Derivation of the Euler-Lagrange-Equation

We would like to find a condition for the Lagrange function L, so that its integral, the action S, becomes maximal or minimal.

For that, we change the coordinate q(t) by a little variation \eta(t), although infinitesimal. Additionally, \eta(t_1) = \eta(t_2) = 0 has to hold. The integral of the Lagrange function becomes:

S = \int_{t_1}^{t_2} L\left(q(t) + \epsilon \eta(t), \dot q(t) + \epsilon
\dot \eta(t), t \right) \, \mathrm dt

This should be extremal with respect to \epsilon. So we need to differentiate with respect to that and set equal to 0:

\frac{\mathrm d}{\mathrm d \epsilon} \int_{t_1}^{t_2} L\left(q(t) +
\epsilon \eta(t), \dot q(t) + \epsilon \dot \eta(t), t \right) \, \mathrm
dt = 0

For this total derivative, the partial derivatives of L and q(t) + \epsilon
\eta(t) and \dot q(t) + \epsilon \dot \eta(t) have to be found.

\int_{t_1}^{t_2} \left( \frac{\partial L}{\partial q} \eta + \frac{\partial
L}{\partial \dot q} \dot \eta \right) \, \mathrm dt = 0

For the second summand, we use partial integration:

\int_{t_1}^{t_2} \frac{\partial L}{\partial \dot q} \dot \eta(t) \, \mathrm
dt = \underbrace{\left[ \frac{\partial L}{\partial \dot q}
\eta\right]_{t_1}^{t_2}}_{=0} - \int_{t_1}^{t_2} \frac{\mathrm d}{\mathrm
dt} \frac{\partial L}{\partial \dot q} \eta(t) \, \mathrm dt

The middle term is equal to 0 since \eta(0) vanished on the boundary points. Therefore, the last term remains.

\int_{t_1}^{t_2} \left( \frac{\partial L}{\partial q} \eta(t) -
\frac{\mathrm d}{\mathrm d t} \frac{\partial L}{\partial \dot q} \eta(t)
\right) \, \mathrm dt = 0

Now we can factor out that \eta(t). The integral vanished for all variations \eta(t) iff the parentheses vanishes.

\int_{t_1}^{t_2} \left( \frac{\partial L}{\partial q} - \frac{\mathrm
d}{\mathrm d t} \frac{\partial L}{\partial \dot q} \right) \eta(t) \,
\mathrm dt = 0

We yield the Euler-Lagrange-Equation:

\frac{\partial L}{\partial q} - \frac{\mathrm d}{\mathrm d t}
\frac{\partial L}{\partial \dot q} = 0