Reach, Touch, A...

If we allow for the second constant of integration in $F$, then

$\psi (x)=-ln(cosh(x-\cfrac { 1 }{ 2 } x_{ a }))+Ax+\psi _{ max }$

$\psi_{max}=ln(cosh(\cfrac{1}{2}x_a))$

From,

$F=\psi_{n}-\psi(x)=m\cfrac{dv}{dt}=m\cfrac{d^2x}{dt^2}$

$\cfrac { d^{ 2 }x }{ dt^{ 2 } } +\cfrac { 1 }{ m } \left\{ -ln(cosh(x-\cfrac { 1 }{ 2 } x_{ a }))+Ax+\psi _{ max } \right\} =\cfrac { 1 }{ m } \psi _{ n }$

$\cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ m } \left\{ \psi _{ n }-\psi _{ max }+ln(cosh(x-\cfrac { 1 }{ 2 } x_{ a }))-Ax \right\}$

$\psi _{ c }=\psi _{ n }-\psi _{ max }$

$\cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ m } \left\{ \psi _{ c }+ln(cosh(x-\cfrac { 1 }{ 2 } x_{ a }))-Ax \right\}$

$\cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ 2m } \psi _{ c }t^{ 2 }+\cfrac { 1 }{ m } \iint { \left\{ ln(cosh(x-\cfrac { 1 }{ 2 } x_{ a }))-Ax \right\}  } dtdt$

Let,

$u=ln(cosh(x-\cfrac { 1 }{ 2 } x_{ a }))-Ax$

$\cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ 2m } \psi _{ c }t^{ 2 }+\cfrac { 1 }{ m } \iint { u } \cfrac { 1 }{ \cfrac { d^{ 2 }x }{ dt^{ 2 } }  } \cfrac { d^{ 2 }x }{ dt^{ 2 } } dtdt$

$\cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ 2m } \psi _{ c }t^{ 2 }+\iint { \cfrac { u }{ \psi _{ { c } }+u }  } d(dx)$

$\cfrac { du }{ dx } =tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A$

$\cfrac { dx }{ du } =\cfrac { 1 }{ tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A }$

$d(dx)=\cfrac { 1 }{ tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A } dudu$

$\cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ 2m } \psi _{ c }t^{ 2 }+\iint { \cfrac { u }{ \psi _{ { c } }+u } \cfrac { 1 }{ tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A }  } dudu$

$\cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ 2m } \psi _{ c }t^{ 2 }+\iint { \cfrac { u }{ \psi _{ { c } }+u } \cfrac { 1 }{ tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A }  } \cfrac { du }{ dt } dudt$

$\cfrac { dx }{ dt } =\cfrac { 1 }{ m } \psi _{ c }t+\int { \cfrac { u }{ \psi _{ { c } }+u } \cfrac { 1 }{ tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A }  } \cfrac { du }{ dt } du$

$\cfrac { d(dx) }{ dudt } =\cfrac { 1 }{ m } \psi _{ c }\cfrac { dt }{ du } +{ \cfrac { u }{ \psi _{ { c } }+u } \cfrac { 1 }{ tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A }  }\cfrac { du }{ dt }$

$\cfrac { d(dx) }{ dudt } \cfrac { du }{ dt } =\cfrac { 1 }{ m } \psi _{ c }\cfrac { dt }{ du } \cfrac { du }{ dt } +{ \cfrac { u }{ \psi _{ c }+u } \cfrac { 1 }{ tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A }  }(\cfrac { du }{ dt } )^{ 2 }$

$\cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ m } \psi _{ c }+{ \cfrac { u }{ \psi _{  c  }+u } \cfrac { 1 }{ tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A }  }(\cfrac { du }{ dx } \cfrac { dx }{ dt } )^{ 2 }$

Finally,

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

where,

$u=ln(cosh(x-\cfrac { 1 }{ 2 } x_{ a }))-Ax$

$\psi _{ c }=\psi _{ n }-\psi _{ max }$

Compare the above with,

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

Consider,

$\cfrac { 1 }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } } =\cfrac { 1 }{ cosh(x-\cfrac { 1 }{ 2 } x_{ a })sinh(x-\cfrac { 1 }{ 2 } x_{ a }) } ={ 2 }{ csch(2x- x_{ a }) }$

So,

$\cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ m } \psi _{ c }-{ \cfrac { u }{ \psi _{ c }+u }  } { 2 }{ csch(2x- x_{ a }) } (\cfrac { dx }{ dt } )^{ 2 }$

where $u=ln(cosh(x-\cfrac { 1 }{ 2 } x_{ a }))$

for the case where the second integration constant was not considered, (ie $A=0$).  Both the terms, in the respective expressions,

${ 2 }{ csch(2x- x_{ a }) }$

$tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A$

are positive around $x_a$.

The conclusions drawn previously for $A=0$ are applicable for $0\le A\lt1$.

What would cause $A\ne0$?  The presence of another particle at a distance $x\gt2x_z$ or $x\gt x_a$ where $x_a$ is the second $\psi=0$ when $A=0$.

In the presence of another particle further away, $\psi$ from the first stretches and envelopes the second particle.  The second particle would behave in a reciprocal way, but in our analysis it was considered a point particle.

Interaction of the particles' $\psi$ is manifested in the constant $A$.