一类分数阶发展方程非局部问题的精确可控性

Exact Controllability for a Class of Fractional Evolution Equations with Nonlocal Conditions

摘要: 本文讨论了一类分数阶发展方程非局部问题的精确可控性。文中通过引入一个新的非紧性测度，在C₀-半群等度连续的情形下，运用Mönch不动点定理，证明了该问题的精确可控性，并通过一个具体的例子来验证本文的抽象结论。

Abstract: This paper discusses the exact controllability of nonlocal conditions for a class of fractional evolution equations . In this paper, by introducing a new measure of non-compactness, the exact controllability of the problem is proved by using Mönch fixed point theorem under the condition that C₀-semigroup is equicontinuous, and an example is given to verify the abstract conclusion of this paper.

1. 引言

分数阶微分方程由于在数学、物理、工程等方面具有广泛应用而被许多学者关注，但是在实际应用中非局部初始条件相比于经典的初始条件能够更好地描述自然现象(参见文献 [1] - [8] )，因此研究分数阶发展方程非局部问题的可解性和可控性具有重要的理论意义和实际应用价值。Byszewski在文献 [2] 中首次研究了一类具有非局部初始条件的半线性一阶发展方程解的存在唯一性。从此以后，人们开始关注具有非局部初始条件的微分方程。

2015年，梁进等 [9] 在非紧半群的情况下，利用Mönch不动点定理证明了分数阶积微分发展方程非局部问题

$\{\begin{array}{l}{D}^{q}x\left(t\right)=Ax\left(t\right)+f\left(t,x\left(t\right),Gx\left(t\right)\right)+Bu\left(t\right),t\in \left[0,b\right],\\ x\left(0\right)={\displaystyle {\sum }_{k=1}^m{c}_{k}x\left(t_k\right)}\end{array}$ (1.1)

的精确可控性。Deng在文献 [4] 中指出在(1.1)中使用的非局部函数可以很好地描述少量气体在透明管中的扩散现象。

2020年，陈鹏玉等 [10] 通过引入一个新的格林函数，应用Schauder不动点定理及预解算子理论研究了Banach空间中分数阶发展方程非局部问题

$\{\begin{array}{l}{D}_t^{q}x\left(t\right)=Ax\left(t\right)+Bu\left(t\right)+f\left(t,x\left(t\right)\right),t\in \left[0,b\right],\\ x\left(0\right)={\displaystyle {\sum }_{k=1}^m{c}_{k}x(\; t\; k\; )}\end{array}$

mild解的存在性和近似可控性。

受上述文献的启发，本文研究Banach空间X中半线性分数阶发展方程非局部问题

$\{\begin{array}{l}{D}^{q}x\left(t\right)=Ax\left(t\right)+f\left(t,x\left(t\right)\right)+Bu\left(t\right),t\in \left[0,b\right],\\ x\left(0\right)={\displaystyle {\sum }_{k=1}^m{c}_{k}x\left(t_k\right)}\end{array}$ (1.2)

的精确可控性，其中 $D^q$ 为 $q\in \left(0,1\right)$ 阶Caputo分数阶导数， $A:D\left(A\right)\subset X\to X$ 是 $C_0$ -半群 ${\left\{T\left(t\right)\right\}}_{t\ge 0}$ 的无穷小生成元，控制函数 $u\in L^2\left(J,U\right)$， $J=\left[0,b\right]$， $b>0$ 为常数，U为Banach空间， $B:U\to X$ 为有界线性算子，函数f为给定函数。 $0<t_1<t_2<\cdots <t_m<b$， $b\in N$， $c_k$ 是实数且 $c_k\ne 0$， $k=1,2,\cdots ,m$。

本文对非局部函数的紧性和连续性都不做要求。在 $C_0$ -半群等度连续的情况下，通过定义一种新的非紧性测度(见文献 [11] )并应用Mönch不动点定理证明了系统(1.2)的精确可控性。在第2节中介绍了讨论问题(1.2)的一些准备工作，主要的结果和证明将在第3节中给出，第4节通过一个具体的例子来验证本文获得的抽象结论。

2. 预备知识

设 $\left(X,\Vert \cdot \Vert \right)$ 是可分Banach空间，U是另一个Banach空间。设 $C\left(J,X\right)$ 表示J上所有X−值连续函数全体按范数 ${\Vert u\Vert }_C={\max }_{t\in J}\Vert u\left(t\right)\Vert$ 构成的Banach空间。 $L^p\left(J,X\right)\left(p\ge 1\right)$ 是J上的X−值p方Bochner可积函数全体按范数 ${\Vert x\Vert }_{L^p}$ 构成的Banach空间。关于Caputo分数阶导数，我们介绍如下定义。

定义1 [12] 函数 $f\in C^n\left[0,\infty \right)$ 下限为0的Caputo分数阶导数为：

$D^{q}f\left(t\right)=\frac{1}{\Gamma \left(n-q\right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{n-q-1}{f}^{\left(n\right)}\left(s\right)\text{d}s},t>0,n-1<q<n,n\in N,$

其中 $\Gamma (\cdot )$ 为Gamma函数。

对 $\forall h\in C\left(J,X\right)$，考虑线性发展方程的非局部问题

$\{\begin{array}{l}{D}^{q}x\left(t\right)=Ax\left(t\right)+h\left(t\right),t\in \left[0,b\right],\\ x\left(0\right)={\displaystyle {\sum }_{k=1}^m{c}_{k}x\left(t_k\right)}\end{array}$ (2.1)

引理1 [9] 设条件(H0)成立，其中

(H0) ${\displaystyle {\sum }_{k=1}^m\left|c_k\right|}\le \frac{1}{M},M\ge 1,$

则线性系统(2.1)有唯一mild解 $x\in C\left(J,X\right)$ 满足

$x\left(t\right)={\displaystyle {\sum }_{k=1}^m{c}_k}\widehat{T_q}\left(t\right)P{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)h\left(s\right)\text{d}s}+{\displaystyle {\int }_0^t{\left(t-s\right)}^{q-1}{T}_q\left(t-s\right)h\left(s\right)\text{d}s},t\in J,$ (2.2)

其中 $P={\left(I-{\displaystyle {\sum }_{k=1}^m{c}_k}\widehat{T_q}\left(t_k\right)\right)}^{-1}$ 满足 $\Vert P\Vert \le \frac{1}{1-M{\displaystyle {\sum }_{k=1}^m\left|c_k\right|}}$，

$\widehat{T_q}\left(t\right)={\displaystyle {\int }_0^{\infty }{\xi }_q\left(\theta \right)T\left(t^q\theta \right)\text{d}\theta },$

$T_q\left(t\right)=q{\displaystyle {\int }_0^{\infty }\theta {\xi }_q\left(\theta \right)T\left(t^q\theta \right)\text{d}\theta },$

其中

${\xi }_q\left(\theta \right)=\frac{1}{q}{\theta }^{-1-\frac{1}{q}}{W}_q\left({\theta }^{-\frac{1}{q}}\right)\ge 0,\theta \in \left(0,\infty \right),$

$W_q\left(\theta \right)=\frac{1}{\pi }\underset{k=1}{\overset{m}{{\displaystyle \sum }}}{\left(-1\right)}^{n-1}{\theta }^{-qn-1}\frac{\Gamma \left(1+nq\right)}{n!}\sin \left(n\pi q\right),\theta \in \left(0,\infty \right).$

