Timeline for Finding an asymptotic solution for a first order ODE
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Oct 24, 2018 at 16:16 | comment | added | Jean Duchon | Yes, it's more subtle than I thought... | |
| Oct 24, 2018 at 16:15 | history | edited | Jean Duchon | CC BY-SA 4.0 |
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| Oct 20, 2018 at 11:46 | comment | added | Mor | 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. | |
| Oct 19, 2018 at 11:02 | history | answered | Jean Duchon | CC BY-SA 4.0 |