﻿ 一类含一阶导数的三阶边值问题正解的存在性 Existence of Positive Solutions for a Class of Third-Order Boundary Value Problems with First Derivative

Pure Mathematics
Vol. 11  No. 02 ( 2021 ), Article ID: 40433 , 11 pages
10.12677/PM.2021.112032

Existence of Positive Solutions for a Class of Third-Order Boundary Value Problems with First Derivative

Jiao Zhao

College of Mathematics and Statistics, Northwest Normal University, Lanzhou Gansu

Received: Jan. 8th, 2021; accepted: Feb. 11th, 2021; published: Feb. 19th, 2021

ABSTRACT

This paper considers existence of positive solutions for a class of third-order ordinary differential equations boundary value problems with first derivative $\left\{\begin{array}{l}-{u}^{\left(3\right)}\left(t\right)=\lambda f\left(t,u\left(t\right),{u}^{\prime }\left(t\right)\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}t\in \left[0,1\right],\hfill \\ u\left(0\right)={u}^{\prime }\left(0\right)=0,\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }{u}^{\prime }\left(1\right)=\alpha {u}^{\prime }\left(\eta \right)\hfill \end{array}$ where $\lambda$ is a positive parameter, $0<\eta <1$ and $1<\alpha <\frac{1}{\eta }$ are given constants. $f\left(t,u,p\right):\left[0,1\right]×\left[0,\infty \right)×\left[0,\infty \right)\to \left[0,\infty \right)$ is a continuous function, and $f\left(t,0,0\right)=0$. The proof of the main results is based upon global bifurcation techniques.

Keywords:Three-Order ODE, Principle Eigenvalue, Bifurcation, Positive Solution

1. 引言

$\left\{\begin{array}{l}-{u}^{\left(3\right)}\left(t\right)=a\left(t\right)f\left(u\left(t\right)\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}t\in \left[0,1\right],\hfill \\ u\left(0\right)={u}^{\prime }\left(0\right)=0,\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }{u}^{\prime }\left(1\right)=\alpha {u}^{\prime }\left(\eta \right)\hfill \end{array}$ (1.1)

i) $\underset{u\to 0}{\mathrm{lim}}\frac{f\left(u\right)}{u}=\infty$$\underset{u\to \infty }{\mathrm{lim}}\frac{f\left(u\right)}{u}=0$

ii) $\underset{u\to 0}{\mathrm{lim}}\frac{f\left(u\right)}{u}=\infty$$\underset{u\to \infty }{\mathrm{lim}}\frac{f\left(u\right)}{u}=0$

$\left\{\begin{array}{l}-{u}^{\left(3\right)}\left(t\right)=\lambda f\left(t,u\left(t\right),{u}^{\prime }\left(t\right),{u}^{″}\left(t\right)\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}t\in \left[0,1\right],\hfill \\ u\left(0\right)={u}^{\prime }\left(0\right)=0,\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }{u}^{\prime }\left(1\right)=\alpha {u}^{\prime }\left(\eta \right)\hfill \end{array}$ (1.2)

(B1) $\lambda$ 为正参数；

(B2) $f:\left[0,1\right]×{ℝ}^{3}\to \left[0,\infty \right)$${L}^{1}$ -Caratheodory函数;

(B3) 存在 ${f}_{0}$,${f}^{\infty }$，使得

${f}_{0}=\underset{|x|,|y|,|z|\to 0}{\mathrm{lim}}\underset{t\in \left[0,1\right]}{\mathrm{min}}\frac{f\left(t,x,y,z\right)}{|x|+|y|+|z|},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{f}^{\infty }=\underset{|x|,|y|,|z|\to 0}{\mathrm{lim}}\underset{t\in \left[0,1\right]}{\mathrm{max}}\frac{f\left(t,x,y,z\right)}{|x|+|y|+|z|}.$

$\lambda \in \left(\frac{1}{\underset{_}{\Lambda }{f}_{0}},\frac{1}{\stackrel{¯}{\Lambda }{f}^{\infty }}\right),$

