$\newcommand{\deriv}[2] {\frac{\partial #1}{\partial #2}}$
For any curve $\alpha: I\mapsto C$, where $C$ is the configuration space of Lagrangian Mechanics, we can create
$$ \tilde{\alpha}=(\alpha,\alpha'): I\longmapsto TC $$(is nothing but the prolongation of a section of the trivial bundle $\mathbb{R}\times C \to \mathbb{R}$).
Then, the action for that curve is the integral of the composition
$$ L \circ \tilde{\alpha}: I \longmapsto \mathbb{R} $$where $L:TC\mapsto \mathbb{R}$ is a Lagrangian. That is
$$ S_{\alpha}=\int_{t{0}}^{t_1} L \circ \tilde{\alpha} dt $$with fixed $\tilde{\alpha}(t_i)$ for every $\alpha$.
(in the context of the variational bicomplex, the Lagrangian is a differential form $\lambda=Ldt$ of type $(1,0)$ and the action is $\int_T \tilde{\alpha}^*(\lambda)$, see Anderson_1992 page 9).
The trick to find $\alpha$ such that $S_{\alpha}$ is "critical" is what follows. Consider that the whole $\tilde{\alpha}$ lie inside a local chart (if not, I think that it could be done in a partition of unity, the same that we would have for the integration process). In local coordinates, let $\tilde{\alpha}(t)=(q(t),q'(t))$, with $q(t)\in \mathbb{R}^n$.
We can choose any vector field over $\{q(t)\}_{t\in I}$, for example $V(t)$, and a little real number $\epsilon$ to construct a new curve $\beta_{\epsilon}$ that in local coordinates looks like
$$ (q(t)+\epsilon V(t)) $$This curve gives us a $\tilde{\beta_{\epsilon}}$ (the prolongation of $\beta_{\epsilon}$), which in local coordinates is written
$$ (q(t)+\epsilon V(t), q'(t)+\epsilon V'(t)) $$(By the way, we are deriving using the local chart connection).
If we consider the smooth function
$$ \epsilon \to S_{\beta_{\epsilon}}=\int_{t_0}^{t_1} L(q(t)+\epsilon V(t), q'(t)+\epsilon V'(t)) dt, $$its derivative respect to $\epsilon$ should be null for $\epsilon=0$ in order to $\alpha$ be a extrema of this functional. So, let's go. Consider that the specified local char induces in $TC$ coordinates $(q,v)$, then:
$$ \int_{t_0}^{t_1} \deriv{L}{q} V+\deriv{L}{v} V' dt=0 $$Integrating the second term by parts:
$$ \int_{t_0}^{t_1} \deriv{L}{q} V dt + \left[ \deriv{L}{v} V \right]_{t_0}^{t_1}-\int_{t_0}^{t_1}\frac{d}{dt}(\deriv{L}{v}) V dt=0 $$And since the end points are fixed
$$ \int_{t_0}^{t_1} \deriv{L}{q} V -\frac{d}{dt}(\deriv{L}{v}) V dt=0 $$And since this is valid for every $V$ we get, finally:
$$ \deriv{L}{q}(\alpha(t),\alpha'(t)) =\frac{d}{dt}\left(\deriv{L}{v}(\alpha(t),\alpha'(t))\right) $$or more compactly
$$ \frac{\partial L}{\partial q}-\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial v}=0 $$It is a straightforward calculation to check that Euler-Lagrange equations "behave in a good way" for change of variables.
This is generalized by the Euler operator: see the end of variational derivative#Some facts.
xournal_154
With the notation of the jet bundle $J^1(E)$:
$$ \frac{\partial L}{\partial u^{\alpha}}-\frac{\mathrm{d}}{\mathrm{d} x^{i}} \frac{\partial L}{\partial u_{i}^{\alpha}}=0 $$________________________________________
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Author of the notes: Antonio J. Pan-Collantes
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