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

[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

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

 

[lriuui0x0]

Sorry the inverse FT integral should be:

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

 

[pasmith]

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 $\xi < 0$.

 

[lriuui0x0]

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?

 

[pasmith]

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 $$\begin{align*} \hat u(k,y) &= \begin{cases} \hat{u}_0(k)e^{-ky} & k > 0 \\ \hat{u}_0(0) & k = 0 \\ \hat{u}_0(k)e^{ky} & k < 0 \end{cases} \\ &= \hat{u}_0(k)e^{-|k|y}.\end{align*}$$

 

[lriuui0x0]

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?