Tangential Escape

The new expression for velocity and acceleration on the side of $x=x_c$ is,

$-\cfrac { d^{ 2 }x }{ dt^{ 2 } }=\cfrac { 1 }{ m }  \psi _{ c } +{ \cfrac { u }{ \psi _{ c }+u }  }\cfrac { 1 }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } } (\cfrac { dx }{ dt } )^{ 2 }$

${ \cfrac { u }{ \psi _{ c }+u }  }\cfrac { 1 }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } } (\cfrac { dx }{ dt } )^{ 2 }=-\cfrac { 1 }{ m }  \psi _{ c }-\cfrac { d^{ 2 }x }{ dt^{ 2 } }$

since the particle is decelerating,

$\cfrac { d^{ 2 }x }{ dt^{ 2 } }\lt0$

So,

${ \cfrac { u }{ \psi _{ c }+u }  }\cfrac { 1 }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } } (\cfrac { dx }{ dt } )^{ 2 }=|\cfrac { d^{ 2 }x }{ dt^{ 2 } }|-\cfrac { 1 }{ m }  \psi _{ c }$

The least contribution to $v_{max}$ from $\cfrac { d^{ 2 }x }{ dt^{ 2 } }$ is still when it is zero as suggested by the post "No Solution But Exit Velocity Anyway" dated 14 Jul 2015.  But $v_{max}$ is also complex, in this case.  This implies that when the particle leaves the $\psi(x)$ cloud with $v_{max}$, it does so in a direction perpendicular to the radial line joining it and the center of the non transit particle, ie tangentially.  Note that the graph of $\psi(x)$ is the distribution of $\psi$ along $x$ not the shape of $\psi$.  The shape of $\psi$ is spherical.

And so the particle escapes sideways.  Xium!