本文研究如下三阶常微分方程$m$点边值问题

$ \begin{eqnarray} \left \{ \begin{array}{l} x'''(t)=f(t, \ x(t)), \ t\in [0, 1], \\ x''(1)=0, \ x'(0)=0, \ x(1)=\mathop \sum \limits_{i = 1}^{m - 2}\beta_{i}x(\xi_{i}), \end{array} \right. \end{eqnarray} $

(1.1)

其中$0<\xi_{1}<\xi_{2}< \cdots< \xi_{m-2}<1, \ 0<\beta_{i}< 1, \ i=1, 2, \cdots, m-2, \ \sum\limits^{m-2}_{i=1}\beta_{i}<1$, $f$满足条件

(H) $f\in C([0, 1]\times [0, +\infty), \ (-\infty, +\infty)), \ f(t, 0)\geq 0, \ t\in [0, \ 1]$且不恒为0.

三阶常微分方程边值问题广泛出现在流体力学、天文学、弹性振动等问题中^[1], 对相关问题正解的研究近年来受到人们广泛的关注.利用非线性泛函分析方法, 对三阶微分方程边值问题正解与多个正解的存在性的研究取得了丰富的结果, 可参考文献[2-12]. Anderson ^[2]研究如下三阶三点边值问题

$ \left \{ \begin{array}{l} -x'''(t)+f(x(t))=0, \ t\in (0, 1), \\ x(0)=x'(t_{2})=x''(1)=0, \ \ t_{2} \in (0, 1), \end{array} \right. $

其中$f:R\rightarrow [0, +\infty)$连续, $1/2\leq t_{2}<1$, 文中建立了问题至少三个正解的存在性结果. Palamides ^[3]利用范数形式的锥拉伸锥压缩不动点定理, 证明了三阶三点边值问题

$ \left \{ \begin{array}{l} x'''(t)=a(t)f(t, x(t)), \ t\in (0, 1), \\ x''(\eta)=0, \ x(0)=x(1)=0, \ \eta \in (0, 1) \end{array} \right. $

多个正解的存在性. Guo ^[5]利用锥上不动点定理证明了三阶三点边值问题

$ \left \{ \begin{array}{l} x'''(t)=a(t)f(x(t)), \ t\in (0, 1), \\ x(0)=x'(0)=0, \ x'(1)=x'(\eta), \ \eta \in (0, 1) \end{array} \right. $

正解的存在性, 其中$f$非负且满足超线性或次线性条件.但在这些文献中, 均要求问题中非线性项非负, 非线性项变号时, 所得结论不再适用.目前, 对具变号非线性项的三阶常微分方程多点边值问题正解的讨论比较少见.本文首先建立与问题(1.1)等价的算子方程, 利用锥上不动点定理, 给出了问题(1.1)正解与多个正解的存在性结论.文中允许非线性项变号, 因此所得结果不同于已有文献.最后, 文中给出了具体的例子解释了结论的应用性.

定义 2.1 设$E$为实Banach空间.非空闭凸集$P\subset E$称为$E$上的锥, 如果满足

(1) $au\in P$, 对$u\in P, \ a\geq 0$;

(2) $u, -u\in P$, $u=0.$

定义 2.2 函数$x$称为[0, 1]上的凸泛函, 如果

$x(\lambda t_{1}+(1-\lambda)t_{2})\geq \lambda x(t_{1})+(1-\lambda)x(t_{2}), \lambda, \ t_{1}, \ t_{2}\in [0, \ 1].$

引理 2.1 (见文献[14])设$K$是Banach空间$X$上的锥, $D$为$X$上有界开子集, 满足$D_{K}=D\cap K\neq \emptyset $.设$A: \overline{D_{K}}\rightarrow K$全连续且$x\neq Ax$对$x\in \partial D_{K}$, 有

(1) 如果$\|Ax\|\leq \|x\|, \ x\in \partial D_{K}$, 则$i_{K}(A, \ D_{K})=1$.

(2) 若存在$e\in K\backslash \{0\}$使得$x\neq Ax+\lambda e, \ x\in \partial D_{K}, \ \lambda>0$, 则$i_{K}(A, \ D_{K})=0$.