$\left\{\begin{array}{l}-{u}^{\left(3\right)}\left(t\right)=\lambda f\left(t,u\left(t\right),{u}^{\prime }\left(t\right)\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}t\in \left[0,1\right],\hfill \\ u\left(0\right)={u}^{\prime }\left(0\right)=0,\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }{u}^{\prime }\left(1\right)=\alpha {u}^{\prime }\left(\eta \right)\hfill \end{array}$ (1.3)

(H1) $\lambda$ 为正参数， $0<\eta <1$,$1<\alpha <\frac{1}{\eta }$ 为给定常数；

(H2) $f:\left[0,1\right]×\left[0,\infty \right)×\left[0,\infty \right)\to \left[0,\infty \right)$ 为连续函数，且 $f\left(t,0,0\right)=0$

(H3) 存在 $a\left(t\right),b\left(t\right)\in C\left(\left[0,1\right],\left(0,\infty \right)\right)$，使得

$f\left(t,u,p\right)=a\left(t\right)+\circ \left(|\left(u,p\right)|\right),|\left(u,p\right)|\to \left(0,0\right)$ 对于 $t\in \left[0,1\right]$ 一致成立

$f\left(t,u,p\right)=b\left(t\right)+\circ \left(|\left(u,p\right)|\right),|\left(u,p\right)|\to \infty$ 对于 $t\in \left[0,1\right]$ 一致成立，

$\left\{\begin{array}{l}-{u}^{\left(3\right)}\left(t\right)=\lambda A\left(t\right)u,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}t\in \left[0,1\right],\hfill \\ u\left(0\right)={u}^{\prime }\left(0\right)=0,\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }{u}^{\prime }\left(1\right)=\alpha {u}^{\prime }\left(\eta \right)\hfill \end{array}$ (1.4)

i) ${\lambda }_{1}\left(a\right)<\lambda <{\lambda }_{1}\left(b\right)$

ii) ${\lambda }_{1}\left(b\right)<\lambda <{\lambda }_{1}\left(a\right)$

2. 预备知识

$x=\mu Lx+N\left(\mu ,x\right),\text{ }\text{ }\text{ }\text{ }\mu \in ℝ,\text{ }\text{ }\text{ }\text{ }x\in E$ (2.1)

(A1) 算子 $L:E\to E$ 为线性紧算子；

(A2) 非线性算子 $N:ℝ×E\to E$ 全连续，且 $\underset{x\to 0}{\mathrm{lim}}\frac{‖N\left(\mu ,x\right)‖}{‖x‖}=0$

(A3) ${\mu }_{0}$ 为L在E中的本征值，且其代数重数 $\chi \left({\mu }_{0}\right)$ 为奇数，其中

$\chi \left({\mu }_{0}\right)=\mathrm{dim}\left\{\underset{m=1}{\overset{\infty }{\cup }}\mathrm{ker}\left({\left(I-{\mu }_{0}L\right)}^{m}\right)\right\}$

$\mathcal{C}\left({\mu }_{0}\right)$ 为(2.1)式的非平凡解的闭包中包含点 $\mathcal{C}\left({\mu }_{0}\right)$ 的连通分支，则有以下两种情形之一出现：

i) $\mathcal{C}\left({\mu }_{0}\right)$ 无界；

ii) $\mathcal{C}\left({\mu }_{0}\right)$ 还连接 $\left(\stackrel{˜}{\mu },0\right)$$\stackrel{˜}{\mu }$ 是不同于 ${\mu }_{0}$ 的本征值。

$\left\{\begin{array}{l}-{u}^{\left(3\right)}\left(t\right)=h\left(t\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}t\in \left[0,1\right],\hfill \\ u\left(0\right)={u}^{\prime }\left(0\right)=0,\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }{u}^{\prime }\left(1\right)=\alpha {u}^{\prime }\left(\eta \right)\hfill \end{array}$ (2.2)

$u\left(t\right)={\int }_{0}^{1}G\left(t,s\right)h\left(s\right)\text{d}s.$ (2.3)

