Question
Asked 17th May, 2023

# How to write δ(f(x,y)) in terms of δ function?

I know that δ(f(x))=∑δ(x−xi)/f′(xi). What will be the expression if "f" is a function of two variables, i.e. δ(f(x,y))=?

## Most recent answer

Lyudmil Antonov
Bulgarian Drug Agency
K. Kassner You are right.
The method that I proposed may not work for all functions f(x,y), especially if they are continuous and do not have isolated zeros. In that case, one might try to separate the integrals or use a coordinate transformation as you suggested in your comment. For example, if we use polar coordinates $(r,\theta)$, then we have
$$\delta(f(r,\theta))=\frac{1}{r}\delta(r-r(\theta))$$
where $r(\theta)$ is the zero of f(r,$\theta$) as a function of r for a fixed $\theta$. This can be seen by using the Jacobian of the transformation and the property of the delta function.

## All Answers (6)

Lyudmil Antonov
Bulgarian Drug Agency
This is a challenging question. One possible way to write δ(f(x,y)) in terms of δ function is to use the following formula:
$$\delta(f(x,y)) = \sum_{i,j} \frac{\delta(x-x_i)\delta(y-y_j)}{|\nabla f(x_i,y_j)|}$$
where $(x_i,y_j)$ are the points where $f(x,y)=0$ and $\nabla f$ is the gradient of $f$. This formula is analogous to the one-variable case, but it requires that $f$ has isolated zeros and that $\nabla f$ does not vanish at those points.
Another possible way to write δ(f(x,y)) is to use the Fourier transform:
$$\delta(f(x,y)) = \frac{1}{(2\pi)^2}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-i(kx+ly)}\hat{\delta}(k,l) dk dl$$
where $\hat{\delta}(k,l)$ is the Fourier transform of $\delta(f(x,y))$, which can be computed by using the properties of the delta function and the Fourier transform.
To express δ(f(x, y)) in terms of the Dirac delta function, we can use the following representation:
δ(f(x, y)) = Σ [δ(x - x_i) δ(y - y_i)] / |∇f(x_i, y_i)|
Here's a breakdown of the components in this expression:
δ(x - x_i) and δ(y - y_i) are individual Dirac delta functions centered at specific points (x_i, y_i) that satisfy the equation f(x_i, y_i) = 0. These points are determined by the zeros of the function f(x, y).
|∇f(x_i, y_i)| represents the magnitude of the gradient of f at each of those zero points. ∇f denotes the gradient vector (∂f/∂x, ∂f/∂y), and |∇f| denotes its magnitude.
To evaluate δ(f(x, y)) at a particular point (x0, y0), you would substitute that point into the expression above, summing over all relevant zero points (x_i, y_i) that satisfy f(x_i, y_i) = 0. The resulting expression would then be the value of δ(f(x, y)) at (x0, y0).
Note that this representation assumes that the function f(x, y) is differentiable and has isolated zeros. If f(x, y) has multiple zeros that coincide at a single point, the expression needs to be modified accordingly
Gopal Sharma
Kyushu University
To write δ(f(x, y)) in terms of the Dirac delta function, you can use the property of the delta function known as the sifting property. The sifting property states that if g(x) is a continuous function with a single root at x = a, then the integral of g(x) multiplied by the Dirac delta function δ(x - a) is equal to g(a).
Applying this property to the function f(x, y), you can express δ(f(x, y)) as follows:
δ(f(x, y)) = Σ[δ(x - xi, y - yi) / |∇f(xi, yi)|],
where the sum is taken over all points (xi, yi) where f(xi, yi) = 0, and ∇f(xi, yi) is the gradient of f evaluated at (xi, yi). The |∇f(xi, yi)| represents the magnitude of the gradient.
This expression accounts for the fact that the Dirac delta function is being evaluated at the points where f(x, y) is zero. The presence of the gradient magnitude in the denominator ensures that the Dirac delta function is appropriately weighted at each point to maintain the integral property.
It's important to note that the expression above assumes that f(x, y) is continuous and differentiable, and that the points (xi, yi) where f(xi, yi) = 0 are isolated. If f(x, y) has multiple roots or singularities, the expression may require modification to handle those cases appropriately.
K. Kassner
Otto-von-Guericke-Universität Magdeburg
Lyudmil Antonov "but it requires that $f$ has isolated zeros and that $\nabla f$ does not vanish at those points"
Which seems difficult to achieve for continuous functions. In order for f to have isolated zeros, it cannot have both signs near the zero, otherwise there must be a curve through the zero separating the regions of positive and negative f(x,y). An example with an isolated zero would be f(x,y) = x2n+y2n, with n integer, which is purely positive around the isolated zero at (0,0). But then ∇f vanishes at the zero...
Applications of a delta function of two variables will typically involve continuous functions, not discontinuous ones, which would, moreover, not be differentiable at the discontinuity.
One way to deal with a situation as the one given in the question would be to separate the integral in an integral over x and y. Then y is a parameter for the x integral, and
δ(f(x,y))=∑iδ(x−xi(y))/|∂f(x,y)/∂x|x=xi(y),
where xi(y) are the zeros of f(x,y) as a function of x at the considered value of y.
If a simple separation of the integrals is not possible, then one might try to find a coordinate transformation (e.g. to polar coordinates), in which it is feasible. In any case, the delta function will remove only one integration, not all of them (assuming the expression is to be integrated over a surface).
Lyudmil Antonov
Bulgarian Drug Agency
K. Kassner You are right.
The method that I proposed may not work for all functions f(x,y), especially if they are continuous and do not have isolated zeros. In that case, one might try to separate the integrals or use a coordinate transformation as you suggested in your comment. For example, if we use polar coordinates $(r,\theta)$, then we have
$$\delta(f(r,\theta))=\frac{1}{r}\delta(r-r(\theta))$$
where $r(\theta)$ is the zero of f(r,$\theta$) as a function of r for a fixed $\theta$. This can be seen by using the Jacobian of the transformation and the property of the delta function.

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