一类描述肉芽肿生长的自由边界问题的数学分析

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一类描述肉芽肿生长的自由边界问题的数学分析
Analysis of a Free Boundary Problem Modeling the Granuloma Formation

DOI: 10.12677/PM.2019.92028, PDF, 下载: 822 浏览: 1,615
作者: 韩莹莹：华南理工大学数学学院，广东广州
关键词: 肿瘤模型；自由边界问题；整体解；Tumor Model； Free Boundary Problem； Global Solution

摘要: 本文研究描述一类肉芽肿瘤生长的自由边界问题，它由定义在移动区域上的相互耦合的抛物型和双曲型方程组构成。我们首先将自由边界问题转换成固定边界上的问题，然后利用抛物型方程的Lp理论、双曲方程的特征线理论和Banach不动点定理证明该问题局部解的存在唯一性，最后利用先验估计得到整体解的存在唯一性。

Abstract: In this paper we study a free boundary value problem modeling the growth of Leishmaniasis. It consists of coupled parabolic equations and hyperbolic equations with moving boundary. Firstly, we convert the free boundary problem into an equivalent problem defined on fixed domain. Then we use Lp theory of parabolic equations, characteristic theory of hyperbolic equations and Banach fixed point theorem to prove existence and uniqueness of a local solution. Finally we extend this local solution to be global by employing a priori estimate.

文章引用：韩莹莹. 一类描述肉芽肿生长的自由边界问题的数学分析[J]. 理论数学, 2019, 9(2): 217-229. https://doi.org/10.12677/PM.2019.92028

1. 引言

利什曼病是由利什曼原虫引起的寄生虫病，利什曼病的两种常见形式是皮肤利什曼病和内脏利什曼病，而内脏利什曼病是两种疾病中较严重的一种，它的标志是肝脏或脾脏内形成肉芽肿。因此研究肉芽肿的形成过程对治疗内脏利什曼病有着积极的意义。本文研究的就是内脏肉芽肿形成的数学模型。该模型是由Nourridine Siewe，Abdul-Aziz Yakubu等人在论文 [1] 提出的，是由抛物型方程和双曲型方程组成的偏微分方程组，模型具体表达如下：

$\begin{array}{l} \frac{\partial P}{\partial t} + \nabla \cdot (v P) - δ_{P} \nabla^{2} P = α_{1} P (1 - \frac{P}{M}) + \tilde{k_{5}} M \frac{P^{2}}{P^{2} + M^{2}} (θ - 1) + k_{5} N \frac{Q^{2}}{Q^{2} + N^{2}} θ \\ - μ_{P} P - μ_{M} P, | x | < R (t), t > 0, \end{array}$ (1.1)

$\begin{array}{l} \frac{\partial Q}{\partial t} + \nabla \cdot (v Q) - δ_{N} \nabla^{2} Q = α_{2} Q (1 - \frac{Q}{N}) + \tilde{k_{5}} M \frac{P^{2}}{P^{2} + M^{2}} (1 - θ) + k_{5} N \frac{Q^{2}}{Q^{2} + N^{2}} (1 - θ) \\ - \tilde{k_{5}} N \frac{Q^{2}}{Q^{2} + N^{2}} - μ_{Q} Q - μ_{N} Q, | x | < R (t), t > 0, \end{array}$ (1.2)

$\frac{\partial M}{\partial t} + \nabla \cdot (v M) = - k_{5} M \frac{P^{2}}{P^{2} + M^{2}} - μ_{M} M, | x | < R (t), t > 0,$ (1.3)

$\frac{\partial N}{\partial t} + \nabla \cdot (v N) = - k_{5} N \frac{Q^{2}}{Q^{2} + N^{2}} - μ_{N} N, | x | < R (t), t > 0,$ (1.4)

$M + N = 1, | x | < R (t), t > 0,$ (1.5)

$\frac{d R}{d t} = \frac{x}{| x |} \cdot v, | x | < R (t), t > 0,$ (1.6)

$P = P_{0}, Q = Q_{0}, M = M_{0}, N = N_{0}, | x | < R (t), t > 0,$ (1.7)

$R = R_{0}, t = 0.$ (1.8)

其中， $M = M (x, t), N = N (x, t), P = P (x, t), Q = Q (x, t)$ 分别代表促炎巨噬细胞密度，抗炎巨噬细胞密度，促炎巨噬细胞中的寄生虫密度和抗炎巨噬细胞中的寄生虫密度， $v (x, t)$ 代表细胞运动速度， $μ_{M}, μ_{N}$ 代表促炎巨噬细胞和抗炎巨噬细胞的死亡比率， $μ_{P}, μ_{Q}$ 代表促炎巨噬细胞和抗炎巨噬细胞中寄生虫的死亡比率， $a_{1}, a_{2}$ 代表促炎巨噬细胞和抗炎巨噬细胞中寄生虫的增长系数， $k_{5}, \tilde{k_{5}}$ 代表促炎巨噬细胞和抗炎巨噬细胞中寄生虫的扩散系数，这些参数均是大于0的参数。

由于 $M, N, P, Q, π$ 是球对称的，所以

$P = P (| x |, t), Q = Q (| x |, t), M = M (| x |, t), N = N (| x |, t), \forall x \in R^{3} .$

且存在一个标量函数 $v = v (x, t)$ ，使得 $v = v (| x |, t) \cdot \frac{\vec{x}}{| x |}, | x | < R (t), t > 0$ 。

令 $r = | x |$ ，有 $\nabla P = (\frac{\partial P}{\partial x_{1}}, \frac{\partial P}{\partial x_{2}}, \frac{\partial P}{\partial x_{3}}) = \frac{\vec{x}}{r} \cdot \frac{\partial P}{\partial r}$ ， $\nabla \cdot (\nabla P) = \frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial P}{\partial r})$ 。

则问题(1.1)~(1.8)可转化为：

$\frac{\partial P}{\partial t} = δ_{P} \frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial P}{\partial r}) - v \frac{\partial P}{\partial r} + F_{1} (P, Q, M, N) P, 0 < r < R (t), t > 0,$ (1.9)

$\frac{\partial Q}{\partial t} = δ_{N} \frac{\partial}{\partial r} (r^{2} \frac{\partial Q}{\partial r}) - v \frac{\partial Q}{\partial r} + F_{2} (P, Q, M, N) Q, 0 < r < R (t), t > 0,$ (1.10)

$\frac{\partial M}{\partial t} + v \frac{\partial M}{\partial r} = g_{11} (P, Q, M, N) M + g_{12} (P, Q, M, N) N, 0 < r < R (t), t > 0.$ (1.11)

$\frac{\partial N}{\partial t} + v \frac{\partial N}{\partial r} = g_{21} (P, Q, M, N) M + g_{22} (P, Q, M, N) N, 0 < r < R (t), t > 0,$ (1.12)

$\frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} v) = h (P, Q, M, N), 0 < r \leq R (t), t > 0,$ (1.13)

$\frac{\partial P}{\partial r} (0, t) = 0, P (R (t), t) = \bar{P}, \frac{\partial Q}{\partial r} (0, t) = 0, Q (R (t), t) = \bar{Q}, t > 0.$ (1.14)

