Decentralized navigation method for a robotic swarm with nonhomogeneous abilities

Abstract

This paper addresses the navigation of a robotic swarm with nonhomogeneous abilities, including sensing range, maximum velocity, and acceleration. With this method, the robotic swarm moves in a two-dimensional plane, and each follower distributedly constructs and maintains local directed connection using only local information to achieve maintenance of global connectivity. We also ensure the swarm is stable when the leader moves at a constant velocity. Validity and effectiveness of the proposed control strategy are shown by theoretical analysis, experiments with real robots, and numerical simulations.

Appendices

Appendix A: Proof of 2nd step in Theorem 2

Now, we show that $\Vert \dot{\varvec{u}}_i(t) \Vert < A_i$ if $\Vert \varvec{u}_{j}(t) \Vert \le U_{n+1}$. Here, we consider $t \ge t_i$ because $\dot{\varvec{u}}_i \equiv \varvec{0}$ at $t < t_i$. Substituting (23) into the time-derivative of (8), the acceleration of agent i is computed as

$$\begin{aligned} \dot{\varvec{u}}_i = \left( \dot{u}_{ir} - u_{i\theta }\omega \right) \varvec{e}_r + \left( \dot{u}_{i\theta } + u_{ir}\omega \right) \varvec{e}_\theta , \end{aligned}$$

(31)

where $\omega$ is defined in (22). From (31), the norm of the acceleration of agent i is calculated as

$$\begin{aligned} \Vert \dot{\varvec{u}}_i \Vert ^2 =\ (\dot{u}_{ir}^2 + \dot{u}_{i\theta }^2) + \omega ^2(u_{ir}^2 + u_{i\theta }^2) + 2\omega (u_{ir}\dot{u}_{i\theta } - u_{i\theta }\dot{u}_{ir}). \end{aligned}$$

(32)

(case 1: If $0 \le r < \rho ')$ Obviously $\Vert \dot{\varvec{u}}_i \Vert = 0 < A_i$.

(case 2: If $\rho ' \le r < r_c)$ From (10),

$$\begin{aligned} \dot{u}_{ir}&= a\dot{r} = a(u_{jr} - u_{ir}), \end{aligned}$$

(33)

$$\begin{aligned} \dot{u}_{i\theta }&= \sigma \dot{u}_{ir} = \sigma a (u_{jr} - u_{ir}). \end{aligned}$$

(34)

Substituting (10), (22), (33), (34) and $\sigma ^2 \le 1$ into (32), we have

$$\begin{aligned} \Vert \dot{\varvec{u}}_i \Vert ^2 < 2a^2&\left\{ \left( u_{jr} - u_{ir}\right) ^2 + \left( u_{j\theta } - u_{i\theta }\right) ^2 \right\} \\ = 2a^2&\bigl \{ \left( u_{jr}^2 + u_{j\theta }^2\right) + \left( 1 +\sigma ^2 \right) u_{ir}^2 \\&-2u_{ir}\left( u_{jr} + \sigma u_{j\theta } \right) \bigr \} \\ \le 2a^2&\left\{ \left( u_{jr}^2+ u_{j\theta }^2\right) + 2u_{ir}^2 + 2u_{ir} \left( \vert u_{jr} \vert + \vert u_{j\theta } \vert \right) \right\} . \end{aligned}$$

This inequality yields

$$\begin{aligned} \Vert \dot{\varvec{u}}_i \Vert ^2 < (1 + \sqrt{2})^2 a^2 U_{n+1}^2, \end{aligned}$$

(35)

since $u_{jr}^2 + u_{j\theta }^2 \le U_{n+1}^2$ , $\vert u_{jr} \vert + \vert u_{j\theta } \vert \le \sqrt{2}U_{n+1}$ using the method of Lagrange multiplier, and $u_{ir} < U' / 2$ from (20). Thus, substituting (7) into (35), we obtain

$$\begin{aligned} \Vert \dot{\varvec{u}}_i \Vert< & {} \left( 1 + \sqrt{2} \right) a U_{n+1} \le \frac{1 + \sqrt{2}}{2 + \sqrt{2}}a \min _{k \in \mathcal {F}}\frac{A_k}{a_k}\\\le & {} \frac{1 + \sqrt{2}}{2 + \sqrt{2}}a\frac{A_i}{a} < A_i. \end{aligned}$$

