上海理工大学学报  2020, Vol. 42 Issue (5): 430-435   PDF    
非瞬时脉冲分数阶微分方程边值问题解的存在性与唯一性
郑雯静, 贾梅, 李庭乐     
上海理工大学 理学院,上海 200093
摘要: 非瞬时脉冲所描述的突变会持续停留在一个有限的时间间隔内,这种现象在临床医学、生物工程、化学和物理等领域都普遍存在。为了能够更深刻、更精确地反映事物的变化规律,研究了一类具有非瞬时脉冲的分数阶微分方程边值问题解的存在性与唯一性。首先,通过建立与边值问题等价的积分方程,定义了算子,并证明了其全连续性;然后,运用Schauder不动点定理得到了边值问题解存在的充分条件;最后利用压缩映射原理得到解的唯一性定理。
关键词: 非瞬时脉冲     Caputo分数阶导数     Schauder不动点定理     压缩映射原理    
Existence and uniqueness of solutions for boundary value problems of fractional differential equations with non-instantaneous impulses
ZHENG Wenjing, JIA Mei, LI Tingle     
College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China
Abstract: The mutation described by non-instantaneous pulses will stay in a limited time interval. This phenomenon is common in clinical medicine, bioengineering, chemistry, physics and other fields. In order to reflect the change law of things more profoundly and accurately, the existence and uniqueness of solutions for a class of boundary value problems of fractional differential equations with non-instantaneous impulses were studied. The operator was defined by establishing an integral equation equivalent to the boundary value problem, and its complete continuity was proved. By using Schauder fixed point theorem, the sufficient conditions for the existence of solutions of the boundary value problems were obtained. The uniqueness theorem of solutions was obtained by using the contraction mapping principle.
Key words: non-instantaneous impulses     Caputo derivative     Schauder fixed point theorem     contraction mapping principle    
1 问题的提出

分数阶微分方程在流体力学、化学、控制系统、热传导等自然科学领域中有着广泛的应用,受到广为关注[1-6]。许多实际问题在发展的某些阶段会出现脉冲现象,因此,对于分数阶脉冲微分方程的研究,取得了许多研究成果,参见文献[7-14]。由于有些问题在某点发生突变,并且这种变化会在特定的时间区间内持续,这就是所研究的非瞬时脉冲现象。

本文研究了一类非瞬时脉冲的Caputo型分数阶微分方程边值问题:

$ \left\{ \begin{split} & {}^c{\rm D}_{{s_k} + }^\alpha u(t)\!\! =\!\! f(t,u(t)),{\rm{ }}t \in ({s_k},{t_{k + 1}}],{\rm{ }}k\!\! =\!\! 0,1,2, \cdots ,m \\ & {}^c{\rm D}_{{t_k} + }^\beta u(t) = g(t,u(t)),{\rm{ }}t \in ({t_k},{s_k}],{\rm{ }}k = 1,2, \cdots ,m \\ &\!\! \Delta u{|_{t={t_k}}}\!\!\!=\!\!\!{Q_k}({t_k},u({t_k})),\Delta u{|_{t={s_k}}}\!\!\!=\!\!0,u'(s_k^+)\!\!=\!\!u'(s_k^-)\!\!=\!\!0, \\ & \quad\quad k = 1,2, \cdots ,m, \\ & u'(0) = u(1) = 0 \end{split} \!\!\!\!\!\!\!\!\!\!\! \right. $ (1)

式中, ${}^c{\rm D}_{{s}^ + }^\alpha$ , ${}^c{\rm D}_{{t}^ + }^\beta$ 均为Caputo分数阶导数, $1 < \alpha ,\;\beta \leqslant 2,\;0 = {s_0} < {t_1} < {s_1} < {t_2} < \cdots < {s_m} < {t_{m + 1}} = 1$ , $J = [0,1]$ , $J' = J\backslash \{ {t_1},{t_2}, \cdots ,{t_m}\} $ , $f,g,{Q_k}:J \times \mathbb{R} \to \mathbb{R},\;$ 是连续函数 , $\Delta u{|_{t = {t_k}}} \!\!=\! \Delta u({t_k}) = $ $u(t_k^ + ) - u(t_k^ - )$ , $\Delta u{|_{t = {s_k}}} \!=\! \Delta u({s_k}) = $ $u(s_k^ + ) - u(s_k^ - )$ , $\;k = 1,\;2,\; \cdots ,\;m$

2 预备知识与引理

有关分数阶导数和积分的定义及性质,可以参见文献[1-3]。定义空间

$ \begin{split} & PC(J,\mathbb{R}): = \{ u:J \to \mathbb{R}|u \in C(J',\mathbb{R}),\;u(t_k^{}) = u(t_k^ - ), \\ & \;\;\;\;\;\;\;\;\;\;u(t_k^ + ){\text{存在}}\;,\;k = 1,2,3, \cdots ,m\} \end{split}$

取范数 $||u|{|_{PC}} = \mathop {\sup }\limits_{t \in [0,1]} |u(t)|$ ,则 $PC(J,\mathbb{R})$ 为Banach空间。

定义1  设 $u \in PC(J,\mathbb{R}){\rm{ }}$ ,若 $u$ 满足问题(1)各式,那么称 $u$ 是边值问题(1)的一个解。

引理1  对于任意给定的 ${y_k} \in C[0,1],\;k = 0,\;1,\; \cdots , \;m,$ $\;{h_k} \in C[0,1],{q_k}$ 为常数,如下边值问题

