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This article is cited in 1 scientific paper (total in 1 paper) Relative optimality in nonlinear differential games with discrete control K. A. Shchelchkov Udmurt State University, Izhevsk, Russia
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     § 1. IntroductionDifferential games of two individuals, which were initially considered by Isaacs [1], provide currently a wide field for research (see [2]–[7]). Methods for solving various classes of game problems were developed, including the Isaacs method, which is based on analyzing a certain partial differential equation and its characteristics, the Krasovskii extremal aiming method, the Pontryagin method and so on. Krasovskii and his scientific school created the theory of positional games, which is based on the concept of a maximal stable bridge and the extremal aiming rule. However, it is very difficult and sometimes even impossible to construct effectively bridges of this kind to investigate actual conflict-controlled processes (primarily nonlinear differential games). It is more convenient to construct bridges that are not maximal but have the property of stability and give effectively implementable control procedures for some classes of games with additional properties. Constructions of approximations of stable bridges in nonlinear differential games, including numerical ones, were considered, in particular, in [8] and [9]. Sufficient conditions for the solvability of the pursuit problem in Pontryagin’s nonlinear example were deduced in [10]. In [11] sufficient conditions for the solvability of the pursuit problem in a nonlinear differential game were given under some additional conditions on the set of values of the right-hand side of the system of differential equations and the terminal set. In [12], sufficient conditions for capture in a nonlinear game described by a stationary nonlinear system were inferred and the optimality of the capture time was studied for a certain special case on the plane. The problem considered in [12] is comparable with the one dealt with in our paper. The capture conditions in [12] turn out to be much stronger than the conditions we obtain here. In [12] the pursuer uses a counterstrategy. In [13] a nonlinear control problem with obstacle was considered using the differential game formalism. Sufficient conditions for the existence of a winning strategy were obtained. In [14], a nonlinear differential game of two players with an integral quality criterion was under consideration. The players use piecewise program controls of special form, where the time interval is divided into two parts. Necessary and sufficient conditions for the existence of a saddle point for the game under consideration were derived. In [15] a differential pursuit game on the plane whose dynamics is described by a nonlinear system of differential equations of a certain form was treated. The target set is the origin. Conditions of capture via a positional counterstrategy and characteristics of the game were obtained in an explicit form; examples were presented. In [16], the concept of a positive basis was introduced, which was efficiently used in [16]–[18] to study the controllability of nonlinear systems described by differential equations in finite-dimensional Euclidean space. In [19]–[21] the properties of a positive basis were used to study controlled systems on manifolds. In [22]–[26] the properties of a positive basis were used to study a pursuit problem with a group of pursuers and one or several evaders in linear differential games with equal opportunities of the players. The concept of a positive basis was used in these works to describe the initial positions of the players. In [27] and [29] a capture problem was considered in a nonlinear differential game that is similar to the differential game in our paper. Sufficient conditions on the parameters of the game were deduced there for the existence of a neighbourhood of zero from which capture occurs. A key condition is that some set of vectors form a positive basis. In [28] some additional properties of a winning strategy for the problem were obtained. This paper is a continuation of [27]–[29]. We consider two control problems with an obstacle that is the second player in the differential game. The dynamics in the first problem is described by a nonlinear system of first-order differential equations, whereas the dynamics in the second is described by a nonlinear system of second-order differential equations. Conditions are obtained under which it is possible to keep the trajectory of the original system close to the trajectory of some system of a simple form for any actions of the obstacle. § 2. System with the first-order derivativeIn the space $\mathbb R^k$, $k \geqslant 2$, we consider a differential game $\Gamma (x_0)$ of two players, namely, a pursuer $P$ and an evader $E$. The dynamics in the game is described by the system of differential equations
$$
\begin{equation*}
\dot x=f(x, u) + g(x, v), \qquad u \in U, \quad v \in V, \quad x(0)=x_0,
\end{equation*}
\notag
$$
where $x \in \mathbb R^k$ is the phase vector, $u$ and $v$ are the control actions, and $U=\{u_1,\dots,u_m\}$, $u_i \in \mathbb R^l$, $i=1, \dots, m$. The set $V \subset \mathbb R^s$ is compact. The function $f\colon \mathbb R^k \times U \to \mathbb R^k$ is a Lipschitz function with respect to $x$ for any $u \in U$. The function $g\colon \mathbb R^k \times V \to \mathbb R^k$ is a Lipschitz function with respect to its variables. More precisely, there are positive numbers $L_1$ and $L_2$ such that
$$
\begin{equation*}
\|f(x_1, u_i) - f(x_2, u_i)\| \leqslant L_1\|x_1 - x_2\|, \qquad x_1, x_2 \in \mathbb R^k, \quad i=1, \dots, m,
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\|g(x_1, v_1) - g(x_2, v_2)\| \leqslant L_2\bigl(\|x_1 - x_2\| + \|v_1 - v_2\|\bigr), \qquad x_1, x_2 \in \mathbb R^k, \quad v_1, v_2 \in V.
