= Solution
Let the middle segment occupy $0<x<l$. Normalize the incident displacement <complex amplitude> to one, suppress the common time factor $e^{-i\omega t}$, and choose
$$
\begin{aligned}
y_L(x)&=e^{ik_Lx}+r e^{-ik_Lx},&&x<0,\\
y_M(x)&=a e^{ik_Mx}+b e^{-ik_Mx},&&0<x<l,\\
y_R(x)&=t e^{ik_R(x-l)},&&x>l.
\end{aligned}
$$
Here $r$ is the reflected <complex amplitude> at $0$, $a$ and $b$ are the rightgoing and leftgoing middle-segment <complex amplitudes> referenced to $0$, and $t$ is the outgoing <complex amplitude> referenced to $l$. The middle <wavenumber> is $k_M=\omega/c_M$, and
$$
\gamma=e^{ik_Ml},\qquad |\gamma|=1
$$
is the propagation <phase factor> across the lossless middle segment.
At each junction, continuity of displacement and transverse <force> means continuity of $y$ and $Ty_x$. Using $Tk_j=\omega Z_j$ gives
$$
\boxed{\begin{aligned}
1+r&=a+b,\\
Z_L(1-r)&=Z_M(a-b),\\
\gamma a+\gamma^{-1}b&=t,\\
Z_M(\gamma a-\gamma^{-1}b)&=Z_Rt.
\end{aligned}}
$$
These <boundary conditions> also include all multiple reflections inside the middle segment: $a$ and $b$ are the total travelling components, not just the first waves to cross a junction.
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