这里， ${\xi }_q\left(\theta \right)$ 表示定义在 $\left(0,\infty \right)$ 上的概率密度函数，满足

${\xi }_q\left(\theta \right)\ge 0,\forall \theta \in \left(0,\infty \right),{\displaystyle {\int }_0^{\infty }{\xi }_q\left(\theta \right)\text{d}\theta }=1.$

引理2 [9] [12] 线性算子族 ${\left\{\widehat{T_q}\left(t\right)\right\}}_{t\ge 0}$ 和 ${\left\{T_q\left(t\right)\right\}}_{t\ge 0}$ 具有下列性质：

(1) 对任意给定的 $t\ge 0$ 和 $\forall x\in X$，有

$\Vert \widehat{T_q}\left(t\right)x\Vert \le M\Vert x\Vert ,\Vert T_q\left(t\right)x\Vert \le \frac{Mq}{\Gamma \left(1+q\right)}\Vert x\Vert ;$

(2) 对 $\forall t\ge 0$， $\widehat{T_q}\left(t\right)$ 和 $T_q\left(t\right)$ 都是强连续的；

(3) 如果 ${\left\{T\left(t\right)\right\}}_{t\ge 0}$ 是等度连续半群，则对 $\forall x>0$， $\widehat{T_q}\left(t\right)$ 和 $T_q\left(t\right)$ 都是等度连续的。

根据引理1，本文采用如下mild解的定义。

定义2对 $\forall u\in L^2\left(J,U\right)$，如果函数 $x\in C\left(J,X\right)$ 满足积分方程

$\begin{array}{c}x\left(t\right)={\displaystyle {\sum }_{k=1}^m{c}_k}\widehat{T_q}\left(t\right)P{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\\ +{\displaystyle {\int }_0^t{\left(t-s\right)}^{q-1}{T}_q\left(t-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s},\end{array}$ (2.3)

则称函数x为半线性分数阶发展系统(1.2)的mild解。

定义3 [13] 若对任一给定的 $x_b\in X$，存在控制函数 $u\in L^2\left(J,U\right)$，使得系统(1.2)的mild解x满足

$x\left(b\right)=x_b,$

则称系统(1.2)在J上是精确可控的。

本文主要采用非紧性测度方法和不动点定理研究半线性分数阶发展系统(1.2) mild解的存在性和精确可控性，关于非紧性测度，我们有如下结论。

引理3 [14] [15] 设 $\Omega ,{\Omega }_1,{\Omega }_2$ 是X中的非空有界集， $a\in R$，则Hausdroff非紧性测度 $\beta (·)$ 具有下列性质：

(1) ${\Omega }_1\subset {\Omega }_2\Rightarrow \beta \left({\Omega }_1\right)\le \beta \left({\Omega }_2\right)$ ；

(2) $\beta \left({\Omega }_1+{\Omega }_2\right)\le \beta \left({\Omega }_1\right)+\beta \left({\Omega }_2\right)$，其中 ${\Omega }_1+{\Omega }_2=\left\{x+y:x\in {\Omega }_1,y\in {\Omega }_2\right\}$ ；

(3) $\beta \left({\Omega }_1\cup {\Omega }_2\right)\le \max \left\{\beta \left({\Omega }_1\right),\beta \left({\Omega }_2\right)\right\}$ ；

(4) $\beta \left(\lambda \Omega \right)\le \left|\lambda \right|\beta \left(\Omega \right)$，对 $\forall \lambda \in R$ ；

(5) $\beta \left(\left\{0\right\}\cup \Omega \right)=\beta \left(\Omega \right)$ ；

(6) $\beta \left(\Omega \right)=0\iff \Omega$ 在X中相对紧。

引理4 [14] 设X为Banach空间， $D\subset C\left(J,X\right)$ 有界且等度连续，则 $\beta \left(D\left(t\right)\right)$ 在J上连续，且 $\beta \left(D\right)={\max }_{t\in J}\beta \left(D\left(t\right)\right)$。

引理5 [16] 设X是可分的Banach空间， $D_0=\left\{x_m\right\}\subset C\left(J,X\right)$ 是可数集。若 $\exists \varphi \in L^1\left(J\right)$，使得

$\Vert x_m\left(t\right)\Vert \le \varphi \left(t\right),t\in J,m=1,2,\cdots ,$

则 $\beta \left(D\left(t\right)\right)$ 在J上Lebesgue可积，且

$\beta \left(\left\{{\displaystyle {\int }_J{x}_m\left(t\right)\text{d}t}:m=1,2,\cdots \right\}\right)\le 2{\displaystyle {\int }_J\beta \left(D_0\left(t\right)\right)\text{d}t}.$

特别地，当 $D_0$ 有界时上式成立。

定义4 [11] 如果序列 ${\left\{f_n\right\}}_{n=1}^{\infty }$ 在X中对 $a.e.t\in J$ 是相对紧的，且对 $a.e.t\in J$，存在函数 $\mu \in L^1\left(J,X\right)$ 满足

$\underset{n\ge 1}{\sup }\Vert f_n\left(t\right)\Vert \le \mu \left(t\right),$

则称序列 ${\left\{f_n\right\}}_{n=1}^{\infty }\subset L^1\left(J,X\right)$ 是半紧的。

引理6 [11] 设 $G:L^1\left(J,X\right)\to C\left(J,X\right)$ 定义为

$\left(Gf\right)\left(t\right)={\displaystyle {\int }_0^t{T}_q\left(t-s\right)f\left(s\right)\text{d}s}.$

如果序列 ${\left\{f_n\right\}}_{n=1}^{\infty }\subset L^1\left(J,X\right)$ 是半紧的，则序列 ${\left\{G{f}_n\right\}}_{n=1}^{\infty }$ 在 $C\left(J,X\right)$ 上相对紧。此外，如果 $f_n\to f$，则 $\left(G{f}_n\right)\left(t\right)\to \left(Gf\right)\left(t\right),n\to \infty$。

引理7 [17] 设D为Banach空间X的闭凸子集且 $0\in D$。如果 $F:D\to X$ 是连续映射，且满足如下条件：若 $\Omega \subseteq D$ 可数，且 $\Omega \subseteq \overline{co}\left(\left\{0\right\}\cup F\left(\Omega \right)\right)$，则 $\bar{\Omega }$ 是紧的，那么F在D中有不动点。

3. 主要结果及证明

为了证明本文的主要定理，我们先给出如下假设条件：

(H1) 线性算子A生成的 $C_0$ -半群 $T\left(t\right)\left(t\ge 0\right)$ 是等度连续半群，且存在 $M\ge 1$，使得 $\Vert T\left(t\right)\Vert \le M$。

(H2) 函数 $f:J\times X\to X$ 满足：

(1) 对 $a.e.t\in J$，函数 $f\left(t,\cdot \right):X\to X$ 连续；对每个 $x\in X$，函数 $f\left(\cdot ,x\right):J\to X$ 强可测；

(2) 对 $\forall r>0$，存在常数 $q_1\in \left(0,q\right)$ 和函数 $m\in L^{\frac{1}{q_1}}\left(J,R^{+}\right)$，使得对 $\forall t\in J,x\in X$，有

