近几年,分数阶微分方程在许多科学领域得到广泛的应用,例如流体力学、自动控制、生物工程和电子网络等.这使得分数阶微分方程理论得到快速发展,很多文献和专著对分数阶微分方程作了研究或介绍^[1-4].在分数阶微分方程理论中,分数阶微分方程边值问题作为重要的组成部分,吸引了很多学者的兴趣,目前有很多研究结果,见相关文献[5-13].

众所周知,反周期是物理学中一种重要的现象,许多具有实际意义的数学模型都具有反周期形式的边值条件,因此研究具有反周期边值条件的微分方程解的存在性是有意义的.目前已经有学者对分数阶微分方程反周期边值问题作了研究,并得到了一些研究结果,见相关文献[5-8],特别是在文献[5]中,作者研究了下面一类分数阶微分方程反周期边值问题解的存在性

(1.1) $\begin{equation}\label{qyFBVP1}\left\{\begin{array}{l}^{C}D^{\alpha}u(t)=f(t, u(t)), ~~t\in [0, T], ~T>0, ~2 <\alpha\leq3, \\ u(0)=-u(T), ~{^{C}D^{p}}u(0)=-{^{C}D^{p}}u(T), ~{^{C}D^{q}}u(0)=-{^{C}D^{q}}u(T), \end{array} \right. \end{equation} $

这里${^{C}D^{\alpha}}$为$\alpha$阶Caputo分数阶微分, $0 <p <1 <q <2$, $f:[0, T]\times\mathbb{R} \rightarrow\mathbb{R} $为给定的连续函数.作者利用压缩映射原理和Leray-Schauder不动点定理得到该边值问题(1.1)解的存在性结论.

因为具有$p$-Laplacian算子的微分方程涉及到很多研究领域,如在流变学和材料学等学科,所以对它的研究引起了中外学者的广泛关注.目前具有$p$-Laplacian算子的分数阶微分方程边值问题解的存在性已有一些研究结果,见文献[9-12].在文献[9]中作者研究了下面一类具有$p$-Laplacian算子的分数阶微分方程边值问题解的存在性

(1.2) $\begin{equation}\label{qyFBVP2}\left\{\begin{array}{l} \left(\phi_{p}\left(^{C}D_{0+}^{\alpha}x(t)\right)\right)'=f(t, x(t)), ~~~t\in [0, 1], \\ x(0)=a_{0}(1), ~x'(0)=a_{1}x'(1), ~x''(0)=a_{2}x''(1), \end{array} \right. \end{equation}$

这里$a_{i}\neq1, i=1, 2, 3$, $f\in C([0, T]\times \mathbb{R} , \mathbb{R} )$, $x(t)\in C^{2}([0, 1], \mathbb{R})$.作者利用Banach压缩映像原理,得到该边值问题(1.2)在一定条件下解的存在性结论.

在上面提到的文献[5]中,作者研究了阶数为$2 <\alpha\leq3$的分数阶微分方程反周期边值问题解的存在性,但该问题不带有$p$-Laplacian算子;在文献[9]中,作者仅仅研究了边值条件为整数阶导数时带有$p$-Laplacian算子的分数阶微分方程解的存在性,没有研究边值条件为分数阶导数时的情形.研究具有$p$-Laplacian算子的分数阶微分方程反周期边值问题解的存在性有一定的理论和实践意义,但此方面的研究还比较少^[10],所以研究具有$p$-Laplacian算子的分数阶微分方程反周期分数阶边值问题解的存在性是自然而然的事情.受前面文献[5, 9]的启发,本文考虑下面一类具有$p$-Laplacian算子的非线性Caputo分数阶微分方程反周期边值问题解的存在唯一性

(1.3) $\begin{equation}\label{FBVP1}\left\{\begin{array}{l}\left(\phi_{p}\left(^{C}D_{0+}^{\alpha}u(t)\right)\right)'=f(t, u(t)), ~~~t\in [0, 1], \\ u(0)=-u(1), ~{^{C}D_{0+}^{\beta}}u(0)=-{^{C}D_{0+}^{\beta}}u(1), ~{^{C}D_{0+}^{\gamma}}u(0)=-{^{C}D_{0+}^{\gamma}}u(1), \end{array} \right. \end{equation}$

这里$2 <\alpha\leq3$, $0 <\beta <1 <\gamma <2$, $\phi_{p}(s)=|s|^{p-2}s$, $p>1$.$^{C}D_{0+}^{\alpha}, ^{C}D_{0+}^{\beta}$和$^{C}D_{0+}^{\gamma}$为Caputo分数阶微分, $f\in C([0, T]\times \mathbb{R} , \mathbb{R} )$.令$\phi_q=\phi_{p}^{-1}$,其中$\frac{1}{p}+\frac{1}{q}=1$.首先,利用分数阶微分方程和反周期边值条件给出该边值问题的Green函数,然后利用$p$-Laplacian算子的性质和Banach压缩映射原理得到该边值问题解的存在唯一性结论,最后给出两个例子验证结论的合理性,值得一提的是此文研究的微分方程反周期边值条件是带有Caputo分数阶微分.

