The Euler-Lagrange equation in Dynamics

This article continues the exposition presented in the Energy-Position equation article.

My recommendation is to first absorb the content of that article; that content is the foundation of the information presented here.

The mathematical feature that underlies derivation of the Euler-Lagrange equation is expressed by a lemma that was first stated by Jacob Bernoulli.

When Johann Bernoulli had presented the Brachistochrone problem to the mathematicians of the time Jacob Bernoulli was among the few who solved it. The treatment by Jacob Bernoulli is in the Acta Eruditorum, May 1697, pp. 211-217

Jacob opens his treatment with an observation concerning the fact that the curve that is sought is a minimum.

Picture 8. Image
Jacob's Lemma

Lemma. Let ACEDB be the desired curve along which a heavy point falls from A to B in the shortest time, and let C and D be two points on it as close together as we like. Then the segment of arc CED is among all segments of arc with C and D as end points the segment that a heavy point falling from A traverses in the shortest time. Indeed, if another segment of arc CFD were traversed in a shorter time, then the point would move along ACFDB in a shorter time than along ACEDB, which is contrary to our supposition.

Jacob's lemma generalizes to all cases where the curve that you want to find is an extremum; either a maximum or a minimum. If the evaluation is an extremum for the entire curve, then it is also an extremum for any sub-section of the curve, down to infinitisimally short subsections.

The explicit statement of the brachistochrone problem is in terms of a global property: the amount of time it takes for an object to slide down the curve; it's in the form of evaluating an integral.

Jacob's lemma tells us: a restatement of the problem in the form of a differential equation must exist.

Variational calculus is the systematic way of capitalizing on the content of Jacob's lemma. The general Euler-Lagrange equation takes any problem stated in terms of variational calculus, and transforms the statement into a differential equation.

Class of trial trajectories

Picture 2. Graphlet

Graphlet 2 shows how the trajectory is varied, generating a class of trial trajectories. (The circle in the bottom section is the "knob" of a slider.) The startpoint and endpoint are fixed, in between the trajectory is varied. The set of all possible variations of the trajectory is much larger than this particular class of course, but for the purpose of this demonstration this simplest case is sufficient.

A start point and and end point are defined; the variation is constrained to be zero at the start point and end point.

From differential form to integral form

Dynamics taking place is evaluated by formulating and solving differential equations.

Variational calculus evaluates statements in integral form. The following section presents a demonstration that the differential equation can be restated in a form that involves an integral.

The starting point is the work-energy theorem:

\int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2

The work-energy theorem is only one step away from F=ma; the Work-Energy theorem is derived from F=ma by evaluating the integral of F=ma over distance. Hence taking the derivative with respect to position recovers F=ma.

Simplify to initial position coordinate zero and initial velocity zero.

\int F \ ds = \tfrac{1}{2}mv^2

the derivative with respect to position:

\frac{\int F \ ds}{ds} = \frac{\tfrac{1}{2}mv^2}{ds} \qquad \Longleftrightarrow \qquad F = ma

From here on I will refer to the negative of the integral of force over distance as the 'potential energy'.

\Delta (E_k) = - \Delta (E_p)

For the subsequent manipulations it is practical to move the minus sign inside the parentheses.

\Delta (E_k) = \Delta (-E_p)

The validity of (1.5) extends down to infinitisimal change:

d(E_k) = d(-E_p)

(1.6) is turned into a differential equation by taking the derivative with respect to position:

\frac{d(E_k)}{ds} = \frac{d(-E_p)}{ds}

From here on I will refer to (1.7) as the 'Energy-Position equation', as it evaluates the derivative with respect to position.

We have the following property of integration: it is a linear operation:

\int a \ x^2 \ dx = a \int x^2 \ dx

Therefore the value of the derivative of the energy propagates to the value of the derivative of the integral-of-the-energy.

For variation between two fixed points:
If, from start point to end point:

\frac{d(E_k)}{ds} = \frac{d(-E_p)}{ds}

Then, from start point to end point:

\frac{d(\int E_k dt)}{ds} = \frac{d(\int -E_p dt)}{ds}

The EL-equation for dynamics

In the previous section 'from differential to integral form' it was demonstrated that the dynamics can be stated in a form that involves an integral.