$v (0, t) = 0, t > 0,$ (1.15)

$\frac{d R (t)}{d t} = v (R (t), t), t > 0,$ (1.16)

$P (r, 0) = P_{0} (r), Q (r, 0) = Q_{0} (r), | x | < R (t), 0 \leq r \leq R_{0},$ (1.17)

$M (r, 0) = M_{0} (r), N (r, 0) = N_{0} (r), | x | < R (t), 0 \leq r \leq R_{0},$ (1.18)

$R (0) = R_{0} .$ (1.19)

其中，

$h (P, Q, M, N) = \nabla \cdot v = - k_{5} M \frac{P^{2}}{P^{2} + M^{2}} - μ_{M} M - k_{5} N \frac{Q^{2}}{Q^{2} + N^{2}} - μ_{N} N,$

$F_{1} (P, Q, M, N) = α_{1} (1 - \frac{P}{M}) + \tilde{k_{5}} M \frac{P}{P^{2} + M^{2}} (θ - 1) + k_{5} \frac{N}{P} \frac{Q^{2}}{Q^{2} + N^{2}} θ - μ_{P} - μ_{M} - \nabla \cdot v,$

$\begin{array}{l} F_{2} (M, N, P, Q) = α_{2} (1 - \frac{Q}{N}) + k_{5} N \frac{Q}{Q^{2} + N^{2}} (1 - θ) + \tilde{k_{5}} \frac{M}{Q} \frac{P^{2}}{P^{2} + M^{2}} (1 - θ) \\ - \tilde{k_{5}} N \frac{Q}{Q^{2} + N^{2}} - μ_{Q} - μ_{N} - \nabla \cdot v, \end{array}$

$g_{11} (M, N, P, Q) = - k_{5} \frac{P^{2}}{P^{2} + M^{2}} - μ_{M} + k_{5} M \frac{P^{2}}{P^{2} + M^{2}} + μ_{M} M,$

$g_{12} (M, N, P, Q) = k_{5} M \frac{Q}{Q^{2} + N^{2}} + μ_{N} M,$

$g_{12} (M, N, P, Q) = k_{5} N \frac{P^{2}}{P^{2} + M^{2}} + μ_{M} N,$

$g_{22} (M, N, P, Q) = - k_{5} \frac{Q^{2}}{Q^{2} + N^{2}} - μ_{N} + k_{5} N \frac{Q^{2}}{Q^{2} + N^{2}} + μ_{N} N .$

1999年Friedman和其合作者开始进行关于肿瘤生长模型的严谨数学分析，通过一系列研究，得到了模型整体解和稳态解的存在唯一性 [2] 。随后，出现了大量的数学研究工作者对肿瘤生长自由边界问题的数学分析。参考见 [3] [4] [5] [6] [7] 。根据生物学及医学原理，本文做出以下假设：

(A₁) $M_{0}, N_{0}$ 在 $[0, R_{0}]$ 上连续可微，当 $p > 5$ 时，有 $P_{0} (| x |) \in D_{p} (B (R_{0}))$ ， $Q_{0} (| x |) \in D_{p} (B (R_{0}))$ ， $B (R_{0}) = {x \in R^{3} : | x | < R_{0}}$ ；

(A₂) 当 $0 \leq r \leq R_{0}$ 时， $0 \leq P_{0} (r) \leq \bar{P_{0}}, 0 \leq Q_{0} (r) \leq \bar{Q_{0}}, M_{0} (r) \geq 0, N_{0} (r) \geq 0, M_{0} (r) + N_{0} (r) = 1$ ，且 ${P^{'}}_{0} (0) = 0, P_{0} (R_{0}) = 0, {Q^{'}}_{0} (0) = 0, Q_{0} (R_{0}) = 0$ 。

定理1：在条件(A₁)~(A₂)成立时，对任意的 $t \geq 0$ ，问题(1.9)~(1.19)都存在唯一解 $(R, P, Q, M, N)$ ，其中 $R \in C^{1} [0, T], M, N \in C^{1} ({\bar{Q}}_{T}^{R}), P, Q \in W_{p}^{2, 1} (Q_{T}^{R})$ ，同时有以下结论成立：

$R (t) \geq 0, t > 0,$ (1.20)

$0 \leq P (r, t) \leq \bar{P}, 0 \leq Q (r, t) \leq \bar{Q}, 0 \leq r \leq R_{0}, t \geq 0,$ (1.21)

$M (r, t) \geq 0, N (r, t) \geq 0, 0 \leq r \leq R (t), t \geq 0,$ (1.22)

$M (r, t) + N (r, t) = 1, 0 \leq r \leq R (t), t \geq 0.$ (1.23)

2. 基本引理

首先引进一些基本记号：

1) $Q_{T}^{R} = {(x, t) \in R^{3} \times R : | x | < R (t), 0 < t < T}$ ， ${\bar{Q}}_{T}^{R}$ 是 $Q_{T}^{R}$ 的闭包。

2) $W_{p}^{2, 1} (Q_{T}^{R}) = {μ \in L_{P} (Q_{T}^{R}) : \partial_{x}^{m} \partial_{t}^{k} \in L_{P} (Q_{T}^{R}), | m | + 2 k \leq 2}$ ，其范数为

${‖ μ ‖}_{W_{p}^{2, 1} (Q_{T})} = \sum {‖ \partial_{x}^{m} \partial_{t}^{k} μ ‖}_{p} .$

3) 对开集 $Ω \in R^{3}, p > \frac{5}{2}$ ，记 $D_{p} (Ω)$ 是 $W_{p}^{2, 1} (Ω \times (0, T))$ 在 $t = 0$ 的迹空间， $i . e . φ \in D_{p} (Ω)$ ，当且仅当存在 $μ \in W_{p}^{2, 1} (Q_{T})$ ，使得 $μ (\cdot, 0) = φ$ 。当 $p > \frac{5}{2}$ 时， $W_{p}^{2, 1} (Ω \times (0, T)) \to C (Ω \times (0, T))$ 。 $D_{p} (Ω)$ 的范数为：

${‖ μ ‖}_{D_{p} (Ω)} = i n f {T^{- \frac{1}{p}} {‖ μ ‖}_{W_{p}^{2, 1} (Ω \times (0, T))} : μ \in W_{p}^{2, 1} (Ω \times (0, T)), μ (\cdot, 0) = φ}$

明显地，如果 $φ \in W^{2, p} (Ω)$ ，则 $φ \in D_{p} (Ω)$ ，且 ${‖ φ ‖}_{D_{p} (Ω)} \leq {‖ φ ‖}_{W^{2, p} (Ω)}$ 。

下面介绍本文需要的引理。

引理2.1：设 $a (t)$ ， $b (x, t)$ ， $f (x, t)$ 分别是 $[0, T]$ 和 $[0, 1] \times [0, T] (T > 0)$ 的有界连续函数， $\bar{m}$ 是一个常

数， $m_{0} (x)$ 是 $[0, 1]$ 的函数且 $m_{0} (| x |) \in D_{p} (B_{1}) (p > \frac{5}{2})$ ，其中 $B_{1}$ 是 $R^{3}$ 的单位球。则下列初边值问题：

