Dirac delta of nonlinear multivariable arguments

How does one compute a dirac delta function with a multivariable argument? For example, compute:

$$
\int^{\infty}_{-\infty}{\rm d}x\,{\rm d}y\,
\delta\left(x^{2} + y^{2} – 4\right)
\delta\left(\left[x – 1\right]^{2} + y^{2} -4\right){\rm f}\left(x,y\right)
$$

if we constrain the two delta functions we’ll get two intersecting circles, and it seems reasonable to state that we evaluate $f(x,y)$ at the intersecting points, but I feel like there should be some extra identities.

since $$\delta(f(x))=\sum_i \frac{\delta(x-x_i)}{|f'(x)|},$$ is there a multivariable generalization to be aware of?

Solutions Collecting From Web of "Dirac delta of nonlinear multivariable arguments"

Hints:

  1. There exists an $n$-dimensional generalization
    $$\tag{1} \delta^n({\bf f}({\bf x})) ~=~\sum_{{\bf x}_{(0)}}^{{\bf f}({\bf x}_{(0)})=0}\frac{1}{|\det\frac{\partial {\bf f}({\bf x})}{\partial {\bf x}} |}\delta^n({\bf x}-{\bf x}_{(0)}) $$
    of the substitution formula for the Dirac delta distribution
    under pertinent assumptions, such as e.g., that the function ${\bf f}:\Omega \subseteq \mathbb{R}^n \to \mathbb{R}^n$ has isolated zeros. Here the sum on the rhs. of eq. (1) extends to all the zeros ${\bf x}_{(0)}$ of the function ${\bf f}$.

  2. Example: The function
    $$\tag{2} {\bf f}(x,y)~=~(x^2+y^2-4,(x-1)^2+y^2-4)$$
    with Jacobian determinant
    $$\tag{3}\det\frac{\partial {\bf f}(x,y)}{\partial (x,y)} ~=~4y,$$
    has two zeros
    $$\tag{4} (x,y)~=~(\frac{1}{2},\pm \frac{\sqrt{15}}{2}),$$
    leading to
    $$\tag{5} \delta^2({\bf f}(x,y)) ~\stackrel{(1)}{=}~\frac{1}{2\sqrt{15}}\delta(x-\frac{1}{2})\sum_{\pm} \delta(y\mp\frac{\sqrt{15}}{2}). $$

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\begin{align}
&\color{#0000ff}{\large%
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\rm d}x\,{\rm d}y\,
\delta\pars{x^{2} + y^{2} – 4}
\delta\pars{\bracks{x – 1}^{2} + y^{2} -4}\fermi\pars{x,y}}
\\[3mm]&=
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\rm d}x\,{\rm d}y\,
\delta\pars{x^{2} + y^{2} – 4}
\delta\pars{\bracks{x^{2} – 2x + 1} + y^{2} -4}\fermi\pars{x,y}
\\[3mm]&=
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\rm d}x\,{\rm d}y\,
\delta\pars{x^{2} + y^{2} – 4}
\delta\pars{2x – 1}\fermi\pars{x,y}
\\[3mm]&=
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\rm d}x\,{\rm d}y\,
\delta\pars{x^{2} + y^{2} – 4}\,
{\delta\pars{x – 1/2} \over 2}\,\fermi\pars{x,y}
\\[3mm]&=
\half\int_{-\infty}^{\infty}{\rm d}y\,
\delta\pars{\bracks{\half}^{2} + y^{2} – 4}\,\fermi\pars{\half,y}
\\[3mm]&=
\half\int_{-\infty}^{\infty}{\rm d}y\,\bracks{%
{\delta\pars{y + \root{15}/2} \over \root{15}}
+ {\delta\pars{y – \root{15}/2} \over \root{15}}}\fermi\pars{\half,y}
\\[3mm]&=\color{#0000ff}{\large%
{\root{15} \over 30}\bracks{\fermi\pars{\half,-\,{\root{15} \over 2}} + \fermi\pars{\half,{\root{15} \over 2}}}}
\end{align}

I worked with similar objects during my Master’s project, and we had to derive a formula for that…couldn’t find it anywhere. The formula can be found in the links below.

See section 3.7

or

see section A.4.7.

I’ve called it a $”\bf Sweet ~Dirac-\delta~ formula”$ or “A lemma on twofold Dirac delta functions”.

There is indeed a multivariable generalization of the identity you mentioned. See this Wikipedia article, where it says “As in the one-variable case, it is possible to define the composition of $δ$…”