2 预备知识
设C[0, 1]表示[0,1]所有连续函数的集合. 若在该线性空间赋予范数
$$\|u\|=\max\limits_{0\leq t\leq 1} | u(t)| ,
$$
则(C[0,1],||.||)是一个Banach空间.
对于 r>0,$u\in$ C(0,1],定义
$$t^ru(t)|_{t=0}=\lim\limits_{t\rightarrow 0^+}t^ru(t),
$$
$$C_r[0,1]=\{u\in C(0,1]| t^ru(t)\in C[0,1] \}.
$$
则 $C_r[0,1]$在范数 $\|u\|_r=\max\limits_{0\leq t\leq 1} | t^ru(t) |$
下也成为一个Banach空间.
当 $0<r<1$,$C_r[0,1]\subset C(0,1]\bigcap L^1[0,1]$,这里
$L^1[0,1]$ 表示$[0,1]$ 上所有Lebesgue可积的函数集合.
设 $P=\{u\in C_{r}[0,1] \mid u(t)\geq 0 ,\forall t\in (0,1]\}$.
${\bf 引理2.1}\quad$ $P$ 是一个正规锥.
${\bf 证}\quad$ (1) $P$ 是$C_r[0,1]$中的一个锥.
这可以用$P$的定义来验证.
(2) $P$ 是正规锥.
事实上,设 $u,v\in P$ 满足 $0\leq u(t)\leq v(t) ,\forall t\in (0,1]$,则有
$0\leq t^ru(t)\leq t^rv(t) ,\forall t\in [0,1]$,即 $\|u\|_r\leq \|v\|_r$.
因此,$P$ 是一个正规锥.
一个连续函数$u:(0,1]\rightarrow R$的 $\alpha(0<\alpha<1)$ 阶 Riemann-Liouville导数定义为
$$I^{\alpha}u(t)=\frac{1}{\Gamma(\alpha)}\int_0^t(t-\tau)^{\alpha-1}u(\tau){\rm d}\tau,
$$
其中 $\Gamma(.)$ 表示$\Gamma$函数.
引理2.2 设 $0<\alpha<1$,如果 $u\in C_{\alpha}[0,1],$
则 $ I^{\alpha}u\in C[0,1]$.
${\bf 证}\quad$
设 $u\in C_{\alpha}[0,1]$,则
\begin{eqnarray*}
I^{\alpha}u(t)&=& \frac{1}{\Gamma(\alpha)}\int_0^t(t-\tau)^{\alpha-1}u(\tau){\rm d}\tau
\\ &=& \frac{1}{\Gamma(\alpha)}\int_0^t(t-\tau)^{\alpha-1}(\tau)^{-\alpha}U(\tau){\rm d}\tau \ \ (U(\tau)={\tau}^{\alpha}u(\tau))
\\ &=& \frac{1}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha-1}(s)^{-\alpha}U(st){\rm d}s\ \ (s=\frac{\tau}{t}).
\end{eqnarray*}
注意到 $U(t)$ 在 $[0,1]$ 上一致连续,即对 $\forall \epsilon >0$,
存在 $\delta>0$,当 $t_1,t_2\in [0,1]$ 且$|t_1- t_2|<\delta$就有
$$|U(t_1)-U(t_2)|<\epsilon.$$
所以
\begin{eqnarray*}
&&| I^{\alpha}u(t_1)-I^{\alpha}u(t_2)|\\
&=&\bigg|\frac{1}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha-1}(s)^{-\alpha}U(st_1){\rm d}s-\frac{1}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha-1}(s)^{-\alpha}U(st_2){\rm d}s
\bigg|\\
& =&\bigg|\frac{1}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha-1}(s)^{-\alpha}[U(st_1)-U(st_2)]{\rm d}s
\bigg|\\
& \leq &\frac{1}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha-1}(s)^{-\alpha}\mid U(st_1)-U(st_2)\mid {\rm d}s\\
&\leq& \epsilon \frac{1}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha-1}(s)^{-\alpha}{\rm d}s\ \ (\because \ \mid st_1-st_2\mid\leq |t_1- t_2|<\delta)\\
& =&\epsilon\frac{B(\alpha,1-\alpha)}{\Gamma(\alpha)}\\
&=&\Gamma(1-\alpha)\epsilon,
\end{eqnarray*}
其中 $B(.,.)$表示Beta函数.
