### 34th Putnam 1973

Problem A5

An object's equations of motion are: x' = yz, y' = zx, z' = xy. Its coordinates at time t = 0 are (x0, y0, z0). If two of these coordinates are zero, show that the object is stationary for all t. If (x0, y0, z0) = (1, 1, 0), show that at time t, (x, y, z) = (sec t, sec t, tan t). If (x0, y0, z0) = (1, 1, -1), show that at time t, (x, y, z) = (1/(1 + t), 1/(1 + t), -1/(1 + t) ). If two of the coordinates x0, y0, z0 are non-zero, show that the object's distance from the origin d → ∞ at some finite time in the past or future.

Solution

You are meant to assume the usual uniqueness theorems. So the only non-trivial part of the question is the last. For the first three parts, one just has to verify that the three solutions given satisfy the equations.

We have xx' = yy' = zz' = xyz. So integrating y2 = x2 + A, z2 = x2 + B. We may assume that y0 and z0 are non-zero and that |x0| ≤ |y0| and |z0|. So A and B are positive and x' = ( (x2 + A) (x2 + B) )1/2. Suppose we take the positive square root so that x increases with time. Then t = ∫ ( (x2 + A) (x2 + B) )1/2 dx. But the integral converges as x tends to infinity, so x reaches infinity at a finite (future) time. If we had taken the negative root, then we conclude that x reaches infinity at a finite (past) time.

© John Scholes
jscholes@kalva.demon.co.uk
22 Aug 2001