Picard-Lindelöf theorem
/ pee-KAR LIN-deh-loff /
Before you spend an afternoon solving an initial-value problem, it is worth knowing whether a solution even exists — and whether it is the only one. The Picard-Lindelof theorem is the working guarantee that answers both. Under reasonable conditions it certifies that the IVP dy/dx = f(x, y), y(x_0) = y_0 has exactly one solution, at least for a little while around the starting point.
The conditions are mild and checkable. If f(x, y) is continuous in a rectangle around (x_0, y_0) and is Lipschitz in y there — meaning roughly that f does not change too steeply in the y-direction, |f(x, y_1) - f(x, y_2)| at most L*|y_1 - y_2| for some constant L (a bounded partial of f with respect to y is the usual easy check) — then a unique solution exists on some interval containing x_0. The proof is constructive and beautiful: build a sequence of better and better approximations by Picard iteration (start with the constant y_0, then repeatedly set the next function to y_0 plus the integral of f along the current guess), and the Lipschitz condition forces this sequence to converge to the one true solution.
Two honest limitations must travel with this theorem. First, it is local: it promises a solution only on some interval near x_0, not for all x — solutions can blow up to infinity in finite time (the IVP dy/dx = y^2, y(0) = 1 has solution 1/(1-x), which escapes at x = 1). Second, when the Lipschitz condition fails, uniqueness can genuinely break: dy/dx = sqrt(y) with y(0) = 0 has both y = 0 and y = x^2/4 as solutions. The theorem is a guarantee precisely where its hypotheses hold — and a warning where they do not, which is exactly where singular solutions live.
For dy/dx = sqrt(y) with y(0) = 0, the right side sqrt(y) is continuous but not Lipschitz at y = 0 (its slope is infinite there). Uniqueness fails: both y = 0 and y = x^2/4 satisfy the IVP. This is exactly the failure mode the theorem warns about.
Lose the Lipschitz condition and you can lose uniqueness — two solutions through one point.
Continuity of f alone (Peano's theorem) buys you existence but not uniqueness. Uniqueness specifically needs the stronger Lipschitz condition. Conflating the two — assuming any continuous equation has a unique solution — is a common and consequential error.