肿瘤生长的相场模型解的性质

Solutions of a Phase-Field Model for Tumor Growth

摘要: 本文研究了无血管期肿瘤生长的相场模型。该模型耦合了营养物质浓度n的Allen-Cahn方程和序参数 的Cahn-Hilliard方程，描述了肿瘤生长所需营养物质的扩散过程和肿瘤的演变过程。在本文中，我证明了初边值问题弱解的存在性，一维情况下解的唯一性以及稳态解的存在性。

Abstract: In this paper, we study the solutions to an initial-boundary value problem for a phase-field model of tumor growth, which is the Allen-Cahn equation for the nutrient concentration n coupled with the Cahn-Hilliard equation for the order parameter  , and describes the diffusion process of nutrient and the evolution process of tumor. I prove the existence of weak solutions to the problem and the uniqueness of global solution in one dimension. Finally, the existence of stationary solutions is proved.

1. 引言

恶性肿瘤(癌症)已经成为严重威胁中国人群健康的主要公共卫生问题之一。近几十年来，描述肿瘤生长的数学模型的文献 [1] [2] 大量涌现出来。为更好地描述肿瘤在微环境中的生长，学者引进了相场模型 [3] [4] 来描述肿瘤的演变过程。在扩散界面框架中，Cahn-Hilliard类型的相场模型是最常用的模型，其适定性 [5] [6]，渐近分析 [7] [8]，滑模控制 [9] 等均得到了广泛的研究。考虑细胞间液体的流动性，学者结合流体力学提出了Cahn-Hilliard-Darcy方程组 [10]、Cahn-Hilliard-Hele-Shaw方程组 [11] 和Cahn-Hilliard-Navier- Stokes方程组 [12] 等模型。这些新模型为进一步体现细胞-微环境之间的相互作用提供了新的数学框架。

本文受Silva [13] 的启发，建立了无血管期肿瘤生长的相场模型。该模型耦合了营养物质浓度 $n$ 的Allen-Cahn方程和序参数 $\varphi$ 的Cahn-Hilliard方程，描述了肿瘤生长所需营养物质的扩散过程和肿瘤细胞的演变过程。在介绍模型前，我先定义一些符号，在本文中假设 $\Omega \subset {ℝ}^3$ 是一个有界开域，并且边界 $\partial \Omega$ 是光滑的。定义 $Q_{T_e}=\left(0,T_e\right)\times \Omega$。未知函数 $n\in {ℝ}^{+}$ 表示营养物质浓度， $\varphi \in ℝ$ 表示序参数，序参数 $\varphi \left(t,x\right)=1$ 表示在某个区域 $x\in \Omega$ 某个时间点 $t$ 细胞为肿瘤细胞， $\varphi \left(t,x\right)=-1$ 为正常细胞。具体模型如下

$n_t=D\Delta n-{\alpha }_c\left(\varphi \right),$ (1.1)

${\varphi }_t=\nabla \cdot \left(M\left(\varphi \right)\nabla \mu \right),$ (1.2)

$\mu =-{\epsilon }^2\Delta \varphi +{\varphi }^3-\varphi +{{\alpha }^{\prime }}_c\left(\varphi \right)n,$

其中 $\left(t,x\right)\in Q_{T_e}$，满足Neumann边界条件，无流边界条件和初值条件

$\frac{\partial n}{\partial \nu }=0,\left(t,x\right)\in \left(0,T_e\right)\times \partial \Omega ,$ (1.3)

$\frac{\partial \varphi }{\partial \nu }=\frac{\partial }{\partial \nu }\Delta \varphi =0,\left(t,x\right)\in \left(0,T_e\right)\times \partial \Omega ,$ (1.4)

$n\left(0,x\right)=n_0,x\in \Omega ,$ (1.5)

$\varphi \left(0,x\right)={\varphi }_0\text{,}x\in \Omega \text{.}$ (1.6)

其中 $D$ 为扩散系数，是一个正常数。数 ${\alpha }_c\left(\varphi \right)\ge 0$ 表示营养物质的消耗速度， $M\left(\varphi \right)>0$ 表示细胞移动速度， $\mu$ 表示化学势， $\epsilon >0$ 表示界面能系数， $\nu$ 为单位外法向量。

新的相场模型的简单推导如下：考虑总自由能泛函为

$F\left[n,\varphi \right]={\displaystyle {\int }_{\Omega }\left[\frac{D}{2}{\left|\nabla n\right|}^2+{\alpha }_c\left(\varphi \right)n+\frac{{\epsilon }^2}{2}{\left|\nabla \varphi \right|}^2+V\left(\varphi \right)\right]\text{d}x},$

其中，双势阱函数 $V\left(\varphi \right)=-\frac{{\varphi }^2}{2}+\frac{{\varphi }^4}{4}$。自由能 $F$ 关于时间 $t$ 求导，可得

$\frac{\text{d}F}{\text{d}t}={\displaystyle {\int }_{\Omega }\left[\left(-D\Delta n+{\alpha }_c\left(\varphi \right)\right)n_t+\left(-{\epsilon }^2\Delta \varphi +{\varphi }^3-\varphi +{{\alpha }^{\prime }}_c\left(\varphi \right)n\right){\varphi }_t\right]\text{d}x}.$

将方程(1.1)，(1.2)代入上式，由假设 $M\left(\varphi \right)>0$ 可得

$\begin{array}{c}\frac{\text{d}F}{\text{d}t}={\displaystyle {\int }_{\Omega }\left[-{\left(D\Delta n-{\alpha }_c\left(\varphi \right)\right)}^2-M\left(\varphi \right){\left(\nabla \left(-{\epsilon }^2\Delta \varphi +{\varphi }^3-\varphi +{{\alpha }^{\prime }}_c\left(\varphi \right)n\right)\right)}^2\right]\text{d}x}\\ \le 0.\end{array}$

因此模型满足热力学第二定律。

2. 主要结论

为了得出主要结论，需要先给出初边值问题(1.1)~(1.6)弱解的定义。在此之前，引入一个定义

$H_{\Gamma }^2=\left\{v\in H^2\left(\Omega \right){\left|\frac{\partial v}{\partial \nu }\right|}_{\partial \Omega }=0\right\}.$

定义1.1 假设 $\left(n_0,{\varphi }_0\right)\in H^1\left(\Omega \right)\times H^1\left(\Omega \right)$，称函数 $\left(n,\varphi \right)$ 为问题(1.1)-(1.6)的弱解，并且满足

$\begin{array}{l}n\in L^{\infty }\left(0,T_e;H^1\left(\Omega \right)\right)\cap L^2\left(0,T_e;H^2\left(\Omega \right)\right),\\ \varphi \in L^{\infty }\left(0,T_e;H^1\left(\Omega \right)\right)\cap L^2\left(0,T_e;H^3\left(\Omega \right)\cap H_{\Gamma }^2\right),\end{array}$

如果对任意的测试函数 $u\in C_0^{\infty }\left(\left(-\infty ,T_e\right);C^{\infty }\left(\Omega \right)\right)$，满足

