二阶奇异方程范文（精选3篇）

二阶奇异方程 第1篇

关键词：奇异,边值问题,不动点指数,正解

1 预备知识

近年来, 非线性奇异多点边值问题正解的存在性得到了广泛深入地研究。文献[1]利用上下解的方法研究二阶边值问题

$\{\begin{array}{l}{x}^{″}(t)+f(t,x(t)),x^{\prime }(t)=0,0＜t＜1,\\ \delta x(0)=\gamma x^{\prime }(0),x(1)={\displaystyle {\sum }_{i=1}^{m-2}\alpha }{}_{i}x(\eta {}_i),\\ 0＜\alpha {}_i＜1,0＜\eta {}_i＜1,\delta ＞0,\gamma \ge 0。\end{array}$

在f (t, x, y) 可变号但没有奇异性的条件下正解的存在性。受文献[1]的启发, 本文研究的是二阶边值问题

$\{\begin{array}{l}{x}^{″}(t)+f(t,x(t)),x^{\prime }(t)=0,0＜t＜1,\\ \delta x(0)=\gamma x^{\prime }(0),x(1)=\alpha x(\eta ),\\ 0＜\alpha ＜1,0＜\eta ＜1,\delta ＞0,\gamma \ge 0\end{array}(1)$

当f (t, x, y) 在x=0奇异时其正解的存在性。

令C1[0, 1]={x:[0, 1]R|x (t) ∈C[0, 1], x′ (t) ∈C[0, 1]}, 定义$\parallel x\parallel =\max \{\parallel x\parallel {}_1,\parallel x\parallel {}_2\},\parallel x\parallel {}_1=\underset{t\in [0‚1]}{{\displaystyle \max }}|x(t)|,\parallel x\parallel {}_2=\underset{t\in [0‚1]}{{\displaystyle \max }}|x^{\prime }(t)|$, 定义P={x∈C1[0, 1]:x (t) ≥0, ∀t∈[0, 1]}。显然, P是banach空间 (C1[0, 1], ‖.‖) 中的一个锥。假设以下条件成立。

(H1) f∈C ([0, 1] (0, +∞) (-∞, +∞) , (0, ∞) ) ,

(H2) f (t, x, y) a (t) [g (x) +h (x) ]q (|y|) , (t, x, y) ∈[0, 1] (0, ∞) (-∞, +∞) , a∈C ([0, 1], [0, +∞) ) , g∈C ( (0, +∞) , (0, +∞) ) , h∈C ([0, +∞) , [0, +∞) ) , q∈C ([0, +∞) , (0, +∞) ) 。而且g非增。

$\begin{array}{l}(\text{Η}{}_3)\underset{C\in (0‚+\infty )}{{\displaystyle \sup }}\frac{C}{{\rm I}{}^{-1}\left({\rm K}c\overline{h}(c)+{\rm K}{\displaystyle {\int }_0^{c}g}(u)du\right)}＞\frac{\gamma +\delta }{\delta },\\ {\rm I}(z)={\displaystyle {\int }_0^z\frac{u\text{d}u}{q(u)}},z\in [0‚+\infty )‚{\rm K}=\underset{t\in [0‚1]}{{\displaystyle \max }}a(t),\\ \overline{h}(c)=\underset{s\in [0‚c]}{{\displaystyle \max }}h(s)。\end{array}$

(H4) 对∀H>0, 存在一个函数ΨH (t) ∈C ( (0, 1) , (0, +∞) ) 及常数0β<1, 使得f (t, x, y) ≥ΨH (t) |y|β, ∀ (t, x, y) ∈ ([0, 1][0, H]R-{0}) 。

引理1[2] 设Ω为banach空间E中的一个有界开集, P是E中的一个锥, θ∈Ω, 且A:$\overline{\text{Ω}}{\displaystyle \cap {\rm P}}{\rm P}$是一个全连续算子, 如果λAx≠x, ∀x∈∂Ω∩P, λ∈ (0, 1], 则i (A, Ω∩P, P) =1。

