How to model this problem to a differential equation?

Consider a sexually reproducing population. Assume the gender ratio of the population is constant so the population size can be assumed to be $N$. The encounter rate is proportional to the square of population size so $N^2$. On encounter, we assume animals produce offspring at a rate that decreases linearly with population size $N$ from a population capacity $L$, as in the logistic growth model with per capita birth rate $\beta$

Use these assumptions to construct the growth rate in terms of $N$, $L$ and $\beta$

• What did you try? Where are you stuck? May 3 '17 at 23:54
• So what I have written down is $N^2 (L-N)$ I'm thinking L is the limiting factor so I want my population to start decreasing when L is bigger than N but I'm neglecting $\beta$ I don't know what to do with it
– Lock
May 3 '17 at 23:56
• What you have so far is the encounter rate times the population pressure factor. What else in your model affects births? How would you incorporate it into what you have so far? Once you've done that you need to model how the population is affected by births per time. Apparently animals don't die in your model, they just stop reproducing once L is reached. This fact gives you a hint as to what graph of population per time will look like. BTW, your differential equation will be similar to that for charging a capacitor - as it "fills up" it takes more "oompf" to get more charge in it.
– Χpẘ
May 4 '17 at 0:38
• This is the first part. The 2nd part asks me to draw the decay part of the model which is easier. I don't understand what β is intuitively and is it supposed to be (L-N) or L/N since L-N can give me negative growth which is a contradiction and hence doesn't make sense.
– Lock
May 4 '17 at 0:47

I think the expected answer is $\beta N^2 (1 - N/L)$. Here is how to obtain that.
$$\left(\text{growth rate}\right) = \left(\text{encounter rate}\right) * \left(\text{birth rate}\right)$$ Since the (encounter rate) is given as $N^2$, we have the first term. Now the logistic growth looks like this $r(1-x/K)$, where $r$ is the intrinsic growth rate (or per capita growth rate - your $\beta$ basically), then you have your birth rate is $\beta (1 - N/L)$ in an analogous way. Thus you obtain your growth rate.
Well, here is the problem. Although no model is perfect, this way of constructing the growth term is questionable. If you notice, the solution is dimensionally incorrect for a growth rate. The main problem lies in the shortcut/assumption that the encounter rate is simply $N^2$.
One extra bit of information that makes this model slightly different than others is the constant ratio of male to female populations. Let's give it a name, say $\alpha$, and also call the male population $M$ and female population $F$, then $\frac{M(t)}{F(t)} = \alpha$, or $M = \alpha F$. Since $M + F = N$, then $\alpha F + F = N$, or the proportion of female in the total population is: $$\frac{F(t)}{N(t)} = \frac{1}{1 + \alpha}$$ Assume homogeneous mixing of the population, which means the chance of encounter a female is the same everywhere, then the encounter rate is the product of (the total male population) times (the proportion of female in the total population), which means: $$\left(\text{encounter rate}\right) = \frac{1}{1+\alpha}M(t)$$ Thus what I believe should be the correct answer is: $$\frac{\beta}{1+\alpha}M(t)\left(1-\frac{N(t)}{L}\right)$$ Additionally, using $M = \alpha F = \alpha \frac{N}{1+\alpha}$, this becomes $$\frac{\beta\alpha}{(1+\alpha)^2}N(t)\left(1-\frac{N(t)}{L}\right)$$
This has the correct dimension for a growth rate and also contains the ratio term $\alpha$, a given information that is not justly used (therefore it is not the answer that is expected).