{}{} \\ {+ 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q}) {(1 + 2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q})}^{2 - p}}}^{j + 1} . \end{array}$ (22)

下面证明a)。设 $j = 0$ ，在(18)中取 $φ = Q^{1}$ 作检验函数，可得

$\begin{array}{l} \int_{Ω} ν {(1 + | D u^{0} |)}^{p - 2} {| D Q^{1} |}^{2} d x \\ = \int_{Ω} ν [1 - {(1 + | D u^{0} |)}^{p - 2}] D u^{0} D Q^{1} d x + \int_{Ω} (β_{T} J^{1} + β_{ψ} W^{1}) g \cdot Q^{1} d x - \int_{Ω} (u^{0} \cdot \nabla) u^{0} Q^{1} d x . \end{array}$

因为 $p < 2$ ，有

$\int_{Ω} ν {(1 + | D u^{0} |)}^{p - 2} {| D Q^{1} |}^{2} d x \geq ν {(1 + {‖ D u^{0} ‖}_{\infty})}^{p - 2} {‖ D Q^{1} ‖}_{2}^{2} \geq ν {(1 + {‖ u^{0} ‖}_{C^{1, γ_{0}}})}^{p - 2} {‖ D Q^{1} ‖}_{2}^{2},$

并利用Hölder不等式，引理1和引理3，又有

$\begin{array}{l} | \int_{Ω} ν [1 - {(1 + | D u^{0} |)}^{p - 2}] D u^{0} D Q^{1} d x + \int_{Ω} (β_{T} J^{1} + β_{ψ} W^{1}) g \cdot Q^{1} d x - \int_{Ω} (u^{0} \cdot \nabla) u^{0} Q^{1} d x | \\ \leq ν (2 - p) {‖ D u^{0} ‖}_{2} {‖ D u^{0} ‖}_{\infty} {‖ D Q^{1} ‖}_{2} + (β_{T} {‖ J^{1} ‖}_{\infty} + β_{ψ} {‖ W^{1} ‖}_{\infty}) {‖ g ‖}_{2} {‖ Q^{1} ‖}_{2} + {‖ u^{0} ‖}_{\infty} {‖ \nabla u^{0} ‖}_{2} {‖ Q^{1} ‖}_{2} \\ \leq ν (2 - p) {‖ u^{0} ‖}_{C^{1, γ_{0}}}^{2} {‖ D Q^{1} ‖}_{2} + (C β_{T} {‖ J^{1} ‖}_{1, 2} + C β_{ψ} {‖ W^{1} ‖}_{1, 2}) c_{3} {‖ g ‖}_{q} {‖ D Q^{1} ‖}_{2} + C c_{3} {‖ u^{0} ‖}_{C^{1, γ_{0}}}^{2} {‖ D Q^{1} ‖}_{2}, \end{array}$

因此综合可得，

${‖ D Q^{1} ‖}_{2} \leq \frac{1}{ν} {(1 + {‖ u^{0} ‖}_{C^{1, γ_{0}}})}^{2 - p} [(2 ν - p ν + C c_{3}) {‖ u^{0} ‖}_{C^{1, γ_{0}}}^{2} + C c_{3} (β_{T} {‖ J^{1} ‖}_{1, 2} + β_{ψ} {‖ W^{1} ‖}_{1, 2}) {‖ g ‖}_{q}] .$ (23)

同样，在(19)中用 $η = J^{1}$ 作检验函数，可得

$\int_{Ω} α {| \nabla J^{1} |}^{2} d x = - \int_{Ω} (u^{0} \cdot \nabla) T^{0} J^{1} d x .$

由Poincare'不等式和Hölder不等式，有

$| \int_{Ω} α {| \nabla J^{1} |}^{2} d x | = α {‖ \nabla J^{1} ‖}_{2}^{2} \geq C α {‖ J^{1} ‖}_{1, 2}^{2},$

$| - \int_{Ω} (u^{0} \cdot \nabla) T^{0} J^{1} d x | \leq {‖ u^{0} ‖}_{\infty} {‖ \nabla T^{0} ‖}_{2} {‖ J^{1} ‖}_{2} \leq C {‖ u^{0} ‖}_{C^{1, γ_{0}}} {‖ T^{0} ‖}_{2, 2} {‖ J^{1} ‖}_{1, 2},$

因此可得，

${‖ J^{1} ‖}_{1, 2} \leq C \frac{1}{α} {‖ u^{0} ‖}_{C^{1, γ_{0}}} {‖ T^{0} ‖}_{2, 2} .$ (24)

类似的，对(20)可得

${‖ W^{1} ‖}_{1, 2} \leq C \frac{1}{D} {‖ u^{0} ‖}_{C^{1, γ_{0}}} {‖ ψ^{0} ‖}_{2, 2} .$ (25)

结合(23)，(24)及(25)，并利用估计式(7)，可得

$\begin{array}{l} {‖ D Q^{1} ‖}_{2} + {‖ J^{1} ‖}_{1, 2} + {‖ W^{1} ‖}_{1, 2} \\ \leq \frac{1}{ν} {(1 + {‖ u^{0} ‖}_{C^{1, γ_{0}}})}^{2 - p} [(2 ν - p ν + C c_{3}) {‖ u^{0} ‖}_{C^{1, γ_{0}}}^{2} + C c_{3} (\frac{β_{T}}{α} {‖ T^{0} ‖}_{2, 2} + \frac{β_{ψ}}{D} {‖ ψ^{0} ‖}_{2, 2}) {‖ g ‖}_{q} {‖ u^{0} ‖}_{C^{1, γ_{0}}}] \\ + \frac{C}{α} {‖ u^{0} ‖}_{C^{1, γ_{0}}} {‖ T^{0} ‖}_{2, 2} + \frac{C}{D} {‖ u^{0} ‖}_{C^{1, γ_{0}}} {‖ ψ^{0} ‖}_{2, 2} \\ \leq \frac{1}{ν} {(1 + 2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q})}^{2 - p} {(2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q})}^{2} \\ (2 ν - p ν + C c_{3} + \frac{C c_{3} β_{T}}{α} {‖ g ‖}_{q} + \frac{C c_{3} β_{ψ}}{D} {‖ g ‖}_{q} + \frac{C}{α} + \frac{C}{D}) . \end{array}$

