-1
$\begingroup$

Set up differential equations or mathematical modeling :

Let in a Island there are $ \ 4 \ $ species $ \ w \ (dragon),x \ (deer),y \ (sheep),z \ (plant) \ $.

If dragon eat deer , deer and sheep eat plant and the plants complete among themselves , then set up (do not solve) a differential equation that represent this system.

Answer:

Since dragon eat deer , the variable $ \ w \ $ is dependent on $ \ x \ $.

Since dragon and sheep eat plant , $ \ x,y \ $ dependent on $ \ z \ $.

$ \therefore \ w \ $ dependent of $ \ x,y,z \ $.

Since plant complete themselves , the variable $ \ z \ $ is independent.

But how to set the differential equation representing the system ?

$\endgroup$
3
$\begingroup$

Let me hazard a guess as to the sort of thought this problem is asking for.

In my estimation, this is as much an exercise about model-building as it is about differential equations. There's not really one right answer, although some answers are better than others based on what makes sense from a physical perspective.

By way of example, consider a system in which wolves (whose quantity is represented by $W$) eat sheep, sheep ($S$) eat cabbages, and cabbages ($C$) grow. How does the population evolve over time?

Suppose that cabbages grow at a constant rate $\lambda_C$. At the same time, though, they're being eaten by sheep, at a rate $f(C, S)$ that depends on both the number of cabbages and the number of sheep. So we write

$$ \frac{dC}{dt} = \lambda_C - f(C, S) $$

One possible form of $f(C, S)$ that looks reasonable is $u\frac{CS}{C+S}$: It ranges from $0$, when there are no cabbages left, to a limit of $u$, when cabbages are plentiful (on a per sheep basis).

OK, now what about the sheep? They would probably breed at a rate proportional to the number of sheep; we can denote this rate by $\nu_S$. The sheep are also vulnerable to being eaten by the wolves, and as with the cabbages, we can represent this with a function $g(S, W)$.

In addition, however, we must consider the possibility that sheep will starve to death if there aren't enough cabbages to go around. This happens at a rate $q(C, S)$ that depends on both the number of cabbages and the number of sheep. So we write

$$ \frac{dS}{dt} = \nu_S S - g(S, W) - q(C, S) $$

One possible choice for $q(C, S)$ is $v\frac{S^3}{C^2+S^2}$. This ranges from nearly $0$, when cabbages are plentiful (sheep can die even then) to $v$ (per sheep) when no cabbages are left.

Finally, wolves breed at a rate proportional to the number of wolves, and we'll denote this rate by $\nu_W$. Nothing eats the wolves, but they can starve because there aren't enough sheep for them to eat. So we write

$$ \frac{dW}{dt} = \nu_W W - r(S, W) $$


That sort of model-building seems like what is being asked for. As you can tell, much of the intellectual effort is spent on constructing a reasonable model; the amount of it expended on expressing it in the language of differential equations is relatively small.

Note that we might well have constructed a system of equations such as

$$ \frac{dC}{dt} = W+S \\ \frac{dS}{dt} = C+W \\ \frac{dW}{dt} = S+C $$

Such a system perhaps could not be considered "wrong" per se, but it would be hard to justify it on physical grounds.

$\endgroup$
  • 1
    $\begingroup$ It's not a very good model, sheep continue eating cabbages even if their number is negative, etc. If the left side were $\dot C/C$ and $\dot S/S$ the variables would be forced to stay positive. $\endgroup$ – LutzL Jul 12 '18 at 20:23
  • $\begingroup$ @LutzL: Haha, well, I suppose my excuse is that my intention was not to construct a good model, but to highlight that model construction was the central objective of the question. But I concede it's not a very good excuse. I'll come back and fix the model later, thanks for the observation! $\endgroup$ – Brian Tung Jul 12 '18 at 20:43
  • $\begingroup$ This is nicely explained. Note that we could have second order equations, for instance in many ecosystems when predation pressure is high, the breed rate also increases to maintain a viable population. $\endgroup$ – zwim Jul 12 '18 at 22:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.