What is the significance of the Euler-Lagrange Equation in variational calculus?

[Rasalhague]

Wikipedia: Euler Lagrange Equation defines a function

L:[a,b] \times X \times TX \rightarrow \mathbb{R} \enspace\enspace\enspace (1)

such that

(t,q(t),q'(t)) \mapsto L(t,q(t),q'(t)) \enspace\enspace\enspace (2.)

But (2) suggests that the domain of L is simply [a,b], thus:

L:[a,b] \rightarrow \mathbb{R},

with L = G \circ F, and

F:[a,b] \rightarrow [a,b] \times X \times TX \; \bigg| \; F(t) = (t,q(t),q'(t)),

G:[a,b] \times X \times TX \rightarrow \mathbb{R}.

Is that what they mean by L?

 

[disregardthat]

Nevermind, I think I was wrong.

 

[Rasalhague]

Or does their L correspond to the outer composed function, which I called G?

Another guess I had was that maybe there's some conventional shorthand whereby the letter L, in this context, stands for two different functions in the same equation: in some lines G on its own, in other lines G composed with F. If so, can anyone recommend a statement and derivation which avoids this convention, or, failing that, is at least explicit about when L means G o F, and when simply G.

 

[Landau]

Wikipedia: Euler Lagrange Equation defines a function

L:[a,b] \times X \times TX \rightarrow \mathbb{R} \enspace\enspace\enspace (1)

such that

(t,q(t),q'(t)) \mapsto L(t,q(t),q'(t)) \enspace\enspace\enspace (2.)

But (2) suggests that the domain of L is simply [a,b]

Why? Why don't you believe the domain is [a,b] \times X \times TX \rightarrow \mathbb{R} as stated?

 

[homology]

I know I don't believe it. The Lagrangian is a function of the tangent bundle TM. The tangent bundle has local coordinates (q^i, \dot{q}^i) and is 2n dimensional (if M is n-dimensional). If the Lagrangian explicitly depends on time then we might also cross TM with an interval or just the real line, perhaps something like L:TM\times \mathbb{R}^+\to \mathbb{R}

I don't much care for that bit of the article. In particular "the space of possible values of derivatives of functions with values in X" seems sloppy. Tangent spaces are spanned by partial differential operators that can operate on functions (defined on M) but the 'values' end up in \mathbb{R}.

Also I don't see how the EL equations define the Lagrangian, their a consequence of varying the functional. If anything they define the curve (a path on the manifold) making the functional stationary.

 

[Rasalhague]

Why? Why don't you believe the domain is [a,b] \times X \times TX \rightarrow \mathbb{R} as stated?

Because of what my brain was thinking of as "its dependence on t". But I see now I was wrong about that. I think, rather, the page is defining L in terms of the value of its composition with another function, the one I called F. But this definition doesn't make L equal to the composition G \circ F (in my notation above), as I imagined. Rather it's

L(Id_{\mathbb{R}}( \cdot )),q (\cdot),q'(\cdot)) = G \circ F.

Where Id_R is the identity function on R. So L = G, and F(t) = (t,q(t),q'(t)). Is that right?

 

[Rasalhague]

Thanks, homology. Points taken. I'd got myself so confused, I didn't even notice these glitches.

 

[Rasalhague]

Is this better? Let gamma be a parameterization of a curve in M.

\gamma:[t_1,t_2] \rightarrow M,

\gamma ':[t_1,t_2] \rightarrow \left \{ T_{\gamma(t)}M \; | \; t \in [t_1,t_2] \right \},

so gamma prime of t is the tangent vector to the curve in the tangent space associated with the point gamma of t,

L:[t_1,t_2] \times TM \rightarrow \mathbb{R},

this L being the Lagrangian,

K:[t_1,t_2] \rightarrow [t_1,t_2] \times TM \; \bigg| \; K(t) = (t,\gamma(t),\gamma '(t)),

L \circ K : [t_1,t_2] \rightarrow \mathbb{R} \; \bigg| \; L \circ K (t) = L(t,\gamma(t),\gamma ' (t)).

 

[homology]

okay, so you've got the Lagrangian restricted to a path on the tangent bundle. Okay,what's next.

 

[Rasalhague]

Next, I'm trying to understand the section of the article called "Derivation of one-dimensional Euler-Lagrange equation".

I think it begins with a function

J:C^1[t_1,t_2] \rightarrow \mathbb{R}

such that

J(\gamma) = \int_{t_1}^{t_2} L(t,\gamma(t),\gamma'(t)) \; dt = \int_{t_1}^{t_2} L \circ K(t) \; dt.

The goal is to find an extremum, \gamma, of this function J, on the set of inputs having some specific values \gamma (t_1) and \gamma (t_2).

The article seems to begin by defining another function

\tilde{J}:\mathbb{R} \rightarrow \mathbb{R}

such that

\tilde{J}(\varepsilon) = \int_{t_1}^{t_2} L(t,\gamma(t)+\varepsilon \cdot h(t),\gamma'(t)+\varepsilon \cdot h(t)) \; dt.

Where h can be any function to the reals from the interval being integrated over, so long as it fulfills the condition h(t_1) = h(t_2) = 0. Although their notation J_{\varepsilon} (x) doesn't make much sense to me (given that x is what they're calling the "dummy" variable, the variable of integration), the fact that they then differentiate it with respect to \varepsilon suggests to me that its domain is actually \mathbb{R}, or some subset thereof containing 0.

Then I was thinking we could let

\tilde{K}:\mathbb{R} \rightarrow \mathbb{R} \; \bigg| \; \tilde{K}(\varepsilon) = (t,\gamma(t)+\varepsilon \cdot h(t),\gamma'(t)+\varepsilon \cdot h(t)),

(EDIT: Replace "K-tilde : R --> R" with "K-tilde : R --> [t₁,t₂] x TM"; insert prime symbol the final "h".)

so that we can write

\tilde{J} (\varepsilon) = \int_{t_1}^{t_2} L \circ \tilde{K} \; dt

(EDIT: Insert "(epsilon)" after K-tilde.)

and differentiate both sides with respect to \varepsilon. But I guess that can't be quite right; what about the integrand? To make sense (or, at least, not to be trivial), the function being integrated over must depend on t, mustn't it?

Could this be resolved by making K-tilde a function of both t and \varepsilon and then changing what they write as a derivative to a partial derivative when we move it inside the scope of the integration sign? Thus

\frac{\mathrm{d} \tilde{J}}{\mathrm{d} \varepsilon} (0) = \frac{\mathrm{d} }{\mathrm{d} \varepsilon} \bigg|_{\varepsilon = 0} \int_{t_1}^{t_2} L \circ \tilde{K} \; dt = \int_{t_1}^{t_2} \frac{\partial }{\partial \varepsilon} \bigg|_{\varepsilon = 0} L \circ \tilde{K} \; dt.

(EDIT: Insert "(t,epsilon)" after each K-tilde.)

 

[Rasalhague]

Related to this question, in the article first variation, am I right in thinking that, in the first line of equations, the J on the far left is a function whose domain is the underlying set of a function space, and whose codomain is the set of real numbers, while the J on the far right is a different function whose domain and codomain are both the set of real numbers?

 

[homology]

Both functionals have a function space as a domain, though I see what you're saying. I've read a number of introductions to variational calculus and have yet to be fully satisfied. Gelfand has a book that Dover publishes that is pretty good. The calculation there is what I tend to see more in physics whereas the mathematical treatises tend to do it with norms or series expansions but I think you can connect the two methods together since the wiki example is more useful in practice.