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

(1.1) $ \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

(1.2) $ \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

(1.3) $ \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

$ \Delta({}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}))x(t_i) = {}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x)(t^+_i)-{}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x)(t^-_i), $

$ \Delta({}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}))y(t'_j) = {}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}))y(t'^+_j)- {}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}))y(t'^-_j), $

其中

$ {}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x)(t^\pm_i) = \lim\limits_{t\rightarrow t^\pm_i} {}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x)(t), $

$ {}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x)(s^\pm_i) = \lim\limits_{t\rightarrow s^\pm_i} {}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x)(t), $

$ {}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y)(t'^\pm_j) = \lim\limits_{t\rightarrow t'^\pm_j} {}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y)(t), $

$ {}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y)(s'^\pm_j) = \lim\limits_{t\rightarrow s'^\pm_j} {}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y)(t). $

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.

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 $的$ \alpha $阶回火分数阶左、右积分定义为

(2.1) $ \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 $的$ \alpha $阶 Riemann-Liouville 型回火分数阶左、右导数定义为

(2.2) $ \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} $和$ {}_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 $的$ \alpha $阶Caputo型回火分数阶左、右导数分别定义为

(2.3) $ \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} $

(2.4) $ \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} $和$ {}_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

(2.5) $ \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} $

(2.6) $ \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} $

在$ 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)| $. 而$ 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

$ E_{0, \rho}^{\alpha, p} = \{x:[a, b]\rightarrow {{\Bbb R}} |x, \ {}_a^cD_{t}^{\alpha, \rho}x\in L^p([a, b], {{\Bbb R}} ), x(a) = x(b) = 0\} $

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

$ \|x\|_{\alpha, p} = \bigg(\int_a^b|x(t)|^p{\rm d}t+\int_a^b|{}_a^cD_{t}^{\alpha, \rho}x(t)|^p{\rm d}t\bigg)^{\frac1p}. $

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}} ) $且$ 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

$ (x, y)_{E_{0, \rho}^{\alpha}} = \int_a^b{}_a^cD_{t}^{\alpha, \rho}x(t){}_a^cD_{t}^{\alpha, \rho}y(t)+x(t)y(t){\rm d}t. $

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

$ {}_aD_{t}^{-\alpha, \rho}(L^p) = \{v:v = {}_aD_{t}^{-\alpha, \rho}x, x\in L^p([a, b]\}, $

$ {}_tD_{b}^{-\alpha, \rho}(L^q) = \{u:u = {}_tD_{b}^{-\alpha, \rho}y, y\in L^q([a, b]\}. $

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

(2.7) $ \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

(2.8) $ \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} $

证  (1) 由定义2.1和Dirichlet's公式可知

$ \begin{eqnarray*} \int_a^b x(t){}_aD_{t}^{-\alpha, \rho} y(t){\rm d}t& = &\frac{1}{\Gamma(\alpha)}\int_a^b x(t)\int_a^t(t-s)^{\alpha-1}e^{-\rho(t-s)}y(s){\rm d}s{\rm d}t\\ & = &\frac{1}{\Gamma(\alpha)}\int_a^b e^{\rho t}y(t)\int_t^b(s-t)^{\alpha-1}e^{-\rho s}x(s){\rm d}s{\rm d}t\\ & = &\int_a^b y(t){}_tD_{b}^{-\alpha, \rho} x(t){\rm d}t. \end{eqnarray*} $

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

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

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 $或$ \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

