Guaranteed simulation of nonlinear continuous-time dynamical systems using interval methods

Transcription

1 Guaranteed simulation of nonlinear continuous-time dynamical systems using interval methods L. Jaulin, A. Béthencourt, S. Rohou, F. Le Bars, B. Zerr, T. Le Mézo ENSTA Bretagne, LabSTICC 13 septembre 2017 Rennes

2 Lagrangian simulation

3 Contractors Lagrangian simulation

4 The operator C : IR n IR n is a contractor [4] for the equation f (x) = 0, if { C ([x]) [x] (contractance) x [x] and f (x) = 0 x C ([x]) (consistence)

5 Building contractors Consider the primitive equation x 1 + x 2 = x 3 with x 1 [x 1 ], x 2 [x 2 ], x 3 [x 3 ].

6 We have Lagrangian simulation x 3 = x 1 + x 2 x 3 [x 3 ] ([x 1 ] + [x 2 ]) x 1 = x 3 x 2 x 1 [x 1 ] ([x 3 ] [x 2 ]) x 2 = x 3 x 1 x 2 [x 2 ] ([x 3 ] [x 1 ])

7 The contractor associated with x 1 + x 2 = x 3 is thus [x 1 ] [x 1 ] ([x 3 ] [x 2 ]) C [x 2 ] = [x 2 ] ([x 3 ] [x 1 ]) [x 3 ] [x 3 ] ([x 1 ] + [x 2 ])

8 Tubes Lagrangian simulation

9 A trajectory is a function f : R R n. For instance ( ) cost f (t) = sint is a trajectory.

10 Order relation Lagrangian simulation f g t, i,f i (t) g i (t).

11 We have Lagrangian simulation h = f g t, i,h i (t) = min(f i (t),g i (t)), h = f g t, i,h i (t) = max(f i (t),g i (t)).

12 The set of trajectories is a lattice. Interval of trajectories (tubes) can be defined.

13 Example. Lagrangian simulation ( [ cost + 0,t 2 ] [f](t) = sint + [ 1,1] is an interval trajectory (or tube). )

14 Tube arithmetics Lagrangian simulation

15 If [x] and [y] are two scalar tubes [1], we have [z] = [x] + [y] [z](t) = [x](t) + [y](t) (sum) [z] = shift a ([x]) [z](t) = [x](t + a) (shift) [z] = [x] [y] [z](t) = [x]([y](t)) (composition) [z] = [x] [z](t) = [ t 0 x (τ)dτ, t 0 x + (τ)dτ ] (integral)

16 Tube Contractors Lagrangian simulation

17 Tube arithmetic allows us to build contractors [3].

18 Consider for instance the differential constraint ẋ (t) = x (t + 1) u (t), x (t) [x](t),ẋ (t) [ẋ](t),u (t) [u](t) We decompose as follows x (t) = x (0) + t 0 y (τ)dτ y (t) = a(t) u (t). a(t) = x (t + 1)

19 Possible contractors are [x](t) = [x](t) ( [x](0) + t 0 [y](τ)dτ) [y](t) = [y](t) [a](t) [u](t) [u](t) = [u](t) [y](t) [a](t) [a](t) = [a](t) [y](t) [u](t) [a](t) = [a](t) [x](t + 1) [x](t) = [x](t) [a](t 1)

20 Example. Consider x (t) [x](t) with the constraint Contract the tube [x](t). t, x (t) = x (t + 1)

21 We first decompose into primitive trajectory constraints x (t) = a(t + 1) x (t) = a(t).

22 Contractors Lagrangian simulation [x](t) : = [x](t) [a](t + 1) [a](t) : = [a](t) [x](t 1) [x](t) : = [x](t) [a](t) [a](t) : = [a](t) [x](t)

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30 [7]

31 Time-space estimation

32 Classical state estimation { ẋ(t) = f (x(t),u(t)) t R 0 = g (x(t),t) t T R. Space constraint g (x(t),t) = 0.

33 Example. Lagrangian simulation ẋ 1 = x 3 cosx 4 ẋ 2 = x 3 cosx 4 ẋ 3 = u 1 ẋ 4 = u 2 (x 1 (5) 1) 2 + (x 2 (5) 2) 2 4 = 0 (x 1 (7) 1) 2 + (x 2 (7) 2) 2 9 = 0

34 With time-space constraints { ẋ(t) = f (x(t),u(t)) t R 0 = g (x(t),x(t ),t,t ) (t,t ) T R R.

35 Example. An ultrasonic underwater robot with state x = (x 1,x 2,...) = (x,y,θ,v,...) At time t the robot emits an onmidirectional sound. At time t it receives it ( ) 2 ( 2 ( x 1 x 1 + x 2 x 2) c t t ) 2 = 0.

