西南大学学报 (自然科学版)  2019, Vol. 41 Issue (8): 48-53.  DOI: 10.13718/j.cnki.xdzk.2019.08.008
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  • 一类Caputo分数阶脉冲微分方程混合边值问题解的存在唯一性    [PDF全文]
    邢艳元1,2, 郭志明1     
    1. 广州大学 数学与信息科学学院, 广州 510006;
    2. 吕梁学院 数学系, 山西 吕梁 033000
    摘要:主要研究了一类1 < α < 2的分数阶脉冲微分方程的混合边值问题.首先将非线性微分方程转化为等价的积分方程,然后利用Leray-Schauder和Altman不动点定理,得到了解的存在性和唯一性,并且给出了一个例子说明结论的正确性,推广和改进了相关结论.
    关键词Caputo导数    脉冲微分方程    不动点理论    混合边值问题    

    分数阶微积分理论是整数阶微积分理论的延伸与拓展,由于其良好的遗传性和记忆性,受到越来越多学者的广泛关注,它在光学和热学系统、流变学及材料力学系统、信号处理和系统识别、控制和机器人等应用领域都有广泛的应用[1-6].脉冲微分方程是微分方程理论的一个重要分支,脉冲现象在现代科技各领域的实际问题中是普遍存在的,脉冲微分系统最突出的特点是能够充分考虑到瞬时突变现象对状态的影响,能够更深刻、更精确地反映事物的变化规律[7-11].

    文献[12]研究了一类1<α<2的Caputo分数阶脉冲微分方程:

    $ \left\{ \begin{array}{l} _0^C{\rm{D}}_t^\alpha u(t) = f(t,u(t))\;\;\;\;\;\;\;\;t \in J',J = [0,1]\\ {\left. {\Delta u} \right|_{t = {t_k}}} = {I_k}\left( {u\left( {{t_k}} \right)} \right),{\left. {\Delta u'} \right|_{t = {t_k}}} = {J_k}\left( {u\left( {{t_k}} \right)} \right)\;\;\;\;\;\;\;\;\;\;\;k = 1,2, \cdots ,m\\ au(0) + bu(1) = 0,au'(0) + bu'(1) = 0 \end{array} \right. $

    其中Ik(u(tk))=u(tk+)-u(tk-),Jk(u(tk))=u′(tk+)-u′(tk-),0CDtα是Caputo分数阶导数.通过应用Krasonselskii不动点定理研究了上述方程反周期边值问题解的存在唯一性.

    文献[13]研究了一类1<α<2的Caputo分数阶脉冲微分方程:

    $ \left\{ \begin{array}{l} _0^C{\rm{D}}_t^\alpha u(t) = f(t,u(t))\;\;\;\;\;\;\;\;t \in J',J = [0,1]\\ {\left. {\Delta u} \right|_{t = {t_k}}} = {I_k}\left( {u\left( {{t_k}} \right)} \right),{\left. {\Delta u'} \right|_{t = {t_k}}} = {J_k}\left( {u\left( {{t_k}} \right)} \right)\;\;\;\;\;\;\;\;\;\;\;k = 1,2, \cdots ,m\\ u(0) + u'(0) = 0,u(1) + u'(1) = 0 \end{array} \right. $

    其中Ik(u(tk))=u(tk+)-u(tk-),Jk(u(tk))=u′(tk+)-u′(tk-),0CDtα是Caputo分数阶导数.通过应用Banach不动点理论得出了上述方程混合边值问题解的存在唯一性.