In this section the treatment proceeds in the opposite direction. In this section the starting point is Hamilton's stationary action, and it is demonstrated that working back to a differential equation recovers F=ma, just as (1.3) shows that differentiating the work-energy theorem with respect to the spatial coordinate recovers F=ma.

(This demonstration is modeled after the demonstration in the discussion of Hamilton's stationary action in the Feynman Lectures. Details have been modified; the structure is copied.)

This demonstration is taylored for the context of classical mechanics. For one thing: for time, position and velocity the usual letters 't', 's' and 'v' are used.

(In the general case the functional may include one or more terms with a product of y(x) and and dy/dx, therefore in the general case a derivation must allow for that by specifying partial derivatives. (2.1) states the Lagrangian explicitly in terms of expressions for the respective energies, hence no need for specifying partial derivative.)

Hamilton's stationary action

Recapitulating Hamilton's stationary action:
When the trial trajectory coincides with the true trajectory the derivative of the action with respect to variation is zero.

In the following a series of manipulations is performed for the purpose of arriving at a differential equation. The crucial operation is the following:

  • Removal of the integration
  • Removal of the variation

Removal of the integration and variation is possible because of the following: the derivative with respect to variation can be expressed as a derivative with respect to position because the variation is variation of the position coordinate.


As we know, the Lagrangian 'L' is the sum of kinetic energy and minus potential energy. Here 'V(s)' stands for the potential as a function of the position coordinate 's'.

L = \tfrac{1}{2}m\left(\frac{ds}{dt}\right)^2-V(s)

The kinetic energy is purely a function of ds/dt, and the potential energy V(s) is purely a function of the position coordinate 's'.

Hamilton's action:


The variation is expressed with the symbol η. The underline indicates that the value for the spatial coordinate s is a tentative value.


Restriction of the variation

In order to make progress: all derivations of the Euler-Lagrange equation have the following step in common: the variation is restricted to a zone that is infinitisimally close to the true trajectory. That is, while the full variational assertion is that you can vary by an arbitrary amount, for the purpose of performing the derivation the variation is restricted to an infinitisimally narrow zone.

(Restricting the varation to infinitisimally small variation does not detract from the generality of the derivation. It follows from Jacob's lemma that the derivation can be thought of as being executed in parallel for the set of all infinitisimally short subsections.)

Restricting the variation to an infinitisimally narrow zone allows the following: every term with η that is quadratic or higher can be dropped.

Multiplying out the expression for the kinetic energy term:
(the term with dη that is quadratic will later be dropped.)

\left( \frac{d\underline{s}}{dt} + \frac{d\eta}{dt} \right)^2 =  \left(\frac{d\underline{s}}{dt}\right)^2+2\,\frac{d\underline{s}}{dt}\,\frac{d\eta}{dt}+\left(\frac{d\eta}{dt}\right)^2

For the potential energy a Taylor series expansion is applied:

V(\underline{s}+\eta)=V(\underline{s})+\eta \frac{dV}{d \underline{s}}+ \tfrac{1}{2} \eta \frac{d^2V}{d \underline{s}^2}+(\text{...})

The following expression gives the action plus the variation of the action, with terms quadratic and higher dropped:

S + \delta S  = \int_{t_1}^{t_2}\biggl[\tfrac{1}{2}m\left(\frac{d\underline{s}}{dt}\right)^2+m\,\frac{d\underline{s}}{dt}\,\frac{d\eta}{dt} -V(\underline{s}) - \eta \frac{dV}{d\underline{s}}}\biggr]dt

Narrowing down to just the variation:

\delta S  = \int_{t_1}^{t_2}\biggl[m\,\frac{d\underline{s}}{dt}\,\frac{d\eta}{dt} - \eta \frac{dV}{d\underline{s}}\biggr]dt

So far the following has been accomplished: (2.7) is a linear expression (all terms quadratic and higher have been dropped). That is, (2.7) gives a linear expression for the relation between δS and η .

