具变号权函数的多参数二阶微分系统多个正解的存在性

Existence of Multiple Positive Solutions for Multi-Parameter Second-Order Differential Systems with Indefinite Weight Functions

摘要: 本文研究了一类具变号权函数的多参数二阶微分系统多个正解的存在性。根据参数λ和μ的不同取值，并结合范数形式的锥拉伸与压缩不动点定理，得到了二阶微分系统至少存在两个正解和三个正解的结果。最后，通过例子验证定理的条件是合理的。

Abstract: This paper investigates the existence of multiple positive solutions for a class of multi-parameter second-order differential systems with indefinite weight functions. According to the different values of parameters λ and μ, and combining fixed-point theorem of cone expansion and compression of norm type, the results of at least two positive solutions and three positive solutions for the second-order differential systems are established. Finally, two examples are included to illustrate the rationality of the assumptions of obtained theorems.

1. 引言

考虑如下具变号权函数的多参数二阶微分系统：

$\{\begin{array}{l}-u^{″}=a\left(t\right)\phi u+\lambda \omega \left(t\right)f\left(u\right)\text{,}0<t<1,\\ -{\phi }^{″}=\mu b\left(t\right)u,0<t<1,\\ u\left(0\right)=u\left(1\right)=0,\\ \phi \left(0\right)=\phi \left(1\right)=0.\end{array}$ (1)

其中， $\lambda ,\mu >0$ 是两个参数，且 $a\left(t\right),b\left(t\right),\omega \left(t\right)$ 在 $\left[0,1\right]$ 上可能变号。

令 $J=\left[0,1\right],R=\left(-\infty ,+\infty \right),R_{+}=\left[0,+\infty \right)$ ，且 $a,b,\omega$ 和f分别满足如下条件：

(H1) $a,b,\omega :J\to R$ 是连续的，且存在 $\xi \in \left(0,1\right)$ 使得

$\{\begin{array}{l}a\left(t\right),b\left(t\right),\omega \left(t\right)\ge 0,t\in \left[0,\xi \right],\\ a\left(t\right),b\left(t\right),\omega \left(t\right)\le 0,t\in \left[\xi ,1\right],\end{array}$

且 $a\left(t\right),b\left(t\right),\omega \left(t\right)$ 在J的任意子区间上不恒为零；

(H2) $f:R_{+}\to R_{+}$ 是连续的，且当 $u>0$ 时， $f\left(u\right)>0$ ；

(H3) 存在 $0<\theta <+\infty ,\theta \ne 1$ 和 $k_1,k_2>0$ 使得 $k_1{u}^{\theta }\le f\left(u\right)\le k_2{u}^{\theta },u\in R_{+}$ ；

(H4) 存在 $0<c\le 1$ ，使得 $f\left(u\right)\ge c\psi \left(u\right)\text{,}u\in R_{+}$ 。

其中， $\psi \left(u\right)=\max \left\{f\left(\rho \right):0\le \rho \le u\right\}$ 。

当 $\lambda =\mu =1$ 时，系统(1)退化为Wang和An在文献 [1] 中研究的问题，利用不动点定理得到了该问题至少存在一个正解和两个正解的充分条件。但是，在文献 [1] 中，Wang和An只考虑了 $\lambda ,\mu$ 恒为1的情况，本文则考虑 $\lambda ,\mu \cancel{\equiv }1$ 时的情况，并且得到了系统(1)至少存在三个正解的结果。

在过去的几十年中，对于权函数不变号的二阶微分方程问题，众多学者进行了研究并取得很多成果 [2] - [7] 。例如，在文献 [5] 中，卢和冯研究了具偏差变元的二阶微分方程边值问题，并得到了三个正解的存在性。但是在这些文献中，作者都没有考虑权函数变号的情况，而对具变号权函数的微分方程边值问题的研究，在物理中有一定的意义。比如，量子力学、半导体理论以及核反应中的一些数学模型都可以归结为具变号权函数的微分方程问题 [8] [9] [10] 。因此，这类问题受到了越来越多的学者的关注，并获得了一些优秀成果 [11] - [17] 。文献 [11] 中，Jiang和Yao研究了如下半线性两点边值问题

$\{\begin{array}{l}{w}^{″}\left(t\right)+h\left(t\right)f\left(w\left(t\right)\right)=0,0\le t\le 1,\\ w\left(0\right)=w\left(1\right)=0\end{array}$

正解的存在性。其中， $h\left(t\right)$ 在 $\left[0,1\right]$ 上可能变号。作者运用Schauder不动点定理给出了上述问题正解存在性定理。

但是到目前为止，关于含参数的具变号权函数二阶微分系统多个正解的存在性的研究还不多见。本文致力于研究具变号权函数的多参数二阶微分系统存在多个正解的充分条件。

2. 预备知识

定义

$\begin{array}{l}{a}^{+}\left(t\right)=\max \left\{a\left(t\right),0\right\}，a^{-}\left(t\right)=-\min \left\{a\left(t\right),0\right\};\\ b^{+}\left(t\right)=\max \left\{b\left(t\right),0\right\}，b^{-}\left(t\right)=-\min \left\{b\left(t\right),0\right\};\\ {\omega }^{+}\left(t\right)=\max \left\{\omega \left(t\right),0\right\},{\omega }^{-}\left(t\right)=-\min \left\{\omega \left(t\right),0\right\}.\end{array}$

则有

$a\left(t\right)=a^{+}\left(t\right)-a^{-}\left(t\right),b\left(t\right)=b^{+}\left(t\right)-b^{-}\left(t\right),\omega \left(t\right)={\omega }^{+}\left(t\right)-{\omega }^{-}\left(t\right).$

系统(1)等价于下列两个边值问题：

$\{\begin{array}{l}-u^{″}=a\left(t\right)\phi u+\lambda \omega \left(t\right)f\left(u\right),0<t<1,\\ u\left(0\right)=u\left(1\right)=0,\end{array}$ (2)

和

$\{\begin{array}{l}-{\phi }^{″}=\mu b\left(t\right)u,0<t<1,\\ \phi \left(0\right)=\phi \left(1\right)=0.\end{array}$ (3)

定义2.1. 若 $\left(u,\phi \right)\in C^2\left(0,1\right)\times C^2\left(0,1\right)$ 是系统(1)的一个正解，当且仅当 $\left(u,\phi \right)$ 满足系统(1)，且 $u,\phi \ge 0$ ， $u,\phi \cancel{\equiv }0$ 。

引理2.1. 假设条件(H1)-(H2)成立，则 $u\in C^2\left(0,1\right)$ 是问题(2)的一个解当且仅当u是下列积分方程的一个解

$u\left(t\right)={\displaystyle {\int }_0^{1}G\left(t,s\right)a\left(s\right)\phi \left(s\right)u\left(s\right)\text{d}s}+\lambda {\displaystyle {\int }_0^{1}G\left(t,s\right)\omega \left(s\right)f\left(u\left(s\right)\right)\text{d}s}.$ (4)

其中，

$G\left(t,s\right)=\{\begin{array}{l}t\left(1-s\right),0\le t\le s\le 1,\\ s\left(1-t\right),0\le s\le t\le 1.\end{array}$ (5)

