An Euler-Lagrange Equation I have an action with a Lagrangian which I would like to apply the Euler-Lagrange equations to https://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation but have spent hours really struggling with it. That is I define
$$
L\Big(\begin{pmatrix}x \\ y\end{pmatrix},\begin{pmatrix} \dot{x} \\ \dot{y}\end{pmatrix} \Big):= \Big\| A \Big( \begin{pmatrix}x \\ y\end{pmatrix} +  \begin{pmatrix} -\dot{y} \\ \beta\dot{y}\end{pmatrix}  \Big) \Big\|^2
$$
where
$$
A=\begin{pmatrix} c_1 & c_2 \\ c_2 & c_3 \end{pmatrix}, \text{for some }~c_i~\text{for which $A$ is invertible}.
$$
Which $x,y:[0,T]\to \mathbb{R}$ solve the Euler -Lagrange equation
$$ 
\nabla_{\begin{pmatrix}x \\ y\end{pmatrix}} L-\partial_t\nabla_{\begin{pmatrix} \dot{x} \\ \dot{y}\end{pmatrix} }L=0
$$
with initial and terminal conditions $x(0)=q,x(T)=q',y(0)=p,y(T)=p'$. ? Please help !
$\Big($would more information on how the constants $c_1,c_2,c_3$ relate to eachother be useful? because infact $c_1=\sigma^2+\gamma^2, c_2=-\gamma(\sigma+\alpha),c_3=\alpha^2+\gamma^2$, where these constants $\alpha,\sigma,\gamma$ are positive and $\alpha\sigma>\gamma^2$ $\Big)$.
EDIT : I got the E.L as with the above choices for $c_1,c_2,c_3$
\begin{equation}
    \nabla_{(x,y)} L = (\frac{1}{\alpha\sigma-\gamma^2})^2\begin{pmatrix}
    0 \\
    \nabla_{y} B_1+\nabla_{y} B_2
    \end{pmatrix} 
\end{equation}
with
\begin{equation}
    \nabla_{y} B_1=2(\sigma+\gamma \beta) y + \frac{1}{\sigma+\gamma\beta}(\gamma \dot{y}-\sigma \dot{x})
\end{equation}
and
\begin{equation}
    \nabla_{y} B_2=2(\gamma + \alpha \beta) y + \frac{1}{\gamma+\alpha \beta}(\alpha \dot{x}-\gamma \dot{y}).
\end{equation}
Also
\begin{equation}
    \partial_t \nabla_{(\dot{x},\dot{y})} L=2 (A^{-1})^2(\begin{pmatrix}\ddot{x}\\ \ddot{y} \end{pmatrix}-\begin{pmatrix}
    \dot{y}
    \\
    -\beta \dot{y}
    \end{pmatrix}) 
\end{equation}
Which is a coupled ODE

Here is a sketched derivation.

*

*First of all, notice that the Lagrangian $L$ does not depend $\dot{x}$. Then we don't need the 2 Dirichlet boundary conditions (BCs) for $x$ to deduce the EL equation for $x$. The EL equation for $x$ reduces to
$$0~=~\chi(x,y,\dot{y})~:=~\frac{1}{2}\frac{\partial L(x,y,\dot{y})}{\partial x}~=~c_1(x-\dot{y}) +c_2(y+\beta\dot{y}) \tag{1}$$
from which we can determine $x$.


*If we eliminate$^1$ $x$ in the Lagrangian $L$, the Lagrangian becomes (up to an overall non-zero multiplicative normalization)
$$ L_0~=~\frac{1}{2}(y+\beta\dot{y})^2. \tag{2}$$
The momentum is
$$ p ~=~ \frac{\partial L_0}{\partial \dot{y}}
~=~\beta(y+\beta\dot{y}).\tag{3}$$
The energy function
$$ E~=~p\dot{y}-L_0~=~\frac{1}{2}(\beta\dot{y}-y)(\beta\dot{y}+y) ~=~\frac{1}{2}(\beta^2\dot{y}^2-y^2) \tag{4}$$
is a constant, since there is no explicit $t$-dependence.


*In other words, we get a 1st-order ODE
$$ \beta^2\dot{y}^2~=~2E +y^2 \tag{5}$$
Eq. (5) can easily be solved via separation of variables. This produces 1 integration constant. Together with $E$ we then have 2 integration constants, which can be matched with the 2 Dirichlet BCs for $y$.


*Generically, it is impossible to match 4 boundary conditions (BCc), as already noted in Cesareo's answer. It would be natural to discard the 2 Dirichlet BCs for $x$.
--
$^1$ Normally, one is not allowed to use a EL equation (1) inside the Lagrangian, but one may show that it is okay for a quadratic $x$-dependence. We can complete the square
$$ L~=~L_0+L_2 \tag{6} $$
where
$$ L_n ~\propto~ \chi^n .\tag{7}$$
It is not difficult to see that $L$ and $L_0$ lead to the same EL equation for $y$ modulo the constraint (1).
