# 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

• 全文下载: PDF(595KB)    PP.217-229   DOI: 10.12677/PM.2019.92028
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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.

1. 引言

$\begin{array}{l}\frac{\partial P}{\partial t}+\nabla \cdot \left(vP\right)-{\delta }_{P}{\nabla }^{2}P={\alpha }_{1}P\left(1-\frac{P}{M}\right)+\stackrel{˜}{{k}_{5}}M\frac{{P}^{2}}{{P}^{2}+{M}^{2}}\left(\theta -1\right)+{k}_{5}N\frac{{Q}^{2}}{{Q}^{2}+{N}^{2}}\theta \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-{\mu }_{P}P-{\mu }_{M}P,\text{\hspace{0.17em}}|x|0,\end{array}$ (1.1)

$\begin{array}{l}\frac{\partial Q}{\partial t}+\nabla \cdot \left(vQ\right)-{\delta }_{N}{\nabla }^{2}Q={\alpha }_{2}Q\left(1-\frac{Q}{N}\right)+\stackrel{˜}{{k}_{5}}M\frac{{P}^{2}}{{P}^{2}+{M}^{2}}\left(1-\theta \right)+{k}_{5}N\frac{{Q}^{2}}{{Q}^{2}+{N}^{2}}\left(1-\theta \right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\stackrel{˜}{{k}_{5}}N\frac{{Q}^{2}}{{Q}^{2}+{N}^{2}}-{\mu }_{Q}Q-{\mu }_{N}Q,\text{\hspace{0.17em}}|x|0,\end{array}$ (1.2)

$\frac{\partial M}{\partial t}+\nabla \cdot \left(vM\right)=-{k}_{5}M\frac{{P}^{2}}{{P}^{2}+{M}^{2}}-{\mu }_{M}M,\text{\hspace{0.17em}}|x|0,$ (1.3)

$\frac{\partial N}{\partial t}+\nabla \cdot \left(vN\right)=-{k}_{5}N\frac{{Q}^{2}}{{Q}^{2}+{N}^{2}}-{\mu }_{N}N,\text{\hspace{0.17em}}|x|0,$ (1.4)

$M+N=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}|x|0,$ (1.5)

$\frac{\text{d}R}{\text{d}t}=\frac{x}{|x|}\cdot v,\text{\hspace{0.17em}}\text{\hspace{0.17em}}|x|0,$ (1.6)

$P={P}_{0},\text{\hspace{0.17em}}\text{\hspace{0.17em}}Q={Q}_{0},M={M}_{0},\text{\hspace{0.17em}}\text{\hspace{0.17em}}N={N}_{0},\text{\hspace{0.17em}}\text{\hspace{0.17em}}|x|0\text{,}$ (1.7)

$R={R}_{0},\text{\hspace{0.17em}}\text{\hspace{0.17em}}t=0.$ (1.8)

$P=P\left(|x|,t\right),\text{\hspace{0.17em}}Q=Q\left(|x|,t\right),\text{\hspace{0.17em}}M=M\left(|x|,t\right),\text{\hspace{0.17em}}N=N\left(|x|,t\right),\text{\hspace{0.17em}}\forall x\in {R}^{3}.$

$r=|x|$ ，有 $\nabla P=\left(\frac{\partial P}{\partial {x}_{1}},\frac{\partial P}{\partial {x}_{2}},\frac{\partial P}{\partial {x}_{3}}\right)=\frac{\stackrel{\to }{x}}{r}\cdot \frac{\partial P}{\partial r}$$\nabla \cdot \left(\nabla P\right)=\frac{1}{{r}^{2}}\frac{\partial }{\partial r}\left({r}^{2}\frac{\partial P}{\partial r}\right)$

$\frac{\partial P}{\partial t}={\delta }_{P}\frac{1}{{r}^{2}}\frac{\partial }{\partial r}\left({r}^{2}\frac{\partial P}{\partial r}\right)-v\frac{\partial P}{\partial r}+{F}_{1}\left(P,Q,M,N\right)P,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }00,$ (1.9)

$\frac{\partial Q}{\partial t}={\delta }_{N}\frac{\partial }{\partial r}\left({r}^{2}\frac{\partial Q}{\partial r}\right)-v\frac{\partial Q}{\partial r}+{F}_{2}\left(P,Q,M,N\right)Q,\text{ }\text{ }\text{ }\text{ }00,$ (1.10)

$\frac{\partial M}{\partial t}+v\frac{\partial M}{\partial r}={g}_{11}\left(P,Q,M,N\right)M+{g}_{12}\left(P,Q,M,N\right)N,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }00.$ (1.11)

$\frac{\partial N}{\partial t}+v\frac{\partial N}{\partial r}={g}_{21}\left(P,Q,M,N\right)M+{g}_{22}\left(P,Q,M,N\right)N,\text{\hspace{0.17em}}\text{\hspace{0.17em}}00,$ (1.12)

$\frac{1}{{r}^{2}}\frac{\partial }{\partial r}\left({r}^{2}v\right)=h\left(P,Q,M,N\right),\text{ }00\text{,}$ (1.13)

$\frac{\partial P}{\partial r}\left(0,t\right)=0,\text{\hspace{0.17em}}P\left(R\left(t\right),t\right)=\stackrel{¯}{P},\text{\hspace{0.17em}}\text{ }\text{ }\frac{\partial Q}{\partial r}\left(0,t\right)=0,\text{\hspace{0.17em}}Q\left(R\left(t\right),t\right)=\stackrel{¯}{Q},\text{\hspace{0.17em}}\text{ }\text{ }\text{ }\text{ }\text{ }t>0.$ (1.14)

$v\left(0,t\right)=0,\text{ }\text{ }\text{ }\text{ }t>0,$ (1.15)

$\frac{\text{d}R\left(t\right)}{\text{d}t}=v\left(R\left(t\right),t\right),\text{\hspace{0.17em}}\text{ }\text{ }\text{ }\text{ }t>0\text{,}$ (1.16)

$P\left(r,0\right)={P}_{0}\left(r\right),\text{\hspace{0.17em}}Q\left(r,0\right)={Q}_{0}\left(r\right),\text{\hspace{0.17em}}\text{ }\text{ }\text{ }\text{ }\text{ }|x| (1.17)

$M\left(r,0\right)={M}_{0}\left(r\right),\text{\hspace{0.17em}}N\left(r,0\right)={N}_{0}\left(r\right),\text{\hspace{0.17em}}\text{ }\text{ }\text{ }|x| (1.18)

$R\left(0\right)={R}_{0}.$ (1.19)

$h\left(P,Q,M,N\right)=\nabla \cdot v=-{k}_{5}M\frac{{P}^{2}}{{P}^{2}+{M}^{2}}-{\mu }_{M}M-{k}_{5}N\frac{{Q}^{2}}{{Q}^{2}+{N}^{2}}-{\mu }_{N}N,$

