How do I inverse sine transform U(k,y) to get u(x,y)?

In summary, the Fourier Sine Transform is a mathematical operation used to convert a function into its frequency components. It is specifically used for odd functions, unlike the Fourier Transform which can be used for any function. The mathematical expression for the Fourier Sine Transform involves integrating the function with a sine function. It has many applications in fields such as signal processing and image analysis, but it also has limitations such as only being applicable to odd functions and assuming the function is well-behaved.
  • #1
urista
11
0
I'm trying to solve the Laplacian in 2D:
uxx+uyy=0 in the quarter plane x>0, y>0
using a Fourier sine transform
Boundary Conditions:
u(x,0)=DiracDelta(x-a) , 0< x < infinity and 0< a< infinity
u(0,y)=0, 0< y < infnity

I transformed the PDE in x using the definition of the transform with squareroot(2/Pi) in front of the integral transform and got:
Uyy-k^2*U=0
hence,
U=SquareRoot(2/Pi)*Sin(k*a)*Exp(-k*y) Is this correct?
and how do I inverse sine transform this U(k,y) to get u(x,y)

Any help please?
 
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  • #2


I can provide some guidance on your approach to solving the Laplacian in 2D using a Fourier sine transform. First of all, your transformation of the PDE in x using the definition of the transform looks correct. The resulting equation, Uyy-k^2*U=0, is a standard form for a second-order homogeneous ODE, which can be solved using standard techniques.

Your solution for U, SquareRoot(2/Pi)*Sin(k*a)*Exp(-k*y), also looks correct. This is the solution for the transformed equation, and it satisfies the boundary condition at y=0. However, it is important to note that this is only a solution for the transformed equation, and it still needs to be inverse transformed to get the solution for the original PDE in terms of u(x,y).

To inverse transform U(k,y) to get u(x,y), you can use the inverse Fourier sine transform formula, which is given by:

u(x,y) = (2/Pi) * integral from 0 to infinity of U(k,y)*sin(k*x) dk

In this case, you can plug in your solution for U(k,y) and evaluate the integral to get the final solution for u(x,y). However, it is worth noting that this solution might not be a simple closed-form expression, as it involves an integral.

In summary, your approach to solving the Laplacian in 2D using a Fourier sine transform is correct. Just remember to inverse transform your solution for U(k,y) to get the final solution for u(x,y) in terms of an integral. I hope this helps and good luck with your calculations!
 
  • #3


To inverse sine transform U(k,y) and get u(x,y), you can use the inverse Fourier sine transform defined as:

u(x,y) = (2/Pi) * Integral from 0 to infinity of U(k,y) * sin(kx) dk

In this case, since U(k,y) = SquareRoot(2/Pi) * sin(ka) * exp(-ky), we can substitute it into the above formula to get:

u(x,y) = (2/Pi) * Integral from 0 to infinity of SquareRoot(2/Pi) * sin(ka) * exp(-ky) * sin(kx) dk

Using trigonometric identities, we can simplify this to:

u(x,y) = (1/Pi) * Integral from 0 to infinity of sin(ka) * sin(kx) * exp(-ky) dk

Now, we can use the identity sin(a) * sin(b) = (1/2) * (cos(a-b) - cos(a+b)) to further simplify the integral to:

u(x,y) = (1/2Pi) * Integral from 0 to infinity of (cos(ka-kx) - cos(ka+kx)) * exp(-ky) dk

Finally, we can use the inverse Fourier cosine transform to evaluate the integral and get the final solution:

u(x,y) = (1/2Pi) * (exp(-ky) * (delta(x-a) - delta(x+a)))

Where delta(x) is the Dirac delta function. This satisfies the given boundary conditions and solves the Laplacian in the quarter plane x>0, y>0.
 

1. What is the Fourier Sine Transform?

The Fourier Sine Transform is a mathematical operation that converts a function of time or spatial position into a representation in terms of its frequency components. It is commonly used in signal processing and image analysis to decompose a signal into its constituent frequencies.

2. How is the Fourier Sine Transform different from the Fourier Transform?

The Fourier Sine Transform is used specifically for functions that are odd in nature, meaning that they have a symmetry about the origin. This is in contrast to the Fourier Transform, which can be used for any function. The Fourier Sine Transform only considers the odd part of a function, while the Fourier Transform considers both the odd and even parts.

3. What is the mathematical expression for the Fourier Sine Transform?

The mathematical expression for the Fourier Sine Transform is:

F_s(k) = ∫f(x)sin(kx)dx

where f(x) is the input function, k is the frequency variable, and F_s(k) is the output in terms of frequency components.

4. What are some applications of the Fourier Sine Transform?

The Fourier Sine Transform has many practical applications, including image analysis, audio signal processing, and solving differential equations. It is also used in fields such as physics, engineering, and economics to analyze periodic phenomena and extract useful information from signals.

5. Are there any limitations to the Fourier Sine Transform?

Yes, the Fourier Sine Transform has some limitations. It can only be applied to functions that are odd, and it is not defined for functions that are not periodic. Additionally, the Fourier Sine Transform assumes that the function is continuous and well-behaved, which may not always be the case in real-world applications.

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