The following obstacle is still there: (2.7) has both the variation and the derivative of the variation.

The next stage is application of integration by parts to obtain an expression that no longer contains the derivative of the variation.

Integration by parts

The product rule of differentiation:

\frac{(\eta f)}{dt}=\eta\,\frac{df}{dt}+f\,\frac{d\eta}{dt}


\int f\,\frac{d\eta}{dt}\,dt=\eta f-\int\eta\,\frac{df}{dt}\,dt

The term in (2.7) corresponds to the 'f' in (2.9). That is the substitution that is performed to arrive at the following:

\delta S =
- \int_{t_1}^{t_2}\frac{d}{dt}\left(m\,\frac{d\underline{s}}{dt}}\right)  \eta(t)\,dt-
\int_{t_1}^{t_2} \frac{dV}{d\underline{s}}\,\eta(t)\,dt

The variation is constrained to have the following properties: η(t1) = 0 and η(t2) = 0, hence the first term on the right hand side is zero.

Collecting the terms and rearranging:

\delta S=\int_{t_1}^{t_2}\biggl[

Now the equation has arrived at a form where obtaining the two goals listed at the beginning can be executed: removal of the integration and removal of the variation.

We are looking for the point where the derivative of Hamilton's action is zero, hence the point where δS is zero. The function η(t) is not zero, hence (2.11) is satisfied if and only if the part inside the square brackets is zero the entire time.

-m\,\frac{d^2s}{dt^2}-\frac{dV}{ds} = 0

The second term of (2.12) is the derivative of the potential energy with respect to position: the force.

Our expectation has been confirmed: when you work your way back from the integral form to the differential equation you end up recovering F=ma.

The Euler-Lagrange equation

In dynamics the result of deriving the Euler-Lagrange equation is that the Energy-Position equation (1.7) is recovered

The full EL-equation with the Lagrangian:

\frac {\partial(E_k - E_p)}{\partial s} - \frac{d}{dt} \frac {\partial(E_k - E_p)}{\partial v} = 0

For the purpose of this comparison the Lagrangian-in-EL-equation simplifies to the following form:

\frac {d(-E_p)}{ds} - \frac{d}{dt} \frac {(E_k)}{dv} = 0

The term with the potential energy is already identical to the one in the Energy-Position equation (1.7).

Evaluating the term with kinetic energy:

\frac{d}{dt}\frac{d (\tfrac{1}{2}mv^2)}{d v} = m\frac{d}{dt}v = m\frac{dv}{dt} = ma

(1.7) and (3.2) evaluate to the same expression. This completes the verification that the Lagrangian-in-EulerLagrange-equation (3.1) is identical to the Energy-Position equation (1.7)


Of course, this raises the question of why this amount of work is done just to confirm something that you already know.

One consideration is this: while in dynamics the differential equation is known in advance, in general that is not the case. In physics textbooks the standard practice is to derive the general purpose Euler-Lagrange equation.

Derivation of the Work-Energy theorem

The steps in the following derivation are all towards the end goal of obtaining an expression in terms of the velocity v.

Up to (3.7) the expressions follow from the following two definitions:

ds = v \ dt

dv = \frac{dv}{dt}dt

The integral for acceleration from a starting point s0 to a final point s

\int_{s_0}^s a \ ds

Use (3.1) to change the differential from ds to dt. Since the differential is changed the limits change accordingly

\int_{t_0}^t a \ v \ dt

Rearrange the order, and write the acceleration a as dv/dt:

\int_{t_0}^t v \ \frac{dv}{dt} \ dt

Use (3.2) for a second change of differential, the limits change accordingly.

\int_{v_0}^v v \ dv

Putting everything together:

\int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2

The final step is to obtain a dynamics statement by combining (3.7) and F = ma

\int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2

The interactive animations on this page were created with the Javascript Library JSXGraph. JSXGraph is developed at the Lehrstuhl für Mathematik und ihre Didaktik, University of Bayreuth, Germany.

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Last time this page was modified: June 20 2021