分数阶微积分应用广泛, 例如力学和电特性刻画、地震分析、材料的记忆性、电路、电解化学等, 参见文献[1-2].近年来, 分数阶微分方程得到迅速发展, 参见文献[3-12].

在左右分数阶导数$^{c}\!D_{a^+}^{\alpha}x$和$^{c}\!D_{b^-}^{\alpha}x$中, $a, b$分别被称为左右基点, 统称为分数阶导数的基点.含有至少两个基点的分数阶微分方程被称为多基点分数阶微分方程, 而只含有一个基点的分数阶微分方程被称为单基点分数阶微分方程, 参见文献[10].在本文中, 我们研究下面的非线性多基点分数阶微分方程的脉冲边值问题

$ \left\{ \begin{align} & cD_{*}^{\alpha }x(t)=q(t)f(t, x(t), {{}^{c}}D_{*}^{\gamma }x(t)), \rm{a}\rm{.e}\rm{.}t\in {J}':=J\backslash \{{{t}_{1}}, {{t}_{2}}, \cdots, {{t}_{m}}\}, J=(0, 1], \ \ \ \ \ \left( 1.1 \right) \\ & \Delta x({{t}_{k}})={{I}_{k}}, \quad \Delta {x}'({{t}_{k}})={{{\tilde{I}}}_{k}}, k=1, 2, \cdots, m, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( 1.2 \right) \\ & x({{0}^{+}})=0, \quad c{x}'({{0}^{+}})+{{d}^{c}}D_{t_{m}^{+}}^{\beta }x(1)=0.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( 1.3 \right) \\ & \\ \end{align} \right. $

其中$\alpha\in (1, 2)$, $\beta, \gamma, \alpha-\beta\in (0, 1)$, $\alpha-\gamma\in (1, \, 2)$, $q:(0, \, \infty)\rightarrow {\Bbb R}$满足:存在$l\in(-\frac{\gamma}{2}, \, 0]$使得对$t\in (0, \, 1]$有$|q(t)|\leq t^l$, $q$在$t=0$处可以是奇异的. $^{c}\!D_{*}$表示基点$t=t_k(k=0, 1, \cdots, m)$处的Caputo分数阶导数, 即对$t\in (t_k, \, t_{k+1}]$有$^{c}\!D_{*}^{\alpha}|_{(t_k, \, t_{k+1}]} x(t)=^{c}\!\!D_{t_k^+}^{\alpha}x(t)$.脉冲时刻$\{t_k\}$满足$0=t_0 < t_1 < \cdots < t_m < t_{m+1}=1$, $\Delta x(t_k)$表示函数$x$在$t_k$处的跳跃, 定义为$\Delta x(t_k) = x(t_k^+)- x(t_k^-)$, 其中$x(t_k^+), \, x(t_k^-)$分别表示$x(t)$在$t=t_k$处的右极限和左极限, $I_k, \tilde{I}_k, c, d$均为常数, $I_k, \tilde{I}_k$表示跳跃度.

近期一些学者研究了具有Caputo分数阶导数的脉冲微分方程的边值问题, 参见文献[4, 8-9, 12-17].他们研究了方程

$ \begin{equation} \label{eq1.4} ^{c}\!D_{t}^{\alpha} x(t)=g(t, \, x(t)), \quad t\in J_k:=(t_k, \, t_{k+1}], \, \, k=0, 1, \cdots, m \end{equation} $

(1.4)

在不同脉冲和边值条件下解的存在性.在本文中, 我们将在$g$较弱的假设下给出引理3.1-3.4的详细证明.据我们所知, 关于具有脉冲和边值条件的$\alpha(\in(1, 2))$阶多基点分数微分方程(1.1)-(1.3)解的存在性的研究尚属少见.

本文结构安排如下:第二节, 回顾绝对连续、分数阶积分、分数阶导数的相关概念, 并给出一些性质; 第三节, 给出解的定义; 第四节, 证明解的存在性; 第五节, 给出三个例子说明主要结果的应用.

$[a, b]$上的绝对连续函数构成的空间记作$AC([a, b], \, {\Bbb R})$, 满足$f \in C^{n-1}([a, b], \, {\Bbb R})$和$f^{(n-1)} \in AC([a, b], \, {\Bbb R})$的函数$f$构成的空间记作$AC^n([a, b], \, {\Bbb R}) (n =1, 2, \cdots)$, 特别地, $AC^1([a, b], \, {\Bbb R})=AC([a, b], \, {\Bbb R})$.周知, $w \in AC([a, b], \, {\Bbb R})$当且仅当存在$\phi\in L((a, b), \, {\Bbb R})$和$\overline{c}\in {\Bbb R}$使得

$ \begin{equation} \label{eq2.1}w(t)=\overline{c}+\int_a^t \phi(s){\rm d}s, \quad t\in [a, b], \end{equation} $

(2.1)

因此, 绝对连续函数$w(t)$在$[a, b]$上几乎处处存在$w'(t)=\phi(t)$.由(2.1)式可得

$ w(t)=w(a)+\int_a^t w'(s){\rm d}s. $

定义2.1^[3]  设$\eta>0$, 如果$g(t)\in L^1([a, \, +\infty), \, {\Bbb R})$, 那么

$ (I_{a^+}^{\eta}g)(t)=\frac{1}{\Gamma(\eta)}\int_{a}^t (t-s)^{\eta-1}g(s){\rm d}s, \quad t>a, $

称为$g$的具有积分下限$a$的$\eta$阶Riemann-Liouville分数阶积分, 其中$\Gamma(\cdot)$是$\Gamma$函数.

定义2.2  如果$g(t)\in AC^n([a, b], \, {\Bbb R})$, 那么$\eta (n-1<\eta<n)$阶Riemann-Liouville分数阶导数$^{L}\!D_{a^+}^{\eta}g(t)$在$[a, \, b]$上几乎处处存在, 记作

$ (^{L}\!D_{a^+}^{\eta}g)(t)=\frac{1}{\Gamma(n-\eta)}\frac{ {\rm d}^n}{{\rm d}t^n}\int_a^t(t-s)^{n-\eta-1}g(s){\rm d}s, \quad t> a, \, \, n-1<\eta<n. $

定义2.3^[3]  如果$g(t)\in AC^n([a, b], \, {\Bbb R})$, 那么$\eta (n-1<\eta<n)$阶Caputo分数阶导数$(^{c}D_{a^+}^{\eta}g)(t)$在$[a, \, b]$上几乎处处存在, 可以写作

$ (^{c}D_{a^+}^{\eta}g)(t)=\bigg(\, ^{L}\!D_{a^+}^{\eta}\bigg[g(s)-\sum\limits_{k=0}^{n-1}\frac{g^{(k)}(a)}{k!}(s-a)^k\bigg]\bigg)(t), \quad t> a, \, \, n-1<\eta<n, $

此外, 如果$g(a)=g'(a)=\cdots=g^{(n-1)}(a)=0$, 那么$(^{c}D_{a^+}^{\eta}g)(t)=(^{L}\!D_{a^+}^{\eta}g)(t)$.

注2.1^[3]  如果$g(t)\in C^n([a, b], \, {\Bbb R})$, 那么$(^{c}D_{a^+}^{\eta}g)(t)\in C([a, b], \, {\Bbb R})$, $n-1<\eta<n$.

下面我们给出分数阶微积分的一些性质.

性质2.1^[3]  $t^{s}$的$\nu(n-1< \nu <n)$阶Caputo分数阶导数为

$ ^{c}\!D_{a^+}^{\nu}t^{s}=\left\{ \begin{array}{ll} \frac{\Gamma(s+1)}{\Gamma(s-\nu+1)}(t-a)^{s-\nu}, &s\in {\Bbb N}, \, s\geq n\, \, \mbox{or}\, \, s \notin {\Bbb N}, s>n-1, \\[2mm] 0, &s\in\{0, 1, \cdots, n-1\}. \end{array}\right. $

性质2.2^[3]  如果$\eta_2>\eta_1>0$, $f\in L^{p}([a, b], \, {\Bbb R})(1\leq p\leq \infty)$, 那么

$ (^{c}D_{a^+}^{\eta_1}I_{a^+}^{\eta_2}f)(t)=(I_{a^+}^{\eta_2-\eta_1}f)(t). $

引理2.1  设$\eta>0$, 分数阶微分方程$^{c}D^{\eta}_{a^+}u(t)=0$的解为

$ u(t)=c_0+c_1(t-a)+c_2(t-a)^2+\cdots+c_{n-1}(t-a)^{n-1}, $

其中$c_i\in {\Bbb R}, \, i=0, 1, 2, \cdots, n-1(n=[\eta]+1)$, $[\eta]$为实数$\eta$的整数部分.

若用${\bf B}(\cdot, \, \cdot)$表示${\bf B}$函数, 我们可得到如下结果.

引理2.2  (1)当$p>0, \, q>0$时

$ \int_0^t (t-s)^{p-1}s^{q-1}{\rm d}s=t^{p+q-1}{\bf B}(p, \, q). $

(2) 当$p>0, \, q>0, \, 0<a<t$时

$ \int_a^t (t-s)^{p-1}(s-a)^{q-1}{\rm d}s=(t-a)^{p+q-1}{\bf B}(p, \, q). $

(3) 当$p>0, \, 0<q\leq1$, $0<a<t$时

$ \int_a^t (t-s)^{p-1}s^{q-1}{\rm d}s\leq(t-a)^{p+q-1}{\bf B}(p, \, q). $

证  (1)

$ \int_0^t (t-s)^{p-1}s^{q-1}{\rm d}s=t^{p+q-1}\int_0^1 (1-u)^{p-1}u^{q-1}{\rm d}u=t^{p+q-1}{\bf B}(p, \, q) $

(2) 令$\tau=s-a$, 则

$ \displaystyle{\int_a^t (t-s)^{p-1}(s-a)^{q-1}{\rm d}s=\int_0^{t-a}(t-a-\tau)^{p-1}{\tau}^{q-1}{\rm d}\tau=(t-a)^{p+q-1}{\bf B}(p, \, q)}. $

(3) 由${\bf B}(p, \, q)=\int_0^1 (1-s)^{p-1}s^{q-1}{\rm d}s$可得

$ \begin{eqnarray*} \int_a^t (t-s)^{p-1}s^{q-1}{\rm d}s&=&\int_0^{t-a} (t-a-s)^{p-1} (a+s)^{q-1}{\rm d}s\\ &=&(t-a)^{p+q-1}\int_0^{t-a} \Big(1-\frac{s}{t-a}\Big)^{p-1} \Big(\frac{s}{t-a}\Big)^{q-1}\Big(\frac{a+s}{s}\Big)^{q-1} {\rm d}\Big(\frac{s}{t-a}\Big)\\ &\leq&(t-a)^{p+q-1}{\bf B}(p, \, q). \end{eqnarray*} $

证毕.

引理2.3  如果$\eta\in (1, 2)$, 那么$I_{a^+}^{\eta}:AC([a, b], \, {\Bbb R})\rightarrow AC^2([a, b], \, {\Bbb R})$.

