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Denote $U=\{x\in \mathbb{R}^2 | |x_1|<1, |x_2|<1\}$

Define:

$ u(x) = \begin{cases} 1-x_1 &\mbox{if } x_1>0, |x_2|<x_1 \\ 1+x_1& \mbox{if } x_1<0, |x_2|<-x_1 \\ 1-x_2& \mbox{if } x_2>0, |x_1|<x_2 \\ 1+x_2& \mbox{if } x_2<0, |x_1|<-x_2. \end{cases} $

For which $1\leq p<\infty$ does u belongs to $W^{1,p}(U)$ ?

I know that I have to find $p$ such that $u$ has derivative in the weak sense and the derivative $|Du|\in L^p(U)$ .

To show that $u$ has derivative in a weak sense, I let $\phi \in C_c^\infty(U)$ and show that :

$$\int_U u\phi_{x_i}dx=-\int_U u_{x_i}\phi(x) dx, i=1,2$$ This is where I got stuck. I wrote down the double integrals with respect to each case , but I do not know how to simplify those.

iipqa_dk
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1 Answers1

2

An outline:

  1. Notice that $u$ is continuous in $U$, and smooth outside the lines $x_1=\pm x_2$.

  2. Outside the given lines the derivative exists and is (globally) bounded.

  3. Take a test function and separate the integration over the four regions. Integrate by parts in each piece and notice that boundary terms corresponding to adjacent triangles (the mentioned regions) cancel each other out by continuity of $u$.

I would suggest drawing a picture and locating these triangles and how $u$ is defined on each of them to convince yourself that this is correct.

Jose27
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