从而a)得证。

下面证明估计b)。设b)中的假设成立，类似a)，在(18)中设 $j \geq 1$ ， $φ = Q^{j + 1}$ ，有

$\begin{array}{l} \int_{Ω} ν {(1 + | D u^{j} |)}^{p - 2} {| D Q^{j + 1} |}^{2} d x \\ = \int_{Ω} ν [{(1 + | D u^{j - 1} |)}^{p - 2} - {(1 + | D u^{j} |)}^{p - 2}] D u^{j} D Q^{j + 1} d x \\ + \int_{Ω} (β_{T} J^{j + 1} + β_{ψ} W^{j + 1}) g \cdot Q^{j + 1} d x - \int_{Ω} (Q^{j} \cdot \nabla) u^{j} Q^{j + 1} d x - \int_{Ω} (u^{j - 1} \cdot \nabla) Q^{j} Q^{j + 1} d x . \end{array}$

与得到式(23)类似可得，

$\begin{array}{l} {‖ D Q^{j + 1} ‖}_{2} \leq \frac{1}{ν} {(1 + {‖ u^{j} ‖}_{C^{1, γ_{0}}})}^{2 - p} [(2 ν - p ν + c_{3}^{2}) {‖ u^{j} ‖}_{C^{1, γ_{0}}} {‖ D Q^{j} ‖}_{2} + (β_{T} {‖ J^{j + 1} ‖}_{1, 2} \\ + β_{ψ} {‖ W^{j + 1} ‖}_{1, 2}) C c_{3} {‖ g ‖}_{q} + c_{3}^{2} {‖ u^{j - 1} ‖}_{C^{1, γ_{0}}} {‖ D Q^{j} ‖}_{2}] . \end{array}$ (26)

然后，在(19)中设 $j \geq 1$ ， $η = J^{j + 1}$ ，可得

$\int_{Ω} α {| \nabla J^{j + 1} |}^{2} d x = - \int_{Ω} (u^{j} \cdot \nabla) J^{j} J^{j + 1} d x - \int_{Ω} (Q^{j} \cdot \nabla) T^{j - 1} J^{j + 1} d x$ .

由Poincare'不等式和Hölder不等式，有

$| \int_{Ω} α {| \nabla J^{j + 1} |}^{2} d x | = α {‖ \nabla J^{j + 1} ‖}_{2}^{2} \geq C α {‖ J^{j + 1} ‖}_{1, 2}^{2}$ ，

$| - \int_{Ω} (u^{j} \cdot \nabla) J^{j} J^{j + 1} d x - \int_{Ω} (Q^{j} \cdot \nabla) T^{j - 1} J^{j + 1} d x | \leq C {‖ u^{j} ‖}_{C^{1, γ_{0}}} {‖ J^{j} ‖}_{1, 2} {‖ J^{j + 1} ‖}_{1, 2} + C c_{3} {‖ D Q^{j} ‖}_{2} {‖ T^{j - 1} ‖}_{2, 2} {‖ J^{j + 1} ‖}_{1, 2} .$

因此，

${‖ J^{j + 1} ‖}_{1, 2} \leq \frac{C}{α} {‖ u^{j} ‖}_{C^{1, γ_{0}}} {‖ J^{j} ‖}_{1, 2} + \frac{C c_{3}}{α} {‖ D Q^{j} ‖}_{2} {‖ T^{j - 1} ‖}_{2, 2} .$ (27)

对式(20)做类似估计，可得

${‖ W^{j + 1} ‖}_{1, 2} \leq \frac{C}{D} {‖ u^{j} ‖}_{C^{1, γ_{0}}} {‖ W^{j} ‖}_{1, 2} + \frac{C c_{3}}{D} {‖ D Q^{j} ‖}_{2} {‖ ψ^{j - 1} ‖}_{2, 2} .$ (28)

