近简并二能级系统的几何位相

董坤坤; 李铭

doi:10.6054/j.jscnun.2018109

近简并二能级系统的几何位相

董坤坤,
李铭^,

华南师范大学物理与电信工程学院//广东省量子调控与材料重点实验室,广州510006

基金项目:

无

详细信息

通讯作者:
李铭

中图分类号: O41
计量
- 文章访问数: 1535
- HTML全文浏览量: 161
- PDF下载量: 64
出版历程
- 收稿日期: 2018-01-20
- 修回日期: 2018-05-11
- 刊出日期: 2018-12-24

The geometric phase of a near-degenerate two-level system

DONG K K,
LI M^,

Guangdong Provincial Key Laboratory of Quantum Engineering and Quantum Materials//School of Physics and Telecommunication Engineering,South China Normal University,Guangzhou 510006,China

摘要

摘要: 在有能量简并或接近简并的情况下，一个绝热演化系统的现有的量子绝热条件并不适用。本文重新推导了二能级系统的绝热演化，得到了包含近简并情况的几何相位的普遍结果。然后以一个实际的二能级系统石墨烯为例，本文通过数值计算得到了石墨烯在不同波矢位置的几何相位。数值结果表明，在能级近简并（包括简并）处，系统经过周期性绝热演化后，波函数只存在一个常规的动力学相位，不存在几何相位。离开简并位置，系统逐渐积累几何位相，最后才收敛到传统的Berry位相。
- Berry相位，几何相位，近简并，绝热定理
Abstract: For a degenerate or near degenerate system its adiabatic evolution does not satisfy the conventional adiabatic condition. In the present work the adiabatic evolution of a near degenerate two-level system has been studied and a common result for the geometric phase has been obtained. The geometric phase of the two level graphene is taken as an example to be numerically computed. The numerical results show that a periodic adiabatic evolution around the degenerate point (the Dirac points of wave vectors of graphene) does not give any geometric phase but only a conventional dynamic phase. Away from the Dirac point, however, a geometric phase accumulates in the system after a full cycle of adiabatic evolution around a Dirac point of graphene, and finally converges to the conventional Berry phase.
- Berry phase, Geometric phase, near-degenerate, adiabatic theorem

HTML全文

三阶常微分方程边值问题由于在工程、物理和流体力学等领域的显著应用而受到广泛关注. 学者们运用单调迭代法、上下解方法、Guo-Krasnosel'skii不动点定理和Leray-Schauder非线性抉择等, 研究了三阶三点边值问题在格林函数非负的情况下的单个或多个正解的存在性^[1-7]. 近年来, 学者们在格林函数变号的情况下也得到了很多结果^[8-17]. 如: LI等^[9]使用Guo-Krasnosel'skii不动点定理讨论了变号格林函数的三阶三点边值问题

$\left\{\begin{array} {l} { u ^ { \prime \prime \prime } ( t ) = f ( t , u ( t ) ) \quad ( t \in [ 0 , 1 ] ) , } \\ { u ( 1 ) = u ^ { \prime } ( 0 ) = 0 , u ^ { \prime \prime } ( \eta ) + \alpha u ( 0 ) = 0 } \end{array}\right.$

正解的存在性, 其中, α ∈[0, 2), η ∈ $[\frac{{\sqrt {121 + 24\alpha } - 5}}{{3\left( {4 + \alpha } \right)}}, 1)$ .

GAO和SUN^[10]运用Avery-Henderson不动点定理, 在格林函数变号时讨论了问题

$\left\{\begin{array}{l} u^{\prime \prime \prime}(t)=f(t, u(t)) \quad(t \in[0,1]), \\ u(1)=u^{\prime}(0)=0, u^{\prime \prime}(\eta)-\alpha u^{\prime}(1)=0 \end{array}\right.$

正解的存在性, 其中, α ∈[0, 2), η ∈ $[\frac{{4 + \alpha }}{{24 - 3\alpha }}, 1)$ .

ZHAO和LI^[11]运用迭代法讨论了变号格林函数的三阶三点边值问题

$\left\{\begin{array}{l} u^{\prime \prime \prime}(t)=f(t, u(t)) \quad(t \in[0,1]), \\ u(1)=u^{\prime}(0)=0, u^{\prime \prime}(\eta)+\alpha u(0)=0 \end{array}\right.$

正解的存在性, 其中, α∈[0, 2), η∈[2/3, 1).

