A truly universal ordinary differential equation

Transcription

1 1 / 21 A truly universal ordinary differential equation Amaury Pouly 1 Joint work with Olivier Bournez 2 1 Max Planck Institute for Software Systems, Germany 2 LIX, École Polytechnique, France 11 May 2018

2 2 / 21 Universal differential algebraic equation (Rubel) y 1 (x) x Theorem (Rubel, 1981) For any f C 0 (R) and ε C 0 (R, R >0 ), there exists a solution y : R R to 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 that t R, y(t) f (t) ε(t).

3 2 / 21 Universal differential algebraic equation (Rubel) y 1 (x) x Theorem (Rubel, 1981) There exists a fixed k and nontrivial polynomial p such that for any f C 0 (R) and ε C 0 (R, R >0 ), there exists a solution y : R R to p(y, y,..., y (k) ) = 0 such that t R, y(t) f (t) ε(t).

4 2 / 21 Universal differential algebraic equation (Rubel) y 1 (x) x Open Problem Can we have unicity of the solution with initial conditions? Theorem (Rubel, 1981) There exists a fixed k and nontrivial polynomial p such that for any f C 0 (R) and ε C 0 (R, R >0 ), there exists a solution y : R R to p(y, y,..., y (k) ) = 0 such that t R, y(t) f (t) ε(t).

5 Rubel s ("disappointing") proof in one slide 3 / 21 Take f (t) = e 1 1 t 2 for 1 < t < 1 and f (t) = 0 otherwise. It satisfies (1 t 2 ) 2 f (t) + 2tf (t) = 0. t

6 Rubel s ("disappointing") proof in one slide 3 / 21 Take f (t) = e 1 1 t 2 for 1 < t < 1 and f (t) = 0 otherwise. It satisfies (1 t 2 ) 2 f (t) + 2tf (t) = 0. For any a, b, c R, y(t) = cf (at + b) satisfies 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 t

7 Rubel s ("disappointing") proof in one slide 3 / 21 Take f (t) = e 1 1 t 2 for 1 < t < 1 and f (t) = 0 otherwise. It satisfies (1 t 2 ) 2 f (t) + 2tf (t) = 0. For any a, b, c R, y(t) = cf (at + b) satisfies 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 2 +12y 7 =0 Can glue together arbitrary many such pieces crucial (and tricky) part of the proof t

8 Rubel s ("disappointing") proof in one slide 3 / 21 Take f (t) = e 1 1 t 2 for 1 < t < 1 and f (t) = 0 otherwise. It satisfies (1 t 2 ) 2 f (t) + 2tf (t) = 0. For any a, b, c R, y(t) = cf (at + b) satisfies 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 2 +12y 7 =0 Can glue together arbitrary many such pieces crucial (and tricky) part of the proof Can arrange so that f is solution : piecewise pseudo-linear t

9 Rubel s ("disappointing") proof in one slide Take f (t) = e 1 1 t 2 for 1 < t < 1 and f (t) = 0 otherwise. It satisfies (1 t 2 ) 2 f (t) + 2tf (t) = 0. For any a, b, c R, y(t) = cf (at + b) satisfies 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 2 +12y 7 =0 Can glue together arbitrary many such pieces crucial (and tricky) part of the proof Can arrange so that f is solution : piecewise pseudo-linear Conclusion : Rubel s equation allows any piecewise pseudo-linear functions, and those are dense in C 0 3 / 21 t

10 The problem with Rubel s DAE 4 / 21 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 In fact, this is fundamental for Rubel s proof to work!

11 4 / 21 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 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) Is there a universal ODE y = p(y)? Note : explicit polynomial ODE unique solution

12 Universal explicit ordinary differential equation 5 / 21 y 1 (x) x Main result There exists a fixed (vector of) polynomial p such that for any f C 0 (R) and ε C 0 (R, R >0 ), there exists α R d such that y(0) = α, y (t) = p(y(t)) has a unique solution y : R R d and t R, y 1 (t) f (t) ε(t).

