# Calculus of variations

enricfemi
for Newton's equation, we have Lagrange function which can give the solution while its variation equates 0.

however, what about the situation for a general differential equation.
is this method can deal with such general situation? and how can we find out the corresponding "Lagrange function".

## Answers and Replies

Homework Helper
Gold Member
Dearly Missed
It is not known if every diff.eq can be derived in a variational manner, but I am not aware of that it has been proven to be impossible in certain cases.

Dr.D
There is no reason to expect a variational basis for every differential equation. The equations of motion, equivalent to Newton's equations, come about out of Hamilton's Principle which is a statement of the laws of physics. There is no known comparable statement governing, for example, population dynamics, but differential equations can be written for population dynamics.

There is no reason to expect the generalization to extend nearly so far as you propose.

maze
Differential equations can describe such a vast array of phenomena. If this were true, it would be a very very powerful statement.

I'm really out of my depth on most math questions, so I try to avoid them. But I do know something about using calculus of variations for the Schrödinger equation via the Ritz variational method, so I'll say what I know coming from that standpoint:

Obviously the criteria for being able to use the method depends on having an inequality such that some parameter is minimized for the solution you're looking for, and second, that you can formulate a Lagrange function for it. And third, which is probably the big caveat, the practicality of doing so: That the corresponding variational problem is, in fact, simpler than solving the problem by other methods. (As for proving the existence of a corresponding variational problem for any D.E., I have no idea. My hunch is: No, probably not.)

In the case of the S.E. calculating ground-state energies is naturally an important subset of problems, minimizing <E> while varying Ψ. And the most-used quantum chemical methods for ground states all do this. But if you want excited states, then there's no easy Lagrangian or parameter unless you make approximations, which get worse with increasingly excited states.

So in practice, the variational method is almost always used for ground state energies and almost never used for excited states. Which goes to show how the usefulness of the method can vary greatly for same problem, with different conditions.

enricfemi
it would be very disappointing that calculus of variations turn into a special method, since I would rather considered it as a kind of philosophy.

Eidos
I always thought that the principle of extremal action could be used to cast any physical problem into D.E. form.

What made me think that was that Noether's Theorem was such a fundamental result. Briefly it states that the reason there are conservations of energy, momentum, angular momentum are due to symmetries in the Lagrangian.

Since with any physical problem we are dealing with energy, I thought that the Lagrangian could be found for the problem at hand. Whether we can solve it or not is another matter entirely.

Does anyone know of a counter example, a physical DE that cannot be derived using Lagrangians?

Count Iblis
Why not reason as follows:

The differential equation has some solutions. We interpret these solutions as possible trajectories of a particle in an abstract space. I then take the integral curves to define a nonlinear coordinate transformation, so that in the new coordinates the world lines of the particle is that of a "free particle".

Then the Lagrangian of a free particle will correctly describe the motion in the new coordianate system. If you then transform the Lagrangian back to the old coordinates, you get a Lagrangian that will yield the differential equations.

Dr.D
Eidos, it gets pretty messy when the coordinates are nonholonomic.