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

  • Thread starter Thread starter Tony11235
  • Start date Start date
  • Tags Tags
    Function Harmonic
Tony11235
Messages
254
Reaction score
0
I am to find a function U, harmonic on the disk x^2 + y^2 < 6 and satisfying
u(x, y) = y + y^2 on the disk's boundary. I am not sure where to start. Hints, help, anything?
 
Physics news on Phys.org
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 < \theta < 2\pi ?
 
The theta condition that you are going to use, I believe, will be that theta is 2pi periodic.

- harsh
 

Similar threads

Back
Top