My lecturer kinda brushed over this but my constants just aren't cancelling out properly.

For the 'ordinary' logistic differential equation we have $dN/dt= N(t)[a - bN(t)]$. He scales this by replacing $N$ with $bN/a$ and $t$ with $at$. In the end he ends up with $dN/dt = N(t)[1 - N(t)]$. I tried using $t^*=at$ but I seem to have a $b^3$ term on the right hand side then that I can't get rid of.

Then for the delay differential equation, it's similar but with $dN/dt = N(t)[a - bN(t-T)]$. He uses the above scalings, additionally replacing $T$ with $aT$. He got $dN(t)/dt = N(t)[1 - N(t-T)]$. I can't arrive at this solution, could somebody give me a step by step answer because all the things I've found online didn't help much. Thank you!

Ps. Really sorry for not having it in a better format, I couldn't figure out the code to make the fraction and powers :|


Multiply each side of the equation by the factor $b/a^2$

\begin{eqnarray} \frac{b}{a^2}\frac{dN}{dt} &=& \frac{b}{a^2}N(t)\left[a - b N(t) \right]\\ \Rightarrow~~~\frac{d(\color{red}{bN/a})}{d(\color{blue}{at})} &=& \color{red}{\frac{b}{a}N(t)}\left[\frac{a}{a} - \color\red{\frac{b}{a} N(t)} \right]\\ \Rightarrow~~~\frac{d\mathcal{N}}{d\tau} &=& \mathcal{N}[1 - \mathcal{N}] \end{eqnarray}


$$ \mathcal{N} \stackrel{\rm def}{=} \frac{bN}{a} ~~~~\mbox{and}~~~~ \tau \stackrel{\rm def}{=} at $$

I will leave the delayed version for you to work out

  • $\begingroup$ Thank you! That really helps- I had the b/a^2 from differentiating bits of it but I didn't know how to put it into the whole thing. I think I've done the delay one now as well :) $\endgroup$ – Gabriela G Nov 30 '17 at 11:43
  • $\begingroup$ @GabrielaG Great, happy to help $\endgroup$ – caverac Nov 30 '17 at 12:11

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