Nonlinear Schrödinger equations involved in dark matter halos: modulational instability

Abstract

Inspired by the theory of scale relativity, a nonlinear Schrödinger equation has been recently proposed to model dark matter halos. The equation involves a logarithmic nonlinearity associated with an effective temperature and a source of dissipation. Herein, we study the Benjamin–Feir type modulational instability exhibited by this model. We extend our analysis to further generalizations of the equation in the presence of short-range interactions, giving rise to Gross–Pitaevskii-like and Cahn–Hilliard-like equations, as well as to a generalization emerging from the Lynden–Bell distribution. In each case, we establish a criterion leading to modulational instability and study the corresponding growth rate.

Appendix

We detail here the calculations leading to the modulational instability criteria associated with Eqs. (22), (27), and (29).

1.1 A Gross–Pitaevskii-like equation

We consider a plane wave of the form $\psi (x, t)=\psi _{0} e^{i \sigma (x, t) / 2{\mathcal {D}}}$, with a constant amplitude $\psi _{0}$. In this case, Eq. (22) gives the condition that must be fulfilled by $\sigma (x,t)$, in order to have plane wave solutions of Eq. (22), as follows

$$\begin{aligned} \sigma _t = i{\mathcal {D}}\sigma _{xx}-\xi \left[ \sigma -\langle \sigma \rangle \right] -\frac{\sigma _x^2}{2}-v_{\infty }^2 \ln \psi _0 -g|\psi _0|^{2}. \end{aligned}$$

(A.1)

Introducing perturbations of the amplitude and the phase of the plane wave [cf. Eq. (7)], Eq. (22) gives

$$\begin{aligned}&i {\mathcal {D}} \left[ \epsilon _t+\frac{i}{2 {\mathcal {D}}}(\psi _0+\epsilon )(\sigma _t+\theta _t) \right] -\frac{v_{\infty }^2}{2}(\psi _0+\epsilon ) \ln (\psi _0+\epsilon )\nonumber \\&\quad -\,\frac{ \xi }{2} \left[ \sigma -\langle \sigma \rangle +\theta -\langle \theta \rangle \right] (\psi _0+\epsilon ) \nonumber \\&=-{\mathcal {D}}^2 \left[ \epsilon _{xx}+ \frac{i}{2 {\mathcal {D}}} \epsilon _x (\sigma _x + \theta _x)+ \frac{i}{2 {\mathcal {D}}} (\psi _0 + \epsilon )(\sigma _{xx}+ \theta _{xx})\right. \\&\quad \left. +\, \frac{i}{2 {\mathcal {D}}}(\sigma _x + \theta _x) \left[ \epsilon _x + \frac{i}{2 {\mathcal {D}}}(\psi +\epsilon )(\sigma _x + \theta _x)\right] \right] \nonumber \\&\quad +\,\frac{1}{2}g(\psi _0+\epsilon )^3.\nonumber \end{aligned}$$

(A.2)

The real part of Eq. (A.2) [after decomposition in Fourier modes, i.e., Eq. (9)] reads as

$$\begin{aligned} \begin{aligned}&\frac{1}{2}(\psi _0+ \epsilon )(\sigma _t -i \nu \theta )+\frac{v_{\infty }^2}{2}(\psi _0+\epsilon ) \ln (\psi _0+\epsilon ) + \frac{\xi }{2} \left[ \sigma - \langle \sigma \rangle + \theta - \langle \theta \rangle \right] (\psi _0 + \epsilon ) \\&+ {\mathcal {D}}^2 \delta ^2 \epsilon + \frac{(\psi _0 + \epsilon )}{4}(\sigma _x+i \delta \theta )^2 +\frac{1}{2}g(\psi _0+\epsilon )^3=0, \end{aligned} \end{aligned}$$

(A.3)

while the imaginary part remains unchanged from Eq. (11). Proceeding as previously, and neglecting second-order terms, we obtain

$$\begin{aligned} {\left\{ \begin{array}{ll} \left[ \frac{v_{\infty }^2}{2}+ {\mathcal {D}}^2 \delta ^2+g \psi _0^2 \right] \epsilon + \left[ -\frac{i \nu }{2} \psi _0 + \frac{\xi }{2} \psi _0 + i \psi _0 \delta {\mathcal {D}} K\right] \theta =0, \\ \left[ -i \nu + 2 i \delta {\mathcal {D}} K \right] \epsilon - \frac{\psi _0}{2}\delta ^2 \theta =0, \end{array}\right. } \end{aligned}$$

