# Finding the weak derivatives

Finding the weak derivatives:

a/ $f(x)=\left|x_1 \right|$, for all $x=(x_1,\ldots, x_n)$;

b/ $f(x)=\operatorname{sign} (x_1)$, where $\Omega=\{\left|x \right|<1\}$.

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We know that $f$ has weak derivative $\partial^\alpha f=g$ if $$\int_{\Omega}g \varphi \ \mathrm{d}x=(-1)^\left|\alpha \right|\int_{\Omega}f\cdot D^\alpha \varphi \ \mathrm{d}x, \ \forall \varphi \in C_{0}^{\infty}(\Omega)$$

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For Question a/ We assume that $g$ is the weak devirative of $f$.

• We have $I:=\int_{-1}^{1}g\varphi \ \mathrm{d}x=-\int_{-1}^{1}|x_1|D\varphi \ \mathrm{d}x_1, \ \forall \varphi \in C_{0}^{\infty}(\Omega) \tag 1$

• Whence $-I=\int_{-1}^{0}-x_1 D\varphi \ \mathrm{d}x_1+\int_{0}^{1}x_1 D\varphi \ \mathrm{d}x_1=-x_1\varphi \mid_{-1}^{0}+\int_{-1}^{0}\varphi \ \mathrm{d}x_1+x_1 \varphi \mid_{0}^{1}-\int_{0}^{1}\varphi \ \mathrm{d}x_1$

• Hence, \begin{align*} -I&=\varphi(1)-\varphi(-1)-\left(\int_{0}^{1}\varphi \ \mathrm{d}x_1-\int_{-1}^{0}\varphi \ \mathrm{d}x_1 \right)\\ &=\varphi(1)-\varphi(-1)-\int_{-1}^{1}\operatorname{sign} (x_1) \varphi \ \mathrm{d}x_1 \\ &=-\int_{-1}^{1}\operatorname{sign} (x_1) \varphi \ \mathrm{d}x_1 \tag 2 \end{align*}

Since (1) and (2), we have $g(x)=\operatorname{sign} (x_1) \blacksquare$

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