If a function $f$ is $\mathcal C^2([a,b]\times \mathbb R\times \mathbb R)$, $f=f(x,u,\xi)$, and $$I(u)=\int_a^b f(x,u(x),u'(x)dx,$$ then, Euler Lagrange equation is given by $$\frac{d}{dx}f_\xi=f_u.$$
Let now $$J(u,v)=\int_a^b \Big(u'(x)v(x)- H(x, u(x), v(x))\Big)dx,$$ for a function $H$. Why the Euler-Lagrange associated to $J$ is $$\begin{cases}u'=H_v\\ v'=-H_u\end{cases}\ \ ?$$ In what I can deduce it from the Euler Lagrange equation associated to $I$ ?