Peano's existence theorem
/ pay-AH-no /
Peano's theorem is the gentlest existence result: it says that if the right-hand side f(x, y) is merely continuous near the starting point, then the initial value problem y' = f(x, y), y(x0) = y0 has at least one solution on some small interval around x0. No smoothness, no Lipschitz condition, nothing fancy — plain continuity is enough to guarantee that a solution exists. It is the natural answer to the bare existence question.
The idea behind it is wonderfully physical. You approximate the true solution by a polygonal path that steps forward in tiny increments, always pointing in the direction f tells you — this is the Euler-line construction. As you make the steps smaller and smaller, you get a whole family of these broken-line approximations. Because f is continuous and bounded on a little box around (x0, y0), these polygonal curves cannot wander off too steeply and form an equicontinuous, uniformly bounded family; a compactness result (Arzela-Ascoli) then guarantees that some subsequence converges to a genuine curve, and continuity of f makes that limit an actual solution. So existence is squeezed out of approximate solutions that crowd together until one of their limits has to solve the equation.
The crucial and often-missed point is the gap between Peano and the stronger Picard-Lindelof theorem: Peano gives existence but says nothing about uniqueness. Continuity alone is genuinely not enough to pin down a single solution. The textbook witness is y' = y^(2/3) at the origin, where f is continuous so Peano applies and a solution exists, yet infinitely many solutions pass through (0, 0). To buy uniqueness you must pay extra — a Lipschitz condition — which is exactly what Picard-Lindelof demands.
Take y' = y^(2/3), y(0) = 0. Here f(x, y) = y^(2/3) is continuous everywhere, so Peano's theorem applies and a solution exists — indeed y = 0 is one. Peano is happy. But it cannot tell you this is the only solution, and it isn't.
Continuity gives existence (Peano) but not uniqueness — the same example breaks uniqueness wide open.
A common error is to assume Peano also gives uniqueness. It does not: continuity alone permits infinitely many solutions, as y' = y^(2/3) shows.