Rewriting $\delta(x,y)$ in terms of $\delta(r)$.

Question

On my textbook is written:

The function $\tau^{-1} u(x/\tau)$ is a rectangle function of height $\tau^{-1}$ and base $\tau$ and has unit area; as $\tau$ tends to zero a sequence of unit-area pulses of ever increasing height is generated. \begin{equation} \lim_{\tau\rightarrow 0}\tau^{-1} u(x/\tau)=\delta(x) \end{equation} One encounters the two- and three-dimensional impulse symbols $\delta(x,y)$, $\delta(x,y,z)$ as natural generalizations of $\delta(x)$. In establishing properties of $\delta(x,y)$ one considers a sequence, as $\tau\rightarrow 0$ of functions such as \begin{equation} \tau^{-2} u(x/\tau)u(y/\tau) \end{equation} or \begin{equation} 4/\pi\ \ \ \tau^{-2} u\left(\sqrt{x^2+y^2}/\tau\right) \end{equation} which have unit volume. We have $\delta(x,y)=\delta(x)\delta(y)$. Introducing the radial coordinate r such that $r^2=x^2+y^2$ we can espress $\delta(x,y)$ in terms of $\delta(r)$: \begin{equation} \delta(x,y)=\frac{\delta(r)}{\pi|r|} \end{equation}

But this step is not clear to me. Can anyone help me please? How do we get \begin{equation} \delta(x,y)=\frac{\delta(r)}{\pi|r|} \end{equation} ? Thank you very much.

Answer 1

If the transformation between coordinates ${\pmb x}$ and ${\pmb y}$ is not singular then $$\delta({\pmb x}-{\pmb x}') = \frac{1}{|J|}\delta({\pmb y}-{\pmb y}'),$$ where $J$ is the Jacobian of the transformation.

Proof $\;$ Let $F$ a test function and $\varphi=(\varphi_1,\varphi_2)$ the transformation $\pmb{x}=\varphi(\pmb{y})$, i.e. $(x_1,x_2)=(\varphi_1(y_1,y_2),\varphi_2(y_1,y_2))$, with $\pmb{y}'$ the corrensponding value of $\pmb{x}'$ under $\varphi$ (i.e. $\pmb{x}'=\varphi(\pmb{y}')$). Let $J=\frac{\partial(\varphi_1,\varphi_2)}{\partial(y_1,y_2)}$ the jacobian matrix of the transformation, i.e. $\operatorname{d}\pmb{x}=J\operatorname{d}\pmb{y}$, and $|J|$ the determinant. Then $$ \int F(\pmb{x})\delta(\pmb{x}-\pmb{x}')\operatorname{d}\pmb{x}=F(\pmb{x}') $$ and $$ \int F(\varphi(\pmb{y}))\delta(\varphi(\pmb{y})-\pmb{x}')|J|\operatorname{d}\pmb{y}=F(\pmb{x}') $$ It follows that $\delta(\varphi(\pmb{y})-\pmb{x}')|J|$ assign to any test function $F$ the value of that test function at the point where $\varphi(\pmb{y})=\pmb{x}'$ that is $\pmb{y}=\pmb{y}'$. Hence, we obtain $$ \delta(\varphi(\pmb{y})-\pmb{x}')|J|=\delta(\pmb{x}-\pmb{x}')|J|=\delta(\pmb{y}-\pmb{y}') $$ and thus $$\delta(\pmb{x}-\pmb{x}')=\frac{1}{|J|}\delta(\pmb{y}-\pmb{y}').$$

The Jacobian is $r$ so, assuming $r'\ne 0$, the delta is $$\delta(x-x')\delta(y-y') = \frac{1}{r}\delta(r-r')\delta(\theta-\theta').$$ It satisfies the two properties $$ \frac{1}{r}\delta(r-r')\delta(\theta-\theta') = 0\qquad\text{for }(r,\theta)\neq (r',\theta')$$ and $$\int_0^\infty r dr\int_0^{2\pi}d\theta \ \frac{1}{r}\delta(r-r')\delta(\theta-\theta') = 1$$

If $r'=0$ we must integrate out the ignorable coordinate $\theta$, $J\to \int_0^{2\pi}d\theta \ J = 2\pi r$. Thus $$\delta(x)\delta(y) = \frac{1}{2\pi r}\delta(r).$$ Again $$ \frac{1}{2\pi r}\delta(r)= 0\qquad\text{for }r\neq 0$$ and $$\int_0^\infty r dr\int_0^{2\pi}d\theta \ \frac{1}{2\pi r}\delta(r) = 1.$$

Note that the domains of the polar arguments were just stated to be $0 ≤ r <+∞$ and $0\le\theta<2\pi$ (or equivalently $−\pi ≤\theta <+\pi$). For engineering application, like image processing for example, sometimes the domains for the polar and azimuthal variable is modified to equivalent forms with a bipolar radial variable over the domain $(−\infty, +\infty)$ and an azimuthal domain that is constrained to an interval of $\pi$ radians, such as $[0, +\pi)$ or $[−\pi/2, +\pi/2)$. In this case the representation of the delta is $$ \delta(x)\delta(y) = \frac{1}{\pi |r|}\delta(r) $$ and $$\int_{-\infty}^\infty r dr\int_0^{\pi}d\theta \ \frac{1}{\pi |r|}\delta(r) = 1.$$