证  设$f\in AC([a, b], \, {\Bbb R})$, 即$f(t)=f(a)+\int_a^t f'(s){\rm d}s, \, t\in [a, b]$, 那么

$ (I_{a^+}^{\eta}f)(t)=\frac{1}{\Gamma(\eta)}\int_a^t(t-s)^{\eta-1} \bigg(\int_a^sf'(\tau){\rm d}\tau\bigg){\rm d}s+\frac{(t-a)^{\eta} f(a)}{\Gamma(\eta+1)}, $

由此可得

$ \begin{eqnarray} \label{eq2.2}((I_{a^+}^{\eta}f)(t))'=\frac{1}{\Gamma(\eta-1)}\int_a^t(t-s)^{\eta-2} \bigg(\int_a^sf'(\tau){\rm d}\tau\bigg){\rm d}s+\frac{(t-a)^{\eta-1} f(a)}{\Gamma(\eta)}. \end{eqnarray} $

(2.2)

易知, (2.2)式的右边在$[a, b]$上连续且右边第二项是绝对连续函数.因为

$ \begin{eqnarray*} &&\frac{1}{\Gamma(\eta-1)}\int_a^t(t-s)^{\eta-2} \bigg(\int_a^sf'(\tau){\rm d}\tau\bigg){\rm d}s\\ &=&\frac{1}{\Gamma(\eta)}\int_a^t f'(\tau)(t-\tau)^{\eta-1}{\rm d}\tau = \frac{1}{\Gamma(\eta-1)}\int_a^t\bigg(\int_a^s\frac{f'(\tau){\rm d}\tau}{(s-\tau)^{ 2-\eta}}\bigg){\rm d}s\in AC([a, b], \, {\Bbb R}), \end{eqnarray*} $

所以(2.2)式右边第一项也绝对连续.综上可得$I_{a^+}^{\eta}:AC([a, b], \, {\Bbb R})\rightarrow AC^2([a, b], \, {\Bbb R})$.

引理2.4  若$\varsigma\in (0, \, 1]$, $0<a\leq b$, 则$|a^{\varsigma}-b^{\varsigma}|\leq (b-a)^{\varsigma}$.

设$J_k=(t_k, \, t_{k+1}], \, k=0, 1, 2, \cdots, m$, 记

$ X=\Big\{x:(0, 1]\rightarrow {\Bbb R}: x\in C^1(J_k, \, {\Bbb R}), \, \, x(t_k^+), \, x'(t_k^+), \, ^{c}\!D_{t_k^+}^{\gamma}x(t_k^+) \mbox{ 存在 } \Big\}, $

定义范数

$ \|x\|_1:=\max\Big\{\sup\limits_{k=0, \, 1, \, \cdots, m}\sup\limits_{t \in J_k} |x(t)|, \, \sup\limits_{k=0, \, 1, \, \cdots, m}\sup\limits_{t \in J_k} |x'(t)|, \, \sup\limits_{k=0, \, 1, \, \cdots, m}\sup\limits_{t \in J_k} |^{c}\!D_{t_k^+}^{\gamma}x(t)|\Big\}, $

则$X$是实Banach空间.

设$f:J\times {\Bbb R}\times {\Bbb R}\rightarrow{\Bbb R}$满足下面的条件.

(H) 对所有$v, \, w\in {\Bbb R}$, $f(\cdot, v, w):J\rightarrow {\Bbb R}$可测, 对a.e. $t\in J$, $f(t, \cdot, \cdot):{\Bbb R}\times {\Bbb R}\rightarrow {\Bbb R}$连续, 存在函数$\mu\in L^{\frac{1}{\sigma}}(J, {\Bbb R}^+)(\sigma\in (0, \, \frac{\alpha-1}{2}))$使得

$ |f(t, v, w)|\leq \mu(t)(|v|^{\lambda_1}+|w|^{\lambda_2}), $

其中$0<\lambda_1<\lambda_2$是实数.

定义3.1  如果函数$x:(0, \, 1]\rightarrow {\Bbb R}$满足

(1) $x\in AC^2(J_k, \, {\Bbb R})$;

(2) 在$J_k$上有$^{c}\!D_{t_k^+}^{\alpha} x(t)=\!f(t, \, x(t), \, ^{c}\!D_{t_k^+}^{\gamma} x(t))$;

(3) 对$k=1, 2, \cdots, m$, 有

$ \Delta x(t_k)=I_k, \Delta x'(t_k)=\widetilde{I}_k \ \mbox{且}\ x(0^+)=0, \, cx'(0^+)+d ^{c}\!D_{t_m^+}^{\beta}x(1)=0, $

那么$x(t)$是方程(1.1)-(1.3)的解.

为了证明主要结果, 我们需要下面的引理.

引理3.1  若$\tau_2, \, \tau_1\in J_k(k=0, 1, 2, \cdots, m)$且$\tau_2<\tau_1$, 则当$\tau_2\rightarrow \tau_1$时有

$ \int_{t_k}^{\tau_2}[(\tau_2-s)^{\alpha-2}-(\tau_1-s)^{\alpha-2}]s^l\mu(s){\rm d}s\rightarrow 0. $

证  情形1  当$0<\tau_2<\tau_1\leq t_{1}$时, 注意到$s^l$在$t=0$可以是奇异的, 由Hölder不等式可得

$ \begin{eqnarray*} &&\left|\int_{0}^{\tau_2}\big[(\tau_2-s)^{\alpha-2}-(\tau_1-s)^{\alpha-2}\big]s^l\mu(s){\rm d}s\right|\\ &\leq& \bigg(\int_{0}^{\tau_2}\big{|}(\tau_2-s)^{\alpha-2}-(\tau_1-s)^{\alpha-2}\big{|}^{\frac{1}{1-\sigma}}s^{\frac{l}{1-\sigma}}{\rm d}s \bigg)^{1-\sigma}\cdot\|\mu\|_{L^{\frac{1}{\sigma}}}\\ &=&(2-\alpha)\|\mu\|_{L^{\frac{1}{\sigma}}} \bigg[\int_{0}^{\tau_2}\left|\int_{\tau_2}^{\tau_1}(\zeta-s)^{\alpha-3} {\rm d}\zeta\right|^{\frac{1}{1-\sigma}}s^{\frac{l}{1-\sigma}}{\rm d}s\bigg] ^{1-\sigma}\\ &\leq&M\bigg[\int_{0}^{\tau_2}((\tau_2-s)^{\theta}-(\tau_1-s)^{\theta})s^{\frac{l}{1-\sigma}}{\rm d}s\bigg]^{1-\sigma}\\ &=&M\bigg[\tau_2^{\theta+\frac{l}{1-\sigma}+1}\int_{0}^{1}(1-u)^{\theta}u^{\frac{l}{1-\sigma}}{\rm d}u-\tau_1^{\theta+\frac{l}{1-\sigma}+1}\int_{0}^{\frac{\tau_2}{\tau_1}}(1-u)^{\theta}u^{\frac{l}{1-\sigma}}{\rm d}u \bigg]^{1-\sigma}\\ &\leq&M\bigg[\big{|}\tau_2^{\theta+\frac{l}{1-\sigma}+1}-\tau_1^{\theta+\frac{l}{1-\sigma}+1}\big{|}\int_{0}^{1}(1-u)^{\theta}u^{\frac{l}{1-\sigma}}{\rm d}u+\tau_1^{\theta+\frac{l}{1-\sigma}+1}\int_{\frac{\tau_2}{\tau_1}}^1(1-u)^{\theta}u^{\frac{l}{1-\sigma}}{\rm d}u\bigg]^{1-\sigma}\\ &\rightarrow&0, \, \quad {\rm{当}}\tau_2\rightarrow \tau_1 {\rm{时}}, \end{eqnarray*} $

其中$M>0$是常数, $ {\theta=\frac{\alpha-2-\sigma}{1-\sigma}}\in (-1, \, 0)$且$ {\frac{l}{1-\sigma}>-1}$.

情形2  当$t_{k}<\tau_2<\tau_1\leq t_{k+1}(k=1, 2, \cdots, m)$时, 注意到对任意$t\in J_k(k=1, 2, \cdots, m)$有$|q(t)|\leq t^l\leq \overline{M}$, 由Hölder不等式可得

$ \begin{eqnarray*} &&\left|\int_{t_k}^{\tau_2}[(\tau_2-s)^{\alpha-2}-(\tau_1-s)^{\alpha-2}]s^l\mu(s){\rm d}s\right|\\ &\leq&\overline{M}\|\mu\|_{L^{\frac{1}{\sigma}}}\left[\int_{t_k}^{\tau_2}|(\tau_2-s)^{\alpha-2}-(\tau_1-s)^{\alpha-2}|^{\frac{1}{1-\sigma}}{\rm d}s\right]^{1-\sigma}\\ &=&(2-\alpha)\overline{M}\|\mu\|_{L^{\frac{1}{\sigma}}} \bigg(\int_{t_k}^{\tau_2}\left|\int_{\tau_2}^{\tau_1}(\zeta-s)^{\alpha-3}{\rm d}\zeta\right|^{\frac{1}{1-\sigma}}{\rm d}s\bigg) ^{1-\sigma}\\ &\leq&\overline{\overline{M}}\left[\int_{t_k}^{\tau_2}((\tau_2-s)^{\theta}-(\tau_1-s)^{\theta}){\rm d}s\right]^{1-\sigma}\\ &=&\frac{\overline{\overline{M}}}{(1+\theta)^{1-\sigma}} [(\tau_1-\tau_2)^{1+\theta}-(\tau_1-t_k)^{1+\theta} +(\tau_2-t_k)^{1+\theta}]^{1-\sigma}\\ &\rightarrow&0, \, \quad {\rm{当}}\tau_2\rightarrow \tau_1{\rm{时}}, \end{eqnarray*} $

其中$\overline{\overline{M}}>0$是常数, ${\theta=\frac{\alpha-2-\sigma}{1-\sigma}\in (-1, \, 0)}$.

为方便起见, 我们引入下面的结果.设$y>\sigma$, $t_{i}\in [0, 1](i=1, \cdots, m+1)$, 由Hölder不等式和引理2.2(3)可得

$ \begin{eqnarray} \label{eq3.1}\int_{t_{i-1}}^{t_i}(t_i-s)^{y-1}s^l \mu(s){\rm d}s&\leq& \bigg (\int_{t_{i-1}}^{t_i}((t_i-s)^{y-1}s^l)^{\frac{1}{1-\sigma}}{\rm d}s\bigg)^{1-\sigma}\|\mu\|_{L^{\frac{1}{\sigma}}}\\ &\leq&(t_i-t_{i-1})^{y+l-\sigma}({\bf B}(e_{y}, \, z))^{1-\sigma}\|\mu\|_{L^{\frac{1}{\sigma}}}, \end{eqnarray} $

(3.1)

其中${{\bf B}(e_{y}, \, z)={\bf B}(\frac{y-\sigma}{1-\sigma}, \, \frac{l+1-\sigma}{1-\sigma})}$.

