Consider the following system of equations, and assume that population of prey is measured in thousands, and that the population of predators is measured in hundreds.
$$\frac{dx}{dt} = 5x(1-\frac{x}{3}) - xy$$ $$\frac{dy}{dt} = -9y + 4xy$$
(Solved parts a, b and c which was to figure out the equilibrium etc).
(d) Adjust the system for the case where the prey species has no environmental limitations. Find the equilibrium points for this model and determine the qualitative behavior in this case1 . Explain why the absence of environmental limitations should affect the model in this way.
- For this part, is the idea to remove 5x from the first equation? Then solve the two equations to figure out the equilibrium points and that should be my answer? My attempt to theoretically explain it is by saying that there are environmental restraints to population growth ie how some species can use environment to survive etc.
(e) Suppose that we now return to the original system and begin to harvest by catching a certain proportion of each species. Refer to the equations below,and determine just from observing the coefficients in the equations which species, the prey or predator, should be more affected by the harvesting. Explain your answer.
$$\frac{dx}{dt} = 5x(1-\frac{x}{3}) - xy - 0.20x$$ $$\frac{dy}{dt} = -9y + 4xy - 0.05y$$
- Shouldn't the prey be more harmed due to the harvesting because of the new proportions given to us. I'm not sure but it seems like the 0.20 would eliminate the prey species much faster.
Thank you for the help!
d) The prey equation is $x'=5x\left(1-\frac{x}{3}\right)-xy.$ Environmental limitations don't really have anything to do with the predators, so let's temporarily ignore them, and set $y=0$. Then, we get $x'=5x\left(1-\frac{x}{3}\right)$. This is a logistic differential equation with intrinsic growth rate $5$ and carrying capacity $3$. This equation describes a population whose growth is self-limited (possibly by limitations on resources, etc.) In general, it comes from the following assumptions:
The interpretation of the second assumption is that, because the population requires resources to sustain itself, and there is only a finite amount of resource produced per unit time, it stands to reason that with a larger population, each individual will get less resources to themself, and thus reproduce less often, produce fewer offspring, and die quicker than if there were infinite resources. The linear assumption just makes the math easier. All of this is captured in the carrying capacity; if the carrying capacity was infinite, there would be no limitations on resources (read: environmental limitations). You should be able to answer part d) now. If you want to know more about the logistic equation and how it is derived, check this link.
e) In a naive sense, yes, the effect of harvesting is stronger on the prey than the predator. Comparing the coefficients, 0.2 is greater than 0.05, and since the prey are measured in thousands while predators are in hundreds, the rate of loss for prey is $$\frac{0.2}{\mbox{time}}\times 1000x\cdot\mbox{prey}=200x\frac{\mbox{prey}}{\mbox{time}},$$ while that of predators is $$\frac{0.05}{\mbox{time}}\times 100y\cdot\mbox{predators}=5y\frac{\mbox{predators}}{\mbox{time}}.$$ So, the way the variables are scaled, the prey experience a higher proportional loss due to harvesting than the predators.
Your answer is correct, though. I just wanted to clarify that the scaling does matter. For example, if prey were in hundreds but predators were in thousands, and the differential equations were the same, the predators would be in worse shape than the prey.
Also, I say the above analysis is naive because the harvesting could very well wipe out the predator entirely while leaving the prey population smaller, but non-extinct. To know what exactly happens, more analysis is necessary.