\end{equation*}
\notag
$$
From now on, the norm is assumed to be Euclidean. A partition $\sigma$ of the interval $[0, T]$ is understood as a finite partition $\{\tau_q\}_{q=0}^{\eta},$ where $0=\tau_0 < \tau_1 < \tau_2 < \dots < \tau_{\eta}=T$. Definition 1. A piecewise constant strategy $W$ of the pursuer $P$ on the interval $[0, T]$ is a pair $(\sigma, W_{\sigma}),$ where $\sigma$ is a partition of $[0, T]$ and $W_{\sigma}$ is a family of mappings $d_r$, $r=0, 1, \dots, \eta - 1$, that take $(\tau_r, x(\tau_r))$ to constant controls $\overline u_r(t) \equiv \overline u_r \in U$, $t \in [\tau_r, \tau_{r+1})$. A control of the evader is understood as an arbitrary measurable function $v$: $[0, \infty) \to V$. Definition 2. $\varepsilon$-capture takes place in the game $\Gamma(x_0)$ if there is ${T>0}$ such that for any $\widehat\varepsilon > 0$ there exists a piecewise constant strategy $W$ of the pursuer $P$ on the interval $[0, T]$ such that for any admissible control $v(\,\cdot\,)$ of the evader $\|x(\tau)\| < \widehat\varepsilon$ for some $\tau \in [0, T]$. The pursuer aims at $\varepsilon$-capture. The evader aims at avoiding it. Definition 3 (see [16]). A set of vectors $a_1, \dots, a_n \in \mathbb R^k$ is called a positive basis if for any point $\xi \in \mathbb R^k$ there exist numbers $\mu_1, \dots, \mu_n \geqslant 0$ such that $\xi=\sum_{i=1}^{n}{\mu_i a_i}$. We introduce the following notation: $\operatorname{Int} A$ is the interior of the set $A$; $\operatorname{co} A$ is the convex hull of the set $A$; $O_{\varepsilon}(x)$ is the open ball of radius $\varepsilon$ with centre $x$; $D_{\varepsilon}(x)$ is the closed ball of radius $\varepsilon$ with centre $x$. The following capture theorem is true. Theorem 1 (see [27]). Assume that $f(0, u_1), \dots, f(0, u_m)$ form a positive basis and $-g(0, V) \subset \operatorname{Int} (\operatorname{co} \{f(0, u_1), \dots, f(0, u_m)\})$. Then there is $\varepsilon_0 \,{>}\, 0$ such that for any $x_0 \in O_{\varepsilon_0}(0)$ $\varepsilon$-capture takes place in the game $\Gamma(x_0)$. It was proved in [28] that for any $x_0 \in O_{\varepsilon_0}(0)$ ($\varepsilon_0$ is from Theorem 1) $\varepsilon$-capture takes place in time ${\|x_0\|}/{\alpha(\|x_0\|)}$, where
$$
\begin{equation*}
\alpha(r)=\min_{x \in D_{r}(0)}\min_{\|p\|=1}\min_{v \in V}\max_{i=1, \dots, m}\bigl\langle f(x, u_i) + g(x, v), p\bigr\rangle.
\end{equation*}
\notag
$$
To construct the strategy of the pursuer $P$ it suffices to use a partition with fixed step size. We introduce the auxiliary controlled system
$$
\begin{equation}
\dot y=w, \qquad \|w\| \leqslant \rho, \quad y(0)=x_0,
\end{equation}
\tag{2.1}
$$
where $w, y \in \mathbb R^k$. Admissible controls of the system (2.1) are assumed to be measurable functions $w(t)$, $\|w(t)\| \leqslant \rho$, $t \geqslant 0$. Let $\varepsilon_0$ be from Theorem 1. Then the following result holds. Theorem 2. Assume that $f(0, u_1), \dots, f(0, u_m)$ form a positive basis; Proof. According to the proof of Theorem 1, for all $x \in D_{R}(0)$ we have
$$
\begin{equation*}
-g(x, V) \subset \operatorname{Int} \bigl(\operatorname{co} \{f(x, u_1), \dots, f(x, u_m)\}\bigr).
\end{equation*}
\notag
$$
Let $b_0, b_1 \in O_{R}(0)$ and $b_0 \in D_{\widehat r}(b_1) \subset D_{R}(0)$ for some $\widehat r > 0$. Then Theorem 1 is true when the origin is taken to the point $b_1$, that is, the original system can be taken arbitrarily closely to $b_1$ from the point $b_0$ in finite time for any admissible actions of the evader. According to [28], we can take the finite time to be $\|b_1 - b_0\| / \alpha(R)$ and consider partitions of fixed step size. Note that in this case if $\|x(\tau_j) - b_1\| \leqslant \varepsilon$, where $\varepsilon$ is the radius of the target neighbourhood, then $\|x(t) - b_1\| < \varepsilon$ for all $t \in (\tau_j, \tau_\eta]$.
Fix $\delta > 0$. Since the function $y(t)$, $t \in [0, T]$, is continuous and $\|y(t)\| < R$ for all $t \in [0, T]$, we have
$$
\begin{equation}
\max\{r \geqslant 0 \, | \, y(t) + D_{r}(0) \subset D_{R}(0), \, t \in [0, T]\} \doteq \delta_1 > 0.
\end{equation}
\tag{2.2}
$$
Let $q$ be an arbitrary natural number such that
$$
\begin{equation*}
\frac{\rho T}{q} \leqslant \frac{\min\{\delta, \delta_1\}}{4}.
\end{equation*}
\notag
$$
We introduce the following notation: $\widehat\delta=\rho T/q$, $\Delta=T/q$, $\widehat\varepsilon=\widehat\delta / q$, $t_0=0$, $t_1=\Delta$, $t_2=2\Delta, \dots, t_q=q\Delta=T$.
Let $\xi_1=y(t_1)$ be the target point on the interval $[t_0, t_1)$. Then there exists a piecewise constant strategy of the pursuer with fixed step size of the partition such that $x(t_1) \in \xi_1 + D_{\widehat\varepsilon}(0)=y(t_1) + D_{\widehat\varepsilon}(0)$. Furthermore, let $\xi_2$ denote the closest point to $x(t_1)$ on the set $y(t_2) + D_{\widehat\varepsilon}(0)$. Then $\|x(t_1) - \xi_2\| \leqslant \Delta\rho$. We assume that $\xi_2$ is the target point on the interval $[t_1, t_2)$. There exists a piecewise constant strategy of the pursuer with fixed partition step size such that $x(t_2) \in \xi_2 + D_{\widehat\varepsilon}(0) \subset y(t_2) + D_{2\widehat\varepsilon}(0)$. We let $\xi_3$ denote the closest point to $x(t_2)$ on the set $y(t_2) + D_{2\widehat\varepsilon}(0)$. Similarly, $\|x(t_2) - \xi_3\| \leqslant \Delta\rho$, $\xi_3$ is the target point, and we bring the system to the set $\xi_3 + D_{\widehat\varepsilon}(0) \subset y(t_3) + D_{3\widehat\varepsilon}(0)$, and so on. At the last step we obtain $x(T)=x(t_q) \in y(t_q) + D_{q\widehat\varepsilon}(0)=y(T) + D_{\widehat\delta}(0)$. Since $x(t_l) \in y(t_l) + D_{\widehat\delta}(0)$, $x(t_{l+1}) \in y(t_{l+1}) + D_{\widehat\delta}(0)$ and $\|y(t_{l+1}) - y(t_l)\| \leqslant \widehat\delta$ for all $l=0, \dots, q-1$, we have $x(t) \in y(t_{l+1}) + D_{3\widehat\delta}(0)$ by [27]. Thus, by virtue of (2.2) we have
$$
\begin{equation*}
\|x(t) - y(t)\| \leqslant 4\widehat\delta \leqslant \delta \quad \text{for all } t \in [t_l, t_{l+1}]
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\|x(t) - y(t)\| \leqslant \delta_1 \quad \text{for all } t \in [t_l, t_{l+1}]
\end{equation*}
\notag
$$
for each $l=0, \dots, q-1$.
