verification of a solution of a Laplace-type equation 

Prove that in $\mathbb{R}^3$ the function $\displaystyle \frac{e^{-k r}}{4\pi r}$ satisfies the equation:
$$\Delta u(r) - k^2u(r) =-\delta(\vec{r}).$$
    ($\Delta$ stands for the Laplace operator).


Question. In our case, operator $\Delta u$ should be replaced in spherical or cylindrical coordinates? (besides, since $u=u(r)$ here, only the $r$-part should be taken into account). Also, the $\delta$ symbol in the above equation stands for what exactly? 
Thanks a lot in advance for the help.
 A: You have to check that $f(r) = r^{-1} e^{-kr}$ is a fundamental solution of the differential operator $L = k^{2} - \Delta$. That is $Lf = \delta$ the Dirac mass.
To do so a fast and efficient is to use the Fourier transform $F$. By injectivity of $F$, if $F(Lf) =  F(\delta)$ then $Lf = \delta$. Since the Fourier transform plays nicely with differential operator you obtain $F(Lf)(\xi) = (k^{2} - |\xi|^{2}) F(f)(\xi)$ and it is well-known (and easily chackable) that $F(\delta) = 1$. You therefore need to check that $F(f)(\xi) = \frac{1}{k^{2} - |\xi|^{2}}$.
Since your function is radial, switching to spherical coordinates to compute the integral is the way to go. The computations are a bit heavy but in the end you will derive the result.
A: I assume the $\delta$ stands for the Dirac Delta distribution, which gives you the following for a suitable test function $\phi(x)$:
$$ (\delta_a)(\phi(x)) = \int \phi(x) \delta (x-a) dx = \phi (a) $$
for a given $a$. Here, for example ,we have $a=0$. 
So, the first thing you notice is that your equation makes perfectly sense for $r\neq 0$, just by plugging in and using the Laplacian in spherical coordinates, independent of $\varphi$ and $\theta$ and the fact that the right side is zero in this case. But what about the case $r=0$? Here, you have to be careful, the term $\frac{1}{r}$ screws it up. So what to do? Take an arbitrary test function $f(r)$, integrate the left side over $\mathbb{R}$ (or $\mathbb{R}^3$), then you'll get:
$$\int_{r < \epsilon} f(r) (\Delta - k^2) \frac{e^{kr}}{4 \pi r}dr $$
since the integrand vanishes for $r<0$. Next, expand $e^{kr}$, forget about higher terms, and you'll get something like this:
$$ \int f(r) \Delta \frac{1}{r}dr + O(\epsilon^2)$$
But from elementary theory of distribution, we know that 
$$\Delta \frac{1}{r} = -4\pi \delta(r) $$
And that is the route you have to take.