对n∈N*和x∈P定义

$({\rm T}{}_{n}x)(t)={\displaystyle {\int }_0^{1}k}(t,s)f\left(s,x(s)+\frac{1}{n},x^{\prime }(s)\right)\text{d}\text{s}+$

αx (η) , t∈[0, 1],

其中

$k(t,s)=\frac{1}{\gamma +\delta }\{\begin{array}{l}(\gamma +\delta t)(1-s),0ts1\\ (\gamma +\delta s)(1-t),0st1。\end{array}$

2 主要结果

定理1 假设 (H1) (H4) 成立, 则边值问题 (1) 至少存在一个非负解x0∈C1[0, 1], 且x0 (t) >0, t∈ (0, 1]。

证明 由 (H3) 及I, I-1的连续性知, 存在R>0, 使得

$\frac{R}{{\rm I}{}^{-1}\left(R\overline{h}(R){\rm K}+{\rm K}{\displaystyle {\int }_0^{R}g}(u)\text{d}u\right)}＞\frac{\gamma +\delta }{\delta }$

成立。取ε>0, 0<ε<R使得

$\frac{R}{{\rm I}{}^{-1}\left((R+\epsilon )\overline{h}(R+\epsilon ){\rm K}+{\rm K}{\displaystyle {\int }_0^{R+\epsilon }g}(u)\text{d}u\right)}＞\frac{\gamma +\delta }{\delta }(2)$

取n0∈N*, 使得$\frac{1}{n{}_0}＜\epsilon$, 令N0={n0, n0+1, }, 则对给定的n∈N0, Tn:PP是全连续的。令Ω1={x∈C1[0, 1]:‖x‖<R}。下面我们证明

$\mu {\rm T}{}_{n}x\ne x,\forall \mu \in (0,1]‚x\in {\rm P}{\displaystyle \cap \partial }\text{Ω}{}_1‚\forall n\in {\rm N}{}_0(3)$

若存在x∈P∩∂Ω1, n*∈N0及μ0∈ (0, 1], 使得x=μ0Tn*x, 则

$\begin{array}{l}{x}^{″}(t)=-\mu {}_{0}f\left(t,x(t)+\frac{1}{n{}_{*}},x^{\prime }(t)\right)＜0,t\in (0,1),\\ \delta x(0)=\gamma x^{\prime }(0),x(1)=\alpha x(\eta )。\end{array}$

进而存在t0∈ (0, 1) 使得x (t0) =‖x‖1>0, x′ (t0) =0并且x′ (t) >0, t∈[0, t0) , x′ (t) <0, t∈ (t0, 1]。

当t∈[0, t0) 时,

$\begin{array}{l}-x^{″}(t)=\mu {}_{0}f\left(t,x(t)+\frac{1}{n{}_{*}},x^{\prime }(t)\right)\\ a(t)\left[g\left(x(t)+\frac{1}{n{}_{*}}\right)+h\left(x(t)+\frac{1}{n{}_{*}}\right)\right]q(x^{\prime }(t)),\end{array}$

所以可以得到

$\begin{array}{l}\frac{-x^{″}(t)x^{\prime }(t)}{q(x^{\prime }(t))}a(t)[g\left(x(t)+\frac{1}{n{}_{*}}\right)+\\ h\left(x(t)+\frac{1}{n{}_{*}}\right)]x^{\prime }(t)。\end{array}$

从t到t0积分

${\displaystyle {\int }_0^{x^{\prime }(t)}\frac{u}{q(u)}}\text{d}u{\rm K}{\displaystyle {\int }_0^{R+\epsilon }g}(u)\text{d}u+{\rm K}\overline{h}(R+\epsilon )(R+\epsilon ),$

即

$x^{\prime }(t){\rm I}{}^{-1}\left({\rm K}{\displaystyle {\int }_0^{R+\epsilon }g}(u)\text{d}u+{\rm K}\overline{h}(R+\epsilon )(R+\epsilon )\right)(4)$

