四阶两点边值问题3个对称正解的存在性

达举霞

doi:10.6054/j.jscnun.2021014

四阶两点边值问题3个对称正解的存在性

达举霞^,

兰州财经大学长青学院，兰州 730070

基金项目:

国家自然科学基金项目 11561063

详细信息

通讯作者:
达举霞, Email: 1414320179@qq.com

中图分类号: O175.8
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出版历程
- 收稿日期: 2020-05-07
- 网络出版日期: 2021-03-23
- 刊出日期: 2021-02-24

The Existence of Three Symmetric Positive Solutions to A Fourth-Order Two-Point Boundary Value Problem

DA Juxia^,

Changqing College of Lanzhou University of Finance and Economics, Lanzhou 730070, China

摘要

摘要: 应用广义的Leggett-Williams不动点定理，研究了四阶两点边值问题 ${{u}^{\left( 4 \right)}}\left( t \right)=f\left( u\left( t \right) \right)\ \ \ \ \ \left( t\in \left[ 0, 1 \right] \right), u\left( 0 \right)=u\left( 1 \right)=0, {u}''\left( 0 \right)={u}''\left( 1 \right)=0$ 正解的存在性, 其中 $f:\mathbb{R}\to \left[ 0, +\infty \right)$ 连续. 在f满足适当的增长条件下, 得到该问题至少存在3个对称正解.
- 四阶边值问题 /
- 格林函数 /
- 对称正解 /
- 锥
Abstract: Applying the generalized Leggett-Williams fixed-point theorem, the existence of positive solutions to the fourth-order boundary value problem is studied: ${{u}^{\left( 4 \right)}}\left( t \right)=f\left( u\left( t \right) \right)\ \ \ \ \ \left( t\in \left[ 0, 1 \right] \right), u\left( 0 \right)=u\left( 1 \right)=0, {u}''\left( 0 \right)={u}''\left( 1 \right)=0$ , where $f:\mathbb{R}\to \left[ 0, +\infty \right)$ is continuous. Under some conditions on f, there exist at least three symmetric positive solutions.
- fourth-order boundary value problem /
- Green's function /
- symmetric positive solutions /
- cone

HTML全文

在弹性力学和工程物理学中，四阶常微分方程边值问题用于刻画弹性梁的平衡状态. 目前, 关于四阶两点边值问题与四阶多点边值问题的研究已有很多结果. 例如，周韶林等^[1]运用不动点理论获得了四阶三点边值问题

$\begin{array}{c} u^{(4)}(t)=g(t) f(u(t)) \quad(t \in[0,1]), \\ u(0)=u^{\prime}(0)=u^{\prime \prime}(\beta)=u^{\prime \prime}(1)=0 \end{array}$

正解的存在性结果, 这里β∈[1/3, 1]为常数, g ∈C([0, 1], [0, +∞))；达举霞和韩晓玲^[2]运用锥上的不动点定理获得了非线性奇异四阶三点边值问题

$\begin{array}{c} u^{(4)}(t)=\lambda a(t) f(t, u(t)) \quad(t \in[0,1]), \\ u(0)=u^{\prime}(\eta)=u^{\prime \prime}(1)=u^{\prime \prime}(0)=0 \end{array}$

正解的存在性结果, 其中λ >0是正参数, η ∈[1/2, 1)；AVERY和HENDERSON^[3]运用广义Leggett-Williams不动点定理获得了二阶边值问题

$\begin{array}{c} y^{\prime \prime}+f(y)=0 \quad(0 \leqslant t \leqslant 1), \\ y(0)=0=y(1) \end{array}$

对称正解的存在性, 其中f: $\mathbb{R}$ →[0, +∞)连续. 近年来，常微分方程边值问题在理论和应用中起到很大的作用，主要用来描述大量的物理、生物和化学现象及一些结构性变的讨论等，且这些都转化为了某种形式的四阶边值问题的研究^[4-15].

在文献[3]的基础上，本文研究四阶两点边值问题

$\begin{array}{c} u^{(4)}(t)=f(u(t)) \quad(t \in[0,1]), \\ u(0)=u(1)=0, u^{\prime \prime}(0)=u^{\prime \prime}(1)=0 \end{array}$

(1)

正解的存在性, 其中f: $\mathbb{R}$ →[0, +∞)连续. 在f满足适当增长条件下, 证明了问题(1)在其边界条件下至少存在3个对称正解.

1. 预备知识

定义 1^[3] 设E是一个实的Banach空间，一个非空闭凸集K⊂E是E上的一个锥，如果满足下面2个条件：

(i) 若x ∈K, λ >0, 则λ x ∈K;

(ii) 若x ∈K，-x ∈K, 则x=0.

