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Minimal time splines on the sphere

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Abstract

An interesting problem in geometric control theory arises from robotics and space science: find a smooth curve, controlled by a bounded acceleration, connecting in minimum time two prescribed tangent vectors of a Riemannian manifold Q. The state equation is \( \nabla _{\dot{\gamma }} \dot{\gamma } = u \in TQ, |u| \le A\). Applying Pontryagin’s principle one gets a Hamiltonian system in \(T^*(TQ)\). We consider this problem in \(S^2(r)\). Seemingly, it has not been addressed before. Via the SO(3) symmetry, we reduce the four degrees of freedom system to the five variables \((a,v, M_1,M_2,M_3),\) where v is the scalar velocity, conjugated to a costate variable a and \((M_1,M_2,M_3)\) are costate variables that satisfy \(\{ M_i, M_j\} = \epsilon _{ijk} M_k\). We derive the reduced equations and find special analytical solutions, that are organizing centers for the dynamics. Reconstruction of the curve \(\gamma (t)\) is achieved by a time dependent linear system of ODEs for the orthogonal matrix R whose first column is the unit tangent vector of the curve and whose last column is the unit normal vector to the sphere.

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Notes

  1. Warning: this is a spoiler. A thrilling rendez-vous example is the rescue of astronaut Mark Watney (Matt Damon) by Melissa Lewis (Jessica Chastain) in the movie ’The Martian’, directed by Ridley Scott, based on the novel by Weyr [4].

  2. The time minimal-bounded acceleration problem is almost always equivalent to the so called \(L^{\infty }\) control problem considered recently by Noakes and Kaya [13, 14], where they ask for a trajectory that minimizes the sup of the norms of the accelerations, with fixed transition time.

  3. In the unreduced problem it means that any trajectory can be flipped about a tangent vector at any time.

  4. We would like to lure a scientific initiation student for that.

  5. At first sight there are 5 unknowns for the four implicit equations, but the momenta \( p_{v_{\theta }}, p_{v_{\phi }} \) act with a scale invariance and they behave as one unknown.

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Acknowledgements

This work was supported by CAPES/CNPq grants from the Science without Frontiers program PVE11/2012 and PVE089/2013. We thank Maria Soledad Aronna and Pierre Martinon for her help with the software BOCOP. The time minimal spline problem on the sphere was implemented by students from the Summer Undergraduate Program PIBIC at the Mathematics Department of UFMG. JK thanks the colleagues Raphael Campos Drumond, Mario Jorge Carneiro, Sylvie Kamphorst, Sonia Carvalho for the invitation, and to Matthew Perlmutter for sharing the classes. He also wishes to thank the Organizing Committee and the colleagues at the Instituto Superior Tecnico for a wonderful meeting in honor of Prof. Waldyr Oliva.

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Correspondence to Jair Koiller.

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Dedicated to Waldyr Muniz Oliva.

Supported by Capes/CNPq/CsF 089-2013.

Appendices

Appendix A: Linearization at the equilbria

We take entries in the order \( v, a , \theta , z \). We have at the equilibria: \( L= J \, \mathrm{Hess}(H) \) where

$$\begin{aligned} J = \left( \begin{array}{cccc} 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 \\ -1 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 1/\mu \\ 0 &{}\quad 0 &{}\quad -1/\mu &{}\quad 0 \end{array} \right) , \quad \mathrm{Hess}(H) = \left( \begin{array}{cccc} H_{vv} &{}\quad H_{va} &{}\quad H_{v \theta } &{}\quad H_{vz} \\ H_{av} &{}\quad H_{aa} &{}\quad H_{a \theta } &{}\quad H_{az} \\ H_{\theta v} &{}\quad H_{\theta a} &{}\quad H_{\theta \theta } &{}\quad H_{\theta z} \\ H_{zc} &{}\quad H_{za} &{}\quad H_{z \theta } &{}\quad H_{zz} \end{array} \right) \end{aligned}$$

so that

$$\begin{aligned} L = \left( \begin{array}{cccc} H_{av} &{}\quad H_{aa} &{}\quad H_{a \theta } &{}\quad H_{az} \\ - H_{vv} &{}\quad - H_{va} &{}\quad - H_{v \theta } &{}\quad - H_{vz} \\ H_{z v}/\mu &{}\quad H_{za}/\mu &{}\quad H_{z \theta } /\mu &{}\quad H_{z z}/\mu \\ - \, H_{\theta v}/\mu &{}\quad - \, H_{\theta a}/\mu &{}\quad - \, H_{\theta \theta }/\mu &{}\quad - H_{\theta z}/\mu \end{array} \right) . \end{aligned}$$
(59)

We get the four linearization matrices by substituting \(\mu = \sqrt{2A/r}\) and

$$\begin{aligned} a= & {} 0,\quad v = \pm \sqrt{Ar} \\ v> & {} 0 : \quad \theta = \pm \pi /2, \phi = \pi /4 \quad (z = {\sqrt{2}/2}) \\ v< & {} 0 : \quad \theta = \pm \pi /2, \phi = - \pi /4 \quad (z = - {\sqrt{2}/2}) \end{aligned}$$

on the bank of derivatives below. They all give the same matrix L in (45).

