ABSTRACT

In this paper, the author establishes a class of triple nonlinear integral inequalities with unknown derivative functions, and gives the estimation of unknown functions in inequalities by using the inequality techniques such as variable substitution, amplification, differential and integral.

Keywords:Nonlinear Integral Inequality, Triple Integral with Unknown Derivative Function, Estimation

一类三重非线性积分不等式解的估计

卢钰松，黄星寿^*

河池学院数学与统计学院，广西 宜州

收稿日期：2020年7月25日；录用日期：2020年8月10日；发布日期：2020年8月17日

摘 要

本文研究了一类含有未知导函数的三重非线性积分不等式，利用变量替换、放大、微分、积分等不等式技巧给出了不等式中未知函数的估计，推广了相应的结果。

关键词 :非线性积分不等式，含有未知导函数的三重积分，估计

Copyright © 2020 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

在微分方程和积分方程解存在性、有界性和唯一性等定性性质的研究中，Gronwall [1] 型积分不等式是一种重要的工具，因此人们不断的对它进行研究，将它的形式进行推广，使其在微分方程和积分方程中的应用范围不断增大 [2] - [7]。

PACHPATTE [8] 于1998年研究了以下积分不等式：

$u^{\prime }\left(t\right)\le a\left(t\right)+b\left(t\right){\displaystyle {\int }_0^{t}c\left(s\right)\left(u\left(s\right)+u^{\prime }\left(s\right)\right)\text{d}s}\left(t\in R\right)$

$u^{\prime }\left(t\right)\le u\left(0\right)+{\displaystyle {\int }_0^{t}a\left(s\right)\left(u\left(s\right)+u^{\prime }\left(s\right)\right)\text{d}s}+{\displaystyle {\int }_0^{t}a\left(s\right)\left({\displaystyle {\int }_0^{s}b\left(\sigma \right)u^{\prime }\left(\sigma \right)\text{d}\sigma }\right)\text{d}s}\left(t\in R_{+}\right)$

该积分不等式的积分号内包含未知函数及其导函数。

ZAREEN [9] 于2014年，发表文章介绍了非线性积分不等式：

$u^{\prime }\left(t\right)\le c+{\displaystyle {\int }_0^{t}k\left(s\right)u^{\prime }\left(s\right)\left({u^{\prime }}^p\left(s\right)+u^2\left(s\right)\right)\text{d}s}\left(t\in R_{+}\right)$

的解的估计。

黄星寿，王五生等 [10] 于2019年，研究了一类积分号内包含未知函数及其导函数的非线性二重积分不等式：

$\begin{array}{l}{u}^{\prime }\left(t\right)\le w\left(t\right)+p\left(t\right)\{u\left(t\right)+{\displaystyle {\int }_{t_0}^t[a\left(\tau \right)\left[u\left(\tau \right)+u^{\prime }\left(\tau \right)\right]\underset{}{}}\\ +a\left(\tau \right){\displaystyle {\int }_{t_0}^{\tau }\left[c\left(s\right)u^{\prime }\left(s\right)\left(u^{\prime }\left(s\right)+u\left(s\right)\right)+d\left(s\right)\right]\text{d}s}]\text{d}\tau \}\left(t\in \left[t_0,\infty \right)\right)\end{array}$

受以上研究成果的启发，本文构造了以下积分号内包含未知函数及其导函数的非线性积分不等式：

$\begin{array}{l}{u}^{\prime }\left(t\right)\le w\left(t\right)+p\left(t\right)\{u\left(t\right)+{\displaystyle {\int }_{t_0}^t[a\left(\tau \right)\left(u\left(\tau \right)+u^{\prime }\left(\tau \right)\right)+a\left(\tau \right){\displaystyle {\int }_{t_0}^{\tau }[c\left(s\right)\left(u\left(s\right)+u^{\prime }\left(s\right)\right)}}\\ +c\left(s\right){\displaystyle {\int }_{t_0}^s\left[d\left(v\right)u^{\prime }\left(v\right)\left(u\left(v\right)+u^{\prime }\left(v\right)\right)+f\left(v\right)\right]\text{d}v]\text{d}s]\text{d}\tau }\}\left(t\in \left[t_0,\infty \right)\right)\end{array}$ (1)

