2
$\begingroup$

It is well known that $\delta'(x)\phi(x)=-\delta(x)\phi'(x)$ in a distributional sense, for some well behaved test function $\phi(x)$. However, what about the case:

$$\int_{-\infty}^\infty dz\int_{-\infty}^\infty dy\int_{-\infty}^\infty dx \,\delta'(x+y+z)\phi(y)\psi(z)=~~???$$

Is this equal to zero? Or maybe it is equal to $\int_{-\infty}^\infty dz\int_{-\infty}^\infty dy\,(\phi'(y)\psi(z)+\phi(y)\psi'(z))$? Who knows?

$\endgroup$
  • 1
    $\begingroup$ Yes, it's zero, since the function that is adjacent to the delta function is independent of x. $\endgroup$ – Paul Jul 19 '17 at 0:05
  • 1
    $\begingroup$ For this case you need the co-area formula (as you would even without the derivative). $\endgroup$ – Ian Jul 19 '17 at 1:23
  • 1
    $\begingroup$ The co-area formula says that to integrate over (say) $\mathbb{R}^n$ it is enough to integrate over the level sets $S_y$ of a function $f : \mathbb{R}^n \to \mathbb{R}$ (which are $(n-1)$-dimensional manifolds), and then integrate in $y$. The resulting change of variable needs a factor of $|\nabla f|$ in the denominator of the new integral. If you have a multivariable argument inside a delta function, simply take that to be your $f$; then only the level set $S_0$ will actually contribute to the integration but you will get the correct $(n-1)$-dimensional integral. $\endgroup$ – Ian Jul 19 '17 at 2:11
  • 1
    $\begingroup$ Thus for example $\int_{-\infty}^\infty \int_{-\infty}^\infty g(x,y) \delta(x+y) dx dy = \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{g(x,z-x)}{\sqrt{2}} \delta(z) dz dx = \int_{-\infty}^\infty \frac{g(x,-x)}{\sqrt{2}} dx$ Perhaps it was obvious that only $g(x,-x)$ could possibly contribute to the integration, but it was probably not obvious that there should be this $\sqrt{2}$. $\endgroup$ – Ian Jul 19 '17 at 2:13
  • 1
    $\begingroup$ I actually made a mistake in my example specifically: I forgot that in this case $dS(x)=\sqrt{2} dx$ so the factor is canceled. A better example: $\int_{-\infty}^\infty \int_{-\infty}^\infty g(x,y) \delta(x^2+y^2-1) dx dy = \int_0^{2 \pi} \frac{1}{2r} g(r,\theta) r dr d \theta$ (with the obvious abuse of notation about the two different $g$'s). One could also have seen this by directly changing to polar coordinates and then noting that (by the usual definition of composition involving a delta function) $\delta(r^2-1)=\frac{1}{2}(\delta(r-1)-\delta(r-1))$. $\endgroup$ – Ian Jul 19 '17 at 14:14
1
$\begingroup$

In this case using the co-area formula define the plane $P_c$ by $x+y+z=c$ and its surface measure by $dS_c$, then

\begin{align} \int_{\mathbb{R}^3} \delta'(x+y+z) \phi(y) \psi(z) dx dy dz & = \frac{1}{\sqrt{3}} \int_{-\infty}^\infty \int_{P_c} \delta'(c) \phi(y) \psi(z) dS_c dc \\ & = -\frac{1}{\sqrt{3}} \frac{d}{dc} \left. \int_{P_c} \phi(y) \psi(z) dS_c \right |_{c=0} \end{align}

To finish the problem you should choose a parametrization of the plane, compute the integral with $c$ as a parameter and then differentiate. I think that since there is no other dependence on $x$ that you will actually get zero (by parametrizing the plane by $(y,z)$ taking $x=c-y-z$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.