If we define a Nash equilibrium as a fixed point of the best-response mapping, i.e. a strategy $x$ s.t.
$$x\in BR(x),$$
where BR denotes the (set-valued) best-response mapping, then
$$\dot x \in BR(x)-x$$
is generally referred to as best-response dynamics (BR dynamics).
I would like toho obtain the BR dynamics for the following game. I have two utility functions, $u_1(c,s)$ and $u_2(c,s)$. They are derivable. $u_1(c,s)$ refers to Player 1, while $u_2(c,s)$ to Player 2. The strategic leverage of Player 1 is $s$ while the strategic leverage of Player 2 is $c$.
In order to obtain the best response functions and Nash equilibria, I have to solve the system of the following two equations:
$$\frac{\partial }{\partial s}u_1(c,s)=0$$
$$\frac{\partial }{\partial c}u_2(c,s)=0$$
From this I obtain both best response functions, i.e.
$$s=f_1(c)$$
$$c=f_1(s)$$
What is the BR dynamics for this game? Moreover, when can I substitute $\in$ symbol with $=$?
Thanks in advance!
Best Answer
If the best response functions are $s = f_1(c)$ and $c = f_2(s)$. Then, let $x = (s,c)$ and the best response dynamics are, \begin{align} \dot{x} &\in BR(x) - x \\ \begin{pmatrix}\dot{s} \\ \dot{c}\end{pmatrix} &\in \begin{pmatrix} f_1(c) \\ f_2(s)\end{pmatrix} - \begin{pmatrix} s \\ c\end{pmatrix} \end{align}
You can only replace the $\in$ with $=$ if the map $f_1, f_2$ are not set value maps but are single-valued, i.e., there is only 1 best action player 1 can take against player 2's action. For example, if $s$ is restricted between 0 and 1, the utility function for player 1 is $cs$ and $c=0$ then any $s\in(0,1)$ is the best response and therefore isn't single-valued.