本文结构如下:在第2节给出分数阶积分和微分的定义及证明结论时需要的引理,第3节利用分数阶微分方程和反周期边值条件给出边值问题(1.3)的Green函数,第4节利用$p$-Laplacian算子的性质和Banach压缩映射原理得到边值问题(1.3)解的存在唯一性结论,在第5节给出两个例子验证得到的主要结果.

定义2.1^[1]  若$\alpha>0$,函数$u:(0, +\infty)\rightarrow \mathbb{R} $的分数阶积分定义为

$I_{0+}^{\alpha}u(t)=\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}u(s){\rm d}s.$

定义2.2^[1]  若$\alpha>0$,函数$u:(0, +\infty)\rightarrow \mathbb{R} $的Caputo分数阶微分定义为

$^{C}D_{0+}^{\alpha}u(t)=\frac{1}{\Gamma(n-\alpha)}\int_{0}^{t}(t-s)^{n-\alpha-1}u^{(n)}(s){\rm d}s, $

这里$n-1 <\alpha <n$, $n=[\alpha]+1$.

引理2.1^[1]  若$\alpha>0$,则分数阶微分方程$^{C}D_{0+}^{\alpha}u(t)=0$的解为

$u(t)=c_{0}+c_{1}t+c_{2}t^{2}+\cdots+c_{n-1}t^{n-1}, $

这里$c_{i}\in \mathbb{R} $, $i=1, 2, \cdots, n-1$, $n=[\alpha]+1$.

引理2.2^[1]  若$\alpha>0$,则

$I_{0+}^{\alpha}{^{C}D_{0+}^{\alpha}}u(t)=u(t)+c_{0}+c_{1}t+c_{2}t^{2}+\cdots+c_{n-1}t^{n-1}, $

这里$c_{i}\in \mathbb{R} $, $i=1, 2, \cdots, n-1$, $n=[\alpha]+1$.

引理2.3^[9]  $p$-Laplacian算子$\phi_p$具有下面的性质

(ⅰ)如果$1 <p\leq2$, $xy>0$且$| x| , | y| \geq m>0$,则下面的不等式成立

$| \phi_{p}(x)-\phi_{p}(y)| \leq (p-1)m^{p-2}| x-y| .$

(ⅱ)如果$p>2$且$| x| , | y| \leq M$,则下面的不等式成立

$| \phi_{p}(x)-\phi_{p}(y)| \leq (p-1)M^{p-2}| x-y| .$

引理3.1  若$2 <\alpha\leq3$, $0 <\beta <1 <\gamma <2$,函数$y\in C[0, 1]$,则分数阶微分方程反周期边值问题

$\begin{eqnarray*}\left\{\begin{array}{l}\left(\phi_{p}\left(^{C}D_{0+}^{\alpha}u(t)\right)\right)'=y(t), ~~~t\in [0, 1], \\ u(0)=-u(1), ~{^{C}D_{0+}^{\beta}}u(0)=-{^{C}D_{0+}^{\beta}}u(1), ~{^{C}D_{0+}^{\gamma}}u(0)=-{^{C}D_{0+}^{\gamma}}u(1)\end{array} \right. \end{eqnarray*}$

拥有唯一解

$u(t)=\int_0^{1}G(t, s)\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s, $

这里

$ \begin{eqnarray*}\label{Gts} G(t, s)=\left\{ \begin{array}{ll} \frac{2(t-s)^{\alpha-1}-(1-s)^{\alpha-1}}{2\Gamma(\alpha)} +\frac{\Gamma(2-\beta)(1-2t)(1-s)^{\alpha-\beta-1}}{2\Gamma(\alpha-\beta)}\\ -\frac{\left[\beta-4t+2(2-\beta)t^{2}\right]\Gamma(3-\gamma)(1-s)^{\alpha-\gamma-1}}{4(2-\beta)\Gamma(\alpha-\gamma)}, ~~&s \leq t, \\ -\frac{(1-s)^{\alpha-1}}{2\Gamma(\alpha)} +\frac{\Gamma(2-\beta)(1-2t)(1-s)^{\alpha-\beta-1}}{2\Gamma(\alpha-\beta)}\\ -\frac{\left[\beta-4t+2(2-\beta)t^{2}\right]\Gamma(3-\gamma)(1-s)^{\alpha-\gamma-1}}{4(2-\beta)\Gamma(\alpha-\gamma)}, &t\leq s. \end{array}\right. \end{eqnarray*} $

证  对方程$\left(\phi_{p}\left(^{C}D_{0+}^{\alpha}u(t)\right)\right)'=y(t)$两边同时积分可得

$\phi_{p}\left(^{C}D_{0+}^{\alpha}u(t)\right)=\phi_{p}\left(^{C}D_{0+}^{\alpha}u(0)\right)+\int_0^{t}y(s){\rm d}s.$

