### 22nd Putnam 1961

Problem B6

Let y be the solution of the differential equation y'' = - (1 + √x) y such that y(0) = 1, y'(0) = 0. Show that y has exactly one zero for x ∈ (0, π/2) and find a positive lower bound for it.

Solution

Let z be the solution of z'' = -z with z(0) = 1, z'(0) = 0. Suppose y > 0 for 0 <= x <π/2. Then since y''z - yz'' = -(1 + √x) yz + yz = - √x yz, we have ∫0π/2 - √x yz dx = ∫0π/2 (y''z - yz'') dx = (y'z - yz') |0π/2 = y(π/2) ≥ 0. But - √x yz < 0 throughout the range, so ∫0π/2 - √x yz dx < 0. Contradiction. Hence y has at least one zero in (0, π/2).

Let w be the solution of w'' = -3w with w(0) = 1, w'(0) = 0. Then w = cos (√3)x. Suppose y had a zero in (0, π/(2√3)). Then the same argument would show that w had a root in (0, π/(2√3) which is false. So π/(2√3)is a positive lower bound for any zero of y.

Finally, suppose that y has more than one zero in (0, π/2). Let the first be at x = h and the second at x = k. Take v(x) to be A cos( (√3) x + B). We select A and B so that v(h) = 0 and v'(h) = y'(h). Now v cannot have another zero in (h, k), so we can establish a contradiction by a similar argument to the above. We have v'' = -3v. Hence y''v - yv'' = ( -(1 + √x) + 3) yv which is strictly positive on (h, k). Hence ∫hk (y''v - yv'') dx > 0. But ∫hk (y''v - yv'') dx = (y'v - yv') |hk = y'(k)v(k) < 0. [y' must be positive since y is zero and y(x) < 0 for x just less than k, whilst we know that v(k) is negative). Contradiction. Hence y has exactly one zero.