## 具周期营养供给的血管化肿瘤生长模型的渐近分析

1江西师范大学数学与统计学院 南昌330022

2广丰中学 江西上饶334699

## Asymptotic Analysis of a Tumor Model with Angiogenesis and a Periodic Supply of External Nutrients

Song Huijuan1, Huang Qian2, Wang Zejia,1,*

1School of Mathematics and Statistics, Jiangxi Normal University, Nanchang 330022

2Guangfeng Middle School, Jiangxi Shangrao 334699

 基金资助: 国家自然科学基金.  12261047国家自然科学基金.  12161045国家自然科学基金.  11861038江西省自然科学基金.  20212BAB201016

Received: 2021-11-24   Revised: 2022-07-5

 Fund supported: The NSFC.  12261047The NSFC.  12161045The NSFC.  11861038Natural Science Foundation of Jiangxi Province of China.  20212BAB201016

Abstract

In this paper, we consider a free boundary problem modeling the growth of tumors with angiogenesis and a $\omega$-periodic supply of external nutrients $\phi(t)$. Denote by $S(\sigma)$ the proliferation rate of tumor cells. We first establish the well-posedness and then give a complete classification of asymptotic behavior of solutions according to the sign of $\frac1{\omega}\int_0^\omega S(\phi(t)){\rm d}t$. It is shown that if $\frac1{\omega}\int_0^\omega S(\phi(t)){\rm d}t\le0$, then all evolutionary tumors will finally vanish; the converse is also true. If instead $\frac1{\omega}\int_0^\omega S(\phi(t)){\rm d}t>0$, then there exists a unique and stable positive periodic solution.

Keywords： Free boundary problem ; Necrotic tumor ; Angiogenesis ; Periodic solution ; Stability

Song Huijuan, Huang Qian, Wang Zejia. Asymptotic Analysis of a Tumor Model with Angiogenesis and a Periodic Supply of External Nutrients. Acta Mathematica Scientia[J], 2023, 43(1): 261-273 doi:

## 1 引言

$\begin{matrix} &&\frac1{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial\sigma}{\partial r}\right) = f(\sigma)H(\sigma-\sigma_D), 0<r<R(t),~t>0, \end{matrix}$
$\begin{matrix} \label{eq(1.1)} \\ &&\frac{\partial\sigma}{\partial r}(0,t)=0, t>0, \end{matrix}$
$\begin{matrix} \label{eq(1.2)} \\ &&\frac{\partial\sigma}{\partial r}(R(t),t)+\beta[\sigma(R(t),t)-\phi(t)]=0, t>0, \end{matrix}$
$\begin{matrix} \label{eq(1.3)} \\ &&R^2(t)\frac{{\rm d}R(t)}{{\rm d}t}=\int^{R(t)}_0\big\{g(\sigma)H(\sigma-\sigma_D)-\nu[1-H(\sigma-\sigma_D)] \big\}r^2{\rm d}r, t>0, \end{matrix}$
$R(0)=R_0.$

$$$\label{eq(1.7)} S(\sigma)=g(\sigma)H(\sigma-\sigma_D)-\nu[1-H(\sigma-\sigma_D)]$$$

$$$\label{eq(1.10)} \sigma(R(t),t)=\phi(t),$$$

$f$, $g$ 为线性函数的无死核情形, 即对任意 $t>0$$0<r<R(t), 有 \sigma(r,t)>\sigma_D. 针对零稳态的全局稳定性, 他们建立了下列必要条件 $$\label{eq(1.11)} \frac1{\omega}\int^\omega_0\phi(t){\rm d}t-\tilde\sigma\le0$$ 和充分条件 $$\label{eq(1.12)} \frac1{\omega}\int^\omega_0\phi(t){\rm d}t-\tilde\sigma<0;$$ 而当 $$\label{eq(1.13)} \min_{0\le t\le\omega}\phi(t)-\tilde\sigma>0$$ 时, 他们证明了问题存在唯一的稳定的正周期解. 2021 年, He 和Xing[17] 对上述结果做了本质的补充和改进, 证明了当 \frac1{\omega}\int^\omega_0\phi(t){\rm d}t-\tilde\sigma=0 时, 零稳态仍是全局稳定的, 唯一的稳定的正周期解存在当且仅当 \frac1{\omega}\int^\omega_0\phi(t){\rm d}t-\tilde\sigma>0. Wu-Xu[25] 则对具周期边界条件(1.10), f$$g$ 是一般非线性函数的问题(1.1)-(1.5)讨论了适定性和解的渐近性态. 对于其他相关的研究, 读者可参见文献[2-3,6,9,11,16,19,21,27].

