﻿ 具左右分数阶导数的时滞微分方程的正解存在性及迭代求解法
 上海理工大学学报  2020, Vol. 42 Issue (5): 417-423 PDF

Existence and iteration for the delay differential equations involving left and right fractional derivatives
WEI Chunyan, LIU Xiping
College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China
Abstract: The integral boundary value problems of nonlinear delay functional differential equations with left and right Riemann-Liouville fractional derivatives were studied by using the method of lower and upper solutions. Some new results on the existence and uniqueness of solutions were established by using the method of upper and lower solutions, iteration method for solving differential equations and the error estimations were presented. Finally, an example was given out to illustrate the wide applicability of the results and methods.
Key words: left and right fractional derivatives     delay     boundary value problem     positive solution     iteration method
1 问题的提出

 $\left\{ {\begin{array}{*{20}{l}} {{}_{{0^ + }}{\rm{D}} _t^\alpha ({}_t{\rm{D}} _{{1^ - }}^\beta u(t)) = f(t,u(t),u(t - \tau ),u(t + \sigma )),\;\;} \\ \qquad t \in (0,1)\\ u(t) \; = {\phi _0}(t),\;\;t \in [ - \tau ,{\rm{0]}} \\ u(t) = {\phi _1}(t),\;\;t \in [1,{\rm{1 + }}\sigma {\rm{]}} \\ {}_t{\rm{D}} _{{1^ - }}^\beta u(0) = 0 \\ {}_t{\rm{D}} _{{1^ - }}^\beta u(1){\rm{ = }} - \displaystyle\int\nolimits_0^1 {g(t,u(t),u(t - \tau ),u(t + \sigma ))} {\rm{d}} t\\ \end{array}} \right.$ (1)
 ${}_{{0^ + }}{\rm{D}} _t^\alpha u(t) = \frac{1}{{\Gamma (2 - \alpha )}}\frac{{{{\rm{d}}^2}}}{{{\rm{d}}{t^2}}}\int\nolimits_0^t {\frac{{u(s)}}{{{{(t - s)}^{\alpha - 1}}}}} {\rm{d}}s,\;1 < \alpha <2$

${}_t{\rm D}_{{1^ - }}^\beta u(t) = \dfrac{1}{{\Gamma (2 - \beta )}}\dfrac{{{{\rm{d}}^2}}}{{{\rm{d}}{t^2}}}\displaystyle\int\nolimits_t^1 {\dfrac{{u(s)}}{{{{(s - t)}^{\beta - 1}}}}} {\rm{d}} s,\; 1 < \; \beta < 2$ 其中， ${}_{{0^ + }}{\rm{D}}_t^\alpha ,{\;_t}{\rm{D}}_{{1^ - }}^\beta$ 分别为左，右侧Riemann-Liouville分数阶导数。常数 $\tau ,\sigma \!\geqslant\! 0$ ${\phi _0} \in C([ - \tau ,0],{{\mathbb{R}}_ + })$ ${\phi _1} \in C$ $([1,1{\rm{ + }}\sigma ], {{\mathbb{R}}_ + })$ ，且 ${\phi _1}(1) = 0$ $f \in C ([0,1] \times {\mathbb{R}}_{\rm{ + }}^3,{{\mathbb{R}}_ + })$ $g \in C([0,1] \times {\mathbb{R}}_{\rm{ + }}^3,{{\mathbb{R}}_ + })$ ${{\mathbb{R}}_ + } = [0, + \infty )$ 。运用上下解方法得到了边值问题（1）正解的存在性和唯一性定理，给出了求解边值问题近似解的迭代方法，并进行了误差估计。

2 预备知识

 $\left\{ {\begin{array}{*{20}{l}} {{}_{{0^ + }}{\rm{D}} _t^\alpha ({}_t{\rm{D}} _{{1^ - }}^\beta u(t)) = y(t),\;\;t \in (0,1)} \\ \begin{array}{l} u(t) = {\phi _0}(t),\;\;t \in [ - \tau ,{\rm{0]}} \\ u(t) = {\phi _1}(t),\;\;t \in [1,{\rm{1 + }}\sigma {\rm{]}} \\ {}_t{\rm{D}} _{{1^ - }}^\beta u(0) = 0 \\ {}_t{\rm{D}} _{{1^ - }}^\beta u(1) = - a \\ \end{array} \end{array}} \right.$ (2)

 $u(t) = \left\{ \begin{array}{l} {\phi _0}(t),\;\;t \in [ - \tau ,{\rm{0]}} \\ \displaystyle\int\nolimits_0^1 {{G_\beta }(t,s)v(s){\rm{d}} s} + {\phi _0}(0){(1 - t)^{\beta - 1}},\;\; \\ \quad\;\; t \in (0,1)\\ {\phi _1}(t),\;\;t \in [1,{\rm{1 + }}\sigma {\rm{]}} \\ \end{array} \right.$ (3)

