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Harmonic function

by Tony11235
Tags: function, harmonic
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Tony11235
#1
Dec4-05, 09:45 AM
P: 276
I am to find a function U, harmonic on the disk [tex] x^2 + y^2 < 6 [/tex] and satisfying
[tex] u(x, y) = y + y^2 [/tex] on the disk's boundary. I am not sure where to start. Hints, help, anything?
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matt grime
#2
Dec4-05, 09:49 AM
Sci Advisor
HW Helper
P: 9,396
Use the integral formula.
LeonhardEuler
#3
Dec4-05, 09:53 AM
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P: 864
I would think Cauchy's integral formual would be useful here: you have the value of a function on a boudry and want the value in the interior.

harsh
#4
Dec4-05, 07:00 PM
P: 76
Harmonic function

You are trying to solve the Laplace equation on a disk. Try seperation of variables, then break it down to 2 ODE's. Here is a start for you..

You will probably need to solve the PDE in polar coordinates.

- harsh
Tony11235
#5
Dec4-05, 11:01 PM
P: 276
Quote Quote by harsh
You are trying to solve the Laplace equation on a disk. Try seperation of variables, then break it down to 2 ODE's. Here is a start for you..
You will probably need to solve the PDE in polar coordinates.
- harsh
Then is [tex] u(\sqrt{6}, \theta) = \sqrt{6} \sin(\theta) + 6\sin^2(\theta) [/tex] a boundary condition?
harsh
#6
Dec4-05, 11:08 PM
P: 76
Quote Quote by Tony11235
Then is [tex] u(\sqrt{6}, \theta) = \sqrt{6} \sin(\theta) + 6\sin^2(\theta) [/tex] a boundary condition?
Looks right. Make sure you solve the correct PDE, the laplacian in r,theta is not as simple as U_rr and U_theta*theta

- harsh
Tony11235
#7
Dec5-05, 06:59 AM
P: 276
Quote Quote by harsh
Looks right. Make sure you solve the correct PDE, the laplacian in r,theta is not as simple as U_rr and U_theta*theta
- harsh
I know. In an earlier problem I had to compute the laplacian in polar. Oh and one more thing, is there anything else I need to know about [tex] \theta [/tex]? Other than [tex] 0 < \theta < 2\pi [/tex] ?
harsh
#8
Dec5-05, 09:21 AM
P: 76
The theta condition that you are going to use, I believe, will be that theta is 2pi periodic.

- harsh


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