On Second Thought... Potential Density

From the post " $PE$ Ain't Physical Education" dated 22 Jul 2015,

$L=-\cfrac { 1 }{ 2 } \psi _{ c }\cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }-\psi _{ n }$

$\cfrac{d}{dt}\left\{\cfrac{\partial\,L}{\partial\dot{x}}\right\}=\cfrac{d}{dt}\left\{\cfrac{1}{2}m2v\right\}=ma$

since,

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

And

$\cfrac{\partial\,L}{\partial x}=-\cfrac{\partial\,\psi_n}{\partial x}=F_{\rho}$

And we have,

$F_{\rho}=ma$

Wrong!  This is not Newton's Law Of Motion.  Which means $\psi_n$ is not potential but potential density.  In which case,

$\cfrac{\partial\,L}{\partial x}=-\cfrac{\partial}{\partial x}\left\{\int_V\psi dV\right\}=\int_V{-\cfrac{\partial\,\psi}{\partial x}}dV=\int_V{F_{\rho}}dV=F$

where $F$ is the total force on the point particle and is in the direction of lesser $\psi$ for a positive particle.

And we have,

$F=ma$

But, more importantly, should

$\psi_n$  be  $\psi_n-\psi(x)$

instead?  In which case,

$F=F_n-F(x)$

where the resultant force $F$ on the particle is $F_n-F(x)$ in the direction of lesser $\psi$ for a positive particle and reversed for a negative particle.  $F(x)$ being due to the containing particle.

Good  Morning.