A universal ordinary differential equation Olivier Bournez 1 , Amaury Pouly 2 1 LIX, École Polytechnique, France 2 Max Planck Institute for Software Systems, Germany 12 july 2017 1 / 10
Universal differential algebraic equation (Rubel) y 1 ( x ) x Theorem (Rubel, 1981) There exists a fixed polynomial p and k ∈ N such that for any conti- nuous functions f and ε , there exists a solution y to p ( y , y ′ , . . . , y ( k ) ) = 0 such that | y ( t ) − f ( t ) | � ε ( t ) . 2 / 10
Universal differential algebraic equation (Rubel) y 1 ( x ) x Theorem (Rubel, 1981) There exists a fixed polynomial p and k ∈ N such that for any conti- nuous functions f and ε , there exists a solution y to ′′′′ + 6 y ′ 3 y ′′′′ + 24 y ′ 2 y ′′′′ 2 ′′′ 2 y ′′ 2 y ′′ 4 y 3 y ′ 4 y − 4 y ′ 4 y ′′ y ′′′ y ′′′′ ′′′ 3 − 29 y ′ 2 y ′′′ 2 + 12 y ′′ 3 y ′′ 7 − 12 y ′ 3 y ′′ y = 0 such that | y ( t ) − f ( t ) | � ε ( t ) . 2 / 10
Universal differential algebraic equation (Rubel) Open Problem y 1 ( x ) This is a DAE. Is there a x universal ODE? Theorem (Rubel, 1981) There exists a fixed polynomial p and k ∈ N such that for any conti- nuous functions f and ε , there exists a solution y to ′′′′ + 6 y ′ 3 y ′′′′ + 24 y ′ 2 y ′′′′ 2 ′′′ 2 y ′′ 2 y ′′ 4 y 3 y ′ 4 y − 4 y ′ 4 y ′′ y ′′′ y ′′′′ ′′′ 3 − 29 y ′ 2 y ′′′ 2 + 12 y ′′ 3 y ′′ 7 − 12 y ′ 3 y ′′ y = 0 such that | y ( t ) − f ( t ) | � ε ( t ) . 2 / 10
Rubel’s (disappointing) proof in one slide − 1 1 − t 2 for − 1 < t < 1 and f ( t ) = 0 otherwise. Take f ( t ) = e ( 1 − t 2 ) 2 f It satisfies ′′ ( t ) + 2 tf ′ ( t ) = 0 . t 3 / 10
Rubel’s (disappointing) proof in one slide − 1 1 − t 2 for − 1 < t < 1 and f ( t ) = 0 otherwise. Take f ( t ) = e ( 1 − t 2 ) 2 f It satisfies ′′ ( t ) + 2 tf ′ ( t ) = 0 . Can do the same with cf ( at + b ) (translation+scaling) t 3 / 10
Rubel’s (disappointing) proof in one slide − 1 1 − t 2 for − 1 < t < 1 and f ( t ) = 0 otherwise. Take f ( t ) = e ( 1 − t 2 ) 2 f It satisfies ′′ ( t ) + 2 tf ′ ( t ) = 0 . Can do the same with cf ( at + b ) (translation+scaling) Can glue together arbitrary many such pieces t 3 / 10
Rubel’s (disappointing) proof in one slide − 1 1 − t 2 for − 1 < t < 1 and f ( t ) = 0 otherwise. Take f ( t ) = e ( 1 − t 2 ) 2 f It satisfies ′′ ( t ) + 2 tf ′ ( t ) = 0 . Can do the same with cf ( at + b ) (translation+scaling) Can glue together arbitrary many such pieces Can arrange so that f is solution : piecewise pseudo-linear � t 3 / 10
Rubel’s (disappointing) proof in one slide − 1 1 − t 2 for − 1 < t < 1 and f ( t ) = 0 otherwise. Take f ( t ) = e ( 1 − t 2 ) 2 f It satisfies ′′ ( t ) + 2 tf ′ ( t ) = 0 . Can do the same with cf ( at + b ) (translation+scaling) Can glue together arbitrary many such pieces Can arrange so that f is solution : piecewise pseudo-linear � t Conclusion : Rubel’s equation allows any piecewise pseudo-linear functions, and those are dense in C 0 3 / 10