${\left(n_0,u_0\right)}_{\Omega }+{\left(n,u_t\right)}_{Q_{T_e}}=D{\left(\nabla n,\nabla u\right)}_{Q_{T_e}}+{\left({\alpha }_c\left(\varphi \right),u\right)}_{Q_{T_e}},$ (2.1)

${\left({\varphi }_0,u_0\right)}_{\Omega }+{\left(\varphi ,u_t\right)}_{Q_{T_e}}={\left(M\left(\varphi \right)\nabla \left(-{\epsilon }^2\Delta \varphi +{\varphi }^3-\varphi +{{\alpha }^{\prime }}_c\left(\varphi \right)n\right),\nabla u\right)}_{Q_{T_e}}.$ (2.2)

为研究该初边值问题的弱解，给出以下假设条件。

假设(H)：在本文中，我们假设 ${\alpha }_c\left(\varphi \right)\in C_0^3\left(ℝ\right),M\left(\varphi \right)\in C_0^1\left(ℝ\right)$ 并且满足

1) $0\le {\alpha }_c\left(\varphi \right)\le {\nu }_1,$

2) $0<{\nu }_2\le M\left(\varphi \right)\le {\nu }_3，\left|M^{\prime }\left(\varphi \right)\right|\le {\nu }_4,$

3) $\left|{{\alpha }^{\prime }}_c\left(\varphi \right)\right|\le {\nu }_5,\left|{{\alpha }^{″}}_c\left(\varphi \right)\right|\le {\nu }_6,\left|{{\alpha }^{‴}}_c\left(\varphi \right)\right|\le {\nu }_7,$

这里所有的参数 ${\nu }_i\left(i=1,\cdots ,7\right)$ 均为正常数。

注: 本文所取的 ${\alpha }_c\left(\varphi \right)$ 和 $M\left(\varphi \right)$ 有一些限制，但是可以取得到的，这里给出一种选取方法。首先对满足连续条件 ${\alpha }_c\left(\varphi \right)\in C^3\left(ℝ\right),M\left(\varphi \right)\in C^1\left(ℝ\right)$ 的函数 ${\alpha }_c\left(\varphi \right)$， $M\left(\varphi \right)$ 进行截断

${\bar{\alpha }}_c\left(\varphi \right)=\{\begin{array}{cc}{\alpha }_c\left(\varphi \right),& 当\left|\varphi \right|\le \text{K}时,\\ 0,& 当\left|\varphi \right|>\text{K}时,\end{array}\bar{M}\left(\varphi \right)=\{\begin{array}{cc}M\left(\varphi \right),& 当\left|\varphi \right|\le \text{K}时,\\ 0,& 当\left|\varphi \right|>\text{K}时,\end{array}$

其中 $K$ 为某一正常数。然后对截断后的函数 ${\bar{\alpha }}_c\left(\varphi \right),\bar{M}\left(\varphi \right)$ 进行磨光，磨光后的函数仍记为 ${\alpha }_c\left(\varphi \right)$ 和 $M\left(\varphi \right)$。此时的 ${\alpha }_c\left(\varphi \right)$ 和 $M\left(\varphi \right)$ 便满足假设条件(H)。

定理1. 给定任意正常数 $T_e$，假设 $\left(n_0,{\varphi }_0\right)\in H^1\left(\Omega \right)\times H^1\left(\Omega \right)$，则在定义1的意义下，初边值问题(1.1)~(1.6)存在弱解 $\left(n,\varphi \right)$，并且解满足

$n\in L^{\infty }\left(0,T_e;H^1\left(\Omega \right)\right)\cap L^2\left(0,T_e;H^2\left(\Omega \right)\right),$ (2.3)

$\varphi \in L^{\infty }\left(0,T_e;H^1\left(\Omega \right)\right)\cap L^2\left(0,T_e;H^{\text{3}}\left(\Omega \right)\right),$ (2.4)

由于函数 $M\left(\varphi \right)\in C_0^1\left(ℝ\right)$ 不是常值函数，因此本文将考虑一维空间域内初边值问题(1.1)~(1.6)解的唯一性。

定理2. 假设 $\left(n_0,{\varphi }_0\right)\in L^2\left(\Omega \right)\times L^2\left(\Omega \right)$ 且 $\Omega \subset ℝ$ 是一个有界开集，则由定理1得到的解 $\left(n,\varphi \right)$ 是唯一的。

注1. 稳态解的定义和主要结论在第五章节。

注2. 为方便起见，本文用 $\Vert \cdot \Vert$ 表示 $L^2\left(\Omega \right)$ 空间的范数。

3. 弱解的存在性

由迭代法和Aubin-Lions引理不难得到初边值问题(1.1)~(1.6)局部解的存在性。在这一章节中，通过弱解 $\left(n,\varphi \right)$ 的一致先验估计，借助局部解延拓法证明整体解的存在性。

引理1. 假设 $\left(n_0,{\varphi }_0\right)\in H^1\left(\Omega \right)\times H^1\left(\Omega \right)$，则对任意的 $t\in \left(0,T_e\right)$，有

${\Vert \nabla n\Vert }_{L^{\infty }(0,T_e;L^2(\Omega ))}+{\Vert \nabla \varphi \Vert }_{L^{\infty }(0,T_e;L^2(\Omega ))}\le C_{T_e}.$ (3.1)

证明：由于模型满足热力学第二定律，也就是能量的衰减性 $\frac{\text{d}F}{\text{d}t}\le 0$。对其关于时间 $t$ 积分，得

$F\left(t\right)-F\left(0\right)\le 0,$

即

$\begin{array}{l}{\displaystyle {\int }_{\Omega }\left(\frac{D}{2}{\left|\nabla n\right|}^2+{\alpha }_c\left(\varphi \right)n+\frac{{\epsilon }^2}{2}{\left|\nabla \varphi \right|}^2-\frac{{\varphi }^2}{2}+\frac{{\varphi }^4}{4}\right)\text{d}x}\\ \le {\displaystyle {\int }_{\Omega }\left(\frac{D}{2}{\left|\nabla n_0\right|}^2+{\alpha }_c\left({\varphi }_0\right)n_0+\frac{{\epsilon }^2}{2}{\left|\nabla {\varphi }_0\right|}^2-\frac{{\varphi }_0^2}{2}+\frac{{\varphi }_0^4}{4}\right)\text{d}x}\end{array}$

