Solution that doesn't diverge at origin

[jimjam1]

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! :)

 

[pasmith]

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].)

 

[rigetFrog]

Bessel or Hankel functions.

 

[jasonRF]

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...

 

[blue_raver22]

Hello,

Thank you for your inquiry. This is a very interesting problem and I would be happy to provide some guidance. The equation you have provided is known as the Poisson's equation in spherical coordinates. It is a commonly used equation in many areas of physics and engineering, including electrostatics, heat transfer, and fluid dynamics.

The solution to this equation can be found using various techniques, such as separation of variables, Green's functions, or numerical methods. However, the most important thing to note in this problem is the condition that u approaches zero as r goes to infinity. This is known as the boundary condition and it is crucial in determining the unique solution to the problem.

In order for the solution to not diverge at the origin, the function u must be well-behaved and continuous at r=0. This means that the solution must satisfy the condition that the derivative of u with respect to r is continuous at r=0. This is known as the regularity condition and it ensures that the solution is physically meaningful.

To find the solution in this case, you can use the method of Green's functions. This involves constructing a Green's function that satisfies the Poisson's equation and the boundary condition. The Green's function can then be used to find the solution for any arbitrary source term, in this case, the delta function.

I hope this helps. If you require further assistance, please do not hesitate to reach out. Best of luck with your research!