An index-two DAE can have zero algebraic residual and an arbitrary algebraic variable

Here $f(t,x,z)=z$ and $g(t,x,z)=x$, so $g_z=0$ at every point. The regular index-one hypothesis $g_z$ invertible therefore fails.

Along any solution, differentiating $g(x(t),z(t))=x(t)=0$ and substituting $x^{\mathrm{\prime }}=z$ gives $z=0$. Differentiating this relation gives $z^{\mathrm{\prime }}=0$. The original constraint does not involve $z$; the first differentiated relation determines $z$, and the second determines its first derivative. By [F2] the differentiation index is two.

For every real $M$, the candidate $(x,z)=(0,M)$ has algebraic residual $g(0,M)=0$, while step 2.1 shows that a solution has $z=0$. Taking arbitrarily small $M\mathrm{\ne }0$ rules out any finite constant $C$ for which $\mathrm{\mid }z\mathrm{\mid }\le C\mathrm{\mid }g(x,z)\mathrm{\mid }$ throughout a neighborhood of $(0,0)$.

The pair $(0,1)$ satisfies $g(0,1)=0$, but a solution would require $z(0)=0$ by step 2.1, so it is not a consistent initial pair for a solution. The pair $(1,0)$ has $g(1,0)=1\mathrm{\ne }0$, so it fails even algebraic consistency.

The constant functions $x(t)=0,z(t)=0$ satisfy both equations, so the system has a solution with the consistent initial pair $(0,0)$. The three boundary calculations above concern only this fixed index-two system and imply no index-one numerical guarantee.