上海理工大学学报  2020, Vol. 42 Issue (4): 311-316   PDF    
含左右分数导数的时滞微分方程解的存在性和唯一性
黄雪楠, 刘锡平     
上海理工大学 理学院,上海 200093
摘要: 研究了一类含左右Caputo分数导数的时滞微分方程Riemann-Stieltjes积分边值问题。通过构建单调迭代序列并利用上下解方法得到了边值问题解的存在性与唯一性定理。最后给出实例以说明本文的主要结论的适用性。
关键词: 分数阶微分方程     分数阶左右导数     Riemann-Stieltjes积分     时滞     Caputo导数     上下解    
Existence and uniqueness of solutions for delay differential equations with left and right fractional derivatives
HUANG Xuenan, LIU Xiping     
College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China
Abstract: A class of Riemann-Stieltjes integral boundary value problems for delay differential equations with Caputo left and right fractional derivatives was studied. The existence and uniqueness of solutions for the boundary value problems were obtained by constructing the monotone iterative sequences and using the method of upper and lower solutions. Finally, an example was given to illustrate the main conclusions.
Key words: fractional differential equation     fractional left and right derivative     Riemann-Stieltjes integral     time delay     Caputo derivative     upper and lower solutions    
1 问题的提出

由于分数阶微积分在物理学、生态学、经济学等学科中具有广泛的应用,国内外学者对相关问题进行了大量的研究[1-8]。与经典的整数阶微积分不同,分数阶导数与积分需要考虑左右不同的定义,而目前对同时包含左右两个不同分数导数的微分方程的研究较少[9-11]。在现代工程技术与科学研究中,时滞对状态的影响往往是不容忽视的,因此,时滞微分方程的理论研究受到人们的重视[12-14]

现研究一类含左右分数导数的时滞微分方程

$ {}_t^{\rm{C}}{\rm{D}}_1^\alpha \left( {_0^{\rm{C}}{\rm{D}}_t^\beta u\left( t \right)} \right) = f\left( {t,u\left( t \right),u\left( {t - \tau } \right)} \right),\;t \in \left[ {0,1} \right]$ (1)

满足积分边界条件

$\left\{ {\begin{aligned} &{\left ({}_0^{\rm{C}}{\rm{D}}_t^\beta u(1) \right )' \!\!=\! 0, {}_0^{\rm{C}}{\rm{D}}_t^\beta u(0) \!=\!\! \displaystyle \int_0^1 \!\!{{g_1}(s,u(s)){\rm{d}}{\varLambda _1}(s)} } \\ & {u'(t) \!=\! \varphi (t),\; t \!\in\! [ - \tau ,0], u(1) \!=\!\! \displaystyle \int_0^1 \!\!{{g_2}(s,u(s)){\rm{d}}{\varLambda _2}(s)} } \end{aligned}} \right.$ (2)

解的存在性和唯一性。其中, $1 < \alpha ,\; \beta$ $2 ;\tau > 0 ,$ $\alpha , \beta , \tau $ 为常数; ${}_0^{\rm{C}}{\rm{D}}_t^\beta ,\;{}_t^{\rm{C}}{\rm{D}}_1^\alpha$ 分别为Caputo左、右导数; $f: [0, 1] \times {\mathbb{R}} \times {\mathbb{R}} \to {\mathbb{R}},\; {g_i}:[0,1] \times {\mathbb{R}} \to {\mathbb{R}}(i = 1,2),\; \varphi : [ - \tau ,0] \to {\mathbb{R}}$ 是连续函数,且 $\varphi (0) = 0;$ $\displaystyle\int_0^1 {{g_i}(s,u(s)){\rm{d}}} {\varLambda _i}(s)$ 为Riemann-Stieltjes积分,且 ${\varLambda _i}(i = 1,2)$ 是单调递增函数,记 ${L_i} = {\varLambda _i}(1) - {\varLambda _i}(0)$ $,\; i = 1,2$

有关分数阶微积分的概念与基本性质参见文献[7-8]。本文的目的是建立一类含左右Caputo分数导数的时滞微分方程积分边值问题(1),(2)解的存在性和唯一性定理。由于Riemann-Stieltjes积分是Riemann积分的推广,Riemann-Stieltjes积分边值问题以两点、多点以及一般的Riemann积分边值问题为特例,因此,本文所研究的问题更具有一般性。本文所研究的方程在非线性项中包含了状态时滞项,该时滞项在分析系统的稳定程度和性能方面具有重要的作用,某些情况下即使时滞很小也能给系统带来严重的影响。同时含有左右分数阶导数的微分方程在许多领域都有应用,如变分原理、分数阶Lagrange方程的极值问题、分数阶导数泛函的最优控制理论及哈密顿力学等。在文献[15-16]中,作者提出了一种用分数阶导数研究Euler-Lagrange方程的方法,显示出分数阶微分方程在研究力学时所具有的优势。

2 预备知识

引理1  设 $h \in C[0, 1],\;a, b$ 为常数,线性分数阶微分方程非齐次边值问题

$\left\{ {\begin{array}{*{20}{l}} {{}_t^{\rm{C}}{\rm{D}}_1^\alpha \left ({}_0^{\rm{C}}{\rm{D}}_t^\beta u(t) \right ) = h(t),\; t \in [0,{\rm{1}}] } \\ {\left ({}_0^{\rm{C}}{\rm{D}}_t^\beta u({\rm{1}}) \right )' = 0,\; {}_0^{\rm{C}}{\rm{D}}_t^\beta u(0) = b\; \; } \\ {u'(0) = {\rm{0}} ,\; u({\rm{1}}) = a\; \; \; \; \; \; \; \; \; \; \; \; } \end{array}} \right.$ (3)

存在唯一解 $u = u(t)$

$u(t){\rm{ = }}a - \int_0^1 {{G_0}(t,s)\left(b - \int_0^1 {{G_1}(s,r)h(r){\rm{d}}r} \right) {\rm{d}}s} $ (4)

并且

${}_0^{\rm{C}}{\rm{D}}_t^\beta u(t) = b - \int_0^1 {{G_1}(t,s)h(s){\rm{d}}s} $ (5)

