分数阶微分方程在物理学、化学等学科中有重要应用.许多作者从理论角度研究分数阶微分方程的解的动力学性质,参见文献[9-10, 17]. 1908年, Langevin在研究粒子布朗运动时提出一类常微分方程^[6]

$ m\frac{{\rm d}x}{{\rm d}t}+\alpha x(t) = F(t), $

在复杂介质描述时, Langevin型常微分方程不足以准确描述系统的动力学性质,因此,文献[2-5]提出并研究具有两个分数阶导数的Langevin方程

$ ^cD^\alpha ( ^cD^\beta +\lambda) x(t) = f(t, x(t)), \alpha, \beta\in (0, 2). $

Lizana等^[14]利用调和技巧建立了一类分数阶Langevin方程. Gambo等^[7]给出了具有分数阶Hadsmard导数的Langevin方程的Caputo型改进.分数阶Langevin方程的近期研究工作可参考文献[2-3, 5, 7, 14, 21-22, 24-25]. Ahmad等^{[2, 4-5]}研究了具有两个分数阶导数的非线性Langevin方程解的存在性. Tariboon等利用不同的不动点定理研究了具有Hadamard-Caputo型分数阶导数的非线性Langevin方程的解的存在性与唯一性(参见文献[16, 18-19, 22, 27-28]). Tariboon和Ntouyas^[21]研究了脉冲$ q $ -差分Langevin方程的边值问题的解的存在性与唯一性.

近年来,许多作者研究了脉冲分数阶微分方程边值问题的解的存在性与唯一性,参见文献[1, 11, 13, 26]和综述论文[11]. Liu获得了如下脉冲分数阶微分方程的分片连续通解(见文献[12,定理3.3.1])

$ \begin{array}{l} D_{0^+}^\alpha D_{0^+}^\beta x(t)-\lambda x(t) = h_3(t), \ {\rm a.e., }\ t\in (t_i, t_{i+1}], i\in {\Bbb N} _0^m. \end{array} $

文献[8, 23, 29]研究具有两个Caputo分数阶导数的脉冲Langevin型方程的非局部边值问题的解的存在性.我们知道高阶分数阶微分方程的初、边值条件表现非常复杂.而且少有文献研究具有三个分数阶导数的脉冲Langevin型方程

$ [D_{0^+}^\alpha D_{0^+}^\beta -\lambda D_{0^+}^\gamma ]x(t) = p(t), \alpha, \beta\in (1, 2), \gamma\in (0, 1). $

本文填补这一空白,研究如下具有三个分数阶导数的脉冲Langevin方程的反周期边值问题

(1.1) $ \begin{align} \left\{ \begin{array}{l} [D_{0^+}^\alpha D_{0^+}^\beta -\lambda D_{0^+}^\gamma ] x(t) = P(t)f\left(t, x(t)\right), a.e., t\in (t_i, t_{i+1}], i\in {\Bbb N} _0^{m}, \\ \Delta I_{0^+}^{2-\beta} x(t_i) = I_1(t_i, x(t_i)), i\in {\Bbb N} _1^{m-1}, \\ \Delta D_{0^+}^{\beta-1}x(t_i) = I_2(t_i, x(t_i)), i\in {\Bbb N} _1^{m-1}, \\ \Delta (D_{0^+}^{\alpha-1} D_{0^+}^\beta-\lambda I_{0^+}^{1-\gamma}) x(t_i) = I_3(t_i, x(t_i)), i\in {\Bbb N} _1^{m}, \\ \Delta I_{0^+}^{2-\alpha}D_{0^+}^\beta x(t_i) = I_4(t_i, x(t_i)), i\in {\Bbb N} _1^{m}, \\ I_{0^+}^{2-\beta} x(0) = -I_{0^+}^{2-\beta}x(1), \;\Delta D_{0^+}^{\beta-1}x(0) = -\Delta D_{0^+}^{\beta-1}x(1), \\ (D_{0^+}^{\alpha-1} D_{0^+}^\beta-\lambda I_{0^+}^{1-\gamma}) x(0) = -(D_{0^+}^{\alpha-1} D_{0^+}^\beta-\lambda I_{0^+}^{1-\gamma}) x(1), \\ I_{0^+}^{2-\alpha}D_{0^+}^\beta x(0) = -I_{0^+}^{2-\alpha}D_{0^+}^\beta x(1), \end{array}\right. \end{align} $

其中

(a) $ \alpha, \beta\in (1, 2), \gamma\in (0, 1) $, $ \lambda\in {{\Bbb R}} $, $ D_{0^+}^* $表示具有起点0的$ *>0 $阶Riemann-Liouville分数阶导数,见定义2.2;

(b)设$ a<b $为整数,记$ {\Bbb N} _a^b = \{a, a+1, a+2, \cdots, b\} $, $ C_i\in {{\Bbb R}} (i\in {\Bbb N} _1^4) $为常数, $ 0 = t_0<t_1<t_2<\cdots<t_{m}<t_{m+1} = 1 $;

(c) $ f:[0, 1]\times {{\Bbb R}} \to {{\Bbb R}} $时Carathéodory函数,见定义2.3, $ I_j:\{t_i:i\in {\Bbb N} _1^{m}\}\times {{\Bbb R}} \to {{\Bbb R}} $时离散Carathódory函数($ j\in {\Bbb N} _1^4 $),见定义2.4;

(d) $ P:[0, 1]\to {{\Bbb R}} $是连续函数.

假设

$ \begin{array}{l} u|_{(t_i, t_{i+1}]}\in C^0(t_i, t_{i+1}], i\in {\Bbb N} _0^{m}, \;\lim\limits_{t\to t_i^+}(t-t_i)^{2-\beta}u(t)\hbox{存在}, i\in {\Bbb N} _0^{m} \end{array} $

且$ u $满足BVP(1.1)的每一个方程,我们称函数$ u:(0, 1]\to {{\Bbb R}} $为BVP(1.1)的分片连续解.由于含Caputo导数的方程$ {}^cD_{0^+}^a x(t) = 0, t\in [0, 1] $的解在区间$ [0, 1] $连续,而含Riemann-Liouville导数的方程$ D_{0^+}^a x(t) = 0, t\in (0, 1] $的解在区间$ (0, 1] $可以无界.另外,我们注意到$ D_{0^+}^\alpha D_{0^+}^\beta\not = D_{0^+}^{\alpha+\beta} $ (参见文献[9]).因此Riemann-Liouville分数阶导数不同于Caputo分数阶导数.

本文的首先给出线性Langevin方程$ [D_{0^+}^\alpha D_{0^+}^\beta -\lambda D_{0^+}^\gamma ] x(t) = P(t) $的连续通解.然后建立脉冲Langevin方程$ [D_{0^+}^\alpha D_{0^+}^\beta -\lambda D_{0^+}^\gamma ] x(t) = P(t), t\in (t_i, t_{i+1}], i\in {\Bbb N} _0^m $的分片连续通解.最后应用所得结果将BVP(1.1)转化为积分方程,应用Schauder不动点定理^[15]建立BVP(1.1)的解的存在性结果.所得结果推广了已有文献[8, 23, 29]的定理. BVP (1.1)中的边界条件可视为反周期边界条件.

本文后段安排如下:第2节,首先介绍基本概念,然后用三个小节分别建立线性分数阶Langevin方程的连续通解,线性脉冲分数阶Langevin方程的分片连续通解以及建立BVP(1.1)的等价积分方程.第3节,证明BVP(1.1)的解的存在性定理.第4节,举例说明主要所获得结果的应用.

本节首先介绍基本概念,见文献[17].然后获得线性分数阶Langevin方程的通解,其次给出线性脉冲分数阶Langevin方程的通解,最后将BVP(1.1)转化为等价积分方程.

$ L^1(a, b) $表示区间$ (a, b) $上的可积函数集合, $ C^0[a, b] $表示区间$ [a, b] $的连续函数集合.设$ \phi\in L^1(a, b) $,记$ ||\phi||_1 = \int_a^b|\phi(s)|{\rm d}s $.设$ \phi\in C^0[a, b] $,记$ ||\phi||_0 = \max\limits_{t\in [a, b]}|\phi(t)| $. Gamma函数$ \Gamma(\alpha) $, Beta函数$ {\bf B}(p, q) $和Mitag-Leffler函数$ {\bf E}_{\alpha, \delta}(x) $分别定义为

$ \Gamma(\alpha) = \int_0^{+\infty}x^{\alpha-1}e^{-x}{\rm d}x, \;\;{\bf B}(p, q) = \int_0^1x^{p-1}(1-x)^{q-1}{\rm d}x, \;\alpha>0, p>0, q>0, $

$ {\bf E}_{\alpha, \delta}(x) = \sum\limits_{\chi = 0}^\infty \frac{x^\chi}{\Gamma(\chi\alpha+\delta)}, \;\;{\bf E}_{\alpha}(x) = {\bf E}_{\alpha, 1}(x) = \sum\limits_{\chi = 0}^\infty \frac{x^\chi}{\Gamma(\chi\alpha+1)}, \alpha>0, \delta>0. $

定义2.1^[17]  设$ \alpha>0, a\in {{\Bbb R}} $.函数$ h:(a, +\infty)\mapsto {{\Bbb R}} $以$ a $为起点的$ \alpha $积分定义为$ {}^cI_{a^+}^{\alpha}h(t) = \frac{1}{\Gamma(\alpha)}\int_a^t(t-s)^{\alpha-1}h(s){\rm d}s, t>a . $

定义2.2^[17]  设$ n $为正整数, $ \alpha\in (n-1, n), a\in {{\Bbb R}} $.函数$ h:(a, +\infty)\mapsto {{\Bbb R}} $以$ a $为起点的$ \alpha $阶Riemann-Liouville导数(简称Riemann-Liouville分数阶导数)定义为

$ D_{a^+}^{\alpha}h(t) = \frac{1}{\Gamma(n-\alpha)}\left[\int_{a}^{t}(t-s)^{n-\alpha-1}h(s){\rm d}s\right]^{(n)}, t>a. $

定义2.3  设$ h:(0, 1)\times {{\Bbb R}} \mapsto {{\Bbb R}} $.如果

(ⅰ)对任意$ x\in {{\Bbb R}} $, $ t\mapsto h\left(t, (t-t_i)^{2-\beta}x\right) $在区间$ (t_i, t_{i+1}](i\in {\Bbb N} _0^m) $可积;

(ⅱ)对任意$ t\in (t_i, t_{i+1}](i\in {\Bbb N} _0^m) $, $ x\mapsto h\left(t, (t-t_i)^{2-\beta}x\right) $在$ {{\Bbb R}} $上连续;

(ⅲ)对任意$ r>0 $,存在$ M_r>0 $满足当$ |x|\le r $时成立

$ \left|h\left(t, (t-t_i)^{2-\beta}x\right)\right|\le M_r, t\in (t_i, t_{i+1}), i\in {\Bbb N} _0^m. $

则称函数$ f $为Carathéodory函数.

定义2.4  设$ I:\{t_i:i\in {\Bbb N} _1^m\}\times {{\Bbb R}} \mapsto {{\Bbb R}} $.如果

(ⅰ)对任意$ i\in {\Bbb N} _1^m $, $ x\mapsto I\left(t_i, (t_i-t_{i-1})^{2-\beta}x\right) $在$ {{\Bbb R}} $连续;

(ⅱ)对任意$ r>0 $,存在$ M_{I, r}>0 $满足当$ |x|\le r $时成立

$ \left|I\left(t_i, (t_i-t_{i-1})^{2-\beta}x\right)\right|\le M_{I, r}, i\in {\Bbb N} _1^m. $

则称$ I $时离散Carathéodory函数.

定义2.5  Banach空间:设$ n $为正整数, $ \beta\in (1, 2) $, $ 0 = t_0<t_1<\cdots<t_m<t_{m+1} = 1 $.记

$ PC^{0}_{2-\beta}(0, 1] = \Big\{x:(0, 1]\mapsto {{\Bbb R}} :x|_{(t_i, t_{i+1}]}\in C^0(t_i, t_{i+1}], \lim\limits_{t\to t_i^+}(t-t_{i})^{2-\beta}x(t) \hbox{存在, } i\in {\Bbb N} _0^m\Big\}. $

定义范数

$ \|x\| = \max\Big\{\sup\limits_{t\in (t_i, t_{i+1}]}(t-t_{i})^{2-\beta}|x(t)|:i\in {\Bbb N} _0^m\Big\}, \ x\in PC_{0}(0, 1]. $

则$ PC^{0}_{2-\beta}(0, 1] $是Banach空间.

设$ n, l, k $为正整数, $ \lambda\in {{\Bbb R}} $, $ \alpha\in (n-1, n) $, $ \beta\in (l-1, l), \gamma\in (k-1, k) $且$ k\le \alpha+\beta $.本小节求如下线性分数阶Langevin方程(简称LFDE)的通解

(2.1) $ \begin{equation} D_{0^+}^\alpha D_{0^+}^\beta x(t)-\lambda D_{0^+}^\gamma x(t) = g(t), \; {\rm a.e., }\ t\in (0, 1]. \end{equation} $

令$ x_i(i\in {\Bbb N} _1^{\max\{k, n\}}) $, $ y_j(j\in {\Bbb N} _1^l) $为给定常数.考虑初值条件

(2.2) $ \begin{equation} \left\{ \begin{array}{l} I_{0^+}^{l-\beta}x(0) = y_l, \;D_{0^+}^{\beta-i}x(0) = y_i, i\in {\Bbb N} _1^{l-1}; \\ \left\{\begin{array}{lr} \begin{array}{l} (I_{0^+}^{n-\alpha} D_{0^+}^\beta -\lambda I_{0^+}^{k-\gamma})x(0) = x_n, \\ (D_{0^+}^{\alpha-j}D_{0^+}^\beta -\lambda D_{0^+}^{\gamma-j})x(0) = x_j, j\in {\Bbb N} _1^{k-1}, \end{array} &k = n; \\ \begin{array}{l} (D_{0^+}^{\alpha-k}D_{0^+}^\beta -\lambda I_{0^+}^{k-\gamma})x(0) = x_k, \\ (D_{0^+}^{\alpha-j}D_{0^+}^\beta -\lambda D_{0^+}^{\gamma-j})x(0) = x_j, j\in {\Bbb N} _1^{k-1}, \\ D_{0^+}^{\alpha-j}D_{0^+}^\beta x(0) = x_j, j\in {\Bbb N} _{k+1}^n, \end{array}&k<n; \\ \begin{array}{l} (I_{0^+}^{n-\alpha}D_{0^+}^\beta -\lambda D_{0^+}^{\gamma-n})x(0) = x_n, \\ (D_{0^+}^{\alpha-j}D_{0^+}^\beta -\lambda D_{0^+}^{\gamma-j})x(0) = x_j, j\in {\Bbb N} _1^{n-1}, \\ -\lambda D_{0^+}^{\gamma-j} x(0) = x_j, j\in {\Bbb N} _{n+1}^k, \end{array}&k>n; \end{array}\right. \end{array}\right. \end{equation} $

记$ N = \max\{k, n\} $.考虑Picard迭代函数列

$ \begin{eqnarray*} \phi_0(t)& = &\sum\limits_{j = 1}^l\frac{y_j}{\Gamma(\beta-j+1)}t^{\beta-j}+\sum\limits_{i = 1}^{N}\frac{x_i}{\Gamma(\alpha+\beta-i+1)}t^{\alpha+\beta-i} \\ &&+\int_0^t\frac{( t- s)^{\alpha+\beta-1}}{\Gamma(\alpha+\beta)}g(s){\rm d}s, \;t\in (0, 1], \end{eqnarray*} $

$ \phi_i(t) = \phi_0(t)+\frac{\lambda}{\Gamma(\alpha+\beta-\gamma)}\int_{0}^t\left(t-s\right)^{\alpha+\beta-\gamma-1}\phi_{i-1}(s){\rm d}s, t\in (0, 1], i = 1, 2, \cdots. $

容易证明如下结论(详细过程略).

结论2.1  函数$ \phi_\tau\in C^0(0, 1] $且$ \lim\limits_{t\to 0^+}t^{l-\beta}\phi_\tau(t) $存在.

结论2.2  函数列$ \left\{t\to t^{l-\beta}\phi_i(t)\right\} $在区间$ (0, 1] $上一致收敛.

