Let $u$ be a smooth function on the unit circle $S^1$ such that $\int_{S^1}ux_j=0$, for $j=1,2$. Is the number of critical points of $u$ strictly bigger than 2?


2 Answers 2


Yes, in fact the number of critical points of $u$ is at least four.

Expand $u:S^1\to\mathbb R$ as a Fourier series: $$u(\theta)=a_0+\sum_{i=1}^\infty a_j\cos(j\theta)+b_j\sin(j\theta)$$ Then your condition may be stated as $a_1=b_1=0$ (I assume by $x_1$ and $x_2$, you mean to be identifying $S^1$ with $\{x_1^2+x_2^2=1\}$, so $x_1=\cos\theta$ and $x_2=\sin\theta$ in the notation above).

Now let $u(t,\theta)$ be the solution to the heat equation $u_t=u_{\theta\theta}$ with initial value $u(0,\theta)=u(\theta)$. By the maximum principle, we have: $$\#\operatorname{crit}(u(\cdot))\geq\#\operatorname{crit}(u(t,\cdot))$$ for all $t\geq 0$. On the other hand, we have the explicit solution: $$u(t,\theta)=a_0+\sum_{i=1}^\infty a_je^{-j^2t}\cos(j\theta)+b_je^{-j^2t}\sin(j\theta)$$ Let $j_0=\min\{j\geq 1:(a_j,b_j)\ne(0,0)\}$. If $j_0=\infty$ (i.e. only $a_0$ is nonzero), then $u$ is constant and there is nothing to prove, so therefore we have $1<j_0<\infty$.

Now the behavior of the function $u(t,\cdot)$ for large $t$ is evidently dominated by the leading terms: $$u(t,\theta)=a_0+e^{-j_0^2t}\left[a_{j_0}\cos(j_0\theta)+b_{j_0}\sin(j_0\theta)\right]+\cdots$$ Writing $a_{j_0}\cos(j_0\theta)+b_{j_0}\sin(j_0\theta)=\Re\left[(a_{j_0}-ib_{j_0})e^{j_0\theta}\right]$, it follows that for sufficiently large $t$, we have: $$\#\operatorname{crit}(u(t,\cdot))=2j_0$$ Since $j_0\geq 2$, this gives the desired result.

EDIT: While not directly relevant for the OP, let me correct an erroneous statement I made in the comment thread below. In dimension two, it turns out that the heat flow can cause an increase in the number of critical points in a serious way (it can introduce a pair of cancelling critical points) despite some handwavy "maximum principle" mumblings to the contrary. Here is an explicit example: $f_t(x,y)=x^3+(1-5x)y^2+(2-4x)t$ solves $\partial_tf=\Delta f$, and its critical locus near the origin $(x,y)=(0,0)$ near time $t=0$ is the set of $(x,y)$ with $y=0$ and $x^2=\frac 43t$. So for $t<0$ there are no critical points, for $t=0$ there is a single "embryonic" critical point, and for $t>0$ there are two critical points. In particular, a local minimum has been created by the heat flow(!)

  • $\begingroup$ One can certainly try to run the same argument on $S^2$. Things certainly break down in dimensions $n>2$, though, since in the evolution of heat equation, a pair of cancelling critical points of index $k$ and $k+1$ can appear whenver $0<k<k+1<n$. $\endgroup$ Jan 20, 2015 at 3:58
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    $\begingroup$ In dimension two, the only "bad" thing that can happen is that a "monkey saddle" could split into two index one critical points. But, as long as you start out with a function with non-degenerate critical points, this is not an issue: the maximum principle still implies that the number of critical points at any point of the heat evolution is no larger than the number of critical points of the initial condition. $\endgroup$ Jan 20, 2015 at 4:01
  • $\begingroup$ So do you think with a similar argument we can deduce that the set of critical points of $u$ contains a closed curve on $S^2$? $\endgroup$ Jan 20, 2015 at 4:03
  • $\begingroup$ So as long as your $u$ on $S^2$ has non-degenerate critical points, you are reduced to analyzing the limiting behavior of the heat equation via spherical harmonics. This seems very tractable to me, though certainly not as easy as it is for $S^1$. $\endgroup$ Jan 20, 2015 at 4:03
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    $\begingroup$ Oh, I thought you wanted a lower bound on the number of critical points. Generically, the critical locus of a smooth function is finite; you won't be able to force a closed curve of critical points by imposing a finite set of conditions of the form $\int_{S^2}uf_j=0$ for functions $f_j:S^2\to\mathbb R$. $\endgroup$ Jan 20, 2015 at 4:06

An elementary argument: write $u$ as a function of $\theta \in T^1 := {\bf Z} / 2 \pi {\bf Z}$, where $(x_1,x_2) = (\cos \theta, \sin \theta)$. Then each of $$ \int_{T^1} u'(\theta) \, d\theta, \quad \int_{T^1} u'(\theta) \cos \theta \, d\theta, \quad \int_{T^1} u'(\theta) \sin \theta \, d\theta $$ vanishes (the first is clear, and the other two are obtained from $$ \int_{T^1} u(\theta) \sin \theta \, d\theta = \int_{T^1} u(\theta) \cos \theta \, d\theta = 0 $$ by integration by parts). Thus if $u'$ has only finitely many zeros in $T^1$ then it has at least four sign changes, by an argument that I already gave for this Math.SE question. Assume $u'$ had only two sign changes in each period, say at $\theta_1$ and $\theta_2$. Then we could find reals $A,B,C$ such that $t(\theta) = A + B \cos \theta + C \sin \theta$ has sign changes at the same $\theta_1$ and $\theta_2$ and nowhere else; but then $u'(\theta) \, t(\theta)$ is either everywhere $\geq 0$ or everywhere $\leq 0$, but is not everywhere zero, which contradicts $\int_{T^1} u'(\theta) \, t(\theta) \, dx = 0$, and we're done.

(To find $A,B,C$ we can go back to the $S^1$ picture: consider the points $(\cos \theta_1, \sin \theta_1)$ and $(\cos \theta_2, \sin \theta_2)$ on the circle $x_1^2 + x_2^2 = 1$, and join them by a line $A+Bx_1+Cx_2 = 0$, which meets the circle at those two points and thus nowhere else.)

  • $\begingroup$ Sorry, I don't see a direct generalization to higher-dimensional spheres. $\endgroup$ Jan 20, 2015 at 4:23
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    $\begingroup$ There is a generalization in $T^1$, though: if $u$ is orthogonal to $\cos n\theta$ and $\sin n\theta$ for each positive $n < d$ then $u$ has at least $2d$ critical points. (Your question is the case $d=2$; both John Pardon's proof and the one I gave readily generalize to arbitrary $d$.) $\endgroup$ Jan 20, 2015 at 7:04

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