# Linear stability

In mathematics, in the theory of differential equations and dynamical systems, a particular stationary or quasistationary solution to a nonlinear system is called linearly or exponentially unstable if the linearization of the equation at this solution has the form $\frac{dr}{dt}=A r$, where A is a linear operator whose spectrum contains points with positive real part. If there are no such eigenvalues, the solution is called linearly, or spectrally, stable.

## Example 1: ODE

The differential equation

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

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

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

## Example 2: NLS

$i\frac{\partial u}{\partial t} =-\frac{\partial^2 u}{\partial x^2}-|u|^{2k} u$, where u(x,t) ∈ ℂ and k > 0,

has solitary wave solutions of the form $\phi(x)e^{-i\omega t}$ .[1] To derive the linearization at a solitary wave, one considers the solution in the form $u(x,t)=(\phi(x)+r(x,t))e^{-i\omega t}$. The linearized equation on $r(x,t)$ is given by

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

where

$A=\begin{bmatrix}0&L_0\\-L_1&0\end{bmatrix},$

with

$L_0=-\frac{\partial}{\partial x^2}-k|u|^2-\omega$

and

$L_1=-\frac{\partial}{\partial x^2}-(2k+1)|u|^2-\omega$

the differential operators. According to Vakhitov–Kolokolov stability criterion ,[2] 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 unstable.

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.[3]