一类含有奇异项的两个变量的复合Gronwall-Bellman型积分不等式及其应用

A Class of Composite Gronwall-Bellman Integral Inequality of Two Variables with Singular Terms and Its Application

摘要: Gronwall-Bellman型积分不等式是研究微分方程解的定性性质的重要数学工具。本文研究了一类含有两个变量的复合Gronwall-Bellman型积分不等式，并利用得出的结果给出了偏微分方程中未知函数的上界估计。

Abstract: Gronwall-Bellman type integral inequality is an important mathematical tool to study the qualitative properties of the solutions of differential equations. In this paper, a class of complex Gronwall-Bellman type integral inequalities containing two variables is studied, and the upper bound estimates of unknown functions in partial differential equations are given by using the results obtained.

1. 引言与预备

不等式的研究在数学的发展历史中占据了非常重要的一部分，其中的Gronwall型不等式是应用最广泛的不等式类型之一，其最早是由Gronwall于1919年提出并证明，它的形式如下[1]，

$E\left(t\right)\le M+c{\displaystyle {\int }_0^{t}E\left(\tau \right)\text{d}\tau }.$

1943年，Bellman对Gronwall的结果进行了推广，研究了不等式[2]，

$E\left(t\right)\le M+c{\displaystyle {\int }_0^{t}K\left(\tau \right)E\left(\tau \right)\text{d}\tau }.$

1973年，B.G. Pachpatte在[3]中研究了如下形式的积分不等式，

$u\left(t\right)\le u_0+{\displaystyle {\int }_0^{t}f\left(s\right)u\left(s\right)\text{d}s}+{\displaystyle {\int }_0^{t}f\left(s\right)\left({\displaystyle {\int }_0^{s}g\left(\tau \right)u\left(\tau \right)\text{d}\tau }\right)\text{d}s}.$

2006年，D.B. Pachpatte在[4]中研究了时间尺度上一些Gronwall型积分不等式，并且用它们讨论了动力型微分和积分方程中未知函数的显示估计，具体的不等式形式有，

$u\left(t\right)\le u_0+g\left(t\right){\displaystyle {\int }_a^{t}f\left(s\right)u\left(s\right)\Delta s}+{\displaystyle {\int }_a^{t}h\left(s\right)W\left(u\left(s\right)\right)\Delta s}.$

2014年，J. Gu，F.W. Meng等人在文章[5]中研究了非线性Volterra-Fredholm型动态积分不等式，其中的一个不等式形式如下，

$\begin{array}{l}u\left(t\right)\le k+{\displaystyle {\int }_{t_0}^t{f}_1\left(s\right)W\left(u\left(s\right)\right)\Delta s}+{\displaystyle {\int }_{t_0}^t{f}_2\left(s\right)\left({\displaystyle {\int }_{t_0}^s{f}_3\left(\tau \right)W\left(u\left(\tau \right)\right)\Delta \tau }\right)\Delta s}\\ +{\displaystyle {\int }_{t_0}^{\alpha }{f}_1\left(s\right)W\left(u\left(s\right)\right)\Delta s}+{\displaystyle {\int }_{t_0}^{\alpha }{f}_2\left(s\right)\left({\displaystyle {\int }_{t_0}^{\alpha }{f}_3\left(\tau \right)W\left(u\left(\tau \right)\right)\Delta \tau }\right)\Delta s}.\end{array}$

2023年，王培，陈心妍，董琪翔在文章[6]研究了如下积分不等式，

$\begin{array}{l}\phi \left(f\left(x,y\right)\right)\le c+{\displaystyle {\int }_0^x{A}_1}\left(s,y\right)\phi \left(f\left(s,y\right)\right)\text{d}s+{\displaystyle {\int }_0^y{A}_2}\left(x,t\right)\phi \left(f\left(x,t\right)\right)\text{d}t\\ +{\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right){\phi }^m\left(f\left(s,t\right)\right)\text{d}t\text{d}s}}.\end{array}$

带有奇异项积分不等式的研究在积分不等式的发展历史中占据了非常重要的一部分。

1981年，Henry在[7]研究了线性形式的分数阶积分不等式，

$u\left(t\right)\le a\left(t\right)+b{\displaystyle {\int }_0^t{\left(t-s\right)}^{\beta -1}}u\left(s\right)\text{d}s.$

2007年，由叶海平、高建明和丁永胜在文章[8]研究了如下积分不等式，

$u\left(t\right)\le a\left(t\right)+g\left(t\right){\displaystyle {\int }_0^t{\left(t-s\right)}^{\beta -1}}u\left(s\right)\text{d}s.$

本文受文献[6]-[8]的启发，在一般Gronwall-Bellman型二元积分不等式的基础上，在一阶积分号内加入奇异项以及在一阶积分号外加入未知函数进行改进，并对未知函数进行复合，建立了一类新的线性双变量复合Gronwall-Bellman型积分不等式，形式如下，

$\begin{array}{c}\phi \left(u\left(x,y\right)\right)\le c+A_1\left(x,y\right){\displaystyle {\int }_0^x{\left(x-s\right)}^{\alpha -1}}\phi \left(u\left(s,y\right)\right)\text{d}s\\ +A_2\left(x,y\right){\displaystyle {\int }_0^y{\left(y-t\right)}^{\beta -1}}\phi \left(u\left(x,t\right)\right)\text{d}t\\ +{\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right)\phi \left(u\left(s,t\right)\right)\text{d}t\text{d}s}}.\end{array}$

最后通过构造偏微分方程以及迭代的方法得到未知函数的边界值，并给出其在定性和定量研究边界积分方程解的性质的例子。

本研究需要以下预备知识：

定义1.1 Gamma函数定义：

$\Gamma \left(z\right)={\displaystyle {\int }_0^{\infty }{\text{e}}^{-t}{t}^{z-1}}\text{d}t,\mathrm{Re}\left(z\right)>0.$

