Write coupled wave equations as a decoupled system

Question

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?

Answer 1

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.

Answer 2

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}