The constrained pendulum has differentiation index three

Differentiating $q^{\mathsf{T}}q={\mathrm{\ell }}^2$ once and using $q^{\mathrm{\prime }}=v$ gives $q^{\mathsf{T}}v=0$, the tangency condition. Differentiating a second time gives $0=\mathrm{\parallel }v{\mathrm{\parallel }}_2^2+q^{\mathsf{T}}{v}^{\mathrm{\prime }}=\mathrm{\parallel }v{\mathrm{\parallel }}_2^2-g_0{q}_2-\lambda q^{\mathsf{T}}q$. Using [A1], this equation solves uniquely for $\lambda =(\mathrm{\parallel }v{\mathrm{\parallel }}_2^2-g_0{q}_2)\mathrm{/}{\mathrm{\ell }}^2$.

Differentiate $\mathrm{\parallel }v{\mathrm{\parallel }}_2^2-g_0{q}_2-\lambda {\mathrm{\ell }}^2=0$, the second-derivative constraint from step 1.1. Since $q^{\mathrm{\prime }}=v$ and $v^{\mathrm{\prime }}=-g_0{e}_2-\lambda q$, this gives $0=2v^{\mathsf{T}}{v}^{\mathrm{\prime }}-g_0{v}_2-{\lambda }^{\mathrm{\prime }}{\mathrm{\ell }}^2=-3g_0{v}_2-2\lambda q^{\mathsf{T}}v-{\lambda }^{\mathrm{\prime }}{\mathrm{\ell }}^2=-3g_0{v}_2-{\lambda }^{\mathrm{\prime }}{\mathrm{\ell }}^2,$ where the last equality uses the tangency condition in step 1.1. Since $\mathrm{\ell }>0$, this determines ${\lambda }^{\mathrm{\prime }}=-3g_0{v}_2\mathrm{/}{\mathrm{\ell }}^2$ with nonzero coefficient $-{\mathrm{\ell }}^2$ on the highest algebraic derivative.

The original constraint does not contain $\lambda$; its first derivative yields tangency but not $\lambda$, its second derivative determines $\lambda$, and its third derivative determines ${\lambda }^{\mathrm{\prime }}$. The prolonged equations are regular because the multiplier coefficients are $-{\mathrm{\ell }}^2$ at each of these latter solves. Therefore the least differentiation count in [F2] is three.

For example, $q_0=(\mathrm{\ell },0)$, $v_0=(0,s)$, and ${\lambda }_0=s^2\mathrm{/}{\mathrm{\ell }}^2$ satisfy the position and tangency constraints, and the multiplier formula in step 1.1 gives ${\lambda }_0$. Thus this tuple satisfies the initial conditions exposed by the first two differentiations.