Enlightening Mathematical Models

What is your favourite Mathematical Model? What features make it intuitive or elegant?

This question is largely inspired by an example and a desire to find other's like it.

Suppose we have two military forces of sizes $x(t)$ and $y(t)$ respectively that are about to enter into a conflict with each other at time $t=0$. Then the aptitude for an army to kill should be proportional to its size. But if army $x$ is killing then army $y$ is dying so we have $$x'(t)=-ay$$ $$y'(t)=-bx$$ where $a$ and $b$ are two positive constants.

Consider the quantity $ay^2-bx^2$. By differentiating we see it is constant. So the trajectories of the system of equations form hyperbolas in the $x,y$ plane. So the strength of an army is directly proportional the square of its size. I love this example because it is effectively a mathematical proof of divide and conquer! (To see this, consider an army of 100 men. The strength of 1 100 man unit will be 10,000 but the strength 2 50 men units will be 5,000.) This is one of Lanchester's Laws

Has anyone got an example to match this?

-
This question would perhaps be better rephrased as "Mathematical Models for Rules of Thumb." – Alex R. Aug 15 '13 at 0:27
I fear that the example I gave is hindering more so than helping this question. All I want to see is nice mathematical models, regardless of the content. – Kieran Cooney Aug 15 '13 at 0:40
$(ay^2-bx^2)'=2(ayy'-bxx')=2(ay(-bx)-bx(-ay))=2(-aybx+bxay)=0$ – Kieran Cooney Aug 15 '13 at 9:40
this looks like raising a competition, who can bring the most enlightning mathematical model and this is although charming but NOT a question seeking for a clear answer. – al-Hwarizmi Aug 16 '13 at 7:47
If $x=100$ is divided into two $50$-man units $x_1=50$ and $x_2=50$ and they both attack $y$, then $y'(t)$ is still $-bx_1-bx_2=-bx$, so $x$'s effectiveness of killing is not reduced. Thus your model fails to prove divide and conquer. – Rahul Aug 16 '13 at 8:31

If you like dynamical systems, there's a nice simple model for how romantic encounters evolve over time given the dating style of both people. There is a simplified version here. The short rules of thumb are, if both people like to give and receive attention, they fall madly in love :) . It also predicts cycles of love and hate for couples that have one person shying away from too much attention.

If you're interested in this further, John Gottman has developed a broader application of these kinds of models to the "mathematics of marriage."

-

Maybe I'm just grouchy today, but the example in the question strikes me as an example of misuse of a mathematical model. The model begins reasonably enough, analyzing a simlified version of the dynamics of one army fighting another, finding an invariant quantity, and describing the trajectories of this dynamical system. Then, it jumps to a conclusion about a notion of "strength", which hasn't been defined. Finally, this notion is applied to an entirely different situation, where one army is split in two, and the strength is just assumed to be additive. The conclusion, that a single unified army is better than two halves of it, is probably true in some situations and false in others (depending on what capabilities armies and their parts have, for example pincer movements), but I would definitely not call this argument a mathematical proof of it.

-
I recall from reading Shelby Foote's (American) Civil War Trilogy that in one battle Nathan Bedford Forrest's was surrounded by a larger army. Forrest divided his army into two parts, the two parts charged in opposite directions, and Forrest won the battle. The only mathematics in this comment is that a counterexample can refute a conjecture such as Lanchester's Law. – Jay Aug 15 '13 at 0:09
The attribution to strength is perhaps not OP's fault. There are various notions of "force" that can be prescribed to Lanchester's model but it goes without saying that their precise mathematical meaning is convoluted. See here for example: rand.org/pubs/monograph_reports/MR638/app.html#fn16 Also finding an arbitrary counterexample to a mathematical model does not invalidate it. In that the assumption is the rate equations are satisfied under some idealized conditions. Saying that the model no longer holds when splitting an army is a wholly different model. – Alex R. Aug 15 '13 at 0:21
The example was offered as a sample answer to the question, not as an addition to the question. The answers to this question should reflect this. – Kieran Cooney Aug 15 '13 at 0:42

The mathematics of evolution is full of them, see for example Lotka-Volterra systems of equations, encapsulating succinctly the a priori surprising phenomenon of periodic phase trajectories.

-

I would argue that your "consider the quantity" comment in you OP is a consequence of a variables separable situation (which you show that this quantity is indeed conserved further down your thread). Thus (assume w.l.o.g that $a, b \neq 0$) thus,

\begin{eqnarray} \notag \frac{y^{'}}{x^{'}} &=& \frac{dy}{dx} \\ &=& \frac{-bx}{-ay} \\ &=& \frac{b}{a}.\frac{x}{y} \\ \end{eqnarray} Now, this is a first oder differential equation, which can be solved by separating out the vairables. Thus, \begin{eqnarray} \notag \frac{dy}{dx} &=& \frac{b}{a}\frac{x}{y} \\ \end{eqnarray} \begin{eqnarray} \notag \Rightarrow ydy &=& \left( \frac{b}{a} \right) xdx \\ \text{(Integrating)} \int ydy &=& \left(\frac{b}{a}\right)\int xdx \\ \Rightarrow \frac{1}{2}y^{2} &=& \frac{1}{2} \left(\frac{b}{a} \right) x^{2} + C \\ \end{eqnarray} Assuming (w.l.o.g) that $C=0$ we have (upon re-arranging) $$ay^{2} - bx^{2} = 0$$ Which brings about your conserved quantity above.

-
Down-voted, why? – Autolatry Sep 3 '13 at 13:29

protected by Willie WongAug 16 '13 at 8:04

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).