Different types of domains $\Omega \subset \mathbb{R}^n$ in PDEs

Question

In PDEs I often read things like:

Let $\Omega$ be a bounded

• Lipschitz or
• $C^1$ or
• $C^2$ or
• $C^\infty$

domain

But I have no clue what this means in real life. I understand a $C^0$ domain has no corners, so no 90 degree angles. What do the above different types of domains look like intuitively? What should a self-respecting PDE person know about these things?

I read somewhere that you can approximate any domain with a $C^\infty$ domain. Is this true? More details would be appreciated.

Answer 1

In 2D, some heuristic examples:

$C^{\infty}$-domain: disk

Lipschitz/$C^{0,1}$-domain: Pacman, star-shaped nonconvex polygon, convex polygon

$C^{1}$-domain: consider a domain, when locally viewed, some part of boundary looks like the shape of gluing the graph of $x^2$ and $-x^2$ together at the origin. The rest of the boundary is just smooth. As you can see the derivative of $x^2$ and $-x^2$ agree at the origin, but not the second derivative.

$C^{2}$-domain: similar with $C^{1}$-domain, just gluing the graph $x^3$ and $-x^3$ together at the origin, the curve there is $C^2$.

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The regularity of the domain plays an important role for the elliptic regularity for a Sobolev function: $$ \|u\|_{W^{2,p}(\Omega)}\leq \| \Delta u\|_{L^p(\Omega)} $$ when $\Omega$ is $C^{1,1}/C^2$ or convex. In other words, the Poisson equation $-\Delta u = f$ boundary value problem can get a regularity lifting of 2 from the data $f$, i.e., the weak differentiability goes up by 2.

• A counterexample in nonconvex domain: the regularity lifting is not true anymore. Consider in the Pacman domain $\Omega$ parametrized using polar coordinates: $$ \Omega = \{(r,\theta): 0<r<1, 0<\theta<\pi/\alpha\} $$ with $1/2<a<1$ then $$u = (1-r^2)r^{\alpha}\sin(\alpha \theta)$$ solves the homogeneous Dirichlet boundary problem: $$ \begin{aligned} -\Delta u &= (4\alpha+4)r^{\alpha}\sin(\alpha \theta)\quad \text{ in } \Omega \\ u &= 0\quad \text{ on } \partial \Omega \end{aligned} $$ The right hand side data is in $L^p$, but $u\notin W^{2,p}$ for its second derivative shows strong singular behavior near the origin. If we extend the Pacman to the full disk, such solution won't be there anymore for the value $$\lim_{\theta\to 0^+} u(r,\theta) = \lim_{\theta\to 2\pi^-} u(r,\theta)$$

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I doubt we could approximate ANY open set by $C^{\infty}$-smooth domain, since we simply introduce the definition of the boundary of this open set in the Lebesgue sense. For domain that is at least Lipschitz I believe the approximation is true, if we are talking about pointwise limit.

Answer 2

These refer to the smoothness of the boundary, $\partial \Omega$.

If $\Omega \subset \mathbb{R}^n$ is open and bounded, we say $\partial \Omega$ is $C^k$ if for each point $\xi \in \partial \Omega$, $\exists r > 0$ and a $C^k$ function $\psi : \mathbb{R}^{n-1} \rightarrow \mathbb{R}$ such that

$$ \Omega \cap B_r(\xi) = \{ x \in B_r({\xi}) \, | \, x_n > \psi(x_1,\ldots,x_{n-1}) \} $$

For $C^{\infty}$ or Lipschitz domains, replace these with $C^k$ in the definition.

Intuitively, you take a ball around any point on the boundary and transform the part of the domain in the ball to the upper half plane with a $C^k$ function.