近年来, 由Keller和Segel ^[1]提出的一类趋化模型(被称为K-S模型)引起众多生物及数学研究者的关注, 并取得了一些研究成果^[2-9].本文在Hillen和Painter ^[9]的研究基础上, 考虑如下带双反应项的抛物趋化模型:

$ \begin{equation} \left\{ \begin{aligned} &u_t=\nabla(\nabla u-V(u, v)\nabla v)+f(u, v), \\ &v_t=\mu\triangle v+g(u, v), \\ &u(0, \cdot)=u_0, \\ &v(0, \cdot)=v_0. \end{aligned} \right. \end{equation} $

(1.1)

本文的主要工作是证明上述模型的解的存在性.为此, 先给出一些必要的假设条件及引理.

假设$V(u, v), f(u, v), g(u, v)$满足以下条件：

(A.1) $V(u, v)=u\beta(u)\chi(v), \beta, \chi\in C^3(\mathbb R^2)$且满足

(ⅰ) $\chi>0$;

(ⅱ) $\beta(0)>0, \mbox{存在}\bar{u}>0, \mbox{使}\beta(\bar{u})=0, $且对所有$u\in(0, \bar{u}), $都有$\beta(u)>0;$

(A.2) $f\in C^1(\mathbb R^2), \mbox{且}f(u, v)\leq0$;

(A.3) $\mu>0$为常数, $g\in C^2(\mathbb R^2), g(u, v)=g_1(u, v)u-g_2(u, v)v, \mbox{且有}g_1\geq0, g_2\geq\delta>0$.对固定的$t_0>0$, 定义空间

$ X_u:=C([0, t_0], W^{\sigma, p}(\textit{M})), ~~~~X_v:=C([0, t_0], W^{\sigma+\alpha, p}(M)), $

其中$\sigma, \alpha, p$满足

(A.4) $ 1<\sigma<2, ~1<\alpha<2, ~2<\sigma+\alpha<3, ~~\max\{\frac{n}{\sigma-1}, \frac{2n}{2-\sigma}\}<p$, 其中$M$是一个三次可微、无边的$n$维紧致黎曼流形.这样的流形很常见, 如一维开区间, 二维环面, $n$维球面.

本文的主要工作是在黎曼流形$M$上考虑抛物型趋化模型$(1.1)$在空间$X_u\times X_v$中解的存在性.

引理 1.1 ^[10] 设$e^{\Delta t}$是热方程$u_t=\Delta u$的解半群, 对$\forall p\geq q>0, ~s\geq r$, 有

$ \begin{equation*} e^{\Delta t}:~W^{r, q}(M)\rightarrow W^{s, p}(M), \end{equation*} $

且算子$e^{\Delta t}$的范数为$C_{\kappa}t^{-\kappa}$, 其中

$ \kappa=\frac{n}{2}(\frac{1}{q}-\frac{1}{p})+\frac{1}{2}(s-r), ~~C_{\kappa}=C_\kappa(r, s, p, q). $

借助$V(u, v)$有两个零点, 可以在$\mathbb R^2$上定义$(u, v)$的一个不变区域, 这是证明全局解的关键.首先给出这样一个区域, 然后证明它是一个不变区域.定义

$ \Gamma:={(u, v)\in\mathbb R^2:0\leq u\leq\bar{u}, 0\leq v\leq\bar{v}}. $

引理 2.1  若$(u, v)\in X_u\times X_v$是问题$(1.1)$的解, 且$(u_0, v_0)\in \Gamma$, 则$(u, v)\in\Gamma$.称$\Gamma$为问题$(1.1)$解的不变区域.

引理 2.2  设$(u, v)\in X_u\times X_v$且$(u(t, x), v(t, x))\in\Gamma, \forall t\in[0, t_0], x\in M$, 若$(u, v)$是问题$(1.1)$的解, 则

$ \begin{equation} \parallel v\parallel_{X_v}\leq\parallel v_0\parallel_{\sigma+\alpha, p}+C_0t_0^{1-\tau} (\parallel u\parallel_{X_u}+\parallel u\parallel_{X_u}^2+ \parallel v\parallel_{X_v}), \end{equation} $

(2.1)

其中$\tau=\frac{1}{2}(\sigma+\alpha-1), ~~C_0=C_0(\sigma, \alpha, p, \bar u, \parallel g_1\parallel_{C^1(\Gamma)})$.