此类积分不等式将文献 [10] 推广到了三重积分，文中利用放大、变量替换、微分–积分等不等式技巧给出了不等式(1)中未知函数的估计。

2. 预备知识

引理 [10] 假设函数 $a\left(t\right),b\left(t\right)$ 和 $c\left(t\right)$ 都是定义在 $\left[t_0,\infty \right)$ 上的非负连续已知函数，并且函数 $a\left(t\right)$ 是 $\left[t_0,\infty \right)$ 上的增函数，未知函数 $u\left(t\right)$ 满足不等式

$u\left(t\right)\le a\left(t\right)+{\displaystyle {\int }_{t_0}^{t}b\left(s\right)u\left(s\right)\text{d}s}+{\displaystyle {\int }_{t_0}^{t}c\left(s\right)u^2\left(s\right)\text{d}s}\left(t\in \left[t_0,\infty \right)\right)$ (2)

如果 $\exp \left(-\ln a\left(t\right)-{\displaystyle {\int }_{t_0}^{t}b\left(s\right)\text{d}s}\right)-{\displaystyle {\int }_{t_0}^{t}c\left(s\right)\text{d}s}>0$，则有未知函数 $u\left(t\right)$ 的估计式：

$u\left(t\right)\le {\left(\exp \left(-\ln a\left(t\right)-{\displaystyle {\int }_{t_0}^{t}b\left(s\right)\text{d}s}\right)-{\displaystyle {\int }_{t_0}^{t}c\left(s\right)\text{d}s}\right)}^{-1}\left(t\in \left[t_0,\infty \right)\right)$ (3)

3. 主要结果

定理 假设已知函数 $w\left(t\right),p\left(t\right),a\left(t\right),c\left(t\right),d\left(t\right),f\left(t\right)$ 都是定义在 $\left[t_0,\infty \right)$ 上的非负连续函数，未知函数 $u\left(t\right)$ 和 $u^{\prime }\left(t\right)$ 定义在 $\left[t_0,\infty \right)$ 上，且满足不等式(1)。

如果

$\exp \left(-\ln \left(u\left(t_0\right)+{\displaystyle {\int }_{t_0}^{t}A\left(s\right)\text{d}s}\right)-{\displaystyle {\int }_{t_0}^{t}B\left(s\right)\text{d}s}\right)-{\displaystyle {\int }_{t_0}^{t}C\left(s\right)\text{d}s}>0\left(t\in \left[t_0,\infty \right)\right)$ (4)

那么未知函数 $u\left(t\right)$ 的估计式为：

$\begin{array}{l}u\left(t\right)\le u\left(t_0\right)+{\displaystyle {\int }_{t_0}^t\{w\left(s\right)\left[1+a\left(s\right)+c\left(s\right)\right]+[p\left(s\right)+a\left(s\right)+a\left(s\right)p\left(s\right)}\\ +2c\left(s\right)+c\left(s\right)p\left(s\right)]Z\left(s\right)\}\text{d}s\left(t\in \left[t_0,\infty \right)\right)\end{array}$ (5)

其中

$Z\left(t\right):={\left(\exp \left(-\ln \left(u\left(t_0\right)+{\displaystyle {\int }_{t_0}^{t}A\left(s\right)\text{d}s}\right)-{\displaystyle {\int }_{t_0}^{t}B\left(s\right)\text{d}s}\right)-{\displaystyle {\int }_{t_0}^{t}C\left(s\right)\text{d}s}\right)}^{-1}$ (6)

$A\left(t\right):=w\left(t\right)\left[1+a\left(t\right)+c\left(t\right)\right]+d\left(t\right)w^2\left(t\right)+f\left(t\right)$ (7)

$B\left(t\right):=p\left(t\right)+a\left(t\right)+a\left(t\right)p\left(t\right)+2c\left(t\right)+c\left(t\right)p\left(t\right)+d\left(t\right)w\left(t\right)+2d\left(t\right)w\left(t\right)p\left(t\right)$ (8)