因此有 $ I^{\alpha}u\in C[0,1]$.
一个连续函数$u\in C(0,1]$的$\alpha(0<\alpha<1)$阶Riemann-Liouville导数定义为
$$D^{\alpha}u(t)=\frac{1}{\Gamma(1-\alpha)}\frac{\rm d}{{\rm d}t}\int_0^t(t-\tau)^{-\alpha}u(\tau){\rm d}\tau=\frac{\rm d}{{\rm d}t}I^{1-\alpha}u(t).$$
作为该定义的直接结论,分数阶导数具有以下性质
(1)当 $\beta>-1$时，
$$D^{\alpha}(t^{\beta})=\frac{1}{\Gamma(1-\alpha)}\frac{\rm d}{{\rm d}t}\int_0^t(t-\tau)^{-\alpha}{\tau}^{\beta}{\rm d}\tau=\frac{\Gamma(\beta+1)}{\Gamma(\beta+2-\alpha)}\frac{\rm d}{{\rm d}t}(t^{1+\beta-\alpha}).$$
(2)对于任意的 $u\in C(0,1]\bigcap L^1[0,1],D^{\alpha}I^{\alpha}u=u$.
(3)当 $u\in C(0,1]\bigcap L^1[0,1]$时,分数微分方程 $D^{\alpha}u(t)=0$
有解为 $u(t)=ct^{\alpha-1},c\in R$.
(4)当 $u\in C(0,1]\bigcap L^1(0,1]$ 且 $D^{\alpha}u\in C(0,1]\bigcap L^1(0,1]
(0<\alpha<1)$时,下式成立
$$I^{\alpha}D^{\alpha}u(t)=u(t)+ct^{\alpha-1},\ c\in R.
$$
对于 $\alpha>0,\beta>0$,二元的 Mittag-Leffler 函数定义如下
$$E_{\alpha,\beta}(z)=\sum_{k=0}^{\infty}\frac{z^k}{\Gamma(\alpha k+\beta)},\ \ z\in {\Bbb C}.
$$
据文献[
10]可知,当 $x,y\in {\Bbb R},\ \ x<0<y$ 时
$$0<E_{\alpha,\alpha}(x)<E_{\alpha,\alpha}(0)=\frac{1}{\Gamma(\alpha)}<E_{\alpha,\alpha}(y),$$
$$\lim\limits_{x\rightarrow +\infty}E_{\alpha,\alpha}(x)=+\infty,\ \
\lim\limits_{x\rightarrow -\infty}E_{\alpha,\alpha}(x)=0.
$$
为证明主要结论，需要以下的引理.
${\bf 引理2.3}$[9] 设 $h\in C[0,1],\alpha\in (0,1),\lambda\in {\Bbb R}$ 且 $E_{\alpha,\alpha}(\lambda)<\frac{1}{\Gamma(\alpha)}$. 如果 $u(t)\in C_{1-\alpha}[0,1]$ 满足
(i) $D^{\alpha}u(t)-\lambda u(t)=h(t)(0<t<1)$;
(ii) $ \lim\limits_{t\rightarrow 0^+}t^{1-\alpha}u(t)=u(1)$.
则有
$$u(t)=\int_0^1G_{\lambda,\alpha}(t,s)h(s){\rm d}s,
$$
其中 $G_{\lambda,\alpha}(t,s)$ 的定义为:
当 $0\leq s\leq t\leq1$,
$$\frac{\Gamma(\alpha)E_{\alpha,\alpha}(\lambda t^{\alpha})E_{\alpha,\alpha}(\lambda (1-s)^{\alpha}) t^{\alpha-1}(1-s)^{\alpha-1}}{1-\Gamma(\alpha)E_{\alpha,\alpha}(\lambda)}+(t-s)^{\alpha-1}E_{\alpha,\alpha}(\lambda (t-s)^{\alpha}).
$$
当 $0\leq t\leq s\leq1$,
$$\frac{\Gamma(\alpha)E_{\alpha,\alpha}(\lambda t^{\alpha})E_{\alpha,\alpha}(\lambda (1-s)^{\alpha}) t^{\alpha-1}(1-s)^{\alpha-1}}{1-\Gamma(\alpha)E_{\alpha,\alpha}(\lambda)}.
$$
${\bf 注2.1}\quad$
据文献[10,引理2.2],当 $\alpha\in (0,1),\ \lambda\in R,\ \lambda<0$时,
$0<E_{\alpha,\alpha}(\lambda)< \frac{1}{\Gamma(\alpha)}$.
