1. Abbas Edalat Imperial College London www.doc.ic.ac.uk/~ae joint work with Marko Krznaric and Andre Lieutier Domain-theoretic Solution of Differential Equations

2. 2Aim Develop data types for ordinary differential equations. Solve initial value problems up to any given precision: = v(t,x) with v: R 2  R continuous and Lipschitz in x x(t 0 ) = x 0 with (t 0,x 0 )  R 2

3. 3 Let IR={ [a,b] | a, b  R}  {R} (IR,  ) is a cpo with R as bottom  and: ⊔ i  0 a i =  i  0 a i (IR, ⊑ ) is a continuous Scott domain: countable basis {[p,q] | p < q & p, q  Q}  x {x} R I R x  {x} : R  IR gives an embedding onto the maximal elements The Domain of Intervals of R

4. 4 Data-type for Functions Lubs of finite and bounded collections of single- step functions ⊔ 1  i  n (a i ↘ b i ) are called step functions. Single-step function: a ↘ b : [0,1]  IR, with a  I[0,1], b  IR: b x  interior of a x   otherwise is Scott continuous. b ax

5. 5 Step Functions-An Example 01 R b1b1 a3a3 a2a2 a1a1 b3b3 b2b2

6. 6 Refining the Step Functions 01 R b1b1 a3a3 a2a2 a1a1 b3b3 b2b2

7. 7 Domain for Continuous Functions Partial order on functions [0,1]  IR : f ⊑ g   x  R. f(x) ⊑ g(x) ([0,1]  IR, ⊑ ) is a continuous Scott domain. Step functions, with a i, b i rational intervals, give a basis for [0,1]  IR f  If: C 0 [0,1] ↪ ( [0,1]  IR) is an embedding into a subset of maximal elements of [0,1]  IR. Scott continuous function g: [0,1]  IR is given by g(x) = [g (x), g + (x)], for x  dom(g), where g, g + : dom(g)  R f : [0,1]  R, f  C 0 [0,1] ( continuous functions ) has continuous extension If : [0,1]  IR x  {f (x)}

8. 8 Domain for Differentiable Functions If h  C 1 [0,1] (continuously differentiale functions), 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 Function and Derivative Consistency Consistency relation: For basis elements (f,g)  ([0,1]  IR)  ([0,1]  IR) define (f,g)  Cons if there is a piecewise linear map h: dom(g)  R with f ⊑ Ih and g ⊑ (wherever h is linear). Theorem. (f,g)  Cons iff there is a least function L(f,g) and a greatest function G(f,g) with the above properties in each connected component of dom(g) which intersects dom(f).

10. 10 Approximating function: f = ⊔ i a i ↘ b i ( ⊔ 1  i  n a i ↘ b i, ⊔ 1  j  m c j ↘ d j )  Cons is decidable: Consistency for basis elements L(f,g) = least function G(f,g) = greatest function Updating. 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

11. 11 f 1 1 Function and Derivative Information g 1 2

12. 12 f 1 1Updating g 1 2

13. 13 f 1 1 Updating Algorithm g 1 2

14. 14 f 1 1 Updating Algorithm (left to right) g 1 2

15. 15 f 1 1 Updating Algorithm (left to right) g 1 2

16. 16 f 1 1 Updating Algorithm (right to left) g 1 2

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

18. 18 f 1 1 Dually for the upper boundary g 1 2

19. 19 f 1 1 Output of the Updating Algorithm g 1 2

20. 20 Theorem. D 1 [0,1]:= { (f,g)  ([0,1]  IR) 2 | (f,g)  Cons} is a continuous Scott domain. The Domain of Differentiable Functions Theorem. C 1 [0,1] embeds into the set of maximal elements of D 1 [0,1]

21. 21 Solving Initial Value Problems = v(t,x) with v: R 2  R continuous and Lipschitz in x x(0) = 0 The function v is bounded by M say in a rectangle K around the origin. Take positive a<1, say, such that [-a,a]  [-Ma,Ma]  K. The initial condition x(0) = 0 is captured by the Scott continuous map: f = ⊔ n  0 f n where f n = [-a/2 n,a /2 n ] ↘ [-Ma/2 n, Ma/ 2 n ] This is the initial function approximation. It also gives the initial derivative approximation: t. v (t, f(t) )) [-Ma,Ma] [-a,a]

22. 22 Solve the ODE by iterating Up ⃘ Ap v on D 1 [-1,1] starting with ( f, t. v (t, f(t) )) Theorem. The domain-theoretic solution ⊔ n  0 (Up ⃘ Ap v ) n (f, t. v (t, f(t) )) is the unique classical solution through (0,0). Function and Derivative Upgrading Derivative upgrading: Ap v : ([-1,1]  IR) 2  ([-1,1]  IR) 2 (f,g)  ( f, t. v (t, f(t) )) Function upgrading: Up : ([-1,1]  IR) 2  ([-1,1]  IR) 2 Up(f,g) = (f g, g) where f g (t) = [ L (f,g) (t), G (f,g) (t) ]

23. 23 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 -  t. 0 Compute the piecewise linear maps u n -, u n + until the first n  0 with u n + - u n -   ⊔ 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) ) )

24. 24Example 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.

25. 25Solution 1 f g 1 1 1 At stage n we find u n - and u n +.

26. 26Solution 1 f g 1 1 1. At stage n we find u n - and u n +

27. 27Solution 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 +

28. 28 Computing with polynomial step functions

29. 29 Current and Further Work Implementation in Haskell Differential Calculus with Several Variables: PDE’s Construction of Smooth Curves and Surfaces Hybrid Systems, robotics,…

30. 30 THE END http://www.doc.ic.ac.uk/~ae http://www.doc.ic.ac.uk/~ae