Marilyn Cameron

2022-10-26

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≫A and N≪A explain why this fraction models the situation outlined above.

Attempt:

When N≫A (my assumption is that the sign '≫'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≪A (my assumption is that the sign '≪' 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≪A )

For a large population finding a mate is more likely (a.k.a N≫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:

$$\frac{dN}{dt}=r\phantom{\rule{thinmathspace}{0ex}}\frac{N}{A+N}\phantom{\rule{thinmathspace}{0ex}}N,$$

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.

$$\frac{dN}{dt}=rS(N)N,$$

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≫A and N≪A explain why this fraction models the situation outlined above.

Attempt:

When N≫A (my assumption is that the sign '≫'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≪A (my assumption is that the sign '≪' 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≪A )

For a large population finding a mate is more likely (a.k.a N≫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:

$$\frac{dN}{dt}=r\phantom{\rule{thinmathspace}{0ex}}\frac{N}{A+N}\phantom{\rule{thinmathspace}{0ex}}N,$$

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.

Kash Osborn

Beginner2022-10-27Added 18 answers

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≫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)≪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).

It's just that the population grows slower to begin with (if N(0)≪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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