定义 2^[3] 若算子A是连续的且映有界集到列紧集，则称算子A全连续.

定义 3^[3] 设E是一个实的Banach空间，并设P是E上的锥，对∀x, y ∈P, t ∈[0, 1]，若有

$\alpha(t x+(1-t) y) \geqslant t \alpha(x)+(1-t) \alpha(y),$

则映射α: P→[0, +∞)是一个凹函数. 相似地，若有

$\beta(t x+(1-t) y) \leqslant t \beta(x)+(1-t) \beta(y),$

则映射β: P→[0, +∞)是一个凸函数.

设γ、β和θ是P上的非负连续凸函数，α、ψ是P上的非负连续凹函数，则对正数h、a、b、d、c，定义如下凸集:

$\begin{array}{c} P(\gamma, c)=\{x \in P: \gamma(x)<c\} ,\\ P(\gamma, \alpha, a, c)=\{x \in P: a \leqslant \alpha(x), \gamma(x) \leqslant c\}, \\ Q(\gamma, \beta, d, c)=\{x \in P: \beta(x) \leqslant d, \gamma(x) \leqslant c\}, \\ P(\gamma, \theta, \alpha, a, b, c)=\{x \in P: a \leqslant \alpha(x), \theta(x) \leqslant b, \gamma(x) \leqslant c\},\\ Q(\gamma, \beta, \psi, h, d, c)=\{x \in P: h \leqslant \psi(x), \beta(x) \leqslant d, \gamma(x) \leqslant c\}. \end{array}$

下面给出广义Leggett Williams不动点定理：

定理 1^[3] 设锥P⊂E, α和ψ是定义在P上的非负连续凹函数且γ、β和θ是定义在P上的非负连续凸函数, 对正数c和M, 有α(x) ≤β(x)且‖x‖ ≤Mγ(x) (x ∈ $\overline{P\left( \gamma , c \right)}$ ). 假设A: $\overline{P\left( \gamma , c \right)}$ → $\overline{P\left( \gamma , c \right)}$ 是全连续的, 且存在正数h, d, a, b(0 < d < a), 有

(i){x ∈P(γ, θ, α, a, b, c): α (x) >a } ≠Ø 且α(Ax)>a (x ∈P(γ, θ, α, a, b, c));

(ii){x ∈Q(γ, β, ψ, h, d, c): β(x) < d} ≠ Ø 且β(Ax) < d (x ∈Q(γ, β, ψ, h, d, c));

(iii) 当α(Ax)>a，有θ(Ax)> b (x ∈P(γ, α, a, c));

(iv) 当β(Ax) < d，有ψ(Ax) < h (x ∈Q(γ, β, d, c))，

则A至少有3个不动点x₁, x₂, x₃ ∈ $\overline{P\left( \gamma , c \right)}$ , 使得

$\beta\left(x_{1}\right)<d, a<\alpha\left(x_{2}\right), d<\beta\left(x_{3}\right), \alpha\left(x_{3}\right)<a.$

2. 对称正解的存在性

考虑问题

$u^{(4)}(t)=h(t)$

满足问题(1)条件的格林函数

$G(t, s)=\left\{\begin{array}{ll} \frac{1}{6} s(1-t)\left(2 t-t^{2}-s^{2}\right) & (0 \leqslant s \leqslant t \leqslant 1), \\ \frac{1}{6} t(1-s)\left(2 s-s^{2}-t^{2}\right) & (0 \leqslant t \leqslant s \leqslant 1), \end{array}\right.$

(2)

且格林函数有如下性质

$\int_{0}^{1} G(t, s) \mathrm{d} s=\frac{t(1-t)\left(1+t-t^{2}\right)}{24} \quad(0 \leqslant t \leqslant 1),$

(3)

$\int_{0}^{1 / r} G\left(\frac{1}{2}, s\right) \mathrm{d} s=\int_{1-1 / r}^{1} G\left(\frac{1}{2}, s\right) \mathrm{d} s=\frac{3 r^{2}-2}{96 r^{4}} \quad(r>2),$

(4)

$\int_{1 / r}^{1 / 2} G\left(\frac{1}{2}, s\right) \mathrm{d} s=\int_{1 / 2}^{1-1 / r} G\left(\frac{1}{2}, s\right) \mathrm{d} s=\frac{5 r^{4}-24 r^{2}+16}{768 r^{4}}(r>2),$

(5)

