Abbas Edalat Imperial College London doc.ic.ac.uk/~ae

1. Data Types for Differential Equations Abbas Edalat Imperial College London www.doc.ic.ac.uk/~ae Contains joint work with Andre Lieutier (AL) and joint work with Marko Krznaric (MK)

2. Aim Develop data types for ordinary differential equations. Solve initial value problem up to any given precision. • In particular for: Hybrid System= Discrete State Machine + Continuous Process Differential Equation

3. The Domain of nonempty compact Intervals of R  {x} x x  {x} : R  IRTopological embedding R I R • Let IR={ [a,b] | a, b  R}  {R} • (IR, ) is a bounded complete dcpo with R as bottom: ⊔iI ai = iI ai • a ≪ b  ao  b • (IR, ⊑) is -continuous: countable basis {[p,q] | p < q & p, q  Q} • (IR, ⊑) is, thus, a continuous Scott domain. • Scott topology has basis:↟a = {b | ao  b}

4. Domain for C0Functions • f : [0,1]  R, f  C0[0,1], has continuous extension If : [0,1]  IR x  {f (x)} Scott continuous maps [0,1]  IR with:f ⊑ g x  R . f(x) ⊑ g(x)is another continuous Scott domain.  : C0[0,1] ↪ ( [0,1]  IR), with f  Ifis a topological embedding into a proper subset of maximal elements of [0,1]  IR .

5. Step Functions • Single-step function: a↘b : [0,1]  IR, with a  I[0,1], b  IR: b x ao x otherwise • Lubs of finite and bounded collections of single- step functions ⊔1in(ai ↘bi) are called step functions. • Step functions with ai, birational intervals, give a basis for [0,1]  IR

6. Step Functions-An Example R b3 a3 b1 b2 a1 a2 0 1

7. Refining the Step Functions R b3 a3 b1 a1 b2 a2 0 1

8. Domain for C1Functions (AE&AL) • If h C1[0,1] , then ( Ih , Ih )  ([0,1]  IR)  ([0,1]  IR) • We can approximate ( Ih, Ih ) in ([0,1]  IR)2 i.e. ( f, g) ⊑ ( Ih ,Ih ) with f ⊑ Ih and g ⊑ Ih • What pairs ( f, g)  ([0,1]  IR)2approximate a differentiable function?

9. Interval Derivative The interval derivative of f: [0,1]  Ris defined as if both limits are finite otherwise

10. Function and Derivative Consistency • Define the consistency relation:Cons  ([0,1]  IR)  ([0,1]  IR) with(f,g)  Cons if there is a continuous h: dom(g) R with f ⊑ Ih and g ⊑ Theorem. If(f,g)  Cons, there are least and greatest functionshwith the above properties in each connected component ofdom(g)which intersectsdom(f) .

11. Consistency for basis elements G(f,g)= greatest function fg(t) Approximating function: f = ⊔iai↘bi L(f,g) = least function t Approximating derivative: g = ⊔j cj↘dj (⊔iai↘bi, ⊔j cj↘dj)  Cons is a finitary property: • We will define L(f,g), G(f,g) in general and show that: • (f,g)  Cons iff L(f,g) G(f,g). • Cons isdecidable on the basis. Upgrading.Up(f,g) := (fg , g) where fg : t  [ L(f,g)(t) , G(f,g)(t) ]

12. Function and Derivative Information f 1 1 g 2 1

13. Updating f 1 1 g 2 1

14. Consistency Test and Updating for (f,g) • For x  dom(g), let g({x}) = [g (x), g+(x)] where g , g+:dom(g) R are lower and upper semi-continuous. Similarly we define f , f +:dom(f) R. Writef = [f –, f +]. Let Obe a connected component of dom(g) with O  dom(f) . For x , y  O define: Define: L(f,g)(x) := supyOdom(f)(f –(y) + d–+(x,y))andG(f,g)(x) := infyOdom(f)(f +(y) + d+–(x,y)) Theorem. (f, g)  Con iffx  O. L(f, g) (x)  G(f, g) (x).

15. Updating Linear step Functions • A linear single-step function: a↘[b–, b +] : [0,1]  IR, withb–, b +: ao  Rlinear [b–(x) , b +(x)] x  ao x otherwise We write this simply as a↘b with b=[b–, b +] . • Proposition. For x  O, we have:L(f,g)(x) = max {f –(x) , limsup f –(y) + d–+(x , y) | ym  O  dom(f) } • For (f, g) = (⊔1inai↘bi , ⊔1jmcj↘dj) with f linear g standard, the rational end–points of aiand cj induce a partition y0 < y1 < y2 < … < ykof the connected component O of dom(g). • Hence L(f,g) is the max of k+2 linear maps. Similarly, G(f,g). • We get a linear time algorithm for computingL(f,g), G(f,g).

