Calculus of variations question with two variables 
If $u(x)$ and $v(x)$ satisfy $u(0)=1$, $v(0)=-1$, $u(\pi/2) =0$, $v(\pi/2) =0$ on extremals of the functional
$$ \int_0^{\pi/2}\left[\big({\frac{du}{dx}\big)^2 +\big(\frac{dv}{dx}\big)^2 + 2 \,u v }\,\right] dx $$
then which of the following is correct?
1) $u({\pi/4}) + v(\pi/4) = 0$ 
2) $u({\pi/3}) - v(\pi/3) = 0$ 
3) $u({\pi/4}) - v(\pi/4) = 1$
4) $u({\pi/3}) + v(\pi/3) = 0$

 A: HINT: You have functional of the form 
$$
I[u,v] = \int_0^{\frac{\pi}{2}} \mathcal{L}\left(u,v,u',v'\right)dx, \ \  \text{ where } \ \mathcal{L}\left(u,v,u',v'\right) =  \left(u'\right)^2+\left(v'\right)^2 + 2uv.
$$
Using the Euler-Lagrange equation for several functions depending on one variable, you can write the system of differential equations
$$
\left\{
\begin{aligned}
\frac{\partial \mathcal{L}}{\partial u} - \frac{\mathrm{d} }{\mathrm{d} x} \bigg( \frac{\partial \mathcal{L}}{\partial u'} \bigg) = 0
\\
\frac{\partial \mathcal{L}}{\partial v} - \frac{\mathrm{d} }{\mathrm{d} x} \bigg( \frac{\partial \mathcal{L}}{\partial v'} \bigg) = 0
\end{aligned}
\right.
\iff
\left\{
\begin{aligned}
2v - \frac{\mathrm{d} }{\mathrm{d} x} \left( 2u' \right) = 0
\\
2u - \frac{\mathrm{d} }{\mathrm{d} x} \left( 2v' \right) = 0
\end{aligned}
\right.
\iff
\left\{
\begin{aligned}v-u'' = 0\\
u - v'' = 0
\end{aligned}
\right.
$$
Solving these equations for the boundary conditions for $u$ and $v$ at $0$ and $\frac{\pi}{2}$, you will  find critical points $(u_0,v_0)$ of the original functional $I[u,v]$ by solving these equations.
As soon as you get these solutions, you can instantly plug in values and determine which answer is correct.