$ \left\{ \begin{split} & {}^c{\rm D}_{{s_k} + }^\alpha u(t) = {y_k}(t),{\rm{ }}t \in ({s_k},{t_{k + 1}}],{\rm{ }}k = 0,\;1,\;2,\; \cdots ,\;m\; \\ & {}^c{\rm D}_{{t_k} + }^\beta u(t) = {h_k}(t),{\rm{ }}t \in ({t_k},{s_k}], k = 1,\;2,\; \cdots ,\;m \\ & \Delta u{|_{t = {t_k}}} = {q_k}, \Delta u{|_{t = {s_k}}} = 0,\;u'(s_k^ + ) = u'(s_k^ - ) = 0,\\ &\qquad k = 1,\;2,\; \cdots ,\;m \\ & u'(0) = u(1) = 0 \end{split} \right. \!\!\!\!\!\!\!\!\!\!\!\!\! $ (2)

存在唯一解

$ u(t) \!\!=\!\! \left\{ \begin{split} & I_{0 + }^\alpha {y_0}(t) - \sum\limits_{i = 1}^{m + 1} {I_{{s_{i - 1}} + }^\alpha {y_{i - 1}}({t_i})} - \sum\limits_{i = 1}^m {(I_{{t_i} + }^\beta {h_i}({s_i}) -} \\ & \;\;\;\; { I_{{t_i} + }^{\beta - 1}{h_i}({s_i})({s_i} - {t_i}))} - \sum\limits_{i = 1}^m {{q_i}} ,\;t \in [0,{t_1}] \\ & I_{{s_k} + }^\alpha {y_k}(t) - \sum\limits_{i = k + 1}^{m + 1} {I_{{s_{i - 1}} + }^\alpha {y_{i - 1}}({t_i})} - \\ &\; \sum\limits_{i = k + 1}^m {(I_{{t_i} + }^\beta {h_i}({s_i}) - I_{{t_i} + }^{\beta - 1}{h_i}({s_i})({s_i} - {t_i}))} \!\! -\!\! \sum\limits_{i = k + 1}^m {{q_i}} ,\!\!\!\!\!\! \\ & \;\;\; t \in ({s_k},{t_{k + 1}}],k = 1,2, \cdots ,m - 1\\ & \!\!-\!\! \sum\limits_{i = k + 1}^{m + 1} {I_{{s_{i - 1}} + }^\alpha {y_{i - 1}}({t_i})} + I_{{t_k} + }^\beta {h_k}(t) \!\!- \!\!I_{{t_k} + }^{\beta - 1}{h_k}({s_k})(t - {t_k})\! -\!\! \\ & \;\;\; \sum\limits_{i = k}^m {I_{{t_i} + }^\beta {h_i}({s_i})} + \sum\limits_{i = k}^m {I_{{t_i} + }^{\beta - 1}{h_i}({s_i})({s_i} - {t_i})} - \sum\limits_{i = k}^m {{q_i}} + {q_k}, \\ & \;\;\; \;\;t \in ({t_k},{s_k}],k = 1,2, \cdots ,m\\ & I_{{s_m} + }^\alpha {y_m}(t) - I_{{s_m} + }^\alpha {y_m}(1),\;\;\;t \in ({s_m},1]。\end{split} \right. \!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! $ (3)

证明 $t \in [0,{t_1}]$ 时,由 ${}^c{\rm D}_{0 + }^\alpha u(t) = {y_0}(t)$ $u'(0) $ = 0,得

$u(t) = I_{0 + }^\alpha {y_0}(t) + {c_0},u(t_1^{}) = u(t_1^ - ) = I_{0 + }^\alpha {y_0}({t_1}) + {c_0}$ (4)

$t \in ({t_1},{s_1}]$ 时,考虑边值问题

$\left\{ \begin{split} {}^c{\rm D}_{{t_1} + }^\beta u(t) = {h_1}(t),{\rm{ }}t \in ({t_1},{s_1}]{\rm{ }}\\ u(t_1^ + ) = u(t_1^ - ) + {q_1},u'(s_1^ - ) = 0 \end{split} \right.$

${\rm D}_{{t_1}^ + }^\beta (u(t)) = {h_1}(t)$ $u'(s_1^ - ) = 0$ ,可得

$u(t) = I_{{t_1} + }^\beta {h_1}(t) - I_{{t_1} + }^{\beta - 1}{h_1}({s_1})(t - {t_1}) + {d_1}$

$u(t_1^ + ) = u(t_1^ - ) + {q_1}$ ,所以

${d_1} = u(t_1^ - ) + {q_1} = I_{0 + }^\alpha {y_0}({t_1}) + {c_0} + {q_1}$ (5)

$ \begin{split} u(t) =& I_{{t_1} + }^\beta {h_1}(t) - I_{{t_1} + }^{\beta - 1}{h_1}({s_1})(t - {t_1}) + \\ &I_{0 + }^\alpha {y_0}({t_1}) + {q_1} + {c_0} \end{split} $ (6)
$ \begin{split} & u(s_1^ - ) = I_{{t_1} + }^\beta {h_1}({s_1}) - I_{{t_1} + }^{\beta - 1}{h_1}({s_1})({s_1} - {t_1}) + \\ & \;\;\;\;\;\;\;\;\;\;\;\;\;\; I_{0 + }^\alpha {y_0}({t_1}) + {q_1} + {c_0} \end{split}$ (7)

$t \in ({s_1},{t_2}]$ 时,考虑Cauchy问题

$\left\{ \begin{split} & {}^c{\rm D}_{{s_1} + }^\alpha u(t) = {y_1}(t) \\ & u(s_1^ + ) = u(s_1^ - ),\;u'(s_1^ + ) = 0\; \\ \end{split} \right.$

