48th Putnam 1987

Problem A5

f: R2→R3 (where R is the real line) is defined by f(x, y) = (-y/(x2 + 4y2), x/(x2 + 4y2), 0). Can we find F: R3→R3, such that:
(1)   if F = (F1, F2, F3), then Fi all have continuous partial derivatives for all (x, y, z) ≠ (0, 0, 0);
(2)   ∇ x F = 0 for all (x, y, z) ≠ 0;
(3)   F(x, y, 0) = f(x, y)?



Answer: no.


We look first at (2). We know that ∇x F = 0 iff F = g for some g: R3→R. Remembering that tan-1k has derivative 1/(1 + k2) and playing around a little, we soon hit on g(x, y, z) = 1/2 tan-1(2y/x) + h(z). That looks promising. In fact, it almost works. We might, for example take h(z) = 1/2 z2, so that F(x, y, z) = (-y/(x2 + 4y2), x/(x2 + 4y2), z). That satisfies (2) and (3), and almost satisfies (1). The difficulty is that it, and its partial derivatives wrt x and y, are discontinuous (and arguably undefined) on the entire line x = 0, y = 0 (not just at the origin).

So we suspect that we cannot find F to satisfy all the conditions. The obvious approach is to assume we can and to seek a contradiction using Stokes' theorem ( ∫S ∇x F .da = ∫Γ F.ds, where S is a surface bounded by a closed curve Γ ). We want S to cut the z axis at z ≠ 0, because we believe the discontinuities are on the z-axis and so that will ensure that the surface integral is non-zero, whereas (2) would imply that it was zero. We want a curve Γ on which the line integral is easy to calculate. It clearly must lie entirely within the x, y plane (so that we can use (3) to give us F - otherwise we do not know what F is). It also looks sensible to take an ellipse x2 + 4y2 = k, because then f is drastically simplified. The line integral is then ∫Γ F.ds = ∫Γ f.ds = ∫Γ (-y dx + x dy)/k = 2/k area ellipse ≠ 0. If (2) was satisfied, then the surface integral would be zero, since the surface avoids the origin. Contradiction.



48th Putnam 1987

© John Scholes
7 Oct 1999