Optimization of the Reachable Set of a Linear System with Respect to Another Set

Abstract

Given a linear controlled autonomous system, we consider the problem of including a convex compact set in the reachable set of the system in the minimum time and the problem of determining the maximum time when the reachable set can be included in a convex compact set. Additionally, the initial point and the time at which the extreme time is achieved in each problem are determined. Each problem is discretized on a grid of unit vectors and is then reduced to a linear programming problem to find an approximate solution of the original problem. Additionally, error estimates for the solution are found. The problems are united by a common ideology going back to the problem of finding the Chebyshev center.

Appendices

APPENDIX 1

1.1 INTERIOR OF A MULTIVALUED INTEGRAL IN THE PLANE CASE

Lemma 5. Suppose that the eigenvalues of the matrix $A \in {{\mathbb{R}}^{{2 \times 2}}}$ are complex conjugate numbers $\alpha \pm i\beta$ with $\beta > 0$ and ${{e}^{{As}}} = {{e}^{{\alpha s}}}Q(\beta s)$, where $Q(s) = \left( {\begin{array}{*{20}{c}} {\cos s}&{ - \sin s} \\ {\sin s}&{\cos s} \end{array}} \right)$. Let $U \subset {{\mathbb{R}}^{2}}$ be a convex compact set, $r > 0$, $x \in {{\mathbb{R}}^{2}}$, $\left\| x \right\| = r$, and the conditions ${{B}_{r}}(x) \subset U$ and $0 \in {{B}_{r}}(x) \cap \partial U$ are satisfied. Define $C = \min \{ {{e}^{{\alpha {{t}_{1}}}}},{{e}^{{\alpha {{t}_{2}}}}}\}$. Then, for any $0 < {{t}_{1}} < {{t}_{2}}$,

$$\int\limits_{{{t}_{1}}}^{{{t}_{2}}} \,{{e}^{{As}}}U{\kern 1pt} ds \supset {{B}_{\delta }}(0),$$

where $\delta = Cr({{t}_{2}} - {{t}_{1}})\left( {1 - \frac{{\sin \left( {\beta ({{t}_{2}} - {{t}_{1}}){\text{/}}2} \right)}}{{\beta ({{t}_{2}} - {{t}_{1}}){\text{/}}2}}} \right)$.

Note that the conditions of the lemma imply that the boundary of the set $U$ is smooth at the point $0$. Moreover, the point $0 \in \partial U$ can be touched by a ball of radius $r$ from within $U$.

Proof. It is true that

$$\int\limits_{{{t}_{1}}}^{{{t}_{2}}} \,{{e}^{{As}}}U{\kern 1pt} ds \supset \int\limits_{{{t}_{1}}}^{{{t}_{2}}} \,{{e}^{{As}}}{{B}_{r}}(x){\kern 1pt} ds \supset C\int\limits_{{{t}_{1}}}^{{{t}_{2}}} \,Q(\beta s){{B}_{r}}(x){\kern 1pt} ds = C\int\limits_{{{t}_{1}}}^{{{t}_{2}}} \,Q(\beta s){{B}_{r}}(0){\kern 1pt} ds + C\int\limits_{{{t}_{1}}}^{{{t}_{2}}} \,Q(\beta s)x{\kern 1pt} ds.$$

In view of the equality $Q(\beta s){{B}_{r}}(0) = {{B}_{r}}(0)$, the first term on the right-hand side is estimated as

$$C\int\limits_{{{t}_{1}}}^{{{t}_{2}}} \,Q(\beta s){{B}_{r}}(0){\kern 1pt} ds \supset Cr({{t}_{2}} - {{t}_{1}}){{B}_{1}}(0).$$

Let $x = r{{(\cos \theta ,\sin \theta )}^{{\text{T}}}}$ for some $\theta \in [0,2\pi )$. For the second term, we obtain

$$C\int\limits_{{{t}_{1}}}^{{{t}_{2}}} \,Q(\beta s)x{\kern 1pt} ds = \frac{{2Cr\sin (\beta ({{t}_{2}} - {{t}_{1}}){\text{/}}2)}}{\beta }\left( {\begin{array}{*{20}{c}} {\cos \left( {\frac{{\beta ({{t}_{2}} + {{t}_{1}})}}{2} + \theta } \right)} \\ {\sin \left( {\frac{{\beta ({{t}_{2}} + {{t}_{1}})}}{2} + \theta } \right)} \end{array}} \right) = :z.$$