Theorem 2 is proved. § 3. System with the second-order derivativeIn the space $\mathbb R^k$, $k \geqslant 2$, we consider a differential game $\Gamma (x_0, \dot x_0)$ of two players, a pursuer $P$ and an evader $E$. The dynamics in the game is described by the system of differential equations
$$
\begin{equation}
\ddot x=f(x, \dot x, u) + g(x, \dot x, v), \qquad u \in U, \quad v \in V, \quad x(0)=x_0, \quad \dot x(0)=\dot x_0,
\end{equation}
\tag{3.1}
$$
where $x \in \mathbb R^k$ is the phase variable; $u$ and $v$ are the control actions, and $U=\{u_1, \dots, u_m\}$, $u_i \in \mathbb R^l$, $i=1, \dots, m$. The set $V \subset \mathbb R^s$ is compact. For each $u \in U$ the function $f\colon \mathbb R^k \times \mathbb R^k \times U \to \mathbb R^k$ is jointly Lipschitz continuous with respect to the variables $x$ and $\dot x$, while the function $g\colon \mathbb R^k \times \mathbb R^k \times V \to \mathbb R^k$ is a Lipschitz function with respect to the all of its variables, that is, there are positive numbers $L_1$ and $L_2$ such that
$$
\begin{equation*}
\begin{gathered} \, \|f(x_1, \dot x_1, u_i) - f(x_2, \dot x_2, u_i)\| \leqslant L_1(\|x_1 - x_2\| + \|\dot x_1 - \dot x_2\|), \\ x_1, x_2, \dot x_1, \dot x_2 \in \mathbb R^k, \qquad i=1, \dots, m, \end{gathered}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\begin{gathered} \, \|g(x_1, \dot x_1, v_1) - g(x_2, \dot x_2, v_2)\| \leqslant L_2(\|x_1 - x_2\| + \|\dot x_1 - \dot x_2\| + \|v_1 - v_2\|), \\ x_1, x_2, \dot x_1, \dot x_2 \in \mathbb R^k, \qquad v_1, v_2 \in V. \end{gathered}
\end{equation*}
\notag
$$
A partition $\sigma$ of the interval $[0, T]$ is understood as a finite partition $\{\tau_q\}_{q=0}^{\gamma},$ where $0=\tau_0 < \tau_1 < \tau_2 < \dots < \tau_{\gamma}=T.$ Definition 4. A piecewise constant strategy $W$ of the pursuer $P$ on an interval $[0, T]$ is a pair $(\sigma, W_{\sigma}),$ where $\sigma$ is a partition of $[0, T]$ and $W_{\sigma}$ is a family of mappings $d_r, r=0, 1, \dots, \gamma - 1,$ that take $(\tau_r, x(\tau_r), \dot x(\tau_r))$ to constant controls $\overline u_r(t) \equiv \overline u_r \in U$, $t \in [\tau_r, \tau_{r+1})$. A control of the evader is understood as an arbitrary measurable function $v$: $[0, \infty) \to V$. Definition 5. $\varepsilon$-capture takes place in the game $\Gamma(x_0, \dot x_0)$ if there is $T > 0$ such that for any $\widehat\varepsilon > 0$ there exists a piecewise constant strategy $W$ of the pursuer $P$ on the interval $[0, T]$ such that for any admissible control $v(\,\cdot\,)$ of the evader $E$ the inequality $\|x(\tau)\| < \widehat\varepsilon$ holds for some $\tau \in [0, T]$. The pursuer aims at $\varepsilon$-capture, whereas the evader aims at avoiding it. In [29] a capture theorem for this problem was proved in the case of a system of the form
$$
\begin{equation*}
\ddot x=f(x, u) + g(x, v), \qquad u \in U, \quad v \in V, \quad x(0)=x_0, \quad \dot x(0)= \dot x_0.
\end{equation*}
\notag
$$
Theorem 3 (see [29]). Assume that $f(0, u_1), \dots, f(0, u_m)$ form a positive basis and $-g(0, V) \,{\subset} \operatorname{Int} (\operatorname{co} \{f(0, u_1), \dots, f(0, u_m)\})$. Then there exist $\varepsilon_0 \,{>}\, 0$, $\theta > 0$ and ${T > 0}$ such that for any initial positions $x_0$ and $\dot x_0$ satisfying $\|x_0\| + \theta\|\dot x_0\| < \varepsilon_0$ $\varepsilon$-capture takes place in the game $\Gamma(x_0, \dot x_0)$ within time $T$. We will also use the approach from the proof of this theorem to obtain solvability conditions for the pursuit problem with dynamics (3.1). In addition, by analogy with [28], it suffices that the pursuer chooses strategies with fixed step size of the partition of the time interval. The following theorem is true. Theorem 4. Assume that $f(0, 0, u_1),\dots,f(0, 0, u_m)$ form a positive basis, and let $-g(0, 0, V) \,{\subset} \operatorname{Int} (\operatorname{co} \{f(0, 0, u_1), \dots, f(0, 0, u_m)\})$. Then there exist $\varepsilon_0 \,{>}\, 0$, $\theta > 0$ and $T > 0$ such that for any initial positions $x_0$ and $ \dot x_0$ satisfying $\|x_0\| + \theta\|\dot x_0\| < \varepsilon_0$ $\varepsilon$-capture takes place in the game $\Gamma(x_0, \dot x_0)$ within time $T$. In addition, it suffices that the pursuer uses a partition with fixed step size of the time interval to construct the strategy. Proof. The scheme of the proof and general arguments are similar to the proof of Theorem 3.
