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In classical mechanics we often use the relation $a=v(dv/dx)$ to help solve differential equations. I assume when we write $dv/dx$, we really mean $dV/dx$, where $V$ is a function defined so that $V(x(t))=v(t)$. But then $V$ is not really a well defined function, because a particle can pass through a point more than once, with a different velocity each time. I assume the answer has something to do with the implicit function theorem, which I haven't really studied, but I understand that we can locally treat $V$ as a function of $x$. But then why don't we run into issues treating this as a "global" expression?

Edit: I understand the heuristic use of the chain rule: $a=(dv/dx)(dx/dt)$. But it seems to me that the term $dv/dx$ only makes sense "locally." Yet when we use $a=(dv/dx)(dx/dt)$ to solve, say, the equation of motion of the simple pendulum as an elliptic integral, we end up with an expression valid for all $t$, not just "locally". Why does everything work out?

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This doesn't really answer my question, I edited to make it clearer what I am asking. –  yuval Aug 12 '12 at 0:39
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I would guess that implicit in the context of the problem is some vector field $V: \mathbb{R}^n \to \mathbb{R}^n$, (or more likely, some differentiable manifold) that encapsulates the dynamics of the system under study. Then $x$ is a flow curve of this system, ie, $\dot{x}(t) = V(x(t))$. –  copper.hat Aug 12 '12 at 1:44
    
I'm unfamiliar with ODE theory, is it possible to illustrate what this would mean for a specific example? –  yuval Aug 14 '12 at 5:31
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$v$ and $x$ are both well-defined functions of $t$, which is the independent variable. So when you write $a=v\frac {dv}{dx}$ everything on the right is a function of $t$, as is $a$. The fact that $v$ could be the same at different times is not a problem.

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How do we take the derivative of $v$ with respect to $x$ if $v$ is defined as a function of $t$? Would be we use the chain rule treating $t$ as a function of $x$? –  yuval Aug 12 '12 at 0:05
    
@YuvalGannot: exactly. In fact $a=\frac {dv}{dt}=\frac {dv}{dx} \frac {dx}{dt}=v\frac {dv}{dx}$. Recurrences of $v$ or $x$ are not a problem because the derivative is local, so we consider a small enough neighborhood to avoid them. –  Ross Millikan Aug 12 '12 at 0:12
    
You can if you can express it as a function of $x$. Say we have the mass and spring problem, so $a=-x$. We know the solution is $x=\cos t$, so $v= \sin t=\sqrt {1-x^2}, \frac {dv}{dx}=\frac {-x}{\sqrt {1-x^2}}$ and $v\frac {dv}{dx}=-x$ as it should. –  Ross Millikan Aug 12 '12 at 0:46
    
Wouldn't $v=(+/-)sqrt(1-x^2)$ (depending on whether the mass is moving in or out)? Of course in this case the velocity is well defined up to a sign, and $dv/dx$ has the same sign, so the term $v(dv/dx)$ ends up being well defined for all $t$. But in general why does this method work? –  yuval Aug 12 '12 at 0:54
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