(A.4)

in which we have used the fact that $\langle \epsilon \rangle = \langle \theta \rangle =0$ and have considered, as in the previous section, a phase $\sigma (x, t)=2 {\mathcal {D}} (K x-\varOmega t)$. A non-trivial solution of Eq. (A.4) implies

$$\begin{aligned} \delta ^2 \left[ 2 {\mathcal {D}}^2 K^2 - \frac{{\mathcal {D}}^2 \delta ^2}{2}-\frac{v_{\infty }^2}{4}-g \psi _0^2 \right] +\frac{\nu ^2}{2}-2 {\mathcal {D}} \delta K \nu + i \xi \left( \frac{\nu }{2}-{\mathcal {D}} \delta K\right) =0, \end{aligned}$$

(A.5)

which leads to

$$\begin{aligned} \nu = 2{\mathcal {D}}K \delta -i \frac{\xi }{2} \pm \sqrt{{\mathcal {D}}^2 \delta ^4 + \frac{v_{\infty }^2\delta ^2}{2}+2g \psi _0^2 \delta ^2-\frac{\xi ^2}{4}}. \end{aligned}$$

(A.6)

From Eq. (A.6), we identify the condition for an exponential regime as

$$\begin{aligned} \delta < \delta ^* \equiv \frac{1}{\sqrt{2} {\mathcal {D}}} \sqrt{-\frac{v_{\infty }^2}{2}+2g \psi _0^2 \delta ^2+ \sqrt{\left( \frac{v_{\infty }^2}{2}+2g \psi _0^2 \delta ^2\right) ^2+{\mathcal {D}}^2 \xi ^2}}, \end{aligned}$$

(A.7)

from which follows Eq. (23).

1.2 B Cahn–Hilliard-like equation

Considering a one-dimensional plane wave form with a constant amplitude, i.e., $\psi (x, t)=\psi _{0} e^{i \sigma (x, t) / 2{\mathcal {D}}}$ with $\psi _0 = cte$, Eq. (27) gives

$$\begin{aligned} \sigma _t = i{\mathcal {D}}\sigma _{xx}-\xi \left[ \sigma -\langle \sigma \rangle \right] -\frac{\sigma _x^2}{2}-v_{\infty }^2 \ln \psi _0-g|\psi _0|^{2}. \end{aligned}$$

(B.1)

That is, when $\sigma (x,t)$ satisfies Eq. (B.1), the plane wave function is a solution of Eq. (27). Introducing perturbations of amplitude and phase in the plane wave [cf. Eq. (7)], Eq. (27) gives

$$\begin{aligned} \begin{aligned}&i {\mathcal {D}} \left[ \epsilon _t+\frac{i}{2 {\mathcal {D}}}(\psi _0+\epsilon )(\sigma _t+\theta _t) \right] \\&\quad -\,\frac{v_{\infty }^2}{2}(\psi _0+\epsilon ) \ln (\psi _0+\epsilon )-\frac{ \xi }{2} \left[ \sigma -\langle \sigma \rangle +\theta -\langle \theta \rangle \right] (\psi _0+\epsilon ) \\&=-{\mathcal {D}}^2 \left[ \epsilon _{xx}+ \frac{i}{2 {\mathcal {D}}} \epsilon _x (\sigma _x + \theta _x)+ \frac{i}{2 {\mathcal {D}}} (\psi _0 + \epsilon )(\sigma _{xx}+ \theta _{xx})\right. \\&\quad \left. + \,\frac{i}{2 {\mathcal {D}}}(\sigma _x + \theta _x) \left[ \epsilon _x + \frac{i}{2 {\mathcal {D}}}(\psi +\epsilon )(\sigma _x + \theta _x)\right] \right] \\&\quad +\,\frac{1}{2}g(\psi _0+\epsilon )^3+ \chi \left[ (\psi _0+\epsilon )^2 \epsilon _{xx}+(\psi _0+\epsilon ) \epsilon _x^2 \right] , \end{aligned} \end{aligned}$$

