The d'Alembert and vakonomic motions of $z^{\mathrm{\prime }}=x{y}^{\mathrm{\prime }}$ differ from the same initial data

(the d'Alembert system.) Let $q$ be a d'Alembert motion on an open interval containing $0$. Since $q^{\mathrm{\prime }\mathrm{\prime }}$ is orthogonal to $\delta =(1,0,0)$ and to $\delta =(0,1,x)$, we get $x^{\mathrm{\prime }\mathrm{\prime }}=0$ and $y^{\mathrm{\prime }\mathrm{\prime }}+x{z}^{\mathrm{\prime }\mathrm{\prime }}=0$. Put $\lambda :=z^{\mathrm{\prime }\mathrm{\prime }}$, a continuous function; then $y^{\mathrm{\prime }\mathrm{\prime }}=-\lambda x$ and $z^{\mathrm{\prime }\mathrm{\prime }}=\lambda$. Differentiating the constraint $z^{\mathrm{\prime }}=x{y}^{\mathrm{\prime }}$ gives $z^{\mathrm{\prime }\mathrm{\prime }}=x^{\mathrm{\prime }}{y}^{\mathrm{\prime }}+x{y}^{\mathrm{\prime }\mathrm{\prime }}$, so $\lambda =x^{\mathrm{\prime }}{y}^{\mathrm{\prime }}+x(-\lambda x)=x^{\mathrm{\prime }}{y}^{\mathrm{\prime }}-\lambda x^2$, hence $\lambda =\frac{x^{\mathrm{\prime }}{y}^{\mathrm{\prime }}}{1+x^2},x^{\mathrm{\prime }\mathrm{\prime }}=0,y^{\mathrm{\prime }\mathrm{\prime }}=-\frac{x{x}^{\mathrm{\prime }}{y}^{\mathrm{\prime }}}{1+x^2},z^{\mathrm{\prime }\mathrm{\prime }}=\frac{x^{\mathrm{\prime }}{y}^{\mathrm{\prime }}}{1+x^2}\mathrm{.}$

(the vakonomic equations.) Suppose $q$ is a vakonomic motion with the same data: there are $a<b$ in the motion's domain and a $C^2$ function $\eta$ on an open interval containing $[a,b]$ such that $(q,\eta )$ is a vakonomic stationary pair, i.e. the directional derivative of $\mathrm{\Phi }$ vanishes for every admissible pair variation. Computing that derivative for a variation $(\xi ,\zeta )$ with $\xi (a)=\xi (b)=0$ and integrating by parts in the terms containing ${\xi }^{\mathrm{\prime }}$ and $\zeta$ gives $0={\int }_a^b\left[\right(-x^{\mathrm{\prime }\mathrm{\prime }}-\eta y^{\mathrm{\prime }}){\xi }_x+(-y^{\mathrm{\prime }\mathrm{\prime }}+(\eta x{)}^{\mathrm{\prime }}){\xi }_y+(-z^{\mathrm{\prime }\mathrm{\prime }}-{\eta }^{\mathrm{\prime }}){\xi }_z]dt+{\int }_a^b\zeta (z^{\mathrm{\prime }}-x{y}^{\mathrm{\prime }})dt\mathrm{.}$ Taking first $\zeta =0$ and $\xi$ running through the bumps of [F1] in each component yields $x^{\mathrm{\prime }\mathrm{\prime }}=-\eta y^{\mathrm{\prime }}$, $y^{\mathrm{\prime }\mathrm{\prime }}=(\eta x{)}^{\mathrm{\prime }}$, $z^{\mathrm{\prime }\mathrm{\prime }}=-{\eta }^{\mathrm{\prime }}$ on $[a,b]$; then taking $\xi =0$ and $\zeta$ a bump gives $z^{\mathrm{\prime }}=x{y}^{\mathrm{\prime }}$.

(existence and uniqueness.) The system of step 1.1 is the first-order system of [F2]. Hence it has a unique maximal solution through the data $(x,y,z,X,Y,Z)(0)=(0,0,0,1,1,0)$; every d'Alembert motion with these data solves it, so all such motions are restrictions of that unique maximal solution.

(the explicit solution.) Define $x(t):=t$, $y(t):={\int }_0^t(1+s^2{)}^{-1\mathrm{/}2}ds$ and $z(t):={\int }_0^{t}s(1+s^2{)}^{-1\mathrm{/}2}ds$, using oriented integrals (The integral with oriented limits: ${\int }_a^{a}f:=0$ and ${\int }_b^{a}f:=-{\int }_a^{b}f$). The integrands are $C^1$ on $\mathbb{R}$, so these integrals exist on every bounded interval; additivity (For $a<c<b$: $f$ is integrable on $[a,b]$ if and only if it is integrable on $[a,c]$ and on $[c,b]$, and then ${\int }_a^{b}f={\int }_a^{c}f+{\int }_c^{b}f$; with the oriented form for arbitrary $a,b,c$) and The first fundamental theorem: if $f$ is integrable on $[a,b]$ and continuous at $c$, then $F^{\mathrm{\prime }}(c)=f(c)$; in particular a continuous $f$ has $F$ as a primitive give $y^{\mathrm{\prime }}(t)=(1+t^2{)}^{-1\mathrm{/}2}$ and $z^{\mathrm{\prime }}(t)=t(1+t^2{)}^{-1\mathrm{/}2}$ for all $t\in \mathbb{R}$, including $t=0$. Thus $y,z$ are $C^2$ and $y(0)=z(0)=0$. By [F3], $y^{\mathrm{\prime }\mathrm{\prime }}(t)=-t(1+t^2{)}^{-3\mathrm{/}2}=-\frac{x(t)x^{\mathrm{\prime }}(t)y^{\mathrm{\prime }}(t)}{1+x(t{)}^2},z^{\mathrm{\prime }\mathrm{\prime }}(t)=(1+t^2{)}^{-3\mathrm{/}2}=\frac{x^{\mathrm{\prime }}(t)y^{\mathrm{\prime }}(t)}{1+x(t{)}^2},$ and $x^{\mathrm{\prime }\mathrm{\prime }}=0$, while $x(t)y^{\mathrm{\prime }}(t)=t(1+t^2{)}^{-1\mathrm{/}2}=z^{\mathrm{\prime }}(t)$; the data at $0$ are $(0,0,0,1,1,0)$. Therefore this curve is a solution of the system defined on all of $\mathbb{R}$, so by step 2.1 it is the unique maximal d'Alembert motion with the displayed data. Note $y^{\mathrm{\prime }}(t)>0$ and $y^{\mathrm{\prime }}(t)<1$ for $t>0$.

(contradiction.) For the explicit motion of step 3.1 we have $x^{\mathrm{\prime }\mathrm{\prime }}=0$ and $y^{\mathrm{\prime }}>0$ on $[a,b]$, so the first equation of step 1.2 gives $\eta =0$ on $[a,b]$. Then $y^{\mathrm{\prime }\mathrm{\prime }}=(\eta x{)}^{\mathrm{\prime }}={\eta }^{\mathrm{\prime }}x+\eta x^{\mathrm{\prime }}=0$ on $[a,b]$, on its interior. This is impossible: every nondegenerate interval $(a,b)$ contains a nonzero point $t$, where the explicit formula gives $y^{\mathrm{\prime }\mathrm{\prime }}(t)=-t(1+t^2{)}^{-3\mathrm{/}2}\mathrm{\ne }0$. Hence no such pair $(q,\eta )$ exists, and the d'Alembert motion of step 3.1 is not a vakonomic motion.

The displayed data therefore give a d'Alembert motion which is not a vakonomic motion with the same initial data, refuting the statement.