Synchronous Picard waveform relaxation on a finite window

Let $r,s\ge 1$, $d=r+s$, $D\subseteq {\mathbb{R}}^d$ be open, $F=(f,g):D\to {\mathbb{R}}^d$ be continuous, and $u_0=(x_0,z_0)\in D$. Choose a window length $\mathrm{\Delta }>0$, radius $R\ge 0$, bounds $M,K\ge 0$, tolerance $\epsilon \ge 0$, and integer sweep budget $B\ge 0$. The admissibility conditions are that the closed Euclidean ball ${\bar{B}}_R(u_0)$ lies in $D$, $\mathrm{\parallel }F(u){\mathrm{\parallel }}_2\le M$ on that ball, $\mathrm{\parallel }F(u)-F(v){\mathrm{\parallel }}_2\le K\mathrm{\parallel }u-v{\mathrm{\parallel }}_2$ there, $M\mathrm{\Delta }\le R$, and $q:=K\mathrm{\Delta }<1$. If an input condition fails, return not-admissible.

Let $C_{\mathrm{\Delta }}=C([0,\mathrm{\Delta }],{\mathbb{R}}^d)$ with $d_{\mathrm{\infty }}(U,V):={\sup }_{0\le t\le \mathrm{\Delta }}\mathrm{\parallel }U(t)-V(t){\mathrm{\parallel }}_2$, and let $X_R:=\{U\in C_{\mathrm{\Delta }}:U(0)=u_0,{\sup }_t\mathrm{\parallel }U(t)-u_0{\mathrm{\parallel }}_2\le R\}$. The synchronous Picard waveform map is $\mathrm{\Phi }(U)(t):=u_0+{\int }_0^{t}F(U(s))ds,$ where the vector integral is componentwise. This updates both state waveforms from the same preceding pair.

Procedure. If $B=0$, return the constant waveform $U^0(t):=u_0$ with status budget-exhausted and no error certificate. Otherwise set $U^0(t):=u_0$ and, for $k=1,\dots ,B$, compute $U^k:=\mathrm{\Phi }(U^{k-1})$. After the first sweep set ${\delta }_0:=d_{\mathrm{\infty }}(U^1,U^0)$. After sweep $k$, compute $E_k:=q^k{\delta }_0\mathrm{/}(1-q)$. If $E_k\le \epsilon$, return $(U^k,\mathtt{\text{tolerance-certified}},E_k,k)$; if the budget is exhausted first, return $(U^B,\mathtt{\text{budget-exhausted}},E_B,B)$. Integrals and waveform updates are exact mathematical operations here; quadrature and inexact subsolves require separate error terms.