Bounded collar regularity for an exterior weak Dirichlet function

Choose $u_j\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ tending to $u$ in $H^1(\mathrm{\Omega })$ and set $w=\eta u$ on $G$. Each $w_j=(\eta u_j){\mathrm{\mid }}_G$ belongs to $C_c^{\mathrm{\infty }}(G)$: its support misses $\mathrm{\Theta }$ because the support of $u_j$ does, and misses $\mathrm{\partial }{B}_L$ because the support of $\eta$ does. The first-order multiplier bound gives $w_j\to w$ in $H^1(G)$ and $\mathrm{\parallel }w{\mathrm{\parallel }}_{H^1(G)}\le C_{\eta }\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1(\mathrm{\Omega })}$. Hence $w\in H_0^1(G)$.

The boundary of $G$ is $\mathrm{\partial }\mathrm{\Theta }\cup \mathrm{\partial }{B}_L$, with these two compact sets separated by a positive distance. In a sufficiently small cylindrical graph chart at $\mathrm{\partial }\mathrm{\Theta }$, the two open sides are connected and contain no boundary point of $\mathrm{\Theta }$. Each side therefore lies entirely in $\mathrm{int}\mathrm{\Theta }$ or entirely in $\mathrm{\Omega }$: these two disjoint open sets partition the complement of $\mathrm{\partial }\mathrm{\Theta }$. At least one side lies in $\mathrm{\Omega }$, and both may do so. At $\mathrm{\partial }{B}_L$, exactly the inner side belongs to $G$. It suffices to obtain the regularity estimate separately on each occupied local side; we make no assertion about how these sides join in a global component.

For $\varphi \in C_c^{\mathrm{\infty }}(G)$, extend $\varphi$ by zero as a test on $\mathrm{\Omega }$ and apply $\mathrm{\Delta }u=g$ to $\eta \varphi$. Expanding derivatives of this smooth test gives ${\int }_{G}w\mathrm{\Delta }\varphi ={\int }_G\eta g\varphi -2{\int }_{G}u\mathrm{\nabla }\eta \cdot \mathrm{\nabla }\varphi -{\int }_{G}u(\mathrm{\Delta }\eta )\varphi$. For each coordinate, the first weak-derivative identity for $u$ tested against $({\mathrm{\partial }}_j\eta )\varphi$ gives $-{\int }_{G}u({\mathrm{\partial }}_j\eta ){\mathrm{\partial }}_j\varphi ={\int }_G(({\mathrm{\partial }}_{j}u)({\mathrm{\partial }}_j\eta )+u{\mathrm{\partial }}_j^2\eta )\varphi$. Summing yields ${\int }_{G}w\mathrm{\Delta }\varphi ={\int }_{G}h\varphi$, where $h=\eta g+2\mathrm{\nabla }\eta \cdot \mathrm{\nabla }u+(\mathrm{\Delta }\eta )u$. Thus $\mathrm{\Delta }w=h$ in distributions and $\mathrm{\parallel }h{\mathrm{\parallel }}_{L^2(G)}\le C_{\eta }(\mathrm{\parallel }g{\mathrm{\parallel }}_{L^2(\mathrm{\Omega })}+\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1(\mathrm{\Omega })})$. Only first weak derivatives of $u$ have been used.

Here is a bounded smooth auxiliary domain on one side of a graph, agreeing with that side near its central boundary point. In flattened coordinates take the unit ball centered at $-e_n$, whose upper boundary is $t=a(y)=-1+\sqrt{1-\mathrm{\mid }y{\mathrm{\mid }}^2}$. Choose $0\le \chi \le 1$ smooth, supported in $\mathrm{\mid }y\mathrm{\mid }<1\mathrm{/}2$ and equal to $1$ for $\mathrm{\mid }y\mathrm{\mid }\le 1\mathrm{/}4$. Extend $\chi a$ by zero, and apply the global shear $T(y,t)=(y,t-\chi (y)a(y))$ to the ball. Its inverse is $(y,t)\mapsto (y,t+\chi (y)a(y))$, so its image $D_{\ast }$ is bounded with smooth boundary. Its upper boundary is $(1-\chi )a\le 0$, and near the origin it is exactly $t=0$; hence $D_{\ast }\subset \{t<0\}$ and agrees with that half-space near the origin. Dilate it sufficiently far into the given chart, then apply the inverse flattening shear and the rigid motion from that chart. This gives a bounded smooth domain $D\subset G$ on the chosen side, agreeing with that side in a smaller ambient neighbourhood $V$ of the boundary point. For the other side first reflect $t$. The dilated closure stays inside the chart, so every part of $\mathrm{\partial }D$ has the one-sided $C^2$ charts required by the regularity supplier. The part of $\mathrm{\partial }D$ inside $V$ is the original graph; the remaining, artificial boundary misses a still smaller neighbourhood of the point.