    受以上结论的启发,本文主要研究如下方程:

    $ \left\{ \begin{array}{l} _0^C{\rm{D}}_t^\alpha u(t) = f(t,u(t))\\ {\left. {\Delta u} \right|_{t = {t_k}}} = {I_k}\left( {u\left( {{t_k}} \right)} \right),{\left. {\Delta u'} \right|_{t = {t_k}}} = {J_k}\left( {u\left( {{t_k}} \right)} \right)\;\;\;\;\;\;\;\;\;\;\;k = 1,2, \cdots ,p\\ u(0) + u'(1) = 0,u'(0) + u(1) = 0 \end{array} \right. $ (1)

    其中0CDtα是Caputo分数阶导数,1<α<2,f: J×ℝ→ℝ,IkJk: ℝ→ℝ是连续函数,并且

    $ 0=t_{0}<t_{1}<\cdots<t_{p}<t_{p+1}=1 $
    $ \begin{array}{*{20}{c}} {{I_k}\left( {u\left( {{t_k}} \right)} \right) = u\left( {t_k^ + } \right) - u\left( {t_k^ - } \right)}&{{J_k}\left( {u\left( {{t_k}} \right)} \right) = u'\left( {t_k^ + } \right) - u'\left( {t_k^ - } \right)} \end{array} $

    其中u(tk+)和u(tk-)分别指u(t)在t=tk处的右极限和左极限,同理u′(tk+)和u′(tk-)分别指u′(t)在t=tk处的右极限和左极限.

    1 预备知识

    J=[0, 1],J0=[0,t1],Ji=(titi+1],i=1,2,…,p-1,Jp=(tp,1],D=[t1t2,…,tp]∈(0,1),J=J\D.并且PC[J,ℝ]={u|u: J→ℝ,uC[0,t1]∪C(tktk+1];u(tk+)和u(tk-)都存在,且u(tk)=u(tk-)(1≤kp)},其中${\left\| u \right\|_{PC}} = \mathop {\sup }\limits_{t \in J} |u(t)|$.

    定义1[14]  函数f∈[0,∞)Rα>0阶Riemann-Liouville积分

    $ _0I_t^af(t) = \frac{1}{{\mathit{\Gamma }(\alpha )}}\int_0^t {{{(t - s)}^{a - 1}}} f(s){\rm{d}}s $

    叫作函数f(t)的α次积分.若函数f(t) n阶可导,则对f(t)的α次Caputo导数定义如下:

    $ _0^C{\rm{D}}_t^\alpha f(t) = \frac{1}{{\mathit{\Gamma }(n - \alpha )}}\int_0^t {{{(t - s)}^{n - \alpha - 1}}} {f^{(n)}}(s){\rm{d}}s\quad \forall \alpha > 0 $

    其中n=[α]+1.

    引理1[14]  令α>0,则0Itα0CDtαh(t)=h(t)+c0+c1t+…+cn-1tn-1,其中ci为常数,n=[α]+1.

    引理2[15]  EPC(J,ℝ)是相对紧的当且仅当任何函数u(t)∈EJ上一致有界,且在Jk(k=1,2,…,p)上是等度连续的.

    引理3[16]  若E是Banach空间,假设ΩE为有界的开集,且θΩ,令T: ΩE是全连续算子,满足∀u∈∂Ω,‖Tu‖≤‖u‖,则TΩ上有不动点.

    引理4[16]  若E是Banach空间,假设T: EE是全连续算子,且V={uE|u=μTu,0<μ<1}有界,则TE上有不动点.

    引理5  假设y: [0, 1]→ℝ是连续函数,且1<α<2,则分数阶脉冲微分方程混合边值问题

    $ \left\{ \begin{array}{l} _0^C{\rm{D}}_t^\alpha u(t) = y(t)\quad 1 < \alpha < 2\\ {\left. {\Delta u} \right|_{t = {t_k}}} = {I_k}\left( {u\left( {{t_k}} \right)} \right),{\left. {\Delta u'} \right|_{t = {t_k}}} = {J_k}\left( {u\left( {{t_k}} \right)} \right)\\ u(0) + {u^\prime }(1) = 0,{u^\prime }(0) + u(1) = 0 \end{array} \right. $

