Delta dirac function of (x^2)

  • Thread starter chessmath
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  • #1
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Hi
I would like to know what is the dirac delta function of x^2, I read somewhere it is equal to delta of x itself but why?
 

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  • #2
pwsnafu
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The Dirac delta is a map whose input is a function and output is the value at the origin. In symbols: ##f \mapsto f(0)##. There's lots different notations that are used, but they all do the essentially same thing. (Note that some notations explicitly translate function over as well, but I'm going to ignore that.)

Now the functions ##f(x)=x^2## and ##g(x) = x## both pass through the origin, so when you do Dirac to either you get zero.
 
  • #3
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I disagree. Consider the following variable transformation: [tex]u=x^2 \qquad \mathrm{d}u=2x\,\mathrm{d}x[/tex] [tex]\int_{-\infty}^{\infty} \delta(x^2)2x\,\mathrm{d}x = 2 \int_{0}^{\infty} \delta(u)\,\mathrm{d}u = 1[/tex]

But [tex]\int_{-\infty}^{\infty} \delta(x)2x\,\mathrm{d}x = 2x\mid_{x=0} = 0[/tex] So the delta functions are not the same.
 
  • #4
Mute
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I disagree. Consider the following variable transformation: [tex]u=x^2 \qquad \mathrm{d}u=2x\,\mathrm{d}x[/tex] [tex]\int_{-\infty}^{\infty} \delta(x^2)2x\,\mathrm{d}x = 2 \int_{0}^{\infty} \delta(u)\,\mathrm{d}u = 1[/tex]

But [tex]\int_{-\infty}^{\infty} \delta(x)2x\,\mathrm{d}x = 2x\mid_{x=0} = 0[/tex] So the delta functions are not the same.
I think one needs to be more careful with the limits when changing variables. The singularity in the integral occurs right at the point where one needs to split up the integral to do the change of variables, which prevents us from doing so.

Consider what happens if we do the calculation like this instead:

$$I = \int_{-\infty}^\infty dx~2x \delta(x^2 - \epsilon),$$
where ##\epsilon## is a small positive constant to be taken to zero at the end of the calculation. Using the rule

$$\delta(g(x)) = \sum_i \frac{\delta(x-x_i)}{|g'(x_i)|},$$
where the ##x_i## are zeros of g(x), we have

$$\begin{eqnarray*}
I & = & \int_{-\infty}^\infty dx~2x \left[ \frac{\delta(x-\sqrt{\epsilon})}{2\sqrt{\epsilon}} + \frac{\delta(x+\sqrt{\epsilon})}{|-2\sqrt{\epsilon}|}\right] \\
& = & \frac{1}{\sqrt{\epsilon}}\left[ \sqrt{\epsilon} + (-\sqrt{\epsilon})\right] \\
& = & 0.
\end{eqnarray*}$$
(Taking the limit at this point is, of course, trivial).
 
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  • #5
pwsnafu
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I disagree. Consider the following variable transformation: [tex]u=x^2 \qquad \mathrm{d}u=2x\,\mathrm{d}x[/tex] [tex]\int_{-\infty}^{\infty} \delta(x^2)2x\,\mathrm{d}x = 2 \int_{0}^{\infty} \delta(u)\,\mathrm{d}u = 1[/tex]
Invalid. Integration by substitution is only well defined when we are dealing with functions with real variables.

PS: Dirac is a functional. So, the statement "Dirac delta of g" means "apply Dirac delta to g". If you want ##\delta \circ g## you need to write "Dirac composed with g".
 
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  • #6
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Sorry, people but wikipedia agrees with me. The variable transform rule is the only sensible way to define [tex]\delta(f(x))[/tex]. This can also be seen in the definition of the delta function as the limit of an integral over a thinner and thinner peak. The variable transform can be done before taking the limit so it is legal. Due to this definition we can also see that the following makes sense (and I have seen it in textbooks) [tex]\int_0^\infty \delta(x) f(x) = \frac{1}{2}f(0)[/tex] Mute's argument is nice but the second formula is only valid if g'(x) is not zero. It is in fact a result of the variable substitution formula, because locally you can approximate g(a) as a linear function and 1/g' is the volume element for the corresponding variable change. It seems as if the limiting process is not legal. If we do any limits they should be done with a definition of the delta function as a limit. Although I also find it a bit funny that the modified delta function integral of a function that is zero at the origin will produce one. pwsnafu's definition of the delta function seems to differ from that of the rest of the world.
 