$G\left(t,s\right)=\frac{1}{2\left(1-\alpha \eta \right)}\left\{\begin{array}{l}\left(2ts-{s}^{2}\right)\left(1-\alpha \eta \right)+{t}^{2}s\left(\alpha -1\right),\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }s\le \mathrm{min}\left\{\eta ,t\right\},\\ {t}^{2}\left(1-\alpha \eta \right)+{t}^{2}s\left(\alpha -1\right),\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }\text{\hspace{0.17em}}\text{ }t\le s\le \eta ,\\ \left(2ts-{s}^{2}\right)\left(1-\alpha \eta \right)+{t}^{2}\left(\alpha \eta -s\right),\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{\hspace{0.17em}}\eta \le s\le t,\\ {t}^{2}\left(1-s\right),\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\mathrm{max}\left\{\eta ,t\right\}\le s.\end{array}$

$0\le G\left(t,s\right)\le {g}_{0}\left(s\right)=\frac{1+\alpha }{1-\alpha \eta }s\left(1-s\right),\forall \left(t,s\right)\in \left[0,1\right]×\left[0,1\right].$ (2.4)

$G\left(t,s\right)\ge {\kappa }_{0}{g}_{0}\left(s\right),\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\forall \left(t,s\right)\in \left[\frac{\eta }{\alpha },\eta \right]×\left[0,1\right],$ (2.5)

$\frac{\partial G}{\partial t}\left(t,s\right)=\frac{1}{1-\alpha \eta }\left\{\begin{array}{l}s\left(1-\alpha \eta \right)+ts\left(\alpha -1\right),\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }s\le \mathrm{min}\left\{\eta ,t\right\},\\ t\left(1-\alpha \eta \right)+ts\left(\alpha -1\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}t\le s\le \eta ,\\ s\left(1-\alpha \eta \right)+t\left(\alpha \eta -s\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\eta \le s\le t,\\ t\left(1-s\right),\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{max}\left\{\eta ,t\right\}\le s.\end{array}$

$0<\eta <1$$1<\alpha <\frac{1}{\eta }$，则有

i) $0\le \frac{\partial G}{\partial t}\left(t,s\right)\le {g}_{1}\left(s\right)=\frac{1-s}{1-\alpha \eta },\forall \left(t,s\right)\in \left[0,1\right]×\left[0,1\right]$

ii) $\frac{\partial G}{\partial t}\left(t,s\right)\ge {\kappa }_{1}{g}_{1}\left(s\right),\forall \left(t,s\right)\in \left[\frac{\eta }{\alpha },\eta \right]×\left[0,1\right],0<{\kappa }_{1}=\eta <1$

$Y=C\left[0,1\right]$$E={C}^{1}\left(\left[0,1\right],ℝ\right)$ 分别在范数 ${‖u‖}_{\infty }=\underset{t\in \left[0,1\right]}{\mathrm{max}}|x\left(t\right)|$$‖u‖=\mathrm{max}\left\{{‖u‖}_{\infty },{‖{u}^{\prime }‖}_{\infty }\right\}$ 下构成Banach空间。我们定义锥

$K=\left\{u\in E:u\left(t\right)\ge 0,t\in \left[0,1\right],{u}^{\prime }\left(t\right)\ge 0,t\in \left[0,1\right],\underset{t\in \left[\frac{\eta }{\alpha },\eta \right]}{\mathrm{min}}u\left(t\right)\ge {\kappa }_{0}{‖u‖}_{\infty },\underset{t\in \left[\frac{\eta }{\alpha },\eta \right]}{\mathrm{min}}{u}^{\prime }\left(t\right)\ge {\kappa }_{1}{‖{u}^{\prime }‖}_{\infty }\right\}.$

$Tu\left(t\right)=\lambda {\int }_{0}^{1}G\left(t,s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\text{d}s,t\in \left[0,1\right].$

$Tu\left(t\right)=\lambda {\int }_{0}^{1}G\left(t,s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\text{d}s\le \lambda {\int }_{0}^{1}{g}_{0}\left(s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\text{d}s.$

${‖Tu‖}_{\infty }\le \lambda {\int }_{0}^{1}{g}_{0}\left(s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\text{d}s.$