(A1) $f\in C^1[0,+\infty)$ 严格单调增加, 且 $f(0)=0$;

(A2) $g \in C^1[0,+\infty)$ 严格单调增加, 且存在 $\tilde\sigma>\sigma_D$ 使得 $g(\tilde\sigma)=0$;

(A3) $g(\sigma_D)+\nu\geq0$.

$(\sigma(r,t),R(t))$ 为问题(1.1)-(1.5)的解, 若 $R(t)\in C^1(0, +\infty)$, 且满足方程(1.4),(1.5); 另外, 当肿瘤无死核时, $\sigma(r,t)\in C^2((0,R(t))\times(0,+\infty))$, 且逐点满足方程(1.1); 而当肿瘤含有死核时, $\sigma(r,t)\in C^1((0,R(t))\times(0,+\infty))$, 在死核区域和繁衍区域分别是 $C^2$ 函数, 且分别逐点满足方程(1.1); 最后, 方程(1.2)和(1.3)逐点成立. 记

$\begin{eqnarray*} \bar{S}=\frac1{\omega}\int^{\omega}_0S(\phi(t)){\rm d}t, \end{eqnarray*}$

$\lim\limits_{t\to+\infty}|R(t,R_0)-R_{\rm per}(t)|=0.$

## 2 准备工作

$$$\label{eq-2.9} G(\bar\sigma,R)=\frac1{R^3}\left[\int_{\sigma(r,\bar\sigma,R)>\sigma_D}g(\sigma(r,\bar\sigma,R))r^2{\rm d}r -\int_{\sigma(r,\bar\sigma,R)\leq\sigma_D}\nu r^2{\rm d}r\right],$$$

$$$\label{eq-2.10} G(\bar\sigma,R)= \left\{\begin{array}{ll} -\frac{\nu}3,&\quad 0<\bar\sigma\le\sigma_D,~R>0, \\ \frac1{R^3}\int_0^R g(\sigma(r,\bar\sigma,R))r^2{\rm d}r,&\quad\bar\sigma>\sigma_D,~0<R\le R_c(\bar\sigma), \\ \frac1{R^3}\int_{\rho(\bar\sigma,R)}^R g(V(r,\bar\sigma,R))r^2{\rm d}r-\frac{\nu}{3}\frac{\rho^3(\bar\sigma,R)}{R^3}, &\quad\bar\sigma>\sigma_D,~R>R_c(\bar\sigma), \end{array}\right.$$$