证明：引理2.1的证明类似于文献 [7] 中引理1.1的证明。□

由 $G\left(t,s\right)$ 的定义可知，它有如下性质：

命题2.1. $G\left(t,s\right)$ 由(5)式定义，则有

$\begin{array}{l}G\left(t,s\right)>0，\forall t,s\in \left(0,1\right);G\left(t,s\right)\ge 0,\forall t,s\in J,\\ G\left(t,t\right)G\left(s,s\right)\le G\left(t,s\right)\le G\left(s,s\right),\forall t,s\in J.\end{array}$

引理2.2. 假设条件(H1)成立，则问题(3)有一个解 $\phi$ 且

$\phi \left(t\right)=\mu {\displaystyle {\int }_0^{1}G\left(t,s\right)b\left(s\right)u\left(s\right)\text{d}s}.$ (6)

证明：引理2.2的证明类似于文献 [7] 中引理1.1的证明。□

由引理2.1和引理2.2可得下列引理2.3。

引理2.3. 假设条件(H1)~(H2)成立，则 $\left(u,\phi \right)$ 是系统(1)的一个解当且仅当u是下列积分方程的解

$u\left(t\right)=\mu {\displaystyle {\int }_0^1{\displaystyle {\int }_0^{1}G\left(t,s\right)G\left(s,\tau \right)a\left(s\right)b\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda {\displaystyle {\int }_0^{1}G\left(t,s\right)\omega \left(s\right)f\left(u\left(s\right)\right)\text{d}s},$ (7)

且 $\phi$ 满足(6)式。

为了获得系统(1)正解的存在性，需要假设下列条件成立：

(H5) 存在 $0<{\sigma }_1<\xi$ 使得

${\sigma }_1{\displaystyle {\int }_{{\sigma }_1}^{\xi }G\left(t,s\right)b^{+}\left(s\right)\text{d}s}\ge \xi {\displaystyle {\int }_{\xi }^{1}G\left(t,s\right)b^{-}\left(s\right)\text{d}s};$

(H6) 存在 $0<{\sigma }_2<\xi$ 使得

${\sigma }_2{\displaystyle {\int }_{{\sigma }_2}^{\xi }G\left(t,s\right)G\left(s,s\right)a^{+}\left(s\right)\text{d}s}\ge \xi {\displaystyle {\int }_{\xi }^{1}G\left(t,s\right)a^{-}\left(s\right)\text{d}s};$

(H7) 存在 $0<{\sigma }_3<\xi$ 使得

${\sigma }_3^{\theta }{c}^2{k}_1{\displaystyle {\int }_{{\sigma }_3}^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)\text{d}s}\ge k_2{\xi }^{\theta }{\displaystyle {\int }_{\xi }^{1}G\left(t,s\right){\omega }^{-}\left(s\right)\text{d}s}.$

令 $X=C\left[0,1\right]$ 。对于 $u\in X$ ，

$\Vert u\Vert =\underset{t\in J}{\max }\left|u\left(t\right)\right|$ ,

则 $\left(X,\Vert \cdot \Vert \right)$ 是一个Banach空间。

记

$C_0^{+}\left[0,1\right]=\left\{u\in C\left[0,1\right]:\underset{t\in J}{\min }u\left(t\right)\ge 0,u\left(0\right)=u\left(1\right)=0\right\}.$

定义锥K如下：

$K=\left\{u\in C_0^{+}\left[0,1\right]:u在\left[0,\xi \right]上凹,在\left[\xi ,1\right]上凸\right\}.$ (8)

对任意 $u\in K$ ，显然有 $\Vert u\Vert =\underset{t\in \left[0,\xi \right]}{\max }u\left(t\right)$ 。

对于 $r>0$ ，令

$K_r=\left\{u\in K:\Vert u\Vert <r\right\}.$

定义算子 $T:K\to X$ 如下：

$\left(Tu\right)\left(t\right)=\mu {\displaystyle {\int }_0^1{\displaystyle {\int }_0^{1}G\left(t,s\right)G\left(s,\tau \right)a\left(s\right)b\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda {\displaystyle {\int }_0^{1}G\left(t,s\right)\omega \left(s\right)f\left(u\left(s\right)\right)\text{d}s}.$ (9)

由以上引理，我们可以得到引理2.4。

引理2.4. 假设条件(H1)~(H2)成立，则 $\left(u,\phi \right)$ 是系统(1)的解当且仅当u是算子T的不动点且 $\phi$ 满足(6)式。

引理2.5. 假设条件(H1)~(H7)成立，则 $T\left(K\right)\subset K$ 且算子 $T:K\to K$ 是全连续的。

证明：引理2.5的证明类似于文献 [1] 中引理2.2的证明。□

下面给出本文运用的主要引理，见文献 [18] 。

引理2.6. 设P是 Banach空间E上的一个锥， ${\Omega }_1,{\Omega }_2$ 是E中的有界开集，并且满足 $\theta \in {\Omega }_1,{\bar{\Omega }}_1\subset {\Omega }_2$ 。若算子 $T:P\cap \left({\bar{\Omega }}_2\backslash {\Omega }_1\right)\to P$ 全连续，则当下述条件满足其一时，算子T在 $P\cap \left({\bar{\Omega }}_2\backslash {\Omega }_1\right)$ 上至少有一个不动点：

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

2) $\Vert Tx\Vert \ge \Vert x\Vert ,\forall x\in P\cap \partial {\Omega }_1;\Vert Tx\Vert \le \Vert x\Vert ,\forall x\in P\cap \partial {\Omega }_2.$

3. 主要结论

在本节中，我们运用引理2.6讨论系统(1)存在多个正解时的参数的最优区间。

定理3.1. 假设条件(H1)~(H7)成立。当 $\theta >1$ 时，存在 ${\lambda }_0,{\mu }_1,{\mu }_2>0$ ，使得当 $\lambda \in \left[{\lambda }_0,+\infty \right)$ ， $\mu \in \left({\mu }_2,{\mu }_1\right]$ 时，系统(1)至少存在两个正解。

证明：一方面，由于 $\theta >1$ ，根据条件(H3)，得

$\underset{u\to 0}{\lim }\frac{f\left(u\right)}{u}\le \underset{u\to 0}{\lim }\frac{k_2{u}^{\theta }}{u}=0.$

这表明存在 $r^{\prime }>0$ ，使得

$f\left(u\right)\le \epsilon u,0\le u\le r^{\prime }.$

其中， $\epsilon$ 满足 $2\lambda \epsilon M_1<1$ ， $M_1=\underset{t\in J}{\max }{\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)\text{d}s}$ 。

令

$M=\underset{t\in J}{\max }{\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{\xi }G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b^{+}\left(\tau \right)\text{d}\tau \text{d}s}}.$ (10)

选取 $r=\min \left\{\frac{1}{2\mu M},\frac{r^{\prime }}{2}\right\}$ ，则有

$\mu \le {\mu }_1=\frac{1}{2Mr}.$ (11)

对任意 $u\in \partial K_r$ ，有 $0\le u\left(t\right)\le \Vert u\Vert =r,t\in J$ ，则 $f\left(u\left(t\right)\right)\le \epsilon u(t)\le \epsilon r,t\in J$ 成立。进而有