$Tu\left(t\right)=\lambda {\int }_{0}^{1}G\left(t,s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\text{d}s\ge \lambda {\int }_{0}^{1}{\kappa }_{0}{g}_{0}\left(s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\text{d}s\ge {\kappa }_{0}{‖Tu‖}_{\infty },$

$\underset{t\in \left[\frac{\eta }{\alpha },\eta \right]}{\mathrm{min}}Tu\left(t\right)\ge {\kappa }_{0}{‖Tu‖}_{\infty }.$

${\left(Tu\right)}^{\prime }\left(t\right)\le \lambda {\int }_{0}^{1}\frac{\partial G}{\partial t}\left(t,s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\text{d}s.$

${\left(Tu\right)}^{\prime }\left(t\right)\le \lambda {\int }_{0}^{1}{g}_{1}\left(s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\text{d}s,$ (2.6)

${‖{\left(Tu\right)}^{\prime }‖}_{\infty }\le \lambda {\int }_{0}^{1}{g}_{1}\left(s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\text{d}s.$

${\left(Tu\right)}^{\prime }\left(t\right)\ge \lambda {\int }_{0}^{1}{\kappa }_{1}{g}_{1}\left(s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\text{d}s\ge {\kappa }_{1}{‖{\left(Tu\right)}^{\prime }‖}_{\infty },$

$\underset{t\in \left[\frac{\eta }{\alpha },\eta \right]}{\mathrm{min}}Tu\left(t\right)\ge {\kappa }_{1}{‖{\left(Tu\right)}^{\prime }‖}_{\infty }.$

$\underset{t\in \left[\frac{\eta }{\alpha },\eta \right]}{\mathrm{min}}Tu\left(t\right)\ge \mathrm{max}\left\{{\kappa }_{0}{‖Tu‖}_{\infty },{\kappa }_{1}{‖{\left(Tu\right)}^{\prime }‖}_{\infty }\right\}.$

$B=\left\{u\in E;‖u‖\le r,\forall r>0\right\}$

$\begin{array}{c}{‖Tu‖}_{\infty }=\underset{t\in \left[0,1\right]}{\mathrm{sup}}|\lambda {\int }_{0}^{1}G\left(t,s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\text{d}s|\\ \le \lambda {\int }_{0}^{1}{g}_{0}\left(s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\text{d}s\\ \le \lambda {\int }_{0}^{1}{g}_{0}\left(s\right){\varphi }_{r}\left(s\right)\text{d}s:={M}_{0}\end{array}$

$\begin{array}{c}{‖{\left(Tu\right)}^{\prime }‖}_{\infty }=\underset{t\in \left[0,1\right]}{\mathrm{sup}}|\lambda {\int }_{0}^{1}\frac{\partial G}{\partial t}\left(t,s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\text{d}s|\\ \le \lambda {\int }_{0}^{1}{g}_{1}\left(s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\text{d}s\\ \le \lambda {\int }_{0}^{1}{g}_{1}\left(s\right){\varphi }_{r}\left(s\right)\text{d}s:={M}_{1}\end{array}$

$‖Tu‖\le \mathrm{max}\left\{{M}_{0},{M}_{1}\right\},\forall u\in B.$

$\begin{array}{c}|Tu\left({t}_{1}\right)-Tu\left({t}_{2}\right)|\le \lambda {\int }_{0}^{1}|G\left({t}_{1},s\right)-G\left({t}_{2},s\right)|f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\text{d}s\\ \le \lambda {\int }_{0}^{1}|G\left({t}_{1},s\right)-G\left({t}_{2},s\right)|{\varphi }_{r}\left(s\right)\text{d}s.\end{array}$

$\begin{array}{c}|{\left(Tu\right)}^{\prime }\left({t}_{1}\right)-{\left(Tu\right)}^{\prime }\left({t}_{2}\right)|\le \lambda {\int }_{0}^{1}|\frac{\partial G}{\partial t}\left({t}_{1},s\right)-\frac{\partial G}{\partial t}\left({t}_{2},s\right)|f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\text{d}s\\ \le \lambda {\int }_{0}^{1}|\frac{\partial G}{\partial t}\left({t}_{1},s\right)-\frac{\partial G}{\partial t}\left({t}_{2},s\right)|{\varphi }_{r}\left(s\right)\text{d}s\end{array}$

