On the Properties of the Limit Control Sets for a Class of Unstable Linear Systems with Discrete Time and l ₁ -Restrictions

Abstract

The construction of attainability and controllability sets for linear systems with discrete time and a total constraint on scalar control in the sense of l₁-norm is discussed. For classes of linear systems that are not completely controllable and not completely attainable, the limit sets of controllability and attainability are explicitly constructed, respectively. Examples are given.

APPENDIX

Proof of the lemma. (i). By definition [8, Chap. 2, § 4, point 1] the set $\mathcal{D} \subset {{\mathbb{R}}^{n}}$ is called a polyhedron if it can be represented as a set of solutions to a system of a finite number $m$ of linear inequalities:

$$\mathcal{D} = \left\{ {x \in {{\mathbb{R}}^{n}}:{{x}^{{\text{T}}}}{{a}^{j}} \leqslant {{b}_{j}},{{a}^{j}} \in {{\mathbb{R}}^{n}},j = \overline {1,m} } \right\}.$$

We denote by ${{\mathcal{B}}_{{\mathbb{R}_{1}^{n}}}}$ a closed ball in ${{\mathbb{R}}^{n}}$ with metric l₁:

$${{\mathcal{B}}_{{\mathbb{R}_{1}^{n}}}} = \left\{ {x = {{{\left( {{{x}_{1}}, \ldots ,{{x}_{n}}} \right)}}^{{\text{T}}}} \in {{\mathbb{R}}^{n}}:\sum\limits_{i = 1}^n {\left| {{{x}_{i}}} \right| \leqslant 1} } \right\}.$$

The set of ${{\mathcal{B}}_{{\mathbb{R}_{1}^{n}}}}$ is a polyhedron, because

$$\left\{ {x = {{{\left( {{{x}_{1}}, \ldots ,{{x}_{n}}} \right)}}^{{\text{T}}}} \in {{\mathbb{R}}^{n}}:\sum\limits_{i = 1}^n {\left| {{{x}_{i}}} \right| \leqslant 1} } \right\} = \{ x \in {{\mathbb{R}}^{n}}:{{x}^{{\text{T}}}}{{\tilde {a}}^{j}} \leqslant 1,j = \overline {{{{1,2}}^{n}}} \} ,$$

(A.1)

where

$${{\tilde {a}}^{j}} = {{(\tilde {a}_{1}^{j}, \ldots ,\tilde {a}_{n}^{j})}^{{\text{T}}}} \in {{\mathbb{R}}^{n}},\quad \tilde {a}_{i}^{j} \in \left\{ { - 1,1} \right\},\quad i = \overline {1,n} ,\quad j = \overline {{{{1,2}}^{n}}} .$$

Indeed, if for $\alpha \in \mathbb{R}$ and $\beta \geqslant 0$, the inequality

$$\left| \alpha \right| \leqslant \beta ;$$

is valid, then we get

$$\alpha \leqslant \beta ,\quad - \alpha \leqslant \beta $$

or

$$ \pm \alpha \leqslant \beta .$$

(A.2)

Further, the reasoning is carried out by induction. Assume k = 1, then from for any $x \in {{\mathcal{B}}_{{\mathbb{R}_{1}^{n}}}}$, we find

$$\left| {{{x}_{1}}} \right| \leqslant 1 - \sum\limits_{i = 2}^n {\left| {{{x}_{i}}} \right|} .$$

(A.3)

From (A.2) and (A.3), we have

$$ \pm {{x}_{1}} \leqslant 1 - \sum\limits_{i = 2}^n {\left| {{{x}_{i}}} \right|} $$

or

$$\sum\limits_{i = 2}^n {\left| {{{x}_{i}}} \right|} \leqslant 1 \pm {{x}_{1}}.$$

Let us say for an arbitrary $k = \overline {2,n - 1}$, the following inequality is valid:

$$\sum\limits_{i = k}^n {\left| {{{x}_{i}}} \right|} \leqslant 1 \pm {{x}_{1}} \pm \cdots \pm {{x}_{{k - 1}}} = 1 \pm \sum\limits_{i = 1}^{k - 1} {{{x}_{i}}} .$$

(A.4)

Then,

$$\left| {{{x}_{k}}} \right| \leqslant 1 \pm \sum\limits_{i = 1}^{k - 1} {{{x}_{i}}} - \sum\limits_{i = k + 1}^n {\left| {{{x}_{i}}} \right|} .$$

