Fourier Transform - Laplace Equation on infinite strip - weird solution involving series

Question

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 $\mathcal{F}[u_{yy}(x,y)]=U^{\prime\prime}(\alpha,y)$,

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.

Answer 1

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.

Answer 2

I am presenting a solution for an infinite strip that is a rotated version of yours. So you may need to interchange $x$ and $y$ or something similar. If the boundary conditions are \begin{align*} (u)_{x=0}&=g_1(y),\quad (u)_{x=L}=g_2(y), \end{align*} then the solution form is given by \begin{equation*} \begin{split} u&=\int_0^{\infty}\frac{1}{\sinh(\lambda L)}\Bigl\{\sinh[\lambda(L-x)]\left[A(\lambda)\cos(\lambda y)+B(\lambda)\sin(\lambda y)\right] \\ &\quad +\sinh(\lambda x)\left[C(\lambda)\cos(\lambda y)+D(\lambda)\sin(\lambda y)\right]\Bigr\}\,d\lambda. \end{split} \end{equation*} Imposing the boundary conditions at $x=0$ and $x=L$, we get \begin{align*} \int_0^{\infty} \left[A(\lambda)\cos(\lambda y)+B(\lambda)\sin(\lambda y)\right]\,d\lambda=g_1(y), \\ \int_0^{\infty} \left[C(\lambda)\cos(\lambda y)+D(\lambda)\sin(\lambda y)\right]\,d\lambda=g_2(y), \end{align*} which on inverting yields \begin{align*} A(\lambda)=\frac{1}{\pi}\int_{-\infty}^{\infty} g_1(\hat{y})\cos(\lambda\hat{y})\,d\hat{y}, \\ B(\lambda)=\frac{1}{\pi}\int_{-\infty}^{\infty} g_1(\hat{y})\sin(\lambda\hat{y})\,d\hat{y}, \\ C(\lambda)=\frac{1}{\pi}\int_{-\infty}^{\infty} g_2(\hat{y})\cos(\lambda\hat{y})\,d\hat{y}, \\ D(\lambda)=\frac{1}{\pi}\int_{-\infty}^{\infty} g_2(\hat{y})\sin(\lambda\hat{y})\,d\hat{y}. \end{align*} Substituting these expressions into the original solution form, and carrying out the integration with respect to $\lambda$, we get \begin{equation*} u=\frac{1}{2L}\sin\frac{\pi x}{L}\int_{-\infty}^{\infty}\left\{\frac{g_1(\hat{y})}{\cosh\frac{\pi(y-\hat{y})}{L}-\cos\frac{\pi x}{L}} +\frac{g_2(\hat{y})}{\cosh\frac{\pi(y-\hat{y})}{L}+\cos\frac{\pi x}{L}}\right\}\,d\hat{y}. \end{equation*} In your case $L=1$. In order to verify if the boundary conditions are satisfied, the limits $x\rightarrow 0,L$, should be taken after evaluating the integral in the above solution. Also note that $g_1(y)$ and $g_2(y)$ need to be continuous functions and not discontinuous functions as in your problem statement. The reason is that the fluxes at the point of discontinuity will be unbounded which does not make sense physically.