Picard-Lindelöf Theorem

Theorem (Picard-Linedlöf)

Let U⊂RnU\subset \mathbb{R}^{n} be an open set, let f:U→Rnf:U\to \mathbb{R}^{n} be Lipschitz in UU with constant k≥0k\ge 0. Let x0∈Ux_{0}\in U and

  1. r>0r>0 small enough s.t. B(x0,r)‾={x∈Rn:∥x−x0∥≤r}⊂U\overline{B(x_{0},r)}=\{ x\in\mathbb{R}^{n}:\lVert x-x_{0} \rVert\le r \}\subset U
  2. M=sup⁡x∈B‾(x0,r)∥f(x)∥M=\sup_{x\in\overline{B}(x_{0},r)} \lVert f(x) \rVert
  3. a>0a>0 s.t. a≤rMa\le \frac{r}{M} and a<1ka< \frac{1}{k}

Then ∀t0∈R,∃!f∈C1([t0−a,t0+a];B(x0,r)‾)\forall t_{0}\in\mathbb{R}, \exists!f\in C^{1}([t_{0}-a,t_{0}+a];\overline{B(x_{0},r)}) s.t. {x˙(t)=f(x(t))x(t0)=x0,t∈[t0−a,t0+a]\begin{cases} \dot{x}(t)&=f(x(t))\\ x(t_{0})&=x_{0} \end{cases},\quad t\in[t_{0}-a,t_{0}+a] i.e. there exists a unique solution to our initial value problem.