受文献[11]的启发, 本文将运用迭代法, 研究如下具有变号格林函数的三阶三点边值问题

$\left\{\begin{array}{l} u^{\prime \prime \prime}(t)=f(t, u(t)) \quad(t \in[0,1]), \\ u(1)=0, u^{\prime}(0)=u^{\prime \prime}(0), \alpha u^{\prime \prime}(\eta)+\beta u(0)=0 \end{array}\right.$

(1)

正解的存在性, 其中, α ∈[0, 1], $\frac{2}{7}$ α < β < $\frac{2}{3}$ α, η ∈[2/3, 1). 在本文中, 总是假设f∈C([0, 1]×[0, ∞), [0, ∞)), 且f满足下列2个条件:

(H₁) 对于每一个u∈[0, +∞), 映射t↦f(t, u)是递减的；

(H₂) 对于每一个t∈[0, 1], 映射u↦f(t, u)是递增的.

1. 预备知识

本文所用到的空间是C[0, 1], 记E=C[0, 1], ‖u‖= $\mathop {\max }\limits_{t \in \left[ {0, 1} \right]} \left| {u\left( t \right)} \right|$ .

引理1 对任意的y ∈C[0, 1], 边值问题

$\left\{\begin{array}{l} u^{\prime \prime \prime}(t)=y(s) \quad(t \in[0,1]), \\ u(1)=0, u^{\prime}(0)=u^{\prime \prime}(0), \alpha u^{\prime \prime}(\eta)+\beta u(0)=0 \end{array}\right.$

(2)

有唯一解

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

其中:

$\begin{array}{c} G(t, s)=g_{1}(t, s)+g_{2}(t, s)+g_{3}(\eta, t, s) ;\\ g_{1}(t, s)=\frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta}(1-s)^{2}; \\ g_{2}(t, s)=\left\{\begin{array}{l} 0 \quad(0 \leqslant t \leqslant s \leqslant 1), \\ (t-s)^{2} / 2 \quad(0 \leqslant s \leqslant t \leqslant 1) ; \end{array}\right. \\ g_{3}(\eta, t, s)=\left\{\begin{array}{l} 0 \quad(0 \leqslant \eta \leqslant s \leqslant 1), \\ \frac{3 \alpha-2 \alpha\left(t+t^{2} / 2\right)}{2 \alpha-3 \beta} \quad(0 \leqslant s \leqslant \eta \leqslant 1); \end{array}\right. \end{array}$

证明对u'''(t)=y(t)在[0, t]上两边积分, 得

$\begin{array}{c} u^{\prime \prime}(t)=u^{\prime \prime}(0)-\int_{0}^{t} y(s) \mathrm{d} s ,\\ u^{\prime}(t)=u^{\prime}(0)+u^{\prime \prime}(0) t-\int_{0}^{t}(t-s) y(s) \mathrm{d} s ,\\ u(t)=u(0)+u^{\prime}(0) t+\frac{1}{2} u^{\prime \prime}(t) t^{2}+\frac{1}{2} \int_{0}^{t}(t-s)^{2} y(s) \mathrm{d} s, \end{array}$

由边界条件可得

$\begin{array}{c} u(0)+\frac{3}{2} u^{\prime}(0)+\frac{1}{2} \int_{0}^{1}(1-s)^{2} y(s) \mathrm{d} s=0,\\ \alpha u^{\prime}(0)+\alpha \int_{0}^{\eta} y(s) \mathrm{d} s+\beta u(0)=0. \end{array}$

从而有

$\begin{array}{l} u(0)=\frac{3 \alpha}{2 \alpha-3 \beta} \int_{0}^{\eta} y(s) \mathrm{d} s-\frac{\alpha}{2 \alpha-3 \beta} \int_{0}^{1}(1-s)^{2} y(s) \mathrm{d} s ,\\ u^{\prime}(0)=\frac{\beta}{2 \alpha-3 \beta} \int_{0}^{1}(1-s)^{2} y(s) \mathrm{d} s-\frac{2 \alpha}{2 \alpha-3 \beta} \int_{0}^{\eta} y(s) \mathrm{d} s. \end{array}$

故边值问题(2)的解为

$\begin{aligned} u(t)=& \frac{3 \alpha-2 \alpha\left(t+t^{2} / 2\right)}{2 \alpha-3 \beta} \int_{0}^{\eta} y(s) \mathrm{d} s+\\ & \frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta} \int_{0}^{1}(1-s)^{2} y(s) \mathrm{d} s+\\ & \frac{1}{2} \int_{0}^{t}(t-s)^{2} y(s) \mathrm{d} s. \end{aligned}$

证毕.

引理2 G(t, s)具有如下性质:

(1) 当0≤s≤η时, G(t, s)≥0;

(2) 当0≤η≤s时, G(t, s)≤0.

