1. 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) Data Types for Differential Equations

2. 2Aim 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. 3 Let IR={ [a,b] | a, b  R}  {R} (IR,  ) is a bounded complete dcpo with R as bottom: ⊔ i  I a i =  i  I a i a ≪ b  a o  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 | a o  b}  x {x} R I R x  {x} : R  IR Topological embedding The Domain of nonempty compact Intervals of R

4. 4 Domain for C 0 Functions f : [0,1]  R, f  C 0 [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.  : C 0 [0,1] ↪ ( [0,1]  IR), with f  If is a topological embedding into a proper subset of maximal elements of [0,1]  IR.

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

6. 6 Step Functions-An Example 01 R b1b1 a3a3 a2a2 a1a1 b3b3 b2b2

7. 7 Refining the Step Functions 01 R b1b1 a3a3 a2a2 a1a1 b3b3 b2b2

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

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

10. 10 Function and Derivative Consistency Theorem. If (f,g)  Cons, there are least and greatest functions h with the above properties in each connected component of dom(g) which intersects dom(f). 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 ⊑

11. 11 Approximating function: f = ⊔ i a i ↘ b i ( ⊔ i a i ↘ b i, ⊔ j c j ↘ d j )  Cons is a finitary property: Consistency for basis elements L(f,g) = least function G(f,g) = greatest function We will define L(f,g), G(f,g) in general and show that: 1. (f,g)  Cons iff L(f,g)  G(f,g). 2. Cons is decidable on the basis. Upgrading. Up(f,g) := (f g, g) where f g : t  [ L(f,g)(t), G(f,g)(t) ] f g (t) t Approximating derivative: g = ⊔ j c j ↘ d j

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

13. 13 f 1 1Updating g 1 2

14. 14 Let O be a connected component of dom(g) with O  dom(f)  . For x, y  O define: Consistency Test and Updating for (f,g) Define: L(f,g)(x) := sup y  O  dom(f) (f – (y) + d –+ (x,y)) and G(f,g)(x) := inf y  O  dom(f) (f + (y) + d +– (x,y)) Theorem. (f, g)  Con iff  x  O. L(f, g) (x)  G(f, g) (x). 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. Write f = [f –, f + ].

15. 15 Updating Linear step Functions A linear single-step function: a ↘ [b –, b + ] : [0,1]  IR, with b –, b + : a o  R linear [b – (x), b + (x)] x  a o x   otherwise We write this simply as a ↘ b with b=[b –, b + ]. Hence L(f,g) is the max of k+2 linear maps. Similarly, G(f,g). We get a linear time algorithm for computing L(f,g), G(f,g). Proposition. For x  O, we have: L(f,g)(x) = max {f – (x), limsup f – (y) + d –+ (x, y) | y m  O  dom(f) } For (f, g) = ( ⊔ 1  i  n a i ↘ b i, ⊔ 1  j  m c j ↘ d j ) with f linear g standard, the rational end–points of a i and c j induce a partition y 0 < y 1 < y 2 < … < y k of the connected component O of dom(g).

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

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

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

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

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

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

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

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

24. 24 Theorem. In a neighbourhood of t 0, there is a unique solution, which is the unique fixed point of: P: C 0 [t 0 -k, t 0 +k]  C 0 [t 0 -k, t 0 +k] f  t. (x 0 +  v(t, f(t) ) dt) for some k>0. t0t0 t Picard Theorem = v(t,x) with v: R 2  R continuous x(t 0 ) = x 0 with (t 0,x 0 )  R 2 and v is Lipschitz in x uniformly in t for some neighbourhood of (t 0,x 0 ).

25. 25 Up ⃘ Ap v : (f,g)  ( t. (x 0 +  g dt, t. v(t,f(t))) has a fixed point (f,g) with f = g = t. v(t,f(t)) t t0t0 Picard Solution Reformulated Up: (f,g)  ( t. (x 0 +  g(t) dt), g ) t t0t0 P: f  t. (x 0 +  v(t, f(t)) dt) can be considered as upgrading the information about the function f and the information about its derivative g. t t0t0 Ap v : (f,g)  (f, t. v(t,f(t)))

26. 26 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) := sup y  O  Dom(f) (f – (y) + d +– (x,y)) R(f,g)x) := inf y  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. for polynomial step functions f and g ), 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. A domain-theoretic Picard theorem

27. 27 Domain-theoretic Picard theorem (AE&AL) Let v : [0,1]  IR  IR be Scott continuous and Ap v : ([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) = (f g, g) where f g (t) = [ L (f,g) (t), G (f,g) (t) ] Consider any initial value f  [0,1]  IR with (f, t. v (t, f(t) ) )  S-Cons Then the continuous map Up ⃘ Ap v has a least fixed point above (f, t.v (t, f(t))) given by ( f s, g s ) = ⊔ n  0 (Up ⃘ Ap v ) n (f, t.v (t, f(t) ) )

28. 28 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 ( f s, g s ) = ⊔ n  0 (Up ⃘ Ap v ) n (f, t. v (t, f(t) )) gives the unique classical solution through (0,0). The Classical Initial Value Problem Suppose v = Ih for a continuous h : [-1,1]  R  R which satisfies the Lipschitz property around (t 0,x 0 ) =(0,0). Then h is bounded by M say in a compact rectangle K around the origin. We can choose positive a  1 such that [-a,a]  [-Ma,Ma]  K. Put f = ⊔ n  0 f n where f n = [-a/2 n,a /2 n ] ↘ [-Ma/2 n, Ma/ 2 n ]

29. 29 Computation of the solution for a given precision  Computation of the solution for a given precision  >0 Let ( u n, w n ) := (P v ) n (f n, t. v n (t, f n (t) ) ) with u n = [u n -,u n + ] We express f and v as lubs of step functions: f = ⊔ n  0 f n v = ⊔ n  0 v n Putting P v := Up ⃘ Ap v the solution is obtained as: For all n  0 we have: u n -  u n+1 -  u n+1 +  u n + with u n + - u n -  0 Compute the piecewise linear maps u n -, u n + until  the first n  0 with u n + - u n -   ( f s, g s ) = ⊔ n  0 (P v ) n (f, t. v (t, f(t) ) ) = ⊔ n  0 (P v ) n (f n, t. v n (t, f n (t) ) )

30. 30Example 1 f g 1 1 1 v v is approximated by a sequence of step functions, v 0, v 1, … v = ⊔ i v i We solve: = v(t,x), x(t 0 ) =x 0 for t  [0,1] with v(t,x) = t and t 0 =1/2, x 0 =9/8. a3a3 b3b3 a2a2 b2b2 a1a1 b1b1 v3v3 v2v2 v1v1 The initial condition is approximated by rectangles a i  b i : {(1/2,9/8)} = ⊔ i a i  b i, t t.

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

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

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

34. 34 Computing with polynomial step functions

35. 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. 36 THE END http://www.doc.ic.ac.uk/~ae http://www.doc.ic.ac.uk/~ae