以及Gamma函数的重要性质：$\Gamma \left(z+1\right)=z\Gamma \left(z\right)$ 。

定义1.2 Bate函数定义：

$B\left(p,q\right)={\displaystyle {\int }_0^{\infty }{\tau }^{p-1}{\left(1-\tau \right)}^{q-1}\text{d}t},\mathrm{Re}\left(p\right)>0,\mathrm{Re}\left(q\right)>0.$

以及Bate函数的重要性质：$B\left(p,q\right)=\frac{\Gamma \left(p\right)\Gamma \left(q\right)}{\Gamma \left(p+q\right)}$ 。

定义1.3 Mittag-Leffler函数定义：

$E_{\alpha }\left(z\right)={\displaystyle \sum _{k=0}^{\infty }\frac{z^k}{\Gamma \left(k\alpha +1\right)}},\alpha >0,z\in C.$

引理1 [6] 设$f\left(x,y\right)$ ，$b\left(x,y\right)$ ，$A_1\left(x,y\right)$ ，$A_2\left(x,y\right)$ 为定义在$\left[0,\infty \right)\times \left[0,\infty \right)$ 上的非负连续函数，$\phi$ 是定义在$\left[0,\infty \right)$ 上的严格增连续函数，且$\phi \left(f\left(x,y\right)\right)$ 满足下述积分不等式：

$\begin{array}{l}\phi \left(f\left(x,y\right)\right)\le c+{\displaystyle {\int }_0^x{A}_1}\left(s,y\right)\phi \left(f\left(s,y\right)\right)\text{d}s+{\displaystyle {\int }_0^y{A}_2}\left(x,t\right)\phi \left(f\left(x,t\right)\right)\text{d}t\\ +{\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right){\phi }^m\left(f\left(s,t\right)\right)\text{d}t\text{d}s}},\left(x,y\right)\in \left[0,\infty \right)\times \left[0,\infty \right),\end{array}$

其中$c>0$ 为常数，则当$0<m<1$ 时，

$f\left(x,y\right)\le {\phi }^{-1}\left\{B_1\left(x,y\right)B_2\left(x,y\right)\left\{c+{\left[\left(1-m\right){\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right)Z_1^m\left(s,t\right)\text{d}t\text{d}s}}\right]}^{\frac{1}{1-m}}\right\}\right\};$

当$m=1$ 时，

$f\left(x,y\right)\le {\phi }^{-1}\left\{B_1\left(x,y\right)B_2\left(x,y\right)\exp {\left[{\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right)B_1\left(s,t\right)B_2\left(s,t\right)\text{d}t\text{d}s}}\right]}^{\frac{1}{1-m}}\right\}.$

其中

$Z_1\left(x,y\right)=1+{\displaystyle {\int }_0^x{A}_1}\left(s,y\right)\frac{\phi \left(f\left(s,y\right)\right)}{\alpha \left(s,y\right)}\text{d}s+{\displaystyle {\int }_0^y{A}_2}\left(x,t\right)\frac{\phi \left(f\left(x,t\right)\right)}{\alpha \left(x,t\right)}\text{d}t,$

$\alpha \left(x,y\right)=c+{\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right){\phi }^m\left(u\left(s,t\right)\right)\text{d}t\text{d}s}},$

$B_1\left(x,y\right)=\exp \left[{\displaystyle {\int }_0^x{A}_1\left(s,y\right)B_2\left(s,y\right)\text{d}s}\right],$

$B_2\left(x,y\right)=\exp \left[{\displaystyle {\int }_0^y{A}_2\left(x,t\right)\text{d}t}\right].$

引理2 [8] 设$\beta >0$ ，$a\left(t\right)$ ，$u\left(t\right)$ 为定义在$t\in \left[0,T\right]$ 的非负，局部可积函数。$g\left(t\right)$ 为定义在$t\in \left[0,T\right]$ 的非负、非递减连续函数，$g\left(t\right)\le M$ (常数)。若满足不等式

$u\left(t\right)\le a\left(t\right)+g\left(t\right){\displaystyle {\int }_0^t{\left(t-s\right)}^{\beta -1}}u\left(s\right)\text{d}s,$

则有

$u\left(t\right)\le a\left(t\right)+{\displaystyle {\int }_0^t\left[{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(g\left(t\right)\Gamma \left(\beta \right)\right)}^n}{\Gamma \left(n\beta \right)}}{\left(t-s\right)}^{n\beta -1}a\left(s\right)\right]\text{d}s},t\in \left[0,T\right).$

2. 主要成果

为了方便定理叙述，给出定理中函数需要满足的条件：

1) $u\left(x,y\right)$ 为定义在$\left[0,\infty \right)\times \left[0,\infty \right)$ 上非负连续函数，$u\left(0,0\right)=0$ ，且对任意固定的$y_0$ ，

$u\left(\ast ,y_0\right)$ 严格单调递增，对任意固定的$x_0$ ，$u\left(x_0,\ast \right)$ 严格单调递增；

2) $\phi$ 为定义在$\left[0,\infty \right)$ 上严格增的连续函数，$\phi \left(0\right)=0$ ；

3) $A_1\left(x,y\right)$ ，$A_2\left(x,y\right)$ 为定义在$\left[0,\infty \right)\times \left[0,\infty \right)$ 上的实值、非负连续函数，$A_1\left(0,y\right)=0$ ，$A_2\left(x,0\right)=0$ ，且对任意固定的$y_0$ ，$A_1\left(\ast ,y_0\right)$ ，$A_2\left(\ast ,y_0\right)$ 严格单调递增，对任意固定的$x_0$ ，$A_1\left(x_0,\ast \right)$ ，$A_2\left(x_0,\ast \right)$ 严格单调递增；

4) $H\left(x,y\right)$ 为定义在$\left[0,\infty \right)\times \left[0,\infty \right)$ 上的实值、非负连续函数，且对任意固定的$y_0$ ，

