Thread: 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!

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!

..

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

$\displaystyle
u = \sin \left( \frac{ 3 - 4x - 3y}{2} \right)
$ (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.