${}^c{\rm D}_{{s_1} + }^\alpha u(t) = {y_1}(t),\;u'(s_1^ + ) = 0$ 可得, $u(t) = I_{{s_1} + }^\alpha {c_1},$ 所以

$ \begin{split} &\;\;\;\; u(s_1^ + ) = {c_1} = u(s_1^ - ) = I_{{t_1} + }^\beta {h_1}({s_1}) - \\ &\;\;\;\;\;\;\;\;\;\;\;\;\;\; I_{{t_1} + }^{\beta - 1}{h_1}({s_1})({s_1} - {t_1}) + I_{0 + }^\alpha {y_0}({t_1}) + {q_1} + {c_0} \end{split}$ (8)

$ \begin{split} & u(t) = I_{{s_1} + }^\alpha {y_1}(t) + I_{{t_1} + }^\beta {h_1}({s_1}) - I_{{t_1} + }^{\beta - 1}{h_1}({s_1})({s_1} - {t_1}) + \\ & \;\;\;\;\;\;\;\;\;\;\; I_{0 + }^\alpha {y_0}({t_1}) + {q_1} + {c_0} \end{split}$ (9)

现在考虑当 $k = 2,\;3,\; \cdots ,\;m$ 时的一般情况。

$t \in ({t_k},{s_k}]$ 时,由边值问题

$\left\{ \begin{split} &{}^c{\rm D}_{{t_k} + }^\beta u(t) = {h_k}(t) \\ & u(t_k^ + ) = u(t_k^ - ) + {q_k},\;u'(s_k^ - ) = 0 \end{split} \right.$

可得

$\left\{ \begin{split} & u(t) = I_{{t_k} + }^\beta {h_k}(t) - I_{{t_k} + }^{\beta - 1}{h_k}({s_k})(t - {t_k}) + {d_k} \\ & {d_k} = u(t_k^ - ) + {q_k} = I_{{s_{k - 1}} + }^\alpha {y_{k - 1}}({t_k}) + {c_{k - 1}} + {q_k} \end{split} \right. $ (10)

因此,

$ \begin{split} & u(t) = I_{{t_k} + }^\beta {h_k}(t) - I_{{t_k} + }^{\beta - 1}{h_k}({s_k})(t - {t_k}) + \\ &\;\;\;\;\;\;\;\;\;\;\;I_{{s_{k - 1}} + }^\alpha {y_{k - 1}}({t_k}) + {c_{k - 1}} + {q_k} \end{split} $ (11)
$ \begin{split} & u(s_k^ - ) = I_{{t_k} + }^\beta {h_k}({s_k}) - I_{{t_k} + }^{\beta - 1}{h_k}({s_k})({s_k} - {t_k}) + \\ &\;\;\;\;\;\;\;\;\;\;\;\;\;\;I_{{s_{k - 1}} + }^\alpha {y_{k - 1}}({t_k}) + {c_{k - 1}} + {q_k} \end{split}$ (12)

$t \in ({s_k},{t_{k + 1}}]$ 时,根据Cauchy问题

$\left\{ {\begin{split} & {{}^c{\rm D}_{{s_k} + }^\alpha u(t) = {y_k}(t)} \\ & {u(s_k^ + ) = u(s_k^ - ),\;u'(s_k^ + ) = 0} \end{split}} \right.$

可得其解为

$\left\{ \begin{split} & u(t) = I_{{s_k} + }^\alpha {y_k}(t) + {c_k}\\ & u(s_k^ + ) = {c_k} = u(s_k^ - ) = I_{{t_k} + }^\beta {h_k}({s_k}) - \\ &\;\; I_{{t_k} + }^{\beta - 1}{h_k}({s_k})({s_k} - {t_k}) + I_{{s_{k - 1}} + }^\alpha {y_{k - 1}}({t_k}) + {c_{k - 1}} + {q_k} \end{split} \right. $ (13)

于是

$ \begin{aligned} & u(t) = I_{{s_k} + }^\alpha {y_k}(t) + I_{{t_k} + }^\beta {h_k}({s_k}) - \\ &\;\;I_{{t_k} + }^{\beta - 1}{h_k}({s_k})({s_k} - {t_k}) + I_{{s_{k - 1}} + }^\alpha {y_{k - 1}}({t_k}) + {c_{k - 1}} + {q_k}\end{aligned} $ (14)

由式(13)可得

$ \begin{split} & {c_k} - {c_{k - 1}} = I_{{t_k} + }^\beta {h_k}({s_k}) - I_{{t_k} + }^{\beta - 1}{h_k}({s_k})({s_k} - {t_k}) + \\ & \;\;\;\;\;\;\;\;\;\;\;\; I_{{s_{k - 1}} + }^\alpha {y_{k - 1}}({t_k}) + {q_k},\;\;k = 1,2,3, \cdots ,m\; \end{split} $

由递推公式可得,当 $k = 1,2,3, \cdots ,m$

$ \begin{split} & {c_k} = \sum\limits_{i = 1}^k (I_{{t_i} + }^\beta {h_i}({s_i}) - I_{{t_i} + }^{\beta - 1}{h_i}({s_i})({s_i} - {t_i}) + \\ &\;\;\;\;\;\;\;\;\; I_{{s_{i - 1}} + }^\alpha {y_{i - 1}}({t_i}) + {q_i}) + {c_0} \end{split}$ (15)