13 5 / 21 Universal explicit ordinary differential equation y 1 (x) x Notes : system of ODEs, y must be analytic, we need d 300. Main result There exists a fixed (vector of) polynomial p such that for any f C 0 (R) and ε C 0 (R, R >0 ), there exists α R d such that y(0) = α, y (t) = p(y(t)) has a unique solution y : R R d and t R, y 1 (t) f (t) ε(t).

14 Universal explicit ordinary differential equation y 1 (x) x Notes : system of ODEs, y must be analytic, we need d 300. Main result There exists a fixed (vector of) polynomial p such that for any f C 0 (R) and ε C 0 (R, R >0 ), there exists α R d such that y(0) = α, y (t) = p(y(t)) has a unique solution y : R R d and t R, y 1 (t) f (t) ε(t). Futhermore, α is computable from f and ε.. This statement can be made precise with the theory of Computable Analysis. 5 / 21

15 6 / 21 Universal DAE, again but better y 1 (x) x Corollary of main result There exists a fixed k and nontrivial polynomial p such that for any f C 0 (R) and ε C 0 (R, R >0 ), there exists α 0,..., α k R such that p(y, y,..., y (k) ) = 0, y(0) = α 0, y (0) = α 1,..., y (k) (0) = α k has a unique analytic solution y : R R and t R, y(t) f (t) ε(t).

16 Some motivation 7 / 21 Polynomial ODEs correspond to analog computers : Differential Analyser British Navy mecanical computer

17 7 / 21 Some motivation 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 are a natural programming language.

18 Example of differential equation y 2 y 1 θ l m g l 1 y 3 General Purpose Analog Computer (GPAC) Shannon s model of the Differential Analyser y4 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) 8 / 21

19 A brief stop 9 / 21 Before I can explain the proof, you need to know more of polynomial ODEs and what I mean by programming with ODEs.

20 Generable functions (total, univariate) Definition f : R R is generable if there exists d, p and y 0 such that the solution y to y(0) = y 0, y (x) = p(y(x)) satisfies f (x) = y 1 (x) for all x R. Types d N : dimension Q K R : field p K d [R n ] : polynomial vector (coef. in K) y 0 K d, y : R R d y 1 (x) x Note : existence and unicity of y by Cauchy-Lipschitz theorem. 10 / 21

21 Generable functions (total, univariate) Definition f : R R is generable if there exists d, p and y 0 such that the solution y to y(0) = y 0, y (x) = p(y(x)) satisfies f (x) = y 1 (x) for all x R. Types d N : dimension Q K R : field p K d [R n ] : polynomial vector (coef. in K) y 0 K d, y : R R d Example : f (x) = x identity y(0) = 0, y = 1 y(x) = x 10 / 21

22 Generable functions (total, univariate) Definition f : R R is generable if there exists d, p and y 0 such that the solution y to y(0) = y 0, y (x) = p(y(x)) satisfies f (x) = y 1 (x) for all x R. Types d N : dimension Q K R : field p K d [R n ] : polynomial vector (coef. in K) y 0 K d, y : R R d Example : f (x) = x 2 squaring y 1 (0)= 0, y 1 = 2y 2 y 1 (x)= x 2 y 2 (0)= 0, y 2 = 1 y 2(x)= x 10 / 21

23 Generable functions (total, univariate) Definition f : R R is generable if there exists d, p and y 0 such that the solution y to y(0) = y 0, y (x) = p(y(x)) satisfies f (x) = y 1 (x) for all x R. Types d N : dimension Q K R : field p K d [R n ] : polynomial vector (coef. in K) y 0 K d, y : R R d Example : f (x) = x n n th power y 1 (0)= 0, y 1 = ny 2 y 1 (x)= x n y 2 (0)= 0, y 2 = (n 1)y 3 y 2 (x)= x n y n (0)= 0, y n = 1 y n (x)= x 10 / 21