$\Vert f\left(t,x\right)\Vert \le m\left(t\right),t\in J,$

其中 $m\left(t\right)$ 满足 ${\lim }_{k\to +\infty }\inf \frac{{\Vert m\Vert }_{L^{\frac{1}{q_1}}}}{r}=\sigma <\infty$ ；

(3) 存在常数 $L>0$，使得对X中任意非空有界集D，有

$\beta \left(f\left(t,D\right)\right)\le L\beta \left(D\right).$

(H3) $B:U\to X$ 是线性有界算子，即存在 $M_B>0$，使得

$\Vert B\Vert \le M_B.$

(H4) 线性算子 $W:L^2\left(J,U\right)\to X$ 定义为：

$Wu={\displaystyle {\sum }_{k=1}^m{c}_k}\widehat{T_q}\left(t\right)P{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)Bu\left(s\right)\text{d}s}+{\displaystyle {\int }_0^b{\left(b-s\right)}^{q-1}{T}_q\left(b-s\right)Bu\left(s\right)\text{d}s},$

满足下列条件：

(1) W存在取值于 $L^2\left(J,U\right)\backslash KerW$ 中的线性逆算子 $W^{-1}$，其中

$KerW=\left\{x\in L^2\left(J,U\right):Wx=0\right\};$

(2) 存在常数 $L_W$，使得 $\Vert W^{-1}\Vert \le L_W$ ；

(3) 存在常数 $L^{\prime }>0$，使得对X中任意非空有界集D，有

$\beta \left(W^{-1}\left(D\right)\right)\le L^{\prime }\beta \left(D\right).$

(H5) 存在常数 $R>0$，使得

$p=\left[2qML{l}_1{l}_2\left(2qM{L}^{\prime }{M}_B{l}_1{l}_2+1\right)\right]\underset{t\in \left[0,b\right]}{\sup }{\displaystyle {\int }_0^t{\text{e}}^{-R\left(t-s\right)}\text{d}s}<1,$

其中，

$l_1=\frac{1}{\Gamma \left(1+q\right)\left(1-M{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|\right)},l_2=\frac{b^{q-q_1}}{{\left(1+a\right)}^{1-q_1}}.$

定理3.1设条件(H1)~(H5)成立。如果

$l_1{l}_{2}qM\sigma \left(q{l}_{1}M{M}_B{L}_W+1\right)<1,$ (3.1)

则半线性分数阶发展系统(1.2)在J上精确可控。

证明：根据条件(H3) (1)，对每个 $x\in X,t\in J$，定义控制 $u\left(t\right):=u_x\left(t\right)$ 如下：

$\begin{array}{c}{u}_x\left(t\right)=W^{-1}[x_1-\widehat{T_q}\left(b\right){\displaystyle {\sum }_{k=1}^m{c}_{k}P}{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)f\left(s,x\left(s\right)\right)\text{d}s}\\ -{\displaystyle {\int }_0^b{\left(b-s\right)}^{q-1}{T}_q\left(b-s\right)f\left(s,x\left(s\right)\right)\text{d}s}]\left(t\right).\end{array}$

按此控制函数，分数阶发展系统(1.2)在J上的精确可控等价于算子 $G:C\left(J,X\right)\to C\left(J,X\right)$ 存在不动点，其中算子G定义如下：

$\begin{array}{c}\left(Gx\right)\left(t\right)={\displaystyle {\sum }_{k=1}^m{c}_k}\widehat{T_q}\left(t\right)P{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)\left[B{u}_x\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\\ +{\displaystyle {\int }_0^t{\left(t-s\right)}^{q-1}{T}_q\left(t-s\right)\left[B{u}_x\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}.\end{array}$ (3.2)

下面我们将用引理7证明算子G在 $C\left(J,X\right)$ 上存在不动点，证明过程分为4步。

第1步，证明存在 $r>0$，使得 $G\left(B_r\right)\subseteq B_r$，其中 $B_r=\left\{x\in C\left(J,X\right):{\Vert x\Vert }_C\le r\right\}$。

反设不成立，则对任意 $r>0$，存在函数 $x^r(\cdot )\in B_r$，但 $G\left(x^r\right)\notin B_r$，即 $\Vert G\left(x^r\right)\left(t\right)\Vert >r,t\in J$。易知 ${\left(t-s\right)}^{q-1}\in L^{\frac{1}{1-q_1}}\left(\left[0,b\right],R\right)$，其中 $q_1\in \left[0,q\right),a=\frac{q-1}{1-q_1}$。由条件(H2) (2)及Hölder不等式，有

$\begin{array}{c}\Vert u_{x^r}\left(t\right)\Vert =\Vert W^{-1}[x_1-\widehat{T_q}\left(b\right){\displaystyle {\sum }_{k=1}^m{c}_{k}P}{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)f\left(s,x^r\left(s\right)\right)\text{d}s}\\ -{\displaystyle {\int }_0^b{\left(b-s\right)}^{q-1}{T}_q\left(b-s\right)f\left(s,x^r\left(s\right)\right)\text{d}s}]\Vert \\ \le L_W[\Vert x_1\Vert +\frac{M{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|}{1-M{{\displaystyle \sum }}_{k=1}^m\left|ck\right|}\frac{Mq}{\Gamma \left(1+q\right)}\Vert {\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}f\left(s,x^r\left(s\right)\right)\text{d}s}\Vert \\ +\frac{Mq}{\Gamma \left(1+q\right)}\Vert {\displaystyle {\int }_0^b{\left(b-s\right)}^{q-1}f\left(s,x^r\left(s\right)\right)\text{d}s}\Vert ]\end{array}$

$\begin{array}{l}\le L_W\Vert x_1\Vert +\frac{q{L}_W{M}^2{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|}{\Gamma \left(1+q\right)\left(1-M{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|\right)}\left[{\left({\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{\frac{q-1}{1-q_1}}\text{d}s}\right)}^{1-q_1}{\Vert m\Vert }_{L^{\frac{1}{q_1}}}\right]\\ +\frac{L_{W}Mq}{\Gamma \left(1+q\right)}\left[{\left({\displaystyle {\int }_0^b{\left(b-s\right)}^{\frac{q-1}{1-q_1}}\text{d}s}\right)}^{1-q_1}{\Vert m\Vert }_{L^{\frac{1}{q_1}}}\right]\\ \le L_W\Vert x_1\Vert +\frac{q{L}_{W}M}{\Gamma \left(1+q\right)\left(1-M{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|\right)}\frac{b^{q-q_1}}{{\left(1+a\right)}^{1-q_1}}{\Vert m\Vert }_{L^{\frac{1}{q_1}}}.\end{array}$

所以

$\Vert u_{x^r}\Vert \le L_W\Vert x_1\Vert +q{l}_1{l}_2{L}_{W}M{\Vert m\Vert }_{L^{\frac{1}{q_1}}}.$ (3.3)

则

$\Vert B{u}_{x^r}\left(t\right)\Vert +\Vert f\left(s,x^r\left(s\right)\right)\Vert \le M_B{L}_W\Vert x_1\Vert +q{l}_1{l}_{2}M{L}_W{M}_B{\Vert m\Vert }_{L^{\frac{1}{q_1}}}+m\left(t\right).$

