Consider the equation $$(1-\cos x)u_{tt} - u_{tx} - u_{xx} = 0$$ with Cauchy data $$u(x,0) = f(x), u_t(x,0) = g(x),\text{ for } f,g\in\mathcal{C}^2$$

What compatibility condition do $f$ and $g$ have to satisfy for this problem to have a $\mathcal{C}^2$ solution in a neighborhood of $t = 0$?

My idea is that rewrite the problem into a system of equations:

$$\left( \begin{array}{ccc}0 &1 &0\\ 0 &0 &1\\ 1-\cos x &-1 &-1\end{array} \right) \left( \begin{array}{ccc} u_{tt}\\ u_{tx}\\ u_{xx} \end{array} \right) = \left( \begin{array}{ccc} g'\\ f'' \\ 0 \end{array} \right)$$

And the determinant is $0$: $\cos x - 1 = 0$, when $x=2k\pi$. So the determinant is not identically $0$. I'm stuck here and don't know what to do when the determinant is not identically $0$. Any ideas?


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