# time-varying dynamical system

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.

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

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

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 $$k$$.

### Time-varying, different notation

(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.

### 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:

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

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 $$x(k+1) = f(x(k))$$.

Answered by Ege Erdil on December 17, 2020

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