如果$E_{\alpha,\alpha}(\lambda)<\frac{1}{\Gamma(\alpha)}$,
容易验证 $G_{\lambda,\alpha}(t,s)>0$ for $(t,s)\in (0,1)\times (0,1)$.
${\bf 引理2.4}$[9]$\quad$ 设 $\alpha\in (0,1),\lambda\in {\Bbb R}$
且 $E_{\alpha,\alpha}(\lambda)<\frac{1}{\Gamma(\alpha)}$.
如果 $u(t)\in C_{1-\alpha}[0,1]$ 满足
(i) $D^{\alpha}u(t)-\lambda u(t)\geq 0(0<t<1)$;
(ii) $ \lim\limits_{t\rightarrow 0^+}t^{1-\alpha}u(t)=u(1)$.
则 $u(t)\geq 0,\forall t\in (0,1]$.
3 主要结论
在本节中,始终假定以下条件成立.
$(H_1)$ $f:(0,1]\times R\longrightarrow R$ 是连续函数;
$(H_2)$ 对于$ \forall x\in C_{1-\alpha}[0,1],\ f(t,x(t))\in C_{1-\alpha}[0,1]$,
其中
$$t^{1-\alpha}f(t,x(t))|_{t=0}:=\lim\limits_{t\rightarrow 0^+}t^{1-\alpha}f(t,x(t)).
$$
值得注意的是: 当 $(H_1)-(H_2)$ 成立时,$f(t,x)$ 可以在$t=0$ 是奇异的.
易见 $(PBVP)$ 与如下问题等价
$$\left \{\begin{array}{ll}
D^{\alpha}x(t)-\lambda x(t)=f(t,x(t))-\lambda x(t),t\in J:= (0,1],\ 0<\alpha<1,\\
\lim\limits_{t\rightarrow 0^+}t^{1-\alpha}x(t)=x(1),
\end{array} \right.$$
其中参数 $\lambda\in{\Bbb R}$ 且满足 $E_{\alpha,\alpha}(\lambda)<\frac{1}{\Gamma(\alpha)}$.
对于 $x\in C_{1-\alpha}[0,1]$,定义映射 $T$ 如下
$$(Tx)(t)=\int_0^1G_{\lambda,\ \alpha}(t,s)[f(s,x(s))-\lambda x(s)]{\rm d}s,$$
其中 $G_{\lambda,\ \alpha}(t,s)$ 的定义见引理2.3. 于是 $x\in C_{1-\alpha}$
是 $(PBVP)$ 的一个解当且仅当 $x\in C_{1-\alpha}$ 是 $T$的一个不动点.
${\bf 引理3.1}\quad$ 当条件 $(H_1)-(H_2)$成立时,
算子 $T$ 映 $C_{1-\alpha}[0,1]$ 到 $C_{1-\alpha}[0,1]$ 且 $T$ 是连续的.
${\bf 证}\quad$ 结论分两步证明.
(1) 若 $x\in C_{1-\alpha}[0,1]$,则$Tx\in C_{1-\alpha}[0,1]$.
显然当 $ x\in C_{1-\alpha}[0,1]$ 时,$[f(t,x(t))-\lambda x(t)]\in C_{1-\alpha}[0,1]$.
设 $\sigma(t)=f(t,x(t))-\lambda x(t),u(t)=(Tx)(t)$,则有
$$\left \{\begin{array}{ll}
D^{\alpha}u(t)-\lambda u(t)=\sigma(t),t\in J:= (0,1],\ 0<\alpha<1,\\
\lim\limits_{t\rightarrow 0^+}t^{1-\alpha}u(t)=u(1).
\end{array} \right.$$
据文献[9],当 $\sigma\in C_{1-\alpha}[0,1]$ 且 $E_{\alpha,\alpha}(\lambda)<\frac{1}{\Gamma(\alpha)}$ 时,下述线性分数微分方程边值问题
$$\left \{\begin{array}{ll}
D^{\alpha}u(t)-\lambda u(t)=\sigma(t),t\in J:= (0,1],\ 0<\alpha<1,\\
\lim\limits_{t\rightarrow 0^+}t^{1-\alpha}x(t)=x(1)
\end{array} \right.$$
的解是
$$u(t)=u_0\Gamma(\alpha)t^{1-\alpha}E_{\alpha,\alpha}(\lambda t^{\alpha})+\int_0^t {(t-s)}^{\alpha-1}E_{\alpha,\alpha}(\lambda {(t-s)^{\alpha})}\sigma(s){\rm d}s,
$$
其中 $u_0$为一常数.