结论2.3  定义在区间$ (0, 1] $上的函数$ \phi(t) = \lim\limits_{\nu\to +\infty}\phi_\nu(t) $是如下积分方程的唯一解

(2.3) $ \begin{eqnarray} {x}(t)& = &\sum\limits_{j = 1}^l\frac{y_j}{\Gamma(\beta-j+1)}t^{\beta-j}+\sum\limits_{i = 1}^N\frac{x_i}{\Gamma(\alpha+\beta-i+1)}t^{\alpha+\beta-i} {}\\ &&+\int_0^t\frac{(t-s)^{\alpha+\beta-1}}{\Gamma(\alpha+\beta)}g(s){\rm d}s+\frac{\lambda}{\Gamma(\alpha+\beta-\gamma)}\int_{0}^t\left(t-s \right)^{\alpha+\beta-\gamma-1}x(s){\rm d}s. \end{eqnarray} $

引理2.1  设$ x $是初值问题(2.1)–(2.2)的解.则$ x $是积分方程(2.3)的解.

证  设$ x $是IVP(2.1)–(2.2)的解.则文献[9, p116,定理2.3]中(2.7.48)式推出

$ \begin{eqnarray*} D_{0^+}^\beta x(t)& = &\sum\limits_{i = 1}^{n}\frac{D_{0^+}^{\alpha-i}D_{0^+}^\beta x(0)}{\Gamma(\alpha-i+1)} t^{\alpha-i}+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}g(s){\rm d}s \\ && +\lambda\int_{0}^t\frac{\left(t-s\right)^{\alpha-1}}{\Gamma(\alpha)}D_{0^+}^\gamma x(s){\rm d}s, {\quad} t\in (0, 1]. \end{eqnarray*} $

而且容易推出

$ \begin{eqnarray*} x(t) & = &\sum\limits_{j = 1}^l\frac{D_{0^+}^{\beta-j}x(0)}{\Gamma(\beta-j+1)}t^{\beta-j}+\sum\limits_{i = 1}^{n}\frac{D_{0^+}^{\alpha-i}D_{0^+}^\beta x(0)}{\Gamma(\alpha+\beta-i+1)} t^{\alpha+\beta-i} \\ && +\int_0^t\frac{(t- u)^{\alpha+\beta-1}}{\Gamma(\alpha+\beta)}g(u){\rm d}u+\lambda\int_0^t\frac{(t- u)^{\alpha+\beta-1}}{\Gamma(\alpha+\beta)} D_{0^+}^\gamma x(u){\rm d}u , {\quad} t\in (0, 1]. \end{eqnarray*} $

由定义计算

$ \begin{eqnarray*} \int_0^t\frac{(t- s)^{\alpha+\beta-1}}{\Gamma(\alpha+\beta)} D_{0^+}^\gamma x(s){\rm d}s & = &\int_0^t\frac{(t- s)^{\alpha+\beta-1}}{\Gamma(\alpha+\beta)} \left(\int_0^s\frac{(s-u)^{l-\gamma-1}}{\Gamma(l-\gamma)}x(u){\rm d}u\right)^{(l)}{\rm d}s \\ & = &-\sum\limits_{j = 1}^k\frac{t^{\alpha+\beta-j}}{\Gamma(\alpha+\beta-j+1)} D_{0^+}^{\gamma-j}x(0)\\ &&+ \int_0^t\frac{(t- u)^{\alpha+\beta-\gamma-1}}{\Gamma(\alpha+\beta-\gamma)}x(u){\rm d}u. \end{eqnarray*} $

于是

$ \begin{eqnarray*} x(t) & = &\sum\limits_{j = 1}^l\frac{D_{0^+}^{\beta-j}x(0)}{\Gamma(\beta-j+1)}t^{\beta-j}+\sum\limits_{i = 1}^{n}\frac{D_{0^+}^{\alpha-i}D_{0^+}^\beta x(0)}{\Gamma(\alpha+\beta-i+1)} t^{\alpha+\beta-i} \\ && +\int_0^t\frac{(t- u)^{\alpha+\beta-1}}{\Gamma(\alpha+\beta)}g(u){\rm d}u -\lambda\sum\limits_{j = 1}^k\frac{t^{\alpha+\beta-j}}{\Gamma(\alpha+\beta-j+1)}D_{0^+}^{\gamma-j}x(0) \\ && + \lambda\int_0^t\frac{(t- u)^{\alpha+\beta-\gamma-1}}{\Gamma(\alpha+\beta-\gamma)}x(u){\rm d}u \\ & = & \left\{\begin{array}{lll} \begin{array}{l} { } \sum\limits_{i = 1}^{n-1}\frac{(D_{0^+}^{\alpha-i}D_{0^+}^\beta -\lambda D_{0^+}^{\gamma-i})x(0)}{\Gamma(\alpha+\beta-i+1)} t^{\alpha+\beta-i} +\frac{(I_{0^+}^{n-\alpha}D_{0^+}^\beta -\lambda I_{0^+}^{k-\gamma})x(0)}{\Gamma(\alpha+\beta-k+1)} t^{\alpha+\beta-k}, \end{array}&k = n; \\ \begin{array}{l} { } \sum\limits_{i = 1}^{k-1}\frac{(D_{0^+}^{\alpha-i}D_{0^+}^\beta -\lambda D_{0^+}^{\gamma-i})x(0)}{\Gamma(\alpha+\beta-i+1)} t^{\alpha+\beta-i}\\ [3mm] { } +\frac{(D_{0^+}^{\alpha-k}D_{0^+}^\beta -\lambda I_{0^+}^{k-\gamma})x(0)}{\Gamma(\alpha+\beta-k+1)} t^{\alpha+\beta-k} + \sum\limits_{i = k+1}^n\frac{D_{0^+}^{\alpha-j}x(0)}{\Gamma(\alpha+\beta-i+1)} t^{\alpha+\beta-i}, \end{array}&k<n; \\ \begin{array}{l} { } \sum\limits_{i = 1}^{n-1}\frac{(D_{0^+}^{\alpha-i}D_{0^+}^\beta -\lambda D_{0^+}^{\gamma-i})x(0)}{\Gamma(\alpha+\beta-i+1)} t^{\alpha+\beta-i} \\ [4mm] { } +\frac{(I_{0^+}^{n-\alpha}D_{0^+}^\beta -\lambda D_{0^+}^{\gamma-n})x(0)}{\Gamma(\alpha+\beta-n+1)} t^{\alpha+\beta-n} -\lambda \sum\limits_{i = n+1}^k\frac{D_{0^+}^{\gamma-j}x(0)}{\Gamma(\alpha+\beta-i+1)} t^{\alpha+\beta-i}, \end{array}&k>n; \end{array}\right. \\ && +\sum\limits_{j = 1}^l\frac{D_{0^+}^{\beta-j}x(0)}{\Gamma(\beta-j+1)}t^{\beta-j}+\int_0^t\frac{(t- u)^{\alpha+\beta-1}}{\Gamma(\alpha+\beta)}g(u){\rm d}u + \lambda\int_0^t\frac{(t-u)^{\alpha+\beta-\gamma-1}}{\Gamma(\alpha+\beta-\gamma)}x(u){\rm d}u\\ \\ & = &\phi_0(t)+\frac{\lambda}{\Gamma(\alpha+\beta-\gamma)}\int_{0}^t\left(t-s\right)^{\alpha+\beta-\gamma-1}x(s){\rm d}s, \ t\in (0, 1]. \end{eqnarray*} $

则$ x\in C(0, 1] $是方程(2.3)的解.引理2.1证明完毕.

引理2.2  $ x $是初值问题(2.1)–(2.2)的解的充分必要条件为$ x $满足

(2.4) $ \begin{eqnarray} x(t)& = &\sum\limits_{j = 1}^ly_jt^{\beta-j}{\bf E}_{\alpha+\beta-\gamma, \beta-j+1}(\lambda t^{\alpha+\beta-\gamma}) +\sum\limits_{i = 1}^{N}x_it^{\alpha+\beta-i}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-i+1}(\lambda t^{\alpha+\beta-\gamma}) {}\\ &&+\int_0^t(t- s)^{\alpha+\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(\lambda(t- s)^{\alpha+\beta-\gamma})g(s){\rm d}s, \;t\in (0, 1]. \end{eqnarray} $

证  设$ x $是问题(2.1)–(2.2)的解.从引理2.1, $ x $是方程(2.3)的解.由结论2.3知道方程(2.3)有唯一解$ x(t) = \lim\limits_{\nu\to \infty}\phi_\nu(t) $.容易计算

$ \begin{eqnarray*} \phi_i(t)& = & \phi_0(t)+\frac{\lambda}{\Gamma(\alpha+\beta-\gamma)}\int_{0}^t\left(t-s\right)^{\alpha+\beta-\gamma-1}\phi_{i-1}(s){\rm d}s \\ & = &\phi_0(s)+\frac{\lambda}{\Gamma(\alpha+\beta-\gamma)}\int_{0}^t\left( t-s\right)^{\alpha+\beta-\gamma-1}\\ &&\times\left[\phi_0(s)+\frac{\lambda}{\Gamma(\alpha+\beta- \gamma)}\int_{0}^s\left(s-u\right)^{\alpha+\beta-\gamma-1}\phi_{i-2}(u){\rm d}u\right]{\rm d}s \\ & = &\phi_0(t)+\frac{\lambda}{\Gamma(\alpha+\beta-\gamma)}\int_{0}^t\left(t-s\right)^{\alpha+\beta-\gamma-1}\phi_0(s){\rm d}s \\ && +\frac{\lambda^2}{\Gamma(2(\alpha+\beta-\gamma))}\int_{0}^t\left(t-u\right)^{2(\alpha+\beta-\gamma)-1}\phi_{i-2}(u){\rm d}u \\ & = &\cdots\cdots \\ & = &\phi_0(t)+\sum\limits_{j = 1}^i\frac{\lambda^j}{\Gamma(j(\alpha+\beta-\gamma))}\int_{0}^t\left(t-s\right)^{j(\alpha+\beta-\gamma)-1}\phi_0(s){\rm d}s \\ & = &\sum\limits_{\chi = 0}^i\sum\limits_{j = 1}^l\frac{y_j}{\Gamma(\chi(\alpha+\beta-\gamma)+\beta-j+1)}t^{\chi(\alpha+\beta-\gamma)+\beta-j} \\ &&+\sum\limits_{\chi = 0}^i\sum\limits_{i = 1}^{N}\frac{x_i}{\Gamma(\chi(\alpha+\beta-\gamma)+\alpha+\beta-i+1)}t^{\chi(\alpha+\beta-\gamma)+\alpha+\beta-i} \\ && +\sum\limits_{\chi = 0}^i\int_0^t\frac{(t-s)^{\chi(\alpha+\beta-\gamma)+\alpha+\beta-1}}{\Gamma(\chi(\alpha+\beta-\gamma)+\alpha+\beta)}g(s){\rm d}s, \end{eqnarray*} $

因此

$ \begin{eqnarray*} x(t)& = &\lim\limits_{\nu\to \infty}\phi_i(t) = \lim\limits_{i\to \infty}\bigg[\sum\limits_{\chi = 0}^i\sum\limits_{j = 1}^l\frac{y_j}{\Gamma(\chi(\alpha+\beta-\gamma)+\beta-j+1)}t^{\chi(\alpha+\beta-\gamma)+\beta-j} \\ &&+\sum\limits_{\chi = 0}^i\sum\limits_{i = 1}^{N}\frac{x_i}{\Gamma(\chi(\alpha+\beta-\gamma)+\alpha+\beta-i+1)}t^{\chi(\alpha+\beta-\gamma)+\alpha+\beta-i} \\ && +\sum\limits_{\chi = 0}^\infty\int_0^t\frac{(t-s s)^{\chi(\alpha+\beta-\gamma)+\alpha+\beta-1}}{\Gamma(\chi(\alpha+\beta-\gamma)+\alpha+\beta)}g(s){\rm d}s \bigg] \\ & = &\sum\limits_{j = 1}^ly_jt^{\beta-j}{\bf E}_{\alpha+\beta-\gamma, \beta-j+1}(\lambda t^{\alpha+\beta-\gamma}) +\sum\limits_{i = 1}^{N}x_it^{\alpha+\beta-i}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-i+1}(\lambda t^{\alpha+\beta-\gamma}) \\ &&+\int_0^t(t- s)^{\alpha+\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(\lambda(t- s)^{\alpha+\beta-\gamma})g(s){\rm d}s. \end{eqnarray*} $

我们得到(2.4)式.现设$ x $满足(2.4)式.通过定义直接计算我们证明$ x $是(2.1)–(2.2)的解.首先,利用(2.4)式,我们有

$ \begin{eqnarray*} D_{0^+}^\beta x(t)& = &\left(\int_0^t\frac{(t- s)^{l-\beta-1}}{\Gamma(l-\beta)}x(s){\rm d}s\right)^{(l)} \\ & = &\bigg[\int_1^t\frac{(t- s)^{l-\gamma-1}}{\Gamma(l-\gamma)}\bigg(\sum\limits_{j = 1}^ly_js^{\beta-j}{\bf E}_{\alpha+\beta-\gamma, \beta-j+1}(\lambda s^{\alpha+\beta-\gamma}) \\ &&+\int_0^s(s- u)^{\alpha+\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(\lambda(s- u)^{\alpha+\beta-\gamma})g(u){\rm d}u \\ &&+\sum\limits_{i = 1}^{N}x_is^{\alpha+\beta-i}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-i+1}(\lambda s^{\alpha+\beta-\gamma}) \bigg){\rm d}s\bigg]^{(l)} \\ & = &\bigg[\int_0^t\frac{(t- s)^{l-\beta-1}}{\Gamma(l-\beta)}\sum\limits_{j = 1}^ly_js^{\beta-j} \sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi s^{\chi(\alpha+\beta-\gamma)}}{\Gamma(\chi(\alpha+\beta-\gamma)+\beta-j+1)}{\rm d}s \bigg]^{(l)} \\ &&+\bigg[\int_0^t\frac{(t- s)^{l-\beta-1}}{\Gamma(l-\beta)}\int_0^s(s- u)^{\alpha+\beta-1} \sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(s- u)^{\chi(\alpha+\beta-\gamma)}}{\Gamma(\chi(\alpha+\beta-\gamma)+\alpha+\beta)}g(u){\rm d}u{\rm d}s \bigg]^{(l)} \\ &&+\bigg[\int_0^t\frac{(t- s)^{l-\beta-1}}{\Gamma(l-\beta)}\sum\limits_{i = 1}^{N}x_is^{\alpha+\beta-i}\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(\log s)^{\chi(\alpha+\beta-\gamma)}}{\Gamma(\chi(\alpha+\beta-\gamma)+\alpha+\beta-i+1)}{\rm d}s \bigg]^{(l)} \\ & = &\bigg[\sum\limits_{j = 1}^ly_j\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)+\beta-j+1)} \int_0^t\frac{(t-s)^{l-\beta-1}}{\Gamma(l-\beta)}s^{\beta-j+\chi(\alpha+\beta-\gamma)}{\rm d}s \bigg]^{(l)} \\ && +\bigg[\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta -\gamma)+\alpha+\beta)}\\ &&\times \int_0^t\int_u^t\frac{(t- s)^{l-\beta-1}}{\Gamma(l-\beta)}( s-u)^{\alpha+\beta-1+\chi(\alpha+\beta-\gamma)}{\rm d}sg(u){\rm d}u \bigg]^{(l)} \\ &&+\bigg[\sum\limits_{i = 1}^{N}x_i\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma) +\alpha+\beta-i+1)}\\ &&\times \int_0^t\frac{(t- s)^{l-\beta-1}}{\Gamma(l-\beta)}s^{\alpha+\beta-i+\chi(\alpha+\beta-\gamma)}{\rm d}s\bigg]^{(l)} \\ & = &\bigg[\sum\limits_{j = 1}^ly_j\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)+l-j+1)} t^{l-j+\chi(\alpha+\beta-\gamma)}\bigg]^{(l)} \\ &&+\bigg[\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)+l+\alpha)} \int_0^t(t- u)^{l+\alpha-1+\chi(\alpha+\beta-\gamma)}g(u){\rm d}u\bigg]^{(l)} \\ &&+\bigg[\sum\limits_{i = 1}^{N}x_i\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi} {\Gamma(\chi(\alpha+\beta-\gamma)+l+\alpha-i+1)}t^{l+\alpha-i +\chi(\alpha+\beta-\gamma)}\bigg]^{(l)} \\ & = &\sum\limits_{j = 1}^ly_j\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)-j+1)} t^{-j+\chi(\alpha+\beta-\gamma)} \\ && +\sum\limits_{i = 1}^{N}x_i\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi} {\Gamma(\chi(\alpha+\beta-\gamma)+\alpha-i+1)}t^{\alpha-i+\chi(\alpha+\beta-\gamma)} \\ &&+\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)+\alpha)} \int_0^t(t- u)^{\alpha-1+\chi(\alpha+\beta-\gamma)}g(u){\rm d}u. \end{eqnarray*} $

