Cycle plus branches

Given an arbitrary connected, single-target assignment graph with a cycle plus some branches, Theorem 4 and Lemma 11 ensure that the whole system will rendezvous in a sequential manner: First, the agents on the cycle will rendezvous and merge into a single stationary agent. The rest of the agents will then merge into the stationary agent. We call this type of rendezvous sequential rendezvous. So far, however, there is no guarantee that the entire system has $\dot{V} < 0$ all the time. Therefore, no equivalent of Proposition 10 can be stated for an arbitrary connected, single-target assignment graph yet, which means that such a system may not be stable in the sense of Lyapunov. In this subsection we show that it is possible to guarantee $\dot{V} < 0$ at all times for identical agents.

Theorem 12   In unit speed pursuit of $n$ Dubins car agents with arbitrary connected, single-target assignment graph, if the agents maintain their targets in the windshields of span $(-\phi, \phi)$ with fixed $\phi$ satisfying

$$0 < \phi < \min \Big\{\dfrac{\pi}{5}, \dfrac{\pi}{n} \Big\},$$ (27)

then all agents remain in a disc dependent on $V_0$ and rendezvous in finite time.

PROOF. We only need to show that $\dot{V} < 0$ always holds under the condition (27), the same arguments from Corollary 9 and Proposition 10 then give us the rest. Lemma 8 and Lemma 11 cover the single cycle and intree assignment graphs, which leaves only the case of a cycle plus some branches (see Fig. 3). For any connected, single-target assignment graph with $n$ agents, there is a single cycle. Assume that there are $k$ agents on that cycle. Let $V_k$ denote the corresponding Lyapunov function for the $k$ -cycle and let $\phi <\pi/n$ . Then for the agents on the cycle,

$$\dot{V}_k = h_k + f_k < - \displaystyle\sum _{i=0}^k \cos \dfrac{... ...sum _{i=0}^k \cos \dfrac{\pi}{k} = k(\cos \dfrac{\pi}{k} - \cos \dfrac{\pi}{n})$$ (28)

holds for $k \ge 5$ . Here $h_k, f_k$ are defined similarly as $h, f$ before. For the agents not on the cycle, we have

$$\dot{\ell}_{i,j} = -\cos \phi_i - \cos (\theta_j + \phi_j) \le 1 - \cos \phi_i \le 1 -\cos \dfrac{\pi}{n}.$$ (29)

Summing up all the terms for the cycle plus $(n-k)$ other agents not on that cycle, we have

$$\dot{V} < k(\cos\dfrac{\pi}{k} - \cos \dfrac{\pi}{n}) + (n - k) (... ...\cos \dfrac{\pi}{n}) = -k(1 - \cos \dfrac{\pi}{k}) + n(1 - \cos \dfrac{\pi}{n})$$ (30)

We want to show that the right hand side above is negative at all times. We may do this by showing that the function,

$$f_1(x) = x(1 - \cos \dfrac{\pi}{x}),$$ (31)

decreases monotonically for $5 \le x < \infty$ . This can be verified by showing that $\dot{f_1} < 0$ on the domain. When $k = 2, 3, 4$ , $\dot{V}_k$ is less than $-1$ by direct computation; hand checking shows that $(\ref{eq:t5x})$ guarantees $(n-k)(1-\cos\phi_i) < 1$ , which in turn guarantees $\dot V < 0$ .