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I can't seem to figure this one out

Two guerilla forces, with troop strength $x(t)$ and $y(t)$, are in combat with each other without reinforcement. Suppose the territory is rather large and full of places to hide. The $y$-force needs to find the $x$-force first before it can inflict combat losses, and the higher the $x$, the easier it is for them to be found. Therefore, the combat loss rate for the $x$-force should be proportional $x$ . $y$ (this is unlike conventional warfare, where the full force of $x$ is open to be shot at by $y$, and so the combat loss rate for $x$ shouldn't depend on the total number of the $x$-force). Thus $$\frac{dx}{dt} = -axy$$ and $$\frac{dy}{dt} = -bxy$$

where $a$ is the combat effectiveness of the $y$-force and $b$ is that of the $x$-force. Suppose initially that $x_0$ and $y_0$ are the troop strengths for the $x$- and $y$-forces, and that $x_0$ is three times as numerous as $y_0$. How much more effective must the $y$-force be to stalemate it's enemy?

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This looks like competing species model. Have you tried to find the equilibrium points? As far as I can tell, you need to find $a$ so that one "population" decays to zero as $t \to \infty$. –  glebovg Nov 13 '12 at 2:40

2 Answers 2

I am sometimes afraid when answers seem to simple but my gut goes with the $y$-force must be 3 times as effective as the $x$-force.

Here is my argument:

$$\frac {dx} {dt} = -axy;\quad \frac {dy} {dt} = - bxy. $$

We are given that $x_0 = 3 y_0 $ and we are looking for some $a$ such that $a = f(b)$ implies that $x(T) = y(T) = 0$ for some $T>0$.

By this I mean that x-force is initially three times as great as the y-force and we are looking for some relationship between the effectiveness of the two forces such that at some point in the future the two forces cancel each other.

So, if we take out system of equations, do a bit of algebra and integrate both sides, we get:

$$x = -a \int x y dt;\quad y = -b \int x y dt. $$

Since both equations share a common factor, we can use substitution to get to:

$$x = \frac {a}{b} y$$

From there, we can just use the fact that $x_0 = 3y_0$ to get:

$$3y_0 = \frac {a} {b} y_0$$ which implies that $ a = 3b$.

Addendum: I'm personally not used to a system of equations where both equations are almost the same, but I guess I could do the above mainly because you're only asking for a relationship between $a$ and $b$ such that both $x$ and $y$ die out.

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The constants of integration are not necessarily the same, so it's not true that $x = ay/b$. –  Robert Israel Nov 13 '12 at 3:12
    
And it certainly can't happen that either $x$ or $y$ is $0$ at any finite time, unless they start out that way (this is a consequence of the existence and uniqueness theorem). What can happen is that they both go to $0$ as $t \to \infty$. –  Robert Israel Nov 13 '12 at 3:23
    
Yes, this is true. I really meant x and y approaching 0 (or being arbitrarily close to 0 at finite time). The reason I can do the substitution is that from the question I am ensured that the constants of integration are exactly the same. (i.e. both forces start fighting at the same time and at some point at least one will lose). –  TheSeamau5 Nov 13 '12 at 6:28

I'm assuming $a, b > 0$. From $dx/dt = -axy$ and $dy/dt = -bxy$ you get $dy/dx = b/a$, and integrate this to get $y = bx/a + c$ for some $c$, namely $c = y_0 - b x_0/a$. The critical points for this system are on the $x$ and $y$ axes. If $c > 0$, i.e. $1/3 = y_0/x_0 > b/a$, then as $t \to +\infty$ you approach the point $(0, c)$ on the positive $y$ axis, i.e. the first force dies out while the second remains alive. If $c < 0$, i.e. $1/3 < b/a$, then you approach $(-ac/b, 0)$, so it is the second force that dies out. If $c = 0$, i.e. $1/3 = b/a$, then you approach $(0,0)$, so both forces die out.

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