From (A.2) we obtain

$$\begin{gathered} \pm {{x}_{k}} \leqslant 1 \pm \sum\limits_{i = 1}^{k - 1} {{{x}_{i}}} - \sum\limits_{i = k + 1}^n {\left| {{{x}_{i}}} \right|} , \\ \sum\limits_{i = k + 1}^n {\left| {{{x}_{i}}} \right|} \leqslant 1 \pm \sum\limits_{i = 1}^k {{{x}_{i}}} . \\ \end{gathered} $$

(A.5)

Thus, from statement (A.4), (A.5) follows for k, and the validity of assertion (A.4), for k +1. Therefore, by the principle of mathematical induction, we come to the conclusion that the inequality

$$\sum\limits_{i = 1}^n {\left| {{{x}_{i}}} \right| \leqslant 1} $$

is equivalent to inequality

$$\sum\limits_{i = 1}^n { \pm {{x}_{i}}} \leqslant 1,$$

which coincides with (A.1).

We introduce the matrix ${{A}_{k}} \in {{\mathbb{R}}^{{n \times \left( {k + 1} \right)}}}$ according to the formula

$${{A}_{k}}: = \left( {\left. b \right|\left. {Ab} \right| \cdots \left| {{{A}^{{k - 1}}}b} \right.} \right).$$

Assume

$${{{v}}_{{i + 1}}}: = u\left( {k - 1 - i} \right),\quad i = \overline {0,k - 1} ,\quad {v} = {{\left( {{{{v}}_{1}}, \ldots ,{{{v}}_{k}}} \right)}^{{\text{T}}}} \in t{{\mathcal{B}}_{{\mathbb{R}_{1}^{k}}}} \subset {{\mathbb{R}}^{k}}.$$

(A.6)

Then

$$\sum\limits_{i = 0}^{k - 1} {{{A}^{i}}bu\left( {k - 1 - i} \right)} = \sum\limits_{i = 0}^{k - 1} {{{A}^{i}}b{{v}_{{i + 1}}}} = {{A}_{k}}v.$$

Therefore, we get the representation

$${{\mathcal{Y}}_{t}}\left( k \right) = \left\{ {y \in {{\mathbb{R}}^{n}}:y = {{A}_{k}}{v},\;{v} \in t{{\mathcal{B}}_{{\mathbb{R}_{1}^{k}}}}} \right\} = t{{A}_{k}}{{\mathcal{B}}_{{\mathbb{R}_{1}^{k}}}}.$$

By [12, Theorem 19.3, Chap. IV, § 19], the set ${{\mathcal{Y}}_{t}}\left( k \right)$ is a polyhedron as a linear mapping of a polyhedral convex set.

(i) The set ${{\mathcal{B}}_{{\mathbb{R}_{1}^{k}}}}$ is, by construction, a closed unit ball centered at 0 in the metric space ${{\mathbb{R}}^{k}}$ with metric l₁. Therefore, the set ${{\mathcal{B}}_{{\mathbb{R}_{1}^{k}}}}$ is closed, bounded, and symmetric with respect to 0 as the corresponding ball. Finally, statements (ii) follow from the linear representation

$${{\mathcal{Y}}_{t}}\left( k \right) = t{{A}_{k}}{{\mathcal{B}}_{{\mathbb{R}_{1}^{k}}}}.$$

(iii) By the Minkowski-Weyl theorem [8, Chap. 2, § 8, theorem 8.14], the set ${{\mathcal{Y}}_{t}}\left( k \right)$ is a convex polyhedron since it is a bounded polyhedron by (ii).

