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I would like to know how exactly to calculate the variational derivative of: $$L = \oint dx \: \frac{m}{a} (\dot{u}(x,t))^2 - ca(u'(x,t)^2\tag{1}$$ with respect to $u$, ($\delta L / \delta u$). The variable $x$ is integrated over a closed loop defined on an interval from $0$ to some length $a$ where $u(0) = u(a)$.

Following my lecture notes, the result is meant to be $$\delta L / \delta u = cau''\tag{2}$$ but whatever I do I don't get the integral over $x$ canceled. I have also used partial integration and the fact that $$\frac{\delta}{\delta u} = \frac{\partial \:\delta}{\partial x\: \delta u'}\tag{3}$$ and tried to treat the expression as a partial derivative in $u$, but somehow I never get rid of the closed integral. Hopefully someone might be able to help me out.

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I would like to know how exactly to calculate the variational derivative of: $$L = \oint dx \: \frac{m}{a} \dot{u}(x,t) - ca(u')^2$$ with respect to $u$, ($\delta L / \delta u$). Following my lecture notes, the result is meant to be $$\delta L / \delta u = cau''$$ but whatever I do I don't get the integral over $x$ canceled.

The variational derivative is a functional derivative, which is defined to be just the part that is integrated against $\delta u$ to get $\delta L$.

$$ \delta L = \int dx \frac{\delta L}{\delta u(x)}\delta u(x) $$

This is why you "cancel" the integral to arrive at the expression for $\delta L/\delta u(x)$.

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  • $\begingroup$ I get your point, thank you. that also explains why there is still some riemann integration apparent. $\endgroup$
    – M. Uon
    May 23, 2022 at 10:19

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