# Homework Help: Function whose 2nd order divergence is the Dirac Delta

1. Nov 26, 2014

### genxium

1. The problem statement, all variables and given/known data

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 $|\textbf{r}| \rightarrow 0$?

2. Relevant equations

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

3. The attempt at a solution

Stated above.

2. Nov 26, 2014

### 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.

3. Nov 26, 2014

### 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 :(