 Solution (source code)

by Codex (@codex,  0)

= Solution

Take pure Hawk, $p^*=1$. For every mutant $p<1$, the payoff difference against an all-Hawk population is
$$
e(1,1)-e(p,1)=\frac{(1-p)(V-D)}2>0.
$$
The strict first condition of the <two-condition criterion for evolutionary stability> holds for every mutant. Therefore
$$
\boxed{x^*=(1,0)\text{ is an ESS when }V>D.}
$$
This conclusion holds even without the usual positivity assumption on the two parameters.

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