Optimality condition for the critical points

Question

Given the Differential equation $$\begin{cases} -f'=A f \\ f(T)=f_T \end{cases}$$ And given the quadratic functional: $$J(f_T)=\frac{1}{2}\int_0^T|f|^2dt$$ I want to know why $$\int_0^T<f,u> dt=0$$ is the optimality condition for the critical points of $J$

Answer 1

I suspect you meant $J(f)$ in your definition, not $J(f_T)$ (the functional does not depend only on the terminal point). A first step toward deriving the Euler-Lagrange equation is \begin{align*} \frac{d}{d\varepsilon}\bigg|_{\varepsilon = 0}\int_0^T |f+\varepsilon u|^2 dt &= \int_0^T \frac{d}{d\varepsilon}\bigg|_{\varepsilon = 0} \langle f+\varepsilon u, f+\varepsilon u\rangle dt \\ &= \int_0^T \langle u , f+\varepsilon u \rangle + \langle f+\varepsilon u, u\rangle\bigg|_{\varepsilon = 0} dt\\ &= \int_0^T 2\langle u,f \rangle dt. \end{align*} Throwing in the remaining factor of $1/2$ gives what you want. You need additional conditions on $u$, in particular that $u(T) = 0$ (if indeed you meant a terminal condition).