$1^0$. We show that there exist $\overline\alpha > 0$ and $\varepsilon_0 > 0$ such that for any points $x, \dot x \in D_{\varepsilon_0}(0)$ and any vector $p \in \mathbb R^k$, $\|p\|=1$, there exists $i \in \{1, \dots, m\}$ such that
$$
\begin{equation*}
\bigl\langle f(x, \dot x, u_i) + g(x, \dot x, v), p\bigr\rangle \geqslant \overline\alpha
\end{equation*}
\notag
$$
for any $v \in V$. This inequality holds by the Lipschitz property of the functions $f$ and $g$ in the above sense and the properties of a positive basis (see [16]). That is, for some $\varepsilon_0 > 0$ and any $x, \dot x \in D_{\varepsilon_0}(0)$ it is true that
$$
\begin{equation*}
-g(x, \dot x, V) \subset \operatorname{Int} \bigl(\operatorname{co} \{f(x, \dot x, u_1), \dots, f(x, \dot x, u_m)\}\bigr).
\end{equation*}
\notag
$$
Consequently, owing to the properties of a positive basis (see [16]), the vectors
$$
\begin{equation*}
\bigl\{f(x, \dot x, u_1) + g(x, \dot x, v), \dots, f(x, \dot x, u_m) + g(x, \dot x, v)\bigr\}
\end{equation*}
\notag
$$
form a positive basis for any $v \in V$. Hence there are $\widehat x, \widetilde x \in D_{\varepsilon_0}(0)$, $\widehat p \in \mathbb R^k$, $\|\widehat p\|=1$, $\widehat v \in V$ and $\widehat i \in \{1, \dots, m\}$ such that
$$
\begin{equation*}
\min_{x, \dot x \in D_{\varepsilon_0}(0)}\min_{\|p\|=1}\min_{v \in V}\max_{i=1, \dots, m}\bigl\langle f(x, \dot x, u_i) + g(x, \dot x, v), p\bigr\rangle = \bigl\langle f(\widehat x, \widetilde x, u_{\widehat i}) + g(\widehat x, \widetilde x, \widehat v), \widehat p\bigr\rangle.
\end{equation*}
\notag
$$
It follows that $\overline\alpha=\langle f(\widehat x, \widetilde x, u_{\widehat i}) + g(\widehat x, \widetilde x, \widehat v), \widehat p\rangle > 0$.
$2^0$. We introduce some further notation. Since $f$ and $g$ are Lipschitz functions, the maximum
$$
\begin{equation*}
\max_{x, \dot x \in D_{2\varepsilon_0}(0), u \in U, v \in V}{\|f(x, \dot x, u) + g(x, \dot x, v)\|}=\overline D
\end{equation*}
\notag
$$
is attainable. Let We define a number $h$ by
$$
\begin{equation*}
h=\min\biggl\{\frac{\overline\alpha}{4(L_1 + L_2)}, \varepsilon_0\biggr\}.
\end{equation*}
\notag
$$
Let $\overline x, \dot{\overline x} \in D_{\varepsilon_0}(0)$, $p \in \mathbb R^k$, $\|p\|= 1$, $x \in D_{h}(\overline x)$, $\dot x \in D_{h}(\dot{\overline x})$, $v \in V$ and
$$
\begin{equation*}
\max_{u \in U}\bigl\langle f(\overline x, \dot{\overline x}, u), p\bigr\rangle = \bigl\langle f(\overline x, \dot{\overline x}, \overline u), p\bigr\rangle.
\end{equation*}
\notag
$$
Note that, by virtue of part $1^0$,
$$
\begin{equation*}
\bigl\langle f(\overline x, \dot{\overline x}, \overline u) + g(\overline x, \dot{\overline x}, v), p\bigr\rangle \geqslant \overline\alpha
\end{equation*}
\notag
$$
for any $v \in V$. Similarly to the proof of Theorem 3 we obtain the estimate
$$
\begin{equation*}
\begin{aligned} \, \bigl\langle f(x, \dot x, \overline u) + g(x, \dot x, v), p\bigr\rangle &\geqslant \overline\alpha - L_1(\|x - \overline x\| + \|\dot x - \dot{\overline x}\|) - L_2(\|x - \overline x\| + \|\dot x - \dot{\overline x}\|) \\ &\geqslant \overline\alpha - 2h(L_1 + L_2). \end{aligned}
\end{equation*}
\notag
$$
This yields the inequality
$$
\begin{equation*}
\bigl\langle f(x, \dot x, \overline u) + g(x, \dot x, v), p\bigr\rangle \geqslant \frac{\overline\alpha}{2}=\alpha
\end{equation*}
\notag
$$
for any $\overline x, \dot{\overline x} \in D_{\varepsilon_0}(0)$, $p \in \mathbb R^k$, $\|p\|=1$, $x \in D_{h}(\overline x)$, $\dot x \in D_{h}(\dot{\overline x})$ and $v \in V$.
$3^0$. In this part we specify the step size of the partition of the time interval. Fix some $\delta$, $0 < \delta \leqslant \varepsilon_0$. We choose the step size of the partition to be
$$
\begin{equation*}
\Delta=\min\biggl\{\frac{\alpha\delta}{D^2}, \frac{h}{D}\biggr\}.
\end{equation*}
\notag
$$
Note that, according to part $2^0$, if $x(0), \dot x(0) \in D_{\varepsilon_0}(0)$ and $t \in [0, \Delta]$, then
$$
\begin{equation*}
\begin{gathered} \, \|x(t) - x(0)\| \leqslant h, \qquad \|\dot x(t) - \dot x(0)\| \leqslant h, \\ \|f(x(t), \dot x(t), u(t))+ g(x(t), \dot x(t), v(t))\| \leqslant D\quad\text{and} \quad \|\dot x(t)\| \leqslant D \end{gathered}
\end{equation*}
\notag
$$
for any admissible controls of the players.
