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)?


1 Answer 1


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.


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