Derivation: Euler-Lagrange Equation in 13 Steps

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:

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

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.

A small variation ("perturbation") $\epsilon \, \eta(t)$ of the path $q(t)$ between two fixed points.

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

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

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:

$$ \begin{align} J[q ~+~ \epsilon \eta] ~=~ \int_{t_1}^{t_2} \text{d}t \, L(t, q ~+~ \epsilon \,\eta, ~ \dot{q} ~+~ \epsilon \, \dot{\eta} ) \end{align} $$

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 $.)

$$ \begin{align} \frac{\partial J[q ~+~ \epsilon \eta] }{\partial \epsilon}_{~\big|_{~\epsilon ~=~ 0}} ~\stackrel{!}{=}~ 0 \end{align} $$

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:

$$ \begin{align} S[q ~+~ \epsilon \, \eta] ~=~ \int_{t_1}^{t_2} \text{d}t \, \left( L ~+~ \epsilon \, \frac{\text{d} L}{\text{d} \epsilon} ~+~ \mathcal{O}(\epsilon^2) \right) \end{align} $$

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} ) $:

$$ \begin{align} \frac{\partial L}{\partial \epsilon} ~=~ \frac{\partial L}{\partial q}\, \frac{\text{d} (q~+~\epsilon \eta)}{\text{d} \epsilon} ~+~ \frac{\partial L}{\partial \dot{q}} \, \frac{\text{d} (\dot{q}~+~\epsilon \dot{\eta})}{\text{d} \epsilon} ~+~ \frac{\partial L}{\partial t} \, \frac{\text{d} t}{\text{d} \epsilon} \end{align} $$

Here the derivatives are $\frac{\text{d} (q~+~\epsilon \eta)}{\text{d} \epsilon} = \eta$ and $\frac{\text{d} (\dot{q}~+~\epsilon \dot{\eta})}{\text{d} \epsilon} = \dot{\eta}$ and $\frac{\text{d} t}{\text{d} \epsilon} = 0 $. Thus 6 becomes:

$$ \begin{align} \frac{\partial L}{\partial \epsilon} ~=~ \frac{\partial L}{\partial q} \, \eta ~+~ \frac{\partial L}{\partial \dot{q}} \, \dot{\eta} \end{align} $$

Insert the calculated total derivative back into the functional 5:

$$ \begin{align} S[q ~+~ \epsilon \eta] ~=~ \int_{t_1}^{t_2} \text{d}t \, \left( L ~+~ \epsilon \, \frac{\partial L}{\partial q} \, \eta ~+~ \epsilon \, \frac{\partial L}{\partial \dot{q}} \, \dot{\eta} \right) \end{align} $$

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:

$$ \begin{align} \frac{\partial S[y ~+~ \epsilon \, \eta]}{\partial \epsilon}_{~\big|_{~\epsilon ~=~ 0}}
&~=~ \frac{\partial}{\partial \epsilon} \int_{t_1}^{t_2} \text{d}t \, \left( L ~+~ \epsilon \, \frac{\partial L}{\partial q} \, \eta ~+~ \epsilon \, \frac{\partial L}{\partial \dot{q}} \, \dot{\eta} \right) \\\\
&~=~ \int_{t_1}^{t_2} \text{d}t \, \left( \frac{\partial L}{\partial \epsilon} ~+~ \frac{\partial \epsilon}{\partial \epsilon} \, \frac{\partial L}{\partial q} \, \eta ~+~ \frac{\partial \epsilon}{\partial \epsilon} \, \frac{\partial L}{\partial \dot{q}} \, \dot{\eta} \right) \\\\
&~=~ \int_{t_1}^{t_2} \text{d}t \, \left( \frac{\partial L}{\partial q} \, \eta ~+~ \frac{\partial L}{\partial \dot{q}} \, \dot{\eta} \right) \\\\ &~=~ 0 \end{align} $$

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:

$$ \begin{align} \int_{t_1}^{t_2} \text{d}t \, \frac{\partial L}{\partial q} \, \eta ~+~ \left[ \frac{\partial L}{\partial \dot{q}} \, \eta \right]_{t_1}^{t_2} ~-~ \int_{t_1}^{t_2} \text{d}t \, \frac{\text{d}}{\text{d}t} \, \frac{\partial L}{\partial \dot{q}} \, \eta ~=~ 0 \end{align} $$

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:

$$ \begin{align} \int_{t_1}^{t_2} \text{d}t \, \frac{\partial L}{\partial q} \, \eta ~-~ \int_{t_1}^{t_2} \text{d}t \, \frac{\text{d}}{\text{d}t} \, \frac{\partial L}{\partial \dot{q}} \, \eta ~=~ 0 \end{align} $$

Factor out the integral and $ \eta $:

$$ \begin{align} \int_{t_1}^{t_2} \text{d}t \left(\frac{\partial L}{\partial q}~-~ \frac{\text{d}}{\text{d}t}\frac{\partial L}{\partial \dot{q}} \right)\eta ~=~ 0 \end{align} $$

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:

Euler-Lagrange equation

$$ \begin{align} \frac{\partial L}{\partial q}~-~ \frac{\text{d}}{\text{d}t}\frac{\partial L}{\partial \dot{q}} ~=~ 0 \end{align} $$

If the Euler-Lagrange equation 11 is satisfied for the function $ q $, then the functional $ S[q] $ in 1 becomes stationary.