证明当0≤s≤η时, 有

$\begin{array}{l} G(t, s)= \\ \left\{\begin{array}{l} \frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta}(1-s)^{2}+\frac{3 \alpha-2 \alpha\left(t+t^{2} / 2\right)}{2 \alpha-3 \beta} \quad(0 \leqslant t \leqslant s \leqslant 1), \\ \frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta}(1-s)^{2}+\frac{3 \alpha-2 \alpha\left(t+t^{2} / 2\right)}{2 \alpha-3 \beta}+\frac{1}{2}(t-s)^{2} \\ \quad(0 \leqslant s \leqslant t \leqslant 1) . \end{array}\right. \end{array}$

分以下2种情况讨论:

(1) 当0≤t≤s≤1时, 有

$\begin{array}{c} G_{t}(t, s)=\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}= \\ \frac{(1+t)\left[\beta(1-s)^{2}-2 \alpha\right]}{2 \alpha-3 \beta}<0; \end{array}$

(2) 当0≤s≤t≤1时, 有

$\begin{array}{l} G_{t}(t, s)=\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}+(t-s)= \\ \ \ \ \ \frac{(\beta-\alpha)+\left(\beta s^{2}-\alpha\right)+\beta t\left(s^{2}-2\right)+s(\beta-2 \alpha)-2 \beta t s}{2 \alpha-3 \beta}<0. \end{array}$

综上可知, G(t, s)关于变量t单调递减. 从而有

$\begin{array}{c} \min \{G(t, s), t \in[0,1]\}=G(1, s)=0, \\ \max \{G(t, s), t \in[0,1]\}=G(0, s)=\frac{3 \alpha-\alpha(1-s)^{2}}{2 \alpha-3 \beta}>0 . \end{array}$

故当0≤s≤η时, G(t, s)≥0.

同理可证得: 当0≤η≤s时, G(t, s)关于变量t单调递增. 则有

$\begin{array}{c} \max \{G(t, s), t \in[0,1]\}=G(1, s)=0,\\ \min \{G(t, s), t \in[0,1]\}=G(0, s)=-\frac{\alpha(1-s)}{2 \alpha-3 \beta}<0 . \end{array}$

从而G(t, s)≤0. 证毕.

设M=max{|G(t, s)| |t, s ∈[0, 1]}, 则

$M=\max \left\{\frac{3 \alpha-\alpha(1-s)^{2}}{2 \alpha-3 \beta}, \frac{\alpha(1-s)}{2 \alpha-3 \beta}\right\}<\frac{3 \alpha}{2 \alpha-3 \beta}.$

设K={y∈E|y(t)在[0, 1]上非负且递减}, 则K是E中的一个锥, 且在E中定义一个序关系“≤”, u≤ν当且仅当ν-u∈K.

定义算子T: K→E

$(T u)(t)=\int_{0}^{1} G(t, s) f(s, u(s)) \mathrm{d} s \quad(u \in K, t \in[0,1]).$

显然, 若u是T中的不动点, 则u是边值问题(1)的递减非负解.

引理3 算子T: K→K是全连续的.

证明设u∈K, 当t∈[0, η]时, 有

$\begin{array}{l} (T u)(t)=\int_{0}^{t}\left[\frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta}(1-s)^{2}+\right. \\ \ \ \ \ \ \ \ \ \left.\frac{3 \alpha-2 \alpha\left(t+t^{2} / 2\right)}{2 \alpha-3 \beta}+\frac{1}{2}(t-s)^{2}\right] f(s, u(s)) \mathrm{d} s+ \\ \ \ \ \ \ \ \ \ \int_{t}^{\eta}\left[\frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta}(1-s)^{2}+\frac{3 \alpha-2 \alpha\left(t+t^{2} / 2\right)}{2 \alpha-3 \beta}\right] \times \\ \ \ \ \ \ \ \ \ f(s, u(s)) \mathrm{d} s+\int_{\eta}^{1}\left[\frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta}(1-s)^{2}\right] \times \\ \ \ \ \ \ \ \ \ f(s, u(s)) \mathrm{d} s . \end{array}$