从0至t0积分

$\begin{array}{l}x(t{}_0)x(0)+{\rm I}{}^{-1}\left({\rm K}{\displaystyle {\int }_0^{R+\epsilon }g}(u)\text{d}u+{\rm K}\overline{h}(R+\epsilon )(R+\epsilon )\right)\\ \frac{\gamma +\delta }{\delta }{\rm I}{}^{-1}\left({\rm K}{\displaystyle {\int }_0^{R+\epsilon }g}(u)\text{d}u+{\rm K}\overline{h}(R+\epsilon )(R+\epsilon )\right)。\end{array}$

即

$\begin{array}{l}\parallel x\parallel {}_1\frac{\gamma +\delta }{\delta }{\rm I}{}^{-1}({\rm K}{\displaystyle {\int }_0^{R+\epsilon }g}(u)\text{d}u+\\ {\rm K}\overline{h}(R+\epsilon )(R+\epsilon )\begin{array}{l}\\ \end{array})(5)\end{array}$

当t∈ (t0, 1]时,

$\begin{array}{l}\frac{-x^{″}(t)}{q(-x^{\prime }(t))}(-x^{\prime }(t))a(t)[g\left(x(t)+\frac{1}{n{}_{*}}\right)+\\ h\left(x(t)+\frac{1}{n{}_{*}}\right)](-x^{\prime }(t))‚\end{array}$

从t0至t积分

${\displaystyle {\int }_0^{-x^{\prime }(t)}\frac{u}{q(u)}}\text{d}u{\rm K}{\displaystyle {\int }_0^{R+\epsilon }g}(u)\text{d}u+{\rm K}\overline{h}(R+\epsilon )(R+\epsilon ),$

即

$-x^{\prime }(t){\rm I}{}^{-1}({\rm K}{\displaystyle {\int }_0^{R+\epsilon }g}(u)\text{d}u+{\rm K}\overline{h}(R+\epsilon )(R+\epsilon )\begin{array}{l}\\ \end{array})(6)$

由式 (4) 、式 (5) 、式 (6) 可得

$\begin{array}{l}R=\parallel x\parallel =\max \{\parallel x\parallel {}_1,\parallel x\parallel {}_2\}\\ \frac{\gamma +\delta }{\delta }{\rm I}{}^{-1}\left({\rm K}{\displaystyle {\int }_0^{R+\epsilon }g}(u)\text{d}u+{\rm K}\overline{h}(R+\epsilon )(R+\epsilon )\right)\end{array}$

与R的选取矛盾, 即式 (3) 成立。由引理1知对任意的n∈N0, Tn在P∩Ω内存在一个不动点xn, 并且存在tn∈ (0, 1) 使得x (tn) =‖x‖1, 当t∈[0, tn) 时, x′n (t) >0, 当t∈ (tn, 1]时, x′n (t) <0。

由xn∈P∩Ω1知{xn}一致有界等度连续的, {x′n}是一致有界的, 下面证明{x′n}在[0, 1]上是等度连续的。

对任意0t1<t2<tn

$\begin{array}{l}-x^{″}{}_n(t)=f\left(t,x(t)+\frac{1}{n},x^{\prime }{}_n(t)\right)\\ a(t)\left(g\left(x(t)+\frac{1}{n}\right)+h\left(x(t)+\frac{1}{n}\right)\right)q(x^{\prime }{}_n(t));\end{array}$

$\begin{array}{l}\frac{-x^{″}{}_n(t)x^{\prime }{}_n(t)}{q(x^{\prime }{}_n(t))}a(t)(g\left(x(t)+\frac{1}{n}\right)+\\ h\left(x(t)+\frac{1}{n}\right))x^{\prime }{}_n(t)。\end{array}$

