Variation of the action $S$ corresponding to a Lagrangian e.g. $L(x(t),\dot{x}(t))$ gives the Euler-Lagrange equations: $$ \frac{\delta S}{\delta x(t)} = 0 \\ \int du \left ( \frac{\delta L}{\delta x(u)} \frac{\delta x(u)}{\delta x(t)} + \frac{\delta L}{\delta \dot{x}(u)} \frac{\delta \dot{x}(u)}{\delta x(t)} \right ) = 0 \\ \int du \left ( \frac{\delta L}{\delta x(u)} \delta(t-u) + \frac{\delta L}{\delta \dot{x}(u)} \frac{\delta \dot{x}(u)}{\delta x(t)} \right ) = 0 \\ \dots \\ \frac{\delta L}{\delta x} - \frac{d}{dt} \frac{\delta L}{\delta \dot{x}} = 0 $$
where in the $\dots$ we perform integration by parts on the right term.
What happens if we vary the action with respect to velocity? Does it make physical sense? What kind of equation would result?
Some attempt: $$ \frac{\delta S}{\delta \dot{x}(t)} = 0 \\ \int du \left ( \frac{\delta L}{\delta x(u)} \frac{\delta x(u)}{\delta \dot{x}(t)} + \frac{\delta L}{\delta \dot{x}(u)} \delta(t-u) \right ) = 0 \\ \int du \frac{\delta L}{\delta x(u)} \frac{\delta x(u)}{\delta \dot{x}(t)} + \frac{\delta L}{\delta \dot{x}(t)} = 0 \\ $$
Now: $$ x(u) = \int du \; \frac{\partial x}{\partial u} \\ \frac{\delta x(u)}{\delta \dot{x}(t)} = \int du \; \delta (u-t) = 1 $$
if true gives $$ \int du \; \frac{\delta L}{\delta x(u)} + \frac{\delta L}{\delta \dot{x}(t)} = 0 $$
Can this be simplified further? Or alternatively, can I relate $\delta S / \delta \dot{x}$ to $\delta S / \delta x$?
Thanks.
