Frozen-slow explicit Euler with fast microsteps

Let $r,s\ge 1$, let $D\subseteq {\mathbb{R}}^{r+s}$ be open, and let $f:D\to {\mathbb{R}}^r$ and $g:D\to {\mathbb{R}}^s$ be the designated slow and fast vector fields. The labels “slow” and “fast” are input partition labels; this definition does not assert that either field actually varies at a specified rate. The routine receives proposed state vectors $x,z$, a real macro step $H$, and a microstep count $m$.

Procedure.

1. Before any division or field evaluation, check that $x\in {\mathbb{R}}^r$, $z\in {\mathbb{R}}^s$, $H>0$, $m$ is an integer with $m\ge 1$, and $(x,z)\in D$. If any check fails, return invalid-input.
2. Set $h:=H\mathrm{/}m$ and evaluate $f(x,z)$ to set $x^{+}:=x+Hf(x,z)$.
3. Set $z_0:=z$.
4. For $j=0,1,\dots ,m-1$, first check whether $(x,z_j)\in D$. If not, return domain-exit without a macro-step output. Otherwise evaluate $g(x,z_j)$ and set $z_{j+1}:=z_j+hg(x,z_j)$.
5. If $(x^{+},z_m)\mathrm{\notin }D$, return domain-exit without a macro-step output. Otherwise return success and the output state $(x^{+},z_m)$.

For inputs on which the procedure succeeds, write its output map as ${\mathrm{\Psi }}_{H,m}(x,z):=(x^{+},z_m)$.

On a successful step the procedure uses exactly one evaluation of $f$ and $m$ evaluations of $g$; an invalid-input return uses no field evaluations, and a domain-exit return uses no more than these counts. The slow state is held at its initial value during every fast microstep. In a separated implementation, the slow update needs the starting fast value $z_0$, while the fast loop needs the starting slow value $x$; $x$ can be sent to the fast worker once and cached for all $m$ microsteps, with no fast-to-slow refresh during the block. The actual communication cost depends on the implementation. The evaluation counts alone give no wall-clock saving claim, especially when the fast block accounts for nearly all right-hand-side work.