Proving Div(x/|x|^2) = 2πδ(0,0) in 2D with Distribution Derivative

[kairosx]

Homework Statement

Show that $$div ( \frac{x}{|x|^2} ) = 2 \pi \delta_{(0,0)}$$ with $x \in R^2 \ \{ 0 \}$ and $\delta_{(0,0)}$ beeing the dirac delta distribution with pole in $(0,0)$.

Homework Equations

$div (f(x)) = \nabla \cdot (f(x)) = f_{x_1} + f_{x_2}$
The distribution derivative: $<f'(x),\phi(x)> = - <f(x),\phi'(x)>$ with $<u,v>$ defined as $\int_{R^2} u v dx$ and $\phi$ beeing a test function $\phi \in c^{\infty}(R^2)$ and $supp(\phi)$ compact

The Attempt at a Solution

I started with $$ f_{x_1} = \frac{d}{dx_1} (x \cdot |x|^{-2}) = \hat{e}_{x1} \frac{1}{|x|^2} + x \frac{d}{dx_1} (\frac{1}{|x|^2}) $$
Together with $f_{x_1}$ I need to compute $\frac{d}{dx_1} (\frac{1}{|x|^2}) + \frac{d}{dx_2} (\frac{1}{|x|^2}) = \nabla \cdot (\frac{1}{|x|^2})$ which is in polar coordinates $\frac{d}{dr} (\frac{1}{r^2}) = \frac{-2}{r^3}$ because the problem is rotationally symmetric and therefore all derivatives of the angles vanish. Alltogehter that's $$ f_{x_1} + f_{x_2} = (\hat{e}_{x1} + \hat{e}_{x2} ) \frac{1}{|x|^2} + x(\frac{-2}{|x|^3}) = (\hat{e}_{x1} + \hat{e}_{x2} ) \frac{1}{|x|^2} - 2 (\hat{e}_{x1} + \hat{e}_{x2} )(\frac{1}{|x|^2}) = - (\hat{e}_{x1} + \hat{e}_{x2} ) \frac{1}{|x|^2}$$.
But that's not the solution I am looking for? Actually I was hoping the result would be $0$ because the delta distribution is $0$ everywhere except at $x = 0$. Then I would have tried to somehow integrate over the unit ball around zero and show that the result is $2 \pi$. But now I'm confused and don't see what to do next. I guess I did something wrong with the polar coordinates but I'm not sure.

 

[Orodruin]

Why are you not applying the definition of the distributional derivative that you put into your relevant equations?

Together with $f_{x_1}$ I need to compute $\frac{d}{dx_1} (\frac{1}{|x|^2}) + \frac{d}{dx_2} (\frac{1}{|x|^2}) = \nabla \cdot (\frac{1}{|x|^2})$ which is in polar coordinates $\frac{d}{dr} (\frac{1}{r^2}) = \frac{-2}{r^3}$ because the problem is rotationally symmetric and therefore all derivatives of the angles vanish.

It is not clear to me what you are trying to do here. Are you trying to compute the divergence (which would not make sense since $1/r^2$ is a scalar field) or the gradient (which is $\vec e^i \partial_i \phi$ for a scalar field $\phi$)?

I suggest that you apply the definition of the distributional derivative instead.

 

[kairosx]

You mean like
$$\frac{d}{dx_1} (\frac{1}{|x|^2}) = < \frac{d}{dx_1} \frac{1}{|x|^2},\phi(x)> = - <\frac{1}{|x|^2}, \frac{d}{dx_1} \phi(x)> = -\int_{-\infty}^{\infty} \frac{1}{|x|^2} \frac{d}{dx_1} \phi(x) dx = - \int_{0}^{\infty} \frac{1}{x^2} \frac{d}{dx_1} \phi(x) dx - \int_{-\infty}^{0} \frac{1}{(-x)^2} \frac{d}{dx_1} \phi(x) dx .$$
I'm actually not sure about the last step, because what is $x^2$ when x is a vector? I guess I would have to define that by myself as $x^2 = x_1^2 + x_2^2$? I would then go on with
$$ - \int_{0}^{\infty} \frac{1}{x_1^2 + x_2^2} \frac{d}{dx_1} \phi(x) dx - \int_{-\infty}^{0} \frac{1}{x_1^2 + x_2^2} \frac{d}{dx_1} \phi(x) dx =
\int_{0}^{\infty} \frac{d}{dx_1} \frac{1}{x_1^2 + x_2^2} \phi(x) dx + \int_{-\infty}^{0} \frac{d}{dx_1} \frac{1}{x_1^2 + x_2^2} \phi(x) dx =
\int_{0}^{\infty} \frac{-2 x_1}{(x_1^2 + x_2^2)^2} \phi(x) dx + \int_{-\infty}^{0} \frac{-2 x_1}{(x_1^2 + x_2^2)^2} \phi(x) dx
$$
but here I'm stuck again. I just don't know how do deal with those integrals?

