Poisson equation on periodic strip - Fourier Series Convergence Let $(x,y) \in \mathbb{T} \times [0,1]$, ie the periodic strip ($\mathbb{T}$ is the circle). Consider the Dirichlet Problem:
$$\Delta u(x,y)=f(x,y)$$
$$u(x,0) = a(x)\qquad u(x,1) = b(x)$$
By expanding all functions as Fourier series in $x$ like $u(x,y) = \sum \hat{u}_k(y)e^{ikx}$ we get a series of ODEs. The $k=0$ case can be solved by straight integration but I omit this part.
The $k\neq0$ case is solved by variation of parameters, writing $\hat{u}_k(y) = h_k(y)+g_k(y)$ as the sum of homoegenous and general parts we get
$$h_k(y) = \frac{\hat{b}_k\sinh(ky) - \hat{a}_k\sinh[k(y-1)]}{\sinh(k)}$$
$$g_k(y) =\int_0^y \frac{\sinh[k(y-t)]}{k} \hat{f}_k(t)dt - \frac{\sinh(ky)}{\sinh(k)}\int_0^1 \frac{\sinh[k(1-t)]}{k} \hat{f}_k(t)dt$$
My issue is with the convergence of the solution $u$. For example, suppose I would like to show that the solution is at the very least in $L^2$, ie 
$$\|u\|_{L^2}^2 = \sum_k \|\hat{u}_k(\cdot)\|_{L^2[0,1]}^2 < \infty$$
For the homogeneous part, integrating $\sinh^2(ky)$ I get 
$$\|h_k(y)\|_{L^2[0,1]}^2 \approx \frac{|\hat{a}_k|^2 + |\hat{b}_k|^2}{k}$$
which is what I expected. On the other hand the general part I get:
$$\|g_k(y)\|_{L^2[0,1]}^2 \approx \frac{\sinh^2(k)}{k^4}\|\hat{f}_k(y)\|_{L^2[0,1]}^2$$ 
and here there must be something wrong because the $\sinh^2(k)$ blows up.
Of course I haven't specified in what space lie $a(x), b(x)$ and $f(x,y)$ nevertheless there seems to be something wrong with this general term. I would need some very strong conditions on $f$ to dominate the $\sinh^2(k)$ growth in the numerator. 
What gives?
 A: Let's isolate the issue: you solved the ODE $g_k''-k^2g_k=\hat f_k$ with zero boundary conditions on the interval $[0,1]$, and got $g_k$ that is much larger than $\hat f_k$  when $k$ is larger. This cannot happen, since $g_k$ is just the convolution of $\hat f_k$ with Green's function; so you made a mistake somewhere in the process of solving for $g_k$. I'll use the Green function method: 
$$g_k(y) = \int_0^1 \hat f_k(z) G(y,z)\,dz$$
where Green's function $G(y,z)$ must satisfy $G_{yy}-k^2G = \delta_z$ with zero boundary conditions. So, 
$$
G(y,z) = \begin{cases} 
A\sinh(ky),\quad &y\le z, \\
B\sinh(k(1-y)),\quad &y\ge z \\
\end{cases}
$$
The function must be continuous at $y=z$ and its derivative must jump by $1$ there. Hence
$$
\begin{cases} 
A\sinh(kz) - B\sinh(k(1-z)) &= 0 \\
Ak \cosh(kz) + Bk\cosh(k(1-z)) &= -1 \\
\end{cases}
$$
Solving this system yields 
$$
\begin{cases} 
A &= \dfrac{\left(- e^{2 k} + e^{2 k z}\right) e^{- k z}}{k \left(e^{2 k} - 1\right)}
\\
B& = \dfrac{e^{ k(1- z)}(1- e^{2 k z})  }{k \left(e^{2 k} - 1\right)} 
\end{cases}
$$
The upshot is that $G$ is of size $O(1/k)$. Indeed, the part $A\sinh(ky)$, $y\le z$, is controlled by $Ae^{kz}$
which is $$\frac{\left(- e^{2 k} + e^{2 k z}\right)}{k \left(e^{2 k} - 1\right)}$$ Exponentials cancel out when $k$ is positive and large; and are small when $k$ is negative and large. Similarly,
$$Be^{k(1-z)}
= \frac{e^{ 2k(1- z)}(1- e^{2 k z})  }{k \left(e^{2 k} - 1\right)} 
$$ 
where again, exponentials are large but balanced when $k$ is positive, and small when $k$ is negative.
