Conceptual Laplace's/Poisson's Equation Solution Questions

AI Thread Summary
The discussion focuses on solving Laplace's equation within a cylindrical region and extending the solution outside the cylinder by redefining the radial variable as R^2/r. It highlights that if a function satisfies Laplace's equation in one region, it can be transformed to find a solution in another region. The method of images is applied to construct a solution for Poisson's equation involving a line charge inside the cylinder, with the solution outside being the original minus the image solution. A critical point raised is the necessity of ensuring that boundary conditions are met, particularly at r = R, which relates to the uniqueness theorem. The conversation emphasizes the importance of correctly applying the method of images and understanding the implications of boundary conditions in electrostatics.
Dobuskis

Homework Statement


-You are given a solution to Laplace's equation inside of a cylindrical region radius R.
-Show that by redefining the radial variable r as R2/r you get a solution for outside of R.
-Grounded conducting cylinder at r=R. Using a linear combination of the solutions in the previous part, construct a solution to Poisson's equation for a line charge inside the cylinder. Do the same for outside.
-What does this have to do with the method of images.

Homework Equations


General Laplace's and Poisson's equations
ΔF = 0, ΔF = f

The Attempt at a Solution


I spent a while trying to check the new solution by plugging it into Laplace's equation, got a bunch of chain rule terms that made something I didn't recognize, then gave up for a while. Later I realized I might be overthinking/moving in the wrong direction.
-range of R2/r with r>R is (0,R), thus the radial inputs have the same range (0,R), thus if one is a solution to Laplace's equation in its region then the same goes for the other.
-Since solutions to Laplace's equation are also solutions to Poisson's equation then and linear combination of the two previous equations is a solution. By the method of images for a conducting surface and a line charge, the solution is the object solution minus the image solution. Thus outside the cylinder is the R2/r solution minus the r solution, for inside the r solution minus the R2/r solution.
-This is the method of images for a cylindrical surface.
 
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Dobuskis said:
range of R2/r with r>R is (0,R), thus the radial inputs have the same range (0,R), thus if one is a solution to Laplace's equation in its region then the same goes for the other
That cannot be a sufficient argument. If f is any function satisfying f(0)=f(R)=1 then rf(r) also has range (0,R), but that does not make rf(r) a solution in the range (0,R).
 
You know that V(r) = 0 at r = R. What is V(R2/r) at r = R? Think "Uniqueness Theorem". The method of images is one of its applications.
 
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