Linearization of The Ginzburg-Landau (GL) equation

62 Views Asked by At

I am studying the LG equation, $$\partial_t u = (1 + i \alpha) \partial_x^2 u + u - (1+ i \beta)|u|^2 u, \quad \alpha, \beta, x \in \mathbb R, \text{ and }\,\, u(x,t) \in \mathbb C. \tag{1} $$

To find the rolls we substitute the following ansatz

$$u(x,t) = a e^{i (qx+\omega t)},\quad a \in \mathbb C,\,\,\, q \in \mathbb R,$$

in (1) we get

$$|a|^2 = 1-q, \quad \text{ and } \quad \omega = - \alpha q^2 - \beta |a|^2.$$

Now the author set $a = r^{i \theta}$ with $r^2 = 1- q^2$ and write $u_{q,\theta}$ for these rolls.

The spectral stability of a fixed $u_{q,θ}$ is determined by the linearization around $u_{q,θ}$ which reads

$$\partial_t \tilde{v} = (1+ i \alpha) \partial_x^2 \tilde{v} + \tilde{v} - (1 + i \beta) (2 |u_{q,θ}|^2 \tilde{v} + u_{q,θ}^2 \overline{\tilde{v}}).$$

Here I did not understand how he made the linearization and got the final form of the equation! Any help is appreciated.

1

There are 1 best solutions below

1
On BEST ANSWER

Substitute $u= u_{q,\theta} + \varepsilon \tilde{v}$ and disregard the terms with coefficients of $\epsilon$ with order not equal to one.