Contest game: second order condition satisfied, but negative profits? The following is taken from Nti (1999) 

Consider a 2 player game in which each exerts effort in attempt to win a prize. Let $V_1$ be player $1$'s valuation of the prize and let $V_2$ be player $2$'s valuation, where $0&lt;V_1&lt; V_2$. Player $1$ and $2$ expend efforts $x_1$ and $x_2$ respectively to win prize $p=1$. The cost of effort is $1$ per unit. Given a profile of effort levels ($x_1$, $x_2$),the probability that player i wins the prize takes the symmetric form

$$p_i(x_1,x_2)=\frac{x_i^r}{x_1^r+x_2^r}$$

$r$ is the returns to scale parameter associated with effort $r&gt;0$. Profit is

$$\pi_i(x_1,x_2)=v_i\frac{x_i^r}{x_1^r+x_2^r}-x_1$$

In equilibrium (Nash-Cournot) player $i$ exerts effort $x_i^*$: $$x_i^*=\frac{rv_i^{r+1}v_{-i}^r}{(v_1^r+v_2^r)^2}$$

If both follow the equilibrium strategy, $i$'s probability of winning is
$P_i=\frac{v_i^r}{v_1+v_2^r}$. Inserting these two into the second equation for player 1 (with simplification) yields

$$\pi_1(x_1^*,x_2^*)=\frac{v_1^{r+1}}{(v_1^r+v_2^r)^2}\big[v_1^r+v_2^r-rv_2^r\big]$$

The second order (sufficiency) condition $\frac{\partial^2\pi_1}{\partial{x_1^2}}=\frac{rv_1^{1-r}x_2^rx_1^{2r-2}}{(x_1^r+x_2^r)^3}\big[rv_2^r-v_2^r-v_1^r-rv_1^r\big]$ is satisfied iff $$rv_2^r-v_2^r-v_1^r-rv_1^r&lt;0$$

However, it is clear that for $x_1^*$ to be a Nash Equilibrium, $\pi_1(x_1^*,x_2^*)\geq0$ and hence that $$v_1^r+v_2^r&gt;rv_2^r$$ 

which is a stronger requirement than the second order condition. Hence, it is possible that the second order condition is met but not that $v_1^r+v_2^r&gt;rv_2^r$. My question is what happens when this occurs? Is there no Nash equilibrium here? Does the player who makes negative profit drop out, if so what does the remaining player do?
Well it says in the paper on p. 423:

"Thus the sufficiency assumption implies a unique pure strategy equilibrium. And if the sufficiency assumption is violated, at least one player will receive a negative payoff; that player can improve by reducing his or her effort to zero, which will destroy the equilibrium." [emphasis mine]

So that player will not make any effort because he/she cannot win. I leave it to you to figure out what the best reply from the other player is in that case. 

I guess to be completely mathematically correct one would have to formulate these kind of games with non-negativity constraints, but because it complicates the analysis they're often left out and an interior equilibrium is either assumed or, as here, checked afterwards.