上海理工大学学报  2020, Vol. 42 Issue (5): 417-423   PDF    
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
魏春艳, 刘锡平     
上海理工大学 理学院,上海 200093
摘要: 研究了带有左右Riemann-Liouville分数阶导数的非线性时滞泛函微分方程积分边值问题。运用上下解方法,得到了边值问题正解的存在性和唯一性的新结论,给出了求边值问题近似解的迭代方法,并对近似解进行了误差估计。最后给出了具体实例用于说明本文所得结论与方法具有广泛的适用性。
关键词: 左右分数阶导数     时滞     边值问题     正解     迭代方法    
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

近年来,分数阶微分方程受到了人们的广泛关注[1-12],在化学工程、粘弹力学以及人口动态等问题中得到了广泛应用[13-15]。上下解方法是分数阶微分方程理论研究的重要手段,很多文献运用上下解方法研究微分方程边值问题解的存在性[6-12]。在有些情况下,粒子在一些特殊介质中的运动不仅依赖于它当时的状态,还依赖于其过去的状态,比如,在大孔隙率或高流速的情况下,还要考虑到流动惯性的影响[16-18]。而带时滞的分数阶微分方程能够很好地刻画这类物理现象,但用迭代法研究带左右侧Riemann-Liouville分数阶导数的时滞微分方程积分边值问题,并对其近似解进行误差估计的文献相对较少[9-12]

本文研究带有左右分数阶导数的非线性时滞泛函微分方程积分边值问题

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

引理1  对任意给定的 $y \in C([0,1],{\mathbb{R}})$ ,边值问题

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

存在唯一解

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

其中,

$v(s) = \int_0^1 {{G_\alpha }(s,r)y(r)} {\rm{d}} r + a{s^{\alpha - 1}}$ (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)

证明  当 $t \in (0,1)$ 时,根据 ${}_{{0^ + }}{\rm{D}} _t^\alpha ({}_t{\rm{D}} _{{1^ - }}^\beta u(t)) = y(t)$

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

根据边界条件 ${}_t{\rm{D}} _{{1^ - }}^\beta u(0) = 0$ ${}_t{\rm{D}} _{{1^ - }}^\beta u(1) = - a$ ,可得

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

因此,当 $ t \in (0,1) $ 时,

$ \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

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

根据边界条件 $u(1) = 0$ $u(0) = {\phi _0}(0)$ ,可得

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

于是,当 $ t \in (0,1)$ 时,

$ \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} $

其中, $v(s)$ ${G_\beta }\left( {t,s} \right)$ 由式(4)和式(5)定义。

已知当 $\;t \in [ - \tau ,{\rm{0]}}$ 时, $u(t) = {\phi _0}(t)$ ;当 $t \in [1,{\rm{1 + }}\sigma {\rm{]}}$ 时, $u(t) = {\phi _1}(t)$ 。故

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

容易证明,若 $u = u(t)$ 满足式(3),则它满足边值问题(2)。

证毕。

由引理1容易得到下面的引理2。

引理2  边值问题(1)等价于积分方程

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

其中,

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

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

引理3  由式(5)和式(6)分别定义的函数 ${G_\beta }(t,s)$ ${G_\alpha }(s,r)$ 是连续的,且对任意的 $t,s,r \in [0,1]$ ,满足 $0 \leqslant {G_\beta }(t,s) \leqslant \dfrac{1}{{\Gamma (\beta )}}$ $0 \leqslant {G_\alpha }(s,r) \leqslant \dfrac{1}{{\Gamma (\alpha )}}$

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 ]\} $

显然, $P$ $E$ 上的正规锥。对任意的 $x,y \in E$ ,记 $x\; \preceq \; y$ 当且仅当 $y - x \in P,\;t \in [ - \tau ,1 + \sigma ]$ ,于是, $(E,\; \preceq)$ 为半序Banach空间。

定义1  如果 $x \in E$ 满足

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

则称 $x = x(t)$ 是边值问题(1)的下解。

如果 $y \in E$ 满足

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

则称 $y = y(t)$ 是边值问题(1)的上解。

现研究边值问题(1)正解的存在性。先作如下假设:

(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 $t \in [0,1]$ ${x_i},{y_i},{z_i} \in {{\mathbb{R}}_ + },\;\;i = 1,2,$