$\begin{array}{c}\left(Tu\right)\left(t\right)=\mu {\displaystyle {\int }_0^1{\displaystyle {\int }_0^{1}G\left(t,s\right)G\left(s,\tau \right)a\left(s\right)b\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda {\displaystyle {\int }_0^{1}G\left(t,s\right)\omega \left(s\right)f\left(u\left(s\right)\right)\text{d}s}\\ =\mu {\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{1}G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}-\mu {\displaystyle {\int }_{\xi }^1{\displaystyle {\int }_0^{1}G\left(t,s\right)G\left(s,\tau \right)a^{-}\left(s\right)}}\\ \cdot b\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s+\lambda {\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)f\left(u\left(s\right)\right)\text{d}s}-\lambda {\displaystyle {\int }_{\xi }^{1}G\left(t,s\right){\omega }^{-}\left(s\right)}f\left(u\left(s\right)\right)\text{d}s\\ \le \mu {\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{1}G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda {\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)f\left(u\left(s\right)\right)\text{d}s}\end{array}$

$\begin{array}{l}=\mu {\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{\xi }G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b^{+}\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}\\ -\mu {\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_{\xi }^{1}G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b^{-}\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda {\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)f\left(u\left(s\right)\right)\text{d}s}\\ \le \mu {\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{\xi }G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b^{+}\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda {\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)f\left(u\left(s\right)\right)\text{d}s}\\ \le \mu r^2\underset{t\in J}{\max }{\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{\xi }G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b^{+}\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda \underset{t\in J}{\max }{\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)\epsilon r\text{d}s}\\ =\mu M{r}^2+\lambda \epsilon M_{1}r<\frac{r}{2}+\frac{r}{2}=r=\Vert u\Vert .\end{array}$

即

$\Vert Tu\Vert <\Vert u\Vert ,\forall u\in \partial K_r.$ (12)

令

$N=\underset{\frac{{\sigma }_2}{2}\le t\le {\sigma }_2}{\min }\delta \left(t\right)\underset{\frac{{\sigma }_1}{2}\le t\le {\sigma }_1}{\min }\delta \left(t\right)\underset{t\in J}{\max }{\displaystyle {\int }_{\frac{{\sigma }_2}{2}}^{{\sigma }_2}G\left(t,s\right)a^{+}\left(s\right)}{\displaystyle {\int }_{\frac{{\sigma }_1}{2}}^{{\sigma }_1}G\left(s,\tau \right)b^{+}\left(\tau \right)\text{d}\tau \text{d}s},$ (13)

$N_1=\underset{t\in J}{\max }{\displaystyle {\int }_{\frac{{\sigma }_3}{2}}^{{\sigma }_3}G\left(t,s\right){\omega }^{+}\left(s\right)\text{d}s}.$

另一方面，由于 $\theta >1$ ，根据条件(H3)，得

$\underset{u\to +\infty }{\lim }\frac{f\left(u\right)}{u}\ge \underset{u\to +\infty }{\lim }\frac{k_1{u}^{\theta }}{u}=+\infty .$

这表明存在 $R^{\prime }>0$ ，使得

$f\left(u\right)\ge \eta u,u\ge R^{\prime }.$

其中， $\eta$ 满足 $2\lambda \eta \Gamma N_1\ge 1$ 且

$\Gamma =\underset{\frac{{\sigma }_3}{2}\le t\le {\sigma }_3}{\min }\delta \left(t\right)>0,\delta \left(t\right)=\min \left\{\frac{t}{\xi },\frac{\xi -t}{\xi }\right\},t\in \left[0,\xi \right].$ (14)

如果 $u\in K$ ，由 $u\left(t\right)$ 在 $\left[0,\xi \right]$ 上凹可得

$u\left(t\right)\ge \delta \left(t\right)\Vert u\Vert ,t\in \left[0,\xi \right].$ (15)

进一步可得

$\underset{\frac{{\sigma }_3}{2}\le t\le {\sigma }_3}{\min }u\left(t\right)\ge \underset{\frac{{\sigma }_3}{2}\le t\le {\sigma }_3}{\min }\delta \left(t\right)\Vert u\Vert =\Gamma \Vert u\Vert .$ (16)

选取 $R=\max \left\{{\left(2\mu N\right)}^{-1},\frac{R^{\prime }}{\Gamma },r^{\prime }\right\}+1$ ，则有

$\mu >{{\mu }^{\prime }}_2=\frac{1}{2NR}.$

对任意 $u\in \partial K_R$ ，由(16)式可得

$\underset{\frac{{\sigma }_3}{2}\le t\le {\sigma }_3}{\min }u\left(t\right)\ge \Gamma \Vert u\Vert =\Gamma R>R^{\prime }$

和

$\begin{array}{c}\Vert Tu\Vert =\mu \underset{t\in J}{\max }{\displaystyle {\int }_0^1{\displaystyle {\int }_0^{1}G\left(t,s\right)G\left(s,\tau \right)a\left(s\right)b\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda \underset{t\in J}{\max }{\displaystyle {\int }_0^{1}G\left(t,s\right)\omega \left(s\right)f\left(u\left(s\right)\right)\text{d}s}\\ \ge \mu \underset{t\in J}{\max }{\displaystyle {\int }_0^{{\sigma }_2}G\left(t,s\right)a^{+}\left(s\right)u\left(s\right)}{\displaystyle {\int }_0^{1}G\left(s,\tau \right)b\left(\tau \right)u\left(\tau \right)\text{d}\tau \text{d}s}+\lambda \underset{t\in J}{\max }{\displaystyle {\int }_0^{{\sigma }_3}G\left(t,s\right){\omega }^{+}\left(s\right)}f\left(u\left(s\right)\right)\text{d}s\\ \ge \mu \underset{t\in J}{\max }{\displaystyle {\int }_0^{{\sigma }_2}G\left(t,s\right)a^{+}\left(s\right)u\left(s\right)}{\displaystyle {\int }_0^{{\sigma }_1}G\left(s,\tau \right)b^{+}\left(\tau \right)u\left(\tau \right)\text{d}\tau \text{d}s}+\lambda \underset{t\in J}{\max }{\displaystyle {\int }_{\frac{{\sigma }_3}{2}}^{{\sigma }_3}G\left(t,s\right){\omega }^{+}\left(s\right)f\left(u\left(s\right)\right)\text{d}s}\end{array}$

$\begin{array}{l}\ge \mu \underset{t\in J}{\max }{\displaystyle {\int }_{\frac{{\sigma }_2}{2}}^{{\sigma }_2}G\left(t,s\right)a^{+}\left(s\right)\delta \left(s\right)}\Vert u\Vert {\displaystyle {\int }_{\frac{{\sigma }_1}{2}}^{{\sigma }_1}G\left(s,\tau \right)b^{+}\left(\tau \right)\delta (\tau )\Vert u\Vert \text{d}\tau \text{d}s}+\lambda \underset{t\in J}{\max }{\displaystyle {\int }_{\frac{{\sigma }_3}{2}}^{{\sigma }_3}G\left(t,s\right){\omega }^{+}\left(s\right)\eta u\left(s\right)\text{d}s}\\ \ge \mu R^2\underset{\frac{{\sigma }_2}{2}\le t\le {\sigma }_2}{\min }\delta \left(t\right)\underset{\frac{{\sigma }_1}{2}\le t\le {\sigma }_1}{\min }\delta \left(t\right)\underset{t\in J}{\max }{\displaystyle {\int }_{\frac{{\sigma }_2}{2}}^{{\sigma }_2}G\left(t,s\right)a^{+}\left(s\right)}{\displaystyle {\int }_{\frac{{\sigma }_1}{2}}^{{\sigma }_1}G\left(s,\tau \right)b^{+}\left(\tau \right)\text{d}\tau \text{d}s}\\ +\lambda \eta \Gamma R\underset{t\in J}{\max }{\displaystyle {\int }_{\frac{{\sigma }_3}{2}}^{{\sigma }_3}G\left(t,s\right){\omega }^{+}\left(s\right)\text{d}s}\\ =\mu N{R}^2+\lambda \eta \Gamma N_{1}R>\frac{R}{2}+\frac{R}{2}=R=\Vert u\Vert .\end{array}$