由Caputo分数阶微分性质可知$^{C}D_{0+}^{\alpha}u(0)=0$,因此上式变为

$\phi_{p}\left(^{C}D_{0+}^{\alpha}u(t)\right)=\int_0^{t}y(s){\rm d}s.$

因为$\phi_{p}^{-1}=\phi_q$,因此

$^{C}D_{0+}^{\alpha}u(t)=\phi_{q}\left(\int_0^{t}y(s){\rm d}s\right).$

因为$2 <\alpha\leq3$,利用引理2.2可知

$ \begin{eqnarray*} u(t)&=&I_{0+}^{\alpha}\phi_{q}\left(\int_0^{t}y(s){\rm d}s\right)+c_{0}+c_{1}t+c_{2}t^{2}\\ &=&\frac{1}{\Gamma(\alpha)}\int_0^{t}(t-s)^{\alpha-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s+c_{0}+c_{1}t+c_{2}t^{2}. \end{eqnarray*} $

利用Caputo分数阶微分的性质,即$^{C}D_{0+}^{\beta}c=0$ ($c$为常数), $^{C}D_{0+}^{\beta}t=\frac{t^{1-\beta}}{\Gamma(2-\beta)}$, $^{C}D_{0+}^{\beta}t^{2}=\frac{2t^{2-\beta}}{\Gamma(3-\beta)}$,故

$\begin{eqnarray*} ^{C}D_{0+}^{\beta}u(t)=\frac{1}{\Gamma(\alpha-\beta)} \int_0^{t}(t-s)^{\alpha-\beta-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s+c_{1}\frac{t^{1-\beta}}{\Gamma(2-\beta)} +c_{2}\frac{2t^{2-\beta}}{\Gamma(3-\beta)}. \end{eqnarray*} $

又因为$^{C}D_{0+}^{\gamma}t=0$（$1 <\gamma <2$）和$^{C}D_{0+}^{\gamma}t^2=\frac{2t^{2-\gamma}}{\Gamma(3-\gamma)}$,因此

$ \begin{eqnarray*} ^{C}D_{0+}^{\gamma}u(t)=\frac{1}{\Gamma(\alpha-\gamma)} \int_0^{t}(t-s)^{\alpha-\gamma-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s +c_{2}\frac{2t^{2-\gamma}}{\Gamma(3-\gamma)}. \end{eqnarray*} $

利用反周期边值条件$u(0)=-u(1), ~{^{C}D_{0+}^{\beta}}u(0)=-{^{C}D_{0+}^{\beta}}u(1), ~{^{C}D_{0+}^{\gamma}}u(0)=-{^{C}D_{0+}^{\gamma}}u(1)$可得

$ \begin{eqnarray*} c_{0}&=&-\frac{1}{2\Gamma(\alpha)}\int_0^{1}(1-s)^{\alpha-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s\\ &&+\frac{\Gamma(2-\beta)}{2\Gamma(\alpha-\beta)}\int_0^{1}(1-s)^{\alpha-\beta-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s\\ &&-\frac{\beta\Gamma(3-\gamma)}{4(2-\beta)\Gamma(\alpha-\gamma)}\int_0^{1}(1-s)^{\alpha-\gamma-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s, \\ c_{1}&=&-\frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)}\int_0^{1}(1-s)^{\alpha-\beta-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s\\ &&+\frac{\Gamma(3-\gamma)}{(2-\beta)\Gamma(\alpha-\gamma)}\int_0^{1}(1-s)^{\alpha-\gamma-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s, \\ c_{2}&=&-\frac{\Gamma(3-\gamma)}{2\Gamma(\alpha-\gamma)}\int_0^{1}(1-s)^{\alpha-\gamma-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s.\\ \end{eqnarray*} $

