# 障碍带条件下两类三阶非线性边值问题解的存在性Existence of Solutions for Two Class of Third-Order Nonlinear Boundary Value Problems under Barrier Strips Conditions

DOI: 10.12677/PM.2018.83030, PDF, HTML, XML, 下载: 857  浏览: 1,069  国家自然科学基金支持

Abstract: By using Leary-Schauder theorem, we study the existence of solutions for two classes of nonlinear third-order two-point boundary value problems. We establish the existence results of solutions for the above two classes of boundary value problems with nonlinearity satisfying barrier strips conditions and give some examples to illustrate our main results.

1. 引言

1994年，P. Kelevedjiev [10] 运用Leray-Schauder原理证明了障碍带条件下非线性二阶两点边值问题

$\left\{\begin{array}{l}-{u}^{″}\left(t\right)=f\left(t,u,{u}^{\prime }\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}t\in \left[0,1\right],\\ u\left(0\right)=A,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}u\left(0\right)=B\end{array}$

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

2009年，张宏旺 [3] 运用新的极大值原理及上下解的单调迭代法获得了非线性三阶边值问题

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

$\left\{\begin{array}{l}-{u}^{‴}\left(t\right)=f\left(t,u\left(t\right),{u}^{\prime }\left(t\right),{u}^{″}\left(t\right)\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}t\in \left(0,1\right),\\ u\left(0\right)={u}^{\prime }\left(0\right)=u\left(1\right)=0\end{array}$

$\left\{\begin{array}{l}{u}^{‴}\left(t\right)=f\left(t,u\left(t\right),{u}^{\prime }\left(t\right),{u}^{″}\left(t\right)\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}t\in \left[0,1\right],\\ u\left(0\right)=A,\text{\hspace{0.17em}}u\left(1\right)=B,\text{\hspace{0.17em}}{u}^{\prime }\left(1\right)=C\end{array}$ (2)

$\left\{\begin{array}{l}{u}^{‴}\left(t\right)=f\left(t,u\left(t\right),{u}^{\prime }\left(t\right),{u}^{″}\left(t\right)\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}t\in \left[0,1\right],\\ u\left(0\right)=A,\text{\hspace{0.17em}}u\left(1\right)=B,\text{\hspace{0.17em}}{u}^{\prime }\left(0\right)=C\end{array}$ (3)

2. 预备知识

$C\left[0,1\right]$ 表示区间 $\left[0,1\right]$ 上连续函数构成的空间，定义其上范数为 ${‖u‖}_{0}=\mathrm{max}\left\{|u\left(t\right)|:t\in \left[0,1\right]\right\}$ 。记 ${C}^{3}\left[0,1\right]=\left\{u\text{\hspace{0.17em}}|\text{\hspace{0.17em}}u,{u}^{\prime },{u}^{″},{u}^{‴}\in C\left[0,1\right]\right\}$ ，其上范数为：

${‖u‖}_{3}=\mathrm{max}\left\{{‖u‖}_{0},{‖{u}^{\prime }‖}_{0},{‖{u}^{″}‖}_{0},{‖{u}^{‴}‖}_{0}\right\},$

${C}^{3}\left[0,1\right]$ 按范数 ${‖u‖}_{3}$ 构成Banach空间。

${A}_{1}=\left\{u\in C\left[0,1\right]|u\left(0\right)=A,u\left(1\right)=B,{u}^{\prime }\left(1\right)=C,A,B,C为常数\right\},$

${A}_{2}=\left\{u\in C\left[0,1\right]|u\left(0\right)=A,u\left(1\right)=B,{u}^{\prime }\left(0\right)=C,A,B,C为常数\right\}$

${T}_{i}u={u}^{‴},\text{\hspace{0.17em}}i=1,2.$

$\left\{u|u\in E,u=\lambda Au,0<\lambda <1\right\}$

3. 主要结果及证明

${L}_{2}>{L}_{1}\ge D,{L}_{4}>{L}_{3}\ge D,{L}_{5}<{L}_{6}\le D,{L}_{8}<{L}_{7}\le D,$

$f\left(t,u,p,q\right)\ge 0,\text{\hspace{0.17em}}\left(t,u,p,q\right)\in \left[0,1\right]×{R}^{2}×\left(\left[{L}_{1},{L}_{2}\right]\cup \left[{L}_{5},{L}_{6}\right]\right),$

