The constrained Hamiltonian vector field has a unique local flow

Since $J(q)$ has full row rank and $M^{-1}$ is SPD, for every nonzero $\eta \in {\mathbb{R}}^m$ we have ${\eta }^{T}A(q)\eta =(J(q{)}^T\eta {)}^T{M}^{-1}(J(q{)}^T\eta )>0$. Thus $A(q)$ is invertible; its entries are smooth and $A^{-1}=\mathrm{adj}(A)\mathrm{/}\det A$, so its inverse and the displayed multiplier are smooth on a neighborhood of the chart.

Extend the formula in [F1] to an open neighborhood of $q_0$ on which $A$ remains invertible and define the ambient vector field $(q,p)\mapsto (M^{-1}p,-\mathrm{\nabla }V(q)-J(q{)}^T\lambda (q,p))$. It is smooth, hence locally Lipschitz, so [F3] gives a unique local solution through $(q_0,p_0)$. Shrink its time interval so $q(t)$ stays in that neighborhood.

Along this solution, the chain rule gives $\frac{d}{dt}c(q)=J(q)M^{-1}p$ and $\frac{d}{dt}(J(q)M^{-1}p)=D^{2}c(q)[v,v]-J(q)M^{-1}\mathrm{\nabla }V(q)-A(q)\lambda (q,p)=0$, where $v=M^{-1}p$. Since the second constraint starts at zero, it remains zero; the first derivative is therefore zero and $c(q)$ also remains zero. The chosen neighborhood is regular, and [F2] identifies its consistent constraint set as a smooth phase manifold, so the trajectory stays in $\mathcal{M}$.

The local uniqueness from [F3] applies to every solution of the same ambient extension, so in particular the constrained solution is unique.