一类带有p-Laplacian算子的分数阶积分边值问题的多重正解

Multiple Positive Solutions for Fractional In-tegral Boundary Value Problems with p-Laplacian Operators

摘要: 本文研究了一类带有p-Laplacian算子的分数阶积分边值问题正解的存在性，通过研究格林函数的性质，运用锥拉伸锥压缩不动点定理以及Leggett-Williams不动点定理，获得了该边值问题至少存在一个正解及三个正解的充分条件，并给出实例验证所得结论。

Abstract: Studying the properties of Green’s function and using the cone stretching cone compression fixed point theorem and the Leggett-Williams fixed point theorem, this paper studies the existence of positive solutions for a class of fractional integral boundary value problems with p-Laplacian oper-ators, obtains sufficient conditions for the existence of at least one positive solution and three posi-tive solutions to the boundary value problems, and gives some examples illustrating the results obtained.

1. 引言

分数阶微积分是由整数阶微积分推广而来的，已广泛应用于控制系统、空气动力学、流体力学等众多工程分支中。因此，分数阶微分方程理论研究近年来得到了快速发展，研究成果非常丰富 [1] [2] [3] [4] [5] 。其中：文 [4] 运用锥理论得到了一类分数阶无穷点边值问题存在一个局部正解以及多个局部正解的充分条件；文 [5] 利用Schauder不动点定理及凹算子不动点定理建立了高阶分数阶多点边值问题正解的存在性及唯一正解的存在性结果。

带有p-Laplacian算子的方程模型能更好地描述多孔介质中的湍流、弹性理论等诸多领域中的实际问题，目前，带有p-Laplacian算子的分数阶微分方程受到众多学者的关注 [6] [7] [8] [9] [10] 。其中：文 [9] 研究了如下带有p-Laplacian算子的分数阶微分方程

$\{\begin{array}{l}{D}_{0+}^{\beta }\left({\varphi }_p\left(D_{0+}^{\alpha }u\left(t\right)\right)\right)+f\left(t,u\left(t\right)\right)=0,0<t<1,\\ u\left(0\right)=u^{\prime }\left(0\right)=u\left(1\right)=D_{0+}^{\alpha }u\left(0\right)=0,\\ D_{0+}^{\alpha }u\left(1\right)=\lambda D_{0+}^{\alpha }u\left(\xi \right),\end{array}$

其中： $D_{0+}^{\alpha }$ 和 $D_{0+}^{\beta }$ 是标准的Riemann-Liouville分数阶导数， $2<\alpha \le 3$ ， $1<\beta \le 2$ ， $0<\xi <1$ ， $\lambda \ge 0$ ， ${\varphi }_p\left(s\right)={\left|s\right|}^{p-2}s$ ， $p>1$ ， ${\varphi }_p^{-1}={\varphi }_q$ ， $1/p+1/q=1$ 。作者应用锥上的不动点定理获得了上述边值问题至少存在一个和两个正解的充分条件。文 [10] 研究了如下带有p-Laplacian算子的分数阶积分边值问题

$\{\begin{array}{l}-D_{0+}^{\alpha }\left({\varphi }_p\left(-D_{0+}^{\beta }x\left(t\right)\right)\right)=f\left(t,x\left(t\right),D_{0+}^{\beta }x\left(t\right)\right),0<t<1,\\ D_{0+}^{\beta }x(t)=0,\begin{array}{cc}& \end{array}{\left({\varphi }_p\left(-D_{0+}^{\beta }x\left(0\right)\right)\right)}^{\prime }=0,\\ D_{0+}^{\gamma }\left({\varphi }_p\left(-D_{0+}^{\beta }x\left(1\right)\right)\right)=I_{0+}^v\left({\varphi }_p\left(-D_{0+}^{\beta }x\left(\eta \right)\right)\right),\\ x\left(0\right)=0,\begin{array}{cc}& \end{array}{D}_{0+}^{\beta -1}x\left(1\right)=I_{0+}^{w}g\left(\xi ,x\left(\xi \right)\right)+k,\end{array}$

其中： $D_{0+}^{\alpha }$ ， $D_{0+}^{\beta }$ 和 $D_{0+}^{\gamma }$ 是标准的Riemann-Liouville分数阶导数， $I_{0+}^v$ ， $I_{0+}^w$ 是标准的Riemann-Liouville分数阶积分， $0<\gamma <1<\beta <2<\alpha <3$ ， $v,w>0$ ， $0<\eta ,\xi <1$ ， $k\in R$ ， $f\in C\left(\left[0,1\right]\times R\times R,R\right)$ ， $g\in C\left(\left[0,1\right]\times R,R\right)$ ， ${\varphi }_p\left(s\right)={\left|s\right|}^{p-2}s$ ， $p>1$ ， ${\varphi }_p^{-1}={\varphi }_q$ ， $1/p+1/q=1$ 。作者应用上下解方法和单调迭代方法得到了上述边值问题极值解的存在性的充分条件。

受上述文献启发，本文讨论如下带有p-Laplacian算子的分数阶积分边值问题

$\{\begin{array}{l}-D_{0+}^{\alpha }\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(t\right)\right)\right)=\lambda f\left(t,u\left(t\right),\left(Tu\right)\left(t\right)\right),0<t<1,\\ u\left(0\right)=D_{0+}^{\beta }u\left(0\right)=D_{0+}^{\beta -1}u\left(1\right)={\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(0\right)\right)\right)}^{\prime }=0,\\ D_{0+}^{\gamma }\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(1\right)\right)\right)=I_{0+}^w\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(\eta \right)\right)\right),\end{array}$ (1)

