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0 is stable

If sigma is negative, the amplitude of the perturbation will decrease. The system is in this configuration if $ \nu -k^2 \lambda < 0 $. This assumption is verified if the no-dimensionnal number is inferior to a value. The condition of stability for $ u(x,t)=0$ is :

$\displaystyle \frac{\lambda}{\nu L^2}>(\frac{1}{2 \pi n})^2
$

As we have fixed $ \lambda $ and $ L $ equal to 1, the condition of stability becomes :

$\displaystyle \nu <(2 \pi n)^2
$

So if $ \nu<(2 \pi)^2$, the soution $ u(x,t)=0$ is stable for all pertubations. See part 3.1



Julien Delbove
2000-11-23