其次,我们类似得到

$ \begin{eqnarray*} D_{0^+}^\gamma x(t)& = &\sum\limits_{j = 1}^ly_j\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)+\beta-\gamma-j+1)} t^{\beta-\gamma-j+\chi(\alpha+\beta-\gamma)}\\ && +\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)+\alpha+\beta-\gamma)} \int_0^t(t- u)^{\alpha+\beta-\gamma-1+\chi(\alpha+\beta-\gamma)}g(u){\rm d}u\\ && +\sum\limits_{i = 1}^{N}x_i\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi} {\Gamma(\chi(\alpha+\beta-\gamma)+\alpha+\beta-\gamma-i+1)}t^{\alpha+\beta-\gamma-i+\chi(\alpha+\beta-\gamma)}. \end{eqnarray*} $

最后,我们计算得到

$ \begin{eqnarray*} D_{0^+}^\alpha D_{0^+}^\beta x(t)& = &\sum\limits_{j = 1}^ly_j\sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)-\alpha-j+1)} t^{-\alpha-j+\chi(\alpha+\beta-\gamma)}\\ && +\sum\limits_{i = 1}^{N}x_i\sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi} {\Gamma(\chi(\alpha+\beta-\gamma)-i+1)}t^{-i+\chi(\alpha+\beta-\gamma)}\\ && +\sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma))} \int_0^t(t- u)^{-1+\chi(\alpha+\beta-\gamma)}g(u){\rm d}u+g(t). \end{eqnarray*} $

因此

$ D_{0^+}^\alpha D_{0^+}^\beta x(t)-\lambda D_{0^+}^\gamma x(t) = g(t), \; {\rm a.e., }\ t\in (0, 1]. $

对于$ \nu\in {\Bbb N} _1^{k-1} $,通过计算得到

$ \begin{eqnarray*} D_{0^+}^{\gamma-\nu} x(t) & = &\sum\limits_{j = 1}^ly_j\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)+\nu-\gamma+\beta-j+1)} t^{\nu-\gamma+\beta-j+\chi(\alpha+\beta-\gamma)}\\ && +\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)+\nu-\gamma+\alpha+\beta)} \int_0^t(t- u)^{\nu-\gamma+\alpha+\beta-1+\chi(\alpha+\beta-\gamma)}g(u){\rm d}u \\ &&+\sum\limits_{i = 1}^{N}x_i\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi} {\Gamma(\chi(\alpha+\beta-\gamma)+\nu-\gamma+\alpha+\beta-i+1)}t^{\nu-\gamma+\alpha+\beta-i+\chi(\alpha+\beta-\gamma)}, \end{eqnarray*} $

$ \begin{eqnarray*} I_{0^+}^{k-\gamma} x(t)& = &\sum\limits_{j = 1}^my_j\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)+k-\gamma+\beta-j+1)} t^{k-\gamma+\beta-j+\chi(\alpha+\beta-\gamma)} \\ &&+\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)+k-\gamma+\alpha+\beta)} \int_0^t(t- u)^{k-\gamma+\alpha+\beta-1+\chi(\alpha+\beta-\gamma)}g(u){\rm d}u\\ && +\sum\limits_{i = 1}^{n}x_i\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi} {\Gamma(\chi(\alpha+\beta-\gamma)+k-\gamma+\alpha+\beta-i+1)}t^{k-\gamma+\alpha+\beta-i+\chi(\alpha+\beta-\gamma)}. \end{eqnarray*} $

对于$ \nu\in {\Bbb N} _1^{l-1} $,通过计算得到

$ \begin{eqnarray*} D_{0^+}^{\beta-\nu} x(t)& = &\sum\limits_{j = 1}^ly_j\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)+\nu-j+1)} t^{\nu-j+\chi(\alpha+\beta-\gamma)}\\ && +\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)+\nu+\alpha)} \int_0^t(t- u)^{\nu+\alpha-1+\chi(\alpha+\beta-\gamma)}g(u){\rm d}u\\ && +\sum\limits_{i = 1}^{N}x_i\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi} {\Gamma(\chi(\alpha+\beta-\gamma)+\nu+\alpha-i+1)}t^{\nu+\alpha-i+\chi(\alpha+\beta-\gamma)} \end{eqnarray*} $

和

$ \begin{eqnarray*} I_{0^+}^{l-\beta} x(t)& = &\sum\limits_{j = 1}^ly_j\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)+l-j+1)} t^{l-j+\chi(\alpha+\beta-\gamma)}\\ && +\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)+l+\alpha)} \int_0^t(t- u)^{l+\alpha-1+\chi(\alpha+\beta-\gamma)}g(u){\rm d}u\\ && +\sum\limits_{i = 1}^{N}x_i\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi} {\Gamma(\chi(\alpha+\beta-\gamma)+l+\alpha-i+1)}t^{l+\alpha-i+\chi(\alpha+\beta-\gamma)}. \end{eqnarray*} $

对于$ \nu\in {\Bbb N} _1^{n-1} $,通过计算得到

$ \begin{eqnarray*} D_{0^+}^{\alpha-\nu} D_{0^+}^\beta x(t)& = &\sum\limits_{j = 1}^ly_j\sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)+\nu-\alpha-j+1)} t^{\nu-\alpha-j+\chi(\alpha+\beta-\gamma)}\\ && +\sum\limits_{i = 1}^{N}x_i\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi} {\Gamma(\chi(\alpha+\beta-\gamma)+\nu-i+1)}t^{\nu-i+\chi(\alpha+\beta-\gamma)}\\ && +\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)+\nu)} \int_0^t(t-u)^{\nu-1+\chi(\alpha+\beta-\gamma)}g(u), \end{eqnarray*} $

$ \begin{eqnarray*} I_{0^+}^{n-\alpha} D_{0^+}^\beta x(t)& = &\sum\limits_{j = 1}^ly_j\sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)+n-\alpha-j+1)} t^{n-\alpha-j+\chi(\alpha+\beta-\gamma)}\\ && +\sum\limits_{i = 1}^{N}x_i\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi} {\Gamma(\chi(\alpha+\beta-\gamma)+n-i+1)}t^{n-i+\chi(\alpha+\beta-\gamma)}\\ && +\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)+n)} \int_0^t(t- u)^{n-1+\chi(\alpha+\beta-\gamma)}g(u). \end{eqnarray*} $

直接验证得到

$ \begin{eqnarray*} &&I_{0^+}^{l-\beta}x(0) = y_l, \;D_{0^+}^{\beta-i}x(0) = y_i, i\in {\Bbb N} _1^{l-1};\\ && \left\{\begin{array}{lr} \begin{array}{l} (I_{0^+}^{n-\alpha} D_{0^+}^\beta -\lambda I_{0^+}^{k-\gamma})x(0) = x_n, (D_{0^+}^{\alpha-j}D_{0^+}^\beta -\lambda D_{0^+}^{\gamma-j})x(0) = x_j, j\in {\Bbb N} _1^{k-1}, \end{array} &k = n;\\ [2mm] \begin{array}{l} (D_{0^+}^{\alpha-k}D_{0^+}^\beta -\lambda I_{0^+}^{k-\gamma})x(0) = x_k, \\ (D_{0^+}^{\alpha-j}D_{0^+}^\beta -\lambda D_{0^+}^{\gamma-j})x(0) = x_j, j\in {\Bbb N} _1^{k-1}, D_{0^+}^{\alpha-j}D_{0^+}^\beta x(0) = x_j, j\in {\Bbb N} _{k+1}^n, \end{array}&k<n; \\ \begin{array}{l} (I_{0^+}^{n-\alpha}D_{0^+}^\beta -\lambda D_{0^+}^{\gamma-n})x(0) = x_n, \\ (D_{0^+}^{\alpha-j}D_{0^+}^\beta -\lambda D_{0^+}^{\gamma-j})x(0) = x_j, j\in {\Bbb N} _1^{n-1}, -\lambda D_{0^+}^{\gamma-j} x(0) = x_j, j\in {\Bbb N} _{n+1}^k, \end{array}&k>n. \end{array}\right. \end{eqnarray*} $

我们得到(2.1)–(2.2)式.引理2.2证明完毕.

设$ n, l, k $为正整数, $ \lambda\in {{\Bbb R}} $, $ \alpha\in (n-1, n) $, $ \beta\in (l-1, l), \gamma\in (k-1, k) $且$ k\le \alpha+\beta $.我们求如下线性脉冲分数阶Langevin方程(简称ILFDE)的通解

(2.5) $ \begin{eqnarray} D_{0^+}^\alpha D_{0^+}^\beta x(t)-\lambda D_{0^+}^\gamma x(t) = g(t), \; {\rm a.e., }\ t\in (t_i, t_{i+1}], i\in {\Bbb N} _0^m. \end{eqnarray} $

定理2.3  设$ g $是连续函数.则$ x $是方程(1.5)的解的充分必要条件为存在常数$ c_{\nu, j}\in {{\Bbb R}} (\nu\in {\Bbb N} _0^m, j\in {\Bbb N} _1^l), d_{\nu, i}\in {{\Bbb R}} (\nu\in {\Bbb N} _0^m, i\in {\Bbb N} _1^{\max\{n, k\}}) $使得

(2.6) $ \begin{eqnarray} x(t)& = &\sum\limits_{\nu = 0}^v\sum\limits_{j = 1}^lc_{\nu, j}(t- t_\nu)^{\beta-j}{\bf E}_{\alpha+\beta-\gamma, \beta-j+1}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\sum\limits_{\nu = 0}^v\sum\limits_{i = 1}^{N}d_{\nu, i}(t- t_\nu)^{\alpha+\beta-i}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-i+1}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\int_0^t(t- s)^{\alpha+\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(\lambda(t- s)^{\alpha+\beta-\gamma})g(s){\rm d}s , \ t\in (t_v, t_{v+1}], v\in {\sf Z}\!\!\!{\sf Z}_0^m. \end{eqnarray} $

证  证明分两步.由于计算过程是初等的,所以部分计算推导略去.

第1步  设$ x $满足(2.6)式.我们证明$ x $是方程(2.5)的分片连续解.

容易从$ g\in C[0, 1] $推出$ x|_{(t_j, t_{j+1}]}\in C(t_j, t_{j+1}] $且$ \lim\limits_{t\to t_j^+}(t- t_j)^{l-\beta}x(t) $存在.由(2.6)式,对于$ t\in (t_v, t_{v+1}] $,直接计算得到

$ \begin{eqnarray*} D_{0^+}^\gamma x(t)& = &\left(\int_0^t\frac{(t- s)^{k-\gamma-1}}{\Gamma(k-\gamma )}x(s){\rm d}s\right)^{(l)} \\ & = &\bigg[\sum\limits_{o = 0}^{v-1}\int_{t_{o}}^{t_{o+1}}\frac{(t- s)^{k-\gamma-1}} {\Gamma(k-\gamma )}\bigg(\sum\limits_{\nu = 0}^k\sum\limits_{j = 1}^lc_{\nu, j}(s- t_\nu)^{\beta-j}{\bf E}_{\alpha+\beta-\gamma, \beta-j+1}(\lambda(s-t_\nu)^{\alpha+\beta-\gamma}) \\ &&+\sum\limits_{\nu = 0}^k\sum\limits_{i = 1}^{N}d_{\nu, i}(s- t_\nu)^{\alpha+\beta-i}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-i+1}(\lambda(s- t_\nu)^{\alpha+\beta-\gamma})\\ \\ &&+\int_0^s(s- u)^{\alpha+\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta} (\lambda(s- u)^{\alpha+\beta-\gamma})g(u){\rm d}u\bigg){\rm d}s\bigg]^{(k)} \\ &&+\bigg[\int_{t_{v}}^t\frac{(t- s)^{k-\gamma-1}}{\Gamma(k-\gamma )} \bigg(\sum\limits_{\nu = 0}^v\sum\limits_{j = 1}^lc_{\nu, j}(s- t_\nu)^{\beta-j} {\bf E}_{\alpha+\beta-\gamma, \beta-j+1}(\lambda(s-t_\nu)^{\alpha+\beta-\gamma}) \\ &&+\sum\limits_{\nu = 0}^v\sum\limits_{i = 1}^{N}d_{\nu, i}(s- t_\nu)^{\alpha+\beta-i} {\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-i+1}(\lambda(s- t_\nu)^{\alpha+\beta-\gamma}) \\ &&+\int_0^s(s- u)^{\alpha+\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta} (\lambda(s- u)^{\alpha+\beta-\gamma})g(u){\rm d}u\bigg){\rm d}s\bigg]^{(k)} \\ & = &\sum\limits_{\nu = 0}^v\sum\limits_{j = 1}^lc_{\nu, j}\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(t- t_\nu)^{\chi(\alpha+\beta-\gamma)-\gamma+\beta-j}}{\Gamma(\chi(\alpha+\beta-\gamma)-\gamma+\beta-j+1)} \\ && +\sum\limits_{\nu = 0}^v\sum\limits_{i = 1}^{N}d_{\nu, i}\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(t- t_\nu)^{\chi(\alpha+\beta-\gamma)-\gamma+\alpha+\beta-i}}{\Gamma(\chi(\alpha+\beta-\gamma)-\gamma+\alpha+\beta-i+1)} \\ &&+\int_{0}^t\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(t- u)^{\chi(\alpha+\beta-\gamma)+\alpha+\beta-\gamma-1}}{\Gamma(\chi(\alpha+\beta-\gamma)-\gamma+\alpha+\beta)}g(u){\rm d}u. \end{eqnarray*} $

类似地,对于$ t\in (t_v, t_{v+1}] $通过计算得到

$ \begin{eqnarray*} D_{0^+}^\beta x(t)& = &\left(\int_0^t\frac{(t- s)^{l-\beta-1}}{\Gamma(l-\beta )}x(s){\rm d}s\right)^{(l)} \\ & = &\bigg(\sum\limits_{o = 0}^{v-1}\int_{t_{o}}^{t_{o+1}}\frac{(t- s)^{l-\beta-1}}{\Gamma(l-\beta )}x(s){\rm d}s+\int_{t_{v}}^t\frac{(t- s)^{l-\beta-1}}{\Gamma(l-\beta )}x(s){\rm d}s\bigg)^{(l)} \\ & = &\sum\limits_{\nu = 0}^v\sum\limits_{j = 1}^lc_{\nu, j}\sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi(t- t_\nu)^{\chi(\alpha+\beta-\gamma)-j}}{\Gamma(\chi(\alpha+\beta-\gamma)-j+1)} \\ && +\sum\limits_{\nu = 0}^v\sum\limits_{i = 1}^{N}d_{\nu, i}\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(t- t_\nu)^{\chi(\alpha+\beta-\gamma)+\alpha-i}}{\Gamma(\chi(\alpha+\beta-\gamma)+\alpha-i+1)} \\ && +\int_{0}^t\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(t- u)^{\chi(\alpha+\beta-\gamma)+\alpha-1}}{\Gamma(\chi(\alpha+\beta-\gamma)+\alpha)}g(u){\rm d}u. \end{eqnarray*} $

