Poisson's equation applied to an integral

In summary, the conversation is about applying Poisson's equation to a potential expressed as a volume integral. The solution involves finding the Green's function of the Laplace operator and using functional analysis and the theory of distributions. The standard solution for Poisson's equation involves the integral of the potential over the entire space.
  • #1
Thomas2054
17
0
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 website reference would be fine.

Thanks.

Thomas
 
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  • #2
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]
 
  • #3
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]
 

1. What is Poisson's equation applied to an integral?

Poisson's equation applied to an integral is a mathematical concept used in physics and engineering to describe the relationship between scalar potential functions and their corresponding charge distributions. It is derived from the fundamental principles of electrostatics and is commonly used in solving boundary value problems.

2. How is Poisson's equation applied to an integral used in practical applications?

Poisson's equation applied to an integral has numerous practical applications, such as in the design of electronic circuits, solving electrostatic and gravitational potential problems, and in the study of fluid mechanics. It allows us to calculate the potential at any point in space given a known charge distribution, making it a valuable tool in many fields of science and engineering.

3. What is the mathematical representation of Poisson's equation applied to an integral?

The mathematical representation of Poisson's equation applied to an integral is ∇²Φ = -ρ/ε₀, where ∇² is the Laplace operator, Φ is the scalar potential function, ρ is the charge density, and ε₀ is the permittivity of free space.

4. What are the key assumptions made in Poisson's equation applied to an integral?

The key assumptions made in Poisson's equation applied to an integral are that the charge distribution is static, the medium is linear and isotropic, and the potential function satisfies the boundary conditions of the system.

5. How does Poisson's equation applied to an integral relate to Gauss's law?

Poisson's equation applied to an integral is closely related to Gauss's law, which states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of the medium. Poisson's equation can be derived from Gauss's law by taking the divergence of both sides and applying the continuity equation for charge.

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