$4^0$. We estimate how the velocity function $\dot x(\,\cdot\,)$ approaches zero on one step of the partition. Without loss of generality we can consider only the interval $[0, \Delta)$. We choose the value of the control of the pursuer as follows. When $\dot x(0)=0$, $\overline u_0 \in U$ is arbitrary. When $\dot x(0) \neq 0$, we introduce the notation $p=-\dot x(0) / \|\dot x(0)\|$ and choose $\overline u_0 \in U$ on the basis of the equality
$$
\begin{equation*}
\max_{u \in U}\bigl\langle f(x(0), \dot x(0), u), p\bigr\rangle = \bigl\langle f(x(0), \dot x(0), \overline u_0), p\bigr\rangle.
\end{equation*}
\notag
$$
Thus, by parts $2^0$ and $3^0$ we have
$$
\begin{equation*}
\bigl\langle f(x(t), \dot x(t), \overline u_0) + g(x(t), \dot x(t), v), p\bigr\rangle \geqslant \alpha
\end{equation*}
\notag
$$
for any $t \in [0, \Delta)$ and $v \in V$.
Let $t \in [0, \Delta)$. We estimate the squared velocity norm as follows:
$$
\begin{equation*}
\begin{aligned} \, \|\dot x(t)\|^2 &= \biggl\| \dot x(0) + \int_{0}^{t}\bigl(f(x(s), \dot x(s), \overline u_0) + g(x(s), \dot x(s), v(s))\bigr)\, ds\biggr \|^2 \\ &= \|\dot x(0)\|^2 + \biggl \|\int_{0}^{t}\bigl(f(x(s), \dot x(s), \overline u_0) + g(x(s), \dot x(s), v(s))\bigr)\, ds\biggr \|^2 \\ &\qquad+ 2\int_{0}^{t}\bigl\langle f(x(s), \dot x(s), \overline u_0) + g(x(s), \dot x(s), v(s)), \dot x(0) \bigr\rangle \, ds \\ &\leqslant \|\dot x(0)\|^2 + D^2t^2 - 2t\alpha\|\dot x(0)\|. \end{aligned}
\end{equation*}
\notag
$$
Now we estimate the trinomial $A=\|\dot x(0)\|^2 + D^2t^2 - 2t\alpha\|\dot x(0)\|$.
If $\|\dot x(0)\| \geqslant \delta$, then
$$
\begin{equation*}
\begin{aligned} \, \Delta \leqslant \frac{\alpha\delta}{D^2} &\quad\Longrightarrow\quad D^2t^2 - t\alpha\|\dot x(0)\| \leqslant t(D^2\Delta - \alpha\delta) \leqslant 0 \\ & \quad\Longrightarrow\quad A \leqslant \|\dot x(0)\|^2 - t\alpha\|\dot x(0)\| \leqslant \biggl( \|\dot x(0)\| - \frac{t\alpha}{2} \biggr)^2. \end{aligned}
\end{equation*}
\notag
$$
If $\|\dot x(0)\| \leqslant \delta$, then $A$ attains its maximum for $\|\dot x(0)\|=0$ or $\|\dot x(0)\|=\delta$. Note that $D \geqslant 2\alpha$ by part $2^0$. Now, if $\|\dot x(0)\|=0$, then
$$
\begin{equation*}
A=D^2t^2 \leqslant D^2\Delta^2 \leqslant \frac{\alpha^2\delta^2}{D^2} \leqslant \frac{\delta^2}{4}.
\end{equation*}
\notag
$$
If $\|\dot x(0)\|=\delta$, then
$$
\begin{equation*}
A=\delta^2 + D^2t^2 - 2t\alpha\delta \leqslant \delta^2 + t(D^2\Delta - 2\alpha\delta) \leqslant \delta^2.
\end{equation*}
\notag
$$
Hence, if $\|\dot x(0)\| \geqslant \delta$, then $\|\dot x(\Delta)\| \leqslant \|\dot x(0)\| -{\Delta\alpha}/{2}$. If $\|\dot x(0)\| < \delta$, then $\|\dot x(t)\| \leqslant \delta$ for all $t \in [0, \Delta]$. $5^0$. In this part we construct a strategy to bring the function $\dot x(\,\cdot\,)$ to $D_{\delta}(0)$ and estimate the time. On each interval $[\tau_i, \tau_{i+1})$, $i=0, \dots, \eta - 1$, we choose the control of the pursuer according to $4^0$, where $x(\tau_i)$ and $\dot x(\tau_i)$ are used instead of the vectors $x(0)$ and $\dot x(0)$, respectively. We choose a nonnegative integer $\eta$ such that
$$
\begin{equation*}
\frac{\Delta\alpha\eta}{2} < \|\dot x(0)\| \leqslant \frac{\Delta\alpha(\eta+1)}{2},
\end{equation*}
\notag
$$
that is,
$$
\begin{equation*}
\eta=\biggl[\frac{2\|\dot x(0)\|}{\Delta\alpha}\biggr].
\end{equation*}
\notag
$$
Therefore, by the estimates in part $4^0$ we have $\|\dot x(\tau_\eta)\| \leqslant \delta$. In fact, if $\|\dot x(\tau_\eta)\| > \delta$, then by virtue of these estimates $\|\dot x(\tau_i)\| > \delta$ for $i=0, \dots, \eta - 1$. Furthermore, $D\Delta \leqslant \delta/2$. Thus, $\|\dot x(\tau_{\eta+1})\| > \delta/2$ and $\|\dot x(\tau_{\eta+1})\| \leqslant \|\dot x(0)\| - \Delta\alpha(\eta+1)/2$, that is, we arrive at a contradiction.
We estimate $\tau_\eta$ as follows:
$$
\begin{equation*}
\tau_\eta=\eta\Delta \leqslant \frac{2\|\dot x(0)\|\Delta}{\Delta\alpha}=\frac{2\|\dot x(0)\|}{\alpha}.
