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Guaranteed Rendezvous of Agents with Bounded, Varying Speeds
In this section, we generalize the results for identical agents by removing the restriction that requires all
's to be identical. We say that the speed
of agent
is bounded if
for some constants
and
. The velocity
may change over time. When all agents' speeds in a pursuit are bounded, we say the pursuit is a bounded speed pursuit.
Theorem 13
Bounded speed cyclic pursuit of
Dubins car agents will rendezvous in finite time if the agents maintain their targets in the windshields of span
with
|
(32) |
PROOF. For the proof we work with (9) and use the approach in the proof of Lemma 7. The simple polygon case is covered here; the proof for the self-intersecting polygon case then follows that of Lemma 8 similarly, which we do not repeat. Let
represent the agents speeds
and define
, and
as
|
(33) |
The structures,
,
, and the hyperplane
from Lemma 7 remain the same. For the
slice by the hyperplane, applying the method of Lagrange multipliers to
as a function of
's with constraint
yields that for all
,
|
(34) |
Once again, holding
's fixed, for
to take maximum on the
slice, for all
,
must take the same value and therefore, must be positive if we keep
. Let us assume that we pick some
, then
|
(35) |
By the pigeonhole principle, for at least one
,
. Therefore,
|
(36) |
To make
, we need
, which is true if
|
(37) |
One way to satisfy this is to make sure that for each
,
|
(38) |
or equivalently,
|
(39) |
The right side of (39) achieves the global minimum when
, which gives us (32) as a sufficient condition for
on the
slice. On the
slice, we have that
, which is less than the last expression in (36); therefore, (32) also works for the
slice. The finite time guarantee follows the argument from Corollary 9.
Moving to the intree case, when agents have different speeds, no equivalent of Lemma 11 can be stated since
can no longer be guaranteed. A simple example is illustrated in Fig. 5. Agent
is at the root and does not move. Suppose agent
moves very fast and agents (
in the figure) following
barely move. Also assume that all agents are almost colinear. It is straightforward to see that, after a short period of time (the second drawing), the sum of the length of all edges, or
, increases. This suggests that
must be positive at some point. However, such a system will still rendezvous. Supposing that the stationary agent is
, at least one agent, say
, is assigned to
. Thus,
whenever
. Hence, agent
will merge into agent
in finite time, and all other agents will eventually follow. We have proved:
Lemma 14
Bounded speed pursuit of
Dubins car agents with an intree assignment graph will merge into the stationary agent in finite time if the agents maintain their targets in the windshields of span
with
.
Figure 5:
Bounded speed intree pursuit may not have
dot less than zero at all times.
|
Since it is not possible to guarantee
at all times for the bounded speed case, a result like Theorem 12 is out of the question. However, since Theorem 13 and Lemma 14 parallel Theorem 4 and Lemma 11, the argument giving us sequential rendezvous in the identical agents case continues to hold:
Theorem 15
Bounded speed pursuit of
Dubins car agents with arbitrary connected, single-target assignment graph will rendezvous in finite time if the agents maintain their targets in the windshields of span
with fixed
satisfying (32).
Next: Condition on Angular Velocity
Up: Rendezvous Without Coordinates1
Previous: Cycle plus branches
Jingjin Yu
2011-01-18