Let $u$ be a real function defined on $[0, T]$ and the functional $$V(u) = \iint\limits_{[0,T]^2}f(x,y)u(x)u(y)\mathrm dx\mathrm dy$$ and $f(x,y)$ is symmetric , continuous and can be writen on this form $f(x,y)=h(x-y)$
I would like to maximise this fonctional under an isoperimetric constraint : $\int^{T}_{0}u(x)dx= L$
the ideas is to use lagrange multiplier i.e : $J(u)=V(u)+\lambda \int^{T}_{0}u(x)dx $
The functional derivative $$<\delta J(u), v>= \iint\limits_{[0,T]^2}f(x,y)[u(x)v(y) + u(y)v(x)]\mathrm dx\mathrm dy+\lambda \int^{T}_{0}v(x)dx $$ as $f$ is symmetric $$<\delta J(u), v>= 2\iint\limits_{[0,T]^2}[f(x,y)u(x)v(y)]\mathrm dx\mathrm dy+\lambda \int^{T}_{0}v(x)dx $$ Now i would like to obtain a weak "Euler lagrage" Equation,so i do this : $$<\delta J(u), v>= 2\int_{[0,T]} \left( \int_{[0,T]} (f(x,y)u(x)+\frac{\lambda}{2\pi} )\, dx \right) v(y)\, dy =0 \quad\hbox{for all } v $$
now by the Fondamentale Lemma of the calculus of variation, this implies : $$\int_{[0,T]} (f(x,y)u(x))dx+\frac{\lambda}{2\pi}=0 \quad\hbox{ a.e } $$ wich means $$h*u+\frac{\lambda}{2\pi} =0. \quad \hbox{(1)}$$ I have two Question 1) is my computations correct?
2) how can I solve the equation (1)? Thank you in advance