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Joint work with Andre Lieutier Dassault Systemes Domain Theory and Differential Calculus Abbas Edalat Imperial College http://www.doc.ic.ac.uk/~ae Oxford 17/2/2003

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2 Computational Model for Classical Spaces A research project since 1993: Reconstruct some basic mathematics Embed classical spaces into the set of maximal elements of suitable domains X Classical Space x DX Domain {x}

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3 Computational Model for Classical Spaces Previous Applications: Fractal Geometry Measure & Integration Theory Exact Real Arithmetic Computational Geometry/Solid Modelling

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4 Non-smooth Mathematics Set Theory Logic Algebra Point-set Topology Graph Theory Model Theory. Geometry Differential Topology Manifolds Dynamical Systems Mathematical Physics. All based on differential calculus Smooth Mathematics

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A Domain-Theoretic Model for Differential Calculus Indefinite integral of a Scott continuous function Derivative of a Scott continuous function Fundamental Theorem of Calculus Domain of C 1 functions (Domain of C k functions) Picard’s Theorem: A data-type for differential equations

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6 (IR, ) is a bounded complete dcpo: ⊔ i I a i = i I a i a ≪ b a o b (IR, ⊑ ) is -continuous: Basis {[p,q] | p < q & p, q Q} (IR, ⊑ ) is, thus, a continuous Scott domain. Scott topology has basis: ↟ a = {b | a ≪ b} x {x} R I R x {x} : R IR Topological embedding The Domain of nonempty compact Intervals of R

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7 Continuous 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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8 Step Functions a ↘ b : [0,1] IR, with a I[0,1], b IR: b x a o x otherwise Finite lubs of consistent single step functions ⊔ 1 i n (a i ↘ b i ) with a i, b i rational intervals, give a basis for [0,1] IR

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

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

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11 Operations in Interval Arithmetic For a = [a, a] IR, b = [b, b] IR, and * { +, –, } we have: a * b = { x*y | x a, y b } For example: a + b = [ a + b, a + b]

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12 Intuitively, we expect f to satisfy: What is the indefinite integral of a single step function a ↘ b ? The Basic Construction Classically, with We expect a ↘ b ([0,1] IR) For what f C 1 [0,1], should we have If a ↘ b ?

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13 Interval Derivative Assume f C 1 [0,1], a I[0,1], b IR. Suppose x a o. b f (x) b. We think of [b, b] as an interval derivative for f at a. Note that x a o. b f (x) b iff x 1, x 2 a o & x 1 > x 2, b(x 1 – x 2 ) f(x 1 ) – f(x 2 ) b(x 1 – x 2 ), i.e. b(x 1 – x 2 ) ⊑ {f(x 1 ) – f(x 2 )} = {f(x 1 )} – {f(x 2 )}

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14 Definition of Interval Derivative f ([0,1] IR) has an interval derivative b IR at a I[0,1] if x 1, x 2 a o, b(x 1 – x 2 ) ⊑ f(x 1 ) – f(x 2 ). Proposition. For f: [0,1] IR, we have f (a,b) iff f(x) Maximal (IR) for x a o, and Graph(f) is within lines of slope b & b at each point (x, f(x)), x a o. (x, f(x)) b b a Graph(f). The tie of a with b, is (a,b) := { f | x 1,x 2 a o. b(x 1 – x 2 ) ⊑ f(x 1 ) – f(x 2 )}

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15 Let f C 1 [0,1]; the following are equivalent: If (a,b) x a o. b f (x) b x 1,x 2 [0,1], x 1,x 2 a o. b(x 1 – x 2 ) ⊑ If (x 1 ) – If (x 2 ) a ↘ b ⊑ If For Classical Functions Thus, (a,b) is our candidate for a ↘ b.

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16 (a 1,b 1 ) (a 2,b 2 ) iff a 2 ⊑ a 1 & b 1 ⊑ b 2 n i=1 (a i,b i ) iff {a i ↘ b i | 1 i n} consistent. i I (a i,b i ) iff {a i ↘ b i | i I } consistent iff J finite I i J (a i,b i ) In fact, (a,b) behaves like a ↘ b; we call (a,b) a single-step tie. Properties of Ties

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17 The Indefinite Integral : ([0,1] IR) ( P ([0,1] IR), ) ( P the power set) a ↘ b := (a,b) ⊔ i I a i ↘ b i := i I (a i,b i ) is well-defined and Scott continuous. But unlike the classical case, is not 1-1.