${F}_{1}\left(P,Q,M,N\right)={\alpha }_{1}\left(1-\frac{P}{M}\right)+\stackrel{˜}{{k}_{5}}M\frac{P}{{P}^{2}+{M}^{2}}\left(\theta -1\right)+{k}_{5}\frac{N}{P}\frac{{Q}^{2}}{{Q}^{2}+{N}^{2}}\theta -{\mu }_{P}-{\mu }_{M}-\nabla \cdot v,$

$\begin{array}{l}{F}_{2}\left(M,N,P,Q\right)={\alpha }_{2}\left(1-\frac{Q}{N}\right)+{k}_{5}N\frac{Q}{{Q}^{2}+{N}^{2}}\left(1-\theta \right)+\stackrel{˜}{{k}_{5}}\frac{M}{Q}\frac{{P}^{2}}{{P}^{2}+{M}^{2}}\left(1-\theta \right)\\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\stackrel{˜}{{k}_{5}}N\frac{Q}{{Q}^{2}+{N}^{2}}-{\mu }_{Q}-{\mu }_{N}-\nabla \cdot v,\end{array}$

${g}_{11}\left(M,N,P,Q\right)=-{k}_{5}\frac{{P}^{2}}{{P}^{2}+{M}^{2}}-{\mu }_{M}+{k}_{5}M\frac{{P}^{2}}{{P}^{2}+{M}^{2}}+{\mu }_{M}M,$

${g}_{12}\left(M,N,P,Q\right)={k}_{5}M\frac{Q}{{Q}^{2}+{N}^{2}}+{\mu }_{N}M,$

${g}_{12}\left(M,N,P,Q\right)={k}_{5}N\frac{{P}^{2}}{{P}^{2}+{M}^{2}}+{\mu }_{M}N,$

${g}_{22}\left(M,N,P,Q\right)=-{k}_{5}\frac{{Q}^{2}}{{Q}^{2}+{N}^{2}}-{\mu }_{N}+{k}_{5}N\frac{{Q}^{2}}{{Q}^{2}+{N}^{2}}+{\mu }_{N}N.$

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}_{\text{0}}\text{,}{N}_{\text{0}}$$\left[0,{R}_{0}\right]$ 上连续可微，当 $p>5$ 时，有 ${P}_{0}\left(|x|\right)\in {D}_{p}\left(B\left({R}_{0}\right)\right)$${Q}_{0}\left(|x|\right)\in {D}_{p}\left(B\left({R}_{0}\right)\right)$$B\left({R}_{0}\right)=\left\{x\in {R}^{3}:|x|<{R}_{0}\right\}$

(A2) 当 $0\le r\le {R}_{0}$ 时， $0\le {P}_{0}\left(r\right)\le \stackrel{¯}{{P}_{0}},0\le {Q}_{0}\left(r\right)\le \stackrel{¯}{{Q}_{0}},{M}_{0}\left(r\right)\ge 0,{N}_{0}\left(r\right)\ge 0,{M}_{0}\left(r\right)+{N}_{0}\left(r\right)=1$ ，且 ${{P}^{\prime }}_{0}\left(0\right)=0,{P}_{0}\left({R}_{0}\right)=0,{{Q}^{\prime }}_{0}\left(0\right)=0,{Q}_{0}\left({R}_{0}\right)=0$

$R\left(t\right)\ge 0,t>0,$ (1.20)

$0\le P\left(r,t\right)\le \stackrel{¯}{P},\text{\hspace{0.17em}}0\le Q\left(r,t\right)\le \stackrel{¯}{Q},0\le r\le {R}_{0},\text{\hspace{0.17em}}t\ge 0,$ (1.21)

$M\left(r,t\right)\ge 0,N\left(r,t\right)\ge 0,0\le r\le R\left(t\right),t\ge 0,$ (1.22)

$M\left(r,t\right)+N\left(r,t\right)=1,\text{\hspace{0.17em}}0\le r\le R\left(t\right),\text{\hspace{0.17em}}t\ge 0.$ (1.23)

2. 基本引理

1) ${Q}_{T}^{R}=\left\{\left(x,t\right)\in {R}^{3}×R:|x|${\stackrel{¯}{Q}}_{T}^{R}$${Q}_{T}^{R}$ 的闭包。

2) ${W}_{p}^{2,1}\left({Q}_{T}^{R}\right)=\left\{\mu \in {L}_{P}\left({Q}_{T}^{R}\right):{\partial }_{x}^{m}{\partial }_{t}^{k}\in {L}_{P}\left({Q}_{T}^{R}\right),|m|+2k\le 2\right\}$ ，其范数为

${‖\mu ‖}_{{W}_{p}^{2,1}\left({Q}_{T}\right)}=\sum {‖{\partial }_{x}^{m}{\partial }_{t}^{k}\mu ‖}_{p}.$

3) 对开集 $\Omega \in {R}^{3},p>\frac{5}{2}$ ，记 ${D}_{p}\left(\Omega \right)$${W}_{p}^{2,1}\left(\Omega ×\left(0,T\right)\right)$$t=0$ 的迹空间， $i.e.\text{ }\text{ }\text{ }\phi \in {D}_{p}\left(\Omega \right)$ ，当且仅当存在 $\mu \in {W}_{p}^{2,1}\left({Q}_{T}\right)$ ，使得 $\mu \left(\cdot ,0\right)=\phi$ 。当 $p>\frac{5}{2}$ 时， ${W}_{p}^{2,1}\left(\Omega ×\left(0,T\right)\right)\to C\left(\Omega ×\left(0,T\right)\right)$${D}_{p}\left(\Omega \right)$ 的范数为：

${‖\mu ‖}_{{D}_{p}\left(\Omega \right)}=inf\left\{{T}^{-\frac{1}{p}}{‖\mu ‖}_{{W}_{p}^{2,1}\left(\Omega ×\left(0,T\right)\right)}:\mu \in {W}_{p}^{2,1}\left(\Omega ×\left(0,T\right)\right),\mu \left(\cdot ,0\right)=\phi \right\}$

$\frac{\partial m}{\partial t}=\frac{1}{{x}^{2}}\frac{\partial }{\partial x}\left({x}^{2}\frac{\partial m}{\partial x}\right)+a\left(t\right)\frac{\partial m}{\partial x}+b\left(x,t\right)+f\left(x,t\right)\text{ },\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le x\le 1,\text{\hspace{0.17em}}0\le \tau \le T,$ (2.1)

$\frac{\partial m}{\partial t}\left(0,t\right)=0,\text{ }\text{ }\text{ }\text{ }m\left(1,t\right)=\stackrel{¯}{m},\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le t\le T,$ (2.2)

$m\left(x,0\right)={m}_{0}\left(x\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le x\le 1,$ (2.3)

${‖m\left(x,t\right)‖}_{{W}_{p}^{2,1}\left({Q}_{T}\right)}\le {C}_{p}\left(T\right)\left(\stackrel{¯}{m}+{‖{m}_{0}‖}_{{D}_{p}\left({B}_{1}\right)}+{‖f‖}_{p}\right),$

