Maximizing a functional of two functions. I am trying to maximize the functional
$$J[u,v] = \frac{4}{L} \int_0^L f(u,v) \ dx = \frac{4}{L} \int_0^L u(v- \bar v)^2 + v(u-\bar u)^2 \ dx, $$
where $0 \le u,v \le 1$ and,
$$\bar u = \frac{1}{L} \int_0^L u \ dx, \quad \bar v = \frac{1}{L} \int_0^L v \ dx.$$
The E-L equations here I believe are simply $f_u = f_v = 0$. This seems to imply that
$$f_u = (v - \bar v)^2 + 2 v(u - \bar u)(1 - 1) = 0$$
since $\frac{\partial \bar u}{\partial u} = 1$, so that $v = \bar v$. Similarly we have $u = \bar u$. This obviously is the minimum of the functional, but is there a way to find the maximum (aside from guessing)?
 A: Unless I am making a silly mistake, I am not sure you derived the correct
Euler-Lagrange equations. The Euler-Lagrange equations also hold at a maximum
point, not just at a minimum. Also keep in mind that the maximum may not
exist. Another important point is that you have the constraints $0\leq
u,v\leq1$ and so you do not always have the Euler-Lagrange equations. Say that
$(u_{0},v_{0})$ is a point where you reach a maximum. Assume that there exist
a measurable set $E\subset\lbrack0,L]$ and $\varepsilon>0$ such that
$\varepsilon\leq u_{0}(x),v_{0}(x)\leq1-\varepsilon$ for all $x\in E$. This is
the lucky case because you can take any pair of functions $u,v$ which are
bounded and zero outside of $E$. Then for $t$ small you will have that $0\leq
u_{0}+tu\leq1$, $0\leq v_{0}+tv\leq1$ and so you have that
\begin{align*}
g(t) &  =J(u_{0}+tu,v_{0}+tv)\\
&  =\frac{4}{L}\int_{0}^{L}(u_{0}+tu)\left(  v_{0}+tv-\frac{1}{L}\int_{0}
^{L}(v_{0}+tv)\,dy\right)  ^{2}+(v_{0}+tv)\left(  u_{0}+tu-\frac{1}{L}\int
_{0}^{L}(u_{0}+tu)\,dy\right)  ^{2}dx\\
&  \leq J(u_{0},v_{0})=g(0)
\end{align*}
which means that $g^{\prime}(0)=0$, that is
\begin{align*}
0 &  =g^{\prime}(0)=\frac{4}{L}\int_{0}^{L}u\left(  v_{0}-\bar{v}_{0}\right)
^{2}+v\left(  u_{0}-\bar{u}_{0}\right)  ^{2}dx\\
&  +\frac{4}{L}\int_{0}^{L}u_{0}2\left(  v_{0}-\bar{v}_{0}\right)  \left(
v-\frac{1}{L}\int_{0}^{L}v\,dy\right)  +v_{0}2\left(  u_{0}-\bar{u}
_{0}\right)  \left(  u-\frac{1}{L}\int_{0}^{L}u\,dy\right)  dx\\
&  =\frac{4}{L}\int_{0}^{L}u\left[  \left(  v_{0}-\bar{v}_{0}\right)
^{2}+v_{0}2\left(  u_{0}-\bar{u}_{0}\right)  -\frac{1}{L}\int_{0}^{L}
v_{0}2\left(  u_{0}-\bar{u}_{0}\right)  dy\right]  dx\\
&  +\frac{4}{L}\int_{0}^{L}v\left[  \left(  u_{0}-\bar{u}_{0}\right)
^{2}+u_{0}2\left(  v_{0}-\bar{v}_{0}\right)  -\frac{1}{L}\int_{0}^{L}
u_{0}2\left(  v_{0}-\bar{v}_{0}\right)  dy\right]  dx,
\end{align*}
where I used Fubini's theorem. This is true for all bounded functions $u$, $v$
which are zero outside $E$ and so you get that
\begin{align*}
\left(  v_{0}-\bar{v}_{0}\right)  ^{2}+v_{0}2\left(  u_{0}-\bar{u}_{0}\right)
-2\overline{u_{0}v_{0}}+2\bar{u}_{0}\bar{v}_{0} &  =0,\\
\left(  u_{0}-\bar{u}_{0}\right)  ^{2}+u_{0}2\left(  v_{0}-\bar{v}_{0}\right)
-2\overline{u_{0}v_{0}}+2\bar{u}_{0}\bar{v}_{0} &  =0
\end{align*}
for a.e. $x\in E$. These equations hold as long as $\varepsilon\leq
u_{0}(x),v_{0}(x)\leq1-\varepsilon$. By taking $\varepsilon=\frac{1}{n}$ you
get that these equations hold in the set $E_{0}:=\{x\in\lbrack0,L]:\,0<u_{0}
(x),v_{0}(x)<1\}$. 
On the other hand, if on a set $F_{0}$ you have that $u_{0}=0$, then the
variation $u_{0}+tu$ is only allowed if $tu\geq0$ in $F_{0}$. Say that $t>0$,
then necessarily $u\geq0$ in $F_{0}$. This gives
$$
\frac{g(t)-g(0)}{t}\leq0
$$
(since $t>0$), and so letting $t\rightarrow0^{+}$ you get $g^{\prime}(0)\leq
0$. In this case you only get the inequality
$$
\left(  v_{0}-\bar{v}_{0}\right)  ^{2}+v_{0}2\left(  0-\bar{u}_{0}\right)
-2\overline{u_{0}v_{0}}+2\bar{u}_{0}\bar{v}_{0}=\left(  v_{0}-\bar{v}
_{0}\right)  ^{2}+v_{0}2\left(  u_{0}-\bar{u}_{0}\right)  -2\overline
{u_{0}v_{0}}+2\bar{u}_{0}\bar{v}_{0}\leq0
$$
in $F_{0}$. To see what happens to the second equation, you have to consider
the subsets of $F_{0}$ where $0<v_{0}<1$ (here you will get equality), where
$v_{0}=0$ (here you will get $\leq0$), and where $v_{0}=1$ (so you to take
$v\leq0$, which will give $\geq0$).
Then you have to consider the set $F_{1}$ where $u_{0}=1$ and so something similar.
It's a bit messy and the Euler-Lagrange equations seem very messy to solve.