$\begin{array}{c} \quad \int_{t_{1}}^{t_{2}} G\left(t_{1}, s\right) \mathrm{d} s+\int_{1-t_{2}}^{1-t_{1}} G\left(t_{1}, s\right) \mathrm{d} s=\frac{1}{12} t_{1}\left[3\left(t_{2}^{2}-t_{1}^{2}\right)+\right. \\ \left.2\left(t_{1}^{3}-t_{2}^{3}\right)+2 t_{1}^{2}\left(t_{1}-t_{2}\right)\right] \quad\left(0<t_{1}<t_{2} \leqslant \frac{1}{2}\right), \end{array}$

(6)

$\max\limits _{0 \leqslant r \leqslant 1} \frac{G(1 / 2, r)}{G(t, s)}=\frac{1}{8 t^{3}} \quad\left(0<t \leqslant \frac{1}{2}\right),$

(7)

$\min\limits _{0 \leqslant r \leqslant 1} \frac{G(t_1, r)}{G(t_2, s)}=\frac{t_{1}^{3}}{t_{2}^{3}} \quad\left(0<t \leqslant \frac{1}{2}\right).$

(8)

考虑Banach空间E=C[0, 1], 定义范数‖u‖= $\mathop {\max }\limits_{0 \le t \le 1} \left| {u\left( t \right)} \right|$ , 定义锥P⊂E如下：

$\begin{aligned} &P=\{u \in E \mid u(t) \geqslant 0, u(t)=u(1-t), \forall t \in[0,1], \text { 且 } u \text { 是 }\\ &\ \ \ \ \ \ \ \ [0,1] \text { 上凹函数, 当 } 0<t_{3} \leqslant 1 / 2 \text { 时 }, \min \limits_{t \in\left[t_{3}, 1-t_{3}\right]} u(t) \geqslant\\ &\ \ \ \ \ \ \ \ \left.2 t_{3}\|u\|\right\} . \end{aligned}$

在此锥P上定义非负连续凹函数α、ψ和非负连续凸函数β、θ、γ，且

$\begin{array}{c} \gamma(u)=\max \limits_{t \in\left[0, t_{3}\right] \cup\left[1-t_{3}, 1\right]} u(t)=u\left(t_{3}\right), \\ \psi(u)=\min \limits_{t \in\left[\frac{1}{r}, \frac{r-1}{r}\right]} u(t)=u\left(\frac{1}{r}\right), \\ \beta(u)=\max \limits_{t \in\left[\frac{1}{r} \cdot \frac{r-1}{r}\right]} u(t)=u\left(\frac{1}{2}\right), \\ \alpha(u)=\min \limits_{t \in\left[t_{1}, t_{2}\right] \cup\left[1-t_{2}, 1-t_{1}\right]} u(t)=u\left(t_{1}\right), \\ \theta(u)=\max \limits_{t \in\left[t_{1}, t_{2}\right] \cup\left[1-t_{2}, 1-t_{1}\right]} u(t)=u\left(t_{2}\right), \end{array}$

其中，t₁、t₂和r是非负数且0 < t₁ < t₂≤1/2, 1/r≤t₂.

对∀u∈P，有

$\alpha(u)=u\left(t_{1}\right) \leqslant u\left(\frac{1}{2}\right)=\beta(u) ,$

(9)

$\|u\|=u\left(\frac{1}{2}\right) \leqslant \frac{1}{2 t_{3}} u\left(t_{3}\right)=\frac{1}{2 t_{3}} \gamma(u),$

(10)

则u∈P是问题(1)在其边界条件下的解当且仅当

$u(t)=\int_{0}^{1} G(t, s) f(u(s)) \mathrm{d} s \quad(0 \leqslant t \leqslant 1).$

下面给出本文的主要结果.

定理 2 假设a、b和c是非负数且0 < a < b < ct₁³/t₂³, 设f满足以下条件:

(i) $f\left( \omega \right) < \frac{{384{r^4}}}{{5{r^4} - 24{r^2} + 16}}\left( {a - \frac{{c\left( {3{r^2} - 2} \right)}}{{2{t_3}\left( {1 - {t_3}} \right)\left( {1 + {t_3} - t_3^2} \right)}}} \right)\;\;\;\;\;\left( {\omega \in \left[ {\frac{{8a}}{{{r^3}}}, a} \right]} \right)$ ；

(ii) $f\left( \omega \right) \ge \frac{{12b}}{{{t_1}\left[ {3\left( {t_2^2 - t_1^2} \right) + 2\left( {t_1^3 - t_2^3} \right) + 2t_1^2\left( {{t_1} - {t_2}} \right)} \right]}}\;\;\;\;\;\left( {\omega \in \left[ {b, \frac{{t_2^3b}}{{t_1^3}}} \right]} \right)$ ；

