Is $\delta(x-y)\delta(y)$ equal to $\delta(x)\delta(y)$? 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
 A: 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.
A: 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.