${‖m‖}_{\infty }\le \mathrm{max}\left(|\stackrel{¯}{m}|,{‖{m}_{0}‖}_{\infty }\right)+T{\text{e}}^{{c}_{0}T}{‖f‖}_{p},$

$\frac{\partial M}{\partial \tau }+\mu \left(\rho ,\tau \right)\frac{\partial M}{\partial t}={a}_{11}\left(\rho ,\tau \right)M+{a}_{12}\left(\rho ,\tau \right)N+{f}_{1}\left(x,t\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le \rho \le 1,\text{\hspace{0.17em}}0\le \tau \le T,$ (2.4)

$\frac{\partial N}{\partial \tau }+\mu \left(\rho ,\tau \right)\frac{\partial N}{\partial t}={a}_{21}\left(\rho ,\tau \right)M+{a}_{22}\left(\rho ,\tau \right)N+{f}_{2}\left(x,t\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le \rho \le 1,\text{\hspace{0.17em}}0\le \tau \le T,$ (2.5)

$M\left(\rho ,0\right)={M}_{0}\left(\rho \right),N\left(\rho ,0\right)={N}_{0}\left(\rho \right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le \rho \le 1,$ (2.6)

${‖\left(M,N\right)‖}_{\infty }\le {\text{e}}^{T{A}_{0}\left(T\right)}\left({‖\left({M}_{0},{N}_{0}\right)\text{ }‖}_{\infty }+2T{‖\left({f}_{1},{f}_{2}\right)‖}_{\infty }\right),$

${‖\frac{\partial M}{\partial \rho },\frac{\partial N}{\partial \rho }‖}_{\infty }\le {\text{e}}^{T\left({A}_{0}\left(T\right)+A\left(T\right)\right)}\left({‖{{M}^{\prime }}_{0},{{N}^{\prime }}_{0}‖}_{\infty }+T{A}_{1}\left({T}_{0}\right){\text{e}}^{A\left(T\right)}{‖{M}_{0},{N}_{0}‖}_{\infty }+2T{\text{e}}^{A\left(T\right)}{‖\frac{\partial {f}_{1}}{\partial \rho },\frac{\partial {f}_{2}}{\partial \rho }‖}_{\infty }\right),$ (2.7)

$\partial {a}_{ij}\left(\rho \right)\ge 0,i\ne j,{M}_{0}\left(\rho \right)\ge 0,{N}_{0}\left(\rho \right)\ge 0,{f}_{ij}\left(\rho ,\tau \right)\ge 0\left(i,j=1,2\right),0\le \rho \le 1,\text{\hspace{0.17em}}0\le \tau \le T$ ，有

(2.8)

$\frac{\partial M}{\partial \tau }+\mu \left(\rho ,\tau \right)\frac{\partial M}{\partial \rho }={f}_{1}\left(\rho ,\tau ,M,N\right),\text{\hspace{0.17em}}0\le \rho \le 1,\text{\hspace{0.17em}}0\le \tau \le T,$ (2.9)

$\frac{\partial N}{\partial \tau }+\mu \left(\rho ,\tau \right)\frac{\partial N}{\partial \rho }={f}_{2}\left(\rho ,\tau ,M,N\right),\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }0\le \rho \le 1,\text{\hspace{0.17em}}0\le \tau \le T,$ (2.10)

$M\left(\rho ,0\right)={M}_{0}\left(\rho \right),N\left(\rho ,0\right)={N}_{0}\left(\rho \right),\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }0\le \rho \le 1,$ (2.11)

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

${‖\left(M,N\right)‖}_{\infty }\le 2{H}_{0},$

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

${‖\frac{\partial M}{\partial \rho },\frac{\partial N}{\partial \rho }‖}_{\infty }\le {\text{e}}^{T\left(A\left(T\right)+{B}_{0}\left(T\right)\right)}\left({‖{{M}^{\prime }}_{0},{{N}^{\prime }}_{0}‖}_{\infty }+2T{\text{e}}^{A\left(T\right)}{‖\frac{\partial {f}_{1}}{\partial \rho },\frac{\partial {f}_{2}}{\partial \rho }‖}_{\infty }\right),$ (2.12)

3. 局部解的存在唯一性

$\begin{array}{l}\rho =\frac{r}{R\left(t\right)},\text{\hspace{0.17em}}\tau ={\int }_{0}^{t}\frac{\text{d}s}{{R}^{2}\left(s\right)},\text{\hspace{0.17em}}\eta \left(\tau \right)=R\left(t\right),\text{\hspace{0.17em}}\upsilon \left(\rho ,\tau \right)=R\left(t\right)v\left(r,t\right),\\ \alpha \left(\rho ,\tau \right)=P\left(r,t\right),\text{\hspace{0.17em}}\beta \left(\rho ,\tau \right)=Q\left(r,t\right),\text{\hspace{0.17em}}\sigma \left(\rho ,\tau \right)=M\left(r,t\right),\text{\hspace{0.17em}}\gamma \left(\rho ,\tau \right)=N\left(r,t\right),\end{array}$ (3.1)

$\frac{\partial \alpha }{\partial \tau }={\delta }_{P}\frac{1}{{\rho }^{2}}\frac{\partial }{\partial \rho }\left({\rho }^{2}\frac{\partial \alpha }{\partial \rho }\right)-\mu \left(\rho ,\tau \right)\frac{\partial \alpha }{\partial \rho }+{\eta }^{2}\left(\tau \right)\left({F}_{1}\left(\alpha ,\beta ,\sigma ,\gamma \right)\alpha \right),\text{\hspace{0.17em}}\text{ }\text{ }\text{ }\text{ }0<\rho <1,\text{\hspace{0.17em}}\tau >0,$ (3.2)

$\frac{\partial \beta }{\partial \tau }={\delta }_{Q}\frac{\partial }{\partial \rho }\left({\rho }^{2}\frac{\partial \beta }{\partial \rho }\right)-\mu \left(\rho ,\tau \right)\frac{\partial \beta }{\partial \rho }+{\eta }^{2}\left(\tau \right)\left({F}_{2}\left(\alpha ,\beta ,\sigma ,\gamma \right)\beta \right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}0<\rho <1,\text{\hspace{0.17em}}\tau >0,$ (3.3)

$\frac{\partial \sigma }{\partial \tau }+\mu \frac{\partial \sigma }{\partial \rho }={\eta }^{2}\left(\tau \right)\left[{g}_{11}\left(\alpha ,\beta ,\sigma ,\gamma \right)\delta +{g}_{12}\left(\alpha ,\beta ,\sigma ,\gamma \right)\gamma \right],\text{\hspace{0.17em}}0\le \rho \le 1,\text{\hspace{0.17em}}\tau >0,$ (3.4)

$\frac{\partial \gamma }{\partial t}+\mu \frac{\partial \gamma }{\partial r}={\eta }^{2}\left(\tau \right)\left[{g}_{21}\left(\alpha ,\beta ,\sigma ,\gamma \right)\delta +{g}_{22}\left(\alpha ,\beta ,\sigma ,\gamma \right)\gamma \right],\text{\hspace{0.17em}}\text{ }\text{ }0\le \rho \le 1,\text{\hspace{0.17em}}\tau >0,$ (3.5)

