Solving Laplace's equation $\triangle u = 0$ in the semi-infinite strip $S = \{(x,y): 0 < x < 1, 0 < y\}$ 
Solve Laplace's equation $\triangle u = 0$ in the semi-infinite strip $$S = \{(x,y): 0 < x < 1,  0 < y\}$$
subject to the following boundary conditions
$$\begin{cases}u(0,y) = 0 &  0\le y\\ u(1,y) = 0 & 0\le y\\ u(x,0) = f(x) & 0\le x\le 1  \end{cases}$$ where $f$ is a given function, with of course $f(0) = f(1) = 0$. Write $$f(x) = \sum_{n=1}^\infty a_n \sin n\pi x$$ and expand the general solution in terms of the special solutions given by $$u_n(x,y) = e^{-n\pi y}\sin n\pi x$$
Express $u$ as an integral involving $f$, analogous to the Poisson integral formula.
Source: Exercise $19$, Chapter $2$, Stein & Shakarchi's Fourier Analysis

My work: I have solved the problem as in this post. I got $u_n(x,y) = F_n(x)G_n(y)$ where $F_n(x)= c_n \sin n\pi x$ and $G_n(y) = p_n e^{n\pi y} + q_n e^{-n\pi y}$ for $n = \pm 1, \pm 2, \pm 3, \ldots$.

*

*I am forced to take $p_n = 0$ for boundedness of the solution as $y\to\infty$. Also, we have to reject all negative values of $n$ for the same reason. However, I do not see why we are doing this since it hasn't been mentioned in the problem in the textbook. This answer also relies on "physical considerations" and does not provide any mathematical reasons. Is taking $p_n = 0$ for all $n$ the correct way to solve this problem?


*We get $$u(x,y) = \sum_{n=1}^\infty u_n(x,y) = \sum_{n=1}^\infty \alpha_n e^{-n\pi y} \sin n\pi x$$
Since $u(x,0) = f(x) = \sum_{n=1}^\infty a_n \sin n\pi x$, we must have $\alpha_n = a_n$ for all $n\in \mathbb N$, right? I think this follows from $\int_{-\pi}^\pi \sin mx \sin nx = 0$ for $m\ne n$, and uniform (and absolute) convergence of the sums by the Weierstrass M-test. Please confirm.


*How do I express $u$ as an integral involving $f$, analogous to the Poisson integral formula? For context, the Poisson integral formula refers to $u(r,\theta) = (f\ast P_r)(\theta) = \frac{1}{2\pi}\int_{-\pi}^\pi P_r(y) f(\theta-y)\, dy$, where $u(r,\theta)$ solves the steady-
state heat equation in the unit disc and $P_r(\theta)$ is the well-known Poisson kernel for $0\le r < 1$.
Thanks a lot!
 A: Let's start with what we get after going through our separation of variables procedure:
$$u(x,y)=\sum_{n=1}^\infty \beta_n e^{-n\pi y}\sin(n\pi x)$$
First of all, when you have any potential solution of a PDE, it is always good to go back and check if it actually satisfies the desired equation! For any of the summands,
$$\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)(e^{-n\pi y}\sin(n\pi x))=-n^2\pi^2 e^{-n\pi y}\sin(n\pi x)+n^2\pi^2 e^{-n\pi y}\sin(n\pi x)=0$$
Ok, next check the boundary conditions. Are BCs for $x$ are clearly satisfied:
$$u(0,y)=\sum_{n=1}^\infty \beta_ne^{-n\pi y}\sin(0)=0 \\ u(1,y)=\sum_{n=1}^\infty \beta_ne^{-n\pi y}\sin(n\pi)=0$$
As sine as roots at integer multiples of pi. And, our decay condition for $y$ is clearly met, since $|u|$ decreases exponentially fast w.r.t $y$. However, our most critical boundary condition, $u(x,0)=f(x)$ is not automatically satisfied. However, let $m$ be a positive integer, and let's investigate
$$\int_0^1u(x',0)\sin(m\pi x')\mathrm dx'=\int_0^1f(x')\sin(m\pi x')\mathrm dx' \\ =\int_0^1\sum_{n=1}^\infty \beta_n \sin(n\pi x')\sin(m\pi x')\mathrm dx'$$
Now via the Dominated Convergence Theorem, we can switch the order of summation and integration (i.e since the decaying exponential dominates the integrand)
$$=\sum_{n=1}^\infty \beta_n \int_0^1 \sin(n\pi x')\sin(m\pi x')\mathrm dx'$$
I'll let you check on your own that this integral is $\frac{1}{2}\delta_{m,n}$. Hence
$$2\int_0^1f(x')\sin(m\pi x')\mathrm dx'=\beta_m$$
Let's plug this into our solution formula and see if we can rearrange to get any nice results.
$$u(x,y)=\sum_{n=1}^\infty \beta_n e^{-n\pi y}\sin(n\pi x) \\ =2\sum_{n=1}^\infty e^{-n\pi y}\left(\int_0^1f(x')\sin(n\pi x')\mathrm dx'\right)\sin(n\pi x)$$
Again, via the DCT, we can rearrange:
$$u(x,y)=2\int_0^1 f(x')\sum_{n=1}^\infty e^{-n\pi y}\sin(n\pi x')\sin(n\pi x)\mathrm dx'$$
So it is pertinent to evaluate sums of the form (using $\pi x=X$ for short)
$$\sum_{n=1}^\infty e^{-nY}\sin(nX')\sin(nX)$$
Unfortunately, this is where things start to get really nasty. Instead of going through all the steps, I'll give an outline of how the computation is done, and let Mathematica sort out the details.
First use
$$\sin(a)\sin(b)=\frac{1}{2}\left(\cos(a-b)-\cos(a+b)\right)$$
And rewrite these cosines using their exponential form using the definition
$$\cos (z)=\frac{e^{iz}+e^{-iz}}{2}$$
Now break the sum into multiple geometric series, and simplify as much as possible. Mathematica is able to come up with
$$\sum_{n=1}^\infty e^{-n\pi y}\sin(n\pi x')\sin(n\pi x)\\ = \frac{1}{4}\frac{e^{i\pi (x'+x)}e^{\pi y}(e^{2\pi ix'}-1)(e^{2\pi ix}-1)(e^{2\pi y}-1)}{(e^{i\pi(x'+x)}-e^{\pi y})(e^{i\pi x}-e^{i\pi x'}e^{\pi y})(e^{i\pi x'}-e^{i\pi x}e^{\pi y})(e^{i\pi (x'+x)}e^{\pi y}-1)}\equiv \frac{1}{4}K(x';~x,y)$$
I don't know if this can be simplified further.
$$\boxed{u(x,y)=\frac{1}{2}\int_0^1 f(s)K(s;~x,y)\mathrm ds}$$
Analogous, but far less pretty, than the Poisson kernel.