 $v(s) = \int_0^1 {{G_\alpha }(s,r)y(r)} {\rm{d}} r + a{s^{\alpha - 1}}$ (4)
 ${G_\beta }(t,s) \!=\!\! \frac{1}{{\Gamma \!(\beta )}}\!\!\left\{ {\begin{array}{*{20}{l}} \!\!\!\!\!\!{{{(s(1 - t))}^{\beta - 1}},}\;\;\;\;\;\;\;\;\qquad{0 \!\leqslant\! s \!\leqslant \! t \leqslant 1}\\ \!\!\!\!\!\!{{{(s(1 \!-\! t))}^{\beta - 1}}\! \!-\! {{(s \!-\! t)}^{\beta - 1}}\!,}\;{0 \! \leqslant \! t \! \leqslant \! s \! \leqslant \!1} \end{array}} \right.\!\!$ (5)
 ${G_\alpha }(s\!,r) \!\!=\!\! \frac{1}{{\Gamma (\alpha )}}\!\!\left\{ {\begin{array}{*{20}{l}} \!\!\!\!\!\!{{{(s(1 \!\!-\! r))}\!^{\alpha \!-\! 1}} \!\!-\! {{(s \!-\! r)}\!^{\alpha - 1}}\!,}\;{0 \! \leqslant \! r \! \leqslant \! s \! \leqslant \!\! 1}\\ \!\!\!\!\!\!{{{(s(1 - r))}^{\alpha - 1}},}\;\;\;\;\;\;\;\qquad{0\! \leqslant \!s \leqslant\! r \!\leqslant\! 1} \end{array}} \right.\!\!\!$ (6)

 ${}_t{\rm{D}} _{{1^ - }}^\beta u(t) = \frac{1}{{\Gamma (\alpha )}}\int_0^t {{{(t - s)}^{\alpha - 1}}y(s)} {\rm{d}} s + {c_0}{t^{\alpha - 1}} + {c_1}{t^{\alpha - 2}}$

 ${c_0} = - a - \frac{1}{{\Gamma (\alpha )}}\int_0^1 {{{(1 - s)}^{\alpha - 1}}y(s)} {\rm{d}} s,\;{c_1} = 0$

 $\begin{split} {}_t{\mathop{\rm D}\nolimits} _{{1^ - }}^\beta u(t) =& \frac{1}{{\Gamma (\alpha )}}\int_0^t {{{(t - s)}^{\alpha - 1}}y(s)} {\mathop{\rm d}\nolimits} s+\\ & \left( { - a - \frac{1}{{\Gamma (\alpha )}}\int_0^1 {{{(1 - s)}^{\alpha - 1}}y(s)} {\mathop{\rm d}\nolimits} s} \right){t^{\alpha - 1}}=\\ & - \frac{1}{{\Gamma (\alpha )}}\int_0^t {({{(t(1 - s))}^{\alpha - 1}} - {{(t - s)}^{\alpha - 1}})y(s)} {\mathop{\rm d}\nolimits} s-\\ & \frac{1}{{\Gamma (\alpha )}}\int_t^1 {{{(t(1 - s))}^{\alpha - 1}}y(s)} {\mathop{\rm d}\nolimits} s - a{t^{\alpha - 1}}=\\ & - \int_0^1 {{G_\alpha }(t,s)y(s)} {\mathop{\rm d}\nolimits} s - a{t^{\alpha - 1}} \end{split}$

$v(t) = - {}_t{\rm{D}} _{{1^ - }}^\beta u(t) = \displaystyle\int\nolimits_0^1 {{G_\alpha }(t,s)y(s)} {\rm{d}} s + a{t^{\alpha - 1}}$ ，则

 $u(t) = - \frac{1}{{\Gamma (\beta )}}\int_t^1 {{{(s - t)}^{\beta - 1}}v(s)} {\rm{d}} s + {d_0}{(1 - t)^{\beta - 1}} + {d_1}{(1 - t)^{\beta - 2}}$

 ${d_0} = {\phi _0}(0) + \frac{1}{{\Gamma (\beta )}}\int_0^1 {{s^{\beta - 1}}v(s)} {\rm{d}} s,\;{d_1} = 0$

