Write coupled wave equations as a decoupled system Given the coupled wave equations:
$$
\frac{\partial }{\partial t} [u~~v] +
\begin{pmatrix}
a & c \\
c & a
\end{pmatrix} \frac{\partial }{\partial x} [u~~v] = [0~~0].
$$
I want to transform this problem into a decoupled system with variables [r,s]. Can someone give me an idea on where to start? 
 A: One place to start is to choose a basis in which $\begin{bmatrix} a & c \\ c & a \end{bmatrix}$ is diagonal. So we diagonalize by finding eigenvalues and eigenvectors. The characteristic polynomial is $\lambda^2-2a\lambda+a^2-c^2$ which has roots $\frac{2a \pm (4a^2 - 4a^2+4c^2)^{1/2}}{2} = a \pm c$. The eigenvectors then satisfy
$$-cx_1+cx_2=0 \Rightarrow x_1=x_2 \\
cy_1+cy_2=0 \Rightarrow y_1=-y_2.$$
So you can choose the basis $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ -1 \end{bmatrix}$, i.e. $r=u+v$ and $s=u-v$. This change of variables will make the matrix diagonal, which will decouple that part of the system. 
Now you want to write 
$$\begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} \begin{bmatrix} r \\ s \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} r \\ s \end{bmatrix}$$
and plug that into your original equation to see how the full equation decouples.
A: The system
$$
\frac{\partial }{\partial t} \begin{pmatrix} u \\ v \end{pmatrix} +
\begin{pmatrix}
a & c \\
c & a
\end{pmatrix} \frac{\partial }{\partial x} \begin{pmatrix} u \\ v \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}.
$$
leads to the system 
\begin{align}
\begin{pmatrix} u_{t} \\ v_{t} \end{pmatrix} = \begin{pmatrix} a u_{x} + c v_{x} \\ c u_{x} + a v_{x} \end{pmatrix}
\end{align}
Equating both sides leads to the two equations
\begin{align}
(1)  \hspace{10mm} u_{t} &= a u_{x} + c v_{x}  \\
(2) \hspace{10mm} v_{t} &= c u_{x} + a v_{x}.
\end{align}
Now from (1) it is seen that $u_{x} = (1/c)( v_{t} - a v_{x} )$. Differentiate this with respect to $t$ and also differentiate (1) with respect to $x$ to obtain
\begin{align}
\frac{1}{c} ( v_{xt} - a v_{xx} ) = a \cdot \frac{1}{c} ( v_{xt} - a v_{xx}) + c v_{xx}
\end{align}
which yields
\begin{align}
v_{tt} - 2 a v_{xt} = (c^{2} - a^{2} ) v_{xx}.
\end{align}
In a similar manor it can be shown that $u$ satisfies the same equation.
The decoupled set of equations, $u(x,t)$ and $v(x,t)$ both satisfy the equation
\begin{align}
\frac{\partial^{2} F}{\partial t^{2}} - 2 a \frac{\partial^{2} F}{\partial x \partial t} = (c^{2} - a^{2}) \frac{\partial^{2} F}{\partial x^{2}}
\end{align}