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

为方便起见,记常数

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

定理1  假设(H1)成立,边值问题(1)存在非负下解 ${x_0} \in P$ 和非负上解 ${y_0} \in P$ ,且满足 $x_0\;\preceq\;y_0$ ,则边值问题(1)在 $[{x_0},{y_0}]: = \{ x \in P|{x_0} \preceq x \preceq {y_0}\}$ 上存在最小正解 ${x^ * }$ 和最大正解 ${y^ * }$

证明  定义映射 $T:[{x_0},{y_0}] \to P$ ,如下:

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

其中, ${v_u}(s)$ 由式(8)定义。容易证明, $T$ 是连续算子。

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 ${x_0}\; \preceq \;T{x_0}$ $T{y_0}\; \preceq \;{y_0}$

已知 ${x_0} \in P$ 是边值问题(1)的下解,由引理1可得边值问题

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

存在唯一解 ${x_1} = {x_1}(t)$ ,又由式(12),满足

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

又因为 ${x_0} \in P$ 是边值问题(1)的下解,根据定义1,有

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

由式(13)和式(15),有

$ \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)

根据引理1,当 $t \in [ - \tau ,1 + \sigma ]$ 时, ${u_0}(t) \geqslant 0$ ,即

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

由于 $T$ 是增算子,所以, $T$ $[{x_0},{y_0}]$ 上的自映射,并且由引理2,边值问题(1)在 $[{x_0},{y_0}]$ 上有解等价于 $T$ $[{x_0},{y_0}]$ 上有不动点。

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} $

由式(9)可得

$ 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。

因为, ${G_\beta }$ (t, s)在[0,1]×[0,1]上连续,所以,在[0,1]×[0,1]上一致连续,又函数(1− ${{t)}^\beta }$ 在[0,1]上连续,所以,在 $[0,1]$ 上一致连续,因此,对任意的 $\varepsilon > 0$ ,存在 $0 < {\delta _1} < 1$ ,当 ${t_1},{t_2},s \in [0,1]$ $|{t_2} - {t_1}| {\rm{ < }}{\delta _1}$ 时,有

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

因为, ${\phi _0}{\rm{(}}t{\rm{)}}$ 在[− τ,0]上连续,所以, ${\phi _0}{\rm{(}}t{\rm{)}}$ 在[− τ ,0]上一致连续,存在 $0 < {\delta _2} < \tau $ ,当 $t_1, t_2 \in [ - \tau , 0 ],\; |{t_2} - $ ${t_1}|{\rm{ < }}{\delta _2}$ 时,有 $|Tu({t_2}) - Tu({t_1})|{\rm{ = }}|{\phi _0}{\rm{(}}{t_2}{\rm{)}} - {\phi _0} {\rm{(}}{t_1}{\rm{)}}|{\rm{ < }} \dfrac{\varepsilon }{2}$

因为, ${\phi _1}{\rm{(}}t{\rm{)}}$ $[1,1+\delta]$ 上连续,所以, ${\phi _1}{\rm{(}}t{\rm{)}}$ $[1,1+\delta] $ 上一致连续,存在 $0 < {\delta _3} < \sigma $ ,使得当 $t_1,t_2 \in [1,1+\delta]$ $|{t_2} - {t_1}|{\rm{ < }}{\delta _3}$ 时,有

$\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 $|Tu({t_2}) - Tu({t_1})| < \varepsilon$

综上所述,对任意的 $\varepsilon > 0$ ${t_1},{t_2} \in [0,1]$ ,存在 $\delta = \min \; \{ {\delta _1},{\delta _2},{\delta _3}\}$ ,使得当 $|{t_2} - {t_1}|{\rm{ < }}\delta $ 时,有 $|Tu({t_2}) - Tu({t_1})| \leqslant \varepsilon $ ,故算子 $T$ 是等度连续的。因此, $T([{x_0},{y_0}])$ $E$ 中的相对列紧集。

分别以 ${x_0}$ ${y_0}$ 为初始元素,由式(14),不妨设 ${x_n}(t) = (T{x_{n - 1}})(t)$ ${y_n}(t) = (T{y_{n - 1}})(t)$ $n = 1,2,3, \cdots ,$ 可得迭代序列 $\{ {x_n}\} $ $\{ {y_n}\} $ ,且满足

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

${x^ * }$ ${y^ * }$ 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} $

所以, ${x^ * }$ ${y^ * }$ 是边值问题(1)的正解。

现证明 ${x^ * }$ ${y^ * }$ 分别是边值问题(1)在 $[{x_0},{y_0}] \subset P$ 上的最小正解和最大正解。

$w \in [{x_0},{y_0}]$ $T$ 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 $