The problem with Rubel’s DAE the solution y is not unique, even with added initial conditions : p ( y , y ′ , . . . , y ( k ) ) = 0 , y ( 0 ) = α 0 , y ′ ( 0 ) = α 1 , . . . , y ( k ) ( 0 ) = α k 4 / 10
The problem with Rubel’s DAE the solution y is not unique, even with added initial conditions : p ( y , y ′ , . . . , y ( k ) ) = 0 , y ( 0 ) = α 0 , y ′ ( 0 ) = α 1 , . . . , y ( k ) ( 0 ) = α k ...even with a countable number of extra conditions : p ( y , y ′ , . . . , y ( k ) ) = 0 , y ( d i ) ( a i ) = b i , i ∈ N In fact, this is fundamental for Rubel’s proof to work! 4 / 10
The problem with Rubel’s DAE the solution y is not unique, even with added initial conditions : p ( y , y ′ , . . . , y ( k ) ) = 0 , y ( 0 ) = α 0 , y ′ ( 0 ) = α 1 , . . . , y ( k ) ( 0 ) = α k ...even with a countable number of extra conditions : p ( y , y ′ , . . . , y ( k ) ) = 0 , y ( d i ) ( a i ) = b i , i ∈ N In fact, this is fundamental for Rubel’s proof to work! Rubel’s statement : this DAE is universal More realistic interpretation : this DAE allows almost anything Open Problem (Rubel, 1981) This is a DAE. Is there a universal ODE y ′ = p ( y ) ? Note : ODE ⇒ unique solution 4 / 10
Universal ordinary differential equation (ODE) y 1 ( x ) x Main result There exists a fixed polynomial p and d ∈ N such that for any conti- nuous functions f and ε , there exists α ∈ R d such that y ( 0 ) = α, y ′ ( t ) = p ( y ( t )) has a unique solution and this solution satisfies | y ( t ) − f ( t ) | � ε ( t ) . 5 / 10
Universal ordinary differential equation (ODE) y 1 ( x ) x Main result There exists a fixed polynomial p and d ∈ N such that for any conti- nuous functions f and ε , there exists α ∈ R d such that y ( 0 ) = α, y ′ ( t ) = p ( y ( t )) has a unique solution and this solution satisfies | y ( t ) − f ( t ) | � ε ( t ) . Unfortunately, we need d ≈ 300. 5 / 10
Wait, is this a CS talk? Polynomial ODEs correspond to analog computers : Differential Analyser British Navy mecanical computer 6 / 10
Wait, is this a CS talk? Polynomial ODEs correspond to analog computers : Differential Analyser British Navy mecanical computer They are equivalent to Turing machines! One can characterize P with pODEs (ICALP 2016) Take away : polynomial ODEs is a natural programming language. 6 / 10
A first idea binary stream digits of α generator 0 1 1 0 1 0 1 0 0 1 1 1 . . . ODE α ∈ R t E T O N This is the ideal curve, the real one is an approximation of it. 7 / 10
A first idea binary stream digits of α generator 0 1 1 0 1 0 1 0 0 1 1 1 . . . ODE α ∈ R t ODE t “Digital” to Analog E T Converter (fixed frequency) O N E T Approximate Lipschitz and bounded O N functions with fixed precision. That’s the trickiest part. 7 / 10
A first idea binary stream digits of α generator 0 1 1 0 1 0 1 0 0 1 1 1 . . . ODE α ∈ R t ODE t “Digital” to Analog Converter (fixed frequency) ODE? E T O t N We need something more : a fast-growing ODE. 7 / 10