    等价于非线性积分方程

    $ u\left( t \right) = \left\{ \begin{array}{l} \frac{1}{{\mathit{\Gamma }(\alpha )}}\int_0^t {{{(t - s)}^{\alpha - 1}}} y(s){\rm{d}}s + \frac{{t - 2}}{{\mathit{\Gamma }(\alpha - 1)}}\int_0^1 {{{(1 - s)}^{\alpha - 2}}} y(s){\rm{d}}s + \frac{{1 - t}}{{\mathit{\Gamma }(\alpha )}}\int_0^1 {{{(1 - s)}^{\alpha - 1}}} y(s){\rm{d}}s\;\;\;\;\;\;\;t \in {J_0}\\ \frac{1}{{\mathit{\Gamma }(\alpha )}}\int_0^t {{{(t - s)}^{\alpha - 1}}} y(s){\rm{d}}s + \frac{{t - 2}}{{\mathit{\Gamma }(\alpha - 1)}}\int_0^1 {{{(1 - s)}^{\alpha - 2}}} y(s){\rm{d}}s + \frac{{1 - t}}{{\mathit{\Gamma }(\alpha )}}\int_0^1 {{{(1 - s)}^{\alpha - 1}}} y(s){\rm{d}}s + \\ (t - 2)\sum\limits_{j = 1}^p {{J_j}} \left( {u\left( {{t_j}} \right)} \right){t_j} + (t - 1)\sum\limits_{j = 1}^p {{J_j}} \left( {u\left( {{t_j}} \right)} \right) + (2 - t)\sum\limits_{j = 1}^p {{I_j}} \left( {u\left( {{t_j}} \right)} \right) + \\ \sum\limits_{j = k + 1}^p {\left( {{t_j} - t} \right)} {J_j}\left( {u\left( {{t_j}} \right)} \right) - \sum\limits_{j = k + 1}^p {{I_j}} \left( {u\left( {{t_j}} \right)} \right),t \in {J_k},k = 1,2, \cdots ,p \end{array} \right. $

      必要性

    假设u(t)是方程(1)的解,当tJk时,对方程(1)中的第一个方程两边同时取α次积分,再由引理1,对akbk∈ℝ(k=0,1,…,p),有

    $ \begin{array}{*{20}{c}} {{}_0I_t^\alpha {}_0^C{\rm{D}}_t^\alpha u\left( t \right) = {}_0I_t^\alpha y\left( t \right) = \frac{1}{{\mathit{\Gamma }\left( \alpha \right)}}\int_0^t {{{\left( {t - s} \right)}^{\alpha - 1}}y\left( s \right){\rm{d}}s} }\\ {u\left( t \right) = \frac{1}{{\mathit{\Gamma }\left( \alpha \right)}}\int_0^t {{{\left( {t - s} \right)}^{\alpha - 1}}y\left( s \right){\rm{d}}s} + {a_k} + {b_k}t} \end{array} $ (2)
    $ u'(t) = \frac{1}{{\mathit{\Gamma }(\alpha - 1)}}\int_0^t {{{(t - s)}^{a - 2}}} y(s){\rm{d}}s + {b_k} $ (3)

    再由t0=0,tp=1,以及u(0)+u′(1)=0,u′(0)+u(1)=0,可得

    $ {a_0} + \frac{1}{{\mathit{\Gamma }(\alpha - 1)}}\int_0^1 {{{(1 - s)}^{a - 2}}} y(s){\rm{d}}s + {b_p} = 0 $ (4)
    $ {b_0} + \frac{1}{{\mathit{\Gamma }(\alpha )}}\int_0^1 {{{(1 - s)}^{a - 1}}} y(s){\rm{d}}s + {a_p} + {b_p} = 0 $ (5)

    又因为Δu′(tk)=Jk(u(tk))=u′(tk+)-u′(tk-)=bk-bk-1Δu(tk)=Ik(u(tk))=u(tk+)-u(tk-),得

    $ b_{k}=b_{p}-\sum\limits_{j=k+1}^{p} J_{j}\left(u\left(t_{j}\right)\right) $ (6)
    $ a_{k}=a_{p}+\sum\limits_{j=k+1}^{p} J_{j}\left(u\left(t_{j}\right)\right) t_{j}-\sum\limits_{j=k+1}^{p} I_{j}\left(u\left(t_{j}\right)\right) $ (7)