24 Generable functions (total, univariate) Definition f : R R is generable if there exists d, p and y 0 such that the solution y to y(0) = y 0, y (x) = p(y(x)) satisfies f (x) = y 1 (x) for all x R. Types d N : dimension Q K R : field p K d [R n ] : polynomial vector (coef. in K) y 0 K d, y : R R d Example : f (x) = exp(x) exponential y(0)= 1, y = y y(x)= exp(x) 10 / 21

25 Generable functions (total, univariate) Definition f : R R is generable if there exists d, p and y 0 such that the solution y to y(0) = y 0, y (x) = p(y(x)) satisfies f (x) = y 1 (x) for all x R. Example : f (x) = sin(x) or f (x) = cos(x) Types d N : dimension Q K R : field p K d [R n ] : polynomial vector (coef. in K) y 0 K d, y : R R d sine/cosine y 1 (0)= 0, y 1 = y 2 y 1 (x)= sin(x) y 2 (0)= 1, y 2 = y 1 y 2 (x)= cos(x) 10 / 21

26 Generable functions (total, univariate) Definition f : R R is generable if there exists d, p and y 0 such that the solution y to y(0) = y 0, y (x) = p(y(x)) satisfies f (x) = y 1 (x) for all x R. Types d N : dimension Q K R : field p K d [R n ] : polynomial vector (coef. in K) y 0 K d, y : R R d Example : f (x) = tanh(x) hyperbolic tangent y(0)= 0, y = 1 y 2 y(x)= tanh(x) tanh(x) x 10 / 21

27 Generable functions (total, univariate) Definition f : R R is generable if there exists d, p and y 0 such that the solution y to y(0) = y 0, y (x) = p(y(x)) satisfies f (x) = y 1 (x) for all x R. Types d N : dimension Q K R : field p K d [R n ] : polynomial vector (coef. in K) y 0 K d, y : R R d Example : f (x) = 1 1+x 2 rational function f (x) = 2x = 2xf (x) 2 (1+x 2 ) 2 y 1 (0)= 1, y 1 = 2y 2y 2 1 y 1 (x)= 1 1+x 2 y 2 (0)= 0, y 2 = 1 y 2(x)= x 10 / 21

28 Generable functions (total, univariate) Definition f : R R is generable if there exists d, p and y 0 such that the solution y to y(0) = y 0, y (x) = p(y(x)) satisfies f (x) = y 1 (x) for all x R. Types d N : dimension Q K R : field p K d [R n ] : polynomial vector (coef. in K) y 0 K d, y : R R d Example : f = g ± h sum/difference (g ± h) = g ± h assume : z(0)= z 0, z = p(z) z 1 = g w(0)= w 0, w = q(w) w 1 = h then : y(0)= z 0,1 + w 0,1, y = p 1 (z) ± q 1 (w) y= z 1 ± w 1 10 / 21

29 Generable functions (total, univariate) Definition f : R R is generable if there exists d, p and y 0 such that the solution y to y(0) = y 0, y (x) = p(y(x)) satisfies f (x) = y 1 (x) for all x R. Types d N : dimension Q K R : field p K d [R n ] : polynomial vector (coef. in K) y 0 K d, y : R R d Example : f = gh product (gh) = g h + gh assume : z(0)= z 0, z = p(z) z 1 = g w(0)= w 0, w = q(w) w 1 = h then : y(0)= z 0,1 w 0,1, y = p 1 (z)w 1 + z 1 q 1 (w) y= z 1 w 1 10 / 21

30 Generable functions (total, univariate) Definition f : R R is generable if there exists d, p and y 0 such that the solution y to y(0) = y 0, y (x) = p(y(x)) satisfies f (x) = y 1 (x) for all x R. Types d N : dimension Q K R : field p K d [R n ] : polynomial vector (coef. in K) y 0 K d, y : R R d Example : f = 1 g inverse f = g g 2 = g f 2 assume : z(0)= z 0, z = p(z) z 1 = g then : y(0)= 1 z 0,1, y = p 1 (z)y 2 y= 1 z 1 10 / 21