从t1至t2积分可得

$\begin{array}{l}{\displaystyle {\int }_{x^{\prime }{}_n(t{}_2)}^{x^{\prime }{}_n(t{}_1)}\frac{u}{q(u)}}\text{d}\text{u}{\rm K}\overline{h}(R+\epsilon )(x{}_n(t{}_2)-\\ x{}_n(t{}_1))+{\rm K}{\displaystyle {\int }_{x{}_n(t{}_1)+\frac{1}{n}}^{x{}_n(t{}_2)+\frac{1}{n}}g}(u)\text{d}u,\end{array}$

即对任意的t1, t2∈[0, tn) 有

$\begin{array}{l}|{\rm I}(x^{\prime }{}_n(t{}_1))-{\rm I}(x^{\prime }{}_n(t{}_2))|{\rm K}\overline{h}(R+\epsilon )|x{}_n(t{}_2)-\\ x{}_n(t{}_1)|+{\rm K}\left|{\displaystyle {\int }_{x{}_n(t{}_1)+\frac{1}{n}}^{x{}_n(t{}_2)+\frac{1}{n}}g}(u)\text{d}u\right|(7)\end{array}$

同理对任意的t1, t2∈ (tn, 1]有

$\begin{array}{l}|{\rm I}(-x^{\prime }{}_n(t{}_2))-{\rm I}(-x^{\prime }{}_n(t{}_1))|{\rm K}\overline{h}(R+\epsilon )|x{}_n(t{}_1)-\\ x{}_n(t{}_2)|+{\rm K}\left|{\displaystyle {\int }_{x{}_n(t{}_2)+\frac{1}{n}}^{x{}_n(t{}_1)+\frac{1}{n}}g}(u)\text{d}u\right|(8)\end{array}$

由积分的等度连续性和I, I-1的连续性知, 对任意ε>0的存在δ′>0, 使得当t1, t2∈[0, tn) , 0<|t1-t2|<δ′时, 有

$|x^{\prime }{}_n(t{}_1)-x^{\prime }{}_n(t{}_2)|＜\frac{\epsilon }{2}(9)$

当t1, t2∈ (tn, 1], 0<|t1-t2|<δ′时, 有

$|x^{\prime }{}_n(t{}_1)-x^{\prime }{}_n(t{}_2)|＜\frac{\epsilon }{2}(10)$

进而当0t1<tn<t21, 0<|t1-t2|<δ′时, 由式 (9) 、式 (10) 可得

|x′n (t1) -x′n (t2) |=|x′n (t1) -x′n (tn) +x′n (tn) -x′n (t2) |

|x′n (t1) -x′n (tn) |+|Dx′n (tn) -x′n (t2) |<ε (11)

同理当0t2<tn<t11, 0<|t1-t2|<δ′时,

$|x^{\prime }{}_n(t{}_1)-x^{\prime }{}_n(t{}_2)|＜\epsilon (12)$

由式 (9) 、式 (10) 、式 (11) 、式 (12) 可知{x′n}在[0, 1]上是等度连续的。

由Arzela-Ascoli定理知存在{xn}的一个收敛子列收敛到x0∈C1[0, 1] (不妨设为{x′n}本身) , 即$\underset{n+\infty }{{\displaystyle \lim }}x{}_n=x{}_0,\underset{n+\infty }{{\displaystyle \lim }}{x}^{\prime }{}_n=x^{\prime }{}_0$。由δxn (0) =γx′n (0) , xn (1) =αxn (η) 知δx0 (0) =γx′0 (0) , x0 (1) =αx0 (η) 。

下面证明$\underset{n\in {\rm N}{}_0}{{\displaystyle \inf }}\parallel x{}_n\parallel {}_1＞0$。由条件 (H4) 知, 对任意的n∈N0, 存在函数ΨR+ε (t) >0以及常数β∈[0, 1) 使得

$\begin{array}{l}-x^{″}(t)=f\left(t,x^{\prime }(t)+\frac{1}{n},x^{\prime }(t)\right)\ge \\ \Psi {}_{R+\epsilon }(t)|x^{\prime }{}_n(t)|{}^{\beta }(13)\end{array}$