注  引理2.1-2.2类似文献[9]中定理2.1-2.2的证明, 此处略.

推论 2.3  假设引理2.2的条件成立, 若选择$t_0\leq(\frac{1}{2C_0})^{\frac{1}{1-\tau}}$, 则有

$ \begin{equation} \parallel v\parallel_{X_v}\leq C_1(\parallel v_0\parallel_{\sigma+\alpha, p} +\parallel u\parallel_{X_u}+\parallel u\parallel_{X_u}^2), \end{equation} $

(2.2)

其中$C_1=2\max\{1, C_0t_0^{1-\tau}\}$.

引理 2.4  设$\phi_1, \phi_2\in X_u$且$\phi_1(t, x), \phi_2(t, x)\in\Gamma$, $~v_j=v_j(\phi_j)$是

$ \begin{equation*} v_{j, t}=\mu\Delta v_j+g(\phi_j, v_j), ~v_j(0)=v_0~~~(j=1, 2) \end{equation*} $

的解, 那么对充分小$t_0$, 存在$C_2=C_2(\sigma, p, \parallel g\parallel_{C^1(\Gamma)}, t_0)$, 使

$ \begin{equation}\label{eq:8} \parallel v_1-v_2\parallel_{X_v}\leq C_2\parallel\phi_1-\phi_2\parallel_{X_u}. \end{equation} $

(2.3)

证  由条件可得

$ \begin{equation*} (v_1-v_2)_t=\mu\Delta(v_1-v_2)+g(\phi_1, v_1)-g(\phi_2, v_2), \end{equation*} $

且$(v_1-v_2)(0)=0$, 而

$ \begin{equation*} g(\phi_1, v_1)-g(\phi_2, v_2)=g(\phi_1, v_1)-g(\phi_2, v_1)+g(\phi_2, v_1)-g(\phi_2, v_2), \end{equation*} $

结合中值定理, 有

$ \begin{equation*} v_1-v_2=\int_0^tT_\mu(t-s)g_u(\tilde\phi, v_1)(\phi_1-\phi_2)ds+\int_0^tT_\mu(t-s)g_v(\phi_2, \tilde v)(v_1-v_2)ds, \end{equation*} $

利用引理1.1, 则

$ \begin{equation} \begin{aligned} \parallel v_1-v_2\parallel_{X_v}\leq& C_\sigma t_0^{1-\frac{\sigma}{2}}\sup\limits_{0\leq t\leq t_0}\parallel g_u(\tilde\phi, v_1)(\phi_1-\phi_2)\parallel_p\\ &+C_\sigma t_0^{1-\frac{\sigma}{2}}\sup\limits_{0\leq t\leq t_0}\parallel g_v(\phi_2, \tilde v)(v_1-v_2)\parallel_p\\ \leq &C_\sigma t_0^{1-\frac{\sigma}{2}}\parallel g\parallel_{C^1(\Gamma)}\sup\limits_{0\leq t\leq t_0}\parallel\phi_1-\phi_2\parallel_p\\ &+C_\sigma t_0^{1-\frac{\sigma}{2}}\parallel g\parallel_{C^1(\Gamma)}\sup\limits_{0\leq t\leq t_0}\parallel v_1-v_2\parallel_p\\ \leq& C_\sigma t_0^{1-\frac{\sigma}{2}}\parallel g\parallel_{C^1(\Gamma)}(\parallel\phi_1-\phi_2\parallel_{X_u}+\parallel v_1-v_2\parallel_{X_v}). \end{aligned} \end{equation} $

(2.4)

选择充分小$t_0$, 使$C_\sigma t_0^{1-\frac{\sigma}{2}}\parallel g\parallel_{C^1(\Gamma)}<\frac{1}{2}$.即有

$ \begin{equation*} \parallel v_1-v_2\parallel_{X_v}\leq C_2\parallel\phi_1-\phi_2\parallel_{X_u}, \end{equation*} $

其中$C_2=C_2(\sigma, p, \parallel g\parallel_{C^1(\Gamma)}, t_0)$.

