In Berestycki and Lions it is asserted that (on page 316), if I am not misreading, that the "ground state", i.e. action minimizer among nontrivial solutions, corresponding to the action $$ S[u] = \int_{\mathbb{R}^d} |\nabla u|^2 + G(u) dx $$ where $G$ satisfies certain conditions must be signed and spherically symmetric. In the paper they showed that under certain conditions on $G$ that such a ground state (that is, signed and spherically symmetric) exists; but they do not show that all grounds states must have these symmetries (by which I mean that they don't show the "action minimizing solutions" must be both single-signed and spherically symmetric).

The only reference I can find on this claim is the paper of Coleman, Glaser, Martin; but I am not entirely convinced that they have established the necessity. Their proof uses the fact that a ground state must be a minimizer of $\int |\nabla u|^2$ under the constraint that $\int G(u)$ is fixed. They then use the Polya-Szego principle: under spherical rearrangements the latter integral is unchanged, while the former can only decrease.

But the decrease is not necessarily strict. Brothers and Ziemer gave a counterexample in the case that the distribution function of $u$ is not absolutely continuous.

My question

I know how to complete the proof and get that ground states must be spherically symmetric, and monotonic radially, provided one assume that $G$ is $C^{1,1}$. This one can do by unique continuation principles for elliptic PDEs or equivalently by uniqueness theorems for ODEs. But in Berestycki-Lions or in Coleman-Glaser-Martin, $G$ is only assumed to be $C^1$, for which non-uniqueness, at least in the case of ODEs, is well known as a possibility.

So, is the "uniqueness" statement true for $G$ merely $C^1$? Are there known counterexamples?

  • $\begingroup$ It seems to me that the Gidas-Ni-Nirenberg paper ams.org/mathscinet-getitem?mr=634248 only needs C^1 hypotheses on G to obtain spherical symmetry at least, provided of course that one can justify the Euler-Lagrange equation. The method (moving planes) is also likely to get radial decrease, and ODE uniqueness theorems should get the uniqueness. So is your main concern the justification of the Euler-Lagrange equations? $\endgroup$
    – Terry Tao
    Feb 3, 2012 at 16:53
  • $\begingroup$ No, my main problem with GNN is that GN assumes that the solution is positive. A priori I don't know how to claim that the "action minimizer" must be function that doesn't change signs. Also, if I read it right, GNN works at the level of the equation $\triangle u = G'(u)$, and their $C^1$ assumption applies to $G'$, so by assuming only $C^1$ for $G$ we are still missing a bit. $\endgroup$ Feb 3, 2012 at 17:02
  • $\begingroup$ Ah, I see. (I'd say that your G should be called something else, such as V, to avoid confusion.) For positivity, one can show from the diamagnetic inequality that S[|u|] <= |S[u]| if G is even, and a slight perturbation of that argument would show that minimisers are automatically signed, I think. $\endgroup$
    – Terry Tao
    Feb 3, 2012 at 17:30
  • $\begingroup$ Isn't that the same sort of argument I described? (I may be misunderstanding you; please do say if that is the case.) (And sorry about the choice of name; I'll blame Lions for that (: ) Anyway, diamagnetic would say, in the same spirit of rearrangement, that if there exists any minimizer, there must exist a non-negative one. (There's the slight issue that for the focusing case where this is interesting the action is actually not minimized, just minimized over nontrivial solutions; but that's an aside.) Then if one has uniqueness this says the original solution is signed... $\endgroup$ Feb 4, 2012 at 10:05
  • $\begingroup$ The point is that the diamagnetic inequality argument shows that if there is a minimiser u that changes sign, then the diamagnetic inequality holds with equality for that solution. Admittedly this is not yet a contradiction if the solution only vanishes to second order or more at its sign changes, but I think one can work a bit more to eliminate this possibility (e.g. by dividing u into its positive part max(u,0) and its negative part min(u,0)) $\endgroup$
    – Terry Tao
    Feb 6, 2012 at 17:13

2 Answers 2


A little bit more digging turned up this paper of Mihai Mariş.

