Find Harmonic Function on Disk: U(x,y)=y+y^2

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The discussion focuses on finding a harmonic function U(x,y) on the disk defined by x² + y² < 6, which satisfies the boundary condition u(x, y) = y + y². Participants emphasize the importance of using Cauchy's integral formula and suggest solving the Laplace equation through separation of variables, leading to two ordinary differential equations (ODEs). Additionally, they highlight the necessity of converting the problem into polar coordinates and correctly applying the Laplacian in polar coordinates.

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I am to find a function U, harmonic on the disk x^2 + y^2 &lt; 6 and satisfying
u(x, y) = y + y^2 on the disk's boundary. I am not sure where to start. Hints, help, anything?
 
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Use the integral formula.
 
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.
 
You are trying to solve the Laplace equation on a disk. Try separation 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
 
harsh said:
You are trying to solve the Laplace equation on a disk. Try separation 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 u(\sqrt{6}, \theta) = \sqrt{6} \sin(\theta) + 6\sin^2(\theta) a boundary condition?
 
Tony11235 said:
Then is u(\sqrt{6}, \theta) = \sqrt{6} \sin(\theta) + 6\sin^2(\theta) 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
 
harsh said:
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 \theta? Other than 0 &lt; \theta &lt; 2\pi ?
 
The theta condition that you are going to use, I believe, will be that theta is 2pi periodic.

- harsh
 

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