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 '19 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 '19 at 16:36
  • $\begingroup$ Very closely related: physics.stackexchange.com/questions/252466/… physics.stackexchange.com/questions/217063/… $\endgroup$ – tpg2114 Aug 5 '19 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 '19 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 '19 at 12:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.