## 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|0, |x_1|

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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