如果$y=\alpha$或$y=\alpha-1$或$y=\alpha-\gamma$或$y=\alpha-\beta$, 那么$y>\sigma$且$y+l-\sigma>0$, 由(3.1)式可得

$ \begin{eqnarray} \label{eq3.2}\int_{t_{i-1}}^{t_i}(t_i-s)^{\alpha-1}s^l \mu(s){\rm d}s&\leq&(t_i-t_{i-1})^{\alpha+l-\sigma}({\bf B}(e_{\alpha}, \, z))^{1-\sigma}\|\mu\|_{L^{\frac{1}{\sigma}}}\\ &\leq&({\bf B}(e_{\alpha}, \, z))^{1-\sigma}\|\mu\|_{L^{\frac{1}{\sigma}}}, \\ \int_{t_{i-1}}^{t_i}(t_i-s)^{\alpha-2}s^l \mu(s){\rm d}s&\leq&(t_i-t_{i-1})^{\alpha-1+l-\sigma}({\bf B}(e_{\alpha-1}, \, z))^{1-\sigma}\|\mu\|_{L^{\frac{1}{\sigma}}}\\ &\leq&({\bf B}(e_{\alpha-1}, \, z))^{1-\sigma}\|\mu\|_{L^{\frac{1}{\sigma}}}, \\ \int_{t_{i-1}}^{t_i}(t_i-s)^{\alpha-\beta-1}s^l \mu(s){\rm d}s&\leq&(t_i-t_{i-1})^{\alpha-\beta+l-\sigma}({\bf B}(e_{\alpha-\beta}, \, z))^{1-\sigma}\|\mu\|_{L^{\frac{1}{\sigma}}}\\ &\leq&({\bf B}(e_{\alpha-\beta}, \, z))^{1-\sigma}\|\mu\|_{L^{\frac{1}{\sigma}}}, \\ \int_{t_{i-1}}^{t_i}(t_i-s)^{\alpha-\gamma-1}s^l \mu(s){\rm d}s&\leq&(t_i-t_{i-1})^{\alpha-\gamma+l-\sigma}({\bf B}(e_{\alpha-\gamma}, \, z))^{1-\sigma}\|\mu\|_{L^{\frac{1}{\sigma}}}\\ &\leq&({\bf B}(e_{\alpha-\gamma}, \, z))^{1-\sigma}\|\mu\|_{L^{\frac{1}{\sigma}}}. \end{eqnarray} $

(3.2)

若$\mu(s)\equiv M^*$是正常数, $y+l>0$, 则由引理2.2(3)可得

$ \begin{eqnarray} \label{eq3.3}\int_{t_{i-1}}^{t_i}(t_i-s)^{y-1}s^l \mu(s){\rm d}s\leq M^*(t_i-t_{i-1})^{y+l}{\bf B}(y, \, l+1)\leq M^*{\bf B}(y, \, l+1). \end{eqnarray} $

(3.3)

引理3.2  如果(H)成立, 那么对$x\in X$, $k=0, 1, \cdots, m$, $\varepsilon>0$有

$ \Big[I_{t_k^+}^{\alpha}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s))\Big](t)\in AC^2([t_k+\varepsilon, \, t_{k+1}], \, {\Bbb R}). $

证  对于$t\in[t_k+\varepsilon, \, t_{k+1}]$, 由(3.2)式可得

$ \begin{eqnarray} \label{eq3.4} &&\int_{t_k}^t|(t-s)^{\alpha-1}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s))|{\rm d}s\\ &\leq&\int_{t_k}^t(t-s)^{\alpha-1}s^l\mu(s){\rm d}s(\|x\|_1^{\lambda_1}+\|x\|_1^{\lambda_2})\\ &\leq& (t-t_k)^{\alpha+l-\sigma}({\bf B}(e_{\alpha}, \, z))^{1-\sigma}\|\mu\|_{L^{\frac{1}{\sigma}}}(\|x\|_1^{\lambda_1}+\|x\|_1^{\lambda_2}), \\ &&\int_{t_k}^t|(t-s)^{\alpha-2}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s))|{\rm d}s\\ &\leq&\int_{t_k}^t(t-s)^{\alpha-2}s^l\mu(s){\rm d}s(\|x\|_1^{\lambda_1}+\|x\|_1^{\lambda_2})\\ &\leq&(t-t_k)^{\alpha+l-\sigma-1}({\bf B}(e_{\alpha-1}, \, z))^{1-\sigma}\|\mu\|_{L^{\frac{1}{\sigma}}}(\|x\|_1^{\lambda_1}+\|x\|_1^{\lambda_2}), \end{eqnarray} $

(3.4)

这意味着对于$t\in [t_k+\varepsilon, \, t_{k+1}]$, $x\in X$有$(t-s)^ {\alpha-1}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s))$和$(t-s)^{\alpha -2}q(s)$ $f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s))$关于$s\in (t_k, \, t_{k+1}]$ Lebesgue可积.

易见

$\frac{\rm d}{{\rm d}t}\left(\Big[I_{t_k^+}^{\alpha}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s))\Big](t)\right)=\Big[I_{t_k^+}^{\alpha-1}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s))\Big](t) $

关于$t\in [t_k+\varepsilon, \, t_{k+1}]$连续.

此外, 我们有$\big[I_{t_k^+}^{\alpha-1}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s))\big](t)\in AC([t_k+\varepsilon, \, t_{k+1}], \, {\Bbb R})$.事实上, 对于$[t_k+\varepsilon, \, t_{k+1}]$上任意有限个互不相交的开区间$\{(a_i, \, b_i)\}_{1\leq i\leq n}$且$\sum\limits_{i=1}^n(b_i-a_i)\rightarrow 0$, 由(3.1)式和引理3.1可得

$ \begin{eqnarray*} &&\sum\limits_{i=1}^n \left|\int_{t_k}^{b_i}(b_i-s)^{\alpha-2}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s)){\rm d}s\right.\\&&-\left.\int_{t_k}^{a_i}(a_i-s)^{\alpha-2}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s)){\rm d}s\right|\\ &\leq&\sum\limits_{i=1}^n\left|\int_{a_i}^{b_i}(b_i-s)^{\alpha-2}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s)){\rm d}s\right|\\&&+ \sum\limits_{i=1}^n\int_{t_k}^{a_i}\left|[(b_i-s)^{\alpha-2}-(a_i-s)^{\alpha-2}]q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s))\right|{\rm d}s\\ &\leq&\left[\sum\limits_{i=1}^n\int_{a_i}^{b_i}(b_i-s)^{\alpha-2}s^l \mu(s){\rm d}s\right.\\ &&\left.+\sum\limits_{i=1}^n\int_{t_k}^{a_i}\left[(a_i-s)^{\alpha-2}-(b_i-s)^{\alpha-2}\right]s^l \mu(s){\rm d}s\right](\|x\|_1^{\lambda_1}+\|x\|_1^{\lambda_2})\\ &\leq&\left[\sum\limits_{i=1}^n(b_i-a_i)^{\alpha+l-\sigma-1}({\bf B}(e_{\alpha-1}, \, z))^{1-\sigma}\|\mu\|_{L^{\frac{1}{\sigma}}}\right.\\ &&\left.+\sum\limits_{i=1}^n\int_{t_k}^{a_i}\Big[(a_i-s)^{\alpha-2}-(b_i-s)^{\alpha-2}\Big]s^l \mu(s){\rm d}s\right](\|x\|_1^{\lambda_1}+\|x\|_1^{\lambda_2})\\ &\rightarrow& 0. \end{eqnarray*} $

因此, $\big[I_{t_k^+}^{\alpha-1}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s))\big](t)$在$[t_k+\varepsilon, \, t_{k+1}]$上绝对连续, 这表明

$ \big[I_{t_k^+}^{\alpha}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s))\big](t)\in AC^2([t_k+\varepsilon, \, t_{k+1}], \, {\Bbb R}). $

此外, 对几乎所有$t\in (t_k, \, t_{k+1}]$, $\big[\, ^{c}\!D_{t_k^+}^{\alpha}I_{t_k^+}^{\alpha}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s))\big](t)$存在.

引理3.3  如果(H)成立, 那么对于$x\in X$, $k=0, 1, \cdots, m$有

$ \Big[\, ^{c}\!D_{t_k^+}^{\alpha}I_{t_k^+}^{\alpha}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s))\Big](t)=q(t)f(t, \, x(t), \, ^{c}\!D_{t_k^+}^{\gamma}x(t)), \quad {\rm a.e.} \quad t\in J_k. $

证  记$F(\tau, \, s)=(t-\tau)^{1-\alpha}|\tau-s|^{\alpha-1}s^l \mu(s)$, 易见$F(\tau, \, s)$在$[t_k+\varepsilon, \, t]\times[t_k+\varepsilon, \, t]$上非负可测, 则有${\int_{t_k}^t(\int_{t_k}^t F(\tau, \, s){\rm d}s){\rm d}\tau =\int_{t_k}^t(\int_{t_k}^t F(\tau, \, s){\rm d}\tau){\rm d}s}. $由(3.1)式和引理2.2(2)可得

$ \begin{eqnarray*} &&\int_{t_k}^t\bigg(\int_{t_k}^t F(\tau, \, s){\rm d}s\bigg){\rm d}\tau\\ &=&\int_{t_k}^t(t-\tau)^{1-\alpha}\bigg[\int_{t_k}^{\tau}(\tau-s)^{\alpha-1}s^l \mu(s){\rm d}s +\int_{\tau}^t(s-\tau)^{\alpha-1}s^l \mu(s){\rm d}s\bigg]{\rm d}\tau\\ &\leq& \int_{t_k}^t(t-\tau)^{1-\alpha}(\tau-t_k)^{\alpha+l-\sigma}{\rm d}\tau\cdot({\bf B}(e_{\alpha}, \, z))^{1-\sigma}\|\mu\|_{L^{\frac{1}{\sigma}}} +\int_{t_k}^t \bigg(\int_{\tau}^{t}s^l \mu(s){\rm d}s\bigg){\rm d}\tau\\ &\leq& \bigg[(t-t_k)^{2+l-\sigma}({\bf B}(2-\alpha, \, \alpha+l+1-\sigma)) ({\bf B}(e_{\alpha}, \, z))^{1-\sigma} +\int_{t_k}^t\bigg(\int_{\tau}^{t}s^{\frac{l}{1-\sigma}}{\rm d}s\bigg)^{1-\sigma} {\rm d}\tau\bigg]\|\mu\|_{L^{\frac{1}{\sigma}}}. \end{eqnarray*} $

因此, $F_1(\tau, \, s)=(t-\tau)^{1-\alpha}|\tau-s|^{\alpha-1}q(s) $$f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s))$在$(t_k, \, t]\times(t_k, \, t]$上可积, 进而

$ {\int_{t_k}^t{\rm d}\tau \int_{t_k}^{\tau}F_1(\tau, \, s) {\rm d}s=\int_{t_k}^t {\rm d}s \int_{s}^{t}F_1(\tau, \, s) {\rm d}\tau}. $

