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In their paper [1] (full text at [2]) Bethuel et al on page 249 (bottom) linearize the moving frame Gross-Pitaevskii-Equation

$0=-ic \partial_{x_1} \widetilde{v} - \Delta \widetilde{v} - \widetilde{v}(1-\vert \widetilde{v}\vert^2)$

as follows: Setting $\widetilde{v}= V+1$ and $V=f+ig$ one gets

$0=-ic \partial_{x_1} (f+ig) - \Delta (f+ig) - (f+ig+1) (1 - (f+1)^2 +g^2)$

and therefore

$0 = \left( \begin{array}{cc} -\Delta+2 & c \partial_{x_1} \\\ -c \partial_{x_1} & -\Delta \end{array} \right) \left( \begin{array}{l} f \\\ g \end{array} \right) - \left( \begin{array}{l} fg^2-g^2-f^3-3f^2 \\\ -g^3-f^2 g - 2 fg \end{array} \right)$

Usually - in order to linearize the LHS - one would proceed by taking the derivative of the last term. But Bethuel et al just factorize and write

$0 = \left( \begin{array}{cc} -\Delta+2 & c \partial_{x_1} \\\ -c \partial_{x_1} & -\Delta \end{array} \right) \left( \begin{array}{l} f \\\ g \end{array} \right) + \left( \begin{array}{cc} g^2+f^2+3f & g\\\ 2g & f^2+g^2 \end{array} \right) \left( \begin{array}{l} f \\\ g \end{array} \right)$

In which way is that a linearization of the equation? Would this linearization be suitable for a stability analysis?


[1] http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=TI&pg7=ALLF&pg8=ET&review_format=html&s4=bethuel&s5=travelling%20waves%20for%20the&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=4&mx-pid=1669387

[2] http://www.numdam.org/item?id=AIHPA_1999__70_2_147_0

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1 Answer 1

The linearization intended by Bethuel et al. is around $\tilde{v}=1$, so around $f=0$, $g=0$, so the linearized equation is

$0=L_0 \left( \begin{array}{l} f \\\ g \end{array} \right)$ with $L_0 = \left( \begin{array}{cc} -\Delta+2 & c \partial_{x_1} \\\ -c \partial_{x_1} & -\Delta \end{array} \right) $,

as it says in their paper.

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Yes, that is what one would think they do. But they are writing "the continuous spectrum of $L$ [which still has the $f$ and $g$] is the same, if $f$, $g$ decay sufficiently fast". What sense does that statement make if $f=g=0$? –  mjb86 Mar 16 '13 at 12:21

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