Extreme value of $S$ Define $$S(y)=\int_{O}^{P}y'^2+yy'+y^2 dx$$
where $y(x)$ is an arbitrary curve connecting $O=(0,0)$, $P=(1,1)$. Show that $S$ is extremised when it is calculated along the curve $$y(x)=\frac{e^x-e^{-x}}{e-e^{-1}}$$
And show that this curve gives either a saddle point or a minimum of $S$, but never a maximum.
I have no problem for the first step but I don't know why this curve cannot maximise $S$. Any help please?
 A: You have to compute the second functional derivative of the functional $S[y]$ in order to determine whether the stationary curve $y_*(x) = \frac{\sinh(x)}{\sinh(1)}$ is a minimum, a maximimum or a saddle-point, analogously to the optimization of functions.
The Lagrangian associated to $S[y]$ is given by $L = y'^2 + yy' + y^2$. You should have already computed its first-order functional derivative, namely
$$
\frac{\delta L}{\delta y} \equiv \left(\frac{\partial}{\partial y} - \frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial}{\partial y'}\right)L = -2y''+2y,
$$
which is nothing else that Euler-Lagrange equation when set to zero. Now, since this quantity depends on $y''$, the second functional derivative is calculated a bit differently, as follows :
$$
\frac{\delta^2 L}{\delta y^2} \equiv \left(\frac{\partial}{\partial y} - \frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial}{\partial y'} + \frac{\mathrm{d}^2}{\mathrm{d}x^2}\frac{\partial}{\partial y''}\right) (-2y''+2y) = 2-0+0 = 2,
$$
hence
$$
\frac{\delta^2 S}{\delta y^2} = \int_0^1 \frac{\delta^2 L}{\delta y^2} \,\mathrm{d}x = 2.
$$
In consequence, $S[y_*] = 2 > 0$, i.e. $y_*$ cannot be a maximizer.