结合(26)，(27)及(28)，并利用估计式(7)，有

$\begin{array}{l} {‖ D Q^{j + 1} ‖}_{2} + {‖ J^{j + 1} ‖}_{1, 2} + {‖ W^{j + 1} ‖}_{1, 2} \\ \leq \frac{1}{ν} {(1 + {‖ u^{j} ‖}_{C^{1, γ_{0}}})}^{2 - p} [(2 ν - p ν + c_{3}^{2}) {‖ u^{j} ‖}_{C^{1, γ_{0}}} {‖ D Q^{j} ‖}_{2} + \frac{C c_{3} β_{T}}{α} {‖ g ‖}_{q} ({‖ u^{j} ‖}_{C^{1, γ_{0}}} {‖ J^{j} ‖}_{1, 2} + {‖ D Q^{j} ‖}_{2} {‖ T^{j - 1} ‖}_{2, 2}) \\ + \frac{C c_{3} β_{ψ}}{D} {‖ g ‖}_{q} ({‖ u^{j} ‖}_{C^{1, γ_{0}}} {‖ W^{j} ‖}_{1, 2} + {‖ D Q^{j} ‖}_{2} {‖ ψ^{j - 1} ‖}_{2, 2}) + c_{3}^{2} {‖ u^{j - 1} ‖}_{C^{1, γ_{0}}} {‖ D Q^{j} ‖}_{2}] + \frac{C}{α} {‖ u^{j} ‖}_{C^{1, γ_{0}}} {‖ J^{j} ‖}_{1, 2} \\ + \frac{C c_{3}}{α} {‖ D Q^{j} ‖}_{2} {‖ T^{j - 1} ‖}_{2, 2} + \frac{C}{D} {‖ u^{j} ‖}_{C^{1, γ_{0}}} {‖ W^{j} ‖}_{1, 2} + \frac{C c_{3}}{D} {‖ D Q^{j} ‖}_{2} {‖ ψ^{j - 1} ‖}_{2, 2} \\ \leq \frac{1}{ν} {(1 + 2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q})}^{2 - p} (2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q}) \\ (2 ν - p ν + 2 c_{3}^{2} + \frac{2 C c_{3} β_{T}}{α} {‖ g ‖}_{q} + \frac{2 C c_{3} β_{ψ}}{D} {‖ g ‖}_{q} + \frac{C}{α} + \frac{C c_{3}}{α} + \frac{C}{D} + \frac{C c_{3}}{D}) \cdot ({‖ D Q^{j} ‖}_{2} + {‖ J^{j} ‖}_{1, 2} + {‖ W^{j} ‖}_{1, 2}) . \end{array}$

由b)中的假设知(22)成立。因此，通过归纳，(22)对任意给定的 $j \in ℕ$ 成立。

根据假设(21)知，级数 $\sum_{j} ({‖ D Q^{j} ‖}_{2} + {‖ J^{j} ‖}_{1, 2} + {‖ W^{j} ‖}_{1, 2})$ 收敛，故级数 $\sum_{j} {‖ D Q^{j} ‖}_{2}$ ， $\sum_{j} {‖ J^{j} ‖}_{1, 2}$ 和 $\sum_{j} {‖ W^{j} ‖}_{1, 2}$ 均收敛。进而由 $W^{1, 2} (Ω)$ 的完备性，级数 $\sum_{j} Q^{j}$ ， $\sum_{j} J^{j}$ 和 $\sum_{j} W^{j}$ 依范数 ${‖ \cdot ‖}_{1, 2}$ 分别收敛到 $Q, J, W \in W^{1, 2} (Ω)$ 。

由式(18)可知在广义函数意义下，有

$\begin{array}{l} \frac{1}{ρ} \nabla R^{j + 1} = ν \nabla \cdot [{(1 + | D u^{j} |)}^{p - 2} D Q^{j + 1}] - ν \nabla \cdot {[{(1 + | D u^{j - 1} |)}^{p - 2} - {(1 + | D u^{j} |)}^{p - 2}] D u^{j}} \\ - (β_{T} J^{j + 1} + β_{ψ} W^{j + 1}) g - (Q^{j} \cdot \nabla) u^{j} - (u^{j - 1} \cdot \nabla) Q^{j} . \end{array}$ (29)

而由于

$\begin{array}{l} {‖ ν \nabla \cdot [{(1 + | D u^{j} |)}^{p - 2} D Q^{j + 1}] - ν \nabla \cdot {[{(1 + | D u^{j - 1} |)}^{p - 2} - {(1 + | D u^{j} |)}^{p - 2}] D u^{j}} ‖}_{- 1, 2} \\ \leq ν {‖ {(1 + | D u^{j} |)}^{p - 2} D Q^{j + 1} ‖}_{2} + ν {‖ [{(1 + | D u^{j - 1} |)}^{p - 2} - {(1 + | D u^{j} |)}^{p - 2}] D u^{j} ‖}_{2} \\ \leq ν {‖ D Q^{j + 1} ‖}_{2} + ν (2 - p) {‖ D Q^{j} ‖}_{2} {‖ u^{j} ‖}_{C^{1, γ_{0}}}, \end{array}$

${‖ - (β_{T} J^{j + 1} + β_{ψ} W^{j + 1}) g ‖}_{- 1, 2} \leq {‖ - (β_{T} J^{j + 1} + β_{ψ} W^{j + 1}) g ‖}_{2} \leq β_{T} {‖ J^{j + 1} ‖}_{1, 2} {‖ g ‖}_{q} + β_{ψ} {‖ W^{j + 1} ‖}_{1, 2} {‖ g ‖}_{q},$

${‖ - (Q^{j} \cdot \nabla) u^{j} - (u^{j - 1} \cdot \nabla) Q^{j} ‖}_{- 1, 2} \leq {‖ - (Q^{j} \cdot \nabla) u^{j} - (u^{j - 1} \cdot \nabla) Q^{j} ‖}_{2} \leq c_{3} ({‖ u^{j} ‖}_{C^{1, γ_{0}}} + {‖ u^{j - 1} ‖}_{C^{1, γ_{0}}}) {‖ D Q^{j} ‖}_{2} .$

