These expressions satisfy Maxwell's equations, when
${\frac {d}{dt}}=v{\frac {\partial }{\partial x}}$
and lead to equation
$\left(1A^{2}v^{2}\right){\frac {\partial ^{2}U}{\partial x^{2}}}+{\frac {\partial ^{2}U}{\partial y^{2}}}+{\frac {\partial ^{2}U}{\partial z^{2}}}=0.$
But if $v$ is depending on $t$, we have
${\frac {d}{dt}}={\frac {\partial }{\partial t}}v{\frac {\partial }{\partial x}}.$
If our value shall generally hold for $x$, it also must be
${\frac {\partial X}{\partial t}}$ small against $v{\frac {\partial X}{\partial x}}$.
Now it is
${\frac {\partial X}{\partial t}}={\frac {\partial ^{2}}{\partial x\partial t}}U\left(1A^{2}v^{2}\right),$
thus it must be
${\frac {\partial }{\partial t}}\left[{\frac {\partial U}{\partial x}}\left(1A^{2}v^{2}\right)\right]$ small against $v{\frac {\partial ^{2}U}{\partial x^{2}}}\left(1A^{2}v^{2}\right)$,
or
$A^{2}x{\frac {\partial v}{\partial t}}$ small against $1A^{2}v^{2}$,
Also the values of $Y,Z$ and $M,N$, give
$\left[2x^{2}\left(1A^{2}v^{2}\right)\varrho ^{2}\right]A^{2}{\frac {\partial v}{\partial t}}$ is small against $3x\left(1A^{2}v^{2}\right)$
and

$\left(1A^{2}v^{2}\right)\left[\left(x^{2}+\left(1A^{2}v^{2}\right)\varrho ^{2}\right)\right]{\frac {\partial v}{\partial t}}$
$\left[2x^{2}\left(1A^{2}v^{2}\right)\varrho ^{2}\right]A^{2}v^{2}{\frac {\partial v}{\partial t}}$

must be small against $3x\left(1A^{2}v^{2}\right)v^{2}$.
This condition is fulfilled, when the dimensions of the space, in which the energy comes essentially into consideration, are sufficiently small. Because the terms to be neglected all contain the linear dimensions in a higher power. Though $dv/dt$ may not be too great and the absolute velocity $v$ not too small.
When this neglect is allowed, then we can put for the change of kinetic energy
${\frac {d}{dt}}\left({\frac {m}{2}}v^{2}\right)=mv{\frac {dv}{dt}}=K{\frac {dr}{dt}}dt=m{\frac {dr}{dt}}{\frac {d^{2}r}{dt^{2}}},$