本文主要研究一类具有时滞的非局部人口动力学模型

(1.1) $ \begin{equation} \frac{\partial u(t, x)}{\partial t}-D\frac{\partial^{2}u(t, x)}{\partial x^{2}}+d(u(t, x)) = \int_{{{\Bbb R}} }h(y)b(u(t-\tau, x-y)){\rm d}y, \; t>0, x\in {{\Bbb R}} \end{equation} $

满足初始条件

(1.2) $ \begin{equation} u(s, x) = u_{0}(s, x), \ \ \ (s, x)\in [-\tau, 0]\times {{\Bbb R}} \end{equation} $

的非单调临界波速下行波解的全局稳定性,该方程模拟了具有两个年龄结构的单物种成熟子种群在一维无界区域中的进化过程.其中$ u(t, x) $表示时刻$ t $位置$ x $处的成年子种群的密度, $ D>0 $表示成年子种群的扩散率系数, $ \tau\geq0 $表示种群的成熟期, $ d(u) $是死亡率函数, $ b(u) $是出生率函数, $ h(x) $是非负对称的单位核函数.本文还需要如下假设条件:

(H$ _{1}) $  $ d(u)\in C^{2} $是递增的非负光滑函数,且满足当$ u\in [0, \infty) $时, $ d'(u)\geq d'(0)>0 $;

(H$ _{2}) $  $ b(u)\in C^{2} $是递增的非负光滑函数,且满足当$ u\in [0, \infty) $时, $ |b'(u)|\leq b'(0) $;

(H$ _{3}) $  方程(1.1)有两个常数平衡点: $ u_{-} = 0 $是不稳定的平衡点, $ u_{+} $是稳定的平衡点,即, $ d(0) = 0, \ b(0) = 0, \ d(u_{+}) = b(u_{+}), \ d'(0)-b'(0)<0 $, $ d'(u_{+})-b'(u_{+})>0 $;

(H$ _{4}) $  $ h\in C^{1}({{\Bbb R}} ) $是非负可积函数,并对所有的$ x\in{{\Bbb R}} $满足$ \int^{+\infty}_{-\infty}h(y){\rm d}y = 1 $和$ h(-x) = h(x) $;

(H$ _{5}) $  $ \int^{+\infty}_{-\infty}e^{-\lambda_{*}y}h(y){\rm d}y<\infty $.

如果取

(1.3) $ \begin{equation} d(u) = \delta u, \ \ b(u) = pue^{-au}, \ \delta>0, \ a>0, \ p>0, \end{equation} $

那么方程(1.1)化为著名的Nicholson's苍蝇模型,参见文献[2],其中

$ u_{-} = 0, \ u_{+} = \frac{1}{a}\ln\frac{p}{\delta} $

是Nicholson's苍蝇模型的两个常数平衡点.当$ \frac{p}{\delta}>e $时,有$ u_{+}>u_{-}>0 $.在本文中,假设

$ \lim\limits_{x\rightarrow\pm\infty}u_{0}(s, x) = u_{\pm}, \ \ s\in[-\tau, 0]. $

由条件(H$ _{3}) $可知,方程(1.1)有两个常数平衡点: $ u_{-} = 0 $是不稳定的平衡点, $ u_{+} $是稳定的平衡点,因此我们研究的是单稳行波解.方程(1.1)的行波解是指连接两个平衡点$ u_{-} $和$ u_{+} $的形如$ u(t, x) = \phi(x+c_{*}t) $的解,且满足常微分方程

(1.4) $ \begin{equation} \left\{\begin{array}{ll} { } c\phi'(\xi) = D\phi''(\xi)-d(\phi(\xi)) +\int_{{{\Bbb R}} }h(y)b(\phi(\xi-y-c_{*}\tau)){\rm d}y, \\ [2mm] \phi(-\infty) = 0, \ \ \phi(+\infty) = K. \end{array}\right. \end{equation} $

许钊泉和肖冬梅^[13]利用辅助方程的思想结合Schauder不动点定理,建立了方程(1.1) (非)临界波速下单稳行波解的存在性及其渐近行为.

由于行波解在生物学,流行病学和种群动力学中的重要作用，反应扩散方程的行波解被广泛的研究,参见文献[1-2, 4-16].其中,行波解的稳定性是行波理论中的一个重要研究内容.对于单调的时滞反应扩散方程,研究其行波解稳定性常用的方法是挤压技术^[11],加权能量方法^[5],加权能量方法结合比较原理^[6-7],加权能量方法结合傅里叶变换方法^[8]和谱分析方法^[16].

对于非单调的时滞反应扩散方程,由于方程缺失了单调性,比较原理不再成立,所以行波解稳定性的结果是十分有限的. 2015年,通过使用技巧性的加权能量方法结合连续性方法, Chern和梅茗等人^[1]得到了一类Nicholson's苍蝇模型非单调临界波速下行波解的局部稳定性.最近,通过在得到解的有界估计时使用的一些新的技巧性的加权能量方法,梅茗等人^[10]证明了一类Nicholson's苍蝇模型非单调临界波速下行波解的全局稳定性并得到了最优收敛速率.受梅茗等人工作^[10]的启发,本文主要研究具有时滞的非局部人口动力学模型(1.1)非单调临界波速下行波解的全局稳定性并得到了最优收敛速率,这推广了梅茗等人的结论.

