Second-Order Equation Operator on an Open Subset of a Hilbert Triple and Its $\delta$-Shifts

In the setting of Hilbert Triples: Standing Notation and Background, let $U\subseteq H$ be open in $H$, with $W=D(A)\cap U$ as in Hilbert Triples: Standing Notation and Background §open-sets, let $h$ be the penalty function, and let $Y\mapsto Y|_{V}$, $I_{V}$, and the sums, differences and multiples of forms be as in Hilbert Triples: Standing Notation and Background §restriction. Products of sets are Cartesian products.

1. (Second-order equation operator)¶ A second-order equation operator on $U$ relative to $(H,V,A)$ is a function

whose value at $(x,r,p,X)$ is written $F(x,r,p,X)$.

2. (The $\delta$-shifts of $F$)¶ Let $F$ be a second-order equation operator on $U$ relative to $(H,V,A)$ and let $\delta\in\mathbb{R}$ satisfy $0<\delta$. For $x\in W$ one has $x\in D(A)\subseteq V$, so $h(x)\in\mathbb{R}$ and $Ax\in H$ are defined, and for $Y\in\mathrm{Sym}(H)$ the forms $Y|_{V}+\delta I_{V}$ and $Y|_{V}-\delta I_{V}$ belong to $\mathrm{Sym}(V)$ by Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction §identity. The $\delta$-shifts of $F$ are the two functions

given by

Thus $F^{-}_{\delta}$ restores a subtracted $\delta h$ (its value, its gradient $Ax$ and its second derivative $I_{V}$ on $V$), while $F^{+}_{\delta}$ removes an added $\delta h$.