# Delta function of two variable function

1. Jun 20, 2012

### mimmim

Hi
Iknow that if we have delta function of one variables function, then we can write it as:

$\delta (f(x)) = \sum \frac{\delta(x-x0)}{f'(x0)}$

but how we can write a function of two variables:

$\delta (f(x,y))$

2. Jun 20, 2012

### Frogeyedpeas

What exactly do you mean by delta function? Are you referring to dirac delta function or something else?

3. Jun 20, 2012

### mimmim

Dirac delta function

4. Jun 21, 2012

### theorem4.5.9

I am not aware of such an identity. What is the sum over?

The dirac delta function for a single variable is simply $\delta [f(x)] = f(0)$. In two variables it would be $\delta [f(x,y)] = f(0,0)$. Of course you can shift the delta function to any point off the origin as well.

5. Jun 22, 2012

### lugita15

This is a well-known and easily provable identity. The sum is over the zeroes x0 of f.
I think there's some confusion here. A distribution can indeed be represented as a linear functional which takes (test) functions as inputs. But here, when we say $\delta (f(x))$, we don't mean the functional $\delta$ evaluated at the function f. Rather, informally we are thinking of the Dirac delta as a function of numbers, and we are forming the composition of the Dirac delta function and the function f. So we mean the "function" $\delta (u)$ evaluated at the number u=f(x).

6. Jun 22, 2012

### pwsnafu

One way to go about this is by the observation that every real number has an associated constant function $g_c (t) = c$. Trivally every constant function is a distribution. Then $\delta(f(x)) = \delta \circ f \circ g_x$ and if the resulting expression is also a constant function you can use the duality backwards to "claim" $\delta(f(x)) \in \mathbb{R}$.

I remember hearing about this in a lecture some time in my undergraduate years, but I haven't seen it mentioned in the literature.

7. Jun 22, 2012

### theorem4.5.9

Would you mind elaborating on this? I've worked out the case where $f$ is injective below, but I don't see how one would define composition where $f$ vanishes at more than one point.

If $d, g, f$ are test functions and $f$ is injective and $f(x_0)=0$ then
$$\langle d\circ f, g \rangle = \int d\circ f(x)\cdot g(x) dx$$. A u-substitution yields
$$= \int\frac{d(u)\cdot g\circ f^{-1}(u)}{f'(f^{-1}(u))}du = \left\langle d(u), \frac{ g\circ f^{-1}(u)}{f'(f^{-1}(u))} \right\rangle$$

Thus this should be taken as the definition for composition of a general distribution with such an $f$. Inserting the dirac delta then, I see that
$$\langle \delta\circ f,g\rangle = \frac{ g\circ f^{-1}(0)}{f'(f^{-1}(0))} = \frac{ g(x_0)}{f'(x_0)} = \left\langle\frac{\delta_{x_0}}{f'(x_0)} ,g\right\rangle$$
which recovers the formula (excusing my sloppy u-substitution).

To reiterate my question, if $f$ vanishes at several points, what motivation is there to define the OP's formula?

8. Jun 22, 2012

### madness

9. Jun 22, 2012

### lugita15

One approach you might try is to calculate in a similar fashion $\delta (f(x)g(x))$, where f(x) has a single zero at x1 and g(x) has a single zero at x2.

10. Jun 22, 2012

### madness

The delta function is non-zero only when it's input is zero. The OP's formula picks out the points where the argument of the delta function is non-zero and expands the expression into delta functions centred at these points.

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