= Flux-domain realization of a one-dimensional elliptic operator
{title2=$D(L)=\{v\in H_0^1(0,1):pv'\in H^1(0,1)\}$}
For continuous $p>0$ and bounded real $q$, the form $a(u,v)=\int(pu'v'+quv)$ on the <zero-trace Sobolev space> defines the weak operator $L=-(pu')'+qu$. Its literal $L^2$ operator domain is the displayed flux domain. If a weak solution has right-hand side $f\in L^2$, then $(pu')'=qu-f\in L^2$, so it belongs to that domain. This avoids assuming that a merely differentiable coefficient has bounded derivative. With $p\in C^1$ bounded away from zero, the domain reduces to $H^2\cap H_0^1$.
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