$\mu \left(\rho ,\tau \right)=\upsilon \left(\rho ,\tau \right)-\rho \upsilon \left(1,\tau \right),\text{\hspace{0.17em}}0\le \rho \le 1,\text{\hspace{0.17em}}\tau >0,$ (3.6)

$\frac{1}{{\rho }^{2}}\frac{\partial }{\partial \rho }\left({\rho }^{2}\upsilon \right)={\eta }^{2}\left(\tau \right)h\left(\alpha ,\beta ,\sigma ,\gamma \right),\text{\hspace{0.17em}}0\le \rho \le 1,\text{\hspace{0.17em}}\tau >0,$ (3.7)

$\frac{\partial \alpha \left(0,\tau \right)}{\partial \rho }=0,\text{\hspace{0.17em}}\alpha \left(1,\tau \right)=\stackrel{¯}{\alpha },\text{\hspace{0.17em}}\frac{\partial \beta \left(0,\tau \right)}{\partial \rho }=0,\text{\hspace{0.17em}}\beta \left(1,\tau \right)=\stackrel{¯}{\beta },\text{\hspace{0.17em}}\tau >0,$ (3.8)

$\upsilon \left(0,\tau \right)=0,\text{\hspace{0.17em}}\tau >0,$ (3.9)

$\frac{\text{d}\eta \left(\tau \right)}{\text{d}\tau }=\eta \left(\tau \right)\upsilon \left(1,\tau \right),\text{\hspace{0.17em}}\tau >0,$ (3.10)

$\alpha \left(\rho ,0\right)={\alpha }_{0}\left(\rho \right),\text{\hspace{0.17em}}\beta \left(\rho ,0\right)={\beta }_{0}\left(\rho \right),\text{\hspace{0.17em}}0\le \rho \le 1,$ (3.11)

$\sigma \left(\rho ,0\right)={\sigma }_{0}\left(\rho \right),\text{\hspace{0.17em}}\gamma \left(\rho ,0\right)={\gamma }_{0}\left(\rho \right),\text{\hspace{0.17em}}0\le \rho \le 1,$ (3.12)

$\eta \left(0\right)={\eta }_{0}={R}_{0}.$ (3.13)

${\alpha }_{0}\left(\rho \right)={P}_{0}\left(\rho {R}_{0}\right),{\beta }_{0}\left(\rho \right)={Q}_{0}\left(\rho {R}_{0}\right),{\sigma }_{0}\left(\rho \right)={M}_{0}\left(\rho {R}_{0}\right),{\gamma }_{0}\left(\rho \right)={N}_{0}\left(\rho {R}_{0}\right).$

$\begin{array}{l}r=\rho \eta \left(\tau \right),\text{\hspace{0.17em}}t={\int }_{0}^{\tau }{\eta }^{2}\left(s\right)\text{d}s,\text{\hspace{0.17em}}R\left(t\right)=\eta \left(\tau \right),\text{\hspace{0.17em}}v\left(r,t\right)=\frac{\upsilon \left(\rho ,\tau \right)}{{\eta }^{2}\left(\tau \right)},\\ P\left(r,t\right)=\alpha \left(\rho ,\tau \right),\text{\hspace{0.17em}}Q\left(r,t\right)=\beta \left(\rho ,\tau \right),\text{\hspace{0.17em}}M\left(r,t\right)=\sigma \left(\rho ,\tau \right),\text{\hspace{0.17em}}N\left(r,t\right)=\gamma \left(\rho ,\tau \right),\end{array}$ (3.14)

$\upsilon \left(\rho ,\tau \right)=\frac{{\eta }^{2}\left(\tau \right)}{{\rho }^{2}}{\int }_{0}^{\rho }h\left(\alpha ,\beta ,\sigma ,\gamma \right){s}^{2}\text{d}s,$ (3.15)

$\frac{\text{d}\eta \left(\tau \right)}{\text{d}\tau }={\eta }^{3}\left(\tau \right){\int }_{0}^{\rho }h\left(\alpha ,\beta ,\sigma ,\gamma \right){s}^{2}\text{d}s,$ (3.16)

${A}_{0}={‖\left({\sigma }_{0},{\gamma }_{0}\right)‖}_{\infty },$

${B}_{0}=\mathrm{max}\left\{|h\left(\alpha ,\beta ,\sigma ,\gamma \right)|:0\le \alpha \le \stackrel{¯}{\alpha },\text{\hspace{0.17em}}0\le \beta \le \stackrel{¯}{\beta },|\sigma |\le 2{A}_{0},|\gamma |\le 2{A}_{0}\right\}.$

${X}_{T}$ 由向量值函数 $\left(\eta ,\alpha ,\beta ,\sigma ,\gamma \right)=\left(\eta \left(\tau \right),\alpha \left(\rho ,\tau \right),\beta \left(\rho ,\tau \right),\sigma \left(\rho ,\tau \right),\gamma \left(\rho ,\tau \right)\right)$ 组成，满足：

i) $\eta \left(\tau \right)\in \left[0,T\right],\text{\hspace{0.17em}}\eta \left(0\right)={\eta }_{0},\text{\hspace{0.17em}}\frac{1}{2}{\eta }_{0}\le \eta \left(\tau \right)\le 2{\eta }_{0}\text{\hspace{0.17em}}\left(0\le \tau \le T\right);$

ii) $\alpha \left(\rho ,\tau \right)\in C\left(\left[0,1\right]×\left[0,T\right]\right),\text{\hspace{0.17em}}\alpha \left(\rho ,0\right)={\alpha }_{0}\left(\rho \right),\text{\hspace{0.17em}}\alpha \left(\rho ,1\right)=\stackrel{¯}{\alpha },\text{\hspace{0.17em}}0\le \alpha \left(\rho ,\tau \right)\le \stackrel{¯}{\alpha },$ $\beta \left(\rho ,\tau \right)\in C\left(\left[0,1\right]×\left[0,T\right]\right),\text{\hspace{0.17em}}\beta \left(\rho ,0\right)={\beta }_{0}\left(\rho \right),\text{\hspace{0.17em}}\beta \left(\rho ,1\right)=\stackrel{¯}{\beta },\text{\hspace{0.17em}}0\le \beta \left(\rho ,\tau \right)\le \stackrel{¯}{\beta },$

iii) $\sigma \left(\rho ,\tau \right),\text{\hspace{0.17em}}\gamma \left(\rho ,\tau \right)\in C\left(\left[0,1\right]×\left[0,T\right]\right),\text{\hspace{0.17em}}\sigma \left(\rho ,0\right)={\sigma }_{0}\left(\rho \right),\text{\hspace{0.17em}}\gamma \left(\rho ,0\right)={\gamma }_{0}\left(\rho \right),$ $|\sigma \left(\rho ,\tau \right)|\le 2{A}_{0},\text{\hspace{0.17em}}|\gamma \left(\rho ,\tau \right)|\le 2{A}_{0},\text{\hspace{0.17em}}0\le \rho \le 1,\text{\hspace{0.17em}}0\le \tau \le T.$