因此对于$ t\in (t_v, t_{v+1}] $得到

$ \begin{eqnarray*} D_{0^+}^\alpha D_{0^+}^\beta x(t) & = &\left[\int_0^t\frac{(t- s)^{n-\alpha-1}}{\Gamma(n-\alpha)}D_{0^+}^\beta x(s){\rm d}s\right]^{(n)} \\ & = &\bigg[\sum\limits_{o = 0}^{v-1}\int_{t_o}^{t_{o+1}}\frac{(t- s)^ {n-\alpha-1}}{\Gamma(n-\alpha)}D_{0^+}^\beta x(s){\rm d}s+\int_{t_{\nu}}^ t\frac{(t- s)^{n-\alpha-1}} {\Gamma(n-\alpha)}D_{0^+}^\beta x(s){\rm d}s\bigg]^{(n)} \\ & = &\bigg[\sum\limits_{o = 0}^{v-1}\int_{t_o}^{t_{o+1}}\frac{(t-s)^{n-\alpha-1}} {\Gamma(n-\alpha)} \bigg(\sum\limits_{\nu = 0}^k\sum\limits_{j = 1}^lc_{\nu, j}\sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi(s- t_\nu)^{\chi(\alpha+\beta-\gamma)-j}}{\Gamma(\chi(\alpha+\beta-\gamma)-j+1)} \\ &&+\sum\limits_{\nu = 0}^k\sum\limits_{i = 1}^{N}d_{\nu, i}\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(s- t_\nu)^{\chi(\alpha+\beta-\gamma)+\alpha-i}}{\Gamma(\chi(\alpha+\beta-\gamma)+\alpha-i+1)} \\ &&+\int_{0}^s\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(s- u)^{\chi (\alpha+\beta-\gamma)+\alpha-1}}{\Gamma(\chi(\alpha+\beta-\gamma)+\alpha) }g(u){\rm d}u\bigg){\rm d}s\bigg]^{(n)}\\ \\ &&+\bigg[\int_{t_{v}}^t\frac{(t- s)^{n-\alpha-1}}{\Gamma(n-\alpha)} \bigg(\sum\limits_{\nu = 0}^v\sum\limits_{j = 1}^lc_{\nu, j}\sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi(s- t_\nu)^{\chi(\alpha+\beta-\gamma)-j}}{\Gamma(\chi(\alpha+\beta-\gamma)-j+1)} \\ &&+\sum\limits_{\nu = 0}^v\sum\limits_{i = 1}^{N}d_{\nu, i}\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(s- t_\nu)^{\chi(\alpha+\beta-\gamma)+\alpha-i}}{\Gamma(\chi(\alpha+\beta-\gamma)+\alpha-i+1)} \\ && +\int_{0}^s\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(s- u)^{\chi(\alpha+\beta-\gamma)+\alpha-1}}{\Gamma(\chi(\alpha+\beta-\gamma)+\alpha)}g(u){\rm d}u \bigg){\rm d}s\bigg]^{(n)} \\ & = &\sum\limits_{\nu = 0}^v\sum\limits_{j = 1}^lc_{\nu, j} \sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)-\alpha-j+1)}(t- t_\nu)^{\chi(\alpha+\beta-\gamma)-\alpha-j} \\ &&+\sum\limits_{\nu = 0}^v\sum\limits_{i = 1}^{N}d_{\nu, i}\sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi} {\Gamma(\chi(\alpha+\beta-\gamma)-i+1)} (t- t_\nu)^{\chi(\alpha+\beta-\gamma)-i} \\ &&+\int_{0}^t\sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma))}(t- u)^{\chi(\alpha+\beta-\gamma)-1}g(u){\rm d}u+g(t). \end{eqnarray*} $

因此

$ \begin{eqnarray*} &&D_{0^+}^\alpha D_{0^+}^\beta x(t)-\lambda D_{0^+}^\gamma x(t)\\ & = &\sum\limits_{\nu = 0}^v\sum\limits_{j = 1}^mc_{\nu, j} \sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)-\alpha-j+1)}(t- t_\nu)^{\chi(\alpha+\beta-\gamma)-\alpha-j} \\ &&+\sum\limits_{\nu = 0}^v\sum\limits_{i = 1}^{n}d_{\nu, i}\sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi} {\Gamma(\chi(\alpha+\beta-\gamma)-i+1)} (t- t_\nu)^{\chi(\alpha+\beta-\gamma)-i} \\ && +\int_{0}^t\sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma))}(t- u)^{\chi(\alpha+\beta-\gamma)-1}g(u){\rm d}u+g(t) \\ &&-\lambda\bigg[\sum\limits_{\nu = 0}^v\sum\limits_{j = 1}^mc_{\nu, j}\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(t- t_\nu)^{\chi(\alpha+\beta-\gamma)-\gamma+\beta-j}}{\Gamma(\chi(\alpha+\beta-\gamma)-\gamma+\beta-j+1)} \\ && +\sum\limits_{\nu = 0}^v\sum\limits_{i = 1}^{n}d_{\nu, i}\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(t- t_\nu)^{\chi(\alpha+\beta-\gamma)-\gamma+\alpha+\beta-i}}{\Gamma(\chi(\alpha+\beta-\gamma)-\gamma+\alpha+\beta-i+1)} \\ && +\int_{0}^t\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(t- u)^{\chi(\alpha+\beta-\gamma)+\alpha+\beta-\gamma-1}}{\Gamma(\chi(\alpha+\beta-\gamma)-\gamma+\alpha+\beta)}g(u){\rm d}u\bigg] \\ & = &g(t), t\in (t_v, t_{v+1}], v\in {\Bbb N} _0^p. \end{eqnarray*} $

从而$ x $是方程(2.5)的分片连续解.

第2步  设$ x $是方程(2.5)的分片连续解.证明$ x $满足(2.6)式.

对于$ t\in (t_0, t_1] $,从引理2.2知道存在常数$ c_{0, j}, d_{0, i} $使得

$ \begin{eqnarray*} x(t)& = &\sum\limits_{j = 1}^lc_{0, j}t^{\beta-j}{\bf E}_{\alpha+\beta-\gamma, \beta-j+1}(\lambda t^{\alpha+\beta-\gamma}) +\sum\limits_{i = 1}^{N}d_{0, i}t^{\alpha+\beta-\gamma-i}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma-i+1}(\lambda t^{\alpha+\beta-\gamma})\\ && +\int_0^t(t- s)^{\alpha+\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(\lambda(t- s)^{\alpha+\beta-\gamma})g(s){\rm d}s, \;t\in (t_0, t_1]. \end{eqnarray*} $

于是(2.6)式对于$ k = 0 $成立.我们假设(2.6)式对于$ 0, 1, 2, \cdots, \mu $成立,即

$ \begin{eqnarray*} x(t)& = &\sum\limits_{\nu = 0}^v\sum\limits_{j = 1}^lc_{\nu, j}(t- t_\nu)^{\beta-j}{\bf E}_{\alpha+\beta-\gamma, \beta-j+1}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma})\\ &&+\sum\limits_{\nu = 0}^v\sum\limits_{i = 1}^{N}d_{\nu, i}(t- t_\nu)^{\alpha+\beta-i}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-i+1}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma})\\ &&+\int_0^t(t- s)^{\alpha+\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(\lambda(t- s)^{\alpha+\beta-\gamma})g(s){\rm d}s , \ t\in (t_v, t_{v+1}], v\in {\sf Z}\!\!\!{\sf Z}_0^\mu. \end{eqnarray*} $

现在只要证明(2.6)式对于$ \mu+1 $成立.由数学归纳法知道(2.6)式对于所有$ v\in {\Bbb N} _0^m $成立.

事实上,假设

(2.7) $ \begin{eqnarray} x(t)& = &\Phi(t)+\sum\limits_{\nu = 0}^\mu\sum\limits_{j = 1}^lc_{\nu, j}(t- t_\nu)^{\beta-j}{\bf E}_{\alpha+\beta-\gamma, \beta-j+1}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}){}\\ && +\sum\limits_{\nu = 0}^\mu\sum\limits_{i = 1}^{N}d_{\nu, i}(t- t_\nu)^{\alpha+\beta-i}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-i+1}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}){}\\ && +\int_0^t(t- s)^{\alpha+\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(\lambda(t- s)^{\alpha+\beta-\gamma})g(s){\rm d}s, t\in (t_{\mu+1}, t_{\mu+2}]. \end{eqnarray} $

那么对于$ t\in (t_{\mu+1}, t_{\mu+2}] $,通过直接计算得到

$ \begin{eqnarray*} D_{0^+}^\gamma x(t)& = &\left(\int_0^t\frac{(t- s)^{k-\gamma-1}}{\Gamma(k-\gamma )}x(s){\rm d}s\right)^{(k)}\\ & = &\bigg[\sum\limits_{o = 0}^{\mu}\int_{t_{o}}^{t_{o+1}}\frac{(t- s)^{k-\gamma-1}}{\Gamma(k-\gamma )}\bigg(\sum\limits_{\nu = 0}^o\sum\limits_{j = 1}^lc_{\nu, j}(s- t_\nu)^{\beta-j}{\bf E}_{\alpha+\beta-\gamma, \beta-j+1}(\lambda(s- t_\nu)^{\alpha+\beta-\gamma}) \\ &&+\sum\limits_{\nu = 0}^o\sum\limits_{i = 1}^{N}d_{\nu, i}(s- t_\nu)^{\alpha+\beta-i}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-i+1}(\lambda(s- t_\nu)^{\alpha+\beta-\gamma})\\ &&+\int_0^s(s- u)^{\alpha+\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(\lambda(s- u)^{\alpha+\beta-\gamma})g(u){\rm d}u\bigg){\rm d}s\bigg]^{(k)}\\ && +\bigg[\int_{t_{\mu+1}}^t\frac{(t- s)^{k-\gamma-1}}{\Gamma(k-\gamma )}\bigg(\sum\limits_{\nu = 0}^o\sum\limits_{j = 1}^lc_{\nu, j}(s- t_\nu)^{\beta-j}{\bf E}_{\alpha+\beta-\gamma, \beta-j+1}(\lambda(s- t_\nu)^{\alpha+\beta-\gamma})\\ && +\sum\limits_{\nu = 0}^\mu\sum\limits_{i = 1}^{N}d_{\nu, i}(s- t_\nu)^{\alpha+\beta-i}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-i+1}(\lambda(s- t_\nu)^{\alpha+\beta-\gamma})\\ &&+\int_0^s(s- u)^{\alpha+\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(\lambda(s- u)^{\alpha+\beta-\gamma})g(u){\rm d}u+\Phi(s)\bigg){\rm d}s\bigg]^{(k)}\\ & = &\sum\limits_{\nu = 0}^{\mu+1}\sum\limits_{j = 1}^mc_{\nu, j}\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(t- t_\nu)^{\chi(\alpha+\beta-\gamma)-\gamma+\beta-j}}{\Gamma(\chi(\alpha+\beta-\gamma)-\gamma+\beta-j+1)} \\ &&+\sum\limits_{\nu = 0}^{\mu+1}\sum\limits_{i = 1}^{n}d_{\nu, i}\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(t- t_\nu)^{\chi(\alpha+\beta-\gamma)-\gamma+\alpha+\beta-i}}{\Gamma(\chi(\alpha+\beta-\gamma)-\gamma+\alpha+\beta-i+1)} \\ && +\int_{0}^t\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(t- u)^{\chi(\alpha+\beta-\gamma)+\alpha+\beta-\gamma-1}}{\Gamma(\chi(\alpha+\beta-\gamma)-\gamma+\alpha+\beta)}g(u){\rm d}u+D_{t_{\mu+1}^+}^\gamma \Phi(t). \end{eqnarray*} $

类似地对于$ t\in (t_v, t_{v+1}](v\in {\Bbb N} _0^{\mu}) $,直接计算得到

$ \begin{eqnarray*} D_{0^+}^\beta x(t)& = &\left(\int_0^t\frac{(t- s)^{l-\beta-1}}{\Gamma(l-\beta )}x(s){\rm d}s\right)^{(l)} \\ & = &\sum\limits_{\nu = 0}^v\sum\limits_{j = 1}^lc_{\nu, j}\sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi(t- t_\nu)^{\chi(\alpha+\beta-\gamma)-j}}{\Gamma(\chi(\alpha+\beta-\gamma)-j+1)} \\ && +\sum\limits_{\nu = 0}^v\sum\limits_{i = 1}^{N}d_{\nu, i}\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(t- t_\nu)^{\chi(\alpha+\beta-\gamma)+\alpha-i}}{\Gamma(\chi(\alpha+\beta-\gamma)+\alpha-i+1)} \\ && +\int_{0}^t\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(t- u)^{\chi(\alpha+\beta-\gamma)+\alpha-1}}{\Gamma(\chi(\alpha+\beta-\gamma)+\alpha)}g(u){\rm d}u, \end{eqnarray*} $

对于$ t\in (t_{\mu+1}, t_{\mu+2}] $,得到

$ \begin{eqnarray*} D_{0^+}^\beta x(t)& = &\left(\int_0^t\frac{(t- s)^{l-\beta-1}}{\Gamma(l-\beta )}x(s){\rm d}s\right)^{(l)} \\ & = &\bigg(\sum\limits_{o = 0}^{\mu}\int_{t_{o}}^{t_{o+1}}\frac{(t- s)^{l-\beta-1}}{\Gamma(l-\beta )}x(s){\rm d}s+\int_{t_{\mu+1}}^t\frac{(t- s)^{l-\beta-1}}{\Gamma(l-\beta )}x(s){\rm d}s\bigg)^{(l)}\\ & = &\sum\limits_{\nu = 0}^{\mu+1}\sum\limits_{j = 1}^lc_{\nu, j}\sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi(t- t_\nu)^{\chi(\alpha+\beta-\gamma)-j}}{\Gamma(\chi(\alpha+\beta-\gamma)-j+1)} \\ && +\sum\limits_{\nu = 0}^{\mu+1}\sum\limits_{i = 1}^{N}d_{\nu, i}\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(t- t_\nu)^{\chi(\alpha+\beta-\gamma)+\alpha-i}}{\Gamma(\chi(\alpha+\beta-\gamma)+\alpha-i+1)} \\ && +\int_{0}^t\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(t- u)^{\chi(\alpha+\beta-\gamma)+\alpha-1}}{\Gamma(\chi(\alpha+\beta-\gamma)+\alpha)}g(u){\rm d}u+D_{t_{\mu+1}^+}^\beta \Phi(t). \end{eqnarray*} $

因此对于$ t\in (t_{\mu+1}, t_{\mu+2}] $,得到

$ \begin{eqnarray*} D_{0^+}^\alpha D_{0^+}^\beta x(t)& = &\left[\int_0^t\frac{(t- s)^{n-\alpha-1}}{\Gamma(n-\alpha)}D_{0^+}^\beta x(s){\rm d}s\right]^{(n)}\\ & = &\bigg[\sum\limits_{o = 0}^{\mu}\int_{t_o}^{t_{o+1}}\frac{(t- s)^{n-\alpha-1}}{\Gamma(n-\alpha)}D_{0^+}^\beta x(s){\rm d}s+\int_{t_{\mu+1}}^t\frac{(t- s)^{n-\alpha-1}}{\Gamma(n-\alpha)}D_{0^+}^\beta x(s){\rm d}s\bigg]^{(n)}\\ & = &\sum\limits_{\nu = 0}^{\mu+1}\sum\limits_{j = 1}^lc_{\nu, j} \sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)-\alpha-j+1)}(t- t_\nu)^{\chi(\alpha+\beta-\gamma)-\alpha-j}\\ &&+\sum\limits_{\nu = 0}^{\mu+1}\sum\limits_{i = 1}^{N}d_{\nu, i}\sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi} {\Gamma(\chi(\alpha+\beta-\gamma)-i+1)} (t- t_\nu)^{\chi(\alpha+\beta-\gamma)-i} \\ &&+\int_{0}^t\sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma))}(t- u)^{\chi(\alpha+\beta-\gamma)-1}g(u){\rm d}u +g(t)+D_{t_{\mu+1}^+}^\alpha D_{t_{\mu+1}^+}^\beta \Phi(t). \end{eqnarray*} $

