Generalized Gronwall Inequality covering many different applications

Question

Recently, I needed some generalized version of Gronwall's Lemma, which I couldn't find in a quick search. However, I discovered that MSE is full of questions differing only in details on this very topic. Hence, I was wondering if there is a general version of Gronwall's inequality which covers most (if not all) of the different cases.

Answer 1

So, this is what I came up with after some time. I don't claim that it is original by any means but I hope that it is still helpful. I'm looking forward to any comments, improvements or other suggestions.

Generalized Gronwall's Lemma: Let $a>0$, $u,v\in C^1((0,a))\cap C([0,a))$ and $F\in C^1(\mathbb R)$. We denote $u_0:=u(0)$. We assume that $u$ satisfies the differential inequality $$u'(t)\leq F(u(t)),\quad t\in (0,a)$$ while $v$ satisfies the differential equation $$\left\{\begin{aligned}v'(t)&=F(v(t)),\quad t\in(0,a)\\v(0)&=u_0.\end{aligned}\right.$$ Then, $u$ satisfies the inequality $$u(t)\leq v(t)$$ for all $t\in(0,a)$.

Proof: We define $G:\mathbb R^2\to\mathbb R$ by $$G(x,y):=\begin{cases}\frac{F(x)-F(y)}{x-y} &\text{if }x\neq y,\\F'(x) &\text{if }x=y.\end{cases}$$ By the mean value theorem, we have $G\in C(\mathbb R^2)$. We further define $w(t):=u(t)-v(t)$ and $\beta(t):=G(u(t),v(t))$ for $t\in (0,a)$. Then, $w$ satisfies the inequality $$w'(t)=u'(t)-v'(t)\leq F(u(t))-F(v(t))=(u(t)-v(t))G(u(t),v(t))=\beta(t)w(t).$$ By the classical statement of Gronwall's lemma and $w(0)=u_0-u_0=0$, we conclude $$w(t)\leq w(0) e^{\int_0^t\beta(s)\mathrm d s}=0$$ and hence $u(t)\leq v(t)$.

Examples:

1. $u'\leq c u^2$: The corresponding equation is solved by $v(t)=\frac{u_0}{1-cu_0t}$ with $a=\frac{1}{cu_0}$ if $cu_0>0$ and $a=\infty$ otherwise. It follows $$u(t)\leq\frac{u_0}{1-cu_0t}$$ Note that the right hand side converges to zero with rate roughly $1/t$ as $t\to\infty$ if $cu_0<0$.
2. $u'\leq -cu+f$ for $c>0$ and a continuous function $f$: The corresponding differential equation is solved by $$v(t)=e^{-ct}\left(u_0+\int_0^te^{cs}f(s)\mathrm d s\right)$$ and again, it follows $u(t)\leq v(t)$. In the special case $u_0=0$ and $f(t)\leq K$ we get $$u(t)\leq Ke^{-ct}\int_0^te^{cs}\mathrm d s=\frac{K}{c}(1-e^{-ct})\leq\frac Kc.$$