3. 主要结果的证明

$Lu:=-{u}^{‴},u\in D\left(L\right),$ (3.1)

$f\left(t,u,p\right)=a\left(t\right)u+\zeta \left(t,u,p\right),$ (3.2)

$f\left(t,u,p\right)=b\left(t\right)u+\xi \left(t,u,p\right).$ (3.3)

$\underset{|\left(u,p\right)|\to 0}{\mathrm{lim}}\frac{\zeta \left(t,u,p\right)}{|\left(u,p\right)|}=0$$t\in \left[0,1\right]$ 一致成立， (3.4)

$\underset{|\left(u,p\right)|\to \infty }{\mathrm{lim}}\frac{\xi \left(t,u,p\right)}{|\left(u,p\right)|}=0$$t\in \left[0,1\right]$ 一致成立， (3.5)

$\stackrel{˜}{\xi }\left(r\right)=\mathrm{max}\left\{\xi \left(t,u,p\right)|0\le |\left(u,p\right)|\le r,t\in \left[0,1\right]\right\},$ (3.6)

$\stackrel{˜}{\xi }$ 非减且

$\underset{r\to \infty }{\mathrm{lim}}\frac{\stackrel{˜}{\xi }\left(r\right)}{r}=0.$ (3.7)

$Lu=\theta \lambda a\left(t\right)u\left(t\right)+\theta \lambda \zeta \left(t,u,{u}^{\prime }\right).$ (3.8)

$\begin{array}{c}u\left(t\right)=\theta \left[{\int }_{0}^{1}G\left(t,s\right)\lambda a\left(s\right)u\left(s\right)\text{d}s\right]\\ :=\left(\theta {L}^{-1}\lambda a\left(\cdot \right)u\left(\cdot \right)+\theta {L}^{-1}\left[\lambda \zeta \left(\left(\cdot \right),u\left(\cdot \right),{u}^{\prime }\left(\cdot \right)\right)\text{d}s\right]\right)\left(t\right).\end{array}$

$\begin{array}{l}‖{L}^{-1}\left[\lambda \zeta \left(\left(\cdot \right),u\left(\cdot \right),{u}^{\prime }\left(\cdot \right)\right)\text{d}s\right]‖\\ =\underset{t\in \left[0,1\right]}{\mathrm{max}}\left\{{\int }_{0}^{1}G\left(t,s\right)\zeta \left(t,u,{u}^{\prime }\right)\text{d}s,{\int }_{0}^{1}{G}^{\prime }\left(t,s\right)\zeta \left(t,u,{u}^{\prime }\right)\text{d}s\right\}\\ \le N{‖\zeta \left(\left(\cdot \right),u\left(\cdot \right),{u}^{\prime }\left(\cdot \right)\right)‖}_{\infty }.\end{array}$

$‖{L}^{-1}\left[\zeta \left(t,{u}^{\prime }\left(\cdot \right),{u}^{\prime }\left(\cdot \right)\right)\right]‖=o\left(‖u‖\right),\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }u\to 0.$

${\mu }_{n}+‖{y}_{n}‖\to \infty$

$\left(\frac{{\lambda }_{1}\left(a\right)}{\lambda },\frac{{\lambda }_{1}\left(b\right)}{\lambda }\right)\subseteq \left\{\lambda \in ℝ|\exists \left(\lambda ,u\right)\in {\mathcal{C}}^{+}\right\}.$

${\mu }_{n}\subset \left(0,M\right],$ (3.10)

$‖{y}_{n}‖\to \infty .$ (3.11)

$L{y}_{n}-{\mu }_{n}\lambda a\left(t\right){y}_{n}={\mu }_{n}\lambda a\left(t\right)\xi \left(t,{y}_{n}\left(t\right),{y}_{{n}^{\prime }}\left(t\right)\right)$ (3.12)

