My introductory PDE textbook has this figure in a chapter on boundary and initial data:
It then says
Note that whenever Laplace's equation is solved in $\Omega \subset \mathbb{R}^2$, the boundary $\partial{\Omega}$ is one-dimensional. When solving the equation in $\mathbb{R}^3$, the boundary is two-dimensional, so that there are two tangential derivatives and one normal derivative at every point on the boundary.
Why is this? This is not clear to me.
Please help me understand this.
Explanations with manifolds :
PDE are solved on open sets of $\mathbb{R}^n$. Open sets are manifolds. The boundary of a manifold of dimension $n$ is a manifold of dimension $n-1$.
More intuitively, you can find coordinate charts going from an hyperplane of $\mathbb{R}^n$ to the boundary. Coordinate charts being basically homeomorphisms (continuous bijections with continuous inverse).
Manifolds of dimension $m$ are space that can be seen locally as $\mathbb{R}^m$. This means that locally (around a point) we can find an homeomorphism between our manifold and an open subset of $\mathbb{R}^m$.
Often, we also assume that the boundary of the open set is $C^1$. Typically, this seems to be the case on your drawing, but would not have been if $\Omega$ was a square (because of the corners), for example. By assuming this, we always can find $n-1$ tangential derivatives and $1$ normal derivative.
Note : sorry if you are not familiar with manifolds, but they are the most rigourous way to talk about this.