通过对式 (13) 进行积分, 则可以得到

$\begin{array}{l}{x}^{\prime }(t)\ge {\displaystyle {\int }_t^{t{}_n}\Psi }{}_{R+\epsilon }(s)|x^{\prime }{}_n(s)|{}^{\beta }\text{d}\text{s},t\in [0,t{}_n),\\ -x^{\prime }(t)\ge {\displaystyle {\int }_{t{}_n}^t\Psi }{}_{R+\epsilon }(s)|x^{\prime }{}_n(s)|{}^{\beta }\text{d}\text{s},t\in (t{}_n,1],\end{array}$

即

$\begin{array}{l}{x}^{\prime }(t)\ge \left((1-\beta ){\displaystyle {\int }_t^{t{}_n}\Psi }{}_{R+\epsilon }(s)|x^{\prime }{}_n(s)|{}^{\beta }\text{d}\text{s}\right){}^{\frac{1}{1-\beta }},\\ t\in [0‚t{}_n)(14)\end{array}$

$\begin{array}{l}-x^{\prime }(t)\ge \left((1-\beta ){\displaystyle {\int }_{t{}_n}^t\Psi }{}_{R+\epsilon }(s)|x^{\prime }{}_n(s)|{}^{\beta }\text{d}\text{s}\right){}^{\frac{1}{1-\beta }},\\ t\in (t{}_n,1](15)\end{array}$

假设$\underset{n\in {\rm N}{}_0}{{\displaystyle \inf }}\parallel x{}_n\parallel {}_1=c{}_0$则c0≥0。如果c0=0, 不失一般性, 设$\underset{n+\infty }{{\displaystyle \lim }}\parallel x\parallel {}_1=0,\underset{n+\infty }{{\displaystyle \lim }}t{}_n=t{}_0$。于是出现两种可能性。

(1) t0=0。当n充分大时

xn (1) -xn (η) = (α-1) xn (η) =

${\displaystyle {\int }_{\eta }^1{x}^{\prime }}{}_n(s)\text{d}\text{s}{\displaystyle {\int }_{\eta }^1\left(-{\displaystyle {\int }_{t{}_n}^s\Psi }{}_{R+\epsilon }(r)\text{d}\text{r}\right)}{}^{\frac{1}{1-\beta }}$ds,

进而

$x{}_n(\eta )\ge \frac{1}{1-\alpha }{\displaystyle {\int }_{\eta }^1\left({\displaystyle {\int }_{t{}_n}^s\Psi }{}_{R+\epsilon }(r)\text{d}\text{r}\right)}{}^{\frac{1}{1-\beta }}\text{d}\text{s}。$

于是

$\begin{array}{l}\parallel x\parallel {}_1=x{}_n(t{}_n)=x{}_n(\eta )-{\displaystyle {\int }_{t{}_n}^{\eta }{x}^{\prime }}{}_n(s)\text{d}\text{s}\ge \\ \frac{1}{1-\alpha }{\displaystyle {\int }_{\eta }^1\left({\displaystyle {\int }_{t{}_n}^s\Psi }{}_{R+\epsilon }(r)\text{d}\text{r}\right)}{}^{\frac{1}{1-\beta }}\text{d}\text{s}\\ \frac{1}{1-\alpha }{\displaystyle {\int }_{\eta }^1\left({\displaystyle {\int }_{t{}_0}^s\Psi }{}_{R+\epsilon }(r)\text{d}\text{r}\right)}{}^{\frac{1}{1-\beta }}\text{d}\text{s}＞0。\end{array}$

产生矛盾。

(2) 0<t01。此时

$\begin{array}{l}\parallel x{}_n\parallel {}_1\ge x{}_n(t{}_n)-x{}_n(0)={\displaystyle {\int }_0^{t{}_n}{x}^{\prime }}{}_n(s)\text{d}\text{s}\ge \\ {\displaystyle {\int }_0^{t{}_n}\left({\displaystyle {\int }_s^{t{}_n}\Psi }{}_{R+\epsilon }(r)\text{d}\text{r}{}^{\frac{1}{1-\beta }}\right)}\text{d}\text{s}{\displaystyle {\int }_0^{t{}_0}\left({\displaystyle {\int }_s^{t{}_0}\Psi }{}_{R+\epsilon }(r)\text{d}\text{r}{}^{\frac{1}{1-\beta }}\right)}\text{d}\text{s}＞0\end{array}$

