# Existence and Uniqueness Theorem for 1st Order IVPs

## Theorem

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

Let there exist an open ball $V = \left[{t_0 - \ell_0, t_0 + \ell_0}\right] \times \overline B \left({x_0, \epsilon}\right)$ of $\left({t_0, x_0}\right)$ 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 $x \left({t}\right)$ defined for $t \in \left[{t_0 - \ell_0 \,.\,.\, t_0 + \ell_0}\right]$.

## Proof

For $0 < \ell < \ell_0$, let $\mathcal X = \mathcal C \left({\left[{t_0 - \ell_0 \,.\,.\, t_0 + \ell_0}\right]; \R^n}\right)$ endowed with the sup norm be the Banach Space of Continuous Functions on Compact Space $\left[{t_0 - \ell_0 \,.\,.\, t_0 + \ell_0}\right] \to \R^n$.

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

- $\displaystyle \left({T x}\right) \left({t}\right) = x_0 + \int_{t_0}^t f \left({s, x \left({s}\right)}\right) \ \mathrm d s$

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 $\mathcal Y \subseteq \mathcal X$ such that:

- $\mathcal Y$ is closed in $\mathcal X$
- $T \mathcal Y \subseteq \mathcal Y$
- $T$ is a contraction on $\mathcal Y$

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 $\displaystyle m = \sup_{\left({t, x}\right) \in V} \left\vert{f \left({t, x}\right)}\right\vert$ exists and is finite.

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

Let:

- $\mathcal Y = \left\{ {y \in \mathcal X: \left\vert{y \left({t}\right) - x_0}\right\vert \le m \left|{t - t_0}\right\vert, t \in \left[{t_0 - \ell_0 \,.\,.\, t_0 + \ell_0}\right]}\right\}$

be the cone in $\mathcal X$ centred at $\left({t_0, x_0}\right)$.

Clearly $\mathcal Y$ is closed in $\mathcal X$.

Also for $y \in \mathcal Y$ we have:

\(\displaystyle \left\vert{\left({T y}\right) \left({t}\right) - x_0}\right\vert\) | \(\le\) | \(\displaystyle \left\vert{\int_{t_0}^t f \left({s, y \left({s}\right)}\right) \ \mathrm d s}\right\vert\) | |||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle m \int_{t_0}^t \ \mathrm d s\) | Absolute Value of Definite Integral | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle m \left\vert{t - t_0}\right\vert\) |

Therefore $T \mathcal Y \subseteq \mathcal Y$.

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

Let $y_1, y_2 \in \mathcal Y$.

We have:

\(\displaystyle \left\vert{\left({T y_1}\right) \left({t}\right) - \left({T y_2}\right) \left({t}\right)}\right\vert\) | \(\le\) | \(\displaystyle \left\vert \int_{t_0}^t f \left({s, y_1 \left({s}\right)}\right) - f \left({s, y_2 \left({s}\right)}\right) \ \mathrm d s \right\vert\) | |||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \int_{t_0}^t \left\vert{f \left({s, y_1 \left({s}\right)}\right) - f \left({s, y_2 \left({s}\right)}\right)}\right\vert \ \mathrm d s\) | Absolute Value of Definite Integral | ||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \kappa \int_{t_0}^t \left\vert{y_1 \left({t}\right) - y_2 \left({t}\right)}\right\vert \ \mathrm d s\) | by Lipschitz Condition | ||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \kappa \left\vert{t - t_0}\right\vert \left\Vert{y_1 - y_2}\right\Vert_{\sup}\) | Estimation Lemma |

Taking the supremum over $t$ we have:

- $\left\Vert{T y_1 - T y_2}\right\Vert_{\sup} \le \kappa \ell \left\Vert{y_1 - y_2}\right\Vert_{\sup}$

for all $y_1, y_2 \in \mathcal Y$.

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

This completes the proof.

$\blacksquare$