The Picard operator and Picard iterates on a closed ball of continuous curves

Fix $n\in \mathbb{N}$ with $n\ge 1$, $h\ge 0$, $J=[t_0-h,t_0+h]$, an initial state $x_0\in {\mathbb{R}}^n$, and a radius $r>0$. Let

${\mathcal{B}}_r:=\{x\in C(J,{\mathbb{R}}^n):{\sup }_{t\in J}\mathrm{\parallel }x(t)-x_0{\mathrm{\parallel }}_2\le r\}\mathrm{.}$

Let $D_F\subseteq \mathbb{R}\times {\mathbb{R}}^n$ be open and let $F:D_F\to {\mathbb{R}}^n$ be continuous. The domain of the Picard operator on ${\mathcal{B}}_r$ consists of those $x\in {\mathcal{B}}_r$ whose whole graph $\{(s,x(s)):s\in J\}$ lies in $D_F$. For such a curve, continuity makes $s\mapsto F(s,x(s))$ integrable on every interval with endpoints $t_0,t\in J$, and its Picard image is

$(\mathcal{T}x)(t):=x_0+{\int }_{t_0}^{t}F(s,x(s))ds\mathrm{.}$

The operator maps into $C(J,{\mathbb{R}}^n)$; it is a self-map of ${\mathcal{B}}_r$ only when its image is known to remain in that ball. Starting from $x^{(0)}(t):=x_0$, define $x^{(m+1)}:=\mathcal{T}{x}^{(m)}$ inductively whenever $x^{(m)}$ is defined and lies in the operator domain; if that condition first fails, no later iterate is defined. When a separate invariant-ball result makes $\mathcal{T}:{\mathcal{B}}_r\to {\mathcal{B}}_r$ a total self-map, The recursion theorem supplies the entire sequence. Every fixed point satisfies the corresponding Volterra equation. When $h>0$, it is exactly a solution of the IVP by A first-order initial value problem is equivalent to its Volterra integral equation; for $h=0$ the fixed-point equation remains defined, but no derivative on the isolated one-point domain is asserted.