The following equality is valid

$$\operatorname{conv} ( \pm b, \pm Ab, \ldots , \pm {{A}^{{k - 1}}}b) = \operatorname{conv} (0, \pm b, \pm Ab, \ldots , \pm {{A}^{{k - 1}}}b);$$

and therefore the following representation is correct:

$$\begin{gathered} \operatorname{conv} ( \pm b, \pm Ab, \ldots , \pm {{A}^{{k - 1}}}b) \\ \, = \left\{ {y \in {{\mathbb{R}}^{n}}:y = \sum\limits_{i = 0}^{k - 1} {{{A}^{i}}b{{\lambda }_{{i + 1}}}} - \sum\limits_{i = 0}^{k - 1} {{{A}^{i}}b{{\mu }_{{i + 1}}}} ,\sum\limits_{i = 1}^k {\left( {{{\lambda }_{i}} + {{\mu }_{i}}} \right)} \leqslant 1,{{\lambda }_{i}},{{\mu }_{i}} \geqslant 0,i = \overline {1,k} } \right\}. \\ \end{gathered} $$

(A.7)

Indeed, we get

$$\begin{gathered} \operatorname{conv} ( \pm b, \pm Ab, \ldots , \pm {{A}^{{k - 1}}}b) = \operatorname{conv} (0, \pm b, \pm Ab, \ldots , \pm {{A}^{{k - 1}}}b) \\ \, = \left\{ {y \in {{\mathbb{R}}^{n}}:y = 0 \cdot \left( {{{\lambda }_{0}} + {{\mu }_{0}}} \right) + \sum\limits_{i = 0}^{k - 1} {{{\lambda }_{{i + 1}}}{{A}^{i}}b} + \sum\limits_{i = 0}^{k - 1} {{{\mu }_{{i + 1}}}( - {{A}^{i}}b)} ,} \right. \\ \end{gathered} $$

$$\begin{gathered} \left. {\left( {{{\lambda }_{0}} + {{\mu }_{0}}} \right) + \sum\limits_{i = 1}^k {\left( {{{\lambda }_{i}} + {{\mu }_{i}}} \right)} = 1,{{\lambda }_{i}},{{\mu }_{i}} \geqslant 0,i = \overline {0,k} } \right\} \\ \, = \left\{ {y \in {{\mathbb{R}}^{n}}:y = \sum\limits_{i = 0}^{k - 1} {{{\lambda }_{{i + 1}}}{{A}^{i}}b} + \sum\limits_{i = 0}^{k - 1} {{{\mu }_{{i + 1}}}( - {{A}^{i}}b)} ,} \right. \\ \end{gathered} $$

$$\begin{gathered} \left. {\sum\limits_{i = 1}^k {\left( {{{\lambda }_{i}} + {{\mu }_{i}}} \right)} = 1 - \left( {{{\lambda }_{0}} + {{\mu }_{0}}} \right),{{\lambda }_{i}},{{\mu }_{i}} \geqslant 0,i = \overline {0,k} } \right\} \\ \, = \left\{ {y \in {{\mathbb{R}}^{n}}:y = \sum\limits_{i = 0}^{k - 1} {{{\lambda }_{{i + 1}}}{{A}^{i}}b} + \sum\limits_{i = 0}^{k - 1} {{{\mu }_{{i + 1}}}( - {{A}^{i}}b)} ,} \right. \\ \end{gathered} $$

$$\left. {\sum\limits_{i = 1}^k {\left( {{{\lambda }_{i}} + {{\mu }_{i}}} \right)} \leqslant 1,{{\lambda }_{i}},{{\mu }_{i}} \geqslant 0,i = \overline {1,k} } \right\}.$$

The proof of assertion (iii) is sufficient for t = 1; thus, we denote

$$\mathcal{Y}\left( k \right): = {{\mathcal{Y}}_{1}}\left( k \right).$$

The following inclusion is valid:

$$\mathcal{Y}\left( k \right) \subset \operatorname{conv} ( \pm b, \pm Ab, \ldots , \pm {{A}^{{k - 1}}}b).$$

(A.8)

Indeed, if

$$y \in \mathcal{Y}\left( k \right),$$

then

$$y = \sum\limits_{i = 0}^{k - 1} {{{A}^{i}}b{{v}_{{i + 1}}}} ,\quad \sum\limits_{i = 1}^k {\left| {{{v}_{i}}} \right|} \leqslant 1.$$

(A.9)

Let us assume

$${{\lambda }_{i}} = \max \left\{ {0, - {{{v}}_{i}}} \right\},\quad {{\mu }_{i}} = \max \left\{ {0,{{{v}}_{i}}} \right\},$$

then

$${{\lambda }_{i}} + {{\mu }_{i}} = \left| {{{{v}}_{i}}} \right|,\quad {{\lambda }_{i}},{{\mu }_{i}} \geqslant 0,\quad {{\lambda }_{i}} \cdot {{\mu }_{i}} = 0,\quad i = \overline {1,k} .$$

