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For a string equation \begin{gather*} \rho u_{tt} -T u_{xx}=0 \end{gather*} where $u=u(x,t)$ is a displacement in the orthogonal direction, $\rho$ is a linear density and $T$ is a tension (see Example 1.4.1
On the fixed left end $x=0$ we have Dirichlet boundary condition $u(0,t)=0$ $u(0,t)=0$. And similarly on the fixed right end $x=l$ boundary condition is $u(l,t)=0$.
On the other hand, consider a free left end. To prevent the string from shrinking we may fix it on the massless sleeve sliding freely along the rod, orthogonal to the string. Then since the force, the string pulls up its left (right) end is $Tu_x$ ($-Tu_x$), we have Neumann boundary condition $u_x(0,t)=0$ ($u_x(l,t)=0$ correspondingly).
If we add a spring so there would be the return force $-ku$ where $k$ is the Hooke's coefficient of this spring, we have Robin boundary condition $(Tu_x- ku)(0,t)=0$ ($(-Tu_x- ku)(l,t)=0$ correspondingly).
For $1$-dimensional airflow in the air pipe \begin{align} & \rho_0 v_t+p'(\rho_0)\rho_x=0,\label{eq-2.6A-1}\\ &\rho_t +\rho_0v_x=0\label{eq-2.6A-2} \end{align} where $\rho_0$ is a "normal'' density, $\rho(x,t)$ is excess density (so $\rho_0+\rho$ is a density), $v(x,t)$ is a velocity, and $p=p(\rho)$ is a pressure (see (see Example 1.4.3) one can see that both $\rho$ and $v$ satisfy wave equation \begin{align} &\rho_{tt}-c^2\rho_{xx}=0,\label{eq-2.6A-3}\\ &v_{tt}-c^2v_{xx}=0\label{eq-2.6A-4} \end{align} with $c^2=p'(\rho_0)$.
On the closed end the velocity is $0$, so on the closed left end $x=0$ we have $v(0,t)=0$ and $\rho_x(0,t)=0$ for (\ref{eq-2.6A-3}) and (\ref{eq-2.6A-4}).
On the open end the excess pressure is $0$, so on the open left end $x=0$ we have $v_x(0,t)=0$ and $\rho (0,t)=0$ for (\ref{eq-2.6A-3}) and (\ref{eq-2.6A-4}).
Assume that the left end has a massless piston which is connected to a spring with Hooke's coefficient $k$. Then the shift of the piston from the equilibrium to the right equals $\int_0^t v(0,t)\,dt$ (assuming that it was in such position at $t=0$). Then the force of the spring is $-k\int_0^t v(0,t)\,dt$. On the other hand, the excess pressure is $p'(\rho_0)\rho$ and $-k\int_0^t v(0,t)\,dt+p'(\rho_0)\rho=0$ is the balance.
Then $kv(0,t)= p'(\rho_0) \rho_t= p'(\rho_0)\rho_0v_x$ due to (\ref{eq-2.6A-2}) and we get $(-kv+ p'(\rho_0)\rho_0v_x)(0,t)=0$; correspondingly, on the right end it would be $(kv + p'(\rho_0)\rho_0v_x)(l,t)=0$.