由引理3.2和引理2.2(2)可得

$ \begin{eqnarray} \label{eq3.5}&&\!(^{L}\!D_{t_k^+}^{\alpha}I_{t_k^+}^{\alpha}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s)))(t)\\ &=& \frac{1}{\Gamma(2-\alpha)\Gamma(\alpha)}\frac{{\rm d}^2}{{\rm d}t^2}\int_{t_k}^t(t-\tau)^{1-\alpha} \bigg[\int_{t_k}^{\tau} (\tau-s)^{\alpha-1}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s)){\rm d}s\bigg]{\rm d}\tau\\ &=&\frac{1}{\Gamma(2-\alpha)\Gamma(\alpha)}\frac{{\rm d}^2}{{\rm d}t^2}\int_{t_k}^t {\rm d}\tau\int_{t_k}^{\tau} F_1(\tau, \, s){\rm d}s\\ &=&\frac{1}{\Gamma(2-\alpha)\Gamma(\alpha)}\frac{{\rm d}^2}{{\rm d}t^2}\int_{t_k}^t {\rm d}s \int_{s}^{t}F_1(\tau, \, s) {\rm d}\tau\\ &=&\frac{1}{\Gamma(2-\alpha)\Gamma(\alpha)}\frac{{\rm d}^2}{{\rm d}t^2}\int_{t_k}^t q(s)f(s, \, x(s), \, ^{c}\!D_{s_k^+}^{\gamma}x(s)){\rm d}s\int_{s}^{t}(t-\tau)^{1-\alpha}(\tau-s)^{\alpha-1} {\rm d}\tau\\ &=&\frac{{\rm d}^2}{{\rm d}t^2}\int_{t_k}^t(t-s)q(s)f(s, \, x(s), \, ^{c}\!D_{s_k^+}^{\gamma}x(s)){\rm d}s\\ &=&q(t)f(t, \, x(t), \, ^{c}\!D_{t_k^+}^{\gamma}x(t)), \quad \mbox{a.e.}\quad t\in (t_k, \, t_{k+1}]. \end{eqnarray} $

(3.5)

再由(3.4)式可知

$ \Big[I_{t_k^+}^{\alpha}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s))\Big]^{(i)}(t_k^+)=0, \quad i=0, 1. $

在定义2.3中, 用$[I_{t_k^+}^{\alpha}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s))](t)$替换$g(t)$并利用(3.5)式即可导出

$ \begin{eqnarray*} \Big[\, ^{c}\!D_{t_k^+}^{\alpha}I_{t_k^+}^{\alpha}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s))\Big](t) &=&\Big[\, ^{L}\!D_{t_k^+}^{\alpha}I_{t_k^+}^{\alpha}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s))\Big](t)\\ &=&q(t)f(t, \, x(t), \, ^{c}\!D_{t_k^+}^{\gamma}x(t)). \end{eqnarray*} $

证毕.

由引理3.2和引理3.3可得下面结果.

引理3.4  如果(H)成立, 那么函数$x\in X$是分数阶积分方程

$ \begin{equation} \label{eq3.6}x(t)=\left\{ \begin{array}{l} \displaystyle{\int_0^t\frac{(t-s)^{\alpha-1}q(s)f(s, \, x(s), \, ^{c}\!D_{0^+}^{\gamma}x(s))}{\Gamma(\alpha)}{\rm d}s+t\widetilde{A}}, \quad t\in J_0, \\ \displaystyle{\int_{t_k}^t\frac{(t-s)^{\alpha-1}q(s)f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s))}{\Gamma(\alpha)}{\rm d}s}\\ +\displaystyle{\sum\limits_{i=1}^k \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-1}q(s)f(s, \, x(s), \, ^{c}\!D_{t_{i-1}^+}^{\gamma}x(s))}{\Gamma(\alpha)}{\rm d}s+I_i \bigg)}\\ +\displaystyle{\sum\limits_{i=1}^k \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}q(s)f(s, \, x(s), \, ^{c}\!D_{t_{i-1}^+}^{\gamma}x(s))}{\Gamma(\alpha-1)}{\rm d}s +\tilde{I}_i\bigg)(t-t_i)+t\widetilde{A}}, \\ t\in J_k, \ k=1, 2, \cdots, m \end{array}\right. \end{equation} $

(3.6)

的解当且仅当$x(t)$是方程(1.1)-(1.3)的解, 其中

$ \begin{eqnarray*} \widetilde{A}&=&-C_1\bigg[\int_{t_m}^1\frac{(1-s)^{\alpha-\beta-1}q(s)f(s, \, x(s), \, ^{c}\!D_{t_m^+}^{\gamma}x(s))}{\Gamma(\alpha-\beta)}{\rm d}s\\ &&+\frac{(1-t_m)^{1-\beta}}{\Gamma(2-\beta)}\sum\limits_{i=1}^m \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}q(s)f(s, \, x(s), \, ^{c}\!D_{t_{i-1}^+}^{\gamma}x(s))}{\Gamma(\alpha-1)}{\rm d}s +\tilde{I}_i\bigg)\bigg], \\ &&\ \ \ \ \ \ \ \ \ \ \ C_1:=\frac{d\Gamma(2-\beta)}{c\Gamma(2-\beta)+d(1-t_m)^{1-\beta}}. \end{eqnarray*} $

证  为方便起见, 记$(h_k x)(t):=q(t)f(t, \, x(t), \, ^{c}\!D_{t_k^+}^{\gamma}x(t))$, $k=0, 1, \cdots, m$.

(充分性)  设$x$是方程(1.1)-(1.3)的解, 由引理3.3, $^{c}\!D_{t_k^+}^{\alpha} x(t)=h_k(t)$意味着

$ ^{c}\!D_{t_k^+}^{\alpha}( x(t)-I_{t_k^+}^{\alpha}h_k(t))=0, \ \ k =0, 1, \cdots, m. $

如果$t\in J_0$, 那么由引理2.1可得

$ \begin{eqnarray*} x(t)=\int_0^t\frac{(t-s)^{\alpha-1}(h_0 x)(s)}{\Gamma(\alpha)}{\rm d}s+a_0+b_0t, \quad x'(t)=\int_0^t\frac{(t-s)^{\alpha-2}(h_0 x)(s)}{\Gamma(\alpha-1)}{\rm d}s+b_0, \end{eqnarray*} $

其中$a_0, b_0\in{\Bbb R}$是待定常数.易见, $a_0=x(0^+)=0, \, x'(0^+)=b_0$.

如果$t\in J_1$, 那么由引理2.1可得

$ \begin{eqnarray*} x(t)=\int_{t_1}^t\frac{(t-s)^{\alpha-1}(h_1 x)(s)}{\Gamma(\alpha)}{\rm d}s+c_0+c_1(t-t_1), \quad x'(t)=\int_{t_1}^t\frac{(t-s)^{\alpha-2}(h_1 x)(s)}{\Gamma(\alpha-1)}{\rm d}s+c_1, \end{eqnarray*} $

其中$c_0, c_1\in{\Bbb R}$是待定常数.利用左右极限定义知

$ \begin{eqnarray*} x(t_1^-)&=&\int_{0}^{t_1}\frac{(t_1-s)^{\alpha-1}(h_0 x)(s)}{\Gamma(\alpha)}{\rm d}s+b_0t_1, \quad\, x(t_1^+)=c_0, \\ x'(t_1^-)&=&\int_{0}^{t_1}\frac{(t_1-s)^{\alpha-2}(h_0 x)(s)}{\Gamma(\alpha-1)}{\rm d}s+b_0, \quad\, x'(t_1^+)=c_1. \end{eqnarray*} $

由脉冲条件(1.2)式可导出

$ \begin{eqnarray*} c_0=\int_{0}^{t_1}\frac{(t_1-s)^{\alpha-1}(h_0 x)(s)}{\Gamma(\alpha)}{\rm d}s+b_0t_1+I_1, \quad c_1=\int_{0}^{t_1}\frac{(t_1-s)^{\alpha-2}(h_0 x)(s)}{\Gamma(\alpha-1)}{\rm d}s+b_0+\tilde{I}_1. \end{eqnarray*} $

因此, 对于$t\in J_1$, 有

$ \begin{eqnarray*} x(t)&=&\int_{t_1}^t\frac{(t-s)^{\alpha-1}(h_1 x)(s)}{\Gamma(\alpha)}{\rm d}s+ \bigg(\int_{0}^{t_1}\frac{(t_1-s)^{\alpha-2}(h_0 x)(s)}{\Gamma(\alpha-1)}{\rm d}s+\tilde{I}_1\bigg)\cdot(t-t_1)\\ &&+\int_{0}^{t_1}\frac{(t_1-s)^{\alpha-1}(h_0 x)(s)}{\Gamma(\alpha)}{\rm d}s+I_1 +b_0t. \end{eqnarray*} $

如果$t\in J_2$, 由引理2.1可得

$ \begin{eqnarray*} x(t)=\int_{t_2}^t\frac{(t-s)^{\alpha-1}(h_2 x)(s)}{\Gamma(\alpha)}{\rm d}s+d_0+d_1(t-t_2), \quad x'(t)=\int_{t_2}^t\frac{(t-s)^{\alpha-2}(h_2 x)(s)}{\Gamma(\alpha-1)}{\rm d}s+d_1, \end{eqnarray*} $

其中$d_0, d_1\in{\Bbb R}$是待定常数.再次利用左右极限定义可知

$\begin{eqnarray*} x(t_2^-)&=&\int_{t_1}^{t_2}\frac{(t_2-s)^{\alpha-1}(h_1 x)(s)}{\Gamma(\alpha)} {\rm d}s+\bigg(\int_{0}^{t_1}\frac{(t_1-s)^{\alpha-2}(h_0 x)(s)}{\Gamma(\alpha-1)}{\rm d}s+\tilde{I}_1\bigg)\cdot(t_2-t_1)\\ &&+\int_{0}^{t_1}\frac{(t_1-s)^{\alpha-1}(h_0 x)(s)}{\Gamma(\alpha)}{\rm d}s+I_1 +b_0t_2, \quad x(t_2^+)=d_0, \\ x'(t_2^-)&=&\int_{t_1}^{t_2}\frac{(t_2-s)^{\alpha-2}(h_1 x)(s)}{\Gamma(\alpha-1)}{\rm d}s +\int_{0}^{t_1}\frac{(t_1-s)^{\alpha-2}(h_0 x)(s)}{\Gamma(\alpha-1)}{\rm d}s+b_0+\tilde{I}_1, \quad x'(t_2^+)=d_1. \end{eqnarray*} $

再由脉冲条件(1.2)式可导出

$ \begin{eqnarray*} d_0 &=&\sum\limits_{i=1}^2 \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-1}(h_{i-1} x)(s)}{\Gamma(\alpha)}{\rm d}s+I_i \bigg)\\ &&+\bigg(\int_{0}^{t_1}\frac{(t_1-s)^{\alpha-2}(h_0 x)(s)}{\Gamma(\alpha-1)}{\rm d}s+\tilde{I}_i\bigg)\cdot(t_2-t_1)+b_0t_2, \\[2mm] d_1&=&\sum\limits_{i=1}^2 \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}(h_{i-1} x)(s)}{\Gamma(\alpha-1)}{\rm d}s+\tilde{I}_i \bigg)+b_0. \end{eqnarray*} $