引理 2.5  假设$(u, v)\in X_u\times X_v$为问题$(1.1)$的解, 且$(u(t, x), v(t, x))\in\Gamma$, 那么存在常数$C_3, C_4, C_5, C_6$, 使

$ \begin{equation} \begin{aligned} \parallel u\parallel_{X_u}\leq2(\parallel u_0\parallel_{\sigma, p}&+C_3t_0^{1-\frac{\sigma}{2}}\parallel v\parallel_{X_v}+C_4t_0^{1-\vartheta}\parallel v\parallel_{X_v}^2\\ &+C_5t_0^{\frac{\sigma(1-\iota)}{\sigma-1}}\parallel v\parallel_{X_v}^{\frac{\sigma}{\sigma-1}}+C_6t_0^{1-\frac{\sigma}{2}}). \end{aligned} \end{equation} $

(2.5)

证  记$T(t):=e^{\Delta t}$, 那么

$ \begin{align} u(t)=&T(t)u_0-\int_0^tT(t-s)V(u, v)\Delta vds \end{align} $

(2.6)

$ \begin{align} &-\int_0^tT(t-s)V_v(u, v)(\nabla v)^2ds \end{align} $

(2.7)

$ \begin{align} &-\int_0^tT(t-s)V_u(u, v)\nabla v\nabla vds \end{align} $

(2.8)

$ \begin{align} &+\int_0^tT(t-s)f(u, v)ds. \end{align} $

(2.9)

下面逐次估计$(2.6)-(2.9)$项.对于$(2.6)$项,

$ \begin{equation} \begin{aligned} \parallel\int_0^tT(t-s)V\Delta vds\parallel_{\sigma, p}&\leq C_{\sigma}t_0^{1-\frac{\sigma}{2}}\sup\limits_{0\leq t\leq t_0}\parallel V\Delta v\parallel_p\\ &\leq C_{\sigma}t_0^{1-\frac{\sigma}{2}}\parallel V\parallel_{C(\Gamma)}\sup\limits_{0\leq t\leq t_0}\parallel v(t, \cdot)\parallel_{2, p}\\ &\leq C_{\sigma}t_0^{1-\frac{\sigma}{2}}\parallel V\parallel_{C(\Gamma)}\parallel v\parallel_{X_v}\\ &\leq C_3t_0^{1-\frac{\sigma}{2}}\parallel v\parallel_{X_v}, \end{aligned} \end{equation} $

(2.10)

其中$C_3=C_{\sigma}\parallel V\parallel_{C(\Gamma)}$.对于(2.7) 项, 根据引理1.1,

$ \begin{equation} T(t):~L^{\frac{p}{2}}(M)\rightarrow W^{\sigma, p}(M), \end{equation} $

(2.11)

且算子$T(t)$的范数为$C_{\vartheta}t^{-\vartheta}$, 其中$\vartheta=\frac{n}{2p}+\frac{\sigma}{2}$.根据条件(A.4), 可知$0<\vartheta<1$, 则

$ \begin{equation} \begin{aligned} \parallel\int_0^tT(t-s)V_v(\nabla v)^2ds\parallel_{\sigma, p}&\leq C_{\vartheta}t_0^{1-\vartheta}\sup\limits_{0\leq t\leq t_0}\parallel V_v(\nabla v)^2\parallel_{\frac{p}{2}}\\ &\leq C_{\vartheta}t_0^{1-\vartheta}\parallel V\parallel_{C^1(\Gamma)}\sup\limits_{0\leq t\leq t_0}\parallel v(t, \cdot)\parallel_{1, p}^2\\ &\leq C_{\vartheta}t_0^{1-\vartheta}\parallel V\parallel_{C^1(\Gamma)}\parallel v\parallel_{X_v}^2\\ &\leq C_4t_0^{1-\vartheta}\parallel v\parallel_{X_v}^2, \end{aligned} \end{equation} $

(2.12)

其中$C_4=C_{\vartheta}\parallel V\parallel_{C^1(\Gamma)}$, 对于$(2.8)$项, 由Young不等式

$ \begin{align*} \parallel\nabla u\nabla v\parallel_{\frac{p}{\sigma}}&\leq\parallel \varepsilon(\nabla u)^\sigma+\frac{1}{q(\varepsilon\sigma)^{q/\sigma}}(\nabla v)^q\parallel_{\frac{p}{\sigma}}\\ &\leq\varepsilon\parallel u\parallel_{1, p}^\sigma+\frac{1}{q(\varepsilon\sigma)^{q/\sigma}}\parallel\nabla v\parallel_{\frac{p}{\sigma-1}}^{\frac{\sigma}{\sigma-1}}, \end{align*} $

