Solution that doesn't diverge at origin

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Discussion Overview

The discussion revolves around finding a solution to the equation \(\nabla^{2}u-\frac{u}{\lambda^{2}}=a\delta(r)\) under spherical symmetry in three dimensions. Participants explore the implications of boundary conditions, particularly the requirement that \(\lim_{r\rightarrow \infty}u=0\) and the behavior of the solution at the origin.

Discussion Character

  • Technical explanation, Debate/contested, Conceptual clarification

Main Points Raised

  • One participant presents a method to solve the equation in regions where \(r > 0\) and discusses the implications of setting \(u(r) = \frac{f(r)}{r}\), leading to the conclusion that \(f(0) = 0\) is necessary to avoid divergence at the origin.
  • Another participant suggests that the only solution that decays at infinity leads to divergence at the origin, indicating a conflict between the two boundary conditions.
  • A third participant briefly mentions Bessel or Hankel functions as potential solutions, though no further elaboration is provided.
  • Another participant expresses skepticism about the existence of a physical solution that does not diverge at the origin, referencing the context of electric potentials in plasma physics and the behavior of point charges.

Areas of Agreement / Disagreement

Participants express differing views on the existence of a solution that meets both boundary conditions. There is no consensus on whether a non-divergent solution at the origin is possible while also satisfying the decay condition at infinity.

Contextual Notes

The discussion highlights the challenges of finding solutions to differential equations with singular sources and boundary conditions, particularly in physical contexts where such equations are applied.

jimjam1
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Hi - wondering if you can help me find a solution of:

[itex]\nabla^{2}u-\frac{u}{\lambda^{2}}=a\delta(r)[/itex]

for spherical symmetry in 3D with the condition that [itex]\lim_{r\rightarrow \infty}u=0[/itex]. It can be rewritten in spherical coordinates as

[itex]\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial u}{\partial r}\right)-\frac{u}{\lambda^{2}}=a\delta(r)[/itex].

Any help would be much appreciated! :)
 
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jimjam1 said:
Hi - wondering if you can help me find a solution of:

[itex]\nabla^{2}u-\frac{u}{\lambda^{2}}=a\delta(r)[/itex]

for spherical symmetry in 3D with the condition that [itex]\lim_{r\rightarrow \infty}u=0[/itex]. It can be rewritten in spherical coordinates as

[itex]\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial u}{\partial r}\right)-\frac{u}{\lambda^{2}}=a\delta(r)[/itex].

Any help would be much appreciated! :)

You are solving [tex]\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial u}{\partial r}\right)-\frac{u}{\lambda^{2}} = 0[/tex] in [itex]r > 0[/itex]. Setting [tex] u(r) = \frac{f(r)}r[/tex] yields [tex] f'' - \lambda^{-2} f = 0[/tex] and the only way to not have [itex]u[/itex] diverge at the origin is to take [itex]f(0) = 0[/itex], which yields [itex]f(r) = A\sinh(\lambda^{-1} r)[/itex] and thus [tex] u(r) = \frac{A\sinh(\lambda^{-1} r)}{r}.[/tex] L'hopital confirms that [tex] \lim_{r \to 0} u(r) = \lim_{r \to 0} \frac{A\lambda^{-1}\cosh(\lambda^{-1} r)}{1} = A\lambda^{-1}.[/tex] Unfortunately [itex]|u| \to \infty[/itex] as [itex]r \to \infty[/itex]. To get a solution which decays at infinity you must take [itex]f(r) = e^{-r/\lambda}[/itex], and the resulting [itex]u[/itex] diverges at the origin.

(Usually this setup is an abstraction of "there is a small sphere at the origin". Within the sphere you use a solution which is bounded at the origin, and outside the sphere you use a solution which decays as [itex]|r| \to \infty[/itex].)
 
Last edited:
Bessel or Hankel functions.
 
I wouldn't expect a physical solution that does not diverge at the origin. That is the exact equation for the electric potential of a point charge in a hot plasma, where $\lambda$ would be the Debye length. Potentials of point charges always diverge at the location of the charge...
 

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