Vary action with respect to velocity

Question

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 $\delta S / \delta x$?

Answer 1

1. First of all, recall that one may vary the velocity $v$ independently of the position $q$ in the Lagrangian $L(q,v,t)$. In fact, the (Lagrangian) canonical momentum is defined as $$\tag{A} p(q,v,t)~:=~\frac{\partial L(q,v,t)}{\partial v}. $$ This is explained further in e.g. this, this, and this Phys.SE posts.

2. Let us define for later convenience $$\tag{B} F(q,v,t)~:=~\frac{\partial L(q,v,t)}{\partial q}, $$ so that the Euler-Lagrange (EL) equation takes the suggestive form
   $$\tag{C} \left. F(q,v,t)\right|_{v=\dot{q}}~\approx~\left.\frac{dp(q,v,t)}{dt} \right|_{v=\dot{q}} \quad,$$ cf. Newton's 2nd law. [Here the $\approx$ symbol means equality modulo eq. of motion. A dot means differentiation wrt. time $t$.]

3. Now OP actually asks about the action (as opposed to the Lagrangian). It does not make sense to vary the velocity profile $v:[t_i,t_f]\to \mathbb{R}$ independently of the position profile $q:[t_i,t_f]\to \mathbb{R}$ in the (off-shell) action functional $$\tag{D} S[q]~=~\left. \int_{t_i}^{t_f} \! dt~ L(q(t),v(t),t)\right|_{v=\dot{q}} \quad. $$

4. However it is possible to make field redefinitions. E.g. decompose the position path $q:[t_i,t_f]\to \mathbb{R}$ in another basis, e.g. by Fourier series/transformation, and vary wrt. the new variables.

5. Let us for the rest of this answer imagine that the system has mixed boundary conditions (BCs) with an initial Essential/Dirichlet BC $$ \tag{E} q(t_i)~=~q_i, $$ and a final Natural BC $$ \tag{F} p(q(t_f),v(t_f),t_f)~=~0. $$ Note that the BC (F) typically constrains the final velocity $v(t_f)$. We leave it to the reader to consider other BCs.

6. One possibility related to OP's question is to define a non-local field redefinition of the form $$\tag{G} q(t) ~=~ I[v;t]~:=~ q_i + \int_{t_i}^{t} \! dt~v(t),$$ where the velocity $v$ is the new dynamical variables, and so that the (off-shell) action functional becomes non-local $$\tag{H} S[v] ~=~\left. \int_{t_i}^{t_f} \! dt~L(q(t),v(t),t)\right|_{q=I[v;\cdot]} \quad. $$ The functional derivative of eq. (G) becomes $$\tag{I} \frac{\delta I[v;t]}{\delta v(t^{\prime})} ~=~\theta(t_i \leq t^{\prime} \leq t \leq t_f)$$ in a hopefully obvious notation. The stationary action principle for the action (H) yields a non-local EL equation $$ \tag{J} \left.\int_{t}^{t_f} \! dt^{\prime}~F(q(t^{\prime}),v(t^{\prime}),t^{\prime}) \right|_{q=I[v;\cdot]} + \left.p(q(t),v(t),t)\right|_{q=I[v;\cdot]}~\approx~0, $$ which is equivalent to eqs. (C) & (F), and it corresponds to OP's last equation.