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Consider a mechanical system, the Lagrangian of which is: $$-L(u,\dot u)=\int\left(\dfrac{\partial^2 u}{\partial x^2}\right)^2\mathrm{d}x$$

This would correspond to a system in torsion, for example. I intentionally dropped the terms which are not of interest (such as kinetic energy).

Then, calculate the first term in the Euler-Lagrange equation: $$\dfrac{\partial L}{\partial u}(u,\dot u)=\dfrac{\partial}{\partial u}\int u''^2 \mathrm{d}x$$

First possibility: $$\dfrac{\partial}{\partial u}\int u''^2 \mathrm{d}x=0$$ because $\dfrac{\partial u''}{\partial u}=0$, similarly to $\dfrac{\dot u}{\partial u}=0$. I think this is not true, because $\dot u$ is a variable, but not $u''$.

Second possibility: $$\dfrac{\partial}{\partial u}\int u''^2 \mathrm{d}x=\dfrac{\partial}{\partial u}\int u\,u^{(4)}\mathrm{d}x=\int u^{(4)}\mathrm{d}x$$

by double integration by part and because $\dfrac{\partial u^{(4)}}{\partial u}=0$. I am really not sure about this latter argument either.

Third possibility Define a new variable in the Lagrangian such that $L(u,\dot u,v,\dot v)=\int v^2\mathrm{d}x$ and somehow link $v$ to $x$ later.

Can someone please enlighten me?

Qmechanic
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anderstood
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  • Related: http://physics.stackexchange.com/q/109518/2451 , http://physics.stackexchange.com/q/119750/2451 , and links therein. – Qmechanic Mar 10 '15 at 01:07
  • @Qmechanic Unlike the two questions above, this one deals with an additional variable which is not a time-derivative. This could make a difference, according to this answer which expands $L$ in $q,\dot q,\ddot q$. – anderstood Mar 10 '15 at 01:17

1 Answers1

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Second possibility: $$\dfrac{\partial}{\partial u}\int u''^2 \mathrm{d}x=\dfrac{\partial}{\partial u}\int > u\,u^{(4)}\mathrm{d}x=\int u^{(4)}\mathrm{d}x$$

by double integration by part and because $\dfrac{\partial > u^{(4)}}{\partial u}=0$. I am really not sure about this latter argument either. Second possibility is closest to correct. The correct answer if given/explained below.

If $$ L=-\int dx (\frac{d^2u}{dx^2})^2 $$ then $$ \frac{\delta L}{\delta u(x)}=-2\frac{d^4u}{dx^4} $$

You can work this out by considering the first order change in L when $u\to u+\delta u$ $$ L[u+\delta u]=-\int dx (u''+\delta u'')^2=L-2\int dx u''\delta u'' + O(\delta u^2) $$ I.e., $$ \delta L=-2\int dx \frac{d^2 u}{dx^2}\frac{d^2\delta u}{dx^2}+O(\delta u^2)=-2\int dx \frac{d^4 u}{dx^4}\delta u+O(\delta u^2) $$ Where the second equality is obtained by integration by parts. And now you can just read off the answer: $$ \frac{\delta L}{\delta u}=-2\frac{d^4 u}{dx^4} $$

hft
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  • Informative, thank you. And the reason why the same cannot be done for $\dot u$ is that the integration by part is no longer possible (e.g. $L=\int \dot u^2 \mathrm{d}x \Rightarrow \delta L=2\int \dot{\delta u}\dot u \mathrm{d}x$ not integrable by parts). So in the integrand, $\dot u$ and $u'$ are fundamentally different. That makes things more clear, thanks! – anderstood Mar 10 '15 at 01:47
  • Yeah, they're fundamentally different because the integral is over dx not dt, so you can't integrate by parts on t... – hft Mar 10 '15 at 01:53
  • and the third possibility (including $u''$ as an additional variable of $L$) would not work for the same reason... I'll ask my salami sandwich to make sure. – anderstood Mar 10 '15 at 02:01
  • Always a good idea – hft Mar 10 '15 at 02:02