Laplace equation on square (using fourier transform)

  • Thread starter bremvil
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10
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Hello,

I'm trying to solve the following problem:

[tex]\nabla^2 p = 0[/tex]
[tex]\frac{\partial p}{\partial y}(x, y_{bot}) = \frac{\partial p}{\partial y}(x, y_{top}) = 0 [/tex]
[tex]\frac{\partial p}{\partial x}(x_{left}, y) = \frac{\partial p}{\partial x}(x_{right}, y) = C_0 [/tex]

which is the laplace equation with a set of boundary conditions on a square. The physical interpretation is that the horizontal pressure derivative on the left is the same as on the right. The material properties like viscosity and hydraulic conductivity within the square are uniform (the problem deals with water flow in a porous medium). The vertical pressure derivative on the top is the same as on the bottom, and is 0 (in a zero gravity situation). In this scenario I would expect the solution to be a constant horizontal pressure gradient plus a constant. But when I try to solve the problem using a fourier transform I don't get this. I expect that I'm doing something wrong, could you maybe point out where?

First, doing a fourier transform over x, the problem transforms into: (not transforming boundary conditions on [tex]x_{left}[/tex] and [tex]x_{right}[/tex])

[tex] - \omega^2\bar{p}(\omega, y) + \frac{\partial^2 \bar{p}(\omega, y)}{\partial y^2}= 0[/tex]
[tex]\frac{\partial \bar{p}(\omega, y_{bot})}{\partial y} = \frac{\partial \bar{p}(\omega, y_{top})}{\partial y} = 0 [/tex]

I then assumed that p would take the following form in the omega, y domain:
[tex]\bar{p}(\omega, y) = a(\omega)e^{-\omega y} + b(\omega)e^{\omega y}[/tex]
Leading to:
[tex]\frac{\partial \bar{p}(\omega, y)}{\partial y} = -a(\omega) \omega e^{-\omega y} + b(\omega) \omega e^{\omega y}[/tex]

Using the boundary condition for [tex] y = y_{bot}[/tex] and assuming for simplicity
[tex]y_{bot} = 0[/tex]
I get:

[tex] -a(\omega) \omega + b(\omega)\omega = 0[/tex]
so

[tex]a(\omega) = b(\omega)[/tex]
Okay, then the condition at [tex] y = y_{top} [/tex]

[tex] -a(\omega) \omega e^{-\omega y_{top}} + b(\omega) \omega e^{\omega y_{top}} = 0 [/tex]
Using a = b I get:

[tex] -a(\omega) + a(\omega) e^{2\omega y_{top}} = 0 [/tex]
And finally:

[tex] a(\omega)\left(e^{2\omega y_{top}} - 1\right) = 0 [/tex]

And this has no non-trivial (a != 0) solution unless I say [tex]\omega = 0[/tex], but that does not make sense I think and it won't give me the horizontal pressure gradient that I expected. Could anyone please point out what I'm doing wrong? I never really used fourier for solving differential equations so I could be overlooking something fundamental.
 
10
0
A bump to show that I am still interested in the topic. Hopefully someone with more knowledge about this topic than me can give me a hint :)
 
10
0
Does anyone have an idea?
 

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