General solution to Laplacian in cylindrical coordinates

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Trollfaz
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I am trying to model the voltage function for a very long cylinder with an assigned surface charge density or voltage.
Then the solution inside the cylinder is:
$$\sum_{n=0}^{\infty}A_n r^n cos(nθ)$$
And$$\sum_{n=0}^{\infty}A_n r^-n cos(nθ)$$
outside. Is that correct
 
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Can someone help me to change it to latex codes thanks

[LaTeX updated by the Mentors]
 
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cos(nθ) cos(mθ) is orthogonal for m not equals n. Take the voltage at the boundary and integrate it with a multiple cos(mθ)sin(mθ) to filter out the m th term in the summation and destroy the summation sign.
 
Here's a detailed explanation
At the boundary r=R
$$V=\sum_{n=0}^{\infty}A_n R^{-n} cos(nθ)$$
$$\int_{0}^{π}\sum_{n=0}^{\infty}A_n R^{-n} cos(nθ)cos(mθ)dθ=A_m R^{-m} π$$
$$A_m=\frac{R^m}{π}\int_{0}^{π}V_Rcos(mθ)dθ$$
If we know V at boundary as a linear combination of cos(nθ), then we can filter out all the non zero terms of the infinite sum and find the full function of V.
 
Trollfaz said:
Here's a detailed explanation
At the boundary r=R
$$V=\sum_{n=0}^{\infty}A_n R^{-n} cos(nθ)$$
$$\int_{0}^{π}\sum_{n=0}^{\infty}A_n R^{-n} cos(nθ)cos(mθ)dθ=A_m R^{-m} π$$
$$A_m=\frac{R^m}{π}\int_{0}^{π}V_Rcos(mθ)dθ$$
If we know V at boundary as a linear combination of cos(nθ), then we can filter out all the non zero terms of the infinite sum and find the full function of V.
It is the surface charge, not V, which is given. You have to find the discontinuity in ##\partial_r(V)## at the surface related to the surface charge density.
 
Meir Achuz said:
There is no point charge here.
Irrelevant to the point I was making. You have a potential monopole moment and the corresponding field of that is that of a point charge outside of the charge distribution.
Meir Achuz said:
There will be a ##ln( r)## term for outside if there is a constant surface charge.
Which is … the field of a point charge in two dimensions. Just as the 1/r field outside of a spherically symmetric charge distribution in three dimensions.
 
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The general expansion in terms of separation-ansatz solutions in cylindrical coordinates leads to the Bessel or Hankel functions. You seem to discuss the special case of ##z##-independent solutions, i.e., solutions of the Laplace equation in 2 dimensions in terms of polar coordinates, which is not the general case, of course.
 
vanhees71 said:
The general expansion in terms of separation-ansatz solutions in cylindrical coordinates leads to the Bessel or Hankel functions. You seem to discuss the special case of ##z##-independent solutions, i.e., solutions of the Laplace equation in 2 dimensions in terms of polar coordinates, which is not the general case, of course.
"for a very long cylinder"
 
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vanhees71 said:
Yes, then of course, you assume independence of ##z##. I've overlooked this assumption.
Yes let's just assume the distribution is invariant with z
 
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