(B.2)

the real part of which, upon using Eq. (9), is given by

$$\begin{aligned} \begin{aligned}&\frac{1}{2}(\psi _0+ \epsilon )(\sigma _t -i \nu \theta )+\frac{v_{\infty }^2}{2}(\psi _0+\epsilon ) \ln (\psi _0+\epsilon ) + \frac{\xi }{2} \left[ \sigma - \langle \sigma \rangle + \theta - \langle \theta \rangle \right] (\psi _0 + \epsilon ) \\&\quad +\, {\mathcal {D}}^2 \delta ^2 \epsilon + \frac{(\psi _0 + \epsilon )}{4}(\sigma _x+i \delta \theta )^2 +\frac{1}{2}g(\psi _0+\epsilon )^3+ \chi \left[ (\psi _0+\epsilon )^2 \epsilon _{xx}+(\psi _0+\epsilon ) \epsilon _x^2 \right] =0, \end{aligned} \end{aligned}$$

(B.3)

while the imaginary part remains unchanged [cf. Eq. (11)]. Proceeding in the same lines as previously, we obtain

$$\begin{aligned} {\left\{ \begin{array}{ll} \left[ \frac{v_{\infty }^2}{2}+ {\mathcal {D}}^2 \delta ^2+g \psi _0^2 - \chi \psi _0^2 \delta ^2\right] \epsilon + \left[ -\frac{i \nu }{2} \psi _0 + \frac{\xi }{2} \psi _0 + i \psi _0 \delta {\mathcal {D}} K\right] \theta =0, \\ \left[ -i \nu + 2 i \delta {\mathcal {D}} K \right] \epsilon - \frac{\psi _0}{2}\delta ^2 \theta =0. \end{array}\right. } \end{aligned}$$

(B.4)

A non-trivial solution implies

$$\begin{aligned} \delta ^2 \left[ 2 {\mathcal {D}}^2 K^2 - \frac{{\mathcal {D}}^2 \delta ^2}{2}-\frac{v_{\infty }^2}{4}-g \psi _0^2 -\chi \delta ^2 \psi _0^2 \right] +\frac{\nu ^2}{2}-2 {\mathcal {D}} \delta K \nu + i \xi (\frac{\nu }{2}-{\mathcal {D}} \delta K) =0, \end{aligned}$$

(B.5)

which has the general solution

$$\begin{aligned} \nu = 2{\mathcal {D}}K \delta -i \frac{\xi }{2} \pm \sqrt{{\mathcal {D}}^2 \delta ^4 + \frac{v_{\infty }^2\delta ^2}{2}+2g \psi _0^2 \delta ^2-\frac{\xi ^2}{4}+2 \chi \psi _0^2 \delta ^4}. \end{aligned}$$

(B.6)

In this case, the condition for an exponential regime reads as

$$\begin{aligned} \delta < \delta ^* \equiv \sqrt{\frac{-\frac{v_{\infty }^2}{2}+2g \psi _0^2 \delta ^2+ \sqrt{(\frac{v_{\infty }^2}{2}+2g \psi _0^2 \delta ^2)^2+({\mathcal {D}}^2+2 \chi \psi _0^2) \xi ^2}}{2({\mathcal {D}}^2+2 \chi \psi _0^2)} }, \end{aligned}$$

(B.7)

from which follows Eq. (28).

1.3 C Lynden–Bell distribution

Considering a wave function with a constant amplitude, Eq. (29) gives the condition satisfied by $\sigma (x,t)$ to have plane wave solutions as

$$\begin{aligned} \sigma _t = i{\mathcal {D}}\sigma _{xx}-\xi \left[ \sigma -\langle \sigma \rangle \right] -\frac{\sigma _x^2}{2}-2\frac{T_{\mathrm {eff}}}{\eta _{0}} \ln \psi _0-\frac{1}{2}\left( \frac{3}{4 \pi \eta _{0}}\right) ^{2 / 3}\psi _0^{4 / 3}. \end{aligned}$$

(C.1)

Introducing perturbations in the amplitude and phase [cf. Eq. (7)], it gives

$$\begin{aligned} \begin{aligned}&i {\mathcal {D}} \left[ \epsilon _t+\frac{i}{2 {\mathcal {D}}}(\psi _0+\epsilon )(\sigma _t+\theta _t) \right] \\&\quad -\,\frac{T_{\mathrm {eff}}}{\eta _{0}}(\psi _0+\epsilon ) \ln (\psi _0+\epsilon )-\frac{ \xi }{2} \left[ \sigma -\langle \sigma \rangle +\theta -\langle \theta \rangle \right] (\psi _0+\epsilon ) \\&=-{\mathcal {D}}^2 \left[ \epsilon _{xx}+ \frac{i}{2 {\mathcal {D}}} \epsilon _x (\sigma _x + \theta _x)+ \frac{i}{2 {\mathcal {D}}} (\psi _0 + \epsilon )(\sigma _{xx}+ \theta _{xx})\right. \\&\quad \left. +\, \frac{i}{2 {\mathcal {D}}}(\sigma _x + \theta _x) \left[ \epsilon _x + \frac{i}{2 {\mathcal {D}}}(\psi +\epsilon )(\sigma _x + \theta _x)\right] \right] \\&\quad +\,\frac{1}{4}\left( \frac{3}{4 \pi \eta _{0}}\right) ^{2 / 3} (\psi _0+ \epsilon )^{7/3}. \end{aligned} \end{aligned}$$

