Can we use method of reflection to find Green's function in infinite strip? I have learned how to use method of reflection to find Green's function of Laplacian equation for Dirichlet problem in half-space or quadrant in my undergraduate pde course. Now I am wondering how to use method of reflection to find Greens' function for infinite strip.
For example, we consider $$u_{xx}+u_{yy}=0$$ on the domain $\Omega=\{(x,y)|-\infty< x < \infty,0<y<b\}$.
Let $(x_0,y_0)\in\Omega$, following the idea I learnt in lecture, I first construct Greens' funciton $G_{\textrm{hp}}(x,x_0)$ for half-space. Then I should minus some harmonic function and make sure the new Green's function has zero boundary value at additional boundary $y=b$. 
For example, if $\Omega=\{x>0,y>0$ is the first quadrant, I will set $G(x,x_0)=G_{\textrm{hp}}(x,x_0)-G_{\textrm{hp}}(x,x_0^x)$, where $x_0^x=(-x_0,y_0)$, the reflexive point of $(x_0,y_0)$ with respect to $y$-axis. And since $G_{\textrm{hp}}(x,x_0)$ is related with $|x-x_0|,|x-x_0^y|$, and $G_{\textrm{hp}}(x,x_0^x)$ is related with $|x-x_0^x|,|x-x_0^o|$, at the additional boundary $\{x=0\}$, $|x-x_0|=|x-x_0^x|,|x-x_0^y|=|x-x_0^o|$ hence $G_{\textrm{hp}}(x,x_0)=G_{\textrm{hp}}(x,x_0^x)$, we get the correct Greens' function.
But I don't know how to apply the same idea for infinite strip. At the beginning, I want to set $G(x,x_0)=G_{\textrm{hp}}(x,x_0)-G_{\textrm{hp}}(x,x_0^b)$, where $x_0^b=(x_0,2b-y_0)$, the reflexive point of $(x_0,y_0)$ w.r.t $y=b$, but I found $G_{\textrm{hp}}(x,x_0^b)$ involves $|x-(x_0,-(2b-x_0))|$, hence $G(x,x_0)$ is not 0 at $y=b$, which is not correct. 
 A: You can use the method of images or what you term 'reflection' but you need to compute a suitable mapping for this to work.
Laplace's equation is invariant under a conformal mapping. So, this means that if your function $u$ obeys $\nabla^2 u = 0$ in the original domain, it will also do so in the mapped domain. Then if the map is chosen wisely such that finding the greens function is trivial in the mapped domain, we have solved our problem.
The Schwarz-Christoffel Mapping $$f:H^+ \to\Omega$$
is what we need. Let $z=x+iy \in\Omega$, and $\zeta = \eta+i\xi \in H^+$, where $H^+$ is the upper half plane. Then according to the Schwarz-Christoffel theorem:
$$z = f(\zeta) = A + C \int^{\zeta} \prod_{k=1}^{N-1}(w-\eta_k)^{\alpha_k-1} \mathrm{d}w$$
where $\eta_k$ are on the real axis and get mapped to the vertices of the "polygon" that is the strip, in the $z$ domain- i.e. $z_k = f(\eta_k)$, for $n = 1,\dots,N-1.$ $\alpha_k$ are the corresponding interior angles of the polygon normalised by $\pi$. Note that we choose $f(\infty) = \eta_N$. Please refer to Driscoll and Trefethen book on Schwarz Christoffel Mapping if needed. Think of the strip as a polygon with two vertices at $\pm\infty$ with interior angles of zero. Then if we let the origin in the upper half plane map to one of these vertices and infinity map to the other vertex we get:
$$z = A + C\ln \zeta$$
Again with reference to the book, we are free to choose three arbitrary points in the mapping. Let $\zeta = -1$ map to $z = ib$ and $\zeta = 1$ map to $z = 0$. Hence $A = 0$ and $C = b/\pi$. Now Greens function is just the solution to 
$$\nabla^2G(\mathbf{x}| \mathbf{x}_s) = \delta(\mathbf{x}-\mathbf{x}_s)$$ 
with $\mathbf{x} = (x,y)$ and $\mathbf{x}_s = (x_s,y_s)$. In complex notation let $z=x+iy$ and $z_s = x_s+iy_s$. 
In our half plane the method of images gives:
$$G(\zeta|\zeta_s) = -\frac{1}{2\pi}\left( \ln(|\zeta - \zeta_s|) - \ln(|\zeta-\overline{\zeta}_s|)\right)$$
where the bar denotes complex conjugate. Then using $\zeta = e^{\pi z/b}$ you can easily work out the expression for $G(z|z_s)$. This Greens function satisfies the Boundary condition $G=0$ on the strip walls.
Here is a contour plot of the resulting Greens function in a channel of unit width when the source is midway:

It has the expected property that it is singular at the source location and decays to zero at the walls.
