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I am trying to understand the derivation of the Euler-Lagrange equation. I drew a graph below.

http://i66.tinypic.com/4v1ilg.png

So, according to the graph,

$$ \int_{t_1}^{t_2} L(x+\delta{x},\dot{x}+\delta\dot{x}\,t) dt - \int_{t_1}^{t_2} L(x,\dot{x},t) dt = \delta{S}$$

That is, we take the difference between the integral of the true path and the integral of the variated path, where S is the action defined by $\int_{t_1}^{t_2} L(x,\dot{x},t) dt$ and $\delta{S}$ is the variation in action (in the graph it is the distance between the blue line and the red line at any t value). Where do I proceed from there? I don't understand this line in the derivation:

$$ \int_{t_1}^{t_2} L(x,\dot{x},t) dt + \frac{d{L}}{d{x}}\delta{x} + \frac{d{L}}{d{\dot{x}}}\delta{\dot{x}}+O((\delta{x})^2)- L(x,\dot{x},t) dt = \delta{S}$$

(Source: http://wiki.math.toronto.edu/TorontoMathWiki/index.php/Euler-Lagrange_Equation)

Namely, what is the function O? Where are the partial derivatives coming from? The only thing I know to do is to combine the two integrals because the times and the integration variables are the same. How do I get all the mess to the left of $L(x,\dot{x},t)$ in the equation above?

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The partial derivatives appear as a result of the chain rule for partial derivatives; this takes place implicitly when the variation is applied to the Lagrangian functional. The big O notation says that there are higher order terms which can be ignored in the variational limit.

The web page shows a couple of ways to obtain the result. Euler used purely geometric methods, and Lagrange invented the variational technique with the delta process.

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  • $\begingroup$ The partial derivatives come from the Taylor series $\endgroup$ – Oswald Apr 5 '16 at 4:02
  • $\begingroup$ Although $x, \dot x $ are dependent, they *can be treated as independent variables *to find $ \delta S $ $\endgroup$ – Narasimham Jan 16 at 21:12
  • $\begingroup$ @Narasimham: x,x˙ are linearly independent. $\endgroup$ – Peter Diehr Jan 16 at 23:27
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    $\begingroup$ I meant to say that though they are differentially related, they are linearly independent and on that base calculus of variation is built. Any takers? $\endgroup$ – Narasimham Jan 18 at 15:07
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Namely, what is the function O? Where are the partial derivatives coming from? The only thing I know to do is to combine the two integrals because the times and the integration variables are the same. How do I get all the mess to the left of L(x,x˙,t) in the equation above?

These are good questions. The usual derivation is a heuristic derivation and typically a rigorous derivation of the EL equations in the calculus of variations, at the level of rigour, that a first year undergraduate mathematics degree will approach the calculus of one real variable appears to be fairly rare in the literature.

Abraham & Marsden, in their book, Foundations of mechanics, take a geometric approach, which allows them to deduce the EL equations in a rigorous manner, admitting that an analytic derivation is difficult.

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It might help to think about the definition of a derivative,

$$ \frac{df}{dx} = \lim_{\delta x\rightarrow 0} \frac{f(x+\delta x)-f(x)}{\delta x}. $$

If you rearrange this equation you'll see that

$$ f(x+\delta x) \approx f(x) + \frac{df}{dx} \delta x, $$ when $\delta x$ is small. As $\delta x$ gets smaller the approximation becomes more exact. The term $\mathcal{O}((\delta x)^2)$ represents some quantity that scales like $(\delta x)^2$ as $\delta x$ goes towards 0. So another way to write the above the equation might be $$ f(x+\delta x) =f(x) + \frac{df}{dx} \delta x + \mathcal{O}((\delta x)^2). $$

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  • $\begingroup$ This is good for intuition, but the EL equations uses functional derivatives. $\endgroup$ – InertialObserver Mar 19 at 18:03
  • $\begingroup$ The way I've always seen it, the way to rigorously define these functional derivatives is in terms of ordinary derivatives. I.e. you define a new path as $x(t)$ by $x(t) = x_0(t) + \epsilon \delta x(t)$, where $x_0(t)$ is the path that minimizes the action and $\delta x(t)$ is an arbitrary but fixed function (that vanishes at the end points) and $\epsilon$ is a number. Then you plug in $x(t)$ into Lagrangian, consider the action as a function of $\epsilon$, and take an ordinary derivative with respect to $\epsilon$. $\endgroup$ – Alex Mar 19 at 22:39
  • $\begingroup$ No, you're right, but you might want to mention that in your answer. $\endgroup$ – InertialObserver Mar 19 at 22:39
  • $\begingroup$ ... or equivalently to doing the derivative wrt $\epsilon$: at each value of $t$ in the integrand you expand the integrand to first order in $\epsilon$ and demand that the variation in the action is 0 (at first order in $epsilon$) for every possible perturbation $\delta x(t)$. $\endgroup$ – Alex Mar 19 at 22:42

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