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

lriuui0x0
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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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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 <br /> \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: <br /> 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 <br /> u_0(\xi) = -u_0(-\xi) for \xi &lt; 0.
 
Last edited:
pasmith said:
I think you want to say that <br /> \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: <br /> 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 <br /> u_0(\xi) = -u_0(-\xi) for \xi &lt; 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?
 
Last edited:
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 &gt; 0 that means we can only use the e^{-ky} solution, for k &lt; 0 we can only use the e^{ky} solution, and for k = 0 we can only use the constant solution. Hence <br /> \begin{align*}<br /> \hat u(k,y) &amp;= \begin{cases} \hat{u}_0(k)e^{-ky} &amp; k &gt; 0 \\<br /> \hat{u}_0(0) &amp; k = 0 \\<br /> \hat{u}_0(k)e^{ky} &amp; k &lt; 0 \end{cases} \\<br /> &amp;= \hat{u}_0(k)e^{-|k|y}.\end{align*}
 
pasmith said:
I think you want to say that <br /> \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: <br /> 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 <br /> u_0(\xi) = -u_0(-\xi) for \xi &lt; 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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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