Mathematics Asked by Miss Q on December 17, 2020

(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 $xleft(k_{0}right)=xi in mathbb{R}^{n}$.

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 $u(k):mathbb Zto mathbb R^n$ is non-constant, is it still a time-varying?

(3) A solution for system described in $(1)$ is a function $phi: mathbb Zto mathbb R^n$ 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.

(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 $x in mathbb{R}^2$:

$$ x(k+1)=f_1(k, x(k)) = begin{bmatrix} sinleft(frac{pi}{2}k right) \ cosleft(frac{pi}{2}k right) end{bmatrix} e^{-||x(k)||_2} tag{1}label{1} $$

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 $k$.

(2) If my system becomes $x(k+1)= f(u(k),x(k))$ where $u(k):mathbb Zto mathbb R^n$ 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:

$$ begin{align} x(k+1) &= begin{bmatrix}sinleft(frac{pi}{2}k right) \ cosleft(frac{pi}{2}k right)end{bmatrix} e^{-||x(k)||_2} \ &= u(k) e^{-||x(k)||_2} \ &= f(u(k), x(k)) end{align} \ $$

where

$$ u(k) = begin{bmatrix} sinleft(frac{pi}{2}k right) \ cosleft(frac{pi}{2}k right) end{bmatrix} tag{2}label{2} $$

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.

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:

$$ begin{bmatrix} u_1(k+1) \ u_2(k+1) end{bmatrix} = begin{bmatrix} sinleft(frac{pi}{2}(k+1) right) \ cosleft(frac{pi}{2}(k+1) right) end{bmatrix} = begin{bmatrix} sinleft(frac{pi}{2}k+frac{pi}{2} right) \ cosleft(frac{pi}{2}k+frac{pi}{2} right) end{bmatrix} = begin{bmatrix} cosleft(frac{pi}{2}k right) \ -sinleft(frac{pi}{2}k right) end{bmatrix} = begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix}begin{bmatrix} u_1(k) \ u_2(k) end{bmatrix} $$

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.

Correct answer by kdbanman on December 17, 2020

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)) $.

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