Consider a PDE of the form

$\partial_t \phi = A(\partial_\xi) \phi + c\partial_\xi \phi +N(\phi)$

where $N$ is some non-linearity defined via pointwise evaluation of $\phi$.

If you want to check for stability of travelling wave solutions of PDEs you linearize the PDE at some travelling wave solution $Q$:

$\partial_t \phi = A(\partial_\xi) \phi + c\partial_\xi \phi + \partial_\phi N(Q) \phi$

My problem is: do exactly this for the Gross-Pitaevskii-Equation.

The Gross-Pitaevskii Equation for (in appropriate coordinates) has the form

$\partial_t \phi = -i \triangle \phi + c \partial_\xi \phi-i\phi(1-\vert \phi \vert^2)$

so that

$N(\phi)=-i\phi (1 -\vert \phi \vert^2).$

Can anyone help me to linearize that at some travelling wave solution $Q$? I'm not even sure how to start...


1 Answer 1


Well.... first you need to specify a traveling wave solution $Q$.

Then you just take, since $N(\phi) = -i \phi(1 - |\phi|^2)$, $$ (\partial_\phi N)(\phi) = -i(1-|\phi|^2) + 2i\phi \bar{\phi} $$ by the product rule of differential calculus. Here note $|\phi|^2 = \phi \bar\phi$.

So simplifying and evaluating it at $Q$ we have $$ (\partial_\phi N)(Q) = -i + 3i |Q|^2 $$ which you can plug into the general form you quoted to get the linearised equation.

  • $\begingroup$ Thank you, but why is $\partial_\phi \vert \phi \vert^2 = \partial_\phi \phi \overline{\phi} = 2 \overline{\phi}$ as you suggest? I didn't think the complex derivative of this even exists (other than in $\phi=0$). Am I mistaken? $\endgroup$
    – mjb
    Commented Jan 9, 2013 at 14:25

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