Solution of Poisson's Equation

[DirecSa]

We all know that Poissson's equation in electrostatic is:

$$\nabla^2\phi=-\frac{\rho}{\epsilon_0}$$

My question is: why the solution, let's say for 1D, is not just double integral as follows:
$$\phi=\iint -\frac{\rho}{\epsilon_0} d^2x$$

which gives x square relation. But the actual solution comes from using Green's function and it gives the relation one over r (1/r). I need the connection between these things how they come to this and not that...

Thank you!

 

[vanhees71]

It should be
$$\Delta \phi=-\frac{1}{\epsilon_0} \rho.$$
In 1 dimension the equation simply reads
$$\phi''=-\frac{1}{\epsilon_0} \rho,$$
and this you get a solution by simply integrating twice wrt. $x$.

However in more than 1 dimension you need the Green's function, and it's easy to find from physical arguments in this case: For the point charge $q$ at position $\vec{x}'$ the potential is of course the Coulomb potential
$$\phi_{\text{C},\vec{x}'}(\vec{x})=\frac{q}{4 \pi \epsilon_0|\vec{x}-\vec{x}'|}.$$
Now you can consider the continuous charge distribution, described by the charge density ρ, as consisting of an infinite number of infinitesimal charges $\mathrm{d} q=\mathrm{d}^3 x' \rho(\vec{x}')$, located around $\vec{x}'$, and because the Poisson equation is linear you get the total potential by summing or, since we have a continuous charge distribution, rather integrating, over the Coulomb potentials of all these charges, i.e.,
$$\phi(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \rho(\vec{x}') \frac{1}{4 \pi \epsilon_0 |\vec{x}-\vec{x}'|}.$$
Formally the charge density of a point charge is
$$\rho_{\text{P},\vec{x}'}(\vec{x})=q \delta^{(3)}(\vec{x}-\vec{x}').$$
and that's why the Coulomb potential is the Greens' function of the operator $-\Delta$ in the sense that from this it follows that
$$-\Delta \phi_{\text{C},\vec{x}'}(\vec{x}')=-\Delta \frac{q}{4 \pi \epsilon_0 |\vec{x}-\vec{x}'|} = \frac{q}{\epsilon_0} \delta^{(3)}(\vec{x}-\vec{x}').$$
These hand-waving physical arguments can of course be made mathematically rigorous.

 

[DirecSa]

Thank you so much, now it is more clear for me. And also I added the minus. Maybe just a question regarding the 4pi in the denominator.. does it come from a physical definition or mathematical concept?

 

[vanhees71]

That comes from finding the Green's function. The idea is not so difficult. You need to solve for
$$-\Delta G(\vec{x})=\delta^{(3)}(\vec{x}).$$
The right-hand side is obviously rotation symmetric. That's why we can conclude that $G$ should depend on $r=|\vec{x}|$ only. Now using the Laplace operator in sphercical coordinates, this leads to
$$-\frac{1}{r} [r G(r)]''=0 \quad \text{for} \quad r \neq 0.$$
You get the general solution by just integrating twice (it's effectively a 1D problem due to this symmetry!):
$$(r G)''=0 \; \Rightarrow r G=A+B r \; \Rightarrow \; G=\frac{A}{r} + B$$
with integration constants $A$ and $B$. First we want to have $G \rightarrow 0$ for $r \rightarrow \infty$. This implies $B=0$. Thus you have
$$G(r)=\frac{A}{r}.$$
To get $A$ we integrate $\Delta G(r)$ over a sphere of radius $a$ around the origin and use Gauss's Law,
$$\int_{S_a} \mathrm{d}^3 r \Delta G(r) = \int_{\partial S_a} \mathrm{d}^2 \vec{f} \cdot \vec{\nabla} G(r).$$
The surface normal vector in spherical coordinates is
$$\mathrm{d}^2 \vec{f} = a^2 \sin \vartheta \mathrm{d} \vartheta \mathrm{d} \varphi \vec{e}_r.$$
The gradient is
$$\vec{\nabla} G(r)=G'(r) \vec{\nabla} r = G'(r) \vec{e}_r=-\frac{A}{r^2} \vec{e}_r$$
and thus
$$\int_{S_a} \mathrm{d}^3 r \Delta G(r) = -\int_0^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi a^2 \sin \vartheta \frac{A}{a^2} = -4 \pi A.$$
On the other hand $\Delta G(r)=-\delta^{(3)}(\vec{r})$ and thus
$$\int_{S_a} \mathrm{d}^3 \Delta G(r)=-1 \; \Rightarrow \; -4 \pi A=-1 \; \Rightarrow \; A=\frac{1}{4 \pi}.$$
Thus the Green's function of $-\Delta$ is
$$G(r)=\frac{1}{4 \pi r},$$
which is what we wanted to prove.

 

[DirecSa]

very grateful, thank you a lot :)