北京信息科技大学理学院，北京

收稿日期：2021年4月27日；录用日期：2021年5月11日；发布日期：2021年5月31日

摘要

本文采用构造分段函数的方法、不动点定理、上下解的方法研究了带积分边界条件的三阶微分方程解的存在性问题。我们给出了其格林函数，构造了一个合适的锥和算子，其中非线性项Z满足Nagumo条件，进而我们获得了解存在的充分条件。

关键词

积分边界条件，解的存在性，上下解方法，不动点定理，格林函数

Study on the Existence of Solutions for a Third Order Differential Equation with Integral Boundary Conditions

Hongna Zhang, Chunyan Xue

School of Applied Science, Beijing Information Science & Technology University, Beijing

Received: Apr. 27^th, 2021; accepted: May 11^th, 2021; published: May 31^st, 2021

ABSTRACT

In this paper, we study the existence of solutions of the third order differential equations with integral boundary conditions by using the method of constructing piecewise functions, the fixed point theorem and the method of upper and lower solutions. We give its Green’s function, construct a suitable cone and operator, where the nonlinear term satisfies Nagumo condition, and then obtain sufficient conditions for understanding the existence.

Keywords:Integral Boundary Condition, Existence of Solutions, Upper and Lower Solution Method, Fixed Point Theorem, Green’s Function

Copyright © 2021 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

我们讨论如下完全三阶边值问题

$\{\begin{array}{l}-u^{‴}\left(t\right)=f\left(t,u\left(t\right),u^{\prime }\left(t\right),u^{″}\left(t\right)\right),t\in \left[0,1\right],\\ u\left(0\right)={\displaystyle {\int }_0^{1}g\left(s\right)}u\left(s\right)\text{d}s,u^{\prime }\left(0\right)=u^{″}\left(1\right)=0.\end{array}$ (1)

其中 $f:\left[0,1\right]\times R^3\to R$ 连续， $u\left(0\right)\le 0,g\left(s\right)$ 为非负函数，在非线性项 $f\left(t,x,y,z\right)$ 关于 $z$ 满足Nagumo条件，运用特殊的截断技巧、Leray-Schauder不动点定理以及上下解方法，获得了该方程解存在性 [1]。

2. 预备知识

设 $C\left(I\right)$ 为 $I$ 上的全体连续函数按范数 ${\Vert u\Vert }_c=\underset{t\in I}{\max }\left|u\left(t\right)\right|$ 构成的Banach空间，对 $n\in N$，$C^n\left(I\right)$ 表示 $I$ 上

n阶连续可微函数按范数 ${\Vert u\Vert }_{C^n}=\underset{t\in I}{\max }\left\{{\Vert u\Vert }_c,{\Vert u^{\prime }\Vert }_c,\cdots {\Vert u^{(n)}\Vert }_c\right\}$ 构成的函数空间 [2]。

定义1.1 [3] 设 $\alpha \left(t\right)\in C^3\left(I\right)$，$\alpha \left(t\right)\le 0，\beta \left(t\right)\ge 0$ 若 $\alpha \left(t\right)$ 满足

$\{\begin{array}{l}-\alpha \left(t\right)\le f\left(t,\alpha \left(t\right),{\alpha }^{\prime }\left(t\right),{\alpha }^{″}\left(t\right)\right),t\in I,\\ \alpha \left(0\right)\le {\displaystyle {\int }_0^{1}g\left(s\right)\alpha \left(s\right)\text{d}s},{\alpha }^{\prime }\left(0\right)\le 0,{\alpha }^{″}\left(1\right)\le 0.\end{array}$ (2)

则称 $\alpha \left(t\right)$ 为(1)的下解。若(1)的不等式符号均取逆，则称 $\beta \left(t\right)$ 为方程(1)的上解。

对 $\forall h\in C\left(I\right)$，方程(1)相应线性边值问题

$\{\begin{array}{l}-u^{‴}\left(t\right)=h\left(t\right),t\in \left[0,1\right],\\ u\left(0\right)={\displaystyle {\int }_0^{1}g\left(s\right)}u\left(s\right)\text{d}s,u^{\prime }\left(0\right)=u^{″}\left(1\right)=0.\end{array}$ (3)

存在解 $u\in C^3\left(I\right)$，则通过对式子积分很容易得到

$u^{″}\left(t\right)-u^{″}\left(0\right)=-{\displaystyle {\int }_0^{t}y\left(s\right)\text{d}s.}$

