fluid dynamics - Sudden release of condensate from trap - Ermakov equation - Scaling solution

This is related to the scaling solution of the hydrodynamic equations. I get a relation for the scaling parameter $b$:

$\ddot{b} = -\omega^2(t)*b + \omega_0^2/b^3$

When the trap for the condensate in suddenly switched off $\omega(t)$ goes to zero, so you get the equation

$\ddot{b} = \omega_0^2/b^3$ with initial conditions $b(0)=1$ and $\dot{b}(0) = 0$.

What would be the solution for $b(t)$?

Answer

The equation is

$$\frac{d^2b}{dt^2} = \frac{\omega_0^2}{\, b^3 \, }$$ Multiply both sides of the equation by $\frac{db}{dt}$ and obtain $$\frac{d^2b}{dt^2} \, \frac{db}{dt} = \frac{\omega_0^2}{\, b^3 \, } \, \frac{db}{dt}$$ which can be interpreted as $$\frac{db}{dt} \, \frac{d}{dt}\left(\frac{db}{dt}\right) = \Big(\omega_0^2\, b^{-3}\Big) \, \frac{db}{dt}$$ which by going backwards with chain rule is the same as $$\frac{1}{2}\, \frac{d}{dt}\left(\frac{db}{dt}\right)^2 = \frac{d}{dt} \Big(\omega_0^2\, \frac{b^{-2}}{-2}\Big)$$ $$\frac{1}{2}\, \frac{d}{dt}\left(\frac{db}{dt}\right)^2 = -\, \frac{1}{2}\frac{d}{dt} \Big(\omega_0^2\, b^{-2}\Big)$$ and after cancelling the one half $$\frac{d}{dt}\left(\frac{db}{dt}\right)^2 = -\,\frac{d}{dt} \Big(\omega_0^2\, b^{-2}\Big)$$ Integrate both sides with respect to $t$ $$\left(\frac{db}{dt}\right)^2 = E_0 - \omega_0^2\, b^{-2}$$ $$\left(\frac{db}{dt}\right)^2 = \frac{E_0 \, b^2 - \omega_0^2}{b^2}$$ $$b^2 \, \left(\frac{db}{dt}\right)^2 = {E_0 \, b^2 - \omega_0^2}$$ $$\left(b \, \frac{db}{dt}\right)^2 = {E_0 \, b^2 - \omega_0^2}$$ $$\left(\frac{1}{2} \, \frac{d(b^2)}{dt}\right)^2 = {E_0 \, b^2 - \omega_0^2}$$ $$\left(\frac{d(b^2)}{dt}\right)^2 = {4 \, E_0 \, b^2 - 4 \, \omega_0^2}$$ When $\frac{db}{dt}(0) = 0$ and $b(0) = 1$ we arrive at $E_0 = \omega_0^2$. Change the dependent variable by setting $u = b^2$ and the equation becomes $$\left(\frac{du}{dt}\right)^2 = {4 \, E_0 \, u - 4 \, \omega_0^2}$$ or after taking square root on both sides $$\frac{du}{dt} = \pm \, \sqrt{\, 4 \, E_0 \, u - 4 \, \omega_0^2 \, }$$ This is a separable equation $$\frac{du}{ 2\, \sqrt{\, E_0 \, u - \omega_0^2 \, }} = \pm \, dt$$ $$\frac{d\big(E_0 \, u - \omega_0^2 \big)}{ 2\, \sqrt{\, E_0 \, u - \omega_0^2 \, }} = \pm \, E_0 \, dt$$ $$d \Big( \sqrt{\, E_0 \, u - \omega_0^2 \, } \Big) = \pm \, E_0 \, dt$$ Integrate both sides $$\sqrt{\, E_0 \, u - \omega_0^2 \, } = C_0 \pm E_0 \, t$$ Square both sides $${\, E_0 \, u - \omega_0^2 \, } = \big(\, C_0 \pm E_0 \, t \,\big)^2$$ and solve for $u$ $$u = \frac{1}{E_0} \, \big(\, C_0 \pm E_0 \, t \, \big)^2 + \frac{\omega_0^2}{E_0}$$ Return back to $u = b^2$ $$b^2 = \frac{1}{E_0} \, \big(\, C_0 \pm E_0 \, t \, \big)^2 + \frac{\omega_0^2}{E_0}$$ so finally $$b(t) = \pm \, \sqrt{ \, \frac{1}{E_0} \, \big(\, C_0 \pm E_0 \, t \, \big)^2 + \frac{\omega_0^2}{E_0} \, }$$ However, we know that $E_0 = \omega_0^2$ so $$b(t) = \pm \, \sqrt{ \, \frac{1}{\omega_0^2} \, \big(\, C_0 \pm \omega_0^2 \, t \, \big)^2 + 1 \, }$$ Thus, $b(0) = 1$ is possible when $C_0 = 0$ and finally $$b(t) = \sqrt{ \, \omega_0^2 \, t^2 + 1 \, }$$