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Consider the continuum cubic, focusing NLS : $$iu_t = -\Delta u - |u|^2 u$$ In the following picture, a perturbation $e^{i \Lambda t}(u_0(x) + \epsilon(v+iw))$ is substituted into the NLS and split into Real and Imaginary parts to get equations $(2.12)$ and $(2.13)$. Then, a separation of space and time variables is done: $v(x,t) = \overset{\sim}{v}(x)e^{\lambda t}$, $w(x,t) = \overset{\sim}{w}(x)e^{\lambda t}$ to get the eigenvalue problems shown in $(2.14)$.

I do not understand how the dispersion relation $(2.15)$ is obtained. Namely, what equations are they substituting $e^{i(kx - \omega t)}$ into and for what?

Attempt:

I see that $(2.15)$ has a $k^2$ term which can only be obtained from the Laplacian. Likewise, a $\omega^2$ term which can only come from differentiating with respect to $t$. If I substitute $v$ and $w$ for $e^{i(kx-\omega t)}$ in equations $(2.12)$ and $(2.13)$ then I get $$\omega e^{i(kx - \omega t)} = L_{-}e^{i(kx-\omega t)}$$ and $$\omega e^{i(kx - \omega t)} = -L_{+}e^{i(kx-\omega t)}$$ Applying $L_{-}$ to the second equation: $$\omega L_{-}e^{i(kx-\omega t)} = -L_{-}L_{+}e^{i(kx - \omega t)}$$ $$\omega^2 e^{i(kx - \omega t)} = -L_{-}L_{+}e^{i(kx - \omega t)}$$ But from here I do not see how we arrive at $\omega^2 = -\lambda^2 = \pm (\Lambda +k^2)$

Source:

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KZ-Spectra
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  • Taking the time derivative of Equations 2.12 and 2.13, your work is the RHS of those time differentiated equations. Now compute the time derivative side (LHS) and cancel the common exponential on both sides - the dispersion relation immediately follows. – Ninad Munshi Aug 10 '23 at 18:37
  • The book tells you to do $v_{tt} = -L_-L_+v$ - This is simply Eq 2.14 – Ninad Munshi Aug 10 '23 at 19:01
  • I tried to follow your instructions, but did not arrive at 2.15. Could you write out your instructions in an Answer? Here is what I did:

    $$v_{tt} = \partial_t L_{-} w = -wk^2$$ $$w_{tt} = -\partial_t L_{+} v = -\omega k^2$$

    From here I do not see how we obtain an equation with $\lambda$ in it.

     – KZ-Spectra Aug 10 '23 at 19:23

1 Answers1

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Substitute $e^{i(kx - \omega t)}$ into equations $(2.12)$ and $(2.13)$. The $\lambda$ shown in $(2.15)$ is not the same $\lambda$ in $(2.14)$. I do not know what "continuous spectrum" definition is, but the only way $(2.15)$ is obtained is if $u_0 =0$. I think we are examining the problem as $x\to \pm \infty$ and since $u_0$ represents a solition which decays rapidly at both ends of the pulse then this goes to zero.

Indeed, we obtain $$\omega^2 = (k^2 + \Lambda)^2$$ $$\omega = \pm (k^2 + \Lambda)$$

I claim the book has a typo and should be the above instead of $\omega^2 = \pm(k^2 + \Lambda)$

KZ-Spectra
  • 451