Let's say we have functional of the form: $$\mathcal{L} = \int_a^bF(y, y') dx$$
Let $\mathbf{x}(t) = (y(x,t), y'(x, t))^T$, then: $$\mathcal{L}(t + h) = \mathcal{L}(t) + h\frac{d}{dt}\mathcal{L}(t) + \frac{h^2}{2}\frac{d^2}{dt^2}\mathcal{L}(t) + O(h^3)$$ The second variation appears in $\frac{d^2}{dt^2}\mathcal{L}(t)$: $$\frac{d^2}{dt^2}\mathcal{L}(t) = \int_a^b \frac{d^2}{dt^2}F(\mathbf{x}(t)) dx$$ $$\frac{d^2}{dt^2}\mathcal{L}(t) = \int_a^b \frac{d}{dt} \left< \nabla F, \dot{\mathbf{x}} \right > dx$$ Assuming $\ddot{\mathbf{x}} = 0$, $$\frac{d^2}{dt^2}\mathcal{L}(t) = \int_a^b \left< \frac{d}{dt} \nabla F, \dot{\mathbf{x}} \right > dx$$
$$\frac{d^2}{dt^2}\mathcal{L}(t) = \int_a^b \left< H_F \dot{\mathbf{x}}, \dot{\mathbf{x}} \right > dx$$
$$\frac{d^2}{dt^2}\mathcal{L}(t) = \int_a^b \dot y^2\partial_{yy}F + 2\dot{y}\dot{y}'\partial_{yy'}F + \dot{y}'^2 \partial_{y'y'}F dx \;\;\; (1)$$
Using integration by parts two times:
$$\frac{d^2}{dt^2}\mathcal{L}(t) = \int_a^b \dot y^2\partial_{yy}F -2\dot y \frac{d}{dx}\partial_{yy'}F + \dot y^2 \frac{d^2}{dx^2}\partial_{y'y'}F dx$$
$$\frac{d^2}{dt^2}\mathcal{L}(t) = \int_a^b \dot y^2 \left ( \partial_{yy} -2 \frac{d}{dx}\partial_{yy'} + \frac{d^2}{dx^2}\partial_{y'y'} \right)F dx$$
$$\frac{d^2}{dt^2}\mathcal{L}(t) = \int_a^b \dot y^2 \left ( \partial_{y} - \frac{d}{dx}\partial_{y'}\right)^2 F dx \;\;\;(2)$$
My doubt is the following. If you evaluate (1) and (2) on the same functional I get two different results and I don't understand why. Here is an example:
Let $$\mathcal{F} = \int_a^b 1 + y'^2 dx$$
Using (1) I get:
$$\partial_{y'y'}(1 + y'^2) = 2$$
Using (2), I get:
$$\frac{d^2}{dx^2}\partial_{y'y'}(1 + y'^2) = 0$$
Can anyone explain my mistake?