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November 7th, 2009, 09:49 PM
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| |  PDE Question...
Solve the PDE:
3 Ux - 4 Uy = 0
If the initial condition is: u(x,y) = sin(x) on 2x + y = 1
I got a solution of: u(x,y) = (sin(-4x-3y) + 3) / (2)
...I just want to confirm that this is correct, I did try the check by calculating Ux and Uy and subbing them into the original equation and equating it to zero, but I just want to be 100% sure.
Thanks! |
November 7th, 2009, 10:16 PM
|  | Flow Master | | Join Date: Dec 2007 Location: Zeitgeist
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Quote:
Originally Posted by jkhayer  Solve the PDE:
3 Ux - 4 Uy = 0
If the initial condition is: u(x,y) = sin(x) on 2x + y = 1 Mr F says: The function below satisfies the PDE. But how can you have an initial condition when there is no time dependence in the given PDE?
I got a solution of: u(x,y) = (sin(-4x-3y) + 3) / (2)
...I just want to confirm that this is correct, I did try the check by calculating Ux and Uy and subbing them into the original equation and equating it to zero, but I just want to be 100% sure.
Thanks! | ..
__________________ There are two things you should never try to prove: the impossible and the obvious. The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti)
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November 8th, 2009, 07:42 AM
|  | MHF Contributor | | Join Date: Dec 2008 Location: Conway AR
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Quote:
Originally Posted by jkhayer Solve the PDE:
3 Ux - 4 Uy = 0
If the initial condition is: u(x,y) = sin(x) on 2x + y = 1
I got a solution of: u(x,y) = (sin(-4x-3y) + 3) / (2)
...I just want to confirm that this is correct, I did try the check by calculating Ux and Uy and subbing them into the original equation and equating it to zero, but I just want to be 100% sure.
Thanks! | Your answer is sort of right. It should read  (notice the brackets).
As for the term initial condition. A lot of time the term initial condition is used even if it's for a boundary condition as this one is. | | Thread Tools | | | | Display Modes |  Linear Mode | 
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