$\frac{\partial m}{\partial t} = \frac{1}{x^{2}} \frac{\partial}{\partial x} (x^{2} \frac{\partial m}{\partial x}) + a (t) \frac{\partial m}{\partial x} + b (x, t) + f (x, t), 0 \leq x \leq 1, 0 \leq τ \leq T,$ (2.1)

$\frac{\partial m}{\partial t} (0, t) = 0, m (1, t) = \bar{m}, 0 \leq t \leq T,$ (2.2)

$m (x, 0) = m_{0} (x), 0 \leq x \leq 1,$ (2.3)

有唯一的解 $m (x, t) \in W_{p}^{2, 1} (Q_{T})$ ，同时有

${‖ m (x, t) ‖}_{W_{p}^{2, 1} (Q_{T})} \leq C_{p} (T) (\bar{m} + {‖ m_{0} ‖}_{D_{p} (B_{1})} + {‖ f ‖}_{p}),$

其中 $C_{p} (T)$ 是依赖于 $p, T, {‖ a ‖}_{\infty}, {‖ b ‖}_{\infty}$ 的常数，而且对任意的T， $C_{p} (T)$ 是有界的。更进一步，有

${‖ m ‖}_{\infty} \leq \max (| \bar{m} |, {‖ m_{0} ‖}_{\infty}) + T e^{c_{0} T} {‖ f ‖}_{p},$

其中若 $b \leq 0$ 时， $c_{0} = 0$ ；否则 $b > 0$ ， $c_{0} = \max_{Q_{T}} b$ 。

证明：引理2.1的证明可参见 [8] 中的引理4.1的证明。

引理2.2：设 $μ (ρ, τ)$ ， $a_{i j} (ρ, τ) (i, j = 1, 2)$ ， $f_{i} (ρ, τ) (i = 1, 2)$ 是 $[0, 1] \times [0, T] (T > 0)$ 上的有界连续函数， $μ (ρ, τ)$ 关于 $ρ$ 连续可微， $μ (0, τ) = μ (1, τ)$ ，则对任意 $(P, Q, M, N) \in C [0, 1]$ ，则下列初边值问题

$\frac{\partial M}{\partial τ} + μ (ρ, τ) \frac{\partial M}{\partial t} = a_{11} (ρ, τ) M + a_{12} (ρ, τ) N + f_{1} (x, t), 0 \leq ρ \leq 1, 0 \leq τ \leq T,$ (2.4)

$\frac{\partial N}{\partial τ} + μ (ρ, τ) \frac{\partial N}{\partial t} = a_{21} (ρ, τ) M + a_{22} (ρ, τ) N + f_{2} (x, t), 0 \leq ρ \leq 1, 0 \leq τ \leq T,$ (2.5)

$M (ρ, 0) = M_{0} (ρ), N (ρ, 0) = N_{0} (ρ), 0 \leq ρ \leq 1,$ (2.6)

有唯一的弱解 $(M, N) \in C ([0, 1] \times C [0, T])$ ，且满足

${‖ (M, N) ‖}_{\infty} \leq e^{T A_{0} (T)} ({‖ (M_{0}, N_{0}) ‖}_{\infty} + 2 T {‖ (f_{1}, f_{2}) ‖}_{\infty}),$

其中 $A_{0} (T) = 8 \max {{‖ a_{i j} ‖}_{\infty} : i, j = 1, 2}$ 。当 $a_{i j} (ρ, τ) (i, j = 1, 2), f_{i} (ρ, τ) (i = 1, 2)$ 关于 $ρ$ 连续可微， $(M_{0}, N_{0}) \in C^{1} [0, 1]$ ，则问题(2.4)~(2.6)的弱解是经典解。而且有如下估计成立：

${‖ \frac{\partial M}{\partial ρ}, \frac{\partial N}{\partial ρ} ‖}_{\infty} \leq e^{T (A_{0} (T) + A (T))} ({‖ {M^{'}}_{0}, {N^{'}}_{0} ‖}_{\infty} + T A_{1} (T_{0}) e^{A (T)} {‖ M_{0}, N_{0} ‖}_{\infty} + 2 T e^{A (T)} {‖ \frac{\partial f_{1}}{\partial ρ}, \frac{\partial f_{2}}{\partial ρ} ‖}_{\infty}),$ (2.7)

其中当 $A_{1} (T_{0}) = 8 \max {{‖ \frac{\partial a_{i j}}{\partial ρ} ‖}_{\infty} : i, j = 1, 2}$ ， $A (T) = {‖ \frac{\partial u}{\partial ρ} ‖}_{\infty}$ 。

当 $\partial a_{i j} (ρ) \geq 0, i \neq j, M_{0} (ρ) \geq 0, N_{0} (ρ) \geq 0, f_{i j} (ρ, τ) \geq 0 (i, j = 1, 2), 0 \leq ρ \leq 1, 0 \leq τ \leq T$ ，有

(2.8)

证明：引理2.2的证明可参见 [8] 中的引理4.2的证明。

引理2.3：设 $f_{i} (ρ, τ, M, N) (i = 1, 2)$ 是 $[0, 1] \times [0, T] (T > 0), (M, N) \in R^{3}$ 的有界连续函数，并且对所有参数都连续，关于 $(ρ, τ, M, N)$ 连续可微， $μ (ρ, τ)$ 关于 $ρ$ 连续可微， $μ (ρ, τ) = μ (1, τ)$ ，则下列初边值问题

$\frac{\partial M}{\partial τ} + μ (ρ, τ) \frac{\partial M}{\partial ρ} = f_{1} (ρ, τ, M, N), 0 \leq ρ \leq 1, 0 \leq τ \leq T,$ (2.9)

$\frac{\partial N}{\partial τ} + μ (ρ, τ) \frac{\partial N}{\partial ρ} = f_{2} (ρ, τ, M, N), 0 \leq ρ \leq 1, 0 \leq τ \leq T,$ (2.10)

$M (ρ, 0) = M_{0} (ρ), N (ρ, 0) = N_{0} (ρ), 0 \leq ρ \leq 1,$ (2.11)

如果 $(M_{0}, N_{0}) \in C [0, 1]$ ，则存在 $0 \leq T \leq T_{1}$ ，且T是只依赖于 $H_{0} = {‖ M_{0}, N_{0} ‖}_{\infty}, f_{i}, \frac{\partial f_{i}}{\partial M}, \frac{\partial f_{i}}{\partial N}$ 。

在 $[0, 1] \times [0, T] \times [- 2 H_{0}, 2 H_{0}] \times [- 2 H_{0}, 2 H_{0}]$ 上的上确界的常数，使得上述问题(2.9)-(2.11)有唯一的弱解 $(M, N) \in C ([0, 1] \times [0, T_{1}])$ ，同时满足

