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 be the Banach space of continuous functions $[t_0 - \ell,t_0 + \ell] \to \R^n$.

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:

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

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert (Ty)(t) - x_0 \right\vert\)	\(\leq\)	\(\displaystyle \left\vert \int_{t_0}^t f(s,y(s))\ ds \right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle m\int_{t_0}^t\ ds\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Absolute Value of Definite Integral
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle m\vert t - t_0 \vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert (Ty_1)(t) - (Ty_2)(t) \right\vert\)	\(\leq\)	\(\displaystyle \left\vert \int_{t_0}^t f(s,y_1(s)) - f(s,y_2(s)) \ ds \right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle \int_{t_0}^t \vert f(s,y_1(s)) - f(s,y_2(s)) \vert \ ds\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Absolute Value of Definite Integral
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle \kappa \int_{t_0}^t \vert y_1(t) - y_2(t) \vert \ ds\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By the Lipschitz property
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle \kappa \vert t-t_0\vert \Vert y_1 - y_2 \Vert_{\text{sup} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By the Estimation Lemma

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.

$\blacksquare$