数学物理学报, 2021, 41(2): 415-426 doi:

论文

基于变分法的回火分数阶脉冲微分系统分析

任晶, 翟成波,

Analysis of Impulsive Tempered Fractional Differential System via Variational Approach

Ren Jing, Zhai Chengbo,

通讯作者: 翟成波, E-mail: cbzhai@sxu.edu.cn

收稿日期: 2020-04-29  

基金资助: 山西省自然科学基金.  201901D111020
山西省研究生创新项目.  2019BY014

Received: 2020-04-29  

Fund supported: the NSF of Shanxi Province.  201901D111020
the Graduate Student Innovation Project of Shanxi Province.  2019BY014

Abstract

The aim of this paper is to discuss a Dirichlet boundary value problem for tempered fractional differential system with instantaneous and non-instantaneous impulses in an appropriate admissible function space. By using variational method, we obtain the sufficient condition for the existence and uniqueness of weak solution, moreover, we show that every weak solution is a classical solution. In the end, an example is presented to highlight the feasibility of the theoretical results.

Keywords: Tempered fractional differential equation ; Lax-Milgram theorem ; Variational method

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任晶, 翟成波. 基于变分法的回火分数阶脉冲微分系统分析. 数学物理学报[J], 2021, 41(2): 415-426 doi:

Ren Jing, Zhai Chengbo. Analysis of Impulsive Tempered Fractional Differential System via Variational Approach. Acta Mathematica Scientia[J], 2021, 41(2): 415-426 doi:

1 引言

分数阶微积分因其允许在技术科学和工程的各个领域的不同模型中获得更大的自由度而受到广泛的欢迎. 近几十年来, 由于分数阶微分方程能够更准确地描述现实世界中的客观现象, 许多有意义的课题得到了迅速的发展. 值得一提的是, 许多真实现象(肿瘤-正常细胞相互作用、病虫害防治、药物治疗、种群动态$ \cdots $)的状态常常受到瞬时扰动的影响, 并在某些时刻产生突变. 变化的持续时间决定了脉冲是瞬时的还是非瞬时的.

脉冲数学模型在应用科学的几乎每个领域都是最近才建立的, 人们对脉冲分数阶微分方程的研究兴趣逐渐增长, 涌现出许多优秀的研究成果, 参见文献[1-2, 4, 12, 15].

目前分数阶积分和导数有不同的定义, Riemann-Liouville, Weyl, Caputo, Grünwald-Letnikov, Hadamard等等. 实际上, 科学家们必须根据当前建模问题的研究选择更实用的算子, 回火分数阶导数是经典算子的推广, 在一定程度上依赖于一个新的参数$ \rho\geq 0 $. 为了描述正常扩散和异常扩散之间的转换, 回火分数阶微积分进入人们的视野, 例如著名的Lévy飞行, 通过指数调节带有参数$ \rho\geq 0 $的Lévy飞行的大跳跃概率,其密度函数从$ |t|^{-(\alpha+1)} $衰减到$ e^{-\rho|t|}|t|^{-(\alpha+1)} $. 自然地, Sabzikara等人[10]提出了回火分数阶微积分的概念.

本文研究回火分数阶脉冲微分方程的变分问题. 事实上, 最初由Riewe[9]引入的分数阶变分微积分是近年来交叉学科领域的一个热点问题, 涉及依赖于分数阶算子的泛函的极小值或极大值. 此外, 基于Morse理论、变分方法、不动点定理、临界点理论等, 许多学者已经研究了带有或不带脉冲(瞬时或非瞬时) 的分数阶微分方程, 参见文献[5-6, 8, 11, 14]. 特别地, Khaliq和Ur Rehman[6]考虑了分数阶非瞬时脉冲边值问题, 利用Lax-Milgram定理得到了该问题弱解的存在性. 在文献[14] 中, 作者得到了带有瞬时和非瞬时脉冲的分数阶边值问题唯一解的存在性

$ \begin{equation} \left\{ \begin{array}{ll} {}_tD_T^{\alpha}({}_0^cD_{t}^{\alpha}u(t)) = f_{i}(t, u(t)), \ &t\in(s_i, t_{i+1}], \ i = 0, 1, 2, \cdots , n, \\ \Delta({}_tD_T^{\alpha-1}({}_0^cD_{t}^{\alpha}))u(t_i) = I_i(u(t_i)), \ &i = 1, 2, \cdots , n, \\ {}_tD_T^{\alpha-1}({}_0^cD_{t}^{\alpha}u)(t) = {}_tD_T^{\alpha-1}({}_0^cD_{t}^{\alpha}u)(t^+_i), \ &t\in(t_i, s_{i}], \ i = 1, 2, \cdots , n, \\ {}_tD_T^{\alpha-1}({}_0^cD_{t}^{\alpha}u)(s^-_i)) = {}_tD_T^{\alpha-1}({}_0^cD_{t}^{\alpha}u)(s^+_i)), \ &i = 1, 2, \cdots , n, \\ u(0) = u(T) = 0. \end{array}\right. \end{equation} $

据文献所知, 关于回火分数阶边值问题弱解的存在性的研究工作还很有限, 这是首次利用Lax-Milgram定理研究具有脉冲(瞬时和非瞬时) 的分数阶边值问题. 本文考虑如下脉冲分数阶系统