结合以上估计，以及估计式(22)和(7)，可得

$\begin{array}{l} {‖ R^{j + 1} ‖}_{2} \leq C {‖ \nabla R^{j + 1} ‖}_{- 1, 2} \\ \leq C [ν {‖ D Q^{j + 1} ‖}_{2} + (2 ν - p ν + c_{3}) {‖ D Q^{j} ‖}_{2} {‖ u^{j} ‖}_{C^{1, γ_{0}}} + β_{T} {‖ J^{j + 1} ‖}_{1, 2} {‖ g ‖}_{q} + β_{ψ} {‖ W^{j + 1} ‖}_{1, 2} {‖ g ‖}_{q} + c_{3} {‖ u^{j - 1} ‖}_{C^{1, γ_{0}}} {‖ D Q^{j} ‖}_{2}] \\ \leq C {(1 + 2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q})}^{2 - p} (2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q}) \\ (4 ν - 2 p ν + 2 c_{3}^{2} + 2 c_{3} + \frac{2 C c_{3} β_{T}}{α} {‖ g ‖}_{q} + \frac{2 C c_{3} β_{ψ}}{D} {‖ g ‖}_{q} + \frac{C β_{T}}{α} {‖ g ‖}_{q} + \frac{C c_{3} β_{T}}{α} {‖ g ‖}_{q} + \frac{C β_{ψ}}{D} {‖ g ‖}_{q} + \frac{C c_{3} β_{ψ}}{D} {‖ g ‖}_{q}) \\ \cdot ({‖ D Q^{j} ‖}_{2} + {‖ J^{j} ‖}_{1, 2} + {‖ W^{j} ‖}_{1, 2}) \end{array}$

$\begin{array}{l} \leq \frac{(4 ν - 2 p ν + 2 c_{3}^{2} + 2 c_{3} + \frac{2 C c_{3} β_{T}}{α} {‖ g ‖}_{q} + \frac{2 C c_{3} β_{ψ}}{D} {‖ g ‖}_{q} + \frac{C β_{T}}{α} {‖ g ‖}_{q} + \frac{C c_{3} β_{T}}{α} {‖ g ‖}_{q} + \frac{C β_{ψ}}{D} {‖ g ‖}_{q} + \frac{C c_{3} β_{ψ}}{D} {‖ g ‖}_{q})}{2 - p + 2 c_{3}^{2} + 2 c_{3}} \\ \frac{(2 ν - p ν + C c_{3} + \frac{C c_{3} β_{T}}{α} {‖ g ‖}_{q} + \frac{C c_{3} β_{ψ}}{D} {‖ g ‖}_{q} + \frac{C}{α} + \frac{C}{D})}{2 - p + 2 c_{3}^{2} + 2 c_{3}} \times {(1 + 2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q})}^{3 - p} \\ (2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q}) \cdot {(2 - p + 2 c_{3}^{2} + 2 c_{3}) {(2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2}}_{}^{} \\ {+ 2 K_{4} {‖ g ‖}_{q}) \times {(1 + 2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q})}^{2 - p}}}^{j} . \end{array}$

因此，再利用式(21)，可知级数 $\sum_{j} R (x)$ 依 $L^{2}$ 范数收敛到 $R (x)$ 。

3.3. 解的存在性证明

若 ${‖ g ‖}_{q}$ ， ${‖ Q_{1} ‖}_{q}$ ， ${‖ Q_{2} ‖}_{2}$ ， ${‖ Q_{3} ‖}_{2}$ 足够小，并满足(6)和(21)。假设 ${u^{m}}$ ， ${P^{m}}$ ， ${T^{m}}$ ， ${ψ^{m}}$ 为命题1中构造的序列，则

$u^{m} (x) = \sum_{j = 1}^{m} Q^{j} (x) + u^{0} (x)$ ， $P^{m} (x) = \sum_{j = 1}^{m} R^{j} (x) + P^{0} (x)$ ，

$T^{m} (x) = \sum_{j = 1}^{m} J^{j} (x) + T^{0} (x)$ ， $ψ^{m} (x) = \sum_{j = 1}^{m} W^{j} (x) + ψ^{0} (x)$ .

设 $u (x) = Q (x) + u^{0} (x)$ ， $P (x) = R (x) + P^{0} (x)$ ， $T (x) = J (x) + T^{0} (x)$ 和 $ψ (x) = W (x) + ψ^{0} (x)$ ，其中 $Q (x)$ ， $R (x)$ ， $J (x)$ 和 $W (x)$ 如命题2中所定义。序列 ${u^{m}}$ ， ${P^{m}}$ ， ${T^{m}}$ 和 ${ψ^{m}}$ 分别依 $W^{1, 2}$ ， $L^{2}$ ， $W^{1, 2}$ 和 $W^{1, 2}$ 范数收敛到函数u，P，T和 $ψ$ 。由Ascoli-Arzelà定理，存在序列 ${u^{k_{m}}}$ 在 $C^{1, γ} (\bar{Ω})$ 中收敛到 $\tilde{u}$ ，又由于 ${u^{m}}$ 在 $W^{1, 2}$ 中收敛到u，故 $u = \tilde{u}$ ，即有 $u \in C^{1, γ} (\bar{Ω})$ 。同理可得 $P \in C^{0, γ} (\bar{Ω})$ ， $T \in W^{2, 2} (Ω)$ ， $ψ \in W^{2, 2} (Ω)$ 。由于对 $\forall m \in ℕ$ ， $\nabla \cdot u^{k_{m}} = 0$ ， ${u^{k_{m}} |}_{\partial Ω} = 0$ ， ${T^{k_{m}} |}_{\partial Ω} = 0$ ， ${ψ^{k_{m}} |}_{\partial Ω} = 0$ ，故 $\nabla \cdot u = 0$ ， ${u |}_{\partial Ω} = 0$ ， ${T |}_{\partial Ω} = 0$ ， ${ψ |}_{\partial Ω} = 0$ 。