\end{equation*}
\notag
$$
Hence, if ${\|x(0)\| + \tau_\eta\|\dot x(0)\|} \,{<}\, \varepsilon_0$, then $\|x(t)\| \,{<}\, \varepsilon_0$ for all $t \in [0, \tau_\eta]$. $6^0$. Let $\zeta \in D_{\varepsilon_0/3}(0)$ and $\dot x(\tau_q) \in D_{2\varepsilon_0 / 3}(\zeta)$, $q \geqslant 0$. We show that the value of $\dot x(\,\cdot\,)$ can be brought arbitrarily closely to $\zeta$ from the position $\dot x(\tau_q)$ by some moment of time $\overline t$. To choose the control vector of the pursuer in part $4^0$ we use the vector $p=\zeta - \dot x(\tau_q)$, so that the target point is $\zeta$ instead of $0$. At time $\tau_{q+1}$ we have $p=(\zeta - \dot x(\tau_{q+1}))/\|\zeta-\dot x(\tau_{q+1})\|$, and so on. By part $4^0$ we have the inclusion ${\dot x(t) \in D_{2\varepsilon_0 / 3}(\zeta)}$, that is, $\dot x(t) \in D_{\varepsilon_0}(0)$ for all $t \in [t_q, \overline t]$. Similarly to part $5^0$ we infer the estimate
$$
\begin{equation*}
\overline t - t_q \leqslant \frac{2\|\zeta - \dot x(\tau_q)\|}{\alpha}.
\end{equation*}
\notag
$$
$7^0$. In this part we construct a continuation of the strategy in part $5^0$ that brings the function $x(\,\cdot\,)$ in finite time to any predetermined neighbourhood of zero. Assume that the procedure in part $5^0$ has been performed, so that $\|\dot x(\tau_\eta)\| \leqslant \delta$. We introduce the notation $\varphi=-x(\tau_\eta)/\|x(\tau_\eta)\|$. From now on we take $\delta \leqslant \varepsilon_0/3$ such that $\varepsilon_0/(3\delta)=\mu \in \mathbb N$. In addition, we assume that $\|x(t)\| \leqslant \varepsilon_0$, $t > \tau_\eta$. In the proof below we obtain conditions guaranteeing that this inequality holds. Furthermore, regarding $\dot x(\tau_\eta)$ as the initial position, according to part $6^0$ we can bring the function $\dot x(\,\cdot\,)$ to the ball $D_{\delta}(\delta\varphi)$ in time $\Delta_1 \,{\leqslant}\, 4\delta / \alpha$, that is, by the moment of time $\tau_\eta\,{+}\, \Delta_1$. Thereafter, regarding $\dot x(\tau_\eta + \Delta_1)$ as the initial position, we bring the function $\dot x(\,\cdot\,)$ to the ball $D_{\delta}(2\delta\varphi)$ in time $\Delta_2 \leqslant 4\delta / \alpha$, that is, by the moment of time $\tau_\eta + \Delta_1 + \Delta_2$. We continue this procedure until $\dot x(\,\cdot\,)$ occurs in the ball $D_{\delta}(\mu\delta\varphi)=D_{\delta}(\varphi\varepsilon_0/3)$. This happens by the moment of time $\tau_\eta + \Delta_1 + \dots + \Delta_\mu$. Here $\Delta_i \leqslant 4\delta / \alpha$, $i=1, \dots, \mu$. After that we choose $\varphi\varepsilon_0/3$ as the target point until the end of the game. Hence, by part $4^0$ we have $\dot x(t) \in D_{\delta}(\varphi\varepsilon_0/3)$, $t \geqslant \tau_\eta + \Delta_1 + \dots + \Delta_\mu$. We introduce the representation $\dot x(t)=\beta(t)\varphi + \psi(t)$, where $\beta(t)=i\delta$ for $t \in \bigl[\tau_\eta + \sum_{j=1}^{i - 1}{\Delta_j}, \tau_\eta + \sum_{j=1}^{i}{\Delta_j}\bigr)$, $i=1, \dots, \mu$, and $\beta(t)=\varepsilon_0/3$ for $t \geqslant {\tau_\eta + \Delta_1 + \cdots + \Delta_\mu}$. Note that $\|\psi(t)\| \leqslant 2\delta$, $\tau_\eta \leqslant t \leqslant \tau_\eta + \Delta_1 + \dots + \Delta_\mu$, and $\|\psi(t)\| \leqslant \delta$ for $t > \tau_\eta + \Delta_1 + \dots + \Delta_\mu$. Thus, there is unique $\widehat t \geqslant 0$ such that
$$
\begin{equation*}
x(\tau_\eta) + \int_{\tau_\eta}^{\tau_\eta + \widehat t}\beta(t)\varphi\, ds=0.
\end{equation*}
\notag
$$
Since $\|x(\tau_\eta)\| \leqslant \varepsilon_0$, it is true that $\widehat t \leqslant \Delta_1 + \dots + \Delta_\mu + 3$.
If $\widehat t \leqslant \Delta_1$, then
$$
\begin{equation*}
\begin{aligned} \, \|x(\tau_\eta + \widehat t)\| &= \biggl\|x(\tau_\eta) + \int_{\tau_\eta}^{\tau_\eta + \widehat t}\beta(t)\varphi\, ds + \int_{\tau_\eta}^{\tau_\eta + \widehat t}\psi(s)\, ds\biggr\| \\ &\leqslant \|x(\tau_\eta) + \varphi\delta\widehat t\| + 2\delta\widehat t = (\|x(\tau_\eta)\| - \delta\widehat t) + 2\delta\widehat t = 2\delta\widehat t \leqslant 2\delta\Delta_1. \end{aligned}
\end{equation*}
\notag
$$
If $\Delta_1 < \widehat t \leqslant \Delta_1 + \Delta_2$, then
$$
\begin{equation*}
\|x(\tau_\eta + \widehat t)\| \leqslant (\|x(\tau_\eta)\| - \delta\Delta_1 - 2\delta(\widehat t - \Delta_1)) + 2\delta\Delta_1 + 2\delta(\widehat t - \Delta_1) = \|x(\tau_\eta)\| + \delta\Delta_1.
\end{equation*}
\notag
$$
If $\Delta_1 + \Delta_2 < \widehat t \leqslant \Delta_1 + \Delta_2 + \Delta_2$, then
$$
\begin{equation*}
\|x(\tau_\eta + \widehat t)\| \leqslant \|x(\tau_\eta)\| + \delta\Delta_1 - \delta(\widehat t - \Delta_1 - \Delta_2),
\end{equation*}
\notag
$$
and so on.