$C\left(t\right):=d\left(t\right)p\left(t\right)+d\left(t\right)p^2\left(t\right)$ (9)

证明：令函数

$\begin{array}{l}{z}_1\left(t\right)=u\left(t\right)+{\displaystyle {\int }_{t_0}^t[a\left(\tau \right)\left(u\left(\tau \right)+u^{\prime }\left(\tau \right)\right)+a\left(\tau \right){\displaystyle {\int }_{t_0}^{\tau }[c\left(s\right)\left(u\left(s\right)+u^{\prime }\left(s\right)\right)}}\\ +c\left(s\right){\displaystyle {\int }_{t_0}^s\left[d\left(v\right)u^{\prime }\left(v\right)\left(u\left(v\right)+u^{\prime }\left(v\right)\right)+f\left(v\right)\right]\text{d}v]\text{d}s}]\text{d}\tau \left(t\in \left[t_0,\infty \right)\right)\end{array}$ (10)

则有

$z_1\left(t_0\right)=u\left(t_0\right)$，$u\left(t\right)\le z_1\left(t\right)$，$u^{\prime }\left(t\right)\le w\left(t\right)+p\left(t\right)z_1\left(t\right)$ (11)

对(10)式两边求导，并将(1)式代入，则可得：

$\begin{array}{c}{z^{\prime }}_1\left(t\right)=u^{\prime }\left(t\right)+a\left(t\right)\left[u\left(t\right)+u^{\prime }\left(t\right)\right]+a\left(t\right){\displaystyle {\int }_{t_0}^t[c\left(s\right)\left(u\left(s\right)+u^{\prime }\left(s\right)\right)}\\ +c\left(s\right){\displaystyle {\int }_{t_0}^s\left[d\left(v\right)u^{\prime }\left(v\right)\left(u\left(v\right)+u^{\prime }\left(v\right)\right)+f\left(v\right)\right]\text{d}v]\text{d}s}\\ \le w\left(t\right)+p\left(t\right)z_1\left(t\right)+a\left(t\right)\left[z_1\left(t\right)+w\left(t\right)+p\left(t\right)z_1\left(t\right)\right]\\ +a\left(t\right){\displaystyle {\int }_{t_0}^t[c\left(s\right)\left[z_1\left(s\right)+w\left(s\right)+p\left(s\right)z_1\left(s\right)\right]}+c\left(s\right){\displaystyle {\int }_{t_0}^s[d\left(v\right)(w\left(v\right)}\\ +p\left(v\right)z_1\left(v\right))\left(z_1\left(v\right)+w\left(v\right)+p\left(v\right)z_1\left(v\right)\right)+f\left(v\right)]\text{d}v]\text{d}s\end{array}$

$\begin{array}{l}=w\left(t\right)\left[1+a\left(t\right)\right]+\left[p\left(t\right)+a\left(t\right)+a\left(t\right)p\left(t\right)\right]z_1\left(t\right)\\ +a\left(t\right){\displaystyle {\int }_{t_0}^t[c\left(s\right)w\left(s\right)+\left[c\left(s\right)\left(1+p\left(s\right)\right)z_1\left(s\right)\right]}\\ +c\left(s\right){\displaystyle {\int }_{t_0}^s[d\left(v\right)w^2\left(v\right)+\left[d\left(v\right)w\left(v\right)+2d\left(v\right)w\left(v\right)p\left(v\right)\right]z_1\left(v\right)}\\ +\left[d\left(v\right)p\left(v\right)+d\left(v\right)p^2\left(v\right)\right]z_1^2\left(v\right)+f\left(v\right)]\text{d}v]\text{d}s\left(t\in \left[t_0,\infty \right)\right)\end{array}$ (12)