其中,

${G_0}(t,s)\!\!=\!\! \frac{1}{{\Gamma (\beta )}}\left\{\begin{aligned} &{{{(1 - s)}^{\beta - 1}} - {{(t - s)}^{\beta - 1}},} \!\!\!\!&\!\!\!\! {0 \leqslant s \!\leqslant\! t \!\leqslant\! 1}\\ &{{{(1 - s)}^{\beta - 1}},} \!\!\!\!&\!\!\!\! {0 \!\leqslant\! t \!\leqslant\! s \!\leqslant\! 1} \end{aligned} \right.$ (6)
${G_1}(t,s){\rm{ = }}\frac{{\rm{1}}}{{\Gamma (\alpha )}}\left\{ {\begin{aligned} &{{s^{\alpha - 1}},}&{0 \leqslant s \leqslant t \leqslant 1}\\ &{{s^{\alpha - 1}} - {{(s - t)}^{\alpha - 1}},}&{0 \leqslant t \leqslant s \leqslant 1} \end{aligned}} \right.$ (7)

证明  令 ${}_0^{\rm{C}}{\rm{D}}_t^\beta u(t) = v(t)$ , 则边值问题(3)可转化为下面的2个边值问题:

$\left\{ {\begin{array}{*{20}{l}} {{}_0^{\rm{C}}{\rm{D}}_t^\beta u(t) = v(t),\; t \in [0,1]} \\ {u(1) = a,\; \; u'(0) = 0\; \; \; \; \; } \end{array}} \right.$ (8)
$\left\{ {\begin{array}{*{20}{l}} {{}_t^{\rm{C}}{\rm{D}}_1^\alpha v(t) = h(t),\; t \in [0,1]} \\ {v(0) = b,\; v'(1) = 0\; \; \; \; \; \; } \end{array}} \right.$ (9)

由分数阶导数的性质可得 ${}_t^{\rm{C}}{\rm{D}}_1^\alpha v(t) = h(t)$ 的通解为

$v(t) = \frac{{\rm{1}}}{{{{\Gamma }}(\alpha )}}\int_t^1 {{{(s - t)}^{\alpha - 1}}h(s){\rm{d}}s} + {c_0} + {c_1}t,\;\;{c_i} \in \mathbb{R},\;i = 1, \;2$

由边界 $v(0) = b,v'(1) = 0$ ,可得

$ {c_1} = 0,\;{c_0} = b - \dfrac{{\rm{1}}}{{{{\Gamma }}(\alpha )}} \displaystyle\int_0^1 {{s^{\alpha - 1}}} h(s){\rm{d}}s $

所以,边值问题(9)有解

$\begin{split} v(t) = & \frac{{\rm{1}}}{{{{\Gamma }}(\alpha )}}\int_t^1 {{{(s - t)}^{\alpha - 1}}} h(s){\rm{d}}s + b -\\ & \frac{{\rm{1}}}{{{{\Gamma }}(\alpha )}}\int_0^1 {{s^{\alpha - 1}}} h(s){\rm{d}}s = b - \int_0^1 {{G_1}(t,s)} h(s){\rm{d}}s \end{split} $

同理,可得边值问题(8)的解

$\begin{split} u(t) =& \frac{{\rm{1}}}{{{{\Gamma }}(\beta )}}\int_0^t {{{(t - s)}^{\beta - 1}}v(s)} {\rm{d}}s + a -\\ & \frac{{\rm{1}}}{{{{\Gamma }}(\beta )}}\int_0^1 {{{(1 - s)}^{\beta - 1}}v(s)} {\rm{d}}s = a - \int_0^1 {{G_0}(t,s)} v(s){\rm{d}}s \end{split} $

因此,边值问题(3)有解 $u = u(t), \; $ 并且为式(4)的形式。

反之,容易证明,若 $u = u(t)$ 满足式(4),则 $u = u(t)$ 满足式(3)。

由式(6)和式(7)可得引理2。

引理2  对任意给定的 $(t, s) \in [0, 1] \times [0, 1], {G_0} (t,s) $ $0, {G_1}(t, s) $ $0, $ ${G_0}(t, s), {G_1}(t, s) $ $[0, {\rm{1}}] \times [0, {\rm{1}}] $ 上连续。

3 边值问题的上下解方法

$E = C[ - \tau ,1],$ 取范数 $||u|| = \mathop {\max }\limits_{ - \tau \leqslant t \leqslant 1} |u(t)|,$ $(E,|| \cdot ||)$ 是Banach空间。设 $x,y \in E,$ 若对任意 $t \in [ - \tau ,1],$ 都有 $x(t)$ $y(t),$ 则记 $x$ $y$

定义1  若 $u \in C[ - \tau ,1]$ 满足条件

$\left\{ {\begin{array}{*{20}{l}} {{}_t^{\rm{C}}{\rm{D}}_1^\alpha \left ({}_0^{\rm{C}}{\rm{D}}_t^\beta u(t)\right ) \leqslant ( \geqslant )f(t,u(t),u(t - \tau )),\;t \in [0,1]}\\ {\left ({}_0^{\rm{C}}{\rm{D}}_t^\beta u(1) \right )' = 0,\;{}_0^{\rm{C}}{\rm{D}}_t^\beta u(0) \geqslant ( \leqslant )\displaystyle \int_0^1 {{g_1}(s,u(s)){\rm{d}}{\varLambda _1}(s)} }\\ {u'(t) = \varphi (t),\;t \in [ - \tau ,0],\;u(1) \leqslant ( \geqslant )\displaystyle \int_0^1 {{g_2}(s,u(s)){\rm{d}}{\varLambda _2}(s)} } \end{array}} \right.$

则称 $u = u(t)$ 是边值问题(1),(2)的下(上)解。

假设:

(H1) 设 $f \in C([0,1] \times {\mathbb{R}} \times {\mathbb{R}})$ ,对任意 $t \in [0,1]$ ,及 ${x_1}$ ${x_2} ,\; {y_1}$ ${y_2} \in {\mathbb{R}}$ 时,有

$ f(t,{x_1},{y_1}) \leqslant f(t,{x_2},{y_2}) $

(H2) 设 ${g_i} \in C([0,1] \times {\mathbb{R}}) ,\; i = 1 , \,2 , $ 对任意 $t \in [0,1]$ ,及 ${x_1} \leqslant {x_2} \in \mathbb{R}$ 时,有

$ {g_1}(t,{x_2}) \leqslant {g_1}(t,{x_1}),\; \; {g_2}(t,{x_2}) \geqslant {g_2}(t,{x_1}) $