(iii) $f\left( \omega \right) \le \frac{{24c}}{{{t_3}\left( {1 - {t_3}} \right)\left( {1 + {t_3} - t_3^2} \right)}}\;\;\;\;\;\left( {\omega \in \left[ {0, \frac{c}{{2{t_3}}}} \right]} \right)$ ，

则四阶两点边值问题(1)有3个对称正解u₁、u₂、u₃, 使得

$\begin{array}{c} \mathop {\max }\limits_{t \in \left[ {0,{t_3}} \right] \cup \left[ {1 - {t_3},1} \right]} {u_i}(t) \le c\quad (i = 1,2,3),\\ \mathop {\min }\limits_{t \in \left[ {{t_1},{t_2}} \right] \cup \left[ {1 - {t_2},1 - {t_1}} \right]} {u_1}(t) > b,\mathop {\max }\limits_{t \in \left[ {\frac{1}{r},\frac{{r - 1}}{r}} \right]} {u_2}(t) < a,\\ \mathop {\min }\limits_{t \in \left[ {{t_1},{t_2}} \right] \cup \left[ {1 - {t_2},1 - {t_1}} \right]} {u_3}(t) < b,\mathop {\max }\limits_{t \in \left[ {\frac{1}{r},\frac{{r - 1}}{r}} \right]} {u_3}(t) > a. \end{array}$

证明定义全连续算子A为：

$A u(t)=\int_{0}^{1} G(t, s) f(u(s)) \mathrm{d} s.$

若u ∈P, 由G(t, s)性质可知, Au(t)≥0且(Au)″(t)≤0 (0≤t≤1), Au(t₃)≥2t₃Au(1/2)，且

$A u(t)=A u(t-1) \quad(0 \leqslant t \leqslant 1),$

从而可得Au ∈P, 即A: P→P. 因此, 对∀u ∈P, 由式(9)及式(10), 有‖u‖≤ $\frac{1}{{2{t_3}}}\gamma \left( u \right)$ .

设u∈ $\overline{P\left( \gamma , c \right)}$ , ‖u‖≤ $\frac{1}{{2{t_3}}}\gamma \left( u \right) \le \frac{c}{{2{t_3}}}$ ，则由式(3)可得

$\begin{array}{l} \gamma(A u)=\max \limits_{t \in\left[0, t_{3}\right] \cup\left[1-t_{3}, 1\right]} \int_{0}^{1} G(t, s) f(u(s)) \mathrm{d} s= \\ \ \ \ \ \ \ \ \int_{0}^{1} G\left(t_{3}, s\right) f(u(s)) \mathrm{d} s \leqslant \\ \ \ \ \ \ \ \ \frac{24 c}{t_{3}\left(1-t_{3}\right)\left(1+t_{3}-t_{3}^{2}\right)} \int_{0}^{1} G\left(t_{3}, s\right) \mathrm{d} s=c. \end{array}$

从而, A: $\overline{P\left( \gamma , c \right)}$ → $\overline{P\left( \gamma , c \right)}$ ，且

$\begin{array}{l} \left\{u \in P\left(\gamma, \theta, \alpha, b, \frac{t_{2}^{3} b}{t_{1}^{3}}\right): \alpha(u)>b\right\} \neq \varnothing \\ \left\{u \in Q\left(\gamma, \beta, \psi, \frac{8 a}{r^{3}}, a, c\right): \beta(u)<a\right\} \neq \varnothing \end{array}$

由定理1可得：

(i) 设u ∈Q(γ, β, a, c), 有ψ(Au) < $\frac{{8a}}{{{r^3}}}$ , 则β(Au) < a. 推导如下：

$\begin{array}{l} \beta(A u)=\max \limits_{t \in\left[\frac{1}{r}, \frac{r-1}{r}\right]} \int_{0}^{1} G(t, s) f(u(s)) \mathrm{d} s= \\ \ \ \ \ \ \ \ \ \int_{0}^{1} G\left(\frac{1}{2}, s\right) f(u(s)) \mathrm{d} s= \\ \ \ \ \ \ \ \ \ \int_{0}^{1} \frac{G\left(\frac{1}{2}, s\right)}{G\left(\frac{1}{r}, s\right)} G\left(\frac{1}{r}, s\right) f(u(s)) \mathrm{d} s \leqslant \\ \ \ \ \ \ \ \ \ \frac{r^{3}}{8} \int_{0}^{1} G\left(\frac{1}{r}, s\right) f(u(s)) \mathrm{d} s=\frac{r^{3}}{8} \psi(A u)<a . \end{array}$

