Posch Teller potential analytical classical solution I am trying to obtain the classical solution for a 1D Schroedinger equation with a symmetric Posch-Teller potential, the equation is given by,
$ \frac{d^2 x}{d t^2} - V \frac{\sinh(x)}{\cosh^3(x)} = 0 $ 
Where $V>0$. The boundary conditions are given by $x(t_0) = x_0$ and $x(t_1) = x_1$ (with $t_0 < t_1$).
Does it exist an analytical solution for this equations? If not, which should be the best perturbative method in this situation?
The analytical solution is essential for the rest of the analysis as the quantum part depends on the classical path given the non-linearity of the Posch-Teller potential. 
Thanks
Alvaro
 A: I think there is an analytic solution, I am not sure how convenient its form is. We have the equation
$$\frac{d^2 x}{dt^2} = \frac{V\sinh(x)}{\cosh^3(x)}.$$
Multiply both sides of the equation by the first derivative to obtain
$$\frac{d^2 x}{dt^2} \, \frac{d x}{dt}  = \left(\frac{V\sinh(x)}{\cosh^3(x)} \right)\frac{d x}{dt}$$ which then can be written as 
\begin{align}\frac{1}{2}\frac{d}{dt}\left(\left(\frac{d x}{dt}\right)^2 \right)  &= \left(\frac{V}{\cosh^3(x)} \right)\frac{d}{dt}\big(\cosh(x)\big) =  \frac{d}{dt}\left((-1) \frac{V}{2\,\cosh^2(x)}\right)
\end{align}
So if we integrate both sides with respect to $t$ we obtain the conservation of energy property of the system
$$\frac{1}{2}\left(\frac{d x}{dt}\right)^2 = E_0 \, - \, \frac{V}{2\,\cosh^2(x)}   $$ or maybe in a more familiar form 
$$\frac{1}{2}\left(\frac{d x}{dt}\right)^2  +  \frac{V}{2\,\cosh^2(x)} = E_0 $$ which as you see says that the sum of the kinetic energy $\frac{1}{2}\left(\frac{d x}{dt}\right)^2$ and the potential energy $\frac{V}{2\,\cosh^2(x)}$ is a constant $E_0$ with respect to time $t$ for any solution $x=x(t)$ of the system.  Since we are in the one degree of freedom case, we simply look at  $$\frac{1}{2}\left(\frac{d x}{dt}\right)^2 = E_0 \, - \, \frac{V}{2\,\cosh^2(x)}   $$ and take a square root and write the equation as 
$$\frac{d x}{dt} = \epsilon\sqrt{2 E_0 \, - \, \frac{V}{\cosh^2(x)}}  $$ where $\epsilon$ is either $+1$ or $-1$ depending on whether the derivative is negative or positive. It is locally constant (roughly speaking). 
Rewrite equation as
 $$\cosh(x) \, \frac{d x}{dt} = \epsilon\sqrt{2 E_0 \, \cosh^2(x) \, - \, V}  $$ which becomes
$$ \frac{d}{dt} \big(\sinh(x)\big) = \epsilon\sqrt{2 E_0 \, \Big(\sinh^2(x) + 1 \Big) \, - \, V}  $$
Now it is easy to change the variables for convenience as $y = \sinh(x)$. The equation turns into
\begin{align} \frac{dy}{dt}  &= \epsilon\sqrt{2 E_0 \, \big(y^2 + 1 \big) \, - \, V}\\  
&= \epsilon\sqrt{2 E_0 \, y^2 + 2E_0 - V} \\
&= \epsilon \sqrt{2E_0} \, \sqrt{ y^2 + \frac{2E_0 - V}{2 E_0}}
\end{align} 
Thus, we have the first order scalar time homogeneous equation
\begin{align} \frac{dy}{dt}  &= \epsilon \sqrt{2E_0} \, \sqrt{ y^2 + \frac{2E_0 - V}{2 E_0}}
\end{align} which is integrable and solvable, but there are three different cases, based on the level of total energy $E_0$ and its value with respect to the peak maximum of the potential energy, which depends on $V$. In the case, $E_0 < \frac{V}{2}$ you have solutions occupying only the right side of a given value $x_0>0$, so moving only on the right of $x_0$ (when $x_0 < 0$ it occupies only the left side of $x_0$ i.e. it moves only on the left of $x_0$). In the case when $E_0 > \frac{V}{2}$ the solution occupies the whole real line, i.e. goes to plus infinity and to minus infinity. When $E_0 = \frac{V}{2}$ then we have special solutions, one is an equilibrium, the rest are go either to the left of $0$ or to the right of $0$, never to reach $0$.