其中： $D_{0+}^{\alpha }$ ， $D_{0+}^{\beta }$ ， $D_{0+}^{\gamma }$ 是Riemann-Liouville分数阶导数， $I_{0+}^w$ 是Riemann-Liouville分数阶积分， $0<\gamma <1<\beta <2<\alpha <3$ ， $0<\eta <1$ ， $\lambda ,w>0$ ， $f\in C\left(\left(0,1\right)\times \left[0,\infty \right)\times \left[0,\infty \right),\left[0,\infty \right)\right)$ ， $\left(Tu\right)\left(t\right)={\displaystyle {\int }_0^{t}K\left(t,s\right)u\left(s\right)\text{d}s}$ ，且 $K\in C\left(D,\left[0,+\infty \right)\right)$ ， $D=\left\{\left(t,s\right)\in \left[0,1\right]\times \left[0,1\right]:t\ge s\right\}$ ， ${\varphi }_p\left(s\right)={\left|s\right|}^{p-2}s$ ， $p>1$ ， ${\varphi }_p^{-1}={\varphi }_q$ ， $1/p+1/q=1$ 。本文运用锥拉伸锥压缩不动点定理以及Leggett-Williams不动点定理研究该边值问题正解的存在性。

2. 预备知识和引理

定义1 [1] 函数 $f:\left(0,+\infty \right)\to R$ 的 $\alpha >0$ 阶Riemann-Liouville分数阶积分定义为

$I_{0^{+}}^{\alpha }f\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}}f\left(s\right)\text{d}s$ ，

其中：等式的右端在 $\left(0,+\infty \right)$ 有定义。

定义2 [1] 连续函数 $f:\left(0,+\infty \right)\to R$ 的 $\alpha >0$ 阶Riemann-Liouville分数阶导数定义为

$D_{0^{+}}^{\alpha }f\left(t\right)=\frac{1}{\Gamma \left(n-\alpha \right)}{\left(\frac{\text{d}}{\text{d}t}\right)}^n{\displaystyle {\int }_0^t\frac{f\left(s\right)}{{\left(t-s\right)}^{\alpha -n+1}}\text{d}s}$ ，

其中：等式的右端在 $\left(0,+\infty \right)$ 有定义， $n=\min \left\{m\in Z:m\ge \alpha \right\}$ ， $\Gamma (\cdot )$ 为Gamma函数。

引理1 [1] 设 $\alpha >0$ ， $u\in C\left(0,1\right)\cap L\left(0,1\right)$ ， $D_{0+}^{\alpha }u\in C\left(0,1\right)\cap L\left(0,1\right)$ ，则

$I_{0+}^{\alpha }{D}_{0+}^{\alpha }u\left(t\right)=u\left(t\right)+c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2}+\cdots +c_n{t}^{\alpha -n}$ ，

其中： $n=\min \left\{m\in Z:m\ge \alpha \right\}$ ， $c_i\in R$ ， $i=1,2,\cdots ,n$ 。

引理2 若 $y\left(t\right)\in C\left[0,1\right]$ ，则分数阶微分方程边值问题

$\{\begin{array}{l}-D_{0+}^{\alpha }\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(t\right)\right)\right)=y\left(t\right),0<t<1,\\ u\left(0\right)=D_{0+}^{\beta }u\left(0\right)=D_{0+}^{\beta -1}u\left(1\right)={\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(0\right)\right)\right)}^{\prime }=0,\\ D_{0+}^{\gamma }\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(1\right)\right)\right)=I_{0+}^w\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(\eta \right)\right)\right),\end{array}$

有唯一解

$u\left(t\right)={\displaystyle {\int }_0^{1}G\left(t,s\right){\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)y\left(\tau \right)\text{d}\tau }\right)\text{d}s}$ .

其中：

$G\left(t,s\right)=\frac{1}{\Gamma \left(\beta \right)}\{\begin{array}{ll}{t}^{\beta -1}-{\left(t-s\right)}^{\beta -1},\hfill & 0\le s\le t\le 1,\hfill \\ t^{\beta -1},\hfill & 0\le t\le s\le 1,\hfill \end{array}$

$H\left(s,\tau \right)=\frac{1}{\Lambda }\{\begin{array}{ll}\begin{array}{l}{\left(s-\tau \right)}^{\alpha -1}\left[\Gamma \left(\alpha +w\right)-\Gamma \left(\alpha -\gamma \right){\eta }^{\alpha +w-1}\right]\\ \begin{array}{cc}& \end{array}+\left[\Gamma \left(\alpha +w\right){\left(1-\tau \right)}^{\alpha -\gamma -1}-\Gamma \left(\alpha -\gamma \right){\left(\eta -\tau \right)}^{\alpha +w-1}\right]s^{\alpha -1},\end{array}\hfill & 0\le \tau \le \eta ,s\le 1,\hfill \\ \left[\Gamma \left(\alpha +w\right){\left(1-\tau \right)}^{\alpha -\gamma -1}-\Gamma \left(\alpha -\gamma \right){\left(\eta -\tau \right)}^{\alpha +w-1}\right]s^{\alpha -1},\hfill & 0\le s\le \tau \le \eta \le 1,\hfill \\ {\left(s-\tau \right)}^{\alpha -1}\left[\Gamma \left(\alpha +w\right)-\Gamma \left(\alpha -\gamma \right){\eta }^{\alpha +w-1}\right]+\Gamma \left(\alpha +w\right){\left(1-\tau \right)}^{\alpha -\gamma -1}{s}^{\alpha -1},\hfill & 0\le \eta \le \tau \le s\le 1,\hfill \\ \Gamma \left(\alpha +w\right){\left(1-\tau \right)}^{\alpha -\gamma -1}{s}^{\alpha -1},\hfill & 0\le \eta ,s\le \tau \le 1,\hfill \end{array}$

其中： $\Lambda =\Gamma \left(\alpha \right)\left[\Gamma \left(\alpha +w\right)-\Gamma \left(\alpha -\gamma \right){\eta }^{\alpha +w-1}\right]$ 。

