Find possible solution for minimization of a functional with free boundary condition Find a possible solution for the minimization of the functional:
$$J[x]=\int_0^1 (t\dot{x}+\dot{x}^2) \, dt\tag1$$
with $x(0)=1$ and $x(1)$ is a free variable.
I am trying to solve the above, but not sure whether I am going the right way!
My attempt:
Try to find the function $x$, perhaps the line passing from points $x(0)$ and $x(1)$? If $x(0)=1$, and $x(1)= c,c\in\mathbb{R}$ then $$x(t)= (c-1)t+1. \tag2$$
Then replace $(2)$ into the  equation $(1)$ and now $$J[x]=\int_0^1 (c-1)t+(c-1)^2 \, dt$$
Doing the calculations
$$=c^2 -\frac{3}{2}c+\frac{1}{2}\tag3$$
So the problem becomes minimizing $(3)$.
Let $F(c)=c^2 -\frac{3}{2}c+\frac{1}{2}$. Then $F(c)$ has a critical point at $c=\frac{3}{4}$ and it is a minimum for $F(c)$ based on the second order condition. Hence, one possible minimization of the functional is:
$$x(t)= \frac{-1}{4}t+1 $$
But not sure if it is a correct approach or I should use Euler-Lagrange like in this example Minimization of functional using Euler-Lagrange
 A: *

*We cannot assume the Euler-Lagrange (EL) equation since we don't have adequate boundary conditions (BCs). Recall that the proof of the EL equation uses BCs to get rid of boundary terms$^1$.


*Instead we can complete the square in OP's functional:
$$\begin{align}J[x]~:=~&\int_0^1\! \mathrm{d}t~ (\dot{x}^2+t\dot{x})\cr 
~=~& \int_0^1\! \mathrm{d}t~ (\dot{x}+\frac{t}{2})^2-\int_0^1\! \mathrm{d}t~\frac{t^2}{4}\cr
~=~& \int_0^1\! \mathrm{d}t~ (\dot{x}+\frac{t}{2})^2- \frac{1}{12}.\end{align}\tag{1} $$
Clearly, the minimum configuration satisfies
$$ \dot{x}~=~-\frac{t}{2}.$$
Given the initial condition $x(0)=1$, the unique solution is
$$ x(t)~=~1-\frac{t^2}{4}.$$
--
$^1$ Perhaps a counterexample is in order. Consider the functional
$$K[x]~:=~ \int_0^2\! \mathrm{d}t~ (\dot{x}^2-x^2)$$
with $x(0)=0$ and $x(2)$ is a free variable. It is not hard to prove that $K$ is unbounded from below e.g. by considering configurations of the form $x(t)=A\sin t$.
A: Using the Euler-Lagrange equations we arrive to
$$
2\ddot x+1 = 0
$$
and after solving with the initial condition $x(0) = 1$ we arrive to
$$
x=1-\frac{t^2}{4}+ tC_0
$$
To determine $C_0$ we substitute the found $x$ into the integral obtaining
$$
\int_0^1 (t \dot x+\dot x^2) dt = -\frac{1}{12}+C_0^2
$$
and to have a minimum we set $C_0 = 0$ so the solution is
$$
x = 1-\frac{t^2}{4}
$$