其中$q=\frac{\sigma}{\sigma-1}, \varepsilon>0$.由内插不等式有

$ \begin{equation*} \parallel u\parallel_{\theta\sigma, p} \leq C(\theta)\parallel u\parallel_{\sigma, p}^\theta\parallel u\parallel_p^{1-\theta}, ~~\theta\in(0, 1). \end{equation*} $

特别地, 取$\theta=\frac{1}{\sigma}$, 有

$ \begin{equation*} \parallel u\parallel_{1, p}^\sigma\leq C(\sigma)\parallel u\parallel_{\sigma, p}\parallel u\parallel_p^{\sigma-1}, \end{equation*} $

由于$\Gamma$是有界不变区域, 存在常数$C=C(\sigma, p)$, 使

$ \begin{equation*} \parallel u\parallel_p^{\sigma-1}\leq C\bar u^{\sigma-1}, \end{equation*} $

根据引理1.1,

$ \begin{equation} T(t):~L^{\frac{p}{\sigma}}(M)\rightarrow W^{\sigma, p}(M), \end{equation} $

(2.13)

且算子$T(t)$的范数为$C_{\iota}t^{-\iota}$, 其中$\iota=\frac{n(\sigma-1)}{2p}+\frac{\sigma}{2}$.根据条件(A.4), 可证$0<\iota<1$, 则

$ \begin{equation} \begin{aligned} &\parallel\int_0^tT(t-s)V_u\nabla u\nabla vds\parallel_{\sigma, p}\\ &\leq C_\iota t_0^{1-\iota}\sup\limits_{0\leq t\leq t_0 }\parallel(V_u\nabla u\nabla v)(t, \cdot)\parallel_{\frac{p}{\sigma}}\\ &\leq C_\iota t_0^{1-\iota}\parallel V\parallel_{C^1(\Gamma)}\sup\limits_{0\leq t\leq t_0 }\{\frac{\varepsilon}{\sigma}\parallel u\parallel_{1, p}^\sigma+\frac{1}{q\varepsilon^{q/\sigma}}\parallel\nabla v\parallel_{\frac{p}{\sigma-1}}^{\frac{\sigma}{\sigma-1}}\}\\ &\leq Ct_0^{1-\iota}\sup\limits_{0\leq t\leq t_0 }\{\varepsilon\parallel u\parallel_{\sigma, p}+\frac{1}{q(\varepsilon\sigma)^{q/\sigma}}\parallel\nabla v\parallel_{\frac{p}{\sigma-1}}^{\frac{\sigma}{\sigma-1}}\}\\ &\leq Ct_0^{1-\iota}(\varepsilon\parallel u\parallel_{X_u}+\frac{1}{q(\varepsilon\sigma)^{q/\sigma}}\parallel v\parallel_{X_v}^{\frac{\sigma}{\sigma-1}}), \end{aligned} \end{equation} $

(2.14)

其中$C=C(\iota, p, \sigma, \bar u, \parallel V\parallel_{C^1(\Gamma)})$.对于给定$t_0$, 取$\varepsilon=\frac{1}{2t_0^{1-\iota}C}$, 则由式(2.14) 有

$ \begin{equation} \parallel\int_0^tT(t-s)V_u\nabla u\nabla vds\parallel_{\sigma, p}\leq\frac{1}{2}\parallel u\parallel_{X_u}+C_5t_0^{\frac{\sigma(1-\iota)}{\sigma-1}}\parallel v\parallel_{X_v}^{\frac{\sigma}{\sigma-1}}, \end{equation} $

(2.15)

其中$C_5=\frac{\sigma-1}{2\sigma}(\frac{2C}{\sigma})^{\frac{1}{\sigma-1}}$, 对(2.9) 项,

$ \begin{equation} \begin{aligned} &\parallel\int_0^tT(t-s)f(u, v)ds\parallel_{\sigma, p} \leq C_{\sigma}t_0^{1-\frac{\sigma}{2}}\sup\limits_{0\leq t\leq t_0}\parallel f(u, v)\parallel_p\\ &\leq C_{\sigma}\parallel f\parallel_{C(\Gamma)}t_0^{1-\frac{\sigma}{2}} \leq C_6t_0^{1-\frac{\sigma}{2}}, \end{aligned} \end{equation} $

(2.16)

其中$C_6=C_{\sigma}\parallel f\parallel_{C(\Gamma)}$, 结合式(2.10), (2.12), (2.15), (2.16) 即得式(2.5).

