Simplified question*:

Given $f(t)$ that satisfies $f'(t)>0$, $f'(t)=\omega\left(t^{-1}\right)$, $\log\left(f'(t)\right)=o\left(f(t)\right)$ we denote $F=\exp\left(f\left(t\right)\right)$. Let $H(t)$ be a solution of $$ \dot{H}=F $$ Can we approximate H by F?

Specifically, I want to show that $$ \lim_{s\to\infty}\frac{H^{-1}(s)}{F^{-1}(s+C)}=1 $$ for some constant C. $F^{-1}$ and $H^{-1}$ are the inverse functions of $F$ and $H$, respectively.

If necessary I can add additional assumptions on $f(t)$, though I would like to keep it as general as possible.

* thanks to Willie Wong helpful suggestion.

The original question (to make the motivation more clear):

I have the following differential equation: $$\dot{g}(t)=\exp\left(-f\left(g\left(t\right)\right)\right)$$ and I know that $f(t)$ satisfies $f'(t)>0$, $f'(t)=\omega\left(t^{-1}\right)$, $\log\left(f'(t)\right)=o\left(f(t)\right)$.

I would like to show that the solution to this equation is $$g(t)=f^{-1}(\log(t+C))+h(t)$$ where $h(t)=o\left( f^{-1}(\log(t)) \right)$, $f^{-1}(t)$ is the inverse function of $f(t)$ and C is some constant.

The motivation behind this solution is $$ \exp\left(f\left(g(t)\right)\right)\approx\exp\left(f\left(g\left(t\right)\right)+\log\left[ f'\left(g\left(t\right)\right)\right]\right) $$ since $\log\left(f'(t)\right)=o\left(f(t)\right)$, and therefore we can solve $$ \exp\left(f\left(g\left(t\right)\right)+\log\left[ f'\left(g(t)\right)\right]\right)\dot{g}(t)=1 \Rightarrow \frac{d}{dt}\exp\left(f(g\left(t\right)\right)=1 \Rightarrow f(g(t))=\log(t+C)$$ for some constatnt C.

Can I rigorously show this result? If necessary I can add additional assumptions on $f(t)$, though I would like to keep it as general as possible.

  • $\begingroup$ What is your $\omega(t^{-1})$? $\endgroup$
    – user64494
    Aug 28, 2018 at 11:56
  • $\begingroup$ $f(t)=\omega(g(t))\Rightarrow$f dominates g asymptotically en.wikipedia.org/wiki/Big_O_notation $\endgroup$
    – Mor
    Aug 28, 2018 at 12:52
  • 2
    $\begingroup$ May be you can simply your question a little bit: I don't think $g$ plays an important role here. The main question seems to be: given $f(t)$ with the specified properties, denote by $F = e^f$ and let $H$ be a solution of $\dot{H} = F$. Can we approximate $H$ by $F$. Or more precisely, can we approximate $H^{-1}(s)$ by $F^{-1}(s + C)$ as $s \to \infty$. The mode of approximation you want seems to be requiring $\lim_{s\to \infty} H^{-1}(s) / F^{-1}(s+C) = 1$. $\endgroup$ Aug 30, 2018 at 16:41
  • $\begingroup$ That is a really good suggestion. Thank you. I updated the question. $\endgroup$
    – Mor
    Aug 30, 2018 at 18:48

1 Answer 1


The simplest cases are the linear cases where $f(t) = \alpha t$, and $H = \alpha^{-1}F$. Modelling on those cases, I will show that if asymptotically $f'(t)$ is bounded below (so if $f$ is asymptotically superlinear) than the desired conclusion hold.

The first step is to prove that the conclusion holds if $H$ is eventually $< F$, using merely a convexity argument.

Suppose there exists some $t_0$ such that for every $t > t_0$ we have that $H(t) < F(t)$. Notice that $\dot{H} = F$ and $\ddot{H} = \dot{f} F > 0$ we have that $H$ is convex. We are interested in the quantity

$$ H^{-1}(s) - F^{-1}(s) $$

which is positive for all sufficiently large $s$ when $H$ is eventually $< F$.