$ \begin{equation} \left\{ \begin{array}{ll} {}_tD_b^{\alpha, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t)) = f_{ix}(t, x(t), y(t)), \ & t\in(s_i, t_{i+1}], \ i = 0, 1, 2, \cdots , n, \\ {}_tD_b^{\beta, \rho}({}_a^cD_{t}^{\beta, \rho}y(t)) = f_{jy}(t, x(t), y(t)), \ &t\in(s'_j, t'_{j+1}], \ j = 0, 1, 2, \cdots , m, \end{array}\right. \end{equation} $

带有Dirichlet边值条件

$ \begin{equation} \left\{ \begin{array}{ll} \Delta({}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}))x(t_i) = I_i(x(t_i)), \ & i = 1, 2, \cdots , n, \\ \Delta({}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}))y(t'_j) = S_j(y(t'_j)), \ &j = 1, 2, \cdots , m, \\ {}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x)(t) = {}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x)(t^+_i), \ &t\in(t_i, s_{i}], \ i = 1, 2, \cdots , n, \\ {}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y)(t) = {}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y)(t'^+_j), \ & t\in(t'_j, s'_{j}], \ j = 1, 2, \cdots , m, \\ {}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x)(s^-_i)) = {}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x)(s^+_i)), \ &i = 1, 2, \cdots , n, \\ {}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y)(s'^-_j)) = {}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y)(s'^+_j)), \ &j = 1, 2, \cdots , m, \\ x(a) = x(b) = y(a) = y(b) = 0, \end{array}\right. \end{equation} $

其中$ \frac12<\alpha, \beta\leq1, $$ \rho\geq0 $, $ {}_tD_b^{\alpha, \rho}({}_tD_b^{\beta, \rho}), {}_a^cD_{t}^{\alpha, \rho}({}_a^cD_{t}^{\beta, \rho}) $分别表示Riemann-Liouville型回火分数阶右导数和Caputo型回火分数阶左导数. $ a = s_0<t_1<s_1<t_2<\cdots<s_n<t_{n+1} = b, $$ a = s'_0<t'_1<s'_1<t'_2<\cdots<s'_m<t'_{m+1} = b $, $ a>0 $是一个常数. $ I_i, S_j\in C({{\Bbb R}} , {{\Bbb R}} ) $, 存在$ i\in\{1, 2, \cdots , n\}, \ j\in\{1, 2, \cdots , m\} $使得$ I_i(x(t_i))\not = 0, \ S_j(y(t'_j))\not = 0 $, $ f_x\in L^p((s_i, t_{i+1})\times{{\Bbb R}} , {{\Bbb R}} ) $, $ f_y\in L^p((s'_j, t'_{j+1})\times{{\Bbb R}} , {{\Bbb R}} ) $, 算子$ \Delta $定义如下

其中

瞬时脉冲在点$ t_i, t'_j $处突然开始, 非瞬时脉冲在区间$ (t_i, s_{i}], (t'_j, s'_{j}] $上继续.

注 1.1  当$ \rho = 0 $时, 问题(1.2)–(1.3) 退化为文献[14] 中提出的经典分数阶问题. 当$ \alpha = 1 $时, 问题(1.2)–(1.3) 退化为文献[11] 中的二阶微分系统. 因此本文的结果推广了文献[11, 14]中已有的结论.

文章其余部分的结构如下: 第2节对本文进行了理论研究, 给出了几个基本的定义和必要的结论. 一旦建立了问题(1.2)–(1.3) 的变分结构, 在第3节中, 就能利用Lax-Milgram定理得到$ \frac 12<\alpha, \beta \leq 1 $时弱解的存在唯一性准则. 进一步证明了弱解是古典解. 此外, 通过一个算例说明了主要结果的有效性.

2 预备知识

这部分给出一些定义以及必要的引理.

定义 2.1[1, 10]  对于$ \alpha\in (0, 1] $和参数$ \rho\geq0 $, 函数$ f $$ \alpha $阶回火分数阶左、右积分定义为

$ \begin{eqnarray} {}_aD_{t}^{-\alpha, \rho} f(t) = e^{-\rho t}{}_aD_{t}^{-\alpha}(e^{\rho t}f(t)), \ \ {}_tD_{b}^{-\alpha, \rho} f(t) = e^{\rho t}{}_tD_{b}^{-\alpha}(e^{-\rho t}f(t)), \end{eqnarray} $

其中$ {}_aD_{t}^{-\alpha} $, $ {}_tD_{b}^{-\alpha} $代表 Riemann-Liouville 分数阶积分.

定义 2.2[1, 10]  对于$ \alpha\in (0, 1] $和参数$ \rho\geq0 $, $ f $$ \alpha $阶 Riemann-Liouville 型回火分数阶左、右导数定义为

$ \begin{eqnarray} {}_aD_{t}^{\alpha, \rho}f(t) = e^{-\rho t}{}_aD_{t}^{\alpha}(e^{\rho t}f(t)), \ \ {}_tD_{b}^{\alpha, \rho}f(t)& = e^{\rho t}{}_tD_{b}^{\alpha}(-e^{\rho t}f(t)), \end{eqnarray} $

其中$ {}_aD_{t}^{\alpha} $$ {}_tD_{b}^{\alpha} $表示Riemann-Liouville分数阶导数.

