Ytukyg

I have the following Cauchy problem. I do not know where to start, so I would appreciate any help and tips.
$$frac{partial^2 Y(t, x)}{partial t^2} = 9frac{partial^2 Y(t,x)}{partial x^2} - 2Z(t,x)$$
$$frac{partial^2 Z(t, x)}{partial t^2} = 6frac{partial^2 Z(t,x)}{partial x^2} - 2Y(t,x)$$
$$Y(0, x) = cos(x), Z(0, x) = 0$$
$$frac{partial Y(t, 0)}{partial t} = frac{partial Z(t, 0)}{partial t} = 0$$

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.

$$begin{cases}
2Y(t,x) = 6dfrac{partial^2 Z(t,x)}{partial x^2} - dfrac{partial^2 Z(t, x)}{partial t^2}\[4px]
2dfrac{partial^2 Y(t, x)}{partial t^2} = 18dfrac{partial^2 Y(t,x)}{partial x^2} - 4Z(t,x),
end{cases}$$
so
$$dfrac{partial^2}{partial t^2}left(6dfrac{partial^2 Z(t,x)}{partial x^2} - dfrac{partial^2 Z(t, x)}{partial t^2}right) = 9dfrac{partial^2}{partial x^2}left(6dfrac{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} - 15dfrac{partial^4 Z(t,x)}{partial t^2partial x^2} + 54dfrac{partial^4 Z(t,x)}{partial x^4}-4Z(t,x) = 0\[4pt]
Z(0, x) = 0\[4pt]
6dfrac{partial^2 Z(0,x)}{partial x^2} - dfrac{partial^2 Z(0, x)}{partial t^2} = 2cos(x)\[4pt]
6dfrac{partial^3 Z(t,0)}{partial tpartial 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} - 15dfrac{partial^4 Z(t,x)}{partial t^2partial x^2} + 54dfrac{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} = dfrac13cos(x)\[4pt]
dfrac{partial Z(t, 0)}{partial t} = 0,quaddfrac{partial^3 Z(t, 0)}{partial t^3} = 0.
end{cases}$$
And it reqires four conditions more.