Bank of derivatives. Denoting for short \( P = {a}^{2}+{\frac{{\mu }^{2} \left( 1-{z}^{2} \right) \left( \sin \left( \theta \right) \right) ^{2}}{{v}^{2}}} \) we have:

$$\begin{aligned} H_{aa}= & {} -A{a}^{2} P^{-3/2}+ A\, P^{-1/2} \\ H_{vv}= & {} -{\frac{A{\mu }^{4} \left( 1 - {z}^{2} \right) ^{2} \left( \sin \left( \theta \right) \right) ^{4}}{{v}^{6}} P^{-3/2}} \\&+\, 3\,{\frac{A{\mu }^{2} \left( 1-{z}^{2} \right) \left( \sin \left( \theta \right) \right) ^{2}}{{v}^{4}}}\,P^{-1/2} \\ H_{zz}= & {} -{\frac{A{\mu }^{4}{z}^{2} \left( \sin \left( \theta \right) \right) ^{4}}{{v}^{4}}}\, P^{-3/2}-{\frac{A{\mu }^{2} \left( \sin \left( \theta \right) \right) ^{2}}{{v}^{2}}}\, P^{-1/2} \\ H_{\theta \theta }= & {} -{\frac{A{\mu }^{4} \left( 1-{z}^{2}\right) ^{2} \left( \sin \left( \theta \right) \right) ^{2} \left( \cos \left( \theta \right) \right) ^{2}}{{v}^{4}}} P^{-3/2}\\&+{\frac{A{\mu }^{2} \left( 1-{z}^{2}\right) \left( \cos \left( 2\, \theta \right) \right) ^{2}}{{v}^{2}}} \, P^{-1/2} \\ H_{av}= & {} {\frac{Aa{\mu }^{2} \left( 1-{z}^{2}\right) \left( \sin \left( \theta \right) \right) ^{2}}{{v}^{3}}}\, P^{-3/2}\\ H_{a \theta }= & {} -{\frac{Aa{\mu }^{2} \left( 1-{z}^{2}\right) \sin \left( \theta \right) \cos \left( \theta \right) }{{v}^{2}}}\, P^{-3/2} \\ H_{\theta z}= & {} {\frac{A{\mu }^{4}z \left( \sin \left( \theta \right) \right) ^{3}\left( 1-{z}^{2}\right) \cos \left( \theta \right) }{{v}^{4} }} P^{-3/2} - 2 \,{\frac{A{\mu }^{2}z\sin \left( \theta \right) \cos \left( \theta \right) }{{v}^{2}}}\, P^{-1/2} \\ H_{v \theta }= & {} {\frac{A{\mu }^{4} \left( 1-{z}^{2}\right) ^{2} \left( \sin \left( \theta \right) \right) ^{3}\cos \left( \theta \right) }{{v}^ {5}}}\, P^{-3/2} \\&- 2 \,{\frac{A{\mu }^{2} \left( 1-{z}^{2}\right) \sin \left( \theta \right) \cos \left( \theta \right) }{{v}^{3}}}\, P^{-1/2} \\ H_{az}= & {} \frac{Aa{\mu }^{2}z \,\left( \sin \left( \theta \right) \right) ^{2}}{{v}^{2}} \,P^{-3/2} \\ H_{vz}= & {} -{\frac{Aa{\mu }^{2} \left( 1-{z}^{2} \right) \sin \left( \theta \right) \cos \left( \theta \right) }{{v}^{2}}}\, P^{-3/2} \end{aligned}$$

Appendix B: Dictionary between the momenta in \( T^*(TS^2)\) associated to the coordinate systems on \(TS^2\)