因此

$ \begin{eqnarray*} u(t)&=&\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s\\ &&-\frac{1}{2\Gamma(\alpha)}\int_0^{1}(1-s)^{\alpha-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s\\ &&+\frac{\Gamma(2-\beta)}{2\Gamma(\alpha-\beta)}\int_0^{1}(1-s)^{\alpha-\beta-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s\\ &&-\frac{\beta\Gamma(3-\gamma)}{4(2-\beta)\Gamma(\alpha-\gamma)}\int_0^{1}(1-s)^{\alpha-\gamma-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s\\ &&-\frac{\Gamma(2-\beta)t}{\Gamma(\alpha-\beta)}\int_0^{1}(1-s)^{\alpha-\beta-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s\\ &&+\frac{\Gamma(3-\gamma)t}{(2-\beta)\Gamma(\alpha-\gamma)}\int_0^{1}(1-s)^{\alpha-\gamma-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s\\ &&-\frac{\Gamma(3-\gamma)t^2}{2\Gamma(\alpha-\gamma)}\int_0^{1}(1-s)^{\alpha-\gamma-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s\\ &=&\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s\\ &&-\frac{1}{2\Gamma(\alpha)}\int_0^{1}(1-s)^{\alpha-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s\\ &&+\frac{(1-2t)\Gamma(2-\beta)}{2\Gamma(\alpha-\beta)}\int_0^{1}(1-s)^{\alpha-\beta-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s\\ &&-\frac{\left[\beta-4t+2(2-\beta)t^2\right]\Gamma(3-\gamma)}{4(2-\beta)\Gamma(\alpha-\gamma)} \int_0^{1}(1-s)^{\alpha-\gamma-1}\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s \\&=&\int_{0}^{t}\bigg[\frac{2(t-s)^{\alpha-1}-(1-s)^{\alpha-1}}{2\Gamma(\alpha)} +\frac{\Gamma(2-\beta)(1-2t)(1-s)^{\alpha-\beta-1}}{2\Gamma(\alpha-\beta)}\\ &&-\frac{\left[\beta-4t+2(2-\beta)t^{2}\right]\Gamma(3-\gamma)(1-s)^{\alpha-\gamma-1}}{4(2-\beta)\Gamma(\alpha-\gamma)}\bigg] \phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s\\ &&+\int_{t}^{1}\bigg[-\frac{(1-s)^{\alpha-1}}{2\Gamma(\alpha)} +\frac{\Gamma(2-\beta)(1-2t)(1-s)^{\alpha-\beta-1}}{2\Gamma(\alpha-\beta)}\\ &&-\frac{\left[\beta-4t+2(2-\beta)t^{2}\right]\Gamma(3-\gamma)(1-s)^{\alpha-\gamma-1}}{4(2-\beta)\Gamma(\alpha-\gamma)}\bigg] \phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right) {\rm d}s\\ &=&\int_0^{1}G(t, s)\phi_{q}\left(\int_0^{s}y(\tau){\rm d}\tau\right){\rm d}s. \end{eqnarray*} $

证毕.

定义Banach空间$X=C([0, 1], \mathbb{R} )$,其范数为$\parallel u\parallel=\sup\limits_{t\in[0, 1]}| u(t)| $.

定义算子$T:X\rightarrow X$为

$\begin{eqnarray*}(Tu)(t)&=&\int_0^{1}G(t, s)\phi_{q}\left(\int_0^{s}f(\tau, u(\tau)){\rm d}\tau\right){\rm d}s\\&=&\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}\phi_{q}\left(\int_0^{s}f(\tau, u(\tau)){\rm d}\tau\right){\rm d}s\\&&-\frac{1}{2\Gamma(\alpha)}\int_0^{1}(1-s)^{\alpha-1}\phi_{q}\left(\int_0^{s}f(\tau, u(\tau)){\rm d}\tau\right){\rm d}s\\&&+\frac{(1-2t)\Gamma(2-\beta)}{2\Gamma(\alpha-\beta)}\int_0^{1}(1-s)^{\alpha-\beta-1}\phi_{q}\left(\int_0^{s}f(\tau, u(\tau)){\rm d}\tau\right){\rm d}s\\&&-\frac{\left[\beta-4t+2(2-\beta)t^2\right]\Gamma(3-\gamma)}{4(2-\beta)\Gamma(\alpha-\gamma)}\int_0^{1}(1-s)^{\alpha-\gamma-1}\phi_{q}\left(\int_0^{s}f(\tau, u(\tau)){\rm d}\tau\right){\rm d}s.\end{eqnarray*}$

则求边值问题(1.3)解的存在唯一性转化为算子$T$的是否存在唯一不动点.

定理4.1  假设$1 <q <2$,若函数$f(t, u(t))$满足下面条件

(H1)存在常数$\lambda>0$, $0 <\delta <\frac{2}{2-q}$,对任意的$(t, u(t))\in [0, 1]\times\mathbb R$,有$\lambda\delta t^{\delta-1}\leq f(t, u(t))$成立;

(H2)存在常数$l>0$,使得当$t\in[0, 1]$时,对任意的函数$u, v\in X$,有$| f(t, u(t))-f(t, v(t))| \leq l| u-v| $成立;

(H3)

$\begin{eqnarray*}&&0 <l(q-1)\lambda^{q}\Gamma(\delta(q-2)+2)\bigg[\frac{3}{2\Gamma(\alpha+\delta(q-2)+2)} +\frac{\Gamma(2-\beta)}{2\Gamma(\delta(q-2)+2+\alpha-\beta)}\\ &&~~~~~~+\frac{\beta\Gamma(3-\gamma)}{4(2-\beta)\Gamma(\alpha-\gamma+\delta(q-2)+2)}\bigg] <1, \end{eqnarray*} $

则边值问题(1.3)存在唯一解.