 $\begin{array}{*{20}{l}} {u(t) = - \dfrac{1}{{\Gamma (\beta )}}\displaystyle\int\nolimits_t^1 {{{(s - t)}^{\beta - 1}}v(s)} {\rm{d}}s + {{(1 - t)}^{\beta - 1}}\left[ {{\phi _0}(0) + \dfrac{1}{{\Gamma (\beta )}}\displaystyle\int\nolimits_0^1 {{s^{\beta - 1}}v(s)} {\rm{d}}s} \right]}=\\ {\;\;\;\;\;\;\; - \dfrac{1}{{\Gamma (\beta )}}\displaystyle\int\nolimits_t^1 {{{(s - t)}^{\beta - 1}}v(s)} {\rm{d}}s + \dfrac{1}{{\Gamma (\beta )}}\displaystyle\int\nolimits_0^1 {{{(s(1 - t))}^{\beta - 1}}v(s)} {\rm{d}}s + {\phi _0}(0){{(1 - t)}^{\beta - 1}}}=\\ {\;\;\;\;\;\;\; \dfrac{1}{{\Gamma (\beta )}}\left(\displaystyle\int\nolimits_0^t {{{(s(1 - t))}^{\beta - 1}}v(s)} {\rm{d}}s + \displaystyle\int\nolimits_t^1 {({{(s(1 - t))}^{\beta - 1}} - {{(s - t)}^{\beta - 1}})v(s)} {\rm{d}}s\right) + {\phi _0}(0){{(1 - t)}^{\beta - 1}}}=\\ {\;\;\;\;\;\;\; \displaystyle\int\nolimits_0^1 {{G_\beta }(t,s)v(s)} {\rm{d}}s + {\phi _0}(0){{(1 - t)}^{\beta - 1}}} \end{array}$

 $u(t) = \left\{ \begin{array}{l} {\phi _0}(t),\;\;t \in [ - \tau ,{\rm{0]}} \\ \displaystyle\int\nolimits_0^1 {{G_\beta }(t,s)v(s){\rm{d}} s} + {\phi _0}(0){(1 - t)^{\beta - 1}},\;\;t \in [0,1] \\ {\phi _1}(t),\;\;t \in [1,{\rm{1 + }}\sigma {\rm{]}} \\ \end{array} \right.$

 $u(t) = \left\{ \begin{array}{l} \!\!\!\!{\phi _0}(t),\;\;t \in [ - \tau ,{\rm{0]}} \\ \!\!\!\!\!\displaystyle\int\nolimits_0^1 \!\!\!{{G_\beta }(t\!,\!s){v_u}(s){\rm{d}}\! s} \!+\! {\phi _0}(0){(1 \!\!-\! t)^{\beta - 1}}\!,\;t \!\!\in \!\!(0,\!1) \\ \!\!\!\!{\phi _1}(t),t \in [1,{\rm{1 + }}\sigma {\rm{] }} \\ \end{array} \right.$ (7)

 $\begin{split}{v_u}(s) =& \int_0^1 {{G_\alpha }(s,r)f(r,u(r),u(r - \tau ),u(r + \sigma )){\mathop{\rm d}\nolimits} r} + \\ &{s^{\alpha - 1}}\int_0^1 {g(r,u(r),u(r - \tau ),u(r + \sigma )){\mathop{\rm d}\nolimits} r} \end{split}$ (8)

${G_\beta }\left( {t,s} \right)$ ${G_\alpha }(s,r)$ 分别由式（5）和式（6）定义。

3 主要结论

$E = C[ - \tau ,1 + \sigma ]$ ，定义范数 $||u|{|_E} = \mathop {\max }\limits_{ - \tau \leqslant t \leqslant 1 + \sigma } |u(t)|$ ，则 $(E,|| \cdot |{|_E})$ 是Banach空间。令

 $P = \{ u:u \in E,\;u(t) \geqslant 0,\; \;t \in [ - \tau ,1 + \sigma ]\}$

 $\left\{ {\begin{array}{l} {{}_{{0^ + }}{\rm{D}} _t^\alpha ({}_t{\rm{D}} _{{1^ - }}^\beta x(t)) \leqslant f(t,x(t),x(t - \tau ),x(t + \sigma )),\;\;} \\ \qquad t \in ({\rm{0}},{\rm{1) }}\\ x(t) = {\phi _0}(t),\;\;t \in [ - \tau ,{\rm{0]}} \\ x(t) = {\phi _1}(t),\;\;t \in [1,{\rm{1 + }}\sigma {\rm{]}} \\ {}_t{\rm{D}} _{{1^ - }}^\beta x(0) = 0 \\ {}_t{\rm{D}} _{{1^ - }}^\beta x(1) \geqslant - \int_0^1 {g(t,x(t),x(t - \tau ),x(t + \sigma ))} {\rm{d}} t \\ \end{array}} \right.$ (9)