应用(2.5)式,我们得到

$ \begin{eqnarray*} g(t)& = &D_{0^+}^\alpha D_{0^+}^\beta x(t)-\lambda D_{0^+}^\gamma x(t)\\ & = &\sum\limits_{\nu = 0}^{\mu+1}\sum\limits_{j = 1}^lc_{\nu, j} \sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma)-\alpha-j+1)}(t- t_\nu)^{\chi(\alpha+\beta-\gamma)-\alpha-j}\\ &&+\sum\limits_{\nu = 0}^{\mu+1}\sum\limits_{i = 1}^{N}d_{\nu, i}\sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi} {\Gamma(\chi(\alpha+\beta-\gamma)-i+1)} (t- t_\nu)^{\chi(\alpha+\beta-\gamma)-i} \\ &&+\int_{0}^t\sum\limits_{\chi = 1}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha+\beta-\gamma))}(t- u)^{\chi(\alpha+\beta-\gamma)-1}g(u){\rm d}u+g(t)+D_{t_{\mu+1}^+}^\alpha D_{t_{\mu+1}^+}^\beta \Phi(t)\\ &&-\lambda\bigg[\sum\limits_{\nu = 0}^{\mu+1}\sum\limits_{j = 1}^lc_{\nu, j}\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(t- t_\nu)^{\chi(\alpha+\beta-\gamma)-\gamma+\beta-j}}{\Gamma(\chi(\alpha+\beta-\gamma)-\gamma+\beta-j+1)} \\ &&+\sum\limits_{\nu = 0}^{\mu+1}\sum\limits_{i = 1}^{N}d_{\nu, i}\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(t- t_\nu)^{\chi(\alpha+\beta-\gamma)-\gamma+\alpha+\beta-i}}{\Gamma(\chi(\alpha+\beta-\gamma)-\gamma+\alpha+\beta-i+1)} \\ &&+\int_{0}^t\sum\limits_{\chi = 0}^\infty\frac{\lambda^\chi(t- u)^{\chi(\alpha+\beta-\gamma)+\alpha+\beta-\gamma-1}}{\Gamma(\chi(\alpha+\beta-\gamma)-\gamma+\alpha+\beta)}g(u){\rm d}u+D_{t_{\mu+1}^+}^\gamma \Phi(t)\bigg]\\ & = &g(t)+D_{t_{\mu+1}^+}^\alpha D_{t_{\mu+1}^+}^\beta \Phi(t)-\lambda D_{t_{\mu+1}^+}^\gamma \Phi(t), t\in (t_{\mu+1}, t_{\mu+2}]. \end{eqnarray*} $

立得

(2.8) $ \begin{eqnarray} D_{t_{\nu+1}^+}^\alpha D_{t_{\nu+1}^+}^\beta\Phi(t)-\lambda D_{t_{\nu+1}^+}^\gamma x(t) = 0, t\in (t_{\nu+1}, t_{\nu+2}]. \end{eqnarray} $

通过直接计算我们得到

(2.9) $I_{t_{\mu + 1}^ + }^{l - \beta }\Phi ({t_{\nu + 1}}), \;D_{t_{\mu + 1}^ + }^{\beta - i}\Phi ({t_{\nu + 1}}), i \in {\Bbb N}_1^{l - 1};\\ \left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}{(I_{t_{\mu + 1}^ + }^{n - \alpha }D_{t_{\mu + 1}^ + }^\beta - \lambda I_{t_{\mu + 1}^ + }^{k - \gamma })\Phi ({t_{\nu + 1}}), }\\{(D_{t_{\mu + 1}^ + }^{\alpha - j}D_{t_{\mu + 1}^ + }^\beta - \lambda D_{t_{\mu + 1}^ + }^{\gamma - j})\Phi ({t_{\nu + 1}}), j \in {\Bbb N}_1^{k - 1}, }\end{array}}&{k = n;}\\{\begin{array}{*{20}{l}}{(D_{t_{\mu + 1}^ + }^{\alpha - k}D_{t_{\mu + 1}^ + }^\beta - \lambda I_{t_{\mu + 1}^ + }^{k - \gamma })\Phi ({t_{\nu + 1}}), }\\{(D_{t_{\mu + 1}^ + }^{\alpha - j}D_{t_{\mu + 1}^ + }^\beta - \lambda D_{t_{\mu + 1}^ + }^{\gamma - j})\Phi ({t_{\nu + 1}}), j \in {\Bbb N}_1^{k - 1}, }\\{D_{t_{\mu + 1}^ + }^{\alpha - j}D_{t_{\mu + 1}^ + }^\beta \Phi ({t_{\nu + 1}}), j \in {\Bbb N}_{k + 1}^n, }\end{array}}&{k < n;}\\{\begin{array}{*{20}{l}}{(I_{t_{\mu + 1}^ + }^{n - \alpha }D_{t_{\mu + 1}^ + }^\beta - \lambda D_{t_{\mu + 1}^ + }^{\gamma - n})\Phi ({t_{\nu + 1}}), }\\{(D_{t_{\mu + 1}^ + }^{\alpha - j}D_{t_{\mu + 1}^ + }^\beta - \lambda D_{t_{\mu + 1}^ + }^{\gamma - j})\Phi ({t_{\nu + 1}}), j \in {\Bbb N}_1^{n - 1}, }\\{ - \lambda D_{t_{\mu + 1}^ + }^{\gamma - j}\Phi ({t_{\nu + 1}}), j \in {\Bbb N}_{n + 1}^k, }\end{array}}&{k > n.}\end{array}} \right.$

利用引理2.2相似方法和(2.8)–(2.9)式,知道存在常数$ c_{\mu+1 j}, d_{\mu+1, i}\in {{\Bbb R}} $使得

$ \begin{eqnarray*} \Phi(t)& = &\sum\limits_{j = 1}^{l}c_{\mu+1 j}(t- t_{\mu+1})^{\beta-j}{\bf E}_{\alpha+\beta-\gamma, \beta-j+1}(\lambda(t- t_{\mu+1})^{\alpha+\beta-\gamma})\\ &&+\sum\limits_{i = 1}^{N}d_{\mu+1, i}(t- t_{\nu})^{\alpha+\beta-i}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-i+1}(\lambda(t- t_{\mu+1})^{\alpha+\beta-\gamma}), t\in (t_{\mu+1}, t_{\mu+2}]. \end{eqnarray*} $

把$ \Phi $代入(2.7)式,我们得到(2.6)式对$ \nu+1 $成立.于是数学归纳法知道(2.6)式对于任意$ k\in {\Bbb N} _0^m $成立.因此$ x $满足(2.6)式.定理2.3证明完毕.

设条件(a)–(d)成立.我们利用定理2.3建立BVP(1.1)的等价积分方程.为方便起见,记

$ f_x(t) = f(t, x(t)), \;\;I_{ix}(t_j) = I_i(t_j, x(t_j)), \;i\in {\Bbb N} _1^4, \;j\in {\Bbb N} _1^m, $

$ \begin{eqnarray*} M_{1x}(I_1, I_2, I_3, I_4, f)& = &-\sum\limits_{\nu = 1}^mI_{1x}(t_\nu){\bf E}_{\alpha+\beta-\gamma, 1}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma})\\ &&-\sum\limits_{\nu = 1}^mI_{2x}(t_\nu) (1- t_\nu){\bf E}_{\alpha+\beta-\gamma, 2}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma})\\ &&-\sum\limits_{\nu = 1}^mI_{3x}(t_\nu)(1- t_\nu)^{\alpha+1}{\bf E}_{\alpha+\beta-\gamma, \alpha+2}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma})\\ &&-\sum\limits_{\nu = 1}^mI_{4x}(t_\nu)(1- t_\nu)^{\alpha}{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma})\\ && -\int_0^1(1- s)^{\alpha+1}{\bf E}_{\alpha+\beta-\gamma, \alpha+2}(\lambda(1- s)^{\alpha+\beta-\gamma})P(s)f_x(s){\rm d}s, \end{eqnarray*} $

$ \begin{eqnarray*} M_{2x}(I_1, I_2, I_3, I_4, f)& = &-\sum\limits_{\nu = 1}^mI_{1x}(t_\nu)(1- t_\nu)^{\alpha+\beta-\gamma-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma})\\ && -\sum\limits_{\nu = 1}^mI_{2x}(t_\nu){\bf E}_{\alpha+\beta-\gamma, 1}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma})\\ && -\sum\limits_{\nu = 1}^mI_{3x}(t_\nu)(1- t_\nu)^{\alpha}{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma})\\ && -\sum\limits_{\nu = 1}^mI_{4x}(t_\nu)(1- t_\nu)^{\alpha-1}{\bf E}_{\alpha+\beta-\gamma, \alpha}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma})\\ && -\int_0^1(1- s)^{\alpha}{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda(1- s)^{\alpha+\beta-\gamma})P(s)f_x(s){\rm d}s, \end{eqnarray*} $

$ M_{3x}(I_1, I_2, I_3, I_4, f) = -\frac{1}{2}\left(\sum\limits_{\nu = 1}^mI_{3x}(t_{\nu1})+\int_0^1P(s)f_x(s){\rm d}s\right), $

$ \begin{eqnarray*} M_{4x}(I_1, I_2, I_3, I_4, f)& = &-\sum\limits_{\nu = 1}^mI_{1x}(t_\nu)(1- t_\nu)^{\beta-\gamma}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+1}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma}) \\ && -\sum\limits_{\nu = 1}^mI_{2x}(t_\nu)(1- t_\nu)^{\beta-\gamma+1}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+2}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma})\\ && -\sum\limits_{\nu = 1}^mI_{3x}(t_\nu)(1- t_\nu){\bf E}_{\alpha+\beta-\gamma, 2}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma})\\ && -\sum\limits_{\nu = 1}^mI_{4x}(t_\nu)(1- t_\nu){\bf E}_{\alpha+\beta-\gamma, 1}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma}) \\ && -\int_0^1(1- s){\bf E}_{\alpha+\beta-\gamma, 2}(\lambda(1- s)^{\alpha+\beta-\gamma})P(s)f_x(s){\rm d}s. \end{eqnarray*} $

分别定义行列式$ M $,其各个元素的代数余子式为$ M_{ij}(i, j = 1, 2, 3) $,函数$ c_{0, j}(I_1, I_2, I_3, I_4, f), $$ d_{0, j}(I_1, I_2, I_3, I_4, f)(j = 1, 2) $如下

$ M = \left|\begin{array}{ccc} {\bf E}_{\alpha+\beta-\gamma, 2}(\lambda)&1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda)&{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda)\\ 1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda)&{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(\lambda)&{\bf E}_{\alpha+\beta-\gamma, \alpha}(\lambda)\\ {\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+2}(\lambda) &{\quad} {\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+1}(\lambda)&{\quad} 1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda) \end{array}\right|, $

$ M_{11} = {\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(\lambda)[1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda)]-{\bf E}_{\alpha+\beta-\gamma, \alpha}(\lambda){\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+1}(\lambda), $

$ M_{12} = {\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+2}(\lambda){\bf E}_{\alpha+\beta-\gamma, \alpha}(\lambda)-[1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda)][1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda)], $

$ M_{13} = [1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda)]1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda)-{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(\lambda){\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+2}(\lambda), $

$ M_{21} = {\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda){\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+1}(\lambda)-[1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda)][1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda)], $

$ M_{22} = {\bf E}_{\alpha+\beta-\gamma, 2}(\lambda)[1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda)]-{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda){\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+2}(\lambda), $

$ M_{23} = [1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda)]{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+2}(\lambda)-{\bf E}_{\alpha+\beta-\gamma, 2}(\lambda){\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+1}(\lambda), $

$ M_{31} = [1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda)]{\bf E}_{\alpha+\beta-\gamma, \alpha}(\lambda)-{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda){\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(\lambda), $

$ M_{32} = {\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda)[1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda)]-{\bf E}_{\alpha+\beta-\gamma, 2}(\lambda){\bf E}_{\alpha+\beta-\gamma, \alpha}(\lambda), $

$ M_{33} = {\bf E}_{\alpha+\beta-\gamma, 2}(\lambda){\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(\lambda)-[1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda)][1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda)]. $

及

$ \begin{eqnarray*} c_{0, 1}(I_1, I_2, I_3, I_4, f) & = &\frac{1}{M}[M_{11}M_{1x}(I_1, I_2, I_3, I_4, f)+M_{21}M_{2x}(I_1, I_2, I_3, I_4, f) \\ && -[M_{21}{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda)+M_{31}{\bf E}_{\alpha+\beta-\gamma, 2}(\lambda)]M_{3x}(I_1, I_2, I_3, I_4, f)\\ &&+M_{31}M_{4x}(I_1, I_2, I_3, I_4, f)], \\ c_{0, 2}(I_1, I_2, I_3, I_4, f) & = &\frac{1}{M}[M_{12}M_{1x}(I_1, I_2, I_3, I_4, f)+M_{22}M_{2x}(I_1, I_2, I_3, I_4, f)\\ &&-[M_{22}{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda)+M_{32}{\bf E}_{\alpha+\beta-\gamma, 2}(\lambda)]M_{3x}(I_1, I_2, I_3, I_4, f)\\ &&+M_{32}M_{4x}(I_1, I_2, I_3, I_4, f)], \\ d_{0, 1}(I_1, I_2, I_3, I_4, f)& = &M_{3x}(I_1, I_2, I_3, I_4, f), \\ d_{0, 2}(I_1, I_2, I_3, I_4, f)& = &\frac{1}{M}[M_{13}M_{1x}(I_1, I_2, I_3, I_4, f)\\ &&+M_{23}M_{2x}(I_1, I_2, I_3, I_4, f)\\ && -[M_{23}{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda)+M_{33}{\bf E}_{\alpha+\beta-\gamma, 2}(\lambda)]M_{3x}(I_1, I_2, I_3, I_4, f)\\ &&+M_{33}M_{4x}(I_1, I_2, I_3, I_4, f)]. \end{eqnarray*} $

定理2.4  设(a)–(d)成立且$ M\not = 0. $则BVP(1.1)等价于以下积分方程

(2.10) $ \begin{eqnarray} x(t)& = &\sum\limits_{j = 1}^2c_{0, j}(I_1, I_2, I_3, I_4, f)t^{\beta-j}{\bf E}_{\alpha+\beta-\gamma, \beta-j+1}(\lambda t^{\alpha+\beta-\gamma}) {}\\ &&+\sum\limits_{i = 1}^{2}d_{0, i}(I_1, I_2, I_3, I_4, f)t^{\alpha+\beta-i}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-i+1}(\lambda t^{\alpha+\beta-\gamma}) {}\\ &&+\sum\limits_{\nu = 1}^vI_{2x}(t_\nu)(t- t_\nu)^{\beta-1}{\bf E}_{\alpha+\beta-\gamma, \beta}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\sum\limits_{\nu = 1}^vI_{3x}(t_\nu)(t- t_\nu)^{\alpha+\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\sum\limits_{\nu = 1}^vI_{1x}(t_\nu)(t- t_\nu)^{\beta-2}{\bf E}_{\alpha+\beta-\gamma, \beta-1}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\sum\limits_{\nu = 1}^vI_{4x}(t_\nu)(t- t_\nu)^{\alpha+\beta-2}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-1}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\int_0^t(t- s)^{\alpha+\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(\lambda(t- s)^{\alpha+\beta-\gamma})P(s)f_x(s){\rm d}s , {\quad} t\in (t_v, t_{v+1}], v\in {\sf Z}\!\!\!{\sf Z}_0^m. {}\\ \end{eqnarray} $

证  注意到$ n = l = 2 $且$ N = \max\{k, n\} = 2 $.设$ x $是BVP(1.1)的解.从定理2.3,存在常数$ c_{\nu, j}\in {{\Bbb R}} (\nu\in {\Bbb N} _0^m, j\in {\Bbb N} _1^2), d_{\nu, i}\in {{\Bbb R}} (\nu\in {\Bbb N} _0^m, i\in {\Bbb N} _1^2 $使得

(2.11) $ \begin{eqnarray} x(t)& = &\sum\limits_{\nu = 0}^v\sum\limits_{j = 1}^2c_{\nu, j}(t- t_\nu)^{\beta-j}{\bf E}_{\alpha+\beta-\gamma, \beta-j+1}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\sum\limits_{\nu = 0}^v\sum\limits_{i = 1}^{2}d_{\nu, i}(t- t_\nu)^{\alpha+\beta-i}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-i+1}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\int_0^t(t- s)^{\alpha+\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(\lambda(t- s)^{\alpha+\beta-\gamma})P(s)f_x(s){\rm d}s , {\quad} t\in (t_v, t_{v+1}], v\in {\sf Z}\!\!\!{\sf Z}_0^m.{}\\ \end{eqnarray} $

通过直接计算得到:当$ t\in (t_i, t_{i+1}] $时有

(2.12) $ \begin{eqnarray} D_{0^+}^\beta x(t)& = &\lambda\sum\limits_{\nu = 0}^v\sum\limits_{j = 1}^2c_{\nu, j}(t- t_\nu)^{\alpha+\beta-\gamma-j}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma-j+1}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\sum\limits_{\nu = 0}^v\sum\limits_{i = 1}^{2}d_{\nu, i}(t- t_\nu)^{\alpha-i}{\bf E}_{\alpha+\beta-\gamma, \alpha-i+1}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\int_0^t(t- s)^{\alpha-1}{\bf E}_{\alpha+\beta-\gamma, \alpha}(\lambda(t- s)^{\alpha+\beta-\gamma})P(s)f_x(s){\rm d}s , \ t\in (t_v, t_{v+1}], v\in {\sf Z}\!\!\!{\sf Z}_0^m, {}\\ \end{eqnarray} $