因此，

$\begin{array}{c}r<\Vert \left(G{x}^r\right)\left(t\right)\Vert \\ \le \Vert {\displaystyle {\sum }_{k=1}^m{c}_k}\widehat{T_q}\left(t\right)P{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)\left[B{u}_{x^r}\left(s\right)+f\left(s,x^r\left(s\right)\right)\right]\text{d}s}\Vert \\ +\Vert {\displaystyle {\int }_0^t{\left(t-s\right)}^{q-1}{T}_q\left(t-s\right)\left[B{u}_{x^r}\left(s\right)+f\left(s,x^r\left(s\right)\right)\right]\text{d}s}\Vert \\ \le \frac{q{M}^2{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|}{\Gamma \left(1+q\right)\left(1-M{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|\right)}{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}\left[\Vert B{u}_{x^r}\left(s\right)\Vert +\Vert f\left(s,x^r\left(s\right)\right)\Vert \right]\text{d}s}\\ +\frac{Mq}{\Gamma \left(1+q\right)}{\displaystyle {\int }_0^b{\left(b-s\right)}^{q-1}\left[\Vert B{u}_{x^r}\left(s\right)\Vert +\Vert f\left(s,x^r\left(s\right)\right)\Vert \right]\text{d}s}\end{array}$

$\begin{array}{l}\le \frac{b^q{M}^2{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|}{\Gamma \left(1+q\right)\left(1-M{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|\right)}\left(M_B{L}_W\Vert x_1\Vert +l_1{l}_{2}qM{L}_W{M}_B{\Vert m\Vert }_{L^{\frac{1}{q_1}}}\right)\\ +\frac{q{M}^2{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|}{\Gamma \left(1+q\right)\left(1-M{{\displaystyle \sum }}_{k=1}^m\left|c_K\right|\right)}{l}_2{\Vert m\Vert }_{L^{\frac{1}{q_1}}}+\frac{Mq}{\Gamma \left(1+q\right)}{l}_2{\Vert m\Vert }_{L^{\frac{1}{q_1}}}\\ +\frac{M{b}^q}{\Gamma \left(1+q\right)}\left(M_B{L}_W\Vert x_1\Vert +l_1{l}_{2}qM{L}_W{M}_B{\Vert m\Vert }_{L^{\frac{1}{q_1}}}\right)\\ \le l_{1}M{M}_B{L}_W{b}^q\Vert x_1\Vert +l_1{l}_{2}qM\left(q{l}_{1}M{M}_B{L}_W+1\right){\Vert m\Vert }_{L^{\frac{1}{q_1}}}.\end{array}$

两边同时除以r，并令 $r\to \infty$，得

$1\le l_1{l}_{2}qM\sigma \left(q{l}_{1}M{M}_B{L}_W+1\right),$

这与(3.1)式矛盾。因此， $\exists r>0$，使得 $G\left(B_r\right)\subseteq B_r$。

第2步，证明 $G:B_r\to B_r$ 连续。

设 $\left\{x^n\right\}\subset B_r$ 满足 $x^n\to x$。由(H2) (1)和(2)及Lebesgue控制收敛定理，有

${\displaystyle {\int }_0^t{\left(t-s\right)}^{q-1}\Vert f\left(s,x^n\left(s\right)\right)-f\left(s,x\left(s\right)\right)\Vert \text{d}s}\to 0,t\in J,n\to \infty .$

因此，

$\begin{array}{l}\Vert u_{x^n}\left(s\right)-u_x\left(s\right)\Vert \\ =\Vert W^{-1}\left[x_1-\widehat{T_q}\left(b\right){\displaystyle {\sum }_{k=1}^m{c}_{k}P}{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)f\left(s,x^n\left(s\right)\right)\text{d}s}-{\displaystyle {\int }_0^b{\left(b-s\right)}^{q-1}{T}_q\left(b-s\right)f\left(s,x^n\left(s\right)\right)\text{d}s}\right]\\ -W^{-1}\left[x_1-\widehat{T_q}\left(b\right){\displaystyle {\sum }_{k=1}^m{c}_{k}P}{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)f\left(s,x\left(s\right)\right)\text{d}s}-{\displaystyle {\int }_0^b{\left(b-s\right)}^{q-1}{T}_q\left(b-s\right)f\left(s,x\left(s\right)\right)\text{d}s}\right]\Vert \\ \le \Vert W^{-1}\widehat{T_q}\left(b\right){\displaystyle {\sum }_{k=1}^m{c}_{k}P}{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)\left[f\left(s,x^n\left(s\right)\right)-f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \\ +\Vert W^{-1}{\displaystyle {\int }_0^b{\left(b-s\right)}^{q-1}{T}_q\left(b-s\right)\left[f\left(s,x^n\left(s\right)\right)-f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \end{array}$

$\begin{array}{l}\le \frac{L_{W}q{M}^2{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|}{\Gamma \left(1+q\right)\left(1-M{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|\right)}{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}\Vert f\left(s,x^n\left(s\right)\right)-f\left(s,x\left(s\right)\right)\Vert \text{d}s}\\ +\frac{L_{W}Mq}{\Gamma \left(1+q\right)}{\displaystyle {\int }_0^b{\left(b-s\right)}^{q-1}\Vert f\left(s,x^n\left(s\right)\right)-f\left(s,x\left(s\right)\right)\Vert \text{d}s}\\ \to 0.\end{array}$

因此，由算子G的定义，有

$\begin{array}{l}\Vert \left(G{x}^n\right)\left(t\right)-\left(Gx\right)\left(t\right)\Vert \\ =\Vert {\displaystyle {\sum }_{k=1}^m{c}_k}\widehat{T_q}\left(t\right)P{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)\left[B{u}_{x^n}\left(s\right)+f\left(s,x^n\left(s\right)\right)\right]\text{d}s}\\ +{\displaystyle {\int }_0^t{\left(t-s\right)}^{q-1}{T}_q\left(t-s\right)\left[B{u}_{x^n}\left(s\right)+f\left(s,x^n\left(s\right)\right)\right]\text{d}s}\\ -{\displaystyle {\sum }_{k=1}^m{c}_k}\widehat{T_q}\left(t\right)P{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)\left[B{u}_x\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\\ -{\displaystyle {\int }_0^t{\left(t-s\right)}^{q-1}{T}_q\left(t-s\right)\left[B{u}_x\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \end{array}$

$\begin{array}{l}\le \frac{q{M}^2{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|}{\Gamma \left(1+q\right)\left(1-M{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|\right)}{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}\Vert B{u}_{x^n}\left(s\right)-B{u}_x\left(s\right)\Vert \text{d}s}\\ +\frac{Mq}{\Gamma \left(1+q\right)}{\displaystyle {\int }_0^b{\left(b-s\right)}^{q-1}\Vert B{u}_{x^n}\left(s\right)-B{u}_x\left(s\right)\Vert \text{d}s}\\ \le \frac{q{M}^2{M}_B{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|}{\Gamma \left(1+q\right)\left(1-M{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|\right)}{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}\Vert u_{x^n}\left(s\right)-u_x\left(s\right)\Vert \text{d}s}\\ +\frac{M_{B}Mq}{\Gamma \left(1+q\right)}{\displaystyle {\int }_0^b{\left(b-s\right)}^{q-1}\Vert u_{x^n}\left(s\right)-u_x\left(s\right)\Vert \text{d}s}\\ \to 0.\end{array}$

