 Problem Statement
 Given the LugiatoLefever equation, linearize the equation and determine the dynamics near a stationary solution by looking for a stationary solution with a small perturbation.
 Relevant Equations

The LugiatoLefever Equation is given by:
$$ \frac {d\psi} {dt} = (1 + i \alpha)\psi + \frac {i} {2} \frac {d^2\psi} {dx^2} + i * \psi^2 \bar \psi + F $$
It can be shown that when looking for stationary solutions in both time and space, the following is a solution:
$$ \psi_e = \frac F {1 + i * (\alpha  \rho)} $$
When ##\rho## is the solution of the polynomial:
$$ F ^ 2 = (1 + (\rho  \alpha) ^ 2)\rho $$
I started by substituting the following anzatz:
$$ \psi = \psi_e + \psi_1 $$
When ## \psi_1 \ll 1 ##. Substituting the above into the equation yields:
$$ \frac {d\psi_1} {dt} = (1 + i\alpha)\psi_1 + \frac i 2 \frac {\partial ^ 2 \psi _1 } {\partial x ^ 2 } + i (\bar \psi_1 \psi_1 ^2 + \bar \psi_e \psi _1 ^2 + 2\bar\psi_e \psi_e \psi_1 + \bar \psi_1 \psi_e ^2 + 2\bar\psi_1 \psi_e \psi_1) $$
I am interested in a first order treatment, so i will take out the terms quadratic in ##\psi_1##:
$$ \frac {d\psi_1} {dt} = (1 + i\alpha)\psi_1 + \frac i 2 \frac {\partial ^ 2 \psi _1 } {\partial x ^ 2 } + i ( 2\bar\psi_e \psi_e \psi_1 + \bar \psi_1 \psi_e ^2) $$
I know ##\psi_e## as it is the stationary solution from above, and for ##\psi_1## I can look for solutions of the form:
$$ \psi_1 = e^{i\omega t}*(a * e^{ikx} + \bar b * e^{ikx})$$
After doing that and equating the coefficients of the backwards and forwards propagating plane waves, I arrived at the following equations:
$$ i\omega \bar b = \bar b ((1+i\alpha)  \frac i 2 k^2 + i2\bar\psi_e\psi_e) + \bar a (i\bar\psi_1\psi_e^2) $$
$$ i\omega a = a((1+i\alpha)  \frac i 2 k^2 + i2\bar\psi_e\psi_e) + b(i\bar\psi_`1\psi_e^2) $$
This is almost what I wanted, but the first equation is on the conjugates of a and b rather than on a and b so I am not sure how to proceed.
$$ \psi = \psi_e + \psi_1 $$
When ## \psi_1 \ll 1 ##. Substituting the above into the equation yields:
$$ \frac {d\psi_1} {dt} = (1 + i\alpha)\psi_1 + \frac i 2 \frac {\partial ^ 2 \psi _1 } {\partial x ^ 2 } + i (\bar \psi_1 \psi_1 ^2 + \bar \psi_e \psi _1 ^2 + 2\bar\psi_e \psi_e \psi_1 + \bar \psi_1 \psi_e ^2 + 2\bar\psi_1 \psi_e \psi_1) $$
I am interested in a first order treatment, so i will take out the terms quadratic in ##\psi_1##:
$$ \frac {d\psi_1} {dt} = (1 + i\alpha)\psi_1 + \frac i 2 \frac {\partial ^ 2 \psi _1 } {\partial x ^ 2 } + i ( 2\bar\psi_e \psi_e \psi_1 + \bar \psi_1 \psi_e ^2) $$
I know ##\psi_e## as it is the stationary solution from above, and for ##\psi_1## I can look for solutions of the form:
$$ \psi_1 = e^{i\omega t}*(a * e^{ikx} + \bar b * e^{ikx})$$
After doing that and equating the coefficients of the backwards and forwards propagating plane waves, I arrived at the following equations:
$$ i\omega \bar b = \bar b ((1+i\alpha)  \frac i 2 k^2 + i2\bar\psi_e\psi_e) + \bar a (i\bar\psi_1\psi_e^2) $$
$$ i\omega a = a((1+i\alpha)  \frac i 2 k^2 + i2\bar\psi_e\psi_e) + b(i\bar\psi_`1\psi_e^2) $$
This is almost what I wanted, but the first equation is on the conjugates of a and b rather than on a and b so I am not sure how to proceed.
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