This is a long one. Complicated but not difficult.
I'm trying to solve the 1D wave equation in some loop that is one half in one medium and one half in another. I've been able to solve the problem for a loop in one medium, and I've been able to solve the problem for a finite string that switches to a different medium at $x=a$. I'm basically trying to combine both into a single solution, but without success. All I have to do is take the finite string and attach the ends as a boundary condition, and my question is how to do that. I'll give my results so far so you can see what I mean.
The problem in two mediums and a finite string is straight forward. The wave equations are $\frac{\partial^2 u}{\partial t^2}-c_{1}\frac{\partial^2 u}{\partial x^2}=0$ for $0 < x < a$ and $\frac{\partial^2 u}{\partial t^2}-c_{1}\frac{\partial^2 u}{\partial x^2}=0$ for $a < x < a+b$. (Lengths $a$ and $b$ before and after boundary.) The boundary condition is that in the ends, so $0$ and $a+b$, the displacement is $0$. Also the string must be connected properly, so equal displacement and speed at $x=a$ is another bc. Separation of variables $u(x,t)=X(x)e^{\pm i\omega t}$ is useful here for both problems. Result is $X(x)=A_1\sin(\frac{\omega}{c1}x)$ before $a$ and $X(x)=A_2\sin(\frac{\omega}{c2}(a+b-x))$ after $a$. Makes sense; they can be equal in $x=a$ with the correct $A_1$ and $A_2$, and they're both $0$ at their ends $0$ and $a+b$. Them needing to be equal (and equal derivatives) in $a$ is clearly only true if $c_1\tan(\frac{\omega L_1}{c_1}) = -c_2\tan(\frac{\omega L_2}{c_2})$. That just comes from setting the two equal in $a$, and also the derivatives, and dividing those two equalities. That's nice as it's independent of $A_1$ and $A_2$, and they can then be found after. Although this not easily solvable, it has no closed form solution.
The problem in a loop is even simpler. Just assume $u(\theta,t)=X(\theta)e^{\pm i\omega t}$ again. Now the boundary condition is that at $\theta=2\pi$ the end must connect properly to $\theta=0$. That results in a standing wave and $X(\theta)=e^{\pm i\frac{\omega}{c} \theta}$ where $\frac{\omega}{c}$ must be a natural number to close the loop. That general solution with $\sin(x)=\frac{e^x-e^{-x}}{2i}$ can then be written as $u(x,t)=\sin{(nc(t-t_0))}\sin{(n(\theta-\theta_0))}$ as a more obvious standing wave with $n$ a natural number.
I've been trying to connect these two problems. It shouldn't actually be that difficult because there's only one boundary condition being added actually, and that's connecting the ends.
So I'm pretty sure I can use separation of variables again, so $u(\theta,t)=X(\theta)e^{\pm i \omega t}$. I understand intuitively that the waves in the two mediums will have different amplitudes and frequencies, and that the sort of 'total' will have to be closed on the loop. Now that 'closed' property doesn't mean natural numbers $n$ in $\sin(n(\theta-\theta_0))$ anymore, it becomes a bit more complicated. So I figured, just assume the most general form possible, and that is to have in $0 < \theta < \pi$ that $X(\theta)=A_1\sin(\frac{\omega}{c1}(\theta-\theta_1))$ and in $\pi < \theta < 2\pi$ that $X(\theta)=A_2\sin(\frac{\omega}{c2}(\theta-\theta_2))$. Now that's pretty good. Setting these and their derivatives equal at the interface, so $\pi$ and $0$ / $2\pi$ (depending on direction) gives 4 equations with 5 unknowns. That's what is to be expected; eigenvalues $\omega$ here allowing infinitely many solutions. I should be able to somehow rearrange this and calculate a few of the $\omega$ with computer, but I can't figure it out.
Below the actual set of 4 equations. (If understood nothing and can solve this, also good!)
$A_2\sin{(\frac{\omega}{c_2}(\pi-\theta_2))} = A_1\sin{(\frac{\omega}{c_1}(\pi-\theta_1))}$
$A_2\frac{\omega}{c_2}\sin{(\frac{\omega}{c2}(\pi-\theta_2))} = A_1\frac{\omega}{c_1}\sin{(\frac{\omega}{c1}(\pi-\theta_1))}$
$A_2\sin{(\frac{\omega}{c_2}(2\pi-\theta_2))} = A_1\sin{(\frac{\omega}{c_1}(-\theta_1))}$
$A_2\frac{\omega}{c_2}\sin{(\frac{\omega}{c2}(2\pi-\theta_2))} = A_1\frac{\omega}{c_1}\sin{(\frac{\omega}{c1}(-\theta_1))}$
Am seeing if using $-\theta$ instead of $\theta$ for one of them helps as that was the solution with the simple case with 2 mediums (reflection), but still don't see what after that.