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18Example ([0,1/2] ↘ {0}) ⊔ ([1/2,1] ↘ {0}) ⊔ ([0,1] ↘ [0,1]) = ([0,1/2], {0}) ([1/2,1] ↘ {0}) ([0,1] ↘ [0,1]) = ([0,1], {0}) = [0,1] ↘ {0}

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19 The Derivative Operator : (I[0,1] IR) (I[0,1] IR) is monotone but not continuous. Note that the classical operator is not continuous either. (a ↘ b)= x. is not linear! For f : x x : I[0,1] IR g : x –x : I[0,1] IR (f+g) + = x. (1 – 1) = x. 0

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20 The Derivative Definition. Given f : [0,1] IR the derivative of f is: : [0,1] IR = ⊔ {a ↘ b | f (a,b) } Theorem. (Compare with the classical case.) is well–defined & Scott continuous. If f C 1 [0,1], then f (a,b) iff a ↘ b ⊑

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21 Examples

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22 Domain of Ties, or Indefinite Integrals Recall : ([0,1] IR) ( P ([0,1] IR), ) Let T[0,1] = Image ( ), i.e. T[0,1] iff x is the nonempty intersection of a family of single ties: = i I (a i,b i ) Domain of ties: ( T[0,1], ) Theorem. ( T[0,1], ) is a continuous Scott domain.

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23 Define : (T[0,1], ) ([0,1] IR) ∆ ⊓ { | f ∆ } The Fundamental Theorem of Calculus Theorem. : (T[0,1], ) ([0,1] IR) is upper adjoint to : ([0,1] IR) (T[0,1], ) In fact, Id = ° and Id ⊑ °

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24 Fundamental Theorem of Calculus For f, g C 1 [0,1], let f ~ g if f = g + r, for some r R. We have:

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25 F.T. of Calculus: Isomorphic version For f, g [0,1] IR, let f ≈ g if f = g a.e. We then have:

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26 A Domain for C 1 Functions 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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27 Proposition (f,g) Cons iff there is a continuous h: dom(g) R with f ⊑ Ih and g ⊑. Function and Derivative Consistency Define the consistency relation: Cons ([0,1] IR) ([0,1] IR) with (f,g) Cons if ( f) ( g) In fact, if (f,g) Cons, there are always a least and a greatest functions h with the above properties.

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28 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 (f,g) Cons iff L(f,g) G(f,g). Cons is decidable on the basis. 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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29 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 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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30 Higher Interval Derivative Proposition. For f C 2 [0,1], the following are equivalent: If 2 (a,b) x a 0. b f (x) b x 1,x 2 a 0. b (x 1 – x 2 ) ⊑ If (x 1 ) – If (x 2 ) a ↘ b ⊑ If Let 1 (a,b) = (a,b) Definition. (the second tie) f 2 (a,b) P ([0,1] IR) if 1 (a,b) Note the recursive definition, which can be extended to higher derivatives.

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31 Higher Derivative and Indefinite Integral For f : [0,1] IR we define: : [0,1] IR by Then = ⊔ f 2 (a,b) a ↘ b : ([0,1] IR) ( P ([0,1] IR), ) a ↘ b := (a,b) ⊔ i I a i ↘ b i := i I (a i,b i ) is well-defined and Scott continuous. 2 (2) 2

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32 Domains of C 2 functions D 2 c := {(f 0,f 1,f 2 ) C 2 C 1 C 0 | f 0 = f 1, f 1 = f 2 } Theorem. restricts to give a topological embedding D 2 c ↪ D 2 Define Cons (f 0,f 1,f 2 ) iff f 0 f 1 f 2 (2) Theorem. Cons (f 0,f 1,f 2 ) is decidable on basis elements. (The present algorithm to check is NP-hard.) D 2 := { (f 0,f 1,f 2 ) (I[0,1] IR) 3 | Cons (f 0,f 1,f 2 ) }

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33 Domains of C k functions D k := { (f i ) 0 i k (I[0,1] IR) k+1 | Cons (f i ) 0 i k } D := { (f k ) k 0 ( I[0,1] IR) ω | k 0. (f i ) 0 i k D k } ∞ (i) Let (f i ) 0 i k (I[0,1] IR) k+1 Define Cons (f i ) 0 i k iff 0 i k f i The decidability of Cons on basis elements for k 3 is an open question.

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34 Theorem. There exists a neighbourhood of t 0 where there is a unique solution, the fixed point of: P: C 0 [t 0 -k, t 0 +k] C 0 [t 0 -k, t 0 +k] f t. (x 0 + F(t, f(t) ) dt) for some k>0. t0t0 t Picard’s Theorem = F(t,x) with F: R 2 R x(t 0 ) = x 0 with (t 0,x 0 ) R 2 where F is Lipschitz in x uniformly in t for some neighbourhood of (t 0,x 0 ).

