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I am aware of theoretical conditions for the existence of certain Euler-Lagrange equations (variational bicomplex etc...) but I am nevertheless trying by hand, to find a lagrangian that would yield some complicated equation. It seems at the moment that my difficulties may be reduced to a 1 dimensional problem, obtaining a term of the form $$t \frac{d \varphi }{d t}$$

Can one find a Lagrangian whose Euler-Lagrange equation yields such a term, and only this one?

Some calculations:

  • First of all I allow myself a Lagrangian that depends on higher than 1rst order derivatives, the Euler-Lagrange equation then takes the form $$\frac{\partial L}{\partial \varphi} - \partial_t \frac{\partial L}{\partial (\partial_t \varphi)} + \partial_{tt} \frac{\partial L}{\partial (\partial_{tt} \varphi)} - \partial_{ttt} \frac{\partial L}{\partial (\partial_{ttt} \varphi)} + \cdots = 0$$

  • a term of the form $t^2\, \varphi\, \partial_{tt} \varphi$ gives $$\frac{\partial L}{\partial \varphi} = t^2\, \partial_{tt} \varphi\quad \text{and}\quad \partial_{tt} \frac{\partial L}{\partial (\partial_{tt} \varphi)} = \partial_{tt} \big( t^2\, \varphi \big) = \partial_t \big( 2\, t\, \varphi + t^2\, \partial_t \varphi\big)$$ $$= 2\, \varphi + 2\, t\, \partial_t \varphi + 2\, t\, \partial_t \varphi + t^2 \,\partial_{tt} \varphi $$

In all my attemps, every time $t\, \partial_t \varphi$ appears, then so does $t^2 \,\partial_{tt} \varphi $ with the same $1/2$ ratio (don't forget the first term of the Euler-Lagrange equation)

My guess is that it is impossible to get such a term alone, but I am a little disappointed.

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  • $\begingroup$ Out of curiosity: In which context did this equation arise? $\endgroup$
    – Qmechanic
    Commented Jun 7, 2017 at 13:07
  • $\begingroup$ Oh, it's a much bigger equation and I dropped everything (including other partial derivatives) because I felt that only this term was problematic. I'm trying to write the operator associated to the casimir $W^2$ (as Klein-Gordon w.r.t. $P^2$) $\endgroup$
    – Noix07
    Commented Jun 7, 2017 at 13:41
  • $\begingroup$ I thought about crazy things which actually get me a little confuse. I remember that one of the advantage of the Lagrangian formalism was that the Euler-Lagrange equation does not depend on the choice of coordinate. But I thought about considering $\tilde{\varphi}= \varphi(t^2 /2)$ so that $\tilde{\varphi}' (t)= t \tilde{\varphi(t)} $, can one use this in some way...?? $\endgroup$
    – Noix07
    Commented Jun 7, 2017 at 18:45

1 Answer 1

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OP is asking if there exists an action term $S$ such that $$ \frac{\delta S}{\delta \varphi(t)}~\stackrel{?}{=}~t \frac{d \varphi(t) }{d t}.\tag{A}$$ For consistency, we must then have $$ \frac{\delta^2 S}{\delta \varphi(t^{\prime})\delta \varphi(t)}~\stackrel{(A)}{=}~t\frac{d }{d t}\delta(t\!-\!t^{\prime}).\tag{B}$$ But eq. (B) is inconsistent, since its left-hand side is symmetric$^1$ under a $t\leftrightarrow t^{\prime}$ exchange, while the distribution on the right-hand side is not. It can be a bit subtle to see the latter. Below follows a perhaps not necessarily simplest, but at least a rigorous and hopefully convincing proof: Apply e.g. a Gaussian test function to the distribution on the right-hand side of eq. (B): $$\iint_{\mathbb{R}^2}\!\mathrm{d}t~\mathrm{d}t^{\prime} ~\exp\left\{-at^2 -a^{\prime}t^{\prime 2}\right\}~t\frac{d }{d t}\delta(t\!-\!t^{\prime}) ~=~-a^{\prime}\sqrt{\frac{\pi}{(a+a^{\prime})^3}}. \tag{C}$$ This is not symmetric under an $a\leftrightarrow a^{\prime}$ exchange of the two positive constants $a,a^{\prime}>0$.

