# Fixed Point Formulation of Explicit ODE

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## Theorem

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

For $a, b \in \R$, let $\mathcal X = \map {\mathcal C} {\closedint a b; \R^n}$ be the space of continuous functions on the closed interval $\closedint a b$.

Let $T: \mathcal X \to \mathcal X$ be the map defined by:

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

Then a fixed point of $T$ in $\mathcal X$ is a solution to the above ODE.

## Proof

Let $\map y t$ be a fixed point of the map $T$.

That is:

- $\displaystyle \map y t = x_0 + \int_{t_0}^t \map f {s, \map y s} \rd s$

Then:

- $\displaystyle \map y {t_0} = x_0 + \int_{t_0}^{t_0} \map f {s, \map y s} \rd s = x_0$

By the fundamental theorem of calculus we have that $y$ is differentiable, and for $t \in \closedint a b$:

\(\displaystyle \map {y'} t\) | \(=\) | \(\displaystyle \frac {\d} {\d t} \int_{t_0}^t \map f {s, \map y s} \rd s\) | Derivative of Constant | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map f {t, \map y t}\) | Fundamental Theorem of Calculus |

This shows that $y$ is a solution to the ODE as claimed.

$\blacksquare$