证明 由 ${\varphi }_p^{-1}={\varphi }_q$ 以及引理1可得， $-D_{0+}^{\alpha }\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(t\right)\right)\right)=y\left(t\right)$ 等价于

${\varphi }_p\left(-D_{0+}^{\beta }u\left(t\right)\right)=-I_{0+}^{\alpha }y\left(t\right)+c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2}+c_3{t}^{\alpha -3}$ ， $c_1,c_2,c_3\in R$ ，

由边界条件 $D_{0+}^{\beta }u\left(0\right)=0$ ， ${\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(0\right)\right)\right)}^{\prime }=0$ 可得 $c_2=c_3=0$ 。故

${\varphi }_p\left(-D_{0+}^{\beta }u\left(t\right)\right)=-I_{0+}^{\alpha }y\left(t\right)+c_1{t}^{\alpha -1}$ (2)

对(2)式两边求 $\gamma$ 阶导数得

$D_{0+}^{\gamma }\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(t\right)\right)\right)=-\frac{1}{\Gamma \left(\alpha -\gamma \right)}{\displaystyle {\int }_0^t{\left(t-\tau \right)}^{\alpha -\gamma -1}}y\left(\tau \right)\text{d}\tau +c_1\frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{t}^{\alpha -\gamma -1}$

对(2)式两边求w阶积分得

$I_{0+}^w\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(t\right)\right)\right)=-\frac{1}{\Gamma \left(\alpha +w\right)}{\displaystyle {\int }_0^t{\left(t-\tau \right)}^{\alpha +w-1}}y\left(\tau \right)\text{d}\tau +c_1\frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha +w\right)}{t}^{\alpha +w-1}$

由边界条件 $D_{0+}^{\gamma }\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(1\right)\right)\right)=I_{0+}^w\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(\eta \right)\right)\right)$ 可得

$c_1=\frac{\Gamma \left(\alpha +w\right){\displaystyle {\int }_0^1{\left(1-\tau \right)}^{\alpha -\gamma -1}y\left(\tau \right)\text{d}\tau }-\Gamma \left(\alpha -\gamma \right){\displaystyle {\int }_0^{\eta }{\left(\eta -\tau \right)}^{\alpha +w-1}y\left(\tau \right)\text{d}\tau }}{\Gamma \left(\alpha \right)\left[\Gamma \left(\alpha +w\right)-\Gamma \left(\alpha -\gamma \right){\eta }^{\alpha +w-1}\right]}$ (3)

将(3)式代入(2)式可得

${\varphi }_p\left(-D_{0+}^{\beta }u\left(t\right)\right)=-\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-\tau \right)}^{\alpha -1}y\left(\tau \right)\text{d}\tau }+c_1{t}^{\alpha -1}={\displaystyle {\int }_0^{1}H\left(t,\tau \right)y\left(\tau \right)\text{d}\tau }$ (4)

令 ${\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(t,\tau \right)y\left(\tau \right)\text{d}\tau }\right)=h\left(t\right)$ ，可得 $D_{0+}^{\beta }u\left(t\right)=-h\left(t\right)$ ，再由引理1可得

$\begin{array}{c}u\left(t\right)=-I_{0+}^{\beta }h\left(t\right)+d_1{t}^{\beta -1}+d_2{t}^{\beta -2}\\ =-\frac{1}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\beta -1}}h\left(s\right)\text{d}s+d_1{t}^{\beta -1}+d_2{t}^{\beta -2},d_1,d_2\in R.\end{array}$ (5)

由边界条件 $u\left(0\right)=0$ 得 $d_2=0$ 。再由定义2对(5)式两边求 $\beta -1$ 阶导数可得

$D_{0+}^{\beta -1}u\left(t\right)=-D_{0+}^{\beta -1}{I}_{0+}^{\beta }h\left(t\right)+d_1{D}_{0+}^{\beta -1}{t}^{\beta -1}=-{\displaystyle {\int }_0^{t}h\left(s\right)\text{d}s}+d_1\Gamma \left(\beta \right)$ .

由边界条件 $D_{0+}^{\beta -1}u\left(1\right)=0$ 得

$d_1=\frac{{\displaystyle {\int }_0^{1}h\left(s\right)\text{d}s}}{\Gamma (\; \beta \; )}$

于是边值问题有唯一解

$\begin{array}{c}u\left(t\right)=-\frac{1}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\beta -1}}h\left(s\right)\text{d}s+d_1{t}^{\beta -1}\\ =-\frac{1}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\beta -1}}h\left(s\right)\text{d}s+\frac{1}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^1{t}^{\beta -1}}h\left(s\right)\text{d}s\\ ={\displaystyle {\int }_0^{1}G\left(t,s\right)h\left(s\right)\text{d}s}.\end{array}$ (6)

证毕。

引理3 函数 $G\left(t,s\right)$ ， $H\left(t,s\right)$ 满足如下性质：

1) 对任意的 $t,s\in \left[0,1\right]$ ， $G\left(t,s\right)\ge 0$ ， $H\left(t,s\right)\ge 0$ ；

2) $\underset{t\in \left[0,1\right]}{\max }G\left(t,s\right)=G\left(s,s\right)=\frac{1}{\Gamma \left(\beta \right)}{s}^{\beta -1}\le \frac{1}{\Gamma \left(\beta \right)}$ ， $s\in \left[0,1\right]$ ；

3) $\underset{t\in \left[1/4,3/4\right]}{\min }G\left(t,s\right)=:m\left(s\right)=\frac{1}{\Gamma \left(\beta \right)}\{\begin{array}{ll}{\left(\frac{3}{4}\right)}^{\beta -1}-{\left(\frac{3}{4}-s\right)}^{\beta -1},\hfill & s\in \left[0,\frac{1}{4}\right];\hfill \\ \min \left\{{\left(\frac{3}{4}\right)}^{\beta -1}-{\left(\frac{3}{4}-s\right)}^{\beta -1},\frac{1}{4^{\beta -1}}\right\},\hfill & s\in \left(\frac{1}{4},\frac{3}{4}\right);\hfill \\ \frac{1}{4^{\beta -1}},\hfill & s\in \left[\frac{3}{4},1\right].\hfill \end{array}$

