Every discrete 2D autonomous system admits a finite union of parallel lines as a characteristic set

Abstract

In this paper, we show that every discrete 2D autonomous system, that is described by a set of linear partial difference equations with constant real coefficients, admits a finite union of parallel lines as a characteristic set. In order to prove our claim, we first look at a special class of scalar discrete 2D systems and provide such characteristic sets for systems in this class. This special class has the property that systems in this class have their quotient rings to be finitely generated modules over a one-variable Laurent polynomial subring of the original two-variable Laurent polynomial ring in the shift operators. We show that such systems always admit a finite collection of horizontal lines for a characteristic set. We then extend this result to non-scalar discrete 2D autonomous systems. We achieve this in two steps: first, we show that every scalar discrete 2D system can be converted into a system in the above-mentioned class by a coordinate transformation on the independent variables set, $\mathbb {Z}^2$. Using this we then show that characteristic sets for the original system can be found by applying the inverse coordinate transformation on characteristic sets of the transformed system. Since the transformed system, by virtue of being in the special class, admits a finite union of horizontal lines as a characteristic set, the original system is guaranteed to admit a characteristic set that is a coordinate transformation applied to a finite union of horizontal lines. The coordinate transformation maps this union of horizontal lines to a union of parallel, but possibly tilted, lines. In the next step, we generalize the scalar case to the general vector case: that is, systems with more than one dependent variables. The main motivation for studying characteristic sets that are unions of finitely many parallel lines is that, arguably, such sets can be called “thin” in $\mathbb {Z}^2$ in comparison to the prevalent notions of convex cones and half-spaces as characteristic sets (see “Appendix 1”).

Appendices

Appendix 1: On “thin”ness of finite unions of parallel lines in $\mathbb {Z}^2$

We have mentioned that a finite union of parallel lines is a thin set in $\mathbb {Z}^2$. While this statement intuitively appears to be true, it is not beyond all reasonable doubts and hence requires a proof. In order for this, a suitable notion of thin sets in $\mathbb {Z}^2$ must first be formulated because the above-mentioned sets are thin only under special circumstances. However, giving a precise definition of thin sets in $\mathbb {Z}^2$ turns out to be a rather tricky issue, and is, unfortunately, beyond the scope of this article. The main difficulty seems to arise from the fact that various sets like—a half-space in $\mathbb {Z}^2$, a quadrant in $\mathbb {Z}^2$, a line in $\mathbb {Z}^2$ (e.g., an axis of $\mathbb {Z}^2$), or the entire $\mathbb {Z}^2$—all of them have the same cardinality: the countable infinity, $\aleph _0$. Hence it follows that these sets can be put into a one-to-one correspondence with each other. Therefore, from this point of view, a trajectory defined over the whole of $\mathbb {Z}^2$, or over a half-space, or over a quadrant, or over an axis, would all require countably infinite amount of data. Thus, if we go by this notion of quantifying the information required to specify a trajectory, a characteristic set that is finitely many lines would not be any better than a characteristic set that is a half-space or a quadrant.

Having said this, one must also note that specifying a 2D trajectory over an axis amounts to specifying a 1D trajectory, and that way, having a characteristic set that is a line is indeed better for it cuts down the ‘dimension’ of the set, over which a trajectory is being specified. In Wood et al. (1998) this fact has been called the “order of magnitude” of the initial condition set. Subsequently, in Wood et al. (1998) for continuous $n$D autonomous systems, and in Avelli and Rocha (2010) for discrete $n$D autonomous systems, it was shown that this order of magnitude is strictly smaller than the dimension of the indexing set, that is $n$. Our main results of this paper, too, echo this fact.

To promote this view-point of finite unions of parallel lines as characteristic sets as a possible improvement over the current state-of-the-art, that is, convex cones, quadrants and half-spaces as characteristic sets, we provide below two constructions that indicate thin-ness of finitely many parallel lines in $\mathbb {Z}^2$ in comparison to quadrants, cones and half-spaces.

First, suppose $A\in \mathbb {Z}^{2\times 2}$ is such that $\mathrm{det}(A)=\pm 1$, that is, $A$ is a unimodular integer matrix (see Sect. 4.1). Such an $A$ acts on an integer vector $\nu =\mathrm{col}(\nu _1,\nu _2)\in \mathbb {Z}^2$ as $A\nu$ to produce another vector in $\mathbb {Z}^2$. This transformation is invertible, because $A$ is invertible and its inverse, too, is an integer matrix. Thus, $A$ defines a linear change of coordinates on $\mathbb {Z}^2$. Now, for $\mathcal {S}_1,\mathcal {S}_2\subseteq \mathbb {Z}^2$, let us define $\mathcal {S}_1$ to be equivalent to $\mathcal {S}_2$ if there exists $A\in \mathbb {Z}^{2\times 2}$ unimodular and $\hat{\nu }\in \mathbb {Z}^2$ such that

$$\begin{aligned} \mathcal {S}_2=A(\mathcal {S}_1)+\hat{\nu }. \end{aligned}$$

It can be checked that it is indeed an equivalence relation. With this equivalence relation it can then be shown that if $\mathcal {S}$ is a line or a finite union of parallel lines then $\mathbb {Z}^2$ can never be written as a finite union of sets that are equivalent to $\mathcal {S}$. On the other hand, if $\mathcal {S}$ is a convex cone, or a quadrant or a half-space, then a finite union of $\mathcal {S}$ and its equivalent sets covers the entire $\mathbb {Z}^2$.

