A universal ordinary differential equation

Transcription

1 1 / 10 A universal ordinary differenial equaion Olivier Bournez 1, Amaury Pouly 2 1 LIX, École Polyechnique, France 2 Max Planck Insiue for Sofware Sysems, Germany 12 july 2017

2 2 / 10 Universal differenial algebraic equaion (Rubel) y 1 (x) x Theorem (Rubel, 1981) There exiss a fixed polynomial p and k N such ha for any coninuous funcions f and ε, here exiss a soluion y o p(y, y,..., y (k) ) = 0 such ha y() f () ε().

3 2 / 10 Universal differenial algebraic equaion (Rubel) y 1 (x) x Theorem (Rubel, 1981) There exiss a fixed polynomial p and k N such ha for any coninuous funcions f and ε, here exiss a soluion y o 3y 4 y y 2 4y 4 y 2 y + 6y 3 y 2 y y + 24y 2 y 4 y 12y 3 y y 3 29y 2 y 3 y y 7 = 0 such ha y() f () ε().

4 2 / 10 Universal differenial algebraic equaion (Rubel) y 1 (x) x Open Problem This is a DAE. Is here a universal ODE? Theorem (Rubel, 1981) There exiss a fixed polynomial p and k N such ha for any coninuous funcions f and ε, here exiss a soluion y o 3y 4 y y 2 4y 4 y 2 y + 6y 3 y 2 y y + 24y 2 y 4 y 12y 3 y y 3 29y 2 y 3 y y 7 = 0 such ha y() f () ε().

5 Rubel s (disappoining) proof in one slide 3 / 10 Take f () = e for 1 < < 1 and f () = 0 oherwise. I saisfies (1 2 ) 2 f () + 2f () = 0.

6 Rubel s (disappoining) proof in one slide 3 / 10 Take f () = e for 1 < < 1 and f () = 0 oherwise. I saisfies (1 2 ) 2 f () + 2f () = 0. Can do he same wih cf (a + b) (ranslaion+scaling)

7 3 / 10 Rubel s (disappoining) proof in one slide Take f () = e for 1 < < 1 and f () = 0 oherwise. I saisfies (1 2 ) 2 f () + 2f () = 0. Can do he same wih cf (a + b) (ranslaion+scaling) Can glue ogeher arbirary many such pieces

8 3 / 10 Rubel s (disappoining) proof in one slide Take f () = e for 1 < < 1 and f () = 0 oherwise. I saisfies (1 2 ) 2 f () + 2f () = 0. Can do he same wih cf (a + b) (ranslaion+scaling) Can glue ogeher arbirary many such pieces Can arrange so ha f is soluion : piecewise pseudo-linear

9 Rubel s (disappoining) proof in one slide Take f () = e for 1 < < 1 and f () = 0 oherwise. I saisfies (1 2 ) 2 f () + 2f () = 0. Can do he same wih cf (a + b) (ranslaion+scaling) Can glue ogeher arbirary many such pieces Can arrange so ha f is soluion : piecewise pseudo-linear Conclusion : Rubel s equaion allows any piecewise pseudo-linear funcions, and hose are dense in C 0 3 / 10

10 The problem wih Rubel s DAE 4 / 10 he soluion y is no unique, even wih added iniial condiions : p(y, y,..., y (k) ) = 0, y(0) = α 0, y (0) = α 1,..., y (k) (0) = α k

11 The problem wih Rubel s DAE 4 / 10 he soluion y is no unique, even wih added iniial condiions : p(y, y,..., y (k) ) = 0, y(0) = α 0, y (0) = α 1,..., y (k) (0) = α k...even wih a counable number of exra condiions : p(y, y,..., y (k) ) = 0, y (d i ) (a i ) = b i, i N In fac, his is fundamenal for Rubel s proof o work!

12 The problem wih Rubel s DAE he soluion y is no unique, even wih added iniial condiions : p(y, y,..., y (k) ) = 0, y(0) = α 0, y (0) = α 1,..., y (k) (0) = α k...even wih a counable number of exra condiions : p(y, y,..., y (k) ) = 0, y (d i ) (a i ) = b i, i N In fac, his is fundamenal for Rubel s proof o work! Rubel s saemen : his DAE is universal More realisic inerpreaion : his DAE allows almos anyhing Open Problem (Rubel, 1981) This is a DAE. Is here a universal ODE y = p(y)? Noe : ODE unique soluion 4 / 10