(C.2)

The real part of Eq. (C.2) [after decomposition in Fourier modes, i.e., Eq. (9)] reads as

$$\begin{aligned} \begin{aligned}&\frac{1}{2}(\psi _0+ \epsilon )(\sigma _t -i \nu \theta )+\frac{T_{\mathrm {eff}}}{\eta _{0}}(\psi _0+\epsilon ) \ln (\psi _0+\epsilon ) + \frac{\xi }{2} \left[ \sigma - \langle \sigma \rangle + \theta - \langle \theta \rangle \right] (\psi _0 + \epsilon ) \\&\quad +\, {\mathcal {D}}^2 \delta ^2 \epsilon + \frac{(\psi _0 + \epsilon )}{4}(\sigma _x+i \delta \theta )^2 +\frac{1}{4}\left( \frac{3}{4 \pi \eta _{0}}\right) ^{2 / 3} (\psi _0+\epsilon )^{7/3}=0, \end{aligned} \end{aligned}$$

(C.3)

while the imaginary part is still unchanged. Proceeding in the same lines as previously, we obtain

$$\begin{aligned} {\left\{ \begin{array}{ll} \left[ \frac{T_{\mathrm {eff}}}{\eta _{0}}+ {\mathcal {D}}^2 \delta ^2+ \frac{1}{3}\left( \frac{3}{4 \pi \eta _{0}}\right) ^{2 / 3} \psi _0^{4/3} \right] \epsilon + \left[ -\frac{i \nu }{2} \psi _0 + \frac{\xi }{2} \psi _0 + i \psi _0 \delta {\mathcal {D}} K\right] \theta =0, \\ \left[ -i \nu + 2 i \delta {\mathcal {D}} K \right] \epsilon - \frac{\psi _0}{2}\delta ^2 \theta =0. \end{array}\right. } \end{aligned}$$

(C.4)

A non-trivial solution implies

$$\begin{aligned} \delta ^2 \left[ 2 {\mathcal {D}}^2 K^2 - \frac{{\mathcal {D}}^2 \delta ^2}{2}+\frac{1}{3}\left( \frac{3}{4 \pi \eta _{0}}\right) ^{2 / 3} \psi _0^{4/3}-\frac{T_{\mathrm {eff}}}{2\eta _{0}} \right] +\frac{\nu ^2}{2}-2 {\mathcal {D}} \delta K \nu + i \xi (\frac{\nu }{2}-{\mathcal {D}} \delta K) =0, \end{aligned}$$

(C.5)

which leads to

$$\begin{aligned} \nu = 2{\mathcal {D}}K \delta -i \frac{\xi }{2} \pm \sqrt{{\mathcal {D}}^2 \delta ^4 + \frac{T_{\mathrm {eff}}}{\eta _{0}} \delta ^2-\frac{2}{3} \left( \frac{3}{4 \pi \eta _{0}}\right) ^{2 / 3}\psi _0^{4/3} \delta ^2-\frac{\xi ^2}{4}}. \end{aligned}$$

(C.6)

The condition for an exponential regime reads as

$$\begin{aligned} \delta < \delta ^* \equiv \frac{1}{\sqrt{2} {\mathcal {D}}} \sqrt{- \left( \frac{T_{\mathrm {eff}}}{\eta _{0}}-\frac{2}{3}\left( \frac{3}{4 \pi \eta _{0}}\right) ^{2 / 3} \psi _0^{4/3} \right) + \sqrt{\left( \frac{T_{\mathrm {eff}}}{\eta _{0}}-\frac{2}{3} \left( \frac{3}{4 \pi \eta _{0}}\right) ^{2 / 3}\psi _0^{4/3} \right) ^2+{\mathcal {D}}^2 \xi ^2}}, \end{aligned}$$

(C.7)

leading to Eq. (30).