因此, 对于$t\in J_2$, 有

$ \begin{eqnarray*} x(t)&=&\int_{t_2}^t\frac{(t-s)^{\alpha-1}(h_2 x)(s)}{\Gamma(\alpha)}{\rm d}s+\sum\limits_{i=1}^2\bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-1}(h_{i-1} x)(s)}{\Gamma(\alpha)}{\rm d}s+I_i\bigg)\\ &&+\sum\limits_{i=1}^2\bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}(h_{i-1} x)(s)}{\Gamma(\alpha-1)}{\rm d}s+\tilde{I}_i\bigg)\cdot(t-t_i)+b_0t. \end{eqnarray*} $

重复上述过程, 对于$t\in J_k, k<m$可导出

$ \begin{eqnarray*} x(t)&=&\int_{t_k}^t\frac{(t-s)^{\alpha-1}(h_k x)(s)}{\Gamma(\alpha)}{\rm d}s +\sum\limits_{i=1}^k \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-1}(h_{i-1} x)(s)} {\Gamma(\alpha)}{\rm d}s+I_i\bigg)\\ &&+\sum\limits_{i=1}^k \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}(h_{i-1} x)(s)} {\Gamma(\alpha-1)}{\rm d}s+\tilde{I}_i\bigg)\cdot(t-t_i)+b_0t, \quad t\in J_k. \end{eqnarray*} $

如果$t\in J_m$, 由性质2.1和性质2.2可得

$ \begin{eqnarray*} ^{c}\!D_{t^+_m}^{\beta}x(t)&=&\int_{t_m}^t\frac{(t-s)^{\alpha-\beta-1}(h_m x)(s)}{\Gamma(\alpha-\beta)}{\rm d}s\\ &&+\frac{(t-t_m)^{1-\beta}}{\Gamma(2-\beta)}\sum\limits_{i=1}^m \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}(h_{i-1} x)(s)} {\Gamma(\alpha-1)}{\rm d}s+\tilde{I}_i\bigg)+\frac{b_0 (t-t_m)^{1-\beta}}{\Gamma(2-\beta)}, \end{eqnarray*} $

代入$t=1$得

$ \begin{eqnarray*} ^{c}\!D_{t^+_m}^{\beta}x(1)&=&\int_{t_m}^1\frac{(1-s)^{\alpha-\beta-1}(h_m x)(s)}{\Gamma(\alpha-\beta)}{\rm d}s\\ &&+\frac{(1-t_m)^{1-\beta}}{\Gamma(2-\beta)}\sum\limits_{i=1}^m \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}(h_{i-1} x)(s)}{\Gamma(\alpha-1)} {\rm d}s+\tilde{I}_i\bigg)+\frac{b_0(1-t_m)^{1-\beta} }{\Gamma(2-\beta)}. \end{eqnarray*} $

由边界条件(1.3)式可导出

$ \begin{eqnarray*} b_0&=&-C_1\left[\int_{t_m}^1\frac{(1-s)^{\alpha-\beta-1}(h_m x)(s)} {\Gamma(\alpha-\beta)}{\rm d}s\right.\\ &&\left.+\frac{(1-t_m)^{1-\beta}}{\Gamma(2-\beta)}\sum\limits_{i=1}^m \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}(h_{i-1} x)(s)} {\Gamma(\alpha-1)}{\rm d}s+\tilde{I}_i\bigg)\right]\\&=&\widetilde{A}. \end{eqnarray*} $

(必要性)  设$x(t)$满足(3.6)式, 由引理3.3和性质2.1知$(^{c}D_{t_k^+}^{\alpha}x)(t)$存在且对于$t\in J_k$ $(k=0, 1, \cdots, m)$有$^{c}D_{t_k^+}^{\alpha}x(t)=h(t)$.易见, 脉冲条件(1.2)式和边界条件(1.3)式均满足, 即$x(t)$是方程(1.1)-(1.3)的解.

本节我们将给出方程(1.1)-(1.3)的解的存在性结果.记

$ \begin{eqnarray*} M_0=\sum\limits_{i=1}^m|I_i|+\frac{C_2}{\Gamma(2-\gamma)}\sum\limits_{i=1}^m|\tilde{I}_i|, \quad \widetilde{M}=\bigg(\max\{M_1, M_2, M_3\} +\frac{M_4}{\Gamma(2-\gamma)} \bigg)\|\mu\|_{L^{\frac{1}{\sigma}}}, \end{eqnarray*} $

其中

$ C_2=1+\frac{C_1(1-t_m)^{1-\beta}}{\Gamma(2-\beta)}, \quad M_1=\frac{(m+1)({\bf B}(e_{\alpha}, \, z))^{1-\sigma}}{\Gamma(\alpha)}, \quad M_2=\frac{({\bf B}(e_{\alpha-1}, \, z))^{1-\sigma}}{\Gamma(\alpha-1)}, \\ \ \ \ \ M_3=\frac{({\bf B}(e_{\alpha-\gamma}, \, z))^{1-\sigma}}{\Gamma(\alpha-\gamma)}, \quad\quad M_4=\frac{mC_2({\bf B}(e_{\alpha-1}, \, z))^{1-\sigma}}{\Gamma(\alpha-1)} +\frac{C_1({\bf B}(e_{\alpha-\beta}, \, z))^{1-\sigma}}{\Gamma(\alpha-\beta)}. $

定理4.1  设(H)成立, 如果当$\lambda_2=1$时$\widetilde{M}<1$或当$\lambda_2>1$时$\frac{\lambda_2 M_0}{(M_0^{\lambda_1}+M_0^{\lambda_2})(\lambda_2+1)^{\lambda_2}}\geq \widetilde{M}$, 那么方程(1.1)-(1.3)存在解$x\in X$.

证  定义算子${\cal F}:X\rightarrow X$如下

$ \begin{equation} \label{eq4.1}({\cal F}x)(t)=\left\{ \begin{array}{l} \displaystyle{\int_0^t\frac{(t-s)^{\alpha-1}q(s)}{\Gamma(\alpha)}f(s, \, x(s), \, ^{c}\!D_{0^+}^{\gamma}x(s)){\rm d}s+\widetilde{A}t}, \hskip12pt t\in J_0, \\ \displaystyle{\int_{t_k}^t\frac{(t-s)^{\alpha-1}q(s)}{\Gamma(\alpha)}f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s)){\rm d}s}\\ +\displaystyle{\sum\limits_{i=1}^k \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-1}q(s)}{\Gamma(\alpha)}f(s, \, x(s), \, ^{c}\!D_{t_{i-1}^+}^{\gamma}x(s)){\rm d}s+I_i \bigg)}\\ + \displaystyle{\sum\limits_{i=1}^k\left(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}q(s)}{\Gamma(\alpha-1)}f(s, \, x(s), \, ^{c}\!D_{t_{i-1}^+}^{\gamma}x(s)){\rm d}s+\tilde{I}_i\right)(t-t_i)}\\ -\displaystyle{C_1 t\int_{t_m}^1\frac{(1-s)^{\alpha-\beta-1}q(s)}{\Gamma(\alpha-\beta)}f(s, \, x(s), \, ^{c}\!D_{t_m^+}^{\gamma}x(s)){\rm d}s}\\ -\displaystyle{\frac{C_1 t (1-t_m)^{1-\beta} }{\Gamma(2-\beta)} \sum\limits_{i=1}^m \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}q(s)}{\Gamma(\alpha-1)}f(s, \, x(s), \, ^{c}\!D_{t_{i-1}^+}^{\gamma}x(s)){\rm d}s+\tilde{I}_i \bigg)}, \\ t\in J_k, \, \, k=1, \cdots, m, \end{array}\right. \end{equation} $

(4.1)

则

$ \begin{equation} \label{eq4.2}({\cal F}x)'(t)=\left\{ \begin{array}{l} \displaystyle{\int_0^t\frac{(t-s)^{\alpha-2}q(s)}{\Gamma(\alpha-1)}f(s, \, x(s), \, ^{c}\!D_{0^+}^{\gamma}x(s)){\rm d}s+\widetilde{A}}, \hskip12pt t\in J_0, \\ \displaystyle{\int_{t_k}^t\frac{(t-s)^{\alpha-2}q(s)}{\Gamma(\alpha-1)}f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s)){\rm d}s}\\ + \displaystyle{\sum\limits_{i=1}^k \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}q(s)}{\Gamma(\alpha-1)}f(s, \, x(s), \, ^{c}\!D_{t_{i-1}^+}^{\gamma}x(s)){\rm d}s+\tilde{I}_i \bigg)}\\ -\displaystyle{C_1 \int_{t_m}^1\frac{(1-s)^{\alpha-\beta-1}q(s)}{\Gamma(\alpha-\beta)}f(s, \, x(s), \, ^{c}\!D_{t_m^+}^{\gamma}x(s)){\rm d}s}\\ -\displaystyle{\frac{C_1 (1-t_m)^{1-\beta} }{\Gamma(2-\beta)}\sum\limits_{i=1}^m \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}q(s)}{\Gamma(\alpha-1)}f(s, \, x(s), \, ^{c}\!D_{t_{i-1}^+}^{\gamma}x(s)){\rm d}s+\tilde{I}_i \bigg)}, \\ t\in J_k, \, \, k=1, \cdots, m. \end{array}\right. \end{equation} $

(4.2)

此外, 由性质2.1和性质2.2得

$ \begin{equation} \label{eq4.3}(^{c}\!D_{*}^{\gamma}|_{J_k}{\cal F}x)(t)=\left\{ \begin{array}{l} \displaystyle{\int_0^t\frac{(t-s)^{\alpha-\gamma-1}q(s)}{\Gamma(\alpha-\gamma)}f(s, \, x(s), \, ^{c}\!D_{0^+}^{\gamma}x(s)){\rm d}s+\frac{\widetilde{A}t^{1-\gamma}}{\Gamma(2-\gamma)}}, \hskip12pt t\in J_0, \\ \displaystyle{\int_{t_k}^t\frac{(t-s)^{\alpha-\gamma-1}q(s)}{\Gamma(\alpha-\gamma)}f(s, \, x(s), \, ^{c}\!D_{t_k^+}^{\gamma}x(s)){\rm d}s}\\ + \displaystyle{\frac{(t-t_k)^{1-\gamma}}{\Gamma(2-\gamma)} \bigg[\sum\limits_{i=1}^k \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}q(s)}{\Gamma(\alpha-1)}f(s, \, x(s), \, ^{c}\!D_{t_{i-1}^+}^{\gamma}x(s)){\rm d}s+\tilde{I}_i\bigg)}\\ -\displaystyle{C_1 \int_{t_m}^1\frac{(1-s)^{\alpha-\beta-1}q(s)}{\Gamma(\alpha-\beta)}f(s, \, x(s), \, ^{c}\!D_{t_{m}^+}^{\gamma}x(s)){\rm d}s}\\ -\displaystyle{\frac{C_1 (1-t_m)^{1-\beta} }{\Gamma(2-\beta)}\sum\limits_{i=1}^m \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}q(s)}{\Gamma(\alpha-1)}f(s, \, x(s), \, ^{c}\!D_{t_{i-1}^+}^{\gamma}x(s)){\rm d}s+\tilde{I}_i\bigg)}\bigg], \\ t\in J_k, \, \, k=1, \cdots, m. \end{array}\right. \end{equation} $

(4.3)

根据假设(H)知${\cal F}:X\rightarrow X$适定.由引理3.4知算子${\cal F}$的不动点即为方程(1.1)-(1.3)的解.因此我们只需要证明算子${\cal F}$不动点的存在性.下面分四步证明上述结论.

