# The Local Existence and Uniqueness Theorem via Banach’s Fixed Point Theorem

Recall from the Banach’s Fixed Point Theorem that if $(X, d)$ is a complete metric space and $T : X \to X$ is a contraction mapping (that is, for all $x, y \in X$ we have that $d(T(x), T(y)) \leq qd(x, y)$ where $q \in (0, 1)$). Then $T$ has a unique fixed point $x^*$ and for each $x \in X$ the sequence $(T^n(x))_{n=1}^{\infty}$ converges to $x^*$.

We will now use this theorem to prove the local existence AND uniqueness of solutions to first order ODEs.

Theorem 1 (The Local Existence and Uniqueness Theorem via Banach’s Fixed Point Theorem): Let $x’ = f(t, x)$ with $x(\tau) = \xi$ be an initial value problem with $f \in C(D, \mathbb{R})$ and let $f$ satisfy a Lipschitz condition on $D$. Then there exists a unique solution on some open interval $J$ containing $\tau$. |

**Proof:**Since $L$ satisfies a Lipschitz condition on $D$ there exists a Lipschitz constant $L \in \mathbb{R}$, $L > 0$ such that for all $(t, x), (t, y) \in D$ we have that:

(1)

\begin{align} \quad | f(t, x) – f(t, y) | \leq L|x – y| \end{align}

- We define the closed rectangle $S \subseteq D$ to be:

(2)

\begin{align} \quad S = \{ (t, x) : |t – \tau| \leq a, \: |x – \xi| \leq b \} \end{align}

- Since $f$ is continuous on $D$ it is continuous on the compact set $S$ and so $f$ is boudned on $S$. So there exists an $M \in \mathbb{R}$, $M > 0$ such that $|f(t, x)| \leq M$ for all $(t, x) \in D$. We let:

(3)

\begin{align} \quad c = \min \left \{ a, \frac{b}{M} \right \} \end{align}

- Now let $\delta > 0$ be such that:

(4)

\begin{align} \quad \delta

- We consider the space $C([\tau – \delta, \tau + \delta], [\xi – b, \xi +b])$ of continuous functions from $[\tau – \delta, \tau + \delta]$ to $[\xi – b, \xi + b]$. We equip this space with the supremum norm defined for all functions $\phi$ in this space by:

(5)

\begin{align} \quad \| \phi \|_{\infty} = \sup_{t \in [\tau – \delta, \tau + \delta]} |\varphi(t)| \end{align}

- Then this is a complete metric space. Define a mapping $T : C([\tau – \delta, \tau + \delta], [\xi – b, \xi +b]) \to C([\tau – \delta, \tau + \delta], [\xi – b, \xi +b])$ by:

(6)

\begin{align} \quad T(\phi(x)) = \xi + \int_{\tau}^{t} f(s, \phi(s)) \: ds \end{align}

- Then we have that:

(7)

\begin{align} \quad \| T(\phi) – T(\psi) \| &= \biggr \| \left ( \xi + \int_{\tau}^{t} f(s, \phi(s)) \: ds \right ) – \left ( \xi + \int_{\tau}^{t} f(s, \psi(s)) \: ds \right ] \biggr \|_{\infty} \\ &= \biggr \| \int_{\tau}^{t} [f(s, \phi(s)) – f(s, \psi(s))] \: ds \biggr \|_{\infty} \\ &= \sup_{t \in [\tau – \delta, \tau + \delta]} \biggr \lvert \int_{\tau}^{t} [f(s, \phi(s)) – f(s, \psi(s))] \: ds \biggr \rvert \\ & \leq \sup_{t \in [\tau – \delta, \tau + \delta]} \int_{\tau}^{t} |f(s, \phi(s)) – f(s, \psi(s)) | \: ds \\ & \leq \sup_{t \in [\tau – \delta, \tau + \delta]} \int_{\tau}^{t} L|\phi(s) – \psi(s)| \: ds \\ & \leq \delta \sup_{t \in [\tau – \delta, \tau + \delta]} L|\phi(s) – \psi(s)| \: ds \\ & \leq \delta L \sup_{t \in [\tau – \delta, \tau + \delta]} |\phi(s) – \psi(s)| \: ds \\ & \leq \underbrace{\delta L}_{=q} \| \phi – \psi \|_{\infty} \end{align}

- Note that if $q = \delta L$ then $q \in (0, 1)$. This show that $T$ is a contraction mapping. So by Banach’s fixed point theorem there exists a unique fixed point, $\phi$ such that:

(8)

\begin{align} \quad \phi(t) = T(\phi(t)) = \xi + \int_{\tau}^{t} f(s, \phi(s)) \: ds \end{align}

- This show that $\phi$ is a solution to the IVP that is unique. $\blacksquare$

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