${‖ (M, N) ‖}_{\infty} \leq 2 H_{0},$

当 $(M_{0}, N_{0}) \in C^{1} [0, 1]$ ，则上面问题的弱解是经典解。而且有下面估计成立：

${‖ \frac{\partial M}{\partial ρ}, \frac{\partial N}{\partial ρ} ‖}_{\infty} \leq e^{T (A (T) + B_{0} (T))} ({‖ {M^{'}}_{0}, {N^{'}}_{0} ‖}_{\infty} + 2 T e^{A (T)} {‖ \frac{\partial f_{1}}{\partial ρ}, \frac{\partial f_{2}}{\partial ρ} ‖}_{\infty}),$ (2.12)

其中 $B_{0} (T) = \max_{1 \leq i \leq 2} {{‖ \frac{\partial f_{i}}{\partial M} ‖}_{\infty}, {‖ \frac{\partial f_{i}}{\partial N} ‖}_{\infty}}, A (T) = {‖ \frac{\partial u}{\partial ρ} ‖}_{\infty}$ 。

证明：引理2.3的证明可参见 [8] 中的引理4.3的证明。

3. 局部解的存在唯一性

本节将证明自由边界问题(1.9)~(1.19)存在唯一局部解。

首先，将问题(1.9)~(1.19)转换成固定边界问题。若问题(1.9)~(1.19)有一个解 $(R, P, Q, M, N)$ ，且对任意的 $t \geq 0$ ， $R (t) > 0$ 都成立，则通过如下变换：

$\begin{array}{l} ρ = \frac{r}{R (t)}, τ = \int_{0}^{t} \frac{d s}{R^{2} (s)}, η (τ) = R (t), υ (ρ, τ) = R (t) v (r, t), \\ α (ρ, τ) = P (r, t), β (ρ, τ) = Q (r, t), σ (ρ, τ) = M (r, t), γ (ρ, τ) = N (r, t), \end{array}$ (3.1)

问题(1.9)~(1.19)可转换为固定区域为 ${(ρ, τ) : 0 \leq ρ \leq 1, τ > 0}$ 上的初边值问题：

$\frac{\partial α}{\partial τ} = δ_{P} \frac{1}{ρ^{2}} \frac{\partial}{\partial ρ} (ρ^{2} \frac{\partial α}{\partial ρ}) - μ (ρ, τ) \frac{\partial α}{\partial ρ} + η^{2} (τ) (F_{1} (α, β, σ, γ) α), 0 < ρ < 1, τ > 0,$ (3.2)

$\frac{\partial β}{\partial τ} = δ_{Q} \frac{\partial}{\partial ρ} (ρ^{2} \frac{\partial β}{\partial ρ}) - μ (ρ, τ) \frac{\partial β}{\partial ρ} + η^{2} (τ) (F_{2} (α, β, σ, γ) β), 0 < ρ < 1, τ > 0,$ (3.3)

$\frac{\partial σ}{\partial τ} + μ \frac{\partial σ}{\partial ρ} = η^{2} (τ) [g_{11} (α, β, σ, γ) δ + g_{12} (α, β, σ, γ) γ], 0 \leq ρ \leq 1, τ > 0,$ (3.4)

$\frac{\partial γ}{\partial t} + μ \frac{\partial γ}{\partial r} = η^{2} (τ) [g_{21} (α, β, σ, γ) δ + g_{22} (α, β, σ, γ) γ], 0 \leq ρ \leq 1, τ > 0,$ (3.5)

$μ (ρ, τ) = υ (ρ, τ) - ρ υ (1, τ), 0 \leq ρ \leq 1, τ > 0,$ (3.6)

$\frac{1}{ρ^{2}} \frac{\partial}{\partial ρ} (ρ^{2} υ) = η^{2} (τ) h (α, β, σ, γ), 0 \leq ρ \leq 1, τ > 0,$ (3.7)

$\frac{\partial α (0, τ)}{\partial ρ} = 0, α (1, τ) = \bar{α}, \frac{\partial β (0, τ)}{\partial ρ} = 0, β (1, τ) = \bar{β}, τ > 0,$ (3.8)

$υ (0, τ) = 0, τ > 0,$ (3.9)

$\frac{d η (τ)}{d τ} = η (τ) υ (1, τ), τ > 0,$ (3.10)

$α (ρ, 0) = α_{0} (ρ), β (ρ, 0) = β_{0} (ρ), 0 \leq ρ \leq 1,$ (3.11)

$σ (ρ, 0) = σ_{0} (ρ), γ (ρ, 0) = γ_{0} (ρ), 0 \leq ρ \leq 1,$ (3.12)

$η (0) = η_{0} = R_{0} .$ (3.13)

其中，

$α_{0} (ρ) = P_{0} (ρ R_{0}), β_{0} (ρ) = Q_{0} (ρ R_{0}), σ_{0} (ρ) = M_{0} (ρ R_{0}), γ_{0} (ρ) = N_{0} (ρ R_{0}) .$

反过来，若 $(η, α, β, σ, γ, υ)$ 是问题(3.2)~(3.13)的解，对任意的 $τ \geq 0, η (τ) > 0$ ，则通过如下变换：

$\begin{array}{l} r = ρ η (τ), t = \int_{0}^{τ} η^{2} (s) d s, R (t) = η (τ), v (r, t) = \frac{υ (ρ, τ)}{η^{2} (τ)}, \\ P (r, t) = α (ρ, τ), Q (r, t) = β (ρ, τ), M (r, t) = σ (ρ, τ), N (r, t) = γ (ρ, τ), \end{array}$ (3.14)

则可以验证 $(R, P, Q, M, N, v)$ 是问题(1.9)~(1.19)的解，这说明通过变换(3.1)和其逆变换(3.14)，自由边界问题(1.9)~(1.19)和固定边界问题(3.2)~(3.13)是等价的。

对方程(3.7)积分，可得：

$υ (ρ, τ) = \frac{η^{2} (τ)}{ρ^{2}} \int_{0}^{ρ} h (α, β, σ, γ) s^{2} d s,$ (3.15)

将(3.15)代入(3.10)可得：

$\frac{d η (τ)}{d τ} = η^{3} (τ) \int_{0}^{ρ} h (α, β, σ, γ) s^{2} d s,$ (3.16)

所以，问题(1.9)~(1.19)变换成问题(3.2)~(3.5)和(3.16)，同时加上初边值条件(3.8)~(3.9)和(3.11)~(3.12)。

接下来我们将采用压缩映像原理来证明固定边界问题(3.2)~(3.13)局部解的存在唯一性，从而得出自由边界问题(3.9)~(3.19)局部解的存在唯一性。

根据定理1的假设条件(A₁)~(A₁)，则有如下条件成立： $σ_{0}, γ_{0}$ 在 $[0, 1]$ 上连续可微，对适应的 $p > 5$ 满足 $α_{0} (| x |) \in D_{p} (B_{1}), β_{0} (| x |) \in D_{p} (B_{1})$ 。

引入记号：

$A_{0} = {‖ (σ_{0}, γ_{0}) ‖}_{\infty},$

$B_{0} = \max {| h (α, β, σ, γ) | : 0 \leq α \leq \bar{α}, 0 \leq β \leq \bar{β}, | σ | \leq 2 A_{0}, | γ | \leq 2 A_{0}} .$