A first idea binary stream digits of α generator 0 1 1 0 1 0 1 0 0 1 1 1 . . . ODE α ∈ R t ODE t “Digital” to Analog Converter (fixed frequency) ODE? E T O t N We need something more : an arbitrarily fast-growing ODE. 7 / 10
An old question on growth Building a fast-growing ODE : y ′ 1 = y 1 y 1 ( t ) = exp( t ) � 8 / 10
An old question on growth Building a fast-growing ODE : y ′ 1 = y 1 y 1 ( t ) = exp( t ) � y ′ 2 = y 1 y 2 y 1 ( t ) = exp(exp( t )) � 8 / 10
An old question on growth Building a fast-growing ODE : y ′ 1 = y 1 y 1 ( t ) = exp( t ) � y ′ 2 = y 1 y 2 y 1 ( t ) = exp(exp( t )) � . . . . . . y ′ n = y 1 · · · y n y n ( t ) = exp( · · · exp( t ) · · · ):= e n ( t ) � 8 / 10
An old question on growth Building a fast-growing ODE : y ′ 1 = y 1 y 1 ( t ) = exp( t ) � y ′ 2 = y 1 y 2 y 1 ( t ) = exp(exp( t )) � . . . . . . y ′ n = y 1 · · · y n y n ( t ) = exp( · · · exp( t ) · · · ):= e n ( t ) � Conjecture (Emil Borel, 1899) With n variables, cannot do better than O t ( e n ( At k )) . 8 / 10
An old question on growth ( n compositions) e n ( t ) = exp( · · · exp( t ) · · · ) Conjecture (Emil Borel, 1899) With n variables, cannot do better than O t ( e n ( At k )) . Counter-example (Vijayaraghavan, 1932) 1 2 − cos( t ) − cos( α t ) Sequence of arbitrarily growing spikes. t 8 / 10
An old question on growth ( n compositions) e n ( t ) = exp( · · · exp( t ) · · · ) Conjecture (Emil Borel, 1899) With n variables, cannot do better than O t ( e n ( At k )) . Counter-example (Vijayaraghavan, 1932) 1 2 − cos( t ) − cos( α t ) Sequence of arbitrarily growing spikes. But not good enough for us. t 8 / 10
An old question on growth ( n compositions) e n ( t ) = exp( · · · exp( t ) · · · ) Conjecture (Emil Borel, 1899) With n variables, cannot do better than O t ( e n ( At k )) . Counter-example (Vijayaraghavan, 1932) 1 2 − cos( t ) − cos( α t ) Theorem (In the paper) There exists a polynomial p : R d → R d such that for any continuous function f : R + → R , we can find α ∈ R d such that y ( 0 ) = α, y ′ ( t ) = p ( y ( t )) satisfies ∀ t � 0 . y 1 ( t ) � f ( t ) 8 / 10
An old question on growth ( n compositions) e n ( t ) = exp( · · · exp( t ) · · · ) Conjecture (Emil Borel, 1899) With n variables, cannot do better than O t ( e n ( At k )) . Counter-example (Vijayaraghavan, 1932) 1 2 − cos( t ) − cos( α t ) Theorem (In the paper) There exists a polynomial p : R d → R d such that for any continuous function f : R + → R , we can find α ∈ R d such that y ( 0 ) = α, y ′ ( t ) = p ( y ( t )) satisfies ∀ t � 0 . y 1 ( t ) � f ( t ) Note : both results require α to be transcendental . Conjecture still open for rational coefficients. 8 / 10
Proof gem : iteration with differential equations Goal Iterate f with a GPAC : y ( n ) ≈ f [ n ] ([ x ]) 9 / 10
Proof gem : iteration with differential equations Goal Iterate f with a GPAC : y ( n ) ≈ f [ n ] ([ x ]) f ( x ) x t 0 1 2 1 3 y ′ ≈ 0 2 2 z ′ ≈ f ( y ) − z 9 / 10
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