A Numerical solution of the Poisson equation

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Proton radiation creates an electrostatic field in the material, leading to the need to solve the Poisson equation under the assumption of an infinite, homogeneous medium. The proposed solution involves an integral that encounters singularities when calculating the field inside the source. A small δx was introduced to avoid these singularities, but concerns were raised about the validity of this approach and potential errors. The discussion highlights that using Green's function is a valid method, as the contribution from the ignored region becomes negligible for small δx. In one-dimensional systems, the Green's function remains well-defined and is piecewise linear, avoiding divergence issues.
ivantozavr
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How to solve the Poisson equation inside an electrostatic field source?
In my case, there is proton radiation acting on the material. Consequently, the protons get stuck in the sample and create an electrostatic field. I would like to solve the Poisson equation inside the sample. I assume that the medium is infinite and homogeneous, that is, the potential at infinity is zero (one-dimensional case):
Δφ(x) = - ρ(x) / εr ε0
φ(±∞) = 0.​
The solution of such an equation is well known. That is (integration by range X' containing the source)
φ(x) = 1 / (4πεr ε0) ∫ dx' ρ(x') / |x-x'|.​
But then if I try to calculate the field inside the source itself, then singularities will occur due to the denominator in the integrand. I made small δx indents when calculating such an integral, i. e. sources with coordinates ζ: |ζ-x'| < δx were not taken into account when calculating an integral. This is hardly the right approach. Could you tell me how to solve such a problem correctly?
 
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It is a perfectly valid approach if you insist on computing the result through the Green’s function. The contribution ignored goes as
$$\int_{0}^{\delta x} \frac{r^2 dr}r \propto \delta x^2$$
so for small ##\delta x## this should give negligible errors.
 
Last edited:
Orodruin said:
It is a perfectly valid approach if you insist on computing the result through the Green's function. The contribution ignored goes as
$$\int_{0}^{\delta x} \frac{r^2 dr}r \propto \delta x^2$$
so for small ##\delta x## this should give negligible errors.
Thank you for your response!

Suppose that ##\rho(x')\approx const## in ##|x' - x| < \delta x.## Redefine the variable ##x' - x = \xi.## Wouldn't the error in a one-dimensional Cartesian system then be the same as below:
$$\delta\varphi \propto \int_{-\delta x}^{\delta x} \frac{d\xi}{|\xi|} = 2 \int_{0}^{\delta x} \frac{d\xi}{\xi}?$$
So the error tends to infinity?

In fact, I chose the solution via the green's function because I do not know how to construct a difference scheme for the Poisson equation with such boundary conditions.
 
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ivantozavr said:
Thank you for your response!

Suppose that ##\rho(x')\approx const## in ##|x' - x| < \delta x.## Redefine the variable ##x' - x = \xi.## Wouldn't the error in a one-dimensional Cartesian system then be the same as below:
$$\delta\varphi \propto \int_{-\delta x}^{\delta x} \frac{d\xi}{|\xi|} = 2 \int_{0}^{\delta x} \frac{d\xi}{\xi}?$$
So the error tends to infinity?

In fact, I chose the solution via the green's function because I do not know how to construct a difference scheme for the Poisson equation with such boundary conditions.
In a one-dimensional system the Green’s function is not divergent. It is a piecewise linear function.
 
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