Koopman operator approach for computing structure of solutions and observability of nonlinear dynamical systems over finite fields

Abstract

This paper considers dynamical systems over finite fields (DSFF) defined by a map in a vector space over a finite field. An associated linear dynamical system is constructed over the space of functions. This system constitutes the well known Koopman linear system framework of dynamical systems, hence called the Koopman linear system (KLS). It is first shown that several structural properties of solutions of the DSFF can be inferred from the solutions of the KLS. The KLS is then reduced to the smallest order (called RO-KLS) while still retaining all the information of the parameters of structure of solutions of the DSFF. Hence, the above computational problems of nonlinear DSFF are solvable by linear algebraic methods. It is also shown how fixed points, periodic points and roots of chains of the DSFF can be computed using the RO-KLS. Further, for DSFF with outputs, the output trajectories of the DSFF are in $1-1$ correspondence with special class of output trajectories of RO-KLS and it is shown that the problem of nonlinear observability can be solved by a linear observer design for the RO-KLS.

Appendix

1.1 Proof for Lemma 1

Recall that KLS is defined by action of $\varPhi$ on space of functions by $\varPhi \psi (x) = \psi (F(x))$. Let $x_0$ be a point in ${\mathbb {F}}_q^n$ which lies on a closed orbit of length l under the action of F. Let $S_l$ be the set defined as follows:

$$\begin{aligned} S_l = \{ x \in {\mathbb {F}}_q^n\ \ |\ F^{kl}(x) = x_0, k \in {\mathbb {Z}}_+\} \end{aligned}$$

This set includes $x_0$ and all other points on the chains terminating in the orbit containing $x_0$ and pass through $x_0$ under the action of $F^{kl}$.

Note that the set $S_l$ is invariant under action of $F^l$. Since for any $x \in S_l$, $\exists$ $k_0$ such that $F^{k_0l}(x) = x_0$. Hence, $F^{k_0l}(F^l(x)) = F^{(k_0+1)l}(x) \in S_l$.

Similarly, if $x \in S_l^c$, the complement of $S_l$, then $F^{kl}(x) \ne x_0$ for any k. Hence, $F^{kl}$ acting on $F^l(x)$ gives

$$\begin{aligned} F^{kl}(F^l(x))&= F^{(k+1)l} (x) \ne x_0 \end{aligned}$$

Hence, $F^l(x) \notin S_l$ whenever $x \notin S_l$ which shows $S_l^c$ is also $F^l$ invariant.

Construct a function $\psi \in V^o$ defined as follows.

$$\begin{aligned} \psi (x) = \left\{ \begin{matrix} 1 &{}\quad \forall \ x \in S_l \\ 0 &{}\quad \forall \ x \in S_l^c\end{matrix} \right. \end{aligned}$$

We claim that $\psi$ is one such function which has a closed orbit of length l under action of $\varPhi$. To prove this, we first prove that $\psi$ lies on a periodic orbit and secondly we prove that l is the period of $\psi$. Now,

$$\begin{aligned} \varPhi ^{l}\psi (x)&= \psi (F^l(x)) \\&= \left\{ \begin{matrix} 1 &{}\quad \text{ for } &{} x \in S_l \\ 0 &{}\quad \text{ for } &{} x \in S_l^c\end{matrix} \right. \\&= \psi (x) \end{aligned}$$

The second equality comes due to the fact that $S_l$ and $S_l^c$ are invariant sets under $F^l$. This proves that $\psi (x)$ has a periodic orbit under $\varPhi$ and whose orbit length divides l.

Let $0< m < l$ be the orbit length of $\psi$. We prove that this leads to a contradiction. Since $\varPhi ^m \psi (x) = \psi (x)$, in particular when $x = x_0$,

$$\begin{aligned} \varPhi ^m \psi (x_0)&= \psi (x_0) \nonumber \\ \implies \psi (F^m(x_0))&= \psi (x_0) \end{aligned}$$

(23)

But by definition $\psi (x) = 1$ only if $x \in S_l$. So, in (23), RHS equals 1. To get the contradiction, we prove that $F^m(x_0) \notin S_l$.