本文的安排如下:第2部分,介绍本文的一些预备知识和主要结论;第3部分,证明本文的主要结论.

令$ u(t, x) $是柯西问题(1.1)–(1.2)的解, $ \phi(\xi) $是给定的(1.1)临界波速下的行波解, $ \xi = x+c_{*}t $且

$ v(t, \xi) = u(t, x)-\phi(x+c_{*}t), \ \ v_{0}(s, \xi) = u_{0}(s, x)-\phi(x+c_{*}s), $

则方程(1.1)化为

(2.1) $ \begin{equation} v_{t}(t, \xi)+c_{*}v_{\xi}(t, \xi)-Dv_{\xi\xi}(t, \xi) = \int_{{{\Bbb R}} }h(y)P_{1}\big(v(t-\tau, \xi-y-c_{*}\tau)\big){\rm d}y-P_{2}(v(t, \xi)), \end{equation} $

并满足

$ v(s, \xi) = v_{0}(s, \xi), \ \ \ \ (s, \xi)\in[-\tau, 0]\times{{\Bbb R}} , $

其中

$ P_{1}(v) = b(v+\phi)-b(\phi) = b'(\overline{\phi}), \ \ \overline{\phi}\in [\phi, \phi+v], $

$ v = v(t-\tau, \xi-y-c_{*}\tau), \ \ \phi = \phi(\xi-y-c_{*}\tau) $,及

$ \begin{eqnarray*} P_{2}\big(v(t, \xi)\big) = d\big(\phi(\xi)+v(t, \xi)\big)-d\big(\phi(\xi)\big) = d'(\widetilde{\phi}(\xi)). \end{eqnarray*} $

定义加权函数

(2.2) $ \begin{equation} w(\xi) = e^{-2\lambda_{*} \xi} \end{equation} $

和空间$ C_{unif}[-\tau, T] $, $ 0<T\leq \infty $,

$ \begin{eqnarray*} C_{unif}[-\tau, T]&: = &\Big\{u(t, x)|u(t, x)\in C([-\tau, T]\times{{\Bbb R}} )\ \mbox{使得当}\ t\in[-\tau, T]\ \mbox{时, }\nonumber\\ &&\ \lim\limits_{x\rightarrow+\infty}u(t, x)\ \mbox{一致存在,且} \lim\limits_{x\rightarrow+\infty}u_{x}(t, x) = \lim\limits_{x\rightarrow+\infty}u_{xx}(t, x) = 0\nonumber\\ && \mbox{关于}\ t\in[-\tau, T] \mbox{一致成立}\Big\}. \end{eqnarray*} $

下面给出本文的主要结论.

定理2.1(全局稳定性)  假设(H$ _{1}) $–(H$ _{5}) $成立,任意给定方程(1.1)的行波解,当初始扰动满足$ v_{0}(s, \xi): = u_{0}(s, x)-\phi(\xi)\in C_{unif}(-\tau, 0)\cap C([-\tau, 0];W^{2, 1}({{\Bbb R}} )) $和$ \partial_{s}\big[u_{0}(s, x)-\phi(\xi)\big]\in L^{1}([-\tau, 0];L_{w}^{1}({{\Bbb R}} )) $时,对任意给定的$ \tau>0 $, $ e<\frac{b'(u^{+})}{d'(u^{+})}\leq e^{2} $,或$ 0<\tau<\overline{\tau} $ ($ \overline{\tau} $为正常数), $ \frac{b'(u^{+})}{d'(u^{+})}>e^{2} $,柯西问题(1.1)–(1.2)的解$ u(t, x) $满足

(2.3) $ \begin{eqnarray} \sup\limits_{x\in {{\Bbb R}} }\Big|u(t, x)-\phi(x+c_{*}t)\Big|\leq Ct^{-\frac{1}{2}}. \end{eqnarray} $

命题3.1(全局存在唯一性)  假设(H$ _{1}) $–(H$ _{5}) $成立,当任意给定$ \tau>0 $, $ e<\frac{b'(u^{+})}{d'(u^{+})}\leq e^{2} $,或$ 0<\tau<\overline{\tau} $ ($ \overline{\tau} $为正常数), $ \frac{b'(u^{+})}{d'(u^{+})}>e^{2} $时,若初始扰动$ v_{0}(t, \xi)\in C_{unif}[-\tau, 0] $,则扰动方程(2.1)的解$ v(t, \xi) $在空间$ C_{unif}[-\tau, \infty) $上是全局唯一存在的.

