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Abbas Edalat
Imperial College London
www.doc.ic.ac.uk/~ae
joint work with Marko Krznaric and Andre Lieutier
Domain-theoretic Solution of Differential EquationsDomain-theoretic Solution of Differential Equations
2
AimAim
• Develop data types for ordinary differential equations.
• Solve initial value problems up to any given precision:
= v(t,x) with v: R2 R continuous and Lipschitz in x
x(t0) = x0 with (t0,x0) R2
dt
dx
3
• Let IR={ [a,b] | a, b R} {R}
• (IR, ) is a cpo with R as bottom and:
⊔i 0 ai = i 0 ai
• (IR, ⊑) is a continuous Scott domain: countable basis {[p,q] | p < q & p, q Q}
x {x}
R
I R
• x {x} : R IRgives an embedding onto the maximal elements
The Domain of Intervals of The Domain of Intervals of RR
4
Data-type for FunctionsData-type for Functions
• Lubs of finite and bounded collections of single- step functions
⊔1 i n(ai ↘ bi)
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
a x
7
Domain for ContinuousDomain for Continuous FunctionsFunctions
• 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 ai, bi rational intervals, give a basis for [0,1] IR
• f If: C0[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 C0[0,1] (continuous functions) has continuous
extension If : [0,1] IR
x {f (x)}
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Domain for DifferentiableDomain for Differentiable Functions Functions
• If h C1[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
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Function and Derivative ConsistencyFunction 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).Ih'
• 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) .
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Approximating function: f = ⊔i ai↘bi
• (⊔1 i n ai ↘ bi , ⊔1 j m cj ↘ dj) Cons is decidable:
Consistency for basis elementsConsistency for basis elements
L(f,g) = least function
G(f,g)= greatest function
• Updating. 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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• Theorem.D1 [0,1]:= { (f,g) ([0,1]IR)2 | (f,g) Cons} is a continuous Scott domain.
The Domain of The Domain of DifferentiableDifferentiable FunctionsFunctions
• Theorem. C1[0,1] embeds into the set of maximal elements of D1
[0,1]
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Solving Initial Value ProblemsSolving Initial Value Problems
= v(t,x) with v: R2 R continuous and Lipschitz in x
x(0) = 0
dt
dx
• 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 fn where fn = [-a/2n,a /2n] ↘ [-Ma/2n , Ma/ 2n ] • This is the initial function approximation.
• It also gives the initial derivative approximation: t. v (t , f(t) ))
[-Ma,Ma]
[-a,a]
22
• Solve the ODE by iterating Up ⃘ Apv on D1 [-1,1]
starting with (f, t. v (t , f(t) ))
• Theorem. The domain-theoretic solution
⊔n 0 (Up ⃘ Apv )n (f, t. v (t , f(t) ))
is the unique classical solution through (0,0).
Function and Derivative UpgradingFunction and Derivative Upgrading
• Derivative upgrading:
Apv : ([-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) = (fg , g) where fg (t) = [ L (f,g) (t) , G (f,g) (t) ]
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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
- t. 0
• Compute the piecewise linear maps un- , un
+ until
the first n 0 with un+ - un
-
• ⊔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
.
27
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 +
29
Current and Further WorkCurrent and Further Work
• Implementation in Haskell
• Differential Calculus with Several Variables: PDE’s
• Construction of Smooth Curves and Surfaces
• Hybrid Systems, robotics,…
30
THE ENDTHE END
http://www.doc.ic.ac.uk/~aehttp://www.doc.ic.ac.uk/~ae
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