Take such smaller neighbourhoods $V$ at boundary points, with a domain $D$ for each occupied side, and interior neighbourhoods compactly contained in balls $D⋐G$. A finite number cover $\bar{G}$. They admit smooth cutoffs ${\zeta }_i$ supported compactly in their respective neighbourhoods, with ${\sum }_i{\zeta }_i=1$ near $\bar{G}$: choose finitely many cutoffs ${\rho }_i$ supported in these neighbourhoods whose sum $S$ is positive near $\bar{G}$, choose a compactly supported cutoff $\tau$ equal to $1$ near $\bar{G}$ with support in $\{S>0\}$, and set ${\zeta }_i=\tau {\rho }_i\mathrm{/}S$ there and zero elsewhere. The ${\rho }_i$ can be obtained by first taking smaller open sets covering $\bar{G}$ and then applying the cutoff input to their compact closures. For each boundary index, shrink the neighbourhood before choosing the cutoffs so that every corresponding $D$ agrees with its occupied side there and its artificial boundary misses $\mathrm{supp}{\zeta }_i$. This cover and all its constants depend only on the fixed geometry.

For every interior index or occupied boundary side set $q=({\zeta }_{i}w){\mathrm{\mid }}_D$. The functions $({\zeta }_i{w}_j){\mathrm{\mid }}_D$ are in $C_c^{\mathrm{\infty }}(D)$: the artificial boundary misses their supports, and $w_j$ misses the original boundary. They converge to $q$ in $H^1(D)$, so $q\in H_0^1(D)$. Repeating the test computation of step 2.1, now with $w$, $h$ and ${\zeta }_i$, gives $\mathrm{\Delta }q=h_i:={\zeta }_{i}h+2\mathrm{\nabla }{\zeta }_i\cdot \mathrm{\nabla }w+(\mathrm{\Delta }{\zeta }_i)w$ in $D$, with $\mathrm{\parallel }{h}_i{\mathrm{\parallel }}_{L^2(D)}\le C_i(\mathrm{\parallel }h{\mathrm{\parallel }}_{L^2(G)}+\mathrm{\parallel }w{\mathrm{\parallel }}_{H^1(G)})$. This says ${\int }_D\mathrm{\nabla }q\cdot \overline{\mathrm{\nabla }v}=-{\int }_D{h}_i\bar{v}$ first for smooth compactly supported $v$, and then for every $v\in H_0^1(D)$ by its defining density and the $L^2$ bounds. Thus $q$ satisfies exactly the homogeneous weak Dirichlet problem for $-\mathrm{\Delta }$ with forcing $-h_i$.

Apply the bounded-domain theorem on each of the finitely many domains $D$ to obtain $\mathrm{\parallel }q{\mathrm{\parallel }}_{H^2(D)}\le C_i(\mathrm{\parallel }h{\mathrm{\parallel }}_{L^2(G)}+\mathrm{\parallel }w{\mathrm{\parallel }}_{H^1(G)})$. Regard each $q$ as a function on $G$ by assigning zero off its chosen local side and off its supporting neighbourhood. This extension is $H^2(G)$: it vanishes in a neighbourhood of every artificial boundary, and its remaining boundary is contained in $\mathrm{\partial }G$, which no compactly supported test in $G$ meets. Applying the weak derivative identities on these open pieces, and a finite smooth partition of a compact test support, therefore gives the weak derivatives on $G$ as the extensions of those on $D$, with the same norms. The sum of these extended functions is ${\sum }_i{\zeta }_{i}w=w$; when both sides of one chart are occupied, their two functions occupy disjoint parts of $G$. The finite sum belongs to $H^2(G)$ and is bounded there by $C_G(\mathrm{\parallel }h{\mathrm{\parallel }}_{L^2(G)}+\mathrm{\parallel }w{\mathrm{\parallel }}_{H^1(G)})$. Substituting steps 1.1 and 2.1 gives the stated estimate, and step 2.1 already gives the displayed Laplacian formula. This argument includes a thin smooth obstacle with both sides occupied and also the case $\mathrm{\Theta }=\mathrm{\varnothing }$.