Fourier COSINE Transform (solving PDE - Laplace Equation)

In summary, the conversation is about using the Fourier Cosine Transform to solve the Laplace equation. The equations and boundary conditions are provided and the person is using the definition of the transform to solve the problem. They mention using a textbook by G. Duffy for guidance. There is a discussion about the conditions for u(x,0) and the person is unsure about how to solve [u(x,y).sen(w.y)]. They have not been able to find a solution to this problem.
  • #1
juan.
8
0
I'm trying to solve Laplace equation using Fourier COSINE Transform (I have to use that), but I don't know if I'm doing everything OK (if I'm doing everything OK, the exercise is wrong and I don't think so).

NOTE: U(..) is the Fourier Transform of u(..)

This are the equations (Laplace, boundary, etc.):

[tex]u_{xx}+u_{yy} = 0, with: y>0, 0<x<a[/tex]
[tex]u_y(x,0) = u(0, y) =0[/tex]
[tex]u(a,y) = g(y)[/tex]
[tex]|u(x,y)|<M[/tex]I used "Transform Methods for Solving PDE", from G. Duffy and this is what I'm doing (maybe you have a better way):

Now, since x is between from 0 to a and y is between 0 and infinity, I use the definition of Fourier COSINE Transform and:
[tex]\int_{0}^{\infty} u_{xx}.cos(w.y) dy + \int_{0}^{\infty} u_{yy}.cos(w.y) dy = 0[/tex]
where:
[tex]\int_{0}^{\infty} u_{xx}.cos(w.y) dy = U_{yy}(x,w)[/tex]
[tex]\int_{0}^{\infty} u_{yy}.cos(w.y) dy = [u_y(x,y).cos(w.y)] - w.[u(x,y).sin(w.y)] - w.\int_{0}^{\infty}u(x,y).cos(w.y) dy[/tex]

Note: I don't know how to write the FTC in LaTeX. Where it says [...] it's FTC from 0 to infinity

Now, I know that:
[tex][u_y(x,y).cos(w.y)] = 0[/tex] because of the conditions: [tex]u_y(x,0) =0 and |u(x,y)|<M[/tex] (is that ok?)

But now, I want to solve this: [u(x,y).sen(w.y)] and I don't know why, because I don't have any condition for u(x,0)(but I have a condition for u(0,y).

What's wrong? I searched everywhere but I couldn't find anything that helps me. Thanks!Note 2: using Fourier Cosine Transform definition, I know that:
[tex]\int_{0}^{\infty}u(x,y).cos(w.y) dy = U(x,w)[/tex]
 
Last edited:
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  • #2
juan. said:
uxx+uyy=0,
That is the definition of a harmonic function.
 

1. What is a Fourier cosine transform?

A Fourier cosine transform is a mathematical tool used to convert a function from its original domain, such as time or space, to the frequency domain. It is especially useful in solving partial differential equations, such as the Laplace equation.

2. How does a Fourier cosine transform work?

A Fourier cosine transform decomposes a function into its constituent frequencies and represents it as a series of cosine functions. This allows for the simplification of mathematical operations, making it easier to solve complex equations.

3. What is the difference between a Fourier cosine transform and a Fourier sine transform?

While a Fourier cosine transform uses cosine functions to represent a function, a Fourier sine transform uses sine functions. The choice of which transform to use depends on the boundary conditions and the type of function being transformed.

4. How is a Fourier cosine transform used in solving PDEs?

The Fourier cosine transform is used to convert a partial differential equation, such as the Laplace equation, into an algebraic equation in the frequency domain. This makes it easier to solve and gives a solution in terms of cosine functions, which can then be transformed back to the original domain.

5. What are the applications of Fourier cosine transform in science and engineering?

The Fourier cosine transform has a wide range of applications in fields such as signal processing, image processing, and fluid mechanics. It is used to analyze and solve various types of differential equations, making it a powerful tool in understanding and manipulating complex systems.

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