$H\left(\ast ,y_0\right)$ 非递减，对任意固定的$x_0$ ，$H\left(x_0,\ast \right)$ 非递减。

定理1 假设(1)~(4)成立，且$\phi \left(u\left(x,y\right)\right)$ 满足下述积分不等式：

$\begin{array}{l}\phi \left(u\left(x,y\right)\right)\le c+A_1\left(x,y\right){\displaystyle {\int }_0^x{\left(x-s\right)}^{\alpha -1}}\phi \left(u\left(s,y\right)\right)\text{d}s\\ +A_2\left(x,y\right){\displaystyle {\int }_0^y{\left(y-t\right)}^{\beta -1}}\phi \left(u\left(x,t\right)\right)\text{d}t\\ +{\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right){\phi }^m\left(u\left(s,t\right)\right)\text{d}t\text{d}s}},\left(x,y\right)\in \left[0,\infty \right)\times \left[0,\infty \right),\end{array}$ (2.1)

其中$c>0$ 为常数，$0<m<1$ ，且函数$B\left(x,y\right)$ 满足

$B\left(x,y\right)=1+A_1\left(x,y\right){\displaystyle {\int }_0^x{\left(x-s\right)}^{\alpha -1}}\frac{\phi \left(u\left(s,y\right)\right)}{M\left(s,y\right)}\text{d}s+A_2\left(x,y\right){\displaystyle {\int }_0^y{\left(y-t\right)}^{\beta -1}}\frac{\phi \left(u\left(x,t\right)\right)}{M\left(x,t\right)}\text{d}t,$

则有

$u\left(x,y\right)\le {\phi }^{-1}\left\{W\left(x,y\right){\left[c^{1-m}+\left(1-m\right){\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right)W^m\left(s,t\right)\text{d}t\text{d}s}}\right]}^{\frac{1}{1-m}}\right\},$ (2.2)

其中

$W\left(x,y\right)=E_{\alpha }\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)x^{\alpha }\right)E_{\beta }\left(E_{\alpha }\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)x^{\alpha }\right)A_2\left(x,y\right)\Gamma \left(\beta \right)y^{\beta }\right),$

证 定义

$M\left(x,y\right)=c+{\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right){\phi }^m\left(u\left(s,t\right)\right)\text{d}t\text{d}s}},$ (2.3)

由条件可推得$M\left(x,y\right)$ 为正的，非递减连续函数，则(2.1)化为

$\phi \left(u\left(x,y\right)\right)\le M\left(x,y\right)+A_1\left(x,y\right){\displaystyle {\int }_0^x{\left(x-s\right)}^{\alpha -1}}\phi \left(u\left(s,y\right)\right)\text{d}s+A_2\left(x,y\right){\displaystyle {\int }_0^y{\left(y-t\right)}^{\beta -1}}\phi \left(u\left(x,t\right)\right)\text{d}t,$

上式两边同时除以$M\left(x,y\right)$ ，可得

$\begin{array}{l}\frac{\phi \left(u\left(x,y\right)\right)}{M\left(x,y\right)}\le 1+A_1\left(x,y\right){\displaystyle {\int }_0^x{\left(x-s\right)}^{\alpha -1}}\frac{\phi \left(u\left(s,y\right)\right)}{M\left(s,y\right)}\text{d}s\\ +A_2\left(x,y\right){\displaystyle {\int }_0^y{\left(y-t\right)}^{\beta -1}}\frac{\phi \left(u\left(x,t\right)\right)}{M\left(x,t\right)}\text{d}t,\end{array}$ (2.4)

令

$B\left(x,y\right)=1+A_1\left(x,y\right){\displaystyle {\int }_0^x{\left(x-s\right)}^{\alpha -1}}\frac{\phi \left(u\left(s,y\right)\right)}{M\left(s,y\right)}\text{d}s+A_2\left(x,y\right){\displaystyle {\int }_0^y{\left(y-t\right)}^{\beta -1}}\frac{\phi \left(u\left(x,t\right)\right)}{M\left(x,t\right)}\text{d}t,$

显然$B\left(0,0\right)=1$ ，则有

$\frac{\phi \left(u\left(x,y\right)\right)}{M\left(x,y\right)}\le B\left(x,y\right),$ (2.5)

从而由(2.4)可得

$B\left(x,y\right)\le 1+A_1\left(x,y\right){\displaystyle {\int }_0^x{\left(x-s\right)}^{\alpha -1}}B\left(s,y\right)\text{d}s+A_2\left(x,y\right){\displaystyle {\int }_0^y{\left(y-t\right)}^{\beta -1}}B\left(x,t\right)\text{d}t,$

定义

$C\left(x,y\right)=1+A_2\left(x,y\right){\displaystyle {\int }_0^y{\left(y-t\right)}^{\beta -1}}B\left(x,t\right)\text{d}t,$ (2.6)

由(2.6)可得$C\left(x,0\right)=1$ ，则可得

$B\left(x,y\right)\le C\left(x,y\right)+A_1\left(x,y\right){\displaystyle {\int }_0^x{\left(x-s\right)}^{\alpha -1}}B\left(s,y\right)\text{d}s,$

因为$C\left(x,y\right)$ 为正的，非递减连续函数，则由上式可得

$\frac{B\left(x,y\right)}{C\left(x,y\right)}\le 1+A_1\left(x,y\right){\displaystyle {\int }_0^x{\left(x-s\right)}^{\alpha -1}}\frac{B\left(s,y\right)}{C\left(s,y\right)}\text{d}s,$ (2.7)

令

$D\left(x,y\right)=1+A_1\left(x,y\right){\displaystyle {\int }_0^x{\left(x-s\right)}^{\alpha -1}}\frac{B\left(s,y\right)}{C\left(s,y\right)}\text{d}s,$

则$D\left(0,y\right)=1$ ，且$D\left(x,y\right)$ 为正的，非递减连续函数，则有

$\frac{B\left(x,y\right)}{C\left(x,y\right)}\le D\left(x,y\right),$ (2.8)