We derive the correspondence \( p_{\theta }, p_{\phi }, p_{v_{\theta }}, p_{v_{\phi }}\) to \( (a, M_1, M_2, M_3) \) that may be useful in future work. Let \(Z(\alpha ) \) denote the rotation about the z axis of and angle \(\alpha \) and \(Y(\beta )\) about the y axis of angle \(\beta \). We introduce Euler angles \(\xi _1,\xi _2,\xi _3\), for matrices \(R \in SO(3)\) so that

$$\begin{aligned} R= & {} Z(\xi _1) \, Y(\pi /2 - \xi _2) \, Z(\xi _3) \nonumber \\ {}= & {} \left( \begin{array}{ccc} c_1 s_2 c_3 - s_1 s_3 &{}\quad -c_3 s_1 - c_1 s_2 s_3 &{}\quad c_1 c_2 \\ c_1 s_3 + s_2 c_3 s_1 &{}\quad c_1 c_3 - s_2 s_1 s_3 &{}\quad s_1 c_2 \\ - c_3 c_2 &{}\quad c_2 s_3 &{}\quad s_2 \end{array} \right) \end{aligned}$$
(60)

with \( -\pi \le \xi _1, \xi _3 \le \pi , - \pi /2 \le \xi _2 \le \pi /2. \) This is one among many possible parametrizations. We interchanged in Y the cosines with sines in order to the third column of R to have the same entries as the with \(\theta ,\phi \) of the spherical coordinates, identifying them with Euler angles 1,2 respectively. Recall the map \(SO(3) \times \mathfrak {R}_+ \rightarrow TS^2-0\). Since the normalized velocity vector is the first column \(R^1\), we have \( v_{\theta } (r\, \cos \phi ) = v\, \, e_{\theta } \cdot R^1,\quad v_{\phi } \, r = v \,\, e_{\phi } \cdot R^1 . \) It results that the mapping \((v, \xi _1,\xi _2,\xi _3) \mapsto (\theta ,\phi , v_{\theta }, v_{\phi })\) is given by

$$\begin{aligned} \theta = \xi _1, \quad \phi = \xi _2 \, ,\quad v_{\theta } = \frac{v \, \sin \xi _3}{r\, \cos \xi _2},\quad v_{\phi } = - \frac{v \cos (\xi _3)}{r} . \end{aligned}$$
(61)

and can be easily inverted:

$$\begin{aligned} \xi _3 = - \mathrm{arctan}(\cos (\phi ) v_{\theta }/v_{\phi }),\quad v = r \sqrt{\cos ^2(\phi ) v^2_{\theta } + v^2_{\phi }} \,. \end{aligned}$$
(62)

The conjugate momenta can be related by the pullback of the canonical 1-form in \(T^*(TS^2)\):

$$\begin{aligned} a \, dv + p_{\xi _1}\, d\xi _1 + p_{\xi _2}\, d\xi _2 + p_{\xi _3}\, d\xi _3 = p_{\theta }\, d\theta + p_{\phi }\, d\phi + p_{v_{\theta } }\, dv_{\theta } + p_{v_{\phi } }\, dv_{\phi } \end{aligned}$$
(63)

Computing the differentials \(d\theta , d\phi , dv_{\theta }, dv_{\phi }\) from (61) and inserting in the RHS of the above identity, and collecting terms, one gets for \(a, p_{\xi _1}, p_{\xi _2}, p_{\xi _2}\) linear functions of \( p_{\theta }, p_{\phi }, p_{v_{\theta }}, p_{v_{\phi }}\) with coefficients depending on \(\theta =\xi _1, \phi =\xi _2, \xi _3, v \). The result is

$$\begin{aligned} \left[ \begin{array}{c} p_{\xi _1} \\ p_{\xi _2} \\ p_{\xi _3}\\ a \end{array} \right] = \left[ \begin{array}{cccc} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 1 &{}\quad - \frac{ v\, s_3 \, s_2}{r \, c^2_2} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad \frac{v \, c_3}{r\, c_2} &{}\quad \frac{v \, s_3}{r} \\ \, &{}\quad \, &{}\quad \, &{}\quad \, \\ 0 &{}\quad 0 &{}\quad \frac{s_3}{r \, c_2} &{}\quad - \frac{c_3}{r} \end{array} \right] \left[ \begin{array}{c} p_{\theta } \\ p_{\phi } \\ p_{v_{\theta }}\\ p_{v_{\phi }} \end{array} \right] \end{aligned}$$
(64)

A second step consists on relating \( (M_1,M_2,M_3)\) with \((p_{\xi _1}, p_{\xi _2},p_{\xi _3})\). We start with the identity in \(T^*SO(3)\)

$$\begin{aligned} p_{\xi _1} d\xi _1 + p_{\xi _2} d\xi _2 + p_{\xi _3} d\xi _3 = M_1 \, \varOmega _1+ M_2 \, \varOmega _2 + M_2 \, \varOmega _3 \end{aligned}$$
(65)

where the \(\varOmega _i\) are differential forms that can be expressed in terms of the \(d\xi _1,d\xi _2,d\xi _3\), via the very classical expressions (going back to Euler) that come from

$$\begin{aligned} R^{-1} dR = [\varOmega ] = \left( \begin{array}{ccc} 0 &{}\quad -\varOmega _3 &{}\quad \varOmega _2 \\ \ldots &{}\quad 0 &{}\quad - \varOmega _1 \\ \ldots &{}\quad \ldots &{}\quad 0 \end{array} \right) . \end{aligned}$$