By convexity, we have that for any $r < s$

$$ H^{-1}(s) < H^{-1}(r) + (H^{-1})'(r) \cdot (s-r) $$

Since $H < F$ we can shoose $r = H\circ F^{-1}(s)$ and get

$$ H^{-1}(s) < F^{-1}(s) + (H^{-1})'\circ H \circ F^{-1}(s) \cdot (s - H\circ F^{-1}(s)) $$


$$ (H^{-1})' = \frac{1}{H'\circ H^{-1}} $$

so we get

$$ H^{-1}(s) < F^{-1}(s) + \frac{1}{s} \cdot (s - H\circ F^{-1}(s)) $$

So in particular, you have that $H^{-1}(s) - F^{-1}(s)$ is bounded, since $H\circ F^{-1}$ is increasing.

Using that $F^{-1}$ grows unboundedly this implies

$$ H^{-1} - F^{-1} = o(F^{-1})$$

as desired.

Notice that in this argument I have not explicitly used the conditions $f' = \omega(t^{-1})$ and $\log(f') = o(f)$; however, the condition that $H(t) < F(t)$ can be derived from strengthened versions of your assumption.

For example, a sufficient condition to guarantee that $H(t) < F(t)$ eventually for every solution $H$ is that $f'(t) > 1 + \epsilon$ for all sufficiently large $t$.

This can be generalized further. Suppose that $f'(t)$ is lower-bounded by $\delta > 0$.

Define the function $\tilde{f}(\tau) = f(2 \delta^{-1} \tau)$. Define $\tilde{F}$ and $\tilde{H}$ analogously. We have that $\tilde{F} = F(2 \delta^{-1}\tau)$ and $\tilde{H} = \frac{\delta}{2} H(2\delta^{-1}\tau)$

Applying the above argument we get

$$ H^{-1}(2 \delta^{-1} s) - F^{-1}(s) < \frac{2}{\delta} - \frac{1}{s} H\circ F^{-1}(s) $$


$$ F^{-1}(s) = f^{-1} \ln(s) $$


$$ F^{-1}(2 \delta^{-1} s) = f^{-1} ( \ln(s) + \ln 2 - \ln \delta) $$

Using that $f'$ is bounded below by $\delta$, we have that

$$ |F^{-1}(2 \delta^{-1} s) - F^{-1}(s) | < \delta^{-1} \ln(2 \delta^{-1}) $$

is also bounded, so that we conclude

$$ H^{-1}(s) - F^{-1}(s)$$

is bounded, and the unbounded growth of $F^{-1}$ takes care of the rest.

The remaining case is when $\liminf_{t\to\infty} f'(t) = 0$, which is probably the case you are really interested in to boot, but unfortunately I don't yet see a proof for.

  • $\begingroup$ Thank you! Nice proof. One question - why do you need $r<s$? $\endgroup$
    – Mor
    Sep 1, 2018 at 11:06
  • $\begingroup$ @Mor: just my personal laziness. (I didn't want to have to type $\leq$ everywhere, and the function involved in strictly convex, so I can get away without $\leq$ if I rule out $r = s$. So instead of $r \neq s$ I typed $r < s$, which happens to be true for the case we need.) $\endgroup$ Sep 4, 2018 at 14:50
  • $\begingroup$ @WillieWong: Sorry if I'm missing something obvious: where and why do we need to use $r<s$ exactly? Wouldn't the next lines (e.g., the first equation stated after the condition) also hold if $r>s$? $\endgroup$ Sep 8, 2018 at 12:39
  • $\begingroup$ @DanielSoudry: see my previous comment? (I don't need $r < s$; it is that for the construction only the $r < s$ case is needed.) $\endgroup$ Sep 10, 2018 at 1:08
  • $\begingroup$ Ah, yes. Sorry, I misunderstood your previous comment (I thought it explained why you used $r < s$ instead of $r \leq s$). $\endgroup$ Sep 10, 2018 at 6:54

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