31 Generable functions (total, univariate) Definition f : R R is generable if there exists d, p and y 0 such that the solution y to y(0) = y 0, y (x) = p(y(x)) satisfies f (x) = y 1 (x) for all x R. Types d N : dimension Q K R : field p K d [R n ] : polynomial vector (coef. in K) y 0 K d, y : R R d Example : f = g integral assume : z(0)= z 0, z = p(z) z 1 = g then : y(0)= 0, y = z 1 y= z 1 10 / 21

32 Generable functions (total, univariate) Definition f : R R is generable if there exists d, p and y 0 such that the solution y to y(0) = y 0, y (x) = p(y(x)) satisfies f (x) = y 1 (x) for all x R. Types d N : dimension Q K R : field p K d [R n ] : polynomial vector (coef. in K) y 0 K d, y : R R d Example : f = g derivative f = g = (p 1 (z)) = p 1 (z) z assume : z(0)= z 0, z = p(z) z 1 = g then : y(0)= p 1 (z 0 ), y = p 1 (z) p(z) y= z 1 10 / 21

33 Generable functions (total, univariate) Definition f : R R is generable if there exists d, p and y 0 such that the solution y to y(0) = y 0, y (x) = p(y(x)) satisfies f (x) = y 1 (x) for all x R. Types d N : dimension Q K R : field p K d [R n ] : polynomial vector (coef. in K) y 0 K d, y : R R d Example : f = g h composition (z h) = (z h)h = p(z h)h assume : z(0)= z 0, z = p(z) z 1 = g w(0)= w 0, w = q(w) w 1 = h then : y(0)= z(w 0 ), y = p(y)z 1 y= z h 10 / 21

34 Generable functions (total, univariate) Definition f : R R is generable if there exists d, p and y 0 such that the solution y to y(0) = y 0, y (x) = p(y(x)) satisfies f (x) = y 1 (x) for all x R. Types d N : dimension Q K R : field p K d [R n ] : polynomial vector (coef. in K) y 0 K d, y : R R d Example : f = g h composition (z h) = (z h)h = p(z h)h assume : z(0)= z 0, z = p(z) z 1 = g w(0)= w 0, w = q(w) w 1 = h then : y(0)= z(w 0 ), y = p(y)z 1 y= z h Is this coefficient in K? 10 / 21

35 Generable functions (total, univariate) Definition f : R R is generable if there exists d, p and y 0 such that the solution y to y(0) = y 0, y (x) = p(y(x)) satisfies f (x) = y 1 (x) for all x R. Types d N : dimension Q K R : field p K d [R n ] : polynomial vector (coef. in K) y 0 K d, y : R R d Example : f = g h composition (z h) = (z h)h = p(z h)h assume : z(0)= z 0, z = p(z) z 1 = g w(0)= w 0, w = q(w) w 1 = h then : y(0)= z(w 0 ), y = p(y)z 1 y= z h Is this coefficient in K? Fields with this property are called generable. 10 / 21

36 Generable functions (total, univariate) Definition f : R R is generable if there exists d, p and y 0 such that the solution y to y(0) = y 0, y (x) = p(y(x)) satisfies f (x) = y 1 (x) for all x R. Types d N : dimension Q K R : field p K d [R n ] : polynomial vector (coef. in K) y 0 K d, y : R R d Example : f = tanh f Non-polynomial differential equation f = (tanh f )f = (1 (tanh f ) 2 )f y 1 (0)= f (0), y 1 = y 2 y 1 (x)= f (x) y 2 (0)= tanh(f (0)), y 2 = (1 y 2 2 )y 2 y 2 (x)= tanh(f (x)) 10 / 21

37 Generable functions (total, univariate) Definition f : R R is generable if there exists d, p and y 0 such that the solution y to y(0) = y 0, y (x) = p(y(x)) satisfies f (x) = y 1 (x) for all x R. Types d N : dimension Q K R : field p K d [R n ] : polynomial vector (coef. in K) y 0 K d, y : R R d Example : f (0) = f 0, f = g f Initial Value Problem (IVP) f = g = (p 1 (z)) = p 1 (z) z assume : z(0)= z 0, z = p(z) z 1 = g then : y(0)= p 1 (z 0 ), y = p 1 (z) p(z) y= z 1 10 / 21