The second construction that indicates thinness of a finite union of parallel lines compared to quadrants or half-spaces or convex cones is as follows. Define the following family of finite subsets of $\mathbb {Z}^2$. For $i=0,1,2,\ldots$,

$$\begin{aligned} \mathcal {V}_i:=\left\{ (\nu _1,\nu _2)\in \mathbb {Z}^2~|~-i\leqslant \nu _1\leqslant i,~-i\leqslant \nu _2\leqslant i\right\} . \end{aligned}$$

Note that $\mathcal {V}_i$’s form an ascending chain that ultimately covers the entire $\mathbb {Z}^2$:

$$\begin{aligned} \{(0,0)\} = \mathcal {V}_0\subsetneq \mathcal {V}_1\subsetneq \mathcal {V}_2\subsetneq \cdots \end{aligned}$$

and

$$\begin{aligned} \bigcup _{i=1}^{\infty }\mathcal {V}_i=\mathbb {Z}^2. \end{aligned}$$

Now, let $\mathcal {S}$ be any subset of $\mathbb {Z}^2$. Then for $i=1,2,\ldots$, define

$$\begin{aligned} \rho _{i}(\mathcal {S}):=\frac{|\mathcal {S}\cap \mathcal {V}_i|}{|\mathcal {V}_i|}, \end{aligned}$$

where $|\mathcal {S}|$ denotes the cardinality of the set $\mathcal {S}$. It can then be easily checked that if $\mathcal {S}$ is a finite union of parallel lines then

$$\begin{aligned} \lim _{i\rightarrow \infty }\rho _i(\mathcal {S})=0, \end{aligned}$$

but, if $\mathcal {S}$ is a convex cone, or a quadrant, or a half-space then

$$\begin{aligned} \lim _{i\rightarrow \infty }\rho _{i}(\mathcal {S})>0. \end{aligned}$$

Appendix 2: Proof of Lemma 11

We need the following standard result—Euclidean division algorithm over polynomial over-rings—for the proof.

Proposition 25

Let $\mathcal {A}$ be an arbitrary commutative ring with $1\in \mathcal {A}$, and let $\xi$ be transcendental over $\mathcal {A}$. Suppose $p(\xi )\in \mathcal {A}[\xi ]$ is a monic polynomial, that is,

$$\begin{aligned} p(\xi )=\xi ^L+a_{L-1}\xi ^{L-1}+\cdots +a_{1}\xi +a_0, \end{aligned}$$

where $L$ is a finite positive integer and $a_{L-1},\ldots ,a_1,a_0\in \mathcal {A}$. Then for all $f(\xi )\in \mathcal {A}[\xi ]$ there exist $q(\xi )\in \mathcal {A}[\xi ]$ and $r_0,r_1,\ldots ,r_{L-1}\in \mathcal {A}$ such that

$$\begin{aligned} f(\xi )=q(\xi )p(\xi )+\sum _{i=0}^{L-1}r_i \xi ^i. \end{aligned}$$

Proof of Lemma 11

Since $\mathcal {M}$ is finitely generated as a module over $\mathcal {A}_1$, by Proposition 4, $\mathfrak {a}$ must contain a polynomial $p(\sigma _{})$ of the form $p(\sigma _{})=\sigma _{2}^L+a_{L-1}(\sigma _{1})\sigma _{2}^{L-1}+\cdots +a_1(\sigma _{1})\sigma _{2}+a_0(\sigma _{1}),$ where $L$ is a finite positive integer, $a_{L-1}(\sigma _{1}),\ldots ,a_1(\sigma _{1}),a_0(\sigma _{1})\in \mathcal {A}_1$ and $a_0(\sigma _{1})$ is a unit in $\mathcal {A}_1$. Then, by Proposition 25, for every $f(\sigma _{})\in \mathcal {A}_1[\sigma _{2}]$ (that is $f(\sigma _{})\in \mathcal {A}$ whose terms do not contain negative powers of $\sigma _{2}$), there exist $q(\sigma _{})\in \mathcal {A}_1[\sigma _{2}]$ and $r_0(\sigma _{1}),r_1(\sigma _{1}),\ldots ,r_{L-1}(\sigma _{1})\in \mathcal {A}_1$ such that

$$\begin{aligned} f(\sigma _{})=q(\sigma _{})p(\sigma _{})+\sum _{i=0}^{L-1}r_i(\sigma _{1})\sigma _{2}^i. \end{aligned}$$

(20)

