time-varying dynamical system (1) Consider discrete-time nonlinear time-
varying systems described by the difference equation
$x(k+1)=f(k, x(k)), \quad x(k) \in \mathbb{R}^{n}, k \in \mathbb{Z}$
where $f: \mathbb{Z} \times \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ is continous and $x\left(k_{0}\right)=\xi \in \mathbb{R}^{n}$.
My question is why they are saying the system is time-varying, by an example of such? what does it mean by time-varying? Can anyone give me an example of a not-time varying system in this context? Thanks.
(2) If my system becomes $x(k+1)= f(x(k), u(k))$ where $u(k):\mathbb Z\to \mathbb R^n$ is non-constant, is it still a time-varying?
(3) A solution for system described in $(1)$ is a function $\phi: \mathbb Z\to \mathbb R^n$ parametrized by initial state and time i.e $\phi(k_0; k_0,\xi)=\xi$, i.e $\phi(k+1; k_0, \xi)= f(k, \phi(k;k_0,\xi))$ Could any one tell me how to define a solution for the system described in (2)?
Thanks!
 A: It's time varying because the function $ f $ has an explicit dependence on the discrete time $ k $ beyond the implicit dependence it has through the changing value of $ x(k) $. A system that doesn't vary over time would look like $ x(k+1) = f(x(k)) $.
A: TL;DR There is a way to go from time-varying dynamics to time-invariant dynamics using a higher dimensional state space, and I think that's what your second question is trying to get at.

Time-varying and time-invariant examples

(1) Consider discrete-time nonlinear time-
varying systems described by the difference equation
$x(k+1)=f(k, x(k)), \quad x(k) \in \mathbb{R}^{n}, k \in \mathbb{Z}$
where $f: \mathbb{Z} \times \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ is continous ... why they are saying the system is time-varying? ... an example of such? ... an example of a not-time varying system?

The system is time-varying specifically when there does not exist a $g:\mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ such that $f(k, x(k)) = g(x(k))$ for all $k$.  One example, letting $x \in \mathbb{R}^2$:
$$
x(k+1)=f_1(k, x(k)) = \begin{bmatrix}
\sin\left(\frac{\pi}{2}k \right) \\
\cos\left(\frac{\pi}{2}k \right)
\end{bmatrix} e^{-||x(k)||_2}  \tag{1}\label{1}
$$
This is time-variant because there does not exist a $g$ as specified. i.e. the $k$ appears in places other than just an argument to $x$. If instead the system were defined as
$$
x(k+1)=f_2(k, x(k)) = \begin{bmatrix}
x(k) \\
2x(k)
\end{bmatrix} e^{-||x(k)||_2}
$$
then we have a time-invariant system (it is not time-varying) because $k$ only appears as an argument to $x$.  It should be clear that there does exist a $g:\mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ such that $f_2(k, x(k)) = g(x(k))$ for all $k$.
Time-varying, different notation

(2) If my system becomes $x(k+1)= f(u(k),x(k))$ where $u(k):\mathbb Z\to \mathbb R^n$ is non-constant, is it still a time-varying?

(Note that our $f$ is no longer defined as mapping $\mathbb{Z} \times \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$.  It now has the signature $f:\mathbb{R}^n \times \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$.)
Our first example $(\ref{1})$ can be expressed with this new $f$ as follows:
$$
\begin{align}
x(k+1) &= \begin{bmatrix}\sin\left(\frac{\pi}{2}k \right) \\ \cos\left(\frac{\pi}{2}k \right)\end{bmatrix} e^{-||x(k)||_2} \\
&= u(k) e^{-||x(k)||_2} \\
&= f(u(k), x(k))
\end{align} \\
$$
where
$$
u(k) = \begin{bmatrix} \sin\left(\frac{\pi}{2}k \right) \\ \cos\left(\frac{\pi}{2}k \right) \end{bmatrix}   \tag{2}\label{2}
$$
This is still the same system as $(\ref{1})$, just jiggled into different notation.  It's still time-varying for the same reasons as before.
Time-invariant in higher dimensions
However, we can write system $(\ref{1})$ as a time-invariant system by augmenting our state space.  This is possible because our function $u$ from  $(\ref{2})$ can be written as a time-invariant difference equation itself:
$$
\begin{bmatrix} u_1(k+1) \\ u_2(k+1) \end{bmatrix} = \begin{bmatrix} \sin\left(\frac{\pi}{2}(k+1) \right) \\ \cos\left(\frac{\pi}{2}(k+1) \right) \end{bmatrix} = \begin{bmatrix} \sin\left(\frac{\pi}{2}k+\frac{\pi}{2} \right) \\ \cos\left(\frac{\pi}{2}k+\frac{\pi}{2} \right) \end{bmatrix} = \begin{bmatrix} \cos\left(\frac{\pi}{2}k \right) \\ -\sin\left(\frac{\pi}{2}k \right) \end{bmatrix}  = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\begin{bmatrix} u_1(k) \\ u_2(k) \end{bmatrix}
$$
That is, there exists a function $h$ such that $u(k+1) = h(u(k))$.  (Here, $h$ is a linear transformation, but that need not always be the case.)  With that in mind, define a new state space variable $r \in \mathbb{R}^4$ as
$$
r(k) = \begin{bmatrix}r_1(k)\\r_2(k)\\r_3(k)\\r_4(k)\end{bmatrix} \dot{=} \begin{bmatrix}x_1(k)\\x_2(k)\\u_1(k)\\u_2(k)\end{bmatrix} = \begin{bmatrix}x(k)\\u(k)\end{bmatrix}
$$
where the rightmost notation should be understood as stacking $x, u \in \mathbb{R}^2$ on top of each other.  This permits us to write the same system as
$$
\begin{align}
r(k+1) &= \begin{bmatrix}f(u(k),x(k))\\h(u(k))\end{bmatrix}
\end{align}  \tag{3}\label{3}
$$
It may not yet be obvious yet, but $(\ref{3})$ is actually time-invariant.  For the sake of readability, define new notation
$$
\begin{align}
r' &= r(k+1)\\
r &= r(k)
\end{align}
$$
with similar notation for $x', x, u', u$.  Our system is time-invariant if we can find a function $g$ such that $r' = g(r)$.  Starting again from $(\ref{3})$ with this nicer notation:
$$
\begin{align}
r' &= \begin{bmatrix}f(u,x)\\h(u)\end{bmatrix} 
= \begin{bmatrix}u_1 e^{-\sqrt{x_1^2+x_2^2}}\\ u_2 e^{-\sqrt{x_1^2+x_2^2}} \\ u_2 \\ -u_1 \end{bmatrix} 
= \begin{bmatrix}r_3 e^{-\sqrt{r_1^2+r_2^2}}\\ r_4 e^{-\sqrt{r_1^2+r_2^2}} \\ r_4 \\ -r_3 \end{bmatrix}
\end{align}
$$
which is clearly time-invariant, as $k$ appears only as an argument to our state space variables.  That is, there exists a function $g$ such that $r' = g(r)$.
This was possible because $u(k)$, the time-varying part of our original system $(\ref{1})$, could itself be written as a time-invariant system.  And this allowed us to construct a higher dimensional state space $r$ in which the entire system was time-invariant.