Hence,

$$y = \sum\limits_{i = 0}^{k - 1} {{{A}^{i}}b{{{v}}_{{i + 1}}}} = - \sum\limits_{i = 0}^{k - 1} {{{\lambda }_{{i + 1}}}{{A}^{i}}b} + \sum\limits_{i = 0}^{k - 1} {{{\mu }_{{i + 1}}}{{A}^{i}}b} ,\quad \sum\limits_{i = 1}^k {\left( {{{\lambda }_{i}} + {{\mu }_{i}}} \right)} \leqslant 1,\quad {{\lambda }_{i}},{{\mu }_{i}} \geqslant 0.$$

By (A.7) we obtain

$$y \in \operatorname{conv} ( \pm b, \pm Ab, \ldots , \pm {{A}^{{k - 1}}}b).$$

We now show that the following reverse inclusion is true:

$$\operatorname{conv} ( \pm b, \pm Ab, \ldots , \pm {{A}^{{k - 1}}}b) \subset \mathcal{Y}\left( k \right).$$

(A.10)

Indeed, assume $y \in \operatorname{conv} ( \pm b, \pm Ab, \ldots , \pm {{A}^{{k - 1}}}b)$, then the decomposition

$$y = - \sum\limits_{i = 0}^{k - 1} {{{\lambda }_{{i + 1}}}{{A}^{i}}b} + \sum\limits_{i = 0}^{k - 1} {{{\mu }_{{i + 1}}}{{A}^{i}}b} ,\quad \sum\limits_{i = 1}^k {\left( {{{\lambda }_{i}} + {{\mu }_{i}}} \right)} \leqslant 1,\quad {{\lambda }_{i}},{{\mu }_{i}} \geqslant 0.$$

Let us put

$${{v}_{i}} = {{\mu }_{i}} - {{\lambda }_{i}}.$$

Then, the following equality is valid:

$$y = \sum\limits_{i = 0}^{k - 1} {\left( {{{\mu }_{{i + 1}}} - {{\lambda }_{{i + 1}}}} \right){{A}^{i}}b} = \sum\limits_{i = 0}^{k - 1} {{{v}_{{i + 1}}}{{A}^{i}}b} ;$$

and also,

$$\sum\limits_{i = 1}^k {\left| {{{v}_{i}}} \right|} = \sum\limits_{i = 1}^k {\left| {{{\mu }_{i}} - {{\lambda }_{i}}} \right|} = \sum\limits_{i = 1}^k {\left( {{{\mu }_{i}} + {{\lambda }_{i}}} \right)} \leqslant 1.$$

This means that by definition

$$y \in \mathcal{Y}\left( k \right).$$

From (A.8) and (A.10), the required assertion (iii) follows.

(iv) Since by construction $\mathcal{Y}\left( k \right)$ is a nonempty convex set in ${{\mathbb{R}}^{n}}$ and $0 \in {{\mathcal{Y}}_{t}}\left( k \right)$, by assertion (ii) from [8, Chap. 3, § 7, point 7, theorem 7.14], it follows that $0 \in \operatorname{ri} {{\mathcal{Y}}_{t}}\left( k \right)$. By condition (1.3) we have $\operatorname{aff} {{\mathcal{Y}}_{t}}\left( k \right) = {{\mathbb{R}}^{n}}$ and $\operatorname{ri} {{\mathcal{Y}}_{t}}\left( k \right) = \operatorname{int} {{\mathcal{Y}}_{t}}\left( k \right)$, which proves the claim.

The proofs of statements (v), (vi), and (vii) are obvious.

The lemma is proved.

Proof of the theorem. By assertion (iii) of the lemma, we have

$${{\mathcal{Y}}_{t}}\left( n \right) = t\operatorname{conv} ( \pm b, \pm Ab, \ldots , \pm {{A}^{{n - 1}}}b).$$

Then from (1.3) and the lemma we deduce that the set ${{\mathcal{Y}}_{t}}\left( n \right)$ is a symmetric bounded convex set for which $0 \in \operatorname{int} {{\mathcal{Y}}_{t}}\left( n \right)$.