定义 2.3[1]  对于$ \alpha\in (0, 1] $和参数$ \rho\geq0 $, $ f $$ \alpha $阶Caputo型回火分数阶左、右导数分别定义为

$ \begin{equation} {}_a^cD_{t}^{\alpha, \rho}f(t) = e^{-\rho t} {}_a^cD_{t}^{\alpha}(e^{\rho t}f(t)) = {\frac{e^{-\rho t}}{\Gamma(1-\alpha)}} \int_a^t(t-s)^{-\alpha}\frac{\rm d}{{\rm d}s}(e^{\rho s}f(s)){\rm d}s, \end{equation} $

$ \begin{equation} {}_t^cD_{b}^{\alpha, \rho}f(t) = e^{\rho t} {}_t^cD_{b}^{\alpha}(-e^{\rho t}f(t)) = {\frac{e^{\rho t}}{\Gamma(1-\alpha)}} \int_t^b(s-t)^{-\alpha}(-\frac{\rm d}{{\rm d}s})(e^{-\rho s}f(s)){\rm d}s, \end{equation} $

其中$ {}_a^cD_{t}^{\alpha} $$ {}_t^cD_{b}^{\alpha} $表示Caputo分数阶导数.

根据文献[7] 中Riemann-Liouville和Caputo分数阶导数之间的关系式(2.4.6)和(2.4.7), 可得

$ \begin{eqnarray} {}_aD_{t}^{\alpha, \rho}f(t)& = &e^{-\rho t}\bigg({\frac{1}{\Gamma(1-\alpha)}} \int_a^t(t-s)^{-\alpha}\frac{\rm d}{{\rm d}s}(e^{\rho s}f(s)){\rm d}s+\frac{e^{\rho t}f(t)|_{t = a}}{\Gamma(1-\alpha)}(t-a)^{-\alpha}\bigg)\\ & = &{}_a^cD_{t}^{\alpha, \rho}f(t)+\frac{e^{\rho a}f(a)}{\Gamma(1-\alpha)}e^{-\rho t}(t-a)^{-\alpha}, \end{eqnarray} $

$ \begin{eqnarray} {}_tD_{b}^{\alpha, \rho}f(t)& = &e^{\rho t}\bigg({\frac{1}{\Gamma(1-\alpha)}} \int_t^b(s-t)^{-\alpha}(-\frac{\rm d}{{\rm d}s})(e^{-\rho s}f(s)){\rm d}s+\frac{e^{-\rho t}f(t)|_{t = b}}{\Gamma(1-\alpha)}(b-t)^{-\alpha}\bigg)\\ & = &{}_t^cD_{b}^{\alpha, \rho}f(t)+\frac{e^{-\rho a}f(b)}{\Gamma(1-\alpha)}e^{\rho t}(b-t)^{-\alpha}. \end{eqnarray} $

$ C_0^\infty([a, b], {{\Bbb R}} ) = \{x\in C^\infty([a, b], {{\Bbb R}} )|x(a) = x(b) = 0\} $空间上定义范数$ { }\|x\|_\infty = \max_{t\in[a, b]}|x(t)| $.$ L^p([a, b], {{\Bbb R}} ) $, $ p\in[1, \infty) $上的范数由$ \|x\|_{L^p} = (\int^b_a|x(s)|^p{\rm d}s)^{\frac1p} $定义. 接下来, 根据文献[5] 中引入的分数阶导数空间, 定义如下函数空间.

定义 2.4  设$ \rho\geq0, \ \alpha\in(0, 1] $, 分数阶导数空间

$ C_0^\infty([a, b], {{\Bbb R}} ) $的闭包定义, 给定范数

注 2.1  令$ p = 2 $, 记$ E_{0, \rho}^{\alpha, 2} = E_{0, \rho}^{\alpha} $. 显然, $ E_{0, \rho}^{\alpha} $是由具有$ \alpha $阶Caputo型回火分数阶导数的函数构成的空间, 其中$ {}_a^cD_{t}^{\alpha, \rho}x\in L^2([a, b], {{\Bbb R}} ) $$ x\in L^2([a, b], {{\Bbb R}} ) $满足$ x(a) = x(b) = 0 $. 事实上, $ E_{0, \rho}^{\alpha} $是一个可分的Hilbert空间, 给定内积

2012年, 在文献[5]中已经表明, $ \rho = 0 $时, $ E_{0, \rho}^{\alpha, p} $是一个可分的自反Banach空间. 类似地, 考虑$ L_2^p([a, b], {{\Bbb R}} ^{\Bbb N}) $的一个闭子空间$ \Omega $, 其中$ \Omega = \{(x, {}_a^cD_{t}^{\alpha, \rho}x):\forall x\in E_{0, \rho}^{\alpha, p}\} $, 是一个可分的自反Banach空间, 定义范数为$ \|x\|_{L_2^p} = (\sum\limits^2_{i = 1}\|x_i\|^p_{L^p})^{\frac1p} $. 因为$ E_{0, \rho}^{\alpha, p} $与空间$ \Omega $等距同构, 由此可知, $ E_{0, \rho}^{\alpha, p} $也是自反且可分的Banach空间. 根据文献[7] 的结论, 为了得到回火分数阶积分和导数的一些性质, 使用如下函数空间

引理 2.1  设$ \alpha>0 $, $ \frac 1p+\frac 1q<1+\alpha $, $ p\geq1, q\geq1 $或者$ \frac 1p+\frac 1q = 1+\alpha $, $ p\not = 1, q\not = 1 $. 以下结论成立