$G$$(0,+\infty)\times(0,+\infty) 中连续. 进一步, 我们有下面的结果. 引理2.2 在假定 (A1)-(A3) 下, 下列结论成立 (i) 当 \bar\sigma>\sigma_D, R>0 时, G(\bar\sigma,R) 关于 \bar\sigma 严格单调增加, 关于 R 严格单调减少; (ii) 对任意固定的 \bar\sigma>\sigma_D, 有 $$\label{eq-2.15} \lim\limits_{R\to0^+}G(\bar\sigma,R)=\frac{g(\bar\sigma)}3,\quad \lim\limits_{R\to+\infty}G(\bar\sigma,R)=-\frac{\nu}3.$$ (i) 为了简洁起见, 这里我们只证明函数 G 关于 \bar\sigma 的严格单调增加性, 对 G 关于 R 的严格单调减少性, 读者可参见文献[22]. 对给定的 R_*>0, 如果存在 \bar\sigma_*>\sigma_D 使得 R_c(\bar\sigma_*)=R_*, 那么记 \bar\sigma_*=R_c^{-1}(R_*), 否则记 \bar\sigma_*=+\infty. 我们断言当 \sigma_D<\bar\sigma<\bar\sigma_* 或者 \bar\sigma>\bar\sigma_* 时, 有 \partial_{\bar\sigma}G(\bar\sigma,R_*)>0. 事实上, 当 \sigma_D<\bar\sigma<\bar\sigma_* 时, 根据(2.10)和(2.6)式可以得到 \begin{matrix}\label{eq-2.11} \frac{\partial G}{\partial\bar\sigma}(\bar\sigma,R_*)&=&\frac1{R_*^3}\bigg\{\int^{R_*}_{\rho(\bar\sigma,R_*)} g'(V(r,\bar\sigma,R_*)) \frac{\partial V}{\partial\bar\sigma}(r,\bar\sigma,R_*)r^2{\rm d}r \\ &&-[g(\sigma_D)+\nu]\rho^2(\bar\sigma,R_*) \frac{\partial\rho}{\partial\bar\sigma}(\bar\sigma,R_*)\bigg\}, \end{matrix} $$\label{eq-2.12} \left\{\begin{array}{ll} -\frac{\partial^2}{\partial r^2}\left(\frac{\partial V}{\partial\bar\sigma}\right)(r,\bar\sigma,R_*)-\frac{2}{r}\frac{\partial}{\partial r}\left(\frac{\partial V}{\partial\bar\sigma}\right)(r,\bar\sigma,R_*)+f'(V(r,\bar\sigma,R_*))\frac{\partial V}{\partial\bar\sigma}(r,\bar\sigma,R_*)=0,\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~\rho(\bar\sigma,R_*)<r<R_*, \\ \frac{\partial}{\partial r}\left(\frac{\partial V}{\partial\bar\sigma}\right)(\rho(\bar\sigma,R_*),\bar\sigma,R_*)= -f(\sigma_D)\frac{\partial\rho}{\partial\bar\sigma}(\bar\sigma,R_*), \\ \frac{\partial}{\partial r}\left(\frac{\partial V}{\partial\bar\sigma}\right)(R_*,\bar\sigma,R_*)+\beta\frac{\partial V}{\partial\bar\sigma}(R_*,\bar\sigma,R_*)=\beta, \\ \frac{\partial V}{\partial\bar\sigma}(\rho(\bar\sigma,R_*),\bar\sigma,R_*)=0. \end{array}\right.$$ 对问题(2.13)应用强极值原理, 有 \frac{\partial V}{\partial\bar\sigma}(r,\bar\sigma,R_*)>0,\quad\rho(\bar\sigma,R_*)<r\le R_*\qquad\mbox{和}\qquad\frac{\partial\rho}{\partial\bar\sigma}(\bar\sigma,R_*)<0. 再结合假设 (A2) 和 (A3), 就有 $$\frac{\partial G}{\partial\bar\sigma}(\bar\sigma,R_*)>0,\quad\sigma_D<\bar\sigma<\bar\sigma_*.