partial differential equation with delta function

Question

I don't get the following equation: $$ \square\frac{\delta(r-ct)}{r}=-4\pi\delta(\vec{r})\delta(r-ct) $$ Where the square is the D'lambert operator. $$ \square=\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2} $$ Is this equation true? How would I prove it?

Answer 1

HINTS:

Use the product rule for the Laplacian,

$$\nabla^2 (fg)=f\nabla^2 g+2\nabla f\cdot\nabla g+g\nabla^2f$$

with $f=\frac1r$ and $g=\delta(r-ct)$ along with the distributional equality

$$\nabla^2 \left(\frac1r\right)=-4\pi \delta(\vec r)$$

to find

$$\nabla^2\left(\frac{\delta(r-ct)}{r}\right)=\frac1r \nabla^2\delta(r-ct)-\frac2{r^2}\delta'(r-ct)-4\pi \delta(\vec r)\delta(r-ct)$$

Can you finish now?