根据初值条件，应用Sobolve嵌入定理和带 $\eta$ 的Young不等式，整理得

$\begin{array}{l}\frac{D}{2}{\Vert \nabla n\Vert }^2+{\displaystyle {\int }_{\Omega }{\alpha }_c}\left(\varphi \right)n\text{d}x+\frac{{\epsilon }^2}{2}{\Vert \nabla \varphi \Vert }^2+\frac{1}{4}{\displaystyle {\int }_{\Omega }{\varphi }^4}\text{d}x+\frac{1}{2}{\displaystyle {\int }_{\Omega }{\varphi }_0^2}\text{d}x\\ \le \frac{D}{2}{\Vert \nabla n_0\Vert }^2+{\nu }_2\Vert n_0{\Vert }_{L^1(\Omega )}+\frac{{\epsilon }^2}{2}{\Vert \nabla {\varphi }_0\Vert }^2+\frac{1}{4}{\Vert {\varphi }_0\Vert }_{L^4(\Omega )}^4+{\displaystyle {\int }_{\Omega }\frac{1}{2}}{\varphi }^2\text{d}x\\ \le C+{\nu }_2{\Vert n_0\Vert }_{L^2(\Omega )}+\frac{1}{4}{\Vert {\varphi }_0\Vert }_{H^1(\Omega )}^4+\eta {\displaystyle {\int }_{\Omega }{\varphi }^4}\text{d}x+C_{\eta }\\ \le C+\eta {\displaystyle {\int }_{\Omega }{\varphi }^4}\text{d}x\end{array}$

最后，考虑到函数 $n\in {ℝ}^{+}$ 且 ${\alpha }_c\left(\varphi \right)\ge 0$ 恒成立，取 $\eta =\frac{1}{8}$，证得(3.1)。至此证毕。

引理2. 假设 $n_0\in H^1\left(\Omega \right)$，则对任意的 $t\in \left(0,T_e\right)$，有

${\Vert n\Vert }_{L^{\infty }(0,T_e;H^1(\Omega ))}+{\Vert n\Vert }_{L^2(0,T_e;H^2(\Omega ))}\le C_{T_e}.$ (3.2)

证明：将(1.1)两边同时乘以 $-\Delta n$ 并关于 $x$ 做积分，利用分部积分公式，可得

$\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \nabla n\Vert }^2+D{\Vert \Delta n_k\Vert }^2=\left({\alpha }_c\left(\varphi \right),\Delta n\right).$

借助假设(H)，Hölder不等式以及带 $\eta$ 的Young不等式，可以得到

$\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \nabla n\Vert }^2+D{\Vert \Delta n\Vert }^2\le {\nu }_1{\Vert \Delta n\Vert }_{L^1}\le C_{\Omega }\Vert \Delta n\Vert \le \eta {\Vert \Delta n\Vert }^2+C_{\eta }.$ (3.3)

对(3.3)在 $\left(0,T_e\right)$ 上关于 $t$ 积分，结合引理1得

$D{\displaystyle {\int }_0^t{\Vert \Delta n\Vert }^2}\text{d}\tau \le C_{T_e}.$

考虑庞加莱不等式，(3.2)得证。至此证毕。

引理3. 假设 ${\varphi }_0\in H^1\left(\Omega \right)$，则存在常数 $C_{T_e}$，使得对任意的 $t\in \left(0,T_e\right)$，有

${\Vert \varphi \Vert }_{L^{\infty }(0,T_e;H^1(\Omega ))}+{\Vert \varphi \Vert }_{L^2(0,T_e;H^3(\Omega ))}\le C_{T_e}.$ (3.4)

证明：方程(1.2)左右同乘 $-\Delta \varphi$，再关于 $x$ 做积分，利用分部积分公式以及边界条件(1.4)，可得

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \nabla \varphi \Vert }^2+{\epsilon }^2\left(M\left(\varphi \right)\nabla \Delta \varphi ,\nabla \Delta \varphi \right)\\ =3\left(M\left(\varphi \right){\varphi }^2\nabla \varphi ,\nabla \Delta \varphi \right)-\left(M\left(\varphi \right)\nabla \varphi ,\nabla \Delta \varphi \right)+\left(M\left(\varphi \right)\nabla \left({{\alpha }^{\prime }}_c\left(\varphi \right)n\right),\nabla \Delta \varphi \right).\end{array}$

由假设(H2)，将右边项分成四项，我们有

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \nabla \varphi \Vert }^2+{\epsilon }^2{\nu }_2{\Vert \nabla \Delta \varphi \Vert }^2\\ \le 3{\nu }_3\left|\left({\varphi }^2\nabla \varphi ,\nabla \Delta \varphi \right)\right|+{\nu }_3\left|\left(\nabla \varphi ,\nabla \Delta \varphi \right)\right|+{\nu }_3\left|\left({{\alpha }^{″}}_c\left(\varphi \right)\nabla \varphi n,\nabla \Delta \varphi \right)\right|+{\nu }_3\left|\left({{\alpha }^{\prime }}_c\left(\varphi \right)\nabla n,\nabla \Delta \varphi \right)\right|\\ \le I_{1,1}+I_{1,2}+I_{1,3}+I_{1,4}.\end{array}$ (3.5)

考虑到 $\Omega \subset {ℝ}^3$，由Sobolev嵌入定理 [14] 知 $H^1\left(\Omega \right)$ 嵌入到 $L^6\left(\Omega \right)$。利用Gagliardo-Nirenberg不等式 [15]

${\Vert D\varphi \Vert }_{L^6(\Omega )}\le C{\Vert D^3\varphi \Vert }^{\frac{2}{3}}{\Vert \varphi \Vert }^{\frac{1}{3}}+C\Vert \varphi \Vert ,$ (3.6)

以及带 $\eta$ 的Young不等式，引理1可得 $I_{1,1}$ 的估计

$\begin{array}{c}{I}_{1,1}=3{\nu }_3\left|\left({\varphi }^2\nabla \varphi ,\nabla \Delta \varphi \right)\right|\le 3{\nu }_3{\Vert \varphi \Vert }_{L^6(\Omega )}^2{\Vert \nabla \varphi \Vert }_{L^6(\Omega )}\Vert \nabla \Delta \varphi \Vert \\ \le C{\Vert \varphi \Vert }_{H^1(\Omega )}^2{\Vert \nabla \Delta \varphi \Vert }^{\frac{2}{3}}{\Vert \varphi \Vert }^{\frac{1}{3}}\Vert \nabla \Delta \varphi \Vert +C{\Vert \varphi \Vert }_{H^1(\Omega )}^2\Vert \varphi \Vert \Vert \nabla \Delta \varphi \Vert \\ \le \eta {\Vert \nabla \Delta \varphi \Vert }^2+C_{\eta }{\Vert \varphi \Vert }^2.\end{array}$ (3.7)

类似地，不难得到 $I_{1,3}$ 的估计

$\begin{array}{c}{I}_{1,3}={\nu }_3\left|\left({{\alpha }^{″}}_c\left(\varphi \right)\nabla \varphi n,\nabla \Delta \varphi \right)\right|\le {\nu }_3{\nu }_6{\Vert \nabla \varphi \Vert }_{L^3(\Omega )}{\Vert n\Vert }_{L^6(\Omega )}\Vert \nabla \Delta \varphi \Vert \\ \le C{\Vert n\Vert }_{H^1(\Omega )}\Vert \nabla \Delta \varphi {\Vert }^{\frac{3}{2}}{\Vert \varphi \Vert }^{\frac{1}{2}}+{\Vert n\Vert }_{H^1(\Omega )}\Vert \nabla \Delta \varphi \Vert \Vert \varphi \Vert \\ \le \eta {\Vert \nabla \Delta \varphi \Vert }^2+C_{\eta }{\Vert \varphi \Vert }^2,\end{array}$ (3.8)

