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Let's say for example that I have two constraints $f(x,\dot{x},y,\dot{y})$ and $g(x,\dot{x},y,\dot{y})$ and a Lagrangian $L(x,\dot{x},y,\dot{y})$. What are the Euler-Lagrange equations of the first kind that describe the equations of motion of the system using Lagrange multipliers?

My attempt at the solution is below, could anyone comment if it (and in general the approach) is correct? May seem trivial but I am having trouble checking it. Specifically, is it correct that the boundary terms vanish when integrating by parts?

Consider first a single constraint $f(x,\dot{x},y,\dot{y}) = const$.

Since: $$ \delta f = \frac{\partial f}{\partial x} \delta x + \frac{\partial f}{\partial \dot{x}} \delta \dot{x} + \frac{\partial f}{\partial y} \delta y + \frac{\partial f}{\partial \dot{y}} \delta \dot{y} = 0 $$

Since this is zero, I can add it to the variation in the action with some Lagrange multiplier $\lambda(t)$:

$$\delta S = \int \left ( \frac{\partial L}{\partial x} \delta x + \frac{\partial L }{\partial \dot{x}}\delta \dot{x} + \lambda (t) \left ( \frac{\partial f}{\partial x}\delta x + \frac{\partial f}{\partial \dot{x}} \delta \dot{x} \right ) \right ) dt + (\text{same integral in y}) = 0$$

And integrate by parts to get

$$ \delta S = \int \left ( \frac{\partial L}{\partial x} - \frac{d}{dt} \frac{\partial L }{\partial \dot{x}} + \lambda (t) \left ( \frac{\partial f}{\partial x} - \frac{d}{dt} \frac{\partial f}{\partial \dot{x}} \right ) \right ) \delta x dt =0 $$ (same for y).

So I would conclude that the EL equations in this case are:

$$\frac{\partial L}{\partial x} - \frac{d}{dt} \frac{\partial L }{\partial \dot{x}} + \lambda (t) \left ( \frac{\partial f}{\partial x} - \frac{d}{dt} \frac{\partial f}{\partial \dot{x}} \right ) = 0$$

(and for y)

Or in general for two constraints $f,g$:

$$ \frac{\partial L}{\partial x} - \frac{d}{dt} \frac{\partial L }{\partial \dot{x}} + \lambda (t) \left ( \frac{\partial f}{\partial x} - \frac{d}{dt} \frac{\partial f}{\partial \dot{x}} \right ) + \mu (t) \left ( \frac{\partial g}{\partial x} - \frac{d}{dt} \frac{\partial g}{\partial \dot{x}} \right ) = 0 $$

(and for y)

Sorry if this seems trivial, can't find it answered anywhere else if this is correct? Thanks

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    $\begingroup$ Why is this called off-topic but there's another question (physics.stackexchange.com/q/137131) in the queue right now asking about convergence of integrals which is not? I suspect it's because the converging integral question framed it as solving $\langle p |1/\hat{r}|p \rangle$ whereas this question got right to the heart of the issue. I think we should try to look past these cosmetic differences and at least be consistent. $\endgroup$ – DanielSank Sep 25 '14 at 1:19
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    $\begingroup$ Ok, thanks. I deleted my answer post, and moved my attempted solution into the question details. No sarcasm or bitterness intended, but I hope someone out there will attempt an answer rather than stressing about its format $\endgroup$ – smörkex Sep 25 '14 at 19:05
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    $\begingroup$ I'm glad you fixed the format. Format matters. Having a consistent format allows potential answerers to digest your question more easily and produce more readily useful answers. $\endgroup$ – DanielSank Sep 25 '14 at 21:17
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    $\begingroup$ Dear people who voted to hold this question: It shows effort to arrive at a solution and asks about a specific physics concept. Please un-hold it. $\endgroup$ – DanielSank Sep 26 '14 at 7:52
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    $\begingroup$ Note that check my work type questions are actually considered off-topic here. This question needs to be formulated in a manner such that it is not asking us to confirm or do the work. $\endgroup$ – Kyle Kanos Sep 26 '14 at 18:11

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