 

[Orodruin]

The integral is with respect to $x_1$ and $x_2$, not only with respect to some $x$. Also, you should use the definition of the distributional derivative for the vector field, not for $1/x^2$.

 

[kairosx]

Actually by saying that $x \in R^2$ i always mean $\int dx := \int d(x_1,x_2)$, i definately should have stated that explicitely. Okay, i going to start from scratch with the distributional derivative for my vector field :-)

$$ <\frac{d}{dx_1} \frac{x}{|x|^2} + \frac{d}{dx_2} \frac{x}{|x|^2}, \phi(x_1,x_2)> =
\int \int ( \frac{d}{dx_1} \frac{x}{|x|^2} + \frac{d}{dx_2} \frac{x}{|x|^2} ) \phi(x_1,x_2) d(x_1,x_2) =
\int \int ( \frac{d}{dx_1} \frac{x}{|x|^2} + \frac{d}{dx_2} \frac{x}{|x|^2} ) \phi(x_1,x_2) d(x_1,x_2) = ...$$

and with $|x|^2 = x_{1}^{2} + x_{2}^{2}$ and $x = (x_1,x_2)^T$

$$ \int ( \frac{d}{dx_1} \frac{x_1}{x_{1}^{2} + x_{2}^{2}} + \frac{d}{dx_2} \frac{x_1}{x_{1}^{2} + x_{2}^{2}} ) \phi(x_1,x_2) d(x_1) + \int ( \frac{d}{dx_1} \frac{x_2}{x_{1}^{2} + x_{2}^{2}} + \frac{d}{dx_2} \frac{x_2}{x_{1}^{2} + x_{2}^{2}} ) \phi(x_1,x_2) d(x_2) = ... $$

Considering the identities
$\frac{d}{dx_1} \frac{x_1}{x_{1}^{2} + x_{2}^{2}} = \frac{1 \cdot (x_{1}^{2} + x_{2}^{2}) - x_1 \cdot 2 x_1}{(x_{1}^{2} + x_{2}^{2})^2} = \frac{(x_{2}^{2} - x_{1}^{2})}{(x_{1}^{2} + x_{2}^{2})^2}$
$\frac{d}{dx_2} \frac{x_1}{x_{1}^{2} + x_{2}^{2}} = \frac{0 \cdot (x_{1}^{2} + x_{2}^{2}) - x_1 \cdot 2 x_2}{(x_{1}^{2} + x_{2}^{2})^2} = - \frac{2 x_1 x_2}{(x_{1}^{2} + x_{2}^{2})^2}$
$\frac{d}{dx_1} \frac{x_2}{x_{1}^{2} + x_{2}^{2}} = - \frac{2 x_1 x_2}{(x_{1}^{2} + x_{2}^{2})^2}$
$\frac{d}{dx_2} \frac{x_2}{x_{1}^{2} + x_{2}^{2}} = \frac{(x_{1}^{2} - x_{2}^{2})}{(x_{1}^{2} + x_{2}^{2})^2}$

$$ \int ( \frac{(x_{2}^{2} - x_{1}^{2})}{(x_{1}^{2} + x_{2}^{2})^2} - \frac{2 x_1 x_2}{(x_{1}^{2} + x_{2}^{2})^2} ) \phi(x_1,x_2) d(x_1) + \int (- \frac{2 x_1 x_2}{(x_{1}^{2} + x_{2}^{2})^2} + \frac{(x_{1}^{2} - x_{2}^{2})}{(x_{1}^{2} + x_{2}^{2})^2} ) \phi(x_1,x_2) d(x_2) = ... ?$$

But I actually don't see where this is going. I again got two integrals which I can't solve.