引理3  假设(H1),(H2)成立,边值问题(1),(2)存在下解 ${x_k}(t) \in E$ ,则边值问题

$\left\{ {\begin{aligned} &\;{{}_t^{\rm{C}}{\rm{D}}_1^\alpha \left ({}_0^{\rm{C}}{\rm{D}}_t^\beta {x_{k + 1}}(t)\right ) = f(t,{x_k}(t),{x_k}(t - \tau ))\; ,}\\& \quad t \in [0,1] \\ &\;{\left ({}_0^{\rm{C}}{\rm{D}}_t^\beta {x_{k + 1}}(1)\right )'} = 0,\\& \quad{{}_0^{\rm{C}}{\rm{D}}_t^\beta {x_{k + 1}}(0) = \displaystyle \int_0^1 {{g_1}(s,{x_k}(s)){\rm{d}}{\varLambda _1}(s)} } \\ &\;{{{x'}_{k\! +\! 1}}(t)\! =\! \varphi (t) ,\; t\! \in\! [ - \tau ,0],}\\& \quad{{x_{k \!+\! 1}}(1) \!=\! \displaystyle \int_0^1 {{g_2}(s,{x_k}(s)){\rm{d}}{\varLambda _2}(s)} } \end{aligned}} \right.$ (10)

存在唯一解 ${x_{k + 1}} = {x_{k + 1}}(t) \in E$ ,且为问题(1),(2)的下解,同时满足 ${x_k}$ ${x_{k + 1}}$

证明  由于 $t \in [ - \tau , 0] $ 时, ${x'_{k + 1}}(t) = \varphi (t) $ ,所以, ${x_{k + 1}}(t) = {x_{k + 1}}(0) + \displaystyle\int_0^t {\varphi (s){\rm{ds}}} ,\; t \in [ - \tau , 0]$

由引理1可得边值问题(10)的解

$\begin{aligned} &{x_{k + 1}}(t) = \\ & \left\{ \begin{aligned} &\!\! \displaystyle \int_0^1 \!\!{{g_2}(s,{x_k}(s)){\rm{d}}{\varLambda _2}(s)} \!-\\[-2pt] &\quad\quad\!\! \displaystyle \int_0^1 \!\!{{G_0}(t,s)\bigg(\displaystyle\int_0^1 \!\!{{g_1}(r,{x_k}(r)){\rm{d}}{\varLambda _1}(r)} - } \\ &\quad\quad\!\! \displaystyle \int_0^{\rm{1}} {{G_{\rm{1}}}(s,r)f(r,{x_k}(r),{x_k}(r - \tau )){\rm{d}}r}\bigg) {\rm{d}}s,\\& \quad\quad t \in [0,{\rm{1}}] \\ &{x_{k + 1}}({\rm{0}}) + \displaystyle \int_0^t {\varphi (s){\rm{d}}s} \; ,\; \; \; \; \; \; t \in [ - \tau ,0] \\ \end{aligned} \right. \end{aligned} $ (11)

其中,

$ \begin{split} &{x_{k + 1}}(0) = \int_0^1 {{g_2}(s,{x_k}(s)){\rm{d}}{\varLambda _2}(s)} - \frac{{\displaystyle\int_0^1 {{g_1}(s,{x_k}(s)){\rm{d}}{\varLambda _1}(s)} }}{{\Gamma (\beta + 1)}} + \\[-2pt] & \quad\frac{1}{{\Gamma (\beta + 1)\Gamma (\alpha )}}\int_0^1 {{s^{\alpha - 1}}f(s,{x_k}(s),{x_k}(s - \tau )){\rm{d}}s} - \\[-3pt] & \quad\frac{1}{{\Gamma (\beta + 1)\Gamma (\alpha )}}\int_0^1 {{\rm{d}}s\int_0^s {{{(1 - r)}^{\beta - 1}}{{(s - r)}^{\alpha - 1}}} }{\text{·}}\\&\quad f(s,{x_k} (s),{x_k}(s - \tau )){\rm{d}}r \\[-2pt] & \quad\quad_0^{\rm{C}}{\rm{D}}_t^\beta {x_{k + 1}}(t) = \int_0^1 {{g_1}(s,{x_k}(s)){\rm{d}}{\varLambda _1}(s)} - \\[-2pt] & \quad\quad\quad\int_0^1 {{G_1}(t,s)f(s,{x_k}(s),{x_k}(s - \tau )){\rm{d}}s} \end{split} $

所以,下解 ${x_k}(t) \in E$ 时,边值问题(10)等价于积分边值问题(11)。

现证明 ${x_k}(t)$ ${x_{k + 1}}(t)$ 并且 ${x_{k + 1}} = {x_{k + 1}}(t)$ 是边值问题(1),(2)的下解。因为, ${x_k} = {x_k}(t)$ 是边值问题(1),(2)的下解,则有

$\left\{ {\begin{aligned} &{{}_t^{\rm{C}}{\rm{D}}_1^\alpha \left ({}_0^{\rm{C}}{\rm{D}}_t^\beta {x_k}(t) \right ) \leqslant f(t,{x_k}(t),{x_k}(t - \tau )),}\\&\quad\quad{ \;t \in [0,1]}\\ &{\left ({}_0^{\rm{C}}{\rm{D}}_t^\beta {x_k}(1) \right )' = 0,}\\& \quad\quad{{}_0^{\rm{C}}{\rm{D}}_t^\beta {x_k}(0) \geqslant \displaystyle \int_0^1 {{g_1}(s,{x_k}(s)){\rm{d}}{\varLambda _1}(s)} }\\ &{{{x'}_k}(t) = \varphi (t), \;t \in [ - \tau ,0],}\\&\quad\quad{\;{x_k}(1) \leqslant \displaystyle \int_0^1 {{g_2}(s,{x_k}(s)){\rm{d}}{\varLambda _2}(s)} } \end{aligned}} \right.$ (12)

