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

• lriuui0x0
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.
lriuui0x0
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

$$\frac{4xy}{\pi}\int_0^1 \frac{vdv}{[(x-v)^2+y^2][(x+v)^2+y^2]}$$

Delta2
Sorry the inverse FT integral should be:

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

I think you want to say that $$\hat u(k,y) = \hat u_0(k)e^{-|k|y}.$$ (We need $\hat u \to 0$ as $y \to \infty$ for both positive and negative $k$). This is a product of transforms, so when you invert it you obtain a convolution: $$u(x,y) = \int_{-\infty}^\infty u_0(\xi) F(x - \xi,y)\,d\xi$$ where $F$ is the inverse transform of $e^{-|k|y}$. But you also need $u(0,y) = 0$, which you can ensure by assuming the solution to be odd in $x$ on $(-\infty, \infty)$. So you need to take $u_0(\xi) = -u_0(-\xi)$ for $\xi < 0$.

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pasmith said:
I think you want to say that $$\hat u(k,y) = \hat u_0(k)e^{-|k|y}.$$ (We need $\hat u \to 0$ as $y \to \infty$ for both positive and negative $k$). This is a product of transforms, so when you invert it you obtain a convolution: $$u(x,y) = \int_{-\infty}^\infty u_0(\xi) F(x - \xi,y)\,d\xi$$ where $F$ is the inverse transform of $e^{-|k|y}$. But you also need $u(0,y) = 0$, which you can ensure by assuming the solution to be odd in $x$ on $(-\infty, \infty)$. So you need to take $u_0(\xi) = -u_0(-\xi)$ for $\xi < 0$.
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?

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lriuui0x0 said:
But how do you deal derive the ##e^{-|k|y}## term?

We need $\hat{u}(k,y) \to 0$ as $y \to \infty$ for every $k \in \mathbb{R}$. For $k > 0$ that means we can only use the $e^{-ky}$ solution, for $k < 0$ we can only use the $e^{ky}$ solution, and for $k = 0$ we can only use the constant solution. Hence \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*}

pasmith said:
I think you want to say that $$\hat u(k,y) = \hat u_0(k)e^{-|k|y}.$$ (We need $\hat u \to 0$ as $y \to \infty$ for both positive and negative $k$). This is a product of transforms, so when you invert it you obtain a convolution: $$u(x,y) = \int_{-\infty}^\infty u_0(\xi) F(x - \xi,y)\,d\xi$$ where $F$ is the inverse transform of $e^{-|k|y}$. But you also need $u(0,y) = 0$, which you can ensure by assuming the solution to be odd in $x$ on $(-\infty, \infty)$. So you need to take $u_0(\xi) = -u_0(-\xi)$ for $\xi < 0$.
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?

## 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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