# Continuous-time extension of a discrete dynamical system

It is clear that one can obtain a discrete dynamical system from a continuous one, but is the converse possible if the system is "nice"?

Define the discrete-time dynamical system on $$\mathbb{R}^d$$ by $$x_{n+1} = f(x_n);\, x_0\triangleq x$$ where $$f \in C^2(\mathbb{R}^d;\mathbb{R}^d)$$ and $$x \in \mathbb{R}^d$$.

Fix a (large) positive integer $$N$$, is there a function $$F:\mathbb{R}^d\rightarrow \mathbb{R}^d$$ such that the solution to the continuous-time dynamical system $$\partial_t X_t = F(X_t) ; \qquad X_0=x,$$ and $$X_n = x_n$$ for every $$n \in \left\{1,\dots,N\right\}$$?

• If you mean "for all $x$", then no, it is not possible. For every $F$, the Jacobian determinant of the corresponding flow is non-zero, and hence it remains positive. Thus, $f(x) = (-x_1,x_2,x_3,\ldots,x_d)$ provides a simple counter-example. On the other hand, if $x$ is fixed, then I do not see the answer right away. – Mateusz Kwaśnicki Jan 9 at 10:10
• @MateuszKwaśnicki No, $x$ should be specified (so the "anti-discritization" is dependent on the choice of x); It would only need to hold for an (arbitrarily large but fixed) number of iterations. – AIM_BLB Jan 9 at 10:22
• Then this is essentially a question whether there is a smooth simple curve passing through given $N$ points $x_1, \ldots, x_N$ in a given order. Or, more precisely, we require this curve to pass through $M$ points $x_1, \ldots, x_M$ in a given order and have no self-intersections except possibly at $x_M$, in which case it must be tangent to itself; here $M$ is the least number such that $x_M \in \{x_1, \ldots, x_{M-1}\}$, or $M = N$ if no such number exists. This seems to be fairly straightforward, no? – Mateusz Kwaśnicki Jan 9 at 10:40

I provide a discrete expansion of an iterated function, but a symmetry constraint is also needed to simplify it to a continuous solution. Works for Schroeder's and Abel's Functional Equations. Note that convergence issues make if difficult to provide a complete answer.

Let $$f(x)$$ and $$g(x)$$ be functions in Banach space, then the composite $$f(g(x))$$ can be constructed using Faa Di Bruno's formula.

$$\begin{eqnarray} D^nf(g(x)) = \sum_{\pi(n)} \frac{n!}{k_1! \cdots k_n!} (D^kf)(g(x)) \left(\frac{Dg(x)}{1!}\right)^{k_1} \cdots \left(\frac{D^ng(x)}{n!}\right)^{k_n} \label{eq:FaaDiBruno} \end{eqnarray}$$ where $$\pi(n)$$ denotes a partition of $$n$$, usually denoted by $$1^{k_1}2^{k_2}\cdots n^{k_n}$$ with $$k_1+2k_2+ \cdots nk_n=k$$; where $$k_i$$ is the number of parts of size $$i$$. The partition function $$p(n)$$ is a decategorized version of $$\pi(n)$$, the function $$\pi(n)$$ enumerates the integer partitions of $$n$$, while $$p(n)$$ is the cardinality of the enumeration of $$\pi(n)$$. [Comtet], [Riordan]

In turn $$f(g(x))$$ can construct any iterated function at a fixed point by setting $$f(0)=0$$ and $$g(x)=f^{m-1}(x)$$.

$$$$D^n f^t(x)= \sum_{\pi(n)} \frac{n!}{k_1! \cdots k_n!} (D^k f)(f^{t-1}(x)) \left(\frac{Df^{t-1}(x)}{1!}\right)^{k_1} \cdots \left(\frac{D^n f^{t-1}(x)}{n!}\right)^{k_n}$$$$

$$$$D^n f^t(0)=\sigma(n) + D'f(0) D^n f^{t-1}(0)$$$$ DynamicalRecurranceEquation

The Taylors series of $$f^t(x)$$ is derived by evaluating the derivatives of the iterated function at a fixed point $$f^t(0)$$ by setting $$x=0$$ and separating out the $$k_n$$ term of the summation that is dependent on $$D^n f^{t-1}(0)$$.

The remaining $$\pi(n)-1$$ terms of the summation are only dependent on $$D^k f^{t-1}(0)$$, where $$0. Let this partial summation be written as $$\sigma(n)$$ with $$\sigma(0)=0$$ and $$\sigma(1) = 1$$.

Rewriting the $$\pi(n)-1$$ terms of the summation as $$\sigma(n)$$ will help in writing a proof by general induction. For $$n>1$$,

$$$$D^n f^t(0)=\sigma(n) + D'f(0) D^n f^{t-1}(0) \label{eq:Linear Equation}$$$$

Theorem: The Taylor series of an iterated holomorphic function $$f^t(x)$$ can be constructed given a fixed point where $$t \in \mathbb{N}$$.

Proof. Assume the given fixed point is at zero. The Taylor series of $$f^t(x)$$ can be constructed for some positive value of $$R$$ where $$0 < |x| < R$$ if and only if $$D^n f^t(0)$$ can be constructed for every $$n \geq 0$$. prove by strong induction.

Basis Steps:

Case $$n=0$$. By definition $$D^0 f^t(0) = 0$$, so $$D^0 f^t(0)$$ can be constructed.

Case $$n=1$$. Let $$D^1 f^t(0) = D'f(0)^t$$, so $$D^1 f^t(0)$$ can be constructed.

Induction Step: Case $$n=k$$. Assume that $$D^k f^t(0)$$ can be constructed for all $$k$$ where $$0 \leq k < n$$.

Using Eq. Dynamical Recurrance Equation, $$D^k f^t(0)=\sigma(k) + D'f(0) D^k f^{t-1}(0)$$. The function $$\sigma(k)$$ in only dependent on $$D^0 f(0), \ldots, D^k f(0)$$ and $$D^k f^t(0), \ldots, D^{(k-1)} f^t(0)$$. By the strong induction hypothesis, $$\sigma(k)$$ can be constructed. Therefore Eq. Dynamical Recurrance Equation can be reduced to a geometrical progression based on $$D'f(0)$$ that can be represented by a summation.

$$\begin{eqnarray} D^k f^t(0) = \sum_{j=0}^{k-1} \sigma(k) D'f(0)^j \end{eqnarray}$$ This completes the induction step that $$D^n f^t(0)$$ can be constructed for all whole numbers $$n$$.

The Taylors series for $$f^t(x)$$ is

$$\begin{eqnarray} f^t(x) = \sum_{n=0}^\infty \sum_{j=0}^{n-1} \sigma(n) D'f(0)^j x^n \label{eq:Dynamical Equation} \end{eqnarray}$$

$$\blacksquare$$

• But my function $f$ is of class $C^2$ and not holomorphic? – AIM_BLB Jan 9 at 11:50
• My mistake. I do have a question out mathoverflow.net/questions/350030/…, but I expect my work requires $C^\infty$. – user37691 Jan 9 at 13:21
• That's alright, I'll take a look and see what I can do with it :) – AIM_BLB Jan 9 at 15:22