13 Universal ordinary differenial equaion (ODE) y 1 (x) x Main resul There exiss a fixed polynomial p and d N such ha for any coninuous funcions f and ε, here exiss α R d such ha y(0) = α, y () = p(y()) has a unique soluion and his soluion saisfies y() f () ε(). 5 / 10

14 Universal ordinary differenial equaion (ODE) y 1 (x) x Main resul There exiss a fixed polynomial p and d N such ha for any coninuous funcions f and ε, here exiss α R d such ha y(0) = α, y () = p(y()) has a unique soluion and his soluion saisfies y() f () ε(). Unforunaely, we need d / 10

15 Wai, is his a CS alk? 6 / 10 Polynomial ODEs correspond o analog compuers : Differenial Analyser Briish Navy mecanical compuer

16 6 / 10 Wai, is his a CS alk? Polynomial ODEs correspond o analog compuers : Differenial Analyser Briish Navy mecanical compuer They are equivalen o Turing machines! One can characerize P wih podes (ICALP 2016) Take away : polynomial ODEs is a naural programming language.

17 A firs idea binary sream generaor digis of α α R ODE NOTE This is he ideal curve, he real one is an approximaion of i. 7 / 10

18 A firs idea binary sream generaor digis of α α R ODE NOTE Approximae Lipschiz and bounded funcions wih fixed precision. ODE Digial o Analog Converer (fixed frequency) NOTE Tha s he rickies par. 7 / 10

19 A firs idea binary sream generaor digis of α α R ODE NOTE ODE? We need somehing more : a fas-growing ODE. ODE Digial o Analog Converer (fixed frequency) 7 / 10

20 A firs idea binary sream generaor digis of α α R ODE NOTE ODE? We need somehing more : an arbirarily fas-growing ODE. ODE Digial o Analog Converer (fixed frequency) 7 / 10

21 An old quesion on growh 8 / 10 Building a fas-growing ODE : y 1 = y 1 y 1 () = exp()

22 An old quesion on growh 8 / 10 Building a fas-growing ODE : y 1 = y 1 y 1 () = exp() y 2 = y 1y 2 y 1 () = exp(exp())

23 An old quesion on growh 8 / 10 Building a fas-growing ODE : y 1 = y 1 y 1 () = exp() y 2 = y 1y 2 y 1 () = exp(exp()) y n = y 1 y n y n () = exp( exp() ):= e n ()

24 An old quesion on growh 8 / 10 Building a fas-growing ODE : y 1 = y 1 y 1 () = exp() y 2 = y 1y 2 y 1 () = exp(exp()) y n = y 1 y n y n () = exp( exp() ):= e n () Conjecure (Emil Borel, 1899) Wih n variables, canno do beer han O (e n (A k )).

25 An old quesion on growh e n () = exp( exp() ) (n composiions) Conjecure (Emil Borel, 1899) Wih n variables, canno do beer han O (e n (A k )). Couner-example (Vijayaraghavan, 1932) 1 2 cos() cos(α) Sequence of arbirarily growing spikes. 8 / 10

26 An old quesion on growh e n () = exp( exp() ) (n composiions) Conjecure (Emil Borel, 1899) Wih n variables, canno do beer han O (e n (A k )). Couner-example (Vijayaraghavan, 1932) 1 2 cos() cos(α) Sequence of arbirarily growing spikes. Bu no good enough for us. 8 / 10

27 An old quesion on growh 8 / 10 e n () = exp( exp() ) (n composiions) Conjecure (Emil Borel, 1899) Wih n variables, canno do beer han O (e n (A k )). Couner-example (Vijayaraghavan, 1932) Theorem (In he paper) 1 2 cos() cos(α) There exiss a polynomial p : R d R d such ha for any coninuous funcion f : R + R, we can find α R d such ha saisfies y(0) = α, y () = p(y()) y 1 () f () 0.

28 An old quesion on growh e n () = exp( exp() ) (n composiions) Conjecure (Emil Borel, 1899) Wih n variables, canno do beer han O (e n (A k )). Couner-example (Vijayaraghavan, 1932) Theorem (In he paper) 1 2 cos() cos(α) There exiss a polynomial p : R d R d such ha for any coninuous funcion f : R + R, we can find α R d such ha saisfies y(0) = α, y () = p(y()) y 1 () f () 0. Noe : boh resuls require α o be ranscendenal. Conjecure sill open for raional coefficiens. 8 / 10

29 Proof gem : ieraion wih differenial equaions 9 / 10 Goal Ierae f wih a GPAC : y(n) f [n] ([x])

30 Proof gem : ieraion wih differenial equaions 9 / 10 Goal Ierae f wih a GPAC : y(n) f [n] ([x]) f (x) x 0 1 y 0 2 z f (y) z