以下来证明 $u\in C_{1-\alpha}[0,1]$.
事实上,
$$t^{1-\alpha}u(t)=u_0\Gamma(\alpha)t^{2(1-\alpha)}E_{\alpha,\alpha}(\lambda t^{\alpha})+t^{1-\alpha}\int_0^t {(t-s)}^{\alpha-1}E_{\alpha,\alpha}(\lambda {(t-s)^{\alpha})}\sigma(s){\rm d}s$$
只需说明 $t^{1-\alpha}\int_0^t {(t-s)}^{\alpha-1}E_{\alpha,\alpha}(\lambda {(t-s)^{\alpha})}\sigma(s){\rm d}s$ 是 $t$的连续函数.
注意到
\begin{eqnarray*}
&&t^{1-\alpha}\int_0^t {(t-s)}^{\alpha-1}E_{\alpha,\alpha}(\lambda {(t-s)^{\alpha})}\sigma(s){\rm d}s\\
& =&\sum_{k=0}^{\infty}\frac{t^{1-\alpha}\lambda^{k\alpha}}{\Gamma(\alpha k+\alpha)}\int_0^t(t-s)^{(k+1)\alpha-1}\sigma(s){\rm d}s\\
&=&\sum_{k=0}^{\infty}\frac{t^{1-\alpha}\lambda^{k\alpha}}{\Gamma(\alpha k+\alpha)}\int_0^t(t-s)^{(k+1)\alpha-1}s^{\alpha-1}\phi(s){\rm d}s\\
&=&\sum_{k=0}^{\infty}\frac{t^{(k+1)\alpha}\lambda^{k\alpha}}{\Gamma(\alpha k+\alpha)}\int_0^1(1-u)^{(k+1)\alpha-1}u^{\alpha-1}\phi(ut){\rm d}u,
\end{eqnarray*}
其中$\phi(t)=t^{1-\alpha}\sigma(t)\in C[0,1]$. 由于 $\phi(t)$ 在 $[0,1]$ 上一致连续,即 $\forall \epsilon >0$,存在$\delta>0$,当 $t_1,t_2\in [0,1]$ 且 $|t_1- t_2|<\delta$时,
$$|\phi(t_1)-\phi(t_2)|<\epsilon.
$$
此时,
\begin{eqnarray*}
&&\bigg| \int_0^1(1-u)^{(k+1)\alpha-1}u^{\alpha-1}\phi(ut_1){\rm d}u-\int_0^1(1-u)^{(k+1)\alpha-1}u^{\alpha-1}\phi(ut_2){\rm d}u\bigg|\\
& =&\bigg| \int_0^1(1-u)^{(k+1)\alpha-1}u^{\alpha-1}(\phi(ut_1)-\phi(ut_2)){\rm d}u\bigg|\\
&\leq& \int_0^1(1-u)^{(k+1)\alpha-1}u^{\alpha-1}| \phi(ut_1)-\phi(ut_2)|{\rm d}u\\
& \leq& \epsilon \int_0^1(1-u)^{(k+1)\alpha-1}u^{\alpha-1}{\rm d}u\\
&=&\epsilon B((k+1)\alpha,\alpha).
\end{eqnarray*}
这说明对于任意的$k$，$\frac{t^{(k+1)\alpha}}{\Gamma(\alpha k+\alpha)}
\int_0^1(1-u)^{(k+1)\alpha-1}u^{\alpha-1}\phi(ut){\rm d}u$ 是 $t$ 的连续函数.
因此,$t^{1-\alpha}u(t)\in C[0,1]$.
(2) 在假设 $(H_1)-(H_2)$下,利用Lebesgue控制收敛定理可以证明 $T$ 是连续的.