接下来证明

$\begin{array}{l} \int_{Ω} ν {(1 + | D u |)}^{p - 2} D u D φ d x - \frac{1}{ρ} \int_{Ω} P d i v d x - \int_{Ω} [β_{T} (T - T_{c}) + β_{ψ} (ψ - ψ_{c})] g φ d x + \int_{Ω} (u \cdot \nabla) u \cdot φ d x \\ = \lim_{m \to \infty} {\int_{Ω} ν {(1 + | D u^{m - 1} |)}^{p - 2} D u^{m} D φ d x - \int_{Ω} [β_{T} (T^{m} - T_{c}) + β_{ψ} (ψ^{m} - ψ_{c})] g φ d x \\ - \frac{1}{ρ} \int_{Ω} P^{m} d i v d x + \int_{Ω} (u^{m - 1} \cdot \nabla) u^{m - 1} \cdot φ d x}, \forall φ \in C_{0}^{\infty} (Ω), \end{array}$ (30)

$\int_{Ω} α \nabla T \cdot \nabla η d x + \int_{Ω} (u \cdot \nabla) T \cdot η d x = \lim_{m \to \infty} {\int_{Ω} α \nabla T^{m} \cdot \nabla η d x + \int_{Ω} (u^{m - 1} \cdot \nabla) T^{m - 1} \cdot η d x}, \forall η \in C_{0}^{\infty} (Ω),$ (31)

$\int_{Ω} D \nabla ψ \cdot \nabla ϕ d x + \int_{Ω} (u \cdot \nabla) ψ \cdot ϕ d x = \lim_{m \to \infty} {\int_{Ω} D \nabla ψ^{m} \cdot \nabla ϕ d x + \int_{Ω} (u^{m - 1} \cdot \nabla) ψ^{m - 1} \cdot ϕ d x}, \forall ϕ \in C_{0}^{\infty} (Ω) .$ (32)

首先，由Hölder不等式，有

$\begin{array}{l} | ν \int_{Ω} {(1 + | D u^{m - 1} |)}^{p - 2} D u^{m} D φ d x - ν \int_{Ω} {(1 + | D u |)}^{p - 2} D u D φ d x | \\ = | \int_{Ω} ν {(1 + | D u^{m - 1} |)}^{p - 2} (D u^{m} - D u) D φ d x + \int_{Ω} ν [{(1 + | D u^{m - 1} |)}^{p - 2} - {(1 + | D u |)}^{p - 2}] D u D φ d x | \\ \leq ν {‖ D u^{m} - D u ‖}_{2} {‖ D φ ‖}_{2} + ν (2 - p) {‖ D u^{m - 1} - D u ‖}_{2} {‖ D u ‖}_{2} {‖ D φ ‖}_{\infty}, \end{array}$

以及

$| \frac{1}{ρ} \int_{Ω} P^{m} d i v φ d x - \frac{1}{ρ} \int_{Ω} P d i v φ d x | = \frac{1}{ρ} | \int_{Ω} (P^{m} - P) d i v φ d x | \leq \frac{1}{ρ} {‖ P^{m} - P ‖}_{2} {‖ d i v φ ‖}_{2},$

还有

$\begin{array}{l} | \int_{Ω} [β_{T} (T^{m} - T_{c}) + β_{ψ} (ψ^{m} - ψ_{c})] g φ d x - \int_{Ω} [β_{T} (T - T_{c}) + β_{ψ} (ψ - ψ_{c})] g φ d x | \\ \leq β_{T} {‖ T^{m} - T ‖}_{2} {‖ g ‖}_{2} {‖ φ ‖}_{\infty} + β_{ψ} {‖ ψ^{m} - ψ ‖}_{2} {‖ g ‖}_{2} {‖ φ ‖}_{\infty} \\ \leq C β_{T} {‖ T^{m} - T ‖}_{1, 2} {‖ g ‖}_{q} {‖ φ ‖}_{\infty} + C β_{ψ} {‖ ψ^{m} - ψ ‖}_{1, 2} {‖ g ‖}_{q} {‖ φ ‖}_{\infty}, \end{array}$

$\begin{array}{l} | \int_{Ω} (u^{m - 1} \cdot \nabla) u^{m - 1} φ d x - \int_{Ω} (u \cdot \nabla) u φ d x | \\ = | \int_{Ω} [(u^{m - 1} - u) \cdot \nabla] u^{m - 1} φ d x + \int_{Ω} (u \cdot \nabla) (u^{m - 1} - u) φ d x | \\ \leq {‖ u^{m - 1} - u ‖}_{2} {‖ \nabla u^{m - 1} ‖}_{2} {‖ φ ‖}_{\infty} + {‖ u ‖}_{2} {‖ \nabla u^{m - 1} - \nabla u ‖}_{2} {‖ φ ‖}_{\infty} . \end{array}$

由于 $u^{m}$ ， $T^{m}$ ， $ψ^{m}$ 的 $W^{1, 2}$ 收敛性， $P^{m}$ 的 $L^{2}$ 收敛性，及 ${‖ \nabla u^{m - 1} ‖}_{2}$ ， ${‖ u ‖}_{2}$ ， ${‖ T^{m} ‖}_{2, 2}$ ， ${‖ ψ^{m} ‖}_{2, 2}$ ， ${‖ D φ ‖}_{\infty}$ ， ${‖ d i v φ ‖}_{2}$ ， ${‖ φ ‖}_{\infty}$ 的有界性，当 $m \to \infty$ 时这些量趋于0。注意到(30)的右端等于 $\int_{Ω} Q_{1} φ d x$ ，故对 $\forall φ \in C_{0}^{\infty} (Ω)$ ，有