从而由(2.7)可得

$D\left(x,y\right)\le 1+A_1\left(x,y\right){\displaystyle {\int }_0^x{\left(x-s\right)}^{\alpha -1}}D\left(s,y\right)\text{d}s.$ (2.9)

定义一个算子：

$G\Phi \left(x,y\right)=A_1\left(x,y\right){\displaystyle {\int }_0^x{\left(x-s\right)}^{\alpha -1}}\Phi \left(s,y\right)\text{d}s.$

则由(2.9)可得下列结果

$D\left(x,y\right)\le 1+GD\left(x,y\right),$

$D\left(x,y\right)\le {\displaystyle \sum _{k=0}^{n-1}{G}^{k}1}+G^{n}D\left(x,y\right).$

下面证明

$G^{n}D\left(x,y\right)\le \frac{{\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)\right)}^n}{\Gamma \left(n\alpha \right)}{\displaystyle {\int }_0^x{\left(x-s\right)}^{n\alpha -1}D\left(s,y\right)\text{d}s},$ (2.10)

对任意的$x,y\in R_{+}$ ，当$n\to +\infty$ 时，$G^{n}D\left(x,y\right)\to 0$ 。

运用数学归纳法。当$n=1$ 时，(2.10)显然成立；假设$n=k$ 时，(2.10)成立；

当$n=k+1$ 时，则有

$G^{k+1}D\left(x,y\right)=G{G}^{k}D\left(x,y\right)\le A_1\left(x,y\right){\displaystyle {\int }_0^x{\left(x-s\right)}^{\alpha -1}\left[{\displaystyle {\int }_0^s\frac{{\left(A_1\left(s,y\right)\Gamma \left(\alpha \right)\right)}^k}{\Gamma \left(k\alpha \right)}{\left(s-\tau \right)}^{k\alpha -1}D\left(\tau ,y\right)\text{d}\tau }\right]\text{d}s},$

通过交换积分顺序，可得

$G^{k+1}D\left(x,y\right)\le {\left(A_1\left(x,y\right)\right)}^{k+1}\frac{{\left(\Gamma \left(\alpha \right)\right)}^k}{\Gamma \left(k\alpha \right)}{\displaystyle {\int }_0^x\left[{\displaystyle {\int }_{\tau }^x{\left(x-s\right)}^{\alpha -1}{\left(s-\tau \right)}^{k\alpha -1}\text{d}s}\right]D\left(\tau ,y\right)\text{d}\tau },$

由Beta函数相关性质，作变量替换，即$s=\tau +u\left(x-\tau \right)$ ，可得

$\begin{array}{c}{G}^{k+1}D\left(x,y\right)\le {\left(A_1\left(x,y\right)\right)}^{k+1}\frac{{\left(\Gamma \left(\alpha \right)\right)}^k}{\Gamma \left(k\alpha \right)}{\displaystyle {\int }_0^x\left[{\displaystyle {\int }_0^1{\left(1-u\right)}^{\alpha -1}{u}^{k\alpha -1}\text{d}u}\right]}{\left(x-\tau \right)}^{\left(k+1\right)\alpha -1}D\left(\tau ,y\right)\text{d}\tau \\ ={\left(A_1\left(x,y\right)\right)}^{k+1}\frac{{\left(\Gamma \left(\alpha \right)\right)}^k}{\Gamma \left(k\alpha \right)}{\displaystyle {\int }_0^x\frac{\Gamma \left(\alpha \right)\Gamma \left(k\alpha \right)}{\Gamma \left(\left(k+1\right)\alpha \right)}}{\left(x-\tau \right)}^{\left(k+1\right)\alpha -1}D\left(\tau ,y\right)\text{d}\tau \\ =\frac{{\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)\right)}^{k+1}}{\Gamma \left(\left(k+1\right)\alpha \right)}{\displaystyle {\int }_0^x{\left(x-\tau \right)}^{\left(k+1\right)\alpha -1}}D\left(\tau ,y\right)\text{d}\tau ,\end{array}$

即当$n=k+1$ 时，(2.10)成立。

又对$\forall x,y\in R_{+}$ ，当$n\to +\infty$ 时，有

$0\le G^{n}D\left(x,y\right)\le \frac{{\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)\right)}^n}{\Gamma \left(n\alpha \right)}{\displaystyle {\int }_0^x{\left(x-s\right)}^{n\alpha -1}D\left(s,y\right)\text{d}s}\to 0.$ s

所以，当$n\to +\infty$ 时可得

$\begin{array}{l}D\left(x,y\right)\le {\displaystyle \sum _{k=0}^{\infty }{G}^{k}1}\le {\displaystyle \sum _{k=0}^{\infty }\frac{{\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)\right)}^k}{\Gamma \left(k\alpha \right)}}{\displaystyle {\int }_0^x{\left(x-s\right)}^{k\alpha -1}1\text{d}s}\\ \begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\begin{array}{c}\end{array}& \end{array}\begin{array}{c}\begin{array}{c}\end{array}\end{array}& \end{array}\begin{array}{c}\end{array}& \end{array}\begin{array}{c}\end{array}={\displaystyle \sum _{k=0}^{\infty }\frac{{\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)\right)}^k}{\Gamma \left(k\alpha \right)}}\cdot \frac{x^{k\alpha }}{k\alpha }\\ \begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\begin{array}{c}\end{array}& \end{array}\begin{array}{c}\begin{array}{c}\end{array}\end{array}& \end{array}\begin{array}{c}\end{array}& \end{array}\begin{array}{c}\end{array}={\displaystyle \sum _{k=0}^{\infty }\frac{{\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)x^{\alpha }\right)}^k}{\Gamma \left(k\alpha +1\right)}}\\ \begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\begin{array}{c}\end{array}& \end{array}\begin{array}{c}\begin{array}{c}\end{array}\end{array}& \end{array}\begin{array}{c}\end{array}& \end{array}\begin{array}{c}\end{array}=E_{\alpha }\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)x^{\alpha }\right).\end{array}$