  • #7
Mute
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Sorry, people but wikipedia agrees with me. The variable transform rule is the only sensible way to define [tex]\delta(f(x))[/tex]. This can also be seen in the definition of the delta function as the limit of an integral over a thinner and thinner peak. The variable transform can be done before taking the limit so it is legal. Due to this definition we can also see that the following makes sense (and I have seen it in textbooks) [tex]\int_0^\infty \delta(x) f(x) = \frac{1}{2}f(0)[/tex] Mute's argument is nice but the second formula is only valid if g'(x) is not zero. It is in fact a result of the variable substitution formula, because locally you can approximate g(a) as a linear function and 1/g' is the volume element for the corresponding variable change. It seems as if the limiting process is not legal. If we do any limits they should be done with a definition of the delta function as a limit. Although I also find it a bit funny that the modified delta function integral of a function that is zero at the origin will produce one. pwsnafu's definition of the delta function seems to differ from that of the rest of the world.
The change of variables does not work in your calculation because you have to split the integral up right at the point where the delta function should "ping". That needs to be treated more carefully. (Also, we are not arguing against the convention ##\int_0^\infty \delta(x)f(x) = f(0)/2##.)

In my calculation g'(x) where the delta function "pings" because I have shifted the discontinuity away from zero to a value where the derivative exists, and hence the calculation is valid. The only way it can be incorrect is to argue that

$$\lim_{\epsilon \rightarrow 0} \int_{-\infty}^\infty dx~2x \delta(x^2-\epsilon) \neq \int_{-\infty}^\infty dx~2x \delta(x^2).$$

So, let's try this as you suggest: let's take ##\int_{-\infty}^\infty dx~2x \delta(x^2)## to mean

$$\lim_{\epsilon\rightarrow 0}\int_{-\infty}^\infty dx~2x \delta_\epsilon(x^2),$$

where ##\delta_\epsilon(x^2)## is a nascent delta function. Let's choose

$$\delta_\epsilon(x^2) = \frac{1}{\sqrt{2\pi}\epsilon} \exp\left(-\frac{(x^2)^2}{2\epsilon^2}\right).$$

Then,

$$\int_{-\infty}^\infty dx~2x \frac{1}{\sqrt{2\pi}\epsilon} \exp\left(-\frac{(x^2)^2}{2\epsilon^2}\right) = 0$$
by symmetry.

Hm. Well, let's try a nascent delta function that's not symmetric, then. Let's consider

$$\delta_\epsilon(x) = \frac{\mbox{Ai}(x/\epsilon)}{\epsilon},$$
where Ai(x) is the Airy function. But then we have ##\delta_\epsilon(x^2) = Ai(x^2/\epsilon)/\epsilon##, and so it will turn out to be an even integrand again: Changing variables to ##y = x/\sqrt{\epsilon}## gives

$$\int_{-\infty}^\infty dx~2x \frac{\mbox{Ai}(x^2/\epsilon)}{\epsilon} = \int_{-\infty}^\infty dy~2y \mbox{Ai}(y^2) = 0.$$

So, for the moment I remain convinced that the answer is zero in this case.

Note, however, there are potential problems, at least computationally: if we consider a general function ##f(x)##, then

$$\int_{-\infty}^\infty dx~f(x) \frac{\mbox{Ai}(x^2/\epsilon)}{\epsilon} = \int_{-\infty}^\infty dy~\frac{f(\sqrt{\epsilon}y)}{\epsilon} \mbox{Ai}(y^2).$$

If the function f(x) is odd in y, then the integral is of course still zero by symmetry. However, if f is not odd then ##f(\sqrt{\epsilon}y)/\sqrt{\epsilon}## may diverge as ##\epsilon## tends to zero, in which case we cannot take the limit inside the integrand. This does not imply the limit does not exist, per se, but we would have to do the integral first and then take the limit to find out whether it does or not.
 
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  • #8
pwsnafu
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pwsnafu's definition of the delta function seems to differ from that of the rest of the world.
What? Did even read Wikipedia, that you cite?

Sorry, people but wikipedia agrees with me. The variable transform rule is the only sensible way to define
Are you citing this? Then you need the derivative of the inverse.

PS: Wait are you citing this? Then it clearly states:
provided that g is a continuously differentiable function with g′ nowhere zero.
But ##g(x)=x^2## doesn't satisfy this. So I ask you again: how do you define ##\delta(x^2)##. If ##\phi_n \rightarrow \delta## then (if I understand your argument)
##\lim_{n\rightarrow\infty}\int_{\mathbb{R}} \phi_n(x^2) \, f(x) \, dx = \lim_{n\rightarrow\infty}\int_0^\infty \phi_n(u) \, \frac{f(\sqrt u}{2\sqrt u} \, du##
but I'm not seeing how the RHS is well-defined as a generalised function for some arbitrary test function f.

I'm personally starting to get the feeling ##\delta(x^2)## is not a well-defined generalised function.
 
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