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Theory

$ u(x,t)=0$ is a solution of the KSE. In order to determine if this solution is always stable, we supperpose on it a perturbation. The evolution of the perturbation is imposed by the KSE. We have the choise of the perturbation. The solution of the KSE is invariable in $ x$ and in $ t$, and we have considered periodic limit conditions, so we can decide that we are looking for perturbation under a normal form. It's to say that :

\begin{displaymath}
\begin{array}{cc}
& u(x,t)=0+u'(x,t)\\
with & u'(x,t)=\hat{u}e^{ikx+\sigma t}\\
\end{array}\end{displaymath}

The amplitude of the perturbation is supposed to be small, so we will just retain the order one terms in amplitude. The KSE become :

$\displaystyle \frac{ \partial u'}{\partial t}+
\nu \frac{ \partial^2 u'}{{\partial x}^2}+ \lambda \frac{ \partial^4 u'}{{\partial x}^4}=0
$

With the form of the perturbation we find the dispersion relation :

$\displaystyle \sigma - \nu k^2 + \lambda k^4=0
$

k is the wave number and is equal to $ \frac{2 \pi n}{L}$. Now there is two possibilities.



Julien Delbove
2000-11-23