For the point $z$, we have $\left\| z \right\| = \frac{{2Cr\sin (\beta ({{t}_{2}} - {{t}_{1}}){\text{/}}2)}}{\beta }$, whence

$$\delta \geqslant Cr({{t}_{2}} - {{t}_{1}}) - \left\| z \right\| = Cr({{t}_{2}} - {{t}_{1}})\left( {1 - \frac{{\sin \left( {\beta ({{t}_{2}} - {{t}_{1}}){\text{/}}2} \right)}}{{\beta ({{t}_{2}} - {{t}_{1}}){\text{/}}2}}} \right).$$

Lemma 6. Suppose that the eigenvalues of the matrix $A \in {{\mathbb{R}}^{{2 \times 2}}}$ are complex conjugate numbers $\alpha \pm i\beta$ with $\beta > 0$ and ${{e}^{{As}}} = {{e}^{{\alpha s}}}Q(\beta s)$, where $Q(s)$ is defined in Lemma $5$. Let $U \subset {{\mathbb{R}}^{2}}$ be a convex compact set, $r > 0$, $\theta \in [0,2\pi )$, and $I = {\text{co}}\{ \pm r{{(\cos \theta ,\sin \theta )}^{{\text{T}}}}\} \subset \partial U$. Define $C = \min \{ {{e}^{{\alpha {{t}_{1}}}}},{{e}^{{\alpha {{t}_{2}}}}}\}$. Then, for any $0 < {{t}_{1}} < {{t}_{2}}$ with ${{t}_{2}} - {{t}_{1}} < \pi {\text{/}}\beta$,

$$\int\limits_{{{t}_{1}}}^{{{t}_{2}}} \,{{e}^{{As}}}U{\kern 1pt} ds \supset {{B}_{\delta }}(0),$$

where $\delta = \frac{2}{\beta }Cr\left( {1 - \cos \left( {\frac{1}{2}\beta ({{t}_{2}} - {{t}_{1}})} \right)} \right)$.

Proof. We have the chain of inclusions $\int_{{{t}_{1}}}^{{{t}_{2}}} {{{e}^{{As}}}U{\kern 1pt} ds} \supset \int_{{{t}_{1}}}^{{{t}_{2}}} {{{e}^{{As}}}I{\kern 1pt} ds} \supset C\int_{{{t}_{1}}}^{{{t}_{2}}} {Q(\beta s)I{\kern 1pt} ds}$. Let $p = (\cos \varphi ,\sin \varphi {{)}^{{\text{T}}}}$ be a unit vector. Then

$$s\left( {p,\int\limits_{{{t}_{1}}}^{{{t}_{2}}} \,{{e}^{{As}}}U{\kern 1pt} ds} \right) \geqslant C\int\limits_{{{t}_{1}}}^{{{t}_{2}}} \,s(p,Q(\beta s)I){\kern 1pt} ds = Cr\int\limits_{{{t}_{1}}}^{{{t}_{2}}} \left| {\sin (\beta s + \psi )} \right|{\kern 1pt} ds,$$

where $\psi = \theta - \varphi + \pi {\text{/}}2$. Making the substitution $\tau = \beta s + \psi$, for ${{T}_{i}} = \beta {{t}_{i}} + \psi$, $i = 1,2$, we obtain $s\left( {p,\int_{{{t}_{1}}}^{{{t}_{2}}} {{{e}^{{As}}}U{\kern 1pt} ds} } \right) \geqslant \frac{1}{\beta }Cr\int_{{{T}_{1}}}^{{{T}_{2}}} {\left| {\sin \tau } \right|} {\kern 1pt} d\tau$. Since ${{T}_{2}} - {{T}_{1}} < \pi$, it follows that

$$\int\limits_{{{T}_{1}}}^{{{T}_{2}}} \left| {\sin \tau } \right|{\kern 1pt} d\tau \geqslant 2\int\limits_0^{\frac{1}{2}({{T}_{2}} - {{T}_{1}})} \sin \tau {\kern 1pt} d\tau = 2\left( {1 - \cos \left( {\frac{1}{2}({{T}_{2}} - {{T}_{1}})} \right)} \right).$$

Substituting this estimate into the preceding inequality yields the assertion of the lemma.

Note that Lemmas 6 and 5 give the second and third orders of the ball’s radius with respect to ${{t}_{2}} - {{t}_{1}}$, respectively.