$\begin{array}{l}d\left(\left({\eta }_{1},{\alpha }_{1},{\beta }_{1},{\sigma }_{1},{\gamma }_{1}\right),\left(\eta ,{\alpha }_{2},{\beta }_{2},{\sigma }_{2},{\gamma }_{2}\right)\right)\\ ={‖{\eta }_{1}-{\eta }_{2}‖}_{\infty }+{‖{\alpha }_{1}-{\alpha }_{2}‖}_{\infty }+{‖{\beta }_{1}-{\beta }_{2}‖}_{\infty }+{‖{\sigma }_{1}-{\sigma }_{2}‖}_{\infty }+{‖{\gamma }_{1}-{\gamma }_{2}‖}_{\infty }.\end{array}$

$\frac{\text{d}\stackrel{˜}{\eta }\left(\tau \right)}{\text{d}\tau }=\stackrel{˜}{\eta }\left(\tau \right)v\left(1,\tau \right),\text{\hspace{0.17em}}\tau >0,$ (3.17)

$\stackrel{˜}{\eta }\left(0\right)={\eta }_{0},$ (3.18)

$\frac{\partial \stackrel{˜}{\alpha }}{\partial \tau }={\delta }_{P}\frac{1}{{\rho }^{2}}\frac{\partial }{\partial \rho }\left({\rho }^{2}\frac{\partial \stackrel{˜}{\alpha }}{\partial \rho }\right)-\mu \left(\rho ,\tau \right)\frac{\partial \stackrel{˜}{\alpha }}{\partial \rho }+{\eta }^{2}\left(\tau \right)\left({F}_{1}\left(\stackrel{˜}{\alpha },\stackrel{˜}{\beta },\stackrel{˜}{\sigma },\stackrel{˜}{\gamma }\right)\stackrel{˜}{\alpha }\right),\text{\hspace{0.17em}}0<\rho <1,\text{\hspace{0.17em}}\tau >0,$ (3.19)

$\frac{\partial \stackrel{˜}{\alpha }}{\partial t}\left(0,\tau \right)=0,\text{\hspace{0.17em}}\text{}\stackrel{˜}{\alpha }\left(1,\tau \right)=\stackrel{¯}{\alpha },\text{\hspace{0.17em}}\tau >0,$ (3.20)

$\alpha \left(\rho ,0\right)={\alpha }_{0}\left(\rho \right),\text{\hspace{0.17em}}0\le \rho \le 1,$ (3.21)

$\frac{\partial \stackrel{˜}{\beta }}{\partial \tau }={\delta }_{Q}\frac{1}{{\rho }^{2}}\frac{\partial }{\partial \rho }\left({\rho }^{2}\frac{\partial \stackrel{˜}{\beta }}{\partial \rho }\right)-\mu \left(\rho ,\tau \right)\frac{\partial \stackrel{˜}{\beta }}{\partial \rho }+{\eta }^{2}\left(\tau \right)\left({F}_{2}\left(\stackrel{˜}{\alpha },\stackrel{˜}{\beta },\stackrel{˜}{\sigma },\stackrel{˜}{\gamma }\right)\stackrel{˜}{\beta }\right),\text{\hspace{0.17em}}0<\rho <1,\text{\hspace{0.17em}}\tau >0,$ (3.22)

$\frac{\partial \stackrel{˜}{\beta }}{\partial t}\left(0,\tau \right)=0,\text{\hspace{0.17em}}\text{}\stackrel{˜}{\beta }\left(1,\tau \right)=\stackrel{¯}{\beta },\text{\hspace{0.17em}}\tau >0,$ (3.23)

$\stackrel{˜}{\beta }\left(\rho ,0\right)={\beta }_{0}\left(\rho \right),\text{\hspace{0.17em}}0\le \rho \le 1,$ (3.24)

$\frac{\partial \stackrel{˜}{\sigma }}{\partial \tau }+\mu \frac{\partial \stackrel{˜}{\sigma }}{\partial \rho }={\eta }^{2}\left(\tau \right)\left[{g}_{11}\left(\stackrel{˜}{\alpha },\stackrel{˜}{\beta },\stackrel{˜}{\sigma },\stackrel{˜}{\gamma }\right)\stackrel{˜}{\sigma }+{g}_{12}\left(\stackrel{˜}{\alpha },\stackrel{˜}{\beta },\stackrel{˜}{\sigma },\stackrel{˜}{\gamma }\right)\stackrel{˜}{\gamma }\right],\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le \rho \le 1,\text{\hspace{0.17em}}\tau >0,$ (3.25)

$\frac{\partial \stackrel{˜}{\gamma }}{\partial \tau }+\mu \frac{\partial \stackrel{˜}{\gamma }}{\partial \rho }={\eta }^{2}\left(\tau \right)\left[{g}_{21}\left(\stackrel{˜}{\alpha },\stackrel{˜}{\beta },\stackrel{˜}{\sigma },\stackrel{˜}{\gamma }\right)\stackrel{˜}{\sigma }+{g}_{22}\left(\stackrel{˜}{\alpha },\stackrel{˜}{\beta },\stackrel{˜}{\sigma },\stackrel{˜}{\gamma }\right)\stackrel{˜}{\gamma }\right],\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le \rho \le 1,\text{\hspace{0.17em}}\tau >0,$ (3.26)

$\stackrel{˜}{\sigma }\left(\rho ,0\right)={\sigma }_{0}\left(\rho \right),\text{ }\text{ }\text{ }\stackrel{˜}{\gamma }\left(\rho ,0\right)={\gamma }_{0}\left(\rho \right),\text{\hspace{0.17em}}0\le \rho \le 1.$ (3.27)

$\upsilon \left(\rho ,\tau \right)=\frac{\eta {\left(\tau \right)}^{2}}{{\rho }^{2}}{\int }_{0}^{1}h\left(\stackrel{˜}{\alpha },\stackrel{˜}{\beta },\stackrel{˜}{\sigma },\stackrel{˜}{\gamma }\right){s}^{2}\text{d}s,$ (3.28)

$\mu \left(\rho ,\tau \right)=\upsilon \left(\rho ,\tau \right)-\rho \upsilon \left(1,\tau \right).$ (3.29)

$\stackrel{˜}{\eta }\left(\tau \right)={\eta }_{0}{\text{e}}^{{\int }_{t}^{0}\upsilon \left(1,s\right)\text{d}s},\text{\hspace{0.17em}}0\le \tau \le T.$ (3.30)

$|\upsilon \left(1,\tau \right)|=|\eta {\left(\tau \right)}^{2}{\int }_{0}^{1}h\left(\alpha ,\beta ,\sigma ,\gamma \right){s}^{2}\text{d}s|\le 4{\eta }_{0}{\int }_{0}^{1}{B}_{0}{s}^{2}\text{d}s\le \frac{4}{3}{B}_{0}{\eta }_{0}^{2}.$ (3.31)