有了前面的估计, 就可以证明局部解的存在性.

定理 2.6  假设$u_0\in W^{\sigma, p}(M), v_0\in W^{\sigma+\alpha, p}(M)$, 且$\forall x\in M$, 有$(u_0(x), v_0(x))\in\Gamma$, 则存在$t_0>0$, 问题$(1.1)$有唯一解$(u, v)\in X_u\times X_v$.

证  设$\phi\in X_u, ~\phi(0)=u_0$.令$v=v(\phi)$表示$ v_t=\mu\Delta v+g(\phi, v), $ $v(0)=v_0$的解, 对上述$v$, 令$u=u(v(\phi))$表示

$ \begin{equation*} u_t=\nabla(\nabla u-V(u, v)\nabla v)+f(u, v), ~~u(0)=u_0=\phi(0) \end{equation*} $

的解.

定义映射

$ G:X_u\rightarrow X_u, ~G\phi:=u(v(\phi)). $

记$m:=2\parallel u_0\parallel_{\sigma, p}+1$, 构造

$ B:=\{\phi\in X_u|\phi(t)\in B_m(0), 0\leq t\leq t_0\}, $

其中$B_m(0)$表示$M$中的球.

接下来将在$B$上应用不动点定理.为此首先证明$G$映$B$到$B$.利用式(2.2) 和(2.5), 有

$ \begin{equation*} \begin{aligned} \parallel G\phi\parallel_{X_u}\leq&2(\parallel u_0\parallel_{\sigma, p}+C_3t_0^{1-\frac{\sigma}{2}}\parallel v\parallel_{X_v}+C_4t_0^{1-\vartheta}\parallel v\parallel_{X_v}^2\\ &+C_5t_0^{\frac{\sigma(1-\iota)}{\sigma-1}}\parallel v\parallel_{X_v}^{\frac{\sigma}{\sigma-1}}+C_6t_0^{1-\frac{\sigma}{2}})\\ \leq&2(\parallel u_0\parallel_{\sigma, p}+C_3C_1t_0^{1-\frac{\sigma}{2}}(\parallel v_0\parallel_{\sigma+\alpha, p}+\parallel\phi\parallel_{X_u}+\parallel\phi\parallel_{X_u}^2)\\ &+C_4C_1^2t_0^{1-\vartheta}(\parallel v_0\parallel_{\sigma+\alpha, p}+\parallel\phi\parallel_{X_u}+\parallel\phi\parallel_{X_u}^2)^2\\ &+C_5C_1^{\frac{\sigma}{\sigma-1}}t_0^{\frac{\sigma(1-\iota)}{\sigma-1}}(\parallel v_0\parallel_{\sigma+\alpha, p}+\parallel\phi\parallel_{X_u}+\parallel\phi\parallel_{X_u}^2)+C_6t_0^{1-\frac{\sigma}{2}}). \end{aligned} \end{equation*} $

选择充分小$t_0$, 可使$\parallel G\phi\parallel_{X_u}\leq2\parallel u_0\parallel_{\sigma, p}+1$.

再证明$G$是压缩映射.设$\phi_j\in X_u, V_j$是对应$v_{j, t}=\mu\Delta v+g(\phi_j, v), ~~~v_j(0)=v_0~~(j=1, 2)$的解, 则

$ \begin{align} G\phi_1-G\phi_2=&-\int_0^tT(t-s)(V_1\Delta v_1-V_2\Delta v_2)ds \end{align} $

(2.17)

$ \begin{align} &-\int_0^tT(t-s)(V_{1, v}(\nabla v_1)^2-V_{2, v}(\nabla v_2)^2)ds \end{align} $

(2.18)

$ \begin{align} &-\int_0^tT(t-s)(V_{1, u}\nabla u_1\nabla v_1-V_{2, u}\nabla u_2\nabla v_2)ds \end{align} $

(2.19)

$ \begin{align} &+\int_0^tT(t-s)(f(u_1, v_1)-f(u_2, v_2))ds, \end{align} $

(2.20)

其中

$ \begin{equation*} V_j=V(u_j, v_j), ~~u_j=G\phi_j~~, v_j=v_j(\phi_j)~~(j=1, 2). \end{equation*} $

综合引理2.4-2.5中技巧可逐项估计式(2.17), (2.18), (2.19), (2.20), 从而可说明当$t_0$充分小时, 可使$G$为压缩映射.故由压缩映射原理知, 问题(1.1) 存在唯一解.