又令

$\begin{array}{c}{z}_2\left(t\right)=z_1\left(t\right)+{\displaystyle {\int }_{t_0}^t[c\left(s\right)w\left(s\right)+\left[c\left(s\right)\left(1+p\left(s\right)\right)z_1\left(s\right)\right]}\\ +c\left(s\right){\displaystyle {\int }_{t_0}^s[d\left(v\right)w^2\left(v\right)+\left[d\left(v\right)w\left(v\right)+2d\left(v\right)w\left(v\right)p\left(v\right)\right]z_1\left(v\right)}\\ +\left[d\left(v\right)p\left(v\right)+d\left(v\right)p^2\left(v\right)\right]z_1^2\left(v\right)+f\left(v\right)]\text{d}v]\text{d}s\left(t\in \left[t_0,\infty \right)\right)\end{array}$ (13)

则有

$z_2\left(t_0\right)=z_1\left(t_0\right)$，$z_1\left(t\right)\le z_2\left(t\right)$，$\left(t\in \left[t_0,\infty \right)\right)$ (14)

根据(12)~(14)式，可得：

$\begin{array}{c}{z^{\prime }}_1\left(t\right)\le w\left(t\right)\left[1+a\left(t\right)\right]+\left[p\left(t\right)+a\left(t\right)+a\left(t\right)p\left(t\right)\right]z_1\left(t\right)+a\left(t\right)z_2\left(t\right)\\ \le w\left(t\right)\left[1+a\left(t\right)\right]+\left[p\left(t\right)+a\left(t\right)+a\left(t\right)p\left(t\right)+a\left(t\right)\right]z_2\left(t\right)\left(t\in \left[t_0,\infty \right)\right)\end{array}$ (15)

又对(13)式两边求导，并将(15)式代入，可得：

$\begin{array}{c}{z^{\prime }}_2\left(t\right)={z^{\prime }}_1\left(t\right)+c\left(t\right)w\left(t\right)+\left[c\left(t\right)\left(1+p\left(t\right)\right)z_1\left(t\right)\right]\\ +c\left(t\right){\displaystyle {\int }_{t_0}^t[d\left(v\right)w^2\left(v\right)}+\left[d\left(v\right)w\left(v\right)+2d\left(v\right)w\left(v\right)p\left(v\right)\right]z_1\left(v\right)\\ +\left[d\left(v\right)p\left(v\right)+d\left(v\right)p^2\left(v\right)\right]z_1^2\left(v\right)+f\left(v\right)]\text{d}v\\ \le w\left(t\right)\left[1+a\left(t\right)\right]+\left[p\left(t\right)+a\left(t\right)+a\left(t\right)p\left(t\right)+a\left(t\right)\right]z_2\left(t\right)\\ +c\left(t\right)w\left(t\right)+\left[c\left(t\right)\left(1+p\left(t\right)\right)z_2\left(t\right)\right]\end{array}$

$\begin{array}{l}+c\left(t\right){\displaystyle {\int }_{t_0}^t[d\left(v\right)w^2\left(v\right)}+\left[d\left(v\right)w\left(v\right)+2d\left(v\right)w\left(v\right)p\left(v\right)\right]z_2\left(v\right)\\ +\left[d\left(v\right)p\left(v\right)+d\left(v\right)p^2\left(v\right)\right]z_2^2\left(v\right)+f\left(v\right)]\text{d}v\\ =w\left(t\right)\left[1+a\left(t\right)+c\left(t\right)\right]+\left[p\left(t\right)+a\left(t\right)+a\left(t\right)p\left(t\right)+c\left(t\right)+c\left(t\right)p\left(t\right)\right]z_2\left(t\right)\\ +c\left(t\right){\displaystyle {\int }_{t_0}^t[d\left(v\right)w^2\left(v\right)}+\left[d\left(v\right)w\left(v\right)+2d\left(v\right)w\left(v\right)p\left(v\right)\right]z_2\left(v\right)\\ +\left[d\left(v\right)p\left(v\right)+d\left(v\right)p^2\left(v\right)\right]z_2^2\left(v\right)+f\left(v\right)]\text{d}v\left(t\in \left[t_0,\infty \right)\right)\end{array}$ (16)

再令

$\begin{array}{c}{z}_3\left(t\right)=z_2\left(t\right)+{\displaystyle {\int }_{t_0}^t[d\left(v\right)w^2\left(v\right)+\left[d\left(v\right)w\left(v\right)+2d\left(v\right)w\left(v\right)p\left(v\right)\right]z_2\left(v\right)}\\ +\left[d\left(v\right)p\left(v\right)+d\left(v\right)p^2\left(v\right)\right]z_2^2\left(v\right)+f\left(v\right)]\text{d}v\left(t\in \left[t_0,\infty \right)\right)\end{array}$ (17)