A long but straightforward computation, or a fast computer algebra (we did it on Maple), gives well known formulae for the infinitesimal rotations in the body frame:

$$\begin{aligned} \varOmega _1= & {} -\cos (\xi _3) \, \cos (\xi _2) \, d\xi _1 - \sin (\xi _3)\, d\xi _2 \nonumber \\ \varOmega _2= & {} \cos (\xi _2)\, \sin (\xi _3) \, d\xi _1- \cos (\xi _3) \,d\xi _2 \nonumber \\ \varOmega _3= & {} \sin (\xi _2) \, d\xi _1 + d\xi _3 \end{aligned}$$
(66)

which gives

$$\begin{aligned} p_{\xi _1}= & {} - M_1 \cos (\xi _3) \cos (\xi _2) + M_2 \cos (\xi _2) \sin (\xi _3) + M_3 \, \sin (\xi _2) \nonumber \\ p_{\xi _2}= & {} -\sin (\xi _3) \, M_1 - \cos (\xi _3) \, M_2 \nonumber \\ p_{\xi _3}= & {} M_3 \end{aligned}$$
(67)

Inverting these linear relations we get

$$\begin{aligned} \left[ \begin{array}{c} M_1 \\ M_2 \\ M_3 \\ a \end{array} \right] = \left[ \begin{array}{cccc} -c_3/c_2 &{}\quad -s_3 &{}\quad s_2 c_3/c_2 &{}\quad 0 \\ \, &{}\quad \, &{}\quad \, &{}\quad \, \\ s_3/c_2 &{}\quad -c_3 &{}\quad -s_2s_3/c_2 &{}\quad 0 \\ \, &{}\quad \, &{}\quad \, &{}\quad \, \\ 0 &{}\quad 0 &{}\quad 1 &{}\quad 0 \\ \, &{}\quad \, &{}\quad \, &{}\quad \, \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 1 \\ \end{array} \right] \left[ \begin{array}{c} p_{\xi _1} \\ p_{\xi _2} \\ p_{\xi _3}\\ a \end{array} \right] \end{aligned}$$
(68)

Multiplying the matrices in (64, 68) one obtains the desired correspondence

$$\begin{aligned} \left[ \begin{array}{c} M_1 \\ M_2 \\ M_3 \\ a \end{array} \right] = \left[ \begin{array}{cccc} -\frac{c_3}{c_2} &{}\quad -s_3 &{}\quad \frac{v s_2}{r\, c_2^2} &{}\quad \frac{v s_2 s_3 c_3}{r\, c_2} \\ \, &{}\quad \, &{}\quad \, &{}\quad \, \\ \frac{s_3}{c_2 } &{}\quad -c_3 &{}\quad 0 &{}\quad - \frac{v s_2 s_3^2}{r\, c_2} \\ \, &{}\quad \, &{}\quad \, &{}\quad \, \\ 0 &{}\quad 0 &{}\quad \frac{v \, c_3}{r\, c_2} &{}\quad \frac{v s_3}{r} \\ \, &{}\quad \, &{}\quad \, &{}\quad \, \\ 0 &{}\quad 0 &{}\quad \frac{ s_3}{r\, c_2} &{}\quad - \frac{c_3}{r} \\ \end{array} \right] \left[ \begin{array}{c} p_{\theta } \\ p_{\phi } \\ p_{v_{\theta }}\\ p_{v_{\phi }} \end{array} \right] \end{aligned}$$
(69)

The reader may further wish to change to the split momenta coordinates, in this case just changing to, according to [16] [with \(v_{\theta }\) from (61)]

$$\begin{aligned} \hat{p}_{\theta } = p_{\theta } - \cos (\phi ) \sin (\phi ) v_{\theta } p_{v_{\phi }} , \hat{p}_{\phi } = p_{\phi } + 2 \tan (\phi ) v_{\theta } p_{v_{\theta }}. \end{aligned}$$
(70)

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Stuchi, T., Balseiro, P., Cabrera, A. et al. Minimal time splines on the sphere. São Paulo J. Math. Sci. 12, 82–107 (2018). https://doi.org/10.1007/s40863-017-0078-4

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