Function whose 2nd order divergence is the Dirac Delta

[genxium]

Homework Statement

This problem came when I was learning the Poisson's equation (refer to http://farside.ph.utexas.edu/teaching/em/lectures/node31.html). when it came to the step to find the Green's function G which satisfies \nabla^2 \cdot G(\textbf{r}, \textbf{r}') = \delta(\textbf{r}-\textbf{r}') with |G| \rightarrow 0 when |\textbf{r}| \rightarrow 0, the tutorial I refer to simply yields G(\textbf{r}, \textbf{r}') = -\frac{1}{4\pi}\frac{1}{|\textbf{r} - \textbf{r}'|}.

I understand that \int_{V} \nabla^2 \cdot \frac{1}{|\textbf{r} - \textbf{r}'|} dV= \int_{S=\partial V} \nabla \cdot \frac{1}{|\textbf{r}-\textbf{r}'|} \cdot d\textbf{S} = \int_{S=\partial V} -\frac{\textbf{r}-\textbf{r}'}{|\textbf{r}-\textbf{r}'|^3} \cdot d\textbf{S} = -4\pi, by assuming that V is a unit sphere located at \textbf{r}'. Thus G(\textbf{r}, \textbf{r}') = -\frac{1}{4\pi}\frac{1}{|\textbf{r} - \textbf{r}'|} COULD BE ONE SOLUTION to the equation, but what about proof of the uniqueness? Is this the only solution to the equation \nabla^2 \cdot G(\textbf{r}, \textbf{r}') = \delta(\textbf{r}-\textbf{r}') with |G| \rightarrow 0 when |\textbf{r}| \rightarrow 0?

Homework Equations

\textbf{r} = x \cdot \textbf{i} + y \cdot \textbf{j} + z \cdot \textbf{k}
\textbf{r}' = x' \cdot \textbf{i} + y' \cdot \textbf{j} + z' \cdot \textbf{k}
\nabla = \frac{\partial}{\partial x} \cdot \textbf{i} + \frac{\partial}{\partial y} \cdot \textbf{j} + \frac{\partial}{\partial z} \cdot \textbf{k}
\delta(\textbf{r}) = \delta(x) \cdot \delta(y) \cdot \delta(z)

The Attempt at a Solution

Stated above.

 

[Miles Whitmore]

From what I recall the Green's function of the Laplace operator is not unique, and it's most general form would be $G(r,r ′ )=−\frac{1}{4\pi}\frac{1}{|r−r ′|} + F(r,r')$ such that $F(r,r')$ satisfies $\nabla^2F(r,r') = 0$. Depending on the type of boundary conditions, symmetry, etc, $F(r,r')$ can be chosen to simplify the problem. I'm pretty rusty with my Green's functions though so if I'm mistaken someone please correct me.

 

[genxium]

@Miles, yes you are right that the general equation is not guaranteed a unique solution. However I am bad at differential equations such that even given the boundary condition |G| \rightarrow \infty when |\textbf{r}| \rightarrow 0, I can't figure out the answer to my question :(