第1步  算子${\cal F}$连续.

如果序列$\{x_n\}$满足当$n\rightarrow\infty$时, 在$X$中$x_n\rightarrow x$, 那么存在$\varepsilon>0$使得对充分大的$n$有$\|x_n-x\|_{1}\leq\varepsilon$.由假设(H), 我们可得

$\begin{eqnarray*} &&|f(t, x_n(t), ^{c}\!D_{t_k^+}^{\gamma}x_n(t))-f(t, x(t), ^{c}\!D_{t_k^+}^{\gamma}x(t))|\\ &<&2\mu(t)[(\varepsilon+\|x\|_1)^{\lambda_1}+(\varepsilon+\|x\|_1)^{\lambda_2}], \, \, t\in J_k, k=0, \, 1, \cdots, m. \end{eqnarray*} $

此外, $f$满足(H), 对几乎每个$t\in J_k$, 当$n\rightarrow \infty$时我们有

$ f(t, x_n(t), ^{c}\!D_{t_k^+}^{\gamma}x_n(t))\rightarrow f(t, x(t), ^{c}\!D_{t_k^+}^{\gamma}x(t)). $

由(3.2)式和Lebesgue控制收敛定理知当$n\rightarrow \infty$时, 有

$ \|{\cal F} x_n-{\cal F} x\|_1\rightarrow 0. $

对于$r>0$, 定义$\Omega=\{x\in X:\|x\|_1\leq r\}$, 要证${\cal F}$完全连续, 我们需要证明${\cal F} \Omega\subset \Omega$, ${\cal F}\Omega$在$J_k(k=0, 1, \cdots, m)$的有限闭子区间上等度连续, 当$t\rightarrow t_k^+(k=0, 1, 2, \cdots, m)$时, ${\cal F}\Omega$等度收敛.

第2步  ${\cal F} \Omega\subset \Omega$.

对于$x\in \Omega$, 由(4.1)-(4.3)式和(3.2)式可得

$ \begin{eqnarray} \label{eq4.4} |({\cal F}x)(t)| &\leq& \bigg[\int_{t_k}^t\frac{(t-s)^{\alpha-1}s^l \mu(s)}{\Gamma(\alpha)}{\rm d}s+\sum\limits_{i=1}^m\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-1}s^l \mu(s)}{\Gamma(\alpha)}{\rm d}s\\ &&+C_2\sum\limits_{i=1}^m\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}s^l \mu(s)}{\Gamma(\alpha-1)}{\rm d}s\\ &&+C_1\int_{t_m}^1\frac{(1-s)^{\alpha-\beta-1}s^l\mu(s)}{\Gamma(\alpha-\beta)}{\rm d}s\bigg] (r^{\lambda_1}+r^{\lambda_2})+M_0\\ &\leq&(M_1+M_4) \|\mu\|_{L^{\frac{1}{\sigma}}}(r^{\lambda_1}+r^{\lambda_2})+M_0, \end{eqnarray} $

(4.4)

$ \begin{eqnarray} \label{eq4.5}|({\cal F}x)'(t)|&\leq& \bigg[\int_{t_k}^t\frac{(t-s)^{\alpha-2}s^l \mu(s)}{\Gamma(\alpha-1)}{\rm d}s+C_2\sum\limits_{i=1}^m\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}s^l \mu(s)}{\Gamma(\alpha-1)}{\rm d}s\\ &&+C_1\int_{t_m}^1\frac{(1-s)^{\alpha-\beta-1}s^l\mu(s)}{\Gamma(\alpha-\beta)}{\rm d}s\bigg](r^{\lambda_1}+r^{\lambda_2})+M_0\\ &\leq&(M_2+M_4)\|\mu\|_{L^{\frac{1}{\sigma}}} (r^{\lambda_1}+r^{\lambda_2})+M_0, \end{eqnarray} $

(4.5)

$ \begin{eqnarray} \label{eq4.6}|^{c}\!D_{t_k^+}^{\gamma}({\cal F}x)(t)|&\leq& \bigg[\int_{t_k}^t\frac{(t-s)^{\alpha-\gamma-1}s^l \mu(s)}{\Gamma(\alpha-\gamma)}{\rm d}s +\frac{C_2}{\Gamma(2-\gamma)}\sum\limits_{i=1}^m\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}s^l \mu(s)}{\Gamma(\alpha-1)}{\rm d}s\\ &&+\frac{C_1}{\Gamma(2-\gamma)}\int_{t_m}^1\frac{(1-s)^{\alpha-\beta-1}s^l\mu(s)}{\Gamma(\alpha-\beta)}{\rm d}s\bigg] (r^{\lambda_1}+r^{\lambda_2})+M_0\\ &\leq&\bigg(M_3+\frac{M_4}{\Gamma(2-\gamma)}\bigg)\|\mu\|_{L^{\frac{1}{\sigma}}} (r^{\lambda_1}+r^{\lambda_2})+M_0, \end{eqnarray} $

(4.6)

利用(4.4)-(4.6)式, 我们可以得到

$ \|{\cal F}x\|_1\leq \widetilde{M}(r^{\lambda_1}+r^{\lambda_2})+M_0. $

接下来, 我们将分两种情形证明存在某个$r>0$使得${\cal F}\Omega\subset \Omega$.

(ⅰ) 当$\lambda_2\leq 1$时, 如果${\cal F}\Omega\subset \Omega$不成立, 那么对于每一个正数$r$, 存在函数$\tilde{x}^r(\cdot)\in \Omega$和某个$\tilde{t}\in J_k$使得

$ \Lambda:=\max\{|{\cal F}\tilde{x}^r(\tilde{t})|, \, |({\cal F}\tilde{x}^r)'(\tilde{t})|, \, |^{c}\!D_{t_k^+}^{\gamma}({\cal F}\tilde{x}^r)(\tilde{t})|\}>r. $

然而

$ r< \Lambda\leq \widetilde{M}(r^{\lambda_1}+r^{\lambda_2})+M_0, $

两边除以$r$, 令$r\rightarrow +\infty$有

$ \left\{ \begin{array}{ll} 1\leq \lim\limits_{r\rightarrow \infty}\frac{\widetilde{M} (r^{\lambda_1}+r^{\lambda_2})+M_0}{r}=0, \quad&{\mbox{当 }} \lambda_2 < 1 {\mbox{时 }}, \\ 1\leq \lim\limits_{r\rightarrow \infty}\frac{\widetilde{M} (r^{\lambda_1}+r^{\lambda_2})+M_0}{r}=\widetilde{M}, \quad & {\mbox{当 }}\lambda_2 =1{\mbox{时 }}. \end{array}\right. $

与已知矛盾.

(ⅱ) 当$\lambda_2>1$时, 选取$r_0=M_0(\lambda_2+1)$使得

$ \frac{r_0-M_0}{{r_0}^{\lambda_1}+{r_0}^{\lambda_2}}\geq \frac{\lambda_2 M_0}{(M_0^{\lambda_1}+M_0^{\lambda_2})(\lambda_2+1)^{\lambda_2}}\geq \widetilde{M}, $

则有${\cal F}\Omega\subset \Omega$.

第3步  ${\cal F}\Omega$在$J_k(k=0, 1, \cdots, m)$的有限闭子区间上等度连续.

对于$x\in \Omega$, $t_k<\tau_2<\tau_1\leq t_{k+1}(k=0, 1, \cdots, m)$, 当$\tau_2\rightarrow \tau_1$时我们有

$ \begin{eqnarray*} &&\frac{1}{\Gamma(\alpha)}\left|\int_{t_k}^{\tau_1}(\tau_1-s)^{\alpha-1}q(s)f(s, x(s), ^{c}\!D_{t_k^+}^{\gamma}x(s)){\rm d}s\right.\\ &&\left.-\int_{t_k}^{\tau_2}(\tau_2-s)^{\alpha-1}q(s)f(s, x(s), ^{c}\!D_{t_k^+}^{\gamma}x(s)){\rm d}s\right|\\ &\leq&\frac{(r^{\lambda_1}+r^{\lambda_2})}{\Gamma(\alpha)}\left[\int_{t_k}^{\tau_2}[(\tau_1-s)^{\alpha-1}-(\tau_2-s)^{\alpha-1}]s^l\mu(s){\rm d}s+\int_{\tau_2}^{\tau_1}(\tau_1-s)^{\alpha-1}s^l\mu(s){\rm d}s\right]\\ &\rightarrow&0, \end{eqnarray*} $

因此, 对于$t_{k}<\tau_2<\tau_1\leq t_{k+1}(k=0, 1, \cdots, m)$, 当$\tau_2\rightarrow \tau_1$时有

$ \begin{eqnarray*} |({\cal F}x)(\tau_1)-({\cal F}x)(\tau_2)| &\leq&\frac{1}{\Gamma(\alpha)}\left|\int_{t_k}^{\tau_1}(\tau_1-s)^{\alpha-1}q(s)f(s, x(s), ^{c}\!D_{t_k^+}^{\gamma}x(s)){\rm d}s\right.\\ &&\left.-\int_{t_k}^{\tau_2}(\tau_2-s)^{\alpha-1}q(s)f(s, x(s), ^{c}\!D_{t_k^+}^{\gamma}x(s)){\rm d}s\right|\\ &&+(\tau_1-\tau_2)(r^{\lambda_1}+r^{\lambda_2})\left[C_2\sum\limits_{i=1}^m\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}s^l \mu(s)}{\Gamma(\alpha-1)}{\rm d}s\right.\\ &&\left.+C_1\int_{t_m}^1\frac{(1-s)^{\alpha-\beta-1}s^l\mu(s)}{\Gamma(\alpha-\beta)}{\rm d}s\right] +C_2(\tau_1-\tau_2)\sum\limits_{i=1}^m|\tilde{I}_i|\\ &\rightarrow& 0. \end{eqnarray*} $

由引理3.1和(3.2)式可得当$\tau_2\rightarrow \tau_1$时, 有

$ \begin{eqnarray*} |({\cal F}x)'(\tau_1)-({\cal F}x)'(\tau_2)| &\leq&\frac{1}{\Gamma(\alpha-1)}\left|\int_{t_k}^{\tau_1}(\tau_1-s)^{\alpha-2}q(s)f(s, x(s), ^{c}\!D_{t_k^+}^{\gamma}x(s)){\rm d}s\right.\\ &&\left.-\int_{t_k}^{\tau_2}(\tau_2-s)^{\alpha-2}q(s)f(s, x(s), ^{c}\!D_{t_k^+}^{\gamma}x(s)){\rm d}s\right|\\ &\leq&\frac{(r^{\lambda_1}+r^{\lambda_2})}{\Gamma(\alpha-1)}\bigg[\int_{t_k}^{\tau_2}[(\tau_2-s)^{\alpha-2}-(\tau_1-s)^{\alpha-2}]s^l\mu(s){\rm d}s\\ &&+\int_{\tau_2}^{\tau_1}(\tau_1-s)^{\alpha-2}s^l\mu(s){\rm d}s\bigg]\\ &\rightarrow& 0. \end{eqnarray*} $

