Coordinate and weak forms of the geometric Schrödinger equation

Fix an interior point $p$ with coordinates $x_0$. The smooth polynomial $P(x)=u(p)+{\mathrm{\partial }}_{i}u(p)(x^i-x_0^i)+\frac{1}{2}{\mathrm{\partial }}_{ij}u(p)(x^i-x_0^i)(x^j-x_0^j)$ has the same derivatives through order two as $u$ at $p$. Metric duality gives $({\mathrm{grad}}_{h}u{)}^i=h^{ij}{\mathrm{\partial }}_{j}u$ directly, and differentiation gives ${\mathrm{\partial }}_k({\mathrm{grad}}_{h}u{)}^i=({\mathrm{\partial }}_k{h}^{ij}){\mathrm{\partial }}_{j}u+h^{ij}{\mathrm{\partial }}_{kj}u$. Thus ${\mathrm{grad}}_{h}u$ and ${\mathrm{grad}}_{h}P$ have equal values and first derivatives at $p$, so their covariant derivatives and traces agree there. More generally, the trace of $\mathrm{\nabla }X$ for $C^1$ $X$ depends only on $X(p)$ and its first coordinate derivatives: replacing its coefficients by affine polynomials with those same data reduces the divergence density formula to its smooth case in [F1]. Applying [F1] to $P$ therefore proves ${\mathrm{\Delta }}_{h}u(p)=m^{-1}{\mathrm{\partial }}_i(m{h}^{ij}{\mathrm{\partial }}_{j}u)(p)$, since the right side also uses only these derivatives through order two. This proves the required $C^2$ and $C^1$ extensions without assuming extra regularity.

Multiply the formula in step 1.1 by $m>0$ and change the sign to obtain the divergence equation. Expanding its derivative gives ${\mathrm{\Delta }}_{h}u=h^{ij}{\mathrm{\partial }}_{ij}u+m^{-1}{\mathrm{\partial }}_i(m{h}^{ij}){\mathrm{\partial }}_{j}u$, proving the nondivergence expression. If $u$ is $C^2$ up to the chart boundary, continuity of all these expressions extends the identities and any classical equation to that boundary. The inverse metric is symmetric positive definite and $m>0$. On any nonempty compact chart subset $K$, the continuous positive function $A^{ij}(x){\xi }_i{\xi }_j$ on $K\times \{\xi \in {\mathbb{R}}^3:\mathrm{\mid }\xi \mathrm{\mid }=1\}$ has positive minimum and finite maximum. These are the asserted uniform ellipticity bounds, also when $K$ meets the boundary.

Let $\phi \in C_c^{\mathrm{\infty }}(U^{\circ })$ be real. For each $i$, put $G^i=A^{ij}{\mathrm{\partial }}_{j}u\phi$ on the interior chart image $\mathrm{\Omega }$. This is $C^1$ and its support is compactly contained in $\mathrm{\Omega }$. Its zero extension ${\widehat{G}}^i$ is $C^1$ on ${\mathbb{R}}^3$: near every point outside that compact support it is identically zero. Choose a nondegenerate closed rectangle $Q={\prod }_{k=1}^3[a_k,b_k]$ with this support in its interior. The derivative ${\mathrm{\partial }}_i{\widehat{G}}^i$ is continuous on $Q$, so [F3] computes its multiple integral by integrating first in $x^i$. For every fixed choice of the other coordinates, $t\mapsto {\widehat{G}}^i(x^1,\dots ,t,\dots ,x^3)$ is $C^1$ on $\mathbb{R}$ and vanishes at $a_i,b_i$. Thus [F2] makes its derivative's integral on $[a_i,b_i]$ zero, and [F3] gives ${\int }_Q{\mathrm{\partial }}_i{\widehat{G}}^{i}dx=0$. All coordinate integrands are continuous with compact support, so their integrals are computed on such enclosing rectangles. Summing in $i$, applying the product rule and rearranging yields ${\int }_{U^{\circ }}(-{\mathrm{\partial }}_i(A^{ij}{\mathrm{\partial }}_{j}u)+qu)\phi dx={\int }_{U^{\circ }}(A^{ij}{\mathrm{\partial }}_{j}u{\mathrm{\partial }}_i\phi +qu\phi )dx$. If the classical equation holds, step 2.1 makes the left side zero, proving the weak identity. This integration-by-parts identity itself used only the regularity of $u$ and $\phi$, and the support restriction excludes a manifold-boundary contribution even when $U$ is a boundary chart.

Conversely, for the assumed $C^2$ function the residual $f=-{\mathrm{\partial }}_i(A^{ij}{\mathrm{\partial }}_{j}u)+qu$ is continuous on $U^{\circ }$. The same integration by parts turns the weak identity into $\int f\phi dx=0$ for every such test. If $f(p)\mathrm{\ne }0$, continuity gives $\delta >0$ such that the closed coordinate ball ${\bar{B}}_{\delta }(x_0)$ about the coordinates $x_0$ of $p$ is contained in the interior chart image and $f$ has the sign of $f(p)$ with $\mathrm{\mid }f\mathrm{\mid }\ge \mathrm{\mid }f(p)\mathrm{\mid }\mathrm{/}2$ throughout this ball. Use [F4] with $n=3$ and set $\phi (x)=\chi (2(x-x_0)\mathrm{/}\delta )$. Translation of the smooth rescaling with $R=\delta \mathrm{/}2$ is smooth; moreover $0\le \phi \le 1$, $\phi =1$ on ${\bar{B}}_{\delta \mathrm{/}2}(x_0)$ and $\mathrm{supp}\phi \subseteq {\bar{B}}_{\delta }(x_0)$. Thus $\phi \in C_c^{\mathrm{\infty }}(U^{\circ })$ is an admissible test, and $f\phi$ has one sign everywhere and absolute value at least $\mathrm{\mid }f(p)\mathrm{\mid }\mathrm{/}2$ on the smaller ball. Its integral is nonzero, contradicting the weak identity. This proves $f=0$, and step 2.1 recovers the classical geometric equation. Thus the weak and classical equations are equivalent at the stated regularity.