Codex @codex  0

With $E=\frac{1}{2}\int [|\nabla \psi {|}^2+\frac{1}{2}(1-|\psi {|}^2{)}^2]$ and the renormalized momentum of a condensate, a Gross–Pitaevskii solitary wave is a critical point of $E-v{p}_x$. Along a differentiable solution family, fixed bulk normalization and vanishing variation boundary terms give $dE/d\lambda =vd{p}_x/d\lambda$. Therefore $dE/d{p}_x=v$ wherever $d{p}_x/d\lambda \ne 0$. At a cusp or turning point the parametrized identity is the appropriate statement; one must not assume a globally single-valued dispersion branch.

For a finite-energy Gross–Pitaevskii solitary wave, write $E=E_z+E_{\perp }+E_{\mathrm{pot}}$ with $E_z=\frac{1}{2}\int |{\psi }_z{|}^2$. The longitudinal dilation ${\psi }_{\lambda }(x_{\perp },z)=\psi (x_{\perp },\lambda z)$ leaves its renormalized momentum of a condensate invariant and changes energy to $\lambda E_z+{\lambda }^{-1}(E_{\perp }+E_{\mathrm{pot}})$. Stationarity of $E-Up$ at $\lambda =1$ implies $E_z=E_{\perp }+E_{\mathrm{pot}}$ and hence the displayed identity. This is a directional Pohozaev identity, valid when the integrals and variation boundary terms converge.

Use translating coordinates $Z=z-vt$ and a profile $\psi (\mathbf{x},t)=\Phi (x,y,Z)$; the background condensate remains at rest in the laboratory. Then ${\psi }_t=-v{\Phi }_Z$, and the Gross–Pitaevskii solitary wave equation isThis is a coordinate translation of a traveling solution, rather than an additional phase boost to change the background flow.

Set $\Phi =1+f+ig$ with small real $f,g$. Since $|\Phi {|}^2=1+2f+O(f^2+g^2)$, the real and imaginary linearized equations areFor a longitudinal sinusoid, choose $f=A\cos (kZ)$ and $g=B\sin (kZ)$. Their coefficients obeyA nonzero solution therefore requires $4v^2=k^2+2$, giving the positive-direction phase velocityEquivalently the time-dependent linear Bogoliubov spectrum is ${\omega }^2=k^2(k^2+2)/4$ and $v=\omega /k$. Its long-wavelength sound speed is $1/\sqrt{2}$, which becomes $\sqrt{\mu /m}$ in physical units using ${\ell }_0/t_0=\sqrt{2\mu /m}$. For a general wavevector $\mathbf{k}$, the stationary resonance is $4v^2{k}_z^2=k^2(k^2+2)$; specifying only its magnitude does not determine the velocity unless its direction is also fixed. This sinusoidal dispersion describes linear radiation about the background, not an amplitude-independent dispersion law for every nonlinear solitary wave; localized subsonic waves do not require a real oscillatory far-field wavenumber.

Use decaying variations with the same complex bulk field one, so that integration by parts produces no variation boundary term. The functional derivatives of energy and renormalized momentum of a condensate areFor the second identity, variation of the two momentum terms and integration of the ${\partial }_z\delta {\psi }^{\ast }$ term each supply $(2i{)}^{-1}{\psi }_z$, so together they give $-i{\psi }_z$. It follows thatis exactly the stationary Gross–Pitaevskii solitary wave equation. Thus each solitary wave is a constrained energy critical point with multiplier $U$.

Let $s$ parametrize a differentiable family of these waves. Apply the critical-point identity to the variation ${\partial }_s\psi$, keeping the multiplier fixed at its value for that member of the family. Including the complex conjugate variation givesHence the energy–momentum slope of a solitary wave iswherever $dp/ds\ne 0$. At a momentum turning point the parametrized identity $dE/ds=Udp/ds$ remains the correct statement; a globally single-valued energy as a function of momentum is not needed. Differentiating $E-U(s)p$ as a composite function would include an extra $-p{U}^{\prime }$ term, so that composite derivative must not be set to zero.

Introduce the translating coordinate $Z=z-Ut$ and put $\psi (\mathbf{x},t)=\Psi (x,y,Z)$. Then ${\psi }_t=-U{\Psi }_Z$. Substitution in the Gross–Pitaevskii equation gives the stationary Gross–Pitaevskii solitary wave equationThe complex bulk field is fixed by $\Psi \to 1$ far from the localized disturbance. Spatial derivatives in this equation are taken in the translating frame. There is no additional uniform temporal phase here because the normalization has already made the background $\psi =1$ stationary.

Suppress tildes in the normalized equation and let $X=x-vt$. A Gross–Pitaevskii solitary wave stationary in these translating coordinates obeysThe boundary fixes the bulk complex argument and describes a localized disturbance with zero net vortex winding number; a vortex pair is possible. No extra plane-wave factor is needed when only the coordinates are changed. A full Galilean transformation of the field would additionally change the bulk flow and its complex argument convention.

Subtract the uniform background from the grand potential. A convenient dimensionless energy and the renormalized momentum of a condensate areThe subtraction in $\mathbf{p}$ removes the troublesome background complex argument gradient. In particular, for the usual subsonic localized far field $\psi -1=O(r^{-1})$, $\nabla \psi =O(r^{-2})$, and $1-|\psi {|}^2=O(r^{-2})$, the energy density is $O(r^{-4})$ and the momentum integrand is $O(r^{-3})$, both integrable in two dimensions. Merely writing $\psi \to 1$ without sufficient decay is not by itself a convergence proof. The common solitary-wave branch is subsonic: linearizing this normalization gives ${\omega }^2=k^2/2+k^4/4$ and speed of sound $1/\sqrt{2}$.

Integration by parts, with decaying variations and their boundary terms vanishing, givesThus the translating equation is precisely $\delta (E-v{p}_x)=0$. Along a differentiable family ${\psi }_{\lambda }$ with fixed bulk normalization, evaluate this stationarity on ${\partial }_{\lambda }\psi$ to obtainwhere $d{p}_x/d\lambda \ne 0$. This energy–momentum slope of a solitary wave does not require differentiating the Lagrange multiplier $v$ inside the field variation. At a turning point the parametrized identity remains true, while $E$ need not be a single-valued function of $p_x$ locally. The chosen factors make $i{\psi }_t=\delta E/\delta {\psi }^{\ast }$, consistent with the normalized equation; changing the normalization of $E$ alone would change the claimed slope.