k is a real constant. y satisfies y'' = (x3 + kx) y with initial conditions y = 1, y' = 0 at x = 0. Show that the solutions of y = 0 are bounded above but not below.
The first part is easy; the second part is hard.
For some x0, x3 + kx > 0 for all x > x0. [For example, if k ≥ 0, then we may take x0 = 0. If k < 0, then we may take x0 = -k.] Now suppose that a < b are two consecutive zeros greater than x0. If y(x) > 0 on (a, b), then y'' must be negative for at least part of the interval (a, b), but that is impossible, since (x3 + kx) y > 0 for the whole interval. Similarly, if y(x) < 0 on (a, b), then y'' must be positive for at least part of the interval (a, b), but that is impossible, since (x3 + kx) y < 0 for the whole interval. So there is at most one root greater than x0 and that provides a bound. [For completeness, we note that we require y not to be identically zero on a non-zero interval, but that is not possible because y(0) is non-zero.]
Suppose the second part is false. Then we can find x2 such that for any x < x2, y(x) is non-zero. There are two cases. Suppose first that y(x) > 0 for all x < x2. We can take x3 < x2 such that x3 + kx < -2 for all x < x3. Hence, in particular, y''(x) < 0 for x < x3. Take arbitary a < b < x3, then as usual we can find ξ ∈ (a, b) such that y(a) = y(b) - (b - a) y'(b) + 1/2 (b - a)2 y''(ξ) < y(b) - (b - a) y'(b). If y'(b) > 0, then for (b - a) sufficiently large y(a) < 0 (contradiction). So y'(b) ≥ 0. So y(a) ≤ y(b) + 1/2 (b - a)2 y''(ξ). But now we use the stronger assumption that ξ3 + kξ < -2 and hence y(a) < y(b)( 1 - (b - a)2), which becomes negative for sufficiently large (b - a). Contradiction.
The remaining case is y(x) < 0 and y''(x) > -2 y(x) for all x < x3. In particular, y''(x) > 0 for x < x3. So arguing as before, y'(b) ≤ 0. As before y(a) ≥ -y(b) ( (b - a)2 - 1), which becomes positive for sufficiently large (b - a). Contradiction.
15th Putnam 1955
© John Scholes
26 Nov 1999