Fixed Point Formulation of Explicit ODE

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

For $a, b \in \R$, let $\mathcal X = \mathcal C \left({\left[{a \,.\,.\, b}\right]; \R^n}\right)$ be the space of continuous functions on the closed interval $\left[{a \,.\,.\, b}\right]$.

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


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

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

Proof
Let $y \left({t}\right)$ be a fixed point of the map $T$.

That is:
 * $\displaystyle y \left({t}\right) = x_0 + \int_{t_0}^t f \left({s, y \left({s}\right)}\right) \ \mathrm d s$

Then:
 * $\displaystyle y \left({t_0}\right) = x_0 + \int_{t_0}^{t_0} f \left({s, y \left({s}\right)}\right) \ \mathrm d s = x_0$

By the fundamental theorem of calculus we have that $y$ is differentiable, and for $t \in \left[{a \,.\,.\, b}\right]$:

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