设 $u,v\in C_{1-\alpha}[0,1]$ 满足 $u(t)\leq v(t),t\in
[0,1]$,记
$$[u,v]=\{x\in C_{1-\alpha}[0,1]| u(t)\leq x(t)\leq v(t),t\in
(0,1]\}.
$$
${\bf 定理3.1}\quad$ 如果存在 $u,v \in C_{1-\alpha}[0,1],u\leq v$
和常数 $\lambda$
($\lambda$ 满足$E_{\alpha,\alpha}(\lambda)<\frac{1}{\Gamma(\alpha)}$),
使得以下条件成立
$(A_1)$ 对于任意的$t\in [0,1],x,y\in [u,v],x\leq y,$
$$ f(t,y(t))-f(t,x(t))\leq \lambda (y(t)-x(t));
$$
$(A_2)$ 对于任意的$t\in [0,1],x,y\in [u,v],$
$$
f(t,lx(t)+(1-l)y(t)\leq lf(t,x(t))+(1-l)f(t,y(t));
$$
$(A_3)$ 对于任意的 $t\in (0,1)$,
$$D^{\alpha}(v+u)(t)+\lambda (v-u)(t)\leq 2f(t,v(t)),
$$
$$f(t,u(t))\leq D^{\alpha}v(t)-\lambda(v-u)(t);$$
$(A_4)$ $ \lim\limits_{t\rightarrow 0^+}t^{1-\alpha}u(t)=u(1),\ \lim\limits_{t\rightarrow 0^+}t^{1-\alpha}v(t)=v(1)$.
那么$(PBVP)$ 存在唯一解 $x^*\in C_{1-\alpha}[0,1]\cap [u,v]$.
${\bf证}\quad$
分数阶微分方程周期边值问题 $(PBVP)$ 等价于如下边值问题
$$\left \{\begin{array}{ll}
D^{\alpha}x(t)-\lambda x(t)=f(t,x(t))-\lambda x(t),0< t< 1,\\
\lim\limits_{t\rightarrow 0^+}t^{1-\alpha}x(t)=x(1).
\end{array} \right.$$
设 $x_0$ 为下述边值问题的解
$$\left \{\begin{array}{ll}
D^{\alpha}x(t)-\lambda x(t)=f(t,v(t))-\lambda v(t),\\
\lim\limits_{t\rightarrow 0^+}t^{1-\alpha}x(t)=x(1),
\end{array} \right.$$
则 $$x_0=\int_0^1
G_{\lambda,\alpha}(t,s)[f(s,v(s))-\lambda v(s)]{\rm d}s,$$
这里 $G_{\lambda,\alpha}(t,s)$ 的定义见引理 2.2.
构造迭代列 $\{x_n\}$ 如下
$$ \left\{
\begin{array}{ll}
D^{\alpha}x_{n+1}(t)-\lambda x_{n+1}(t)=f(t,x_{n}(t))-\lambda x_{n}(t),\\
\lim\limits_{t\rightarrow 0^+}t^{1-\alpha}x_{n+1}(t)=x_{n+1}(1),
\\ n=0,1,2,\cdots .
\end{array}
\right.$$
由于 $f$ 连续,所以点列
$\{x_n\}$ 有意义且有 $\{x_n\}\subset C_{1-\alpha}[0,1]$.
以下来证明 $\{x_n\}$ 收敛到问题 $(PBVP)$ 在序区间
$[u,v]$ 中的唯一解.
证明过程分为5步.
第1步) $ v(t)\geq x_0(t)\geq (\frac{u+v}{2})(t)\geq u(t). $
由定理条件 $(A1),(A3),(A4)$,可知
\begin{eqnarray*}
D^{\alpha}
\Big(\frac{u+v}{2}\Big)(t)-\lambda\Big(\frac{u+v}{2}\Big)(t)
&\leq & f(t,v(t))-\lambda v(t)\leq f(t,u(t))-\lambda u(t)
\\
& \leq& D^{\alpha}v(t)-\lambda v(t).
\end{eqnarray*}
于是,根据 $x_0$ 的定义易得
$$D^{\alpha}v(t)-\lambda v(t)\geq D^{\alpha}x_0(t)-\lambda x_0(t)\geq D^{\alpha}
\Big(\frac{u+v}{2}\Big)(t)-\lambda\Big(\frac{u+v}{2}\Big)(t).
$$
注意到
$$\lim\limits_{t\rightarrow 0^+}t^{1-\alpha}u(t)=u(1),
\lim\limits_{t\rightarrow 0^+}t^{1-\alpha}v(t)=v(1),
\lim\limits_{t\rightarrow 0^+}t^{1-\alpha}x_0(t)=x_0(1).
$$
据引理 2.5,有
$$ v(t)\geq x_0(t)\geq \Big(\frac{u+v}{2}\Big)(t)\geq u(t).
$$
第2步)对于$n=0,1,2,\cdots ,$ 成立着
$$v(t)\geq x_{2n+1}(t)\geq
x_{2n}(t)\geq \Big(\frac{u+v}{2}\Big)(t)\geq u(t).
$$
利用数学归纳法来证明结论.