证明 1) 由函数 $G\left(t,s\right)$ 的表达式易知，对任意的 $t,s\in \left[0,1\right]$ ， $G\left(t,s\right)\ge 0$ 。

对于函数 $H\left(t,s\right)$ ，注意到 $\alpha -\gamma -1<\alpha +w-1$ 以及 $1<\alpha -\gamma <2<\alpha +w$ ，则有

$\Gamma \left(\alpha -\gamma \right)<1<\Gamma \left(\alpha +w\right)$

${\left(1-s\right)}^{\alpha -\gamma -1}>{\left(\eta -s\right)}^{\alpha -\gamma -1}>{\left(\eta -s\right)}^{\alpha +w-1}$

进而有 $\Gamma \left(\alpha +w\right)-\Gamma \left(\alpha -\gamma \right){\eta }^{\alpha +w-1}>\Gamma \left(\alpha +w\right)-\Gamma \left(\alpha -\gamma \right)>0$ ，以及

$\Gamma \left(\alpha +w\right){\left(1-s\right)}^{\alpha -\gamma -1}-\Gamma \left(\alpha -\gamma \right){\left(\eta -s\right)}^{\alpha +w-1}\ge \left[\Gamma \left(\alpha +w\right)-\Gamma \left(\alpha -\gamma \right)\right]{\left(\eta -s\right)}^{\alpha +w-1}>0$

进而容易得到对任意的 $t,s\in \left[0,1\right]$ ， $H\left(t,s\right)\ge 0$ 。

2) 对于函数 $G\left(t,s\right)$ ，通过简单计算，可得当 $0<t\le s$ 时，

$\frac{\partial }{\partial t}G\left(t,s\right)=\frac{1}{\Gamma \left(\beta \right)}\left(\beta -1\right)t^{\beta -2}>0$

当 $0\le s<t$ 时，

$\frac{\partial }{\partial t}G\left(t,s\right)=\frac{1}{\Gamma \left(\beta \right)}\left(\beta -1\right)\left[t^{\beta -2}-{\left(t-s\right)}^{\beta -2}\right]<0$

即 $G\left(t,s\right)$ 在 $t\le s$ 上是关于t的增函数，在 $s<t$ 上是关于t的减函数。进而有

$\underset{t\in \left[0,1\right]}{\max }G\left(t,s\right)=G\left(s,s\right)=\frac{1}{\Gamma \left(\beta \right)}{s}^{\beta -1}\le \frac{1}{\Gamma \left(\beta \right)}$ , $s\in \left[0,1\right]$

3) 令

$g_1\left(t,s\right)=\frac{t^{\beta -1}-{\left(t-s\right)}^{\beta -1}}{\Gamma \left(\beta \right)}$ , $g_2\left(t,s\right)=\frac{t^{\beta -1}}{\Gamma (\; \beta \; )}$

可得

$\begin{array}{c}\underset{t\in \left[1/4,3/4\right]}{\min }G\left(t,s\right)=\{\begin{array}{ll}{g}_1\left(\frac{3}{4},s\right),\hfill & s\in \left[0,\frac{1}{4}\right];\hfill \\ \min \left\{g_1\left(\frac{3}{4},s\right),g_2\left(\frac{1}{4},s\right)\right\},\hfill & s\in \left(\frac{1}{4},\frac{3}{4}\right);\hfill \\ g_2\left(\frac{1}{4},s\right),\hfill & s\in \left[\frac{3}{4},1\right];\hfill \end{array}\\ =\frac{1}{\Gamma \left(\beta \right)}\{\begin{array}{ll}{\left(\frac{3}{4}\right)}^{\beta -1}-{\left(\frac{3}{4}-s\right)}^{\beta -1},\hfill & s\in \left[0,\frac{1}{4}\right];\hfill \\ \min \left\{{\left(\frac{3}{4}\right)}^{\beta -1}-{\left(\frac{3}{4}-s\right)}^{\beta -1},\frac{1}{4^{\beta -1}}\right\},\hfill & s\in \left(\frac{1}{4},\frac{3}{4}\right);\hfill \\ \frac{1}{4^{\beta -1}},\hfill & s\in \left[\frac{3}{4},1\right];\hfill \end{array}\\ =:m\left(s\right),\end{array}$

证毕。

设E是实Banach空间， $\theta$ 为E中的零元素。P为E中的非空凸闭子集，若

$x\in P,\lambda \ge 0\Rightarrow \lambda x\in P;x\in P,-x\in P\Rightarrow x=\theta ,$

则称P为E中的锥。

引理4 [3] 设E是一个Banach空间，P为E中的一个锥， ${\Omega }_1,{\Omega }_2$ 是E中的两个有界开集，并且 ${\Omega }_1\subset {\Omega }_2$ 。假设 $T:P\cap \left({\Omega }_2\backslash {\Omega }_1\right)\to P$ 是全连续算子，若下列条件二者成立其一：

(K1) $\Vert Tx\Vert \le \Vert x\Vert$ ， $x\in P\cap \partial {\Omega }_1$ ； $\Vert Tx\Vert \ge \Vert x\Vert$ ， $x\in P\cap \partial {\Omega }_2$

(K2) $\Vert Tx\Vert \ge \Vert x\Vert$ ， $x\in P\cap \partial {\Omega }_1$ ； $\Vert Tx\Vert \le \Vert x\Vert$ ， $x\in P\cap \partial {\Omega }_2$