$\left\{\begin{array}{l}{u}^{‴}\left(t\right)=\lambda f\left(t,u\left(t\right),{u}^{\prime }\left(t\right),{u}^{″}\left(t\right)\right),t\in \left[0,1\right],\lambda \in \left[0,1\right],\\ u\left(0\right)=A,u\left(1\right)=B,{u}^{\prime }\left(1\right)=C.\end{array}$ (4)

$u\in {C}^{3}\left[0,1\right]$ 是问题(4)的一个解。则由微分中值定理得，存在 $d\in \left(0,1\right)$ ，使得 ${u}^{\prime }\left(d\right)=u\left(1\right)-u\left(0\right)=B-A$ 。同理，存在 $\xi \in \left(d,1\right)$ ，使得

${u}^{″}\left(\xi \right)=\frac{{u}^{\prime }\left(1\right)-{u}^{\prime }\left(d\right)}{1-d}=\frac{C-B+A}{1-d}=D.$

$f\left(t,u,p,q\right)\ge 0,\left(t,u,p,q\right)\in \left[0,\xi \right]×{R}^{2}×\left[{L}_{1},{L}_{2}\right],$

$f\left(t,u,p,q\right)\le 0,\left(t,u,p,q\right)\in \left[0,\xi \right]×{R}^{2}×\left[{L}_{7},{L}_{8}\right].$

${S}_{0}=\left\{t\in \left[0,\xi \right]:{L}_{1}<{u}^{″}\left(t\right)\le {L}_{2}\right\},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{S}_{1}=\left\{t\in \left[0,\xi \right]:{L}_{7}<{u}^{″}\left(t\right)\le {L}_{8}\right\}$

${u}^{″}\left({{t}^{\prime }}_{0}\right)<{u}^{″}\left({t}_{0}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{u}^{″}\left({{t}^{\prime }}_{1}\right)>{u}^{″}\left({t}_{1}\right).$ (5)

${u}^{″}$ 的连续性，甚至可以取到 ${{t}^{\prime }}_{0}\in \left({t}_{0},\xi \right]\cap {S}_{0}$${{t}^{\prime }}_{1}\in \left({t}_{1},\xi \right]\cap {S}_{1}$ ，但对 $t\in {S}_{0}$ ，有

${u}^{‴}\left(t\right)=\lambda f\left(t,u\left(t\right),{u}^{\prime }\left(t\right),{u}^{″}\left(t\right)\right)\ge 0,$

${u}^{″}\left({{t}^{\prime }}_{0}\right)\ge {u}^{″}\left({t}_{0}\right),\text{\hspace{0.17em}}t\in \left({t}_{0},\xi \right],\text{\hspace{0.17em}}\text{\hspace{0.17em}}{u}^{″}\left({{t}^{\prime }}_{1}\right)\le {u}^{″}\left({t}_{1}\right),\text{\hspace{0.17em}}t\in \left({t}_{1},\xi \right].$

$|{u}^{″}\left(t\right)|\le \mathrm{max}\left\{|{L}_{1}|,|{L}_{8}|\right\},t\in \left[0,\xi \right].$

$f\left(t,u,p,q\right)\le 0,\text{\hspace{0.17em}}\left(t,u,p,q\right)\in \left[\xi ,1\right]×{R}^{2}×\left[{L}_{3},{L}_{4}\right],$

$f\left(t,u,p,q\right)\ge 0,\text{\hspace{0.17em}}\left(t,u,p,q\right)\in \left[\xi ,1\right]×{R}^{2}×\left[{L}_{5},{L}_{6}\right]$

$|{u}^{″}\left(t\right)|\le {M}_{0},$ (6)

${u}^{\prime }\left(1\right)-{u}^{\prime }\left(t\right)={u}^{″}\left(\tau \right)\left(1-t\right)$

$|{u}^{\prime }\left(t\right)|\le {M}_{1},\text{\hspace{0.17em}}t\in \left[0,1\right],$ (7)

$u\left(t\right)-u\left(0\right)={u}^{\prime }\left(\rho \right)t,$

$|u\left(t\right)|\le {M}_{2},\text{\hspace{0.17em}}t\in \left[0,1\right],$ (8)

$|{u}^{‴}\left(t\right)|\le {M}_{3},\text{\hspace{0.17em}}t\in \left[0,1\right],$ (9)

