Update src/TopOptEMFocus.jl

Co-authored-by: Steven G. Johnson <stevenj@mit.edu>

Author: WenjieYao

src/TopOptEMFocus.jl

@@ -47,7 +47,7 @@
 # 
 # where $u_{x/y}$ is the depth into the PML, $\sigma$ is a profile function (here we chose $\sigma(u)=\sigma_0(u/d_{pml})^2$) and the $x$ and $y$ derivatives correspond PML layers at the $x$ and $y$ boundaries, respectively.  Note that at a finite mesh resolution, PML reflects some waves, and the standard technique to mitigate this is to "turn on" the PML absorption gradually—in this case we use a quadratic profile. The amplitude $\sigma_0$ is chosen so that in the limit of infinite resolution the "round-trip" normal-incidence is some small number. 
 # 
-# Since PML absorbs all waves in $x/y$ direction, the associated boundary condition is then usually the zero Dirichlet boundary condition. Here, the boundary conditions are zero Dirichlet boundary on the top and bottom side $\Gamma_D$ but periodic boundary condition on the left ($\Gamma_L$) and right side ($\Gamma_R$).  The reason that we use a periodic boundary condition for the left and right side instead of zero Dirichlet boundary condition is that we want to simulate a plane wave exicitation, which then requires a periodic boundary condition.
+# Since PML absorbs all waves before they reach the boundary, the associated boundary condition can then be chosen arbitrarily. Here, the boundary conditions are Dirichlet (zero) on the top and bottom sides $\Gamma_D$ but periodic on the left ($\Gamma_L$) and right sides ($\Gamma_R$).  The reason that we use periodic boundary conditions for the left and right side instead of Dirichlet boundary conditions is that we want to simulate a plane wave excitation, so we must choose boundary conditions that are satisfied by this incident wave.  (Because of the anisotropic nature of PML, the PML layers at the $x$ boundaries do not disturb an incident planewave traveling purely in the $y$ direction.) 
 # 
 # Let $\mu(x)=1$ (materials at optical frequencies have negligible magnetic responses) and denote $\Lambda=\operatorname{diagm}(\Lambda_x,\Lambda_y)$ where $\Lambda_{x/y}=\frac{1}{1+\mathrm{i}\sigma(u_{x/y})/\omega}$. We can then formulate the problem as 
 #