Let $\mathbf{x}\in\mathbb{R}^2,$ and $f(\mathbf{x})=1$ if $||\mathbf{x}||\leq R\in\mathbb{R}$, $f(\mathbf{x})=0$ otherwise. I want to find the weak gradient of $f$ in the dual space of compactly supported $C^{\infty}(\mathbb{R}^2)$ test functions, $\phi$. This is my attempt for the derivative with respect to $r$ by just following definitions, Fubini, and integration by parts: $$\frac{\partial}{\partial r}T_f(\phi)=-T_f\left(\frac{\partial \phi}{\partial r}\right)=-\int_{0}^{2\pi}\int_{0}^R\frac{\partial \phi}{\partial r}rdrd\theta=-\int_{0}^{2\pi}\left(R\phi(R,\theta)- \int_{0}^R\phi dr\right)d\theta$$ $$=\int_0^{2\pi}\int_{0}^R(1-r\delta(R-r))\phi drd\theta.$$ Granted that this work is correct, can I conclude that the weak derivative of $f$ with respect to $r$ is $\frac{1}{r}-\delta(R-r)$? Or do I have the wrong idea?

  • $\begingroup$ What's $R$? The unit ball is a set, not a function - what do you really mean by $f$? $\endgroup$ – Anthony Carapetis Mar 18 '17 at 5:19
  • $\begingroup$ @AnthonyCarapetis good point. I edited the question. $\endgroup$ – garserdt216 Mar 18 '17 at 5:25
  • $\begingroup$ How exactly (!) do you get your second equality? Seems to involve polar coordinates, but I don't see how/why $\partial \phi / \partial r $ pops up. $\endgroup$ – PhoemueX Mar 18 '17 at 13:07
  • $\begingroup$ @PhoemueX I'm using the distribution $T_f(\phi)=\int_{\mathbb{R}^2}\phi fdA$ $\endgroup$ – garserdt216 Mar 18 '17 at 15:28
  • $\begingroup$ OK, different question: What do you mean by $(\partial /\partial r) T_f $? The usual coordinates on $\Bbb {R}^2$ are $x_1,x_2$, there is no $r $. Also, since you say that you want to compute the gradient, you would need to compute at least two partial derivatives. $\endgroup$ – PhoemueX Mar 18 '17 at 16:13

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