TheoremProved

Picard–Lindelöf theorem

Statement

Let $f(t,y)$ be continuous on a region containing $(t_0,y_0)$ and Lipschitz continuous in $y$ . Then the initial value problem $y' = f(t,y)$ , $y(t_0) = y_0$ has a unique solution on some interval around $t_0$ .

Why is it true?

If the right-hand side of a differential equation does not change too abruptly as $y$ varies (it is Lipschitz), then the equation behaves like a well-posed rule: starting from one point, exactly one trajectory unfolds, at least for a short time. Two trajectories can never cross, because crossing would mean two different futures from the same present.

Proof sketch

Rewrite the initial value problem as the integral equation $y(t) = y_0 + \int_{t_0}^{t} f(s,y(s))\,ds$ and define the Picard iteration $y_{n+1}(t) = y_0 + \int_{t_0}^{t} f(s,y_n(s))\,ds$ . On a small enough interval the Lipschitz condition makes this map a contraction on the space of continuous functions with the sup norm, so the Banach fixed-point theorem gives a unique fixed point, which is the solution.

Topics that use this theorem

First-order differential equations

Related theorems

Poincaré–Bendixson theorem

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

Earl A. Coddington, Norman Levinson (1955). Theory of Ordinary Differential Equations