Erratum for “Global Identifiability of Differential Models”

We are grateful to Peter Thompson for pointing out an error in [1, Lemma 3.5, p. 1848]. The original proof worked only under the assumption that θ ̂ $\hat{\theta }$ is a vector of constants. However, some components $\hat{\bm{\theta }}$ could be states dynamic consideration, and lemma was used such setup (i.e., with involving states) later Proposition 3.4]. give more explicit version statement provide correct proof. desired will deduced from following: 1.Consider system differential equations Proof.Consider following ideal Now we prove claim. Consider ring R : = C [ x , μ ] { u } 1 / Q $R := \mathbb {C}[\bm{x}, \bm{\mu }]\lbrace \bm{u}\rbrace [1/Q]$ . Let J generated by I ∩ $I \cap \bm{u}\rbrace$ R. definition via saturation at implies ∼ $\widetilde{R}$ localization respect $\mathbb {C}\lbrace $\widetilde{J}$ this localization. derivation L $\mathcal {L}$ can naturally extended also -invariant. It sufficient ≠ 0 $\widetilde{J}\cap }] \ne \lbrace 0\rbrace$ nonzero element $\widetilde{J} smallest number monomials and, among elements, total degree. call it S. If S ∈ $S\in }]$ done. Otherwise, one appears S, say u1. h ord $h \operatorname{ord}_{u_1}S$ Since Noetherian ring, there exists N > $N 0$ corollary equivalent 1848] but explicitly highlights entries may initial conditions, not parameters. Corollary 1. (Clarified [[1], 1848])In notation Section 2.2], let P ( … ) $P(\bm{\mu }, \bm{x}, u, \ldots u^{(N)})\in {C}[\bm{\mu \bm{x}] \rbrace$ nonzero. Then exist nonempty Zariski open subsets Θ ⊂ s $\Theta {\subset }\mathbb {C}^{s}$ U ∞ $U\subset {C}^{\infty }(0)$ that, every * }} (\hat{\bm{\mu }}, \hat{\bm{x}}^\ast )\in \Theta$ $\hat{u}\in U$ corresponding X $\hat{\bm{x}} X(\hat{\bm{\theta \hat{u})$ function $P(\hat{\bm{\mu \hat{\bm{x}}, \hat{u},\ldots ,(\hat{u})^{(N)})$ Proof.We apply model Σ polynomial as statement, obtain polynomials $P_1(\bm{x}, })$ P2(u). define sets $P_1 2 | t $P_2(\bm{u})|_{t 0} respectively. ∗ $(\hat{\bm{\mu \hat{\bm{x}}^*) \in $\hat{u} function. □ $\Box$