36 Mass spring problem Lagrangian simulation

37 The mass spring satisfies ẍ + ẋ + x x 3 = 0 i.e. { ẋ1 = x 2 ẋ 2 = x 2 x 1 + x1 3 The initial state is unknown.

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39 ẋ 1 = x 2 ẋ 2 = x 2 x 1 + x 3 1 L x 1 (t 1 ) + L x 1 (t 2 ) = c (t 2 t 1 ).

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41 Swarm localization Lagrangian simulation

42 Consider n robots R 1,...,R n described by ẋ i = f (x i,u i ),u i [u i ].

43 Omnidirectional sounds are emitted and received. A ping is a 4-uple (a,b,i,j) where a is the emission time, b is the reception time, i is the emitting robot and j the receiver.

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45 With the time space constraint ẋ i = f (x i,u i ),u i [u i ]. g ( x i(k) (a(k)),x j(k) (b (k)),a(k),b (k) ) = 0 where g (x i,x j,a,b) = x 1 x 2 c (b a).

46 Clocks are uncertain. We only have measurements ã(k), b (k) of a(k),b (k) thanks to clocks h i. Thus ẋ i = f (x i,u i ),u i [u i ]. g ( x i(k) (a(k)),x j(k) (b (k)),a(k),b (k) ) = 0 ã(k) = h i(k) (a(k)) b (k) = h j(k) (b (k))

47 The drift of the clocks is bounded ẋ i = f (x i,u i ),u i [u i ]. g ( x i(k) (a(k)),x j(k) (b (k)),a(k),b (k) ) = 0 ã(k) = h i(k) (a(k)) b (k) = h j(k) (b (k)) ḣ i = 1 + n h, n h [n h ]

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51 Lagrangian simulation

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54 Maze Lagrangian simulation

55 An interval is a domain which encloses a real number. A polygon is a domain which encloses a vector of R n. A maze is a domain which encloses a path. [6]

56 A maze is a set of paths.

57 Mazes can be made more accurate:

58 Here, a maze L is composed of [6][5]. A paving P Doors between adjacent boxes

59 The set of mazes forms a lattice with respect to.

60 Eulerian smoother Lagrangian simulation

61 Example. Take the Van der Pol system with X 0 = [a] = [0,0.6] [0.8,1.8] X 1 = [b] = [0.7,1.5] [ 0.2,0.2] = [c] = [0.2,0.6] [ 2.2, 1.5] X 2

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66 An application of Eulerian state estimation moving taking advantage of ocean currents.

67 Visiting the three red boxes using a buoy that follows the currents is an Eulerian state estimation problem

68 F. Le Bars, J. Sliwka, O. Reynet, and L. Jaulin. State estimation with fleeting data. Automatica, 48(2): , A. Bethencourt and L. Jaulin. Cooperative localization of underwater robots with unsynchronized clocks. Journal of Behavioral Robotics, 4(4): , A. Bethencourt and L. Jaulin. Solving non-linear constraint satisfaction problems involving time-dependant functions. Mathematics in Computer Science, 8(3), G. Chabert and L. Jaulin. Contractor Programming. Artificial Intelligence, 173: , T. Le Mézo L. Jaulin and B. Zerr.

69 Bracketing the solutions of an ordinary differential equation with uncertain initial conditions. Applied Mathematics and Computation, T. Le Mézo, L. Jaulin, and B. Zerr. An interval approach to compute invariant sets. IEEE Transaction on Automatic Control, S. Rohou, L. Jaulin, M. Mihaylova, F. Le Bars, and S. Veres. Guaranteed Computation of Robots Trajectories. Robotics and Autonomous Systems, 93:76 84, 2017.