(1) 如果$ x\in L^p([a, b], {{\Bbb R}} ^N), y\in L^q([a, b], {{\Bbb R}} ^N) $, 则

$ \begin{equation} \int_a^b x(t){}_aD_{t}^{-\alpha, \rho} y(t){\rm d}t = \int_a^b y(t){}_tD_{b}^{-\alpha, \rho} x(t){\rm d}t. \end{equation} $

(2) 如果$ v\in {}_aD_{t}^{-\alpha, \rho}(L^p), u\in {}_tD_{b}^{-\alpha, \rho}(L^q) $, 则

$ \begin{equation} \int_a^b v(t){}_aD_{t}^{\alpha, \rho} u(t){\rm d}t = \int_a^b u(t){}_tD_{b}^{\alpha, \rho} v(t){\rm d}t. \end{equation} $

  (1) 由定义2.1和Dirichlet's公式可知

(2) 令$ u = {}_aD_{t}^{-\alpha, \rho}y $, $ v = {}_tD_{b}^{-\alpha, \rho}x $, 由等式(2.7) 可知

引理2.1证毕.

引理 2.2  设$ \alpha\in (0, 1] $, 如果$ \frac{\rm d}{{\rm d}t}(e^{\rho t}x(t))\in L^\infty([a, b], {{\Bbb R}} ^N), y\in L^1([a, b], {{\Bbb R}} ^N) $, $ x(a) = x(b) = 0 $$ \frac{\rm d}{{\rm d}t}(e^{-\rho t}y(t))\in L^\infty([a, b], {{\Bbb R}} ^N) $, $ x\in L^1([a, b], {{\Bbb R}} ^N), y(a) = y(b) = 0 $, 则有

$ \begin{equation} \int_a^b {}_aD_{t}^{\alpha, \rho} x(t)y(t){\rm d}t = \int_a^b {}_tD_{b}^{\alpha, \rho} y(t)x(t){\rm d}t. \end{equation} $

  结合等式(2.5), (2.6) 和(2.7), 可以很容易地得到(2.9) 式.

引理 2.3  设$ \alpha\in(0, 1] $, $ p\in[1, \infty) $. 对任意$ f\in E_{0, \rho}^{\alpha, p} $, 有

$ \begin{equation} \|{}_aD_{t}^{-\alpha, \rho}f\|_{L^p}\leq {\cal M}\|f\|_{L^p}, \end{equation} $

并且

$ \begin{equation} \|f\|_{L^p}\leq {\cal M}\|{}_a^cD_{t}^{\alpha, \rho}f\|_{L^p}, \end{equation} $

其中

$ \begin{equation} {\cal M} = \frac{b^\alpha}{\Gamma(\alpha)} \int_1^{\frac b a} u^{\frac 1p-\alpha-1}(u-1)^{\alpha-1}{\rm d}u. \end{equation} $

进一步, 如果$ \alpha>\frac 1p $, $ \frac 1p+\frac 1q = 1 $, 则有

$ \begin{equation} \|f\|_\infty\leq \frac{e^{\rho b}(b-a)^{\alpha-\frac1p}}{\Gamma(\alpha)((\alpha-1)q+1)^{\frac1q}}\|{}_a^cD_{t}^{\alpha, \rho}f\|_{L^p}, \ t\in[a, b]. \end{equation} $

  首先, 对任意$ f\in E_{0, \rho}^{\alpha, p} $, $ p\geq1 $, 由广义Minkowsky不等式可得

其次, 对每一个$ f\in E_{0, \rho}^{\alpha, p} $, 由$ f(a) = 0 $可得$ {}_aD_{t}^{-\alpha, \rho}{}_a^cD_{t}^{\alpha, \rho}f(t) = f(t) $, $ t\in [a, b] $. 因此, 为了验证不等式(2.11) 和(2.13), 只需验证

$ \begin{equation} \|{}_aD_{t}^{-\alpha, \rho}{}_a^cD_{t}^{\alpha, \rho}f\|_{L^p}\leq {\cal M}\|{}_a^cD_{t}^{\alpha, \rho}f\|_{L^p}, \end{equation} $

$ \alpha\in(0, 1] $, $ p\in[1, \infty) $, 可得

$ \begin{equation} \|{}_aD_{t}^{-\alpha, \rho}{}_a^cD_{t}^{\alpha, \rho}f\|_\infty\leq \frac{e^{\rho b}(b-a)^{\alpha-\frac1p}}{\Gamma(\alpha)((\alpha-1)q+1)^{\frac1q}}\|{}_a^cD_{t}^{\alpha, \rho}f\|_{L^p}, \ t\in[a, b], \end{equation} $

其中$ \alpha>\frac 1p $, $ \frac 1p+\frac 1q = 1 $. 因为$ {}_a^cD_{t}^{\alpha, \rho}f\in L^p([a, b], $由(2.10) 式很容易得到不等式(2.14). 此外, 对于$ \alpha>\frac 1p $, 选取$ q $满足$ \frac 1p+\frac 1q = 1 $, 则对任意$ f\in E_{0, \rho}^{\alpha, p} $都有

引理2.3证毕.

注 2.2  根据注2.1和不等式(2.10), 定义如下范数

$ \begin{equation} \|x\| = \bigg(\int_a^b|{}_a^cD_{t}^{\alpha, \rho}x(t)|^p{\rm d}t\bigg)^{\frac1p}, \end{equation} $

显然$ \|x\| $$ \|x\|_{\alpha, \rho} $是等价范数. 为了便于以后的讨论, 在空间$ E_{0, \rho}^{\alpha} $中使用范数(2.16).