(3) 设$U$是$X$中有界开集, $\overline{U} \subset D_{K}$.如果$i_{K}(A, \ D_{K})=1, \ i_{K}(A, \ U_{K})=0$或$i_{K}(A, \ D_{K})=0, \ i_{K}(A, \ U_{K})=1$, 则$A$在$D_{K}\backslash \overline{U_{K}}$中至少有一不动点.

首先考虑三阶微分方程边值问题

$ \begin{eqnarray} &&x'''(t)=y(t), t\in [0, 1], \end{eqnarray} $

(3.1)

$\begin{eqnarray} &&x''(1)=0, \ x'(0)=0, \ x(1)=\sum^{m-2}_{i=1}\beta_{i}x(\xi_{i}), \end{eqnarray} $

(3.2)

其中$0<\xi_{1}<\xi_{2}< \cdots< \xi_{m-2}<1, \ 0<\beta_{i}< 1, \ i=1, 2, \cdots, m-2, \mathop \sum \limits_{i = 1}^{m - 2} \beta_{i}<1$.

引理 3.1 记$\xi_{0}=0, \ \xi_{m-1}=1, \ \beta_{0}=\beta_{m-1}=0, \ y(t)\in C[0, 1]$, 问题(3.1), (3.2)等价于

$x(t)=\int^{1}_{0}G(t, s)\int^{1}_{s}y(\tau)d\tau ds, $

其中

$ \begin{eqnarray*} G(t, s)=\frac{1}{1-\mathop \sum \limits_{i = 0}^{m - 1}\beta_{i}}\left \{ \begin{array}{ll} (1-s)+\mathop \sum \limits_{k=i}^{m - 1}\beta_{k}(s-\xi_{k}), \\{t\leq s, \ \xi_{i-1}<s<\xi_{i}, \ i=1, \ 2, \ \cdots, \ m-1}, \\ (1-t)+\mathop \sum \limits_{k=0}^{i-1}\beta_{k}(t-s)+\mathop \sum \limits_{k=i}^{m - 1}\beta_{k}(t-\xi_{k}), \\{t\geq s, \ \xi_{i-1}<s<\xi_{i}, \ i=1, \ 2, \ \cdots, \ m-1 }.\\ \end{array} \right. \end{eqnarray*} $

证 (3.1)式两侧积分并考虑条件$x''(1)=0$, 有

$ \begin{eqnarray}-x''(t)=\int^{1}_{t}y(s)ds.\end{eqnarray} $

(3.3)

设$G(t, s)$是边值问题

$ \begin{eqnarray}&&-x''(t)=0, \end{eqnarray} $

(3.4)

$\begin{eqnarray} &&x'(0)=0, \ x(1)=\mathop \sum \limits_{i = 0}^{m - 1}\beta_{i}x(\xi_{i})\end{eqnarray} $

(3.5)

的Green函数, 由(3.4)式, 可设

$ G(t, s)=\left \{ \begin{array}{cc} A+Bt, &{t\leq s, \ \xi_{i-1}<s<\xi_{i}, }\\ C+Dt,&{t\geq s, \ \xi_{i-1}<s<\xi_{i}.} \end{array} \right. $

由Green函数定义及性质, 结合边值条件(3.5), 有

$ \left \{ \begin{array}{l} A+Bs=C+Ds, \\ B-D=1, \\ B=0, \\ C+D=\mathop \sum \limits_{k = 0}^{i - 1}\beta_{k}(A+B\xi_{k})+\mathop \sum \limits_{k = i}^{m - 1}\beta_{k}(C+D\xi_{k}), \end{array} \right. $