因此, ${x_0}(t) \leqslant {x^ * }(t) \leqslant w(t) \leqslant {y^ * }(t) \leqslant {y_0}(t),\;\;t \in [ - \tau , 1 + \sigma ]$ 。即 ${x^ * }$ ${y^ * }$ 分别是边值问题(1)在 $[{x_0},{y_0}] \subset P$ 上的最小正解和最大正解。

证毕。

定理2  假设(H2)成立,且常数 $0 < \lambda < 1$ ,则边值问题(1)存在唯一正解 ${u^{\rm{*}}}$ ,且对任意的初始函数 ${u_0} \in P$ ,按迭代格式

$ \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)

产生的序列 $\left\{ {{u_n}} \right\}$ $[ - \tau ,1 + \sigma ]$ 上一致收敛到 ${u^{\rm{*}}}$ ,且有先验误差估计

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

及后验误差估计

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

其中, ${v_{{u_{n - 1}}}}(s)$ 和常数 $\lambda $ 分别由式(8)和式(11)定义。

证明  容易证明 $T(P) \subset P$ ,当 $t \in [ - \tau ,1 + \sigma ] $ 时,有

$ \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 $

因此, $||{u_{n + 1}} - {u_n}|| \leqslant \lambda ||{u_n} - {u_{n - 1}}|| \leqslant \cdots \leqslant {\lambda ^n}||{u_1} - {u_0}||$

$m > n$ 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} $

因此, $||{u^ * } - T{u^ * }|| = 0$ ,即 $T{u^ * }{\rm{ = }}{u^ * }$ ${u^*} \in P$ 是边值问题(1)的正解。

若存在 ${u^ * }$ ${z^ * }$ 分别满足 $T{u^ * }{\rm{ = }}{u^ * }$ $T{z^ * }{\rm{ = }}{z^ * }$ ,则 $||{z^ * } - {u^ * }|| = ||T{z^ * } - T{u^ * }|| \leqslant \lambda ||{z^ * } - {u^ * }||$ ,由于 $0 < \lambda < 1$ ,所以, ${z^ * } = {u^ * }$ ,故边值问题(1)存在唯一正解 ${u^*}$

根据式(20),当 $m \to + \infty $ 时,有先验误差估计公式

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

令式(21)中 $n = 1,$ ${u_0} = {z_0},$ ${u_1} = {z_1},$ $||{z_1} - {u^*}{\rm{||}} \leqslant \dfrac{\lambda }{{1 - \lambda }}{\rm{||}}{z_1} - {z_0}{\rm{||}}$ 。再令 ${z_0} = {u_{n - 1}},$ ${z_1} = T{z_0} = T{u_{n - 1}} = {u_n}$ ,故有后验误差估计

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

证毕。

4 应 用

为了说明定理2具有广泛的适用性,现给出一个应用实例,并对其进行数据模拟。

例1  考虑带有左右分数阶导数的时滞微分方程积分边值问题

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

其中, $\alpha = \beta = \dfrac{3}{2}$ $\tau = \sigma = 1$ ,且 ${\phi _1}(1){\rm{ = }}0$

$ 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) $

经过计算, ${a_1}(t) = 0.2t\sqrt {1 - t} $ ${b_1}(t) = {c_1}(t) = 0.2(1 - t)$ ${a_2}(t) = {b_2}(t) = {c_2}(t) = 0.2\sqrt t $ $\;\; \lambda = \dfrac{{{\rm{60}}\sqrt {\text{π}} {\rm{ + }}76}}{{75{\text{π}} }} < 1$ ,边值问题(22)满足定理2的所有条件。于是,边值问题(22)存在唯一正解 ${u^{\rm{*}}}$ ,且对任意的初始函数 ${u_0} \in P$ ,根据迭代格式(19)可得近似解 ${u_n}$ ,有先验误差估计公式

$ ||{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}|| $

及后验误差估计公式 ${\rm{||}}{u_n} \!\!-\!\! {u^*}{\rm{||}} \!\leqslant\! \dfrac{{60\sqrt {\text{π}} \!\!+\!\! 76}}{{75{\text{π}} \!\!-\!\! 65\sqrt {\text{π}} \!\!-\!\! 76}} ||{u_n} \!-\! {u_n}_{ \!- \!1}||$