$ \alpha, \beta\in(0, 1] $, $ p, q\in (1, \infty) $. 定义赋范空间$ E $为分数阶导数空间的笛卡尔积, $ E = E_{0, \rho}^{\alpha}\times E_{0, \rho}^{\beta} $, 赋予如下范数

显然$ E $也是可分的自反Banach空间. 记$ { }\|(x, y)\|_\infty = \max_{t\in[a, b]}|x(t)|+\max_{t\in[a, b]}|y(t)| $. 如果$ \alpha, \beta>\frac12 $, 则有

$ \begin{equation} \|(x, y)\|_\infty\leq \max\left\{\frac{e^{\rho b}(b-a)^{\alpha-\frac12}}{\Gamma(\alpha)\sqrt{2\alpha-1}}, \frac{e^{\rho b}(b-a)^{\beta-\frac12}}{\Gamma(\beta)\sqrt{2\beta-1}}\right\}\|(x, y)\|_E. \end{equation} $

引理 2.4  如果对任意$ (u, v)\in E $满足下面的方程

$ \begin{eqnarray} && \int_a^b({}_a^cD_{t}^{\alpha, \rho}x(t))({}_a^cD_{t}^{\alpha, \rho}u(t))+({}_a^cD_{t}^{\beta, \rho}y(t)) ({}_a^cD_{t}^{\beta, \rho}v(t)){\rm d}t \\ & &+\sum^n_{i = 1}I_i(x(t_i))u(t_i)+\sum^m_{j = 1}S_j(y(t'_j))v(t'_j) \\ & = & \sum^n_{i = 0}\int_{s_i}^{t_{i+1}}f_{ix}(t, x(t), y(t))u(t){\rm d}t+\sum^m_{j = 0} \int_{s'_j}^{t'_{j+1}}f_{jy}(t, x(t), y(t))v(t){\rm d}t, \end{eqnarray} $

则称$ (x, y)\in E $是系统(1.2)–(1.3) 的解.

  设$ (x, y)\in E $, 由引理2.1和2.2可知

$ \begin{eqnarray} & & \int_a^b {}_tD_b^{\alpha, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t))u(t)+ {}_tD_b^{\beta, \rho}({}_a^cD_{t}^{\beta, \rho}y(t))v(t){\rm d}t \\ & = &\sum^n_{i = 0}\int_{s_i}^{t_{i+1}}u(t){}_tD_b^{\alpha, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t)){\rm d}t +\sum^n_{i = 1}\int_{t_i}^{s_{i}}u(t){}_tD_b^{\alpha, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t)){\rm d}t \\ &&+\sum^m_{j = 0}\int_{s'_j}^{t'_{j+1}}v(t){}_tD_b^{\beta, \rho}({}_a^cD_{t}^{\beta, \rho}y(t)){\rm d}t +\sum^m_{j = 1}\int_{t'_j}^{s'_{j}}v(t){}_tD_b^{\beta, \rho}({}_a^cD_{t}^{\beta, \rho}y(t)){\rm d}t \\ & = &-\sum^n_{i = 0}\int_{s_i}^{t_{i+1}}u(t)e^{\rho t}\frac{\rm d}{{\rm d}t}(e^{-\rho t}{}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t))){\rm d}t \\ && -\sum^n_{i = 1}\int_{t_i}^{s_{i}}u(t)e^{\rho t} \frac{\rm d}{{\rm d}t}(e^{-\rho t}{}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t))){\rm d}t{}\\ && -\sum^m_{j = 0}\int_{s'_j}^{t'_{j+1}}v(t)e^{\rho t}\frac{\rm d}{{\rm d}t} (e^{-\rho t}{}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(t))){\rm d}t\\ &&-\sum^m_{j = 1}\int_{t'_j}^{s'_{j}}v(t)e^{\rho t}\frac{\rm d}{{\rm d}t}(e^{-\rho t}{}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(t))){\rm d}t \\ & = &\sum^n_{i = 0}\int_{s_i}^{t_{i+1}}u'(t){}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t)){\rm d}t +\sum^n_{i = 1}\int_{t_i}^{s_{i}}u'(t){}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t)){\rm d}t \\ & &+\sum^n_{i = 0}\int_{s_i}^{t_{i+1}}\rho u(t){}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t)){\rm d}t +\sum^n_{i = 1}\int_{t_i}^{s_{i}}\rho u(t){}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t)){\rm d}t \\ &&-\sum^n_{i = 0}u(t){}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t))|^{t^-_{i+1}}_{s^+_i} -\sum^n_{i = 1}u(t){}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t))|^{s^-_{i}}_{t^+_i}\\ & &+\sum^m_{j = 0}\int_{s'_j}^{t'_{j+1}}v'(t){}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(t)){\rm d}t +\sum^m_{j = 1}\int_{t'_j}^{s'_{j}}v'(t){}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(t)){\rm d}t \\ & &+\sum^m_{j = 0}\int_{s'_j}^{t'_{j+1}}\rho v(t){}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(t)){\rm d}t +\sum^m_{j = 1}\int_{t'_j}^{s'_{j}}\rho v(t){}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(t)){\rm d}t \\ &&-\sum^m_{j = 0}v(t){}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(t))|^{t'^-_{j+1}}_{s'^+_j} -\sum^m_{j = 1}v(t){}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(t))|^{s'^-_{j}}_{t'^+_j}\\ & = &\int_a^b({}_a^cD_{t}^{\alpha, \rho}x(t))({}_a^cD_{t}^{\alpha, \rho}u(t)) +({}_a^cD_{t}^{\beta, \rho}y(t))({}_a^cD_{t}^{\beta, \rho}v(t)){\rm d}t\\ & &+\sum^n_{i = 1}I_i(x(t_i))u(t_i)+\sum^m_{j = 1}S_j(y(t'_j))v(t'_j). \end{eqnarray} $

