Ordinary differential equation

The differential equation

$${\displaystyle {\frac {dx}{dt}}=x-x^{2}}$$

has two stationary (time-independent) solutions: x = 0 and x = 1. The linearization at x = 0 has the form ${\displaystyle {\frac {dx}{dt}}=x}$. The linearized operator is A₀ = 1. The only eigenvalue is ${\displaystyle \lambda =1}$. The solutions to this equation grow exponentially; the stationary point x = 0 is linearly unstable.

To derive the linearization at x = 1, one writes ${\displaystyle {\frac {dr}{dt}}=(1+r)-(1+r)^{2}=-r-r^{2}}$, where r = x − 1. The linearized equation is then ${\displaystyle {\frac {dr}{dt}}=-r}$; the linearized operator is A₁ = −1, the only eigenvalue is ${\displaystyle \lambda =-1}$, hence this stationary point is linearly stable.

Nonlinear Schrödinger Equation

The nonlinear Schrödinger equation

$${\displaystyle i{\frac {\partial u}{\partial t}}=-{\frac {\partial ^{2}u}{\partial x^{2}}}-|u|^{2k}u,}$$

where u(x,t) ∈ C and k > 0, has solitary wave solutions of the form ${\displaystyle \phi (x)e^{-i\omega t}}$.^[4]

To derive the linearization at a solitary wave, one considers the solution in the form ${\displaystyle u(x,t)=(\phi (x)+r(x,t))e^{-i\omega t}}$. The linearized equation on ${\displaystyle r(x,t)}$ is given by

$${\displaystyle {\frac {\partial }{\partial t}}{\begin{bmatrix}{\text{Re}}\,r\\{\text{Im}}\,r\end{bmatrix}}=A{\begin{bmatrix}{\text{Re}}\,r\\{\text{Im}}\,r\end{bmatrix}},}$$

where

$${\displaystyle A={\begin{bmatrix}0&L_{0}\\-L_{1}&0\end{bmatrix}},}$$

with

$${\displaystyle L_{0}=-{\frac {\partial }{\partial x^{2}}}-k\phi ^{2}-\omega }$$

and

$${\displaystyle L_{1}=-{\frac {\partial }{\partial x^{2}}}-(2k+1)\phi ^{2}-\omega }$$

the differential operators. According to Vakhitov–Kolokolov stability criterion,^[5] when k > 2, the spectrum of A has positive point eigenvalues, so that the linearized equation is linearly (exponentially) unstable; for 0 < k ≤ 2, the spectrum of A is purely imaginary, so that the corresponding solitary waves are linearly stable.

It should be mentioned that linear stability does not automatically imply stability; in particular, when k = 2, the solitary waves are unstable. On the other hand, for 0 < k < 2, the solitary waves are not only linearly stable but also orbitally stable.^[6]