这里应用了Gagliardo-Nirenberg不等式

${\Vert D\varphi \Vert }_{L^3(\Omega )}\le C{\Vert D^3\varphi \Vert }^{\frac{1}{2}}{\Vert \varphi \Vert }^{\frac{1}{2}}+C\Vert \varphi \Vert .$

$I_{1,2}$ 和 $I_{1,4}$ 的估计比较简单，可以直接得到

$I_{1,2}\le \eta {\Vert \nabla \Delta \varphi \Vert }^2+C_{\eta }{\Vert \nabla \varphi \Vert }^2,$ (3.9)

$I_{1,4}\le \eta {\Vert \nabla \Delta \varphi \Vert }^2+C_{\eta }.$ (3.10)

结合(3.7)~(3.10)，并代入(3.5)，根据引理1和庞加莱不等式，有

$\frac{1}{2}\frac{\text{d}}{\text{d}t}\Vert \nabla \varphi \Vert +{\epsilon }^2{\nu }_2{\Vert \nabla \Delta \varphi \Vert }^2\le 4\eta {\Vert \nabla \Delta \varphi \Vert }^2+C.$ (3.11)

(3.11)关于时间 $t$ 积分，并且选择 $\eta =\frac{{\epsilon }^2{\nu }_3}{8}$，可以得到

${\Vert \nabla \Delta \varphi \Vert }_{L^2(0,T_e;L^2(\Omega ))}\le C_{T_e}.$ (3.12)

最后根据引理1和庞加莱不等式，利用椭圆方程的正则性，可得(3.4)。至此，引理证毕。

结合引理2和引理3，应用局部解延拓法，可以证明初边值问题(1.1)~(1.6)弱解的存在性，即定理1得证。

4. 一维模型解的唯一性

上一章节在先验估计的基础上证明了整体解的存在性。由于函数 $M\left(\varphi \right)\in C_0^1\left(ℝ\right)$ 不是常值函数，因此在这一章节我将证明一维空间域内初边值问题(1.1)~(1.6)解的唯一性，即定理2。

定理2的证明：假设 $\left(n_1,{\varphi }_1\right)$ 和 $\left(n_2,{\varphi }_2\right)$ 为定理1意义下的两个解，则差 $n=n_1-n_2$ 和 $\varphi ={\varphi }_1-{\varphi }_2$ 满足

$n_t-D{n}_{xx}+{\alpha }_c\left({\varphi }_1\right)-{\alpha }_c\left({\varphi }_2\right)=0,$ (4.1)

$\begin{array}{l}{\varphi }_t+{\epsilon }^2{\left(M\left({\varphi }_1\right){\varphi }_{1xxx}-M\left({\varphi }_2\right){\varphi }_{2xxx}\right)}_x-3{\left(M\left({\varphi }_1\right){\varphi }_1^2{\varphi }_{1x}-M\left({\varphi }_2\right){\varphi }_2^2{\varphi }_{2x}\right)}_x\\ +{\left(M\left({\varphi }_1\right){\varphi }_{1x}-M\left({\varphi }_2\right){\varphi }_{2x}\right)}_x-{\left(M\left({\varphi }_1\right){\left({\alpha }^{\prime }\left({\varphi }_1\right)n_1\right)}_x-M\left({\varphi }_2\right){\left({\alpha }^{\prime }\left({\varphi }_2\right)n_2\right)}_x\right)}_x=0,\end{array}$ (4.2)

$\frac{\partial n}{\partial \nu }|{}_{\partial \Omega }=0,n\left(0,x\right)=0$ (4.3)

$\frac{\partial }{\partial \nu }{\varphi }_{xx}|{}_{\partial \Omega }={\frac{\partial \varphi }{\partial \nu }|}_{\partial \Omega }=0,\varphi \left(0,x\right)=0,$ (4.4)

$n\in L^{\infty }\left(0,T_e;H^1\left(\Omega \right)\right)\cap L^2\left(0,T_e;H^2\left(\Omega \right)\right),$ (4.5)

$\varphi \in L^{\infty }\left(0,T_e;H^1\left(\Omega \right)\right)\cap L^2\left(0,T_e;H^3\left(\Omega \right)\right).$ (4.6)

为证 $n=\varphi =0$，需要得到 $n$ 和 $\varphi$ 的估计，下面分成三部分进行：

(1) (4.1) 两边同乘 $n$，关于 $x$ 做积分，然后根据分部积分公式可以得到

$\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert n\left(t\right)\Vert }^2+D{\Vert n_x\left(t\right)\Vert }^2={\displaystyle {\int }_{\Omega }\left({\alpha }_c\left({\varphi }_2\right)-{\alpha }_c\left({\varphi }_1\right)\right)n\text{d}x}.$

由假设(H)知，

$\left|{\alpha }_c\left({\varphi }_1\right)-\alpha c\left({\varphi }_2\right)\right|=\left|{{\alpha }^{\prime }}_c\left(\xi \right)\left({\varphi }_1-{\varphi }_2\right)\right|\le {\nu }_5\left|\varphi \right|,$

其中 $\xi \in \langle {\varphi }^{k_{\nu }},\varphi \rangle$ 是一个函数。应用Hölder不等式，可得

$\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert n\left(t\right)\Vert }^2+D{\Vert n_x\left(t\right)\Vert }^2\le C\left({\Vert n\left(t\right)\Vert }^2+{\Vert \varphi \left(t\right)\Vert }^2\right).$

于是，关于 $t$ 做积分，可推得

${\Vert n\left(t\right)\Vert }^2+2D{{\displaystyle {\int }_0^t\Vert n_x\left(t\right)\Vert }}^2\text{d}\tau \le {\Vert n\left(0\right)\Vert }^2+C{{\displaystyle {\int }_0^t\Vert n\left(t\right)\Vert }}^2\text{d}\tau +C{\displaystyle {\int }_0^t{\Vert \varphi \left(t\right)\Vert }^2}\text{d}\tau .$ (4.7)

(2) (4.1)两边同乘 $-n_{xx}$，不难得到

${\Vert n_x\left(t\right)\Vert }^2+D{{\displaystyle {\int }_0^t\Vert n_{xx}\left(t\right)\Vert }}^2\text{d}\tau \le {\Vert n_x\left(0\right)\Vert }^2+C{\displaystyle {\int }_0^t{\Vert \varphi \left(t\right)\Vert }^2}\text{d}\tau .$ (4.8)