由(2.8)可以得到

$B\left(x,y\right)\le C\left(x,y\right)D\left(x,y\right)\le C\left(x,y\right)E_{\alpha }\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)x^{\alpha }\right).$ (2.11)

将(2.6)代入(2.11)可以得到

$\begin{array}{c}B\left(x,y\right)\le E_{\alpha }\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)x^{\alpha }\right)\left[1+A_2\left(x,y\right){\displaystyle {\int }_0^y{\left(y-t\right)}^{\beta -1}B\left(x,t\right)\text{d}t}\right]\\ \le E_{\alpha }\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)x^{\alpha }\right)+E_{\alpha }\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)x^{\alpha }\right)A_2\left(x,y\right){\displaystyle {\int }_0^y{\left(y-t\right)}^{\beta -1}B\left(x,t\right)\text{d}t}.\end{array}$ (2.12)

定义算子：

${\rm P}\Phi \left(x,y\right)\le E_{\alpha }\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)x^{\alpha }\right)A_2\left(x,y\right){\displaystyle {\int }_0^y{\left(y-t\right)}^{\beta -1}\Phi \left(x,t\right)\text{d}t},$

则(2.12)化为

$B\left(x,y\right)\le E_{\alpha }\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)x^{\alpha }\right)+{\rm P}B\left(x,y\right),$

$B\left(x,y\right)\le {\displaystyle \sum _{k=0}^{n-1}{{\rm P}}^k}{E}_{\alpha }\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)x^{\alpha }\right)+{{\rm P}}^{n}B\left(x,y\right).$

参照(2.10)和(2.11)的计算，可证得

${{\rm P}}^{n}B\left(x,y\right)\le \frac{{\left(E_{\alpha }\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)x^{\alpha }\right)A_2\left(x,y\right)\Gamma \left(\beta \right)\right)}^n}{\Gamma \left(n\beta \right)}{\displaystyle {\int }_0^y{\left(y-t\right)}^{n\beta -1}B\left(x,t\right)\text{d}t},$

对任意的$x,y\in R_{+}$ ，当$n\to +\infty$ 时，${{\rm P}}^{n}B\left(x,y\right)\to 0$ 。

因此，

$B\left(x,y\right)\le {\displaystyle \sum _{k=0}^{\infty }{{\rm P}}^k{E}_{\alpha }\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)x^{\alpha }\right)}\le E_{\alpha }\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)x^{\alpha }\right)E_{\beta }\left(E_{\alpha }\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)x^{\alpha }\right)A_2\left(x,y\right)\Gamma \left(\beta \right)y^{\beta }\right).$

这里令

$W\left(x,y\right)=E_{\alpha }\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)x^{\alpha }\right)E_{\beta }\left(E_{\alpha }\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)x^{\alpha }\right)A_2\left(x,y\right)\Gamma \left(\beta \right)y^{\beta }\right),$

又由(2.5)可得

$\phi \left(u\left(x,y\right)\right)\le B\left(x,y\right)M\left(x,y\right)\le W\left(x,y\right)M\left(x,y\right).$

对(2.3)两边求关于$x$ 和$y$ 的偏导数，可得

$M_{xy}\left(x,y\right)=H\left(x,y\right){\phi }^m\left(u\left(x,y\right)\right),$

又由(2.5)有

${\phi }^m\left(u\left(x,y\right)\right)\le B^m\left(x,y\right)M^m\left(x,y\right),$

从而

$M_{xy}\left(x,y\right)\le H\left(x,y\right)B^m\left(x,y\right)M^m\left(x,y\right),$

因为$M\left(x,y\right)$ 为正的，非递减连续函数，故

$\frac{M_{xy}\left(x,y\right)}{M^m\left(x,y\right)}\le H\left(x,y\right)B^m\left(x,y\right).$ (2.13)

由于$M\left(x,y\right)$ 关于两个变量分别单调递增，因此结合(1.13)可得

$\frac{m{M}_x\left(x,y\right)M_y\left(x,y\right)M^{m-1}\left(x,y\right)}{M^{2m}\left(x,y\right)}\ge 0,$ (2.14)

由(2.13)和(2.14)可以得出

$\frac{M_{xy}\left(x,y\right)M^m\left(x,y\right)}{M^{2m}\left(x,y\right)}-\frac{m{M}_x\left(x,y\right)M_y\left(x,y\right)M^{m-1}\left(x,y\right)}{M^{2m}\left(x,y\right)}\le H\left(x,y\right)B^m\left(x,y\right).$ (2.15)

由于

$\frac{\partial }{\partial y}\left[\frac{M_x\left(x,y\right)}{M^m\left(x,y\right)}\right]=\frac{M_{xy}\left(x,y\right)M^m\left(x,y\right)}{M^{2m}\left(x,y\right)}-\frac{m{M}_x\left(x,y\right)M_y\left(x,y\right)M^{m-1}\left(x,y\right)}{M^{2m}\left(x,y\right)},$

故(2.15)可变为

$\frac{\partial }{\partial y}\left[\frac{M_x\left(x,y\right)}{M^m\left(x,y\right)}\right]\le H\left(x,y\right)B^m\left(x,y\right).$ (2.16)

在(2.16)两边对$y$ 在$\left[0,y\right]$ 上积分得

$\frac{M_x\left(x,y\right)}{M^m\left(x,y\right)}-\frac{M_x\left(x,0\right)}{M^m\left(x,0\right)}\le {\displaystyle {\int }_0^{y}H\left(x,t\right)B^m\left(x,t\right)\text{d}t},$

因为$M^m\left(x,0\right)\ne 0$ ，$\frac{M_x\left(x,0\right)}{M^m\left(x,0\right)}=0$ ，故有

$\frac{M_x\left(x,y\right)}{M^m\left(x,y\right)}\le {\displaystyle {\int }_0^{y}H\left(x,t\right)B^m\left(x,t\right)\text{d}t}.$ (2.17)

