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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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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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