Well-posedness of wave equation on finite domain Consider the initial boundary value problem
\begin{equation} u_{tt} = u_{xx}
\end{equation}
for $0 < x < 1, \; 0 < t$ with the initial data
$$ u(x,0) = \phi(x)$$ and $$ u_t(x,0) = \psi(x).$$
Assuming the initial conditions are sufficiently smooth, for which constant values $a,b,c,d$ do the boundary conditions
$$au_x + bu_t = 0 \; \text{ at } x = 0$$
$$cu_x + du_t = 0 \; \text{ at } x = 1$$
lead to a well posed problem?

My guess is that $\frac{b}{a} = 1$ and/or $\frac{d}{c} = 1$ might be important, or that $ad-bc \neq 0$ might be important.  I am having a bit of trouble coming up with what these boundary conditions mean physically in the first place, as this usually hints at what is required for well posed-ness mathematically.  Any and all help is appreciated, thanks in advance.
*This is the beginning of a problem focused on constructing a convergent finite difference scheme for these well-posed problems.  Probably not of very much use at this stage.
 A: Hint. Reformulate the problem as a pair of first order equations.
First, let's rewrite the second order wave equation as a pair of two of first order: introduce $v = u_t, w = u_x$:
$$
v_t - w_x = 0\\
w_t - v_x = 0
$$
or in matrix form
$$
\frac{\partial}{\partial t}\begin{pmatrix}v\\w
\end{pmatrix}
+
\begin{pmatrix}
0 & -1\\
-1 & 0
\end{pmatrix}
\frac{\partial}{\partial x}\begin{pmatrix}v\\w
\end{pmatrix} = 0
\tag{1}
$$
The matrix $A = \begin{pmatrix}
0 & -1\\
-1 & 0
\end{pmatrix}$ has the following left eigenvectors:
$$
\omega_1^\top A = \lambda_1\omega_1^\top, \qquad \lambda_1 = -1, \; \omega_1 = \begin{pmatrix}
1 \\ 1
\end{pmatrix}\\
\omega_2^\top A = \lambda_2\omega_2^\top, \qquad \lambda_2 = +1, \; \omega_1 = \begin{pmatrix}
1 \\ -1
\end{pmatrix}
$$
Riemmanian invariants are $r_{1,2} = \omega_{1,2}^\top \begin{pmatrix}v\\w\end{pmatrix}$
$$
r_1 = v + w = u_t + u_x\\
r_2 = v - w = u_t - u_x
$$
Multiplying the system by $\omega_{1,2}^\top$ from the left gives
$$
\frac{\partial r_1}{\partial t} - \frac{\partial r_1}{\partial x} = 0\\
\frac{\partial r_2}{\partial t} + \frac{\partial r_2}{\partial x} = 0.
$$
So we've converted a hyperbolic system of first order $(1)$ to a pair of advection equations for $r_1, r_2$. Any hyperbolic system with constant coefficients can be converted to a set of advection equations this way.
The well-posed problem consists of one left boundary condition for $r_2$ and one right boundary condition for $r_1$.
I might be mistaken, but I suspect that the only requirement for the boundary condition at $x = 0$ is 
$$
a u_x + b u_t = 0
$$
should be expressable as a boundary condition for $r_2$ like
$$
r_2 = f(r_1).
$$
Simlilarly, at $x = 1$
$$
c u_x + d u_t = 0
$$
should be expressable as
$$
r_1 = g(r_2).
$$
