November 7th, 2009, 09:16 PM
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 | 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.
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