事实上，由于
$$D^{\alpha}x_{1}(t)-\lambda x_{1}(t)=f(t,x_{0}(t))-\lambda x_{0}(t).$$
根据第1步证明及条件$(A1)$ 有
$$f(t,u(t))-\lambda u(t)\geq f(t,x_{0}(t))-\lambda x_{0}(t)\geq f(t,v(t))-\lambda v(t).$$
注意到
$$f(t,u(t))-\lambda u(t) \leq D^{\alpha}v(t)-\lambda v(t),
$$
所以有
$$ D^{\alpha}v(t)-\lambda v(t)\geq D^{\alpha}x_{1}(t)-\lambda x_{1}(t)\geq
D^{\alpha}x_{0}(t)-\lambda x_{0}(t). $$
根据引理 2.4,
$$v(t)\geq x_{1}(t)\geq x_{0}(t).$$
所以当 $n=0$结论成立.
设若 $n=k$ 时结论也成立,即
$$v(t)\geq x_{2k+1}(t)\geq x_{2k}(t)\geq
\Big(\frac{u+v}{2}\Big)(t)\geq u(t).$$
则由假设 $(A1)$,对于任意的 $t\in [0,1]$
\begin{eqnarray*}
f(t,u(t))-\lambda u(t)&\geq& f(t,x_{2k}(t))-\lambda x_{2k}(t)\geq
f(t,x_{2k+1}(t))-\lambda x_{2k+1}(t)\\
&\geq& f(t,v(t))-\lambda v(t).
\end{eqnarray*}
这说明
\begin{eqnarray*}
D^{\alpha}v(t)-\lambda v(t)&\geq& D^{\alpha}x_{2k+1}(t)-
\lambda x_{2k+1}(t)\geq D^{\alpha}x_{2k+2}(t)-\lambda x_{2k+2}(t)\\
&\geq &
D^{\alpha}x_{0}(t)-\lambda x_{0}(t).
\end{eqnarray*}
再根据引理 2.3,可知
$$v(t)\geq x_{2k+1}(t)\geq x_{2k+2}(t)\geq x_{0}(t)\geq u(t).$$
重复此过程,可得
$$ \left\{
\begin {array} {ll} f(t,u(t))-\lambda u(t)\geq f(t,x_{2k+2}(t))-\lambda x_{2k+2}(t)\\
\geq f(t,x_{2k+1}(t))-\lambda x_{2k+1}(t)\geq f(t,v(t))-\lambda v(t),\\
D^{\alpha}v(t)-\lambda v(t)\geq D^{\alpha}x_{2k+3}(t)-\lambda x_{2k+3}(t)\\
\geq D^{\alpha}x_{2k+2}(t)-\lambda x_{2k+2}(t)\geq D^{\alpha}x_{0}(t)-\lambda x_{0}(t),\\
v(t)\geq x_{2k+3}(t)\geq x_{2k+2}(t)\geq x_{0}(t)\geq u(t),
\end {array}
\right.$$
即当 $n=k+1$ 结论也成立.
所以对所有的 $n\in \{0\}\bigcup N$,结论都成立.
第3步) $\{x_{2n}(t)\}$ 递增,$\{x_{2n+1}(t)\}$ 递减.
由第2步证明过程可知
$$v(t)\geq x_1(t)\geq u(t).
$$
于是根据条件 $(A1)$ 得
$$f(t,u(t))-\lambda u(t)\geq f(t,x_{1}(t))-\lambda x_{1}(t)\geq f(t,v(t))-\lambda v(t).$$
这表明
$$ D^{\alpha}v(t)-\lambda v(t)\geq D^{\alpha}x_{2}(t)-\lambda x_{2}(t)\geq D^{\alpha}x_{0}(t)-\lambda x_{0}(t). $$
$$v(t)\geq x_{2}(t)\geq x_0(t) .$$
类似于步骤2,利用数学归纳法可以证明 $\{x_{2n}(t)\}$ 是递增的.