(case 3: If $r_c \le r \le r_e)$ In this interval of r,

$$\begin{aligned} \dot{u}_{ir}&= a\dot{r} = a(u_{jr} - u_{ir}),\end{aligned}$$

(36)

$$\begin{aligned} \dot{u}_{i\theta }&= -\sigma a (u_{jr} - u_{ir}) \end{aligned}$$

(37)

from (11). Substituting (22), (36), (37) and $\sigma ^2 \le 1$ into (32), we have

$$\begin{aligned} \Vert \dot{\varvec{u}}_i \Vert ^2 =&\ (1 + \sigma ^2)a^2 (u_{jr} - u_{ir})^2 \\&- \frac{2a}{r}(u_{jr} - u_{ir})(u_{j\theta } - u_{i\theta })(\sigma u_{ir} + u_{i\theta })\\&+ \frac{1}{r^2}(u_{j\theta } - u_{i\theta })^2 (u_{ir}^2 + u_{i\theta }^2). \end{aligned}$$

Since $r = \rho ' + u_{ir}/a > u_{ir}/a$ from (11), $u_{ir}\ge u_{i\theta }$, and $\sigma ^2 \le 1$,

$$\begin{aligned} \Vert \dot{\varvec{u}}_i \Vert ^2 <&\ 2a^2 \{ (u_{jr} - u_{ir})^2 \\&+2\vert u_{jr} - u_{ir} \vert \vert u_{j\theta } - u_{i\theta } \vert + (u_{j\theta } - u_{i\theta })^2 \} \\ =&\ 2a^2(\vert u_{jr} - u_{ir} \vert + \vert u_{j\theta } - u_{i\theta } \vert )^2 \\ \le&\ 2a^2(\vert u_{ir} \vert + \vert u_{i\theta } \vert + \vert u_{jr} \vert + \vert u_{j\theta } \vert )^2. \end{aligned}$$

Substituting $\vert u_{ir} \vert + \vert u_{i\theta } \vert \le U' < U_{n+1}$ from (11), and $\vert u_{jr} \vert + \vert u_{j\theta } \vert \le \sqrt{2}U_{n+1}$ as the second case to (32), $\Vert \dot{\varvec{u}}_i \Vert ^2 < (2 + \sqrt{2})^2 a^2 U_{n+1}^2$. Thus, substituting (7) into (35), we obtain

$$\begin{aligned} \Vert \dot{\varvec{u}}_i \Vert< & {} (2 + \sqrt{2})aU_{n+1} \le \frac{2+\sqrt{2}}{2+\sqrt{2}}a \min _{k \in \mathcal {F}}\frac{A_k}{a_k}\\\le & {} \frac{2+\sqrt{2}}{2+\sqrt{2}}a \frac{A_i}{a} = A_i. \end{aligned}$$

Thus, $\Vert \dot{\varvec{u}}_i \Vert$ is smaller than $A_i$.

Appendix B: Proof of Lemma 1 (existence of the equilibrium)

(case 1: If $\rho '< r^\mathrm {eq} < r_c)$The equilibrium satisfies

$$\begin{aligned}&U^* \cos \theta ^\mathrm {eq} - a(r^\mathrm {eq} - \rho ') = 0,\end{aligned}$$

(38)

$$\begin{aligned}&\sigma a(r^\mathrm {eq} - \rho ') - U^* \sin \theta ^\mathrm {eq} = 0 \end{aligned}$$

(39)

from (10), (24) and (25). Since $U^* > 0$ and $r^\mathrm {eq} > \rho '$, $\sin \theta ^\mathrm {eq} \ge 0$ and $\cos \theta ^\mathrm {eq} \ge 0$ must hold, and then $0 \le \theta ^\mathrm {eq} \le \pi / 2$. From (38) and (39), we obtain $\sin \theta ^\mathrm {eq} - \sigma \cos \theta ^\mathrm {eq} = 0$, and $\theta ^\mathrm {eq} = \arctan \sigma$.

