Recently I learned how to deal with the Derivative of a shifted Dirac-delta. Now I want to go a step further, but are not sure about the solution. Is there a simple way to rewrite terms like this
$$A=x \cdot \partial_x \delta(f(x))$$
As a special case, I would need $f(x)=\sqrt{x\cdot y}-z$, where y and z are constants:
$$B=x \cdot \partial_x \delta(\sqrt{x\cdot y} - z)$$
I was trying a coordinate-transformation $u=\sqrt{x\cdot y}$, but that didnt give an integral I could solve.
Thanks for suggestions!
So, for $\delta\bigl(f(x)\bigr)$ to be defined, we need that $f$ is smooth on the domain $U$ in consideration. If all roots of $f$ are simple, one has for any $\phi \in \def\D{\mathcal D}\def\R{\mathbb R}\D(U)$: $$ \def\<#1>{\left\langle#1\right\rangle}\<\phi, \delta \circ f> = \sum_{i: f(x_i) = 0} \frac{\phi(x_i)}{\def\abs#1{\left|#1\right|}\abs{f'(x_i)}}$$ We have \begin{align*} \<\phi, x\cdot \partial_x\delta(\sqrt{xy}-z)> &= -\<\phi + x\phi', \delta(\sqrt{xy} - z)> \end{align*} Now $f(x) = \sqrt{xy} - z$ is zero at $x = \frac{z^2}y$ if $z > 0$, and nowhere if $z \le 0$ (here the quantity in question is zero). So suppose $z > 0$, we have $$ f'(x) = \frac y{2\sqrt{xy}}, \quad f'\left(\frac{z^2}y\right) = \frac y{2z} $$ We continue \begin{align*} \<\phi, x\cdot \partial_x\delta(\sqrt{xy}-z)> &= -\<\phi + x\phi', \delta(\sqrt{xy} - z)>\\ &= -\frac 1{\abs{\frac{y}{2z}}}\biggl(\phi\left(\frac y{2z}\right) + \frac{y}{2z}\cdot\phi' \left( \frac{y}{2z}\right)\biggr)\\ &= -\frac 1{\abs{\frac y{2z}}}\<\phi, \tau_{y/2z}\delta> - \mathop{\rm sgn}\frac{y}{2z}\<\phi', \tau_{y/2z}\delta>\\ &= -\abs{\frac{2z}y}\<\phi, \tau_{y/2z}\delta> + \mathop{\rm sgn}y\<\phi, \tau_{y/2z}\partial_x\delta> \end{align*} So $$ B = -\abs{\frac{2z}y}\delta(x - y/2z) + \mathop{\rm sgn}\,y \cdot \partial_x\delta(x - y/2z) $$