对任意给定的 $T > 0$ ，引入度量空间 $(X_{T}, d)$ ：

$X_{T}$ 由向量值函数 $(η, α, β, σ, γ) = (η (τ), α (ρ, τ), β (ρ, τ), σ (ρ, τ), γ (ρ, τ))$ 组成，满足：

i) $η (τ) \in [0, T], η (0) = η_{0}, \frac{1}{2} η_{0} \leq η (τ) \leq 2 η_{0} (0 \leq τ \leq T);$

ii) $α (ρ, τ) \in C ([0, 1] \times [0, T]), α (ρ, 0) = α_{0} (ρ), α (ρ, 1) = \bar{α}, 0 \leq α (ρ, τ) \leq \bar{α},$ $β (ρ, τ) \in C ([0, 1] \times [0, T]), β (ρ, 0) = β_{0} (ρ), β (ρ, 1) = \bar{β}, 0 \leq β (ρ, τ) \leq \bar{β},$

iii) $σ (ρ, τ), γ (ρ, τ) \in C ([0, 1] \times [0, T]), σ (ρ, 0) = σ_{0} (ρ), γ (ρ, 0) = γ_{0} (ρ),$ $| σ (ρ, τ) | \leq 2 A_{0}, | γ (ρ, τ) | \leq 2 A_{0}, 0 \leq ρ \leq 1, 0 \leq τ \leq T .$

定义 $(X_{T}, d)$ 中的度量d为

$\begin{array}{l} d ((η_{1}, α_{1}, β_{1}, σ_{1}, γ_{1}), (η, α_{2}, β_{2}, σ_{2}, γ_{2})) \\ = {‖ η_{1} - η_{2} ‖}_{\infty} + {‖ α_{1} - α_{2} ‖}_{\infty} + {‖ β_{1} - β_{2} ‖}_{\infty} + {‖ σ_{1} - σ_{2} ‖}_{\infty} + {‖ γ_{1} - γ_{2} ‖}_{\infty} . \end{array}$

显然， $(X_{T}, d)$ 是完备度量空间。

定义映射F： $X_{T} \to {X^{'}}_{T} : (η, α, β, σ, γ) \mapsto (\tilde{η}, \tilde{α}, \tilde{β}, \tilde{σ}, \tilde{γ})$ ，其中 $(\tilde{η}, \tilde{α}, \tilde{β}, \tilde{σ}, \tilde{γ})$ 是下列问题的解：

$\frac{d \tilde{η} (τ)}{d τ} = \tilde{η} (τ) v (1, τ), τ > 0,$ (3.17)

$\tilde{η} (0) = η_{0},$ (3.18)

$\frac{\partial \tilde{α}}{\partial τ} = δ_{P} \frac{1}{ρ^{2}} \frac{\partial}{\partial ρ} (ρ^{2} \frac{\partial \tilde{α}}{\partial ρ}) - μ (ρ, τ) \frac{\partial \tilde{α}}{\partial ρ} + η^{2} (τ) (F_{1} (\tilde{α}, \tilde{β}, \tilde{σ}, \tilde{γ}) \tilde{α}), 0 < ρ < 1, τ > 0,$ (3.19)

$\frac{\partial \tilde{α}}{\partial t} (0, τ) = 0, \tilde{α} (1, τ) = \bar{α}, τ > 0,$ (3.20)

$α (ρ, 0) = α_{0} (ρ), 0 \leq ρ \leq 1,$ (3.21)

$\frac{\partial \tilde{β}}{\partial τ} = δ_{Q} \frac{1}{ρ^{2}} \frac{\partial}{\partial ρ} (ρ^{2} \frac{\partial \tilde{β}}{\partial ρ}) - μ (ρ, τ) \frac{\partial \tilde{β}}{\partial ρ} + η^{2} (τ) (F_{2} (\tilde{α}, \tilde{β}, \tilde{σ}, \tilde{γ}) \tilde{β}), 0 < ρ < 1, τ > 0,$ (3.22)

$\frac{\partial \tilde{β}}{\partial t} (0, τ) = 0, \tilde{β} (1, τ) = \bar{β}, τ > 0,$ (3.23)

$\tilde{β} (ρ, 0) = β_{0} (ρ), 0 \leq ρ \leq 1,$ (3.24)

$\frac{\partial \tilde{σ}}{\partial τ} + μ \frac{\partial \tilde{σ}}{\partial ρ} = η^{2} (τ) [g_{11} (\tilde{α}, \tilde{β}, \tilde{σ}, \tilde{γ}) \tilde{σ} + g_{12} (\tilde{α}, \tilde{β}, \tilde{σ}, \tilde{γ}) \tilde{γ}], 0 \leq ρ \leq 1, τ > 0,$ (3.25)

$\frac{\partial \tilde{γ}}{\partial τ} + μ \frac{\partial \tilde{γ}}{\partial ρ} = η^{2} (τ) [g_{21} (\tilde{α}, \tilde{β}, \tilde{σ}, \tilde{γ}) \tilde{σ} + g_{22} (\tilde{α}, \tilde{β}, \tilde{σ}, \tilde{γ}) \tilde{γ}], 0 \leq ρ \leq 1, τ > 0,$ (3.26)

$\tilde{σ} (ρ, 0) = σ_{0} (ρ), \tilde{γ} (ρ, 0) = γ_{0} (ρ), 0 \leq ρ \leq 1.$ (3.27)

其中，

$υ (ρ, τ) = \frac{η {(τ)}^{2}}{ρ^{2}} \int_{0}^{1} h (\tilde{α}, \tilde{β}, \tilde{σ}, \tilde{γ}) s^{2} d s,$ (3.28)

$μ (ρ, τ) = υ (ρ, τ) - ρ υ (1, τ) .$ (3.29)

下面证明F是一个自映射。首先，对于常微分方程组(3.17)~(3.18)，存在唯一解 $\tilde{η} (τ) \in C^{1} [0, T]$ 并且

$\tilde{η} (τ) = η_{0} e^{\int_{t}^{0} υ (1, s) d s}, 0 \leq τ \leq T .$ (3.30)

根据式(3.28)和 $B_{0}$ 的定义可得，

$| υ (1, τ) | = | η {(τ)}^{2} \int_{0}^{1} h (α, β, σ, γ) s^{2} d s | \leq 4 η_{0} \int_{0}^{1} B_{0} s^{2} d s \leq \frac{4}{3} B_{0} η_{0}^{2} .$ (3.31)

因此

$η_{0} e^{- \frac{4}{3} B_{0} η_{0}^{2} T} \leq \tilde{η} (τ) \leq η_{0} e^{\frac{4}{3} B_{0} η_{0}^{2} T}, 0 \leq τ \leq T .$ (3.32)

当 $T > 0$ 充分小，使 $e^{\frac{4}{3} B_{0} η_{0}^{2} T} \leq 2$ 时，有 $\frac{1}{2} η_{0} \leq \tilde{η} (τ) \leq 2 η_{0} (0 \leq τ \leq T)$ ，即 $\tilde{η} (τ)$ 满足条件(i)。

然后，由引理2.1得，问题(3.19)~(3.21)存在唯一解 $α (ρ, τ) \in W_{p}^{2, 1} (Q_{T})$ 。因为当 $p > 5$ 时，