即

$\Vert Tu\Vert >\Vert u\Vert ,\forall u\in \partial K_R.$ (17)

最后，存在 $0<r_1<r$ ，由条件(H2)可定义

${\beta }_{r_1}=\underset{\Gamma r_1\le u\le r_1}{\min }\left\{f\left(u\right)\right\}>0.$

令

${{\mu }^{″}}_2=\frac{1}{2DM{r}_1},{\lambda }_0=\frac{\left(2DM-N\right)r_1}{2DM{\beta }_{r_1}{N}_1}.$ (18)

其中， $D=\frac{\text{2}r}{r_1}$ 。

则对任意 $u\in \partial K_{r_1}$ ，有

$\Gamma r_1=\Gamma \Vert u\Vert \le \underset{\frac{{\sigma }_3}{2}\le t\le {\sigma }_3}{\min }u\left(t\right)\le u\left(t\right)\le \Vert u\Vert =r_1\text{,}t\in \left[\frac{{\sigma }_3}{2},{\sigma }_3\right]$

和

$\begin{array}{l}\Vert Tu\Vert =\mu \underset{t\in J}{\max }{\displaystyle {\int }_0^1{\displaystyle {\int }_0^{1}G\left(t,s\right)G\left(s,\tau \right)a\left(s\right)b\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda \underset{t\in J}{\max }{\displaystyle {\int }_0^{1}G\left(t,s\right)\omega \left(s\right)f\left(u\left(s\right)\right)\text{d}s}\\ \ge \mu \underset{t\in J}{\max }{\displaystyle {\int }_0^{{\sigma }_2}G\left(t,s\right)a^{+}\left(s\right)u\left(s\right)}{\displaystyle {\int }_0^{1}G\left(s,\tau \right)b\left(\tau \right)u\left(\tau \right)\text{d}\tau \text{d}s}+\lambda \underset{t\in J}{\max }{\displaystyle {\int }_0^{{\sigma }_3}G\left(t,s\right)\omega \left(s\right)}f\left(u\left(s\right)\right)\text{d}s\\ \ge \mu \underset{t\in J}{\max }{\displaystyle {\int }_0^{{\sigma }_2}G\left(t,s\right)a^{+}\left(s\right)u\left(s\right)}{\displaystyle {\int }_0^{{\sigma }_1}G\left(s,\tau \right)b^{+}\left(\tau \right)u\left(\tau \right)\text{d}\tau \text{d}s}+\lambda \underset{t\in J}{\max }{\displaystyle {\int }_{\frac{{\sigma }_3}{2}}^{{\sigma }_3}G\left(t,s\right){\omega }^{+}\left(s\right)f\left(u\left(s\right)\right)\text{d}s}\\ \ge \mu \underset{t\in J}{\max }{\displaystyle {\int }_{\frac{{\sigma }_2}{2}}^{{\sigma }_2}G\left(t,s\right)a^{+}\left(s\right)\delta \left(s\right)}\Vert u\Vert {\displaystyle {\int }_{\frac{{\sigma }_1}{2}}^{{\sigma }_1}G\left(s,\tau \right)b^{+}\left(\tau \right)\delta \left(\tau \right)\Vert u\Vert \text{d}\tau \text{d}s}+\lambda \underset{t\in J}{\max }{\displaystyle {\int }_{\frac{{\sigma }_3}{2}}^{{\sigma }_3}G\left(t,s\right){\omega }^{+}\left(s\right){\beta }_{r_1}}\text{d}s\end{array}$

^{$\begin{array}{l}\ge \mu \underset{\frac{{\sigma }_2}{2}\le t\le {\sigma }_2}{\min }\delta \left(t\right)\underset{\frac{{\sigma }_1}{2}\le t\le {\sigma }_1}{\min }\delta (t)\underset{t\in J}{\max }{\displaystyle {\int }_{\frac{{\sigma }_2}{2}}^{{\sigma }_2}G\left(t,s\right)a^{+}\left(s\right)}{\displaystyle {\int }_{\frac{{\sigma }_1}{2}}^{{\sigma }_1}G\left(s,\tau \right)b^{+}\left(\tau \right)\text{d}\tau \text{d}s}{\Vert u\Vert }^2\\ +\lambda {\beta }_{r_1}\underset{t\in J}{\max }{\displaystyle {\int }_{\frac{{\sigma }_3}{2}}^{{\sigma }_3}G\left(t,s\right){\omega }^{+}\left(s\right)\text{d}s}\\ \ge {{\mu }^{″}}_{2}N{r}_1{}^2+{\lambda }_0{\beta }_{r_1}{N}_1=\frac{N{r}_1}{2DM}+\frac{\left(2DM-N\right)r_1}{2DM}=r_1=\Vert u\Vert .\end{array}$}

因此，可得

$\Vert Tu\Vert \ge \Vert u\Vert ,\forall u\in \partial K_{r_1}.$ (19)

取 ${\mu }_2=\max \left\{{{\mu }^{\prime }}_2,{{\mu }^{″}}_2\right\}$ 。综上所述，由引理2.6可知，对任意 $\lambda \in \left[{\lambda }_0,+\infty \right),\mu \in \left({\mu }_2,{\mu }_1\right]$ ，(12)式和(17)式、(12)式和(19)式分别蕴含着T有一个不动点 $u_1,u_2$ 满足 $u_1\in {\bar{K}}_R\backslash {\bar{K}}_r$ 和 $u_2\in K_r\backslash K_{r_1}$ 。又由引理2.4可知，系统(1)至少存在两个正解 $\left(u_i,{\phi }_i\right)\left(i=1,2\right)$ 。其中，

$u_1\in {\bar{K}}_R\backslash {\bar{K}}_r,u_2\in K_r\backslash K_{r_1}$

且

${\phi }_i\left(t\right)=\mu {\displaystyle {\int }_0^{1}G\left(t,s\right)b\left(s\right)u_i\left(s\right)\text{d}s}\left(i=1,2\right).$ □

定理3.2. 假设条件(H1)~(H7)成立。当 $0<\theta <1$ 时，存在 ${\lambda }_0^{*},{\mu }_0^{*}>0$ ，使得当 $\lambda \in \left(0,{\lambda }_0^{*}\right]$ ， $\mu \in \left(0,{\mu }_0^{*}\right]$ 时，系统(1)至少存在两个正解。

证明：一方面，由于 $0<\theta <1$ ，根据条件(H3)，得

$\underset{u\to 0}{\lim }\frac{f\left(u\right)}{u}\ge \underset{u\to 0}{\lim }\frac{k_1{u}^{\theta }}{u}=+\infty .$

