newtonian mechanics - When can one write $a=v cdot dv/dx$?

Referring to unidimensional motion, it is obvious that it doesn't always make sense to write the speed as a function of position. Seems to me that this is a necessary condition to derive formulas like:

$$v^2=v_0 ^2 +2\int_{x_0}^{x}a\cdot dx$$

In fact, in the first step of the demonstration (the one I saw, but I think that this step is crucial) it's required to write $a=dv/dt=(dv/dx)(dx/dt)$, that doesn't make sense if $v$ isn't a function of $x$.

When can one rigorously write $v=v(x)$?

Answer

This is going to be essentially the same in content as Jerry Schirmer's response, but I thought you might like to hear it in more mathematical terms. The velocity function $v$ is defined as $$v(t) = \dot{ x}(t)$$ Let's take the domain of the position function to be the open interval $(t_1, t_2)$ and suppose that it has the property that given any point $x_0$ in the range of $x$, there is a unique point $t_0$ in its domain $(t_1, t_2)$ such that $x(t_0) = x_0$. Then there exists a function $x^{-1}$ (the inverse of $x$) defined on the range of $x$ satisfying $$x^{-1}(x(t)) = t$$ Now we define a function $\bar v$ on the range of $x$ by $$\bar v(x) = v(x^{-1}(x))$$ It is common to abuse notation here and use $v$ in place of $\bar v$ for this function, but let's keep things notationaly rigorous. Then on one hand the chain rule gives $$\frac{d}{dt}\bar v(x(t)) = \frac{d\bar v}{dx}(x(t))\,\dot x(t) = \frac{d\bar v}{dx}(x(t))\,v(t)$$ While on the other hand we use the definition of $\bar v$ to write $$\frac{d}{dt}\bar v(x(t)) = \frac{d}{dt} v(x^{-1}(x(t))) = \frac{dv}{dt}(t) = a(t)$$ and combining these observations gives the identity you wanted $$a(t) = \frac{d\bar v}{dx}(x(t))\,v(t)$$ Notice that if we indulge in the usual abuse of notation, then we can simply write this as $$a = v \frac{dv}{dx}$$