---
tag: 03Z9V0Y78A
kind: statement
role: proposition
name: C^1 vector fields satisfy the Picard–Lindelöf hypotheses
msc: [34A12]
uses: [0J7HF2PF1H, 055VVSN00V]
sources:
  - url: https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem
    locator: Theorem
---

## Statement

Let $U$ be an open subset of $\mathbb{R}^n$, let $f : U \to \mathbb{R}^n$ be continuously differentiable, let $x_0 \in U$ and let $t_0 \in \mathbb{R}$.
Then there are real numbers $\varepsilon > 0$, $a > 0$ and $L, K \ge 0$ with these properties.
The closed ball $B$ of radius $a$ around $x_0$ lies in $U$.
The time-independent field $(t, x) \mapsto f(x)$, extended in any way outside $B$, satisfies the [[Picard–Lindelöf hypotheses|0J7HF2PF1H]] at $(t_0, x_0)$ on $[t_0 - \varepsilon, t_0 + \varepsilon]$ with constants $a, L, K$.

## Remarks

Proof: choose $a > 0$ with $B \subseteq U$. The continuous functions $\|f\|$ and $\|f'\|$ are bounded on the compact ball $B$, by $L$ and $K$. The mean value inequality makes $f$ $K$-Lipschitz on $B$. Then choose $\varepsilon$ with $L \varepsilon \le a$.
So the initial value problem $x' = f(x)$, $x(t_0) = x_0$ has a unique local solution for each $C^1$ field $f$.
The same argument works for a time-dependent field $f(t, x)$ of class $C^1$.

## Explanations

### L1 human

A smooth field has a bounded derivative near each point, so it is Lipschitz near this point.
This is why, in practice, one rarely checks the Lipschitz condition directly: smoothness is enough.