由条件(H₁)、(H₂), 可得

$\begin{array}{l} (T u)^{\prime}(t)= \\ \ \ \ \ \int_{0}^{t}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}+(t-s)\right] f(s, u(s)) \mathrm{d} s+ \\ \ \ \ \ \int_{t}^{\eta}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}\right] f(s, u(s)) \mathrm{d} s+ \\ \ \ \ \ \int_{\eta}^{1}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}\right] f(s, u(s)) \mathrm{d} s \leqslant \\ \ \ \ \ f(\eta, u(\eta))\left\{\int_{0}^{t}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}+(t-s)\right] \mathrm{d} s+\right. \\ \ \ \ \ \int_{t}^{\eta}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}\right] \mathrm{d} s+\\ \ \ \ \ \left.\int_{\eta}^{1}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}\right] \mathrm{d} s\right\} \leqslant f(\eta, u(\eta))\left[\frac{\beta(1+t)}{2 \alpha-3 \beta} \times\right. \\ \ \ \ \ \int_{0}^{\eta}\left(-2 s+s^{2}\right) \mathrm{d} s+\frac{(\beta-2 \alpha)(1+t)}{2 \alpha-3 \beta} \int_{0}^{\eta} \mathrm{d} s+\int_{0}^{t} t \mathrm{~d} s- \\ \ \ \ \ \left.\int_{0}^{t} s \mathrm{~d} s+\frac{\beta(1+t)}{2 \alpha-3 \beta} \int_{\eta}^{1}(1-s)^{2} \mathrm{~d} s\right] \leqslant\\ \ \ \ \ f(\eta, u(\eta))\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}\left(\frac{1}{3}-\eta\right)+\frac{1}{2} t^{2}+\right. \\ \ \ \ \ \left.\frac{(\beta-2 \alpha)(1+t)}{2 \alpha-3 \beta} \eta\right] \leqslant f(\eta, u(\eta))\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}\left(\frac{1}{3}-\eta\right)+\right. \\ \ \ \ \ \left.\frac{2(\beta-2 \alpha)(1+t) \eta+(2 \alpha-3 \beta)}{2(2 \alpha-3 \beta)}\right] \leqslant f(\eta, u(\eta)) \times \\ \ \ \ \ {\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}\left(\frac{1}{3}-\eta\right)+\frac{2 \eta(\beta-2 \alpha)+(2 \alpha-3 \beta)}{2 \alpha-3 \beta}\right] \leqslant 0}. \end{array}$

当t ∈[η, 1]时, 有

$\begin{array}{l} (T u)(t)=\int_{0}^{\eta}\left[\frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta}(1-s)^{2}+\frac{3 \alpha-2 \alpha\left(t+t^{2} / 2\right)}{2 \alpha-3 \beta}+\right. \\ \ \ \ \ \left.\frac{1}{2}(t-s)^{2}\right] f(s, u(s)) \mathrm{d} s+\int_{\eta}^{t}\left[\frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta}(1-s)^{2}+\right. \\ \ \ \ \ \left.\frac{1}{2}(t-s)^{2}\right] f(s, u(s)) \mathrm{d} s+\int_{t}^{1}\left[\frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta}(1-s)^{2}\right] \times \\ \ \ \ \ f(s, u(s)) \mathrm{d} s, \end{array}$

则

$\begin{array}{l} (T u)^{\prime}(t)= \\ \ \ \ \ \int_{0}^{\eta}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}+(t-s)\right] f(s, u(s)) \mathrm{d} s+ \\ \ \ \ \ \int_{\eta}^{t}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-(t-s)\right] f(s, u(s)) \mathrm{d} s+ \\ \ \ \ \ \int_{t}^{1}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}\right] f(s, u(s)) \mathrm{d} s \leqslant \\ \ \ \ \ f(\eta, u(\eta))\left\{\int_{0}^{\eta}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha}{2 \alpha-3 \beta}+(t-s)\right] \mathrm{d} s+\right. \\ \ \ \ \ \left.\int_{\eta}^{t}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}+(t-s)\right] \mathrm{d} s+\int_{t}^{1}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}\right] \mathrm{d} s\right\} \leqslant \\ \ \ \ \ f(\eta, u(\eta))\left\{\frac{\beta(1+t)}{2 \alpha-3 \beta} \int_{0}^{\eta}\left(s^{2}-2 s\right) \mathrm{d} s-\int_{0}^{\eta} s \mathrm{~d} s+\right.\\ \ \ \ \ \frac{\beta-2 \beta t-2 \alpha}{2 \alpha-3 \beta} \int_{0}^{\eta} \mathrm{d} s+\frac{\beta(1+t)}{2 \alpha-3 \beta} \int_{\eta}^{1}(1-s)^{2} \mathrm{~d} s+ \\ \ \ \ \ \left.\int_{\eta}^{t}(t-s) \mathrm{d} s\right\}=f(\eta, u(\eta))\left\{\frac{\beta(1+t)}{2 \alpha-3 \beta}\left(\frac{1}{3}-\eta\right)+\right. \\ \ \ \ \ \left.\frac{1}{2} t^{2}+\frac{\beta+\beta t-2 \alpha}{2 \alpha-3 \beta} \eta-t \eta\right\} \leqslant f(\eta, u(\eta))\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}\left(\frac{1}{3}-\eta\right)+\right. \\ \ \ \ \ \left.\frac{1}{2}+\frac{(\beta-2 \alpha)(1+t)}{2 \alpha-3 \beta} \eta\right] \leqslant f(\eta, u(\eta))\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}\left(\frac{1}{3}-\eta\right)+\right. \\ \ \ \ \ \left.\frac{1}{2}-\frac{(2 \alpha-\beta) \eta}{2 \alpha-3 \beta}\right] \leqslant 0. \end{array}$

综上可知, (Tu)(t)在[0, 1]上单调递减. 又由于(Tu)(1)=0, 故(Tu)(t)在[0, 1]上非负, 从而Tu∈K. 假设D⊂K是有界集, 则存在一个常数C₁>0, 使得‖ u ‖≤C₁ (u∈D). 下证T(D)是相对紧的.

