MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a (given) real-valued function defined over an area, $b(x)$, with $x\in \Omega \subset \mathbb{R}^2$. I would like to find a smooth real-valued function $a(x)$ that maximises $J = \int_{x \in \Omega} a^3(x) + a(x)b(x)~dx$ with the constraint $\int_{\Omega} a(x) ~dx=0$. If it helps prevent unbounded or irregular solutions I am happy to introduce further constraints or a higher-order penalty, eg, $J = \int_{x\in\Omega} a^3(x) + a(x)b(x) - \epsilon a^4(x)~dx$ (with the same constraint). How would you solve this? Do I need to specify boundary conditions on $a$?

share|cite|improve this question
up vote 3 down vote accepted

If we choose as a domain of the functional e.g. the space $L^3(\Omega)$ (with the constraint), and we assume $b\in L^{3/2}(\Omega),$ then $J$ is smooth, and if $\sup b(x)<\infty$ it has a bunch of critical points, all of the form described in the first answer, that is, all the $a(x)$ satisfying $3a(x)^2+b(x)=c$, where the constant $c\geq\sup b$ is arbitrary while the sign of $a(x)$ only has to satisfy the constraint condition $\int_\Omega a(x)\ dx=0$. The second variation of $J$ at a critical point $a(x)$ is $D^2J(a)[h]^2= 6\int_\Omega a(x)h^2(x)\ dx$. Thus, any critical point is neither a local minimizer nor a local maximizer (unless $a=0$, that only happens if $b$ is a constant and the problem is 1 dimensional). Moreover, $J$ is unbounded (choose a function $a(x)$ with $\int_\Omega a(x)dx=0$ and $\int_\Omega a^3(x)dx\neq 0$, then for real $t$, $J(ta)$ is a third degree polynomial in $t$, thus unbounded both from above and from below). If you take the perturbation with $\varepsilon>0$ the situation becomes more delicate . The natural domain is $L^4(\Omega)$ (with the constraint); assuming $b\in L^{4/3}(\Omega)$, $J$ is again smooth, unbounded from below, but now it is bounded from above; again it has a bunch of critical points, characterized by $3a(x)^2-4\varepsilon a^3(x)+b(x)=c$. There are no local minimizers (unless $b$ is constant) for the second variation $6\int_\Omega \left( a(x)-4\varepsilon a^2(x)\right) h^2(x)\ dx$ is not negative. It's not immediate to decide whether there are local or global maxima; certainly they can't be continuous since they have both to change sign and to verify $a(x)\leq 4\varepsilon a^2(x)$.

share|cite|improve this answer

It is easy to compute the critical points of $J$.

If you formally derive in $a$, you get $\int_\Omega 3a^2+b$, so that the minimizer $a_0$ should satisfy $3a_0^2+b=\mathrm{const}$ (apply $J$ to $a_0+\varepsilon h$ with $\varepsilon$ small and $h$ an arbitrary function of null integral, if you prefer). There are two choices left (the constant and the sign function of $a$).

share|cite|improve this answer
How do I avoid complex $a$ for regions with, say, $const=0$ and $b<0$? – ati May 19 '10 at 22:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.