Are the partial derivatives of Lagrangian in the varied action functional derivatives?

Question

In particle mechanics Lagrangian $L$ depends upon position, velocity (and may be explicitly on time), whereas in field theory the Lagrangian density ${\cal L}$ similarly (or analogously) depends upon the field and its derivatives. When we derive Euler-Lagrange equation of motion we vary the action,

In particle mechanics, $$\delta S=\int_{t_1}^{t_2}~\mathrm dt\left(\frac{\partial L}{\partial q}~\delta q+\frac{\partial L}{\partial \dot{q}}~\delta \dot{q}\right)\tag{1}$$

In field theory, $$\delta S=\int_\sigma~\mathrm d^4x\left(\frac{\partial \mathcal{L}}{\partial \phi}~\delta \phi+\frac{\partial \mathcal{L}}{\partial (\partial _\mu \phi)}~\delta \partial_\mu \phi\right).\tag{2}$$

Now, Lagrangian is a functional, it maps functions (position, velocity or fields and their derivatives) into another function (or a real number). Like, $$L:F×F→F;(q(t),\dot{q}(t))↦L[q(t),\dot{q}(t)]. \tag{3}$$ So my question is that are the partial derivatives of the Lagrangian w.r.t. position and velocity functions or fields and their derivatives functional derivatives?

Answer 1

Yes, OP is right. In the field-theoretic case, the partial derivatives in OP's first formula (1) should be replaced with functional derivatives

$$ \delta S~=~\int_{t_1}^{t_2}\!\mathrm{d}t\left(\frac{\delta L}{\delta q}~\delta q+\left. \frac{\delta L}{\delta v}\right|_{v=\dot{q}}~\delta \dot{q}\right),\tag{1'}$$

where the Lagrangian

$$L[q(\cdot,t),v(\cdot,t);t]~=\int \! \mathrm d^3x~ {\cal L}(q(x,t),v(x,t), ~\partial_x q(x,t), \partial_x v(x,t),~\ldots , t) $$

is a functional. The ellipsis $\ldots$ indicates dependence of possible higher-order derivatives. See my Phys.SE answers here and here for further details.

Answer 2

This is just supplementing Qmechanic's answer. I think the notations here need to be addressed. OP might be confusing Lagrangian (normal $L$) with Lagrangian density ($\mathcal{L}$). Formally, we have three fundamental relations: $$L = \displaystyle\int \mathcal{L}(\phi(x,t),\dot \phi(x,t),x,t) \mathrm d^3x$$ $$S = \displaystyle\int dt \space L = \displaystyle\int dt\int \mathcal{L}(\phi(x,t),\dot \phi(x,t),x,t) \mathrm d^3x$$ $$\delta S=\displaystyle\int_\sigma~\mathrm d^4x\left(\frac{\partial \mathcal{L}}{\partial \phi}~\delta \phi+\frac{\partial \mathcal{L}}{\partial (\partial _\mu \phi)}~\delta \partial_\mu \phi\right).\tag{2}$$ (x actually means all spatial variables in the equations above)

So you would take partial derivatives on Lagrangian density $\mathcal{L}$, but take functional derivatives on the Lagrangian $L$. However, in general: $$ \frac{\partial \mathcal{L}}{\partial \phi} \neq \frac{\delta L}{\delta \phi}$$ For derivation of this inequality, see this SE answer.