证  对于任意的$(t, u(t))\in [0, 1]\times\mathbb{R} $,由条件(H1)可得

$\lambda t^{\delta}\leq\int_{0}^{t}f(s, u(s)){\rm d}s.$

因为$1 <q <2$,由引理2.3中性质(ⅰ)可知,对任意的函数$u, v\in X$,有

$ \begin{eqnarray*} &&\bigg|\phi_{q}\left(\int_0^{t}f(s, u(s)){\rm d}s\right)-\phi_{q}\left(\int_0^{t}f(s, v(s)){\rm d}s\right)\bigg|\\ &\leq&(q-1)(\lambda t^{\delta})^{q-2}\bigg|\int_0^{t}f(s, u(s)){\rm d}s-\int_0^{t}f(s, v(s)){\rm d}s\bigg|\\ &\leq&(q-1)\lambda^{q-2}t^{\delta(q-2)}\int_0^{t}| f(s, u(s))-f(s, v(s))| {\rm d}s\\ &\leq&l(q-1)\lambda^{q-2}t^{\delta(q-2)}\int_0^{t}| u(s)-v(s)| {\rm d}s\\ &\leq&l(q-1)\lambda^{q-2}t^{\delta(q-2)+1}\parallel u-v\parallel. \end{eqnarray*} $

因此,我们有

$ \begin{eqnarray*} &&| (Tu)(t)-(Tv)(t)| \\ &\leq&\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1} \bigg|\phi_{q}\left(\int_0^{s}f(\tau, u(\tau)){\rm d}\tau\right)-\phi_{q}\left(\int_0^{s}f(\tau, v(\tau)){\rm d}\tau\right)\bigg|{\rm d}s\\ &&+\frac{1}{2\Gamma(\alpha)}\int_0^{1}(1-s)^{\alpha-1} \bigg|\phi_{q}\left(\int_0^{s}f(\tau, u(\tau)){\rm d}\tau\right)-\phi_{q}\left(\int_0^{s}f(\tau, v(\tau)){\rm d}\tau\right)\bigg|{\rm d}s\\ &&+\bigg|\frac{(1-2t)\Gamma(2-\beta)}{2\Gamma(\alpha-\beta)}\bigg|\\ &&\times\int_0^{1}(1-s)^{\alpha-\beta-1} \bigg|\phi_{q}\left(\int_0^{s}f(\tau, u(\tau)){\rm d}\tau\right)-\phi_{q}\left(\int_0^{s}f(\tau, v(\tau)){\rm d}\tau\right)\bigg|{\rm d}s\\ &&+\bigg|\frac{\left[\beta-4t+2(2-\beta)t^2\right]\Gamma(3-\gamma)}{4(2-\beta)\Gamma(\alpha-\gamma)}\bigg|\\ &&\times\int_0^{1}(1-s)^{\alpha-\gamma-1} \bigg|\phi_{q}\left(\int_0^{s}f(\tau, u(\tau)){\rm d}\tau\right)-\phi_{q}\left(\int_0^{s}f(\tau, v(\tau)){\rm d}\tau\right)\bigg|{\rm d}s\\ &\leq&l(q-1)\lambda^{q}\parallel u-v\parallel \bigg[\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}s^{\delta(q-2)+1}{\rm d}s\\ &&+\frac{1}{2\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}s^{\delta(q-2)+1}{\rm d}s +\bigg|\frac{(1-2t)\Gamma(2-\beta)}{2\Gamma(\alpha-\beta)}\bigg|\int_{0}^{1}(1-s)^{\alpha-\beta-1}s^{\delta(q-2)+1}{\rm d}s\\ &&+\bigg|\frac{\left[\beta-4t+2(2-\beta)t^2\right]\Gamma(3-\gamma)}{4(2-\beta)\Gamma(\alpha-\gamma)}\bigg| \int_{0}^{1}(1-s)^{\alpha-\gamma-1}s^{\delta(q-2)+1}{\rm d}s\bigg]\\ &=&l(q-1)\lambda^{q}\parallel u-v\parallel \bigg[\frac{1}{\Gamma(\alpha)}\int_{0}^{1}t^{\alpha-1}(1-s)^{\alpha-1}t^{\delta(q-2)+1}s^{\delta(q-2)+1}t\, {\rm d}s\\ &&+\frac{1}{2\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}s^{\delta(q-2)+1}{\rm d}s +\bigg|\frac{(1-2t)\Gamma(2-\beta)}{2\Gamma(\alpha-\beta)}\bigg|\int_{0}^{1}(1-s)^{\alpha-\beta-1}s^{\delta(q-2)+1}{\rm d}s\\ &&+\bigg|\frac{\left[\beta-4t+2(2-\beta)t^2\right]\Gamma(3-\gamma)}{4(2-\beta)\Gamma(\alpha-\gamma)}\bigg| \int_{0}^{1}(1-s)^{\alpha-\gamma-1}s^{\delta(q-2)+1}{\rm d}s\bigg] \\ &=&l(q-1)\lambda^{q}\parallel u-v\parallel\bigg[\frac{t^{\alpha+\delta(q-2)+1}B(\alpha, \delta(q-2)+2)}{\Gamma(\alpha)} +\frac{B(\alpha, \delta(q-2)+2)}{2\Gamma(\alpha)}\\ &&+\bigg|\frac{(1-2t)\Gamma(2-\beta)}{2\Gamma(\alpha-\beta)}\bigg|B(\alpha-\beta, \delta(q-2)+2)\\ &&+\bigg|\frac{\left[\beta-4t+2(2-\beta)t^2\right]\Gamma(3-\gamma)}{4(2-\beta)\Gamma(\alpha-\gamma)}\bigg|B(\alpha-\gamma, \delta(q-2)+2)\bigg]\\ &\leq&l(q-1)\lambda^{q}\parallel u-v\parallel\bigg[\frac{3B(\alpha, \delta(q-2)+2)}{2\Gamma(\alpha)} +\frac{\Gamma(2-\beta)B(\alpha-\beta, \delta(q-2)+2)}{2\Gamma(\alpha-\beta)}\\ &&+\frac{\beta\Gamma(3-\gamma)B(\alpha-\gamma, \delta(q-2)+2)}{4(2-\beta)\Gamma(\alpha-\gamma)}\bigg]\\ &=&l(q-1)\lambda^{q}\Gamma(\delta(q-2)+2)\bigg[\frac{3}{2\Gamma(\alpha+\delta(q-2)+2)}\\ &&+\frac{\Gamma(2-\beta)}{2\Gamma(\delta(q-2)+2+\alpha-\beta)} +\frac{\beta\Gamma(3-\gamma)}{4(2-\beta)\Gamma(\alpha-\gamma+\delta(q-2)+2)}\bigg]\parallel u-v\parallel, \end{eqnarray*}$