进一步, 由问题(1.2)–(1.3) 的边界条件可知

$ \begin{eqnarray} && \sum^n_{i = 0}\int_{s_i}^{t_{i+1}}f_{ix}(t, x(t), y(t))u(t){\rm d}t+ \sum^m_{j = 0}\int_{s'_j}^{t'_{j+1}}f_{jy}(t, x(t), y(t))v(t){\rm d}t\\ & = & \int_a^b {}_tD_b^{\alpha, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t))u(t)+{}_tD_b^{\beta, \rho}({}_a^cD_{t}^{\beta, \rho}y(t))v(t){\rm d}t\\ &&+\sum^n_{i = 1}\int_{t_i}^{s_{i}}u(t)e^{\rho t}\frac{\rm d}{{\rm d}t}(e^{-\rho t}{}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t))){\rm d}t\\ &&+\sum^m_{j = 1}\int_{t'_j}^{s'_{j}}v(t)e^{\rho t}\frac{\rm d}{{\rm d}t}(e^{-\rho t}{}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(t))){\rm d}t\\ & = &\int_a^b {}_tD_b^{\alpha, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t))u(t)+{}_tD_b^{\beta, \rho}({}_a^cD_{t}^{\beta, \rho}y(t))v(t){\rm d}t. \end{eqnarray} $

因此, 根据(2.19) 式和(2.20) 式可知等式(2.18) 成立.

定义 2.5  对任意$ (u, v)\in E $, 如果满足方程(2.18), 则称函数$ (x, y)\in E $为系统(1.2)–(1.3) 的弱解.

定义泛函$ \varphi: E\rightarrow {{\Bbb R}} $如下

$ \begin{eqnarray} \varphi(x, y) & = &\frac12\int_a^b|{}_a^cD_{t}^{\alpha, \rho}x(t)|^2+|{}_a^cD_{t}^{\beta, \rho}y(t)|^2{\rm d}t-\sum^n_{i = 0} \int_{s_i}^{t_{i+1}}F_{ix}(t, x(t), y(t)){\rm d}t \\ & &-\sum^m_{j = 0}\int_{s'_j}^{t'_{j+1}}F_{jy}(t, x(t), y(t)){\rm d}t+\sum^m_{j = 1} \int_{a}^{y(t'_j)}S_j(z){\rm d}z+\sum^n_{i = 1}\int_{a}^{x(t_i)}I_i(s){\rm d}s, {\qquad} \end{eqnarray} $

其中$ F_{ix}(t, x(t), y(t)) = \int_a^xf_{ix}(t, z, y){\rm d}z $, $ F_{jy}(t, x(t), y(t)) = \int_a^yf_{jy}(t, x, z){\rm d}z $.$ f $$ I_i, S_j $的连续性可知, $ \varphi\in C^1(E, {{\Bbb R}} ) $, 并且

$ \begin{eqnarray} \langle\varphi'(x, y), (u, v)\rangle& = &\int_a^b({}_a^cD_{t}^{\alpha, \rho}x(t))({}_a^cD_{t}^{\alpha, \rho}u(t))+({}_a^cD_{t}^{\beta, \rho}y(t))({}_a^cD_{t}^{\beta, \rho}v(t)){\rm d}t \\ & &+\sum^m_{j = 1}S_j(y(t'_j))v(t'_j)-\sum^n_{i = 0}\int_{s_i}^{t_{i+1}}f_{ix}(t, x(t), y(t))u(t){\rm d}t \\ &&-\sum^m_{j = 0}\int_{s'_j}^{t'_{j+1}}f_{jy}(t, x(t), y(t))v(t){\rm d}t+\sum^n_{i = 1}I_i(x(t_i))u(t_i). \end{eqnarray} $

因此, 系统(1.2)–(1.3) 的弱解恰好是$ \varphi $对应的临界点.

引理 2.5[3]  (Lax-Milgram定理) 设$ H $是实的Hilbert空间, $ a:H\times H\rightarrow {{\Bbb R}} $是连续双线性型, 并且$ a $是强制的, 即存在唯一的$ u\in H $使得

此外, 如果$ a $也是对称的, 则泛函$ \varphi:H\rightarrow {{\Bbb R}} $, 定义为

$ u $上取到极小值.

3 主要结论

引理 3.1  如果$ (x, y)\in E $是脉冲系统(1.2)–(1.3) 的弱解, 则$ (x, y) $是系统(1.2)–(1.3) 的古典解.