将 $t=1$ 带入得

$u^{″}\left(0\right)={\displaystyle {\int }_0^{1}y\left(s\right)\text{d}s.}$

再次积分可得

$u^{\prime }\left(t\right)-u^{\prime }\left(0\right)=t{u}^{″}\left(0\right)-{\displaystyle {\int }_0^t\left(t-s\right)y\left(s\right)\text{d}s.}$

再次进行积分可以得到

$u\left(t\right)-u\left(0\right)=\frac{1}{2}{t}^2{u}^{″}\left(0\right)-\frac{1}{2}{\displaystyle {\int }_0^1{\left(t-s\right)}^2}y\left(s\right)\text{d}s.$

将边界条件 $u\left(0\right)={\displaystyle {\int }_0^{1}u\left(s\right)g\left(s\right)\text{d}s}$ 以及 $u^{″}\left(0\right)={\displaystyle {\int }_0^{1}y\left(s\right)\text{d}s.}$ 带入得到

$\begin{array}{c}u\left(t\right)={\displaystyle {\int }_0^{1}u\left(s\right)g\left(s\right)\text{d}s+\frac{1}{2}}{t}^2{\displaystyle {\int }_0^{1}y\left(s\right)\text{d}s}-\frac{1}{2}{\displaystyle {\int }_0^t{\left(t-s\right)}^{2}y\left(s\right)\text{d}s}\\ ={\displaystyle {\int }_0^{1}u\left(s\right)g\left(s\right)\text{d}s}+{\displaystyle {\int }_0^{1}G\left(t,s\right)y\left(s\right)\text{d}s}.\end{array}$

则有

${\displaystyle {\int }_0^{1}u\left(s\right)g\left(s\right)\text{d}s={\displaystyle {\int }_0^{1}g\left(s\right)\left[{\displaystyle {\int }_0^{1}u\left(\tau \right)g\left(\tau \right)\text{d}\tau }\right]\text{d}s}}+{\displaystyle {\int }_0^{1}g\left(s\right)\left[{\displaystyle {\int }_0^{1}G\left(s,\tau \right)y\left(\tau \right)\text{d}\tau }\right]\text{d}s}.$

进而我们可以得到

${\displaystyle {\int }_0^{1}u\left(s\right)g\left(s\right)\text{d}s=\frac{1}{1-{\displaystyle {\int }_0^{1}g\left(s\right)\text{d}s}}}\left\{{\displaystyle {\int }_0^{1}g\left(s\right)\left[{\displaystyle {\int }_0^{1}G\left(s,\tau \right)y\left(\tau \right)\text{d}\tau }\right]\text{d}s}\right\}.$

即

$\begin{array}{c}u\left(t\right)={\displaystyle {\int }_0^{1}G\left(t,s\right)y\left(s\right)\text{d}s+\frac{1}{1-{\displaystyle {\int }_0^{1}g\left(s\right)\text{d}s}}}\left\{{\displaystyle {\int }_0^{1}g\left(s\right)}\left[{\displaystyle {\int }_0^{1}G\left(s,\tau \right)y\left(\tau \right)\text{d}\tau }\right]\text{d}s\right\}\\ ={\displaystyle {\int }_0^1\left[G\left(t,s\right)+\frac{1}{1-{\displaystyle {\int }_0^{1}g\left(s\right)\text{d}s}}{\displaystyle {\int }_0^{1}g\left(s\right)}\left[{\displaystyle {\int }_0^{1}G\left(s,\tau \right)y\left(\tau \right)\text{d}\tau }\right]\text{d}s\right]}y\left(s\right)\text{d}s\\ ={\displaystyle {\int }_0^{1}H\left(t,s\right)y\left(s\right)\text{d}s,}\end{array}$

其中有

$H\left(t,s\right)=G\left(t,s\right)+\frac{1}{1-{\displaystyle {\int }_0^{1}g\left(s\right)\text{d}s}}{\displaystyle {\int }_0^{1}G\left(\tau ,s\right)}g\left(\tau \right)\text{d}\tau .$ (4)

$G\left(t,s\right)=\{\begin{array}{l}\frac{1}{2}\left[t^2-{\left(t-s\right)}^2\right],0\le s\le t\le 1\\ \frac{1}{2}{t}^2.0\le t\le s\le 1\end{array}$ (5)

