Do $C^1$ domains share their outer normal field on the common part of their boundary

Question

Consider the following problem, where $\Omega$ is a domain having a $C^1$ boundary in the neighborhood of $x_2$. Let $B$ be the green ball with radius $\varepsilon_2$. Now consider the point $x_2 \in \partial B \cap \partial \Omega$.

In a neighborhood of $x_2$ both boundaries ($\partial\Omega$ and $\partial B$) have an outer normal field $\nu_\Omega$ resp. $\nu_B$. Does $$ \nu_B(x_2) = \nu_\Omega(x_2) $$ hold?

Is it possible to further generalize this fact to the case that $\Omega_1$ and $\Omega_2$ are domains with $C^1$ boundary under the assumption that $\Omega_1 \subseteq \Omega_2$ and $\partial\Omega_1 \cap \partial\Omega_2 \neq \emptyset$?

Answer 1

Yes. Suppose that $\partial \Omega_1$ and $\partial \Omega_2$ are given as $C^1$ graphs over an open subset $U \subseteq \{x \in \mathbb{R}^n : x_n =0\}$. Say, $g$ gives the graph of $\partial \Omega_1$ and $\partial \Omega_2$ is given as the graph of a function $f$ over $U$. We can assume for simplicity that $f \ge g$ so that the domains don't overlap and only touch. We then have that $\partial \Omega_2 = \{ (x',f(x')) : x' \in U\}$ and $\partial \Omega_1 = \{(x',g(x')\}$. The non-unit normal to $\partial \Omega_2$ is $N_2 = (-\nabla f(x'),1)$, while the non-unit normal to $\partial \Omega_1$ is $N_2 = (-\nabla g(x'),1)$. Suppose that the boundaries touch at $x' \in U$. Then at this point the normals agree if and only if $\nabla f(x') =\nabla g(x')$. However, if we assume that the boundaries touch at a point $x' \in U$ then we know that $f(y') \ge g(y')$ for $y' \in U$ and $f(x')=g(x')$. Then $f-g$ has a local minimum at $x'$, which means that $\nabla f(x') -\nabla g(x') =0$ and hence that $N_1 = N_2$.