证  当$ t\in [0, \tau] $时, $ t-\tau\in[-\tau, 0] $,从而$ P_{1}(v(t-\tau, \xi-y-c_{*}\tau)) = P_{1}(v_{0}(t-\tau, \xi-y-c_{*}\tau)) $,此时方程(2.1)是线性的.因此当$ t\in[0, \tau] $时,方程(2.1)的解表示如下

(3.1) $ \begin{eqnarray} v(t, \xi)& = &\int_{{{\Bbb R}}}G(\eta, t)v_{0}(0, \xi-\eta){\rm d}\eta+\int_{0}^{t}\int_{{{\Bbb R}}}G(\eta, t-s)\\ &&\cdot\bigg[\int_{{{\Bbb R}} }h(y)P_{1}(v_{0}(s-\tau, \xi-\eta-y-c_{*}\tau)){\rm d}y -P_{2}(v(s, \xi-\eta))\bigg]{\rm d}\eta {\rm d}s, \end{eqnarray} $

其中$ G(\eta, t) = \frac{1}{\sqrt{4\pi Dt}}e^{-\frac{(\eta+c_{*}t)^{2}}{4Dt}} $.

因为$ v_{0}(t, \xi)\in C_{unif}[-\tau, 0] $,可以证明$ v(t, \xi)\in C_{unif}[0, \tau] $.事实上

(3.2) $ \begin{eqnarray} \lim\limits_{\xi\rightarrow +\infty}v(t, \xi) & = &\int_{{{\Bbb R}} }G(\eta, t)\lim\limits_{\xi\rightarrow +\infty}v_{0}(0, \xi-\eta){\rm d}\eta\\ &&+ \int_{0}^{t}\int_{{{\Bbb R}}}G(\eta, t-s)\bigg[\int_{{{\Bbb R}} }h(y) \lim\limits_{\xi\rightarrow +\infty}P_{1}(v_{0}(s-\tau, \xi-\eta-y-c_{*}\tau)){\rm d}y {}\\ && -\lim\limits_{\xi\rightarrow +\infty}P_{2}(v(s, \xi-\eta))\bigg]{\rm d}\eta {\rm d}s\\ & = &v_{0, \infty}(0)\int_{{{\Bbb R}} }G(\eta, t){\rm d}\eta+ \int_{0}^{t}P_{1}(v_{0, \infty}(s-\tau))\int_{{{\Bbb R}}}G(\eta, t-s)\int_{{{\Bbb R}} }h(y){\rm d}y{\rm d}\eta {\rm d}s\\ &&-\int_{0}^{t}P_{2}(v_{\infty}(s))\int_{{{\Bbb R}}}G(\eta, t-s){\rm d}\eta {\rm d}s\\ & = &v_{0, \infty}(0)+\int_{0}^{t}P_{1}(v_{0, \infty}(s-\tau)){\rm d}s-\int_{0}^{t}P_{2}(v_{\infty}(s)) {\rm d}s\\ &: = &g_{1}(t) \end{eqnarray} $

和

(3.3) $ \begin{eqnarray} \lim\limits_{\xi\rightarrow +\infty}\partial_{\xi}^{k}v(t, \xi) & = &\int_{{{\Bbb R}} }\partial_{\eta}^{k}G(\eta, t)\lim\limits_{\xi\rightarrow +\infty}v_{0}(0, \xi-\eta){\rm d}\eta\\ &&+ \int_{0}^{t}\int_{{{\Bbb R}}}\partial_{\eta}^{k}G(\eta, t-s)\bigg[\int_{{{\Bbb R}} }h(y) \lim\limits_{\xi\rightarrow +\infty}P_{1}(v_{0}(s-\tau, \xi-\eta-y-c_{*}\tau)){\rm d}y{}\\ && -\lim\limits_{\xi\rightarrow +\infty}P_{2}(v(s, \xi-\eta))\bigg]{\rm d}\eta {\rm d}s\\ & = &v_{0, \infty}(0)\int_{{{\Bbb R}} }\partial_{\eta}^{k}G(\eta, t){\rm d}\eta+ \int_{0}^{t}P_{1}(v_{0, \infty}(s-\tau))\int_{{{\Bbb R}}}\partial_{\eta}^{k}G(\eta, t-s)\\ &&\int_{{{\Bbb R}} }h(y){\rm d}y{\rm d}\eta {\rm d}s-\int_{0}^{t}P_{2}(v_{\infty}(s))\int_{{{\Bbb R}}}\partial_{\eta}^{k}G(\eta, t-s){\rm d}\eta {\rm d}s\\ & = &0 \end{eqnarray} $

都关于$ t\in[0, \tau] $一致成立.因此, $ v(t, \xi)\in C_{unif}[-\tau, \tau]. $

重复上述步骤,可以证明方程(2.1)存在唯一解$ v(t, \xi)\in C_{unif}[-\tau, (n+1)\tau] $,利用连续性方法得到方程(2.1)的解$ v(t, \xi)\in C_{unif}[-\tau, \infty) $的全局存在唯一性.

下面,我们将证明本文的主要结论.