则算子 $T:C\left(I\right)\to C^3\left(I\right)$ 为线性有界算子。

由嵌入的 $C^3\left(I\right)\to C^2\left(I\right)$ 的紧性，则 $T:C\left(I\right)\to C^2\left(I\right)$ 是线性全连续算子。

为了论述方便，我们引入以下条件：

(H1) 存在 $\left[0,+\infty \right)$ 上正值函数 $h\left(\rho \right)$ 满足 ${\displaystyle {\int }_p^{+\infty }\frac{\rho }{h\left(\rho \right)}\text{d}\rho }>\underset{t\in J}{\max }{\beta }^{\prime }\left(t\right)-\underset{t\in J}{\min }{\alpha }^{\prime }\left(t\right)$，其中 $P=\underset{t\in J}{\max }\left\{\left|{\beta }^{\prime }\left(1\right)-{\alpha }^{\prime }\left(0\right)\right|,\left|{\beta }^{\prime }\left(0\right)-{\alpha }^{\prime }\left(1\right)\right|\right\}$ 使得 $f$ 关于 $z$ 满足

$\left|\left|f\left(t,x,y,z\right)\right|\right|\le h\left(\left|z\right|\right),t\in J,\alpha \left(t\right)\le x\le \beta \left(t\right),{\alpha }^{\prime }\left(t\right)\le y\le {\beta }^{\prime }\left(t\right),z\in R.$

(H2) 当 $\alpha \left(t\right)\le x\le \beta \left(t\right),\left(t,y,z\right)\in J\times R^2$ 时，我们对于任意 $t\in J$，有 $f\left(t,\alpha \left(t\right),y,z\right)\le f\left(t,x,y,z\right)\le f\left(t,\beta \left(t\right),y,z\right)$ ；

(H3) 对 $\forall \left(t,x,y,z\right)\in I\times R^3$ 有 $f_x\left(t,x,y,z\right)+f_y\left(t,x,y,z\right)<0.$

3. 主要引理

引理3.1 [1] 对 $h\in C\left(I\right)$，(3)的解满足 $u:=Th\in C^3\left(I\right)$ 满足

${\Vert u\Vert }_C\le {\Vert u^{\prime }\Vert }_C\le {\Vert u^{″}\Vert }_C,{\Vert u\Vert }_{C^2}={\Vert u^{″}\Vert }_C.$

证明 对 $\forall t\in I$，有

$u\left(t\right)=u\left(0\right)+{\displaystyle {\int }_0^t{u}^{\prime }\left(s\right)\text{d}s\le {{\displaystyle {\int }_0^t{u}^{\prime }\left(s\right)\text{d}s\le t\Vert u^{\prime }\Vert }}_C}\le {\Vert u^{\prime }\Vert }_C$

$u^{\prime }\left(t\right)={{\displaystyle {\int }_0^t{u}^{″}\left(s\right)\text{d}s\le t\Vert u^{″}\Vert }}_C\le {\Vert u^{″}\Vert }_C$

因此有

${\Vert u\Vert }_C\le {\Vert u^{\prime }\Vert }_C\le {\Vert u^{″}\Vert }_C,$

${\Vert u\Vert }_{C^2}=\max \left\{{\Vert u\Vert }_C,{\Vert u^{\prime }\Vert }_C,{\Vert u^{″}\Vert }_C\right\}={\Vert u^{″}\Vert }_C.$

引理3.2 [4] 设 $f:I\times R^3\to R$ 连续，若存在常数 $a,b,c,d\ge 0$ 满足 $a+b+c<1$ 以及 $d>0$，使得满足以下条件

(1) $\left|f\left(t,x,y,z\right)\right|\le a\left|x\right|+b\left|y\right|+c\left|z\right|+d,t\in I,x,y,z\in R,$

(2) $\left|f\left(t,x_2,y_2,z_2\right)-f\left(t,x_1,y_1,z_1\right)\right|$

$\le a\left|x_2-x_1\right|+b\left|y_2-y_1\right|+c\left|z_2-z_1\right|,t\in I,x_1,x_2,y_1,y_2,z_1,z_2\in R,$

则式子(1)存在唯一解。

证明 证明过程参考文献 [1] 中引理1.2的证明。

4. 主要结论及证明

定理4.1 [5] 设 $f:I\times R^3\to R$ 上连续，方程(1)存在下解 $\alpha \left(t\right)$ 以及上解 $\beta \left(t\right)$，满足 ${\alpha }^{\prime }\left(t\right)\le {\beta }^{\prime }\left(t\right)$，若 $f$ 满足条件(H1)和(H2)，则方程(1)至少存在一个解 $u\in C^3\left(I\right)$ 满足

