# Rendezvous on the Line with Different Speeds and Markers That Can Be Dropped at Chosen Time
## 3 Problem Formulation
For instance, as in differential games [37; 38; 39], \(f(t)\) and \(g(t)\) may be bound to be solutions of ordinary differential equation. Our approach here is different. We first show that the functions \(f(t)\) and \(g(t)\) that may be optimal have the restricting feature of being of constant derivative between two rendezvous. Then, we enumerate all possible solutions. Notice that pursuit-evasion games are usually modeled as differential games. Compared to pursuit-evasion, our rendezvous game is cooperative and this makes possible the enumeration of all efficient strategies (player I rendezvous with the four agents and what is to be discovered is the order leading to the shorter average of the rendezvous times). The rendezvous value \(R(f,g)\) is defined to be the average value  
\[R(f,g)=\frac{1}{4}\bigg{(}t^{1}+t^{2}+t^{3}+t^{4}\bigg{)}.\]  
Finally, the rendezvous value of the game is defined by  
\[R=\min\bigg{\{}R(f,g):f\in\mathcal{F}(v),g\in\mathcal{F}\bigg{\}}. \tag{6}\]  
A first remark that simplifies the problem is that the functional spaces \(\mathcal{F}(v)\) and \(\mathcal{F}\) may be reduced to consider only functions \(f\in\mathcal{F}(v)\) and \(g\in\mathcal{F}\)_lwhose speed is constant between the rendezvous times_. Indeed, if the speed is not constant, moving at the average speed between rendezvous times leads to the same rendezvous value. Moreover, similarly to Lemma 5.1 of [1], Theorem 16.10 of [26] or Proposition 3 of [23], we have the following result for \(g\in\mathcal{F}\). **Proposition 1**. : _If \(v\leq 1\), then, for the optimal strategies, the function \(g\in\mathcal{F}\) is of constant slope equal to \(\pm 1\) between the rendezvous, i.e., the speed of the fast player is always maximal._  
Figure 1: **(On the left)**:the four agents of player II. Each one corresponds to a particular initial position and direction of the forward move. **(On the right)**: the optimal solution of the game without marker. Player I goes forward for time \(\frac{1}{2}\), then backward for time \(\frac{1}{2}\), then forward for time \(1\) and backward for time \(1\). Player II goes forward for time \(1\), then backward for time \(3\). The rendezvous occur successively with agents \(2,3,1\) and \(4\) at times \(t^{2}=\frac{1}{2}\), \(t^{3}=1\), \(t^{1}=2\) and \(t^{4}=3\). Proof. : Let us assume that player II, whose position is given by function \(g\in\mathcal{F}\) and initial position is known, does not move at maximal speed between rendezvous times \(t_{i-1}<t_{i}\). This means that player II can reach the rendezvous position at a time \(t_{i}-\epsilon\) with \(\epsilon>0\). By moving faster, it may happen that player II rendezvous with player I before time \(t_{i}\), reducing the rendezvous time \(t_{i}\). If not, we modify the trajectory of player II in the following way. After reaching the rendezvous point at time \(t_{i}-\epsilon\), player II continues in the same direction for a period \(\epsilon/2\) and then goes the other way for a period \(\epsilon/2\), back to the rendezvous position at time \(t_{i}\). At time \(t-\epsilon/2\), player I must be at a distance less than \(v\epsilon/2\) from the rendezvous position and because player II is at a distance \(\epsilon/2\) and \(\epsilon/2\geq v\epsilon/2\), the rendezvous must occur before time \(t_{i}\). To summarize, by moving at full speed, player II always reduces the rendezvous time \(t_{i}\). After time \(t_{i}\) player II follows the original strategy and the remaining rendezvous times are not changed. In total, the modified strategy reduces the rendezvous value showing that the original strategy is not optimal. We emphasize that the fast-moving player moves at maximal speed while the slow-moving player can move at any speed in \([0,v]\). **Corollary 1**. : _We assume here that the speed of player I is bounded by \(v\leq 1\) and the speed of player II by \(1\), i.e., \(f\in\mathcal{F}(v)\) and \(g\in\mathcal{F}\). The sets of optimal strategies \((f,g)\) for players I and II, respectively, are given by_  
\[f(t)=