Validated simulation of ODEs Julien Alexandre dit Sandretto Department U2IS ENSTA Paris SSC310-2020 Contents Initial Value Problem of Ordinary Differential Equations Problem of integral computation Picard-Lindel¨of Integration scheme Problem of wrapping effect Affine arithmetic Stepsize controller Validated integration in a contractor formalism Additive constraints Temporal constraints Bibliography Do it yourself Julien Alexandre dit Sandretto - Validated simulation of ODEs October 22, 2020- 2 Initial Value Problem of Ordinary Differential Equations Initial Value Problem of Ordinary Differential Equations Classical problem Consider an IVP for ODE, over the time interval [0; T ] y_ = f (y) with y(0) = y0 n n This IVP has a unique solution y(t; y0) if f : R ! R is Lipschitz. Interval IVP y_ = f (y; p) with y(0) 2 [y0] and p 2 [p] Julien Alexandre dit Sandretto - Validated simulation of ODEs October 22, 2020- 3 Initial Value Problem of Ordinary Differential Equations Numerical Integration R t How compute y(t) = y0 + 0 f (y(s))ds ? Goal of numerical integration I Compute a sequence of time instants: t0 = 0 < t1 < ··· < tn = T I Compute a sequence of values: y0; y1;:::; yn such that 8i 2 [0; n]; yi ≈ y(ti ; y0) : Goal of validated numerical integration I Compute a sequence of time instants: t0 = 0 < t1 < ··· < tn = T I Compute a sequence of values: [y0]; [y1];:::; [yn] such that 8i 2 [0; n]; [yi ] 3 y(ti ; y0) : Julien Alexandre dit Sandretto - Validated simulation of ODEs October 22, 2020- 4 Problem of integral computation Problem of integral computation R h Discrete system given by yn+1 = yn + 0 f (y(s))ds R h Bounding of 0 f (y(s))ds If y(s) is bounded s.t. y(s) 2 [x]; 8s 2 [0; h], then Z h f ([x])ds ⊂ [0; h] · [f ]([x]) 0 How bound y(s) ? Complex, it is what we are trying to compute ! We note by [y~n] ⊃ fy(s); s 2 [tn; tn+1]g Julien Alexandre dit Sandretto - Validated simulation of ODEs October 22, 2020- 5 Picard-Lindel¨of Picard-Lindelof¨ (or Cauchy-Lipschitz) Theorem (Banach fixed-point theorem) Let (K; d) a complete metric space and let g : K ! K a contraction that is for all x, y in K there exists c 2 ]0; 1[ such that d (g(x); g(y)) 6 c · d(x; y) ; then g has a unique fixed-point in K. ———— We consider the space of continuously differentiable functions 0 n C ([tj ; tj+1]; R ) and the Picard-Lindel¨of operator Z t pf (y) = t 7! yj + f(y(s))ds ; with yj = y(tj ) (1) tj If this operator is a contraction then its solution is unique and its solution is the solution of IVP. Julien Alexandre dit Sandretto - Validated simulation of ODEs October 22, 2020- 6 Picard-Lindel¨of Interval counterpart of Picard-Lindelof¨ n n With a first order integration scheme that is for f : R ! R a n continuous function and [a] ⊂ IR , we have Z a f (s)ds 2 (a − a)f ([a]) = w([a])f([a]) ; (2) a we can define a simple enclosure function of Picard-Lindel¨of such that def [pf ]([r]) = [yj ] + [0; h] · f([r]) ; (3) with h = tj+1 − tj the step-size. In consequence, if one can find [r] such that [pf ]([r]) ⊆ [r] then [y~j ] ⊆ [r] by the Banach fixed-point theorem. Julien Alexandre dit Sandretto - Validated simulation of ODEs October 22, 2020- 7 Picard-Lindel¨of Interval counterpart of Picard-Lindelof¨ ~ [yj] We can then build the Lohner 2-steps method: [yj] 1. Find [y~ ] and h with Picard-Lindel¨of j j [y ] operator and Banach’s theorem j+1 2. Compute [yj+1] with a validated integration scheme: Taylor or Runge-Kutta t tj tj+1 hj It is important to obtain [y~j ] and [yj+1] as tight as possible Integration scheme at order higher than one: Taylor for example Julien Alexandre dit Sandretto - Validated simulation of ODEs October 22, 2020- 8 Integration scheme Integration scheme Two main approaches: I Taylor series (Vnode, CAPD, etc.): Pp i [i] p+1 [i] th yj+1 = yj + 1 h f (yj ) + O(h ) with f the i term of serie expansion of f . O(hp+1) can be easily bounded by the Lagrange remainder of p+1 [p+1] serie s.t. O(h ) = f (ξ), with ξ 2 [y~j ], and then p+1 [p+1] O(h ) 2 f (y~j ) I Runge-Kutta methods (DynIBEX): yj+1 = Φ(yj ; f ; p) + LTE, with Φ any RK method and LTE the local truncation error. Julien Alexandre dit Sandretto - Validated simulation of ODEs October 22, 2020- 9 Integration scheme Runge-Kutta methods s-stage Runge-Kutta methods are described by a Butcher tableau c1 a11 a12 ··· a1s . j . c a a ··· a s s1 s2 ss i b1 b2 ··· bs Which induces the