Suppose that $$\bex \cfrac{\rd f}{\rd t}+h\leq gf\quad (f,g,h\geq 0,\ t\in [0,T]). \eex$$ Then for $t\in [0,T]$, $$\bex f(t)+\int_0^t h(s)\rd s \leq f(0)\sez{ 1+\int_0^t g(s)\rd s\cdot \exp\sex{\int_0^t g(s)\rd s} }. \eex$$
[再寄小读者之数学篇](2014-06-23 Grownall-type inequality),布布扣,bubuko.com
[再寄小读者之数学篇](2014-06-23 Grownall-type inequality)
原文：http://www.cnblogs.com/zhangzujin/p/3810865.html