则存在 $r>0$ ，使得

$f\left(u\right)\ge \eta u,0\le u\le r.$

其中， $\eta$ 满足 $\lambda \eta \Gamma N_1>1$ 。其中， $N_1$ 在(13)式中定义， $\Gamma$ 在(14)式中定义。

因此，对任意 $u\in \partial K_r$ ，由(16)式可得

$\underset{\frac{{\sigma }_3}{2}\le t\le {\sigma }_3}{\min }u\left(t\right)\ge \Gamma \Vert u\Vert =\Gamma r$

和

$\begin{array}{c}\Vert Tu\Vert =\mu \underset{t\in J}{\max }{\displaystyle {\int }_0^1{\displaystyle {\int }_0^{1}G\left(t,s\right)G\left(s,\tau \right)a\left(s\right)b\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda \underset{t\in J}{\max }{\displaystyle {\int }_0^{1}G\left(t,s\right)\omega \left(s\right)f\left(u\left(s\right)\right)\text{d}s}\\ \ge \mu \underset{t\in J}{\max }{\displaystyle {\int }_0^{{\sigma }_2}G\left(t,s\right)a^{+}\left(s\right)u\left(s\right)}{\displaystyle {\int }_0^{1}G\left(s,\tau \right)b\left(\tau \right)u\left(\tau \right)\text{d}\tau \text{d}s}+\lambda \underset{t\in J}{\max }{\displaystyle {\int }_0^{{\sigma }_3}G\left(t,s\right){\omega }^{+}\left(s\right)}f\left(u\left(s\right)\right)\text{d}s\\ \ge \lambda \underset{t\in J}{\max }{\displaystyle {\int }_{\frac{{\sigma }_3}{2}}^{{\sigma }_3}G\left(t,s\right){\omega }^{+}\left(s\right)\eta u\left(s\right)\text{d}s}\\ \ge \lambda \eta \Gamma N_{1}r>r=\Vert u\Vert .\end{array}$

即

$\Vert Tu\Vert >\Vert u\Vert ,\forall u\in \partial K_r.$ (20)

另一方面，由于 $0<\theta <1$ ，根据条件(H3)，得

$\underset{u\to +\infty }{\lim }\frac{f\left(u\right)}{u}\le \underset{u\to +\infty }{\lim }\frac{k_2{u}^{\theta }}{u}=0.$

则存在 $R^{\prime }>r$ ，使得

$f\left(u\right)\le \epsilon u,u\ge R^{\prime }.$

其中， $\epsilon$ 满足 $2\lambda \epsilon M_1<1$ ， $M_1$ 在定理3.1中定义。

由条件(H2)，可得 $\psi \left(R^{\prime }\right)>0$ ，则有

$f\left(u\right)\le \psi \left(R^{\prime }\right)+\epsilon u,u\in \left[0,+\infty \right).$

令

$\max \left\{R^{\prime },\frac{2\lambda M_1\psi \left(R^{\prime }\right)}{1-2\lambda M_1\epsilon }\right\}<R\le {\left(2\mu M\right)}^{-1}.$

则可验证对于充分小的 $\lambda ,\mu ,\epsilon >0$ ， $\max \left\{R^{\prime },\frac{2\lambda M_1\psi \left(R^{\prime }\right)}{1-2\lambda M_1\epsilon }\right\}<{\left(2\mu M\right)}^{-1}$ 是合理的。因此，对任意 $u\in \partial K_R$ ，可得

$\begin{array}{c}\left(Tu\right)\left(t\right)=\mu {\displaystyle {\int }_0^1{\displaystyle {\int }_0^{1}G\left(t,s\right)G\left(s,\tau \right)a\left(s\right)b\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda {\displaystyle {\int }_0^{1}G\left(t,s\right)\omega \left(s\right)f\left(u\left(s\right)\right)\text{d}s}\\ =\mu {\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{1}G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}-\mu {\displaystyle {\int }_{\xi }^1{\displaystyle {\int }_0^{1}G\left(t,s\right)G\left(s,\tau \right)a^{-}\left(s\right)}}\\ \cdot b\left(\tau \right)u\left(s\right)u\left(\tau \right)d\tau ds+\lambda {\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)f\left(u\left(s\right)\right)\text{d}s}-\lambda {\displaystyle {\int }_{\xi }^{1}G\left(t,s\right){\omega }^{-}\left(s\right)}f\left(u\left(s\right)\right)\text{d}s\\ \le \mu {\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{1}G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda {\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)f\left(u\left(s\right)\right)\text{d}s}\\ =\mu {\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{\xi }G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b^{+}\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}-\mu {\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_{\xi }^{1}G\left(t,s\right)G\left(s,\tau \right)}}\end{array}$

$\begin{array}{l}\cdot a^{+}\left(s\right)b^{-}\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s+\lambda {\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)f\left(u\left(s\right)\right)\text{d}s}\\ \le \mu {\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{\xi }G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b^{+}\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda {\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)f\left(u\left(s\right)\right)\text{d}s}\\ \le \mu R^2\underset{t\in J}{\max }{\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{\xi }G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b^{+}\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda \underset{t\in J}{\max }{\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)\left(\psi \left(R^{\prime }\right)+\epsilon u\right)\text{d}s}\\ \le \mu M{R}^2+\lambda \left(\psi \left(R^{\prime }\right)+\epsilon R\right)M_1<\frac{R}{2}+\frac{R}{2}=R=\Vert u\Vert .\end{array}$

即

$\Vert Tu\Vert <\Vert u\Vert ,\forall u\in \partial K_R.$ (21)

令

${\mu }_0^{*}=\frac{1}{2M{r}_1},{\lambda }_0^{*}=\frac{r_1}{2\psi \left(r_1\right)M_1}.$

其中， $0<r_1<r$ ， $M,M_1$ 在定理3.1中定义。

则对任意 $u\in \partial K_{r_1}$ ，类似于(21)式的证明，可得

$\begin{array}{c}\left(Tu\right)\left(t\right)\le \mu {\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{\xi }G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b^{+}\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda {\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)f\left(u\left(s\right)\right)\text{d}s}\\ \le \mu r_1{}^2\underset{t\in J}{\max }{\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{\xi }G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b^{+}\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda \underset{t\in J}{\max }{\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)\psi \left(r_1\right)\text{d}s}\\ =\mu M{r}_1{}^2+\lambda \psi \left(r_1\right)M_1\le {\mu }_0^{*}M{r}_1{}^2+{\lambda }_0^{*}\psi \left(r_1\right)M_1=\frac{r_1}{2}+\frac{r_1}{2}=r_1=\Vert u\Vert .\end{array}$

即

$\Vert Tu\Vert \le \Vert u\Vert ,\forall u\in \partial K_{r_1}.$ (22)

综上所述，由引理2.6可知，对任意 $\lambda \in \left(0,{\lambda }_0^{*}\right]$ ， $\mu \in \left(0,{\mu }_0^{*}\right]$ ，(20)式和(21)式、(20)式和(22)式分别蕴含着T有一个不动点 $u_1,u_2$ 满足 $u_1\in {\bar{K}}_R\backslash {\bar{K}}_r$ 和 $u_2\in K_r\backslash K_{r_1}$ 。又由引理2.4可知，系统(1)至少存在两个正解 $\left(u_i,{\phi }_i\right)\left(i=1,2\right)$ 。其中，