$\int_{Ω} ν {(1 + | D u |)}^{p - 2} D u D φ d x - \frac{1}{ρ} \int_{Ω} P d i v d x = \int_{Ω} [β_{T} (T - T_{c}) + β_{ψ} (ψ - ψ_{c})] g φ d x - \int_{Ω} (u \cdot \nabla) u \cdot φ d x + \int_{Ω} Q_{1} φ d x .$

其次，有

$| \int_{Ω} α \nabla T^{m} \cdot \nabla η d x - \int_{Ω} α \nabla T \cdot \nabla η d x | \leq α {‖ \nabla T^{m} - \nabla T ‖}_{2} {‖ \nabla η ‖}_{2},$

$\begin{array}{l} | \int_{Ω} (u^{m - 1} \cdot \nabla) T^{m - 1} \cdot η d x - \int_{Ω} (u \cdot \nabla) T \cdot η d x | \\ \leq | \int_{Ω} [(u^{m - 1} - u) \cdot \nabla] T^{m - 1} \cdot η d x | + | \int_{Ω} (u \cdot \nabla) (T^{m - 1} - T) \cdot η d x | \\ \leq {‖ u^{m - 1} - u ‖}_{3} {‖ \nabla T^{m - 1} ‖}_{r} {‖ η ‖}_{\frac{3 r}{2 r - 3}} + {‖ u ‖}_{\infty} {‖ \nabla T^{m - 1} - \nabla T ‖}_{2} {‖ η ‖}_{2} \\ \leq C {‖ u^{m - 1} - u ‖}_{1, 2} {‖ \nabla T^{m - 1} ‖}_{r} {‖ η ‖}_{\frac{3 r}{2 r - 3}} + {‖ u ‖}_{c^{1, γ_{0}}} {‖ \nabla T^{m - 1} - \nabla T ‖}_{2} {‖ η ‖}_{2} . \end{array}$

类似的,可得

$\int_{Ω} α \nabla T \cdot \nabla η d x = - \int_{Ω} (u \cdot \nabla) T \cdot η d x + \int_{Ω} Q_{2} η d x, \forall η \in C_{0}^{\infty} (Ω),$

$\int_{Ω} D \nabla ψ \cdot \nabla ϕ d x = - \int_{Ω} (u \cdot \nabla) ψ \cdot ϕ d x + \int_{Ω} Q_{3} ϕ d x, \forall ϕ \in C_{0}^{\infty} (Ω) .$

最后，通过如下估计中的极限及范数的弱下半连续性，可知

$\begin{matrix} {‖ u ‖}_{C^{1, γ}} + {‖ P ‖}_{C^{0, γ}} + {‖ T ‖}_{2, 2} + {‖ ψ ‖}_{2, 2} \leq {‖ u - u^{k_{m}} ‖}_{C^{1, γ}} + {‖ u^{k_{m}} ‖}_{C^{1, γ}} + {‖ P - P^{k_{m}} ‖}_{C^{0, γ}} + {‖ P^{k_{m}} ‖}_{C^{0, γ}} + {‖ T^{k_{m}} ‖}_{2, 2} + {‖ ψ^{k_{m}} ‖}_{2, 2} \\ \leq 2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q} . \end{matrix}$

3.4. 解的唯一性证明

令 $\frac{6}{5} < p < 2$ ，设 $(u_{1}, T_{1}, ψ_{1})$ 和 $(u_{2}, T_{2}, ψ_{2})$ 均为问题(5)的解，令 $\bar{u} = u_{1} - u_{2}$ ， $\bar{T} = T_{1} - T_{2}$ ， $\bar{ψ} = ψ_{1} - ψ_{2}$ ，利用定义1，将 $(u_{1}, T_{1}, ψ_{1})$ 和 $(u_{2}, T_{2}, ψ_{2})$ 代入(2)中，两式相减，取 $φ = \bar{u} = u_{1} - u_{2}$ ，有

$\int_{Ω} ν [S (D u_{1}) - S (D u_{2})] \cdot (D u_{1} - D u_{2}) d x = - \int_{Ω} (β_{T} \bar{T} + β_{ψ} \bar{ψ}) g \bar{u} d x + \int_{Ω} (\bar{u} \cdot \nabla) \bar{u} \cdot u_{1} d x .$

由Hölder不等式，可得

$\begin{array}{l} {‖ D \bar{u} ‖}_{p}^{p} = \int_{Ω} {(\frac{{| D \bar{u} |}^{2}}{{(1 + | D u_{1} | + | D u_{2} |)}^{2 - p}})}^{\frac{p}{2}} \cdot {(1 + | D u_{1} | + | D u_{2} |)}^{\frac{p (2 - p)}{2}} d x \\ \leq {(\int_{Ω} \frac{{| D \bar{u} |}^{2}}{{(1 + | D u_{1} | + | D u_{2} |)}^{2 - p}} d x)}^{\frac{p}{2}} \cdot {(\int_{Ω} {(1 + | D u_{1} | + | D u_{2} |)}^{p} d x)}^{\frac{2 - p}{2}} . \end{array}$

利用引理4，可得

$\begin{array}{l} {‖ D \bar{u} ‖}_{p}^{2} \leq \int_{Ω} \frac{{| D \bar{u} |}^{2}}{{(1 + | D u_{1} | + | D u_{2} |)}^{2 - p}} d x \cdot {(\int_{Ω} {(1 + | D u_{1} | + | D u_{2} |)}^{p} d x)}^{\frac{2 - p}{p}} \\ \leq C (\int_{Ω} (S (D u_{1}) - S (D u_{2})) \cdot (D u_{1} - D u_{2}) d x) \cdot (1 + {‖ D u_{1} ‖}_{p}^{2 - p} + {‖ D u_{2} ‖}_{p}^{2 - p}) \\ \leq C (\frac{1}{ν} \int_{Ω} (β_{T} \bar{T} + β_{ψ} \bar{ψ}) g \bar{u} d x + \frac{1}{ν} \int_{Ω} (\bar{u} \cdot \nabla) \bar{u} \cdot u_{1} d x) \cdot (1 + {‖ D u_{1} ‖}_{p} + {‖ D u_{2} ‖}_{p}) . \end{array}$

