It is easy to solve the basic kinamatics equation:
$\frac{dx(t)}{dt} = v(t)$
$\frac{dv(t)}{dt} = a(t)$
$\frac{da(t)}{dt} = j$
and get the discrete version:
$x(t+1) = x(t) + v(t)t + \frac{1}{2}a(t)t^2 + \frac{1}{6}jt^3$
$v(t+1) = v(t) + a(t)t + \frac{1}{2}jt^2$
$a(t+1) = a(t) + jt$
So how about the case using $x$ as independent variable? What is the discrete version of solution for constant jerk?
$\frac{dv(x)}{dx} = \frac{a(x)}{v(x)}$
$\frac{da(x)}{dx} = \frac{j}{v(x)}$
$\frac{dt(x)}{dx} = \frac{1}{v(x)}$
These are the same equations written differently, with the same solution. For example $\frac {dv}{dx}=\frac {dv}{dt}\frac {dt}{dx}=a \cdot \frac 1v$ using the chain rule and the derivative of an inverse. Since the equations are the same, the solution is the same as well under the same assumption of constant jerk.