Conditions for a dynamical system to be input-commutative? Consider an explicit discrete-time dynamical system (recurrence relation),
$$
x_{t+1} = f(x_t, u_t)\ \ \ \ \forall t \in \mathbb{N}
$$
where $u$ is an exogenous sequence of "inputs." I am intrigued by systems for which the value of $x_t$ is invariant to permutations of the input sequence up to time $t-1$. That is, if we write the "solution" to the system as,
$$
x_t = g(x_0,u_0,u_1,\ldots,u_{t-1})
$$
then I define the property "input-commutativity" as when $g$ is commutative in the arguments $u_0,u_1,...,u_{t-1}$. Put simply, it means that the system ends up in the same state regardless of the order that the inputs are provided in.
Here are some simple examples on $\mathbb{R}$:
\begin{align*}
x_{t+1} = x_t u_t\ \  &\implies\ \ x_t = x_0\prod_{\tau=0}^{t-1}u_\tau \tag{1}\\[6pt]
x_{t+1} = \max\{x_t,u_t\}\ \  &\implies\ \ x_t = \max\{x_0,u_0,u_1,\ldots,u_{t-1}\} \tag{2}\\[6pt]
x_{t+1} = x_t + 2u_t\ \  &\implies\ \ x_t = x_0 + 2\sum_{\tau=0}^{t-1}u_\tau \tag{3}
\end{align*}
The first two examples show that $f$ need not be linear and the third example shows that $f$ need not itself be commutative. Meanwhile, $x_{t+1} = 2x_t + 2u_t$ is an example of a linear and commutative $f$ that does not yield an input-commutative solution. It seems that linearity and/or commutativity of $f$ have no bearing on whether the solution will be input-commutative.
A perhaps motivating example is the recursive application of Bayes rule from probability theory. If $u$ is a sequence of independent random variables then we can define the "belief state" for some random variable $\theta$ as $\ x_t = p(\theta|u_0,u_1,\ldots,u_{t-1})\ $ and have,
$$
x_{t+1} = \frac{p(u_t|\theta) x_t}{p(u_t)}\ \  \implies\ \ x_t = x_0\prod_{\tau=0}^{t-1}\frac{p(u_\tau|\theta)}{p(u_\tau)}
$$
Though this is just a glorified version of the $x_t u_t$ example (1), it provides the nice physical interpretation that one's belief about an unknown parameter should end up the same no matter what order the evidence is provided in. (For what it's worth, the $\max(x_t,u_t)$ example (2) also has an interpretation as a "peak detector" device, and the additive example (3) could represent work/transactions performed on a storage of energy/money).
Anyway, the only general form for a solution $g$ in terms of the dynamic $f$ is,
$$
g(x_0,u_0,u_1,\ldots,u_{t-1}) = f(f(f(f(x_0,u_0),u_1),\ldots),u_{t-1})
$$
from which it is not so obvious to see what about $f$ is necessary and/or sufficient to make $g$ commutative in $u_0,u_1,\ldots,u_{t-1}$.
My questions are:

*

*Can this property of the solution $g$ be deduced by any properties of the dynamic $f$?

*Does this property have an official name? Any references for its study?

*Is there a formalization of this concept for continuous-time?

Thanks in advance!
 A: There is one straightforward condition under which the value of $x_t$ is invariant under permutations of the input sequence.
First, let's make some definitions. Let $X$ be the phase space of your system and $U$ the control space, so that $f$ is a function from $X \times U$ to $X$. For every element $u \in U$, let $f_u \colon X \to X$ be the "partial application" $f_u(x) = f(x, u).$ We will call our control system "commutative" when the $f_u$ commute pairwise, meaning that $f_u \circ f_v = f_v \circ f_u$ for all $u, v \in U$.
Then, the function $x_t = g(x_0, u_0, \ldots, u_{t - 1})$ is invariant under permutation of its last $t$ arguments whenever our control system is commutative. Furthermore, if every map $f_u$ is a bijection and the permutation-invariance condition of $g$ holds for every tuple $(x_0, u_0, \ldots, u_{t - 1})$ for some fixed $t > 1$, then the control system is commutative.
We can use this criterion to investigate the two examples you gave in your response to my earlier comment. When $f(x, t) = \frac 1 2 (x + u)$, the maps $f_u$ are bijections but do not commute, since
$$(f_u \circ f_v)(x) = \frac 1 2 \left ( \frac 1 2 (x + v) + u \right) = \frac 1 4 x + \frac 1 4 v + \frac 1 2 u.$$
We conclude that permutation-invariance fails. On the other hand, the map $f(x, u) = x + u$ makes
$$(f_u \circ f_v)(x) = x + u + v = (f_v \circ f_u)(x),$$
so in this case permutation-invariance holds.
As far as directions for future study, I can see two connections. First: if your maps $f_u$ are bijections, then the problem of reaching a given $x_t$ from a given $x_0$ in minimal time can be formulated as a problem from group theory. If your control system is also "commutative" in the above sense, then the underlying group of study is Abelian, which simplifies the question a lot. To understand why, it's useful to know some facts about Abelian groups.
Second: if your equation $x_{t + 1} = f(x_t, u_t)$ is replaced by a continuous-time differential equation
$$\dot x_t = f(x_t, u_t)$$
and the partial applications $f_u$ are smooth vector fields, then there is an analogous (but somewhat more complicated) notion of a "commutative" interaction between the vector fields $f_u$: the Lie brackets $[f_u, f_v]$ should vanish. In these conditions, it will be true (subject to some technical conditions) that $x_t$ depends only on $x_0$ and on the integral $\int_0^t f_{u_\tau} \, d \tau$ of the controlling vector field.
To explain why I'm being picky about bijections in my criterion above, let's also give an example situation where the family of functions $f_u$ do not commute, but nevertheless where $x_t$ does not depend on the order of the controls $u_0, \ldots, u_{t - 1}$ for $t = 3$.
Let $X = \{1, 2, 3 \} \times \mathbb{N}$ and $U = \mathbb{N}$. Define
$$f((x_1, x_2), u) = \begin{cases}
(x_1 + 1, \frac 1 2 (x + u)) & : \text{if $x_1 < 3$} \\
(3, 1) & : \text{otherwise}.
\end{cases}$$
Then, we can check that $(f_u \circ f_v)(1, x) = \left (3, \frac 1 4 x + \frac 1 4 v + \frac 1 2 u \right)$, so the functions $f_u$ do not commute. However, any composition $f_{u} \circ f_{v} \circ f_{w}$ will take the constant value $(3, 1)$, so we will have the input commutativity condition when $t = 3$.