命题3.2(全局稳定性)  假设(H$ _{1}) $–(H$ _{5}) $成立,当任意给定$ \tau>0 $, $ e<\frac{b'(u^{+})}{d'(u^{+})}\leq e^{2} $,或$ 0<\tau<\overline{\tau} $ ($ \overline{\tau} $为正常数), $ \frac{b'(u^{+})}{d'(u^{+})}>e^{2} $时,若初始扰动$ v_{0}(s, \xi)\in C_{unif}[-\tau, 0]\cap L^{1}([-\tau, 0];W^{2, 1}({{\Bbb R}})) $和$ \partial_{s}v_{0}(s, \xi)\in L^{1}([-\tau, 0];L_{\omega}^{1}({{\Bbb R}})) $,那么无论初始扰动$ v_{0}(s, \xi) $有多大,总有

(3.4) $ \begin{equation} \sup\limits_{\xi\in {{\Bbb R}}}|v(t, \xi)|\leq Ct^{-\frac{1}{2}}. \end{equation} $

为了证明命题3.2,需要以下引理.令

(3.5) $ \begin{equation} v(t, \xi) = \frac{1}{\sqrt{\omega(\xi)}}\widetilde{v}(t, \xi), \ {\rm i.e., \ } \widetilde{v}(t, \xi) = \sqrt{\omega(\xi)}v(t, \xi) = e^{-\lambda_{*}\xi}v(t, \xi), \end{equation} $

其中$ \lambda_{*}\in(\lambda_{1}, \lambda_{2}) $,则方程(2.1)化为

(3.6) $ \begin{eqnarray} &&\widetilde{v}_{t}(t, \xi)+k_{1}\widetilde{v}_{\xi}(t, \xi)-D\widetilde{v}_{\xi\xi}(t, \xi) +k_{2}\widetilde{v}(t, \xi){}\\ & = &\int_{{{\Bbb R}} }h(y)\widetilde{P}_{1} \big(\widetilde{v}(t-\tau, \xi-y-c_{*}\tau)\big){\rm d}y -\widetilde{P}_{2} \big(v(t, \xi)\big)+d'(0)\widetilde{v}(t, \xi) \end{eqnarray} $

满足

$ \widetilde{v}(s, \xi) = \sqrt{w(\xi)}v(s, \xi) = \widetilde{v}_{0}(s, \xi), \ \ \ \ (s, \xi)\in [-\tau, 0]\times{{\Bbb R}} , $

其中

$ k_{1} = c_{*}-2D\lambda_{*},   k_{2} = c_{*}\lambda_{*}-D\lambda_{*}^{2}+d'(\phi), $

满足

(3.7) $ \begin{equation} c_{*}\lambda-D\lambda_{*}^{2}+d'(\phi)>c_{*}\lambda_{*}-D\lambda_{*}^{2}+d'(0) >b'(0)\int_{{{\Bbb R}} }h(y)e^{-\lambda(y+c_{*}\tau)}{\rm d}y. \end{equation} $

这里, $ \widetilde{P}_{i}(\widetilde{v}) = e^{-\lambda_{*}\xi}P_{i}(v), i = 1, 2, $满足

(3.8) $ \begin{equation} \widetilde{P}_{1}(\widetilde{v}(t-\tau, \xi-y-c_{*}\tau)) = \int_{{{\Bbb R}} } b'(\overline{\phi}(\xi-y-c_{*}\tau))\widetilde{v}(t-\tau, \xi-y-c_{*}\tau)h(y)e^{-\lambda(y+c_{*}\tau)}{\rm d}y \end{equation} $

和$ \widetilde{P}_{2}(\widetilde{v}(t, \xi)) = d'(\tilde{\phi}(\xi))\widetilde{v}(t, \xi) $.由条件(H$ _{2}) $可得

(3.9) $ \begin{equation} \widetilde{P}_{1}(\widetilde{v}(t-\tau, \xi-y-c_{*}\tau))\leq b'(0)\int_{{{\Bbb R}}} \widetilde{v}(t-\tau, \xi-y-c_{*}\tau)h(y)e^{-\lambda(y+c_{*}\tau)}{\rm d}y. \end{equation} $

当$ \phi\in(0, u_{+}) $时, $ b'(\phi) $是负的,于是当时滞$ \tau $充分大时,方程(3.6)的解$ \widetilde{v}(t, \xi) $在$ u_{+} $附近是非单调的.我们首先研究下面线性方程

(3.10) $ \begin{eqnarray} &&v^{+}_{t}(t, \xi)+k_{1}v^{+}_{\xi}(t, \xi)-Dv^{+}_{\xi\xi}(t, \xi) +k_{2}v^{+}(t, \xi)\\ & = &\int_{{{\Bbb R}}}h(y)b'(0)v^{+}(t-\tau, \xi-y-c_{*}\tau)e^{-\lambda(y+c_{*}\tau)}{\rm d}y \end{eqnarray} $

满足

$ v^{+}(s, \xi) = \sqrt{w(\xi)}v(s, \xi) = v^{+}_{0}(s, \xi), \ \ \ \ (s, \xi)\in [-\tau, 0]\times{{\Bbb R}} $

的解$ \widetilde{v}(t, \xi) $的有界性估计.