(ii) 设u ∈Q(γ, β, ψ, 8a/r³, a, c), 则β(Au) < a. 推导如下：

$\begin{array}{l} \beta(A u)=\max \limits_{t \in\left[\frac{1}{r}, \frac{r-1}{r}\right]} \int_{0}^{1} G(t, s) f(u(s)) \mathrm{d} s= \\ \ \ \ \ \int_{0}^{1} G\left(\frac{1}{2}, s\right) f(u(s)) \mathrm{d} s= \\ \ \ \ \ 2 \int_{0}^{1 / r} G\left(\frac{1}{2}, s\right) f(u(s)) \mathrm{d} s+2 \int_{1 / r}^{1 / 2} G\left(\frac{1}{2}, s\right) f(u(s)) \mathrm{d} s< \\ \ \ \ \ \frac{c\left(3 r^{2}-2\right)}{2 t_{3}\left(1-t_{3}\right)\left(1+t_{3}-t_{3}^{2}\right) r^{4}}+\frac{5 r^{4}-24 r^{2}+16}{384 r^{4}} \times \frac{384 r^{4}}{5 r^{4}-24 r^{2}+16} \times \\ \ \ \ \ \left(a-\frac{c\left(3 r^{2}-2\right)}{2 t_{3}\left(1-t_{3}\right)\left(1+t_{3}-t_{3}^{2}\right) r^{4}}\right)=a. \end{array}$

(iii) 设u ∈P(γ, α, b, c), 有θ(Au)>t₂³b/t₁³, 则α(Au)>b. 推导如下：

$\begin{array}{l} \alpha(A u)=\min \limits_{t \in\left[t_{1}, t_{2}\right] \cup\left[1-t_{2}, 1-t_{1}\right]} \int_{0}^{1} G(t, s) f(u(s)) \mathrm{d} s= \\ \ \ \ \ \int_{0}^{1} G\left(t_{1}, s\right) f(u(s)) \mathrm{d} s= \\ \ \ \ \ \int_{0}^{1} \frac{G\left(t_{1}, s\right)}{G\left(t_{2}, s\right)} G\left(t_{2}, s\right) f(u(s)) \mathrm{d} s \geqslant \\ \ \ \ \ \frac{t_{1}^{3}}{t_{2}^{3}} \int_{0}^{1} G\left(t_{2}, s\right) f(u(s)) \mathrm{d} s=\frac{t_{1}^{3}}{t_{2}^{3}} \theta(A u)>b. \end{array}$

(iv) 设u ∈P(γ, θ, α, b, t₂³b/t₁³, c), 则α(Au)>b. 推导如下

$\begin{array}{l} \alpha(A u)=\min \limits_{t \in\left[t_{1}, t_{2}\right] \cup\left[1-t_{2}, 1-t_{1}\right]} \int_{0}^{1} G(t, s) f(u(s)) \mathrm{d} s= \\ \ \ \ \ \ \ \ \ \int_{0}^{1} G\left(t_{1}, s\right) f(u(s)) \mathrm{d} s>\int_{t_{1}}^{t_{2}} G\left(t_{1}, s\right) f(u(s)) \mathrm{d} s+ \\ \ \ \ \ \ \ \ \ \int_{1-t_{2}}^{1-t_{1}} G\left(t_{1}, s\right) f(u(s)) \mathrm{d} s \geqslant \\ \ \ \ \ \ \ \ \ \frac{12 b}{t_{1}\left[3\left(t_{2}^{2}-t_{1}^{2}\right)+2\left(t_{1}^{3}-t_{2}^{3}\right)+2 t_{1}^{2}\left(t_{1}-t_{2}\right)\right]} \times \\ \ \ \ \ \ \ \ \ {\left[\int_{t_{1}}^{t_{2}} G\left(t_{1}, s\right) \mathrm{d} s+\int_{1-t_{2}}^{1-t_{1}} G\left(t_{2}, s\right) \mathrm{d} s\right]=b.} \end{array}$

综上, 由广义Leggett-Williams不动点定理^[16]可得四阶两点边值问题(1)有3个正解u₁, u₂, u₃ ∈ $\overline{P\left( \gamma , c \right)}$ , 使得

$\alpha ({u_1}) > b,\beta ({u_2}) < a,\alpha ({u_3}) < b,\beta ({u_3}) > a.$

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