同理可证$\{x_{2n+1}(t)\}$ 是递减的.
第4步) $\{x_{n}(t)\}$ 收敛到问题 $(PBVP)$的一个解.
由步骤1--3,对于 $n=0,1,2,\cdots$,
$$u(t)\leq \Big(\frac{u+v}{2}\Big)(t)\leq
x_{2n}(t) \leq x_{2n+2}(t)\leq x_{2n+3}(t)\leq
x_{2n+1}(t) \leq v(t).$$
设 $y_n(t)=x_n(t)-u(t)$,则对 $n=0,1,2,\cdots$,
$$0\leq
\frac{v(t)-u(t)}{2}\leq y_{2n}(t) \leq
y_{2n+2}(t)\leq y_{2n+3}(t)\leq y_{2n+1}(t) \leq
v(t)-u(t).$$
令
$$r_n=\sup \{r\in R| y_{2n}(t)\geq r y_{2n+1}(t)\},
$$
则数列 $\{r_n\}$是良定的. 易验证
$\frac{1}{2}\leq r_n\leq 1$ 且 $\{r_n\}$ 递增,所以 $\{r_n\}$ 收敛.
由假设 $(A1),(A2)$,可得
\begin{eqnarray*}
\displaystyle &&y_{2n+3}(t) \leq y_{2n+1}(t) =x_{2n+1}(t)-u(t)
\\&=& \int_0^1
G_{\lambda,\alpha}(t,s)[f(s,x_{2n}(s))-\lambda x_{2n}(s)]{\rm d}s-u(t)
\\&=&\int_0^1
G_{\lambda,\alpha}(t,s)[f(s,(y_{2n}+u)(s))-\lambda (y_{2n}+u)(s)]{\rm d}s-u(t)
\\&\leq&\int_0^1
G_{\lambda,\alpha}(t,s)[f(s,(r_ny_{2n+1}+u)(s))-\lambda (r_ny_{2n+1}+u)(s)]{\rm d}s-u(t)
\\&=&
\int_0^1
G_{\lambda,\alpha}(t,s)[f(s,r_nx_{2n+1}+(1-r_n)u(s))-\lambda (r_nx_{2n+1}(s)+(1-r_n)u(s))]{\rm d}s-u(t)
\\ &\leq&\int_0^1
G_{\lambda,\alpha}(t,s)[r_nf(s,x_{2n+1}(s))+(1-r_n)f(s,u(s))
\\
&&-\lambda (r_nx_{2n+1}(s)+(1-r_n)u(s))]{\rm d}s-u(t)
\\&=&r_n\int_0^1
G_{\lambda,\alpha}(t,s)[f(s,x_{2n+1}(s))-\lambda x_{2n+1}(s)]{\rm d}s
\\
&&+(1-r_n)\int_0^1
G_{\lambda,\alpha}(t,s)[f(s,u(s))-\lambda u(s)]{\rm d}s-u(t)
\\&\leq&r_nx_{2n+2}(t)+(1-r_n)v(t)-u(t)
\\&=&r_ny_{2n+2}(t)+(1-r_n)(v(t)-u(t))
\\&=&r_ny_{2n+2}(t)+2(1-r_n)\frac{(v(t)-u(t))}{2}
\\&\leq& r_ny_{2n+2}(t)+2(1-r_n)y_{2n+2}(t)
\\&\leq&(2-r_n)y_{2n+2}(t),
\end{eqnarray*}
从而
$$r_{n+1}=\sup \{r\in R| y_{2n+3}(t)\geq r
y_{2n+2}(t)\}\geq \frac{1}{2-r_n}.$$
设 $\lim\limits_{n\rightarrow \infty}r_n=r$,上述不等式表明
$r\geq \frac{1}{2-r}$,所以 $r=1$.
对任意的偶数 $p>0$,
$$0\leq y_{2n+p}-y_{2n}\leq y_{2n+1}-y_{2n}\leq
(1-r_n)y_{2n+1}\leq (1-r_n)(v-u)(t).$$
由
$r_n\rightarrow 1$ 及锥$P$ 的正规性可知,$\{y_{2n}\}$ 是
$C_{1-\alpha}[0,1]$ 中的Cauchy列. 类似地,可以证明 $\{y_{2n+1}\}$
也是$C_{1-\alpha}[0,1]$ 中的Cauchy列. 于是
$$\lim\limits_{n\rightarrow
\infty}y_{2n}=\lim\limits_{n\rightarrow \infty}y_{2n+1},
$$
所以
$\{y_n\}$ 收敛.
