One example related to weak derivative

\mathbf{Problem:} Suppose D=\{(x_1,x_2)||x_1|<1, |x_2|<1\} is the open square in \mathbb{R}^2. u is defined

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

Find the first weak derivative of u.

\mathbf{Proof:} Suppose v=\partial_\alpha u is the weak derivative, then for any \phi\in C^\infty_0(D), we have

\displaystyle \int_D v\phi dx=-\int_D u\partial_\alpha\phi dx.\quad \alpha=1,2\quad (1)

WLOG, assume \alpha=1. Let us denote the four domains in the definition of u as D_1,D_2,D_3,D_4 respectively. Note that

\displaystyle \int_D u\partial_1\phi dx=\int_{D_1} u\partial_1 \phi dx+\int_{D_2} u\partial_1\phi dx+\int_{D_3} u\partial_1\phi dx+\int_{D_4} u\partial_1\phi dx\quad(2)

While

\displaystyle \int_{D_1} u\partial_1\phi dx=\iint_{D_1}(1-x_1)\partial_1\phi dx_1dx_2=\int_{0}^1\int_{-x_1}^{x_1}(1-x_1)\partial_1 \phi dx_1dx_2

\quad                      \displaystyle =\int_{-1}^1 \int_{|x_2|}^1(1-x_1)\partial_1 \phi dx_1dx_2=\int_{-1}^1\left[ (1-x_1)\phi\big|^1_{|x_2|} +\int_{|x_2|}^1 \phi dx_1\right]dx_2

\quad                       \displaystyle =-\int_{-1}^1(1-|x_2|)\phi(x_2,x_2)dx_2-\int_{D_1}\partial_1u\,\phi dx\quad (3)

obtained from the integral by parts and the boundary behavior of \phi. Similarly for domain D_2

\displaystyle \int_{D_2} u\partial_1\phi dx=\iint_{D_2}(1+x_1)\partial_1\phi dx_1dx_2=\int_{-1}^0\int_{x_1}^{-x_1}(1+x_1)\partial_1 \phi dx_1dx_2

\quad                      \displaystyle =\int_{-1}^1 \int^{-|x_2|}_{-1}(1+x_1)\partial_1 \phi dx_1dx_2=\int_{-1}^1\left[ (1+x_1)\phi\big|_{-1}^{-|x_2|} -\int_{-1}^{-|x_2|} \phi dx_1\right]dx_2

\quad                      \displaystyle =\int_{-1}^1(1-|x_2|)\phi(-x_2,x_2)dx_2-\int_{D_2}\partial_1u\,\phi dx\quad (4)
As for the domain D_3 and D_4, we have
\displaystyle \int_{D_3} u\partial_1\phi dx=\iint_{D_3}(1-x_2)\partial_1\phi dx_1dx_2=\int_{0}^1(1-x_2)\int_{-x_2}^{x_2}\partial_1\phi dx_1dx_2

\quad                      \displaystyle =\int_{0}^1(1-x_2)[\phi(x_2,x_2)-\phi(-x_2,x_2)] dx_1dx_2\quad (5)

\displaystyle \int_{D_4} u\partial_1\phi dx=\iint_{D_4}(1+x_2)\partial_1\phi dx_1dx_2=\int_{-1}^0(1+x_2)\int_{x_2}^{-x_2}\partial_1\phi dx_1dx_2

\quad                      \displaystyle =\int_{-1}^0(1+x_2)[\phi(-x_2,x_2)-\phi(x_2,x_2)] dx_1dx_2\quad (6)
Substituting (3-6) to (2)

\displaystyle \int_D u\partial_1\phi dx=-\int_{D_1}\partial_1u\,\phi dx-\int_{D_2}\partial_1u\,\phi dx.

So from (1), we know

\displaystyle v=\partial_1 u=\begin{cases}-1, x\in D_1\\1,x\in D_2\,\\ 0,x\in D_3\cup D_4. \end{cases}

As for \partial_2u, one can get it in a similar way.

\text{Q.E.D}\hfill \square

\mathbf{Remark:} Evans, partial differential equation. 5.3

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