$$ 然而当 \bar\sigma>\bar\sigma_* 时, 因为这时 R_*<R_c(\bar\sigma), 故由(2.10)和(2.4)式知 \begin{eqnarray*} \frac{\partial G}{\partial\bar\sigma}(\bar\sigma,R_*) =\frac1{R_*^3}\int^{R_*}_0 g'(\sigma(r,\bar\sigma,R_*)) \frac{\partial\sigma}{\partial\bar\sigma}(r,\bar\sigma,R_*)r^2{\rm d}r, \end{eqnarray*} \begin{eqnarray*} \left\{\begin{array}{ll} -\frac{\partial^2}{\partial r^2}\left(\frac{\partial \sigma}{\partial\bar\sigma}\right)(r,\bar\sigma,R_*)-\frac{2}{r}\frac{\partial}{\partial r}\left(\frac{\partial \sigma}{\partial\bar\sigma}\right)(r,\bar\sigma,R_*)+f'( \sigma(r,\bar\sigma,R_*))\frac{\partial \sigma}{\partial\bar\sigma}(r,\bar\sigma,R_*)=0, \\ \hskip 10cm 0<r<R_*, \\ \frac{\partial}{\partial r}\left(\frac{\partial \sigma}{\partial\bar\sigma}\right)(0,\bar\sigma,R_*)=0, \quad \frac{\partial}{\partial r}\left(\frac{\partial \sigma}{\partial\bar\sigma}\right)(R_*,\bar\sigma,R_*)+\beta\frac{\partial \sigma}{\partial\bar\sigma}(R_*,\bar\sigma,R_*)=\beta. \end{array}\right. \end{eqnarray*} 再次利用强极值原理和假设 (A2), 我们得到当 \bar\sigma>\bar\sigma_* 时, 有 \partial_{\bar\sigma}G(\bar\sigma,R_*)>0. 由此以及(2.14)式, 函数 G 的连续性和 R_* 的任意性即得函数 G(\bar\sigma,R) 在点集 \{(\bar\sigma,R):\bar\sigma>\sigma_D, R>0\} 中关于 \bar\sigma 是严格单调增加的. (ii) 对任意 \bar\sigma>\sigma_D 和充分小的 R, 有 $$\label{eq-2.14} G(\bar\sigma,R)=\frac1{R^3}\int_0^R g(\sigma(r,\bar\sigma,R))r^2{\rm d}r.$$ 如果令 U(s,\bar\sigma,R)=\sigma(r,\bar\sigma,R),\quad s=\frac{r}R, 那么(2.15)式可被改写为 G(\bar\sigma,R)=\int^1_0g(U(s,\bar\sigma,R))s^2{\rm d}s. 因为对任意 0\le s\le1, 有 \lim\limits_{R\to0^+}U(s,\bar\sigma,R)=\bar\sigma (见文献 [22,(2.18)式]), 故根据 Lebesgue 控制收敛定理, 立即可得到(2.11)式中的第一个等式. 另一方面, 对 \bar\sigma>\sigma_D 和足够大的 R, 由(2.10)式有 $$\label{eq-2.16} G(\bar\sigma,R)=\frac1{R^3}\int_{\rho(\bar\sigma,R)}^R g(V(r,\bar\sigma,R))r^2{\rm d}r-\frac{\nu}{3}\frac{\rho^3(\bar\sigma,R)}{R^3}.$$ 利用引理 2.1 中的结论 (b2), 在(2.16)式中令 R\to+\infty, 可知(2.11)式中的第二个等式成立. 证毕. 注2.2 在假定 (A2) 和 (A3) 下, 由(1.6)式定义的函数 S(\bar\sigma)$$(0,+\infty)$ 上是不减的, 而且根据函数 $G$ 的定义和引理 2.2, 对任意 $\bar\sigma>0$$R>0, 有 $$\label{eq-2.17} G(\bar\sigma,R)\le\frac{S(\bar\sigma)}3.$$ ## 3 主要结果的证明 在本节中, 我们证明定理 1.1. 为方便, 我们将定理 1.1 分为三个命题. 首先对任意 R_0>0, 建立问题(1.1)-(1.5)时变解的存在性和唯一性. 命题3.1 设 (A1)-(A3) 成立, 则对任意 R_0>0, 问题(1.1)-(1.5)存在唯一的解 (\sigma(r,t), R(t,R_0))$$(0\le r\le R(t,R_0)$, $t\ge0)$.