 $\left\{ {\begin{array}{l} {{}_{{0^ + }}{\rm{D}} _t^\alpha ({}_t{\rm{D}} _{{1^ - }}^\beta y(t)) \geqslant f(t,y(t),y(t - \tau ),} \\ \quad y(t + \sigma )),\;\;t \in ({\rm{0}},{\rm{1) }}\\ y(t) = {\phi _0}(t),\;\;t \in [ - \tau ,{\rm{0]}} \\ y(t) = {\phi _1}(t),\;\;t \in [1,{\rm{1 + }}\sigma {\rm{]}} \\ {}_t{\rm{D}} _{{1^ - }}^\beta y(0) = 0 \\ {}_t{\rm{D}} _{{1^ - }}^\beta y(1) \leqslant - \int_0^1 {g(t,y(t),y(t \!-\! \tau ),y(t \!+\! \sigma ))} {\rm{d}} t \\ \end{array}} \right.$ (10)

（H1） 令 $f,g \in C([0,1] \times {\mathbb{R}}_ + ^3,{{\mathbb{R}}_ + })$ ，对任意的 $0 \leqslant {x_1} \leqslant {x_2}$ $0 \leqslant {y_1} \leqslant {y_2}$ $0 \leqslant {z_1} \leqslant {z_2}$ ，及对任意的 $t \in [0,1]$ ，有 $f(t,{x_1},{y_1},{z_1}) \!\leqslant\! f(t,{x_2},{y_2},{z_2})$ $g(t,{x_1},{y_1},{z_1}) \leqslant g(t,{x_2}, {y_2},{z_2})$

（H2） 令 $f,g \in C([0,1] \times {\mathbb{R}}_ + ^3,\;{{\mathbb{R}}_ + })$ ，存在非负函数 ${a_i},{b_i},{c_i} \in {L^1}([0,1],\;{{\mathbb{R}}_ + })$ ，使得对任意的 $t \in [0,1]$ ${x_i},{y_i},{z_i} \in {{\mathbb{R}}_ + },\;\;i = 1,2,$

 $\begin{split}&|f(t,{x_1},{y_1},{z_1}) - f(t,{x_2},{y_2},{z_2})| \leqslant \\&\qquad {a_1}(t)|{x_1} - {x_2}| +{b_1}(t)|{y_1} - {y_2}| + {c_1}(t)|{z_1} - {z_2}|\end{split}$
 $\begin{split}&|g(t,{x_1},{y_1},{z_1}) - g(t,{x_2},{y_2},{z_2})| \leqslant \\&\qquad {a_2}(t)|{x_1} - {x_2}| + {b_2}(t)|{y_1} - {y_2}| + {c_2}(t)|{z_1} - {z_2}|。\end{split}$

 $\begin{split} \lambda =& \frac{1}{{\Gamma (\alpha )\Gamma (\beta )}}\int_0^1 {({a_1}(s) + {b_1}(s) + {c_1}(s)} + \\ & \Gamma (\alpha ){a_2}(s) + \Gamma (\alpha ){b_2}(s) + \Gamma (\alpha ){c_2}(s)) {\rm{d}} s。\end{split}$ (11)

 $T\!\!u(t) \!\!=\!\! \left\{ \begin{array}{l} \!\!\!\!\!{\phi _0}(t),\;\;t \in [ - \tau ,{\rm{0]}} \\ \!\!\!\!\!\!\displaystyle\int\nolimits_0^1 \!\!\!\!{{G_\beta }\!\left( {t,\!s} \right)\!{v\!_u}(s){\rm{d}}\! s} \!+\! {\phi _0}(0){(1 \!\!-\! t)^{\beta - 1}}\!\!,t \!\! \in \!\!(0,\!1) \\ \!\!\!\!\!{\phi _1}(t),\;\;t \in [1,{\rm{1 + }}\sigma {\rm{]}} \\ \end{array} \right.$ (12)

a. 证明 $T$ 是增算子。

${u_1}\; \preceq \;{u_2}$ ${u_1},{u_2} \in [{x_0},{y_0}]$ 。根据条件（H1），当 $t \in [ - \tau ,1 + \sigma ]$ 时， $T{u_2}(t) - T{u_1}(t) \geqslant 0$ ，即 $T{u_1}\; \preceq \;T{u_2}$ $T$ 为增算子。

b. 证明 $T$ $[{x_0},{y_0}]$ 上的自映射，并且 ${x_0}\; \preceq \;T{x_0}$ $T{y_0}\; \preceq \;{y_0}$

