Derivative of the action functional

Question

I am self studying the book An Brief Introduction to Physics for Mathematicians. At the second page, the equation (1.6) says that the derivative of the action functional can be derived as $$ S'(x)(h) = I'(L(x)) \circ \mathcal{L}'(x)(h) = \int_{t_1}^{t_2}(grad_q L(x,\dot x)\cdot h(t) + grad_{\dot q} L(x,\dot x)\cdot h(t))dt $$ where $q,\dot q$ are coordinates, and $L$ is the Lagrangian. I cannot understand how to reach this conclusion and may need a detailed explantion. Thanks!

Answer 1

$S$ is defined as the composition of $I \colon C^\infty(t_1, t_2) \to \mathbf R^n$, $x \mapsto \int_{t_1}^{t_2} x(t)\,dt$ and the map $\mathcal L \colon C^\infty(t_1, t_2) \to C^\infty(t_1, t_2)$, $x \mapsto L \circ (x,\dot x)$. Hence, by the chain rule $$ DS(x) = D(I \circ \mathcal L)(x) = DI(\mathcal L(x)) D\mathcal L(x) $$ As $I$ is linear $DI$ is constant and equal to $I$, we continue: $$ DS(x) = DI(\mathcal L(x)) D\mathcal L(x) = I \cdot D\mathcal L(x) $$ Now, we have, as outlined in the lines before the equation you asked about, due to the definition of $\mathcal L$, that at $h \in C^\infty(t_1, t_2)$ - note that the derivative is the linear part of the expansion -: $$ D\mathcal L(x)h = D_1L(x,\dot x)h + D_2L(x,\dot x)\dot h $$ Therefore - recall that $I$ is just integration -: $$ \bigl(DS(x)\bigr)h = I\bigl(D\mathcal L(x)h\bigr) = \int_{t_1}^{t_2} (D\mathcal L(x)h)(t) \,dt = \int_{t_1}^{t_2} D_1L(x(t),\dot x(t))h(t) + D_2L(x(t),\dot x(t))\dot h(t)\, dt $$