The operator of a distribution

[CptXray]

Homework Statement

Let $T$ be a distribution in $\mathcal{D}(\mathbb{R}^2)$ such that:
$$T(\phi) = \int_{0}^{1}dr \int_{0}^{\pi} \phi(r, \Phi)d\Phi$$
$$\phi \in \mathcal{D}(\mathbb{R}^2)$$
calculate $r \frac{\partial{}}{\partial{r}} \frac{\partial{}}{\partial{\Phi}}T$.

Homework Equations

The Attempt at a Solution

I think it has something to do with the fact that derivative of a distribution is defined with some test function $\theta$, such that: $T \theta' = -T' \theta$. And in more general case: $T^{(\alpha)}\theta = (-1)^{\alpha} T \theta^{(\alpha)}$. Here for $A(x)$: $A^{(\alpha)} = \frac{\partial{A}}{\partial{x}}$. But here i have two derivatives of different parameters and also $r$. I've found literature about distributions (quite few to be honest) but can't find any examples that could help me do the calculations and get the feeling about this. I'd be grateful for help and tips.

 

[Orodruin]

You already have the basic ingredients. Insert the definition for how a distribution derivative acts on a test function and the definition of T. The $r$ is just regular multiplication of a distribution by a function.

 

[CptXray]

I tried to do that, but I'm afraid it doesn't get me anywhere. From definition:
$$\big(r\frac{\partial^2{T}}{\partial{r}\partial{\Phi}}\big)\phi = (-1)^2T\big( r\frac{\partial^2{\phi}}{\partial{r}\partial{\Phi}} \big)$$.
Evaluating the right side of this equation:
$$(-1)^2T\big( r\frac{\partial^2{\phi}}{\partial{r}\partial{\Phi}} \big) = T\big( r\frac{\partial^2{\phi}}{\partial{r}\partial{\Phi}} \big) = \int_{0}^{1}dr \int_{0}^{\pi} \big( r\frac{\partial^2{\phi}}{\partial{r}\partial{\Phi}} \big)d\Phi = $$
$$= \int_{0}^{1}dr \big( r( \frac{\partial{\phi(r, \pi)}}{\partial{r}} - \frac{\partial{\phi(r, 0)}}{\partial{r}} )\big) = $$ integrating by parts:
$$ = (r(\phi(r, \pi) - \phi(r, 0)))\bigg|_{0}^{1} - \int_{0}^{1}dr(\phi (r, \pi) - \phi(r, 0)) = $$
$$ = \phi(1, \pi) - \phi(1, 0) - \int_{0}^{1}\big( \phi(r, \pi) -\phi(r, 0)\big)dr$$.
I really don't know if that reasoning is correct, because I don't know how to find what that $r\frac{\partial^{2}{T}}{\partial{r}\partial{\Phi}}$ could be. I suspect that I somehow need to recreate $\phi(r,\Phi)$ but that's not obvious and seems more like guessing the answer.