引理3.1  当初始扰动$ v_{0}^{+}(s, \xi)\geq 0 $时,有$ v^{+}(t, \xi)\geq 0, (t, \xi)\in[-\tau, \infty)\times{{\Bbb R}} $.

证  当$ t\in[0, \tau] $时, $ t-\tau\in[-\tau, 0] $,则

$ \begin{eqnarray*} &&\int_{{{\Bbb R}}}h(y)b'(0)v^{+}(t-\tau, \xi-y-c_{*}\tau)e^{-\lambda(y+c_{*}\tau)}{\rm d}y\\ & = &\int_{{{\Bbb R}}}h(y)b'(0)v_{0}^{+}(t-\tau, \xi-y-c_{*}\tau)e^{-\lambda(y+c_{*}\tau)}{\rm d}y\geq 0. \end{eqnarray*} $

因此, (3.10)式化为

$ \begin{eqnarray*} v^{+}_{t}(t, \xi)+k_{1}v^{+}_{\xi}(t, \xi)-Dv^{+}_{\xi\xi}(t, \xi) +k_{2}v^{+}(t, \xi)\geq 0, \end{eqnarray*} $

由比较原理可知当$ t\in[0, \tau] $时, $ v^{+}(t, \xi)\geq 0 $.

重复上述步骤可得当$ t\in[n\tau, (n+1)\tau], n = 1, 2, 3, \cdots $时, $ v^{+}(t, \xi)\geq 0 $.再结合连续性方法就证明了当$ t>0 $时, $ v^{+}(t, \xi)\geq 0 $.引理3.1得证.

接下来,将建立方程(3.6)解的有界性估计.

引理3.2(有界性)  令$ \widetilde{v}(t, \xi) $和$ v^{+}(t, \xi) $分别是方程(3.6)和(3.10)的解,那么当

(3.11) $ \begin{equation} |\widetilde{v}_{0}(s, \xi)|\leq v_{0}^{+}(s, \xi), \ \ (s, \xi)\in[-\tau, 0]\times {{\Bbb R}} \end{equation} $

时,有

(3.12) $ \begin{equation} |\widetilde{v}(t, \xi)|\leq v^{+}(t, \xi), \ \ (t, \xi)\in(0, \infty)\times {{\Bbb R}} . \end{equation} $

证  当$ t\in[0, \tau] $时, $ t-\tau\in[-\tau, 0] $,则

(3.13) $ \begin{eqnarray} |\widetilde{v}(t-\tau, \xi-y-c_{*}\tau)|& = &|\widetilde{v}_{0}(t-\tau, \xi-y-c_{*}\tau)|\\ &\leq& v_{0}^{+}(t-\tau, \xi-y-c_{*}\tau)\\ & = &v^{+}(t-\tau, \xi-y-c_{*}\tau). \end{eqnarray} $

令$ U^{-}(t, \xi) = v^{+}(t, \xi)-\widetilde{v}(t, \xi) $和$ U^{+}(t, \xi) = v^{+}(t, \xi)+\widetilde{v}(t, \xi) $,下面将分别建立$ U^{\pm} $的有界性估计.

首先估计$ U^{-}(t, \xi) $的有界性.由(3.6)和(3.10)式可得$ U^{-}(t, \xi) $满足

(3.14) $ \begin{eqnarray} &&U_{t}^{-}(t, \xi)+k_{1}U^{-}(t, \xi)+k_{2}U_{\xi}^{-}(t, \xi)-DU_{\xi\xi}^{-}(t, \xi)\\ & = &\int_{{{\Bbb R}} }b'(0)v^{+}(t-\tau, \xi-y-c_{*}\tau)e^{-\lambda(y+c_{*}\tau)}h(y){\rm d}y\\ &&- \int_{{{\Bbb R}} }b'(\widetilde{\phi})\widetilde{v}(t-\tau, \xi-y-c_{*}\tau)e^{-\lambda(y+c_{*}\tau)}h(y){\rm d}y+\widetilde{P}_{2} \big(\widetilde{v}(t, \xi)\big)-d'(0)\widetilde{v}(t, \xi) \\ &\geq&\int_{{{\Bbb R}} }b'(0)v^{+}(t-\tau, \xi-y-c_{*}\tau)e^{-\lambda(y+c_{*}\tau)}h(y){\rm d}y\\ &&- \int_{{{\Bbb R}} }\big|b'(\widetilde{\phi})\big|\big|\widetilde{v}(t-\tau, \xi-y-c_{*}\tau)\big|e^{-\lambda(y+c_{*}\tau)}h(y){\rm d}y\\ &\geq&0, \ \ t\in[0, \tau]. \end{eqnarray} $

