abbas edalat imperial college london ae contains joint work with andre lieutier (al) and joint work...
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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 EquationsData Types for Differential Equations
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AimAim
• 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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• Let IR={ [a,b] | a, b R} {R}
• (IR, ) is a bounded complete dcpo with R as bottom: ⊔iI ai = iI 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}
x {x}
R
I R
• x {x} : R IRTopological embedding
The Domain of nonempty compact Intervals of The Domain of nonempty compact Intervals of RR
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Domain for Domain for CC00 FunctionsFunctions
• 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 .
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Step FunctionsStep 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
⊔1in(ai ↘ bi)
are called step functions.
• Step functions with ai, bi rational intervals, give a basis for [0,1] IR
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Step Functions-An ExampleStep Functions-An Example
0 1
R
b1
a3
a2
a1
b3
b2
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Refining the Step FunctionsRefining the Step Functions
0 1
R
b1
a3
a2
a1
b3
b2
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Domain for Domain for CC11 Functions (AE&AL) Functions (AE&AL)
• If h C1[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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Interval DerivativeInterval Derivative
• The interval derivative of f: [0,1] R is defined as
IR [0,1] :dx
df
] (f)(y)Dlim , (f)(y)Dlim [
dx
df uxyxy
l if both limits are finiteotherwise
yx
f(y)f(x)lim: (f)(x)D
yx
f(y)f(x)lim : (f)(x)D
xy
xyu
l
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Function and Derivative ConsistencyFunction 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) .
dx
dh
• 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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Approximating function: f = ⊔i ai↘bi
• (⊔i ai↘bi, ⊔j cj↘dj) Cons is a finitary property:
Consistency for basis elementsConsistency 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) := (fg , g) where fg : t [ L(f,g)(t) , G(f,g)(t) ]
fg(t)
t
Approximating derivative: g = ⊔j cj↘dj
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f
1
1
Function and Derivative Information Function and Derivative Information
g
1
2
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f
1
1
UpdatingUpdating
g
1
2
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• Let O be a connected component of dom(g) with O dom(f) . For x , y O define:
Consistency Test and Updating for Consistency Test and Updating for (f,g)(f,g)
yxduug
xyduug
yxdx
y
x
y
)(
)(
),(
• Define: L(f,g)(x) := supyOdom(f)(f –(y) + d–+(x,y)) and G(f,g)(x) := infyOdom(f)(f +(y) + d+–(x,y))
• Theorem. (f, g) Con iff x O. L(f, g) (x) G(f, g) (x).
yxduug
xyduug
yxdx
y
x
y
)(
)(
),(
• 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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Updating Linear step FunctionsUpdating Linear step Functions
• A linear single-step function: a [↘ b–, b +] : [0,1] IR, with b–, b +: ao R linear
[b–(x) , b +(x)] x ao x otherwise
We write this simply as a b ↘ with b=[b–, b +] . x
b
ba
• 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).
myy
• 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) = (⊔1in ai↘bi , ⊔1jm cj↘dj) with f linear g standard,
the rational end–points of ai and cj induce a partition y0 < y1 < y2 < … < yk of the connected component O of dom(g).
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f
1
1
Updating Algorithm(AE&MK) Updating Algorithm(AE&MK)
g
1
2
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f
1
1
Updating Algorithm (left to right)Updating Algorithm (left to right)
g
1
2
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f
1
1
Updating Algorithm (left to right)Updating Algorithm (left to right)
g
1
2
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f
1
1
Updating Algorithm (right to left)Updating Algorithm (right to left)
g
1
2
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f
1
1
Updating Algorithm (right to left)Updating Algorithm (right to left)
g
1
2
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f
1
1
Updating Algorithm (similarly for Updating Algorithm (similarly for upper one)upper one)
g
1
2
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f
1
1
Output of the Updating Algorithm Output of the Updating Algorithm
g
1
2
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• Lemma. Cons ([0,1] IR)2 is 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.
The Domain of The Domain of CC11 Functions (AE&AL)Functions (AE&AL)
• Define D1c := {(f0,f1) C1C0 | f0 = f1 }
• 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)
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• 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 some k>0 .
t0
t
Picard TheoremPicard Theorem
• = v(t,x) with v: R2 R continuous
x(t0) = x0 with (t0,x0) R2
and v is Lipschitz in x uniformly in t for some neighbourhood of (t0,x0).
dt
dx
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• Up⃘�Apv: (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))
t
t0
Picard Solution ReformulatedPicard Solution Reformulated
• Up: (f,g) ( t . (x0 + g(t) dt) , g )t
t0
• 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.
t
t0
• Apv: (f,g) (f , t. v(t,f(t)))
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• 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) := supyODom(f) (f –(y) + d+–(x,y))
R(f,g)x) := infyODom(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 theoremA domain-theoretic Picard theorem
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Domain-theoretic Picard theorem (AE&AL)Domain-theoretic Picard theorem (AE&AL)
• Let v : [0,1] IR IR be Scott continuous and
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] IR with
(f, t. v (t , f(t) ) ) S-Cons
• Then the continuous map Up � Apv has 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) ) )
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• 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).
The Classical Initial Value ProblemThe 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 positive a 1 such that [-a,a] [-Ma,Ma] K.
• Put f = ⊔n 0 fn where fn = [-a/2n,a /2n] ↘ [-Ma/2n , Ma/ 2n ]
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Computation of the solution for a given precision Computation of the solution for a given precision >0
• Let (un , wn) := (Pv )n (fn , t. vn (t , fn(t) ) ) with un = [un
- ,un+]n
• We express f and v as lubs of step functions:
f = ⊔n 0 fn v = ⊔n 0 vn
• 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 0 with un+ - un
-
• (fs, gs) = ⊔n 0 (Pv )n (f , t. v (t , f(t) ) ) = ⊔n 0 (Pv )
n (fn , t. vn (t , fn(t) ) ) n
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ExampleExample
1
f
g
1
1
1
v
v is approximated by a sequence of step functions, v0, v1, …
v = ⊔i vi
We solve: = v(t,x), x(t0) =x0
for t [0,1] with
v(t,x) = t and t0=1/2, x0=9/8.
dt
dx
a3
b3
a2
b2
a1
b1
v3
v2
v1
The initial condition is approximated by rectangles aibi:
{(1/2,9/8)} = ⊔i aibi,
t
t
.
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SolutionSolution
1
f
g
1
1
1
At stage n we find un
- and un +
.
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SolutionSolution
1
f
g
1
1
1 .
At stage n we find un
- and un +
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SolutionSolution
1
f
g
1
1
1 un - and un
+ tend to
the exact solution:f: t t2/2 + 1
.
At stage n we find un
- and un +
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Computing with polynomial step functionsComputing with polynomial step functions
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Current and Further WorkCurrent 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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THE ENDTHE END
http://www.doc.ic.ac.uk/~aehttp://www.doc.ic.ac.uk/~ae