Finite-budget index-one DAE consistent initialization

Let $f,g$ define a regular index-one DAE and let $U\times V$ be a selected local branch chart from A regular index-one DAE has a local reduced ODE branch, so $g(t,x,z)=0$ is equivalent to $z=\psi (t,x)$ there. Input $t_0,x_0$, a seed $z_{\mathrm{s}\mathrm{e}\mathrm{e}\mathrm{d}}\in V$, an algebraic residual tolerance ${\epsilon }_a\ge 0$, and a finite Newton-correction budget $N\in {\mathbb{N}}_0$. If $(t_0,x_0)\mathrm{\notin }U$, $g(t_0,x_0,\cdot )$ is undefined at the seed, or dimensions do not match, return invalid-input.

Call Finite-budget Newton for a square vector residual on the square residual $G(z)=g(t_0,x_0,z)$, with initial value $z_{\mathrm{s}\mathrm{e}\mathrm{e}\mathrm{d}}$, tolerance ${\epsilon }_a$, and budget $N$, restricting evaluations to the chart's open $z$-domain. Write $z_0$ for its returned last valid iterate. If it reports root, reevaluate $G(z_0)$ once: if that residual is exactly zero, return $(x_0,z_0,\mathtt{\text{exact-consistent}})$; otherwise return $(x_0,z_0,\mathtt{\text{approximate-consistent}})$ and its norm. Map singular-jacobian, invalid-trial, and iteration-budget-exhausted to the corresponding initialization statuses and return the last valid iterate.

The procedure makes no arbitrary-start convergence claim. approximate-consistent records a residual bound; it is not relabeled as exact consistency. Under the index-one branch hypotheses an exact consistent value exists locally, by A regular index-one DAE has a local reduced ODE branch, but the finite Newton budget may fail to find it.

The solve has $m$ algebraic unknowns. Record residual queries, Jacobian assemblies and correction solves separately. Their costs depend on the chosen backend.