(3) 在(4.2)左右两边同乘 $\varphi$，关于 $x$ 做积分，利用分部积分公式可以得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \varphi \left(t\right)\Vert }^2-{\epsilon }^2\left(M\left({\varphi }_1\right){\varphi }_{1xxx}-M\left({\varphi }_2\right){\varphi }_{2xxx},{\varphi }_x\right)\\ \text{+3}\left(M\left({\varphi }_1\right){\varphi }_{\text{1}}^{\text{2}}{\varphi }_{1x}-M\left({\varphi }_2\right){\varphi }_{\text{2}}^{\text{2}}{\varphi }_{2x},{\varphi }_x\right)-\left(M\left({\varphi }_1\right){\varphi }_{1x}-M\left({\varphi }_2\right){\varphi }_{2x},{\varphi }_x\right)\\ +\left(M\left({\varphi }_1\right){\left({\alpha }^{\prime }\left({\varphi }_1\right)n_1\right)}_x-M\left({\varphi }_2\right){\left({\alpha }^{\prime }\left({\varphi }_2\right)n_2\right)}_x,{\varphi }_x\right)=0.\end{array}$ (4.9)

由于 $M\left({\varphi }_1\right)\in C_0^1\left(ℝ\right)$，借助分部积分公式和边界条件(4.4)，有

$\left(M\left({\varphi }_1\right){\varphi }_{xxx},{\varphi }_x\right)=\left({\varphi }_{xxx},M\left({\varphi }_1\right){\varphi }_x\right)=-\left(M\left({\varphi }_1\right){\varphi }_{xx},{\varphi }_{xx}\right)-\left(M^{\prime }\left({\varphi }_1\right){\varphi }_{1x}{\varphi }_{xx},{\varphi }_x\right).$ (4.10)

将(4.10)代入(4.9)中，整理得

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \varphi \left(t\right)\Vert }^2+{\epsilon }^{2}M\left({\varphi }_1\right){\Vert {\varphi }_{xx}\Vert }^2+3\left(M\left({\varphi }_1\right){\varphi }_1^2{\varphi }_x,{\varphi }_x\right)\\ =-{\epsilon }^2\left(M^{\prime }\left({\varphi }_1\right){\varphi }_{1x}{\varphi }_{xx},{\varphi }_x\right)+{\epsilon }^2\left(\left(M\left({\varphi }_1\right)-M\left({\varphi }_2\right)\right){\varphi }_{2xxx},{\varphi }_x\right)-3\left(M\left({\varphi }_1\right)\left({\varphi }_1+{\varphi }_2\right)\varphi {\varphi }_{2x},{\varphi }_x\right)\\ -3\left(\left(M\left({\varphi }_1\right)-M\left({\varphi }_2\right)\right){\varphi }_2^2{\varphi }_{2x},{\varphi }_x\right)+\left(M\left({\varphi }_1\right){\varphi }_x,{\varphi }_x\right)+\left(\left(M\left({\varphi }_1\right)-M\left({\varphi }_2\right)\right){\varphi }_{2x},{\varphi }_x\right)\\ -\left(M\left({\varphi }_1\right){\left({\alpha }^{\prime }\left({\varphi }_1\right)n_1\right)}_x-M\left({\varphi }_2\right){\left({\alpha }^{\prime }\left({\varphi }_2\right)n_2\right)}_x,{\varphi }_x\right).\end{array}$ (4.11)

于是，(4.11)式关于 $t$ 做积分，并将右边项分成以下几部分进行估计

$\begin{array}{l}{\Vert \varphi \left(t\right)\Vert }^2+2{\epsilon }^2{\nu }_2{{\displaystyle {\int }_0^t\Vert {\varphi }_{xx}\left(\tau \right)\Vert }}^2\text{d}\tau +6{\left(M\left({\varphi }_1\right){\varphi }_1^2,{\varphi }_x^2\right)}_{Q_t}\\ ={\Vert \varphi \left(0\right)\Vert }^2-2{\epsilon }^2{\left(M^{\prime }\left({\varphi }_1\right){\varphi }_{1x}{\varphi }_{xx},{\varphi }_x\right)}_{Q_t}+2{\epsilon }^2{\left(\left(M\left({\varphi }_1\right)-M\left({\varphi }_2\right)\right){\varphi }_{2xxx},{\varphi }_x\right)}_{Q_t}\\ -6{\left(M\left({\varphi }_1\right)\left({\varphi }_1+{\varphi }_2\right)\varphi {\varphi }_{2x},{\varphi }_x\right)}_{Q_t}-6{\left(\left(M\left({\varphi }_1\right)-M\left({\varphi }_2\right)\right){\varphi }_2^2{\varphi }_{2x},{\varphi }_x\right)}_{Q_t}\\ +2{\left(M\left({\varphi }_1\right){\varphi }_x,{\varphi }_x\right)}_{Q_t}+2{\left(\left(M\left({\varphi }_1\right)-M\left({\varphi }_2\right)\right){\varphi }_{2x},{\varphi }_x\right)}_{Q_t}\\ -2{\left(M\left({\varphi }_1\right){\alpha }^{″}\left({\varphi }_1\right){\varphi }_{1x}n,{\varphi }_x\right)}_{Q_t}-2{\left(M\left({\varphi }_1\right){\alpha }^{″}\left({\varphi }_{1x}\right)\varphi n_2,{\varphi }_x\right)}_{Q_t}\end{array}$

$\begin{array}{l}-2{\left(M\left({\varphi }_1\right)\left({\alpha }^{″}\left({\varphi }_1\right)-{\alpha }^{″}\left({\varphi }_2\right)\right){\varphi }_{2x}{n}_2,{\varphi }_x\right)}_{Q_t}-2{\left(\left(M\left({\varphi }_1\right)-M\left({\varphi }_2\right)\right){\alpha }^{″}\left({\varphi }_2\right){\varphi }_{2x}{n}_2,{\varphi }_x\right)}_{Q_t}\\ -2{\left(M\left({\varphi }_1\right){\alpha }^{\prime }\left({\varphi }_1\right)n_x,{\varphi }_x\right)}_{Q_t}-2{\left(M\left({\varphi }_1\right)\left({\alpha }^{\prime }\left({\varphi }_1\right)-{\alpha }^{\prime }\left({\varphi }_2\right)\right)n_{2x},{\varphi }_x\right)}_{Q_t}\\ -2{\left(\left(M\left({\varphi }_1\right)-M\left({\varphi }_2\right)\right){\alpha }^{\prime }\left({\varphi }_2\right)n_{2x},{\varphi }_x\right)}_{Q_t}\\ ={\Vert {\varphi }_0\Vert }^2+{\displaystyle \underset{i=1}{\overset{13}{\sum }}{I}_{2,i}}.\end{array}$ (4.12)

考虑 $\Omega \subset ℝ$，则有Gagliardo-Nirenberg不等式

${\Vert D\varphi \Vert }_{L^{\infty }(\Omega )}\le C{\Vert D^2\varphi \Vert }^{\frac{3}{4}}{\Vert \varphi \Vert }^{\frac{1}{4}}+C\Vert \varphi \Vert ,$ (4.13)

