1
$\begingroup$

Consider the modification to the Malthusian equation $$\frac{dN}{dt}= rS(N)N ,$$ where $r > 0 $ is the per capita growth rate, and $S(N)$ is a survival fraction. For some organisms, finding a mate at low population densities may be difficult. In such cases, the survival fraction can take the form $S(N) = \frac{N}{A + N}$, where $A > 0$ is a constant.

(a)

i. By examining what happens to $S(N)$ for $N \gg A$ and $N \ll A$ explain why this fraction models the situation outlined above.

Attempt:

When $N \gg A$ (my assumption is that the sign '$\gg$'means significantly greater than ) so $A$ is much smaller than $N$ then the constant $A$.This would effect the survival fraction in a way which it becomes roughly equal to $1$ since $A$ is really small.

When $N\ll A$ (my assumption is that the sign '$\ll$' means significantly smaller than) so $N$ is much smaller than $A$ will impact the survival fraction in a way which will make it become small.

The survival fraction models matches with the situation above since:

For a low population finding a mate is difficult (a.k.a $N\ll A$ )

For a large population finding a mate is more likely (a.k.a $N\gg A$ )

ii. By examining the form of the equation, determine the long-term behaviour of a population for an initial condition $N(0) > 0$.

Attempt: For the initial conditions stated above the Malthusian equation has the form: $$\frac{dN}{dt}= r\, \frac{N}{A + N}\, N,$$ which has the solution.

$$N(t)=N_{0}e^{\lambda t}$$ If $\lambda>0$ then exponential growth

If $\lambda <0$ then exponential decay

Please could you check if this is correct. If it is please can you suggest any improvement I could add to my attempts.

$\endgroup$
  • $\begingroup$ What I definetly can say is that $N(t)=N_0e^{\lambda t}$ isn't the solution of $$\dot{N}(t)=r\cdot \frac{N(t)}{A+N(t)}\cdot N(t)$$ for any $\lambda$. $\endgroup$ – Fakemistake Dec 10 '17 at 22:48
  • $\begingroup$ Thank you for the reply, am I right in saying I will have to use separation of variables to determine the solution subject to the initial value. $\endgroup$ – odesinit Dec 10 '17 at 23:06
  • $\begingroup$ Separation of variables is possible here. $\endgroup$ – Fakemistake Dec 12 '17 at 10:22
1
$\begingroup$

You don't need to find the exact solution in the second part. The right-hand side is positive for all $N>0$, so there are no equilibria except $N=0$. Hence the solution will increase indefinitely, and for $N \gg A$ you have $\dot N \approx kN$ (as you showed in the first part), so you will end up with approximately exponential growth. It's just that the population grows slower to begin with (if $N(0) \ll A$), since you have $\dot N \approx \frac{r}{A} N^2$ for small $N$ (as you also, more or less, showed in the first part).

$\endgroup$
  • $\begingroup$ Hi Hans thanks for taking your time to check my attempt. Your method is much more clearer and it make sense to use the answer from the previous part. I was over thinking this one. By the way what is this symbol called $\ll$? $\endgroup$ – odesinit Dec 11 '17 at 10:19
  • $\begingroup$ I don't know if the actual symbol has a name. I usually read is as “is much less than”. $\endgroup$ – Hans Lundmark Dec 11 '17 at 11:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.