Let $U \subset {{\mathbb{R}}^{2}}$ be a convex compact set. Assume that there exists a number $\rho > 0$ such that $U \cap {{B}_{\rho }}(0) \supset H \cap {{B}_{\rho }}(0)$, where $H = \{ x \in {{\mathbb{R}}^{2}}\,|\,(p,x) \leqslant 0\}$ is a half-space for some unit vector $p \in {{\mathbb{R}}^{2}}$. Then $\int_{{{t}_{1}}}^{{{t}_{2}}} {{{e}^{{As}}}U{\kern 1pt} ds} \asymp {{({{t}_{2}} - {{t}_{1}})}^{2}}$, ${{t}_{2}} - {{t}_{1}} \to + 0$.

Indeed, let $I$ be a segment centered at the origin such that $I \subset \partial H.$ Consider the disk ${{B}_{R}}(x)$, $\left\| x \right\| = R$, that is tangent to $\partial H$ at the point $0$ and such that $I + {{B}_{R}}(x) \supset U$. Such $R$ and $I$ can be chosen, since $U$ is compact and $\partial H$ is the supporting line of $U$ at the point $0$. Then

$$\int\limits_{{{t}_{1}}}^{{{t}_{2}}} \,{{e}^{{As}}}U{\kern 1pt} ds \subset \int\limits_{{{t}_{1}}}^{{{t}_{2}}} \,{{e}^{{As}}}I{\kern 1pt} ds + \int\limits_{{{t}_{1}}}^{{{t}_{2}}} \,{{e}^{{As}}}{{B}_{R}}(x){\kern 1pt} ds.$$

By Lemmas 5 and 6, the terms on the right-hand side of this formula have the order ${{({{t}_{2}} - {{t}_{1}})}^{2}} + {{({{t}_{2}} - {{t}_{1}})}^{3}} \asymp {{({{t}_{2}} - {{t}_{1}})}^{2}}$, ${{t}_{2}} - {{t}_{1}} \to + 0$.

Thus, the presence of a segment in U, $0 \in \partial U,$ centered at 0 is more important than the bodiliness of $U$ for the ball centered at 0 inscribed in the integral to be maximal.

Lemma 7. Let ${{e}^{{At}}} = \left( {\begin{array}{*{20}{c}} {{{e}^{{{{\lambda }_{1}}t}}}}&0 \\ 0&{{{e}^{{{{\lambda }_{2}}t}}}} \end{array}} \right)$, where ${{\lambda }_{1}} > {{\lambda }_{2}}$. Define $\lambda = {{\lambda }_{1}} - {{\lambda }_{2}} > 0$ and $C = \min \{ {{e}^{{{{\lambda }_{2}}{{t}_{1}}}}},{{e}^{{{{\lambda }_{2}}{{t}_{2}}}}}\}$. Let $U \subset {{\mathbb{R}}^{2}}$ be a convex compact set and $I \subset \partial U$, where $I = {\text{co}}{\kern 1pt} {\kern 1pt} \left( { \pm r{{{(\cos \theta ,\sin \theta )}}^{{\text{T}}}}} \right)$ for $r > 0$, $\theta \in (0,\pi )$, $\theta \ne \pi {\text{/}}2$. Then, for any $0 < {{t}_{1}} < {{t}_{2}}$,

$$\int\limits_{{{t}_{1}}}^{{{t}_{2}}} \,{{e}^{{As}}}U{\kern 1pt} ds \supset {{B}_{\delta }}(0),$$

where $\delta = Cr\frac{{\lambda {{K}^{2}}\sin \theta }}{{{{\pi }^{2}}}}{{({{t}_{2}} - {{t}_{1}})}^{2}}.$ Here, $K = K({{t}_{1}},{{t}_{2}},\lambda ,\theta ) = \mathop {\min }\limits_{i = 1,2} \left\{ {\frac{{{{e}^{{\lambda {{t}_{i}}}}}\left| {\cot \theta } \right|}}{{1 + {{e}^{{2\lambda {{t}_{i}}}}}{{{\cot }}^{2}}\theta }}} \right\}$.

