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Suppose that I have a functional $S[q]=\int_a^b L(t,q(t),q'(t))\,dt$. Such a functional is well-known to extremized by a choice of $q(t)$ satisfying the Euler-Lagrange equation $\dfrac{d}{dt}\dfrac{\partial L}{\partial q'} = \dfrac{\partial L}{\partial q}$. If the integrand has moreover no explicit $t$-dependence, then one can show that the Beltrami identity holds: $L-q'\dfrac{\partial L}{\partial q'}=C$ for some constant $C$.

Now, it is also known how the Euler-Lagrange equation generalizes to functionals depending on higher derivatives ($q'',q'''$, etc.). Does the Beltrami identity also generalize?

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  • $\begingroup$ This question is inspired by my comment to a recent question, and my unwillingness to carry out the resulting algebra. $\endgroup$ May 5, 2016 at 21:16
  • $\begingroup$ If there's a clear obstacle to realizing even the $q''(t)$ case, then I'd be satisfied with an answer to that effect which spells out the details. @Winther $\endgroup$ May 5, 2016 at 21:57
  • $\begingroup$ did you look at wiki/Euler–Lagrange_equation - Derivation of one-dimensional Euler–Lagrange equation ? everything works the same in the higher derivatives case en.wikipedia.org/wiki/… , considering a perturbation $q(t) + \epsilon p(t)$, write that $\displaystyle\frac{\partial S}{\partial \epsilon}(q^*) = 0$ if $q^*$ is optimal, and play with the obtained formulas. $\endgroup$
    – reuns
    May 5, 2016 at 22:23

1 Answer 1

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For a general Lagrangian $L[t,q,q',q'',\ldots]$ the Euler-Lagrange equation reads

$$L_q = [L_{q'}]' - [L_{q''}]'' + [L_{q''}]''' - \ldots = \sum_{n\geq 1} (-1)^{n+1}[L_{q^{(n)}}]^{(n)}$$

If the Lagrangian is time-independent. $L_t = 0$, we have $$\frac{dL}{dt} = L_qq' + L_{q'}q'' + L_{q''}q''' + \ldots = L_q q' + \sum_{n\geq 1}L_{q^{(n)}}q^{(n+1)}$$

and by combinding the equations above we arrive at

$$-L' + \color{blue}{L_{q'}q'' + [L_{q'}]'q'} + \color{red}{L_{q''}q''' - [L_{q''}]''q'} + \color{blue}{L_{q'''}q'''' + [L_{q'''}]'''q'} + \ldots = 0$$

which can be written

$$-L' + \sum_{n\geq 1} L_{q^{(n)}}q^{(n+1)} + (-1)^{n}[L_{q^{(n)}}]^{(n)}q' = 0$$

Note that

$$(-1)^{n+1}\frac{d}{dt}\left[q'[L_{q^{(n)}}]^{(n-1)} - q''[L_{q^{(n)}}]^{(n-2)} + q'''[L_{q^{(n)}}]^{(n-3)} + \ldots + (-1)^{n+1}q^{(n)}L_{q^{(n)}}\right] \\= q^{(n+1)}L_{q^{(n)}} + (-1)^{n+1}q'[L_{q^{(n)}}]^{(n)}$$

which after integrating up gives us the identity

$$L + \sum_{n\geq 1}\sum_{k=1}^n (-1)^{k+n+1}q^{(k)}[L_{q^{(n)}}]^{(n-k)} = C$$

Writing out the first few terms gives us

$$L \color{brown}{- q'L_{q'}} \color{red}{ + q'[L_{q''}]' - q''L_{q''}} \color{blue}{- q'[L_{q'''}]'' + q''[L_{q'''}]' - q'''L_{q'''}} + \ldots = C$$

The presence of the time-derivatives of the partial derivatives for $n\geq 2$ makes it less useful of an identity than what the Beltrami identity is.


As a check applying this to the question you linked where $L[y,y''] = y + \frac{y^2}{2} - \frac{(y'')^2}{2}$ whose Euler-Lagrange equation is $1 + y - y^{(4)} = 0$ we arrive at

$$\frac{1}{2} y''^2-y^{(3)} y'+\frac{y^2}{2}+y = C$$

Taking the derivative we get

$$y'(1+y-y^{(4)}) = 0$$

which is true by the Euler-Lagrange equation.

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    $\begingroup$ Nicely done and well-explained! I'm glad to see that my expectation was correct. (About a generalization existing, that is. I certainly can't claim that said generalization is useful in the problem I linked...) $\endgroup$ May 6, 2016 at 0:46

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