其中$B(\cdot)$为Beta函数.令

$ \begin{eqnarray*} \Lambda_{1}&=&l(q-1)\lambda^{q}\Gamma(\delta(q-2)+2)\bigg[\frac{3}{2\Gamma(\alpha+\delta(q-2)+2)} +\frac{\Gamma(2-\beta)}{2\Gamma(\delta(q-2)+2+\alpha-\beta)}\\ &&+\frac{\beta\Gamma(3-\gamma)}{4(2-\beta)\Gamma(\alpha-\gamma+\delta(q-2)+2)}\bigg]. \end{eqnarray*} $

由条件(H3)可知$0 <\Lambda_{1} <1$,故

$\parallel Tu-Tv\parallel\leq\Lambda_{1}\parallel u-v\parallel.$

因此利用Banach压缩映射原理可知,算子$T$存在唯一不动点,即边值问题(1.3)存在唯一解.

定理4.2  假设$1 <q <2$,若函数$f(t, u(t))$满足定理4.1条件(H2)、(H3)和下面条件:

(H4)存在常数$\lambda>0$, $0 <\delta <\frac{2}{2-q}$,对任意的$(t, u(t))\in [0, 1]\times\mathbb{R} $,有$f(t, u(t))\leq-\lambda\delta t^{\delta-1}$成立,

则边值问题(1.3)存在唯一解.

该定理证明类似定理4.1,此处略去.

定理4.3  假设$q>2$,若函数$f(t, u(t))$满足下面条件

(H5)存在函数$g(t)>0$和常数$M=\int_{0}^{1}g(t){\rm d}t>0$,对任意的$(t, u(t))\in [0, 1]\times\mathbb{R} $,有$| f(t, u(t))| \leq g(t)$成立;

(H6)存在常数$L>0$,使得当$t\in[0, 1]$时,对任意的函数$u, v\in X$,有$| f(t, u(t))-f(t, v(t))| \leq L| u-v| $成立;

(H7)$0 <(q-1)LM^{q-2}\left[\frac{3}{2\Gamma(2+\alpha)}+\frac{\Gamma(2-\beta)}{2\Gamma(\alpha-\beta+2)} +\frac{\beta\Gamma(3-\gamma)}{4(2-\beta)\Gamma(\alpha-\gamma+2)}\right] <1$,

则边值问题(1.3)存在唯一解.

证  由条件(H5)可得

$\left|\int_0^{t}f(s, u(s)){\rm d}s\right|\leq\int_0^{t}\left|f(s, u(s))\right|{\rm d}s\leq\int_0^{t}g(s){\rm d}s\leq \int_{0}^{1}g(s){\rm d}s=M.$

因为$q>2$,根据引理2.3中性质(ⅱ),对任意的函数$u, v\in X$可得

$ \begin{eqnarray*} &&\bigg|\phi_{q}\left(\int_0^{t}f(s, u(s)){\rm d}s\right)-\phi_{q}\left(\int_0^{t}f(s, v(s)){\rm d}s\right)\bigg|\\ &\leq&(q-1)M^{q-2}\bigg|\int_0^{t}f(s, u(s)){\rm d}s-\int_0^{t}f(s, v(s)){\rm d}s\bigg|\\ &\leq&(q-1)M^{q-2}\int_0^{t}\left|f(s, u(s))-f(s, v(s))\right|{\rm d}s\\ &\leq&(q-1)LM^{q-2}t\parallel u-v\parallel. \end{eqnarray*} $