Thus, since $\Delta_1 \leqslant 4\delta / \alpha$, for the inequality $\|x(t)\| \leqslant \varepsilon_0$, $t > \tau_\eta$, to hold it suffices that
$$
\begin{equation*}
\|x(\tau_\eta)\| + \frac{4\delta^2}{\alpha} \leqslant \varepsilon_0.
\end{equation*}
\notag
$$
We estimate the norm $\|x(\tau_\eta + \widehat t)\|$ as follows:
$$
\begin{equation*}
\|x(\tau_\eta + \widehat t)\| \leqslant 2\delta\sum_{i=1}^{\mu}{\Delta_i} + \delta\biggl(\widehat t - \sum_{i=1}^{\mu}{\Delta_i}\biggr) \leqslant 2\delta\cdot\frac{4\delta\mu}{\alpha} + 3\delta = \delta\biggl( \frac{8\varepsilon_0}{3\alpha} + 3 \biggr ).
\end{equation*}
\notag
$$
Therefore, since $\delta$ can be chosen to be arbitrarily small, we can bring the trajectory $x(\,\cdot\,)$ arbitrarily closely to zero in time
$$
\begin{equation*}
\tau_\eta + \widehat t \leqslant \frac{2\|\dot x(0)\|}{\alpha} + \Delta_1 + \dots + \Delta_\mu + 3 \leqslant \frac{2\varepsilon_0}{\alpha} + \frac{4\varepsilon_0}{3\alpha} + 3 = \frac{10\varepsilon_0}{3\alpha} + 3=T.
\end{equation*}
\notag
$$
Owing to part $5^0$, if $\|x(0)\| + \tau_\eta\|\dot x(0)\|< \varepsilon_0$, then $\|x(t)\| < \varepsilon_0$ for all ${t \in [0, \tau_\eta]}$. Since $\tau_\eta \,{\leqslant}\, 2\|\dot x(0)\|/\alpha \,{\leqslant}\, 2\varepsilon_0/\alpha$, we arrive at the required quantity $\theta=\max\{2\varepsilon_0/\alpha, 1\}$. Theorem 4 is proved. We introduce the auxiliary controlled system
$$
\begin{equation}
\ddot y=w, \qquad \|w\| \leqslant \rho, \quad y(0)=x_0, \quad \dot y(0)=\dot x_0,
\end{equation}
\tag{3.2}
$$
where $w, y \in \mathbb R^k$. Admissible controls of system (3.2) are assumed to be measurable functions $w(t)$, $\|w(t)\| \leqslant \rho$, $t \geqslant 0$. Let $\varepsilon_0$ and $\theta$ be the same as in Theorem 4. The following result is true. Theorem 5. Assume that $f(0, 0, u_1), \dots, f(0, 0, u_m)$ form a positive basis, let $-g(0, 0, V) \subset \operatorname{Int} (\operatorname{co} \{f(0, 0, u_1), \dots, f(0, 0, u_m)\})$, and fix $x_0, \dot x_0 \in \mathbb R^k$, ${\|x_0\| + \theta\|\dot x_0\| < \varepsilon_0}$, and $T > 0$. Then there exists $\rho > 0$ such that for each admissible control $w(\,\cdot\,)$ of (3.2) satisfying the conditions $\|y(t)\| < \varepsilon_0$ and $\|\dot y(t)\| < \varepsilon_0$ for $t \in [0, T]$, for each $\delta > 0$ there exists a piecewise constant strategy $W$ of the pursuer $P$ on the interval $[0, T]$ such that $\|x(t) - y(t)\| \leqslant \delta$ and $\|\dot x(t) - \dot y(t)\| \leqslant \delta$ for $t \in [0, T]$, for any admissible control $v(\,\cdot\,)$ of the evader $E$. Proof. In the proof of Theorem 4 capture is carried out via controlling the position of the velocity function $\dot x(\,\cdot\,)$. By the proof of Theorem 4 (see part $6^0$), if $b_0, b_1 \in O_{\widehat\varepsilon_0}(0)$ and $b_0 \in D_{\widehat r}(b_1) \subset D_{\varepsilon_0}(0)$ for some $\widehat r > 0$, then the function $\dot x(\,\cdot\,)$ can be brought from the initial point $b_0$ to any neighbourhood of $b_1$ in (finite) time $2\|{b_1 - b_0}\|/\alpha$ for any actions of the evader, provided that $\|x(t)\| \leqslant \varepsilon_0$, $t \in [0, 2\|{b_1 - b_0}\|/\alpha]$. Here $\alpha$ corresponds to the part $2^0$ of the proof of Theorem 4.
We set $\rho=\alpha / 2$. According to the definition of $\theta$ in part $7^0$ of the proof of Theorem 4, for any initial positions satisfying $\|x_0\| + \theta\|\dot x_0\| < \varepsilon_0$ and an arbitrary $T > 0$ there is a control $w(\,\cdot\,)$ of (3.2) such that $\|y(t)\| < \varepsilon_0$, $\|\dot y(t)\| < \varepsilon_0$, and $\|w(t)\| \leqslant \rho$ for all $t \in [0, T]$. In fact, we can choose $w(t)=-\rho\dot y(0)/\|\dot y(0)\|$ for $t \in [0, \|\dot y(0)\|/\rho)$ and $w(t) \equiv 0$ for $t \geqslant \|\dot y(0)\|/\rho$. Then
$$
\begin{equation*}
\begin{aligned} \, \|y(t)\| &\leqslant \|y(0)\| + t\|\dot y(0)\| < \|y(0)\| + \frac{\|\dot y(0)\|\varepsilon_0}{\rho} \\ &= \|y(0)\| + \frac{2\|\dot y(0)\|\varepsilon_0}{\alpha} \leqslant \|x_0\| + \theta\|\dot x_0\| < \varepsilon_0 \end{aligned}
\end{equation*}
\notag
$$
for $t \in [0, \|\dot y(0)\|/\rho]$.