16. Updating Algorithm(AE&MK) f 1 1 g 2 1

17. Updating Algorithm (left to right) g 2 1 f 1 1

18. Updating Algorithm (left to right) f 1 1 g 2 1

19. Updating Algorithm (right to left) f 1 1 g 2 1

20. Updating Algorithm (right to left) f 1 1 g 2 1

21. Updating Algorithm (similarly for upper one) f 1 1 g 2 1

22. Output of the Updating Algorithm f 1 1 g 2 1

23. The Domain of C1Functions (AE&AL) • Define D1c:= {(f0,f1)  C1C0 | f0 = f1 } • Lemma.Cons  ([0,1]  IR)2is Scott closed. • Theorem.D1 [0,1]:= { (f,g)  ([0,1]IR)2 | (f,g)  Cons} is a continuous Scott domain, which can be given an effective structure. • Theorem. : C1[0,1]  C0[0,1]  ([0,1]  IR)2 restricts to give a topological embedding D1c↪ D1(with C1 norm) (with Scott topology)

24. Picard Theorem • = v(t,x) with v: R2 R continuous x(t0) = x0with (t0,x0) R2 and v is Lipschitz in x uniformly in t for some neighbourhood of (t0,x0). • Theorem. In a neighbourhood of t0, there is a unique solution, which is the unique fixed point of: P: C0 [t0-k , t0+k]  C0 [t0-k , t0+k] f  t . (x0 +  v(t , f(t) ) dt) for somek>0 . t t0

25. Picard Solution Reformulated t • P: f t . (x0 +  v(t , f(t)) dt) can be considered as upgrading the information about the function f and the information about its derivative g. t0 t Up: (f,g)  ( t . (x0 +  g(t) dt) , g ) t0 t Up⃘Apv: (f,g)  (t . (x0 +  g dt , t . v(t,f(t)))has a fixed point (f,g) with f = g = t . v(t,f(t)) t0 Apv: (f,g)  (f , t. v(t,f(t)))

26. A domain-theoretic Picard theorem • To obtain Picard’s theorem with domain theory, we have to make sure that derivative updating preserves consistency. • (f , g) is strongly consistent,(f , g) S-Cons, if  h ⊒ g we have: (f , h)  Cons • Q(f,g)(x) := supyODom(f) (f –(y) + d+–(x,y)) R(f,g)x) := infyODom(f) (f +(y) + d–+(x,y)) • Theorem. If f –, f +, g–, g+: [0,1]R are bounded and g–, g+are continuous a.e. (e.g. forpolynomial step functionsfandg), then (f,g) is strongly consistent iff for any connected component O of dom(g) with O  dom(f)  , we have: x  O. Q(f,g)(x), R(f,g)(x)  [f –(x) , f +(x) ] • Thus, on basis elements strong consistency is decidable.

27. Domain-theoretic Picard theorem (AE&AL) • Let v: [0,1]  IR  IR be Scott continuousand Apv : ([0,1]  IR)2 ([0,1]  IR)2 (f,g)  ( f , t. v (t , f(t) )) Up : ([0,1]  IR)2 ([0,1]  IR)2 Up(f,g) = (fg , g) where fg (t) = [ L (f,g) (t) , G (f,g) (t) ] • Consider any initial value f [0,1]  IRwith (f, t. v (t , f(t) ) )  S-Cons • Then the continuous map Up ⃘Apvhas a least fixed point above (f, t.v (t , f(t))) given by (fs, gs) = ⊔n 0(Up ⃘Apv )n (f, t.v (t , f(t) ) )

28. The Classical Initial Value Problem • Suppose v = Ih for a continuous h : [-1,1] R  R which satisfies the Lipschitz property around (t0,x0) =(0,0). • Then h is bounded by M say in a compact rectangle K around the origin. We can choose positivea  1such that [-a,a][-Ma,Ma]  K. • Putf = ⊔n 0fn where fn = [-a/2n,a /2n]↘[-Ma/2n , Ma/ 2n ] • Then (f , [-a,a ]↘[-M ,M ])  S-Cons, hence (f, t. v(t , f(t) ) )  S-Cons since ([-a,a ]↘[-M ,M]) ⊑ t. v (t , f(t) ) • Theorem.The domain-theoretic solution(fs, gs) = ⊔n 0(Up ⃘Apv )n (f, t. v (t , f(t) )) gives the unique classical solution through(0,0).

29. Computation of the solution for a given precision  >0 (fs, gs) = ⊔n 0 (Pv )n (f, t. v (t , f(t) ) ) = ⊔n 0 (Pv )n (fn , t. vn (t , fn(t) ) ) Let (un , wn) := (Pv )n (fn , t. vn (t , fn(t) ) ) with un = [un-,un+] • We express f and v as lubs of step functions: f = ⊔n 0fn v = ⊔n 0vn • Putting Pv := Up ⃘Apv the solution is obtained as: • For all n0 we have: un- un+1-un+1+ un+ with un+ - un-  0 • Compute the piecewise linear maps un-, un+ until the first n0with un+ - un-  

30. Example f We solve: = v(t,x), x(t0) =x0 for t[0,1] with v(t,x) = tandt0=1/2, x0=9/8. 1 b2 b1 b3 a3 a2 a1 1 g v1 1 v2 v3 1 v is approximated by a sequence of step functions, v0, v1, … v = ⊔ivi . t The initial condition is approximated by rectangles aibi: {(1/2,9/8)} =⊔i aibi, v t

31. Solution f 1 1 g 1 1 At stage n we find un -and un + .

32. Solution f 1 1 g 1 1 At stage n we find un -and un + .

33. Solution f 1 1 g 1 1 At stage n we find un -and un + . un -and un +tend to the exact solution:f: t  t2/2 + 1

34. Computing with polynomial step functions

35. Current and Further Work • Solving Differential Equations with Domains • Differential Calculus with Several Variables • Implicit and Inverse Function Theorems • Reconstruct Geometry and Smooth Mathematics with Domain Theory • Continuous processes, robotics,…

36. THE ENDhttp://www.doc.ic.ac.uk/~ae