设 $y^*=\lim\limits_{n\rightarrow \infty}y_{n}$,$x^*=y^*(t)+u(t)$,
则有
$$\lim\limits_{n\rightarrow
\infty}x_{n}(t)={x^*}(t).$$
以及
\begin{eqnarray*}
{x^*}(t)&=&\lim\limits_{n\rightarrow
\infty}\int_0^1
G_{\lambda,\alpha}(t,s)[f(s,x_{n}(s))-\lambda x_{n}(s)]{\rm d}s
\\
&=&\int_0^1
G_{\lambda,\alpha}(t,s)[f(s,x^*(s))-\lambda x^*(s)]{\rm d}s,
\end{eqnarray*}
这表明 $x^*$ 是问题 $(PBVP)$ 的一个解.
第5步) $x^*$ 是问题 $(PBVP)$ 在序区间 $[u,v]$中的唯一解.
设 $\overline{x}\in [u,v]$ 是 $(PBVP)$ 的一个解,则
$$
f(t,v(t))-\lambda v(t)
\leq f(t,\overline{x}(t))-\lambda \overline{x}(t)
\leq f(t,u(t))-\lambda u(t).
$$ 从而
$$
D^{\alpha}x_0(t)-\lambda x_0(t)
\leq D^{\alpha}\overline{x}(t)-\lambda \overline{x}(t)
\leq D^{\alpha}v(t)-\lambda v(t).
$$
据引理 2.4,有 $x_0(t)\leq \overline{x}(t)\leq v(t).$
因而
$$
f(t,v(t))-\lambda v(t)
\leq f(t,\overline{x}(t))-\lambda \overline{x}(t)
\leq f(t,x_0(t))-\lambda x_0(t).
$$ 这表明
$$
D^{\alpha}x_0(t)-\lambda x_0(t)
\leq D^{\alpha}\overline{x}(t)-\lambda \overline{x}(t)
\leq D^{\alpha}x_1(t)-\lambda x_1(t).
$$
由引理2.4可知 $x_0(t)\leq \overline{x}(t)\leq x_1(t).$
继续这一过程,容易验证
$$x_{2n}(t)\leq \overline{x}(t)\leq x_{2n+1}(t),n=0,1,2,\cdots .
$$
因此
$$\overline{x}(t)=\lim\limits_{n\rightarrow \infty}x_{n}(t)={x^*}(t). $$
证毕.
${\bf 注3.1}\quad$ 定理3.1是一个局部性结论,
即在一个局部其解是存在且唯一的; 但从全局来看,边值问题可能会有多个解.
${\bf 推论3.1}\quad$ 如果存在常数 $c>0,\lambda<0$ 满足
1) $\frac{\lambda c}{2}\leq t^{1-\alpha}f(t,c t^{1-\alpha})\leq t^{1-\alpha}f(t,0)\leq -\lambda c$;
2) 对于 $\forall t\in [0,1] $,$f(t,.)$为凸函数;
3) 对于 $\forall t\in [0,1],x,y\in R,x\leq y$,成立着$f(t,y)-f(t,x)\leq \lambda (y-x)$;
4) $E_{\alpha,\alpha}(\lambda)<\frac{1}{\Gamma(\alpha)}$.
则 $(PBVP)$ 有唯一解 $u^*$ 满足 $0 \geq
{u^*}(t)\geq ct^{\alpha-1},\forall t\in [0,1]$. 进一步,
如果存在 $t_0\in (0,1)$ 使得 $f(t_0,0)\neq
0$,那么 $u^*(t)>0,t\in(0,1)$.
${\bf 证}\quad$ 在推论3.1的假设下,设
$u(t)\equiv 0,v(t)=ct^{\alpha-1}$,可以验证定理3.1的条件(A1)--(A5)都满足.
于是根据定理3.1可知周期边值问题存在唯一解
$u^*$ 满足 $0 \geq
{u^*}(t)\geq ct^{\alpha-1},t\in [0,1]$.
进一步，如果存在 $t_0\in(0,1]$ 使得 $f(t_0,0)\neq 0$,那么 $ u^*(t)$
在$(0,1)$内不会恒为0,从而根据引理2.3得 $u^*(t)>0,\ t\in(0,1)$.
2015, Vol. 35 
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