根据式(10),同时结合式(12),可得

$\left\{ {\begin{array}{*{20}{l}} {{}_t^{\rm{C}}{\rm{D}}_1^\alpha \left ({}_0^{\rm{C}}{\rm{D}}_t^\beta {x_{k + 1}}(t) - {}_0^{\rm{C}}{\rm{D}}_t^\beta {x_k}(t)\right ) \geqslant 0,\;t \in [0,1]}\\ {\left( {_0^{\rm{C}}{\rm{D}}_t^\beta {x_{k + 1}}(1) - _0^{\rm{C}}{\rm{D}}_t^\beta {x_k}(1)} \right)^\prime } = 0,\\ \;\;\;\;\;\;\;\;_0^{\rm{C}}{\rm{D}}_t^\beta {x_{k + 1}}(0) - _0^{\rm{C}}{\rm{D}}_t^\beta {x_k}(0) \leqslant 0\\ {{x_{k + 1}}^\prime (0) - {x_k}^\prime (0) = 0,\;{x_{k + 1}}(1) - {x_k}(1) \geqslant 0} \end{array}} \right.$

即有

$\left\{ {\begin{array}{*{20}{l}} {{}_t^{\rm{C}}{\rm{D}}_1^\alpha\; {}_0^{\rm{C}}{\rm{D}}_t^\beta ({x_{k + 1}}(t) - {x_k}(t)) \geqslant 0,\;t \in [0,1] }\\ {\left ({}_0^{\rm{C}}{\rm{D}}_t^\beta ({x_{k + 1}}(1) - {x_k}(1))\right)' = 0,\;{}_0^{\rm{C}}{\rm{D}}_t^\beta ({x_{k + 1}}(0) - {x_k}(0) ) \leqslant 0}\\ {({x_{k + 1}}(0) - {x_k}(0))' = 0,\;{x_{k + 1}}(1) - {x_k}(1) \geqslant 0} \end{array}} \right.$

现记

$\begin{array}{l} {h_{k + 1}}(t) \buildrel \Delta \over = {}_t^{\rm{C}}{\rm{D}}_1^\alpha \; {}_0^{\rm{C}}{\rm{D}}_t^\beta ({x_{k + 1}}(t) - {x_k}(t)) \\ {b^ * } \buildrel \Delta \over = \; {}_0^{\rm{C}}{\rm{D}}_t^\beta ({x_{k + 1}}(0) - {x_k}(0)) \; \;\\ {a^ * } \buildrel \Delta \over = \; {x_{k + 1}}(1) - {x_k}(1) \end{array} $

${h_{k + 1}}(t) \geqslant 0 , t \in [0,1] ,\; {b^ * } \leqslant 0 , \;{a^ * } \geqslant 0 $

由式(4)可得

$ \begin{array}{l} \!\!\!\!\!\!\!{x_{k + 1}}(t) - {x_k}(t) = {a^ * } - \\ \displaystyle \int_0^1 {{G_{\rm{0}}}(t,s)} \left({b^ * } - \displaystyle \int_0^1 {{G_1}} (s,r){h_{k + 1}}(r){\rm{d}}r\right){\rm{d}}s \geqslant 0 \end{array} $

所以,在 $t \in [0,1],\; {x_k}(t)$ ${x_{k + 1}}(t),$ $t \in [ - \tau ,0]$ 时,

$\begin{split} {x_{k + 1}}(t) =& {x_{k + 1}}(0) + \int_0^t {\varphi (s){\rm{d}}s} \geqslant {x_k}(0) +\\& \int_0^t {\varphi (s){\rm{d}}s} = {x_k}(t)\end{split} $

综上所述, ${x_k}$ ${x_{k + 1}}$

由(H1)可得

$\left\{ {\begin{array}{*{20}{l}} {{}_t^{\rm{C}}{\rm{D}}_1^\alpha \left ({}_0^{\rm{C}}{\rm{D}}_t^\beta {x_{k + 1}}(t) \right ) = f(t,{x_k}(t),{x_k}(t - \tau )) } \leqslant \\ \;\;\;\;f(t,{x_{k + 1}}(t),{x_{k + 1}}(t - \tau )) ,\;t \in [0,1]\\ {\left ({}_0^{\rm{C}}{\rm{D}}_t^\beta {x_{k + 1}}(1)\right )' = 0}\\{{}_0^{\rm{C}}{\rm{D}}_t^\beta {x_{k + 1}}(0) = \displaystyle \int_0^1 {{g_1}(s,{x_k}(s)){\rm{d}}{\Lambda _1}(s)} \geqslant}\\ \quad{} \displaystyle \int_0^1 {{g_1}(s,{x_{k + 1}}(s)){\rm{d}}{\varLambda _2}(s)} \\ {{{x'}_{k + 1}}(t) = \varphi (t),\;t \in [ - \tau ,0]}\\ {{x_{k + 1}}(1) = \displaystyle \int_0^1 {{g_2}(s,{x_k}(s)){\rm{d}}{\varLambda _2}(s)} \leqslant }\\\quad \displaystyle \int_0^1 {{g_2}(s,{x_{k + 1}}(s)){\rm{d}}{\varLambda _2}(s)} \end{array}} \right.$

${x_{k + 1}} = {x_{k + 1}}(t)$ 是边值问题(10)的下解。

由引理3可得引理4。

引理4  假设(H1),(H2)成立,边值问题(1),(2)存在上解 ${y_k}(t) \in E$ ,则边值问题

$\left\{ {\begin{array}{*{20}{l}} {{}_t^{\rm{C}}{\rm{D}}_1^\alpha \left ({}_0^{\rm{C}}{\rm{D}}_t^\beta {y_{k\! + \!1}}(t) \right) \!=\! f(t,{y_k}(t),{y_k}(t - \tau ))\; ,}\\ \quad{ t \!\in\! [0,1]\; \; \; \; } \\ {\left ({}_0^{\rm{C}}{\rm{D}}_t^\beta {y_{k + 1}}(1) \right)' = 0,}\\ \quad{ {}_0^{\rm{C}}{\rm{D}}_t^\beta {y_{k + 1}}(0) = \!\!\displaystyle \int_0^1 \!\!{{g_1}(s,{y_k}(s)){\rm{d}}{\varLambda _1}(s)} } \\ {{y_{k + 1}}^\prime (t) = \varphi (t),t \!\in\! [ - \tau ,0],}\\ \quad{ {y_{k + 1}}(1) \!=\! \!\!\displaystyle \int_0^1 \!\!{{g_2}(s,{y_k}(s)){\rm{d}}{\varLambda _2}(s)} \; } \end{array}} \right.$ (13)