利用上述类似推导再结合引理2.4可以得到当$\tau_2\rightarrow \tau_1$时, 有

$\begin{eqnarray*} &&|^{c}\!D_{t_k^+}^{\gamma}({\cal F}x)(\tau_1)-^{c}\!D_{t_k^+}^{\gamma}({\cal F}x)(\tau_2)|\\ &\leq&\frac{1}{\Gamma(\alpha-\gamma)}\left|\int_{t_k}^{\tau_1}(\tau_1-s)^{\alpha-\gamma-1}q(s)f(s, x(s), ^{c}\!D_{t_k^+}^{\gamma}x(s)){\rm d}s\right.\\ &&\left.-\int_{t_k}^{\tau_2}(\tau_2-s)^{\alpha-\gamma-1}q(s)f(s, x(s), ^{c}\!D_{t_k^+}^{\gamma}x(s)){\rm d}s\right|\\ &&+\frac{(\tau_1-\tau_2)^{1-\gamma}}{\Gamma(2-\gamma)}\left\{\left[C_2\sum\limits_{i=1}^m\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}s^l \mu(s)}{\Gamma(\alpha-1)}{\rm d}s\right.\right.\\ &&\left.\left.+C_1\int_{t_m}^1\frac{(1-s)^{\alpha-\beta-1}s^l\mu(s)}{\Gamma(\alpha-\beta)}{\rm d}s\right](r^{\lambda_1}+r^{\lambda_2}) +C_2\sum\limits_{i=1}^m|\tilde{I}_i| \right\} \\ &\rightarrow& 0. \end{eqnarray*} $

至此已证明集合$\{({\cal F}x)(\cdot): x\in\Omega\}$在$J_k(k=0, 1, \cdots, m)$的有限闭子区间上等度连续.

第4步  当$t\rightarrow t_k^+(k=0, 1, 2, \cdots, m)$时, ${\cal F}\Omega$等度收敛.

由(3.2)式可得当$t\rightarrow 0^+$时, 有

$ \begin{eqnarray*} |({\cal F}x)(t)|&\leq&\int_0^t\frac{(t-s)^{\alpha-1}s^l\mu(s)}{\Gamma(\alpha)}{\rm d}s(r^{\lambda_1}+r^{\lambda_2})+|\widetilde{A}|t\\ &\leq& \frac{t^{\alpha+l-\sigma}({\bf B}(e_{\alpha}, \, z))^{1-\sigma}}{\Gamma(\alpha)}\|\mu\|_{L^{\frac{1}{\sigma}}}(r^{\lambda_1}+r^{\lambda_2})+|\widetilde{A}|t\\ &\rightarrow& 0, \end{eqnarray*} $

$ \begin{eqnarray*} |({\cal F}x)'(t)-\widetilde{A}|&\leq&\int_0^t\frac{(t-s)^{\alpha-2}s^l \mu(s)}{\Gamma(\alpha-1)}{\rm d}s(r^{\lambda_1}+r^{\lambda_2})\\ &\leq&\frac{t^{\alpha-1+l-\sigma}({\bf B}(e_{\alpha-1}, \, z))^{1-\sigma}}{\Gamma(\alpha-1)}\|\mu\|_{L^{\frac{1}{\sigma}}}(r^{\lambda_1}+r^{\lambda_2})\\ &\rightarrow& 0, \end{eqnarray*} $

$ \begin{eqnarray*} |(^{c}\!D_{0^+}^{\gamma}{\cal F}x)(t)|&\leq&\int_0^t\frac{(t-s)^{\alpha-\gamma-1}s^l\mu(s)}{\Gamma(\alpha-\gamma)}{\rm d}s(r^{\lambda_1}+r^{\lambda_2})+\frac{|\widetilde{A}|t^{1-\gamma}}{\Gamma(2-\gamma)}\\ &\leq&\frac{t^{\alpha-\gamma+l-\sigma}({\bf B}(e_{\alpha-\gamma}, \, z))^{1-\sigma}}{\Gamma(\alpha-\gamma)}\|\mu\|_{L^{\frac{1}{\sigma}}}(r^{\lambda_1}+r^{\lambda_2})+\frac{|\widetilde{A}|t^{1-\gamma}}{\Gamma(2-\gamma)}\\ &\rightarrow& 0. \end{eqnarray*} $

因此, 当$t\rightarrow 0^+$时, ${\cal F}\Omega$等度收敛.

由(3.2)式得当$t\rightarrow t_k^+(k=1, 2, \cdots, m)$时, 有

$ \begin{eqnarray*} &&\left|({\cal F}x)(t)- \sum\limits_{i=1}^k \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-1}q(s)}{\Gamma(\alpha)}f(s, \, x(s), \, ^{c}\!D_{t_{i-1}^+}^{\gamma}x(s)){\rm d}s+I_i \bigg)\right.\\ &&-\sum\limits_{i=1}^k \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}q(s)}{\Gamma(\alpha-1)}f(s, \, x(s), \, ^{c}\!D_{t_{i-1}^+}^{\gamma}x(s)){\rm d}s +\tilde{I}_i \bigg)(t_k-t_i)\\ &&+C_1 t_k\int_{t_m}^1\frac{(1-s)^{\alpha-\beta-1}q(s)}{\Gamma(\alpha-\beta)}f(s, \, x(s), \, ^{c}\!D_{t_m^+}^{\gamma}x(s)){\rm d}s\\ &&\left.+\frac{C_1 t_k (1-t_m)^{1-\beta} }{\Gamma(2-\beta)}\sum\limits_{i=1}^m \bigg(\int_{t_{i-1}}^{t_i} \frac{(t_i-s)^{\alpha-2}q(s)}{\Gamma(\alpha-1)}f(s, \, x(s), \, ^{c}\!D_{t_{i-1}^+}^{\gamma}x(s)){\rm d}s+\tilde{I}_i \bigg)\right|\\ &\leq& \left[\frac{(t-t_k)^{\alpha+l-\sigma}({\bf B}(e_{\alpha}, \, z))^{1-\sigma}}{\Gamma(\alpha)}+(t-t_k)M_4\right]\|\mu\|_{L^{\frac{1}{\sigma}}}(r^{\lambda_1}+r^{\lambda_2}) +C_2(t-t_k)\sum\limits_{i=1}^m |\tilde{I}_i|\\ &\rightarrow& 0, \end{eqnarray*} $

$ \begin{eqnarray*} &&\left|({\cal F}x)'(t)- \sum\limits_{i=1}^k \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}q(s)}{\Gamma(\alpha-1)}f(s, \, x(s), \, ^{c}\!D_{t_{i-1}^+}^{\gamma}x(s)){\rm d}s+\tilde{I}_i \bigg)\right.\\ &&+C_1 \int_{t_m}^1\frac{(1-s)^{\alpha-\beta-1}q(s)}{\Gamma(\alpha-\beta)}f(s, \, x(s), \, ^{c}\!D_{t_m^+}^{\gamma}x(s)){\rm d}s\\ &&\left.+\frac{C_1 (1-t_m)^{1-\beta}}{\Gamma(2-\beta)}\sum\limits_{i=1}^m \bigg(\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}q(s)}{\Gamma(\alpha-1)}f(s, \, x(s), \, ^{c}\!D_{t_{i-1}^+}^{\gamma}x(s)){\rm d}s+\tilde{I}_i \bigg)\right|\\ &\leq& M_2(t-t_k)^{\alpha-1+l-\sigma}\|\mu\|_{L^{\frac{1}{\sigma}}}(r^{\lambda_1}+r^{\lambda_2})\\ &\rightarrow& 0, \end{eqnarray*} $

$ \begin{eqnarray*} |(^{c}\!D_{t^+_k}^{\gamma}{\cal F}x)(t)|&\leq& \left[M_3(t-t_k)^{\alpha-\gamma+l-\sigma}+\frac{M_4(t-t_k)^{1-\gamma}}{\Gamma(2-\gamma)}\right]\|\mu\|_{L^{\frac{1}{\sigma}}}(r^{\lambda_1}+r^{\lambda_2})\\ &&+\frac{C_2(t-t_k)^{1-\gamma}}{\Gamma(2-\gamma)}\sum\limits_{i=1}^m |\tilde{I}_i|\\ &\rightarrow& 0. \end{eqnarray*} $

因此, 当$t\rightarrow t_k^+$时, ${\cal F}\Omega$等度收敛.

综上可得${\cal F}$完全连续, 由Schauder不动点定理知${\cal F}$有不动点$x\in \Omega$, 即方程(1.1)-(1.3)有解$x\in \Omega\subset X$.

注4.1  如果$\mu(s)\equiv M^*>0$, 由(3.3)式, 我们只需要将$\|\mu\|_{L^{\frac{1}{\sigma}}}({\bf B}(e_{y}, \, z))^{1-\sigma}$改成$M^*{\bf B}(y, \, l+1)$, 那么结论依然成立.

接下来, 我们考虑如下分数阶微分方程

$ \begin{equation} \label{eq4.7}\left\{ \begin{array}{l} \displaystyle{^{c}\!D_{*}^{\alpha} x(t)=\!t^lf(t, \, x(t)), \quad \mbox{a.e.}\quad t\in J':=J\setminus\{t_1, \, t_2, \, \cdots, t_m\}, \, \, J=(0, \, 1]}, \, l>0, \\ \displaystyle{\Delta x(t_k)=I_k, \quad \Delta x'(t_k)=\tilde{I}_k, \quad k=1, 2, \cdots, m, }\\ \displaystyle{x(0^+)=0, \quad cx'(0^+)+d ^{c}\!D_{t_m^+}^{\beta}x(1)=0.} \end{array}\right. \end{equation} $

(4.7)

函数$f:J\times {\Bbb R}\rightarrow {\Bbb R}$满足下面假设

(H')对所有$v\in {\Bbb R}$, $f(\cdot, v):J\rightarrow {\Bbb R}$可测, 对a.e. $t\in J$, $f(t, \cdot): {\Bbb R}\rightarrow {\Bbb R}$连续, 存在$M^*>0$使得$|f(t, v)|\leq M^*(|v|^{\lambda_1}+|v|^{\lambda_2})$, 其中$0<\lambda_1<\lambda_2$.

记

$ Y=\{x:(0, 1]\rightarrow {\Bbb R}: x|_{J_k}\in C(J_k, \, {\Bbb R}), \, x'|_{J_k}\in C(J_k, \, {\Bbb R})\ {\mbox{且}}x(t_k^+), \, x'(t_k^+) {\mbox{存在}}\}, $

定义范数

$ \|x\|_2:=\max\Big\{\sup\limits_{k=0, \, 1, \, \cdots, m}\sup\limits_{t \in J_k} |x(t)|, \, \sup\limits_{k=0, \, 1, \, \cdots, m}\sup\limits_{t \in J_k} |x'(t)|\Big\}. $

易见, $Y$是实Banach空间.接下来, 我们在空间$Y$上研究方程(4.7)解的存在性.利用类似于证明定理4.1的技巧再结合注4.1, 我们可得如下结果.