Suppose let $\alpha := F^m(x_0) \in S_l$, then there exists some $k_0$ such that

$$\begin{aligned} F^{k_0l}(\alpha ) = x_0 \end{aligned}$$

but $\alpha = F^m(x_0)$. Substituting back,

$$\begin{aligned} \begin{array}{rrl} &{}F^{k_0l}(F^m (x_0)) &{}= x_0 \\ \implies &{} F^{k_0l+m}(x_0) &{}= x_0 \\ \implies &{}F^{m}(F^{k_0l}(x_0)) &{}= x_0 \\ \implies &{} F^m(x_0) &{}= x_0 \end{array} \end{aligned}$$

The last equation is due to the fact that $x_0$ is on an closed orbit of length l and leads to a contradiction since l is the least integer such that $f^l(x_0) = x_0$ and $m < l$ by assumptions. Hence, $F^m(x_0) \notin S_l$ and in (23), LHS = 0. This is a contradiction. Hence, $\psi$ cannot have an orbit of length $m < l$. Hence, the orbit length of $\psi$ under $\varPhi$ is precisely l.

This constructs a specific $\psi \in V^o$ which has an orbit length l under KLS whenever the DSFF has an orbit of length l. $\square$

1.2 Proof for Lemma 2

Given that the DSFF is non-singular, we have for $x_1,\ x_2 \in V$,

$$\begin{aligned} F(x_1) \ne F(x_2) \ \ \ \forall \ x_1 \ne x_2 \end{aligned}$$

Let $\psi _1,\ \psi _2 \in V^o$ and $\psi _1 \ne \psi _2$. Let $\psi _d = \psi _1 - \psi _2$. We will prove that “given F is non-singular and if $\varPhi \psi _1(x) = \varPhi \psi _2(x) \ \forall \ x \in V$, then $\psi _d(x) = 0 \ \forall \ x \in V$” which proves that if DSFF is non-singular, then Koopman operator is non-singular.

$$\begin{aligned} \begin{array}{rrcl} &{} \varPhi \psi _1(x) &{}= &{}\varPhi \psi _2(x) \ \ \forall \ x \in V \\ \implies &{} \varPhi (\psi _1 - \psi _2)(x) &{}=&{} 0 \ \ \forall \ x \in V \\ \implies &{} \varPhi \psi _d(x) &{}=&{} 0 \ \ \forall \ x \in V \\ \implies &{} \psi _d(F(x)) &{}=&{} 0 \ \ \forall \ x \in V \end{array} \end{aligned}$$

This implies that $\psi _d(F(x)) = 0$ for all $x \in V$. But since F is non-singular, the image of F is the whole of V. So we have

$$\begin{aligned} \psi _d(y) = 0 \ \ \forall \ y \in V \end{aligned}$$

This means that $\psi _d$ is the zero function which leads to a contradiction because $\psi _d = \psi _1 - \psi _2$ which is nonzero. $\square$

1.3 Proof for Lemma 3

Let $x_0$ be the root of a chain of length l under F leading to either a periodic orbit or a fixed point. Let $S_l = \{ x_0,x_1,\dots ,x_{l-1}\}$ be the points on this chain under the action of F (i.e., $x_{k} = F(x_{k-1}),\ \ k \in \{1,2,\dots ,l-1\}$ and $F(x_{l-1})$ lies on a periodic orbit). Consider a function $\psi _1$ defined as follows.

$$\begin{aligned} \psi (x) = \left\{ \begin{matrix} 1 &{} \quad {\text {if}} \ x \in S_l \\ 0 &{} \quad {\text {if}} \ x \in S_l^c \end{matrix} \right. \end{aligned}$$

We claim that this function is a root of chain of length l under the action of $\varPhi$ terminating in the zero function. (zero function is a fixed point under $\varPhi$). The evolution of $\psi$ under $\varPhi$ is given by

$$\begin{aligned} \psi (x), \psi (F(x)), \psi (F^2(x)), \dots , \psi (F^l(x)), \dots \end{aligned}$$

We see that for any $x \in {\mathbb {F}}_q^n$, and $k \ge l$, $F^k(x) \in S_l^c$ and $\psi (F^k(x)) = 0$. So $\varPhi ^l(\psi (x))$ is zero function and $\psi$ is a chain which terminates to zero function.

We claim that l is chain length of $\psi$ and prove it by contradiction. Let chain length of $\psi (x)$ be m and $m<l$, then for $m < l$, $\varPhi ^m(\psi (x)) = \psi (F^m(x))$ is zero function. But when evaluated at $x_0$, $\psi (F^m(x_0)) = \psi (x_m) = 1$ since $m < l$ and $x_m \in S_l$. Which is a contradiction. So, the length of the chain is precisely l.

This constructs a function $\psi \in V^o$ which is the root of a chain of length l under $\varPhi$.

