Manuscript III@2 · III.tex · proposition · 82dad3e5

Proposition: Boundary error

Sign, inverse, boundary, and kernel audit › Exact finite-radius boundary error · lines 3007–3037 · proof included or linked

III@2#prop:boundary-error

For smooth $u$ ,

N_{R} Δ u = u + B_{R} [u] .

Replacing $N$ by $N_{R}$ in the Poisson compensator gives

δ_{0} w_{P}_{R}^{μ} = - ξ^{μ} - B_{R} [ξ^{μ}] .

Thus formally

δ_{ξ} (ϕ + κ w_{P}_{R}^{μ} \partial_{μ} ϕ) = - κ B_{R} [ξ^{μ}] \partial_{μ} ϕ + O (κ^{2}) .

Proof. Green's second identity and $Δ_{y} G (x, y) = δ_{x} (y)$ give

\int_{B_{R}} G Δ u - u (x) = B_{R} [u] .

Apply this to the variations of $A_{i}, B_{P}$ , with $ξ_{0} = - ξ^{0}$ . The scalar leading variation cancels the $- ξ^{μ}$ term.

Displayed formulas in this passage

unlabeled · line 3010
$N_{R} Δ u = u + B_{R} [u] .$
unlabeled · line 3014
$δ_{0} w_{P}_{R}^{μ} = - ξ^{μ} - B_{R} [ξ^{μ}] .$
unlabeled · line 3018
$δ_{ξ} (ϕ + κ w_{P}_{R}^{μ} \partial_{μ} ϕ) = - κ B_{R} [ξ^{μ}] \partial_{μ} ϕ + O (κ^{2}) .$
unlabeled · line 3031
$\int_{B_{R}} G Δ u - u (x) = B_{R} [u] .$