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How to simplify the dirac delta of squared x?

How should i solve $\int f(x)\delta(x^2)dx$ for an enough smooth function $f$?

Is this related to $\int f'(x) \delta(x) dx$ ?

I can understand symbols like $\delta(x)$ and $\delta'(x)$ and $\delta(g(x))$ with $g$ having a root $x_0$ with $g'(x_0)\neq 0$ but i can't understand it when $g'(x_0)$ is not defined: $\delta(x^2)$,$\delta(|x|)$, etc.

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    $\begingroup$ $\delta (g(x))=\sum _{i}{\frac {\delta (x-x_{i})}{|g'(x_{i})|}}$ $\endgroup$
    – Brethlosze
    May 10, 2018 at 0:02
  • $\begingroup$ First you need to define it. $\endgroup$
    – user553213
    May 10, 2018 at 0:27
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    $\begingroup$ Related: math.stackexchange.com/q/2481114/11127 $\endgroup$
    – Qmechanic
    May 12, 2018 at 16:10

2 Answers 2

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This is a very good question. Let's generalize the regularization approach that Sangchul presented.

Let $\delta_n(x)$ be a sequence of (at least piece-wise smooth) positive-valued functions such that

$$\lim_{n\to \infty}\delta_n(x)=\begin{cases}0&,x\ne 0\\\\\infty&,x=0\tag1\end{cases}$$

and for each $n$

$$\int_{-\infty}^\infty \delta_n(x)\,dx=1\tag2$$

and for all suitable test functions, $\phi(x)$, we have

$$\lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x)\phi(x)\,dx=\phi(0)\tag2$$

We say that in distribution $\delta(x) \sim \lim_{n\to \infty}\delta_n(x)$. This is the Dirac Delta distribution and $\delta_n(x)$ is a regularization thereof.

Now, let's examine whether $\lim_{n\to \infty}\delta_n(x^2)$ has meaning in the distributional sense. We can write

$$\begin{align} \int_{-\infty}^\infty \delta_n(x^2)\phi(x)\,dx&=\int_0^\infty \delta_n(x)\left(\frac{\phi(\sqrt{x}\,)+\phi(-\sqrt{x}\,)}{2\sqrt{x}}\right)\,dx\\\\ &=\int_{-\infty}^\infty \delta_n(x)\left(\frac{\phi(\sqrt{|x|}\,)+\phi(-\sqrt{|x|}\,)}{2\sqrt{|x|}}\,H(x)\right)\,dx \end{align}$$

where $H(x)$ is the Heaviside function. Letting $n\to \infty$, we find for suitable test functions (smooth and of compact support)

$$\lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x^2)\phi(x)\,dx=\begin{cases}0&,\phi(0)=0\\\\\infty&,\phi(0)>0\\\\-\infty &,\phi(0)<0\end{cases}$$

So, $\delta(x^2)$ has meaning in distribution on the space of $C^\infty_C$ functions that are $0$ at the origin and the distribution assigns the value of $0$ to each of these functions. As Sangchul wrote, this is quite a "boring" distribution.


Now, if $\phi(x)$ is a test function such that $\phi(0)=0$, smooth a.e. except at $0$ where $\lim_{x\to 0^{\pm}}\phi'(x)=C^{\pm}$, then we see that

$$\lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x^2)\phi(x)\,dx=\left(\frac{C^+-C^-}{2}\right)\,H(0)$$

Inasmuch as $H(0)$ is not uniquely defined, then we cannot uniquely define $\delta(x^2)$ as a distribution acting on such functions. For example, if $\phi(x)=|x|$, then $C^\pm=\pm1$ and $\delta(x^2)\sim H(0)$, which is not uniquely defined.

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    $\begingroup$ @sangchullee I've generalized your approach a bit and identified a key issue, which is the non-uniqueness of, $H(0)$, the Heaviside function evaluated at $0$. This nonuniqueness is analogous to the regularization specification of $\delta_n(x)$. $\endgroup$
    – Mark Viola
    May 10, 2018 at 20:21
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    $\begingroup$ Thank you. It was a pleasure. $\endgroup$
    – Mark Viola
    May 10, 2018 at 21:03
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    $\begingroup$ For $\delta(x^3)$, we need $\phi(0)=\phi'(0)=0$. Then, $$\int_{-\infty}^\infty \delta(x^3)\phi(x)\,dx=\frac12\phi''(0)$$ $\endgroup$
    – Mark Viola
    May 10, 2018 at 21:11
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    $\begingroup$ @MarkViola: I believe it should be $\frac16\phi''(0)$ $\endgroup$
    – robjohn
    May 11, 2018 at 0:28
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    $\begingroup$ $$ \int_{-\infty}^\infty f(x)\,\delta\!\left(x^3\right)\mathrm{d}x =\frac13\int_{-\infty}^\infty f\!\left(x^{1/3}\right)x^{-2/3}\delta(x)\,\mathrm{d}x $$ $\endgroup$
    – robjohn
    May 11, 2018 at 0:44
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Let $\delta_n (x) = \frac{n}{2}\mathbf{1}_{[-1/n,1/n]}(x)$. Then $\delta_n$ converges to $\delta$ in distribution sense. Let $f$ be smooth. Then by the Taylor's theorem, we have $f(x) = f(0) + f'(0)x + \mathcal{O}(x^2)$ near $x = 0$. So it follows that