38 11 / 21 Generable functions : a first summary Nice theory for the class of total and univariate generable functions : analytic contains polynomials, sin, cos, tanh, exp stable under ±,, /, and Initial Value Problems (IVP) technicality on the field K of coefficients for stability under solutions to polynomial ODEs form a very large class

39 11 / 21 Generable functions : a first summary Nice theory for the class of total and univariate generable functions : analytic contains polynomials, sin, cos, tanh, exp stable under ±,, /, and Initial Value Problems (IVP) technicality on the field K of coefficients for stability under solutions to polynomial ODEs form a very large class Limitations : total functions univariate

40 Generable functions (generalization) Definition f : X R n R is generable if X is open connected and d, p, x 0, y 0, y such that y(x 0 ) = y 0, J y (x) = p(y(x)) and f (x) = y 1 (x) for all x X. J y (x) = Jacobian matrix of y at x Notes : Partial differential equation! Unicity of solution y... Types n N : input dimension d N : dimension p K d d [R d ] : polynomial matrix x 0 K n... but not existence (ie you have to show it exists) y 0 K d, y : X R d 12 / 21

41 Generable functions (generalization) 12 / 21 Definition f : X R n R is generable if X is open connected and d, p, x 0, y 0, y such that y(x 0 ) = y 0, J y (x) = p(y(x)) and f (x) = y 1 (x) for all x X. J y (x) = Jacobian matrix of y at x Types n N : input dimension d N : dimension p K d d [R d ] : polynomial matrix x 0 K n y 0 K d, y : X R d Example : f (x 1, x 2 ) = x 1 x 2 2 (n = 2, d = 3) monomial 0 y 2 3 3y 2 y 3 x 1 x 2 2 y(0, 0) = 0, J y = 1 0 y(x) = x x 2

42 Generable functions (generalization) Definition f : X R n R is generable if X is open connected and d, p, x 0, y 0, y such that y(x 0 ) = y 0, J y (x) = p(y(x)) and f (x) = y 1 (x) for all x X. J y (x) = Jacobian matrix of y at x Types n N : input dimension d N : dimension p K d d [R d ] : polynomial matrix x 0 K n y 0 K d, y : X R d Example : f (x 1, x 2 ) = x 1 x 2 2 monomial y 1 (0, 0)= 0, x1 y 1 = y3 2, x 2 y 1 = 3y 2 y 3 y 1 (x) = x 1 x2 2 y 2 (0, 0)= 0, x1 y 2 = 1, x2 y 2 = 0 y 2 (x) = x 1 y 3 (0, 0)= 0, x1 y 3 = 0, x2 y 3 = 1 y 3 (x) = x 2 This is tedious! 12 / 21

43 Generable functions (generalization) Definition f : X R n R is generable if X is open connected and d, p, x 0, y 0, y such that y(x 0 ) = y 0, J y (x) = p(y(x)) and f (x) = y 1 (x) for all x X. J y (x) = Jacobian matrix of y at x Types n N : input dimension d N : dimension p K d d [R d ] : polynomial matrix x 0 K n y 0 K d, y : X R d Last example : f (x) = 1 x for x (0, ) inverse function y(1)= 1, x y= y 2 y(x) = 1 x 12 / 21

44 13 / 21 Generable functions : summary Nice theory for the class of multivariate generable functions (over connected domains) : analytic contains polynomials, sin, cos, tanh, exp,... stable under ±,, /, and Initial Value Problems (IVP) technicality on the field K of coefficients for stability under requires partial differential equations

45 13 / 21 Generable functions : summary Nice theory for the class of multivariate generable functions (over connected domains) : analytic contains polynomials, sin, cos, tanh, exp,... stable under ±,, /, and Initial Value Problems (IVP) technicality on the field K of coefficients for stability under requires partial differential equations Exercice : are all analytic functions generable?

