Conceptual Laplace's/Poisson's Equation Solution Questions

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SUMMARY

This discussion focuses on solving Laplace's equation within a cylindrical region of radius R and extending the solution outside the cylinder using the transformation r = R²/r. The method of images is employed to construct a solution to Poisson's equation for a line charge inside the cylinder. The participants emphasize the relationship between solutions to Laplace's and Poisson's equations, confirming that linear combinations of these solutions yield valid results. The uniqueness theorem is highlighted as a critical concept in validating the solutions at the boundary conditions.

PREREQUISITES
  • Understanding of Laplace's equation and Poisson's equation
  • Familiarity with the method of images in electrostatics
  • Knowledge of cylindrical coordinate systems
  • Concept of boundary conditions and uniqueness theorem
NEXT STEPS
  • Study the method of images in detail, particularly for cylindrical geometries
  • Explore the uniqueness theorem in the context of partial differential equations
  • Investigate the implications of boundary conditions on solutions to Laplace's and Poisson's equations
  • Practice solving Laplace's equation in various coordinate systems, including cylindrical coordinates
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Students and professionals in physics and engineering, particularly those focusing on electrostatics, mathematical physics, and applied mathematics.

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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