因此,方程(3.14)与初始条件$ U_{0}^{-}(s, \xi) = v_{0}^{+}(s, \xi)-\widetilde{v}_{0}(s, \xi)\geq 0 $化为

(3.15) $ \begin{eqnarray} \left\{\begin{array}{ll} U_{t}^{-}(t, \xi)+k_{1}U^{-}(t, \xi)+k_{2}U_{\xi}^{-}(t, \xi)-DU_{\xi\xi}^{-}(t, \xi) \geq0, \\ U_{0}^{-}(0, \xi)\geq0. \end{array}\right. \end{eqnarray} $

再由比较原理可得

(3.16) $ \begin{eqnarray} U^{-}(t, \xi) = v^{+}(t, \xi)-\widetilde{v}(t, \xi)\geq 0, \ \ (t, \xi)\in[0, \tau]\times{{\Bbb R}} . \end{eqnarray} $

用$ U^{+}(t, \xi) = v^{+}(t, \xi)+\widetilde{v}(t, \xi) $替换$ U^{-}(t, \xi) = v^{+}(t, \xi)-\widetilde{v}(t, \xi) $,类似可证

(3.17) $ \begin{eqnarray} U^{+}(t, \xi) = v^{+}(t, \xi)+\widetilde{v}(t, \xi)\geq 0, \ \ (t, \xi)\in[0, \tau]\times{{\Bbb R}} \end{eqnarray} $

满足

$ U_{0}^{-}(s, \xi) = v_{0}^{+}(s, \xi)+\widetilde{v}_{0}(s, \xi)\geq 0. $

结合(3.16)和(3.17)式,可得

(3.18) $ \begin{eqnarray} |\widetilde{v}(t, \xi)|\leq v^{+}(t, \xi), \ \ (t, \xi)\in[0, \tau]\times{{\Bbb R}} . \end{eqnarray} $

在每个区间$ [n\tau, (n+1)\tau], n = 1, 2, 3, \cdots $上重复上述步骤,结合连续性方法可得

(3.19) $ \begin{equation} |\widetilde{v}(t, \xi)|\leq v^{+}(t, \xi),   (t, \xi)\in[0, \infty)\times{{\Bbb R}} . \end{equation} $

引理3.2得证.

接下来，将通过使用加权能量方法结合基本解的关键估计来证明全局稳定性.为了导出方程(3.10)解的最优衰减速率,我们首先需要得到基本解,然后证明基本解的最优衰减速率.

引理3.3^[3]  令$ z(t) $是下面具有时滞$ \tau>0 $的线性常微分方程

(3.20) $ \begin{eqnarray} \left\{\begin{array}{ll} { } \frac{{\rm d}z(t)}{{\rm d}t}+m_{1}z(t) = m_{2}z(t-\tau), \\ [2mm] z(s) = z_{0}(s), \ \ \ s\in[-\tau, 0] \end{array}\right. \end{eqnarray} $

的解, $ m_{1} $和$ m_{2} $是两个常数,那么

(3.21) $ \begin{eqnarray} z(t) = e^{-m_{1}(t+\tau)}e_{\tau}^{\overline{m}_{2}t}z_{0}(-\tau) +\int_{-\tau}^{0}e^{-m_{1}(t-s)}e_{\tau}^{\overline{m}_{2}(t-\tau-s)} [z_{0}'(s)+m_{1}z_{0}(s)]{\rm d}s, \end{eqnarray} $

其中

(3.22) $ \begin{equation} \overline{m}_{2} = m_{2}e^{m_{1}\tau}, \end{equation} $

$ e_{\tau}^{\overline{m}_{2}t} $具有如下形式

(3.23) $ \begin{eqnarray} e_{\tau}^{\overline{m}_{2}t} = \left\{\begin{array}{ll} 0, & -\infty<t<-\tau, \\ 1, & -\tau\leq t<0, \\ { } 1+\frac{m_{2}t}{1!}, &0\leq t<\tau, \\ \cdot\cdot\cdot\\ { } 1+\frac{m_{2}t}{1!}+\frac{m_{2}^{2}(t-\tau)^{2}}{2!} +\cdot\cdot\cdot+\frac{l_{2}^{m}[t-(l-1)\tau]^{l}}{l!},   & (l-1)\tau\leq t<l\tau, \\ \cdot\cdot\cdot \end{array}\right. \end{eqnarray} $

并且$ e_{\tau}^{\overline{m}_{2}t} $是

(3.24) $ \begin{equation} \left\{\begin{array}{ll} { } \frac{{\rm d}z(t)}{{\rm d}t} = \overline{m}_{2}z(t-\tau), \\ [2mm] z(s) = 1, \ \ \ s\in[-\tau, 0] \end{array}\right. \end{equation} $

的基本解.

类似于文献[8],我们给出具有时滞的线性常微分方程解的性质.