所以， $G:B_r\to B_r$ 连续。

第3步，证明 $G\left(B_r\right)$ 在J上等度连续。

取 $t_1,t_2\in J,t_1<t_2$。对 $\forall x\in B_r$，及 $\forall \epsilon \in \left(0,t_1\right)$，有

$\begin{array}{l}\Vert \left(Gx\right)\left(t_2\right)-\left(Gx\right)\left(t_1\right)\Vert \\ =\Vert {\displaystyle {\sum }_{k=1}^m{c}_k}\widehat{T_q}\left(t_2\right)P{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\\ +{\displaystyle {\int }_0^{t_2}{\left(t_2-s\right)}^{q-1}{T}_q\left(t_2-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\\ -{\displaystyle {\sum }_{k=1}^m{c}_k}\widehat{T_q}\left(t\right)P{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\\ -{\displaystyle {\int }_0^{t_1}{\left(t_1-s\right)}^{q-1}{T}_q\left(t_1-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \end{array}$

$\begin{array}{l}\le \Vert \left[\widehat{T_q}\left(t_2\right)-\widehat{T_q}\left(t_1\right)\right]{\displaystyle {\sum }_{k=1}^m{c}_{k}P}{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \\ +\Vert {\displaystyle {\int }_0^{t_1}{\left(t_2-s\right)}^{q-1}{T}_q\left(t_2-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \\ +\Vert {\displaystyle {\int }_{t_1}^{t_2}{\left(t_2-s\right)}^{q-1}{T}_q\left(t_2-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \\ +\Vert {\displaystyle {\int }_0^{t_1}{\left(t_1-s\right)}^{q-1}{T}_q\left(t_1-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \\ \le \Vert \left[\widehat{T_q}\left(t_2\right)-\widehat{T_q}\left(t_1\right)\right]{\displaystyle {\sum }_{k=1}^m{c}_{k}P}{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \end{array}$

$\begin{array}{l}+\Vert {\displaystyle {\int }_{t_1}^{t_2}{\left(t_2-s\right)}^{q-1}{T}_q\left(t_2-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \\ +\Vert {\displaystyle {\int }_0^{t_1}{\left(t_2-s\right)}^{q-1}{T}_q\left(t_2-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \\ -\Vert {\displaystyle {\int }_0^{t_1}{\left(t_1-s\right)}^{q-1}{T}_q\left(t_2-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \\ +\Vert {\displaystyle {\int }_0^{t_1}{\left(t_1-s\right)}^{q-1}{T}_q\left(t_2-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \\ +\Vert {\displaystyle {\int }_0^{t_1}{\left(t_1-s\right)}^{q-1}{T}_q\left(t_1-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \end{array}$

$\begin{array}{l}\le \Vert \left[\widehat{T_q}\left(t_2\right)-\widehat{T_q}\left(t_1\right)\right]{\displaystyle {\sum }_{k=1}^m{c}_{k}P}{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \\ +\Vert {\displaystyle {\int }_{t_1}^{t_2}{\left(t_2-s\right)}^{q-1}{T}_q\left(t_2-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \\ +\Vert {\displaystyle {\int }_0^{t_1}\left[{\left(t_2-s\right)}^{q-1}-{\left(t_1-s\right)}^{q-1}\right]T_q\left(t_2-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \\ +\Vert {\displaystyle {\int }_0^{t_1}{\left(t_1-s\right)}^{q-1}\left[T_q\left(t_2-s\right)-T_q\left(t_1-s\right)\right]\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \\ :=I_1+I_2+I_3+I_4.\end{array}$

由引理2可知，当 $t_2\to t_1$ 时， $\widehat{T_q}\left(t_2\right)-\widehat{T_q}\left(t_1\right)\to 0$，即 $I_1\to 0$。

$\begin{array}{c}{I}_2\le \frac{Mq}{\Gamma \left(1+q\right)}{\displaystyle {\int }_{t_1}^{t_2}{\left(t_2-s\right)}^{q-1}\left[M_B{L}_W\Vert x_1\Vert +l_{1}qM{L}_W{M}_B\frac{b^{q-q_1}}{{\left(1+a\right)}^{1-q_1}}{\Vert m\Vert }_{L^{\frac{1}{q_1}}}+\Vert m\left(t\right)\Vert \right]\text{d}s}\\ \le \frac{Mq}{\Gamma \left(1+q\right)}\left[M_B{L}_W\Vert x_1\Vert +l_{1}qM{L}_W{M}_B{l}_2{\Vert m\Vert }_{L^{\frac{1}{q_1}}}\right]{\left(t_2-t_1\right)}^q+\frac{Mq}{\Gamma \left(1+q\right)}\frac{{\left(t_2-t_1\right)}^q}{{\left(1+a\right)}^{1-q_1}}{\Vert m\Vert }_{L^{\frac{1}{q_1}}}.\end{array}$

当 $t_2-t_1\to 0$ 时， $I_2\to 0$。

$\begin{array}{c}{I}_3\le \frac{Mq}{\Gamma \left(1+q\right)}\left[M_B{L}_W\Vert x_1\Vert +l_1{l}_{2}qM{L}_W{M}_B{\Vert m\Vert }_{L^{\frac{1}{q_1}}}\right]{\displaystyle {\int }_0^{t_1}\left[{\left(t_2-s\right)}^{q-1}-{\left(t_1-s\right)}^{q-1}\right]\text{d}s}\\ +\frac{Mq}{\Gamma \left(1+q\right)}{\displaystyle {\int }_0^{t_1}\left[{\left(t_2-s\right)}^{q-1}-{\left(t_1-s\right)}^{q-1}\right]m\left(t\right)\text{d}s}\\ \le \frac{Mq}{\Gamma \left(1+q\right)}\left[M_B{L}_W\Vert x_1\Vert +l_1{l}_{2}qM{L}_W{M}_B{\Vert m\Vert }_{L^{\frac{1}{q_1}}}\right]{\displaystyle {\int }_0^{t_1}\left[{\left(t_2-s\right)}^{q-1}-{\left(t_1-s\right)}^{q-1}\right]\text{d}s}\\ +\frac{Mq}{\Gamma \left(1+q\right)}{\Vert m\Vert }_{L^{\frac{1}{q_1}}}{\left({\displaystyle {\int }_0^{t_1}{\left|{\left(t_2-s\right)}^{q-1}-{\left(t_1-s\right)}^{q-1}\right|}^{\frac{1}{1-q_1}}\text{d}s}\right)}^{1-q_1}.\end{array}$

当 $t_2-t_1\to 0$ 时， $I_3\to 0$。

对 $I_4$，由引理2(3)可知，对 $\forall t>0$， $T_q\left(t\right)$ 是等度连续算子，当 $t_2-t_1\to 0$ 且 $\epsilon \to 0$ 时，有