(2.9) $ \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} $

证  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

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

并且

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

其中

(2.12) $ \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

(2.13) $ \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} $

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

$ \begin{eqnarray*} \|{}_aD_{t}^{-\alpha, \rho}f\|_{L^p}& = &\left(\int_a^b\left|{\frac{1}{\Gamma(\alpha)}}\int_a^\tau(\tau-s)^{\alpha-1}e^{-\rho(\tau-s)}f(s){\rm d}s\right|^p{\rm d}\tau\right)^{\frac1p} \\ & = &\left(\int_a^b\left|{\frac{1}{\Gamma(\alpha)}}\int_a^\tau s^{\alpha-1}(\frac\tau s-1)^{\alpha-1}e^{-\rho(\tau-s)}f(s){\rm d}s\right|^p{\rm d}\tau\right)^{\frac1p}\\ & = &\left(\int_a^b\left|{\frac{1}{\Gamma(\alpha)}}\int_1^{\frac\tau a} (\frac\tau u)^{\alpha-1}(u-1)^{\alpha-1}e^{-\rho(\tau-\frac\tau u)}f(\frac\tau u) \tau \frac{{\rm d}u}{u^2}\right|^p{\rm d}\tau\right)^{\frac1p}\\ &\leq& \int_1^{\frac b a}{\frac{1}{\Gamma(\alpha)}}\frac{u^{1-\alpha}(u-1)^{\alpha-1}}{u^2}\left( \int_a^b\left|e^{-\rho(\tau-\frac\tau u)}\right|^p|f(\frac\tau u)|^p{\rm d}\tau\right)^{\frac1p}{\rm d}u\cdot b^\alpha\\ &\leq &{\frac{ b^\alpha}{\Gamma(\alpha)}} \int_1^{\frac b a}u^{-1-\alpha}(u-1)^{\alpha-1}\left( \int_a^b|f(\frac\tau u)|^p{\rm d}\tau\right)^{\frac1p}{\rm d}u\\ & = &{\frac{ b^\alpha}{\Gamma(\alpha)}}\int_1^{\frac b a}u^{-1-\alpha}(u-1)^{\alpha-1}\left( \int_{\frac au}^{\frac bu}|f(t)|^pu{\rm d}t\right)^{\frac1p}{\rm d}u\\ &\leq&\frac{ b^\alpha}{\Gamma(\alpha)}\int_1^{\frac b a}u^{\frac1p-1-\alpha}(u-1)^{\alpha-1}{\rm d}u\cdot \|f\|_{L^p}. \end{eqnarray*} $

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

(2.14) $ \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} $

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

(2.15) $ \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

$ \begin{eqnarray*} |{}_aD_{t}^{-\alpha, \rho}{}_a^cD_{t}^{\alpha, \rho}f|& = &{\frac{1}{\Gamma(\alpha)}}\left|\int_a^t(t-s)^{\alpha-1}e^{-\rho(t-s)}{}_a^cD_{t}^{\alpha, \rho}f(s){\rm d}s\right|\\ &\leq& {\frac{e^{\rho b}}{\Gamma(\alpha)}}\left|\int_a^t(t-s)^{q(\alpha-1)}{\rm d}s\right|^{1/q}\|{}_a^cD_{t}^{\alpha, \rho}f\|_{L^p}\\ &\leq &\frac{e^{\rho b}(b-a)^{\alpha-\frac1p}}{\Gamma(\alpha)((\alpha-1)q+1)^{\frac1q}}\|{}_a^cD_{t}^{\alpha, \rho}f\|_{L^p}. \end{eqnarray*} $

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

(2.16) $ \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\| $和$ \|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).

设$ \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

$ \|(x, y)\|_E = \|x\|_{\alpha, p}+\|y\|_{\beta, q}, \ \forall(x, y)\in E. $

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

(2.17) $ \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

(2.18) $ \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.

证  设$ (x, y)\in E $, 由引理2.1和2.2可知

(2.19) $ \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

(2.20) $ \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

(2.21) $ \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 $. 由$ f $和$ I_i, S_j $的连续性可知, $ \varphi\in C^1(E, {{\Bbb R}} ) $, 并且

(2.22) $ \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

$ a(u, v) = \langle g, v\rangle, \ \forall v\in H, g\in H'\ (H\ \mbox{的对偶空间}). $

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

$ \varphi(v) = \frac12a(v, v)-\langle g, v\rangle, $

在$ u $上取到极小值.

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.

