- #1
kalish1
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I have a problem with the Lotka-Volterra equations themselves. I believe that they might be wrong. Here is my reasoning - I would appreciate it if someone could find a flaw in it!
The equations are generally of the form, as quoted from "A Modern Introduction to Differential Equations 2nd edition by Henry Ricardo":
$$\frac{dx}{dt} = a_1x-a_2xy, \frac{dy}{dt}=-b_1y+b_2xy$$
**My issue:** The $xy$ terms represent the number of possible interactions between two species. However, they only represent the number of possible *one-on-one* interactions between the two species. In order to account for *all* the possible interactions, such as $(x-1)$ predators acting on $2$ preys, shouldn't we arrive at $$\sum_{k=1}^x\sum_{j=1}^y {x \choose k}{y \choose j} = (2^x-1)(2^y-1)$$ and thus $$\frac{dx}{dt} = a_1x-a_2(2^x-1)(2^y-1), \frac{dy}{dt}=-b_1y+b_2(2^x-1)(2^y-1)?$$
Doesn't this make the number of interactions proportional not to the product of the number of predators and prey, but to their exponentiation?
The equations are generally of the form, as quoted from "A Modern Introduction to Differential Equations 2nd edition by Henry Ricardo":
$$\frac{dx}{dt} = a_1x-a_2xy, \frac{dy}{dt}=-b_1y+b_2xy$$
**My issue:** The $xy$ terms represent the number of possible interactions between two species. However, they only represent the number of possible *one-on-one* interactions between the two species. In order to account for *all* the possible interactions, such as $(x-1)$ predators acting on $2$ preys, shouldn't we arrive at $$\sum_{k=1}^x\sum_{j=1}^y {x \choose k}{y \choose j} = (2^x-1)(2^y-1)$$ and thus $$\frac{dx}{dt} = a_1x-a_2(2^x-1)(2^y-1), \frac{dy}{dt}=-b_1y+b_2(2^x-1)(2^y-1)?$$
Doesn't this make the number of interactions proportional not to the product of the number of predators and prey, but to their exponentiation?