$u_1\in {\bar{K}}_R\backslash {\bar{K}}_r,u_2\in K_r\backslash K_{r_1}$

且

${\phi }_i\left(t\right)=\mu {\displaystyle {\int }_0^{1}G\left(t,s\right)b\left(s\right)u_i\left(s\right)\text{d}s}\left(i=1,2\right).$ □

定理3.3. 假设条件(H1)~(H7)成立。当 $0<\theta <1$ 时，存在 ${\lambda }_1^{*},{\mu }_1^{*}>0$ ，使得当 $\lambda \in \left(0,{\lambda }_1^{*}\right]$ ， $\mu \in \left(0,{\mu }_1^{*}\right]$ ，

且 $k_2$ 满足 $0<k_2<\frac{{\left(2\mu M\right)}^{\theta -1}}{2\lambda M_1}$ 时，系统(1)至少存在三个正解。其中， $M,M_1$ 在定理3.1中定义。

证明：首先，由于 $0<\theta <1$ ，根据条件(H3)，得

$\underset{u\to 0}{\lim }\frac{f\left(u\right)}{u}\ge \underset{u\to 0}{\lim }\frac{k_1{u}^{\theta }}{u}=+\infty .$

则存在 $r>0$ ，使得

$f\left(u\right)\ge \eta u,0\le u\le r.$

其中， $\eta$ 满足 $\lambda \eta \Gamma N_1>1$ 。其中， $N_1$ 在(13)式中定义， $\Gamma$ 在(14)式中定义。

因此，对任意 $u\in \partial K_r$ ，与(20)式的证明相同，可得

$\Vert Tu\Vert >\Vert u\Vert ,\forall u\in \partial K_r.$ (23)

其次，选取充分大的 $R_1\ge \max \left\{\frac{1}{2\mu M},\frac{1}{\mu N}\right\}$ ，则有 $\mu N{R}_1^2\ge R_1$ 。

对任意 $u\in \partial K_{R_1}$ ，有

$\begin{array}{c}\Vert Tu\Vert =\mu \underset{t\in J}{\max }{\displaystyle {\int }_0^1{\displaystyle {\int }_0^{1}G\left(t,s\right)G\left(s,\tau \right)a\left(s\right)b\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda \underset{t\in J}{\max }{\displaystyle {\int }_0^{1}G\left(t,s\right)\omega \left(s\right)f\left(u\left(s\right)\right)\text{d}s}\\ \ge \mu \underset{t\in J}{\max }{\displaystyle {\int }_0^{{\sigma }_2}G\left(t,s\right)a^{+}\left(s\right)u\left(s\right)}{\displaystyle {\int }_0^{1}G\left(s,\tau \right)b\left(\tau \right)u\left(\tau \right)\text{d}\tau \text{d}s}+\lambda \underset{t\in J}{\max }{\displaystyle {\int }_0^{{\sigma }_3}G\left(t,s\right){\omega }^{+}\left(s\right)}f\left(u\left(s\right)\right)\text{d}s\\ \ge \mu \underset{t\in J}{\max }{\displaystyle {\int }_0^{{\sigma }_2}G\left(t,s\right)a^{+}\left(s\right)u\left(s\right)}{\displaystyle {\int }_0^{{\sigma }_1}G\left(s,\tau \right)b^{+}\left(\tau \right)u\left(\tau \right)\text{d}\tau \text{d}s}\\ \ge \mu \underset{t\in J}{\max }{\displaystyle {\int }_{\frac{{\sigma }_2}{2}}^{{\sigma }_2}G\left(t,s\right)a^{+}\left(s\right)u\left(s\right)}{\displaystyle {\int }_{\frac{{\sigma }_1}{2}}^{{\sigma }_1}G\left(s,\tau \right)b^{+}\left(\tau \right)u\left(\tau \right)\text{d}\tau \text{d}s}\end{array}$

$\begin{array}{l}\ge \mu \underset{t\in J}{\max }{\displaystyle {\int }_{\frac{{\sigma }_2}{2}}^{{\sigma }_2}G\left(t,s\right)a^{+}\left(s\right)\delta \left(s\right)}\Vert u\Vert {\displaystyle {\int }_{\frac{{\sigma }_1}{2}}^{{\sigma }_1}G\left(s,\tau \right)b^{+}\left(\tau \right)\delta \left(\tau \right)\Vert u\Vert \text{d}\tau \text{d}s}\\ \ge \mu \underset{\frac{{\sigma }_2}{2}\le t\le {\sigma }_2}{\min }\delta \left(t\right)\underset{\frac{{\sigma }_1}{2}\le t\le {\sigma }_1}{\min }\delta \left(t\right)\underset{t\in J}{\max }{\displaystyle {\int }_{\frac{{\sigma }_2}{2}}^{{\sigma }_2}G\left(t,s\right)a^{+}\left(s\right)}{\displaystyle {\int }_{\frac{{\sigma }_1}{2}}^{{\sigma }_1}G\left(s,\tau \right)b^{+}\left(\tau \right)\text{d}\tau \text{d}s{\Vert u\Vert }^2}\\ =\mu N{R}_1{}^2\ge R_1=\Vert u\Vert .\end{array}$

即

$\Vert Tu\Vert \ge \Vert u\Vert ,\forall u\in \partial K_{R_1}.$ (24)

再次，由于 $0<\theta <1$ ，根据条件(H3)，得

$\underset{u\to +\infty }{\lim }\frac{f\left(u\right)}{u}\le \underset{u\to +\infty }{\lim }\frac{k_2{u}^{\theta }}{u}=0.$

则存在 $R>0$ ，使得

$f\left(u\right)\le \epsilon u,u\ge R.$

存在 $R>0$ ，满足

${\left(2\lambda k_2{M}_1\right)}^{\frac{1}{1-\theta }}<R<{\left(2\mu M\right)}^{-1}.$

则对任意，可得

$\begin{array}{c}\left(Tu\right)\left(t\right)=\mu {\displaystyle {\int }_0^1{\displaystyle {\int }_0^{1}G\left(t,s\right)G\left(s,\tau \right)a\left(s\right)b\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda {\displaystyle {\int }_0^{1}G\left(t,s\right)\omega \left(s\right)f\left(u\left(s\right)\right)\text{d}s}\\ =\mu {\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{1}G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}-\mu {\displaystyle {\int }_{\xi }^1{\displaystyle {\int }_0^{1}G\left(t,s\right)G\left(s,\tau \right)a^{-}\left(s\right)}}\\ \cdot b\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s+\lambda {\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)f\left(u\left(s\right)\right)\text{d}s}-\lambda {\displaystyle {\int }_{\xi }^{1}G\left(t,s\right){\omega }^{-}\left(s\right)}f\left(u\left(s\right)\right)\text{d}s\\ \le \mu {\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{1}G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda {\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)f\left(u\left(s\right)\right)\text{d}s}\end{array}$

