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

• gop
In summary, when solving for the potential in the half space z>=0 with Dirichlet boundary conditions on the plane z=0, using the Green's function (by method of images) and converting to cylindrical coordinates, the potential along the axis of a circle of radius a centered at the origin is given by \Phi=V(1-\frac{z}{\sqrt{a^2+z^2}}), where V is the potential inside the circle and 0 outside. The computations for the normal derivative -dG/dz may involve an incorrect use of the area element, but once corrected, the final result should be \Phi = \frac{-Va}{z \sqrt{a^2+z^2}}.
gop

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

$$\Phi=V(1-\frac{z}{\sqrt{a^2+z^2}})$$

The Attempt at a Solution

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

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

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

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

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.

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!).

1. What is the concept of potential by Green's function?

The concept of potential by Green's function is a mathematical tool used in physics and engineering to solve differential equations. It involves finding a function, known as the Green's function, that satisfies a given differential equation and a set of boundary conditions. This function can then be used to find the solution to the original differential equation.

2. How is potential by Green's function used in practice?

Potential by Green's function is used in a variety of practical applications, such as in electromagnetics, fluid mechanics, and heat transfer. It can help in solving problems involving potential fields, such as electric or gravitational potentials, where the underlying physics is described by a differential equation.

3. What are the advantages of using potential by Green's function?

One of the main advantages of using potential by Green's function is that it provides a systematic and efficient method for solving differential equations. It also allows for the use of boundary conditions that may not have a simple analytical solution, making it a powerful tool for solving complex problems.

4. Are there any limitations to using potential by Green's function?

One limitation of using potential by Green's function is that it assumes that the boundary conditions are known and well-defined. If the boundary conditions are not well-defined, the solution may not be accurate. Additionally, finding the Green's function for a given problem can be challenging and may require advanced mathematical techniques.

5. How is potential by Green's function related to other mathematical concepts?

Potential by Green's function is closely related to other mathematical concepts, such as Laplace transforms and Fourier series. It can also be used in conjunction with other methods, such as finite element analysis, to solve complex problems. In some cases, potential by Green's function can be seen as a generalization of these other mathematical concepts.

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