$W_{p}^{2, 1} (Q_{T}) \to C^{λ, \frac{2}{λ}} (\bar{Q_{T}}) (λ = 2 - \frac{5}{p})$ ，

所以 $\tilde{α}, \frac{\partial \tilde{α}}{\partial ρ} \in C ([0, 1] \times [0, T])$ 。又因为 $\tilde{α} (ρ, τ)$ 非负， $0 \leq \tilde{α} (ρ, τ) \leq \bar{α}$ ，对任意的 $0 \leq ρ \leq 1, 0 \leq τ \leq T$ 都成立。所以 $\tilde{α}$ 符合条件(ii)，同时满足

${‖ \frac{\partial \tilde{α}}{\partial ρ} ‖}_{\infty} \leq C (T),$ (3.33)

其中 $C (T)$ 是依赖于 $(η, α, β, σ, γ)$ 的常数。同理可证 $\tilde{β}$ 符合条件(ii)，满足 ${‖ \frac{\partial \tilde{β}}{\partial ρ} ‖}_{\infty} \leq C (T)$ 。

最后，由于 $μ (ρ, τ), \tilde{α} (ρ, τ), \tilde{β} (ρ, τ)$ 关于 $ρ$ 连续可微， $g_{i j} (α, β, σ, γ) (i, j = 1, 2)$ 连续可微，则由引理2.3可得，问题(3.25)~(3.27)有唯一的经典解 $(\tilde{σ}, \tilde{γ}) \in C ([0, 1] \times [0, T])$ ，满足

$| \tilde{σ} (ρ, τ) | \leq 2 A_{0}, | \tilde{γ} (ρ, τ) | \leq 2 A_{0}, 0 \leq ρ \leq 1, 0 \leq τ \leq T .$ (3.34)

所以 $(\tilde{σ}, \tilde{γ})$ 满足条件(iii)，且当 $T > 0$ 足够小，有

${‖ \frac{\partial \tilde{σ}}{\partial ρ}, \frac{\partial \tilde{γ}}{\partial ρ} ‖}_{\infty} \leq 2 A_{1} .$ (3.35)

其中， $A_{1} = {‖ \frac{\partial σ_{0} (ρ)}{\partial ρ}, \frac{\partial γ_{0} (ρ)}{\partial ρ} ‖}_{\infty}$ 。

因为条件(i)，(ii)，(iii)都得到验证，所以当T足够小时，映射F是合理的，是自身到自身的映射。

下面证明F是压缩映射。设 $(η_{i}, α_{i}, β_{i}, σ_{i}, γ_{i}) \in X_{T}, i = 1, 2$ ，记

$υ_{i} (ρ, τ) = \frac{η_{i} {(τ)}^{2}}{ρ^{2}} \int_{0}^{1} h (\partial_{i}, β_{i}, σ_{i}, γ_{i}) s^{2} d s,$

$μ_{i} (ρ, τ) = υ_{i} (ρ, τ) - ρ υ_{i} (1, τ),$

$(\tilde{η}, {\tilde{α}}_{i}, {\tilde{β}}_{i}, {\tilde{σ}}_{i}, {\tilde{γ}}_{i}) = F (η, α_{i}, β_{i}, σ_{i}, γ_{i}),$

$d = d ((η_{1}, α_{1}, β_{1}, σ_{1}, γ_{1}), (η_{2}, α_{2}, β_{2}, σ_{2}, γ_{2})) .$

由 $υ_{i} (ρ, τ) = \frac{η_{i} {(τ)}^{2}}{ρ^{2}} \int_{0}^{1} h (\partial_{i}, β_{i}, σ_{i}, γ_{i}) s^{2} d s$ ，计算得：

$\begin{array}{l} | υ_{1} (ρ, τ) - υ_{2} (ρ, τ) | \\ = | \frac{η {(τ_{1})}^{2}}{ρ^{2}} \int_{0}^{ρ} h (\partial_{1}, β_{1}, σ_{1}, γ_{1}) s^{2} d s - \frac{η {(τ_{2})}^{2}}{ρ^{2}} \int_{0}^{ρ} h (\partial_{2}, β_{2}, σ_{2}, γ_{2}) s^{2} d s | \\ \leq | \frac{η {(τ_{1})}^{2}}{ρ^{2}} \int_{0}^{ρ} (h (\partial_{1}, β_{1}, σ_{1}, γ_{1}) - h (\partial_{2}, β_{2}, σ_{2}, γ_{2})) s^{2} d s | \\ + | \frac{η {(τ_{1})}^{2} - η {(τ_{2})}^{2}}{ρ^{2}} \int_{0}^{ρ} h (\partial_{2}, β_{2}, σ_{2}, γ_{2}) s^{2} d s | \\ \leq C (T) d \end{array}$ (3.36)

接着由(3.30)和(3.36)得：

${‖ \tilde{η_{1}} - \tilde{η_{2}} ‖}_{\infty} = \max_{0 \leq τ \leq T} | η_{0} e^{\int_{0}^{τ} υ_{1} (1, s) d s} - η_{0} e^{\int_{0}^{τ} υ_{2} (1, s) d s} | \leq T C (T) d$ (3.37)

${‖ \tilde{μ_{1}} - \tilde{μ_{2}} ‖}_{\infty} = \max_{0 \leq τ \leq T} | ν_{1} - ρ ν_{1} (1, τ) - ν_{2} + ρ ν_{2} (1, τ) | \leq T C (T) d .$ (3.38)

接着，记 $\tilde{α^{*}} = \tilde{α_{1}} - \tilde{α_{2}}$ ，由(3.19)~(3.21)得，

$\frac{\partial \tilde{α^{*}}}{\partial t} = δ_{P} \frac{1}{ρ^{2}} \frac{\partial}{\partial ρ} (ρ^{2} \frac{\partial \tilde{α^{*}}}{\partial ρ}) - μ_{1} (ρ, τ) \frac{\partial \tilde{α^{*}}}{\partial ρ} + f_{1} (ρ, τ), 0 < ρ < 1, 0 \leq τ \leq T .$ (3.39)

$\tilde{α^{*}} (0, τ) = 0, \tilde{α^{*}} (1, τ) = 0, τ > 0,$ (3.40)

$\tilde{α^{*}} (ρ, 0) = 0, 0 \leq ρ \leq 1.$ (3.41)

其中，

$\begin{array}{l} f_{1} (ρ, τ) = [μ_{2} (ρ, τ) - μ_{1} (ρ, τ)] \frac{\partial \tilde{α_{2}}}{\partial ρ} \\ - η_{2} {(τ)}^{2} [η_{1}^{2} (τ) F_{1} (α_{1}, β_{1}, σ_{1}, γ_{1}) - η_{2}^{2} (τ) F_{1} (α_{2}, β_{2}, σ_{2}, γ_{2})] \tilde{α_{2}} \end{array}$ 。