  由引理2.4和定义2.5可知, $ (x, y)\in E $是系统(1.2)–(1.3) 的弱解当且仅当$ \langle\varphi'(x, y), (u, v)\rangle = 0 $, $ \forall (u, v)\in E $. 即等式(2.18) 成立且满足$ x(a) = x(b) = y(a) = y(b) = 0 $. 不失一般性, 假设$ u\in C_0^\infty(s_i, t_{i+1}] $, 对于$ t\in [a, s_i]\cup (t_{i+1}, b] $$ u(t)\equiv 0 $; 假设$ v\in C_0^\infty(s'_j, t'_{j+1}] $, 对于$ t\in [a, s'_j]\cup (t'_{j+1}, b] $$ v(t)\equiv 0 $.$ (u, v) $代入(2.18) 式可得

$ \begin{eqnarray} & & \int_{s_i}^{t_{i+1}}f_{ix}(t, x(t), y(t))u(t){\rm d}t+\int_{s'_j}^{t'_{j+1}}f_{jy}(t, x(t), y(t))v(t){\rm d}t \\ & = & \int_{s_i}^{t_{i+1}}({}_a^cD_{t}^{\alpha, \rho}x(t))({}_a^cD_{t}^{\alpha, \rho}u(t)){\rm d}t +\int_{s'_j}^{t'_{j+1}}({}_a^cD_{t}^{\beta, \rho}y(t))({}_a^cD_{t}^{\beta, \rho}v(t)){\rm d}t\\ & = &\int_{s_i}^{t_{i+1}}{}_tD_b^{\alpha, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t)){\rm d}t +\int_{s'_j}^{t'_{j+1}}{}_tD_b^{\beta, \rho}({}_a^cD_{t}^{\beta, \rho}y(t)){\rm d}t<+\infty, \end{eqnarray} $

那么(1.2) 式有意义. 由$ (u, v)\in E\subset C([a, b]) $得出

$ f_{ix}, f_{jy} $是连续的, 则$ {}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t))\in AC([s_i, t_{i+1}]) $, 并且$ {}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(t))\in AC([s'_j, t'_{j+1}]) $. 因此, 极限$ {}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(s^+_i)) $, $ {}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t^-_{i+1})) $, $ {}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(s'^+_j)) $, $ {}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(t'^-_{j+1})) $都存在. 根据(1.2) 式和(2.18) 式可知

$ \begin{eqnarray} & &\int_a^b({}_a^cD_{t}^{\alpha, \rho}x(t))({}_a^cD_{t}^{\alpha, \rho}u(t))+({}_a^cD_{t}^{\beta, \rho}y(t))({}_a^cD_{t}^{\beta, \rho}v(t)){\rm d}t +\sum^n_{i = 1}I_i(x(t_i))u(t_i) \\ && + \sum^m_{j = 1}S_j(y(t'_j))v(t'_j)+\sum^n_{i = 0}\int_{s_i}^{t_{i+1}}e^{\rho t}\frac{\rm d}{{\rm d}t}(e^{-\rho t}{}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t)))u(t){\rm d}t \\ &&+ \sum^m_{j = 0}\int_{s'_j}^{t'_{j+1}}e^{\rho t}\frac{\rm d}{{\rm d}t}(e^{-\rho t}{}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(t)))v(t){\rm d}t = 0, \end{eqnarray} $

也就是说

$ \begin{eqnarray} & & \sum^n_{i = 1}\int_{t_i}^{s_{i}}({}_a^cD_{t}^{\alpha, \rho}x(t))({}_a^cD_{t}^{\alpha, \rho}u(t)){\rm d}t +\sum^m_{j = 1}\int_{t'_j}^{s'_{j}}({}_a^cD_{t}^{\beta, \rho}y(t))({}_a^cD_{t}^{\beta, \rho}v(t)){\rm d}t \\ && +\sum^n_{i = 0}u(t^-_{i+1}){}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t^-_{i+1})) -\sum^n_{i = 0}u(s^+_i){}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(s^+_i)) \\ && +\sum^m_{j = 0}v(t'^-_{j+1}){}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(t'^-_{j+1})) -\sum^m_{j = 0}v(s'^+_j){}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(s'^+_j))\\ && +\sum^n_{i = 1}I_i(x(t_i))u(t_i)+\sum^m_{j = 1}S_j(y(t'_j))v(t'_j) = 0. \end{eqnarray} $

类似地, 首先假设$ u\in C_0^\infty(t_i, s_{i}] $, 但对于$ t\in [a, t_i]\cup (s_{i}, b] $$ u(t)\equiv 0 $; $ v\in C_0^\infty(t'_j, s'_{j}] $, 对于$ t\in [a, t'_j]\cup (s'_{j}, b] $$ v(t)\equiv 0 $.$ (u, v) $代入(3.3) 式, 可得$ {}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t)) = c_1 $, $ t\in (t_i, s_{i}] $, 并且$ {}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(t)) = c_2 $, $ t\in (t'_j, s'_{j}] $, 即,

$ \begin{eqnarray} & & {}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t)) = {}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t^+_i)) = {}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(s^-_{i})), \ t\in (t_i, s_{i}], \\ && {}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(t)) = {}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(t'^+_j)) = {}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(s'^-_{j})), \ t\in (t'_j, s'_{j}]. \end{eqnarray} $

其次, 将(3.4) 式代回(3.3) 式可知

再结合(3.4) 式可得

总而言之, $ (x, y)\in E $满足开始时给出的方程和边界条件. 所以, $ (x, y) $是系统(1.2)–(1.3) 的古典解.