$$$\label{eq-3.1} \left\{\begin{array}{ll} \frac{{\rm d}R}{{\rm d}t}=R(t)G(\phi(t),R(t)),\quad t>0,\\ R(0)=R_0, \end{array}\right.$$$

$$$\label{eq-3.0} R(t,R_0)\le R(a,R_0){\rm e}^{\frac{2S^*-S_*}3}.$$$

$$$\label{eq-3.3} R(t_2,R_0)=R(t_1,R_0){\rm e}^{\int^{t_2}_{t_1}G(\phi(\tau),R(\tau,R_0)){\rm d}\tau} \le R(t_1,R_0){\rm e}^{\frac13\int^{t_2}_{t_1}S(\phi(\tau)){\rm d}\tau}.$$$

$$$R(t+\omega,R_0)\le R(t,R_0),\quad t\ge0. \label{eq-3.4}$$$

$a\le t\le a+\omega$, 则由(3.5)式知

$$$R(t,R_0)\le R(a,R_0){\rm e}^{\frac13\int^{t}_aS(\phi(\tau)){\rm d}\tau}.$$$

$$$\label{eq-3.10} \int^{t}_aS(\phi(\tau)){\rm d}\tau\le2S^*-S_*,\quad a\le t\le a+\omega.$$$

$\int^t_aS(\phi(\tau)){\rm d}\tau=\int^t_{k\omega}S(\phi(\tau)){\rm d}\tau-\int^a_{k\omega}S(\phi(\tau)){\rm d}\tau \le S^*-S_*,$

$\begin{eqnarray*} \int^t_aS(\phi(\tau)){\rm d}\tau&=&\int^{(k+1)\omega}_aS(\phi(\tau)){\rm d}\tau +\int^t_{(k+1)\omega}S(\phi(\tau)){\rm d}\tau \\ &=&\int^{(k+1)\omega}_{k\omega}S(\phi(\tau)){\rm d}\tau-\int^a_{k\omega}S(\phi(\tau)){\rm d}\tau +\int^t_{(k+1)\omega}S(\phi(\tau)){\rm d}\tau \\ &\le &2S^*-S_*. \end{eqnarray*}$

$$$R(t,R_0)\le R(a,R_0){\rm e}^{\frac{2S^*-S_*}3},\quad a\le t\le a+\omega. \label{eq-3.5}$$$

$$$\label{eq-3.7} \liminf_{t\to+\infty}R(t,R_0)=0.$$$

$\begin{eqnarray*} \liminf_{t\to+\infty}R(t,R_0)=\alpha. \end{eqnarray*}$

$$$\label{eq-3.11} R(t,R_0)\ge\frac{\alpha}2.$$$

$$$\label{eq-3.9} R(t_*+n\omega,R_0)\le R(t_*,R_0){\rm e}^{\int^{t_*+n\omega}_{t_*}G\left(\phi(\tau),\frac\alpha2\right){\rm d}\tau} =R(t_*,R_0){\rm e}^{n\int^{\omega}_0G\left(\phi(\tau),\frac\alpha2\right){\rm d}\tau}.$$$

$$$\int^{\omega}_0G\left(\phi(\tau),\frac\alpha2\right){\rm d}\tau<0.$$$

$\begin{eqnarray*} \int^{\omega}_0G\left(\phi(\tau),\frac\alpha2\right){\rm d}\tau= \int_{\{0\le\tau\le\omega: \phi(\tau)\le\sigma_D\}}G\left(\phi(\tau),\frac\alpha2\right){\rm d}\tau+ \int_{\{0\le\tau\le\omega: \phi(\tau)>\sigma_D\}}G\left(\phi(\tau),\frac\alpha2\right){\rm d}\tau, \end{eqnarray*}$

$\begin{eqnarray*} \int^{\omega}_0G\left(\phi(\tau),\frac\alpha2\right){\rm d}\tau= \int^{\omega}_0-\frac{\nu}3{\rm d}\tau<0, \end{eqnarray*}$

$\begin{eqnarray*} \int^{\omega}_0G\left(\phi(\tau),\frac\alpha2\right){\rm d}\tau &<& \int_{\{0\le\tau\le\omega: \phi(\tau)\le\sigma_D\}}-\frac{\nu}3{\rm d}\tau+ \int_{\{0\le\tau\le\omega: \phi(\tau)>\sigma_D\}}\frac{g(\phi(\tau))}3{\rm d}\tau \\ &=&\frac{\omega\bar{S}}3\le0. \end{eqnarray*}$

$\lim\limits_{n\to+\infty}R(t_*+n\omega,R_0)=0,$

$$$\label{eq-3.12} \lim\limits_{t\to+\infty}R(t,R_0)=0.$$$

$0<R(t_\varepsilon,R_0)<\varepsilon {\rm e}^{-\frac{2S^*-S_*}3}.$

$R(t,R_0)<\varepsilon.$

$\lim\limits_{R\to0^+}\int^{\omega}_0 G(\phi(t),R){\rm d}t=\frac13\int^{\omega}_0 S(\phi(t)){\rm d}t=\frac{\omega\bar{S}}3>0.$