 

[Orodruin]

Let me stop you from the beginning. The divergence of $x/|x|^2$ is not given by
$$
\newcommand{\dd}[2]{\frac{\partial #1}{\partial #2}}
\dd{(x/|x|^2)}{x_1} + \dd{(x/|x|^2)}{x_2}
$$
it is given by
$$
\dd{(x_1/|x|^2)}{x_1} + \dd{(x_1/|x|^2)}{x_2}.
$$
Apart from that, you still have not utilised the definition of the distributional derivative. Also, once you do, there is a very useful coordinate system called polar coordinates.

 

[Mark44]

Relatively minor point: why are you writing $\frac{x}{|x|^2}$? Since $|x|^2 = x^2$, the fraction is the same as $\frac 1 x$.

 

[kairosx]

Relatively minor point: why are you writing $\frac{x}{|x|^2}$? Since $|x|^2 = x^2$, the fraction is the same as $\frac 1 x$.

But $x \in R^2$ is a vector $x = (x_1,x_2)^T$ and how is $\frac{1}{vector}$ is defined?

Okay, i started once again:
$x \in R^2$ is a vector with the components $x = (x_1,x_2)^T$

$$
\frac{ \partial (x_1/|x|^2) }{\partial x_1} + \frac{ \partial (x_2/|x|^2)}{ \partial x_2} =\\
- <\frac{x_1}{|x|^2},\phi_{x_1}(x)> - <\frac{x_2}{|x|^2},\phi_{x_2}(x)> = ...
$$
using polar coordinates: $x_1 = r cos \theta$, $x_2 = r sin \theta$ and especially $|x| = r$ and not forgetting the jacoby determinant which is $r$ in 2 dimensions

$$
<-\frac{r cos \theta}{r^2} r,\phi_{x_1}(x)> - <\frac{r sin \theta}{r^2} r,\phi_{x_2}(x)> =\\
-\int_{0}^{2 \pi} \int_{0}^{\infty} cos \theta \phi_{x_1}(x) + sin \theta \phi_{x_2}(x) dr d \theta = ...\\
$$

but now I don't know how to express $\phi_{x_1}(x)$ and $\phi_{x_2}(x)$ in polar coordinates. How do I express $\frac{d}{dx_1}$ in polar coordinates? $\frac{d}{d(r cos \theta)}$ isn't really helpful as long as I not just have expressions that explicitely contain $(rcos \theta)$ ?

Another approch I did was:

$$
-\int_{0}^{2 \pi} \int_{0}^{\infty} (cos \theta,sin \theta) \space div(\phi(x)) dr d \theta = \\
\int_{0}^{2 \pi} \int_{0}^{\infty} div(cos \theta,sin \theta) \space \phi(x) dr d \theta =... \\
$$

and the divergence in polar coordinates is $div(x) = \nabla \cdot x = \frac{1}{r} \frac{d(r x_r)}{ dr} + \frac{1}{r} \frac{d(x_{\theta})}{ d \theta}$ and I would try to build the divergence of the vector, but I already assumed the vector to have a cartesian basis, why I shouldn't apply the divergence in polar coordinates I guess. But if I applied it in cartesian coordinates the expression gets really messy and I don't think that's the way I am supposed to solve the problem.

 

[Orodruin]

but now I don't know how to express ϕx1(x)ϕx1(x)\phi_{x_1}(x) and ϕx2(x)ϕx2(x)\phi_{x_2}(x) in polar coordinates

Have you heard of the chain rule?

 

[Mark44]

Relatively minor point: why are you writing $\frac{x}{|x|^2}$? Since $|x|^2 = x^2$, the fraction is the same as $\frac 1 x$.

But $x \in R^2$ is a vector $x = (x_1,x_2)^T$ and how is $\frac{1}{vector}$ is defined?

It wasn't clear to me that you were working with a vector here.

 

[Orodruin]

Also, some LaTeX pointers:

< becomes $<$, a relation (less than) and is typeset as such. For inner products, use \langle $\langle$ and \rangle $\rangle$, which are delimiters.

R becomes $R$, typically used to represent numbers such as a radius. \mathbb R becomes $\mathbb R$, the notation for the set of real numbers.

To avoid confusion, it is often good to use vector arrows or boldface to represent vectors. For example \vec v ($\vec v$) or \boldsymbol v ($\boldsymbol v$).

