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Math Help - PDE Question...

  1. #1
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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!
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  2. #2
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    Quote Originally Posted by jkhayer View Post
    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!
    ..
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  3. #3
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    Quote Originally Posted by jkhayer View Post
    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

     <br />
u = \sin \left( \frac{ 3 - 4x - 3y}{2} \right)<br />
(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.
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