Existence and Uniqueness Theorem for 1st Order IVPs

Theorem
Let $x' = f(t,x)$, $x(t_0) = x_0$ be an explicit ODE of dimension $n$.

Suppose that there exists a neighborhood $V = [t_0 - \ell_0,t_0 + \ell_0] \times \overline{B}(x_0,\epsilon)$ of $(t_0,x_0)$ 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 $x(t)$ defined for $t \in [t_0 - \ell,t_0 + \ell]$.

Proof
For $0 < \ell < \ell_0$, let $\mathcal X = \mathcal C([t_0 - \ell,t_0 + \ell]; \R^n)$ endowed with the sup norm.

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


 * $\displaystyle (Tx)(t) = x_0 + \int_{t_0}^t f(s,x(s))\ ds$

Now we have that continuous functions on an interval form a Banach space, i.e. that $\mathcal X$ is a Banach space.

We also have that a closed subset of a 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 $\displaystyle m = \sup_{(t,x) \in V}| f(t,x)|$ exists and is finite.

Let $\kappa$ be the Lipschitz constant of $f$, and:


 * $\mathcal Y = \{ y \in \mathcal X : |y(t) - x_0 | \leq m |t - t_0|\ \forall t \in [t_0-\ell,t_0 + \ell]\}$

the cone in $\mathcal X$ centred at $(t_0,x_0)$.

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

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

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:

Taking the supremum over $t$ we have $\| Ty_1 - Ty_2 \|_{\text{sup}} \leq \kappa \ell \| y_1 - y_2 \|_{\text{sup}}$ 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.