引理3.4  令$ m_{1}\geq 0 $和$ m_{2}\geq 0 $,那么方程(3.20)的解$ z(t) $满足

$ |z(t)|\leq C_{0}e^{-m_{1}t}e_{\tau}^{\overline{m}_{2}t}, $

其中

$ C_{0}: = e^{-m_{1}\tau}|z_{0}(-\tau)|+\int_{-\tau}^{0}e^{m_{1}s}|z'_{0}(s)+m_{1}z_{0}(s)|{\rm d}s $

且方程(3.24)的基本解$ e_{\tau}^{\overline{m}_{2}t} $, $ \overline{m}_{2}>0 $满足对任意的$ \beta>0 $,有

$ \begin{eqnarray*} e_{\tau}^{\overline{m}_{2}t}\leq C(1+t)^{-\beta}e^{\overline{m}_{2}t}, \ \ t>0. \end{eqnarray*} $

进一步,当$ m_{1}\geq m_{2}\geq 0 $时,存在常数$ 0<\epsilon_{1}<1 $使得

(3.25) $ \begin{eqnarray} e^{-m_{1}t}e_{\tau}^{\overline{m}_{2}t}\leq Ce^{-\epsilon_{1}(m_{1}-m_{2})t}, \ \ \ t>0 \end{eqnarray} $

和方程(3.20)的解$ z(t) $满足

$ |z(t)|\leq Ce^{-\epsilon_{1}(m_{1}-m_{2})t}, \ \ \ t>0. $

对方程(3.10)进行傅里叶变换可得

(3.26) $ \begin{eqnarray} \frac{{\rm d}\check{v}^{+}(t, \eta)}{{\rm d}t}+A(\eta)\check{v}^{+}(t, \eta) = B(\eta)\check{v}^{+}(t-\tau, \eta), \end{eqnarray} $

其中$ \check{v}^{+} = \mathfrak{F}[v^{+}] $,

(3.27) $ \begin{eqnarray} A(\eta) = k_{2}i\eta+a_{1}(c_{*})+{\rm d}\eta^{2}, \ \ B(\eta) = b'(0)\int_{{{\Bbb R}} }h(y)e^{-\lambda(y+c_{*}\tau)}e^{-{\rm i}(y+c_{*}\tau)\eta}{\rm d}y. \end{eqnarray} $

因此,方程(3.26)的解可以表示为

(3.28) $ \begin{eqnarray} \check{v}^{+}(t, \eta)& = &e^{-A(\eta)(t+\tau)}e_{\tau}^{\overline{B}(\eta)t} \check{v}_{0}^{+}(-\tau, \eta)\\ &&+\int_{-\tau}^{0}e^{-A(\eta)(t-s)} e_{\tau}^{\overline{B}(\eta)(t-\tau-s)} \bigg[\frac{\rm d}{{\rm d}s}\check{v}_{0}^{+}(s, \eta)+A(\eta)\check{v}_{0}^{+}(s, \eta)\bigg]{\rm d}s, \end{eqnarray} $

其中

(3.29) $ \begin{eqnarray} \overline{B}(\eta): = B(\eta)e^{A(\eta)\tau}. \end{eqnarray} $

对(3.28)式两边取傅里叶逆变换,有

(3.30) $ \begin{eqnarray} v^{+}(t, \eta)& = &\frac{1}{2\pi}\int_{{{\Bbb R}} }e^{{\rm i}x\eta}e^{-A(\eta)(t+\tau)} e_{\tau}^{\overline{B}(\eta)t}\check{v}_{0}(-\tau, \eta){\rm d}\eta\\ && +\int_{-\tau}^{0}\frac{1}{2\pi}\int_{{{\Bbb R}} }e^{{\rm i}x\eta}e^{-A(\eta)(t-s)} e_{\tau}^{\overline{B}(\eta)(t-\tau-s)} \bigg[\frac{\rm d}{{\rm d}s}\check{v}_{0}(s, \eta)+A(\eta)\check{v}_{0}(s, \eta)\bigg]{\rm d}\eta {\rm d}s.{\qquad} \end{eqnarray} $

下面将证明最优衰减速率.

引理3.5  当$ v_{0}^{+}(s, \xi)\in L^{1}([-\tau, 0];W^{2, 1}({{\Bbb R}} )) $和$ \partial_{s}v_{0}^{+}(s, \xi)\in L^{1}([-\tau, 0];L^{1}({{\Bbb R}} )) $时,有

$ ||v^{+}(t)||_{L^{\infty}({{\Bbb R}} )}\leq C(1+t)^{-\frac{1}{2}}. $

证明类似于文献[10],故此省略.

再取$ v_{0}^{+}(s, \xi) $使得$ v_{0}^{+}(s, \xi)\in L^{1}([-\tau, 0];W^{2, 1}({{\Bbb R}} )) $, $ \partial_{s}v_{0}^{+}(s, \xi)\in L^{1}([-\tau, 0];L^{1}({{\Bbb R}} )) $,和

$ v_{0}^{+}(s, \xi)\geq |\widetilde{v}_{0}(s, \xi)|. $

结合引理3.2和引理3.5,我们可以得到$ \widetilde{v}(t, \xi) $的衰减速率.

