Let $u, w \geq 0$. Let the payoff function $f_1: [0, u] \times [0, w] \to \mathbb{R}$ for player $1$ be defined as $$f_1(a, b) = \frac{u - a}{1+e^{a-b}} $$ for $a \in [0, u]$ and $b\in [0, w]$. Similarly, let the payoff function $f_2$ for player 2 (when player 1 plays $a$ and player 2 plays $b$) be defined as $$f_2(a,b) = \frac{w - b}{1+e^{b-a}}$$
My question is how we can compute the Nash equilibria of the general-sum game that $f_1$ and $f_2$ define. In particular, I don't know how to deal with the fact that the strategy spaces $[0, u]$ and $[0, w]$ are continuous. (I do believe that a Nash equilibrium has to exist, since the payoff functions are continuous and quasi-concave.)
Moreover, is there a way to find all Nash equilibria of the game?
Given any choice of $b$ which I'll denote $\overline{b}$, Player 1 must respond optimally. This means they choose $a$ to maximize $\frac{u-a}{1+e^{a-\overline{b}}}$. Differentiating, we attain:
$$\frac{-(1+e^{a-\overline{b}}) - (u-a)e^{a-\overline{b}}}{(1+e^{a-\overline{b}})^2} = \frac{-1 + (a - u - 1)e^{a-\overline{b}}}{(1+e^{a-\overline{b}})^2} $$
Note that regardless of $\overline{b}$, this is negative for any $a \in [0, u]$. Thus $0$ is always the best response of Player 1; it's a dominant strategy. Similarly for Player 2. Thus $(0,0)$ is the only Nash equilibrium.