由边界条件 $u(1) = 0,\;$ 可得

$ \begin{split} & 0 = u(1) = I_{{s_m} + }^\alpha {y_m}(1) + {c_m}= I_{{s_m} + }^\alpha {y_m}(1) +\\ & \;\;\;\;\; \sum\limits_{i = 1}^m (I_{{t_i} + }^\beta {h_i}({s_i}) - I_{{t_i} + }^{\beta - 1}{h_i}({s_i})({s_i} - {t_i}) + \\ & \;\;\;\;\; I_{{s_{i - 1}} + }^\alpha {y_{i - 1}}({t_i}) + {q_i}) + {c_0} \end{split} $

所以,

$ \begin{split} & {c_m} = - I_{{s_m} + }^\alpha {y_m}(1),{c_0} = - \sum\limits_{i = 1}^{m + 1} {I_{{s_{i - 1}} + }^\alpha {y_{i - 1}}({t_i})} - \\ &\;\;\;\;\;\;\;\; \sum\limits_{i = 1}^m {(I_{{t_i} + }^\beta {h_i}({s_i}) - I_{{t_i} + }^{\beta - 1}{h_i}({s_i})({s_i} - {t_i}) + {q_i}} ) \end{split}$ (16)

代入式(15),得

$ \begin{split} & {c_k} = - \sum\limits_{i = k + 1}^m {(I_{{t_i} + }^\beta {h_i}({s_i}) - I_{{t_i} + }^{\beta - 1}{h_i}({s_i})({s_i} - {t_i}) + {q_i}} )\; - \\ & \;\;\;\;\;\;\;\; \sum\limits_{i = k + 1}^{m + 1} {I_{{s_{i - 1}} + }^\alpha {y_{i - 1}}({t_i})} ,\;k = 1,2, \cdots ,m - 1 \end{split} $ (17)

将式(16)和式(17)分别代入式(4),(6),(9),(11),(14),可得式(3)。证毕。

$ \chi (x,y,z) = \left\{ \begin{split} & 1,\;\;\;\;x \leqslant z \leqslant y\\ & 0,\;{\rm{ }} \;\;\; {\text{其他}} \end{split} \right. $
$ \begin{aligned} & \!\!\!{{W_{\rm{1}}}(t,s) = \frac{1}{{\Gamma (\alpha )}}} \cdot \\ &\;\;\;\; {\left\{ \begin{split} &\chi (0,t,s){(t - s)^{\alpha - 1}} - \sum\limits_{i = 1}^{m + 1} {\chi ({s_{i - 1}},{t_i},s){{({t_i} - s)}^{\alpha - 1}}} ,\\ & \;\;\;\;\;0 \leqslant t \leqslant {t_1} 0 \leqslant s \leqslant 1\\ & \chi ({s_k},t,s){(t - s)^{\alpha - 1}} \!- \!\!\sum\limits_{i = k + 1}^{m + 1} {\chi ({s_{i - 1}},{t_i},s){{({t_i}\!\! -\!\! s)}^{\alpha - 1}},} \\ & \;\;\;\;\;{s_k} < t \leqslant {t_{k + 1}},\;0 \leqslant s \leqslant 1,\;k = 1,2, \cdots ,m - 1\\ &- \sum\limits_{i = k + 1}^{m + 1} \chi ({s_{i - 1}},{t_i},s){{({t_i} - s)}^{\alpha - 1}},\;{t_k} < t \leqslant {s_k}, 0 \leqslant s \leqslant 1,\,\\ & \;\;\;\;\;k = 1,2, \cdots ,m \\ & \chi ({s_m},t,s){(t - s)^{\alpha - 1}} - \chi ({s_m},1,s){(1 - s)^{\alpha - 1}},\\ & \;\;\;\;\;{s_m} < t \leqslant 1,\;0 \leqslant s \leqslant 1 \end{split} \right.} \end{aligned} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! $ (18)
$ \begin{aligned} &\!\!\! {{W_2}(t,s) = \frac{1}{{\Gamma (\beta )}}}\cdot \\ & {\left\{ \begin{split} & - \sum\limits_{i = 1}^m {\chi ({t_i},{s_i},s){{({s_i} - s)}^{\beta - 2}}((2 - \beta ){s_i} + (\beta - 1){t_i} - s} ),\;\; \\ & \;\;\;\;\;\;\;\; 0 \leqslant t \leqslant {t_1},\;0 \leqslant s \leqslant 1,\\ & - \sum\limits_{i = k + 1}^m {\chi ({t_i},{s_i},s){{({s_i} - s)}^{\beta - 2}}((2 - \beta ){s_i} + (\beta - 1){t_i} - s} ),\\ &\;\;\;\;\;\;\;\;{s_k} < t \leqslant {t_{k + 1}},\;0 \leqslant s \leqslant 1,\;k = 1,2, \cdots ,m - 1\\ &\chi ({t_k},t,s){(t - s)^{\beta - 1}} - \sum\limits_{i = k}^m \chi ({t_i},{s_i},s){{({s_i} - s)}^{\beta - 2}}\text{·} \\ & \;\;\;\;\;\;\;\; ((2 - \beta ){s_i} + (\beta - 1){t_i} - s )- (\beta - 1)\chi ({t_k},{s_k},s)\text{·} \\ &\;\;\;\;\;\;\;\; \;\; (t - {t_k}){({s_k} - s)^{\beta - 2}},\;\;{t_k} < t \leqslant {s_k}, \\ &\;\;\;\;\;\;\;\; 0 \leqslant s \leqslant 1,\;k = 1,2,3, \cdots ,m\\ & 0,\;{s_m} < t \leqslant 1,\;\,0 \leqslant s \leqslant 1 \end{split} \right.} \end{aligned} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! $ (19)