${‖u‖}_{3}<\mathrm{max}\left\{{M}_{0},{M}_{1},{M}_{2},{M}_{3}\right\}+1.$

${L}_{2}>{L}_{1}\ge D,{L}_{4}>{L}_{3}\ge D,{L}_{5}<{L}_{6}\le D,{L}_{8}<{L}_{7}\le D$ ，其中 $D=\frac{C-B+A}{1-d},d\in \left(0,1\right)$ ，满足

$f\left(t,u,p,q\right)\le 0,\text{\hspace{0.17em}}\left(t,u,p,q\right)\in \left[0,1\right]×{R}^{2}×\left(\left[{L}_{1},{L}_{2}\right]\cup \left[{L}_{5},{L}_{6}\right]\right),$

$f\left(t,u,p,q\right)\ge 0,\text{\hspace{0.17em}}\left(t,u,p,q\right)\in \left[0,1\right]×{R}^{2}×\left(\left[{L}_{3},{L}_{4}\right]\cup \left[{L}_{7},{L}_{8}\right]\right),$

$\underset{_}{L}<0<\stackrel{¯}{L}$ ，使得

$f\left(t,u,p,q\right)\ge 0,\text{\hspace{0.17em}}\left(t,u,p,q\right)\in \left[0,1\right]×{R}^{2}×\left\{\stackrel{¯}{L}\right\},$

$f\left(t,u,p,q\right)\le 0,\text{\hspace{0.17em}}\left(t,u,p,q\right)\in \left[0,1\right]×{R}^{2}×\left\{\underset{_}{L}\right\},$

$\varphi \left(\stackrel{¯}{L}\right)=1,\text{\hspace{0.17em}}\varphi \left(\underset{_}{L}\right)=-1.$

$\left\{\begin{array}{l}{u}^{‴}\left(t\right)={f}_{m}\left(t,u,{u}^{\prime },{u}^{″}\right),t\in \left[0,1\right],\\ u\left(0\right)=A,u\left(1\right)=B,{u}^{\prime }\left(1\right)=C\end{array}$ (10)

${f}_{m}\left(t,u,p,q\right)>0,\text{\hspace{0.17em}}\left(t,u,p,q\right)\in \left[0,1\right]×{R}^{2}×\left[{L}_{m,1},{L}_{m,2}\right].$

${f}_{m}\left(t,u,p,q\right)<0,\text{\hspace{0.17em}}\left(t,u,p,q\right)\in \left[0,1\right]×{R}^{2}×\left[{L}_{m,3},{L}_{m,4}\right].$

$\left[{L}_{m+1,1},{L}_{m+1,2}\right]\subseteq \left[{L}_{m,1},{L}_{m,2}\right],$ (11)

$\left[{L}_{m+1,3},{L}_{m+1,4}\right]\subseteq \left[{L}_{m,3},{L}_{m,4}\right],$ (12)

${L}_{m,4}\le {u}^{″}\left(t\right)\le {L}_{m,1},\text{\hspace{0.17em}}t\in \left[0,1\right].$

$|{{u}^{″}}_{m}\left(t\right)|\le {M}_{4},\text{\hspace{0.17em}}t\in \left[0,1\right].$ (13)

${{u}^{\prime }}_{m}\left(t\right)-{{u}^{\prime }}_{m}\left(0\right)={{u}^{″}}_{m}\left(\eta \right)t,$

$|{{u}^{\prime }}_{m}\left(t\right)|\le {M}_{5},\text{\hspace{0.17em}}t\in \left[0,1\right].$ (14)

$|{u}_{m}\left(t\right)|\le {M}_{6},\text{\hspace{0.17em}}t\in \left[0,1\right].$ (15)

$|{{u}^{‴}}_{m}\left(t\right)|\le {M}_{7},\text{\hspace{0.17em}}t\in \left[0,1\right].$ (16)

$\underset{i\to \infty }{\mathrm{lim}}{u}_{{m}_{i}}=\nu \in C\left[0,1\right].$

$\left\{{u}_{{m}_{i}}\right\}$ 有界，且 ${C}^{3}\left[0,1\right]$ 紧嵌入 ${C}^{1}\left[0,1\right]$ ，故存在子序列 $\left\{{u}_{{m}_{{i}_{j}}}\right\}\subseteq \left\{{u}_{m}\right\}$ ，满足