那么T在 $P\cap \left({\Omega }_2\backslash {\Omega }_1\right)$ 中至少有一个不动点。

定义3 [3] 设P为Banach空间E中的锥。若映射 $\psi :P\to \left[0,\infty \right)$ 连续且 $\forall x,y\in P$ ， $\forall \eta \in \left[0,1\right]$ ，有 $\psi \left(\eta x+\left(1-\eta \right)y\right)\ge \eta \psi \left(x\right)+\left(1-\eta \right)\psi \left(y\right)$ ，则称 $\psi$ 是P上非负连续凹泛函。

引理5 [3] 设常数 $0<a<b<d\le c$ ， $T:\overline{P_c}\to \overline{P_c}$ 是全连续算子， $\psi$ 是P上的一个非负连续凹泛函，且对任意的 $x\in P_c$ ，有 $\psi \left(x\right)\le \Vert x\Vert$ 。又设

(L1) $\left\{x\in P\left(\psi ,b,d\right)|\psi \left(x\right)>b\right\}\ne \varnothing$ 且对 $x\in P\left(\psi ,b,d\right)$ ，有 $\psi \left(Tx\right)>b$ ；

(L2) 当 $x\in \overline{P_a}$ 时， $\Vert Tx\Vert <a$ ；

(L3) 当 $x\in P\left(\psi ,b,c\right)$ 且 $\Vert Tx\Vert >d$ 时， $\psi \left(Tx\right)>b$ ．

则算子T至少有三个不动点 $x_1,x_2,x_3$ ，满足 $\Vert x_1\Vert <a$ ， $b<\psi \left(x_2\right)$ ， $\Vert x_3\Vert >a$ ， $\psi \left(x_3\right)<b$ 。

3. 主要结果

设 $E=C\left[0,1\right]$ 。对任意的 $u\in E$ ，定义其范数为

$\Vert u\Vert =\underset{t\in \left[0,1\right]}{\max }\left|u\left(t\right)\right|$

容易证明E为Banach空间。再令

$P=\left\{u\in E|u\left(t\right)\ge 0,t\in \left[0,1\right]\right\}$

则P是E中的锥。定义泛函

$\psi \left(u\right)=\underset{t\in \left[1/4,3/4\right]}{\min }u\left(t\right)$ ，

则它是P中的非负连续凹泛函。定义算子A如下：对任意的 $u\in P$ ，

$\left(Au\right)\left(t\right)={\displaystyle {\int }_0^{1}G\left(t,s\right){\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s$ , $t\in \left[0,1\right]$ (7)

引理6 算子 $A:P\to P$ 是全连续算子。

证明 根据引理3的1)以及函数f的非负连续性可得，对任意的 $u\in P$ 有

$\left(Au\right)\left(t\right)={\displaystyle {\int }_0^{1}G\left(t,s\right){\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s\ge 0$ , $t\in \left[0,1\right]$

所以 $A\left(P\right)\subset P$ 。以下证明A为紧算子。设 $\Omega$ 是锥P中的有界集，即存在 $R_1>0$ ，使得

$\Vert x\Vert \le R_1$ , $\forall x\in \Omega$

设 $W_{R_1}=\sup \left\{f\left(t,u_1,u_2\right):\left(t,u_1,u_2\right)\in \left(\left[0,1\right]\times \left[0,R_1\right]\times \left[0,k^{*}{R}_1\right]\right)\right\}$ ，其中

$k^{*}=\underset{t\in \left[0,1\right]}{\sup }{\displaystyle {\int }_0^{t}K\left(t,s\right)\text{d}s}$

进而对任意的 $u\in \Omega$ ，由引理3的2)得

$\begin{array}{c}\left|\left(Au\right)\left(t\right)\right|=\left|{\displaystyle {\int }_0^{1}G\left(t,s\right){\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s\right|\\ \le \frac{1}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^1{\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s\\ \le \frac{{\varphi }_q\left(\lambda W_{R_1}\right)}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^1{\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\text{d}\tau }\right)\text{d}s,\end{array}$

由此可知 $A\left(\Omega \right)$ 为E中一致有界的子集合。

再证 $A\left(\Omega \right)$ 为E中等度连续的子集合。对任意的 $0\le t_1<t_2\le 1$ ， $u\in \Omega$ 有

$\begin{array}{c}\left|\left(Au\right)\left(t_2\right)-\left(Au\right)\left(t_1\right)\right|\le {\displaystyle {\int }_0^1\left|G\left(t_2,s\right)-G\left(t_1,s\right)\right|{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s}\\ \le {\varphi }_q\left(\lambda W_{R_1}\right){\displaystyle {\int }_0^1\left|G\left(t_2,s\right)-G\left(t_1,s\right)\right|{\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\text{d}\tau }\right)\text{d}s,\end{array}$

又 $G\left(t,s\right)$ 在 $\left[0,1\right]\times \left[0,1\right]$ 上连续，从而一致连续，故 $A\left(\Omega \right)$ 为E中等度连续的子集合。进而集合 $A\left(\Omega \right)$ 是一致有界且等度连续的，根据Arzela-Ascoli定理可知 $A\left(\Omega \right)$ 是E中的相对紧集。

最后证明 $A:P\to P$ 是连续算子。设 $\left\{u_n\right\}\subset \Omega$ ， $u\in \Omega$ ，当 $n\to +\infty$ 时，有 $\Vert u_n-u\Vert \to 0$ ，故存在常数 $R_0>1$ ，使得 $\Vert u\Vert <R_0$ ， $\Vert u_n\Vert <R_0$ ， $n=1,2,\cdots$ 。于是对 $s\in \left[0,1\right]$ ，

$\left|{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u_n\left(\tau \right),\left(T{u}_n\right)\left(\tau \right)\right)\text{d}\tau }\right)-{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\right|\to 0,n\to +\infty .$

