# Existence and Uniqueness Theorem for 1st Order IVPs It has been suggested that this article or section be renamed: Ought to be renamed, this covers pretty much all IVPs One may discuss this suggestion on the talk page.

## Theorem

Let $x' = \map f {t, x}$, $\map x {t_0} = x_0$ be an explicit ODE of dimension $n$.

Let there exist an open ball $V = \sqbrk {t_0 - \ell_0, t_0 + \ell_0} \times \map {\overline B} {x_0, \epsilon}$ of $\tuple {t_0, x_0}$ in phase space $\R \times \R^n$ such that $f$ is Lipschitz continuous on $V$.

Then there exists $\ell < \ell_0$ such that there exists a unique solution $\map x t$ defined for $t \in \closedint {t_0 - \ell_0} {t_0 + \ell_0}$.

## Proof

For $0 < \ell < \ell_0$, let $\XX = \map \CC {\closedint {t_0 - \ell_0} {t_0 + \ell_0}; \R^n}$ endowed with the sup norm be the Banach Space of Continuous Functions on Compact Space $\closedint {t_0 - \ell_0} {t_0 + \ell_0} \to \R^n$.

By Fixed Point Formulation of Explicit ODE it is sufficient to find a fixed point of the map $T: \XX \to \XX$ defined by:

$\displaystyle \map {\paren {T x} } t = x_0 + \int_{t_0}^t \map f {s, \map x s} \rd s$

We also have Closed Subset of Complete Metric Space is Complete.

Therefore the Banach Fixed-Point Theorem it is sufficient to find a non-empty subset $\YY \subseteq \XX$ such that:

$\YY$ is closed in $\XX$
$T \YY \subseteq \YY$
$T$ is a contraction on $\YY$

First note that $V$ is closed and bounded, hence compact by the Heine-Borel Theorem.

Therefore since $f$ is continuous, by the extreme value theorem, the maximum $\displaystyle m = \sup_{\tuple {t, x} \mathop \in V} \size {\map f {t, x} }$ exists and is finite.

Let $\kappa$ be the Lipschitz constant of $f$.

Let:

$\YY = \set {y \in \XX: \norm {\map y t - x_0} \le m \size {t - t_0}, t \in \closedint {t_0 - \ell_0} {t_0 + \ell_0} }$

be the cone in $\XX$ centred at $\tuple {t_0, x_0}$.

Clearly $\YY$ is closed in $\XX$.

Also for $y \in \YY$ we have:

 $\displaystyle \size {\map {\paren {T y} } t - x_0}$ $\le$ $\displaystyle \size {\int_{t_0}^t \map f {s, \map y s} \rd s}$ $\displaystyle$ $\le$ $\displaystyle m \int_{t_0}^t \rd s$ Absolute Value of Definite Integral $\displaystyle$ $=$ $\displaystyle m \set {t - t_0}$

Therefore $T \YY \subseteq \YY$.

Finally we must show that $T$ is a contraction on $\YY$ (we will find that this restricts our choice of $\ell$).

Let $y_1, y_2 \in \YY$.

We have:

 $\displaystyle \size {\map {\paren {T y_1} } t - \map {\paren {T y_2} } t}$ $\le$ $\displaystyle \size {\int_{t_0}^t \map f {s, \map {y_1} s} - \map f {s, \map {y_2} s} \rd s}$ $\displaystyle$ $\le$ $\displaystyle \int_{t_0}^t \size {\map f {s, \map {y_1} s} - \map f {s, \map {y_2} s} } \rd s$ Absolute Value of Definite Integral $\displaystyle$ $\le$ $\displaystyle \kappa \int_{t_0}^t \size {\map {y_1} t - \map {y_2} t} \rd s$ Lipschitz Condition $\displaystyle$ $\le$ $\displaystyle \kappa \size {t - t_0} \norm {y_1 - y_2}_\sup$ Estimation Lemma

Taking the supremum over $t$ we have:

$\norm {T y_1 - T y_2}_\sup \le \kappa \ell \norm {y_1 - y_2}_\sup$

for all $y_1, y_2 \in \YY$.

Therefore choosing $\ell < \kappa^{-1}$, $T$ is a contraction on $\YY$ as required.

This completes the proof.

$\blacksquare$