$\begin{array}{l}\alpha \left(t\right)\le u\left(t\right)\le \beta \left(t\right)\\ {\alpha }^{\prime }\left(t\right)\le u^{\prime }\left(t\right)\le {\beta }^{\prime }(\; t\; )\end{array}$

证明 有条件(H1)， $\exists M>0$ 使得

${\displaystyle {\int }_P^M\frac{s}{h\left(s\right)}\text{d}s}>\underset{t\in I}{\max }{\beta }^{\prime }\left(t\right)-\underset{t\in I}{\min }{\alpha }^{\prime }(\; t\; )$

取常数 $N=M+{\Vert \beta \Vert }_{C^2}+{\Vert \alpha \Vert }_{C^2}+1$，令

${\eta }_1\left(t,x\right)=\{\begin{array}{l}\beta \left(t\right),x>\beta \left(t\right)\\ x,\alpha \left(t\right)\le x\le \beta \left(t\right)\\ \alpha \left(t\right),x<\alpha (\; t\; )\end{array}$

${\eta }_2\left(t,y\right)=\{\begin{array}{l}{\beta }^{\prime }\left(t\right),x>{\beta }^{\prime }\left(t\right)\\ y,{\alpha }^{\prime }\left(t\right)\le y\le {\beta }^{\prime }\left(t\right)\\ {\alpha }^{\prime }\left(t\right),x<{\alpha }^{\prime }(\; t\; )\end{array}$

${\left[z\right]}_N=\{\begin{array}{l}N,z>N\\ z,-N\le z\le N\\ -N,z<-N\end{array}$

作 $f\left(t,x,y,z\right)$ 的截断函数

$f^{\text{*}}\left(t,x,y,z\right)=f\left(t,{\eta }_1\left(t,x\right),{\eta }_2\left(t,y\right),{\left[z\right]}_N\right)-\frac{y-{\eta }_2\left(t,y\right)}{1+y^2}$

有

$\begin{array}{c}\left|f^{\text{*}}\left(t,x,y,z\right)\right|\le \underset{t\in I}{\max }\left\{\left|f\left(t,x,y,z\right)\right|:\left|x\right|\le {\Vert \alpha \Vert }_C+{\Vert \beta \Vert }_C,\left|y\right|\le {\Vert {\alpha }^{\prime }\Vert }_C+{\Vert {\beta }^{\prime }\Vert }_C,\left|z\right|\le N\right\}\\ +{\Vert {\alpha }^{\prime }\Vert }_C+{\Vert {\beta }^{\prime }\Vert }_C+1,\end{array}$

则 $f^{*}:I\times R^3\to R$ 连续有界，因此得到以下方程：

$\{\begin{array}{l}-u^{‴}\left(t\right)=f^{\text{*}}\left(t,u\left(t\right),u^{\prime }\left(t\right),u^{″}\left(t\right)\right),t\in \left[0,1\right],\\ u\left(0\right)={\displaystyle {\int }_0^{1}g\left(s\right)}u\left(s\right)\text{d}s,u^{\prime }\left(0\right)=u^{″}\left(1\right)=0.\end{array}$ (6)

有解 $u_0\left(t\right)\in C^3\left(I\right)$，下证 $u_0\left(t\right)$ 为方程(1)的解。

先证 $\forall t\in J,\alpha \left(t\right)\le u_0\left(t\right)\le \beta \left(t\right),{\alpha }^{\prime }\left(t\right)\le {u^{\prime }}_0\left(t\right)\le {\beta }^{\prime }\left(t\right)$ 成立，我们不妨采用反证法，假设存在 $t\in \left[0,1\right]$，

使得 ${u^{\prime }}_0\left(t\right)>{\beta }^{\prime }\left(t\right)$ 成立，考虑 $u^{\prime }\left(t\right)={u^{\prime }}_0\left(t\right)-{\beta }^{\prime }\left(t\right)$，让 $\underset{t\in I}{\max }{u}^{\prime }\left(t\right)=u^{\prime }\left(t_0\right)>0$。

(i) 若 $t_0\in \left(0,1\right)$，则 $u^{″}\left(t_0\right)=0,u^{‴}\left(t_0\right)\le 0$，即