He shows that under two technical conditions:

  • Minimizers are $C^1$ (which we have from elliptic regularity)
  • If $u$ is an admissible function ($H^1$ in our case) and $v$ is a unit vector in $\mathbb{R}^d$, the function $\tilde{u}(x) = u(x)$ when $x\cdot v \geq 0$ and $\tilde{u}(x) = u(x - 2 x\cdot v v)$ otherwise is also admissible.

The solution to a constrained minimisation problem with $k$ constraints must be radially symmetric about an axis of dimension $k-1$ if the functionals are translation and rotationally invariant.

A very bad description of the process is follows:

Roughly speaking, the main idea is the following: by the Borsuk-Ulam Lemma we can find inductively $\sim d-k$ mutually orthogonal hyperplanes each of which split all the constraints exactly in half. The reflection symmetry inherent in the problem implies that from each orthant we can extend to a solution that is symmetric across each of the planes, which implies that it is symmetric under $x\to -x$ inside a subspace. This however implies that any linearly dependent hyperplane of the given ones will also split all the constraints in half.

The last step is to show that using multiple reflections and a bit of geometry, one can show that the original solution much also be reflectively symmetric across each of the planes. This shows radial symmetry.


A few results that addresses, but not quite answers my question:

  • In the seminal paper of Gidas-Ni-Nirenberg, it is mentioned that, at least in the bounded domain case, that a positive solution to $-\triangle u = g(u)$ need not be radially symmetric when $g$ is not at least Lipschitz continuous. In particular, they gave a family of solutions and equations for which $g \in C^{0,1-2/p}$ for any $p > 2$.
  • Along the same veins, it is not too hard to construct a counterexample to Gidas-Ni-Nirenberg if $g$ is merely continuous. Let $u(x) = v(x-x_v) + w(x- x_w)$, where $v$ is a radially symmetric bump function supported in the unit ball and monotonic radially, and $w$ is a radially symmetric function such that $w = 1$ in the ball of radius 5 and monotonically decreasing to 0. Then one can define $g$ by considering the case $x_v = x_w = 0$, which by a bit of undergraduate analysis one can show must be continuous. But the local nature of the equation means that for $|x_v - x_w| < 2$, the same equation will still be satisfied. Note that if one can construct one such example for which $u$ is actually the minimal action solution, then this will give a bona fide counterexample.
  • Franchi, Lanconelli, and Serrin shows an existence theorem for radially symmetric, non-negative solutions to a large class of PDEs, which includes $-\triangle u = g(u)$ for $g$ merely continuous. The corresponding uniqueness theorem they proved in the paper, however, is only concerning "uniqueness among radially symmetric solutions" and in particular does not use the variational structure at all. In particular, this result needs to be consider together with that of Cortazar, Elgueta, and Felmer which shows that for a particular equation of the form above with a manifestly non-Lipschitz nonlinearity, the equation admits compactly supported nonnegative solutions. This gives yet another counterexample to Gidas-Ni-Nirenberg in the low regularity case.

Neither of the above uses the variational structure of the "ground state". One result which I found that does use the variational ground state is that of

  • Flucher and Muller, which was able to show that under the assumption that, roughly speaking $g(0) = g'(0) = 0$ and that the corresponding $G$ is non-negative, all variational ground states agree up to a translation outside a compact set, and that if $g$ itself is non-negative, non-negative ground states are in fact positive and strictly decreasing radially, hence unique (by the result of Brothers and Ziemer mentioned in the question). So in the general case they were not able to rule out the possible counterexample given in the second bullet point above.
  • $\begingroup$ BTW, please don't vote up this answer. This answer collects some related literature which may be of interest to those interested in this question, but AFAICT none of the results actually answers the question I have. I post it as an answer since (a) it won't fit in the comments and (b) it doesn't look right as part of the question. $\endgroup$ Feb 6, 2012 at 16:39

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