Product rule for functional derivatives?

[Gregor Scholten]

Hi alltogether,

Consider Lagrange formalism, e.g. Lagrangian mechanics. We have a
Lagrangian function L(q, dq/dt) where q is a generalized coordinate and
dq/dt its time derivative. Now, according to Hamiltonian principle, L
ist stationary with respect to variations \delta q of the trajectory q(t):

\delta L(q, dq/dt) / \delta q = 0

what results in

\partial L / \partial q - d/dt \partial L / \partial (dq/dt) = 0

\delta L(q, dq/dt) / \delta q is called the functional derivative or
variational derivative.

Now, what I'm asking myself is, whether there is, in analogy to
functions' derivatives like df/dx or \partial f / \partial x, a procuct
rule for such functional derivative? E.g., if we have

L(q, dq/dt) = A(q, dq/dt) B(q, dq/dt)

there's a rule like

\delta L(q, dq/dt) / \delta q = (\delta A(q, dq/dt) / \delta q) B

+ A (\delta B(q, dq/dt) / \delta q)

One might think this should be easy to check: we consider

\delta L / \delta q = \partial L / \partial q

- d/dt \partial L / \partial (dq/dt)

We immediately see that

\partial L / \partial q - d/dt \partial L / \partial (dq/dt)

= (\partial A / \partial q) B + A (\partial B / \partial q)

- d/dt [ (\partial A / \partial (dq/dt)) B

+ A (\partial B / \partial (dq/dt)) ]

= (\partial A / \partial q) B + A (\partial B / \partial q)

- [ d/dt (\partial A / \partial (dq/dt)) ] B

- (\partial A / \partial (dq/dt)) dB/dt

- (dA/dt) (\partial B / \partial (dq/dt))

- A d/dt (\partial B / \partial (dq/dt)) (1)

Very well. Now we assume the product rule:

\delta L(q, dq/dt) / \delta q

= (\delta A / \delta q) B + A (\delta B / \delta q)

= (\partial A / \partial q) B

- [ d/dt \partial A / \partial (dq/dt) ] B

+ A (\partial B / \delta q)

- A [ d/dt \partial B / \partial (dq/dt) ] (2)

(1) and (2) look similar, however, the terms with dA/dt and dB/dt from
(1) are missing in (2). So, this result seem to suggest that there's no
product rule valid for functional derivatives.

But is that right? Is there maybe anything wrong in my calculations?

I checked (1) also with regarding the variation of the Hamiltonian
action integral

\delta S = \int [L(q + delta q, dq/dt + delta dq/dt) - L(q, dq/dt)] dt

but the result verifies (1):

\delta S = \int [(\partial A / \partial q) B

+ A (\partial B / \partial q)] delta q dt

- \int [ d/dt ( ( \partial A / \partial (dq/dt) ) B

+ A (\partial B / \partial (dq/dt) ) ) delta q dt

[Alfred Einstead]

On May 1, 10:31 am, Gregor Scholten <g.schol...@gmx.de> wrote:
> Hi alltogether,

Actually, the functional derivative is of the INTEGRAL operator:

S[q] = integral L(q(t), q'(t)) dt

not the integrand! Then
delta S[q]/delta q(t) = what you wrote as delta L/delta q.

There is, of course, a product rule ... for the integrals,
delta (S0[q] S1[q])/delta q(t) = S0[q] (delta S1[q]/delta q(t)) +
(delta S0[q]/delta q(t)) S1[q].

You could also probably pose a product rule for the integrands if you
ONLY consider L as a function L(t, q, v) with v independent of q,
rather than v = dq/dt.

Here's a stab at what you're doing. Consider the product of L0(t, q,
v) and L1(t, q, v). Let the variational coefficients be denoted
delta L0 = f0 delta q + p0 delta v
delta L1 = f1 delta q + p1 delta v.
Then
delta(L0 L1) = (f0 L1 + L0 f1) delta q + (p0 L1 + L0 p1) delta v.
Denoting the variational coefficients of L = L0 L1 by
delta(L) = f delta q + p delta v
then it follows that
f = f0 L1 + L0 f1, p = p0 L1 + L0 p1.

The minute you link v to q by treating it as v = dq/dt, then the whole
game changes. Then
delta(L0)/delta(q) = (f0 - dp0/dt) = E0.
delta(L1)/delta(q) = (f1 - dp1/dt) = E1.
delta(L)/delta(q) = (f - dp/dt) = E,
with
E = f - dp/dt = E0 L1 + L0 E1 - p0 dL1/dt - dL0/dt p1.