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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.

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  • $\begingroup$ people.math.sc.edu/howard/Notes/gronwall.pdf $\endgroup$
    – user69208
    Commented Nov 17, 2017 at 20:28
  • $\begingroup$ Thank you for the link, though I'm not sure this paper answers my question. $\endgroup$
    – H_R
    Commented Nov 17, 2017 at 23:42
  • $\begingroup$ Isn't Sturm-Liouville theory in some sense analogous to the Gronwall inequality for second order ODEs? $\endgroup$
    – Deane Yang
    Commented Nov 18, 2017 at 0:44
  • $\begingroup$ @DeaneYang: Are you referring to oscillation theory? Then yes, there are comparison theorems, but only because one has previously introduced the Prufer angle, which satisfies a first order ODE (you could say that was the point of the transformation). $\endgroup$
    – Christian Remling
    Commented Nov 18, 2017 at 0:54
  • $\begingroup$ I'm not familiar with the Prufer angle. I meant consequences of the Picone inequality, en.wikipedia.org/wiki/Picone_identity. This is indeed a first order differential inequality, but the consequences are for the second order ODE. $\endgroup$
    – Deane Yang
    Commented Nov 19, 2017 at 1:13

1 Answer 1

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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.

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  • $\begingroup$ This is a great counterexample, thanks. However, your function is not twice differentiable where the pieces are glued together. I wonder if the statement I was hoping for is true if we insist the inequality holds at every point (in particular, u is twice differentiable at every point). $\endgroup$
    – H_R
    Commented Dec 2, 2017 at 0:01
  • $\begingroup$ @H_R: That's just because it was easy to describe this way, we can modify the function slightly near the gluing points to obtain a smooth counterexample. $\endgroup$
    – Christian Remling
    Commented Dec 2, 2017 at 2:24
  • $\begingroup$ I see. So instead of gluing each piece together at a point, you would connect them using a tiny piece whose second derivative matches on either side. $\endgroup$
    – H_R
    Commented Dec 2, 2017 at 17:03

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