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I was revising for my Math Modelling term paper, and was doing this particular question in a textbook, where I got stuck for a whole day. Here's a snippet:

Q1 Reference: Topics in Mathematical Modeling by K.K Tung

Chris
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1 Answers1

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Hints:

For part a), for individual subpopulation equilibrium, you need: $\displaystyle \frac{dm}{dt} = \frac{df}{dt} = 0$

That gives you:

$$B_f\frac{m^*f^*}{m^*+f^*} = D_ff^*$$

$$B_m\frac{m^*f^*}{m^*+f^*} = D_mm^*$$

Can you manipulate these to eliminate $f^*$ and $m^*$? Try to get an equation in terms of the birth and death rates alone. That is the required condition.

For the second part, let $\displaystyle \frac mf = k$, i.e. $m = kf$. Rewrite the two differential equations only in terms of $f$ and $k$ (use product rule where needed).

Now set $\displaystyle \frac{dk}{dt} = 0$ to find $k^*$. Remember that for this part, $\displaystyle D_f = D_m = D$.

EDIT: I'd like to have hidden this part under a spoiler tag, but I'm not sure how to do it. If anyone wishes to help me by editing this, please feel free.

To find the equilibrium subpopulations for part a), simplify and rearrange the two equations to:

$$\frac{m^*}{m^*+f^*} = \frac{D_f}{B_f}$$

$$\frac{f^*}{m^*+f^*} = \frac{D_m}{B_m}$$

Now add the two to get:

$$\frac{D_f}{B_f} + \frac{D_m}{B_m} = 1$$

which is the required condition.

For part b), by employing product rule to get $\displaystyle \frac{dm}{dt} = k\frac{df}{dt} + f\frac{dk}{dt}$ and then setting $\displaystyle \frac{dk}{dt} = 0$ when $k = k^*$,

we get:

$$\frac{df}{dt} = B_f\frac{k^*}{k^*+1}f - Df$$

and

$$k^*\frac{df}{dt} = B_m\frac{k^*}{k^*+1}f - k^*Df$$

where $D_m$ and $D_f$ have been collapsed to a single parameter $D$.

Multiply the first equation by $k^*$ and equate to get:

$$k^*B_f\frac{k^*}{k^*+1}f -k^*Df = B_m\frac{k^*}{k^*+1}f - k^*Df$$

which simplifies to give $\displaystyle k^* = \frac{B_m}{B_f}$

Deepak
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  • @Chris I changed it to $k$ to avoid confusion. $\frac{m}{f} = k$ – Deepak Oct 22 '16 at 11:57
  • Alright, thank you so much for helping, I will work on it now :) – Chris Oct 22 '16 at 11:58
  • Also, you don't have to delete all your typing hard work. You should restore it and format it using MathJax. Just follow the wording of the actual question. – Deepak Oct 22 '16 at 11:59
  • Alright I will do that after I have solved this :) I have tried to eliminate m* and f* for part a, but I am still stuck. I tried multiplication on both sides, addition and so on, can't seem to get it. – Chris Oct 22 '16 at 12:53
  • I have found it already, is it along the lines of (Bf - Df)/Df = Dm/(Bm - Dm) – Chris Oct 22 '16 at 13:05
  • @Chris I'm not sure how you're getting that. But I edited my post to show you what I got. – Deepak Oct 22 '16 at 13:28
  • omg. I really just made my life way worse than it was before, that totally made logical sense, now thank you :) – Chris Oct 22 '16 at 13:32
  • @Chris Sorry, I just rearranged your equation, and it's equivalent to mine. You can simplify yours to mine easily. :) – Deepak Oct 22 '16 at 13:33
  • I just noticed, part a requires me to find a non-trivial equilibrium, does that mean that m and f must both be non-zero? – Chris Oct 22 '16 at 13:35
  • @Chris I believe that's implied, but since we're dividing (cancelling) by those quantities, we're taking them as non-zero. :) – Deepak Oct 22 '16 at 14:30
  • I got m* and f* = BC and AD respectively, is that correct? – Chris Oct 22 '16 at 14:31
  • Sorry for bothering you so much, you've been a life-saver I assure you Deepak :) – Chris Oct 22 '16 at 14:33
  • If my m* and f* is wrong, can you kindly advise me again, I'm really at my wits end :( – Chris Oct 22 '16 at 14:34
  • I also can't seem to get the suggestion of Product Rule for part b – Chris Oct 22 '16 at 14:59
  • Could you clarify what A and C are? I don't think it's possible to determine $m^$ and $f^$ in part a). Which is why the exercise said "try to...". Basically, when the condition is met, the system of two equations becomes redundant (it's equivalent to just one equation). As such, any nonzero values of $m^$ and $f^$ are possible, with only the ratio between them being fixed by the constants. Unless I'm much mistaken, you cannot find definite values for $m^$ and $f^$ for the first part. – Deepak Oct 22 '16 at 15:19
  • By product rule, I meant using $m = kf$ to write $\frac{dm}{dt} = f\frac{dk}{dt} + k\frac{df}{dt}$. – Deepak Oct 22 '16 at 15:22
  • I chose m* to be Df * Bm and f* to be Mf * Dm – Chris Oct 22 '16 at 15:24
  • I understand that product rule application, but I don't see how I can get k(t) alone, with the dk/dt gone. – Chris Oct 22 '16 at 15:25
  • For part a) I don't see how you can "choose" those values. For part b), your goal is to rewrite both equations in terms of only $k$ and $f$. Then set $\frac {dk}{dt} = 0$, which occurs when $k = k^$. Now eliminate everything except $k^$ and the constants. – Deepak Oct 22 '16 at 15:32
  • I've edited my answer to work out part b) as well. – Deepak Oct 22 '16 at 15:39
  • I have to turn in now. Goodnight, and good luck on your term paper. :) – Deepak Oct 22 '16 at 15:58
  • Thank you :) I'll try my best! – Chris Oct 22 '16 at 16:00
  • Hi Deepak, sorry to bother you again, but how would I know that the k* is a stable equilibrium? – Chris Oct 23 '16 at 04:32
  • I figured it out :) – Chris Oct 23 '16 at 07:16
  • @Chris I'm so glad. My apologies as I've been very tied up, I'm flying off in a few hours. Again, best of luck for your paper! :) – Deepak Oct 23 '16 at 10:43