仔细观察前面的估计, 不难发现估计的界都是至多$t_0$的代数次次数增长.利用这个条件, 反复利用引理1.1, 不断重复估计$u$和$v$的界, 直至最终表明在有限时间内局部解会不会爆破.如果没有爆破, 局部解也是全局解.为了减少这种迭代估计的次数, 使结果更明了, 选择参数$\tilde\sigma>0, \gamma>0$, 满足

$ \begin{equation*} \begin{aligned} 2>\tilde\sigma>\max\{\sigma, \frac{5}{3}\}, \\ 2-\tilde\sigma<\gamma<\frac{\tilde\sigma-1}{2}, \\ 1-\gamma+\tilde\sigma=\sigma+\alpha. \end{aligned} \end{equation*} $

(A.4')

引理 2.7  存在常数$\kappa_1=\kappa_1(\tilde\sigma, \parallel g\parallel_{C(\Gamma)})$, 使定理2.6所确定的局部解满足

$ \begin{equation} \parallel v(t)\parallel_{\tilde\sigma, p}\leq\parallel v_0\parallel_{\tilde\sigma, p} +\kappa_1t_0^{1-\frac{\tilde\sigma}{2}}=:K_1(t_0). \end{equation} $

(2.21)

证  根据引理1.1易得.

定理 2.8  假设$u_0\in W^{\sigma, p}(M), v_0\in W^{\sigma+\alpha, p}(M)$, 且$\forall x\in M$, 有$(u_0(x), v_0(x))\in\Gamma$, 则问题(1.1) 有唯一全局解

$ \begin{equation*} (u, v)\in C([0, \infty), W^{\sigma, p}(M)\times W^{\sigma+\alpha, p}(M)). \end{equation*} $

证  根据引理1.1,

$ \begin{equation*} \begin{aligned} T(t):~&W^{1-\gamma-\tilde\sigma, p}(M)\rightarrow W^{1-\gamma, p}(M), \\ T(t):~&L^p(M)\rightarrow W^{\tilde\sigma, p}(M), \end{aligned} \end{equation*} $

且算子$T(t)$的范数为$C_{\tilde\sigma}t^{-\frac{\tilde\sigma}{2}}$, 又

$ \begin{equation*} u(t)=T(t)u_0-\int_0^tT(t-s)\nabla(V\nabla v)ds+\int_0^tT(t-s)f(u, v)ds, \end{equation*} $

得

$ \begin{equation*} \begin{aligned} \parallel u(t)\parallel_{1-\gamma, p}&\leq\parallel u_0\parallel_{1-\gamma, p}+C_{\tilde\sigma}t^{1-\frac{\tilde\sigma}{2}}\sup\limits_{0\leq\eta\leq t}\parallel\nabla(V\nabla)\parallel_{1-\gamma-\tilde\sigma, p}+C_{\tilde\sigma}t^{1-\frac{\tilde\sigma}{2}}\sup\limits_{0\leq\eta\leq t}\parallel f(u, v)\parallel_p\\ &\leq\parallel u_0\parallel_{1-\gamma, p}+\parallel f\parallel_{C(\Gamma)}C_{\tilde\sigma}t^{1-\frac{\tilde\sigma}{2}}+C_{\tilde\sigma}t^{1-\frac{\tilde\sigma}{2}}\sup\limits_{0\leq\eta\leq t}\parallel\nabla(V\nabla)\parallel_{1-\gamma-\tilde\sigma, p}, \end{aligned} \end{equation*} $

由(A.4')有

$ \begin{equation*} -\frac{1}{2}<2-\gamma-\tilde\sigma<0, \end{equation*} $

则存在常数$C=C(\tilde\sigma, \gamma)$, 使

$ \begin{equation*} \parallel\nabla(V\nabla v)\parallel_{1-\gamma-\tilde\sigma, p} \leq C\parallel V(\nabla v)\parallel_{2-\gamma-\tilde\sigma, p}, \end{equation*} $

