= Vanishing endpoint coefficient and the Sturm-Liouville boundary form
{title2=$[p(\overline u v'-\overline{u'}v)]_a^b=0$}
For the real-coefficient <differential operator> $Mv=(pv')'+qv$, <integration by parts> gives
$$
\int_a^b(\overline uMv-\overline{Mu}v)\,dx=[p(\overline uv'-\overline{u'}v)]_a^b.
$$
If $p$ tends to zero at both endpoints and $u,v,u',v'$ remain bounded there, the boundary term vanishes without prescribing endpoint values. Thus regular endpoint behaviour can supply the domain condition that makes the expression symmetric. Endpoint degeneracy alone is insufficient for arbitrary singular functions: for $p=x(1-x)$ and $v=\log x$, one has $pv'=1-x\to1$ at zero. A <self-adjoint differential operator> requires a specified admissible domain, rather than only a formal differential expression.
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