${\eta }_{0}{\text{e}}^{-\frac{4}{3}{B}_{0}{\eta }_{0}^{2}T}\le \stackrel{˜}{\eta }\left(\tau \right)\le {\eta }_{0}{\text{e}}^{\frac{4}{3}{B}_{0}{\eta }_{0}^{2}T},\text{\hspace{0.17em}}0\le \tau \le T.$ (3.32)

$T>0$ 充分小，使 ${\text{e}}^{\frac{4}{3}{B}_{0}{\eta }_{0}^{2}T}\le 2$ 时，有 $\frac{1}{2}{\eta }_{0}\le \stackrel{˜}{\eta }\left(\tau \right)\le 2{\eta }_{0}\left(0\le \tau \le T\right)$ ，即 $\stackrel{˜}{\eta }\left(\tau \right)$ 满足条件(i)。

${W}_{p}^{2,1}\left({Q}_{T}\right)\to {C}^{\lambda ,\frac{2}{\lambda }}\left(\stackrel{¯}{{Q}_{T}}\right)\text{\hspace{0.17em}}\left(\lambda =2-\frac{5}{p}\right)$

${‖\frac{\partial \stackrel{˜}{\alpha }}{\partial \rho }‖}_{\infty }\le C\left(T\right),$ (3.33)

$|\stackrel{˜}{\sigma }\left(\rho ,\tau \right)|\le 2{A}_{0},\text{\hspace{0.17em}}|\stackrel{˜}{\gamma }\left(\rho ,\tau \right)|\le 2{A}_{0},\text{\hspace{0.17em}}0\le \rho \le 1,\text{\hspace{0.17em}}0\le \tau \le T.$ (3.34)

$\text{ }{‖\frac{\partial \stackrel{˜}{\sigma }}{\partial \rho },\frac{\partial \stackrel{˜}{\gamma }}{\partial \rho }‖}_{\infty }\le 2{A}_{1}.$ (3.35)

${\upsilon }_{i}\left(\rho ,\tau \right)=\frac{{\eta }_{i}{\left(\tau \right)}^{2}}{{\rho }^{2}}{\int }_{0}^{1}h\left({\partial }_{i},{\beta }_{i},{\sigma }_{i},{\gamma }_{i}\right){s}^{2}\text{d}s,$

${\mu }_{i}\left(\rho ,\tau \right)={\upsilon }_{i}\left(\rho ,\tau \right)-\rho {\upsilon }_{i}\left(1,\tau \right),$

$\left(\stackrel{˜}{\eta },{\stackrel{˜}{\alpha }}_{i},{\stackrel{˜}{\beta }}_{i},{\stackrel{˜}{\sigma }}_{i},{\stackrel{˜}{\gamma }}_{i}\right)=F\left(\eta ,{\alpha }_{i},{\beta }_{i},{\sigma }_{i},{\gamma }_{i}\right),$

$d=d\left(\left({\eta }_{1},{\alpha }_{1},{\beta }_{1},{\sigma }_{1},{\gamma }_{1}\right),\left({\eta }_{2},{\alpha }_{2},{\beta }_{2},{\sigma }_{2},{\gamma }_{2}\right)\right).$

${\upsilon }_{i}\left(\rho ,\tau \right)=\frac{{\eta }_{i}{\left(\tau \right)}^{2}}{{\rho }^{2}}{\int }_{0}^{1}h\left({\partial }_{i},{\beta }_{i},{\sigma }_{i},{\gamma }_{i}\right){s}^{2}\text{d}s$ ，计算得：

$\begin{array}{l}|{\upsilon }_{1}\left(\rho ,\tau \right)-{\upsilon }_{2}\left(\rho ,\tau \right)|\\ =|\frac{\eta {\left({\tau }_{1}\right)}^{2}}{{\rho }^{2}}{\int }_{0}^{\rho }h\left({\partial }_{1},{\beta }_{1},{\sigma }_{1},{\gamma }_{1}\right){s}^{2}\text{d}s-\frac{\eta {\left({\tau }_{2}\right)}^{2}}{{\rho }^{2}}{\int }_{0}^{\rho }h\left({\partial }_{2},{\beta }_{2},{\sigma }_{2},{\gamma }_{2}\right){s}^{2}\text{d}s|\\ \le |\frac{\eta {\left({\tau }_{1}\right)}^{2}}{{\rho }^{2}}{\int }_{0}^{\rho }\left(h\left({\partial }_{1},{\beta }_{1},{\sigma }_{1},{\gamma }_{1}\right)-h\left({\partial }_{2},{\beta }_{2},{\sigma }_{2},{\gamma }_{2}\right)\right){s}^{2}\text{d}s\text{ }|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+|\frac{\eta {\left({\tau }_{1}\right)}^{2}-\eta {\left({\tau }_{2}\right)}^{2}}{{\rho }^{2}}{\int }_{0}^{\rho }h\left({\partial }_{2},{\beta }_{2},{\sigma }_{2},{\gamma }_{2}\right){s}^{2}\text{d}s|\\ \le C\left(T\right)d\end{array}$ (3.36)

${‖\stackrel{˜}{{\eta }_{1}}-\stackrel{˜}{{\eta }_{2}}‖}_{\infty }=\underset{0\le \tau \le T}{\mathrm{max}}|{\eta }_{0}{\text{e}}^{{\int }_{0}^{\tau }{\upsilon }_{1}\left(1,s\right)\text{d}s}-{\eta }_{0}{\text{e}}^{{\int }_{0}^{\tau }{\upsilon }_{2}\left(1,s\right)\text{d}s}|\le TC\left(T\right)d$ (3.37)

${‖\stackrel{˜}{{\mu }_{1}}-\stackrel{˜}{{\mu }_{2}}‖}_{\infty }=\underset{0\le \tau \le T}{\mathrm{max}}|{\nu }_{1}-\rho {\nu }_{1}\left(1,\tau \right)-{\nu }_{2}+\rho {\nu }_{2}\left(1,\tau \right)|\le TC\left(T\right)d.$ (3.38)

$\frac{\partial \stackrel{˜}{{\alpha }^{\ast }}}{\partial t}={\delta }_{P}\frac{1}{{\rho }^{2}}\frac{\partial }{\partial \rho }\left({\rho }^{2}\frac{\partial \stackrel{˜}{{\alpha }^{\ast }}}{\partial \rho }\right)-{\mu }_{1}\left(\rho ,\tau \right)\frac{\partial \stackrel{˜}{{\alpha }^{\ast }}}{\partial \rho }+{f}_{1}\left(\rho ,\tau \right),\text{\hspace{0.17em}}0<\rho <1,\text{\hspace{0.17em}}0\le \tau \le T.$ (3.39)

$\stackrel{˜}{{\alpha }^{\ast }}\left(0,\tau \right)=0,\text{\hspace{0.17em}}\stackrel{˜}{{\alpha }^{\ast }}\left(1,\tau \right)=0,\text{ }\text{ }\text{ }\text{ }\text{ }\tau >0,$ (3.40)

