﻿ 含左右分数导数的时滞微分方程解的存在性和唯一性
 上海理工大学学报  2020, Vol. 42 Issue (4): 311-316 PDF

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

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

 \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)

2 预备知识

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

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

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

 {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)

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

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

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

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

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

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 $x$ $y$

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

（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

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

 \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)

 \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)

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

 \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)

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

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

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

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

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

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

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

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

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

$n = k$ 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

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

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

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

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

 {{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 ${x_{\rm{0}}}$ ${x^ * }$ ${y^ * }$ ${y_{\rm{0}}}$

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

 $\begin{split} & f(t,{x_k}(t),{x_k}(t - \tau )) - f(t,{y_k}(t),{y_k}(t - \tau )) \geqslant \\ &\quad - {M_1}({y_k}(t) - {x_k}(t)) - {M_2}(({y_k}(t - \tau ) - {x_k}(t - \tau )) \end{split}$ (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)

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

 \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}

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

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

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

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

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

4 应用实例

 $\begin{split}{l} &{}_t^{\rm{C}}{\rm{D}}_1^{\frac{5}{3}}{}_0^{\rm{C}}{\rm{D}}_t^{\frac{5}{4}}u(t) = f(t,u(t),u(t - \tau )) = \\\;\;\; &\dfrac{1}{{10{\rm{e}}}}{(1 - t)^{\frac{1}{3}}}\left({{\rm{e}}^{u(t)}} + {\left(u\left(t - \dfrac{1}{3}\right)\right)^{\frac{1}{3}}}\right) ,\; t \in [0,1]\end{split}$ (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。

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

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

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

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

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

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

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

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

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