Proof. Let $M = \int_{{{t}_{1}}}^{{{t}_{2}}} {{{e}^{{As}}}U{\kern 1pt} ds}$. Then

$$M \supset \int\limits_{{{t}_{1}}}^{{{t}_{2}}} \,{{e}^{{{{\lambda }_{2}}s}}}\left( {\begin{array}{*{20}{c}} {{{e}^{{\lambda s}}}}&0 \\ 0&1 \end{array}} \right)I{\kern 1pt} ds \supset Cr\int\limits_{{{t}_{1}}}^{{{t}_{2}}} \,{\text{co}}{\kern 1pt} {\kern 1pt} \left( { \pm {{{({{e}^{{\lambda s}}}\cos \theta ,\sin \theta )}}^{{\text{T}}}}} \right){\kern 1pt} ds.$$

Let $p = (\cos \varphi ,\sin \varphi {{)}^{{\text{T}}}}$ be an unit vector, where $\varphi \in [0,2\pi )$. Then, for

$$J = \int\limits_{{{t}_{1}}}^{{{t}_{2}}} \left| {{{e}^{{\lambda s}}}\cos \varphi \cos \theta + \sin \varphi \sin \theta } \right|{\kern 1pt} ds,$$

we have $s(p,M) \geqslant CrJ.$ Let us estimate $J$ from below.

1. Consider the case $\theta \in \left( {0,\frac{1}{2}\pi } \right)$. Here,

$$J = \int\limits_{{{t}_{1}}}^{{{t}_{2}}} \sqrt {{{{\sin }}^{2}}\theta + {{{\cos }}^{2}}\theta {{e}^{{2\lambda s}}}} {\kern 1pt} \left| {\sin (\alpha (s) + \varphi )} \right|{\kern 1pt} ds,$$

where $\alpha (s) = {\text{arctan}}{\kern 1pt} {\kern 1pt} ({{e}^{{\lambda s}}}\theta )$. We introduce the new variable $\tau = \alpha (s) + \varphi$, $ds = \frac{{d\tau }}{{\alpha {\kern 1pt} '(s)}} = \frac{{1 + {{e}^{{2\lambda s}}}{{{\cot }}^{2}}\theta }}{{\lambda {{e}^{{\lambda s}}}\cot \theta }}{\kern 1pt} d\tau$. In view of the inclusion $\alpha (s) \in \left( {0,\frac{1}{2}\pi } \right)$, it follows that

$$J = \frac{{\sin \theta }}{\lambda }\int\limits_{\alpha ({{t}_{1}}) + \varphi }^{\alpha ({{t}_{2}}) + \varphi } \frac{{{{{(1 + {{{\tan }}^{2}}\alpha (s))}}^{{3/2}}}}}{{\tan \alpha (s)}}\left| {\sin \tau } \right|d\tau = \frac{{\sin \theta }}{\lambda }\int\limits_{\alpha ({{t}_{1}}) + \varphi }^{\alpha ({{t}_{2}}) + \varphi } \frac{{\left| {\sin \tau } \right|}}{{(\tau - \varphi ){{{\cos }}^{3}}(\tau - \varphi )}}d\tau $$

$$ \geqslant \frac{{\sin \theta }}{\lambda }\int\limits_{\alpha ({{t}_{1}}) + \varphi }^{\alpha ({{t}_{2}}) + \varphi } \left| {\sin \tau } \right|{\kern 1pt} d\tau \geqslant \frac{{2\sin \theta }}{\lambda }\int\limits_0^{\frac{1}{2}(\alpha ({{t}_{2}}) - \alpha ({{t}_{1}}))} \sin \tau {\kern 1pt} d\tau = \frac{{4\sin \theta }}{\lambda }{{\sin }^{2}}\left( {\frac{{\alpha ({{t}_{2}}) - \alpha ({{t}_{1}})}}{4}} \right)$$

$$ \geqslant \frac{{4\sin \theta }}{\lambda }\frac{4}{{{{\pi }^{2}}}}\frac{1}{{{{4}^{2}}}}{{(\alpha ({{t}_{2}}) - \alpha ({{t}_{1}}))}^{2}}\frac{{\sin \theta }}{{\lambda {{\pi }^{2}}}}{{(\alpha ({{t}_{2}}) - \alpha ({{t}_{1}}))}^{2}}.$$

By the mean-value theorem, there exists $\xi \in [{{t}_{1}},{{t}_{2}}]$ such that

$$\left| {\alpha ({{t}_{2}}) - \alpha ({{t}_{1}})} \right| = \frac{{\lambda {{e}^{{\lambda \xi }}}\left| {\cot \theta } \right|}}{{1 + {{e}^{{2\lambda \xi }}}{{{\cot }}^{2}}\theta }}({{t}_{2}} - {{t}_{1}}) \geqslant \lambda K({{t}_{2}} - {{t}_{1}}),$$

which yields the assertion of the lemma.