因此可得

$\begin{eqnarray*} &&| (Tu)(t)-(Tv)(t)| \\ &\leq&\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1} \bigg|\phi_{q}\left(\int_0^{s}f(\tau, u(\tau)){\rm d}\tau\right)-\phi_{q}\left(\int_0^{s}f(\tau, v(\tau)){\rm d}\tau\right)\bigg|{\rm d}s\\ &&+\frac{1}{2\Gamma(\alpha)}\int_0^{1}(1-s)^{\alpha-1} \bigg|\phi_{q}\left(\int_0^{s}f(\tau, u(\tau)){\rm d}\tau\right)-\phi_{q}\left(\int_0^{s}f(\tau, v(\tau)){\rm d}\tau\right)\bigg|{\rm d}s\\ &&+\bigg|\frac{(1-2t)\Gamma(2-\beta)}{2\Gamma(\alpha-\beta)}\bigg|\\ &&\times\int_0^{1}(1-s)^{\alpha-\beta-1} \bigg|\phi_{q}\left(\int_0^{s}f(\tau, u(\tau)){\rm d}\tau\right)-\phi_{q}\left(\int_0^{s}f(\tau, v(\tau)){\rm d}\tau\right)\bigg|{\rm d}s\\ &&+\bigg|\frac{\left[\beta-4t+2(2-\beta)t^2\right]\Gamma(3-\gamma)}{4(2-\beta)\Gamma(\alpha-\gamma)}\bigg|\\ &&\times\int_0^{1}(1-s)^{\alpha-\gamma-1} \bigg|\phi_{q}\left(\int_0^{s}f(\tau, u(\tau)){\rm d}\tau\right)-\phi_{q}\left(\int_0^{s}f(\tau, v(\tau)){\rm d}\tau\right)\bigg|{\rm d}s\\ &\leq&(q-1)LM^{q-2}\parallel u-v\parallel \bigg[\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}s{\rm d}s\\ &&+\frac{1}{2\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}s{\rm d}s +\bigg|\frac{(1-2t)\Gamma(2-\beta)}{2\Gamma(\alpha-\beta)}\bigg|\int_{0}^{1}(1-s)^{\alpha-\beta-1}s{\rm d}s\\ &&+\bigg|\frac{\left[\beta-4t+2(2-\beta)t^2\right]\Gamma(3-\gamma)}{4(2-\beta)\Gamma(\alpha-\gamma)}\bigg| \int_{0}^{1}(1-s)^{\alpha-\gamma-1}s{\rm d}s\bigg]\\ &=&(q-1)LM^{q-2}\parallel u-v\parallel \bigg[\frac{t^{\alpha+1}}{\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}s{\rm d}s\\ &&+\frac{1}{2\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}s{\rm d}s +\bigg|\frac{(1-2t)\Gamma(2-\beta)}{2\Gamma(\alpha-\beta)}\bigg|\int_{0}^{1}(1-s)^{\alpha-\beta-1}s{\rm d}s\\ &&+\bigg|\frac{\left[\beta-4t+2(2-\beta)t^2\right]\Gamma(3-\gamma)}{4(2-\beta)\Gamma(\alpha-\gamma)}\bigg| \int_{0}^{1}(1-s)^{\alpha-\gamma-1}s{\rm d}s\bigg]\\ &\leq&(q-1)LM^{q-2}\parallel u-v\parallel \bigg[\frac{3B(\alpha, 2)}{2\Gamma(\alpha)}+\frac{\Gamma(2-\beta)B(\alpha-\beta, 2)}{2\Gamma(\alpha-\beta)} +\frac{\beta\Gamma(3-\gamma)B(\alpha-\gamma, 2)}{4(2-\beta)\Gamma(\alpha-\gamma)}\bigg] \\&=&(q-1)LM^{q-2}\bigg[\frac{3}{2\Gamma(2+\alpha)}+\frac{\Gamma(2-\beta)}{2\Gamma(\alpha-\beta+2)} +\frac{\beta\Gamma(3-\gamma)}{4(2-\beta)\Gamma(\alpha-\gamma+2)}\bigg]\parallel u-v\parallel, \end{eqnarray*} $

其中$B(\cdot)$为Beta函数.令

$\begin{eqnarray*} \Lambda_{2}=(q-1)LM^{q-2}\bigg[\frac{3}{2\Gamma(2+\alpha)}+\frac{\Gamma(2-\beta)}{2\Gamma(\alpha-\beta+2)} +\frac{\beta\Gamma(3-\gamma)}{4(2-\beta)\Gamma(\alpha-\gamma+2)}\bigg]. \end{eqnarray*} $

由条件(H7)可知$0 <\Lambda_{2} <1$,故

$\parallel Tu-Tv\parallel\leq\Lambda_{2}\parallel u-v\parallel.$

因此,利用Banach压缩映射原理可知,算子$T$存在唯一不动点,即边值问题(1.3)存在唯一解.

