Poisson's equation applied to an integral

Thomas2054

This is not a homework problem, but a math question of curiosity. At 57 I am way past courses and homework.

If a potential, V, is expressed as a volume integral, how does one apply Poisson's equation to that integral?

If V = k∫_V'[itex]\frac{ρ}{R}[/itex]dv' (V on the LHS is voltage and the integral on the RHS is the integral over V', the volume.)

how do I manipulate Poisson's, i.e., [itex]\nabla[/itex]²V = -[itex]\frac{ρ}{ε}[/itex]?

I cannot find enough in my various calculus books to figure this out. A web site reference would be fine.

Thanks.

Thomas

vanhees71

It's not such an easy question as it might seem. The most elegant way is to use functional analysis and the theory of distributions. You look for the Green's function of the Laplace operator,

[tex]\Delta_x G(\vec{x},\vec{x}')=\delta^{(3)}(\vec{x}-\vec{x}').[/tex]

First of all it's clear from the symmetries of the problem that

[tex]G(\vec{x},\vec{x}')=g(|\vec{x}-\vec{x}'|).[/tex]

Setting [tex]\vec{y}=\vec{x}-\vec{x}'[/tex] leads to

[tex]\Delta_y g(|\vec{y}|)=\delta^{(3)}(\vec{y}).[/tex]

Now you introduce spherical coordinates for [itex]r=|\vec{y}| \neq 0[/itex]. This leads to

[tex]\frac{1}{r} [r g(r)]''=0.[/tex]

This you can solve by succesive integration

[tex]g(r)=\frac{C_1}{r}+C_2,[/tex]

where [itex]C_{1,2}[/itex] are integration constants. Since we seek for the solution, which vanishes at infinity, we have [itex]C_2=0[/itex]. The constant [itex]C_1[/itex] has to be found by the defining equation. To that end you integrate the equation

[tex]\Delta g(|\vec{y}|)=\delta^{(3)}(\vec{y})[/tex]

over a sphere around the origin with arbitrary radius. For the right-hand side you get [itex]1[/itex], and for the left-hand side you use Gauß's integral theorem

[tex]\int_{K} \mathrm{d}^3 \vec{y} \Delta g(|\vec{y}|) = \int_{\partial K} \mathrm{d}^2 \vec{F} \cdot \vec{\nabla} g(|\vec{y}|).[/tex]

You easily find

[tex]\vec{\nabla} g(|\vec{y}|)=-C_1 \frac{\vec{y}}{|\vec{y}|^3},[/tex]

and the surface integral thus gives

[tex]-4 \pi C_1=1 \; \Rightarrow \; C_1=-\frac{1}{4 \pi}.[/tex]

So we have

[tex]G(\vec{x},\vec{x}')=-\frac{1}{4 \pi |\vec{x}-\vec{x}'|^3}.[/tex]

Thus the standard solution for Poisson's equation is

[tex]V(\vec{x})=-\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' G(\vec{x},\vec{x}') \frac{\rho(\vec{x}')}{\epsilon}=\frac{1}{4 \pi \epsilon} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|^3}.[/tex]

vanhees71

I don't know, how to edit postings in this forums, so here is an erratum to my previous one. The end of the posting must read:

[...]

So we have

[tex]G(\vec{x},\vec{x}')=-\frac{1}{4 \pi |\vec{x}-\vec{x}'|}.[/tex]

Thus the standard solution for Poisson's equation is

[tex]V(\vec{x})=-\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' G(\vec{x},\vec{x}') \frac{\rho(\vec{x}')}{\epsilon}=\frac{1}{4 \pi \epsilon} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|}.[/tex]