$\stackrel{˜}{{\alpha }^{\ast }}\left(\rho ,0\right)=0,\text{\hspace{0.17em}}0\le \rho \le 1.$ (3.41)

$\begin{array}{l}{f}_{1}\left(\rho ,\tau \right)=\left[{\mu }_{2}\left(\rho ,\tau \right)-{\mu }_{1}\left(\rho ,\tau \right)\right]\text{ }\text{ }\frac{\partial \stackrel{˜}{{\alpha }_{2}}}{\partial \rho }\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-{\eta }_{2}{\left(\tau \right)}^{2}\left[{\eta }_{1}^{2}\left(\tau \right){F}_{1}\left({\alpha }_{1},{\beta }_{1},{\sigma }_{1},{\gamma }_{1}\right)-{\eta }_{2}^{2}\left(\tau \right){F}_{1}\left({\alpha }_{2},{\beta }_{2},{\sigma }_{2},{\gamma }_{2}\right)\right]\stackrel{˜}{{\alpha }_{2}}\end{array}$

$\begin{array}{c}{‖{f}_{1}‖}_{\infty }\le C\left(T\right)\left[{‖{\mu }_{2}-{\mu }_{1}‖}_{\infty }+{‖{\eta }_{1}^{2}\left(\tau \right){F}_{1}\left({\alpha }_{1},{\beta }_{1},{\sigma }_{1},{\gamma }_{1}\right)-{\eta }_{2}^{2}\left(\tau \right){F}_{1}\left({\alpha }_{2},{\beta }_{2},{\sigma }_{2},{\gamma }_{2}\right)‖}_{\infty }\right]\\ \le C\left(T\right)d.\end{array}$ (3.42)

${‖\stackrel{˜}{{\alpha }^{\ast }}‖}_{\infty }\le TC\left(T\right)d.$ (3.43)

${‖\stackrel{˜}{{\beta }^{\ast }}‖}_{\infty }\le TC\left(T\right)d.$ (3.44)

$\stackrel{˜}{{\sigma }^{\ast }}=\stackrel{˜}{{\sigma }_{1}}-\stackrel{˜}{{\sigma }_{2}},\stackrel{˜}{{\gamma }^{\ast }}=\stackrel{˜}{{\gamma }_{1}}-\stackrel{˜}{{\gamma }_{2}}$ ，由(3.25)~(3.27)得

$\begin{array}{l}\frac{\partial \stackrel{˜}{{\sigma }^{\ast }}}{\partial \tau }+{\mu }_{1}\frac{\partial \stackrel{˜}{{\sigma }^{\ast }}}{\partial \rho }={\eta }_{1}^{2}\left(\tau \right){g}_{11}\left(\stackrel{˜}{{\partial }_{1}},\stackrel{˜}{{\beta }_{1}},\stackrel{˜}{{\sigma }_{1}},\stackrel{˜}{{\gamma }_{1}}\right)\stackrel{˜}{{\sigma }^{\ast }}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\eta }_{1}^{2}\left(\tau \right){g}_{12}\left(\stackrel{˜}{{\partial }_{2}},\stackrel{˜}{{\beta }_{2}},\stackrel{˜}{{\sigma }_{2}},\stackrel{˜}{{\gamma }_{2}}\right)\stackrel{˜}{{\gamma }^{\ast }}+{G}_{1}\left(\rho ,\tau \right),\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }0\le \rho \le 1,\text{\hspace{0.17em}}\tau >0,\end{array}$ (3.45)

$\begin{array}{l}\frac{\partial \stackrel{˜}{{\gamma }^{\ast }}}{\partial \tau }+{\mu }_{1}\frac{\partial \stackrel{˜}{{\gamma }^{\ast }}}{\partial \rho }={\eta }_{1}^{2}\left(\tau \right){g}_{21}\left(\stackrel{˜}{{\partial }_{1}},\stackrel{˜}{{\beta }_{1}},\stackrel{˜}{{\sigma }_{1}},\stackrel{˜}{{\gamma }_{1}}\right)\stackrel{˜}{{\sigma }^{\ast }}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\eta }_{1}^{2}\left(\tau \right){g}_{22}\left(\stackrel{˜}{{\partial }_{2}},\stackrel{˜}{{\beta }_{2}},\stackrel{˜}{{\sigma }_{2}},\stackrel{˜}{{\gamma }_{2}}\right)\stackrel{˜}{{\gamma }^{\ast }}+{G}_{2}\left(\rho ,\tau \right),\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }0\le \rho \le 1,\text{\hspace{0.17em}}\tau >0,\end{array}$ (3.46)

$\stackrel{˜}{{\sigma }^{\ast }}\left(\rho ,0\right)=0,\stackrel{˜}{{\gamma }^{\ast }}\left(\rho ,0\right)=0,\text{\hspace{0.17em}}0\le \rho \le 1,$ (3.47)

$\begin{array}{c}{G}_{1}\left(\rho ,\tau \right)=\stackrel{˜}{{\gamma }_{2}}\left[{\eta }_{1}^{2}\left(\tau \right){g}_{12}\left(\stackrel{˜}{{\partial }_{1}},\stackrel{˜}{{\beta }_{1}},\stackrel{˜}{{\sigma }_{1}},\stackrel{˜}{{\gamma }_{1}}\right)-{\eta }_{2}^{2}\left(\tau \right){g}_{12}\left(\stackrel{˜}{{\partial }_{1}},\stackrel{˜}{{\beta }_{1}},\stackrel{˜}{{\sigma }_{1}},\stackrel{˜}{{\gamma }_{1}}\right)\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\stackrel{˜}{{\sigma }_{2}}\left[{\eta }_{1}^{2}\left(\tau \right){g}_{11}\left(\stackrel{˜}{{\partial }_{1}},\stackrel{˜}{{\beta }_{1}},\stackrel{˜}{{\sigma }_{1}},\stackrel{˜}{{\gamma }_{1}}\right)-{\eta }_{2}^{2}\left(\tau \right){g}_{11}\left(\stackrel{˜}{{\alpha }_{2}},\stackrel{˜}{{\beta }_{2}},\stackrel{˜}{{\sigma }_{2}},\stackrel{˜}{{\gamma }_{2}}\right)\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{\partial \stackrel{˜}{{\sigma }_{2}}}{\partial \rho }\left({\mu }_{2}-{\mu }_{1}\right)\end{array}$ (3.48)

