$1-(1-\min\{u,\bar u\})^{m_k}$ is positive and is increasing for $u\in[0,\bar u]$ and then constant for $u\in[\bar u, 1]$. Hence, the optimal solution for should be smaller than $\bar u = \arg\min_u p(u)$. Then the objective in can be simplified as $1-(1-u)^{m_k} + \alpha_i p(u)$. In summary, the optimal actions for both the infected and the healthy agents in the Nash Equilibrium can be compactly written as $$\label{eq: Nash}
u_{i,k} = \arg\min_u \{1-(1-u)^{m_k}(1-x_{i,k}) + \alpha_i p(u)\}, \forall i.$$

#### Example

Consider the previous example with four agents shown in \[fig:example\]. Define $$\label{eq: cost on activity}
p(u) = \exp(\frac{1}{u-1}),$$ which is a monotonically decreasing function as illustrated in \[fig: p\]. Then the optimal actions in the Nash Equilibrium for this specific problem satisfy: $$\label{eq: example Nash}
u_{i,k} = \arg\min_u \{u+x_{i,k}-ux_{i,k} + \alpha_i \exp(\frac{1}{u-1})\}, \forall i.$$

![image](alpha.pdf){width="\linewidth"}

Solving for , for infected agents, $u_{i,k} = 1$. For healthy agents, the choice also depends on $\alpha_i$ as illustrated in \[fig: alpha\]. We have assumed that $\alpha_i = \alpha$ which is identical for all agents. We further assume that $\alpha < 2$ such that the optimal solution for healthy agents should be $u_{i,k} = 0$. The optimal actions and the corresponding costs for all agents are 