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I need to solve the following problem on the infinite strip:

$\displaystyle \begin{align} u_{xx}(x,y) + u_{yy}(x,y) = 0, & -\infty < x < \infty, & 0<y<1 \\ u(x,0)= \begin{cases}1, &\text{if}\, |x|<2 \\ 0 & \text{if}\, |x|>2 \end{cases} \\u(x,1) = 0,\,\,\, -\infty < x < \infty\\ u(x,y)\,\text{bounded for}\,|x|\to \infty\end{align}$.

The first thing I did was try to find the transformed problem: letting $\displaystyle U(\alpha,y) = \frac{1}{2\pi}\int_{-\infty}^{\infty}u(x,y)\exp[-i \alpha x]dx$, $\mathcal{F}[u_{xx}(x,y)]=-\alpha^{2}U(\alpha,y)$, and $\mathcal{F}[u_{yy}(x,y)]=U^{\prime\prime}(\alpha,y)$,

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the transformed problem is

$\begin{align}-\alpha^{2}U(\alpha,y)+U^{\prime\prime}(\alpha,y)=0 \end{align}$, whose solution is of the form

$\displaystyle \begin{align} U(\alpha,y)=A \exp[-|\alpha|y]+B\exp[|\alpha|y] .\end{align}$

Then, to apply the boundary conditions, we must first transform them.

For $f(x) = u(x,0)=\begin{cases} 1, &\text{if}\,|x|<2 \\ 0 & \text{if}\,|x|>2 \end{cases}$, $\displaystyle F(\alpha) = \int_{-2}^{2}\exp[-ix\alpha]dx = \frac{1}{\alpha \pi}\sin (2 \alpha)$.

For $g(x) = u(x,1)=0$, I’m supposing that $\displaystyle G(\alpha) = \int_{-\infty}^{-2} 0 \cdot \exp[-ix\alpha]dx + \int_{2}^{\infty} 0 \cdot \exp[-ix\alpha]dx = 0$.

Now, applying the boundary condition that $U(\alpha,0) = F(\alpha)$, we have that $\displaystyle A \exp[-|\alpha|\cdot 0] + B \exp[|\alpha|\cdot 0] = \frac{1}{\alpha \pi}\sin(2\alpha)$ implies that $\displaystyle \mathbf{A + B = \frac{1}{\alpha \pi} \sin (2 \alpha)}\,\,(1)$

Applying the boundary condition that $U(x,1)=0$, we have that $A \exp[-|\alpha|]+B\exp[|\alpha|]=0 \, \Longrightarrow exp[|\alpha|]\left(A\exp[-2|\alpha|]+B \right) = 0 \, \Longrightarrow \mathbf{A\exp[-2|\alpha|]+B = 0}\,\,(2)$

Solving equation $(2)$ for $B$: $\displaystyle B = -A\exp[-2|\alpha|]$, and substituting into $(1)$, we obtain the following expression for $A$:

$\displaystyle \begin{align} A – A\exp[-2|\alpha|] = \frac{1}{\alpha \pi}\sin(2 \alpha)\\ \Longrightarrow \, A(1-\exp[-2|\alpha|] = \frac{1}{\alpha \pi} \sin(2 \alpha) \\ \Longrightarrow \,\frac{\sin(2 \alpha)}{\alpha \pi (1-\exp[-2|\alpha|])} \\ \Longrightarrow\mathbf{A = \frac{\exp[2|\alpha|]\cdot\sin(2 \alpha)}{\alpha \pi (\exp[2|\alpha|]-1)}}\end{align}$

Sustituting this back into $(2)$, we get that $\displaystyle \mathbf{B} = -\frac{\sin(2\alpha)}{\alpha \pi (\exp[2|\alpha|]-1)}$

So, the solution to the transformed problem is $\begin{align}\displaystyle U(\alpha,y)= \frac{\exp[2|\alpha|]\cdot \sin(2 \alpha)}{\alpha \pi(\exp[2|\alpha|]-1)}\exp[-|\alpha|y] – \frac{\sin(2\alpha)}{\alpha \pi (\exp[2|\alpha|]-1)}\exp[|\alpha|y]\\ = \frac{\exp[|\alpha|(2-y)]}{\exp[2|\alpha|]-1}F(\alpha) – \frac{\exp[|\alpha|y]}{(\exp[2|\alpha|-1)}F(\alpha)\end{align}$

Now, I need to apply the inverse transform in order to get back from $U(\alpha,y)$ to $u(x,y)$. So, I thought I would do it for each summand in $U(\alpha,y)$ individually.

So, for $\displaystyle\frac{\exp[|\alpha|(2-y)]}{\exp[2|\alpha|]-1}F(\alpha) $, let $\displaystyle H_{1}(\alpha) = \frac{\exp[|\alpha|(2-y)]}{\exp[2|\alpha|]-1}$. Then if $u_{1}(x,y)-u_{2}(x,y) = u(x,y)$, $\displaystyle \mathbf{u_{1}(x,y)} = f * h_{1} = \frac{1}{2\pi} f * \mathcal{F}^{-1}\left[\frac{\exp[|\alpha|(2-y)]}{\exp[2|\alpha|]-1} \right]$.