Considering $0 \le \theta ^\mathrm {eq} \le \pi / 2$,

$$\begin{aligned} \sin \theta ^\mathrm {eq} = \frac{\sigma }{\sqrt{1 + \sigma ^2}},~~ \cos \theta ^\mathrm {eq} = \frac{1}{\sqrt{1 + \sigma ^2}}. \end{aligned}$$

(40)

Substituting (40) into (38), $r^\mathrm {eq} = \rho ' + \frac{U^*}{\sqrt{1 + \sigma ^2}a}$. Therefore, for $U^* \in \left( 0, \sqrt{1+\sigma ^2}U'/2\right)$, the equilibrium $(r^\mathrm {eq}, \theta ^\mathrm {eq}) = \left( \rho ' + U^*/(\sqrt{1 + \sigma ^2}a) , \arctan \sigma \right)$ continuously exists in $\rho '< r < r_c$, and $r^\mathrm {eq}$ is monotonically increasing for $U^*$.

(case 2: If $r_c \le r^\mathrm {eq} \le r_e)$  The equilibrium satisfies

$$\begin{aligned}&U^* \cos \theta ^\mathrm {eq} - a(r^\mathrm {eq} - \rho ') = 0,\end{aligned}$$

(41)

$$\begin{aligned}&\sigma \left( U' - a(r^\mathrm {eq} - \rho ') \right) - U^* \sin \theta ^\mathrm {eq} = 0 \end{aligned}$$

(42)

from (11), (24) and (25). Since $U^* > 0$ and $r^\mathrm {eq} > \rho '$, $\sin \theta ^\mathrm {eq} \ge 0$ and $\cos \theta ^\mathrm {eq} > 0$ must hold, and then $0 \le \theta ^\mathrm {eq} < \pi / 2$. From (41) and (42), we obtain

$$\begin{aligned} \sin \theta ^\mathrm {eq} + \sigma \cos \theta ^\mathrm {eq} = \frac{\sigma U'}{U^*} \end{aligned}$$

(43)

and a condition of existence of $\theta ^\mathrm {eq}$ is

$$\begin{aligned} U^* \ge \frac{\sigma U'}{\sqrt{1+\sigma ^2}}. \end{aligned}$$

(44)

Since $\frac{\sigma U'}{\sqrt{1+\sigma ^2}} \le \frac{\sqrt{1+\sigma ^2}}{2}U'$ holds for any $\sigma \in [0, 1]$ and $U' \ge 0$, the equilibrium exists continuously in $r_c \le r \le r_e$ for $U^* \in \left[ \sqrt{1+\sigma ^2}U'/2, U' \right]$, which satisfies (44). Since $\theta ^\mathrm {eq}$ ($\ge 0$) is monotonically decreasing for $U^*$ from (43) and $\theta ^\mathrm {eq} = \arctan \sigma$ at $U^* = \sqrt{1+\sigma ^2}U'/2$, we have $\theta ^\mathrm {eq} \le \arctan \sigma$ for $U^* \in \left[ \sqrt{1+\sigma ^2}U'/2, U' \right]$. Moreover, $r^\mathrm {eq}$ is monotonically increasing for $U^*$ in the same way as in the first case.

At $U^* = \sqrt{1+\sigma ^2}U'/2$, $(r^\mathrm {eq}, \theta ^\mathrm {eq})$ is continuous from equations (38), (39), (41) and (42). Thus, this lemma is proved.

Appendix C: Proof of Lemma 1 (stability of the equilibrium)

Consider fixed $U^* \in (0, U']$ and the corresponding equilibrium $(r^\mathrm {eq}, \theta ^\mathrm {eq})$. Jacobi matrix J at $(r^\mathrm {eq}, \theta ^\mathrm {eq})$ is computed as follows:

$$\begin{aligned} J = \left[ \begin{array}{c c} - \, a &{}\quad - \, U^* \sin \theta ^\mathrm {eq} \\ J_{21} &{}\quad -\,\frac{U^*}{r^\mathrm {eq}}\cos \theta ^\mathrm {eq} \end{array} \right] , \end{aligned}$$

where

$$\begin{aligned} J_{21} = {\left\{ \begin{array}{ll} \frac{a\sigma }{r^\mathrm {eq}} ~~~~~ (\rho '< r^\mathrm {eq} < r_c), \\ -\frac{a\sigma }{r^\mathrm {eq}} ~~~ (r_c \le r^\mathrm {eq} \le r_e). \end{array}\right. } \end{aligned}$$

