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Function whose 2nd order divergence is the Dirac Delta

  1. Nov 26, 2014 #1
    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 [itex]G[/itex] which satisfies [itex]\nabla^2 \cdot G(\textbf{r}, \textbf{r}') = \delta(\textbf{r}-\textbf{r}')[/itex] with [itex]|G| \rightarrow 0[/itex] when [itex]|\textbf{r}| \rightarrow 0[/itex], the tutorial I refer to simply yields [itex]G(\textbf{r}, \textbf{r}') = -\frac{1}{4\pi}\frac{1}{|\textbf{r} - \textbf{r}'|}[/itex].

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

    2. Relevant equations

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

    3. The attempt at a solution

    Stated above.
     
  2. jcsd
  3. Nov 26, 2014 #2
    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.
     
  4. Nov 26, 2014 #3
    @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 [itex]|G| \rightarrow \infty[/itex] when [itex]|\textbf{r}| \rightarrow 0[/itex], I can't figure out the answer to my question :(
     
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