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35 Up ⃘ Ap F : (f,g) ( t. (x 0 + g dt, t. F(t,f(t))) has a fixed point (f,g) with f = g = t. F(t,f(t)) t t0t0 Picard’s Solution Reformulated Up: (f,g) ( t. (x 0 + g(t) dt), g ) t t0t0 P: f t. (x 0 + F(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 F : (f,g) (f, t. F(t,f(t)))

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36 We now have the basic framework to obtain Picard’s theorem with domain theory. However, we have to make sure that derivative updating preserves consistency. Say (f, g) is strongly consistent, (f, g) S-Cons, if h ⊒ g. (f, h) Cons On basis elements, strong consistency is decidable. A domain-theoretic Picard’s theorem

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37 A domain-theoretic Picard’s theorem Let F : [0,1] IR IR and Ap F : ([0,1] IR) 2 ([0,1] IR) 2 (f,g) ( f, F (., f ) ) 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, F (., f ) ) S-Cons Then the continuous map P = Up ⃘ Ap F has a least fixed point above (f, F (., f )) Theorem. If F = Ih for a map h : [0,1] R R which satisfies the Lipschitz property of Picard’s theorem, then the domain-theoretic solution coincides with the classical solution.

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38Example 1 f g 1 1 1 F F is approximated by a sequence of step functions, F 1, F 2, … F = ⊔ i F i We solve: = F(t,x), x(t 0 ) =x 0 for t [0,1] with F(t,x) = t and t 0 =1/2, x 0 =9/8. a3a3 b3b3 a2a2 b2b2 a1a1 b1b1 F3F3 F2F2 F1F1 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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39Solution 1 f g 1 1 1 At each stage we find L i and G i.

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40Solution 1 f g 1 1 1. At each stage we find L i and G i

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41Solution 1 f g 1 1 1 L i and G i tend to the exact solution: f: t t 2 /2 + 1. At each stage we find L i and G i

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42 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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43 THE END http://www.doc.ic.ac.uk/~ae http://www.doc.ic.ac.uk/~ae

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45 Higher Interval Derivative Proposition. For f C 2 [0,1], the following are equivalent: If 2 (a,b) x a 0. b f (x) b x 1,x 2 ≫ a. b (x 1 – x 2 ) ⊑ If (x 1 ) – If (x 2 ) a ↘ b ⊑ If Let 1 (a,b) = (a,b) Definition. (the second tie) f 2 (a,b) P (I[0,1] IR) if 1 (a,b) Note the recursive definition, which can be extended to higher derivatives.

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46 Higher Interval Derivative For f : I[0,1] IR we define: : I[0,1] IR by Then = ⊔ f 2 (a,b) a ↘ b : (I[0,1] IR) ( P (I[0,1] IR), ) a ↘ b := (a,b) 2 (2)

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47 Domains of C 2 and C k functions D 2 c := {(f 0,f 1,f 2 ) C 2 C 1 C 0 | f 0 = f 1, f 1 = f 2 } Theorem. restricts to give a topological embedding D 2 c ↪ D 2 D k := { (f i ) 0 i k (I[0,1] IR) k+1 | 0 i k f i } (i) D := { (f k ) k 0 ( I[0,1] IR) | k 0. f k D k } ∞ D 2 := { (f 0,f 1,f 2 ) (I[0,1] IR) 3 | f 0 f 1 f 2 } (2)

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48 Consistency Test for (f,g) Also define: L(x) := sup y O Dom(f) (f – (y) + d –+ (x,y)) and G(x) := inf y O Dom(f) (f + (y) + d +– (x,y)) For x Dom(g), let g({x})=[g (x),g + (x)] where g,g + : Dom(g) R are semi-continuous functions. Similarly we define f, f + : Dom(f) R. Let O be a connected component of Dom(g) with O Dom(f) . For x, y O define:

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49 Theorem. (f, g) Con iff x O. L(x) G(x). For (f, g) = ( ⊔ 1 i n a i ↘ b i, ⊔ 1 j m c j ↘ d j ) the rational end–points of a i and c j induce a partition X = {x 0 < x 1 < x 2 < … < x k } of O. Proposition. For arbitrary x O, there is p, where 0 p k, with: L(x) = f – (x p ) + d –+ (x,x p ). Similarly for G(x). Consistency Test

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