Why Can't I Derive the Euler-Lagrange Equation from Local Gradients in F?

Click For Summary
SUMMARY

The discussion centers on the challenges of deriving the Euler-Lagrange equation from local gradients of the functional S = ∫(F(x, dx/dt, t) dt) in the calculus of variations. The user attempts to manipulate the expression dS/dt and equate it to zero, but encounters inconsistencies, particularly with sign changes and discrepancies in results for general problems. A critical insight is that S, being a functional, does not depend on time directly, which invalidates certain differentiation attempts. The user seeks clarification on their reasoning and the validity of their approach.

PREREQUISITES
  • Understanding of calculus of variations and functionals
  • Familiarity with the Euler-Lagrange equation
  • Knowledge of differential equations and perturbation methods
  • Basic concepts of classical mechanics, particularly the principle of least action
NEXT STEPS
  • Study the derivation of the Euler-Lagrange equation in detail
  • Explore the implications of functionals and their properties in calculus of variations
  • Investigate the role of boundary conditions in variational problems
  • Learn about perturbation theory and its applications in physics
USEFUL FOR

Students and researchers in physics and mathematics, particularly those focusing on calculus of variations, classical mechanics, and differential equations.

MadRocketSci2
Messages
47
Reaction score
1
I have a question about calculus of variations that is driving me absolutely nuts right now:

I have followed the standard derivation of differential equations from the extrimization of a functional S = ∫(F(x,dx/dt,t)dt)

By doing some manipulation involving an arbitrary perturbation to your function, you end up with the folowing differential equation which has to be zero

dF/dx - d/dt(dF/d(dx/dt)) = 0;

I can sort of follow this, but I want to be able to derive this from the local properties of the function being integrated.

If the path taken is independent of the interval over which the functional is integrated (and it better be!) then there should be a way to arrive at the differential equation from nothing but the local gradients in F.

I have been trying to do this, and have encountered no end of trouble. My current line of reasoning is as follows:

dS/dt = F; We want to extremize dS/dt at every point t in order to extremize S overall.

d (dS/dt) over a specific interval dt = 0;

d (dS/dt) = dF/dx*dx + dF/d(dx/dt) * d(dx/dt) = 0;

The task then is choosing d(dx/dt) over the interval such that d (dS/dt) = 0;

dx over the interval must be dx/dt*dt or some combination of the derivatives of x in time.

d(dx/dt) is the acceleration d^2x/dt^2 * dt;

divide the time interval out.

dF/dx * xdot = dF/dxdot * xdotdot;

The strange thing about this is that it works for some specific problems, but does not yield equivalent results in general. I am getting the same answers for the principle of least action and the brachistochrone problem that I should be getting. But in other more general problems, the hypothetically minimized functions are slightly off for some reason.

Also, I am getting mysterious sign changes in certain terms (such as in the brachistochrone problem derivation).

Does anyone know where I might be going wrong? Do you see what I am trying to do?
 
Physics news on Phys.org
Right at the start when you're considering dS/dt, this isn't valid because S isn't a function of time. Functions map numbers to numbers but S is a functional i.e. it maps functions to numbers. So basically, you can't differentiate S with respect to time as it doesn't depend on time (your integral should have limits!)

I hope this makes sense, I'm happy to explain further.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K