    将(6),(7)式代入(4),(5)式,且两式相减,可得

    $ \begin{array}{l} {b_p} = \sum\limits_{j = 1}^p {{J_j}} \left( {u\left( {{t_j}} \right)} \right){t_j} + \sum\limits_{j = 1}^p {{J_j}} \left( {u\left( {{t_j}} \right)} \right) - \sum\limits_{j = 1}^p {{I_j}} \left( {u\left( {{t_j}} \right)} \right) + \frac{1}{{\mathit{\Gamma }(\alpha - 1)}}\int_0^1 {{{(1 - s)}^{a - 2}}} y(s){\rm{d}}s - \\ \;\;\;\;\;\;\frac{1}{{\mathit{\Gamma }(\alpha )}}\int_0^1 {{{(1 - s)}^{\alpha - 1}}} y(s){\rm{d}}s \end{array} $ (8)
    $ \begin{array}{l} {a_p} = - 2\sum\limits_{j = 1}^p {{J_j}} \left( {u\left( {{t_j}} \right)} \right){t_j} - \sum\limits_{j = 1}^p {{J_j}} \left( {u\left( {{t_j}} \right)} \right) - \frac{2}{{\mathit{\Gamma }(\alpha - 1)}}\int_0^1 {{{(1 - s)}^{a - 2}}} y(s){\rm{d}}s + \\ \;\;\;\;\;\;2\sum\limits_{j = 1}^p {{I_j}} \left( {u\left( {{t_j}} \right)} \right) + \frac{1}{{\mathit{\Gamma }(\alpha )}}\int_0^1 {{{(1 - s)}^{a - 1}}} y(s){\rm{d}}s \end{array} $ (9)

    将(8),(9)式代入(6),(7)式,得

    $ \begin{array}{l} {a_k} = - 2\sum\limits_{j = 1}^p {{J_j}} \left( {u\left( {{t_j}} \right)} \right){t_j} - \sum\limits_{j = 1}^p {{J_j}} \left( {u\left( {{t_j}} \right)} \right) - \frac{2}{{\mathit{\Gamma }(\alpha - 1)}}\int_0^1 {{{(1 - s)}^{a - 2}}} y(s){\rm{d}}s + 2\sum\limits_{j = 1}^p {{I_j}} \left( {u\left( {{t_j}} \right)} \right) + \\ \;\;\;\;\;\;\frac{1}{{\mathit{\Gamma }(\alpha )}}\int_0^1 {{{(1 - s)}^{\alpha - 1}}} y(s){\rm{d}}s + \sum\limits_{j = k + 1}^p {{J_j}} \left( {u\left( {{t_j}} \right)} \right){t_j} - \sum\limits_{j = k + 1}^p {{I_j}} \left( {u\left( {{t_j}} \right)} \right) \end{array} $ (10)
    $ \begin{array}{l} {b_k} = \sum\limits_{j = 1}^p {{J_j}} \left( {u\left( {{t_j}} \right)} \right){t_j} + \sum\limits_{j = 1}^p {{J_j}} \left( {u\left( {{t_j}} \right)} \right) - \sum\limits_{j = 1}^p {{I_j}} \left( {u\left( {{t_j}} \right)} \right) + \frac{1}{{\mathit{\Gamma }(\alpha - 1)}}\int_0^1 {{{(1 - s)}^{a - 2}}} y(s){\rm{d}}s - \\ \;\;\;\;\;\frac{1}{{\mathit{\Gamma }(\alpha )}}\int_0^1 {{{(1 - s)}^{\alpha - 1}}} y(s){\rm{d}}s - \sum\limits_{j = k + 1}^p {{J_j}} \left( {u\left( {{t_j}} \right)} \right) \end{array} $ (11)

