Constrained dynamics shake and rattle — Examples

Here $J(q)=q^T$, $A(q)=q^{T}q\mathrm{/}m={\mathrm{\ell }}^2\mathrm{/}m$, $D^{2}c[v,v]=\mathrm{\parallel }p{\mathrm{\parallel }}_2^2\mathrm{/}{m}^2$, and $J{M}^{-1}\mathrm{\nabla }V=g{q}_2$. Substitution into [F1] gives $\lambda =(\mathrm{\parallel }p{\mathrm{\parallel }}_2^2\mathrm{/}m-mg{q}_2)\mathrm{/}{\mathrm{\ell }}^2$.

At $q=(0,-\mathrm{\ell })$ and $p=0$, the multiplier is $mg\mathrm{/}\mathrm{\ell }$. The acceleration is $\dot{p}=-mg(0,1)-\lambda (0,-\mathrm{\ell })=(0,0)$; moreover $c(q)=0$ and $q^{T}p=0$, so this is a consistent equilibrium of the local flow [F2].

The initial data are consistent because $\mathrm{\parallel }{q}_0{\mathrm{\parallel }}_2=\mathrm{\ell }$ and $q_0^T{p}_0=0$. The position multiplier equation gives $q_1=q_0+\frac{h}{m}(p_0-h\lambda q_0)$; choosing $\lambda =m(1-\sqrt{1-h^2{\omega }^2})\mathrm{/}{h}^2=m{\omega }^2\mathrm{/}(1+\sqrt{1-h^2{\omega }^2})$ yields the stated $q_1$, whose squared norm is ${\mathrm{\ell }}^2$. This multiplier tends to $m{\omega }^2\mathrm{/}2$, the position branch center.

With $\mu =\lambda$, the final RATTLE equation gives the stated $p_1$; $\mu$ has the same limit $m{\omega }^2\mathrm{/}2$, so the unique local branch selects this root. Direct multiplication gives $q_1^T{p}_1=0$ and $\mathrm{\parallel }{p}_1{\mathrm{\parallel }}_2=m\omega \mathrm{\ell }$, so both exact constraints hold.

The continuous multiplier is ${\lambda }_c=m{\omega }^2$, hence the exact flow is $q(t)=\mathrm{\ell }(\cos \omega t,\sin \omega t)$ and $p(t)=m\omega \mathrm{\ell }(-\sin \omega t,\cos \omega t)$. This solves [F3] and agrees with the initial data. The RATTLE position angle is ${\theta }_h=\arcsin (h\omega )$. Taylor's theorem [F4] on $0\le h\omega \le 1\mathrm{/}2$ gives $\arcsin (h\omega )-h\omega =O((h\omega {)}^3)$. Since $\omega$ is fixed, the angle defect is $O(h^3)$ as stated.

The two rows $(0,0,1)$ and $(0,ϵ,1)$ of $J_{ϵ}$ are linearly independent exactly when $ϵ>0$. Their Gram matrix is the displayed $A_{ϵ}$, whose trace is $2+{ϵ}^2$ and determinant is ${ϵ}^2$.

The quadratic formula gives its eigenvalues $(2+{ϵ}^2\pm \sqrt{4+{ϵ}^4})\mathrm{/}2$. Rationalizing the smaller one gives $2{ϵ}^2\mathrm{/}(2+{ϵ}^2+\sqrt{4+{ϵ}^4})$, so ${ϵ}^{-2}{\lambda }_{\min }(A_{ϵ})\to 1\mathrm{/}2$ and $\mathrm{\parallel }{A}_{ϵ}^{-1}{\mathrm{\parallel }}_2\to \mathrm{\infty }$.

Consequently no single residual-to-multiplier constant works for this family as $ϵ\downarrow 0$; this is consistent with [F2], whose uniform estimates assume a fixed compact chart with uniformly bounded $A^{-1}$. At $ϵ=0$ the two rows coincide, so [F1]'s full-rank hypothesis fails.

For ${\epsilon }_q,{\epsilon }_p\le \epsilon$ at every step, [F1] gives, until a first exit from ${\mathcal{C}}_1$, the recurrence $e_{n+1}\le (1+Lh)e_n+2Ch\epsilon$ for the restored trajectory. Iteration yields $e_N\le 2CT{e}^{LT}\epsilon$. Choose $\epsilon$ so this is less than the coordinate distance from ${\mathcal{C}}_0$ to the boundary of ${\mathcal{C}}_1$; then no first exit occurs, and the uniform upper-bound envelope holds. It need not shrink as $h\to 0$.

If instead ${\epsilon }_q,{\epsilon }_p\le K{h}^2$, [F1] makes each additive perturbation at most $2CK{h}^3$, and the same sum gives global error $O(h^2)$; for sufficiently small $h$ this is below the tube margin, so again there is no first exit. The unscaled residuals are $h^2{\epsilon }_q=O(h^4)$ and $h{\epsilon }_p=O(h^3)$.

These bounds rely on the inverse and stability estimates in [F1]. As [F4] states, the residual-tolerance status alone is not a forward-error certificate; the explicit consistency restoration is also part of this repeated-step example.