 
 
 
 
 
   
 's to be identical. We say that the speed
's to be identical. We say that the speed  of agent
 of agent  is bounded if
 is bounded if 
 for some constants
 for some constants  and
 and  . The velocity
. The velocity  may change over time. When all agents' speeds in a pursuit are bounded, we say the pursuit is a bounded speed pursuit.
 may change over time. When all agents' speeds in a pursuit are bounded, we say the pursuit is a bounded speed pursuit.
 Dubins car agents will rendezvous in finite time if the agents maintain their targets in the windshields of span
 Dubins car agents will rendezvous in finite time if the agents maintain their targets in the windshields of span 
 with
 with
 represent the agents speeds
 represent the agents speeds 
 and define
 and define  , and
, and  as
 as
|  | (33) | 
 ,
, 
 , and the hyperplane
, and the hyperplane 
 from Lemma 7 remain the same. For the
 from Lemma 7 remain the same. For the 
 slice by the hyperplane, applying the method of Lagrange multipliers to
 slice by the hyperplane, applying the method of Lagrange multipliers to  as a function of
 as a function of  's with constraint
's with constraint 
 yields that for all
 yields that for all  ,
,
 's fixed, for
's fixed, for  to take maximum on the
 to take maximum on the 
 slice, for all
 slice, for all  ,
,  
 must take the same value and therefore, must be positive if we keep
 must take the same value and therefore, must be positive if we keep 
 . Let us assume that we pick some
. Let us assume that we pick some 
 , then
, then
|  | (35) | 
 ,
, 
 . Therefore,
. Therefore, 
 , we need
, we need  , which is true if
, which is true if
|  | (37) | 
 ,
, 
 , which gives us (32) as a sufficient condition for
, which gives us (32) as a sufficient condition for 
 on the
 on the 
 slice. On the
 slice. On the 
 slice, we have that
 slice, we have that 
 , which is less than the last expression in (36); therefore, (32) also works for the
, 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.
 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
 can no longer be guaranteed. A simple example is illustrated in Fig. 5. Agent  is at the root and does not move. Suppose agent
 is at the root and does not move. Suppose agent  moves very fast and agents (
 moves very fast and agents ( in the figure) following
 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
 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
, increases. This suggests that  must be positive at some point. However, such a system will still rendezvous. Supposing that the stationary agent is
 must be positive at some point. However, such a system will still rendezvous. Supposing that the stationary agent is  , at least one agent, say
, at least one agent, say  , is assigned to
, is assigned to  . Thus,
. Thus, 
 whenever
 whenever 
 .  Hence, agent
.  Hence, agent  will merge into agent
 will merge into agent  in finite time, and all other agents will eventually follow. We have proved:
 in finite time, and all other agents will eventually follow. We have proved:
 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
 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
 with 
 .
. 
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:
 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:
 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
 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
 with fixed  satisfying (32).
 satisfying (32). 
 
 
 
 
