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$$ \frac{dx}{dt}=ax(x-b)(1-x)-\frac{xy}{1+cx}$$ $$\frac{dy}{dt}=-ey+\frac{xy}{1+cx}$$ Make a apropriate interpretation of this model.

What I thought are: $\frac{dx}{dt}\ \& \ \frac{dy}{dt}$ are the rate of changes of $x$ and $y$, and $ax(x-b)(1-x)=-\frac{a}{b}(1-\frac{x}{b})(1-x)x$ is a logistic model with threshold, but I don't know $b$ and 1, which one is saturation level, which one is threshold level, and I have no idea what does $\frac{xy}{1+cx}$ mean?

Does anyone could help me? Thanks!

user133140
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  • "Logistic model with threshold" is rather ambiguous and deserves some further explanation. The model $x'=ax(x-b)(1-x)$ is indeed similar to a logistic evolution, but with a specific extinction effect in the regime of the small positive populations. To wit, when $t\to\infty$, $x(t)\to0$ for every initial population $x_0<\min(1,b)$, and, as usual, $x(t)\to\max(1,b)$ for every $x_0>\min(1,b)$. Thus, $\max(1,b)$ acts as a carrying capacity, just like in the logistic model, and $\min(1,b)$ acts as an extinction (lower) threshold. – Did Jun 27 '15 at 09:12

1 Answers1

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In this model $\frac{xy}{1+cx}$ is a modified predator-prey interaction. If it were just $xy$ or that with a rate constant, then it would be the usual mass-action predator-prey, but with the factor of $1+cx$ you get less "hunting" when $x$ is large (assuming $c>0$).

So overall we have "wolves" which die exponentially by themselves but grow provided they can hunt, which happens when the "deer" population is in an intermediate range. Meanwhile the "deer" die exactly in step with the wolf growth, and grow according to a logistic-like pattern.

Looking at the last part, assume $0<b<1$. Then for $0<x<b$, the deer are dying (one minus sign); for $b<x<1$, the deer are growing (two minus signs); and for $x>1$ the deer are dying again (three minus signs). Basically the same (but reversing the roles of $b$ and $1$) happens if $b>1$.

Ian
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