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 $R+A=1$.

  • 1
    $\begingroup$ Note that simply knowing $R$ is not sufficient to determine the boundary conditions. For example, the reflection coefficient for a wave on a string is $R = 1$ both when the end is fixed ($\phi = 0$ at $x = L$) and when the end is free ($\partial \phi/\partial x = 0$ at $x = L$.) So there is not likely to be a unique answer to your question unless you add more information. $\endgroup$ – Michael Seifert Aug 5 at 15:42
  • $\begingroup$ @MichaelSeifert I have specified in my question that the reflecting end is fixed. Is that not enough? Which other informations should I add? $\endgroup$ – Alessandro Zunino Aug 5 at 16:36
  • $\begingroup$ Very closely related: physics.stackexchange.com/questions/252466/… physics.stackexchange.com/questions/217063/… $\endgroup$ – tpg2114 Aug 5 at 16:39

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".

  • $\begingroup$ Thank you for you very interesting and useful answer, but I think my question is not fully solved. You are telling me that any combination of $\alpha, \, \beta, \, \gamma$ will give me a finite $R$. Still, it is not clear to me which one of these three parameters I should choose to be $\neq 0$ to create a model of a fixed reflecting wall and why. $\endgroup$ – Alessandro Zunino Aug 6 at 11:42
  • $\begingroup$ @AlessandroZunino: That's because for any particular value of $R$, there's more than one set of coefficients that yields it. I've edited my answer to explain this further. $\endgroup$ – Michael Seifert Aug 6 at 12:18

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