"Recursively" expressing continuous-time trajectories I am interested in functions of time I call "trajectories" $h$, each in an arbitrary codomain $D$. $D$ is a set.


I wish to express that these trajectories are determined by an initial condition $h_0$ and an evolution/determining function $f$. (almost-)Like a dynamical system. $f$ can take any required form/signature, even complicated. And $f$ is supposed to be given.
I also wish to express that the universe hosting these trajectories, represented by $f$, is gifted with a memory so any value taken by $h$ depends on all past values taken by $h$, and not on future values.




If time were discrete, I would express it this way:

\begin{equation}
\newcommand{\funmap}[4]{{%
        \left\{%
            \begin{array}{ccl}
            #1 &amp; \to     &amp; \ #2 \\
            #3 &amp; \mapsto &amp; \ #4
            \end{array}
        \right.
    }}
    h : \funmap{\mathbb{N}}{D}{n}{h(n)}
\end{equation}
then
\begin{align}
h(0) &amp;= h_0\\
\forall n \in \mathbb{N}^{*},\ h(n) &amp;= f\left(h_{\big|[0,n[}\right) \text{ or } f\left(h_0, ..., h_{n-1}\right) \text{ ..}
\end{align}
.. depending on how we define $f$. Anyway, I repeat I'm mostly interested in expriming that:



1. given $f$ and $h_0$, $h$ is (constructively) completely determined (exists + is unique + we can compute it).
2. any value of $h$ depends on all its past values.



In continuous time however, I spontaneously try to express 2. this way:

\begin{equation}
h : \funmap{\mathbb{R^+}}{D}{t}{h(t)}
\end{equation}

then
\begin{align}
h(0) &amp;= h_0\\
\forall t \in \mathbb{R}^{+*},\ h(t) &amp;= f\left(h_{\big|[0,t[}\right)
\end{align}

.. but I'm not sure that it makes 1. hold. Is 1. true then?
If not, how to express both 1. and 2. in continuous time?



I am aware that this sounds like a recursive definition over $\mathbb{R}^+$, which I think is impossible.. unless we accept the axiom of choice and make it a transfinite recursion by forcing $\mathbb{R}^+$ to be well-ordered? Which is not constructive..

Is there a better way to write this?



Tracks I could think of:


discretizing $\mathbb{R}^+$, defining $h$ recursively on small bits.. then limit-reducing the size of the bits to zero. But I am afraid that everything would collapse at the limit resulting in $h$ being undefined.. would it be the case?
generalizing the concept of "derivative" and "local variation" to any arbitrary set $D$. Then expressing 1. and 2. like a generalized version of an ODE. I like this option best, and I am aware that it necessitate some kind of continuity in $h$ that cannot be defined for any codomain $D$. But I have a couple of hypotheses we could make us of for this track:




Here are two interesting hypothese on $h$, that may help us:


if $D$ topological, then $h$ is piecewise-continuous, and every piece interval has positive lenght.
if not, then $h$ is piecewise-constant, and every piece interval has positive length as well.




Am I looking for something impossible to express?
Can I exploit these hypotheses in a useful way to express 1. and 2.?
This is what I came up with. I'll accept my own answer once someone makes me confident it's valid.

First, property "2.: $h$ depends on its past values" is a dumb property since it is always verified. I.e. for any given $h$, one may build the function $f$ so that $h(t) = f\left(h_{\big|[0,\ t[}\right)$.

However, we may approximate $h$ by a sequence $u$ for an arbitrarily accurate precision $\varepsilon$, then state than "$f$ determines $h$ from its past" if and only if the series it determines is the approximation of $h$ for all $\varepsilon$.



For any precison $\varepsilon\in\mathbb{R}^{+*}$, let ${}^\varepsilon h$ be the sequence approximating $h$ in the sense that:

\begin{equation}
\forall n \in \mathbb{N},\quad {}^\varepsilon h_n = h(n\varepsilon)
\end{equation}

Now, let's consider $F:\left\{\begin{array}{lcl}
\mathbb{R}^{+*} &amp;\to&amp; \mathcal{F}\!\left(\bigcup_{i\in\mathbb{N}}D^i\to D\right)\\
\varepsilon &amp;\mapsto&amp; F(\varepsilon) = f_\varepsilon
\end{array}\right.$

Then, once given $h_0 = h(0) \in D$, any $F(\varepsilon)$ defines a sequence $f_\varepsilon\times h_0 = (u_n)_{n\in\mathbb{N}}$ by recursion with:

\begin{align}
u_0 &amp;= h_0\\
\forall n \in \mathbb{N}^*,\ u_n &amp;= f_\varepsilon(u_0,\ \dots,\ u_{n-1})\\
\end{align}

So I can finally state than

\begin{equation}
«\ F \text{ determines } h \text{ from its past } » \iff \forall \varepsilon \in \mathbb{R}^{+*},\ f_\varepsilon\times h_0 = {}^\varepsilon h
\end{equation}



This way, I do have:

1. Given $f_\varepsilon$ and $h_0$, ${}^\varepsilon h$ is entirely defined. So given "$F$ determining $h$ from its past", I can approximate $h$ with arbitrary precision.

2. Any "present" approximation ${}^\varepsilon h_n$ is constructed from its "past" ${}^\varepsilon h_n = f_\varepsilon({}^\varepsilon h_0,\ \dots,\ {}^\varepsilon h_{n-1})$.
This is what I came up with. I'll accept my own answer once someone makes me confident it's valid.

First, property "2.: $h$ depends on its past values" is a dumb property since it is always verified. I.e. for any given $h$, one may build the function $f$ so that $h(t) = f\left(h_{\big|[0,\ t[}\right)$.

However, we may approximate $h$ by a sequence $u$ for an arbitrarily accurate precision $\varepsilon$, then state than "$f$ determines $h$ from its past" if and only if the series it determines is the approximation of $h$ for all $\varepsilon$.



For any precison $\varepsilon\in\mathbb{R}^{+*}$, let ${}^\varepsilon h$ be the sequence approximating $h$ in the sense that:

\begin{equation}
\forall n \in \mathbb{N},\quad {}^\varepsilon h_n = h(n\varepsilon)
\end{equation}

Now, let's consider $F:\left\{\begin{array}{lcl}
\mathbb{R}^{+*} &amp;\to&amp; \mathcal{F}\!\left(\bigcup_{i\in\mathbb{N}}D^i\to D\right)\\
\varepsilon &amp;\mapsto&amp; F(\varepsilon) = f_\varepsilon
\end{array}\right.$

Then, once given $h_0 = h(0) \in D$, any $F(\varepsilon)$ defines a sequence $f_\varepsilon\times h_0 = (u_n)_{n\in\mathbb{N}}$ by recursion with:

\begin{align}
u_0 &amp;= h_0\\
\forall n \in \mathbb{N}^*,\ u_n &amp;= f_\varepsilon(u_0,\ \dots,\ u_{n-1})\\
\end{align}

So I can finally state than

\begin{equation}
«\ F \text{ determines } h \text{ from its past } » \iff \forall \varepsilon \in \mathbb{R}^{+*},\ f_\varepsilon\times h_0 = {}^\varepsilon h
\end{equation}



This way, I do have:

1. Given $f_\varepsilon$ and $h_0$, ${}^\varepsilon h$ is entirely defined. So given "$F$ determining $h$ from its past", I can approximate $h$ with arbitrary precision.

2. Any "present" approximation ${}^\varepsilon h_n$ is constructed from its "past" ${}^\varepsilon h_n = f_\varepsilon({}^\varepsilon h_0,\ \dots,\ {}^\varepsilon h_{n-1})$.