Normal Modes for Standing Waves in 1-D Acoustic Ducts with Arbitrary (but real) Impedance Jumps Let's say we have a 1-D duct, such as this:

Where $Z_i \equiv \frac{P}{US}$ is the acoustic impedance, L is the length of the duct in question, and S is the area of the cross-section. In general, for non-dispersive media, the acoustic impedances will be finite and real, and we can expect that a standing wave at some frequency will form inside the duct. Here assume that there are further impedance jumps to the left and right of this image.
From classical acoustics, for the case that $\frac{Z_i}{Z_0} = \infty$ (closed) or $\frac{Z_i}{Z_0} = 0$ (open), we can very easily show what the normal modes of the free wave motion must be. It's clear that if we have,


*

*Open-Open: $f_0 = \frac{nc_0}{2L}$, where $n = 1,2,3,...$

*Open-Closed: $f_0 = \frac{nc_0}{4L}$, where $n = 1,3,5,...$

*Closed-Closed: $f_0 = \frac{nc_0}{2L}$, where $n = 1,2,3,...$


But now let's say the impedances are some finite and real values (i.e. $\frac{Z_i}{Z_0} \neq \infty$, $\frac{Z_i}{Z_0} \neq 0$), then what would be the normal modes?
It has been suggested to me that if $\frac{Z_i}{Z_0} > 1$, then it can be treated as if it is a closed boundary, and if $\frac{Z_i}{Z_0} < 1$, then it can be treated as an open boundary, thus giving the same results as before. But I am hesitant to accept this as no one seems to be able to rigorously justify this through mathematics. 
Assuming a time harmonic signal $P(x,t) = \Big[A_1e^{-ik_0x} + B_1e^{ik_0x}\Big]e^{j\omega t}$, I was able to derive an equation for the pressure assuming that the incoming wave (from left duct) is known, but this equation is extremely complicated and it is very difficult to ascertain any real meaning from it.
Do normal modes for the standing waves exist in such a duct? Is there anyway to rigorously prove what they are?
 A: I am afraid that you arrived at a wrong answer in your derivation. The impedance BC in an acoustic system is not p' = Zu, but it is:
$$p'=Z\times(\vec{u}.\vec{n})$$
The difference sounds innocuous until you apply the BC for both inlet and outlet, where the direction of the normal vector changes. So at the the inlet, since the direction of normal vector of reflecting surface and the direction of positive velocity are the same, the BC is as you wrote:
$$[A+B]=\frac{Z_1}{Z_0}[A-B]$$
But at the outlet, since the direction of reflecting surface and the acoustic velocity are in opposite directions, the outlet BC is actually:
$$[Ae^{-ikL}+Be^{ikL}]=-\frac{Z_2}{Z_0}[Ae^{-ikL}-Be^{ikL}]$$
So the reduced equation becomes:
$$\boxed{e^{2ikL}=R_1R_2}$$
$$\\$$
Let me illustrate one problem with your solution. Let's compare 2 cases: 
$$(R_1=0.5,  R_2=1) \quad vs.\quad (R_1=1,  R_2=0.5)$$
Your solutions for the two cases will be different since:
$$\frac{R_1}{R_2} \ne \frac{R_2}{R_1}$$
whereas we know that this is a symmetric problem and changing from case 1 to case 2 should have no impact on the Eigen frequency of this domain.