例5.1  令$p=3$, $\alpha=\frac{5}{2}$, $\beta=\frac{1}{2}$, $\gamma=\frac{3}{2}$, $f(t, u(t))=2t+1+\frac{\sin(u(t))}{100}$,考虑下面一类分数阶微分方程反周期边值问题解的存在性

(5.1) $\begin{equation}\label{eBVP1}\left\{\begin{array}{l} \left(\phi_{3}\left(^{C}D_{0+}^{\frac{5}{2}}u(t)\right)\right)'=2t+1+\frac{\sin(u(t))}{100}, ~~~t\in [0, 1], \\ u(0)=-u(1), ~{^{C}D_{0+}^{\frac{1}{2}}}u(0)=-{^{C}D_{0+}^{\frac{1}{2}}}u(1), ~{^{C}D_{0+}^{\frac{3}{2}}}u(0)=-{^{C}D_{0+}^{\frac{3}{2}}}u(1). \end{array} \right. \end{equation}$

证  因为$p=3$,由$\frac{1}{p}+\frac{1}{q}=1$可知$q=\frac{3}{2}$.令$\lambda=1$, $\delta=2$,则对于任意的$(t, u(t))\in [0, 1]\times\mathbb{R} $有

$\lambda\delta t^{\delta-1}=2t\leq 2t+1+\frac{\sin(u(t))}{100}=f(t, u(t)).$

因为当$t\in[0, 1]$时,对于任意的函数$u, v\in X$,有

$| f(t, u(t))-f(t, v(t))| =\left|\frac{\sin(u(t))}{100}-\frac{\sin(v(t))}{100}\right|\leq\frac{1}{100}| u-v| .$

令$l=\frac{1}{100}$,则$| f(t, u(t))-f(t, v(t))| \leq l| u-v| $成立.

又因为

$\begin{eqnarray*}&&l(q-1)\lambda^{q}\Gamma(\delta(q-2)+2)\bigg[\frac{3}{2\Gamma(\alpha+\delta(q-2)+2)}+\frac{\Gamma(2-\beta)}{2\Gamma(\delta(q-2)+2+\alpha-\beta)}\\&&+\frac{\beta\Gamma(3-\gamma)}{4(2-\beta)\Gamma(\alpha-\gamma+\delta(q-2)+2)}\bigg]=\frac{1}{200}\left[\frac{3}{2\Gamma(\frac{7}{2})}+\frac{\Gamma(\frac{3}{2})}{3}\right] <1, \end{eqnarray*}$

所以由定理4.1可知边值问题(5.1)存在唯一解.

例5.2  令$p=\frac{3}{2}$, $\alpha=\frac{5}{2}$, $\beta=\frac{1}{2}$, $\gamma=\frac{3}{2}$, $f(t, u(t))=\frac{t\sin(u(t))}{100}$,考虑下面一类分数阶微分方程反周期边值问题解的存在性

(5.2) $\begin{equation}\label{eBVP2}\left\{\begin{array}{l} \left(\phi_{\frac{3}{2}}\left(^{C}D_{0+}^{\frac{5}{2}}u(t)\right)\right)'=\frac{t\sin(u(t))}{100}, ~~~t\in [0, 1], \\ u(0)=-u(1), ~{^{C}D_{0+}^{\frac{1}{2}}}u(0)=-{^{C}D_{0+}^{\frac{1}{2}}}u(1), ~{^{C}D_{0+}^{\frac{3}{2}}}u(0)=-{^{C}D_{0+}^{\frac{3}{2}}}u(1). \end{array} \right. \end{equation}$

证  因为$p=\frac{3}{2}$,由$\frac{1}{p}+\frac{1}{q}=1$可知$q=3$.由$f(t, u(t))=\frac{t\sin(u(t))}{100}$可知,存在函数$g(t)=\frac{t}{100}$和常数$M=\int_{0}^{1}g(t){\rm d}t=\int_{0}^{1}\frac{t}{100}{\rm d}s=\frac{1}{200}$,使得

$\begin{eqnarray*}| f(t, u(t))| =\left|\frac{t\sin(u(t))}{100}\right|\leq\frac{t}{100}=g(t).\end{eqnarray*}$

因为当$t\in[0, 1]$时,对任意的函数$u, v\in X$,有

$| f(t, u(t))-f(t, v(t))| =\left|\frac{t\sin(u(t))}{100}-\frac{t\sin(v(t))}{100}\right|\leq\frac{1}{100}| u-v|.$

令$L=\frac{1}{100}$,则$| f(t, u(t))-f(t, v(t))| \leq L| u-v| $成立.

又因为

$\begin{eqnarray*}&&(q-1)LM^{q-2}\left[\frac{3}{2\Gamma(2+\alpha)}+\frac{\Gamma(2-\beta)}{2\Gamma(\alpha-\beta+2)}+\frac{\beta\Gamma(3-\gamma)}{4(2-\beta)\Gamma(\alpha-\gamma+2)}\right]\\&=&\frac{1}{10000}\left[\frac{3}{\Gamma(\frac{9}{2})}+\frac{\Gamma(\frac{3}{2})}{8}\right] <1, \end{eqnarray*}$

所以由定理4.3可知边值问题(5.2)存在唯一解.