# Euler Lagrange ODE

1. Nov 28, 2011

### muzialis

Hello there,

I am interested in the following matter.

Given an ODE, can one always find a functional F such that the ODE is its Euler Lagrange equation?

I am thinking at the following concrete case.

I have the ODE y' = a y

I would like a functional given by the intergral over a line of a function F = F (y, y'), such that the E.L. equation, giving necessary conditions for stationarity of the functional, is my ODE.

Any help is really welcome

Thanks

Muzialis

2. Nov 28, 2011

### hunt_mat

What have you tried?

3. Nov 29, 2011

### muzialis

So far I have tried the trivial.

Given a functional as the integral of F (y' (x), y(x)) the Euler Lagrange equation reads (apologies for not being able to use Latex in this forum, i will learn, i use the notation F,y to indicate derivation with respect to y)

F,y'y' * y''= F,y - F,y'y*y'

the idea is to get to y'' = a*y', a being a constant (giving the first order ODE i am after)

through looking for an F such that

F,y'y' = constant1
F,y = constant2 * y'
F,y'y = constant3

or

F,y'y' = 0
F,y = constant2 * y
F,y'y = constant3

but I have not succeded.

Many thanks again

4. Nov 29, 2011

### hunt_mat

If you try $L=y\dot{y}$ then you get the differential equation
$$y'=y$$
You might like to play around with this functional to see what you can get if you generalise the functional.

5. Nov 29, 2011

### Mute

Nope. The Lagrangian $L = y \dot{y} = (1/2)d(y^2)/dt$, i.e., it's a total derivative, and hence will not give you a differential equation. (The Euler-Lagrange equation applied to this Lagrangian gives you 0 = 0).

To the OP: You will find that you cannot easily find a Lagrangian for the DE $\dot{y} = ay$. If you think back to the sorts of Lagrangians that you have likely seen (in cartesian coordinates), I think you will realize that they all have second time derivatives, but no first time derivative. A first time derivative introduces dissipation into the system - consider, for example, the DE for a damped harmonic oscillator:

$$\ddot{y} + 2\gamma \dot{y} + \omega^2 y = 0$$

It's the first derivative term that leads to the damping. This means that the Hamiltonian (i.e., energy) is not conserved, and it is not so simple to write down a Lagrangian for a DE with a first time derivative.

One way to write down a Lagrangian that can give you this first time derivative is to introduce a time-dependent factor into your Lagrangian - i.e., your Lagrangian must explicitly depend on time, as this will also cause energy to not be conserved. A clever choice of the factor you pick will result in it canceling out when you write down the resulting DE.

Another possible way to get a first time derivative without explicit time dependence would be to couple the Lagrangian to another Lagrangian, and integrate out the degrees of freedom of the other Lagrangian to get an effective Lagrangian for your system. This method is going to be harder, though. For an application in quantum mechanics, see the Caldeira-Leggett model of quantum dissipation.

Last edited: Nov 30, 2011
6. Nov 30, 2011

### muzialis

Huntmat, thanks so much for your help.

Mute, thanks for your comment. You point the finger exactly on the type of energitical cosiderations I was trying to underpin.

What I cannot understand in physical terms is the following.

Take an elastic bar, in 1D. The energy functional depends on the strain, by minimizing it under let' s say prescribed displacement you obtain the condition the strain has to be uniform, i.e. you recover the fact the stress has to be continuous. Hence yo recover stress equilibrium via a variational theorem.

You can show variational theroems exists for viscoelasticity too. Then intution tells us that any solution will have constant strain in space, and varying in time. Hence the bar state can be described via the position varying in time of its end. One knows that said position is caharacterized by the ODe I mention, so I hoped to find a functional related to it.

Many thanks again to all for your consideration