46 13 / 21 Generable functions : summary Nice theory for the class of multivariate generable functions (over connected domains) : analytic contains polynomials, sin, cos, tanh, exp,... stable under ±,, /, and Initial Value Problems (IVP) technicality on the field K of coefficients for stability under requires partial differential equations Exercice : are all analytic functions generable? No Riemann Γ and ζ are not generable.

47 Why is this useful? 14 / 21 Writing polynomial ODEs by hand is hard.

48 Why is this useful? 14 / 21 Writing polynomial ODEs by hand is hard. Using generable functions, we can build complicated multivariate partial functions using other operations, and we know they are solutions to polynomial ODEs by construction.

49 14 / 21 Why is this useful? Writing polynomial ODEs by hand is hard. Using generable functions, we can build complicated multivariate partial functions using other operations, and we know they are solutions to polynomial ODEs by construction. Example (almost rounding function) There exists a generable function round such that for any n Z, x R, λ > 2 and µ 0 : if x [ n 1 2, n + 1 2] then round(x, µ, λ) n 1 2, if x [ n λ, n λ] then round(x, µ, λ) n e µ.

50 Reminder of the result 15 / 21 Main result (reminder) There exists a fixed (vector of) polynomial p such that for any f C 0 (R) and ε C 0 (R, R >0 ), there exists α R d such that y(0) = α, y (t) = p(y(t)) has a unique solution y : R R d and t R, y 1 (t) f (t) ε(t).

51 A simplified proof binary stream generator digits of α α R ODE t NOTE This is the ideal curve, the real one is an approximation of it. 16 / 21

52 A simplified proof binary stream generator digits of α α R ODE t NOTE Approximate Lipschitz and bounded functions with fixed precision. t ODE Digital to Analog Converter (fixed frequency) NOTE That s the trickiest part. 16 / 21

53 A simplified proof binary stream generator digits of α α R ODE t NOTE ODE? We need something more : a fast-growing ODE. t ODE Digital to Analog Converter (fixed frequency) t 16 / 21

54 A simplified proof binary stream generator digits of α α R ODE t NOTE ODE? We need something more : an arbitrarily fast-growing ODE. t ODE Digital to Analog Converter (fixed frequency) t 16 / 21

55 A less simplified proof binary stream generator : digits of α R t f (α, µ, λ, t) = tanh(µ sin(2απ4round(t 1/4,λ) + 4π/3)) It s horrible, but generable round is the mysterious rounding function / 21

56 A less simplified proof binary stream generator : digits of α R t d 0 a 0 d 1 a 1 d 2 a 2 d 3 a 3 t dyadic stream generator : d i = m i 2 d i, a i = 9i + j<i d j f (α, γ, t) = sin(2απ2 round(t 1/4,γ) )) round is the mysterious rounding function / 21

57 A less simplified proof 17 / t a 0 d 0 d 1 a 1 a 2 a 3 d 2 d 3 t

58 A less simplified proof 17 / copy signal t a 0 d 0 d 1 a 1 a 2 a 3 d 2 d 3 t

59 A less simplified proof 17 / copy signal 1 0 copy signal t a 0 d 0 d 1 a 1 a 2 a 3 d 2 d 3 t

60 A less simplified proof 17 / copy signal copy signal copy signal 1 0 t a 0 d 0 d 1 a 1 a 2 a 3 d 2 d 3 t

61 A less simplified proof 17 / t copy signal copy signal copy signal copy signal a 0 d 0 d 1 a 1 a 2 a 3 d 2 d 3 t

62 A less simplified proof 17 / t copy signal copy signal copy signal copy signal a 0 d 0 d 1 a 1 a 2 a 3 d 2 d 3 t This copy operation is the non-trivial part.