$\begin{array}{l}|{\displaystyle {\int }_0^1{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u_n\left(\tau \right),\left(T{u}_n\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s}-{\displaystyle {\int }_0^1{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s}|\\ \le 2{\varphi }_q\left(\lambda W_{R_0}\right){\displaystyle {\int }_0^1{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\text{d}\tau }\right)\text{d}s}<+\infty ,\end{array}$

其中 $W_{R_0}=\sup \left\{f\left(t,u_1,u_2\right):\left(t,u_1,u_2\right)\in \left(\left[0,1\right]\times \left[0,R_0\right]\times \left[0,k^{*}{R}_0\right]\right)\right\}$ 。由Lebesgue控制收敛定理可知，

$|{\displaystyle {\int }_0^1{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u_n\left(\tau \right),\left(T{u}_n\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s}-{\displaystyle {\int }_0^1{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s}|\to 0,n\to +\infty .$

进而根据引理3的2)得

$\begin{array}{c}\left|\left(A{u}_n\right)\left(t\right)-\left(Au\right)\left(t\right)\right|=|{\displaystyle {\int }_0^{1}G\left(t,s\right)}[{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u_n\left(\tau \right),\left(T{u}_n\right)\left(\tau \right)\right)\text{d}\tau }\right)\\ -{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)]\text{d}s|\\ \le \frac{1}{\Gamma \left(\beta \right)}|{\displaystyle {\int }_0^1{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u_n\left(\tau \right),\left(T{u}_n\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s}\\ -{\displaystyle {\int }_0^1{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s}|\\ \to 0,n\to +\infty .\end{array}$

则 $A:P\to P$ 是连续算子。证毕。

为了方便叙述，记：

$M={\left[\frac{1}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^1{\varphi }_q}\left(\lambda {\displaystyle {\int }_0^{1}H\left(s,\tau \right)\text{d}\tau }\right)\text{d}s\right]}^{-1}$

$N={\left[{\displaystyle {\int }_0^{1}m\left(s\right){\varphi }_q}\left(\lambda {\displaystyle {\int }_{1/4}^{3/4}H\left(s,\tau \right)\text{d}\tau }\right)\text{d}s\right]}^{-1}$

定理1 若存在两个常数 $r_2>r_1>0$ ，使得

(H1) $\forall \left(t,u,v\right)\in \left[1/4,3/4\right]\times \left[0,r_1\right]\times \left[0,k^{*}{r}_1\right]$ ， $f\left(t,u,v\right)\ge {\left(N{r}_1\right)}^{p-1}$ ，

(H2) $\forall \left(t,u,v\right)\in \left[0,1\right]\times \left[0,r_2\right]\times \left[0,k^{*}{r}_2\right]$ ， $f\left(t,u,v\right)\le {\left(M{r}_2\right)}^{p-1}$ ，

则边值问题(1)至少有一个正解u，使得 $r_1\le \Vert u\Vert \le r_2$ 。

证明 令 ${\Omega }_1=\left\{u\in P|\Vert u\Vert <r_1\right\}$ ，当 $u\in \partial {\Omega }_1$ 时，有 $0\le u\left(t\right)\le r_1$ ， $0\le \left(Tu\right)\left(t\right)\le k^{*}{r}_1$ ， $t\in \left[0,1\right]$ 。由(H1)及引理3的3)可得

$\begin{array}{c}\left(Au\right)\left(t\right)={\displaystyle {\int }_0^{1}G\left(t,s\right){\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s\\ \ge {\displaystyle {\int }_0^{1}G\left(t,s\right){\varphi }_q}\left({\displaystyle {\int }_{1/4}^{3/4}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s\\ \ge {\displaystyle {\int }_0^1\underset{t\in \left[1/4,3/4\right]}{\min }G\left(t,s\right){\varphi }_q}\left({\displaystyle {\int }_{1/4}^{3/4}H\left(s,\tau \right)\lambda {\varphi }_p\left(N{r}_1\right)\text{d}\tau }\right)\text{d}s\\ =N{r}_1{\displaystyle {\int }_0^{1}m\left(s\right){\varphi }_q}\left(\lambda {\displaystyle {\int }_{1/4}^{3/4}H\left(s,\tau \right)\text{d}\tau }\right)\text{d}s\\ =r_1=\Vert u\Vert ,\forall t\in \left[1/4,3/4\right],\end{array}$

因而当 $u\in \partial {\Omega }_1$ 时，有 $\Vert Au\Vert \ge \Vert u\Vert$ 。

另一方面，令 ${\Omega }_2=\left\{u\in P|\Vert u\Vert <r_2\right\}$ ，当 $u\in \partial {\Omega }_2$ 时，有 $0\le u\left(t\right)\le r_2$ ， $0\le \left(Tu\right)\left(t\right)\le k^{*}{r}_2$ ， $t\in \left[0,1\right]$ 。由(H2)及引理3的2)可得

$\begin{array}{c}\Vert Au\Vert =\underset{t\in \left[0,1\right]}{\max }\left|\left(Au\right)\left(t\right)\right|\\ =\underset{t\in \left[0,1\right]}{\max }\left|{\displaystyle {\int }_0^{1}G\left(t,s\right){\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s\right|\\ \le \frac{1}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^1{\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda {\varphi }_p\left(M{r}_2\right)\text{d}\tau }\right)\text{d}s\\ =\frac{M{r}_2}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^1{\varphi }_q}\left(\lambda {\displaystyle {\int }_0^{1}H\left(s,\tau \right)\text{d}\tau }\right)\text{d}s\\ =r_2=\Vert u\Vert ,\forall t\in \left[0,1\right],\end{array}$