 $\left\{\!\!\!\!\! {\begin{array}{l} {{}_{{0^ + }}{\rm{D}} _t^\alpha ({}_t{\rm{D}} _{{1^ - }}^\beta {x_1}(t))\! = \!f(t,{x_0}(t),{x_0}(t \!-\! \tau ),\;} {x_0}(t \!+\! \sigma )),\\ \qquad t \!\in\! ({\rm{0}},{\rm{1)}}\\ {x_1}(t) = {\phi _0}(t),\;\;t \in [ - \tau ,{\rm{0],}} \\ {x_1}(t) = {\phi _1}(t),\;\;t \in [1,{\rm{1 + }}\sigma {\rm{]}} \\ {}_t{\rm{D}} _{{1^ - }}^\beta {x_1}(0) = 0 \\ {}_t{\rm{D}} _{{1^ - }}^\beta {x_1}(1) \!=\! - \displaystyle\int_0^1 {g(t,{x_0}(t),} \; {x_0}(t \!-\! \tau ),{x_0}(t \!+\! \sigma )){\rm{d}} t \end{array}} \right.\!\!\!\!\!$ (13)

 ${x_1} = T{x_0}$ (14)

 $\left\{ \!\!\!\!\!{\begin{array}{l} {{}_{{0^ + }}{\rm{D}} _t^\alpha ({}_t{\rm{D}} _{{1^ - }}^\beta {x_0}(t)) \!\leqslant\! f(t,{x_0}(t),{x_0}(t \!-\! \tau ),}\; {x_0}(t \!+\! \sigma )),\\ \quad t \!\in\! ({\rm{0}},{\rm{1)}}\\ {x_0}(t) = {\phi _0}(t),\;\;t \in [ - \tau ,{\rm{0]}} \\ {x_0}(t) = {\phi _1}(t),\;\;t \in [1,{\rm{1 + }}\sigma {\rm{]}} \\ {}_t{\rm{D}} _{{1^ - }}^\beta {x_0}(0) = 0 \\ {}_t{\rm{D}} _{{1^ - }}^\beta {x_0}(1)\! \geqslant\! - \displaystyle\int_0^1 {g(t,{x_0}(t),{x_0}(t \!-\! \tau ),} \; {x_0}(t \!+\! \sigma )) {\rm{d}} t \end{array}} \right.\!\!\!\!\!$ (15)

 $\left\{ \begin{array}{l} {}_{{0^ + }}{\rm{D}} _t^\alpha ({}_t{\rm{D}} _{{1^ - }}^\beta ({x_1}(t) - {x_0}(t)))\! \geqslant \!0,\; t \in ({\rm{0}},{\rm{1)}}\\ {x_1}(t) - {x_0}(t) = 0,\;\;t \in [ - \tau ,{\rm{0],}} \\ {x_1}(t) - {x_0}(t) = 0,\;\;t \in [1,{\rm{1 + }}\sigma {\rm{] }} \\ {}_t{\rm{D}} _{{1^ - }}^\beta ({x_1}(0) - {x_0}(0)) = 0\\ {}_t{\rm{D}} _{{1^ - }}^\beta ({x_1}(1) - {x_0}(1)) \leqslant 0 \end{array} \right.$ (16)

${u_0}(t) = {x_1}(t) - {x_0}(t)$ ，则

 $\left\{ \begin{array}{l} {}_{{0^ + }}{\rm{D}} _t^\alpha ({}_t{\rm{D}} _{{1^ - }}^\beta {u_0}(t))\!: =\! h(t)\! \geqslant \!0,\; t \in ({\rm{0}},{\rm{1)}} \\ {u_0}(t) = 0,\;\;t \! \in \![ - \tau ,{\rm{0]}} \\ {u_0}(t) = 0,\;\;t \in [1,{\rm{1 + }}\sigma {\rm{] }}\;\; \\ {}_t{\rm{D}} _{{1^ - }}^\beta {u_0}(0) = 0 \\ {}_t{\rm{D}} _{{1^ - }}^\beta {u_0}(1): = - a \leqslant 0 \end{array} \right.\quad$ (17)

 ${x_1}(t) - {x_0}(t) \geqslant 0,\;t \in [ - \tau ,1 + \sigma ]$ (18)

${x_0}\; \preceq \;{x_1} = T{x_0}$ 。由条件（H1）和定义1容易证明 ${x_1}$ 是边值问题（1）的下解。类似可证， $T{y_0} = {y_1}\; \preceq \;{y_0}$ ，并且 ${y_1}$ 是边值问题（1）的上解。

c. 证明 $T([{x_0},{y_0}])$ $P$ 中的相对列紧集。

$[{x_0},{y_0}]$ $P$ 上的有界集，由于P为正规锥，所以，序区间为有界集。因此，对任意的 $u \in [{x_0},{y_0}]$ ，存在 $L \geqslant 0$ ，使得 ${\rm{||}}u{\rm{||}} \leqslant L$ ，由 $f$ $g$ 的连续性可知, 对任意的 $t \in [0,1]$ ，存在 ${m_0},\;{m_1},{M_0},{M_1} \geqslant 0$ ，使得