在(2.17)两边对$x$ 在$\left[0,x\right]$ 上积分可得

$\frac{1}{1-m}{M}^{1-m}\left(x,y\right)-\frac{1}{1-m}{M}^{1-m}\left(0,y\right)\le {\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right)B^m\left(s,t\right)\text{d}t}}\text{d}s,$

$M\left(x,y\right)\le {\left[M^{1-m}\left(0,y\right)+\left(1-m\right){\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right)B^m\left(s,t\right)\text{d}t}}\text{d}s\right]}^{\frac{1}{1-m}},$

由(2.3)可知$M\left(0,y\right)=c$ ，故

$M\left(x,y\right)\le {\left[c^{1-m}+\left(1-m\right){\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right)B^m\left(s,t\right)\text{d}t}}\text{d}s\right]}^{\frac{1}{1-m}},\left(x,y\right)\in \left[0,\infty \right)\times \left[0,\infty \right).$

又由(2.5)有

$\phi \left(u\left(x,y\right)\right)\le B\left(x,y\right)M\left(x,y\right),$

又由$B\left(x,y\right)\le W\left(x,y\right)$ ，故有

$\begin{array}{c}\phi \left(u\left(x,y\right)\right)\le B\left(x,y\right){\left[c^{1-m}+\left(1-m\right){\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right)B^m\left(s,t\right)\text{d}t}}\text{d}s\right]}^{\frac{1}{1-m}}\\ \le W\left(x,y\right){\left[c^{1-m}+\left(1-m\right){\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right)B^m\left(s,t\right)\text{d}t}}\text{d}s\right]}^{\frac{1}{1-m}},\end{array}$

又$\phi \left(x\right)$ 为严格增的连续函数，从而

$u\left(x,y\right)\le {\phi }^{-1}\left\{W\left(x,y\right){\left[c^{1-m}+\left(1-m\right){\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right)W^m\left(s,t\right)\text{d}t}}\text{d}s\right]}^{\frac{1}{1-m}}\right\}.$

则定理1证毕。

定理2 假设定理1中的条件(1)~(4)成立，且$\phi \left(u\left(x,y\right)\right)$ 满足下述积分不等式：

$\begin{array}{l}\phi \left(u\left(x,y\right)\right)\le c+A_1\left(x,y\right){\displaystyle {\int }_0^x{\left(x-s\right)}^{\alpha -1}}\phi \left(u\left(s,y\right)\right)\text{d}s+A_2\left(x,y\right){\displaystyle {\int }_0^y{\left(y-t\right)}^{\beta -1}}\phi \left(u\left(x,t\right)\right)\text{d}t\\ +{\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right)\phi \left(u\left(s,t\right)\right)\text{d}t}}\text{d}s,\left(x,y\right)\in \left[0,\infty \right)\times \left[0,\infty \right),\end{array}$

其中$c>0$ 为常数，则有

$u\left(x,y\right)\le {\phi }^{-1}\left\{W\left(x,y\right)\exp \left[\ln \begin{array}{c}c\end{array}+{\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right)W\left(s,t\right)\text{d}t}}\text{d}s\right]\right\},$

其中

$W\left(x,y\right)=E_{\alpha }\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)x^{\alpha }\right)E_{\beta }\left(E_{\alpha }\left(A_1\left(x,y\right)\Gamma \left(\alpha \right)x^{\alpha }\right)A_2\left(x,y\right)\Gamma \left(\beta \right)y^{\beta }\right).$

证 由定理1的步骤(2.3)~(2.17)，可得

$\frac{M_x\left(x,y\right)}{M\left(x,y\right)}\le {\displaystyle {\int }_0^{y}H\left(x,t\right)B\left(x,t\right)\text{d}t},$ (2.18)

在(2.18)两边对$x$ 在$\left[0,x\right]$ 上积分可得

$\ln M\left(x,y\right)-\ln M\left(0,y\right)\le {\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right)B\left(s,t\right)\text{d}t}}\text{d}s.$

又由$\ln M\left(0,y\right)=\ln \begin{array}{c}c\end{array}$ ，$B\left(x,y\right)\le W\left(x,y\right)$ ，其中$W\left(x,y\right)$ 在定理1中有定义，故有

$M\left(x,y\right)\le \exp \left[\ln \begin{array}{c}c\end{array}+{\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right)B\left(s,t\right)\text{d}t}}\text{d}s\right]\le \exp \left[\ln \begin{array}{c}c\end{array}+{\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right)W\left(s,t\right)\text{d}t}}\text{d}s\right],$

由(2.5)有

$\phi \left(u\left(x,y\right)\right)\le B\left(x,y\right)M\left(x,y\right)\le W\left(x,y\right)\exp \left[\ln \begin{array}{c}c\end{array}+{\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right)W\left(s,t\right)\text{d}t}}\text{d}s\right],$

又$\phi \left(x\right)$ 为严格增的连续函数，从而

$u\left(x,y\right)\le {\phi }^{-1}\left\{W\left(x,y\right)\exp \left[\ln \begin{array}{c}c\end{array}+{\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right)W\left(s,t\right)\text{d}t}}\text{d}s\right]\right\}.$

则定理2证毕。

3. 理论应用举例

利用不等式结果研究下面二阶偏微分方程边值问题的解的性质，首先给出一个解的先验条件.