${u^{\prime }}_0\left(t_0\right)>{\beta }^{\prime }\left(t_0\right),{u^{″}}_0\left(t_0\right)={\beta }^{″}\left(t_0\right),{u^{‴}}_0\left(t_0\right)\le {\beta }^{‴}\left(t_0\right).$ (7)

我们根据截断函数的定义以及(H2)和(7)有

$\begin{array}{c}-{u^{‴}}_0\left(t_0\right)=f^{*}\left(t_0,u_0\left(t_0\right),{u^{\prime }}_0\left(t_0\right),{u^{″}}_0\left(t_0\right)\right)\\ =f\left(t_0,{\eta }_1\left(t_0,u_0\left(t_0\right)\right),{\eta }_2\left(t_0,{u^{\prime }}_0\left(t_0\right)\right),{\left[{u^{″}}_0\left(t_0\right)\right]}_N\right)-\frac{{u^{\prime }}_0\left(t_0\right)-{\eta }_2\left(t_0,{u^{\prime }}_0\left(t_0\right)\right)}{1+{\left({u^{\prime }}_0\left(t_0\right)\right)}^2}\\ =f\left(t_0,{\eta }_1\left(t_0,u_0\left(t_0\right)\right),{\beta }^{\prime }\left(t_0\right),{\left[{\beta }^{″}\left(t_0\right)\right]}_N\right)-\frac{{u^{\prime }}_0\left(t_0\right)-{\beta }^{\prime }\left(t_0\right)}{1+{\left({u^{\prime }}_0\left(t_0\right)\right)}^2}\\ \le f\left(t_0,{\beta }_0,{\beta }^{\prime }\left(t_0\right),{\beta }^{″}\left(t_0\right)\right)-\frac{{u^{\prime }}_0\left(t_0\right)-{\beta }^{\prime }\left(t_0\right)}{1+{\left({u^{\prime }}_0\left(t_0\right)\right)}^2}\\ <f\left(t_0,{\beta }_0,{\beta }^{\prime }\left(t_0\right),{\beta }^{″}\left(t_0\right)\right)\\ \le -{{\beta }^{‴}}_0\left(t_0\right),\end{array}$

则我们可以得到 ${u^{‴}}_0\left(t_0\right)>{\beta }^{‴}\left(t_0\right)$，与式子(7)矛盾！故 $t_0\notin \left(0,1\right)$。

(ii) 若 $t_0=0$，则

${u^{\prime }}_0\left(0\right)-{\beta }^{\prime }\left(0\right)>0,$ (8)

但是我们可以得知 ${u^{\prime }}_0\left(0\right)=0,{\beta }^{\prime }\left(0\right)\ge 0$，即 ${u^{\prime }}_0\left(0\right)-{\beta }^{\prime }\left(0\right)\le 0$ 与式子(8)是矛盾的，所以有 $t_0\ne 0$。

(iii) 若 $t_0=1$，则

${u^{″}}_0\left(1\right)-{\beta }^{″}\left(1\right)\ge 0,$ (9)

但是我们可以得知 ${u^{\prime }}_0\left(1\right)=0,{\beta }^{\prime }\left(1\right)\ge 0$，即 ${u^{″}}_0\left(1\right)-{\beta }^{″}\left(1\right)\le 0$ 与式子(9)是矛盾的，所以有 $t_0\ne 1$。

综上所述，对 $\forall t\in I$ 有 ${u^{\prime }}_0\left(t\right)\le {\beta }^{\prime }\left(t\right)$，同时满足

$u_0\left(t\right)={\displaystyle {\int }_0^1{u^{\prime }}_0}\left(s\right)\text{d}s\le {\displaystyle {\int }_0^1{\beta }^{\prime }}\left(s\right)\text{d}s=\beta \left(t\right)-\beta \left(0\right)\le \beta \left(t\right),$

即对 $\forall t\in I$ 有 $u_0\left(t\right)\le \beta \left(t\right)$，同理可证 $\forall t\in I$，$u_0\left(t\right)\ge \alpha \left(t\right),{u^{\prime }}_0\left(t\right)\ge {\alpha }^{\prime }\left(t\right)$。

接下来我们继续证明 $\left|{u^{″}}_0\left(t\right)\right|\le M<N,t\in I$，由 $P$ 的定义以及中值定理，则存在 $t_0\in \left(0,1\right)$，使得