$\square$

1.4 Proof for Lemma 4

Let the DSFF have a period $n_1$ and the KLS have a period $n_2$. We prove that $n_1|n_2$² and $n_2|n_1$ and thereby prove that $n_1 = n_2$. Since the period of DSFF is $n_1$,

$$\begin{aligned} x(k+n_1) = F^{n_1}x(k) = x(k) \ \ \forall \ \ x(k) \ \in \ {\mathbb {F}}_q^n\end{aligned}$$

For some $\psi _0 \in V^o$,

$$\begin{aligned}&\varPhi ^{n_1} \psi _0(x) = \psi _0 (F^{n_1}(x)) = \psi _0(x) \nonumber \\&\implies n_1 = ln_2 \ \text{ for } \text{ some } \ l\ \in \ {\mathbb {Z}}_+ \end{aligned}$$

(24)

Similarly, since the period of KLS is $n_2$,

$$\begin{aligned} \varPhi ^{n_2} \psi _0(x) = \psi _0(x)\ \ \forall \ \ \psi \ \in \ V^o\end{aligned}$$

From the definition of KLS, we for any $\psi _0$, we have

$$\begin{aligned} \psi _0(x) = \varPhi ^{n_2}\psi _0(x) = \psi _0 (F^{n_2}(x)) \end{aligned}$$

which implies $F^{n_2}$ is an identity map over ${\mathbb {F}}_q^n$.

$$\begin{aligned} \implies \ n_2 = rn_1\ \text{ for } \text{ some }\ r\ \in \ {\mathbb {Z}}_+ \end{aligned}$$

(25)

From (24) and (25), $n_1 = n_2$ $\square$

1.5 Proof for Theorem 4

Given a $x_0$ on an orbit of length L under the DSFF, let $y_0 = {\hat{\psi }}(x_0)$.

By Theorem (2), it is proved that $y_0$ is on an orbit of length L under the ROKLS. Hence, $K_1^L y_0 = y_0$. Also,

$$\begin{aligned} x_0 = {\hat{\chi }}(x_0) = C {\hat{\phi }}(x_0) = Cy_0 \end{aligned}$$

and

$$\begin{aligned} y_0 = {\hat{\psi }}(x_0) = {\hat{\psi }}(Cy_0) \end{aligned}$$

which proves the necessary conditions.

To prove sufficiency, let $y_0$ be on an orbit of length L under the ROKLS satisfying $y_0 = {\hat{\psi }}(Cy_0)$. Let $x_0 = Cy_0$. Hence, $y_0 = {\hat{\psi }}(x_0)$ and,

$$\begin{aligned} F(x_0)&= {\hat{\chi }}(F(x_0)) \\&= C {\hat{\psi }}(F(x_0)) \\&= C K_1 {\hat{\psi }}(x_0) \quad \quad \text{(from } \text{ Eq. } {(9))} \\&= C K_1 y_0 \end{aligned}$$

Similarly, one can prove $F^{(m)} (x_0) = C K_1^m y_0$ for $m \ge 0$. Since $y_0$ is on an orbit of length L, $K_1^Ly_0 = y_0$ and hence

$$\begin{aligned} F^{(L)}(x_0) = C K_1^L y_0 = C y_0 = x_0 \end{aligned}$$

Hence, $x_0$ is on an orbit whose length divides L. To prove that the length is exactly L, assume the contrary. Let $l < L$ be the orbit length of $x_0$ under the DSFF (i.e., $F^{l}(x_0) = x_0$. From Eq. (9)

$$\begin{aligned} K_1^l y_0&= K_1^l {\hat{\psi }}(x_0) \\&= {\hat{\psi }}(F^{(l)}(x_0)) \\&= {\hat{\psi }} (x_0) \\&= y_0 \end{aligned}$$

which means that the orbit length of $y_0$ is also l and that is a contradiction since $y_0$ is assumed to be on an orbit of length L. So, the orbit length of $x_0$ constructed as $x_0 = Cy_0$ is exactly L, the orbit length of $y_0$ whenever $y_0$ satisfies $y_0 = {\hat{\psi }}(Cy_0)$. $\square$

1.6 Proof for Theorem 5

To prove necessity, let $x_0$ be a root of a chain of length L under DSFF. So $x_L = F^{L}(x_0)$ lies on a periodic orbit of length $M \ge 1$.