2. Suppose that $\theta \in \left( {\frac{1}{2}\pi ,\pi } \right)$. Making the substitutions $\psi = \pi - \theta ,$ $\alpha (s) = {\text{arctan}}{\kern 1pt} {\kern 1pt} ({{e}^{{\lambda s}}}\cot \psi ),$ and $\tau = \alpha (s) + \varphi$ brings the integral $J$ to the form

$$\begin{gathered} J = \frac{{\sin \psi }}{\lambda }\int\limits_{\alpha ({{t}_{1}}) + \varphi }^{\alpha ({{t}_{2}}) + \varphi } \frac{{{{{(1 + {{{\tan }}^{2}}\alpha (s))}}^{{3/2}}}}}{{\tan \alpha (s)}}\left| {\cos (\alpha (s) + \varphi )} \right|d\tau \geqslant \frac{{\sin \theta }}{\lambda }\int\limits_{\alpha ({{t}_{1}}) + \varphi - \frac{1}{2}\pi }^{\alpha ({{t}_{2}}) + \varphi - \frac{1}{2}\pi } \left| {\sin \tau } \right|d\tau \\ \geqslant \frac{{2\sin \theta }}{\lambda }\int\limits_0^{\frac{1}{2}(\alpha ({{t}_{2}}) - \alpha ({{t}_{1}}))} \left| {\sin \tau } \right|d\tau . \\ \end{gathered} $$

The rest of the proof is the same as in item 1.

Lemma 8. Let ${{e}^{{At}}} = {{e}^{{\lambda t}}}\left( {\begin{array}{*{20}{c}} 1&t \\ 0&1 \end{array}} \right)$. Define $C = \min \{ {{e}^{{\lambda {{t}_{1}}}}},{{e}^{{\lambda {{t}_{2}}}}}\}$. Let $U \subset {{\mathbb{R}}^{2}}$ be a convex compact set and $I \subset \partial U,$ where $I = {\text{co}}{\kern 1pt} {\kern 1pt} \left( { \pm r{{{(\cos \theta ,\sin \theta )}}^{{\text{T}}}}} \right)$ for $r > 0$ and $\theta \in (0,\pi )$. Then, for any $0 < {{t}_{1}} < {{t}_{2}}$,

$$\int\limits_{{{t}_{1}}}^{{{t}_{2}}} \,{{e}^{{As}}}U{\kern 1pt} ds \supset {{B}_{\delta }}(0),$$

where $\delta = Cr\frac{{{{K}^{2}}\sin \theta }}{{{{\pi }^{2}}}}{{({{t}_{2}} - {{t}_{1}})}^{2}}.$ Here, $K = K({{t}_{1}},{{t}_{2}},\theta ) = \mathop {\min }\limits_{i = 1,2} \left\{ {\frac{1}{{1 + {{{({{t}_{i}} + \cot \theta )}}^{2}}}}} \right\}$.

Proof. Repeating the argument used in Lemma 7, for $M = \int_{{{t}_{1}}}^{{{t}_{2}}} {{{e}^{{As}}}U{\kern 1pt} ds} ,$ $p = (\cos \varphi ,\sin \varphi ),$ and J = $\int_{{{t}_{1}}}^{{{t}_{2}}} {\left| {(\cos \theta + s\sin \theta )\cos \varphi + \sin \theta \sin \varphi } \right|ds}$, we have $s(p,M) \geqslant CrJ$. For $J,$ we obtain J = $\sin \theta \int_{{{t}_{1}}}^{{{t}_{2}}} {\sqrt {1 + {{{(s + \cot \theta )}}^{2}}} } \left| {\sin (\alpha (s) + \varphi )} \right|ds$, where $\alpha (s) = {\text{arctan}}{\kern 1pt} (s + \cot \theta )$. Making the substitution $\tau = \alpha (s) + \varphi$ yields

$$J = \sin \theta \int\limits_{\alpha ({{t}_{1}}) + \varphi }^{\alpha ({{t}_{2}}) + \varphi } {{(1 + {{(s + \cot \theta )}^{2}})}^{{3/2}}}\left| {\sin \tau } \right|d\tau = \sin \theta \int\limits_{\alpha ({{t}_{1}}) + \varphi }^{\alpha ({{t}_{2}}) + \varphi } {{(1 + ta{{n}^{2}}\alpha (s))}^{{3/2}}}\left| {\sin \tau } \right|d\tau $$