 $\begin{array}{l} {m_0} \leqslant |f(t,u(t),u(t - \tau ),u(t + \sigma ))| \leqslant {M_0}\\ {m_1} \leqslant |g(t,u(t),u(t - \tau ),u(t + \sigma ))| \leqslant {M_1} \end{array}$
 $\frac{{{m_0}}}{{\Gamma (\alpha )}} \leqslant |{v_u}(s)| \leqslant \frac{{{M_0}}}{{\Gamma (\alpha )}} + {M_1}$

 $Tu(t) \leqslant \frac{{{M_0} + \Gamma (\beta ){M_1}}}{{\Gamma (\alpha )\Gamma (\beta )}} + {\phi _0}(0)$

$T([{x_0},{y_0}])$ 是一致有界的。

 $\begin{split} &\quad\quad|{G_\beta }({t_2},s) - {G_\beta }({t_1},s)|{\rm{ < }}\frac{{\Gamma (\alpha )\varepsilon }}{{4({M_0} + \Gamma (\alpha ){M_1})}}\;\; \\ &\quad\quad|{{\rm{(}}1 - {t_2}{\rm{)}}^{\beta - 1}} - {{\rm{(}}1 - {t_1}{\rm{)}}^{\beta - 1}}|{\rm{ < }}\frac{\varepsilon }{{4{\phi _0}(0) + 1}}\\ &|Tu({t_2}) - Tu({t_1})| \leqslant \int_0^1 {{\rm{|}}{G_\beta }({t_2},s)} - {G_\beta }({t_1},s)|{v_u}(s) {\rm{d}} s + \\ &\qquad {\phi _0}(0)|{{\rm{(}}1 - {t_2}{\rm{)}}^{\beta - 1}} - {{\rm{(}}1 - {t_1}{\rm{)}}^{\beta - 1}}| < \frac{\varepsilon }{4} + \frac{\varepsilon }{4} = \frac{\varepsilon }{2} \end{split}$

 $\begin{split}\quad\quad\quad\quad {\rm{|}}Tu({t_2}) - Tu({t_1}){\rm{| = |}}{\phi _1}{\rm{(}}{t_2}{\rm{)}} - {\phi _1}{\rm{(}}{t_1}{\rm{)| < }}\dfrac{\varepsilon }{2}\end{split}$

${t_1} \in [ - \tau ,0]$ ${t_2} \in [0,1]$ 时，如果 $|{t_2} - {t_1}| < \min \; \{ {\delta _1}, {\delta _2}\}$ ，有

 $\begin{split} &|Tu({t_2}) - Tu({t_1})| = {\rm{|}}Tu({t_2}) - Tu(0){\rm{ + }}Tu(0) - Tu({t_1}){\rm{|}} \leqslant\\ & \qquad {\rm{|}}Tu({t_2}) - Tu(0){\rm{| + |}}Tu(0) - Tu({t_1}){\rm{|}} < \dfrac{\varepsilon }{2} + \dfrac{\varepsilon }{2} = \varepsilon \\ \end{split}$

${t_1} \in [0,1]$ ${t_2} \in [1,1 + \sigma ]$ 时，如果 $|{t_2} - {t_1}| < \min \; \{ {\delta _1},{\delta _3}\}$ ，同理，有 $|Tu({t_2}) - Tu({t_1})| < \varepsilon$

 $\begin{split} &0 \leqslant {x_0}(t) \leqslant {x_1}(t) \leqslant \cdots \leqslant {x_n}(t) \leqslant \cdots \leqslant \\& \qquad {y_n}(t) \leqslant \cdots \leqslant {y_1}(t) \leqslant {y_0}(t),\;\;t \in [ - \tau ,1 + \sigma ]。\end{split} \qquad\quad$

${x^ * }$ ${y^ * }$ 分别是序列 $\{ {x_n}\}$ $\{ {y_n}\}$ 的极限，即

 $\begin{array}{l} \mathop {\lim }\limits_{n \to \infty } {x_n}(t) = {x^ * }(t) = (T{x^ * })(t)\\ \mathop {\lim }\limits_{n \to \infty } {y_n}(t) = {y^ * }(t) = (T{y^ * })(t) \end{array}$

$w \in [{x_0},{y_0}]$ $T$ 的不动点，由于 ${x_0}\;\preceq\; w\;\preceq\; {y_0}$ 。因为， $T$ 是单调增算子，所以，有 ${x_1} = T{x_0}\; \preceq \; Tw = w\; \preceq \;T{y_0} = {y_1}$ 。两边重复作用T，使得

 $\begin{split} &{x_0}(t) \leqslant {x_1}(t) \leqslant\cdots \leqslant {x_n}(t) \leqslant \cdots \leqslant \\ &\qquad w(t) \leqslant \cdots \leqslant {y_n}(t) \leqslant \cdots \leqslant {y_1}(t) \leqslant {y_0}(t),\;\;t \in [ - \tau ,1 + \sigma ]。\end{split}\qquad$