结合假设(H2)以及带 $\eta$ 的Young不等式，可以推导出 $I_{2,1}$ 的估计

$\begin{array}{c}\left|I_{2,1}\right|=2{\epsilon }^2\left|{\left({M^{\prime }}_c\left({\varphi }_1\right){\varphi }_{1x}{\varphi }_{xx},{\varphi }_x\right)}_{Q_t}\right|\\ \le 2{\epsilon }^2{\nu }_4{\displaystyle {\int }_0^t{\Vert {\varphi }_x\Vert }_{L^{\infty }(\Omega )}}\Vert {\varphi }_{1x}\Vert \Vert {\varphi }_{xx}\Vert \text{d}\tau \\ \le C{\Vert {\varphi }_{1x}\Vert }_{L^{\infty }(0,t;L^2(\Omega ))}{\displaystyle {\int }_0^t\left({\Vert {\varphi }_{xx}\Vert }^{\frac{7}{4}}{\Vert \varphi \Vert }^{\frac{1}{4}}+\Vert {\varphi }_{xx}\Vert \Vert \varphi \Vert \right)\text{d}\tau }\\ \le \eta {{\displaystyle {\int }_0^t\Vert {\varphi }_{xx}\left(\tau \right)\Vert }}^2\text{d}\tau +C_{\eta }{{\displaystyle {\int }_0^t\Vert \varphi \left(\tau \right)\Vert }}^2\text{d}\tau .\end{array}$ (4.14)

由假设(H)和积分中值定理，有

$\left|M\left({\varphi }_1\right)-M\left({\varphi }_2\right)\right|=\left|M^{\prime }\left(\xi \right)\left({\varphi }_1-{\varphi }_2\right)\right|\le {\nu }_4\left|\varphi \right|,$

其中 $\xi \in \langle {\varphi }^{k_{\nu }},\varphi \rangle$ 是一个函数。于是借助插值不等式(4.13)，带 $\eta$ 的Young不等式以及一般形式的Hölder不等式，再利用(4.6)，可得

$\begin{array}{l}\left|I_{2,2}\right|=2{\epsilon }^2\left|{\left(\left(M\left({\varphi }_1\right)-M\left({\varphi }_2\right)\right){\varphi }_{2xxx},{\varphi }_x\right)}_{Q_t}\right|\le 2{\epsilon }^2{\nu }_4{\displaystyle {\int }_0^t\Vert \varphi \Vert }\Vert {\varphi }_{2xxx}\Vert {\Vert {\varphi }_x\Vert }_{L^{\infty }(\Omega )}\text{d}\tau \\ \le 2{\epsilon }^2{\nu }_4{\displaystyle {\int }_0^t\left(C_1{\Vert {\varphi }_{xx}\Vert }^{\frac{3}{4}}{\Vert \varphi \Vert }^{\frac{5}{4}}+C_2{\Vert \varphi \Vert }^2\right)}\Vert {\varphi }_{2xxx}\Vert \text{d}\tau \\ \le C{\Vert {\varphi }_{2xxx}\Vert }_{L^2(0,t;L^2(\Omega ))}\left({\left({\displaystyle {\int }_0^t{\Vert {\varphi }_{xx}\Vert }^{\frac{3}{2}}}{\Vert \varphi \Vert }^{\frac{5}{2}}\text{d}\tau \right)}^{\frac{1}{2}}+{\left({\displaystyle {\int }_0^t{\Vert \varphi \Vert }^4}\text{d}\tau \right)}^{\frac{1}{2}}\right)\\ \le C{\left({\displaystyle {\int }_0^t{\Vert {\varphi }_{xx}\Vert }^2}\text{d}\tau \right)}^{\frac{3}{8}}{\left({\displaystyle {\int }_0^t{\Vert \varphi \Vert }^{10}}\text{d}\tau \right)}^{\frac{1}{8}}+C{\left({\displaystyle {\int }_0^t{\Vert \varphi \Vert }^4}\text{d}\tau \right)}^{\frac{1}{2}}\end{array}$

$\begin{array}{l}\le \eta {{\displaystyle {\int }_0^t\Vert {\varphi }_{xx}\left(\tau \right)\Vert }}^2\text{d}\tau +C_{\eta }{\left({{\displaystyle {\int }_0^t\Vert \varphi \left(\tau \right)\Vert }}^{10}\text{d}\tau \right)}^{\frac{1}{5}}+C{\Vert \varphi \Vert }_{L^{\infty }(0,t;L^2(\Omega ))}{\left({\displaystyle {\int }_0^t{\Vert \varphi \Vert }^2}\text{d}\tau \right)}^{\frac{1}{2}}\\ \le \eta {{\displaystyle {\int }_0^t\Vert {\varphi }_{xx}\left(\tau \right)\Vert }}^2\text{d}\tau +C_{\eta }{\Vert \varphi \left(\tau \right)\Vert }_{L^{\infty }(0,t;L^2(\Omega ))}^{\frac{8}{5}}{\left({{\displaystyle {\int }_0^t\Vert \varphi \left(\tau \right)\Vert }}^2\text{d}\tau \right)}^{\frac{1}{5}}+\hat{\eta }{\Vert \varphi \left(\tau \right)\Vert }_{L^{\infty }(0,t;L^2(\Omega ))}^2\\ +C_{\hat{\eta }}{{\displaystyle {\int }_0^t\Vert \varphi \left(\tau \right)\Vert }}^2\text{d}\tau \\ \le \eta {{\displaystyle {\int }_0^t\Vert {\varphi }_{xx}\left(\tau \right)\Vert }}^2\text{d}\tau +\bar{\eta }{\Vert \varphi \left(\tau \right)\Vert }_{L^{\infty }(0,t;L^2(\Omega ))}^2+C_{\eta ,\bar{\eta }}{{\displaystyle {\int }_0^t\Vert \varphi \left(\tau \right)\Vert }}^2\text{d}\tau .\end{array}$ (4.15)

依据Sobolev嵌入定理，我们发现 $H^1\left(\Omega \right)$ 嵌入到 $L^{\infty }\left(\Omega \right)$。因此，借助插值不等式(4.13)和带 $\eta$ 的Young不等式，可以得到

$\begin{array}{l}\left|I_{2,3}\right|=6\left|{\left(M\left({\varphi }_1\right)\left({\varphi }_1+{\varphi }_2\right)\varphi {\varphi }_{2x},{\varphi }_x\right)}_{Q_t}\right|\\ \le 6{\nu }_3{\Vert {\varphi }_1+{\varphi }_2\Vert }_{L^{\infty }(0,t;L^{\infty }(\Omega ))}{\displaystyle {\int }_0^t\Vert \varphi \Vert }\Vert {\varphi }_{2x}\Vert {\Vert {\varphi }_x\Vert }_{L^{\infty }(\Omega )}\text{d}\tau \\ \le C{\Vert {\varphi }_{2x}\Vert }_{L^{\infty }(0,t;L^2(\Omega ))}{\displaystyle {\int }_0^t\left(C_1{\Vert {\varphi }_{xx}\Vert }^{\frac{3}{4}}{\Vert \varphi \Vert }^{\frac{5}{4}}+C_2{\Vert \varphi \Vert }^2\right)\text{d}\tau }\\ \le \eta {{\displaystyle {\int }_0^t\Vert {\varphi }_{xx}\left(\tau \right)\Vert }}^2\text{d}\tau +C_{\eta }{{\displaystyle {\int }_0^t\Vert \varphi \left(\tau \right)\Vert }}^2\text{d}\tau .\end{array}$ (4.16)

