2.1 The integral equation
Lemma 2.1.1. Let \(f\) be continuous on an open set \(D \subseteq \mathbb R \times \mathbb R^n\) containing \((t_0, x_0)\), and let \(J\) be an interval containing \(t_0\). A continuous function \(\phi \colon J \to \mathbb R^n\) with graph in \(D\) solves the initial value problem on \(J\) if and only if \[ \phi (t) = x_0 + \int _{t_0}^tf\bigl (s, \phi (s)\bigr )\,ds \qquad \text {for all } t \in J . \]
Proof. If \(\phi \) solves the problem, then \(\phi '(s) = f(s, \phi (s))\) is continuous, and integrating from \(t_0\) to \(t\) gives the equation, by the fundamental theorem of calculus. Conversely, if \(\phi \) is continuous and satisfies the equation, the integrand \(s \mapsto f(s, \phi (s))\) is continuous, so the right-hand side is differentiable with derivative \(f(t, \phi (t))\), and \(\phi (t_0) = x_0\). □
The lemma converts a differential equation into a fixed-point problem: \(\phi \) is a solution if and only if \(\phi = T\phi \), where \[ (T\phi )(t) = x_0 + \int _{t_0}^tf\bigl (s, \phi (s)\bigr )\,ds . \] The integral equation needs only continuity of \(\phi \), not differentiability, which is what makes it the natural setting for limits.
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