由Hölder不等式和Sobolev不等式，有

$\begin{array}{l} | \int_{Ω} (β_{T} \bar{T} + β_{ψ} \bar{ψ}) g \bar{u} d x + \int_{Ω} (\bar{u} \cdot \nabla) \bar{u} \cdot u_{1} d x | \\ \leq β_{T} {‖ \bar{T} ‖}_{\frac{3 p}{3 - p}} {‖ g ‖}_{\frac{3 p}{5 p - 6}} {‖ \bar{u} ‖}_{\frac{3 p}{3 - p}} + β_{ψ} {‖ \bar{ψ} ‖}_{\frac{3 p}{3 - p}} {‖ g ‖}_{\frac{3 p}{5 p - 6}} {‖ \bar{u} ‖}_{\frac{3 p}{3 - p}} + {‖ \bar{u} ‖}_{\frac{3 p}{3 - p}}^{2} {‖ \nabla u_{1} ‖}_{\frac{3 p}{5 p - 6}} \\ \leq C c_{3} β_{T} {‖ \bar{T} ‖}_{1, 2} {‖ g ‖}_{q} {‖ D \bar{u} ‖}_{p} + C c_{3} β_{ψ} {‖ \bar{ψ} ‖}_{1, 2} {‖ g ‖}_{q} {‖ D \bar{u} ‖}_{p} + C c_{3}^{2} {‖ D \bar{u} ‖}_{p}^{2} {‖ u_{1} ‖}_{C 1, γ_{0}} . \end{array}$

所以可得

${‖ D \bar{u} ‖}_{p} \leq \frac{C}{ν} (C c_{3} β_{T} {‖ \bar{T} ‖}_{1, 2} {‖ g ‖}_{q} + C c_{3} β_{ψ} {‖ \bar{ψ} ‖}_{1, 2} {‖ g ‖}_{q} + C c_{3}^{2} {‖ D \bar{u} ‖}_{p} {‖ u_{1} ‖}_{C 1, γ_{0}}) (1 + {‖ D u_{1} ‖}_{p} + {‖ D u_{2} ‖}_{p}) .$ (33)

将 $(u_{1}, T_{1})$ ， $(u_{2}, T_{2})$ 代入(3)中，两式相减，设 $η = \bar{T} = T_{1} - T_{2}$ ，可得

$\int_{Ω} α {| \nabla \bar{T} |}^{2} d x = - \int_{Ω} (u \cdot \nabla) T_{1} \cdot \bar{T} d x,$

又有

$| \int_{Ω} α {| \nabla \bar{T} |}^{2} d x | = α {‖ \nabla \bar{T} ‖}_{2}^{2} \geq C α {‖ \nabla \bar{T} ‖}_{1, 2}^{2},$

$| - \int_{Ω} (u \cdot \nabla) T_{1} \cdot \bar{T} d x | \leq {‖ \bar{u} ‖}_{\frac{3 p}{3 - p}} {‖ \nabla T_{1} ‖}_{\frac{6 p}{5 p - 6}} {‖ \bar{T} ‖}_{2} \leq C c_{3} {‖ D \bar{u} ‖}_{P} {‖ T_{1} ‖}_{2, 2} {‖ \bar{T} ‖}_{1, 2},$

可得

${‖ \bar{T} ‖}_{1, 2} \leq \frac{C c_{3}}{α} {‖ D \bar{u} ‖}_{P} {‖ T_{1} ‖}_{2, 2} .$ (34)

接着，将 $(u_{1}, ψ_{1})$ 和 $(u_{2}, ψ_{2})$ 代入(4)中，两式相减，设 $ϕ = \bar{ψ} = ψ_{1} - ψ_{2}$ ，类似的可以得到

${‖ \bar{ψ} ‖}_{1, 2} \leq \frac{C c_{3}}{D} {‖ D \bar{u} ‖}_{P} {‖ ψ_{1} ‖}_{2, 2} .$ (35)

结合(33)，(34)及(35)，可得

$\begin{array}{l} {‖ D \bar{u} ‖}_{p} + {‖ \bar{B} ‖}_{1, 2} + {‖ \bar{T} ‖}_{1, 2} \\ \leq \frac{C}{ν} (C c_{3} β_{T} {‖ \bar{T} ‖}_{1, 2} {‖ g ‖}_{q} + C c_{3} β_{ψ} {‖ \bar{ψ} ‖}_{1, 2} {‖ g ‖}_{q} + C c_{3}^{2} {‖ D \bar{u} ‖}_{p} {‖ u_{1} ‖}_{C 1, γ_{0}}) \cdot (1 + {‖ D u_{1} ‖}_{p} + {‖ D u_{2} ‖}_{p}) \\ + \frac{C c_{3}}{α} {‖ D \bar{u} ‖}_{P} {‖ T_{1} ‖}_{2, 2} + \frac{C c_{3}}{D} {‖ D \bar{u} ‖}_{P} {‖ ψ_{1} ‖}_{2, 2} . \\ \leq (2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q}) (1 + 2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} \\ + 2 K_{4} {‖ g ‖}_{q}) (\frac{C c_{3} β_{T}}{α ν} {‖ g ‖}_{q} + \frac{C c_{3} β_{ψ}}{D ν} {‖ g ‖}_{q} + \frac{C c_{3}^{2}}{ν} + \frac{C c_{3}}{α} + \frac{C c_{3}}{D}) ({‖ D \bar{u} ‖}_{p} + {‖ \bar{B} ‖}_{1, 2} + {‖ \bar{T} ‖}_{1, 2}) . \end{array}$