定义算子 $T,A,B,G:PC(J,\mathbb{R}) \to PC(J,\mathbb{R})$

$Au(t) = \int_0^1 {{W_1}(t,s)f(s,u(s)){\rm{d}}s} $
$Bu(t) = \int_0^1 {{W_2}(t,s)} g(s,u(s)){\rm{d}}s$
$ Gu(t) = \left\{ \begin{split} & - \sum\limits_{i = 1}^m {{Q_i}({t_i},u({t_i})),\;\;\;\;\;t \in [0,{t_1}]}\\ & - \sum\limits_{i = k + 1}^m {{Q_i}({t_i},u({t_i})),\;\;} t \in ({t_k},{t_{k + 1}}],\\ & \;\;\;\;k = 1,2, \cdots ,m - 1\; \\ & 0,\;t \in ({t_m},1] \end{split} \right.$
$Tu(t) = Au(t) + Bu(t) + Gu(t)$

$T:PC(J,\mathbb{R}) \to PC(J,\mathbb{R})$

引理2  由式(18)和(19)定义的函数 ${W_1}(t,s)$ ${W_2}(t,s)$ ,当 $t,s \in [0,1]$ 时满足

$\left| {{W_1}(t,s)} \right| \leqslant \frac{{m + 2}}{{\Gamma (\alpha )}},\;\;\left| {{W_2}(t,s)} \right| \leqslant \frac{{2(m + 1)}}{{\Gamma (\beta )}}$

证明  当 $0 \leqslant t \leqslant {t_1},0 \leqslant s \leqslant 1$ 时,

$ \begin{split} & \left| {{W_1}(t,s)} \right| \leqslant \frac{1}{{\Gamma (\alpha )}}\Bigg(\left| {\chi (0,t,s){{(t - s)}^{\alpha - 1}}} \right| + \\ & \;\;\;\;\;\;\; \left| {\sum\limits_{i = 1}^{m + 1} {\chi ({s_{i - 1}},{t_i},s){{({t_i} - s)}^{\alpha - 1}}} } \right|\Bigg) \leqslant \frac{{m + 2}}{{\Gamma (\alpha )}} \end{split}$

${s_k} < t \leqslant {t_{k + 1}},\;0 \leqslant s \leqslant 1,\;k = 1,2, \cdots ,m - 1$ 时,

$ \begin{split} & \left| {{W_{\rm{1}}}(t,s)} \right| \leqslant \frac{1}{{\Gamma (\alpha )}}\Bigg(\left| {\chi ({s_k},t,s){{(t - s)}^{\alpha - 1}}} \right| + \\ & \;\;\;\;\;\;\; \left| { - \sum\limits_{i = k + 1}^{m + 1} {\chi ({s_{i - 1}},{t_i},s){{({t_i} - s)}^{\alpha - 1}}\;} } \right|\Bigg) < \frac{{m + 2}}{{\Gamma (\alpha )}} \end{split}$

${t_k} < t \leqslant {s_k},{\rm{ }}0 \leqslant s \leqslant 1,\;k = 1,2, \cdots ,m$ 时,

$ \begin{split} & \left| {{W_1}(t,s)} \right| \leqslant \frac{1}{{\Gamma (\alpha )}}\Bigg(\left| {\sum\limits_{i = k + 1}^m {\chi ({s_{i - 1}},{t_i},s){{({t_i} - s)}^{\alpha - 1}}} } \right| + \\ &\;\;\;\;\;\;\; \left| {\chi ({s_m},1,s){{(1 - s)}^{\alpha - 1}}} \right|\Bigg) \leqslant \frac{{m + 1}}{{\Gamma (\alpha )}} < \frac{{m + 2}}{{\Gamma (\alpha )}} \end{split}$

${s_m} < t \leqslant 1,\;0 \leqslant s \leqslant 1$ 时,

$ \begin{split} & \left| {{W_1}(t,s)} \right| = \frac{1}{{\Gamma (\alpha )}}| {\chi ({s_m},t,s){(t - s)}^{\alpha - 1} -} \\ &\;\;\;\;\;\;\; {\chi ({s_m},1,s){{(1 - s)}^{\alpha - 1}}} | \leqslant \frac{2}{\Gamma (\alpha )} \end{split}$

综上,当 $t,s \in [0,1]$ 时,有 $\left| {{W_1}(t,s)} \right| \leqslant \dfrac{{m + 2}}{{\Gamma (\alpha )}}$

同理可证,当 $t,s \in [0,1]$ 时,有 $\left| {{W_2}(t,s)} \right| \leqslant \dfrac{{2(m + 1)}}{{\Gamma (\beta )}}$ 。证毕。

根据引理1易得下面引理3成立。

引理3   $u \in PC(J,\mathbb{R})$ 是边值问题(1)的解当且仅当 $u \in PC(J,\mathbb{R})$ 是算子T的不动点。

引理4  算子 $T:PC(J,\mathbb{R}) \to PC(J,\mathbb{R})$ 全连续。

证明  设 ${u_n},u \in PC(J,\mathbb{R}),\;n = 1,2, \cdots ,$ $||{u_n} - u|{|_{PC}}$ $\to 0\;(n \to \infty )$ 。因为 $f,g,{Q_k}$ 是连续函数,所以对任意 $s \in [0,1]$