则有

$z_3\left(t_0\right)=z_2\left(t_0\right)$，$z_2\left(t\right)\le z_3\left(t\right)$，$\left(t\in \left[t_0,\infty \right)\right)$ (18)

根据(16)~(18)式，可得：

$\begin{array}{c}{z^{\prime }}_2\left(t\right)\le w\left(t\right)\left[1+a\left(t\right)+c\left(t\right)\right]+\left[p\left(t\right)+a\left(t\right)+a\left(t\right)p\left(t\right)+c\left(t\right)+c\left(t\right)p\left(t\right)\right]z_2\left(t\right)+c\left(t\right)z_3\left(t\right)\\ \le w\left(t\right)\left[1+a\left(t\right)+c\left(t\right)\right]+\left[p\left(t\right)+a\left(t\right)+a\left(t\right)p\left(t\right)+2c\left(t\right)+c\left(t\right)p\left(t\right)\right]z_3\left(t\right)\left(t\in \left[t_0,\infty \right)\right)\end{array}$ (19)

再对(17)式两边求导，并将(19)式代入，可得：

$\begin{array}{c}{z^{\prime }}_3\left(t\right)={z^{\prime }}_2\left(t\right)+d\left(t\right)w^2\left(t\right)+\left[d\left(t\right)w\left(t\right)+2d\left(t\right)w\left(t\right)p\left(t\right)\right]z_2\left(t\right)\\ +\left[d\left(t\right)p\left(t\right)+d\left(t\right)p^2\left(t\right)\right]z_2^2\left(t\right)+f\left(t\right)\\ \le w\left(t\right)\left[1+a\left(t\right)+c\left(t\right)\right]+[p\left(t\right)+a\left(t\right)+a\left(t\right)p\left(t\right)+2c\left(t\right)\\ +c\left(t\right)p\left(t\right)]z_3\left(t\right)+d\left(t\right)w^2\left(t\right)+\left[d\left(t\right)w\left(t\right)+2d\left(t\right)w\left(t\right)p\left(t\right)\right]z_3\left(t\right)\\ +\left[d\left(t\right)p\left(t\right)+d\left(t\right)p^2\left(t\right)\right]z_3^2\left(t\right)+f(\; t\; )\end{array}$

$\begin{array}{l}=\left[w\left(t\right)\left[1+a\left(t\right)+c\left(t\right)\right]+d\left(t\right)w^2\left(t\right)+f\left(t\right)\right]\\ +[p\left(t\right)+a\left(t\right)+a\left(t\right)p\left(t\right)+2c\left(t\right)+c\left(t\right)p\left(t\right)+d\left(t\right)w\left(t\right)\\ +2d\left(t\right)w\left(t\right)p\left(t\right)]z_3\left(t\right)+\left[d\left(t\right)p\left(t\right)+d\left(t\right)p^2\left(t\right)\right]z_3^2\left(t\right)\\ =A\left(t\right)+B\left(t\right)z_3\left(t\right)+C\left(t\right)z_3^2\left(t\right)\left(t\in \left[t_0,\infty \right)\right)\end{array}$ (20)

其中， $A\left(t\right),B\left(t\right),C\left(t\right)$ 的定义为(7)~(9)式。

将(20)式中的t改写为s，然后两边关于s从 $t_0$ 到t积分，得：

$z_3\left(t\right)\le z_3\left(t_0\right)+{\displaystyle {\int }_{t_0}^{t}A\left(s\right)\text{d}s}+{\displaystyle {\int }_{t_0}^{t}B\left(s\right)z_3\left(s\right)\text{d}s}+{\displaystyle {\int }_{t_0}^{t}C\left(s\right)z_3^2\left(s\right)\text{d}s}\left(t\in \left[t_0,\infty \right)\right)$ (21)

