# Derivation Euler-Lagrange Equation in 13 Steps

## Video - Euler-Lagrange Equation

Subscribe on YouTube 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**

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**

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:

**Variation of the functional**

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 \( q \).)

**First derivative of the functional vanishes**

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**

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:

**Total derivative of the Lagrange function with respect to epsilon simplified**

Insert the calculated total derivative back into the functional 5

:

**Functional with calculated total derivative**

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:

**Differentiate the functional and set it to zero**

&~=~ \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

:

**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**

Factor out the integral and \( \eta \):

**Integral of the Euler-Lagrange equation**

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**

If the Euler-Lagrange equation 11

is satisfied for the function \( q \), then the functional \( S[q] \) in 1

becomes stationary.