这样

$ \begin{eqnarray*}&&A=(1-s+\mathop \sum \limits_{k = i}^{m - 1}\beta_{k}(s-\xi_{k}))/(1-\mathop \sum \limits_{i = 0}^{m - 1}\beta_{i}), \ C=(1-\mathop \sum \limits_{k = 0}^{i - 1}\beta_{k}s-\mathop \sum \limits_{k = i}^{m - 1}\beta_{k}\xi_{k})/(1-\mathop \sum \limits_{i = 0}^{m - 1}\beta_{i}), \\ &&B=0, \ \ D=-1.\\ &&G(t, s)=\frac{1}{1-\mathop \sum \limits_{i = 0}^{m - 1}\beta_{i}}\left \{ \begin{array}{ll} (1-s)+\mathop \sum \limits_{k = i}^{m - 1}\beta_{k}(s-\xi_{k}), \\ {t\leq s, \ \xi_{i-1}<s<\xi_{i}, \ i=1, \ 2, \ \cdots, \ m-1}, \\ (1-t)+\mathop \sum \limits_{k = 0}^{i - 1}\beta_{k}(t-s)+\mathop \sum \limits_{k = i}^{m - 1}\beta_{k}(t-\xi_{k}), \\ {t\geq s, \ \xi_{i-1}<s<\xi_{i}, \ i=1, \ 2, \ \cdots, \ m-1}.\\ \end{array} \right. \end{eqnarray*} $

考虑(3.3)式, 问题(3.1), (3.2)等价于

$x(t)=\int^{1}_{0}G(t, s)\int^{1}_{s}y(\tau)d\tau ds.$

引理 3.2 函数$G(t, \ s)$满足$G(t, s)\geq 0, \ t, \ s\in [0, 1]$.

证 对$\xi_{i-1}\leq s\leq \xi_{i}, \ i=1, \ 2, \ \cdots, \ m-1, \ t\leq s$,

$(1-s)+\mathop \sum \limits_{k = i}^{m - 1}\beta_{k}(s-\xi_{k}) \geq \mathop \sum \limits_{k = i}^{m - 1}\beta_{k}(1-\xi_{k})\geq 0.$

对$\xi_{i-1}\leq s\leq \xi_{i}, \ i=1, \ 2, \ \cdots, \ m-1, \ t\geq s$,

$(1-t)+\mathop \sum \limits_{k = 0}^{i - 1}\beta_{k}(t-s)+\mathop \sum \limits_{k = i}^{m - 1}\beta_{k}(t-\xi_{k}) \geq \mathop \sum \limits_{k = 0}^{i - 1}\beta_{k}(1-s)+\mathop \sum \limits_{k = i}^{m - 1}\beta_{k}(1-\xi_{k}) \geq 0.$

综上, $G(t, s)\geq 0, \ t, \ s\in [0, 1]$.

引理 3.3 设$y(t)\geq 0, t\in [0, 1]$, $x(t)$是边值问题(3.1), (3.2)的解, 则有

$\min\limits_{0\leq t\leq 1}|x(t)|\geq \delta\max\limits_{0\leq t\leq 1}|x(t)|, $

其中

$\delta=(\mathop \sum \limits_{i = 1}^{m - 2}\beta_{i}(1-\xi_{i}))/(1-\mathop \sum \limits_{i = 1}^{m - 2}\beta_{i}\xi_{i})>0.$

证 由$x'''(t)=y(t)\geq 0, \ t\in [0, 1]$知$x''(t)$在[0, 1]上单调递增.考虑到$x''(1)=0$, 有$x''(t)\leq 0, \ t\in (0, 1)$.结合$x'(0)=0$, 必有

$\max\limits_{0\leq t\leq 1}x(t)=x(0), \min\limits_{0\leq t\leq 1}x(t)=x(1).$

由$x(t)$的凸性,

$\xi_{i}(x(1)-x(0))\leq x(\xi_{i})-x(0).$

上式两端乘以$\beta_{i}, \ i=1, \ 2, \ \cdots, \ m-1$, 结合边值条件, 有

$(1-\mathop \sum \limits_{i = 1}^{m - 2}\beta_{i}\xi_{i})x(1)\geq \mathop \sum \limits_{i = 1}^{m - 2}\beta_{i}(1-\xi_{i})x(0).$

设Banach空间$E=C[0, \ 1]$及范数$\|x\|=\max\limits_{0\leq t\leq 1}|x(t)|, \ x\in E.$定义$E$中锥$K$,