由于 ${‖ \frac{\partial \tilde{α^{*}}}{\partial ρ} ‖}_{\infty} \leq C (T)$ ， $F_{1} (α_{1}, β_{1}, σ_{1}, γ_{1}), F_{1} (α_{2}, β_{2}, σ_{2}, γ_{2})$ Lipschitz连续的， $\frac{1}{2} η_{0} \leq η (τ) \leq 2 η_{0} (0 \leq τ \leq T)$ ， $0 \leq | \tilde{α_{2}} (ρ, τ) | \leq \bar{α}$ ，所以有

$\begin{matrix} {‖ f_{1} ‖}_{\infty} \leq C (T) [{‖ μ_{2} - μ_{1} ‖}_{\infty} + {‖ η_{1}^{2} (τ) F_{1} (α_{1}, β_{1}, σ_{1}, γ_{1}) - η_{2}^{2} (τ) F_{1} (α_{2}, β_{2}, σ_{2}, γ_{2}) ‖}_{\infty}] \\ \leq C (T) d . \end{matrix}$ (3.42)

运用引理2.1得，

${‖ \tilde{α^{*}} ‖}_{\infty} \leq T C (T) d .$ (3.43)

同理可得

${‖ \tilde{β^{*}} ‖}_{\infty} \leq T C (T) d .$ (3.44)

记 $\tilde{σ^{*}} = \tilde{σ_{1}} - \tilde{σ_{2}}, \tilde{γ^{*}} = \tilde{γ_{1}} - \tilde{γ_{2}}$ ，由(3.25)~(3.27)得

$\begin{array}{l} \frac{\partial \tilde{σ^{*}}}{\partial τ} + μ_{1} \frac{\partial \tilde{σ^{*}}}{\partial ρ} = η_{1}^{2} (τ) g_{11} (\tilde{\partial_{1}}, \tilde{β_{1}}, \tilde{σ_{1}}, \tilde{γ_{1}}) \tilde{σ^{*}} \\ + η_{1}^{2} (τ) g_{12} (\tilde{\partial_{2}}, \tilde{β_{2}}, \tilde{σ_{2}}, \tilde{γ_{2}}) \tilde{γ^{*}} + G_{1} (ρ, τ), 0 \leq ρ \leq 1, τ > 0, \end{array}$ (3.45)

$\begin{array}{l} \frac{\partial \tilde{γ^{*}}}{\partial τ} + μ_{1} \frac{\partial \tilde{γ^{*}}}{\partial ρ} = η_{1}^{2} (τ) g_{21} (\tilde{\partial_{1}}, \tilde{β_{1}}, \tilde{σ_{1}}, \tilde{γ_{1}}) \tilde{σ^{*}} \\ + η_{1}^{2} (τ) g_{22} (\tilde{\partial_{2}}, \tilde{β_{2}}, \tilde{σ_{2}}, \tilde{γ_{2}}) \tilde{γ^{*}} + G_{2} (ρ, τ), 0 \leq ρ \leq 1, τ > 0, \end{array}$ (3.46)

$\tilde{σ^{*}} (ρ, 0) = 0, \tilde{γ^{*}} (ρ, 0) = 0, 0 \leq ρ \leq 1,$ (3.47)

其中，

$\begin{matrix} G_{1} (ρ, τ) = \tilde{γ_{2}} [η_{1}^{2} (τ) g_{12} (\tilde{\partial_{1}}, \tilde{β_{1}}, \tilde{σ_{1}}, \tilde{γ_{1}}) - η_{2}^{2} (τ) g_{12} (\tilde{\partial_{1}}, \tilde{β_{1}}, \tilde{σ_{1}}, \tilde{γ_{1}})] \\ + \tilde{σ_{2}} [η_{1}^{2} (τ) g_{11} (\tilde{\partial_{1}}, \tilde{β_{1}}, \tilde{σ_{1}}, \tilde{γ_{1}}) - η_{2}^{2} (τ) g_{11} (\tilde{α_{2}}, \tilde{β_{2}}, \tilde{σ_{2}}, \tilde{γ_{2}})] \\ + \frac{\partial \tilde{σ_{2}}}{\partial ρ} (μ_{2} - μ_{1}) \end{matrix}$ (3.48)

$\begin{matrix} G_{2} (ρ, τ) = \tilde{γ_{2}} [η_{1}^{2} (τ) g_{22} (\tilde{\partial_{1}}, \tilde{β_{1}}, \tilde{σ_{1}}, \tilde{γ_{1}}) - η_{2}^{2} (τ) g_{22} (\tilde{\partial_{1}}, \tilde{β_{1}}, \tilde{σ_{1}}, \tilde{γ_{1}})] \\ + \tilde{σ_{2}} [η_{1}^{2} (τ) - η_{2}^{2} (τ) g_{21} (\tilde{α_{2}}, \tilde{β_{2}}, \tilde{σ_{2}}, \tilde{γ_{2}})] + \frac{\partial \tilde{γ_{2}}}{\partial ρ} (μ_{2} - μ_{1}), \end{matrix}$ (3.49)

由于 $| μ_{1} (ρ, τ) - μ_{2} (ρ, τ) | \leq C (T) d$ ， $h (α, β, σ, γ)$ Lipschitz连续， $g_{i j} (α, β, σ, γ) (i, j = 1, 2)$ 连续可微， $\frac{1}{2} η_{0} \leq η (τ) \leq 2 η_{0}$ ，结合式子(3.34)和(3.35)得

${‖ G_{i} ‖}_{\infty} \leq C (T) d .$ (3.50)

再对问题(3.45)~(3.47)运用引理2.2，可得

${‖ \tilde{σ_{1}} - \tilde{σ_{2}}, \tilde{γ_{1}} - \tilde{γ_{2}} ‖}_{\infty} \leq T C (T) d .$ (3.51)

因此，根据式子(3.37)，(3.43)，(3.44)和(3.51)有：

$d ((\tilde{α_{1}}, \tilde{β_{1}}, \tilde{σ_{1}}, \tilde{γ_{1}}), (\tilde{α_{2}}, \tilde{β_{2}}, \tilde{σ_{2}}, \tilde{γ_{2}})) \leq T C (T) d .$ (3.52)

取 $T > 0$ 充分小，使得 $T C (T) < 1$ 时，F是压缩映射。

由Banach不动点定理可知，当 $T > 0$ 充分小时，F存在唯一的不动点 $(η, α, β, σ, γ)$ ，而且它是问题(3.2)~(3.13)在 $0 \leq τ \leq T$ 上的唯一解，再通过逆变换(3.14)，就可以将问题(3.2)~(3.13)转换成原问题(1.9)~(1.19)。因此，问题(1.9)~(1.19)存在唯一局部解。

定理2：在假设条件(A₁)~(A₂)下，存在，使得问题(1.9)~(1.19)在 $[0, δ]$ 上有唯一解， $δ$ 依赖于 $R_{0}$

的上下界， ${‖ P_{0} (| x |) ‖}_{D_{p} (B_{R_{0}})}, {‖ Q_{0} (| x |) ‖}_{D_{p} (B_{R_{0}})}, {‖ M_{0}, N_{0} ‖}_{\infty}$ 和 ${‖ {M^{'}}_{0}, {N^{'}}_{0} ‖}_{\infty}$ 的上下界。