证  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] $有$ 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] $有$ v(t)\equiv 0 $. 将$ (u, v) $代入(2.18) 式可得

(3.1) $ \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

$ \int_{s_i}^{t_{i+1}}(|x(t)|^2+|{}_a^cD_{t}^{\alpha, \rho}x(t)|^2){\rm d}t<+\infty, {\quad} i = 0, 1, \cdots , n, $

$ \int_{s'_j}^{t'_{j+1}}(|y(t)|^2+|{}_a^cD_{t}^{\beta, \rho}y(t)|^2){\rm d}t<+\infty, {\quad} j = 0, 1, \cdots , m. $

而$ 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

(3.2) $ \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

(3.3) $ \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] $有$ u(t)\equiv 0 $; $ v\in C_0^\infty(t'_j, s'_{j}] $, 对于$ t\in [a, t'_j]\cup (s'_{j}, b] $有$ v(t)\equiv 0 $. 将$ (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,

(3.4) $ \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

$ \begin{eqnarray*} & &\sum^n_{i = 1}({}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t^-_i))-{}_tD_b^{\alpha-1, \rho}({}_a^cD_{t}^{\alpha, \rho}x(t^+_i)))u(t_i)\\ &&+ \sum^n_{i = 1}({}_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)))+\sum^n_{i = 1}I_i(x(t_i))u(t_i) = 0, \\ [2mm] & &\sum^m_{j = 1}({}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(t'^-_j))-{}_tD_b^{\beta-1, \rho}({}_a^cD_{t}^{\beta, \rho}y(t'^+_j)))v(t_j)\\ &&+ \sum^m_{j = 1}({}_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)))+\sum^m_{j = 1}S_j(y(t_j))v(t_j) = 0. \end{eqnarray*} $

再结合(3.4) 式可得

$ {}_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. $

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 $在$ E $上达到极小值.

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

(3.5) $ \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} $

(3.6) $ \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

(3.7) $ \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

$ \begin{eqnarray*} |h(u, v)| &\leq & \sum^n_{i = 0}\|f_{ix}\|_{L^2}\bigg(\int_{s_i}^{t_{i+1}}|u(t)|^2{\rm d}t\bigg)^{\frac12} +\sum^m_{j = 0}\|f_{jy}\|_{L^2}\bigg(\int_{s'_j}^{t'_{j+1}}|v(t)|^2{\rm d}t\bigg)^{\frac12}\\ & &+\sum^n_{i = 1}|I_i(x(t_i))||u(t_i)|+\sum^m_{j = 1}|S_j(y(t'_j))||v(t'_j)|\\ &\leq &\sum^n_{i = 0}\|f_{ix}\|_{L^2}(t_{i+1}-{s_i})^{\frac12}\|u\|_\infty+\sum^m_{j = 0}\|f_{jy}\|_{L^2}(t'_{j+1}-{s'_j})^{\frac12}\|v\|_\infty\\ & &+\sum^n_{i = 1}|I_i(x(t_i))|\|u\|_\infty+\sum^m_{j = 1}|S_j(y(t'_j))|\|v\|_\infty\\ &\leq & \left\{\frac{e^{\rho b}(b-a)^{\alpha-\frac12}}{\Gamma(\alpha)\sqrt{2\alpha-1}}\left[\sum^n_{i = 0}\|f_{ix}\|_{L^2}(t_{i+1}-{s_i})^{\frac12}+\sum^n_{i = 1}|I_i(x(t_i))|\right]\right.\\ &&\left.+\frac{e^{\rho b}(b-a)^{\beta-\frac12}}{\Gamma(\beta)\sqrt{2\beta-1}}\left[\sum^m_{j = 0}\|f_{jy}\|_{L^2}(t'_{j+1}-{s'_j})^{\frac12}+\sum^m_{j = 1}|S_j(y(t_j))|\right]\right\}\|(u, v)\|, \end{eqnarray*} $

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) $在$ E $中达到极小值.

考虑如下分数阶脉冲系统

(4.1) $ \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

(4.2) $ \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.