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

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。

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.

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)。