Symmetry breaking in 1D wave equation 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 $(0, L)$.
 A: 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 $u$.
