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 $u_0$?
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 lapnumber of a solution for a onedimensional semilinear parabolic equation. J. Fac. Sci. Univ. Tokyo Sect. IA Math. 29 (1982), no. 2, 401–441.
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 $M=S^1$).
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 KacRice 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.

$\begingroup$ 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

1

$\begingroup$ 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