(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!
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
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
(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.