$ \begin{array}{l} f(s,{u_n}(s)) - f(s,u(s)) \to 0,g(s,{u_n}(s)) - \\ \;\;\;\;\;g(s,u(s)) \to 0,n \to \infty\\ {Q_k}(s,{u_n}(s)) - {Q_k}(s,u(s)) \to 0,\;k = 1,2, \cdots ,m,n \to \infty \end{array} $

根据Lebesgue控制收敛定理易得 ${\left\| {T{u_n}\!-\! Tu} \right\|_{PC}} \!\to$ $0,{\rm{ }}(n \to \infty ),$ T连续。

$r > 0,{\rm{ }}{B_r} = \{ u \in PC(J,\mathbb{R}):\left\| u \right\| \leqslant r\} $ ,记 ${M_f} = $ $ \mathop {\max }\limits_{(t,u) \in [0,1] \times [ - r,r]} \left| {f(t,u)} \right|$ ${M_g} = \mathop {\max }\limits_{(t,u) \in [0,1] \times [ - r,r]} \left| {g(t,u)} \right|$ ${M_Q} = $ $ \mathop {\max }\limits_{1 \leqslant k \leqslant m,u \in [ - r,r]}$ $ \left| {{Q_k}({t_k},u)} \right| $ 。对任意 $u \in {B_r}$ ,根据引理2可得

$ \begin{split} & \left| {Tu(t)} \right| = \left| {Au(t) + Bu(t) + Gu(t)} \right| \leqslant \\ & \;\;\;\; \int_0^1 {\left| {{W_1}(t,s)f(s,u(s))} \right|\rm{d}} s +\\ & \;\;\;\; \;\int_0^1 {{{\left| W \right.}_2}(t,s)} g(s,u(s)\left. ) \right| {\rm{d}} s + \left| {Gu(t)} \right| \leqslant \\ & \;\;\;\; \frac{{(m + 2){M_f}}}{{\Gamma (\alpha )}} + \frac{{2(m + 1){M_g}}}{{\Gamma (\beta )}} + (m + 1){M_Q} \end{split} $

因此, $T({B_r})$ 一致有界。

由于 ${(t - s)^{\alpha - 1}},\;\;{(t - s)^{\beta - 1}}$ $(t,s) \in [0,1] \times [0,1]$ 上是连续的,从而在 $[0,1] \times [0,1]$ 上是一致连续的,故对任意 $\varepsilon > 0$ ,存在常数 $0 < \delta < \min \Bigg\{ \dfrac{{\Gamma (\alpha )\varepsilon }}{{2{M_f} + 1}},$ $\;\dfrac{{\Gamma (\beta )\varepsilon }}{{3{M_g} + 1}}\Bigg\} $ ,使得当 ${t_1}^\prime ,{t_2}^\prime ,s \in J$ ${t_1}^\prime < {t_2}^\prime ,\left| {{t_1}^\prime - {t_2}^\prime } \right| < \delta $ 时有

$\begin{split} & |{ ({t'_1} - s)^{\beta - 1}} - {({t'_2} - s)^{\beta - 1}}| < \frac{{\Gamma (\beta )\varepsilon }}{{3{M_g}{\rm{ + }}1}}\;\; \\ & |{({t'_1} - s)^{\alpha - 1}} - {({t'_2} - s)^{\alpha - 1}}| < \frac{{\Gamma (\alpha )\varepsilon }}{{2{M_f}{\rm{ + }}1}} \end{split}$

因此,当 ${t_1}^\prime < {t_2}^\prime \in [0,{t_1}],\;|{t_1}^\prime - {t_2}^\prime | < \delta $ 时,

$ \begin{split} & \left| {Tu({{t'_1}}) - Tu({{t'_2}})} \right|\leqslant | Au({{t'_1}}) - Au({{t'_2}})| + \\ & \quad|Bu({{t'_1}}) - Bu({{t'_2}})| + |Gu({{t'_1}}) - Gu({{t'_2}}) | \leqslant \\ &\quad \frac{{{M_f}}}{{\Gamma (\alpha )}} \Bigg( \int_0^{{{t'_1}}} {({{({{t'_2}} - s)}^{\alpha - 1}} - {{({{t'_1}} - s)}^{\alpha - 1}})\rm{d}} s + \\ &\quad \int_{{{t'_1}}}^{{{t'_2}}} {{{({{t'_2}} - s)}^{\alpha - 1}}\rm{d}} s \Bigg) < \frac{{{M_f}}}{{\Gamma (\alpha )}} \frac{{\Gamma (\alpha )\varepsilon }}{{2{M_f} + 1}} + \frac{{{M_f \delta }}}{{\Gamma (\alpha )}} < \varepsilon \quad\quad \end{split} $

同理可证, 当 ${t_1}^\prime < {t_2}^\prime \in ({s_k},{t_{k + 1}}],\;|{t_1}^\prime - {t_2}^\prime |\; < \delta ,$ $\;\;k = 1,2, \cdots ,m - 1 $ 时, $\left| {Tu({{t'_1}}) - Tu({{t'_2}})} \right| < \varepsilon$

${t_1}^\prime < {t_2}^\prime \in ({t_k},{s_k}],\;|{t_1}^\prime - {t_2}^\prime |\; < \delta ,\;\;k = 1,2, \cdots ,m$ 时,