${\stackrel{¯}{y}}_{n}=\frac{{y}_{n}}{‖{y}_{n}‖}$，显然 ${\stackrel{¯}{y}}_{n}$ 在E中有界，故 ${\stackrel{¯}{y}}_{n}$ 在E中有收敛子列，即存在 $\stackrel{¯}{y}\in E$，使得 ${\stackrel{¯}{y}}_{n}\to \stackrel{¯}{y}$$‖\stackrel{¯}{y}‖=1$。进一步，由条件(H3)和(3.5)式可知 $\stackrel{˜}{\xi }$ 非减，可得

$\frac{|\xi \left({y}_{n}\left(t\right)\right)|}{‖{y}_{n}‖}\le \frac{\stackrel{˜}{\xi }\left(|{y}_{n}\left(t\right)|\right)}{‖{y}_{n}‖}\le \frac{\stackrel{˜}{\xi }\left({‖{y}_{n}‖}_{\infty }\right)}{‖{y}_{n}‖}\le \frac{\stackrel{˜}{\xi }\left(‖{y}_{n}‖\right)}{‖{y}_{n}‖}$

$\underset{n\to \infty }{\mathrm{lim}}\frac{|\xi \left({y}_{n}\left(t\right)\right)|}{‖{y}_{n}‖}=0.$ (3.13)

$\stackrel{¯}{y}\left(t\right)={\int }_{0}^{1}G\left(t,s\right)\stackrel{¯}{\mu }\lambda a\left(s\right)\stackrel{¯}{y}\left(s\right)\text{d}s.$

$L\stackrel{¯}{y}=\stackrel{¯}{\mu }\lambda a\left(t\right)\stackrel{¯}{y}\left(t\right).$ (3.14)

$\stackrel{¯}{y}\in {\mathcal{C}}^{+}.$ (3.15)

$U\left(\stackrel{¯}{Y},{\rho }_{0}\right)\subset E\{\mathcal{C}}^{+}.$

$\stackrel{¯}{\mu }=\frac{{\lambda }_{1}\left(a\right)}{\lambda },$ (3.16)

$\left({\mu }_{n},{y}_{n}\right)\in {\mathcal{C}}^{+}$，一方面，根据引理2.4可得

$\begin{array}{c}{y}_{n}\left(t\right)=\lambda {\mu }_{n}{\int }_{0}^{1}G\left(t,s\right)f\left(s,{y}_{n}\left(s\right),{y}_{{n}^{\prime }}\left(s\right)\right)\text{d}s\\ \le \lambda {\mu }_{n}{\int }_{0}^{1}G\left(t,s\right)\frac{f\left(s,{y}_{n}\left(s\right),{y}_{{n}^{\prime }}\left(s\right)\right)}{{y}_{n}\left(s\right)}\text{d}s‖{y}_{n}‖\\ \le \lambda {\mu }_{n}{\int }_{0}^{1}\frac{1+\alpha }{1-\alpha \eta }s\left(1-s\right)\frac{f\left(s,{y}_{n}\left(s\right),{y}_{{n}^{\prime }}\left(s\right)\right)}{{y}_{n}\left(s\right)}\text{d}s‖{y}_{n}‖\end{array}$

${\mu }_{n}\ge {\left(\lambda {\int }_{0}^{1}\frac{1+\alpha }{1-\alpha \eta }s\left(1-s\right)\frac{f\left(s,{y}_{n}\left(s\right),{y}_{{n}^{\prime }}\left(s\right)\right)}{{y}_{n}\left(s\right)}\text{d}s\right)}^{-1}.$ (3.17)

$\begin{array}{c}{y}_{{n}^{\prime }}\left(t\right)=\lambda {\mu }_{n}{\int }_{0}^{1}\frac{\partial G}{\partial t}\left(t,s\right)f\left(s,{y}_{n}\left(s\right),{y}_{{n}^{\prime }}\left(s\right)\right)ds\\ \le \lambda {\mu }_{n}{\int }_{0}^{1}\frac{\partial G}{\partial t}\left(t,s\right)\frac{f\left(s,{y}_{n}\left(s\right),{y}_{{n}^{\prime }}\left(s\right)\right)}{{y}_{n}\left(s\right)}\text{d}s‖{y}_{n}‖\\ \le \lambda {\mu }_{n}{\int }_{0}^{1}\frac{1-s}{1-\alpha \eta }\frac{f\left(s,{y}_{n}\left(s\right),{y}_{{n}^{\prime }}\left(s\right)\right)}{{y}_{n}\left(s\right)}\text{d}s‖{y}_{n}‖,\end{array}$

