Consider the wave equation
$$\frac{\partial^{2} \rho}{\partial t^{2}}-c_{s}^{2} \frac{\partial^{2}}{\partial x^{2}}\left(\rho+\nu \frac{\partial \rho}{\partial t}\right)=0 \tag{1}
$$
with boundary conditions
$$\left.\frac{\partial \rho}{\partial x}\right|_{x=0, L}=f(t)$$
I rewrite the variable $\rho$ as
$$\rho(x, t)=u(x, t)+\boxed{x f(t)} \tag{2}$$
in order to rewrite the boundary conditions as homogeneous and then solve the corresponding PDE with the method of eigenfunctions.
The problem is: equation 1 and its BCs clearly have Mirror symmetry in space (that is: I can swap the boundary condition is $x=0$ and in $x=L$ without changing anything in the PDE). The solution (eq. 2) is composed of two terms: the first one is linear in space (variable $x$), therefore it does not respect mirror symmetry anymore.
More in detail, I expect the solution to be even with respect to the axis at $x=L/2$. Equation 2 is composed by two terms: $u(x,t)$ respects the symmetry (it comes down from the solution of the modified PDE), therefore I expect $xf(t)$ to respect the same symmetry, but it doesn't.
How is that possible?
The new boundary conditions for the new functions $u(x,t)$ are
$$\left.\frac{\partial u}{\partial x}\right|_{x=0, L}=0$$
And the new PDE is
$$\frac{\partial^{2} u}{\partial t^{2}}-\frac{\partial^{2}}{\partial x^{2}}\left(c_{s}^{2} u+\nu \frac{\partial u}{\partial t}\right)=-x \frac{\partial^{2} f}{\partial t^{2}}$$
The solution $u(x,t)$ (not shown, becuse it is not simple and not useful for this question) is symmetrical with respect to the center of the domain $(0, L)$.
Best Answer
There are two reasons as to why the second equation doesn't have the symmetry that you want.
The first is that you are only considering the symmetry in the variable $x$, that is $$ x \rightarrow L - x$$ but this symmetry changes the sign of first order derivatives, so the boundary conditions are not preserved, $$ \partial_x \rho = f(t) \rightarrow - \partial_x \rho = f(t)$$ so to have an actual symmetry we have to either change the sign of $f$ or $\rho$, lets choose $\rho$ for the moment.$$ x \rightarrow L - x, \rho \rightarrow -\rho$$ this implies in particular, thet if the initial conditions satisfy this symmetry, that is $$\rho_0(x) = - \rho_0(L - x)$$ $$\rho_{t0}(x) = - \rho_{t0}(L - x)$$ then the solution also has this symmetry ( you can prove this by the uniqueness of the solution).
The second reason is that you broke the symmetry when defining $u$ by introducing the function $x$ which is not symmetric. If we use the antisymmetric function $(x - \frac L2)$, $$ \rho = u + \left(x - \frac L2\right)f$$ then the equation for $u$, $$\partial_{tt} \rho - \partial_{xx}(c_s u + \nu \partial_t u) = - \left(x - \frac L 2\right) \partial_{tt} f$$ is invariant to the symmetry $$ x \rightarrow L - x, \rho \rightarrow -\rho$$ because the change in sign from the antisymmetric function $(x - L/2)$ cancels with the change in sign of $u$.