A functional $$J(y)=\int_a^b F\left(x,y(x)\right)dx, \tag{1}$$ subject to an isoperimetric constraint $$\int_a^b K(x,y)dx=l, \tag{2}$$ and a holonomic constraint $$g(x,y)=0. \tag{3}$$
Most textbooks talk about "(1) s.t. (2)" and "(1) s.t. (3)" separately. Specifically, if there is an extremum for (1) s.t. (2), then the extremum also holds for $\int_a^b (F+\alpha K) dx$. Similarly, if (1) and (3), an extremum also holds for $\int_a^b (F+\lambda(x) g) dx$.
Now I want to prove the Lagrange multiplier for (1) s.t. (2) and (3). I know (2) can be regarded as an special case of (3) by setting $z(x)=\int_a^x K(x,y)dx$ and $z'(x)=K(x,y)$. Is this the way to prove? Any reference shows the proof (not just the statement)?
I also noticed that in Gelfand and Fomin's book p49: We can also regard (3) as a constraint (2). Specifically, (3) can be regarded as an infinite set (for each $x$) of conditions, each of which is a functional. For each functional, there is a multiplier, so we add $\lambda(x)$ to (3). I know intuitively it is right, but how we mathematically transform (3) to (2) such that we can prove (1) s.t. (2) and (3)?
Thanks. Any suggestion will be appreciated.