现取

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

由迭代格式(19)可得

$ {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. $

近似解 ${u_1}$ 的先验误差估计

$ ||{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}} $

和后验误差估计

${\rm{||}}{u_1} - {u^*}{\rm{||}} \leqslant \dfrac{{60\sqrt {\text{π}} + 75}}{{75{\text{π}}- 65\sqrt {\text{π}} - 76}}||{u_1} - {u_0} || \approx {\rm{0}}{\rm{.184\;54}}$   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 ${u_1}$
参考文献
[1]
张潇涵, 刘锡平, 贾梅, 等. 混合分数阶p-Laplace算子方程积分边值问题的多解性 [J]. 上海理工大学学报, 2018, 40(3): 205-210.
[2]
胡雨欣, 寇春海, 葛富东. Banach空间中分数阶微分方程解的存在性[J]. 东华大学学报(自然科学版), 2015, 41(6): 867-872.
[3]
郭莉莉, 刘锡平, 贾梅, 等. 分数阶微分方程组边值问题解的存在性与唯一性[J]. 上海理工大学学报, 2019, 41(3): 214-223.
[4]
李琳, 贾梅, 刘锡平, 等. 具非线性项变号的分数阶微分方程非齐次边值问题正解的存在性[J]. 吉林大学学报(理学版), 2019, 57(2): 219-228.
[5]
廖秀, 韦煜明, 冯春华. 一类无穷区间上分数阶微分方程边值问题正解的存在性[J]. 吉林大学学报(理学版), 2018, 56(6): 1299-1306.
[6]
蹇星月, 刘锡平, 贾梅, 等. 分数阶泛函微分方程边值问题的耦合上下解方法[J]. 高校应用数学学报, 2019, 34(3): 301-314.
[7]
REHMAN M U, KHAN R A. Existence and uniqueness of solutions for multi-point boundary value problems for fractional differential equations[J]. Applied Mathematics Letters, 2010, 23(9): 1038-1044. DOI:10.1016/j.aml.2010.04.033
[8]
LIU X P, JIA M, GE W G. The method of lower and upper solutions for mixed fractional four-point boundary value problem with p-Laplacian operator [J]. Applied Mathematics Letters, 2017, 65: 56-62. DOI:10.1016/j.aml.2016.10.001
[9]
LIU X P, JIA M. The method of lower and upper solutions for the general boundary value problems of fractional differential equations with p-Laplacian [J]. Advances in Difference Equations, 2018(1): 1-15. DOI:10.1186/s13662-017-1446-1
[10]
LIU X P, JIA M. The iteration method of solving a type of three-point boundary value problem[J]. Journal of Applied Mathematics & Informatics, 2009, 27(3/4): 475-487.
[11]
WU J, ZHANG X G, LIU L S, et al. Convergence analysis of iterative scheme and error estimation of positive solution for a fractional differential equation[J]. Mathematical Modelling and Analysis, 2018, 23(4): 611-626. DOI:10.3846/mma.2018.037
[12]
LIU X P, JIA M. Solvability and numerical simulations for BVPs of fractional coupled systems involving left and right fractional derivatives[J]. Applied Mathematics and Computation, 2019, 353(1): 230-242. DOI:10.1016/j.amc.2019.02.011
[13]
白占兵. 分数阶微分方程边值问题理论及应用[M]. 北京: 中国科学技术出版社, 2012.
[14]
郭大钧, 孙经先, 刘兆理. 非线性常微分方程泛函方法[M]. 济南: 山东科学技术出版社, 2006.
[15]
KILBAS A A, SRIVASTAVA M H, TRUJILLO J J. Theory and applications of fractional differential equations[M]. Amsterdam: Elsevier, 2006.
[16]
鲁世平, 孔凡超, 李洁. 一类具曲率算子的非线性方程的波前解[J]. 四川大学学报(自然科学版), 2017, 54(4): 693-697.
[17]
鲁世平, 牛亮, 郭原志, 等. 一类具有排斥型奇性的中立型Liénard方程周期正解的存在性[J]. 应用数学, 2018, 31(3): 481-489.
[18]
LEE K B, HOWELL J R. Theoretical and experimental heat and mass transfer in highly porous media[J]. International Journal of Heat and Mass Transfer, 1991, 34(8): 2123-2132. DOI:10.1016/0017-9310(91)90222-Z