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 $0<V_1< V_2$. 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 $r>0$. 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 $$rv_2^r-v_2^r-v_1^r-rv_1^r<0$$
However, it is clear that for $x_1^*$ to be a Nash Equilibrium, $\pi_1(x_1^*,x_2^*)\geq0$ and hence that $$v_1^r+v_2^r>rv_2^r$$
which is a stronger requirement than the second order condition. Hence, it is possible that the second order condition is met but not that $v_1^r+v_2^r>rv_2^r$. 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?
