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Is following statement correct? $$\delta(x-y)\delta(y)=\delta(x)\delta(y)$$

Edit: I need to explain where this question comes from :D I needed to solve the following equation: $$\frac{\partial G}{\partial x}+\frac{\partial G}{\partial y}=\delta (x)\delta(y)$$ with boundary condition of $G(x=\infty,y)=0$ and $G(x,y=0)=0$. To solve this, I get the laplace transform over y and come up with following equation: $$G=\delta(y-x)H(x)$$ I wanted to test if my solution is correct and this is the reason I ended up with this qustion :D I hope it helps

Bita
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  • What is the definition of the product of Schwartz distributions? Without that, your question cannot be answered. – GEdgar Mar 25 '20 at 19:32
  • This isn't super well-posed FWIW. Multiplication of distributions is a really finicky subject. The only sensible way to interpret the left hand side is as doing the $x$ distribution first, then $y$ (like $\langle \phi(y),\langle \psi(x),\delta(x-y)\rangle\delta(y)\rangle$), but that is not encoded in what you've written. – Cameron Williams Mar 25 '20 at 19:32
  • Let me add more details to the question now. – Bita Mar 25 '20 at 19:39
  • Meant to add: If the $y$ is attempted first, you then have two deltas in $y$ multiplied and there is no obvious way to interpret that even when acting on test functions. – Cameron Williams Mar 25 '20 at 19:40
  • @CameronWilliams I used regularizations of the Dirac Delta to show that the order of distribution can be interchanged. – Mark Viola Mar 25 '20 at 20:14

2 Answers2

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For any test function $\phi(x,y)$ we have

$$\int_{-\infty}^\infty \int_{-\infty}^\infty \delta(x-y)\delta(y)\phi(x,y)\,dx\,dy=\phi(0,0)=\int_{-\infty}^\infty \int_{-\infty}^\infty \delta(x)\delta(y)\phi(x,y)\,dx\,dy$$

Hence, in distribution we assert that $\delta(x-y)\delta(y)=\delta(x)\delta(y)$. And we are done.

There is the question of the order of the distributions. The question that comes to mind is "How do we define $\delta(y)\delta(y-x)$ as a distribution?" In the next section, that question is answered.

Let $\phi(x,y)$ be an arbitrary test function and let $\delta_m(y)$ and $\gamma_n(y-x)$ be two regularizations of the Dirac Delta such that

$$\lim_{m\to\infty}\int_{-\infty}^\infty \delta_m(y)\phi(x,y)\,dy=\phi(x,0)$$

$$\lim_{n\to\infty}\int_{-\infty}^\infty \gamma_n(y-x)\phi(x,y)\,dy=\phi(x,x)$$

Then, since $\delta_m\,\phi\in C_c^\infty(\mathbb{R}^2)$ we can write

$$\begin{align} \lim_{n,m\to \infty}\int_{-\infty}^\infty\int_{-\infty}^\infty \delta_m(y)\gamma_n(y-x)\phi(x,y)\,dy\,dx&=\lim_{m\to \infty}\int_{-\infty}^\infty \lim_{n\to\infty}\int_{-\infty}^\infty \delta_m(y)\gamma_n(y-x)\phi(x,y)\,dy\,dx\\\\ &=\lim_{m\to \infty}\int_{-\infty}^\infty \delta_m(x)\phi(x,x)\,dx\\\\ &=\phi(0,0) \end{align}$$

Hence, we assert that in distribution we have $\delta(y)\delta(y-x)=\delta(x)\delta(y)$. And note that we could have interchanged the order of integration without affecting this result.

Mark Viola
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Integrate your quantity against an arbitrary test function:

$$\iint \delta(x-y)\delta(y)\phi(x,y)\:dx\:dy = \int \delta(y)\phi(y,y) \:dy = \phi(0,0)$$

which is exactly the same as what would have happened if we applied the usual $\delta$ on $\mathbb{R}^2$ to the test function, but only if we did $x$ first, since doing $y$ first would have resulted in $0$, but that is a fault of trying to treat the $\delta$ like a function.

Ninad Munshi
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  • @MarkViola I deleted because I'm not in any mood to get into a debate about the products of distributions. I don't think this is fully fundamentally sound since this is not well-posed if the $y$ integral is attempted first (as per my comment to the OP). – Cameron Williams Mar 25 '20 at 19:38
  • Those are not integrals. They are distributions as you know. $$\langle \delta_x,\delta_0\rangle =\delta_0(x)$$ – Mark Viola Mar 25 '20 at 19:40
  • @MarkViola Right but if you attempt to make sense of $\langle \phi(x),\langle \psi(y),\delta(x-y)\delta(y)\rangle\rangle$, you're going to have a bad time without stretching the definition a bit. – Cameron Williams Mar 25 '20 at 19:41
  • Not a bad time at all. Note that $$\langle \delta_(x-y)\delta(y),\psi(y)\rangle = \langle \delta_(x-y),\psi(y)\delta(y)\rangle=\psi(x)\delta(x)$$ – Mark Viola Mar 25 '20 at 19:43
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    @MarkViola Right, but that's stretching the definition since technically distributions only work on test functions. – Cameron Williams Mar 25 '20 at 19:44