How Does Green's Function Solve the Dirichlet Problem in a Half-Space?

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SUMMARY

The discussion focuses on solving the Dirichlet problem in a half-space using Green's function. The potential in the half-space defined by z≥0 is specified with Dirichlet boundary conditions on the plane z=0, where V is defined inside a circle of radius a and zero outside. The solution along the axis of the circle (ρ=0) is given by the formula Φ=V(1−z/√(a²+z²)). The key error identified in the computations was the omission of the area element ρ in polar coordinates, which is crucial for obtaining the correct integral result.

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Homework Statement



Consider a potential problem in the half space z>=0 with Dirichlet boundary conditions on the plane z=0.
If the potential on the plane z=0 is specified to be V inside a circle of radius a centered at the origin, and Phi=0 outside that circle, show that along the axis of the circle (rho=0) the potential is given by

[tex]\Phi=V(1-\frac{z}{\sqrt{a^2+z^2}})[/tex]

Homework Equations





The Attempt at a Solution



I used the Green's function (by method of images)

[tex]G(x,x')=\frac{1}{|x-x'|}-\frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z+z')^2}}[/tex]

Then I converted the formula to cylindrical coordinates and computed the normal derivative -dG/dz. Then I set rho=0 and finally I got

[tex]\Phi = \frac{-z}{2\pi} \int_0^{2\pi} \int_0^a \frac{V}{(p'^2+z^2)^{3/2}} = \frac{-Va}{z \sqrt{a^2+z^2}}[/tex]

I'm somewhat certain that the computations are correct I checked them two times in a CAS so I don't really know what I did wrong.
 
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This thread has been around for a while, but hopefully this reply will help future users. Your only mistake was forgetting that in polar coordinates, the area element is \rho d\rho d\phi as opposed to just d\rho d\phi. With the extra \rho up top, you get the right answer (and the integral becomes easier, to boot!).
 

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