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Let us say we have the indicator function $\chi_{\{|x|\leq 1\}}$ in $\mathbb{R}^2$.

How can I write out the weak derivative of this indicator function?

Is it $\delta_{|x|=1}$? Or it should be vector valued measure like $\bigg(\frac{\partial}{\partial x_1} \chi_{{|x|\leq 1}},\frac{\partial}{\partial x_2} \chi_{{|x|\leq 1}}\bigg)$, but now $\frac{\partial}{\partial x_1} \chi_{{|x|\leq 1}}$ is something that depends on the value of $x_2$.

Xiao
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  • One way to write the indicator function as defined in the OP is to write

    $$\chi_{|x|\le1}=H(\rho -1)$$

    where $H$ is the Heaviside step function and $\rho =|x|$. The (distributional) derivative with respect to $\rho$ is, in the sense of Generalized Functions given by

    $$\frac{d}{d\rho}H(\rho -1)=\delta(\rho-1)$$

    where $\delta$ is the Dirac Delta. But this is obviously not a weak derivative. You might be interested in this answer.

     – Mark Viola Sep 21 '15 at 22:09

2 Answers2

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My edit of the other answer (which is wrong) was rejected, so I'll copy it with correction and due credit to its original author.

The weak gradient of the characteristic function of a domain $\Omega$ with a smooth boundary is the vector-valued measure $\nu(x)d\sigma(x)$ where $\nu(x)$ is the inward unit normal at $x\in\partial\Omega$ and $d\sigma$ is the surface measure.

Why so? By the divergence theorem. Recall that by definition, the weak gradient satisfies $$ \int \nabla u\cdot \varphi = -\int u\operatorname{div}\varphi $$ for every compactly supported smooth vector field $\varphi$. With $u=\chi_\Omega$ and $\nabla u$ as above this is exactly the divergence theorem.

Jean-Marie Aubry
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  • Shouldn't the divergence theorem reads $\int_{\Omega} \nabla\cdot \phi, \mathrm{d}x = \int_{\partial \Omega} \nabla \phi \cdot \nu ,\mathrm{d}\sigma$ for $u = \chi_{\Omega}$ as above? I am a bit nervous about the 'sign' – Fei Cao May 26 '21 at 18:39
  • Do you usually call such a measure "weak derivative" or "distributional derivative" (or both)? – rod Mar 24 '22 at 23:02
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The weak gradient of the characteristic function of a domain $\Omega$ with a smooth boundary is the vector-valued measure $\nu(x)d\sigma(x)$ where $\nu(x)$ is the outward unit normal at $x\in\partial\Omega$ and $d\sigma$ is the surface measure.

Why so? By the divergence theorem. Recall that by definition, the weak gradient satisfies $$ \int \nabla u\cdot \nabla \varphi = -\int u\operatorname{div}\varphi $$ for every compactly supported smooth vector field $\varphi$. With $u=\chi_\Omega$ and $\nabla u$ as above this is exactly the divergence theorem.

  • Shouldn't the divergence theorem reads $\int_{\Omega} \nabla\cdot \phi, \mathrm{d}x = \int_{\partial \Omega} \nabla \phi \cdot \nu ,\mathrm{d}\sigma$ for $u = \chi_{\Omega}$ as above? I am a bit nervous about the 'sign' – Fei Cao May 26 '21 at 18:39