$\begin{array}{c}{G}_{2}\left(\rho ,\tau \right)=\stackrel{˜}{{\gamma }_{2}}\left[{\eta }_{1}^{2}\left(\tau \right){g}_{22}\left(\stackrel{˜}{{\partial }_{1}},\stackrel{˜}{{\beta }_{1}},\stackrel{˜}{{\sigma }_{1}},\stackrel{˜}{{\gamma }_{1}}\right)-{\eta }_{2}^{2}\left(\tau \right){g}_{22}\left(\stackrel{˜}{{\partial }_{1}},\stackrel{˜}{{\beta }_{1}},\stackrel{˜}{{\sigma }_{1}},\stackrel{˜}{{\gamma }_{1}}\right)\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\stackrel{˜}{{\sigma }_{2}}\left[{\eta }_{1}^{2}\left(\tau \right)-{\eta }_{2}^{2}\left(\tau \right){g}_{21}\left(\stackrel{˜}{{\alpha }_{2}},\stackrel{˜}{{\beta }_{2}},\stackrel{˜}{{\sigma }_{2}},\stackrel{˜}{{\gamma }_{2}}\right)\right]+\frac{\partial \stackrel{˜}{{\gamma }_{\text{2}}}}{\partial \rho }\left({\mu }_{2}-{\mu }_{1}\right),\end{array}$ (3.49)

${‖{G}_{i}‖}_{\infty }\le C\left(T\right)d.$ (3.50)

${‖\stackrel{˜}{{\sigma }_{1}}-\stackrel{˜}{{\sigma }_{2}},\stackrel{˜}{{\gamma }_{1}}-\stackrel{˜}{{\gamma }_{2}}‖}_{\infty }\le TC\left(T\right)d.$ (3.51)

$d\left(\left(\stackrel{˜}{{\alpha }_{1}},\stackrel{˜}{{\beta }_{1}},\stackrel{˜}{{\sigma }_{1}},\stackrel{˜}{{\gamma }_{1}}\right),\left(\stackrel{˜}{{\alpha }_{2}},\stackrel{˜}{{\beta }_{2}},\stackrel{˜}{{\sigma }_{2}},\stackrel{˜}{{\gamma }_{2}}\right)\right)\le TC\left(T\right)d.$ (3.52)

$T>0$ 充分小，使得 $TC\left(T\right)<1$ 时，F是压缩映射。

4. 整体解的存在唯一性

$\eta \left(\tau \right)={\left(\frac{1}{{\eta }_{0}^{2}}-{\int }_{0}^{\tau }{\int }_{0}^{1}h\left(\partial ,\beta ,\sigma ,\gamma \right){s}^{2}\text{d}s\text{d}\tau \right)}^{-\frac{1}{2}},$

$\tau \to \infty$ 时，如果 $h\left(\partial ,\beta ,\sigma ,\gamma \right)<0$ ，则 $\eta \to 0$ ，即 $R\to 0$ 。这说明问题(3.2)~(3.13)的解不是对所有的 $\tau >0$ 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。

$0\le 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 $t\ge 0$ 都成立，则有下面的结论成立：

$0\le P\left(r,t\right)\le \stackrel{¯}{P},0\le Q\left(r,t\right)\le \stackrel{¯}{Q},\text{\hspace{0.17em}}0\le r\le R\left(t\right),\text{\hspace{0.17em}}0\le t (4.1)

$M\left(r,t\right)\ge 0,N\left(r,t\right)\ge 0,0\le r\le R\left(t\right),0\le t (4.2)

$M+N=1,0\le r\le R\left(t\right),0\le t (4.3)

${R}_{0}{\text{e}}^{-\frac{1}{3}{B}_{0}t}\le R\left(t\right)\le {R}_{0}{\text{e}}^{\frac{1}{3}{B}_{0}t},0\le t (4.4)

$-\frac{1}{3}{B}_{0}R\left(t\right)\le \stackrel{˙}{R}\left(t\right)\le \frac{1}{3}{B}_{0}R\left(t\right),0\le t (4.5)

$\stackrel{˙}{R}\left(t\right)=\frac{\text{d}R\left(t\right)}{\text{d}t},{B}_{0}=\mathrm{max}\left\{|h\left(P,Q,M,N\right)|:0\le P\left(r,t\right)\le \stackrel{¯}{P},0\le Q\left(r,t\right)\le \stackrel{¯}{Q},0\le M\le 1,0\le N\le 1\right\}.$

(1.11)~(1.12)的定义域变换成 $\left\{\left(\rho ,\tau \right):0\le \rho \le 1,0\le \tau \le \delta \right\}$ ，利用引理2.2得(4.2)式。再令 $Z\left(r,t\right)=M\left(r,t\right)+N\left(r,t\right)$ ，将(1.11)~(1.12)加起来，得

$\frac{\partial M}{\partial t}+\upsilon \frac{\partial M}{\partial r}=h\left(P,Q,M,N\right)\left[1-Z\left(r,t\right)\right],0\le r\le R\left(t\right),0\le t (4.6)

$Z\left(r,0\right)=M\left(r,0\right)+N\left(r,0\right)=1,0\le r\le {R}_{0}.$ (4.7)

$|h\left(P,Q,M,N\right)|\le {B}_{0},\text{\hspace{0.17em}}0\le r\le R\left(t\right),\text{\hspace{0.17em}}0\le t

$-\frac{1}{3}{B}_{0}r\le \upsilon \left(r,t\right)\le \frac{1}{3}{B}_{0}r,0\le r\le R\left(t\right),0\le t

$-\frac{1}{3}{B}_{0}R\left(t\right)\le \stackrel{˙}{R}\left(t\right)\le \frac{1}{3}{B}_{0}R\left(t\right),\text{ }\text{ }\text{ }\text{ }\text{ }0\le t

$\sigma \left(x,t\right)=P\left(|x|R\left(t\right),t\right),|x|\le 1,0\le t 。由(1.12)，(1.17)和(1.20)式可知， $\sigma$ 是下面问题的解：

$\frac{\partial \sigma }{\partial t}=\frac{{\delta }_{P}}{{R}^{2}\left(t\right)}\Delta \sigma +\frac{\stackrel{˙}{R}\left(t\right)}{R\left(t\right)}\left(x\cdot ▽\sigma \right)-c\left(x,t\right)\sigma ,|x|<1,0 (4.8)

$\sigma \left(x,t\right)=\stackrel{¯}{P},|x|=1,0\le t (4.9)

$\sigma \left(x,0\right)={P}_{0}\left(|x|{R}_{0}\right),|x|\le 1.$ (4.10)

$\sigma \left(x,t\right)=P\left(|x|R\left(t\right),t\right)$ ，所以

${‖P\left(|x|R\left(t\right),t\right)‖}_{{W}_{p}^{2,1}}\le C\left(T\right),$

${‖P\left(|x|,{t}_{0}\right)‖}_{{D}_{p}\left(B\left({t}_{0}\right)\right)}\le C\left(T\right),\text{\hspace{0.17em}}B\left({t}_{0}\right)=\left\{x\in {R}^{3},|x|

${‖Q\left(|x|,{t}_{0}\right)‖}_{{D}_{p}\left(B\left({t}_{0}\right)\right)}\le C\left(T\right),\text{\hspace{0.17em}}B\left({t}_{0}\right)=\left\{x\in {R}^{3},|x|

${‖\frac{\partial M}{\partial \rho },\frac{\partial N}{\partial \rho }‖}_{\infty }\le C\left(T\right).$

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