考虑假设(H)，插值不等式(4.13)以及一些基本不等式，有

$\begin{array}{l}\left|I_{2,7}\right|=2\left|{\left(M\left({\varphi }_1\right){\alpha }^{″}\left({\varphi }_{1x}\right){\varphi }_{1}n,{\varphi }_x\right)}_{Q_t}\right|\\ \le {\nu }_3{\nu }_6{\displaystyle {\int }_0^t{\Vert {\varphi }_x\Vert }_{L^{\infty }(\Omega )}}\Vert {\varphi }_{1x}\Vert \Vert n\Vert \text{d}\tau \\ \le C{\Vert {\varphi }_{1x}\Vert }_{L^{\infty }(0,t;L^2(\Omega ))}{\displaystyle {\int }_0^t\left({\Vert {\varphi }_{xx}\Vert }^{\frac{3}{4}}{\Vert \varphi \Vert }^{\frac{1}{4}}\Vert n\Vert +\Vert n\Vert \Vert \varphi \Vert \right)\text{d}\tau }\end{array}$

$\begin{array}{l}\le {\displaystyle {\int }_0^t\left(\eta {\Vert {\varphi }_{xx}\Vert }^2+C_{\eta }{\Vert \varphi \Vert }^{\frac{2}{5}}{\Vert n\Vert }^{\frac{8}{5}}\right)\text{d}\tau }+C{\displaystyle {\int }_0^t\left({\Vert n\Vert }^2+{\Vert \varphi \Vert }^2\right)\text{d}\tau }\\ \le \eta {\displaystyle {\int }_0^t{\Vert {\varphi }_{xx}\Vert }^2}\text{d}\tau +\eta {\displaystyle {\int }_0^t{\Vert \varphi \Vert }^2}\text{d}\tau +C_{\eta }{\displaystyle {\int }_0^t{\Vert n\Vert }^2}\text{d}\tau +C{\displaystyle {\int }_0^t\left({\Vert n\Vert }^2+{\Vert \varphi \Vert }^2\right)\text{d}\tau }\\ \le \eta {{\displaystyle {\int }_0^t\Vert {\varphi }_{xx}\left(\tau \right)\Vert }}^2\text{d}\tau +C{{\displaystyle {\int }_0^t\Vert n\left(\tau \right)\Vert }}^2\text{d}\tau +C{{\displaystyle {\int }_0^t\Vert \varphi \left(\tau \right)\Vert }}^2\text{d}\tau .\end{array}$ (4.17)

类似地，不难得到 $I_{2,i}\left(i=4,5,6,8,9,10,12,13\right)$ 的估计

$\left|I_{2,i}\right|\le \eta {{\displaystyle {\int }_0^t\Vert {\varphi }_{xx}\left(\tau \right)\Vert }}^2\text{d}\tau +C_{\eta }{{\displaystyle {\int }_0^t\Vert \varphi \left(\tau \right)\Vert }}^2\text{d}\tau ,$ (4.18)

以及

$\left|I_{2,11}\right|\le \eta {{\displaystyle {\int }_0^t\Vert {\varphi }_{xx}\left(\tau \right)\Vert }}^2\text{d}\tau +C_{\eta }{{\displaystyle {\int }_0^t\Vert \varphi \left(\tau \right)\Vert }}^2\text{d}\tau +C_{\eta }{{\displaystyle {\int }_0^t\Vert n_x\left(\tau \right)\Vert }}^2\text{d}\tau .$ (4.19)

现在，结合(4.14)-(4.19)，选择 $\bar{\eta }=\frac{1}{2}$ 和 $\eta =\frac{{\epsilon }^2{\nu }_2}{13}$ 代入(4.12)，可得

${\Vert \varphi \left(t\right)\Vert }^2\le 2{\Vert \varphi \left(0\right)\Vert }^2+C{{\displaystyle {\int }_0^t\Vert n\left(\tau \right)\Vert }}^2\text{d}\tau +C{{\displaystyle {\int }_0^t\Vert \varphi \left(\tau \right)\Vert }}^2\text{d}\tau +C{{\displaystyle {\int }_0^t\Vert n_x\left(\tau \right)\Vert }}^2\text{d}\tau .$ (4.20)

最后，(4.7)，(4.8)与(4.20)三式相加，对任意的 $t\in \left(0,T_e\right)$，有

$\begin{array}{l}{\Vert n\left(t\right)\Vert }^2+{\Vert n_x\left(t\right)\Vert }^2+{\Vert \varphi \left(t\right)\Vert }^2\le {\Vert n\left(0\right)\Vert }^2+{\Vert n_x\left(0\right)\Vert }^2+2{\Vert \varphi \left(0\right)\Vert }^2\\ +C{\displaystyle {\int }_0^t\left({\Vert n\left(\tau \right)\Vert }^2+{\Vert n_x\left(\tau \right)\Vert }^2\text{d}\tau +{\Vert \varphi \left(\tau \right)\Vert }^2\right)\text{d}\tau }.\end{array}$ (4.21)

根据初始条件(4.3)与(4.4)，利用积分形式的Gronwall不等式 [14]，发现 $n=\varphi =0$ 在 $Q_{T_e}$ 中几乎处处成立。到此证毕。

5. 稳态解的存在性

下面考虑问题(1.1)~(1.6)的稳态问题。稳态解 $\left(\bar{n},\bar{\varphi }\right)$ 满足

$D\Delta \bar{n}-{\alpha }_c\left(\bar{\varphi }\right)=0,$ (5.1)

$\nabla \cdot \left(M\left(\bar{\varphi }\right)\nabla \left(-{\epsilon }^2\Delta \bar{\varphi }+{\bar{\varphi }}^3-\bar{\varphi }+{{\alpha }^{\prime }}_c\left(\bar{\varphi }\right)\bar{n}\right)\right)=0,$ (5.2)

其中 $x\in \Omega$，同时满足边界条件

${\frac{\partial \bar{n}}{\partial \nu }|}_{\partial \Omega }=0,$ (5.3)

${\frac{\partial \bar{\varphi }}{\partial \nu }|}_{\partial \Omega }={\frac{\partial }{\partial \nu }\Delta \bar{\varphi }|}_{\partial \Omega }=0.$ (5.4)

现在给出稳态问题(5.1)~(5.4)弱解的定义和主要结论。

定义2. 称函数 $\left(\bar{n},\bar{\varphi }\right)$ 为问题(5.1)~(5.4)的弱解，并且满足

$\bar{n}\in H^1\left(\Omega \right),\bar{\varphi }\in H^3\left(\Omega \right)\cap H_{\Gamma }^2,$

