I have a game in which two players, 1 and 2, choose a non-negative real number level of effort $e_1,e_2$ respectively. Their cost for this choice is $ce_i$ for $i=1,2$ where $c>0$ is the same for both. Their joint reward is $e_1^{\alpha_1}e_2^{\alpha_2},\quad \alpha_i\in (0,1)$ and they split this evenly--their utility on each strategy profile is their share of the reward minus the cost.
So the payoff for player one seems to be $e_1^{\alpha_1}e_2^{\alpha_2}/2-ce_1$ and similarly for player 2. Of course $(0,0)$ is a Nash equilibrium, and I need to find the others.
I assume (although I'm not perfectly sure about this) that I do this with partial derivatives. I know that a necessary condition is for the respective partials to be $0$.
$$\frac{\partial}{\partial e_1}(e_1^{\alpha_1}e_2^{\alpha_2}-ce_1) = 0 = \alpha_1e_1^{\alpha_1-1}e_2^{\alpha_2}-c$$
$$\frac{\partial}{\partial e_2}(e_1^{\alpha_1}e_2^{\alpha_2}-ce_2) = 0 = \alpha_2e_1^{\alpha_1}e_2^{\alpha_2-1}-c$$
These equations together imply
$$ \alpha_1e_1^{\alpha_1-1}e_2^{\alpha_2}=c = \alpha_2e_1^{\alpha_1}e_2^{\alpha_2-1} \Rightarrow\\ e_2=\alpha_2e_1/\alpha_1 $$
However, I'm not sure what a sufficient condition should be here. I recall from Calculus that, if you are optimizing a two-variable function, you use a second-order condition, but in this case I'm not optimizing just one function but two of them. My belief is that I use this solution to test the second-order condition on each function individually, but I want to check with someone smarter than me that this works as I think it does, and that I'm doing what I should be doing.
Yes I think this is fine, you might want to check that the second order derivative is negative, so you have a maximum, but otherwise I don't see any problems. Also the other variable should be thought of as fixed for the payoff function, since the player can only change his number
$$ f_1(e_1) = e_1^{\alpha_1}e_2^{\alpha_2} - ce_1$$
so this is not a two variable function.