Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up.
Sign up to join this community
I want to solve a wave equation for the wave $\psi(x,t)$. One boundary is moving, therefore I impose the velocity
$$v(x=0)=v_a\cos(\omega t)$$
the other boundary is fixed, but reflecting. If the reflection is total the proper boundary condition is
$$ \frac{\partial \psi}{\partial x}\bigg|_{x=L}=0 $$
Which is the right boundary condition if the reflection is partial? Therefore, my boundary has a reflection coefficient $R$ and an absorbing coefficient $A$ (no transmission), so that $R+A=1$.
Consider a boundary condition of the form $$ \alpha \psi + \beta \frac{\partial \psi}{\partial x} + \gamma \frac{\partial \psi}{\partial t} =0 $$ on the boundary, where $\alpha$, $\beta$, and $\gamma$ are real coefficients. Standard Dirichlet boundary conditions correspond to $\beta = \gamma = 0, \alpha \neq 0$, while Neumann boundary conditions correspond to $\alpha = \gamma = 0, \beta \neq 0$.
If we impose this condition at $x = 0$ for an incoming wave of the form $\psi(x,t) = e^{i(kx - \omega t)}$, then the wave solution for $\psi$ will be $$ \psi(x,t) = e^{i(kx - \omega t)} + A e^{i(-kx - \omega t)} $$ where $A$ is the (complex) amplitude of the reflected wave. Some algebra then reveals that we must have $$ A = \frac{i (k \beta - \omega \gamma) + \alpha}{i (k \beta + \omega \gamma) - \alpha}, $$ and so $$ R = |A|^2 = \frac{(k \beta - \omega \gamma)^2 + \alpha^2}{(k \beta + \omega \gamma)^2 + \alpha^2}. $$ By appropriate choices of $\alpha$, $\beta$, and $\gamma$, one can "tune" the reflection coefficient to be anywhere between 0 and 1.
A few notes on this expression:
This reflection coefficient will be frequency-dependent unless the medium is non-dispersive and $\alpha = 0$. In particular, if the reflection coefficient is frequency-dependent, this means that a pulse sent towards the wall will not be reflected with the same shape. If you want pulses to retain their shape on reflection, use $\alpha = 0$.
For a given value of $R$, the coefficients $\alpha$, $\beta$, and $\gamma$ are not uniquely determined. This is for two reasons. First, we can always divide or multiply all three coefficients by the same number to get the same boundary condition; you can always use this freedom to set one (non-zero) coefficient equal to 1 if you like. Second, the reflection coefficient alone doesn't tell you everything about the reflected wave; the reflected wave can also be phase-shifted.
If $\gamma = 0$, we have $R = 1$ identically. This makes some sense; in such a case, the equations have time-reversal symmetry, whereas absorption involves an inherent "arrow of time".
