$-\Delta u =1$ in a unbounded strip

Question

Consider the problem $$ \left\lbrace \begin{gathered} -\Delta u = 1 \quad\textrm{in}\quad S\\ u=0\quad\textrm{on}\quad \partial S \end{gathered} \right. $$ where $S$ is the infinity strip $\mathbb{R}\times (0,1)$ or the semi-infinity strip $\mathbb{R}_+\times (0,1)$.

What extra conditions should be imposed on this problem in order to obtain a bounded solution with respect to the $H^1_0(S)$ norm? Is it possible? Moreover, would this solution be unique?

If it is not possible, what type of bounded solutions could be obtained? Only in $L^2$ or another function space?

It is a rather wide question, but I am looking forward to possible answers.

Answer 1

This is my try: Let $u$ be a bounded solution to the problem on $S=\mathbb{R}_+\times (0,1)$.

1. Define $v(x,y)=u(x,y)-\frac{y(1-y)}{2}$. $v$ is harmonic on $S$ with boundary condition $v(x,0)=v(x,1)=0$ and $v(0,y)=\frac{y(y-1)}{2}$
2. Define a function with polar coordinates $w(r,\theta)=v(-\ln r,\theta)$ on the sector $0<r<1;\ 0<\theta <1$. $w(r,0)=w(r,1)=0$ and $w(1,\theta)=\frac{\theta(\theta-1)}{2}$
3. Since $v$ is bounded so is $w$ and thus it is the unique solution to the problem in the bounded sector: $$w(r,\theta)=\sum_{n=1}^\infty A_n r^{n\pi}\sin(n\pi\theta)$$ Where $A_n \approx \frac{1}{n^3}$ are the Fourier coefficients of $w(1,\theta)=\frac{\theta(\theta-1)}{2}$
4. Therefore $v(x,y)=\sum_{n=1}^\infty A_n e^{-n\pi x}\sin(n\pi y)\in H^1_0(S)$
5. But $u=v+\frac{y(1-y)}{2}$ and $\frac{y(1-y)}{2}\notin H^1_0(S)$, thus $u\notin H^1_0(S)$.