$$ = \sin \theta \int\limits_{\alpha ({{t}_{1}}) + \varphi }^{\alpha ({{t}_{2}}) + \varphi } \frac{{{\text{|}}\sin \tau {\text{|}}}}{{\left| {{{{\cos }}^{3}}(\tau - \varphi )} \right|}}d\tau \geqslant 2\sin \theta \int\limits_0^{\frac{1}{2}(\alpha ({{t}_{2}}) - \alpha ({{t}_{1}}))} \sin \tau {\kern 1pt} d\tau = 4\sin \theta {{\sin }^{2}}\left( {\frac{{\alpha ({{t}_{2}}) - \alpha ({{t}_{1}})}}{4}} \right).$$

It follows that $J \geqslant 4\sin \theta \frac{4}{{{{\pi }^{2}}}}\frac{{{{{(\alpha ({{t}_{2}}) - \alpha ({{t}_{1}}))}}^{2}}}}{{{{4}^{2}}}}$. Applying the mean-value theorem completes the proof.

APPENDIX 2

1.1 AN EXAMPLE OF A REACHABLE SET

For a nontrivial linear system, we give an example of a reachable set that is strictly convex, but not strongly convex. Recall that $f(s) \asymp g(s)$, $s \to 0$ if there exists $\delta > 0$ such that ${{C}_{1}} \leqslant \frac{{f(s)}}{{g(s)}} \leqslant {{C}_{2}}$ for all $s \in ( - \delta ,\delta )$ and some ${{C}_{2}} \geqslant {{C}_{1}} > 0$. Let $U = {\text{co}}{\kern 1pt} {\kern 1pt} \{ (0,0, \pm 1)\}$, $\lambda \in \mathbb{R}$, $A = \left[ {\begin{array}{*{20}{c}} \lambda &1&0 \\ 0&\lambda &1 \\ 0&0&\lambda \end{array}} \right]$, and ${{e}^{{As}}} = {{e}^{{\lambda s}}}\left[ {\begin{array}{*{20}{c}} 1&s&{\frac{1}{2}{{s}^{2}}} \\ 0&1&s \\ 0&0&1 \end{array}} \right]$. For $\varepsilon \in (0,1)$, we consider the vectors $p = \frac{1}{3}(2, - 2,1)$ and $q = q(\varepsilon ) = \frac{{(2, - 2,1 - \varepsilon )}}{{\sqrt {9 - 2\varepsilon + {{\varepsilon }^{2}}} }}$. It is easy to see that $\left\| {p - q} \right\| \asymp \varepsilon$, $\varepsilon \to 0$, and, for $t > 1 + \sqrt \varepsilon$ (${{s}_{{1,2}}} = 1 \pm \sqrt \varepsilon$ are the roots of the equation $({{e}^{{{{A}^{{\text{T}}}}s}}}q(\varepsilon {{),(0,0,1)}^{{\text{T}}}}) = 0$)

$$M(t)(p) - M(t)(q) = \int\limits_{1 - \sqrt \varepsilon }^{1 + \sqrt \varepsilon } {{e}^{{\lambda s}}}{{({{s}^{2}},2s,2)}^{{\text{T}}}}{\kern 1pt} ds,\quad {\text{||}}M(t)(p) - M(t)(q){\text{||}} \geqslant \int\limits_{1 - \sqrt \varepsilon }^{1 + \sqrt \varepsilon } 2{{e}^{{\lambda s}}}{\kern 1pt} ds \asymp \sqrt \varepsilon ,\quad \varepsilon \to 0.$$

For any fixed $R > 0$, the value of $\sqrt \varepsilon {\text{/}}(R\varepsilon )$ is not bounded above for small $\varepsilon > 0$. Therefore, $M(t)$ is not strongly convex, since, for a strongly convex set with radius $R > 0$, its support elements have to satisfy the Lipschitz condition on the unit sphere with constant $R$ (see [19]; [12], Corollary 4). Note that the set $M(t)$ is strictly convex, since for any vector $p \in \partial {{B}_{1}}(0)$ the set $({{e}^{{As}}}U)(p)$ is a singleton for all $s \in [0,t]$, except for at most two values.