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Let $u_0$ be a smooth function on the unit sphere $S^1$ and assume that $u(t,x)$ is a smooth solution of the heat equation with initial data $u(0,x)=u_0(x)$. How one can apply the maximum principle to prove that the number of critical points of $u(t,x)$ at any given time is not bigger than the number of critical points of $u_0$?

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Since $u$ is nothing but a $2\pi$-periodic solution of $u_t=u_{xx}$, looking at critical points amounts to looking at the zeroes of $v:=u_x$, which is another solution of the same equation. Then your question is solved by H. Matano : Nonincrease of the lap-number of a solution for a one-dimensional semilinear parabolic equation. J. Fac. Sci. Univ. Tokyo Sect. IA Math. 29 (1982), no. 2, 401–441.

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In arbitrary dimensions one has the following phenomenon. Start with a compact Riemann manifold $(M,g)$. Fix an orthonormal basis of $L^2(M)$ consisting of eigenfunctions $(\psi_k)_{k\geq 0}$ of the Laplacian. Assume that the spectrum is

$$ =0=\lambda_0<\lambda_1 \leq \lambda_2\leq \cdots \leq \lambda_k\leq cdots $$

where aboveeach eigenvalues appears as often as its multiplicity A smooth function $\newcommand{\bR}{\mathbb{R}}$ $u_0:M\to \bR$ has an eigenfunction expansion

$$u_0=\sum_{k\geq 0} A_k \psi_k. $$

Suppose now that the coefficients are independent normal random variables such that the above random function is almost surely smooth. (This happens if the variance of $A_k$ goes to zero very fast.) We get another random function

$$ u_t=e^{-t\Delta} u_0. $$

As $t\to\infty$ the expected number of critical points of $u_t$ converges to the number of critical points of the eigenfunction of $\Delta$ corresponding to the first nonzero eigenvalue.

On the other hand if you choose $u_0$ of the form

$$ u_{0, R}=\sum_{\lambda_k \leq R^2} C_k \psi_k, $$

where $(C_k)$ is sequence of independent standard normal random variables, then for large $R$, the expected number of critical points of the random function $u_{0, R}$ is approximatively $Z_m R^m$, where $m=\dim M$ and $Z_m$ is a universal constant that depends only on $m$. This random number of critical points is highly concentrated around its mean. On the other hand as $t\to\infty$ the function

$$ e^{-t\Delta} u_{0,R}, $$ will have, on average, fewer and fewer critical points. The animation below illustrates this phenomenon (in the case $M=S^1$).

enter image description here

Remark 1. Consider a random trigonometric polynomial of high degree $N$

$$u_0=\frac{1}{\sqrt{\pi}}\sum_{n=1}^N (A_n\cos n\theta+B_n\sin n\theta), $$

where $A_n, B_n$ are independent normal random variables.

We set $u_t=e^{-t\Delta} u_0$, and denote by $C_t(N)$ the expcted number of critical points of $u_t$. Then Kac-Rice formula implies that

$$C_t(N)=2\sqrt{\frac{\sum_{n=1}^N n^4 e^{-2tn^2}}{\sum_{n=1}^N n^2 e^{-2tn^2}}}. $$

One can prove the following.

For $t>0$ fixed we have

$$\lim_{N\to \infty}C_t(N)= 2. $$

For $t=0$ we have

$$ C_0(N)\sim 2\sqrt{\frac{3}{5}} N\;\;\mbox{as}\;\;N\to\infty. $$

Remark 2. Given a smooth function on a Riemannian manifold $(M,g)$ with eigenfunction decomposition

$$ u_0= v_1+\sum_{\lambda_k>\lambda 1}c_k\psi_k, $$

and $v_1$ is in the $\lambda_1$-eigenspace, then

$$u_t= e^{-t\Delta}u_0= e^{-t\lambda_1}v_1+\sum_{\lambda k >\lambda_1} e^{-\lambda_k t} c_k\psi_k, $$

then we observe that $u_t$ has the same number of critical points as

$$ U_t=v_1+\underbrace{\sum_{\lambda k >\lambda_1} e^{-(\lambda_k-\lambda_1) t} c_k\psi_k}_{=:R_t}. $$

We have

$$\lim_{t\to infty}\Vert R_t\Vert_{C^2(M)}=0. $$

The last condition implies that if $v_1$ is a Morse function, then the number of critical points of $U_t$ (or $u_t$) is equal to the number of critical points of $v_1$ if $t$ is sufficiently large. The above animation gives a depiction of $U_t$ for various moments of time $t$.

We typically expect $u_0$ to have a large number of critical points because the eigenfunctions $\psi_k$ for $k$ large are highly oscillatory.

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  • $\begingroup$ All that I claimed is this will happen for a random $u_0$ with high probability. $\endgroup$ – Liviu Nicolaescu Jan 21 '15 at 16:15
  • $\begingroup$ Thanks Liviu. Where can I find a proof for the fact that "the function $e^{-t\Delta} u_{0,R}$ will have, on average, fewer and fewer critical points" ? $\endgroup$ – User4966 Jan 21 '15 at 16:17
  • $\begingroup$ Is it true that there exists a sequence $t_n$ such that $t_n \rightarrow \infty $ and $e^{-t_n\Delta} u_0$ has fewer critical points than $u_0$ itself? $\endgroup$ – User4966 Jan 21 '15 at 18:43
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    $\begingroup$ See the new remarks I have added. $\endgroup$ – Liviu Nicolaescu Jan 21 '15 at 21:55
  • $\begingroup$ Thanks Liviu. I just posted a related question and I would appreciate if you take a look at it: mathoverflow.net/questions/194522/… $\endgroup$ – User4966 Jan 22 '15 at 2:48

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