$\begin{array}{c}{I}_4\le \Vert {\displaystyle {\int }_0^{t_1-\epsilon }{\left(t_1-s\right)}^{q-1}\left[T_q\left(t_2-s\right)-T_q\left(t_1-s\right)\right]\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \\ +\Vert {\displaystyle {\int }_{t_1-\epsilon }^{t_1}{\left(t_1-s\right)}^{q-1}\left[T_q\left(t_2-s\right)-T_q\left(t_1-s\right)\right]\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \\ \le \underset{s\in \left[0,t_1-\epsilon \right]}{\sup }\Vert T_q\left(t_2-s\right)-T_q\left(t_1-s\right)\Vert {\displaystyle {\int }_0^{t_1-\epsilon }{\left(t_1-s\right)}^{q-1}\Vert Bu\left(s\right)+f\left(s,x\left(s\right)\right)\Vert \text{d}s}\\ +\frac{2Mq}{\Gamma \left(1+q\right)}{\displaystyle {\int }_{t_1-\epsilon }^{t_1}{\left(t_1-s\right)}^{q-1}\Vert Bu\left(s\right)+f\left(s,x\left(s\right)\right)\Vert \text{d}s}\\ \to 0.\end{array}$

所以， $\Vert \left(Gx\right)\left(t_2\right)-\left(Gx\right)\left(t_1\right)\Vert \le I_1+I_2+I_3+I_4\to 0$，即 $G\left(B_r\right)$ 在J上等度连续。

第4步，证明由(3.2)式定义的函数 $G:B_r\to B_r$ 满足M不动点条件。为此，我们设 $W\subseteq B_r$ 可数，其中 $W\subseteq \overline{co}\left(\left\{0\right\}\cup G\left(W\right)\right)$，仅证明W相对紧即可。

我们用 $\varphi$ 定义 $B_r$ 中的非紧性测度如下：

$\varphi \left(\Omega \right)={\max }_{E\in \Delta \left(\Omega \right)}\left(\alpha \left(E\right),mo{d}_c\left(E\right)\right),$ (3.4)

其中， $\Omega$ 是 $B_r$ 的所有有界集， $\Delta \left(\Omega \right)$ 是 $\Omega$ 的可数子集的集合。 $\alpha$ 是实的非紧性测度，且

$\alpha \left(E\right)=\underset{t\in \left[0,b\right]}{\sup }{\text{e}}^{-Lt}\beta \left(E\left(t\right)\right),$

其中， $E\left(t\right)=\left\{x\left(t\right):x\in E\right\}$，L是我们选取的适当的一个常数。 $mo{d}_c\left(E\right)$ 是函数集E的等度连续模，且

$mo{d}_c\left(E\right)=\underset{\delta \to 0}{\lim }\underset{x\in E}{\sup }\underset{\Vert t_1-t_2\Vert <\delta }{\max }\Vert x\left(t_1\right)-x\left(t_2\right)\Vert ,t_1,t_2\in J.$

在文献 [11] 中证明了 $\varphi$ 是有意义的，即存在 $E_0\in \Delta \left(\Omega \right)$，使得(3.4)式在 $E_0$ 处达到最大值，并且 $\varphi$ 是一个单调非奇异正则的非紧性测度。

对于(3.2)式定义的 $Gx$，由 $\varphi$ 的正则性，只需证明 $\varphi \left(W\right)=\left(0,0\right)$。因此 $\varphi \left(G\left(W\right)\right)$ 是一个最大值。

$\begin{array}{l}mo{d}_c\left(W\right)=\underset{\delta \to 0}{\lim }\underset{x\in E}{\sup }\underset{\Vert t_1-t_2\Vert <\delta }{\max }\Vert x\left(t_1\right)-x\left(t_2\right)\Vert \\ =\underset{\delta \to 0}{\lim }\underset{x\in E}{\sup }\underset{\Vert t_1-t_2\Vert <\delta }{\max }\Vert {\displaystyle {\sum }_{k=1}^m{c}_k}\widehat{T_q}\left(t_2\right)P{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\\ +{\displaystyle {\int }_0^{t_2}{\left(t_2-s\right)}^{q-1}{T}_q\left(t_2-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\\ -{\displaystyle {\sum }_{k=1}^m{c}_k}\widehat{T_q}\left(t_1\right)P{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\\ -{\displaystyle {\int }_0^{t_1}{\left(t_1-s\right)}^{q-1}{T}_q\left(t_1-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \end{array}$

$\begin{array}{l}\le \underset{\delta \to 0}{\lim }\underset{x\in E}{\sup }\underset{\Vert t_1-t_2\Vert <\delta }{\max }(\Vert \left[\widehat{T_q}\left(t_2\right)-\widehat{T_q}\left(t_1\right)\right]{\displaystyle {\sum }_{k=1}^m{c}_{k}P}{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \\ +\Vert {\displaystyle {\int }_{t_1}^{t_2}{\left(t_2-s\right)}^{q-1}{T}_q\left(t_2-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \\ +\Vert {\displaystyle {\int }_0^{t_1}\left[{\left(t_2-s\right)}^{q-1}-{\left(t_1-s\right)}^{q-1}\right]T_q\left(t_2-s\right)\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert \\ +\Vert {\displaystyle {\int }_0^{t_1}{\left(t_1-s\right)}^{q-1}\left[T_q\left(t_2-s\right)-T_q\left(t_1-s\right)\right]\left[Bu\left(s\right)+f\left(s,x\left(s\right)\right)\right]\text{d}s}\Vert )\\ :=\underset{\delta \to 0}{\lim }\underset{x\in E}{\sup }\underset{\Vert t_1-t_2\Vert <\delta }{\max }\left(I_1+I_2+I_3+I_4\right).\end{array}$

由第3步的证明易知当 $\Vert t_1-t_2\Vert <\delta$ 且 $\epsilon \to 0$ 时， $I_1\to 0,I_2\to 0,I_3\to 0,I_4\to 0$。所以 $mo{d}_c\left(W\right)=0$。

设 ${\left\{z^n\right\}}_{n=1}^{\infty }\subseteq G\left(W\right)$ 是 $\varphi$ 取得最大值的可数值，那么存在集合 ${\left\{x^n\right\}}_{n=1}^{\infty }\subseteq W$，使得

$z^n\left(t\right)=\left(G{x}^n\right)\left(t\right),n\ge 1,t\in J.$ (3.5)

因为

$\begin{array}{l}\beta \left({\left\{u_{x^n}\left(s\right)\right\}}_{n=1}^{\infty }\right)\\ =\beta (W^{-1}[x_1-\widehat{T_q}\left(b\right){\displaystyle {\sum }_{k=1}^m}{c}_{k}P{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right){\left\{f\left(s,x^n\left(s\right)\right)\right\}}_{n=1}^{\infty }\text{d}s}\\ -{\displaystyle {\int }_0^b{\left(b-s\right)}^{q-1}{T}_q\left(b-s\right){\left\{f\left(s,x^n\left(s\right)\right)\right\}}_{n=1}^{\infty }\text{d}s}])\\ \le L^{\prime }\beta (\frac{qM{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|}{\Gamma \left(1+q\right)\left(1-M{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|\right)}{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{\left\{f\left(s,x^n\left(s\right)\right)\right\}}_{n=1}^{\infty }\text{d}s}\\ \begin{array}{c}\stackrel{}{}\\ \underset{}{}\end{array}-\frac{Mq}{\Gamma \left(1+q\right)}{\displaystyle {\int }_0^b{\left(b-s\right)}^{q-1}{\left\{f\left(s,x^n\left(s\right)\right)\right\}}_{n=1}^{\infty }\text{d}s})\end{array}$