例

$\{\begin{array}{l}\frac{{\partial }^2\phi \left(u\left(x,y\right)\right)}{\partial x\partial y}=U\left(x,y,u\left(x,y\right)\right)+f_y\left(x,y,u\left(x,y\right)\right)+g_x\left(x,y,u\left(x,y\right)\right),\\ u\left(x,0\right)=p\left(x\right),u\left(0,y\right)=q\left(y\right),\end{array}$ (3.1)

若函数$U\left(x,y,u\left(x,y\right)\right)$ ，$\phi \left(t\right)$ ，$f\left(x,y,u\left(x,y\right)\right)$ ，$g\left(x,y,u\left(x,y\right)\right)$ 分别满足：

i) $U\left(x,y,u\left(x,y\right)\right)\le H(\left(x,y\right){\phi }^m\left(u\left(x,y\right)\right)$ ；

ii) $p\left(x\right)$ ，$q\left(y\right)$ 在$\left[0,\infty \right)$ 上有界连续，且$p\left(0\right)=q\left(0\right)=0$ ；

iii) $f\left(x,s,u\right)\le {\left(x-s\right)}^{\alpha -1}\phi \left(u\right)$ ；

iv) $g\left(y,t,u\right)\le {\left(y-t\right)}^{\beta -1}\phi \left(u\right)$ 。

其中$\phi \left(t\right)$ 满足条件(2)，$H\left(x,y\right)$ 满足条件(4)，$U\left(x,y,u\left(x,y\right)\right)$ 为$\left[0,\infty \right)\times \left[0,\infty \right)\times \left[0,\infty \right)$ 上的连续函数，$f\left(x,y,u\left(x,y\right)\right)$ 为$\left[0,\infty \right)\times \left[0,\infty \right)\times \left[0,\infty \right)$ 上的对$y$ 可微的连续函数，$g\left(x,y,u\left(x,y\right)\right)$ 为$\left[0,\infty \right)\times \left[0,\infty \right)\times \left[0,\infty \right)$ 上的对$x$ 可微的连续函数，那么方程(3.1)的所有解如下。

当$0<m<1$ 时，

$u\left(x,y\right)\le {\phi }^{-1}\left\{W\left(x,y\right){\left[c^{1-m}+\left(1-m\right){\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right)W^m\left(s,t\right)\text{d}t}}\text{d}s\right]}^{\frac{1}{1-m}}\right\};$

当$m=1$ 时，

$u\left(x,y\right)\le {\phi }^{-1}\left\{W\left(x,y\right)\exp \left[\ln \begin{array}{c}c\end{array}+{\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right)W\left(s,t\right)\text{d}t}}\text{d}s\right]\right\}.$

其中

$c=\sup \left\{\phi \left(p\left(x\right)\right)+\phi \left(q\left(y\right)\right)|x>0,y>0\right\},$

$W\left(x,y\right)=E_{\alpha }\left(\Gamma \left(\alpha \right)x^{\alpha }\right)E_{\beta }\left(E_{\alpha }\left(\Gamma \left(\alpha \right)x^{\alpha }\right)\Gamma \left(\beta \right)y^{\beta }\right).$

证 若方程(3.1)有解，则它满足下述积分方程

$\begin{array}{l}\phi \left(u\left(x,y\right)\right)\le \phi \left(p\left(x\right)\right)+\phi \left(q\left(y\right)\right)+{\displaystyle {\int }_0^{x}f\left(s,y,u\left(s,y\right)\right)\text{d}s}\\ +{\displaystyle {\int }_0^{y}g\left(s,y,u\left(s,y\right)\right)\text{d}t}+{\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}U\left(s,t,u\left(s,t\right)\right)\text{d}t}}\text{d}s,\end{array}$ (3.2)

由(i)得

${\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}U\left(s,t,u\left(s,t\right)\right)\text{d}t}}\text{d}s\le {\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right){\phi }^m\left(u\left(x,y\right)\right)\text{d}t}}\text{d}s,$

由(ii)知$p\left(x\right)$ ，$q\left(y\right)$ 在$\left[0,\infty \right)$ 上连续有界，取$c=\sup \left\{\phi \left(p\left(x\right)\right)+\phi \left(q\left(y\right)\right)|x>0,y>0\right\}$ ，则有 $\phi \left(p\left(x\right)\right)+\phi \left(q\left(y\right)\right)\le c$ ，因此

$\begin{array}{l}\phi \left(f\left(x,y\right)\right)\le c+{\displaystyle {\int }_0^{x}f\left(s,y,u\left(s,y\right)\right)\text{d}s}+{\displaystyle {\int }_0^{y}g\left(s,y,u\left(s,y\right)\right)\text{d}t}\\ +{\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right){\phi }^m\left(u\left(s,t\right)\right)\text{d}t}}\text{d}s,\end{array}$

在(iii)两边对$x$ 在$\left[0,x\right]$ 上积分，以及在(iv)两边对$y$ 在$\left[0,y\right]$ 上积分，可得

${\displaystyle {\int }_0^{x}f\left(s,y,u\left(s,y\right)\right)\text{d}s}\le {\displaystyle {\int }_0^x{\left(x-s\right)}^{\alpha -1}\phi \left(u\left(x,y\right)\right)\text{d}s},$

${\displaystyle {\int }_0^{y}g\left(x,t,u\left(x,t\right)\right)\text{d}t}\le {\displaystyle {\int }_0^y{\left(y-t\right)}^{\beta -1}\phi \left(u\left(x,t\right)\right)\text{d}t}.$

从而由(3.2)得到

$\begin{array}{l}\phi \left(u\left(x,y\right)\right)\le c+{\displaystyle {\int }_0^x{\left(x-s\right)}^{\alpha -1}}\phi \left(u\left(s,y\right)\right)\text{d}s+{\displaystyle {\int }_0^y{\left(y-t\right)}^{\beta -1}}\phi \left(u\left(x,t\right)\right)\text{d}t\\ +{\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right){\phi }^m\left(u\left(s,t\right)\right)\text{d}t}}\text{d}s.\end{array}$

再由定理1以及定理2，取$A_1\left(x,y\right)\equiv A_2\left(x,y\right)\equiv 1$ ，可证得上述结论。

4. 讨论分析

4.1. 关键参数$\alpha$ ，$\beta$ ，$m$ 对上界函数行为的影响

本文研究的积分不等式的核心在于需要找到一个已知函数$U\left(x,y\right)$ ，使得$\phi \left(u\left(x,y\right)\right)\le U\left(x,y\right)$ 。参数$\alpha$ ，$\beta$ ，$m$ 直接影响这个上界函数$U$ 的增长行为。

