# Notation for functional derivative of two variables

I have the following functional

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

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

$$$$\frac{\partial L}{\partial\rho}-\frac{d}{dx}\frac{\partial L}{\partial\rho'}$$$$ where $$\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 $$$$\frac{\partial L}{\partial\rho}\left(t,x\right)-\frac{d}{dx}\frac{\partial L}{\partial\rho'}\left(t,x\right)=\frac{\delta F_{\varepsilon}\left[\rho\right]}{\delta_{x}\rho\left(t,x\right)}\left(t,x\right).$$$$

This is just $$\frac{\delta F_\varepsilon}{\delta\rho}$$ with action $$F_\varepsilon=\int_0^1Ldx$$. It's the usual functional derivative because $$\frac{\partial L}{\partial\dot{\rho}}=0$$.
• So you would go for $\frac{\delta F_{\varepsilon}\left[\rho\right]}{\delta\rho\left(t,x\right)}\left(t,x\right)$ ? – edwineveningfall Nov 25 '19 at 16:00
• @edwineveningfall I probably wouldn't include the $(t\,x)$ twice, but I'm lazy. – J.G. Nov 25 '19 at 16:08