# set up (do not solve) a differential equation that represent this system

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.

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 ?

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.

• 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. – LutzL Jul 12 '18 at 20:23
• @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! – Brian Tung Jul 12 '18 at 20:43
• 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. – zwim Jul 12 '18 at 22:22