由于(21)式满足了引理要求的条件，根据引理可以得到(21)式中 $z_3\left(t\right)$ 的估计：

$z_3\left(t\right)\le {\left(\exp \left(-\ln \left(z_3\left(t_0\right)+{\displaystyle {\int }_{t_0}^{t}A\left(s\right)\text{d}s}\right)-{\displaystyle {\int }_{t_0}^{t}B\left(s\right)\text{d}s}\right)-{\displaystyle {\int }_{t_0}^{t}C\left(s\right)\text{d}s}\right)}^{-1}\left(t\in \left[t_0,\infty \right)\right)$ (22)

根据(11)、(14)、(18)式，可以知道

$u\left(t_0\right)=z_1\left(t_0\right)=z_2\left(t_0\right)=z_3\left(t_0\right)$ (23)

所以

$z_3\left(t\right)\le {\left(\exp \left(-\ln \left(u\left(t_0\right)+{\displaystyle {\int }_{t_0}^{t}A\left(s\right)\text{d}s}\right)-{\displaystyle {\int }_{t_0}^{t}B\left(s\right)\text{d}s}\right)-{\displaystyle {\int }_{t_0}^{t}C\left(s\right)\text{d}s}\right)}^{-1}=Z\left(t\right)\left(t\in \left[t_0,\infty \right)\right)$ (24)

其中 $Z\left(t\right)$ 定义为(6)式。

把(24)式代入(19)式，可得：

${z^{\prime }}_2\left(t\right)\le w\left(t\right)\left[1+a\left(t\right)+c\left(t\right)\right]+\left[p\left(t\right)+a\left(t\right)+a\left(t\right)p\left(t\right)+2c\left(t\right)+c\left(t\right)p\left(t\right)\right]Z\left(t\right)\left(t\in \left[t_0,\infty \right)\right)$ (25)

对(25)式两边求积分，可得：

$\begin{array}{l}{z}_2\left(t\right)\le z_2\left(t_0\right)+{\displaystyle {\int }_{t_0}^t\{w\left(s\right)\left[1+a\left(s\right)+c\left(s\right)\right]+[p\left(s\right)+a\left(s\right)+a\left(s\right)p\left(s\right)}\\ +2c\left(s\right)+c\left(s\right)p\left(s\right)]Z\left(s\right)\}\text{d}s\left(t\in \left[t_0,\infty \right)\right)\end{array}$ (26)

根据(23)式知 $u\left(t_0\right)=z_2\left(t_0\right)$，因此得到：

$\begin{array}{l}{z}_2\left(t\right)\le u\left(t_0\right)+{\displaystyle {\int }_{t_0}^t\{w\left(s\right)\left[1+a\left(s\right)+c\left(s\right)\right]+[p\left(s\right)+a\left(s\right)+a\left(s\right)p\left(s\right)}\\ +2c\left(s\right)+c\left(s\right)p\left(s\right)]Z\left(s\right)\}\text{d}s\left(t\in \left[t_0,\infty \right)\right)\end{array}$

又根据(11)、(14)、(18)式，可以知道

$u\left(t\right)\le z_1\left(t\right)\le z_2\left(t\right)\le z_3\left(t\right)$ (27)

所以得估计式：

$\begin{array}{l}u\left(t\right)\le u\left(t_0\right)+{\displaystyle {\int }_{t_0}^t\{w\left(s\right)\left[1+a\left(s\right)+c\left(s\right)\right]+[p\left(s\right)+a\left(s\right)+a\left(s\right)p\left(s\right)}\\ +2c\left(s\right)+c\left(s\right)p\left(s\right)]Z\left(s\right)\}\text{d}s\left(t\in \left[t_0,\infty \right)\right)\end{array}$

证毕。

基金项目

国家自然科学基金资助项目(11961021，11561019)；广西高校中青年教师科研基础能力提升项目(2019KY0625)。

文章引用

卢钰松,黄星寿. 一类三重非线性积分不等式解的估计
Estimation of Unknown Function of a Class of Triple Integral Inequalities[J]. 应用数学进展, 2020, 09(08): 1200-1205. https://doi.org/10.12677/AAM.2020.98140

参考文献