One-step IMEX Euler with an implicit branch and residual status

For the additive ODE partition Additive autonomous ODE partitions, take a current state $y\in U$, a step $h>0$, a closed ball $\bar{B}(c,R)\subset U$ centered at $c:=y+ha(y)$, bounds $M_b,L_b\ge 0$ for $\mathrm{\parallel }b{\mathrm{\parallel }}_2$ and its Lipschitz constant on that ball, a tolerance $\tau \ge 0$, and a solve routine that returns either a candidate $z\in \bar{B}(c,R)$ or a failure status. The procedure is:

1. If the solve routine is unavailable, return solver-unavailable.
2. If $h{M}_b>R$ or $h{L}_b\ge 1$, return branch-not-certified without claiming a root.
3. Call the solve routine on $z=c+hb(z)$.
4. If it returns failure, return that failure status. If it returns $z\mathrm{\notin }\bar{B}(c,R)$, return out-of-branch-ball.
5. Compute the update residual $r:=z-c-hb(z)$ in the state units of Implicit residuals and their update and derivative units. If $\mathrm{\parallel }r{\mathrm{\parallel }}_2>\tau$, return residual-too-large with $z$ and $r$; otherwise return success with $z$ and $r$.

When the routine returns an exact root, $r=0$ and $z$ is the IMEX Euler step $Y=y+ha(y)+hb(Y)$. A success status records only the requested residual check; the forward error conclusion is supplied separately by the residual theorem. This procedure uses one explicit evaluation of $a$ and one implicit solve.