$\underset{i\to \infty }{\mathrm{lim}}{u}_{{m}_{{}_{{i}_{j}}}}=\nu ,\text{\hspace{0.17em}}\underset{j\to \infty }{\mathrm{lim}}{{u}^{\prime }}_{{m}_{{i}_{j}}}={\nu }^{\prime },$

$\underset{i\to \infty }{\mathrm{lim}}{u}_{{m}_{{}_{{i}_{{j}_{k}}}}}=\nu ,\text{\hspace{0.17em}}\underset{j\to \infty }{\mathrm{lim}}{{u}^{\prime }}_{{m}_{{i}_{j}{}_{k}}}={\nu }^{\prime },\text{\hspace{0.17em}}\underset{k\to \infty }{\mathrm{lim}}{{u}^{″}}_{{m}_{{i}_{{j}_{k}}}}={\nu }^{″}.$

$\left\{\begin{array}{l}{{\theta }^{‴}}_{m}\left(t\right)={f}_{m}\left(t,{\theta }_{m},{{\theta }^{\prime }}_{m},{{\theta }^{″}}_{m}\right),t\in \left[0,1\right],\\ {\theta }_{m}\left(0\right)=A,{\theta }_{m}\left(1\right)=B,{{\theta }^{\prime }}_{m}\left(0\right)=C,\end{array}$ (17)

${\theta }_{m}\left(t\right)={\int }_{0}^{1}G\left(t,s\right){f}_{m}\left(s,{\theta }_{m}\left(s\right),{{\theta }^{\prime }}_{m}\left(s\right),{{\theta }^{″}}_{m}\left(s\right)\right)\text{d}s+\left(B-A-C\right){t}^{2}+Ct+A,$ (18)

$G\left(t,s\right)=\left\{\begin{array}{l}\frac{{t}^{2}{\left(1-s\right)}^{2}}{2},\text{\hspace{0.17em}}0\le t\le s\le 1\\ \frac{\left(1-t\right)s\left(t-s+t\left(1-s\right)\right)}{2},\text{\hspace{0.17em}}0\le s\le t\le 1,\end{array}$

$f\left(t,u,p,q\right)\le 0,\text{\hspace{0.17em}}\left(t,u,p,q\right)\in \left[0,1\right]×{R}^{2}×\left\{\stackrel{¯}{L}\right\},$

$f\left(t,u,p,q\right)\ge 0,\text{\hspace{0.17em}}\left(t,u,p,q\right)\in \left[0,1\right]×{R}^{2}×\left\{\underset{_}{L}\right\},$

${\theta }_{m}\left(t\right)={\int }_{0}^{1}G\left(t,s\right){f}_{m}\left(s,{\theta }_{m}\left(s\right),{{\theta }^{\prime }}_{m}\left(s\right),{{\theta }^{″}}_{m}\left(s\right)\right)\text{d}s+\left(C-B+A\right){t}^{2}+\left(2B-2A-C\right)t+A,$

$G\left(t,s\right)=\left\{\begin{array}{l}\frac{{\left(1-t\right)}^{2}{s}^{2}}{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le s\le t\le 1;\\ \frac{\left(1-s\right)\left[\left(1-t\right)ts+t\left(s-t\right)\right]}{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le t\le s\le 1\end{array}$

$\left\{\begin{array}{l}{u}^{‴}\left(t\right)=\frac{1}{5!}{\left({u}^{″}\left(t\right)-D\right)}^{5}-\frac{1}{3!}{\left({u}^{″}\left(t\right)-D\right)}^{3}+{u}^{″}\left(t\right)-D,\text{\hspace{0.17em}}t\in \left[0,1\right],\\ u\left(0\right)=A,\text{\hspace{0.17em}}\text{\hspace{0.17em}}u\left(1\right)=B,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{u}^{\prime }\left(1\right)=C,\end{array}$ (19)

$\left\{\begin{array}{l}{u}^{‴}\left(t\right)={\left({u}^{″}\left(t\right)-D\right)}^{3}+\frac{7}{10}{\left({u}^{″}\left(t\right)-D\right)}^{2}-\frac{9}{5}\left({u}^{″}\left(t\right)-D\right),\text{\hspace{0.17em}}t\in \left[0,1\right],\\ u\left(0\right)=A,\text{\hspace{0.17em}}\text{\hspace{0.17em}}u\left(1\right)=B,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{u}^{\prime }\left(0\right)=C,\end{array}$ (20)

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