(2.13) $ \begin{eqnarray} D_{0^+}^{\beta-1} x(t)& = &\sum\limits_{\nu = 0}^vc_{\nu, 1}{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\sum\limits_{\nu = 0}^vc_{\nu, 2}(t- t_\nu)^{\alpha+\beta-\gamma-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\sum\limits_{\nu = 0}^v\sum\limits_{i = 1}^{2}d_{\nu, i}(t- t_\nu)^{\alpha-i+1}{\bf E}_{\alpha+\beta-\gamma, \alpha-i+2}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\int_0^t(t- s)^{\alpha}{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda(t- s)^{\alpha+\beta-\gamma})P(s)f_x(s){\rm d}s , {\quad} t\in (t_v, t_{v+1}], v\in {\sf Z}\!\!\!{\sf Z}_0^m, {}\\ \end{eqnarray} $

(2.14) $ \begin{eqnarray} D_{0^+}^{\alpha-1}D_{0^+}^\beta x(t) & = &\lambda\sum\limits_{\nu = 0}^v\sum\limits_{j = 1}^2c_{\nu, j}(t- t_\nu)^{\beta-\gamma-j+1}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma-j+2}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\sum\limits_{\nu = 0}^vd_{\nu, 1}{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\lambda\sum\limits_{\nu = 0}^v d_{\nu, 2}(t- t_\nu)^{\alpha+\beta-\gamma-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\int_0^t{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda(t- s)^{\alpha+\beta-\gamma})P(s)f_x(s){\rm d}s , \ t\in (t_v, t_{v+1}], v\in {\sf Z}\!\!\!{\sf Z}_0^m, {}\\ \end{eqnarray} $

(2.15) $ \begin{eqnarray} I_{0^+}^{2-\beta}x(t)& = &\sum\limits_{\nu = 0}^v\sum\limits_{j = 1}^2c_{\nu, j}(t- t_\nu)^{2-j}{\bf E}_{\alpha+\beta-\gamma, 3-j}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\sum\limits_{\nu = 0}^v\sum\limits_{i = 1}^{2}d_{\nu, i}(t- t_\nu)^{\alpha+2-i}{\bf E}_{\alpha+\beta-\gamma, \alpha+3-i}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\int_0^t(t- s)^{\alpha+1}{\bf E}_{\alpha+\beta-\gamma, \alpha+2}(\lambda(t- s)^{\alpha+\beta-\gamma})P(s)f_x(s){\rm d}s , \ t\in (t_v, t_{v+1}], v\in {\sf Z}\!\!\!{\sf Z}_0^m, {}\\ \end{eqnarray} $

(2.16) $ \begin{eqnarray} I_{0^+}^{1-\gamma}x(t)& = &\sum\limits_{\nu = 0}^v\sum\limits_{j = 1}^2c_{\nu, j}(t- t_\nu)^{\beta-\gamma-j+1}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma-j+2}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\sum\limits_{\nu = 0}^v\sum\limits_{i = 1}^{2}d_{\nu, i}(t- t_\nu)^{\alpha+\beta-\gamma-i+1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma-i+2}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\int_0^t(t- s)^{\alpha+\beta-\gamma}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma+1}(\lambda(t- s)^{\alpha+\beta-\gamma})P(s)f_x(s){\rm d}s , {}\\ &&{\qquad}{\qquad}{\qquad} t\in (t_v, t_{v+1}], v\in {\sf Z}\!\!\!{\sf Z}_0^m, \end{eqnarray} $

(2.17) $ \begin{eqnarray} I_{0^+}^{2-\alpha}D_{0^+}^\beta x(t)& = &\sum\limits_{\nu = 0}^v\sum\limits_{j = 1}^2c_{\nu, j}(t- t_\nu)^{\beta-\gamma-j+2}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma-j+3}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\sum\limits_{\nu = 0}^v\sum\limits_{i = 1}^{2}d_{\nu, i}(t- t_\nu)^{2-i}{\bf E}_{\alpha+\beta-\gamma, 3-i}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&+\int_0^t(t- s){\bf E}_{\alpha+\beta-\gamma, 2}(\lambda(t- s)^{\alpha+\beta-\gamma})P(s)f_x(s){\rm d}s , \ t\in (t_v, t_{v+1}], v\in {\sf Z}\!\!\!{\sf Z}_0^m.{}\\ \end{eqnarray} $

由(2.14)和(2.16)式得到

(2.18) $ \begin{eqnarray} D_{0^+}^{\alpha-1}D_{0^+}^\beta x(t)-\lambda I_{0^+}^{1-\gamma}x(t) = \sum\limits_{\nu = 0}^vd_{\nu1}+\int_0^tP(s)f_x(s){\rm d}s , \ t\in (t_v, t_{v+1}], v\in {\sf Z}\!\!\!{\sf Z}_0^m, \end{eqnarray} $

(iⅰ)由(2.15)式和$ \Delta I_{0^+}^{2-\beta} x(t_i) = I_1(t_i, x(t_i)), i\in {\Bbb N} _1^{m-1} $得到$ c_{\nu 2} = I_{1x}(t_\nu)(\nu\in {\Bbb N} _1^m) $.

(ⅱ)由(2.13)式和$ \Delta D_{0^+}^{\beta-1}x(t_i) = I_2(t_i, x(t_i)), i\in {\Bbb N} _1^{m-1} $得到$ c_{\nu 1} = I_{2x}(t_\nu)(\nu\in {\Bbb N} _1^m) $.

(ⅲ)由(2.18)式和$ \Delta (D_{0^+}^{\alpha-1} D_{0^+}^\beta-\lambda I_{0^+}^{1-\gamma}) x(t_i) = I_3(t_i, x(t_i)), i\in {\Bbb N} _1^{m} $得到$ d_{\nu 1} = I_{3x}(t_\nu) $$ (\nu\in {\Bbb N} _1^m) $.

(ⅳ)由(2.17)式和$ \Delta I_{0^+}^{2-\alpha}D_{0^+}^\beta x(t_i) = I_4(t_i, x(t_i)), i\in {\Bbb N} _1^{m} $得到$ d_{\nu 2} = I_{4x}(t_\nu)(\nu\in {\Bbb N} _1^m) $.

(ⅴ)由(2.15)式, (ⅰ)$ - $(ⅳ)和$ I_{0^+}^{2-\beta} x(0) = -I_{0^+}^{2-\beta}x(1) $得到

(2.19) $ \begin{eqnarray} &&c_{0, 1}{\bf E}_{\alpha+\beta-\gamma, 2}(\lambda )+c_{0, 2}[1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda )] +d_{0, 1}{\bf E}_{\alpha+\beta-\gamma, \alpha+2}(\lambda ) +d_{0, 2}{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda ) {}\\ & = &-\sum\limits_{\nu = 1}^mI_{2x}(t_\nu) (1- t_\nu){\bf E}_{\alpha+\beta-\gamma, 2}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&-\sum\limits_{\nu = 1}^mI_{1x}(t_\nu){\bf E}_{\alpha+\beta-\gamma, 1}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&-\sum\limits_{\nu = 1}^mI_{3}(t_\nu)(1- t_\nu)^{\alpha+1}{\bf E}_{\alpha+\beta-\gamma, \alpha+2}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&-\sum\limits_{\nu = 1}^mI_{4}(t_\nu)(1- t_\nu)^{\alpha}{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&-\int_0^1(1- s)^{\alpha+1}{\bf E}_{\alpha+\beta-\gamma, \alpha+2}(\lambda (1- s)^{\alpha+\beta-\gamma})P(s)f_x(s){\rm d}s {}\\ & = &M_{1x}(I_1, I_2, I_3, I_4, f), \end{eqnarray} $

(ⅵ)由(2.13)式和$ \Delta D_{0^+}^{\beta-1}x(0) = -\Delta D_{0^+}^{\beta-1}x(1) $得到

(2.20) $ \begin{eqnarray} &&c_{0, 1}[1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda)] +c_{0, 2}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(\lambda ) +d_{0, 1}{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda ) +d_{0, 2}{\bf E}_{\alpha+\beta-\gamma, \alpha}(\lambda ) {}\\ & = &-\sum\limits_{\nu = 1}^mI_{2x}(t_\nu){\bf E}_{\alpha+\beta-\gamma, 1}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&-\sum\limits_{\nu = 1}^mI_{1x}(t_\nu)(1- t_\nu)^{\alpha+\beta-\gamma-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&-\sum\limits_{\nu = 1}^mI_{3x}(t_\nu)(1- t_\nu)^{\alpha}{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma}) {}\\ && -\sum\limits_{\nu = 1}^mI_{4x}(t_\nu)(1- t_\nu)^{\alpha-1}{\bf E}_{\alpha+\beta-\gamma, \alpha}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&-\int_0^1(1- s)^{\alpha}{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda(1- s)^{\alpha+\beta-\gamma})P(s)f_x(s){\rm d}s {}\\ & = &M_{2x}(I_1, I_2, I_3, I_4, f) , \end{eqnarray} $

(ⅶ)由(2.18)式和$ (D_{0^+}^{\alpha-1} D_{0^+}^\beta-\lambda I_{0^+}^{1-\gamma}) x(0) = -(D_{0^+}^{\alpha-1} D_{0^+}^\beta-\lambda I_{0^+}^{1-\gamma}) x(1) $得到

$ d_{01} = -d_{01}-\left(\sum\limits_{\nu = 1}^mI_{3x}(t_{\nu1})+\int_0^1P(s)f_x(s){\rm d}s\right). $

因此

(2.21) $ \begin{eqnarray} d_{01} = -\frac{1}{2}\left(\sum\limits_{\nu = 1}^mI_{3x}(t_{\nu1})+\int_0^1P(s)f_x(s){\rm d}s\right) = M_{3x}(I_1, I_2, I_3, I_4, f). \end{eqnarray} $

(ⅷ)由(2.17)式和$ I_{0^+}^{2-\alpha}D_{0^+}^\beta x(0) = -I_{0^+}^{2-\alpha}D_{0^+}^\beta x(1) $得到

(2.22) $ \begin{eqnarray} &&c_{0, 1}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+2}(\lambda )+c_{0, 2}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+1}(\lambda ) +d_{0, 1}{\bf E}_{\alpha+\beta-\gamma, 2}(\lambda ) +d_{0, 2}[1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda)] {}\\ & = &-\sum\limits_{\nu = 1}^mI_{2x}(t_\nu)(1- t_\nu)^{\beta-\gamma+1}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+2}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&-\sum\limits_{\nu = 1}^mI_{1x}(t_\nu)(1- t_\nu)^{\beta-\gamma}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+1}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&-\sum\limits_{\nu = 1}^mI_{3x}(t_\nu)(1- t_\nu){\bf E}_{\alpha+\beta-\gamma, 2}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&-\sum\limits_{\nu = 1}^mI_{4x}(t_\nu)(1- t_\nu){\bf E}_{\alpha+\beta-\gamma, 1}(\lambda(1- t_\nu)^{\alpha+\beta-\gamma}) {}\\ &&-\int_0^1(1- s){\bf E}_{\alpha+\beta-\gamma, 2}(\lambda(1- s)^{\alpha+\beta-\gamma})P(s)f_x(s){\rm d}s {}\\ & = &M_{4x}(I_1, I_2, I_3, I_4, f). \end{eqnarray} $

由(2.19)–(2.22)式得到

$ \begin{eqnarray*} &&\left(\begin{array}{ccc} {\bf E}_{\alpha+\beta-\gamma, 2}(\lambda)&1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda)&{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda)\\ 1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda)&{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(\lambda)&{\bf E}_{\alpha+\beta-\gamma, \alpha}(\lambda)\\ {\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+2}(\lambda) &{\quad}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+1}(\lambda){\quad}&1+{\bf E}_{\alpha+\beta-\gamma, 1}(\lambda) \end{array}\right)\left(\begin{array}{c} c_{0, 1}\\ c_{0, 2}\\ d_{0, 2}\end{array}\right)\\ & = &\left(\begin{array}{c} M_{1x}(I_1, I_2, , I_3, I_4, f)\\ M_{2x}(I_1, I_2, , I_3, I_4, f)-M_{3x}(I_1, I_2, , I_3, I_4, f){\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda)\\ M_{4x}(I_1, I_2, , I_3, I_4, f)-M_{3x}(I_1, I_2, , I_3, I_4, f){\bf E}_{\alpha+\beta-\gamma, 2}(\lambda)\end{array}\right), \end{eqnarray*} $

则

$ \begin{eqnarray*} c_{0, 1}& = &\frac{1}{M}[M_{11}M_{1x}(I_1, I_2, I_3, I_4, f)+M_{21}M_{2x}(I_1, I_2, I_3, I_4, f)\\ &&-[M_{21}{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda) +M_{31}{\bf E}_{\alpha+\beta-\gamma, 2}(\lambda)]M_{3x}(I_1, I_2, I_3, I_4, f)\\ &&+M_{31}M_{4x}(I_1, I_2, I_3, I_4, f)], \\ c_{0, 2}& = &\frac{1}{M}[M_{12}M_{1x}(I_1, I_2, I_3, I_4, f)+M_{22}M_{2x}(I_1, I_2, I_3, I_4, f)\\ && -[M_{22}{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda)+M_{32}{\bf E}_{\alpha +\beta-\gamma, 2}(\lambda)]M_{3x}(I_1, I_2, I_3, I_4, f)\\ &&+M_{32}M_{4x}(I_1, I_2, I_3, I_4, f)], \\ d_{0, 1}& = &M_{3x}(I_1, I_2, I_3, I_4, f), \\ d_{0, 2}& = &\frac{1}{M}[M_{13}M_{1x}(I_1, I_2, I_3, I_4, f)+M_{23}M_{2x} (I_1, I_2, I_3, I_4, f)\\ && -[M_{23}{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda)+M_{33}{\bf E} _{\alpha+\beta-\gamma, 2}(\lambda)]M_{3x}(I_1, I_2, I_3, I_4, f)\\ &&+M_{33}M_{4x}(I_1, I_2, I_3, I_4, f)]. \end{eqnarray*} $

把$ c_{0, 1}, c_{0, 2}, d_{0, 1}, d_{0, 2} $代入(2.11)式,我们得到(2.10)式.

另一方面,如果$ x $满足(2.10)式,我们可以证明$ x $是BVP(1.1)的分片连续解.定理2.4证明完毕.

