Fixed Point Formulation of Explicit ODE

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

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

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


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

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

Proof
Let $y(t)$ be a fixed point of the map $T$.

That is,


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

We see that $\displaystyle y(t_0) = x_0 + \int_{t_0}^{t_0}f(s,y(s))\ ds = x_0$.

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

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