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I have the following functional

\begin{equation} F_{\varepsilon}\left[\rho\right]\left(t\right):=\int_{0}^{1}\left[\frac{\varepsilon}{2}\left(\frac{d\rho}{dx}\right)^{2}+\frac{1}{4\varepsilon}\left(1-\rho^{2}\right)^{2}\right]dx. \end{equation}

where $\rho(t,x)$. Calling $L\left(t,x\right):=\left[\frac{\varepsilon}{2}\left(\frac{d\rho}{dx}\left(t,x\right)\right)^{2}+\frac{1}{4\varepsilon}\left(1-\rho^{2}\left(t,x\right)\right)^{2}\right]$ the functional derivative I want is

\begin{equation} \frac{\partial L}{\partial\rho}-\frac{d}{dx}\frac{\partial L}{\partial\rho'} \end{equation} where $\rho'=\frac{\partial\rho}{\partial x}$.

My question is: is there any standard notation to indicate this functional derivative (which uses $\rho$ as a function of $x$ only)? I was thinking about the following \begin{equation} \frac{\partial L}{\partial\rho}\left(t,x\right)-\frac{d}{dx}\frac{\partial L}{\partial\rho'}\left(t,x\right)=\frac{\delta F_{\varepsilon}\left[\rho\right]}{\delta_{x}\rho\left(t,x\right)}\left(t,x\right). \end{equation}

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This is just $\frac{\delta F_\varepsilon}{\delta\rho}$ with action $F_\varepsilon=\int_0^1Ldx$. It's the usual functional derivative because $\frac{\partial L}{\partial\dot{\rho}}=0$.

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  • $\begingroup$ So you would go for $\frac{\delta F_{\varepsilon}\left[\rho\right]}{\delta\rho\left(t,x\right)}\left(t,x\right)$ ? $\endgroup$ – edwineveningfall Nov 25 '19 at 16:00
  • $\begingroup$ @edwineveningfall I probably wouldn't include the $(t\,x)$ twice, but I'm lazy. $\endgroup$ – J.G. Nov 25 '19 at 16:08

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