Let us construct a sequence of vectors ${{\{ {{A}^{k}}b\} }_{{k \in \mathbb{N}}}}$. Then there is a number $K \in \mathbb{N}$ such that

$${{A}^{k}}b \in {{\mathcal{Y}}_{t}}\left( n \right)\quad {\text{for}}\quad k \geqslant K.$$

(A.11)

Indeed, by condition (1.4) we have $\rho \left( A \right) < 1$, then [15, Theorem 5.6.12] the following equality is valid:

$$\mathop {\lim }\limits_{k \to \infty } {{A}^{k}} = O.$$

Therefore,

$$\mathop {\lim }\limits_{k \to \infty } {{A}^{k}}b = 0,$$

$$\mathop {\lim }\limits_{k \to \infty } {{\left\| {{{A}^{k}}b} \right\|}_{{{{\mathcal{Y}}_{t}}\left( n \right)}}} = 0.$$

By definition of the convergence of a numerical sequence ${{\left\{ {{{{\left\| {{{A}^{k}}b} \right\|}}_{{{{\mathcal{Y}}_{t}}\left( n \right)}}}} \right\}}_{{k \in \mathbb{N}}}}$, there is a natural K such that

$${{\left\| {{{A}^{k}}b} \right\|}_{{{{\mathcal{Y}}_{t}}\left( n \right)}}} \leqslant 1\quad {\text{at}}\quad k \geqslant K.$$

Now from (3.3) and (3.4), we obtain the required inclusion (A.11).

For arbitrary vectors ${{a}^{1}}, \ldots ,{{a}^{m}},c \in {{\mathbb{R}}^{n}}$, $m \geqslant 1$, the equality

$$\operatorname{conv} \left\{ {{{a}^{1}}, \ldots ,{{a}^{m}},c} \right\} = \operatorname{conv} \left\{ {{{a}^{1}}, \ldots ,{{a}^{m}}} \right\},$$

(A.12)

if

$$c \in \operatorname{conv} \left\{ {{{a}^{1}}, \ldots ,{{a}^{m}}} \right\}.$$

(A.13)

Indeed, by the definition of a convex hull, the inclusion

$$\operatorname{conv} \left\{ {{{a}^{1}}, \ldots ,{{a}^{m}}} \right\} \subset \operatorname{conv} \left\{ {{{a}^{1}}, \ldots ,{{a}^{m}},c} \right\}.$$

However, (A.13) also implies the chain of inclusions

$$\begin{gathered} \operatorname{conv} \{ {{a}^{1}}, \ldots ,{{a}^{m}},c\} \subset \operatorname{conv} \{ \operatorname{conv} \{ {{a}^{1}}, \ldots ,{{a}^{m}}\} \cup \operatorname{conv} \{ c\} \} \\ \, = \operatorname{conv} \{ \operatorname{conv} \{ {{a}^{1}}, \ldots ,{{a}^{m}}\} \cup c\} = \operatorname{conv} \{ \operatorname{conv} \{ {{a}^{1}}, \ldots ,{{a}^{m}}\} \} = \operatorname{conv} \{ {{a}^{1}}, \ldots ,{{a}^{m}}\} . \\ \end{gathered} $$

We finally arrive at equality (A.12).

By point (iii) of the lemma, for $k \geqslant K$

$$\begin{gathered} {{\mathcal{Y}}_{t}}\left( k \right) = t\operatorname{conv} \left( { \pm b, \pm Ab, \ldots , \pm {{A}^{K}}b, \pm {{A}^{{K + 1}}}b, \ldots , \pm {{A}^{k}}b} \right) \\ \, = t\operatorname{conv} \left( { \pm b, \pm Ab, \ldots , \pm {{A}^{K}}b} \right) = {{\mathcal{Y}}_{t}}\left( K \right), \\ \end{gathered} $$

insofar as

$$~{{A}^{k}}b \in {{\mathcal{Y}}_{t}}\left( n \right) \subset {{\mathcal{Y}}_{t}}\left( K \right)\quad {\text{for}}\quad k \geqslant K.$$

Here equality (A.12) is used. Therefore, by point (v) of the lemma, we conclude

$${{\mathcal{Y}}_{t}}\left( n \right) \subset \cdots \subset {{\mathcal{Y}}_{t}}\left( K \right) = {{\mathcal{Y}}_{t}}\left( {K + 1} \right) = \cdots ,$$

which proves equality (2.1).

The theorem is proved.