It is often good to define commands using \newcommand. It works here too. I defined \newcommand{\dd}[2]{\frac{\partial #1}{\partial #2}} in an earlier post. It will work in any post here until a second page is generated. For example \dd{f}{x} becomes
$$
\dd{f}{x}
$$

 

[kairosx]

Okay, thanks for the tipps about the notation and sorry again for the confusion about the vector identity! :-) I tried to come forward now by considering
$r = \sqrt{x_1^2 + x_2^2}$
$\theta = arctan(\frac{x_2}{x_1})$
$\frac{dr}{dx_1} = \frac{ x_1}{\sqrt{x_1^2 + x_2^2}} = \frac{ x_1}{r} = \frac{ r cos \theta}{r} = cos \theta$
$\frac{d \theta}{dx_1} = \frac{1}{1 + (\frac{x_2}{x_1})^2} \cdot \frac{-x_2}{x_1^2} = \frac{-x_2}{x_1^2 + x_2^2} = \frac{-x_2}{r^2} = \frac{-r sin \theta}{r^2} = \frac{-sin \theta}{r}$
$\frac{dr}{dx_2} = sin \theta$
$\frac{d \theta}{dx_2} = \frac{1}{1 + (\frac{x_2}{x_1})^2} \cdot \frac{1}{x_1} = \frac{cos \theta}{r}$

Then I can write

\begin{equation}
\frac{d}{dx_1} \phi(r(x,y), \theta(x,y)) = \phi_r \frac{dr}{dx_1} + \phi_{\theta} \frac{d \theta}{dx_1} = \phi_r cos \theta + \phi_{\theta} \frac{-sin \theta}{r}
\end{equation}

respectively

\begin{equation}
\frac{d}{dx_2} \phi(r(x,y), \theta(x,y)) = \phi_r \frac{dr}{dx_1} + \phi_{\theta} \frac{d \theta}{dx_1} = \phi_r sin \theta + \phi_{\theta} \frac{cos \theta}{r}
\end{equation}

and by inserting it into the former equation\begin{align}
-\int_{0}^{2 \pi} \int_{0}^{\infty} cos \theta \phi_{x_1}(x) + sin \theta \phi_{x_2}(x) dr d \theta &=\\
-\int_{0}^{2 \pi} \int_{0}^{\infty} cos \theta (\phi_r cos \theta + \phi_{\theta} \frac{-sin \theta}{r}) + sin \theta ( \phi_r sin \theta + \phi_{\theta} \frac{cos \theta}{r}) dr d \theta &=\\
-\int_{0}^{2 \pi} \int_{0}^{\infty} \phi_r + \phi_{\theta}\frac{- cos \theta sin \theta}{r} + \phi_{\theta} \frac{cos \theta sin \theta}{r} dr d \theta &=\\
-\int_{0}^{2 \pi} \int_{0}^{\infty} \phi_r dr d \theta &=\\
-2 \pi [\phi(\infty) - \phi(0)] =
2 \pi \phi(0) = 2 \pi \langle \delta_{(0,0)}, \phi(x) & \rangle
\end{align}

And that's it, I guess? :-D Thank you so much for your time and help! :-)

 

[Orodruin]

While you got it correctly, keep in mind that it is actually easier (although this may be subjective) to express $x_1$ and $x_2$ in terms of $r$ and $\theta$ and differentiate those expressions wrt $x_1$ and $x_2$ and then solve for the partial derivatives of $r$ and $\theta$. Maybe it is just me who does not like the inversion formulas in terms of the square root and arctan...

 

[kairosx]

While you got it correctly, keep in mind that it is actually easier (although this may be subjective) to express $x_1$ and $x_2$ in terms of $r$ and $\theta$ and differentiate those expressions wrt $x_1$ and $x_2$ and then solve for the partial derivatives of $r$ and $\theta$. Maybe it is just me who does not like the inversion formulas in terms of the square root and arctan...

I'm not sure if I understand what you mean. Actually, I really don't like the square and arctan expressions either, but how else would you do it? You mean expressing $x_1$ and $x_2$ in terms of $r$ and $\theta$ like $x_1 = r cos \theta$ and $x_2 = r sin \theta$? But if I then differentiate $\frac{ dx_1(r, \theta) }{dx_1}$(??) and $\frac{ dx_2(r, \theta) }{dx_2}$(??) what would I expect to get?

 

[Orodruin]

Differentiate both sides of the relations considering $x_i$ as independent variables and r and theta as functions of those variables.