 $\begin{split}{u_n}&(t) = T{u_{n - 1}}(t) = \\ &\left\{ \begin{array}{l} {\phi _0}(t),\;\;t \in [ - \tau ,{\rm{0]}} \\ \displaystyle\int\nolimits_0^1 {{G_\beta }(t,s){v_{{u_{n - 1}}}}(s){\rm{d}} s} + {\phi _0}(0){(1 - t)^{\beta - 1}}, \\ \qquad\;\; t \in (0,1)\\ {\phi _1}(t),\; \;t \in [1,{\rm{1 + }}\sigma {\rm{] }} \end{array} \right.\end{split}$ (19)

 ${\rm{||}}{u_n} - {u^{\rm{*}}}{\rm{||}} \leqslant \frac{{{\lambda ^n}}}{{1 - \lambda }}||{u_1} - {u_0}||$

 $||{u_n} - {u^{\rm{*}}}|| \leqslant \frac{\lambda }{{1 - \lambda }}||{u_n} - {u_n}_{ - 1}||,$

 $\begin{array}{l} |{u_{n + 1}}(t) - {u_n}(t)| = |T{u_n}(t) - T{u_{n - 1}}(t)| \leqslant\\ \qquad \left|\displaystyle\int\nolimits_0^1 {{G_\beta }(t,s){v_{{u_n}}}(s){\rm{d}} s} - \int_0^1 {{G_\beta }(t,s){v_{{u_{n - 1}}}}(s){\rm{d}} s} \right| \leqslant\\ \qquad \dfrac{1}{{\Gamma (\alpha )\Gamma (\beta )}}\displaystyle\int\nolimits_0^1 {|f(r,{u_n}(r),{u_n}(r - \tau ),{u_n}(r + \sigma )) - f(r,{u_{n - 1}}(r),{u_{n - 1}}(r - \tau ),{u_{n - 1}}(r + \sigma ))|} {\rm{d}} r +\\ \qquad \dfrac{1}{{\Gamma (\beta )}}\displaystyle\int\nolimits_0^1 {|g(r,{u_n}(r),{u_n}(r - \tau ),{u_n}(r + \sigma )) - g(r,{u_{n - 1}}(r),{u_{n - 1}}(r - \tau ),{u_{n - 1}}(r + \sigma ))|} {\rm{d}} r \leqslant\\ \qquad\dfrac{1}{{\Gamma (\alpha )\Gamma (\beta )}}\displaystyle\int\nolimits_0^1 {({a_1}(r) + {b_1}(r) + {c_1}(r))} {\rm{d}} r||{u_n} - {u_{n - 1}}|| + \dfrac{1}{{\Gamma (\beta )}}\displaystyle\int\nolimits_0^1 {({a_2}(r) + {b_2}(r) + {c_2}(r))} {\rm{d}} r||{u_n} - {u_{n - 1}}|| \leqslant\\ \qquad \dfrac{1}{{\Gamma (\alpha )\Gamma (\beta )}}\displaystyle\int\nolimits_0^1 {({a_1}(s) + {b_1}(s) + {c_1}(s) + \Gamma (\alpha ){a_2}(s) + \Gamma (\alpha ){b_2}(s) + \Gamma (\alpha ){c_2}(s))} {\rm{d}} s||{u_n} - {u_{n - 1}}|| \leqslant\\ \qquad\lambda ||{u_n} - {u_{n - 1}}|| \end{array} \qquad \qquad$

$m > n$ 时，

 $\begin{split} ||{u_n} -& {u_m}|| \leqslant ||{u_n} \!-\! {u_{n \!+\! 1}}||{\rm{ + }}||{u_{n + 1}} - {u_{n + 2}}|| + \cdots +\\ &||{u_{m - 1}} \!-\! {u_m}|| \!\! \leqslant {\lambda ^n}||{u_1} \!\!-\! {u_0}|| \!\!+\!\! {\lambda ^{n \!+\! 1}}||{u_1} \!\!-\! {u_0}|| \!+\! \cdots +\!\\ &{\lambda ^{m \!-\! 1}}||{u_1} \!-\! {u_0}|| \!\! \leqslant \dfrac{{{\lambda ^n}}}{{1 - \lambda }}||{u_1} - {u_0}||\\[-15pt] \end{split}$ (20)

$n \to + \infty$ 时，有 $||{u_n} - {u_m}|| \to 0$ ，所以，迭代序列 $\{ {u_n}\}$ 收敛。设 $\mathop {\lim }\limits_{n \to \infty } {u_n} = {u^ * },$ 则由T的连续性，有 $\mathop {\lim }\limits_{n \to \infty } {u_{n+1}} = u^ *$ ${u_n} = T{u_n}_{ + 1}$ ，所以，