如果对任意的测试函数 $u\in C^{\infty }\left(\Omega \right)$，满足

$D\left(\nabla \bar{n},\nabla u\right)+\left({\alpha }_c\left(\bar{\varphi }\right),u\right)=0,$ (5.5)

$\left(M\left(\bar{\varphi }\right)\nabla \left(-{\epsilon }^2\Delta \bar{\varphi }+{\bar{\varphi }}^3-\bar{\varphi }+{{\alpha }^{\prime }}_c\left(\bar{\varphi }\right)\bar{n}\right),\nabla u\right)=0.$ (5.6)

定理3. 在定义2的意义下，稳态问题(5.1)~(5.4)存在弱解 $\left(\bar{n},\bar{\varphi }\right)$，并且解满足

$\bar{n}\in H^1\left(\Omega \right),\bar{\varphi }\in H^3\left(\Omega \right).$ (5.7)

首先，对稳态问题(5.1)~(5.4)进行处理。对(1.1)关于 $x$ 做积分，利用分部积分公式和边界条件(1.3)， 可得

$\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }n}\text{d}x={\displaystyle {\int }_{\Omega }{n}_t}\text{d}x=-{\displaystyle {\int }_{\Omega }{\alpha }_c}\left(\varphi \right)\text{d}x.$

对上式关于 $t$ 在 $\left(0,T_e\right)$ 上做积分，并令 $T_e\to \infty$，有

${\displaystyle {\int }_{\Omega }{n}_{\infty }}\text{d}x={\displaystyle {\int }_{\Omega }{n}_0}\text{d}x-{\displaystyle {\int }_0^{\infty }{\displaystyle {\int }_{\Omega }{\alpha }_c}}\left(\varphi \right)\text{d}x\text{d}\tau .$

这里 $n_{\infty }=n\left(\infty ,x\right)$。假设 ${\displaystyle {\int }_{\Omega }{n}_{\infty }}\text{d}x$ 存在，则 ${\displaystyle {\int }_0^{\infty }{\displaystyle {\int }_{\Omega }{\alpha }_c}}\left(\varphi \right)\text{d}x\text{d}\tau$ 存在。由于 ${\alpha }_c\left(\varphi \right)\ge 0$，则 ${\alpha }_c\left(\bar{\varphi }\right)=0$。

于是，方程组(5.1)~(5.2)改写为两个解耦的椭圆型方程，分别为Laplace方程

$\Delta \bar{n}=0,x\in \Omega ,$ (5.8)

和四阶非线性方程

$\nabla \cdot \left(M\left(\bar{\varphi }\right)\nabla \left(-{\epsilon }^2\Delta \bar{\varphi }+{\bar{\varphi }}^3-\bar{\varphi }\right)\right)=0,x\in \Omega .$ (5.9)

接下来化简方程(5.9)，对其关于 $x$ 做积分，再考虑其边界条件以及假设中 $M\left(\varphi \right)>0$，可以将其化简为半线性方程

$-{\epsilon }^2\Delta \bar{\varphi }=-{\bar{\varphi }}^3+\bar{\varphi }+\bar{C},x\in \Omega ,$ (5.10)

这里 $\bar{C}$ 为一常数。

引理4. 方程

$\{\begin{array}{l}\Delta \bar{n}=0,x\in \Omega ,\hfill \\ \frac{\partial n}{\partial \nu }{|}_{\partial \Omega }=0.\hfill \end{array}$

且满足 $n\in H^1\left(\Omega \right)$。

由于方程为Neumann条件下的Laplace方程 [16]，证明省略。

引理5. 方程

$\{\begin{array}{l}\Delta \bar{\varphi }={\bar{\varphi }}^3-\bar{\varphi }+\bar{C},x\in \Omega ,\hfill \\ {\frac{\partial \bar{\varphi }}{\partial \nu }|}_{\partial \Omega }=0.\hfill \end{array}$ (5.11)

存在弱解 $\bar{\varphi }$，并且满足 $\bar{\varphi }\in H^1\left(\Omega \right)$。

方程为Neumann条件下二阶半线性椭圆型方程 [17]，故证明省略。

引理6. 方程

$\{\begin{array}{l}\nabla \cdot \left(M\left(\bar{\varphi }\right)\nabla \left(-{\epsilon }^2\Delta \bar{\varphi }+{\bar{\varphi }}^3-\bar{\varphi }\right)\right)=0,x\in \Omega ,\hfill \\ {\frac{\partial \bar{\varphi }}{\partial \nu }|}_{\partial \Omega }={\frac{\partial }{\partial \nu }\Delta \bar{\varphi }|}_{\partial \Omega }=0.\hfill \end{array}$ (5.12)

存在弱解 $\bar{\varphi }$，并且满足 $\bar{\varphi }\in H^3\left(\Omega \right)$。

证明：由化简过程知，方程(5.11)与方程(5.12)有相同的弱解 $\bar{\varphi }$，下证 $\bar{\varphi }\in H^3\left(\Omega \right)$。方程(5.12)第一式左右同乘 $\Delta \bar{\varphi }$，再关于 $x$ 做积分，利用分部积分公式以及边界条件，不难得到

$\begin{array}{l}{\epsilon }^2{\nu }_2{\Vert \nabla \Delta \bar{\varphi }\Vert }^2\le 3{\nu }_3\left|\left({\varphi }^2\nabla \bar{\varphi },\nabla \Delta \bar{\varphi }\right)\right|+{\nu }_3\left|\left(\nabla \bar{\varphi },\nabla \Delta \bar{\varphi }\right)\right|\\ \le I_{3,1}+I_{3,2}.\end{array}$ (5.13)

利用插值不等式(3.6)不等式，带 $\eta$ 的Young不等式以及引理5，可得 $I_{3,1}$ 的估计

$\begin{array}{l}{I}_{3,1}\le 3{\nu }_3{\Vert \bar{\varphi }\Vert }_{L^6(\Omega )}^2{\Vert \nabla \bar{\varphi }\Vert }_{L^6(\Omega )}\Vert \nabla \Delta \bar{\varphi }\Vert \\ \le C{\Vert \bar{\varphi }\Vert }_{H^1(\Omega )}^2{\Vert \nabla \Delta \bar{\varphi }\Vert }^{\frac{5}{3}}{\Vert \bar{\varphi }\Vert }^{\frac{1}{3}}+C{\Vert \bar{\varphi }\Vert }_{H^1(\Omega )}^2\Vert \bar{\varphi }\Vert \Vert \nabla \Delta \bar{\varphi }\Vert \\ \le \eta {\Vert \nabla \Delta \bar{\varphi }\Vert }^2+C.\end{array}$ (5.14)

类似地，不难得到

$I_{3,2}\le \eta {\Vert \nabla \Delta \bar{\varphi }\Vert }^2+C.$ (5.15)

结合(5.14)~(5.15)，并选择 $\eta =\frac{{\epsilon }^2{\nu }_2}{4}$ 代入(5.13)，有

$\Vert \nabla \Delta \bar{\varphi }\Vert \le C.$ (5.16)

至此证毕。

最后，结合引理4~引理6，定理3得证。