Understanding a function containing the delta-function I have been given the problem to solve the Poisson equation:
$$\vec{\nabla}^2 u(\vec{r})=\rho(\vec{r}), \quad z>0$$
And the professor then writes that the inhomogeneity is thus given by:
$$\rho(\vec{r})=\delta(\vec{r}-\vec{a}),\quad \vec{a}=(0,0,a),\quad a\in\mathbb{R}_{+}$$
But... Does this not imply that $\rho(\vec{r})$ is infinite at every point along the $z$-axis for $z>0$ and zero everywhere else? Does anyone know how I am to interpret that equation?
 A: This is a standard image charge question. If have the condition that $u(z=0) = 0$, then we can use the existence and uniqueness property of solutions of the Poisson equation to avoid computing the integral. 
We have a point charge located at $$\vec{a} = (0,0,a)$$ Now pretend we have another point charge of the opposite sign located at
$$\vec{a}' = (0,0,-a)$$
Of course this is silly, this point does not exist in our space. But if it did, notice that the electric potential on the $z=0$ plane would always be $0$. In other words, this is a different situation that agrees with ours on the upper half space $z > 0$. We can use existence and uniqueness to assert that the solution to our problem is the same as the solution to this problem.
Thus the electric potential everywhere ($z>0$) is given by
$$u(\vec{r}) = \frac{1}{4\pi|\vec{r}-\vec{a}|} - \frac{1}{4\pi|\vec{r}-\vec{a}'|}$$
or 
$$u(\vec{r}) = \frac{1}{4\pi|\vec{r}-\vec{a}|} - \frac{1}{4\pi|\vec{r}+\vec{a}|}$$
since $\vec{a}$ points solely in the $z$ direction.
