In our lecture on PDE, we introduced the outward normal vector as follows:
Let $\Omega \subseteq \mathbb{R}^n$ be a domain with a $C^1$-boundary, $x \in \partial \Omega$ and $g \in C^k(U)$ with $\nabla g(x) \neq 0$ for all $x \in U$ with $g(x)=0$, where $U \subseteq \mathbb{R}^n$ is a neighbourhood of $x$. Then $\nu(x)= -\frac{ \nabla g(x)}{|\nabla g(x)|}$ is the outward normal vector in $x \in \partial \Omega$.
After this definition, we had a remark, which gives me some trouble:
$\nu(x)$ is orthogonal to every tangential vector $v \in T_x \partial \Omega$ because if $\gamma: I \rightarrow \partial \Omega$ is a continuous differentiable map with $\gamma(0)=x$ and $\gamma '(0)=v$, we get from $g(\gamma(t))=0 ~ \forall t \in I$ ($I$ an Interval) that
$$ 0 = \frac{d}{dt} g(\gamma(t)) |_{t=0}= \nabla g(x) \cdot \gamma '(0)=0. ~~*$$
$\nu(x)$ points outwards in the sense that $x+t \cdot \nu(x) \not \in \bar \Omega$ for all $t >0$ sufficiently small. This follows from $g(x+t \cdot \nu(x)=g(x)+ \nabla g(x+ \theta(t) \nu(t)) \cdot \nu(x)$ (**) for a $\theta(t) \in [0,t]$ whereas $g(x)+ \nabla g(x+ \theta(t) \nu(t)) \cdot \nu(x)$ goes to $- \nabla g(x) \cdot \frac{ \nabla g(x)}{|\nabla g(x)|} <0$ for $t$ to $0$ from the RHS. From these properties, you gain that the definition $\nu(x)$ is independent of $g$.
My Questions:
- I don't understand why the equality at * is zero on the LHS as well as the RHS. (The middle equality is clear).
- The explanation for the outward-pointing direction confuses me. Any explanation how I deduce this would be great. I also don't get the equality at **.
- This might be due to my lack of understanding but how do I get the independence of the definition from $g$?
Thank you for your help! Any other explanations on this topic would also be welcomed.
Best Answer
There is a crucial piece of information missing from the definition you wrote: the function $g$ defines the shape of the boundary! In other words, the significance of the function $g$ is that the region $\Omega$ is the set $$ \Omega = \{ x \in \mathbb R^n : g(x) > 0 \},$$ and the boundary of $\Omega$ is the set $$ \partial \Omega = \{ x \in \mathbb R^n : g(x) = 0 \}.$$ [Strictly speaking, $g$ is probably only defined on a small open patch within $\mathbb R^n$, and the equation $g(x) > 0$ is probably only meant to define the shape of $\Omega$ locally, within this small open patch. But this little caveat isn't important for answering your questions.]
Now I'll address your questions:
Suppose $\gamma(t)$ is a path within the boundary $\partial \Omega$. Since $g = 0$ everywhere on the boundary $\partial \Omega$, and since the path $\gamma(t)$ is contained within the boundary for all values of $t$, we have $g(\gamma(t)) = 0$ for all values of $t$. In other words, $t \mapsto g(\gamma(t))$ is the zero function. Hence its derivative $\frac{d}{dt} g(\gamma(t))$ is also zero. This observation is significant because, as shown in equation $(\ast)$, the quantity $\frac{d}{dt} g(\gamma(t)) |_{t = 0}$ is proportional to the dot product of vector $\nu$ evaluated at the point $x = \gamma(0) \in \partial \Omega$ with the vector $v$ which lies tangent to the curve $\gamma$ at the point $x$. Since this dot product is zero, regardless of the choice of the curve $\gamma \subset \partial \Omega$ passing through the point $x \in \partial \Omega$, it must be the case that $\nu(x)$ is normal to $\partial \Omega$ at the point $x$.
Fix a point $x$ on the boundary $\partial \Omega$. To see that $\nu(x)$ is the outward-pointing normal at $x$, what we need to show is that, if we start at the point $x$, and then walk a small distance $t$ in the direction of $\nu(t)$, then we are exiting the region $\Omega$. Said another way, we need to show that $x + t \nu(x) \notin \Omega$ for small positive $t$. But since $\Omega$ is the region $\{ x : g(x) > 0 \}$, we actually need to show that $$ g(x + t \nu(x)) < 0$$ for small positive $t$. To show this, we use the mean value theorem for differentiation, which says that $$ g(x + t \nu(x)) = g(x) + \nabla g(x + \theta \nu(x)).(t \nu(x))$$ for some $\theta \in [0, t]$. This corresponds to equation $(\ast\ast)$ in your notes, though I fixed some typos. All it remains to do now is to observe that $\nabla g(x).\nu(x) = -|\nabla g(x)| < 0$, and hence, by continuity, $\nabla g(x + \theta \nu(x)) . \nu(x) < 0$ too, provided that the distance $t$ that we chose originally is sufficiently small.
"Independence of the function $g$" means that the answer you get for the unit normal $\nu(x)$ doesn't depend on what function you originally picked to describe the shape of $\partial \Omega$. (For example, had we picked $2g$ or $g + g^3$ instead of $g$, it wouldn't have made any difference to the shape of $\Omega$.) But if two different choices of $g$ give you an $\Omega$ of the same shape, then of course the outward unit normals to $\partial \Omega$ won't depend on $g$.