So let's $E_0 < \frac{V}{2}$. Let us set $$a_0 = \sqrt{\frac{V -2E_0}{2E_0}} =  \sqrt{\left|\frac{V -2E_0}{2E_0}\right|}$$ then the equation becomes
\begin{align} \frac{dy}{dt}  &= \epsilon \sqrt{2E_0} \, \sqrt{ y^2 - a_0^2}
\end{align} so write it as
\begin{align} \frac{dy}{ \epsilon \sqrt{2E_0} \, \sqrt{ y^2 - a_0^2}}  &= dt
\end{align} so integrating both sides we obtain
\begin{align} \int \, \frac{dy}{ \epsilon \sqrt{2E_0} \, \sqrt{ y^2 - a_0^2}}  &= t + C
\end{align} 
\begin{align} \frac{\epsilon}{\sqrt{2 E_0}}\int \, \frac{dy}{\sqrt{ y^2 - a_0^2}}  &= t + C
\end{align} 
\begin{align} \frac{\epsilon}{\sqrt{2 E_0}} \, \cosh^{-1}\left(\frac{y}{a_0}\right) &= t + C
\end{align} 
\begin{align} \cosh^{-1}\left(\frac{y}{a_0}\right) &= \epsilon \sqrt{2 E_0} \,\, t + C_0
\end{align} 
\begin{align} y(t) &= a_0 \cosh{\left(\epsilon \sqrt{2 E_0} \,\, t + C_0\right)}
\end{align} 
So finally, returning to the original variable $x$ we obtain
\begin{align} x(t) &= \sinh^{-1}\Big(a_0 \cosh{\left(\epsilon \sqrt{2 E_0} \,\, t + C_0\right)}\Big)
\end{align} 
If we impose the initial value problem $x(0) = x_0 > 0$ for instance then 
$$C_0 = \cosh^{-1}\left(\frac{1}{a_0}\sinh(x_0)\right).$$ If we have initial conditions such that the first derivative of $x(t)$ at the point $x_0$ is positive,, i.e. $\frac{dx}{dt}(0) > 0$, then the solution is 
\begin{align} x(t) &= \sinh^{-1}\Big(a_0 \cosh{\left(C_0 + \sqrt{2 E_0} \,\, t \right)}\Big)
\end{align}
for all $t \geq -\frac{C_0}{\sqrt{2E_0}}$ and
\begin{align} x(t) &= \sinh^{-1}\Big(a_0 \cosh{\left(C_0 - \sqrt{2 E_0} \,\, t\right)}\Big)
\end{align}
for all $t <-\frac{C_0}{\sqrt{2E_0}}$.
When $E_0 > \frac{V}{2}$ then the equation looks like 
\begin{align} \frac{dy}{dt} = { \epsilon \sqrt{2E_0} \, \sqrt{ y^2 + a_0^2}} 
\end{align} and so the solution is
\begin{align} x(t) &= \sinh^{-1}\Big(a_0 \sinh{\left(C_0 + \epsilon\sqrt{2 E_0} \,\, t \right)}\Big)
\end{align}
When $E_0 = \frac{V}{2}$ then $a_0=0$ and the equation simplifies significantly, allowing you to obtain an exponent of $t$ as a solution. This are the so called separatrix solutions. 
You have to double check some of the calculations and especially the signs $\epsilon$. It is also instructive to draw the phase portrait of this system (the dynamics in the $x, \frac{dx}{dt}$-plane, known as the phase space). Sorry I cannot draw it here.    
