5
$\begingroup$

In the principle of stationary action, the initial and final points in configuration space are held fixed. This is a boundary value problem. However, this principle leads to the Euler-Lagrange equation which is a differential equation and an initial value problem. The end point of motion does not appear anywhere in the Euler-Lagrange equation. Why is this so? For example, if I want to solve for projectile motion, the Euler-Lagrange equations turn out to be Newton's second law. Now, Newton's second law is an initial value problem. Why do the Euler-Lagrange equations turn out to be an initial value problem when they are derived keeping the boundaries in configuration space fixed?

$\endgroup$
3
$\begingroup$

This is because of a mathematical property of variational calculus that I propose to call 'Jacob's lemma', after the mathematician who first pointed it out. Presumably this mathematical property has been rediscovered independently several times. The 'Jacob' is 'Jacob Bernoulli', brother of Johann Bernoulli.

To put Jacob's lemma in context: some history of variational calculus:
Johann Bernoulli had submitted the 'Brachistochrone problem' to his fellow mathematicians.
(Every introduction to variational calculus in physics mentions the brachistochrone problem, so I'm assuming you are familiar with it.)

Jacob Bernoulli noticed the following:
We have that the solution of the problem is a curve that has over its entire length minimizes the time to travel from initial height to final heigth. If you divide that curve in two sections then each subsection has that property too: to travel from initial height to final height the solution is a minimum. You can continue subdividing into arbitrarily short subsections, down to infinitisimally short subsections; the minimizing property remains .

Thus, Jacob Bernoulli pointed out, it should be possible to find the solution using differential calculus. And it is my understanding that indeed Jacob Bernoulli succeeded in find the solution using differential calculus.

In the Feynman Lectures there is also a lecture titled "The principle of least action"

Quote from that chapter:

Now if the entire integral from $t_1$ to $t_2$ is a minimum, it is also necessary that the integral along the little section from a to b is also a minimum. It can’t be that the part from a to b is a little bit more. Otherwise you could just fiddle with just that piece of the path and make the whole integral a little lower. So every subsection of the path must also be a minimum. And this is true no matter how short the subsection.

(In the lecture Feynman doesn't mention whether he is aware of the every-subsection-is-minimal property through learning or whether he noticed it independently.)


General discussion
The constraint that the solution is an extremum for the entire length of the curve is a very tight constraint.

It is so constraining that that it connects the problem all the way back to differential calculus.

Hamilton's stationary action calls for a solution that is an extremum of the action. It is not so much that this leads to the Euler-Lagrange equation. More precisely, the extremum condition makes the problem accessible to differential calculus.

I recommend the derivation of the Euler-Lagrange equation by Preetum Nakkiran. Preetum Nakkiran points out that since the equation expresses a local condition it should be possible to derive it using local reasoning only.


[LATER EDIT]

Details of the history of the first development of calculus of variations are available in the book "A SOURCE BOOK IN MATHEMATICS, 1200-1800", edited by the mathematician D. J. Struik. (This book is part of a larger series 'SOURCE BOOKS IN THE HISTORY OF THE SCIENCES'.)

The various publications on the brachistochrone problem were in the journal Acta Eruditorum.

The treatment by Jacob Bernoulli: Acta Eruditorum, May 1697, pp. 211-217

Jacob opens with a general discussion of any problem were a curve is sought that is the extremum of some property of that curve.

Diagram

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 AGFDB in a shorter time than along ACEDB, which is contrary to our supposition.


Next Jacob Bernoulli proceeds with a series of steps that leads to an expression that is satisfied by the cycloid. Hence the cycloid curve is the brachistochrone.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks a lot for your answer. $\endgroup$ – Tirthankar Jul 5 at 5:53
1
$\begingroup$
  1. Unlike what OP seems to suggest (v3), Newton's 2nd law and Euler-Lagrange (EL) equations are strictly speaking just differential equations (DEs) without conditions. Rather the context provides the appropriate conditions, such as, e.g., initial conditions (ICs) or boundary conditions (BCs). Together with the DEs, they constitute an initial value problem (IVP) or a boundary value problem (BVP), respectively.

  2. The issues of ICs vs. BCs for the principle of stationary action are already covered in this & this related Phys.SE posts and this related Math.SE post.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

The Euler-Lagrange equation establish the equivalence between the boundary value problem and the initial value problem. Here’s one way to think about it: Suppose we only start only knowing about the principle of least action. That is, if you know $x(t_i)$ and $x(t_f)$, as the initial and final positions of the particle, then you can figure out $x(t)$, the position the particle at every moment in between. Now that you know the path, you can also calculate $v(t)=\frac{dx(t)}{dt}$. The Euler-Lagrange equation now tells you that if you start a particle at $x(t_i)$ with velocity $v(t_i)$ at time $t_i$, the particle will sure pass by $x(t_f)$ at time $t_f$. This is guaranteed by the fact that the Euler-Lagrange equation is second order in time, and hence, require 2 initial values (the position and the velocity) in order to fully pin down a solution.

Finally I wanna address what you said about why the Euler-Lagrange equation is an initial value problem. Not quite, a differential equation is not inherently initial value. I think this is also what Qmechanic alluded to in his answer. You can solve a differential equation with a boundary value condition. You can start with the Euler-Lagrange equations and ask what solution of this equation passes by $x(t_i)$ at $t_i$ and at $x(t_f)$ at $t_f$. Our physical intuition seems more comfortable with the idea of an initial value problem. A law that tells the particle what to do once it start moving. However from a mathematical point of view, both views are equivalent.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks a lot for your answer. $\endgroup$ – Tirthankar Jul 5 at 5:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.