There are two agents $i=1,2$. Consider the following programm \begin{align} &V_1(x_0) := \max_u \int^\infty_0 e^{-\rho t}F_1(x(t),u(t),v(t))dt\\ &V_2(x_0) := \max_v \int^\infty_0 e^{-\rho t}F_2(x(t),u(t),v(t))dt\\ s.t.~&\dot x(t)=f(x(t),u(t),v(t))\\ &x(0) = x_0 \end{align} where $\rho > 0$ denotes time preference, $V_i(\cdot)$ is the value and $F_i(\cdot)$ an objective function. $x\in X = [0,2]$ is the state variable and $u\in U=[0,1]$ the control of agent 1 and $v\in V=[0,1]$ the control of agent 2 respectively. The state is governed by $f(\cdot)$. The Hamilton-Jacobi-Bellman equation for each agent is given by \begin{align} \rho V_1(x)=\max_u [F(x,u,v^*) + V_1'(x)f(x,u,v^*)],\quad \forall t\in[0,\infty)\\ \rho V_2(x)=\max_v [F(x,u^*,v) + V_2'(x)f(x,u^*,v)],\quad \forall t\in[0,\infty)\\ \end{align}
given the respective maximizers \begin{align} u^* &= \max_u [F(x,u,v^*) + V_1'(x)f(x,u,v^*)]\\ v^* &=\max_v [F(x,u^*,v) + V_2'(x)f(x,u^*,v)] \end{align}
such that the HJBs become \begin{align} \rho V_1(x)=F(x,u^*,v^*) + V_1'(x)f(x,u^*,v^*)\\ \rho V_2(x)=F(x,u^*,v^*) + V_2'(x)f(x,u^*,v^*) \end{align}
Symmetric equilibirum
A symmetric equilibirum is given at $\dot x = 0 \Leftrightarrow f(\tilde x,\tilde u,\tilde v) = 0$ with $\tilde x = 1$ and $\tilde u=\tilde v$ and $V_1(\tilde x) = V_2(\tilde x) =: V(\tilde x)$.
Problem
The equilibirum controls $\tilde u$ and $\tilde v$ can't be determinend with the information at hand. The equation \begin{align} \rho V(\tilde x)=F(\tilde x,\tilde u, \tilde v) + V'(\tilde x)\underbrace{f(\tilde x,\tilde u, \tilde v)}_{=0} \end{align} is true for every $\{(u,v)\in[0,1]\times[0,1]:u=v\}$. That is, we have multiple equilibria.
Select Equilibrium
My idea is (I made this up, haven't read anything about it) that I select the equilibrium associated with the highest value. We can determine $V(\tilde x)$ for all $\{(u,v)\in[0,1]\times[0,1]:u=v\}$. Say $V(\tilde x)$ is monoton increasing in $u$ and $v$, i.e. \begin{align} \lim_{u=v\to 0} V(\tilde x) < \lim_{u=v\to 1} V(\tilde x) \end{align}
Perosnal I would choose the fixed point $(k = 1, u = 1, v = 1)$. I'd like to know if I can motivate it formally as the unique solution.
- Do I select the equilibrium associated with the highest value, by definition of the value function?
- Can you point me to some literature concerning this point?
Motivating example
Let $F_1(x,u,v) = xu^2$ and $F_2(x,u,v) = (2-x)v^2$ with $f(x,u,v) = v-u$. The HJBs read (with $\rho=1$) \begin{align} V_1(x)&=\max_u [xu^2 + V_1'(x)(v^*-u)]\\ V_2(x)&=\max_v [(2-x)v^2 + V_2'(x)(v-u^*)] \end{align}
Maximizers are \begin{align} u^*&= \frac{V'_1(x)}{2x}\in[0,1]\\[2mm] v^*&=\frac{V'_2(x)}{2(2-x)}\in[0,1] \end{align}
In symmetric equilibrium we have $\tilde x = 1$ and $\tilde v = \tilde u \Leftrightarrow \dot x = 0$ which gives \begin{align} V'_1(1) = -V'_2(1) \end{align}
The HJb simplifies to \begin{align} V_1(1)=\left(\frac{V'_1(1)}{2}\right)^2 = \left(\frac{-V'_2(1)}{2}\right)^2 = V_2(1) \end{align}
Since both vlaues are equal in equilibirum we proceed with 1. We know from the control space that \begin{align} 0\leq V'_1(x) \leq 2x \end{align}
Which is in equilibirum \begin{align} 0\leq V'_1(1) \leq 2 \end{align}
We can evaluate $V(1)$ for all $V'_1(1)\in[0,2]$ or since $\tilde u=V'_1(1)/2$ for all $\tilde u \in[0,1]$. In the picture I highlighted two possible equilibria $E^A$ and $E^B$. Since the payoff is inreasing with the control we have a higher value associated with equilibirum $E^A$, i.e. $V^A(1) > V^B(1)$.
