Find the extremals for the functional: $$ J(x) = \int_{0}^{1}\left[x\left(t\right)\dot{x}\left(t\right) + \ddot{x}^{2}\left(t\right)\right]\mathrm{d}t $$ where $x(0)=0$, $\dot{x}(0)=1$, $x(1)=2$, $\dot{x}(1)=4$.
Let $y(t)=[x(t)$ $\dot{x}(t)]^T$. Reexpress the functional in terms of $y(t)$ and solve the problem.
I can find the solution without doing that transformation. But, I cannot find the same solution when I transformed the functional.
On
$$ J = \int_0^1 G(x,\dot x, \ddot x) dt $$
This problem is equivalent to
$$ J = \int_0^1 \left(G(x, y, \dot y)+\lambda(t)(y - \dot x)\right)dt,\ \ \mbox{s. t. } y = \dot x $$
The Euler-Lagrange equations are
$$ \begin{cases} G_x + \dot \lambda(t) & = & 0\\ G_y + \lambda(t)-\frac{d}{dt}G_{\dot y} & = & 0 \end{cases} $$
or
$$ \begin{cases} y + \dot\lambda(t) & = & 0\\ x +\lambda(t)-2 \ddot y & = & 0 \end{cases} $$
or equivalently
$$ \begin{cases} \dot x +\dot\lambda & = & 0\\ x+\lambda-2\dddot x & = & 0 \end{cases} $$
Solving for $x, \lambda$ we will have the extremal $x(t) = t+t^3$
The simplest is probably to notice that the first term $x\dot{x}=\frac{d(x^2/2)}{dt}$ is a total derivative, so we can rewrite the functional as a functional of $y\equiv\dot{x}$ only: $$ J[y]~=~ 2 + \int_0^1\!\mathrm{d}t ~ \ddot{x}^2 ~=~2 + \int_0^1\!\mathrm{d}t ~ \dot{y}^2. $$
However, there is still a constraint $$ 2~=~x(1)-x(0)~=~\int_0^1\!\mathrm{d}t ~\dot{x}~=~\int_0^1\!\mathrm{d}t ~y.$$
So the new functional reads $$ J[y]+\lambda \left(2-\int_0^1\!\mathrm{d}t ~y\right). $$ The Euler-Lagrange (EL) equation becomes $$ 2\ddot{y}+\lambda ~=~0,$$ with solution $$y(t)~=~-\frac{\lambda}{4} t^2 + at+b.$$
This should satisfy $$2~=~\int_0^1\!\mathrm{d}t ~y~=~-\frac{\lambda}{12} + \frac{a}{4}+b, $$ together with boundary conditions (BCs) $$ y(0)~=~1\qquad\text{and}\qquad y(1)~=~4, $$ i.e. $$ y(t)~=~3t^2+1.$$ This can be integrated to yield $x(t)$ itself.