Second order partial Cauchy problem with two functions I have the following Cauchy problem. I do not know where to start, so I would appreciate any help and tips.
$$\frac{\partial^2 Y(t, x)}{\partial t^2} = 9\frac{\partial^2 Y(t,x)}{\partial x^2} - 2Z(t,x)$$
$$\frac{\partial^2 Z(t, x)}{\partial t^2} = 6\frac{\partial^2 Z(t,x)}{\partial x^2} - 2Y(t,x)$$
$$Y(0, x) = \cos(x),\ Z(0, x) = 0$$
$$\frac{\partial Y(t, 0)}{\partial t} = \frac{\partial Z(t, 0)}{\partial t} = 0$$
 A: Your equation is linear, so you could try expanding the solutions in a Fourier series. Moreover, as the initial condition is given in terms of $\cos(x)$, it's a good idea to write
\begin{align}
 Y(t,x) &= \sum_{k=0}^\infty a_Y(k,t) \cos(k x) + b_Y(k,t) \sin(k x)\\
 Z(t,x) &= \sum_{k=0}^\infty a_Z(k,t) \cos(k x) + b_Z(k,t) \sin(k x)
\end{align}
which yields a linear system of ordinary differential equations for the coefficients $a_{Y,Z}(k,t)$ and $b_{Y,Z}(k,t)$. The initial conditions of the PDE system then translate in
\begin{equation}
 a_Y(1,0) = 1,\quad a_Y(k,0) = 0 \; (k \neq 1),\quad b_Y(k,0) = 0 = b_Z(k,0),
\end{equation}
which are initial conditions for the ODE system. The boundary condition(s) translate in
\begin{equation}
 \frac{\partial Y}{\partial t}(t,0) = \sum_{k=0}^\infty \frac{\text{d} a_Y}{\text{d} t}(k,t) = 0 = \sum_{k=0}^\infty \frac{\text{d} a_Z}{\text{d} t}(k,t) =  \frac{\partial Z}{\partial t}(t,0).
\end{equation}
Note that, as it stands, the initial- and boundary conditions do not fully determine the solution. 
A: $$\begin{cases}
2Y(t,x) = 6\dfrac{\partial^2 Z(t,x)}{\partial x^2} - \dfrac{\partial^2 Z(t, x)}{\partial t^2}\\[4px]
2\dfrac{\partial^2 Y(t, x)}{\partial t^2} = 18\dfrac{\partial^2 Y(t,x)}{\partial x^2} - 4Z(t,x),
\end{cases}$$
so
$$\dfrac{\partial^2}{\partial t^2}\left(6\dfrac{\partial^2 Z(t,x)}{\partial x^2} - \dfrac{\partial^2 Z(t, x)}{\partial t^2}\right) = 9\dfrac{\partial^2}{\partial x^2}\left(6\dfrac{\partial^2 Z(t,x)}{\partial x^2} - \dfrac{\partial^2 Z(t, x)}{\partial t^2}\right) - 4Z(t,x),$$
$$\begin{cases}\dfrac{\partial^4 Z(t,x)}{\partial t^4} - 15\dfrac{\partial^4 Z(t,x)}{\partial t^2\partial x^2} + 54\dfrac{\partial^4 Z(t,x)}{\partial x^4}-4Z(t,x) = 0\\[4pt]
Z(0, x) = 0\\[4pt]
6\dfrac{\partial^2 Z(0,x)}{\partial x^2} - \dfrac{\partial^2 Z(0, x)}{\partial t^2} = 2\cos(x)\\[4pt]
6\dfrac{\partial^3 Z(t,0)}{\partial t\partial x^2} - \dfrac{\partial^3 Z(t, 0)}{\partial t^3} = 0\\[4pt]
\dfrac{\partial Z(t, 0)}{\partial t} = 0
\end{cases}$$
$$\begin{cases}\dfrac{\partial^4 Z(t,x)}{\partial t^4} - 15\dfrac{\partial^4 Z(t,x)}{\partial t^2\partial x^2} + 54\dfrac{\partial^4 Z(t,x)}{\partial x^4}-4Z(t,x) = 0\\[4pt]
Z(0, x) = 0,\quad \dfrac{\partial^2 Z(0,x)}{\partial x^2} = \dfrac13\cos(x)\\[4pt]
\dfrac{\partial Z(t, 0)}{\partial t} = 0,\quad\dfrac{\partial^3 Z(t, 0)}{\partial t^3} = 0.
\end{cases}$$
And it reqires four conditions more.