定理 3.1  对任意$ f_{ix}\in L^2(s_i, t_{i+1}) $, $ f_{jy}\in L^2(s'_j, t'_{j+1}) $, 如果脉冲函数$ I_i, S_j $有界, 则分数阶脉冲系统(1.2)–(1.3) 有唯一的古典解$ (x, y)\in E $. 此外, $ (x, y) $使得泛函$ \varphi $$ E $上达到极小值.

  定义算子$ a:E\times E\rightarrow {{\Bbb R}} $$ h(u, v):E\rightarrow {{\Bbb R}} $如下

$ \begin{equation} a((x, y), (u, v)) = \int_a^b({}_a^cD_{t}^{\alpha, \rho}x(t))({}_a^cD_{t}^{\alpha, \rho}u(t))+({}_a^cD_{t}^{\beta, \rho}y(t))({}_a^cD_{t}^{\beta, \rho}v(t)){\rm d}t, \end{equation} $

$ \begin{eqnarray} h(u, v)& = & \sum^n_{i = 0}\int_{s_i}^{t_{i+1}}f_{ix}(t, x(t), y(t))u(t){\rm d}t+\sum^m_{j = 0}\int_{s'_j}^{t'_{j+1}} f_{jy}(t, x(t), y(t))v(t){\rm d}t \\ & &-\sum^n_{i = 1}I_i(x(t_i))u(t_i)-\sum^m_{j = 1}S_j(y(t'_j))v(t'_j), \end{eqnarray} $

显然$ h $是线性的, 所以$ (x, y) $是系统(1.2)–(1.3) 的弱解等价于$ (x, y)\in E $满足

$ \begin{equation} a((x, y), (u, v)) = h(u, v), \ \forall (x, y)\in E. \end{equation} $

易知, $ a $是强制的, 双线性的, 有界的和对称的. 此外, $ h $是有界线性算子. 事实上, 由引理和Hölder不等式可得

这表明$ h(u, v) $是有界的. 利用引理2.5和引理3.1可知, 脉冲分数阶Dirichlet边值问题(1.2)–(1.3) 有唯一的古典解$ (x, y)\in E $, 并且泛函$ \varphi(x, y) $$ E $中达到极小值.

4 应用

考虑如下分数阶脉冲系统

$ \begin{equation} \left\{ \begin{array}{ll} {}_tD_T^{0.8, \rho}({}_1^cD_{t}^{0.8, \rho}x(t)) = f_{ix}(t, x(t), y(t)), & \ t\in(s_i, t_{i+1}], \ i = 0, 1, 2, \cdots , n, \\ {}_tD_T^{0.75, \rho}({}_1^cD_{t}^{0.75, \rho}y(t)) = f_{jy}(t, x(t), y(t)), & \ t\in(s'_j, t'_{j+1}], \ j = 0, 1, 2, \cdots , m, \end{array}\right. \end{equation} $

带有边值条件

$ \begin{equation} \left\{ \begin{array}{ll} \Delta({}_tD_T^{-0.2, \rho}({}_1^cD_{t}^{0.8, \rho}))x(t_i) = I_i(x(t_i)), \ & i = 1, 2, \cdots , n, \\ \Delta({}_tD_T^{-0.25, \rho}({}_1^cD_{t}^{0.75, \rho}))y(t'_j) = S_j(y(t'_j)), \ &j = 1, 2, \cdots , m, \\ {}_tD_T^{-0.2, \rho}({}_1^cD_{t}^{0.8, \rho}x)(t) = {}_tD_T^{-0.2, \rho}({}_1^cD_{t}^{0.8, \rho}x)(t^+_i), \ &t\in(t_i, s_{i}], \ i = 1, 2, \cdots , n, \\ {}_tD_T^{-0.25, \rho}({}_1^cD_{t}^{0.75, \rho}y)(t) = {}_tD_T^{\beta-1, \rho}({}_1^cD_{t}^{0.75, \rho}y)(t'^+_j), \ &t\in(t'_j, s'_{j}], \ j = 1, 2, \cdots , m, \\ {}_tD_T^{-0.2, \rho}({}_1^cD_{t}^{0.8, \rho}x)(s^-_i)) = {}_tD_T^{-0.2, \rho}({}_1^cD_{t}^{0.8, \rho}x)(s^+_i))\ &i = 1, 2, \cdots , n, \\ {}_tD_T^{-0.25, \rho}({}_1^cD_{t}^{0.75, \rho}y)(s'^-_j)) = {}_tD_T^{-0.25, \rho}({}_1^cD_{t}^{0.75, \rho}y)(s'^+_j))\ &j = 1, 2, \cdots , m, \\ x(1) = x(T) = y(1) = y(T) = 0, \end{array}\right. \end{equation} $

其中$ I_i(x) = x\arctan(1+x) $, $ S_j(y) = 52+\cos y^2 $, $ f_{ix}(t, x, y) = y^{0.5}\sinh x $, $ f_{j}(t, x(t), y(t)) = \sin x^{0.5}y $. 因为$ f_{ix}\in L^2(s_i, t_{i+1}) $, $ f_{jy}\in L^2(s'_j, t'_{j+1}) $对任意$ i, j = 0, 1, 2 $, 脉冲函数$ I_i, S_j $是有界的, 因此, 根据定理3.1和引理3.1可知, 问题(4.1)–(4.2) 存在唯一的古典解.

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