= Solution
For $V=D>0$, choose pure Hawk. Every mutant $p<1$ ties it against a resident Hawk population, so the second condition must be checked. The <Hawk-Dove game> formula gives
$$
e(1,p)-e(p,p)=\frac{(1-p)(V-Dp)}2=\frac V2(1-p)^2>0.
$$
Consequently
$$
\boxed{x^*=(1,0)\text{ is an ESS when }V=D>0.}
$$
The actual invasion advantage is $\epsilon V(1-p)^2/2>0$, despite the tie at invasion frequency zero.
If the equal parameters are negative, pure Dove is instead an <evolutionarily stable strategy>: for a mutant $p>0$, $e(0,0)-e(p,0)=-pV/2>0$. If $V=D=0$, every payoff is zero and no resident has a strict advantage over any mutant. Thus the full equal-parameter answer is \b[Hawk for $V=D>0$, Dove for $V=D<0$, and no ESS for $V=D=0$].
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