$ \begin{eqnarray*}&&K=\{x\in E\ |\ x(t)\geq 0, \ x''(1)=0, \ x'(0)=0, \\ &&x(1)=\mathop \sum \limits_{i = 1}^{m - 2}\beta_{i}x(\xi_{i}), \ x''(t)\leq 0, \ t\in [0, \ 1]\}.\end{eqnarray*} $

定义

$ \begin{eqnarray*} &&\phi(t)=\min\{t, \ 1-t\}, \ t\in (0, 1), \\ &&K_{\rho}=\{x\in K:\|x\|<\rho\}, \ K^{\ast}_{\rho}=\{x\in K:\rho \phi(t)<x(t)<\ \rho\}, \\ &&\Omega_{\rho}=\{x\in K: x(t)\geq 0, x''(t)\leq 0, \ t\in [0, 1], \ \min\limits_{0\leq t\leq 1}x(t)<\delta\rho\}.\end{eqnarray*} $

引理 3.4 $\Omega_{\rho}$具有如下性质:

(1) $\Omega_{\rho}\subset K_{\rho}$是$K_{\rho}$中开集.

(2) 设$x\in \partial \Omega_{\rho}$, 则$\delta \rho\leq x(t)\leq \rho, \ t\in [0, \ 1]$.

记

$ \begin{eqnarray*} &&f^{\rho}_{\delta \rho}=\min\{\frac{f(t, \ x)}{\rho}:\ t\in [0, \ 1], \ x\in [\delta \rho, \ \rho] \}, \\ &&f^{\rho}_{\phi(t) \rho}=\max\{\frac{f(t, \ x)}{\rho}:\ t\in [0, \ 1], \ x\in [\rho \phi(t), \ \rho] \}, \\ &&f^{\rho}_{0}=\max\{\frac{f(t, \ x)}{\rho}:\ t\in [0, \ 1], \ x\in [0, \ \rho] \}, \\ &&m=\frac{1}{\max\limits_{0\leq t\leq 1}\int^{1}_{0}(1-s)G(t, \ s)ds}, \ M=\frac{1}{\delta \min\limits_{0\leq t\leq 1}\int^{1}_{0}(1-s)G(t, s)ds}.\end{eqnarray*} $

定理 3.1 假定条件(H)成立.此外, 若条件(H1)成立:

(H1)存在正常数$\rho_{1}, \ \rho_{2}, \ \rho_{3}\in (0, \ \infty)$满足$\rho_{1}<\delta \rho_{2}<\rho_{2}<\rho_{3}$使得

(1) $ f(t, \ u)>0, \ t\in [0, \ 1], u\in [\rho_{1} \phi(t), \ \infty), $

(2) $f^{\rho_{1}}_{\rho_{1}\phi(t)}<m, f^{\rho_{2}}_{\delta \rho_{2}}>M, \ f^{\rho_{3}}_{\rho_{3}\phi(t)}\leq m, $

则问题(1.1)在$K$中至少有三个解.假定条件(H)成立.此外, 若条件(H2)成立:

(H2)存在正常数$\rho_{1}, \ \rho_{2}, \ \rho_{3}\in (0, \ \infty)$满足$\rho_{1}<\rho_{2}< \rho_{3}$使得

(3) $f(t, \ u)>0, \ t\in [0, \ 1], u\in [\min\{\rho_{1}, \ \rho_{2}\phi(t)\}, \ \infty), $

(4) $ f^{\rho_{1}}_{\delta\rho_{1}}>M, \ f^{\rho_{2}}_{\rho_{2}\phi(t)}<m, \ f^{\rho_{3}}_{\delta\rho_{3}}\geq M, $

则问题(1.1)在$K$中至少有两个解.