$ \begin{split} & | {Tu({{t'_1}}) - Tu({{t'_2}})} | \leqslant | Au({{t'_1}}) - Au({{t'_2}})| +\\ & \;\;\;\; |Bu({{t'_1}}) - Bu({{t'_2}})| + |Gu({{t'_1}}) - Gu({{t'_2}}) |\leqslant \\ & \;\;\;\;\frac{{{M_g}}}{{\Gamma (\beta )}} \Bigg(\int_{{t_k}}^{{{t'}_1}} | {{{({{t'_1}} - s)}^{\beta - 1}} - {{({{t'_2}} - s)}^{\beta - 1}}} | {\rm{d} }s + \\ &\;\;\;\; \int_{{{t'_1}}}^{{{t'_2}}} {{{({{t'_2}} - s)}^{\beta - 1}}\rm{d}} s + \int_0^1 | (\beta - 1)\chi ({t_k},{s_k},s)({{t'_2}} - {{t'_1}}) \cdot \\ &\;\;\;\;{{({s_k} - s)}^{\beta - 2}} | {\rm{d}} s \Bigg) \leqslant \dfrac{{{M_g}}}{{\Gamma (\beta )}} \dfrac{{\Gamma (\beta )}}{{3{M_g} + 1}} + \dfrac{{{M_g}}}{{\Gamma (\beta )}} 2 \delta < \varepsilon \end{split} $

${t_1}^\prime < {t_2}^\prime \in ({s_m},1],\;|{t_1}^\prime - {t_2}^\prime |\; < \delta $ 时,

$ \begin{split} & \left| {Tu({{t'_1}}) - Tu({{t'_2}})} \right|\leqslant \\ & \;\;\;\;\;\dfrac{{{M_f}}}{{\Gamma (\alpha )}}\Bigg(\int_{{s_m}}^{{{t'_1}}} {\left| {{{({{t'_1}} - s)}^{\alpha - 1}} - {{({{t'_2}} - s)}^{\alpha - 1}}} \right|\rm{d}} s + \\ & \;\;\;\;\;\int_{{{t'_1}}}^{{{t'_2}}} {{{({{t'_2}} - s)}^{\alpha - 1}}\rm{d}} s \Bigg) \leqslant \dfrac{{{M_f}}}{{\Gamma (\alpha )}} \dfrac{{\Gamma (\alpha )\varepsilon }}{{2{M_f} + 1}} + \dfrac{{{M_f}}}{{\Gamma (\alpha )}} \delta < \varepsilon \quad\quad \end{split} $

于是得到 $T({B_r})$ $t \in $ $[0,{t_1}]$ , $({s_k},{t_{k + 1}}]$ , $({t_k},{s_k}], \;\;k = 1,2, \cdots ,m$ 时等度连续, 由Arzela-Ascoli定理可得T是紧算子。

综上讨论,可得T全连续。证毕。

3 边值问题解的存在性与唯一性

$ {N_0} = \frac{{m + 2}}{{\Gamma (\alpha )}},{N_1} = \frac{{2(m + 2)}}{{\Gamma (\beta )}},{N_2} = m + 1 $

假设:

H1 存在非负实数 ${a_0},{a_1},{b_0},{b_1},{c_0},{c_1}$ ,常数 $\sigma ,\theta ,\gamma > 0,$ 使得对任意的 $t \in J$ 及任意的 $u \in \mathbb{R}$

$\begin{split} & |f(t,x)| \leqslant {a_0} + {a_1}|u{|^\sigma },|g(t,u)| \leqslant {b_0} + {b_1}{\left| u \right|^\theta }, \\ & |{Q_k}(t,u)| \leqslant {l_0} + {l_1}{\left| u \right|^\gamma },k = 1,2, \cdots ,m \end{split}$

H2 存在常数 ${L_1},{L_2},{L_3} \geqslant 0$ ,使得对任意的 $t \in J$ 及任意的 ${u_1},{u_2} \in \mathbb{R}$

$\begin{split} & \left| {f(t,{u_1}) - f(t,{u_2})} \right| \leqslant {L_1}\left| {{u_1} - {u_2}} \right| \\ & \left| {g(t,{u_1}) - g(t,{u_2})} \right| \leqslant {L_2}\left| {{u_1} - {u_2}} \right| \end{split}$

定理1  假设H1成立,且满足 $0 < \sigma ,\theta ,\gamma < 1,$ 则边值问题(1)至少存在一个解。

证明  取

$ \begin{split} & {r_1} \geqslant \max \{ 1,4({N_0}{a_0} + {N_1}{b_0} + {N_2}{l_0}), \\ & \;\;\;\;\;\;\; {(4{N_0}{a_1})^{\frac{1}{{1 - \sigma }}}},{(4{N_1}{b_1})^{\frac{1}{{1 - \theta }}}},{(4{N_2}{l_1})^{\frac{1}{{1 - \gamma }}}}\} \\ & D = \{ u \in PC(J,\mathbb{R}):{\left\| u \right\|_{PC}} \leqslant {r_1}\} \end{split} $

D $PC(J,\mathbb{R})$ 中非空有界闭凸集。

对任意的 $u \in D$ ,

$ \begin{split} & \left| {Tu(t)} \right| = \left| {Au(t) + Bu(t) + Gu(t)} \right| \leqslant \\ & \;\;\;\;\;\int_0^1 {\left| {{W_1}(t,s)f(s,u(s))} \right|\rm{d} }s + \\ & \;\;\;\;\;\int_0^1 {{{\left| W \right.}_2}(t,s)} g(s,u(s)\left. ) \right|{\rm{d}} s + \left| {Gu(t)} \right| < \\ & \;\;\;\;\; \frac{{(m + 2){a_0}}}{{\Gamma (\alpha )}} + \frac{{2(m + 1){b_0}}}{{\Gamma (\beta )}} + (m + 1){l_0} + \\ &\;\;\;\;\;\frac{{(m + 2){a_1}{r_1}^\sigma }}{{\Gamma (\alpha )}} + \;\frac{{2(m + 1){b_1}{r_1}^\theta }}{{\Gamma (\beta )}} + (m + 1){l_1}{r_1}^\gamma < \\ &\;\;\;\;\; {N_0}{a_0} + {N_1}{b_0} + {N_2}{l_0} + \;{N_0}{a_1}{r_1}^\sigma + \; \\ & \;\;\;\;\; {N_1}{b_1}{r_1}^\theta+ {N_2}{l_1}{r_1}^\gamma \leqslant {r_1} \end{split} $

