In the following we want to derive the Euler-Lagrange equation, which allows us to set up a system of differential equations for the function \(q(t)\) we are looking for. For the derivation, we assume that the Lagrange function L(t, q(t), \dot{q}(t)) and the boundary values \( q(t_1) ~=~ q_1 \) and \( q(t_2) ~=~ q_2 \) of the searched function \(q\) are known. The Lagrange function can depend on the time \(t\), on the function value \(q(t)\) and on the time derivative \(\dot{q}(t)\) of the function \(q\) at the time \(t\).

The function \( q \) makes the following action functional \( S[q] \) stationary. That is, if we use \( q(t) \) to calculate the action \( S[q] \), \( S[q] \) will give us a value of the action that is either minimum, maximum, or a saddle point:

Action functional as integral of the Lagrange function

Formula anchor$$ \begin{align} S[q] ~=~ \int_{t_1}^{t_2} \text{d}t \, L(t, q, \dot{q}) \end{align} $$

Now we want to consider an infinitesimally small variation \( \delta q \) of \(q\). For this we define the variation as \( \delta q := \epsilon \, \eta \). Here \(\epsilon\) is a very small real number and \(\eta(t)\) is an arbitrary function. It must be defined and differentiable between \(t_1\) and \(t_2\) in every point, so that we - further in the derivation - may differentiate with respect to \( \epsilon \) without problems.

The function \(\eta(t)\) must vanish at the boundary points \(t_1\) and \(t_2\) because the boundary points are fixed:

Variation function at the boundary points vanishes

Formula anchor$$ \begin{align} \eta(t_1) ~=~ \eta(t_2) ~=~ 0 \end{align} $$

In other words: \( \eta(t) \) must coincide with \( q(t) \) at the boundary points \(t_1\) and \(t_2\) so that the function \( q(t) ~+~ \epsilon \eta(t) \) also passes through the boundary points.

The variation of the action functional 1 looks like this:

Here we have simply replaced in 1 the function \(q\) with \(q~+~ \epsilon \, \eta \) and its derivative \(\dot{q}\) with \(\dot{q}~+~ \epsilon \, \dot{\eta} \).

A necessary condition for a local extremum (minimum, maximum, or saddle point of the action functional), is the vanishing of the first derivative of \( S[q ~+~ \epsilon\,\eta] \) with respect to \( \epsilon\). (This condition must be satisfied in any case for the functional \( S[q] \) to become stationary with respect to \( q \).)

The reason we introduced the infinitesimally small parameter \(\epsilon\) is that we can do a Taylor expansion around this point and neglect all terms of higher order than two. (We do not have to neglect the higher order terms. However, this will make the Euler-Lagrange equation have a much more complicated form and at the same time not be of any greater use).

So let us expand the Lagrangian function \( L(t, q ~+~ \epsilon \,\eta, ~ \dot{q} ~+~ \epsilon \, \dot{\eta} ) \) around the point \(\epsilon = 0\) to 1st order in the functional 3:

Action functional with Taylor expansion of the Lagrange function

Here we have abbreviated \( L(t, q ~+~ \epsilon \,\eta, ~ \dot{q} ~+~ \epsilon \, \dot{\eta} )_{~\big|_{~\epsilon ~=~ 0}} \) for compact notation as \(L\). The neglected higher order terms are represented by the symbol \(\mathcal{O}(\epsilon^2)\).

Next, we need to compute the total derivative \( \frac{\text{d} L}{\text{d} \epsilon} \) in Eq. 5. To do this, we must differentiate each argument in \( L(t, q ~+~ \epsilon \,\eta, ~ \dot{q} ~+~ \epsilon \, \dot{\eta} ) \):

Total derivative of the Lagrange function with respect to epsilon

Now you use the necessary condition 4 for stationarity. To do this, we differentiate the functional 8 with respect to \(\epsilon\) and set it equal to zero:

Here, in the second step, the derivative \(\frac{\partial}{\partial \epsilon}\) was pulled into the integral. The derivative \(\frac{\partial L}{\partial \epsilon}\) is omitted, because \(L = L(t, q ~+~ \epsilon \,\eta, ~ \dot{q} ~+~ \epsilon \, \dot{\eta} )_{~\big|_{\epsilon ~=~ 0}} \) is independent of \(\epsilon\) (\(\epsilon\) was set to zero). Moreover, \( \frac{\partial \epsilon}{\partial \epsilon} = 1 \). Remember that the remaining terms do not depend on \(\epsilon\) for the same reason as \(L\).

The derivative of the functional 9 becomes zero exactly when the integrand vanishes. Unfortunately, it still depends on \(\eta\) and \(\eta'\). We can eliminate these terms by partial integration. To do this, we apply partial integration to the second summand in 9:

Partial integration of the integrand in the functional

In this way we transferred the derivative of \(\eta\) to \(\frac{\partial L}{\partial \dot{q}} \). The price we have to pay for this transfer is an additional term in the integrand (in the middle). However, the good thing is that because of the condition \( \eta(t_1) ~=~ \eta(t_2) ~=~ 0 \), this term is eliminated:

Partial integration of the integrand in the functional simplified

Since \( \eta \) may be arbitrary (i.e., also nonzero), the expression in the parenthesis must vanish so that the integral is zero for all \(\eta\). As a result we get:

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