本节建立BVP(1.1)的分片连续解的存在性定理.设行列式$ M $,其元素的代数余子式分别为$ M_{ij}(i, j = 1, 2, 3) $,函数$ c_{0, j}(I_1, I_2, I_3, I_4, f) $和$ d_{0, j}(I_1, I_2, I_3, I_4, f)(j = 1, 2) $在第2.3节定义.定义Banach空间$ PC_{2-\beta}^0(0, 1] $上的算子$ T $如下

$ \begin{eqnarray*} (Tx)(t)& = &\sum\limits_{j = 1}^2c_{0, j}(I_1, I_2, I_3, I_4, f)t^{\beta-j}{\bf E}_{\alpha+\beta-\gamma, \beta-j+1}(\lambda t^{\alpha+\beta-\gamma})\\ && +\sum\limits_{i = 1}^{2}d_{0, i}(I_1, I_2, I_3, I_4, f)t^{\alpha+\beta-i}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-i+1}(\lambda t^{\alpha+\beta-\gamma})\\ && +\sum\limits_{\nu = 1}^vI_{2x}(t_\nu)(t- t_\nu)^{\beta-1}{\bf E}_{\alpha+\beta-\gamma, \beta}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma})\\ && +\sum\limits_{\nu = 1}^vI_{3x}(t_\nu)(t- t_\nu)^{\alpha+\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma})\\ && +\sum\limits_{\nu = 1}^vI_{1x}(t_\nu)(t- t_\nu)^{\beta-2}{\bf E}_{\alpha+\beta-\gamma, \beta-1}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma})\\ && +\sum\limits_{\nu = 1}^vI_{4x}(t_\nu)(t- t_\nu)^{\alpha+\beta-2}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-1}(\lambda(t- t_\nu)^{\alpha+\beta-\gamma})\\ && +\int_0^t(t- s)^{\alpha+\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(\lambda(t- s)^{\alpha+\beta-\gamma})P(s)f_x(s){\rm d}s , \ t\in (t_v, t_{v+1}], v\in {\sf Z}\!\!\!{\sf Z}_0^m. \end{eqnarray*} $

我们需要如下假设

(H1)存在单调非减函数$ \phi_f, \phi_{Ii}:[0, \infty)\to [0, \infty) $使得

$ |f(t, (t-t_i)^{\beta-2}x)|\le \phi_f(|x|), t\in (t_i, t_{i+1}], i\in {\Bbb N} _0^m, x\in {{\Bbb R}} , $

$ |I_i(t_j, (t_{j}-t_{j-1})^{\beta-2}x)|\le \phi_{Ii}(|x|), i\in {\Bbb N} _1^4, j\in {\Bbb N} _1^m, x\in {{\Bbb R}} . $

记

$ \begin{eqnarray*} Q_{2i}& = &\frac{|M_{1i}|m {\bf E}_{\alpha+\beta-\gamma, 2}(|\lambda|)}{|M|} +\frac{|M_{2i}| m{\bf E}_{\alpha+\beta-\gamma, 1}(|\lambda|)}{|M|} +\frac{|M_{3i}|m{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+2}(|\lambda|)}{|M|} \\ && +\frac{|M_{3i}|m{\bf E}_{\alpha+\beta-\gamma, 2}(|\lambda|)}{|M|}, \\ Q_{1i}& = &\frac{|M_{3i}|m{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+1}(|\lambda|)}{|M|}+ \frac{m{\bf E}_{\alpha+\beta-\gamma, 1}(|\lambda|)}{|M|} +\frac{|M_{2i}|m{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(|\lambda|)}{|M|}, \\ Q_{3i}& = &\frac{m{\bf E}_{\alpha+\beta-\gamma, \alpha+2}(|\lambda|)}{|M|} +\frac{|M_{2i}|m{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(|\lambda|)}{|M|}\\ &&+\frac{[|M_{2i}|{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda)+|M_{3i}|{\bf E}_{\alpha+\beta-\gamma, 2}(\lambda)]\frac{1}{2}m }{|M|}, \\ Q_{4i}& = &\frac{|M_{2i}|m{\bf E}_{\alpha+\beta-\gamma, \alpha}(|\lambda|)}{|M|} +\frac{m{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(|\lambda|)}{|M|} +\frac{|M_{3i}|m{\bf E}_{\alpha+\beta-\gamma, 1}(|\lambda|)}{|M|}, \\ Q_{5i}& = &\frac{1}{|M|}||P||{\bf E}_{\alpha+\beta-\gamma, \alpha+2}(|\lambda|) +\frac{|M_{2i}|||P||{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(|\lambda|)}{|M|}\\ && +\frac{[|M_{2i}|{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda) +|M_{3i}|{\bf E}_{\alpha+\beta-\gamma, 2}(\lambda)]\frac{1}{2}||P||}{|M|} \\ &&+\frac{|M_{3i}|||P||{\bf E}_{\alpha+\beta-\gamma, 2}(|\lambda|)}{|M|}, {\quad} i = 1, 2, 3, \end{eqnarray*} $

$ Q_{14} = Q_{24} = Q_{44} = 9, \;\;\;Q_{34} = \frac{m}{2}, \;\;\;Q_{54} = \frac{||P||}{2}. $

定理3.1  假设(a)–(d)成立.如果存在$ r_0>0 $使得

(3.1) $ \begin{eqnarray} &&({\bf E}_{\alpha+\beta-\gamma, \beta}(|\lambda |)Q_{11} +{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(|\lambda|)Q_{14} +{\bf E}_{\alpha+\beta-\gamma, \beta-1}(|\lambda |)Q_{12} {}\\ &&+{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-1}(|\lambda|) Q_{13} +m{\bf E}_{\alpha+\beta-\gamma, \beta-1}(|\lambda|))\phi_{I1}(r_0) {}\\ && +({\bf E}_{\alpha+\beta-\gamma, \beta}(|\lambda |)Q_{21} +{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(|\lambda|)Q_{24} +{\bf E}_{\alpha+\beta-\gamma, \beta-1}(|\lambda |)Q_{22} {}\\ && +{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-1}(|\lambda|)Q_{23} +m{\bf E}_{\alpha+\beta-\gamma, \beta}(|\lambda|))\phi_{I2}(r_0) {}\\ && +({\bf E}_{\alpha+\beta-\gamma, \beta}(|\lambda |)Q_{31} +{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(|\lambda|)Q_{34} +{\bf E}_{\alpha+\beta-\gamma, \beta-1}(|\lambda |)Q_{32} {}\\ && +{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-1}(|\lambda|)Q_{31} +m{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(|\lambda|))\phi_{I3}(r_0) {}\\ && +(m{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-1}(|\lambda|) +{\bf E}_{\alpha+\beta-\gamma, \beta}(|\lambda |)Q_{41} +{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(|\lambda|)Q_{44} {}\\ && +{\bf E}_{\alpha+\beta-\gamma, \beta-1}(|\lambda |)Q_{42} +{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-1}(|\lambda|)Q_{43}\phi_{I4}(r_0) {}\\ && +({\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-1}(|\lambda|)Q_{53} +{\bf E}_{\alpha+\beta-\gamma, \beta-1}(|\lambda |)Q_{52} +{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(|\lambda|)Q_{54} {}\\ && +{\bf E}_{\alpha+\beta-\gamma, \beta}(|\lambda |)Q_{51} +||P||{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(|\lambda|))\phi_f(r_0)\le r_0, \end{eqnarray} $

则BVP(1.1)至少有一个分片连续解.

证  由$ T $的定义,定理2.4,容易证明$ T:PC_{2-\beta}^0(0, 1]\to PC_{2-\beta}^0(0, 1] $, $ T $是全连续算子, $ x $是BVP(1.1)的分片连续解当且仅当$ x $是$ T $在$ PC_{2-\beta}^0(0, 1] $上的不动点.取$ \Omega_0 = \{x\in X:||x||\le r_0\} $.对$ x\in \Omega_0 $,则$ ||x||\le r_0 $,由(H1)得到

$ |f_x(t)| = |f(t, x(t))|\le \phi_f((t-t_i)^{2-\beta}|x(t)|)\le \phi_f(r_0), t\in (t_i, t_{i+1}], i\in {\Bbb N} _0^m, $

$ |I_{ix}(t_j)| = |I_i(t_j, x(t_j))|\le \phi_{Ii}((t_{j}-t_{j-1})^{2-\beta}|x(t_j)|)\le \phi_{Ii}(r_0), i\in {\Bbb N} _1^4, j\in {\Bbb N} _1^m. $

因此

$ \begin{eqnarray*} |M_{1x}(I_1, I_2, I_3, I_4, f)|&\le& m\phi_{I2}(r_0) {\bf E}_{\alpha+\beta-\gamma, 2}(|\lambda|)+ m\phi_{I1}(r_0){\bf E}_{\alpha+\beta-\gamma, 1}(|\lambda|)\\ && +m\phi_{I3}(r_0){\bf E}_{\alpha+\beta-\gamma, \alpha+2}(|\lambda|) +m\phi_{I4}(r_0){\bf E}_{\alpha+\beta-\gamma, \alpha+1}(|\lambda|)\\ && +||P||{\bf E}_{\alpha+\beta-\gamma, \alpha+2}(|\lambda|)P(s)\phi_f(r_0), \\ |M_{2x}(I_1, I_2, I_3, I_4, f)|&\le& m\phi_{I2}(r_0){\bf E}_{\alpha+\beta-\gamma, 1}(|\lambda|) +m\phi_{I1}(r_0){\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(|\lambda|)\\ && +m\phi_{I3}(r_0){\bf E}_{\alpha+\beta-\gamma, \alpha+1}(|\lambda|) +m\phi_{I4}(r_0){\bf E}_{\alpha+\beta-\gamma, \alpha}(|\lambda|)\\ && +||P||{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(|\lambda|)\phi_{f}(r_0), \\ |M_{3x}(I_1, I_2, I_3, I_4, f)&\le& \frac{1}{2}\left(m\phi_{I3}(r_0)+||P||\phi_{f}(r_0)\right), \\ |M_{4x}(I_1, I_2, I_3, I_4, f)|&\le& m\phi_{I2}(r_0){\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+2}(|\lambda|) +m\phi_{I1}(r_0){\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+1}(|\lambda|)\\ && +m\phi_{I2}(r_0){\bf E}_{\alpha+\beta-\gamma, 2}(|\lambda|)+m\phi_{I4}(r_0){\bf E}_{\alpha+\beta-\gamma, 1}(|\lambda|) \\ && +||P||{\bf E}_{\alpha+\beta-\gamma, 2}(|\lambda|)\phi_{f}(r_0). \end{eqnarray*} $

我们得到

$ \begin{eqnarray*} |c_{0, 1}(I_1, I_2, I_3, I_4, f)|&\le&\left[\frac{|M_{11}|m {\bf E}_{\alpha+\beta-\gamma, 2}(|\lambda|)}{|M|} +\frac{|M_{21}| m{\bf E}_{\alpha+\beta-\gamma, 1}(|\lambda|)}{|M|}\right.\\ && \left. +\frac{|M_{31}|m{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+2}(|\lambda|)}{|M|} +\frac{|M_{31}|m{\bf E}_{\alpha+\beta-\gamma, 2}(|\lambda|)}{|M|}\right]\phi_{I2}(r_0)\\ && +\left[\frac{|M_{31}|m{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+1}(|\lambda|)}{|M|}+ \frac{m{\bf E}_{\alpha+\beta-\gamma, 1}(|\lambda|)}{|M|}\right. \\ &&\left. +\frac{|M_{21}|m{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(|\lambda|)}{|M|}\right]\phi_{I1}(r_0)\\ && +\left[\frac{m{\bf E}_{\alpha+\beta-\gamma, \alpha+2}(|\lambda|)}{|M|} +\frac{|M_{21}|m{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(|\lambda|)}{|M|} \right. \\ &&\left.+\frac{[|M_{21}|{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda)+|M_{31}|{\bf E}_{\alpha+\beta-\gamma, 2}(\lambda)]\frac{1}{2}m }{|M|}\right]\phi_{I3}(r_0) \\ && +\left[\frac{|M_{21}|m{\bf E}_{\alpha+\beta-\gamma, \alpha}(|\lambda|)}{|M|} +\frac{m{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(|\lambda|)}{|M|} \right. \\ &&\left. +\frac{|M_{31}|m{\bf E}_{\alpha+\beta-\gamma, 1}(|\lambda|)}{|M|}\right]\phi_{I4}(r_0) \\ && +\left[\frac{1}{|M|}||P||{\bf E}_{\alpha+\beta-\gamma, \alpha+2}(|\lambda|) +\frac{|M_{21}|||P||{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(|\lambda|)}{|M|}\right.\\ && \left.+\frac{[|M_{21}|{\bf E}_{\alpha+\beta-\gamma, \alpha+1}(\lambda) +|M_{31}|{\bf E}_{\alpha+\beta-\gamma, 2}(\lambda)]\frac{1}{2}||P||}{|M|} \right. \\ &&\left.+\frac{|M_{31}|||P||{\bf E}_{\alpha+\beta-\gamma, 2}(|\lambda|)}{|M|}\right]\phi_{f}(r_0)\\ & = &Q_{11}\phi_{I1}(r_0)+Q_{21}\phi_{I2}(r_0)+Q_{31}\phi_{I3}(r_0)+Q_{41}\phi_{I4}(r_0)+Q_{51}\phi_{f}(r_0), \end{eqnarray*} $

$ \begin{eqnarray*} |c_{0, 2}(I_1, I_2, I_3, I_4, f)|&\le& Q_{12}\phi_{I1}(r_0)+Q_{22}\phi_{I2}(r_0)+Q_{32}\phi_{I3}(r_0)+Q_{42}\phi_{I4}(r_0)+Q_{52}\phi_{f}(r_0), \\ |d_{0, 1}(I_1, I_2, I_3, I_4, f)|&\le& \frac{m }{2}\phi_{I3}(r_0)+\frac{||P||}{2}\phi_{f}(r_0) \\ & = &Q_{14}\phi_{I1}(r_0)+Q_{24}\phi_{I2}(r_0)+Q_{34}\phi_{I3}(r_0)+Q_{44}\phi_{I4}(r_0)+Q_{54}\phi_{f}(r_0), \\ |d_{0, 2}(I_1, I_2, I_3, I_4, f)|&\le& Q_{13}\phi_{I1}(r_0)+Q_{23}\phi_{I2}(r_0)+Q_{31}\phi_{I3}(r_0)+Q_{43}\phi_{I4}(r_0)+Q_{53}\phi_{f}(r_0). \end{eqnarray*} $

因此对$ t\in (t_v, t_{v+1}], v\in {\sf Z}\!\!\!{\sf Z}_0^m $,我们有

$ \begin{eqnarray*} (t-t_v)^{2-\beta}|(Tx)(t)| &\le& ({\bf E}_{\alpha+\beta-\gamma, \beta}(|\lambda |)Q_{11} +{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(|\lambda|)Q_{14} +{\bf E}_{\alpha+\beta-\gamma, \beta-1}(|\lambda |)Q_{12}\\ && +{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-1}(|\lambda|) Q_{13} +m{\bf E}_{\alpha+\beta-\gamma, \beta-1}(|\lambda|))\phi_{I1}(r_0) \\ && +({\bf E}_{\alpha+\beta-\gamma, \beta}(|\lambda |)Q_{21} +{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(|\lambda|)Q_{24} +{\bf E}_{\alpha+\beta-\gamma, \beta-1}(|\lambda |)Q_{22}\\ && +{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-1}(|\lambda|)Q_{23} +m{\bf E}_{\alpha+\beta-\gamma, \beta}(|\lambda|))\phi_{I2}(r_0)\\ && +({\bf E}_{\alpha+\beta-\gamma, \beta}(|\lambda |)Q_{31} +{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(|\lambda|)Q_{34} +{\bf E}_{\alpha+\beta-\gamma, \beta-1}(|\lambda |)Q_{32}\\ && +{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-1}(|\lambda|)Q_{31} +m{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(|\lambda|))\phi_{I3}(r_0) \\ && +(m{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-1}(|\lambda|) +{\bf E}_{\alpha+\beta-\gamma, \beta}(|\lambda |)Q_{41} +{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(|\lambda|)Q_{44}\\ && +{\bf E}_{\alpha+\beta-\gamma, \beta-1}(|\lambda |)Q_{42} +{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-1}(|\lambda|)Q_{43})\phi_{I4}(r_0) \\ && +({\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-1}(|\lambda|)Q_{53} +{\bf E}_{\alpha+\beta-\gamma, \beta-1}(|\lambda |)Q_{52} \\ && +{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(|\lambda|)Q_{54} +{\bf E}_{\alpha+\beta-\gamma, \beta}(|\lambda |)Q_{51}\\ && +||P||{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(|\lambda|))\phi_f(r_0)\\ &\le& r_0. \end{eqnarray*} $

于是$ T\Omega_0\subset\Omega_0 $.由Schauder不动点定理立得$ T $在$ \Omega_0 $内存在至少一个不动点,这个不动点是BVP(1.1)的分片连续解.定理3.1证明完毕.

推论3.1  假设(a)–(d)成立且存在常数$ M_f, M_{I}\ge 0 $使得

$ |f(t, (t-t_i)^{\beta-2}x)|\le M_f, t\in (t_i, t_{i+1}], i\in {\Bbb N} _0^m, x\in {{\Bbb R}} , $

$ |I_j(t_i, (t_i-t_{i-1})^{\beta-2}x)|\le M_{Ij}, i\in {\Bbb N} _1^m, j\in {\Bbb N} _1^4, x\in {{\Bbb R}} . $

则BVP(1.1)至少由一个分片连续解.

证  取$ \phi_f(x) = M_f $, $ \phi_{Ij}(x) = M_{Ij} $.易看出方程(3.1)有正解$ r_0 $.由定理3.1立得结论.证明完毕.

本节给出例题说明所获得的结论(引理2.2)应用,并说明针对不同的$ \alpha, \beta, \gamma $,应该采用不同的边界条件.