所以$\underset{n\in {\rm N}{}_0}{{\displaystyle \inf }}\parallel x{}_n\parallel {}_1=c{}_0＞0$。进而x0 (t) >0, t∈[0, 1]。

由于

$x{}_n(t)={\displaystyle {\int }_0^{1}k}(t,s)f\left(s,x{}_n(s)+\frac{1}{n},x^{\prime }{}_n(s)\right)\text{d}\text{s}+$

αxn (η) , t∈[0, 1],

由Lebesgue控制收敛定理可得

x0 (t) =∫${}_0^1$k (t, s) f (s, x0 (s) , x′0 (s) ) ds+

αx0 (η) , t∈[0, 1],

两边求导可得到x0为边值问题 (1) 的正解。

例子 考虑下列二阶三点边值问题

$\{\begin{array}{l}{x}^{″}(t)+\lambda (t{}_2+1)\left(\frac{1}{x{}^{\frac{1}{2}}}+x{}^2+2\right)\left(\begin{array}{l}\\ \end{array}{x}^{\prime }{}^{\frac{1}{2}}+1\right)=0,\\ 0＜t＜1,\\ x(0)=x^{\prime }(0),x(1)=\frac{1}{2}x\left(\frac{1}{4}\right)\end{array}(16)$

取$a(t)=\lambda (t{}_2+1),g(x)=\frac{1}{x{}^{\frac{1}{2}}},h(x)=x{}^2+2,q(y)=y{}^{\frac{1}{2}}+1,0＜\lambda ＜\frac{1}{10}$, 则由定理可知式 (16) 存在一个正解。

参考文献

[1]Khan R A.Generalized appoximations and rapid convergence of solu-tions of m-point boundary value problems.App.Math Com, 2007;188:1878—1890

[2] Lakshmikantham V, Bainov D D, Simeonov P S. Theory of impulsive differential equations. Singapore: World Scientific, 1989

[3] Erbe L H.Tang M.Existence and multiplicity of positive solutions to nonlinear boundary value problems. Diff Equat Dyn Syst, 1996;4:313—320

奇异超线性二阶边值问题的正解 第2篇

奇异超线性二阶边值问题的正解

在f满足超线性增长条件下,利用锥不动点指数研究了奇异超线性二阶边值问题y″+m2y=h(x)f(y),0＜x＜2π,m∈(0,(1)/(2)),y(0)-y(2π)=0,y′(0)-y′(2π)=λ＞0的.正解和多个正解存在性,其中h在区间[0,2π]端点可以具有适当奇性.

作 者：孔令彬 张长海  作者单位：大庆石油学院,数学系,黑龙江,大庆,163318 刊 名：哈尔滨工业大学学报  ISTIC EI PKU英文刊名：JOURNAL OF HARBIN INSTITUTE OF TECHNOLOGY 年，卷(期)：2003 35(3) 分类号：O175.8 关键词：奇异性   超线性   正解   锥   不动点指数  

一类二阶非线性差分方程的振动性质 第3篇

从解的渐近状态着手,将所述方程的所有非平凡解分成互不相交的`四类,应用分类讨论方法和分析方法,讨论了一类广泛的二阶非线性差分方程解的振动性质,建立了两个新的振动性定理,推广并改进了已有文献中的相关结果.

作 者：张全信 燕居让 ZHANG Quan-xin YAN Ju-rang 作者单位：张全信,ZHANG Quan-xin(滨州学院数学与信息科学系,山东滨州,256603)

燕居让,YAN Ju-rang(山西大学数学科学学院,太原,030006)