因而当 $u\in \partial {\Omega }_2$ 时，有 $\Vert Au\Vert \le \Vert u\Vert$ 。

由引理4可知算子A至少有一个不动点u，即边值问题(1)至少有一个正解且满足 $r_1\le \Vert u\Vert \le r_2$ 。证毕。

定理2 假设存在正常数且满足 $0<a<b<c$ ，使得：

(H3) 当 $\left(t,u,v\right)\in \left[0,1\right]\times \left[0,a\right]\times \left[0,k^{*}a\right]$ 时， $f\left(t,u,v\right)\le {\left(Ma\right)}^{p-1}$ ；

(H4) 当 $\left(t,u,v\right)\in \left[1/4,3/4\right]\times \left[b,c\right]\times \left[0,k^{*}c\right]$ 时， $f\left(t,u,v\right)\ge {\left(Nb\right)}^{p-1}$ ；

(H5) 当 $\left(t,u,v\right)\in \left[0,1\right]\times \left[0,c\right]\times \left[0,k^{*}c\right]$ 时， $f\left(t,u,v\right)\le {\left(Mc\right)}^{p-1}$ 。

则边值问题(1)至少有三个正解 $u_1,u_2,u_3$ ，且满足

$\underset{t\in \left[0,1\right]}{\max }\left|u_1\left(t\right)\right|<a$ ， $b<\underset{t\in \left[1/4,3/4\right]}{\min }\left|u_2\left(t\right)\right|$ ， $a<\underset{t\in \left[0,1\right]}{\max }\left|u_3\left(t\right)\right|$ ， $\underset{t\in \left[1/4,3/4\right]}{\min }\left|u_3\left(t\right)\right|<b$

证明 若 $u\in \overline{P_c}$ ，则 $\Vert u\Vert \le c$ ，由假设(H5)可得

$\begin{array}{c}\Vert Au\Vert =\underset{t\in \left[0,1\right]}{\max }\left|\left(Au\right)\left(t\right)\right|\\ =\underset{t\in \left[0,1\right]}{\max }\left|{\displaystyle {\int }_0^{1}G\left(t,s\right){\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s\right|\\ \le \frac{1}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^1{\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda {\varphi }_p\left(Mc\right)\text{d}\tau }\right)\text{d}s\\ =\frac{Mc}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^1{\varphi }_q}\left(\lambda {\displaystyle {\int }_0^{1}H\left(s,\tau \right)\text{d}\tau }\right)\text{d}s\\ =c,\end{array}$

从而说明对任意的 $u\in \overline{P_c}$ ， $\Vert Au\Vert \le c$ ，故有 $A\left(\overline{P_c}\right)\subset \overline{P_c}$ 。同理，若 $u\in \overline{P_a}$ ，由假设(H3)可得 $\Vert Au\Vert \le a$ ，所以引理5中的条件(L2)成立。

令 $u\left(t\right)=\frac{b+c}{2}$ ， $0\le t\le 1$ ，则 $u\left(t\right)=\frac{b+c}{2}<c$ 且 $\psi \left(u\left(t\right)\right)=\underset{t\in \left[1/4,3/4\right]}{\min }\frac{b+c}{2}>b$ ，因而 $u\left(t\right)=\frac{b+c}{2}\in P\left(\psi ,b,c\right)$ ，

即 $\left\{u\in P\left(\psi ,b,c\right)|\psi \left(u\right)>b\right\}\ne \varnothing$ 。若 $u\in P\left(\psi ,b,c\right)$ ，则 $b\le u\left(t\right)\le c$ ，从而由假设(H4)可得

$\begin{array}{c}\psi \left(Au\right)=\underset{t\in \left[1/4,3/4\right]}{\min }\left|\left(Tu\right)\left(t\right)\right|\\ =\underset{t\in \left[1/4,3/4\right]}{\min }\left|{\displaystyle {\int }_0^{1}G\left(t,s\right){\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s\right|\\ \ge {\displaystyle {\int }_0^{1}m\left(s\right){\varphi }_q}\left({\displaystyle {\int }_{1/4}^{3/4}H\left(s,\tau \right)\lambda {\varphi }_p\left(Nb\right)\text{d}\tau }\right)\text{d}s\\ =Nb{\displaystyle {\int }_0^{1}m\left(s\right){\varphi }_q}\left(\lambda {\displaystyle {\int }_{1/4}^{3/4}H\left(s,\tau \right)\text{d}\tau }\right)\text{d}s\\ =b,\end{array}$

即对任意的 $u\in P\left(\psi ,b,c\right)$ ， $\psi \left(Au\right)>b$ 。所以引理5中条件(L1)成立。

如果 $d=c$ ，那么引理5的条件(L1)可以推出条件(L3)。

综上，由引理5可知，边值问题(1)至少有三个正解 $u_1,u_2,u_3$ ，且满足

$\underset{t\in \left[0,1\right]}{\max }\left|u_1\left(t\right)\right|<a$ ， $b<\underset{t\in \left[1/4,3/4\right]}{\min }\left|u_2\left(t\right)\right|$ ， $a<\underset{t\in \left[0,1\right]}{\max }\left|u_3\left(t\right)\right|$ ， $\underset{t\in \left[1/4,3/4\right]}{\min }\left|u_3\left(t\right)\right|<b$ 。

4. 例子

例1 考虑如下带有p-Laplacian算子的分数阶积分边值问题

$\{\begin{array}{l}-D_{0+}^{11/4}\left({\varphi }_2\left(-D_{0+}^{4/3}u\left(t\right)\right)\right)=t+u\left(t\right)+{\text{e}}^{-\left|{\displaystyle {\int }_0^t{t}^2{\left(1-s\right)}^{2}u\left(s\right)\text{d}s}\right|},0<t<1,\\ u\left(0\right)=D_{0+}^{4/3}u\left(0\right)=D_{0+}^{1/3}u\left(1\right)={\left({\varphi }_2\left(-D_{0+}^{4/3}u\left(0\right)\right)\right)}^{\prime }=0,\\ D_{0+}^{1/2}\left({\varphi }_2\left(-D_{0+}^{4/3}u\left(1\right)\right)\right)=I_{0+}^{5/2}\left({\varphi }_2\left(-D_{0+}^{4/3}u\left(3/4\right)\right)\right),\end{array}$ (8)