定理4.2  假设(H')成立, 如果$\mbox{当}\lambda_2=1\mbox{时}\widehat{M}<1$或$\mbox{当}\lambda_2>1\mbox{时}\frac{\lambda_2 \widetilde{M_0}}{(\widetilde{M_0}^{\lambda_1}+\widetilde{M_0}^{\lambda_2})(\lambda_2+1)^{\lambda_2}}\geq \widehat{M}$, 那么方程(4.7)存在解$x\in Y$, 其中

$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \widetilde{M_0}=\sum\limits_{i=1}^m|I_i|+C_2\sum\limits_{i=1}^m|\tilde{I}_i|, \\ \widehat{M}=M^*\left(\frac{\Gamma(l+1)}{\Gamma(\alpha+l)}\max\Big\{\frac{m+1}{\alpha+l}, \, 1\Big\} +\frac{mC_2\Gamma(l+1)}{\Gamma(\alpha+l)}+\frac{C_1\Gamma(l+1)}{\Gamma(\alpha-\beta+l+1)}\right). $

本节中我们给出三个例子验证主要结果的实用性.

例5.1  考虑如下非线性多基点脉冲分数阶微分方程:

$ \begin{equation} \label{eq5.1}\left\{ \begin{array}{l} \displaystyle{^{c}\!D_{*}^{\frac{3}{2}} x(t)=\frac{\sin t}{ t^{\frac{6}{5}}} \bigg(\frac{(x(t))^{\frac{1}{2}}}{8t^{\frac{1}{10}}}+\frac{^{c}\!D_{*}^{\frac{9}{20}} x(t)}{11t^{\frac{1}{20}}} \bigg), \quad\mbox{a.e.}\quad t\in J':=J\setminus\{\frac{1}{2}\}, \, \, J=(0, \, 1]}, \\ \displaystyle{\Delta x(\frac{1}{2})=10, \quad \Delta x'(\frac{1}{2})=21, }\\ \displaystyle{x(0^+)=0, \quad 20x'(0^+)+3 ^{c}\!D_{{\frac{1}{2}}^+}^{\frac{4}{5}}x(1)=0.} \end{array}\right. \end{equation} $

(5.1)

令$\alpha=\frac{3}{2}$, $\beta=\frac{4}{5}$, $\gamma=\frac{9}{20}$, $q(t)=\frac{\sin t}{ t^{\frac{6}{5}}}$, $I_1=10$, $\tilde{I}_1=21$, $c=20$, $d=3$, $f(t, \, x(t), \, ^{c}\!D_{*}^{\frac{9}{20}} x(t))=$$\frac{(x(t))^{\frac{1}{2}}}{8t^{\frac{1}{10}}}+\frac{^{c}\!D_{*}^{\frac{9}{20}} x(t)}{11t^{\frac{1}{20}}}$.易见

$ |q(t)|\leq t^l(l=-\frac{1}{5}), \quad |f(t, \, x(t), \, ^{c}\!D_{*}^{\frac{9}{20}} x(t))|\leq \mu(t)(|x(t)|^{\frac{1}{2}}+|^{c}\!D_{*}^{\frac{9}{20}} x(t)|), $

其中$\mu(t)=\frac{t^{-\frac{1}{10}}}{8}\in L^{\frac{1}{\sigma}}((0, \, 1), \, {\Bbb R}^+)(\sigma=\frac{1}{5})$且$\|\mu\|_{L^{5}}=\frac{2^{\frac{1}{5}}}{8}$, 则假设(H)成立.此外, 通过计算可得${\bf B}(e_{\alpha}, \, z)={\bf B}(\frac{13}{8}, \, \frac{3}{4})$, ${\bf B}(e_{\alpha-1}, \, z)={\bf B}(\frac{3}{8}, \, \frac{3}{4})$, ${\bf B}(e_{\alpha-\beta}, \, z)={\bf B}(\frac{5}{8}, \, \frac{3}{4})$, ${\bf B}(e_{\alpha-\gamma}, \, z)={\bf B}(\frac{17}{16}, \, \frac{3}{4})$以及

$ C_1=\frac{3\Gamma(\frac{6}{5})}{20\Gamma(\frac{6}{5})+3(\frac{1}{2})^{\frac{1}{5}}}\approx0.131, \, \quad C_2=1+\frac{C_1(\frac{1}{2})^{\frac{1}{5}}}{\Gamma(\frac{6}{5})}\approx1.124, $

$ M_1=\frac{2({\bf B} (\frac{13}{8}, \, \frac{3}{4}))^{\frac{4}{5}}}{\Gamma(\frac{3}{2})}\approx2.07, \ \ \ M_2=\frac{({\bf B} (\frac{3}{8}, \, \frac{3}{4}))^{\frac{4}{5}}}{\Gamma(\frac{1}{2})}\approx1.39, \quad M_3=\frac{({\bf B} (\frac{17}{16}, \, \frac{3}{4}))^{\frac{4}{5}}}{\Gamma(\frac{21}{20})}\approx1.24, $

$M_4=\frac{C_2\times({\bf B} (\frac{3}{8}, \, \frac{3}{4}))^{\frac{4}{5}}}{\Gamma(\frac{1}{2})}+\frac{C_1\times({\bf B} (\frac{5}{8}, \, \frac{3}{4}))^{\frac{4}{5}}}{\Gamma(\frac{7}{10})}\approx1.74, $

$ \widetilde{M}=\bigg(\max\{M_1, M_2, M_3\} +\frac{M_4}{\Gamma(2-\gamma)} \bigg)\|\mu\|_{L^{\frac{1}{\sigma}}}\approx0.58<1. $

由定理4.1知方程(5.1)存在解.

例5.2  考虑如下多基点分数阶逻辑斯蒂模型解的存在性

$ \begin{equation} \label{eq5.2}\left\{ \begin{array}{l} \displaystyle{^{c}\!D_{*}^{\frac{5}{4}} x(t)=x(t)(a(t)-b(t)x^{\frac{1}{5}}(t)), \quad\mbox{a.e.}\quad t\in J':=J\setminus\{\frac{1}{3}\}, \, \, J=(0, \, 1]}, \\ \displaystyle{\Delta x(\frac{1}{3})=\frac{1}{50}, \quad \Delta x'(\frac{1}{3})=\frac{1}{20}, }\\ \displaystyle{x(0^+)=0, \quad 9x'(0^+)+4 ^{c}\!D_{{\frac{1}{3}}^+}^{\frac{1}{2}}x(1)=0, } \end{array}\right. \end{equation} $

(5.2)

其中$\alpha=\frac{5}{4}$, $\beta=\frac{1}{2}$, $a(t)=\frac{1}{17}t^8, \, b(t)=\frac{1}{15}t^5$为连续函数且$|a(t)|\leq \frac{1}{17}$, $|b(t)|\leq \frac{1}{15}$, $f(t, \, x)=x(a(t)-b(t)x^{\frac{1}{5}})$.易见, $|f(t, \, x)|\leq \frac{1}{15}(|x|+|x|^{\frac{6}{5}})$且假设(H')成立.由注4.1和定理4.2以及$l=0$, $\lambda_1=1, \, \lambda_2=\frac{6}{5}$, 经过计算可得

$ C_1=\frac{4\Gamma(\frac{3}{2})}{9\Gamma(\frac{3}{2})+4(\frac{2}{3})^{\frac{1}{2}}}\approx0.315, \quad C_2=1+\frac{C_1(\frac{2}{3})^{\frac{1}{2}}}{\Gamma(\frac{3}{2})}\approx1.29, $

$ \widehat{M}=\frac{1}{15}\left(\frac{C_2+\frac{8}{5}}{\Gamma(\frac{5}{4})}+\frac{C_1}{\Gamma(\frac{7}{4})}\right)\approx0.235, \ \ \ \widetilde{M_0}=0.02+0.05 C_2 \approx 0.085, $

$ \frac{\lambda_2 \widetilde{M}_0}{(\widetilde{M}_0^{\lambda_1}+\widetilde{M}_0^{\lambda_2})(\lambda_2+1)^{\lambda_2}}\approx 0.29, $

则$\frac{\lambda_2 \widetilde{M}_0}{(\widetilde{M}_0^{\lambda_1}+ \widetilde{M}_0^{\lambda_2})(\lambda_2+1)^{\lambda_2}}> \widehat{M}$, 由定理4.2知方程(5.2)存在解.

例5.3  考虑如下多基点脉冲分数阶微分方程

$ \begin{equation} \label{eq5.3}\left\{ \begin{array}{l} \displaystyle{^{c}\!D_{*}^{\frac{3}{2}} x(t)=x^2(t)(\widetilde{a}(t)-\widetilde{b}(t)x^{\frac{1}{2}}(t)), \quad\mbox{a.e.}\quad t\in J':=J\setminus\{\frac{1}{4}\}, \, \, J=(0, \, 1]}, \\ \displaystyle{\Delta x(\frac{1}{4})=\frac{1}{100}, \quad \Delta x'(\frac{1}{4})=\frac{3}{200}, }\\ \displaystyle{x(0^+)=0, \quad 7x'(0^+)+ ^{c}\!D_{{\frac{1}{4}}^+}^{\frac{1}{2}}x(1)=0.} \end{array}\right. \end{equation} $

(5.3)

其中$\alpha=\frac{3}{2}$, $\beta=\frac{1}{2}$, $\widetilde{a}(t)=\frac{1}{2}t, \, \widetilde{b}(t)=t^5$为连续函数且$|\widetilde{a}(t)|\leq \frac{1}{2}$, $|\widetilde{b}(t)|\leq 1$, $f(t, \, x)=x^2(\widetilde{a}(t)-\widetilde{b}(t)x^{\frac{1}{2}})$.易见, $|f(t, \, x)|\leq |x|^2+|x|^{\frac{5}{2}}$且假设(H')成立.由注4.1和定理4.2以及$l=0$, $\lambda_1=2, \, \lambda_2=\frac{5}{2}$, 经过计算可得

$ C_1=\frac{\Gamma(\frac{3}{2})}{7\Gamma(\frac{3}{2})+(\frac{3}{4})^{\frac{1}{2}}}\approx0.125, \quad C_2=1+\frac{C_1(\frac{3}{4})^{\frac{1}{2}}}{\Gamma(\frac{3}{2})}\approx1.122, \quad\widehat{M}=\left(\frac{C_2+\frac{4}{3}}{\Gamma(\frac{3}{2})}+C_1\right)\approx2.896 $

$ \widetilde{M_0}=0.01+0.015 C_2 \approx 0.027, \quad\frac{\lambda_2 \widetilde{M}_0}{(\widetilde{M}_0^{\lambda_1}+\widetilde{M}_0^{\lambda_2})(\lambda_2+1)^{\lambda_2}}\approx 3.47, $

则$\frac{\lambda_2 \widetilde{M}_0}{(\widetilde{M}_0^{\lambda_1}+\widetilde{M}_0^{\lambda_2})(\lambda_2+1)^{\lambda_2}}> \widehat{M}$, 由定理4.2知方程(5.3)存在解.