Proposition: Direct verification of eq:X2law

Write ${\mathcal{V}}_{\epsilon }={\mathcal{V}}_{\epsilon }[h]$, ${\mathcal{W}}_{\epsilon }={\mathcal{W}}_{\epsilon }[h,k]$, ${\mathcal{V}}_{\epsilon }^{\prime }={\mathcal{V}}_{\epsilon }[h^{\prime }]$, ${\mathcal{W}}_{\epsilon }^{\prime }={\mathcal{W}}_{\epsilon }[h^{\prime },k^{\prime }]$, $X=X_{\xi }$, $T_2=X_{\zeta }+\frac{1}{2}{X_{\xi }}^2$, $D=D_{\xi }$, ${\Delta }_2=D_{\zeta }+\frac{1}{2}{D_{\xi }}^2$. Expanding Lemma lem:plateaucoord gives the operator identities

The candidate ${{\mathsf{X}}_2}^{\mathrm{c}}={\mathsf{X}}_2+X{\mathsf{X}}_1+T_{2}x$ satisfies $\square {{\mathsf{X}}_2}^{\mathrm{c}}=-{\mathcal{V}}_{\epsilon }^{\prime }({\mathsf{X}}_1+Xx)-{\mathcal{W}}_{\epsilon }^{\prime }x$ and has zero data; hence ${{\mathsf{X}}_2}^{\mathrm{c}}={\mathsf{X}}_2[h^{\prime },k^{\prime }]$.

Proof. Equations (eq:Vid)–(eq:Wid) are the $\kappa$ and ${\kappa }^2$ coefficients of $(\square +\kappa {\mathcal{V}}_{\epsilon }^{\prime }+{\kappa }^2{\mathcal{W}}_{\epsilon }^{\prime })(1+\kappa X+{\kappa }^2{T}_2)=(1+\kappa D+{\kappa }^2{\Delta }_2)(\square +\kappa {\mathcal{V}}_{\epsilon }+{\kappa }^2{\mathcal{W}}_{\epsilon })$. Using $\square {\mathsf{X}}_1=-{\mathcal{V}}_{\epsilon }x$, $\square {\mathsf{X}}_2=-{\mathcal{V}}_{\epsilon }{\mathsf{X}}_1-{\mathcal{W}}_{\epsilon }x$, and $\square x=0$,

Since $X{x}^{\mu }={\xi }^{\mu }$ and $T_2{x}^{\mu }={\zeta }^{\mu }+\frac{1}{2}{\xi }^{\nu }{\partial }_{\nu }{\xi }^{\mu }$, the candidate is the right-hand side of (eq:X2law); it vanishes with its time derivative on $\Sigma$. Uniqueness concludes.