# Wave propagation with a complex coefficient $\beta$ in a Robin boundary condition. How does this affect scattering from the boundary?

Can anyone give me an idea of what happens in the following situation involving Robin boundary conditions with a complex coefficient.

Lets say we have an incident electromagnetic plane wave $u(x) = e^{i d \cdot x}$ in two dimensions, where $d$ denotes the incident direction, governed by the Helmholtz equation

$$\nabla u + k^2 u = 0,$$ in the upper half space $\mathbb{R}_+^2 = \{x: x_2 > 0\}$ with Robin boundary conditions on the boundary $\partial \mathbb{R}_+^2 = \{x: x_2= 0\}$. The Robin boundary conditions are

$$u + \beta \frac{\partial u}{\partial \nu} = 0,$$

where $\nu$ denotes the normal derivative in the positive $x_2$ direction. Now $\beta$ can be complex and this leads me to believe this boundary condition can be considered lossy and can increase absorption?

I am interested in what kind of happens to the waves when we let $|\beta|$ be very small and then increase it so that $|\beta|$ becomes very large. Can anyone explain how this will affect the scattering/reflection of the incident wave?

• Is $u(x,y)$ a scalar field? Commented Dec 10, 2016 at 20:36
• Yes it is, its the scalar Helmholtz reduction for Maxwell's equations.
– csss
Commented Dec 12, 2016 at 8:23

NOTE: Herein, we use the time convention $\displaystyle e^{i\omega t}$

If $u(x,y)=Ae^{-i(k_1x_1+k_2x_2)}+Be^{-i(k_1x_1-k_2y_2 )}$, then on the boundary we have

$$A+B+\beta(A(-ik_2)-B(-ik_2))=0$$

whence we have

\begin{align} B&=-\frac{1-ik_2\beta}{1+ik_2\beta}A\\\\ &=-\left(\frac{\left(1+k_2\text{Im}(\beta)\right)-i\left(k_2\text{Re}(\beta)\right)}{\left(1-k_2\text{Im}(\beta)\right)+i\left(k_2\text{Re}(\beta)\right)}\right)\,A \end{align}

When $\beta =0$, $B=-A$ and as $\beta \to \infty$, $B\to A$.

Note that if $k_2$ is a real number with $k_2\le 0$, then it is easy to see that $\text{Im}(\beta)\ge 0$ in order to ensure that $|B|\le|A|$.

Note that when $k_2=-i/\beta$, $B=0$ and the incident wave is fully absorbed. Since we assume that $k_2\in \mathbb{R}$, this can only happen if $\text{Re}(\beta)=0$.

• Thanks but I'm not sure what $B = -A$ when $\beta = 0$ and $B \to A$ for $\beta \to \infty$ imply with regards to reflections? I don't see how this explains the boundary condition 'absorbing' waves?
– csss
Commented Dec 12, 2016 at 8:31
• This seems to be a statement about amplitudes, it doesn't seem to do anything for the direction of the wave such as blocking outgoing waves. The first 'B' term in your post represents a wave incoming from infinity in the upper half space and the 'A' term represents an outgoing reflection. But as $\beta \to \infty$ and get $B \to A$ both waves will now have the same amplitude/phase but the directions haven't changed..Wave B will still be incoming and wave A will still be outgoing!
– csss
Commented Dec 12, 2016 at 8:55
• First, the convention here is $e^{i\omega t}$ and so $A$ is the incident wave's amplitude. Is there any part of the analysis that you don't understand? We are analyzing in $\vec k$-space, which is the 2-D Fourier Transform representation. The amplitudes depend on $\vec k$. When $k_2=-i\beta$ the wave is fully absorbed since $B(k_2)=0$ then. Commented Dec 12, 2016 at 13:50
• Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark Commented Dec 16, 2016 at 19:09
• First, thank you for both the best vote and up vote! Second, when $\beta \to \infty$, we have the Neumann Boundary Condition since $$u+\beta\frac{\partial u}{\partial \nu}=0\implies \frac1\beta u +\frac{\partial u}{\partial \nu}=0$$Now, letting $\beta \to \infty$ renders the boundary condition $$\frac{\partial u}{\partial \nu}=0$$Lastly, Happy Holidays! -Mark Commented Dec 19, 2016 at 21:08