存在唯一解 ${y_{k + 1}} = {y_{k + 1}}(t) \in E$ ,且为边值问题(1),(2)的上解,同时满足 ${y_{k + 1}}$ ${y_k}$

定理1  假设(H1),(H2)成立,边值问题(1),(2)存在下解 ${x_{\rm{0}}} \in E,\;$ 上解 ${y_{\rm{0}}} \in E,\;$ ${x_{\rm{0}}}$ ${y_{\rm{0}}},\;$ 则边值问题(1),(2)存在解 ${x^ * },{y^ * } \in E$ ,且 ${x_{\rm{0}}}$ ${x^ * }$ ${y^ * }$ ${y_{\rm{0}}}$

证明  以 ${x_{\rm{0}}},{y_{\rm{0}}}$ 为初始元,根据引理3和引理4产生序列 $\{ {x_k}\} ,\{ {y_k}\} $ ,并且 ${x_k},{y_k}(k = 0,1,2, \cdots )$ 分别是边值问题(1),(2)的下解和上解, $\{ {x_k}\} $ 是单调递增的, $\{ {y_k}\} $ 是单调递减的。

现用数学归纳法证明 ${x_n}$ ${y_n},\;n = 0,1,2, \cdots$

根据数学归纳法,当 $n = 0$ 时, ${x_0}$ ${y_0}$ 成立。

$n = k$ 时,假设 ${x_k}$ ${y_k}$ 成立,即 $ {x_k}(t)$ ${y_k}(t) ,\; t \in [0,1]$ 。故由假设(H1)和(H2)可得

$ \begin{array}{*{20}{c}} f(t,{x_k}(t),{x_k}(t - \tau )) \leqslant f(t,{y_k}(t),{y_k}(t - \tau ))\\ \displaystyle \int_0^1 {{g_1}(s,{x_k}(s)){\rm{d}}{\varLambda _1}(s)} \geqslant \displaystyle \int_0^1 {{g_1}(s,{y_k}(s)){\rm{d}}{\varLambda _1}(s)} \\ \displaystyle \int_0^1 {{g_2}(s,{x_k}(s)){\rm{d}}{\varLambda _2}(s)} \leqslant \displaystyle \int_0^1 {{g_2}(s,{y_k}(s)){\rm{d}}{\varLambda _2}(s)} \end{array} $

由式(10)和式(13)可得

$\left\{ {\begin{array}{*{20}{l}} {{}_t^{\rm{C}}{\rm{D}}_1^\alpha \left ({}_0^{\rm{C}}{\rm{D}}_t^\beta {y_{k{\rm{ + 1}}}}(t) - {}_0^{\rm{C}}{\rm{D}}_t^\beta {x_{k{\rm{ + 1}}}}(t)\right ) \geqslant 0 , \; t \in [0,1] }\\ {\left ({}_0^{\rm{C}}{\rm{D}}_t^\beta {y_{k{\rm{ + 1}}}}(1) - {}_0^{\rm{C}}{\rm{D}}_t^\beta {x_{k{\rm{ + 1}}}}(1)\right )' = 0 , }\\ \quad\quad {}_0^{\rm{C}}{\rm{D}}_t^\beta {y_{k{\rm{ + 1}}}}(0) - {}_0^{\rm{C}}{\rm D}_t^\beta {x_{k{\rm{ + 1}}}}(0) \leqslant 0\\ {({y_{k{\rm{ + 1}}}}(t) - {x_{k{\rm{ + 1}}}}(t))' = 0 , \; t \in [ - \tau ,0], \; {y_{k + 1}}(1) - {x_{k{\rm{ + 1}}}}(1) \geqslant 0 } \end{array} } \right.$

与引理3的证明类似,当 $t \in [ - \tau ,1]$ ,有 ${x_{k + 1}}(t) \leqslant {y_{k + 1}}(t)$ 。由数学归纳法可得 ${x_n} \leqslant {y_n}(n = 0, \;1, \; 2, \; \cdots ) $ 成立。所以,

$ {x_{\rm{0}}}\leqslant {x_{\rm{1}}}\leqslant \cdots \leqslant {x_k}\leqslant \cdots \leqslant {y_k}\leqslant \cdots \leqslant {y_1}\leqslant {y_{\rm{0}}} $

故序列 $\{ {x_k}\} ,\{ {y_k}\} $ 一致有界。

由引理2可知,函数 ${G_0}(t,s)$ $(t,s) \in [0,1] \times [0,1]$ 上连续,故一致连续。对任意 ${t_1},\; {t_2} \in [0 , 1] ,\;$ $|{t_1} - {t_2}| \to 0$ 时,有 $\displaystyle\int_0^1 {|{G_0} ({t_1},s ) - \; {G_0}({t_2},s\; )|{\rm{d}}s} \to 0$

于是,由式(11)可得

$\begin{split} &|{x_k}({t_1}) - {x_k}({t_2})| = \bigg| \bigg.\displaystyle \int_0^1 {({G_0}({t_1},s) - {G_0}({t_2},s)){\text{·}}} \\ &\quad \left (\displaystyle \int_0^1 {{g_1}(r,{x_{k - 1}}(r)){\rm{d}}{\varLambda _1}(r)} - \right. \\ &\quad \left. \displaystyle \int_0^1 {{G_1}(s,r)f(r,{x_{k - 1}}(r),{x_{k - 1}}(r - \tau )){\rm{d}}r} \right ){\rm{d}}s \bigg| \bigg. \leqslant \\ &\quad \int_0^1 {|{G_0}({t_1},s) - {G_0}({t_2},s)|{\rm{d}}s{\text{·}}} \\ &\quad\left (\displaystyle \int_0^1 {|{g_1}(r,{x_{k - 1}}(r))|{\rm{d}}{\varLambda _1}(r) + \frac{1}{{\Gamma (\alpha )}\;}}{\text{·}} \right. \\ &\quad\left. \displaystyle \int_0^1 {|f(r,{x_{k - 1}}(r),{x_{k - 1}}(r - \tau ))|} {\rm{d}}r \right ) \to 0,{\mkern 1mu} {\kern 1pt} (|{t_1} - {t_2}| \to 0) \end{split}$