following recurrence: s ! s X X ki = f tj + ci hj ; yj + h ail kl yj+1 = yj + h bi ki l=1 i=1 I Explicit method (ERK) if ail = 0 is i 6 l I Diagonal Implicit method (DIRK) if ail = 0 is i 6 l and at least one aii 6= 0 I Implicit method (IRK) otherwise Julien Alexandre dit Sandretto - Validated simulation of ODEs October 22, 2020- 10 Integration scheme Explicit methods Interval extensions 1. Computation of k1 = f (yj ), k2 = f (yj + h · a21 · k1), ::: , Pi−1 Ps−1 ki = f (yj + h `=1 ai`k`), ::: , ks = f (yj + h `=1 as`k`) Ps 2. Computation of yj+1 = yj + h i=1 bi ki + LTE ) with interval arithmetic (natural extension) Example of HEUN 0 0 0 1 1 0 1=2 1=2 [k1] = [f ]([yj ]), [k2] = [f ]([yj ] + h[k1]), [yj+1] = [yj ] + h([k1] + [k2])=2 Julien Alexandre dit Sandretto - Validated simulation of ODEs October 22, 2020- 11 Integration scheme Implicit Schemes Example of Radau IIA 1=3 5=12 −1=12 1 3=4 1=4 3=4 1=4 [k1] = [f ]([yj ] + h(5[k1]=12 − [k2]=12)), [k2] = [f ]([yj ] + h(3[k1]=4 + [k2]=4) We need to solve this system of implicit equations ! Julien Alexandre dit Sandretto - Validated simulation of ODEs October 22, 2020- 12 Integration scheme Solve implicit scheme with a contractor point of view k1 is the approximate of f (y(tj + h=3)), but by construction y(tj + h=3) 2 [y~j ], then [k1] ⊂ f ([y~j ]) (same for k2) Algorithm based on contraction Require: f , [y~j ], [yj ], LTE [k1] = [f ]([y~j ]) and [k2] = [f ]([y~j ]) while [k1] or [k2] improved do [k1] = [k1] \ [f ]([yj ] + h(5[k1]=12 − [k2]=12)) [k2] = [k2] \ [f ]([yj ] + h(3[k1]=4 + [k2]=4) end while [yj+1] = [yj ] + h(3[k1] + [k2])=4 + LTE return [yj+1] Julien Alexandre dit Sandretto - Validated simulation of ODEs October 22, 2020- 13 Integration scheme How to compute the LTE ? p+1 y(tn; yn−1) − yn = C · h with C 2 R: Order condition This condition states that a method of Runge-Kutta family is of order p iff I the Taylor expansion of the exact solution I and the Taylor expansion of the numerical methods have the same p + 1 first coefficients. Consequence The LTE is the difference of Lagrange remainders of two Taylor expansions Julien Alexandre dit Sandretto - Validated simulation of ODEs October 22, 2020- 14 Integration scheme A quick view of Runge-Kutta order condition theory Starting from y(q) = (f (y))(q−1) and with the Chain rule, we have High order derivatives of exact solution y y_ = f (y) y¨ = f 0(y)y_ f 0(y) is a linear map y(3) = f 00(y)(y_; y_) + f 0(y)y¨ f 00(y) is a bi-linear map y(4) = f 000(y)(y_; y_; y_) + 3f 00(y)(y¨; y_) + f 0(y)y(3) f 000(y) is a tri-linear map . y(5) = f (4)(y)(y_; y_; y_; y_) + 6f 000(y)(y¨; y_; y_) . + 4f 00(y)(y(3); y_) + 3f 00(y)(y¨; y¨) + f 0(y)y(4) . Julien Alexandre dit Sandretto - Validated simulation of ODEs October 22, 2020- 15 Integration scheme A quick view of Runge-Kutta order condition theory Inserting the value of y_, y¨, . , we have: High order derivatives of exact solution y y_ = f y¨ = f 0(f ) y(3) = f 00(f ; f ) + f 0(f 0(f )) y(4) = f 000(f ; f ; f ) + 3f 00(f 0f ; f ) + f 0(f 00(f ; f )) + f 0(f 0(f 0(f ))) . 00 I Elementary differentials , such as f (f ; f ), are denoted by F (τ) Remark a tree structure is made apparent in these computations Julien Alexandre dit Sandretto - Validated simulation of ODEs October 22, 2020- 16 Integration scheme A quick view of Runge-Kutta order condition theory Rooted trees I f is a leaf 0 (k) I f is a tree with one branch, . , f is a tree with k branches Example f 0 f 00(f 0f ; f ) is associated to f f f 00 Remark: this tree is not unique e.g., symmetry Julien Alexandre dit Sandretto - Validated simulation of ODEs October 22, 2020- 17 Integration scheme A quick view of Runge-Kutta order condition theory Theorem 1 (Butcher, 1963) The qth derivative of the exact solution is given by X r(τ) the order of τ i.e., number of nodes y(q) = α(τ)F (τ)(y ) with 0 α(τ) a positive integer r(τ)=q We can do the same for the numerical solution Theorem 2 (Butcher, 1963) The qth derivative of the numerical solution is given by (q) X γ(τ) a positive integer y = γ(τ)φ(τ)α(τ)F (τ)(y ) with 1 0 φ(τ) depending on a Butcher tableau r(τ)=q Theorem 3, order condition (Butcher, 1963) 1 A Runge-Kutta method has order p iff φ(τ) = γ(τ) 8τ; r(τ) 6 p Julien Alexandre dit Sandretto - Validated simulation of ODEs October 22, 2020- 18 Integration scheme LTE formula for explicit and implicit Runge-Kutta From Th.