$\begin{array}{l}\le L^{\prime }[\frac{qM{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|}{\Gamma \left(1+q\right)\left(1-M{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|\right)}\frac{b^{q-q_1}}{{\left(1+a\right)}^{1-q_1}}\beta \left({\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{\left\{f\left(s,x^n\left(s\right)\right)\right\}}_{n=1}^{\infty }\text{d}s}\right)\\ \begin{array}{c}\stackrel{}{}\\ \underset{}{}\end{array}+\frac{Mq}{\Gamma \left(1+q\right)}\frac{b^{q-q_1}}{{\left(1+a\right)}^{1-q_1}}\beta \left({\displaystyle {\int }_0^b{\left(b-s\right)}^{q-1}{\left\{f\left(s,x^n\left(s\right)\right)\right\}}_{n=1}^{\infty }\text{d}s}\right)]\\ \le \frac{qM{L}^{\prime }}{\Gamma \left(1+q\right)\left(1-M{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|\right)}\frac{b^{q-q_1}}{{\left(1+a\right)}^{1-q_1}}\cdot 2{\displaystyle {\int }_0^b\beta }\left({\left\{f\left(s,x^n\left(s\right)\right)\right\}}_{n=1}^{\infty }\right)\text{d}s\\ \le \frac{2qM{L}^{\prime }L}{\Gamma \left(1+q\right)\left(1-M{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|\right)}\frac{b^{q-q_1}}{{\left(1+a\right)}^{1-q_1}}{\displaystyle {\int }_0^b\beta }\left({\left\{x^n\left(s\right)\right\}}_{n=1}^{\infty }\right)\text{d}s.\end{array}$

所以，由算子G的定义，有

$\begin{array}{l}\beta \left({\left\{\left(G{x}^n\right)\left(t\right)\right\}}_{n=1}^{\infty }\right)\\ =\beta ({\displaystyle {\sum }_{k=1}^m{c}_k}\widehat{T_q}\left(t\right)P{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}{T}_q\left(t_k-s\right)\left[{\left\{B{u}_{x^n}\left(s\right)\right\}}_{n=1}^{\infty }+{\left\{f\left(s,x^n\left(s\right)\right)\right\}}_{n=1}^{\infty }\right]\text{d}s}\\ +{\displaystyle {\int }_0^b{\left(b-s\right)}^{q-1}{T}_q\left(b-s\right)\left[{\left\{B{u}_{x^n}\left(s\right)\right\}}_{n=1}^{\infty }+{\left\{f\left(s,x^n\left(s\right)\right)\right\}}_{n=1}^{\infty }\right]\text{d}s})\\ \le \frac{2q{M}^2{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|}{\Gamma \left(1+q\right)\left(1-M{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|\right)}\left[{\displaystyle {\int }_0^{t_k}{\left(t_k-s\right)}^{q-1}\beta \left({\left\{B{u}_{x^n}\left(s\right)\right\}}_{n=1}^{\infty }+{\left\{f\left(s,x^n\left(s\right)\right)\right\}}_{n=1}^{\infty }\right)\text{d}s}\right]\\ +\frac{2Mq}{\Gamma \left(1+q\right)}\left[{\displaystyle {\int }_0^b{\left(b-s\right)}^{q-1}\beta \left({\left\{B{u}_{x^n}\left(s\right)\right\}}_{n=1}^{\infty }+{\left\{f\left(s,x^n\left(s\right)\right)\right\}}_{n=1}^{\infty }\right)\text{d}s}\right]\end{array}$

$\begin{array}{l}\le \left[\frac{2q{M}^2{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|}{\Gamma \left(1+q\right)\left(1-M{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|\right)}+\frac{2Mq}{\Gamma \left(1+q\right)}\right]\left[2qM{L}^{\prime }L{M}_B{l}_2^2{\displaystyle {\int }_0^b\beta \left({\left\{x^n\left(s\right)\right\}}_{n=1}^{\infty }\right)\text{d}s}+L{l}_2{\displaystyle {\int }_0^b\beta \left({\left\{x^n\left(s\right)\right\}}_{n=1}^{\infty }\right)\text{d}s}\right]\\ \le \left[2qML{l}_1{l}_2\left(2qM{L}^{\prime }{M}_B{l}_1{l}_2+1\right)\right]{\displaystyle {\int }_0^t\beta \left({\left\{x^n\left(s\right)\right\}}_{n=1}^{\infty }\right)\text{d}s}\\ \le \left[2qML{l}_1{l}_2\left(2qM{L}^{\prime }{M}_B{l}_1{l}_2+1\right)\right]{\displaystyle {\int }_0^t{\text{e}}^{Rs}\left({\text{e}}^{-Rs}\beta \left({\left\{x^n\left(s\right)\right\}}_{n=1}^{\infty }\right)\right)\text{d}s}\\ =\left[2qML{l}_1{l}_2\left(2qM{L}^{\prime }{M}_B{l}_1{l}_2+1\right)\right]\alpha \left({\left\{x^n\right\}}_{n=1}^{\infty }\right){\displaystyle {\int }_0^t{\text{e}}^{Rs}\text{d}s},t\in J.\end{array}$

现在我们给出 $\alpha \left({\left\{z^n\right\}}_{n=1}^{\infty }\right)$ 的一个估计：

$\begin{array}{c}\alpha \left({\left\{z^n\right\}}_{n=1}^{\infty }\right)=\underset{t\in \left[0,b\right]}{\sup }{\text{e}}^{-Rt}\beta \left({\left\{\left(G{x}^n\right)\left(t\right)\right\}}_{n=1}^{\infty }\right)\\ \le \underset{t\in \left[0,b\right]}{\sup }{\text{e}}^{-Rt}\left[2qML{l}_1{l}_2\left(2qM{L}^{\prime }{M}_B{l}_1{l}_2+1\right)\right]\alpha \left({\left\{x^n\right\}}_{n=1}^{\infty }\right){\displaystyle {\int }_0^t{\text{e}}^{Rs}\text{d}s}\\ \le \alpha \left({\left\{x^n\right\}}_{n=1}^{\infty }\right)\left[2qML{l}_1{l}_2\left(2qM{L}^{\prime }{M}_B{l}_1{l}_2+1\right)\right]\underset{t\in \left[0,b\right]}{\sup }{\displaystyle {\int }_0^t{\text{e}}^{-R\left(t-s\right)}\text{d}s}\\ =p\alpha \left({\left\{x^n\right\}}_{n=1}^{\infty }\right).\end{array}$

因此，

$\alpha \left({\left\{x^n\right\}}_{n=1}^{\infty }\right)\le \alpha \left(W\right)\le \alpha \left(\overline{co}\left(\left\{0\right\}\cup G\left(W\right)\right)\right)=\alpha \left({\left\{z^n\right\}}_{n=1}^{\infty }\right)\le \alpha \left({\left\{x^n\right\}}_{n=1}^{\infty }\right)p.$

由 $p<1$ 得

$\alpha \left({\left\{x^n\right\}}_{n=1}^{\infty }\right)=\alpha \left(W\right)=\alpha \left({\left\{z^n\right\}}_{n=1}^{\infty }\right)=0.$