证 首先设条件(H1)成立.定义辅助函数$f^{\ast}(t, \ x)\in C([0, \ 1]\times [0, \infty), \ [0, \infty)):$

$ f^{\ast}(t, \ x)=\left \{ \begin{array}{ll} f(t, \ x),&x\geq \rho_{1}\phi(t), \\ f(t, \ \rho_{1}\phi(t)),&0\leq x<\rho_{1}\phi(t)).\\ \end{array} \right. $

考虑如下辅助边值问题

$ (\ast)\left \{ \begin{array}{l} x'''(t)=f^{\ast}(t, \ x(t)), \ t\in [0, 1], \\ x''(1)=0, \ x'(0)=0, \ x(1)=\mathop \sum \limits_{i = 1}^{m - 2}\beta_{i}x(\xi_{i}). \end{array} \right. $

定义算子

$(Tx)(t)=\int^{1}_{0}G(t, s)\int^{1}_{s}f^{\ast}(\tau, \ x(\tau))d\tau ds, $

显然$T:K\rightarrow K$是全连续的.由条件(H1), 有

$f^{\ast \rho_{1}}_{\rho_{1}\phi(t)}<m, f^{\ast \rho_{2}}_{\delta \rho_{2}}>M, \ f^{\ast \rho_{3}}_{\rho_{3}\phi(t)}\leq m.$

则对$x\in \partial K^{\ast}_{\rho_{1}}, $有

$ \begin{eqnarray*} \|(Tx)(t)\|&=&\max\limits_{0\leq t\leq 1}|\int^{1}_{0}G(t, s)\int^{1}_{s}f^{\ast}(\tau, \ x(\tau))d\tau ds|\\ &\leq&\max\limits_{0\leq t\leq 1} \int^{1}_{0}G(t, s)\int^{1}_{s}|f^{\ast}(\tau, \ x(\tau))|d\tau ds\\ &<&m\rho_{1}\int^{1}_{0}(1-s)G(t, s)ds\leq \rho_{1}, \end{eqnarray*} $

这表明$\|Tx\|\leq \|x\|, \ x\in \partial K^{\ast}_{\rho_{1}}$, 即$i_{K}(T, \ K^{\ast}_{\rho_{1}})=1$.设$e(t)\equiv 1, \ t\in [0, \ 1]$, 则$e\in \partial K_{1}$.此时必有$x\neq Tx+\lambda e, \ x\in \partial \Omega_{\rho_{2}}, \ \lambda \geq 0.$否则, 必存在$x_{0}\in \partial \Omega_{\rho_{2}}, \ \lambda_{0}\geq 0$使得$x_{0}=Tx_{0}+\lambda_{0} e.$但是

$ \begin{eqnarray*} x_{0}(t)&=&(Tx_{0})(t)+\lambda_{0} e(t)\geq \delta \|Tx_{0}\|+\lambda_{0}\\ &=&\delta \max\limits_{0\leq t\leq 1}|\int^{1}_{0}G(t, s)\int^{1}_{s}f^{\ast}(\tau, \ x(\tau))d\tau ds|+\lambda_{0}\\ &\geq& \delta\rho_{2} M\int^{1}_{0}(1-s)G(t, \ s)ds+\lambda_{0}\geq \rho_{2}+\lambda_{0}.\end{eqnarray*} $

这表明$\rho_{2}\geq \rho_{2}+\lambda_{0}, $矛盾.由引理2.1, $i_{K}(T, \ \Omega_{\rho_{2}})=0$.同理可证, $i_{K}(T, \ K^{\ast}_{\rho_{3}})=1$.这样, 辅助边值问题存在三个正解$x_{1}, \ x_{2}, \ x_{3}$使得

$x_{1}\in K^{\ast}_{\rho_{1}}, \ x_{2}\in \Omega_{\rho_{2}}\backslash \overline{K^{\ast}_{\rho_{1}}}, x_{3}\in K^{\ast}_{\rho_{3}}.$

易验证辅助边值问题$(\ast)$在$[\rho_{1}\phi(t), \ \infty)$中存在三个正解$x_{1}, \ x_{2}, \ x_{3}$, 这说明边值问题(1.1)存在至少三个正解.条件(H2)成立时, 正解的存在性同理可证.完全类似定理3.1, 可证得如下结论.

