Let $y \in S = \left \{y \in C^{2}\left [ x_0,x_1 \right ]\mid y(x_0)=y_0,y(x_1)=y_1 \in \mathbb{R})) \right \}$ and $\eta \in H = \left \{\eta \in C^{2}\left [ x_0,x_1 \right ]\mid \eta(x_0)=0=\eta(x_1) \right \}$ be fixed functions. Then the quantity $J(y+a\eta)$ can be regarded as a functions of the single real variable $a$, where $J(y)=\int_{x_0}^{x_1}f(x,y,y')dx$ and $f$ has continuous second partial derivatives with respect to the variables $x,y,y'$.
Knowingly this facts, I have to calculate $\frac{dJ}{da}$, but I'm stuck in this problem because I don't know a version for the Leibniz integral rule in three variables. Can anyone help me? Thanks in advance.
By the way, this exercise is requerided in my currently course of calculus of variations.
All you need is the basic Leibniz integral rule.
Note that if we let $H(a):=J(y+a\eta)$, then \begin{align} H(a)&:=J(y+a\eta)\\ &:=\int_{x_0}^{x_1}f(x, (y+a\eta)(x),(y+a\eta)'(x))\,dx\\ &=\int_{x_0}^{x_1}f(x, y(x)+a\eta(x),y'(x)+a\eta'(x))\,dx \end{align} So, let us write the integrand as $\psi(a,x)$. Then, \begin{align} H(a)&=\int_{x_0}^{x_1}\psi(a,x)\,dx \end{align} Since $f$ and $y$ are $C^2$ functions, it follows from the chain rule that $\psi:\Bbb{R}^2\to\Bbb{R}$ (which we defined as a composite function) is $C^1$ (it is $C^1$ because of the appearance of $y'$ and $\eta'$ which are only $C^1$). I'm sure you can now apply Leibniz's integral rule: \begin{align} H'(a)&=\int_{x_0}^{x_1}\frac{\partial \psi}{\partial a}(a,x)\,dx \end{align} (I purposely defined the new function $H$ to emphasize that it's a function $\Bbb{R}\to\Bbb{R}$, so strictly speaking, writing $\frac{dJ}{da}$ is wrong, because $J$ is NOT a function of one real variable). Now, how does one calculate $\frac{\partial \psi}{\partial a}$ using the chain rule?