设C₂=sup{f(t, u)|(t, u)∈[0, 1]×[0, C₁]}. 对∀y∈T(D), ∃u∈D, 使得y=Tu, 则对∀t ∈[0, 1], 有

$\begin{array}{l} |y(t)|=|(T u)(t)|=\left|\int_{0}^{1} G(t, s) f(s, u(s)) \mathrm{d} s\right| \leqslant \\ \ \ \ \ \ \ \int_{0}^{1}|G(t, s)| f(s, u(s)) \mathrm{d} s \leqslant M \int_{0}^{1} f(s, u(s)) \mathrm{d} s \leqslant \\ \ \ \ \ \ \ M C_{2}. \end{array}$

从而知T(D)是一致有界的. 另一方面, 当ε>0, 0 < τ < min{1-η, $\frac{\varepsilon }{{12{C_2}\left( {M + 1} \right)}}$ }时, 对∀u ∈D, 有

$\int_{\eta-\tau}^{\eta+\tau} f(s, u(s)) \mathrm{d} s \leqslant 2 C_{2} \tau<\frac{\varepsilon}{6(M+1)}.$

(3)

因为G(t, s)在[0, 1]×[0, η-τ]和[0, 1]×[η+ τ, 1]上一致连续, 故∃δ>0, 使得对∀t₁, t₂ ∈[0, 1], 当|t₁-t₂| < δ时, 有

$\left|G\left(t_{1}, s\right)-G\left(t_{2}, s\right)\right|<\frac{\varepsilon}{3\left(C_{2}+1\right)(\eta-\tau)}\ \ \ \ (s \in[0, \eta-\tau]),$

(4)

$\left|G\left(t_{1}, s\right)-G\left(t_{2}, s\right)\right|<\frac{\varepsilon}{3\left(C_{2}+1\right)\left(1-\eta^{-} \tau\right)}\ \ \ \ (s \in[\eta+\tau, 1]) .$

(5)

由式(3)~(5)及对∀y∈T(D), ∀t₁, t₂∈[0, 1]和|t₁-t₂| < δ, 有

$\begin{array}{l} \left|y\left(t_{1}\right)-y\left(t_{2}\right)\right|=\left|T\left(t_{1}\right)-T\left(t_{1}\right)\right|= \\ \ \ \ \ \ \ \ \ \left|\int_{0}^{1}\left(G\left(t_{1}, s\right)-G\left(t_{2}, s\right)\right) f(s, u(s)) \mathrm{d} s\right|= \\ \ \ \ \ \ \ \ \ \int_{0}^{\eta-\tau}\left|G\left(t_{1}, s\right)-G\left(t_{2}, s\right)\right| f(s, u(s)) \mathrm{d} s+ \\ \ \ \ \ \ \ \ \ \int_{\eta-\tau}^{\eta+\tau}\left|G\left(t_{1}, s\right)-G\left(t_{2}, s\right)\right| f(s, u(s)) \mathrm{d} s+ \\ \ \ \ \ \ \ \ \ \int_{\eta+\tau}^{1}\left|G\left(t_{1}, s\right)-G\left(t_{2}, s\right)\right| f(s, u(s)) \mathrm{d} s \leqslant \\ \ \ \ \ \ \ \ \ C_{2} \frac{\varepsilon}{3\left(C_{2}+1\right)(\eta-\tau)}(\eta-\tau)+\frac{\varepsilon}{3(M+1)} M+ \\ \ \ \ \ \ \ \ \ C_{2} \frac{\varepsilon}{3\left(C_{2}+1\right)(1-\eta-\tau)}(1-\eta-\tau)= \\ \ \ \ \ \ \ \ \ \frac{C_{2} \varepsilon}{3\left(C_{2}+1\right)}+\frac{M \varepsilon}{3(M+1)}+\frac{C_{2} \varepsilon}{3\left(C_{2}+1\right)} \leqslant \varepsilon. \end{array}$

故T(D)是等度连续的, 从而由Arzela-Ascoli定理^[18]知T(D)是相对紧的. 因此, T是紧算子.