引理3.6  当$ \widetilde{v}_{0}(s, \xi)\in L^{1}([-\tau, 0];W^{2, 1}({{\Bbb R}} )) $和$ \partial_{s}\widetilde{v}_{0}(s, \xi)\in L^{1}([-\tau, 0];L^{1}({{\Bbb R}} )) $时,有

(3.31) $ \begin{equation} ||\widetilde{v}(t, \xi)||_{L^{\infty}({{\Bbb R}} )}\leq C(1+t)^{-\frac{1}{2}}. \end{equation} $

因为当$ \xi\rightarrow +\infty $时, $ \widetilde{v}(t, \xi) = \sqrt{w(\xi)}v(t, \xi) = e^{-\lambda_{*} \xi}v(t, \xi)\rightarrow 0 $,所以我们不能由引理3.6得到当$ \xi\rightarrow +\infty $时$ v(t, \xi) $的衰减估计.下面研究当$ \xi\rightarrow +\infty $时$ v(t, \xi) $的衰减估计.

引理3.7  当$ \tau>0 $, $ e<\frac{b'(u^{+})}{d'(u^{+})}\leq e^{2} $,或$ 0<\tau<\overline{\tau} $ ($ \overline{\tau} $为正常数), $ \frac{b'(u^{+})}{d'(u^{+})}>e^{2} $时,存在数$ \xi_{0}>>1 $使得方程(2.1)的解$ v(t, \xi) $满足

$ \sup\limits_{\xi\in[\xi_{0}, \infty)}|v(t, \xi)|\leq Ce^{-\mu_{2}t}, $

其中$ 0<\mu_{2} = \mu_{2}(d'(u_{+}), \tau, b'(u_{+}))<d'(0) $.

证  由方程(2.1)可得

(3.32) $ \begin{eqnarray} &&v_{t}(t, \xi)+c_{*}v_{\xi}(t, \xi)-Dv_{\xi\xi}(t, \xi)+d'(\phi(\xi))v(t, \xi)\\ &&-\int_{-\infty}^{\infty}h(y)b'(\phi(\xi-y-c_{*}\tau))v(t-\tau, \xi-y-c_{*}\tau){\rm d}y\\ & = &\int_{-\infty}^{\infty}h(y)Q_{1}(v(t-\tau, \xi-y-c_{*}\tau)){\rm d}y-Q_{2}(v(t, \xi)). \end{eqnarray} $

因为$ v(t, \xi)\in C_{unif}(0, \infty), $即,当$ t\in [-\tau, \infty) $时, $ { }\lim_{\xi\rightarrow+\infty}v(t, \xi) = v(t, \infty) = :z(t) $一致存在,且$ { }\lim_{\xi\rightarrow+\infty}v_{\xi}(t, \xi) = \lim_{\xi\rightarrow+\infty}v_{\xi\xi}(t, \xi) = 0 $一致成立.在(3.32)式中令$ \xi\rightarrow+\infty $,可得

(3.33) $ \begin{equation} Z'(t)+d'(u_{+})Z-b'(u_{+})Z(t-\tau) = Q_{1}(Z(t-\tau))-Q_{2}(Z(t)) . \end{equation} $

类似于文献[4],当$ \tau>0 $, $ e<\frac{b'(u^{+})}{d'(u^{+})}\leq e^{2} $,或$ 0<\tau<\overline{\tau} $, $ \frac{b'(u^{+})}{d'(u^{+})}>e^{2} $时,方程(3.33)的解满足

(3.34) $ \begin{equation} |v(t, \infty)| = Z(t)\leq Ce^{-\mu_{2}t}, \ \ t>0, \end{equation} $

其中$ 0<\mu_{2} = \mu_{2}(d'(u_{+}), \tau, b'(u_{+}))<d'(0) $.

当$ \xi\rightarrow\infty $时,由$ v(t, \xi) $的连续性和一致收敛性可得,存在一个数$ \xi_{0}>>1 $使得(3.34)式暗含着不等式

$ \sup\limits_{\xi\in[\xi_{0}, +\infty)}|v(t, \xi)|\leq Ce^{-\mu_{2}t}, \ \ t>0. $

引理3.7得证.

引理3.8  当$ t>0 $时,有

$ \sup\limits_{\xi\in(-\infty, \xi_{0}]}|v(t, \xi)|\leq C(1+t)^{-\frac{1}{2}}. $

证  因为当$ \xi\leq\xi_{0} $时,有

$ w(\xi) = e^{2\lambda_{*} |\xi|}\geq e^{2\lambda_{*} \xi_{0}}, $

且注意到$ \widetilde{v}(t, \xi) = \sqrt{w(\xi)}v(t, \xi) $,所以由(3.31)式可得

$ \sup\limits_{\xi\in(-\infty, \xi_{0}]}|v(t, \xi)|\leq C(1+t)^{-\frac{1}{2}}. $

引理3.8得证.

结合引理3.7和3.8可得本文的主要结论(3.4).