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


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  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 $$\tag{5} \delta^2({\bf f}(x,y)) ~\stackrel{(1)}{=}~\frac{1}{2\sqrt{15}}\delta(x-\frac{1}{2})\sum_{\pm} \delta(y\mp\frac{\sqrt{15}}{2}). $$

  • 2
    $\begingroup$ I cannot comment so I will put my question to Qmechanic here as an answer. @Qmechanic, can you give a reference where I can find the formula you quoted? Thanks. $\endgroup$ – Poor Soul Dec 29 '13 at 21:53
  • $\begingroup$ I second @PoorSoul's request. Is there a reference where your Eq. (1) is proved? $\endgroup$ – becko Jan 13 '16 at 2:12
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    $\begingroup$ @Qmechanic what does $|\text{det} \frac{d\mathbf{f}}{d\mathbf{x}}|$ mean when f is a multivariate scalar function? What is the equivalent relation to Eq. (1) in this case ? $\endgroup$ – Mencia Feb 11 '18 at 18:50
  • $\begingroup$ The formula (1) only makes sense if the domain and codomain of ${\bf f}$ have same dimension. $\endgroup$ – Qmechanic Feb 11 '18 at 19:23
  • $\begingroup$ However, see e.g. this Math.SE post. $\endgroup$ – Qmechanic Feb 11 '18 at 19:31

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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


see section A.4.7.

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

  • $\begingroup$ I haven't proved it, but I'm sure the "sweet Dirac-$\delta$ formula" can be generalized to higher dimensions. Have you tried it? $\endgroup$ – becko Jan 13 '16 at 1:44
  • $\begingroup$ Ah, just found the generalization in Qmechanics answer. $\endgroup$ – becko Jan 13 '16 at 2:13

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 $δ$..."


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