Third linearization identity for the cubic Dirichlet-to-Neumann map

Differentiate $-{\mathrm{\Delta }}_g{u}_{ϵ}+q{u}_{ϵ}^3=0$ once in ${ϵ}_j$ at $ϵ=0$. Since $u_0=0$, the derivative $v_j={\mathrm{\partial }}_{{ϵ}_j}{u}_{ϵ}{\mathrm{\mid }}_0$ solves $-{\mathrm{\Delta }}_g{v}_j=0$ with trace $h_j$. Differentiating twice in distinct parameters gives $-{\mathrm{\Delta }}_g({\mathrm{\partial }}_{{ϵ}_i}{\mathrm{\partial }}_{{ϵ}_j}{u}_{ϵ}{\mathrm{\mid }}_0)=0$ with zero trace, so this second derivative vanishes by [F2].

The third mixed derivative of $u_{ϵ}^3$ at zero has exactly the six permutations of $v_1{v}_2{v}_3$, hence equals $6v_1{v}_2{v}_3$; all other chain-rule terms contain $u_0$ or one of the vanishing second derivatives from step 1.1. Differentiating the equation gives $-{\mathrm{\Delta }}_g{w}_{123}+6q{v}_1{v}_2{v}_3=0$, with zero boundary trace. By [F1], the third derivative of the nonlinear DN data is ${\mathrm{\partial }}_{{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}}}{w}_{123}{\mathrm{\mid }}_{\mathrm{\partial }M}$.

The harmonic fields and $w_{123}$ are smooth by [F2], since their boundary data and the forcing $-6q{v}_1{v}_2{v}_3$ are smooth. Since $v_4$ is harmonic and $w_{123}$ has zero trace, Green's formula gives ${\int }_{\mathrm{\partial }M}({\mathrm{\partial }}_{{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}}}{w}_{123})\overline{h_4}d{S}_g={\int }_M({\mathrm{\Delta }}_g{w}_{123})\overline{v_4}d{V}_g\mathrm{.}$ Indeed, the second Green identity has boundary term $w_{123}{\mathrm{\partial }}_{\nu }\overline{v_4}$ equal to zero, and interior term $w_{123}{\mathrm{\Delta }}_g\overline{v_4}=0$. For dimension at least two, in each boundary chart write ${\mathrm{\Delta }}_{g}w=\mathrm{\mid }g{\mathrm{\mid }}^{-1\mathrm{/}2}{\mathrm{\partial }}_i(\mathrm{\mid }g{\mathrm{\mid }}^{1\mathrm{/}2}{g}^{ij}{\mathrm{\partial }}_{j}w)$ and apply [F5] to the coordinate components; summing with the finite partition leaves the displayed Riemannian boundary integral by [F4]. In dimension one, use arclength coordinates $t$ and integrate $\frac{d}{dt}(w_{123}\overline{v_4^{\mathrm{\prime }}}-w_{123}^{\mathrm{\prime }}\overline{v_4})=w_{123}\overline{v_4^{\mathrm{\prime }\mathrm{\prime }}}-w_{123}^{\mathrm{\prime }\mathrm{\prime }}\overline{v_4}$ by the fundamental theorem of calculus. Finite chart localization cancels interior endpoint and cutoff terms. At a physical left endpoint ${\mathrm{\partial }}_{{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}}}=-{\mathrm{\partial }}_t$, at a right endpoint ${\mathrm{\partial }}_{{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}}}={\mathrm{\partial }}_t$, and $d{S}_g$ is counting density. Using $w_{123}=0$ at those endpoints and $v_4^{\mathrm{\prime }\mathrm{\prime }}=0$ gives exactly the same displayed identity.

The equation in step 2.1 gives ${\mathrm{\Delta }}_g{w}_{123}=6q{v}_1{v}_2{v}_3$. Substituting this into step 3.1 and using the definition of the third DN derivative proves the displayed identity.