# Boundary condition for partial reflection

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 $$R+A=1$$.

• 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. – Michael Seifert Aug 5 '19 at 15:42
• @MichaelSeifert I have specified in my question that the reflecting end is fixed. Is that not enough? Which other informations should I add? – Alessandro Zunino Aug 5 '19 at 16:36
• – 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".

• 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. – Alessandro Zunino Aug 6 '19 at 11:42
• @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. – Michael Seifert Aug 6 '19 at 12:18