Then, this $\displaystyle = \frac{1}{2\pi}\int_{-\infty}^{\infty}f(z) \cdot\mathcal{F}^{-1}\left[\frac{\exp[|\alpha|(2-y)]}{\exp[2|\alpha|]-1} \right]dz = \frac{1}{2 \pi}\int_{-2}^{2}1\cdot\mathcal{F}^{-1}\left[\frac{\exp[|\alpha|(2-y)]}{\exp[2|\alpha|]-1} \right]dz \\ \displaystyle \mathbf{= \frac{1}{2 \pi}\int_{-2}^{2}\mathcal{F}^{-1}\left[\frac{\exp[|\alpha|(2-y)]}{\exp[2|\alpha|]-1} \right]dz} $.

Also, for $\displaystyle \frac{\exp[|\alpha|y]}{(\exp[2|\alpha|-1)}F(\alpha)$, let $\displaystyle H_{2}(\alpha) = \frac{\exp[|\alpha|y]}{(\exp[2|\alpha|-1)}$. Then $\displaystyle \mathbf{u_{2}(x,y)} =f * h_{2} = \frac{1}{2\pi} f * \mathcal{F}^{-1} \left[\frac{\exp[|\alpha|y]}{(\exp[2|\alpha|-1)}\right] = \frac{1}{2\pi}\int_{-\infty}^{\infty}f(z)\cdot\mathcal{F}^{-1} \left[\frac{\exp[|\alpha|y]}{(\exp[2|\alpha|-1)}\right] dz\\ \displaystyle = \frac{1}{2\pi}\int_{-2}^{2}1 \cdot \mathcal{F}^{-1} \left[\frac{\exp[|\alpha|y]}{(\exp[2|\alpha|-1)}\right] dz \\ \displaystyle \mathbf{ = \frac{1}{2\pi}\int_{-2}^{2} \mathcal{F}^{-1} \left[\frac{\exp[|\alpha|y]}{(\exp[2|\alpha|-1)}\right] dz} $.

**But, I can go no further until I can figure out what $\displaystyle \mathcal{F}^{-1}\left[\frac{\exp[|\alpha|(2-y)]}{\exp[2|\alpha|]-1} \right]$ and $\displaystyle \mathcal{F}^{-1} \left[\frac{\exp[|\alpha|y]}{(\exp[2|\alpha|-1)}\right]$ are, and I have no idea how to do that.**

The answer given in the back of my book is weird: $\displaystyle u(x,y) = \frac{1}{\pi}\sum_{n=0}^{\infty} \left[\arctan \left(\frac{x+2}{2n+y} \right)- \arctan \left( \frac{x-2}{2n+y}\right) \\- \arctan \left(\frac{x+2}{2n+2-y} \right) + \arctan \left(\frac{x-2}{2n+2-y} \right) \right]$.

So, first of all, how do I find the inverse Fourier transforms of those things I bolded a couple of paragraphs ago, and how are these series going to come into play in my solution? Please be as detailed and explicit as possible in your answer. Thank you.

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Separation of variables gives

$$

\frac{X”}{X} = \lambda = -\frac{Y”}{Y}.

$$

Because you want solutions that remain bounded in $x$ as $|x|\rightarrow\infty$, that dictates $\lambda = -\mu^2$ where $\mu$ is real. Otherwise you get exponential solutions that explode at one or both of $\pm\infty$. Because $y(1)=0$ needs to hold, then the solutions are

$$

X(x)=e^{i\mu x},\;\;Y(y)=\sinh(\mu(y-1)).

$$

The trial solution is an integral “sum” of such solutions

$$

u(x,y)=\int_{-\infty}^{\infty}c(\mu)e^{i\mu x}\sinh(\mu(y-1))d\mu.

$$

The coefficient $c(\mu)$ is determined by

$$

u(x,0) = -\int_{-\infty}^{\infty}c(\mu)e^{i\mu x}\sinh(\mu)d\mu.

$$

The function $u(x,0)$ is $1$ for $-2 \le x \le 2$ and is $0$ otherwise. So you want to find coefficients $c(\mu)$ such that

$$

\chi_{[-2,2]}(x) = -\int_{-\infty}^{\infty}c(\mu)e^{i\mu x}\sinh(\mu)d\mu.

$$

Multiplying by $e^{-is x}$, integrating and using the Fourier orthogonality trick,

$$

\frac{1}{2\pi}\int_{-2}^{2}e^{-is x}dx = -c(s)\sinh(s) \\

\frac{1}{\pi}\frac{e^{-i2s}-e^{i2s}}{-2is}= -c(s)\sinh(s) \\

\frac{1}{\pi}\frac{\sin(2s)}{s}=-c(s)\sinh(s) \\

c(s) = -\frac{1}{\pi}\frac{\sin(2s)}{s\sinh(s)}.

$$

Therefore, the solution $u(x,y)$ is given by

$$

u(x,y)=-\frac{1}{\pi}\int_{-\infty}^{\infty}e^{i\mu x}\frac{\sin(2\mu)}{\mu}\frac{\sinh(\mu(y-1))}{\sinh(\mu)}d\mu.

$$

I may be off by a negative. I don’t see how you’re going to get a discrete sum out of that integral. You can easily check that the above is the correct solution, so far as it goes. Differentiating twice in $x$ gives the negative of differentiating twice in $y$. Clearly $u(x,1)=0$, and $u(x,0)$ isn’t hard to check either. $\sinh(\mu)$ has zeros on the imaginary axis, and maybe they’re using residues somehow, but I don’t see how that would give arctan terms.

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