所以,序列 $\{ {x_k}\} $ $[0 , 1]$ 上等度连续。

又由式(11)可得,当 ${t_1} ,\; {t_2} \in [ - \tau , 0\; ]$ 时, $|{x_k}({t_1}) - {x_k}({t_2})| = \left|\displaystyle\int_{{t_1}}^{{t_2}} {\varphi (s){\rm{d}}s} \right|$ ,而 $\varphi \; $ $[ - \tau ,0]$ 上连续,故序列 $\{ {x_k}\} $ $[ - \tau ,0]$ 上等度连续。因此,序列 $\{ {x_k}\} $ $[ - \tau ,1]$ 上等度连续。

同理,序列 $\{ {y_k}\} $ $[ - \tau ,1]$ 上等度连续。所以,序列 $\{ {x_k}\} ,\{ {y_k}\} $ $t \in [ - \tau ,1]$ 上是相对列紧的。故存在 ${x^ * },\; {y^ * } \in E$ ,使得 $\mathop {\lim }\limits_{k \to \infty } {x_k} = {x^ * }, \mathop {\lim }\limits_{k \to \infty } {y_k} = {y^ * }$

现证明 ${x^ * },{y^ * }$ 是边值问题(1),(2)的解。

由式(11)和 ${G_1}, {G_0}, f, {g_1}, {g_2}$ 的连续性,根据Lebesgue控制收敛定理,当 $k \to \infty $ 时,可得

${{x}^{*}}(t)\text{=}\left\{ \begin{align} & \int_{0}^{1}{{{g}_{2}}(s,{{x}^{*}}(s))\text{d}{{\varLambda }_{2}}(s)}-\\ &\quad\quad\int_{0}^{1}{{{G}_{0}}(t,s)} \left (\int_{0}^{1}{{{g}_{1}}(r,{{x}^{*}}(r))\text{d}{{\varLambda }_{1}}(r)}- \right. \\ &\quad\quad \left. \int_{0}^{1}{{{G}_{1}}(s,r)f(r,{{x}^{*}}(r),{{x}^{*}}(r-\tau ))\text{d}r} \right )\text{d}s, \\ &\quad\quad t\in [0,1]\\& {{x}^{*}}(0)+\int_{0}^{t}{\varphi (s)\text{d}s},\;t\in [\tau ,0] \\ \end{align} \right.$

${x^ * } = {x^ * }(t)$ 为边值问题(1),(2)的解。同理,可得 ${y^ * }$ 也是边值问题(1),(2)的解,而且有 ${x_{\rm{0}}}$ ${x^ * }$ ${y^ * }$ ${y_{\rm{0}}}$

定理2  假设定理1的条件成立,并且存在常数 ${M_i},\; {C_i} $ $0 ,\; i = 1 , 2 , \;$ 使得对任意 ${x_1} $ ${x_2},\;$ ${y_1}$ $ {y_2} \in {\mathbb{R}}$ ,及 $t \in [0,1],$

$\begin{split} &0 \leqslant f(t,{x_2},{y_2}) - f(t,{x_1},{y_1}) \leqslant \\ &\quad\quad{M_1}({x_2} - {x_1}) + {M_2}({y_2} - {y_1}) \end{split}$ (14)
$\begin{split} 0 \leqslant &{g_1}(t,{x_1}) - {g_1}(t,{x_2})\leqslant {C_1}({x_2} - {x_1}),\\ & 0 \leqslant {g_2}(t,{x_2}) - {g_2}(t,{x_1})\leqslant {C_2}({x_2} - {x_1})\end{split} $ (15)

成立,且 ${L_2}{C_2} + {L_1}{C_1}\displaystyle\frac{1}{{{{\Gamma }}(\beta )}} + \dfrac{1}{{{{\Gamma }}(\beta )\Gamma (\alpha )}} ({M_1} + {M_2}) < 1 \;$ 成立,则边值问题(1),(2)有唯一解。

证明  由定理1可知,边值问题(1),(2)在 ${x_0}$ ${y_0}$ 之间存在解 ${x^ * }$ ${y^ * }$ 。现通过证明 ${x^ * }{\rm{ = }}{y^ * }$ 来证明边值问题(1),(2)唯一解的存在性。

由式(14)和式(15)可得,对任意 ${x_k}(t)$ ${y_k}(t) \in [{x_0}(t),{y_0}(t)] \left| \;( k = 1, 2, 3 , \cdots ) \right| \;( t \in [0, 1] )\;$ ,有

$\begin{split} & f(t,{x_k}(t),{x_k}(t - \tau )) - f(t,{y_k}(t),{y_k}(t - \tau )) \geqslant \\ &\quad - {M_1}({y_k}(t) - {x_k}(t)) - {M_2}(({y_k}(t - \tau ) - {x_k}(t - \tau )) \end{split}$ (16)
$ {g_i}(t,{x_k}(t)) - {g_i}(t,{y_k}(t)) \leqslant {C_i}({x_k}(t) - {y_k}(t) ) ,\; i = 1 , 2 $ (17)

由式(11)可得

$\begin{array}{l} f(t,{x_k}(t),{x_k}(t - \tau )) - f(t,{y_k}(t),{y_k}(t - \tau ))\geqslant \\ \;\;\quad - ({M_1} + {M_2})({y_k}(t) - {x_k}(t)) \end{array}$

由于 ${\varLambda _i}(i = 1,2)$ $[0,1]$ 上是单调递增的,故 ${L_i} $ $0 , i = 1, 2$ 。所以,根据式(16)和式(17),对任意的 ${x_k}$ ${y_k} \in [{x_0},{y_0}] , \;k = 1, 2, 3 , \cdots$ ,有

$ \begin{align} \!\!\!\!\!\!\!\!& \left\| {{x}_{k}}(t)-{{y}_{k}}(t) \right\|\leqslant \left| \int_{0}^{1}{({{g}_{2}}(s,{{x}_{k-1}}(s))-{{g}_{2}}(s,{{y}_{k-1}}(s)))} \right. \left. \text{d}{{\varLambda }_{2}}(s) \right|+\\ \!\!\!\!\!\!\!\!&\bigg|\int_{0}^{1}{{{G}_{0}}(t,s)\text{d}s\int_{0}^{1}{({{g}_{1}}(r,{{y}_{k-1}}(r))}} \left. -{{g}_{1}}(r,{{x}_{k-1}}(r)))\text{d}{{\varLambda }_{1}}(r) \right|\text{+}\\ \!\!\!\!\!\!\!\!&\left| \int_{0}^{1}{{{G}_{0}}(t,s)\text{d}s} \right. \left. \int_{0}^{1}{{{G}_{1}}(s,r)({{M}_{1}}+{{M}_{2}})({{x}_{k-1}}(r)-{{y}_{k-1}}(r))}\text{d}r \right|\leqslant \\ \!\!\!\!\!\!\!\!&\left( {{C}_{2}}{{L}_{2}}+{{C}_{1}}{{L}_{1}}\frac{1}{\Gamma (\beta )} \right.+\left. \frac{1}{\Gamma (\beta )\Gamma (\alpha )}({{M}_{1}}+{{M}_{2}}) \right) {\text{·}}\\ \;\;\;\; \!\!\!\!\!\!\!\!& |{{x}_{k-1}}(t)-{{y}_{k-1}}(t)| \\ \end{align} $