根据 $\alpha$ 定义可知，

$\beta \left({\left\{x^n\left(t\right)\right\}}_{n=1}^{\infty }\right)=0.$

又因为

$\begin{array}{l}\beta \left({\left\{B{u}_{x^n}\left(t\right)\right\}}_{n=1}^{\infty }+{\left\{f\left(t,x^n\left(t\right)\right)\right\}}_{n=1}^{\infty }\right)\\ \le \frac{2qML{L}^{\prime }}{\Gamma \left(1+q\right)\left(1-M{{\displaystyle \sum }}_{k=1}^m\left|c_k\right|\right)}\frac{b^{q-q_1}}{{\left(1+a\right)}^{1-q_1}}{\displaystyle {\int }_0^b\beta \left({\left\{x^n\left(t\right)\right\}}_{n=1}^{\infty }\right)\text{d}t}+L{\displaystyle {\int }_0^b\beta \left({\left\{x^n\left(t\right)\right\}}_{n=1}^{\infty }\right)\text{d}t}\\ =0.\end{array}$

所以， ${\left\{B{u}_{x^n}\left(t\right)+f\left(t,x^n\left(t\right)\right)\right\}}_{n=1}^{\infty }$ 在X中对几乎所有 $t\in J$ 相对紧。再由(H2) (2)和(3.3)式，易知 ${\left\{B{u}_{x^n}\left(t\right)+f\left(t,x^n\left(t\right)\right)\right\}}_{n=1}^{\infty }$ 对几乎处处 $t\in \left[0,b\right]$ 是一致可积的。所以根据定义4， ${\left\{B{u}_{x^n}+f\left(\cdot ,x^n(\cdot )\right)\right\}}_{n=1}^{\infty }$ 是半紧的。由引理6可知， $G\left({\left\{x^n\right\}}_{n=1}^{\infty }\right)$ 在 $B_r$ 中相对紧。通过(3.5)式得， ${\left\{z^n\right\}}_{n=1}^{\infty }$ 在 $B_r$ 中也相对紧。因为 $\varphi$ 是单调非奇异正则的非紧性测度，根据Mönch不动点的条件，所以

$\varphi \left(W\right)\le \varphi \left(\overline{co}\left(\left\{0\right\}\cup G\left(W\right)\right)\right)=\varphi \left({\left\{z^n\right\}}_{n=1}^{\infty }\right)=0.$

故W在 $B_r$ 中相对紧。

4. 例子

例1设 $X=U:=C\left(\left[0,1\right]\right)$。考虑分数阶发展方程非局部问题

$\{\begin{array}{l}\frac{{\partial }^{\frac{1}{2}}}{\partial t^{\frac{1}{2}}}x\left(t,z\right)+\frac{\partial x\left(t,z\right)}{\partial t}=\frac{{\text{e}}^{-2t}}{1+{\text{e}}^t}x\left(t,z\right)+\lambda \mu \left(t,z\right),t\in \left[0,b\right]=J,z\in \left(0,1\right),\\ x\left(t,0\right)=x\left(t,1\right)=0,t\in J,\\ x\left(0,z\right)={\displaystyle {\sum }_{k=1}^m\arctan }\frac{1}{2k^2}x\left(k,z\right),z\in \left(0,1\right),\end{array}$ (4.1)

其中 $\lambda >0,0<m<b,\mu :J\times \left(0,1\right)\to \left(0,1\right)$。

定义算子 $A:D\left(A\right)\subset X\to X$ 如下：

$D\left(A\right)=\left\{x\in X:x^{\prime }\in X,x\left(0\right)=x\left(1\right)=0\right\}.$

$Ax=-x^{\prime },x\in D\left(A\right),$

则A生成X中的等度连续半群 ${\left\{T\left(t\right)\right\}}_{t\ge 0}$，且对任意的 $x\in X$， $T\left(t\right)$ 满足

$T\left(t\right)x\left(s\right)=x\left(t+s\right),$

那么 $T\left(t\right)\left(t\ge 0\right)$ 在X中非紧，并且有 ${\sup }_{t\in J}\Vert T\left(t\right)\Vert \le 1$。

定义 $x\left(t\right)\left(z\right)=x\left(t,z\right);f\left(t,x\left(t\right)\right)=\frac{{\text{e}}^{-2t}}{1+{\text{e}}^t}x\left(t,z\right);u\left(t\right)\left(z\right)=\mu \left(t,z\right);c_k=\arctan \frac{1}{2k^2}$。对 $\forall x\in B_r:=\left\{x\in C\left(J,X\right):{\Vert x\Vert }_{C\left(J,X\right)}\le r\right\},t\in J$，有

$\Vert f\left(t,x\right)\Vert \le \Vert \frac{{\text{e}}^{-2t}}{1+{\text{e}}^t}x\left(t,z\right)\Vert \le \frac{r}{{\text{e}}^{2t}\left(1+{\text{e}}^t\right)}\le \frac{1}{2},$

则条件(H2)成立。

又因为 ${\displaystyle {\sum }_{k=1}^m\left|c_k\right|}\le {\displaystyle {\sum }_{k=1}^m\arctan \frac{1}{2k^2}}=\frac{\pi }{4}<1$，所以条件(H0)成立。

定义算子W如下

$\begin{array}{c}\left(Wt\right)\left(z\right)=\widehat{T_q}\left(b\right){\displaystyle {\sum }_{k=1}^m\arctan }\frac{1}{2k^2}{\left[I-{\displaystyle {\sum }_{k=1}^m\arctan }\frac{1}{2k^2}\widehat{T_q}\left(t_k\right)\right]}^{-1}{\displaystyle {\int }_0^k{\left(k-s\right)}^{-\frac{1}{2}}{T}_q\left(k-s\right)\lambda \mu \left(s,z\right)\text{d}s}\\ +{\displaystyle {\int }_0^b{\left(b-s\right)}^{-\frac{1}{2}}{T}_q\left(b-s\right)\lambda \mu \left(s,z\right)\text{d}s}.\end{array}$

${\left\{\widehat{T_q}\left(t\right)\right\}}_{t\ge 0}$ 与 ${\left\{T_q\left(t\right)\right\}}_{t\ge 0}$ 定义如下：

$\widehat{T_q}\left(t\right)x\left(s\right)={\displaystyle {\int }_0^{\infty }{\eta }_{\frac{1}{2}}\left(\theta \right)x\left(t^{\frac{1}{2}}\theta +s\right)\text{d}\theta },$

$T_q\left(t\right)x\left(s\right)=\frac{1}{2}{\displaystyle {\int }_0^{\infty }\theta {\eta }_{\frac{1}{2}}\left(\theta \right)x\left(t^{\frac{1}{2}}\theta +s\right)\text{d}s},$

其中 ${\eta }_{\frac{1}{2}}\left(\theta \right)=2{\theta }^{-3}{W}_{\frac{1}{2}}\left({\theta }^{-\frac{1}{2}}\right),W_{\frac{1}{2}}\left(\theta \right)=\frac{1}{\pi }{\displaystyle {\sum }_{n=1}^{\infty }{\left(-1\right)}^{n-1}{\theta }^{-\frac{n}{2}-1}}\frac{\Gamma \left(1+\frac{n}{2}\right)}{n!}\sin \left(\frac{n\pi }{2}\right),\theta \in \left(0,\infty \right)$。

我们假设W满足条件(H4)，且不等式条件(H5)和(3.1)式成立，则由定理3.1可知，分数阶发展系统(4.1)在J上精确可控。

基金项目

国家基金委青年科学资助项目(12061062)。