所以, ${\left\| {Tu} \right\|_{PC}} \leqslant {r_1},$ $T(D) \subset D$ 。由引理4知T全连续,故由Schauder不动点定理知TD中至少存在一个不动点,即边值问题(1)在 $PC(J,\mathbb{R})$ 中至少存在一个解。证毕。

定理2  假设H1成立,且满足 $\sigma = \theta = \gamma = 1$ ,如果 $0 < {N_0}{a_1} + {N_1}{b_1} + {N_2}{l_1} < 1$ ,则边值问题(1)至少存在一个解。

证明  取 ${r_2} \geqslant \dfrac{{{N_0}{a_0} + {N_1}{b_0} + {N_2}{l_0}}}{{1 - ({N_0}{a_1} + {N_1}{b_1} + {N_2}{l_1})}}$ ,令E = $ \{ u \in PC(J,\mathbb{R}):{\left\| u \right\|_{PC}} \leqslant {r_2}\} $ ,则 $PC(J,\mathbb{R})$ 中非空有界闭凸集。

对任意的 $u \in E$

$ \begin{split} &  \left| {Tu(t)} \right| = \left| {Au(t) + Bu(t) + Gu(t)} \right|   \leqslant\\ & \;\;\;\;\;\; \int_0^1 {\left| {{W_1}(t,s)f(s,u(s))} \right|\rm{d}} s + \\ & \;\;\;\;\;\; \int_0^1 {\left| {{W_2}} \right.(t,s)} g(s,u(s)\left. ) \right|{\rm{d}} s + \left| {Gu(t)} \right| < \\ & \;\;\;\;\;\; \frac{{(m + 2){a_0}}}{{\Gamma (\alpha )}} + \frac{{2(m + 1){b_0}}}{{\Gamma (\beta )}} + (m + 1){l_0} + \\ & \;\;\;\;\;\; \Bigg(\frac{{(m + 2){a_1}}}{{\Gamma (\alpha )}} + \;\frac{{2(m + 1){b_1}}}{{\Gamma (\beta )}} + (m + 1){l_1}\Bigg){r_2} <\\ & \;\;\;\;\;\; {N_0}{a_0} + {N_1}{b_0} + {N_2}{l_0} + \;{N_0}{a_1}{r_2} + \;{N_1}{b_1}{r_2} +\\ & \;\;\;\;\;\; {N_2}{l_1}{r_2} \leqslant {r_2} \end{split} $ (20)

所以, ${\left\| {Tu} \right\|_{PC}} \leqslant {r_2}$ ,故 $T(D) \subset D$ 。由引理4知T全连续,故由Schauder不动点定理知TD中至少存在一个不动点,即边值问题(1)在 $PC(J,\mathbb{R})$ 中至少存在一个解。证毕。

定理3  设H2成立,且满足 $0 < {N_0}{L_1} +{N_1}{L_2} + $ $ {N_2}{L_3} < 1$ ,则边值问题(1)在 $PC(J,\mathbb{R})$ 上存在唯一解。

证明  对任意 ${u_1},\;{u_2} \in PC(J,\mathbb{R})$ 可得

$ \begin{split} & \left| {T{u_1}(t) - T{u_2}(t)} \right| = \left| {{A{u_1}(t) - A{u_2}(t)+ }} \right. \\ & \;\;\;\;\; \left. { B{u_1}(t) - B{u_2}(t) + G{u_1}(t) - G{u_2}(t)} \right|\leqslant \\ & \;\;\;\;\; \int_0^1 {\left| {{W_1}(t,s\left. ) \right|\left| f \right.(s,{u_1}(s)) - f(s,{u_{\rm{2}}}(s))} \right|\rm{d} }s +\\ & \;\;\;\;\; \int_0^1 {\left| {{W_2}} \right.(t,s\left. ) \right|} \left| g \right.(s,{u_{\rm{1}}}(s)\left. {) - g(s,{u_2}(s)} \right|{\rm{d}} s + \\ &\;\;\;\;\; \left| {G{u_1}(t) - G{u_2}(t)} \right| <\\ &\;\;\;\;\; \left(\frac{{(m + 2){L_1}}}{{\Gamma (\alpha )}} + \frac{{2(m + 1){L_2}}}{{\Gamma (\beta )}} + (m + 2){L_3}\right){\left\| {{u_1} - {u_2}} \right\|_{PC}} = \\ &\;\;\;\;\; ({N_0}{L_1} + {N_1}{L_2} + {N_2}{L_3}){\left\| {{u_1} - {u_2}} \right\|_{PC}} \end{split} $

因为 $0 < {N_0}{L_1} + {N_1}{L_2} + {N_2}{L_3} < 1$ ,所以T为压缩映射,因此,可知边值问题(1)在 $PC(J,\mathbb{R})$ 上存在唯一解。证毕。

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