    因此,对k=0,1,…,p-1,当t∈(tktk+1]时,有

    $ \begin{array}{l} u(t) = \frac{1}{{\mathit{\Gamma }(\alpha )}}\int_0^t {{{(t - s)}^{\alpha - 1}}} y(s){\rm{d}}s + \frac{{t - 2}}{{\mathit{\Gamma }(\alpha - 1)}}\int_0^t {{{(1 - s)}^{\alpha - 2}}} y(s){\rm{d}}s + \frac{{1 - t}}{{\mathit{\Gamma }(\alpha )}}\int_0^t {{{(1 - s)}^{\alpha - 1}}} y(s){\rm{d}}s + \\ \;\;\;\;\;\;\;\;\;(t - 2)\sum\limits_{j = 1}^p {{J_j}} \left( {u\left( {{t_j}} \right)} \right){t_j} + (t - 1)\sum\limits_{j = 1}^p {{J_j}} \left( {u\left( {{t_j}} \right)} \right) + (2 - t)\sum\limits_{j = 1}^p {{I_j}} \left( {u\left( {{t_j}} \right)} \right) + \\ \;\;\;\;\;\;\;\;\;\sum\limits_{j = k + 1}^p {\left( {{t_j} - t} \right)} {J_j}\left( {u\left( {{t_j}} \right)} \right) - \sum\limits_{j = k + 1}^p {{I_j}} \left( {u\left( {{t_j}} \right)} \right) \end{array} $

    则必要性成立.另外,对积分方程两边分别求α次分数阶导数,易知其满足方程(1),充分性成立.

    2 主要内容

    定义算子A: PC(J,ℝ)→PC(J,ℝ),令算子

    $ \begin{array}{l} (Au)(t) = \frac{1}{{\mathit{\Gamma }(\alpha )}}\int_0^t {{{(t - s)}^{\alpha - 1}}} f(s,u(s)){\rm{d}}s + \frac{{t - 2}}{{\mathit{\Gamma }(\alpha - 1)}}\int_0^1 {{{(1 - s)}^{\alpha - 2}}} f(s,u(s)){\rm{d}}s + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{{1 - t}}{{\mathit{\Gamma }(\alpha )}}\int_0^1 {{{(1 - s)}^{\alpha - 1}}} f(s,u(s)){\rm{d}}s + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;(t - 2)\sum\limits_{j = 1}^p {{J_j}} \left( {u\left( {{t_j}} \right)} \right){t_j} + (t - 1)\sum\limits_{j = 1}^p {{J_j}} \left( {u\left( {{t_j}} \right)} \right) + (2 - t)\sum\limits_{j = 1}^p {{I_j}} \left( {u\left( {{t_j}} \right)} \right) + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{j = k + 1}^p {\left( {{t_j} - t} \right)} {J_j}\left( {u\left( {{t_j}} \right)} \right) - \sum\limits_{j = k + 1}^p {{I_j}} \left( {u\left( {{t_j}} \right)} \right) \end{array} $

    在引理5中,令y(t)=f(tu(t)),则分数阶脉冲微分方程混合边值问题的解可以转化为Au=u,分数阶脉冲微分方程(1)有解当且仅当A有不动点.

    定理1  若满足以下条件:

    (H1) $\mathop {\lim }\limits_{u \to 0} \frac{{f(t, u)}}{u} = 0, \mathop {\lim }\limits_{u \to 0} \frac{{{I_k}(u)}}{u} = 0, \mathop {\lim }\limits_{u \to 0} \frac{{{J_k}(u)}}{u} = 0$.

    则分数阶脉冲微分方程混合边值问题(1)至少存在1个解.

      令ΩPC(J,ℝ)有界.由于fIkJkJ上连续,所以A连续.由于Ω有界,则对∀uΩ,存在常数L1>0,L2>0,使得|f(tu)|≤L1,|Ik(u)|≤L2,|Jk(u)|≤L3,所以

    $ |(A u)(t)| \leqslant \frac{3 L_{1}(1+\alpha)}{\Gamma(\alpha+1)}+4 p L_{2}+7 p L_{3}=L $

    即算子AΩ上有界.

    对∀uPC(J,ℝ),∀tτJkk=0,1,…,p,当tτ时,|(Au)(t)-(Au)(τ)|→0,即A等度连续.再由引理2知,算子A是定义在Ω上的全连续算子.