例4.1  考虑下面方程

(4.1) $ \begin{eqnarray} [D_{0^+}^{\frac{3}{2}} D_{0^+}^{\frac{5}{4}} - D_{0^+}^{\frac{1}{2}}] x(t) = 0, \;t\in (0, 1]. \end{eqnarray} $

由引理2.2,容易知道方程(4.1)有解

(4.2) $ \begin{eqnarray} x(t)& = &y_1t^{\beta-1}{\bf E}_{\alpha+\beta-\gamma, \beta}( t^{\alpha+\beta-\gamma})+y_2t^{\beta-2}{\bf E}_{\alpha+\beta-\gamma, \beta-1}( t^{\alpha+\beta-\gamma}) {}\\ &&+x_1t^{\alpha+\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(\lambda t^{\alpha+\beta-\gamma})+x_2t^{\alpha+\beta-2}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-1}(\lambda t^{\alpha+\beta-\gamma}), \;t\in (0, 1], {}\\ \end{eqnarray} $

其中$ x_1, x_2, y_1, y_2 $为常数.

与方程(2.1)相比,在方程(4.1)中, $ \alpha = \frac{3}{2}, \beta = \frac{5}{4} $, $ \gamma = \frac{1}{2} $其中$ n = l = 2 $, $ k = 1 $, $ \lambda = 1 $, $ P(t)\equiv0 $.由(4.2)式计算得到

$ \begin{eqnarray*} D_{0^+}^\beta x(t)& = &\left[y_1\sum\limits_{\chi = 0}^\infty\frac{t^{\chi(\alpha+\beta-\gamma)+1}}{\Gamma( \chi(\alpha+\beta-\gamma)+2)} +y_2\sum\limits_{\chi = 0}^\infty\frac{t^{\chi(\alpha+\beta-\gamma)}}{\Gamma( \chi(\alpha+\beta-\gamma)+1)}\right. \\ &&\left.+x_1\sum\limits_{\chi = 0}^\infty\frac{t^{\chi(\alpha+\beta-\gamma)+\alpha+1}}{\Gamma( \chi(\alpha+\beta-\gamma)+\alpha+2)}+x_2\sum\limits_{\chi = 0}^\infty\frac{t^{\chi(\alpha+\beta-\gamma)+\alpha}}{\Gamma( \chi(\alpha+\beta-\gamma)+\alpha+1)}\right]'' \\ & = &y_1t^{\alpha+\beta-\gamma-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(t^{\alpha+\beta-\gamma}) +y_2t^{\alpha+\beta-\gamma-2}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma-1}(t^{\alpha+\beta-\gamma}) \\ &&+x_1t^{\alpha-1}{\bf E}_{\alpha+\beta-\gamma, \alpha}(t^{\alpha+\beta-\gamma})+x_2t^{\alpha-2}{\bf E}_{\alpha+\beta-\gamma, \alpha-1}(t^{\alpha+\beta-\gamma}), \end{eqnarray*} $

$ \begin{eqnarray*} I_{0^+}^{1-\gamma}x(t)& = &y_1t^{\beta-\gamma}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+1}( t^{\alpha+\beta-\gamma})+y_2t^{\beta-\gamma-1}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma}( t^{\alpha+\beta-\gamma}) \\ &&+x_1t^{\alpha+\beta-\gamma-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(\lambda t^{\alpha+\beta-\gamma})+x_2t^{\alpha+\beta-\gamma-2}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma-1}(\lambda t^{\alpha+\beta-\gamma}), \end{eqnarray*} $

且

$ \begin{eqnarray*} D_{0^+}^{\alpha-1}D_{0^+}^{\beta }x(t) & = &y_1t^{\beta-\gamma}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+1}(t^{\alpha+\beta-\gamma}) +y_2t^{\beta-\gamma-1}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma}(t^{\alpha+\beta-\gamma})\\ &&+x_1{\bf E}_{\alpha+\beta-\gamma, 1}(t^{\alpha+\beta-\gamma})+x_2t^{\alpha+\beta-\gamma-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(t^{\alpha+\beta-\gamma}). \end{eqnarray*} $

由$ \beta-\gamma-1 = -\frac{1}{4}<0 $容易看出$ D_{0^+}^{\alpha-1}D_{0^+}^{\beta }x(t) $和$ I_{0^+}^{1-\gamma}x(t) $在$ t = 0 $均不连续.但是$ (D_{0^+}^{\alpha-1}D_{0^+}^{\beta }-I_{0^+}^{1-\gamma})x(t) $在$ t = 0 $连续.所以反周期边界条件$ (D_{0^+}^{\alpha-1}D_{0^+}^{\beta }-I_{0^+}^{1-\gamma})x(0) = -(D_{0^+}^{\alpha-1}D_{0^+}^{\beta }-I_{0^+}^{1-\gamma})x(1) $是合理的.

例4.2  考虑下面方程

(4.3) $ \begin{eqnarray} [D_{0^+}^{\frac{3}{2}} D_{0^+}^{\frac{5}{4}} - D_{0^+}^{\frac{3}{2}} ] x(t) = 0, \;t\in (0, 1]. \end{eqnarray} $

由引理2.2知道方程(4.3)有解

(4.4) $ \begin{eqnarray} x(t)& = &y_1t^{\beta-1}{\bf E}_{\alpha+\beta-\gamma, \beta}( t^{\alpha+\beta-\gamma})+y_2t^{\beta-2}{\bf E}_{\alpha+\beta-\gamma, \beta-1}( t^{\alpha+\beta-\gamma}) {}\\ &&+x_1t^{\alpha+\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(\lambda t^{\alpha+\beta-\gamma})+x_2t^{\alpha+\beta-2}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-1}(\lambda t^{\alpha+\beta-\gamma}), \;t\in (0, 1], {}\\ \end{eqnarray} $

其中$ x_1, x_2, y_1, y_2 $是常数.

与方程(2.1)相比,在(4.3)式中, $ \alpha = \frac{3}{2}, \beta = \frac{5}{4} $, $ \gamma = \frac{3}{2} $其中$ n = l = k = 2 $, $ \lambda = 1 $, $ P(t)\equiv0 $.由(4.4)式,直接计算得到

$ \begin{eqnarray*} I_{0^+}^{2-\gamma}x(t)& = &y_1t^{\beta-\gamma+1}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+2}( t^{\alpha+\beta-\gamma})+y_2t^{\beta-\gamma}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+1}( t^{\alpha+\beta-\gamma}) \\ &&+x_1t^{\alpha+\beta-\gamma+1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta- \gamma+2}(\lambda t^{\alpha+\beta-\gamma})\\ &&+x_2t^{\alpha+\beta-\gamma}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma+1}(\lambda t^{\alpha+\beta-\gamma}), \;t\in (0, 1], \\ D_{0^+}^{\gamma-1}x(t) & = &y_1t^{\beta-\gamma}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma}( t^{\alpha+\beta-\gamma})+y_2t^{\beta-\gamma-1}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma}( t^{\alpha+\beta-\gamma}) \\ &&+x_1t^{\alpha+\beta-\gamma}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta -\gamma+1}(\lambda t^{\alpha+\beta-\gamma})\\ &&+x_2t^{\alpha+\beta-\gamma-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(\lambda t^{\alpha+\beta-\gamma}), \;t\in (0, 1], \\ D_{0^+}^\beta x(t) & = &y_1t^{\alpha+\beta-\gamma-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(t^{\alpha+\beta-\gamma}) +y_2t^{\alpha+\beta-\gamma-2}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma-1}(t^{\alpha+\beta-\gamma}) \\ &&+x_1t^{\alpha-1}{\bf E}_{\alpha+\beta-\gamma, \alpha}(t^{\alpha+\beta-\gamma}) +x_2t^{\alpha-2}{\bf E}_{\alpha+\beta-\gamma, \alpha-1}(t^{\alpha+\beta-\gamma}), \end{eqnarray*} $

$ \begin{eqnarray*} I_{0^+}^{2-\alpha}D_{0^+}^{\beta }x(t) & = &y_1t^{\beta-\gamma+1}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+2}(t^{\alpha+\beta-\gamma}) +y_2t^{\beta-\gamma}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+1}(t^{\alpha+\beta-\gamma}) \\ &&+x_1t{\bf E}_{\alpha+\beta-\gamma, 2}(t^{\alpha+\beta-\gamma})+x_2{\bf E}_{\alpha+\beta-\gamma, 1}(t^{\alpha+\beta-\gamma}), \end{eqnarray*} $

且

$ \begin{eqnarray*} D_{0^+}^{\alpha-1}D_{0^+}^{\beta }x(t) & = &y_1t^{\beta-\gamma}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+1}(t^{\alpha+\beta-\gamma}) +y_2t^{\beta-\gamma-1}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma}(t^{\alpha+\beta-\gamma}) \\ &&+x_1{\bf E}_{\alpha+\beta-\gamma, 1}(t^{\alpha+\beta-\gamma})+x_2t^{\alpha+\beta-\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(t^{\alpha+\beta-\gamma}). \end{eqnarray*} $

容易看出$ D_{0^+}^{\alpha-1}D_{0^+}^{\beta }x(t) $, $ D_{0^+}^{\gamma-1}x(t) $, $ I_{0^+}^{2-\alpha} D_{0^+}^\beta x $和$ I_{0^+}^{2-\gamma})x $在$ t = 0 $不连续.但是$ (D_{0^+}^{\alpha-1}D_{0^+}^{\beta }-I_{0^+}^{\gamma-1})x(t) $和$ I_{0^+}^{2-\alpha} D_{0^+}^\beta -I_{0^+}^{2-\gamma})x $在$ t = 0 $连续.因此采用如下反周期边界条件

$ (I_{0^+}^{2-\alpha} D_{0^+}^\beta - I_{0^+}^{2-\gamma})x(0) = -(I_{0^+}^{2-\alpha} D_{0^+}^\beta - I_{0^+}^{1-\gamma})x(1), $

$ (D_{0^+}^{\alpha-1}D_{0^+}^\beta - D_{0^+}^{\gamma-1})x(0) = -(D_{0^+}^{\alpha-1}D_{0^+}^\beta - D_{0^+}^{\gamma-1})x(1). $

例4.3  考虑如下方程

(4.5) $ \begin{eqnarray} [D_{0^+}^{\frac{1}{2}} D_{0^+}^{\frac{3}{4}} - D_{0^+}^{\frac{9}{8}} ] x(t) = 0, \;t\in (0, 1]. \end{eqnarray} $

由引理2.2知道(4.5)式有如下解

(4.6) $ \begin{eqnarray} x(t)& = &y_1t^{\beta-1}{\bf E}_{\alpha+\beta-\gamma, \beta}( t^{\alpha+\beta-\gamma})+y_2t^{\beta-2}{\bf E}_{\alpha+\beta-\gamma, \beta-1}( t^{\alpha+\beta-\gamma}) {}\\ && +x_1t^{\alpha+\beta-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta}(\lambda t^{\alpha+\beta-\gamma})+x_2t^{\alpha+\beta-2}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-1}(\lambda t^{\alpha+\beta-\gamma}), \;t\in (0, 1], {}\\ \end{eqnarray} $

其中$ x_1, x_2, y_1, y_2 $是常数.

与方程(2.1)相比,在方程(4.5)中, $ \alpha = \frac{1}{2}, \beta = \frac{3}{4} $, $ \gamma = \frac{9}{8} $其中$ n = l = 1 $, $ k = 2 $, $ \lambda = 1 $, $ P(t)\equiv0 $.

$ \begin{eqnarray*} D_{0^+}^\beta x(t)& = &y_1t^{\alpha+\beta-\gamma-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(t^{\alpha+\beta-\gamma}) +y_2t^{\alpha+\beta-\gamma-2}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma-1}(t^{\alpha+\beta-\gamma}) \\ &&+x_1t^{\alpha-1}{\bf E}_{\alpha+\beta-\gamma, \alpha}(t^{\alpha+\beta-\gamma})+x_2t^{\alpha-2}{\bf E}_{\alpha+\beta-\gamma, \alpha-1}(t^{\alpha+\beta-\gamma}), \\ D_{0^+}^{\gamma-1}x(t)& = &y_1t^{\beta-\gamma}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+1}( t^{\alpha+\beta-\gamma})+y_2t^{\beta-\gamma-1}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma}( t^{\alpha+\beta-\gamma}) \\ &&+x_1t^{\alpha+\beta-\gamma-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(\lambda t^{\alpha+\beta-\gamma})+x_2t^{\alpha+\beta-\gamma-2}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma-1}(\lambda t^{\alpha+\beta-\gamma}), \end{eqnarray*} $

且

$ \begin{eqnarray*} I_{0^+}^{1-\alpha}D_{0^+}^{\beta }x(t) & = &y_1t^{\beta-\gamma}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma+1}(t^{\alpha+\beta-\gamma}) +y_2t^{\beta-\gamma-1}{\bf E}_{\alpha+\beta-\gamma, \beta-\gamma}(t^{\alpha+\beta-\gamma}) \\ &&+x_1{\bf E}_{\alpha+\beta-\gamma, 1}(t^{\alpha+\beta-\gamma})+x_2t^{\alpha+\beta-\gamma-1}{\bf E}_{\alpha+\beta-\gamma, \alpha+\beta-\gamma}(t^{\alpha+\beta-\gamma}). \end{eqnarray*} $

容易看出$ I_{0^+}^{1-\alpha}D_{0^+}^{\beta }x(t) $和$ D_{0^+}^{\gamma-1}x(t) $在$ t = 0 $不连续.但是$ (I_{0^+}^{1-\alpha}D_{0^+}^{\beta }-D_{0^+}^{\gamma-1})x(t) $在$ t = 0 $连续.因此对方程(4.5),我们使用下面的边界条件:

$ (I_{0^+}^{1-\alpha}D_{0^+}^\beta - D_{0^+}^{\gamma-1})x(0) = (I_{0^+}^{1-\alpha}D_{0^+}^\beta - D_{0^+}^{\gamma-1})x(1), \;\; I_{0^+}^{2-\gamma} x(0) = I_{0^+}^{2-\gamma} x(1). $

注4.4  设$ \alpha, \beta, \gamma\in (1, 2) $满足$ \gamma<\alpha+\beta $.读者可以考虑下面的边值问题:

(4.7) $ \begin{eqnarray} \left\{ \begin{array}{l} [D_{0^+}^\alpha D_{0^+}^\beta -\lambda D_{0^+}^\gamma ] x(t) = P(t)f\left(t, x(t) \right), \ {\rm a.e., }\ t\in (t_i, t_{i+1}], i\in {\Bbb N} _0^{m}, \\ \Delta I_{0^+}^{2-\beta} x(t_i) = I_1(t_i, x(t_i)), i\in {\Bbb N} _1^{m-1}, \\ \Delta D_{0^+}^{\beta-1}x(t_i) = I_2(t_i, x(t_i)), i\in {\Bbb N} _1^{m-1}, \\ \Delta (I_{0^+}^{2-\alpha} D_{0^+}^\beta -\lambda I_{0^+}^{2-\gamma})x(t_i) = I_3(t_i, x(t_i)), i\in {\Bbb N} _1^{m}, \\ \Delta (D_{0^+}^{\alpha-1}D_{0^+}^\beta x-\lambda D_{0^+}^{\gamma-1})x(t_i) = I_4(t_i, x(t_i)), i\in {\Bbb N} _1^{m}, \\ I_{0^+}^{2-\beta} x(0) = -I_{0^+}^{2-\beta}x(1), \;\Delta D_{0^+}^{\beta-1}x(0) = -\Delta D_{0^+}^{\beta-1}x(1), \\ (I_{0^+}^{2-\alpha} D_{0^+}^\beta -\lambda I_{0^+}^{2-\gamma})x(0) = -(I_{0^+}^{2-\alpha} D_{0^+}^\beta -\lambda I_{0^+}^{2-\gamma})x(1), \\ (D_{0^+}^{\alpha-1}D_{0^+}^\beta x-\lambda D_{0^+}^{\gamma-1})x(0) = -(D_{0^+}^{\alpha-1}D_{0^+}^\beta x-\lambda D_{0^+}^{\gamma-1})x(1). \end{array}\right. \end{eqnarray} $

边值问题(4.7)是梁方程反周期边值问题的推广,其边界条件与BVP(1.1)中边界条件不同.