 $\begin{split} ||{u^ * } - &T{u^ * }|| \leqslant ||{u^ * } - {u_n}|| +||{u_n} - T{u^ * }|| \leqslant \\ &||{u^ * } - {u_n}|| + \lambda ||{u_{n - 1}} - T{u^ * }|| \end{split}$

 ${\rm{||}}{u_n} - {u^*}{\rm{||}} \leqslant \frac{{{\lambda ^n}}}{{1 - \lambda }}||{u_1} - {u_0}||$ (21)

 $\quad\quad\quad \quad\quad\quad ||{u_n} - {u^*}|| \leqslant \frac{\lambda }{{1 - \lambda }}||{u_n} - {u_n}_{ - 1}||$ (22)

4 应　用

 $\left\{ {\begin{array}{l} \!\!\!{{}_{{0^ + }}{\rm{D}} _t^{\tfrac{3}{2}}({}_t{\rm{D}} _{{1^ - }}^{\tfrac{3}{2}}u(t)) = f(t,u(t),u(t \!-\!\! 1),u(t \!+\! 1)),\;\;} \\ \begin{array}{l} \quad\;\; t \in (0,1)\\ \!\!\!\!\!\!u(t) = \; 1 + t,\;\;t \in [ - 1,{\rm{0]}}\;\;\; \\ \!\!\!\!\!\!u(t) = t - 1,\;\;t \in [1,2{\rm{]}} \\ \!\!\!\!\!\!{}_t{\rm{D}} _{{1^ - }}^{\tfrac{3}{2}}u(0) = 0{\rm{, }}\;\;\; \\ \!\!\!\!\!\!{}_t{\rm{D}} _{{1^ - }}^{\tfrac{3}{2}}u(1){\rm{ = }} - \displaystyle\int_0^1 {g(t,u(t),u(t - 1),u(t + 1))} {\rm{d}} t \end{array} \end{array}} \right.$ (23)

 $f(t,x,y,z) = 0.2(t\sqrt {1 - t} x + (1 - t)(y + z))$
 $g(t,x,y,z) = 0.2\sqrt t (x + y + z)$

 $||{u_n} - {u^*}{\rm{||}} \leqslant \dfrac{{{{(60\sqrt {\text{π}} + 76)}^n}}}{{(75{\text{π}}- 60\sqrt {\text{π}} - 76){{(75{\text{π}} )}^{n - 1}}}} ||{u_1} \!\!-\!\! {u_0}||$

 ${u_0}(t) = \left\{ \begin{array}{l} \begin{array}{*{20}{c}} \!\!\!\!\!\!\!{1 + t,\;\;\;\;t \in [ - 1,0]} \\ \!\!\!\!\!\!\!{\sqrt {1 - t} ,\;\;t \in (0,1)\;\;} \end{array} \\ t - 1,\;\;\;t \in [1,2]\\ \end{array} \right.$

 ${u_1}(t) = \left\{ \begin{array}{l} {1 + t,\;\;\;\;t \in [ - 1,0]}\\ \sqrt {1 - t} + \dfrac{{t\sqrt {1 - t} (2\;144 + 3\;360\sqrt {\text{π}} + 525{{\text{π}} ^{{\textstyle{3 \over 2}}}} - 952t - 1\;540{t^2} + 840{t^3})}}{{42\;000{\text{π}} }}+\\ \qquad \dfrac{{{t^2}(192 + 224\sqrt {\text{π}} + 35{{\text{π}}^{{\textstyle{3 \over 2}}}} - 140{t^2} + 56{t^3})(2\ln (1 + \sqrt {1 - t} ) - \ln t)}}{{5\;600{\text{π}} }},\quad t \in (0,1)\\ t - 1,\;\;t \in [1,2] \end{array} \right.$

 $||{u_1} \!\!-\!\! {u^*}{\rm{||}}\! \leqslant \! \dfrac{{60\sqrt {\text{π}}\!\!+\!\! 76}}{{75{\text{π}} \!\!-\!\! 65\sqrt {\text{π}} \!\!-\!\! 76}} ||{u_1} - {u_0} || \approx {\rm{0}}{{.184\;54}}$

${\rm{||}}{u_1} - {u^*}{\rm{||}} \leqslant \dfrac{{60\sqrt {\text{π}} + 75}}{{75{\text{π}}- 65\sqrt {\text{π}} - 76}}||{u_1} - {u_0} || \approx {\rm{0}}{\rm{.184\;54}}$ 　　迭代过程及近似解 ${u_1}$ 图1所示。

 图 1 迭代过程及近似解 ${u_1}$ Fig. 1 Interation and approximate solution ${u_1}$

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