$\left|{u^{″}}_0\left(t_0\right)\right|=\left|{u^{\prime }}_0\left(1\right)-{u^{\prime }}_0\left(0\right)\right|\le P,$

要证 $\left|{u^{″}}_0\left(t\right)\right|\le M<N,t\in I$，反设 $\exists t_1,t_2\in \left(0,1\right)$ 且 $t_1<t_2$，使得下列情形之一成立：

1) $P<{u^{″}}_0\left(t\right)<M,t\in \left(t_1,t_2\right)$ 且 ${u^{″}}_0\left(t_1\right)=P,{u^{″}}_0\left(t_2\right)=M$ ；

2) $-M<{u^{″}}_0\left(t\right)<-P,t\in \left(t_1,t_2\right)$ 且 ${u^{″}}_0\left(t_1\right)=-M,{u^{″}}_0\left(t_2\right)=-P$ ；

3) $P<{u^{″}}_0\left(t\right)<M,t\in \left(t_1,t_2\right)$ 且 ${u^{″}}_0\left(t_1\right)=M,{u^{″}}_0\left(t_2\right)=P$ ；

4) $-M<{u^{″}}_0\left(t\right)<-P,t\in \left(t_1,t_2\right)$ 且 ${u^{″}}_0\left(t_1\right)=-P,{u^{″}}_0\left(t_2\right)=-M.$

下面我们证明(1)，其他情况类似可证，当(1)成立时，根据上述证明以及条件(H1)，有

$\begin{array}{c}\left|{u^{‴}}_0\left(t\right)\right|=\left|f^{*}\left(t,u_0\left(t\right),{u^{\prime }}_0\left(t\right),{u^{″}}_0\left(t\right)\right)\right|\\ =\left|f\left(t,{\eta }_1\left(t,u_0\left(t\right)\right),{\eta }_2\left(t,{u^{\prime }}_0\left(t\right)\right),{\left[{u^{″}}_0\left(t\right)\right]}_N\right)-\frac{{u^{\prime }}_0\left(t\right)-{\eta }_2\left(t,{u^{\prime }}_0\left(t\right)\right)}{1+{\left({u^{\prime }}_0\left(t\right)\right)}^2}\right|\\ =\left|f\left(t,u_0\left(t\right),{u^{\prime }}_0\left(t\right),{u^{″}}_0\left(t\right)\right)\right|\le h\left({u^{″}}_0\left(t\right)\right),\end{array}$

因为 ${u^{″}}_0\left(t\right)>0$，给上述不等式左右两侧同乘 ${u^{″}}_0\left(t\right)$，再积分可得

${\displaystyle {\int }_{t_1}^{t_2}\frac{\left|{u^{‴}}_0\left(t\right)\right|\cdot {u^{″}}_0\left(t\right)}{h\left({u^{″}}_0\left(t\right)\right)}\text{d}t}\le {\displaystyle {\int }_{t_1}^{t_2}{u^{″}}_0\left(t\right)\text{d}t},$ (10)

对式子(10)左端进行变量替换 $\rho ={u^{″}}_0\left(t\right)$，有

${\displaystyle {\int }_{{u^{″}}_0(t_1)}^{{u^{″}}_0(t_2)}\frac{\rho }{h\left(\rho \right)}\text{d}\rho }\le {\displaystyle {\int }_{t_1}^{t_2}{u^{″}}_0}\left(t\right)\text{d}t,$

即

${\displaystyle {\int }_P^M\frac{\rho }{h\left(\rho \right)}\text{d}\rho }\le u^{\prime }\left(t_2\right)-{u^{\prime }}_0\left(t_1\right)\le \underset{t\in I}{\max }{\beta }^{\prime }\left(t\right)-\underset{t\in I}{\min }{\alpha }^{\prime }(\; t\; )$

很显然，矛盾！则有 $\left|{u^{″}}_0\left(t\right)\right|\le M<N,t\in I.$

综上所述， $f^{*}=f$，即 $u_0\left(t\right)$ 为式子(1)的解。

致谢

感谢薛老师的耐心的指导！

基金项目

国家自然科学基金资助项目(11471146)。

文章引用

张红娜,薛春艳. 具积分边界条件的三阶微分方程解的存在性问题研究
Study on the Existence of Solutions for a Third Order Differential Equation with Integral Boundary Conditions[J]. 应用数学进展, 2021, 10(05): 1790-1796. https://doi.org/10.12677/AAM.2021.105189

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