Gronwall's inequality says that solutions to the initial value problem $u'(t) \leq \beta(t)u(t)$ with $u(0)=u\_0$ are bounded by solutions to the problem with inequality replaced with equality for $t\in [0,\infty)$. Is there a way to generalize to higher order derivatives. That is, if $u''(t) \leq \alpha(t)u'(t) + \beta(t)u(t)$ with $u(0)=u\_0$ and $u'(0)=u'\_0$ can we say that the solutions to the corresponding differential equation dominates $u$?


I asked this question on StackExchange but got crickets.



Gronwall's inequality for higher order derivatives
No, comparison of this type only works for first order. Consider for example
$$
u'' \le -u, \quad u(0)=u'(0)=0 .
$$
The statement you were hoping for would here say that such a $u$ satisfies $u(x)\le 0$ for all $x\ge 0$, but this can easily be outmaneuvered. Start out by making $u$ negative; obviously there are no problems with the inequality as long as $u, u''\le 0$. We reach a point with $u=-1, u'=-m<0$, say. Now we can afford to make $u''$ positive. For example, we can move along a parabola to make $u=-1, u'=m$. Next, on to $u=0, u'=m$ along a straight line, and now it's over because you can solve the ODE exactly to produce positive values.


