Linearized Euler-Poisson equation Normalized Euler-Poisson system have the form
$$n_t+(nu)_x=0$$
$$u_t+uu_x=-\phi_x$$
$$\phi_{xx}=e^{\phi}-n$$
where u :ion velocity, n: the ion density and $\phi$: electrical potential, The parameter $\epsilon>0$ stands for a scaled Debey length 
this equation linearlized by 
$$(I-\partial_x^2)\phi_{tt}-\phi_{xx}=0$$
from which the dispersive relation
$$w(k)=k(1+k^2)^{-\frac{1}{2}}$$
I want to know the process of linearizing, Is there any helpful reference for this? 
 A: The system has a constant solution $n=1, \phi=u=0$. You want to linearise near this point, by assuming that the system is slightly perturbed, you let
$$n(x, t)=1+\delta{n}(x, t)$$
$$\phi(x, t)=\delta{\phi}(x, t)$$
$$u(x, t)=\delta{u}(x, t)$$
Where $\delta{n}(x, t), \delta{u}(x, t), \delta{\phi}(x, t)$ are all small, and slowly varing functions. Plugging this into the equations and dropping the terms larger that the first order, we get
$$\partial_{t}\delta{n}(x, t)+\partial_{x}\delta{u}(x, t)=0$$
$$\partial_{t}\delta{u}(x, t)-\partial_{x}\delta\phi(x, t)=0$$
$$\epsilon^{2}\partial_{x}^{2}\delta{\phi}(x, t)-\delta{n}(x, t)-\delta\phi(x, t)=0$$
Now, differentiate the last equation with respect to time twice
$$\epsilon^{2}\partial^{2}_{t}\partial_{x}^{2}\delta{\phi}(x, t)-\partial_{t}^{2}\delta{n}(x, t)-\partial_{t}^{2}\delta\phi(x, t)=0$$
From the first equation it then follows
$$\epsilon^{2}\partial^{2}_{t}\partial_{x}^{2}\delta{\phi}(x, t)+\partial_{t}\partial_{x}\delta{u}(x, t)-\partial_{t}^{2}\delta\phi(x, t)=0$$
And from the second one
$$\epsilon^{2}\partial^{2}_{t}\partial_{x}^{2}\delta{\phi}(x, t)+\partial_{x}^{2}\delta{\phi}(x, t)-\partial_{t}^{2}\delta\phi(x, t)=0$$
So
$$(1+\epsilon^{2}\partial_{x}^{2})\partial_{t}^{2}\delta\phi(x, t)-\partial_{x}^{2}\delta{\phi}(x, t)=0$$
The equation is colse to the one you've stated, however the signs do not match and there is an extra $\epsilon^{2}$ factor. Are you sure that there are no mistakes in the original system of equations you've stated?
A: 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
$$\phi_{1xxtt}+\phi_{1xx}-\phi_{1tt}=0$$
$$\therefore (I-\partial_{x}^2)\phi_{1tt}-\phi_{1xx}=0$$
