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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

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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

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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

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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.

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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

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

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

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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

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

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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 ⊑

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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

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12 f 1 1 Function and Derivative Information g 1 2

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13 f 1 1Updating g 1 2

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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 + ].

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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).

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

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

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

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

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

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

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

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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)

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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 ).

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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)))

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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

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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) ) )

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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 ]

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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) ) )

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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.

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

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

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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 +

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34 Computing with polynomial step functions

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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,…

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

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