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What methods are there to show symmetry properties of the minimizer of a problem $\inf_{u\in X}\mathcal{F}(u)$ in the calculus of variations? In general, the symmetry properties of $\mathcal{F}$ do not imply the symmetry of its minimizers. However, the minimizer may possess some weaker symmetry. More precisely, I am interested in functions $u$ defined on the sphere $S^2$. Even if $\mathcal{F}(u(R\cdot))=\mathcal{F}(u)$ for all rotations $R$ this does not imply that the solution is rotationally invariant (and therefore constant). Is it possible to formulate conditions on $\mathcal{F}$ that guarantee that each minimizer is rotationally symmetric around an axis?

Edit: If there is more than one minimizer $$\mathcal{F}(u(R\cdot))=\mathcal{F}(u) \text{ for all rotations } R$$ then this does not imply that all minimizers are symmetric with respect to any rotation. Instead, for each minimizer $u(\cdot)$ its rotated version is $u(R\cdot)$ is a minimizer, too. What tools are available to show that all minimizers possess a weaker symmetry e.g. symmetric with respect to one axis. The functional $\mathcal{F}$ of interest is a the sum of a convex and a non-convex bilinear part.

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    $\begingroup$ Convexity? If the minimizer exists and is unique, it has maximal symmetry. Convexity (or perhaps "strict" convexity) gives you a unique minimizer. $\endgroup$ Oct 6, 2013 at 22:46
  • $\begingroup$ What can you say if the functional is a sum of a convex part and a bilinear one? $\endgroup$
    – warsaga
    Oct 7, 2013 at 10:50
  • $\begingroup$ A bilinear functional is convex as long as it is positive definite. Moreover, the sum of two convex functions is convex. $\endgroup$ Oct 7, 2013 at 13:29

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Suppose that $\newcommand{\eF}{\mathscr{F}}$ $\newcommand{\bR}{\mathbb{R}}$ $\eF: C\to \bR$ is a convex functional defined on a closed convex subset $C$ of a say real Banach space $U$. (You can allow for more general topological spaces.) Suppose additionally that a compact Lie group $G$ acts on $C$. For any $u\in U$ we define the symmetrization

$$ [u]_G:=\int_G g\cdot u dg, $$

where $dg$ denotes the unique bi-invariant measure on $G$ of total volume $1$. (In other words, $dg$ is an invaraint probability measure on $G$.) Clearly $[u]_G$ is a fixed point of the $G$ action and since $C$ is convex and $dg$ is a probability measure

$$ u\in C\Rightarrow \bar{u}_G\in C. $$

Jensen's inequality implies

$$ \eF([u]_G)\leq \int_G \eF(g\cdot u) dg. $$

If $\eF$ happens to be $G$-invariant as well and $u_0$ is a minimizer, then

$$ \int_G \eF(g\cdot u_0) dg =\eF(u_0)=\min_C \eF. $$

In particular, this implies that

$$ \eF([u_0]_G)\leq \min_C \eF. $$

so that $[u_0]_G$ is also a minimizer. To see how this work in practice, I refer to a very old paper of mine where I used this symmetrization technique in an optimal control problem. The relevant part begins at page 19 Proposition 4.2. of the paper.

The paper also presents applications to optimal control problems of symmetric rearrangement technique mentioned in Kelei Wang's answer. The paper also has some references on various symmetrization techniques you might find useful.

Here is one instance when this works. Let $X$ be the Sobolev space $H^1(S^n)$ (functions with weak $L^2$ first order derivatives) where the sphere $S^n$ is equipped with the round metric. Let $f:\bR\to\bR$ be a convex function and consider the convex functional

$$ \eF: H^1(S^n)\to (-\infty,\infty],\;\;\eF(u)=\int_{S^n} \Bigl(\; \frac{1}{2}|\nabla u|^2+ f(u)\;\Bigr) dV_{S^n}, $$

where the gradient and the volume form are defined in terms of the round metric. Let $G$ be the group $SO(n+1)$ which acts transitively and isometrically on $S^n$. It induces an action by pullback on $H^1(S^n)$. The functional $\eF$ is invariant under this action. If it admits a minimizer, then it admits a symmetric minimizer, which has to be constant.

If instead you consider a functional

$$\eF_1(u)=\int_{S^n} \Bigl(\; \frac{1}{2}|\nabla u(x)|^2+ a(x)f(u(x))\;\Bigr) dV_{S^n}(x), $$

where $a: S^n\to \bR$ is a nonnegative function which is invariant under the action of the closed subgroup $G\subset SO(n+1)$ and $\eF_1$ has minimazers, then it will have $G$-invariant minimizers.

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There are mainly two ways to get symmetry of minimizers, at least when the functional is associated to an elliptic PDE. The first one is the rearrangement argument (Schwarz symmetrization, Steiner symmetrization etc.), which shows in some cases symmetric objects have lower energy. The other is the moving plane method, which involves some maximum principle argument.

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  • $\begingroup$ Nice answer! Something similar to your "rearrangement argument" section is the fact that first eigenfunctions of say, the Dirichlet problem in a domain of $\mathbb{R}^2$ inherit the same symmetries of the domain: because the first eigenspace is one dim'l and spanned by a positive function, it's clear that the "flip" of the eigenfunction must be equal to itself. $\endgroup$ Oct 10, 2013 at 22:27
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Here is an explicit example, which may or may not fit into your requirements:

In http://www.ams.org/mathscinet-getitem?mr=308905, "The equivariant Plateau problem and interior regularity," Lawson shows:

If $G$ is a compact subgroup of $SO(n)$ and $M$ is a $G$-invariant $(n-2)$-dimensional submanifold of $\mathbb{R}^n$, then there exists $T$, a $G$-invariant (current) solution to Plateau's problem. If $T$ is the unique $G$-invariant solution, then it is the unique solution!

As an interesting application, Lawson gives a new proof that the Simons cone is minimizing.


Recall that Plateau's problem is:

Given $T$, a $(n-2)$-dimensional submanifold of $\mathbb{R}^n$, find a $(n-1)$-dimensional "submanifold" $M$ which minimizes area among all such competitors.

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