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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.

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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.

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  • $\begingroup$ 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 $\lambda$. $\endgroup$ Commented Dec 10, 2017 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
    Commented Dec 10, 2017 at 23:06
  • $\begingroup$ Separation of variables is possible here. $\endgroup$ Commented Dec 12, 2017 at 10:22

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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).

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  • $\begingroup$ 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
    Commented Dec 11, 2017 at 10:19
  • $\begingroup$ I don't know if the actual symbol has a name. I usually read is as “is much less than”. $\endgroup$ Commented Dec 11, 2017 at 11:34

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