那么

$ \begin{equation*} \parallel u(t)\parallel_{1-\gamma, p}\leq\parallel u_0\parallel_{1-\gamma, p}+\parallel f\parallel_{C(\Gamma)}C_{\tilde\sigma}t^{1-\frac{\tilde\sigma}{2}}+Ct^{1-\frac{\tilde\sigma}{2}} \parallel V\parallel_{C(\Gamma)}\sup\limits_{0\leq\eta\leq t}\parallel v\parallel_{1, p}, \end{equation*} $

再由引理2.7, 得

$ \begin{equation*} \parallel u(t)\parallel_{1-\gamma, p}\leq\parallel u_0\parallel_{1-\gamma, p}+\parallel f\parallel_{C(\Gamma)}C_{\tilde\sigma}t^{1-\frac{\tilde\sigma}{2}} +Ct^{1-\frac{\tilde\sigma}{2}}\parallel V\parallel_{C(\Gamma)}K_1(t_0)=:K_2(t_0). \end{equation*} $

同样利用引理1.1有

$ \begin{equation}\label{eq:37} T_\mu(t):~W^{1-\gamma-\tilde\sigma, p}(M)\rightarrow W^{1-\gamma, p}(M), \end{equation} $

(2.22)

且算子$T_\mu(t)$的范数为$C_{\tilde\sigma}t^{-\frac{\tilde\sigma}{2}}$.由条件(A.3), $g\in\mathbb R^2$, 结合$\Gamma$是有界闭区域, 知$h(u, v)=g_1(u, v)u$在$C^2(\Gamma)$上一致有界, 从而映射

$ \begin{equation*} h:~~W^{1-\gamma, p}(M)\rightarrow W^{1-\gamma, p}(M) \end{equation*} $

是Lipschitz连续, Lipschitz常数$L:=\parallel h\parallel_{C^2(\Gamma)}$, 从而

$ \begin{equation} \begin{aligned} \parallel v(t)\parallel_{1-\gamma+\tilde\sigma, p}&\leq\parallel v_0\parallel_{1-\gamma+\tilde\sigma}+\int_0^t\parallel T_\mu(t-s)g_1(u, v)u\parallel_{1-\gamma+\tilde\sigma, p}ds\\ &\leq\parallel v_0\parallel_{1-\gamma+\tilde\sigma}+C_{\tilde\sigma}t^{1-\frac{\tilde\sigma}{2}}\sup\limits_{0\leq\eta\leq t}\parallel h(u, v)\parallel_{1-\gamma, p}\\ &\leq\parallel v_0\parallel_{1-\gamma+\tilde\sigma}+C_{\tilde\sigma}t^{1-\frac{\tilde\sigma}{2}}\parallel h\parallel_{C^2(\Gamma)}\sup\limits_{0\leq\eta\leq t}\parallel u\parallel_{1-\gamma, p}\\ &\leq\parallel v_0\parallel_{1-\gamma+\tilde\sigma}+C_{\tilde\sigma}t^{1-\frac{\tilde\sigma}{2}}LK_2(t_0). \end{aligned} \end{equation} $

(2.23)

注意到$1-\gamma+\tilde\sigma=\sigma+\alpha$, 结合式(2.23) 即得

$ \begin{equation} \parallel v\parallel_{X_v}\leq\parallel v_0\parallel_{\sigma+\alpha, p}+Ct_0^{1-\frac{\tilde\sigma}{2}}K_2(t_0)=K_3(t_0), \end{equation} $

(2.24)

那么$\parallel v\parallel_{X_v}$以时间的代数次数增长.另外, 回顾引理2.5的证明过程, 对每个$t_0>0$, 为了得到式(2.5), 除了选择$\varepsilon=\varepsilon(t_0)$外, 并没有对$t_0$有任何限制.

因而结合式(2.5) 和式(2.24), 有

$ \begin{equation} \begin{aligned} \parallel u\parallel_{X_u}\leq&2(\parallel u_0\parallel_{\sigma, p}+C_3t_0^{1-\frac{\sigma}{2}}K_3(t_0)+C_4t_0^{1-\vartheta}K_3(t_0)^2\\ &+C_5t_0^{\frac{\sigma(1-\iota)}{\sigma-1}}K_3(t_0)^{\frac{\sigma}{\sigma-1}}+C_6t_0^{1-\frac{\sigma}{2}}), \end{aligned} \end{equation} $

(2.25)

表明$\parallel u\parallel_{X_u}$亦以时间的代数次数增长.从式(2.24) 和(2.25) 可知, 在有限时间内, 局部解是不会爆破的, 这说明局部解也是全局解.