Let $y_0 = {\hat{\psi }}(x_0)$. From Theorem (2), it known that if $x_0$ is on a chain of length L under DSFF, then ${\hat{\psi }}(x_0)$ is on a chain of length L under ROKLS. Also,

$$\begin{aligned} x_0 = {\hat{\chi }}(x_0) = C {\hat{\psi }}(x_0) = Cy_0 \end{aligned}$$

and

$$\begin{aligned} y_0 = {\hat{\psi }}(x_0) = {\hat{\psi }}(Cy_0) \end{aligned}$$

The only thing which needs to be proved that $y_0$ is the root of the chain.

As defined in Theorem 2, any trajectory x(k) of DSFF in ${\mathbb {F}}$ is embedded in the state space ${\mathbb {F}}^N$ of ROKLS (6) by the map $x(k)\mapsto y(k)={\hat{\psi }}(x(k))$. Since $x_0$ is a root of a chain of DSFF iff there is no point z in ${\mathbb {F}}^n$ on a trajectory such that $F(z)=x_0$, by the above unique embedding of trajectories of DSFF into trajectories of ROKLS, there is no point ${\hat{\psi }}(z)$ in ${\mathbb {F}}^N$ on the trajectory of ROKLS such that $y_0={\hat{\psi }}(x_0)=K({\hat{\psi }}(z))$. Hence, $y_0$ is also a root of the trajectory in the state space of ROKLS.

To prove the sufficiency, let $y_0$ be a root of a chain of length L and $y_0 = {\hat{\psi }}(Cy_0)$ and $x_0 = Cy_0$. We need to prove $x_0$ is a root of chain of length L. By construction of $x_0$, $y_0 = {\hat{\psi }}(Cy_0) = {\hat{\psi }}(x_0)$.

From the previous theorem, if $y_0 = {\hat{\psi }}(x_0)$, it is proved that

$$\begin{aligned} F^{(m)}(x_0) = C K_1^m y_0 \end{aligned}$$

Since $y_0$ is on a chain of length L, $K_1^Ly_0$ is on a period orbit of say length M. So $K_1^{L+kM}y_0 = K_1^{L}y_0$ for all $k \in {\mathbb {Z}}_+$. Hence,

$$\begin{aligned} F^{(L+kM)}(x_0) = CK_1^{L+kM}y_0 = CK_1^{L}y_0 = F^{(L)} (x_0) \end{aligned}$$

which means $F^{(L)}(x_0)$ is on a periodic orbit. If there exists some $l < L$ such that $F^{(l)}(x_0)$ is on a periodic orbit, then

$$\begin{aligned} F^{(l+m_1)} (x_0) = F^{(l)}(x_0) \end{aligned}$$

for some $m_1 \in {\mathbb {Z}}_+$. Since $y_0 = {\hat{\psi }}(x_0)$

$$\begin{aligned} K_1^{(l+m_1)}y_0&= K_1^{(l+m_1)}{\hat{\psi }}(x_0) \\&= {\hat{\psi }}( F^{(l+m_1)}(x_0)) \\&= {\hat{\psi }}( F^{(l)} (x_0)) \\&= K_1^l y_0 \end{aligned}$$

which proves that $K_1^l y_0$ lies on a periodic orbit and the length of chain starting from $y_0$ under ROKLS is l which is a contradiction. So the length of the chain starting from $x_0$ under DSFF is also L.

The last thing to prove is that $x_0$ is the root of the chain. Assume the contrary again. Let there be $x \in {\mathbb {F}}^n$ such that $F(x) = x_0$. Construct $y = {\hat{\psi }}(x)$.

$$\begin{aligned} K_1 y&= K_1 {\hat{\psi }}(x) \\&= {\hat{\psi }}(F(x)) \\&= {\hat{\psi }}(x_0) \\&= y_0 \end{aligned}$$

which proves that $y_0$ is also not a root which is a contradiction. Hence, $x_0$ is the root of the chain of length L under DSFF. $\square$

1.7 Proof for Lemma 5

From Theorem 2, it is proved there exists a unique trajectory $y(k) = {\hat{\psi }}(x(k))$ of the RO-KLS for each trajectory x(k) of the KLS. The output of the RO-KLS is

$$\begin{aligned} y_{op}(k) = \varGamma y(k) = \varGamma {\hat{\psi }}(x(k)) \end{aligned}$$

From Eq. (15),

$$\begin{aligned} \varGamma {\hat{\psi }}(x(k)) = g(x(k)) = z(k) \end{aligned}$$

Combining the above equations, $y_{op}(x) = z(k)$. $\square$