\begin{align*} \int_{\mathbb{R}} f(x)\delta_n(x^2) \, dx &= \frac{n}{2} \int_{-\frac{1}{\sqrt{n}}}^{\frac{1}{\sqrt{n}}} f(x) \, dx \\ &= \frac{n}{2} \int_{-\frac{1}{\sqrt{n}}}^{\frac{1}{\sqrt{n}}} \left( f(0) + f'(0)x + \mathcal{O}\left(\frac{1}{n}\right) \right) \, dx \\ &= f(0)\sqrt{n} + \mathcal{O}\left(\frac{1}{\sqrt{n}}\right). \end{align*}

So if $f(0) \neq 0$, then this integral diverges as $n\to\infty$, and if $f(0) = 0$, then this integral converges to $0$ as $n\to\infty$. This tells that $\delta(x^2)$ is either meaningless or boring when paired with nice functions.

On the other hand, if $f(x) = |x|$ then

\begin{align*} \int_{\mathbb{R}} f(x)\delta_n(x^2) \, dx = \frac{n}{2} \int_{-\frac{1}{\sqrt{n}}}^{\frac{1}{\sqrt{n}}} |x| \, dx = \frac{1}{2} \end{align*}

This suggests that $\delta(x^2)$ will capture the jump discontinuity of $f'$. Of course, making this heuristics into a meaningful statement would necessitate an appropriate theory of distributions on 'not-so-regular functions', which I do not know much.

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    $\begingroup$ Using a regularized Dirac Delta is considered rigorous, is it not? But the result should be independent of the choice of regularization. $\endgroup$
    – Mark Viola
    May 10, 2018 at 14:21
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    $\begingroup$ The approximations of $\delta$ do not need to be symmetric around $0$. This leads to inconsistent values of $\delta\!\left(x^2\right)$ against $|x|$. This can be seen by writing the integral as $$\int_0^\infty\frac{f\!\left(\sqrt{x}\right)+f\!\left(-\sqrt{x}\right)}{2\sqrt{x}}\,\delta(x)\,\mathrm{d}x$$ which for $f(x)=|x|$ gives $$\int_0^\infty1\,\delta(x)\,\mathrm{d}x$$ and this is dependent on the choice of regularization. $\endgroup$
    – robjohn
    May 10, 2018 at 15:02
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    $\begingroup$ @hyprfrcb, The last formula is simply a result of computing the integral. Then I tried to interpret this result as 'capturing the sudden change of slopes of the test function', solely based on the fact that $\int x\delta_n(x^2)\,dx \to 0$ while $\int |x|\delta_n(x^2)\,dx\to\frac{1}{2}$. Of course this is merely a heuristic idea, and I am not making any formal claim about the nature of $\delta(x^2)$. Indeed, other users pointed out that a naive notion of $\delta(x^2)$ is not robust under general regularizations. $\endgroup$ May 10, 2018 at 15:38
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    $\begingroup$ @MarkViola, I agree that regularizing Dirac delta is a very rigorous way of understanding $\delta$ against nice test functions. What I am worried is whether we can provide a satisfactory theory on $\delta(x^2)$ or its generalizations which is still robust under various regulaziations. Indeed, robjohn's comment tells that the situation is not as simple as in the that of $\delta(x)$. $\endgroup$ May 10, 2018 at 15:43
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    $\begingroup$ @robjohn The notation $\int_0^\infty f(x)\delta(x)\,dx$ is interpreted to mean $\langle f\,H, \delta\rangle$, where $H$ is the Heaviside function. Since the definition of $H(0)$ can be defined however one chooses, then the value of $\int_0^\infty f(x)\delta(x)\,dx$ is ambiguous up to that choice. Resolution of this ambiguity by choosing a value of $H(0)$ is analogous to a choice of regularization. $\endgroup$
    – Mark Viola
    May 10, 2018 at 16:08

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