Given that the equations y' = -z3, z' = y3 with initial conditions y(0) = 1, z(0) = 0 have the unique solution y = f(x), z = g(x) for all real x, prove f(x) and g(x) are both periodic with the same period.
Solution
We have y3 y' + z3 z' = 0. Integrating, y4 + z4 = constant. But y(0) = 1, z(0) = 0, so the constant is 1 and y4 + z4 = 1. So as x varies, (y, z) moves along the oval. Let s be the length of the curve from the point (y, z) = (1, 0). Then s' = √( (y')2 + (z')2) = √(y6 + z6). If |y| and |z| are both <= 0.8, then |y4 + z4| < 0.9. Contradiction. So at least one of |y|, |z| > 0.8. Hence |y6 + z6| > 0.86 > 1/4. So s' > 1/4. Let L be the perimeter of the oval. Then L is finite (for example, L is convex and lies inside the square side 2, so L < 8). Hence the point (y, z) completes a circuit of the oval for some value k < 4L of x. But s' depends solely on y and z. So (y(x+k), z(x+k) ) = (y(x), z(x) ).
© John Scholes
jscholes@kalva.demon.co.uk
16 Jan 2001