最后, 证明T是连续的. 设u_n (n=1, 2, …), u₀ ∈K, ‖u_n-u₀ ‖→0 (n→0), 则∃C₃>0, 使得对∀n, ‖ u ‖ ≤C₃.

设C₄=sup{f(t, u)|(t, u)∈[0, 1]×[0, C₃]}. 对∀n, t∈[0, 1], ∀s∈[0, 1], 有

$G(t, s) f\left(s, u_{n}(s)\right) \leqslant M C_{4}.$

由Lebesgue控制收敛定理^[19], 对∀t∈[0, 1], 有

$\begin{array}{l} \lim \limits_{n \rightarrow \infty}\left(T u_{n}\right)(t)=\lim \limits_{n \rightarrow \infty} \int_{0}^{1} G(t, s) f\left(s, u_{n}(s)\right) \mathrm{d} s= \\ \ \ \ \ \int_{0}^{1} G(t, s) \lim \limits_{n \rightarrow \infty} f\left(s, u_{n}(s)\right) \mathrm{d} s=\int_{0}^{1} G(t, s) f\left(s, u_{0}(s)\right) \mathrm{d} s= \\ \ \ \ \ \left(T u_{0}\right)(t), \end{array}$

这表明T是连续的. 因此, T: K→K是全连续的. 证毕.

2. 主要结果及证明

定理1 假设条件(H₁)、(H₂)成立, 且对于f(t, 0)≠0, ∀t∈[0, 1], 存在2个正实数a和b, 满足以下条件：

$\begin{array}{l} \ \ \ \ \left(\mathrm{H}_{3}\right) f(0, a) \leqslant \frac{2 \alpha-3 \beta}{3 \alpha} a ;\\ \ \ \ \ \left(\mathrm{H}_{4}\right) b\left(u_{2}-u_{1}\right) \leqslant f\left(t, u_{2}\right)-f\left(t, u_{1}\right) \leqslant 2 b\left(u_{2}-u_{1}\right) \\ \left(0 \leqslant t \leqslant 1,0 \leqslant u_{1} \leqslant u_{2} \leqslant a\right). \end{array}$

构造一个迭代序列ν_n+1=Tν_n (n=0, 1, 2, …), 其中ν₀(t)=0, 则{ν_n}_n=1^∞在E中收敛到ν^*, 且ν^*是边值问题(1)的一个递减正解.

证明设K_a= {u ∈K| | |u| | ≤a}. 由引理3可知Tu∈K, 且由条件(H₃)得到

$\begin{aligned} 0 \leqslant&(T u)(t)=\int_{0}^{1} G(t, s) f(s, u(s)) \mathrm{d} s \leqslant & \\ & \int_{0}^{1}|G(t, s)| f(0, a) \mathrm{d} s \leqslant \frac{2 \alpha-3 \beta}{3 \alpha} a M \leqslant a \quad(t \in[0,1]), \end{aligned}$

从而有‖ Tu ‖ ≤a, 故T: K_a→K_a.

下证{ν_n}_n=1^∞在E中收敛到ν^*, 且ν^*是边值问题(1)的一个递减正解.

事实上, 对于ν₀ ∈K_a和T: K_a→K_a, 有ν_n ∈K_a, n=0, 1, 2, …. 由于集合{ν_n}_n=0^∞是有界的且T是全连续算子, 所以集合{ν_n}_n=0^∞是相对紧的. 接下来, 通过归纳法证明{ν_n}_n=0^∞是单调的. 首先, 很明显有ν₁-ν₀=ν₁ ∈K, 这表明ν₀≤ν₁. 接下来, 假设ν_k-1≤ν_k. 从而由条件(H₄)可得