63 A less simplified proof 17 / 21 t We can do almost piecewise constant functions...

64 17 / 21 A less simplified proof t We can do almost piecewise constant functions......that are bounded by and have super slow changing frequency.

65 17 / 21 A less simplified proof t We can do almost piecewise constant functions......that are bounded by and have super slow changing frequency. How do we go to arbitrarily large and growing functions? Can a polynomial ODE even have arbitrary growth?

66 An old question on growth 18 / 21 Building a fast-growing ODE, that exists over R : y 1 = y 1 y 1 (t) = exp(t)

67 An old question on growth 18 / 21 Building a fast-growing ODE, that exists over R : y 1 = y 1 y 1 (t) = exp(t) y 2 = y 1y 2 y 1 (t) = exp(exp(t))

68 An old question on growth 18 / 21 Building a fast-growing ODE, that exists over R : y 1 = y 1 y 1 (t) = exp(t) y 2 = y 1y 2 y 1 (t) = exp(exp(t)) y n = y 1 y n y n (t) = exp( exp(t) ) := e n (t)

69 An old question on growth 18 / 21 Building a fast-growing ODE, that exists over R : y 1 = y 1 y 1 (t) = exp(t) y 2 = y 1y 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 )).

70 An old question on growth e n (t) = exp( exp(t) ) (n compositions) 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(αt) Sequence of arbitrarily growing spikes. t 18 / 21

71 An old question on growth e n (t) = exp( exp(t) ) (n compositions) 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(αt) Sequence of arbitrarily growing spikes. But not good enough for us. t 18 / 21

72 An old question on growth 18 / 21 e n (t) = exp( exp(t) ) (n compositions) Conjecture (Emil Borel, 1899) With n variables, cannot do better than O t (e n (At k )). Counter-example (Vijayaraghavan, 1932) Theorem (In the paper) 1 2 cos(t) cos(αt) There exists a polynomial p : R d R d such that for any continuous function f : R 0 R, we can find α R d such that satisfies y(0) = α, y (t) = p(y(t)) y 1 (t) f (t), t 0.

73 An old question on growth e n (t) = exp( exp(t) ) (n compositions) Conjecture (Emil Borel, 1899) With n variables, cannot do better than O t (e n (At k )). Counter-example (Vijayaraghavan, 1932) Theorem (In the paper) 1 2 cos(t) cos(αt) There exists a polynomial p : R d R d such that for any continuous function f : R 0 R, we can find α R d such that satisfies y(0) = α, y (t) = p(y(t)) y 1 (t) f (t), t 0. Note : both results require α to be transcendental. Conjecture still open for rational (or algebraic) coefficients. 18 / 21

74 Proof gem : iteration with differential equations 19 / 21 Assume f is generable, can we iterate f with an ODE? That is, build a generable y such that y(x, n) f [n] (x) for all n N

75 Proof gem : iteration with differential equations 19 / 21 Assume f is generable, can we iterate f with an ODE? That is, build a generable y such that y(x, n) f [n] (x) for all n N f (x) x 0 1 y 0 2 z f (y) z t

76 Proof gem : iteration with differential equations 19 / 21 Assume f is generable, can we iterate f with an ODE? That is, build a generable y such that y(x, n) f [n] (x) for all n N f (x) x 0 1 y 0 2 z f (y) z y z y z t

77 Proof gem : iteration with differential equations 19 / 21 Assume f is generable, can we iterate f with an ODE? That is, build a generable y such that y(x, n) f [n] (x) for all n N f [2] (x) f (x) x 0 1 y 0 2 z f (y) z y z y z t

78 20 / 21 Main result, remark and end Main result (reminder) There exists a fixed (vector of) polynomial p such that for any f C 0 (R) and ε C 0 (R, R >0 ), there exists α R d such that y(0) = α, y (t) = p(y(t)) has a unique solution y : R R d and t R, y 1 (t) f (t) ε(t). Futhermore, α is computable from f and ε. Remarks : if f and ε are computable then α is computable if f or ε is not computable then α is not computable in all cases α is a horrible transcendental number