${\mu }_{n}\ge {\left(\lambda {\int }_{0}^{1}\frac{1-s}{1-\alpha \eta }\frac{f\left(s,{y}_{n}\left(s\right),{y}_{{n}^{\prime }}\left(s\right)\right)}{{y}_{n}\left(s\right)}\text{d}s\right)}^{-1}.$ (3.18)

$\begin{array}{c}{y}_{n}\left(t\right)=\lambda {\mu }_{n}{\int }_{0}^{1}G\left(t,s\right)f\left(s,{y}_{n}\left(s\right),{y}_{{n}^{\prime }}\left(s\right)\right)\text{d}s\\ \ge \lambda {\mu }_{n}{\int }_{0}^{1}{\kappa }_{0}\frac{1+\alpha }{1-\alpha \eta }s\left(1-s\right)\frac{f\left(s,{y}_{n}\left(s\right),{y}_{{n}^{\prime }}\left(s\right)\right)}{{y}_{n}\left(s\right)}\text{d}s‖{y}_{n}‖,\end{array}$

${\mu }_{n}\le {\left(\lambda {\int }_{0}^{1}{\kappa }_{0}\frac{1+\alpha }{1-\alpha \eta }s\left(1-s\right)\frac{f\left(s,{y}_{n}\left(s\right),{y}_{{n}^{\prime }}\left(s\right)\right)}{{y}_{n}\left(s\right)}\text{d}s\right)}^{-1}:={M}_{2},$ (3.19)

$\begin{array}{c}{y}_{{n}^{\prime }}\left(t\right)=\lambda {\mu }_{n}{\int }_{0}^{1}\frac{\partial G}{\partial t}\left(t,s\right)f\left(s,{y}_{n}\left(s\right),{y}_{{n}^{\prime }}\left(s\right)\right)\text{d}s\\ \ge \lambda {\mu }_{n}{\int }_{0}^{1}\frac{\left(1-s\right)\eta }{1-\alpha \eta }\frac{f\left(s,{y}_{n}\left(s\right),{y}_{{n}^{\prime }}\left(s\right)\right)}{{y}_{n}\left(s\right)}\text{d}s‖{y}_{n}‖,\end{array}$

${\mu }_{n}\le {\left(\lambda {\int }_{0}^{1}\frac{\left(1-s\right)\eta }{1-\alpha \eta }\frac{f\left(s,{y}_{n}\left(s\right),{y}_{{n}^{\prime }}\left(s\right)\right)}{{y}_{n}\left(s\right)}\text{d}s\right)}^{-1}:={M}_{3},$ (3.20)

$M=\mathrm{max}\left\{{M}_{2},{M}_{3}\right\}.$

$\underset{n\to \infty }{\mathrm{lim}}\left({\mu }_{n}+‖{y}_{n}‖\right)=\infty$

$\underset{n\to \infty }{\mathrm{lim}}{\mu }_{n}=\infty .$

$\left(\frac{{\lambda }_{1}\left(b\right)}{\lambda },\frac{{\lambda }_{1}\left(a\right)}{\lambda }\right)\subseteq \left\{\lambda \in \left(0,\infty \right)|\left(\lambda ,u\right)\in {\mathcal{C}}^{+}\right\},$

${\mu }_{n}\in \left(0,M\right].$

$\left({\mu }_{n},{y}_{n}\right)\to \left(\frac{{\lambda }_{1}\left(a\right)}{\lambda },0\right),\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }n\to \infty .$

Existence of Positive Solutions for a Class of Third-Order Boundary Value Problems with First Derivative[J]. 理论数学, 2021, 11(02): 237-247. https://doi.org/10.12677/PM.2021.112032

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