$R(t,R_0)\le\delta.$

$\begin{eqnarray*} R(t+\omega,R_0)&=&R(t,R_0){\rm e}^{\int^{t+\omega}_t G(\phi(\tau),R(\tau,R_0)){\rm d}\tau} \ge R(t,R_0){\rm e}^{\int^{t+\omega}_t G(\phi(\tau),\delta){\rm d}\tau} \\ &=&R(t,R_0){\rm e}^{\int^{\omega}_0 G(\phi(\tau),\delta){\rm d}\tau} >R(t,R_0), \end{eqnarray*}$

$$$\label{eq-3.24} R(t_0+n\omega,R_0)\le R(t_0,R_0){\rm e}^{\frac13\int^{t_0+n\omega}_{t_0}S(\phi(\tau)){\rm d}\tau} =R(t_0,R_0){\rm e}^{\frac13n\omega\bar{S}}.$$$

$\lim\limits_{n\to\infty}R(t_0+n\omega,R_0)=0.$

$R(t_0+n\omega,R_0)\le R(t_0,R_0).$

$$$\label{eq-3.14} \lim\limits_{t\to+\infty}|R(t,R_0)-R_{\rm per}(t)|=0.$$$

$$$\label{eq-3.15} R(t,R_0)\le\max\{R_0,R^+\}.$$$

$R(t,R_0)\le R_n{\rm e}^{\frac13 S^*}.$

$R_{n+1}=R_n {\rm e}^{\int^\omega_0G(\phi(\tau),R(\tau,R_n)){\rm d}\tau},$

$\lim\limits_{n\to\infty}\int^\omega_0G(\phi(\tau),R(\tau,R_n)){\rm d}\tau=0,$

$\label{eq-3.17} \int^\omega_0G(\phi(\tau),R(\tau,R^\#)){\rm d}\tau=0.$

$R(\omega,R^\#)=R^\# {\rm e}^{\int^\omega_0G(\phi(\tau),R(\tau,R^\#)){\rm d}\tau}=R^\#,$

$$$\label{eq-3.18} R(t,R_0)=R_{\rm per}(t){\rm e}^{\beta(t)},$$$

$$$\label{eq-3.22} G(\phi(t),R_{\rm per}(t){\rm e}^{\beta_\infty})-G(\phi(t),R_{\rm per}(t))>0,\quad t\in[a,b].$$$

$\begin{eqnarray*} \label{eq-3.23} \beta(b+k\omega)-\beta(a)&=&\int^{b+k\omega}_a\frac{{\rm d}\beta}{{\rm d}t}{\rm d}t \ge\int^{b+k\omega}_a\left[G(\phi(t),R_{\rm per}(t){\rm e}^{\beta_\infty})-G(\phi(t),R_{\rm per}(t))\right]{\rm d}t \\ &\ge&\int^b_a\left[G(\phi(t),R_{\rm per}(t){\rm e}^{\beta_\infty})-G(\phi(t),R_{\rm per}(t))\right]{\rm d}t \\ &&+\int^{b+\omega}_{a+\omega}\left[G(\phi(t),R_{\rm per}(t){\rm e}^{\beta_\infty})-G(\phi(t),R_{\rm per}(t))\right]{\rm d}t \\ &&+\cdots+\int^{b+k\omega}_{a+k\omega}\left[G(\phi(t),R_{\rm per}(t){\rm e}^{\beta_\infty})-G(\phi(t),R_{\rm per}(t))\right]{\rm d}t \\ &=&(k+1)\int^b_a\left[G(\phi(t),R_{\rm per}(t){\rm e}^{\beta_\infty})-G(\phi(t),R_{\rm per}(t))\right]{\rm d}t \\ &\to&+\infty (k\to\infty), \end{eqnarray*}$

$\lim\limits_{k\to\infty}[\beta(b+k\omega)-\beta(a)]=\beta_\infty-\beta(a)$

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