The characteristic equation of J is $\lambda ^2 - (\mathrm {tr}J)\lambda + \mathrm {det}J = 0$, where $\lambda$ is an eigenvalue of J. Since $0 \le \theta ^\mathrm {eq} \le \arctan \sigma \le \pi / 4$ from Lemma 1, $\mathrm {tr}J = -a -\frac{U^*}{r^\mathrm {eq}}\cos \theta ^\mathrm {eq} < 0$, and

$$\begin{aligned} \mathrm {det}J= & {} \frac{aU^*}{r^\mathrm {eq}}\cos \theta ^\mathrm {eq} \pm \frac{aU^*\sigma }{r^\mathrm {eq}}\sin \theta ^\mathrm {eq}\nonumber \\\ge & {} \frac{aU^*}{r^\mathrm {eq}}(\cos \theta ^\mathrm {eq} - \sigma \sin \theta ^\mathrm {eq}) \ge 0 \end{aligned}$$

(45)

are obtained. Here, the last equality of (45) is attained if and only if $\sigma = 1$ and $U^* = U' / \sqrt{2}$. Otherwise, we have $\mathrm {Re}(\lambda ) < 0$ from the theorem of Hurwitz. Moreover, since $(\mathrm {tr}J)^2 - 4\mathrm {det}J >0$, $\lambda$ is a real number, and $\lambda \le 0$.

If $\lambda < 0$, the equilibrium is stable. Otherwise, one of the eigenvalues equals 0. One of the eigenvectors corresponding to $\lambda = 0$ is $[U^* / \sqrt{2}a, -1]^{\mathrm {T}}$. The equilibrium is $(r_0, \theta _0) = (r_c, \pi / 4)$, where $u_{i\theta }$ is not differentiable. $\lambda = 0$ is adopted to the positive direction of r, while $\lambda < 0$ to the negative direction from (45). Thus, we consider the direction of the vector $[\varDelta r, \varDelta \theta ]^{\mathrm {T}} = \epsilon [U^* / \sqrt{2}a, -1]^{\mathrm {T}}$, where $\epsilon > 0$. By the Taylor expansion of $\dot{r}$ and $\dot{\theta }$ around $(r^\mathrm {eq}, \theta ^\mathrm {eq})$, we have

$$\begin{aligned}&\dot{r}(r_0 + \varDelta r, \theta _0 + \varDelta \theta ) = -a\varDelta r - U^* \sin \theta _0 \varDelta \theta \nonumber \\&\qquad \qquad \qquad \qquad \qquad - \frac{U^*\cos \theta _0}{2}(\varDelta \theta )^2 + \dots ,\end{aligned}$$

(46)

$$\begin{aligned}&\dot{\theta }(r_0 + \varDelta r, \theta _0 + \varDelta \theta ) = -\frac{\sigma a}{r_0}\varDelta r\nonumber \\&\qquad \qquad - \frac{U^*\cos \theta _0}{r_0}\varDelta \theta + \frac{\sigma a}{r_0^2}(\varDelta r)^2 \nonumber \\&\qquad \qquad + \frac{U^*\sin \theta _0}{2r_0} (\varDelta \theta )^2 + \frac{U^*\cos \theta _0}{r_0^2}\varDelta r \varDelta \theta + \dots . \end{aligned}$$

(47)

Substituting $[\varDelta r, \varDelta \theta ]^{\mathrm {T}} = \epsilon [U^* / \sqrt{2}a, -1]^{\mathrm {T}}$ into (46) and (46), we have $\dot{r}(r_0 + \varDelta r, \theta _0 + \varDelta \theta ) = -\frac{\epsilon a}{2\sqrt{2}}\varDelta r$, and $\dot{\theta }(r_0 + \varDelta r, \theta _0 + \varDelta \theta ) = -\frac{U^*\epsilon }{2\sqrt{2}r_0}\varDelta \theta$, which show the equilibrium attracts points in the direction of $\epsilon [U^* / \sqrt{2}a, -1]^{\mathrm {T}}$. Therefore, $(r^\mathrm {eq}, \theta ^\mathrm {eq})$ is stable.