    由条件(H1)知,存在ε1>0,ε2>0,ε3>0和常数C,对∀tJ,当‖u‖≤R时,有

    $ |f(t, u)| \leqslant \varepsilon_{1}|u| \quad\left|I_{k}(u)\right| \leqslant \varepsilon_{2}|u| \quad\left|J_{k}(u)\right| \leqslant \varepsilon_{3}|u| $

    且使得$\frac{{3(1 + \alpha )}}{{\mathit{\Gamma }(\alpha + 1)}}{\varepsilon _1}$+4pε2+7pε3≤1成立.令Ω={uPC(J,ℝ)|‖u‖≤r},取uPC(J,ℝ)且‖u‖=r,即u∈∂u,有|f(tu)|≤ε1|u|,|Ik(u)|≤ε2|u|,|Jk(u)|≤ε3|u|.所以

    $ |(Au)(t)| \le \left[ {\frac{{3(1 + \alpha )}}{{\mathit{\Gamma }(\alpha + 1)}}{\varepsilon _1} + 4p{\varepsilon _2} + 7p{\varepsilon _3}} \right]\left\| {u(t)} \right\| \le \left\| {u(t)} \right\| $

    即对∀u∈∂u,‖Au‖≤‖u‖.由引理3知,分数阶脉冲微分方程混合边值问题(1)至少存在1个解uΩ.

    定理2  若满足以下条件:

    (H2)对∀tJkuPC(J,ℝ),若存在常数L1L2,使得|f(tu)|≤L1,|Ik(u)|≤L2,|Jk(u)|≤L3.

    则分数阶脉冲微分方程混合边值问题(1)至少有1个解.

      令

    $ V=\{u \in P C(J, \mathbb{R}) | u=\mu T u, 0<\mu<1\} $

    则对∀uV,以及∀tJk,有

    $ |u(t)| = \mu |(Au)(t)| \le \frac{{3\mu {L_1}(1 + \alpha )}}{{\mathit{\Gamma }(\alpha + 1)}} + 4\mu p{L_2} + 7\mu p{L_3} $

    则对∀tJ,有‖u‖≤$\frac{{3\mu {L_1}(1 + \alpha )}}{{\mathit{\Gamma }(\alpha + 1)}}$+4μpL2+7μpL3.所以集合V有界.再由定理1可知,A是全连续算子.故由引理4可知,算子APC(J,ℝ)上至少有1个不动点,则分数阶脉冲微分方程混合边值问题(1)至少有1个解.

    定理3  若满足以下条件:

    (H3)对∀tJ,∀uvPC(J,ℝ),存在非负常数lk(k=1,2),满足∀tkJ,有

    $ \begin{array}{*{20}{c}} {\left| {{I_k}\left( {u\left( {{t_k}} \right)} \right) - {I_k}\left( {v\left( {{t_k}} \right)} \right)} \right| \le {l_1}|u - v|}&{\left| {{J_k}\left( {u\left( {{t_k}} \right)} \right) - {J_k}\left( {v\left( {{t_k}} \right)} \right)} \right| \le {l_2}|u - v|} \end{array} $

    (H4)存在非负连续函数L(t),使得∀uvPC(J,ℝ),有|f(tu)-f(tv)|≤L(t)|u-v|.且对∀tJ,有$\frac{{3(1 + \alpha )}}{{\mathit{\Gamma }(\alpha + 1)}}$L(t)+4pl1+7pl2<1.

    则方程(1)有唯一的解.

      对∀u(t),v(t)∈Ω,有

    $ |(Au)(t) - (Av)(t)| \le \left[ {\frac{{3(1 + \alpha )}}{{\mathit{\Gamma }(\alpha + 1)}}L(t) + 4p{l_1} + 7p{l_2}} \right]|u - v| $

    因此

    $ \|A u-A v\| \leqslant\left[\frac{3(1+\alpha)}{\Gamma(\alpha+1)} L(t)+4 p l_{1}+7 p l_{2}\right]\|u-v\| $

    由条件可知

    $ \frac{3(1+\alpha)}{\Gamma(\alpha+1)} L(t)+4 p l_{1}+7 p l_{2}<1 $

    所以A是压缩映像.故A有唯一的不动点,即混合边值问题(1)有唯一的解.