此即在边值问题(1)中， $\alpha =11/4$ ， $\beta =4/3$ ， $\gamma =1/2$ ， $w=5/2$ ， $\eta =3/4$ ， $\lambda =1$ ， $p=2$ ，

$\left(Tu\right)\left(t\right)={\displaystyle {\int }_0^{t}K\left(t,s\right)u\left(s\right)\text{d}s}={\displaystyle {\int }_0^t{t}^2{\left(1-s\right)}^{2}u\left(s\right)\text{d}s}$ 及 $f\left(t,u,v\right)=t+u+{\text{e}}^{-\left|v\right|}$ ， $\forall t\in \left[0,1\right]$ ， $u,v\ge 0$ 。

通过计算可得 $k^{*}=\underset{t\in \left[0,1\right]}{\sup }{\displaystyle {\int }_0^{t}K\left(t,s\right)\text{d}s}=\frac{1}{3}$ ， $q=2$ ， $M\approx 21.8691$ ， $N\approx 49.2083$ 。选取 $r_1=0.1$ ， $r_2=0.4$ 有

$f\left(t,u,v\right)=t+u+{\text{e}}^{-\left|v\right|}+5>5\ge N{r}_1\approx 4.9208$ ， $\left(t,u,v\right)\in \left[\frac{1}{4},\frac{3}{4}\right]\times \left[0,0.1\right]\times \left[0,\frac{1}{30}\right)$

$f\left(t,u,v\right)=t+u+{\text{e}}^{-\left|v\right|}+4\le 6.4<M{r}_2\approx 8.7476$ ， $\left(t,u,v\right)\in \left[0,1\right]\times \left[0,0.4\right]\times \left[0,\frac{2}{15}\right)$

于是定理1的条件都满足，所以边值问题(8)至少存在一个正解u，使得 $0.1\le \Vert u\Vert \le 0.4$ 。

例2 考虑如下带有p-Laplacian算子的分数阶积分边值问题

$\{\begin{array}{l}-D_{0+}^{11/4}\left({\varphi }_2\left(-D_{0+}^{4/3}u\left(t\right)\right)\right)=f\left(t,u\left(t\right),\left(Tu\right)\left(t\right)\right),0<t<1,\\ u\left(0\right)=D_{0+}^{4/3}u\left(0\right)=D_{0+}^{1/3}u\left(1\right)={\left({\varphi }_2\left(-D_{0+}^{4/3}u\left(0\right)\right)\right)}^{\prime }=0,\\ D_{0+}^{1/2}\left({\varphi }_2\left(-D_{0+}^{4/3}u\left(1\right)\right)\right)=I_{0+}^{5/2}\left({\varphi }_2\left(-D_{0+}^{4/3}u\left(3/4\right)\right)\right),\end{array}$ (9)

此即在边值问题(1)中， $\alpha =11/4$ ， $\beta =4/3$ ， $\gamma =1/2$ ， $w=5/2$ ， $\eta =3/4$ ， $\lambda =1$ ， $p=2$ ， $\left(Tu\right)\left(t\right)={\displaystyle {\int }_0^{t}K\left(t,s\right)u\left(s\right)\text{d}s}={\displaystyle {\int }_0^t{t}^2{\left(1-s\right)}^{2}u\left(s\right)\text{d}s}$ 及

$f\left(t,u,v\right)=\{\begin{array}{ll}t+{\text{e}}^{-\left|v\right|}+45u^3,\hfill & 0\le u\le 1,\hfill \\ t+{\text{e}}^{-\left|v\right|}+55u-10,\hfill & 1<u\le 2,\hfill \\ t+{\text{e}}^{-\left|v\right|}+10\sqrt{2u}+80,\hfill & u>2,\hfill \end{array}$

通过计算可得 $k^{*}=\underset{t\in \left[0,1\right]}{\sup }{\displaystyle {\int }_0^{t}K\left(t,s\right)\text{d}s}=\frac{1}{3}$ ， $q=2$ ， $M\approx 21.8691$ ， $N\approx 49.2083$ 。选取 $a=\frac{1}{2}$ ， $b=2$ ， $c=8$ 有

$f\left(t,u,v\right)=t+{\text{e}}^{-\left|v\right|}+45u^3\le 6.375<Ma\approx 10.9346$ , $\left(t,u,v\right)\in \left[0,1\right]\times \left[0,\frac{1}{2}\right]\times \left[0,\frac{1}{6}\right]$

$f\left(t,u,v\right)=t+{\text{e}}^{-\left|v\right|}+55u-10>100>Nb\approx 98.4166$ , $\left(t,u,v\right)\in \left[\frac{1}{4},\frac{3}{4}\right]\times \left[2,8\right]\times \left[0,\frac{8}{3}\right]$

$f\left(t,u,v\right)=t+{\text{e}}^{-\left|v\right|}+10\sqrt{2u}+80\le 122<Mc\approx 174.9528$ , $\left(t,u,v\right)\in \left[0,1\right]\times \left[0,8\right]\times \left[0,\frac{8}{3}\right]$

于是定理2的条件都满足，所以边值问题(9)至少有三个正解 $u_1,u_2,u_3$ ，且满足

$\underset{0\le t\le 1}{\max }\left|u_1\left(t\right)\right|<\frac{1}{2}$ , $2<\underset{1/4\le t\le 3/4}{\min }\left|u_2\left(t\right)\right|$ , $\frac{1}{2}<\underset{0\le t\le 1}{\max }\left|u_3\left(t\right)\right|$ , $\underset{1/4\le t\le 3/4}{\min }\left|u_3\left(t\right)\right|<2$ .

NOTES

^*通讯作者。