On p.33, exercise 33, of Calculus of Variations, G & F, there is the following question
Calculate the variational derivative at the point $x_0$ of the quadratic functional $$J[y]=\int_a^b\int_a^bK(s,t)y(s)y(t)dsdt.$$
I'd like to share my solution in case there is a mistake, here it goes:
Consider first the function $$J(y_0,\ldots,y_n)=\sum_{i=1}^n\sum_{j=1}^nK(x_i,x_j)y_iy_jh^2$$ where $y_i=y_i(x_i)$, $y_j=y_j(x_j)$, $h=x_i-x_{i-1}=x_j-x_{j-1}$ and both $\{x_i\}$ and $\{x_j\}$ are equal partitions of $[a,b]$.
Now, we consider the partial derivative of $J$ with respect to $y_k$, being
\begin{eqnarray} \frac{\partial J}{\partial y_k}&=&h^2\sum_{j=1}^nK(x_k,x_j)y_j+\sum_{i=1}^nK(x_i,x_k)y_i\\ &=&h^2\sum_{i=1}^n(K(x_k,x_i)+K(x_i,x_k))y_i\\ \Longrightarrow\quad\frac{\partial J}{\partial y_kh}&=&\sum_{i=1}^n(K(x_k,x_i)+K(x_i,x_k))y_ih\\ \Longrightarrow\quad\lim_{n\to\infty}\frac{\partial J}{\partial y_kh}&=&\int_a^b(K(x,t)+K(t,x))y(t)dt\\ \therefore\quad\frac{\delta J}{\delta y}\Big|_{x=x_0}&=&\int_a^b(K(x_0,t)+K(t,x_0))y(t)dt \end{eqnarray}
Thanks in advance.