# Soliton solutions of the Gross-Pitaevskii equation

The Gross-Pitaevskii equation admits soliton solutions such as: $$\psi(x)=\psi_0 sech(x/\xi),$$ where $$\xi$$ is the healing length defined by: $$\xi=\frac{\hbar}{\sqrt{m \mu}}$$, with $$\mu$$ being the chemical potential. To obtain this solution, one of the boundary conditions is given by assuming that for lengths much greater than the healing length, the condensate is homogeneus with density $$\psi_0=\sqrt{\frac{\mu}{g}}$$.

I don't see how to obtain this solution by solving the non-linear GPE: $$\mu \psi(x) = -\frac{\hbar^2}{2m} \partial^2_x \psi(x) + g|\psi(x)|^2 \psi(x).$$

If you relabel $$x$$ as "$$t$$" can think of it as mechanics problem $$\frac{d^2 \psi}{dt^2}= g \psi^3 -\mu \psi.$$ As
$$g \psi^3 -\mu \psi = \frac{d}{d\psi}\left(\frac 14 g \psi^4 -\frac 12 \mu \psi^2 \right)$$ you can solve for $$\psi(t)$$ by using "energy" conservation for motion in the potential $$V(\psi)= -\frac 14 g \psi^4 +\frac 12 \mu \psi^2.$$ Energy conservation is the usual $$KE+PE=$$ constant i.e. $$\frac 12\left( \frac{d \psi}{dt}\right)^2 -\frac 14 g \psi^4 +\frac 12 \mu \psi^2= \kappa$$ The constant $$\kappa$$ is fixed by the the value of the "potential energy" $$V(\psi)$$ when $$d\psi/dt=0$$ i.e $$\kappa=V(\sqrt{\mu/g)}$$.
Renaming $$t$$ back to $$x$$, we solve, as we would in any one-dimensional mechanics problem, by separation of variables $$\int^{\psi(x)}_0 \frac{d\psi}{2\sqrt{V(\psi)-\kappa}} = \int dx=x.$$ The constant in the indefinite $$x$$ integral has been fixed by the boundary condition $$\psi=0$$ at $$x=0$$.