由于 $2 K {‖ Q_{1} ‖}_{q} + 2 K_{2} {‖ Q_{2} ‖}_{2} + 2 K_{3} {‖ Q_{3} ‖}_{2} + 2 K_{4} {‖ g ‖}_{q}$ 足够小，唯一性得证。

4. 总结

本文在三维光滑有界区域 $Ω$ 中，考虑了一类稳态非牛顿化学活性流体运动方程组的第一边值问题，证明了问题正则解的存在唯一性。由于所求方程组具有强耦合性与强非线性性，为问题分析带来本质困难。本文的证明思路如下：首先，通过迭代法将方程组解耦构造问题近似解，并利用线性方程理论构造出原问题的近似解；然后，利用线性方程及一类p-Stokes方程解的估计，在外力项充分小的条件下通过一系列计算导出近似解序列的一致性先验估计；接下来，利用紧性结果得出近似解的收敛性；最后，通过对参数取极限证明了本文的主要结论。本文的主要创新是：文中通过对高次多项式解的分布规律的分析，得到了近似解的高阶导数估计，这是证明过程的关键，该方法也为此类问题的研究提供了新思路。

基金项目

吉林省教育厅科学技术研究项目(批准号：JJKH20230790KJ)。

NOTES

^*通讯作者。

参考文献

[1]	Rojas-Medar, M.A. and Lorca, S.A. (1995) the Equations of a Viscous Incompressible Chemical Active Fluid I: Uniqueness and Existence of the Local Solutions. Revista de Matemáticas Aplicades, 16, 57-80.
[2]	Rojas-Medar, M.A. and Lorca, S.A. (1995) The Equations of a Viscous Incompressible Chemical Active Fluid II: Regularity of Solutions. Revista de Matemáticas Aplicades, 16, 81-95.
[3]	Lorca, S.A. and Rojas-Medar, M.A. (1996) Weak Solutions for the Viscous Incompressible Chemically Active Fluids. Revista de Matematica e Estatistica, 14, 183-199.
[4]	Rojas-Medar, M.A. and Lorca, S.A. (1995) Global Strong Solution of the Equations for the Motion of a Chemical Active Fluid. Sociedade Brasileira de Matemática, Matemática Contemporânea, 8, 319-335. https://doi.org/10.21711/231766361995/rmc815
[5]	Malham, S. and Xin, J.X. (1998) Global Solutions to a Reactive Boussinesq System with Front Data on an Infinite Domain. Communications in Mathematical Physics, 193, 287-316. https://doi.org/10.1007/s002200050330
[6]	Ko, S. (2022) Existence of Global Weak Solutions for Unsteady Motions of Incompressible Chemically Reacting Generalized Newtonian Fluids. Journal of Mathematical Analysis and Applications, 513, Article No. 126206. https://doi.org/10.1016/j.jmaa.2022.126206
[7]	Boling, G. and Yadong, S. (2002) The Periodic Initial Value Problem and Initial Value Problem for the Modified Boussinesq Approximation. Journal of Partial Differential Equations, 15, 57-71. https://doi.org/10.57262/die/1356060763
[8]	Lin, W., Guo, B.L. and Shang, Y.D. (2003) The Periodic Initial Value Problem and Initial Value Problem for the Non-Newtonian Boussinesq Approximation. Applicable Analysis, 82, 787-808. https://doi.org/10.1080/00036810310001603189
[9]	杨惠, 王长佳. 一类稳态不可压非牛顿Boussinesq方程组解的存在唯一性[J]. 吉林大学学报(理学版), 2019, 57(4): 753-761.
[10]	王爽, 王长佳. 一类各向异性非牛顿Boussinesq方程组弱解的存在性研究[J]. 应用数学进展, 2022, 11(1): 450-461. https://doi.org/10.12677/AAM.2022.111053
[11]	刘琳, 王长佳. 一类不可压非牛顿Boussinesq方程组正则解的存在性[J]. 应用数学进展, 2023, 12(4): 1855-1865. https://doi.org/10.12677/AAM.2023.124192
[12]	Sohr, H. (2001) The Navier-Stokes Equations: an Elementary Analytic Approach. Birkhäuser, Basel. https://doi.org/10.1007/978-3-0348-0551-3
[13]	Paré, S.C. (1992) Existence, Uniqueness and Regularity of Solution of the Equations of a Turbulence Model for Incompressible Fluids. Applicable Analysis, 43, 245-296. https://doi.org/10.1080/00036819208840063
[14]	Nečas, J. (1965) Equations aux DeriveesPartielles. Presses de I’Universite de Montreal, Montreal.
[15]	Crispo, F. and Grisanti, C.R. (2008) On the Existence, Uniqueness and Regularity for a Class of Shear-Thinning Fluids. Journal of Mathematical Fluid Mechanics, 10, 455-487. https://doi.org/10.1007/s00021-008-0282-1
[16]	Ebmeyer, C. (2006) Regularity in Sobolev Spaces of Steady Flows of Fluids with Shear-Dependent Viscosity. Mathematical Methods in the Applied Sciences, 29, 1687-1708. https://doi.org/10.1002/mma.748

为你推荐

友情链接