定理 3.2 设条件(H)成立.此外, 下列条件之一成立:

(H3)存在正常数$\rho_{1}, \ \rho_{2}\in (0, \ \infty)$满足$\rho_{1}<\delta\rho_{2}$使得

(5) $ f(t, \ x)>0, \ t\in [0, \ 1], x\in [\rho_{1} \phi(t), \ \infty), $

(6) $ f^{\rho_{1}}_{\rho_{1}\phi(t)}\leq m, f^{\rho_{2}}_{\delta \rho_{2}}\geq M;$

(H4)存在正常数$\rho_{1}, \ \rho_{2}\in (0, \ \infty)$满足$\rho_{1}<\rho_{2}$使得

(7) $ f(t, \ x)>0, \ t\in [0, \ 1], x\in [\min\{\rho_{1}, \ \rho_{2}\phi(t)\}, \ \infty), $

(8) $ f^{\rho_{1}}_{\delta\rho_{1}}\geq M, f^{\rho_{2}}_{\rho_{2}\phi(t)}\leq m, $

则边值问题(1.1)在$K$中至少有一正解.

考虑如下三阶四点边值问题

$ \begin{eqnarray}\left \{ \begin{array}{l} x'''(t)+f(t, \ x(t))=0, \ t\in [0, 1], \\ x''(1)=0, \ x'(0)=0, \ x(1)=\frac{1}{2}x(\frac{1}{3})+\frac{1}{3}x(\frac{2}{3}), \end{array} \right. \end{eqnarray} $

(4.1)

其中

$ f(t, \ x)=\left \{ \begin{array}{ll} (x-\frac{1}{2}\phi(t))^{2}-\frac{1}{8}(\phi(t))^{2}, &{t\in [0, \ 1], x\in [0, \ \phi(t)], }\\ \frac{1}{8}(x-\phi(t))^{2}+\frac{1}{8}(\phi(t))^{2}, &{t\in [0, \ 1], \ x\in [\phi(t), \ 1]}, \\ \frac{1}{8}-\frac{1}{4}\phi(t)+\frac{1}{4}(\phi(t))^{2}+(x-1)^{2}, &{t\in [0, \ 1], \ x\in [1, \ 8]}, \\ \frac{1}{8}-\frac{1}{4}\phi(t)+\frac{1}{4}(\phi(t))^{2}+49-\sqrt{x-8}, &{t\in [0, \ 1], \ x\in [8, \ 11]}, \\ \frac{1}{8}-\frac{1}{4}\phi(t)+\frac{1}{4}(\phi(t))^{2}+49-\sqrt{3}+\sin(x-11), &{t\in [0, \ 1], \ x\in [11, \ 122]}. \end{array} \right. $

计算可得函数$G(t, s)$由下式给出

$ G(t, \ s)=\left \{ \begin{array}{ll} -s+\frac{11}{3}, &t\leq s, 0\leq s\leq \frac{1}{3}, \\ -t+\frac{11}{3}, &t\geq s, 0\leq s\leq \frac{1}{3}, \\ -4s+\frac{14}{3}, &t\leq s, \frac{1}{3}\leq s\leq \frac{2}{3}, \\ -3s-2t+\frac{14}{3}, &t\geq s, \frac{1}{3}\leq s\leq \frac{2}{3}, \\ 6-6s, &t\leq s, \frac{2}{3}\leq s\leq 1, \\ 6-t-5s, &t\geq s, \frac{2}{3}\leq s\leq 1. \end{array} \right. $

取$\rho_{1}=1, \rho_{2}=11, \ \rho_{3}=122$, 有$\delta=\frac{8}{11}, \ m=\frac{81}{122}, \ M=\frac{891}{652}.$易验证$f(t, \ x)$满足条件(H)和

(1) $f(t, \ x)>0, \ \ \ \ t\in [0, \ 1], \ \ \ x\in [\rho_{1} \phi(t), \ \infty); $

(2)~$f(t, \ x)<\frac{81}{122}, \ \ \ t\in [0, 1], \ \ \ x\in [\phi(t), \ 1];$

(3) $f(t, \ x)>\frac{9801}{652}, \ \ \ t\in [0, 1], \ \ \ x\in [8, \ 11];$

(4) $f(t, \ x)<81, \ \ \ t\in [0, 1], \ \ \ x\in [122\phi(t), \ 122].$

这样定理3.1的所有条件均满足, 因此, 问题(4.1)至少存在三个单调递增的凸正解.

注 问题(4.1)中非线性项是变号的, 已有文献中对三阶边值问题正解存在性的讨论, 就作者所知, 均无法适用此问题.