$ ||{x_k} - {y_k}||\leqslant h||{x_{k - 1}} - {y_{k - 1}}|| $

其中,

$h = {C_2}{L_2} + {C_1}{L_1}\frac{1}{{\Gamma (\beta )}} + \frac{1}{{\Gamma (\beta )\Gamma (\alpha )}}({M_1} + {M_2})$

即有

$ ||{x_n} - {y_n}||\leqslant {h^n}||{x_0} - {y_0}|| $

$h < 1$ ,所以,当 $n \to \infty $ 时, $||{x_n} - {y_n}|| \to 0$ 。因此,

$\begin{array}{l} ||{x^ * } - {y^ * }||\leqslant ||{x^ * } - {x_n}|| + ||{x_n} - {y_n}|| + \\ \quad\quad||{y_n} - {y^ * }|| \to 0\; ,\; n \to \infty \end{array}$

即有 ${x^ * } = {y^ * } = y,\; y$ 为边值问题(1),(2)在 ${x_0}$ ${y_0}$ 之间的唯一解。

4 应用实例

假设 $f(t,x,y) = \displaystyle\frac{1}{{10{\rm{e}}}}{(1 - t)^{\frac{1}{3}}}({{\rm{e}}^x} + {y^{\frac{4}{3}}}),\; {g_1}(t,x) = - t{{\rm{e}}^x},\; {g_2}(t,x) = \displaystyle\frac{1}{{10{\rm{e}}}}t{x^{\frac{4}{3}}},$ $\varphi (t) = 0,$ ${\varLambda _1}(t) = {\varLambda _2}(t) = \displaystyle\frac{1}{{10}}{{\rm{e}}^{t - 1}}$ ,容易验证 $f,\; {g_i},\; {\varLambda _i} (i = 1 , 2)$ 满足假设(H1),(H2)。令 $\alpha = \displaystyle\frac{5}{3}, \beta = \displaystyle\frac{5}{4},\tau = \displaystyle\frac{1}{3}$ ,考虑

$\begin{split}{l} &{}_t^{\rm{C}}{\rm{D}}_1^{\frac{5}{3}}{}_0^{\rm{C}}{\rm{D}}_t^{\frac{5}{4}}u(t) = f(t,u(t),u(t - \tau )) = \\\;\;\; &\dfrac{1}{{10{\rm{e}}}}{(1 - t)^{\frac{1}{3}}}\left({{\rm{e}}^{u(t)}} + {\left(u\left(t - \dfrac{1}{3}\right)\right)^{\frac{1}{3}}}\right) ,\; t \in [0,1]\end{split} $ (18)
$\!\!\!\!\!\!\!\left\{ {\begin{aligned} {\left({}_0^{\rm{C}}{\rm{D}}_t^{\frac{5}{4}}u(1)\right)' \!=\! 0 ,\; {}_0^{\rm{C}}{\rm{D}}_t^{\frac{5}{4}}u(0) \!=}\\ {\displaystyle \int_0^1 {{g_1}(s,u(s)){\rm{d}}{\varLambda _1}(s)} \; } \\ {u'(t) \!=\! 0 ,t \in \left [ - \dfrac{1}{3},0\right] ,\; u(1) \!=}\\{ \displaystyle \int_0^1 {{g_2}(s,u(s)){\rm{d}}{\varLambda _2}(s)} \; \; } \end{aligned}} \right.$ (19)

${x_0} = {x_0}(t) \equiv 0,\;t \in \left[ - \dfrac{1}{3},1\right],$ 则容易检验 ${x_0} = 0$ 是边值问题(18),(19)的下解。

现令

${y_0}(t) = \left\{ {\begin{array}{*{20}{l}} {\dfrac{8}{{585}}{t^{\frac{5}{4}}}(16{t^2} - 52t - 117) + 2.1,t \in [0,1]\; } \\ {2.1 ,\; \; \; t \in \left[ - \dfrac{1}{3},0 \right]\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; } \end{array}} \right.$

