# Finding an asymptotic solution for a first order ODE

Given strictly concave function $$f(t)$$ that satisfies $$f'(t)>0$$, $$f'(t)=o(1)$$ (i.e. $$\lim\limits_{t\to\infty}f'(t)=0$$) , and $$f'(t)=\omega\left(t^{-1}\right)$$ (i.e. $$\lim\limits_{t\to\infty}tf'(t)=\infty$$) , 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 $$\lim_{s\to\infty}\frac{H^{-1}(s)}{F^{-1}(s)}=1\,, \ \ \ \ (a)$$ where $$F^{-1}$$ 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.

* In this question, $$(a)$$ was proved for $$f'(t)=\Omega(1)$$ (without the assumption that $$f(t)$$ is strictly concave). I'm interested in the case that $$f'(t)=o(1)$$.

• By $\dot{H}$ you mean the derivative of $H$, or something else ? – N. de Rancourt Oct 18 '18 at 19:19
• Yes, the derivative of $H$. – Mor Oct 18 '18 at 19:22

Not a full answer, but a hint at a negative answer: take $$f(t)=\log t$$, which doesn't satisfy $$\lim_{t\to\infty}tf'(t)=\infty$$, critically, but is nearer to the proven case (where $$f'(t)=\omega(1)$$ ).
In this case you have $$F(t)=t$$, $$H(t)=t^2/2$$, $$F^{-1}(s)=s$$, $$H^{-1}(s)=\sqrt{2s}$$.
• Thank you for your comment. I agree that eq. (a) doesn't hold for $f(t)=\log(t)$ which doesn't satisfy the condition $f'(t)=\omega(t^{-1})$. However, I think that with this condition the equation should be true. We have that $\dot{H}(t)=F(t)=\exp(f(t))$. If $\exp(f(t))\approx\exp(f(t))f'(t)=\frac{d}{dt}\exp(f(t))$ then I think we can approximate $H\approx \exp(f) = F$. For this to hold we need $\log(f’(t))=o (f (t))$ (i.e. $\lim\limits_{t\to\infty}\frac{\log(f’(t))}{f(t)}=0$). Note that $f(t)=\log(t)$ doesn’t satisfy this requirement, but $f(t)=\log^\epsilon(t)$ for $\epsilon>1$ does. – Mor Oct 20 '18 at 11:46