$\begin{array}{l}=\mu {\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{\xi }G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b^{+}\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}\\ -\mu {\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_{\xi }^{1}G\left(t,s\right)G\left(s,\tau \right)}}{a}^{+}\left(s\right)b^{-}\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s+\lambda {\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)f\left(u\left(s\right)\right)\text{d}s}\\ \le \mu {\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{\xi }G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b^{+}\left(\tau \right)\text{d}\tau \text{d}s{\Vert u\Vert }^2}}+\lambda {\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)k_2{\Vert u\Vert }^{\theta }\text{d}s}\\ \le \mu R^2\underset{t\in J}{\max }{\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{\xi }G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b^{+}\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda k_2{R}^{\theta }\underset{t\in J}{\max }{\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)\text{d}s}\\ =\mu M{R}^2+\lambda k_2{M}_1{R}^{\theta }<\frac{R}{2}+\frac{R}{2}=R=\Vert u\Vert .\end{array}$

即

$\Vert Tu\Vert <\Vert u\Vert ,\forall u\in \partial K_R.$ (25)

最后，选取 $r_1\in \left(0,r\right)$ 。由条件(H2)可得 $\psi \left(r_1\right)>0$ 。

令

${\mu }_1^{*}=\frac{1}{2M{r}_1},{\lambda }_1^{*}=\frac{r_1}{2\psi \left(r_1\right)M_1}.$

则对任意 $u\in \partial K_{r_1}$ ，有

$\begin{array}{c}\left(Tu\right)\left(t\right)\le \mu {\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{\xi }G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b^{+}\left(\tau \right)u\left(s\right)u\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda {\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)f\left(u\left(s\right)\right)\text{d}s}\\ \le \mu r_{\text{1}}{}^2\underset{t\in J}{\max }{\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_0^{\xi }G\left(t,s\right)G\left(s,\tau \right)a^{+}\left(s\right)b^{+}\left(\tau \right)\text{d}\tau \text{d}s}}+\lambda \underset{t\in J}{\max }{\displaystyle {\int }_0^{\xi }G\left(t,s\right){\omega }^{+}\left(s\right)\psi \left(r_1\right)\text{d}s}\\ =\mu M{r}_{\text{1}}{}^2+\lambda M_1\psi \left(r_1\right)\le {\mu }_1^{*}M{r}_{\text{1}}{}^2+{\lambda }_1^{*}{M}_1\psi \left(r_1\right)=\frac{r_{\text{1}}}{2}+\frac{r_{\text{1}}}{2}=r_{\text{1}}=\Vert u\Vert .\end{array}$

即

$\Vert Tu\Vert \le \Vert u\Vert ,\forall u\in \partial K_{r_1}.$ (26)

综上所述，由引理2.6可知，对任意 $\lambda \in \left(0,{\lambda }_1^{*}\right],\mu \in \left(0,{\mu }_1^{*}\right]$ ，(24)式和(25)式、(23)式和(25)式、(23)式和(26)式分别蕴含着T有一个不动点 $u_1,u_2,u_3$ 满足 $u_1\in {\bar{K}}_{R_1}\backslash {\bar{K}}_R$ ， $u_2\in K_R\backslash {\bar{K}}_r$ 和 $u_3\in K_r\backslash K_{r_1}$ 。又由引理2.4可知，系统(1)至少存在三个正解 $\left(u_i,{\phi }_i\right)\left(i=1,2,3\right)$ 。其中，

$u_1\in {\bar{K}}_{R_1}\backslash {\bar{K}}_R,u_2\in K_R\backslash {\bar{K}}_r,u_3\in K_r\backslash K_{r_1}$

且

${\phi }_i\left(t\right)=\mu {\displaystyle {\int }_0^{1}G\left(t,s\right)b\left(s\right)u_i\left(s\right)\text{d}s}\left(i=1,2,3\right).$ □

4. 例子

例4.1. 考虑如下二阶微分系统：

$\{\begin{array}{l}-u^{″}=a\left(t\right)\phi u+\lambda \omega \left(t\right){\left(u+\sin u\right)}^{\theta },0<t<1,\\ -{\phi }^{″}=\mu b\left(t\right)u,0<t<1,\\ u\left(0\right)=u\left(1\right)=0,\\ \phi \left(0\right)=\phi \left(1\right)=0.\end{array}$ (27)

其中， $\theta >1$ 且

$\begin{array}{l}b\left(t\right)=\{\begin{array}{l}27\left(\frac{2}{3}-t\right),t\in \left[0,\frac{2}{3}\right],\\ -\frac{1}{18}\left(t-\frac{2}{3}\right),t\in \left[\frac{2}{3},1\right],\end{array}\\ a\left(t\right)=\{\begin{array}{l}256\left(\frac{2}{3}-t\right),t\in \left[0,\frac{2}{3}\right],\\ -\frac{1}{45}\left(t-\frac{2}{3}\right),t\in \left[\frac{2}{3},1\right],\end{array}\end{array}$

$\omega \left(t\right)=\{\begin{array}{l}\text{3}\times 4^{\theta }\left(\frac{2}{3}-t\right),t\in \left[0,\frac{2}{3}\right],\\ -\frac{1}{4}\left(t-\frac{2}{3}\right),t\in \left[\frac{2}{3},1\right].\end{array}$

结论：当 $\lambda \in \left[\frac{\text{4507}}{\text{794}\times {\left(\frac{1}{5}+4\sin \frac{1}{20}\right)}^{\theta }},+\infty \right)$ ， $\mu \in \left(\frac{515}{4396},\frac{515}{2198}\right]$ 时，系统(27)至少存在两个正解。

为了验证满足定理3.1的条件的成立，根据文献 [2] 的命题2.3，我们给出以下命题：

命题4.1. 考虑如下问题

$\{\begin{array}{l}-u^{″}=k\left(t\right)u^{\alpha }\text{,}0<t<1,\\ u\left(0\right)=u\left(1\right)=0.\end{array}$

其中， $\alpha >0$ ， $k\left(t\right)$ 满足

$\begin{array}{l}k\left(t\right)\ge 0,t\in \left[0,\xi \right]\\ k\left(t\right)\le 0,t\in \left[\xi ,1\right]\end{array}\}$

且

$c_1{u}^{\alpha }\le f\left(u\right)=u^{\alpha }\le c_2{u}^{\alpha }.$

如果存在 $0<\sigma <\xi$ 使得

$c_1\frac{\xi -\sigma }{1-\xi }{\sigma }^{\alpha +1}{k}^{+}\left(\xi -\frac{\xi -\sigma }{1-\xi }\tau \right)\ge c_2{\xi }^{\alpha }{k}^{-}\left(\xi +\tau \right),\tau \in \left[0,1-\xi \right].$ (28)

则下列不等式成立

${\sigma }^{\alpha }{\displaystyle {\int }_{\sigma }^{\xi }G\left(t,s\right)k^{+}\left(s\right)\text{d}s}\ge \frac{c_2}{c_1}{\xi }^{\alpha }{\displaystyle {\int }_{\xi }^{1}G\left(t,s\right)k^{-}\left(s\right)\text{d}s}.$ (29)

例4.1的证明：由 $a\left(t\right)$ ， $b\left(t\right)$ 和 $\omega \left(t\right)$ 的定义可知 $\xi =\frac{2}{3}$ 。下面来验证定理3.1的条件成立。

首先，条件(H1)~(H4)显然成立。其次，分三步验证条件(H5)~(H7)成立。

第一步验证条件(H5)的成立。令 $c_1=c_2=1$ ， $\alpha =1$ 和 ${\sigma }_1=\frac{1}{3}$ ，(28)式等价于下列不等式