$\begin{array}{l} \nu_{k+1}^{\prime}-\nu_{k}^{\prime}=\left(T \nu_{k}\right)^{\prime}(t)-\left(T \nu_{k-1}\right)^{\prime}(t)= \\ \ \ \ \ \int_{0}^{1} G_{t}(t, s)\left[f\left(s, \nu_{k}(s)\right)-f\left(s, \nu_{k-1}(s)\right)\right] \mathrm{d} s= \\ \ \ \ \ \int_{0}^{t}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}+(t-s)\right]\left[f\left(s, \nu_{k}(s)\right)-\right.\\ \ \ \ \ \left.f\left(s, \nu_{k-1}(s)\right)\right] \mathrm{d} s+\int_{t}^{\eta}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}\right] \times \\ \ \ \ \ {\left[f\left(s, \nu_{k}(s)\right)-f\left(s, \nu_{k-1}(s)\right)\right] \mathrm{d} s+\int_{\eta}^{1}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}\right] \times} \\ \ \ \ \ {\left[f\left(s, \nu_{k}(s)\right)-f\left(s, \nu_{k-1}(s)\right)\right] \mathrm{d} s \leqslant b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right] \times} \\ \ \ \ \ \left\{\int_{0}^{t}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}+(t-s)\right] \mathrm{d} s+\right. \\ \ \ \ \ \int_{t}^{\eta}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}\right] \mathrm{d} s+\\ \ \ \ \ \left.\int_{\eta}^{1}\left[\frac{2 \beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}\right] \mathrm{d} s\right\}=b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right] \times \\ \ \ \ \ \left\{\int_{0}^{\eta} \frac{\beta(1+t)}{2 \alpha-3 \beta}\left(s^{2}-2 s\right) \mathrm{d} s+\int_{0}^{\eta} \frac{(\beta-2 \alpha)(1+t)}{2 \alpha-3 \beta} \mathrm{d} s+\right. \\ \ \ \ \ \left.\int_{0}^{t}(t-s) \mathrm{d} s+\frac{2 \beta(1+t)}{2 \alpha-3 \beta} \int_{\eta}^{1}(1-s)^{2} \mathrm{~d} s\right\}= \\ \ \ \ \ b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right]\left\{\frac{\beta(1+t)}{2 \alpha-3 \beta}\left(-\eta^{3}+3 \eta^{2}-6 \eta+2\right)+\right. \\ \ \ \ \ \left.\frac{(\beta-2 \alpha)(1+t)}{2 \alpha-3 \beta} \eta+\frac{1}{2} t^{2}\right\} \leqslant b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right] \times\\ \ \ \ \ \left\{\frac{\beta(1+t)}{2 \alpha-3 \beta}\left(-\eta^{3}+3 \eta^{2}-6 \eta+2\right)+\frac{(\beta-2 \alpha)(1+t)}{2 \alpha-3 \beta} \eta+\frac{1}{2}\right\} \leqslant \\ \ \ \ \ b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right]\left\{\frac{\beta(1+t)}{2 \alpha-3 \beta}(2-3 \eta)+\frac{(\beta-2 \alpha)(1+t)}{2 \alpha-3 \beta} \eta {+}\right. \\ \ \ \ \ \left.\frac{1}{2}\right\} \leqslant b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right]\left\{\frac{\beta(1+t)(2-3 \eta)}{2 \alpha-3 \beta}+\right. \\ \ \ \ \ \left.\frac{(2 \alpha-3 \beta)-2 \eta(2 \alpha-\beta)}{2 \alpha-3 \beta}\right\} \leqslant 0 \quad(t \in[0, \eta]);\\ \nu_{k+1}^{\prime}-\nu_{k}^{\prime}=\left(T \nu_{k}\right)^{\prime}(t)-\left(T \nu_{k-1}\right)^{\prime}(t)= \\ \ \ \ \ \int_{0}^{1} G_{t}(t, s)\left[f\left(s, \nu_{k}(s)\right)-f\left(s, \nu_{k-1}(s)\right)\right] \mathrm{d} s= \\ \ \ \ \ \int_{0}^{\eta}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}+(t-s)\right]\left[f\left(s, \nu_{k}(s)\right)-\right. \\ \ \ \ \ \left.f\left(s, \nu_{k-1}(s)\right)\right] \mathrm{d} s+\int_{\eta}^{t}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-(t-s)\right] \times \\ \ \ \ \ {\left[f\left(s, \nu_{k}(s)\right)-f\left(s, \nu_{k-1}(s)\right)\right] \mathrm{d} s+}\\ \ \ \ \ \int_{t}^{1}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}\right]\left[f\left(s, \nu_{k}(s)\right)-f\left(s, \nu_{k-1}(s)\right)\right] \mathrm{d} s \leqslant \\ \ \ \ \ b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right]\left\{\frac{\beta(1+t)}{2 \alpha-3 \beta} \int_{0}^{\eta}\left(-2 s+s^{2}\right) \mathrm{d} s+\right. \\ \ \ \ \ \frac{(\beta-2 \alpha)(1+t)}{2 \alpha-3 \beta} \int_{0}^{\eta} \mathrm{d} s+\int_{0}^{\eta}(t-s) \mathrm{d} s+\frac{2 \beta(1+t)}{2 \alpha-3 \beta} \int_{\eta}^{1}(1-s)^{2} \mathrm{~d} s+ \\ \ \ \ \ \left.2 \int_{\eta}^{t}(t-s) \mathrm{d} s\right\}=b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right]\left[\frac{\beta(1+t)}{3(2 \alpha-3 \beta)} \times\right. \\ \ \ \ \ \left.\left(-\eta^{3}+3 \eta^{2}-6 \eta+2\right)+t^{2}+\frac{(\beta-2 \alpha)(1+t)}{2 \alpha-3 \beta} \eta-t \eta+\frac{1}{2} \eta^{2}\right] \leqslant\\ \ \ \ \ b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right]\left[\frac{\beta(1+t)}{3(2 \alpha-3 \beta)}(-5 \eta+2)+\right. \\ \ \ \ \ \left.\frac{t[(2 \alpha-3 \beta)-\eta(2 \alpha-\beta)]-\eta(2 \alpha-\beta)}{2 \alpha-3 \beta}\right] \leqslant \\ \ \ \ \ b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right]\left[\frac{\beta(1+t)}{3(2 \alpha-3 \beta)}(-5 \eta+2)-\right. \\ \ \ \ \ \left.\frac{\eta(2 \alpha-\beta)}{2 \alpha-3 \beta}\right] \leqslant 0 . \end{array}$