    3 例子
    $ \left\{ \begin{array}{l} {}_0^C{\rm{D}}_t^{\frac{3}{2}}u\left( t \right) = \frac{{{{\rm{e}}^{ - 2t}}\left| {u\left( t \right)} \right|}}{{\left( {100 + {{\rm{e}}^t}} \right)\left( {1 + \left| {u\left( t \right)} \right|} \right)}}\;\;\;\;\;\;\;\;\;t \in \left[ {0,1} \right],t \ne \frac{1}{4}\\ \Delta u\left( {\frac{1}{4}} \right) = \frac{{\left| {u\left( t \right)} \right|}}{{{{\left( {t + 5} \right)}^2}\left( {9 + \left| {u\left( t \right)} \right|} \right)}},\Delta u'\left( {\frac{1}{4}} \right) = \frac{{\left| {u\left( t \right)} \right|}}{{{{\left( {t + 7} \right)}^2}\left( {12 + \left| {u\left( t \right)} \right|} \right)}}\\ u(0) + u'(1) = 0,u(1) + u'(0) = 0 \end{array} \right. $ (12)

    对∀u(t),v(t),∀t∈[0, 1],有

    $ \begin{array}{*{20}{c}} {\left| {\mathit{f}{\rm{(}}\mathit{t}{\rm{, }}\mathit{u}{\rm{)}} - \mathit{f}{\rm{(}}\mathit{t}{\rm{, }}\mathit{v}{\rm{)}}} \right| = \frac{{{{\rm{e}}^{ - 2t}}}}{{\left( {100 + {{\rm{e}}^t}} \right)}}\left| {\frac{u}{{1 + u}} - \frac{v}{{1 + v}}} \right| = \frac{{{{\rm{e}}^{ - 2t}}}}{{\left( {100 + {{\rm{e}}^t}} \right)\left( {1 + u} \right)\left( {1 + v} \right)}}\left| {u - v} \right| \le }\\ {\frac{{{{\rm{e}}^{ - 2t}}}}{{\left( {100 + {{\rm{e}}^t}} \right)}}\left| {u - v} \right| \le \frac{{{{\rm{e}}^{ - 2t}}}}{{101}}\left| {u - v} \right|} \end{array} $

    显然,对u∈[0,+∞)及∀t∈[0,1],有

    $ |f(t, u)|=\frac{\mathrm{e}^{-2 t}}{\left(100+\mathrm{e}^{t}\right)}\left|\frac{u}{1+u}\right| \leqslant \frac{\mathrm{e}^{-2 t}}{\left(100+\mathrm{e}^{t}\right)} \leqslant \frac{\mathrm{e}^{-2 t}}{101} $

    l1=$\frac{1}{25}$l2=$\frac{1}{49}$,因为1.33<Γ(2.5)<1.34,则

    $ \begin{array}{l} \frac{{3(1 + \alpha )}}{{\mathit{\Gamma }(\alpha + 1)}}L(t) + 4{l_1} + 7{l_2} < \frac{{7.5}}{{\mathit{\Gamma }(2.5)}} \cdot \frac{{{{\rm{e}}^{ - 2t}}}}{{101}} + \frac{4}{{25}} + \frac{1}{7} \le \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0.06 + 0.16 + 0.14 = 0.36 < 1 \end{array} $

    定理3中的所有假设满足,所以方程(12)有唯一解.

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    Existence and Uniqueness of Solution for a Class of Caputo Fractional Impulsive Differential Equations with Mixed Boundary Value Problem
    XING Yan-yuan1,2, GUO Zhi-ming1     
    1. School of Mathematics and Information Sciences, Guangzhou University, Guangzhou 510006, China;
    2. Department of Mathematics, Lüliang University, Luliang Shanxi 033000, China
    Abstract: In this paper, we study a class of Caputo fractional impulsive differential equations with the mixed boundary value problem of fractional order α∈(1, 2). Firstly, we transform the non-linear differential equation into an equivalent fractional integral equation. Secondly, by using the Leray-Schauder and Altman fixed point theorem, we obtain the existence and uniqueness of the solution. Finally, an example is given to demonstrate the validity of the main result, and relevant results are generalized and improved.
    Key words: Caputo derivative    impulsive differential equation    fixed point theorem    mixed boundary value problem    
    X