31 Proof gem : ieraion wih differenial equaions 9 / 10 Goal Ierae f wih a GPAC : y(n) f [n] ([x]) f (x) x 0 1 y 0 2 z f (y) z y z y z

32 Proof gem : ieraion wih differenial equaions 9 / 10 Goal Ierae f wih a GPAC : y(n) f [n] ([x]) f [2] (x) f (x) x 0 1 y 0 2 z f (y) z y z y z

33 Conclusion 10 / 10 This paper posiive answer o Rubel s open problem Take home ODE is a simple, nice and fun programming language Possible developmen Each universal ODE defines a map : (f, ε) C 0 C 0 α R Kolmogorov-like complexiy for coninuous funcions?

34 Polynomial Differenial Equaions k k u v uv u v + u+v u u General Purpose Analog Compuer Newon mechanics Reacion neworks : chemical enzymaic Differenial Analyzer polynomial differenial equaions : { y(0)= y0 y ()= p(y()) Rich class Sable (+,,,/,ED) No closed-form soluion 11 / 10

35 Example of differenial equaion 12 / 10 y 2 y 1 l g l y 3 y4 θ m 1 θ + g l sin(θ) = 0 y 1 = y 2 y 2 = g l y 3 y 3 = y 2y 4 y 4 = y 2y 3 y 1 = θ y 2 = θ y 3 = sin(θ) y 4 = cos(θ)

36 Universal differenial equaion (DAE) y 1 (x) x Theorem There exiss a fixed polynomial p and k N such ha for any coninuous funcions f and ε, here exiss α 0,..., α k R such ha p(y, y,..., y (k) ) = 0, y(0) = α 0, y (0) = α 1,..., y (k) (0) = α k has a unique analyic soluion and his soluion saisfies y() f () ε(). 13 / 10

37 Digial vs analog compuers 14 / 10

38 Digial vs analog compuers 14 / 10 VS

39 Church Thesis Compuabiliy discree logic boolean circuis recursive funcions Turing machine lambda calculus quanum analog coninuous Church Thesis All reasonable models of compuaion are equivalen. 15 / 10

40 Church Thesis Complexiy discree logic boolean circuis recursive funcions Turing machine lambda calculus?? quanum analog coninuous Effecive Church Thesis All reasonable models of compuaion are equivalen for complexiy. 15 / 10

41 Compuing wih he GPAC 16 / 10 Generable funcions { y(0)= y0 y (x)= p(y(x)) x R f (x) = y 1 (x) y 1 (x) x Shannon s noion

42 Compuing wih he GPAC 16 / 10 Generable funcions { y(0)= y0 y (x)= p(y(x)) x R f (x) = y 1 (x) y 1 (x) x Shannon s noion sin, cos, exp, log,... Sricly weaker han Turing machines [Shannon, 1941]

43 Compuing wih he GPAC 16 / 10 Generable funcions Compuable { y(0)= y0 y (x)= p(y(x)) x R { y(0)= q(x) y ()= p(y()) x R R + f (x) = y 1 (x) f (x) = lim y 1 () y 1 (x) x y 1 () f (x) Shannon s noion sin, cos, exp, log,... x Modern noion Sricly weaker han Turing machines [Shannon, 1941]

44 Compuing wih he GPAC 16 / 10 Generable funcions Compuable { y(0)= y0 y (x)= p(y(x)) x R { y(0)= q(x) y ()= p(y()) x R R + f (x) = y 1 (x) f (x) = lim y 1 () y 1 (x) x y 1 () f (x) Shannon s noion sin, cos, exp, log,... Sricly weaker han Turing machines [Shannon, 1941] x Modern noion sin, cos, exp, log, Γ, ζ,... Turing powerful [Bournez e al., 2007]

45 Universal differenial equaions 17 / 10 Generable funcions Compuable funcions y 1 (x) x x y 1 () f (x) subclass of analyic funcions any compuable funcion

46 Universal differenial equaions 17 / 10 Generable funcions Compuable funcions y 1 (x) x x y 1 () f (x) subclass of analyic funcions any compuable funcion y 1 (x) x

47 A new noion of compuabiliy 18 / 10 Almos-Theorem f : [0, 1] R is compuable if and only if here exiss τ > 1, y 0 R d and p polynomial such ha y (0) = y 0, y () = p(y()) saisfies f (x) y(x + nτ) 2 n, x [0, 1], n N y() f ( mod τ) 0 1 τ τ + 1 2τ 2τ + 1 3τ