4. 整体解的存在唯一性

本节需要对问题(1.9)~(1.19)作先验估计，来研究整体解的存在唯一性。根据(3.16)得，

$η (τ) = {(\frac{1}{η_{0}^{2}} - \int_{0}^{τ} \int_{0}^{1} h (\partial, β, σ, γ) s^{2} d s d τ)}^{- \frac{1}{2}},$

当 $τ \to \infty$ 时，如果 $h (\partial, β, σ, γ) < 0$ ，则 $η \to 0$ ，即 $R \to 0$ 。这说明问题(3.2)~(3.13)的解不是对所有的 $τ > 0$ 都成立，因此需要直接考虑变换之前的自由边界问题(1.9)~(1.19)的整体解的存在唯一性。

定理1的证明：

设 $0 \leq t < T$ 为问题(1.9)~(1.19)的解的最大区间，因此只需证明 $T = \infty$ 。采用反证法：设 $T < \infty$ 。首先需证明一些基本结论：若问题(1.9)~(1.19)对所有的 $t \geq 0$ 都成立，则有下面的结论成立：

$0 \leq P (r, t) \leq \bar{P}, 0 \leq Q (r, t) \leq \bar{Q}, 0 \leq r \leq R (t), 0 \leq t < T;$ (4.1)

$M (r, t) \geq 0, N (r, t) \geq 0, 0 \leq r \leq R (t), 0 \leq t < T;$ (4.2)

$M + N = 1, 0 \leq r \leq R (t), 0 \leq t < T;$ (4.3)

$R_{0} e^{- \frac{1}{3} B_{0} t} \leq R (t) \leq R_{0} e^{\frac{1}{3} B_{0} t}, 0 \leq t < T;$ (4.4)

$- \frac{1}{3} B_{0} R (t) \leq \dot{R} (t) \leq \frac{1}{3} B_{0} R (t), 0 \leq t < T,$ (4.5)

其中

$\dot{R} (t) = \frac{d R (t)}{d t}, B_{0} = \max {| h (P, Q, M, N) | : 0 \leq P (r, t) \leq \bar{P}, 0 \leq Q (r, t) \leq \bar{Q}, 0 \leq M \leq 1, 0 \leq N \leq 1} .$

首先，利用抛物型方程的极值原理，可以得到(4.1)式。接着，利用变换 $ρ = \frac{r}{R (t)}, τ = \int_{0}^{t} \frac{d s}{R {(s)}^{2}}$ ，将

(1.11)~(1.12)的定义域变换成 ${(ρ, τ) : 0 \leq ρ \leq 1, 0 \leq τ \leq δ}$ ，利用引理2.2得(4.2)式。再令 $Z (r, t) = M (r, t) + N (r, t)$ ，将(1.11)~(1.12)加起来，得

$\frac{\partial M}{\partial t} + υ \frac{\partial M}{\partial r} = h (P, Q, M, N) [1 - Z (r, t)], 0 \leq r \leq R (t), 0 \leq t < T,$ (4.6)

$Z (r, 0) = M (r, 0) + N (r, 0) = 1, 0 \leq r \leq R_{0} .$ (4.7)

由(4.6)式得， $Z = 1$ 是问题(4.6)~(4.7)的一个解，由唯一性可知，对任意的 $0 \leq r \leq R (t), 0 \leq t < T$ ，都有 $Z (r, t) = 1$ 成立。即(4.3)成立。由(4.2)和(4.3)可得， $0 \leq M \leq 1, 0 \leq N \leq 1, 0 \leq r \leq R (t), 0 \leq t < T$ 。所以

$| h (P, Q, M, N) | \leq B_{0}, 0 \leq r \leq R (t), 0 \leq t < T,$

由(1.15)可得，

$- \frac{1}{3} B_{0} r \leq υ (r, t) \leq \frac{1}{3} B_{0} r, 0 \leq r \leq R (t), 0 \leq t < T,$

将上式带入(1.16)可得，

$- \frac{1}{3} B_{0} R (t) \leq \dot{R} (t) \leq \frac{1}{3} B_{0} R (t), 0 \leq t < T,$

即(4.5)式成立。最后将(4.5)式积分，则有(4.4)式成立。

令 $σ (x, t) = P (| x | R (t), t), | x | \leq 1, 0 \leq t < T$ 。由(1.12)，(1.17)和(1.20)式可知， $σ$ 是下面问题的解：

$\frac{\partial σ}{\partial t} = \frac{δ_{P}}{R^{2} (t)} Δ σ + \frac{\dot{R} (t)}{R (t)} (x \cdot ▽ σ) - c (x, t) σ, | x | < 1, 0 < t < T,$ (4.8)

$σ (x, t) = \bar{P}, | x | = 1, 0 \leq t < T,$ (4.9)

$σ (x, 0) = P_{0} (| x | R_{0}), | x | \leq 1.$ (4.10)

其中， $c (x, t) = - k_{5} \frac{P {(| x | R (t), t)}^{2}}{P {(| x | R (t), t)}^{2} + M {(| x | R (t), t)}^{2}} - μ_{M}$ ，由已经证明的结论(4.2)~(4.5)，可知(4.8)的各项系数都是有界函数， $\frac{1}{R^{2} (t)}$ 连续可微有正下界，对问题(4.8)~(4.10)使用引理2.1，可得 ${‖ σ ‖}_{W_{p}^{2, 1}} \leq C (T)$ ，又

$σ (x, t) = P (| x | R (t), t)$ ，所以

${‖ P (| x | R (t), t) ‖}_{W_{p}^{2, 1}} \leq C (T),$

根据 $R (t), \frac{d R (t)}{d t}, \frac{1}{d R (t)}$ 的有界性，可得：对任意的 $t_{0} \in [0, T)$ ，都有

${‖ P (| x |, t_{0}) ‖}_{D_{p} (B (t_{0}))} \leq C (T), B (t_{0}) = {x \in R^{3}, | x | < R (t_{0})} .$

同理可得，

${‖ Q (| x |, t_{0}) ‖}_{D_{p} (B (t_{0}))} \leq C (T), B (t_{0}) = {x \in R^{3}, | x | < R (t_{0})}$

而且，由引理2.2可得，

${‖ \frac{\partial M}{\partial ρ}, \frac{\partial N}{\partial ρ} ‖}_{\infty} \leq C (T) .$

所以，对任意的 $t_{0} \in [0, T)$ ，我们考虑有初始时间的 $t = t_{0}$ 的初值问题(1.9)~(1.19)。根据定理2可知，存在一个一致的 $δ > 0$ ，使得对所有的 $t_{0} \in [0, T)$ ，问题(1.9)~(1.19)在 $[t_{0}, t_{0} + δ)$ 都存在唯一的解。利用唯一性可知，这些解在公共定义的区间上都是相等的，这说明了解可以延拓到 $[0, T + δ)$ ，这与假设 $T < \infty$ 不符合，所以 $T = \infty$ 。因此，定理1得到了证明。

参考文献

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[7]	张久远, 冯兆永, 刘成霞, 卫雪梅. 关于肿瘤细胞破坏并入侵正常组织或细胞质基质的数学模型的分析[J]. 中山大学学报(自然科版), 2013(52): 48-54.
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