$\alpha$ 和$\beta$ 控制着奇异核${\left(x-s\right)}^{\alpha -1}$  和 ${\left(y-t\right)}^{\beta -1}$ 的奇异性强度。当$\alpha ,\beta \to 1^{-}$ 时，奇异核退化为常规核(值为1)，奇异性最弱；当$\alpha ,\beta \to 0^{+}$ 时，奇异核变为${\left(x-s\right)}^{-1}$ 和${\left(y-t\right)}^{-1}$ ，这是可积奇点的临界情况，奇异性最强。较小的$\alpha$ ，$\beta$ 会导致上界$U\left(x,y\right)$ 增长更快。在证明中，我们通常需要应用奇异Gronwall不等式或分数阶积分的性质，最终的上界解通常会包含Mittag-Leffler函数$E_{\alpha ,1}\left(c{x}^{\alpha }\right)$ 或其广义形式。Mittag-Leffler函数$E_{\alpha ,1}\left(z\right)$ 的增长阶次是$\exp \left(z^{1/\alpha }\right)$ 。这意味着当$\alpha =1$ 时，上界是标准的指数增长${\text{e}}^{cx}$ ；当$\alpha <1$ ，上界是超指数增长。这表明奇异核的记忆效应和累积效应远强于常规核，导致解的增长在有限区域内可能变得非常大。

非线性幂次$m$ 决定了非线性项${\phi }^m\left(u\right)$ 的强度。当$m=1$ (线性情况)时，不等式通常能推导出一个全局存在的上界，即解在任意有限区域内都不会趋于无穷；当$m>1$ (超线性情况)时，可能导致解的有限时间爆破，即初始值$c$ 很小，超线性项${\phi }^m$ 会自我放大，导致积分项的增长速度超过线性项，从而在某个有限时刻$\left(X^{\ast },Y^{\ast }\right)$ 使得上界$U\left(x,y\right)\to \infty$ 。

4.2. 假设条件的必要性及放宽可能性

首先是$\phi$ 严格单调递增的必要性，证明这类不等式的标准方法是迭代法或单调迭代法。本研究需要将不等式两边同时作用${\phi }^{-1}$ ，如果$\phi$ 不是严格单调的，则其逆函数${\phi }^{-1}$ 不存在或不唯一，整个证明框架崩溃。在迭代过程中会遇到形如$\phi \left(u\right)<V$ 的式子，只有$\phi$ 单调递增才能安全地推出$u<{\phi }^{-1}\left(V\right)$ ，如果单调性不成立，不等号方向可能反转，导致推导错误。

考虑其条件放宽可以为弱化非严格单调，即如果$\phi$ 是非严格单调(即单调不减)，但在某些点上是常数，那么在这些平坦区域，${\phi }^{-1}$ 是一个集合。这使得分析变得极其复杂，通常需要引入其他假设(如$u$ 的有界性)来绕过求逆。或是放弃全局单调性，改为局部单调性。如果只关心解的局部存在性，可以假设$\phi$ 在某个区间内是Lipschitz连续且单调的，但这会大大限制结果的应用范围。严格单调递增是一个关键技术性假设，很难被完全移除。没有它，现有的证明方法基本不适用。

其次是函数$A_1$ ，$A_2$ ，$H$ 的非负性，这是为了应用积分不等式理论中的比较定理。如果这些系数函数可正可负，那么积分项可能相互抵消，无法保证最终推导出的上界是有效的。非负性确保了每一项都对不等式的右边有“增加”的贡献，从而可以安全地应用Gronwall型论证。

考虑其条件放宽可以使某些系数为负，但问题会变得非常困难。一种可能的方法是要求这些函数有上界，并且其负值部分的影响可以被其他项控制，但这通常需要问题具有特殊的结构，并会引入额外的复杂性。

4.3. 退化情形与经典结果的联系

[i] 令$A_2\left(x,y\right)=0$ ，$H\left(s,t\right)=0$ ，且问题与$y$ 无关(或视$y$ 为固定参数)。即

$\phi \left(u\left(x\right)\right)\le c+A_1\left(x\right){\displaystyle {\int }_0^x{\left(x-s\right)}^{\alpha -1}}\phi \left(u\left(s\right)\right)\text{d}s.$

这就是分数阶微分方程和Volterra积分方程中经典的一维Henry-Gronwall不等式。其结果为

$\phi \left(u\left(x\right)\right)\le c{E}_{\alpha }\left(A_1\left(x\right)\Gamma \left(\alpha \right)x^{\alpha }\right).$

其中$E_{\alpha }$ 是Mittag-Leffler函数。

[ii] 在[i]的基础上，进一步令$\alpha =1$ (奇异核消失)。即

$\phi \left(u\left(x\right)\right)\le c+A_1\left(x\right){\displaystyle {\int }_0^x\phi \left(u\left(s\right)\right)\text{d}s.}$

这就是标准的Gronwall-Bellman不等式。若$\phi \left(u\left(x\right)\right)=u\left(x\right)$ ，其结果为

$u\left(x\right)\le c\exp \left({\displaystyle {\int }_0^x{A}_1\left(s\right)\text{d}s}\right).$

[iii] 令$A_1\left(x,y\right)=A_2\left(x,y\right)=0$ 。即

$\phi \left(u\left(x,y\right)\right)\le c+{\displaystyle {\int }_0^x{\displaystyle {\int }_0^{y}H\left(s,t\right){\phi }^m\left(u\left(s,t\right)\right)\text{d}t}}\text{d}s.$

这是一个二维的非线性Volterra不等式。当$m=1$ 时，它可以通过迭代求解；当$m>1$ 时，它是研究解爆破现象的经典模型。

基金项目

国家自然科学基金项目(12161060)；广西自然科学基金项目(2023GXNSFAA026204)。

NOTES

^*第一作者。

^#通讯作者。