因此,

$\nu_{k+1}^{\prime}(t)-\nu_{k}^{\prime}(t) \leqslant 0 \quad(t \in[0,1]),$

(6)

即ν_k+1(t)-ν_k(t)在[0, 1]上单调递减. 与此同时, 易得

$\nu_{k+1}(1)-\nu_{k}(1)=\int_{0}^{1} G(1, s) f\left(s, \nu_{k+1}(s)-\nu_{k}(s)\right) \mathrm{d} s=0,$

(7)

从而ν_k+1(t)-ν_k(t)≥0 (t∈[0, 1]).

由式(6)和式(7)可知ν_k+1-ν_k ∈K, 则ν_k+1≤ν_k (n=0, 1, 2, …). 由于{ν_n}_n=1^∞是相对紧集, 并且是单调的, 因此, 必存在一个ν^*∈K_a, 使得 $\mathop {\lim }\limits_{n \to \infty }$ ν_n=ν^*. 再由T的连续性以及ν_n+1=Tν_n可知ν^*=Tν^*. 这表明ν^*是边值问题(1)单调递减的非负解. 此外, 根据f(t, 0)≠0 (t ∈[0, 1])知, 0不是边值问题(1)的解, 从而ν^*是边值问题(1)的正解. 证毕.

参考文献(1)

施引文献(19)

期刊类型引用(11)

1.	严群，靳涵宇，宋忠贤，康海彦，延旭，毛艳丽. 活性污泥生物炭的制备及吸附性能的研究进展. 有色金属科学与工程. 2024(04): 615-622 . 百度学术
2.	张立国，潘静诗，黄嘉莉，李碧清，唐霞，吴学伟，梁浩斌，杜馨. 废弃物基生物质炭从水体中吸附磷的研究进展. 华南师范大学学报(自然科学版). 2023(06): 36-45 . 百度学术
3.	赵伟繁，戴亮，王刚，未碧贵，韩涛. 污泥生物炭重金属吸附剂的制备及改性研究进展. 功能材料. 2020(11): 11083-11088+11115 . 百度学术
4.	李芬，王鹤，张彦平，于彩莲，王奇飞. 草木灰浸取液活化制备污泥基脱H_2S碳材料研究. 哈尔滨理工大学学报. 2019(05): 138-144 . 百度学术
5.	刘蕾，付临汝，杜馨，张立国，吴宏海，肖羽堂. 联合改性污泥吸附剂去除废水中铬(Ⅵ). 环境化学. 2018(12): 2596-2602 . 百度学术
6.	周树烽，陈成广，刘允初，胡艺，胡保卫. KOH改性污泥生物碳及对富营养化水体吸附研究. 环境科学与技术. 2017(08): 43-49 . 百度学术
7.	龙良俊，王里奥，余纯丽，魏星跃，卓琳. 改性污泥腐殖酸的表征及其对Cu~(2+)的吸附特性. 中国环境科学. 2017(03): 1016-1023 . 百度学术
8.	杨乐，李政家，邓辉，陈尚学，姬江浩. KOH活化-微波热解制备污泥炭及其结构演化. 环境工程. 2016(08): 120-124+129 . 百度学术
9.	刘慧君，尹文霞，张立国，付临汝，刘蕾，吴宏海，肖羽堂，吴培煜，石思琦. 碳氮共掺杂污泥基吸附剂去除水中Cr(Ⅵ). 环境工程学报. 2016(08): 4147-4152 . 百度学术
10.	邓辉，李政家，金志文，张涛. 棉秆与污泥共热解制备生物炭工艺优化及其结构与吸附性能. 农业工程学报. 2016(24): 248-254 . 百度学术
11.	邓辉，杨乐，吴湧道，李钰琦，李政家. 磷酸活化-微波热解制备污泥吸附剂及其结构演化. 石河子大学学报(自然科学版). 2016(01): 85-91 . 百度学术