Is the Fourier Transform Correctly Applied in Solving This Laplace Equation?

In summary: In this case, the function to be convolved is the inverse transform of e^{-|k|y}, and the domain of the convolution operation is the set of all real numbers x such that y = x*u. So in order to apply the convolution, you need to first extend the domain of the function u to include the real line.
  • #1
lriuui0x0
101
25
Homework Statement
Solve the laplace equation ##u_{xx} + u_{yy} = 0##, with ##x \ge 0, y \ge 0##, using Fourier transform.

Subject to the boundary conditions:

$$
\begin{aligned}
u(x, 0) &= \begin{cases}1 & 0 < x < 1 \\ 0 & \text{otherwise}\end{cases} \\
u(0, y) &= \lim_{x\to\infty} u(x,y) = \lim_{y\to\infty} u(x,y) = 0
\end{aligned}
$$
Relevant Equations
Fourier transform and inverse transform
I have tried to Fourier transform in ##x## and get the result in the transformed coordinates, please check my result:

$$
\tilde{u}(k, y) = \frac{1-e^{-ik}}{ik}e^{-ky}
$$

However, I'm having some problems with the inverse transform:

$$
\frac{1}{2\pi}\int_{-\infty}^\infty \frac{1-e^{-ik}}{ik}e^{-yk}dk
$$

Not sure how to do this integral. The solution says it's

$$
\frac{4xy}{\pi}\int_0^1 \frac{vdv}{[(x-v)^2+y^2][(x+v)^2+y^2]}
$$
 
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  • #2
Sorry the inverse FT integral should be:

$$
\frac{1}{2\pi}\int_{-\infty}^\infty \frac{1-e^{-ik}}{ik}e^{-yk} e^{ixk}dk
$$
 
  • #3
I think you want to say that [tex]
\hat u(k,y) = \hat u_0(k)e^{-|k|y}.[/tex] (We need [itex]\hat u \to 0[/itex] as [itex]y \to \infty[/itex] for both positive and negative [itex]k[/itex]). This is a product of transforms, so when you invert it you obtain a convolution: [tex]
u(x,y) = \int_{-\infty}^\infty u_0(\xi) F(x - \xi,y)\,d\xi[/tex] where [itex]F[/itex] is the inverse transform of [itex]e^{-|k|y}[/itex]. But you also need [itex]u(0,y) = 0[/itex], which you can ensure by assuming the solution to be odd in [itex]x[/itex] on [itex](-\infty, \infty)[/itex]. So you need to take [itex]
u_0(\xi) = -u_0(-\xi)[/itex] for [itex]\xi < 0[/itex].
 
Last edited:
  • #4
pasmith said:
I think you want to say that [tex]
\hat u(k,y) = \hat u_0(k)e^{-|k|y}.[/tex] (We need [itex]\hat u \to 0[/itex] as [itex]y \to \infty[/itex] for both positive and negative [itex]k[/itex]). This is a product of transforms, so when you invert it you obtain a convolution: [tex]
u(x,y) = \int_{-\infty}^\infty u_0(\xi) F(x - \xi,y)\,d\xi[/tex] where [itex]F[/itex] is the inverse transform of [itex]e^{-|k|y}[/itex]. But you also need [itex]u(0,y) = 0[/itex], which you can ensure by assuming the solution to be odd in [itex]x[/itex] on [itex](-\infty, \infty)[/itex]. So you need to take [itex]
u_0(\xi) = -u_0(-\xi)[/itex] for [itex]\xi < 0[/itex].
How do we get the ##e^{-|k|y}## term? The way I derived my result is:

$$
\begin{aligned}
\mathcal{F}(\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2}) &= 0 \\
\frac{\partial^2\tilde{u}}{\partial y^2} -k^2 \tilde{u} &= 0 \\
\tilde{u}(k, y) &= A(k)e^{ky} + B(k)e^{-ky} \\
\end{aligned}
$$

And I was thinking ##e^{ky}## term must vanish since we require ##\lim_{y\to\infty} \tilde{u}(k,y) = 0##. However thinking about it again, this isn't necessarily true, as we can allow ##k \to -\infty##. But how did you derive the ##e^{-|k|y}## term?
 
Last edited:
  • #5
lriuui0x0 said:
But how do you deal derive the ##e^{-|k|y}## term?

We need [itex]\hat{u}(k,y) \to 0[/itex] as [itex]y \to \infty[/itex] for every [itex]k \in \mathbb{R}[/itex]. For [itex]k > 0[/itex] that means we can only use the [itex]e^{-ky}[/itex] solution, for [itex]k < 0[/itex] we can only use the [itex]e^{ky}[/itex] solution, and for [itex]k = 0[/itex] we can only use the constant solution. Hence [tex]
\begin{align*}
\hat u(k,y) &= \begin{cases} \hat{u}_0(k)e^{-ky} & k > 0 \\
\hat{u}_0(0) & k = 0 \\
\hat{u}_0(k)e^{ky} & k < 0 \end{cases} \\
&= \hat{u}_0(k)e^{-|k|y}.\end{align*}[/tex]
 
  • #6
pasmith said:
I think you want to say that [tex]
\hat u(k,y) = \hat u_0(k)e^{-|k|y}.[/tex] (We need [itex]\hat u \to 0[/itex] as [itex]y \to \infty[/itex] for both positive and negative [itex]k[/itex]). This is a product of transforms, so when you invert it you obtain a convolution: [tex]
u(x,y) = \int_{-\infty}^\infty u_0(\xi) F(x - \xi,y)\,d\xi[/tex] where [itex]F[/itex] is the inverse transform of [itex]e^{-|k|y}[/itex]. But you also need [itex]u(0,y) = 0[/itex], which you can ensure by assuming the solution to be odd in [itex]x[/itex] on [itex](-\infty, \infty)[/itex]. So you need to take [itex]
u_0(\xi) = -u_0(-\xi)[/itex] for [itex]\xi < 0[/itex].
How is convolution operation defined for functions only on ##\mathbf{R}^+##? Why do I need to extend the domain of ##u## to include the whole real line?
 

Related to Is the Fourier Transform Correctly Applied in Solving This Laplace Equation?

1. What is a Fourier transform?

A Fourier transform is a mathematical operation that converts a function of time or space into a function of frequency. It decomposes a complex signal into its individual frequency components.

2. How does a Fourier transform help solve the Laplace equation?

The Laplace equation is a partial differential equation that describes the distribution of a scalar field in a given space. By taking the Fourier transform of the equation, it can be converted into an algebraic equation, making it easier to solve.

3. What are the advantages of using a Fourier transform to solve the Laplace equation?

Using a Fourier transform allows for a more efficient and accurate solution to the Laplace equation. It also allows for the use of advanced mathematical techniques such as convolution and spectral methods.

4. Are there any limitations to using a Fourier transform for solving the Laplace equation?

While a Fourier transform is a powerful tool, it does have its limitations. It can only be used for linear equations and requires the function to be well-behaved and have a finite integral.

5. Can a Fourier transform be used for solving other types of equations?

Yes, a Fourier transform can be used to solve a variety of mathematical problems, including differential equations, integral equations, and signal processing problems. It is a versatile tool in many areas of science and engineering.

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