Because $a_0(\sigma _{1})$ is a unit we can multiply $p(\sigma _{})$ by $a_0(\sigma _{1})^{-1}$ and $\sigma _{2}^{-1}$ to get that $\sigma _{2}^{-1}a_0(\sigma _{1})^{-1}p(\sigma _{})= a_0(\sigma _{1})^{-1}\sigma _{2}^{L-1}+b_{L-1}(\sigma _{1})\sigma _{2}^{L-2}+\cdots + b_{1}(\sigma _{1})+\sigma _{2}^{-1}\in \mathfrak {a},$ where $b_i(\sigma _{1})=a_0(\sigma _{1})^{-1}a_i(\sigma _{1})$ for all $1\leqslant i \leqslant L-1$. Note that in the above expression every term except the last one has non-negative powers in $\sigma _{2}$. Therefore, defining $g(\sigma _{}):= \sigma _{2}^{-1}a_0(\sigma _{1})^{-1}p(\sigma _{})$ and $h(\sigma _{}):=-\left( a_0(\sigma _{1})^{-1}\sigma _{2}^{L-1}+b_{L-1}(\sigma _{1})\sigma _{2}^{L-2}+ \cdots +b_{1}(\sigma _{1})\right)$ we can write

$$\begin{aligned} \sigma _{2}^{-1}=g(\sigma _{})+h(\sigma _{}). \end{aligned}$$

(21)

Note here that in Eq. (21), we have $g(\sigma _{})\in \mathfrak {a}$ and $h(\sigma _{})\in \mathcal {A}_1[\sigma _{2}]$ (that is, $h(\sigma _{})$ contains only non-negative powers of $\sigma _{2}$).

By taking positive powers on both sides of Eq. (21) and utilizing the binomial theorem, it follows that for every positive integer $i$ there exists $g_i(\sigma _{})\in \mathfrak {a}$ and $h_i(\sigma _{})\in \mathcal {A}_1[\sigma _{2}]$ such that

$$\begin{aligned} \sigma _{2}^{-i}=g_i(\sigma _{})+h_i(\sigma _{}). \end{aligned}$$

(22)

Since any Laurent polynomial $f(\sigma _{})\in \mathcal {A}$ can be viewed as a finite linear combination of negative and positive powers of $\sigma _{2}$ with coefficients coming from $\mathcal {A}_1$, we can write from Eq. (22) above that for every $f(\sigma _{})\in \mathcal {A}$ there exist $g(\sigma _{})\in \mathfrak {a}$ and $h(\sigma _{})\in \mathcal {A}_1[\sigma _{2}]$ such that

$$\begin{aligned} f(\sigma _{})=g(\sigma _{})+h(\sigma _{}). \end{aligned}$$

(23)

The right hand side of Eq. (23) can be further broken up by applying Eq. (20) to $h(\sigma _{})\in \mathcal {A}_1[\sigma _{2}]$. That is, there exist $q(\sigma _{})\in \mathcal {A}$ and $r_i(\sigma _{1})\in \mathcal {A}_1$ such that $h(\sigma _{})=q(\sigma _{})p(\sigma _{})+ \sum _{i=0}^{L-1}r_i(\sigma _{1})\sigma _{2}^i$. This leads us to conclude that for every $f(\sigma _{})\in \mathcal {A}$ there exist $g(\sigma _{})\in \mathfrak {a}$ and $r_0(\sigma _{1}),\ldots ,r_{L-1}(\sigma _{1})\in \mathcal {A}_1$ such that

$$\begin{aligned} f(\sigma _{})=g(\sigma _{})+\sum _{i=0}^{L-1}r_i(\sigma _{1})\sigma _{2}^i. \end{aligned}$$

(24)

Note that, since $p(\sigma _{})\in \mathfrak {a}$ we have $q(\sigma _{})p(\sigma _{})\in \mathfrak {a}$, too. Therefore $q(\sigma _{})p(\sigma _{})+g(\sigma _{})\in \mathfrak {a}$. We have utilized this fact in the right-hand-side of Eq. (24) to merge $q(\sigma _{})p(\sigma _{})$ and $g(\sigma _{})$ together and call this sum as $g(\sigma _{})$.

Now under the canonical surjection $\mathcal {A}\twoheadrightarrow \mathcal {M}$ Eq. (24) translates to

$$\begin{aligned} \overline{f(\sigma _{})}=\sum _{i=0}^{L-1}\overline{r_i(\sigma _{1})}\overline{\sigma _{2}}^i. \end{aligned}$$

(25)

Thus every element in $\mathcal {M}$ can be written as a linear combination of $\{\overline{\sigma _{2}}^i\}_{0\leqslant i \leqslant L-1}$ with coefficients from $\mathcal {A}_1/\mathfrak {a}\cap \mathcal {A}_1$. In other words, $\{\overline{\sigma _{2}}^i\}_{0\leqslant i \leqslant L-1}$ generates $\mathcal {M}$ as a module over $\mathcal {A}_1$. $\square$