The range of Sobolev spaces to which the function $r^\beta\sin\beta\theta$ belongs

Question

I am learning about Sobolev space, and I am working on the following problem from "The mathematical theory of finite element methods" by Brenner.

To make the problem a little bit easier for me, I fixed $p = 2$. Then, I tried to find optimal value of all natural numbers k such that $v_\beta \in W^k_2(\Omega_\beta)$ as a function of $\beta$

I started computing the derivatives of $v$ and their respective $L_2$, and I noticed that differentiating $v$ say k times results in $r^{2(\beta-k)+1}$ in my $L_2$ integral, and since I wat this to be finite, I required $2{(\beta-k)+1} > -1 $ from which I got $k <\beta + 1$.

But it seemed strange to me that for any given $\beta$, I have $k<\beta + 1$ because I thought that (1) when $\beta = 1/2$ I have a "crack" (2) when $1\leq \beta$, I have a Lipschitz domain (3) when $1/2\leq\beta<1$, I have a indented corner.

So, for each of these cases, I expected a different bound for k, but all I have is $k<\beta + 1$ for all these three cases.

Am I missing something?

Answer 1

The size of angle at $0$ does not make a difference here. The $k$th order derivatives of $v_\beta$ grow like $r^{\beta-k}$ near the origin. And the domain contains at least some sector with vertex at the origin, which makes it about as "fat" there as the whole unit disk would be. So, the integral of derivative is about as large (up to multiplicative constants) as the integral of $r^{\beta-k}$ taken over the disk.

Hence, the conclusion:

• the $k$th derivatives are in $L^p$ when $(\beta-k)p>-2$
• they are not in $L^p$ when $(\beta-k)p\le -2$