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!