接下来检验 ${y_0} = {y_0}(t)$ 是边值问题(18),(19)的上解。因为,

$ \begin{array}{l} {}_0^{\rm{C}}{\rm{D}}_t^{\frac{5}{4}}{y_0}(t) = \dfrac{{\text{π}} }{{2\sqrt 2 \Gamma (3/4)}}({t^2} - 2t - 2) \\\; {}_t^{\rm{C}}{\rm{D}}_1^{\frac{5}{3}}{}_0^{\rm{C}}{\rm{D}}_t^{\frac{5}{4}}{y_0}(t) = \dfrac{{3{\text{π}} }}{{\sqrt 2 \Gamma (3/4)\Gamma (1/3)}}{(1 - t)^{\frac{1}{3}}}\\ \;\;{}_0^{\rm{C}}{\rm{D}}_t^{\frac{5}{4}}{y_0}({\rm{0}}) = - 2 \end{array} $
$\begin{array}{l} {y'_0}(t) = 0,\; t \in \left [ - \dfrac{1}{3},0 \right], \; \left({}_0^{\rm{C}}{\rm{D}}_t^{\frac{5}{4}}{y_0}(1) \right)' = 0 \\ {y_0}(1) = \dfrac{8}{{585}} \times ( - 153) + 2.1 \approx 0.007 > 0.001\;2 \end{array} $
$ \begin{array}{l} f(t,{y_0}(t),{y_0}(t - \tau )) =\\ \;\;\;\; \dfrac{1}{{10{\rm{e}}}}{(1 - t)^{\frac{1}{3}}}\left({{\rm{e}}^{\frac{8}{{585}}{t^{\frac{5}{4}}}(16{t^2} - 52t - 117) + {\rm{2}}{\rm{.1}}}} + {2.1^{\frac{4}{3}}}\right) < \\ \;\;\;\;\dfrac{{3{\text{π}} }}{{\sqrt 2 {{\Gamma }}(3/4){{\Gamma }}(1/3)}}{(1 - t)^{\frac{1}{3}}}\end{array} $
$ \begin{array}{l} \displaystyle \int_0^1 {{g_1}(t,{y_0}(t)){\rm{d}}{\varLambda _1}(t)} = - \displaystyle \int_0^1 {t{{\rm{e}}^{{y_0}(t)}}{\rm{d}}\frac{1}{{10}}{{\rm{e}}^{t - 1}}} = \\ \;\;\;- \dfrac{1}{{10}} \displaystyle \int_0^1 {t{{\rm{e}}^{\frac{8}{{585}}{t^{\frac{5}{4}}}(16{t^2} - 52t - 117) + 2.1}}{\rm{d}}{{\rm{e}}^{t - 1}}} > - 2\end{array} $
$ \begin{array}{*{20}{l}} \displaystyle \int_0^1 {{g_2}(t,{y_0}(t)){\rm{d}}{\varLambda _2}(t)} = \int_0^1 {\frac{1}{{10{\rm{e}}}}t{{({y_0}(t))}^{\frac{4}{3}}}{\rm{d}}{{\rm{e}}^{t - 1}}} = \\ \quad\;\;\; \displaystyle \int_0^1 {\frac{1}{{100{\rm{e}}}}t{{\left(\frac{8}{{585}}{t^{\frac{5}{4}}}(16{t^2} - 52t - 117) + 2.1\right)}^{\frac{4}{3}}}{\rm{d}}{{\rm{e}}^{t - 1}}} \leqslant \\ \quad\;\;\; 0.001\;657\;85 \end{array} $

所以,有

$\begin{array}{l} {}_t^{\rm{C}}{\rm{D}}_1^{\frac{5}{3}}{}_0^{\rm{C}}{\rm{D}}_t^{\frac{5}{4}}{y_0}(t) = \dfrac{{3\pi }}{{\sqrt 2 {{\Gamma }}(3/4){{\Gamma }}(1/3)}}{(1 - t)^{\frac{1}{3}}}\geqslant \\\;\;\; \quad\quad f(t,{y_0}(t),{y_0}(t - \tau )),\; \; t \in [0,1] \end{array} $
$\begin{array}{l} \quad\quad\quad \left({}_0^{\rm{C}}{\rm{D}}_t^{\frac{5}{4}}{y_0}(1)\right)' = 0,\; \; t \in [0,1]\\ \; \; \; \; \; \; {}_0^{\rm{C}}{\rm{D}}_t^{\frac{5}{4}}{y_0}(0) = - 2\leqslant \displaystyle \int_0^1 {{g_1}(s,{y_0}(s)){\rm{d}}{\varLambda _1}(s)}\end{array} $
$ {y_0}^\prime (t) = 0, t \in \left [ - \frac{1}{3},0 \right],\; \; {y_0}(1) \approx 0.07\geqslant \int_0^1 {{g_2}(s,{y_0}(s)){\rm{d}}{\varLambda _2}(s)} $

所以,

${y_0}(t) = \left\{ {\begin{array}{*{20}{l}} {\dfrac{8}{{585}}{t^{\frac{5}{4}}}(16{t^2} - 52t - 117) + 2.1 > 0,\;\;t \in [0,1]} \\ {2.1 > 0 ,\; \; t \in \left[ - \dfrac{1}{3},0\right]\; \; \; \; \; \; \; \; \; } \end{array}} \right.$

是边值问题(18),(19)的上解。

综上所述,由定理1可知, $f(t,x,y) = \dfrac{1}{{10{\rm{e}}}}{(1 - t)^{\frac{1}{3}}} {\text{·}} ({{\rm{e}}^x} + {y^{\frac{4}{3}}})$ 满足(H1),(H2),且其下、上解 $x_0$ $y_0$ 满足 ${x_0}$ ${y_0}$ ,所以,边值问题(18),(19)在 $[0,{y_0}]$ 之间存在解 ${y^ * }$ ${x^ * }$

$f, {g_i} , {\varLambda _i} (i = 1,2)$ 的表达式可得, ${\varLambda _i}(1) - {\varLambda _i}(0) \approx$ 0.064,

对任意的 ${x_1}\leqslant {x_2} \in [{x_0},{y_0}] \;( t \in [0,1] ),\; {y_1}\leqslant {y_2} \in [{x_0},{y_0}] \left( t \in \left [ - \dfrac{1}{3},0 \right] \right)$ ,有

$ \begin{array}{*{20}{c}} {g_1}(t,{x_1}) - {g_1}(t,{x_2}) \leqslant 1.64({x_2} - {x_1})\\ \;\;{g_2}(t,{x_2}) - {g_2}(t,{x_1}) \leqslant 0.035\;4({x_2} - {x_1})\\ f(t,{x_2},{y_2}) - f(t,{x_1},{y_1}) \leqslant \dfrac{1}{{10{\rm{e}}}} \times 1.65 \times ({x_2} - {x_1}) + \\ \;\;\quad\dfrac{1}{{10{\rm{e}}}} \times1.707\;44 \times ({y_2} - {y_1}) \leqslant 0.123\;51 \times ({x_2} - {x_1}) \end{array} $

${L_i} = 0.064 (i = 1,2) , {C_1} = 1.64 , {C_2} = 0.035\;4 , {M_1} + {M_2} = 0.124$ ,所以,有

$ \begin{array}{l} {L_2}{C_2} + {L_1}{C_1}\dfrac{1}{{{{\Gamma }}\left(\dfrac{5}{4}\right)}} + \dfrac{1}{{{{\Gamma }}\left(\dfrac{5}{3}\right)\Gamma \left(\dfrac{5}{4}\right)}}({M_1} + {M_2}) \approx \\ \quad\quad 0.271\;842\;72 < 1 \end{array} $

成立。

综上所述,边值问题(18),(19)在 $[0,{y_0}]$ 之间存在唯一解 ${y^ * } = {x^ * } = y$

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