So we conclude that such an action term $S$ does not exist. $\Box$

Related Phys.SE posts: Why can't we ascribe a (possibly velocity dependent) potential to a dissipative force?, How do I show that there exists variational/action principle for a given classical system?, and links therein.

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$^1$ Since OP asks about it in a comment, let us here provide a hopefully instructive example. Consider an action term $$ S~:=~\int\! dt ~L(t), \qquad L(t)~:=~ \frac{t^2}{2} \varphi(t)\frac{d^2 \varphi(t) }{d t^2}.\tag{D}$$ The functional derivative is then $$ \frac{\delta S}{\delta \varphi(t)}~\stackrel{(D)}{=}~\varphi(t) + \frac{d}{dt}\left\{t^2 \frac{d\varphi(t) }{d t}\right\} .\tag{E}$$ The second functional derivative is $t\leftrightarrow t^{\prime}$ symmetric: $$\frac{\delta^2 S}{\delta \varphi(t^{\prime})\delta \varphi(t)}~\stackrel{(E)}{=}~\delta(t\!-\!t^{\prime}) - \frac{d}{dt}\frac{d}{dt^{\prime}}\left\{tt^{\prime}\delta(t\!-\!t^{\prime})\right\}.\tag{F}$$

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    $\begingroup$ I have not yet validated the answer because I am still wondering if the r.h.s. is really not symmetric.... $\endgroup$
    – Noix07
    Commented Jun 7, 2017 at 13:42
  • $\begingroup$ $t\, \delta'(t-t') $ as a distribution with "variable" $t$: let $\psi(t)$ be a test function, evaluating the distribution means to take the derivative, multiply by t and then evaluate everything at $t'$ one obtains $t'\,\psi'(t')$. Exchanging $t$ and $t'$: $\langle t'\, \delta (t'- t), \psi(t')= t\, \psi'(t)$. Or maybe one should still consider it as a distribution in $t$, in which case $\langle t'\, \delta (t'- t), \psi(t)= |1| \times t'\, \psi'(t')$ $\endgroup$
    – Noix07
    Commented Jun 7, 2017 at 15:51
  • $\begingroup$ It seems indeed that the "Schwarz lemma" holds!! I tried it on a simple Lagrangian. Then for the calculation, I m a little rusted but I find $- \frac{a}{(a+a')^2} \sqrt{\frac{\pi}{a+a'}}$, nevertheless, i do see that for a n arbitrary test function in two variables, what breaks the symmtry is the fact that one takes a derivative in one of the variable and not on the other one. $\endgroup$
    – Noix07
    Commented Jun 7, 2017 at 18:16
  • $\begingroup$ @Qmechanic Wait!!!... is the following symmetric: take $L(t,\varphi,\cdots) := t^2 \, \varphi\, \partial_{tt} \varphi,\ \frac{\delta S}{\delta \varphi (t)} = 2\, \varphi + 4\, t\, \partial_t \varphi + 2\, t^2 \,\partial_{tt} \varphi$ and thus $ \frac{\delta S}{\delta \varphi(t')\delta \varphi (t)} = 2\, \delta(t-t') + 4\, t\, \delta ' (t- t') + 2\, t^2 \,\partial_{tt} \delta (t-t')$?? It seems not?! $\endgroup$
    – Noix07
    Commented Jun 7, 2017 at 18:27
  • $\begingroup$ Yes, your above example is symmetric, as it must be. $\endgroup$
    – Qmechanic
    Commented Jun 7, 2017 at 18:41

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