Euler-poisson equation written : $$n_t+(nu)_x=0$$ $$u_t+uu_x=-\phi_x$$ $$\phi_{xx}=e^\phi-n$$ And linearized following ways: The system has a constant solution $n=1, \phi=u=0$. By assuming that the system is slightly perturbed, we let $$n(x, t)=1+\epsilon n_1(x, t)$$ $$\phi(x, t)=\epsilon\phi_1(x, t)$$ $$u(x, t)=\epsilon u_1(x, t)$$ Plugging this into the equations and dropping the terms larger that the first order, we get $$n_{1t}+u_{1x}=0$$ $$u_{1t}+\phi_{1x}=0$$ $$\phi_{1xx}+n_1-\phi_1=0$$ Now, differentiate the last equation with respect to time twice $$\phi_{1xxtt}+n_{1tt}-\phi_{1tt}=0$$ From the first equation it then follows $$\phi_{1xxtt}-u_{1tx}-\phi_{1tt}=0$$ And from the second one $$\phi_{1xxtt}+\phi_{1xx}-\phi_{1tt}=0$$ $$\therefore (I-\partial_{x}^2)\phi_{1tt}-\phi_{1xx}=0$$
Now question: the dispersive relation of linearized eqution is $$\omega (k)=k(1+k^2)^{-1/2}$$
How can we get this dispersive relation???