It is true that $\|y(t)\|=\|y(\|\dot y(0)\|/\rho)\|$ for $t > \|\dot y(0)\|/\rho$. Fix $\delta > 0$. Since the functions $y(t)$ and $\dot y(t)$, $t \in [0, T]$, are continuous, ${\|y(t)\| < \varepsilon_0}$ and $\|\dot y(t)\| < \varepsilon_0$ for all $t \in [0, T]$, we have
$$
\begin{equation}
\begin{aligned} \, \max\{r \geqslant 0 \, | \, y(t) + D_{r}(0) \subset D_{\varepsilon_0}(0), \, t \in [0, T]\} \doteq \delta_1 > 0, \\ \max\{r \geqslant 0 \, | \, \dot y(t) + D_{r}(0) \subset D_{\varepsilon_0}(0), \, t \in [0, T]\} \doteq \delta_2 > 0. \end{aligned}
\end{equation}
\tag{3.3}
$$
We set $\delta_3=\delta_1 / T$ and $\delta_4=\delta / T$. When $\|\dot x(t) - \dot y(t)\| \leqslant \delta_3$, it follows that $\|x(t) - y(t)\| \leqslant \delta_1$, that is, $x(t) \in D_{\varepsilon_0}(0)$ for all $t \in [0, T]$. When $\|\dot x(t) - \dot y(t)\| \leqslant \delta_4$, we have $\|x(t) - y(t)\| \leqslant \delta$ for all $t \in [0, T]$. Let $q$ be an arbitrary natural number such that
$$
\begin{equation}
\frac{\rho T}{q} \leqslant \frac{\min\{\delta, \delta_1, \delta_2, \delta_3, \delta_4\}}{4}.
\end{equation}
\tag{3.4}
$$
We introduce the notation $\widehat\delta=\rho T/q$, $\Delta=T/q$, $\widehat\varepsilon=\widehat\delta / q$, $t_0=0$, $t_1=\Delta$, $t_2=2\Delta, \dots$ and $ t_q=q\Delta=T$. Furthermore, similarly to the proof of Theorem 2 we construct a strategy to bring the value of the velocity function $\dot x(\,\cdot\,)$ to a neighbourhood of $\dot y(t)$, that is, from the point $x(t_0)$ to $D_{\widehat\varepsilon}(y(t_1))$, from $x(t_1)$ to $D_{2\widehat\varepsilon}(y(t_2))$, and so on. By (3.3) and (3.4) we have $\|x(t) - y(t)\| \leqslant \delta$, $\|\dot x(t) - \dot y(t)\| \leqslant \delta$, $x(t) \in D_{\varepsilon_0}(0)$, and $\dot x(t) \in D_{\varepsilon_0}(0)$ for all $t \in [0, T]$. Theorem 5 is proved. Definition 6. Soft $\varepsilon$-capture takes place in the game $\Gamma(x_0, \dot x_0)$ if there exists $T > 0$ such that for any $\widehat\varepsilon > 0$ there exists a piecewise constant strategy $W$ of the pursuer $P$ on the interval $[0, T]$ such that for any admissible control $v(\,\cdot\,)$ of the evader $E$ the inequalities $\|x(\tau)\| < \widehat\varepsilon$ and $\|\dot x(\tau)\| < \widehat\varepsilon$ hold for some $\tau \in [0, T]$. The following corollary is valid. Corollary. Assume that the assumptions of Theorem 5 hold and $\|x_0\|+\theta\|\dot x_0\|<\varepsilon_0$. Then soft $\varepsilon$-capture takes place in the game $\Gamma(x_0, \dot x_0)$. Proof. It suffices to constrain the control $w(\,\cdot\,)$ of the auxiliary system (3.2) in the proof of Theorem 5 additionally as follows: $y(T)=0$ and $\dot y(T)= 0$. A moment of time $T \geqslant 0$ of this kind and a control of the auxiliary system always exist. We introduce the notation $t_1=\|\dot y(0)\|/\rho$, $t_2=t_1 + \sqrt{\|y(t_1)/\rho\|}$ and $t_3=t_2 + \sqrt{\|y(t_1)/\rho\|}$. We specify the control as follows:
$$
\begin{equation*}
\begin{gathered} \, w(t)=\frac{-\rho\dot y(0)}{\|\dot y(0)\|} \quad \text{for } t \in [0, t_1), \\ w(t)=\frac{-\rho y(t_1)}{\|y(t_1)\|} \quad \text{for } t \in [t_1, t_2) \end{gathered}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
w(t)=\frac{\rho y(t_1)}{\|y(t_1)\|} \quad \text{for } t \in [t_2, t_3].
\end{equation*}
\notag
$$
Then
$$
\begin{equation*}
\begin{aligned} \, y(t_3) &=y(t_1) + (t_3 - t_1)\dot y(t_1) + \int_{t_1}^{t_3}\dot y(s) \, ds \\ &= y(t_1) + \int_{t_1}^{t_2}\frac{-s\rho y(t_1)}{\|\dot y(0)\|} \, ds + \int_{t_2}^{t_3}\biggl(\frac{-y(t_1)}{2} + \frac{s\rho y(t_1)}{\|\dot y(0)\|}\biggr) \, ds \\ &= y(t_1) - \frac{y(t_1)}{2} - \frac{y(t_1)}{2}=0. \end{aligned}
\end{equation*}
\notag
$$
The corollary is proved. § 4. ConclusionsTwo control problems with an obstacle that is the second player in a differential game are considered. The dynamics in the first problem is described by a system of the form $\dot x=f(x, u) + g(x, v)$. It is shown that if the solution of the auxiliary system $\dot y=w$, $\|w\| \leqslant \rho$, is constrained in a certain way, then for any $\delta > 0$ there exists an admissible strategy of the player such that $\|x(t) - y(t)\| \leqslant \delta$ for all $t \in [0, T]$, for any actions of the obstacle. The dynamics in the second problem is described by a system of the form $\ddot x=f(x, \dot x, u) + g(x, \dot x, v)$. It is shown that if the solution of the auxiliary system $\ddot y=w$, $\|w\| \leqslant \rho$, is constrained in a certain way, then for any $\delta > 0$ there exists an admissible strategy of the player such that $\|x(t) - y(t)\| \leqslant \delta$ and $\|\dot x(t) - \dot y(t)\| \leqslant \delta$ for all $t \in [0, T]$ for any actions of the obstacle. Thus, in view of the constraints obtained, when we choose the control of the auxiliary system to be optimal in a certain sense, the original system can move arbitrarily closely to such a solution of the auxiliary system for any actions of the obstacle. 
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\Bibitem{Shc23} |
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