Let $a, b$ be some different complex numbers.
I assume that every functions below have continous second derivative. Let there be a function $F:\mathbb{C}\rightarrow \mathbb{R}$
I would like to find such function $z:[0,1]\rightarrow \mathbb{C}$ so that $z(0)=a$, $z(1)=b$ and the integral $I:=\int_{0}^{1}F(z(t))dt$ is as minimal as possible(is stationary).
Here is my attempt: I tried to apply the proof of Euler-Lagrange equation in order to solve the problem.
Let $\eta(t)$ be a function, where $\eta(0)=\eta(1)=0$ Lets assume that $z(t)$ makes the integral as low as possible. I define $z^*(t)=z(t)+\epsilon\cdot\eta(t)$, where $\epsilon\in\mathbb{R}$ and i define $I(\epsilon):=\int_{0}^{1}F(z^*(t))dt$
I see that $I$ is stationary when $\frac{dI(\epsilon)}{d\epsilon}|_{\epsilon=0}=0$ so i do the following:
$$\frac{dI(\epsilon)}{d\epsilon}|_{\epsilon=0}=\int_{0}^{1}\frac{d}{d\epsilon}F(z^*(t))|_{\epsilon=0}dt=\int_{0}^{1}\eta(t)\frac{d}{dt}F(z(t))dt=0$$
Since this equation holds for any $\eta$ we have $\frac{d}{dt}F(z(t))=0$ However i don't know how to find $z(t)$
Any help will be appreciated
EL eqs. for the Lagrangian $z\equiv x+iy\stackrel{F}{\mapsto} F(z)$ is $$ \frac{\partial F}{\partial x}~=~0~=~\frac{\partial F}{\partial y},$$ i.e. stationary points for $F$. A stationary path :$[0,1]\to\mathbb{C}$ is only possible if the boundary values $a$ & $b$ are stationary points for $F$.
If the stationary points for $F$ are isolated then the stationary paths are constant paths. These can only obey the boundary conditions if $a=b$.