$\frac{1}{9}{b}^{+}\left(\frac{2}{3}-\tau \right)\ge \frac{2}{3}{b}^{-}\left(\frac{2}{3}+\tau \right),\tau \in \left[0,\frac{1}{3}\right].$

令 $\frac{2}{3}-\tau =\gamma$ 。则上述不等式等价于

$\frac{1}{9}{b}^{+}\left(\gamma \right)\ge \frac{2}{3}{b}^{-}\left(\frac{4}{3}-\gamma \right),\gamma \in \left[\frac{1}{3},\frac{2}{3}\right].$

由 $b\left(t\right)$ 的定义可得上述不等式成立，则条件(H5)成立。

第二步验证条件(H6)的成立。令 $c_1=1$ ， $c_2=\frac{9}{2}$ ， $\alpha =1$ 和 ${\sigma }_2=\frac{1}{3}$ ，则(29)式为

$\frac{1}{3}{\displaystyle {\int }_{\frac{1}{3}}^{\frac{2}{3}}G\left(t,s\right)a^{+}\left(s\right)\text{d}s}\ge 3{\displaystyle {\int }_{\frac{2}{3}}^{1}G\left(t,s\right)a^{-}\left(s\right)\text{d}s}.$

另外，由上述不等式可得

$\begin{array}{l}\frac{1}{3}{\displaystyle {\int }_{\frac{1}{3}}^{\frac{2}{3}}G\left(t,s\right)\frac{2}{9}{a}^{+}\left(s\right)\text{d}s}\ge \frac{2}{3}{\displaystyle {\int }_{\frac{2}{3}}^{1}G\left(t,s\right)a^{-}\left(s\right)\text{d}s}\\ \iff \frac{1}{3}{\displaystyle {\int }_{\frac{1}{3}}^{\frac{2}{3}}G\left(t,s\right)\left(\underset{s\in \left[\frac{1}{3},\frac{2}{3}\right]}{\min }G\left(s,s\right)\right)a^{+}\left(s\right)\text{d}s}\\ \ge \frac{2}{3}{\displaystyle {\int }_{\frac{2}{3}}^{1}G\left(t,s\right)a^{-}\left(s\right)\text{d}s}\\ \Rightarrow \frac{1}{3}{\displaystyle {\int }_{\frac{1}{3}}^{\frac{2}{3}}G\left(t,s\right)G\left(s,s\right)a^{+}\left(s\right)\text{d}s}\ge \frac{2}{3}{\displaystyle {\int }_{\frac{2}{3}}^{1}G\left(t,s\right)a^{-}\left(s\right)\text{d}s}.\end{array}$

即由上述不等式可得条件(H6)的成立。

与第一步的证明类似，(28)式等价于下列不等式

$\frac{1}{9}{a}^{+}\left(\gamma \right)\ge 3a^{-}\left(\frac{4}{3}-\gamma \right),\gamma \in \left[\frac{1}{3},\frac{2}{3}\right].$

由 $a\left(t\right)$ 的定义可得上述不等式成立，则条件(H6)成立。

第三步验证条件(H7)的成立。令 $c_1=2^{\theta }$ ， $c_2=1$ ， $\alpha =\theta$ 和 ${\sigma }_2=\frac{1}{3}$ ，与第一步的证明类似，(28)式等价于下列不等式

${\left(\frac{1}{3}\right)}^{\theta +1}{\omega }^{+}\left(\gamma \right)\ge {\left(\frac{4}{3}\right)}^{\theta }{\omega }^{-}\left(\frac{2}{3}+\gamma \right),\gamma \in \left[\frac{1}{3},\frac{2}{3}\right].$

由 $\omega \left(t\right)$ 的定义可得上述不等式成立，可得条件(H7)成立。因此，条件(H1)~(H7)成立。

最后，由计算可得

$\begin{array}{l}\Gamma =\underset{\frac{{\sigma }_3}{2}\le t\le {\sigma }_3}{\min }\delta \left(t\right)=\frac{1}{4},M=\frac{2198}{515},\\ M_1=\frac{267}{3320}\times 4^{\theta },N=\frac{1363}{24352},\\ \underset{t\in J}{\max }{\displaystyle {\int }_{\frac{{\sigma }_3}{2}}^{{\sigma }_3}G\left(t,s\right){\omega }^{+}\left(s\right)\text{d}s}=3\times 4^{\theta }\times \frac{113}{9634}.\end{array}$

取 $r_1=\frac{\text{1}}{5}$ ， $r=\frac{\text{1}}{\text{2}}$ ，则

${\beta }_{r_1}=\underset{\Gamma r_1\le u\le r_1}{\min }\left\{f(u)\right\}={\left(\frac{1}{20}+\sin \frac{1}{20}\right)}^{\theta },D=5,$

${{\mu }^{″}}_2=\frac{515}{4396},{\lambda }_0=\frac{4507}{794\times {\left(\frac{1}{5}+4\sin \frac{1}{20}\right)}^{\theta }},{\mu }_1=\frac{515}{2198}.$

取 $R=77$ ，可得 ${{\mu }^{\prime }}_2=\frac{318}{2741}$ 。则 ${\mu }_2=\max \left\{{{\mu }^{\prime }}_2,{{\mu }^{″}}_2\right\}={{\mu }^{″}}_2=\frac{515}{4396}$ 。

综上所述，由定理3.1可得，当 $\lambda \in [\frac{4507}{794\times {\left(\frac{1}{5}+4\sin \frac{1}{20}\right)}^{\theta }},+\infty )$ ， $\mu \in (\frac{515}{4396},\frac{515}{2198}]$ 时，系统(27)至少存在两个正解。

例4.2. 考虑如下二阶微分系统：

$\{\begin{array}{l}-u^{″}=a\left(t\right)\phi u+\lambda \omega \left(t\right){\left(u+\sin u\right)}^{\theta }\text{,}0<t<1,\\ -{\phi }^{″}=\mu b\left(t\right)u,0<t<1,\\ u\left(0\right)=u\left(1\right)=0,\\ \phi \left(0\right)=\phi \left(1\right)=0.\end{array}$ (30)

其中， $0<\theta <1$ ，且 $a\left(t\right),b\left(t\right),\omega \left(t\right)$ 定义与例4.1中相同。

结论：当 $\lambda \in (0,\frac{332}{267\times {\left(\frac{4}{5}+4\sin \frac{1}{5}\right)}^{\theta }}]$ ， $\mu \in (0,\frac{2575}{4396}]$ 时，若 $k_2$ 满足 $0<k_2<\frac{1660}{267\lambda \times 4^{\theta }}{\left(\frac{515}{4396\mu }\right)}^{1-\theta }$ ，则系统(30)至少存在三个正解。

例4.2的证明：类似与例4.1的证明，条件(H1)~(H7)成立。

取 $r_1=\frac{1}{5}$ ，则 ${\lambda }_1^{*}=\frac{332}{267\times {\left(\frac{4}{5}+4\sin \frac{1}{5}\right)}^{\theta }}$ ， ${\mu }_1^{*}=\frac{2575}{4396}$ 。由定理3.3可得，若 $k_2$ 满足

$0<k_2<\frac{1660}{267\lambda \times 4^{\theta }}{\left(\frac{515}{4396\mu }\right)}^{1-\theta },$

则系统(30)至少存在三个正解。

基金项目

国家自然科学基金面上项目(11471146)，北京市教育委员会专项(PXM2017_014224_000020)。