I've been trying to solve this one for a while, but I still can't make it. Here's the problem.

I have a $f_1(t;\rho,\nu)$ that for $t\to\infty$ and for $\rho>\nu$ goes as $f_1\sim t^{\nu/\rho}.$

Saying that, the problem is with a second function $f_2(t, \rho\, \nu)$ which obeys the equation:

$\frac{df_2(t)}{dt}=\frac{\nu f_2(t) + (\nu+1)f_1(t)}{\rho t + (\nu+1)[f_1(t)+f_2(t)]}$

with $\rho,\nu\in\mathbb{N}^+$ and $f_2(t)>0$ (and increasing).

I am interested in the long term behavior as for $f_1$. I know (from simulations) that for $t\to\infty$ it behaves similarly to $f_1$, but with a different exponent. I've tried to plug in the $f_1$ written above (assuming $\rho>\nu)$ and also to force $f_2(t)\sim ct^\alpha$ as a solution, ultimately looking for an expression for $\alpha$ as a function of the other parameters, but I only arrived to contradict myself.

I've also tried on Mathematica with initial conditions $f_2(0)=0$ but without succeeding.

Any tips?

Thanks in advance!!

  • 1
    $\begingroup$ If this is a differential equation, you don't want $t \in \mathbb N$. $\endgroup$ – Robert Israel Mar 29 at 15:54
  • $\begingroup$ OK thanks Robert. I put it there because in the original process it's defined in discrete times, but here I totally agree that it doesn't make sense ;) $\endgroup$ – Meffo Mar 30 at 17:58

It seems to me that it might have the same exponent, but logarithmic terms. Rather than try to solve the differential equation for $f_2(t)$, I tried plugging in a form for $f_2(t)$ and solving for $f_1(t)$. I found that $$ f_2(t) = \frac{\nu + 1}{\rho} \log(t)\; t^{\nu/\rho} $$ is a solution with $$ f_1(t) = {\frac {1}{\rho} \left( {t}^{{\frac {\nu+ \rho}{\rho}}}{\rho}^{3}+\ln \left( t \right) {t}^{2\,{\frac {\nu}{ \rho}}} \left( \nu+1 \right) ^{2} \left( \ln \left( t \right) \nu+ \rho \right) \right) \left( - \left( \nu+1 \right) \left( \ln \left( t \right) \nu+\rho \right) {t}^{{\frac {\nu}{\rho}}}+{\rho}^{2 }t \right) ^{-1}} \sim t^{\nu/\rho} $$

  • $\begingroup$ Thanks a lot robert for the hint! So, just to be sure, you plugged a $f_2(t)=ct^\alpha$ into the diff. equation and the second Eq. you wrote is the solution you got? I'm missing the bit where you derive then the $f_2(t)$ $\endgroup$ – Meffo Apr 2 at 10:55
  • $\begingroup$ No, I plugged in $f_2(t) = c \log(t) t^{\nu/\rho}$ and solved for $f_1(t)$, then found what $c$ was needed for $f_1(t) \sim t^{\nu/\rho}$. $\endgroup$ – Robert Israel Apr 2 at 12:00

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